Mathematical Methods for Physics and Engineering - Matematica.NET

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Mathematical Methods for Physics and EngineeringThe third edition of this highly acclaimed undergraduate textbook is suitablefor teaching all the mathematics ever likely to be needed for an undergraduatecourse in any of the physical sciences. As well as lucid descriptions of all thetopics covered and many worked examples, it contains more than 800 exercises.A number of additional topics have been included and the text has undergonesignificant reorganisation in some areas. New stand-alone chapters:• give a systematic account of the ‘special functions’ of physical science• cover an extended range of practical applications of complex variables includingWKB methods and saddle-point integration techniques• provide an introduction to quantum operators.Further tabulations, of relevance in statistics and numerical integration, havebeen added. In this edition, all 400 odd-numbered exercises are provided withcomplete worked solutions in a separate manual, available to both students andtheir teachers; these are in addition to the hints and outline answers given inthe main text. The even-numbered exercises have no hints, answers or workedsolutions and can be used for unaided homework; full solutions to them areavailable to instructors on a password-protected website.Ken Riley read mathematics at the University of Cambridge and proceededto a Ph.D. there in theoretical and experimental nuclear physics. He became aresearch associate in elementary particle physics at Brookhaven, and then, havingtaken up a lectureship at the Cavendish Laboratory, Cambridge, continued thisresearch at the Rutherford Laboratory and Stanford; in particular he was involvedin the experimental discovery of a number of the early baryonic resonances. Aswell as having been Senior Tutor at Clare College, where he has taught physicsand mathematics for over 40 years, he has served on many committees concernedwith the teaching and examining of these subjects at all levels of tertiary andundergraduate education. He is also one of the authors of 200 Puzzling PhysicsProblems.Michael Hobson read natural sciences at the University of Cambridge, specialisingin theoretical physics, and remained at the Cavendish Laboratory tocomplete a Ph.D. in the physics of star-formation. As a research fellow at TrinityHall, Cambridge and subsequently an advanced fellow of the Particle Physicsand Astronomy Research Council, he developed an interest in cosmology, andin particular in the study of fluctuations in the cosmic microwave background.He was involved in the first detection of these fluctuations using a ground-basedinterferometer. He is currently a University Reader at the Cavendish Laboratory,his research interests include both theoretical and observational aspects of cosmology,and he is the principal author of General Relativity: An Introduction for

Physicists. He is also a Director of Studies in Natural Sciences at Trinity Hall andenjoys an active role in the teaching of undergraduate physics and mathematics.Stephen Bence obtained both his undergraduate degree in Natural Sciencesand his Ph.D. in Astrophysics from the University of Cambridge. He then becamea Research Associate with a special interest in star-formation processes and thestructure of star-forming regions. In particular, his research concentrated on thephysics of jets and outflows from young stars. He has had considerable experienceof teaching mathematics and physics to undergraduate and pre-universtiystudents.ii

Mathematical Methodsfor Physics and EngineeringThird EditionK.F. RILEY, M.P. HOBSON and S.J. BENCE

cambridge university pressCambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São PauloCambridge University PressThe Edinburgh Building, Cambridge cb2 2ru, UKPublished in the United States of America by Cambridge University Press, New Yorkwww.cambridge.orgInformation on this title: www.cambridge.org/9780521861533© K. F. Riley, M. P. Hobson and S. J. Bence 2006This publication is in copyright. Subject to statutory exception and to the provision ofrelevant collective licensing agreements, no reproduction of any part may take placewithout the written permission of Cambridge University Press.First published in print format 2006isbn-13 978-0-511-16842-0 eBook (EBL)isbn-10 0-511-16842-x eBook (EBL)isbn-13 978-0-521-86153-3 hardbackisbn-10 0-521-86153-5 hardbackisbn-13 978-0-521-67971-8 paperbackisbn-10 0-521-67971-0 paperbackCambridge University Press has no responsibility for the persistence or accuracy of urlsfor external or third-party internet websites referred to in this publication, and does notguarantee that any content on such websites is, or will remain, accurate or appropriate.

ContentsPreface to the third editionPreface to the second editionPreface to the first editionpage xxxxiiixxv1 Preliminary algebra 11.1 Simple functions and equations 1Polynomial equations; factorisation; properties of roots1.2 Trigonometric identities 10Single angle; compound angles; double- and half-angle identities1.3 Coordinate geometry 151.4 Partial fractions 18Complications and special cases1.5 Binomial expansion 251.6 Properties of binomial coefficients 271.7 Some particular methods of proof 30Proof by induction; proof by contradiction; necessary and sufficient conditions1.8 Exercises 361.9 Hints and answers 392 Preliminary calculus 412.1 Differentiation 41Differentiation from first principles; products; the chain rule; quotients;implicit differentiation; logarithmic differentiation; Leibnitz’ theorem; specialpoints of a function; curvature; theorems of differentiationv

CONTENTS2.2 Integration 59Integration from first principles; the inverse of differentiation; by inspection;sinusoidal functions; logarithmic integration; using partial fractions;substitution method; integration by parts; reduction formulae; infinite andimproper integrals; plane polar coordinates; integral inequalities; applicationsof integration2.3 Exercises 762.4 Hints and answers 813 Complex numbers and hyperbolic functions 833.1 The need for complex numbers 833.2 Manipulation of complex numbers 85Addition and subtraction; modulus and argument; multiplication; complexconjugate; division3.3 Polar representation of complex numbers 92Multiplication and division in polar form3.4 de Moivre’s theorem 95trigonometric identities; finding the nth roots of unity; solving polynomialequations3.5 Complex logarithms and complex powers 993.6 Applications to differentiation and integration 1013.7 Hyperbolic functions 102Definitions; hyperbolic–trigonometric analogies; identities of hyperbolicfunctions; solving hyperbolic equations; inverses of hyperbolic functions;calculus of hyperbolic functions3.8 Exercises 1093.9 Hints and answers 1134 Series and limits 1154.1 Series 1154.2 Summation of series 116Arithmetic series; geometric series; arithmetico-geometric series; the differencemethod; series involving natural numbers; transformation of series4.3 Convergence of infinite series 124Absolute and conditional convergence; series containing only real positiveterms; alternating series test4.4 Operations with series 1314.5 Power series 131Convergence of power series; operations with power series4.6 Taylor series 136Taylor’s theorem; approximation errors; standard Maclaurin series4.7 Evaluation of limits 1414.8 Exercises 1444.9 Hints and answers 149vi

CONTENTS5 Partial differentiation 1515.1 Definition of the partial derivative 1515.2 The total differential and total derivative 1535.3 Exact and inexact differentials 1555.4 Useful theorems of partial differentiation 1575.5 The chain rule 1575.6 Change of variables 1585.7 Taylor’s theorem for many-variable functions 1605.8 Stationary values of many-variable functions 1625.9 Stationary values under constraints 1675.10 Envelopes 1735.11 Thermodynamic relations 1765.12 Differentiation of integrals 1785.13 Exercises 1795.14 Hints and answers 1856 Multiple integrals 1876.1 Double integrals 1876.2 Triple integrals 1906.3 Applications of multiple integrals 191Areas and volumes; masses, centres of mass and centroids; Pappus’ theorems;moments of inertia; mean values of functions6.4 Change of variables in multiple integrals 199Change∫of variables in double integrals; evaluation of the integral I =∞−∞ e−x2 dx; change of variables in triple integrals; general properties ofJacobians6.5 Exercises 2076.6 Hints and answers 2117 Vector algebra 2127.1 Scalars and vectors 2127.2 Addition and subtraction of vectors 2137.3 Multiplication by a scalar 2147.4 Basis vectors and components 2177.5 Magnitude of a vector 2187.6 Multiplication of vectors 219Scalar product; vector product; scalar triple product; vector triple productvii

CONTENTS7.7 Equations of lines, planes and spheres 2267.8 Using vectors to find distances 229Point to line; point to plane; line to line; line to plane7.9 Reciprocal vectors 2337.10 Exercises 2347.11 Hints and answers 2408 Matrices and vector spaces 2418.1 Vector spaces 242Basis vectors; inner product; some useful inequalities8.2 Linear operators 2478.3 Matrices 2498.4 Basic matrix algebra 250Matrix addition; multiplication by a scalar; matrix multiplication8.5 Functions of matrices 2558.6 The transpose of a matrix 2558.7 The complex and Hermitian conjugates of a matrix 2568.8 The trace of a matrix 2588.9 The determinant of a matrix 259Properties of determinants8.10 The inverse of a matrix 2638.11 The rank of a matrix 2678.12 Special types of square matrix 268Diagonal; triangular; symmetric and antisymmetric; orthogonal; Hermitianand anti-Hermitian; unitary; normal8.13 Eigenvectors and eigenvalues 272Of a normal matrix; of Hermitian and anti-Hermitian matrices; of a unitarymatrix; of a general square matrix8.14 Determination of eigenvalues and eigenvectors 280Degenerate eigenvalues8.15 Change of basis and similarity transformations 2828.16 Diagonalisation of matrices 2858.17 Quadratic and Hermitian forms 288Stationary properties of the eigenvectors; quadratic surfaces8.18 Simultaneous linear equations 292Range; null space; N simultaneous linear equations in N unknowns; singularvalue decomposition8.19 Exercises 3078.20 Hints and answers 3149 Normal modes 3169.1 Typical oscillatory systems 3179.2 Symmetry and normal modes 322viii

CONTENTS9.3 Rayleigh–Ritz method 3279.4 Exercises 3299.5 Hints and answers 33210 Vector calculus 33410.1 Differentiation of vectors 334Composite vector expressions; differential of a vector10.2 Integration of vectors 33910.3 Space curves 34010.4 Vector functions of several arguments 34410.5 Surfaces 34510.6 Scalar and vector fields 34710.7 Vector operators 347Gradient of a scalar field; divergence of a vector field; curl of a vector field10.8 Vector operator formulae 354Vector operators acting on sums and products; combinations of grad, div andcurl10.9 Cylindrical and spherical polar coordinates 35710.10 General curvilinear coordinates 36410.11 Exercises 36910.12 Hints and answers 37511 Line, surface and volume integrals 37711.1 Line integrals 377Evaluating line integrals; physical examples; line integrals with respect to ascalar11.2 Connectivity of regions 38311.3 Green’s theorem in a plane 38411.4 Conservative fields and potentials 38711.5 Surface integrals 389Evaluating surface integrals; vector areas of surfaces; physical examples11.6 Volume integrals 396Volumes of three-dimensional regions11.7 Integral forms for grad, div and curl 39811.8 Divergence theorem and related theorems 401Green’s theorems; other related integral theorems; physical applications11.9 Stokes’ theorem and related theorems 406Related integral theorems; physical applications11.10 Exercises 40911.11 Hints and answers 41412 Fourier series 41512.1 The Dirichlet conditions 415ix

CONTENTS12.2 The Fourier coefficients 41712.3 Symmetry considerations 41912.4 Discontinuous functions 42012.5 Non-periodic functions 42212.6 Integration and differentiation 42412.7 Complex Fourier series 42412.8 Parseval’s theorem 42612.9 Exercises 42712.10 Hints and answers 43113 Integral transforms 43313.1 Fourier transforms 433The uncertainty principle; Fraunhofer diffraction; the Dirac δ-function;relation of the δ-function to Fourier transforms; properties of Fouriertransforms; odd and even functions; convolution and deconvolution; correlationfunctions and energy spectra; Parseval’s theorem; Fourier transforms in higherdimensions13.2 Laplace transforms 453Laplace transforms of derivatives and integrals; other properties of Laplacetransforms13.3 Concluding remarks 45913.4 Exercises 46013.5 Hints and answers 46614 First-order ordinary differential equations 46814.1 General form of solution 46914.2 First-degree first-order equations 470Separable-variable equations; exact equations; inexact equations, integratingfactors; linear equations; homogeneous equations; isobaric equations;Bernoulli’s equation; miscellaneous equations14.3 Higher-degree first-order equations 480Equations soluble for p; forx; fory; Clairaut’s equation14.4 Exercises 48414.5 Hints and answers 48815 Higher-order ordinary differential equations 49015.1 Linear equations with constant coefficients 492Finding the complementary function y c (x); finding the particular integraly p (x); constructing the general solution y c (x) +y p (x); linear recurrencerelations; Laplace transform method15.2 Linear equations with variable coefficients 503The Legendre and Euler linear equations; exact equations; partially knowncomplementary function; variation of parameters; Green’s functions; canonicalform for second-order equationsx

CONTENTS15.3 General ordinary differential equations 518Dependent variable absent; independent variable absent; non-linear exactequations; isobaric or homogeneous equations; equations homogeneous in xor y alone; equations having y = Ae x as a solution15.4 Exercises 52315.5 Hints and answers 52916 Series solutions of ordinary differential equations 53116.1 Second-order linear ordinary differential equations 531Ordinary and singular points16.2 Series solutions about an ordinary point 53516.3 Series solutions about a regular singular point 538Distinct roots not differing by an integer; repeated root of the indicialequation; distinct roots differing by an integer16.4 Obtaining a second solution 544The Wronskian method; the derivative method; series form of the secondsolution16.5 Polynomial solutions 54816.6 Exercises 55016.7 Hints and answers 55317 Eigenfunction methods for differential equations 55417.1 Sets of functions 556Some useful inequalities17.2 Adjoint, self-adjoint and Hermitian operators 55917.3 Properties of Hermitian operators 561Reality of the eigenvalues; orthogonality of the eigenfunctions; constructionof real eigenfunctions17.4 Sturm–Liouville equations 564Valid boundary conditions; putting an equation into Sturm–Liouville form17.5 Superposition of eigenfunctions: Green’s functions 56917.6 A useful generalisation 57217.7 Exercises 57317.8 Hints and answers 57618 Special functions 57718.1 Legendre functions 577General solution for integer l; properties of Legendre polynomials18.2 Associated Legendre functions 58718.3 Spherical harmonics 59318.4 Chebyshev functions 59518.5 Bessel functions 602General solution for non-integer ν; general solution for integer ν; propertiesof Bessel functionsxi

CONTENTS18.6 Spherical Bessel functions 61418.7 Laguerre functions 61618.8 Associated Laguerre functions 62118.9 Hermite functions 62418.10 Hypergeometric functions 62818.11 Confluent hypergeometric functions 63318.12 The gamma function and related functions 63518.13 Exercises 64018.14 Hints and answers 64619 Quantum operators 64819.1 Operator formalism 648Commutators19.2 Physical examples of operators 656Uncertainty principle; angular momentum; creation and annihilation operators19.3 Exercises 67119.4 Hints and answers 67420 Partial differential equations: general and particular solutions 67520.1 Important partial differential equations 676The wave equation; the diffusion equation; Laplace’s equation; Poisson’sequation; Schrödinger’s equation20.2 General form of solution 68020.3 General and particular solutions 681First-order equations; inhomogeneous equations and problems; second-orderequations20.4 The wave equation 69320.5 The diffusion equation 69520.6 Characteristics and the existence of solutions 699First-order equations; second-order equations20.7 Uniqueness of solutions 70520.8 Exercises 70720.9 Hints and answers 71121 Partial differential equations: separation of variablesand other methods 71321.1 Separation of variables: the general method 71321.2 Superposition of separated solutions 71721.3 Separation of variables in polar coordinates 725Laplace’s equation in polar coordinates; spherical harmonics; other equationsin polar coordinates; solution by expansion; separation of variables forinhomogeneous equations21.4 Integral transform methods 747xii

CONTENTS21.5 Inhomogeneous problems – Green’s functions 751Similarities to Green’s functions for ordinary differential equations; generalboundary-value problems; Dirichlet problems; Neumann problems21.6 Exercises 76721.7 Hints and answers 77322 Calculus of variations 77522.1 The Euler–Lagrange equation 77622.2 Special cases 777F does not contain y explicitly; F does not contain x explicitly22.3 Some extensions 781Several dependent variables; several independent variables; higher-orderderivatives; variable end-points22.4 Constrained variation 78522.5 Physical variational principles 787Fermat’s principle in optics; Hamilton’s principle in mechanics22.6 General eigenvalue problems 79022.7 Estimation of eigenvalues and eigenfunctions 79222.8 Adjustment of parameters 79522.9 Exercises 79722.10 Hints and answers 80123 Integral equations 80323.1 Obtaining an integral equation from a differential equation 80323.2 Types of integral equation 80423.3 Operator notation and the existence of solutions 80523.4 Closed-form solutions 806Separable kernels; integral transform methods; differentiation23.5 Neumann series 81323.6 Fredholm theory 81523.7 Schmidt–Hilbert theory 81623.8 Exercises 81923.9 Hints and answers 82324 Complex variables 82424.1 Functions of a complex variable 82524.2 The Cauchy–Riemann relations 82724.3 Power series in a complex variable 83024.4 Some elementary functions 83224.5 Multivalued functions and branch cuts 83524.6 Singularities and zeros of complex functions 83724.7 Conformal transformations 83924.8 Complex integrals 845xiii

CONTENTS24.9 Cauchy’s theorem 84924.10 Cauchy’s integral formula 85124.11 Taylor and Laurent series 85324.12 Residue theorem 85824.13 Definite integrals using contour integration 86124.14 Exercises 86724.15 Hints and answers 87025 Applications of complex variables 87125.1 Complex potentials 87125.2 Applications of conformal transformations 87625.3 Location of zeros 87925.4 Summation of series 88225.5 Inverse Laplace transform 88425.6 Stokes’ equation and Airy integrals 88825.7 WKB methods 89525.8 Approximations to integrals 905Level lines and saddle points; steepest descents; stationary phase25.9 Exercises 92025.10 Hints and answers 92526 Tensors 92726.1 Some notation 92826.2 Change of basis 92926.3 Cartesian tensors 93026.4 First- and zero-order Cartesian tensors 93226.5 Second- and higher-order Cartesian tensors 93526.6 The algebra of tensors 93826.7 The quotient law 93926.8 The tensors δ ij and ɛ ijk 94126.9 Isotropic tensors 94426.10 Improper rotations and pseudotensors 94626.11 Dual tensors 94926.12 Physical applications of tensors 95026.13 Integral theorems for tensors 95426.14 Non-Cartesian coordinates 95526.15 The metric tensor 95726.16 General coordinate transformations and tensors 96026.17 Relative tensors 96326.18 Derivatives of basis vectors and Christoffel symbols 96526.19 Covariant differentiation 96826.20 Vector operators in tensor form 971xiv

CONTENTS26.21 Absolute derivatives along curves 97526.22 Geodesics 97626.23 Exercises 97726.24 Hints and answers 98227 Numerical methods 98427.1 Algebraic and transcendental equations 985Rearrangement of the equation; linear interpolation; binary chopping;Newton–Raphson method27.2 Convergence of iteration schemes 99227.3 Simultaneous linear equations 994Gaussian elimination; Gauss–Seidel iteration; tridiagonal matrices27.4 Numerical integration 1000Trapezium rule; Simpson’s rule; Gaussian integration; Monte Carlo methods27.5 Finite differences 101927.6 Differential equations 1020Difference equations; Taylor series solutions; prediction and correction;Runge–Kutta methods; isoclines27.7 Higher-order equations 102827.8 Partial differential equations 103027.9 Exercises 103327.10 Hints and answers 103928 Group theory 104128.1 Groups 1041Definition of a group; examples of groups28.2 Finite groups 104928.3 Non-Abelian groups 105228.4 Permutation groups 105628.5 Mappings between groups 105928.6 Subgroups 106128.7 Subdividing a group 1063Equivalence relations and classes; congruence and cosets; conjugates andclasses28.8 Exercises 107028.9 Hints and answers 107429 Representation theory 107629.1 Dipole moments of molecules 107729.2 Choosing an appropriate formalism 107829.3 Equivalent representations 108429.4 Reducibility of a representation 108629.5 The orthogonality theorem for irreducible representations 1090xv

CONTENTS29.6 Characters 1092Orthogonality property of characters29.7 Counting irreps using characters 1095Summation rules for irreps29.8 Construction of a character table 110029.9 Group nomenclature 110229.10 Product representations 110329.11 Physical applications of group theory 1105Bonding in molecules; matrix elements in quantum mechanics; degeneracy ofnormal modes; breaking of degeneracies29.12 Exercises 111329.13 Hints and answers 111730 Probability 111930.1 Venn diagrams 111930.2 Probability 1124Axioms and theorems; conditional probability; Bayes’ theorem30.3 Permutations and combinations 113330.4 Random variables and distributions 1139Discrete random variables; continuous random variables30.5 Properties of distributions 1143Mean; mode and median; variance and standard deviation; moments; centralmoments30.6 Functions of random variables 115030.7 Generating functions 1157Probability generating functions; moment generating functions; characteristicfunctions; cumulant generating functions30.8 Important discrete distributions 1168Binomial; geometric; negative binomial; hypergeometric; Poisson30.9 Important continuous distributions 1179Gaussian; log-normal; exponential; gamma; chi-squared; Cauchy; Breit–Wigner; uniform30.10 The central limit theorem 119530.11 Joint distributions 1196Discrete bivariate; continuous bivariate; marginal and conditional distributions30.12 Properties of joint distributions 1199Means; variances; covariance and correlation30.13 Generating functions for joint distributions 120530.14 Transformation of variables in joint distributions 120630.15 Important joint distributions 1207Multinominal; multivariate Gaussian30.16 Exercises 121130.17 Hints and answers 1219xvi

CONTENTS31 Statistics 122131.1 Experiments, samples and populations 122131.2 Sample statistics 1222Averages; variance and standard deviation; moments; covariance and correlation31.3 Estimators and sampling distributions 1229Consistency, bias and efficiency; Fisher’s inequality; standard errors; confidencelimits31.4 Some basic estimators 1243Mean; variance; standard deviation; moments; covariance and correlation31.5 Maximum-likelihood method 1255ML estimator; transformation invariance and bias; efficiency; errors andconfidence limits; Bayesian interpretation; large-N behaviour; extendedML method31.6 The method of least squares 1271Linear least squares; non-linear least squares31.7 Hypothesis testing 1277Simple and composite hypotheses; statistical tests; Neyman–Pearson; generalisedlikelihood-ratio; Student’s t; Fisher’sF; goodness of fit31.8 Exercises 129831.9 Hints and answers 1303Index 1305xvii

CONTENTSI am the very Model for a Student MathematicalI am the very model for a student mathematical;I’ve information rational, and logical and practical.I know the laws of algebra, and find them quite symmetrical,And even know the meaning of ‘a variate antithetical’.I’m extremely well acquainted, with all things mathematical.I understand equations, both the simple and quadratical.About binomial theorems I’m teeming with a lot o’news,With many cheerful facts about the square of the hypotenuse.I’m very good at integral and differential calculus,And solving paradoxes that so often seem to rankle us.In short in matters rational, and logical and practical,I am the very model for a student mathematical.I know the singularities of equations differential,And some of these are regular, but the rest are quite essential.I quote the results of giants; with Euler, Newton, Gauss, Laplace,And can calculate an orbit, given a centre, force and mass.I can reconstruct equations, both canonical and formal,And write all kinds of matrices, orthogonal, real and normal.I show how to tackle problems that one has never met before,By analogy or example, or with some clever metaphor.I seldom use equivalence to help decide upon a class,But often find an integral, using a contour o’er a pass.In short in matters rational, and logical and practical,I am the very model for a student mathematical.When you have learnt just what is meant by ‘Jacobian’ and ‘Abelian’;When you at sight can estimate, for the modal, mean and median;When describing normal subgroups is much more than recitation;When you understand precisely what is ‘quantum excitation’;When you know enough statistics that you can recognise RV;When you have learnt all advances that have been made in SVD;And when you can spot the transform that solves some tricky PDE,You will feel no better student has ever sat for a degree.Your accumulated knowledge, whilst extensive and exemplary,Will have only been brought down to the beginning of last century,But still in matters rational, and logical and practical,You’ll be the very model of a student mathematical.xixKFR, with apologies to W.S. Gilbert

Preface to the third editionAs is natural, in the four years since the publication of the second edition ofthis book we have somewhat modified our views on what should be includedand how it should be presented. In this new edition, although the range of topicscovered has been extended, there has been no significant shift in the general levelof difficulty or in the degree of mathematical sophistication required. Further, wehave aimed to preserve the same style of presentation as seems to have been wellreceived in the first two editions. However, a significant change has been madeto the format of the chapters, specifically to the way that the exercises, togetherwith their hints and answers, have been treated; the details of the change areexplained below.The two major chapters that are new in this third edition are those dealing with‘special functions’ and the applications of complex variables. The former presentsa systematic account of those functions that appear to have arisen in a moreor less haphazard way as a result of studying particular physical situations, andare deemed ‘special’ for that reason. The treatment presented here shows that,in fact, they are nearly all particular cases of the hypergeometric or confluenthypergeometric functions, and are special only in the sense that the parametersof the relevant function take simple or related values.The second new chapter describes how the properties of complex variables canbe used to tackle problems arising from the description of physical situationsor from other seemingly unrelated areas of mathematics. To topics treated inearlier editions, such as the solution of Laplace’s equation in two dimensions, thesummation of series, the location of zeros of polynomials and the calculation ofinverse Laplace transforms, has been added new material covering Airy integrals,saddle-point methods for contour integral evaluation, and the WKB approach toasymptotic forms.Other new material includes a stand-alone chapter on the use of coordinate-freeoperators to establish valuable results in the field of quantum mechanics; amongstxx

PREFACE TO THE THIRD EDITIONthe physical topics covered are angular momentum and uncertainty principles.There are also significant additions to the treatment of numerical integration.In particular, Gaussian quadrature based on Legendre, Laguerre, Hermite andChebyshev polynomials is discussed, and appropriate tables of points and weightsare provided.We now turn to the most obvious change to the format of the book, namelythe way that the exercises, hints and answers are treated. The second edition ofMathematical Methods for Physics and Engineering carried more than twice asmany exercises, based on its various chapters, as did the first. In its preface wediscussed the general question of how such exercises should be treated but, inthe end, decided to provide hints and outline answers to all problems, as in thefirst edition. This decision was an uneasy one as, on the one hand, it did notallow the exercises to be set as totally unaided homework that could be used forassessment purposes but, on the other, it did not give a full explanation of howto tackle a problem when a student needed explicit guidance or a model answer.In order to allow both of these educationally desirable goals to be achieved,we have, in this third edition, completely changed the way in which this matteris handled. A large number of exercises have been included in the penultimatesubsections of the appropriate, sometimes reorganised, chapters. Hints and outlineanswers are given, as previously, in the final subsections, but only for the oddnumberedexercises. This leaves all even-numbered exercises free to be set asunaided homework, as described below.For the four hundred plus odd-numbered exercises, complete solutions areavailable, to both students and their teachers, in the form of a separate manual,Student Solutions Manual for Mathematical Methods for Physics and Engineering(Cambridge: Cambridge University Press, 2006); the hints and outline answersgiven in this main text are brief summaries of the model answers given in themanual. There, each original exercise is reproduced and followed by a fullyworked solution. For those original exercises that make internal reference to thistext or to other (even-numbered) exercises not included in the solutions manual,the questions have been reworded, usually by including additional information,so that the questions can stand alone.In many cases, the solution given in the manual is even fuller than one thatmight be expected of a good student that has understood the material. This isbecause we have aimed to make the solutions instructional as well as utilitarian.To this end, we have included comments that are intended to show how theplan for the solution is fomulated and have given the justifications for particularintermediate steps (something not always done, even by the best of students). Wehave also tried to write each individual substituted formula in the form that bestindicates how it was obtained, before simplifying it at the next or a subsequentstage. Where several lines of algebraic manipulation or calculus are needed toobtain a final result, they are normally included in full; this should enable thexxi

PREFACE TO THE THIRD EDITIONstudent to determine whether an incorrect answer is due to a misunderstandingof principles or to a technical error.The remaining four hundred or so even-numbered exercises have no hints oranswers, outlined or detailed, available for general access. They can therefore beused by instructors as a basis for setting unaided homework. Full solutions tothese exercises, in the same general format as those appearing in the manual(though they may contain references to the main text or to other exercises), areavailable without charge to accredited teachers as downloadable pdf files on thepassword-protected website http://www.cambridge.org/9780521679718. Teacherswishing to have access to the website should contact solutions@cambridge.orgfor registration details.In all new publications, errors and typographical mistakes are virtually unavoidable,and we would be grateful to any reader who brings instances toour attention. Retrospectively, we would like to record our thanks to ReinhardGerndt, Paul Renteln and Joe Tenn for making us aware of some errors inthe second edition. Finally, we are extremely grateful to Dave Green for hisconsiderable and continuing advice concerning L A TEX.Ken Riley, Michael Hobson,Cambridge, 2006xxii

Preface to the second editionSince the publication of the first edition of this book, both through teaching thematerial it covers and as a result of receiving helpful comments from colleagues,we have become aware of the desirability of changes in a number of areas.The most important of these is that the mathematical preparation of currentsenior college and university entrants is now less thorough than it used to be.To match this, we decided to include a preliminary chapter covering areas suchas polynomial equations, trigonometric identities, coordinate geometry, partialfractions, binomial expansions, necessary and sufficient condition and proof byinduction and contradiction.Whilst the general level of what is included in this second edition has notbeen raised, some areas have been expanded to take in topics we now feel werenot adequately covered in the first. In particular, increased attention has beengiven to non-square sets of simultaneous linear equations and their associatedmatrices. We hope that this more extended treatment, together with the inclusionof singular value matrix decomposition, will make the material of more practicaluse to engineering students. In the same spirit, an elementary treatment of linearrecurrence relations has been included. The topic of normal modes has been givena small chapter of its own, though the links to matrices on the one hand, and torepresentation theory on the other, have not been lost.Elsewhere, the presentation of probability and statistics has been reorganised togive the two aspects more nearly equal weights. The early part of the probabilitychapter has been rewritten in order to present a more coherent developmentbased on Boolean algebra, the fundamental axioms of probability theory andthe properties of intersections and unions. Whilst this is somewhat more formalthan previously, we think that it has not reduced the accessibility of these topicsand hope that it has increased it. The scope of the chapter has been somewhatextended to include all physically important distributions and an introduction tocumulants.xxiii

PREFACE TO THE SECOND EDITIONStatistics now occupies a substantial chapter of its own, one that includes systematicdiscussions of estimators and their efficiency, sample distributions and t-and F-tests for comparing means and variances. Other new topics are applicationsof the chi-squared distribution, maximum-likelihood parameter estimation andleast-squares fitting. In other chapters we have added material on the followingtopics: curvature, envelopes, curve-sketching, more refined numerical methodsfor differential equations and the elements of integration using Monte Carlotechniques.Over the last four years we have received somewhat mixed feedback aboutthe number of exercises at the ends of the various chapters. After consideration,we decided to increase the number substantially, partly to correspond to theadditional topics covered in the text but mainly to give both students andtheir teachers a wider choice. There are now nearly 800 such exercises, many withseveral parts. An even more vexed question has been whether to provide hints andanswers to all the exercises or just to ‘the odd-numbered’ ones, as is the normalpractice for textbooks in the United States, thus making the remainder moresuitable for setting as homework. In the end, we decided that hints and outlinesolutions should be provided for all the exercises, in order to facilitate independentstudy while leaving the details of the calculation as a task for the student.In conclusion, we hope that this edition will be thought by its users to be‘heading in the right direction’ and would like to place on record our thanks toall who have helped to bring about the changes and adjustments. Naturally, thosecolleagues who have noted errors or ambiguities in the first edition and broughtthem to our attention figure high on the list, as do the staff at The CambridgeUniversity Press. In particular, we are grateful to Dave Green for continued L A TEXadvice, Susan Parkinson for copy-editing the second edition with her usual keeneye for detail and flair for crafting coherent prose and Alison Woollatt for onceagain turning our basic L A TEX into a beautifully typeset book. Our thanks goto all of them, though of course we accept full responsibility for any remainingerrors or ambiguities, of which, as with any new publication, there are bound tobe some.On a more personal note, KFR again wishes to thank his wife Penny for herunwavering support, not only in his academic and tutorial work, but also in theirjoint efforts to convert time at the bridge table into ‘green points’ on their record.MPH is once more indebted to his wife, Becky, and his mother, Pat, for theirtireless support and encouragement above and beyond the call of duty. MPHdedicates his contribution to this book to the memory of his father, RonaldLeonard Hobson, whose gentle kindness, patient understanding and unbreakablespirit made all things seem possible.Ken Riley, Michael HobsonCambridge, 2002xxiv

Preface to the first editionA knowledge of mathematical methods is important for an increasing number ofuniversity and college courses, particularly in physics, engineering and chemistry,but also in more general science. Students embarking on such courses come fromdiverse mathematical backgrounds, and their core knowledge varies considerably.We have therefore decided to write a textbook that assumes knowledge only ofmaterial that can be expected to be familiar to all the current generation ofstudents starting physical science courses at university. In the United Kingdomthis corresponds to the standard of Mathematics A-level, whereas in the UnitedStates the material assumed is that which would normally be covered at juniorcollege.Starting from this level, the first six chapters cover a collection of topicswith which the reader may already be familiar, but which are here extendedand applied to typical problems encountered by first-year university students.They are aimed at providing a common base of general techniques used inthe development of the remaining chapters. Students who have had additionalpreparation, such as Further Mathematics at A-level, will find much of thismaterial straightforward.Following these opening chapters, the remainder of the book is intended tocover at least that mathematical material which an undergraduate in the physicalsciences might encounter up to the end of his or her course. The book is alsoappropriate for those beginning graduate study with a mathematical content, andnaturally much of the material forms parts of courses for mathematics students.Furthermore, the text should provide a useful reference for research workers.The general aim of the book is to present a topic in three stages. The firststage is a qualitative introduction, wherever possible from a physical point ofview. The second is a more formal presentation, although we have deliberatelyavoided strictly mathematical questions such as the existence of limits, uniformconvergence, the interchanging of integration and summation orders, etc. on thexxv

PREFACE TO THE FIRST EDITIONgrounds that ‘this is the real world; it must behave reasonably’. Finally a workedexample is presented, often drawn from familiar situations in physical scienceand engineering. These examples have generally been fully worked, since, inthe authors’ experience, partially worked examples are unpopular with students.Only in a few cases, where trivial algebraic manipulation is involved, or whererepetition of the main text would result, has an example been left as an exercisefor the reader. Nevertheless, a number of exercises also appear at the end of eachchapter, and these should give the reader ample opportunity to test his or herunderstanding. Hints and answers to these exercises are also provided.With regard to the presentation of the mathematics, it has to be accepted thatmany equations (especially partial differential equations) can be written morecompactly by using subscripts, e.g. u xy for a second partial derivative, instead ofthe more familiar ∂ 2 u/∂x∂y, and that this certainly saves typographical space.However, for many students, the labour of mentally unpacking such equationsis sufficiently great that it is not possible to think of an equation’s physicalinterpretation at the same time. Consequently, wherever possible we have decidedto write out such expressions in their more obvious but longer form.During the writing of this book we have received much help and encouragementfrom various colleagues at the Cavendish Laboratory, Clare College, Trinity Halland Peterhouse. In particular, we would like to thank Peter Scheuer, whosecomments and general enthusiasm proved invaluable in the early stages. Forreading sections of the manuscript, for pointing out misprints and for numeroususeful comments, we thank many of our students and colleagues at the Universityof Cambridge. We are especially grateful to Chris Doran, John Huber, GarthLeder, Tom Körner and, not least, Mike Stobbs, who, sadly, died before the bookwas completed. We also extend our thanks to the University of Cambridge andthe Cavendish teaching staff, whose examination questions and lecture hand-outshave collectively provided the basis for some of the examples included. Of course,any errors and ambiguities remaining are entirely the responsibility of the authors,and we would be most grateful to have them brought to our attention.We are indebted to Dave Green for a great deal of advice concerning typesettingin L A TEX and to Andrew Lovatt for various other computing tips. Our thanksalso go to Anja Visser and Graça Rocha for enduring many hours of (sometimesheated) debate. At Cambridge University Press, we are very grateful to our editorAdam Black for his help and patience and to Alison Woollatt for her experttypesetting of such a complicated text. We also thank our copy-editor SusanParkinson for many useful suggestions that have undoubtedly improved the styleof the book.Finally, on a personal note, KFR wishes to thank his wife Penny, not only fora long and happy marriage, but also for her support and understanding duringhis recent illness – and when things have not gone too well at the bridge table!MPH is indebted both to Rebecca Morris and to his parents for their tirelessxxvi

PREFACE TO THE FIRST EDITIONsupport and patience, and for their unending supplies of tea. SJB is grateful toAnthony Gritten for numerous relaxing discussions about J. S. Bach, to SusannahTicciati for her patience and understanding, and to Kate Isaak for her calminglate-night e-mails from the USA.Ken Riley, Michael Hobson and Stephen BenceCambridge, 1997xxvii

1Preliminary algebraThis opening chapter reviews the basic algebra of which a working knowledge ispresumed in the rest of the book. Many students will be familiar with much, ifnot all, of it, but recent changes in what is studied during secondary educationmean that it cannot be taken for granted that they will already have a masteryof all the topics presented here. The reader may assess which areas need furtherstudy or revision by attempting the exercises at the end of the chapter. The mainareas covered are polynomial equations and the related topic of partial fractions,curve sketching, coordinate geometry, trigonometric identities and the notions ofproof by induction or contradiction.1.1 Simple functions and equationsIt is normal practice when starting the mathematical investigation of a physicalproblem to assign an algebraic symbol to the quantity whose value is sought, eithernumerically or as an explicit algebraic expression. For the sake of definiteness, inthis chapter we will use x to denote this quantity most of the time. Subsequentsteps in the analysis involve applying a combination of known laws, consistencyconditions and (possibly) given constraints to derive one or more equationssatisfied by x. These equations may take many forms, ranging from a simplepolynomial equation to, say, a partial differential equation with several boundaryconditions. Some of the more complicated possibilities are treated in the laterchapters of this book, but for the present we will be concerned with techniquesfor the solution of relatively straightforward algebraic equations.1.1.1 Polynomials and polynomial equationsFirstly we consider the simplest type of equation, a polynomial equation, inwhicha polynomial expression in x, denoted by f(x), is set equal to zero and thereby1

PRELIMINARY ALGEBRAforms an equation which is satisfied by particular values of x, called the roots ofthe equation:f(x) =a n x n + a n−1 x n−1 + ···+ a 1 x + a 0 =0. (1.1)Here n is an integer > 0, called the degree of both the polynomial and theequation, and the known coefficients a 0 ,a 1 ,...,a n are real quantities with a n ≠0.Equations such as (1.1) arise frequently in physical problems, the coefficients a ibeing determined by the physical properties of the system under study. What isneeded is to find some or all of the roots of (1.1), i.e. the x-values, α k ,thatsatisfyf(α k )=0;herek is an index that, as we shall see later, can take up to n differentvalues, i.e. k =1, 2,...,n. The roots of the polynomial equation can equally wellbe described as the zeros of the polynomial. When they are real, they correspondto the points at which a graph of f(x) crosses the x-axis. Roots that are complex(see chapter 3) do not have such a graphical interpretation.For polynomial equations containing powers of x greater than x 4 generalmethods do not exist for obtaining explicit expressions for the roots α k . Evenfor n = 3 and n = 4 the prescriptions for obtaining the roots are sufficientlycomplicated that it is usually preferable to obtain exact or approximate valuesby other methods. Only for n = 1 and n = 2 can closed-form solutions be given.These results will be well known to the reader, but they are given here for thesake of completeness. For n = 1, (1.1) reduces to the linear equationa 1 x + a 0 = 0; (1.2)the solution (root) is α 1 = −a 0 /a 1 .Forn = 2, (1.1) reduces to the quadraticequationa 2 x 2 + a 1 x + a 0 = 0; (1.3)the two roots α 1 and α 2 are given by√α 1,2 = −a 1 ± a 2 1 − 4a 2a 0. (1.4)2a 2When discussing specifically quadratic equations, as opposed to more generalpolynomial equations, it is usual to write the equation in one of the two notationsax 2 + bx + c =0, ax 2 +2bx + c =0, (1.5)with respective explicit pairs of solutionsα 1,2 = −b ± √ b 2 − 4ac2a, α 1,2 = −b ± √ b 2 − ac. (1.6)aOf course, these two notations are entirely equivalent and the only importantpoint is to associate each form of answer with the corresponding form of equation;most people keep to one form, to avoid any possible confusion.2

1.1 SIMPLE FUNCTIONS AND EQUATIONSIf the value of the quantity appearing under the square root sign is positivethen both roots are real; if it is negative then the roots form a complex conjugatepair, i.e. they are of the form p ± iq with p and q real (see chapter 3); if it haszero value then the two roots are equal and special considerations usually arise.Thus linear and quadratic equations can be dealt with in a cut-and-dried way.We now turn to methods for obtaining partial information about the roots ofhigher-degree polynomial equations. In some circumstances the knowledge thatan equation has a root lying in a certain range, or that it has no real roots at all,is all that is actually required. For example, in the design of electronic circuitsit is necessary to know whether the current in a proposed circuit will breakinto spontaneous oscillation. To test this, it is sufficient to establish whether acertain polynomial equation, whose coefficients are determined by the physicalparameters of the circuit, has a root with a positive real part (see chapter 3);complete determination of all the roots is not needed for this purpose. If thecomplete set of roots of a polynomial equation is required, it can usually beobtained to any desired accuracy by numerical methods such as those describedin chapter 27.There is no explicit step-by-step approach to finding the roots of a generalpolynomial equation such as (1.1). In most cases analytic methods yield onlyinformation about the roots, rather than their exact values. To explain the relevanttechniques we will consider a particular example, ‘thinking aloud’ on paper andexpanding on special points about methods and lines of reasoning. In moreroutine situations such comment would be absent and the whole process brieferand more tightly focussed.Example: the cubic caseLet us investigate the roots of the equationg(x) =4x 3 +3x 2 − 6x − 1 = 0 (1.7)or, in an alternative phrasing, investigate the zeros of g(x).Wenotefirstofallthat this is a cubic equation. It can be seen that for x large and positive g(x)will be large and positive and, equally, that for x large and negative g(x) willbe large and negative. Therefore, intuitively (or, more formally, by continuity)g(x) must cross the x-axis at least once and so g(x) = 0 must have at least onereal root. Furthermore, it can be shown that if f(x) isannth-degree polynomialthen the graph of f(x) must cross the x-axis an even or odd number of timesas x varies between −∞ and +∞, according to whether n itself is even or odd.Thus a polynomial of odd degree always has at least one real root, but one ofeven degree may have no real root. A small complication, discussed later in thissection, occurs when repeated roots arise.Having established that g(x) = 0 has at least one real root, we may ask how3

PRELIMINARY ALGEBRAmany real roots it could have. To answer this we need one of the fundamentaltheorems of algebra, mentioned above:An nth-degree polynomial equation has exactly n roots.It should be noted that this does not imply that there are n real roots (only thatthere are not more than n); some of the roots may be of the form p + iq.To make the above theorem plausible and to see what is meant by repeatedroots, let us suppose that the nth-degree polynomial equation f(x) = 0, (1.1), hasr roots α 1 ,α 2 ,...,α r , considered distinct for the moment. That is, we suppose thatf(α k )=0fork =1, 2,...,r,sothatf(x) vanishes only when x is equal to one ofthe r values α k . But the same can be said for the functionF(x) =A(x − α 1 )(x − α 2 ) ···(x − α r ), (1.8)in which A is a non-zero constant; F(x) can clearly be multiplied out to form apolynomial expression.We now call upon a second fundamental result in algebra: that if two polynomialfunctions f(x) andF(x) have equal values for all values of x, then theircoefficients are equal on a term-by-term basis. In other words, we can equatethe coefficients of each and every power of x in the two expressions (1.8) and(1.1); in particular we can equate the coefficients of the highest power of x. Fromthis we have Ax r ≡ a n x n and thus that r = n and A = a n .Asr is both equalto n and to the number of roots of f(x) = 0, we conclude that the nth-degreepolynomial f(x) =0hasn roots. (Although this line of reasoning may make thetheorem plausible, it does not constitute a proof since we have not shown that itis permissible to write f(x) in the form of equation (1.8).)We next note that the condition f(α k )=0fork =1, 2,...,r, could also be metif (1.8) were replaced byF(x) =A(x − α 1 ) m1 (x − α 2 ) m2 ···(x − α r ) mr , (1.9)with A = a n . In (1.9) the m k are integers ≥ 1 and are known as the multiplicitiesof the roots, m k being the multiplicity of α k . Expanding the right-hand side (RHS)leads to a polynomial of degree m 1 + m 2 + ···+ m r . This sum must be equal to n.Thus, if any of the m k is greater than unity then the number of distinct roots, r,is less than n; the total number of roots remains at n, but one or more of the α kcounts more than once. For example, the equationF(x) =A(x − α 1 ) 2 (x − α 2 ) 3 (x − α 3 )(x − α 4 )=0has exactly seven roots, α 1 being a double root and α 2 a triple root, whilst α 3 andα 4 are unrepeated (simple) roots.We can now say that our particular equation (1.7) has either one or three realroots but in the latter case it may be that not all the roots are distinct. To decidehow many real roots the equation has, we need to anticipate two ideas from the4

1.1 SIMPLE FUNCTIONS AND EQUATIONSφ 1 (x)φ 2 (x)β 2xβ 1 β 1β 2xFigure 1.1 Two curves φ 1 (x) andφ 2 (x), both with zero derivatives at thesame values of x, but with different numbers of real solutions to φ i (x) =0.next chapter. The first of these is the notion of the derivative of a function, andthe second is a result known as Rolle’s theorem.The derivative f ′ (x) of a function f(x) measures the slope of the tangent tothe graph of f(x) at that value of x (see figure 2.1 in the next chapter). Forthe moment, the reader with no prior knowledge of calculus is asked to acceptthat the derivative of ax n is nax n−1 , so that the derivative g ′ (x) ofthecurveg(x) =4x 3 +3x 2 − 6x − 1 is given by g ′ (x) =12x 2 +6x − 6. Similar expressionsfor the derivatives of other polynomials are used later in this chapter.Rolle’s theorem states that if f(x) has equal values at two different values of xthen at some point between these two x-values its derivative is equal to zero; i.e.the tangent to its graph is parallel to the x-axis at that point (see figure 2.2).Having briefly mentioned the derivative of a function and Rolle’s theorem, wenow use them to establish whether g(x) has one or three real zeros. If g(x) =0does have three real roots α k ,i.e.g(α k )=0fork =1, 2, 3, then it follows fromRolle’s theorem that between any consecutive pair of them (say α 1 and α 2 )theremust be some real value of x at which g ′ (x) = 0. Similarly, there must be a furtherzero of g ′ (x) lying between α 2 and α 3 . Thus a necessary condition for three realroots of g(x) = 0 is that g ′ (x) = 0 itself has two real roots.However, this condition on the number of roots of g ′ (x) = 0, whilst necessary,is not sufficient to guarantee three real roots of g(x) = 0. This can be seen byinspecting the cubic curves in figure 1.1. For each of the two functions φ 1 (x) andφ 2 (x), the derivative is equal to zero at both x = β 1 and x = β 2 . Clearly, though,φ 2 (x) = 0 has three real roots whilst φ 1 (x) = 0 has only one. It is easy to see thatthe crucial difference is that φ 1 (β 1 )andφ 1 (β 2 ) have the same sign, whilst φ 2 (β 1 )and φ 2 (β 2 ) have opposite signs.It will be apparent that for some equations, φ(x) =0say,φ ′ (x) equals zero5

PRELIMINARY ALGEBRAat a value of x for which φ(x) is also zero. Then the graph of φ(x) justtouchesthe x-axis. When this happens the value of x so found is, in fact, a double realroot of the polynomial equation (corresponding to one of the m k in (1.9) havingthe value 2) and must be counted twice when determining the number of realroots.Finally, then, we are in a position to decide the number of real roots of theequationg(x) =4x 3 +3x 2 − 6x − 1=0.The equation g ′ (x) =0,withg ′ (x) =12x 2 +6x − 6, is a quadratic equation withexplicit solutions § β 1,2 = −3 ± √ 9+7212,so that β 1 = −1 andβ 2 = 1 2 . The corresponding values of g(x) areg(β 1)=4andg(β 2 )=− 114 , which are of opposite sign. This indicates that 4x3 +3x 2 −6x−1 =0has three real roots, one lying in the range −1

1.1 SIMPLE FUNCTIONS AND EQUATIONS(iii) The equation f ′ (x) = 0 can be written as x(7x 5 +30x 4 +4x 2 − 3x +2)=0and thus x = 0 is a root. The derivative of f ′ (x), denoted by f ′′ (x), equals42x 5 + 150x 4 +12x 2 − 6x +2. That f ′ (x) is zero whilst f ′′ (x) is positiveat x = 0 indicates (subsection 2.1.8) that f(x) has a minimum there. This,together with the facts that f(0) is negative and f(∞) =∞, implies thatthe total number of real roots to the right of x = 0 must be odd. Sincethe total number of real roots must be odd, the number to the left mustbe even (0, 2, 4 or 6).This is about all that can be deduced by simple analytic methods in this case,although some further progress can be made in the ways indicated in exercise 1.3.There are, in fact, more sophisticated tests that examine the relative signs ofsuccessive terms in an equation such as (1.1), and in quantities derived fromthem, to place limits on the numbers and positions of roots. But they are notprerequisites for the remainder of this book and will not be pursued furtherhere.We conclude this section with a worked example which demonstrates that thepractical application of the ideas developed so far can be both short and decisive.◮For what values of k, ifany,doesf(x) =x 3 − 3x 2 +6x + k =0have three real roots?Firstly we study the equation f ′ (x) =0,i.e.3x 2 − 6x + 6 = 0. This is a quadratic equationbut, using (1.6), because 6 2 < 4 × 3 × 6, it can have no real roots. Therefore, it followsimmediately that f(x) has no maximum or minimum; consequently f(x) = 0 cannot havemore than one real root, whatever the value of k. ◭1.1.2 Factorising polynomialsIn the previous subsection we saw how a polynomial with r given distinct zerosα k could be constructed as the product of factors containing those zeros:f(x) =a n (x − α 1 ) m1 (x − α 2 ) m2 ···(x − α r ) mr= a n x n + a n−1 x n−1 + ···+ a 1 x + a 0 , (1.10)with m 1 + m 2 + ···+ m r = n, the degree of the polynomial. It will cause no loss ofgenerality in what follows to suppose that all the zeros are simple, i.e. all m k =1and r = n, and this we will do.Sometimes it is desirable to be able to reverse this process, in particular whenone exact zero has been found by some method and the remaining zeros are tobe investigated. Suppose that we have located one zero, α; it is then possible towrite (1.10) asf(x) =(x − α)f 1 (x), (1.11)7

PRELIMINARY ALGEBRAwhere f 1 (x) is a polynomial of degree n−1. How can we find f 1 (x)? The procedureis much more complicated to describe in a general form than to carry out foran equation with given numerical coefficients a i . If such manipulations are toocomplicated to be carried out mentally, they could be laid out along the lines ofan algebraic ‘long division’ sum. However, a more compact form of calculationis as follows. Write f 1 (x) asf 1 (x) =b n−1 x n−1 + b n−2 x n−2 + b n−3 x n−3 + ···+ b 1 x + b 0 .Substitution of this form into (1.11) and subsequent comparison of the coefficientsof x p for p = n, n − 1, ..., 1, 0 with those in the second line of (1.10) generatesthe series of equationsb n−1 = a n ,b n−2 − αb n−1 = a n−1 ,b n−3 − αb n−2 = a n−2 ,..b 0 − αb 1 = a 1 ,−αb 0 = a 0 .These can be solved successively for the b j , starting either from the top or fromthe bottom of the series. In either case the final equation used serves as a check;if it is not satisfied, at least one mistake has been made in the computation –or α is not a zero of f(x) = 0. We now illustrate this procedure with a workedexample.◮Determine by inspection the simple roots of the equationf(x) =3x 4 − x 3 − 10x 2 − 2x +4=0and hence, by factorisation, find the rest of its roots.From the pattern of coefficients it can be seen that x = −1 is a solution to the equation.We therefore writef(x) =(x +1)(b 3 x 3 + b 2 x 2 + b 1 x + b 0 ),whereb 3 =3,b 2 + b 3 = −1,b 1 + b 2 = −10,b 0 + b 1 = −2,b 0 =4.These equations give b 3 =3,b 2 = −4,b 1 = −6,b 0 = 4 (check) and sof(x) =(x +1)f 1 (x) =(x + 1)(3x 3 − 4x 2 − 6x +4).8

1.1 SIMPLE FUNCTIONS AND EQUATIONSWe now note that f 1 (x) =0ifx is set equal to 2. Thus x − 2isafactoroff 1 (x), whichtherefore can be written asf 1 (x) =(x − 2)f 2 (x) =(x − 2)(c 2 x 2 + c 1 x + c 0 )withc 2 =3,c 1 − 2c 2 = −4,c 0 − 2c 1 = −6,−2c 0 =4.These equations determine f 2 (x) as3x 2 +2x − 2. Since f 2 (x) = 0 is a quadratic equation,its solutions can be written explicitly asx = −1 ± √ 1+6.3Thus the four roots of f(x) =0are−1, 2, 1 (−1+√ 7) and 1 (−1 − √ 7). ◭3 31.1.3 Properties of rootsFrom the fact that a polynomial equation can be written in any of the alternativeformsf(x) =a n x n + a n−1 x n−1 + ···+ a 1 x + a 0 =0,f(x) =a n (x − α 1 ) m1 (x − α 2 ) m2 ···(x − α r ) mr =0,f(x) =a n (x − α 1 )(x − α 2 ) ···(x − α n )=0,it follows that it must be possible to express the coefficients a i in terms of theroots α k . To take the most obvious example, comparison of the constant terms(formally the coefficient of x 0 ) in the first and third expressions shows thata n (−α 1 )(−α 2 ) ···(−α n )=a 0 ,or, using the product notation,n∏α k =(−1) n a 0. (1.12)a nk=1Only slightly less obvious is a result obtained by comparing the coefficients ofx n−1 in the same two expressions of the polynomial:n∑α k = − a n−1. (1.13)a nk=1Comparing the coefficients of other powers of x yields further results, thoughthey are of less general use than the two just given. One such, which the readermay wish to derive, isn∑n∑j=1 k>jα j α k = a n−2a n. (1.14)9

PRELIMINARY ALGEBRAIn the case of a quadratic equation these root properties are used sufficientlyoften that they are worth stating explicitly, as follows. If the roots of the quadraticequation ax 2 + bx + c = 0 are α 1 and α 2 thenα 1 + α 2 = − b a ,α 1 α 2 = c a .If the alternative standard form for the quadratic is used, b is replaced by 2b inboth the equation and the first of these results.◮Find a cubic equation whose roots are −4, 3 and 5.From results (1.12) – (1.14) we can compute that, arbitrarily setting a 3 =1,3∑3∑ 3∑3∏−a 2 = α k =4, a 1 = α j α k = −17, a 0 =(−1) 3 α k =60.k=1j=1 k>jThus a possible cubic equation is x 3 +(−4)x 2 +(−17)x + (60) = 0. Of course, any multipleof x 3 − 4x 2 − 17x + 60 = 0 will do just as well. ◭k=11.2 Trigonometric identitiesSo many of the applications of mathematics to physics and engineering areconcerned with periodic, and in particular sinusoidal, behaviour that a sure andready handling of the corresponding mathematical functions is an essential skill.Even situations with no obvious periodicity are often expressed in terms ofperiodic functions for the purposes of analysis. Later in this book whole chaptersare devoted to developing the techniques involved, but as a necessary prerequisitewe here establish (or remind the reader of) some standard identities with which heor she should be fully familiar, so that the manipulation of expressions containingsinusoids becomes automatic and reliable. So as to emphasise the angular natureof the argument of a sinusoid we will denote it in this section by θ rather than x.1.2.1 Single-angle identitiesWe give without proof the basic identity satisfied by the sinusoidal functions sin θand cos θ, namelycos 2 θ +sin 2 θ =1. (1.15)If sin θ and cos θ have been defined geometrically in terms of the coordinates ofa point on a circle, a reference to the name of Pythagoras will suffice to establishthis result. If they have been defined by means of series (with θ expressed inradians) then the reader should refer to Euler’s equation (3.23) on page 93, andnote that e iθ has unit modulus if θ is real.10

1.2 TRIGONOMETRIC IDENTITIESy ′yRPMTNBx ′OAxFigure 1.2 Illustration of the compound-angle identities. Refer to the maintext for details.Other standard single-angle formulae derived from (1.15) by dividing throughby various powers of sin θ and cos θ are1+tan 2 θ =sec 2 θ, (1.16)cot 2 θ +1=cosec 2 θ. (1.17)1.2.2 Compound-angle identitiesThe basis for building expressions for the sinusoidal functions of compoundangles are those for the sum and difference of just two angles, since all othercases can be built up from these, in principle. Later we will see that a study ofcomplex numbers can provide a more efficient approach in some cases.To prove the basic formulae for the sine and cosine of a compound angleA + B in terms of the sines and cosines of A and B, we consider the constructionshown in figure 1.2. It shows two sets of axes, Oxy and Ox ′ y ′ , with a commonorigin but rotated with respect to each other through an angle A. The pointP lies on the unit circle centred on the common origin O and has coordinatescos(A + B), sin(A + B) with respect to the axes Oxy and coordinates cos B,sin Bwith respect to the axes Ox ′ y ′ .Parallels to the axes Oxy (dotted lines) and Ox ′ y ′ (broken lines) have beendrawn through P . Further parallels (MR and RN) totheOx ′ y ′ axes have been11

PRELIMINARY ALGEBRAdrawn through R, the point (0, sin(A + B)) in the Oxy system. That all the anglesmarked with the symbol • are equal to A follows from the simple geometry ofright-angled triangles and crossing lines.We now determine the coordinates of P in terms of lengths in the figure,expressing those lengths in terms of both sets of coordinates:(i) cos B = x ′ = TN + NP = MR + NP= OR sin A + RP cos A = sin(A + B)sinA +cos(A + B)cosA;(ii) sin B = y ′ = OM − TM = OM − NR= OR cos A − RP sin A = sin(A + B)cosA − cos(A + B)sinA.Now, if equation (i) is multiplied by sin A and added to equation (ii) multipliedby cos A, the result issin A cos B +cosA sin B = sin(A + B)(sin 2 A +cos 2 A)=sin(A + B).Similarly, if equation (ii) is multiplied by sin A and subtracted from equation (i)multiplied by cos A, the result iscos A cos B − sin A sin B =cos(A + B)(cos 2 A +sin 2 A)=cos(A + B).Corresponding graphically based results can be derived for the sines and cosinesof the difference of two angles; however, they are more easily obtained by settingB to −B in the previous results and remembering that sin B becomes − sin Bwhilst cos B is unchanged. The four results may be summarised bysin(A ± B) =sinA cos B ± cos A sin B (1.18)cos(A ± B) =cosA cos B ∓ sin A sin B. (1.19)Standard results can be deduced from these by setting one of the two anglesequal to π or to π/2:sin(π − θ) =sinθ, cos(π − θ) =− cos θ, (1.20)sin ( 12 π − θ) =cosθ, cos ( 12 π − θ) =sinθ, (1.21)From these basic results many more can be derived. An immediate deduction,obtained by taking the ratio of the two equations (1.18) and (1.19) and thendividing both the numerator and denominator of this ratio by cos A cos B, istan A ± tan Btan(A ± B) =1 ∓ tan A tan B . (1.22)One application of this result is a test for whether two lines on a graphare orthogonal (perpendicular); more generally, it determines the angle betweenthem. The standard notation for a straight-line graph is y = mx + c, inwhichmis the slope of the graph and c is its intercept on the y-axis. It should be notedthat the slope m is also the tangent of the angle the line makes with the x-axis.12

1.2 TRIGONOMETRIC IDENTITIESConsequently the angle θ 12 between two such straight-line graphs is equal to thedifference in the angles they individually make with the x-axis, and the tangentof that angle is given by (1.22):tan θ 12 = tan θ 1 − tan θ 21+tanθ 1 tan θ 2= m 1 − m 21+m 1 m 2. (1.23)For the lines to be orthogonal we must have θ 12 = π/2, i.e. the final fraction onthe RHS of the above equation must equal ∞, andsom 1 m 2 = −1. (1.24)A kind of inversion of equations (1.18) and (1.19) enables the sum or differenceof two sines or cosines to be expressed as the product of two sinusoids; theprocedure is typified by the following. Adding together the expressions given by(1.18) for sin(A + B) and sin(A − B) yieldssin(A + B)+sin(A − B) =2sinA cos B.If we now write A + B = C and A − B = D, this becomes( ) C + Dsin C +sinD =2sin cos2( C − DIn a similar way each of the following equations can be derived:2). (1.25)( ) ( )C + D C − Dsin C − sin D =2cos sin , (1.26)22( ) ( )C + D C − Dcos C +cosD =2cos cos , (1.27)22( ) ( )C + D C − Dcos C − cos D = −2sin sin . (1.28)22The minus sign on the right of the last of these equations should be noted; it mayhelp to avoid overlooking this ‘oddity’ to recall that if C>Dthen cos C

PRELIMINARY ALGEBRAand use made of equation (1.15), the following results are obtained:sin 2θ =2sinθ cos θ, (1.29)cos 2θ =cos 2 θ − sin 2 θ=2cos 2 θ − 1=1− 2sin 2 θ, (1.30)tan 2θ =2tanθ1 − tan 2 θ . (1.31)A further set of identities enables sinusoidal functions of θ to be expressed interms of polynomial functions of a variable t = tan(θ/2). They are not used intheir primary role until the next chapter, but we give a derivation of them herefor reference.If t = tan(θ/2), then it follows from (1.16) that 1+t 2 =sec 2 (θ/2) and cos(θ/2) =(1 + t 2 ) −1/2 , whilst sin(θ/2) = t(1 + t 2 ) −1/2 . Now, using (1.29) and (1.30), we maywrite:sin θ =2sin θ 2 cos θ 2 = 2t1+t 2 , (1.32)cos θ =cos 2 θ 2 − θ sin2 2 = 1 − t21+t 2 , (1.33)tan θ =2t1 − t 2 . (1.34)It can be further shown that the derivative of θ with respect to t takes thealgebraic form 2/(1 + t 2 ). This completes a package of results that enablesexpressions involving sinusoids, particularly when they appear as integrands, tobe cast in more convenient algebraic forms. The proof of the derivative propertyand examples of use of the above results are given in subsection (2.2.7).We conclude this section with a worked example which is of such a commonlyoccurring form that it might be considered a standard procedure.◮Solve for θ the equationa sin θ + b cos θ = k,where a, b and k are given real quantities.To solve this equation we make use of result (1.18) by setting a = K cos φ and b = K sin φfor suitable values of K and φ. We then havek = K cos φ sin θ + K sin φ cos θ = K sin(θ + φ),withK 2 = a 2 + b 2 and φ =tan −1 b a .Whether φ lies in 0 ≤ φ ≤ π or in −π

1.3 COORDINATE GEOMETRYwith K and φ as given above. Notice that the inverse sine yields two values in the range 0to 2π and that there is no real solution to the original equation if |k| > |K| =(a 2 +b 2 ) 1/2 . ◭1.3 Coordinate geometryWe have already mentioned the standard form for a straight-line graph, namelyy = mx + c, (1.35)representing a linear relationship between the independent variable x and thedependent variable y. Theslopem is equal to the tangent of the angle the linemakes with the x-axis whilst c is the intercept on the y-axis.An alternative form for the equation of a straight line isto which (1.35) is clearly connected byax + by + k =0, (1.36)m = − a band c = − k b .This form treats x and y on a more symmetrical basis, the intercepts on the twoaxes being −k/a and −k/b respectively.A power relationship between two variables, i.e. one of the form y = Ax n ,canalso be cast into straight-line form by taking the logarithms of both sides. Whilstit is normal in mathematical work to use natural logarithms (to base e, writtenln x), for practical investigations logarithms to base 10 are often employed. Ineither case the form is the same, but it needs to be remembered which has beenused when recovering the value of A from fitted data. In the mathematical (basee) form, the power relationship becomesln y = n ln x +lnA. (1.37)Now the slope gives the power n, whilst the intercept on the ln y axis is ln A,which yields A, either by exponentiation or by taking antilogarithms.The other standard coordinate forms of two-dimensional curves that studentsshould know and recognise are those concerned with the conic sections – so calledbecause they can all be obtained by taking suitable sections across a (double)cone. Because the conic sections can take many different orientations and scalingstheir general form is complex,Ax 2 + By 2 + Cxy + Dx + Ey + F =0, (1.38)but each can be represented by one of four generic forms, an ellipse, a parabola, ahyperbola or, the degenerate form, a pair of straight lines. If they are reduced totheir standard representations, in which axes of symmetry are made to coincide15

PRELIMINARY ALGEBRAwith the coordinate axes, the first three take the forms(x − α) 2 (y − β)2a 2 +b 2 = 1 (ellipse), (1.39)(y − β) 2 =4a(x − α) (parabola), (1.40)(x − α) 2 (y − β)2a 2 −b 2 = 1 (hyperbola). (1.41)Here, (α, β) gives the position of the ‘centre’ of the curve, usually taken asthe origin (0, 0) when this does not conflict with any imposed conditions. Theparabola equation given is that for a curve symmetric about a line parallel tothe x-axis. For one symmetrical about a parallel to the y-axis the equation wouldread (x − α) 2 =4a(y − β).Of course, the circle is the special case of an ellipse in which b = a and theequation takes the form(x − α) 2 +(y − β) 2 = a 2 . (1.42)The distinguishing characteristic of this equation is that when it is expressed inthe form (1.38) the coefficients of x 2 and y 2 are equal and that of xy is zero; thisproperty is not changed by any reorientation or scaling and so acts to identify ageneral conic as a circle.Definitions of the conic sections in terms of geometrical properties are alsoavailable; for example, a parabola can be defined as the locus of a point thatis always at the same distance from a given straight line (the directrix) asitisfrom a given point (the focus). When these properties are expressed in Cartesiancoordinates the above equations are obtained. For a circle, the defining propertyis that all points on the curve are a distance a from (α, β); (1.42) expresses thisrequirement very directly. In the following worked example we derive the equationfor a parabola.◮Find the equation of a parabola that has the line x = −a as its directrix and the point(a, 0) as its focus.Figure 1.3 shows the situation in Cartesian coordinates. Expressing the defining requirementthat PN and PF are equal in length gives(x + a) =[(x − a) 2 + y 2 ] 1/2 ⇒ (x + a) 2 =(x − a) 2 + y 2which, on expansion of the squared terms, immediately gives y 2 =4ax. This is (1.40) withα and β both set equal to zero. ◭Although the algebra is more complicated, the same method can be used toderive the equations for the ellipse and the hyperbola. In these cases the distancefrom the fixed point is a definite fraction, e, known as the eccentricity, ofthedistance from the fixed line. For an ellipse 0

1.3 COORDINATE GEOMETRYyNP(x, y)OF(a, 0)xx = −aFigure 1.3 Construction of a parabola using the point (a, 0) as the focus andthe line x = −a as the directrix.The values of a and b (with a ≥ b) in equation (1.39) for an ellipse are relatedto e throughe 2 = a2 − b 2a 2and give the lengths of the semi-axes of the ellipse. If the ellipse is centred onthe origin, i.e. α = β = 0, then the focus is (−ae, 0) and the directrix is the linex = −a/e.For each conic section curve, although we have two variables, x and y, theyarenot independent, since if one is given then the other can be determined. However,determining y when x is given, say, involves solving a quadratic equation on eachoccasion, and so it is convenient to have parametric representations of the curves.A parametric representation allows each point on a curve to be associated witha unique value of a single parameter t. The simplest parametric representationsfor the conic sections are as given below, though that for the hyperbola useshyperbolic functions, not formally introduced until chapter 3. That they do givevalid parameterizations can be verified by substituting them into the standardforms (1.39)–(1.41); in each case the standard form is reduced to an algebraic ortrigonometric identity.x = α + a cos φ, y = β + b sin φ (ellipse),x = α + at 2 , y = β +2at (parabola),x = α + a cosh φ, y = β + b sinh φ (hyperbola).As a final example illustrating several topics from this section we now prove17

PRELIMINARY ALGEBRAthe well-known result that the angle subtended by a diameter at any point on acircle is a right angle.◮Taking the diameter to be the line joining Q =(−a, 0) and R =(a, 0) and the point P tobe any point on the circle x 2 + y 2 = a 2 , prove that angle QP R is a right angle.If P is the point (x, y), the slope of the line QP ism 1 = y − 0x − (−a) = yx + a .That of RP ism 2 = y − 0x − (a) = yx − a .Thusm 1 m 2 =y2x 2 − a . 2But, since P is on the circle, y 2 = a 2 − x 2 and consequently m 1 m 2 = −1. From result (1.24)this implies that QP and RP are orthogonal and that QP R is therefore a right angle. Notethat this is true for any point P on the circle. ◭1.4 Partial fractionsIn subsequent chapters, and in particular when we come to study integrationin chapter 2, we will need to express a function f(x) that is the ratio of twopolynomials in a more manageable form. To remove some potential complexityfrom our discussion we will assume that all the coefficients in the polynomialsare real, although this is not an essential simplification.The behaviour of f(x) is crucially determined by the location of the zeros ofits denominator, i.e. if f(x) is written as f(x) =g(x)/h(x) where both g(x) andh(x) are polynomials, § then f(x) changes extremely rapidly when x is close tothose values α i that are the roots of h(x) = 0. To make such behaviour explicit,we write f(x) as a sum of terms such as A/(x − α) n ,inwhichA is a constant, α isone of the α i that satisfy h(α i )=0andn is a positive integer. Writing a functionin this way is known as expressing it in partial fractions.Suppose, for the sake of definiteness, that we wish to express the functionf(x) =4x +2x 2 +3x +2§ It is assumed that the ratio has been reduced so that g(x) andh(x) do not contain any commonfactors, i.e. there is no value of x that makes both vanish at the same time. We may also assumewithout any loss of generality that the coefficient of the highest power of x in h(x) has been madeequal to unity, if necessary, by dividing both numerator and denominator by the coefficient of thishighest power.18

1.4 PARTIAL FRACTIONSin partial fractions, i.e. to write it asf(x) = g(x) 4x +2=h(x) x 2 +3x +2 = A 1(x − α 1 ) + A 2n1 (x − α 2 ) + ··· . n2 (1.43)The first question that arises is that of how many terms there should be onthe right-hand side (RHS). Although some complications occur when h(x) hasrepeated roots (these are considered below) it is clear that f(x) only becomesinfinite at the two values of x, α 1 and α 2 , that make h(x) = 0. Consequently theRHS can only become infinite at the same two values of x and therefore containsonly two partial fractions – these are the ones shown explicitly. This argumentcan be trivially extended (again temporarily ignoring the possibility of repeatedroots of h(x)) to show that if h(x) is a polynomial of degree n then there should ben terms on the RHS, each containing a different root α i of the equation h(α i )=0.A second general question concerns the appropriate values of the n i .Thisisanswered by putting the RHS over a common denominator, which will clearlyhave to be the product (x − α 1 ) n1 (x − α 2 ) n2 ···. Comparison of the highest powerof x in this new RHS with the same power in h(x) showsthatn 1 + n 2 + ···= n.This result holds whether or not h(x) = 0 has repeated roots and, although wedo not give a rigorous proof, strongly suggests the following correct conclusions.• The number of terms on the RHS is equal to the number of distinct roots ofh(x) = 0, each term having a different root α i in its denominator (x − α i ) ni .• If α i is a multiple root of h(x) = 0 then the value to be assigned to n i in (1.43) isthat of m i when h(x) is written in the product form (1.9). Further, as discussedon p. 23, A i has to be replaced by a polynomial of degree m i − 1. This is alsoformally true for non-repeated roots, since then both m i and n i are equal tounity.Returning to our specific example we note that the denominator h(x) has zerosat x = α 1 = −1 andx = α 2 = −2; these x-values are the simple (non-repeated)roots of h(x) = 0. Thus the partial fraction expansion will be of the form4x +2x 2 +3x +2 = A 1x +1 + A 2x +2 . (1.44)We now list several methods available for determining the coefficients A 1 andA 2 . We also remind the reader that, as with all the explicit examples and techniquesdescribed, these methods are to be considered as models for the handling of anyratio of polynomials, with or without characteristics that make it a special case.(i) The RHS can be put over a common denominator, in this case (x+1)(x+2),and then the coefficients of the various powers of x can be equated in the19

PRELIMINARY ALGEBRAnumerators on both sides of the equation. This leads to4x +2=A 1 (x +2)+A 2 (x +1),4=A 1 + A 2 2=2A 1 + A 2 .Solving the simultaneous equations for A 1 and A 2 gives A 1 = −2 andA 2 =6.(ii) A second method is to substitute two (or more generally n) differentvalues of x into each side of (1.44) and so obtain two (or n) simultaneousequations for the two (or n) constantsA i . To justify this practical way ofproceeding it is necessary, strictly speaking, to appeal to method (i) above,which establishes that there are unique values for A 1 and A 2 valid forall values of x. It is normally very convenient to take zero as one of thevalues of x, but of course any set will do. Suppose in the present case thatwe use the values x = 0 and x = 1 and substitute in (1.44). The resultingequations are22 = A 11 + A 22 ,66 = A 12 + A 23 ,which on solution give A 1 = −2 andA 2 = 6, as before. The reader caneasily verify that any other pair of values for x (except for a pair thatincludes α 1 or α 2 ) gives the same values for A 1 and A 2 .(iii) The very reason why method (ii) fails if x is chosen as one of the rootsα i of h(x) = 0 can be made the basis for determining the values of the A icorresponding to non-multiple roots without having to solve simultaneousequations. The method is conceptually more difficult than the other methodspresented here, and needs results from the theory of complex variables(chapter 24) to justify it. However, we give a practical ‘cookbook’ recipefor determining the coefficients.(a) To determine the coefficient A k , imagine the denominator h(x)written as the product (x − α 1 )(x − α 2 ) ···(x − α n ), with any m-foldrepeated root giving rise to m factors in parentheses.(b) Now set x equal to α k and evaluate the expression obtained afteromitting the factor that reads α k − α k .(c) Divide the value so obtained into g(α k ); the result is the requiredcoefficient A k .For our specific example we find that in step (a) that h(x) =(x +1)(x +2)and that in evaluating A 1 step (b) yields −1 + 2, i.e. 1. Since g(−1) =4(−1) + 2 = −2, step (c) gives A 1 as (−2)/(1), i.e in agreement with ourother evaluations. In a similar way A 2 is evaluated as (−6)/(−1) = 6.20

1.4 PARTIAL FRACTIONSThus any one of the methods listed above shows that4x +2x 2 +3x +2 = −2x +1 + 6x +2 .The best method to use in any particular circumstance will depend on thecomplexity, in terms of the degrees of the polynomials and the multiplicities ofthe roots of the denominator, of the function being considered and, to someextent, on the individual inclinations of the student; some prefer lengthy butstraightforward solution of simultaneous equations, whilst others feel more athome carrying through shorter but more abstract calculations in their heads.1.4.1 Complications and special casesHaving established the basic method for partial fractions, we now show, throughfurther worked examples, how some complications are dealt with by extensionsto the procedure. These extensions are introduced one at a time, but of course inany practical application more than one may be involved.The degree of the numerator is greater than or equal to that of the denominatorAlthough we have not specifically mentioned the fact, it will be apparent fromtrying to apply method (i) of the previous subsection to such a case, that if thedegree of the numerator (m) is not less than that of the denominator (n) then theratio of two polynomials cannot be expressed in partial fractions.To get round this difficulty it is necessary to start by dividing the denominatorh(x) into the numerator g(x) to obtain a further polynomial, which we will denoteby s(x), together with a function t(x) thatis a ratio of two polynomials for whichthe degree of the numerator is less than that of the denominator. The functiont(x) can therefore be expanded in partial fractions. As a formula,f(x) = g(x)r(x)= s(x)+t(x) ≡ s(x)+h(x) h(x) . (1.45)It is apparent that the polynomial r(x) istheremainder obtained when g(x) isdivided by h(x), and, in general, will be a polynomial of degree n − 1. It is alsoclear that the polynomial s(x) will be of degree m − n. Again, the actual divisionprocess can be set out as an algebraic long division sum but is probably moreeasily handled by writing (1.45) in the formor, more explicitly, asg(x) =s(x)h(x)+r(x) (1.46)g(x) =(s m−n x m−n + s m−n−1 x m−n−1 + ···+ s 0 )h(x)+(r n−1 x n−1 + r n−2 x n−2 + ···+ r 0 )(1.47)and then equating coefficients.21

PRELIMINARY ALGEBRAWe illustrate this procedure with the following worked example.◮Find the partial fraction decomposition of the functionf(x) = x3 +3x 2 +2x +1.x 2 − x − 6Since the degree of the numerator is 3 and that of the denominator is 2, a preliminarylong division is necessary. The polynomial s(x) resulting from the division will have degree3 − 2 = 1 and the remainder r(x) will be of degree 2 − 1 = 1 (or less). Thus we writex 3 +3x 2 +2x +1=(s 1 x + s 0 )(x 2 − x − 6) + (r 1 x + r 0 ).From equating the coefficients of the various powers of x on the two sides of the equation,starting with the highest, we now obtain the simultaneous equations1=s 1 ,3=s 0 − s 1 ,2=−s 0 − 6s 1 + r 1 ,1=−6s 0 + r 0 .These are readily solved, in the given order, to yield s 1 =1,s 0 =4,r 1 =12andr 0 = 25.Thus f(x) can be written as12x +25f(x) =x +4+x 2 − x − 6 .The last term can now be decomposed into partial fractions as previously. The zeros ofthe denominator are at x =3andx = −2 and the application of any method from theprevious subsection yields the respective constants as A 1 =12 1 and A 5 2 = − 1 . Thus the5final partial fraction decomposition of f(x) isx +4+615(x − 3) − 15(x +2) . ◭Factors of the form a 2 + x 2 in the denominatorWe have so far assumed that the roots of h(x) = 0, needed for the factorisation ofthe denominator of f(x), can always be found. In principle they always can butin some cases they are not real. Consider, for example, attempting to express inpartial fractions a polynomial ratio whose denominator is h(x) =x 3 −x 2 +2x−2.Clearly x = 1 is a zero of h(x), and so a first factorisation is (x − 1)(x 2 +2).However we cannot make any further progress because the factor x 2 + 2 cannotbe expressed as (x − α)(x − β) foranyrealα and β.Complex numbers are introduced later in this book (chapter 3) and, when thereader has studied them, he or she may wish to justify the procedure set outbelow. It can be shown to be equivalent to that already given, but the zeros ofh(x) are now allowed to be complex and terms that are complex conjugates ofeach other are combined to leave only real terms.Since quadratic factors of the form a 2 +x 2 that appear in h(x) cannot be reducedto the product of two linear factors, partial fraction expansions including themneed to have numerators in the corresponding terms that are not simply constants22

1.4 PARTIAL FRACTIONSA i but linear functions of x, i.e.oftheformB i x + C i . Thus, in the expansion,linear terms (first-degree polynomials) in the denominator have constants (zerodegreepolynomials) in their numerators, whilst quadratic terms (second-degreepolynomials) in the denominator have linear terms (first-degree polynomials) intheir numerators. As a symbolic formula, the partial fraction expansion ofg(x)(x − α 1 )(x − α 2 ) ···(x − α p )(x 2 + a 2 1 )(x2 + a 2 2 ) ···(x2 + a 2 q )should take the formA 1x − α 1+ A 2x − α 2+ ···+A p+ B 1x + C 1x − α p x 2 + a 2 + B 2x + C 21x 2 + a 2 2+ ···+ B qx + C qx 2 + a 2 .qOf course, the degree of g(x) must be less than p +2q; if it is not, an initialdivision must be carried out as demonstrated earlier.Repeated factors in the denominatorConsider trying (incorrectly) to expandf(x) =x − 4(x +1)(x − 2) 2in partial fraction form as follows:x − 4(x +1)(x − 2) 2 = A 1x +1 + A 2(x − 2) 2 .Multiplying both sides of this supposed equality by (x +1)(x − 2) 2 produces anequation whose LHS is linear in x, whilst its RHS is quadratic. This is clearlywrong and so an expansion in the above form cannot be valid. The correction wemust make is very similar to that needed in the previous subsection, namely thatsince (x − 2) 2 is a quadratic polynomial the numerator of the term containing itmust be a first-degree polynomial, and not simply a constant.The correct form for the part of the expansion containing the doubly repeatedroot is therefore (Bx+ C)/(x − 2) 2 . Using this form and either of methods (i) and(ii) for determining the constants gives the full partial fraction expansion asx − 4(x +1)(x − 2) 2 = − 5 5x − 16+9(x +1) 9(x − 2) 2 ,as the reader may verify.Since any term of the form (Bx + C)/(x − α) 2 can be written asB(x − α)+C + Bα(x − α) 2 = Bx − α + C + Bα(x − α) 2 ,and similarly for multiply repeated roots, an alternative form for the part of thepartial fraction expansion containing a repeated root α isD 1x − α + D 2(x − α) 2 + ···+ D p(x − α) p . (1.48)23

PRELIMINARY ALGEBRAIn this form, all x-dependence has disappeared from the numerators but at theexpense of p−1 additional terms; the total number of constants to be determinedremains unchanged, as it must.When describing possible methods of determining the constants in a partialfraction expansion, we noted that method (iii), p. 20, which avoids the need tosolve simultaneous equations, is restricted to terms involving non-repeated roots.In fact, it can be applied in repeated-root situations, when the expansion is putin the form (1.48), but only to find the constant in the term involving the largestinverse power of x − α, i.e.D p in (1.48).We conclude this section with a more protracted worked example that containsall three of the complications discussed.◮Resolve the following expression F(x) into partial fractions:F(x) = x5 − 2x 4 − x 3 +5x 2 − 46x + 100.(x 2 +6)(x − 2) 2We note that the degree of the denominator (4) is not greater than that of the numerator(5), and so we must start by dividing the latter by the former. It follows, from the differencein degrees and the coefficients of the highest powers in each, that the result will be a linearexpression s 1 x + s 0 with the coefficient s 1 equal to 1. Thus the numerator of F(x) mustbeexpressible as(x + s 0 )(x 4 − 4x 3 +10x 2 − 24x + 24) + (r 3 x 3 + r 2 x 2 + r 1 x + r 0 ),where the second factor in parentheses is the denominator of F(x) written as a polynomial.Equating the coefficients of x 4 gives −2 =−4+s 0 and fixes s 0 as 2. Equating the coefficientsof powers less than 4 gives equations involving the coefficients r i as follows:−1 =−8+10+r 3 ,5=−24+20+r 2 ,−46 = 24 − 48 + r 1 ,100 = 48 + r 0 .Thus the remainder polynomial r(x) can be constructed and F(x) written asF(x) =x +2+ −3x3 +9x 2 − 22x +52≡ x +2+f(x).(x 2 +6)(x − 2) 2The polynomial ratio f(x) can now be expressed in partial fraction form, noting that itsdenominator contains both a term of the form x 2 + a 2 and a repeated root. Thusf(x) = Bx + Cx 2 +6 + D 1x − 2 + D 2(x − 2) . 2We could now put the RHS of this equation over the common denominator (x 2 +6)(x−2) 2and find B,C,D 1 and D 2 by equating coefficients of powers of x. It is quicker, however,to use methods (iii) and (ii). Method (iii) gives D 2 as (−24 + 36 − 44 + 52)/(4 + 6) = 2.We choose to evaluate the other coefficients by method (ii), and setting x =0,x =1and24

1.5 BINOMIAL EXPANSIONx = −1 gives respectively5224 = C 6 − D 12 + 2 4 ,367 = B + C − D 1 +2,78663 = C − B − D 17 3 + 2 9 .These equations reduce to4C − 12D 1 =40,B + C − 7D 1 =22,−9B +9C − 21D 1 =72,with solution B =0,C =1,D 1 = −3.Thus, finally, we may rewrite the original expression F(x) in partial fractions asF(x) =x +2+ 1x 2 +6 − 3x − 2 + 2(x − 2) . ◭ 21.5 Binomial expansionEarlier in this chapter we were led to consider functions containing powers ofthe sum or difference of two terms, e.g. (x − α) m . Later in this book we will findnumerous occasions on which we wish to write such a product of repeated factorsas a polynomial in x or, more generally, as a sum of terms each of which containspowers of x and α separately, as opposed to a power of their sum or difference.To make the discussion general and the result applicable to a wide variety ofsituations, we will consider the general expansion of f(x) =(x + y) n ,wherex andy may stand for constants, variables or functions and, for the time being, n is apositive integer. It may not be obvious what form the general expansion takesbut some idea can be obtained by carrying out the multiplication explicitly forsmall values of n. Thus we obtain successively(x + y) 1 = x + y,(x + y) 2 =(x + y)(x + y) =x 2 +2xy + y 2 ,(x + y) 3 =(x + y)(x 2 +2xy + y 2 )=x 3 +3x 2 y +3xy 2 + y 3 ,(x + y) 4 =(x + y)(x 3 +3x 2 y +3xy 2 + y 3 )=x 4 +4x 3 y +6x 2 y 2 +4xy 3 + y 4 .This does not establish a general formula, but the regularity of the terms inthe expansions and the suggestion of a pattern in the coefficients indicate that ageneral formula for power n will have n + 1 terms, that the powers of x and y inevery term will add up to n and that the coefficients of the first and last termswill be unity whilst those of the second and penultimate terms will be n.25

PRELIMINARY ALGEBRAIn fact, the general expression, the binomial expansion for power n, is given by∑k=n(x + y) n =n C k x n−k y k , (1.49)k=0where n C k is called the binomial coefficient and is expressed in terms of factorialfunctions by n!/[k!(n − k)!]. Clearly, simply to make such a statement does notconstitute proof of its validity, but, as we will see in subsection 1.5.2, (1.49) canbe proved using a method called induction. Before turning to that proof, weinvestigate some of the elementary properties of the binomial coefficients.1.5.1 Binomial coefficientsAs stated above, the binomial coefficients are defined by( )n n! nC k ≡k!(n − k)! ≡ for 0 ≤ k ≤ n, (1.50)kwhere in the second identity we give a common alternative notation for n C k .Obvious properties include(i) n C 0 = n C n =1,(ii) n C 1 = n C n−1 = n,(iii) n C k = n C n−k .We note that, for any given n, the largest coefficient in the binomial expansion isthe middle one (k = n/2) if n is even; the middle two coefficients (k = 1 2(n ± 1))are equal largest if n is odd. Somewhat less obvious is the resultn C k + n n!C k−1 =k!(n − k)! + n!(k − 1)!(n − k +1)!n![(n +1− k)+k]=k!(n +1− k)!(n +1)!=k!(n +1− k)! = n+1 C k . (1.51)An equivalent statement, in which k has been redefined as k +1,isn C k + n C k+1 = n+1 C k+1 . (1.52)1.5.2 Proof of the binomial expansionWe are now in a position to prove the binomial expansion (1.49). In doing so, weintroduce the reader to a procedure applicable to certain types of problems andknown as the method of induction. The method is discussed much more fully insubsection 1.7.1.26

1.6 PROPERTIES OF BINOMIAL COEFFICIENTSWe start by assuming that (1.49) is true for some positive integer n = N.Wenowproceed to show that this implies that it must also be true for n = N+1, as follows:(x + y) N+1 =(x + y)k=0N∑N C k x N−k y kk=0N∑N∑=N C k x N+1−k y k +N C k x N−k y k+1=k=0N∑N+1∑N C k x N+1−k y k +N C j−1 x (N+1)−j y j ,k=0where in the first line we have used the assumption and in the third line havemoved the second summation index by unity, by writing k +1 = j. Wenowseparate off the first term of the first sum, N C 0 x N+1 , and write it as N+1 C 0 x N+1 ;we can do this since, as noted in (i) following (1.50), n C 0 =1foreveryn. Similarly,the last term of the second summation can be replaced by N+1 C N+1 y N+1 .The remaining terms of each of the two summations are now written together,with the summation index denoted by k in both terms. Thus(x + y) N+1 = N+1 C 0 x N+1 += N+1 C 0 x N+1 +j=1N∑ ( NC k + N )C k−1 x (N+1)−k y k + N+1 C N+1 y N+1k=1N∑N+1 C k x (N+1)−k y k + N+1 C N+1 y N+1k=1N+1∑=N+1 C k x (N+1)−k y k .k=0In going from the first to the second line we have used result (1.51). Now weobserve that the final overall equation is just the original assumed result (1.49)but with n = N + 1. Thus it has been shown that if the binomial expansion isassumed to be true for n = N, thenitcanbeproved to be true for n = N +1. Butit holds trivially for n = 1, and therefore for n = 2 also. By the same token it isvalid for n =3, 4,..., and hence is established for all positive integers n.1.6 Properties of binomial coefficients1.6.1 Identities involving binomial coefficientsThere are many identities involving the binomial coefficients that can be deriveddirectly from their definition, and yet more that follow from their appearance inthe binomial expansion. Only the most elementary ones, given earlier, are worthcommitting to memory but, as illustrations, we now derive two results involvingsums of binomial coefficients.27

PRELIMINARY ALGEBRAThe first is a further application of the method of induction. Consider theproposal that, for any n ≥ 1andk ≥ 0,∑n−1k+s C k = n+k C k+1 . (1.53)s=0Notice that here n, the number of terms in the sum, is the parameter that varies,k is a fixed parameter, whilst s is a summation index and does not appear on theRHS of the equation.Now we suppose that the statement (1.53) about the value of the sum of thebinomial coefficients k C k , k+1 C k ,..., k+n−1 C k is true for n = N. Wenextwritedowna series with an extra term and determine the implications of the supposition forthe new series:N+1−1∑N−1∑k+s C k =k+s C k + k+N C ks=0s=0= N+k C k+1 + N+k C k= N+k+1 C k+1 .But this is just proposal (1.53) with n now set equal to N + 1. To obtain the lastline, we have used (1.52), with n set equal to N + k.It only remains to consider the case n = 1, when the summation only containsone term and (1.53) reduces tok C k = 1+k C k+1 .This is trivially valid for any k since both sides are equal to unity, thus completingthe proof of (1.53) for all positive integers n.The second result, which gives a formula for combining terms from two setsof binomial coefficients in a particular way (a kind of ‘convolution’, for readerswho are already familiar with this term), is derived by applying the binomialexpansion directly to the identity(x + y) p (x + y) q ≡ (x + y) p+q .Written in terms of binomial expansions, this readsp∑q∑∑p+qp C s x p−s y s q C t x q−t y t =p+q C r x p+q−r y r .s=0t=0We now equate coefficients of x p+q−r y r on the two sides of the equation, notingthat on the LHS all combinations of s and t such that s + t = r contribute. Thisgives as an identity thatr∑r∑p C q r−t C t = p+q C r =p C q t C r−t . (1.54)r=0t=0t=028

1.6 PROPERTIES OF BINOMIAL COEFFICIENTSWe have specifically included the second equality to emphasise the symmetricalnature of the relationship with respect to p and q.Further identities involving the coefficients can be obtained by giving x and yspecial values in the defining equation (1.49) for the expansion. If both are setequal to unity then we obtain (using the alternative notation so as to producefamiliarity with it)( ( ( ( n n n n+ + + ···+ =20)1)2)n)n , (1.55)whilst setting x = 1 and y = −1 yields( ( ( ( )n n n n− + − ···+(−1)0)1)2)n n=0. (1.56)1.6.2 Negative and non-integral values of nUp till now we have restricted n in the binomial expansion to be a positiveinteger. Negative values can be accommodated, but only at the cost of an infiniteseries of terms rather than the finite one represented by (1.49). For reasons thatare intuitively sensible and will be discussed in more detail in chapter 4, veryoften we require an expansion in which, at least ultimately, successive terms inthe infinite series decrease in magnitude. For this reason, if x>ywe consider(x + y) −m ,wherem itself is a positive integer, in the form(x + y) n =(x + y) −m = x −m ( 1+ y x) −m.Since the ratio y/x is less than unity, terms containing higher powers of it will besmall in magnitude, whilst raising the unit term to any power will not affect itsmagnitude. If y>xthe roles of the two must be interchanged.We can now state, but will not explicitly prove, the form of the binomialexpansion appropriate to negative values of n (n equal to −m):∑∞ ( y) k(x + y) n =(x + y) −m = x −m −m C k , (1.57)xwhere the hitherto undefined quantity −m C k , which appears to involve factorialsof negative numbers, is given by−m k m(m +1)···(m + k − 1) k (m + k − 1)!C k =(−1) =(−1) =(−1) k m+k−1 C k .k!(m − 1)!k!(1.58)The binomial coefficient on the extreme right of this equation has its normalmeaning and is well defined since m + k − 1 ≥ k.Thus we have a definition of binomial coefficients for negative integer valuesof n in terms of those for positive n. The connection between the two may not29k=0

PRELIMINARY ALGEBRAbe obvious, but they are both formed in the same way in terms of recurrencerelations. Whatever the sign of n, the series of coefficients n C k can be generatedby starting with n C 0 = 1 and using the recurrence relationn C k+1 = n − k n C k . (1.59)k +1The difference is that for positive integer n the series terminates when k = n,whereas for negative n there is no such termination – in line with the infiniteseries of terms in the corresponding expansion.Finally we note that, in fact, equation (1.59) generates the appropriate coefficientsfor all values of n, positive or negative, integer or non-integer, with theobvious exception of the case in which x = −y and n is negative. For non-integern the expansion does not terminate, even if n is positive.1.7 Some particular methods of proofMuch of the mathematics used by physicists and engineers is concerned withobtaining a particular value, formula or function from a given set of data andstated conditions. However, just as it is essential in physics to formulate the basiclaws and so be able to set boundaries on what can or cannot happen, so itis important in mathematics to be able to state general propositions about theoutcomes that are or are not possible. To this end one attempts to establishtheorems that state in as general a way as possible mathematical results thatapply to particular types of situation. We conclude this introductory chapter bydescribing two methods that can sometimes be used to prove particular classesof theorems.The two general methods of proof are known as proof by induction (whichhas already been met in this chapter) and proof by contradiction. They sharethe common characteristic that at an early stage in the proof an assumptionis made that a particular (unproven) statement is true; the consequences ofthat assumption are then explored. In an inductive proof the conclusion isreached that the assumption is self-consistent and has other equally consistentbut broader implications, which are then applied to establish the general validityof the assumption. A proof by contradiction, however, establishes an internalinconsistency and thus shows that the assumption is unsustainable; the naturalconsequence of this is that the negative of the assumption is established as true.Later in this book use will be made of these methods of proof to explore newterritory, e.g. to examine the properties of vector spaces, matrices and groups.However, at this stage we will draw our illustrative and test examples from earliersections of this chapter and other topics in elementary algebra and number theory.30

1.7 SOME PARTICULAR METHODS OF PROOF1.7.1 Proof by inductionThe proof of the binomial expansion given in subsection 1.5.2 and the identityestablished in subsection 1.6.1 have already shown the way in which an inductiveproof is carried through. They also indicated the main limitation of the method,namely that only an initially supposed result can be proved. Thus the methodof induction is of no use for deducing a previously unknown result; a putativeequation or result has to be arrived at by some other means, usually by noticingpatterns or by trial and error using simple values of the variables involved. Itwill also be clear that propositions that can be proved by induction are limitedto those containing a parameter that takes a range of integer values (usuallyinfinite).For a proposition involving a parameter n, the five steps in a proof usinginduction are as follows.(i) Formulate the supposed result for general n.(ii) Suppose (i) to be true for n = N (or more generally for all values ofn ≤ N; seebelow),whereN is restricted to lie in the stated range.(iii) Show, using only proven results and supposition (ii), that proposition (i)is true for n = N +1.(iv) Demonstrate directly, and without any assumptions, that proposition (i) istrue when n takes the lowest value in its range.(v) It then follows from (iii) and (iv) that the proposition is valid for all valuesof n in the stated range.(It should be noted that, although many proofs at stage (iii) require the validityof the proposition only for n = N, some require it for all n less than or equal to N– hence the form of inequality given in parentheses in the stage (ii) assumption.)To illustrate further the method of induction, we now apply it to two workedexamples; the first concerns the sum of the squares of the first n natural numbers.◮Prove that the sum of the squares of the first n natural numbers is given byn∑r 2 = 1 n(n + 1)(2n +1). (1.60)6r=1As previously we start by assuming the result is true for n = N. Then it follows that∑N+1N∑r 2 = r 2 +(N +1) 2r=1r=1= 1 N(N + 1)(2N +1)+(N +1)26= 1 (N +1)[N(2N +1)+6N +6]6= 1 (N + 1)[(2N +3)(N +2)]6= 1 (N +1)[(N +1)+1][2(N +1)+1].631

PRELIMINARY ALGEBRAThis is precisely the original assumption, but with N replaced by N + 1. To complete theproof we only have to verify (1.60) for n = 1. This is trivially done and establishes theresult for all positive n. The same and related results are obtained by a different methodin subsection 4.2.5. ◭Our second example is somewhat more complex and involves two nested proofsby induction: whilst trying to establish the main result by induction, we find thatwe are faced with a second proposition which itself requires an inductive proof.◮Show that Q(n) =n 4 +2n 3 +2n 2 + n is divisible by 6 (without remainder) for all positiveinteger values of n.Again we start by assuming the result is true for some particular value N of n, whilstnoting that it is trivially true for n = 0. We next examine Q(N + 1), writing each of itsterms as a binomial expansion:Q(N +1)=(N +1) 4 +2(N +1) 3 +2(N +1) 2 +(N +1)=(N 4 +4N 3 +6N 2 +4N +1)+2(N 3 +3N 2 +3N +1)+2(N 2 +2N +1)+(N +1)=(N 4 +2N 3 +2N 2 + N)+(4N 3 +12N 2 +14N +6).Now, by our assumption, the group of terms within the first parentheses in the last lineis divisible by 6 and clearly so are the terms 12N 2 and 6 within the second parentheses.Thus it comes down to deciding whether 4N 3 +14N is divisible by 6 – or equivalently,whether R(N) =2N 3 +7N is divisible by 3.To settle this latter question we try using a second inductive proof and assume thatR(N) is divisible by 3 for N = M, whilst again noting that the proposition is trivially truefor N = M = 0. This time we examine R(M +1):R(M +1)=2(M +1) 3 +7(M +1)=2(M 3 +3M 2 +3M +1)+7(M +1)=(2M 3 +7M)+3(2M 2 +2M +3)By assumption, the first group of terms in the last line is divisible by 3 and the secondgroup is patently so. We thus conclude that R(N) is divisible by 3 for all N ≥ M, andtaking M = 0 shows that it is divisible by 3 for all N.We can now return to the main proposition and conclude that since R(N) =2N 3 +7Nis divisible by 3, 4N 3 +12N 2 +14N + 6 is divisible by 6. This in turn establishes that thedivisibility of Q(N + 1) by 6 follows from the assumption that Q(N) divides by 6. SinceQ(0) clearly divides by 6, the proposition in the question is established for all values of n. ◭1.7.2 Proof by contradictionThe second general line of proof, but again one that is normally only useful whenthe result is already suspected, is proof by contradiction. The questions it canattempt to answer are only those that can be expressed in a proposition thatis either true or false. Clearly, it could be argued that any mathematical resultcan be so expressed but, if the proposition is no more than a guess, the chancesof success are negligible. Valid propositions containing even modest formulaeare either the result of true inspiration or, much more normally, yet anotherreworking of an old chestnut!32

1.7 SOME PARTICULAR METHODS OF PROOFThe essence of the method is to exploit the fact that mathematics is requiredto be self-consistent, so that, for example, two calculations of the same quantity,starting from the same given data but proceeding by different methods, must givethe same answer. Equally, it must not be possible to follow a line of reasoning anddraw a conclusion that contradicts either the input data or any other conclusionbased upon the same data.It is this requirement on which the method of proof by contradiction is based.The crux of the method is to assume that the proposition to be proved isnot true, and then use this incorrect assumption and ‘watertight’ reasoning todraw a conclusion that contradicts the assumption. The only way out of theself-contradiction is then to conclude that the assumption was indeed false andtherefore that the proposition is true.It must be emphasised that once a (false) contrary assumption has been made,every subsequent conclusion in the argument must follow of necessity. Proof bycontradiction fails if at any stage we have to admit ‘this may or may not bethe case’. That is, each step in the argument must be a necessary consequence ofresults that precede it (taken together with the assumption), rather than simply apossible consequence.It should also be added that if no contradiction can be found using soundreasoning based on the assumption then no conclusion can be drawn about eitherthe proposition or its negative and some other approach must be tried.We illustrate the general method with an example in which the mathematicalreasoning is straightforward, so that attention can be focussed on the structureof the proof.◮A rational number r is a fraction r = p/q in which p and q are integers with q positive.Further, r is expressed in its lowest terms, any integer common factor of p and q havingbeen divided out.Prove that the square root of an integer m cannot be a rational number, unless the squareroot itself is an integer.We begin by supposing that the stated result is not true and that we can write an equation√ m = r =pqfor integers m, p, q with q ≠1.It then follows that p 2 = mq 2 .But,sincer is expressed in its lowest terms, p and q, andhence p 2 and q 2 , have no factors in common. However, m is an integer; this is only possibleif q =1andp 2 = m. This conclusion contradicts the requirement that q ≠1andsoleadsto the conclusion that it was wrong to suppose that √ m can be expressed as a non-integerrational number. This completes the proof of the statement in the question. ◭Our second worked example, also taken from elementary number theory,involves slightly more complicated mathematical reasoning but again exhibits thestructure associated with this type of proof.33

PRELIMINARY ALGEBRA◮The prime integers p i are labelled in ascending order, thus p 1 =1, p 2 =2, p 5 =7, etc.Show that there is no largest prime number.Assume, on the contrary, that there is a largest prime and let it be p N .Considernowthenumber q formed by multiplying together all the primes from p 1 to p N and then addingone to the product, i.e.q = p 1 p 2 ···p N +1.By our assumption p N is the largest prime, and so no number can have a prime factorgreater than this. However, for every prime p i , i =1, 2,...,N, the quotient q/p i has theform M i +(1/p i )withM i an integer and 1/p i non-integer. This means that q/p i cannot bean integer and so p i cannot be a divisor of q.Since q is not divisible by any of the (assumed) finite set of primes, it must be itself aprime. As q is also clearly greater than p N , we have a contradiction. This shows that ourassumption that there is a largest prime integer must be false, and so it follows that thereis no largest prime integer.It should be noted that the given construction for q does not generate all the primesthat actually exist (e.g. for N =3,q = 7 rather than the next actual prime value of 5, isfound), but this does not matter for the purposes of our proof by contradiction. ◭1.7.3 Necessary and sufficient conditionsAs the final topic in this introductory chapter, we consider briefly the notionof, and distinction between, necessary and sufficient conditions in the contextof proving a mathematical proposition. In ordinary English the distinction iswell defined, and that distinction is maintained in mathematics. However, inthe authors’ experience students tend to overlook it and assume (wrongly) that,having proved that the validity of proposition A implies the truth of propositionB, it follows by ‘reversing the argument’ that the validity of B automaticallyimplies that of A.As an example, let proposition A be that an integer N is divisible withoutremainder by 6, and proposition B be that N is divisible without remainder by2. Clearly, if A is true then it follows that B is true, i.e. A is a sufficient conditionfor B; it is not however a necessary condition, as is trivially shown by taking Nas 8. Conversely, the same value of N shows that whilst the validity of B is anecessary condition for A to hold, it is not sufficient.An alternative terminology to ‘necessary’ and ‘sufficient’ often employed bymathematicians is that of ‘if’ and ‘only if’, particularly in the combination ‘if andonly if’ which is usually written as IFF or denoted by a double-headed arrow⇐⇒ . The equivalent statements can be summarised byA if B A is true if B is true or B =⇒ A,B is a sufficient condition for A B =⇒ A,A only if B A is true only if B is true or A =⇒ B,B is a necessary consequence of A A =⇒ B,34

1.7 SOME PARTICULAR METHODS OF PROOFA IFF B A is true if and only if B is true or B ⇐⇒ A,A and B necessarily imply each other B ⇐⇒ A.Although at this stage in the book we are able to employ for illustrative purposesonly simple and fairly obvious results, the following example is given as a modelof how necessary and sufficient conditions should be proved. The essential pointis that for the second part of the proof (whether it be the ‘necessary’ part or the‘sufficient’ part) one needs to start again from scratch; more often than not, thelines of the second part of the proof will not be simply those of the first writtenin reverse order.◮Prove that (A) a function f(x) is a quadratic polynomial with zeros at x =2and x =3if and only if (B) the function f(x) has the form λ(x 2 − 5x +6) with λ a non-zero constant.(1) Assume A, i.e.thatf(x) is a quadratic polynomial with zeros at x =2andx =3.Letits form be ax 2 + bx + c with a ≠ 0. Then we have4a +2b + c =0,9a +3b + c =0,and subtraction shows that 5a + b =0andb = −5a. Substitution of this into the first ofthe above equations gives c = −4a − 2b = −4a +10a =6a. Thus, it follows thatf(x) =a(x 2 − 5x +6) with a ≠0,and establishes the ‘A only if B’ part of the stated result.(2) Now assume that f(x) has the form λ(x 2 − 5x +6)withλ a non-zero constant. Firstlywe note that f(x) is a quadratic polynomial, and so it only remains to prove that itszeros occur at x =2andx =3.Considerf(x) = 0, which, after dividing through by thenon-zero constant λ, givesx 2 − 5x +6=0.We proceed by using a technique known as completing the square, for the purposes ofillustration, although the factorisation of the above equation should be clear to the reader.Thus we writex 2 − 5x +( 5 2 )2 − ( 5 2 )2 +6=0,(x − 5 2 )2 = 1 , 4x − 5 = ± 1 .2 2The two roots of f(x) = 0 are therefore x =2andx =3;thesex-values give the zerosof f(x). This establishes the second (‘A if B’) part of the result. Thus we have shownthat the assumption of either condition implies the validity of the other and the proof iscomplete. ◭It should be noted that the propositions have to be carefully and preciselyformulated. If, for example, the word ‘quadratic’ were omitted from A, statementB would still be a sufficient condition for A but not a necessary one, since f(x)could then be x 3 − 4x 2 + x + 6 and A would not require B. Omitting the constantλ from the stated form of f(x) inB has the same effect. Conversely, if A were tostate that f(x) =3(x − 2)(x − 3) then B would be a necessary condition for A butnot a sufficient one.35

PRELIMINARY ALGEBRA1.8 ExercisesPolynomial equations1.1 Continue the investigation of equation (1.7), namelyg(x) =4x 3 +3x 2 − 6x − 1,as follows.(a) Make a table of values of g(x) for integer values of x between −2 and2.Useit and the information derived in the text to draw a graph and so determinethe roots of g(x) = 0 as accurately as possible.(b) Find one accurate root of g(x) = 0 by inspection and hence determine precisevalues for the other two roots.(c) Show that f(x) =4x 3 +3x 2 − 6x − k = 0 has only one real root unless−5 ≤ k ≤ 7 . 41.2 Determine how the number of real roots of the equationg(x) =4x 3 − 17x 2 +10x + k =0depends upon k. Are there any cases for which the equation has exactly twodistinct real roots?1.3 Continue the analysis of the polynomial equationf(x) =x 7 +5x 6 + x 4 − x 3 + x 2 − 2=0,investigated in subsection 1.1.1, as follows.(a) By writing the fifth-degree polynomial appearing in the expression for f ′ (x)in the form 7x 5 +30x 4 + a(x − b) 2 + c, show that there is in fact only onepositive root of f(x) =0.(b) By evaluating f(1), f(0) and f(−1), and by inspecting the form of f(x) fornegative values of x, determine what you can about the positions of the realroots of f(x) =0.1.4 Given that x =2isonerootofg(x) =2x 4 +4x 3 − 9x 2 − 11x − 6=0,use factorisation to determine how many real roots it has.1.5 Construct the quadratic equations that have the following pairs of roots:(a) −6, −3; (b) 0, 4; (c) 2, 2;(d)3+2i, 3 − 2i, wherei 2 = −1.1.6 Use the results of (i) equation (1.13), (ii) equation (1.12) and (iii) equation (1.14)to prove that if the roots of 3x 3 − x 2 − 10x +8=0areα 1 ,α 2 and α 3 then(a) α1 −1 + α −12+ α −13=5/4,(b) α 2 1 + α2 2 + α2 3 =61/9,(c) α 3 1 + α3 2 + α3 3 = −125/27.(d) Convince yourself that eliminating (say) α 2 and α 3 from (i), (ii) and (iii) doesnot give a simple explicit way of finding α 1 .1.7 Prove thatby consideringTrigonometric identitiescos π 12 = √3+12 √ 236

1.8 EXERCISES(a) the sum of the sines of π/3 andπ/6,(b) the sine of the sum of π/3 andπ/4.1.8 The following exercises are based on the half-angle formulae.(a) Use the fact that sin(π/6) = 1/2 to prove that tan(π/12) = 2 − √ 3.(b) Use the result of (a) to show further that tan(π/24) = q(2 − q) whereq 2 =2+ √ 3.1.9 Find the real solutions of(a) 3 sin θ − 4cosθ =2,(b) 4 sin θ +3cosθ =6,(c) 12 sin θ − 5cosθ = −6.1.10 If s = sin(π/8), prove that8s 4 − 8s 2 +1=0,and hence show that s =[(2− √ 2)/4] 1/2 .1.11 Find all the solutions ofsin θ +sin4θ =sin2θ +sin3θthat lie in the range −π

PRELIMINARY ALGEBRA1.16 Express the following in partial fraction form:(a) 2x3 − 5x +1x 2 − 2x − 8 , (b) x2 + x − 1x 2 + x − 2 .1.17 Rearrange the following functions in partial fraction form:x − 6(a)x 3 − x 2 +4x − 4 , (b) x3 +3x 2 + x +19.x 4 +10x 2 +91.18 Resolve the following into partial fractions in such a way that x does not appearin any numerator:(a)2x 2 + x +1(x − 1) 2 (x +3) , (b) x 2 − 2x 3 +8x 2 +16x , (c) x 3 − x − 1(x +3) 3 (x +1) .Binomial expansion1.19 Evaluate those of the following that are defined: (a) 5 C 3 ,(b) 3 C 5 ,(c) −5 C 3 ,(d)−3 C 5 .1.20 Use a binomial expansion to evaluate 1/ √ 4.2 to five places of decimals, andcompare it with the accurate answer obtained using a calculator.Proof by induction and contradiction1.21 Prove by induction thatn∑n∑r = 1 n(n +1) and r 3 = 1 2 4 n2 (n +1) 2 .r=11.22 Prove by induction that1+r + r 2 + ···+ r k + ···+ r n = 1 − rn+1.1 − r1.23 Prove that 3 2n +7,wheren is a non-negative integer, is divisible by 8.1.24 If a sequence of terms, u n , satisfies the recurrence relation u n+1 =(1− x)u n + nx,with u 1 = 0, show, by induction, that, for n ≥ 1,r=1u n = 1 x [nx − 1+(1− x)n ].1.25 Prove by induction thatn∑( )1 θ2 tan = 1 ( ) θ r 2 r 2 cot − cot θ.n 2 n r=11.26 The quantities a i in this exercise are all positive real numbers.(a) Show that( ) 2 a1 + a 2a 1 a 2 ≤.2(b) Hence prove, by induction on m, that( ) p a1 + a 2 + ···+ a pa 1 a 2 ···a p ≤,pwhere p =2 m with m a positive integer. Note that each increase of m byunity doubles the number of factors in the product.38

1.9 HINTS AND ANSWERS1.27 Establish the values of k for which the binomial coefficient p C k is divisible by pwhen p is a prime number. Use your result and the method of induction to provethat n p − n is divisible by p for all integers n and all prime numbers p. Deducethat n 5 − n isdivisibleby30foranyintegern.1.28 An arithmetic progression of integers a n is one in which a n = a 0 + nd, wherea 0and d are integers and n takes successive values 0, 1, 2,....(a) Show that if any one term of the progression is the cube of an integer thenso are infinitely many others.(b) Show that no cube of an integer can be expressed as 7n + 5 for some positiveinteger n.1.29 Prove, by the method of contradiction, that the equationx n + a n−1 x n−1 + ···+ a 1 x + a 0 =0,in which all the coefficients a i are integers, cannot have a rational root, unlessthat root is an integer. Deduce that any integral root must be a divisor of a 0 andhence find all rational roots of(a) x 4 +6x 3 +4x 2 +5x +4=0,(b) x 4 +5x 3 +2x 2 − 10x +6=0.Necessary and sufficient conditions1.30 Prove that the equation ax 2 + bx + c =0,inwhicha, b and c are real and a>0,has two real distinct solutions IFF b 2 > 4ac.1.31 For the real variable x, show that a sufficient, but not necessary, condition forf(x) =x(x + 1)(2x + 1) to be divisible by 6 is that x is an integer.1.32 Given that at least one of a and b, and at least one of c and d, are non-zero,show that ad = bc is both a necessary and sufficient condition for the equationsax + by =0,cx + dy =0,to have a solution in which at least one of x and y is non-zero.1.33 The coefficients a i in the polynomial Q(x) =a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x are allintegers. Show that Q(n) is divisible by 24 for all integers n ≥ 0 if and only if allof the following conditions are satisfied:(i) 2a 4 + a 3 is divisible by 4;(ii) a 4 + a 2 is divisible by 12;(iii) a 4 + a 3 + a 2 + a 1 is divisible by 24.1.9 Hints and answers1.1 (b) The roots are 1, 1 (−7+√ 33) = −0.1569, 1 (−7 − √ 33) = −1.593. (c) −5 and8 87are the values of k that make f(−1) and f( 1 ) equal to zero.4 21.3 (a) a =4,b= 3 23and c = are all positive. Therefore 8 16 f′ (x) > 0forallx>0.(b) f(1) = 5, f(0) = −2 andf(−1) = 5, and so there is at least one root in eachof the ranges 0

PRELIMINARY ALGEBRA1.11 Show that the equation is equivalent to sin(5θ/2) sin(θ)sin(θ/2) = 0.Solutions are −4π/5, −2π/5, 0, 2π/5, 4π/5,π.Thesolutionθ = 0 has multiplicity3.1.13 (a) A circle of radius 5 centred on (−3, −4).(b) A hyperbola with ‘centre’ (3, −2) and ‘semi-axes’ 2 and 3.(c) The expression factorises into two lines, x +2y − 3=0and2x + y +2=0.(d) Write the expression as (x+y) 2 =8(x−y) to see that it represents a parabolapassing through the origin, with the line x + y = 0 as its axis of symmetry.51.15 (a)7(x − 2) + 97(x +5) , (b) − 43x + 43(x − 3) .1.17 (a) x +2x 2 +4 − 1x +1, (b)x − 1 x 2 +9 + 2x 2 +1 .1.19 (a) 10, (b) not defined, (c) −35, (d) −21.1.21 Look for factors common to the n = N sum and the additional n = N +1term,so as to reduce the sum for n = N +1toasingleterm.1.23 Write 3 2n as 8m − 7.1.25 Use the half-angle formulae of equations (1.32) to (1.34) to relate functions ofθ/2 k to those of θ/2 k+1 .1.27 Divisible for k =1, 2,...,p− 1. Expand (n +1) p as n p + ∑ p−11p C k n k + 1. Applythe stated result for p = 5. Note that n 5 − n = n(n − 1)(n +1)(n 2 + 1); the productof any three consecutive integers must divide by both 2 and 3.1.29 By assuming x = p/q with q ≠ 1, show that a fraction −p n /q isequaltoaninteger a n−1 p n−1 + ···+ a 1 pq n−2 + a 0 q n−1 . This is a contradiction, and is onlyresolved if q = 1 and the root is an integer.(a) The only possible candidates are ±1, ±2, ±4. None is a root.(b) The only possible candidates are ±1, ±2, ±3, ±6. Only −3 isaroot.1.31 f(x) can be written as x(x +1)(x +2)+x(x +1)(x − 1). Each term consists ofthe product of three consecutive integers, of which one must therefore divide by2 and (a different) one by 3. Thus each term separately divides by 6, and sotherefore does f(x). Note that if x is the root of 2x 3 +3x 2 + x − 24 = 0 that liesnear the non-integer value x =1.826, then x(x + 1)(2x + 1) = 24 and thereforedivides by 6.1.33 Note that, e.g., the condition for 6a 4 + a 3 to be divisible by 4 is the same as thecondition for 2a 4 + a 3 to be divisible by 4.For the necessary (only if) part of the proof set n =1, 2, 3 and take integercombinations of the resulting equations.For the sufficient (if) part of the proof use the stated conditions to prove theproposition by induction. Note that n 3 − n is divisible by 6 and that n 2 + n iseven.40

2Preliminary calculusThis chapter is concerned with the formalism of probably the most widely usedmathematical technique in the physical sciences, namely the calculus. The chapterdivides into two sections. The first deals with the process of differentiation and thesecond with its inverse process, integration. The material covered is essential forthe remainder of the book and serves as a reference. Readers who have previouslystudied these topics should ensure familiarity by looking at the worked examplesin the main text and by attempting the exercises at the end of the chapter.2.1 DifferentiationDifferentiation is the process of determining how quickly or slowly a functionvaries, as the quantity on which it depends, its argument, is changed. Morespecifically it is the procedure for obtaining an expression (numerical or algebraic)for the rate of change of the function with respect to its argument. Familiarexamples of rates of change include acceleration (the rate of change of velocity)and chemical reaction rate (the rate of change of chemical composition). Bothacceleration and reaction rate give a measure of the change of a quantity withrespect to time. However, differentiation may also be applied to changes withrespect to other quantities, for example the change in pressure with respect to achange in temperature.Although it will not be apparent from what we have said so far, differentiationis in fact a limiting process, that is, it deals only with the infinitesimal change inone quantity resulting from an infinitesimal change in another.2.1.1 Differentiation from first principlesLet us consider a function f(x) that depends on only one variable x, together withnumerical constants, for example, f(x) =3x 2 or f(x) =sinx or f(x) =2+3/x.41

PRELIMINARY CALCULUSf(x +∆x)AP∆ff(x)θx∆xx +∆xFigure 2.1 The graph of a function f(x) showing that the gradient or slopeof the function at P ,givenbytanθ, is approximately equal to ∆f/∆x.Figure 2.1 shows an example of such a function. Near any particular point,P , the value of the function changes by an amount ∆f, say, as x changesby a small amount ∆x. The slope of the tangent to the graph of f(x) at Pis then approximately ∆f/∆x, and the change in the value of the function is∆f = f(x +∆x) − f(x). In order to calculate the true value of the gradient, orfirst derivative, of the function at P , we must let ∆x become infinitesimally small.We therefore define the first derivative of f(x) asf ′ (x) ≡ df(x)dx≡ lim∆x→0f(x +∆x) − f(x), (2.1)∆xprovided that the limit exists. The limit will depend in almost all cases on thevalue of x. If the limit does exist at a point x = a then the function is said to bedifferentiable at a; otherwise it is said to be non-differentiable at a. Theformalconcept of a limit and its existence or non-existence is discussed in chapter 4; forpresent purposes we will adopt an intuitive approach.In the definition (2.1), we allow ∆x to tend to zero from either positive ornegative values and require the same limit to be obtained in both cases. Afunction that is differentiable at a is necessarily continuous at a (there must beno jump in the value of the function at a), though the converse is not necessarilytrue. This latter assertion is illustrated in figure 2.1: the function is continuousat the ‘kink’ A but the two limits of the gradient as ∆x tends to zero frompositive or negative values are different and so the function is not differentiableat A.It should be clear from the above discussion that near the point P we may42

2.1 DIFFERENTIATIONapproximate the change in the value of the function, ∆f, that results from a smallchange ∆x in x by∆f ≈ df(x) ∆x. (2.2)dxAs one would expect, the approximation improves as the value of ∆x is reduced.In the limit in which the change ∆x becomes infinitesimally small, we denote itby the differential dx, and (2.2) readsdf = df(x) dx. (2.3)dxThis equality relates the infinitesimal change in the function, df, to the infinitesimalchange dx that causes it.So far we have discussed only the first derivative of a function. However, wecan also define the second derivative as the gradient of the gradient of a function.Again we use the definition (2.1) but now with f(x) replaced by f ′ (x). Hence thesecond derivative is defined byf ′′ f ′ (x +∆x) − f ′ (x)(x) ≡ lim, (2.4)∆x→0 ∆xprovided that the limit exists. A physical example of a second derivative is thesecond derivative of the distance travelled by a particle with respect to time. Sincethe first derivative of distance travelled gives the particle’s velocity, the secondderivative gives its acceleration.We can continue in this manner, the nth derivative of the function f(x) beingdefined byf (n) f (n−1) (x +∆x) − f (n−1) (x)(x) ≡ lim. (2.5)∆x→0 ∆xIt should be noted that with this notation f ′ (x) ≡ f (1) (x), f ′′ (x) ≡ f (2) (x), etc., andthat formally f (0) (x) ≡ f(x).All this should be familiar to the reader, though perhaps not with such formaldefinitions. The following example shows the differentiation of f(x) =x 2 from firstprinciples. In practice, however, it is desirable simply to remember the derivativesof standard functions; the techniques given in the remainder of this section canbe applied to find more complicated derivatives.43

PRELIMINARY CALCULUS◮Find from first principles the derivative with respect to x of f(x) =x 2 .Using the definition (2.1),f ′ f(x +∆x) − f(x)(x) = lim∆x→0 ∆x(x +∆x) 2 − x 2= lim∆x→0 ∆x2x∆x +(∆x) 2= lim∆x→0 ∆x= lim (2x +∆x).∆x→0As ∆x tends to zero, 2x +∆x tends towards 2x, hencef ′ (x) =2x. ◭Derivatives of other functions can be obtained in the same way. The derivativesof some simple functions are listed below (note that a is a constant):ddx (xn ) = nx n−1 ,dd(sin ax) = a cos ax,dxddx (tan ax) = a sec2 ax,ddx (eax ) = ae ax ,d(cos ax) = −a sin ax,dxddx (cot ax) = −a cosec2 ax,d(cos −1 x )−1= √dx a a2 − x ,2ddx (ln ax) = 1 x ,(sec ax) = a sec ax tan ax,dxd(cosec ax) = −a cosec ax cot ax,dxd(sin −1 x )1= √dx a a2 − x ,2d(tan −1 x )a=dx a a 2 + x 2 .Differentiation from first principles emphasises the definition of a derivative asthe gradient of a function. However, for most practical purposes, returning to thedefinition (2.1) is time consuming and does not aid our understanding. Instead, asmentioned above, we employ a number of techniques, which use the derivativeslisted above as ‘building blocks’, to evaluate the derivatives of more complicatedfunctions than hitherto encountered. Subsections 2.1.2–2.1.7 develop the methodsrequired.2.1.2 Differentiation of productsAs a first example of the differentiation of a more complicated function, weconsider finding the derivative of a function f(x) that can be written as theproduct of two other functions of x, namely f(x) =u(x)v(x). For example, iff(x) = x 3 sin x then we might take u(x) = x 3 and v(x) = sinx. Clearly the44

2.1 DIFFERENTIATIONseparation is not unique. (In the given example, possible alternative break-upswould be u(x) =x 2 , v(x) =x sin x, orevenu(x) =x 4 tan x, v(x) =x −1 cos x.)The purpose of the separation is to split the function into two (or more) parts,of which we know the derivatives (or at least we can evaluate these derivativesmore easily than that of the whole). We would gain little, however, if we didnot know the relationship between the derivative of f and those of u and v.Fortunately, they are very simply related, as we shall now show.Since f(x) is written as the product u(x)v(x), it follows thatf(x +∆x) − f(x) =u(x +∆x)v(x +∆x) − u(x)v(x)= u(x +∆x)[v(x +∆x) − v(x)] + [u(x +∆x) − u(x)]v(x).From the definition of a derivative (2.1),dfdx = lim f(x +∆x) − f(x)∆x→0{∆x= lim u(x +∆x)∆x→0 ∆x[ v(x +∆x) − v(x)]+[ u(x +∆x) − u(x)∆x] }v(x) .In the limit ∆x → 0, the factors in square brackets become dv/dx and du/dx(by the definitions of these quantities) and u(x +∆x) simply becomes u(x).Consequently we obtaindfdx = d [u(x)v(x)] = u(x)dv(x)dx dx+ du(x) v(x). (2.6)dxIn primed notation and without writing the argument x explicitly, (2.6) is statedconcisely asf ′ =(uv) ′ = uv ′ + u ′ v. (2.7)This is a general result obtained without making any assumptions about thespecific forms f, u and v, other than that f(x) =u(x)v(x). In words, the resultreads as follows. The derivative of the product of two functions is equal to thefirst function times the derivative of the second plus the second function times thederivative of the first.◮Find the derivative with respect to x of f(x) =x 3 sin x.Using the product rule, (2.6),ddx (x3 sin x) =x 3 d d(sin x)+dx dx (x3 )sinx= x 3 cos x +3x 2 sin x. ◭The product rule may readily be extended to the product of three or morefunctions. Considering the functionf(x) =u(x)v(x)w(x) (2.8)45

PRELIMINARY CALCULUSand using (2.6), we obtain, as before omitting the argument,dfdx = u d du(vw) +dx dx vw.Using (2.6) again to expand the first term on the RHS gives the complete resultddw(uvw) =uvdx dx + u dvdx w + du vw (2.9)dxor(uvw) ′ = uvw ′ + uv ′ w + u ′ vw. (2.10)It is readily apparent that this can be extended to products containing any numbern of factors; the expression for the derivative will then consist of n terms withthe prime appearing in successive terms on each of the n factors in turn. This isprobably the easiest way to recall the product rule.2.1.3 The chain ruleProducts are just one type of complicated function that we may encounter indifferentiation. Another is the function of a function, e.g. f(x) =(3+x 2 ) 3 = u(x) 3 ,where u(x) =3+x 2 .If∆f, ∆u and ∆x are small finite quantities, it follows that∆f∆x = ∆f ∆u∆u ∆x ;As the quantities become infinitesimally small we obtaindfdx = df dudu dx . (2.11)This is the chain rule, which we must apply when differentiating a function of afunction.◮Find the derivative with respect to x of f(x) =(3+x 2 ) 3 .Rewriting the function as f(x) =u 3 ,whereu(x) =3+x 2 , and applying (2.11) we finddf du d=3u2dx dx =3u2 dx (3 + x2 )=3u 2 × 2x =6x(3 + x 2 ) 2 . ◭Similarly, the derivative with respect to x of f(x) =1/v(x) may be obtained byrewriting the function as f(x) =v −1 and applying (2.11):df dv= −v−2dx dx = − 1 dvv 2 dx . (2.12)The chain rule is also useful for calculating the derivative of a function f withrespect to x when both x and f are written in terms of a variable (or parameter),say t.46

2.1 DIFFERENTIATION◮Find the derivative with respect to x of f(t) =2at, wherex = at 2 .We could of course substitute for t and then differentiate f as a function of x, but in thiscase it is quicker to usedfdx = df dtdt dx =2a 12at = 1 t ,where we have used the fact that( ) −1dt dxdx = . ◭dt2.1.4 Differentiation of quotientsApplying (2.6) for the derivative of a product to a function f(x) =u(x)[1/v(x)],we may obtain the derivative of the quotient of two factors. Thus( u) ( ) ′ ′ ( ) )1 1f ′ = = u + u ′ = u(− v′v v v v 2 + u′v ,where (2.12) has been used to evaluate (1/v) ′ . This can now be rearranged intothe more convenient and memorisable form( u) ′f ′ vu ′ − uv ′= =v v 2 . (2.13)This can be expressed in words as the derivative of a quotient is equal to the bottomtimes the derivative of the top minus the top times the derivative of the bottom, allover the bottom squared.◮Find the derivative with respect to x of f(x) =sinx/x.Using (2.13) with u(x) =sinx, v(x) =x and hence u ′ (x) =cosx, v ′ (x) = 1, we findf ′ (x) =x cos x − sin xx 2= cos xx− sin x . ◭x 22.1.5 Implicit differentiationSo far we have only differentiated functions written in the form y = f(x).However, we may not always be presented with a relationship in this simpleform. As an example consider the relation x 3 − 3xy + y 3 = 2. In this case it isnot possible to rearrange the equation to give y as a function of x. Nevertheless,by differentiating term by term with respect to x (implicit differentiation), we canfind the derivative of y.47

PRELIMINARY CALCULUS◮Find dy/dx if x 3 − 3xy + y 3 =2.Differentiating each term in the equation with respect to x we obtainddx (x3 ) − ddx (3xy)+ ddx (y3 )= ddx (2),(⇒ 3x 2 − 3x dy )dx +3y +3y 2 dydx =0,where the derivative of 3xy has been found using the product rule. Hence, rearranging fordy/dx,dydx = y − x2y 2 − x .Note that dy/dx is a function of both x and y and cannot be expressed as a function of xonly. ◭2.1.6 Logarithmic differentiationIn circumstances in which the variable with respect to which we are differentiatingis an exponent, taking logarithms and then differentiating implicitly is the simplestway to find the derivative.◮Find the derivative with respect to x of y = a x .To find the required derivative we first take logarithms and then differentiate implicitly:ln y =lna x 1 dy= x ln a ⇒y dx =lna.Now, rearranging and substituting for y, we finddydx = y ln a = ax ln a. ◭2.1.7 Leibnitz’ theoremWe have discussed already how to find the derivative of a product of two or morefunctions. We now consider Leibnitz’ theorem, which gives the correspondingresults for the higher derivatives of products.Consider again the function f(x) =u(x)v(x). We know from the product rulethat f ′ = uv ′ + u ′ v. Using the rule once more for each of the products, we obtainf ′′ =(uv ′′ + u ′ v ′ )+(u ′ v ′ + u ′′ v)= uv ′′ +2u ′ v ′ + u ′′ v.Similarly, differentiating twice more givesf ′′′ = uv ′′′ +3u ′ v ′′ +3u ′′ v ′ + u ′′′ v,f (4) = uv (4) +4u ′ v ′′′ +6u ′′ v ′′ +4u ′′′ v ′ + u (4) v.48

2.1 DIFFERENTIATIONThe pattern emerging is clear and strongly suggests that the results generalise tof (n) =n∑r=0n!r!(n − r)! u(r) v (n−r) =n∑n C r u (r) v (n−r) , (2.14)where the fraction n!/[r!(n − r)!] is identified with the binomial coefficient n C r(see chapter 1). To prove that this is so, we use the method of induction as follows.Assume that (2.14) is valid for n equal to some integer N. Thenf (N+1) ===N∑r=0N d (C r u (r) v (N−r))dxr=0N∑N C r [u (r) v (N−r+1) + u (r+1) v (N−r) ]r=0N∑N+1∑N C s u (s) v (N+1−s) +N C s−1 u (s) v (N+1−s) ,s=0where we have substituted summation index s for r in the first summation, andfor r + 1 in the second. Now, from our earlier discussion of binomial coefficients,equation (1.51), we haves=1N C s + N C s−1 = N+1 C sand so, after separating out the first term of the first summation and the lastterm of the second, obtainf (N+1) = N C 0 u (0) v (N+1) +N∑N+1 C s u (s) v (N+1−s) + N C N u (N+1) v (0) .s=1But N C 0 =1= N+1 C 0 and N C N =1= N+1 C N+1 , and so we may writef (N+1) = N+1 C 0 u (0) v (N+1) +N∑N+1 C s u (s) v (N+1−s) + N+1 C N+1 u (N+1) v (0)s=1N+1∑=N+1 C s u (s) v (N+1−s) .s=0This is just (2.14) with n set equal to N + 1. Thus, assuming the validity of (2.14)for n = N implies its validity for n = N + 1. However, when n =1equation(2.14) is simply the product rule, and this we have already proved directly. Theseresults taken together establish the validity of (2.14) for all n and prove Leibnitz’theorem.49

PRELIMINARY CALCULUSf(x)QABCSFigure 2.2 A graph of a function, f(x), showing how differentiation correspondsto finding the gradient of the function at a particular point. Points B,Q and S are stationary points (see text).x◮Find the third derivative of the function f(x) =x 3 sin x.Using (2.14) we immediately findf ′′′ (x) =6sinx +3(6x)cosx +3(3x 2 )(− sin x)+x 3 (− cos x)=3(2− 3x 2 )sinx + x(18 − x 2 )cosx. ◭2.1.8 Special points of a functionWe have interpreted the derivative of a function as the gradient of the function atthe relevant point (figure 2.1). If the gradient is zero for some particular value ofx then the function is said to have a stationary point there. Clearly, in graphicalterms, this corresponds to a horizontal tangent to the graph.Stationary points may be divided into three categories and an example of eachis shown in figure 2.2. Point B is said to be a minimum since the function increasesin value in both directions away from it. Point Q is said to be a maximum sincethe function decreases in both directions away from it. Note that B is not theoverall minimum value of the function and Q is not the overall maximum; rather,they are a local minimum and a local maximum. Maxima and minima are knowncollectively as turning points.The third type of stationary point is the stationary point of inflection, S. Inthis case the function falls in the positive x-direction and rises in the negativex-direction so that S is neither a maximum nor a minimum. Nevertheless, thegradient of the function is zero at S, i.e. the graph of the function is flat there,and this justifies our calling it a stationary point. Of course, a point at which the50

2.1 DIFFERENTIATIONgradient of the function is zero but the function rises in the positive x-directionand falls in the negative x-direction is also a stationary point of inflection.The above distinction between the three types of stationary point has beenmade rather descriptively. However, it is possible to define and distinguish stationarypoints mathematically. From their definition as points of zero gradient,all stationary points must be characterised by df/dx = 0. In the case of theminimum, B, the slope, i.e. df/dx, changes from negative at A to positive at Cthrough zero at B. Thus df/dx is increasing and so the second derivative d 2 f/dx 2must be positive. Conversely, at the maximum, Q, we must have that d 2 f/dx 2 isnegative.It is less obvious, but intuitively reasonable, that at S, d 2 f/dx 2 is zero. This maybe inferred from the following observations. To the left of S the curve is concaveupwards so that df/dx is increasing with x and hence d 2 f/dx 2 > 0. To the rightof S, however, the curve is concave downwards so that df/dx is decreasing withx and hence d 2 f/dx 2 < 0.In summary, at a stationary point df/dx = 0 and(i) for a minimum, d 2 f/dx 2 > 0,(ii) for a maximum, d 2 f/dx 2 < 0,(iii) for a stationary point of inflection, d 2 f/dx 2 = 0 and d 2 f/dx 2 changes signthrough the point.In case (iii), a stationary point of inflection, in order that d 2 f/dx 2 changes signthrough the point we normally require d 3 f/dx 3 ≠ 0 at that point. This simplerule can fail for some functions, however, and in general if the first non-vanishingderivative of f(x) at the stationary point is f (n) then if n is even the point is amaximum or minimum and if n is odd the point is a stationary point of inflection.This may be seen from the Taylor expansion (see equation (4.17)) of the functionabout the stationary point, but it is not proved here.◮Find the positions and natures of the stationary points of the functionf(x) =2x 3 − 3x 2 − 36x +2.The first criterion for a stationary point is that df/dx = 0, and hence we setdfdx =6x2 − 6x − 36 = 0,from which we obtain(x − 3)(x +2)=0.Hence the stationary points are at x =3andx = −2. To determine the nature of thestationary point we must evaluate d 2 f/dx 2 :d 2 f=12x − 6.dx2 51

PRELIMINARY CALCULUSf(x)GxFigure 2.3 The graph of a function f(x) that has a general point of inflectionat the point G.Now, we examine each stationary point in turn. For x =3,d 2 f/dx 2 = 30. Since this ispositive, we conclude that x = 3 is a minimum. Similarly, for x = −2, d 2 f/dx 2 = −30 andso x = −2 is a maximum. ◭So far we have concentrated on stationary points, which are defined to havedf/dx = 0. We have found that at a stationary point of inflection d 2 f/dx 2 isalso zero and changes sign. This naturally leads us to consider points at whichd 2 f/dx 2 is zero and changes sign but at which df/dx is not, in general, zero. Suchpoints are called general points of inflection or simply points of inflection. Clearly,a stationary point of inflection is a special case for which df/dx is also zero.At a general point of inflection the graph of the function changes from beingconcave upwards to concave downwards (or vice versa), but the tangent to thecurve at this point need not be horizontal. A typical example of a general pointof inflection is shown in figure 2.3.The determination of the stationary points of a function, together with theidentification of its zeros, infinities and possible asymptotes, is usually sufficientto enable a graph of the function showing most of its significant features to besketched. Some examples for the reader to try are included in the exercises at theend of this chapter.2.1.9 Curvature of a functionIn the previous section we saw that at a point of inflection of the functionf(x), the second derivative d 2 f/dx 2 changes sign and passes through zero. Thecorresponding graph of f shows an inversion of its curvature at the point ofinflection. We now develop a more quantitative measure of the curvature of afunction (or its graph), which is applicable at general points and not just in theneighbourhood of a point of inflection.As in figure 2.1, let θ be the angle made with the x-axis by the tangent at a52

2.1 DIFFERENTIATIONf(x)C∆θρPQθθ +∆θxFigure 2.4 Two neighbouring tangents to the curve f(x) whose slopes differby ∆θ. The angular separation of the corresponding radii of the circle ofcurvature is also ∆θ.point P on the curve f = f(x), with tan θ = df/dx evaluated at P . Now consideralso the tangent at a neighbouring point Q on the curve, and suppose that itmakes an angle θ +∆θ with the x-axis, as illustrated in figure 2.4.It follows that the corresponding normals at P and Q, which are perpendicularto the respective tangents, also intersect at an angle ∆θ. Furthermore, their pointof intersection, C in the figure, will be the position of the centre of a circle thatapproximates the arc PQ, at least to the extent of having the same tangents atthe extremities of the arc. This circle is called the circle of curvature.For a finite arc PQ, the lengths of CP and CQ will not, in general, be equal,as they would be if f = f(x) were in fact the equation of a circle. But, as Qis allowed to tend to P ,i.e.as∆θ → 0, they do become equal, their commonvalue being ρ, the radius of the circle, known as the radius of curvature. It followsimmediately that the curve and the circle of curvature have a common tangentat P and lie on the same side of it. The reciprocal of the radius of curvature, ρ −1 ,defines the curvature of the function f(x) at the point P .The radius of curvature can be defined more mathematically as follows. Thelength ∆s of arc PQ is approximately equal to ρ∆θ and, in the limit ∆θ → 0, thisrelationship defines ρ as∆sρ = lim∆θ→0 ∆θ = dsdθ . (2.15)It should be noted that, as s increases, θ may increase or decrease according towhether the curve is locally concave upwards (i.e. shaped as if it were near aminimum in f(x)) or concave downwards. This is reflected in the sign of ρ, whichtherefore also indicates the position of the curve (and of the circle of curvature)53

PRELIMINARY CALCULUSrelative to the common tangent, above or below. Thus a negative value of ρindicates that the curve is locally concave downwards and that the tangent liesabove the curve.We next obtain an expression for ρ, not in terms of s and θ but in termsof x and f(x). The expression, though somewhat cumbersome, follows from thedefining equation (2.15), the defining property of θ that tan θ = df/dx ≡ f ′ andthe fact that the rate of change of arc length with x is given by[ ( ) ] 2 1/2ds dfdx = 1+. (2.16)dxThis last result, simply quoted here, is proved more formally in subsection 2.2.13.From the chain rule (2.11) it follows thatρ = dsdθ = ds dxdx dθ . (2.17)Differentiating both sides of tan θ = df/dx with respect to x givessec 2 θ dθdx = d2 fdx 2 ≡ f′′ ,from which, using sec 2 θ =1+tan 2 θ =1+(f ′ ) 2 , we can obtain dx/dθ asdxdθ = 1+tan2 θf ′′ = 1+(f′ ) 2f ′′ . (2.18)Substituting (2.16) and (2.18) into (2.17) then yields the final expression for ρ,ρ =[1+(f ′ ) 2] 3/2f ′′ . (2.19)It should be noted that the quantity in brackets is always positive and that itsthree-halves root is also taken as positive. The sign of ρ is thus solely determinedby that of d 2 f/dx 2 , in line with our previous discussion relating the sign towhether the curve is concave or convex upwards. If, as happens at a point ofinflection, d 2 f/dx 2 is zero then ρ is formally infinite and the curvature of f(x) iszero. As d 2 f/dx 2 changes sign on passing through zero, both the local tangentand the circle of curvature change from their initial positions to the opposite sideof the curve.54

2.1 DIFFERENTIATION◮Show that the radius of curvature at the point (x, y) on the ellipsex 2a + y22 b =1 2has magnitude (a 4 y 2 + b 4 x 2 ) 3/2 /(a 4 b 4 ) and the opposite sign to y. Check the special caseb = a, for which the ellipse becomes a circle.Differentiating the equation of the ellipse with respect to x gives2xa + 2y dy2 b 2 dx =0and sodydx = − b2 xa 2 y .A second differentiation, using (2.13), then yields( ) ( )d 2 ydx = − b2 y − xy′= − b4 y22 a 2 y 2 a 2 y 3 b + x2= − b42 a 2 a 2 y , 3where we have used the fact that (x, y) lies on the ellipse. We note that d 2 y/dx 2 , and henceρ, has the opposite sign to y 3 and hence to y. Substituting in (2.19) gives for the magnitudeof the radius of curvature[ 1+b 4 x 2 /(a 4 y 2 ) ] 3/2|ρ| =∣ −b 4 /(a 2 y 3 ) ∣ = (a4 y 2 + b 4 x 2 ) 3/2.a 4 b 4For the special case b = a, |ρ| reduces to a −2 (y 2 + x 2 ) 3/2 and, since x 2 + y 2 = a 2 ,thisinturn gives |ρ| = a, as expected. ◭The discussion in this section has been confined to the behaviour of curvesthat lie in one plane; examples of the application of curvature to the bending ofloaded beams and to particle orbits under the influence of a central forces can befound in the exercises at the ends of later chapters. A more general treatment ofcurvature in three dimensions is given in section 10.3, where a vector approach isadopted.2.1.10 Theorems of differentiationRolle’s theoremRolle’s theorem (figure 2.5) states that if a function f(x) is continuous in therange a ≤ x ≤ c, is differentiable in the range a

PRELIMINARY CALCULUSf(x)a b cxFigure 2.5 The graph of a function f(x), showing that if f(a) =f(c) thenatone point at least between x = a and x = c the graph has zero gradient.f(x)f(c)Cf(a)Aa b cxFigure 2.6 The graph of a function f(x); at some point x = b it has the samegradient as the line AC.Mean value theoremThe mean value theorem (figure 2.6) states that if a function f(x) is continuousin the range a ≤ x ≤ c and differentiable in the range a

2.1 DIFFERENTIATIONand hence the difference between the curve and the line ish(x) =f(x) − g(x) =f(x) − f(a) − (x − a)f(c) − f(a).c − aSince the curve and the line intersect at A and C, h(x) = 0 at both of these points.Hence, by an application of Rolle’s theorem, h ′ (x) = 0 for at least one point bbetween A and C. Differentiating our expression for h(x), we findand hence at b, whereh ′ (x) =0,h ′ (x) =f ′ (x) −f ′ (b) =f(c) − f(a),c − af(c) − f(a).c − aApplications of Rolle’s theorem and the mean value theoremSince the validity of Rolle’s theorem is intuitively obvious, given the conditionsimposed on f(x), it will not be surprising that the problems that can be solvedby applications of the theorem alone are relatively simple ones. Nevertheless wewill illustrate it with the following example.◮What semi-quantitative results can be deduced by applying Rolle’s theorem to the followingfunctions f(x), witha and c chosen so that f(a) =f(c) =0?(i)sinx, (ii) cos x,(iii)x 2 − 3x +2, (iv) x 2 +7x +3, (v) 2x 3 − 9x 2 − 24x + k.(i) If the consecutive values of x that make sin x =0areα 1 ,α 2 ,... (actually x = nπ, forany integer n) then Rolle’s theorem implies that the derivative of sin x, namelycosx, hasat least one zero lying between each pair of values α i and α i+1 .(ii) In an exactly similar way, we conclude that the derivative of cos x, namely− sin x,has at least one zero lying between consecutive pairs of zeros of cos x. Thesetworesultstaken together (but neither separately) imply that sin x and cos x have interleavingzeros.(iii) For f(x) =x 2 − 3x +2, f(a) =f(c) =0ifa and c are taken as 1 and 2 respectively.Rolle’s theorem then implies that f ′ (x) =2x − 3=0hasasolutionx = b with b in therange 1

PRELIMINARY CALCULUSIn each case, as might be expected, the application of Rolle’s theorem does no more thanfocus attention on particular ranges of values; it does not yield precise answers. ◭Direct verification of the mean value theorem is straightforward when it isapplied to simple functions. For example, if f(x) =x 2 , it states that there is avalue b in the interval a

2.2 INTEGRATIONf(x)abxFigure 2.7An integral as the area under a curve.2.2 IntegrationThe notion of an integral as the area under a curve will be familiar to the reader.In figure 2.7, in which the solid line is a plot of a function f(x), the shaded arearepresents the quantity denoted byI =∫ baf(x) dx. (2.21)This expression is known as the definite integral of f(x) betweenthelower limitx = a and the upper limit x = b, andf(x) is called the integrand.2.2.1 Integration from first principlesThe definition of an integral as the area under a curve is not a formal definition,but one that can be readily visualised. The formal definition of I involvessubdividing the finite interval a ≤ x ≤ b into a large number of subintervals, bydefining intermediate points ξ i such that a = ξ 0

PRELIMINARY CALCULUSf(x)ax 1 x 2 x 3 x 4 x 5ξ 1 ξ 2 ξ 3 ξ 4bxFigure 2.8 The evaluation of a definite integral by subdividing the intervala ≤ x ≤ b into subintervals.◮Evaluate from first principles the integral I = ∫ b0 x2 dx.We first approximate the area under the curve y = x 2 between 0 and b by n rectangles ofequal width h. If we take the value at the lower end of each subinterval (in the limit of aninfinite number of subintervals we could equally well have chosen the value at the upperend) to give the height of the corresponding rectangle, then the area of the kth rectanglewill be (kh) 2 h = k 2 h 3 . The total area is thus∑n−1A = k 2 h 3 =(h 3 ) 1 n(n − 1)(2n − 1),6k=0where we have used the expression for the sum of the squares of the natural numbersderived in subsection 1.7.1. Now h = b/n and so( ) (b3 n b3A = (n − 1)(2n − 1) = 1 − 1 )(2 − 1 ).n 3 6 6 n nAs n →∞, A → b 3 /3, which is thus the value I of the integral. ◭Some straightforward properties of definite integrals that are almost self-evidentare as follows:∫ ba∫ ca∫ ba0 dx =0,f(x) dx =∫ b[f(x)+g(x)] dx =a∫ aaf(x) dx +∫ ba60f(x) dx =0, (2.23)∫ cbf(x) dx +f(x) dx, (2.24)∫ bag(x) dx. (2.25)

2.2 INTEGRATIONCombining (2.23) and (2.24) with c set equal to a shows that∫ bf(x) dx = −∫ aabf(x) dx. (2.26)2.2.2 Integration as the inverse of differentiationThe definite integral has been defined as the area under a curve between twofixed limits. Let us now consider the integralF(x) =∫ xaf(u) du (2.27)in which the lower limit a remains fixed but the upper limit x is now variable. Itwill be noticed that this is essentially a restatement of (2.21), but that the variablex in the integrand has been replaced by a new variable u. It is conventional torename the dummy variable in the integrand in this way in order that the samevariable does not appear in both the integrand and the integration limits.It is apparent from (2.27) that F(x) is a continuous function of x, but at firstglance the definition of an integral as the area under a curve does not connect withour assertion that integration is the inverse process to differentiation. However,by considering the integral (2.27) and using the elementary property (2.24), weobtainF(x +∆x) ==∫ x+∆xa∫ xa= F(x)+f(u) duf(u) du +x∫ x+∆xRearranging and dividing through by ∆x yieldsF(x +∆x) − F(x)∆xx= 1∆x∫ x+∆xf(u) du.∫ x+∆xxf(u) duf(u) du.Letting ∆x → 0 and using (2.1) we find that the LHS becomes dF/dx, whereasthe RHS becomes f(x). The latter conclusion follows because when ∆x is smallthe value of the integral on the RHS is approximately f(x)∆x, and in the limit∆x → 0 no approximation is involved. ThusdF(x)= f(x), (2.28)dxor, substituting for F(x) from (2.27),[∫d x]f(u) du = f(x).dxa61

PRELIMINARY CALCULUSFrom the last two equations it is clear that integration can be considered asthe inverse of differentiation. However, we see from the above analysis that thelower limit a is arbitrary and so differentiation does not have a unique inverse.Any function F(x) obeying (2.28) is called an indefinite integral of f(x), thoughany two such functions can differ by at most an arbitrary additive constant. Sincethe lower limit is arbitrary, it is usual to writeF(x) =∫ xf(u) du (2.29)and explicitly include the arbitrary constant only when evaluating F(x). Theevaluation is conventionally written in the form∫f(x) dx = F(x)+c (2.30)where c is called the constant of integration. It will be noticed that, in the absenceof any integration limits, we use the same symbol for the arguments of both fand F. This can be confusing, but is sufficiently common practice that the readerneeds to become familiar with it.We also note that the definite integral of f(x) between the fixed limits x = aand x = b can be written in terms of F(x). From (2.27) we have∫ baf(x) dx =∫ bx 0f(x) dx −∫ ax 0f(x) dx= F(b) − F(a), (2.31)where x 0 is any third fixed point. Using the notation F ′ (x) =dF/dx, wemayrewrite (2.28) as F ′ (x) =f(x), and so express (2.31) as∫ baF ′ (x) dx = F(b) − F(a) ≡ [F] b a.In contrast to differentiation, where repeated applications of the product ruleand/or the chain rule will always give the required derivative, it is not alwayspossible to find the integral of an arbitrary function. Indeed, in most real physicalproblems exact integration cannot be performed and we have to revert tonumerical approximations. Despite this cautionary note, it is in fact possible tointegrate many simple functions and the following subsections introduce the mostcommon types. Many of the techniques will be familiar to the reader and so aresummarised by example.2.2.3 Integration by inspectionThe simplest method of integrating a function is by inspection. Some of the moreelementary functions have well-known integrals that should be remembered. Thereader will notice that these integrals are precisely the inverses of the derivatives62

2.2 INTEGRATIONfound near the end of subsection 2.1.1. A few are presented below, using the formgiven in (2.30):∫∫adx= ax + c, ax n dx = axn+1n +1 + c,∫∫e ax dx = eaxaa + c, dx = a ln x + c,x∫∫a sin bx−a cos bxa cos bx dx = + c, a sin bx dx = + c,bb∫∫a tan bx dx =−a ln(cos bx)b+ c,a(a 2 + x 2 dx x)=tan−1 + c,a∫−1( √a2 − x dx x)2=cos−1 + c,a∫∫a cos bx sin n bx dx = a sinn+1 bxb(n +1)a sin bx cos n bx dx = −a cosn+1 bxb(n +1)∫+ c,+ c,1( √a2 − x dx x)2=sin−1 + c,awhere the integrals that depend on n are valid for all n ≠ −1 andwherea and bare constants. In the two final results |x| ≤a.2.2.4 Integration of sinusoidal functionsIntegrals of the type ∫ sin n xdx and ∫ cos n xdx may be found by using trigonometricexpansions. Two methods are applicable, one for odd n and the other foreven n. They are best illustrated by example.◮Evaluate the integral I = ∫ sin 5 xdx.Rewriting the integral as a product of sin x and an even power of sin x, and then usingthe relation sin 2 x =1− cos 2 x yields∫I = sin 4 x sin xdx∫= (1 − cos 2 x) 2 sin xdx∫= (1 − 2cos 2 x +cos 4 x)sinxdx∫= (sin x − 2sinx cos 2 x +sinx cos 4 x) dx= − cos x + 2 3 cos3 x − 1 5 cos5 x + c,where the integration has been carried out using the results of subsection 2.2.3. ◭63

PRELIMINARY CALCULUS◮Evaluate the integral I = ∫ cos 4 xdx.Rewriting the integral as a power of cos 2 x and then using the double-angle formulacos 2 x = 1 (1+cos2x) yields2∫∫ ( ) 2 1+cos2xI = (cos 2 x) 2 dx =dx2∫1= (1+2cos2x 4 +cos2 2x) dx.Using the double-angle formula again we may write cos 2 2x = 1 (1+cos4x), and hence2∫ [I = 1 + 1 cos 2x + 1 (1+cos4x)] dx4 2 8= 1 x + 1 sin 2x + 1 x + 1 sin 4x + c4 4 8 32= 3 x + 1 1sin 2x + sin 4x + c. ◭8 4 322.2.5 Logarithmic integrationIntegrals for which the integrand may be written as a fraction in which thenumerator is the derivative of the denominator may be evaluated using∫ f ′ (x)dx =lnf(x)+c. (2.32)f(x)This follows directly from the differentiation of a logarithm as a function of afunction (see subsection 2.1.3).◮Evaluate the integral∫I =6x 2 +2cosxx 3 +sinxdx.We note first that the numerator can be factorised to give 2(3x 2 +cosx), and then thatthe quantity in brackets is the derivative of the denominator. Hence∫ 3x 2 +cosxI =2x 3 +sinx dx =2ln(x3 +sinx)+c. ◭2.2.6 Integration using partial fractionsThe method of partial fractions was discussed at some length in section 1.4, butin essence consists of the manipulation of a fraction (here the integrand) in sucha way that it can be written as the sum of two or more simpler fractions. Againwe illustrate the method by an example.64

2.2 INTEGRATION◮Evaluate the integral∫I =1x 2 + x dx.We note that the denominator factorises to give x(x + 1). Hence∫1I =x(x +1) dx.We now separate the fraction into two partial fractions and integrate directly:∫ ( 1I =x − 1 )( ) xdx =lnx − ln(x +1)+c =ln + c. ◭x +1x +12.2.7 Integration by substitutionSometimes it is possible to make a substitution of variables that turns a complicatedintegral into a simpler one, which can then be integrated by a standardmethod. There are many useful substitutions and knowing which to use is a matterof experience. We now present a few examples of particularly useful substitutions.◮Evaluate the integral∫I =1√1 − x2 dx.Making the substitution x =sinu, we note that dx =cosudu, and hence∫∫∫1I = √1 − sin 2 u cos udu= 1√cos2 u cos udu= du = u + c.Now substituting back for u,I =sin −1 x + c.This corresponds to one of the results given in subsection 2.2.3. ◭Another particular example of integration by substitution is afforded by integralsof the form∫∫1I =a + b cos x dx or I = 1dx. (2.33)a + b sin xIn these cases, making the substitution t = tan(x/2) yields integrals that canbe solved more easily than the originals. Formulae expressing sin x and cos x interms of t were derived in equations (1.32) and (1.33) (see p. 14), but before wecan use them we must relate dx to dt as follows.65

PRELIMINARY CALCULUSSincethe required relationship isdtdx = 1 x 2 sec2 2 = 1 (1+tan 2 x )= 1+t2 ,2 2 2dx = 2 dt. (2.34)1+t2 ◮Evaluate the integral∫I =21+3cosx dx.Rewriting cos x in terms of t and using (2.34) yields∫I =∫=∫=∫=21+3 [ (1 − t 2 )(1 + t 2 ) −1] ( 21+t 2 )dt2(1 + t 2 )1+t 2 +3(1− t 2 )∫22 − t dt = 2( 21+t 2 )dt2( √ 2 − t)( √ 2+t) dt(1 1√ √ + 1 )√ dt2 2 − t 2+t= − 1 √2ln( √ 2 − t)+1 √2ln( √ 2+t)+c[= √ 1√ ]2+tan(x/2)ln √ + c. ◭2 2 − tan (x/2)Integrals of a similar form to (2.33), but involving sin 2x, cos2x, tan 2x, sin 2 x,cos 2 x or tan 2 x instead of cos x and sin x, should be evaluated by using thesubstitution t =tanx. Inthiscasesin x =t√1+t2 , cos x = 1√1+t2and dx = dt1+t 2 . (2.35)A final example of the evaluation of integrals using substitution is the methodof completing the square (cf. subsection 1.7.3).66

2.2 INTEGRATION◮Evaluate the integral∫I =1x 2 +4x +7 dx.We can write the integral in the form∫I =1(x +2) 2 +3 dx.Substituting y = x + 2, we find dy = dx and hence∫1I =y 2 +3 dy,Hence, by comparison with the table of standard integrals (see subsection 2.2.3)√ ( ) √ ( ) 3 y√3 3 x +2I =3 tan−1 + c =3 tan−1 √ + c. ◭32.2.8 Integration by partsIntegration by parts is the integration analogy of product differentiation. Theprinciple is to break down a complicated function into two functions, at least oneof which can be integrated by inspection. The method in fact relies on the resultfor the differentiation of a product. Recalling from (2.6) thatd dv(uv) =udx dx + dudx v,where u and v are functions of x, we now integrate to find∫uv = u dv ∫ dudx dx + dx vdx.Rearranging into the standard form for integration by parts gives∫u dv ∫ dudx dx = uv − vdx. (2.36)dxIntegration by parts is often remembered for practical purposes in the formthe integral of a product of two functions is equal to {the first times the integral ofthe second} minus the integral of {the derivative of the first times the integral ofthe second}. Here, u is ‘the first’ and dv/dx is ‘the second’; clearly the integral vof ‘the second’ must be determinable by inspection.◮Evaluate the integral I = ∫ x sin x dx.In the notation given above, we identify x with u and sin x with dv/dx. Hence v = − cos xand du/dx = 1 and so using (2.36)∫I = x(− cos x) − (1)(− cos x) dx = −x cos x +sinx + c. ◭67

PRELIMINARY CALCULUSThe separation of the functions is not always so apparent, as is illustrated bythe following example.◮Evaluate the integral I = ∫ x 3 e −x2 dx.Firstly we rewrite the integral as∫I =x 2 (xe −x2) dx.Now, using the notation given above, we identify x 2 with u and xe −x2 with dv/dx. Hencev = − 1 2 e−x2 and du/dx =2x, sothat∫I = − 1 2 x2 e −x2 − (−x)e −x2 dx = − 1 2 x2 e −x2 − 1 2 e−x2 + c. ◭A trick that is sometimes useful is to take ‘1’ as one factor of the product, asis illustrated by the following example.◮Evaluate the integral I = ∫ ln xdx.Firstly we rewrite the integral as∫I = (ln x)1dx.Now, using the notation above, we identify ln x with u and 1 with dv/dx. Hence we havev = x and du/dx =1/x, andso∫ ( ) 1I =(lnx)(x) − xdx= x ln x − x + c. ◭xIt is sometimes necessary to integrate by parts more than once. In doing so,we may occasionally re-encounter the original integral I. In such cases we canobtain a linear algebraic equation for I that can be solved to obtain its value.◮Evaluate the integral I = ∫ e ax cos bx dx.Integrating by parts, taking e ax as the first function, we find( ) ∫ ( )sin bxsin bxI = e ax − ae ax dx,bbwhere, for convenience, we have omitted the constant of integration. Integrating by partsa second time,( ) ( ) ∫ ( )sin bx − cos bx− cos bxI = e ax − ae ax + a 2 e ax dx.bb 2 b 2Notice that the integral on the RHS is just −a 2 /b 2 times the original integral I. Thus( 1I = e ax b sin bx + a )b cos bx − a22 b I. 268

2.2 INTEGRATIONRearranging this expression to obtain I explicitly and including the constant of integrationwe findI =eax(b sin bx + a cos bx)+c. (2.37)a 2 + b2 Another method of evaluating this integral, using the exponential of a complex number,is given in section 3.6. ◭2.2.9 Reduction formulaeIntegration using reduction formulae is a process that involves first evaluating asimple integral and then, in stages, using it to find a more complicated integral.◮Using integration by parts, find a relationship between I n and I n−1 whereI n =∫ 10(1 − x 3 ) n dxand n is any positive integer. Hence evaluate I 2 = ∫ 10 (1 − x3 ) 2 dx.Writing the integrand as a product and separating the integral into two we findI n ==∫ 10∫ 10(1 − x 3 )(1 − x 3 ) n−1 dx(1 − x 3 ) n−1 dx −∫ 10x 3 (1 − x 3 ) n−1 dx.The first term on the RHS is clearly I n−1 and so, writing the integrand in the second termon the RHS as a product,Integrating by parts we findI n = I n−1 +I n = I n−1 −∫ 10(x)x 2 (1 − x 3 ) n−1 dx.[ x] 1∫ 13n (1 − x3 ) n −0 013n (1 − x3 ) n dx= I n−1 +0− 13n I n,which on rearranging givesI n =3n3n +1 I n−1.We now have a relation connecting successive integrals. Hence, if we can evaluate I 0 ,wecan find I 1 , I 2 etc. Evaluating I 0 is trivial:I 0 =∫ 10(1 − x 3 ) 0 dx =∫ 10dx = [x] 1 0 =1.Hence(3 × 1)I 1 =(3 × 1) + 1 × 1= 3 4 , I (3 × 2)2 =(3 × 2) + 1 × 3 4 = 9 14 .Although the first few I n could be evaluated by direct multiplication, this becomes tediousfor integrals containing higher values of n; these are therefore best evaluated using thereduction formula. ◭69

PRELIMINARY CALCULUS2.2.10 Infinite and improper integralsThe definition of an integral given previously does not allow for cases in whicheither of the limits of integration is infinite (an infinite integral) or for casesin which f(x) is infinite in some part of the range (an improper integral), e.g.f(x) = (2 − x) −1/4 near the point x = 2. Nevertheless, modification of thedefinition of an integral gives infinite and improper integrals each a meaning.In the case of an integral I = ∫ baf(x) dx, the infinite integral, in which b tendsto ∞, is defined byI =∫ ∞af(x) dx = lim∫ bb→∞ af(x) dx = limb→∞F(b) − F(a).As previously, F(x) is the indefinite integral of f(x) and lim b→∞ F(b) meansthelimit (or value) that F(b) approaches as b →∞; it is evaluated after calculatingthe integral. The formal concept of a limit will be introduced in chapter 4.◮Evaluate the integral∫ ∞xI =0 (x 2 + a 2 ) dx. 2Integrating, we find F(x) =− 1 2 (x2 + a 2 ) −1 + c and so]I = lim−b→∞[−12(b 2 + a 2 )( −12a 2 )= 12a 2 . ◭For the case of improper integrals, we adopt the approach of excluding theunbounded range from the integral. For example, if the integrand f(x) is infiniteat x = c (say), a ≤ c ≤ b then∫ b∫ c−δ∫ bf(x) dx = lim f(x) dx + lim f(x) dx.aδ→0 aɛ→0 c+ɛ◮Evaluate the integral I = ∫ 20 (2 − x)−1/4 dx.Integrating directly,[I = lim −4(2 − x)3/4] 2−ɛ [= limɛ→03 0 −4ɛ3/4] + 4ɛ→03 3 23/4 = ( )43 2 3/4 . ◭2.2.11 Integration in plane polar coordinatesIn plane polar coordinates ρ, φ, a curve is defined by its distance ρ from theorigin as a function of the angle φ between the line joining a point on the curveto the origin and the x-axis, i.e. ρ = ρ(φ). The area of an element is given by70

2.2 INTEGRATIONyCρ(φ + dφ)dAρ(φ)ρdφBOxFigure 2.9 Finding the area of a sector OBC defined by the curve ρ(φ) andthe radii OB, OC, at angles to the x-axis φ 1 , φ 2 respectively.dA = 1 2 ρ2 dφ, as illustrated in figure 2.9, and hence the total area between twoangles φ 1 and φ 2 is given by∫ φ21A =2 ρ2 dφ. (2.38)φ 1An immediate observation is that the area of a circle of radius a is given byA =∫ 2π012 a2 dφ = [ 12 a2 φ ] 2π= πa 2 .0◮The equation in polar coordinates of an ellipse with semi-axes a and b is1ρ = cos2 φ+ sin2 φ.2 a 2 b 2Find the area A of the ellipse.Using (2.38) and symmetry, we have∫ 2πA = 1 a 2 b 2∫ π/22 0 b 2 cos 2 φ + a 2 sin 2 φ dφ =2a2 b 2 0To evaluate this integral we write t =tanφ and use (2.35):∫ ∞A =2a 2 b 20∫1∞b 2 + a 2 t dt 2 =2b201b 2 cos 2 φ + a 2 sin 2 φ dφ.1(b/a) 2 + t 2 dt.Finally, from the list of standard integrals (see subsection 2.2.3),A =2b 2 [ 1(b/a) tan−1] ∞t=2ab(b/a)0( π2 − 0 )= πab. ◭71

PRELIMINARY CALCULUS2.2.12 Integral inequalitiesConsider the functions f(x), φ 1 (x) andφ 2 (x) such that φ 1 (x) ≤ f(x) ≤ φ 2 (x) forall x in the range a ≤ x ≤ b. It immediately follows that∫ baφ 1 (x) dx ≤∫ baf(x) dx ≤∫ baφ 2 (x) dx, (2.39)which gives us a way of estimating an integral that is difficult to evaluate explicitly.◮Show that the value of the integrallies between 0.810 and 0.882.I =∫ 101dx(1 + x 2 + x 3 )1/2We note that for x in the range 0 ≤ x ≤ 1, 0 ≤ x 3 ≤ x 2 .Hence(1 + x 2 ) 1/2 ≤ (1 + x 2 + x 3 ) 1/2 ≤ (1+2x 2 ) 1/2 ,and so1(1 + x 2 ) ≥ 11/2 (1 + x 2 + x 3 ) ≥ 11/2 (1+2x 2 ) . 1/2Consequently,∫ 10∫11(1 + x 2 ) dx ≥ 1/2from which we obtain[ln(x + √ ] 11+x 2 ) ≥ I ≥00∫11(1 + x 2 + x 3 ) dx ≥ 1/2[√12ln0.8814 ≥ I ≥ 0.81050.882 ≥ I ≥ 0.810.01dx,(1+2x 2 )1/2( √ )] 11x + + 2 x2 0In the last line the calculated values have been rounded to three significant figures,one rounded up and the other rounded down so that the proved inequality cannot beunknowingly made invalid. ◭2.2.13 Applications of integrationMean value of a functionThe mean value m of a function between two limits a and b is defined bym = 1 ∫ bf(x) dx. (2.40)b − a aThe mean value may be thought of as the height of the rectangle that has thesame area (over the same interval) as the area under the curve f(x). This isillustrated in figure 2.10.72

2.2 INTEGRATIONf(x)mabxFigure 2.10The mean value m of a function.◮Find the mean value m of the function f(x) =x 2 between the limits x =2and x =4.Using (2.40),m = 14 − 2∫ 42x 2 dx = 1 [ x32 3] 42= 1 2( 433 − 233)= 283 . ◭Finding the length of a curveFinding the area between a curve and certain straight lines provides one exampleof the use of integration. Another is in finding the length of a curve. If a curveis defined by y = f(x) then the distance along the curve, ∆s, that corresponds tosmall changes ∆x and ∆y in x and y is given by∆s ≈ √ (∆x) 2 +(∆y) 2 ; (2.41)this follows directly from Pythagoras’ theorem (see figure 2.11). Dividing (2.41)through by ∆x and letting ∆x → 0weobtain §√( )dsdy 2dx = 1+ .dxClearly the total length s of the curve between the points x = a and x = b is thengiven by integrating both sides of the equation:s =∫ ba√1+( ) 2 dydx. (2.42)dx§ Instead of considering small changes ∆x and ∆y and letting these tend to zero, we could havederived (2.41) by considering infinitesimal changes dx and dy from the start. After writing (ds) 2 =(dx) 2 +(dy) 2 , (2.41) may be deduced by using the formal device of dividing through by dx. Althoughnot mathematically rigorous, this method is often used and generally leads to the correct result.73

PRELIMINARY CALCULUSf(x)y = f(x)∆s∆x∆yxFigure 2.11 The distance moved along a curve, ∆s, corresponding to thesmall changes ∆x and ∆y.In plane polar coordinates,ds = √ (dr) 2 +(rdφ) 2 ⇒ s =∫ r2r 1√1+r 2 ( dφdr) 2dr.(2.43)◮Find the length of the curve y = x 3/2 from x =0to x =2.Using (2.42) and noting that dy/dx = 3 2√ x, the length s of the curve is given by∫ 2s =[=023= 827√1+ 9 4 xdx( 4)(9 1+9[ ( 112] x) 3/2240) ] 3/2− 1 . ◭[ (1+ ]= 8 9x) 3/2227 40Surfaces of revolutionConsider the surface S formed by rotating the curve y = f(x) about the x-axis(see figure 2.12). The surface area of the ‘collar’ formed by rotating an elementof the curve, ds, about the x-axis is 2πy ds, and hence the total surface area isS =∫ ba2πy ds.Since (ds) 2 =(dx) 2 +(dy) 2 from (2.41), the total surface area between the planesx = a and x = b is√∫ b( ) 2 dyS = 2πy 1+ dx. (2.44)dxa74

2.2 INTEGRATIONydsf(x)adxVbxSFigure 2.12The surface and volume of revolution for the curve y = f(x).◮Find the surface area of a cone formed by rotating about the x-axis the line y =2xbetween x =0and x = h.Using (2.44), the surface area is given byS ===∫ h0∫ h0(2π)2x√1+[ ddx (2x) ] 2dx4πx ( 1+2 2) ∫ h1/2dx = 4 √ 5πx dx[2 √ 5πx 2 ] h0=2 √ 5π(h 2 − 0) = 2 √ 5πh 2 . ◭0We note that a surface of revolution may also be formed by rotating a lineabout the y-axis. In this case the surface area between y = a and y = b isS =∫ ba√( ) dx 22πx 1+ dy. (2.45)dyVolumes of revolutionThe volume V enclosed by rotating the curve y = f(x) about the x-axis can alsobe found (see figure 2.12). The volume of the disc between x and x + dx is givenby dV = πy 2 dx. Hence the total volume between x = a and x = b isV =∫ ba75πy 2 dx. (2.46)

PRELIMINARY CALCULUS◮Find the volume of a cone enclosed by the surface formed by rotating about the x-axisthe line y =2x between x =0and x = h.Using (2.46), the volume is given byV =∫ h0π(2x) 2 dx =∫ h04πx 2 dx= [ 43 πx3] h0 = 4 3 π(h3 − 0) = 4 3 πh3 . ◭As before, it is also possible to form a volume of revolution by rotating a curveabout the y-axis. In this case the volume enclosed between y = a and y = b isV =∫ baπx 2 dy. (2.47)2.3 Exercises2.1 Obtain the following derivatives from first principles:(a) the first derivative of 3x +4;(b) the first, second and third derivatives of x 2 + x;(c) the first derivative of sin x.2.2 Find from first principles the first derivative of (x+3) 2 and compare your answerwith that obtained using the chain rule.2.3 Find the first derivatives of(a) x 2 exp x, (b)2sinx cos x, (c)sin2x, (d)x sin ax,(e) (exp ax)(sin ax)tan −1 ax, (f)ln(x a + x −a ),(g) ln(a x + a −x ), (h) x x .2.4 Find the first derivatives of(a) x/(a + x) 2 ,(b)x/(1 − x) 1/2 ,(c)tanx, assinx/ cos x,(d) (3x 2 +2x +1)/(8x 2 − 4x +2).2.5 Use result (2.12) to find the first derivatives of(a) (2x +3) −3 ,(b)sec 2 x,(c)cosech 3 3x, (d)1/ ln x, (e)1/[sin −1 (x/a)].2.6 Show that the function y(x) =exp(−|x|) defined by⎧⎪⎨ exp x for x0,is not differentiable at x = 0. Consider the limiting process for both ∆x >0and∆x

2.3 EXERCISES2.10 The function y(x) is defined by y(x) =(1+x m ) n .(a) Use the chain rule to show that the first derivative of y is nmx m−1 (1 + x m ) n−1 .(b) The binomial expansion (see section 1.5) of (1 + z) n is(1 + z) n =1+nz +n(n − 1)z 2 + ···+2!n(n − 1) ···(n − r +1)z r + ··· .r!Keeping only the terms of zeroth and first order in dx, apply this result twiceto derive result (a) from first principles.(c) Expand y in a series of powers of x before differentiating term by term.Show that the result is the series obtained by expanding the answer givenfor dy/dx in (a).2.11 Show by differentiation and substitution that the differential equation4x 2 d2 y dy− 4xdx2 dx +(4x2 +3)y =0has a solution of the form y(x) =x n sin x, and find the value of n.2.12 Find the positions and natures of the stationary points of the following functions:(a) x 3 − 3x +3;(b)x 3 − 3x 2 +3x; (c)x 3 +3x +3;(d) sin ax with a ≠0;(e)x 5 + x 3 ;(f)x 5 − x 3 .2.13 Show that the lowest value taken by the function 3x 4 +4x 3 − 12x 2 +6is−26.2.14 By finding their stationary points and examining their general forms, determinethe range of values that each of the following functions y(x) can take. In eachcase make a sketch-graph incorporating the features you have identified.(a) y(x) =(x − 1)/(x 2 +2x +6).(b) y(x) =1/(4+3x − x 2 ).(c) y(x) =(8sinx)/(15 + 8 tan 2 x).2.15 Show that y(x) =xa 2x exp x 2 has no stationary points other than x =0,ifexp(− √ 2)

PRELIMINARY CALCULUSOcCρr +∆rrρQpp +∆pPFigure 2.13 The coordinate system described in exercise 2.20.2.20 A two-dimensional coordinate system useful for orbit problems is the tangentialpolarcoordinate system (figure 2.13). In this system a curve is defined by r, thedistance from a fixed point O to a general point P of the curve, and p, theperpendicular distance from O to the tangent to the curve at P . By proceedingas indicated below, show that the radius of curvature, ρ, atP canbewritteninthe form ρ = r dr/dp.Consider two neighbouring points, P and Q, on the curve. The normals to thecurve through those points meet at C, with (in the limit Q → P ) CP = CQ = ρ.Apply the cosine rule to triangles OPC and OQC to obtain two expressions forc 2 ,oneintermsofr and p and the other in terms of r +∆r and p +∆p. Byequating them and letting Q → P deduce the stated result.2.21 Use Leibnitz’ theorem to find(a) the second derivative of cos x sin 2x,(b) the third derivative of sin x ln x,(c) the fourth derivative of (2x 3 +3x 2 + x +2)exp2x.2.22 If y =exp(−x 2 ), show that dy/dx = −2xy and hence, by applying Leibnitz’theorem, prove that for n ≥ 1y (n+1) +2xy (n) +2ny (n−1) =0.2.23 Use the properties of functions at their turning points to do the following:(a) By considering its properties near x = 1, show that f(x) =5x 4 − 11x 3 +26x 2 − 44x + 24 takes negative values for some range of x.(b) Show that f(x) =tanx − x cannot be negative for 0 ≤ x

2.3 EXERCISES2.26 Use the mean value theorem to establish bounds in the following cases.(a) For − ln(1 − y), by considering ln x in the range 0 < 1 − y

PRELIMINARY CALCULUS(c) [(x − a)/(b − x)] 1/2 .2.38 Determine whether the following integrals exist and, where they do, evaluatethem: ∫ ∞∫ ∞x(a) exp(−λx) dx; (b)0−∞ (x 2 + a 2 ) dx;∫ 2 ∞∫11(c)1 x +1 dx; (d) 10 x dx;∫ 2 π/2∫ 1x(e) cot θdθ; (f)dx.00 (1 − x 2 )1/22.39 Use integration∫by parts to evaluate the following:y∫ y(a) x 2 sin xdx; (b) x ln xdx;(c)0∫ y0sin −1 xdx;(d)1∫ y1ln(a 2 + x 2 )/x 2 dx.2.40 Show, using the following methods, that the indefinite integral of x 3 /(x +1) 1/2 isJ = 235 (5x3 − 6x 2 +8x − 16)(x +1) 1/2 + c.(a) Repeated integration by parts.(b) Setting x +1=u 2 and determining dJ/du as (dJ/dx)(dx/du).2.41 The gamma function Γ(n) is defined for all n>−1 byΓ(n +1)=∫ ∞0x n e −x dx.Find a recurrence relation connecting Γ(n +1)andΓ(n).(a) Deduce (i) the value of Γ(n +1) when n is a non-negative integer, and (ii)the value of Γ ( ) (72 ,giventhatΓ 1) √2 = π.(b) Now, ( ) taking factorial m for any m to be defined by m! =Γ(m + 1), evaluate−32 !.2.42 Define J(m, n), for non-negative integers m and n, by the integralJ(m, n) =∫ π/20cos m θ sin n θdθ.(a) Evaluate J(0, 0), J(0, 1), J(1, 0), J(1, 1), J(m, 1), J(1,n).(b) Using integration by parts, prove that, for m and n both > 1,J(m, n) = m − 1n − 1J(m − 2,n) and J(m, n) = J(m, n − 2).m + n m + n(c) Evaluate (i) J(5, 3), (ii) J(6, 5) and (iii) J(4, 8).2.43 By integrating by parts twice, prove that I n as defined in the first equality belowfor positive integers n has the value given in the second equality:∫ π/2I n = sin nθ cos θdθ= n − sin(nπ/2) .0n 2 − 12.44 Evaluate the following definite integrals:∫ ∞[(x 3 +1)/(x 4 +4x +1) ] dx;(a)(c)0 xe−x dx; (b) ∫ 10∫ π/2[a +(a − 1) cos θ] −1 dθ with a> 1 ; (d) ∫ ∞0 2 −∞ (x2 +6x +18) −1 dx.80

2.4 HINTS AND ANSWERS2.45 If J r is the integral∫ ∞0x r exp(−x 2 ) dxshow that(a) J 2r+1 =(r!)/2,(b) J 2r =2 −r (2r − 1)(2r − 3) ···(5)(3)(1) J 0 .2.46 Find positive constants a, b such that ax ≤ sin x ≤ bx for 0 ≤ x ≤ π/2. Usethis inequality to find (to two significant figures) upper and lower bounds for theintegral∫ π/2I = (1+sinx) 1/2 dx.0Use the substitution t =tan(x/2) to evaluate I exactly.2.47 By noting that for 0 ≤ η ≤ 1, η 1/2 ≥ η 3/4 ≥ η, prove that23 ≤ 1 ∫ a(a 2 − x 2 ) 3/4 dx ≤ π a 5/2 04 .2.48 Show that the total length of the astroid x 2/3 + y 2/3 = a 2/3 , which can beparameterised as x = a cos 3 θ, y = a sin 3 θ,is6a.2.49 By noting that sinh x< 1 2 ex < cosh x, andthat1+z 2 < (1 + z) 2 for z>0, showthat, for x>0, the length L of the curve y = 1 2 ex measured from the originsatisfies the inequalities sinh x

PRELIMINARY CALCULUSy2aπa2πaxFigure 2.14 The solution to exercise 2.17.2.21 (a) 2(2 − 9cos 2 x)sinx; (b)(2x −3 − 3x −1 )sinx − (3x −2 +lnx)cosx; (c)8(4x 3 +30x 2 +62x + 38) exp 2x.2.23 (a) f(1) = 0 whilst f ′ (1) ≠0,andsof(x) mustbenegativeinsomeregionwithx = 1 as an endpoint.(b) f ′ (x) =tan 2 x>0andf(0) = 0; g ′ (x) =(− cos x)(tan x − x)/x 2 , which isnever positive in the range.2.25 The false result arises because tan nx is not differentiable at x = π/(2n), whichlies in the range 0 0:(i) ∆ −1 ln[(2ax + b − ∆)/(2ax + b +∆)]+k;(ii) 2∆ ′−1 tan −1 [(2ax + b)/∆ ′ ]+k;(iii) −2(2ax + b) −1 + k.2.35 f ′ (x)=(1+sinx)/ cos 2 x = f(x)secx; J =ln(f(x)) + c =ln(secx +tanx)+c.2.37 Note that dx =2(b − a)cosθ sin θdθ.(a) π; (b)π(b − a) 2 /8; (c) π(b − a)/2.2.39 (a) (2 − y 2 )cosy + 2y sin y − 2; (b) [(y 2 ln y)/2] + [(1 − y 2 )/4];(c) y sin −1 y +(1− y 2 ) 1/2 − 1;(d) ln(a 2 +1)− (1/y)ln(a 2 + y 2 )+(2/a)[tan −1 (y/a) − tan −1 (1/a)].2.41 Γ(n +1)=nΓ(n); (a) (i) n!, (ii) 15 √ π/8; (b) −2 √ π.2.43 By integrating twice, recover a multiple of I n .2.45 J 2r+1 = rJ 2r−1 and 2J 2r =(2r − 1)J 2r−2 .2.47 Set η =1− (x/a) 2 throughout, and x = a sin θ in one of the bounds.2.49 L = ∫ x0(1+14 exp 2x) 1/2dx.82

3Complex numbers andhyperbolic functionsThis chapter is concerned with the representation and manipulation of complexnumbers. Complex numbers pervade this book, underscoring their wide applicationin the mathematics of the physical sciences. The application of complexnumbers to the description of physical systems is left until later chapters andonly the basic tools are presented here.3.1 The need for complex numbersAlthough complex numbers occur in many branches of mathematics, they arisemost directly out of solving polynomial equations. We examine a specific quadraticequation as an example.Consider the quadratic equationEquation (3.1) has two solutions, z 1 and z 2 , such thatz 2 − 4z +5=0. (3.1)(z − z 1 )(z − z 2 )=0. (3.2)Using the familiar formula for the roots of a quadratic equation, (1.4), thesolutions z 1 and z 2 , written in brief as z 1,2 ,arez 1,2 = 4 ± √ (−4) 2 − 4(1 × 5)√2−4=2±2 . (3.3)Both solutions contain the square root of a negative number. However, it is nottrue to say that there are no solutions to the quadratic equation. The fundamentaltheorem of algebra states that a quadratic equation will always have two solutionsand these are in fact given by (3.3). The second term on the RHS of (3.3) iscalled an imaginary term since it contains the square root of a negative number;83

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSf(z)543211234 zFigure 3.1 The function f(z) =z 2 − 4z +5.the first term is called a real term. The full solution is the sum of a real termand an imaginary term and is called a complex number. A plot of the functionf(z) =z 2 − 4z + 5 is shown in figure 3.1. It will be seen that the plot does notintersect the z-axis, corresponding to the fact that the equation f(z) =0hasnopurely real solutions.The choice of the symbol z for the quadratic variable was not arbitrary; theconventional representation of a complex number is z, wherez is the sum of areal part x and i times an imaginary part y, i.e.z = x + iy,where i is used to denote the square root of −1. The real part x and the imaginarypart y are usually denoted by Re z and Im z respectively. We note at this pointthat some physical scientists, engineers in particular, use j instead of i. However,for consistency, we will use i throughout this book.In our particular example, √ −4=2 √ −1=2i, and hence the two solutions of(3.1) arez 1,2 =2± 2i =2± i.2Thus, here x = 2 and y = ±1.For compactness a complex number is sometimes written in the formz =(x, y),where the components of z may be thought of as coordinates in an xy-plot. Sucha plot is called an Argand diagram and is a common representation of complexnumbers; an example is shown in figure 3.2.84

3.2 MANIPULATION OF COMPLEX NUMBERSIm zyz = x + iyxRe zFigure 3.2The Argand diagram.Our particular example of a quadratic equation may be generalised readily topolynomials whose highest power (degree) is greater than 2, e.g. cubic equations(degree 3), quartic equations (degree 4) and so on. For a general polynomial f(z),of degree n, the fundamental theorem of algebra states that the equation f(z) =0will have exactly n solutions. We will examine cases of higher-degree equationsin subsection 3.4.3.The remainder of this chapter deals with: the algebra and manipulation ofcomplex numbers; their polar representation, which has advantages in manycircumstances; complex exponentials and logarithms; the use of complex numbersin finding the roots of polynomial equations; and hyperbolic functions.3.2 Manipulation of complex numbersThis section considers basic complex number manipulation. Some analogy maybe drawn with vector manipulation (see chapter 7) but this section stands aloneas an introduction.3.2.1 Addition and subtractionThe addition of two complex numbers, z 1 and z 2 , in general gives anothercomplex number. The real components and the imaginary components are addedseparately and in a like manner to the familiar addition of real numbers:z 1 + z 2 =(x 1 + iy 1 )+(x 2 + iy 2 )=(x 1 + x 2 )+i(y 1 + y 2 ),85

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSIm zz 2z 1 + z 2z 1Re zFigure 3.3The addition of two complex numbers.or in component notationz 1 + z 2 =(x 1 ,y 1 )+(x 2 ,y 2 )=(x 1 + x 2 ,y 1 + y 2 ).The Argand representation of the addition of two complex numbers is shown infigure 3.3.By straightforward application of the commutativity and associativity of thereal and imaginary parts separately, we can show that the addition of complexnumbers is itself commutative and associative, i.e.z 1 + z 2 = z 2 + z 1 ,z 1 +(z 2 + z 3 )=(z 1 + z 2 )+z 3 .Thus it is immaterial in what order complex numbers are added.◮Sum the complex numbers 1+2i, 3 − 4i, −2+i.Summing the real terms we obtain1+3− 2=2,and summing the imaginary terms we obtain2i − 4i + i = −i.Hence(1 + 2i)+(3− 4i)+(−2+i) =2− i. ◭The subtraction of complex numbers is very similar to their addition. As in thecase of real numbers, if two identical complex numbers are subtracted then theresult is zero.86

3.2 MANIPULATION OF COMPLEX NUMBERSIm zy|z|xarg zRe zFigure 3.4The modulus and argument of a complex number.3.2.2 Modulus and argumentThe modulus of the complex number z is denoted by |z| and is defined as|z| = √ x 2 + y 2 . (3.4)Hence the modulus of the complex number is the distance of the correspondingpoint from the origin in the Argand diagram, as may be seen in figure 3.4.The argument of the complex number z is denoted by arg z and is defined asarg z =tan −1 ( yx). (3.5)It can be seen that arg z is the angle that the line joining the origin to z onthe Argand diagram makes with the positive x-axis. The anticlockwise directionis taken to be positive by convention. The angle arg z is shown in figure 3.4.Account must be taken of the signs of x and y individually in determining inwhich quadrant arg z lies. Thus, for example, if x and y are both negative thenarg z lies in the range −π

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS3.2.3 MultiplicationComplex numbers may be multiplied together and in general give a complexnumber as the result. The product of two complex numbers z 1 and z 2 is foundby multiplying them out in full and remembering that i 2 = −1, i.e.z 1 z 2 =(x 1 + iy 1 )(x 2 + iy 2 )= x 1 x 2 + ix 1 y 2 + iy 1 x 2 + i 2 y 1 y 2=(x 1 x 2 − y 1 y 2 )+i(x 1 y 2 + y 1 x 2 ). (3.6)◮Multiply the complex numbers z 1 =3+2i and z 2 = −1 − 4i.By direct multiplication we findz 1 z 2 =(3+2i)(−1 − 4i)= −3 − 2i − 12i − 8i 2=5− 14i. ◭ (3.7)The multiplication of complex numbers is both commutative and associative,i.e.z 1 z 2 = z 2 z 1 , (3.8)(z 1 z 2 )z 3 = z 1 (z 2 z 3 ). (3.9)The product of two complex numbers also has the simple propertiesThese relations are derived in subsection 3.3.1.|z 1 z 2 | = |z 1 ||z 2 |, (3.10)arg(z 1 z 2 )=argz 1 +arg z 2 . (3.11)◮Verify that (3.10) holds for the product of z 1 =3+2i and z 2 = −1 − 4i.From (3.7)We also find|z 1 z 2 | = |5 − 14i| = √ 5 2 +(−14) 2 = √ 221.|z 1 | = √ 3 2 +2 2 = √ 13,|z 2 | = √ (−1) 2 +(−4) 2 = √ 17,and hence|z 1 ||z 2 | = √ 13 √ 17 = √ 221 = |z 1 z 2 |. ◭We now examine the effect on a complex number z of multiplying it by ±1and ±i. These four multipliers have modulus unity and we can see immediatelyfrom (3.10) that multiplying z by another complex number of unit modulus givesa product with the same modulus as z. We can also see from (3.11) that if we88

3.2 MANIPULATION OF COMPLEX NUMBERSizIm zzRe z−z−izFigure 3.5Multiplication of a complex number by ±1 and±i.multiply z by a complex number then the argument of the product is the sumof the argument of z and the argument of the multiplier. Hence multiplyingz by unity (which has argument zero) leaves z unchanged in both modulusand argument, i.e. z is completely unaltered by the operation. Multiplying by−1 (which has argument π) leads to rotation, through an angle π, of the linejoining the origin to z in the Argand diagram. Similarly, multiplication by i or −ileads to corresponding rotations of π/2 or−π/2 respectively. This geometricalinterpretation of multiplication is shown in figure 3.5.◮Using the geometrical interpretation of multiplication by i, find the product i(1 − i).The complex number 1 − i has argument −π/4 and modulus √ 2. Thus, using (3.10) and(3.11), its product with i has argument +π/4 and unchanged modulus √ 2. The complexnumber with modulus √ 2 and argument +π/4 is1+i and soi(1 − i) =1+i,as is easily verified by direct multiplication. ◭The division of two complex numbers is similar to their multiplication butrequires the notion of the complex conjugate (see the following subsection) andso discussion is postponed until subsection 3.2.5.3.2.4 Complex conjugateIf z has the convenient form x + iy then the complex conjugate, denoted by z ∗ ,may be found simply by changing the sign of the imaginary part, i.e. if z = x + iythen z ∗ = x − iy. More generally, we may define the complex conjugate of z asthe (complex) number having the same magnitude as z that when multiplied byz leaves a real result, i.e. there is no imaginary component in the product.89

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSIm zyz = x + iyxRe z−yz ∗ = x − iyFigure 3.6The complex conjugate as a mirror image in the real axis.Inthecasewherez can be written in the form x + iy it is easily verified, bydirect multiplication of the components, that the product zz ∗ gives a real result:zz ∗ =(x + iy)(x − iy) =x 2 − ixy + ixy − i 2 y 2 = x 2 + y 2 = |z| 2 .Complex conjugation corresponds to a reflection of z in the real axis of theArgand diagram, as may be seen in figure 3.6.◮Find the complex conjugate of z = a +2i +3ib.The complex number is written in the standard formz = a + i(2+3b);then, replacing i by −i, weobtainz ∗ = a − i(2+3b). ◭In some cases, however, it may not be simple to rearrange the expression forz into the standard form x + iy. Nevertheless, given two complex numbers, z 1and z 2 , it is straightforward to show that the complex conjugate of their sum(or difference) is equal to the sum (or difference) of their complex conjugates, i.e.(z 1 ± z 2 ) ∗ = z1 ∗ ± z∗ 2 . Similarly, it may be shown that the complex conjugate of theproduct (or quotient) of z 1 and z 2 is equal to the product (or quotient) of theircomplex conjugates, i.e. (z 1 z 2 ) ∗ = z1 ∗z∗ 2 and (z 1/z 2 ) ∗ = z1 ∗/z∗ 2 .Using these results, it can be deduced that, no matter how complicated theexpression, its complex conjugate may always be found by replacing every i by−i. To apply this rule, however, we must always ensure that all complex parts arefirst written out in full, so that no i’s are hidden.90

3.2 MANIPULATION OF COMPLEX NUMBERS◮Find the complex conjugate of the complex number z = w (3y+2ix) ,wherew = x +5i.Although we do not discuss complex powers until section 3.5, the simple rule given abovestill enables us to find the complex conjugate of z.In this case w itself contains real and imaginary components and so must be writtenout in full, i.e.z = w 3y+2ix =(x +5i) 3y+2ix .Now we can replace each i by −i to obtainz ∗ =(x − 5i) (3y−2ix) .It can be shown that the product zz ∗ is real, as required. ◭The following properties of the complex conjugate are easily proved and othersmay be derived from them. If z = x + iy then(z ∗ ) ∗ = z, (3.12)z + z ∗ =2Rez =2x, (3.13)z − z ∗ =2i Im z =2iy, (3.14)(z x 2z ∗ = − y 2 ) ( ) 2xyx 2 + y 2 + ix 2 + y 2 . (3.15)The derivation of this last relation relies on the results of the following subsection.3.2.5 DivisionThe division of two complex numbers z 1 and z 2 bears some similarity to theirmultiplication. Writing the quotient in component form we obtainz 1= x 1 + iy 1. (3.16)z 2 x 2 + iy 2In order to separate the real and imaginary components of the quotient, wemultiply both numerator and denominator by the complex conjugate of thedenominator. By definition, this process will leave the denominator as a realquantity. Equation (3.16) givesz 1= (x 1 + iy 1 )(x 2 − iy 2 )z 2 (x 2 + iy 2 )(x 2 − iy 2 ) = (x 1x 2 + y 1 y 2 )+i(x 2 y 1 − x 1 y 2 )x 2 2 + y2 2= x 1x 2 + y 1 y 2x 2 2 + y2 2+ i x 2y 1 − x 1 y 2x 2 2 + .y2 2Hence we have separated the quotient into real and imaginary components, asrequired.In the special case where z 2 = z ∗ 1 ,sothatx 2 = x 1 and y 2 = −y 1 , the generalresult reduces to (3.15).91

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS◮Express z in the form x + iy, whenz =3 − 2i−1+4i .Multiplying numerator and denominator by the complex conjugate of the denominatorwe obtain(3 − 2i)(−1 − 4i) −11 − 10iz = =(−1+4i)(−1 − 4i) 17= − 1117 − 1017 i. ◭In analogy to (3.10) and (3.11), which describe the multiplication of twocomplex numbers, the following relations apply to division:∣ z 1 ∣∣∣∣ = |z 1|z 2 |z 2 | , (3.17)( )z1arg =argz 1 − arg z 2 . (3.18)z 2The proof of these relations is left until subsection 3.3.1.3.3 Polar representation of complex numbersAlthough considering a complex number as the sum of a real and an imaginarypart is often useful, sometimes the polar representation proves easier to manipulate.This makes use of the complex exponential function, which is defined bye z =expz ≡ 1+z + z22! + z3+ ··· . (3.19)3!Strictly speaking it is the function exp z that is defined by (3.19). The number eis the value of exp(1), i.e. it is just a number. However, it may be shown that e zand exp z are equivalent when z is real and rational and mathematicians thendefine their equivalence for irrational and complex z. For the purposes of thisbook we will not concern ourselves further with this mathematical nicety but,rather, assume that (3.19) is valid for all z. We also note that, using (3.19), bymultiplying together the appropriate series we may show that (see chapter 24)e z1 e z2 = e z1+z2 , (3.20)which is analogous to the familiar result for exponentials of real numbers.92

3.3 POLAR REPRESENTATION OF COMPLEX NUMBERSIm zyz = re iθrθxRe zFigure 3.7The polar representation of a complex number.From (3.19), it immediately follows that for z = iθ, θ real,e iθ =1+iθ − θ22! − iθ3 + ··· (3.21)3!(···)=1− θ22! + θ44! − ···+ i θ − θ33! + θ55! − (3.22)and hence thate iθ =cosθ + i sin θ, (3.23)where the last equality follows from the series expansions of the sine and cosinefunctions (see subsection 4.6.3). This last relationship is called Euler’s equation. Italso follows from (3.23) thate inθ =cosnθ + i sin nθfor all n. From Euler’s equation (3.23) and figure 3.7 we deduce thatre iθ = r(cos θ + i sin θ)= x + iy.Thus a complex number may be represented in the polar formz = re iθ . (3.24)Referring again to figure 3.7, we can identify r with |z| and θ with arg z. Thesimplicity of the representation of the modulus and argument is one of the mainreasons for using the polar representation. The angle θ lies conventionally in therange −π

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSIm zr 1 r 2 e i(θ 1+θ 2 )r 2 e iθ 2r 1 e iθ 1Re zFigure 3.8 The multiplication of two complex numbers. In this case r 1 andr 2 are both greater than unity.The algebra of the polar representation is different from that of the real andimaginary component representation, though, of course, the results are identical.Some operations prove much easier in the polar representation, others much morecomplicated. The best representation for a particular problem must be determinedby the manipulation required.3.3.1 Multiplication and division in polar formMultiplication and division in polar form are particularly simple. The product ofz 1 = r 1 e iθ1 and z 2 = r 2 e iθ2 is given byz 1 z 2 = r 1 e iθ1 r 2 e iθ2= r 1 r 2 e i(θ1+θ2) . (3.25)The relations |z 1 z 2 | = |z 1 ||z 2 | and arg(z 1 z 2 ) = arg z 1 +argz 2 follow immediately.An example of the multiplication of two complex numbers is shown in figure 3.8.Division is equally simple in polar form; the quotient of z 1 and z 2 is given byz 1= r 1e iθ1z 2 r 2 e = r 1e i(θ1−θ2) . (3.26)iθ2 r 2The relations |z 1 /z 2 | = |z 1 |/|z 2 | and arg(z 1 /z 2 ) = arg z 1 − arg z 2 are again94

3.4 DE MOIVRE’S THEOREMIm zr 1 e iθ 1r 2 e iθ 2r 1e i(θ 1−θ 2 )r 2Re zFigure 3.9 The division of two complex numbers. As in the previous figure,r 1 and r 2 are both greater than unity.immediately apparent. The division of two complex numbers in polar form isshown in figure 3.9.3.4 de Moivre’s theoremWe now derive an extremely important theorem. Since ( e iθ) n= e inθ , we have(cos θ + i sin θ) n =cosnθ + i sin nθ, (3.27)where the identity e inθ =cosnθ + i sin nθ follows from the series definition ofe inθ (see (3.21)). This result is called de Moivre’s theorem andisoftenusedinthemanipulation of complex numbers. The theorem is valid for all n whether real,imaginary or complex.There are numerous applications of de Moivre’s theorem but this sectionexamines just three: proofs of trigonometric identities; finding the nth roots ofunity; and solving polynomial equations with complex roots.3.4.1 Trigonometric identitiesThe use of de Moivre’s theorem in finding trigonometric identities is best illustratedby example. We consider the expression of a multiple-angle function interms of a polynomial in the single-angle function, and its converse.95

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS◮Express sin 3θ and cos 3θ in terms of powers of cos θ and sin θ.Using de Moivre’s theorem,cos 3θ + i sin 3θ =(cosθ + i sin θ) 3=(cos 3 θ − 3cosθ sin 2 θ)+i(3 sin θ cos 2 θ − sin 3 θ). (3.28)We can equate the real and imaginary coefficients separately, i.e.andcos 3θ =cos 3 θ − 3cosθ sin 2 θ=4cos 3 θ − 3cosθ (3.29)sin 3θ =3sinθ cos 2 θ − sin 3 θ=3sinθ − 4sin 3 θ. ◭This method can clearly be applied to finding power expansions of cos nθ andsin nθ for any positive integer n.The converse process uses the following properties of z = e iθ ,z n + 1 =2cosnθ,zn (3.30)z n − 1 =2i sin nθ.zn (3.31)These equalities follow from simple applications of de Moivre’s theorem, i.e.z n + 1 z n =(cosθ + i sin θ)n +(cosθ + i sin θ) −n=cosnθ + i sin nθ +cos(−nθ)+i sin(−nθ)=cosnθ + i sin nθ +cosnθ − i sin nθ=2cosnθandz n − 1 z n =(cosθ + i sin θ)n − (cos θ + i sin θ) −n=cosnθ + i sin nθ − cos nθ + i sin nθ=2i sin nθ.In the particular case where n =1,z + 1 z = eiθ + e −iθ =2cosθ, (3.32)z − 1 z = eiθ − e −iθ =2i sin θ. (3.33)96

3.4 DE MOIVRE’S THEOREM◮Find an expression for cos 3 θ in terms of cos 3θ and cos θ.Using (3.32),cos 3 θ = 1 (z + 1 ) 32 3 z(z 3 +3z + 3 z + 1 )z 3= 1 8= 1 8Now using (3.30) and (3.32), we find(z 3 + 1 )+ 3 (z + 1 ).z 3 8 zcos 3 θ = 1 cos 3θ + 3 cos θ. ◭4 4This result happens to be a simple rearrangement of (3.29), but cases involvinglarger values of n are better handled using this direct method than by rearrangingpolynomial expansions of multiple-angle functions.3.4.2 Finding the nth roots of unityThe equation z 2 = 1 has the familiar solutions z = ±1. However, now thatwe have introduced the concept of complex numbers we can solve the generalequation z n = 1. Recalling the fundamental theorem of algebra, we know thatthe equation has n solutions. In order to proceed we rewrite the equation asz n = e 2ikπ ,where k is any integer. Now taking the nth root of each side of the equation wefindz = e 2ikπ/n .Hence, the solutions of z n = 1 arez 1,2,...,n =1, e 2iπ/n , ..., e 2i(n−1)π/n ,corresponding to the values 0, 1, 2,...,n− 1fork. Larger integer values of k donot give new solutions, since the roots already listed are simply cyclically repeatedfor k = n, n +1,n+2,etc.◮Find the solutions to the equation z 3 =1.By applying the above method we findz = e 2ikπ/3 .Hence the three solutions are z 1 = e 0i =1,z 2 = e 2iπ/3 , z 3 = e 4iπ/3 . We note that, as expected,the next solution, for which k =3,givesz 4 = e 6iπ/3 =1=z 1 , so that there are only threeseparate solutions. ◭97

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSIm ze 2iπ/3 2π/32π/31Re ze −2iπ/3Figure 3.10 The solutions of z 3 =1.Not surprisingly, given that |z 3 | = |z| 3 from (3.10), all the roots of unity haveunit modulus, i.e. they all lie on a circle in the Argand diagram of unit radius.The three roots are shown in figure 3.10.The cube roots of unity are often written 1, ω and ω 2 . The properties ω 3 =1and 1 + ω + ω 2 = 0 are easily proved.3.4.3 Solving polynomial equationsA third application of de Moivre’s theorem is to the solution of polynomialequations. Complex equations in the form of a polynomial relationship must firstbe solved for z in a similar fashion to the method for finding the roots of realpolynomial equations. Then the complex roots of z may be found.◮Solve the equation z 6 − z 5 +4z 4 − 6z 3 +2z 2 − 8z +8=0.We first factorise to give(z 3 − 2)(z 2 +4)(z − 1) = 0.Hence z 3 =2orz 2 = −4 orz = 1. The solutions to the quadratic equation are z = ±2i;to find the complex cube roots, we first write the equation in the formz 3 =2=2e 2ikπ ,where k is any integer. If we now take the cube root, we getz =2 1/3 e 2ikπ/3 .98

3.5 COMPLEX LOGARITHMS AND COMPLEX POWERSTo avoid the duplication of solutions, we use the fact that −π

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSWe may use (3.34) to investigate further the properties of Ln z. We have alreadynoted that the argument of a complex number is multivalued, i.e. arg z = θ +2nπ,where n is any integer. Thus, in polar form, the complex number z should strictlybe written asz = re i(θ+2nπ) .Taking the logarithm of both sides, and using (3.34), we findLn z =lnr + i(θ +2nπ), (3.35)where ln r is the natural logarithm of the real positive quantity r and so iswritten normally. Thus from (3.35) we see that Ln z is itself multivalued. To avoidthis multivalued behaviour it is conventional to define another function ln z, theprincipal value of Ln z, which is obtained from Ln z by restricting the argumentof z to lie in the range −π

3.6 APPLICATIONS TO DIFFERENTIATION AND INTEGRATION3.6 Applications to differentiation and integrationWe can use the exponential form of a complex number together with de Moivre’stheorem (see section 3.4) to simplify the differentiation of trigonometric functions.◮Find the derivative with respect to x of e 3x cos 4x.We could differentiate this function straightforwardly using the product rule (see subsection2.1.2). However, an alternative method in this case is to use a complex exponential.Let us consider the complex numberz = e 3x (cos 4x + i sin 4x) =e 3x e 4ix = e (3+4i)x ,where we have used de Moivre’s theorem to rewrite the trigonometric functions as a complexexponential. This complex number has e 3x cos 4x as its real part. Now, differentiatingz with respect to x we obtaindzdx =(3+4i)e(3+4i)x =(3+4i)e 3x (cos 4x + i sin 4x), (3.36)where we have again used de Moivre’s theorem. Equating real parts we then findd (e 3x cos 4x ) = e 3x (3 cos 4x − 4sin4x).dxBy equating the imaginary parts of (3.36), we also obtain, as a bonus,d (e 3x sin 4x ) = e 3x (4 cos 4x +3sin4x). ◭dxIn a similar way the complex exponential can be used to evaluate integralscontaining trigonometric and exponential functions.◮Evaluate the integral I = ∫ e ax cos bx dx.Let us consider the integrand as the real part of the complex numbere ax (cos bx + i sin bx) =e ax e ibx = e (a+ib)x ,where we use de Moivre’s theorem to rewrite the trigonometric functions as a complexexponential. Integrating we find∫e (a+ib)x dx = e(a+ib)xa + ib + c(a − ib)e(a+ib)x=(a − ib)(a + ib) + c= eax (ae ibx − ibe ibx) + c, (3.37)a 2 + b 2where the constant of integration c is in general complex. Denoting this constant byc = c 1 + ic 2 and equating real parts in (3.37) we obtain∫I = e ax cos bx dx =eaxa 2 + b (a cos bx + b sin bx)+c 1,2which agrees with result (2.37) found using integration by parts. Equating imaginary partsin (3.37) we obtain, as a bonus,∫J = e ax sin bx dx =eaxa 2 + b (a sin bx − b cos bx)+c 2. ◭2101

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS3.7 Hyperbolic functionsThe hyperbolic functions are the complex analogues of the trigonometric functions.The analogy may not be immediately apparent and their definitions may appearat first to be somewhat arbitrary. However, careful examination of their propertiesreveals the purpose of the definitions. For instance, their close relationship withthe trigonometric functions, both in their identities and in their calculus, meansthat many of the familiar properties of trigonometric functions can also be appliedto the hyperbolic functions. Further, hyperbolic functions occur regularly, and sogiving them special names is a notational convenience.3.7.1 DefinitionsThe two fundamental hyperbolic functions are cosh x and sinh x, which, as theirnames suggest, are the hyperbolic equivalents of cos x and sin x. They are definedby the following relations:cosh x = 1 2 (ex + e −x ), (3.38)sinh x = 1 2 (ex − e −x ). (3.39)Note that cosh x is an even function and sinh x is an odd function. By analogywith the trigonometric functions, the remaining hyperbolic functions aretanh x = sinh xcosh x = ex − e −xe x , (3.40)+ e−x sech x = 1cosh x = 2e x , (3.41)+ e−x cosech x = 1sinh x = 2e x ,− e−x (3.42)coth x = 1tanh x = ex + e −xe x .− e−x (3.43)All the hyperbolic functions above have been defined in terms of the real variablex. However, this was simply so that they may be plotted (see figures 3.11–3.13);the definitions are equally valid for any complex number z.3.7.2 Hyperbolic–trigonometric analogiesIn the previous subsections we have alluded to the analogy between trigonometricand hyperbolic functions. Here, we discuss the close relationship between the twogroups of functions.Recalling (3.32) and (3.33) we findcos ix = 1 2 (ex + e −x ),sin ix = 1 2 i(ex − e −x ).102

3.7 HYPERBOLIC FUNCTIONS43cosh x21sech x−2 −11 2xFigure 3.11Graphs of cosh x and sechx.4cosech x2sinh x−2 −11 2x−2cosech x−4Figure 3.12Graphs of sinh x and cosechx.Hence, by the definitions given in the previous subsection,cosh x =cosix, (3.44)i sinh x =sinix, (3.45)cos x =coshix, (3.46)i sin x = sinh ix. (3.47)These useful equations make the relationship between hyperbolic and trigono-103

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS4coth x2tanh x−2 −11 2x−2coth x−4Figure 3.13 Graphs of tanh x and coth x.metric functions transparent. The similarity in their calculus is discussed furtherin subsection 3.7.6.3.7.3 Identities of hyperbolic functionsThe analogies between trigonometric functions and hyperbolic functions havingbeen established, we should not be surprised that all the trigonometric identitiesalso hold for hyperbolic functions, with the following modification. Whereversin 2 x occurs it must be replaced by − sinh 2 x, and vice versa. Note that thisreplacement is necessary even if the sin 2 x is hidden, e.g. tan 2 x =sin 2 x/ cos 2 xand so must be replaced by (− sinh 2 x/ cosh 2 x)=− tanh 2 x.◮Find the hyperbolic identity analogous to cos 2 x +sin 2 x =1.Using the rules stated above cos 2 x is replaced by cosh 2 x,andsin 2 x by − sinh 2 x,andsothe identity becomescosh 2 x − sinh 2 x =1.This can be verified by direct substitution, using the definitions of cosh x and sinh x; see(3.38) and (3.39). ◭Some other identities that can be proved in a similar way aresech 2 x =1− tanh 2 x, (3.48)cosech 2 x =coth 2 x − 1, (3.49)sinh 2x =2sinhx cosh x, (3.50)cosh 2x =cosh 2 x + sinh 2 x. (3.51)104

3.7 HYPERBOLIC FUNCTIONS3.7.4 Solving hyperbolic equationsWhen we are presented with a hyperbolic equation to solve, we may proceedby analogy with the solution of trigonometric equations. However, it is almostalways easier to express the equation directly in terms of exponentials.◮Solve the hyperbolic equation cosh x − 5sinhx − 5=0.Substituting the definitions of the hyperbolic functions we obtain12 (ex + e −x ) − 5 2 (ex − e −x ) − 5=0.Rearranging, and then multiplying through by −e x , gives in turn−2e x +3e −x − 5=0and2e 2x +5e x − 3=0.Now we can factorise and solve:(2e x − 1)(e x +3)=0.Thus e x =1/2 ore x = −3. Hence x = − ln 2 or x =ln(−3). The interpretation of thelogarithm of a negative number has been discussed in section 3.5. ◭3.7.5 Inverses of hyperbolic functionsJust like trigonometric functions, hyperbolic functions have inverses. If y =cosh x then x =cosh −1 y, which serves as a definition of the inverse. By usingthe fundamental definitions of hyperbolic functions, we can find closed-formexpressions for their inverses. This is best illustrated by example.◮Find a closed-form expression for the inverse hyperbolic function y =sinh −1 x.First we write x as a function of y, i.e.y =sinh −1 x ⇒ x =sinhy.Now, since cosh y = 1 2 (ey + e −y ) and sinh y = 1 2 (ey − e −y ),e y =coshy +sinhy√= 1+sinh 2 y +sinhye y = √ 1+x 2 + x,and hencey =ln( √ 1+x 2 + x). ◭In a similar fashion it can be shown thatcosh −1 x =ln( √ x 2 − 1+x).105

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS42sech −1 xcosh −1 x123 4x−2−4sech −1 xcosh −1 xFigure 3.14 Graphs of cosh −1 x and sech −1 x.◮Find a closed-form expression for the inverse hyperbolic function y =tanh −1 x.First we write x as a function of y, i.e.y =tanh −1 x ⇒ x =tanhy.Now, using the definition of tanh y and rearranging, we findx = ey − e −ye y + e −y ⇒ (x +1)e −y =(1− x)e y .Thus, it follows thate 2y = 1+x1 − x√1+x⇒ e y =1 − x ,√1+xy =ln1 − x ,tanh −1 x = 1 ( ) 1+x2 ln . ◭1 − xGraphs of the inverse hyperbolic functions are given in figures 3.14–3.16.3.7.6 Calculus of hyperbolic functionsJust as the identities of hyperbolic functions closely follow those of their trigonometriccounterparts, so their calculus is similar. The derivatives of the two basic106

3.7 HYPERBOLIC FUNCTIONS4cosech −1 x2sinh −1 x−2 −11 2xcosech −1 x−2−4Figure 3.15 Graphs of sinh −1 x and cosech −1 x.42tanh −1 xcoth −1 x−2 −11 2xcoth −1 x−2−4Figure 3.16 Graphs of tanh −1 x and coth −1 x.hyperbolic functions are given byd(cosh x) = sinh x, (3.52)dxd(sinh x) =coshx. (3.53)dxThey may be deduced by considering the definitions (3.38), (3.39) as follows.107

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS◮Verify the relation (d/dx)coshx =sinhx.Using the definition of cosh x,cosh x = 1 2 (ex + e −x ),and differentiating directly, we findddx (cosh x) = 1 2 (ex − e −x )=sinhx. ◭Clearly the integrals of the fundamental hyperbolic functions are also definedby these relations. The derivatives of the remaining hyperbolic functions can bederived by product differentiation and are presented below only for completeness.ddx (tanh x) =sech2 x, (3.54)d(sech x) = −sech x tanh x,dx(3.55)d(cosech x) = −cosech x coth x,dx(3.56)ddx (coth x) = −cosech2 x. (3.57)The inverse hyperbolic functions also have derivatives, which are given by thefollowing:d(cosh −1 x )=dx ad(sinh −1 x )=dx ad(tanh −1 x )=dx addx(coth −1 x a1√x2 − a 2 , (3.58)1√x2 + a 2 , (3.59)aa 2 − x 2 , for x2 a 2 . (3.61)These may be derived from the logarithmic form of the inverse (see subsection3.7.5).108

3.8 EXERCISES◮Evaluate (d/dx)sinh −1 x using the logarithmic form of the inverse.From the results of section 3.7.5,d (sinh −1 x ) = ddxdx[ (ln1=x + √ x 2 +11=x + √ x 2 +1=1√x2 +1 . ◭x + √ )]x 2 +1(1+x√x2 +1)( √ )x2 +1+x√x2 +13.8 Exercises3.1 Two complex numbers z and w are given by z =3+4i and w =2− i. OnanArgand diagram, plot(a) z + w, (b)w − z, (c)wz, (d)z/w,(e) z ∗ w + w ∗ z,(f)w 2 ,(g)lnz, (h)(1+z + w) 1/2 .3.2 By considering the real and imaginary parts of the product e iθ e iφ prove thestandard formulae for cos(θ + φ) and sin(θ + φ).3.3 By writing π/12 = (π/3) − (π/4) and considering e iπ/12 , evaluate cot(π/12).3.4 Find the locus in the complex z-plane of points that satisfy the following equations.( ) 1+it(a) z − c = ρ , where c is complex, ρ is real and t is a real parameter1 − itthat varies in the range −∞

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS(a) the equalities of corresponding angles, and(b) the constant ratio of corresponding sides,in the two triangles.By noting that any complex quantity can be expressed asz = |z| exp(i arg z),deduce thata(B − C)+b(C − A)+c(A − B) =0.3.9 For the real constant a find the loci of all points z = x + iy in the complex planethat satisfy{ ( )} z − ia(a) Re ln= c, c>0,z + ia{ ( )} z − ia(b) Im ln= k, 0 ≤ k ≤ π/2.z + iaIdentify the two families of curves and verify that in case (b) all curves passthrough the two points ±ia.3.10 The most general type of transformation between one Argand diagram, in thez-plane, and another, in the Z-plane, that gives one and only one value of Z foreach value of z (and conversely) is known as the general bilinear transformationand takes the formz = aZ + bcZ + d .(a) Confirm that the transformation from the Z-plane to the z-plane is also ageneral bilinear transformation.(b) Recalling that the equation of a circle can be written in the form∣ z − z 1 ∣∣∣∣ = λ, λ ≠1,z − z 2show that the general bilinear transformation transforms circles into circles(or straight lines). What is the condition that z 1 , z 2 and λ must satisfy if thetransformed circle is to be a straight line?3.11 Sketch the parts of the Argand diagram in which(a) Re z 2 < 0, |z 1/2 |≤2;(b) 0 ≤ arg z ∗ ≤ π/2;(c) | exp z 3 |→0as|z| →∞.What is the area of the region in which all three sets of conditions are satisfied?3.12 Denote the nth roots of unity by 1, ω n , ωn, 2 ... , ωn n−1 .(a) Prove that∑n−1∏n−1(i) ωn r =0, (ii) ωn r =(−1) n+1 .r=0(b) Express x 2 + y 2 + z 2 − yz − zx − xy as the product of two factors, each linearin x, y and z, with coefficients dependent on the third roots of unity (andthose of the x terms arbitrarily taken as real).r=0110

3.8 EXERCISES3.13 Prove that x 2m+1 − a 2m+1 ,wherem is an integer ≥ 1, can be written asm∏[( ) ]2πrx 2m+1 − a 2m+1 =(x − a) x 2 − 2ax cos+ a 2 .2m +1r=13.14 The complex position vectors of two parallel interacting equal fluid vorticesmoving with their axes of rotation always perpendicular to the z-plane are z 1and z 2 . The equations governing their motions aredz ∗ 1dt= − iz 1 − z 2,dz ∗ 2dt= − iz 2 − z 1.Deduce that (a) z 1 + z 2 ,(b)|z 1 − z 2 | and (c) |z 1 | 2 + |z 2 | 2 are all constant in time,and hence describe the motion geometrically.3.15 Solve the equationz 7 − 4z 6 +6z 5 − 6z 4 +6z 3 − 12z 2 +8z +4=0,(a) by examining the effect of setting z 3 equal to 2, and then(b) by factorising and using the binomial expansion of (z + a) 4 .Plot the seven roots of the equation on an Argand plot, exemplifying that complexroots of a polynomial equation always occur in conjugate pairs if the polynomialhas real coefficients.3.16 The polynomial f(z) is defined byf(z) =z 5 − 6z 4 +15z 3 − 34z 2 +36z − 48.(a) Show that the equation f(z) = 0 has roots of the form z = λi, whereλ isreal, and hence factorize f(z).(b) Show further that the cubic factor of f(z) can be written in the form(z + a) 3 + b, wherea and b are real, and hence solve the equation f(z) =0completely.3.17 The binomial expansion of (1 + x) n , discussed in chapter 1, can be written for apositive integer n asn∑(1 + x) n =n C r x r ,where n C r = n!/[r!(n − r)!].(a) Use de Moivre’s theorem to show that the sumS 1 (n) = n C 0 − n C 2 + n C 4 − ···+(−1) m n C 2m , n− 1 ≤ 2m ≤ n,has the value 2 n/2 cos(nπ/4).(b) Derive a similar result for the sumS 2 (n) = n C 1 − n C 3 + n C 5 − ···+(−1) m n C 2m+1 , n− 1 ≤ 2m +1≤ n,and verify it for the cases n =6,7and8.3.18 By considering (1 + exp iθ) n , prove thatn∑n C r cos rθ =2 n cos n (θ/2) cos(nθ/2),r=0r=0n∑n C r sin rθ =2 n cos n (θ/2) sin(nθ/2),r=0where n C r = n!/[r!(n − r)!].111

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS3.19 Use de Moivre’s theorem with n = 4 to prove thatcos 4θ =8cos 4 θ − 8cos 2 θ +1,and deduce that(cos π 8 = 2+ √ ) 1/22.43.20 Express sin 4 θ entirely in terms of the trigonometric functions of multiple anglesand deduce that its average value over a complete cycle is 3 . 83.21 Use de Moivre’s theorem to prove thattan 5θ = t5 − 10t 3 +5t5t 4 − 10t 2 +1 ,where t =tanθ. Deduce the values of tan(nπ/10) for n =1,2,3,4.3.22 Prove the following results involving hyperbolic functions.(a) That( ) x + ycosh x − cosh y =2sinh sinh2(b) That, if y =sinh −1 x,(x 2 +1) d2 ydx + x dy2 dx =0.3.23 Determine the conditions under which the equationa cosh x + b sinh x = c, c > 0,( x − yhas zero, one, or two real solutions for x. What is the solution if a 2 = c 2 + b 2 ?3.24 Use the definitions and properties of hyperbolic functions to do the following:(a) Solve cosh x =sinhx +2sechx.(b) Show that the real solution x of tanh x =cosechx canbewrittenintheform x =ln(u + √ u). Find an explicit value for u.(c) Evaluate tanh x when x is the real solution of cosh 2x =2coshx.3.25 Express sinh 4 x in terms of hyperbolic cosines of multiples of x, and hence findthe real solutions of2cosh4x − 8cosh2x +5=0.2).3.26 In the theory of special relativity, the relationship between the position and timecoordinates of an event, as measured in two frames of reference that have parallelx-axes, can be expressed in terms of hyperbolic functions. If the coordinates arex and t in one frame and x ′ and t ′ in the other, then the relationship take theformx ′ = x cosh φ − ct sinh φ,ct ′ = −x sinh φ + ct cosh φ.Express x and ct in terms of x ′ , ct ′ and φ and show thatx 2 − (ct) 2 =(x ′ ) 2 − (ct ′ ) 2 .112

3.9 HINTS AND ANSWERS3.27 A closed barrel has as its curved surface the surface obtained by rotating aboutthe x-axis the part of the curvey = a[2 − cosh(x/a)]lying in the range −b ≤ x ≤ b, whereb0. With these conditions,there are two roots if a 2 >b 2 , but only one if b 2 >a 2 .For a 2 = c 2 + b 2 , x = 1 ln[(a − b)/(a + b)].23.25 Reduce the equation to 16 sinh 4 x = 1, yielding x = ±0.481.113

COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS3.27 Show that ds =(coshx/a) dx;curved surface area = πa 2 [8 sinh(b/a) − sinh(2b/a)] − 2πab.flat ends area = 2πa 2 [4 − 4cosh(b/a)+cosh 2 (b/a)].114

4Series and limits4.1 SeriesMany examples exist in the physical sciences of situations where we are presentedwith a sum of terms to evaluate. For example, we may wish to add the contributionsfrom successive slits in a diffraction grating to find the total light intensity at aparticular point behind the grating.A series may have either a finite or infinite number of terms. In either case, thesum of the first N terms of a series (often called a partial sum) is writtenS N = u 1 + u 2 + u 3 + ···+ u N ,where the terms of the series u n , n =1, 2, 3,...,N are numbers, that may ingeneral be complex. If the terms are complex then S N will in general be complexalso, and we can write S N = X N + iY N ,whereX N and Y N are the partial sums ofthe real and imaginary parts of each term separately and are therefore real. If aseries has only N terms then the partial sum S N is of course the sum of the series.Sometimes we may encounter series where each term depends on some variable,x, say. In this case the partial sum of the series will depend on the value assumedby x. For example, consider the infinite seriesS(x) =1+x + x22! + x33! + ··· .This is an example of a power series; these are discussed in more detail insection 4.5. It is in fact the Maclaurin expansion of exp x (see subsection 4.6.3).Therefore S(x) =expx and, of course, varies according to the value of thevariable x. A series might just as easily depend on a complex variable z.A general, random sequence of numbers can be described as a series and a sumof the terms found. However, for cases of practical interest, there will usually be115

SERIES AND LIMITSsome sort of relationship between successive terms. For example, if the nth termof a series is given byu n = 1 2 n ,for n =1, 2, 3,...,N then the sum of the first N terms will beS N =N∑n=1u n = 1 2 + 1 4 + 1 8 + ···+ 12 N . (4.1)It is clear that the sum of a finite number of terms is always finite, providedthat each term is itself finite. It is often of practical interest, however, to considerthe sum of a series with an infinite number of finite terms. The sum of aninfinite number of terms is best defined by first considering the partial sumof the first N terms, S N . If the value of the partial sum S N tends to a finitelimit, S, asN tends to infinity, then the series is said to converge and its sumis given by the limit S. In other words, the sum of an infinite series is givenbyS = limN→∞S N ,provided the limit exists. For complex infinite series, if S N approaches a limitS = X + iY as N →∞, this means that X N → X and Y N → Y separately, i.e.the real and imaginary parts of the series are each convergent series with sumsX and Y respectively.However, not all infinite series have finite sums. As N →∞, the value of thepartial sum S N may diverge: it may approach +∞ or −∞, or oscillate finitelyor infinitely. Moreover, for a series where each term depends on some variable,its convergence can depend on the value assumed by the variable. Whether aninfinite series converges, diverges or oscillates has important implications whendescribing physical systems. Methods for determining whether a series convergesare discussed in section 4.3.4.2 Summation of seriesIt is often necessary to find the sum of a finite series or a convergent infiniteseries. We now describe arithmetic, geometric and arithmetico-geometric series,which are particularly common and for which the sums are easily found. Othermethods that can sometimes be used to sum more complicated series are discussedbelow.116

4.2 SUMMATION OF SERIES4.2.1 Arithmetic seriesAn arithmetic series has the characteristic that the difference between successiveterms is constant. The sum of a general arithmetic series is writtenN−1∑S N = a +(a + d)+(a +2d)+···+ [a +(N − 1)d] = (a + nd).Rewriting the series in the opposite order and adding this term by term to theoriginal expression for S N , we findS N = N 2 [a + a +(N − 1)d] = N (first term + last term). (4.2)2If an infinite number of such terms are added the series will increase (or decrease)indefinitely; that is to say, it diverges.n=0◮Sum the integers between 1 and 1000 inclusive.This is an arithmetic series with a =1,d =1andN = 1000. Therefore, using (4.2) we findS N = 1000 (1 + 1000) = 500500,2which can be checked directly only with considerable effort. ◭4.2.2 Geometric seriesEquation (4.1) is a particular example of a geometric series, which has thecharacteristic that the ratio of successive terms is a constant (one-half in thiscase). The sum of a geometric series is in general writtenN−1∑S N = a + ar + ar 2 + ···+ ar N−1 = ar n ,where a is a constant and r is the ratio of successive terms, the common ratio. Thesum may be evaluated by considering S N and rS N :n=0S N = a + ar + ar 2 + ar 3 + ···+ ar N−1 ,rS N = ar + ar 2 + ar 3 + ar 4 + ···+ ar N .If we now subtract the second equation from the first we obtainand hence(1 − r)S N = a − ar N ,S N = a(1 − rN ). (4.3)1 − r117

SERIES AND LIMITSFor a series with an infinite number of terms and |r| < 1, we have lim N→∞ r N =0,and the sum tends to the limitS =a1 − r . (4.4)In (4.1), r = 1 2 , a = 1 2,andsoS =1.For|r| ≥1, however, the series either divergesor oscillates.◮Consider a ball that drops from a height of 27 m and on each bounce retains only a thirdof its kinetic energy; thus after one bounce it will return to a height of 9m,aftertwobounces to 3m, and so on. Find the total distance travelled between the first bounce andthe Mth bounce.The total distance travelled between the first bounce and the Mth bounce is given by thesum of M − 1 terms:M−2∑ 9S M−1 =2(9+3+1+···) =23 mfor M>1, where the factor 2 is included to allow for both the upward and the downwardjourney. Inside the parentheses we clearly have a geometric series with first term 9 andcommon ratio 1/3 and hence the distance is given by (4.3), i.e.[9 1 − ( ) ]1 M−1[3S M−1 =2×=27 1 − ( ) ]1 M−1,1 − 1 33where the number of terms N in (4.3) has been replaced by M − 1. ◭m=04.2.3 Arithmetico-geometric seriesAn arithmetico-geometric series, as its name suggests, is a combined arithmeticand geometric series. It has the general formN−1∑S N = a +(a + d)r +(a +2d)r 2 + ···+ [a +(N − 1)d] r N−1 = (a + nd)r n ,and can be summed, in a similar way to a pure geometric series, by multiplyingby r and subtracting the result from the original series to obtain(1 − r)S N = a + rd + r 2 d + ···+ r N−1 d − [a +(N − 1)d] r N .Using the expression for the sum of a geometric series (4.3) and rearranging, wefindS N =a − [a +(N − 1)d] rN1 − r+ rd(1 − rN−1 )(1 − r) 2 .For an infinite series with |r| < 1, lim N→∞ r N = 0 as in the previous subsection,and the sum tends to the limitS =a1 − r + rd(1 − r) 2 . (4.5)As for a geometric series, if |r| ≥1 then the series either diverges or oscillates.118n=0

4.2 SUMMATION OF SERIES◮Sum the seriesS =2+ 5 2 + 8 2 2 + 112 3 + ··· .This is an infinite arithmetico-geometric series with a =2,d =3andr =1/2. Therefore,from (4.5), we obtain S = 10. ◭4.2.4 The difference methodThe difference method is sometimes useful in summing series that are morecomplicated than the examples discussed above. Let us consider the general seriesN∑u n = u 1 + u 2 + ···+ u N .n=1If the terms of the series, u n , can be expressed in the formu n = f(n) − f(n − 1)for some function f(n) then its (partial) sum is given byN∑S N = u n = f(N) − f(0).n=1This can be shown as follows. The sum is given byS N = u 1 + u 2 + ···+ u Nand since u n = f(n) − f(n − 1), it may be rewrittenS N =[f(1) − f(0)] + [f(2) − f(1)] + ···+[f(N) − f(N − 1)].By cancelling terms we see thatS N = f(N) − f(0).◮Evaluate the sumN∑n=11n(n +1) .Using partial fractions we find( 1u n = −n +1 − 1 ).nHence u n = f(n) − f(n − 1) with f(n) =−1/(n +1),andsothesumisgivenbyS N = f(N) − f(0) = − 1N +1 +1= NN +1 . ◭119

SERIES AND LIMITSThe difference method may be easily extended to evaluate sums in which eachterm can be expressed in the formu n = f(n) − f(n − m), (4.6)where m is an integer. By writing out the sum to N terms with each term expressedin this form, and cancelling terms in pairs as before, we findm∑m∑S N = f(N − k +1)− f(1 − k).k=1k=1◮Evaluate the sumN∑n=11n(n +2) .Using partial fractions we find[u n = −12(n +2) − 1 2nHence u n = f(n) − f(n − 2) with f(n) =−1/[2(n +2)],andsothesumisgivenbyS N = f(N)+f(N − 1) − f(0) − f(−1) = 3 4 − 1 ( 12 N +2 + 1 ). ◭N +1In fact the difference method is quite flexible and may be used to evaluatesums even when each term cannot be expressed as in (4.6). The method still relies,however, on being able to write u n in terms of a single function such that mostterms in the sum cancel, leaving only a few terms at the beginning and the end.This is best illustrated by an example.].◮Evaluate the sumN∑n=11n(n +1)(n +2) .Using partial fractions we find1u n =2(n +2) − 1n +1 + 1 2n .Hence u n = f(n) − 2f(n − 1) + f(n − 2) with f(n) =1/[2(n + 2)]. If we write out the sum,expressing each term u n in this form, we find that most terms cancel and the sum is givenbyS N = f(N) − f(N − 1) − f(0) + f(−1) = 1 4 + 1 2( 1N +2 − 1 ). ◭N +1120

4.2 SUMMATION OF SERIES4.2.5 Series involving natural numbersSeries consisting of the natural numbers 1, 2, 3, ..., or the square or cube of thesenumbers, occur frequently and deserve a special mention. Let us first considerthe sum of the first N natural numbers,N∑S N =1+2+3+···+ N = n.This is clearly an arithmetic series with first term a = 1 and common differenced = 1. Therefore, from (4.2), S N = 1 2N(N +1).Next, we consider the sum of the squares of the first N natural numbers:N∑S N =1 2 +2 2 +3 2 + ...+ N 2 = n 2 ,which may be evaluated using the difference method. The nth term in the seriesis u n = n 2 , which we need to express in the form f(n) − f(n − 1) for some functionf(n). Consider the functionf(n) =n(n + 1)(2n +1) ⇒ f(n − 1) = (n − 1)n(2n − 1).For this function f(n) − f(n − 1) = 6n 2 , and so we can writeu n = 1 6[f(n) − f(n − 1)].Therefore, by the difference method,S N = 1 6 [f(N) − f(0)] = 1 6N(N + 1)(2N +1).Finally, we calculate the sum of the cubes of the first N natural numbers,N∑S N =1 3 +2 3 +3 3 + ···+ N 3 = n 3 ,again using the difference method. Consider the functionf(n) =[n(n +1)] 2 ⇒ f(n − 1) = [(n − 1)n] 2 ,for which f(n) − f(n − 1) = 4n 3 . Therefore we can write the general nth term ofthe series asu n = 1 4[f(n) − f(n − 1)],and using the difference method we findS N = 1 4 [f(N) − f(0)] = 1 4 N2 (N +1) 2 .Note that this is the square of the sum of the natural numbers, i.e.(N∑N 2∑n 3 = n).n=1121n=1n=1n=1n=1

SERIES AND LIMITS◮Sum the seriesN∑(n +1)(n +3).n=1The nth term in this series isu n =(n +1)(n +3)=n 2 +4n +3,and therefore we can writeN∑N∑(n +1)(n +3)= (n 2 +4n +3)n=1n=1N∑N∑ N∑= n 2 +4 n + 3n=1n=1n=1= 1 N(N + 1)(2N +1)+4× 1 N(N +1)+3N6 2= 1 6 N(2N2 +15N + 31). ◭4.2.6 Transformation of seriesA complicated series may sometimes be summed by transforming it into afamiliar series for which we already know the sum, perhaps a geometric seriesor the Maclaurin expansion of a simple function (see subsection 4.6.3). Varioustechniques are useful, and deciding which one to use in any given case is a matterof experience. We now discuss a few of the more common methods.The differentiation or integration of a series is often useful in transforming anapparently intractable series into a more familiar one. If we wish to differentiateor integrate a series that already depends on some variable then we may do soin a straightforward manner.◮Sum the seriesS(x) =x43(0!) + x54(1!) + x65(2!) + ··· .Dividing both sides by x we obtainS(x)x = x33(0!) + x44(1!) + x55(2!) + ··· ,which is easily differentiated to give[ ]d S(x)= x2dx x 0! + x31! + x42! + x53! + ··· .Recalling the Maclaurin expansion of exp x given in subsection 4.6.3, we recognise thatthe RHS is equal to x 2 exp x. Having done so, we can now integrate both sides to obtain∫S(x)/x = x 2 exp xdx.122

4.2 SUMMATION OF SERIESIntegrating the RHS by parts we findS(x)/x = x 2 exp x − 2x exp x +2expx + c,where the value of the constant of integration c canbefixedbytherequirementthatS(x)/x =0atx = 0. Thus we find that c = −2 and that the sum is given byS(x) =x 3 exp x − 2x 2 exp x +2x exp x − 2x. ◭Often, however, we require the sum of a series that does not depend on avariable. In this case, in order that we may differentiate or integrate the series,we define a function of some variable x such that the value of this function isequal to the sum of the series for some particular value of x (usually at x =1).◮Sum the seriesS =1+ 2 2 + 3 2 2 + 4 2 3 + ··· .Let us begin by defining the functionf(x)=1+2x +3x 2 +4x 3 + ··· ,so that the sum S = f(1/2). Integrating this function we obtain∫f(x) dx = x + x 2 + x 3 + ··· ,which we recognise as an infinite geometric series with first term a = x and common ratior = x. Therefore, from (4.4), we find that the sum of this series is x/(1 − x). In other words∫f(x) dx =x1 − x ,so that f(x) isgivenbyf(x) = d ( x)1=dx 1 − x (1 − x) . 2The sum of the original series is therefore S = f(1/2) = 4. ◭Aside from differentiation and integration, an appropriate substitution cansometimes transform a series into a more familiar form. In particular, series withterms that contain trigonometric functions can often be summed by the use ofcomplex exponentials.◮Sum the seriesS(θ)=1+cosθ +cos 2θ2!+cos 3θ3!+ ··· .Replacing the cosine terms with a complex exponential, we obtain{}exp 2iθ exp 3iθS(θ) =Re 1+expiθ + + + ···2! 3!}(exp iθ)2 (exp iθ)3=Re{1+expiθ + + + ··· .2! 3!123

SERIES AND LIMITSAgain using the Maclaurin expansion of exp x given in subsection 4.6.3, we notice thatS(θ) = Re [exp(exp iθ)] = Re [exp(cos θ + i sin θ)]=Re{[exp(cos θ)][exp(i sin θ)]} = [exp(cos θ)]Re [exp(i sin θ)]=[exp(cosθ)][cos(sin θ)]. ◭4.3 Convergence of infinite seriesAlthough the sums of some commonly occurring infinite series may be found,the sum of a general infinite series is usually difficult to calculate. Nevertheless,it is often useful to know whether the partial sum of such a series converges toa limit, even if the limit cannot be found explicitly. As mentioned at the end ofsection 4.1, if we allow N to tend to infinity, the partial sumN∑S N =u nn=1of a series may tend to a definite limit (i.e. the sum S of the series), or increaseor decrease without limit, or oscillate finitely or infinitely.To investigate the convergence of any given series, it is useful to have availablea number of tests and theorems of general applicability. We discuss them below;some we will merely state, since once they have been stated they become almostself-evident, but are no less useful for that.4.3.1 Absolute and conditional convergenceLet us first consider some general points concerning the convergence, or otherwise,of an infinite series. In general an infinite series ∑ u n can have complex terms,and in the special case of a real series the terms can be positive or negative. Fromany such series, however, we can always construct another series ∑ |u n | in whicheach term is simply the modulus of the corresponding term in the original series.Then each term in the new series will be a positive real number.If the series ∑ |u n | converges then ∑ u n also converges, and ∑ u n is said to beabsolutely convergent, i.e. the series formed by the absolute values is convergent.For an absolutely convergent series, the terms may be reordered without affectingthe convergence of the series. However, if ∑ |u n | diverges whilst ∑ u n convergesthen ∑ u n is said to be conditionally convergent. For a conditionally convergentseries, rearranging the order of the terms can affect the behaviour of the sumand, hence, whether the series converges or diverges. In fact, a theorem dueto Riemann shows that, by a suitable rearrangement, a conditionally convergentseries may be made to converge to any arbitrary limit, or to diverge, or to oscillatefinitely or infinitely! Of course, if the original series ∑ u n consists only of positivereal terms and converges then automatically it is absolutely convergent.124

4.3 CONVERGENCE OF INFINITE SERIES4.3.2 Convergence of a series containing only real positive termsAs discussed above, in order to test for the absolute convergence of a series∑un , we first construct the corresponding series ∑ |u n | that consists only of realpositive terms. Therefore in this subsection we will restrict our attention to seriesof this type.We discuss below some tests that may be used to investigate the convergence ofsuch a series. Before doing so, however, we note the following crucial consideration.In all the tests for, or discussions of, the convergence of a series, it is not whathappens in the first ten, or the first thousand, or the first million terms (or anyother finite number of terms) that matters, but what happens ultimately.Preliminary testA necessary but not sufficient condition for a series of real positive terms ∑ u nto be convergent is that the term u n tends to zero as n tends to infinity, i.e. werequirelim u n =0.n→∞If this condition is not satisfied then the series must diverge. Even if it is satisfied,however, the series may still diverge, and further testing is required.Comparison testThe comparison test is the most basic test for convergence. Let us consider twoseries ∑ u n and ∑ v n and suppose that we know the latter to be convergent (bysome earlier analysis, for example). Then, if each term u n in the first series is lessthan or equal to the corresponding term v n in the second series, for all n greaterthan some fixed number N that will vary from series to series, then the originalseries ∑ u n is also convergent. In other words, if ∑ v n is convergent andu n ≤ v nfor n>N,then ∑ u n converges.However, if ∑ v n diverges and u n ≥ v n for all n greater than some fixed numberthen ∑ u n diverges.◮Determine whether the following series converges:∞∑ 1n!+1 = 1 2 + 1 3 + 1 7 + 1 + ··· . (4.7)25n=1Let us compare this series with the series∞∑ 1n! = 1 0! + 1 1! + 1 2! + 1 3! + ···=2+ 1 2! + 1 + ··· , (4.8)3!n=0125

SERIES AND LIMITSwhich is merely the series obtained by setting x = 1 in the Maclaurin expansion of exp x(see subsection 4.6.3), i.e.exp(1) = e =1+ 1 1! + 1 2! + 1 3! + ··· .Clearly this second series is convergent, since it consists of only positive terms and has afinite sum. Thus, since each term u n in the series (4.7) is less than the corresponding term1/n! in (4.8), we conclude from the comparison test that (4.7) is also convergent. ◭D’Alembert’s ratio testThe ratio test determines whether a series converges by comparing the relativemagnitude of successive terms. If we consider a series ∑ u n and set( )un+1ρ = lim , (4.9)n→∞ u nthen if ρ1 the series is divergent; if ρ =1then the behaviour of the series is undetermined by this test.To prove this we observe that if the limit (4.9) is less than unity, i.e. ρ

4.3 CONVERGENCE OF INFINITE SERIESRatio comparison testAs its name suggests, the ratio comparison test is a combination of the ratio andcomparison tests. Let us consider the two series ∑ u n and ∑ v n and assume thatwe know the latter to be convergent. It may be shown that ifu n+1u n≤ v n+1v nfor all n greater than some fixed value N then ∑ u n is also convergent.Similarly, ifu n+1≥ v n+1u n v nfor all sufficiently large n, and ∑ v n diverges then ∑ u n also diverges.◮Determine whether the following series converges:∞∑ 1(n!) =1+ 1 2 2 + 1 2 6 + ··· .2n=1In this case the ratio of successive terms, as n tends to infinity, is given by[ ] 2 ( ) 2n!1R = lim= lim ,n→∞ (n +1)! n→∞ n +1which is less than the ratio seen in (4.11). Hence, by the ratio comparison test, the seriesconverges. (It is clear that this series could also be found to be convergent using the ratiotest.) ◭Quotient testThe quotient test may also be considered as a combination of the ratio andcomparison tests. Let us again consider the two series ∑ u n and ∑ v n , and defineρ as the limit( )unρ = lim . (4.12)n→∞ v nThen, it can be shown that:(i) if ρ ≠ 0 but is finite then ∑ u n and ∑ v n either both converge or bothdiverge;(ii) if ρ = 0 and ∑ v n converges then ∑ u n converges;(iii) if ρ = ∞ and ∑ v n diverges then ∑ u n diverges.127

SERIES AND LIMITS◮Given that the series ∑ ∞n=11/n diverges, determine whether the following series converges:∞∑ 4n 2 − n − 3n 3 +2n . (4.13)n=1If we set u n =(4n 2 − n − 3)/(n 3 +2n) andv n =1/n then the limit (4.12) becomes[ ] [ ](4n 2 − n − 3)/(n 3 +2n) 4n 3 − n 2 − 3nρ = lim= lim=4.n→∞ 1/nn→∞ n 3 +2nSince ρ is finite but non-zero and ∑ v n diverges, from (i) above ∑ u n must also diverge. ◭Integral testThe integral test is an extremely powerful means of investigating the convergenceof a series ∑ u n . Suppose that there exists a function f(x) which monotonicallydecreases for x greater than some fixed value x 0 and for which f(n) =u n ,i.e.thevalue of the function at integer values of x is equal to the corresponding termin the series under investigation. Then it can be shown that, if the limit of theintegral∫ Nlim f(x) dxN→∞exists, the series ∑ u n is convergent. Otherwise the series diverges. Note that theintegral defined here has no lower limit; the test is sometimes stated with a lowerlimit, equal to unity, for the integral, but this can lead to unnecessary difficulties.◮Determine whether the following series converges:∞∑ 1(n − 3/2) =4+4+ 4 2 9 + 425 + ··· .n=1Let us consider the function f(x) =(x − 3/2) −2 . Clearly f(n) =u n and f(x) monotonicallydecreases for x>3/2. Applying the integral test, we consider∫ N( )1−1limdx = lim=0.N→∞ (x − 3/2)2 N→∞ N − 3/2Since the limit exists the series converges. Note, however, that if we had included a lowerlimit, equal to unity, in the integral then we would have run into problems, since theintegrand diverges at x =3/2. ◭The integral test is also useful for examining the convergence of the Riemannzeta series. This is a special series that occurs regularly and is of the form∞∑ 1n p .n=1It converges for p>1 and diverges if p ≤ 1. These convergence criteria may bederived as follows.128

4.3 CONVERGENCE OF INFINITE SERIESUsing the integral test, we consider∫ N( )1N1−plim dx = lim ,N→∞ xp N→∞ 1 − pand it is obvious that the limit tends to zero for p>1andto∞ for p ≤ 1.Cauchy’s root testCauchy’s root test may be useful in testing for convergence, especially if the nthterms of the series contains an nth power. If we define the limitρ = lim (u n ) 1/n ,n→∞then it may be proved that the series ∑ u n converges if ρ1 then theseries diverges. Its behaviour is undetermined if ρ =1.◮Determine whether the following series converges:∞∑( ) n 1=1+ 1 n 4 + 1 27 + ··· .n=1Using Cauchy’s root test, we findand hence the series converges. ◭( ) 1ρ = lim =0,n→∞ nGrouping termsWe now consider the Riemann zeta series, mentioned above, with an alternativeproof of its convergence that uses the method of grouping terms. In general thereare better ways of determining convergence, but the grouping method may beused if it is not immediately obvious how to approach a problem by a bettermethod.First consider the case where p>1, and group the terms in the series asfollows:S N = 1 ( 11 p + 2 p + 1 ) ( 13 p +4 p + ···+ 1 )7 p + ··· .Now we can see that each bracket of this series is less than each term of thegeometric seriesS N = 1 1 p + 2 2 p + 4 4 p + ··· .This geometric series has common ratio r = ( 12) p−1;sincep>1, it follows thatr1.129

SERIES AND LIMITSThe divergence of the Riemann zeta series for p ≤ 1 can be seen by firstconsidering the case p = 1. The series isS N =1+ 1 2 + 1 3 + 1 4 + ··· ,which does not converge, as may be seen by bracketing the terms of the series ingroups in the following way:N∑( ( 1 1S N = u n =1+ +2)3 4)+ 1 ( 1+5 + 1 6 + 1 7 8)+ 1 + ··· .n=1The sum of the terms in each bracket is ≥ 1 2and, since as many such groupingscan be made as we wish, it is clear that S N increases indefinitely as N is increased.Now returning to the case of the Riemann zeta series for p 1/n for n>1, p

4.4 OPERATIONS WITH SERIESis always less than u N for all m and u n → 0asn →∞, the alternating seriesconverges. It is clear that an analogous proof can be constructed in the casewhere N is even.◮Determine whether the following series converges:∞∑(−1) n+1 1 n =1− 1 2 + 1 3 − ··· .n=1This alternating series clearly satisfies conditions (i) and (ii) above and hence converges.However, as shown above by the method of grouping terms, the corresponding series withall positive terms is divergent. ◭4.4 Operations with seriesSimple operations with series are fairly intuitive, and we discuss them here onlyfor completeness. The following points apply to both finite and infinite seriesunless otherwise stated.(i) If ∑ u n = S then ∑ ku n = kS where k is any constant.(ii) If ∑ u n = S and ∑ v n = T then ∑ (u n + v n )=S + T .(iii) If ∑ u n = S then a + ∑ u n = a + S. A simple extension of this trivial resultshows that the removal or insertion of a finite number of terms anywherein a series does not affect its convergence.(iv) If the infinite series ∑ u n and ∑ v n are both absolutely convergent thenthe series ∑ w n ,wherew n = u 1 v n + u 2 v n−1 + ···+ u n v 1 ,is also absolutely convergent. The series ∑ w n is called the Cauchy productof the two original series. Furthermore, if ∑ u n converges to the sum Sand ∑ v n converges to the sum T then ∑ w n converges to the sum ST.(v) It is not true in general that term-by-term differentiation or integration ofa series will result in a new series with the same convergence properties.A power series has the form4.5 Power seriesP (x) =a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ··· ,where a 0 ,a 1 ,a 2 ,a 3 etc. are constants. Such series regularly occur in physics andengineering and are useful because, for |x| < 1, the later terms in the series maybecome very small and be discarded. For example the seriesP (x) =1+x + x 2 + x 3 + ··· ,131

SERIES AND LIMITSalthough in principle infinitely long, in practice may be simplified if x happens tohave a value small compared with unity. To see this note that P (x) forx =0.1has the following values: 1, if just one term is taken into account; 1.1, for twoterms; 1.11, for three terms; 1.111, for four terms, etc. If the quantity that itrepresents can only be measured with an accuracy of two decimal places, then allbut the first three terms may be ignored, i.e. when x =0.1 orlessP (x) =1+x + x 2 +O(x 3 ) ≈ 1+x + x 2 .This sort of approximation is often used to simplify equations into manageableforms. It may seem imprecise at first but is perfectly acceptable insofar as itmatches the experimental accuracy that can be achieved.The symbols O and ≈ used above need some further explanation. They areused to compare the behaviour of two functions when a variable upon whichboth functions depend tends to a particular limit, usually zero or infinity (andobvious from the context). For two functions f(x) andg(x), with g positive, theformal definitions of the above symbols are as follows:(i) If there exists a constant k such that |f| ≤kg as the limit is approachedthen f =O(g).(ii) If as the limit of x is approached f/g tends to a limit l, wherel ≠0,thenf ≈ lg. The statement f ≈ g means that the ratio of the two sides tendsto unity.4.5.1 Convergence of power seriesThe convergence or otherwise of power series is a crucial consideration in practicalterms. For example, if we are to use a power series as an approximation, it isclearly important that it tends to the precise answer as more and more terms ofthe approximation are taken. Consider the general power seriesP (x) =a 0 + a 1 x + a 2 x 2 + ··· .Using d’Alembert’s ratio test (see subsection 4.3.2), we see that P (x) convergesabsolutely if∣ ∣ ∣∣∣ a n+1ρ = lim xn→∞ a n∣ = |x| lima n+1 ∣∣∣n→∞ ∣ < 1.a nThus the convergence of P (x) depends upon the value of x, i.e. there is, in general,a range of values of x for which P (x) converges, an interval of convergence. Notethat at the limits of this range ρ = 1, and so the series may converge or diverge.The convergence of the series at the end-points may be determined by substitutingthese values of x into the power series P (x) and testing the resulting series usingany applicable method (discussed in section 4.3).132

4.5 POWER SERIES◮Determine the range of values of x for which the following power series converges:P (x)=1+2x +4x 2 +8x 3 + ··· .By using the interval-of-convergence method discussed above,∣ ∣ ∣∣∣ 2 n+1 ∣∣∣ρ = limn→∞ 2 x = |2x|,nand hence the power series will converge for |x| < 1/2. Examining the end-points of theinterval separately, we findP (1/2)=1+1+1+··· ,P (−1/2) = 1 − 1+1− ··· .Obviously P (1/2) diverges, while P (−1/2) oscillates. Therefore P (x) is not convergent ateither end-point of the region but is convergent for −1

SERIES AND LIMITSr = − exp iθ. Therefore, on the the circle of convergence we have1P (z) =1+expiθ .Unless θ = π this is a finite complex number, and so P (z) converges at all points on thecircle |z| =2exceptatθ = π (i.e. z = −2), where it diverges. Note that P (z) isjustthebinomial expansion of (1 + z/2) −1 , for which it is obvious that z = −2 is a singular point.In general, for power series expansions of complex functions about a given point in thecomplex plane, the circle of convergence extends as far as the nearest singular point. Thisis discussed further in chapter 24. ◭Note that the centre of the circle of convergence does not necessarily lie at theorigin. For example, applying the ratio test to the complex power seriesP (z) =1+ z − 12+(z − 1)24+(z − 1)38+ ··· ,we find that for it to converge we require |(z − 1)/2| < 1. Thus the series convergesfor z lying within a circle of radius 2 centred on the point (1, 0) in the Arganddiagram.4.5.2 Operations with power seriesThe following rules are useful when manipulating power series; they apply topower series in a real or complex variable.(i) If two power series P (x) andQ(x) have regions of convergence that overlapto some extent then the series produced by taking the sum, the difference or theproduct of P (x) andQ(x) converges in the common region.(ii) If two power series P (x) andQ(x) converge for all values of x then oneseries may be substituted into the other to give a third series, which also convergesfor all values of x. For example, consider the power series expansions of sin x ande x given below in subsection 4.6.3,sin x = x − x33! + x55! − x77! + ···e x =1+x + x22! + x33! + x44! + ··· ,both of which converge for all values of x. Substituting the series for sin x intothat for e x we obtaine sin x =1+x + x22! − 3x44! − 8x55! + ··· ,which also converges for all values of x.If, however, either of the power series P (x) andQ(x) has only a limited regionof convergence, or if they both do so, then further care must be taken whensubstituting one series into the other. For example, suppose Q(x) converges forall x, but P (x) onlyconvergesforx within a finite range. We may substitute134

4.5 POWER SERIESQ(x) intoP (x) toobtainP (Q(x)), but we must be careful since the value of Q(x)may lie outside the region of convergence for P (x), with the consequence that theresulting series P (Q(x)) does not converge.(iii) If a power series P (x) converges for a particular range of x then the seriesobtained by differentiating every term and the series obtained by integrating everyterm also converge in this range.This is easily seen for the power seriesP (x) =a 0 + a 1 x + a 2 x 2 + ··· ,which converges if |x| < lim n→∞ |a n /a n+1 |≡k. The series obtained by differentiatingP (x) with respect to x is given bydPdx = a 1 +2a 2 x +3a 3 x 2 + ···and converges if∣ ∣ ∣∣∣ na n ∣∣∣|x| < lim= k.n→∞ (n +1)a n+1Similarly the series obtained by integrating P (x) term by term,converges if∫P (x) dx = a 0 x + a 1x 22+ a 2x 33+ ··· ,∣ ∣ ∣∣∣ (n +2)a n ∣∣∣|x| < lim= k.n→∞ (n +1)a n+1So, series resulting from differentiation or integration have the same interval ofconvergence as the original series. However, even if the original series convergesat either end-point of the interval, it is not necessarily the case that the new serieswill do so. The new series must be tested separately at the end-points in orderto determine whether it converges there. Note that although power series may beintegrated or differentiated without altering their interval of convergence, this isnot true for series in general.It is also worth noting that differentiating or integrating a power series termby term within its interval of convergence is equivalent to differentiating orintegrating the function it represents. For example, consider the power seriesexpansion of sin x,sin x = x − x33! + x55! − x7+ ··· , (4.14)7!which converges for all values of x. If we differentiate term by term, the seriesbecomes1 − x22! + x44! − x66! + ··· ,which is the series expansion of cos x, asweexpect.135

SERIES AND LIMITS4.6 Taylor seriesTaylor’s theorem provides a way of expressing a function as a power series in x,known as a Taylor series, but it can be applied only to those functions that arecontinuous and differentiable within the x-range of interest.4.6.1 Taylor’s theoremSuppose that we have a function f(x) that we wish to express as a power seriesin x − a about the point x = a. We shall assume that, in a given x-range, f(x)is a continuous, single-valued function of x having continuous derivatives withrespect to x, denoted by f ′ (x), f ′′ (x) and so on, up to and including f (n−1) (x). Weshall also assume that f (n) (x) exists in this range.From the equation following (2.31) we may write∫ a+haf ′ (x) dx = f(a + h) − f(a),where a, a + h are neighbouring values of x. Rearranging this equation, we mayexpress the value of the function at x = a + h in terms of its value at a byf(a + h) =f(a)+∫ a+haf ′ (x) dx. (4.15)A first approximation for f(a + h) may be obtained by substituting f ′ (a) forf ′ (x) in (4.15), to obtainf(a + h) ≈ f(a)+hf ′ (a).This approximation is shown graphically in figure 4.1. We may write this firstapproximation in terms of x and a asand, in a similar way,f(x) ≈ f(a)+(x − a)f ′ (a),f ′ (x) ≈ f ′ (a)+(x − a)f ′′ (a),f ′′ (x) ≈ f ′′ (a)+(x − a)f ′′′ (a),and so on. Substituting for f ′ (x) in (4.15), we obtain the second approximation:f(a + h) ≈ f(a)+∫ a+ha≈ f(a)+hf ′ (a)+ h22 f′′ (a).[f ′ (a)+(x − a)f ′′ (a)] dxWe may repeat this procedure as often as we like (so long as the derivativesof f(x) exist) to obtain higher-order approximations to f(a + h); we find the136

4.6 TAYLOR SERIESf(x)QRf(a)Pθhhf ′ (a)aa + hxFigure 4.1 The first-order Taylor series approximation to a function f(x).The slope of the function at P ,i.e.tanθ, equalsf ′ (a). Thus the value of thefunction at Q, f(a + h), is approximated by the ordinate of R, f(a)+hf ′ (a).(n − 1)th-order approximation § to bef(a + h) ≈ f(a)+hf ′ (a)+ h22! f′′ (a)+···+ hn−1(n − 1)! f(n−1) (a). (4.16)As might have been anticipated, the error associated with approximating f(a+h)by this (n − 1)th-order power series is of the order of the next term in the series.This error or remainder can be shown to be given byR n (h) = hnn! f(n) (ξ),for some ξ that lies in the range [a, a + h]. Taylor’s theorem then states that wemay write the equalityf(a + h) =f(a)+hf ′ (a)+ h22! f′′ (a)+···+ h(n−1)(n − 1)! f(n−1) (a)+R n (h).(4.17)The theorem may also be written in a form suitable for finding f(x) giventhe value of the function and its relevant derivatives at x = a, by substituting§ The order of the approximation is simply the highest power of h in the series. Note, though, thatthe (n − 1)th-order approximation contains n terms.137

SERIES AND LIMITSx = a + h in the above expression. It then readsf(x) =f(a)+(x − a)f ′ (a)+where the remainder now takes the form(x − a)2f ′′ (x − a)n−1(a)+···+2!(n − 1)! f(n−1) (a)+R n (x),(4.18)(x − a)nR n (x) = f (n) (ξ),n!and ξ lies in the range [a, x]. Each of the formulae (4.17), (4.18) gives us theTaylor expansion of the function about the point x = a. A special case occurswhen a = 0. Such Taylor expansions, about x = 0, are called Maclaurin series.Taylor’s theorem is also valid without significant modification for functionsof a complex variable (see chapter 24). The extension of Taylor’s theorem tofunctions of more than one variable is given in chapter 5.For a function to be expressible as an infinite power series we require it to beinfinitely differentiable and the remainder term R n to tend to zero as n tends toinfinity, i.e. lim n→∞ R n = 0. In this case the infinite power series will represent thefunction within the interval of convergence of the series.◮Expand f(x) =sinx as a Maclaurin series, i.e. about x =0.We must first verify that sin x may indeed be represented by an infinite power series. It iseasily shown that the nth derivative of f(x) isgivenby(f (n) (x) =sin x + nπ ).2Therefore the remainder after expanding f(x) asan(n − 1)th-order polynomial aboutx = 0 is given byR n (x) =(ξ xnn! sin + nπ ),2where ξ lies in the range [0,x]. Since the modulus of the sine term is always less than orequal to unity, we can write |R n (x)| < |x n |/n!. For any particular value of x, sayx = c,R n (c) → 0asn →∞. Hence lim n→∞ R n (x) = 0, and so sin x can be represented by aninfinite Maclaurin series.Evaluating the function and its derivatives at x =0weobtainf(0) = sin 0 = 0,f ′ (0) = sin(π/2) = 1,f ′′ (0) = sin π =0,f ′′′ (0) = sin(3π/2) = −1,and so on. Therefore, the Maclaurin series expansion of sin x is given bysin x = x − x33! + x55! − ··· .Note that, as expected, since sin x is an odd function, its power series expansion containsonly odd powers of x. ◭138

4.6 TAYLOR SERIESWe may follow a similar procedure to obtain a Taylor series about an arbitrarypoint x = a.◮Expand f(x) =cosx as a Taylor series about x = π/3.As in the above example, it is easily shown that the nth derivative of f(x) isgivenby(f (n) (x) =cos x + nπ ).2Therefore the remainder after expanding f(x) asan(n − 1)th-order polynomial aboutx = π/3 isgivenbyR n (x) =(x − π/3)nn!(cos ξ + nπ ),2where ξ lies in the range [π/3,x]. The modulus of the cosine term is always less than orequal to unity, and so |R n (x)| < |(x−π/3) n |/n!. As in the previous example, lim n→∞ R n (x) =0 for any particular value of x, andsocosx can be represented by an infinite Taylor seriesabout x = π/3.Evaluating the function and its derivatives at x = π/3 weobtainf(π/3) = cos(π/3) = 1/2,f ′ (π/3) = cos(5π/6) = − √ 3/2,f ′′ (π/3) = cos(4π/3) = −1/2,and so on. Thus the Taylor series expansion of cos x about x = π/3 isgivenbycos x = 1 2 − √32(x − π/3)−12( ) 2x − π/3+ ··· . ◭2!4.6.2 Approximation errors in Taylor seriesIn the previous subsection we saw how to represent a function f(x) by an infinitepower series, which is exactly equal to f(x) for all x within the interval ofconvergence of the series. However, in physical problems we usually do not wantto have to sum an infinite number of terms, but prefer to use only a finite numberof terms in the Taylor series to approximate the function in some given rangeof x. In this case it is desirable to know what is the maximum possible errorassociated with the approximation.As given in (4.18), a function f(x) can be represented by a finite (n − 1)th-orderpower series together with a remainder term such thatf(x) =f(a)+(x − a)f ′ (a)+(x − a)2f ′′ (x − a)n−1(a)+···+2!(n − 1)! f(n−1) (a)+R n (x),where(x − a)nR n (x) = f (n) (ξ)n!and ξ lies in the range [a, x]. R n (x) is the remainder term, and represents the errorin approximating f(x) bytheabove(n − 1)th-order power series. Since the exact139

SERIES AND LIMITSvalue of ξ that satisfies the expression for R n (x) is not known, an upper limit onthe error may be found by differentiating R n (x) withrespecttoξ and equatingthe derivative to zero in the usual way for finding maxima.◮Expand f(x) =cosx as a Taylor series about x =0and find the error associated withusing the approximation to evaluate cos(0.5) if only the first two non-vanishing terms aretaken. (Note that the Taylor expansions of trigonometric functions are only valid for anglesmeasured in radians.)Evaluating the function and its derivatives at x = 0, we findf(0) = cos 0 = 1,f ′ (0) = − sin 0 = 0,f ′′ (0) = − cos 0 = −1,f ′′′ (0) = sin 0 = 0.So, for small |x|, we find from (4.18)cos x ≈ 1 − x22 .Note that since cos x is an even function, its power series expansion contains only evenpowers of x. Therefore, in order to estimate the error in this approximation, we mustconsider the term in x 4 , which is the next in the series. The required derivative is f (4) (x)and this is (by chance) equal to cos x. Thus, adding in the remainder term R 4 (x), we findcos x =1− x22 + x4cos ξ,4!where ξ lies in the range [0,x]. Thus, the maximum possible error is x 4 /4!, since cos ξcannot exceed unity. If x =0.5, taking just the first two terms yields cos(0.5) ≈ 0.875 witha predicted error of less than 0.00260. In fact cos(0.5) = 0.87758 to 5 decimal places. Thus,to this accuracy, the true error is 0.00258, an error of about 0.3%. ◭4.6.3 Standard Maclaurin seriesIt is often useful to have a readily available table of Maclaurin series for standardelementary functions, and therefore these are listed below.sin x = x − x33! + x55! − x7+ ···7!for −∞

4.7 EVALUATION OF LIMITSThese can all be derived by straightforward application of Taylor’s theorem tothe expansion of a function about x =0.4.7 Evaluation of limitsThe idea of the limit of a function f(x) asx approaches a value a is fairly intuitive,though a strict definition exists and is stated below. In many cases the limit ofthe function as x approaches a will be simply the value f(a), but sometimes thisis not so. Firstly, the function may be undefined at x = a, as, for example, whenf(x) = sin xx ,which takes the value 0/0 atx = 0. However, the limit as x approaches zerodoes exist and can be evaluated as unity using l’Hôpital’s rule below. Anotherpossibility is that even if f(x) is defined at x = a its value may not be equal to thelimiting value lim x→a f(x). This can occur for a discontinuous function at a pointof discontinuity. The strict definition of a limit is that if lim x→a f(x) =l thenfor any number ɛ however small, it must be possible to find a number η such that|f(x)−l|

SERIES AND LIMITS◮Evaluate the limitslimx→1 (x2 +2x 3 ),lim(x cos x),x→0sin xlimx→π/2 x .Using (a) above,Using (b),Using (c),limx→1 (x2 +2x 3 ) = lim x 2 + lim 2x 3 =3.x→1 x→1lim(x cos x) = lim x lim cos x =0× 1=0.x→0 x→0 x→0sin xlimx→π/2 x= lim x→π/2 sin xlim x→π/2 x = 1π/2 = 2 π . ◭(iv) Limits of functions of x that contain exponents that themselves depend onx can often be found by taking logarithms.◮Evaluate the limit) x 2lim(1 − a2.x→∞ x 2Let us definelim ln y = limx→∞ x→∞) x 2y =(1 − a2x 2and consider the logarithm of the required limit, i.e.[x 2 ln(1 − a2x 2 )].Using the Maclaurin series for ln(1 + x) given in subsection 4.6.3, we can expand thelogarithm as a series and obtainlim ln y = lim[x(− ···)]2 a2x→∞ x→∞ x − a42 2x + = −a 2 .4Therefore, since lim x→∞ ln y = −a 2 it follows that lim x→∞ y =exp(−a 2 ). ◭(v) L’Hôpital’s rule may be used; it is an extension of (iii)(c) above. In caseswhere both numerator and denominator are zero or both are infinite, furtherconsideration of the limit must follow. Let us first consider lim x→a f(x)/g(x),where f(a) =g(a) = 0. Expanding the numerator and denominator as Taylorseries we obtainf(x)g(x) = f(a)+(x − a)f′ (a)+[(x − a) 2 /2!]f ′′ g(a)+(x − a)g ′ (a)+[(x − a) 2 /2!]g ′′ (a)+···.However, f(a) =g(a) =0sof(x)g(x) = f′ (a)+[(x − a)/2!]f ′′ g ′ (a)+[(x − a)/2!]g ′′ (a)+···.142

4.7 EVALUATION OF LIMITSTherefore we findf(x)limx→a g(x) = f′ (a)g ′ (a) ,provided f ′ (a) and g ′ (a) are not themselves both equal to zero. If, however,f ′ (a) andg ′ (a) are both zero then the same process can be applied to the ratiof ′ (x)/g ′ (x) to yieldf(x)limx→a g(x) = f′′ (a)g ′′ (a) ,provided that at least one of f ′′ (a) andg ′′ (a) is non-zero. If the original limit doesexist then it can be found by repeating the process as many times as is necessaryfor the ratio of corresponding nth derivatives not to be of the indeterminate form0/0, i.e.f(x)limx→a g(x) = f(n) (a)g (n) (a) .◮Evaluate the limitsin xlimx→0 x .We first note that if x = 0, both numerator and denominator are zero. Thus we applyl’Hôpital’s rule: differentiating, we obtainlim(sin x/x) = lim(cos x/1) = 1. ◭x→0 x→0So far we have only considered the case where f(a) =g(a) =0.Forthecasewhere f(a) =g(a) =∞ we may still apply l’Hôpital’s rule by writingf(x)limx→a g(x) = lim 1/g(x)x→a 1/f(x) ,whichisnowoftheform0/0 atx = a. Note also that l’Hôpital’s rule is stillvalid for finding limits as x →∞,i.e.whena = ∞. This is easily shown by lettingy =1/x as follows:f(x)limx→∞ g(x) = lim f(1/y)y→0 g(1/y)−f ′ (1/y)/y 2= limy→0 −g ′ (1/y)/y 2f ′ (1/y)= limy→0 g ′ (1/y)f ′ (x)= limx→∞ g ′ (x) .143

SERIES AND LIMITSSummary of methods for evaluating limitsTo find the limit of a continuous function f(x) at a point x = a, simply substitute0the value a into the function noting that∞ = 0 and that ∞ 0= ∞. Theonlydifficulty occurs when either of the expressions 0 0 or ∞ ∞results. In this casedifferentiate top and bottom and try again. Continue differentiating until the topand bottom limits are no longer both zero or both infinity. If the undeterminedform 0 ×∞occurs then it can always be rewritten as 0 0 or ∞ ∞ .4.8 Exercises4.1 Sum the even numbers between 1000 and 2000 inclusive.4.2 If you invest £1000 on the first day of each year, and interest is paid at 5% onyour balance at the end of each year, how much money do you have after 25years?4.3 How does the convergence of the series∞∑ (n − r)!n!depend on the integer r?4.4 Show that for testing the convergence of the seriesn=rx + y + x 2 + y 2 + x 3 + y 3 + ··· ,where 0

4.8 EXERCISES4.8 The N + 1 complex numbers ω m are given by ω m =exp(2πim/N), for m =0, 1, 2,... ,N.(a) Evaluate the following:N∑N∑N∑(i) ω m , (ii) ωm,2 (iii) ω m x m .m=0m=0(b) Use these results to evaluate:N∑[ ( 2πm(i) cosNn=1)− cosm=0( )] 4πm, (ii)Nm=03∑( ) 2πm2 m sin .34.9 Prove thatcos θ +cos(θ + α)+···+cos(θ + nα) = sin 1 (n +1)α2sin 1 α cos(θ + 1 nα). 224.10 Determine whether the following series converge (θ and p are positive realnumbers):∞∑∞2sinnθ(a)n(n +1) , (b) ∑ 2n , (c) ∑ ∞12 2n , 1/2∞∑ (−1) n (n 2 +1) 1/2∞∑ n p(d), (e)n ln nn! .n=2n=14.11 Find the real values of x for which the following series are convergent:∞∑ x n∞(a)n +1 , (b) ∑∞∑(sin x) n , (c) n x ,n=1(d)∞∑e nx ,n=1n=1n=1(e)∞∑(ln n) x .4.12 Determine whether the following series are convergent:∞∑ n 1/2(a)(n +1) , (b) ∑ ∞n 21/2 n! , (c) ∑ ∞(ln n) n∞∑ n n, (d)n n/2 n! .n=1n=1n=1n=14.13 Determine whether the following series are absolutely convergent, convergent oroscillatory:∞∑ (−1) n(a)n , (b) ∑ ∞(−1) n (2n +1)∞∑ (−1) n |x| n, (c),5/2 nn!n=1n=1n=0∞∑ (−1) n(d)n 2 +3n +2 , (e) ∑ ∞(−1) n 2 n.n 1/2 n=0n=14.14 Obtain the positive values of x for which the following series converges:∞∑ x n/2 e −n.nn=1n=2m=0n=1n=1145

SERIES AND LIMITS4.15 Prove that∞∑n=2[ ]n r +(−1) nlnn ris absolutely convergent for r = 2, but only conditionally convergent for r =1.4.16 An extension to the proof of the integral test (subsection 4.3.2) shows that, if f(x)is positive, continuous and monotonically decreasing, for x ≥ 1, and the seriesf(1) + f(2) + ··· is convergent, then its sum does not exceed f(1) + L, whereLis the integral∫ ∞1f(x) dx.Use this result to show that the sum ζ(p) of the Riemann zeta series ∑ n −p ,withp>1, is not greater than p/(p − 1).4.17 Demonstrate that rearranging the order of its terms can make a conditionally∑ convergent series converge to a different limit by considering the series(−1) n+1 n −1 =ln2=0.693. Rearrange the series asS = 1 + 1 − 1 + 1 + 1 − 1 + 1 + 1 − 1 + 1 + ···1 3 2 5 7 4 9 11 6 13and group each set of three successive terms. Show that the series can then bewritten∞∑m=18m − 32m(4m − 3)(4m − 1) ,which is convergent (by comparison with ∑ n −2 ) and contains only positiveterms. Evaluate the first of these and hence deduce that S is not equal to ln 2.4.18 Illustrate result (iv) of section 4.4, concerning Cauchy products, by consideringthedoublesummationS =∞∑n∑n=1 r=11r 2 (n +1− r) 3 .By examining the points in the nr-plane over which the double summation is tobe carried out, show that S can be written asS =∞∑∞∑n=r r=11r 2 (n +1− r) 3 .Deduce that S ≤ 3.4.19 A Fabry–Pérot interferometer consists of two parallel heavily silvered glass plates;light enters normally to the plates, and undergoes repeated reflections betweenthem, with a small transmitted fraction emerging at each reflection. Find theintensity of the emerging wave, |B| 2 ,wherewith r and φ real.B = A(1 − r)146∞∑r n e inφ ,n=0

4.8 EXERCISES4.20 Identify the series∞∑ (−1) n+1 x 2n(2n − 1)! ,n=1and then, by integration and differentiation, deduce the values S of the followingseries:∞∑∞∑(a)(c)n=1∞∑n=1(−1) n+1 n 2, (b)(2n)!n=1(−1) n+1 nπ 2n∞4 n (2n − 1)! , (d) ∑n=0(−1) n+1 n(2n +1)! ,4.21 Starting from the Maclaurin series for cos x, show that(−1) n (n +1).(2n)!(cos x) −2 =1+x 2 + 2x43 + ··· .Deduce the first three terms in the Maclaurin series for tan x.4.22 Find the Maclaurin series for:( 1+x(a) ln1 − x4.23 Writing the nth derivative of f(x) =sinh −1 x as), (b) (x 2 +4) −1 , (c) sin 2 x.f (n) P n (x)(x) =(1 + x 2 ) , n−1/2where P n (x) is a polynomial (of order n − 1), show that the P n (x) satisfytherecurrence relationP n+1 (x) =(1+x 2 )P n(x) ′ − (2n − 1)xP n (x).Hence generate the coefficients necessary to express sinh −1 x as a Maclaurin seriesup to terms in x 5 .4.24 Find the first three non-zero terms in the Maclaurin series for the followingfunctions:(a) (x 2 +9) −1/2 , (b) ln[(2 + x) 3 ], (c) exp(sin x),(d) ln(cos x), (e) exp[−(x − a) −2 ], (f) tan −1 x.4.25 By using the logarithmic series, prove that if a and b are positive and nearlyequal thenln a 2(a − b)≃b a + b .Show that the error in this approximation is about 2(a − b) 3 /[3(a + b) 3 ].4.26 Determine whether the following functions f(x) are (i) continuous, and (ii)differentiable at x =0:(a) f(x) =exp(−|x|);(b) f(x) =(1− cos x)/x 2 for x ≠0,f(0) = 1 ; 2(c) f(x) =x sin(1/x) forx ≠0,f(0) = 0;(d) f(x) =[4− x 2 ], where [y] denotes the integer part of y.4.27 Find the limit as x → 0of[ √ 1+x m − √ 1 − x m ]/x n ,inwhichm and n are positiveintegers.4.28 Evaluate the following limits:147

SERIES AND LIMITSsin 3x(a) limx→0 sinh x ,tan x − x(c) limx→0 cos x − 1 ,tan x − tanh x(b) limx→0 sinh x − x ,( cosec x(d) lim − sinh x ).x→0 x 3 x 54.29 Find the limits of the following functions:x 3 + x 2 − 5x − 2(a), as x → 0, x →∞and x → 2;2x 3 − 7x 2 +4x +4sin x − x cosh x(b), as x → 0;sinh x − x∫ π/2( )y cos y − sin y(c)dy, as x → 0.x y 24.30 Use Taylor expansions to three terms to find approximations to (a) 4√ 17, and(b) 3√ 26.4.31 Using a first-order Taylor expansion about x = x 0 , show that a better approximationthan x 0 to the solution of the equationf(x) =sinx +tanx =2is given by x = x 0 + δ, where2 − f(x 0 )δ =.cos x 0 +sec 2 x 0(a) Use this procedure twice to find the solution of f(x) = 2 to six significantfigures, given that it is close to x =0.9.(b) Use the result in (a) to deduce, to the same degree of accuracy, one solutionof the quartic equationy 4 − 4y 3 +4y 2 +4y − 4=0.4.32 Evaluate[ ( 1lim cosec x − 1x→0 x 3 x − x )].64.33 In quantum theory, a system of oscillators, each of fundamental frequency ν andinteracting at temperature T , has an average energy Ē given by∑ ∞n=0Ē = ∑ nhνe−nx∞ ,n=0 e−nxwhere x = hν/kT , h and k being the Planck and Boltzmann constants, respectively.Prove that both series converge, evaluate their sums, and show that at hightemperatures Ē ≈ kT, whilst at low temperatures Ē ≈ hν exp(−hν/kT ).4.34 In a very simple model of a crystal, point-like atomic ions are regularly spacedalong an infinite one-dimensional row with spacing R. Alternate ions carry equaland opposite charges ±e. The potential energy of the ith ion in the electric fielddue to another ion, the jth, isq i q j,4πɛ 0 r ijwhere q i , q j are the charges on the ions and r ij is the distance between them.Write down a series giving the total contribution V i of the ithiontotheoverallpotential energy. Show that the series converges, and, if V i is written asV i =148αe24πɛ 0 R ,

4.9 HINTS AND ANSWERSfind a closed-form expression for α, the Madelung constant for this (unrealistic)lattice.4.35 One of the factors contributing to the high relative permittivity of water to staticelectric fields is the permanent electric dipole moment, p, of the water molecule.In an external field E the dipoles tend to line up with the field, but they do notdo so completely because of thermal agitation corresponding to the temperature,T , of the water. A classical (non-quantum) calculation using the Boltzmanndistribution shows that the average polarisability per molecule, α, isgivenbyα = p E (coth x − x−1 ),where x = pE/(kT) andk is the Boltzmann constant.At ordinary temperatures, even with high field strengths (10 4 Vm −1 or more),x ≪ 1. By making suitable series expansions of the hyperbolic functions involved,show that α = p 2 /(3kT) to an accuracy of about one part in 15x −2 .4.36 In quantum theory, a certain method (the Born approximation) gives the (socalled)amplitude f(θ) for the scattering of a particle of mass m through an angleθ by a uniform potential well of depth V 0 and radius b (i.e. the potential energyof the particle is −V 0 within a sphere of radius b and zero elsewhere) asf(θ) = 2mV 0(sin Kb − Kbcos Kb). 2 K3 Here is the Planck constant divided by 2π, the energy of the particle is 2 k 2 /(2m)and K is 2k sin(θ/2).Use l’Hôpital’s rule to evaluate the amplitude at low energies, i.e. when k andhence K tend to zero, and so determine the low-energy total cross-section.[ Note: the differential cross-section is given by |f(θ)| 2 and the total crosssectionby the integral of this over all solid angles, i.e. 2π ∫ π0 |f(θ)|2 sin θdθ.]4.9 Hints and answers4.1 Write as 2( ∑ 1000n=1 n − ∑ 499n=1n) = 751500.4.3 Divergent for r ≤ 1; convergent for r ≥ 2.4.5 (a) ln(N + 1), divergent; (b) 1 [1 − 3 (−2)n ], oscillates infinitely; (c) Add 1 S 3 N to theS N series; 3 [1 − 16 (−3)−N ]+ 3 4 N(−3)−N−1 ,convergentto 3 . 164.7 Write the nth term as the difference between two consecutive values of a partialfractionfunction of n. Thesumequals 1 (1 − 2 N−2 ).4.9 Sum the geometric series with rth term exp[i(θ + rα)]. Its real part is{cos θ − cos [(n +1)α + θ] − cos(θ − α)+cos(θ + nα)} /4sin 2 (α/2),which can be reduced to the given answer.4.11 (a) −1 ≤ x

SERIES AND LIMITS4.15 Divide the series into two series, n odd and n even. For r = 2 both are absolutelyconvergent, by comparison with ∑ n −2 .Forr = 1 neither series is convergent,by comparison with ∑ n −1 . However, the sum of the two is convergent, by thealternating sign test or by showing that the terms cancel in pairs.4.17 The first term has value 0.833 and all other terms are positive.4.19 |A| 2 (1 − r) 2 /(1 + r 2 − 2r cos φ).4.21 Use the binomial expansion and collect terms up to x 4 . Integrate both sides ofthe displayed equation. tan x = x + x 3 /3+2x 5 /15 + ···.4.23 For example, P 5 (x) =24x 4 − 72x 2 +9.sinh −1 x = x − x 3 /6+3x 5 /40 − ···.4.25 Set a = D + δ and b = D − δ and use the expansion for ln(1 ± δ/D).4.27 The limit is 0 for m>n,1form = n, and∞ for m

5Partial differentiationIn chapter 2, we discussed functions f of only one variable x, which were usuallywritten f(x). Certain constants and parameters may also have appeared in thedefinition of f, e.g.f(x) =ax+2 contains the constant 2 and the parameter a, butonly x was considered as a variable and only the derivatives f (n) (x) =d n f/dx nwere defined.However, we may equally well consider functions that depend on more than onevariable, e.g. the function f(x, y) =x 2 +3xy, which depends on the two variablesx and y. For any pair of values x, y, the function f(x, y) has a well-defined value,e.g. f(2, 3) = 22. This notion can clearly be extended to functions dependent onmore than two variables. For the n-variable case, we write f(x 1 ,x 2 ,...,x n )fora function that depends on the variables x 1 ,x 2 ,...,x n .Whenn =2,x 1 and x 2correspond to the variables x and y used above.Functions of one variable, like f(x), can be represented by a graph on aplane sheet of paper, and it is apparent that functions of two variables can,with little effort, be represented by a surface in three-dimensional space. Thus,we may also picture f(x, y) as describing the variation of height with positionin a mountainous landscape. Functions of many variables, however, are usuallyvery difficult to visualise and so the preliminary discussion in this chapter willconcentrate on functions of just two variables.5.1 Definition of the partial derivativeIt is clear that a function f(x, y) of two variables will have a gradient in alldirections in the xy-plane. A general expression for this rate of change can befound and will be discussed in the next section. However, we first consider thesimpler case of finding the rate of change of f(x, y) in the positive x- andydirections.These rates of change are called the partial derivatives with respect151

PARTIAL DIFFERENTIATIONto x and y respectively, and they are extremely important in a wide range ofphysical applications.For a function of two variables f(x, y) we may define the derivative with respectto x, for example, by saying that it is that for a one-variable function when y isheld fixed and treated as a constant. To signify that a derivative is with respectto x, but at the same time to recognize that a derivative with respect to y alsoexists, the former is denoted by ∂f/∂x and is the partial derivative of f(x, y) withrespect to x. Similarly, the partial derivative of f with respect to y is denoted by∂f/∂y.To define formally the partial derivative of f(x, y) with respect to x, we have∂f∂x = lim∆x→0f(x +∆x, y) − f(x, y), (5.1)∆xprovided that the limit exists. This is much the same as for the derivative of aone-variable function. The other partial derivative of f(x, y) is similarly definedas a limit (provided it exists):∂f∂y = lim∆y→0f(x, y +∆y) − f(x, y). (5.2)∆yIt is common practice in connection with partial derivatives of functionsinvolving more than one variable to indicate those variables that are held constantby writing them as subscripts to the derivative symbol. Thus, the partial derivativesdefined in (5.1) and (5.2) would be written respectively as( ∂f∂x)yand( ∂f∂yIn this form, the subscript shows explicitly which variable is to be kept constant.A more compact notation for these partial derivatives is f x and f y . However, it isextremely important when using partial derivatives to remember which variablesare being held constant and it is wise to write out the partial derivative in explicitform if there is any possibility of confusion.The extension of the definitions (5.1), (5.2) to the general n-variable case isstraightforward and can be written formally as∂f(x 1 ,x 2 ,...,x n )∂x i[f(x 1 ,x 2 ,...,x i +∆x i ,...,x n ) − f(x 1 ,x 2 ,...,x i ,...,x n )]= lim,∆x i→0∆x iprovided that the limit exists.Just as for one-variable functions, second (and higher) partial derivatives maybe defined in a similar way. For a two-variable function f(x, y) theyare(∂ ∂f∂x ∂x(∂ ∂f∂x ∂y)= ∂2 f∂x 2 = f xx,)= ∂2 f∂x∂y = f xy,152∂∂y∂∂y( ∂f∂y( ∂f∂x).x)= ∂2 f∂y 2 = f yy,)= ∂2 f∂y∂x = f yx.

5.2 THE TOTAL DIFFERENTIAL AND TOTAL DERIVATIVEOnly three of the second derivatives are independent since the relation∂ 2 f∂x∂y = ∂2 f∂y∂x ,is always obeyed, provided that the second partial derivatives are continuousat the point in question. This relation often proves useful as a labour-savingdevice when evaluating second partial derivatives. It can also be shown that fora function of n variables, f(x 1 ,x 2 ,...,x n ), under the same conditions,∂ 2 f=∂2 f.∂x i ∂x j ∂x j ∂x i◮Find the first and second partial derivatives of the functionf(x, y) =2x 3 y 2 + y 3 .The first partial derivatives are∂f∂x =6x2 y 2 ,and the second partial derivatives are∂ 2 f∂x 2 =12xy2 ,∂ 2 f∂y 2 =4x3 +6y,the last two being equal, as expected. ◭∂f∂y =4x3 y +3y 2 ,∂ 2 f∂x∂y =12x2 y,∂ 2 f∂y∂x =12x2 y,5.2 The total differential and total derivativeHaving defined the (first) partial derivatives of a function f(x, y), which give therate of change of f along the positive x- andy-axes, we consider next the rate ofchange of f(x, y) in an arbitrary direction. Suppose that we make simultaneoussmall changes ∆x in x and ∆y in y and that, as a result, f changes to f +∆f.Then we must have∆f = f(x +∆x, y +∆y) − f(x, y)= f(x +∆x, y +∆y) − f(x, y +∆y)+f(x, y +∆y) − f(x, y)[ ] [ ]f(x +∆x, y +∆y) − f(x, y +∆y) f(x, y +∆y) − f(x, y)=∆x +∆y.∆x∆y(5.3)In the last line we note that the quantities in brackets are very similar to thoseinvolved in the definitions of partial derivatives (5.1), (5.2). For them to be strictlyequal to the partial derivatives, ∆x and ∆y would need to be infinitesimally small.But even for finite (but not too large) ∆x and ∆y the approximate formula∆f ≈∂f(x, y)∆x +∂x153∂f(x, y)∆y, (5.4)∂y

PARTIAL DIFFERENTIATIONcan be obtained. It will be noticed that the first bracket in (5.3) actually approximatesto ∂f(x, y +∆y)/∂x but that this has been replaced by ∂f(x, y)/∂x in (5.4).This approximation clearly has the same degree of validity as that which replacesthe bracket by the partial derivative.How valid an approximation (5.4) is to (5.3) depends not only on how small∆x and ∆y are but also on the magnitudes of higher partial derivatives; this isdiscussed further in section 5.7 in the context of Taylor series for functions ofmore than one variable. Nevertheless, letting the small changes ∆x and ∆y in(5.4) become infinitesimal, we can define the total differential df of the functionf(x, y), without any approximation, asdf = ∂f ∂fdx + dy. (5.5)∂x ∂yEquation (5.5) can be extended to the case of a function of n variables,f(x 1 ,x 2 ,...,x n );df = ∂f dx 1 + ∂f dx 2 + ···+ ∂f dx n . (5.6)∂x 1 ∂x 2 ∂x n◮Find the total differential of the function f(x, y) =y exp(x + y).Evaluating the first partial derivatives, we find∂f∂f= y exp(x + y),∂x=exp(x + y)+y exp(x + y).∂yApplying (5.5), we then find that the total differential is given bydf =[y exp(x + y)]dx +[(1+y)exp(x + y)]dy. ◭In some situations, despite the fact that several variables x i , i =1, 2,...,n,appear to be involved, effectively only one of them is. This occurs if there aresubsidiary relationships constraining all the x i to have values dependent on thevalue of one of them, say x 1 . These relationships may be represented by equationsthat are typically of the formx i = x i (x 1 ), i =2, 3,...,n. (5.7)In principle f can then be expressed as a function of x 1 alone by substitutingfrom (5.7) for x 2 ,x 3 ,...,x n , and then the total derivative (or simply the derivative)of f with respect to x 1 is obtained by ordinary differentiation.Alternatively, (5.6) can be used to givedfdx 1= ∂f∂x 1+( ∂f∂x 2) dx2dx 1+ ···+( ∂f∂x n) dxndx 1. (5.8)It should be noted that the LHS of this equation is the total derivative df/dx 1 ,whilst the partial derivative ∂f/∂x 1 forms only a part of the RHS. In evaluating154

5.3 EXACT AND INEXACT DIFFERENTIALSthis partial derivative account must be taken only of explicit appearances of x 1 inthe function f, andno allowance must be made for the knowledge that changingx 1 necessarily changes x 2 ,x 3 ,...,x n . The contribution from these latter changes isprecisely that of the remaining terms on the RHS of (5.8). Naturally, what hasbeen shown using x 1 in the above argument applies equally well to any other ofthe x i , with the appropriate consequent changes.◮Find the total derivative of f(x, y) =x 2 +3xy with respect to x, given that y =sin −1 x.We can see immediately that∂f∂f=2x +3y,∂x∂y =3x,dydx = 1(1 − x 2 ) 1/2and so, using (5.8) with x 1 = x and x 2 = y,dfdx =2x +3y +3x 1(1 − x 2 ) 1/2=2x +3sin −1 3xx +(1 − x 2 ) . 1/2Obviously the same expression would have resulted if we had substituted for y from thestart, but the above method often produces results with reduced calculation, particularlyin more complicated examples. ◭5.3 Exact and inexact differentialsIn the last section we discussed how to find the total differential of a function, i.e.its infinitesimal change in an arbitrary direction, in terms of its gradients ∂f/∂xand ∂f/∂y in the x- andy- directions (see (5.5)). Sometimes, however, we wishto reverse the process and find the function f that differentiates to give a knowndifferential. Usually, finding such functions relies on inspection and experience.As an example, it is easy to see that the function whose differential is df =xdy+ ydx is simply f(x, y) =xy + c, wherec is a constant. Differentials such asthis, which integrate directly, are called exact differentials, whereas those that donot are inexact differentials. For example, xdy+3ydx is not the straightforwarddifferential of any function (see below). Inexact differentials can be made exact,however, by multiplying through by a suitable function called an integratingfactor. This is discussed further in subsection 14.2.3.◮Show that the differential xdy+3ydx is inexact.On the one hand, if we integrate with respect to x we conclude that f(x, y) =3xy + g(y),where g(y) is any function of y. On the other hand, if we integrate with respect to y weconclude that f(x, y) =xy + h(x) whereh(x) is any function of x. These conclusions areinconsistent for any and every choice of g(y) andh(x), and therefore the differential isinexact. ◭It is naturally of interest to investigate which properties of a differential make155

PARTIAL DIFFERENTIATIONit exact. Consider the general differential containing two variables,df = A(x, y) dx + B(x, y) dy.We see that∂f∂f= A(x, y),∂x= B(x, y)∂yand, using the property f xy = f yx , we therefore require∂A∂y = ∂B∂x . (5.9)This is in fact both a necessary and a sufficient condition for the differential tobe exact.◮Using (5.9) show that xdy+3ydx is inexact.In the above notation, A(x, y) =3y and B(x, y) =x and so∂A∂y =3,∂B∂x =1.As these are not equal it follows that the differential is inexact. ◭Determining whether a differential containing many variable x 1 ,x 2 ,...,x n isexact is a simple extension of the above. A differential containing many variablescanbewritteningeneralasand will be exact ifdf =n∑g i (x 1 ,x 2 ,...,x n ) dx ii=1∂g i∂x j= ∂g j∂x ifor all pairs i, j. (5.10)There will be 1 2n(n − 1) such relationships to be satisfied.◮Show thatis an exact differential.(y + z) dx + xdy+ xdzIn this case, g 1 (x, y, z) =y + z, g 2 (x, y, z) =x, g 3 (x, y, z) =x and hence ∂g 1 /∂y =1=∂g 2 /∂x, ∂g 3 /∂x =1=∂g 1 /∂z, ∂g 2 /∂z =0=∂g 3 /∂y; therefore, from (5.10), the differentialis exact. As mentioned above, it is sometimes possible to show that a differential is exactsimply by finding by inspection the function from which it originates. In this example, itcan be seen easily that f(x, y, z) =x(y + z)+c. ◭156

5.4 USEFUL THEOREMS OF PARTIAL DIFFERENTIATION5.4 Useful theorems of partial differentiationSo far our discussion has centred on a function f(x, y) dependent on two variables,x and y. Equally, however, we could have expressed x as a function of f and y,or y as a function of f and x. To emphasise the point that all the variables areof equal standing, we now replace f by z. This does not imply that x, y and zare coordinate positions (though they might be). Since x is a function of y and z,it follows thatdx =( ) ∂xdy +∂yz( ) ∂xdz (5.11)∂zyand similarly, since y = y(x, z),( ) ( )∂y ∂ydy = dx + dz. (5.12)∂xz∂zxWe may now substitute (5.12) into (5.11) to obtain( ) ( ) [ (∂x ) ( ) ( ) ]∂x ∂y∂y ∂xdx =dx ++ dz. (5.13)∂yz∂xz∂yz∂zx∂zyNow if we hold z constant, so that dz = 0, we obtain the reciprocity relation( ) ( ) −1∂x ∂y= ,∂yz∂xzwhich holds provided both partial derivatives exist and neither is equal to zero.Note, further, that this relationship only holds when the variable being keptconstant, in this case z, is the same on both sides of the equation.Alternatively we can put dx = 0 in (5.13). Then the contents of the squarebrackets also equal zero, and we obtain the cyclic relation( ) ( ) ( )∂y ∂z ∂x= −1,∂zx∂xy∂yzwhich holds unless any of the derivatives vanish. In deriving this result we haveused the reciprocity relation to replace (∂x/∂z) −1y by (∂z/∂x) y .5.5 The chain ruleSo far we have discussed the differentiation of a function f(x, y) with respect toits variables x and y. We now consider the case where x and y are themselvesfunctions of another variable, say u. If we wish to find the derivative df/du,we could simply substitute in f(x, y) the expressions for x(u) andy(u) andthendifferentiate the resulting function of u. Such substitution will quickly give thedesired answer in simple cases, but in more complicated examples it is easier tomake use of the total differentials described in the previous section.157

PARTIAL DIFFERENTIATIONFrom equation (5.5) the total differential of f(x, y) is given bydf = ∂f ∂fdx +∂x ∂y dy,but we now note that by using the formal device of dividing through by du thisimmediately impliesdfdu = ∂f dx∂x du + ∂f dy∂y du , (5.14)which is called the chain rule for partial differentiation. This expression providesa direct method for calculating the total derivative of f with respect to u and isparticularly useful when an equation is expressed in a parametric form.◮Given that x(u) =1+au and y(u) =bu 3 , find the rate of change of f(x, y) =xe −y withrespect to u.As discussed above, this problem could be addressed by substituting for x and y to obtainf as a function only of u and then differentiating with respect to u. However, using (5.14)directly we obtaindfdu =(e−y )a +(−xe −y )3bu 2 ,which on substituting for x and y givesdfdu = e−bu3 (a − 3bu 2 − 3bau 3 ). ◭Equation (5.14) is an example of the chain rule for a function of two variableseach of which depends on a single variable. The chain rule may be extended tofunctions of many variables, each of which is itself a function of a variable u, i.e.f(x 1 ,x 2 ,x 3 ,...,x n ), with x i = x i (u). In this case the chain rule givesdfdu =n∑i=1∂f dx i∂x i du = ∂f dx 1∂x 1 du + ∂f dx 2 ∂f dx n+ ···+∂x 2 du ∂x n du . (5.15)5.6 Change of variablesIt is sometimes necessary or desirable to make a change of variables during thecourse of an analysis, and consequently to have to change an equation expressedin one set of variables into an equation using another set. The same situation arisesif a function f depends on one set of variables x i ,sothatf = f(x 1 ,x 2 ,...,x n ) butthe x i are themselves functions of a further set of variables u j and given by theequationsx i = x i (u 1 ,u 2 ,...,u m ). (5.16)158

5.6 CHANGE OF VARIABLESyρφxFigure 5.1The relationship between Cartesian and plane polar coordinates.For each different value of i, x i will be a different function of the u j .Inthiscasethe chain rule (5.15) becomes∂f∂u j=n∑i=1∂f∂x i∂x i∂u j, j =1, 2,...,m, (5.17)and is said to express a change of variables. In general the number of variablesin each set need not be equal, i.e. m need not equal n, but if both the x i and theu i are sets of independent variables then m = n.◮Plane polar coordinates, ρ and φ, and Cartesian coordinates, x and y, arerelatedbytheexpressionsx = ρ cos φ, y = ρ sin φ,as can be seen from figure 5.1. An arbitrary function f(x, y) canbere-expressedasafunction g(ρ, φ). Transform the expression∂ 2 f∂x + ∂2 f2 ∂y 2into one in ρ and φ.We first note that ρ 2 = x 2 + y 2 , φ =tan −1 (y/x). We can now write down the four partialderivatives∂ρ∂x = x(x 2 + y 2 ) =cosφ, ∂φ1/2 ∂x = −(y/x2 )1+(y/x) = − sin φ2 ρ ,∂ρ∂y = y(x 2 + y 2 ) =sinφ, ∂φ1/2 ∂y = 1/x1+(y/x) = cos φ2 ρ .159

PARTIAL DIFFERENTIATIONThus, from (5.17), we may write∂∂x =cosφ ∂∂ρ − sin φ ∂ρ ∂φ ,Now it is only a matter of writing∂ 2 f∂x = ∂ ( ) ∂f= ∂ ( ∂2 ∂x ∂x ∂x ∂x(= cos φ ∂∂ρ − sin φ ∂ρ ∂φ(= cos φ ∂ ∂∂ρ − sin φρ ∂φ)f∂∂y =sinφ ∂∂ρ + cos φρ)(cos φ ∂)(cos φ ∂g=cos 2 φ ∂2 g 2cosφ sin φ ∂g+∂ρ2 ρ 2 ∂φ+ sin2 φ ∂gρ ∂ρ + sin2 φ ∂ 2 gρ 2 ∂φ 2∂ρ − sin φ ∂ρ ∂φ∂ρ − sin φ )∂gρ ∂φ−2cosφ sin φρand a similar expression for ∂ 2 f/∂y 2 ,(∂ 2 f∂y = sin φ ∂2 ∂ρ + cos φ )(∂sin φ ∂ρ ∂φ ∂ρ + cos φρ=sin 2 φ ∂2 g 2cosφ sin φ ∂g−∂ρ2 ρ 2 ∂φ+ cos2 φ ∂gρ ∂ρ + cos2 φ ∂ 2 gρ 2 ∂φ . 2+2cosφ sin φρ∂∂φ .)g∂ 2 g∂φ∂ρ∂∂φ)g∂ 2 g∂φ∂ρWhen these two expressions are added together the change of variables is complete andwe obtain∂ 2 f∂x + ∂2 f2 ∂y = ∂2 g2 ∂ρ + 1 ∂g2 ρ ∂ρ + 1 ∂ 2 gρ 2 ∂φ . ◭ 25.7 Taylor’s theorem for many-variable functionsWe have already introduced Taylor’s theorem for a function f(x) of one variable,in section 4.6. In an analogous way, the Taylor expansion of a function f(x, y) oftwo variables is given byf(x, y) =f(x 0 ,y 0 )+ ∂f ∂f∆x +∂x ∂y ∆y+ 1 [ ∂ 2 ]f2! ∂x 2 (∆x)2 +2 ∂2 f∂x∂y ∆x∆y + ∂2 f∂y 2 (∆y)2 + ··· , (5.18)where ∆x = x − x 0 and ∆y = y − y 0 , and all the derivatives are to be evaluatedat (x 0 ,y 0 ).160

5.7 TAYLOR’S THEOREM FOR MANY-VARIABLE FUNCTIONS◮Find the Taylor expansion, up to quadratic terms in x − 2 and y − 3, off(x, y) =y exp xyabout the point x =2, y =3.We first evaluate the required partial derivatives of the function, i.e.∂f∂x = y2 exp xy,∂f=expxy + xy exp xy,∂y∂ 2 f∂x 2 = y3 exp xy,∂ 2 f∂y 2 =2x exp xy + x2 y exp xy,∂ 2 f∂x∂y =2y exp xy + xy2 exp xy.Using (5.18), the Taylor expansion of a two-variable function, we findf(x, y) ≈ e 6 {3+9(x − 2) + 7(y − 3)+(2!) −1 [ 27(x − 2) 2 + 48(x − 2)(y − 3) + 16(y − 3) 2]} . ◭It will be noticed that the terms in (5.18) containing first derivatives can bewritten as(∂f ∂f∆x +∂x ∂y ∆y = ∆x ∂∂x +∆y ∂ )f(x, y),∂ywhere both sides of this relation should be evaluated at the point (x 0 ,y 0 ). Similarlythe terms in (5.18) containing second derivatives can be written as[1 ∂ 2 ]f2! ∂x 2 (∆x)2 +2 ∂2 f∂x∂y ∆x∆y + ∂2 f∂y 2 (∆y)2 = 1 (∆x ∂2! ∂x +∆y ∂ ) 2f(x, y),∂y(5.19)where it is understood that the partial derivatives resulting from squaring theexpression in parentheses act only on f(x, y) and its derivatives, and not on ∆xor ∆y; again both sides of (5.19) should be evaluated at (x 0 ,y 0 ). It can be shownthat the higher-order terms of the Taylor expansion of f(x, y) can be written inan analogous way, and that we may write the full Taylor series asf(x, y) =∞∑n=0[(1∆x ∂n! ∂x +∆y ∂ ) nf(x, y)]∂yx 0,y 0where, as indicated, all the terms on the RHS are to be evaluated at (x 0 ,y 0 ).The most general form of Taylor’s theorem, for a function f(x 1 ,x 2 ,...,x n )ofnvariables, is a simple extension of the above. Although it is not necessary to doso, we may think of the x i as coordinates in n-dimensional space and write thefunction as f(x), where x is a vector from the origin to (x 1 ,x 2 ,...,x n ). Taylor’s161

PARTIAL DIFFERENTIATIONtheorem then becomesf(x) =f(x 0 )+ ∑ i∂f∂x i∆x i + 1 2!∑ ∑ij∂ 2 f∂x i ∂x j∆x i ∆x j + ··· ,(5.20)where ∆x i = x i − x i0 and the partial derivatives are evaluated at (x 10 ,x 20 ,...,x n0 ).For completeness, we note that in this case the full Taylor series can be writtenin the formf(x) =∞∑n=01 [(∆x · ∇) n f(x) ] n!x=x 0,where ∇ is the vector differential operator del, to be discussed in chapter 10.5.8 Stationary values of many-variable functionsThe idea of the stationary points of a function of just one variable has alreadybeen discussed in subsection 2.1.8. We recall that the function f(x) has a stationarypoint at x = x 0 if its gradient df/dx is zero at that point. A function may haveany number of stationary points, and their nature, i.e. whether they are maxima,minima or stationary points of inflection, is determined by the value of the secondderivative at the point. A stationary point is(i) a minimum if d 2 f/dx 2 > 0;(ii) a maximum if d 2 f/dx 2 < 0;(iii) a stationary point of inflection if d 2 f/dx 2 = 0 and changes sign throughthe point.We now consider the stationary points of functions of more than one variable;we will see that partial differential analysis is ideally suited to the determinationof the position and nature of such points. It is helpful to consider first the caseof a function of just two variables but, even in this case, the general situationis more complex than that for a function of one variable, as can be seen fromfigure 5.2.This figure shows part of a three-dimensional model of a function f(x, y). Atpositions P and B there are a peak and a bowl respectively or, more mathematically,a local maximum and a local minimum. At position S the gradient in anydirection is zero but the situation is complicated, since a section parallel to theplane x = 0 would show a maximum, but one parallel to the plane y =0wouldshow a minimum. A point such as S is known as a saddle point. The orientationof the ‘saddle’ in the xy-plane is irrelevant; it is as shown in the figure solely forease of discussion. For any saddle point the function increases in some directionsaway from the point but decreases in other directions.162

5.8 STATIONARY VALUES OF MANY-VARIABLE FUNCTIONSPSyxBFigure 5.2 Stationary points of a function of two variables. A minimumoccurs at B, a maximum at P and a saddle point at S.For functions of two variables, such as the one shown, it should be clear that anecessary condition for a stationary point (maximum, minimum or saddle point)to occur is that∂f∂x = 0 and ∂f=0. (5.21)∂yThe vanishing of the partial derivatives in directions parallel to the axes is enoughto ensure that the partial derivative in any arbitrary direction is also zero. Thelatter can be considered as the superposition of two contributions, one alongeach axis; since both contributions are zero, so is the partial derivative in thearbitrary direction. This may be made more precise by considering the totaldifferentialdf = ∂f ∂fdx +∂x ∂y dy.Using (5.21) we see that although the infinitesimal changes dx and dy can bechosen independently the change in the value of the infinitesimal function df isalways zero at a stationary point.We now turn our attention to determining the nature of a stationary point ofa function of two variables, i.e. whether it is a maximum, a minimum or a saddlepoint. By analogy with the one-variable case we see that ∂ 2 f/∂x 2 and ∂ 2 f/∂y 2must both be positive for a minimum and both be negative for a maximum.However these are not sufficient conditions since they could also be obeyed atcomplicated saddle points. What is important for a minimum (or maximum) isthat the second partial derivative must be positive (or negative) in all directions,not just in the x- andy- directions.163

PARTIAL DIFFERENTIATIONTo establish just what constitutes sufficient conditions we first note that, sincef is a function of two variables and ∂f/∂x = ∂f/∂y = 0, a Taylor expansion ofthe type (5.18) about the stationary point yieldsf(x, y) − f(x 0 ,y 0 ) ≈ 1 2![(∆x) 2 f xx +2∆x∆yf xy +(∆y) 2 f yy],where ∆x = x − x 0 and ∆y = y − y 0 and where the partial derivatives have beenwritten in more compact notation. Rearranging the contents of the bracket asthe weighted sum of two squares, we findf(x, y) − f(x 0 ,y 0 ) ≈ 1 2( [f xx ∆x + f ) ( )]2xy∆y+(∆y) 2 f yy − f2 xy.f xx f xx(5.22)For a minimum, we require (5.22) to be positive for all ∆x and ∆y, and hencef xx > 0andf yy − (fxy/f 2 xx ) > 0. Given the first constraint, the second can bewritten f xx f yy >fxy. 2 Similarly for a maximum we require (5.22) to be negative,and hence f xx < 0andf xx f yy >fxy 2 . For minima and maxima, symmetry requiresthat f yy obeys the same criteria as f xx . When (5.22) is negative (or zero) for somevalues of ∆x and ∆y but positive (or zero) for others, we have a saddle point. Inthis case f xx f yy

5.8 STATIONARY VALUES OF MANY-VARIABLE FUNCTIONS◮Show that the function f(x, y) =x 3 exp(−x 2 −y 2 ) has a maximum at the point ( √ 3/2, 0),a minimum at (− √ 3/2, 0) and a stationary point at the origin whose nature cannot bedetermined by the above procedures.Setting the first two partial derivatives to zero to locate the stationary points, we find∂f∂x =(3x2 − 2x 4 )exp(−x 2 − y 2 )=0, (5.23)∂f∂y = −2yx3 exp(−x 2 − y 2 )=0. (5.24)For (5.24) to be satisfied we require x =0ory = 0 and for (5.23) to be satisfied we requirex =0orx = ± √ 3/2. Hence the stationary points are at (0, 0), ( √ 3/2, 0) and (− √ 3/2, 0).We now find the second partial derivatives:f xx =(4x 5 − 14x 3 +6x)exp(−x 2 − y 2 ),f yy = x 3 (4y 2 − 2) exp(−x 2 − y 2 ),f xy =2x 2 y(2x 2 − 3) exp(−x 2 − y 2 ).We then substitute the pairs of values of x and y for each stationary point and find thatat (0, 0)f xx =0, f yy =0, f xy =0and at (± √ 3/2, 0)f xx = ∓6 √ 3/2 exp(−3/2), f yy = ∓3 √ 3/2 exp(−3/2), f xy =0.Hence, applying criteria (i)–(iii) above, we find that (0, 0) is an undetermined stationarypoint, ( √ 3/2, 0) is a maximum and (− √ 3/2, 0) is a minimum. The function is shown infigure 5.3. ◭Determining the nature of stationary points for functions of a general numberof variables is considerably more difficult and requires a knowledge of theeigenvectors and eigenvalues of matrices. Although these are not discussed untilchapter 8, we present the analysis here for completeness. The remainder of thissection can therefore be omitted on a first reading.For a function of n real variables, f(x 1 ,x 2 ,...,x n ), we require that, at allstationary points,∂f= 0 for all x i .∂x iIn order to determine the nature of a stationary point, we must expand thefunction as a Taylor series about the point. Recalling the Taylor expansion (5.20)for a function of n variables, we see that∆f = f(x) − f(x 0 ) ≈ 1 ∑ ∑ ∂ 2 f∆x i ∆x j . (5.25)2 ∂x i ∂x j165ij

PARTIAL DIFFERENTIATIONmaximum0.40.20−0.2−0.4−3−2minimum−1 0x1 2 320y−2Figure 5.3 The function f(x, y) =x 3 exp(−x 2 − y 2 ).If we define the matrix M to have elements given bythen we can rewrite (5.25) asM ij =∂2 f∂x i ∂x j,∆f = 1 2 ∆xT M∆x, (5.26)where ∆x is the column vector with the ∆x i as its components and ∆x T is itstranspose. Since M is real and symmetric it has n real eigenvalues λ r and northogonal eigenvectors e r , which after suitable normalisation satisfyMe r = λ r e r , e T r e s = δ rs ,where the Kronecker delta, written δ rs , equals unity for r = s and equals zerootherwise. These eigenvectors form a basis set for the n-dimensional space andwe can therefore expand ∆x in terms of them, obtaining∆x = ∑ ra r e r ,166

5.9 STATIONARY VALUES UNDER CONSTRAINTSwhere the a r are coefficients dependent upon ∆x. Substituting this into (5.26), wefind∑∆f = 1 2 ∆xT M∆x = 1 2λ r a 2 r .Now, for the stationary point to be a minimum, we require ∆f = 1 ∑2 r λ ra 2 r > 0for all sets of values of the a r , and therefore all the eigenvalues of M to begreater than zero. Conversely, for a maximum we require ∆f = 1 ∑2 r λ ra 2 r < 0,and therefore all the eigenvalues of M to be less than zero. If the eigenvalues havemixed signs, then we have a saddle point. Note that the test may fail if some orall of the eigenvalues are equal to zero and all the non-zero ones have the samesign.◮Derive the conditions for maxima, minima and saddle points for a function of two realvariables, using the above analysis.For a two-variable function the matrix M is given by( )fxx fM =xy.f yx f yyTherefore its eigenvalues satisfy the equation∣ f xx − λ f xyf xy f yy − λ ∣ =0.Hence(f xx − λ)(f yy − λ) − fxy 2 =0⇒ f xx f yy − (f xx + f yy )λ + λ 2 − fxy 2 =0√⇒ 2λ =(f xx + f yy ) ± (f xx + f yy ) 2 − 4(f xx f yy − fxy),2which by rearrangement of the terms under the square root gives√2λ =(f xx + f yy ) ± (f xx − f yy ) 2 +4fxy.2Now, that M is real and symmetric implies that its eigenvalues are real, and so for botheigenvalues to be positive (corresponding to a minimum), we require f xx and f yy positiveand also√f xx + f yy > (f xx + f yy ) 2 − 4(f xx f yy − fxy),2⇒ f xx f yy − fxy 2 > 0.A similar procedure will find the criteria for maxima and saddle points. ◭r5.9 Stationary values under constraintsIn the previous section we looked at the problem of finding stationary values ofa function of two or more variables when all the variables may be independently167

PARTIAL DIFFERENTIATIONvaried. However, it is often the case in physical problems that not all the variablesused to describe a situation are in fact independent, i.e. some relationshipbetween the variables must be satisfied. For example, if we walk through a hillylandscape and we are constrained to walk along a path, we will never reachthe highest peak on the landscape unless the path happens to take us to it.Nevertheless, we can still find the highest point that we have reached during ourjourney.We first discuss the case of a function of just two variables. Let us considerfinding the maximum value of the differentiable function f(x, y) subject to theconstraint g(x, y) =c, wherec is a constant. In the above analogy, f(x, y) mightrepresent the height of the land above sea-level in some hilly region, whilstg(x, y) =c is the equation of the path along which we walk.We could, of course, use the constraint g(x, y) =c to substitute for x or y inf(x, y), thereby obtaining a new function of only one variable whose stationarypoints could be found using the methods discussed in subsection 2.1.8. However,such a procedure can involve a lot of algebra and becomes very tedious for functionsof more than two variables. A more direct method for solving such problemsis the method of Lagrange undetermined multipliers, which we now discuss.To maximise f we requiredf = ∂f ∂fdx + dy =0.∂x ∂yIf dx and dy were independent, we could conclude f x =0=f y . However, herethey are not independent, but constrained because g is constant:dg = ∂g ∂gdx + dy =0.∂x ∂yMultiplying dg by an as yet unknown number λ and adding it to df we obtaind(f + λg) =( ∂f∂x + λ ∂g∂x)dx +( ∂f∂y + λ∂g ∂y)dy =0,where λ is called a Lagrange undetermined multiplier. In this equation dx and dyare to be independent and arbitrary; we must therefore choose λ such that∂f∂x + λ ∂g =0, (5.27)∂x∂f∂y + λ∂g =0. (5.28)∂yThese equations, together with the constraint g(x, y) =c, are sufficient to find thethree unknowns, i.e. λ and the values of x and y at the stationary point.168

5.9 STATIONARY VALUES UNDER CONSTRAINTS◮The temperature of a point (x, y) on a unit circle is given by T (x, y) =1+xy. Findthetemperature of the two hottest points on the circle.We need to maximise T (x, y) subject to the constraint x 2 + y 2 = 1. Applying (5.27) and(5.28), we obtainy +2λx =0, (5.29)x +2λy =0. (5.30)These results, together with the original constraint x 2 + y 2 = 1, provide three simultaneousequations that may be solved for λ, x and y.From (5.29) and (5.30) we find λ = ±1/2, which in turn implies that y = ∓x. Rememberingthat x 2 + y 2 = 1, we find thaty = x ⇒ x = ± √ 1 ,2y = ± √ 12y = −x ⇒ x = ∓ √ 1 ,2y = ± √ 1 .2We have not yet determined which of these stationary points are maxima and which areminima. In this simple case, we need only substitute the four pairs of x- andy- values intoT (x, y) =1+xy to find that the maximum temperature on the unit circle is T max =3/2 atthe points y = x = ±1/ √ 2. ◭The method of Lagrange multipliers can be used to find the stationary points offunctions of more than two variables, subject to several constraints, provided thatthe number of constraints is smaller than the number of variables. For example,if we wish to find the stationary points of f(x, y, z) subject to the constraintsg(x, y, z) =c 1 and h(x, y, z) =c 2 ,wherec 1 and c 2 are constants, then we proceedas above, obtaining∂(f + λg + µh) =∂f∂x ∂x + λ ∂g∂x + µ ∂h∂x = 0,∂(f + λg + µh) =∂f∂y ∂y + λ∂g ∂y + µ ∂h∂y= 0, (5.31)∂(f + λg + µh) =∂f∂z ∂z + λ∂g ∂z + µ∂h ∂z = 0.We may now solve these three equations, together with the two constraints, togive λ, µ, x, y and z.169

PARTIAL DIFFERENTIATION◮Find the stationary points of f(x, y, z) =x 3 + y 3 + z 3 subject to the following constraints:(i) g(x, y, z) =x 2 + y 2 + z 2 =1;(ii) g(x, y, z) =x 2 + y 2 + z 2 =1 and h(x, y, z) =x + y + z =0.Case (i). Since there is only one constraint in this case, we need only introduce a singleLagrange multiplier to obtain∂∂x (f + λg) =3x2 +2λx =0,∂∂y (f + λg) =3y2 +2λy =0, (5.32)∂∂z (f + λg) =3z2 +2λz =0.These equations are highly symmetrical and clearly have the solution x = y = z = −2λ/3.Using the constraint x 2 + y 2 + z 2 = 1 we find λ = ± √ 3/2 and so stationary points occuratx = y = z = ± √ 1 . (5.33)3In solving the three equations (5.32) in this way, however, we have implicitly assumedthat x, y and z are non-zero. However, it is clear from (5.32) that any of these values canequal zero, with the exception of the case x = y = z = 0 since this is prohibited by theconstraint x 2 + y 2 + z 2 = 1. We must consider the other cases separately.If x = 0, for example, we require3y 2 +2λy =0,3z 2 +2λz =0,y 2 + z 2 =1.Clearly, we require λ ≠ 0, otherwise these equations are inconsistent. If neither y norz is zero we find y = −2λ/3 =z and from the third equation we require y = z =±1/ √ 2. If y =0,however,thenz = ±1 and, similarly, if z =0theny = ±1. Thus thestationary points having x =0are(0, 0, ±1), (0, ±1, 0) and (0, ±1/ √ 2, ±1/ √ 2). A similarprocedure can be followed for the cases y =0andz = 0 respectively and, in additionto those already obtained, we find the stationary points (±1, 0, 0), (±1/ √ 2, 0, ±1/ √ 2) and(±1/ √ 2, ±1/ √ 2, 0).Case (ii). We now have two constraints and must therefore introduce two Lagrangemultipliers to obtain (cf. (5.31))∂∂x (f + λg + µh) =3x2 +2λx + µ =0, (5.34)∂∂y (f + λg + µh) =3y2 +2λy + µ =0, (5.35)∂∂z (f + λg + µh) =3z2 +2λz + µ =0. (5.36)These equations are again highly symmetrical and the simplest way to proceed is tosubtract (5.35) from (5.34) to obtain3(x 2 − y 2 )+2λ(x − y) =0⇒ 3(x + y)(x − y)+2λ(x − y) =0. (5.37)This equation is clearly satisfied if x = y; then, from the second constraint, x + y + z =0,170

5.9 STATIONARY VALUES UNDER CONSTRAINTSwe find z = −2x. Substituting these values into the first constraint, x 2 + y 2 + z 2 =1,weobtainx = ± √ 1 , y = ± √ 1 , z = ∓ √ 2 . (5.38)6 6 6Because of the high degree of symmetry amongst the equations (5.34)–(5.36), we may obtainby inspection two further relations analogous to (5.37), one containing the variables y, zand the other the variables x, z. Assuming y = z in the first relation and x = z in thesecond, we find the stationary pointsx = ± 1 √6, y = ∓ 2 √6, z = ± 1 √6(5.39)andx = ∓ 2 √6, y = ± 1 √6, z = ± 1 √6. (5.40)We note that in finding the stationary points (5.38)–(5.40) we did not need to evaluate theLagrange multipliers λ and µ explicitly. This is not always the case, however, and in someproblems it may be simpler to begin by finding the values of these multipliers.Returning to (5.37) we must now consider the case where x ≠ y; then we find3(x + y)+2λ =0. (5.41)However, in obtaining the stationary points (5.39), (5.40), we did not assume x = y butonly required y = z and x = z respectively. It is clear that x ≠ y at these stationary points,and it can be shown that they do indeed satisfy (5.41). Similarly, several stationary pointsfor which x ≠ z or y ≠ z have already been found.Thus we need to consider further only two cases, x = y = z, andx, y and z are alldifferent. The first is clearly prohibited by the constraint x + y + z = 0. For the secondcase, (5.41) must be satisfied, together with the analogous equations containing y, z andx, z respectively, i.e.3(x + y)+2λ =0,3(y + z)+2λ =0,3(x + z)+2λ =0.Adding these three equations together and using the constraint x+y +z = 0 we find λ =0.However, for λ = 0 the equations are inconsistent for non-zero x, y and z. Therefore allthe stationary points have already been found and are given by (5.38)–(5.40). ◭The method may be extended to functions of any number n of variablessubject to any smaller number m of constraints. This means that effectively thereare n − m independent variables and, as mentioned above, we could solve bysubstitution and then by the methods of the previous section. However, for largen this becomes cumbersome and the use of Lagrange undetermined multipliers isa useful simplification.171

PARTIAL DIFFERENTIATION◮A system contains a very large number N of particles, each of which can be in any of Renergy levels with a corresponding energy E i , i =1, 2,...,R. The number of particles in theith level is n i and the total energy of the system is a constant, E. Find the distribution ofparticles amongst the energy levels that maximises the expressionN!P =n 1 !n 2 ! ···n R ! ,subject to the constraints that both the number of particles and the total energy remainconstant, i.e.R∑R∑g = N − n i =0 and h = E − n i E i =0.i=1i=1The way in which we proceed is as follows. In order to maximise P , we must minimiseits denominator (since the numerator is fixed). Minimising the denominator is the same asminimising the logarithm of the denominator, i.e.f =ln(n 1 !n 2 ! ···n R !) =ln(n 1 !) +ln(n 2 !) + ···+ln(n R !) .Using Stirling’s approximation, ln (n!) ≈ n ln n − n, we find thatf = n 1 ln n 1 + n 2 ln n 2 + ···+ n R ln n R − (n 1 + n 2 + ···+ n R )( R∑)= n i ln n i − N.i=1It has been assumed here that, for the desired distribution, all the n i are large. Thus, wenow have a function f subject to two constraints, g =0andh = 0, and we can apply theLagrange method, obtaining (cf. (5.31))∂f+ λ ∂g + µ ∂h =0,∂n 1 ∂n 1 ∂n 1∂f+ λ ∂g + µ ∂h =0,∂n 2 ∂n 2 ∂n 2 .∂f+ λ ∂g + µ ∂h =0.∂n R ∂n R ∂n RSince all these equations are alike, we consider the general case∂f+ λ ∂g + µ ∂h =0,∂n k ∂n k ∂n kfor k =1, 2,...,R. Substituting the functions f, g and h into this relation we findn k+lnn k + λ(−1) + µ(−E k )=0,n kwhich can be rearranged to giveln n k = µE k + λ − 1,and hencen k = C exp µE k .172

5.10 ENVELOPESWe now have the general form for the distribution of particles amongst energy levels, butin order to determine the two constants µ, C we recall thatandR∑C exp µE k = Nk=1R∑CE k exp µE k = E.k=1This is known as the Boltzmann distribution and is a well-known result from statisticalmechanics. ◭5.10 EnvelopesAs noted at the start of this chapter, many of the functions with which physicists,chemists and engineers have to deal contain, in addition to constants and oneor more variables, quantities that are normally considered as parameters of thesystem under study. Such parameters may, for example, represent the capacitanceof a capacitor, the length of a rod, or the mass of a particle – quantities thatare normally taken as fixed for any particular physical set-up. The correspondingvariables may well be time, currents, charges, positions and velocities. However,the parameters could be varied and in this section we study the effects of doing so;in particular we study how the form of dependence of one variable on another,typically y = y(x), is affected when the value of a parameter is changed in asmooth and continuous way. In effect, we are making the parameter into anadditional variable.As a particular parameter, which we denote by α, is varied over its permittedrange, the shape of the plot of y against x will change, usually, but not always,in a smooth and continuous way. For example, if the muzzle speed v of a shellfired from a gun is increased through a range of values then its height–distancetrajectories will be a series of curves with a common starting point that areessentially just magnified copies of the original; furthermore the curves do notcross each other. However, if the muzzle speed is kept constant but θ, the angleof elevation of the gun, is increased through a series of values, the correspondingtrajectories do not vary in a monotonic way. When θ has been increased beyond45 ◦ the trajectories then do cross some of the trajectories corresponding to θ45 ◦ all lie within a curve that touches each individualtrajectory at one point. Such a curve is called the envelope to the set of trajectorysolutions; it is to the study of such envelopes that this section is devoted.For our general discussion of envelopes we will consider an equation of theform f = f(x, y, α) = 0. A function of three Cartesian variables, f = f(x, y, α),is defined at all points in xyα-space, whereas f = f(x, y, α) =0isasurface inthis space. A plane of constant α, which is parallel to the xy-plane, cuts such173

PARTIAL DIFFERENTIATIONP 1PP 2yf(x, y, α 1 )=0 f(x, y, α 1 + h) =0xFigure 5.4 Two neighbouring curves in the xy-plane of the family f(x, y, α) =0 intersecting at P .Forfixedα 1 , the point P 1 is the limiting position of P ash → 0. As α 1 is varied, P 1 delineates the envelope of the family (broken line).a surface in a curve. Thus different values of the parameter α correspond todifferent curves, which can be plotted in the xy-plane. We now investigate howthe envelope equation for such a family of curves is obtained.5.10.1 Envelope equationsSuppose f(x, y, α 1 )=0andf(x, y, α 1 + h) = 0 are two neighbouring curves of afamily for which the parameter α differs by a small amount h. Let them intersectat the point P with coordinates x, y, as shown in figure 5.4. Then the envelope,indicated by the broken line in the figure, touches f(x, y, α 1 ) = 0 at the point P 1 ,which is defined as the limiting position of P when α 1 is fixed but h → 0. Thefull envelope is the curve traced out by P 1 as α 1 changes to generate successivemembers of the family of curves. Of course, for any finite h, f(x, y, α 1 + h) =0isone of these curves and the envelope touches it at the point P 2 .We are now going to apply Rolle’s theorem, see subsection 2.1.10, with theparameter α as the independent variable and x and y fixed as constants. In thiscontext, the two curves in figure 5.4 can be thought of as the projections onto thexy-plane of the planar curves in which the surface f = f(x, y, α) = 0 meets theplanes α = α 1 and α = α 1 + h.Along the normal to the page that passes through P ,asα changes from α 1to α 1 + h the value of f = f(x, y, α) will depart from zero, because the normalmeets the surface f = f(x, y, α) =0onlyatα = α 1 and at α = α 1 + h. However,at these end points the values of f = f(x, y, α) will both be zero, and thereforeequal. This allows us to apply Rolle’s theorem and so to conclude that for someθ in the range 0 ≤ θ ≤ 1 the partial derivative ∂f(x, y, α 1 + θh)/∂α is zero. When174

5.10 ENVELOPESh is made arbitrarily small, so that P → P 1 , the three defining equations reduceto two, which define the envelope point P 1 :∂f(x, y, α 1 )f(x, y, α 1 )=0 and=0. (5.42)∂αIn (5.42) both the function and the gradient are evaluated at α = α 1 . The equationof the envelope g(x, y) = 0 is found by eliminating α 1 between the two equations.As a simple example we will now solve the problem which when posed mathematicallyreads ‘calculate the envelope appropriate to the family of straight linesin the xy-plane whose points of intersection with the coordinate axes are a fixeddistance apart’. In more ordinary language, the problem is about a ladder leaningagainst a wall.◮A ladder of length L stands on level ground and can be leaned at any angle against avertical wall. Find the equation of the curve bounding the vertical area below the ladder.We take the ground and the wall as the x- andy-axes respectively. If the foot of the ladderis a from the foot of the wall and the top is b above the ground then the straight-lineequation of the ladder isxa + y b =1,where a and b are connected by a 2 + b 2 = L 2 . Expressed in standard form with only oneindependent parameter, a, the equation becomesf(x, y, a) = x a + y− 1=0. (5.43)(L 2 − a 2 )1/2Now, differentiating (5.43) with respect to a and setting the derivative ∂f/∂a equal tozero gives− x a + ay=0;2 (L 2 − a 2 )3/2from which it follows thatLx 1/3a =and (L 2 − a 2 ) 1/2 Ly 1/3=(x 2/3 + y 2/3 ) 1/2 (x 2/3 + y 2/3 ) . 1/2Eliminating a by substituting these values into (5.43) gives, for the equation of theenvelope of all possible positions on the ladder,x 2/3 + y 2/3 = L 2/3 .This is the equation of an astroid (mentioned in exercise 2.19), and, together with the walland the ground, marks the boundary of the vertical area below the ladder. ◭Other examples, drawn from both geometry and and the physical sciences, areconsidered in the exercises at the end of this chapter. The shell trajectory problemdiscussed earlier in this section is solved there, but in the guise of a questionabout the water bell of an ornamental fountain.175

PARTIAL DIFFERENTIATION5.11 Thermodynamic relationsThermodynamic relations provide a useful set of physical examples of partialdifferentiation. The relations we will derive are called Maxwell’s thermodynamicrelations. They express relationships between four thermodynamic quantities describinga unit mass of a substance. The quantities are the pressure P , the volumeV , the thermodynamic temperature T and the entropy S of the substance. Thesefour quantities are not independent; any two of them can be varied independently,but the other two are then determined.The first law of thermodynamics may be expressed asdU = TdS− PdV, (5.44)where U is the internal energy of the substance. Essentially this is a conservationof energy equation, but we shall concern ourselves, not with the physics, but ratherwith the use of partial differentials to relate the four basic quantities discussedabove. The method involves writing a total differential, dU say, in terms of thedifferentials of two variables, say X and Y , thus( ) ( )∂U∂UdU = dX + dY , (5.45)∂XY∂YXand then using the relationship∂ 2 U∂X∂Y =∂2 U∂Y ∂Xto obtain the required Maxwell relation. The variables X and Y aretobechosenfrom P , V , T and S.◮Show that (∂T/∂V) S = −(∂P/∂S) V .Here the two variables that have to be held constant, in turn, happen to be those whosedifferentials appear on the RHS of (5.44). And so, taking X as S and Y as V in (5.45), wehaveand find directly thatTdS− PdV= dU =( ) ∂U= T and∂SV( ) ∂UdS +∂SV( ) ∂UdV ,∂VS( ) ∂U= −P.∂VSDifferentiating the first expression with respect to V and the second with respect to S, andusing∂ 2 U∂V∂S =∂2 U∂S∂V ,we find the Maxwell relation( ) ( )∂T ∂P= − . ◭∂VS∂SV176

5.11 THERMODYNAMIC RELATIONS◮Show that (∂S/∂V) T =(∂P/∂T) V .Applying (5.45) to dS, with independent variables V and T , we find[( ) ( ) ]∂S∂SdU = TdS− PdV= T dV + dT − PdV.∂VT∂TVSimilarly applying (5.45) to dU, we find( ) ∂UdU =∂VThus, equating partial derivatives,( ) ( )∂U ∂S= T − P∂V ∂VBut, sinceT∂ 2 U∂T∂V =TdV +Tand∂2 U∂V∂T , i.e. ∂∂Tit follows that( ) ∂S∂VTThus finally we get the Maxwell relation( ) ∂S∂V( ) ∂UdT .∂TV( ) ( )∂U ∂S= T .∂TV∂TV( ) ∂U= ∂∂VT∂V( ) ∂U,∂TV( )+ T ∂2 S ∂P∂T∂V − = ∂ [ ( ) ] ∂ST= T ∂2 S∂TV∂V ∂TV T∂V∂T .T=( ) ∂P. ◭∂TVThe above derivation is rather cumbersome, however, and a useful trick thatcan simplify the working is to define a new function, called a potential. Theinternal energy U discussed above is one example of a potential but three othersare commonly defined and they are described below.◮Show that (∂S/∂V) T =(∂P/∂T) V by considering the potential U − ST.We first consider the differential d(U − ST). From (5.5), we obtaind(U − ST)=dU − SdT − TdS = −SdT − PdVwhen use is made of (5.44). We rewrite U − ST as F for convenience of notation; F iscalled the Helmholtz potential. ThusdF = −SdT − PdV,and it follows that( )( )∂F∂F= −S and= −P.∂TV∂VTUsing these results together with∂ 2 F∂T∂V =∂2 F∂V∂T ,we can see immediately that( ) ( )∂S ∂P= ,∂VT∂TVwhich is the same Maxwell relation as before. ◭177

PARTIAL DIFFERENTIATIONAlthough the Helmholtz potential has other uses, in this context it has simplyprovided a means for a quick derivation of the Maxwell relation. The otherMaxwell relations can be derived similarly by using two other potentials, theenthalpy, H = U + PV, and the Gibbs free energy, G = U + PV − ST (seeexercise 5.25).5.12 Differentiation of integralsWe conclude this chapter with a discussion of the differentiation of integrals. Letus consider the indefinite integral (cf. equation (2.30))∫F(x, t) = f(x, t) dt,from which it follows immediately that∂F(x, t)= f(x, t).∂tAssuming that the second partial derivatives of F(x, t) are continuous, we have∂ 2 F(x, t)∂t∂x= ∂2 F(x, t),∂x∂tand so we can write[ ]∂ ∂F(x, t)= ∂ [ ] ∂F(x, t) ∂f(x, t)=∂t ∂x ∂x ∂t ∂x .Integrating this equation with respect to t then gives∫∂F(x, t) ∂f(x, t)= dt. (5.46)∂x ∂xNow consider the definite integralI(x) =∫ t=vt=uf(x, t) dt= F(x, v) − F(x, u),where u and v are constants. Differentiating this integral with respect to x, andusing (5.46), we see thatdI(x) ∂F(x, v)=dx ∂x=u∂F(x, u)∂x−∫ v ∫∂f(x, t) u∂x dt − ∂f(x, t)∂xdt∫ v∂f(x, t)=∂xdt.This is Leibnitz’ rule for differentiating integrals, and basically it states that for178

5.13 EXERCISESconstant limits of integration the order of integration and differentiation can bereversed.In the more general case where the limits of the integral are themselves functionsof x, it follows immediately thatI(x) =∫ t=v(x)t=u(x)which yields the partial derivatives∂I∂I= f(x, v(x)),∂vConsequentlyf(x, t) dt= F(x, v(x)) − F(x, u(x)),( ) ( )dI ∂I dv ∂I dudx = ∂v dx + ∂u dx + ∂I∂x= −f(x, u(x)).∂u= f(x, v(x)) dvdu− f(x, u(x))dx dx + ∂∂x= f(x, v(x)) dv∫du v(x)− f(x, u(x))dx dx + u(x)∫ v(x)u(x)f(x, t)dt∂f(x, t)∂xdt, (5.47)where the partial derivative with respect to x in the last term has been takeninside the integral sign using (5.46). This procedure is valid because u(x) andv(x)are being held constant in this term.◮Find the derivative with respect to x of the integralApplying (5.47), we see thatdIdxI(x) =∫ x 2sin x3 sin x2= (2x) −x 2= 2sinx3x−xsin x2xsin xtdt.t∫ x 2x (1) + [ sin xt+xsin x3 sin x2=3 − 2x x= 1 x (3 sin x3 − 2sinx 2 ). ◭x] x 2xt cos xtdtt5.13 Exercises5.1 Using the appropriate properties of ordinary derivatives, perform the following.179

PARTIAL DIFFERENTIATION(a) Find all the first partial derivatives of the following functions f(x, y):(i) x 2 y, (ii) x 2 + y 2 + 4, (iii) sin(x/y), (iv) tan −1 (y/x),(v) r(x, y, z) =(x 2 + y 2 + z 2 ) 1/2 .(b) For (i), (ii) and (v), find ∂ 2 f/∂x 2 , ∂ 2 f/∂y 2 and ∂ 2 f/∂x∂y.(c) For (iv) verify that ∂ 2 f/∂x∂y = ∂ 2 f/∂y∂x.5.2 Determine which of the following are exact differentials:(a) (3x +2)ydx+ x(x +1)dy;(b) y tan xdx+ x tan ydy;(c) y 2 (ln x +1)dx +2xy ln xdy;(d) y 2 (ln x +1)dy +2xy ln xdx;(e) [x/(x 2 + y 2 )] dy − [y/(x 2 + y 2 )] dx.5.3 Show that the differentialdf = x 2 dy − (y 2 + xy) dxis not exact, but that dg =(xy 2 ) −1 df is exact.5.4 Show thatdf = y(1 + x − x 2 ) dx + x(x +1)dyis not an exact differential.Find the differential equation that a function g(x) mustsatisfyifdφ = g(x)dfis to be an exact differential. Verify that g(x) =e −x is a solution of this equationand deduce the form of φ(x, y).5.5 The equation 3y = z 3 +3xz defines z implicitly as a function of x and y. Evaluateall three second partial derivatives of z with respect to x and/or y. Verifythatzis a solution ofx ∂2 z∂y + ∂2 z2 ∂x =0. 25.6 A possible equation of state for a gas takes the form(PV = RT exp −α ),VRTin which α and R are constants. Calculate expressions for( ) ( ) ( )∂P∂V∂T∂V∂T∂P,Tand show that their product is −1, as stated in section 5.4.5.7 The function G(t) is defined byG(t) =F(x, y) =x 2 + y 2 +3xy,where x(t) =at 2 and y(t) =2at. Use the chain rule to find the values of (x, y) atwhich G(t) has stationary values as a function of t. Do any of them correspondto the stationary points of F(x, y) as a function of x and y?5.8 In the xy-plane, new coordinates s and t are defined bys = 1 (x + y), t 2Transform the equation,P= 1 (x − y).2∂ 2 φ∂x − ∂2 φ2 ∂y =0 2into the new coordinates and deduce that its general solution can be writtenφ(x, y) =f(x + y)+g(x − y),where f(u) andg(v) are arbitrary functions of u and v, respectively.180,V

5.13 EXERCISES5.9 The function f(x, y) satisfies the differential equationy ∂f∂x + x ∂f∂y =0.By changing to new variables u = x 2 − y 2 and v =2xy, show that f is, in fact, afunction of x 2 − y 2 only.5.10 If x = e u cos θ and y = e u sin θ, show that( )∂ 2 φ∂u + ∂2 φ∂ 2 ∂θ 2 =(x2 + y 2 2 f)∂x + ∂2 f,2 ∂y 2where f(x, y) =φ(u, θ).5.11 Find and evaluate the maxima, minima and saddle points of the functionf(x, y) =xy(x 2 + y 2 − 1).5.12 Show thatf(x, y) =x 3 − 12xy +48x + by 2 , b ≠0,has two, one, or zero stationary points, according to whether |b| is less than,equal to, or greater than 3.5.13 Locate the stationary points of the functionf(x, y) =(x 2 − 2y 2 )exp[−(x 2 + y 2 )/a 2 ],where a is a non-zero constant.Sketch the function along the x- andy-axes and hence identify the nature andvalues of the stationary points.5.14 Find the stationary points of the functionf(x, y) =x 3 + xy 2 − 12x − y 2and identify their natures.5.15 Find the stationary values off(x, y) =4x 2 +4y 2 + x 4 − 6x 2 y 2 + y 4and classify them as maxima, minima or saddle points. Make a rough sketch ofthe contours of f in the quarter plane x, y ≥ 0.5.16 The temperature of a point (x, y, z) on the unit sphere is given byT (x, y, z) =1+xy + yz.By using the method of Lagrange multipliers, find the temperature of the hottestpoint on the sphere.5.17 A rectangular parallelepiped has all eight vertices on the ellipsoidx 2 +3y 2 +3z 2 =1.Using the symmetry of the parallelepiped about each of the planes x = 0,y =0,z = 0, write down the surface area of the parallelepiped in terms ofthe coordinates of the vertex that lies in the octant x, y, z ≥ 0. Hence find themaximum value of the surface area of such a parallelepiped.5.18 Two horizontal corridors, 0 ≤ x ≤ a with y ≥ 0, and 0 ≤ y ≤ b with x ≥ 0, meetat right angles. Find the length L of the longest ladder (considered as a stick)that may be carried horizontally around the corner.5.19 A barn is to be constructed with a uniform cross-sectional area A throughoutits length. The cross-section is to be a rectangle of wall height h (fixed) andwidth w, surmounted by an isosceles triangular roof that makes an angle θ with181

PARTIAL DIFFERENTIATIONthe horizontal. The cost of construction is α perunitheightofwallandβ perunit (slope) length of roof. Show that, irrespective of the values of α and β, tominimise costs w should be chosen to satisfy the equationw 4 =16A(A − wh),and θ made such that 2 tan 2θ = w/h.5.20 Show that the envelope of all concentric ellipses that have their axes along thex- andy-coordinate axes, and that have the sum of their semi-axes equal to aconstant L, is the same curve (an astroid) as that found in the worked examplein section 5.10.5.21 Find the area of the region covered by points on the linesxa + y b =1,where the sum of any line’s intercepts on the coordinate axes is fixed and equalto c.5.22 Prove that the envelope of the circles whose diameters are those chords of agiven circle that pass through a fixed point on its circumference, is the cardioidr = a(1 + cos θ).Here a is the radius of the given circle and (r, θ) are the polar coordinates of theenvelope. Take as the system parameter the angle φ between a chord and thepolar axis from which θ is measured.5.23 A water feature contains a spray head at water level at the centre of a roundbasin. The head is in the form of a small hemisphere perforated by many evenlydistributed small holes, through which water spurts out at the same speed, v 0 ,inall directions.(a) What is the shape of the ‘water bell’ so formed?(b) What must be the minimum diameter of the bowl if no water is to be lost?5.24 In order to make a focussing mirror that concentrates parallel axial rays to onespot (or conversely forms a parallel beam from a point source), a parabolic shapeshould be adopted. If a mirror that is part of a circular cylinder or sphere wereused, the light would be spread out along a curve. This curve is known as acaustic and is the envelope of the rays reflected from the mirror. Denoting by θthe angle which a typical incident axial ray makes with the normal to the mirrorat the place where it is reflected, the geometry of reflection (the angle of incidenceequals the angle of reflection) is shown in figure 5.5.Show that a parametric specification of the caustic isx = R cos θ ( 12 +sin2 θ ) , y = R sin 3 θ,where R is the radius of curvature of the mirror. The curve is, in fact, part of anepicycloid.5.25 By considering the differentialdG = d(U + PV − ST),where G is the Gibbs free energy, P the pressure, V the volume, S the entropyand T the temperature of a system, and given further that the internal energy UsatisfiesdU = TdS− PdV,derive a Maxwell relation connecting (∂V/∂T) P and (∂S/∂P) T .182

5.13 EXERCISESyORθθ2θxFigure 5.5 The reflecting mirror discussed in exercise 5.24.5.26 Functions P (V,T), U(V,T)andS(V,T) are related byTdS= dU + PdV,where the symbols have the same meaning as in the previous question. Thepressure P is known from experiment to have the formin appropriate units. IfP = T 43 + T V ,U = αV T 4 + βT,where α, β, are constants (or, at least, do not depend on T or V ), deduce that αmust have a specific value, but that β may have any value. Find the correspondingform of S.5.27 As in the previous two exercises on the thermodynamics of a simple gas, thequantity dS = T −1 (dU + PdV) is an exact differential. Use this to prove that( ∂U∂V)T= T( ∂P∂T)V− P.In the van der Waals model of a gas, P obeys the equationP =RTV − b − aV , 2where R, a and b are constants. Further, in the limit V →∞, the form of Ubecomes U = cT , where c is another constant. Find the complete expression forU(V,T).5.28 The entropy S(H,T), the magnetisation M(H,T) and the internal energy U(H,T)of a magnetic salt placed in a magnetic field of strength H, at temperature T ,are connected by the equationTdS = dU − HdM.183

PARTIAL DIFFERENTIATIONBy considering d(U − TS − HM) prove that( ) ( ∂M ∂S=∂TH∂HFor a particular salt,M(H,T)=M 0 [1 − exp(−αH/T )].Show that if, at a fixed temperature, the applied field is increased from zero to astrength such that the magnetization of the salt is 3 M 4 0, then the salt’s entropydecreases by an amountM 0(3 − ln 4).4α5.29 Using the results of section 5.12, evaluate the integralHence show that5.30 The integralI(y) =∫ ∞0).Te −xy sin xdx.x∫ ∞sin xJ =0 x dx = π 2 .∫ ∞−∞e −αx2 dxhas the value (π/α) 1/2 . Use this result to evaluateJ(n) =∫ ∞−∞x 2n e −x2 dx,where n is a positive integer. Express your answer in terms of factorials.5.31 The function f(x) is differentiable and f(0) = 0. A second function g(y) is definedby∫ yf(x) dxg(y) = √ .0 y − xProve that∫dg ydy = df dx√ .0 dx y − xFor the case f(x) =x n , prove thatd n gdy =2(n!)√ y.n5.32 The functions f(x, t) andF(x) are defined byf(x, t) =e −xt ,F(x) =Verify, by explicit calculation, that∫ x∫dFxdx = f(x, x)+1840f(x, t) dt.0∂f(x, t)dt.∂x

5.14 HINTS AND ANSWERS5.33 If∫ 1I(α) =0x α − 1dx, α > −1,ln xwhat is the value of I(0)? Show thatddα xα = x α ln x,and deduce thatddα I(α) = 1α +1 .Hence prove that I(α) =ln(1+α).5.34 Find the derivative, with respect to x, of the integralI(x) =∫ 3xxexp xt dt.5.35 The function G(t, ξ) is defined for 0 ≤ t ≤ π by{− cos t sin ξ for ξ ≤ t,G(t, ξ) =− sin t cos ξ for ξ>t.Show that the function x(t) defined byx(t) =∫ π0G(t, ξ)f(ξ) dξsatisfies the equationd 2 xdt + x = f(t),2where f(t) canbeany arbitrary (continuous) function. Show further that x(0) =[dx/dt] t=π = 0, again for any f(t), but that the value of x(π) does depend uponthe form of f(t).[ The function G(t, ξ) is an example of a Green’s function, an importantconcept in the solution of differential equations and one studied extensively inlater chapters. ]5.14 Hints and answers5.1 (a) (i) 2xy, x 2 ; (ii) 2x, 2y; (iii) y −1 cos(x/y), (−x/y 2 )cos(x/y);(iv) −y/(x 2 + y 2 ),x/(x 2 + y 2 ); (v) x/r, y/r, z/r.(b) (i) 2y, 0, 2x; (ii) 2, 2, 0; (v) (y 2 + z 2 )r −3 , (x 2 + z 2 )r −3 , −xyr −3 .(c) Both second derivatives are equal to (y 2 − x 2 )(x 2 + y 2 ) −2 .5.3 2x ≠ −2y − x. For g, both sides of equation (5.9) equal y −2 .5.5 ∂ 2 z/∂x 2 =2xz(z 2 + x) −3 , ∂ 2 z/∂x∂y =(z 2 − x)(z 2 + x) −3 , ∂ 2 z/∂y 2 = −2z(z 2 + x) −3 .5.7 (0, 0), (a/4, −a) and(16a, −8a). Only the saddle point at (0, 0).5.9 The transformed equation is 2(x 2 + y 2 )∂f/∂v = 0; hence f does not depend on v.5.11 Maxima, equal to 1/8, at ±(1/2, −1/2), minima, equal to −1/8, at ±(1/2, 1/2),saddle points, equalling 0, at (0, 0), (0, ±1), (±1, 0).5.13 Maxima equal to a 2 e −1 at (±a, 0), minima equal to −2a 2 e −1 at (0, ±a), saddlepoint equalling 0 at (0, 0).5.15 Minimum at (0, 0); saddle points at (±1, ±1). To help with sketching the contours,determine the behaviour of g(x) =f(x, x).5.17 The Lagrange multiplier method gives z = y = x/2, for a maximal area of 4.185

PARTIAL DIFFERENTIATION5.19 The cost always includes 2αh, which can therefore be ignored in the optimisation.With Lagrange multiplier λ, sinθ = λw/(4β) andβ sec θ − 1 λw tan θ = λh, leading2to the stated results.5.21 The envelope of the lines x/a + y/(c−a) − 1 = 0, as a is varied, is √ x+ √ y = √ c.Area = c 2 /6.5.23 (a) Using α =cotθ, whereθ is the initial angle a jet makes with the vertical, theequation is f(z, ρ,α) =z−ρα+[gρ 2 (1+α 2 )/(2v0 2 )], and setting ∂f/∂α = 0 givesα = v0 2/(gρ). The water bell has a parabolic profile z = v2 0 /(2g) − gρ2 /(2v0 2).(b) Setting z = 0 gives the minimum diameter as 2v0 2/g.5.25 Show that (∂G/∂P) T = V and (∂G/∂T) P = −S. From each result, obtain anexpression for ∂ 2 G/∂T∂P and equate these, giving (∂V/∂T) P = −(∂S/∂P) T .5.27 Find expressions for (∂S/∂V) T and (∂S/∂T) V , and equate ∂ 2 S/∂V∂T with5.29∂ 2 S/∂T∂V. U(V,T)=cT − aV −1 .dI/dy = −Im[ ∫ ∞0 y2 ). Integrate dI/dy from 0 to ∞.I(∞) =0andI(0) = J.5.31 Integrate the RHS of the equation by parts, before differentiating with respectto y. Repeated application of the method establishes the result for all orders ofderivative.5.33 I(0) = 0; use Leibnitz’ rule.5.35 Write x(t) =− cos t ∫ tπcos ξf(ξ) dξ and differentiate each0 tterm as a product to obtain dx/dt. Obtaind 2 x/dt 2 in a similar way. Note thatintegrals that have equal lower and upper limits have value zero. The value ofx(π) is ∫ π0186

6Multiple integralsFor functions of several variables, just as we may consider derivatives with respectto two or more of them, so may the integral of the function with respect to morethan one variable be formed. The formal definitions of such multiple integrals areextensions of that for a single variable, discussed in chapter 2. We first discussdouble and triple integrals and illustrate some of their applications. We thenconsider changing the variables in multiple integrals and discuss some generalproperties of Jacobians.6.1 Double integralsFor an integral involving two variables – a double integral – we have a function,f(x, y) say, to be integrated with respect to x and y between certain limits. Theselimits can usually be represented by a closed curve C bounding a region R in thexy-plane. Following the discussion of single integrals given in chapter 2, let usdivide the region R into N subregions ∆R p of area ∆A p , p =1, 2,...,N, and let(x p ,y p ) be any point in subregion ∆R p . Now consider the sumS =N∑f(x p ,y p )∆A p ,p=1and let N →∞as each of the areas ∆A p → 0. If the sum S tends to a uniquelimit, I, then this is called the double integral of f(x, y) over the region R and iswritten∫I = f(x, y) dA, (6.1)Rwhere dA stands for the element of area in the xy-plane. By choosing thesubregions to be small rectangles each of area ∆A =∆x∆y, and letting both ∆x187

MULTIPLE INTEGRALSydSdxdA = dxdyRVdyUcTCa b xFigure 6.1 A simple curve C in the xy-plane, enclosing a region R.and ∆y → 0, we can also write the integral as∫∫I = f(x, y) dx dy, (6.2)Rwhere we have written out the element of area explicitly as the product of thetwo coordinate differentials (see figure 6.1).Some authors use a single integration symbol whatever the dimension of theintegral; others use as many symbols as the dimension. In different circumstancesboth have their advantages. We will adopt the convention used in (6.1) and (6.2),that as many integration symbols will be used as differentials explicitly written.The form (6.2) gives us a clue as to how we may proceed in the evaluationof a double integral. Referring to figure 6.1, the limits on the integration maybewrittenasanequationc(x, y) = 0 giving the boundary curve C. However, anexplicit statement of the limits can be written in two distinct ways.One way of evaluating the integral is first to sum up the contributions fromthe small rectangular elemental areas in a horizontal strip of width dy (as shownin the figure) and then to combine the contributions of these horizontal strips tocover the region R. In this case, we write∫ y=d{∫ x=x2(y)}I =f(x, y) dx dy, (6.3)y=cx=x 1(y)where x = x 1 (y) andx = x 2 (y) are the equations of the curves TSV and TUVrespectively. This expression indicates that first f(x, y) istobeintegratedwithrespect to x (treating y as a constant) between the values x = x 1 (y) andx = x 2 (y)and then the result, considered as a function of y, is to be integrated between thelimits y = c and y = d. Thus the double integral is evaluated by expressing it interms of two single integrals called iterated (or repeated) integrals.188

6.1 DOUBLE INTEGRALSAn alternative way of evaluating the integral, however, is first to sum up thecontributions from the elemental rectangles arranged into vertical strips and thento combine these vertical strips to cover the region R. Wethenwrite∫ x=b{∫ y=y2(x)}I =f(x, y) dy dx, (6.4)x=ay=y 1(x)where y = y 1 (x) andy = y 2 (x) are the equations of the curves STU and SVUrespectively. In going to (6.4) from (6.3), we have essentially interchanged theorder of integration.In the discussion above we assumed that the curve C was such that any lineparallel to either the x- ory-axis intersected C at most twice. In general, providedf(x, y) is continuous everywhere in R and the boundary curve C has this simpleshape, the same result is obtained irrespective of the order of integration. In caseswhere the region R has a more complicated shape, it can usually be subdividedinto smaller simpler regions R 1 , R 2 etc. that satisfy this criterion. The doubleintegral over R is then merely the sum of the double integrals over the subregions.◮Evaluate the double integral∫∫I = x 2 ydxdy,Rwhere R is the triangular area bounded by the lines x =0, y =0and x + y =1. Reversethe order of integration and demonstrate that the same result is obtained.The area of integration is shown in figure 6.2. Suppose we choose to carry out theintegration with respect to y first. With x fixed, the range of y is 0 to 1 − x. Wecantherefore write∫ x=1{∫ y=1−x}I =x 2 ydy dxx=0∫ x=1y=0] y=1−x[ x 2 y 2∫ 1x 2 (1 − x) 2=dx =dx = 1x=0 2y=00 2 60 .Alternatively, we may choose to perform the integration with respect to x first. With yfixed, the range of x is 0 to 1 − y, so we have∫ y=1{∫ x=1−y}I =x 2 ydx dy=y=0∫ y=1y=0[ x 3 y3x=0] x=1−yx=0∫ 1dx =0(1 − y) 3 y3dy = 1 60 .As expected, we obtain the same result irrespective of the order of integration. ◭We may avoid the use of braces in expressions such as (6.3) and (6.4) by writing(6.4), for example, asI =∫ badx∫ y2(x)y 1(x)dy f(x, y),where it is understood that each integral symbol acts on everything to its right,189

MULTIPLE INTEGRALSy1dyRx + y =100dx1xFigure 6.2 The triangular region whose sides are the axes x =0,y =0andthe line x + y =1.and that the order of integration is from right to left. So, in this example, theintegrand f(x, y) is first to be integrated with respect to y and then with respectto x. With the double integral expressed in this way, we will no longer write theindependent variables explicitly in the limits of integration, since the differentialof the variable with respect to which we are integrating is always adjacent to therelevant integral sign.Using the order of integration in (6.3), we could also write the double integral asI =∫ dcdy∫ x2(y)x 1(y)dx f(x, y).Occasionally, however, interchange of the order of integration in a double integralis not permissible, as it yields a different result. For example, difficulties mightarise if the region R were unbounded with some of the limits infinite, though inmany cases involving infinite limits the same result is obtained whichever orderof integration is used. Difficulties can also occur if the integrand f(x, y) has anydiscontinuities in the region R or on its boundary C.6.2 Triple integralsThe above discussion for double integrals can easily be extended to triple integrals.Consider the function f(x, y, z) defined in a closed three-dimensional region R.Proceeding as we did for double integrals, let us divide the region R into Nsubregions ∆R p of volume ∆V p , p =1, 2,...,N,andlet(x p ,y p ,z p ) be any point inthe subregion ∆R p . Now we form the sumS =N∑f(x p ,y p ,z p )∆V p ,p=1190

6.3 APPLICATIONS OF MULTIPLE INTEGRALSand let N →∞as each of the volumes ∆V p → 0. If the sum S tends to a uniquelimit, I, then this is called the triple integral of f(x, y, z) over the region R and iswritten∫I = f(x, y, z) dV , (6.5)Rwhere dV stands for the element of volume. By choosing the subregions to besmall cuboids, each of volume ∆V =∆x∆y∆z, and proceeding to the limit, wecanalsowritetheintegralas∫∫∫I = f(x, y, z) dx dy dz, (6.6)Rwhere we have written out the element of volume explicitly as the product of thethree coordinate differentials. Extending the notation used for double integrals,we may write triple integrals as three iterated integrals, for example,I =∫ x2x 1dx∫ y2(x)y 1(x)dy∫ z2(x,y)z 1(x,y)dz f(x, y, z),where the limits on each of the integrals describe the values that x, y and z takeon the boundary of the region R. As for double integrals, in most cases the orderof integration does not affect the value of the integral.We can extend these ideas to define multiple integrals of higher dimensionalityin a similar way.6.3 Applications of multiple integralsMultiple integrals have many uses in the physical sciences, since there are numerousphysical quantities which can be written in terms of them. We now discuss afew of the more common examples.6.3.1 Areas and volumesMultiple integrals are often used in finding areas and volumes. For example, theintegral∫ ∫∫A = dA = dx dyRRis simply equal to the area of the region R. Similarly, if we consider the surfacez = f(x, y) in three-dimensional Cartesian coordinates then the volume under thissurface that stands vertically above the region R is given by the integral∫ ∫∫V = zdA= f(x, y) dx dy,Rwhere volumes above the xy-plane are counted as positive, and those below asnegative.191R

MULTIPLE INTEGRALSzcdV = dx dy dzdzdxdybyaxFigure 6.3 The tetrahedron bounded by the coordinate surfaces and theplane x/a + y/b + z/c = 1 is divided up into vertical slabs, the slabs intocolumns and the columns into small boxes.◮Find the volume of the tetrahedron bounded by the three coordinate surfaces x =0, y =0and z =0and the plane x/a + y/b + z/c =1.Referring to figure 6.3, the elemental volume of the shaded region is given by dV = zdxdy,and we must integrate over the triangular region R in the xy-plane whose sides are x =0,y =0andy = b − bx/a. The total volume of the tetrahedron is therefore given by∫∫V = zdxdy=R= c= c∫ adx0∫ a0∫ a0∫ b−bx/a0(dy c 1 − y b − x )adx[y − y22b − xy a( bx2dx2a − bx 2 a + b 2] y=b−bx/ay=0)= abc6 . ◭Alternatively, we can write the volume of a three-dimensional region R as∫ ∫∫∫V = dV = dx dy dz, (6.7)RRwhere the only difficulty occurs in setting the correct limits on each of theintegrals. For the above example, writing the volume in this way corresponds todividing the tetrahedron into elemental boxes of volume dx dy dz (as shown infigure 6.3); integration over z then adds up the boxes to form the shaded columnin the figure. The limits of integration are z =0toz = c ( 1 − y/b − x/a ) ,and192

6.3 APPLICATIONS OF MULTIPLE INTEGRALSthe total volume of the tetrahedron is given byV =∫ a0dx∫ b−bx/a0dy∫ c(1−y/b−x/a)0dz, (6.8)which clearly gives the same result as above. This method is illustrated further inthe following example.◮Find the volume of the region bounded by the paraboloid z = x 2 + y 2z =2y.and the planeThe required region is shown in figure 6.4. In order to write the volume of the region inthe form (6.7), we must deduce the limits on each of the integrals. Since the integrationscan be performed in any order, let us first divide the region into vertical slabs of thicknessdy perpendicular to the y-axis, and then as shown in the figure we cut each slab intohorizontal strips of height dz, and each strip into elemental boxes of volume dV = dx dy dz.Integrating first with respect to x (adding up the elemental boxes to get a horizontal strip),the limits on x are x = − √ z − y 2 to x = √ z − y 2 . Now integrating with respect to z(adding up the strips to form a vertical slab) the limits on z are z = y 2 to z =2y. Finally,integrating with respect to y (adding up the slabs to obtain the required region), the limitson y are y =0andy = 2, the solutions of the simultaneous equations z =0 2 + y 2 andz =2y. So the volume of the region isV ==∫ 20∫ 20dy∫ 2yy 2∫ √ z−y 2dz dx =−√z−y 2dy [ 43 (z − y2 ) 3/2] z=2yz=y 2 =∫ 20∫ 20∫ 2ydy dz 2 √ z − y 2y 2dy 4 3 (2y − y2 ) 3/2 .The integral over y may be evaluated straightforwardly by making the substitution y =1+sinu, and gives V = π/2. ◭In general, when calculating the volume (area) of a region, the volume (area)elements need not be small boxes as in the previous example, but may be of anyconvenient shape. The latter is usually chosen to make evaluation of the integralas simple as possible.6.3.2 Masses, centres of mass and centroidsIt is sometimes necessary to calculate the mass of a given object having a nonuniformdensity. Symbolically, this mass is given simply by∫M = dM,where dM is the element of mass and the integral is taken over the extent of theobject. For a solid three-dimensional body the element of mass is just dM = ρdV,where dV is an element of volume and ρ is the variable density. For a laminarbody (i.e. a uniform sheet of material) the element of mass is dM = σdA,whereσ is the mass per unit area of the body and dA is an area element. Finally, fora body in the form of a thin wire we have dM = λds,whereλ is the mass per193

MULTIPLE INTEGRALSzz =2yz = x 2 + y 20 2 ydV = dx dy dzxFigure 6.4 The region bounded by the paraboloid z = x 2 + y 2 and the planez =2y is divided into vertical slabs, the slabs into horizontal strips and thestrips into boxes.unit length and ds is an element of arc length along the wire. When evaluatingthe required integral, we are free to divide up the body into mass elements inthe most convenient way, provided that over each mass element the density isapproximately constant.◮ Find the mass of the tetrahedron bounded by the three coordinate surfaces and the planex/a + y/b + z/c =1, if its density is given by ρ(x, y, z) =ρ 0 (1 + x/a).From (6.8), we can immediately write down the mass of the tetrahedron as∫ (M = ρ 0 1+ x ) ∫ a (dV = dx ρ 0 1+ x ) ∫ b−bx/a ∫ c(1−y/b−x/a)dydz,R a0a 00where we have taken the density outside the integrations with respect to z and y since itdepends only on x. Therefore the integrations with respect to z and y proceed exactly asthey did when finding the volume of the tetrahedron, and we have∫ a (M = cρ 0 dx 1+ x ) ( bx 2a 2a − bx 2 a + b 20). (6.9)We could have arrived at (6.9) more directly by dividing the tetrahedron into triangularslabs of thickness dx perpendicular to the x-axis (see figure 6.3), each of which is ofconstant density, since ρ depends on x alone. A slab at a position x has volume dV =1c(1 − x/a)(b − bx/a) dx and mass dM = ρdV = ρ 2 0(1 + x/a) dV . Integrating over x weagain obtain (6.9). This integral is easily evaluated and gives M = 5 abcρ 24 0. ◭194

6.3 APPLICATIONS OF MULTIPLE INTEGRALSThe coordinates of the centre of mass of a solid or laminar body may also bewritten as multiple integrals. The centre of mass of a body has coordinates ¯x, ȳ,¯z given by the three equations∫¯x∫∫dM =∫xdMȳ dM = ydM∫ ∫¯z dM = zdM,where again dM is an element of mass as described above, x, y, z are thecoordinates of the centre of mass of the element dM and the integrals are takenover the entire body. Obviously, for any body that lies entirely in, or is symmetricalabout, the xy-plane (say), we immediately have ¯z = 0. For completeness, we notethat the three equations above can be written as the single vector equation (seechapter 7)¯r = 1 ∫r dM,Mwhere ¯r is the position vector of the body’s centre of mass with respect to theorigin, r is the position vector of the centre of mass of the element dM andM = ∫ dM is the total mass of the body. As previously, we may divide the bodyinto the most convenient mass elements for evaluating the necessary integrals,provided each mass element is of constant density.We further note that the coordinates of the centroid of a body are defined asthose of its centre of mass if the body had uniform density.◮Find the centre of mass of the solid hemisphere bounded by the surfaces x 2 + y 2 + z 2 = a 2and the xy-plane, assuming that it has a uniform density ρ.Referring to figure 6.5, we know from symmetry that the centre of mass must lie onthe z-axis. Let us divide the hemisphere into volume elements that are circular slabs ofthickness dz parallel to the xy-plane. For a slab at a height z, the mass of the element isdM = ρdV = ρπ(a 2 − z 2 ) dz. Integrating over z, we find that the z-coordinate of the centreof mass of the hemisphere is given by¯z∫ a0ρπ(a 2 − z 2 ) dz =∫ a0zρπ(a 2 − z 2 ) dz.The integrals are easily evaluated and give ¯z =3a/8. Since the hemisphere is of uniformdensity, this is also the position of its centroid. ◭6.3.3 Pappus’ theoremsThe theorems of Pappus (which are about seventeen centuries old) relate centroidsto volumes of revolution and areas of surfaces, discussed in chapter 2, and may beuseful for finding one quantity given another that can be calculated more easily.195

MULTIPLE INTEGRALSza√a2 − z 2dzxaayFigure 6.5 The solid hemisphere bounded by the surfaces x 2 + y 2 + z 2 = a 2and the xy-plane.ydAAyȳxFigure 6.6 An area A in the xy-plane, which may be rotated about the x-axisto form a volume of revolution.If a plane area is rotated about an axis that does not intersect it then the solidso generated is called a volume of revolution. Pappus’ first theorem states that thevolume of such a solid is given by the plane area A multiplied by the distancemoved by its centroid (see figure 6.6). This may be proved by considering thedefinition of the centroid of the plane area as the position of the centre of massif the density is uniform, so thatȳ = 1 ∫ydA.ANow the volume generated by rotating the plane area about the x-axis is given by∫V = 2πy dA =2πȳA,which is the area multiplied by the distance moved by the centroid.196

6.3 APPLICATIONS OF MULTIPLE INTEGRALSydsyȳxFigure 6.7 A curve in the xy-plane, which may be rotated about the x-axisto form a surface of revolution.Pappus’ second theorem states that if a plane curve is rotated about a coplanaraxis that does not intersect it then the area of the surface of revolution so generatedis given by the length of the curve L multiplied by the distance moved by itscentroid (see figure 6.7). This may be proved in a similar manner to the firsttheorem by considering the definition of the centroid of a plane curve,ȳ = 1 L∫yds,and noting that the surface area generated is given by∫S = 2πy ds =2πȳL,which is equal to the length of the curve multiplied by the distance moved by itscentroid.◮ A semicircular uniform lamina is freely suspended from one of its corners. Show that itsstraight edge makes an angle of 23.0 ◦ with the vertical.Referring to figure 6.8, the suspended lamina will have its centre of gravity C verticallybelow the suspension point and its straight edge will make an angle θ =tan −1 (d/a) withthe vertical, where 2a is the diameter of the semicircle and d is the distance of its centreof mass from the diameter.Since rotating the lamina about the diameter generates a sphere of volume 4 3 πa3 , Pappus’first theorem requires that43 πa3 =2πd × 1 2 πa2 .Hence d = 4a4and θ =tan−13π 3π =23.0◦ . ◭197

MULTIPLE INTEGRALSaθdCFigure 6.8Suspending a semicircular lamina from one of its corners.6.3.4 Moments of inertiaFor problems in rotational mechanics it is often necessary to calculate the momentof inertia of a body about a given axis. This is defined by the multiple integral∫I = l 2 dM,where l is the distance of a mass element dM from the axis. We may again choosemass elements convenient for evaluating the integral. In this case, however, inaddition to elements of constant density we require all parts of each element tobe at approximately the same distance from the axis about which the moment ofinertia is required.◮ Find the moment of inertia of a uniform rectangular lamina of mass M with sides a andb about one of the sides of length b.Referring to figure 6.9, we wish to calculate the moment of inertia about the y-axis.We therefore divide the rectangular lamina into elemental strips parallel to the y-axis ofwidth dx. The mass of such a strip is dM = σb dx, whereσ is the mass per unit area ofthe lamina. The moment of inertia of a strip at a distance x from the y-axisissimplydI = x 2 dM = σbx 2 dx. The total moment of inertia of the lamina about the y-axis isthereforeI =∫ a0σbx 2 dx = σba33 .Since the total mass of the lamina is M = σab, wecanwriteI = 1 3 Ma2 . ◭198

6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALSybdM = σb dxdxaxFigure 6.9 A uniform rectangular lamina of mass M with sides a and b canbe divided into vertical strips.6.3.5 Mean values of functionsIn chapter 2 we discussed average values for functions of a single variable. Thisis easily extended to functions of several variables. Let us consider, for example,a function f(x, y) defined in some region R of the xy-plane. Then the averagevalue ¯f of the function is given by∫ ∫¯f dA = f(x, y) dA. (6.10)RThis definition is easily extended to three (and higher) dimensions; if a functionf(x, y, z) is defined in some three-dimensional region of space R then the averagevalue ¯f of the function is given by∫ ∫¯f dV = f(x, y, z) dV . (6.11)RRR◮A tetrahedron is bounded by the three coordinate surfaces and the plane x/a+y/b+z/c =1 and has density ρ(x, y, z) =ρ 0 (1 + x/a). Find the average value of the density.From (6.11), the average value of the density is given by∫ ∫¯ρ dV = ρ(x, y, z) dV .RRNow the integral on the LHS is just the volume of the tetrahedron, which we found insubsection 6.3.1 to be V = 1 5abc, and the integral on the RHS is its mass M = abcρ 6 24 0,calculated in subsection 6.3.2. Therefore ¯ρ = M/V = 5 ρ 4 0. ◭6.4 Change of variables in multiple integralsIt often happens that, either because of the form of the integrand involved orbecause of the boundary shape of the region of integration, it is desirable to199

MULTIPLE INTEGRALSyu =constantv =constantNKMLRCFigure 6.10 A region of integration R overlaid with a grid formed by thefamily of curves u =constantandv = constant. The parallelogram KLMNdefines the area element dA uv .xexpress a multiple integral in terms of a new set of variables. We now considerhow to do this.6.4.1 Change of variables in double integralsLet us begin by examining the change of variables in a double integral. Supposethat we require to change an integral∫∫I = f(x, y) dx dy,Rin terms of coordinates x and y, into one expressed in new coordinates u and v,given in terms of x and y by differentiable equations u = u(x, y) andv = v(x, y)with inverses x = x(u, v) andy = y(u, v). The region R in the xy-plane and thecurve C that bounds it will become a new region R ′ and a new boundary C ′ inthe uv-plane, and so we must change the limits of integration accordingly. Also,the function f(x, y) becomes a new function g(u, v) of the new coordinates.Now the part of the integral that requires most consideration is the area element.In the xy-plane the element is the rectangular area dA xy = dx dy generated byconstructing a grid of straight lines parallel to the x- andy- axes respectively.Our task is to determine the corresponding area element in the uv-coordinates. Ingeneral the corresponding element dA uv will not be the same shape as dA xy , butthis does not matter since all elements are infinitesimally small and the value ofthe integrand is considered constant over them. Since the sides of the area elementare infinitesimal, dA uv will in general have the shape of a parallelogram. We canfind the connection between dA xy and dA uv by considering the grid formed by thefamily of curves u =constantandv = constant, as shown in figure 6.10. Since v200

6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALSis constant along the line element KL, the latter has components (∂x/∂u) du and(∂y/∂u) du in the directions of the x- andy-axes respectively. Similarly, since uis constant along the line element KN, the latter has corresponding components(∂x/∂v) dv and (∂y/∂v) dv. Using the result for the area of a parallelogram givenin chapter 7, we find that the area of the parallelogram KLMN is given by∣ dA uv =∂x∣ ∂u du∂y ∂x ∂y ∣∣∣dv − dv∂v ∂v ∂u du =∂x ∂y∣ ∂u ∂v − ∂x∂y∂v ∂u∣ du dv.Defining the Jacobian of x, y with respect to u, v as∂(x, y)J =∂(u, v) ≡ ∂x ∂y∂u ∂v − ∂x ∂y∂v ∂u ,we havedA uv =∂(x, y)∣ ∂(u, v) ∣ du dv.The reader acquainted with determinants will notice that the Jacobian can alsobe written as the 2 × 2 determinant∣ ∣∣∣∣∣∣∣ ∂x ∂y∂(x, y)J =∂(u, v) = ∂u ∂u∂x ∂y.∣∂v ∂vSuch determinants can be evaluated using the methods of chapter 8.So, in summary, the relationship between the size of the area element generatedby dx, dy and the size of the corresponding area element generated by du, dv isdx dy =∂(x, y)∣ ∂(u, v) ∣ du dv.This equality should be taken as meaning that when transforming from coordinatesx, y to coordinates u, v, the area element dx dy should be replaced by theexpression on the RHS of the above equality. Of course, the Jacobian can, andin general will, vary over the region of integration. We may express the doubleintegral in either coordinate system as∫∫∫∫I = f(x, y) dx dy = g(u, v)∂(x, y)∣RR ∂(u, v) ∣ du dv. (6.12)′When evaluating the integral in the new coordinate system, it is usually advisableto sketch the region of integration R ′ in the uv-plane.201

MULTIPLE INTEGRALS◮Evaluate the double integral∫∫I =(a + √ )x 2 + y 2 dx dy,where R is the region bounded by the circle x 2 + y 2 = a 2 .00RIn Cartesian coordinates, the integral may be written∫ a ∫ √ a 2 −x 2I = dx−a − √ dy(a + √ )x 2 + y 2 ,a 2 −x 2and can be calculated directly. However, because of the circular boundary of the integrationregion, a change of variables to plane polar coordinates ρ, φ is indicated. The relationshipbetween Cartesian and plane polar coordinates is given by x = ρ cos φ and y = ρ sin φ.Using (6.12) we can therefore write∫∫I = (a + ρ)∂(x, y)∣R ′ ∂(ρ, φ) ∣ dρ dφ,where R ′ is the rectangular region in the ρφ-plane whose sides are ρ =0,ρ = a, φ =0and φ =2π. The Jacobian is easily calculated, and we obtain∣ ∂(x, y) ∣∣∣J =∂(ρ, φ) = cos φ sin φ−ρ sin φ ρcos φ ∣ = ρ(cos2 φ +sin 2 φ)=ρ.So the relationship between the area elements in Cartesian and in plane polar coordinates isdx dy = ρdρdφ.Therefore, when expressed in plane polar coordinates, the integral is given by∫∫I = (a + ρ)ρdρdφR∫ ′ 2π ∫ a[ ] aρ2 a= dφ dρ (a + ρ)ρ =2π2 + ρ3= 5πa3303 . ◭6.4.2 Evaluation of the integral I = ∫ ∞−∞ e−x2 dxBy making a judicious change of variables, it is sometimes possible to evaluatean integral that would be intractable otherwise. An important example of thismethod is provided by the evaluation of the integralI =∫ ∞−∞e −x2 dx.Its value may be found by first constructing I 2 , as follows:I 2 =∫ ∞−∞e −x2 dx∫ ∞−∞∫ ∞∫ ∞e −y2 dy = dx dy e −(x2 +y 2 )−∞ −∞∫∫= e −(x2 +y 2) dx dy,202R

6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALSya−aax−aFigure 6.11 The regions used to illustrate the convergence properties of theintegral I(a) = ∫ a−a e−x2 dx as a →∞.where the region R is the whole xy-plane. Then, transforming to plane polarcoordinates, we find∫∫∫ 2π ∫ ∞I 2 = e −ρ2 ρdρdφ= dφ dρ ρe −ρ2 =2π[− 1 2 e−ρ2] ∞= π.R ′ 0 00Therefore the original integral is given by I = √ π. Because the integrand is aneven function of x, it follows that the value of the integral from 0 to ∞ is simply√ π/2.We note, however, that unlike in all the previous examples, the regions ofintegration R and R ′ are both infinite in extent (i.e. unbounded). It is thereforeprudent to derive this result more rigorously; this we do by considering theintegral∫ aI(a) = e −x2 dx.−aWe then have∫∫I 2 (a) = e −(x2 +y 2) dx dy,Rwhere R is the square of side 2a centred on the origin. Referring to figure 6.11,since the integrand is always positive the value of the integral taken over thesquare lies between the value of the integral taken over the region bounded bythe inner circle of radius a and the value of the integral taken over the outercircle of radius √ 2a. Transforming to plane polar coordinates as above, we may203

MULTIPLE INTEGRALSzRTu = c 1v = c 2SPQw = c 3CyxFigure 6.12 A three-dimensional region of integration R, showing an elementof volume in u, v, w coordinates formed by the coordinate surfacesu =constant,v =constant,w =constant.evaluate the integrals over the inner and outer circles respectively, and we find(π 1 − e −a2) (

6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALSw are constant, and so PQ has components (∂x/∂u) du, (∂y/∂u) du and (∂z/∂u) duin the directions of the x-, y- andz- axes respectively. The components of theline elements PS and ST are found by replacing u by v and w respectively.The expression for the volume of a parallelepiped in terms of the componentsof its edges with respect to the x-, y- andz-axes is given in chapter 7. Using this,we find that the element of volume in u, v, w coordinates is given bydV uvw =∂(x, y, z)∣∂(u, v, w) ∣ du dv dw,where the Jacobian of x, y, z with respect to u, v, w is a short-hand for a 3 × 3determinant:∣ ∣∣∣∣∣∣∣∣∣∣∣ ∂x ∂y ∂z∂u ∂u ∂u∂(x, y, z)∂(u, v, w) ≡ ∂x ∂y ∂z∂v ∂v ∂v.∣∂x∂wSo, in summary, the relationship between the elemental volumes in multipleintegrals formulated in the two coordinate systems is given in Jacobian form bydx dy dz =∂(x, y, z)∣∂(u, v, w) ∣ du dv dw,and we can write a triple integral in either set of coordinates as∫∫∫∫∫∫I = f(x, y, z) dx dy dz = g(u, v, w)∂(x, y, z)∣RR ∂(u, v, w) ∣ du dv dw.′∂y∂w∂z∂w∣◮ Find an expression for a volume element in spherical polar coordinates, and hence calculatethe moment of inertia about a diameter of a uniform sphere of radius a and mass M.Spherical polar coordinates r, θ, φ are defined byx = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ(and are discussed fully in chapter 10). The required Jacobian is therefore∣ ∣∣∣∣∣∂(x, y, z)sin θ cos φ sin θ sin φ cos θJ =∂(r, θ, φ) = r cos θ cos φ rcos θ sin φ −r sin θ−r sin θ sin φ rsin θ cos φ 0 ∣ .The determinant is most easily evaluated by expanding it with respect to the last column(see chapter 8), which givesJ =cosθ(r 2 sin θ cos θ)+r sin θ(r sin 2 θ)= r 2 sin θ(cos 2 θ +sin 2 θ)=r 2 sin θ.Therefore the volume element in spherical polar coordinates is given bydV =∂(x, y, z)∂(r, θ, φ) dr dθ dφ = r2 sin θdrdθdφ,205

MULTIPLE INTEGRALSwhich agrees with the result given in chapter 10.If we place the sphere with its centre at the origin of an x, y, z coordinate system thenits moment of inertia about the z-axis (which is, of course, a diameter of the sphere) isI =∫ (x 2 + y 2) dM = ρ∫ (x 2 + y 2) dV ,where the integral is taken over the sphere, and ρ is the density. Using spherical polarcoordinates, we can write this as∫∫∫(I = ρ r 2 sin 2 θ ) r 2 sin θdrdθdφ= ρV∫ 2π0dφ∫ π0dθ sin 3 θ= ρ × 2π × 4 × 1 3 5 a5 = 8 15 πa5 ρ.Since the mass of the sphere is M = 4 3 πa3 ρ, the moment of inertia can also be written asI = 2 5 Ma2 . ◭∫ a0dr r 46.4.4 General properties of JacobiansAlthough we will not prove it, the general result for a change of coordinates inan n-dimensional integral from a set x i to a set y j (where i and j both run from1ton) isdx 1 dx 2 ···dx n =∂(x 1 ,x 2 ,...,x n )∣ ∂(y 1 ,y 2 ,...,y n ) ∣ dy 1 dy 2 ···dy n ,where the n-dimensional Jacobian can be written as an n × n determinant (seechapter 8) in an analogous way to the two- and three-dimensional cases.For readers who already have sufficient familiarity with matrices (see chapter 8)and their properties, a fairly compact proof of some useful general propertiesof Jacobians can be given as follows. Other readers should turn straight to theresults (6.16) and (6.17) and return to the proof at some later time.Consider three sets of variables x i , y i and z i , with i running from 1 to n foreach set. From the chain rule in partial differentiation (see (5.17)), we know that∂x in∑ ∂x i ∂y k=. (6.13)∂z j ∂y k ∂z jk=1Now let A, B and C be the matrices whose ijth elements are ∂x i /∂y j , ∂y i /∂z j and∂x i /∂z j respectively. We can then write (6.13) as the matrix productn∑c ij = a ik b kj or C = AB. (6.14)k=1We may now use the general result for the determinant of the product of twomatrices, namely |AB| = |A||B|, and recall that the JacobianJ xy = ∂(x 1,...,x n )= |A|, (6.15)∂(y 1 ,...,y n )206

6.5 EXERCISESand similarly for J yz and J xz . On taking the determinant of (6.14), we thereforeobtainJ xz = J xy J yzor, in the usual notation,∂(x 1 ,...,x n )∂(z 1 ,...,z n ) = ∂(x 1,...,x n ) ∂(y 1 ,...,y n )∂(y 1 ,...,y n ) ∂(z 1 ,...,z n ) . (6.16)As a special case, if the set z i istakentobeidenticaltothesetx i ,andtheobvious result J xx = 1 is used, we obtainJ xy J yx =1or, in the usual notation,[ ] −1∂(x 1 ,...,x n )∂(y 1 ,...,y n ) = ∂(y1 ,...,y n ). (6.17)∂(x 1 ,...,x n )The similarity between the properties of Jacobians and those of derivatives isapparent, and to some extent is suggested by the notation. We further note from(6.15) that since |A| = |A T |,whereA T is the transpose of A, we can interchange therows and columns in the determinantal form of the Jacobian without changingits value.6.5 Exercises6.1 Identify the curved wedge bounded by the surfaces y 2 =4ax, x + z = a andz = 0, and hence calculate its volume V .6.2 Evaluate the volume integral of x 2 + y 2 + z 2 over the rectangular parallelepipedbounded by the six surfaces x = ±a, y = ±b and z = ±c.6.3 Find the volume integral of x 2 y over the tetrahedral volume bounded by theplanes x =0,y =0,z =0,andx + y + z =1.6.4 Evaluate the surface integral of f(x, y) over the rectangle 0 ≤ x ≤ a, 0≤ y ≤ bfor the functionsx(a) f(x, y) =x 2 + y , (b) f(x, y) =(b − y + 2 x)−3/2 .6.5 Calculate the volume of an ellipsoid as follows:(a) Prove that the area of the ellipsex 2a + y22 b =1 2is πab.(b) Use this result to obtain an expression for the volume of a slice of thicknessdz of the ellipsoidx 2a + y22 b + z22 c =1. 2Hence show that the volume of the ellipsoid is 4πabc/3.207

MULTIPLE INTEGRALS6.6 The function(Ψ(r) =A 2 − Zr )e −Zr/2aagives the form of the quantum-mechanical wavefunction representing the electronin a hydrogen-like atom of atomic number Z, when the electron is in its firstallowed spherically symmetric excited state. Here r is the usual spherical polarcoordinate, but, because of the spherical symmetry, the coordinates θ and φ donot appear explicitly in Ψ. Determine the value that A (assumed real) must haveif the wavefunction is to be correctly normalised, i.e. if the volume integral of|Ψ| 2 over all space is to be equal to unity.6.7 In quantum mechanics the electron in a hydrogen atom in some particular stateis described by a wavefunction Ψ, which is such that |Ψ| 2 dV is the probability offinding the electron in the infinitesimal volume dV . In spherical polar coordinatesΨ=Ψ(r, θ, φ) anddV = r 2 sin θdrdθdφ. Two such states are described by( ) 1/2 ( ) 3/2 1 1Ψ 1 =2e −r/a 0,4π a 0( ) 1/2 ( ) 3/2 31Ψ 2 = − sin θe iφ re −r/2a 0√ .8π2a 0 a 0 3(a) Show that each Ψ i is normalised, i.e. the integral over all space ∫ |Ψ| 2 dV isequal to unity – physically, this means that the electron must be somewhere.(b) The (so-called) dipole matrix element between the states 1 and 2 is given bythe integral∫p x = Ψ ∗ 1qr sin θ cos φ Ψ 2 dV ,where q is the charge on the electron. Prove that p x has the value −2 7 qa 0 /3 5 .6.8 A planar figure is formed from uniform wire and consists of two equal semicirculararcs, each with its own closing diameter, joined so as to form a letter ‘B’. Thefigure is freely suspended from its top left-hand corner. Show that the straightedge of the figure makes an angle θ with the vertical given by tan θ =(2+π) −1 .6.9 A certain torus has a circular vertical cross-section of radius a centredonahorizontal circle of radius c (> a).(a) Find the volume V and surface area A of the torus, and show that they canbe written asV = π24 (r2 o − ri 2 )(r o − r i ), A = π 2 (ro 2 − ri 2 ),where r o and r i are, respectively, the outer and inner radii of the torus.(b) Show that a vertical circular cylinder of radius c, coaxial with the torus,divides A in the ratioπc +2a : πc − 2a.6.10 A thin uniform circular disc has mass M and radius a.(a) Prove that its moment of inertia about an axis perpendicular to its planeand passing through its centre is 1 2 Ma2 .(b) Prove that the moment of inertia of the same disc about a diameter is 1 4 Ma2 .208

6.5 EXERCISESThis is an example of the general result for planar bodies that the moment ofinertia of the body about an axis perpendicular to the plane is equal to the sumof the moments of inertia about two perpendicular axes lying in the plane; in anobvious notation∫ ∫I z = r 2 dm =∫(x 2 + y 2 ) dm =∫x 2 dm +y 2 dm = I y + I x .6.11 In some applications in mechanics the moment of inertia of a body about asingle point (as opposed to about an axis) is needed. The moment of inertia, I,about the origin of a uniform solid body of density ρ is given by the volumeintegral∫I = (x 2 + y 2 + z 2 )ρdV.VShow that the moment of inertia of a right circular cylinder of radius a, length2b and mass M about its centre is( a2M2 + b236.12 The shape of an axially symmetric hard-boiled egg, of uniform density ρ 0 ,isgiven in spherical polar coordinates by r = a(2 − cos θ), where θ is measuredfrom the axis of symmetry.(a) Prove that the mass M of the egg is M = 40 πρ 3 0a 3 .(b) Prove that the egg’s moment of inertia about its axis of symmetry is 342175 Ma2 .6.13 In spherical polar coordinates r, θ, φ the element of volume for a body thatis symmetrical about the polar axis is dV =2πr 2 sin θdrdθ, whilst its elementof surface area is 2πr sin θ[(dr) 2 + r 2 (dθ) 2 ] 1/2 . A particular surface is defined byr =2a cos θ, wherea is a constant and 0 ≤ θ ≤ π/2. Find its total surface areaand the volume it encloses, and hence identify the surface.6.14 By expressing both the integrand and the surface element in spherical polarcoordinates, show that the surface integral∫x 2x 2 + y dS 2over the surface x 2 + y 2 = z 2 ,0≤ z ≤ 1, has the value π/ √ 2.6.15 By transforming to cylindrical polar coordinates, evaluate the integral∫ ∫ ∫I = ln(x 2 + y 2 ) dx dy dzover the interior of the conical region x 2 + y 2 ≤ z 2 ,0≤ z ≤ 1.6.16 Sketch the two families of curvesy 2 =4u(u − x), y 2 =4v(v + x),where u and v are parameters.By transforming to the uv-plane, evaluate the integral of y/(x 2 + y 2 ) 1/2 overthe part of the quadrant x>0, y>0 that is bounded by the lines x =0,y =0and the curve y 2 =4a(a − x).6.17 By making two successive simple changes of variables, evaluate∫ ∫ ∫I = x 2 dx dy dz).209

MULTIPLE INTEGRALSover the ellipsoidal regionx 2a + y22 b + z22 c ≤ 1. 26.18 Sketch the domain of integration for the integralI =∫ 1 ∫ 1/y0x=yy 3x exp[y2 (x 2 + x −2 )] dx dyand characterise its boundaries in terms of new variables u = xy and v = y/x.Show that the Jacobian for the change from (x, y) to(u, v) isequalto(2v) −1 ,andhence evaluate I.6.19 Sketch the part of the region 0 ≤ x, 0≤ y ≤ π/2 that is bounded by the curvesx =0,y =0,sinhx cos y =1andcoshx sin y = 1. By making a suitable changeof variables, evaluate the integral∫ ∫I = (sinh 2 x +cos 2 y)sinh2x sin 2ydxdyover the bounded subregion.6.20 Define a coordinate system u, v whose origin coincides with that of the usualx, y system and whose u-axis coincides with the x-axis, whilst the v-axis makesan angle α with it. By considering the integral I = ∫ exp(−r 2 ) dA, wherer is theradial distance from the origin, over the area defined by 0 ≤ u

6.6 HINTS AND ANSWERS(a) Let R be a real positive number and define K m byK m =∫ R−R(R 2 − x 2) mdx.Show, using integration by parts, that K m satisfies the recurrence relation(2m +1)K m =2mR 2 K m−1 .(b) For integer n, define I n = K n and J n = K n+1/2 . Evaluate I 0 and J 0 directlyand hence prove that(c)I n = 22n+1 (n!) 2 R 2n+1and J n =(2n +1)!A sequence of functions V n (R) is defined byV 0 (R) =1,∫ R (√ )V n (R) = V n−1 R2 − x 2−Rπ(2n +1)!R2n+22 2n+1 n!(n +1)! .dx, n ≥ 1.Prove by induction thatV 2n (R) = πn R 2n, V 2n+1 (R) = πn 2 2n+1 n!R 2n+1.n!(2n +1)!(d) For interest,(i) show that V 2n+2 (1)

7Vector algebraThis chapter introduces space vectors and their manipulation. Firstly we deal withthe description and algebra of vectors, then we consider how vectors may be usedto describe lines and planes and finally we look at the practical use of vectors infinding distances. Much use of vectors will be made in subsequent chapters; thischapter gives only some basic rules.7.1 Scalars and vectorsThe simplest kind of physical quantity is one that can be completely specified byits magnitude, a single number, together with the units in which it is measured.Such a quantity is called a scalar and examples include temperature, time anddensity.A vector is a quantity that requires both a magnitude (≥ 0) and a direction inspace to specify it completely; we may think of it as an arrow in space. A familiarexample is force, which has a magnitude (strength) measured in newtons and adirection of application. The large number of vectors that are used to describethe physical world include velocity, displacement, momentum and electric field.Vectors are also used to describe quantities such as angular momentum andsurface elements (a surface element has an area and a direction defined by thenormal to its tangent plane); in such cases their definitions may seem somewhatarbitrary (though in fact they are standard) and not as physically intuitive as forvectors such as force. A vector is denoted by bold type, the convention of thisbook, or by underlining, the latter being much used in handwritten work.This chapter considers basic vector algebra and illustrates just how powerfulvector analysis can be. All the techniques are presented for three-dimensionalspace but most can be readily extended to more dimensions.Throughout the book we will represent a vector in diagrams as a line togetherwith an arrowhead. We will make no distinction between an arrowhead at the212

7.2 ADDITION AND SUBTRACTION OF VECTORSabb + aa + bbaFigure 7.1 Addition of two vectors showing the commutation relation. Wemake no distinction between an arrowhead at the end of the line and onealong the line’s length, but rather use that which gives the clearer diagram.end of the line and one along the line’s length but, rather, use that which gives theclearer diagram. Furthermore, even though we are considering three-dimensionalvectors, we have to draw them in the plane of the paper. It should not be assumedthat vectors drawn thus are coplanar, unless this is explicitly stated.7.2 Addition and subtraction of vectorsThe resultant or vector sum of two displacement vectors is the displacement vectorthat results from performing first one and then the other displacement, as shownin figure 7.1; this process is known as vector addition. However, the principleof addition has physical meaning for vector quantities other than displacements;for example, if two forces act on the same body then the resultant force actingon the body is the vector sum of the two. The addition of vectors only makesphysical sense if they are of a like kind, for example if they are both forcesacting in three dimensions. It may be seen from figure 7.1 that vector addition iscommutative, i.e.a + b = b + a. (7.1)The generalisation of this procedure to the addition of three (or more) vectors isclear and leads to the associativity property of addition (see figure 7.2), e.g.a +(b + c) =(a + b)+c. (7.2)Thus, it is immaterial in what order any number of vectors are added.The subtraction of two vectors is very similar to their addition (see figure 7.3),that is,a − b = a +(−b)where −b is a vector of equal magnitude but exactly opposite direction to vector b.213

VECTOR ALGEBRAabcb + cbcab + ca +(b + c)aba + ba + b c(a + b)+cFigure 7.2Addition of three vectors showing the associativity relation.a−ba − babFigure 7.3Subtraction of two vectors.The subtraction of two equal vectors yields the zero vector, 0, which has zeromagnitude and no associated direction.7.3 Multiplication by a scalarMultiplication of a vector by a scalar (not to be confused with the ‘scalarproduct’, to be discussed in subsection 7.6.1) gives a vector in the same directionas the original but of a proportional magnitude. This can be seen in figure 7.4.The scalar may be positive, negative or zero. It can also be complex in someapplications. Clearly, when the scalar is negative we obtain a vector pointingin the opposite direction to the original vector. Multiplication by a scalar isassociative, commutative and distributive over addition. These properties may besummarised for arbitrary vectors a and b and arbitrary scalars λ and µ by(λµ)a = λ(µa) =µ(λa), (7.3)λ(a + b) =λa + λb, (7.4)(λ + µ)a = λa + µa. (7.5)214

7.3 MULTIPLICATION BY A SCALARλ aaFigure 7.4Scalar multiplication of a vector (for λ>1).BµbpPλAOaFigure 7.5 An illustration of the ratio theorem. The point P divides the linesegment AB in the ratio λ : µ.Having defined the operations of addition, subtraction and multiplication by ascalar, we can now use vectors to solve simple problems in geometry.◮A point P divides a line segment AB in the ratio λ : µ (see figure 7.5). If the positionvectors of the points A and B are a and b, respectively, find the position vector of thepoint P .As is conventional for vector geometry problems, we denote the vector from the point Ato the point B by AB. If the position vectors of the points A and B, relative to some originO, area and b, it should be clear that AB = b − a.Now, from figure 7.5 we see that one possible way of reaching the point P from O isfirst to go from O to A and to go along the line AB for a distance equal to the the fractionλ/(λ + µ) of its total length. We may express this in terms of vectors asOP = p = a +λλ + µ AB= a + λ (b − a)=λ + µ(1 − λλ + µ= µλ + µ a +)a +λλ + µ bλ b, (7.6)λ + µwhich expresses the position vector of the point P in terms of those of A and B. Wewould,of course, obtain the same result by considering the path from O to B and then to P . ◭215

VECTOR ALGEBRACEAGFDacBbOFigure 7.6 The centroid of a triangle. The triangle is defined by the points A,B and C that have position vectors a, b and c. The broken lines CD, BE, AFconnect the vertices of the triangle to the mid-points of the opposite sides;these lines intersect at the centroid G of the triangle.Result (7.6) is a version of the ratio theorem and we may use it in solving morecomplicated problems.◮The vertices of triangle ABC have position vectors a, b and c relative to some origin O(see figure 7.6). Find the position vector of the centroid G of the triangle.From figure 7.6, the points D and E bisect the lines AB and AC respectively. Thus fromthe ratio theorem (7.6), with λ = µ =1/2, the position vectors of D and E relative to theorigin ared = 1 a + 1 b,2 2e = 1 a + 1 c.2 2Using the ratio theorem again, we may write the position vector of a general point on theline CD that divides the line in the ratio λ :(1− λ) asr =(1− λ)c + λd,=(1− λ)c + 1 λ(a + b), (7.7)2where we have expressed d in terms of a and b. Similarly, the position vector of a generalpoint on the line BE can be expressed asr =(1− µ)b + µe,=(1− µ)b + 1 µ(a + c). (7.8)2Thus, at the intersection of the lines CD and BE we require, from (7.7), (7.8),(1 − λ)c + 1 λ(a + b) =(1− µ)b + 1 µ(a + c).2 2By equating the coefficents of the vectors a, b, c we find1λ = µ, λ =1− µ, 1 − λ = 1 µ.2 2216

7.4 BASIS VECTORS AND COMPONENTSThese equations are consistent and have the solution λ = µ =2/3. Substituting thesevalues into either (7.7) or (7.8) we find that the position vector of the centroid G is givenbyg = 1 (a + b + c). ◭37.4 Basis vectors and componentsGiven any three different vectors e 1 , e 2 and e 3 , which do not all lie in a plane,it is possible, in a three-dimensional space, to write any other vector in terms ofscalar multiples of them:a = a 1 e 1 + a 2 e 2 + a 3 e 3 . (7.9)The three vectors e 1 , e 2 and e 3 are said to form a basis (for the three-dimensionalspace); the scalars a 1 , a 2 and a 3 , which may be positive, negative or zero, arecalled the components of the vector a with respect to this basis. We say that thevector has been resolved into components.Most often we shall use basis vectors that are mutually perpendicular, for easeof manipulation, though this is not necessary. In general, a basis set must(i) have as many basis vectors as the number of dimensions (in more formallanguage, the basis vectors must span the space) and(ii) be such that no basis vector may be described as a sum of the others, or,more formally, the basis vectors must be linearly independent. Putting thismathematically, in N dimensions, we requirec 1 e 1 + c 2 e 2 + ···+ c N e N ≠ 0,for any set of coefficients c 1 ,c 2 ,...,c N except c 1 = c 2 = ···= c N =0.In this chapter we will only consider vectors in three dimensions; higher dimensionalitycan be achieved by simple extension.If we wish to label points in space using a Cartesian coordinate system (x, y, z),we may introduce the unit vectors i, j and k, which point along the positive x-,y- andz- axes respectively. A vector a may then be written as a sum of threevectors, each parallel to a different coordinate axis:a = a x i + a y j + a z k. (7.10)A vector in three-dimensional space thus requires three components to describefully both its direction and its magnitude. A displacement in space may bethought of as the sum of displacements along the x-, y- andz- directions (seefigure 7.7). For brevity, the components of a vector a with respect to a particularcoordinate system are sometimes written in the form (a x ,a y ,a z ). Note that the217

VECTOR ALGEBRAkaja z ka y ja x iiFigure 7.7 A Cartesian basis set. The vector a is the sum of a x i, a y j and a z k.basis vectors i, j and k may themselves be represented by (1, 0, 0), (0, 1, 0) and(0, 0, 1) respectively.We can consider the addition and subtraction of vectors in terms of theircomponents. The sum of two vectors a and b is found by simply adding theircomponents, i.e.a + b = a x i + a y j + a z k + b x i + b y j + b z k=(a x + b x )i +(a y + b y )j +(a z + b z )k, (7.11)and their difference by subtracting them,a − b = a x i + a y j + a z k − (b x i + b y j + b z k)=(a x − b x )i +(a y − b y )j +(a z − b z )k. (7.12)◮Two particles have velocities v 1 = i +3j +6k and v 2 = i − 2k, respectively. Find thevelocity u of the second particle relative to the first.The required relative velocity is given byu = v 2 − v 1 =(1− 1)i +(0− 3)j +(−2 − 6)k= −3j − 8k. ◭7.5 Magnitude of a vectorThe magnitude of the vector a is denoted by |a| or a. In terms of its componentsin three-dimensional Cartesian coordinates, the magnitude of a is given by√a ≡|a| = a 2 x + a 2 y + a 2 z. (7.13)Hence, the magnitude of a vector is a measure of its length. Such an analogy isuseful for displacement vectors but magnitude is better described, for example, by‘strength’ for vectors such as force or by ‘speed’ for velocity vectors. For instance,218

7.6 MULTIPLICATION OF VECTORSbOθ b cos θaFigure 7.8 The projection of b onto the direction of a is b cos θ. The scalarproduct of a and b is ab cos θ.in the previous example, the speed of the second particle relative to the first isgiven byu = |u| = √ (−3) 2 +(−8) 2 = √ 73.A vector whose magnitude equals unity is called a unit vector. The unit vectorin the direction a is usually notated â and may be evaluated asâ = a|a| . (7.14)The unit vector is a useful concept because a vector written as λâ then has magnitudeλ and direction â. Thus magnitude and direction are explicitly separated.7.6 Multiplication of vectorsWe have already considered multiplying a vector by a scalar. Now we considerthe concept of multiplying one vector by another vector. It is not immediatelyobvious what the product of two vectors represents and in fact two productsare commonly defined, the scalar product and the vector product. As their namesimply, the scalar product of two vectors is just a number, whereas the vectorproduct is itself a vector. Although neither the scalar nor the vector productis what we might normally think of as a product, their use is widespread andnumerous examples will be described elsewhere in this book.7.6.1 Scalar productThe scalar product (or dot product) of two vectors a and b is denoted by a · band is given bya · b ≡|a||b| cos θ, 0 ≤ θ ≤ π, (7.15)where θ is the angle between the two vectors, placed ‘tail to tail’ or ‘head to head’.Thus, the value of the scalar product a · b equals the magnitude of a multipliedby the projection of b onto a (see figure 7.8).219

VECTOR ALGEBRAFrom (7.15) we see that the scalar product has the particularly useful propertythata · b = 0 (7.16)is a necessary and sufficient condition for a to be perpendicular to b (unless eitherof them is zero). It should be noted in particular that the Cartesian basis vectorsi, j and k, being mutually orthogonal unit vectors, satisfy the equationsi · i = j · j = k · k =1, (7.17)i · j = j · k = k · i =0. (7.18)Examples of scalar products arise naturally throughout physics and in particularin connection with energy. Perhaps the simplest is the work done F · r inmoving the point of application of a constant force F through a displacement r;notice that, as expected, if the displacement is perpendicular to the direction ofthe force then F · r = 0 and no work is done. A second simple example is affordedby the potential energy −m · B of a magnetic dipole, represented in strength andorientation by a vector m, placed in an external magnetic field B.As the name implies, the scalar product has a magnitude but no direction. Thescalar product is commutative and distributive over addition:a · b = b · a (7.19)a · (b + c) =a · b + a · c. (7.20)◮Four non-coplanar points A, B, C, D are positioned such that the line AD is perpendicularto BC and BD is perpendicular to AC. Show that CD is perpendicular to AB.Denote the four position vectors by a, b, c, d. As none of the three pairs of lines actuallyintersect, it is difficult to indicate their orthogonality in the diagram we would normallydraw. However, the orthogonality can be expressed in vector form and we start by notingthat, since AD ⊥ BC, it follows from (7.16) that(d − a) · (c − b) =0.Similarly, since BD ⊥ AC,(d − b) · (c − a) =0.Combining these two equations we find(d − a) · (c − b) =(d − b) · (c − a),which, on mutliplying out the parentheses, givesd · c − a · c − d · b + a · b = d · c − b · c − d · a + b · a.Cancelling terms that appear on both sides and rearranging yieldsd · b − d · a − c · b + c · a =0,which simplifies to give(d − c) · (b − a) =0.From (7.16), we see that this implies that CD is perpendicular to AB. ◭220

7.6 MULTIPLICATION OF VECTORSIf we introduce a set of basis vectors that are mutually orthogonal, such as i, j,k, we can write the components of a vector a, with respect to that basis, in termsof the scalar product of a with each of the basis vectors, i.e. a x = a·i, a y = a·j anda z = a · k. In terms of the components a x , a y and a z the scalar product is given bya · b =(a x i + a y j + a z k) · (b x i + b y j + b z k)=a x b x + a y b y + a z b z , (7.21)where the cross terms such as a x i · b y j are zero because the basis vectors aremutually perpendicular; see equation (7.18). It should be clear from (7.15) thatthe value of a · b has a geometrical definition and that this value is independentof the actual basis vectors used.◮Find the angle between the vectors a = i +2j +3k and b =2i +3j +4k.From (7.15) the cosine of the angle θ between a and b is given bycos θ = a · b|a||b| .From (7.21) the scalar product a · b has the valuea · b =1× 2+2× 3+3× 4=20,and from (7.13) the lengths of the vectors areThus,|a| = √ 1 2 +2 2 +3 2 = √ 14 and |b| = √ 2 2 +3 2 +4 2 = √ 29.cos θ =20√14√29≈ 0.9926 ⇒ θ =0.12 rad. ◭We can see from the expressions (7.15) and (7.21) for the scalar product that ifθ is the angle between a and b thencos θ = a x b xa b + a y b ya b + a z b za bwhere a x /a, a y /a and a z /a are called the direction cosines of a, since they give thecosine of the angle made by a with each of the basis vectors. Similarly b x /b, b y /band b z /b are the direction cosines of b.If we take the scalar product of any vector a with itself then clearly θ = 0 andfrom (7.15) we havea · a = |a| 2 .Thus the magnitude of a can be written in a coordinate-independent form as|a| = √ a · a.Finally, we note that the scalar product may be extended to vectors withcomplex components if it is redefined asa · b = a ∗ xb x + a ∗ yb y + a ∗ zb z ,where the asterisk represents the operation of complex conjugation. To accom-221

VECTOR ALGEBRAa × bθbaFigure 7.9set.The vector product. The vectors a, b and a×b form a right-handedmodate this extension the commutation property (7.19) must be modified toreada · b =(b · a) ∗ . (7.22)In particular it should be noted that (λa) · b = λ ∗ a · b, whereas a · (λb) =λa · b.However, the magnitude of a complex vector is still given by |a| = √ a · a, sincea · a is always real.7.6.2 Vector productThe vector product (or cross product) of two vectors a and b is denoted by a × band is defined to be a vector of magnitude |a||b| sin θ in a direction perpendicularto both a and b;|a × b| = |a||b| sin θ.The direction is found by ‘rotating’ a into b through the smallest possible angle.The sense of rotation is that of a right-handed screw that moves forward in thedirection a × b (see figure 7.9). Again, θ is the angle between the two vectorsplaced ‘tail to tail’ or ‘head to head’. With this definition a, b and a × b form aright-handed set. A more directly usable description of the relative directions ina vector product is provided by a right hand whose first two fingers and thumbare held to be as nearly mutually perpendicular as possible. If the first finger ispointed in the direction of the first vector and the second finger in the directionof the second vector, then the thumb gives the direction of the vector product.The vector product is distributive over addition, but anticommutative and nonassociative:(a + b) × c =(a × c)+(b × c), (7.23)b × a = −(a × b), (7.24)(a × b) × c ≠ a × (b × c). (7.25)222

7.6 MULTIPLICATION OF VECTORSPFθRrOFigure 7.10 The moment of the force F about O is r×F. The cross representsthe direction of r × F, which is perpendicularly into the plane of the paper.From its definition, we see that the vector product has the very useful propertythat if a × b = 0 then a is parallel or antiparallel to b (unless either of them iszero). We also note thata × a = 0. (7.26)◮Show that if a = b + λc, for some scalar λ, thena × c = b × c.From (7.23) we havea × c =(b + λc) × c = b × c + λc × c.However, from (7.26), c × c = 0 and soa × c = b × c. (7.27)We note in passing that the fact that (7.27) is satisfied does not imply that a = b. ◭An example of the use of the vector product is that of finding the area, A, ofa parallelogram with sides a and b, using the formulaA = |a × b|. (7.28)Another example is afforded by considering a force F acting through a point R,whose vector position relative to the origin O is r (see figure 7.10). Its momentor torque about O is the strength of the force times the perpendicular distanceOP, which numerically is just Fr sin θ, i.e. the magnitude of r × F. Furthermore,the sense of the moment is clockwise about an axis through O that pointsperpendicularly into the plane of the paper (the axis is represented by a crossin the figure). Thus the moment is completely represented by the vector r × F,in both magnitude and spatial sense. It should be noted that the same vectorproduct is obtained wherever the point R is chosen, so long as it lies on the lineof action of F.Similarly, if a solid body is rotating about some axis that passes through theorigin, with an angular velocity ω then we can describe this rotation by a vectorω that has magnitude ω and points along the axis of rotation. The direction of ω223

VECTOR ALGEBRAis the forward direction of a right-handed screw rotating in the same sense as thebody. The velocity of any point in the body with position vector r is then givenby v = ω × r.Since the basis vectors i, j, k are mutually perpendicular unit vectors, forminga right-handed set, their vector products are easily seen to bei × i = j × j = k × k = 0, (7.29)i × j = −j × i = k, (7.30)j × k = −k × j = i, (7.31)k × i = −i × k = j. (7.32)Using these relations, it is straightforward to show that the vector product of twogeneral vectors a and b is given in terms of their components with respect to thebasis set i, j, k, bya × b =(a y b z − a z b y )i +(a z b x − a x b z )j +(a x b y − a y b x )k. (7.33)For the reader who is familiar with determinants (see chapter 8), we record thatthis can also be written as∣ i j k ∣∣∣∣∣a × b =a x a y a z .∣ b x b y b zThat the cross product a × b is perpendicular to both a and b can be verifiedin component form by forming its dot products with each of the two vectors andshowing that it is zero in both cases.◮Find the area A of the parallelogram with sides a = i +2j +3k and b =4i +5j +6k.The vector product a × b is given in component form bya × b =(2× 6 − 3 × 5)i +(3× 4 − 1 × 6)j +(1× 5 − 2 × 4)k= −3i +6j − 3k.Thus the area of the parallelogram isA = |a × b| = √ (−3) 2 +6 2 +(−3) 2 = √ 54. ◭7.6.3 Scalar triple productNow that we have defined the scalar and vector products, we can extend ourdiscussion to define products of three vectors. Again, there are two possibilities,the scalar triple product and the vector triple product.224

7.6 MULTIPLICATION OF VECTORSvPOφθcbaFigure 7.11The scalar triple product gives the volume of a parallelepiped.The scalar triple product is denoted by[a, b, c] ≡ a · (b × c)and, as its name suggests, it is just a number. It is most simply interpreted as thevolume of a parallelepiped whose edges are given by a, b and c (see figure 7.11).The vector v = a × b is perpendicular to the base of the solid and has magnitudev = ab sin θ, i.e. the area of the base. Further, v · c = vc cos φ. Thus, since c cos φ= OP is the vertical height of the parallelepiped, it is clear that (a × b) · c =areaof the base × perpendicular height = volume. It follows that, if the vectors a, band c are coplanar, a · (b × c) =0.Expressed in terms of the components of each vector with respect to theCartesian basis set i, j, k the scalar triple product isa · (b × c) =a x (b y c z − b z c y )+a y (b z c x − b x c z )+a z (b x c y − b y c x ),(7.34)which can also be written as a determinant:∣ a x a y a z ∣∣∣∣∣a · (b × c) =b x b y b z .∣ c x c y c zBy writing the vectors in component form, it can be shown thata · (b × c) =(a × b) · c,so that the dot and cross symbols can be interchanged without changing the result.More generally, the scalar triple product is unchanged under cyclic permutationof the vectors a, b, c. Other permutations simply give the negative of the originalscalar triple product. These results can be summarised by[a, b, c] =[b, c, a] =[c, a, b] =−[a, c, b] =−[b, a, c] =−[c, b, a]. (7.35)225

VECTOR ALGEBRA◮Find the volume V of the parallelepiped with sides a = i +2j +3k, b =4i +5j +6k andc =7i +8j +10k.We have already found that a × b = −3i +6j − 3k, in subsection 7.6.2. Hence the volumeof the parallelepiped is given byV = |a · (b × c)| = |(a × b) · c|= |(−3i +6j − 3k) · (7i +8j +10k)|= |(−3)(7) + (6)(8) + (−3)(10)| =3. ◭Another useful formula involving both the scalar and vector products is Lagrange’sidentity (see exercise 7.9), i.e.(a × b) · (c × d) ≡ (a · c)(b · d) − (a · d)(b · c). (7.36)7.6.4 Vector triple productBy the vector triple product of three vectors a, b, c we mean the vector a × (b × c).Clearly, a × (b × c) is perpendicular to a and lies in the plane of b and c and socan be expressed in terms of them (see (7.37) below). We note, from (7.25), thatthe vector triple product is not associative, i.e. a × (b × c) ≠(a × b) × c.Two useful formulae involving the vector triple product area × (b × c) =(a · c)b − (a · b)c, (7.37)(a × b) × c =(a · c)b − (b · c)a, (7.38)which may be derived by writing each vector in component form (see exercise 7.8).It can also be shown that for any three vectors a, b, c,a × (b × c)+b × (c × a)+c × (a × b) =0.7.7 Equations of lines, planes and spheresNow that we have described the basic algebra of vectors, we can apply the resultsto a variety of problems, the first of which is to find the equation of a line invector form.7.7.1 Equation of a lineConsider the line passing through the fixed point A with position vector a andhaving a direction b (see figure 7.12). It is clear that the position vector r of ageneral point R on the line can be written asr = a + λb, (7.39)226

7.7 EQUATIONS OF LINES, PLANES AND SPHERESRAbraOFigure 7.12 The equation of a line. The vector b is in the direction AR andλb is the vector from A to R.since R can be reached by starting from O, going along the translation vectora to the point A on the line and then adding some multiple λb of the vector b.Different values of λ give different points R on the line.Taking the components of (7.39), we see that the equation of the line can alsobewrittenintheformx − a xb x= y − a yb y= z − a zb z= constant. (7.40)Taking the vector product of (7.39) with b and remembering that b × b = 0 givesan alternative equation for the line(r − a) × b = 0.We may also find the equation of the line that passes through two fixed pointsA and C with position vectors a and c. SinceAC is given by c − a, the positionvector of a general point on the line isr = a + λ(c − a).7.7.2 Equation of a planeThe equation of a plane through a point A with position vector a and perpendicularto a unit position vector ˆn (see figure 7.13) is(r − a) · ˆn =0. (7.41)This follows since the vector joining A to a general point R with position vectorr is r − a; r will lie in the plane if this vector is perpendicular to the normal tothe plane. Rewriting (7.41) as r · ˆn = a · ˆn, we see that the equation of the planemay also be expressed in the form r · ˆn = d, or in component form aslx + my + nz = d, (7.42)227

VECTOR ALGEBRAˆnARadrOFigure 7.13 The equation of the plane is (r − a) · ˆn =0.where the unit normal to the plane is ˆn = li + mj + nk and d = a · ˆn is theperpendicular distance of the plane from the origin.The equation of a plane containing points a, b and c isr = a + λ(b − a)+µ(c − a).This is apparent because starting from the point a in the plane, all other pointsmay be reached by moving a distance along each of two (non-parallel) directionsin the plane. Two such directions are given by b − a and c − a. It can be shownthat the equation of this plane may also be written in the more symmetrical formwhere α + β + γ =1.r = αa + βb + γc,◮Find the direction of the line of intersection of the two planes x +3y − z = 5 and2x − 2y +4z =3.The two planes have normal vectors n 1 = i +3j − k and n 2 =2i − 2j +4k. Itisclearthat these are not parallel vectors and so the planes must intersect along some line. Thedirection p of this line must be parallel to both planes and hence perpendicular to bothnormals. Thereforep = n 1 × n 2= [(3)(4) − (−2)(−1)] i +[(−1)(2) − (1)(4)] j + [(1)(−2) − (3)(2)] k=10i − 6j − 8k. ◭7.7.3 Equation of a sphereClearly, the defining property of a sphere is that all points on it are equidistantfrom a fixed point in space and that the common distance is equal to the radius228

7.8 USING VECTORS TO FIND DISTANCESof the sphere. This is easily expressed in vector notation as|r − c| 2 =(r − c) · (r − c) =a 2 , (7.43)where c is the position vector of the centre of the sphere and a is its radius.◮Find the radius ρ of the circle that is the intersection of the plane ˆn · r = p and the sphereof radius a centred on the point with position vector c.The equation of the sphere is|r − c| 2 = a 2 , (7.44)and that of the circle of intersection is|r − b| 2 = ρ 2 , (7.45)where r is restricted to lie in the plane and b is the position of the circle’s centre.As b lies on the plane whose normal is ˆn, thevectorb − c must be parallel to ˆn, i.e.b − c = λˆn for some λ. Further, by Pythagoras, we must have ρ 2 + |b − c| 2 = a 2 . Thusλ 2 = a 2 − ρ 2 .Writing b = c + √ a 2 − ρ 2 ˆn and substituting in (7.45) gives(r 2 − 2r · c + √ )a 2 − ρ 2 ˆn + c 2 +2(c · ˆn) √ a 2 − ρ 2 + a 2 − ρ 2 = ρ 2 ,whilst, on expansion, (7.44) becomesr 2 − 2r · c + c 2 = a 2 .Subtracting these last two equations, using ˆn · r = p and simplifying yieldsp − c · ˆn = √ a 2 − ρ 2 .On rearrangement, this gives ρ as √ a 2 − (p − c · ˆn) 2 , which places obvious geometricalconstraints on the values a, c, ˆn and p can take if a real intersection between the sphereand the plane is to occur. ◭7.8 Using vectors to find distancesThis section deals with the practical application of vectors to finding distances.Some of these problems are extremely cumbersome in component form, but theyall reduce to neat solutions when general vectors, with no explicit basis set,are used. These examples show the power of vectors in simplifying geometricalproblems.7.8.1 Distance from a point to a lineFigure 7.14 shows a line having direction b that passes through a point A whoseposition vector is a. To find the minimum distance d of the line from a point Pwhose position vector is p, we must solve the right-angled triangle shown. We seethat d = |p − a| sin θ; so, from the definition of the vector product, it follows thatd = |(p − a) × ˆb|.229

VECTOR ALGEBRAPp − apdAbθaOFigure 7.14The minimum distance from a point to a line.◮Find the minimum distance from the point P with coordinates (1, 2, 1) to the line r = a+λb,where a = i + j + k and b =2i − j +3k.Comparison with (7.39) shows that the line passes through the point (1, 1, 1) and hasdirection 2i − j +3k. The unit vector in this direction isˆb = √ 1 (2i − j +3k).14The position vector of P is p = i +2j + k and we find(p − a) × ˆb = √ 1 [ j × (2i − 3j +3k)]14= √ 1 (3i − 2k).14Thus the minimum distance from the line to the point P is d = √ 13/14. ◭7.8.2 Distance from a point to a planeThe minimum distance d from a point P whose position vector is p to the planedefined by (r − a) · ˆn = 0 may be deduced by finding any vector from P to theplane and then determining its component in the normal direction. This is shownin figure 7.15. Consider the vector a − p, which is a particular vector from P tothe plane. Its component normal to the plane, and hence its distance from theplane, is given byd =(a − p) · ˆn, (7.46)where the sign of d depends on which side of the plane P is situated.230

7.8 USING VECTORS TO FIND DISTANCESPpˆndaOFigure 7.15The minimum distance d from a point to a plane.◮Find the distance from the point P with coordinates (1, 2, 3) to the plane that contains thepoints A, B and C having coordinates (0, 1, 0), (2, 3, 1) and (5, 7, 2).Let us denote the position vectors of the points A, B, C by a, b, c. Two vectors in theplane areb − a =2i +2j + k and c − a =5i +6j +2k,and hence a vector normal to the plane isn =(2i +2j + k) × (5i +6j +2k) =−2i + j +2k,and its unit normal isˆn = n|n| = 1 (−2i + j +2k).3Denoting the position vector of P by p, the minimum distance from the plane to P isgiven byd =(a − p) · ˆn=(−i − j − 3k) · 1 (−2i + j +2k)3= 2 − 1 − 2 = − 5 .3 3 3If we take P to be the origin O, then we find d = 1 , i.e. a positive quantity. It follows from3this that the original point P with coordinates (1, 2, 3), for which d was negative, is on theopposite side of the plane from the origin. ◭7.8.3 Distance from a line to a lineConsider two lines in the directions a and b, as shown in figure 7.16. Since a × bis by definition perpendicular to both a and b, the unit vector normal to boththese lines isˆn = a × b|a × b| .231

VECTOR ALGEBRAQbqˆnPpaOFigure 7.16The minimum distance from one line to another.If p and q are the position vectors of any two points P and Q on different linesthen the vector connecting them is p − q. Thus, the minimum distance d betweenthe lines is this vector’s component along the unit normal, i.e.d = |(p − q) · ˆn|.◮A line is inclined at equal angles to the x-, y- and z-axes and passes through the origin.Another line passes through the points (1, 2, 4) and (0, 0, 1). Find the minimum distancebetween the two lines.The first line is given byr 1 = λ(i + j + k),and the second byr 2 = k + µ(i +2j +3k).Hence a vector normal to both lines isn =(i + j + k) × (i +2j +3k) =i − 2j + k,and the unit normal isˆn = √ 1 (i − 2j + k).6A vector between the two lines is, for example, the one connecting the points (0, 0, 0)and (0, 0, 1), which is simply k. Thus it follows that the minimum distance between thetwo lines isd = √ 1 |k · (i − 2j + k)| = √ 1 . ◭6 67.8.4 Distance from a line to a planeLet us consider the line r = a + λb. This line will intersect any plane to which itis not parallel. Thus, if a plane has a normal ˆn then the minimum distance from232

7.9 RECIPROCAL VECTORSthe line to the plane is zero unlessin which case the distance, d, will bewhere r is any point in the plane.b · ˆn =0,d = |(a − r) · ˆn|,◮A line is given by r = a + λb, wherea = i +2j +3k and b =4i +5j +6k. Findthecoordinates of the point P at which the line intersects the planex +2y +3z =6.A vector normal to the plane isn = i +2j +3k,from which we find that b · n ≠ 0. Thus the line does indeed intersect the plane. To findthe point of intersection we merely substitute the x-, y- andz- values of a general pointon the line into the equation of the plane, obtaining1+4λ +2(2+5λ)+3(3+6λ) =6 ⇒ 14 + 32λ =6.This gives λ = − 1 , which we may substitute into the equation for the line to obtain4x =1− 1 (4) = 0, y =2− 1 (5) = 3 and z =3− 1 (6) = 3 . Thus the point of intersection is4 4 4 4 2(0, 3 4 , 3 2 ). ◭ 7.9 Reciprocal vectorsThe final section of this chapter introduces the concept of reciprocal vectors,which have particular uses in crystallography.The two sets of vectors a, b, c and a ′ , b ′ , c ′ are called reciprocal sets ifanda · a ′ = b · b ′ = c · c ′ = 1 (7.47)a ′ · b = a ′ · c = b ′ · a = b ′ · c = c ′ · a = c ′ · b =0. (7.48)It can be verified (see exercise 7.19) that the reciprocal vectors of a, b and c aregiven bya ′ =b × ca · (b × c) , (7.49)b ′ =c × aa · (b × c) , (7.50)c ′ =a × ba · (b × c) , (7.51)where a · (b × c) ≠ 0. In other words, reciprocal vectors only exist if a, b and c are233

VECTOR ALGEBRAnot coplanar. Moreover, if a, b and c are mutually orthogonal unit vectors thena ′ = a, b ′ = b and c ′ = c, so that the two systems of vectors are identical.◮Construct the reciprocal vectors of a =2i, b = j + k, c = i + k.First we evaluate the triple scalar product:a · (b × c) =2i · [(j + k) × (i + k)]=2i · (i + j − k) = 2.Now we find the reciprocal vectors:a ′ = 1 (j + k) × (i + k) = 1 (i + j − k),2 2b ′ = 1 (i + k) × 2i = j,2c ′ = 1 (2i) × (j + k) = −j + k.2It is easily verified that these reciprocal vectors satisfy their defining properties (7.47),(7.48). ◭We may also use the concept of reciprocal vectors to define the components of avector a with respect to basis vectors e 1 , e 2 , e 3 that are not mutually orthogonal.If the basis vectors are of unit length and mutually orthogonal, such as theCartesian basis vectors i, j, k, then (see the text preceeding (7.21)) the vector acanbewrittenintheforma =(a · i)i +(a · j)j +(a · k)k.If the basis is not orthonormal, however, then this is no longer true. Nevertheless,we may write the components of a with respect to a non-orthonormal basise 1 , e 2 , e 3 in terms of its reciprocal basis vectors e ′ 1 , e′ 2 , e′ 3 , which are defined as in(7.49)–(7.51). If we leta = a 1 e 1 + a 2 e 2 + a 3 e 3 ,then the scalar product a · e ′ 1 is given bya · e ′ 1 = a 1 e 1 · e ′ 1 + a 2 e 2 · e ′ 1 + a 3 e 3 · e ′ 1 = a 1 ,where we have used the relations (7.48). Similarly, a 2 = a·e ′ 2 and a 3 = a·e ′ 3 ;sonowa =(a · e ′ 1)e 1 +(a · e ′ 2)e 2 +(a · e ′ 3)e 3 . (7.52)7.10 Exercises7.1 Which of the following statements about general vectors a, b and c are true?(a) c · (a × b) =(b × a) · c.(b) a × (b × c) =(a × b) × c.(c) a × (b × c) =(a · c)b − (a · b)c.(d) d = λa + µb implies (a × b) · d =0.(e) a × c = b × c implies c · a − c · b = c|a − b|.(f) (a × b) × (c × b) =b[b · (c × a)].234

7.10 EXERCISES7.2 A unit cell of diamond is a cube of side A, with carbon atoms at each corner, atthe centre of each face and, in addition, at positions displaced by 1 A(i + j + k)4from each of those already mentioned; i, j, k are unit vectors along the cube axes.One corner of the cube is taken as the origin of coordinates. What are the vectorsjoining the atom at 1 A(i + j + k) to its four nearest neighbours? Determine the4angle between the carbon bonds in diamond.7.3 Identify the following surfaces:(a) |r| = k; (b)r · u = l; (c)r · u = m|r| for −1 ≤ m ≤ +1;(d) |r − (r · u)u| = n.Here k, l, m and n are fixed scalars and u is a fixed unit vector.7.4 Find the angle between the position vectors to the points (3, −4, 0) and (−2, 1, 0)and find the direction cosines of a vector perpendicular to both.7.5 A, B, C and D are the four corners, in order, of one face of a cube of side 2units. The opposite face has corners E,F,G and H, withAE, BF, CG and DHas parallel edges of the cube. The centre O ofthecubeistakenastheoriginand the x-, y- andz-axes are parallel to AD, AE and AB, respectively. Find thefollowing:(a) the angle between the face diagonal AF and the body diagonal AG;(b) the equation of the plane through B that is parallel to the plane CGE;(c) the perpendicular distance from the centre J of the face BCGF to the planeOCG;(d) the volume of the tetrahedron JOCG.7.6 Use vector methods to prove that the lines joining the mid-points of the oppositeedges of a tetrahedron OABC meet at a point and that this point bisects each ofthe lines.7.7 The edges OP, OQ and OR of a tetrahedron OPQR are vectors p, q and r,respectively, where p =2i +4j, q =2i − j +3k and r =4i − 2j +5k. Show thatOP is perpendicular to the plane containing OQR. Express the volume of thetetrahedron in terms of p, q and r and hence calculate the volume.7.8 Prove, by writing it out in component form, that(a × b) × c =(a · c)b − (b · c)a,and deduce the result, stated in equation (7.25), that the operation of formingthe vector product is non-associative.7.9 Prove Lagrange’s identity, i.e.(a × b) · (c × d) =(a · c)(b · d) − (a · d)(b · c).7.10 For four arbitrary vectors a, b, c and d, evaluate(a × b) × (c × d)in two different ways and so prove thata[b, c, d] − b[c, d, a]+c[d, a, b] − d[a, b, c] =0.Show that this reduces to the normal Cartesian representation of the vector d,i.e. d x i + d y j + d z k,ifa, b and c are taken as i, j and k, the Cartesian base vectors.7.11 Show that the points (1, 0, 1), (1, 1, 0) and (1, −3, 4) lie on a straight line. Give theequation of the line in the formr = a + λb.235

VECTOR ALGEBRA7.12 The plane P 1 contains the points A, B and C, which have position vectorsa = −3i +2j, b =7i +2j and c =2i +3j +2k, respectively. Plane P 2 passesthrough A and is orthogonal to the line BC, whilst plane P 3 passes throughB and is orthogonal to the line AC. Find the coordinates of r, the point ofintersection of the three planes.7.13 Two planes have non-parallel unit normals ˆn and ˆm and their closest distancesfrom the origin are λ and µ, respectively. Find the vector equation of their lineof intersection in the form r = νp + a.7.14 Two fixed points, A and B, in three-dimensional space have position vectors aand b. Identify the plane P given by(a − b) · r = 1 2 (a2 − b 2 ),where a and b are the magnitudes of a and b.Show also that the equation(a − r) · (b − r) =0describes a sphere S of radius |a − b|/2. Deduce that the intersection of P and Sis also the intersection of two spheres, centred on A and B, and each of radius|a − b|/ √ 2.7.15 Let O, A, B and C be four points with position vectors 0, a, b and c, and denoteby g = λa + µb + νc the position of the centre of the sphere on which they all lie.(a) Prove that λ, µ and ν simultaneously satisfy(a · a)λ +(a · b)µ +(a · c)ν = 1 2 a2and two other similar equations.(b) By making a change of origin, find the centre and radius of the sphere onwhich the points p =3i+j−2k, q =4i+3j−3k, r =7i−3k and s =6i+j−kall lie.7.16 The vectors a, b and c are coplanar and related byλa + µb + νc =0,where λ, µ, ν are not all zero. Show that the condition for the points with positionvectors αa, βb and γc to be collinear isλα + µ β + ν γ =0.7.17 Using vector methods:(a) Show that the line of intersection of the planes x +2y +3z = 0 and3x +2y + z = 0 is equally inclined to the x- andz-axes and makes an anglecos −1 (−2/ √ 6) with the y-axis.(b) Find the perpendicular distance between one corner of a unit cube and themajor diagonal not passing through it.7.18 Four points X i , i =1, 2, 3, 4, taken for simplicity as all lying within the octantx, y, z ≥ 0, have position vectors x i . Convince yourself that the direction ofvector x n lies within the sector of space defined by the directions of the otherthree vectors if[ ]xi · x jmin ,over j |x i ||x j |considered for i =1, 2, 3, 4 in turn, takes its maximum value for i = n, i.e.n equalsthat value of i for which the largest of the set of angles which x i makes withthe other vectors, is found to be the lowest. Determine whether any of the four236

7.10 EXERCISESabcdaFigure 7.17A face-centred cubic crystal.points with coordinatesX 1 =(3, 2, 2), X 2 =(2, 3, 1), X 3 =(2, 1, 3), X 4 =(3, 0, 3)lies within the tetrahedron defined by the origin and the other three points.7.19 The vectors a, b and c are not coplanar. The vectors a ′ , b ′ and c ′ are theassociated reciprocal vectors. Verify that the expressions (7.49)–(7.51) define a setof reciprocal vectors a ′ , b ′ and c ′ with the following properties:(a) a ′ · a = b ′ · b = c ′ · c =1;(b) a ′ · b = a ′ · c = b ′ · a etc = 0;(c) [a ′ , b ′ , c ′ ]=1/[a, b, c];(d) a =(b ′ × c ′ )/[a ′ , b ′ , c ′ ].7.20 Three non-coplanar vectors a, b and c, have as their respective reciprocal vectorsthe set a ′ , b ′ and c ′ . Show that the normal to the plane containing the pointsk −1 a, l −1 b and m −1 c is in the direction of the vector ka ′ + lb ′ + mc ′ .7.21 In a crystal with a face-centred cubic structure, the basic cell can be taken as acube of edge a with its centre at the origin of coordinates and its edges parallelto the Cartesian coordinate axes; atoms are sited at the eight corners and at thecentre of each face. However, other basic cells are possible. One is the rhomboidshown in figure 7.17, which has the three vectors b, c and d as edges.(a) Show that the volume of the rhomboid is one-quarter that of the cube.(b) Show that the angles between pairs of edges of the rhomboid are 60 ◦ and thatthe corresponding angles between pairs of edges of the rhomboid defined bythe reciprocal vectors to b, c, d are each 109.5 ◦ . (This rhomboid can be usedas the basic cell of a body-centred cubic structure, more easily visualised asa cube with an atom at each corner and one at its centre.)(c) In order to use the Bragg formula, 2d sin θ = nλ, for the scattering of X-raysby a crystal, it is necessary to know the perpendicular distance d betweensuccessive planes of atoms; for a given crystal structure, d has a particularvalue for each set of planes considered. For the face-centred cubic structurefind the distance between successive planes with normals in the k, i + j andi + j + k directions.237

VECTOR ALGEBRA7.22 In subsection 7.6.2 we showed how the moment or torque of a force about an axiscould be represented by a vector in the direction of the axis. The magnitude ofthe vector gives the size of the moment and the sign of the vector gives the sense.Similar representations can be used for angular velocities and angular momenta.(a) The magnitude of the angular momentum about the origin of a particle ofmass m moving with velocity v on a path that is a perpendicular distance dfrom the origin is given by m|v|d. Show that if r is the position of the particlethen the vector J = r × mv represents the angular momentum.(b) Now consider a rigid collection of particles (or a solid body) rotating aboutan axis through the origin, the angular velocity of the collection beingrepresented by ω.(i) Show that the velocity of the ith particle isv i = ω × r iand that the total angular momentum J isJ = ∑ im i [r 2 i ω − (r i · ω)r i ].(ii) Show further that the component of J along the axis of rotation canbe written as Iω, whereI, the moment of inertia of the collectionabout the axis or rotation, is given byI = ∑ im i ρ 2 i .Interpret ρ i geometrically.(iii) Prove that the total kinetic energy of the particles is 1 2 Iω2 .7.23 By proceeding as indicated below, prove the parallel axis theorem, whichstatesthat, for a body of mass M, the moment of inertia I about any axis is related tothe corresponding moment of inertia I 0 about a parallel axis that passes throughthe centre of mass of the body byI = I 0 + Ma 2 ⊥,where a ⊥ is the perpendicular distance between the two axes. Note that I 0 canbe written as∫(ˆn × r) · (ˆn × r) dm,where r is the vector position, relative to the centre of mass, of the infinitesimalmass dm and ˆn is a unit vector in the direction of the axis of rotation. Write asimilar expression for I in which r is replaced by r ′ = r − a, wherea is the vectorposition of any point on the axis to which I refers. Use Lagrange’s identity andthe fact that ∫ r dm = 0 (by the definition of the centre of mass) to establish theresult.7.24 Without carrying out any further integration, use the results of the previousexercise, the worked example in subsection 6.3.4 and exercise 6.10 to prove thatthe moment of inertia of a uniform rectangular lamina, of mass M and sides aand b, about an axis perpendicular to its plane and passing through the point(αa/2,βb/2), with −1 ≤ α, β ≤ 1, isM12 [a2 (1 + 3α 2 )+b 2 (1 + 3β 2 )].238

7.10 EXERCISESV 0 cos ωtV 4V 1V 2R 1 =50ΩI 2I 1I 3LC =10µFV 3R 2Figure 7.18 An oscillatory electric circuit. The power supply has angularfrequency ω =2πf = 400π s −1 .7.25 Define a set of (non-orthogonal) base vectors a = j + k, b = i + k and c = i + j.(a) Establish their reciprocal vectors and hence express the vectors p =3i−2j+k,q = i +4j and r = −2i + j + k in terms of the base vectors a, b and c.(b) Verify that the scalar product p · q has the same value, −5, when evaluatedusing either set of components.7.26 Systems that can be modelled as damped harmonic oscillators are widespread;pendulum clocks, car shock absorbers, tuning circuits in television sets and radios,and collective electron motions in plasmas and metals are just a few examples.In all these cases, one or more variables describing the system obey(s) anequation of the formẍ +2γẋ + ω0x 2 = P cos ωt,where ẋ = dx/dt, etc. and the inclusion of the factor 2 is conventional. In thesteady state (i.e. after the effects of any initial displacement or velocity have beendamped out) the solution of the equation takes the formx(t) =A cos(ωt + φ).By expressing each term in the form B cos(ωt+ɛ), and representing it by a vectorof magnitude B making an angle ɛ with the x-axis, draw a closed vector diagram,at t = 0, say, that is equivalent to the equation.(a) Convince yourself that whatever the value of ω (> 0) φ must be negative(−π

VECTOR ALGEBRAof vector plots for potential differences and currents (they could all be on thesame plot if suitable scales were chosen), determine all unknown currents andpotential differences and find values for the inductance of L and the resistanceof R 2 .[ Scales of 1 cm = 0.1 V for potential differences and 1 cm = 1 mA for currentsare convenient. ]7.11 Hints and answers7.1 (c), (d) and (e).7.3 (a) A sphere of radius k centred on the origin; (b) a plane with its normal in thedirection of u and at a distance l from the origin; (c) a cone with its axis parallelto u and of semiangle cos −1 m; (d) a circular cylinder of radius n with its axisparallel to u.7.5 (a) cos √ −1 2/3; (b) z − x =2;(c)1/ √ 2; (d) 1 1[ (c × g) · j = 1 .3 2 37.7 Show that q × r is parallel to p; volume = 1 1 (q × r) · p] = 5 .3 2 37.9 Note that (a × b) · (c × d) =d · [(a × b) × c] and use the result for a triple vectorproduct to expand the expression in square brackets.7.11 Show that the position vectors of the points are linearly dependent; r = a + λbwhere a = i + k and b = −j + k.7.13 Show that p must have the direction ˆn× ˆm and write a as xˆn+y ˆm. By obtaining apair of simultaneous equations for x and y, prove that x =(λ−µˆn· ˆm)/[1−(ˆn· ˆm) 2 ]and that y =(µ − λˆn · ˆm)/[1 − (ˆn · ˆm) 2 ].7.15 (a) Note that |a − g| 2 = R 2 = |0 − g| 2 , leading to a · a =2a · g.(b) Make p the new origin and solve the three simultaneous linear equations toobtain λ =5/18, µ =10/18, ν = −3/18, giving g =2i − k and a sphere ofradius √ 5 centred on (5, 1, −3).7.17 (a) Find two points on both planes, say (0, 0, 0) and (1, −2, 1), and hence determinethe direction cosines of the line of intersection; (b) ( 2 3 )1/2 .7.19 For (c) and (d), treat (c × a) × (a × b) as a triple vector product with c × a as oneof the three vectors.7.21 (b) b ′ = a −1 (−i + j + k), c ′ = a −1 (i − j + k), d ′ = a −1 (i + j − k); (c) a/2 fordirectionk; successive planes through (0, 0, 0) and (a/2, 0,a/2) give a spacing of a/ √ 8fordirection i + j; successive planes through (−a/2, 0, 0) and (a/2, 0, 0) give a spacingof a/ √ 3fordirectioni + j + k.7.23 Note that a 2 − (ˆn · a) 2 = a 2 ⊥ .7.25 p = −2a +3b, q = 3 a − 3 b + 5 c and r =2a − b − c. Remember that a · a = b · b =2 2 2c · c =2anda · b = a · c = b · c =1.7.27 With currents in mA and potential differences in volts:I 1 =(7.76, −23.2 ◦ ), I 2 =(14.36, −50.8 ◦ ), I 3 =(8.30, 103.4 ◦ );V 1 =(0.388, −23.2 ◦ ), V 2 =(0.287, −50.8 ◦ ), V 4 =(0.596, 39.2 ◦ );L = 33 mH, R 2 =20Ω.240

8Matrices and vector spacesIn the previous chapter we defined a vector as a geometrical object which hasboth a magnitude and a direction and which may be thought of as an arrow fixedin our familiar three-dimensional space, a space which, if we need to, we defineby reference to, say, the fixed stars. This geometrical definition of a vector is bothuseful and important since it is independent of any coordinate system with whichwe choose to label points in space.In most specific applications, however, it is necessary at some stage to choosea coordinate system and to break down a vector into its component vectors inthe directions of increasing coordinate values. Thus for a particular Cartesiancoordinate system (for example) the component vectors of a vector a will be a x i,a y j and a z k and the complete vector will bea = a x i + a y j + a z k. (8.1)Although we have so far considered only real three-dimensional space, we mayextend our notion of a vector to more abstract spaces, which in general canhave an arbitrary number of dimensions N. We may still think of such a vectoras an ‘arrow’ in this abstract space, so that it is again independent of any (Ndimensional)coordinate system with which we choose to label the space. As anexample of such a space, which, though abstract, has very practical applications,we may consider the description of a mechanical or electrical system. If the stateof a system is uniquely specified by assigning values to a set of N variables,which could be angles or currents, for example, then that state can be representedby a vector in an N-dimensional space, the vector having those values as itscomponents.In this chapter we first discuss general vector spaces and their properties. Wethen go on to discuss the transformation of one vector into another by a linearoperator. This leads naturally to the concept of a matrix, a two-dimensional arrayof numbers. The properties of matrices are then discussed and we conclude with241

MATRICES AND VECTOR SPACESa discussion of how to use these properties to solve systems of linear equations.The application of matrices to the study of oscillations in physical systems istakenupinchapter9.8.1 Vector spacesA set of objects (vectors) a, b, c, ... is said to form a linear vector space V if:(i) the set is closed under commutative and associative addition, so thata + b = b + a, (8.2)(a + b)+c = a +(b + c); (8.3)(ii) the set is closed under multiplication by a scalar (any complex number) toform a new vector λa, the operation being both distributive and associativeso thatλ(a + b) =λa + λb, (8.4)(λ + µ)a = λa + µa, (8.5)λ(µa) =(λµ)a, (8.6)where λ and µ are arbitrary scalars;(iii) there exists a null vector 0 such that a + 0 = a for all a;(iv) multiplication by unity leaves any vector unchanged, i.e. 1 × a = a;(v) all vectors have a corresponding negative vector −a such that a+(−a) =0.It follows from (8.5) with λ = 1 and µ = −1 that−a is the same vector as(−1) × a.We note that if we restrict all scalars to be real then we obtain a real vectorspace (an example of which is our familiar three-dimensional space); otherwise,in general, we obtain a complex vector space. We note that it is common to use theterms ‘vector space’ and ‘space’, instead of the more formal ‘linear vector space’.The span of a set of vectors a, b,...,s is defined as the set of all vectors thatmay be written as a linear sum of the original set, i.e. all vectorsx = αa + βb + ···+ σs (8.7)that result from the infinite number of possible values of the (in general complex)scalars α,β,...,σ.Ifx in (8.7) is equal to 0 for some choice of α,β,...,σ (not allzero), i.e. ifαa + βb + ···+ σs = 0, (8.8)then the set of vectors a, b,...,s, issaidtobelinearly dependent. Insuchasetat least one vector is redundant, since it can be expressed as a linear sum ofthe others. If, however, (8.8) is not satisfied by any set of coefficients (other than242

8.1 VECTOR SPACESthe trivial case in which all the coefficients are zero) then the vectors are linearlyindependent, and no vector in the set can be expressed as a linear sum of theothers.If, in a given vector space, there exist sets of N linearly independent vectors,but no set of N + 1 linearly independent vectors, then the vector space is said tobe N-dimensional. (In this chapter we will limit our discussion to vector spaces offinite dimensionality; spaces of infinite dimensionality are discussed in chapter 17.)8.1.1 Basis vectorsIf V is an N-dimensional vector space then any set of N linearly independentvectors e 1 , e 2 ,...,e N forms a basis for V .Ifx is an arbitrary vector lying in V thenthe set of N + 1 vectors x, e 1 , e 2 ,...,e N , must be linearly dependent and thereforesuch thatαe 1 + βe 2 + ···+ σe N + χx = 0, (8.9)where the coefficients α,β,...,χ are not all equal to 0, and in particular χ ≠0.Rearranging (8.9) we may write x as a linear sum of the vectors e i as follows:x = x 1 e 1 + x 2 e 2 + ···+ x N e N =N∑x i e i , (8.10)for some set of coefficients x i that are simply related to the original coefficients,e.g. x 1 = −α/χ, x 2 = −β/χ, etc. Since any x lying in the span of V can beexpressed in terms of the basis or base vectors e i , the latter are said to forma complete set. The coefficients x i are the components of x with respect to thee i -basis. These components are unique, since if boththeni=1i=1N∑N∑x = x i e i and x = y i e i ,i=1N∑(x i − y i )e i = 0, (8.11)i=1which, since the e i are linearly independent, has only the solution x i = y i for alli =1, 2,...,N.From the above discussion we see that any set of N linearly independentvectors can form a basis for an N-dimensional space. If we choose a different sete ′ i , i =1,...,N then we can write x asx = x ′ 1e ′ 1 + x ′ 2e ′ 2 + ···+ x ′ Ne ′ N =243N∑x ′ ie ′ i. (8.12)i=1

MATRICES AND VECTOR SPACESWe reiterate that the vector x (a geometrical entity) is independent of the basis– it is only the components of x that depend on the basis. We note, however,that given a set of vectors u 1 , u 2 ,...,u M ,whereM ≠ N, inanN-dimensionalvector space, then either there exists a vector that cannot be expressed as alinear combination of the u i or, for some vector that can be so expressed, thecomponents are not unique.8.1.2 The inner productWe may usefully add to the description of vectors in a vector space by definingthe inner product of two vectors, denoted in general by 〈a|b〉, which is a scalarfunction of a and b. The scalar or dot product, a · b ≡|a||b| cos θ, ofvectorsin real three-dimensional space (where θ is the angle between the vectors), wasintroduced in the last chapter and is an example of an inner product. In effect thenotion of an inner product 〈a|b〉 is a generalisation of the dot product to moreabstract vector spaces. Alternative notations for 〈a|b〉 are (a, b), or simply a · b.The inner product has the following properties:(i) 〈a|b〉 = 〈b|a〉 ∗ ,(ii) 〈a|λb + µc〉 = λ〈a|b〉 + µ〈a|c〉.We note that in general, for a complex vector space, (i) and (ii) imply that〈λa + µb|c〉 = λ ∗ 〈a|c〉 + µ ∗ 〈b|c〉, (8.13)〈λa|µb〉 = λ ∗ µ〈a|b〉. (8.14)Following the analogy with the dot product in three-dimensional real space,two vectors in a general vector space are defined to be orthogonal if 〈a|b〉 =0.Similarly, the norm of a vector a is given by ‖a‖ = 〈a|a〉 1/2 and is clearly ageneralisation of the length or modulus |a| of a vector a in three-dimensionalspace. In a general vector space 〈a|a〉 can be positive or negative; however, weshall be primarily concerned with spaces in which 〈a|a〉 ≥0 and which are thussaid to have a positive semi-definite norm. Insuchaspace〈a|a〉 = 0 implies a = 0.Let us now introduce into our N-dimensional vector space a basis ê 1 , ê 2 ,...,ê Nthat has the desirable property of being orthonormal (the basis vectors are mutuallyorthogonal and each has unit norm), i.e. a basis that has the property〈ê i |ê j 〉 = δ ij . (8.15)Here δ ij is the Kronecker delta symbol (of which we say more in chapter 26) andhas the properties{1 for i = j,δ ij =0 for i ≠ j.244

8.1 VECTOR SPACESIn the above basis we may express any two vectors a and b asN∑N∑a = a i ê i and b = b i ê i .i=1Furthermore, in such an orthonormal basis we have, for any a,N∑N∑〈ê j |a〉 = 〈ê j |a i ê i 〉 = a i 〈ê j |ê i 〉 = a j . (8.16)i=1Thus the components of a are given by a i = 〈ê i |a〉. Note that this is not trueunless the basis is orthonormal. We can write the inner product of a and b interms of their components in an orthonormal basis as〈a|b〉 = 〈a 1 ê 1 + a 2 ê 2 + ···+ a N ê N |b 1 ê 1 + b 2 ê 2 + ···+ b N ê N 〉N∑N∑ N∑= a ∗ i b i 〈ê i |ê i 〉 + a ∗ i b j 〈ê i |ê j 〉=i=1N∑a ∗ i b i,i=1i=1j≠iwhere the second equality follows from (8.14) and the third from (8.15). This isclearly a generalisation of the expression (7.21) for the dot product of vectors inthree-dimensional space.We may generalise the above to the case where the base vectors e 1 , e 2 ,...,e Nare not orthonormal (or orthogonal). In general we can define the N 2 numbersi=1i=1G ij = 〈e i |e j 〉. (8.17)Then, if a = ∑ Ni=1 a ie i and b = ∑ Ni=1 b ie i , the inner product of a and b is given by〈 ∣∑ N ∣∣∣∣∣ 〉∑N〈a|b〉 = a i e i b j e j==N∑i=1i=1 j=1N∑i=1 j=1j=1N∑a ∗ i b j 〈e i |e j 〉N∑a ∗ i G ij b j . (8.18)We further note that from (8.17) and the properties of the inner product werequire G ij = G ∗ ji . This in turn ensures that ‖a‖ = 〈a|a〉 is real, since thenN∑ N∑N∑ N∑〈a|a〉 ∗ = a i G ∗ ija ∗ j = a ∗ j G ji a i = 〈a|a〉.i=1 j=1j=1 i=1245

MATRICES AND VECTOR SPACES8.1.3 Some useful inequalitiesFor a set of objects (vectors) forming a linear vector space in which 〈a|a〉 ≥0forall a, the following inequalities are often useful.(i) Schwarz’s inequality is the most basic result and states that|〈a|b〉| ≤ ‖a‖‖b‖, (8.19)where the equality holds when a is a scalar multiple of b, i.e.whena = λb.It is important here to distinguish between the absolute value of a scalar,|λ|, andthenorm of a vector, ‖a‖. Schwarz’s inequality may be proved byconsidering‖a + λb‖ 2 = 〈a + λb|a + λb〉= 〈a|a〉 + λ〈a|b〉 + λ ∗ 〈b|a〉 + λλ ∗ 〈b|b〉.If we write 〈a|b〉 as |〈a|b〉|e iα then‖a + λb‖ 2 = ‖a‖ 2 + |λ| 2 ‖b‖ 2 + λ|〈a|b〉|e iα + λ ∗ |〈a|b〉|e −iα .However, ‖a + λb‖ 2 ≥ 0 for all λ, sowemaychooseλ = re −iα and requirethat, for all r,0 ≤‖a + λb‖ 2 = ‖a‖ 2 + r 2 ‖b‖ 2 +2r|〈a|b〉|.This means that the quadratic equation in r formed by setting the RHSequal to zero must have no real roots. This, in turn, implies that4|〈a|b〉| 2 ≤ 4‖a‖ 2 ‖b‖ 2 ,which, on taking the square root (all factors are necessarily positive) ofboth sides, gives Schwarz’s inequality.(ii) The triangle inequality states that‖a + b‖ ≤‖a‖ + ‖b‖ (8.20)and may be derived from the properties of the inner product and Schwarz’sinequality as follows. Let us first consider‖a + b‖ 2 = ‖a‖ 2 + ‖b‖ 2 +2Re〈a|b〉 ≤‖a‖ 2 + ‖b‖ 2 +2|〈a|b〉|.Using Schwarz’s inequality we then have‖a + b‖ 2 ≤‖a‖ 2 + ‖b‖ 2 +2‖a‖‖b‖ =(‖a‖ + ‖b‖) 2 ,which, on taking the square root, gives the triangle inequality (8.20).(iii) Bessel’s inequality requires the introduction of an orthonormal basis ê i ,i =1, 2,...,N into the N-dimensional vector space; it states that‖a‖ 2 ≥ ∑ i|〈ê i |a〉| 2 , (8.21)246

8.2 LINEAR OPERATORSwhere the equality holds if the sum includes all N basis vectors. If notall the basis vectors are included in the sum then the inequality results(though of course the equality remains if those basis vectors omitted allhave a i = 0). Bessel’s inequality can also be written〈a|a〉 ≥ ∑ i|a i | 2 ,where the a i are the components of a in the orthonormal basis. From (8.16)these are given by a i = 〈ê i |a〉. The above may be proved by considering∣ a − ∑ i∣ ∣∣∣ ∣∣∣ 〈〈ê i |a〉ê i2= a − ∑ i∣ ∣∣a ∑〈ê i |a〉ê i − 〈ê j |a〉ê j〉.Expanding out the inner product and using 〈ê i |a〉 ∗ = 〈a|ê i 〉, we obtain∣ a − ∑ ∣ ∣∣∣ ∣∣∣2〈ê i |a〉ê i = 〈a|a〉−2 ∑ 〈a|ê i 〉〈ê i |a〉 + ∑ ∑〈a|ê i 〉〈ê j |a〉〈ê i |ê j 〉.iii jNow 〈ê i |ê j 〉 = δ ij , since the basis is orthonormal, and so we find0 ≤∣ a − ∑ ∣ ∣∣∣ ∣∣∣2〈ê i |a〉ê i = ‖a‖ 2 − ∑ |〈ê i |a〉| 2 ,iiwhich is Bessel’s inequality.We take this opportunity to mention also(iv) the parallelogram equality‖a + b‖ 2 + ‖a − b‖ 2 =2 ( ‖a‖ 2 + ‖b‖ 2) , (8.22)which may be proved straightforwardly from the properties of the innerproduct.j8.2 Linear operatorsWe now discuss the action of linear operators on vectors in a vector space. Alinear operator A associates with every vector x another vectory = A x,in such a way that, for two vectors a and b,A (λa + µb) =λA a + µA b,where λ, µ are scalars. We say that A ‘operates’ on x to give the vector y. Wenote that the action of A is independent of any basis or coordinate system and247

MATRICES AND VECTOR SPACESmay be thought of as ‘transforming’ one geometrical entity (i.e. a vector) intoanother.If we now introduce a basis e i , i =1, 2,...,N, into our vector space then theaction of A on each of the basis vectors is to produce a linear combination ofthe latter; this may be written asA e j =N∑A ij e i , (8.23)i=1where A ij is the ith component of the vector A e j in this basis; collectively thenumbers A ij are called the components of the linear operator in the e i -basis. Inthis basis we can express the relation y = A x in component form asy =⎛ ⎞N∑N∑y i e i = A ⎝ x j e j⎠ =i=1and hence, in purely component form, in this basis we havey i =j=1N∑j=1x jN ∑i=1A ij e i ,N∑A ij x j . (8.24)j=1If we had chosen a different basis e ′ i , in which the components of x, y and Aare x ′ i , y′ i and A ′ ij respectively then the geometrical relationship y = A x would berepresented in this new basis byy ′ i =N∑A ′ ijx ′ j.j=1We have so far assumed that the vector y is in the same vector space asx. If,however,y belongs to a different vector space, which may in general beM-dimensional (M ≠ N) then the above analysis needs a slight modification. Byintroducing a basis set f i , i =1, 2,...,M, into the vector space to which y belongswe may generalise (8.23) asA e j =M∑A ij f i ,i=1where the components A ij of the linear operator A relate to both of the bases e jand f i .248

8.3 MATRICES8.2.1 Properties of linear operatorsIf x is a vector and A and B are two linear operators then it follows that(A + B )x = A x + B x,(λA )x = λ(A x),(AB)x = A (B x),where in the last equality we see that the action of two linear operators insuccession is associative. The product of two linear operators is not in generalcommutative, however, so that in general ABx ≠ BAx. In an obvious way wedefine the null (or zero) and identity operators byO x = 0 and I x = x,for any vector x in our vector space. Two operators A and B are equal ifA x = B x for all vectors x. Finally, if there exists an operator A −1 such thatAA −1 = A −1 A = Ithen A −1 is the inverse of A . Some linear operators do not possess an inverseand are called singular, whilst those operators that do have an inverse are termednon-singular.8.3 MatricesWe have seen that in a particular basis e i both vectors and linear operatorscan be described in terms of their components with respect to the basis. Thesecomponents may be displayed as an array of numbers called a matrix. Ingeneral,if a linear operator A transforms vectors from an N-dimensional vector space,for which we choose a basis e j , j =1, 2,...,N, into vectors belonging to anM-dimensional vector space, with basis f i , i =1, 2,...,M, then we may representthe operator A by the matrix⎛⎞A 11 A 12 ... A 1NA 21 A 22 ... A 2NA = ⎜⎝. ⎟. . .. . ⎠ . (8.25)A M1 A M2 ... A MNThe matrix elements A ij are the components of the linear operator with respectto the bases e j and f i ; the component A ij of the linear operator appears in theith row and jth column of the matrix. The array has M rows and N columnsand is thus called an M × N matrix. If the dimensions of the two vector spacesare the same, i.e. M = N (for example, if they are the same vector space) then wemay represent A by an N × N or square matrix of order N. The component A ij ,which in general may be complex, is also denoted by (A) ij .249

MATRICES AND VECTOR SPACESIn a similar way we may denote a vector x in terms of its components x i in abasis e i , i =1, 2,...,N, by the array⎛ ⎞x 1x 2x = ⎜ ⎟⎝. ⎠ ,x Nwhich is a special case of (8.25) and is called a column matrix (or conventionally,and slightly confusingly, a column vector or even just a vector – strictly speakingthe term ‘vector’ refers to the geometrical entity x). The column matrix x can alsobe written asx =(x 1 x 2 ··· x N ) T ,which is the transpose of a row matrix (see section 8.6).We note that in a different basis e ′ i the vector x would be represented by adifferent column matrix containing the components x ′ i in the new basis, i.e.⎛x ′ = ⎜⎝Thus, we use x and x ′ to denote different column matrices which, in different basese i and e ′ i , represent the same vector x. In many texts, however, this distinction isnot made and x (rather than x) is equated to the corresponding column matrix; ifwe regard x as the geometrical entity, however, this can be misleading and so weexplicitly make the distinction. A similar argument follows for linear operators;the same linear operator A is described in different bases by different matrices Aand A ′ , containing different matrix elements.x ′ 1x ′ 2..x ′ N⎞⎟⎠ .8.4 Basic matrix algebraThe basic algebra of matrices may be deduced from the properties of the linearoperators that they represent. In a given basis the action of two linear operatorsA and B on an arbitrary vector x (see the beginning of subsection 8.2.1), whenwritten in terms of components using (8.24), is given by∑(A + B) ij x j = ∑ A ij x j + ∑ B ij x j ,jjj∑(λA) ij x j = λ ∑ A ij x j ,jj∑(AB) ij x j = ∑ A ik (Bx) k = ∑ ∑A ik B kj x j .jkj k250

8.4 BASIC MATRIX ALGEBRANow, since x is arbitrary, we can immediately deduce the way in which matricesare added or multiplied, i.e.(A + B) ij = A ij + B ij , (8.26)(λA) ij = λA ij , (8.27)(AB) ij = ∑ A ik B kj .k(8.28)We note that a matrix element may, in general, be complex. We now discussmatrix addition and multiplication in more detail.8.4.1 Matrix addition and multiplication by a scalarFrom (8.26) we see that the sum of two matrices, S = A + B, is the matrix whoseelements are given byS ij = A ij + B ijfor every pair of subscripts i, j, with i =1, 2,...,M and j =1, 2,...,N. Forexample, if A and B are 2 × 3 matrices then S = A + B is given by( ) ( )S11 S 12 S 13 A11 A=12 A 13+S 21 S 22 S 23 A 21 A 22 A 23=( )B11 B 12 B 13B 21 B 22 B 23( )A11 + B 11 A 12 + B 12 A 13 + B 13. (8.29)A 21 + B 21 A 22 + B 22 A 23 + B 23Clearly, for the sum of two matrices to have any meaning, the matrices must havethe same dimensions, i.e. both be M × N matrices.From definition (8.29) it follows that A + B = B + A and that the sum of anumber of matrices can be written unambiguously without bracketting, i.e. matrixaddition is commutative and associative.The difference of two matrices is defined by direct analogy with addition. Thematrix D = A − B has elementsD ij = A ij − B ij , for i =1, 2,...,M, j =1, 2,...,N. (8.30)From (8.27) the product of a matrix A with a scalar λ is the matrix withelements λA ij , for example( ) ( )A11 Aλ12 A 13 λA11 λA=12 λA 13. (8.31)A 21 A 22 A 23 λA 21 λA 22 λA 23Multiplication by a scalar is distributive and associative.251

MATRICES AND VECTOR SPACES◮The matrices A, B and C are given by( ) ( ) ( )2 −11 0−2 1A =, B =, C =.3 10 −2−1 1Find the matrix D = A +2B − C.(2 −1D =3 1=) (1 0+20 −2) (−2 1−−1 1(2+2× 1 − (−2) −1+2× 0 − 13+2× 0 − (−1) 1+2× (−2) − 1))=( )6 −2. ◭4 −4From the above considerations we see that the set of all, in general complex,M × N matrices (with fixed M and N) forms a linear vector space of dimensionMN. One basis for the space is the set of M × N matrices E (p,q) with the propertythat E (p,q)ij =1ifi = p and j = q whilst E (p,q)ij = 0 for all other values of i andj, i.e. each matrix has only one non-zero entry, which equals unity. Here the pair(p, q) is simply a label that picks out a particular one of the matrices E (p,q) ,thetotal number of which is MN.8.4.2 Multiplication of matricesLet us consider again the ‘transformation’ of one vector into another, y = A x,which, from (8.24), may be described in terms of components with respect to aparticular basis asy i =N∑A ij x j for i =1, 2,...,M. (8.32)j=1Writing this in matrix form as y = Ax we have⎛ ⎞⎛⎛⎞y 1 A 11 A 12 ... A 1Ny 2⎜ ⎟⎝ . ⎠ = A 21 A 22 ... A 2N⎜⎝.. ⎟. .. . ⎠ ⎜⎝A M1 A M2 ... A MNy Mx 1x 2..x N⎞⎟⎠(8.33)where we have highlighted with boxes the components used to calculate theelement y 2 : using (8.32) for i =2,y 2 = A 21 x 1 + A 22 x 2 + ···+ A 2N x N .All the other components y i are calculated similarly.If instead we operate with A on a basis vector e j having all components zero252

8.4 BASIC MATRIX ALGEBRAexcept for the jth, which equals unity, then we find⎛ ⎞0⎛⎞A 11 A 12 ... A 1N0⎛A 21 A 22 ... A 2N..Ae j = ⎜⎝. ⎟= ⎜. . .. . ⎠1⎝A M1 A M2 ... A⎜ . ⎟MN ⎝ . ⎠0and so confirm our identification of the matrix element A ij as the ith componentof Ae j in this basis.From (8.28) we can extend our discussion to the product of two matricesP = AB, whereP is the matrix of the quantities formed by the operation ofthe rows of A on the columns of B, treating each column of B in turn as thevector x represented in component form in (8.32). It is clear that, for this to bea meaningful definition, the number of columns in A must equal the number ofrows in B. Thus the product AB of an M × N matrix A with an N × R matrix Bis itself an M × R matrix P, whereN∑P ij = A ik B kj for i =1, 2,...,M, j =1, 2,...,R.k=1For example, P = AB may be written in matrix form() ( ) ⎛ P 11 P 12 A11 A=12 A 13 ⎝P 21 P 22 A 21 A 22 A 23whereP 11 = A 11 B 11 + A 12 B 21 + A 13 B 31 ,P 21 = A 21 B 11 + A 22 B 21 + A 23 B 31 ,P 12 = A 11 B 12 + A 12 B 22 + A 13 B 32 ,P 22 = A 21 B 12 + A 22 B 22 + A 23 B 32 .A 1jA 2j.A Mj⎞⎟⎠ ,B 11 B 12⎞B 21 B 22⎠Multiplication of more than two matrices follows naturally and is associative.So, for example,A(BC) ≡ (AB)C, (8.34)provided, of course, that all the products are defined.As mentioned above, if A is an M × N matrix and B is an N × M matrix thentwo product matrices are possible, i.e.P = AB and Q = BA.253

MATRICES AND VECTOR SPACESThese are clearly not the same, since P is an M × M matrix whilst Q is anN × N matrix. Thus, particular care must be taken to write matrix products inthe intended order; P = AB but Q = BA. We note in passing that A 2 means AA,A 3 means A(AA) =(AA)A etc. Even if both A and B are square, in generalAB ≠ BA, (8.35)i.e. the multiplication of matrices is not, in general, commutative.◮Evaluate P = AB and Q = BA where⎛A = ⎝ 3 2 −10 3 21 −3 4⎞ ⎛⎠ , B =⎝ 2 −2 31 1 03 2 1⎞⎠ .As we saw for the 2 × 2 case above, the element P ij of the matrix P = AB is found bymentally taking the ‘scalar product’ of the ith row of A with the jth column of B. Forexample, P 11 =3× 2+2× 1+(−1) × 3=5,P 12 =3× (−2) + 2 × 1+(−1) × 2=−6, etc.Thus⎛P = AB = ⎝ 3 2 −1⎞ ⎛0 3 2 ⎠ ⎝ 2 −2 3⎞ ⎛1 1 0 ⎠ = ⎝ 5 −6 8⎞9 7 2 ⎠ ,1 −3 4 3 2 1 11 3 7and, similarly,⎛Q = BA = ⎝ 2 −2 3⎞ ⎛1 1 0 ⎠ ⎝ 3 2 −1⎞ ⎛0 3 2 ⎠ = ⎝ 9 −11 6⎞3 5 1 ⎠ .3 2 1 1 −3 4 10 9 5These results illustrate that, in general, two matrices do not commute. ◭andThe property that matrix multiplication is distributive over addition, i.e. thatfollows directly from its definition.(A + B)C = AC + BC (8.36)C(A + B) =CA + CB, (8.37)8.4.3 The null and identity matricesBoth the null matrix and the identity matrix are frequently encountered, and wetake this opportunity to introduce them briefly, leaving their uses until later. Thenull or zero matrix 0 has all elements equal to zero, and so its properties areA0 = 0 = 0A,A + 0 = 0 + A = A.254

8.5 FUNCTIONS OF MATRICESThe identity matrix I has the propertyAI = IA = A.It is clear that, in order for the above products to be defined, the identity matrixmust be square. The N × N identity matrix (often denoted by I N )hastheform⎛⎞1 0 ··· 0.I N =0 1 .⎜ .⎝. ⎟. .. 0 ⎠ .0 ··· 0 18.5 Functions of matricesIf a matrix A is square then, as mentioned above, one can define powers of A ina straightforward way. For example A 2 = AA, A 3 = AAA, or in the general caseA n = AA ···A(n times),where n is a positive integer. Having defined powers of a square matrix A, wemay construct functions of A of the formS = ∑ na n A n ,where the a k are simple scalars and the number of terms in the summation maybe finite or infinite. In the case where the sum has an infinite number of terms,the sum has meaning only if it converges. A common example of such a functionis the exponential of a matrix, which is defined byexp A =∞∑n=0A nn! . (8.38)This definition can, in turn, be used to define other functions such as sin A andcos A.8.6 The transpose of a matrixWe have seen that the components of a linear operator in a given coordinate systemcan be written in the form of a matrix A. We will also find it useful, however,to consider the different (but clearly related) matrix formed by interchanging therows and columns of A. The matrix is called the transpose of A and is denotedby A T .255

MATRICES AND VECTOR SPACES◮Find the transpose of the matrix( )3 1 2A =.0 4 1By interchanging the rows and columns of A we immediately obtain⎛A T = ⎝ 3 0⎞1 4 ⎠ . ◭2 1It is obvious that if A is an M × N matrix then its transpose A T is a N × Mmatrix. As mentioned in section 8.3, the transpose of a column matrix is arow matrix and vice versa. An important use of column and row matrices isin the representation of the inner product of two real vectors in terms of theircomponents in a given basis. This notion is discussed fully in the next section,where it is extended to complex vectors.The transpose of the product of two matrices, (AB) T , is given by the productof their transposes taken in the reverse order, i.e.This is proved as follows:(AB) T = B T A T . (8.39)(AB) T ij =(AB) ji = ∑ k= ∑ kA jk B ki(A T ) kj (B T ) ik = ∑ k(B T ) ik (A T ) kj =(B T A T ) ij ,and the proof can be extended to the product of several matrices to give(ABC ···G) T = G T ···C T B T A T .8.7 The complex and Hermitian conjugates of a matrixTwo further matrices that can be derived from a given general M × N matrixare the complex conjugate, denoted by A ∗ ,andtheHermitian conjugate, denotedby A † .The complex conjugate of a matrix A is the matrix obtained by taking thecomplex conjugate of each of the elements of A, i.e.(A ∗ ) ij =(A ij ) ∗ .Obviously if a matrix is real (i.e. it contains only real elements) then A ∗ = A.256

8.7 THE COMPLEX AND HERMITIAN CONJUGATES OF A MATRIX◮Find the complex conjugate of the matrix(1 2 3iA =1+i 1 0).By taking the complex conjugate of each element we obtain immediately( )1 2 −3iA ∗ =. ◭1 − i 1 0The Hermitian conjugate, or adjoint, of a matrix A is the transpose of itscomplex conjugate, or equivalently, the complex conjugate of its transpose, i.e.A † =(A ∗ ) T =(A T ) ∗ .We note that if A is real (and so A ∗ = A) thenA † = A T , and taking the Hermitianconjugate is equivalent to taking the transpose. Following the previous line ofargument for the transpose of the product of several matrices, the Hermitianconjugate of such a product can be shown to be given by(AB ···G) † = G † ···B † A † . (8.40)◮Find the Hermitian conjugate of the matrix(1 2 3iA =1+i 1 0).Taking the complex conjugate of A and then forming the transpose we find⎛A † = ⎝ 1 1− i⎞2 1 ⎠ .−3i 0We obtain the same result, of course, if we first take the transpose of A andthentakethecomplex conjugate. ◭An important use of the Hermitian conjugate (or transpose in the real case)is in connection with the inner product of two vectors. Suppose that in a givenorthonormal basis the vectors a and b may be represented by the column matrices⎛a = ⎜⎝a 1a 2.a N⎞⎟⎠⎛and b = ⎜⎝b 1b 2.b N⎞⎟⎠ . (8.41)Taking the Hermitian conjugate of a, to give a row matrix, and multiplying (on257

MATRICES AND VECTOR SPACESthe right) by b we obtain⎛a † b =(a ∗ 1 a ∗ 2 ··· a ∗ N) ⎜⎝b 1b 2.b N⎞N⎟⎠ = ∑a ∗ i b i , (8.42)which is the expression for the inner product 〈a|b〉 in that basis. We note that forreal vectors (8.42) reduces to a T b = ∑ Ni=1 a ib i .If the basis e i is not orthonormal, so that, in general,〈e i |e j 〉 = G ij ≠ δ ij ,then, from (8.18), the scalar product of a and b in terms of their components withrespect to this basis is given by〈a|b〉 =N∑i=1 j=1i=1N∑a ∗ i G ij b j = a † Gb,where G is the N × N matrix with elements G ij .8.8 The trace of a matrixFor a given matrix A, in the previous two sections we have considered variousother matrices that can be derived from it. However, sometimes one wishes toderive a single number from a matrix. The simplest example is the trace (or spur)of a square matrix, which is denoted by Tr A. This quantity is defined as the sumof the diagonal elements of the matrix,Tr A = A 11 + A 22 + ···+ A NN =N∑A ii . (8.43)It is clear that taking the trace is a linear operation so that, for example,Tr(A ± B) =TrA ± Tr B.A very useful property of traces is that the trace of the product of two matricesis independent of the order of their multiplication; this results holds whether ornot the matrices commute and is proved as follows:N∑N∑ N∑N∑ N∑N∑Tr AB = (AB) ii = A ij B ji = B ji A ij =i=1i=1 j=1i=1 j=1i=1j=1(BA) jj =TrBA.(8.44)The result can be extended to the product of several matrices. For example, from(8.44), we immediately findTr ABC =TrBCA =TrCAB,258

8.9 THE DETERMINANT OF A MATRIXwhich shows that the trace of a multiple product is invariant under cyclicpermutations of the matrices in the product. Other easily derived properties ofthe trace are, for example, Tr A T =TrA and Tr A † =(TrA) ∗ .8.9 The determinant of a matrixFor a given matrix A, the determinant det A (like the trace) is a single number (oralgebraic expression) that depends upon the elements of A. Also like the trace,the determinant is defined only for square matrices. If, for example, A is a 3 × 3matrix then its determinant, of order 3, is denoted bydet A = |A| =∣A 11 A 12 A 13A 21 A 22 A 23A 31 A 32 A 33∣ ∣∣∣∣∣. (8.45)In order to calculate the value of a determinant, we first need to introducethe notions of the minor and the cofactor of an element of a matrix. (Weshall see that we can use the cofactors to write an order-3 determinant as theweighted sum of three order-2 determinants, thereby simplifying its evaluation.)The minor M ij of the element A ij of an N × N matrix A is the determinant ofthe (N − 1) × (N − 1) matrix obtained by removing all the elements of the ithrow and jth column of A; the associated cofactor, C ij , is found by multiplyingthe minor by (−1) i+j .◮Find the cofactor of the element A 23 of the matrix⎛A = ⎝ A ⎞11 A 12 A 13A 21 A 22 A 23⎠ .A 31 A 32 A 33Removing all the elements of the second row and third column of A and forming thedeterminant of the remaining terms gives the minorM 23 =∣ A ∣11 A 12 ∣∣∣.A 31 A 32Multiplying the minor by (−1) 2+3 =(−1) 5 = −1 givesC 23 = −∣ A ∣11 A 12 ∣∣∣. ◭A 31 A 32We now define a determinant as the sum of the products of the elements of anyrow or column and their corresponding cofactors, e.g.A 21 C 21 + A 22 C 22 + A 23 C 23 orA 13 C 13 + A 23 C 23 + A 33 C 33 . Such a sum is called a Laplace expansion. For example,in the first of these expansions, using the elements of the second row of the259

MATRICES AND VECTOR SPACESdeterminant defined by (8.45) and their corresponding cofactors, we write |A| asthe Laplace expansion|A| = A 21 (−1) (2+1) M 21 + A 22 (−1) (2+2) M 22 + A 23 (−1) (2+3) M 23= −A 21∣ ∣∣∣ A 12 A 13A 32 A 33∣ ∣∣∣+ A 22∣ ∣∣∣ A 11 A 13A 31 A 33∣ ∣∣∣− A 23∣ ∣∣∣ A 11 A 12A 31 A 32∣ ∣∣∣.We will see later that the value of the determinant is independent of the rowor column chosen. Of course, we have not yet determined the value of |A| but,rather, written it as the weighted sum of three determinants of order 2. However,applying again the definition of a determinant, we can evaluate each of theorder-2 determinants.◮Evaluate the determinant ∣ ∣∣∣ A 12 A 13A 32 A 33∣ ∣∣∣.By considering the products of the elements of the first row in the determinant, and theircorresponding cofactors, we find∣ A ∣12 A 13 ∣∣∣= AA 32 A 12 (−1) (1+1) |A 33 | + A 13 (−1) (1+2) |A 32 |33= A 12 A 33 − A 13 A 32 ,where the values of the order-1 determinants |A 33 | and |A 32 | are defined to be A 33 and A 32respectively. It must be remembered that the determinant is not the same as the modulus,e.g. det (−2) = |−2| = −2, not 2. ◭We can now combine all the above results to show that the value of thedeterminant (8.45) is given by|A| = −A 21 (A 12 A 33 − A 13 A 32 )+A 22 (A 11 A 33 − A 13 A 31 )− A 23 (A 11 A 32 − A 12 A 31 ) (8.46)= A 11 (A 22 A 33 − A 23 A 32 )+A 12 (A 23 A 31 − A 21 A 33 )+ A 13 (A 21 A 32 − A 22 A 31 ), (8.47)where the final expression gives the form in which the determinant is usuallyremembered and is the form that is obtained immediately by considering theLaplace expansion using the first row of the determinant. The last equality, whichessentially rearranges a Laplace expansion using the second row into one usingthe first row, supports our assertion that the value of the determinant is unaffectedby which row or column is chosen for the expansion.260

8.9 THE DETERMINANT OF A MATRIX◮Suppose the rows of a real 3 × 3 matrix A are interpreted as the components in a givenbasis of three (three-component) vectors a, b and c. Show that one can write the determinantof A as|A| = a · (b × c).If one writes the rows of A as the components in a given basis of three vectors a, b and c,we have from (8.47) that∣ a 1 a 2 a 3 ∣∣∣∣∣|A| =b 1 b 2 b 3 = a 1 (b 2 c 3 − b 3 c 2 )+a 2 (b 3 c 1 − b 1 c 3 )+a 3 (b 1 c 2 − b 2 c 1 ).∣ c 1 c 2 c 3From expression (7.34) for the scalar triple product given in subsection 7.6.3, it followsthat we may write the determinant as|A| = a · (b × c). (8.48)In other words, |A| is the volume of the parallelepiped defined by the vectors a, b andc. (One could equally well interpret the columns of the matrix A as the components ofthree vectors, and result (8.48) would still hold.) This result provides a more memorable(and more meaningful) expression than (8.47) for the value of a 3 × 3 determinant. Indeed,using this geometrical interpretation, we see immediately that, if the vectors a 1 , a 2 , a 3 arenot linearly independent then the value of the determinant vanishes: |A| =0.◭The evaluation of determinants of order greater than 3 follows the same generalmethod as that presented above, in that it relies on successively reducing the orderof the determinant by writing it as a Laplace expansion. Thus, a determinantof order 4 is first written as a sum of four determinants of order 3, whichare then evaluated using the above method. For higher-order determinants, onecannot write down directly a simple geometrical expression for |A| analogous tothat given in (8.48). Nevertheless, it is still true that if the rows or columns ofthe N × N matrix A are interpreted as the components in a given basis of N(N-component) vectors a 1 , a 2 ,...,a N , then the determinant |A| vanishes if thesevectors are not all linearly independent.8.9.1 Properties of determinantsA number of properties of determinants follow straightforwardly from the definitionof det A; their use will often reduce the labour of evaluating a determinant.We present them here without specific proofs, though they all follow readily fromthe alternative form for a determinant, given in equation (26.29) on page 942,and expressed in terms of the Levi–Civita symbol ɛ ijk (see exercise 26.9).(i) Determinant of the transpose. The transpose matrix A T (which, we recall,is obtained by interchanging the rows and columns of A) has the samedeterminant as A itself, i.e.|A T | = |A|. (8.49)261

MATRICES AND VECTOR SPACESIt follows that any theorem established for the rows of A will apply to thecolumns as well, and vice versa.(ii) Determinant of the complex and Hermitian conjugate. It is clear that thematrix A ∗ obtained by taking the complex conjugate of each element of Ahas the determinant |A ∗ | = |A| ∗ . Combining this result with (8.49), we findthat|A † | = |(A ∗ ) T | = |A ∗ | = |A| ∗ . (8.50)(iii) Interchanging two rows or two columns. If two rows (columns) of A areinterchanged, its determinant changes sign but is unaltered in magnitude.(iv) Removing factors. If all the elements of a single row (column) of A havea common factor, λ, then this factor may be removed; the value of thedeterminant is given by the product of the remaining determinant and λ.Clearly this implies that if all the elements of any row (column) are zerothen |A| = 0. It also follows that if every element of the N × N matrix Ais multiplied by a constant factor λ then|λA| = λ N |A|. (8.51)(v) Identical rows or columns. If any two rows (columns) of A are identical orare multiples of one another, then it can be shown that |A| =0.(vi) Adding a constant multiple of one row (column) to another. The determinantof a matrix is unchanged in value by adding to the elements of one row(column) any fixed multiple of the elements of another row (column).(vii) Determinant of a product. IfA and B are square matrices of the same orderthen|AB| = |A||B| = |BA|. (8.52)A simple extension of this property gives, for example,|AB ···G| = |A||B| ···|G| = |A||G| ···|B| = |A ···GB|,which shows that the determinant is invariant under permutation of thematrices in a multiple product.There is no explicit procedure for using the above results in the evaluation ofany given determinant, and judging the quickest route to an answer is a matterof experience. A general guide is to try to reduce all terms but one in a row orcolumn to zero and hence in effect to obtain a determinant of smaller size. Thesteps taken in evaluating the determinant in the example below are certainly notthe fastest, but they have been chosen in order to illustrate the use of most of theproperties listed above.262

8.10 THE INVERSE OF A MATRIX◮Evaluate the determinant|A| =∣1 0 2 30 1 −2 13 −3 4 −2−2 1 −2 −1.∣Taking a factor 2 out of the third column and then adding the second column to the thirdgives1 0 1 31 0 1 30 1 −1 10 1 0 1|A| =23 −3 2 −2=23 −3 −1 −2.∣ −2 1 −1 −1 ∣ ∣ −2 1 0 −1 ∣Subtracting the second column from the fourth gives1 0 1 30 1 0 0|A| =23 −3 −1 1.∣ −2 1 0 −2 ∣We now note that the second row has only one non-zero element and so the determinantmay conveniently be written as a Laplace expansion, i.e.∣ ∣ ∣∣∣∣∣ 1 1 3∣∣∣∣∣ 4 0 4|A| =2× 1 × (−1) 2+2 3 −1 1−2 0 −2 ∣ =2 3 −1 1−2 0 −2 ∣ ,where the last equality follows by adding the second row to the first. It can now be seenthat the first row is minus twice the third, and so the value of the determinant is zero, byproperty (v) above. ◭8.10 The inverse of a matrixOur first use of determinants will be in defining the inverse of a matrix. If wewere dealing with ordinary numbers we would consider the relation P = AB asequivalent to B = P/A, provided that A ≠ 0. However, if A, B and P are matricesthen this notation does not have an obvious meaning. What we really want toknow is whether an explicit formula for B can be obtained in terms of A andP. It will be shown that this is possible for those cases in which |A| ≠0.Asquare matrix whose determinant is zero is called a singular matrix; otherwise itis non-singular. We will show that if A is non-singular we can define a matrix,denoted by A −1 and called the inverse of A, which has the property that if AB = Pthen B = A −1 P.Inwords,B can be obtained by multiplying P from the left byA −1 . Analogously, if B is non-singular then, by multiplication from the right,A = PB −1 .It is clear thatAI = A ⇒ I = A −1 A, (8.53)where I is the unit matrix, and so A −1 A = I = AA −1 . These statements are263

MATRICES AND VECTOR SPACESequivalent to saying that if we first multiply a matrix, B say, by A and thenmultiply by the inverse A −1 , we end up with the matrix we started with, i.e.A −1 AB = B. (8.54)This justifies our use of the term inverse. It is also clear that the inverse is onlydefined for square matrices.So far we have only defined what we mean by the inverse of a matrix. Actuallyfinding the inverse of a matrix A may be carried out in a number of ways. We willshow that one method is to construct first the matrix C containing the cofactorsof the elements of A, as discussed in the last subsection. Then the required inverseA −1 can be found by forming the transpose of C and dividing by the determinantof A. Thus the elements of the inverse A −1 are given by(A −1 ) ik = (C)T ik|A|= C ki|A| . (8.55)That this procedure does indeed result in the inverse may be seen by consideringthe components of A −1 A,i.e.(A −1 A) ij = ∑ k(A −1 ) ik (A) kj = ∑ kC ki|A| A kj = |A||A| δ ij. (8.56)The last equality in (8.56) relies on the property∑C ki A kj = |A|δ ij ; (8.57)kthis can be proved by considering the matrix A ′ obtained from the original matrixA when the ith column of A is replaced by one of the other columns, say the jth.Thus A ′ is a matrix with two identical columns and so has zero determinant.However, replacing the ith column by another does not change the cofactors C kiof the elements in the ith column, which are therefore the same in A and A ′ .Recalling the Laplace expansion of a determinant, i.e.|A| = ∑ kA ki C ki ,we obtain0=|A ′ | = ∑ kA ′ kiC ′ ki = ∑ kA kj C ki , i ≠ j,which together with the Laplace expansion itself may be summarised by (8.57).It is immediately obvious from (8.55) that the inverse of a matrix is not definedif the matrix is singular (i.e. if |A| =0).264

8.10 THE INVERSE OF A MATRIX◮Find the inverse of the matrix⎛A = 1 −2 −2⎝ 2 4 3⎞⎠ .−3 3 2We first determine |A|:|A| =2[−2(2) − (−2)3] + 4[(−2)(−3) − (1)(2)] + 3[(1)(3) − (−2)(−3)]=11. (8.58)This is non-zero and so an inverse matrix can be constructed. To do this we need thematrix of the cofactors, C, and hence C T . We find⎛C = ⎝ 2 4 −3⎞⎛1 13 −18 ⎠ and C T = ⎝ 2 1 −2⎞4 13 7 ⎠ ,−2 7 −8−3 −18 −8and henceA −1 = CT|A| = 111⎛⎝ 4 13 72 1 −2−3 −18 −8⎞⎠ . ◭ (8.59)For a 2 × 2 matrix, the inverse has a particularly simple form. If the matrix is( )A11 AA =12A 21 A 22then its determinant |A| is given by |A| = A 11 A 22 − A 12 A 21 ,andthematrixofcofactors is( )A22 −AC =21.−A 12 A 11Thus the inverse of A is given by( )A −1 = CT|A| = 1A22 −A 12. (8.60)A 11 A 22 − A 12 A 21 −A 21 A 11It can be seen that the transposed matrix of cofactors for a 2 × 2matrixisthesame as the matrix formed by swapping the elements on the leading diagonal(A 11 and A 22 ) and changing the signs of the other two elements (A 12 and A 21 ).This is completely general for a 2 × 2 matrix and is easy to remember.The following are some further useful properties related to the inverse matrix265

MATRICES AND VECTOR SPACESand may be straightforwardly derived.(i) (A −1 ) −1 = A.(ii) (A T ) −1 =(A −1 ) T .(iii) (A † ) −1 =(A −1 ) † .(iv) (AB) −1 = B −1 A −1 .(v) (AB ···G) −1 = G −1 ···B −1 A −1 .◮Prove the properties (i)–(v) stated above.We begin by writing down the fundamental expression defining the inverse of a nonsingularsquare matrix A:AA −1 = I = A −1 A. (8.61)Property (i). This follows immediately from the expression (8.61).Property (ii). Taking the transpose of each expression in (8.61) gives(AA −1 ) T = I T =(A −1 A) T .Using the result (8.39) for the transpose of a product of matrices and noting that I T = I,we find(A −1 ) T A T = I = A T (A −1 ) T .However, from (8.61), this implies (A −1 ) T =(A T ) −1 and hence proves result (ii) above.Property (iii). This may be proved in an analogous way to property (ii), by replacing thetransposes in (ii) by Hermitian conjugates and using the result (8.40) for the Hermitianconjugate of a product of matrices.Property (iv). Using (8.61), we may write(AB)(AB) −1 = I = (AB) −1 (AB),From the left-hand equality it follows, by multiplying on the left by A −1 ,thatA −1 AB(AB) −1 = A −1 I and hence B(AB) −1 = A −1 .Now multiplying on the left by B −1 givesB −1 B(AB) −1 = B −1 A −1 ,and hence the stated result.Property (v). Finally, result (iv) may extended to case (v) in a straightforward manner.For example, using result (iv) twice we find(ABC) −1 = (BC) −1 A −1 = C −1 B −1 A −1 . ◭We conclude this section by noting that the determinant |A −1 | of the inversematrix can be expressed very simply in terms of the determinant |A| of the matrixitself. Again we start with the fundamental expression (8.61). Then, using theproperty (8.52) for the determinant of a product, we find|AA −1 | = |A||A −1 | = |I|.It is straightforward to show by Laplace expansion that |I| = 1, and so we arriveat the useful result|A −1 | = 1|A| . (8.62)266

8.11 THE RANK OF A MATRIX8.11 The rank of a matrixThe rank of a general M × N matrix is an important concept, particularly inthe solution of sets of simultaneous linear equations, to be discussed in the nextsection, and we now discuss it in some detail. Like the trace and determinant,the rank of matrix A is a single number (or algebraic expression) that dependson the elements of A. Unlike the trace and determinant, however, the rank of amatrix can be defined even when A is not square. As we shall see, there are twoequivalent definitions of the rank of a general matrix.Firstly, the rank of a matrix may be defined in terms of the linear independenceof vectors. Suppose that the columns of an M × N matrix are interpreted asthe components in a given basis of N (M-component) vectors v 1 , v 2 ,...,v N ,asfollows:⎛A = ⎝↑ ↑ ↑v 1 v 2 ... v N↓ ↓ ↓Then the rank of A, denoted by rank A or by R(A), is defined as the numberof linearly independent vectors in the set v 1 , v 2 ,...,v N , and equals the dimensionof the vector space spanned by those vectors. Alternatively, we may consider therows of A to contain the components in a given basis of the M (N-component)vectors w 1 , w 2 ,...,w M as follows:⎛A = ⎜⎝← w 1 →← w 2 →.← w M →It may then be shown § that the rank of A is also equal to the number oflinearly independent vectors in the set w 1 , w 2 ,...,w M . From this definition it isshould be clear that the rank of A is unaffected by the exchange of two rows(or two columns) or by the multiplication of a row (or column) by a constant.Furthermore, suppose that a constant multiple of one row (column) is added toanother row (column): for example, we might replace the row w i by w i + cw j .This also has no effect on the number of linearly independent rows and so leavesthe rank of A unchanged. We may use these properties to evaluate the rank of agiven matrix.A second (equivalent) definition of the rank of a matrix may be given and usesthe concept of submatrices. A submatrix of A is any matrix that can be formedfrom the elements of A by ignoring one, or more than one, row or column. It⎞⎠ .⎞⎟⎠ .§ For a fuller discussion, see, for example, C. D. Cantrell, Modern Mathematical Methods for Physicistsand Engineers (Cambridge: Cambridge University Press, 2000), chapter 6.267

MATRICES AND VECTOR SPACESmay be shown that the rank of a general M × N matrix is equal to the size ofthe largest square submatrix of A whose determinant is non-zero. Therefore, if amatrix A has an r × r submatrix S with |S| ≠ 0, but no (r +1)× (r + 1) submatrixwith non-zero determinant then the rank of the matrix is r. From either definitionit is clear that the rank of A is less than or equal to the smaller of M and N.◮Determine the rank of the matrix⎛A = ⎝ 1 1 0 −22 0 2 24 1 3 1⎞⎠ .The largest possible square submatrices of A must be of dimension 3 × 3. Clearly, Apossesses four such submatrices, the determinants of which are given by∣ 1 1 0∣∣∣∣∣ 1 1 −22 0 2∣ 4 1 3 ∣ =0, 2 0 24 1 1 ∣ =0,∣1 0 −22 2 24 3 1∣ ∣∣∣∣∣ 1 0 −2∣ =0, 0 2 21 3 1∣ =0.(In each case the determinant may be evaluated as described in subsection 8.9.1.)The next largest square submatrices of A are of dimension 2 × 2. Consider, for example,the 2 × 2 submatrix formed by ignoring the third row and the third and fourth columnsof A; this has determinant∣ 1 12 0 ∣ =1× 0 − 2 × 1=−2.Since its determinant is non-zero, A is of rank 2 and we need not consider any other 2 × 2submatrix. ◭In the special case in which the matrix A is a square N×N matrix, by comparingeither of the above definitions of rank with our discussion of determinants insection 8.9, we see that |A| = 0 unless the rank of A is N. Inotherwords,A issingular unless R(A) =N.8.12 Special types of square matrixMatrices that are square, i.e. N × N, are very common in physical applications.We now consider some special forms of square matrix that are of particularimportance.8.12.1 Diagonal matricesThe unit matrix, which we have already encountered, is an example of a diagonalmatrix. Such matrices are characterised by having non-zero elements only on the268

8.12 SPECIAL TYPES OF SQUARE MATRIXleading diagonal, i.e. only elements A ij with i = j may be non-zero. For example,⎛1 0 0⎞A = ⎝ 0 2 0 ⎠ ,0 0 −3is a 3 × 3 diagonal matrix. Such a matrix is often denoted by A = diag (1, 2, −3).By performing a Laplace expansion, it is easily shown that the determinant of anN × N diagonal matrix is equal to the product of the diagonal elements. Thus, ifthe matrix has the form A = diag(A 11 ,A 22 ,...,A NN )then|A| = A 11 A 22 ···A NN . (8.63)Moreover, it is also straightforward to show that the inverse of A is also adiagonal matrix given by( )1A −1 1 1= diag , ,..., .A 11 A 22 A NNFinally, we note that, if two matrices A and B are both diagonal then they havethe useful property that their product is commutative:This is not true for matrices in general.AB = BA.8.12.2 Lower and upper triangular matricesA square matrix A is called lower triangular if all the elements above the principaldiagonal are zero. For example, the general form for a 3 × 3 lower triangularmatrix is⎛A 11 0 0⎞A 31 A 32 A 33A = ⎝ A 21 A 22 0 ⎠ ,where the elements A ij may be zero or non-zero. Similarly an upper triangularsquare matrix is one for which all the elements below the principal diagonal arezero. The general 3 × 3 form is thus⎛A 11 A 12 A 13⎞0 0 A 33A = ⎝ 0 A 22 A 23⎠ .By performing a Laplace expansion, it is straightforward to show that, in thegeneral N × N case, the determinant of an upper or lower triangular matrix isequal to the product of its diagonal elements,|A| = A 11 A 22 ···A NN . (8.64)269

MATRICES AND VECTOR SPACESClearly result (8.63) for diagonal matrices is a special case of this result. Moreover,it may be shown that the inverse of a non-singular lower (upper) triangular matrixis also lower (upper) triangular.8.12.3 Symmetric and antisymmetric matricesA square matrix A of order N with the property A = A T is said to be symmetric.Similarly a matrix for which A = −A T is said to be anti- or skew-symmetricand its diagonal elements a 11 ,a 22 ,...,a NN are necessarily zero. Moreover, if A is(anti-)symmetric then so too is its inverse A −1 . This is easily proved by notingthat if A = ±A T then(A −1 ) T =(A T ) −1 = ±A −1 .Any N × N matrix A can be written as the sum of a symmetric and anantisymmetric matrix, since we may writeA = 1 2 (A + AT )+ 1 2 (A − AT )=B + C,where clearly B = B T and C = −C T . The matrix B is therefore called thesymmetric part of A, andC is the antisymmetric part.◮If A is an N × N antisymmetric matrix, show that |A| =0if N is odd.If A is antisymmetric then A T = −A. Using the properties of determinants (8.49) and(8.51), we have|A| = |A T | = |−A| =(−1) N |A|.Thus, if N is odd then |A| = −|A|, which implies that |A| =0.◭8.12.4 Orthogonal matricesA non-singular matrix with the property that its transpose is also its inverse,A T = A −1 , (8.65)is called an orthogonal matrix. It follows immediately that the inverse of anorthogonal matrix is also orthogonal, since(A −1 ) T =(A T ) −1 =(A −1 ) −1 .Moreover, since for an orthogonal matrix A T A = I, we have|A T A| = |A T ||A| = |A| 2 = |I| =1.Thus the determinant of an orthogonal matrix must be |A| = ±1.An orthogonal matrix represents, in a particular basis, a linear operator thatleaves the norms (lengths) of real vectors unchanged, as we will now show.270

8.12 SPECIAL TYPES OF SQUARE MATRIXSuppose that y = A x is represented in some coordinate system by the matrixequation y = Ax; then〈y|y〉 is given in this coordinate system byy T y = x T A T Ax = x T x.Hence 〈y|y〉 = 〈x|x〉, showing that the action of a linear operator represented byan orthogonal matrix does not change the norm of a real vector.8.12.5 Hermitian and anti-Hermitian matricesAn Hermitian matrix is one that satisfies A = A † ,whereA † is the Hermitian conjugatediscussed in section 8.7. Similarly if A † = −A, thenA is called anti-Hermitian.A real (anti-)symmetric matrix is a special case of an (anti-)Hermitian matrix, inwhich all the elements of the matrix are real. Also, if A is an (anti-)Hermitianmatrix then so too is its inverse A −1 ,since(A −1 ) † =(A † ) −1 = ±A −1 .Any N × N matrix A can be written as the sum of an Hermitian matrix andan anti-Hermitian matrix, sinceA = 1 2 (A + A† )+ 1 2 (A − A† )=B + C,where clearly B = B † and C = −C † . The matrix B is called the Hermitian part ofA, andC is called the anti-Hermitian part.8.12.6 Unitary matricesA unitary matrix A is defined as one for whichA † = A −1 . (8.66)Clearly, if A is real then A † = A T , showing that a real orthogonal matrix is aspecial case of a unitary matrix, one in which all the elements are real. We notethat the inverse A −1 of a unitary is also unitary, since(A −1 ) † =(A † ) −1 =(A −1 ) −1 .Moreover, since for a unitary matrix A † A = I, we have|A † A| = |A † ||A| = |A| ∗ |A| = |I| =1.Thus the determinant of a unitary matrix has unit modulus.A unitary matrix represents, in a particular basis, a linear operator that leavesthe norms (lengths) of complex vectors unchanged. If y = A x is represented insome coordinate system by the matrix equation y = Ax then 〈y|y〉 is given in thiscoordinate system byy † y = x † A † Ax = x † x.271

MATRICES AND VECTOR SPACESHence 〈y|y〉 = 〈x|x〉, showing that the action of the linear operator represented bya unitary matrix does not change the norm of a complex vector. The action of aunitary matrix on a complex column matrix thus parallels that of an orthogonalmatrix acting on a real column matrix.8.12.7 Normal matricesA final important set of special matrices consists of the normal matrices, for whichAA † = A † A,i.e. a normal matrix is one that commutes with its Hermitian conjugate.We can easily show that Hermitian matrices and unitary matrices (or symmetricmatrices and orthogonal matrices in the real case) are examples of normalmatrices. For an Hermitian matrix, A = A † and soAA † = AA = A † A.Similarly, for a unitary matrix, A −1 = A † and soAA † = AA −1 = A −1 A = A † A.Finally, we note that, if A is normal then so too is its inverse A −1 ,sinceA −1 (A −1 ) † = A −1 (A † ) −1 =(A † A) −1 =(AA † ) −1 =(A † ) −1 A −1 =(A −1 ) † A −1 .This broad class of matrices is important in the discussion of eigenvectors andeigenvalues in the next section.8.13 Eigenvectors and eigenvaluesSuppose that a linear operator A transforms vectors x in an N-dimensionalvector space into other vectors A x in the same space. The possibility then arisesthat there exist vectors x each of which is transformed by A into a multiple ofitself. Such vectors would have to satisfyA x = λx. (8.67)Any non-zero vector x that satisfies (8.67) for some value of λ is called aneigenvector of the linear operator A ,andλ is called the corresponding eigenvalue.As will be discussed below, in general the operator A has N independenteigenvectors x i , with eigenvalues λ i .Theλ i are not necessarily all distinct.If we choose a particular basis in the vector space, we can write (8.67) in termsof the components of A and x with respect to this basis as the matrix equationAx = λx, (8.68)where A is an N × N matrix. The column matrices x that satisfy (8.68) obviously272

8.13 EIGENVECTORS AND EIGENVALUESrepresent the eigenvectors x of A in our chosen coordinate system. Conventionally,these column matrices are also referred to as the eigenvectors of the matrixA. § Clearly, if x is an eigenvector of A (with some eigenvalue λ) then any scalarmultiple µx is also an eigenvector with the same eigenvalue. We therefore oftenuse normalised eigenvectors, for whichx † x =1(note that x † x corresponds to the inner product 〈x|x〉 in our basis). Any eigenvectorx can be normalised by dividing all its components by the scalar (x † x) 1/2 .As will be seen, the problem of finding the eigenvalues and correspondingeigenvectors of a square matrix A plays an important role in many physicalinvestigations. Throughout this chapter we denote the ith eigenvector of a squarematrix A by x i and the corresponding eigenvalue by λ i . This superscript notationfor eigenvectors is used to avoid any confusion with components.◮A non-singular matrix A has eigenvalues λ i and eigenvectors x i . Find the eigenvalues andeigenvectors of the inverse matrix A −1 .The eigenvalues and eigenvectors of A satisfyAx i = λ i x i .Left-multiplying both sides of this equation by A −1 , we findA −1 Ax i = λ i A −1 x i .Since A −1 A = I, on rearranging we obtainA −1 x i = 1 λ ix i .Thus, we see that A −1 has the same eigenvectors x i as does A, but the correspondingeigenvalues are 1/λ i . ◭In the remainder of this section we will discuss some useful results concerningthe eigenvectors and eigenvalues of certain special (though commonly occurring)square matrices. The results will be established for matrices whose elements maybe complex; the corresponding properties for real matrices may be obtained asspecial cases.8.13.1 Eigenvectors and eigenvalues of a normal matrixIn subsection 8.12.7 we defined a normal matrix A as one that commutes with itsHermitian conjugate, so thatA † A = AA † .§ In this context, when referring to linear combinations of eigenvectors x we will normally use theterm ‘vector’.273

MATRICES AND VECTOR SPACESWe also showed that both Hermitian and unitary matrices (or symmetric andorthogonal matrices in the real case) are examples of normal matrices. We nowdiscuss the properties of the eigenvectors and eigenvalues of a normal matrix.If x is an eigenvector of a normal matrix A with corresponding eigenvalue λthen Ax = λx, or equivalently,(A − λI)x = 0. (8.69)Denoting B = A−λI, (8.69) becomes Bx = 0 and, taking the Hermitian conjugate,we also haveFrom (8.69) and (8.70) we then haveHowever, the product B † B is given by(Bx) † = x † B † = 0. (8.70)x † B † Bx = 0. (8.71)B † B =(A − λI) † (A − λI) =(A † − λ ∗ I)(A − λI) =A † A − λ ∗ A − λA † + λλ ∗ .Now since A is normal, AA † = A † A and soB † B = AA † − λ ∗ A − λA † + λλ ∗ =(A − λI)(A − λI) † = BB † ,and hence B is also normal. From (8.71) we then findx † B † Bx = x † BB † x =(B † x) † B † x = 0,from which we obtainB † x =(A † − λ ∗ I)x = 0.Therefore, for a normal matrix A, the eigenvalues of A † are the complex conjugatesof the eigenvalues of A.Let us now consider two eigenvectors x i and x j of a normal matrix A correspondingto two different eigenvalues λ i and λ j . We then haveMultiplying (8.73) on the left by (x i ) † we obtainHowever, on the LHS of (8.74) we haveAx i = λ i x i , (8.72)Ax j = λ j x j . (8.73)(x i ) † Ax j = λ j (x i ) † x j . (8.74)(x i ) † A =(A † x i ) † =(λ ∗ i x i ) † = λ i (x i ) † , (8.75)where we have used (8.40) and the property just proved for a normal matrix to274

8.13 EIGENVECTORS AND EIGENVALUESwrite A † x i = λ ∗ i xi . From (8.74) and (8.75) we have(λ i − λ j )(x i ) † x j = 0. (8.76)Thus, if λ i ≠ λ j the eigenvectors x i and x j must be orthogonal, i.e.(x i ) † x j = 0.It follows immediately from (8.76) that if all N eigenvalues of a normal matrixA are distinct then all N eigenvectors of A are mutually orthogonal. If, however,two or more eigenvalues are the same then further consideration is required. Aneigenvalue corresponding to two or more different eigenvectors (i.e. they are notsimply multiples of one another) is said to be degenerate. Suppose that λ 1 is k-folddegenerate, i.e.Ax i = λ 1 x i for i =1, 2,...,k, (8.77)but that it is different from any of λ k+1 , λ k+2 , etc. Then any linear combinationof these x i is also an eigenvector with eigenvalue λ 1 ,since,forz = ∑ ki=1 c ix i ,Az ≡ Ak∑c i x i =i=1k∑c i Ax i =i=1k∑c i λ 1 x i = λ 1 z. (8.78)If the x i defined in (8.77) are not already mutually orthogonal then we canconstruct new eigenvectors z i that are orthogonal by the following procedure:i=1z 1 = x 1 ,[z 2 = x 2 − (ẑ 1 ) † x 2] ẑ 1 ,[z 3 = x 3 − (ẑ 2 ) † x 3] [ẑ 2 − (ẑ 1 ) † x 3] ẑ 1 ,.z k = x k −[(ẑ k−1 ) † x k] [ẑ k−1 − ···− (ẑ 1 ) † x k] ẑ 1 .In this procedure, known as Gram–Schmidt orthogonalisation, each new eigenvectorz i is normalised to give the unit vector ẑ i before proceeding to the constructionof the next one (the normalisation is carried out by dividing each element ofthe vector z i by [(z i ) † z i ] 1/2 ). Note that each factor in brackets (ẑ m ) † x n is a scalarproduct and thus only a number. It follows that, as shown in (8.78), each vectorz i so constructed is an eigenvector of A with eigenvalue λ 1 and will remain soon normalisation. It is straightforward to check that, provided the previous neweigenvectors have been normalised as prescribed, each z i is orthogonal to all itspredecessors. (In practice, however, the method is laborious and the example insubsection 8.14.1 gives a less rigorous but considerably quicker way.)Therefore, even if A has some degenerate eigenvalues we can by constructionobtain a set of N mutually orthogonal eigenvectors. Moreover, it may be shown(although the proof is beyond the scope of this book) that these eigenvectorsare complete in that they form a basis for the N-dimensional vector space. As275

MATRICES AND VECTOR SPACESa result any arbitrary vector y can be expressed as a linear combination of theeigenvectors x i :y =N∑a i x i , (8.79)i=1where a i =(x i ) † y. Thus, the eigenvectors form an orthogonal basis for the vectorspace. By normalising the eigenvectors so that (x i ) † x i = 1 this basis is madeorthonormal.◮Show that a normal matrix A can be written in terms of its eigenvalues λ i and orthonormaleigenvectors x i asN∑A = λ i x i (x i ) † . (8.80)i=1The key to proving the validity of (8.80) is to show that both sides of the expression givethesameresultwhenactingonanarbitaryvectory. SinceA is normal, we may expand yin terms of the eigenvectors x i , as shown in (8.79). Thus, we haveN∑N∑Ay = A a i x i = a i λ i x i .Alternatively, the action of the RHS of (8.80) on y is given byN∑N∑λ i x i (x i ) † y = a i λ i x i ,i=1i=1since a i =(x i ) † y. We see that the two expressions for the action of each side of (8.80) on yare identical, which implies that this relationship is indeed correct. ◭i=1i=18.13.2 Eigenvectors and eigenvalues of Hermitian and anti-Hermitian matricesFor a normal matrix we showed that if Ax = λx then A † x = λ ∗ x. However, if A isalso Hermitian, A = A † , it follows necessarily that λ = λ ∗ . Thus, the eigenvaluesof an Hermitian matrix are real, a result which may be proved directly.◮Prove that the eigenvalues of an Hermitian matrix are real.For any particular eigenvector x i , we take the Hermitian conjugate of Ax i = λ i x i to give(x i ) † A † = λ ∗ i (x i ) † . (8.81)Using A † = A, sinceA is Hermitian, and multiplying on the right by x i ,weobtain(x i ) † Ax i = λ ∗ i (x i ) † x i . (8.82)But multiplying Ax i = λ i x i through on the left by (x i ) † gives(x i ) † Ax i = λ i (x i ) † x i .Subtracting this from (8.82) yields0 =(λ ∗ i − λ i )(x i ) † x i .276

8.13 EIGENVECTORS AND EIGENVALUESBut (x i ) † x i is the modulus squared of the non-zero vector x i and is thus non-zero. Henceλ ∗ i must equal λ i and thus be real. The same argument can be used to show that theeigenvalues of a real symmetric matrix are themselves real. ◭The importance of the above result will be apparent to any student of quantummechanics. In quantum mechanics the eigenvalues of operators correspond tomeasured values of observable quantities, e.g. energy, angular momentum, parityand so on, and these clearly must be real. If we use Hermitian operators toformulate the theories of quantum mechanics, the above property guaranteesphysically meaningful results.Since an Hermitian matrix is also a normal matrix, its eigenvectors are orthogonal(or can be made so using the Gram–Schmidt orthogonalisation procedure).Alternatively we can prove the orthogonality of the eigenvectors directly.◮Prove that the eigenvectors corresponding to different eigenvalues of an Hermitian matrixare orthogonal.Consider two unequal eigenvalues λ i and λ j and their corresponding eigenvectors satisfyingAx i = λ i x i , (8.83)Ax j = λ j x j . (8.84)Taking the Hermitian conjugate of (8.83) we find (x i ) † A † = λ ∗ i (xi ) † . Multiplying this on theright by x j we obtain(x i ) † A † x j = λ ∗ i (x i ) † x j ,and similarly multiplying (8.84) through on the left by (x i ) † we find(x i ) † Ax j = λ j (x i ) † x j .Then, since A † = A, the two left-hand sides are equal and, because the λ i are real, onsubtraction we obtain0 =(λ i − λ j )(x i ) † x j .Finally we note that λ i ≠ λ j and so (x i ) † x j = 0, i.e. the eigenvectors x i and x j areorthogonal. ◭In the case where some of the eigenvalues are equal, further justification of theorthogonality of the eigenvectors is needed. The Gram–Schmidt orthogonalisationprocedure discussed above provides a proof of, and a means of achieving,orthogonality. The general method has already been described and we will notrepeat it here.We may also consider the properties of the eigenvalues and eigenvectors of ananti-Hermitian matrix, for which A † = −A and thusAA † = A(−A) =(−A)A = A † A.Therefore matrices that are anti-Hermitian are also normal and so have mutuallyorthogonal eigenvectors. The properties of the eigenvalues are also simplydeduced, since if Ax = λx thenλ ∗ x = A † x = −Ax = −λx.277

MATRICES AND VECTOR SPACESHence λ ∗ = −λ and so λ must be pure imaginary (or zero). In a similar mannerto that used for Hermitian matrices, these properties may be proved directly.8.13.3 Eigenvectors and eigenvalues of a unitary matrixA unitary matrix satisfies A † = A −1 and is also a normal matrix, with mutuallyorthogonal eigenvectors. To investigate the eigenvalues of a unitary matrix, wenote that if Ax = λx thenx † x = x † A † Ax = λ ∗ λx † x,and we deduce that λλ ∗ = |λ| 2 = 1. Thus, the eigenvalues of a unitary matrixhave unit modulus.8.13.4 Eigenvectors and eigenvalues of a general square matrixWhen an N × N matrix is not normal there are no general properties of itseigenvalues and eigenvectors; in general it is not possible to find any orthogonalset of N eigenvectors or even to find pairs of orthogonal eigenvectors (exceptby chance in some cases). While the N non-orthogonal eigenvectors are usuallylinearly independent and hence form a basis for the N-dimensional vector space,this is not necessarily so. It may be shown (although we will not prove it) that anyN × N matrix with distinct eigenvalues has N linearly independent eigenvectors,which therefore form a basis for the N-dimensional vector space. If a generalsquare matrix has degenerate eigenvalues, however, then it may or may not haveN linearly independent eigenvectors. A matrix whose eigenvectors are not linearlyindependent is said to be defective.8.13.5 Simultaneous eigenvectorsWe may now ask under what conditions two different normal matrices can havea common set of eigenvectors. The result – that they do so if, and only if, theycommute – has profound significance for the foundations of quantum mechanics.To prove this important result let A and B be two N × N normal matrices andx i be the ith eigenvector of A corresponding to eigenvalue λ i ,i.e.Ax i = λ i x i for i =1, 2,... ,N.For the present we assume that the eigenvalues are all different.(i) First suppose that A and B commute. Now considerABx i = BAx i = Bλ i x i = λ i Bx i ,where we have used the commutativity for the first equality and the eigenvectorproperty for the second. It follows that A(Bx i )=λ i (Bx i ) and thus that Bx i is an278

8.13 EIGENVECTORS AND EIGENVALUESeigenvector of A corresponding to eigenvalue λ i . But the eigenvector solutions of(A − λ i I)x i = 0 are unique to within a scale factor, and we therefore conclude thatBx i = µ i x ifor some scale factor µ i . However, this is just an eigenvector equation for B andshows that x i is an eigenvector of B, in addition to being an eigenvector of A. Byreversing the roles of A and B, it also follows that every eigenvector of B is aneigenvector of A. Thus the two sets of eigenvectors are identical.(ii) Now suppose that A and B have all their eigenvectors in common, a typicalone x i satisfying bothAx i = λ i x i and Bx i = µ i x i .As the eigenvectors span the N-dimensional vector space, any arbitrary vector xin the space can be written as a linear combination of the eigenvectors,i=1x =N∑c i x i .Now consider bothN∑N∑N∑ABx = AB c i x i = A c i µ i x i = c i λ i µ i x i ,andi=1i=1i=1i=1i=1N∑N∑N∑BAx = BA c i x i = B c i λ i x i = c i µ i λ i x i .It follows that ABx and BAx are the same for any arbitrary x and hence that(AB − BA)x = 0for all x. Thatis,A and B commute.This completes the proof that a necessary and sufficient condition for twonormal matrices to have a set of eigenvectors in common is that they commute.It should be noted that if an eigenvalue of A, say, is degenerate then not all ofits possible sets of eigenvectors will also constitute a set of eigenvectors of B.However, provided that by taking linear combinations one set of joint eigenvectorscan be found, the proof is still valid and the result still holds.When extended to the case of Hermitian operators and continuous eigenfunctions(sections 17.2 and 17.3) the connection between commuting matrices anda set of common eigenvectors plays a fundamental role in the postulatory basisof quantum mechanics. It draws the distinction between commuting and noncommutingobservables and sets limits on how much information about a systemcan be known, even in principle, at any one time.279i=1

MATRICES AND VECTOR SPACES8.14 Determination of eigenvalues and eigenvectorsThe next step is to show how the eigenvalues and eigenvectors of a given N × Nmatrix A are found. To do this we refer to (8.68) and as in (8.69) rewrite it asAx − λIx =(A − λI)x = 0. (8.85)The slight rearrangement used here is to write x as Ix, whereI is the unit matrixof order N. The point of doing this is immediate since (8.85) now has the formof a homogeneous set of simultaneous equations, the theory of which will bedeveloped in section 8.18. What will be proved there is that the equation Bx = 0only has a non-trivial solution x if |B| = 0. Correspondingly, therefore, we musthave in the present case that|A − λI| = 0, (8.86)if there are to be non-zero solutions x to (8.85).Equation (8.86) is known as the characteristic equation for A and its LHS asthe characteristic or secular determinant of A. The equation is a polynomial ofdegree N in the quantity λ. TheN roots of this equation λ i , i =1, 2,...,N, givethe eigenvalues of A. Corresponding to each λ i there will be a column vector x i ,which is the ith eigenvector of A and can be found by using (8.68).It will be observed that when (8.86) is written out as a polynomial equation inλ, the coefficient of −λ N−1 in the equation will be simply A 11 + A 22 + ···+ A NNrelative to the coefficient of λ N . As discussed in section 8.8, the quantity ∑ Ni=1 A iiis the trace of A and, from the ordinary theory of polynomial equations, will beequal to the sum of the roots of (8.86):N∑λ i =TrA. (8.87)i=1This can be used as one check that a computation of the eigenvalues λ i has beendone correctly. Unless equation (8.87) is satisfied by a computed set of eigenvalues,they have not been calculated correctly. However, that equation (8.87) is satisfied isa necessary, but not sufficient, condition for a correct computation. An alternativeproof of (8.87) is given in section 8.16.◮Find the eigenvalues and normalised eigenvectors of the real symmetric matrix⎛A = ⎝ 1 1 31 1 −3⎞⎠ .3 −3 −3Using (8.86),∣1 − λ 1 31 1− λ −33 −3 −3 − λ280∣ =0.

8.14 DETERMINATION OF EIGENVALUES AND EIGENVECTORSExpanding out this determinant gives(1 − λ) [(1 − λ)(−3 − λ) − (−3)(−3)] +1[(−3)(3) − 1(−3 − λ)]+3[1(−3) − (1 − λ)(3)] =0,which simplifies to give(1 − λ)(λ 2 +2λ − 12) + (λ − 6) + 3(3λ − 6) = 0,⇒ (λ − 2)(λ − 3)(λ +6)=0.Hence the roots of the characteristic equation, which are the eigenvalues of A, areλ 1 =2,λ 2 =3,λ 3 = −6. We note that, as expected,λ 1 + λ 2 + λ 3 = −1=1+1− 3=A 11 + A 22 + A 33 =TrA.For the first root, λ 1 = 2, a suitable eigenvector x 1 , with elements x 1 , x 2 , x 3 ,mustsatisfyAx 1 =2x 1 or, equivalently,x 1 + x 2 +3x 3 =2x 1 ,x 1 + x 2 − 3x 3 =2x 2 , (8.88)3x 1 − 3x 2 − 3x 3 =2x 3 .These three equations are consistent (to ensure this was the purpose in finding the particularvalues of λ) andyieldx 3 =0,x 1 = x 2 = k, wherek is any non-zero number. A suitableeigenvector would thus bex 1 = (k k 0) T .If we apply the normalisation condition, we require k 2 + k 2 +0 2 =1ork =1/ √ 2. Hence( ) T 1x 1 1= √2 √2 0 = √ 1 (1 1 0) T .2Repeating the last paragraph, but with the factor 2 on the RHS of (8.88) replacedsuccessively by λ 2 =3andλ 3 = −6, gives two further normalised eigenvectorsx 2 = √ 1 (1 − 1 1) T , x 3 = √ 1 (1 − 1 − 2) T . ◭3 6In the above example, the three values of λ are all different and A is areal symmetric matrix. Thus we expect, and it is easily checked, that the threeeigenvectors are mutually orthogonal, i.e.(x1 ) Tx 2 = ( x 1) Tx 3 = ( x 2) Tx 3 =0.It will be apparent also that, as expected, the normalisation of the eigenvectorshas no effect on their orthogonality.8.14.1 Degenerate eigenvaluesWe return now to the case of degenerate eigenvalues, i.e. those that have two ormore associated eigenvectors. We have shown already that it is always possibleto construct an orthogonal set of eigenvectors for a normal matrix, see subsection8.13.1, and the following example illustrates one method for constructingsuch a set.281

MATRICES AND VECTOR SPACES◮Construct an orthonormal set of eigenvectors for the matrix⎛A = ⎝ 1 0 30 −2 0⎞⎠ .3 0 1We first determine the eigenvalues using |A − λI| =0:1 − λ 0 30=0 −2 − λ 0∣ 3 0 1− λ ∣ = −(1 − λ)2 (2 + λ) + 3(3)(2 + λ)=(4− λ)(λ +2) 2 .Thus λ 1 =4,λ 2 = −2 =λ 3 . The eigenvector x 1 = (x 1 x 2 x 3 ) T is found from⎛⎝ 1 0 3⎞ ⎛0 −2 0 ⎠ ⎝ x ⎞ ⎛1x 2⎠ =4⎝ x ⎞⎛1x 2⎠ ⇒ x 1 = √ 1 ⎝ 1 ⎞0 ⎠ .3 0 1 x 3 x 32 1A general column vector that is orthogonal to x 1 isx = (a b − a) T , (8.89)and it is easily shown that⎛⎞ ⎛ ⎞ ⎛ ⎞Ax = ⎝ 1 0 30 −2 0 ⎠ ⎝a b ⎠ = −2 ⎝a b ⎠ = −2x.3 0 1 −a−aThus x is a eigenvector of A with associated eigenvalue −2. It is clear, however, that thereis an infinite set of eigenvectors x all possessing the required property; the geometricalanalogue is that there are an infinite number of corresponding vectors x lying in theplane that has x 1 as its normal. We do require that the two remaining eigenvectors areorthogonal to one another, but this still leaves an infinite number of possibilities. For x 2 ,therefore, let us choose a simple form of (8.89), suitably normalised, say,x 2 = (0 1 0) T .The third eigenvector is then specified (to within an arbitrary multiplicative constant)by the requirement that it must be orthogonal to x 1 and x 2 ; thus x 3 may be found byevaluating the vector product of x 1 and x 2 and normalising the result. This givesx 3 = 1 √2(−1 0 1) T ,to complete the construction of an orthonormal set of eigenvectors. ◭8.15 Change of basis and similarity transformationsThroughout this chapter we have considered the vector x as a geometrical quantitythat is independent of any basis (or coordinate system). If we introduce a basise i , i =1, 2,...,N, into our N-dimensional vector space then we may writex = x 1 e 1 + x 2 e 2 + ···+ x N e N ,282

8.15 CHANGE OF BASIS AND SIMILARITY TRANSFORMATIONSand represent x in this basis by the column matrixx =(x 1 x 2 ··· x n ) T ,having components x i . We now consider how these components change as a resultof a prescribed change of basis. Let us introduce a new basis e ′ i , i =1, 2,...,N,which is related to the old basis bye ′ j =N∑S ij e i , (8.90)i=1the coefficient S ij being the ith component of e ′ j with respect to the old (unprimed)basis. For an arbitrary vector x it follows thatN∑ N∑N∑x = x i e i = x ′ je ′ j =i=1j=1x ′ jj=1 i=1N∑S ij e i .From this we derive the relationship between the components of x in the twocoordinate systems asN∑x i = S ij x ′ j,which we can write in matrix form asj=1x = Sx ′ (8.91)where S is the transformation matrix associated with the change of basis.Furthermore, since the vectors e ′ j are linearly independent, the matrix S isnon-singular and so possesses an inverse S −1 . Multiplying (8.91) on the left byS −1 we findx ′ = S −1 x, (8.92)which relates the components of x in the new basis to those in the old basis.Comparing (8.92) and (8.90) we note that the components of x transform inverselyto the way in which the basis vectors e i themselves transform. This has to be so,as the vector x itself must remain unchanged.We may also find the transformation law for the components of a linearoperator under the same change of basis. Now, the operator equation y = A x(which is basis independent) can be written as a matrix equation in each of thetwo bases asBut, using (8.91), we may rewrite the first equation asy = Ax, y ′ = A ′ x ′ . (8.93)Sy ′ = ASx ′ ⇒ y ′ = S −1 ASx ′ .283

MATRICES AND VECTOR SPACESComparing this with the second equation in (8.93) we find that the componentsof the linear operator A transform asA ′ = S −1 AS. (8.94)Equation (8.94) is an example of a similarity transformation – a transformationthat can be particularly useful in converting matrices into convenient forms forcomputation.Given a square matrix A, we may interpret it as representing a linear operatorA in a given basis e i . From (8.94), however, we may also consider the matrixA ′ = S −1 AS, for any non-singular matrix S, as representing the same linearoperator A but in a new basis e ′ j , related to the old basis bye ′ j = ∑ iS ij e i .Therefore we would expect that any property of the matrix A that representssome (basis-independent) property of the linear operator A will also be sharedby the matrix A ′ . We list these properties below.(i) If A = I then A ′ = I, since, from (8.94),(ii) The value of the determinant is unchanged:A ′ = S −1 IS = S −1 S = I. (8.95)|A ′ | = |S −1 AS| = |S −1 ||A||S| = |A||S −1 ||S| = |A||S −1 S| = |A|. (8.96)(iii) The characteristic determinant and hence the eigenvalues of A ′ are thesame as those of A: from (8.86),|A ′ − λI| = |S −1 AS − λI| = |S −1 (A − λI)S|= |S −1 ||S||A − λI| = |A − λI|. (8.97)(iv) The value of the trace is unchanged: from (8.87),Tr A ′ = ∑ A ′ ii = ∑ ∑ ∑(S −1 ) ij A jk S kiii j k= ∑ ∑ ∑S ki (S −1 ) ij A jk = ∑ ∑δ kj A jk = ∑ A jji j kj kj=TrA. (8.98)An important class of similarity transformations is that for which S is a unitarymatrix; in this case A ′ = S −1 AS = S † AS. Unitary transformation matricesare particularly important, for the following reason. If the original basis e i is284

8.16 DIAGONALISATION OF MATRICESorthonormal and the transformation matrix S is unitary then〈∑ ∣〈e ′ i|e ′ ∣∣∑ 〉j〉 = S ki e k S rj e r= ∑ k= ∑ kkS ∗ kiS ∗ kir∑S rj 〈e k |e r 〉r∑S rj δ kr = ∑ SkiS ∗ kj =(S † S) ij = δ ij ,rkshowing that the new basis is also orthonormal.Furthermore, in addition to the properties of general similarity transformations,for unitary transformations the following hold.(i) If A is Hermitian (anti-Hermitian) then A ′ is Hermitian (anti-Hermitian),i.e. if A † = ±A then(A ′ ) † =(S † AS) † = S † A † S = ±S † AS = ±A ′ . (8.99)(ii) If A is unitary (so that A † = A −1 )thenA ′ is unitary, since(A ′ ) † A ′ =(S † AS) † (S † AS) =S † A † SS † AS = S † A † AS= S † IS = I. (8.100)8.16 Diagonalisation of matricesSuppose that a linear operator A is represented in some basis e i , i =1, 2,...,N,by the matrix A. Consider a new basis x j given byx j =N∑S ij e i ,where the x j are chosen to be the eigenvectors of the linear operator A ,i.e.i=1A x j = λ j x j . (8.101)In the new basis, A is represented by the matrix A ′ = S −1 AS, which has aparticularly simple form, as we shall see shortly. The element S ij of S is the ithcomponent, in the old (unprimed) basis, of the jth eigenvector x j of A, i.e.thecolumns of S are the eigenvectors of the matrix A:⎛S = ⎝↑ ↑ ↑x 1 x 2 ··· x N↓ ↓ ↓285⎞⎠ ,

MATRICES AND VECTOR SPACESthat is, S ij =(x j ) i . Therefore A ′ is given by(S −1 AS) ij = ∑ ∑(S −1 ) ik A kl S ljk l= ∑ ∑(S −1 ) ik A kl (x j ) lk l= ∑ k= ∑ k(S −1 ) ik λ j (x j ) kλ j (S −1 ) ik S kj = λ j δ ij .So the matrix A ′ is diagonal with the eigenvalues of A as the diagonal elements,i.e.⎛λ 1 0 ··· 0⎞0 ··· 0 λ N A ′ =0 λ 2 .⎜⎝. ⎟. .. 0 ⎠ .Therefore, given a matrix A, if we construct the matrix S that has the eigenvectorsof A as its columns then the matrix A ′ = S −1 AS is diagonal and has theeigenvalues of A as its diagonal elements. Since we require S to be non-singular(|S| ≠ 0), the N eigenvectors of A must be linearly independent and form a basisfor the N-dimensional vector space. It may be shown that any matrix with distincteigenvalues can be diagonalised by this procedure. If, however, a general squarematrix has degenerate eigenvalues then it may, or may not, have N linearlyindependent eigenvectors. If it does not then it cannot be diagonalised.For normal matrices (which include Hermitian, anti-Hermitian and unitarymatrices) the N eigenvectors are indeed linearly independent. Moreover, whennormalised, these eigenvectors form an orthonormal set (or can be made to doso). Therefore the matrix S with these normalised eigenvectors as columns, i.e.whose elements are S ij =(x j ) i ,hastheproperty(S † S) ij = ∑ k(S † ) ik (S) kj = ∑ kS ∗ kiS kj = ∑ k(x i ) ∗ k(x j ) k = (x i ) † x j = δ ij .Hence S is unitary (S −1 = S † ) and the original matrix A can be diagonalised byA ′ = S −1 AS = S † AS.Therefore, any normal matrix A can be diagonalised by a similarity transformationusing a unitary transformation matrix S.286

8.16 DIAGONALISATION OF MATRICES◮Diagonalise the matrixA =⎛⎝ 1 0 0 −2 30⎞⎠ .3 0 1The matrix A is symmetric and so may be diagonalised by a transformation of the formA ′ = S † AS, whereS has the normalised eigenvectors of A as its columns. We have alreadyfound these eigenvectors in subsection 8.14.1, and so we can write straightaway⎛⎞S = √ 1 1√0 −1⎝ 0 2 0 ⎠ .2 1 0 1We note that although the eigenvalues of A are degenerate, its three eigenvectors arelinearly independent and so A can still be diagonalised. Thus, calculating S † AS we obtain⎛⎞ ⎛S † AS = 1 1√0 1⎝ 0 2 0 ⎠ ⎝ 1 0 3⎞ ⎛⎞1√0 −10 −2 0 ⎠ ⎝ 0 2 0 ⎠2−1 0 1 3 0 1 1 0 1⎛= ⎝ 4 0 0⎞0 −2 0 ⎠ ,0 0 −2which is diagonal, as required, and has as its diagonal elements the eigenvalues of A. ◭If a matrix A is diagonalised by the similarity transformation A ′ = S −1 AS, sothat A ′ = diag(λ 1 ,λ 2 ,...,λ N ), then we have immediatelyTr A ′ =TrA =N∑λ i , (8.102)i=1|A ′ | = |A| =N∏λ i , (8.103)since the eigenvalues of the matrix are unchanged by the transformation. Moreover,these results may be used to prove the rather useful trace formulai=1| exp A| = exp(Tr A), (8.104)where the exponential of a matrix is as defined in (8.38).◮Prove the trace formula (8.104).At the outset, we note that for the similarity transformation A ′ = S −1 AS, we have(A ′ ) n =(S −1 AS)(S −1 AS) ···(S −1 AS) =S −1 A n S.Thus, from (8.38), we obtain exp A ′ = S −1 (exp A)S, from which it follows that | exp A ′ | =287

MATRICES AND VECTOR SPACES| exp A|. Moreover, by choosing the similarity transformation so that it diagonalises A, wehave A ′ =diag(λ 1 ,λ 2 ,...,λ N ), and soN∏| exp A| = | exp A ′ | = | exp[diag(λ 1 ,λ 2 ,...,λ N )]| = |diag(exp λ 1 , exp λ 2 ,...,exp λ N )| = exp λ i .Rewriting the final product of exponentials of the eigenvalues as the exponential of thesum of the eigenvalues, we find(N∏N)∑| exp A| = exp λ i =exp λ i =exp(TrA),i=1which gives the trace formula (8.104). ◭i=1i=18.17 Quadratic and Hermitian formsLet us now introduce the concept of quadratic forms (and their complex analogues,Hermitian forms). A quadratic form Q is a scalar function of a real vectorx given byQ(x) =〈x|A x〉, (8.105)for some real linear operator A . In any given basis (coordinate system) we canwrite (8.105) in matrix form asQ(x) =x T Ax, (8.106)where A is a real matrix. In fact, as will be explained below, we need only considerthe case where A is symmetric, i.e. A = A T . As an example in a three-dimensionalspace,( ) ⎛ 1 1 3Q = x T Ax = x 1 x 2 x 3⎝ 1 1 −33 −3 −3⎞ ⎛⎠ ⎝x 1x 2x 3⎞⎠= x 2 1 + x 2 2 − 3x 2 3 +2x 1 x 2 +6x 1 x 3 − 6x 2 x 3 . (8.107)It is reasonable to ask whether a quadratic form Q = x T Mx, whereM is any(possibly non-symmetric) real square matrix, is a more general definition. Thatthis is not the case may be seen by expressing M in terms of a symmetric matrixA = 1 2 (M+MT ) and an antisymmetric matrix B = 1 2 (M−MT ) such that M = A+B.We then haveHowever, Q is a scalar quantity and soQ = x T Mx = x T Ax + x T Bx. (8.108)Q = Q T =(x T Ax) T +(x T Bx) T = x T A T x + x T B T x = x T Ax − x T Bx.(8.109)Comparing (8.108) and (8.109) shows that x T Bx = 0, and hence x T Mx = x T Ax,288

8.17 QUADRATIC AND HERMITIAN FORMSi.e. Q is unchanged by considering only the symmetric part of M. Hence, with noloss of generality, we may assume A = A T in (8.106).From its definition (8.105), Q is clearly a basis- (i.e. coordinate-) independentquantity. Let us therefore consider a new basis related to the old one by anorthogonal transformation matrix S, the components in the two bases of anyvector x being related (as in (8.91)) by x = Sx ′ or, equivalently, by x ′ = S −1 x =S T x. We then haveQ = x T Ax =(x ′ ) T S T ASx ′ =(x ′ ) T A ′ x ′ ,where (as expected) the matrix describing the linear operator A in the newbasis is given by A ′ = S T AS (since S T = S −1 ). But, from the last section, if wechoose as S the matrix whose columns are the normalised eigenvectors of A thenA ′ = S T AS is diagonal with the eigenvalues of A as the diagonal elements. (SinceA is symmetric, its normalised eigenvectors are orthogonal, or can be made so,and hence S is orthogonal with S −1 = S T .)In the new basisQ = x T Ax =(x ′ ) T Λx ′ = λ 1 x ′ 12 + λ 2 x ′ 22 + ···+ λ N x ′ N2 , (8.110)where Λ = diag(λ 1 ,λ 2 ,...,λ N )andtheλ i are the eigenvalues of A. It should benoted that Q contains no cross-terms of the form x ′ 1 x′ 2 .◮Find an orthogonal transformation that takes the quadratic form (8.107) into the formλ 1 x ′ 12 + λ 2 x ′ 22 + λ 3 x ′ 32 .The required transformation matrix S has the normalised eigenvectors of A as its columns.We have already found these in section 8.14, and so we can write immediately⎛ √ √ ⎞3S = √ 1 √ √ 2 1⎝ 3 − 2 −1 ⎠6 √,0 2 −2which is easily verified as being orthogonal. Since the eigenvalues of A are λ =2,3,and−6, the general result already proved shows that the transformation x = Sx ′ will carry(8.107) into the form 2x ′ 12 +3x ′ 22 − 6x ′ 32 . This may be verified most easily by writing outthe inverse transformation x ′ = S −1 x = S T x and substituting. The inverse equations arex ′ 1 =(x 1 + x 2 )/ √ 2,x ′ 2 =(x 1 − x 2 + x 3 )/ √ 3, (8.111)x ′ 3 =(x 1 − x 2 − 2x 3 )/ √ 6.If these are substituted into the form Q =2x ′ 12 +3x ′ 22 − 6x ′ 32 then the original expression(8.107) is recovered. ◭In the definition of Q it was assumed that the components x 1 , x 2 , x 3 and thematrix A were real. It is clear that in this case the quadratic form Q ≡ x T Ax is real289

MATRICES AND VECTOR SPACESalso. Another, rather more general, expression that is also real is the HermitianformH(x) ≡ x † Ax, (8.112)where A is Hermitian (i.e. A † = A) and the components of x may now be complex.It is straightforward to show that H is real, sinceH ∗ =(H T ) ∗ = x † A † x = x † Ax = H.With suitable generalisation, the properties of quadratic forms apply also to Hermitianforms, but to keep the presentation simple we will restrict our discussionto quadratic forms.A special case of a quadratic (Hermitian) form is one for which Q = x T Axis greater than zero for all column matrices x. By choosing as the basis theeigenvectors of A we have Q in the formQ = λ 1 x 2 1 + λ 2 x 2 2 + λ 3 x 2 3.The requirement that Q>0 for all x means that all the eigenvalues λ i of A mustbe positive. A symmetric (Hermitian) matrix A with this property is called positivedefinite. If,instead,Q ≥ 0 for all x then it is possible that some of the eigenvaluesare zero, and A is called positive semi-definite.8.17.1 The stationary properties of the eigenvectorsConsider a quadratic form, such as Q(x) =〈x|A x〉, equation (8.105), in a fixedbasis. As the vector x is varied, through changes in its three components x 1 , x 2and x 3 , the value of the quantity Q also varies. Because of the homogeneousform of Q we may restrict any investigation of these variations to vectors of unitlength (since multiplying any vector x by any scalar k simply multiplies the valueof Q by a factor k 2 ).Of particular interest are any vectors x that make the value of the quadraticform a maximum or minimum. A necessary, but not sufficient, condition for thisis that Q is stationary with respect to small variations ∆x in x, whilst 〈x|x〉 ismaintained at a constant value (unity).In the chosen basis the quadratic form is given by Q = x T Ax and, usingLagrange undetermined multipliers to incorporate the variational constraints, weare led to seek solutions of∆[x T Ax − λ(x T x − 1)] = 0. (8.113)This may be used directly, together with the fact that (∆x T )Ax = x T A ∆x, sinceAis symmetric, to obtainAx = λx (8.114)290

8.17 QUADRATIC AND HERMITIAN FORMSas the necessary condition that x must satisfy. If (8.114) is satisfied for someeigenvector x then the value of Q(x) is given byQ = x T Ax = x T λx = λ. (8.115)However, if x and y are eigenvectors corresponding to different eigenvalues thenthey are (or can be chosen to be) orthogonal. Consequently the expression y T Axis necessarily zero, sincey T Ax = y T λx = λy T x =0. (8.116)Summarising, those column matrices x of unit magnitude that make thequadratic form Q stationary are eigenvectors of the matrix A, and the stationaryvalue of Q is then equal to the corresponding eigenvalue. It is straightforwardto see from the proof of (8.114) that, conversely, any eigenvector of A makes Qstationary.Instead of maximising or minimising Q = x T Ax subject to the constraintx T x = 1, an equivalent procedure is to extremise the functionλ(x) = xT Axx T x .◮Show that if λ(x) is stationary then x is an eigenvector of A and λ(x) is equal to thecorresponding eigenvalue.We require ∆λ(x) = 0 with respect to small variations in x. Now∆λ = 1 [(x T x) ( ∆x T Ax + x T A ∆x ) − x T Ax ( ∆x T x + x T ∆x )](x T x) 2 ( )= 2∆xT Ax x T Ax ∆x T x− 2x T x x T x x T x ,since x T A ∆x =(∆x T )Ax and x T ∆x =(∆x T )x. Thus∆λ = 2x T x ∆xT [Ax − λ(x)x].Hence, if ∆λ =0thenAx = λ(x)x, i.e.x is an eigenvector of A with eigenvalue λ(x). ◭Thus the eigenvalues of a symmetric matrix A are the values of the functionλ(x) = xT Axx T xat its stationary points. The eigenvectors of A lie along those directions in spacefor which the quadratic form Q = x T Ax has stationary values, given a fixedmagnitude for the vector x. Similar results hold for Hermitian matrices.291

MATRICES AND VECTOR SPACES8.17.2 Quadratic surfacesThe results of the previous subsection may be turned round to state that thesurface given byx T Ax = constant = 1 (say) (8.117)and called a quadratic surface, has stationary values of its radius (i.e. origin–surface distance) in those directions that are along the eigenvectors of A. Morespecifically, in three dimensions the quadratic surface x T Ax = 1 has its principalaxes along the three mutually perpendicular eigenvectors of A, andthesquaresof the corresponding principal radii are given by λ −1i , i =1, 2, 3. As well ashaving this stationary property of the radius, a principal axis is characterised bythe fact that any section of the surface perpendicular to it has some degree ofsymmetry about it. If the eigenvalues corresponding to any two principal axes aredegenerate then the quadratic surface has rotational symmetry about the thirdprincipal axis and the choice of a pair of axes perpendicular to that axis is notuniquely defined.◮Find the shape of the quadratic surfacex 2 1 + x 2 2 − 3x 2 3 +2x 1 x 2 +6x 1 x 3 − 6x 2 x 3 =1.If, instead of expressing the quadratic surface in terms of x 1 , x 2 , x 3 , as in (8.107), wewere to use the new variables x ′ 1 , x′ 2 , x′ 3 defined in (8.111), for which the coordinate axesare along the three mutually perpendicular eigenvector directions (1, 1, 0), (1, −1, 1) and(1, −1, −2), then the equation of the surface would take the form (see (8.110))x ′ 212(1/ √ 2) + x′ 22 (1/ √ 3) − x′ 32 (1/ √ 6) =1.2Thus, for example, a section of the quadratic surface in the plane x ′ 3 =0,i.e.x 1 − x 2 −2x 3 = 0, is an ellipse, with semi-axes 1/ √ 2and1/ √ 3. Similarly a section in the planex ′ 1 = x 1 + x 2 = 0 is a hyperbola. ◭Clearly the simplest three-dimensional situation to visualise is that in which allthe eigenvalues are positive, since then the quadratic surface is an ellipsoid.28.18 Simultaneous linear equationsIn physical applications we often encounter sets of simultaneous linear equations.In general we may have M equations in N unknowns x 1 ,x 2 ,...,x N of the formA 11 x 1 + A 12 x 2 + ···+ A 1N x N = b 1 ,A 21 x 1 + A 22 x 2 + ···+ A 2N x N = b 2 ,..A M1 x 1 + A M2 x 2 + ···+ A MN x N = b M ,(8.118)292

8.18 SIMULTANEOUS LINEAR EQUATIONSwhere the A ij and b i have known values. If all the b i are zero then the system ofequations is called homogeneous, otherwise it is inhomogeneous. Depending on thegiven values, this set of equations for the N unknowns x 1 , x 2 , ..., x N may haveeither a unique solution, no solution or infinitely many solutions. Matrix analysismay be used to distinguish between the possibilities. The set of equations may beexpressed as a single matrix equation Ax = b, or, written out in full, as⎛ ⎞⎛⎞ ⎛ ⎞ bA 11 A 12 ... A 1N x 11A 21 A 22 ... A 2Nx 2b ⎜⎝. ⎟ ⎜ ⎟. . .. . ⎠ ⎝ . ⎠ = 2.⎜⎝ . ⎟⎠A M1 A M2 ... A MNx Nb M8.18.1 The range and null space of a matrixAs we discussed in section 8.2, we may interpret the matrix equation Ax = b asrepresenting, in some basis, the linear transformation A x = b of a vector x in anN-dimensional vector space V into a vector b in some other (in general different)M-dimensional vector space W .In general the operator A will map any vector in V into some particularsubspace of W , which may be the entire space. This subspace is called the rangeof A (or A) and its dimension is equal to the rank of A. Moreover, if A (andhence A) issingular then there exists some subspace of V that is mapped ontothe zero vector 0 in W ; that is, any vector y that lies in the subspace satisfiesA y = 0. This subspace is called the null space of A and the dimension of thisnull space is called the nullity of A. We note that the matrix A must be singularif M ≠ N and may be singular even if M = N.The dimensions of the range and the null space of a matrix are related throughthe fundamental relationshiprank A + nullity A = N, (8.119)where N is the number of original unknowns x 1 ,x 2 ,...,x N .◮Prove the relationship (8.119).As discussed in section 8.11, if the columns of an M × N matrix A are interpreted as thecomponents, in a given basis, of N (M-component) vectors v 1 , v 2 ,...,v N then rank A isequal to the number of linearly independent vectors in this set (this number is also equalto the dimension of the vector space spanned by these vectors). Writing (8.118) in termsof the vectors v 1 , v 2 ,...,v N , we havex 1 v 1 + x 2 v 2 + ···+ x N v N = b. (8.120)From this expression, we immediately deduce that the range of A is merely the span ofthe vectors v 1 , v 2 ,...,v N and hence has dimension r =rankA.293

MATRICES AND VECTOR SPACESIf a vector y lies in the null space of A then A y = 0, whichwemaywriteasy 1 v 1 + y 2 v 2 + ···+ y N v N = 0. (8.121)As just shown above, however, only r (≤ N) of these vectors are linearly independent. Byrenumbering, if necessary, we may assume that v 1 , v 2 ,...,v r form a linearly independentset; the remaining vectors, v r+1 , v r+2 ,...,v N , can then be written as a linear superpositionof v 1 , v 2 ,...,v r . We are therefore free to choose the N − r coefficients y r+1 ,y r+2 ,...,y Narbitrarily and (8.121) will still be satisfied for some set of r coefficients y 1 ,y 2 ,...,y r (whichare not all zero). The dimension of the null space is therefore N − r, and this completesthe proof of (8.119). ◭Equation (8.119) has far-reaching consequences for the existence of solutionsto sets of simultaneous linear equations such as (8.118). As mentioned previously,these equations may have no solution, aunique solution or infinitely many solutions.We now discuss these three cases in turn.No solutionThe system of equations possesses no solution unless b lies in the range of A ;inthis case (8.120) will be satisfied for some x 1 ,x 2 ,...,x N . This in turn requires thesetofvectorsb, v 1 , v 2 ,...,v N to have the same span (see (8.8)) as v 1 , v 2 ,...,v N .Interms of matrices, this is equivalent to the requirement that the matrix A and theaugmented matrix⎛⎞A 11 A 12 ... A 1N b 1A 21 A 22 ... A 2N b 1M = ⎜⎝. ⎟... . ⎠A M1 A M2 ... A MN b Mhave the same rank r. If this condition is satisfied then b does lie in the range ofA , and the set of equations (8.118) will have either a unique solution or infinitelymany solutions. If, however, A and M have different ranks then there will be nosolution.A unique solutionIf b lies in the range of A and if r = N then all the vectors v 1 , v 2 ,...,v N in (8.120)are linearly independent and the equation has a unique solution x 1 ,x 2 ,...,x N .Infinitely many solutionsIf b lies in the range of A and if r

8.18 SIMULTANEOUS LINEAR EQUATIONSany vector in the null space of A (i.e. A y = 0) thenA (x + y) =A x + A y = A x + 0 = b,and so x + y is also a solution. Since the null space is (n − r)-dimensional, so toois the space of solutions.We may use the above results to investigate the special case of the solution ofa homogeneous set of linear equations, for which b = 0. Clearly the set always hasthe trivial solution x 1 = x 2 = ··· = x n =0,andifr = N this will be the onlysolution. If r

MATRICES AND VECTOR SPACES◮Show that the set of simultaneous equations2x 1 +4x 2 +3x 3 =4,x 1 − 2x 2 − 2x 3 =0, (8.123)−3x 1 +3x 2 +2x 3 = −7,has a unique solution, and find that solution.The simultaneous equations can be represented by the matrix equation Ax = b, i.e.⎛⎝ 2 4 3⎞ ⎛1 −2 −2 ⎠ ⎝ x ⎞ ⎛1x 2⎠ = ⎝ 4 ⎞0 ⎠ .−3 3 2 x 3 −7As we have already shown that A −1 exists and have calculated it, see (8.59), it follows thatx = A −1 b or, more explicitly, that⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞⎝ x 1x 2⎠ = 1 ⎝ 2 1 −24 13 7 ⎠ ⎝ 4 0 ⎠ = ⎝ 2−3x113−3 −18 −8 −7 4⎠ . (8.124)Thus the unique solution is x 1 =2,x 2 = −3, x 3 =4.◭LU decompositionAlthough conceptually simple, finding the solution by calculating A −1 can becomputationally demanding, especially when N is large. In fact, as we shall nowshow, it is not necessary to perform the full inversion of A in order to solve thesimultaneous equations Ax = b. Rather, we can perform a decomposition of thematrix into the product of a square lower triangular matrix L and a square uppertriangular matrix U, which are such thatA = LU, (8.125)and then use the fact that triangular systems of equations can be solved verysimply.We must begin, therefore, by finding the matrices L and U such that (8.125)is satisfied. This may be achieved straightforwardly by writing out (8.125) incomponent form. For illustration, let us consider the 3 × 3 case. It is, in fact,always possible, and convenient, to take the diagonal elements of L as unity, sowe have⎛A = ⎝⎛= ⎝1 0 0L 21 1 0L 31 L 32 1⎞ ⎛⎠ ⎝U 11 U 12 U 13⎞0 U 22 U 23⎠U 11 U 12 U 13⎞L 21 U 11 L 21 U 12 + U 22 L 21 U 13 + U 23⎠ (8.126)The nine unknown elements of L and U can now be determined by equating296

8.18 SIMULTANEOUS LINEAR EQUATIONSthe nine elements of (8.126) to those of the 3 × 3 matrix A. This is done in theparticular order illustrated in the example below.Once the matrices L and U have been determined, one can use the decompositionto solve the set of equations Ax = b in the following way. From (8.125), we haveLUx = b, but this can be written as two triangular sets of equationsLy = b and Ux = y,where y is another column matrix to be determined. One may easily solve the firsttriangular set of equations for y, which is then substituted into the second set.The required solution x is then obtained readily from the second triangular setof equations. We note that, as with direct inversion, once the LU decompositionhas been determined, one can solve for various RHS column matrices b 1 , b 2 , ... ,with little extra work.◮Use LU decomposition to solve the set of simultaneous equations (8.123).We begin the determination of the matrices L and U by equating the elements of thematrix in (8.126) with those of the matrix⎛A = 1 −2 −2⎝ 2 4 3⎞⎠ .−3 3 2This is performed in the following order:1st row: U 11 =2, U 12 =4, U 13 =31st column: L 21 U 11 =1, L 31 U 11 = −3 ⇒ L 21 = 1 , L 2 31 = − 3 22nd row: L 21 U 12 + U 22 = −2 L 21 U 13 + U 23 = −2 ⇒ U 22 = −4, U 23 = − 7 22nd column: L 31 U 12 + L 32 U 22 =3 ⇒ L 32 = − 9 43rd row: L 31 U 13 + L 32 U 23 + U 33 =2 ⇒ U 33 = − 118Thus we may write the matrix A as⎛1 0 0⎞ ⎛2 4 3⎞⎜ 1A = LU = ⎝ 21 0⎟ ⎜⎠ ⎝ 0 −4 − 7 2⎟⎠ .− 3 2− 9 41 0 0 − 118We must now solve the set of equations Ly = b, whichread⎛⎜⎝⎞ ⎛1 0 01⎟ ⎜1 02 ⎠ ⎝− 3 − 9 12 4y 1y 2y 3⎞ ⎛⎟ ⎜⎠ = ⎝Since this set of equations is triangular, we quickly findy 1 =4, y 2 =0− ( 1 )(4) = −2, y 2 3 = −7 − (− 3 )(4) − (− 9 11)(−2) = − .2 4 2These values must then be substituted into the equations Ux = y, whichread⎛⎞ ⎛ ⎞ ⎛ ⎞2 4 3 x 1 4⎜⎝ 0 −4 − 7 ⎟ ⎜ ⎟ ⎜2 ⎠ ⎝ x 2 ⎠ = ⎝ −2⎟⎠ .0 0 − 11 x83 − 11229740−7⎞⎟⎠ .

MATRICES AND VECTOR SPACESThis set of equations is also triangular, and we easily find the solutionx 1 =2, x 2 = −3, x 3 =4,which agrees with the result found above by direct inversion. ◭We note, in passing, that one can calculate both the inverse and the determinantof A from its LU decomposition. To find the inverse A −1 , one solves the systemof equations Ax = b repeatedly for the N different RHS column matrices b = e i ,i =1, 2,...,N,wheree i is the column matrix with its ith element equal to unityand the others equal to zero. The solution x in each case gives the correspondingcolumn of A −1 . Evaluation of the determinant |A| is much simpler. From (8.125),we have|A| = |LU| = |L||U|. (8.127)Since L and U are triangular, however, we see from (8.64) that their determinantsare equal to the products of their diagonal elements. Since L ii = 1 for all i, wethus findN∏|A| = U 11 U 22 ···U NN = U ii .As an illustration, in the above example we find |A| = (2)(−4)(−11/8) = 11,which, as it must, agrees with our earlier calculation (8.58).Finally, we note that if the matrix A is symmetric and positive semi-definitethen we can decompose it asi=1A = LL † , (8.128)where L is a lower triangular matrix whose diagonal elements are not, in general,equal to unity. This is known as a Cholesky decomposition (in the special casewhere A is real, the decomposition becomes A = LL T ). The reason that we cannotset the diagonal elements of L equal to unity in this case is that we require thesame number of independent elements in L as in A. The requirement that thematrix be positive semi-definite is easily derived by considering the Hermitianform (or quadratic form in the real case)x † Ax = x † LL † x =(L † x) † (L † x).Denoting the column matrix L † x by y, we see that the last term on the RHSis y † y, which must be greater than or equal to zero. Thus, we require x † Ax ≥ 0for any arbitrary column matrix x, andsoA must be positive semi-definite (seesection 8.17).We recall that the requirement that a matrix be positive semi-definite is equivalentto demanding that all the eigenvalues of A are positive or zero. If one ofthe eigenvalues of A is zero, however, then from (8.103) we have |A| = 0 and so Ais singular. Thus, if A is a non-singular matrix, it must be positive definite (rather298

8.18 SIMULTANEOUS LINEAR EQUATIONSthan just positive semi-definite) in order to perform the Cholesky decomposition(8.128). In fact, in this case, the inability to find a matrix L that satisfies (8.128)implies that A cannot be positive definite.The Cholesky decomposition can be applied in an analogous way to the LUdecomposition discussed above, but we shall not explore it further.Cramer’s ruleAn alternative method of solution is to use Cramer’s rule, which also providessome insight into the nature of the solutions in the various cases. To illustratethis method let us consider a set of three equations in three unknowns,A 11 x 1 + A 12 x 2 + A 13 x 3 = b 1 ,A 21 x 1 + A 22 x 2 + A 23 x 3 = b 2 , (8.129)A 31 x 1 + A 32 x 2 + A 33 x 3 = b 3 ,which may be represented by the matrix equation Ax = b. We wish either to findthe solution(s) x to these equations or to establish that there are no solutions.From result (vi) of subsection 8.9.1, the determinant |A| is unchanged by addingto its first column the combinationx 2× (second column of |A|)+ x 3× (third column of |A|).x 1 x 1We thus obtain∣ A 11 A 12 A 13 ∣∣∣∣∣ |A| =A 21 A 22 A 23 =∣ A 31 A 32 A ∣ 33∣A 11 +(x 2 /x 1 )A 12 +(x 3 /x 1 )A 13 A 12 A 13 ∣∣∣∣∣A 21 +(x 2 /x 1 )A 22 +(x 3 /x 1 )A 23 A 22 A 23 ,A 31 +(x 2 /x 1 )A 32 +(x 3 /x 1 )A 33 A 32 A 33which, on substituting b i /x 1 for the ith entry in the first column, yields|A| = 1 x 1∣ ∣∣∣∣∣ b 1 A 12 A 13b 2 A 22 A 23b 3 A 32 A 33∣ ∣∣∣∣∣= 1 x 1∆ 1 .The determinant ∆ 1 is known as a Cramer determinant. Similar manipulations ofthe second and third columns of |A| yield x 2 and x 3 , and so the full set of resultsreadsx 1 = ∆ 1|A| , x 2 = ∆ 2|A| , x 3 = ∆ 3|A| , (8.130)where∣ ∣ b 1 A 12 A 13 ∣∣∣∣∣ A 11 b 1 A 13 ∣∣∣∣∣ ∆ 1 =b 2 A 22 A 23 , ∆ 2 =A 21 b 2 A 23 , ∆ 3 =∣ b 3 A 32 A ∣33 A 31 b 3 A ∣33A 11 A 12 b 1A 21 A 22 b 2A 31 A 32 b 3∣ ∣∣∣∣∣.It can be seen that each Cramer determinant ∆ i is simply |A| but with column ireplaced by the RHS of the original set of equations. If |A| ≠ 0 then (8.130) gives299

MATRICES AND VECTOR SPACESthe unique solution. The proof given here appears to fail if any of the solutionsx i is zero, but it can be shown that result (8.130) is valid even in such a case.◮Use Cramer’s rule to solve the set of simultaneous equations (8.123).Let us again represent these simultaneous equations by the matrix equation Ax = b, i.e.⎛⎝ 2 4 3⎞ ⎛1 −2 −2 ⎠ ⎝ x ⎞ ⎛1x 2⎠ = ⎝ 4 ⎞0 ⎠ .−3 3 2 x 3 −7From (8.58), the determinant of A is given by |A| = 11. Following the discussion givenabove, the three Cramer determinants are4 4 3∆ 1 =0 −2 −2∣ −7 3 2 ∣ , ∆ 2 4 32 =1 0 −2∣ −3 −7 2 ∣ , ∆ 2 4 43 =1 −2 0∣ −3 3 −7 ∣ .These may be evaluated using the properties of determinants listed in subsection 8.9.1and we find ∆ 1 = 22, ∆ 2 = −33 and ∆ 3 = 44. From (8.130) the solution to the equations(8.123) is given byx 1 = 2211 =2, x 2 = −3311 = −3, x 3 = 4411 =4,which agrees with the solution found in the previous example. ◭At this point it is useful to consider each of the three equations (8.129) as representinga plane in three-dimensional Cartesian coordinates. Using result (7.42)of chapter 7, the sets of components of the vectors normal to the planes are(A 11 ,A 12 ,A 13 ), (A 21 ,A 22 ,A 23 )and(A 31 ,A 32 ,A 33 ), and using (7.46) the perpendiculardistances of the planes from the origin are given byd i =b i(A2i1+ A 2 i2 + A2 i3) 1/2for i =1, 2, 3.Finding the solution(s) to the simultaneous equations above corresponds to findingthe point(s) of intersection of the planes.If there is a unique solution the planes intersect at only a single point. Thishappens if their normals are linearly independent vectors. Since the rows of Arepresent the directions of these normals, this requirement is equivalent to |A| ≠0.If b = (0 0 0) T = 0 then all the planes pass through the origin and, since thereis only a single solution to the equations, the origin is that solution.Let us now turn to the cases where |A| = 0. The simplest such case is that inwhich all three planes are parallel; this implies that the normals are all paralleland so A is of rank 1. Two possibilities exist:(i) the planes are coincident, i.e. d 1 = d 2 = d 3 , in which case there is aninfinity of solutions;(ii) the planes are not all coincident, i.e. d 1 ≠ d 2 and/or d 1 ≠ d 3 and/ord 2 ≠ d 3 , in which case there are no solutions.300

8.18 SIMULTANEOUS LINEAR EQUATIONS(a)(b)Figure 8.1 The two possible cases when A is of rank 2. In both cases all thenormals lie in a horizontal plane but in (a) the planes all intersect on a singleline (corresponding to an infinite number of solutions) whilst in (b) there areno common intersection points (no solutions).It is apparent from (8.130) that case (i) occurs when all the Cramer determinantsare zero and case (ii) occurs when at least one Cramer determinant is non-zero.The most complicated cases with |A| = 0 are those in which the normals to theplanes themselves lie in a plane but are not parallel. In this case A has rank 2.Again two possibilities exist and these are shown in figure 8.1. Just as in therank-1 case, if all the Cramer determinants are zero then we get an infinity ofsolutions (this time on a line). Of course, in the special case in which b = 0 (andthe system of equations is homogeneous), the planes all pass through the originand so they must intersect on a line through it. If at least one of the Cramerdeterminants is non-zero, we get no solution.These rules may be summarised as follows.(i) |A| ≠0,b ≠ 0: The three planes intersect at a single point that is not theorigin, and so there is only one solution, given by both (8.122) and (8.130).(ii) |A| ≠0,b = 0: The three planes intersect at the origin only and there isonly the trivial solution, x =0.(iii) |A| =0,b ≠ 0, Cramer determinants all zero: There is an infinity ofsolutions either on a line if A is rank 2, i.e. the cofactors are not all zero,or on a plane if A is rank 1, i.e. the cofactors are all zero.(iv) |A| =0,b ≠ 0, Cramer determinants not all zero: No solutions.(v) |A| =0,b = 0: The three planes intersect on a line through the origingiving an infinity of solutions.8.18.3 Singular value decompositionThere exists a very powerful technique for dealing with a simultaneous set oflinear equations Ax = b, such as (8.118), which may be applied whether or not301

MATRICES AND VECTOR SPACESthe number of simultaneous equations M is equal to the number of unknowns N.This technique is known as singular value decomposition (SVD) and is the methodof choice in analysing any set of simultaneous linear equations.We will consider the general case, in which A is an M × N (complex) matrix.Let us suppose we can write A as the product §A = USV † , (8.131)where the matrices U, S and V have the following properties.(i) The square matrix U has dimensions M × M and is unitary.(ii) The matrix S has dimensions M × N (thesamedimensionsasthoseofA)and is diagonal in the sense that S ij =0ifi ≠ j. We denote its diagonalelements by s i for i =1, 2,...,p,wherep = min(M,N); these elements aretermed the singular values of A.(iii) The square matrix V has dimensions N × N and is unitary.We must now determine the elements of these matrices in terms of the elements ofA. From the matrix A, we can construct two square matrices: A † A with dimensionsN ×N and AA † with dimensions M ×M. Both are clearly Hermitian. From (8.131),and using the fact that U and V are unitary, we findA † A = VS † U † USV † = VS † SV † (8.132)AA † = USV † VS † U † = USS † U † , (8.133)where S † S and SS † are diagonal matrices with dimensions N × N and M × Mrespectively. The first p elements of each diagonal matrix are s 2 i , i =1, 2,...,p,where p = min(M,N), and the rest (where they exist) are zero.These two equations imply that both V −1 A † AV ( = V −1 A † A(V † ) −1) and, bya similar argument, U −1 AA † U, must be diagonal. From our discussion of thediagonalisation of Hermitian matrices in section 8.16, we see that the columns ofV must therefore be the normalised eigenvectors v i , i =1, 2,...,N, of the matrixA † A and the columns of U must be the normalised eigenvectors u j , j =1, 2,...,M,of the matrix AA † . Moreover, the singular values s i must satisfy s 2 i = λ i ,wherethe λ i are the eigenvalues of the smaller of A † A and AA † . Clearly, the λ i arealso some of the eigenvalues of the larger of these two matrices, the remainingones being equal to zero. Since each matrix is Hermitian, the λ i are real and thesingular values s i may be taken as real and non-negative. Finally, to make thedecomposition (8.131) unique, it is customary to arrange the singular values indecreasing order of their values, so that s 1 ≥ s 2 ≥ ···≥ s p .§ The proof that such a decomposition always exists is beyond the scope of this book. For afull account of SVD one might consult, for example, G. H. Golub and C. F. Van Loan, MatrixComputations, 3rd edn (Baltimore MD: Johns Hopkins University Press, 1996).302

8.18 SIMULTANEOUS LINEAR EQUATIONS◮Show that, for i =1, 2,...,p, Av i = s i u i and A † u i = s i v i ,wherep =min(M,N).Post-multiplying both sides of (8.131) by V, and using the fact that V is unitary, we obtainAV = US.Since the columns of V and U consist of the vectors v i and u j respectively and S has onlydiagonal non-zero elements, we find immediately that, for i =1, 2,...,p,Av i = s i u i . (8.134)Moreover, we note that Av i =0fori = p +1,p+2,...,N.Taking the Hermitian conjugate of both sides of (8.131) and post-multiplying by U, weobtainA † U = VS † = VS T ,where we have used the fact that U is unitary and S is real. We then see immediately that,for i =1, 2,...,p,A † u i = s i v i . (8.135)We also note that A † u i =0fori = p +1,p+2,...,M. Results (8.134) and (8.135) are usefulfor investigating the properties of the SVD. ◭The decomposition (8.131) has some advantageous features for the analysis ofsets of simultaneous linear equations. These are best illustrated by writing thedecomposition (8.131) in terms of the vectors u i and v i asp∑A = s i u i (v i ) † ,i=1where p = min(M,N). It may be, however, that some of the singular values s iare zero, as a result of degeneracies in the set of M linear equations Ax = b.Let us suppose that there are r non-zero singular values. Since our convention isto arrange the singular values in order of decreasing size, the non-zero singularvalues are s i , i =1, 2,...,r, and the zero singular values are s r+1 ,s r+2 ,...,s p .Therefore we can write A asr∑A = s i u i (v i ) † . (8.136)i=1Let us consider the action of (8.136) on an arbitrary vector x. This is given byr∑Ax = s i u i (v i ) † x.i=1Since (v i ) † x is just a number, we see immediately that the vectors u i , i =1, 2,...,r,must span the range of the matrix A; moreover, these vectors form an orthonormalbasis for the range. Further, since this subspace is r-dimensional, we haverank A = r, i.e. the rank of A is equal to the number of non-zero singular values.The SVD is also useful in characterising the null space of A. From (8.119),we already know that the null space must have dimension N − r; so,ifA has r303

MATRICES AND VECTOR SPACESnon-zero singular values s i , i =1, 2,...,r, then from the worked example abovewe haveAv i =0 fori = r +1,r+2,...,N.Thus, the N − r vectors v i , i = r +1,r+2,...,N, form an orthonormal basis forthe null space of A.◮ Find the singular value decompostion of the matrix⎛⎞2 2 2 2⎜ 17 1A = ⎝ − 17 − 1 ⎟10 10 10 10 ⎠ . (8.137)3595− 3 5− 9 5The matrix A has dimension 3 × 4(i.e.M =3,N =4),andsowemayconstructfromit the 3 × 3matrixAA † and the 4 × 4matrixA † A (in fact, since A is real, the Hermitianconjugates are just transposes). We begin by finding the eigenvalues λ i and eigenvectors u iof the smaller matrix AA † . This matrix is easily found to be given by⎛⎞16 0 0AA † 12= ⎝⎠ ,02950125and its characteristic equation reads∣ 16 − λ 0 0 ∣∣∣∣∣ 2905∣− λ 125=(16− λ)(36 − 13λ + λ 2 )=0.12 360 − λ 5 5Thus, the eigenvalues are λ 1 = 16, λ 2 =9,λ 3 = 4. Since the singular values of A are givenby s i = √ λ i and the matrix S in (8.131) has the same dimensions as A, we have⎛S = ⎝ 4 0 0 0⎞0 3 0 0 ⎠ , (8.138)0 0 2 0where we have arranged the singular values in order of decreasing size. Now the matrix Uhas as its columns the normalised eigenvectors u i of the 3×3 matrixAA † . These normalisedeigenvectors correspond to the eigenvalues of AA † as follows:λ 1 =16 ⇒ u 1 = (1 0 0) Tand so we obtain the matrix5365λ 2 =9 ⇒ u 2 3=(05λ 3 =4 ⇒ u 3 =(0 − 4 5⎛1 0 0⎜U = ⎝ 03045− 4 5 535⎞45 )T35 )T ,⎟⎠ . (8.139)The columns of the matrix V in (8.131) are the normalised eigenvectors of the 4 × 4matrix A † A, which is given by⎛⎞29 21 3 11A † A = 1 ⎜ 21 29 11 3 ⎟⎝4 3 11 29 21⎠ .11 3 21 29304

8.18 SIMULTANEOUS LINEAR EQUATIONSWe already know from the above discussion, however, that the non-zero eigenvalues ofthis matrix are equal to those of AA † found above, and that the remaining eigenvalue iszero. The corresponding normalised eigenvectors are easily found:λ 1 =16 ⇒ v 1 = 1 (1 1 1 1)T2andsothematrixV is given byλ 2 =9 ⇒ v 2 = 1 (1 1 − 1 − 1)T2λ 3 =4 ⇒ v 3 = 1 (−1 1 1 − 1)T2λ 4 =0 ⇒ v 4 = 1 (1 − 1 1 − 1)T2⎛V = 1 ⎜⎝2⎞1 1 −1 11 1 1 −1 ⎟1 −1 1 1⎠ . (8.140)1 −1 −1 −1Alternatively, we could have found the first three columns of V by using the relation(8.135) to obtainv i = 1 A † u i for i =1, 2, 3.s iThe fourth eigenvector could then be found using the Gram–Schmidt orthogonalisationprocedure. We note that if there were more than one eigenvector corresponding to a zeroeigenvalue then we would need to use this procedure to orthogonalise these eigenvectorsbefore constructing the matrix V.Collecting our results together, we find the SVD of the matrix A:⎛1 0 0A = USV † ⎜= ⎝ 03045− 4 5 535⎛⎞ ⎛⎟⎠ ⎝ 4 0 0 0⎞0 3 0 0 ⎠⎜0 0 2 0 ⎝this can be verified by direct multiplication. ◭1212− 1 212112122− 1 2− 1 21212− 1 212− 1 212− 1 2Let us now consider the use of SVD in solving a set of M simultaneous linearequations in N unknowns, which we write again as Ax = b. Firstly, considerthe solution of a homogeneous set of equations, for which b = 0. As mentionedpreviously, if A is square and non-singular (and so possesses no zero singularvalues) then the equations have the unique trivial solution x = 0. Otherwise,anyof the vectors v i , i = r +1,r+2,...,N, or any linear combination of them, willbe a solution.In the inhomogeneous case, where b is not a zero vector, the set of equationswill possess solutions if b lies in the range of A. To investigate these solutions, itis convenient to introduce the N × M matrix S, which is constructed by takingthe transpose of S in (8.131) and replacing each non-zero singular value s i onthe diagonal by 1/s i . It is clear that, with this construction, SS is an M × Mdiagonal matrix with diagonal entries that equal unity for those values of j forwhich s j ≠ 0, and zero otherwise.Now consider the vectorˆx = VSU † b. (8.141)305⎞⎟⎠ ;

MATRICES AND VECTOR SPACESUsing the unitarity of the matrices U and V, we find thatAˆx − b = USSU † b − b = U(SS − I)U † b. (8.142)The matrix (SS − I) is diagonal and the jth element on its leading diagonal isnon-zero (and equal to −1) only when s j = 0. However, the jth element of thevector U † b is given by the scalar product (u j ) † b;ifb lies in the range of A, thisscalar product can be non-zero only if s j ≠ 0. Thus the RHS of (8.142) mustequal zero, and so ˆx given by (8.141) is a solution to the equations Ax = b. Wemay, however, add to this solution any linear combination of the N − r vectors v i ,i = r+1,r+2,...,N, that form an orthonormal basis for the null space of A; thus,in general, there exists an infinity of solutions (although it is straightforward toshow that (8.141) is the solution vector of shortest length). The only way in whichthe solution (8.141) can be unique is if the rank r equals N, so that the matrix Adoes not possess a null space; this only occurs if A is square and non-singular.If b does not lie in the range of A then the set of equations Ax = b doesnot have a solution. Nevertheless, the vector (8.141) provides the closest possible‘solution’ in a least-squares sense. In other words, although the vector (8.141)does not exactly solve Ax = b, it is the vector that minimises the residualɛ = |Ax − b|,where here the vertical lines denote the absolute value of the quantity theycontain, not the determinant. This is proved as follows.Suppose we were to add some arbitrary vector x ′ to the vector ˆx in (8.141).This would result in the addition of the vector b ′ = Ax ′ to Aˆx − b; b ′ is clearly inthe range of A since any part of x ′ belonging to the null space of A contributesnothing to Ax ′ . We would then have|Aˆx − b + b ′ | = |(USSU † − I)b + b ′ |= |U[(SS − I)U † b + U † b ′ ]|= |(SS − I)U † b + U † b ′ |; (8.143)in the last line we have made use of the fact that the length of a vector is leftunchanged by the action of the unitary matrix U. Now, the jth component of thevector (SS − I)U † b will only be non-zero when s j = 0. However, the jth elementof the vector U † b ′ is given by the scalar product (u j ) † b ′ , which is non-zero only ifs j ≠0,sinceb ′ lies in the range of A. Thus, as these two terms only contribute to(8.143) for two disjoint sets of j-values, its minimum value, as x ′ is varied, occurswhen b ′ = 0; this requires x ′ = 0.◮Find the solution(s) to the set of simultaneous linear equations Ax = b, whereA is givenby (8.137) and b = (1 0 0) T .To solve the set of equations, we begin by calculating the vector given in (8.141),x = VSU † b,306

8.19 EXERCISESwhere U and V are given by (8.139) and (8.140) respectively and S is obtained by takingthe transpose of S in (8.138) and replacing all the non-zero singular values s i by 1/s i . Thus,S reads⎛1⎞0 04 10 03 S = ⎜⎟⎝10 0 ⎠ .20 0 0Substituting the appropriate matrices into the expression for x we findx = 1 (1 1 1 8 1)T . (8.144)It is straightforward to show that this solves the set of equations Ax = b exactly, andso the vector b = (1 0 0) T must lie in the range of A. This is, in fact, immediatelyclear, since b = u 1 . The solution (8.144) is not, however, unique. There are three non-zerosingular values, but N = 4. Thus, the matrix A has a one-dimensional null space, whichis ‘spanned’ by v 4 , the fourth column of V, given in (8.140). The solutions to our set ofequations, consisting of the sum of the exact solution and any vector in the null space ofA, therefore lie along the linex = 1 (1 1 1 8 1)T + α(1 − 1 1 − 1) T ,where the parameter α can take any real value. We note that (8.144) is the point on thisline that is closest to the origin. ◭8.19 Exercises8.1 Which of the following statements about linear vector spaces are true? Where astatement is false, give a counter-example to demonstrate this.(a) Non-singular N × N matrices form a vector space of dimension N 2 .(b) Singular N × N matrices form a vector space of dimension N 2 .(c) Complex numbers form a vector space of dimension 2.(d) Polynomial functions of x form an infinite-dimensional vector space.(e) Series {a 0 ,a 1 ,a 2 ,...,a N } for which ∑ Nn=0 |a n| 2 =1formanN-dimensionalvector space.(f) Absolutely convergent series form an infinite-dimensional vector space.(g) Convergent series with terms of alternating sign form an infinite-dimensionalvector space.8.2 Evaluate the determinants∣ a h g∣∣∣∣∣∣ 1 0 2 3(a)h b f∣ g f c ∣ , (b) 0 1 −2 13 −3 4 −2−2 1 −2 1 ∣andgc ge a + ge gb + ge0 b b b(c)c e e b+ e.∣ a b b+ f b+ d ∣307

MATRICES AND VECTOR SPACES8.3 Using the properties of determinants, solve with a minimum of calculation thefollowing equations for x:x a a 1a x b 1x +2 x +4 x − 3(a)a b x 1=0, (b)x +3 x x+5∣ a b c 1 ∣∣ x − 2 x − 1 x +1 ∣ =0.8.4 Consider the matrices⎛(a) B =⎝ i 0 −i0 −i i−i i 0⎞⎛ √ √ √ ⎞⎠ , (b) C = √ 1 3 −√ 2 − 3⎝ 1 6 −1 ⎠ .82 0 2Are they (i) real, (ii) diagonal, (iii) symmetric, (iv) antisymmetric, (v) singular,(vi) orthogonal, (vii) Hermitian, (viii) anti-Hermitian, (ix) unitary, (x) normal?8.5 By considering the matricesA =(1 00 0) ( )0 0, B =,3 4show that AB = 0 does not imply that either A or B is the zero matrix, but thatit does imply that at least one of them is singular.8.6 This exercise considers a crystal whose unit cell has base vectors that are notnecessarily mutually orthogonal.(a) The basis vectors of the unit cell of a crystal, with the origin O at one corner,are denoted by e 1 , e 2 , e 3 .ThematrixG has elements G ij ,whereG ij = e i · e jand H ij are the elements of the matrix H ≡ G −1 . Show that the vectorsf i = ∑ j H ije j are the reciprocal vectors and that H ij = f i · f j .(b) If the vectors u and v are given byu = ∑ u i e i , v = ∑ v i f i ,iiobtain expressions for |u|, |v|, andu · v.(c) If the basis vectors are each of length a and the angle between each pair isπ/3, write down G and hence obtain H.(d) Calculate (i) the length of the normal from O onto the plane containing thepoints p −1 e 1 , q −1 e 2 , r −1 e 3 , and (ii) the angle between this normal and e 1 .8.7 Prove the following results involving Hermitian matrices:(a) If A is Hermitian and U is unitary then U −1 AU is Hermitian.(b) If A is anti-Hermitian then iA is Hermitian.(c) The product of two Hermitian matrices A and B is Hermitian if and only ifA and B commute.(d) If S is a real antisymmetric matrix then A =(I − S)(I + S) −1 is orthogonal.If A is given by( )cos θ sin θA =− sin θ cos θthen find the matrix S that is needed to express A in the above form.(e) If K is skew-hermitian, i.e. K † = −K, thenV =(I + K)(I − K) −1 is unitary.8.8 A and B are real non-zero 3 × 3 matrices and satisfy the equation(AB) T + B −1 A = 0.(a) Prove that if B is orthogonal then A is antisymmetric.308

8.19 EXERCISES(b) Without assuming that B is orthogonal, prove that A is singular.8.9 The commutator [X, Y] of two matrices is defined by the equation[X, Y] =XY − YX.Two anticommuting matrices A and B satisfyA 2 = I, B 2 = I, [A, B] =2iC.(a) Prove that C 2 = I and that [B, C] =2iA.(b) Evaluate [[[A, B], [B, C]], [A, B]].8.10 The four matrices S x , S y , S z and I are defined by( ) ( )0 10 −iS x =, S1 0y =,i 0( ) ( )1 01 0S z =, I =,0 −10 1where i 2 = −1. Show that S 2 x = I and S x S y = iS z , and obtain similar resultsby permutting x, y and z. Giventhatv is a vector with Cartesian components(v x ,v y ,v z ), the matrix S(v) is defined asS(v) =v x S x + v y S y + v z S z .Prove that, for general non-zero vectors a and b,S(a)S(b) =a · b I + i S(a × b).Without further calculation, deduce that S(a) andS(b) commute if and only if aand b are parallel vectors.8.11 A general triangle has angles α, β and γ and corresponding opposite sides a,b and c. Express the length of each side in terms of the lengths of the othertwo sides and the relevant cosines, writing the relationships in matrix and vectorform, using the vectors having components a, b, c and cos α, cos β,cos γ. Invert thematrix and hence deduce the cosine-law expressions involving α, β and γ.8.12 Given a matrixA =⎛⎝ 1 β α 1 00⎞⎠ ,0 0 1where α and β are non-zero complex numbers, find its eigenvalues and eigenvectors.Find the respective conditions for (a) the eigenvalues to be real and (b) theeigenvectors to be orthogonal. Show that the conditions are jointly satisfied ifand only if A is Hermitian.8.13 Using the Gram–Schmidt procedure:(a) construct an orthonormal set of vectors from the following:x 1 = (0 0 1 1) T , x 2 =(1 0 − 1 0) T ,x 3 = (1 2 0 2) T , x 4 =(2 1 1 1) T ;309

MATRICES AND VECTOR SPACES(b) find an orthonormal basis, within a four-dimensional Euclidean space, forthe subspace spanned by the three vectors (1 2 0 0) T ,(3 − 1 2 0) Tand (0 0 2 1) T .8.14 If a unitary matrix U is written as A + iB, whereA and B are Hermitian withnon-degenerate eigenvalues, show the following:(a) A and B commute;(b) A 2 + B 2 = I;(c) The eigenvectors of A are also eigenvectors of B;(d) The eigenvalues of U have unit modulus (as is necessary for any unitarymatrix).8.15 Determine which of the matrices below are mutually commuting, and, for thosethat are, demonstrate that they have a complete set of eigenvectors in common:( ) ( )6 −21 8A =, B =,C =−2 9(−9 −10−10 5), D =8 −11(14 22 118.16 Find the eigenvalues and a set of eigenvectors of the matrix⎛⎝ 3 4 −21 3 −1⎞⎠ .−1 −2 2Verify that its eigenvectors are mutually orthogonal.8.17 Find three real orthogonal column matrices, each of which is a simultaneouseigenvector ofA =⎛⎝ 0 0 10 1 01 0 0⎞⎛⎠ and B =).⎝ 0 1 11 0 11 1 08.18 Use the results of the first worked example in section 8.14 to evaluate, withoutrepeated matrix multiplication, the expression A 6 x,wherex =(2 4 − 1) T andA is the matrix given in the example.8.19 Given that A is a real symmetric matrix with normalised eigenvectors e i ,obtainthe coefficients α i involved when column matrix x, which is the solution ofAx − µx = v, (∗)is expanded as x = ∑ i α ie i .Hereµ is a given constant and v is a given columnmatrix.⎞⎠ .(a) Solve (∗) whenA =⎛⎝ 2 1 1 2 00⎞⎠ ,0 0 3µ =2andv = (1 2 3) T .(b) Would (∗) have a solution if µ = 1 and (i) v = (1 2 3) T , (ii) v =(2 2 3) T ?310

8.19 EXERCISES8.20 Demonstrate that the matrix⎛⎞A = −6 4 4⎝ 2 0 0⎠3 −1 0is defective, i.e. does not have three linearly independent eigenvectors, by showingthe following:(a) its eigenvalues are degenerate and, in fact, all equal;(b) any eigenvector has the form (µ (3µ − 2ν) ν) T .(c) if two pairs of values, µ 1 ,ν 1 and µ 2 ,ν 2 , define two independent eigenvectorsv 1 and v 2 ,thenany third similarly defined eigenvector v 3 canbewrittenasa linear combination of v 1 and v 2 ,i.e.v 3 = av 1 + bv 2 ,wherea = µ 3ν 2 − µ 2 ν 3µ 1 ν 2 − µ 2 ν 1and b = µ 1ν 3 − µ 3 ν 1.µ 1 ν 2 − µ 2 ν 1Illustrate (c) using the example (µ 1 ,ν 1 )=(1, 1), (µ 2 ,ν 2 )=(1, 2) and (µ 3 ,ν 3 )=(0, 1).Show further that of the formany matrix⎛⎝ 6n − 6 4− 2n 4 − 4n2 0 0⎞⎠3 − 3n n− 1 2nis defective, with the same eigenvalues and eigenvectors as A.8.21 By finding the eigenvectors of the Hermitian matrix( )10 3iH =,−3i 2construct a unitary matrix U such that U † HU = Λ, where Λ is a real diagonalmatrix.8.22 Use the stationary properties of quadratic forms to determine the maximum andminimum values taken by the expressionQ =5x 2 +4y 2 +4z 2 +2xz +2xyon the unit sphere, x 2 + y 2 + z 2 = 1. For what values of x, y and z do they occur?8.23 Given that the matrix⎛A = −1 2 −1⎝ 2 −1 0⎞⎠0 −1 2has two eigenvectors of the form (1 y 1) T , use the stationary property of theexpression J(x) =x T Ax/(x T x) to obtain the corresponding eigenvalues. Deducethe third eigenvalue.8.24 Find the lengths of the semi-axes of the ellipse73x 2 +72xy +52y 2 = 100,and determine its orientation.8.25 The equation of a particular conic section isQ ≡ 8x 2 1 +8x 2 2 − 6x 1 x 2 = 110.Determine the type of conic section this represents, the orientation of its principalaxes, and relevant lengths in the directions of these axes.311

MATRICES AND VECTOR SPACES8.26 Show that the quadratic surface5x 2 +11y 2 +5z 2 − 10yz +2xz − 10xy =4is an ellipsoid with semi-axes of lengths 2, 1 and 0.5. Find the direction of itslongest axis.8.27 Find the direction of the axis of symmetry of the quadratic surface7x 2 +7y 2 +7z 2 − 20yz − 20xz +20xy =3.8.28 For the following matrices, find the eigenvalues and sufficient of the eigenvectorsto be able to describe the quadratic surfaces associated with them:⎛(a) ⎝ 5 1 −1⎞ ⎛1 5 1 ⎠ , (b) ⎝ 1 2 2⎞ ⎛2 1 2 ⎠ , (c) ⎝ 1 2 1⎞2 4 2 ⎠ .−1 1 52 2 11 2 18.29 This exercise demonstrates the reverse of the usual procedure of diagonalising amatrix.(a) Rearrange the result A ′ = S −1 AS of section 8.16 to express the originalmatrix A in terms of the unitary matrix S and the diagonal matrix A ′ . Henceshow how to construct a matrix A that has given eigenvalues and given(orthogonal) column matrices as its eigenvectors.(b) Find the matrix that has as eigenvectors (1 2 1) T , (1 − 1 1) T and(1 0 − 1) T , with corresponding eigenvalues λ, µ and ν.(c) Try a particular case, say λ =3,µ = −2 andν = 1, and verify by explicitsolution that the matrix so found does have these eigenvalues.8.30 Find an orthogonal transformation that takes the quadratic formQ ≡−x 2 1 − 2x 2 2 − x 2 3 +8x 2 x 3 +6x 1 x 3 +8x 1 x 2into the formµ 1 y1 2 + µ 2 y2 2 − 4y3,2and determine µ 1 and µ 2 (see section 8.17).8.31 One method of determining the nullity (and hence the rank) of an M × N matrixA is as follows.• Write down an augmented transpose of A, by adding on the right an N × Nunit matrix and thus producing an N × (M + N) array B.• Subtract a suitable multiple of the first row of B from each of the other lowerrows so as to make B i1 =0fori>1.• Subtract a suitable multiple of the second row (or the uppermost row thatdoes not start with M zero values) from each of the other lower rows so as tomake B i2 =0fori>2.• Continue in this way until all remaining rows have zeros in the first M places.The number of such rows is equal to the nullity of A, andtheN rightmostentries of these rows are the components of vectors that span the null space.They can be made orthogonal if they are not so already.Use this method to show that the nullity of⎛⎞−1 3 2 73 10 −6 17A = ⎜−1 −2 2 −3⎟⎝ 2 3 −4 4 ⎠4 0 −8 −4312

8.19 EXERCISESis 2 and that an orthogonal base for the null space of A is provided by any twocolumn matrices of the form (2 + α i − 2α i 1 α i ) T ,forwhichtheα i (i =1, 2)are real and satisfy 6α 1 α 2 +2(α 1 + α 2 )+5=0.8.32 Do the following sets of equations have non-zero solutions? If so, find them.(a) 3x +2y + z =0, x − 3y +2z =0, 2x + y +3z =0.(b) 2x = b(y + z), x =2a(y − z), x =(6a − b)y − (6a + b)z.8.33 Solve the simultaneous equations2x +3y + z =11,x + y + z =6,5x − y +10z =34.8.34 Solve the following simultaneous equations for x 1 , x 2 and x 3 , using matrixmethods:x 1 +2x 2 +3x 3 =1,3x 1 +4x 2 +5x 3 =2,x 1 +3x 2 +4x 3 =3.8.35 Show that the following equations have solutions only if η = 1 or 2, and findthem in these cases:x + y + z =1,x +2y +4z = η,x +4y +10z = η 2 .8.36 Find the condition(s) on α such that the simultaneous equationsx 1 + αx 2 =1,x 1 − x 2 +3x 3 = −1,2x 1 − 2x 2 + αx 3 = −2have (a) exactly one solution, (b) no solutions, or (c) an infinite number ofsolutions; give all solutions where they exist.8.37 Make an LU decomposition of the matrix⎛A = ⎝ 3 6 9⎞1 0 5⎠2 −2 16and hence solve Ax = b, where(i)b = (21 9 28) T , (ii) b = (21 7 22) T .8.38 Make an LU decomposition of the matrix⎛⎞2 −3 1 3⎜1 4 −3 −3⎟A = ⎝5 3 −1 −1⎠ .3 −6 −3 1Hence solve Ax = b for (i) b =(−4 1 8 −5) T , (ii) b =(−10 0 −3 −24) T .Deduce that det A = −160 and confirm this by direct calculation.8.39 Use the Cholesky separation method to determine whether the following matricesare positive definite. For each that is, determine the corresponding lower diagonalmatrix L:⎛A =⎝ 2 1 31 3 −13 −1 1⎞ ⎛⎠ , B =313⎝ 5 0 √ ⎞3√0 3 0 ⎠ .3 0 3

MATRICES AND VECTOR SPACES8.40 Find the equation satisfied by the squares of the singular values of the matrixassociated with the following over-determined set of equations:2x +3y + z =0x − y − z =12x + y =02y + z = −2.Show that one of the singular values is close to zero. Determine the two largersingular values by an appropriate iteration process and the smallest one byindirect calculation.8.41 Find the SVD of⎛A = ⎝ 0 −1⎞1 1 ⎠ ,−1 0showing that the singular values are √ 3and1.8.42 Find the SVD form of the matrix⎛⎞22 28 −22⎜ 1 −2 −19⎟A = ⎝19 −2 −1⎠ .−6 12 6Use it to determine the best solution x of the equation Ax = b when (i) b =(6 − 39 15 18) T , (ii) b =(9 − 42 15 15) T , showing that (i) has an exactsolution, but that the best solution to (ii) has a residual of √ 18.8.43 Four experimental measurements of particular combinations of three physicalvariables, x, y and z, gave the following inconsistent results:13x +22y − 13z =4,10x − 8y − 10z =44,10x − 8y − 10z =47,9x − 18y − 9z =72.Find the SVD best values for x, y and z. Identify the null space of A and henceobtain the general SVD solution.8.20 Hints and answers8.1 (a) False. O N ,theN × N null matrix, ( is not ) non-singular. ( )1 0 0 0(b) False. Consider the sum ofand.0 0 0 1(c) True.(d) True.(e) False. Consider b n = a n + a n for which ∑ Nn=0 |b n| 2 =4≠ 1, or note that thereis no zero vector with unit norm.(f) True.(g) False. Consider the two series defined bya 0 = 1 , a 2 n =2(− 1 2 )n for n ≥ 1; b n = −(− 1 2 )n for n ≥ 0.Theseriesthatisthesumof{a n } and {b n } does not have alternating signsand so closure does not hold.8.3 (a) x = a, b or c; (b)x = −1; the equation is linear in x.314

8.20 HINTS AND ANSWERS8.5 Use the ( property of the determinant ) of a matrix product.0 − tan(θ/2)8.7 (d) S =.tan(θ/2) 0(e) Note that (I + K)(I − K) =I − K 2 =(I − K)(I + K).8.9 (b) 32iA.8.11 a = b cos γ + c cos β, and cyclic permutations; a 2 = b 2 + c 2 − 2bc cos α, and cyclicpermutations.8.13 (a) 2 −1/2 (0 0 1 1) T , 6 −1/2 (2 0 − 1 1) T ,39 −1/2 (−1 6 − 1 1) T , 13 −1/2 (2 1 2 − 2) T .(b) 5 −1/2 (1 2 0 0) T , (345) −1/2 (14 − 7 10 0) T ,(18 285) −1/2 (−56 28 98 69) T .8.15 C does not commute with the others; A, B and D have (1 − 2) T and (2 1) T ascommon eigenvectors.8.17 For A : (1 0 − 1) T , (1 α 1 1) T , (1 α 2 1) T .For B : (1 1 1) T , (β 1 γ 1 − β 1 − γ 1 ) T , (β 2 γ 2 − β 2 − γ 2 ) T .The α i , β i and γ i are arbitrary.Simultaneous and orthogonal: (1 0 − 1) T , (1 1 1) T , (1 − 2 1) T .8.19 α j =(v · e j∗ )/(λ j − µ), where λ j is the eigenvalue corresponding to e j .(a) x = (2 1 3) T .(b) Since µ is equal to one of A’s eigenvalues λ j , the equation only has a solutionif v · e j∗ = 0; (i) no solution; (ii) x =(1 1 3/2) T .8.21 U = (10) −1/2 (1, 3i;3i, 1), Λ = (1, 0; 0, 11).8.23 J =(2y 2 −4y +4)/(y 2 +2), with stationary values at y = ± √ 2 and correspondingeigenvalues 2 ∓ √ 2. From the trace property of A, the third eigenvalue equals 2.8.25 Ellipse; θ = π/4, a = √ 22; θ =3π/4, b = √ 10.8.27 The direction of the eigenvector having the unrepeated eigenvalue is(1, 1, −1)/ √ 3.8.29 (a) A = SA ′ S † ,whereS is the matrix whose columns are the eigenvectors of thematrix A to be constructed, and A ′ =diag(λ, µ, ν).(b) A =(λ +2µ +3ν, 2λ − 2µ, λ +2µ − 3ν; 2λ − 2µ, 4λ +2µ, 2λ − 2µ;λ +2µ − 3ν, 2λ − 2µ, λ +2µ +3ν).1(c) (1, 5, −2; 5, 4, 5; −2, 5, 1).38.31 The null space is spanned by (2 0 1 0) T and (1 − 2 0 1) T .8.33 x =3,y =1,z =2.8.35 First show that A is singular. η =1,x =1+2z, y = −3z; η =2,x =2z,y =1− 3z.8.37 L =(1, 0, 0; 1 , 1, 0; 2 , 3, 1),3 3U =(3, 6, 9; 0, −2, 2; 0, 0, 4).(i) x =(−1 1 2) T . (ii) x =(−3 2 2) T .8.39 A is not positive definite, as L 33 is calculated to be √ −6.B = LL T , where the non-zero elements of L areL 11 = √ 5,L 31 = √ 3/5, L 22 = √ 3,L 33 = √ 12/5.8.41⎛( )2 1A † A =, U = √ 1 ⎝1 26√ √ ⎞−1 3√ 22 0 2−1 − √ ⎠√, V = 1 (1 1√3 2 2 1 −18.43 The singular values are 12 √ 6, 0, 18 √ 3 and the calculated best solution is x =1.71,y = −1.94,z = −1.71. The null space is the line x = z,y = 0 and the generalSVD solution is x =1.71 + λ, y = −1.94, z= −1.71 + λ.).315

9Normal modesAny student of the physical sciences will encounter the subject of oscillations onmany occasions and in a wide variety of circumstances, for example the voltageand current oscillations in an electric circuit, the vibrations of a mechanicalstructure and the internal motions of molecules. The matrices studied in theprevious chapter provide a particularly simple way to approach what may appear,at first glance, to be difficult physical problems.We will consider only systems for which a position-dependent potential exists,i.e., the potential energy of the system in any particular configuration dependsupon the coordinates of the configuration, which need not be be lengths, however;the potential must not depend upon the time derivatives (generalised velocities) ofthese coordinates. So, for example, the potential −qv · A used in the Lagrangiandescription of a charged particle in an electromagnetic field is excluded. Afurther restriction that we place is that the potential has a local minimum atthe equilibrium point; physically, this is a necessary and sufficient condition forstable equilibrium. By suitably defining the origin of the potential, we may takeits value at the equilibrium point as zero.We denote the coordinates chosen to describe a configuration of the systemby q i , i =1, 2,...,N.Theq i need not be distances; some could be angles, forexample. For convenience we can define the q i so that they are all zero at theequilibrium point. The instantaneous velocities of various parts of the system willdepend upon the time derivatives of the q i , denoted by ˙q i . For small oscillationsthe velocities will be linear in the ˙q i and consequently the total kinetic energy Twill be quadratic in them – and will include cross terms of the form ˙q i˙q j withi ≠ j. The general expression for T can be written as the quadratic formT = ∑ ∑a ij ˙q i ˙q j = ˙q T A˙q, (9.1)i jwhere ˙q is the column vector (˙q 1 ˙q 2 ··· ˙q N ) T and the N × N matrix Ais real and may be chosen to be symmetric. Furthermore, A, like any matrix316

9.1 TYPICAL OSCILLATORY SYSTEMScorresponding to a kinetic energy, is positive definite; that is, whatever non-zeroreal values the ˙q i take, the quadratic form (9.1) has a value > 0.Turning now to the potential energy, we may write its value for a configurationq by means of a Taylor expansion about the origin q = 0,V (q) =V (0)+ ∑ i∂V(0)∂q iq i + 1 2∑ ∑ij∂ 2 V (0)∂q i ∂q jq i q j + ··· .However, we have chosen V (0) = 0 and, since the origin is an equilibrium point,there is no force there and ∂V(0)/∂q i = 0. Consequently, to second order in theq i we also have a quadratic form, but in the coordinates rather than in their timederivatives:V = ∑ ∑b ij q i q j = q T Bq, (9.2)i jwhere B is, or can be made, symmetric. In this case, and in general, the requirementthat the potential is a minimum means that the potential matrix B, like the kineticenergy matrix A, is real and positive definite.9.1 Typical oscillatory systemsWe now introduce particular examples, although the results of this section aregeneral, given the above restrictions, and the reader will find it easy to apply theresults to many other instances.Consider first a uniform rod of mass M and length l, attached by a light stringalso of length l to a fixed point P and executing small oscillations in a verticalplane. We choose as coordinates the angles θ 1 and θ 2 shown, with exaggeratedmagnitude, in figure 9.1. In terms of these coordinates the centre of gravity of therod has, to first order in the θ i , a velocity component in the x-direction equal tol˙θ 1 + 1 2 l˙θ 2 andinthey-direction equal to zero. Adding in the rotational kineticenergy of the rod about its centre of gravity we obtain, to second order in the ˙θ i ,T ≈ 1 2 Ml2 (˙θ 2 1 + 1 4 ˙θ 2 2 + ˙θ 1˙θ2 )+ 1 24 Ml2˙θ2 2= 1 6 Ml2 ( 3˙θ 2 1 +3˙θ 1˙θ2 + ˙θ 2 2)=112 Ml2˙q T ( 6 33 2)˙q, (9.3)where ˙q T =(˙θ 1˙θ2 ). The potential energy is given byso thatV = Mlg [ (1 − cos θ 1 )+ 1 2 (1 − cos θ 2) ] (9.4)( ) 6 0V ≈ 1 4 Mlg(2θ2 1 + θ2)= 2 112 MlgqT q, (9.5)0 3where g is the acceleration due to gravity and q =(θ 1second order in the θ i .317θ 2 ) T ; (9.5) is valid to

NORMAL MODESP P Pθ 1θ 2θ 2lθ 1θ 1θ 2(a) (b) (c)lFigure 9.1 A uniform rod of length l attached to the fixed point P by a lightstring of the same length: (a) the general coordinate system; (b) approximationto the normal mode with lower frequency; (c) approximation to the mode withhigher frequency.With these expressions for T and V we now apply the conservation of energy,d(T + V )=0, (9.6)dtassuming that there are no external forces other than gravity. In matrix form(9.6) becomesddt (˙qT A˙q + q T Bq) =¨q T A˙q + ˙q T A¨q + ˙q T Bq + q T B˙q =0,which, using A = A T and B = B T , gives2˙q T (A¨q + Bq) =0.We will assume, although it is not clear that this gives the only possible solution,that the above equation implies that the coefficient of each ˙q i is separately zero.HenceA¨q + Bq = 0. (9.7)For a rigorous derivation Lagrange’s equations should be used, as in chapter 22.Nowwesearchforsetsofcoordinatesq that all oscillate with the same period,i.e. the total motion repeats itself exactly after a finite interval. Solutions of thisform will satisfyq = x cos ωt; (9.8)the relative values of the elements of x in such a solution will indicate how each318

9.1 TYPICAL OSCILLATORY SYSTEMScoordinate is involved in this special motion. In general there will be N valuesof ω if the matrices A and B are N × N and these values are known as normalfrequencies or eigenfrequencies.Putting (9.8) into (9.7) yields−ω 2 Ax + Bx =(B − ω 2 A)x = 0. (9.9)Our work in section 8.18 showed that this can have non-trivial solutions only if|B − ω 2 A| =0. (9.10)This is a form of characteristic equation for B, except that the unit matrix I hasbeen replaced by A. It has the more familiar form if a choice of coordinates ismade in which the kinetic energy T is a simple sum of squared terms, i.e. it hasbeen diagonalised, and the scale of the new coordinates is then chosen to makeeach diagonal element unity.However, even in the present case, (9.10) can be solved to yield ω 2 k for k =1, 2,...,N,whereN is the order of A and B. The values of ω k can be usedwith (9.9) to find the corresponding column vector x k and the initial (stationary)physical configuration that, on release, will execute motion with period 2π/ω k .In equation (8.76) we showed that the eigenvectors of a real symmetric matrixwere, except in the case of degeneracy of the eigenvalues, mutually orthogonal.In the present situation an analogous, but not identical, result holds. It is shownin section 9.3 that if x 1 and x 2 are two eigenvectors satisfying (9.9) for differentvalues of ω 2 then they are orthogonal in the sense that(x 2 ) T Ax 1 =0 and (x 2 ) T Bx 1 =0.The direct ‘scalar product’ (x 2 ) T x 1 , formally equal to (x 2 ) T Ix 1 ,isnot,ingeneral,equal to zero.Returning to the suspended rod, we find from (9.10)∣)− ω2 Ml 2( 6 00 3( 6 33 2)∣ Mlg∣∣∣∣=0.1212Writing ω 2 l/g = λ, this becomes∣ 6 − 6λ −3λ−3λ 3 − 2λ ∣ =0 ⇒ λ2 − 10λ +6=0,which has roots λ =5± √ 19. Thus we find that the two normal frequencies aregiven by ω 1 =(0.641g/l) 1/2 and ω 2 =(9.359g/l) 1/2 . Putting the lower of the twovalues for ω 2 , namely (5 − √ 19)g/l, into (9.9) shows that for this modex 1 : x 2 =3(5− √ 19) : 6( √ 19 − 4) = 1.923 : 2.153.This corresponds to the case where the rod and string are almost straight out, i.e.they almost form a simple pendulum. Similarly it may be shown that the higher319

NORMAL MODESfrequency corresponds to a solution where the string and rod are moving withopposite phase and x 1 : x 2 =9.359 : −16.718. The two situations are shown infigure 9.1.In connection with quadratic forms it was shown in section 8.17 how to makea change of coordinates such that the matrix for a particular form becomesdiagonal. In exercise 9.6 a method is developed for diagonalising simultaneouslytwo quadratic forms (though the transformation matrix may not be orthogonal).If this process is carried out for A and B in a general system undergoing stableoscillations, the kinetic and potential energies in the new variables η i take theformsT = ∑ iV = ∑ iµ i˙η 2 i = ˙η T M˙η, M = diag (µ 1 ,µ 2 ,...,µ N ), (9.11)ν i η 2 i = η T Nη, N = diag (ν 1 ,ν 2 ...,ν N ), (9.12)and the equations of motion are the uncoupled equationsµ i¨η i + ν i η i =0, i =1, 2,...,N. (9.13)Clearly a simple renormalisation of the η i can be made that reduces all the µ iin (9.11) to unity. When this is done the variables so formed are called normalcoordinates and equations (9.13) the normal equations.When a system is executing one of these simple harmonic motions it is said tobe in a normal mode, and once started in such a mode it will repeat its motionexactly after each interval of 2π/ω i . Any arbitrary motion of the system maybe written as a superposition of the normal modes, and each component modewill execute harmonic motion with the corresponding eigenfrequency; however,unless by chance the eigenfrequencies are in integer relationship, the system willnever return to its initial configuration after any finite time interval.As a second example we will consider a number of masses coupled together bysprings. For this type of situation the potential and kinetic energies are automaticallyquadratic functions of the coordinates and their derivatives, provided theelastic limits of the springs are not exceeded, and the oscillations do not have tobe vanishingly small for the analysis to be valid.◮Find the normal frequencies and modes of oscillation of three particles of masses m, µm,m connected in that order in a straight line by two equal light springs of force constant k.This arrangement could serve as a model for some linear molecules, e.g. CO 2 .The situation is shown in figure 9.2; the coordinates of the particles, x 1 , x 2 , x 3 , aremeasured from their equilibrium positions, at which the springs are neither extended norcompressed.The kinetic energy of the system is simplyT = 1 m ( ẋ 2 2 1 + µ ẋ 2 2 + ẋ3) 2 ,320

9.1 TYPICAL OSCILLATORY SYSTEMSmk µmkmx 1 x 2 x 3Figure 9.2 Three masses m, µm and m connected by two equal light springsof force constant k.(a)(b)(c)Figure 9.3 The normal modes of the masses and springs of a linear moleculesuch as CO 2 .(a)ω 2 =0;(b)ω 2 = k/m; (c)ω 2 =[(µ +2)/µ](k/m).whilst the potential energy stored in the springs isV = 1 k [ (x2 2 − x 1 ) 2 +(x 3 − x 2 ) 2] .The kinetic- and potential-energy symmetric matrices are thus⎛A = m ⎝ 1 0 0⎞⎛0 µ 0 ⎠ , B = k ⎝ 1 −1 0−1 2 −12 0 0 12 0 −1 1From (9.10), to find the normal frequencies we have to solve |B − ω 2 A| =0. Thus, writingmω 2 /k = λ, we have1 − λ −1 0−1 2− µλ −1∣ 0 −1 1− λ ∣ =0,which leads to λ =0,1or1+2/µ. The corresponding eigenvectors are respectively⎛x 1 = √ 1 ⎝ 1 ⎞⎛1 ⎠ , x 2 = √ 1 ⎝ 1 ⎞⎛0 ⎠ , x 3 1= √ ⎝ 1⎞−2/µ ⎠ .3 12 −12+(4/µ2 ) 1The physical motions associated with these normal modes are illustrated in figure 9.3.The first, with λ = ω =0andallthex i equal, merely describes bodily translation of thewhole system, with no (i.e. zero-frequency) internal oscillations.In the second solution the central particle remains stationary, x 2 = 0, whilst the othertwo oscillate with equal amplitudes in antiphase with each other. This motion, which hasfrequency ω =(k/m) 1/2 , is illustrated in figure 9.3(b).321⎞⎠ .

NORMAL MODESThe final and most complicated of the three normal modes has angular frequencyω = {[(µ +2)/µ](k/m)} 1/2 , and involves a motion of the central particle which is inantiphase with that of the two outer ones and which has an amplitude 2/µ timesasgreat.In this motion (see figure 9.3(c)) the two springs are compressed and extended in turn. Wealso note that in the second and third normal modes the centre of mass of the moleculeremains stationary. ◭9.2 Symmetry and normal modesIt will have been noticed that the system in the above example has an obvioussymmetry under the interchange of coordinates 1 and 3: the matrices A and B,the equations of motion and the normal modes illustrated in figure 9.3 are allunaltered by the interchange of x 1 and −x 3 . This reflects the more general resultthat for each physical symmetry possessed by a system, there is at least onenormal mode with the same symmetry.The general question of the relationship between the symmetries possessed bya physical system and those of its normal modes will be taken up more formallyin chapter 29 where the representation theory of groups is considered. However,we can show here how an appreciation of a system’s symmetry properties willsometimes allow its normal modes to be guessed (and then verified), somethingthat is particularly helpful if the number of coordinates involved is greater thantwo and the corresponding eigenvalue equation (9.10) is a cubic or higher-degreepolynomial equation.Consider the problem of determining the normal modes of a system consistingof four equal masses M at the corners of a square of side 2L, each pairof masses being connected by a light spring of modulus k that is unstretchedin the equilibrium situation. As shown in figure 9.4, we introduce Cartesiancoordinates x n ,y n , with n =1, 2, 3, 4, for the positions of the masses and denotetheir displacements from their equilibrium positions R n by q n = x n i + y n j.Thusr n = R n + q n with R n = ±Li ± Lj.The coordinates for the system are thus x 1 ,y 1 ,x 2 ,...,y 4 and the kinetic energymatrix A is given trivially by MI 8 , where I 8 is the 8 × 8 identity matrix.The potential energy matrix B is much more difficult to calculate and involves,for each pair of values m, n, evaluating the quadratic approximation to theexpressionb mn = 1 2 k ( |r m − r n |−|R m − R n | ) 2.Expressing each r i in terms of q i and R i and making the normal assumption that322

9.2 SYMMETRY AND NORMAL MODESy 1 y 2x 2MkMx 1kkkkMx 3ky 3 y 4x 4MFigure 9.4 The arrangement of four equal masses and six equal springsdiscussed in the text. The coordinate systems x n ,y n for n =1, 2, 3, 4 measurethe displacements of the masses from their equilibrium positions.|R m − R n |≫|q m − q n |,weobtainb mn (= b nm ):b mn = 1 2 k [ |(R m − R n )+(q m − q n )|−|R m − R n | ] 2= 1 2 k { [|Rm− R n | 2 +2(q m − q n ) · (R M − R n )+|q m − q n )| 2] 1/2−|Rm − R n |{ [= 1 2 k|R m − R n | 2 1+ 2(q ] 1/2 2m − q n ) · (R M − R n )|R m − R n | 2 + ··· − 1}{ }≈ 1 2 k (qm − q n ) · (R M − R n ) 2.|R m − R n |This final expression is readily interpretable as the potential energy stored in thespring when it is extended by an amount equal to the component, along theequilibrium direction of the spring, of the relative displacement of its two ends.Applying this result to each spring in turn gives the following expressions forthe elements of the potential matrix.m n 2b mn /k1 2 (x 1 − x 2 ) 21 3 (y 1 − y 3 ) 21 412 (−x 1 + x 4 + y 1 − y 4 ) 22 312 (x 2 − x 3 + y 2 − y 3 ) 22 4 (y 2 − y 4 ) 23 4 (x 3 − x 4 ) 2 .} 2323

NORMAL MODESThe potential matrix is thus constructed as⎛⎞3 −1 −2 0 0 0 −1 1−1 3 0 0 0 −2 1 −1−2 0 3 1 −1 −1 0 0B = k 0 0 1 3 −1 −1 0 −2.40 0 −1 −1 3 1 −2 0⎜0 −2 −1 −1 1 3 0 0⎟⎝ −1 1 0 0 −2 0 3 −1 ⎠1 −1 0 −2 0 0 −1 3To solve the eigenvalue equation |B − λA| = 0 directly would mean solvingan eigth-degree polynomial equation. Fortunately, we can exploit intuition andthe symmetries of the system to obtain the eigenvectors and correspondingeigenvalues without such labour.Firstly, we know that bodily translation of the whole system, without anyinternal vibration, must be possible and that there will be two independentsolutions of this form, corresponding to translations in the x- andy- directions.The eigenvector for the first of these (written in row form to save space) isEvaluation of Bx (1) givesx (1) = (1 0 1 0 1 0 1 0) T .Bx (1) = (0 0 0 0 0 0 0 0) T ,showing that x (1) is a solution of (B − ω 2 A)x = 0 corresponding to the eigenvalueω 2 = 0, whatever form Ax may take. Similarly,x (2) = (0 1 0 1 0 1 0 1) Tis a second eigenvector corresponding to the eigenvalue ω 2 =0.The next intuitive solution, again involving no internal vibrations, and, therefore,expected to correspond to ω 2 = 0, is pure rotation of the whole systemabout its centre. In this mode each mass moves perpendicularly to the line joiningits position to the centre, and so the relevant eigenvector isx (3) = 1 √2(1 1 1 − 1 − 1 1 − 1 − 1) T .It is easily verified that Bx (3) = 0 thus confirming both the eigenvector and thecorresponding eigenvalue. The three non-oscillatory normal modes are illustratedin diagrams (a)–(c) of figure 9.5.We now come to solutions that do involve real internal oscillations, and,because of the four-fold symmetry of the system, we expect one of them to be amode in which all the masses move along radial lines – the so-called ‘breathing324

9.2 SYMMETRY AND NORMAL MODES(a) ω 2 =0 (b)ω 2 =0 (c)ω 2 =0 (d) ω 2 =2k/M(e) ω 2 = k/M (f) ω 2 = k/M (g) ω 2 = k/M (h) ω 2 = k/MFigure 9.5 The displacements and frequencies of the eight normal modes ofthe system shown in figure 9.4. Modes (a), (b) and (c) are not true oscillations:(a) and (b) are purely translational whilst (c) is a mode of bodily rotation.Mode (d), the ‘breathing mode’, has the highest frequency and the remainingfour, (e)–(h), of lower frequency, are degenerate.mode’. Expressing this motion in coordinate form gives as the fourth eigenvectorx (4) = 1 √2(−1 1 1 1 − 1 − 1 1 − 1) T .Evaluation of Bx (4) yieldsBx (4) =k4 √ 2 (−8 8 8 8 − 8 − 8 8 − 8)T =2kx (4) ,i.e. a multiple of x (4) , confirming that it is indeed an eigenvector. Further, sinceAx (4) = Mx (4) , it follows from (B − ω 2 A)x = 0 that ω 2 =2k/M forthisnormalmode. Diagram (d) of the figure illustrates the corresponding motions of the fourmasses.As the next step in exploiting the symmetry properties of the system we notethat, because of its reflection symmetry in the x-axis, the system is invariant underthe double interchange of y 1 with −y 3 and y 2 with −y 4 . This leads us to try aneigenvector of the formx (5) =(0 α 0 β 0 − α 0 − β) T .Substituting this trial vector into (B − ω 2 A)x = 0 gives, of course, eight simulta-325

NORMAL MODESneous equations for α and β, but they are all equivalent to just two, namelyα + β =0,5α + β = 4Mω2 α;kthese have the solution α = −β and ω 2 = k/M. The latter thus gives the frequencyof the mode with eigenvectorx (5) =(0 1 0 − 1 0 − 1 0 1) T .Note that, in this mode, when the spring joining masses 1 and 3 is most stretched,the one joining masses 2 and 4 is at its most compressed. Similarly, based onreflection symmetry in the y-axis,x (6) =(1 0 − 1 0 − 1 0 1 0) Tcan be shown to be an eigenvector corresponding to the same frequency. Thesetwo modes are shown in diagrams (e) and (f) of figure 9.5.This accounts for six of the expected eight modes, and the other two could befound by considering motions that are symmetric about both diagonals of thesquare or are invariant under successive reflections in the x- andy- axes. However,since A is a multiple of the unit matrix, and since we know that (x (j) ) T Ax (i) =0ifi ≠ j, we can find the two remaining eigenvectors more easily by requiring themto be orthogonal to each of those found so far.Let us take the next (seventh) eigenvector, x (7) , to be given byx (7) =(a b c d e f g h) T .Then orthogonality with each of the x (n) for n =1, 2,...,6 yields six equationssatisfied by the unknowns a,b,...,h. As the reader may verify, they can be reducedto the six simple equationsa + g =0, d+ f =0, a+ f = d + g,b + h =0, c+ e =0, b+ c = e + h.With six homogeneous equations for eight unknowns, effectively separated intotwo groups of four, we may pick one in each group arbitrarily. Taking a = b =1gives d = e = 1 and c = f = g = h = −1 as a solution. Substitution ofx (7) =(1 1 − 1 1 1 − 1 − 1 − 1) T .into the eigenvalue equation checks that it is an eigenvector and shows that thecorresponding eigenfrequency is given by ω 2 = k/M.We now have the eigenvectors for seven of the eight normal modes and theeighth can be found by making it simultaneously orthogonal to each of the otherseven. It is left to the reader to show (or verify) that the final solution isx (8) =(1 − 1 1 1 − 1 − 1 − 1 1) T326

9.3 RAYLEIGH–RITZ METHODand that this mode has the same frequency as three of the other modes. Thegeneral topic of the degeneracy of normal modes is discussed in chapter 29. Themovements associated with the final two modes are shown in diagrams (g) and(h) of figure 9.5; this figure summarises all eight normal modes and frequencies.Although this example has been lengthy to write out, we have seen that theactual calculations are quite simple and provide the full solution to what isformally a matrix eigenvalue equation involving 8 × 8 matrices. It should benoted that our exploitation of the intrinsic symmetries of the system played acrucial part in finding the correct eigenvectors for the various normal modes.9.3 Rayleigh–Ritz methodWe conclude this chapter with a discussion of the Rayleigh–Ritz method forestimating the eigenfrequencies of an oscillating system. We recall from theintroduction to the chapter that for a system undergoing small oscillations thepotential and kinetic energy are given byV = q T Bq and T = ˙q T A˙q,where the components of q are the coordinates chosen to represent the configurationof the system and A and B are symmetric matrices (or may be chosen to besuch). We also recall from (9.9) that the normal modes x i and the eigenfrequenciesω i are given by(B − ω 2 i A)x i = 0. (9.14)It may be shown that the eigenvectors x i corresponding to different normal modesare linearly independent and so form a complete set. Thus, any coordinate vectorq can be written q = ∑ j c jx j . We now consider the value of the generalisedquadratic form∑λ(x) = xT Bxx T Ax = ∑ m(xm ) T c ∗ mB ∑ i c ix ij (xj ) T c ∗ j A ∑ k c kx k ,which, since both numerator and denominator are positive definite, is itself nonnegative.Equation (9.14) can be used to replace Bx i , with the result thatλ(x) ==∑m (xm ) T c ∗ mA ∑ ∑i ω2 i c ix ij (xj ) T c ∗ j A ∑ k c kx k∑m (xm ) T c ∗ m∑i ω2 i c iAx i∑j (xj ) T c ∗ j A ∑ k c kx k . (9.15)Now the eigenvectors x i obtained by solving (B − ω 2 A)x = 0 are not mutuallyorthogonal unless either A or B is a multiple of the unit matrix. However, it may327

NORMAL MODESbe shown that they do possess the desirable properties(x j ) T Ax i =0 and (x j ) T Bx i =0 ifi ≠ j. (9.16)This result is proved as follows. From (9.14) it is clear that, for general i and j,(x j ) T (B − ω 2 i A)x i = 0. (9.17)But, by taking the transpose of (9.14) with i replaced by j and recalling that Aand B are real and symmetric, we obtain(x j ) T (B − ω 2 j A) =0.Forming the scalar product of this with x i and subtracting the result from (9.17)gives(ω 2 j − ω 2 i )(x j ) T Ax i =0.Thus, for i ≠ j and non-degenerate eigenvalues ωi2 and ωj 2 , we have that(x j ) T Ax i = 0, and substituting this into (9.17) immediately establishes the correspondingresult for (x j ) T Bx i . Clearly, if either A or B is a multiple of the unitmatrix then the eigenvectors are mutually orthogonal in the normal sense. Theorthogonality relations (9.16) are derived again, and extended, in exercise 9.6.Using the first of the relationships (9.16) to simplify (9.15), we find thatλ(x) =∑i |c i| 2 ω 2 i (xi ) T Ax i∑k |c k| 2 (x k ) T Ax k . (9.18)Now, if ω0 2 is the lowest eigenfrequency then ω2 i ≥ ω0 2 for all i and, further, since(x i ) T Ax i ≥ 0 for all i the numerator of (9.18) is ≥ ω02 ∑i |c i| 2 (x i ) T Ax i .Henceλ(x) ≡ xT Bxx T Ax ≥ ω2 0, (9.19)for any x whatsoever (whether x is an eigenvector or not). Thus we are able toestimate the lowest eigenfrequency of the system by evaluating λ for a varietyof vectors x, the components of which, it will be recalled, give the ratios of thecoordinate amplitudes. This is sometimes a useful approach if many coordinatesare involved and direct solution for the eigenvalues is not possible.An additional result is that the maximum eigenfrequency ωm2 may also beestimated. It is obvious that if we replace the statement ‘ωi2 ≥ ω0 2 for all i’ by‘ωi 2 ≤ ωm 2 for all i’, then λ(x) ≤ ω2 m for any x. Thus λ(x) always lies betweenthe lowest and highest eigenfrequencies of the system. Furthermore, λ(x) has astationary value, equal to ωk 2 ,whenx is the kth eigenvector (see subsection 8.17.1).328

9.4 EXERCISES◮Estimate the eigenfrequencies of the oscillating rod of section 9.1.Firstly we recall thatA = Ml212(6 3)3 2andB = Mlg12( )6 0.0 3Physical intuition suggests that the slower mode will have a configuration approximatingthat of a simple pendulum (figure 9.1), in which θ 1 = θ 2 , and so we use this as a trialvector. Takingx =(θ θ) T ,λ(x) = xT Bxx T Ax = 3Mlgθ2 /47Ml 2 θ 2 /6 = 9g14l =0.643 g l ,and we conclude from (9.19) that the lower (angular) frequency is ≤ (0.643g/l) 1/2 .Wehave already seen on p. 319 that the true answer is (0.641g/l) 1/2 and so we have comevery close to it.Next we turn to the higher frequency. Here, a typical pattern of oscillation is not soobvious but, rather preempting the answer, we try θ 2 = −2θ 1 ; we then obtain λ =9g/land so conclude that the higher eigenfrequency ≥ (9g/l) 1/2 . We have already seen that theexact answer is (9.359g/l) 1/2 and so again we have come close to it. ◭A simplified version of the Rayleigh–Ritz method may be used to estimate theeigenvalues of a symmetric (or in general Hermitian) matrix B, the eigenvectorsof which will be mutually orthogonal. By repeating the calculations leading to(9.18), A being replaced by the unit matrix I, it is easily verified that ifλ(x) = xT Bxx T xis evaluated for any vector x thenλ 1 ≤ λ(x) ≤ λ m ,where λ 1 ,λ 2 ...,λ m are the eigenvalues of B in order of increasing size. A similarresult holds for Hermitian matrices.9.4 Exercises9.1 Three coupled pendulums swing perpendicularly to the horizontal line containingtheir points of suspension, and the following equations of motion are satisfied:−mẍ 1 = cmx 1 + d(x 1 − x 2 ),−Mẍ 2 = cMx 2 + d(x 2 − x 1 )+d(x 2 − x 3 ),−mẍ 3 = cmx 3 + d(x 3 − x 2 ),where x 1 , x 2 and x 3 are measured from the equilibrium points; m, M and mare the masses of the pendulum bobs; and c and d are positive constants. Findthe normal frequencies of the system and sketch the corresponding patterns ofoscillation. What happens as d → 0ord →∞?9.2 A double pendulum, smoothly pivoted at A, consists of two light rigid rods, ABand BC, each of length l, which are smoothly jointed at B and carry masses m andαm at B and C respectively. The pendulum makes small oscillations in one plane329

NORMAL MODESunder gravity. At time t, AB and BC make angles θ(t) andφ(t), respectively, withthe downward vertical. Find quadratic expressions for the kinetic and potentialenergies of the system and hence show that the normal modes have angularfrequencies given byω 2 = g [1+α ± √ ]α(1 + α) .lFor α =1/3, show that in one of the normal modes the mid-point of BC doesnot move during the motion.9.3 Continue the worked example, modelling a linear molecule, discussed at the endofsection9.1,forthecaseinwhichµ =2.(a) Show that the eigenvectors derived there have the expected orthogonalityproperties with respect to both A and B.(b) For the situation in which the atoms are released from rest with initialdisplacements x 1 =2ɛ, x 2 = −ɛ and x 3 = 0, determine their subsequentmotions and maximum displacements.9.4 Consider the circuit consisting of three equal capacitors and two different inductorsshown in the figure. For charges Q i on the capacitors and currents I iQ 1 Q 2CCQ 3L 1 C L 2I 1 I 2through the components, write down Kirchhoff’s law for the total voltage changearound each of two complete circuit loops. Note that, to within an unimportantconstant, the conservation of current implies that Q 3 = Q 1 − Q 2 . Express the loopequations in the form given in (9.7), namelyA¨Q + BQ = 0.Use this to show that the normal frequencies of the circuit are given byω 2 =1CL 1 L 2[L1 + L 2 ± (L 2 1 + L 2 2 − L 1 L 2 ) 1/2] .Obtain the same matrices and result by finding the total energy stored in thevarious capacitors (typically Q 2 /(2C)) and in the inductors (typically LI 2 /2).For the special case L 1 = L 2 = L determine the relevant eigenvectors and sodescribe the patterns of current flow in the circuit.9.5 It is shown in physics and engineering textbooks that circuits containing capacitorsand inductors can be analysed by replacing a capacitor of capacitance C by a‘complex impedance’ 1/(iωC) and an inductor of inductance L by an impedanceiωL, whereω is the angular frequency of the currents flowing and i 2 = −1.Use this approach and Kirchhoff’s circuit laws to analyse the circuit shown in330

9.4 EXERCISESthe figure and obtain three linear equations governing the currents I 1 , I 2 and I 3 .Show that the only possible frequencies of self-sustaining currents satisfy eitherPCI 1QLULSI 2 TI 3CCR(a) ω 2 LC =1or(b)3ω 2 LC = 1. Find the corresponding current patterns and,in each case, by identifying parts of the circuit in which no current flows, drawan equivalent circuit that contains only one capacitor and one inductor.9.6 The simultaneous reduction to diagonal form of two real symmetric quadratic forms.Consider the two real symmetric quadratic forms u T Au and u T Bu, whereu Tstands for the row matrix (x y z), and denote by u n those column matricesthat satisfyBu n = λ n Au n ,(E9.1)in which n is a label and the λ n are real, non-zero and all different.(a) By multiplying (E9.1) on the left by (u m ) T , and the transpose of the correspondingequation for u m on the right by u n , show that (u m ) T Au n =0forn ≠ m.(b) By noting that Au n =(λ n ) −1 Bu n , deduce that (u m ) T Bu n =0form ≠ n.(c) It can be shown that the u n are linearly independent; the next step is toconstruct a matrix P whose columns are the vectors u n .(d) Make a change of variables u = Pv such that u T Au becomes v T Cv, andu T Bubecomes v T Dv. Show that C and D are diagonal by showing that c ij =0ifi ≠ j, and similarly for d ij .Thus u = Pv or v = P −1 u reduces both quadratics to diagonal form.To summarise, the method is as follows:(a) find the λ n that allow (E9.1) a non-zero solution, by solving |B − λA| =0;(b) for each λ n construct u n ;(c) construct the non-singular matrix P whose columns are the vectors u n ;(d) make the change of variable u = Pv.9.7 (It is recommended that the reader does not attempt this question until exercise 9.6has been studied.)If, in the pendulum system studied in section 9.1, the string is replaced by asecond rod identical to the first then the expressions for the kinetic energy T andthe potential energy V become(tosecondorderintheθ i )T ≈ Ml ( 2 8 ˙θ 2 3 1 +2˙θ 1˙θ2 + 2 ˙θ 2 3 2),V ≈ Mgl ( 32 θ2 1 + 1 2 2) θ2 .Determine the normal frequencies of the system and find new variables ξ and ηthat will reduce these two expressions to diagonal form, i.e. toa 1˙ξ2 + a 2˙η 2 and b 1 ξ 2 + b 2 η 2 .331

NORMAL MODES9.8 (It is recommended that the reader does not attempt this question until exercise 9.6has been studied.)Find a real linear transformation that simultaneously reduces the quadraticforms3x 2 +5y 2 +5z 2 +2yz +6zx − 2xy,5x 2 +12y 2 +8yz +4zxto diagonal form.9.9 Three particles of mass m are attached to a light horizontal string having fixedends, the string being thus divided into four equal portions each of length a andunder a tension T . Show that for small transverse vibrations the amplitudes x iof the normal modes satisfy =(maω 2 /T )x, whereB is the matrixBx⎛⎝ −1 2 −12 −1 0⎞⎠ .0 −1 2Estimate the lowest and highest eigenfrequencies using trial vectors (3 4 3)(Tand (3 − 4 3) T . Use also the exact vectors 1 √ ) T (2 1 and 1 − √ ) T2 1and compare the results.9.10 Use the Rayleigh–Ritz method to estimate the lowest oscillation frequency of aheavy chain of N links, each of length a (= L/N), which hangs freely from oneend. (Try simple calculable configurations such as all links but one vertical, orall links collinear, etc.)9.5 Hints and answers9.1 See figure 9.6.9.3 (b) x 1 = ɛ(cos ωt +cos √ 2ωt), x 2 = −ɛ cos √ 2ωt, x 3 = ɛ(− cos ωt +cos √ 2ωt).At various times the three displacements will reach 2ɛ, ɛ, 2ɛ respectively. For example,x 1 canbewrittenas2ɛ cos[( √ 2−1)ωt/2] cos[( √ 2+1)ωt/2], i.e. an oscillationof angular frequency ( √ 2+1)ω/2 and modulated amplitude 2ɛ cos[( √ 2−1)ωt/2];the amplitude will reach 2ɛ after a time ≈ 4π/[ω( √ 2 − 1)].9.5 As the circuit loops contain no voltage sources, the equations are homogeneous,and so for a non-trivial solution the determinant of coefficients must vanish.(a) I 1 =0,I 2 = −I 3 ; no current in PQ; equivalent to two separate circuits ofcapacitance C and inductance L.(b) I 1 = −2I 2 = −2I 3 ; no current in TU; capacitance 3C/2 and inductance 2L.9.7 ω =(2.634g/l) 1/2 or (0.3661g/l) 1/2 ; θ 1 = ξ + η, θ 2 =1.431ξ − 2.097η.9.9 Estimated, 10/17

9.5 HINTS AND ANSWERS1 2 3mMm(a) ω 2 = c + d mlabel2kM2kmkM(c) ω 2 = c + 2dM + d mFigure 9.6 The normal modes, as viewed from above, of the coupled pendulumsin example 9.1.333

10Vector calculusIn chapter 7 we discussed the algebra of vectors, and in chapter 8 we consideredhow to transform one vector into another using a linear operator. In this chapterand the next we discuss the calculus of vectors, i.e. the differentiation andintegration both of vectors describing particular bodies, such as the velocity ofa particle, and of vector fields, in which a vector is defined as a function of thecoordinates throughout some volume (one-, two- or three-dimensional). Since theaim of this chapter is to develop methods for handling multi-dimensional physicalsituations, we will assume throughout that the functions with which we have todeal have sufficiently amenable mathematical properties, in particular that theyare continuous and differentiable.10.1 Differentiation of vectorsLet us consider a vector a that is a function of a scalar variable u. By thiswe mean that with each value of u we associate a vector a(u). For example, inCartesian coordinates a(u) =a x (u)i + a y (u)j + a z (u)k, wherea x (u), a y (u) anda z (u)are scalar functions of u and are the components of the vector a(u) inthex-, y-and z- directions respectively. We note that if a(u) is continuous at some pointu = u 0 then this implies that each of the Cartesian components a x (u), a y (u) anda z (u) is also continuous there.Let us consider the derivative of the vector function a(u) withrespecttou.The derivative of a vector function is defined in a similar manner to the ordinaryderivative of a scalar function f(x) given in chapter 2. The small change inthe vector a(u) resulting from a small change ∆u in the value of u is given by∆a = a(u +∆u) − a(u) (see figure 10.1). The derivative of a(u) with respect to u isdefined to bedadu = lim∆u→0a(u +∆u) − a(u), (10.1)∆u334

10.1 DIFFERENTIATION OF VECTORSa(u +∆u)∆a = a(u +∆u) − a(u)a(u)Figure 10.1 A small change in a vector a(u) resulting from a small changein u.assuming that the limit exists, in which case a(u) is said to be differentiable atthat point. Note that da/du is also a vector, which is not, in general, parallel toa(u). In Cartesian coordinates, the derivative of the vector a(u) =a x i + a y j + a z kis given bydadu = da xdu i + da ydu j + da zdu k.Perhaps the simplest application of the above is to finding the velocity andacceleration of a particle in classical mechanics. If the time-dependent positionvector of the particle with respect to the origin in Cartesian coordinates is givenby r(t) =x(t)i + y(t)j + z(t)k then the velocity of the particle is given by the vectorv(t) = drdt = dxdt i + dydt j + dzdt k.The direction of the velocity vector is along the tangent to the path r(t) attheinstantaneous position of the particle, and its magnitude |v(t)| is equal to thespeed of the particle. The acceleration of the particle is given in a similar mannerbya(t) = dvdt = d2 xdt 2 i + d2 ydt 2 j + d2 zdt 2 k.◮The position vector of a particle at time t in Cartesian coordinates is given by r(t) =2t 2 i +(3t − 2)j +(3t 2 − 1)k. Find the speed of the particle at t =1and the component ofits acceleration in the direction s = i +2j + k.The velocity and acceleration of the particle are given byv(t) = dr =4ti +3j +6tk,dta(t) = dv =4i +6k.dt335

VECTOR CALCULUSyê φjê ρρiφxFigure 10.2coordinates.Unit basis vectors for two-dimensional Cartesian and plane polarThespeedoftheparticleatt = 1 is simply|v(1)| = √ 4 2 +3 2 +6 2 = √ 61.The acceleration of the particle is constant (i.e. independent of t), and its component inthe direction s is given bya · ŝ =(4i +6k) · (i +2j + k)√12 +2 2 +1 2 = 5√ 63 . ◭Note that in the case discussed above i, j and k are fixed, time-independentbasis vectors. This may not be true of basis vectors in general; when we arenot using Cartesian coordinates the basis vectors themselves must also be differentiated.We discuss basis vectors for non-Cartesian coordinate systems indetail in section 10.10. Nevertheless, as a simple example, let us now considertwo-dimensional plane polar coordinates ρ, φ.Referring to figure 10.2, imagine holding φ fixed and moving radially outwards,i.e. in the direction of increasing ρ. Let us denote the unit vector in this directionby ê ρ . Similarly, imagine keeping ρ fixed and moving around a circle of fixed radiusin the direction of increasing φ. Let us denote the unit vector tangent to the circleby ê φ . The two vectors ê ρ and ê φ are the basis vectors for this two-dimensionalcoordinate system, just as i and j are basis vectors for two-dimensional Cartesiancoordinates. All these basis vectors are shown in figure 10.2.An important difference between the two sets of basis vectors is that, whilei and j are constant in magnitude and direction, the vectors ê ρ and ê φ haveconstant magnitudes but their directions change as ρ and φ vary. Therefore,when calculating the derivative of a vector written in polar coordinates we mustalso differentiate the basis vectors. One way of doing this is to express ê ρ and ê φ336

10.1 DIFFERENTIATION OF VECTORSin terms of i and j. From figure 10.2, we see thatê ρ =cosφ i +sinφ j,ê φ = − sin φ i +cosφ j.Since i and j are constant vectors, we find that the derivatives of the basis vectorsê ρ and ê φ with respect to t are given bydê ρdtdê φdt= − sin φ dφdt i +cosφdφ dt j = ˙φ ê φ , (10.2)= − cos φ dφ i − sin φdφdt dt j = −˙φ ê ρ , (10.3)where the overdot is the conventional notation for differentiation with respect totime.◮The position vector of a particle in plane polar coordinates is r(t) =ρ(t)ê ρ .Findexpressionsfor the velocity and acceleration of the particle in these coordinates.Using result (10.4) below, the velocity of the particle is given byv(t) =ṙ(t) =˙ρ ê ρ + ρ ˙ê ρ = ˙ρ ê ρ + ρ˙φ ê φ ,where we have used (10.2). In a similar way its acceleration is given bya(t) = d dt (˙ρ ê ρ + ρ˙φ ê φ )= ¨ρ ê ρ + ˙ρ ˙ê ρ + ρ˙φ ˙ê φ + ρ¨φ ê φ + ˙ρ˙φ ê φ= ¨ρ ê ρ + ˙ρ(˙φ ê φ )+ρ˙φ(−˙φ ê ρ )+ρ¨φ ê φ + ˙ρ˙φ ê φ=(¨ρ − ρ˙φ 2 ) ê ρ +(ρ¨φ +2˙ρ˙φ) ê φ . ◭Here we have used (10.2) and (10.3).10.1.1 Differentiation of composite vector expressionsIn composite vector expressions each of the vectors or scalars involved may bea function of some scalar variable u, as we have seen. The derivatives of suchexpressions are easily found using the definition (10.1) and the rules of ordinarydifferential calculus. They may be summarised by the following, in which weassume that a and b are differentiable vector functions of a scalar u and that φis a differentiable scalar function of u:d(φa) =φdadu du + dφ a, (10.4)duddb(a · b) =a ·du du + da · b, (10.5)duddb(a × b) =a ×du du + da × b. (10.6)du337

VECTOR CALCULUSThe order of the factors in the terms on the RHS of (10.6) is, of course, just asimportant as it is in the original vector product.◮A particleofmassm with position vector r relative to some origin O experiences a forceF, which produces a torque (moment) T = r × F about O. The angular momentum of theparticle about O is given by L = r × mv, wherev is the particle’s velocity. Show that therate of change of angular momentum is equal to the applied torque.The rate of change of angular momentum is given bydLdt = d (r × mv).dtUsing (10.6) we obtaindLdt = drdt × mv + r × d dt (mv)= v × mv + r × d dt (mv)= 0 + r × F = T,where in the last line we use Newton’s second law, namely F = d(mv)/dt. ◭If a vector a is a function of a scalar variable s that is itself a function of u, sothat s = s(u), then the chain rule (see subsection 2.1.3) givesda(s)du= ds dadu ds . (10.7)The derivatives of more complicated vector expressions may be found by repeatedapplication of the above equations.One further useful result can be derived by considering the derivativedda(a · a) =2a ·du du ;since a · a = a 2 ,wherea = |a|, weseethata · da =0 ifa is constant. (10.8)duIn other words, if a vector a(u) has a constant magnitude as u varies then it isperpendicular to the vector da/du.10.1.2 Differential of a vectorAs a final note on the differentiation of vectors, we can also define the differentialof a vector, in a similar way to that of a scalar in ordinary differential calculus.In the definition of the vector derivative (10.1), we used the notion of a smallchange ∆a in a vector a(u) resulting from a small change ∆u in its argument. Inthe limit ∆u → 0, the change in a becomes infinitesimally small, and we denote itby the differential da. From (10.1) we see that the differential is given byda = da du. (10.9)du338

10.2 INTEGRATION OF VECTORSNote that the differential of a vector is also a vector. As an example, theinfinitesimal change in the position vector of a particle in an infinitesimal time dt isdr = dr dt = v dt,dtwhere v is the particle’s velocity.10.2 Integration of vectorsThe integration of a vector (or of an expression involving vectors that may itselfbe either a vector or scalar) with respect to a scalar u canberegardedastheinverse of differentiation. We must remember, however, that(i) the integral has the same nature (vector or scalar) as the integrand,(ii) the constant of integration for indefinite integrals must be of the samenature as the integral.For example, if a(u) =d[A(u)]/du then the indefinite integral of a(u) is given by∫a(u) du = A(u)+b,where b is a constant vector. The definite integral of a(u) fromu = u 1 to u = u 2is given by∫ u2u 1a(u) du = A(u 2 ) − A(u 1 ).◮A small particle of mass m orbits a much larger mass M centred at the origin O. Accordingto Newton’s law of gravitation, the position vector r of the small mass obeys the differentialequationm d2 rdt = − GMm ˆr.2 r 2Show that the vector r × dr/dt is a constant of the motion.Forming the vector product of the differential equation with r, weobtainr × d2 rdt = − GM2 r r × ˆr.2Since r and ˆr are collinear, r × ˆr = 0 and therefore we haveHowever,r × d2 r= 0. (10.10)dt2 (dr × dr )= r × d2 rdt dt dt + dr2 dt × drdt = 0,339

VECTOR CALCULUSzCˆnPˆtˆbr(u)OyxFigure 10.3 The unit tangent ˆt, normal ˆn and binormal ˆb to the space curve Cat a particular point P .since the first term is zero by (10.10), and the second is zero because it is the vector productof two parallel (in this case identical) vectors. Integrating, we obtain the required resultr × dr = c, (10.11)dtwhere c is a constant vector.As a further point of interest we may note that in an infinitesimal time dt the changein the position vector of the small mass is dr and the element of area swept out by theposition vector of the particle is simply dA = 1 |r×dr|. Dividing both sides of this equation2by dt, we conclude thatdAdt = 1 2 ∣ r × drdt ∣ = |c|2 ,and that the physical interpretation of the above result (10.11) is that the position vector rof the small mass sweeps out equal areas in equal times. This result is in fact valid formotion under any force that acts along the line joining the two particles. ◭10.3 Space curvesIn the previous section we mentioned that the velocity vector of a particle is atangent to the curve in space along which the particle moves. We now give a morecomplete discussion of curves in space and also a discussion of the geometricalinterpretation of the vector derivative.AcurveC in space can be described by the vector r(u) joining the origin O ofa coordinate system to a point on the curve (see figure 10.3). As the parameter uvaries, the end-point of the vector moves along the curve. In Cartesian coordinates,r(u) =x(u)i + y(u)j + z(u)k,where x = x(u), y = y(u) andz = z(u) aretheparametric equations of the curve.340

10.3 SPACE CURVESThis parametric representation can be very useful, particularly in mechanics whenthe parameter may be the time t. We can, however, also represent a space curveby y = f(x), z = g(x), which can be easily converted into the above parametricform by setting u = x, sothatr(u) =ui + f(u)j + g(u)k.Alternatively, a space curve can be represented in the form F(x, y, z) = 0,G(x, y, z) = 0, where each equation represents a surface and the curve is theintersection of the two surfaces.A curve may sometimes be described in parametric form by the vector r(s),where the parameter s is the arc length along the curve measured from a fixedpoint. Even when the curve is expressed in terms of some other parameter, it isstraightforward to find the arc length between any two points on the curve. Forthe curve described by r(u), let us consider an infinitesimal vector displacementdr = dx i + dy j + dz kalong the curve. The square of the infinitesimal distance moved is then given by(ds) 2 = dr · dr =(dx) 2 +(dy) 2 +(dz) 2 ,fromwhichitcanbeshownthat( ) ds 2= drdu du · drdu .Therefore, the arc length between two points on the curve r(u), given by u = u 1and u = u 2 ,iss =∫ u2u 1√drdu · dr du. (10.12)du◮ Acurvelyinginthexy-plane is given by y = y(x), z =0. Using (10.12), show that thearc length along the curve between x = a and x = b is given by s = ∫ √ba 1+y ′2 dx, wherey ′ = dy/dx.Let us first represent the curve in parametric form by setting u = x, sothatr(u) =ui + y(u)j.Differentiating with respect to u, we finddrdu = i + dydu j,from which we obtaindrdu · dr ( ) 2 dydu =1+ .du341

VECTOR CALCULUSTherefore, remembering that u = x, from (10.12) the arc length between x = a and x = bis given by∫ b√ √drs =a du · dr ∫ b( ) 2 dydu du = 1+ dx.a dxThis result was derived using more elementary methods in chapter 2. ◭If a curve C is described by r(u) then, by considering figures 10.1 and 10.3, wesee that, at any given point on the curve, dr/du is a vector tangent to C at thatpoint, in the direction of increasing u. In the special case where the parameter uis the arc length s along the curve then dr/ds is a unit tangent vector to C and isdenoted by ˆt.The rate at which the unit tangent ˆt changes with respect to s is given byd ˆt/ds, and its magnitude is defined as the curvature κ of the curve C at a givenpoint,∣ ∣ κ =d ˆt∣∣∣ ∣ ds ∣ = d 2 ˆr ∣∣∣ds 2 .We can also define the quantity ρ =1/κ, which is called the radius of curvature.Since ˆt is of constant (unit) magnitude, it follows from (10.8) that it is perpendicularto d ˆt/ds. The unit vector in the direction perpendicular to ˆt is denotedby ˆn and is called the principal normal at the point. We therefore haved ˆt= κ ˆn. (10.13)dsThe unit vector ˆb = ˆt × ˆn, which is perpendicular to the plane containing ˆtand ˆn, is called the binormal to C. The vectors ˆt, ˆn and ˆb form a right-handedrectangular cooordinate system (or triad) at any given point on C (see figure 10.3).As s changes so that the point of interest moves along C, the triad of vectors alsochanges.The rate at which ˆb changes with respect to s is given by d ˆb/ds and is ameasure of the torsion τ of the curve at any given point. Since ˆb is of constantmagnitude, from (10.8) it is perpendicular to d ˆb/ds. We may further show thatd ˆb/ds is also perpendicular to ˆt, as follows. By definition ˆb · ˆt =0,whichondifferentiating yields0= d ( )ˆb · ˆt = d ˆbds ds · ˆt + ˆb · d ˆtds= d ˆbds · ˆt + ˆb · κ ˆn= d ˆbds · ˆt,where we have used the fact that ˆb · ˆn = 0. Hence, since d ˆb/ds is perpendicularto both ˆb and ˆt, we must have d ˆb/ds ∝ ˆn. The constant of proportionality is −τ,342

10.3 SPACE CURVESso we finally obtaind ˆb = −τ ˆn. (10.14)dsTaking the dot product of each side with ˆn, we see that the torsion of a curve isgiven byτ = − ˆn · d ˆbds .We may also define the quantity σ =1/τ, which is called the radius of torsion.Finally, we consider the derivative d ˆn/ds. Since ˆn = ˆb × ˆt we haved ˆnds = d ˆbds × ˆt + ˆb × d ˆtds= −τ ˆn × ˆt + ˆb × κ ˆn= τ ˆb − κ ˆt. (10.15)In summary, ˆt, ˆn and ˆb and their derivatives with respect to s are related to oneanother by the relations (10.13), (10.14) and (10.15), the Frenet–Serret formulae,d ˆtds = κ ˆn,d ˆnds = τ ˆb − κ ˆt,d ˆbds= −τ ˆn. (10.16)◮Show that the acceleration of a particle travelling along a trajectory r(t) is given bya(t) = dvdt ˆt + v2ρ ˆn,where v is the speed of the particle, ˆt is the unit tangent to the trajectory, ˆn is its principalnormal and ρ is its radius of curvature.The velocity of the particle is given byv(t) = drdt = dr dsds dt = dsdt ˆt,where ds/dt is the speed of the particle, which we denote by v, and ˆt is the unit vectortangent to the trajectory. Writing the velocity as v = v ˆt, and differentiating once morewith respect to time t, weobtaina(t) = dvdt = dvdt ˆt + v d ˆtdt ;but we note thatd ˆtdt = ds d ˆtdt ds = vκ ˆn = v ρ ˆn.Therefore, we havea(t) = dvdt ˆt + v2ρ ˆn.This shows that in addition to an acceleration dv/dt along the tangent to the particle’strajectory, there is also an acceleration v 2 /ρ in the direction of the principal normal. Thelatter is often called the centripetal acceleration. ◭343

VECTOR CALCULUSFinally, we note that a curve r(u) representing the trajectory of a particle maysometimes be given in terms of some parameter u that is not necessarily equal tothe time t but is functionally related to it in some way. In this case the velocityof the particle is given byv = drdt = dr dudu dt .Differentiating again with respect to time gives the acceleration asa = dvdt = d ( ) ( ) 2 dr du= d2 r dudt du dt du 2 + dr d 2 udt du dt 2 .10.4 Vector functions of several argumentsThe concept of the derivative of a vector is easily extended to cases where thevectors (or scalars) are functions of more than one independent scalar variable,u 1 ,u 2 ,...,u n . In this case, the results of subsection 10.1.1 are still valid, exceptthat the derivatives become partial derivatives ∂a/∂u i defined as in ordinarydifferential calculus. For example, in Cartesian coordinates,∂a∂u = ∂a x∂u i + ∂a y∂u j + ∂a z∂u k.In particular, (10.7) generalises to the chain rule of partial differentiation discussedin section 5.5. If a = a(u 1 ,u 2 ,...,u n ) and each of the u i is also a functionu i (v 1 ,v 2 ,...,v n ) of the variables v i then, generalising (5.17),∂a∂v i= ∂a∂u 1∂u 1∂v i+ ∂a∂u 2∂u 2∂v i+ ···+ ∂a∂u n∂u n∂v i=n∑j=1∂a∂u j∂u j∂v i. (10.17)A special case of this rule arises when a is an explicit function of some variablev, as well as of scalars u 1 ,u 2 ,...,u n that are themselves functions of v; thenwehavedadv = ∂a∂v +n∑j=1∂a∂u j∂u j∂v . (10.18)We may also extend the concept of the differential of a vector given in (10.9)to vectors dependent on several variables u 1 ,u 2 ,...,u n :da = ∂a du 1 + ∂a du 2 + ···+ ∂an∑ ∂adu n = du j . (10.19)∂u 1 ∂u 2 ∂u n ∂u jAs an example, the infinitesimal change in an electric field E in moving from aposition r to a neighbouring one r + dr is given bydE = ∂E∂xdx +∂E∂y344j=1∂Edy + dz. (10.20)∂z

10.5 SURFACESzT∂r∂uSu = c 1P∂r∂vr(u, v)v = c 2OyxFigure 10.4 The tangent plane T to a surface S at a particular point P ;u = c 1 and v = c 2 are the coordinate curves, shown by dotted lines, that passthrough P . The broken line shows some particular parametric curve r = r(λ)lying in the surface.10.5 SurfacesAsurfaceS in space can be described by the vector r(u, v) joining the origin O ofa coordinate system to a point on the surface (see figure 10.4). As the parametersu and v vary, the end-point of the vector moves over the surface. This is verysimilar to the parametric representation r(u) of a curve, discussed in section 10.3,but with the important difference that we require two parameters to describe asurface, whereas we need only one to describe a curve.In Cartesian coordinates the surface is given byr(u, v) =x(u, v)i + y(u, v)j + z(u, v)k,where x = x(u, v), y = y(u, v) andz = z(u, v) are the parametric equations of thesurface. We can also represent a surface by z = f(x, y) org(x, y, z) =0.Eitherof these representations can be converted into the parametric form in a similarmanner to that used for equations of curves. For example, if z = f(x, y) thenbysetting u = x and v = y the surface can be represented in parametric form byr(u, v) =ui + vj + f(u, v)k.Any curve r(λ), where λ is a parameter, on the surface S can be representedby a pair of equations relating the parameters u and v, for example u = f(λ)and v = g(λ). A parametric representation of the curve can easily be found bystraightforward substitution, i.e. r(λ) =r(u(λ),v(λ)). Using (10.17) for the casewhere the vector is a function of a single variable λ so that the LHS becomes a345

VECTOR CALCULUStotal derivative, the tangent to the curve r(λ) at any point is given bydrdλ = ∂r du∂u dλ + ∂r dv∂v dλ . (10.21)The two curves u =constantandv = constant passing through any point Pon S are called coordinate curves. Forthecurveu = constant, for example, wehave du/dλ = 0, and so from (10.21) its tangent vector is in the direction ∂r/∂v.Similarly, the tangent vector to the curve v = constant is in the direction ∂r/∂u.If the surface is smooth then at any point P on S the vectors ∂r/∂u and∂r/∂v are linearly independent and define the tangent plane T at the point P (seefigure 10.4). A vector normal to the surface at P is given byn = ∂r∂u × ∂r∂v . (10.22)In the neighbourhood of P , an infinitesimal vector displacement dr is writtendr = ∂r ∂rdu +∂u ∂v dv.The element of area at P , an infinitesimal parallelogram whose sides are thecoordinate curves, has magnitude∣ dS =∂r ∂r ∣∣∣∣ du ×∂u ∂v dv =∂r∣∂u × ∂r∂v ∣ du dv = |n| du dv. (10.23)Thus the total area of the surface is∫∫A =∂r∣∂u ∫∫∂v∣ |n| du dv, (10.24)Rwhere R is the region in the uv-plane corresponding to the range of parametervalues that define the surface.◮ Find the element of area on the surface of a sphere of radius a, and hence calculate thetotal surface area of the sphere.We can represent a point r on the surface of the sphere in terms of the two parameters θand φ:r(θ, φ) =a sin θ cos φ i + a sin θ sin φ j + a cos θ k,where θ and φ are the polar and azimuthal angles respectively. At any point P ,vectorstangent to the coordinate curves θ =constantandφ = constant are∂r= a cos θ cos φ i + a cos θ sin φ j − a sin θ k,∂θ∂r= −a sin θ sin φ i + a sin θ cos φ j.∂φ346R

10.6 SCALAR AND VECTOR FIELDSA normal n to the surface at this point is then given by∣ ∣∣∣∣∣∣n = ∂r∂θ × ∂ri j k∂φ = a cos θ cos φ acos θ sin φ −a sin θ−a sin θ sin φ asin θ cos φ 0∣= a 2 sin θ(sin θ cos φ i +sinθ sin φ j +cosθ k),which has a magnitude of a 2 sin θ. Therefore, the element of area at P is, from (10.23),dS = a 2 sin θdθdφ,and the total surface area of the sphere is given byA =∫ πdθ∫ 2π0 0dφ a 2 sin θ =4πa 2 .This familiar result can, of course, be proved by much simpler methods! ◭10.6 Scalar and vector fieldsWe now turn to the case where a particular scalar or vector quantity is definednot just at a point in space but continuously as a field throughout some regionof space R (which is often the whole space). Although the concept of a field isvalid for spaces with an arbitrary number of dimensions, in the remainder of thischapter we will restrict our attention to the familiar three-dimensional case. Ascalar field φ(x, y, z) associates a scalar with each point in R, while a vector fielda(x, y, z) associates a vector with each point. In what follows, we will assume thatthe variation in the scalar or vector field from point to point is both continuousand differentiable in R.Simple examples of scalar fields include the pressure at each point in a fluidand the electrostatic potential at each point in space in the presence of an electriccharge. Vector fields relating to the same physical systems are the velocity vectorin a fluid (giving the local speed and direction of the flow) and the electric field.With the study of continuously varying scalar and vector fields there arises theneed to consider their derivatives and also the integration of field quantities alonglines, over surfaces and throughout volumes in the field. We defer the discussionof line, surface and volume integrals until the next chapter, and in the remainderof this chapter we concentrate on the definition of vector differential operatorsand their properties.10.7 Vector operatorsCertain differential operations may be performed on scalar and vector fieldsand have wide-ranging applications in the physical sciences. The most importantoperations are those of finding the gradient of a scalar field and the divergenceand curl of a vector field. It is usual to define these operators from a strictly347

VECTOR CALCULUSmathematical point of view, as we do below. In the following chapter, however, wewill discuss their geometrical definitions, which rely on the concept of integratingvector quantities along lines and over surfaces.Central to all these differential operations is the vector operator ∇, whichiscalled del (or sometimes nabla) and in Cartesian coordinates is defined by∇≡i ∂∂x + j ∂∂y + k ∂ ∂z . (10.25)The form of this operator in non-Cartesian coordinate systems is discussed insections 10.9 and 10.10.10.7.1 Gradient of a scalar fieldThe gradient of a scalar field φ(x, y, z) is defined bygrad φ = ∇φ = i ∂φ∂x + j∂φ ∂y + k∂φ ∂z . (10.26)Clearly, ∇φ is a vector field whose x-, y- andz- components are the first partialderivatives of φ(x, y, z) with respect to x, y and z respectively. Also note that thevector field ∇φ should not be confused with the vector operator φ∇, which hascomponents (φ ∂/∂x, φ ∂/∂y, φ ∂/∂z).◮Find the gradient of the scalar field φ = xy 2 z 3 .From (10.26) the gradient of φ is given by∇φ = y 2 z 3 i +2xyz 3 j +3xy 2 z 2 k. ◭The gradient of a scalar field φ has some interesting geometrical properties.Let us first consider the problem of calculating the rate of change of φ in someparticular direction. For an infinitesimal vector displacement dr, forming its scalarproduct with ∇φ we obtain(∇φ · dr = i ∂φ)∂x + j∂φ ∂y + k∂φ · (i dx + j dy + k dx) ,∂z= ∂φ ∂φ ∂φdx + dy +∂x ∂y ∂z dz,= dφ, (10.27)which is the infinitesimal change in φ in going from position r to r + dr. Inparticular, if r depends on some parameter u such that r(u) defines a space curve348

10.7 VECTOR OPERATORS∇φaθQPdφdsin the direction aφ =constantFigure 10.5 Geometrical properties of ∇φ. PQ gives the value of dφ/ds inthe direction a.then the total derivative of φ with respect to u along the curve is simplydφ dr= ∇φ ·du du . (10.28)In the particular case where the parameter u is the arc length s along the curve,the total derivative of φ with respect to s along the curve is given bydφds = ∇φ · ˆt, (10.29)where ˆt is the unit tangent to the curve at the given point, as discussed insection 10.3.In general, the rate of change of φ with respect to the distance s in a particulardirection a is given bydφ= ∇φ · â (10.30)dsand is called the directional derivative. Since â is a unit vector we havedφ= |∇φ| cos θdswhere θ is the angle between â and ∇φ as shown in figure 10.5. Clearly ∇φ liesin the direction of the fastest increase in φ, and|∇φ| is the largest possible valueof dφ/ds. Similarly, the largest rate of decrease of φ is dφ/ds = −|∇φ| in thedirection of −∇φ.349

VECTOR CALCULUS◮For the function φ = x 2 y + yz at the point (1, 2, −1), find its rate of change with distancein the direction a = i +2j +3k. At this same point, what is the greatest possible rate ofchange with distance and in which direction does it occur?The gradient of φ is given by (10.26):∇φ =2xyi +(x 2 + z)j + yk,=4i +2k at the point (1, 2, −1).The unit vector in the direction of a is â = √ 114(i +2j +3k), so the rate of change of φwith distance s in this direction is, using (10.30),dφds = ∇φ · â = √ 1 (4+6)= √ 10 .14 14From the above discussion, at the point (1, 2, −1) dφ/ds will be greatest in the directionof ∇φ =4i +2k and has the value |∇φ| = √ 20 in this direction. ◭We can extend the above analysis to find the rate of change of a vectorfield (rather than a scalar field as above) in a particular direction. The scalardifferential operator â · ∇ can be shown to give the rate of change with distancein the direction â of the quantity (vector or scalar) on which it acts. In Cartesiancoordinates it may be written as∂â · ∇ = a x∂x + a ∂y∂y + a ∂z∂z . (10.31)Thus we can write the infinitesimal change in an electric field in moving from rto r + dr given in (10.20) as dE =(dr · ∇)E.A second interesting geometrical property of ∇φ may be found by consideringthe surface defined by φ(x, y, z) =c, wherec is some constant. If ˆt is a unittangent to this surface at some point then clearly dφ/ds = 0 in this directionand from (10.29) we have ∇φ · ˆt = 0. In other words, ∇φ is a vector normal tothe surface φ(x, y, z) =c at every point, as shown in figure 10.5. If ˆn is a unitnormal to the surface in the direction of increasing φ(x, y, z), then the gradient issometimes written∇φ ≡ ∂φ ˆn, (10.32)∂nwhere ∂φ/∂n ≡|∇φ| is the rate of change of φ in the direction ˆn and is calledthe normal derivative.◮Find expressions for the equations of the tangent plane and the line normal to the surfaceφ(x, y, z) =c at the point P with coordinates x 0 ,y 0 ,z 0 . Use the results to find the equationsof the tangent plane and the line normal to the surface of the sphere φ = x 2 + y 2 + z 2 = a 2at the point (0, 0,a).A vector normal to the surface φ(x, y, z) =c at the point P is simply ∇φ evaluated at thatpoint; we denote it by n 0 .Ifr 0 is the position vector of the point P relative to the origin,350

10.7 VECTOR OPERATORSz(0, 0,a)ˆn 0z = aOayφ = x 2 + y 2 + z 2 = a 2xFigure 10.6 The tangent plane and the normal to the surface of the sphereφ = x 2 + y 2 + z 2 = a 2 at the point r 0 with coordinates (0, 0,a).and r is the position vector of any point on the tangent plane, then the vector equation ofthe tangent plane is, from (7.41),(r − r 0 ) · n 0 =0.Similarly, if r is the position vector of any point on the straight line passing through P(with position vector r 0 ) in the direction of the normal n 0 then the vector equation of thisline is, from subsection 7.7.1,(r − r 0 ) × n 0 = 0.For the surface of the sphere φ = x 2 + y 2 + z 2 = a 2 ,∇φ =2xi +2yj +2zk=2ak at the point (0, 0,a).Therefore the equation of the tangent plane to the sphere at this point is(r − r 0 ) · 2ak =0.This gives 2a(z − a) =0orz = a, as expected. The equation of the line normal to thesphere at the point (0, 0,a)is(r − r 0 ) × 2ak = 0,which gives 2ayi − 2axj = 0 or x = y =0,i.e.thez-axis, as expected. The tangent planeand normal to the surface of the sphere at this point are shown in figure 10.6. ◭Further properties of the gradient operation, which are analogous to those ofthe ordinary derivative, are listed in subsection 10.8.1 and may be easily proved.351

VECTOR CALCULUSIn addition to these, we note that the gradient operation also obeys the chainrule as in ordinary differential calculus, i.e. if φ and ψ are scalar fields in someregion R then∇ [φ(ψ)] = ∂φ∂ψ ∇ψ.10.7.2 Divergence of a vector fieldThe divergence of a vector field a(x, y, z) is defined bydiv a = ∇ · a = ∂a x∂x + ∂a y∂y + ∂a z∂z , (10.33)where a x , a y and a z are the x-, y- andz- components of a. Clearly, ∇ · a is a scalarfield. Any vector field a for which ∇ · a = 0 is said to be solenoidal.◮Find the divergence of the vector field a = x 2 y 2 i + y 2 z 2 j + x 2 z 2 k.From (10.33) the divergence of a is given by∇ · a =2xy 2 +2yz 2 +2x 2 z =2(xy 2 + yz 2 + x 2 z). ◭We will discuss fully the geometric definition of divergence and its physicalmeaning in the next chapter. For the moment, we merely note that the divergencecan be considered as a quantitative measure of how much a vector field diverges(spreads out) or converges at any given point. For example, if we consider thevector field v(x, y, z) describing the local velocity at any point in a fluid then ∇ · vis equal to the net rate of outflow of fluid per unit volume, evaluated at a point(by letting a small volume at that point tend to zero).Now if some vector field a is itself derived from a scalar field via a = ∇φ then∇ · a has the form ∇ · ∇φ or, as it is usually written, ∇ 2 φ,where∇ 2 (del squared)is the scalar differential operator∇ 2 ≡ ∂2∂x 2 + ∂2∂y 2 + ∂2∂z 2 . (10.34)∇ 2 φ is called the Laplacian of φ and appears in several important partial differentialequations of mathematical physics, discussed in chapters 20 and 21.◮Find the Laplacian of the scalar field φ = xy 2 z 3 .From (10.34) the Laplacian of φ is given by∇ 2 φ = ∂2 φ∂x + ∂2 φ2 ∂y + ∂2 φ2 ∂z 2 =2xz3 +6xy 2 z. ◭352

10.7 VECTOR OPERATORS10.7.3 Curl of a vector fieldThe curl of a vector field a(x, y, z) is defined by( ∂azcurl a = ∇×a =∂y − ∂a ) (y ∂axi +∂z ∂z − ∂a ) (z ∂ayj +∂x ∂x − ∂a )xk,∂ywhere a x , a y and a z are the x-, y- andz- components of a. The RHS can bewritten in a more memorable form as a determinant:∣ i j k ∣∣∣∣∣∣∣ ∇×a =∂ ∂ ∂, (10.35)∂x ∂y ∂z∣ a x a y a zwhere it is understood that, on expanding the determinant, the partial derivativesin the second row act on the components of a in the third row. Clearly, ∇×a isitself a vector field. Any vector field a for which ∇×a = 0 is said to be irrotational.◮Find the curl of the vector field a = x 2 y 2 z 2 i + y 2 z 2 j + x 2 z 2 k.The curl of a is given by∣ i j k ∣∣∣∣∣∣∣ ∂ ∂ ∂∇φ == −2 [ y 2 zi +(xz 2 − x 2 y 2 z)j + x 2 yz 2 k ] . ◭∂x ∂y ∂z∣x 2 y 2 z 2 y 2 z 2 x 2 z 2For a vector field v(x, y, z) describing the local velocity at any point in a fluid,∇×v is a measure of the angular velocity of the fluid in the neighbourhood ofthat point. If a small paddle wheel were placed at various points in the fluid thenit would tend to rotate in regions where ∇×v ≠ 0, while it would not rotate inregions where ∇×v = 0.Another insight into the physical interpretation of the curl operator is gainedby considering the vector field v describing the velocity at any point in a rigidbody rotating about some axis with angular velocity ω. Ifr is the position vectorof the point with respect to some origin on the axis of rotation then the velocityof the point is given by v = ω × r. Without any loss of generality, we may takeω to lie along the z-axis of our coordinate system, so that ω = ω k. The velocityfield is then v = −ωy i + ωx j. The curl of this vector field is easily found to bei j k∇×v =∂ ∂ ∂∂x ∂y ∂z=2ωk =2ω. (10.36)∣ −ωy ωx 0 ∣353

VECTOR CALCULUS∇(φ + ψ) =∇φ + ∇ψ∇ · (a + b) =∇ · a + ∇ · b∇×(a + b) =∇×a + ∇×b∇(φψ) =φ∇ψ + ψ∇φ∇(a · b) =a × (∇×b)+b × (∇×a)+(a · ∇)b +(b · ∇)a∇ · (φa) =φ∇ · a + a · ∇φ∇ · (a × b) =b · (∇×a) − a · (∇×b)∇×(φa) =∇φ × a + φ∇×a∇×(a × b) =a(∇ · b) − b(∇ · a)+(b · ∇)a − (a · ∇)bTable 10.1 Vector operators acting on sums and products. The operator ∇ isdefined in (10.25); φ and ψ are scalar fields, a and b are vector fields.Therefore the curl of the velocity field is a vector equal to twice the angularvelocity vector of the rigid body about its axis of rotation. We give a fullgeometrical discussion of the curl of a vector in the next chapter.10.8 Vector operator formulaeIn the same way as for ordinary vectors (chapter 7), for vector operators certainidentities exist. In addition, we must consider various relations involving theaction of vector operators on sums and products of scalar and vector fields. Someof these relations have been mentioned earlier, but we list all the most importantones here for convenience. The validity of these relations may be easily verifiedby direct calculation (a quick method of deriving them using tensor notation isgiven in chapter 26).Although some of the following vector relations are expressed in Cartesiancoordinates, it may be proved that they are all independent of the choice ofcoordinate system. This is to be expected since grad, div and curl all have cleargeometrical definitions, which are discussed more fully in the next chapter andwhich do not rely on any particular choice of coordinate system.10.8.1 Vector operators acting on sums and productsLet φ and ψ be scalar fields and a and b be vector fields. Assuming these fieldsare differentiable, the action of grad, div and curl on various sums and productsof them is presented in table 10.1.These relations can be proved by direct calculation.354

10.8 VECTOR OPERATOR FORMULAE◮Show that∇×(φa) =∇φ × a + φ∇×a.The x-component of the LHS is∂∂y (φa z) − ∂ ∂z (φa y)=φ ∂a z∂y + ∂φ∂y a z − φ ∂a y∂z − ∂φ∂z a y,( ∂az= φ∂y − ∂a ) (y ∂φ+∂z ∂y a z − ∂φ )∂z a y ,= φ(∇×a) x +(∇φ × a) x ,where, for example, (∇φ×a) x denotes the x-component of the vector ∇φ×a. Incorporatingthe y- andz- components, which can be similarly found, we obtain the stated result. ◭Some useful special cases of the relations in table 10.1 are worth noting. If r isthe position vector relative to some origin and r = |r|, then∇φ(r) = dφdr ˆr,∇ · [φ(r)r] =3φ(r)+r dφ(r) ,dr∇ 2 φ(r) = d2 φ(r)dr 2 + 2 dφ(r),r dr∇×[φ(r)r] = 0.These results may be proved straightforwardly using Cartesian coordinates butfar more simply using spherical polar coordinates, which are discussed in subsection10.9.2. Particular cases of these results aretogether with∇r = ˆr, ∇ · r =3, ∇×r = 0,( ) 1∇ = − ˆrr r 2 ,( ) ( ˆr1∇ ·r 2 = −∇ 2 r)=4πδ(r),where δ(r) is the Dirac delta function, discussed in chapter 13. The last equation isimportant in the solution of certain partial differential equations and is discussedfurther in chapter 20.10.8.2 Combinations of grad, div and curlWe now consider the action of two vector operators in succession on a scalar orvector field. We can immediately discard four of the nine obvious combinations ofgrad, div and curl, since they clearly do not make sense. If φ is a scalar field and355

VECTOR CALCULUSa is a vector field, these four combinations are grad(grad φ), div(div a), curl(div a)and grad(curl a). In each case the second (outer) vector operator is acting on thewrong type of field, i.e. scalar instead of vector or vice versa. In grad(grad φ), forexample, grad acts on grad φ, which is a vector field, but we know that grad onlyacts on scalar fields (although in fact we will see in chapter 26 that we can formthe outer product of the del operator with a vector to give a tensor, but that neednot concern us here).Of the five valid combinations of grad, div and curl, two are identically zero,namelycurl grad φ = ∇×∇φ = 0, (10.37)div curl a = ∇ · (∇×a) =0. (10.38)From (10.37), we see that if a is derived from the gradient of some scalar functionsuch that a = ∇φ then it is necessarily irrotational (∇ ×a = 0). We also notethat if a is an irrotational vector field then another irrotational vector field isa + ∇φ + c, whereφ is any scalar field and c is a constant vector. This followssince∇×(a + ∇φ + c) =∇×a + ∇×∇φ = 0.Similarly, from (10.38) we may infer that if b is the curl of some vector field asuch that b = ∇×a then b is solenoidal (∇ · b = 0). Obviously, if b is solenoidaland c is any constant vector then b + c is also solenoidal.The three remaining combinations of grad, div and curl arediv grad φ = ∇ · ∇φ = ∇ 2 φ = ∂2 φ∂x 2 + ∂2 φ∂y 2 + ∂2 φ∂z 2 , (10.39)grad div a = ∇(∇ · a),( ∂ 2 a x=∂x 2++ ∂2 a y∂x∂y + ∂2 a z∂x∂z) ( ∂ 2 )a xi +∂y∂x + ∂2 a y∂y 2 + ∂2 a zj∂y∂z( ∂ 2 )a x∂z∂x + ∂2 a y∂z∂y + ∂2 a z∂z 2 k, (10.40)curl curl a = ∇×(∇×a) =∇(∇ · a) −∇ 2 a, (10.41)where (10.39) and (10.40) are expressed in Cartesian coordinates. In (10.41), theterm ∇ 2 a has the linear differential operator ∇ 2 acting on a vector (as opposed toa scalar as in (10.39)), which of course consists of a sum of unit vectors multipliedby components. Two cases arise.(i) If the unit vectors are constants (i.e. they are independent of the values ofthe coordinates) then the differential operator gives a non-zero contributiononly when acting upon the components, the unit vectors being merelymultipliers.356

10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATES(ii) If the unit vectors vary as the values of the coordinates change (i.e. arenot constant in direction throughout the whole space) then the derivativesof these vectors appear as contributions to ∇ 2 a.Cartesian coordinates are an example of the first case in which each componentsatisfies (∇ 2 a) i = ∇ 2 a i . In this case (10.41) can be applied to each componentseparately:[∇×(∇×a)] i =[∇(∇ · a)] i −∇ 2 a i . (10.42)However, cylindrical and spherical polar coordinates come in the second class.For them (10.41) is still true, but the further step to (10.42) cannot be made.More complicated vector operator relations may be proved using the relationsgiven above.◮Show thatwhere φ and ψ are scalar fields.∇ · (∇φ ×∇ψ) =0,From the previous section we have∇ · (a × b) =b · (∇×a) − a · (∇×b).If we let a = ∇φ and b = ∇ψ then we obtain∇ · (∇φ ×∇ψ) =∇ψ · (∇×∇φ) −∇φ · (∇×∇ψ) =0, (10.43)since ∇×∇φ =0=∇×∇ψ, from (10.37). ◭10.9 Cylindrical and spherical polar coordinatesThe operators we have discussed in this chapter, i.e. grad, div, curl and ∇ 2 ,have all been defined in terms of Cartesian coordinates, but for many physicalsituations other coordinate systems are more natural. For example, many systems,such as an isolated charge in space, have spherical symmetry and spherical polarcoordinates would be the obvious choice. For axisymmetric systems, such as fluidflow in a pipe, cylindrical polar coordinates are the natural choice. The physicallaws governing the behaviour of the systems are often expressed in terms ofthe vector operators we have been discussing, and so it is necessary to be ableto express these operators in these other, non-Cartesian, coordinates. We firstconsider the two most common non-Cartesian coordinate systems, i.e. cylindricaland spherical polars, and go on to discuss general curvilinear coordinates in thenext section.10.9.1 Cylindrical polar coordinatesAs shown in figure 10.7, the position of a point in space P having Cartesiancoordinates x, y, z may be expressed in terms of cylindrical polar coordinates357

VECTOR CALCULUSρ, φ, z, wherex = ρ cos φ, y = ρ sin φ, z = z, (10.44)and ρ ≥ 0, 0 ≤ φ

10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATESzê zPê φê ρrkziOjρyφxFigure 10.7 Cylindrical polar coordinates ρ, φ, z.zρdφdzdρφρdφρdφyxFigure 10.8 The element of volume in cylindrical polar coordinates is givenby ρdρdφdz.Factors, such as the ρ in ρdφ, that multiply the coordinate differentials to givedistances are known as scale factors. From (10.52), the scale factors for the ρ-, φ-and z- coordinates are therefore 1, ρ and 1 respectively.The magnitude ds of the displacement dr is given in cylindrical polar coordinatesby(ds) 2 = dr · dr =(dρ) 2 + ρ 2 (dφ) 2 +(dz) 2 ,where in the second equality we have used the fact that the basis vectors areorthonormal. We can also find the volume element in a cylindrical polar system(see figure 10.8) by calculating the volume of the infinitesimal parallelepiped359

VECTOR CALCULUS∇Φ = ∂Φ∂ρ êρ + 1 ∂Φρ ∂φ êφ + ∂Φ∂z êz∇ · a = 1 ∂ρ ∂ρ (ρa ρ)+ 1 ∂a φρ ∂φ + ∂a z∂z∣ ∇×a = 1 ê ρ ρê φ ê z ∣∣∣∣∣∣∣∂ ∂ ∂ρ∂ρ ∂φ ∂z∣ a ρ ρa φ a z∇ 2 Φ = 1 (∂ρ ∂Φ )+ 1 ∂ 2 Φρ ∂ρ ∂ρ ρ 2 ∂φ + ∂2 Φ2 ∂z 2Table 10.2 Vector operators in cylindrical polar coordinates; Φ is a scalarfield and a is a vector field.defined by the vectors dρ ê ρ , ρdφê φ and dz ê z :dV = |dρ ê ρ · (ρdφê φ × dz ê z )| = ρdρdφdz,which again uses the fact that the basis vectors are orthonormal. For a simplecoordinate system such as cylindrical polars the expressions for (ds) 2 and dV areobvious from the geometry.We will now express the vector operators discussed in this chapter in termsof cylindrical polar coordinates. Let us consider a vector field a(ρ, φ, z) andascalar field Φ(ρ, φ, z), where we use Φ for the scalar field to avoid confusion withthe azimuthal angle φ. We must first write the vector field in terms of the basisvectors of the cylindrical polar coordinate system, i.e.a = a ρ ê ρ + a φ ê φ + a z ê z ,where a ρ , a φ and a z are the components of a in the ρ-, φ- andz- directionsrespectively. The expressions for grad, div, curl and ∇ 2 can then be calculatedand are given in table 10.2. Since the derivations of these expressions are rathercomplicated we leave them until our discussion of general curvilinear coordinatesin the next section; the reader could well postpone examination of these formalproofs until some experience of using the expressions has been gained.◮Express the vector field a = yz i − y j + xz 2 k in cylindrical polar coordinates, and hencecalculate its divergence. Show that the same result is obtained by evaluating the divergencein Cartesian coordinates.The basis vectors of the cylindrical polar coordinate system are given in (10.49)–(10.51).Solving these equations simultaneously for i, j and k we obtaini =cosφ ê ρ − sin φ ê φj =sinφ ê ρ +cosφ ê φk = ê z .360

10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATESzê rê φPê θkθriOjyφxFigure 10.9 Spherical polar coordinates r, θ, φ.Substituting these relations and (10.44) into the expression for a we finda = zρ sin φ (cos φ ê ρ − sin φ ê φ ) − ρ sin φ (sin φ ê ρ +cosφ ê φ )+z 2 ρ cos φ ê z=(zρsin φ cos φ − ρ sin 2 φ) ê ρ − (zρsin 2 φ + ρ sin φ cos φ) ê φ + z 2 ρ cos φ ê z .Substituting into the expression for ∇ · a given in table 10.2,∇ · a =2z sin φ cos φ − 2sin 2 φ − 2z sin φ cos φ − cos 2 φ +sin 2 φ +2zρcos φ=2zρcos φ − 1.Alternatively, and much more quickly in this case, we can calculate the divergencedirectly in Cartesian coordinates. We obtain∇ · a = ∂a x∂x + ∂a y∂y + ∂a z=2zx − 1,∂zwhich on substituting x = ρ cos φ yields the same result as the calculation in cylindricalpolars. ◭Finally, we note that similar results can be obtained for (two-dimensional)polar coordinates in a plane by omitting the z-dependence. For example, (ds) 2 =(dρ) 2 + ρ 2 (dφ) 2 , while the element of volume is replaced by the element of areadA = ρdρdφ.10.9.2 Spherical polar coordinatesAs shown in figure 10.9, the position of a point in space P , with Cartesiancoordinates x, y, z, may be expressed in terms of spherical polar coordinatesr, θ, φ, wherex = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, (10.53)361

VECTOR CALCULUSand r ≥ 0, 0 ≤ θ ≤ π and 0 ≤ φ

10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATES∇Φ = ∂Φ∂r êr + 1 ∂Φr ∂θ êθ + 1 ∂Φr sin θ ∂φ êφ∇ · a =∇×a =∇ 2 Φ =1 ∂r 2 ∂r (r2 a r )+ 1 ∂r sin θ ∂θ (sin θa θ)+ 1 ∂a φr sin θ ∂φ∣ ê r rê θ r sin θ ê φ ∣∣∣∣∣∣∣1∂ ∂ ∂r 2 sin θ∂r ∂θ ∂φ∣ a r ra θ r sin θa φ(1 ∂r 2 ∂Φ )+ 1 (∂sin θ ∂Φ )+r 2 ∂r ∂r r 2 sin θ ∂θ ∂θ1r 2 sin 2 θ∂ 2 Φ∂φ 2Table 10.3 Vector operators in spherical polar coordinates; Φ is a scalar fieldand a is a vector field.zdφdrθrdθrdθr sin θdφφdφyr sin θr sin θdφxFigure 10.10 The element of volume in spherical polar coordinates is givenby r 2 sin θdrdθdφ.we can rewrite the first term on the RHS as follows:(1 ∂r 2 r 2 ∂Φ )= 1 ∂ 2∂r ∂r r ∂r 2 (rΦ),which can often be useful in shortening calculations.363

VECTOR CALCULUS10.10 General curvilinear coordinatesAs indicated earlier, the contents of this section are more formal and technicallycomplicated than hitherto. The section could be omitted until the reader has hadsome experience of using its results.Cylindrical and spherical polars are just two examples of what are calledgeneral curvilinear coordinates. In the general case, the position of a point Phaving Cartesian coordinates x, y, z may be expressed in terms of the threecurvilinear coordinates u 1 ,u 2 ,u 3 ,whereand similarlyx = x(u 1 ,u 2 ,u 3 ), y = y(u 1 ,u 2 ,u 3 ), z = z(u 1 ,u 2 ,u 3 ),u 1 = u 1 (x, y, z), u 2 = u 2 (x, y, z), u 3 = u 3 (x, y, z).We assume that all these functions are continuous, differentiable and have asingle-valued inverse, except perhaps at or on certain isolated points or lines,so that there is a one-to-one correspondence between the x, y, z and u 1 ,u 2 ,u 3systems. The u 1 -, u 2 -andu 3 - coordinate curves of a general curvilinear systemare analogous to the x-, y- andz- axes of Cartesian coordinates. The surfacesu 1 = c 1 , u 2 = c 2 and u 3 = c 3 , where c 1 ,c 2 ,c 3 are constants, are called thecoordinate surfaces and each pair of these surfaces has its intersection in a curvecalled a coordinate curve or line (see figure 10.11).If at each point in space the three coordinate surfaces passing through the pointmeet at right angles then the curvilinear coordinate system is called orthogonal.For example, in spherical polars u 1 = r, u 2 = θ, u 3 = φ and the three coordinatesurfaces passing through the point (R,Θ, Φ) are the sphere r = R, the circularcone θ = Θ and the plane φ = Φ, which intersect at right angles at thatpoint. Therefore spherical polars form an orthogonal coordinate system (as docylindrical polars) .If r(u 1 ,u 2 ,u 3 ) is the position vector of the point P then e 1 = ∂r/∂u 1 is a vectortangent to the u 1 -curve at P (for which u 2 and u 3 are constants) in the directionof increasing u 1 . Similarly, e 2 = ∂r/∂u 2 and e 3 = ∂r/∂u 3 are vectors tangent tothe u 2 -andu 3 - curves at P in the direction of increasing u 2 and u 3 respectively.Denoting the lengths of these vectors by h 1 , h 2 and h 3 ,theunit vectors in each ofthese directions are given byê 1 = 1 ∂r, ê 2 = 1 ∂r, ê 3 = 1 ∂r,h 1 ∂u 1 h 2 ∂u 2 h 3 ∂u 3where h 1 = |∂r/∂u 1 |, h 2 = |∂r/∂u 2 | and h 3 = |∂r/∂u 3 |.The quantities h 1 , h 2 , h 3 are the scale factors of the curvilinear coordinatesystem. The element of distance associated with an infinitesimal change du i inone of the coordinates is h i du i . In the previous section we found that the scale364

10.10 GENERAL CURVILINEAR COORDINATESuzê 3 ˆɛ 3u 1 = c 1u 2 = c 2 ˆɛê 21 u 2P ê 2u 13ˆɛ 1ku 3 = c 3iOjyxFigure 10.11General curvilinear coordinates.factors for cylindrical and spherical polar coordinates werefor cylindrical polars h ρ =1, h φ = ρ, h z =1,for spherical polars h r =1, h θ = r, h φ = r sin θ.Although the vectors e 1 , e 2 , e 3 form a perfectly good basis for the curvilinearcoordinate system, it is usual to work with the corresponding unit vectors ê 1 , ê 2 ,ê 3 . For an orthogonal curvilinear coordinate system these unit vectors form anorthonormal basis.An infinitesimal vector displacement in general curvilinear coordinates is givenby, from (10.19),dr = ∂r du 1 + ∂r du 2 + ∂r du 3 (10.55)∂u 1 ∂u 2 ∂u 3= du 1 e 1 + du 2 e 2 + du 3 e 3 (10.56)= h 1 du 1 ê 1 + h 2 du 2 ê 2 + h 3 du 3 ê 3 . (10.57)In the case of orthogonal curvilinear coordinates, where the ê iperpendicular, the element of arc length is given byare mutually(ds) 2 = dr · dr = h 2 1(du 1 ) 2 + h 2 2(du 2 ) 2 + h 2 3(du 3 ) 2 . (10.58)The volume element for the coordinate system is the volume of the infinitesimalparallelepiped defined by the vectors (∂r/∂u i ) du i = du i e i = h i du i ê i ,fori =1, 2, 3.365

VECTOR CALCULUSFor orthogonal coordinates this is given bydV = |du 1 e 1 · (du 2 e 2 × du 3 e 3 )|= |h 1 ê 1 · (h 2 ê 2 × h 3 ê 3 )| du 1 du 2 du 3= h 1 h 2 h 3 du 1 du 2 du 3 .Now, in addition to the set {ê i }, i =1, 2, 3, there exists another useful set ofthree unit basis vectors at P .Since∇u 1 is a vector normal to the surface u 1 = c 1 ,a unit vector in this direction is ˆɛ 1 = ∇u 1 /|∇u 1 |. Similarly, ˆɛ 2 = ∇u 2 /|∇u 2 | andˆɛ 3 = ∇u 3 /|∇u 3 | are unit vectors normal to the surfaces u 2 = c 2 and u 3 = c 3respectively.Therefore at each point P in a curvilinear coordinate system, there exist, ingeneral, two sets of unit vectors: {ê i }, tangent to the coordinate curves, and {ˆɛ i },normal to the coordinate surfaces. A vector a can be written in terms of eitherset of unit vectors:a = a 1 ê 1 + a 2 ê 2 + a 3 ê 3 = A 1ˆɛ 1 + A 2ˆɛ 2 + A 3ˆɛ 3 ,where a 1 , a 2 , a 3 and A 1 , A 2 , A 3 are the components of a in the two systems. Itmay be shown that the two bases become identical if the coordinate system isorthogonal.Instead of the unit vectors discussed above, we could instead work directly withthe two sets of vectors {e i = ∂r/∂u i } and {ɛ i = ∇u i },whicharenot,ingeneral,ofunit length. We can then write a vector a asa = α 1 e 1 + α 2 e 2 + α 3 e 3 = β 1 ɛ 1 + β 2 ɛ 2 + β 3 ɛ 3 ,or more explicitly as∂r ∂r ∂ra = α 1 + α 2 + α 3 = β 1 ∇u 1 + β 2 ∇u 2 + β 3 ∇u 3 ,∂u 1 ∂u 2 ∂u 3where α 1 ,α 2 ,α 3 and β 1 ,β 2 ,β 3 are called the contravariant and covariant componentsof a respectively. A more detailed discussion of these components, inthe context of tensor analysis, is given in chapter 26. The (in general) non-unitbases {e i } and {ɛ i } are often the most natural bases in which to express vectorquantities.◮Show that {e i } and {ɛ i } are reciprocal systems of vectors.Let us consider the scalar product e i · ɛ j ; using the Cartesian expressions for r and ∇, weobtaine i · ɛ j = ∂r · ∇u j∂u i)=( ∂x∂u ii + ∂y∂u ij + ∂z∂u ik= ∂x∂u i∂u j∂x + ∂y∂u i∂u j∂y + ∂z∂u i∂u j∂z = ∂u j∂u i.366( ∂uj·∂x i + ∂u j∂y j + ∂u )j∂z k

10.10 GENERAL CURVILINEAR COORDINATESIn the last step we have used the chain rule for partial differentiation. Therefore e i · ɛ j =1if i = j, ande i · ɛ j = 0 otherwise. Hence {e i } and {ɛ j } are reciprocal systems of vectors. ◭We now derive expressions for the standard vector operators in orthogonalcurvilinear coordinates. Despite the useful properties of the non-unit bases discussedabove, the remainder of our discussion in this section will be in terms ofthe unit basis vectors {ê i }. The expressions for the vector operators in cylindricaland spherical polar coordinates given in tables 10.2 and 10.3 respectively can befound from those derived below by inserting the appropriate scale factors.GradientThe change dΦ in a scalar field Φ resulting from changes du 1 ,du 2 ,du 3 in thecoordinates u 1 ,u 2 ,u 3 is given by, from (5.5),dΦ = ∂Φ du 1 + ∂Φ du 2 + ∂Φ du 3 .∂u 1 ∂u 2 ∂u 3For orthogonal curvilinear coordinates u 1 ,u 2 ,u 3 we find from (10.57), and comparisonwith (10.27), that we can write this aswhere ∇Φ is given bydΦ =∇Φ · dr, (10.59)∇Φ = 1 ∂Φê 1 + 1 ∂Φê 2 + 1 ∂Φê 3 . (10.60)h 1 ∂u 1 h 2 ∂u 2 h 3 ∂u 3This implies that the del operator can be written∇ = ê1h 1∂∂u 1+ ê2h 2∂∂u 2+ ê3h 3∂∂u 3.◮Show that for orthogonal curvilinear coordinates ∇u i = ê i /h i . Hence show that the twosets of vectors {ê i } and {ˆɛ i } are identical in this case.Letting Φ = u i in (10.60) we find immediately that ∇u i = ê i /h i . Therefore |∇u i | =1/h i ,andso ˆɛ i = ∇u i /|∇u i | = h i ∇u i = ê i . ◭DivergenceIn order to derive the expression for the divergence of a vector field in orthogonalcurvilinear coordinates, we must first write the vector field in terms of the basisvectors of the coordinate system:The divergence is then given by∇ · a = 1h 1 h 2 h 3[ ∂∂u 1(h 2 h 3 a 1 )+a = a 1 ê 1 + a 2 ê 2 + a 3 ê 3 .∂∂u 2(h 3 h 1 a 2 )+367∂ ](h 1 h 2 a 3 ) .∂u 3(10.61)

VECTOR CALCULUS◮Prove the expression for ∇ · a in orthogonal curvilinear coordinates.Let us consider the sub-expression ∇ · (a 1 ê 1 ). Now ê 1 = ê 2 × ê 3 = h 2 ∇u 2 × h 3 ∇u 3 . Therefore∇ · (a 1 ê 1 )=∇ · (a 1 h 2 h 3 ∇u 2 ×∇u 3 ),= ∇(a 1 h 2 h 3 ) · (∇u 2 ×∇u 3 )+a 1 h 2 h 3 ∇ · (∇u 2 ×∇u 3 ).However, ∇ · (∇u 2 ×∇u 3 ) = 0, from (10.43), so we obtain( )ê2∇ · (a 1 ê 1 )=∇(a 1 h 2 h 3 ) · × ê3ê 1= ∇(a 1 h 2 h 3 ) · ;h 2 h 3 h 2 h 3letting Φ = a 1 h 2 h 3 in (10.60) and substituting into the above equation, we find∇ · (a 1 ê 1 )= 1 ∂(a 1 h 2 h 3 ).h 1 h 2 h 3 ∂u 1Repeating the analysis for ∇ · (a 2 ê 2 )and∇ · (a 3 ê 3 ), and adding the results we obtain (10.61),as required. ◭LaplacianIn the expression for the divergence (10.61), leta = ∇Φ = 1 ∂Φê 1 + 1 ∂Φê 2 + 1 ∂Φê 3 ,h 1 ∂u 1 h 2 ∂u 2 h 3 ∂u 3where we have used (10.60). We then obtain∇ 2 Φ= 1 [ ( )∂ h2 h 3 ∂Φ+ ∂ ( )h3 h 1 ∂Φ+ ∂ ( )]h1 h 2 ∂Φ,h 1 h 2 h 3 ∂u 1 h 1 ∂u 1 ∂u 2 h 2 ∂u 2 ∂u 3 h 3 ∂u 3which is the expression for the Laplacian in orthogonal curvilinear coordinates.CurlThe curl of a vector field a = a 1 ê 1 + a 2 ê 2 + a 3 ê 3coordinates is given byin orthogonal curvilinear∣ ∣ ∣∣∣∣∣∣∣∣ h 1 ê 1 h 2 ê 2 h 3 ê 3 ∣∣∣∣∣∣∣∣∇×a = 1 ∂ ∂ ∂. (10.62)h 1 h 2 h 3 ∂u 1 ∂u 2 ∂u 3h 1 a 1 h 2 a 2 h 3 a 3◮Prove the expression for ∇×a in orthogonal curvilinear coordinates.Let us consider the sub-expression ∇×(a 1 ê 1 ). Since ê 1 = h 1 ∇u 1 we have∇×(a 1 ê 1 )=∇×(a 1 h 1 ∇u 1 ),= ∇(a 1 h 1 ) ×∇u 1 + a 1 h 1 ∇×∇u 1 .But ∇×∇u 1 =0,soweobtain∇×(a 1 ê 1 )=∇(a 1 h 1 ) × ê1h 1.368

10.11 EXERCISES∇Φ =∇ · a =∇×a =∇ 2 Φ =1 ∂Φê 1 + 1 ∂Φê 2 + 1 ∂Φê 3h 1 ∂u 1 h 2 ∂u 2 h 3 ∂u 3[1 ∂(h 2 h 3 a 1 )+h 1 h 2 h 3 ∂u 1∣ ∣ ∣∣∣∣∣∣ h 1 ê 1 h 2 ê 2 h 3 ê 3 ∣∣∣∣∣∣1 ∂ ∂ ∂h 1 h 2 h 3 ∂u 1 ∂u 2 ∂u 3h 1 a 1 h 2 a 2 h 3 a 3∂∂u 2(h 3 h 1 a 2 )+∂ ](h 1 h 2 a 3 )∂u 3[ ( )1 ∂ h2 h 3 ∂Φ+ ∂ ( )h3 h 1 ∂Φ+ ∂ ( )]h1 h 2 ∂Φh 1 h 2 h 3 ∂u 1 h 1 ∂u 1 ∂u 2 h 2 ∂u 2 ∂u 3 h 3 ∂u 3Table 10.4 Vector operators in orthogonal curvilinear coordinates u 1 ,u 2 ,u 3 .Φ is a scalar field and a is a vector field.Letting Φ = a 1 h 1 in (10.60) and substituting into the above equation, we find∇×(a 1 ê 1 )=ê2 ∂(a 1 h 1 ) −ê3 ∂(a 1 h 1 ).h 3 h 1 ∂u 3 h 1 h 2 ∂u 2The corresponding analysis of ∇×(a 2 ê 2 ) produces terms in ê 3 and ê 1 , whilst that of∇×(a 3 ê 3 ) produces terms in ê 1 and ê 2 . When the three results are added together, thecoefficients multiplying ê 1 , ê 2 and ê 3 are the same as those obtained by writing out (10.62)explicitly, thus proving the stated result. ◭The general expressions for the vector operators in orthogonal curvilinearcoordinates are shown for reference in table 10.4. The explicit results for cylindricaland spherical polar coordinates, given in tables 10.2 and 10.3 respectively, areobtained by substituting the appropriate set of scale factors in each case.A discussion of the expressions for vector operators in tensor form, whichare valid even for non-orthogonal curvilinear coordinate systems, is given inchapter 26.10.11 Exercises10.1 Evaluate the integral∫ [a(ḃ · a + b · ȧ)+ȧ(b · a) − 2(ȧ · a)b − ḃ|a|2] dtin which ȧ, ḃ are the derivatives of a, b with respect to t.10.2 At time t = 0, the vectors E and B are given by E = E 0 and B = B 0 ,wheretheunit vectors, E 0 and B 0 are fixed and orthogonal. The equations of motion aredEdt = E 0 + B × E 0 ,dBdt = B 0 + E × B 0 .Find E and B at a general time t, showing that after a long time the directionsof E and B have almost interchanged.369

VECTOR CALCULUS10.3 The general equation of motion of a (non-relativistic) particle of mass m andcharge q when it is placed in a region where there is a magnetic field B and anelectric field E ism¨r = q(E + ṙ × B);here r is the position of the particle at time t and ṙ = dr/dt, etc. Write this asthree separate equations in terms of the Cartesian components of the vectorsinvolved.For the simple case of crossed uniform fields E = Ei, B = Bj, inwhichtheparticle starts from the origin at t =0withṙ = v 0 k, find the equations of motionand show the following:(a) if v 0 = E/B then the particle continues its initial motion;(b) if v 0 = 0 then the particle follows the space curve given in terms of theparameter ξ byx = mEmE(1 − cos ξ), y =0, z = (ξ − sin ξ).B 2 q B 2 qInterpret this curve geometrically and relate ξ to t. Show that the totaldistance travelled by the particle after time t is given by∫2E tBqt′B ∣sin 0 2m ∣ dt′ .10.4 Use vector methods to find the maximum angle to the horizontal at which a stonemay be thrown so as to ensure that it is always moving away from the thrower.10.5 If two systems of coordinates with a common origin O are rotating with respectto each other, the measured accelerations differ in the two systems. Denotingby r and r ′ position vectors in frames OXY Z and OX ′ Y ′ Z ′ , respectively, theconnection between the two is¨r ′ = ¨r + ˙ω × r +2ω × ṙ + ω × (ω × r),where ω is the angular velocity vector of the rotation of OXY Z with respect toOX ′ Y ′ Z ′ (taken as fixed). The third term on the RHS is known as the Coriolisacceleration, whilst the final term gives rise to a centrifugal force.Consider the application of this result to the firing of a shell of mass m froma stationary ship on the steadily rotating earth, working to the first order inω (= 7.3 × 10 −5 rad s −1 ). If the shell is fired with velocity v at time t =0andonlyreaches a height that is small compared with the radius of the earth, show thatits acceleration, as recorded on the ship, is given approximately by¨r = g − 2ω × (v + gt),where mg is the weight of the shell measured on the ship’s deck.The shell is fired at another stationary ship (a distance s away) and v is suchthat the shell would have hit its target had there been no Coriolis effect.(a) Show that without the Coriolis effect the time of flight of the shell wouldhave been τ = −2g · v/g 2 .(b) Show further that when the shell actually hits the sea it is off-target byapproximately2τg [(g × ω) · v](gτ + v) − (ω × 2 v)τ2 − 1 3 (ω × g)τ3 .(c) Estimate the order of magnitude ∆ of this miss for a shell for which theinitial speed v is 300 m s −1 , firing close to its maximum range (v makes anangle of π/4 with the vertical) in a northerly direction, whilst the ship isstationed at latitude 45 ◦ North.370

10.11 EXERCISES10.6 Prove that for a space curve r = r(s), where s is the arc length measured alongthe curve from a fixed point, the triple scalar product( ) drds × d2 r· d3 rds 2 ds 3at any point on the curve has the value κ 2 τ,whereκ is the curvature and τ thetorsion at that point.10.7 For the twisted space curve y 3 +27axz − 81a 2 y = 0, given parametrically byx = au(3 − u 2 ), y =3au 2 , z = au(3 + u 2 ),show that the following hold:(a) ds/du =3 √ 2a(1 + u 2 ), where s is the distance along the curve measured fromthe origin;(b) the length of the curve from the origin to the Cartesian point (2a, 3a, 4a) is4 √ 2a;(c) the radius of curvature at the point with parameter u is 3a(1 + u 2 ) 2 ;(d) the torsion τ and curvature κ at a general point are equal;(e) any of the Frenet–Serret formulae that you have not already used directlyare satisfied.10.8 The shape of the curving slip road joining two motorways, that cross at rightangles and are at vertical heights z =0andz = h, can be approximated by thespace curve√2hr =π( zπ)√2hln cos i +2h π( zπ)ln sin j + zk.2hShow that the radius of curvature ρ of the slip road is (2h/π)cosec (zπ/h) atheight z and that the torsion τ = −1/ρ. To shorten the algebra, set z =2hθ/πand use θ as the parameter.10.9 In a magnetic field, field lines are curves to which the magnetic induction B iseverywhere tangential. By evaluating dB/ds, wheres is the distance measuredalong a field line, prove that the radius of curvature at any point on a line isgiven byB 3ρ =|B × (B · ∇)B| .10.10 Find the areas of the given surfaces using parametric coordinates.(a) Using the parameterisation x = u cos φ, y = u sin φ, z = u cot Ω, find thesloping surface area of a right circular cone of semi-angle Ω whose base hasradius a. Verifythatitisequalto 1 ×perimeter of the base ×slope height.2(b) Using the same parameterization as in (a) for x and y, and an appropriatechoice for z, find the surface area between the planes z =0andz = Z ofthe paraboloid of revolution z = α(x 2 + y 2 ).10.11 Parameterising the hyperboloidx 2a + y22 b − z22 c =1 2by x = a cos θ sec φ, y = b sin θ sec φ, z = c tan φ, show that an area element onits surface isdS =sec 2 φ [ c 2 sec 2 φ ( b 2 cos 2 θ + a 2 sin 2 θ ) + a 2 b 2 tan 2 φ ] 1/2dθ dφ.371

VECTOR CALCULUSUse this formula to show that the area of the curved surface x 2 + y 2 − z 2 = a 2between the planes z =0andz =2a is(πa 2 6+ √ 1 sinh −1 2 √ )2 .210.12 For the functionz(x, y) =(x 2 − y 2 )e −x2 −y 2 ,find the location(s) at which the steepest gradient occurs. What are the magnitudeand direction of that gradient? The algebra involved is easier if plane polarcoordinates are used.10.13 Verify by direct calculation that∇ · (a × b) =b · (∇×a) − a · (∇×b).10.14 In the following exercises, a, b and c are vector fields.(a) Simplify∇×a(∇ · a) + a × [∇×(∇×a)] + a ×∇ 2 a.(b) By explicitly writing out the terms in Cartesian coordinates, prove that[c · (b · ∇) − b · (c · ∇)] a =(∇×a) · (b × c).(c) Prove that a × (∇×a) =∇( 1 2 a2 ) − (a · ∇)a.10.15 Evaluate the Laplacian of the functionzx 2ψ(x, y, z) =x 2 + y 2 + z 2(a) directly in Cartesian coordinates, and (b) after changing to a spherical polarcoordinate system. Verify that, as they must, the two methods give the sameresult.10.16 Verify that (10.42) is valid for each component separately when a is the Cartesianvector x 2 y i + xyz j + z 2 y k, by showing that each side of the equation is equal toz i +(2x +2z) j + x k.10.17 The (Maxwell) relationship between a time-independent magnetic field B and thecurrent density J (measured in SI units in A m −2 ) producing it,∇×B = µ 0 J,can be applied to a long cylinder of conducting ionised gas which, in cylindricalpolar coordinates, occupies the region ρaand B = B(ρ) forρ

10.11 EXERCISES(a) For cylindrical polar coordinates ρ, φ, z, evaluate the derivatives of the threeunit vectors with respect to each of the coordinates, showing that only ∂ê ρ /∂φand ∂ê φ /∂φ are non-zero.(i) Hence evaluate ∇ 2 a when a is the vector ê ρ , i.e. a vector of unit magnitudeeverywhere directed radially outwards and expressed by a ρ =1,a φ =a z =0.(ii) Note that it is trivially obvious that ∇×a = 0 and hence that equation(10.41) requires that ∇(∇ · a) =∇ 2 a.(iii) Evaluate ∇(∇ · a) and show that the latter equation holds, but that[∇(∇ · a)] ρ ≠ ∇ 2 a ρ .(b) Rework the same problem in Cartesian coordinates (where, as it happens,the algebra is more complicated).10.19 Maxwell’s equations for electromagnetism in free space (i.e. in the absence ofcharges, currents and dielectric or magnetic media) can be written(i) ∇ · B =0, (ii) ∇ · E =0,(iii) ∇×E + ∂B∂t = 0, (iv) ∇×B − 1 ∂Ec 2 ∂t = 0.A vector A is defined by B = ∇×A, andascalarφ by E = −∇φ − ∂A/∂t. Showthat if the condition(v) ∇ · A + 1 ∂φc 2 ∂t =0is imposed (this is known as choosing the Lorentz gauge), then A and φ satisfywave equations as follows:(vi) ∇ 2 φ − 1 ∂ 2 φc 2 ∂t =0, 2(vii) ∇ 2 A − 1 ∂ 2 Ac 2 ∂t = 0. 2The reader is invited to proceed as follows.(a) Verify that the expressions for B and E in terms of A and φ are consistentwith (i) and (iii).(b) Substitute for E in (ii) and use the derivative with respect to time of (v) toeliminate A from the resulting expression. Hence obtain (vi).(c) Substitute for B and E in (iv) in terms of A and φ. Then use the gradient of(v) to simplify the resulting equation and so obtain (vii).10.20 In a description of the flow of a very viscous fluid that uses spherical polarcoordinates with axial symmetry, the components of the velocity field u are givenin terms of the stream function ψ by1 ∂ψu r =r 2 sin θ ∂θ , u θ = −1 ∂ψr sin θ ∂r .Find an explicit expression for the differential operator E defined byEψ = −(r sin θ)(∇×u) φ .The stream function satisfies the equation of motion E 2 ψ = 0 and, for the flow ofa fluid past a sphere, takes the form ψ(r, θ) =f(r)sin 2 θ. Show that f(r) satisfiesthe (ordinary) differential equationr 4 f (4) − 4r 2 f ′′ +8rf ′ − 8f =0.373

VECTOR CALCULUS10.21 Paraboloidal coordinates u, v, φ are defined in terms of Cartesian coordinates byx = uv cos φ, y = uv sin φ, z = 1 2 (u2 − v 2 ).Identify the coordinate surfaces in the u, v, φ system. Verify that each coordinatesurface (u = constant, say) intersects every coordinate surface on which one ofthe other two coordinates (v, say) is constant. Show further that the system ofcoordinates is an orthogonal one and determine its scale factors. Prove that theu-component of ∇×a is given by(1 aφ(u 2 + v 2 ) 1/2 v + ∂a )φ− 1 ∂a v∂v uv ∂φ .10.22 Non-orthogonal curvilinear coordinates are difficult to work with and should beavoided if at all possible, but the following example is provided to illustrate thecontent of section 10.10.In a new coordinate system for the region of space in which the Cartesiancoordinate z satisfies z ≥ 0, the position of a point r is given by (α 1 ,α 2 ,R), whereα 1 and α 2 are respectively the cosines of the angles made by r with the x- andycoordinateaxes of a Cartesian system and R = |r|. The ranges are −1 ≤ α i ≤ 1,0 ≤ R

10.12 HINTS AND ANSWERS(a) Express z and the perpendicular distance ρ from P to the z-axis in terms ofu 1 ,u 2 ,u 3 .(b) Evaluate ∂x/∂u i , ∂y/∂u i , ∂z/∂u i ,fori =1, 2, 3.(c)Find the Cartesian components of û j and hence show that the new coordinatesare mutually orthogonal. Evaluate the scale factors and the infinitesimalvolume element in the new coordinate system.(d) Determine and sketch the forms of the surfaces u i =constant.(e) Find the most general function f of u 1 only that satisfies ∇ 2 f =0.10.12 Hints and answers10.1 Group the term so that they form the total derivatives of compound vectorexpressions. The integral has the value a × (a × b)+h.10.3 For crossed uniform fields, ẍ+(Bq/m) 2 x = q(E −Bv 0 )/m, ÿ =0,mż = qBx+mv 0 ;(b) ξ = Bqt/m; the path is a cycloid in the plane y =0;ds =[(dx/dt) 2 +(dz/dt) 2 ] 1/2 dt.10.5 g = ¨r ′ − ω × (ω × r), where ¨r ′ is the shell’s acceleration measured by an observerfixed in space. To first order in ω, the direction of g is radial, i.e. parallel to ¨r ′ .(a) Note that s is orthogonal to g.(b) If the actual time of flight is T ,use(s +∆)· g =0toshowthatT ≈ τ(1 + 2g −2 (g × ω) · v + ···).In the Coriolis terms, it is sufficient to put T ≈ τ.(c) For this situation (g × ω) · v =0andω × v = 0; τ ≈ 43 s and ∆ = 10–15 mto the East.10.7 (a) Evaluate (dr/du) · (dr/du).(b) Integrate the previous result between u =0andu =1.(c) ˆt =[ √ 2(1 + u 2 )] −1 [(1 − u 2 )i +2uj +(1+u 2 )k]. Use dˆt/ds =(dˆt/du)/(ds/du);ρ −1 = |dˆt/ds|.(d) ˆn =(1+u 2 ) −1 [−2ui +(1− u 2 )j]. ˆb =[ √ 2(1 + u 2 )] −1 [(u 2 − 1)i − 2uj +(1+u 2 )k].Use dˆb/ds =(dˆb/du)/(ds/du) and show that this equals −[3a(1 + u 2 ) 2 ] −1 ˆn.(e) Show that dˆn/ds = τ(ˆb − ˆt) =−2[3 √ 2a(1 + u 2 ) 3 ] −1 [(1 − u 2 )i +2uj].10.9 Note that dB =(dr · ∇)B and that B = Bˆt, withˆt = dr/ds. Obtain(B · ∇)B/B =ˆt(dB/ds)+ˆn(B/ρ) and then take the vector product of ˆt with this equation.10.11 To integrate sec 2 φ(sec 2 φ +tan 2 φ) 1/2 dφ put tan φ =2 −1/2 sinh ψ.10.13 Work in Cartesian coordinates, regrouping the terms obtained by evaluating thedivergence on the LHS.10.15 (a) 2z(x 2 +y 2 +z 2 ) −3 [(y 2 +z 2 )(y 2 +z 2 −3x 2 )−4x 4 ]; (b) 2r −1 cos θ (1−5sin 2 θ cos 2 φ);both are equal to 2zr −4 (r 2 − 5x 2 ).10.17 Use the formulae given in table 10.2.(a) C = −B 0 /(µ 0 a); B(ρ) =B 0 ρ/a.(b) B 0 ρ 2 /(3a) forρa.(c) [B0 2/(2µ 0)][1 − (ρ/a) 2 ].10.19 Recall that ∇×∇φ = 0 for any scalar φ and that ∂/∂t and ∇ act on differentvariables.10.21 Two sets of paraboloids of revolution about the z-axis and the sheaf of planescontaining the z-axis. For constant u, −∞

VECTOR CALCULUS10.23 The tangent vectors are as follows: for u = 0, the line joining (1, 0, 0) and(−1, 0, 0); for v = 0, the line joining (1, 0, 0) and (∞, 0, 0); for v = π/2, the line(0, 0,z); for v = π, the line joining (−1, 0, 0) and (−∞, 0, 0).ψ(u) =2tan −1 e u + c, derivedfrom∂[cosh u(∂ψ/∂u)]/∂u =0.376

11Line, surface and volume integralsIn the previous chapter we encountered continuously varying scalar and vectorfields and discussed the action of various differential operators on them. Inaddition to these differential operations, the need often arises to consider theintegration of field quantities along lines, over surfaces and throughout volumes.In general the integrand may be scalar or vector in nature, but the evaluationof such integrals involves their reduction to one or more scalar integrals, whichare then evaluated. In the case of surface and volume integrals this requires theevaluation of double and triple integrals (see chapter 6).11.1 Line integralsIn this section we discuss line or path integrals, in which some quantity relatedto the field is integrated between two given points in space, A and B, along aprescribed curve C that joins them. In general, we may encounter line integralsof the forms ∫ ∫∫φdr, a · dr, a × dr, (11.1)CCwhere φ is a scalar field and a is a vector field. The three integrals themselves arerespectively vector, scalar and vector in nature. As we will see below, in physicalapplications line integrals of the second type are by far the most common.The formal definition of a line integral closely follows that of ordinary integralsand can be considered as the limit of a sum. We may divide the path C joiningthe points A and B into N small line elements ∆r p , p =1,...,N.If(x p ,y p ,z p )isany point on the line element ∆r p then the second type of line integral in (11.1),for example, is defined as∫∑Na · dr = lim a(x p ,y p ,z p ) · ∆r p ,CN→∞p=1where it is assumed that all |∆r p |→0asN →∞.377C

LINE, SURFACE AND VOLUME INTEGRALSEach of the line integrals in (11.1) is evaluated over some curve C that may beeither open (A and B being distinct points) or closed (the curve C forms a loop,so that A and B are coincident). In the case where C is closed, the line integralis written ∮ Cto indicate this. The curve may be given either parametrically byr(u) =x(u)i + y(u)j + z(u)k or by means of simultaneous equations relating x, y, zfor the given path (in Cartesian coordinates). A full discussion of the differentrepresentations of space curves was given in section 10.3.In general, the value of the line integral depends not only on the end-pointsA and B but also on the path C joining them. For a closed curve we must alsospecify the direction around the loop in which the integral is taken. It is usuallytaken to be such that a person walking around the loop C in this directionalways has the region R on his/her left; this is equivalent to traversing C in theanticlockwise direction (as viewed from above).11.1.1 Evaluating line integralsThe method of evaluating a line integral is to reduce it to a set of scalar integrals.It is usual to work in Cartesian coordinates, in which case dr = dx i + dy j + dz k.The first type of line integral in (11.1) then becomes simply∫ ∫∫∫φdr = i φ(x, y, z) dx + j φ(x, y, z) dy + k φ(x, y, z) dz.CCCThe three integrals on the RHS are ordinary scalar integrals that can be evaluatedin the usual way once the path of integration C has been specified. Note that inthe above we have used relations of the form∫ ∫φ i dx = i φdx,which is allowable since the Cartesian unit vectors are of constant magnitudeand direction and hence may be taken out of the integral. If we had been usinga different coordinate system, such as spherical polars, then, as we saw in theprevious chapter, the unit basis vectors would not be constant. In that case thebasis vectors could not be factorised out of the integral.The second and third line integrals in (11.1) can also be reduced to a set ofscalar integrals by writing the vector field a in terms of its Cartesian componentsas a = a x i + a y j + a z k,wherea x ,a y ,a z are each (in general) functions of x, y, z.The second line integral in (11.1), for example, can then be written as∫ ∫a · dr = (a x i + a y j + a z k) · (dx i + dy j + dz k)CC∫= (a x dx + a y dy + a z dz)C∫ ∫ ∫= a x dx + a y dy + a z dz. (11.2)CC378CC

11.1 LINE INTEGRALSA similar procedure may be followed for the third type of line integral in (11.1),which involves a cross product.Line integrals have properties that are analogous to those of ordinary integrals.In particular, the following are useful properties (which we illustrate using thesecond form of line integral in (11.1) but which are valid for all three types).(i) Reversing the path of integration changes the sign of the integral. If thepath C along which the line integrals are evaluated has A and B as itsend-points then∫ BAa · dr = −∫ ABa · dr.This implies that if the path C is a loop then integrating around the loopin the opposite direction changes the sign of the integral.(ii) If the path of integration is subdivided into smaller segments then the sumof the separate line integrals along each segment is equal to the line integralalong the whole path. So, if P is any point on the path of integration thatlies between the path’s end-points A and B then∫ BAa · dr =∫ PAa · dr +∫ BPa · dr.◮Evaluate the line integral I = ∫ a · dr, wherea =(x + y)i +(y − x)j, along each of theCpaths in the xy-plane shown in figure 11.1, namely(i) the parabola y 2 = x from (1, 1) to (4, 2),(ii) the curve x =2u 2 + u +1, y =1+u 2 from (1, 1) to (4, 2),(iii) the line y = 1 from (1, 1) to (4, 1), followed by the line x = 4 from (4, 1)to (4, 2).Since each of the paths lies entirely in the xy-plane, we have dr = dx i + dy j. Wecantherefore write the line integral as∫ ∫I = a · dr = [(x + y) dx +(y − x) dy]. (11.3)CCWe must now evaluate this line integral along each of the prescribed paths.Case (i). Along the parabola y 2 = x we have 2ydy = dx. Substituting for x in (11.3)and using just the limits on y, weobtainI =∫ (4,2)(1,1)[(x + y) dx +(y − x) dy] =∫ 21[(y 2 + y)2y +(y − y 2 )] dy =11 1 3 .Note that we could just as easily have substituted for y and obtained an integral in x,which would have given the same result.Case (ii). The second path is given in terms of a parameter u. We could eliminate ubetween the two equations to obtain a relationship between x and y directly and proceedas above, but it is usually quicker to write the line integral in terms of the parameter u.Along the curve x =2u 2 + u +1, y =1+u 2 we have dx =(4u +1)du and dy =2udu.379

LINE, SURFACE AND VOLUME INTEGRALSy(i)(4, 2)(ii)(1, 1)(iii)Figure 11.1 Different possible paths between the points (1, 1) and (4, 2).xSubstituting for x and y in (11.3) and writing the correct limits on u, weobtainI ==∫ (4,2)(1,1)∫ 10[(x + y) dx +(y − x) dy][(3u 2 + u + 2)(4u +1)− (u 2 + u)2u] du =10 2 3 .Case (iii). For the third path the line integral must be evaluated along the two linesegments separately and the results added together. First, along the line y = 1 we havedy = 0. Substituting this into (11.3) and using just the limits on x forthissegment,weobtain∫ (4,1)∫ 4[(x + y) dx +(y − x) dy] = (x +1)dx =10 1 . 2(1,1)Next, along the line x = 4 we have dx = 0. Substituting this into (11.3) and using just thelimits on y for this segment, we obtain∫ (4,2)(4,1)[(x + y) dx +(y − x) dy] =1∫ 21(y − 4) dy = −2 1 2 .The value of the line integral along the whole path is just the sum of the values of the lineintegrals along each segment, and is given by I =10 1 2 − 2 1 2 =8.◭When calculating a line integral along some curve C, which is given in termsof x, y and z, we are sometimes faced with the problem that the curve C is suchthat x, y and z are not single-valued functions of one another over the entirelength of the curve. This is a particular problem for closed loops in the xy-plane(and also for some open curves). In such cases the path may be subdivided intoshorter line segments along which one coordinate is a single-valued function ofthe other two. The sum of the line integrals along these segments is then equalto the line integral along the entire curve C. A better solution, however, is torepresent the curve in a parametric form r(u) that is valid for its entire length.380

11.1 LINE INTEGRALS◮Evaluate the line integral I = ∮ xdy,whereC is the circle in the xy-plane defined byCx 2 + y 2 = a 2 , z =0.Adopting the usual convention mentioned above, the circle C is to be traversed in theanticlockwise direction. Taking the circle as a whole means x is not a single-valuedfunction of y. We must therefore divide the path into two parts with x =+ √ a 2 − y 2 forthe semicircle lying to the right of x =0,andx = − √ a 2 − y 2 for the semicircle lying tothe left of x = 0. The required line integral is then the sum of the integrals along the twosemicircles. Substituting for x, itisgivenby∮ ∫ a √∫ −aI = xdy = a2 − y 2 dy +(− √ )a 2 − y 2 dyC−aa∫ a √=4 a2 − y 2 dy = πa 2 .0Alternatively, we can represent the entire circle parametrically, in terms of the azimuthalangle φ, sothatx = a cos φ and y = a sin φ with φ running from 0 to 2π. The integral cantherefore be evaluated over the whole circle at once. Noting that dy = a cos φdφ,wecanrewrite the line integral completely in terms of the parameter φ and obtain∮ ∫ 2πI = xdy = a 2 cos 2 φdφ= πa 2 . ◭C011.1.2 Physical examples of line integralsThere are many physical examples of line integrals, but perhaps the most commonis the expression for the total work done by a force F when it moves its pointof application from a point A to a point B along a given curve C. We allow themagnitude and direction of F to vary along the curve. Let the force act at a pointr and consider a small displacement dr along the curve; then the small amountof work done is dW = F · dr, as discussed in subsection 7.6.1 (note that dW canbe either positive or negative). Therefore, the total work done in traversing thepath C is∫W C = F · dr.CNaturally, other physical quantities can be expressed in such a way. For example,the electrostatic potential energy gained by moving a charge q along a path C inan electric field E is −q ∫ CE · dr. We may also note that Ampère’s law concerningthe magnetic field B associated with a current-carrying wire can be written as∮B · dr = µ 0 I,Cwhere I is the current enclosed by a closed path C traversed in a right-handedsense with respect to the current direction.Magnetostatics also provides a physical example of the third type of line381

LINE, SURFACE AND VOLUME INTEGRALSintegral in (11.1). If a loop of wire C carrying a current I is placed in a magneticfield B then the force dF on a small length dr of the wire is given by dF = Idr×B,and so the total (vector) force on the loop is∮F = I dr × B.C11.1.3 Line integrals with respect to a scalarIn addition to those listed in (11.1), we can form other types of line integral,which depend on a particular curve C but for which we integrate with respectto a scalar du, rather than the vector differential dr. This distinction is somewhatarbitrary, however, since we can always rewrite line integrals containing the vectordifferential dr as a line integral with respect to some scalar parameter. If the pathC along which the integral is taken is described parametrically by r(u) thendr = drdu du,and the second type of line integral in (11.1), for example, can be written as∫C∫a · dr =Ca · drdu du.A similar procedure can be followed for the other types of line integral in (11.1).Commonly occurring special cases of line integrals with respect to a scalar are∫ ∫φds, a ds,Cwhere s is the arc length along the curve C. We can always represent C parametricallyby r(u), and from section 10.3 we have√drds =du · drdu du.The line integrals can therefore be expressed entirely in terms of the parameter uand thence evaluated.◮Evaluate the line integral I = ∫ C (x − y)2 ds, whereC is the semicircle of radius a runningfrom A =(a, 0) to B =(−a, 0) and for which y ≥ 0.The semicircular path from A to B can be described in terms of the azimuthal angle φ(measured from the x-axis) byr(φ) =a cos φ i + a sin φ j,where φ runs from 0 to π. Therefore the element of arc length is given, from section 10.3,byds =√drdφ · drdφ dφ = a(cos2 φ +sin 2 φ) dφ = adφ.382C

11.2 CONNECTIVITY OF REGIONS(a) (b) (c)Figure 11.2 (a) A simply connected region; (b) a doubly connected region;(c) a triply connected region.Since (x − y) 2 = a 2 (1 − sin 2φ), the line integral becomes∫∫ πI = (x − y) 2 ds = a 3 (1 − sin 2φ) dφ = πa 3 . ◭C0As discussed in the previous chapter, the expression (10.58) for the square ofthe element of arc length in three-dimensional orthogonal curvilinear coordinatesu 1 ,u 2 ,u 3 is(ds) 2 = h 2 1 (du 1 ) 2 + h 2 2 (du 2 ) 2 + h 2 3 (du 3 ) 2 ,where h 1 ,h 2 ,h 3 are the scale factors of the coordinate system. If a curve C inthree dimensions is given parametrically by the equations u i = u i (λ) fori =1, 2, 3then the element of arc length along the curve isds =√h 2 1( ) 2 ( ) 2 ( ) 2 du1+ h 2 du22+ h 2 du33dλ.dλ dλ dλ11.2 Connectivity of regionsIn physical systems it is usual to define a scalar or vector field in some region R.In the next and some later sections we will need the concept of the connectivityof such a region in both two and three dimensions.We begin by discussing planar regions. A plane region R is said to be simplyconnected if every simple closed curve within R can be continuously shrunk toa point without leaving the region (see figure 11.2(a)). If, however, the regionR contains a hole then there exist simple closed curves that cannot be shrunkto a point without leaving R (see figure 11.2(b)). Such a region is said to bedoubly connected, since its boundary has two distinct parts. Similarly, a regionwith n − 1 holes is said to be n-fold connected, ormultiply connected (the regionin figure 11.2(c) is triply connected).383

LINE, SURFACE AND VOLUME INTEGRALSydVSRUcTCa b xFigure 11.3 A simply connected region R bounded by the curve C.These ideas can be extended to regions that are not planar, such as generalthree-dimensional surfaces and volumes. The same criteria concerning the shrinkingof closed curves to a point also apply when deciding the connectivity of suchregions. In these cases, however, the curves must lie in the surface or volumein question. For example, the interior of a torus is not simply connected, sincethere exist closed curves in the interior that cannot be shrunk to a point withoutleaving the torus. The region between two concentric spheres of different radii issimply connected.11.3 Green’s theorem in a planeIn subsection 11.1.1 we considered (amongst other things) the evaluation of lineintegrals for which the path C is closed and lies entirely in the xy-plane. Sincethe path is closed it will enclose a region R of the plane. We now discuss how toexpress the line integral around the loop as a double integral over the enclosedregion R.Suppose the functions P (x, y), Q(x, y) and their partial derivatives are singlevalued,finite and continuous inside and on the boundary C of some simplyconnected region R in the xy-plane. Green’s theorem in a plane (sometimes calledthe divergence theorem in two dimensions) then states∮∫∫ ( ∂Q(P dx+ Qdy)=CR ∂x − ∂P )dx dy, (11.4)∂yand so relates the line integral around C to a double integral over the enclosedregion R. This theorem may be proved straightforwardly in the following way.Consider the simply connected region R in figure 11.3, and let y = y 1 (x) and384

11.3 GREEN’S THEOREM IN A PLANEy = y 2 (x) be the equations of the curves STU and SVU respectively. We thenwrite∫∫∫∂Pb ∫ y2(x)R ∂y dx dy = dx dy ∂P ∫ b [ ] y=y2(x)a y 1(x) ∂y = dx P (x, y)ay=y 1(x)∫ b []= P (x, y 2 (x)) − P (x, y 1 (x)) dxa∫ b∫ a∮= − P (x, y 1 (x)) dx − P (x, y 2 (x)) dx = − Pdx.abCIf we now let x = x 1 (y) andx = x 2 (y) be the equations of the curves TSV andTUV respectively, we can similarly show that∫∫∫∂Qd ∫ x2(y)R ∂x dx dy = dy dx ∂Q ∫ d [ ] x=x2(y)c x 1(y) ∂x = dy Q(x, y)cx=x 1(y)∫ d []= Q(x 2 (y),y) − Q(x 1 (y),y) dyc∫ c∫ d∮= Q(x 1 ,y) dy + Q(x 2 ,y) dy = Qdy.dcCSubtracting these two results gives Green’s theorem in a plane.◮Show ∮ that the area of a region R enclosed by a simple closed curve C is given by A =1(xdy−ydx)=∮ xdy = − ∮ ydx. Hence calculate the area of the ellipse x = a cos φ,2 C C Cy = b sin φ.In Green’s theorem (11.4) put P = −y and Q = x; then∮∫∫∫∫(xdy− ydx)= (1+1)dx dy =2 dx dy =2A.CRR∮Therefore the area of the region is A = 1 (xdy−ydx). Alternatively, we could put P =02 Cand Q = x and obtain A = ∮ xdy, or put P = −y and Q = 0, which gives A = − ∮ ydx.C CThe area of the ellipse x = a cos φ, y = b sin φ is given byA = 1 ∮ydx)=2 C(xdy− 1 ∫ 2πab(cos 2 φ +sin 2 φ) dφ2 0= ab ∫ 2πdφ = πab. ◭2It may further be shown that Green’s theorem in a plane is also valid formultiply connected regions. In this case, the line integral must be taken overall the distinct boundaries of the region. Furthermore, each boundary must betraversed in the positive direction, so that a person travelling along it in thisdirection always has the region R on their left. In order to apply Green’s theorem3850

LINE, SURFACE AND VOLUME INTEGRALSyRC 2C 1Figure 11.4 A doubly connected region R bounded by the curves C 1 and C 2 .xto the region R shown in figure 11.4, the line integrals must be taken overboth boundaries, C 1 and C 2 , in the directions indicated, and the results addedtogether.We may also use Green’s theorem in a plane to investigate the path independence(or not) of line integrals when the paths lie in the xy-plane. Let us considerthe line integralI =∫ BA(P dx+ Qdy).For the line integral from A to B to be independent of the path taken, it musthave the same value along any two arbitrary paths C 1 and C 2 joining the points.Moreover, if we consider as the path the closed loop C formed by C 1 − C 2 thenthe line integral around this loop must be zero. From Green’s theorem in a plane,(11.4), we see that a sufficient condition for I = 0 is that∂P∂y = ∂Q∂x , (11.5)throughout some simply connected region R containing the loop, where we assumethat these partial derivatives are continuous in R.It may be shown that (11.5) is also a necessary condition for I = 0 and isequivalent to requiring Pdx+ Qdy to be an exact differential of some functionφ(x, y) such that Pdx+ Qdy = dφ. It follows that ∫ BA(Pdx+ Qdy)=φ(B) − φ(A)and that ∮ C(Pdx+ Qdy) around any closed loop C in the region R is identicallyzero. These results are special cases of the general results for paths in threedimensions, which are discussed in the next section.386

11.4 CONSERVATIVE FIELDS AND POTENTIALS◮Evaluate the line integral∮I = [(e x y +cosx sin y) dx +(e x +sinx cos y) dy] ,around the ellipse x 2 /a 2 + y 2 /b 2 =1.CClearly, it is not straightforward to calculate this line integral directly. However, if we letP = e x y +cosx sin y and Q = e x +sinx cos y,then ∂P/∂y = e x +cosx cos y = ∂Q/∂x, andsoPdx+ Qdy is an exact differential (itis actually the differential of the function f(x, y) =e x y +sinx sin y). From the abovediscussion, we can conclude immediately that I =0.◭11.4 Conservative fields and potentialsSo far we have made the point that, in general, the value of a line integralbetween two points A and B depends on the path C taken from A to B. Intheprevious section, however, we saw that, for paths in the xy-plane, line integralswhose integrands have certain properties are independent of the path taken. Wenow extend that discussion to the full three-dimensional case.For line integrals of the form ∫ Ca · dr, there exists a class of vector fields forwhich the line integral between two points is independent of the path taken. Suchvector fields are called conservative. A vector field a that has continuous partialderivatives in a simply connected region R is conservative if, and only if, any ofthe following is true.(i) The integral ∫ BAa · dr, whereA and B lie in the region R, is independent ofthe path from A to B. Hence the integral ∮ Ca · dr around any closed loopin R is zero.(ii) There exists a single-valued function φ of position such that a = ∇φ.(iii) ∇×a = 0.(iv) a · dr is an exact differential.The validity or otherwise of any of these statements implies the same for theother three, as we will now show.First, let us assume that (i) above is true. If the line integral from A to Bis independent of the path taken between the points then its value must be afunction only of the positions of A and B. We may therefore write∫ BAa · dr = φ(B) − φ(A), (11.6)which defines a single-valued scalar function of position φ. If the points A and Bare separated by an infinitesimal displacement dr then (11.6) becomesa · dr = dφ,387

LINE, SURFACE AND VOLUME INTEGRALSwhich shows that we require a · dr to be an exact differential: condition (iv). From(10.27) we can write dφ = ∇φ · dr, and so we have(a −∇φ) · dr =0.Since dr is arbitrary, we find that a = ∇φ; this immediately implies ∇×a = 0,condition (iii) (see (10.37)).Alternatively, if we suppose that there exists a single-valued function of positionφ such that a = ∇φ then ∇×a = 0 follows as before. The line integral around aclosed loop then becomes∮C∮∮a · dr = ∇φ · dr =CSince we defined φ to be single-valued, this integral is zero as required.Now suppose ∇×a = 0. From Stoke’s theorem, which is discussed in section11.9, we immediately obtain ∮ Ca · dr =0;thena = ∇φ and a · dr = dφ followas above.Finally, let us suppose a · dr = dφ. Then immediately we have a = ∇φ, andtheother results follow as above.◮Evaluate the line integral I = ∫ Ba · dr, wherea A =(xy2 + z)i +(x 2 y +2)j + xk, A is thepoint (c, c, h) and B is the point (2c, c/2,h), along the different paths(i) C 1 ,givenbyx = cu, y = c/u, z = h,(ii) C 2 ,givenby2y =3c − x, z = h.Show that the vector field a is in fact conservative, and find φ such that a = ∇φ.Expanding out the integrand, we haveI =∫ (2c, c/2,h)(c, c, h)dφ.[(xy 2 + z) dx +(x 2 y +2)dy + xdz ] , (11.7)which we must evaluate along each of the paths C 1 and C 2 .(i) Along C 1 we have dx = cdu, dy = −(c/u 2 ) du, dz = 0, and on substituting in (11.7)and finding the limits on u, weobtain∫ 2I = c(h − 2 )du = c(h − 1).u 21(ii) Along C 2 we have 2 dy = −dx, dz = 0 and, on substituting in (11.7) and using thelimits on x, weobtainI =∫ 2cc( 12 x3 − 9 4 cx2 + 9 4 c2 x + h − 1 ) dx = c(h − 1).Hence the line integral has the same value along paths C 1 and C 2 . Taking the curl of a,we have∇×a =(0− 0)i +(1− 1)j +(2xy − 2xy)k = 0,so a is a conservative vector field, and the line integral between two points must be388

11.5 SURFACE INTEGRALSindependent of the path taken. Since a is conservative, we can write a = ∇φ. Therefore, φmust satisfy∂φ∂x = xy2 + z,which implies that φ = 1 2 x2 y 2 + zx + f(y, z) for some function f. Secondly, we require∂φ∂y = x2 y + ∂f∂y = x2 y +2,which implies f =2y + g(z). Finally, since∂φ∂z = x + ∂g∂z = x,we have g =constant=k. It can be seen that we have explicitly constructed the functionφ = 1 2 x2 y 2 + zx +2y + k. ◭The quantity φ that figures so prominently in this section is called the scalarpotential function of the conservative vector field a (which satisfies ∇×a = 0), andis unique up to an arbitrary additive constant. Scalar potentials that are multivaluedfunctions of position (but in simple ways) are also of value in describingsome physical situations, the most obvious example being the scalar magneticpotential associated with a current-carrying wire. When the integral of a fieldquantity around a closed loop is considered, provided the loop does not enclosea net current, the potential is single-valued and all the above results still hold. Ifthe loop does enclose a net current, however, our analysis is no longer valid andextra care must be taken.If, instead of being conservative, a vector field b satisfies ∇ · b =0(i.e.bis solenoidal) then it is both possible and useful, for example in the theory ofelectromagnetism, to define a vector field a such that b = ∇×a. Itmaybeshownthat such a vector field a always exists. Further, if a is one such vector field thena ′ = a + ∇ψ + c, whereψ is any scalar function and c is any constant vector, alsosatisfies the above relationship, i.e. b = ∇×a ′ . This was discussed more fully insubsection 10.8.2.11.5 Surface integralsAs with line integrals, integrals over surfaces can involve vector and scalar fieldsand, equally, can result in either a vector or a scalar. The simplest case involvesentirely scalars and is of the form∫φdS. (11.8)SAs analogues of the line integrals listed in (11.1), we may also encounter surfaceintegrals involving vectors, namely∫ ∫∫φdS, a · dS, a × dS. (11.9)SS389S

LINE, SURFACE AND VOLUME INTEGRALSSVdSSdSC(a)(b)Figure 11.5 (a) A closed surface and (b) an open surface. In each case anormal to the surface is shown: dS = ˆn dS.All the above integrals are taken over some surface S, which may be eitheropen or closed, and are therefore, in general, double integrals. Following thenotation for line integrals, for surface integrals over a closed surface ∫ Sis replacedby ∮ S .The vector differential dS in (11.9) represents a vector area element of thesurface S. It may also be written dS = ˆn dS, whereˆn is a unit normal to thesurface at the position of the element and dS is the scalar area of the element usedin (11.8). The convention for the direction of the normal ˆn to a surface dependson whether the surface is open or closed. A closed surface, see figure 11.5(a),does not have to be simply connected (for example, the surface of a torus is not),but it does have to enclose a volume V , which may be of infinite extent. Thedirection of ˆn is taken to point outwards from the enclosed volume as shown.An open surface, see figure 11.5(b), spans some perimeter curve C. The directionof ˆn is then given by the right-hand sense with respect to the direction in whichthe perimeter is traversed, i.e. follows the right-hand screw rule discussed insubsection 7.6.2. An open surface does not have to be simply connected but forour purposes it must be two-sided (a Möbius strip is an example of a one-sidedsurface).The formal definition of a surface integral is very similar to that of a lineintegral. We divide the surface S into N elements of area ∆S p , p =1, 2,...,N,each with a unit normal ˆn p .If(x p ,y p ,z p ) is any point in ∆S p then the second typeof surface integral in (11.9), for example, is defined as∫∑Na · dS = lim a(x p ,y p ,z p ) · ˆn p ∆S p ,SN→∞p=1where it is required that all ∆S p → 0asN →∞.390

11.5 SURFACE INTEGRALSzkαdSSyxRdAFigure 11.6 A surface S (or part thereof) projected onto a region R in thexy-plane; dS is a surface element.11.5.1 Evaluating surface integralsWe now consider how to evaluate surface integrals over some general surface. Thisinvolves writing the scalar area element dS in terms of the coordinate differentialsof our chosen coordinate system. In some particularly simple cases this is verystraightforward. For example, if S is the surface of a sphere of radius a (or somepart thereof) then using spherical polar coordinates θ, φ on the sphere we havedS = a 2 sin θdθdφ. For a general surface, however, it is not usually possible torepresent the surface in a simple way in any particular coordinate system. In suchcases, it is usual to work in Cartesian coordinates and consider the projections ofthe surface onto the coordinate planes.Consider a surface (or part of a surface) S as in figure 11.6. The surface S isprojected onto a region R of the xy-plane, so that an element of surface area dSprojects onto the area element dA. From the figure, we see that dA = | cos α| dS,where α is the angle between the unit vector k in the z-direction and the unitnormal ˆn to the surface at P . So, at any given point of S, we have simplydS =dA| cos α| = dA|ˆn · k| .Now, if the surface S is given by the equation f(x, y, z) = 0 then, as shown in subsection10.7.1, the unit normal at any point of the surface is given by ˆn = ∇f/|∇f|evaluated at that point, cf. (10.32). The scalar element of surface area then becomesdS =dA |∇f| dA |∇f| dA= =|ˆn · k| ∇f · k ∂f/∂z , (11.10)391

LINE, SURFACE AND VOLUME INTEGRALSwhere |∇f| and ∂f/∂z are evaluated on the surface S. We can therefore expressany surface integral over S as a double integral over the region R in the xy-plane.◮Evaluate the surface integral I = ∫ a · dS, wherea = xi and S is the surface of theShemisphere x 2 + y 2 + z 2 = a 2 with z ≥ 0.The surface of the hemisphere is shown in figure 11.7. In this case dS may be easilyexpressed in spherical polar coordinates as dS = a 2 sin θdθdφ, and the unit normal to thesurface at any point is simply ˆr. On the surface of the hemisphere we have x = a sin θ cos φand soa · dS = x (i · ˆr) dS =(a sin θ cos φ)(sin θ cos φ)(a 2 sin θdθdφ).Therefore, inserting the correct limits on θ and φ, we have∫∫ π/2I = a · dS = a 3 dθ sin 3 θS0∫ 2π0dφ cos 2 φ = 2πa33 .We could, however, follow the general prescription above and project the hemisphere Sonto the region R in the xy-plane that is a circle of radius a centred at the origin. Writingthe equation of the surface of the hemisphere as f(x, y) =x 2 + y 2 + z 2 − a 2 =0andusing(11.10), we have∫ ∫∫|∇f| dAI = a · dS = x (i · ˆr) dS = x (i · ˆr)SSR ∂f/∂z .Now ∇f =2xi +2yj +2zk =2r, soonthesurfaceS we have |∇f| =2|r| =2a. OnS wealso have ∂f/∂z =2z =2 √ a 2 − x 2 − y 2 and i · ˆr = x/a. Therefore, the integral becomes∫∫I =Rx 2√ dx dy.a2 − x 2 − y2 Although this integral may be evaluated directly, it is quicker to transform to plane polarcoordinates:∫∫ρ 2 cos 2 φI = √R ′ a2 − ρ ρdρdφ2=∫ 2π0∫ acos 2 φdφ0Making the substitution ρ = a sin u, we finally obtainI =∫ 2π0cos 2 φdφ∫ π/20ρ 3 dρ√a2 − ρ 2 .a 3 sin 3 udu= 2πa33 . ◭In the above discussion we assumed that any line parallel to the z-axis intersectsS only once. If this is not the case, we must split up the surface into smallersurfaces S 1 , S 2 etc. that are of this type. The surface integral over S is then thesum of the surface integrals over S 1 , S 2 and so on. This is always necessary forclosed surfaces.Sometimes we may need to project a surface S (or some part of it) onto thezx- oryz-plane, rather than the xy-plane; for such cases, the above analysis iseasily modified.392

11.5 SURFACE INTEGRALSSzadSxaCadA = dx dyyFigure 11.7 The surface of the hemisphere x 2 + y 2 + z 2 = a 2 , z ≥ 0.11.5.2 Vector areas of surfacesThe vector area of a surface S is defined as∫S = dS,where the surface integral may be evaluated as above.S◮Find the vector area of the surface of the hemisphere x 2 + y 2 + z 2 = a 2 with z ≥ 0.As in the previous example, dS = a 2 sin θdθdφˆr in spherical polar coordinates. Thereforethe vector area is given by∫∫S = a 2 sin θ ˆr dθ dφ.SNow, since ˆr varies over the surface S, it also must be integrated. This is most easilyachieved by writing ˆr in terms of the constant Cartesian basis vectors. On S we haveˆr =sinθ cos φ i +sinθ sin φ j +cosθ k,so the expression for the vector area becomes( ∫ 2π ∫ ) (π/2∫ 2π ∫ )π/2S = i a 2 cos φdφ sin 2 θdθ + j a 2 sin φdφ sin 2 θdθ0000( ∫ 2π ∫ )π/2+ k a 2 dφ sin θ cos θdθ= 0 + 0 + πa 2 k = πa 2 k.00Note that the magnitude of S is the projected area, of the hemisphere onto the xy-plane,and not the surface area of the hemisphere. ◭393

LINE, SURFACE AND VOLUME INTEGRALSdrCrOFigure 11.8 The conical surface spanning the perimeter C and having itsvertex at the origin.The hemispherical shell discussed above is an example of an open surface. Fora closed surface, however, the vector area is always zero. This may be seen byprojecting the surface down onto each Cartesian coordinate plane in turn. Foreach projection, every positive element of area on the upper surface is cancelledby the corresponding negative element on the lower surface. Therefore, eachcomponent of S = ∮ SdS vanishes.An important corollary of this result is that the vector area of an open surfacedepends only on its perimeter, or boundary curve, C. This may be proved asfollows. If surfaces S 1 and S 2 have the same perimeter then S 1 − S 2 is a closedsurface, for which∮ ∫ ∫dS = dS − dS = 0.S 1 S 2Hence S 1 = S 2 . Moreover, we may derive an expression for the vector area ofan open surface S solely in terms of a line integral around its perimeter C.Since we may choose any surface with perimeter C, we will consider a conewith its vertex at the origin (see figure 11.8). The vector area of the elementarytriangular region shown in the figure is dS = 1 2r × dr. Therefore, the vector areaof the cone, and hence of any open surface with perimeter C, is given by the lineintegralS = 1 ∮r × dr.2CFor a surface confined to the xy-plane, r = xi + yj and dr = dx i + dy j,and we∮obtain for this special case that the area of the surface is given byA = 1 2 C(xdy− ydx), as we found in section 11.3.394

11.5 SURFACE INTEGRALS◮Find the vector area of the surface ∮ of the hemisphere x 2 + y 2 + z 2evaluating the line integral S = 1 r × dr around its perimeter.2 C= a 2 , z ≥ 0, byThe perimeter C of the hemisphere is the circle x 2 + y 2 = a 2 , on which we haver = a cos φ i + a sin φ j, dr = −a sin φdφi + a cos φdφj.Therefore the cross product r × dr is given byi j kr × dr =a cos φ asin φ 0∣ −a sin φdφ acos φdφ 0 ∣ = a2 (cos 2 φ +sin 2 φ) dφ k = a 2 dφ k,and the vector area becomes∫ 2πS = 1 2 a2 k dφ = πa 2 k. ◭011.5.3 Physical examples of surface integralsThere are many examples of surface integrals in the physical sciences. Surfaceintegrals of the form (11.8) occur in computing the total electric charge on asurface or the mass of a shell, ∫ Sρ(r) dS, given the charge or mass density ρ(r).For surface integrals involving vectors, the second form in (11.9) is the mostcommon. For a vector field a, the surface integral ∫ Sa · dS is called the fluxof a through S. Examples of physically important flux integrals are numerous.For example, let us consider a surface S in a fluid with density ρ(r) that has avelocity field v(r). The mass of fluid crossing an element of surface area dS intime dt is dM = ρv · dS dt. Therefore the net total mass flux of fluid crossing Sis M = ∫ Sρ(r)v(r) · dS. As a another example, the electromagnetic flux of energyout of a given volume V bounded by a surface S is ∮ S(E × H) · dS.The solid angle, to be defined below, subtended at a point O by a surface (closedor otherwise) can also be represented by an integral of this form, although it isnot strictly a flux integral (unless we imagine isotropic rays radiating from O).The integral∫ ∫r · dS ˆr · dSΩ=S r 3 =S r 2 , (11.11)gives the solid angle Ω subtended at O by a surface S if r is the position vectormeasured from O of an element of the surface. A little thought will show that(11.11) takes account of all three relevant factors: the size of the element ofsurface, its inclination to the line joining the element to O and the distance fromO. Such a general expression is often useful for computing solid angles when thethree-dimensional geometry is complicated. Note that (11.11) remains valid whenthe surface S is not convex and when a single ray from O in certain directionswould cut S in more than one place (but we exclude multiply connected regions).395

LINE, SURFACE AND VOLUME INTEGRALSIn particular, when the surface is closed Ω = 0 if O is outside S and Ω = 4π if Ois an interior point.Surface integrals resulting in vectors occur less frequently. An example isafforded, however, by the total resultant force experienced by a body immersed ina stationary fluid in which the hydrostatic pressure is given by p(r). The pressureis everywhere inwardly directed and the resultant force is F = − ∮ SpdS, takenover the whole surface.11.6 Volume integralsVolume integrals are defined in an obvious way and are generally simpler thanline or surface integrals since the element of volume dV is a scalar quantity. Wemay encounter volume integrals of the forms∫∫φdV, a dV . (11.12)VClearly, the first form results in a scalar, whereas the second form yields a vector.Two closely related physical examples, one of each kind, are provided by the totalmass of a fluid contained in a volume V , given by ∫ Vρ(r) dV , and the total linearmomentum of that same fluid, given by ∫ Vρ(r)v(r) dV where v(r) is the velocityfield in the fluid. As a slightly more complicated example of a volume integral wemay consider the following.◮Find an expression for the angular momentum of a solid body rotating with angularvelocity ω about an axis through the origin.Consider a small volume element dV situated at position r; its linear momentum is ρdVṙ,where ρ = ρ(r) is the density distribution, and its angular momentum about O is r × ρṙ dV .Thus for the whole body the angular momentum L is∫L = (r × ṙ)ρdV.Putting ṙ = ω × r yields∫∫L = [r × (ω × r)] ρdV =VVVV∫ωr 2 ρdV − (r · ω)rρdV. ◭VThe evaluation of the first type of volume integral in (11.12) has already beenconsidered in our discussion of multiple integrals in chapter 6. The evaluation ofthe second type of volume integral follows directly since we can write∫ ∫ ∫∫a dV = i a x dV + j a y dV + k a z dV , (11.13)VVwhere a x ,a y ,a z are the Cartesian components of a. Of course, we could havewritten a in terms of the basis vectors of some other coordinate system (e.g.spherical polars) but, since such basis vectors are not, in general, constant, they396VV

11.6 VOLUME INTEGRALSdSSVrOFigure 11.9 A general volume V containing the origin and bounded by theclosed surface S.cannot be taken out of the integral sign as in (11.13) and must be included aspart of the integrand.11.6.1 Volumes of three-dimensional regionsAs discussed in chapter 6, the volume of a three-dimensional region V is simplyV = ∫ VdV , which may be evaluated directly once the limits of integration havebeen found. However, the volume of the region obviously depends only on thesurface S that bounds it. We should therefore be able to express the volume Vin terms of a surface integral over S. This is indeed possible, and the appropriateexpression may derived as follows. Referring to figure 11.9, let us suppose thatthe origin O is contained within V . The volume of the small shaded cone isdV = 1 3r · dS; the total volume of the region is thus given byV = 1 ∮r · dS.3 SIt may be shown that this expression is valid even when O is not contained in V .Although this surface integral form is available, in practice it is usually simplerto evaluate the volume integral directly.◮Find the volume enclosed between a sphere of radius a centred on the origin and a circularcone of half-angle α with its vertex at the origin.The element of vector area dS on the surface of the sphere is given in spherical polarcoordinates by a 2 sin θdθdφˆr. Now taking the axis of the cone to lie along the z-axis (fromwhich θ is measured) the required volume is given byV = 1 ∮r · dS = 1 ∫ 2π ∫ αdφ a 2 sin θ r · ˆr dθ33S= 1 30∫ 2π0dφ0∫ α0a 3 sin θdθ= 2πa3 (1 − cos α). ◭3397

LINE, SURFACE AND VOLUME INTEGRALS11.7 Integral forms for grad, div and curlIn the previous chapter we defined the vector operators grad, div and curl in purelymathematical terms, which depended on the coordinate system in which they wereexpressed. An interesting application of line, surface and volume integrals is theexpression of grad, div and curl in coordinate-free, geometrical terms. If φ is ascalar field and a is a vector field then it may be shown that at any point P( ∮ 1∇φ = limV →0 V S( ∮ 1∇ · a = limV →0 V∇×a = limV →0( 1V)φdS)a · dSS∮ )dS × aS(11.14)(11.15)(11.16)where V is a small volume enclosing P and S is its bounding surface. Indeed,we may consider these equations as the (geometrical) definitions of grad, div andcurl. An alternative, but equivalent, geometrical definition of ∇×a at a point P ,which is often easier to use than (11.16), is given by( ∮ )1(∇×a) · ˆn = lim a · dr , (11.17)A→0 A Cwhere C is a plane contour of area A enclosing the point P and ˆn is the unitnormal to the enclosed planar area.It may be shown, in any coordinate system, that all the above equations areconsistent with our definitions in the previous chapter, although the difficulty ofproof depends on the chosen coordinate system. The most general coordinatesystem encountered in that chapter was one with orthogonal curvilinear coordinatesu 1 ,u 2 ,u 3 , of which Cartesians, cylindrical polars and spherical polars are allspecial cases. Although it may be shown that (11.14) leads to the usual expressionfor grad in curvilinear coordinates, the proof requires complicated manipulationsof the derivatives of the basis vectors with respect to the coordinates and is notpresented here. In Cartesian coordinates, however, the proof is quite simple.◮Show that the geometrical definition of grad leads to the usual expression for ∇φ inCartesian coordinates.Consider the surface S of a small rectangular volume element ∆V =∆x ∆y ∆z that has itsfaces parallel to the x, y, andz coordinate surfaces; the point P (see above) is at one corner.We must calculate the surface integral (11.14) over each of its six faces. Remembering thatthe normal to the surface points outwards from the volume on each face, the two faceswith x = constant have areas ∆S = −i ∆y ∆z and ∆S = i ∆y ∆z respectively. Furthermore,over each small surface element, we may take φ to be constant, so that the net contribution398

11.7 INTEGRAL FORMS FOR grad, div AND curlto the surface integral from these two faces is then([(φ +∆φ) − φ]∆y ∆z i = φ + ∂φ )∂x ∆x − φ ∆y ∆z i= ∂φ ∆x ∆y ∆z i.∂xThe surface integral over the pairs of faces with y =constantandz = constant respectivelymay be found in a similar way, and we obtain∮ ( ∂φφdS =S ∂x i + ∂φ∂y j + ∂φ )∂z k ∆x ∆y ∆z.Therefore ∇φ at the point P is given by[ (1 ∂φ∇φ = lim∆x,∆y,∆z→0 ∆x ∆y ∆z ∂x i + ∂φ∂y j + ∂φ ) ]∂z k ∆x ∆y ∆z= ∂φ∂x i + ∂φ∂y j + ∂φ∂z k. ◭We now turn to (11.15) and (11.17). These geometrical definitions may beshown straightforwardly to lead to the usual expressions for div and curl inorthogonal curvilinear coordinates.◮By considering the infinitesimal volume element dV = h 1 h 2 h 3 ∆u 1 ∆u 2 ∆u 3 shown in figure11.10, show that (11.15) leads to the usual expression for ∇·a in orthogonal curvilinearcoordinates.Let us write the vector field in terms of its components with respect to the basis vectorsof the curvilinear coordinate system as a = a 1 ê 1 + a 2 ê 2 + a 3 ê 3 . We consider first thecontribution to the RHS of (11.15) from the two faces with u 1 = constant, i.e. PQRSand the face opposite it (see figure 11.10). Now, the volume element is formed from theorthogonal vectors h 1 ∆u 1 ê 1 , h 2 ∆u 2 ê 2 and h 3 ∆u 3 ê 3 at the point P andsoforPQRS wehave∆S = h 2 h 3 ∆u 2 ∆u 3 ê 3 × ê 2 = −h 2 h 3 ∆u 2 ∆u 3 ê 1 .Reasoning along the same lines as in the previous example, we conclude that the contributionto the surface integral of a · dS over PQRS and its opposite face taken together isgiven by∂(a · ∆S)∆u 1 =∂ (a 1 h 2 h 3 )∆u 1 ∆u 2 ∆u 3 .∂u 1 ∂u 1The surface integrals over the pairs of faces with u 2 = constant and u 3 = constantrespectively may be found in a similar way, and we obtain∮ [ ∂a · dS = (a 1 h 2 h 3 )+∂ (a 2 h 3 h 1 )+∂u 1 ∂u 2STherefore ∇ · a at the point P is given by[1∇ · a = lim∆u 1 ,∆u 2 ,∆u 3 →0 h 1 h 2 h 3 ∆u 1 ∆u 2 ∆u 3∮S= 1 [ ∂(a 1 h 2 h 3 )+h 1 h 2 h 3 ∂u 1∂∂u 3(a 3 h 1 h 2 )∂∂u 2(a 2 h 3 h 1 )+]a · dS]∆u 1 ∆u 2 ∆u 3 .∂ ](a 3 h 1 h 2 ) . ◭∂u 3399

LINE, SURFACE AND VOLUME INTEGRALSzh 1 ∆u 1 ê 1RS TQh 2 ∆u 2 ê 2P h 3 ∆u 3 ê 3yxFigure 11.10 A general volume ∆V in orthogonal curvilinear coordinatesu 1 ,u 2 ,u 3 . PT gives the vector h 1 ∆u 1 ê 1 , PS gives h 2 ∆u 2 ê 2 and PQ givesh 3 ∆u 3 ê 3 .◮By considering the infinitesimal planar surface element PQRS in figure 11.10, show that(11.17) leads to the usual expression for ∇×a in orthogonal curvilinear coordinates.The planar surface PQRS is defined by the orthogonal vectors h 2 ∆u 2 ê 2 and h 3 ∆u 3 ê 3at the point P . If we traverse the loop in the direction PSRQ then, by the right-handconvention, the unit normal to the plane is ê 1 .Writinga = a 1 ê 1 + a 2 ê 2 + a 3 ê 3 , the lineintegral around the loop in this direction is given by∮PSRQa · dr = a 2 h 2 ∆u 2 +[− a 2 h 2 +[ ∂= (a 3 h 3 ) −∂u 2[a 3 h 3 +∂ ](a 3 h 3 )∆u 2 ∆u 3∂u 2∂ ](a 2 h 2 )∆u 3 ∆u 2 − a 3 h 3 ∆u 3∂u 3]∂∂u 3(a 2 h 2 )∆u 2 ∆u 3 .Therefore from (11.17) the component of ∇×a in the direction ê 1 at P is given by[]1(∇×a) 1 = lima · dr∆u 2 ,∆u 3 →0 h 2 h 3 ∆u 2 ∆u 3∮PSRQ= 1 [ ∂(h 3 a 3 ) −∂ ](h 2 a 2 ) .h 2 h 3 ∂u 2 ∂u 3The other two components are found by cyclically permuting the subscripts 1, 2, 3. ◭Finally, we note that we can also write the ∇ 2 operator as a surface integral bysetting a = ∇φ in (11.15), to obtain( ∮ )1∇ 2 φ = ∇ · ∇φ = lim ∇φ · dS .V →0 V400S

11.8 DIVERGENCE THEOREM AND RELATED THEOREMS11.8 Divergence theorem and related theoremsThe divergence theorem relates the total flux of a vector field out of a closedsurface S to the integral of the divergence of the vector field over the enclosedvolume V ; it follows almost immediately from our geometrical definition ofdivergence (11.15).Imagine a volume V , in which a vector field a is continuous and differentiable,to be divided up into a large number of small volumes V i . Using (11.15), we havefor each small volume∮(∇ · a)V i ≈ a · dS,S iwhere S i is the surface of the small volume V i . Summing over i we find thatcontributions from surface elements interior to S cancel since each surface elementappears in two terms with opposite signs, the outward normals in the two termsbeing equal and opposite. Only contributions from surface elements that are alsoparts of S survive. If each V i is allowed to tend to zero then we obtain thedivergence theorem,∫∮∇ · a dV = a · dS. (11.18)VSWe note that the divergence theorem holds for both simply and multiply connectedsurfaces, provided that they are closed and enclose some non-zero volumeV . The divergence theorem may also be extended to tensor fields (see chapter 26).The theorem finds most use as a tool in formal manipulations, but sometimesit is of value in transforming surface integrals of the form ∫ Sa · dS into volumeintegrals or vice versa. For example, setting a = r we immediately obtain∫∫∮∇ · r dV = 3 dV =3V = r · dS,VVwhich gives the expression for the volume of a region found in subsection 11.6.1.The use of the divergence theorem is further illustrated in the following example.◮Evaluate the surface integral I = ∫ S a · dS, wherea =(y − x) i + x2 z j +(z + x 2 ) k and Sis the open surface of the hemisphere x 2 + y 2 + z 2 = a 2 , z ≥ 0.We could evaluate this surface integral directly, but the algebra is somewhat lengthy. Wewill therefore evaluate it by use of the divergence theorem. Since the latter only holdsfor closed surfaces enclosing a non-zero volume V , let us first consider the closed surfaceS ′ = S + S 1 ,whereS 1 is the circular area in the xy-plane given by x 2 + y 2 ≤ a 2 , z =0;S ′then encloses a hemispherical volume V . By the divergence theorem we have∫∮ ∫ ∫∇ · a dV = a · dS = a · dS + a · dS.VS ′ SS 1Now ∇ · a = −1+0+1=0,sowecanwrite∫∫a · dS = − a · dS.SS 1401S

LINE, SURFACE AND VOLUME INTEGRALSyCRdxdrdyˆn dsxFigure 11.11 A closed curve C in the xy-plane bounding a region R. Vectorstangent and normal to the curve at a given point are also shown.The surface integral over S 1 is easily evaluated. Remembering that the normal to thesurface points outward from the volume, a surface element on S 1 is simply dS = −k dx dy.On S 1 we also have a =(y − x) i + x 2 k,sothat∫ ∫∫I = − a · dS = x 2 dx dy,S 1 Rwhere R is the circular region in the xy-plane given by x 2 + y 2 ≤ a 2 . Transforming to planepolar coordinates we have∫∫∫ 2π ∫ aI = ρ 2 cos 2 φ ρ dρ dφ = cos 2 φdφ ρ 3 dρ = πa4R ′ 00 4 . ◭It is also interesting to consider the two-dimensional version of the divergencetheorem. As an example, let us consider a two-dimensional planar region R inthe xy-plane bounded by some closed curve C (see figure 11.11). At any pointon the curve the vector dr = dx i + dy j is a tangent to the curve and the vectorˆn ds = dy i − dx j is a normal pointing out of the region R. If the vector field a iscontinuous and differentiable in R then the two-dimensional divergence theoremin Cartesian coordinates gives∫∫R( ∂ax∂x + ∂a y∂y)∮dx dy =∮a · ˆn ds = (a x dy − a y dx).CLetting P = −a y and Q = a x , we recover Green’s theorem in a plane, which wasdiscussed in section 11.3.11.8.1 Green’s theoremsConsider two scalar functions φ and ψ that are continuous and differentiable insome volume V bounded by a surface S. Applying the divergence theorem to the402

11.8 DIVERGENCE THEOREM AND RELATED THEOREMSvector field φ∇ψ we obtain∮∫φ∇ψ · dS =S∫=VV∇ · (φ∇ψ) dV[φ∇ 2 ψ +(∇φ) · (∇ψ) ] dV . (11.19)Reversing the roles of φ and ψ in (11.19) and subtracting the two equations gives∮∫(φ∇ψ − ψ∇φ) · dS = (φ∇ 2 ψ − ψ∇ 2 φ) dV . (11.20)SEquation (11.19) is usually known as Green’s first theorem and (11.20) as hissecond. Green’s second theorem is useful in the development of the Green’sfunctions used in the solution of partial differential equations (see chapter 21).V11.8.2 Other related integral theoremsThere exist two other integral theorems which are closely related to the divergencetheorem and which are of some use in physical applications. If φ is a scalar fieldand b is a vector field and both φ and b satisfy our usual differentiabilityconditions in some volume V bounded by a closed surface S then∫ ∮∇φdV = φdS, (11.21)VS∫∮∇×b dV = dS × b. (11.22)VS◮Use the divergence theorem to prove (11.21).In the divergence theorem (11.18) let a = φc, wherec is a constant vector. We then have∫∮∇ · (φc) dV = φc · dS.Expanding out the integrand on the LHS we have∇ · (φc) =φ∇ · c + c · ∇φ = c · ∇φ,since c is constant. Also, φc · dS = c · φdS, soweobtain∫∮c · (∇φ) dV = c · φdS.Since c is constant we may take it out of both integrals to give∫∮c · ∇φdV = c · φdS,and since c is arbitrary we obtain the stated result (11.21). ◭VVVEquation (11.22) may be proved in a similar way by letting a = b × c in thedivergence theorem, where c is again a constant vector.403SSS

LINE, SURFACE AND VOLUME INTEGRALS11.8.3 Physical applications of the divergence theoremThe divergence theorem is useful in deriving many of the most important partialdifferential equations in physics (see chapter 20). The basic idea is to use thedivergence theorem to convert an integral form, often derived from observation,into an equivalent differential form (used in theoretical statements).◮For a compressible fluid with time-varying position-dependent density ρ(r,t) and velocityfield v(r,t), in which fluid is neither being created nor destroyed, show that∂ρ+ ∇ · (ρv) =0.∂tFor an arbitrary volume V in the fluid, the conservation of mass tells us that the rate ofincrease or decrease of the mass M of fluid in the volume must equal the net rate at whichfluid is entering or leaving the volume, i.e.∮dM= − ρv · dS,dt Swhere S is the surface bounding V . But the mass of fluid in V is simply M = ∫ ρdV,soVwe have∫ ∮dρdV + ρv · dS =0.dt VSTaking the derivative inside the first integral on the RHS and using the divergence theoremto rewrite the second integral, we obtain∫∂ρV ∂t∫VdV + ∇ · (ρv) dV =∫V[ ∂ρ∂t + ∇ · (ρv) ]dV =0.Since the volume V is arbitrary, the integrand (which is assumed continuous) must beidentically zero, so we obtain∂ρ+ ∇ · (ρv) =0.∂tThis is known as the continuity equation. It can also be applied to other systems, forexample those in which ρ is the density of electric charge or the heat content, etc. For theflow of an incompressible fluid, ρ = constant and the continuity equation becomes simply∇ · v =0.◭In the previous example, we assumed that there were no sources or sinks inthe volume V , i.e. that there was no part of V in which fluid was being createdor destroyed. We now consider the case where a finite number of point sourcesand/or sinks are present in an incompressible fluid. Let us first consider thesimple case where a single source is located at the origin, out of which a quantityof fluid flows radially at a rate Q (m 3 s −1 ). The velocity field is given byv =Qr4πr 3 = Qˆr4πr 2 .Now, for a sphere S 1 of radius r centred on the source, the flux across S 1 is∮v · dS = |v|4πr 2 = Q.S 1404

11.8 DIVERGENCE THEOREM AND RELATED THEOREMSSince v has a singularity at the origin it is not differentiable there, i.e. ∇·v is notdefined there, but at all other points ∇ · v = 0, as required for an incompressiblefluid. Therefore, from the divergence theorem, for any closed surface S 2 that doesnot enclose the origin we have∮ ∫v · dS = ∇ · v dV =0.S 2 VThus we see that the surface integral ∮ Sv · dS has value Q or zero dependingon whether or not S encloses the source. In order that the divergence theorem isvalid for all surfaces S, irrespective of whether they enclose the source, we write∇ · v = Qδ(r),where δ(r) is the three-dimensional Dirac delta function. The properties of thisfunction are discussed fully in chapter 13, but for the moment we note that it isdefined in such a way that∫Vδ(r − a) =0 forr ≠ a,{f(a) if a lies in Vf(r)δ(r − a) dV =0 otherwisefor any well-behaved function f(r). Therefore, for any volume V containing thesource at the origin, we have∫∫∇ · v dV = Q δ(r) dV = Q,Vwhich is consistent with ∮ Sv · dS = Q for a closed surface enclosing the source.Hence, by introducing the Dirac delta function the divergence theorem can bemade valid even for non-differentiable point sources.The generalisation to several sources and sinks is straightforward. For example,if a source is located at r = a and a sink at r = b then the velocity field isv =and its divergence is given byV(r − a)Q (r − b)Q−4π|r − a|34π|r − b| 3∇ · v = Qδ(r − a) − Qδ(r − b).Therefore, the integral ∮ Sv · dS has the value Q if S encloses the source, −Q ifS encloses the sink and 0 if S encloses neither the source nor sink or enclosesthem both. This analysis also applies to other physical systems – for example, inelectrostatics we can regard the sources and sinks as positive and negative pointcharges respectively and replace v by the electric field E.405

LINE, SURFACE AND VOLUME INTEGRALS11.9 Stokes’ theorem and related theoremsStokes’ theorem is the ‘curl analogue’ of the divergence theorem and relates theintegral of the curl of a vector field over an open surface S to the line integral ofthe vector field around the perimeter C bounding the surface.Following the same lines as for the derivation of the divergence theorem, wecan divide the surface S into many small areas S i with boundaries C i and unitnormals ˆn i . Using (11.17), we have for each small area∮(∇×a) · ˆn i S i ≈ a · dr.C iSumming over i we find that on the RHS all parts of all interior boundariesthat are not part of C are included twice, being traversed in opposite directionson each occasion and thus contributing nothing. Only contributions from lineelements that are also parts of C survive. If each S i is allowed to tend to zerothen we obtain Stokes’ theorem,∫∮(∇×a) · dS = a · dr. (11.23)SCWe note that Stokes’ theorem holds for both simply and multiply connected opensurfaces, provided that they are two-sided. Stokes’ theorem may also be extendedto tensor fields (see chapter 26).Just as the divergence theorem (11.18) can be used to relate volume and surfaceintegrals for certain types of integrand, Stokes’ theorem can be used in evaluatingsurface integrals of the form ∮ S(∇×a) · dS as line integrals or vice versa.◮Given the vector field a = y i − x j + z k, verify Stokes’ theorem for the hemisphericalsurface x 2 + y 2 + z 2 = a 2 , z ≥ 0.Let us first evaluate the surface integral∫S(∇×a) · dSover the hemisphere. It is easily shown that ∇×a = −2 k, and the surface element isdS = a 2 sin θdθdφ ˆr in spherical polar coordinates. Therefore∫∫ 2π ∫ π/2(∇×a) · dS = dφ dθ ( −2a 2 sin θ ) ˆr · kS0∫ 2π= −2a 2 dφ∫ 2π= −2a 2 dφ000∫ π/20∫ π/20( z)sin θ dθasin θ cos θdθ= −2πa 2 .We now evaluate the line integral around the perimeter curve C of the surface, which406

11.9 STOKES’ THEOREM AND RELATED THEOREMSis the circle x 2 + y 2 = a 2 in the xy-plane. This is given by∮ ∮a · dr = (y i − x j + z k) · (dx i + dy j + dz k)CC∮= (ydx− xdy).CUsing plane polar coordinates, on C we have x = a cos φ, y = a sin φ so that dx =−a sin φdφ, dy = a cos φdφ, and the line integral becomes∮∫ 2π∫ 2π(ydx− xdy)=−a 2 (sin 2 φ +cos 2 φ) dφ = −a 2 dφ = −2πa 2 .C0Since the surface and line integrals have the same value, we have verified Stokes’ theoremin this case. ◭The two-dimensional version of Stokes’ theorem also yields Green’s theorem ina plane. Consider the region R in the xy-plane shown in figure 11.11, in which avector field a is defined. Since a = a x i+a y j, we have ∇×a =(∂a y /∂x−∂a x /∂y) k,and Stokes’ theorem becomes∫∫R( ∂ay∂x − ∂a x∂y)∮dx dy = (a x dx + a y dy).CLetting P = a x and Q = a y we recover Green’s theorem in a plane, (11.4).011.9.1 Related integral theoremsAs for the divergence theorem, there exist two other integral theorems that areclosely related to Stokes’ theorem. If φ is a scalar field and b is a vector field,and both φ and b satisfy our usual differentiability conditions on some two-sidedopen surface S bounded by a closed perimeter curve C, then∫∮dS ×∇φ = φdr, (11.24)SC∫∮(dS ×∇) × b = dr × b. (11.25)SC◮Use Stokes’ theorem to prove (11.24).In Stokes’ theorem, (11.23), let a = φc, wherec is a constant vector. We then have∫∮[∇×(φc)] · dS = φc · dr. (11.26)SExpanding out the integrand on the LHS we have∇×(φc) =∇φ × c + φ∇×c = ∇φ × c,since c is constant, and the scalar triple product on the LHS of (11.26) can therefore bewritten[∇×(φc)] · dS =(∇φ × c) · dS = c · (dS ×∇φ).407C

LINE, SURFACE AND VOLUME INTEGRALSSubstituting this into (11.26) and taking c out of both integrals because it is constant, wefind∫∮c · dS ×∇φ = c · φdr.SCSince c is an arbitrary constant vector we therefore obtain the stated result (11.24). ◭Equation (11.25) may be proved in a similar way, by letting a = b × c in Stokes’theorem, where c is again a constant vector. We also note that by setting b = rin (11.25) we find∫∮(dS ×∇) × r = dr × r.SCExpanding out the integrand on the LHS gives(dS ×∇) × r = dS − dS(∇ · r) =dS − 3 dS = −2 dS.Therefore, as we found in subsection 11.5.2, the vector area of an open surface Sis given by∫S = dS = 1 ∮r × dr.S 2 C11.9.2 Physical applications of Stokes’ theoremLike the divergence theorem, Stokes’ theorem is useful in converting integralequations into differential equations.◮From Ampère’s law, derive Maxwell’s equation in the case where the currents are steady,i.e. ∇×B − µ 0 J = 0.Ampère’s rule for a distributed current with current density J is∮∫B · dr = µ 0 J · dS,Cfor any circuit C bounding a surface S. Using Stokes’ theorem, the LHS can be transformedinto ∫ (∇×B) · dS; henceS∫S(∇×B − µ 0 J) · dS =0for any surface S. This can only be so if ∇×B − µ 0 J = 0, which is the required relation.Similarly, from Faraday’s law of electromagnetic induction we can derive Maxwell’sequation ∇×E = −∂B/∂t. ◭In subsection 11.8.3 we discussed the flow of an incompressible fluid in thepresence of several sources and sinks. Let us now consider vortex flow in anincompressible fluid with a velocity fieldSv = 1 ρêφ,in cylindrical polar coordinates ρ, φ, z. For this velocity field ∇×v equals zero408

11.10 EXERCISESeverywhere except on the axis ρ =0,wherev has a singularity. Therefore ∮ C v · drequals zero for any path C that does not enclose the vortex line on the axis and2π if C does enclose the axis. In order for Stokes’ theorem to be valid for allpaths C, we therefore set∇×v =2πδ(ρ),where δ(ρ) is the Dirac delta function, to be discussed in subsection 13.1.3. Now,since ∇×v = 0, except on the axis ρ = 0, there exists a scalar potential ψ suchthat v = ∇ψ. It may easily be shown that ψ = φ, the polar angle. Therefore, if Cdoes not enclose the axis then ∮ ∮v · dr = dφ =0,Cand if C does enclose the axis,∮v · dr =∆φ =2πn,Cwhere n is the number of times we traverse C. Thus φ is a multivalued potential.Similar analyses are valid for other physical systems – for example, in magnetostaticswe may replace the vortex lines by current-carrying wires and the velocityfield v by the magnetic field B.11.10 Exercises11.1 The vector field F is defined byF =2xzi +2yz 2 j +(x 2 +2y 2 z − 1)k.Calculate ∇×F and deduce that F can be written F = ∇φ. Determine the formof φ.11.2 The vector field Q is defined byQ = [ 3x 2 (y + z)+y 3 + z 3] i + [ 3y 2 (z + x)+z 3 + x 3] j + [ 3z 2 (x + y)+x 3 + y 3] k.Show that Q is a conservative field, construct its potential function and henceevaluate the integral J = ∫ Q · dr along any line connecting the point A at(1, −1, 1) to B at (2, 1, 2).11.3 F is a vector field xy 2 i+2j+xk, andL is a path parameterised by x = ct, y = c/t,z = d for the range 1 ≤ t ≤ 2. Evaluate (a) ∫ F dt, (b)∫ F dy and (c) ∫ F · dr.L L L11.4 By making an appropriate choice for the functions P (x, y)andQ(x, y) that appearin Green’s theorem in a plane, show that the integral of x − y over the upper halfof the unit circle centred on the origin has the value − 2 . Show the same result3by direct integration in Cartesian coordinates.11.5 Determine the point of intersection P , in the first quadrant, of the two ellipsesx 2a + y22 b =1 and x 22 b + y22 a =1. 2Taking b

LINE, SURFACE AND VOLUME INTEGRALSin section 11.3, evaluate the two remaining line integrals and hence find the totalarea common to the two ellipses.11.6 By using parameterisations of the form x = a cos n θ and y = a sin n θ for suitablevalues of n, find the area bounded by the curvesx 2/5 + y 2/5 = a 2/5 and x 2/3 + y 2/3 = a 2/3 .11.7 Evaluate the line integral∮[I = y(4x 2 + y 2 ) dx + x(2x 2 +3y 2 ) dy ]Caround the ellipse x 2 /a 2 + y 2 /b 2 =1.11.8 Criticise the following ‘proof’ that π =0.(a) Apply Green’s theorem in a plane to the functions P (x, y) =tan −1 (y/x) andQ(x, y) =tan −1 (x/y), taking the region R to be the unit circle centred on theorigin.(b) The RHS of the equality so produced is∫ ∫y − xdx dy,R x 2 + y2 which, either from symmetry considerations or by changing to plane polarcoordinates, can be shown to have zero value.(c) In the LHS of the equality, set x =cosθ and y =sinθ, yielding P (θ) =θand Q(θ) =π/2 − θ. The line integral becomes∫ 2π0[( π)]2 − θ cos θ − θ sin θ dθ,which has the value 2π.(d) Thus 2π = 0 and the stated result follows.11.9 A single-turn coil C of arbitrary shape is placed in a magnetic field B and carriesa current I. Show that the couple acting upon the coil can be written as∫∫M = I (B · r) dr − I B(r · dr).CFor a planar rectangular coil of sides 2a and 2b placed with its plane verticaland at an angle φ to a uniform horizontal field B, show that M is, as expected,4abBI cos φ k.11.10 Find the vector area S of the part of the curved surface of the hyperboloid ofrevolutionx 2a − y2 + z 2=12 b 2that lies in the region z ≥ 0anda ≤ x ≤ λa.11.11 An axially symmetric solid body with its axis AB vertical is immersed in anincompressible fluid of density ρ 0 . Use the following method to show that,whatever the shape of the body, for ρ = ρ(z) in cylindrical polars the Archimedeanupthrust is, as expected, ρ 0 gV, whereV is the volume of the body.Express the vertical component of the resultant force on the body, − ∫ pdS,where p is the pressure, in terms of an integral; note that p = −ρ 0 gz and that foran annular surface element of width dl, n · n z dl = −dρ. Integrate by parts anduse the fact that ρ(z A )=ρ(z B )=0.410C

11.10 EXERCISES11.12 Show that the expression below is equal to the solid angle subtended by arectangular aperture, of sides 2a and 2b, at a point on the normal through itscentre, and at a distance c from the aperture:∫ bacΩ=4dy.0 (y 2 + c 2 )(y 2 + c 2 + a 2 )1/2By setting y =(a 2 + c 2 ) 1/2 tan φ, change this integral into the form∫ φ14ac cos φ0 c 2 + a 2 sin 2 φ dφ,where tan φ 1 = b/(a 2 + c 2 ) 1/2 , and hence show that[]Ω=4tan −1 ab.c(a 2 + b 2 + c 2 ) 1/211.13 A vector field a is given by −zxr −3 i−zyr −3 j+(x 2 +y 2 )r −3 k,wherer 2 = x 2 +y 2 +z 2 .Establish that the field is conservative (a) by showing that ∇×a = 0, and(b)byconstructing its potential function φ.11.14 A vector field a is given by (z 2 +2xy) i +(x 2 +2yz) j +(y 2 +2zx) k. Show thata is conservative and that the line integral ∫ a · dr along any line joining (1, 1, 1)and (1, 2, 2) has the value 11.11.15 A force F(r) acts on a particle at r. In which of the following cases can F berepresented in terms of a potential? Where it can, find the potential.[] )2(x − y)(a) F = F 0 i − j − r exp(− r2;a 2a 2(b) F = F []0zk + (x2 + y 2 − a 2 )r expaa 2(c) F = F 0[k +]a(r × k).r 2(− r2a 2 );11.16 One of Maxwell’s electromagnetic equations states that all magnetic fields Bare solenoidal (i.e. ∇ · B = 0). Determine whether each of the following vectorscould represent a real magnetic field; where it could, try to find a suitable vectorpotential A, i.e. such that B = ∇×A. (Hint: seek a vector potential that is parallelto ∇×B.):B 0 b(a)r [(x − y)z i +(x − y)z j 3 +(x2 − y 2 ) k] in Cartesians with r 2 = x 2 +y 2 +z 2 ;B 0 b 3(b) [cos θ cos φ êr 3r − sin θ cos φ ê θ +sin2θ sin φ ê φ ] in spherical polars;[](c) B 0 b 2 zρ1(b 2 + z 2 ) 2 êρ +b 2 + z 2 êz in cylindrical polars.11.17 The vector field f has components yi−xj+k and γ is a curve given parametricallybyr =(a − c + c cos θ)i +(b + c sin θ)j + c 2 θk, 0 ≤ θ ≤ 2π.Describe the shape of the path γ and show that the line integral ∫ f · dr vanishes.γDoes this result imply that f is a conservative field?11.18 A vector field a = f(r)r is spherically symmetric and everywhere directed awayfrom the origin. Show that a is irrotational, but that it is also solenoidal only iff(r) isoftheformAr −3 .411

LINE, SURFACE AND VOLUME INTEGRALS11.19 Evaluate the surface integral ∫ r·dS, wherer is the position vector, over that partof the surface z = a 2 − x 2 − y 2 for which z ≥ 0, by each of the following methods.(a) Parameterise the surface as x = a sin θ cos φ, y = a sin θ sin φ, z = a 2 cos 2 θ,and show thatr · dS = a 4 (2 sin 3 θ cos θ +cos 3 θ sin θ) dθ dφ.(b) Apply the divergence theorem to the volume bounded by the surface andthe plane z =0.11.20 Obtain an expression for the value φ P at a point P of a scalar function φ thatsatisfies ∇ 2 φ = 0, in terms of its value and normal derivative on a surface S thatencloses it, by proceeding as follows.(a) In Green’s second theorem, take ψ at any particular point Q as 1/r, whereris the distance of Q from P . Show that ∇ 2 ψ = 0, except at r =0.(b) Apply the result to the doubly connected region bounded by S and a smallsphere Σ of radius δ centred on P.(c) Apply the divergence theorem to show that the surface integral over Σinvolving 1/δ vanishes, and prove that the term involving 1/δ 2 has the value4πφ P .(d) Conclude thatφ P = − 1 ∫φ ∂ ( ) 1dS + 1 ∫4π S ∂n r 4π S1 ∂φr ∂n dS.This important result shows that the value at a point P of a function φthat satisfies ∇ 2 φ = 0 everywhere within a closed surface S that encloses Pmay be expressed entirely in terms of its value and normal derivative on S.This matter is taken up more generally in connection with Green’s functionsin chapter 21 and in connection with functions of a complex variable insection 24.10.11.21 Use result (11.21), together with an appropriately chosen scalar function φ, toprove that the position vector ¯r of the centre of mass of an arbitrarily shapedbody of volume V and uniform density can be written¯r = 1 ∮12Vr2 dS.S11.22 A rigid body of volume V and surface S rotates with angular velocity ω. Show thatω = − 1 ∮u × dS,2V Swhere u(x) is the velocity of the point x on the surface S.11.23 Demonstrate the validity of the divergence theorem:(a) by calculating the flux of the vectorαrF =(r 2 + a 2 ) 3/2through the spherical surface |r| = √ 3a;(b) by showing that3αa 2∇ · F =(r 2 + a 2 ) 5/2and evaluating the volume integral of ∇ · F over the interior of the sphere|r| = √ 3a. The substitution r = a tan θ will prove useful in carrying out theintegration.412

11.10 EXERCISES11.24 Prove equation (11.22) and, by taking b = zx 2 i + zy 2 j +(x 2 − y 2 )k, show that thetwo integrals∫∫I = x 2 dV and J = cos 2 θ sin 3 θ cos 2 φdθdφ,both taken over the unit sphere, must have the same value. Evaluate both directlyto show that the common value is 4π/15.11.25 In a uniform conducting medium with unit relative permittivity, charge density ρ,current density J, electric field E and magnetic field B, Maxwell’s electromagneticequations take the form (with µ 0 ɛ 0 = c −2 )(i) ∇ · B = 0, (ii) ∇ · E = ρ/ɛ 0 ,(iii) ∇×E + Ḃ = 0, (iv) ∇×B − (Ė/c2 )=µ 0 J.The density of stored energy in the medium is given by 1 (ɛ 2 0E 2 + µ −10 B2 ). Showthat the rate of change of the total stored energy in a volume V is equal to∫− J · E dV − 1 ∮(E × B) · dS,µ 0Vwhere S is the surface bounding V .[ The first integral gives the ohmic heating loss, whilst the second gives theelectromagnetic energy flux out of the bounding surface. The vector µ −10 (E × B)is known as the Poynting vector. ]11.26 A vector field F is defined in cylindrical polar coordinates ρ, θ, z by( )x cos λz y cos λzF = F 0 i + j +(sinλz)k ≡ F 0ρa aa (cos λz)e ρ + F 0 (sin λz)k,where i, j and k are the unit vectors along the Cartesian axes and e ρ is the unitvector (x/ρ)i +(y/ρ)j.(a) Calculate, as a surface integral, the flux of F through the closed surfacebounded by the cylinders ρ = a and ρ =2a and the planes z = ±aπ/2.(b) Evaluate the same integral using the divergence theorem.11.27 The vector field F is given byF =(3x 2 yz + y 3 z + xe −x )i +(3xy 2 z + x 3 z + ye x )j +(x 3 y + y 3 x + xy 2 z 2 )k.Calculate (a) directly, and (b) by using Stokes’ theorem the value of the lineintegral ∫ F · dr, whereL is the (three-dimensional) closed contour OABCDEOLdefined by the successive vertices (0, 0, 0), (1, 0, 0), (1, 0, 1), (1, 1, 1), (1, 1, 0), (0, 1, 0),(0, 0, 0).11.28 A vector force field F is defined in Cartesian coordinates by[( y3F = F 03a + y ) ( xy23 a exy/a2 +1 i + + x + y )e xy/a2 j + z ]a 3 aa exy/a2 k .Use Stokes’ theorem to calculate∮F · dr,Lwhere L is the perimeter of the rectangle ABCD given by A =(0, 1, 0), B =(1, 1, 0),C =(1, 3, 0) and D =(0, 3, 0).S413

LINE, SURFACE AND VOLUME INTEGRALS11.11 Hints and answers11.1 Show that ∇×F = 0. The potential φ F (r) =x 2 z + y 2 z 2 − z.11.3 (a) c 3 ln 2 i +2j +(3c/2)k; (b)(−3c 4 /8)i − c j − (c 2 ln 2)k; (c)c 4 ln 2 − c.11.5 For P , x = y = ab/(a 2 + b 2 ) 1/2 . The relevant limits are 0 ≤ θ 1 ≤ tan −1 (b/a) andtan −1 (a/b) ≤ θ 2 ≤ π/2. The total common area is 4ab tan −1 (b/a).11.7 Show that, in the notation of section 11.3, ∂Q/∂x − ∂P/∂y =2x 2 ; I = πa 3 b/2.11.9 M = I ∫ r × (dr × B). Show that the horizontal sides in the first term and theCwhole of the second term contribute nothing to the couple.11.11 Note that, if ˆn is the outward normal to the surface, ˆn z · ˆn dl is equal to −dρ.11.13 (b) φ = c + z/r.11.15 (a) Yes, F 0 (x − y)exp(−r 2 /a 2 ); (b) yes, −F 0 [(x 2 + y 2 )/(2a)] exp(−r 2 /a 2 );(c) no, ∇×F ≠ 0.11.17 A spiral of radius c with its axis parallel to the z-direction and passing through(a, b). The pitch of the spiral is 2πc 2 . No, because (i) γ is not a closed loop and(ii) the line integral must be zero for every closed loop, not just for a particularone. In fact ∇×f = −2k ≠ 0 shows that f is not conservative.11.19 (a) dS =(2a 3 cos θ sin 2 θ cos φ i +2a 3 cos θ sin 2 θ sin φ j + a 2 cos θ sin θ k) dθ dφ.(b) ∇ · r = 3; over the plane z =0,r · dS =0.The necessarily common value is 3πa 4 /2.11.21 Write r as ∇( 1 2 r2 ).11.23 The answer is 3 √ 3πα/2 in each case.11.25 Identify the expression for ∇ · (E × B) and use the divergence theorem.11.27 (a) The successive contributions to the integral are:1 − 2e −1 , 0, 2+ 1 2 e, − 7 3 , −1+2e−1 , − 1 2 .(b) ∇×F =2xyz 2 i − y 2 z 2 j + ye x k. Show that the contour is equivalent to thesum of two plane square contours in the planes z =0andx = 1, the latter beingtraversed in the negative sense. Integral = 1 (3e − 5).6414

12Fourier seriesWe have already discussed, in chapter 4, how complicated functions may beexpressed as power series. However, this is not the only way in which a functionmay be represented as a series, and the subject of this chapter is the expressionof functions as a sum of sine and cosine terms. Such a representation is called aFourier series. Unlike Taylor series, a Fourier series can describe functions that arenot everywhere continuous and/or differentiable. There are also other advantagesin using trigonometric terms. They are easy to differentiate and integrate, theirmoduli are easily taken and each term contains only one characteristic frequency.This last point is important because, as we shall see later, Fourier series are oftenused to represent the response of a system to a periodic input, and this responseoften depends directly on the frequency content of the input. Fourier series areused in a wide variety of such physical situations, including the vibrations of afinite string, the scattering of light by a diffraction grating and the transmissionof an input signal by an electronic circuit.12.1 The Dirichlet conditionsWe have already mentioned that Fourier series may be used to represent somefunctions for which a Taylor series expansion is not possible. The particularconditions that a function f(x) must fulfil in order that it may be expanded as aFourier series are known as the Dirichlet conditions, and may be summarised bythe following four points:(i) the function must be periodic;(ii) it must be single-valued and continuous, except possibly at a finite numberof finite discontinuities;(iii) it must have only a finite number of maxima and minima within oneperiod;(iv) the integral over one period of |f(x)| must converge.415

FOURIER SERIESf(x)xLLFigure 12.1 An example of a function that may be represented as a Fourierseries without modification.If the above conditions are satisfied then the Fourier series converges to f(x)at all points where f(x) is continuous. The convergence of the Fourier seriesat points of discontinuity is discussed in section 12.4. The last three Dirichletconditions are almost always met in real applications, but not all functions areperiodic and hence do not fulfil the first condition. It may be possible, however,to represent a non-periodic function as a Fourier series by manipulation of thefunction into a periodic form. This is discussed in section 12.5. An example ofa function that may, without modification, be represented as a Fourier series isshown in figure 12.1.We have stated without proof that any function that satisfies the Dirichletconditions may be represented as a Fourier series. Let us now show why this isa plausible statement. We require that any reasonable function (one that satisfiesthe Dirichlet conditions) can be expressed as a linear sum of sine and cosineterms. We first note that we cannot use just a sum of sine terms since sine, beingan odd function (i.e. a function for which f(−x) =−f(x)), cannot represent evenfunctions (i.e. functions for which f(−x) =f(x)). This is obvious when we tryto express a function f(x) that takes a non-zero value at x = 0. Clearly, sincesin nx = 0 for all values of n, we cannot represent f(x) atx = 0 by a sine series.Similarly odd functions cannot be represented by a cosine series since cosine isan even function. Nevertheless, it is possible to represent all odd functions by asine series and all even functions by a cosine series. Now, since all functions maybewrittenasthesumofanoddandanevenpart,f(x) = 1 2 [f(x)+f(−x)] + 1 2[f(x) − f(−x)]= f even (x)+f odd (x),416

12.2 THE FOURIER COEFFICIENTSwe can write any function as the sum of a sine series and a cosine series.All the terms of a Fourier series are mutually orthogonal, i.e. the integrals, overone period, of the product of any two terms have the following properties:∫ x0+L( ) ( )2πrx 2πpxsin cos dx = 0 for all r and p, (12.1)L Lx 0∫ x0+Lx 0∫ x0+Lx 0( 2πrxcosL( 2πrxsinL)cos)sin⎧( ) 2πpx⎨L for r = p =0,1dx =L ⎩2L for r = p>0,0 for r ≠ p,⎧( ) 2πpx⎨0 for r = p =0,1dx =L ⎩2L for r = p>0,0 for r ≠ p,(12.2)(12.3)where r and p are integers greater than or equal to zero; these formulae are easilyderived. A full discussion of why it is possible to expand a function as a sum ofmutually orthogonal functions is given in chapter 17.The Fourier series expansion of the function f(x) is conventionally writtenf(x) = a ∞02 + ∑[a r cosr=1( 2πrxL)+ b r sin( 2πrxL)], (12.4)where a 0 ,a r ,b r are constants called the Fourier coefficients. These coefficients areanalogous to those in a power series expansion and the determination of theirnumerical values is the essential step in writing a function as a Fourier series.This chapter continues with a discussion of how to find the Fourier coefficientsfor particular functions. We then discuss simplifications to the general Fourierseries that may save considerable effort in calculations. This is followed by thealternative representation of a function as a complex Fourier series, and weconclude with a discussion of Parseval’s theorem.12.2 The Fourier coefficientsWe have indicated that a series that satisfies the Dirichlet conditions may bewritten in the form (12.4). We now consider how to find the Fourier coefficientsfor any particular function. For a periodic function f(x) ofperiodL we will findthat the Fourier coefficients are given bya r = 2 ∫ x0+L( ) 2πrxf(x)cos dx, (12.5)L x 0Lb r = 2 ∫ x0+L( ) 2πrxf(x)sin dx, (12.6)L x 0Lwhere x 0 is arbitrary but is often taken as 0 or −L/2. The apparently arbitraryfactor 1 2which appears in the a 0 term in (12.4) is included so that (12.5) may417

FOURIER SERIESapply for r = 0 as well as r>0. The relations (12.5) and (12.6) may be derivedas follows.Suppose the Fourier series expansion of f(x) can be written as in (12.4),f(x) = a ∞ 02 + ∑[a r cosr=1( 2πrxL)+ b r sin( 2πrxL)].Then, multiplying by cos(2πpx/L), integrating over one full period in x andchanging the order of the summation and integration, we get∫ x0+Lx 0( ) 2πpxf(x)cosLdx = a 02∫ x0+L( ) 2πpxcos dxx 0L∞∑∫ x0+L( 2πrx+ a r cosr=1x 0L∞∑∫ x0+L( 2πrx+ b r sinLr=1x 0)cos)cos( ) 2πpxdxL( ) 2πpxdx.L(12.7)We can now find the Fourier coefficients by considering (12.7) as p takes differentvalues. Using the orthogonality conditions (12.1)–(12.3) of the previous section,we find that when p = 0 (12.7) becomes∫ x0+Lf(x)dx = a 0x 02 L.When p ≠ 0 the only non-vanishing term on the RHS of (12.7) occurs whenr = p, andso∫ x0+L( ) 2πrxf(x)cos dx = a rL 2 L.x 0The other Fourier coefficients b r may be found by repeating the above processbut multiplying by sin(2πpx/L) instead of cos(2πpx/L) (see exercise 12.2).◮Express the square-wave function illustrated in figure 12.2 as a Fourier series.Physically this might represent the input to an electrical circuit that switches between ahigh and a low state with time period T . The square wave may be represented by{−1 for − 1f(t) =T ≤ t

12.3 SYMMETRY CONSIDERATIONSf(t)1− T 20 T2t−1Figure 12.2A square-wave function.following section). To evaluate the coefficients in the sine series we use (12.6). Henceb r = 2 ∫ T/2( ) 2πrtf(t)sin dtT −T/2 T= 4 ∫ T/2( ) 2πrtsin dtT 0 T= 2 πr [1 − (−1)r ] .Thus the sine coefficients are zero if r is even and equal to 4/(πr) ifr is odd. Hence theFourier series for the square-wave function may be written as()f(t) = 4 sin 3ωt sin 5ωtsin ωt + + + ···π3 5, (12.8)where ω =2π/T is called the angular frequency. ◭12.3 Symmetry considerationsThe example in the previous section employed the useful property that since thefunction to be represented was odd, all the cosine terms of the Fourier series wereabsent. It is often the case that the function we wish to express as a Fourier serieshas a particular symmetry, which we can exploit to reduce the calculational labourof evaluating Fourier coefficients. Functions that are symmetric or antisymmetricabout the origin (i.e. even and odd functions respectively) admit particularlyuseful simplifications. Functions that are odd in x have no cosine terms (seesection 12.1) and all the a-coefficients are equal to zero. Similarly, functions thatare even in x have no sine terms and all the b-coefficients are zero. Since theFourier series of odd or even functions contain only half the coefficients requiredfor a general periodic function, there is a considerable reduction in the algebraneeded to find a Fourier series.The consequences of symmetry or antisymmetry of the function about thequarter period (i.e. about L/4) are a little less obvious. Furthermore, the results419

FOURIER SERIESare not used as often as those above and the remainder of this section can beomitted on a first reading without loss of continuity. The following argumentgives the required results.Suppose that f(x) has even or odd symmetry about L/4, i.e. f(L/4 − x) =±f(x − L/4). For convenience, we make the substitution s = x − L/4 and hencef(−s) =±f(s). We can now see thatb r = 2 ∫ x0+L( 2πrsf(s)sinL x 0L+ πr )ds,2where the limits of integration have been left unaltered since f is, of course,periodic in s as well as in x. If we use the expansion( 2πrssinL+ πr ) ( ) 2πrs( πr) ( 2πrs=sin cos +cos2L 2L)sin( πr),2we can immediately see that the trigonometric part of the integrand is an oddfunction of s if r is even and an even function of s if r is odd. Hence if f(s) iseven and r is even then the integral is zero, and if f(s) is odd and r is odd thenthe integral is zero. Similar results can be derived for the Fourier a-coefficientsand we conclude that(i) if f(x) is even about L/4 thena 2r+1 = 0 and b 2r =0,(ii) if f(x) is odd about L/4 thena 2r = 0 and b 2r+1 =0.All the above results follow automatically when the Fourier coefficients areevaluated in any particular case, but prior knowledge of them will often enablesome coefficients to be set equal to zero on inspection and so substantially reducethe computational labour. As an example, the square-wave function shown infigure 12.2 is (i) an odd function of t, sothatalla r = 0, and (ii) even about thepoint t = T/4, so that b 2r = 0. Thus we can say immediately that only sine termsof odd harmonics will be present and therefore will need to be calculated; this isconfirmed in the expansion (12.8).12.4 Discontinuous functionsThe Fourier series expansion usually works well for functions that are discontinuousin the required range. However, the series itself does not produce adiscontinuous function and we state without proof that the value of the expandedf(x) at a discontinuity will be half-way between the upper and lowervalues. Expressing this more mathematically, at a point of finite discontinuity, x d ,the Fourier series converges to12 lim[f(xd + ɛ)+f(x d − ɛ)].ɛ→0At a discontinuity, the Fourier series representation of the function will overshootits value. Although as more terms are included the overshoot moves in position420

12.4 DISCONTINUOUS FUNCTIONS(a)1(b)1− T 2T2− T 2T2−1−1(c)1(d)1δ− T 2T2− T 2T2−1−1Figure 12.3 The convergence of a Fourier series expansion of a square-wavefunction, including (a) one term, (b) two terms, (c) three terms and (d) 20terms. The overshoot δ is shown in (d).arbitrarily close to the discontinuity, it never disappears even in the limit of aninfinite number of terms. This behaviour is known as Gibbs’ phenomenon. Afulldiscussion is not pursued here but suffice it to say that the size of the overshootis proportional to the magnitude of the discontinuity.◮Find the value to which the Fourier series of the square-wave function discussed in section12.2 converges at t =0.It can be seen that the function is discontinuous at t = 0 and, by the above rule, we expectthe series to converge to a value half-way between the upper and lower values, in otherwords to converge to zero in this case. Considering the Fourier series of this function,(12.8), we see that all the terms are zero and hence the Fourier series converges to zero asexpected. The Gibbs phenomenon for the square-wave function is shown in figure 12.3. ◭421

FOURIER SERIES(a)0L(b)0L2L(c)0L2L(d)0L2LFigure 12.4Possible periodic extensions of a function.12.5 Non-periodic functionsWe have already mentioned that a Fourier representation may sometimes be usedfor non-periodic functions. If we wish to find the Fourier series of a non-periodicfunction only within a fixed range then we may continue the function outside therange so as to make it periodic. The Fourier series of this periodic function wouldthen correctly represent the non-periodic function in the desired range. Since weare often at liberty to extend the function in a number of ways, we can sometimesmake it odd or even and so reduce the calculation required. Figure 12.4(b) showsthe simplest extension to the function shown in figure 12.4(a). However, thisextension has no particular symmetry. Figures 12.4(c), (d) show extensions as oddand even functions respectively with the benefit that only sine or cosine termsappear in the resulting Fourier series. We note that these last two extensions givea function of period 2L.In view of the result of section 12.4, it must be added that the continuationmust not be discontinuous at the end-points of the interval of interest; if it isthe series will not converge to the required value there. This requirement thatthe series converges appropriately may reduce the choice of continuations. Thisis discussed further at the end of the following example.◮Find the Fourier series of f(x) =x 2 for 0

12.5 NON-PERIODIC FUNCTIONSf(x) =x 2−2 0 2xLFigure 12.5f(x) =x 2 ,0

FOURIER SERIESconverge to the correct values of f(x) =±4 atx = ±2; it converges, instead, tozero, the average of the values at the two ends of the range.12.6 Integration and differentiationIt is sometimes possible to find the Fourier series of a function by integration ordifferentiation of another Fourier series. If the Fourier series of f(x) isintegratedterm by term then the resulting Fourier series converges to the integral of f(x).Clearly, when integrating in such a way there is a constant of integration that mustbe found. If f(x) is a continuous function of x for all x and f(x) is also periodicthen the Fourier series that results from differentiating term by term converges tof ′ (x), provided that f ′ (x) itself satisfies the Dirichlet conditions. These propertiesof Fourier series may be useful in calculating complicated Fourier series, sincesimple Fourier series may easily be evaluated (or found from standard tables)and often the more complicated series can then be built up by integration and/ordifferentiation.◮Find the Fourier series of f(x) =x 3 for 0

12.7 COMPLEX FOURIER SERIESwhere the Fourier coefficients are given byc r = 1 ∫ x0+L(f(x)exp − 2πirx )dx. (12.10)LLx 0This relation can be derived, in a similar manner to that of section 12.2, by multiplying(12.9) by exp(−2πipx/L) before integrating and using the orthogonalityrelation∫ x0+L(exp − 2πipx ) ( ) {2πirx L for r = p,exp dx =LL0 for r ≠ p.x 0The complex Fourier coefficients in (12.9) have the following relations to the realFourier coefficients:c r = 1 2 (a r − ib r ),c −r = 1 2 (a r + ib r ).(12.11)Note that if f(x) isrealthenc −r = c ∗ r , where the asterisk represents complexconjugation.◮Find a complex Fourier series for f(x) =x in the range −2

FOURIER SERIES12.8 Parseval’s theoremParseval’s theorem gives a useful way of relating the Fourier coefficients to thefunction that they describe. Essentially a conservation law, it states that∫1 x0+L∞∑|f(x)| 2 dx = |c r | 2L x 0r=−∞= ( 12 a ) 20 +12∞∑(a 2 r + b 2 r ). (12.13)In a more memorable form, this says that the sum of the moduli squared ofthe complex Fourier coefficients is equal to the average value of |f(x)| 2 over oneperiod. Parseval’s theorem can be proved straightforwardly by writing f(x) asa Fourier series and evaluating the required integral, but the algebra is messy.Therefore, we shall use an alternative method, for which the algebra is simpleand which in fact leads to a more general form of the theorem.Let us consider two functions f(x) and g(x), which are (or can be made)periodic with period L and which have Fourier series (expressed in complexform)∞∑( ) 2πirxf(x) = c r exp ,Lr=−∞∞∑( ) 2πirxg(x) = γ r exp ,Lr=−∞where c r and γ r are the complex Fourier coefficients of f(x) andg(x) respectively.Thus∞∑( ) 2πirxf(x)g ∗ (x) = c r g ∗ (x)exp .Lr=−∞Integrating this equation with respect to x over the interval (x 0 ,x 0 + L) anddividing by L, we find∫1 x0+L∞∑∫f(x)g ∗ 1 x0+L( ) 2πirx(x) dx = c r g ∗ (x)exp dxL x 0Lr=−∞ x 0L∞∑[ ∫ 1 x0+L( ) ] ∗−2πirx= c r g(x)expdxLr=−∞ x 0L∞∑= c r γr ∗ ,r=−∞where the last equality uses (12.10). Finally, if we let g(x) =f(x) then we obtainParseval’s theorem (12.13). This result can be proved in a similar manner using426r=1

12.9 EXERCISESthe sine and cosine form of the Fourier series, but the algebra is slightly morecomplicated.Parseval’s theorem is sometimes used to sum series. However, if one is presentedwith a series to sum, it is not usually possible to decide which Fourier seriesshould be used to evaluate it. Rather, useful summations are nearly always foundserendipitously. The following example shows the evaluation of a sum by aFourier series method.◮Using Parseval’s theorem and the Fourier series for f(x) =x 2 found in section 12.5,calculate the sum ∑ ∞r=1 r−4 .Firstly we find the average value of [f(x)] 2 over the interval −2

FOURIER SERIESbe better for numerical evaluation? Relate your answer to the relevant periodiccontinuations.12.7 For the continued functions used in exercise 12.6 and the derived correspondingseries, consider (i) their derivatives and (ii) their integrals. Do they give meaningfulequations? You will probably find it helpful to sketch all the functions involved.12.8 The function y(x) =x sin x for 0 ≤ x ≤ π is to be represented by a Fourier seriesof period 2π that is either even or odd. By sketching the function and consideringits derivative, determine which series will have the more rapid convergence. Findthe full expression for the better of these two series, showing that the convergence∼ n −3 and that alternate terms are missing.12.9 Find the Fourier coefficients in the expansion of f(x) =expx over the range−1

12.9 EXERCISESDeduce the value of the sum S of the series1 − 1 3 3 + 1 5 3 − 1 7 3 + ··· .12.15 Using the result of exercise 12.14, determine, as far as possible by inspection, theforms of the functions of which the following are the Fourier series:(a)(b)cos θ + 1 9 cos 3θ + 1 cos 5θ + ··· ;25sin θ + 1 1sin 3θ + sin 5θ + ··· ;27 125(c)[L 23 − 4L2 cos πxπ 2 L − 1 2πxcos4 L+ 1 3πx···]cos9 L − .(You may find it helpful to first set x = 0 in the quoted result and so obtainvalues for S o = ∑ (2m +1) −2 and other sums derivable from it.)12.16 By finding a cosine Fourier series of period 2 for the function f(t) thattakestheform f(t) =cosh(t − 1) in the range 0 ≤ t ≤ 1, prove that∞∑n=11n 2 π 2 +1 = 1e 2 − 1 .Deduce values for the sums ∑ (n 2 π 2 +1) −1 over odd n and even n separately.12.17 Find the (real) Fourier series of period 2 for f(x) =coshx and g(x) =x 2 in therange −1 ≤ x ≤ 1. By integrating the series for f(x) twice, prove that∞∑n=1(−1) n+1n 2 π 2 (n 2 π 2 +1) = 1 2( 1sinh 1 − 5 612.18 Express the function f(x) =x 2 as a Fourier sine series in the range 0

FOURIER SERIES12.21 Find the complex Fourier series for the periodic function of period 2π defined inthe range −π ≤ x ≤ π by y(x) =coshx. By setting x = 0 prove that∞∑ (−1) nn 2 +1 = 1 ( π)2 sinh π − 1 .n=112.22 The repeating output from an electronic oscillator takes the form of a sine wavef(t) =sint for 0 ≤ t ≤ π/2; it then drops instantaneously to zero and startsagain. The output is to be represented by a complex Fourier series of the form∞∑c n e 4nti .n=−∞Sketch the function and find an expression for c n .Verifythatc −n = c ∗ n.Demonstratethat setting t =0andt = π/2 produces differing values for the sum∞∑ 116n 2 − 1 .n=1Determine the correct value and check it using the result of exercise 12.20.12.23 Apply Parseval’s theorem to the series found in the previous exercise and soderive a value for the sum of the series17(15) + 652 (63) + 1452 (143) + ···+ 16n2 +12 (16n 2 − 1) + ··· .212.24 A string, anchored at x = ±L/2, has a fundamental vibration frequency of 2L/c,where c is the speed of transverse waves on the string. It is pulled aside at itscentre point by a distance y 0 and released at time t = 0. Its subsequent motioncan be described by the series∞∑y(x, t) = a n cos nπx nπctcosL L .n=1Find a general expression for a n and show that only the odd harmonics of thefundamental frequency are present in the sound generated by the released string.By applying Parseval’s theorem, find the sum S of the series ∑ ∞0 (2m +1)−4 .12.25 Show that Parseval’s theorem for two real functions whose Fourier expansionshave cosine and sine coefficients a n , b n and α n , β n takes the form∫1 Lf(x)g ∗ (x) dx = 1 L 04 a 0α 0 + 1 ∞∑(a n α n + b n β n ).2n=1(a) Demonstrate that for g(x) =sinmx or cos mx this reduces to the definitionof the Fourier coefficients.(b) Explicitly verify the above result for the case in which f(x) =x and g(x) isthe square-wave function, both in the interval −1 ≤ x ≤ 1.12.26 An odd function f(x) ofperiod2π is to be approximated by a Fourier sine serieshaving only m terms. The error in this approximation is measured by the squaredeviation∫ [2 πm∑E m = f(x) − b n sin nx]dx.−πn=1By differentiating E m with respect to the coefficients b n , find the values of b n thatminimise E m .430

12.10 HINTS AND ANSWERS0 1 0 1 0 1 0 2 4(a) (b) (c) (d)Figure 12.6 Continuations of exp(−x 2 )in0≤ x ≤ 1 to give: (a) cosine termsonly; (b) sine terms only; (c) period 1; (d) period 2.Sketch the graph of the function f(x), where{−x(π + x) for −π ≤ x

FOURIER SERIES12.21 c n =[(−1) n sinh π]/[π(1 + n 2 )]. Having set x = 0, separate out the n =0termand note that (−1) n =(−1) −n .12.23 (π 2 − 8)/16.12.25 (b) All a n and α n are zero; b n =2(−1) n+1 /(nπ) andβ n =4/(nπ). You will needthe result quoted in exercise 12.19.432

13Integral transformsIn the previous chapter we encountered the Fourier series representation of aperiodic function in a fixed interval as a superposition of sinusoidal functions. It isoften desirable, however, to obtain such a representation even for functions definedover an infinite interval and with no particular periodicity. Such a representationis called a Fourier transform and is one of a class of representations called integraltransforms.We begin by considering Fourier transforms as a generalisation of Fourierseries. We then go on to discuss the properties of the Fourier transform and itsapplications. In the second part of the chapter we present an analogous discussionof the closely related Laplace transform.13.1 Fourier transformsThe Fourier transform provides a representation of functions defined over aninfinite interval and having no particular periodicity, in terms of a superpositionof sinusoidal functions. It may thus be considered as a generalisation of theFourier series representation of periodic functions. Since Fourier transforms areoften used to represent time-varying functions, we shall present much of ourdiscussion in terms of f(t), rather than f(x), although in some spatial examplesf(x) will be the more natural notation and we shall use it as appropriate. Our|f(t)| dt is finite.In order to develop the transition from Fourier series to Fourier transforms, wefirst recall that a function of period T may be represented as a complex Fourierseries, cf. (12.9),only requirement on f(t) will be that ∫ ∞−∞f(t) =∞∑c r e 2πirt/T =r=−∞∞∑c r e iωrt , (13.1)r=−∞where ω r =2πr/T. As the period T tends to infinity, the ‘frequency quantum’433

INTEGRAL TRANSFORMSc(ω)expiωt− 2πT0−1 02πT4πTω r1 2 rFigure 13.1 The relationship between the Fourier terms for a function ofperiod T and the Fourier integral (the area below the solid line) of thefunction.∆ω =2π/T becomes vanishingly small and the spectrum of allowed frequenciesω r becomes a continuum. Thus, the infinite sum of terms in the Fourier seriesbecomes an integral, and the coefficients c r become functions of the continuousvariable ω, as follows.We recall, cf. (12.10), that the coefficients c r in (13.1) are given by∫ T/2∫ T/2c r = 1 f(t) e −2πirt/T dt = ∆ω f(t) e −iωrt dt, (13.2)T −T/22π −T/2where we have written the integral in two alternative forms and, for convenience,made one period run from −T/2to+T/2 rather than from 0 to T . Substitutingfrom (13.2) into (13.1) gives∞∑∫∆ω T/2f(t) =f(u) e −iωru du e iωrt . (13.3)2πr=−∞ −T/2At this stage ω r is still a discrete function of r equal to 2πr/T.The solid points in figure 13.1 are a plot of (say, the real part of) c r e iωrt asa function of r (or equivalently of ω r ) and it is clear that (2π/T)c r e iωrt givesthe area of the rth broken-line rectangle. If T tends to ∞ then ∆ω (= 2π/T)becomes infinitesimal, the width of the rectangles tends to zero and, from themathematical definition of an integral,∞∑r=−∞In this particular case∆ω2π g(ω r) e iωrt → 12π∫ ∞−∞∫ T/2g(ω r )= f(u) e −iωru du,−T/2434g(ω) e iωt dω.

13.1 FOURIER TRANSFORMSand (13.3) becomesf(t) = 1 ∫ ∞ ∫ ∞dω e iωt du f(u) e −iωu . (13.4)2π −∞−∞This result is known as Fourier’s inversion theorem.From it we may define the Fourier transform of f(t) byand its inverse by˜f(ω) =1√2π∫ ∞−∞f(t) = 1 √2π∫ ∞−∞f(t) e −iωt dt, (13.5)˜f(ω) e iωt dω. (13.6)Including the constant 1/ √ 2π in the definition of ˜f(ω) (whose mathematicalexistence as T →∞is assumed here without proof) is clearly arbitrary, the onlyrequirement being that the product of the constants in (13.5) and (13.6) shouldequal 1/(2π). Our definition is chosen to be as symmetric as possible.◮ Find the Fourier transform of the exponential decay function f(t) =0for t0).Using the definition (13.5) and separating the integral into two parts,∫1 0˜f(ω) = √ (0) e −iωt dt + A ∫ ∞√ e −λt e −iωt dt2π −∞2π 0=0+ √ A [ ] ∞− e−(λ+iω)t2π λ + iω0A= √ ,2π(λ + iω)which is the required transform. It is clear that the multiplicative constant A does notaffect the form of the transform, merely its amplitude. This transform may be verified byresubstitution of the above result into (13.6) to recover f(t), but evaluation of the integralrequires the use of complex-variable contour integration (chapter 24). ◭13.1.1 The uncertainty principleAn important function that appears in many areas of physical science, eitherprecisely or as an approximation to a physical situation, is the Gaussian ornormal distribution. Its Fourier transform is of importance both in itself and alsobecause, when interpreted statistically, it readily illustrates a form of uncertaintyprinciple.435

INTEGRAL TRANSFORMS◮Find the Fourier transform of the normalised Gaussian distributionf(t) = 1 ( )τ √ 2π exp − t2, −∞

13.1 FOURIER TRANSFORMSis a wavefunction and the distribution of the wave intensity in time is given by|f| 2 (also a Gaussian). Similarly, the intensity distribution in frequency is givenby |˜f| 2 . These two distributions have respective root mean square deviations ofτ/ √ 2and1/( √ 2τ), giving, after incorporation of the above relations,∆E ∆t = /2 and ∆p ∆x = /2.The factors of 1/2 that appear are specific to the Gaussian form, but anydistribution f(t) produces for the product ∆E∆t a quantity λ in which λ isstrictly positive (in fact, the Gaussian value of 1/2 is the minimum possible).13.1.2 Fraunhofer diffractionWe take our final example of the Fourier transform from the field of optics. Thepattern of transmitted light produced by a partially opaque (or phase-changing)object upon which a coherent beam of radiation falls is called a diffraction patternand, in particular, when the cross-section of the object is small compared withthe distance at which the light is observed the pattern is known as a Fraunhoferdiffraction pattern.We will consider only the case in which the light is monochromatic withwavelength λ. The direction of the incident beam of light can then be describedby the wave vector k; the magnitude of this vector is given by the wave numberk =2π/λ of the light. The essential quantity in a Fraunhofer diffraction patternis the dependence of the observed amplitude (and hence intensity) on the angle θbetween the viewing direction k ′ and the direction k of the incident beam. Thisis entirely determined by the spatial distribution of the amplitude and phase ofthe light at the object, the transmitted intensity in a particular direction k ′ beingdetermined by the corresponding Fourier component of this spatial distribution.As an example, we take as an object a simple two-dimensional screen of width2Y on which light of wave number k is incident normally; see figure 13.2. Wesuppose that at the position (0,y) the amplitude of the transmitted light is f(y)per unit length in the y-direction (f(y) may be complex). The function f(y) iscalled an aperture function. Both the screen and beam are assumed infinite in thez-direction.Denoting the unit vectors in the x- andy- directions by i and j respectively,the total light amplitude at a position r 0 = x 0 i + y 0 j, with x 0 > 0, will be thesuperposition of all the (Huyghens’) wavelets originating from the various partsof the screen. For large r 0 (= |r 0 |), these can be treated as plane waves to give §A(r 0 )=∫ Y−Yf(y) exp[ik ′ · (r 0 − yj)]dy. (13.8)|r 0 − yj|§ This is the approach first used by Fresnel. For simplicity we have omitted from the integral amultiplicative inclination factor that depends on angle θ and decreases as θ increases.437

INTEGRAL TRANSFORMSYyk ′k0θx−YFigure 13.2 Diffraction grating of width 2Y with light of wavelength 2π/kbeing diffracted through an angle θ.The factor exp[ik ′ · (r 0 − yj)] represents the phase change undergone by the lightin travelling from the point yj on the screen to the point r 0 , and the denominatorrepresents the reduction in amplitude with distance. (Recall that the system isinfinite in the z-direction and so the ‘spreading’ is effectively in two dimensionsonly.)If the medium is the same on both sides of the screen then k ′ = k cos θ i+k sin θ j,and if r 0 ≫ Y then expression (13.8) can be approximated by∫A(r 0 )= exp(ik′ · r 0 ) ∞f(y) exp(−iky sin θ) dy. (13.9)r 0 −∞We have used that f(y) =0for|y| >Y to extend the integral to infinite limits.The intensity in the direction θ is then given byI(θ) =|A| 2 = 2πr2 |˜f(q)| 2 , (13.10)0where q = k sin θ.◮Evaluate I(θ) for an aperture consisting of two long slits each of width 2b whose centresare separated by a distance 2a, a>b; the slits are illuminated by light of wavelength λ.The aperture function is plotted in figure 13.3. We first need to find ˜f(q):∫1 −a+b˜f(q) = √ e −iqx dx + 1 ∫ a+b√ e −iqx dx2π 2π−a−b] −a+ba−b] a+b= √ 1 [− e−iqx+ 1 [√ − e−iqx2π iq−a−b 2π iqa−b= −1iq √ [e −iq(−a+b) − e −iq(−a−b) + e −iq(a+b) − e −iq(a−b)] .2π438

13.1 FOURIER TRANSFORMSf(y)1−a−a − b −a + ba − baa + bxFigure 13.3The aperture function f(y) for two wide slits.After some manipulation we obtain4cosqa sin qb˜f(q) =q √ .2πNow applying (13.10), and remembering that q =(2π sin θ)/λ, we findI(θ) = 16 cos2 qa sin 2 qb,q 2 r2 0where r 0 is the distance from the centre of the aperture. ◭13.1.3 The Dirac δ-functionBefore going on to consider further properties of Fourier transforms we make adigression to discuss the Dirac δ-function and its relation to Fourier transforms.The δ-function is different from most functions encountered in the physicalsciences but we will see that a rigorous mathematical definition exists; the utilityof the δ-function will be demonstrated throughout the remainder of this chapter.It can be visualised as a very sharp narrow pulse (in space, time, density, etc.)which produces an integrated effect having a definite magnitude. The formalproperties of the δ-function may be summarised as follows.The Dirac δ-function has the property thatδ(t) =0 fort ≠0, (13.11)but its fundamental defining property is∫f(t)δ(t − a) dt = f(a), (13.12)provided the range of integration includes the point t = a; otherwise the integral439

INTEGRAL TRANSFORMSequals zero. This leads immediately to two further useful results:and∫ b−aδ(t) dt = 1 for all a, b > 0 (13.13)∫δ(t − a) dt =1, (13.14)provided the range of integration includes t = a.Equation (13.12) can be used to derive further useful properties of the Diracδ-function:δ(t) =δ(−t), (13.15)δ(at) = 1 δ(t), (13.16)|a|tδ(t) =0. (13.17)◮Prove that δ(bt) =δ(t)/|b|.Let us first consider the case where b>0. It follows that∫ ∞∫ ∞( ) t′f(t)δ(bt) dt = f δ(t ′ ) dt′−∞−∞ b b = 1 b f(0) = 1 ∫ ∞f(t)δ(t) dt,b −∞where we have made the substitution t ′ = bt. Butf(t) is arbitrary and so we immediatelysee that δ(bt) =δ(t)/b = δ(t)/|b| for b>0.Now consider the case where b = −c

13.1 FOURIER TRANSFORMSThe derivative of the delta function, δ ′ (t), is defined by∫ ∞−∞[ ] ∞∫ ∞f(t)δ ′ (t) dt = f(t)δ(t) − f ′ (t)δ(t) dt−∞ −∞= −f ′ (0), (13.19)and similarly for higher derivatives.For many practical purposes, effects that are not strictly described by a δ-function may be analysed as such, if they take place in an interval much shorterthan the response interval of the system on which they act. For example, theidealised notion of an impulse of magnitude J applied at time t 0 can be representedbyj(t) =Jδ(t − t 0 ). (13.20)Many physical situations are described by a δ-function in space rather than intime. Moreover, we often require the δ-function to be defined in more than onedimension. For example, the charge density of a point charge q at a point r 0 maybe expressed as a three-dimensional δ-functionρ(r) =qδ(r − r 0 )=qδ(x − x 0 )δ(y − y 0 )δ(z − z 0 ), (13.21)so that a discrete ‘quantum’ is expressed as if it were a continuous distribution.From (13.21) we see that (as expected) the total charge enclosed in a volume Vis given by∫∫{q if r0 lies in V,ρ(r) dV = qδ(r − r 0 ) dV =VV0 otherwise.Closely related to the Dirac δ-function is the Heaviside or unit step functionH(t), for which{1 for t>0,H(t) =0 for t

INTEGRAL TRANSFORMS◮Prove relation (13.23).Considering the integral∫ ∞−∞[ ] ∞ ∫ ∞f(t)H ′ (t) dt = f(t)H(t) − f ′ (t)H(t) dt−∞ −∞∫ ∞= f(∞) − f ′ (t) dt= f(∞) −0[f(t) = f(0),and comparing it with (13.12) when a = 0 immediately shows that H ′ (t) =δ(t). ◭] ∞013.1.4 Relation of the δ-function to Fourier transformsIn the previous section we introduced the Dirac δ-function as a way of representingvery sharp narrow pulses, but in no way related it to Fourier transforms.We now show that the δ-function can equally well be defined in a way that morenaturally relates it to the Fourier transform.Referring back to the Fourier inversion theorem (13.4), we havef(t) = 1 ∫ ∞ ∫ ∞dω e iωt du f(u) e −iωu=2π−∞∫ ∞−∞du f(u){ 12π−∞∫ ∞−∞}e iω(t−u) dω .Comparison of this with (13.12) shows that we may write the δ-function asδ(t − u) = 1 ∫ ∞e iω(t−u) dω. (13.24)2π −∞Considered as a Fourier transform, this representation shows that a verynarrow time peak at t = u results from the superposition of a complete spectrumof harmonic waves, all frequencies having the same amplitude and all waves beingin phase at t = u. This suggests that the δ-function may also be represented asthe limit of the transform of a uniform distribution of unit height as the widthof this distribution becomes infinite.Consider the rectangular distribution of frequencies shown in figure 13.4(a).From (13.6), taking the inverse Fourier transform,f Ω (t) = √ 1 ∫ Ω1 × e iωt dω2π −Ω= √ 2Ω sin Ωt. (13.25)2π ΩtThis function is illustrated in figure 13.4(b) and it is apparent that, for large Ω, itbecomes very large at t = 0 and also very narrow about t = 0, as we qualitatively442

13.1 FOURIER TRANSFORMS˜fΩ2Ω(2π) 1/2f Ω (t)1−Ω(a)ΩωπΩ(b)tFigure 13.4 (a) A Fourier transform showing a rectangular distribution offrequencies between ±Ω; (b) the function of which it is the transform, whichis proportional to t −1 sin Ωt.expect and require. We also note that, in the limit Ω →∞, f Ω (t), as defined bythe inverse Fourier transform, tends to (2π) 1/2 δ(t) by virtue of (13.24). Hence wemay conclude that the δ-function can also be represented byδ(t) = limΩ→∞( sin Ωtπt). (13.26)Several other function representations are equally valid, e.g. the limiting cases ofrectangular, triangular or Gaussian distributions; the only essential requirementsare a knowledge of the area under such a curve and that undefined operationssuch as dividing by zero are not inadvertently carried out on the δ-function whilstsome non-explicit representation is being employed.We also note that the Fourier transform definition of the delta function, (13.24),shows that the latter is real sinceδ ∗ (t) = 1 ∫ ∞e −iωt dω = δ(−t) =δ(t).2π −∞Finally, the Fourier transform of a δ-function is simply˜δ(ω) = 1 √2π∫ ∞−∞δ(t) e −iωt dt = 1 √2π. (13.27)13.1.5 Properties of Fourier transformsHaving considered the Dirac δ-function, we now return to our discussion of theproperties of Fourier transforms. As we would expect, Fourier transforms havemany properties analogous to those of Fourier series in respect of the connectionbetween the transforms of related functions. Here we list these properties withoutproof; they can be verified by working from the definition of the transform. Aspreviously, we denote the Fourier transform of f(t) by˜f(ω) orF[f(t)].443

INTEGRAL TRANSFORMS(i) Differentiation:F [ f ′ (t) ] = iω˜f(ω). (13.28)This may be extended to higher derivatives, so thatandsoon.(ii) Integration:F [ f ′′ (t) ] = iωF [ f ′ (t) ] = −ω 2˜f(ω),t]F[∫f(s) ds = 1iω ˜f(ω)+2πcδ(ω), (13.29)where the term 2πcδ(ω) represents the Fourier transform of the constantof integration associated with the indefinite integral.(iii) Scaling:F[f(at)] = 1 ( ω). (13.30)a˜fa(iv) Translation:(v) Exponential multiplication:where α may be real, imaginary or complex.F[f(t + a)] = e iaω˜f(ω). (13.31)F [ e αt f(t) ] = ˜f(ω + iα), (13.32)◮Prove relation (13.28).Calculating the Fourier transform of f ′ (t) directly, we obtainF [ f ′ (t) ] = √ 1 ∫ ∞f ′ (t) e −iωt dt2π −∞= √ 1 [e −iωt f(t) + 1 ∫ ∞√ iω e −iωt f(t) dt2π 2π] ∞−∞= iω˜f(ω),if f(t) → 0att = ±∞, asitmustsince ∫ ∞|f(t)| dt is finite. ◭−∞To illustrate a use and also a proof of (13.32), let us consider an amplitudemodulatedradio wave. Suppose a message to be broadcast is represented by f(t).The message can be added electronically to a constant signal a of magnitudesuch that a + f(t) is never negative, and then the sum can be used to modulatethe amplitude of a carrier signal of frequency ω c . Using a complex exponentialnotation, the transmitted amplitude is now−∞g(t) =A [a + f(t)] e iωct . (13.33)444

13.1 FOURIER TRANSFORMSIgnoring in the present context the effect of the term Aa exp(iω c t), which gives acontribution to the transmitted spectrum only at ω = ω c , we obtain for the newspectrum˜g(ω) = √ 1 ∫ ∞A f(t) e iωct e −iωt dt2π −∞= √ 1 ∫ ∞A f(t) e −i(ω−ωc)t dt2π−∞= A˜f(ω − ω c ), (13.34)which is simply a shift of the whole spectrum by the carrier frequency. The useof different carrier frequencies enables signals to be separated.13.1.6 Odd and even functionsIf f(t) is odd or even then we may derive alternative forms of Fourier’s inversiontheorem, which lead to the definition of different transform pairs. Let us firstconsider an odd function f(t) =−f(−t), whose Fourier transform is given by˜f(ω) = √ 1 ∫ ∞f(t) e −iωt dt2π −∞= √ 1 ∫ ∞f(t)(cos ωt − i sin ωt) dt2π −∞= √ −2i ∫ ∞f(t)sinωt dt,2π0where in the last line we use the fact that f(t) andsinωt are odd, whereas cos ωtis even.We note that ˜f(−ω) =−˜f(ω), i.e. ˜f(ω) is an odd function of ω. Hencef(t) = √ 1 ∫ ∞˜f(ω) e iωt dω =2i ∫ ∞√ ˜f(ω)sinωt dω2π −∞2π 0= 2 ∫ ∞{∫ ∞}dω sin ωt f(u)sinωu du .π 00Thus we may define the Fourier sine transform pair for odd functions:√ ∫ 2 ∞˜f s (ω) = f(t) sinωt dt, (13.35)π 0√ ∫ 2 ∞f(t) = ˜fs (ω) sinωt dω. (13.36)π 0Note that although the Fourier sine transform pair was derived by consideringan odd function f(t) defined over all t, the definitions (13.35) and (13.36) onlyrequire f(t) and˜f s (ω) to be defined for positive t and ω respectively. For an445

INTEGRAL TRANSFORMSg(y)(a)(b)(c)(d)0yFigure 13.5 Resolution functions: (a) ideal δ-function; (b) typical unbiasedresolution; (c) and (d) biases tending to shift observations to higher valuesthan the true one.even function, i.e. one for which f(t) =f(−t), we can define the Fourier cosinetransform pair in a similar way, but with sin ωt replaced by cos ωt.13.1.7 Convolution and deconvolutionIt is apparent that any attempt to measure the value of a physical quantity islimited, to some extent, by the finite resolution of the measuring apparatus used.On the one hand, the physical quantity we wish to measure will be in general afunction of an independent variable, x say, i.e. the true function to be measuredtakes the form f(x). On the other hand, the apparatus we are using does not givethe true output value of the function; a resolution function g(y) is involved. Bythis we mean that the probability that an output value y = 0 will be recordedinstead as being between y and y+dy is given by g(y) dy. Some possible resolutionfunctions of this sort are shown in figure 13.5. To obtain good results we wishthe resolution function to be as close to a δ-function as possible (case (a)). Atypical piece of apparatus has a resolution function of finite width, although ifit is accurate the mean is centred on the true value (case (b)). However, someapparatus may show a bias that tends to shift observations to higher or lowervalues than the true ones (cases (c) and(d)), thereby exhibiting systematic error.Given that the true distribution is f(x) and the resolution function of ourmeasuring apparatus is g(y), we wish to calculate what the observed distributionh(z) will be. The symbols x, y and z all refer to the same physical variable (e.g.446

13.1 FOURIER TRANSFORMSf(x)∗ g(y) =12bh(z)2b−aaFigure 13.6x y z−b b−aaThe convolution of two functions f(x) andg(y).length or angle), but are denoted differently because the variable appears in theanalysis in three different roles.The probability that a true reading lying between x and x + dx, and so havingprobability f(x) dx of being selected by the experiment, will be moved by theinstrumental resolution by an amount z − x into a small interval of width dz isg(z − x) dz. Hence the combined probability that the interval dx will give rise toan observation appearing in the interval dz is f(x) dx g(z − x) dz. Adding togetherthe contributions from all values of x thatcanleadtoanobservationintherangez to z + dz, we find that the observed distribution is given byh(z) =∫ ∞−∞f(x)g(z − x) dx. (13.37)The integral in (13.37) is called the convolution of the functions f and g and isoften written f ∗ g. The convolution defined above is commutative (f ∗ g = g ∗ f),associative and distributive. The observed distribution is thus the convolution ofthe true distribution and the experimental resolution function. The result will bethat the observed distribution is broader and smoother than the true one and, ifg(y) has a bias, the maxima will normally be displaced from their true positions.It is also obvious from (13.37) that if the resolution is the ideal δ-function,g(y) =δ(y) thenh(z) =f(z) and the observed distribution is the true one.It is interesting to note, and a very important property, that the convolution ofany function g(y) with a number of delta functions leaves a copy of g(y) attheposition of each of the delta functions.◮Find the convolution of the function f(x) =δ(x + a) +δ(x − a) with the function g(y)plotted in figure 13.6.Using the convolution integral (13.37)h(z) =∫ ∞−∞f(x)g(z − x) dx =∫ ∞−∞[δ(x + a)+δ(x − a)]g(z − x) dx= g(z + a)+g(z − a).This convolution h(z) is plotted in figure 13.6. ◭Let us now consider the Fourier transform of the convolution (13.37); this is447

INTEGRAL TRANSFORMSgiven by∫1 ∞{∫ ∞}˜h(k) = √ dz e −ikz f(x)g(z − x) dx2π −∞−∞= √ 1 ∫ ∞{∫ ∞}dx f(x) g(z − x) e −ikz dz .2π −∞−∞If we let u = z − x in the second integral we have∫1 ∞{∫ ∞}˜h(k) = √ dx f(x) g(u) e −ik(u+x) du2π −∞−∞= √ 1 ∫ ∞∫ ∞f(x) e −ikx dx g(u) e −iku du2π−∞−∞= 1 √2π× √ 2π ˜f(k) × √ 2π˜g(k) = √ 2π ˜f(k)˜g(k). (13.38)Hence the Fourier transform of a convolution f ∗ g is equal to the product of theseparate Fourier transforms multiplied by √ 2π; this result is called the convolutiontheorem.It may be proved similarly that the converse is also true, namely that theFourier transform of the product f(x)g(x) is given byF[f(x)g(x)] = √ 1 ˜f(k) ∗ ˜g(k). (13.39)2π◮Find the Fourier transform of the function in figure 13.3 representing two wide slits byconsidering the Fourier transforms of (i) two δ-functions, at x = ±a, (ii) a rectangularfunction of height 1 and width 2b centred on x =0.(i) The Fourier transform of the two δ-functions is given by∫1 ∞˜f(q) = √ δ(x − a) e −iqx dx + 1 ∫ ∞√ δ(x + a) e −iqx dx2π −∞2π −∞= √ 1 (e −iqa + e iqa) = 2cosqa √ .2π 2π(ii) The Fourier transform of the broad slit is˜g(q) = √ 1 ∫ be −iqx dx = 1 [ ] e−iqx b√2π −b2π −iq−b= −1iq √ 2π (e−iqb − e iqb )= 2sinqbq √ 2π .We have already seen that the convolution of these functions is the required functionrepresenting two wide slits (see figure 13.6). So, using the convolution theorem, the Fouriertransform of the convolution is √ 2π times the product of the individual transforms, i.e.4cosqa sin qb/(q √ 2π). This is, of course, the same result as that obtained in the examplein subsection 13.1.2. ◭448

13.1 FOURIER TRANSFORMSThe inverse of convolution, called deconvolution, allows us to find a truedistribution f(x) given an observed distribution h(z) and a resolution functiong(y).◮An experimental quantity f(x) is measured using apparatus with a known resolution functiong(y) to give an observed distribution h(z). Howmayf(x) be extracted from the measureddistribution?From the convolution theorem (13.38), the Fourier transform of the measured distributionis˜h(k) =√2π ˜f(k)˜g(k),from which we obtain1 ˜h(k)˜f(k) = √2π ˜g(k) .Then on inverse Fourier transforming we find[f(x) = √ 1 ˜h(k)F −1 .2π ˜g(k)In words, to extract the true distribution, we divide the Fourier transform of the observeddistribution by that of the resolution function for each value of k and then take the inverseFourier transform of the function so generated. ◭This explicit method of extracting true distributions is straightforward for exactfunctions but, in practice, because of experimental and statistical uncertainties inthe experimental data or because data over only a limited range are available, itis often not very precise, involving as it does three (numerical) transforms eachrequiring in principle an integral over an infinite range.]13.1.8 Correlation functions and energy spectraThe cross-correlation of two functions f and g is defined byC(z) =∫ ∞−∞f ∗ (x)g(x + z) dx. (13.40)Despite the formal similarity between (13.40) and the definition of the convolutionin (13.37), the use and interpretation of the cross-correlation and of the convolutionare very different; the cross-correlation provides a quantitative measure ofthe similarity of two functions f and g as one is displaced through a distance zrelative to the other. The cross-correlation is often notated as C = f ⊗ g, and, likeconvolution, it is both associative and distributive. Unlike convolution, however,it is not commutative, in fact[f ⊗ g](z) =[g ⊗ f] ∗ (−z). (13.41)449

INTEGRAL TRANSFORMS◮Prove the Wiener–Kinchin theorem,˜C(k) = √ 2π [˜f(k)] ∗˜g(k). (13.42)Following a method similar to that for the convolution of f and g, let us consider theFourier transform of (13.40):˜C(k) = √ 1 ∫ ∞{∫ ∞}dz e −ikz f ∗ (x)g(x + z) dx2π −∞−∞= √ 1 ∫ ∞{∫ ∞}dx f ∗ (x) g(x + z) e −ikz dz .2π −∞−∞Making the substitution u = x + z inthesecondintegralweobtain˜C(k) = √ 1 ∫ ∞{∫ ∞}dx f ∗ (x) g(u) e −ik(u−x) du2π −∞−∞= √ 1 ∫ ∞∫ ∞f ∗ (x) e ikx dx g(u) e −iku du2π −∞−∞= √ 1 × √ 2π [˜f(k)] ∗ × √ 2π ˜g(k) = √ 2π [˜f(k)] ∗˜g(k). ◭2πThus the Fourier transform of the cross-correlation of f and g is equal tothe product of [˜f(k)] ∗ and ˜g(k) multiplied by √ 2π. This a statement of theWiener–Kinchin theorem. Similarly we can derive the converse theoremF [ f ∗ (x)g(x) ] = √ 1 ˜f ⊗ ˜g.2πIf we now consider the special case where g is taken to be equal to f in (13.40)then, writing the LHS as a(z), we havea(z) =∫ ∞−∞f ∗ (x)f(x + z) dx; (13.43)this is called the auto-correlation function of f(x). Using the Wiener–Kinchintheorem (13.42) we see thata(z) = √ 1 ∫ ∞ã(k) e ikz dk2π −∞= √ 1 ∫ ∞ √2π [˜f(k)]∗˜f(k) e ikz dk,2π−∞so that a(z) is the inverse Fourier transform of √ 2π |˜f(k)| 2 , which is in turn calledthe energy spectrum of f.13.1.9 Parseval’s theoremUsing the results of the previous section we can immediately obtain Parseval’stheorem. The most general form of this (also called the multiplication theorem) is450

13.1 FOURIER TRANSFORMSobtained simply by noting from (13.42) that the cross-correlation (13.40) of twofunctions f and g canbewrittenasC(z) =∫ ∞−∞f ∗ (x)g(x + z) dx =∫ ∞−∞[˜f(k)] ∗˜g(k) e ikz dk. (13.44)Then, setting z = 0 gives the multiplication theorem∫ ∞∫f ∗ (x)g(x) dx = [˜f(k)] ∗˜g(k) dk. (13.45)−∞Specialising further, by letting g = f, we derive the most common form ofParseval’s theorem,∫ ∞∫ ∞|f(x)| 2 dx = |˜f(k)| 2 dk. (13.46)−∞−∞When f is a physical amplitude these integrals relate to the total intensity involvedin some physical process. We have already met a form of Parseval’s theorem forFourier series in chapter 12; it is in fact a special case of (13.46).◮The displacement of a damped harmonic oscillator as a function of time is given by{0 for t

INTEGRAL TRANSFORMStwo- or three-dimensional functions of position. For example, in three dimensionswe can define the Fourier transform of f(x, y, z) as∫∫∫1˜f(k x ,k y ,k z )=f(x, y, z) e −ikxx e −ikyy e −ikzz dx dy dz, (13.47)(2π) 3/2and its inverse as∫∫∫1f(x, y, z) =˜f(k(2π) 3/2 x ,k y ,k z ) e ikxx e ikyy e ikzz dk x dk y dk z . (13.48)Denoting the vector with components k x ,k y ,k z by k and that with componentsx, y, z by r, we can write the Fourier transform pair (13.47), (13.48) as∫1˜f(k) = f(r) e −ik·r d 3 r, (13.49)(2π) 3/2 ∫1f(r) = ˜f(k) e ik·r d 3 k. (13.50)(2π) 3/2From these relations we may deduce that the three-dimensional Dirac δ-functioncanbewrittenasδ(r) = 1 ∫(2π) 3 e ik·r d 3 k. (13.51)Similar relations to (13.49), (13.50) and (13.51) exist for spaces of other dimensionalities.◮In three-dimensional space a function f(r) possesses spherical symmetry, so that f(r) =f(r). Find the Fourier transform of f(r) as a one-dimensional integral.Let us choose spherical polar coordinates in which the vector k of the Fourier transformlies along the polar axis (θ =0).Thiswecandosincef(r) is spherically symmetric. Wethen haved 3 r = r 2 sin θdrdθdφ and k · r = kr cos θ,where k = |k|. The Fourier transform is then given by∫1˜f(k) = f(r) e −ik·r d 3 r(2π) 3/2 ∫1 ∞ ∫ π ∫ 2π=dr dθ dφ f(r)r 2 −ikr cos θsin θe(2π) 3/2 0 0 0∫1 ∞ ∫ π=dr 2πf(r)r 2 dθ sin θe −ikr cos θ .(2π) 3/2 00The integral over θ may be straightforwardly evaluated by noting thatddθ (e−ikr cos θ )=ikr sin θe −ikr cos θ .Therefore∫1 ∞[ ] e˜f(k) = dr 2πf(r)r 2 −ikr cos θ θ=π(2π) 3/2 0ikrθ=0∫1 ∞( ) sin kr=4πr 2 f(r) dr. ◭(2π) 3/2 kr0452

13.2 LAPLACE TRANSFORMSA similar result may be obtained for two-dimensional Fourier transforms inwhich f(r) =f(ρ), i.e. f(r) is independent of azimuthal angle φ. In this case, usingthe integral representation of the Bessel function J 0 (x) given at the very end ofsubsection 18.5.3, we find˜f(k) = 1 ∫ ∞2πρf(ρ)J 0 (kρ) dρ. (13.52)2π013.2 Laplace transformsOften we are interested in functions f(t) for which the Fourier transform does notexist because f ↛ 0ast →∞, and so the integral defining ˜f does not converge.This would be the case for the function f(t) =t, which does not possess a Fouriertransform. Furthermore, we might be interested in a given function only for t>0,for example when we are given the value at t = 0 in an initial-value problem.This leads us to consider the Laplace transform, ¯f(s) orL [f(t)], off(t), whichis defined by¯f(s) ≡∫ ∞0f(t)e −st dt, (13.53)provided that the integral exists. We assume here that s is real, but complex valueswould have to be considered in a more detailed study. In practice, for a givenfunction f(t) there will be some real number s 0 such that the integral in (13.53)exists for s>s 0 but diverges for s ≤ s 0 .Through (13.53) we define a linear transformation L that converts functionsof the variable t to functions of a new variable s:L [af 1 (t)+bf 2 (t)] = aL [f 1 (t)] + bL [f 2 (t)] = a ¯f 1 (s)+b ¯f 2 (s). (13.54)◮Find the Laplace transforms of the functions (i) f(t) =1, (ii) f(t) =e at , (iii) f(t) =t n ,for n =0, 1, 2,... .(i) By direct application of the definition of a Laplace transform (13.53), we find∫ ∞[ ] ∞ −1L [1] = e −st dt =0s e−st = 1 , if s>0,0swhere the restriction s>0 is required for the integral to exist.(ii) Again using (13.53) directly, we find∫ ∞∫ ∞¯f(s) = e at e −st dt = e (a−s)t dt00[ ] e(a−s)t ∞= = 1 if s>a.a − s0s − a453

INTEGRAL TRANSFORMS(iii) Once again using the definition (13.53) we have∫ ∞¯f n (s) = t n e −st dt.0Integrating by parts we find[ ] −t¯f n e −st ∞n (s) =+ n ∫ ∞t n−1 e −st dts0s 0=0+ n s ¯f n−1 (s), if s>0.We now have a recursion relation between successive transforms and by calculating¯f 0 we can infer ¯f 1 , ¯f2 ,etc.Sincet 0 = 1, (i) above gives¯f 0 = 1 , if s>0, (13.55)sand¯f 1 (s) = 1 s , 2 ¯f2 (s) = 2!s , ..., 3 ¯fn (s) = n! if s>0.s n+1Thus, in each case (i)–(iii), direct application of the definition of the Laplace transform(13.53) yields the required result. ◭Unlike that for the Fourier transform, the inversion of the Laplace transformis not an easy operation to perform, since an explicit formula for f(t), given ¯f(s),is not straightforwardly obtained from (13.53). The general method for obtainingan inverse Laplace transform makes use of complex variable theory and is notdiscussed until chapter 25. However, progress can be made without having to findan explicit inverse, since we can prepare from (13.53) a ‘dictionary’ of the Laplacetransforms of common functions and, when faced with an inversion to carry out,hope to find the given transform (together with its parent function) in the listing.Such a list is given in table 13.1.When finding inverse Laplace transforms using table 13.1, it is useful to notethat for all practical purposes the inverse Laplace transform is unique § and linearso thatL −1[ a ¯f 1 (s)+b ¯f 2 (s) ] = af 1 (t)+bf 2 (t). (13.56)In many practical problems the method of partial fractions can be useful inproducing an expression from which the inverse Laplace transform can be found.◮Using table 13.1 find f(t) if¯f(s) = s +3s(s +1) .Using partial fractions ¯f(s) may be written¯f(s) = 3 s − 2s +1 .§ This is not strictly true, since two functions can differ from one another at a finite number ofisolated points but have the same Laplace transform.454

13.2 LAPLACE TRANSFORMSf(t) ¯f(s) s0c c/s 0ct n cn!/s n+1 0sin bt b/(s 2 + b 2 ) 0cos bt s/(s 2 + b 2 ) 0e at 1/(s − a) at n e at n!/(s − a) n+1 asinh at a/(s 2 − a 2 ) |a|cosh at s/(s 2 − a 2 ) |a|e at sin bt b/[(s − a) 2 + b 2 ] ae at cos bt (s − a)/[(s − a) 2 + b 2 ] at 1/2 1 2 (π/s3 ) 1/2 0t −1/2 (π/s) 1/2 0δ(t − t 0 ) e −st 0 0{1 for t ≥ t 0H(t − t 0 )=e −st 0/s 00 for ts 0 .Comparing this with the standard Laplace transforms in table 13.1, we find that the inversetransform of 3/s is 3 for s>0 and the inverse transform of 2/(s +1)is 2e −t for s>−1,and sof(t) =3− 2e −t , if s>0. ◭13.2.1 Laplace transforms of derivatives and integralsOne of the main uses of Laplace transforms is in solving differential equations.Differential equations are the subject of the next six chapters and we will returnto the application of Laplace transforms to their solution in chapter 15. Inthe meantime we will derive the required results, i.e. the Laplace transforms ofderivatives.The Laplace transform of the first derivative of f(t) is given byL[ dfdt]=∫ ∞0dfdt e−st dt= [ f(t)e −st] ∫ ∞∞0 + s f(t)e −st dt0= −f(0) + s ¯f(s), for s>0. (13.57)The evaluation relies on integration by parts and higher-order derivatives maybe found in a similar manner.455

INTEGRAL TRANSFORMS◮Find the Laplace transform of d 2 f/dt 2 .Using the definition of the Laplace transform and integrating by parts we obtain[ ] dL2 ∫f ∞d 2 f=dt 2 dt 2 e−st dt0[ ] ∞ ∫ df∞=dt e−st + s0 0= − dfdt (0) + s[s ¯f(s) − f(0)],dfdt e−st dtfor s>0,where (13.57) has been substituted for the integral. This can be written more neatly as[ ] dL2 f= s 2 df¯f(s) − sf(0) −dt 2 dt (0), for s>0. ◭In general the Laplace transform of the nth derivative is given by[ d n ]fLdt n = s n ¯f − s n−1 n−2 dff(0) − sdt (0) − ···− dn−1 f(0), for s>0.dtn−1 (13.58)We now turn to integration, which is much more straightforward. From thedefinition (13.53),[∫ t] ∫ ∞ ∫ tL f(u) du = dt e −st f(u) du000=[− 1 ∫ t] ∞ ∫ ∞1s e−st f(u) du +00 0 s e−st f(t) dt.The first term on the RHS vanishes at both limits, and so[∫ t]L f(u) du = 1 L [f] . (13.59)s013.2.2 Other properties of Laplace transformsFrom table 13.1 it will be apparent that multiplying a function f(t) bye at has theeffect on its transform that s is replaced by s − a. This is easily proved generally:L [ e at f(t) ] ==∫ ∞0∫ ∞0f(t)e at e −st dtf(t)e −(s−a)t dt= ¯f(s − a). (13.60)As it were, multiplying f(t) bye at moves the origin of s by an amount a.456

13.2 LAPLACE TRANSFORMSWe may now consider the effect of multiplying the Laplace transform ¯f(s) bye −bs (b>0). From the definition (13.53),∫ ∞e −bs ¯f(s) = e −s(t+b) f(t) dt=0∫ ∞0e −sz f(z − b) dz,on putting t + b = z. Thus e −bs ¯f(s) is the Laplace transform of a function g(t)defined by{0 for 0 b.In other words, the function f has been translated to ‘later’ t (larger values of t)by an amount b.Further properties of Laplace transforms can be proved in similar ways andare listed below.(i)L [f(at)] = 1 a ¯f( s, (13.61)a)(ii)(iii)L [t n n dn ¯f(s)f(t)] =(−1)ds n , for n =1, 2, 3,..., (13.62)[ ] ∫ f(t) ∞L = ¯f(u) du, (13.63)t sprovided lim t→0 [f(t)/t] exists.Related results may be easily proved.◮Find an expression for the Laplace transform of td 2 f/dt 2 .From the definition of the Laplace transform we have[ ] ∫L t d2 f ∞= e −st t d2 fdt 2 dt dt 20∫ ∞= − d e −st d2 fds 0 dt dt 2= − d ds [s2 ¯f(s) − sf(0) − f ′ (0)]= −s 2 d ¯fds − 2s ¯f + f(0). ◭Finally we mention the convolution theorem for Laplace transforms (which isanalogous to that for Fourier transforms discussed in subsection 13.1.7). If thefunctions f and g have Laplace transforms ¯f(s) andḡ(s) then[∫ t]L f(u)g(t − u) du = ¯f(s)ḡ(s), (13.64)0457

INTEGRAL TRANSFORMSFigure 13.7text).Two representations of the Laplace transform convolution (seewhere the integral in the brackets on the LHS is the convolution of f and g,denoted by f ∗ g. As in the case of Fourier transforms, the convolution definedabove is commutative, i.e. f ∗ g = g ∗ f, and is associative and distributive. From(13.64) we also see thatL −1[ ∫] t¯f(s)ḡ(s) = f(u)g(t − u) du = f ∗ g.0◮Prove the convolution theorem (13.64) for Laplace transforms.From the definition (13.64),¯f(s)ḡ(s) ==∫ ∞0∫ ∞0e −su f(u) dudu∫ ∞0∫ ∞0e −sv g(v) dvdv e −s(u+v) f(u)g(v).Now letting u + v = t changes the limits on the integrals, with the result that¯f(s)ḡ(s) =∫ ∞0du f(u)∫ ∞udt g(t − u) e −st .As shown in figure 13.7(a) the shaded area of integration may be considered as the sumof vertical strips. However, we may instead integrate over this area by summing overhorizontal strips as shown in figure 13.7(b). Then the integral can be written as¯f(s)ḡ(s) ==∫ t0∫ ∞0∫ ∞du f(u) dt g(t − u) e −st0{∫ t}dt e −st f(u)g(t − u) du[∫ t]= L f(u)g(t − u) du . ◭04580

13.3 CONCLUDING REMARKSThe properties of the Laplace transform derived in this section can sometimesbe useful in finding the Laplace transforms of particular functions.◮Find the Laplace transform of f(t) =t sin bt.Although we could calculate the Laplace transform directly, we can use (13.62) to give¯f(s) =(−1) d ds L [sin bt] = − d ( ) b 2bs= , for s>0. ◭ds s 2 + b 2 (s 2 + b 2 )213.3 Concluding remarksIn this chapter we have discussed Fourier and Laplace transforms in some detail.Both are examples of integral transforms, which can be considered in a moregeneral context.A general integral transform of a function f(t) takes the formF(α) =∫ baK(α, t)f(t) dt, (13.65)where F(α) is the transform of f(t) with respect to the kernel K(α, t), and α isthe transform variable. For example, in the Laplace transform case K(s, t) =e −st ,a =0,b = ∞.Very often the inverse transform can also be written straightforwardly andwe obtain a transform pair similar to that encountered in Fourier transforms.Examples of such pairs are(i) the Hankel transformF(k) =f(x) =∫ ∞0∫ ∞0f(x)J n (kx)xdx,F(k)J n (kx)kdk,where the J n are Bessel functions of order n, and(ii) the Mellin transformF(z) =∫ ∞0f(t) = 12πit z−1 f(t) dt,∫ i∞−i∞t −z F(z) dz.Although we do not have the space to discuss their general properties, thereader should at least be aware of this wider class of integral transforms.459

INTEGRAL TRANSFORMS13.4 Exercises13.1 Find the Fourier transform of the function f(t) =exp(−|t|).(a) By applying Fourier’s inversion theorem prove that∫π∞2 exp(−|t|) = cos ωt0 1+ω dω. 2(b) By making the substitution ω =tanθ, demonstrate the validity of Parseval’stheorem for this function.13.2 Use the general definition and properties of Fourier transforms to show thefollowing.(a) If f(x) is periodic with period a then ˜f(k) = 0, unless ka =2πn for integer n.(b) The Fourier transform of tf(t) isid˜f(ω)/dω.(c) The Fourier transform of f(mt + c) ise iωc/mm ˜f( ω).m13.3 Find the Fourier transform of H(x−a)e −bx ,whereH(x) is the Heaviside function.13.4 Prove that the Fourier transform of the function f(t) defined in the tf-plane bystraight-line segments joining (−T,0) to (0, 1) to (T,0), with f(t) = 0 outside|t|

13.4 EXERCISESDetermine the convolution of f with itself and, without further integration,deduce its transform. Deduce that∫ ∞sin 2 ωdω = π,ω 2−∞∫ ∞−∞sin 4 ωω 4 dω = 2π 3 .13.8 Calculate the Fraunhofer spectrum produced by a diffraction grating, uniformlyilluminated by light of wavelength 2π/k, as follows. Consider a grating with 4Nequal strips each of width a and alternately opaque and transparent. The aperturefunction is then{A for (2n +1)a ≤ y ≤ (2n +2)a, −N ≤ n

INTEGRAL TRANSFORMS13.10 In many applications in which the frequency spectrum of an analogue signal isrequired, the best that can be done is to sample the signal f(t) a finite number oftimes at fixed intervals, and then use a discrete Fourier transform F k to estimatediscrete points on the (true) frequency spectrum ˜f(ω).(a) By an argument that is essentially the converse of that given in section 13.1,show that, if N samples f n , beginning at t = 0 and spaced τ apart, are taken,then ˜f(2πk/(Nτ)) ≈ F k τ whereF k = √ 1 ∑N−1f n e −2πnki/N .2π(b) For the function f(t) defined by{1 for 0 ≤ t

13.4 EXERCISES(a) Find the Fourier transform off(γ, p, t) ={e −γt sin pt t > 0,0 t 0) and p are constant parameters.(b) The current I(t) flowing through a certain system is related to the appliedvoltage V (t) by the equationI(t) =∫ ∞−∞K(t − u)V (u) du,whereK(τ) =a 1 f(γ 1 ,p 1 ,τ)+a 2 f(γ 2 ,p 2 ,τ).The function f(γ, p, t) is as given in (a) and all the a i ,γ i (> 0) and p i are fixedparameters. By considering the Fourier transform of I(t), find the relationshipthat must hold between a 1 and a 2 if the total net charge Q passed throughthe system (over a very long time) is to be zero for an arbitrary appliedvoltage.13.14 Prove the equality∫ ∞e −2at sin 2 at dt = 1 ∫ ∞a 20π 0 4a 4 + ω dω. 413.15 A linear amplifier produces an output that is the convolution of its input and itsresponse function. The Fourier transform of the response function for a particularamplifier isiω˜K(ω) = √ .2π(α + iω)2Determine the time variation of its output g(t) when its input is the Heavisidestep function. (Consider the Fourier transform of a decaying exponential functionand the result of exercise 13.2(b).)13.16 In quantum mechanics, two equal-mass particles having momenta p j = k j andenergies E j = ω j and represented by plane wavefunctions φ j =exp[i(k j·r j −ω j t)],j =1, 2, interact through a potential V = V (|r 1 − r 2 |). In first-order perturbationtheory the probability of scattering to a state with momenta and energies p ′ j ,E′ jis determined by the modulus squared of the quantity∫∫∫M = ψf ∗ Vψ i dr 1 dr 2 dt.The initial state, ψ i ,isφ 1 φ 2 and the final state, ψ f ,isφ ′ 1 φ′ 2 .(a) By writing r 1 + r 2 =2R and r 1 − r 2 = r and assuming that dr 1 dr 2 = dR dr,show that M can be written as the product of three one-dimensional integrals.(b) From two of the integrals deduce energy and momentum conservation in theform of δ-functions.(c) Show that M is proportional to the Fourier transform of V ,i.e.toṼ (k)where 2k =(p 2 − p 1 ) − (p ′ 2 − p′ 1 ) or, alternatively, k = p′ 1 − p 1.13.17 For some ion–atom scattering processes, the potential V of the previous exercisemay be approximated by V = |r 1 − r 2 | −1 exp(−µ|r 1 − r 2 |). Show, using the resultof the worked example in subsection 13.1.10, that the probability that the ionwill scatter from, say, p 1 to p ′ 1 is proportional to (µ2 + k 2 ) −2 ,wherek = |k| and kis as given in part (c) of that exercise.463

INTEGRAL TRANSFORMS13.18 The equivalent duration and bandwidth, T e and B e , of a signal x(t) are definedin terms of the latter and its Fourier transform ˜x(ω) byT e = 1 ∫ ∞x(t) dt,x(0) −∞B e = 1 ∫ ∞˜x(ω) dω,˜x(0) −∞where neither x(0) nor ˜x(0) is zero. Show that the product T e B e =2π (this is aform of uncertainty principle), and find the equivalent bandwidth of the signalx(t) =exp(−|t|/T ).For this signal, determine the fraction of the total energy that lies in the frequencyrange |ω| |a|;2464

13.4 EXERCISES(c) L [sinh at cos bt] = a(s 2 − a 2 + b 2 )[(s − a) 2 + b 2 ] −1 [(s + a) 2 + b 2 ] −1 .13.24 Find the solution (the so-called impulse response or Green’s function) of theequationT dxdt + x = δ(t)by proceeding as follows.(a) Show by substitution thatx(t) =A(1 − e −t/T )H(t)is a solution, for which x(0) = 0, ofT dxdt + x = AH(t), (∗)where H(t) is the Heaviside step function.(b) Construct the solution when the RHS of (∗) is replaced by AH(t − τ), withdx/dt = x =0fort0, the function y(t) obeys the differential equationd 2 ydt + a dy2 dt + by = c cos2 ωt,where a, b and c are positive constants. Find ȳ(s) andshowthatsȳ(s) → c/2bas s → 0. Interpret the result in the t-domain.13.26 By writing f(x) as an integral involving the δ-function δ(ξ − x) and taking theLaplace transforms of both sides, show that the transform of the solution of theequationd 4 ydx − y = f(x)4for which y and its first three derivatives vanish at x = 0 can be written as∫ ∞ȳ(s) = f(ξ) e−sξ0 s 4 − 1 dξ.Use the properties of Laplace transforms and the entries in table 13.1 to showthaty(x) = 1 ∫ xf(ξ) [sinh(x − ξ) − sin(x − ξ)] dξ.20465

INTEGRAL TRANSFORMS13.27 The function f a (x) is defined as unity for 0

13.5 HINTS AND ANSWERS13.17 Ṽ (k) ∝ [−2π/(ik)] ∫ {exp[−(µ − ik)r] − exp[−(µ + ik)r]} dr.13.19 Note that the lower limit in the calculation of a(z) is0,forz>0, and |z|, forz

14First-order ordinary differentialequationsDifferential equations are the group of equations that contain derivatives. Chapters14–21 discuss a variety of differential equations, starting in this chapter andthe next with those ordinary differential equations (ODEs) that have closed-formsolutions. As its name suggests, an ODE contains only ordinary derivatives (nopartial derivatives) and describes the relationship between these derivatives ofthe dependent variable, usually called y, with respect to the independent variable,usually called x. The solution to such an ODE is therefore a function of x andis written y(x). For an ODE to have a closed-form solution, it must be possibleto express y(x) in terms of the standard elementary functions such as exp x, lnx,sin x etc. The solutions of some differential equations cannot, however, be writtenin closed form, but only as an infinite series; these are discussed in chapter 16.Ordinary differential equations may be separated conveniently into differentcategories according to their general characteristics. The primary groupingadopted here is by the order of the equation. The order of an ODE is simply theorder of the highest derivative it contains. Thus equations containing dy/dx, butno higher derivatives, are called first order, those containing d 2 y/dx 2 are calledsecond order and so on. In this chapter we consider first-order equations, and inthe next, second- and higher-order equations.Ordinary differential equations may be classified further according to degree.The degree of an ODE is the power to which the highest-order derivative israised, after the equation has been rationalised to contain only integer powers ofderivatives. Hence the ODEd 3 ( )y dy 3/2dx 3 + x + x 2 y =0,dxis of third order and second degree, since after rationalisation it contains the term(d 3 y/dx 3 ) 2 .The general solution to an ODE is the most general function y(x) that satisfiesthe equation; it will contain constants of integration which may be determined by468

14.1 GENERAL FORM OF SOLUTIONthe application of some suitable boundary conditions. For example, we may betold that for a certain first-order differential equation, the solution y(x) isequaltozero when the parameter x is equal to unity; this allows us to determine the valueof the constant of integration. The general solutions to nth-order ODEs, whichare considered in detail in the next chapter, will contain n (essential) arbitraryconstants of integration and therefore we will need n boundary conditions if theseconstants are to be determined (see section 14.1). When the boundary conditionshave been applied, and the constants found, we are left with a particular solutionto the ODE, which obeys the given boundary conditions. Some ODEs of degreegreater than unity also possess singular solutions, which are solutions that containno arbitrary constants and cannot be found from the general solution; singularsolutions are discussed in more detail in section 14.3. When any solution to anODE has been found, it is always possible to check its validity by substitutioninto the original equation and verification that any given boundary conditionsare met.In this chapter, firstly we discuss various types of first-degree ODE and then goon to examine those higher-degree equations that can be solved in closed form.At the outset, however, we discuss the general form of the solutions of ODEs;this discussion is relevant to both first- and higher-order ODEs.14.1 General form of solutionIt is helpful when considering the general form of the solution of an ODE toconsider the inverse process, namely that of obtaining an ODE from a givengroup of functions, each one of which is a solution of the ODE. Suppose themembers of the group can be written asy = f(x, a 1 ,a 2 ,...,a n ), (14.1)each member being specified by a different set of values of the parameters a i .Forexample, consider the group of functionsy = a 1 sin x + a 2 cos x; (14.2)here n =2.Since an ODE is required for which any of the group is a solution, it clearlymust not contain any of the a i . As there are n of the a i in expression (14.1), wemust obtain n + 1 equations involving them in order that, by elimination, we canobtain one final equation without them.Initially we have only (14.1), but if this is differentiated n times, a total of n +1equations is obtained from which (in principle) all the a i can be eliminated, togive one ODE satisfied by all the group. As a result of the n differentiations,d n y/dx n will be present in one of the n + 1 equations and hence in the finalequation, which will therefore be of nth order.469

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSIn the case of (14.2), we havedydx = a 1 cos x − a 2 sin x,d 2 ydx 2 = −a 1 sin x − a 2 cos x.Here the elimination of a 1 and a 2 is trivial (because of the similarity of the formsof y and d 2 y/dx 2 ), resulting ind 2 ydx 2 + y =0,a second-order equation.Thus, to summarise, a group of functions (14.1) with n parameters satisfies annth-order ODE in general (although in some degenerate cases an ODE of lessthan nth order is obtained). The intuitive converse of this is that the generalsolution of an nth-order ODE contains n arbitrary parameters (constants); forour purposes, this will be assumed to be valid although a totally general proof isdifficult.As mentioned earlier, external factors affect a system described by an ODE,by fixing the values of the dependent variables for particular values of theindependent ones. These externally imposed (or boundary) conditions on thesolution are thus the means of determining the parameters and so of specifyingprecisely which function is the required solution. It is apparent that the numberof boundary conditions should match the number of parameters and hence theorder of the equation, if a unique solution is to be obtained. Fewer independentboundary conditions than this will lead to a number of undetermined parametersin the solution, whilst an excess will usually mean that no acceptable solution ispossible.For an nth-order equation the required n boundary conditions can take manyforms, for example the value of y at n different values of x, or the value of anyn − 1ofthen derivatives dy/dx, d 2 y/dx 2 , ..., d n y/dx n together with that of y, allfor the same value of x, or many intermediate combinations.14.2 First-degree first-order equationsFirst-degree first-order ODEs contain only dy/dx equated to some function of xand y, and can be written in either of two equivalent standard forms,dy= F(x, y), A(x, y) dx + B(x, y) dy =0,dxwhere F(x, y) =−A(x, y)/B(x, y), and F(x, y), A(x, y) andB(x, y) are in generalfunctions of both x and y. Which of the two above forms is the more usefulfor finding a solution depends on the type of equation being considered. There470

14.2 FIRST-DEGREE FIRST-ORDER EQUATIONSare several different types of first-degree first-order ODEs that are of interest inthe physical sciences. These equations and their respective solutions are discussedbelow.14.2.1 Separable-variable equationsA separable-variable equation is one which may be written in the conventionalformdy= f(x)g(y), (14.3)dxwhere f(x) andg(y) are functions of x and y respectively, including cases inwhich f(x) org(y) is simply a constant. Rearranging this equation so that theterms depending on x and on y appear on opposite sides (i.e. are separated), andintegrating, we obtain∫ ∫ dyg(y) = f(x) dx.Finding the solution y(x) that satisfies (14.3) then depends only on the ease withwhich the integrals in the above equation can be evaluated. It is also worthnoting that ODEs that at first sight do not appear to be of the form (14.3) cansometimes be made separable by an appropriate factorisation.◮Solvedydx = x + xy.Since the RHS of this equation can be factorised to give x(1 + y), the equation becomesseparable and we obtain∫ ∫dy1+y = xdx.Now integrating both sides separately, we findln(1 + y) = x22 + c,and so( ) x21+y =exp2 + c = A expwhere c and hence A is an arbitrary constant. ◭Solution method. Factorise the equation so that it becomes separable. Rearrangeit so that the terms depending on x and those depending on y appear on oppositesides and then integrate directly. Remember the constant of integration, which canbe evaluated if further information is given.( x22),471

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS14.2.2 Exact equationsAn exact first-degree first-order ODE is one of the formA(x, y) dx + B(x, y) dy = 0 and for which ∂A∂y = ∂B∂x . (14.4)In this case A(x, y) dx + B(x, y) dy is an exact differential, dU(x, y) say (seesection 5.3). In other wordsAdx+ Bdy= dU = ∂U ∂Udx +∂x ∂y dy,from which we obtainA(x, y) = ∂U∂x , (14.5)B(x, y) = ∂U∂y . (14.6)Since ∂ 2 U/∂x∂y = ∂ 2 U/∂y∂x we therefore require∂A∂y = ∂B∂x . (14.7)If (14.7) holds then (14.4) can be written dU(x, y) = 0, which has the solutionU(x, y) =c, wherec is a constant and from (14.5) U(x, y) is given by∫U(x, y) = A(x, y) dx + F(y). (14.8)The function F(y) can be found from (14.6) by differentiating (14.8) with respectto y and equating to B(x, y).◮Solvex dy +3x + y =0.dxRearranging into the form (14.4) we have(3x + y) dx + xdy =0,i.e. A(x, y) =3x + y and B(x, y) =x. Since∂A/∂y =1=∂B/∂x, the equation is exact, andby (14.8) the solution is given by∫U(x, y) = (3x + y) dx + F(y) =c 1 ⇒ 3x22 + yx + F(y) =c 1.Differentiating U(x, y) with respect to y and equating it to B(x, y) =x we obtain dF/dy =0,which integrates immediately to give F(y) =c 2 . Therefore, letting c = c 1 − c 2 ,thesolutionto the original ODE is3x 2+ xy = c. ◭2472

14.2 FIRST-DEGREE FIRST-ORDER EQUATIONSSolution method. Check that the equation is an exact differential using (14.7) thensolve using (14.8). Find the function F(y) by differentiating (14.8) with respect toy and using (14.6).14.2.3 Inexact equations: integrating factorsEquations that may be written in the formA(x, y) dx + B(x, y) dy = 0 but for which ∂A∂y ≠ ∂B∂x(14.9)are known as inexact equations. However, the differential Adx+ Bdy can alwaysbe made exact by multiplying by an integrating factor µ(x, y), which obeys∂(µA)∂y= ∂(µB)∂x . (14.10)For an integrating factor that is a function of both x and y, i.e.µ = µ(x, y), thereexists no general method for finding it; in such cases it may sometimes be foundby inspection. If, however, an integrating factor exists that is a function of eitherx or y alone then (14.10) can be solved to find it. For example, if we assumethat the integrating factor is a function of x alone, i.e. µ = µ(x), then (14.10)readsµ ∂A∂y = µ∂B ∂x + B dµdx .Rearranging this expression we finddµµ = 1 ( ∂AB ∂y − ∂B )dx = f(x) dx,∂xwhere we require f(x) also to be a function of x only; indeed this provides ageneral method of determining whether the integrating factor µ is a function ofx alone. This integrating factor is then given by{∫µ(x) =exp}f(x) dxwheref(x) = 1 B( ∂A∂y − ∂B ). (14.11)∂xSimilarly, if µ = µ(y) then{∫µ(y) =exp}g(y) dywhere473g(y) = 1 A( ∂B∂x − ∂A ). (14.12)∂y

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Solvedydx = − 2 y − 3y2x .Rearranging into the form (14.9), we have(4x +3y 2 ) dx +2xy dy =0, (14.13)i.e. A(x, y) =4x +3y 2 and B(x, y) =2xy. Now∂A∂y =6y, ∂B∂x =2y,so the ODE is not exact in its present form. However, we see that(1 ∂AB ∂y − ∂B )= 2 ∂x x ,a function of x alone. Therefore an integrating factor exists that is also a function of xalone and, ignoring the arbitrary constant of integration, is given by{ ∫ } dxµ(x) =exp 2 =exp(2lnx) =x 2 .xMultiplying (14.13) through by µ(x) =x 2 we obtain(4x 3 +3x 2 y 2 ) dx +2x 3 ydy=4x 3 dx +(3x 2 y 2 dx +2x 3 ydy)=0.By inspection this integrates immediately to give the solution x 4 + y 2 x 3 = c, wherec is aconstant. ◭Solution method. Examine whether f(x) and g(y) are functions of only x or yrespectively. If so, then the required integrating factor is a function of either x ory only, and is given by (14.11) or (14.12) respectively. If the integrating factor isa function of both x and y, then sometimes it may be found by inspection or bytrial and error. In any case, the integrating factor µ must satisfy (14.10). Once theequation has been made exact, solve by the method of subsection 14.2.2.14.2.4 Linear equationsLinear first-order ODEs are a special case of inexact ODEs (discussed in theprevious subsection) and can be written in the conventional formdy+ P (x)y = Q(x). (14.14)dxSuch equations can be made exact by multiplying through by an appropriateintegrating factor in a similar manner to that discussed above. In this case,however, the integrating factor is always a function of x alone and may beexpressed in a particularly simple form. An integrating factor µ(x) must be suchthatµ(x) dyd+ µ(x)P (x)y = [µ(x)y] = µ(x)Q(x), (14.15)dx dx474

14.2 FIRST-DEGREE FIRST-ORDER EQUATIONSwhich may then be integrated directly to give∫µ(x)y = µ(x)Q(x) dx. (14.16)The required integrating factor µ(x) is determined by the first equality in (14.15),i.e.d dy(µy) =µdx dx + dµdx y = µ dydx + µPy,which immediately gives the simple relation{∫dµdx = µ(x)P (x) ⇒ µ(x) =exp}P (x) dx . (14.17)◮Solvedy+2xy =4x.dxThe integrating factor is given immediately by{∫ }µ(x) =exp 2xdx =expx 2 .Multiplying through the ODE by µ(x) =expx 2 and integrating, we have∫y exp x 2 =4 x exp x 2 dx =2expx 2 + c.The solution to the ODE is therefore given by y =2+c exp(−x 2 ). ◭Solution method. Rearrange the equation into the form (14.14) and multiply by theintegrating factor µ(x) given by (14.17). The left- and right-hand sides can then beintegrated directly, giving y from (14.16).14.2.5 Homogeneous equationsHomogeneous equation are ODEs that may be written in the formdy A(x, y)( y)=dx B(x, y) = F , (14.18)xwhere A(x, y) and B(x, y) are homogeneous functions of the same degree. Afunction f(x, y) is homogeneous of degree n if, for any λ, itobeysf(λx, λy) =λ n f(x, y).For example, if A = x 2 y − xy 2 and B = x 3 + y 3 then we see that A and B areboth homogeneous functions of degree 3. In general, for functions of the form ofA and B, we see that for both to be homogeneous, and of the same degree, werequire the sum of the powers in x and y in each term of A and B to be the same475

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS(in this example equal to 3). The RHS of a homogeneous ODE can be written asa function of y/x. The equation may then be solved by making the substitutiony = vx, sothatdydx = v + x dvdx = F(v).This is now a separable equation and can be integrated directly to give∫∫dv dxF(v) − v = x . (14.19)◮Solvedydx = y ( y)x +tan .xSubstituting y = vx we obtainv + x dvdx = v +tanv.Cancelling v on both sides, rearranging and integrating gives∫∫ dxcot vdv=x =lnx + c 1.But∫∫cot vdv=cos vsin v dv =ln(sinv)+c 2,so the solution to the ODE is y = x sin −1 Ax, whereA is a constant. ◭Solution method. Check to see whether the equation is homogeneous. If so, makethe substitution y = vx, separate variables as in (14.19) and then integrate directly.Finally replace v by y/x to obtain the solution.14.2.6 Isobaric equationsAn isobaric ODE is a generalisation of the homogeneous ODE discussed in theprevious section, and is of the formdy A(x, y)=dx B(x, y) , (14.20)where the equation is dimensionally consistent if y and dy are each given a weightm relative to x and dx, i.e. if the substitution y = vx m makes it separable.476

14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS◮Solvedydx = −1 (y 2 + 2 ).2yx xRearranging we have(y 2 + 2 )dx +2yx dy =0.xGiving y and dy the weight m and x and dx the weight 1, the sums of the powers in eachterm on the LHS are 2m +1, 0 and2m + 1 respectively. These are equal if 2m +1=0, i.e.if m = − 1 2 . Substituting y = vxm = vx −1/2 , with the result that dy = x −1/2 dv − 1 2 vx−3/2 dx,we obtainvdv+ dx x =0,which is separable and may be integrated directly to give 1 2 v2 +lnx = c. Replacingv byy √ x we obtain the solution 1 2 y2 x +lnx = c. ◭Solution method. Write the equation in the form Adx+ Bdy =0. Giving y anddy each a weight m and x and dx each a weight 1, write down the sum of powersin each term. Then, if a value of m that makes all these sums equal can be found,substitute y = vx m into the original equation to make it separable. Integrate theseparated equation directly, and then replace v by yx −m to obtain the solution.Bernoulli’s equation has the form14.2.7 Bernoulli’s equationdydx + P (x)y = Q(x)yn where n ≠0or1. (14.21)This equation is very similar in form to the linear equation (14.14), but is in factnon-linear due to the extra y n factor on the RHS. However, the equation can bemade linear by substituting v = y 1−n and correspondingly( )dy yn dvdx = 1 − n dx .Substituting this into (14.21) and dividing through by y n , we finddv+(1− n)P (x)v =(1− n)Q(x),dxwhich is a linear equation and may be solved by the method described insubsection 14.2.4.477

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Solvedydx + y x =2x3 y 4 .If we let v = y 1−4 = y −3 thendydx = − y4 dv3 dx .Substituting this into the ODE and rearranging, we obtaindvdx − 3vx = −6x3 ,which is linear and may be solved by multiplying through by the integrating factor (seesubsection 14.2.4){ ∫exp −3dxx}=exp(−3lnx) = 1 x 3 .This yields the solutionv= −6x + c.x3 Remembering that v = y −3 ,weobtainy −3 = −6x 4 + cx 3 . ◭Solution method. Rearrange the equation into the form (14.21) and make the substitutionv = y 1−n . This leads to a linear equation in v, which can be solved by themethod of subsection 14.2.4. Then replace v by y 1−n to obtain the solution.14.2.8 Miscellaneous equationsThere are two further types of first-degree first-order equation that occur fairlyregularly but do not fall into any of the above categories. They may be reducedto one of the above equations, however, by a suitable change of variable.Firstly, we considerdy= F(ax + by + c), (14.22)dxwhere a, b and c are constants, i.e. x and y only appear on the RHS in the particularcombination ax + by + c and not in any other combination or by themselves. Thisequation can be solved by making the substitution v = ax + by + c, inwhichcasedvdx = a + b dy = a + bF(v), (14.23)dxwhich is separable and may be integrated directly.478

14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS◮Solvedydx =(x + y +1)2 .Making the substitution v = x + y + 1, we obtain, as in (14.23),dvdx = v2 +1,which is separable and integrates directly to give∫ ∫dv1+v = dx ⇒ tan −1 v = x + c 2 1 .So the solution to the original ODE is tan −1 (x + y +1)=x + c 1 ,wherec 1 is a constant ofintegration. ◭Solution method. In an equation such as (14.22), substitute v = ax+by+c to obtaina separable equation that can be integrated directly. Then replace v by ax + by + cto obtain the solution.Secondly, we discussdy ax + by + c=dx ex + fy + g , (14.24)where a, b, c, e, f and g are all constants. This equation may be solved by lettingx = X + α and y = Y + β, whereα and β are constants found fromaα + bβ + c = 0 (14.25)eα + fβ + g =0. (14.26)Then (14.24) can be written asdY aX + bY=dX eX + fY ,which is homogeneous and can be solved by the method of subsection 14.2.5.Note, however, that if a/e = b/f then (14.25) and (14.26) are not independentand so cannot be solved uniquely for α and β. However, in this case, (14.24)reduces to an equation of the form (14.22), which was discussed above.◮Solvedy 2x − 5y +3=dx 2x +4y − 6 .Let x = X + α and y = Y + β, whereα and β obey the relations2α − 5β +3=02α +4β − 6=0,which solve to give α = β = 1. Making these substitutions we finddY 2X − 5Y=dX 2X +4Y ,479

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSwhich is a homogeneous ODE and can be solved by substituting Y = vX (see subsection14.2.5) to obtaindvdX2 − 7v − 4v2= .X(2 + 4v)This equation is separable, and using partial fractions we find∫2+4v2 − 7v − 4v dv = − 4 ∫dv2 3 4v − 1 − 2 ∫ ∫dv dX3 v +2 = X ,which integrates to giveln X + 1 ln(4v − 1) + 2 ln(v +2)=c 3 3 1,orX 3 (4v − 1)(v +2) 2 =exp3c 1 .Remembering that Y = vX, x = X +1 andy = Y + 1, the solution to the original ODEis given by (4y − x − 3)(y +2x − 3) 2 = c 2 ,wherec 2 =exp3c 1 . ◭Solution method. If in (14.24) a/e ≠ b/f then make the substitution x = X + α,y = Y + β, whereα and β are given by (14.25) and (14.26); the resulting equationis homogeneous and can be solved as in subsection 14.2.5. Substitute v = Y/X,X = x − α and Y = y − β to obtain the solution. If a/e = b/f then (14.24) is ofthe same form as (14.22) and may be solved accordingly.14.3 Higher-degree first-order equationsFirst-order equations of degree higher than the first do not occur often inthe description of physical systems, since squared and higher powers of firstorderderivatives usually arise from resistive or driving mechanisms, when anacceleration or other higher-order derivative is also present. They do sometimesappear in connection with geometrical problems, however.Higher-degree first-order equations can be written as F(x, y, dy/dx) =0.Themost general standard form isp n + a n−1 (x, y)p n−1 + ···+ a 1 (x, y)p + a 0 (x, y) =0, (14.27)where for ease of notation we write p = dy/dx. If the equation can be solved forone of x, y or p then either an explicit or a parametric solution can sometimes beobtained. We discuss the main types of such equations below, including Clairaut’sequation, which is a special case of an equation explicitly soluble for y.14.3.1 Equations soluble for pSometimes the LHS of (14.27) can be factorised into the form(p − F 1 )(p − F 2 ) ···(p − F n )=0, (14.28)480

14.3 HIGHER-DEGREE FIRST-ORDER EQUATIONSwhere F i = F i (x, y). We are then left with solving the n first-degree equationsp = F i (x, y). Writing the solutions to these first-degree equations as G i (x, y) =0,the general solution to (14.28) is given by the productG 1 (x, y)G 2 (x, y) ···G n (x, y) =0. (14.29)◮Solve(x 3 + x 2 + x +1)p 2 − (3x 2 +2x +1)yp +2xy 2 =0. (14.30)This equation may be factorised to give[(x +1)p − y][(x 2 +1)p − 2xy] =0.Taking each bracket in turn we have(x +1) dydx − y =0,(x 2 +1) dy − 2xy =0,dxwhich have the solutions y − c(x +1) = 0 and y − c(x 2 + 1) = 0 respectively (seesection 14.2 on first-degree first-order equations). Note that the arbitrary constants inthese two solutions can be taken to be the same, since only one is required for a first-orderequation. The general solution to (14.30) is then given by[y − c(x +1)] [ y − c(x 2 +1) ] =0. ◭Solution method. If the equation can be factorised into the form (14.28) then solvethe first-order ODE p − F i =0for each factor and write the solution in the formG i (x, y) =0. The solution to the original equation is then given by the product(14.29).14.3.2 Equations soluble for xEquations that can be solved for x, i.e. such that they may be written in the formx = F(y, p), (14.31)can be reduced to first-degree first-order equations in p by differentiating bothsides with respect to y, sothatdxdy = 1 p = ∂F∂y + ∂F dp∂p dy .This results in an equation of the form G(y, p) = 0, which can be used togetherwith (14.31) to eliminate p and give the general solution. Note that often a singularsolution to the equation will be found at the same time (see the introduction tothis chapter).481

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Solve6y 2 p 2 +3xp − y =0. (14.32)This equation can be solved for x explicitly to give 3x =(y/p) − 6y 2 p. Differentiating bothsides with respect to y, we find3 dxdy = 3 p = 1 p − y dp dp−p 2 6y2dy dy − 12yp,which factorises to give( ) ( 1+6yp22p + y dp )=0. (14.33)dySetting the factor containing dp/dy equal to zero gives a first-degree first-order equationin p, which may be solved to give py 2 = c. Substituting for p in (14.32) then yields thegeneral solution of (14.32):y 3 =3cx +6c 2 . (14.34)If we now consider the first factor in (14.33), we find 6p 2 y = −1 as a possible solution.Substituting for p in (14.32) we find the singular solution8y 3 +3x 2 =0.Note that the singular solution contains no arbitrary constants and cannot be found fromthe general solution (14.34) by any choice of the constant c. ◭Solution method. Write the equation in the form (14.31) and differentiate bothsides with respect to y. Rearrange the resulting equation into the form G(y, p) =0,which can be used together with the original ODE to eliminate p and so give thegeneral solution. If G(y, p) can be factorised then the factor containing dp/dy shouldbe used to eliminate p and give the general solution. Using the other factors in thisfashion will instead lead to singular solutions.14.3.3 Equations soluble for yEquations that can be solved for y, i.e. are such that they may be written in theformy = F(x, p), (14.35)can be reduced to first-degree first-order equations in p by differentiating bothsides with respect to x, sothatdydx = p = ∂F∂x + ∂F dp∂p dx .This results in an equation of the form G(x, p) = 0, which can be used togetherwith (14.35) to eliminate p and give the general solution. An additional (singular)solution to the equation is also often found.482

14.3 HIGHER-DEGREE FIRST-ORDER EQUATIONS◮Solvexp 2 +2xp − y =0. (14.36)This equation can be solved for y explicitly to give y = xp 2 +2xp. Differentiating bothsides with respect to x, we finddydxwhich after factorising gives= p =2xpdpdx + p2 +2x dpdx +2p,((p +1) p +2x dp )=0. (14.37)dxTo obtain the general solution of (14.36), we consider the factor containing dp/dx. Thisfirst-degree first-order equation in p has the solution xp 2 = c (see subsection 14.3.1), whichwe then use to eliminate p from (14.36). Thus we find that the general solution to (14.36)is(y − c) 2 =4cx. (14.38)If instead, we set the other factor in (14.37) equal to zero, we obtain the very simplesolution p = −1. Substituting this into (14.36) then givesx + y =0,which is a singular solution to (14.36). ◭Solution method. Write the equation in the form (14.35) and differentiate bothsides with respect to x. Rearrange the resulting equation into the form G(x, p) =0,which can be used together with the original ODE to eliminate p and so give thegeneral solution. If G(x, p) can be factorised then the factor containing dp/dx shouldbe used to eliminate p and give the general solution. Using the other factors in thisfashion will instead lead to singular solutions.14.3.4 Clairaut’s equationFinally, we consider Clairaut’s equation, which has the formy = px + F(p) (14.39)and is therefore a special case of equations soluble for y, as in (14.35). It may besolved by a similar method to that given in subsection 14.3.3, but for Clairaut’sequation the form of the general solution is particularly simple. Differentiating(14.39) with respect to x, we finddydx = p = p + x dpdx + dF dpdp dx⇒dp ( ) dFdx dp + x =0. (14.40)Considering first the factor containing dp/dx, we finddpdx = d2 ydx 2 =0 ⇒ y = c 1x + c 2 . (14.41)483

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSSince p = dy/dx = c 1 , if we substitute (14.41) into (14.39) we find c 1 x + c 2 =c 1 x + F(c 1 ). Therefore the constant c 2 is given by F(c 1 ), and the general solutionto (14.39) isy = c 1 x + F(c 1 ), (14.42)i.e. the general solution to Clairaut’s equation can be obtained by replacing pin the ODE by the arbitrary constant c 1 . Now, considering the second factor in(14.40), we also havedF+ x =0, (14.43)dpwhich has the form G(x, p) = 0. This relation may be used to eliminate p from(14.39) to give a singular solution.◮Solvey = px + p 2 . (14.44)From (14.42) the general solution is y = cx + c 2 . But from (14.43) we also have 2p + x =0 ⇒ p = −x/2. Substituting this into (14.44) we find the singular solution x 2 +4y =0.◭Solution method. Write the equation in the form (14.39), then the general solutionis given by replacing p by some constant c, as shown in (14.42). Using the relationdF/dp+x =0to eliminate p from the original equation yields the singular solution.14.4 Exercises14.1 A radioactive isotope decays in such a way that the number of atoms present ata given time, N(t), obeys the equationdNdt = −λN.If there are initially N 0 atoms present, find N(t) atlatertimes.14.2 Solve the following equations by separation of the variables:(a) y ′ − xy 3 =0;(b) y ′ tan −1 x − y(1 + x 2 ) −1 =0;(c) x 2 y ′ + xy 2 =4y 2 .14.3 Show that the following equations either are exact or can be made exact, andsolve them:(a) y(2x 2 y 2 +1)y ′ + x(y 4 +1)=0;(b) 2xy ′ +3x + y =0;(c) (cos 2 x + y sin 2x)y ′ + y 2 =0.14.4 Find the values of α and β that make( 1dF(x, y) =x 2 +2 + α )dx +(xy β +1)dyyan exact differential. For these values solve F(x, y) =0.484

14.4 EXERCISES14.5 By finding suitable integrating factors, solve the following equations:(a) (1 − x 2 )y ′ +2xy =(1− x 2 ) 3/2 ;(b) y ′ − y cot x +cosecx =0;(c) (x + y 3 )y ′ = y (treat y as the independent variable).14.6 By finding an appropriate integrating factor, solvedydx = − 2x2 + y 2 + x.xy14.7 Find, in the form of an integral, the solution of the equationα dydt + y = f(t)for a general function f(t). Find the specific solutions for(a) f(t) =H(t),(b) f(t) =δ(t),(c) f(t) =β −1 e −t/β H(t) withβ

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSFind an expression for the velocity v of the mass as a function of time, giventhat it has an initial velocity v 0 .14.13 Using the results about Laplace transforms given in chapter 13 for df/dt andtf(t), show, for a function y(t) thatsatisfiest dy +(t − 1)y =0dt (∗)with y(0) finite, that ȳ(s) =C(1 + s) −2 for some constant C.Given that∞∑y(t) =t + a n t n ,determine C and show that a n =(−1) n−1 /(n − 1)!. Compare this result with thatobtained by integrating (∗) directly.14.14 Solvedydx = 1x +2y +1 .14.15 Solvedydx = − x + y3x +3y − 4 .14.16 If u =1+tany, calculated(ln u)/dy; hence find the general solution ofdy=tanx cos y (cos y +siny).dx14.17 Solvex(1 − 2x 2 y) dydx + y =3x2 y 2 ,given that y(1) = 1/2.14.18 A reflecting mirror is made in the shape of the surface of revolution generated byrevolving the curve y(x) about the x-axis. In order that light rays emitted from apoint source at the origin are reflected back parallel to the x-axis, the curve y(x)must obeyyx = 2p1 − p , 2where p = dy/dx. By solving this equation for x, find the curve y(x).14.19 Find the curve with the property that at each point on it the sum of the interceptson the x- andy-axes of the tangent to the curve (taking account of sign) is equalto 1.14.20 Find a parametric solution of( ) 2 dyx + dydx dx − y =0as follows.(a) Write an equation for y in terms of p = dy/dx and show thatp = p 2 +(2px +1) dpdx .(b) Using p as the independent variable, arrange this as a linear first-orderequation for x.486n=2

14.4 EXERCISES(c) Find an appropriate integrating factor to obtainx = ln p − p + c ,(1 − p) 2which, together with the expression for y obtained in (a), gives a parameterisationof the solution.(d) Reverse the roles of x and y in steps (a) to (c), putting dx/dy = p −1 ,andshow that essentially the same parameterisation is obtained.14.21 Using the substitutions u = x 2 and v = y 2 , reduce the equation( ) 2 dyxy − (x 2 + y 2 − 1) dy + xy =0dxdxto Clairaut’s form. Hence show that the equation represents a family of conicsand the four sides of a square.14.22 The action of the control mechanism on a particular system for an input f(t) isdescribed, for t ≥ 0, by the coupled first-order equations:ẏ +4z = f(t),ż − 2z = ẏ + 1 y. 2Use Laplace transforms to find the response y(t) of the system to a unit stepinput, f(t) =H(t), given that y(0) = 1 and z(0) = 0.Questions 23 to 31 are intended to give the reader practice in choosing an appropriatemethod. The level of difficulty varies within the set; if necessary, the hints maybe consulted for an indication of the most appropriate approach.14.23 Find the general solutions of the following:(a)dydx + xya 2 + x = x; dy(b) 2 dx = 4y2x − 2 y2 .14.24 Solve the following first-order equations for the boundary conditions given:(a) y ′ − (y/x) =1, y(1) = −1;(b) y ′ − y tan x =1, y(π/4) = 3;(c) y ′ − y 2 /x 2 =1/4, y(1) = 1;(d) y ′ − y 2 /x 2 =1/4, y(1) = 1/2.14.25 An electronic system has two inputs, to each of which a constant unit signal isapplied, but starting at different times. The equations governing the system thustake the formẋ +2y = H(t),ẏ − 2x = H(t − 3).Initially (at t =0),x =1andy = 0; find x(t) atlatertimes.14.26 Solve the differential equationsin x dy +2y cos x =1,dxsubject to the boundary condition y(π/2) = 1.14.27 Find the complete solution of( ) 2 dy− y dydx x dx + A x =0,where A is a positive constant.487

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS14.28 Find the solution of(5x + y − 7) dy =3(x + y +1).dx14.29 Find the solution y = y(x) ofx dydx + y − y2=0,x3/2 subject to y(1) = 1.14.30 Find the solution of(2 sin y − x) dydx =tany,if (a) y(0) = 0, and (b) y(0) = π/2.14.31 Find the family of solutions ofthat satisfy y(0) = 0.( )d 2 2y dydx + + dy2 dx dx =014.5 Hints and answers14.1 N(t) =N 0 exp(−λt).14.3 (a) exact, x 2 y 4 + x 2 + y 2 = c; (b) IF = x −1/2 , x 1/2 (x + y) = c; (c) IF =sec 2 x, y 2 tan x + y = c.14.5 (a) IF = (1 − x 2 ) −2 , y =(1− x 2 )(k +sin −1 x); (b) IF = cosec x, leading toy = k sin x +cosx; (c) exact equation is y −1 (dx/dy) − xy −2 = y, leading tox = y(k + y 2 /2).14.7 y(t) =e ∫ −t/α t α −1 e t′ /α f(t ′ )dt ′ ;(a)y(t) =1− e −t/α ;(b)y(t) =α −1 e −t/α ;(c)y(t) =(e −t/α − e −t/β )/(α − β). It becomes case (b).14.9 Note that, if the angle between the tangent and the radius vector is α, thencos α = dr/ds and sin α = p/r.14.11 Homogeneous equation, put y = vx to obtain (1 − v)(v 2 +2v +2) −1 dv = x −1 dx;write 1 − v as 2 − (1 + v), and v 2 +2v +2as1+(1+v) 2 ;A[x 2 +(x + y) 2 ]=exp { 4tan −1 [(x + y)/x] } .14.13 (1 + s)(dȳ/ds)+2ȳ =0.C = 1; use separation of variables to show directly thaty(t) =te −t .14.15 The equation is of the form of (14.22), set v = x + y; x +3y +2ln(x + y − 2) = A.14.17 The equation is isobaric with weight y = −2; setting y = vx −2 givesv −1 (1 − v) −1 (1 − 2v) dv = x −1 dx; 4xy(1 − x 2 y)=1.14.19 The curve must satisfy y =(1−p −1 ) −1 (1−x+px), which has solution x =(p−1) −2 ,leading to y =(1± √ x) 2 or x =(1± √ y) 2 ; the singular solution p ′ = 0 givesstraight lines joining (θ, 0) and (0, 1 − θ) foranyθ.14.21 v = qu + q/(q − 1), where q = dv/du. General solution y 2 = cx 2 + c/(c − 1),hyperbolae for c>0 and ellipses for c

14.5 HINTS AND ANSWERS14.31 Show that p =(Ce x − 1) −1 , where p = dy/dx; y =ln[C − e −x )/(C − 1)] orln[D − (D − 1)e −x ]orln(e −K +1− e −x )+K.489

15Higher-order ordinary differentialequationsFollowing on from the discussion of first-order ordinary differential equations(ODEs) given in the previous chapter, we now examine equations of second andhigher order. Since a brief outline of the general properties of ODEs and theirsolutions was given at the beginning of the previous chapter, we will not repeatit here. Instead, we will begin with a discussion of various types of higher-orderequation. This chapter is divided into three main parts. We first discuss linearequations with constant coefficients and then investigate linear equations withvariable coefficients. Finally, we discuss a few methods that may be of use insolving general linear or non-linear ODEs. Let us start by considering somegeneral points relating to all linear ODEs.Linear equations are of paramount importance in the description of physicalprocesses. Moreover, it is an empirical fact that, when put into mathematicalform, many natural processes appear as higher-order linear ODEs, most oftenas second-order equations. Although we could restrict our attention to thesesecond-order equations, the generalisation to nth-order equations requires littleextra work, and so we will consider this more general case.A linear ODE of general order n has the forma n (x) dn ydx n + a n−1(x) dn−1 ydx n−1 + ···+ a 1(x) dydx + a 0(x)y = f(x). (15.1)If f(x) = 0 then the equation is called homogeneous; otherwise it is inhomogeneous.The first-order linear equation studied in subsection 14.2.4 is a special case of(15.1). As discussed at the beginning of the previous chapter, the general solutionto (15.1) will contain n arbitrary constants, which may be determined if n boundaryconditions are also provided.In order to solve any equation of the form (15.1), we must first find thegeneral solution of the complementary equation, i.e. the equation formed by setting490

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSf(x) =0:a n (x) dn ydx n + a n−1(x) dn−1 ydx n−1 + ···+ a 1(x) dydx + a 0(x)y =0. (15.2)To determine the general solution of (15.2), we must find n linearly independentfunctions that satisfy it. Once we have found these solutions, the general solutionis given by a linear superposition of these n functions. In other words, if the nsolutions of (15.2) are y 1 (x), y 2 (x),...,y n (x), then the general solution is given bythe linear superpositiony c (x) =c 1 y 1 (x)+c 2 y 2 (x)+···+ c n y n (x), (15.3)where the c m are arbitrary constants that may be determined if n boundaryconditions are provided. The linear combination y c (x) is called the complementaryfunction of (15.1).The question naturally arises how we establish that any n individual solutions to(15.2) are indeed linearly independent. For n functions to be linearly independentover an interval, there must not exist any set of constants c 1 ,c 2 ,...,c n such thatc 1 y 1 (x)+c 2 y 2 (x)+···+ c n y n (x) = 0 (15.4)over the interval in question, except for the trivial case c 1 = c 2 = ···= c n =0.A statement equivalent to (15.4), which is perhaps more useful for the practicaldetermination of linear independence, can be found by repeatedly differentiating(15.4), n − 1 times in all, to obtain n simultaneous equations for c 1 ,c 2 ,...,c n :c 1 y 1 (x)+c 2 y 2 (x)+···+ c n y n (x) =0c 1 y ′ 1 (x)+c 2 y ′ 2 (x)+···+ c n y ′ n (x) =0.(15.5).c 1 y (n−1)1(x)+c 2 y (n−1)2+ ···+ c n y n (n−1) (x) =0,where the primes denote differentiation with respect to x. Referring to thediscussion of simultaneous linear equations given in chapter 8, if the determinantof the coefficients of c 1 ,c 2 ,...,c n is non-zero then the only solution to equations(15.5) is the trivial solution c 1 = c 2 = ···= c n = 0. In other words, the n functionsy 1 (x), y 2 (x),...,y n (x) are linearly independent over an interval if∣∣W (y 1 ,y 2 ,...,y n )=∣y 1 y 2 ... y n..′y 1′y 2... ....y (n−1)1... ... y n(n−1)≠ 0 (15.6)∣over that interval; W (y 1 ,y 2 ,...,y n ) is called the Wronskian of the set of functions.It should be noted, however, that the vanishing of the Wronskian does notguarantee that the functions are linearly dependent.491

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSIf the original equation (15.1) has f(x) = 0 (i.e. it is homogeneous) then ofcourse the complementary function y c (x) in (15.3) is already the general solution.If, however, the equation has f(x) ≠ 0 (i.e. it is inhomogeneous) then y c (x) isonlyone part of the solution. The general solution of (15.1) is then given byy(x) =y c (x)+y p (x), (15.7)where y p (x)istheparticular integral,whichcanbeany function that satisfies (15.1)directly, provided it is linearly independent of y c (x). It should be emphasised forpractical purposes that any such function, no matter how simple (or complicated),is equally valid in forming the general solution (15.7).It is important to realise that the above method for finding the general solutionto an ODE by superposing particular solutions assumes crucially that the ODEis linear. For non-linear equations, discussed in section 15.3, this method cannotbe used, and indeed it is often impossible to find closed-form solutions to suchequations.15.1 Linear equations with constant coefficientsIf the a m in (15.1) are constants rather than functions of x then we haved n ya ndx n + a d n−1 yn−1dx n−1 + ···+ a dy1dx + a 0y = f(x). (15.8)Equations of this sort are very common throughout the physical sciences andengineering, and the method for their solution falls into two parts as discussedin the previous section, i.e. finding the complementary function y c (x) and findingthe particular integral y p (x). If f(x) = 0 in (15.8) then we do not have to finda particular integral, and the complementary function is by itself the generalsolution.15.1.1 Finding the complementary function y c (x)The complementary function must satisfyd n ya ndx n + a d n−1 yn−1dx n−1 + ···+ a dy1dx + a 0y = 0 (15.9)and contain n arbitrary constants (see equation (15.3)). The standard methodfor finding y c (x) is to try a solution of the form y = Ae λx , substituting this into(15.9). After dividing the resulting equation through by Ae λx , we are left with apolynomial equation in λ of order n; thisistheauxiliary equation and readsa n λ n + a n−1 λ n−1 + ···+ a 1 λ + a 0 =0. (15.10)492

15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTSIn general the auxiliary equation has n roots, say λ 1 ,λ 2 ,...,λ n . In certain cases,some of these roots may be repeated and some may be complex. The three maincases are as follows.(i) All roots real and distinct. In this case the n solutions to (15.9) are exp λ m xfor m =1ton. It is easily shown by calculating the Wronskian (15.6)of these functions that if all the λ m are distinct then these solutions arelinearly independent. We can therefore linearly superpose them, as in(15.3), to form the complementary functiony c (x) =c 1 e λ1x + c 2 e λ2x + ···+ c n e λnx . (15.11)(ii) Some roots complex. For the special (but usual) case that all the coefficientsa m in (15.9) are real, if one of the roots of the auxiliary equation (15.10)is complex, say α + iβ, then its complex conjugate α − iβ is also a root. Inthis case we can writec 1 e (α+iβ)x + c 2 e (α−iβ)x = e αx (d 1 cos βx + d 2 sin βx)= Ae αx { sincos}(βx + φ), (15.12)where A and φ are arbitrary constants.(iii) Some roots repeated. If, for example, λ 1 occurs k times (k >1) as a rootof the auxiliary equation, then we have not found n linearly independentsolutions of (15.9); formally the Wronskian (15.6) of these solutions, havingtwo or more identical columns, is equal to zero. We must therefore findk −1 further solutions that are linearly independent of those already foundand also of each other. By direct substitution into (15.9) we find thatxe λ1x , x 2 e λ1x , ..., x k−1 e λ1xare also solutions, and by calculating the Wronskian it is easily shown thatthey, together with the solutions already found, form a linearly independentset of n functions. Therefore the complementary function is given byy c (x) =(c 1 + c 2 x + ···+ c k x k−1 )e λ1x + c k+1 e λk+1x + c k+2 e λk+2x + ···+ c n e λnx .(15.13)If more than one root is repeated the above argument is easily extended.For example, suppose as before that λ 1 is a k-fold root of the auxiliaryequation and, further, that λ 2 is an l-fold root (of course, k>1andl>1).Then, from the above argument, the complementary function readsy c (x) =(c 1 + c 2 x + ···+ c k x k−1 )e λ1x+(c k+1 + c k+2 x + ···+ c k+l x l−1 )e λ2x+ c k+l+1 e λk+l+1x + c k+l+2 e λk+l+2x + ···+ c n e λnx . (15.14)493

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Find the complementary function of the equationd 2 ydx − 2 dy2 dx + y = ex . (15.15)Setting the RHS to zero, substituting y = Ae λx and dividing through by Ae λx we obtainthe auxiliary equationλ 2 − 2λ +1=0.The root λ = 1 occurs twice and so, although e x is a solution to (15.15), we must finda further solution to the equation that is linearly independent of e x .Fromtheabovediscussion, we deduce that xe x is such a solution, so that the full complementary functionis given by the linear superpositiony c (x) =(c 1 + c 2 x)e x . ◭Solution method. Set the RHS of the ODE to zero (if it is not already so), andsubstitute y = Ae λx . After dividing through the resulting equation by Ae λx , obtainan nth-order polynomial equation in λ (the auxiliary equation, see (15.10)). Solvethe auxiliary equation to find the n roots, λ 1 ,λ 2 ,...,λ n , say. If all these roots arereal and distinct then y c (x) is given by (15.11). If, however, some of the roots arecomplex or repeated then y c (x) is given by (15.12) or (15.13), or the extension(15.14) of the latter, respectively.15.1.2 Finding the particular integral y p (x)There is no generally applicable method for finding the particular integral y p (x)but, for linear ODEs with constant coefficients and a simple RHS, y p (x) can oftenbe found by inspection or by assuming a parameterised form similar to f(x). Thelatter method is sometimes called the method of undetermined coefficients. Iff(x)contains only polynomial, exponential, or sine and cosine terms then, by assuminga trial function for y p (x) of similar form but one which contains a number ofundetermined parameters and substituting this trial function into (15.9), theparameters can be found and y p (x) deduced. Standard trial functions are asfollows.(i) If f(x) =ae rx then tryy p (x) =be rx .(ii) If f(x) =a 1 sin rx + a 2 cos rx (a 1 or a 2 may be zero) then tryy p (x) =b 1 sin rx + b 2 cos rx.(iii) If f(x) =a 0 + a 1 x + ···+ a N x N (some a m may be zero) then tryy p (x) =b 0 + b 1 x + ···+ b N x N .494

15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS(iv) If f(x) is the sum or product of any of the above then try y p (x) asthesum or product of the corresponding individual trial functions.It should be noted that this method fails if any term in the assumed trialfunction is also contained within the complementary function y c (x). In such acase the trial function should be multiplied by the smallest integer power of xsuch that it will then contain no term that already appears in the complementaryfunction. The undetermined coefficients in the trial function can now be foundby substitution into (15.8).Three further methods that are useful in finding the particular integral y p (x) arethose based on Green’s functions, the variation of parameters, and a change in thedependent variable using knowledge of the complementary function. However,since these methods are also applicable to equations with variable coefficients, adiscussion of them is postponed until section 15.2.◮Find a particular integral of the equationd 2 ydx − 2 dy2 dx + y = ex .From the above discussion our first guess at a trial particular integral would be y p (x) =be x .However, since the complementary function of this equation is y c (x) =(c 1 + c 2 x)e x (asin the previous subsection), we see that e x is already contained in it, as indeed is xe x .Multiplying our first guess by the lowest integer power of x such that the result does notappear in y c (x), we therefore try y p (x) =bx 2 e x . Substituting this into the ODE, we findthat b =1/2, so the particular integral is given by y p (x) =x 2 e x /2. ◭Solution method. If the RHS of an ODE contains only functions mentioned at thestart of this subsection then the appropriate trial function should be substitutedinto it, thereby fixing the undetermined parameters. If, however, the RHS of theequation is not of this form then one of the more general methods outlined in subsections15.2.3–15.2.5 should be used; perhaps the most straightforward of these isthe variation-of-parameters method.15.1.3 Constructing the general solution y c (x)+y p (x)As stated earlier, the full solution to the ODE (15.8) is found by adding togetherthe complementary function and any particular integral. In order to illustratefurther the material discussed in the last two subsections, let us find the generalsolution to a new example, starting from the beginning.495

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Solved 2 ydx 2 +4y = x2 sin 2x. (15.16)First we set the RHS to zero and assume the trial solution y = Ae λx . Substituting this into(15.16) leads to the auxiliary equationλ 2 +4=0 ⇒ λ = ±2i. (15.17)Therefore the complementary function is given byy c (x) =c 1 e 2ix + c 2 e −2ix = d 1 cos 2x + d 2 sin 2x. (15.18)We must now turn our attention to the particular integral y p (x). Consulting the list ofstandard trial functions in the previous subsection, we find that a first guess at a suitabletrial function for this case should be(ax 2 + bx + c)sin2x +(dx 2 + ex + f)cos2x. (15.19)However, we see that this trial function contains terms in sin 2x and cos 2x, both of whichalready appear in the complementary function (15.18). We must therefore multiply (15.19)by the smallest integer power of x which ensures that none of the resulting terms appearsin y c (x). Since multiplying by x will suffice, we finally assume the trial function(ax 3 + bx 2 + cx)sin2x +(dx 3 + ex 2 + fx)cos2x. (15.20)Substituting this into (15.16) to fix the constants appearing in (15.20), we find the particularintegral to bey p (x) =− x3 x2cos 2x +12 16 sin 2x + x cos 2x. (15.21)32The general solution to (15.16) then readsy(x) =y c (x)+y p (x)= d 1 cos 2x + d 2 sin 2x − x3 x2cos 2x +12 16 sin 2x + x cos 2x. ◭3215.1.4 Linear recurrence relationsBefore continuing our discussion of higher-order ODEs, we take this opportunityto introduce the discrete analogues of differential equations, which are calledrecurrence relations (or sometimes difference equations). Whereas a differentialequation gives a prescription, in terms of current values, for the new value of adependent variable at a point only infinitesimally far away, a recurrence relationdescribes how the next in a sequence of values u n , defined only at (non-negative)integer values of the ‘independent variable’ n, istobecalculated.In its most general form a recurrence relation expresses the way in which u n+1is to be calculated from all the preceding values u 0 ,u 1 ,... ,u n . Just as the mostgeneral differential equations are intractable, so are the most general recurrencerelations, and we will limit ourselves to analogues of the types of differentialequations studied earlier in this chapter, namely those that are linear, have496

15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTSconstant coefficients and possess simple functions on the RHS. Such equationsoccur over a broad range of engineering and statistical physics as well as in therealms of finance, business planning and gambling! They form the basis of manynumerical methods, particularly those concerned with the numerical solution ofordinary and partial differential equations.A general recurrence relation is exemplified by the formulaN−1∑u n+1 = a r u n−r + k, (15.22)r=0where N and the a r are fixed and k is a constant or a simple function of n.Such an equation, involving terms of the series whose indices differ by up to N(ranging from n−N +1 to n), is called an Nth-order recurrence relation. It is clearthat, given values for u 0 ,u 1 ,... ,u N−1 , this is a definitive scheme for generating theseries and therefore has a unique solution.Parallelling the nomenclature of differential equations, if the term not involvingany u n is absent, i.e. k = 0, then the recurrence relation is called homogeneous.The parallel continues with the form of the general solution of (15.22). If v n isthe general solution of the homogeneous relation, and w n is any solution of thefull relation, thenu n = v n + w nis the most general solution of the complete recurrence relation. This is straightforwardlyverified as follows:u n+1 = v n+1 + w n+1∑N−1=r=0N−1=N−1∑a r v n−r +r=0a r w n−r + k∑a r (v n−r + w n−r )+kr=0∑a r u n−r + k.N−1=r=0Of course, if k =0thenw n = 0 for all n is a trivial particular solution and thecomplementary solution, v n , is itself the most general solution.First-order recurrence relationsFirst-order relations, for which N = 1, are exemplified byu n+1 = au n + k, (15.23)with u 0 specified. The solution to the homogeneous relation is immediate,u n = Ca n ,497

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSand, if k is a constant, the particular solution is equally straightforward: w n = Kfor all n, provided K is chosen to satisfyK = aK + k,i.e. K = k(1 − a) −1 . This will be sufficient unless a =1,inwhichcaseu n = u 0 + nkis obvious by inspection.Thus the general solution of (15.23) is{Ca n + k/(1 − a) a ≠1,u n =u 0 + nk a =1.(15.24)If u 0 is specified for the case of a ≠1thenC must be chosen as C = u 0 −k/(1−a),resulting in the equivalent formu n = u 0 a n + k 1 − an1 − a . (15.25)We now illustrate this method with a worked example.◮A house-buyer borrows capital B from a bank that charges a fixed annual rate of interestR%. If the loan is to be repaid over Y years, at what value should the fixed annual paymentsP , made at the end of each year, be set? For a loan over 25 years at 6%, what percentageof the first year’s payment goes towards paying off the capital?Let u n denote the outstanding debt at the end of year n, andwriteR/100 = r. Then therelevant recurrence relation isu n+1 = u n (1 + r) − Pwith u 0 = B. From (15.25) we haveu n = B(1 + r) n 1 − (1 + r)n− P1 − (1 + r) .AstheloanistoberepaidoverY years, u Y = 0 and thusBr(1 + r)YP =(1 + r) Y − 1 .The first year’s interest is rB and so the fraction of the first year’s payment goingtowards capital repayment is (P − rB)/P , which, using the above expression for P ,isequalto (1 + r) −Y . With the given figures, this is (only) 23%. ◭With only small modifications, the method just described can be adapted tohandle recurrence relations in which the constant k in (15.23) is replaced by kα n ,i.e. the relation isu n+1 = au n + kα n . (15.26)As for an inhomogeneous linear differential equation (see subsection 15.1.2), wemay try as a potential particular solution a form which resembles the term thatmakes the equation inhomogeneous. Here, the presence of the term kα n indicates498

15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTSthat a particular solution of the form u n = Aα n should be tried. Substituting thisinto (15.26) givesAα n+1 = aAα n + kα n ,from which it follows that A = k/(α − a) and that there is a particular solutionhaving the form u n = kα n /(α − a), provided α ≠ a. For the special case α = a, thereader can readily verify that a particular solution of the form u n = Anα n is appropriate.This mirrors the corresponding situation for linear differential equationswhen the RHS of the differential equation is contained in the complementaryfunction of its LHS.In summary, the general solution to (15.26) isu n =with C 1 = u 0 − k/(α − a) andC 2 = u 0 .{C1 a n + kα n /(α − a) α ≠ a,C 2 a n + knα n−1 α = a,(15.27)Second-order recurrence relationsWe consider next recurrence relations that involve u n−1 in the prescription foru n+1 and treat the general case in which the intervening term, u n , is also present.A typical equation is thusu n+1 = au n + bu n−1 + k. (15.28)As previously, the general solution of this is u n = v n + w n ,wherev n satisfiesv n+1 = av n + bv n−1 (15.29)and w n is any particular solution of (15.28); the proof follows the same lines asthat given earlier.We have already seen for a first-order recurrence relation that the solution tothe homogeneous equation is given by terms forming a geometric series, and weconsider a corresponding series of powers in the present case. Setting v n = Aλ n in(15.29) for some λ, as yet undetermined, gives the requirement that λ should satisfyAλ n+1 = aAλ n + bAλ n−1 .Dividing through by Aλ n−1 (assumed non-zero) shows that λ could be either ofthe roots, λ 1 and λ 2 ,ofλ 2 − aλ − b =0, (15.30)which is known as the characteristic equation of the recurrence relation.That there are two possible series of terms of the form Aλ n is consistent with thefact that two initial values (boundary conditions) have to be provided before theseries can be calculated by repeated use of (15.28). These two values are sufficientto determine the appropriate coefficient A for each of the series. Since (15.29) is499

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSboth linear and homogeneous, and is satisfied by both v n = Aλ n 1 and v n = Bλ n 2 ,itsgeneral solution isv n = Aλ n 1 + Bλ n 2.If the coefficients a and b are such that (15.30) has two equal roots, i.e. a 2 = −4b,then, as in the analogous case of repeated roots for differential equations (seesubsection 15.1.1(iii)), the second term of the general solution is replaced by Bnλ n 1to givev n =(A + Bn)λ n 1.Finding a particular solution is straightforward if k is a constant: a trivial butadequate solution is w n = k(1 − a − b) −1 for all n. As with first-order equations,particular solutions can be found for other simple forms of k by trying functionssimilar to k itself. Thus particular solutions for the cases k = Cn and k = Dα ncan be found by trying w n = E + Fn and w n = Gα n respectively.◮Find the value of u 16 if the series u n satisfiesu n+1 +4u n +3u n−1 = nfor n ≥ 1, withu 0 =1and u 1 = −1.We first solve the characteristic equation,λ 2 +4λ +3=0,to obtain the roots λ = −1 andλ = −3. Thus the complementary function isv n = A(−1) n + B(−3) n .In view of the form of the RHS of the original relation, we tryas a particular solution and obtainyielding F =1/8 andE =1/32.Thus the complete general solution isw n = E + FnE + F(n +1)+4(E + Fn)+3[E + F(n − 1)] = n,u n = A(−1) n + B(−3) n + n 8 + 132 ,and now using the given values for u 0 and u 1 determines A as 7/8 andB as 3/32. Thusu n = 1 32 [28(−1)n +3(−3) n +4n +1] .Finally, substituting n =16givesu 16 = 4 035 633, a value the reader may (or may not)wish to verify by repeated application of the initial recurrence relation. ◭500

15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTSHigher-order recurrence relationsIt will be apparent that linear recurrence relations of order N>2 do not presentany additional difficulty in principle, though two obvious practical difficulties are(i) that the characteristic equation is of order N and in general will not have rootsthat can be written in closed form and (ii) that a correspondingly large numberof given values is required to determine the N otherwise arbitrary constants inthe solution. The algebraic labour needed to solve the set of simultaneous linearequations that determines them increases rapidly with N. We do not give specificexamples here, but some are included in the exercises at the end of the chapter.15.1.5 Laplace transform methodHaving briefly discussed recurrence relations, we now return to the main topicof this chapter, i.e. methods for obtaining solutions to higher-order ODEs. Onesuch method is that of Laplace transforms, which is very useful for solvinglinear ODEs with constant coefficients. Taking the Laplace transform of such anequation transforms it into a purely algebraic equation in terms of the Laplacetransform of the required solution. Once the algebraic equation has been solvedfor this Laplace transform, the general solution to the original ODE can beobtained by performing an inverse Laplace transform. One advantage of thismethod is that, for given boundary conditions, it provides the solution in justone step, instead of having to find the complementary function and particularintegral separately.In order to apply the method we need only two results from Laplace transformtheory (see section 13.2). First, the Laplace transform of a function f(x) is definedby¯f(s) ≡∫ ∞0e −sx f(x) dx, (15.31)from which we can derive the second useful relation. This concerns the Laplacetransform of the nth derivative of f(x):f (n) (s) =s n¯f(s) − s n−1 f(0) − s n−2 f ′ (0) − ···− sf (n−2) (0) − f (n−1) (0),(15.32)where the primes and superscripts in parentheses denote differentiation withrespect to x. Using these relations, along with table 13.1, on p. 455, which givesLaplace transforms of standard functions, we are in a position to solve a linearODE with constant coefficients by this method.501

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Solved 2 ydx − 3 dy2 dx +2y =2e−x , (15.33)subject to the boundary conditions y(0) = 2, y ′ (0) = 1.Taking the Laplace transform of (15.33) and using the table of standard results we obtains 2 ȳ(s) − sy(0) − y ′ (0) − 3 [sȳ(s) − y(0)] +2ȳ(s) = 2s +1 ,which reduces to(s 2 − 3s +2)ȳ(s) − 2s +5= 2s +1 . (15.34)Solving this algebraic equation for ȳ(s), the Laplace transform of the required solution to(15.33), we obtain2s 2 − 3s − 3ȳ(s) =(s +1)(s − 1)(s − 2) = 13(s +1) + 2s − 1 − 13(s − 2) , (15.35)where in the final step we have used partial fractions. Taking the inverse Laplace transformof (15.35), again using table 13.1, we find the specific solution to (15.33) to bey(x) = 1 3 e−x +2e x − 1 3 e2x . ◭Note that if the boundary conditions in a problem are given as symbols, ratherthan just numbers, then the step involving partial fractions can often involvea considerable amount of algebra. The Laplace transform method is also veryconvenient for solving sets of simultaneous linear ODEs with constant coefficients.◮Two electrical circuits, both of negligible resistance, each consist of a coil having selfinductanceL and a capacitor having capacitance C. The mutual inductance of the twocircuits is M. There is no source of e.m.f. in either circuit. Initially the second capacitoris given a charge CV 0 , the first capacitor being uncharged, and at time t =0aswitchinthe second circuit is closed to complete the circuit. Find the subsequent current in the firstcircuit.Subject to the initial conditions q 1 (0) = ˙q 1 (0) = ˙q 2 (0) = 0 and q 2 (0) = CV 0 = V 0 /G, say,we have to solveL¨q 1 + M¨q 2 + Gq 1 =0,M¨q 1 + L¨q 2 + Gq 2 =0.On taking the Laplace transform of the above equations, we obtain(Ls 2 + G)¯q 1 + Ms 2¯q 2 = sMV 0 C,Ms 2¯q 1 +(Ls 2 + G)¯q 2 = sLV 0 C.Eliminating ¯q 2 and rewriting as an equation for ¯q 1 , we findMV 0 s¯q 1 (s) =[(L + M)s 2 + G ][(L − M)s 2 + G ]= V []0 (L + M)s (L − M)s− .2G (L + M)s 2 + G (L − M)s 2 + G502

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTSUsing table 13.1,q 1 (t) = 1 V 2 0C(cos ω 1 t − cos ω 2 t),where ω1 2(L + M) =G and ω2 2 (L − M) =G. Thus the current is given byi 1 (t) = 1 V 2 0C(ω 2 sin ω 2 t − ω 1 sin ω 1 t). ◭Solution method. Perform a Laplace transform, as defined in (15.31), on the entireequation, using (15.32) to calculate the transform of the derivatives. Then solve theresulting algebraic equation for ȳ(s), the Laplace transform of the required solutionto the ODE. By using the method of partial fractions and consulting a table ofLaplace transforms of standard functions, calculate the inverse Laplace transform.The resulting function y(x) is the solution of the ODE that obeys the given boundaryconditions.15.2 Linear equations with variable coefficientsThere is no generally applicable method of solving equations with coefficientsthat are functions of x. Nevertheless, there are certain cases in which a solution ispossible. Some of the methods discussed in this section are also useful in findingthe general solution or particular integral for equations with constant coefficientsthat have proved impenetrable by the techniques discussed above.15.2.1 The Legendre and Euler linear equationsLegendre’s linear equation has the forma n (αx + β) n dn ydx n + ···+ a 1(αx + β) dydx + a 0y = f(x), (15.36)where α, β and the a n are constants and may be solved by making the substitutionαx + β = e t . We then havedydx = dt dydx dt = α dyαx + β dtd 2 ydx 2 = ddxdydx = α 2 ( d 2 y(αx + β) 2 dt 2 − dy )dtand so on for higher derivatives. Therefore we can write the terms of (15.36) as(αx + β) dydx = αdy dt ,(αx + β) 2 d2 ydx 2 = α2 d dt.(αx + β) n dn ydx n = αn d dt( ddt − 1 )y,( ) ( )d ddt − 1 ···dt − n +1 y.503(15.37)

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSSubstituting equations (15.37) into the original equation (15.36), the latter becomesa linear ODE with constant coefficients, i.e.a n α n d ( ) ( )d ddt dt − 1 ···dt − n +1 y + ···+ a 1 α dy ( e t )dt + a − β0y = f ,αwhich can be solved by the methods of section 15.1.A special case of Legendre’s linear equation, for which α = 1 and β =0,isEuler’s equation,a n x n dn ydx n + ···+ a 1x dydx + a 0y = f(x); (15.38)it may be solved in a similar manner to the above by substituting x = e t .Iff(x) = 0 in (15.38) then substituting y = x λ leads to a simple algebraic equationin λ, which can be solved to yield the solution to (15.38). In the event that thealgebraic equation for λ has repeated roots, extra care is needed. If λ 1 is a k-foldroot (k >1) then the k linearly independent solutions corresponding to this rootare x λ1 ,x λ1 ln x,...,x λ1 (ln x) k−1 .◮Solvex 2 d2 ydx + x dy − 4y = 0 (15.39)2 dxby both of the methods discussed above.First we make the substitution x = e t , which, after cancelling e t , gives an equation withconstant coefficients, i.e.( )d ddt dt − 1 y + dydt − 4y =0 ⇒ d2 y− 4y =0. (15.40)dt2 Using the methods of section 15.1, the general solution of (15.40), and therefore of (15.39),is given byy = c 1 e 2t + c 2 e −2t = c 1 x 2 + c 2 x −2 .Since the RHS of (15.39) is zero, we can reach the same solution by substituting y = x λinto (15.39). This givesλ(λ − 1)x λ + λx λ − 4x λ =0,which reduces to(λ 2 − 4)x λ =0.This has the solutions λ = ±2, so we obtain again the general solutiony = c 1 x 2 + c 2 x −2 . ◭Solution method. If the ODE is of the Legendre form (15.36) then substitute αx +β = e t . This results in an equation of the same order but with constant coefficients,which can be solved by the methods of section 15.1. If the ODE is of the Eulerform (15.38) with a non-zero RHS then substitute x = e t ; this again leads to anequation of the same order but with constant coefficients. If, however, f(x) =0inthe Euler equation (15.38) then the equation may also be solved by substituting504

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTSy = x λ . This leads to an algebraic equation whose solution gives the allowed valuesof λ; the general solution is then the linear superposition of these functions.15.2.2 Exact equationsSometimes an ODE may be merely the derivative of another ODE of one orderlower. If this is the case then the ODE is called exact. The nth-order linear ODEa n (x) dn ydx n + ···+ a 1(x) dydx + a 0(x)y = f(x), (15.41)is exact if the LHS can be written as a simple derivative, i.e. ifa n (x) dn ydx n + ···+ a 0(x)y = d []b n−1 (x) dn−1 ydx dx n−1 + ···+ b 0(x)y . (15.42)It may be shown that, for (15.42) to hold, we requirea 0 (x) − a ′ 1(x)+a ′′2(x) − ···+(−1) n a n (n) (x) =0, (15.43)where the prime again denotes differentiation with respect to x. If (15.43) issatisfied then straightforward integration leads to a new equation of one orderlower. If this simpler equation can be solved then a solution to the originalequation is obtained. Of course, if the above process leads to an equation that isitself exact then the analysis can be repeated to reduce the order still further.◮Solve(1 − x 2 ) d2 y dy− 3x − y =1. (15.44)dx2 dxComparing with (15.41), we have a 2 =1− x 2 , a 1 = −3x and a 0 = −1. It is easily shownthat a 0 − a ′ 1 + a′′ 2 = 0, so (15.44) is exact and can therefore be written in the form[db 1 (x) dy ]dx dx + b 0(x)y =1. (15.45)Expanding the LHS of (15.45) we find( )d dyb 1dx dx + b d 2 y0y = b 1dx 2 +(b′ 1 + b 0 ) dydx + b′ 0y. (15.46)Comparing (15.44) and (15.46) we findb 1 =1− x 2 , b ′ 1 + b 0 = −3x, b ′ 0 = −1.These relations integrate consistently to give b 1 =1− x 2 and b 0 = −x, so (15.44) can bewritten as[d(1 − x 2 ) dy ]dx dx − xy =1. (15.47)Integrating (15.47) gives us directly the first-order linear ODEdy(dx − x)y = x + c 11 − x 2 1 − x , 2which can be solved by the method of subsection 14.2.4 and has the solutiony = c 1 sin −1 x + c 2√1 − x2− 1. ◭505

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSIt is worth noting that, even if a higher-order ODE is not exact in its given form,it may sometimes be made exact by multiplying through by some suitable function,an integrating factor, cf. subsection 14.2.3. Unfortunately, no straightforwardmethod for finding an integrating factor exists and one often has to rely oninspection or experience.◮Solvex(1 − x 2 ) d2 y dy− 3x2 − xy = x. (15.48)dx2 dxIt is easily shown that (15.48) is not exact, but we also see immediately that by multiplyingit through by 1/x we recover (15.44), which is exact and is solved above. ◭Another important point is that an ODE need not be linear to be exact,although no simple rule such as (15.43) exists if it is not linear. Nevertheless, it isoften worth exploring the possibility that a non-linear equation is exact, since itcould then be reduced in order by one and may lead to a soluble equation. Thisis discussed further in subsection 15.3.3.Solution method. For a linear ODE of the form (15.41) check whether it is exactusing equation (15.43). If it is not then attempt to find an integrating factor whichwhen multiplying the equation makes it exact. Once the equation is exact write theLHS as a derivative as in (15.42) and, by expanding this derivative and comparingwith the LHS of the ODE, determine the functions b m (x) in (15.42). Integrate theresulting equation to yield another ODE, of one order lower. This may be solved orsimplified further if the new ODE is itself exact or can be made so.15.2.3 Partially known complementary functionSuppose we wish to solve the nth-order linear ODEa n (x) dn ydx n + ···+ a 1(x) dydx + a 0(x)y = f(x), (15.49)andwehappentoknowthatu(x) is a solution of (15.49) when the RHS isset to zero, i.e. u(x) is one part of the complementary function. By making thesubstitution y(x) =u(x)v(x), we can transform (15.49) into an equation of ordern − 1indv/dx. This simpler equation may prove soluble.In particular, if the original equation is of second order then we obtaina first-order equation in dv/dx, which may be soluble using the methods ofsection 14.2. In this way both the remaining term in the complementary functionand the particular integral are found. This method therefore provides a usefulway of calculating particular integrals for second-order equations with variable(or constant) coefficients.506

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS◮Solved 2 y+ y =cosecx. (15.50)dx2 We see that the RHS does not fall into any of the categories listed in subsection 15.1.2,and so we are at an initial loss as to how to find the particular integral. However, thecomplementary function of (15.50) isy c (x) =c 1 sin x + c 2 cos x,and so let us choose the solution u(x) =cosx (we could equally well choose sin x) andmake the substitution y(x) =v(x)u(x) =v(x)cosx into (15.50). This givescos x d2 v dv− 2sinx =cosecx, (15.51)dx2 dxwhich is a first-order linear ODE in dv/dx and may be solved by multiplying through bya suitable integrating factor, as discussed in subsection 14.2.4. Writing (15.51) asd 2 v dv− 2tanxdx2 dx = cosec xcos x , (15.52)we see that the required integrating factor is given by{ ∫ }exp −2 tan xdx =exp[2ln(cosx)] =cos 2 x.Multiplying both sides of (15.52) by the integrating factor cos 2 x we obtain(dcos 2 x dv )=cotx,dx dxwhich integrates to givecos 2 x dvdx =ln(sinx)+c 1.After rearranging and integrating again, this becomes∫∫v = sec 2 x ln(sin x) dx + c 1 sec 2 xdx=tanx ln(sin x) − x + c 1 tan x + c 2 .Therefore the general solution to (15.50) is given by y = uv = v cos x, i.e.y = c 1 sin x + c 2 cos x +sinx ln(sin x) − x cos x,which contains the full complementary function and the particular integral. ◭Solution method. If u(x) is a known solution of the nth-order equation (15.49) withf(x) =0, then make the substitution y(x) =u(x)v(x) in (15.49). This leads to anequation of order n − 1 in dv/dx, which might be soluble.507

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS15.2.4 Variation of parametersThe method of variation of parameters proves useful in finding particular integralsfor linear ODEs with variable (and constant) coefficients. However, it requiresknowledge of the entire complementary function, not just of one part of it as inthe previous subsection.Suppose we wish to find a particular integral of the equationa n (x) dn ydx n + ···+ a 1(x) dydx + a 0(x)y = f(x), (15.53)and the complementary function y c (x) (the general solution of (15.53) withf(x) = 0) isy c (x) =c 1 y 1 (x)+c 2 y 2 (x)+···+ c n y n (x),where the functions y m (x) are known. We now assume that a particular integral of(15.53) can be expressed in a form similar to that of the complementary function,but with the constants c m replaced by functions of x, i.e.weassumeaparticularintegral of the formy p (x) =k 1 (x)y 1 (x)+k 2 (x)y 2 (x)+···+ k n (x)y n (x). (15.54)This will no longer satisfy the complementary equation (i.e. (15.53) with the RHSset to zero) but might, with suitable choices of the functions k i (x), be made equalto f(x), thus producing not a complementary function but a particular integral.Since we have n arbitrary functions k 1 (x),k 2 (x),...,k n (x), but only one restrictionon them (namely the ODE), we may impose a further n − 1 constraints. Wecan choose these constraints to be as convenient as possible, and the simplestchoice is given byk 1(x)y ′ 1 (x)+k 2(x)y ′ 2 (x)+···+ k n(x)y ′ n (x) =0k 1(x)y ′ 1(x)+k ′ 2(x)y ′ 2(x)+···+ ′ k n(x)y ′ n(x) ′ =0.. (15.55)k 1(x)y ′ (n−2)1(x)+k 2(x)y ′ (n−2)2(x)+···+ k n(x)y ′ n (n−2) (x) =0k 1(x)y ′ (n−1)1(x)+k 2(x)y ′ (n−1)2(x)+···+ k n(x)y ′ n(n−1) (x) = f(x)a n (x) ,where the primes denote differentiation with respect to x. The last of theseequations is not a freely chosen constraint; given the previous n − 1 constraintsand the original ODE, it must be satisfied.This choice of constraints is easily justified (although the algebra is quitemessy). Differentiating (15.54) with respect to x, we obtainy ′ p = k 1 y ′ 1 + k 2 y ′ 2 + ···+ k n y ′ n +[k ′ 1y 1 + k ′ 2y 2 + ···+ k ′ ny n ],where, for the moment, we drop the explicit x-dependence of these functions. Since508

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTSwe are free to choose our constraints as we wish, let us define the expression inparentheses to be zero, giving the first equation in (15.55). Differentiating againwe findy p ′′ = k 1 y 1 ′′ + k 2 y 2 ′′ + ···+ k n y n ′′ +[k 1y ′ 1 ′ + k 2y ′ 2 ′ + ···+ k ny ′ n ′ ].Once more we can choose the expression in brackets to be zero, giving the secondequation in (15.55). We can repeat this procedure, choosing the correspondingexpression in each case to be zero. This yields the first n − 1 equations in (15.55).The mth derivative of y p for m

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSintegration equal to zero, to give k m (x). The general solution to (15.53) is thengiven byn∑y(x) =y c (x)+y p (x) = [c m + k m (x)]y m (x).Note that if the constants of integration are included in the k m (x) then, as wellas finding the particular integral, we redefine the arbitrary constants c m in thecomplementary function.◮Use the variation-of-parameters method to solved 2 y+ y =cosecx, (15.57)dx2 subject to the boundary conditions y(0) = y(π/2) = 0.The complementary function of (15.57) is againy c (x) =c 1 sin x + c 2 cos x.We therefore assume a particular integral of the formy p (x) =k 1 (x)sinx + k 2 (x)cosx,and impose the additional constraints of (15.55), i.e.k 1(x)sinx ′ + k 2(x)cosx ′ =0,k 1(x)cosx ′ − k 2(x)sinx ′ =cosecx.Solving these equations for k 1 ′ (x) andk′ 2 (x) givesk 1(x) ′ =cosx cosec x =cotx,k 2(x) ′ =− sin x cosec x = −1.Hence, ignoring the constants of integration, k 1 (x) andk 2 (x) are given byk 1 (x) = ln(sin x),k 2 (x) =−x.The general solution to the ODE (15.57) is thereforey(x) =[c 1 +ln(sinx)] sin x +(c 2 − x)cosx,which is identical to the solution found in subsection 15.2.3. Applying the boundaryconditions y(0) = y(π/2) = 0 we find c 1 = c 2 =0andsoy(x) = ln(sin x)sinx − x cos x. ◭m=1Solution method. If the complementary function of (15.53) is known then assumea particular integral of the same form but with the constants replaced by functionsof x. Impose the constraints in (15.55) and solve the resulting system of equationsfor the unknowns k ′ 1 (x),k′ 2 ,...,k′ n(x). Integrate these functions, setting constants ofintegration equal to zero, to obtain k 1 (x),k 2 (x),...,k n (x) and hence the particularintegral.510

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS15.2.5 Green’s functionsThe Green’s function method of solving linear ODEs bears a striking resemblanceto the method of variation of parameters discussed in the previous subsection;it too requires knowledge of the entire complementary function in order to findthe particular integral and therefore the general solution. The Green’s functionapproach differs, however, since once the Green’s function for a particular LHSof (15.1) and particular boundary conditions has been found, then the solutionfor any RHS (i.e. any f(x)) can be written down immediately, albeit in the formof an integral.Although the Green’s function method can be approached by considering thesuperposition of eigenfunctions of the equation (see chapter 17) and is alsoapplicable to the solution of partial differential equations (see chapter 21), thissection adopts a more utilitarian approach based on the properties of the Diracdelta function (see subsection 13.1.3) and deals only with the use of Green’sfunctions in solving ODEs.Let us again consider the equationa n (x) dn ydx n + ···+ a 1(x) dydx + a 0(x)y = f(x), (15.58)but for the sake of brevity we now denote the LHS by Ly(x), i.e. as a lineardifferential operator acting on y(x). Thus (15.58) now readsLy(x) =f(x). (15.59)Let us suppose that a function G(x, z) (the Green’s function) exists such that thegeneral solution to (15.59), which obeys some set of imposed boundary conditionsin the range a ≤ x ≤ b, is given byy(x) =∫ baG(x, z)f(z) dz, (15.60)where z is the integration variable. If we apply the linear differential operator Lto both sides of (15.60) and use (15.59) then we obtainLy(x) =∫ ba[LG(x, z)] f(z) dz = f(x). (15.61)Comparison of (15.61) with a standard property of the Dirac delta function (seesubsection 13.1.3), namelyf(x) =∫ baδ(x − z)f(z) dz,for a ≤ x ≤ b, shows that for (15.61) to hold for any arbitrary function f(x), werequire (for a ≤ x ≤ b) thatLG(x, z) =δ(x − z), (15.62)511

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSi.e. the Green’s function G(x, z) must satisfy the original ODE with the RHS setequal to a delta function. G(x, z) may be thought of physically as the response ofa system to a unit impulse at x = z.In addition to (15.62), we must impose two further sets of restrictions onG(x, z). The first is the requirement that the general solution y(x) in (15.60) obeysthe boundary conditions. For homogeneous boundary conditions, in which y(x)and/or its derivatives are required to be zero at specified points, this is mostsimply arranged by demanding that G(x, z) itself obeys the boundary conditionswhen it is considered as a function of x alone; if, for example, we requirey(a) =y(b) = 0 then we should also demand G(a, z) =G(b, z) = 0. Problemshaving inhomogeneous boundary conditions are discussed at the end of thissubsection.The second set of restrictions concerns the continuity or discontinuity of G(x, z)and its derivatives at x = z and can be found by integrating (15.62) with respectto x over the small interval [z − ɛ, z + ɛ] and taking the limit as ɛ → 0. We thenobtainlimn∑∫ z+ɛɛ→0m=0z−ɛa m (x) dm G(x, z)dx m dx = lim∫ z+ɛɛ→0 z−ɛδ(x − z) dx =1. (15.63)Since d n G/dx n exists at x = z but with value infinity, the (n−1)th-order derivativemust have a finite discontinuity there, whereas all the lower-order derivatives,d m G/dx m for m

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS(ii) When considered as a function of x alone G(x, z) obeys the specified(homogeneous) boundary conditions on y(x).(iii) The derivatives of G(x, z) with respect to x up to order n−2 are continuousat x = z, but the (n − 1)th-order derivative has a discontinuity of 1/a n (z)at this point.◮Use Green’s functions to solved 2 y+ y =cosecx, (15.66)dx2 subject to the boundary conditions y(0) = y(π/2) = 0.From (15.62) we see that the Green’s function G(x, z) mustsatisfyd 2 G(x, z)+ G(x, z) =δ(x − z). (15.67)dx 2Now it is clear that for x ≠ z the RHS of (15.67) is zero, and we are left with thetask of finding the general solution to the homogeneous equation, i.e. the complementaryfunction. The complementary function of (15.67) consists of a linear superposition of sin xand cos x and must consist of different superpositions on either side of x = z, since its(n − 1)th derivative (i.e. the first derivative in this case) is required to have a discontinuitythere. Therefore we assume the form of the Green’s function to be{ A(z)sinx + B(z)cosx for xz.Note that we have performed a similar (but not identical) operation to that used in thevariation-of-parameters method, i.e. we have replaced the constants in the complementaryfunction with functions (this time of z).We must now impose the relevant restrictions on G(x, z) in order to determine thefunctions A(z),...,D(z). The first of these is that G(x, z) should itself obey the homogeneousboundary conditions G(0,z)=G(π/2,z) = 0. This leads to the conclusion that B(z) =C(z) = 0, so we now have{ A(z)sinx for xz.The second restriction is the continuity conditions given in equations (15.64), (15.65),namely that, for this second-order equation, G(x, z) is continuous at x = z and dG/dx hasa discontinuity of 1/a 2 (z) = 1 at this point. Applying these two constraints we haveD(z)cosz − A(z)sinz =0−D(z)sinz − A(z)cosz =1.Solving these equations for A(z) andD(z), we findA(z) =− cos z, D(z) =− sin z.Thus we have{ − cos z sin x for xz.Therefore, from (15.60), the general solution to (15.66) that obeys the boundary conditions513

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSy(0) = y(π/2) = 0 is given byy(x) =∫ π/20= − cos xG(x, z)coseczdz∫ x0sin z cosec zdz− sin x= −x cos x +sinx ln(sin x),∫ π/2xcos z cosec zdzwhich agrees with the result obtained in the previous subsections. ◭As mentioned earlier, once a Green’s function has been obtained for a givenLHS and boundary conditions, it can be used to find a general solution for anyRHS; thus, the solution of d 2 y/dx 2 + y = f(x), with y(0) = y(π/2) = 0, is givenimmediately byy(x) =∫ π/20= − cos xG(x, z)f(z) dz∫ x0sin zf(z) dz − sin x∫ π/2xcos zf(z) dz. (15.68)As an example, the reader may wish to verify that if f(x) =sin2x then (15.68)gives y(x) =(− sin 2x)/3, a solution easily verified by direct substitution. Ingeneral, analytic integration of (15.68) for arbitrary f(x) will prove intractable;then the integrals must be evaluated numerically.Another important point is that although the Green’s function method abovehas provided a general solution, it is also useful for finding a particular integralif the complementary function is known. This is easily seen since in (15.68) theconstant integration limits 0 and π/2 lead merely to constant values by whichthe factors sin x and cos x are multiplied; thus the complementary function isreconstructed. The rest of the general solution, i.e. the particular integral, comesto − ∫ x ,andsodropping the constant integration limits, we can find just the particular integral.For example, a particular integral of d 2 y/dx 2 + y = f(x) that satisfies the aboveboundary conditions is given byfrom the variable integration limit x. Therefore by changing ∫ π/2xy p (x) =− cos x∫ xsin zf(z) dz +sinx∫ xcos zf(z) dz.A very important point to realise about the Green’s function method is that aparticular G(x, z) applies to a given LHS of an ODE and the imposed boundaryconditions, i.e. the same equation with different boundary conditions will have adifferent Green’s function. To illustrate this point, let us consider again the ODEsolved in (15.68), but with different boundary conditions.514

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS◮Use Green’s functions to solved 2 y+ y = f(x), (15.69)dx2 subject to the one-point boundary conditions y(0) = y ′ (0) = 0.We again require (15.67) to hold and so again we assume a Green’s function of the form{ A(z)sinx + B(z)cosx for xz.However, we now require G(x, z) to obey the boundary conditions G(0,z)=G ′ (0,z)=0,which imply A(z) =B(z) = 0. Therefore we have{ 0 for xz.Applying the continuity conditions on G(x, z) as before now givesC(z)sinz + D(z)cosz =0,C(z)cosz − D(z)sinz =1,which are solved to giveC(z) =cosz, D(z) =− sin z.So finally the Green’s function is given by{ 0 for xz,and the general solution to (15.69) that obeys the boundary conditions y(0) = y ′ (0) = 0 isy(x) ==∫ ∞0∫ x0G(x, z)f(z) dzsin(x − z)f(z) dz. ◭Finally, we consider how to deal with inhomogeneous boundary conditionssuch as y(a) =α, y(b) =β or y(0) = y ′ (0) = γ, whereα, β, γ are non-zero. Thesimplest method of solution in this case is to make a change of variable such thatthe boundary conditions in the new variable, u say, are homogeneous, i.e. u(a) =u(b) =0oru(0) = u ′ (0) = 0 etc. For nth-order equations we generally requiren boundary conditions to fix the solution, but these n boundary conditions canbe of various types: we could have the n-point boundary conditions y(x m )=y mfor m =1ton, or the one-point boundary conditions y(x 0 )=y ′ (x 0 )=··· =y (n−1) (x 0 )=y 0 , or something in between. In all cases a suitable change of variableisu = y − h(x),where h(x) isan(n − 1)th-order polynomial that obeys the boundary conditions.515

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSFor example, if we consider the second-order case with boundary conditionsy(a) =α, y(b) =β then a suitable change of variable isu = y − (mx + c),where y = mx + c is the straight line through the points (a, α) and(b, β), for whichm =(α − β)/(a − b) andc =(βa − αb)/(a − b). Alternatively, if the boundaryconditions for our second-order equation are y(0) = y ′ (0) = γ then we wouldmake the same change of variable, but this time y = mx + c would be the straightline through (0,γ) with slope γ, i.e.m = c = γ.Solution method. Require that the Green’s function G(x, z) obeys the original ODE,but with the RHS set to a delta function δ(x − z). This is equivalent to assumingthat G(x, z) is given by the complementary function of the original ODE, with theconstants replaced by functions of z; these functions are different for xz. Now require also that G(x, z) obeys the given homogeneous boundary conditionsand impose the continuity conditions given in (15.64) and (15.65). The generalsolution to the original ODE is then given by (15.60). For inhomogeneous boundaryconditions, make the change of dependent variable u = y − h(x), whereh(x) is apolynomial obeying the given boundary conditions.15.2.6 Canonical form for second-order equationsIn this section we specialise from nth-order linear ODEs with variable coefficientsto those of order 2. In particular we consider the equationd 2 ydx 2 + a 1(x) dydx + a 0(x)y = f(x), (15.70)which has been rearranged so that the coefficient of d 2 y/dx 2 is unity. By makingthe substitution y(x) =u(x)v(x) we obtain( ) ( 2uv ′′ ′u+u + a 1 v ′ ′′ + a 1 u ′ )+ a 0 u+v = f uu , (15.71)where the prime denotes differentiation with respect to x. Since (15.71) would bemuch simplified if there were no term in v ′ ,letuschooseu(x) such that the firstfactor in parentheses on the LHS of (15.71) is zero, i.e.2u ′{ ∫u + a 1 =0 ⇒ u(x) =exp − 1 2We then obtain an equation of the form}a 1 (z) dz . (15.72)d 2 v+ g(x)v = h(x), (15.73)dx2 516

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTSwhereg(x) =a 0 (x) − 1 4 [a 1(x)] 2 − 1 2 a′ 1{ ∫ }(x)h(x) =f(x)exp a 1 (z) dz .Since (15.73) is of a simpler form than the original equation, (15.70), it mayprove easier to solve.12◮Solve4x 2 d2 y dy+4xdx2 dx +(x2 − 1)y =0. (15.74)Dividing (15.74) through by 4x 2 , we see that it is of the form (15.70) with a 1 (x) =1/x,a 0 (x) =(x 2 − 1)/4x 2 and f(x) = 0. Therefore, making the substitution( ∫ ) 1y = vu = v exp −2x dx = √ Av , xwe obtaind 2 vdx + v =0. (15.75)2 4Equation (15.75) is easily solved to givev = c 1 sin 1 x + c 2 2 cos 1 x, 2so the solution of (15.74) isy = √ v = c 1 sin 1 x + c 2 2 cos 1 2√ x . ◭x xAs an alternative to choosing u(x) such that the coefficient of v ′ in (15.71) iszero, we could choose a different u(x) such that the coefficient of v vanishes. Forthis to be the case, we see from (15.71) that we would requireu ′′ + a 1 u ′ + a 0 u =0,so u(x) would have to be a solution of the original ODE with the RHS set tozero, i.e. part of the complementary function. If such a solution were known thenthe substitution y = uv would yield an equation with no term in v, which couldbe solved by two straightforward integrations. This is a special (second-order)case of the method discussed in subsection 15.2.3.Solution method. Write the equation in the form (15.70), then substitute y = uv,where u(x) is given by (15.72). This leads to an equation of the form (15.73), inwhich there is no term in dv/dx and which may be easier to solve. Alternatively,if part of the complementary function is known then follow the method of subsection15.2.3.517

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS15.3 General ordinary differential equationsIn this section, we discuss miscellaneous methods for simplifying general ODEs.These methods are applicable to both linear and non-linear equations and insome cases may lead to a solution. More often than not, however, finding aclosed-form solution to a general non-linear ODE proves impossible.15.3.1 Dependent variable absentIf an ODE does not contain the dependent variable y explicitly, but only itsderivatives, then the change of variable p = dy/dx leads to an equation of oneorder lower.◮Solved 2 ydx +2dy =4x (15.76)2 dxThis is transformed by the substitution p = dy/dx to the first-order equationdp+2p =4x. (15.77)dxThe solution to (15.77) is then found by the method of subsection 14.2.4 and readsp = dydx = ae−2x +2x − 1,where a is a constant. Thus by direct integration the solution to the original equation,(15.76), isy(x) =c 1 e −2x + x 2 − x + c 2 . ◭An extension to the above method is appropriate if an ODE contains onlyderivatives of y that are of order m and greater. Then the substitution p = d m y/dx mreduces the order of the ODE by m.Solution method. If the ODE contains only derivatives of y that are of order m andgreater then the substitution p = d m y/dx m reduces the order of the equation by m.15.3.2 Independent variable absentIf an ODE does not contain the independent variable x explicitly, except in d/dx,d 2 /dx 2 etc., then as in the previous subsection we make the substitution p = dy/dx518

15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONSbut also writed 2 ydx 2 = dpdx = dy dpdx dy = p dpdyd 3 ydx 3 = d (p dp )= dy ddx dy dx dy(p dp )( ) 2= p 2 d2 p dpdy dy 2 + p , (15.78)dyand so on for higher-order derivatives. This leads to an equation of one orderlower.◮Solve( ) 21+y d2 y dydx + =0. (15.79)2 dxMaking the substitutions dy/dx = p and d 2 y/dx 2 = p(dp/dy) we obtain the first-orderODE1+yp dpdy + p2 =0,which is separable and may be solved as in subsection 14.2.1 to obtain(1 + p 2 )y 2 = c 1 .Using p = dy/dx we therefore have√p = dydx = ± c 2 1 − y2,y 2which may be integrated to give the general solution of (15.79); after squaring this reads(x + c 2 ) 2 + y 2 = c 2 1. ◭Solution method. If the ODE does not contain x explicitly then substitute p =dy/dx, along with the relations for higher derivatives given in (15.78), to obtain anequation of one order lower, which may prove easier to solve.15.3.3 Non-linear exact equationsAs discussed in subsection 15.2.2, an exact ODE is one that can be obtained bystraightforward differentiation of an equation of one order lower. Moreover, thenotion of exact equations is useful for both linear and non-linear equations, sincean exact equation can be immediately integrated. It is possible, of course, thatthe resulting equation may itself be exact, so that the process can be repeated.In the non-linear case, however, there is no simple relation (such as (15.43) forthe linear case) by which an equation can be shown to be exact. Nevertheless, ageneral procedure does exist and is illustrated in the following example.519

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Solve2y d3 ydx +6dy d 2 y= x. (15.80)3 dx dx2 Directing our attention to the term on the LHS of (15.80) that contains the highest-orderderivative, i.e. 2yd 3 y/dx 3 , we see that it can be obtained by differentiating 2yd 2 y/dx 2 since( )d2y d2 y=2y d3 ydx dx 2 dx +2dy d 2 y3 dx dx . (15.81)2Rewriting the LHS of (15.80) using (15.81), we are left with 4(dy/dx)(d 2 y/dy 2 ), which mayitself be written as a derivative, i.e.[4 dy d 2 ydx dx = d 22 dx( ) ] 2 dy. (15.82)dxSince, therefore, we can write the LHS of (15.80) as a sum of simple derivatives of otherfunctions, (15.80) is exact. Integrating (15.80) with respect to x, and using (15.81) and(15.82), now gives( ) 2 ∫2y d2 y dydx +2 = 2 dxxdx= x22 + c 1. (15.83)Now we can repeat the process to find whether (15.83) is itself exact. Considering the termon the LHS of (15.83) that contains the highest-order derivative, i.e. 2yd 2 y/dx 2 , we notethat we obtain this by differentiating 2ydy/dx, as follows:ddx(2y dydx)=2y d2 ydx 2 +2 ( dydxThe above expression already contains all the terms on the LHS of (15.83), so we canintegrate (15.83) to give2y dydx = x36 + c 1x + c 2 .Integrating once more we obtain the solutiony 2 = x424 + c 1x 22 + c 2x + c 3 . ◭It is worth noting that both linear equations (as discussed in subsection 15.2.2)and non-linear equations may sometimes be made exact by multiplying throughby an appropriate integrating factor. Although no general method exists forfinding such a factor, one may sometimes be found by inspection or inspiredguesswork.Solution method. Rearrange the equation so that all the terms containing y or itsderivatives are on the LHS, then check to see whether the equation is exact byattempting to write the LHS as a simple derivative. If this is possible then theequation is exact and may be integrated directly to give an equation of one orderlower. If the new equation is itself exact the process can be repeated.) 2.520

15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONS15.3.4 Isobaric or homogeneous equationsIt is straightforward to generalise the discussion of first-order isobaric equationsgiven in subsection 14.2.6 to equations of general order n. Annth-order isobaricequation is one in which every term can be made dimensionally consistent upongiving y and dy each a weight m, andx and dx each a weight 1. Then the nthderivative of y with respect to x, for example, would have dimensions m in yand −n in x. In the special case m = 1, for which the equation is dimensionallyconsistent, the equation is called homogeneous (not to be confused with linearequations with a zero RHS). If an equation is isobaric or homogeneous then thechange in dependent variable y = vx m (y = vx in the homogeneous case) followedby the change in independent variable x = e t leadstoanequationinwhichthenew independent variable t is absent except in the form d/dt.◮Solvex 3 d2 ydx 2 − (x2 + xy) dydx +(y2 + xy) =0. (15.84)Assigning y and dy the weight m, andx and dx the weight 1, the weights of the five termson the LHS of (15.84) are, from left to right: m +1, m +1, 2m, 2m, m +1. For theseweights all to be equal we require m = 1; thus (15.84) is a homogeneous equation. Since itis homogeneous we now make the substitution y = vx, which, after dividing the resultingequation through by x 3 ,givesx d2 v dv+(1− v) =0. (15.85)dx2 dxNow substituting x = e t into (15.85) we obtain (after some working)d 2 vdt − v dv =0, (15.86)2 dtwhich can be integrated directly to givedvdt = 1 2 v2 + c 1 . (15.87)Equation (15.87) is separable, and integrates to give∫1t + d dv2 2 =v 2 + d 2 1= 1 ( ) vtan −1 .d 1 d 1Rearranging and using x = e t and y = vx we finally obtain the solution to (15.84) asy = d 1 x tan ( 1d )2 1 ln x + d 1 d 2 . ◭Solution method. Assume that y and dy have weight m, and x and dx weight 1,and write down the combined weights of each term in the ODE. If these weights canbe made equal by assuming a particular value for m then the equation is isobaric(or homogeneous if m =1). Making the substitution y = vx m followed by x = e tleads to an equation in which the new independent variable t is absent except in theform d/dt.521

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS15.3.5 Equations homogeneous in x or y aloneIt will be seen that the intermediate equation (15.85) in the example of theprevious subsection was simplified by the substitution x = e t , in that this led toan equation in which the new independent variable t occurred only in the formd/dt; see (15.86). A closer examination of (15.85) reveals that it is dimensionallyconsistent in the independent variable x taken alone; this is equivalent to givingthe dependent variable and its differential a weight m = 0. For any equation thatis homogeneous in x alone, the substitution x = e t will lead to an equation thatdoes not contain the new independent variable t except as d/dt. Note that theEuler equation of subsection 15.2.1 is a special, linear example of an equationhomogeneous in x alone. Similarly, if an equation is homogeneous in y alone, thensubstituting y = e v leads to an equation in which the new dependent variable, v,occurs only in the form d/dv.◮Solvex 2 d2 ydx 2 + x dydx + 2 y 3 =0.This equation is homogeneous in x alone, and on substituting x = e t we obtaind 2 ydt + 2 2 y =0, 3which does not contain the new independent variable t except as d/dt. Such equationsmay often be solved by the method of subsection 15.3.2, but in this case we can integratedirectly to obtaindydt = √ 2(c 1 +1/y 2 ).This equation is separable, and we find∫dy√2(c1 +1/y 2 ) = t + c 2.By multiplying the numerator and denominator of the integrand on the LHS by y, we findthe solution√c1 y 2 +1√2c1= t + c 2 .Remembering that t =lnx, we finally obtain√c1 y 2 +1√ =lnx + c 2 . ◭2c1Solution method. If the weight of x taken alone is the same in every term in theODE then the substitution x = e t leads to an equation in which the new independentvariable t is absent except in the form d/dt. If the weight of y taken alone is thesame in every term then the substitution y = e v leads to an equation in which thenew dependent variable v is absent except in the form d/dv.522

15.4 EXERCISES15.3.6 Equations having y = Ae x as a solutionFinally, we note that if any general (linear or non-linear) nth-order ODE issatisfied identically by assuming thaty = dydx = ···= dn ydx n (15.88)then y = Ae x is a solution of that equation. This must be so because y = Ae x isa non-zero function that satisfies (15.88).◮Find a solution of(x 2 + x) dy d 2 ydx dx − 2 x2 y dy ( ) 2 dydx − x =0. (15.89)dxSetting y = dy/dx = d 2 y/dx 2 in (15.89), we obtain(x 2 + x)y 2 − x 2 y 2 − xy 2 =0,which is satisfied identically. Therefore y = Ae x is a solution of (15.89); this is easilyverified by directly substituting y = Ae x into (15.89). ◭Solution method. If the equation is satisfied identically by making the substitutionsy = dy/dx = ···= d n y/dx n then y = Ae x is a solution.15.4 Exercises15.1 A simple harmonic oscillator, of mass m and natural frequency ω 0 , experiencesan oscillating driving force f(t) =ma cos ωt. Therefore, its equation of motion isd 2 xdt + 2 ω2 0x = a cos ωt,where x is its position. Given that at t = 0 we have x = dx/dt = 0, find thefunction x(t). Describe the solution if ω is approximately, but not exactly, equalto ω 0 .15.2 Find the roots of the auxiliary equation for the following. Hence solve them forthe boundary conditions stated.(a) d2 fdt +2df2 dt +5f =0, with f(0) = 1,f′ (0) = 0.(b) d2 fdt +2df2 dt +5f = e−t cos 3t, with f(0) = 0,f ′ (0) = 0.15.3 The theory of bent beams shows that at any point in the beam the ‘bendingmoment’ is given by K/ρ,whereK is a constant (that depends upon the beammaterial and cross-sectional shape) and ρ is the radius of curvature at that point.Consider a light beam of length L whose ends, x =0andx = L, are supportedat the same vertical height and which has a weight W suspended from its centre.Verify that at any point x (0 ≤ x ≤ L/2 for definiteness) the net magnitude ofthe bending moment (bending moment = force × perpendicular distance) due tothe weight and support reactions, evaluated on either side of x, isWx/2.523

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSIf the beam is only slightly bent, so that (dy/dx) 2 ≪ 1, where y = y(x) isthedownward displacement of the beam at x, show that the beam profile satisfiesthe approximate equationd 2 ydx = − Wx2 2K .By integrating this equation twice and using physically imposed conditions onyour solution at x =0andx = L/2, show that the downward displacement atthe centre of the beam is WL 3 /(48K).15.4 Solve the differential equationd 2 fdt +6df2 dt +9f = e−t ,subject to the conditions f =0anddf/dt = λ at t =0.Find the equation satisfied by the positions of the turning points of f(t) andhence, by drawing suitable sketch graphs, determine the number of turning pointsthe solution has in the range t>0if(a)λ =1/4, and (b) λ = −1/4.15.5 The function f(t) satisfies the differential equationd 2 fdt +8df2 dt +12f =12e−4t .For the following sets of boundary conditions determine whether it has solutions,and, if so, find them:(a) f(0) = 0, f ′ (0) = 0, f(ln √ 2) = 0;(b) f(0) = 0, f ′ (0) = −2, f(ln √ 2) = 0.15.6 Determine the values of α and β for which the following four functions arelinearly dependent:y 1 (x) =x cosh x +sinhx,y 2 (x) =x sinh x +coshx,y 3 (x) =(x + α)e x ,y 4 (x) =(x + β)e −x .You will find it convenient to work with those linear combinations of the y i (x)that can be written the most compactly.15.7 A solution of the differential equationd 2 ydx +2dy2 dx + y =4e−xtakes the value 1 when x = 0 and the value e −1 when x = 1. What is its valuewhen x =2?15.8 The two functions x(t) andy(t) satisfy the simultaneous equationsdx− 2y = − sin t,dtdy+2x =5cost.dtFind explicit expressions for x(t) andy(t), given that x(0) = 3 and y(0) = 2.Sketch the solution trajectory in the xy-plane for 0 ≤ t

15.4 EXERCISES15.9 Find the general solutions of(a)(b)d 3 y dy− 12 +16y =32x − 8,dx3 ( dx )(d 1 dy1+(2a coth 2ax)dx y dxydydx)=2a 2 ,where a is a constant.15.10 Use the method of Laplace transforms to solve(a)d 2 fdt +5df2 dt +6f =0, f(0) = 1, f′ (0) = −4,(b)d 2 fdt +2df2 dt +5f =0, f(0) = 1, f′ (0) = 0.15.11 The quantities x(t), y(t) satisfy the simultaneous equationsẍ +2nẋ + n 2 x =0,ÿ +2nẏ + n 2 y = µẋ,where x(0) = y(0) = ẏ(0) = 0 and ẋ(0) = λ. Show thaty(t) = ( 1 2 µλt2 1 − 1 nt) exp(−nt).315.12 Use Laplace transforms to solve, for t ≥ 0, the differential equationsẍ +2x + y =cost,ÿ +2x +3y =2cost,which describe a coupled system that starts from rest at the equilibrium position.Show that the subsequent motion takes place along a straight line in the xy-plane.Verify that the frequency at which the system is driven is equal to one of theresonance frequencies of the system; explain why there is no resonant behaviourin the solution you have obtained.15.13 Two unstable isotopes A and B and a stable isotope C have the following decayrates per atom present: A → B, 3s −1 ; A → C, 1s −1 ; B → C, 2s −1 . Initially aquantity x 0 of A is present, but there are no atoms of the other two types. UsingLaplace transforms, find the amount of C present at a later time t.15.14 For a lightly damped (γ < ω 0 ) harmonic oscillator driven at its undampedresonance frequency ω 0 , the displacement x(t) attimet satisfies the equationd 2 x dx+2γdt 2 dt + ω2 0x = F sin ω 0 t.Use Laplace transforms to find the displacement at a general time if the oscillatorstarts from rest at its equilibrium position.(a) Show that ultimately the oscillation has amplitude F/(2ω 0 γ), with a phaselag of π/2 relative to the driving force per unit mass F.(b) By differentiating the original equation, conclude that if x(t) is expanded asa power series in t for small t, then the first non-vanishing term is Fω 0 t 3 /6.Confirm this conclusion by expanding your explicit solution.15.15 The ‘golden mean’, which is said to describe the most aesthetically pleasingproportions for the sides of a rectangle (e.g. the ideal picture frame), is givenby the limiting value of the ratio of successive terms of the Fibonacci series u n ,which is generated byu n+2 = u n+1 + u n ,with u 0 =0andu 1 = 1. Find an expression for the general term of the series and525

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSverify that the golden mean is equal to the larger root of the recurrence relation’scharacteristic equation.15.16 In a particular scheme for numerically modelling one-dimensional fluid flow, thesuccessive values, u n , of the solution are connected for n ≥ 1 by the differenceequationc(u n+1 − u n−1 )=d(u n+1 − 2u n + u n−1 ),where c and d are positive constants. The boundary conditions are u 0 =0andu M = 1. Find the solution to the equation, and show that successive values of u nwill have alternating signs if c>d.15.17 The first few terms of a series u n , starting with u 0 ,are1, 2, 2, 1, 6, −3. The seriesis generated by a recurrence relation of the formu n = Pu n−2 + Qu n−4 ,where P and Q are constants. Find an expression for the general term of theseries and show that, in fact, the series consists of two interleaved series given byu 2m = 2 + 1 3 3 4m ,u 2m+1 = 7 − 1 3 3 4m ,for m =0, 1, 2,... .15.18 Find an explicit expression for the u n satisfyingu n+1 +5u n +6u n−1 =2 n ,given that u 0 = u 1 = 1. Deduce that 2 n − 26(−3) n is divisible by 5 for allnon-negative integers n.15.19 Find the general expression for the u n satisfyingu n+1 =2u n−2 − u nwith u 0 = u 1 =0andu 2 = 1, and show that they can be written in the formu n = 1 ( )5 − 2n/2 3πn√ cos5 4 − φ ,where tan φ =2.15.20 Consider the seventh-order recurrence relationu n+7 − u n+6 − u n+5 + u n+4 − u n+3 + u n+2 + u n+1 − u n =0.Find the most general form of its solution, and show that:(a) if only the four initial values u 0 =0,u 1 =2,u 2 =6andu 3 = 12, are specified,then the relation has one solution that cycles repeatedly through this set offour numbers;(b) but if, in addition, it is required that u 4 = 20, u 5 =30andu 6 = 42 then thesolution is unique, with u n = n(n +1).15.21 Find the general solution ofx 2 d2 ydx − x dy2 dx + y = x,given that y(1) = 1 and y(e) =2e.15.22 Find the general solution of(x +1) 2 d2 y+3(x +1)dydx2 dx + y = x2 .526

15.4 EXERCISES15.23 Prove that the general solution of(x − 2) d2 ydx 2 +3dy dx + 4yx 2 =0is given by[ (1 2y(x) = k(x − 2) 2 3x − 1 ) ]+ cx 2 .215.24 Use the method of variation of parameters to find the general solutions of(a)d 2 ydx 2 − y = xn ,(b) d2 ydx 2 − 2 dydx + y =2xex .15.25 Use the intermediate result of exercise 15.24(a) to find the Green’s function thatsatisfiesd 2 G(x, ξ)− G(x, ξ) =δ(x − ξ) with G(0,ξ)=G(1,ξ)=0.dx 215.26 Consider the equationF(x, y) =x(x +1) d2 ydx +(2− 2 x2 ) dy − (2 + x)y =0.dx(a) Given that y 1 (x) =1/x is one of its solutions, find a second linearly independentone,(i) by setting y 2 (x) =y 1 (x)u(x), and(ii) by noting the sum of the coefficients in the equation.(b) Hence, using the variation of parameters method, find the general solutionofF(x, y) =(x +1) 2 .15.27 Show generally that if y 1 (x) andy 2 (x) are linearly independent solutions ofd 2 y dy+ p(x) + q(x)y =0,dx2 dxwith y 1 (0) = 0 and y 2 (1) = 0, then the Green’s function G(x, ξ) for the interval0 ≤ x, ξ ≤ 1andwithG(0,ξ)=G(1,ξ)=0canbewrittenintheform{y1 (x)y 2 (ξ)/W (ξ) 0

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS15.29 The equation of motion for a driven damped harmonic oscillator can be writtenẍ +2ẋ +(1+κ 2 )x = f(t),with κ ≠0.Ifitstartsfromrestwithx(0) = 0 and ẋ(0) = 0, find the correspondingGreen’s function G(t, τ) and verify that it can be written as a function of t − τonly. Find the explicit solution when the driving force is the unit step function,i.e. f(t) =H(t). Confirm your solution by taking the Laplace transforms of bothit and the original equation.15.30 Show that the Green’s function for the equationd 2 ydx + y 2 4 = f(x),subject to the boundary conditions y(0) = y(π) = 0, is given by{−2cos12G(x, z) =x sin 1 z 0 ≤ z ≤ x,2−2sin 1 x cos 1 z x ≤ z ≤ π.2 215.31 Find the Green’s function x = G(t, t 0 ) that solvesd 2 xdt + α dx2 dt = δ(t − t 0)under the initial conditions x = dx/dt =0att =0.Hencesolved 2 xdt + α dx2 dt = f(t),where f(t) =0fort0).15.32 Consider the equationd 2 y+ f(y) =0,dx2 where f(y) can be any function.(a) By multiplying through by dy/dx, obtain the general solution relating x andy.(b) A mass m, initially at rest at the point x = 0, is accelerated by a force[f(x) =A(x 0 − x) 1+2ln(1 − x )].x 0Its equation of motion is md 2 x/dt 2 = f(x). Find x as a function of time, andshow that ultimately the particle has travelled a distance x 0 .15.33 Solve(2y d3 ydx +2 y +3 dy ) ( ) d 2 2y dy3 dx dx +2 = sinx.2 dx15.34 Find the general solution of the equationx d3 y ydx 3 +2d2 dx = Ax. 215.35 Express the equationd 2 y dy+4xdx2 dx +(4x2 +6)y = e −x2 sin 2xin canonical form and hence find its general solution.528

15.5 HINTS AND ANSWERS15.36 Find the form of the solutions of the equation( )dy d 3 y d 2 2 ( ) 2dx dx − 2 y dy+ =03 dx 2 dxthat have y(0) = ∞.[ You will need the result ∫ z cosech udu= − ln(cosech z +cothz). ]15.37 Consider the equation( ) 2x p y ′′ n +3− 2p p − 2+ x p−1 y ′ + x p−2 y = y n ,n − 1n − 1in which p ≠2andn>−1 but n ≠ 1. For the boundary conditions y(1) = 0 andy ′ (1) = λ, show that the solution is y(x) =v(x)x (p−2)/(n−1) ,wherev(x) isgivenby∫ v(x)0dz[λ2 +2z n+1 /(n +1) ] 1/2 =lnx.15.5 Hints and answers15.1 The function is a(ω0 2 − ω2 ) −1 (cos ωt − cos ω 0 t); for moderate t, x(t) is a sine waveof linearly increasing amplitude (t sin ω 0 t)/(2ω 0 ); for large t it shows beats ofmaximum amplitude 2(ω0 2 − ω2 ) −1 .15.3 Ignore the term y ′2 , compared with 1, in the expression for ρ. y =0atx =0.From symmetry, dy/dx =0atx = L/2.15.5 General solution f(t) = Ae −6t + Be −2t − 3e −4t . (a) No solution, inconsistentboundary conditions; (b) f(t) =2e −6t + e −2t − 3e −4t .15.7 The auxiliary equation has repeated roots and the RHS is contained in thecomplementary function. The solution is y(x) =(A+Bx)e −x +2x 2 e −x . y(2) = 5e −2 .15.9 (a) The auxiliary equation has roots 2, 2, −4; (A+Bx)exp2x+C exp(−4x)+2x+1;(b) ∫ multiply through by sinh 2ax and note thatcosech 2ax dx =(2a) −1 ln(| tanh ax|); y = B(sinh 2ax) 1/2 (| tanh ax|) A .15.11 Use Laplace transforms; write s(s + n) −4 as (s + n) −3 − n(s + n) −4 .15.13 L [C(t)] = x 0 (s +8)/[s(s +2)(s + 4)], yieldingC(t) =x 0 [1 + 1 exp(−4t) − 3 exp(−2t)].2 215.15 The characteristic equation is λ 2 − λ − 1=0.u n =[(1+ √ 5) n − (1 − √ 5) n ]/(2 n√ 5).15.17 From u 4 and u 5 , P =5,Q= −4. u n =3/2 − 5(−1) n /6+(−2) n /4+2 n /12.15.19 The general solution is A+B2 n/2 exp(i3πn/4)+C2 n/2 exp(i5πn/4). The initial valuesimply that A =1/5,B =( √ 5/10) exp[i(π − φ)] and C =( √ 5/10) exp[i(π + φ)].15.21 This is Euler’s equation; setting x =expt produces d 2 z/dt 2 − 2 dz/dt + z =expt,with complementary function (A + Bt)expt and particular integral t 2 (exp t)/2;y(x) =x +[x ln x(1 + ln x)]/2.15.23 After multiplication through by x 2 the coefficients are such that this is anexact equation. The resulting first-order equation, in standard form, needs anintegrating factor (x − 2) 2 /x 2 .15.25 Given the boundary conditions, it is better to work with sinh x and sinh(1 − x)than with e ±x ; G(x, ξ) =−[sinh(1 − ξ)sinhx]/ sinh 1 for xξ.15.27 Follow the method of subsection 15.2.5, but using general rather than specificfunctions.15.29 G(t, τ) =0fortτ. For a unit step input,x(t) =(1+κ 2 ) −1 (1 − e −t cos κt − κ −1 e −t sin κt). Both transforms are equivalent tos[(s +1) 2 + κ 2 )]¯x =1.529

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS15.31 Use continuity and the step condition on ∂G/∂t at t = t 0 to show thatG(t, t 0 )=α −1 {1 − exp[α(t 0 − t)]} for 0 ≤ t 0 ≤ t;x(t) =A(α − a) −1 {a −1 [1 − exp(−at)] − α −1 [1 − exp(−αt)]}.15.33 The LHS of the equation is exact for two stages of integration and then needsan integrating factor exp x; 2yd 2 y/dx 2 +2ydy/dx+2(dy/dx) 2 ;2ydy/dx+ y 2 =d(y 2 )/dx + y 2 ; y 2 = A exp(−x)+Bx + C − (sin x − cos x)/2.15.35 Follow the method of subsection 15.2.6; u(x) =e −x2 and v(x) satisfiesv ′′ +4v =sin 2x, for which a particular integral is (−x cos 2x)/4. The general solution isy(x) =[A sin 2x +(B − 1 4 x)cos2x]e−x2 .15.37 The equation is isobaric, with y of weight m, where m + p − 2 = mn; v(x)satisfies x 2 v ′′ + xv ′ = v n .Setx = e t and v(x) =u(t), leading to u ′′ = u n withu(0) = 0,u ′ (0) = λ. Multiply both sides by u ′ to make the equation exact.530

16Series solutions of ordinarydifferential equationsIn the previous chapter the solution of both homogeneous and non-homogeneouslinear ordinary differential equations (ODEs) of order ≥ 2 was discussed. In particularwe developed methods for solving some equations in which the coefficientswere not constant but functions of the independent variable x. In each case wewere able to write the solutions to such equations in terms of elementary functions,or as integrals. In general, however, the solutions of equations with variablecoefficients cannot be written in this way, and we must consider alternativeapproaches.In this chapter we discuss a method for obtaining solutions to linear ODEsin the form of convergent series. Such series can be evaluated numerically, andthose occurring most commonly are named and tabulated. There is in fact nodistinct borderline between this and the previous chapter, since solutions in termsof elementary functions may equally well be written as convergent series (i.e. therelevant Taylor series). Indeed, it is partly because some series occur so frequentlythat they are given special names such as sin x, cosx or exp x.Since we shall be concerned principally with second-order linear ODEs in thischapter, we begin with a discussion of these equations, and obtain some generalresults that will prove useful when we come to discuss series solutions.16.1 Second-order linear ordinary differential equationsAny homogeneous second-order linear ODE can be written in the formy ′′ + p(x)y ′ + q(x)y =0, (16.1)where y ′ = dy/dx and p(x) andq(x) are given functions of x. From the previouschapter, we recall that the most general form of the solution to (16.1) isy(x) =c 1 y 1 (x)+c 2 y 2 (x), (16.2)531

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSwhere y 1 (x) andy 2 (x) arelinearly independent solutions of (16.1), and c 1 and c 2are constants that are fixed by the boundary conditions (if supplied).A full discussion of the linear independence of sets of functions was givenat the beginning of the previous chapter, but for just two functions y 1 and y 2to be linearly independent we simply require that y 2 is not a multiple of y 1 .Equivalently, y 1 and y 2 must be such that the equationc 1 y 1 (x)+c 2 y 2 (x) =0is only satisfied for c 1 = c 2 = 0. Therefore the linear independence of y 1 (x) andy 2 (x) can usually be deduced by inspection but in any case can always be verifiedby the evaluation of the Wronskian of the two solutions,W (x) =∣ y 1 y 2y 1 ′ y 2′ ∣ = y 1y 2 ′ − y 2 y 1. ′ (16.3)If W (x) ≠ 0 anywhere in a given interval then y 1 and y 2 are linearly independentin that interval.An alternative expression for W (x), of which we will make use later, may bederived by differentiating (16.3) with respect to x to giveW ′ = y 1 y 2 ′′ + y 1y ′ 2 ′ − y 2 y 1 ′′ − y 2y ′ 1 ′ = y 1 y 2 ′′ − y 1y ′′2 .Since both y 1 and y 2 satisfy (16.1), we may substitute for y 1 ′′ and y′′ 2 to obtainW ′ = −y 1 (py 2 ′ + qy 2 )+(py 1 ′ + qy 1 )y 2 = −p(y 1 y 2 ′ − y 1y ′ 2 )=−pW .Integrating, we find{ ∫ x}W (x) =C exp − p(u) du , (16.4)where C is a constant. We note further that in the special case p(x) ≡ 0weobtainW =constant.◮The functions y 1 =sinx and y 2 =cosx are both solutions of the equation y ′′ + y =0. Evaluate the Wronskian of these two solutions, and hence show that they are linearlyindependent.The Wronskian of y 1 and y 2 is given byW = y 1 y 2 ′ − y 2 y 1 ′ = − sin 2 x − cos 2 x = −1.Since W ≠ 0 the two solutions are linearly independent. We also note that y ′′ + y =0isa special case of (16.1) with p(x) = 0. We therefore expect, from (16.4), that W will be aconstant, as is indeed the case. ◭From the previous chapter we recall that, once we have obtained the generalsolution to the homogeneous second-order ODE (16.1) in the form (16.2), thegeneral solution to the inhomogeneous equationy ′′ + p(x)y ′ + q(x)y = f(x) (16.5)532

16.1 SECOND-ORDER LINEAR ORDINARY DIFFERENTIAL EQUATIONScan be written as the sum of the solution to the homogeneous equation y c (x)(the complementary function) and any function y p (x) (the particular integral) thatsatisfies (16.5) and is linearly independent of y c (x). We have thereforey(x) =c 1 y 1 (x)+c 2 y 2 (x)+y p (x). (16.6)General methods for obtaining y p , that are applicable to equations with variablecoefficients, such as the variation of parameters or Green’s functions, were discussedin the previous chapter. An alternative description of the Green’s functionmethod for solving inhomogeneous equations is given in the next chapter. For thepresent, however, we will restrict our attention to the solutions of homogeneousODEs in the form of convergent series.16.1.1 Ordinary and singular points of an ODESo far we have implicitly assumed that y(x) isareal function of a real variablex. However, this is not always the case, and in the remainder of this chapter webroaden our discussion by generalising to a complex function y(z) ofacomplexvariable z.Let us therefore consider the second-order linear homogeneous ODEy ′′ + p(z)y ′ + q(z) =0, (16.7)where now y ′ = dy/dz; this is a straightforward generalisation of (16.1). A fulldiscussion of complex functions and differentiation with respect to a complexvariable z is given in chapter 24, but for the purposes of the present chapter weneed not concern ourselves with many of the subtleties that exist. In particular,we may treat differentiation with respect to z in a way analogous to ordinarydifferentiation with respect to a real variable x.In (16.7), if, at some point z = z 0 , the functions p(z) andq(z) are finite and canbe expressed as complex power series (see section 4.5), i.e.∞∑∞∑p(z) = p n (z − z 0 ) n , q(z) = q n (z − z 0 ) n ,n=0then p(z) andq(z) are said to be analytic at z = z 0 , and this point is called anordinary point of the ODE. If, however, p(z) orq(z), or both, diverge at z = z 0then it is called a singular point of the ODE.Even if an ODE is singular at a given point z = z 0 , it may still possess anon-singular (finite) solution at that point. In fact the necessary and sufficientcondition § for such a solution to exist is that (z − z 0 )p(z) and(z − z 0 ) 2 q(z) areboth analytic at z = z 0 . Singular points that have this property are called regularn=0§ See, for example, H. Jeffreys and B. S. Jeffreys, Methods of Mathematical Physics, 3rdedn(Cambridge:Cambridge University Press, 1966), p. 479.533

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSsingular points, whereas any singular point not satisfying both these criteria istermed an irregular or essential singularity.◮Legendre’s equation has the form(1 − z 2 )y ′′ − 2zy ′ + l(l +1)y =0, (16.8)where l is a constant. Show that z =0is an ordinary point and z = ±1 are regular singularpoints of this equation.Firstly, divide through by 1 − z 2 to put the equation into our standard form (16.7):y ′′ −2z l(l +1)1 − z 2 y′ +1 − z y =0.2Comparing this with (16.7), we identify p(z) andq(z) asp(z) =−2z1 − z = −2zl(l +1) l(l +1), q(z) = =2 (1 + z)(1 − z) 1 − z 2 (1 + z)(1 − z) .By inspection, p(z) andq(z) are analytic at z = 0, which is therefore an ordinary point,but both diverge for z = ±1, which are thus singular points. However, at z =1weseethat both (z − 1)p(z) and(z − 1) 2 q(z) are analytic and hence z = 1 is a regular singularpoint. Similarly, at z = −1 both (z +1)p(z) and(z +1) 2 q(z) are analytic, and it too is aregular singular point. ◭So far we have assumed that z 0 is finite. However, we may sometimes wish todetermine the nature of the point |z| →∞. This may be achieved straightforwardlyby substituting w =1/z into the equation and investigating the behaviour atw =0.◮Show that Legendre’s equation has a regular singularity at |z| →∞.Letting w =1/z, the derivatives with respect to z becomedydz = dy dwdw dz = − 1 dy dy=z 2 −w2dw dw ,d 2 ydz = dw ( ) (d dy= −w 2 −2w dy ) (2 dz dw dzdw − d2 yw2 = w 3 2 dy )dw 2 dw + w d2 y.dw 2If we substitute these derivatives into Legendre’s equation (16.8) we obtain(1 − 1 ) (w 3 2 dy )w 2 dw + w d2 y+2 1 dydw 2 w2 + l(l +1)y =0,w dwwhich simplifies to givew 2 (w 2 − 1) d2 y dy+2w3 + l(l +1)y =0.dw2 dwDividing through by w 2 (w 2 − 1) to put the equation into standard form, and comparingwith (16.7), we identify p(w) andq(w) asp(w) =2wl(l +1), q(w) =w 2 − 1 w 2 (w 2 − 1) .At w =0,p(w) is analytic but q(w) diverges, and so the point |z| →∞is a singular pointof Legendre’s equation. However, since wp and w 2 q are both analytic at w =0,|z| →∞is a regular singular point. ◭534

16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINTEquation Regular Essentialsingularities singularitiesHypergeometricz(1 − z)y ′′ +[c − (a + b +1)z]y ′ − aby =0 0, 1, ∞ —Legendre(1 − z 2 )y ′′ − 2zy ′ + l(l +1)y =0 −1, 1, ∞ —Associated Legendre](1 − z 2 )y ′′ − 2zy ′ +[l(l +1)− m2y =0 −1, 1, ∞ —1 − z 2Chebyshev(1 − z 2 )y ′′ − zy ′ + ν 2 y =0 −1, 1, ∞ —Confluent hypergeometriczy ′′ +(c − z)y ′ − ay =0 0 ∞Besselz 2 y ′′ + zy ′ +(z 2 − ν 2 )y =0 0 ∞Laguerrezy ′′ +(1− z)y ′ + νy =0 0 ∞Associated Laguerrezy ′′ +(m +1− z)y ′ +(ν − m)y =0 0 ∞Hermitey ′′ − 2zy ′ +2νy =0 — ∞Simple harmonic oscillatory ′′ + ω 2 y =0 — ∞Table 16.1 Important second-order linear ODEs in the physical sciences andengineering.Table 16.1 lists the singular points of several second-order linear ODEs thatplay important roles in the analysis of many problems in physics and engineering.A full discussion of the solutions to each of the equations in table 16.1 and theirproperties is left until chapter 18. We now discuss the general methods by whichseries solutions may be obtained.16.2 Series solutions about an ordinary pointIf z = z 0 is an ordinary point of (16.7) then it may be shown that every solutiony(z) of the equation is also analytic at z = z 0 . From now on we will take z 0 as theorigin, i.e. z 0 = 0. If this is not already the case, then a substitution Z = z − z 0will make it so. Since every solution is analytic, y(z) can be represented by a535

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSpower series of the form (see section 24.11)∞∑y(z) = a n z n . (16.9)n=0Moreover, it may be shown that such a power series converges for |z|

16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINTagivena 0 . Alternatively, starting with a 1 , we obtain the odd coefficients a 3 ,a 5 ,etc.Twoindependent solutions of the ODE can be obtained by setting either a 0 =0ora 1 =0.Firstly, if we set a 1 = 0 and choose a 0 = 1 then we obtain the solution∞y 1 (z) =1− z22! + z44! − ···= ∑ (−1) n(2n)! z2n .n=0Secondly, if we set a 0 = 0 and choose a 1 = 1 then we obtain a second, independent, solution∞y 2 (z) =z − z33! + z55! − ···= ∑ (−1) n(2n +1)! z2n+1 .n=0Recognising these two series as cos z and sin z, we can write the general solution asy(z) =c 1 cos z + c 2 sin z,where c 1 and c 2 are arbitrary constants that are fixed by boundary conditions (if supplied).We note that both solutions converge for all z, as might be expected since the ODEpossesses no singular points (except |z| →∞). ◭Solving the above example was quite straightforward and the resulting serieswere easily recognised and written in closed form (i.e. in terms of elementaryfunctions); this is not usually the case. Another simplifying feature of the previousexample was that we obtained a two-term recurrence relation relating a n+2 anda n , so that the odd- and even-numbered coefficients were independent of oneanother. In general, the recurrence relation expresses a n in terms of any numberof the previous a r (0 ≤ r ≤ n − 1).◮Find the series solutions, about z =0,ofy ′′ −2(1 − z) 2 y =0.By inspection, z = 0 is an ordinary point, and therefore we may find two independentsolutions by substituting y = ∑ ∞n=0 a nz n . Using (16.10) and (16.11), and multiplying throughby (1 − z) 2 , we find∞∑∞∑(1 − 2z + z 2 ) n(n − 1)a n z n−2 − 2 a n z n =0,n=0n=0n=0which leads to∞∑∞∑∞∑∞∑n(n − 1)a n z n−2 − 2 n(n − 1)a n z n−1 + n(n − 1)a n z n − 2 a n z n =0.In order to write all these series in terms of the coefficients of z n , we must shift thesummation index in the first two sums, obtaining∞∑∞∑∞∑(n +2)(n +1)a n+2 z n − 2 (n +1)na n+1 z n + (n 2 − n − 2)a n z n =0,n=0which can be written as∞∑(n +1)[(n +2)a n+2 − 2na n+1 +(n − 2)a n ]z n =0.n=0n=0537n=0n=0n=0n=0

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSBy demanding that the coefficients of each power of z vanish separately, we obtain thethree-term recurrence relation(n +2)a n+2 − 2na n+1 +(n − 2)a n =0 forn ≥ 0,which determines a n for n ≥ 2intermsofa 0 and a 1 . Three-term (or more) recurrencerelations are a nuisance and, in general, can be difficult to solve. This particular recurrencerelation, however, has two straightforward solutions. One solution is a n = a 0 for all n, inwhich case (choosing a 0 = 1) we findy 1 (z) =1+z + z 2 + z 3 + ···= 11 − z .The other solution to the recurrence relation is a 1 = −2a 0 , a 2 = a 0 and a n =0forn>2,so that (again choosing a 0 =1)weobtainapolynomial solution to the ODE:y 2 (z) =1− 2z + z 2 =(1− z) 2 .The linear independence of y 1 and y 2 is obvious but can be checked by computing theWronskianW = y 1 y 2 ′ − y 1y ′ 2 = 11 − z [−2(1 − z)] − 1(1 − z) (1 − 2 z)2 = −3.Since W ≠ 0, the two solutions y 1 and y 2 are indeed linearly independent. The generalsolution of the ODE is thereforey(z) = c 11 − z + c 2(1 − z) 2 .We observe that y 1 (and hence the general solution) is singular at z =1,whichisthesingular point of the ODE nearest to z = 0, but the polynomial solution, y 2 , is valid forall finite z. ◭The above example illustrates the possibility that, in some cases, we may findthat the recurrence relation leads to a n =0forn>N, for one or both of thetwo solutions; we then obtain a polynomial solution to the equation. Polynomialsolutions are discussed more fully in section 16.5, but one obvious property ofsuch solutions is that they converge for all finite z. By contrast, as mentionedabove, for solutions in the form of an infinite series the circle of convergenceextends only as far as the singular point nearest to that about which the solutionis being obtained.16.3 Series solutions about a regular singular pointFrom table 16.1 we see that several of the most important second-order linearODEs in physics and engineering have regular singular points in the finite complexplane. We must extend our discussion, therefore, to obtaining series solutions toODEs about such points. In what follows we assume that the regular singularpoint about which the solution is required is at z = 0, since, as we have seen, ifthis is not already the case then a substitution of the form Z = z − z 0 will makeit so.If z = 0 is a regular singular point of the equationy ′′ + p(z)y ′ + q(z)y =0538

16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINTthen at least one of p(z) andq(z) is not analytic at z = 0, and in general weshould not expect to find a power series solution of the form (16.9). We musttherefore extend the method to include a more general form for the solution. Infact, it may be shown (Fuch’s theorem) that there exists at least one solution tothe above equation, of the formy = z σ∞ ∑n=0a n z n , (16.12)where the exponent σ is a number that may be real or complex and where a 0 ≠0(since, if it were otherwise, σ could be redefined as σ +1orσ +2or··· so as tomake a 0 ≠ 0). Such a series is called a generalised power series or Frobenius series.As in the case of a simple power series solution, the radius of convergence of theFrobenius series is, in general, equal to the distance to the nearest singularity ofthe ODE.Since z = 0 is a regular singularity of the ODE, it follows that zp(z) andz 2 q(z)are analytic at z = 0, so that we may write∞∑zp(z) ≡ s(z) = s n z n ,z 2 q(z) ≡ t(z) =n=0∞∑t n z n ,where we have defined the analytic functions s(z) andt(z) for later convenience.The original ODE therefore becomesy ′′ + s(z)z y′ + t(z)z 2 y =0.Let us substitute the Frobenius series (16.12) into this equation. The derivativesof (16.12) with respect to z are given by∞∑y ′ = (n + σ)a n z n+σ−1 , (16.13)n=0y ′′ =n=0n=0∞∑(n + σ)(n + σ − 1)a n z n+σ−2 , (16.14)n=0and we obtain∞∑∞∑∞∑(n + σ)(n + σ − 1)a n z n+σ−2 + s(z) (n + σ)a n z n+σ−2 + t(z) a n z n+σ−2 =0.n=0Dividing this equation through by z σ−2 , we find∞∑[(n + σ)(n + σ − 1) + s(z)(n + σ)+t(z)] a n z n =0. (16.15)n=0539n=0

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSSetting z = 0, all terms in the sum with n>0 vanish, implying that[σ(σ − 1) + s(0)σ + t(0)]a 0 =0,which, since we require a 0 ≠ 0, yields the indicial equationσ(σ − 1) + s(0)σ + t(0) = 0. (16.16)This equation is a quadratic in σ and in general has two roots, the nature ofwhich determines the forms of possible series solutions.The two roots of the indicial equation, σ 1 and σ 2 , are called the indices ofthe regular singular point. By substituting each of these roots into (16.15) inturn and requiring that the coefficients of each power of z vanish separately, weobtain a recurrence relation (for each root) expressing each a n as a function ofthe previous a r (0 ≤ r ≤ n − 1). We will see that the larger root of the indicialequation always yields a solution to the ODE in the form of a Frobenius series(16.12). The form of the second solution depends, however, on the relationshipbetween the two indices σ 1 and σ 2 . There are three possible general cases: (i)distinct roots not differing by an integer; (ii) repeated roots; (iii) distinct rootsdiffering by an integer (not equal to zero). Below, we discuss each of these in turn.Before continuing, however, we note that, as was the case for solutions inthe form of a simple power series, it is always worth investigating whether aFrobenius series found as a solution to a problem is summable in closed formor expressible in terms of known functions. We illustrate this point below, butthe reader should avoid gaining the impression that this is always so or that, ifone worked hard enough, a closed-form solution could always be found withoutusing the series method. As mentioned earlier, this is not the case, and very oftenan infinite series solution is the best one can do.16.3.1 Distinct roots not differing by an integerIf the roots of the indicial equation, σ 1 and σ 2 , differ by an amount that is notan integer then the recurrence relations corresponding to each root lead to twolinearly independent solutions of the ODE:y 1 (z) =z σ1∞ ∑n=0∑∞a n z n , y 2 (z) =z σ2 b n z n ,with both solutions taking the form of a Frobenius series. The linear independenceof these two solutions follows from the fact that y 2 /y 1 is not a constant sinceσ 1 − σ 2 is not an integer. Because y 1 and y 2 are linearly independent, we may usethem to construct the general solution y = c 1 y 1 + c 2 y 2 .We also note that this case includes complex conjugate roots where σ 2 = σ ∗ 1 ,since σ 1 − σ 2 = σ 1 − σ ∗ 1 =2i Im σ 1 cannot be equal to a real integer.540n=0

16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINT◮Find the power series solutions about z =0of4zy ′′ +2y ′ + y =0.Dividing through by 4z to put the equation into standard form, we obtainy ′′ + 1 2z y′ + 1 y =0, (16.17)4zand on comparing with (16.7) we identify p(z) =1/(2z) andq(z) =1/(4z). Clearly z =0is a singular point of (16.17), but since zp(z) =1/2 andz 2 q(z) =z/4 are finite there, itis a regular singular point. We therefore substitute the Frobenius series y = z ∑ σ ∞n=0 a nz ninto (16.17). Using (16.13) and (16.14), we obtain∞∑(n + σ)(n + σ − 1)a n z n+σ−2 + 1 ∞∑(n + σ)a n z n+σ−1 + 1 ∞∑a n z n+σ =0,2z4zn=0n=0which, on dividing through by z σ−2 ,gives∞∑ [(n + σ)(n + σ − 1) +1(n + σ)+ 1 z] a2 4 n z n =0. (16.18)n=0If we set z = 0 then all terms in the sum with n>0 vanish, and we obtain the indicialequationσ(σ − 1) + 1 σ =0, 2which has roots σ =1/2 andσ = 0. Since these roots do not differ by an integer, weexpect to find two independent solutions to (16.17), in the form of Frobenius series.Demanding that the coefficients of z n vanish separately in (16.18), we obtain therecurrence relation(n + σ)(n + σ − 1)a n + 1 (n + σ)a 2 n + 1 a 4 n−1 =0. (16.19)If we choose the larger root, σ =1/2, of the indicial equation then (16.19) becomes(4n 2 +2n)a n + a n−1 =0 ⇒ a n = −a n−12n(2n +1) .Setting a 0 = 1, we find a n =(−1) n /(2n + 1)!, and so the solution to (16.17) is given byy 1 (z) = √ z∞∑n=0= √ z − (√ z) 33!(−1) n(2n +1)! zn+ (√ z) 55!− ···=sin √ z.To obtain the second solution we set σ = 0 (the smaller root of the indicial equation) in(16.19), which gives(4n 2 − 2n)a n + a n−1 =0 ⇒ a n = − a n−12n(2n − 1) .Setting a 0 =1nowgivesa n =(−1) n /(2n)!, and so the second (independent) solution to(16.17) isy 2 (z) =∞∑n=0n=0(−1) n(2n)! zn =1− (√ z) 2+ (√ 4) 4− ···=cos √ z.2! 4!541

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSWe may check that y 1 (z) andy 2 (z) are indeed linearly independent by computing theWronskian as follows:W = y 1 y 2 ′ − y 2 y 1′=sin √ (z − 12 √ z sin √ )z − cos √ ( 1z2 √ z cos √ )z= − 12 √ ( √ sin2z +cos √ 2 z ) = − 1z2 √ z ≠0.Since W ≠0,thesolutionsy 1 (z) andy 2 (z) are linearly independent. Hence, the generalsolution to (16.17) is given byy(z) =c 1 sin √ z + c 2 cos √ z. ◭16.3.2 Repeated root of the indicial equationIf the indicial equation has a repeated root, so that σ 1 = σ 2 = σ, then obviouslyonly one solution in the form of a Frobenius series (16.12) may be found asdescribed above, i.e.y 1 (z) =z σ∞ ∑n=0a n z n .Methods for obtaining a second, linearly independent, solution are discussed insection 16.4.16.3.3 Distinct roots differing by an integerWhatever the roots of the indicial equation, the recurrence relation correspondingto the larger of the two always leads to a solution of the ODE. However, if theroots of the indicial equation differ by an integer then the recurrence relationcorresponding to the smaller root may or may not lead to a second linearlyindependent solution, depending on the ODE under consideration. Note that forcomplex roots of the indicial equation, the ‘larger’ root is taken to be the onewith the larger real part.◮Find the power series solutions about z =0ofz(z − 1)y ′′ +3zy ′ + y =0. (16.20)Dividing through by z(z − 1) to put the equation into standard form, we obtainy ′′ + 3 1(z − 1) y′ + y =0, (16.21)z(z − 1)and on comparing with (16.7) we identify p(z) =3/(z − 1) and q(z) =1/[z(z − 1)]. Weimmediately see that z = 0 is a singular point of (16.21), but since zp(z) =3z/(z − 1) andz 2 q(z) =z/(z−1) are finite there, it is a regular singular point and we expect to find at least542

16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINTone solution in the form of a Frobenius series. We therefore substitute y = z ∑ σ ∞n=0 a nz ninto (16.21) and, using (16.13) and (16.14), we obtain∞∑(n + σ)(n + σ − 1)a n z n+σ−2 + 3 ∞∑(n + σ)a n z n+σ−1z − 1n=0which, on dividing through by z σ−2 ,gives∞∑n=0n=0[(n + σ)(n + σ − 1) + 3z (n + σ)+zz − 1+1z(z − 1)∞∑a n z n+σ =0,n=0]a n z n =0.z − 1Although we could use this expression to find the indicial equation and recurrence relations,the working is simpler if we now multiply through by z − 1togive∞∑[(z − 1)(n + σ)(n + σ − 1) + 3z(n + σ)+z] a n z n =0. (16.22)n=0If we set z = 0 then all terms in the sum with the exponent of z greater than zero vanish,and we obtain the indicial equationσ(σ − 1) = 0,which has the roots σ =1andσ = 0. Since the roots differ by an integer (unity), it may notbe possible to find two linearly independent solutions of (16.21) in the form of Frobeniusseries. We are guaranteed, however, to find one such solution corresponding to the largerroot, σ =1.Demanding that the coefficients of z n vanish separately in (16.22), we obtain therecurrence relation(n − 1+σ)(n − 2+σ)a n−1 − (n + σ)(n + σ − 1)a n +3(n − 1+σ)a n−1 + a n−1 =0,which can be simplified to give(n + σ − 1)a n =(n + σ)a n−1 . (16.23)On substituting σ = 1 into this expression, we obtain( ) n +1a n = a n−1 ,nandonsettinga 0 = 1 we find a n = n + 1; so one solution to (16.21) is given by∞∑y 1 (z) =z (n +1)z n = z(1+2z +3z 2 + ···)n=0z=(1 − z) . (16.24)2If we attempt to find a second solution (corresponding to the smaller root of the indicialequation) by setting σ = 0 in (16.23), we find( n)a n = a n−1 .n − 1But we require a 0 ≠0,soa 1 is formally infinite and the method fails. We discuss how tofind a second linearly independent solution in the next section. ◭One particular case is worth mentioning. If the point about which the solution543

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSis required, i.e. z = 0, is in fact an ordinary point of the ODE rather than aregular singular point, then substitution of the Frobenius series (16.12) leads toan indicial equation with roots σ = 0 and σ = 1. Although these roots differ byan integer (unity), the recurrence relations corresponding to the two roots yieldtwo linearly independent power series solutions (one for each root), as expectedfrom section 16.2.16.4 Obtaining a second solutionWhilst attempting to construct solutions to an ODE in the form of Frobeniusseries about a regular singular point, we found in the previous section that whenthe indicial equation has a repeated root, or roots differing by an integer, we can(in general) find only one solution of this form. In order to construct the generalsolution to the ODE, however, we require two linearly independent solutions y 1and y 2 . We now consider several methods for obtaining a second solution in thiscase.16.4.1 The Wronskian methodIf y 1 and y 2 are two linearly independent solutions of the standard equationy ′′ + p(z)y ′ + q(z)y =0then the Wronskian of these two solutions is given by W (z) =y 1 y 2 ′ − y 2y 1 ′ .Dividing the Wronskian by y1 2 we obtain[ ( )]Wy12 = y′ 2− y′ 1y 1 y12 y 2 = y′ 2 d 1+ y 2 = d ( )y2,y 1 dz y 1 dz y 1which integrates to give∫ zW (u)y 2 (z) =y 1 (z)y1 2 du.(u)Now using the alternative expression for W (z) given in (16.4) with C =1(sincewe are not concerned with this normalising factor), we find∫ z{ ∫1u}y 2 (z) =y 1 (z)y1 2(u) exp − p(v) dv du. (16.25)Hence, given y 1 , we can in principle compute y 2 . Note that the lower limits ofintegration have been omitted. If constant lower limits are included then theymerely lead to a constant times the first solution.◮Find a second solution to (16.21) using the Wronskian method.For the ODE (16.21) we have p(z) =3/(z − 1), and from (16.24) we see that one solution544

16.4 OBTAINING A SECOND SOLUTIONto (16.21) is y 1 = z/(1 − z) 2 . Substituting for p and y 1 in (16.25) we have∫z z((1 − u) 4 ∫ u) 3y 2 (z) =exp −(1 − z) 2 u 2v − 1 dv du∫z z(1 − u) 4=(1 − z) 2 u∫ 2z zu − 1=du(1 − z) 2 u 2=z(1 − z) 2 (ln z + 1 z).exp [−3ln(u − 1)] duBy calculating the Wronskian of y 1 and y 2 it is easily shown that, as expected, the twosolutions are linearly independent. In fact, as the Wronskian has already been evaluatedas W (u) =exp[−3ln(u − 1)], i.e. W (z) =(z − 1) −3 , no calculation is needed. ◭An alternative (but equivalent) method of finding a second solution is simply toassume that the second solution has the form y 2 (z) =u(z)y 1 (z) for some functionu(z) to be determined (this method was discussed more fully in subsection 15.2.3).From (16.25), we see that the second solution derived from the Wronskian isindeed of this form. Substituting y 2 (z) =u(z)y 1 (z) into the ODE leads to afirst-order ODE in which u ′ is the dependent variable; this may then be solved.16.4.2 The derivative methodThe derivative method of finding a second solution begins with the derivation ofa recurrence relation for the coefficients a n in a Frobenius series solution, as in theprevious section. However, rather than putting σ = σ 1 in this recurrence relationto evaluate the first series solution, we now keep σ as a variable parameter. Thismeans that the computed a n are functions of σ and the computed solution is nowa function of z and σ:y(z,σ) =z σ∞ ∑n=0a n (σ)z n . (16.26)Of course, if we put σ = σ 1 in this, we obtain immediately the first series solution,but for the moment we leave σ as a parameter.For brevity let us denote the differential operator on the LHS of our standardODE (16.7) by L, sothatL = d2dz 2 + p(z) d dz + q(z),and examine the effect of L on the series y(z,σ) in (16.26). It is clear that theseries Ly(z,σ) will contain only a term in z σ , since the recurrence relation definingthe a n (σ) is such that these coefficients vanish for higher powers of z. Butthecoefficient of z σ is simply the LHS of the indicial equation. Therefore, if the roots545

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSof the indicial equation are σ = σ 1 and σ = σ 2 then it follows thatLy(z,σ) =a 0 (σ − σ 1 )(σ − σ 2 )z σ . (16.27)Therefore, as in the previous section, we see that for y(z,σ) tobeasolutionofthe ODE Ly =0,σ must equal σ 1 or σ 2 . For simplicity we shall set a 0 = 1 in thefollowing discussion.Let us first consider the case in which the two roots of the indicial equationare equal, i.e. σ 2 = σ 1 . From (16.27) we then haveLy(z,σ) =(σ − σ 1 ) 2 z σ .Differentiating this equation with respect to σ we obtain∂∂σ [Ly(z,σ)] =(σ − σ 1) 2 z σ ln z +2(σ − σ 1 )z σ ,which equals zero if σ = σ 1 . But since ∂/∂σ and L are operators that differentiatewith respect to different variables, we can reverse their order, implying that[ ] ∂L∂σ y(z,σ) =0 atσ = σ 1 .Hence, the function in square brackets, evaluated at σ = σ 1 and denoted by[ ] ∂∂σ y(z,σ) , (16.28)σ=σ 1is also a solution of the original ODE Ly = 0, and is in fact the second linearlyindependent solution that we were looking for.The case in which the roots of the indicial equation differ by an integer isslightly more complicated but can be treated in a similar way. In (16.27), since Ldifferentiates with respect to z we may multiply (16.27) by any function of σ, sayσ − σ 2 , and take this function inside the operator L on the LHS to obtainL [(σ − σ 2 )y(z,σ)] =(σ − σ 1 )(σ − σ 2 ) 2 z σ . (16.29)Therefore the function[(σ − σ 2 )y(z,σ)] σ=σ2is also a solution of the ODE Ly = 0. However, it can be proved § that thisfunction is a simple multiple of the first solution y(z,σ 1 ), showing that it is notlinearly independent and that we must find another solution. To do this wedifferentiate (16.29) with respect to σ and find∂∂σ {L [(σ − σ 2)y(z,σ)]} =(σ − σ 2 ) 2 z σ +2(σ − σ 1 )(σ − σ 2 )z σ+(σ − σ 1 )(σ − σ 2 ) 2 z σ ln z,§ For a fuller discussion see, for example, K. F. Riley, Mathematical Methods for the Physical Sciences(Cambridge: Cambridge University Press, 1974), pp. 158–9.546

16.4 OBTAINING A SECOND SOLUTIONwhich is equal to zero if σ = σ 2 . As previously, since ∂/∂σ and L are operatorsthat differentiate with respect to different variables, we can reverse their order toobtain{ }∂L∂σ [(σ − σ 2)y(z,σ)] =0 atσ = σ 2 ,and so the function{ }∂∂σ [(σ − σ 2)y(z,σ)](16.30)σ=σ 2is also a solution of the original ODE Ly = 0, and is in fact the second linearlyindependent solution.◮Find a second solution to (16.21) using the derivative method.From (16.23) the recurrence relation (with σ as a parameter) is given by(n + σ − 1)a n =(n + σ)a n−1 .Setting a 0 = 1 we find that the coefficients have the particularly simple form a n (σ) =(σ + n)/σ. We therefore consider the function∞∑y(z, σ) =z σ a n (σ)z n = z σ σ + nσ zn .n=0n=0The smaller root of the indicial equation for (16.21) is σ 2 = 0, and so from (16.30) asecond, linearly independent, solution to the ODE is given by{ { [ ∞]}∂∂σ}σ=0[σy(z,σ)] ∂ ∑= z σ (σ + n)z n .∂σn=0σ=0The derivative with respect to σ is given by[ ∞]∂ ∑∞∑∑ ∞z σ (σ + n)z n = z σ ln z (σ + n)z n + z σ z n ,∂σn=0n=0n=0which on setting σ = 0 gives the second solution∞∑ ∞∑y 2 (z) =lnz nz n + z nn=0z=(1 − z) ln z + 12 1 − z(z= ln z + 1 )(1 − z) 2 z − 1 .This second solution is the same as that obtained by the Wronskian method in the previoussubsection except for the addition of some of the first solution. ◭n=0∞∑16.4.3 Series form of the second solutionUsing any of the methods discussed above, we can find the general form of thesecond solution to the ODE. This form is most easily found, however, using the547

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSderivative method. Let us first consider the case where the two solutions of theindicial equation are equal. In this case a second solution is given by (16.28),which may be written as[ ] ∂y(z,σ)y 2 (z) =∂σσ=σ 1∑∞ ∑∞ [ ]=(lnz)z σ1a n (σ 1 )z n dan (σ)+ z σ1dσn=0n=1σ=σ 1z n∑∞= y 1 (z)lnz + z σ1 b n z n , (16.31)n=1where b n =[da n (σ)/dσ] σ=σ1 . One could equally obtain the coefficients b n by directsubstitution of the form (16.31) into the original ODE.In the case where the roots of the indicial equation differ by an integer (notequal to zero), then from (16.30) a second solution is given by{ }∂y 2 (z) =∂σ [(σ − σ 2)y(z,σ)]σ=σ 2[]∑∞=lnz (σ − σ 2 )z σ a n (σ)z n + z σ2n=0σ=σ 2∞ ∑n=0[ ]ddσ (σ − σ 2)a n (σ) z n .σ=σ 2But, as we mentioned in the previous section, [(σ − σ 2 )y(z,σ)] at σ = σ 2 is just amultiple of the first solution y(z,σ 1 ). Therefore the second solution is of the formy 2 (z) =cy 1 (z)lnz + z σ2∞ ∑n=0b n z n , (16.32)where c is a constant. In some cases, however, c might be zero, and so the secondsolution would not contain the term in ln z and could be written simply as aFrobenius series. Clearly this corresponds to the case in which the substitutionof a Frobenius series into the original ODE yields two solutions automatically.In either case, the coefficients b n may also be found by direct substitution of theform (16.32) into the original ODE.16.5 Polynomial solutionsWe have seen that the evaluation of successive terms of a series solution to adifferential equation is carried out by means of a recurrence relation. The formof the relation for a n depends upon n, the previous values of a r (r

16.5 POLYNOMIAL SOLUTIONSis a positive integer or zero, then we are left with a finite polynomial of degreeN ′ = N + σ as a solution of the ODE:N∑y(z) = a n z n+σ . (16.33)n=0In many applications in theoretical physics (particularly in quantum mechanics)the termination of a potentially infinite series after a finite number of termsis of crucial importance in establishing physically acceptable descriptions andproperties of systems. The condition under which such a termination occurs istherefore of considerable importance.◮Find power series solutions about z =0ofy ′′ − 2zy ′ + λy =0. (16.34)For what values of λ does the equation possess a polynomial solution? Find such a solutionfor λ =4.Clearly z = 0 is an ordinary point of (16.34) and so we look for solutions of the formy = ∑ ∞n=0 a nz n . Substituting this into the ODE and multiplying through by z 2 we find∞∑[n(n − 1) − 2z 2 n + λz 2 ]a n z n =0.n=0By demanding that the coefficients of each power of z vanish separately we derive therecurrence relationn(n − 1)a n − 2(n − 2)a n−2 + λa n−2 =0,which may be rearranged to give2(n − 2) − λa n = a n−2 for n ≥ 2. (16.35)n(n − 1)The odd and even coefficients are therefore independent of one another, and two solutionsto (16.34) may be derived. We either set a 1 =0anda 0 =1toobtainy 1 (z) =1− λ z2 z4z6− λ(4 − λ) − λ(4 − λ)(8 − λ) − ··· (16.36)2! 4! 6!or set a 0 =0anda 1 =1toobtainy 2 (z) =z +(2− λ) z3z5z7+(2− λ)(6 − λ) +(2− λ)(6 − λ)(10 − λ)3! 5! 7! + ··· .Now, from the recurrence relation (16.35) (or in this case from the expressions for y 1and y 2 themselves) we see that for the ODE to possess a polynomial solution we requireλ =2(n − 2) for n ≥ 2 or, more simply, λ =2n for n ≥ 0, i.e. λ must be an even positiveinteger. If λ = 4 then from (16.36) the ODE has the polynomial solutiony 1 (z) =1− 4z22! =1− 2z2 . ◭A simpler method of obtaining finite polynomial solutions is to assume asolution of the form (16.33), where a N ≠ 0. Instead of starting with the lowestpower of z, as we have done up to now, this time we start by considering the549

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONScoefficient of the highest power z N ; such a power now exists because of ourassumed form of solution.◮By assuming a polynomial solution find the values of λ in (16.34) for which such a solutionexists.We assume a polynomial solution to (16.34) of the form y = ∑ Nn=0 a nz n . Substituting thisform into (16.34) we findN∑ [n(n − 1)an z n−2 − 2zna n z n−1 + λa n z n] =0.n=0Now, instead of starting with the lowest power of z, we start with the highest. Thus,demanding that the coefficient of z N vanishes, we require −2N + λ =0,i.e.λ =2N, aswefound in the previous example. By demanding that the coefficient of a general power of zis zero, the same recurrence relation as above may be derived and the solutions found. ◭16.6 Exercises16.1 Find two power series solutions about z = 0 of the differential equation(1 − z 2 )y ′′ − 3zy ′ + λy =0.Deduce that the value of λ for which the corresponding power series becomes anNth-degree polynomial U N (z) isN(N + 2). Construct U 2 (z) andU 3 (z).16.2 Find solutions, as power series in z, of the equation4zy ′′ +2(1− z)y ′ − y =0.Identify one of the solutions and verify it by direct substitution.16.3 Find power series solutions in z of the differential equationzy ′′ − 2y ′ +9z 5 y =0.Identify closed forms for the two series, calculate their Wronskian, and verifythat they are linearly independent. Compare the Wronskian with that calculatedfrom the differential equation.16.4 Change the independent variable in the equationd 2 f df+2(z − a) +4f =0dz2 dz (∗)from z to x = z − α, and find two independent series solutions, expanded aboutx = 0, of the resulting equation. Deduce that the general solution of (∗) is∞∑ (−4) m m!f(z,α) =A(z − α)e −(z−α)2 + B(z − α) 2m ,(2m)!m=0with A and B arbitrary constants.16.5 Investigate solutions of Legendre’s equation at one of its singular points asfollows.(a) Verify that z = 1 is a regular singular point of Legendre’s equation and thatthe indicial equation for a series solution in powers of (z − 1) has roots 0and 3.(b) Obtain the corresponding recurrence relation and show that σ = 0 does notgive a valid series solution.550

16.6 EXERCISES(c) Determine the radius of convergence R of the σ = 3 series and relate it tothe positions of the singularities of Legendre’s equation.16.6 Verify that z = 0 is a regular singular point of the equationz 2 y ′′ − 3 2 zy′ +(1+z)y =0,and that the indicial equation has roots 2 and 1/2. Show that the general solutionis given byy(z) =6a 0 z 2+ b 0(∞∑n=0(−1) n (n +1)2 2n z n(2n +3)!z 1/2 +2z 3/2 − z1/24∞∑n=2)(−1) n 2 2n z n.n(n − 1)(2n − 3)!16.7 Use the derivative method to obtain, as a second solution of Bessel’s equationfor the case when ν = 0, the following expression:(∞∑n∑)(−1) n 1 ( z) 2nJ 0 (z)lnz −,(n!) 2 r 2n=1r=1given that the first solution is J 0 (z), as specified by (18.79).16.8 Consider a series solution of the equationzy ′′ − 2y ′ + yz =0 (∗)about its regular singular point.(a) Show that its indicial equation has roots that differ by an integer but thatthe two roots nevertheless generate linearly independent solutionsy 1 (z) =3a 0∞∑∞∑n=1(−1) n+1 2nz 2n+1,(2n +1)!(−1) n+1 (2n − 1)z 2ny 2 (z) =a 0.(2n)!n=0(b) Show that y 1 (z) isequalto3a 0 (sin z − z cos z) by expanding the sinusoidalfunctions. Then, using the Wronskian method, find an expression for y 2 (z) interms of sinusoids. You will need to write z 2 as (z/ sin z)(z sin z) and integrateby parts to evaluate the integral involved.(c) Confirm that the two solutions are linearly independent by showing thattheir Wronskian is equal to −z 2 , as would be expected from the form of (∗).16.9 Find series solutions of the equation y ′′ − 2zy ′ − 2y = 0. Identify one of the seriesas y 1 (z) =expz 2 and verify this by direct substitution. By setting y 2 (z) =u(z)y 1 (z)and solving the resulting equation for u(z), find an explicit form for y 2 (z) anddeduce that∫ x∑ ∞n!e −v2 dv = e −x202(2n +1)! (2x)2n+1 .n=016.10 Solve the equationz(1 − z) d2 y dy+(1− z) + λy =0dz2 dzas follows.(a) Identify and classify its singular points and determine their indices.551

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS(b) Find one series solution in powers of z. Give a formal expression for asecond linearly independent solution.(c) Deduce the values of λ for which there is a polynomial solution P N (z) ofdegree N. Evaluate the first four polynomials, normalised in such a way thatP N (0) = 1.16.11 Find the general power series solution about z =0oftheequationz d2 y dy+(2z − 3)dz2 dz + 4 z y =0.16.12 Find the radius of convergence of a series solution about the origin for theequation (z 2 + az + b)y ′′ +2y = 0 in the following cases:(a) a =5,b=6; (b)a =5,b=7.Show that if a and b are real and 4b >a 2 , then the radius of convergence isalways given by b 1/2 .16.13 For the equation y ′′ + z −3 y = 0, show that the origin becomes a regular singularpoint if the independent variable is changed from z to x =1/z. Hence find aseries solution of the form y 1 (z) = ∑ ∞0 a nz −n .Bysettingy 2 (z) =u(z)y 1 (z) andexpanding the resulting expression for du/dz in powers of z −1 , show that y 2 (z)has the asymptotic form[( )] ln zy 2 (z) =c z +lnz − 1 +O ,2zwhere c is an arbitrary constant.16.14 Prove that the Laguerre equation,z d2 y dy+(1− z) + λy =0,dz2 dzhas polynomial solutions L N (z) ifλ is a non-negative integer N, and determinethe recurrence relationship for the polynomial coefficients. Hence show that anexpression for L N (z), normalised in such a way that L N (0) = N!, isN∑ (−1) n (N!) 2L N (z) =(N − n)!(n!) 2 zn .n=0Evaluate L 3 (z) explicitly.16.15 The origin is an ordinary point of the Chebyshev equation,(1 − z 2 )y ′′ − zy ′ + m 2 y =0,which therefore has series solutions of the form z ∑ σ ∞0 a nz n for σ =0andσ =1.(a) Find the recurrence relationships for the a n in the two cases and show thatthere exist polynomial solutions T m (z):(i) for σ =0,whenm is an even integer, the polynomial having 1 (m +2)2terms;(ii) for σ =1,whenm is an odd integer, the polynomial having 1 (m +1)2terms.(b) T m (z) is normalised so as to have T m (1) = 1. Find explicit forms for T m (z)for m =0, 1, 2, 3.552

16.7 HINTS AND ANSWERS(c)Show that the corresponding non-terminating series solutions S m (z) have astheir first few termsS 0 (z) =a 0(z + 1 3! z3 + 9 )5! z5 + ··· ,S 1 (z) =a 0(1 − 1 2! z2 − 3 )4! z4 − ··· ,(S 2 (z) =a 0 z − 3 3! z3 − 15 )5! z5 − ··· ,S 3 (z) =a 0(1 − 9 2! z2 + 454! z4 + ···16.16 Obtain the recurrence relations for the solution of Legendre’s equation (18.1) ininverse powers of z, i.e.sety(z) = ∑ a n z σ−n ,witha 0 ≠ 0. Deduce that, if l is aninteger, then the series with σ = l will terminate and hence converge for all z,whilst the series with σ = −(l + 1) does not terminate and hence converges onlyfor |z| > 1.).16.7 Hints and answers16.1 Note that z = 0 is an ordinary point of the equation.For σ =0,a n+2 /a n =[n(n +2)− λ]/[(n +1)(n + 2)] and, correspondingly, forσ =1,U 2 (z) =a 0 (1 − 4z 2 )andU 3 (z) =a 0 (z − 2z 3 ).16.3 σ =0and3;a 6m /a 0 =(−1) m /(2m)! and a 6m /a 0 =(−1) m /(2m + 1)!, respectively.y 1 (z) =a 0 cos z 3 and y 2 (z) =a 0 sin z 3 .TheWronskianis±3a 2 0 z2 ≠0.16.5 (b) a n+1 /a n =[l(l +1)− n(n +1)]/[2(n +1) 2 ].(c) R = 2, equal to the distance between z = 1 and the closest singularity atz = −1.(−1) n z 2n16.7 A typical term in the series for y(σ, z) is[(σ +2)(σ +4)···(σ +2n)] . 216.9 The origin is an ordinary point. Determine the constant of integration by examiningthe behaviour of the related functions for small x.y 2 (z) =(expz 2 ) ∫ z0 exp(−x2 ) dx.16.11 Repeated roots σ =2.∞∑y(z) =az 2 − 4az 3 +6bz 3 (n +1)(−2z) n+2 { a}+n! 4 + b [ln z + g(n)] ,n=2whereg(n) = 1n +1 − 1 n − 1n − 1 − ···− 1 2 − 2.16.13 The transformed equation is xy ′′ +2y ′ + y =0;a n =(−1) n (n +1) −1 (n!) −2 a 0 ;du/dz = A[ y 1 (z)] −2 .16.15 (a) (i) a n+2 =[a n (n 2 − m 2 )]/[(n +2)(n + 1)],(ii) a n+2 = {a n [(n +1) 2 − m 2 ]}/[(n +3)(n + 2)]; (b) 1, z, 2z 2 − 1, 4z 3 − 3z.553

17Eigenfunction methods fordifferential equationsIn the previous three chapters we dealt with the solution of differential equationsof order n by two methods. In one method, we found n independent solutionsof the equation and then combined them, weighted with coefficients determinedby the boundary conditions; in the other we found solutions in terms of serieswhose coefficients were related by (in general) an n-term recurrence relation andthence fixed by the boundary conditions. For both approaches the linearity of theequation was an important or essential factor in the utility of the method, andin this chapter our aim will be to exploit the superposition properties of lineardifferential equations even further.We will be concerned with the solution of equations of the inhomogeneousformLy(x) =f(x), (17.1)where f(x) is a prescribed or general function and the boundary conditions tobe satisfied by the solution y = y(x), for example at the limits x = a and x = b,are given. The expression Ly(x) stands for a linear differential operator L actingupon the function y(x).In general, unless f(x) is both known and simple, it will not be possible tofind particular integrals of (17.1), even if complementary functions can be foundthat satisfy Ly = 0. The idea is therefore to exploit the linearity of L by buildingup the required solution y(x) asasuperposition, generally containing an infinitenumber of terms, of some set of functions {y i (x)} that each individually satisfythe boundary conditions. Clearly this brings in a quite considerable complicationbut since, within reason, we may select the set of functions to suit ourselves, wecan obtain sizeable compensation for this complication. Indeed, if the set chosenis one containing functions that, when acted upon by L, produce particularlysimple results then we can ‘show a profit’ on the operation. In particular, if the554

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSset consists of those functions y i for whichLy i (x) =λ i y i (x), (17.2)where λ i is a constant (and which satisfy the boundary conditions), then a distinctadvantage may be obtained from the manoeuvre because all the differentiationwill have disappeared from (17.1).Equation (17.2) is clearly reminiscent of the equation satisfied by the eigenvectorsx i of a linear operator A , namelyA x i = λ i x i , (17.3)where λ i is a constant and is called the eigenvalue associated with x i . By analogy,in the context of differential equations a function y i (x) satisfying (17.2) is calledan eigenfunction of the operator L (under the imposed boundary conditions) andλ i is then called the eigenvalue associated with the eigenfunction y i (x). Clearly,the eigenfunctions y i (x) ofL are only determined up to an arbitrary scale factorby (17.2).Probably the most familiar equation of the form (17.2) is that which describesa simple harmonic oscillator, i.e.Ly ≡− d2 ydt 2 = ω2 y, where L ≡−d 2 /dt 2 . (17.4)Imposing the boundary condition that the solution is periodic with period T ,the eigenfunctions in this case are given by y n (t) =A n e iωnt ,whereω n =2πn/T ,n =0, ±1, ±2,... and the A n are constants. The eigenvalues are ωn 2 = n 2 ω1 2 =n 2 (2π/T) 2 . (Sometimes ω n is referred to as the eigenvalue of this equation, butwe will avoid such confusing terminology here.)We may discuss a somewhat wider class of differential equations by consideringa slightly more general form of (17.2), namelyLy i (x) =λ i ρ(x)y i (x), (17.5)where ρ(x) isaweight function. In many applications ρ(x) is unity for all x, inwhich case (17.2) is recovered; in general, though, it is a function determined bythe choice of coordinate system used in describing a particular physical situation.The only requirement on ρ(x) is that it is real and does not change sign in therange a ≤ x ≤ b, so that it can, without loss of generality, be taken to be nonnegativethroughout; of course, ρ(x) must be the same function for all values ofλ i . A function y i (x) that satisfies (17.5) is called an eigenfunction of the operatorL with respect to the weight function ρ(x).This chapter will not cover methods used to determine the eigenfunctions of(17.2) or (17.5), since we have discussed those in previous chapters, but, rather, willuse the properties of the eigenfunctions to solve inhomogeneous equations of theform (17.1). We shall see later that the sets of eigenfunctions y i (x) of a particular555

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSclass of operators called Hermitian operators (the operator in the simple harmonicoscillator equation is an example) have particularly useful properties and thesewill be studied in detail. It turns out that many of the interesting differentialoperators met within the physical sciences are Hermitian. Before continuing ourdiscussion of the eigenfunctions of Hermitian operators, however, we will considersome properties of general sets of functions.17.1 Sets of functionsIn chapter 8 we discussed the definition of a vector space but concentrated onspaces of finite dimensionality. We consider now the infinite-dimensional spaceof all reasonably well-behaved functions f(x), g(x), h(x), ... on the intervala ≤ x ≤ b. That these functions form a linear vector space is shown by notingthe following properties. The set is closed under(i) addition, which is commutative and associative, i.e.f(x)+g(x) =g(x)+f(x),[f(x)+g(x)] + h(x) =f(x)+[g(x)+h(x)] ,(ii) multiplication by a scalar, which is distributive and associative, i.e.Furthermore, in such a spaceλ [f(x)+g(x)] = λf(x)+λg(x),λ [µf(x)] =(λµ)f(x),(λ + µ)f(x) =λf(x)+µf(x).(iii) there exists a ‘null vector’ 0 such that f(x)+0=f(x),(iv) multiplication by unity leaves any function unchanged, i.e. 1 × f(x) =f(x),(v) each function has an associated negative function −f(x) that is such thatf(x)+[−f(x)] = 0.By analogy with finite-dimensional vector spaces we now introduce a setof linearly independent basis functions y n (x), n = 0, 1,...,∞, such that any‘reasonable’ function in the interval a ≤ x ≤ b (i.e. it obeys the Dirichlet conditionsdiscussed in chapter 12) can be expressed as the linear sum of these functions:∞∑f(x) = c n y n (x).n=0Clearly if a different set of linearly independent basis functions u n (x) is chosenthen the function can be expressed in terms of the new basis,∞∑f(x) = d n u n (x),n=0556

17.1 SETS OF FUNCTIONSwhere the d n are a different set of coefficients. In each case, provided the basisfunctions are linearly independent, the coefficients are unique.We may also define an inner product on our function space by〈f|g〉 =∫ baf ∗ (x)g(x)ρ(x) dx, (17.6)where ρ(x) is the weight function, which we require to be real and non-negativein the interval a ≤ x ≤ b. As mentioned above, ρ(x) is often unity for all x. Twofunctions are said to be orthogonal (with respect to the weight function ρ(x)) onthe interval [a, b] if〈f|g〉 =∫ baf ∗ (x)g(x)ρ(x) dx =0, (17.7)and the norm of a function is defined as[∫ b1/2 [∫ b1/2‖f‖ = 〈f|f〉 1/2 = f ∗ (x)f(x)ρ(x) dx]= |f(x)| 2 ρ(x) dx]. (17.8)aIt is also common practice to define a normalised function by ˆf = f/‖f‖, whichhas unit norm.An infinite-dimensional vector space of functions, for which an inner productis defined, is called a Hilbert space. Using the concept of the inner product, wecan choose a basis of linearly independent functions ˆφ n (x), n =0, 1, 2,... that areorthonormal, i.e. such that〈 ˆφ i | ˆφ j 〉 =∫ baaˆφ ∗ i (x) ˆφ j (x)ρ(x) dx = δ ij . (17.9)If y n (x), n =0, 1, 2,..., are a linearly independent, but not orthonormal, basis forthe Hilbert space then an orthonormal set of basis functions ˆφ n may be produced(in a similar manner to that used in the construction of a set of orthogonaleigenvectors of an Hermitian matrix; see chapter 8) by the following procedure:φ 0 = y 0 ,φ 1 = y 1 − ˆφ 0 〈 ˆφ 0 |y 1 〉,φ 2 = y 2 − ˆφ 1 〈 ˆφ 1 |y 2 〉− ˆφ 0 〈 ˆφ 0 |y 2 〉,.φ n = y n − ˆφ n−1 〈 ˆφ n−1 |y n 〉−···− ˆφ 0 〈 ˆφ 0 |y n 〉,.It is straightforward to check that each φ n is orthogonal to all its predecessorsφ i , i =0, 1, 2,...,n− 1. This method is called Gram–Schmidt orthogonalisation.Clearly the functions φ n form an orthogonal set, but in general they do not haveunit norms.557

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS◮Starting from the linearly independent functions y n (x) =x n , n =0, 1,..., construct threeorthonormal functions over the range −1

17.2 ADJOINT, SELF-ADJOINT AND HERMITIAN OPERATORS17.1.1 Some useful inequalitiesSince for a Hilbert space 〈f|f〉 ≥0, the inequalities discussed in subsection 8.1.3hold. The proofs are not repeated here, but the relationships are listed forcompleteness.(i) The Schwarz inequality states that|〈f|g〉| ≤ 〈f|f〉 1/2 〈g|g〉 1/2 , (17.12)where the equality holds when f(x) is a scalar multiple of g(x), i.e. whenthey are linearly dependent.(ii) The triangle inequality states that‖f + g‖ ≤‖f‖ + ‖g‖, (17.13)where again equality holds when f(x) is a scalar multiple of g(x).(iii) Bessel’s inequality requires the introduction of an orthonormal basis ˆφ n (x)so that any function f(x) can be written as∞∑f(x) = c n ˆφn (x),n=0where c n = 〈 ˆφ n |f〉. Bessel’s inequality then states that〈f|f〉 ≥ ∑ n|c n | 2 . (17.14)The equality holds if the summation is over all the basis functions. If somevalues of n are omitted from the sum then the inequality results (unless,of course, the c n happen to be zero for all values of n omitted, in whichcase the equality remains).17.2 Adjoint, self-adjoint and Hermitian operatorsHaving discussed general sets of functions, we now return to the discussion ofeigenfunctions of linear operators. We begin by introducing the adjoint of anoperator L, denoted by L † , which is defined by∫ baf ∗ (x) [Lg(x)] dx =∫ ba[L † f(x)] ∗ g(x) dx + boundary terms,(17.15)where the boundary terms are evaluated at the end-points of the interval [a, b].Thus, for any given linear differential operator L, the adjoint operator L † can befound by repeated integration by parts.An operator is said to be self-adjoint if L † = L. If, in addition, certain boundaryconditions are met by the functions f and g on which a self-adjoint operator acts,559

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSor by the operator itself, such that the boundary terms in (17.15) vanish, then theoperator is said to be Hermitian over the interval a ≤ x ≤ b. Thus, in this case,∫ baf ∗ (x) [Lg(x)] dx =∫ ba[Lf(x)] ∗ g(x) dx. (17.16)A little careful study will reveal the similarity between the definition of anHermitian operator and the definition of an Hermitian matrix given in chapter 8.◮Show that the linear operator L = d 2 /dt 2 is self-adjoint, and determine the requiredboundary conditions for the operator to be Hermitian over the interval t 0 to t 0 + T .Substituting into the LHS of the definition of the adjoint operator (17.15) and integratingby parts gives∫ t0 +T[f ∗ d2 gt 0dt dt = f ∗ dg ] t0 +T ∫ t0 +Tdf ∗ dg−2 dtt 0 t 0dt dt dt.Integrating the second term on the RHS by parts once more yields∫ t0 +T[f ∗ d2 gt 0dt dt = f ∗ dg ] t0 +T ] t0+[− df∗ +T ∫ t0 +T2 dtt 0dt g + g d2 f ∗t 0 t 0dt dt, 2which, by comparison with (17.15), proves that L is a self-adjoint operator. Moreover,from (17.16), we see that L is an Hermitian operator over the required interval provided[f ∗ dg ] t0 +T [ ] df∗ t0 +Tdt dt g . ◭t 0=We showed in chapter 8 that the eigenvalues of Hermitian matrices are real andthat their eigenvectors can be chosen to be orthogonal. Similarly, the eigenvaluesof Hermitian operators are real and their eigenfunctions can be chosen to beorthogonal (we will prove these properties in the following section). Hermitianoperators (or matrices) are often used in the formulation of quantum mechanics.The eigenvalues then give the possible measured values of an observable quantitysuch as energy or angular momentum, and the physical requirement that suchquantities must be real is ensured by the reality of these eigenvalues. Furthermore,the infinite set of eigenfunctions of an Hermitian operator form a complete basisset over the relevant interval, so that it is possible to expand any function y(x)obeying the appropriate conditions in an eigenfunction series over this interval:∞∑y(x) = c n y n (x), (17.17)n=0where the choice of suitable values for the c n will make the sum arbitrarily closeto y(x). § These useful properties provide the motivation for a detailed study ofHermitian operators.t 0§ The proof of the completeness of the eigenfunctions of an Hermitian operator is beyond the scopeof this book. The reader should refer, for example, to R. Courant and D. Hilbert, Methods ofMathematical Physics (New York: Interscience, 1953).560

17.3 PROPERTIES OF HERMITIAN OPERATORS17.3 Properties of Hermitian operatorsWe now provide proofs of some of the useful properties of Hermitian operators.Again much of the analysis is similar to that for Hermitian matrices in chapter 8,although the present section stands alone. (Here, and throughout the remainderof this chapter, we will write out inner products in full. We note, however, thatthe inner product notation often provides a neat form in which to express results.)17.3.1 Reality of the eigenvaluesConsider an Hermitian operator for which (17.5) is satisfied by at least twoeigenfunctions y i (x) andy j (x), which have corresponding eigenvalues λ i and λ j ,so thatLy i = λ i ρ(x)y i , (17.18)Ly j = λ j ρ(x)y j , (17.19)where we have allowed for the presence of a weight function ρ(x). Multiplying(17.18) by yj ∗ and (17.19) by y∗ i and then integrating gives∫ b∫ byj ∗ Ly i dx = λ i yj ∗ y i ρdx, (17.20)a∫ baa∫ byi ∗ Ly j dx = λ j yi ∗ y j ρdx. (17.21)Remembering that we have required ρ(x) to be real, the complex conjugate of(17.20) becomes∫ bay j (Ly i ) ∗ dx = λ ∗ ia∫ bay ∗ i y j ρdx, (17.22)and using the definition of an Hermitian operator (17.16) it follows that the LHSof (17.22) is equal to the LHS of (17.21). Thus(λ ∗ i − λ j )∫ bay ∗ i y j ρdx=0. (17.23)If i = j then λ i = λ ∗ i (since ∫ ba y∗ i y iρdx ≠ 0), which is a statement that theeigenvalue λ i is real.17.3.2 Orthogonality and normalisation of the eigenfunctionsFrom (17.23), it is immediately apparent that two eigenfunctions y i and y j thatcorrespond to different eigenvalues, i.e. such that λ i ≠ λ j , satisfy∫ bay ∗ i y j ρdx=0, (17.24)561

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSwhich is a statement of the orthogonality of y i and y j .If one (or more) of the eigenvalues is degenerate, however, we have differenteigenfunctions corresponding to the same eigenvalue, and the proof of orthogonalityis not so straightforward. Nevertheless, an orthogonal set of eigenfunctionsmay be constructed using the Gram–Schmidt orthogonalisation method mentionedearlier in this chapter and used in chapter 8 to construct a set of orthogonaleigenvectors of an Hermitian matrix. We repeat the analysis here for completeness.Suppose, for the sake of our proof, that λ 0 is k-fold degenerate, i.e.Ly i = λ 0 ρy i for i =0, 1,...,k− 1, (17.25)but that λ 0 is different from any of λ k , λ k+1 , etc. Then any linear combination ofthese y i is also an eigenfunction with eigenvalue λ 0 since∑k−1∑k−1∑k−1Lz ≡ L c i y i = c i Ly i = c i λ 0 ρy i = λ 0 ρz. (17.26)i=0i=0If the y i defined in (17.25) are not already mutually orthogonal then considerthe new eigenfunctions z i constructed by the following procedure, in which eachof the new functions z i is to be normalised, to give ẑ i , before proceeding to theconstruction of the next one (the normalisation can be carried out by dividingthe eigenfunction z i by ( ∫ ba z∗ i z iρdx) 1/2 ):z 0 = y 0 ,∫ b)z 1 = y 1 −(ẑ 0 ẑ0y ∗ 1 ρdx ,a∫ b) ∫ b)z 2 = y 2 −(ẑ 1 ẑ1y ∗ 2 ρdx −(ẑ 0 ẑ0y ∗ 2 ρdx ,aa.∫ b)∫ b)z k−1 = y k−1 −(ẑ k−2 ẑk−2y ∗ k−1 ρdx − ···−(ẑ 0 ẑ0y ∗ k−1 ρdx .aaEach of the integrals is just a number and thus each new function z i is, as can beshown from (17.26), an eigenvector of L with eigenvalue λ 0 . It is straightforwardto check that each z i is orthogonal to all its predecessors. Thus, by this explicitconstruction we have shown that an orthogonal set of eigenfunctions of anHermitian operator L can be obtained. Clearly the orthogonal set obtained, z i ,isnot unique.In general, since L is linear, the normalisation of its eigenfunctions y i (x) isarbitrary. It is often convenient, however, to work in terms of the normalisedeigenfunctions ŷ i (x), so that ∫ ba ŷ∗ i ŷiρdx= 1. These therefore form an orthonormal562i=0

17.3 PROPERTIES OF HERMITIAN OPERATORSset and we can write∫ bwhich is valid for all pairs of values i, j.aŷ ∗ i ŷjρdx= δ ij , (17.27)17.3.3 Completeness of the eigenfunctionsAs noted earlier, the eigenfunctions of an Hermitian operator may be shownto form a complete basis set over the relevant interval. One may thus expandany (reasonable) function y(x) obeying appropriate boundary conditions in aneigenfunction series over the interval, as in (17.17). Working in terms of thenormalised eigenfunctions ŷ n (x), we may thus writef(x) = ∑ n∫ bŷ n (x)∫ baŷ ∗ n(z)f(z)ρ(z) dz= f(z)ρ(z) ∑ ŷ n (x)ŷn(z) ∗ dz.anSince this is true for any f(x), we must have thatρ(z) ∑ nŷ n (x)ŷ ∗ n(z) =δ(x − z). (17.28)This is called the completeness or closure property of the eigenfunctions. It definesa complete set. If the spectrum of eigenvalues of L is anywhere continuous thenthe eigenfunction y n (x) must be treated as y(n, x) and an integration carried outover n.We also note that the RHS of (17.28) is a δ-function and so is only non-zerowhen z = x; thus ρ(z) on the LHS can be replaced by ρ(x) if required, i.e.ρ(z) ∑ nŷ n (x)ŷ ∗ n(z) =ρ(x) ∑ nŷ n (x)ŷ ∗ n(z). (17.29)17.3.4 Construction of real eigenfunctionsRecall that the eigenfunction y i satisfiesLy i = λ i ρy i (17.30)and that the complex conjugate of this givesLyi ∗ = λ ∗ i ρyi ∗ = λ i ρyi ∗ , (17.31)where the last equality follows because the eigenvalues are real, i.e. λ i = λ ∗ i .Thus, y i and yi∗ are eigenfunctions corresponding to the same eigenvalue andhence, because of the linearity of L, atleastoneofyi ∗ + y i and i(yi ∗ − y i ), which563

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSare both real, is a non-zero eigenfunction corresponding to that eigenvalue. Itfollows that the eigenfunctions can always be made real by taking suitable linearcombinations, though taking such linear combinations will only be necessary incases where a particular λ is degenerate, i.e. corresponds to more than one linearlyindependent eigenfunction.17.4 Sturm–Liouville equationsOne of the most important applications of our discussion of Hermitian operatorsis to the study of Sturm–Liouville equations, which take the general formp(x) d2 y dy+ r(x) + q(x)y + λρ(x)y =0,dx2 dxwhere r(x) =dp(x)dx(17.32)and p, q and r are real functions of x. § A variational approach to the Sturm–Liouville equation, which is useful in estimating the eigenvalues λ for a given setof boundary conditions on y, is discussed in chapter 22. For now, however, weconcentrate on demonstrating that solutions of the Sturm–Liouville equation thatsatisfy appropriate boundary conditions are the eigenfunctions of an Hermitianoperator.It is clear that (17.32) can be writtenLy = λρ(x)y, where L ≡−[p(x) d2dx 2 + r(x) d ]dx + q(x) . (17.33)Using the condition that r(x) =p ′ (x), it will be seen that the general Sturm–Liouville equation (17.32) can also be rewritten as(py ′ ) ′ + qy + λρy =0, (17.34)where primes denote differentiation with respect to x. Using (17.33) this may alsobe written Ly ≡−(py ′ ) ′ − qy = λρy, which defines a more useful form for theSturm–Liouville linear operator, namelyL ≡−[ ddx(p(x) ddx) ]+ q(x) . (17.35)17.4.1 Hermitian nature of the Sturm–Liouville operatorAs we now show, the linear operator of the Sturm–Liouville equation (17.35) isself-adjoint. Moreover, the operator is Hermitian over the range [a, b] provided§ We note that sign conventions vary in this expression for the general Sturm–Liouville equation;some authors use −λρ(x)y on the LHS of (17.32).564

17.4 STURM–LIOUVILLE EQUATIONScertain boundary conditions are met, namely that any two eigenfunctions y i andy j of (17.33) must satisfy[y∗i py j]′ y ∗ x=a i py j′ ]x=bfor all i, j. (17.36)Rearranging (17.36), we can write[ ] x=byi ∗ py j′ = 0 x=a (17.37)as an equivalent statement of the required boundary conditions. These boundaryconditions are in fact not too restrictive and are met, for instance, by the setsy(a) =y(b) =0;y(a) =y ′ (b) =0;p(a) =p(b) = 0 and by many other sets. Itis important to note that in order to satisfy (17.36) and (17.37) one boundarycondition must be specified at each end of the range.◮Prove that the Sturm–Liouville operator is Hermitian over the range [a, b] and under theboundary conditions (17.37).Putting the Sturm–Liouville form Ly = −(py ′ ) ′ − qy into the definition (17.16) of anHermitian operator, the LHS may be written as a sum of two terms, i.e.∫ b∫[ ] b∫ b− y∗i (py j) ′ ′ + yi ∗ qy j dx = − yi ∗ (py j) ′ ′ dx − yi ∗ qy j dx.aThe first term may be integrated by parts to give[ ] b ∫ b− yi ∗ py j′ + (yi ∗ ) ′ py j ′ dx.a aThe boundary-value term in this is zero because of the boundary conditions, and sointegrating by parts again yields[ ] b ∫ b(yi ∗ ) ′ py j − ((yi ∗ ) ′ p) ′ y j dx.aAgain, the boundary-value term is zero, leaving us with−∫ ba[y∗i (py ′ j) ′ + y ∗ i qy j]dx = −∫ baaa[ ]yj (p(yi ∗ ) ′ ) ′ + y j qyi∗ dx,which proves that the Sturm–Liouville operator is Hermitian over the prescribed interval. ◭It is also worth noting that, since p(a) =p(b) = 0 is a valid set of boundaryconditions, many Sturm–Liouville equations possess a ‘natural’ interval [a, b] overwhich the corresponding differential operator L is Hermitian irrespective of theboundary conditions satisfied by its eigenfunctions at x = a and x = b (the onlyrequirement being that they are regular at these end-points).a17.4.2 Transforming an equation into Sturm–Liouville formMany of the second-order differential equations encountered in physical problemsare examples of the Sturm–Liouville equation (17.34). Moreover, any second-order565

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSEquation p(x) q(x) λ ρ(x)Hypergeometric x c (1 − x) a+b−c+1 0 −ab x c−1 (1 − x) a+b−cLegendre 1 − x 2 0 l(l +1) 1Associated Legendre 1 − x 2 −m 2 /(1 − x 2 ) l(l +1) 1Chebyshev (1 − x 2 ) 1/2 0 ν 2 (1 − x 2 ) −1/2Confluent hypergeometric x c e −x 0 −a x c−1 e −xBessel ∗ x −ν 2 /x α 2 xLaguerre xe −x 0 ν e −xAssociated Laguerre x m+1 e −x 0 ν x m e −xHermite e −x2 0 2ν e −x2Simple harmonic 1 0 ω 2 1Table 17.1 The Sturm–Liouville form (17.34) for important ODEs in thephysical sciences and engineering. The asterisk denotes that, for Bessel’s equation,a change of variable x → x/a is required to give the conventionalnormalisation used here, but is not needed for the transformation into Sturm–Liouville form.differential equation of the formp(x)y ′′ + r(x)y ′ + q(x)y + λρ(x)y = 0 (17.38)can be converted into Sturm–Liouville form by multiplying through by a suitableintegrating factor, which is given by{∫ xr(u) − p ′ }(u)F(x) =expdu . (17.39)p(u)It is easily verified that (17.38) then takes the Sturm–Liouville form,[F(x)p(x)y ′ ] ′ + F(x)q(x)y + λF(x)ρ(x)y =0, (17.40)with a different, but still non-negative, weight function F(x)ρ(x). Table 17.1summarises the Sturm–Liouville form (17.34) for several of the equations listedin table 16.1. These forms can be determined using (17.39), as illustrated in thefollowing example.◮Put the following equations into Sturm–Liouville (SL) form:(i) (1 − x 2 )y ′′ − xy ′ + ν 2 y =0 (Chebyshev equation);(ii) xy ′′ +(1− x)y ′ + νy =0 (Laguerre equation);(iii) y ′′ − 2xy ′ +2νy =0 (Hermite equation).(i) From (17.39), the required integrating factor is(∫ x) uF(x) =exp1 − u du =exp [ − 1 ln(1 − 2 2 x2 ) ] =(1− x 2 ) −1/2 .Thus, the Chebyshev equation becomes(1 − x 2 ) 1/2 y ′′ − x(1 − x 2 ) −1/2 y ′ + ν 2 (1 − x 2 ) −1/2 y = [ (1 − x 2 ) 1/2 y ′] ′+ ν 2 (1 − x 2 ) −1/2 y =0,whichisinSLformwithp(x) =(1− x 2 ) 1/2 , q(x) =0,ρ(x) =(1− x 2 ) −1/2 and λ = ν 2 .566

17.4 STURM–LIOUVILLE EQUATIONS(ii) From (17.39), the required integrating factor isx)F(x) =exp(∫−1 du =exp(−x).Thus, the Laguerre equation becomesxe −x y ′′ +(1− x)e −x y ′ + νe −x y =(xe −x y ′ ) ′ + νe −x y =0,which is in SL form with p(x) =xe −x , q(x) =0,ρ(x) =e −x and λ = ν.(iii) From (17.39), the required integrating factor isx)F(x) =exp(∫−2udu =exp(−x 2 ).Thus, the Hermite equation becomese −x2 y ′′ − 2xe −x2 y ′ +2νe −x2 y =(e −x2 y ′ ) ′ +2νe −x2 y =0,which is in SL form with p(x) =e −x2 , q(x) =0,ρ(x) =e −x2 and λ =2ν. ◭From the p(x) entries in table 17.1, we may read off the natural interval overwhich the corresponding Sturm–Liouville operator (17.35) is Hermitian; in eachcase this is given by [a, b], where p(a) =p(b) = 0. Thus, the natural intervalfor the Legendre equation, the associated Legendre equation and the Chebyshevequation is [−1, 1]; for the Laguerre and associated Laguerre equations theinterval is [0, ∞]; and for the Hermite equation it is [−∞, ∞]. In addition, from(17.37), one sees that for the simple harmonic equation one requires only that[a, b] =[x 0 ,x 0 +2π]. We also note that, as required, the weight function in eachcase is finite and non-negative over the natural interval. Occasionally, a little morecare is required when determining the conditions for a Sturm–Liouville operatorof the form (17.35) to be Hermitian over some natural interval, as is illustratedin the following example.◮Express the hypergeometric equation,x(1 − x)y ′′ +[c − (a + b +1)x ]y ′ − aby =0,in Sturm–Liouville form. Hence determine the natural interval over which the resultingSturm–Liouville operator is Hermitian and the corresponding conditions that one must imposeon the parameters a, b and c.As usual for an equation not already in SL form, we first determine the appropriate567

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSintegrating factor. This is given, as in equation (17.39), by[ ∫ xc − (a + b +1)u − 1+2uF(x) =expu(1 − u)[ ∫ x]c − 1 − (a + b − 1)u=expduu(1 − u)[ ∫ x( c − 1=exp1 − u + c − 1u]du− a + b − 1 )]du1 − u=exp[(a + b − c)ln(1− x)+(c − 1) ln x ]= x c−1 (1 − x) a+b−c .When the equation is multiplied through by F(x) ittakestheform[x c (1 − x) a+b−c+1 y ′ ] ′− abx c−1 (1 − x) a+b−c y =0.Now, for the corresponding Sturm–Liouville operator to be Hermitian, the conditionsto be imposed are as follows.(i) The boundary condition (17.37); if c>0anda + b − c +1> 0, this is satisfiedautomatically for 0 ≤ x ≤ 1, which is thus the natural interval in this case.(ii) The weight function x c−1 (1 − x) a+b−c must be finite and not change sign in theinterval 0 ≤ x ≤ 1. This means that both exponents in it must be positive, i.e.c − 1 > 0anda + b − c>0.Putting together the conditions on the parameters gives the double inequality a + b>c>1. ◭Finally, we consider Bessel’s equation,x 2 y ′′ + xy ′ +(x 2 − ν 2 )y =0,which may be converted into Sturm–Liouville form, but only in a somewhatunorthodox fashion. It is conventional first to divide the Bessel equation by xand then to change variables to ¯x = x/α. In this case, it becomes¯xy ′′ (α¯x)+y ′ (α¯x) − ν2¯x y(α¯x)+α2¯xy(α¯x) =0, (17.41)where a prime now indicates differentiation with respect to ¯x. Dropping the barson the independent variable, we thus have[xy ′ (αx)] ′ − ν2x y(αx)+α2 xy(αx) =0, (17.42)whichisinSLformwithp(x) =x, q(x) =−ν 2 /x, ρ(x) =x and λ = α 2 .Itshould be noted, however, that in this case the eigenvalue (actually its squareroot) appears in the argument of the dependent variable.568

17.5 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS17.5 Superposition of eigenfunctions: Green’s functionsWe have already seen that ifLy n (x) =λ n ρ(x)y n (x), (17.43)where L is an Hermitian operator, then the eigenvalues λ n are real and theeigenfunctions y n (x) are orthogonal (or can be made so). Let us assume that weknow the eigenfunctions y n (x) ofL that individually satisfy (17.43) and someimposed boundary conditions (for which L is Hermitian).Now let us suppose we wish to solve the inhomogeneous differential equationLy(x) =f(x), (17.44)subject to the same boundary conditions. Since the eigenfunctions of L form acomplete set, the full solution, y(x), to (17.44) may be written as a superpositionof eigenfunctions, i.e.∞∑y(x) = c n y n (x), (17.45)n=0for some choice of the constants c n . Making full use of the linearity of L, we have( ∞)∑∞∑∞∑f(x) =Ly(x) =L c n y n (x) = c n Ly n (x) = c n λ n ρ(x)y n (x).n=0n=0n=0(17.46)Multiplying the first and last terms of (17.46) by yj ∗ and integrating, we obtain∫ bay ∗ j (z)f(z) dz =∞∑n=0∫ bac n λ n y ∗ j (z)y n (z)ρ(z) dz, (17.47)where we have used z as the integration variable for later convenience. Finally,using the orthogonality condition (17.27), we see that the integrals on the RHSare zero unless n = j, and so obtainc n = 1 ∫ ba y∗ n(z)f(z) dz∫λ b na y∗ n (z)y n(z)ρ(z) dz . (17.48)Thus, if we can find all the eigenfunctions of a differential operator then (17.48)can be used to find the weighting coefficients for the superposition, to give as thefull solutiony(x) =∞∑ 1λ nn=0∫ ba y∗ n(z)f(z) dz∫ ba y∗ n (z)y n(z)ρ(z) dz y n(x). (17.49)If we work with normalised eigenfunctions ŷ n (x), so that∫ baŷ ∗ n(z)ŷ n (z)ρ(z) dz = 1 for all n,569

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSand we assume that we may interchange the order of summation and integration,then (17.49) can be written as∫ {b ∑ ∞ [ ] } 1y(x) =ŷ n (x)ŷ ∗λn(z) f(z) dz.nan=0The quantity in braces, which is a function of x and z only, is usually writtenG(x, z), and is the Green’s function for the problem. With this notation,wherey(x) =G(x, z) =∫ baG(x, z)f(z) dz, (17.50)∞∑n=01λ nŷ n (x)ŷ ∗ n(z). (17.51)We note that G(x, z) is determined entirely by the boundary conditions and theeigenfunctions ŷ n , and hence by L itself, and that f(z) depends purely on theRHS of the inhomogeneous equation (17.44). Thus, for a given L and boundaryconditions we can establish, once and for all, a function G(x, z) that will enableus to solve the inhomogeneous equation for any RHS. From (17.51) we also notethatG(x, z) =G ∗ (z,x). (17.52)We have already met the Green’s function in the solution of second-order differentialequations in chapter 15, as the function that satisfies the equationL[G(x, z)] = δ(x − z) (and the boundary conditions). The formulation givenabove is an alternative, though equivalent, one.◮Find an appropriate Green’s function for the equationy ′′ + 1 y = f(x),4with boundary conditions y(0) = y(π) =0.Hence,solvefor(i) f(x) =sin2x and (ii)f(x) =x/2.One approach to solving this problem is to use the methods of chapter 15 and finda complementary function and particular integral. However, in order to illustrate thetechniques developed in the present chapter we will use the superposition of eigenfunctions,which, as may easily be checked, produces the same solution.The operator on the LHS of this equation is already Hermitian under the given boundaryconditions, and so we seek its eigenfunctions. These satisfy the equationy ′′ + 1 y = λy.4This equation has the familiar solution(√ ) (√ )1y(x) =A sin − λ 1x + B cos − λ x.4 4570

17.5 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS(√ )1Now, the boundary conditions require that B =0andsin − λ π =0,andso4√1− λ = n, where n =0, ±1, ±2,....4Therefore, the independent eigenfunctions that satisfy the boundary conditions arey n (x) =A n sin nx,where n is any non-negative integer, and the corresponding eigenvalues are λ n = 1 − 4 n2 .The normalisation condition further requires∫ π( ) 1/2 2A 2 n sin 2 nx dx =1 ⇒ A n = .0πComparison with (17.51) shows that the appropriate Green’s function is therefore givenbyG(x, z) = 2 ∞∑ sin nx sin nz1π − .n=0 4 n2Case (i). Using (17.50), the solution with f(x) =sin2x is given byy(x) = 2 ∫ (π ∞)∑ sin nx sin nz1π 0− sin 2zdz= 2 ∞∑n=0 4 n2 πn=0Now the integral is zero unless n = 2, in which case it is∫ π0sin 2 2zdz= π 2 .∫sin nx π1− sin nz sin 2zdz.4 n2 0Thusy(x) =− 2 sin 2x ππ 15/4 2 = − 4 sin 2x15is the full solution for f(x) =sin2x. This is, of course, exactly the solution found by usingthe methods of chapter 15.Case (ii). The solution with f(x) =x/2 isgivenby∫ ()π2∞∑ sin nx sin nz zy(x) =10 π − n=0 4 n2 2 dz = 1 ∞∑πn=0The integral may be evaluated by integrating by parts. For n ≠0,∫ [π∫z cos nzπcos nzz sin nz dz = − + dznn0=−π cos nπn] π00[ ] π sin nz+n 2 0∫sin nx π1− z sin nz dz.4 n2 0= − π(−1)n .nFor n = 0 the integral is zero, and thus∞∑y(x) = (−1) n+1 sin nxn ( 1− n2) ,n=14is the full solution for f(x) =x/2. Using the methods of subsection 15.1.2, the solutionis found to be y(x) =2x − 2π sin(x/2), which may be shown to be equal to the abovesolution by expanding 2x − 2π sin(x/2) as a Fourier sine series. ◭571

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS17.6 A useful generalisationSometimes we encounter inhomogeneous equations of a form slightly more generalthan (17.1), given byLy(x) − µρ(x)y(x) =f(x) (17.53)for some Hermitian operator L, with y subject to the appropriate boundaryconditions and µ a given (i.e. fixed) constant. To solve this equation we expandy(x) andf(x) in terms of the eigenfunctions y n (x) of the operator L, which satisfyLy n (x) =λ n ρ(x)y n (x).Working in terms of the normalised eigenfunctions ŷ n (x), we first expand f(x)as follows:∞∑∫ bf(x) = ŷ n (x) ŷn(z)f(z)ρ(z) ∗ dzUsing (17.29) this becomes=f(x) =n=0∫ ba∫ ba= ρ(x)ρ(z)ρ(x)a∞∑ŷ n (x)ŷn(z)f(z) ∗ dz. (17.54)n=0∞∑ŷ n (x)ŷn(z)f(z) ∗ dzn=0∞∑ŷ n (x)n=0∫ baŷ ∗ n(z)f(z) dz. (17.55)Next, we expand y(x)asy = ∑ ∞n=0 c nŷ n (x) and seek the coefficients c n . Substitutingthis and (17.55) into (17.53) we have∞∑∞∑∫ bρ(x) (λ n − µ)c n ŷ n (x) =ρ(x) ŷ n (x) ŷn(z)f(z) ∗ dz,n=0from which we find thatn=0n=0∫ ∞∑ ba ŷ∗ nn=0n=0(z)f(z) dzc n =.λ n − µHence the solution of (17.53) is given by∞∑∞∑∫ŷ n (x) by = c n ŷ n (x) =ŷ ∗λ n − µn(z)f(z) dz =From this we may identify the Green’s function∞∑ ŷ n (x)ŷ ∗G(x, z) =n(z)λ n − µ .an=0572a∫ ba∞∑n=0ŷ n (x)ŷn(z)∗ f(z) dz.λ n − µ

17.7 EXERCISESWe note that if µ = λ n ,i.e.ifµ equals one of the eigenvalues of L, thenG(x, z)becomes infinite and this method runs into difficulty. No solution then existsunless the RHS of (17.53) satisfies the relation∫ baŷ ∗ n(x)f(x) dx =0.If the spectrum of eigenvalues of the operator L is anywhere continuous,the orthonormality and closure relationships of the normalised eigenfunctionsbecome∫ ba∫ ∞0ŷ ∗ n(x)ŷ m (x)ρ(x) dx = δ(n − m),ŷ ∗ n(z)ŷ n (x)ρ(x) dn = δ(x − z).Repeating the above analysis we then find that the Green’s function is given byG(x, z) =∫ ∞0ŷ n (x)ŷ ∗ n(z)λ n − µdn.17.7 Exercises17.1 By considering 〈h|h〉, whereh = f + λg with λ real, prove that, for two functionsf and g,〈f|f〉〈g|g〉 ≥ 1 [〈f|g〉 + 4 〈g|f〉]2 .The function y(x) is real and positive for all x. Its Fourier cosine transform ỹ c (k)is defined by∫ ∞ỹ c (k) = y(x)cos(kx) dx,−∞and it is given that ỹ c (0) = 1. Prove thatỹ c (2k) ≥ 2[ỹ c (k)] 2 − 1.17.2 Write the homogeneous Sturm-Liouville eigenvalue equation for which y(a) =y(b) =0asL(y; λ) ≡ (py ′ ) ′ + qy + λρy =0,where p(x),q(x) andρ(x) are continuously differentiable functions. Show that ifz(x) andF(x) satisfyL(z; λ) =F(x), with z(a) =z(b) =0,then∫ bay(x)F(x) dx =0.Demonstrate the validity of this general result by direct calculation for thespecific case in which p(x) =ρ(x) =1,q(x) =0,a = −1, b =1andz(x) =1− x 2 .17.3 Consider the real eigenfunctions y n (x) of a Sturm–Liouville equation,(py ′ ) ′ + qy + λρy =0, a ≤ x ≤ b,in which p(x), q(x) andρ(x) are continuously differentiable real functions andp(x) does not change sign in a ≤ x ≤ b. Takep(x) as positive throughout the573

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSinterval, if necessary by changing the signs of all eigenvalues. For a ≤ x 1 ≤ x 2 ≤ b,establish the identity(λ n − λ m )∫ x2x 1ρy n y m dx = [ ]y n py m ′ − y m py n′ x2.x 1Deduce that if λ n >λ m then y n (x) must change sign between two successive zerosof y m (x).[ The reader may find it helpful to illustrate this result by sketching the first feweigenfunctions of the system y ′′ + λy =0,withy(0) = y(π) = 0, and the Legendrepolynomials P n (z) forn =2, 3, 4, 5. ]17.4 Show that the equationy ′′ + aδ(x)y + λy =0,with y(±π) =0anda real, has a set of eigenvalues λ satisfyingtan(π √ λ)= 2√ λa .Investigate the conditions under which negative eigenvalues, λ = −µ 2 ,withµreal, are possible.17.5 Use the properties of Legendre polynomials to carry out the following exercises.(a) Find the solution of (1 − x 2 )y ′′ − 2xy ′ + by = f(x), valid in the range−1 ≤ x ≤ 1 and finite at x = 0, in terms of Legendre polynomials.(b) If b =14andf(x) =5x 3 , find the explicit solution and verify it by directsubstitution.[ The first six Legendre polynomials are listed in Subsection 18.1.1. ]17.6 Starting from the linearly independent functions 1, x, x 2 , x 3 , ... , in the range0 ≤ x

17.7 EXERCISESwhere κ is a constant and{x 0 ≤ x ≤ π/2,f(x) =π − x π/2

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS17.14 Express the solution of Poisson’s equation in electrostatics,∇ 2 φ(r) =−ρ(r)/ɛ 0 ,where ρ is the non-zero charge density over a finite part of space, in the form ofan integral and hence identify the Green’s function for the ∇ 2 operator.17.15 In the quantum-mechanical study of the scattering of a particle by a potential,a Born-approximation solution can be obtained in terms of a function y(r) thatsatisfies an equation of the form(−∇ 2 − K 2 )y(r) =F(r).Assuming that y k (r) =(2π) −3/2 exp(ik·r) is a suitably normalised eigenfunction of−∇ 2 corresponding to eigenvalue k 2 , find a suitable Green’s function G K (r, r ′ ). Bytaking the direction of the vector r − r ′ as the polar axis for a k-space integration,show that G K (r, r ′ ) can be reduced to14π 2 |r − r ′ |∫ ∞−∞w sin wdw,w 2 − w02where w 0 = K|r − r ′ |.[ This integral can be evaluated using a contour integration (chapter 24) to give(4π|r − r ′ |) −1 exp(iK|r − r ′ |). ]17.8 Hints and answers17.1 Express the condition 〈h|h〉 ≥0 as a quadratic equation in λ and then apply the∫condition for no real roots, noting that 〈f|g〉 + 〈g|f〉 is real. To put a limit ony cos 2 kx dx, setf = y 1/2 cos kx and g = y 1/2 in the inequality.17.3 Follow an argument similar to that used for proving the reality of the eigenvalues,but integrate from x 1 to x 2 , rather than from a to b. Takex 1 and x 2 as twosuccessive zeros of y m (x) and note that, if the sign of y m is α then the sign of y m(x ′ 1 )is α whilst that of y m(x ′ 2 )is−α. Now assume that y n (x) does not change sign inthe interval and has a constant sign β; show that this leads to a contradictionbetween the signs of the two sides of the identity.17.5 (a) y = ∑ a n P n (x) witha n = n +1/2 ∫ 1f(z)P n (z) dz;b − n(n +1) −1(b) 5x 3 =2P 3 (x)+3P 1 (x), giving a 1 =1/4 anda 3 = 1, leading to y =5(2x 3 −x)/4.17.7 (a) No, ∫ gf ∗′ dx ≠ 0; (b) yes; (c) no, i ∫ f ∗ gdx ≠0;(d)yes.17.9 The normalised eigenfunctions are (2/π) 1/2 sin nx, withn an integer.y(x) =(4/π) ∑ n odd [(−1)(n−1)/2 sin nx]/[n 2 (κ − n 2 )].17.11 λ n =(n +1/2) 2 π 2 ,n=0, 1, 2,... .(a) Since y n (1)y m(1) ′ ≠ 0, the Sturm–Liouville boundary conditions are not satisfiedand the appropriate weight function has to be justified by inspection. Thenormalised eigenfunctions are √ 2e −x/2 sin[(n +1/2)πx], with ρ(x) =e x .(b) y(x) =(−2/π 3 ) ∑ ∞n=0 e−x/2 sin[(n +1/2)πx]/(n +1/2) 3 .17.13 y n (x) = √ 2x −1/2 sin(nπ ln x) withλ n = −n 2 π 2 ;{ ∫ −(nπ)−2 e√a n =1 2x −1 sin(nπ ln x) dx = − √ 8(nπ) −3 for n odd,0 for n even.17.15 Use the form of Green’s function that is the integral over all eigenvalues of the‘outer product’ of two eigenfunctions corresponding to the same eigenvalue, butwith arguments r and r ′ .576

18Special functionsIn the previous two chapters, we introduced the most important second-orderlinear ODEs in physics and engineering, listing their regular and irregular singularpoints in table 16.1 and their Sturm–Liouville forms in table 17.1. Theseequations occur with such frequency that solutions to them, which obey particularcommonly occurring boundary conditions, have been extensively studied andgiven special names. In this chapter, we discuss these so-called ‘special functions’and their properties. In addition, we also discuss some special functions that arenot derived from solutions of important second-order ODEs, namely the gammafunction and related functions. These convenient functions appear in a numberof contexts, and so in section 18.12 we gather together some of their properties,with a minimum of formal proofs.18.1 Legendre functionsLegendre’s differential equation has the form(1 − x 2 )y ′′ − 2xy ′ + l(l +1)y =0, (18.1)and has three regular singular points, at x = −1, 1, ∞. It occurs in numerousphysical applications and particularly in problems with axial symmetry thatinvolve the ∇ 2 operator, when they are expressed in spherical polar coordinates.In normal usage the variable x in Legendre’s equation is the cosine of the polarangle in spherical polars, and thus −1 ≤ x ≤ 1. The parameter l is a given realnumber, and any solution of (18.1) is called a Legendre function.In subsection 16.1.1, we showed that x = 0 is an ordinary point of (18.1), and sowe expect to find two linearly independent solutions of the form y = ∑ ∞n=0 a nx n .Substituting, we find∞∑ [n(n − 1)an x n−2 − n(n − 1)a n x n − 2na n x n + l(l +1)a n x n] =0,n=0577

SPECIAL FUNCTIONSwhich on collecting terms gives∞∑{(n +2)(n +1)a n+2 − [n(n +1)− l(l +1)]a n } x n =0.n=0The recurrence relation is therefore[n(n +1)− l(l +1)]a n+2 = a n , (18.2)(n +1)(n +2)for n =0, 1, 2,... . If we choose a 0 = 1 and a 1 = 0 then we obtain the solutiony 1 (x) =1− l(l +1) x2+(l − 2)l(l +1)(l +3)x4 − ··· , (18.3)2! 4!whereas on choosing a 0 = 0 and a 1 = 1 we find a second solutiony 2 (x) =x − (l − 1)(l +2) x3+(l − 3)(l − 1)(l +2)(l +4)x5 − ··· . (18.4)3! 5!By applying the ratio test to these series (see subsection 4.3.2), we find that bothseries converge for |x| < 1, and so their radius of convergence is unity, which(as expected) is the distance to the nearest singular point of the equation. Since(18.3) contains only even powers of x and (18.4) contains only odd powers, thesetwo solutions cannot be proportional to one another, and are therefore linearlyindependent. Hence, the general solution to (18.1) for |x| < 1isy(x) =c 1 y 1 (x)+c 2 y 2 (x).18.1.1 Legendre functions for integer lIn many physical applications the parameter l in Legendre’s equation (18.1) isan integer, i.e. l =0, 1, 2,... . In this case, the recurrence relation (18.2) gives[l(l +1)− l(l +1)]a l+2 = a l =0,(l +1)(l +2)i.e. the series terminates and we obtain a polynomial solution of order l. Inparticular, if l is even, then y 1 (x) in (18.3) reduces to a polynomial, whereas if l isoddthesameistrueofy 2 (x) in (18.4). These solutions (suitably normalised) arecalled the Legendre polynomials of order l; they are written P l (x) and are validfor all finite x. It is conventional to normalise P l (x) in such a way that P l (1) = 1,and as a consequence P l (−1) = (−1) l . The first few Legendre polynomials areeasily constructed and are given byP 0 (x) =1,P 2 (x) = 1 2 (3x2 − 1),P 4 (x) = 1 8 (35x4 − 30x 2 +3),P 1 (x) =x,P 3 (x) = 1 2 (5x3 − 3x),P 5 (x) = 1 8 (63x5 − 70x 3 +15x).578

18.1 LEGENDRE FUNCTIONS2P 01P 1−1P 2P 3−0.5 0.51x−1−2Figure 18.1The first four Legendre polynomials.The first four Legendre polynomials are plotted in figure 18.1.Although, according to whether l is an even or odd integer, respectively, eithery 1 (x) in (18.3) or y 2 (x) in (18.4) terminates to give a multiple of the correspondingLegendre polynomial P l (x), the other series in each case does not terminate andtherefore converges only for |x| < 1. According to whether l is even or odd, wedefine Legendre functions of the second kind as Q l (x) =α l y 2 (x)orQ l (x) =β l y 1 (x),respectively, where the constants α l and β l are conventionally taken to have thevaluesα l = (−1)l/2 2 l [(l/2)!] 2l!β l = (−1)(l+1)/2 2 l−1 {[(l − 1)/2]!} 2l!for l even, (18.5)for l odd. (18.6)These normalisation factors are chosen so that the Q l (x) obey the same recurrencerelations as the P l (x) (see subsection 18.1.2).The general solution of Legendre’s equation for integer l is thereforey(x) =c 1 P l (x)+c 2 Q l (x), (18.7)579

SPECIAL FUNCTIONSwhere P l (x) is a polynomial of order l, and so converges for all x, andQ l (x) isan infinite series that converges only for |x| < 1. §By using the Wronskian method, section 16.4, we may obtain closed forms forthe Q l (x).◮Use the Wronskian method to find a closed-form expression for Q 0 (x).From (16.25) a second solution to Legendre’s equation (18.1), with l =0,is∫ x(∫ 1u) 2vy 2 (x) =P 0 (x)[P 0 (u)] exp 2 1 − v dv du2∫ x= exp [ − ln(1 − u 2 ) ] du∫ x( )du1+x=(1 − u 2 ) = 1 ln , (18.8)21 − xwhere in the second line we have used the fact that P 0 (x) =1.All that remains is to adjust the normalisation of this solution so that it agrees with(18.5). Expanding the logarithm in (18.8) as a Maclaurin series we obtainy 2 (x) =x + x33 + x55 + ··· .Comparing this with the expression for Q 0 (x), using (18.4) with l = 0 and the normalisation(18.5), we find that y 2 (x) is already correctly normalised, and so( ) 1+xQ 0 (x) = 1 ln .21 − xOf course, we might have recognised the series (18.4) for l = 0, but to do so for larger lwould prove progressively more difficult. ◭Using the above method for l = 1, we find( ) 1+xQ 1 (x) = 1 2 x ln − 1.1 − xClosed forms for higher-order Q l (x) may now be found using the recurrencerelation (18.27) derived in the next subsection. The first few Legendre functionsof the second kind are plotted in figure 18.2.18.1.2 Properties of Legendre polynomialsAs stated earlier, when encountered in physical problems the variable x inLegendre’s equation is usually the cosine of the polar angle θ in spherical polarcoordinates, and we then require the solution y(x) to be regular at x = ±1, whichcorresponds to θ =0orθ = π. For this to occur we require the equation to havea polynomial solution, and so l must be an integer. Furthermore, we also require§ It is possible, in fact, to find a second solution in terms of an infinite series of negative powers ofx that is finite for |x| > 1 (see exercise 16.16).580

18.1 LEGENDRE FUNCTIONS1Q 00.5−1−0.5−0.5−1xQ 10.5 1Q 2Figure 18.2The first three Legendre functions of the second kind.the coefficient c 2 of the function Q l (x) in (18.7) to be zero, since Q l (x) is singularat x = ±1, with the result that the general solution is simply some multiple of therelevant Legendre polynomial P l (x). In this section we will study the propertiesof the Legendre polynomials P l (x) in some detail.Rodrigues’ formulaAs an aid to establishing further properties of the Legendre polynomials we nowdevelop Rodrigues’ representation of these functions. Rodrigues’ formula for theP l (x) isP l (x) = 1 d l2 l l! dx l (x2 − 1) l . (18.9)To prove that this is a representation we let u =(x 2 −1) l ,sothatu ′ =2lx(x 2 −1) l−1and(x 2 − 1)u ′ − 2lxu =0.If we differentiate this expression l + 1 times using Leibnitz’ theorem, we obtain[(x 2 − 1)u (l+2) +2x(l +1)u (l+1) + l(l +1)u (l)] − 2l [ xu (l+1) +(l +1)u (l)] =0,581

SPECIAL FUNCTIONSwhich reduces to(x 2 − 1)u (l+2) +2xu (l+1) − l(l +1)u (l) =0.Changing the sign all through, we recover Legendre’s equation (18.1) with u (l) asthe dependent variable. Since, from (18.9), l is an integer and u (l) is regular atx = ±1, we may make the identificationu (l) (x) =c l P l (x), (18.10)for some constant c l that depends on l. To establish the value of c l we notethat the only term in the expression for the lth derivative of (x 2 − 1) l thatdoes not contain a factor x 2 − 1, and therefore does not vanish at x =1,is(2x) l l!(x 2 − 1) 0 . Putting x = 1 in (18.10) and recalling that P l (1) = 1, thereforeshows that c l =2 l l!, thus completing the proof of Rodrigues’ formula (18.9).◮Use Rodrigues’ formula to show thatI l =∫ 1−1P l (x)P l (x) dx = 22l +1 . (18.11)The result is trivially obvious for l = 0 and so we assume l ≥ 1. Then, by Rodrigues’formula,∫1 1[ ][ ]d l (x 2 − 1) l d l (x 2 − 1) lI l =dx.2 2l (l!) 2 −1 dx l dx lRepeated integration by parts, with all boundary terms vanishing, reduces this to∫ 1I l =(−1)l (x 2 − 1) l d2l2 2l (l!) 2 −1 dx 2l (x2 − 1) l dx= (2l)! ∫ 1(1 − x 2 ) l dx.2 2l (l!) 2 −1If we writeK l =∫ 1−1(1 − x 2 ) l dx,then integration by parts (taking a factor 1 as the second part) givesK l =∫ 1Writing 2lx 2 as 2l − 2l(1 − x 2 )weobtainK l =2l∫ 1−1−12lx 2 (1 − x 2 ) l−1 dx.(1 − x 2 ) l−1 dx − 2l∫ 1−1(1 − x 2 ) l dx=2lK l−1 − 2lK land hence the recurrence relation (2l +1)K l =2lK l−1 . We therefore findK l =2l 2l − 22l +12l − 1 ··· 23 K 0 =2 l 2 l l!l!(2l +1)! 2= 22l+1 (l!) 2(2l +1)! ,which, when substituted into the expression for I l , establishes the required result. ◭582

18.1 LEGENDRE FUNCTIONSMutual orthogonalityIn section 17.4, we noted that Legendre’s equation was of Sturm–Liouville formwith p =1− x 2 , q =0,λ = l(l +1) and ρ = 1, and that its natural intervalwas [−1, 1]. Since the Legendre polynomials P l (x) are regular at the end-pointsx = ±1, they must be mutually orthogonal over this interval, i.e.∫ 1−1P l (x)P k (x) dx =0 ifl ≠ k. (18.12)Although this result follows from the general considerations of the previouschapter, it may also be proved directly, as shown in the following example.◮Prove directly that the Legendre polynomials P l (x) are mutually orthogonal over theinterval −1

SPECIAL FUNCTIONS◮Prove the expression (18.14) for the coefficients in the Legendre polynomial expansionof a function f(x).If we multiply (18.13) by P k (x) and integrate from x = −1 tox = 1 then we obtain∫ 1∞∑∫ 1P k (x)f(x) dx = a l P k (x)P l (x) dx−1l=0∫ 1= a k P k (x)P k (x) dx =2a k−12k +1 ,where we have used the orthogonality property (18.12) and the normalisation property(18.11). ◭Generating functionA useful device for manipulating and studying sequences of functions or quantitieslabelled by an integer variable (here, the Legendre polynomials P l (x) labelled byl) isagenerating function. The generating function has perhaps its greatest utilityin the area of probability theory (see chapter 30). However, it is also a greatconvenience in our present study.The generating function for, say, a series of functions f n (x) forn =0, 1, 2,... isa function G(x, h) containing, as well as x, a dummy variable h such that∞∑G(x, h) = f n (x)h n ,n=0i.e. f n (x) is the coefficient of h n in the expansion of G in powers of h. The utilityof the device lies in the fact that sometimes it is possible to find a closed formfor G(x, h).For our study of Legendre polynomials let us consider the functions P n (x)defined by the equation∞∑G(x, h) =(1− 2xh + h 2 ) −1/2 = P n (x)h n . (18.15)As we show below, the functions so defined are identical to the Legendre polynomialsand the function (1 − 2xh + h 2 ) −1/2 is in fact the generating function forthem. In the process we will also deduce several useful relationships between thevarious polynomials and their derivatives.◮Show that the functions P n (x) defined by (18.15) satisfy Legendre’s equationIn the following dP n (x)/dx will be denoted by P n. ′ Firstly, we differentiate the definingequation (18.15) with respect to x and geth(1 − 2xh + h 2 ) −3/2 = ∑ P nh ′ n . (18.16)Also, we differentiate (18.15) with respect to h to yield(x − h)(1 − 2xh + h 2 ) −3/2 = ∑ nP n h n−1 . (18.17)584−1n=0

18.1 LEGENDRE FUNCTIONSEquation (18.16) can then be written, using (18.15), ash ∑ P n h n =(1− 2xh + h 2 ) ∑ P nh ′ n ,and equating the coefficients of h n+1 we obtain the recurrence relationP n = P n+1 ′ − 2xP n ′ + P n−1. ′ (18.18)Equations (18.16) and (18.17) can be combined as(x − h) ∑ P nh ′ n = h ∑ nP n h n−1 ,from which the coefficent of h n yields a second recurrence relation,xP n ′ − P n−1 ′ = nP n ; (18.19)eliminating P n−1 ′ between (18.18) and (18.19) then gives the further result(n +1)P n = P n+1 ′ − xP n. ′ (18.20)If we now take the result (18.20) with n replaced by n − 1andaddx times (18.19) to itwe obtain(1 − x 2 )P n ′ = n(P n−1 − xP n ). (18.21)Finally, differentiating both sides with respect to x and using (18.19) again, we find(1 − x 2 )P n ′′ − 2xP n ′ = n[(P n−1 ′ − xP n) ′ − P n ]= n(−nP n − P n )=−n(n +1)P n ,andsotheP n defined by (18.15) do indeed satisfy Legendre’s equation. ◭The above example shows that the functions P n (x) defined by (18.15) satisfyLegendre’s equation with l = n (an integer) and, also from (18.15), these functionsare regular at x = ±1. Thus P n must be some multiple of the nth Legendrepolynomial. It therefore remains only to verify the normalisation. This is easilydone at x =1,whenG becomesG(1,h)=[(1− h) 2 ] −1/2 =1+h + h 2 + ··· ,and we can see that all the P n so defined have P n (1) = 1 as required, and are thusidentical to the Legendre polynomials.A particular use of the generating function (18.15) is in representing the inversedistance between two points in three-dimensional space in terms of Legendrepolynomials. If two points r and r ′ are at distances r and r ′ , respectively, fromthe origin, with r ′ r,however,585

SPECIAL FUNCTIONSr and r ′ must be exchanged in (18.22) or the series would not converge. Thisresult may be used, for example, to write down the electrostatic potential at apoint r due to a charge q at the point r ′ . Thus, in the case r ′

18.2 ASSOCIATED LEGENDRE FUNCTIONS18.2 Associated Legendre functionsThe associated Legendre equation has the form](1 − x 2 )y ′′ − 2xy ′ +[l(l +1)− m21 − x 2 y =0, (18.28)which has three regular singular points at x = −1, 1, ∞ and reduces to Legendre’sequation (18.1) when m = 0. It occurs in physical applications involving theoperator ∇ 2 , when expressed in spherical polars. In such cases, −l ≤ m ≤ l andm is restricted to integer values, which we will assume from here on. As was thecase for Legendre’s equation, in normal usage the variable x is the cosine of thepolar angle in spherical polars, and thus −1 ≤ x ≤ 1. Any solution of (18.28) iscalled an associated Legendre function.The point x = 0 is an ordinary point of (18.28), and one could obtain seriessolutions of the form y = ∑ n=0 a nx n in the same manner as that used forLegendre’s equation. In this case, however, it is more instructive to note that ifu(x) is a solution of Legendre’s equation (18.1), theny(x) =(1− x 2 ) |m|/2 d|m| u(18.29)dx |m|is a solution of the associated equation (18.28).◮Prove that if u(x) is a solution of Legendre’s equation, then y(x) given in (18.29) is asolution of the associated equation.For simplicity, let us begin by assuming that m is non-negative. Legendre’s equation for ureads(1 − x 2 )u ′′ − 2xu ′ + l(l +1)u =0,and, on differentiating this equation m times using Leibnitz’ theorem, we obtain(1 − x 2 )v ′′ − 2x(m +1)v ′ +(l − m)(l + m +1)v =0, (18.30)where v(x) =d m u/dx m . On settingy(x) =(1− x 2 ) m/2 v(x),the derivatives v ′ and v ′′ may be written as(v ′ =(1− x 2 ) −m/2 y ′ +v ′′ =(1− x 2 ) −m/2 [y ′′ +mx )1 − x y ,22mx1 − x 2 y′ + m m(m +2)x2y + y1 − x2 (1 − x 2 ) 2Substituting these expressions into (18.30) and simplifying, we obtain](1 − x 2 )y ′′ − 2xy ′ +[l(l +1)− m2y =0,1 − x 2which shows that y is a solution of the associated Legendre equation (18.28). Finally, wenote that if m is negative, the value of m 2 is unchanged, and so a solution for positive mis also a solution for the corresponding negative value of m. ◭From the two linearly independent series solutions to Legendre’s equation given587].

SPECIAL FUNCTIONSin (18.3) and (18.4), which we now denote by u 1 (x) andu 2 (x), we may obtain twolinearly-independent series solutions, y 1 (x) andy 2 (x), to the associated equationby using (18.29). From the general discussion of the convergence of power seriesgiven in section 4.5.1, we see that both y 1 (x) andy 2 (x) will also converge for|x| < 1. Hence the general solution to (18.28) in this range is given byy(x) =c 1 y 1 (x)+c 2 y 2 (x).18.2.1 Associated Legendre functions for integer lIf l and m are both integers, as is the case in many physical applications, thenthe general solution to (18.28) is denoted byy(x) =c 1 P m l (x)+c 2 Q m l (x), (18.31)where Pl m(x) andQm l (x) are associated Legendre functions of the first and secondkind, respectively. For non-negative values of m, these functions are related to theordinary Legendre functions for integer l byP m l (x) =(1− x 2 ) m/2 dm P ldx m ,Qm l (x) =(1− x 2 ) m/2 dm Q ldx m . (18.32)We see immediately that, as required, the associated Legendre functions reduceto the ordinary Legendre functions when m = 0. Since it is m 2 that appears inthe associated Legendre equation (18.28), the associated Legendre functions fornegative m values must be proportional to the corresponding function for nonnegativem. The constant of proportionality is a matter of convention. For thePl m (x) it is usual to regard the definition (18.32) as being valid also for negative mvalues. Although differentiating a negative number of times is not defined, whenP l (x) is expressed in terms of the Rodrigues’ formula (18.9), this problem doesnot occur for −l ≤ m ≤ l. § In this case,Pl−mm (l − m)!(x) =(−1)(l + m)! P l m (x). (18.33)◮Prove the result (18.33).From (18.32) and the Rodrigues’ formula (18.9) for the Legendre polynomials, we havePl m (x) = 12 l l! (1 − x2 m/2 dl+m)dx l+m (x2 − 1) l ,and, without loss of generality, we may assume that m is non-negative. It is convenient to§ Some authors define Pl−m (x) =Pl m(x), and similarly for the Qm l (x), in which case m is replaced by|m| in the definitions (18.32). It should be noted that, in this case, many of the results presented inthis section also require m to be replaced by |m|.588

18.2 ASSOCIATED LEGENDRE FUNCTIONSwrite (x 2 − 1) = (x +1)(x − 1) and use Leibnitz’ theorem to evaluate the derivative, whichyieldsPl m (x) = 12 l l! (1 − ∑l+mx2 ) m/2 (l + m)! d r (x +1) l d l+m−r (x − 1) l.r!(l + m − r)! dx r dx l+m−r r=0Considering the two derivative factors in a term in the summation, we note that the firstis non-zero only for r ≤ l and the second is non-zero for l + m − r ≤ l. Combining theseconditions yields m ≤ r ≤ l. Performing the derivatives, we thus obtainPl m (x) = 1l∑2 l l! (1 − x2 ) m/2 (l + m)! l!(x +1) l−r l!(x − 1) r−mr!(l + m − r)! (l − r)! (r − m)!r=mm/2 l!(l + m)!l∑ (x +1) l−r+ m 2 (x − 1) r− m 2=(−1)2 l r!(l + m − r)!(l − r)!(r − m)! . (18.34)r=mRepeating the above calculation for Pl−m (x) and identifying once more those terms inthe sum that are non-zero, we findPl−m−m/2l!(l − m)!(x) =(−1)∑2 l l−mr=0(x +1) l−r− m 2 (x − 1) r+ m 2r!(l − m − r)!(l − r)!(r + m)!−m/2 l!(l − m)!l∑ (x +1) l−¯r+ m 2 (x − 1)¯r− m 2=(−1)2 l (¯r − m)!(l − ¯r)!(l + m − ¯r)!¯r! , (18.35)¯r=mwhere, in the second equality, we have rewritten the summation in terms of the new index¯r = r + m. Comparing (18.34) and (18.35), we immediately arrive at the required result(18.33). ◭Since P l (x) is a polynomial of order l, we have Pl m (x) =0for|m| >l.Fromits definition, it is clear that Pl m (x) is also a polynomial of order l if m is even,but contains the factor (1 − x 2 ) to a fractional power if m is odd. In either case,Pl m (x) is regular at x = ±1. The first few associated Legendre functions of thefirst kind are easily constructed and are given by (omitting the m = 0 cases)P 1 1 (x) =(1− x2 ) 1/2 , P 1 2 (x) =3x(1 − x2 ) 1/2 ,P 2 2 (x) =3(1− x2 ), P 1 3 (x) = 3 2 (5x2 − 1)(1 − x 2 ) 1/2 ,P 2 3 (x) =15x(1 − x2 ), P 3 3 (x) = 15(1 − x2 ) 3/2 .Finally, we note that the associated Legendre functions of the second kind Q m l (x),like Q l (x), are singular at x = ±1.18.2.2 Properties of associated Legendre functions P m l (x)When encountered in physical problems, the variable x in the associated Legendreequation (as in the ordinary Legendre equation) is usually the cosine of the polarangle θ in spherical polar coordinates, and we then require the solution y(x) tobe regular at x = ±1 (corresponding to θ =0orθ = π). For this to occur, werequire l to be an integer and the coefficient c 2 of the function Q m l (x) in (18.31)589

SPECIAL FUNCTIONSto be zero, since Q m l (x) is singular at x = ±1, with the result that the generalsolution is simply some multiple of one of the associated Legendre functions ofthe first kind, Pl m (x). We will study the further properties of these functions inthe remainder of this subsection.Mutual orthogonalityAs noted in section 17.4, the associated Legendre equation is of Sturm–Liouvilleform (py) ′ + qy + λρy = 0, with p =1− x 2 , q = −m 2 /(1 − x 2 ), λ = l(l +1)and ρ = 1, and its natural interval is thus [−1, 1]. Since the associated Legendrefunctions Pl m (x) are regular at the end-points x = ±1, they must be mutuallyorthogonal over this interval for a fixed value of m, i.e.∫ 1−1P m l (x)P m k (x) dx =0 ifl ≠ k. (18.36)This result may also be proved directly in a manner similar to that used for demonstratingthe orthogonality of the Legendre polynomials P l (x) in section 18.1.2.Note that the value of m must be the same for the two associated Legendrefunctions for (18.36) to hold. The normalisation condition when l = k may beobtained using the Rodrigues’ formula, as shown in the following example.◮Show thatI lm ≡∫ 1−1Pl m (x)Pl m (x) dx = 2 (l + m)!2l +1(l − m)! . (18.37)From the definition (18.32) and the Rodrigues’ formula (18.9) for P l (x), we may write∫1 1][ ]I lm =[(1 − x 2 ) m dl+m (x 2 − 1) l d l+m (x 2 − 1) ldx,2 2l (l!) 2 −1dx l+m dx l+mwhere the square brackets identify the factors to be used when integrating by parts.Performing the integration by parts l + m times, and noting that all boundary termsvanish, we obtain∫ 1[]I lm = (−1)l+m (x 2 − 1) l dl+m(1 − x 2 ) m dl+m (x 2 − 1) ldx.2 2l (l!) 2 −1 dx l+m dx l+mUsing Leibnitz’ theorem, the second factor in the integrand may be written as[]d l+m(1 − x 2 ) m dl+m (x 2 − 1) ldx l+m dx l+m∑l+m=r=0(l + m)! d r (1 − x 2 ) m d 2l+2m−r (x 2 − 1) l.r!(l + m − r)! dx r dx 2l+2m−rConsidering the two derivative factors in a term in the summation on the RHS, wesee that the first is non-zero only for r ≤ 2m, whereas the second is non-zero only for2l +2m − r ≤ 2l. Combining these conditions, we find that the only non-zero term in thesum is that for which r =2m. Thus, we may write∫I lm = (−1)l+m (l + m)! 1(1 − x 2 ) l d2m (1 − x 2 ) m d 2l (1 − x 2 ) ldx.2 2l (l!) 2 (2m)!(l − m)!dx 2m dx 2l−1590

18.2 ASSOCIATED LEGENDRE FUNCTIONSSince d 2l (1 − x 2 ) l /dx 2l =(−1) l (2l)!, and noting that (−1) 2l+2m = 1, we have1 (2l)!(l + m)!I lm =2 2l (l!) 2 (l − m)!We have already shown in section 18.1.2 that∫ 1∫ 1−1(1 − x 2 ) l dx.K l ≡ (1 − x 2 ) l dx = 22l+1 (l!) 2−1(2l +1)! ,and so we obtain the final resultI lm = 2 (l + m)!2l +1(l − m)! . ◭The orthogonality and normalisation conditions, (18.36) and (18.37) respectively,mean that the associated Legendre functions Pl m (x), with m fixed, may beused in a similar way to the Legendre polynomials to expand any reasonablefunction f(x) on the interval |x| < 1 in a series of the formf(x) =∞∑a m+k Pm+k(x), m (18.38)k=0where, in this case, the coefficients are given by2l +1(l − m)!a l = f(x)Pl m (x) dx.2 (l + m)! −1We note that the series takes the form (18.38) because Pl m (x) =0form>l.Finally, it is worth noting that the associated Legendre functions Pl m (x) mustalso obey a second orthogonality relationship. This has to be so because one mayequally well write the associated Legendre equation (18.28) in Sturm–Liouvilleform (py) ′ +qy+λρy = 0, with p =1−x 2 , q = l(l+1), λ = −m 2 and ρ =(1−x 2 ) −1 ;once again the natural interval is [−1, 1]. Since the associated Legendre functionsPl m (x) are regular at the end-points x = ±1, they must therefore be mutuallyorthogonal with respect to the weight function (1 − x 2 ) −1 over this interval for afixed value of l, i.e.∫ 1−1∫ 1P m l (x)P k l (x)(1 − x 2 ) −1 dx =0 if|m| ≠ |k|. (18.39)One may also show straightforwardly that the corresponding normalisation conditionwhen m = k is given by∫ 1Pl m (x)Pl m (x)(1 − x 2 ) −1 (l + m)!dx =−1m(l − m)! .In solving physical problems, however, the orthogonality condition (18.39) is notof any practical use.591

SPECIAL FUNCTIONSGenerating functionThe generating function for associated Legendre functions can be easily derivedby combining their definition (18.32) with the generating function for the Legendrepolynomials given in (18.15). We find that(2m)!(1 − x 2 ) m/2G(x, h) =2 m m!(1 − 2hx + h 2 ) = ∑∞ Pn+m(x)h m n . (18.40)m+1/2n=0◮Derive the expression (18.40) for the associated Legendre generating function.The generating function (18.15) for the Legendre polynomials reads∞∑P n h n =(1− 2xh + h 2 ) −1/2 .n=0Differentiating both sides of this result m times (assumimg m to be non-negative), mutliplyingthrough by (1 − x 2 ) m/2 and using the definition (18.32) of the associated Legendrefunctions, we obtain∞∑Pn m h n =(1− x 2 m/2dm)dx (1 − 2xh + m h2 ) −1/2 .n=0Performing the derivatives on the RHS gives∞∑Pn m h n = 1 · 3 · 5 ···(2m − 1)(1 − x2 ) m/2 h m.(1 − 2xh + h 2 ) m+1/2 n=0Dividing through by h m , re-indexing the summation on the LHS and noting that, quitegenerally,1 · 3 · 5 ···(2r − 1) = 1 · 2 · 3 ···2r2 · 4 · 6 ···2r = (2r)!2 r r! ,we obtain the final result (18.40). ◭Recurrence relationsAs one might expect, the associated Legendre functions satisfy certain recurrencerelations. Indeed, the presence of the two indices n and m means that a much widerrange of recurrence relations may be derived. Here we shall content ourselveswith quoting just four of the most useful relations:Pn m+1 2mx=(1 − x 2 ) P m 1/2 n +[m(m − 1) − n(n +1)]Pn m−1 , (18.41)(2n +1)xPn m =(n + m)Pn−1 m +(n − m +1)Pn+1, m (18.42)(2n + 1)(1 − x 2 ) 1/2 Pn m = Pn+1 m+1 − P n−1 m+1 , (18.43)2(1 − x 2 ) 1/2 (Pn m ) ′ = Pnm+1 − (n + m)(n − m +1)Pn m−1 . (18.44)We note that, by virtue of our adopted definition (18.32), these recurrence relationsare equally valid for negative and non-negative values of m. These relations may592

18.3 SPHERICAL HARMONICSbe derived in a number of ways, such as using the generating function (18.40) orby differentiation of the recurrence relations for the Legendre polynomials P l (x).◮Use the recurrence relation (2n +1)P n = P n+1 ′ − P n−1 ′ for Legendre polynomials to derivethe result (18.43).Differentiating the recurrence relation for the Legendre polynomials m times, we have(2n +1) dm P ndx = dm+1 P n+1− dm+1 P n−1.m dx m+1 dx m+1Multiplying through by (1 − x 2 ) (m+1)/2 and using the definition (18.32) immediately givesthe result (18.43). ◭18.3 Spherical harmonicsThe associated Legendre functions discussed in the previous section occur mostcommonly when obtaining solutions in spherical polar coordinates of Laplace’sequation ∇ 2 u = 0 (see section 21.3.1). In particular, one finds that, for solutionsthat are finite on the polar axis, the angular part of the solution is given byΘ(θ)Φ(φ) =P m l (cos θ)(C cos mφ + D sin mφ),where l and m are integers with −l ≤ m ≤ l. This general form is sufficientlycommon that particular functions of θ and φ called spherical harmonics aredefined and tabulated. The spherical harmonics Yl m (θ, φ) are defined by[ ] 1/2 2l +1Yl m (θ, φ) =(−1) m (l − m)!Pl m (cos θ) exp(imφ). (18.45)4π (l + m)!Using (18.33), we note thatYl−m (θ, φ) =(−1) m [ Yl m (θ, φ) ] ∗,where the asterisk denotes complex conjugation. The first few spherical harmonicsYl mm(θ, φ) ≡ Ylare as follows:√√Y0 0 = 14π , Y 1 0 = 34πcos θ,√√Y ±1301= ∓8πsin θ exp(±iφ), Y2 = 516π (3 cos2 θ − 1),√√Y ±115±2152= ∓8πsin θ cos θ exp(±iφ), Y2=32π sin2 θ exp(±2iφ).Since they contain as their θ-dependent part the solution Plm to the associatedLegendre equation, the Ylm are mutually orthogonal when integrated from −1 to+1 over d(cos θ). Their mutual orthogonality with respect to φ (0 ≤ φ ≤ 2π) iseven more obvious. The numerical factor in (18.45) is chosen to make the Ylm an593

SPECIAL FUNCTIONSorthonormal set, i.e.∫ 1 ∫ 2π−10[Yml (θ, φ) ] ∗Ym ′l (θ, φ) dφ d(cos θ) =δ ′ ll ′δ mm ′. (18.46)In addition, the spherical harmonics form a complete set in that any reasonablefunction (i.e. one that is likely to be met in a physical situation) of θ and φ canbe expanded as a sum of such functions,f(θ, φ) =∞∑l∑l=0 m=−la lm Yl m (θ, φ), (18.47)the constants a lm being given by∫ 1 ∫ 2π[a lm = Yml (θ, φ) ] ∗f(θ, φ) dφ d(cos θ). (18.48)−10This is in exact analogy with a Fourier series and is a particular example of thegeneral property of Sturm–Liouville solutions.Aside from the orthonormality condition (18.46), the most important relationshipobeyed by the Ylm is the spherical harmonic addition theorem. This readsP l (cos γ) =4π2l +1l∑m=−lY ml (θ, φ)[Y ml (θ ′ ,φ ′ )] ∗ , (18.49)where (θ, φ)and(θ ′ ,φ ′ ) denote two different directions in our spherical polar coordinatesystem that are separated by an angle γ. In general, spherical trigonometry(or vector methods) shows that these angles obey the identitycos γ =cosθ cos θ ′ +sinθ sin θ ′ cos(φ − φ ′ ). (18.50)◮Prove the spherical harmonic addition theorem (18.49).For the sake of brevity, it will be useful to denote the directions (θ, φ) and(θ ′ ,φ ′ )byΩandΩ ′ , respectively. We will also denote the element of solid angle on the sphere bydΩ =dφ d(cos θ). We begin by deriving the form of the closure relationship obeyed by thespherical harmonics. Using (18.47) and (18.48), and reversing the order of the summationand integration, we may write∫f(Ω) = dΩ ′ f(Ω ′ ) ∑ Ylm∗ (Ω ′ )Yl m (Ω),4πlmwhere ∑ lmis a convenient shorthand for the double summation in (18.47). Thus we maywrite the closure relationship for the spherical harmonics as∑lmY ml (Ω)Y m∗l (Ω ′ )=δ(Ω − Ω ′ ), (18.51)where δ(Ω − Ω ′ ) is a Dirac delta function with the properties that δ(Ω − Ω ′ )=0ifΩ≠Ω ′and ∫ 4π δ(Ω) dΩ =1. 594

18.4 CHEBYSHEV FUNCTIONSSince δ(Ω − Ω ′ ) can depend only on the angle γ between the two directions Ω and Ω ′ ,we may also expand it in terms of a series of Legendre polynomials of the formδ(Ω − Ω ′ )= ∑ lb l P l (cos γ). (18.52)From (18.14), the coefficients in this expansion are given byb l ==2l +122l +14π∫ 1δ(Ω − Ω ′ )P l (cos γ) d(cos γ)−1∫ 2π ∫ 10−1δ(Ω − Ω ′ )P l (cos γ) d(cos γ) dψ,where, in the second equality, we have introduced an additional integration over anazimuthal angle ψ about the direction Ω ′ (and γ is now the polar angle measured fromΩ ′ to Ω). Since the rest of the integrand does not depend upon ψ, this is equivalentto multiplying it by 2π/2π. However, the resulting double integral now has the form ofa solid-angle integration over the whole sphere. Moreover, when Ω = Ω ′ , the angle γseparating the two directions is zero, and so cos γ = 1. Thus, we find2l +1b l =4π P 2l +1l(1) =4π ,and combining this expression with (18.51) and (18.52) gives∑Yl m (Ω)Ylm∗ (Ω ′ )= ∑ 2l +14π P l(cos γ). (18.53)lmlComparing this result with (18.49), we see that, to complete the proof of the additiontheorem, we now only need to show that the summations in l on either side of (18.53) canbe equated term by term.That such a procedure is valid may be shown by considering an arbitrary rigid rotationof the coordinate axes, thereby defining new spherical polar coordinates ¯Ω on the sphere.Any given spherical harmonic Yl m(¯Ω)in the new coordinates can be written as a linearcombination of the spherical harmonics Yl m (Ω) of the old coordinates, all having the samevalue of l. Thus,l∑Yl m (¯Ω) = D mm′ (Ω),m ′ =−ll Ylm′where the coefficients Dlmm′ depend on the rotation; note that in this expression Ω and ¯Ωrefer to the same direction, but expressed in the two different coordinate systems. If wechoose the polar axis of the new coordinate system to lie along the Ω ′ direction, then from(18.45), with m in that equation set equal to zero, we may write√4πl∑P l (cos γ) =2l +1 Y l 0 (¯Ω) = Cl 0m′ Ylm′ (Ω)m ′ =−lfor some set of coefficients Cl 0m that depend on Ω ′ . Thus, we see that the equality (18.53)does indeed hold term by term in l, thus proving the addition theorem (18.49). ◭18.4 Chebyshev functionsChebyshev’s equation has the form(1 − x 2 )y ′′ − xy ′ + ν 2 y =0, (18.54)595

SPECIAL FUNCTIONSand has three regular singular points, at x = −1, 1, ∞. By comparing it with(18.1), we see that the Chebyshev equation is very similar in form to Legendre’sequation. Despite this similarity, equation (18.54) does not occur very oftenin physical problems, though its solutions are of considerable importance innumerical analysis. The parameter ν is a given real number, but in nearly allpractical applications it takes an integer value. From here on we thus assumethat ν = n, wheren is a non-negative integer. As was the case for Legendre’sequation, in normal usage the variable x is the cosine of an angle, and so−1 ≤ x ≤ 1. Any solution of (18.54) is called a Chebyshev function.The point x = 0 is an ordinary point of (18.54), and so we expect to findtwo linearly independent solutions of the form y = ∑ ∞m=0 a mx m . One could findthe recurrence relations for the coefficients a m in a similar manner to that usedfor Legendre’s equation in section 18.1 (see exercise 16.15). For Chebyshev’sequation, however, it is easier and more illuminating to take a different approach.In particular, we note that, on making the substitution x =cosθ, and consequentlyd/dx =(−1/ sin θ) d/dθ, Chebyshev’s equation becomes (with ν = n)d 2 ydθ 2 + n2 y =0,which is the simple harmonic equation with solutions cos nθ and sin nθ. Thecorresponding linearly independent solutions of Chebyshev’s equation are thusgiven byT n (x) =cos(n cos −1 x) and V n (x) =sin(n cos −1 x). (18.55)It is straightforward to show that the T n (x) arepolynomials of order n, whereasthe V n (x) arenot polynomials◮Find explicit forms for the series expansions of T n (x) and V n (x).Writing x =cosθ, it is convenient first to form the complex superpositionT n (x)+iV n (x) =cosnθ + i sin nθ=(cosθ + i sin θ) n(= x + i √ ) n1 − x 2 for |x| ≤1.Then, on expanding out the last expression using the binomial theorem, we obtainT n (x) =x n − n C 2 x n−2 (1 − x 2 )+ n C 4 x n−4 (1 − x 2 ) 2 − ··· , (18.56)V n (x) = √ 1 − x 2 [n C 1 x n−1 − n C 3 x n−3 (1 − x 2 )+ n C 5 x n−5 (1 − x 2 ) 2 − ···] , (18.57)where n C r = n!/[r!(n−r)!] is a binomial coefficient. We thus see that T n (x) is a polynomialof order n, but V n (x) is not a polynomial. ◭It is conventional to define the additional functionsW n (x) =(1− x 2 ) −1/2 T n+1 (x) and U n (x) =(1− x 2 ) −1/2 V n+1 (x).596(18.58)

18.4 CHEBYSHEV FUNCTIONS1T 0T 2T 30.5T 1−1−0.50.51−0.5−1Figure 18.3The first four Chebyshev polynomials of the first kind.From (18.56) and (18.57), we see immediately that U n (x) isapolynomial of ordern, but that W n (x) isnot a polynomial. In practice, it is usual to work entirely interms of T n (x)andU n (x), which are known, respectively, as Chebyshev polynomialsof the first and second kind. In particular, we note that the general solution toChebyshev’s equation can be written in terms of these polynomials as⎧√⎨c 1 T n (x)+c 2 1 − x2 U n−1 (x) for n =1, 2, 3,...,y(x) =⎩c 1 + c 2 sin −1 x for n =0.The n = 0 solution could also be written as d 1 + c 2 cos −1 x with d 1 = c 1 + 1 2 πc 2.The first few Chebyshev polynomials of the first kind are easily constructedand are given byT 0 (x) =1,T 1 (x) =x,T 2 (x) =2x 2 − 1, T 3 (x) =4x 3 − 3x,T 4 (x) =8x 4 − 8x 2 +1,T 5 (x) =16x 5 − 20x 3 +5x.The functions T 0 (x), T 1 (x), T 2 (x) andT 3 (x) are plotted in figure 18.3. In general,the Chebyshev polynomials T n (x) satisfy T n (−x) =(−1) n T n (x), which is easilydeduced from (18.56). Similarly, it is straightforward to deduce the following597

SPECIAL FUNCTIONS4U 2U 32U 0U 1−1 −0.5 0.5 1−2−4Figure 18.4The first four Chebyshev polynomials of the second kind.special values:T n (1) = 1, T n (−1) = (−1) n , T 2n (0) = (−1) n , T 2n+1 (0) = 0.The first few Chebyshev polynomials of the second kind are also easily foundand readU 0 (x) =1,U 1 (x) =2x,U 2 (x) =4x 2 − 1, U 3 (x) =8x 3 − 4x,U 4 (x) =16x 4 − 12x 2 +1, U 5 (x) =32x 5 − 32x 3 +6x.The functions U 0 (x), U 1 (x), U 2 (x) and U 3 (x) are plotted in figure 18.4. TheChebyshev polynomials U n (x) also satisfy U n (−x) =(−1) n U n (x), which may bededuced from (18.57) and (18.58), and have the special values:U n (1) = n +1, U n (−1) = (−1) n (n +1), U 2n (0) = (−1) n , U 2n+1 (0) = 0.◮Show that the Chebyshev polynomials U n (x) satisfy the differential equation(1 − x 2 )U n ′′ (x) − 3xU n(x)+n(n ′ +2)U n (x) =0. (18.59)From (18.58), we have V n+1 =(1− x 2 ) 1/2 U n and these functions satisfy the Chebyshevequation (18.54) with ν = n +1,namely(1 − x 2 )V n+1 ′′ − xV n+1 ′ +(n +1) 2 V n+1 =0. (18.60)598

18.4 CHEBYSHEV FUNCTIONSEvaluating the first and second derivatives of V n+1 ,weobtainV ′ n+1 =(1− x 2 ) 1/2 U ′ n − x(1 − x 2 ) −1/2 U nV n+1 ′′ =(1− x 2 ) 1/2 U n ′′ − 2x(1 − x 2 ) −1/2 U n ′ − (1 − x 2 ) −1/2 U n − x 2 (1 − x 2 ) −3/2 U n .Substituting these expressions into (18.60) and dividing through by (1 − x 2 ) 1/2 , we find(1 − x 2 )U n ′′ − 3xU n ′ − U n +(n +1) 2 U n =0,which immediately simplifies to give the required result (18.59). ◭18.4.1 Properties of Chebyshev polynomialsThe Chebyshev polynomials T n (x) andU n (x) have their principal applicationsin numerical analysis. Their use in representing other functions over the range|x| < 1 plays an important role in numerical integration; Gauss–Chebyshevintegration is of particular value for the accurate evaluation of integrals whoseintegrands contain factors (1 − x 2 ) ±1/2 . It is therefore worthwhile outlining someof their main properties.Rodrigues’ formulaThe Chebyshev polynomials T n (x) andU n (x) may be expressed in terms of aRodrigues’ formula, in a similar way to that used for the Legendre polynomialsdiscussed in section 18.1.2. For the Chebyshev polynomials, we haveT n (x) = (−1)n√ π(1 − x 2 ) 1/2 d n2 n (n − 1 2 )! dx n (1 − x2 ) n− 1 2 ,(−1) n√ π(n +1) d nU n (x) =2 n+1 (n + 1 2 )!(1 − x2 ) 1/2 dx n (1 − x2 ) n+ 1 2 .These Rodrigues’ formulae may be proved in an analogous manner to that usedin section 18.1.2 when establishing the corresponding expression for the Legendrepolynomials.Mutual orthogonalityIn section 17.4, we noted that Chebyshev’s equation could be put into Sturm–Liouville form with p =(1− x 2 ) 1/2 , q =0,λ = n 2 and ρ =(1− x 2 ) −1/2 , and itsnatural interval is thus [−1, 1]. Since the Chebyshev polynomials of the first kind,T n (x), are solutions of the Chebyshev equation and are regular at the end-pointsx = ±1, they must be mutually orthogonal over this interval with respect to theweight function ρ =(1− x 2 ) −1/2 ,i.e.∫ 1−1T n (x)T m (x)(1 − x 2 ) −1/2 dx =0 ifn ≠ m. (18.61)599

SPECIAL FUNCTIONSThe normalisation, when m = n, is easily found by making the substitutionx =cosθ and using (18.55). We immediately obtain∫ 1−1T n (x)T n (x)(1 − x 2 ) −1/2 dx ={π for n =0,π/2 for n =1, 2, 3,... .(18.62)The orthogonality and normalisation conditions mean that any (reasonable)function f(x) can be expanded over the interval |x| < 1inaseriesoftheformf(x) = 1 2 a 0 +∞∑a n T n (x),n=1where the coefficients in the expansion are given by∫ 1a n = 2 f(x)T n (x)(1 − x 2 ) −1/2 dx.π −1For the Chebyshev polynomials of the second kind, U n (x), we see from (18.58)that (1 − x 2 ) 1/2 U n (x) =V n+1 (x) satisfies Chebyshev’s equation (18.54) with ν =n + 1. Thus, the orthogonality relation for the U n (x), obtained by replacing T i (x)by V i+1 (x) in equation (18.61), reads∫ 1−1U n (x)U m (x)(1 − x 2 ) 1/2 dx =0 ifn ≠ m.The corresponding normalisation condition, when n = m, can again be found bymaking the substitution x =cosθ, as illustrated in the following example.◮Show thatI ≡∫ 1−1U n (x)U n (x)(1 − x 2 ) 1/2 dx = π 2 .From (18.58), we see thatI =∫ 1−1which, on substituting x =cosθ, givesI =∫ 0πV n+1 (x)V n+1 (x)(1 − x 2 ) −1/2 dx,sin(n +1)θ sin(n +1)θ1sin θ (− sin θ) dθ = π 2 . ◭The above orthogonality and normalisation conditions allow one to expandany (reasonable) function in the interval |x| < 1 in a series of the formf(x) =∞∑a n U n (x),n=0600

18.4 CHEBYSHEV FUNCTIONSin which the coefficients a n are given bya n = 2 π∫ 1−1f(x)U n (x)(1 − x 2 ) 1/2 dx.Generating functionsThe generating functions for the Chebyshev polynomials of the first and secondkinds are given, respectively, by1 − xh∞G I (x, h) =1 − 2xh + h 2 = ∑T n (x)h n , (18.63)n=0∞1G II (x, h) =1 − 2xh + h 2 = ∑U n (x)h n . (18.64)These prescriptions may be proved in a manner similar to that used in section18.1.2 for the generating function of the Legendre polynomials. For theChebyshev polynomials, however, the generating functions are of less practicaluse, since most of the useful results can be obtained more easily by takingadvantage of the trigonometric forms (18.55), as illustrated below.Recurrence relationsThere exist many useful recurrence relationships for the Chebyshev polynomialsT n (x) andU n (x). They are most easily derived by setting x =cosθ and using(18.55) and (18.58) to writeT n (x) =T n (cos θ) =cosnθ, (18.65)sin(n +1)θU n (x) =U n (cos θ) = . (18.66)sin θOne may then use standard formulae for the trigonometric functions to derivea wide variety of recurrence relations. Of particular use are the trigonometricidentitiesn=0cos(n ± 1)θ =cosnθ cos θ ∓ sin nθ sin θ, (18.67)sin(n ± 1)θ =sinnθ cos θ ± cos nθ sin θ. (18.68)◮Show that the Chebyshev polynomials satisfy the recurrence relationsT n+1 (x) − 2xT n (x)+T n−1 (x) =0, (18.69)U n+1 (x) − 2xU n (x)+U n−1 (x) =0. (18.70)Adding the result (18.67) with the plus sign to the corresponding result with a minus signgivescos(n +1)θ +cos(n − 1)θ =2cosnθ cos θ.601

SPECIAL FUNCTIONSUsing (18.65) and setting x =cosθ immediately gives a rearrangement of the requiredresult (18.69). Similarly, adding the plus and minus cases of result (18.68) givessin(n +1)θ +sin(n − 1)θ =2sinnθ cos θ.Dividing through on both sides by sin θ and using (18.66) yields (18.70). ◭The recurrence relations (18.69) and (18.70) are extremely useful in the practicalcomputation of Chebyshev polynomials. For example, given the values of T 0 (x)and T 1 (x) at some point x, the result (18.69) may be used iteratively to obtainthe value of any T n (x) at that point; similarly, (18.70) may be used to calculatethe value of any U n (x) at some point x, given the values of U 0 (x) andU 1 (x) atthat point.Further recurrence relations satisfied by the Chebyshev polynomials areT n (x) =U n (x) − xU n−1 (x), (18.71)(1 − x 2 )U n (x) =xT n+1 (x) − T n+2 (x), (18.72)which establish useful relationships between the two sets of polynomials T n (x)and U n (x). The relation (18.71) follows immediately from (18.68), whereas (18.72)follows from (18.67), with n replaced by n + 1, on noting that sin 2 θ =1− x 2 .Additional useful results concerning the derivatives of Chebyshev polynomialsmay be obtained from (18.65) and (18.66), as illustrated in the following example.◮Show thatT n(x) ′ =nU n−1 (x),(1 − x 2 )U n(x) ′ =xU n (x) − (n +1)T n+1 (x).These results are most easily derived from the expressions (18.65) and (18.66) by notingthat d/dx =(−1/ sin θ) d/dθ. Thus,T n(x) ′ =− 1 d(cos nθ) n sin nθ= = nU n−1 (x).sin θ dθ sin θSimilarly, we findU n(x) ′ =− 1 [d sin(n +1)θsin θ dθ sin θwhich rearranges immediately to yield the stated result. ◭]sin(n +1)θ cos θ (n +1)cos(n +1)θ= −sin 3 θsin 2 θ= xU n(x)1 − x − (n +1)T n+1(x),2 1 − x 2Bessel’s equation has the form18.5 Bessel functionsx 2 y ′′ + xy ′ +(x 2 − ν 2 )y =0, (18.73)which has a regular singular point at x = 0 and an essential singularity at x = ∞.The parameter ν is a given number, which we may take as ≥ 0 with no loss of602

18.5 BESSEL FUNCTIONSgenerality. The equation arises from physical situations similar to those involvingLegendre’s equation but when cylindrical, rather than spherical, polar coordinatesare employed. The variable x in Bessel’s equation is usually a multiple of a radialdistance and therefore ranges from 0 to ∞.We shall seek solutions to Bessel’s equation in the form of infinite series. Writing(18.73) in the standard form used in chapter 16, we havey ′′ + 1 )x y′ +(1 − ν2x 2 y =0. (18.74)By inspection, x = 0 is a regular singular point; hence we try a solution of theform y = x σ ∑ ∞n=0 a nx n . Substituting this into (18.74) and multiplying the resultingequation by x 2−σ ,weobtain∞∑ [(σ + n)(σ + n − 1) + (σ + n) − ν2 ] ∞∑a n x n + a n x n+2 =0,n=0which simplifies ton=0n=0∞∑ [(σ + n) 2 − ν 2] ∞∑a n x n + a n x n+2 =0.Considering the coefficient of x 0 , we obtain the indicial equationσ 2 − ν 2 =0,and so σ = ±ν. For coefficients of higher powers of x we find[(σ +1) 2 − ν 2] a 1 =0, (18.75)[(σ + n) 2 − ν 2] a n + a n−2 =0 forn ≥ 2. (18.76)n=0Substituting σ = ±ν into (18.75) and (18.76), we obtain the recurrence relations(1 ± 2ν)a 1 =0, (18.77)n(n ± 2ν)a n + a n−2 =0 forn ≥ 2. (18.78)We consider now the form of the general solution to Bessel’s equation (18.73) fortwo cases: the case for which ν is not an integer and that for which it is (includingzero).18.5.1 Bessel functions for non-integer νIf ν is a non-integer then, in general, the two roots of the indicial equation,σ 1 = ν and σ 2 = −ν, will not differ by an integer, and we may obtain two linearlyindependent solutions in the form of Frobenius series. Special considerations doarise, however, when ν = m/2 form =1, 3, 5,...,andσ 1 − σ 2 =2ν = m is an(odd positive) integer. When this happens, we may always obtain a solution in603

SPECIAL FUNCTIONSthe form of a Frobenius series corresponding to the larger root, σ 1 = ν = m/2,as described above. However, for the smaller root, σ 2 = −ν = −m/2, we mustdetermine whether a second Frobenius series solution is possible by examiningthe recurrence relation (18.78), which readsn(n − m)a n + a n−2 =0 forn ≥ 2.Since m is an odd positive integer in this case, we can use this recurrence relation(starting with a 0 ≠ 0) to calculate a 2 ,a 4 ,a 6 ,... in the knowledge that all theseterms will remain finite. It is possible in this case, therefore, to find a secondsolution in the form of a Frobenius series, one that corresponds to the smallerroot σ 2 .Thus, in general, for non-integer ν we have from (18.77) and (18.78)1a n = −n(n ± 2ν) a n−2 for n =2, 4, 6,...,= 0 for n =1, 3, 5,....Setting a 0 = 1 in each case, we obtain the two solutionsy ±ν (x) =x[1 ±ν x 2−2(2 ± 2ν) + x 4···]2 × 4(2 ± 2ν)(4 ± 2ν) − .It is customary, however, to set1a 0 =2 ±ν Γ(1 ± ν) ,where Γ(x) isthegamma function, described in subsection 18.12.1; it may beregarded as the generalisation of the factorial function to non-integer and/ornegative arguments. § The two solutions of (18.73) are then written as J ν (x) andJ −ν (x), where1 xJ ν (x) =Γ(ν +1)(2∞∑=n=0) ν[1 − 1(−1) nn!Γ(ν + n +1)( x2x) 2 1 1( x) 4+ − ···]ν +1(2 (ν +1)(ν +2) 2! 2) ν+2n; (18.79)replacing ν by −ν gives J −ν (x). The functions J ν (x) andJ −ν (x) are called Besselfunctions of the first kind, of order ν. Since the first term of each series is afinite non-zero multiple of x ν and x −ν , respectively, if ν is not an integer thenJ ν (x) andJ −ν (x) are linearly independent. This may be confirmed by calculatingthe Wronskian of these two functions. Therefore, for non-integer ν the generalsolution of Bessel’s equation (18.73) is given byy(x) =c 1 J ν (x)+c 2 J −ν (x). (18.80)§ In particular, Γ(n +1)=n! forn =0, 1, 2,..., and Γ(n) is infinite if n is any integer ≤ 0.604

18.5 BESSEL FUNCTIONSWe note that Bessel functions of half-integer order are expressible in closed formin terms of trigonometric functions, as illustrated in the following example.◮Find the general solution ofx 2 y ′′ + xy ′ +(x 2 − 1 )y =0.4This is Bessel’s equation with ν =1/2, so from (18.80) the general solution is simplyy(x) =c 1 J 1/2 (x)+c 2 J −1/2 (x).However, Bessel functions of half-integral order can be expressed in terms of trigonometricfunctions. To show this, we note from (18.79) thatJ ±1/2 (x) =x ±1/2∞∑n=0(−1) n x 2n2 2n±1/2 n!Γ(1 + n ± 1 2 ) .Using the fact that Γ(x +1)=xΓ(x) andΓ( 1 2 )=√ π,wefindthat,forν =1/2,J 1/2 (x) = ( 1 2 x)1/2Γ( 3 2 ) − ( 1 2 x)5/21!Γ( 5 2 ) + ( 1 2 x)9/22!Γ( 7 2 ) − ···= ( 1 2 x)1/2( 1 )√ π − ( 1 2 x)5/21!( 3 )( 1 )√ π + ( 1 2 x)9/22!( 5 )( 3 )( 1 )√ π − ···2 2 2 2 2 2= ( 1 2 x)1/2( 1 2 )√ πwhereas for ν = −1/2 weobtain(1 − x23! + x45! − ···)J −1/2 (x) = ( 1 2 x)−1/2Γ( 1 2 ) − ( 1 2 x)3/21!Γ( 3 2 ) + ( 1 2 x)7/22!Γ( 5 2 ) − ···= ( 1 (···)2 x)−1/2√ 1 − x2π 2! + x44! − =Therefore the general solution we require isy(x) =c 1 J 1/2 (x)+c 2 J −1/2 (x) =c 1√2πx sin x + c 2= ( √ 12 x)1/2 sin x 2( 1 )√ π x = sin x,πx2√2cos x.πx√2cos x. ◭πx18.5.2 Bessel functions for integer νThe definition of the Bessel function J ν (x) given in (18.79) is, of course, valid forall values of ν, but, as we shall see, in the case of integer ν the general solution ofBessel’s equation cannot be written in the form (18.80). Firstly, let us consider thecase ν = 0, so that the two solutions to the indicial equation are equal, and weclearly obtain only one solution in the form of a Frobenius series. From (18.79),605

SPECIAL FUNCTIONS1.51J 0J 1J 20.52 4 6 8 10 x−0.5Figure 18.5The first three integer-order Bessel functions of the first kind.this is given byJ 0 (x) =∞∑n=0(−1) n x 2n2 2n n!Γ(1 + n)=1− x22 2 + x42 2 4 2 − x62 2 4 2 6 2 + ··· .In general, however, if ν is a positive integer then the solutions of the indicialequation differ by an integer. For the larger root, σ 1 = ν, we may find a solutionJ ν (x), for ν =1, 2, 3,..., in the form of the Frobenius series given by (18.79).Graphs of J 0 (x), J 1 (x) andJ 2 (x) are plotted in figure 18.5 for real x. Forthesmaller root, σ 2 = −ν, however, the recurrence relation (18.78) becomesn(n − m)a n + a n−2 =0 forn ≥ 2,where m =2ν is now an even positive integer, i.e. m =2, 4, 6,... . Starting witha 0 ≠ 0 we may then calculate a 2 ,a 4 ,a 6 ,..., but we see that when n = m thecoefficient a n is formally infinite, and the method fails to produce a secondsolution in the form of a Frobenius series.In fact, by replacing ν by −ν in the definition of J ν (x) given in (18.79), it canbe shown that, for integer ν,J −ν (x) =(−1) ν J ν (x),606

18.5 BESSEL FUNCTIONSand hence that J ν (x) andJ −ν (x) are linearly dependent. So, in this case, we cannotwrite the general solution to Bessel’s equation in the form (18.80). One thereforedefines the functionY ν (x) = J ν(x)cosνπ − J −ν (x), (18.81)sin νπwhich is called a Bessel function of the second kind of order ν (or, occasionally,a Weber or Neumann function). As Bessel’s equation is linear, Y ν (x) is clearly asolution, since it is just the weighted sum of Bessel functions of the first kind.Furthermore, for non-integer ν it is clear that Y ν (x) is linearly independent ofJ ν (x). It may also be shown that the Wronskian of J ν (x) andY ν (x) is non-zerofor all values of ν. Hence J ν (x) andY ν (x) always constitute a pair of independentsolutions.◮If n is an integer, show that Y n+1/2 (x) =(−1) n+1 J −n−1/2 (x).From (18.81), we haveY n+1/2 (x) = J n+1/2(x)cos(n + 1 )π − J 2 −n−1/2(x)sin(n + 1 )π .2If n is an integer, cos(n + 1 )π = 0 and sin(n + 1 )π 2 2 =(−1)n , and so we immediately obtainY n+1/2 (x) =(−1) n+1 J −n−1/2 (x), as required. ◭The expression (18.81) becomes an indeterminate form 0/0 when ν is aninteger, however. This is so because for integer ν we have cos νπ =(−1) ν andJ −ν (x) =(−1) ν J ν (x). Nevertheless, this indeterminate form can be evaluated usingl’Hôpital’s rule (see chapter 4). Therefore, for integer ν, weset[ ]Jµ (x)cosµπ − J −µ (x)Y ν (x) = lim, (18.82)µ→ν sin µπwhich gives a linearly independent second solution for this case. Thus, we maywrite the general solution of Bessel’s equation, valid for all ν, asy(x) =c 1 J ν (x)+c 2 Y ν (x). (18.83)The functions Y 0 (x), Y 1 (x) andY 2 (x) are plotted in figure 18.6Finally, we note that, in some applications, it is convenient to work withcomplex linear combinations of Bessel functions of the first and second kindsgiven byH (1)ν (x) =J ν (x)+iY ν (x), H ν(2) (x) =J ν (x) − iY ν (x);these are called, respectively, Hankel functions of the first and second kind oforder ν.607

SPECIAL FUNCTIONS10.5Y 0Y 1 Y22 4 6 8 10 x−0.5−1Figure 18.6The first three integer-order Bessel functions of the second kind.18.5.3 Properties of Bessel functions J ν (x)In physical applications, we often require that the solution is regular at x =0,but, from its definition (18.81) or (18.82), it is clear that Y ν (x) is singular atthe origin, and so in such physical situations the coefficient c 2 in (18.83) mustbe set to zero; the solution is then simply some multiple of J ν (x). These Besselfunctions of the first kind have various useful properties that are worthy offurther discussion. Unless otherwise stated, the results presented in this sectionapply to Bessel functions J ν (x) of integer and non-integer order.Mutual orthogonalityIn section 17.4, we noted that Bessel’s equation (18.73) could be put into conventionalSturm–Liouville form with p = x, q = −ν 2 /x, λ = α 2 and ρ = x,provided αx is the argument of y. From the form of p, we see that there is nonatural interval over which one would expect the solutions of Bessel’s equationcorresponding to different eigenvalues λ (but fixed ν) to be automatically orthogonal.Nevertheless, provided the Bessel functions satisfied appropriate boundaryconditions, we would expect them to obey an orthogonality relationship oversome interval [a, b] oftheform∫ baxJ ν (αx)J ν (βx) dx =0 forα ≠ β. (18.84)608

18.5 BESSEL FUNCTIONSTo determine the required boundary conditions for this result to hold, let usconsider the functions f(x) =J ν (αx) andg(x) =J ν (βx), which, as will be provedbelow, respectively satisfy the equationsx 2 f ′′ + xf ′ +(α 2 x 2 − ν 2 )f =0, (18.85)x 2 g ′′ + xg ′ +(β 2 x 2 − ν 2 )g =0. (18.86)◮Show that f(x) =J ν (αx) satisfies (18.85).If f(x) =J ν (αx) and we write w = αx, thendfdx = α dJ ν(w)d 2 fanddwdx = d2 J ν (w)2 α2 .dw 2When these expressions are substituted into (18.85), its LHS becomesx 2 α 2 d2 J ν (w)dw 2+ xα dJ ν(w)dw+(α2 x 2 − ν 2 )J ν (w)= w 2 d2 J ν (w)dw 2+ w dJ ν(w)dw+(w2 − ν 2 )J ν (w).But, from Bessel’s equation itself, this final expression is equal to zero, thus verifying thatf(x) does satisfy (18.85). ◭Now multiplying (18.86) by f(x) and (18.85) by g(x) and subtracting them givesddx [x(fg′ − gf ′ )] = (α 2 − β 2 )xfg, (18.87)wherewehaveusedthefactthatddx [x(fg′ − gf ′ )] = x(fg ′′ − gf ′′ )+(fg ′ − gf ′ ).By integrating (18.87) over any given range x = a to x = b, weobtain∫ b1[bxf(x)g(x) dx =aα 2 − β 2 xf(x)g ′ (x) − xg(x)f (x)] ′ ,awhich, on setting f(x) =J ν (αx) andg(x) =J ν (βx), becomes∫ b1[] bxJ ν (αx)J ν (βx) dx =aα 2 − β 2 βxJ ν (αx)J ν(βx) ′ − αxJ ν (βx)J ν(αx)′ . a(18.88)If α ≠ β, and the interval [a, b] is such that the expression on the RHS of (18.88)equals zero, then we obtain the orthogonality condition (18.84). This happens, forexample, if J ν (αx) andJ ν (βx) vanish at x = a and x = b, orifJ ν(αx) ′ andJ ν(βx)′vanish at x = a and x = b, or for many more general conditions. It should benoted that the boundary term is automatically zero at the point x =0,asonemight expect from the fact that the Sturm–Liouville form of Bessel’s equationhas p(x) =x.If α = β, the RHS of (18.88) takes the indeterminant form 0/0. This may be609

SPECIAL FUNCTIONSevaluated using l’Hôpital’s rule, or alternatively we may calculate the relevantintegral directly.◮Evaluate the integral∫ baJ 2 ν (αx)xdx.Ignoring the integration limits for the moment,∫J 2 ν (αx)xdx= 1 α 2 ∫J 2 ν (u)u du,where u = αx. Integrating by parts yields∫∫I = Jν 2 (u)udu= 1 2 u2 Jν 2 (u) −J ν (u)J ′ ν(u)u 2 du.Now Bessel’s equation (18.73) can be rearranged asu 2 J ν (u) =ν 2 J ν (u) − uJ ν(u) ′ − u 2 J ν ′′ (u),which, on substitution into the expression for I, gives∫I = 1 2 u2 Jν 2 (u) − J ν(u)[ν ′ 2 J ν (u) − uJ ν(u) ′ − u 2 J ν ′′ (u)] du= 1 2 u2 Jν 2 (u) − 1 2 ν2 Jν 2 (u)+ 1 2 u2 [J ν(u)] ′ 2 + c.Since u = αx, the required integral is given by∫ bJν 2 (αx)xdx= 1 ) ] b [(x 2 − ν2J 2a2 α 2 ν (αx)+x 2 [J ν(αx)] ′ 2 , (18.89)awhich gives the normalisation condition for Bessel functions of the first kind. ◭Since the Bessel functions J ν (x) possess the orthogonality property (18.88), wemay expand any reasonable function f(x), i.e. one obeying the Dirichlet conditionsdiscussed in chapter 12, in the interval 0 ≤ x ≤ b as a sum of Bessel functions ofa given (non-negative) order ν,f(x) =∞∑c n J ν (α n x), (18.90)n=0provided that the α n are chosen such that J ν (α n b) = 0. The coefficients c n are thengiven by∫ b2c n =b 2 Jν+1 2 (α f(x)J ν (α n x)xdx. (18.91)nb) 0Theintervalistakentobe0≤ x ≤ b, as then one need only ensure that theappropriate boundary condition is satisfied at x = b, since the boundary conditionat x = 0 is met automatically.610

18.5 BESSEL FUNCTIONS◮Prove the expression (18.91).If we multiply (18.90) by xJ ν (α m x) and integrate from x =0tox = b then we obtain∫ b∞∑∫ bxJ ν (α m x)f(x) dx = c n xJ ν (α m x)J ν (α n x) dx0n=00∫ b= c m Jν 2 (α m x)xdx0= 1 c 2 mb 2 J ′2 ν(α m b)= 1 c 2 mb 2 Jν+1(α 2 m b),where in the last two lines we have used (18.88) with α m = α ≠ β = α n , (18.89), the factthat J ν (α m b) = 0 and (18.95), which is proved below. ◭Recurrence relationsThe recurrence relations enjoyed by Bessel functions of the first kind, J ν (x), canbe derived directly from the power series definition (18.79).◮Prove the recurrence relationddx [xν J ν (x)] = x ν J ν−1 (x). (18.92)From the power series definition (18.79) of J ν (x) weobtainddx [xν J ν (x)] = d ∞∑ (−1) n x 2ν+2ndx 2 ν+2n n!Γ(ν + n +1)=∞∑n=0n=0= x ν ∞∑(−1) n x 2ν+2n−12 ν+2n−1 n!Γ(ν + n)n=0(−1) n x (ν−1)+2n2 (ν−1)+2n n!Γ((ν − 1) + n +1) = xν J ν−1 (x). ◭It may similarly be shown thatddx [x−ν J ν (x)] = −x −ν J ν+1 (x). (18.93)From (18.92) and (18.93) the remaining recurrence relations may be derived.Expanding out the derivative on the LHS of (18.92) and dividing through byx ν−1 , we obtain the relationxJ ν ′ (x)+νJ ν(x) =xJ ν−1 (x). (18.94)Similarly, by expanding out the derivative on the LHS of (18.93), and multiplyingthrough by x ν+1 , we findxJ ν(x) ′ − νJ ν (x) =−xJ ν+1 (x). (18.95)Adding (18.94) and (18.95) and dividing through by x givesJ ν−1 (x) − J ν+1 (x) =2J ν(x). ′ (18.96)611

SPECIAL FUNCTIONSFinally, subtracting (18.95) from (18.94) and dividing by x givesJ ν−1 (x)+J ν+1 (x) = 2ν x J ν(x). (18.97)◮Given that J 1/2 (x) =(2/πx) 1/2 sin x and that J −1/2 (x) =(2/πx) 1/2 cos x, expressJ 3/2 (x)and J −3/2 (x) in terms of trigonometric functions.From (18.95) we haveJ 3/2 (x) = 12x J 1/2(x) − J ′ 1/2 (x)= 1 ( ) 1/2 ( ) 1/2 22sin x − cos x + 12x πxπx2x( ) 1/2 ( )2 1=πx x sin x − cos x .Similarly, from (18.94) we haveJ −3/2 (x) =− 12x J −1/2(x)+J ′ −1/2 (x)= − 1 ( ) 1/2 ( ) 1/2 22cos x − sin x − 12x πxπx2x( ) 1/2 2=(− 1 )πx x cos x − sin x .( ) 1/2 2sin xπx( ) 1/2 2cos xπxWe see that, by repeated use of these recurrence relations, all Bessel functions J ν (x) ofhalfintegerorder may be expressed in terms of trigonometric functions. From their definition(18.81), Bessel functions of the second kind, Y ν (x), of half-integer order can be similarlyexpressed. ◭Finally, we note that the relations (18.92) and (18.93) may be rewritten inintegral form as∫x ν J ν−1 (x) dx = x ν J ν (x),∫x −ν J ν+1 (x) dx = −x −ν J ν (x).If ν is an integer, the recurrence relations of this section may be proved usingthe generating function for Bessel functions discussed below. It may be shownthat Bessel functions of the second kind, Y ν (x), also satisfy the recurrence relationsderived above.Generating functionThe Bessel functions J ν (x), where ν = n is an integer, can be described by agenerating function in a way similar to that discussed for Legendre polynomials612

18.5 BESSEL FUNCTIONSin subsection 18.1.2. The generating function for Bessel functions of integer orderis given by[ ( xG(x, h) =exp h − 1 )] ∞∑= J n (x)h n . (18.98)2 hn=−∞By expanding the exponential as a power series, it is straightfoward to verify thatthe functions J n (x) defined by (18.98) are indeed Bessel functions of the first kind,as given by (18.79).The generating function (18.98) is useful for finding, for Bessel functions ofinteger order, properties that can often be extended to the non-integer case. Inparticular, the Bessel function recurrence relations may be derived.◮Use the generating function to prove, for integer ν, the recurrence relation (18.97), i.e.J ν−1 (x)+J ν+1 (x) = 2ν x J ν(x).Differentiating G(x, h) with respect to h we obtain(1+ 1 )G(x, h) =h 2∂G(x, h)∂h= x 2which can be written using (18.98) again as(1+ 1 )∑ ∞h 2x2n=−∞J n (x)h n =∞∑nJ n (x)h n−1 ,n=−∞∞∑nJ n (x)h n−1 .n=−∞Equating coefficients of h n we obtainx2 [J n(x)+J n+2 (x)] = (n +1)J n+1 (x),which, on replacing n by ν − 1, gives the required recurrence relation. ◭Integral representationsThe generating function (18.98) is also useful for deriving integral representationsof Bessel functions of integer order.◮Show that for integer n the Bessel function J n (x) is given byJ n (x) = 1 π∫ π0cos(nθ − x sin θ) dθ. (18.99)By expanding out the cosine term in the integrand in (18.99) we obtain the integral∫ πI = 1 [cos(x sin θ)cosnθ + sin(x sin θ)sinnθ] dθ. (18.100)π 0Now, we may express cos(x sin θ) and sin(x sin θ) in terms of Bessel functions by settingh =expiθ in (18.98) to give[ x]∞∑exp2 (exp iθ − exp(−iθ)) =exp(ix sin θ) = J m (x)expimθ.613m=−∞

SPECIAL FUNCTIONSUsing de Moivre’s theorem, exp iθ =cosθ + i sin θ, we then obtain∞∑exp (ix sin θ) =cos(x sin θ)+i sin(x sin θ) = J m (x)(cos mθ + i sin mθ).m=−∞Equating the real and imaginary parts of this expression gives∞∑cos(x sin θ) = J m (x)cosmθ,sin(x sin θ) =m=−∞∞∑J m (x)sinmθ.m=−∞Substituting these expressions into (18.100) then yieldsI = 1 ∞∑∫ π[J m (x)cosmθ cos nθ + J m (x)sinmθ sin nθ] dθ.πm=−∞ 0However, using the orthogonality of the trigonometric functions [ see equations (12.1)–(12.3) ], we obtainI = 1 ππ 2 [J n(x)+J n (x)] = J n (x),which proves the integral representation (18.99). ◭Finally, we mention the special case of the integral representation (18.99) forn =0:J 0 (x) = 1 ∫ πcos(x sin θ) dθ = 1 ∫ 2πcos(x sin θ) dθ,π 02π 0since cos(x sin θ) repeats itself in the range θ = π to θ =2π. However, sin(x sin θ)changes sign in this range and so∫ 2π1sin(x sin θ) dθ =0.2π 0Using de Moivre’s theorem, we can therefore writeJ 0 (x) = 12π∫ 2π0exp(ix sin θ) dθ = 12π∫ 2π0exp(ix cos θ) dθ.There are in fact many other integral representations of Bessel functions; theycan be derived from those given.18.6 Spherical Bessel functionsWhen obtaining solutions of Helmholtz’ equation (∇ 2 + k 2 )u = 0 in sphericalpolar coordinates (see section 21.3.2), one finds that, for solutions that are finiteon the polar axis, the radial part R(r) of the solution must satisfy the equationr 2 R ′′ +2rR ′ +[k 2 r 2 − l(l +1)]R =0, (18.101)614

18.6 SPHERICAL BESSEL FUNCTIONSwhere l is an integer. This equation looks very much like Bessel’s equation andcan in fact be reduced to it by writing R(r) =r −1/2 S(r), in which case S(r) thensatisfies[r 2 S ′′ + rS ′ + k 2 r 2 − ( ) ]l + 1 22S =0.On making the change of variable x = kr and letting y(x) =S(kr), we obtainx 2 y ′′ + xy ′ +[x 2 − (l + 1 2 )2 ]y =0,where the primes now denote d/dx. This is Bessel’s equation of order l + 1 2and has as its solutions y(x) =J l+1/2 (x) andY l+1/2 (x). The general solution of(18.101) can therefore be writtenR(r) =r −1/2 [c 1 J l+1/2 (kr)+c 2 Y l+1/2 (kr)],where c 1 and c 2 are constants that may be determined from the boundaryconditions on the solution. In particular, for solutions that are finite at the originwe require c 2 =0.The functions x −1/2 J l+1/2 (x) andx −1/2 Y l+1/2 (x), when suitably normalised, arecalled spherical Bessel functions of the first and second kind, respectively, and aredenoted as follows:j l (x) =√ π2x J l+1/2(x), (18.102)n l (x) =√ π2x Y l+1/2(x). (18.103)For integer l, we also note that Y l+1/2 (x) =(−1) l+1 J −l−1/2 (x), as discussed insection 18.5.2. Moreover, in section 18.5.1, we noted that Bessel functions of thefirst kind, J ν (x), of half-integer order are expressible in closed form in terms oftrigonometric functions. Thus, all spherical Bessel functions of both the first andsecond kinds may be expressed in such a form. In particular, using the results ofthe worked example in section 18.5.1, we find thatj 0 (x) = sin xx , (18.104)n 0 (x) =− cos xx . (18.105)Expressions for higher-order spherical Bessel functions are most easily obtainedby using the recurrence relations for Bessel functions.615

SPECIAL FUNCTIONS◮Show that the lth spherical Bessel function is given by( ) l 1f l (x) =(−1) l x l df 0 (x), (18.106)x dxwhere f l (x) denotes either j l (x) or n l (x).The recurrence relation (18.93) for Bessel functions of the first kind readsJ ν+1 (x) =−x ν d [x −ν J ν (x) ] .dxThus, on setting ν = l + 1 and rearranging, we find2x −1/2 J l+3/2 (x) =−x l d [ x −1/2 ]J l+1/2,dx x lwhich on using (18.102) yields the recurrence relationj l+1 (x) =−x l ddx [x−l j l (x)].We now change l +1→ l and iterate this result:j l (x) =−x l−1 d= −x l−1 ddx=(−1) 2 xlx= ···dx [ x−l+1 j l−1 (x)]{x −l+1 (−1)x l−2 ddxddx{ 1xddx[x −l+2 j l−2 (x) ]}[x −l+2 j l−2 (x) ]}( ) l 1=(−1) l x l dj 0 (x).x dxThis is the expression for j l (x) as given in (18.106). One may prove the result (18.106) forn l (x) in an analogous manner by setting ν = l − 1 in the recurrence relation (18.92) for2Bessel functions of the first kind and using the relationship Y l+1/2 (x) =(−1) l+1 J −l−1/2 (x). ◭Using result (18.106) and the expressions (18.104) and (18.105), one quicklyfinds, for example,j 1 (x) = sin xx 2 − cos x3x , j 2(x) =(x 3 − 1 )sin x − 3cosxxx 2 ,n 1 (x) =− cos xx 2− sin xx , n 2(x) =−( 3x 3 − 1 x)cos x − 3sinxx 2 .Finally, we note that the orthogonality properties of the spherical Bessel functionsfollow directly from the orthogonality condition (18.88) for Bessel functions ofthe first kind.18.7 Laguerre functionsLaguerre’s equation has the formxy ′′ +(1− x)y ′ + νy = 0; (18.107)616

18.7 LAGUERRE FUNCTIONSit has a regular singularity at x = 0 and an essential singularity at x = ∞. Theparameter ν is a given real number, although it nearly always takes an integervalue in physical applications. The Laguerre equation appears in the descriptionof the wavefunction of the hydrogen atom. Any solution of (18.107) is called aLaguerre function.Since the point x = 0 is a regular singularity, we may find at least one solutionin the form of a Frobenius series (see section 16.3):y(x) =∞∑a m x m+σ . (18.108)m=0Substituting this series into (18.107) and dividing through by x σ−1 ,weobtain∞∑[(m + σ)(m + σ − 1) + (1 − x)(m + σ)+νx] a m x m =0.m=0(18.109)Setting x = 0, so that only the m = 0 term remains, we obtain the indicialequation σ 2 = 0, which trivially has σ = 0 as its repeated root. Thus, Laguerre’sequation has only one solution of the form (18.108), and it, in fact, reduces toa simple power series. Substituting σ = 0 into (18.109) and demanding that thecoefficient of x m+1 vanishes, we obtain the recurrence relationa m+1 =m − ν(m +1) 2 a m.As mentioned above, in nearly all physical applications, the parameter ν takesinteger values. Therefore, if ν = n, wheren is a non-negative integer, we see thata n+1 = a n+2 = ···= 0, and so our solution to Laguerre’s equation is a polynomialof order n. It is conventional to choose a 0 = 1, so that the solution is given byL n (x) =[x (−1)n n − n2n! 1! xn−1 + n2 (n − 1) 2]x n−2 − ···+(−1) n n! (18.110)2!n∑= (−1) m n!(m!) 2 (n − m)! xm , (18.111)m=0where L n (x) is called the nth Laguerre polynomial. We note in particular thatL n (0) = 1. The first few Laguerre polynomials are given byL 0 (x) =1, 3!L 3 (x) =−x 3 +9x 2 − 18x +6,L 1 (x) =−x +1, 4!L 4 (x) =x 4 − 16x 3 +72x 2 − 96x +24,2!L 2 (x) =x 2 − 4x +2, 5!L 5 (x) =−x 5 +25x 4 − 200x 3 + 600x 2 − 600x + 120.The functions L 0 (x), L 1 (x), L 2 (x) andL 3 (x) are plotted in figure 18.7.617

SPECIAL FUNCTIONS105L 2L 3L 0L 11 2 3 45 6 7x−5−10Figure 18.7The first four Laguerre polynomials.18.7.1 Properties of Laguerre polynomialsThe Laguerre polynomials and functions derived from them are important inthe analysis of the quantum mechanical behaviour of some physical systems. Wetherefore briefly outline their useful properties in this section.Rodrigues’ formulaThe Laguerre polynomials can be expressed in terms of a Rodrigues’ formulagiven byd nL n (x) = ex (x nn! dx n e −x) , (18.112)which may be proved straightforwardly by calculating the nth derivative explicitlyusing Leibnitz’ theorem and comparing the result with (18.111). This is illustratedin the following example.618

18.7 LAGUERRE FUNCTIONS◮Prove that the expression (18.112) yields the nth Laguerre polynomial.Evaluating the nth derivative in (18.112) using Leibnitz’ theorem, we findn∑L n (x) = ex n d r x n d n−r e −xC rn! dx r dx n−rr=0n∑= ex n! n!n! r!(n − r)! (n − r)! xn−r (−1) n−r e −xr=0n∑= (−1) n−r n!r!(n − r)!(n − r)! xn−r .r=0Relabelling the summation using the index m = n − r, weobtainn∑L n (x) = (−1) m n!(m!) 2 (n − m)! xm ,m=0which is precisely the expression (18.111) for the nth Laguerre polynomial. ◭Mutual orthogonalityIn section 17.4, we noted that Laguerre’s equation could be put into Sturm–Liouville form with p = xe −x , q =0,λ = ν and ρ = e −x , and its natural intervalis thus [0, ∞]. Since the Laguerre polynomials L n (x) aresolutionsoftheequationand are regular at the end-points, they must be mutually orthogonal over thisinterval with respect to the weight function ρ = e −x ,i.e.∫ ∞0L n (x)L k (x)e −x dx =0 ifn ≠ k.This result may also be proved directly using the Rodrigues’ formula (18.112).Indeed, the normalisation, when k = n, is most easily found using this method.◮Show thatI ≡∫ ∞0L n (x)L n (x)e −x dx =1. (18.113)Using the Rodrigues’ formula (18.112), we may writeI = 1 ∫ ∞L n (x) dnn! dx n (xn e −x ) dx = (−1)nn!0∫ ∞0d n L ndx nxn e −x dx,where, in the second equality, we have integrated by parts n times and used the fact that theboundary terms all vanish. When d n L n /dx n is evaluated using (18.111), only the derivativeof the m = n term survives and that has the value [ (−1) n n! n!]/[(n!) 2 0!] = (−1) n . Thuswe haveI = 1 ∫ ∞x n e −x dx =1,n! 0where, in the second equality, we use the expression (18.153) defining the gamma function(see section 18.12). ◭619

SPECIAL FUNCTIONSThe above orthogonality and normalisation conditions allow us to expand any(reasonable) function in the interval 0 ≤ x

18.8 ASSOCIATED LAGUERRE FUNCTIONSwhich trivially rearranges to give the recurrence relation (18.115).To obtain the recurrence relation (18.116), we begin by differentiating the generatingfunction (18.114) with respect to x, which yields∂G∂x = − he−xh/(1−h)= ∑ L ′(1 − h)nh n ,2and thus we have−h ∑ L n h n =(1− h) ∑ L ′ nh n .Equating coefficients of h n on each side then gives−L n−1 = L ′ n − L ′ n−1,which immediately simplifies to give (18.116). ◭18.8 Associated Laguerre functionsThe associated Laguerre equation has the formxy ′′ +(m +1− x)y ′ + ny = 0; (18.118)it has a regular singularity at x = 0 and an essential singularity at x = ∞. Werestrict our attention to the situation in which the parameters n and m are bothnon-negative integers, as is the case in nearly all physical problems. The associatedLaguerre equation occurs most frequently in quantum-mechanical applications.Any solution of (18.118) is called an associated Laguerre function.Solutions of (18.118) for non-negative integers n and m are given by theassociated Laguerre polynomialsL m n (x) =(−1) m dmdx m L n+m(x), (18.119)where L n (x) are the ordinary Laguerre polynomials. §◮Show that the functions L m n (x) defined in (18.119) are solutions of (18.118).Since the Laguerre polynomials L n (x) are solutions of Laguerre’s equation (18.107), wehavexL ′′n+m +(1− x)L ′ n+m +(n + m)L n+m =0.Differentiating this equation m times using Leibnitz’ theorem and rearranging, we findxL (m+2)n+m+(m +1− x)L (m+1)n+m + nL (m)n+m =0.On multiplying through by (−1) m and setting L m n =(−1) m L (m)n+m, in accord with (18.119),we obtainx(L m n ) ′′ +(m +1− x)(L m n ) ′ + nL m n =0,which shows that the functions L m n are indeed solutions of (18.118). ◭§ Note that some authors define the associated Laguerre polynomials as L m n (x) =(d m /dx m )L n (x),which is thus related to our expression (18.119) by L m n (x) =(−1) m L m n+m (x).621

SPECIAL FUNCTIONSIn particular, we note that L 0 n(x) =L n (x). As discussed in the previous section,L n (x) is a polynomial of order n and so it follows that L m n (x) is also. The first fewassociated Laguerre polynomials are easily found using (18.119):L m 0 (x) =1,L m 1 (x) =−x + m +1,2!L m 2 (x) =x 2 − 2(m +2)x +(m +1)(m +2),3!L m 3 (x) =−x 3 +3(m +3)x 2 − 3(m +2)(m +3)x +(m +1)(m +2)(m +3).Indeed, in the general case, one may show straightforwardly, from the definition(18.119) and the expression (18.111) for the ordinary Laguerre polynomials, thatL m n (x) =n∑(−1) k (n + m)!k!(n − k)!(k + m)! xk . (18.120)k=018.8.1 Properties of associated Laguerre polynomialsThe properties of the associated Laguerre polynomials follow directly from thoseof the ordinary Laguerre polynomials through the definition (18.119). We shalltherefore only briefly outline the most useful results here.Rodrigues’ formulaA Rodrigues’ formula for the associated Laguerre polynomials is given byL m n (x) = ex x −m d nn! dx n (xn+m e −x ). (18.121)It can be proved by evaluating the nth derivative using Leibnitz’ theorem (seeexercise 18.7).Mutual orthogonalityIn section 17.4, we noted that the associated Laguerre equation could be transformedinto a Sturm–Liouville one with p = x m+1 e −x , q =0,λ = n and ρ = x m e −x ,and its natural interval is thus [0, ∞]. Since the associated Laguerre polynomialsL m n (x) are solutions of the equation and are regular at the end-points, thosewith the same m but differing values of the eigenvalue λ = n must be mutuallyorthogonal over this interval with respect to the weight function ρ = x m e −x ,i.e.∫ ∞0L m n (x)L m k (x)x m e −x dx =0 ifn ≠ k.This result may also be proved directly using the Rodrigues’ formula (18.121), asmay the normalisation condition when k = n.622

18.8 ASSOCIATED LAGUERRE FUNCTIONS◮Show thatI ≡∫ ∞0L m n (x)L m n (x)x m e −x dx =(n + m)!. (18.122)n!Using the Rodrigues’ formula (18.121), we may writeI = 1 ∫ ∞∫ ∞L m n (x) dnn! 0 dx n (xn+m e −x ) dx = (−1)n d n L m nx n+m e −x dx,n! 0 dx nwhere, in the second equality, we have integrated by parts n times and used the fact thatthe boundary terms all vanish. From (18.120) we see that d n L m n /dx n =(−1) n . Thus we have∫ ∞I = 1 x n+m e −x (n + m)!dx = ,n! 0n!where, in the second equality, we use the expression (18.153) defining the gamma function(see section 18.12). ◭The above orthogonality and normalisation conditions allow us to expand any(reasonable) function in the interval 0 ≤ x

SPECIAL FUNCTIONSwhere, in the second equality, we have expanded the RHS using the binomial theorem. Onequating coefficients of h n , we immediately obtainL m n (0) =(n + m)!. ◭n!m!Recurrence relationsThe various recurrence relations satisfied by the associated Laguerre polynomialsmay be derived by differentiating the generating function (18.123) with respect toeither or both of x and h, or by differentiating with respect to x the recurrencerelations obeyed by the ordinary Laguerre polynomials, discussed in section 18.7.1.Of the many recurrence relations satisfied by the associated Laguerre polynomials,two of the most useful are as follows:(n +1)L m n+1(x) =(2n + m +1− x)L m n (x) − (n + m)L m n−1(x), (18.124)x(L m n ) ′ (x) =nL m n (x) − (n + m)L m n−1(x). (18.125)For proofs of these relations the reader is referred to exercise 18.7.Hermite’s equation has the form18.9 Hermite functionsy ′′ − 2xy ′ +2νy =0, (18.126)and has an essential singularity at x = ∞. The parameter ν is a given realnumber, although it nearly always takes an integer value in physical applications.The Hermite equation appears in the description of the wavefunction of theharmonic oscillator. Any solution of (18.126) is called a Hermite function.Since x = 0 is an ordinary point of the equation, we may find two linearlyindependent solutions in the form of a power series (see section 16.2):∞∑y(x) = a m x m . (18.127)m=0Substituting this series into (18.107) yields∞∑[(m +1)(m +2)a m+2 +2(ν − m)a m ] x m =0.m=0Demanding that the coefficient of each power of x vanishes, we obtain therecurrence relation2(ν − m)a m+2 = −(m +1)(m +2) a m.As mentioned above, in nearly all physical applications, the parameter ν takesinteger values. Therefore, if ν = n, wheren is a non-negative integer, we see that624

18.9 HERMITE FUNCTIONS105−1.5H 0H 1−1−0.5H 2H 3x0.5 1 1.5−5−10Figure 18.8The first four Hermite polynomials.a n+2 = a n+4 = ···= 0, and so one solution of Hermite’s equation is a polynomialof order n. For even n, it is conventional to choose a 0 =(−1) n/2 n!/(n/2)!, whereasfor odd n one takes a 1 =(−1) (n−1)/2 2n!/[ 1 2(n − 1)]!. These choices allow a generalsolution to be written asH n (x) =(2x) n − n(n − 1)(2x) n−1 +[n/2]∑=m=0n(n − 1)(n − 2)(n − 3)(2x) n−4 − ···(18.128)2!(−1) m n!m!(n − 2m)! (2x)n−2m , (18.129)where H n (x) is called the nth Hermite polynomial and the notation [n/2] denotesthe integer part of n/2. We note in particular that H n (−x) =(−1) n H n (x). Thefirst few Hermite polynomials are given byH 0 (x) =1, H 3 (x) =8x 2 − 12x,H 1 (x) =2x, H 4 (x) =16x 4 − 48x 2 +12,H 2 (x) =4x 2 − 2, H 5 (x) =32x 5 − 160x 3 + 120x.The functions H 0 (x), H 1 (x), H 2 (x) andH 3 (x) are plotted in figure 18.8.625

SPECIAL FUNCTIONS18.9.1 Properties of Hermite polynomialsThe Hermite polynomials and functions derived from them are important in theanalysis of the quantum mechanical behaviour of some physical systems. Wetherefore briefly outline their useful properties in this section.Rodrigues’ formulaThe Rodrigues’ formula for the Hermite polynomials is given byH n (x) =(−1) n e x2dx n (e−x2 ). (18.130)This can be proved using Leibnitz’ theorem.◮Prove the Rodrigues’ formula (18.130) for the Hermite polynomials.Letting u = e −x2 and differentiating with respect to x, we quickly find thatu ′ +2xu =0.Differentiating this equation n + 1 times using Leibnitz’ theorem then givesu (n+2) +2xu (n+1) +2(n +1)u (n) =0,which, on introducing the new variable v =(−1) n u (n) , reduces tov ′′ +2xv ′ +2(n +1)v =0. (18.131)Now letting y = e x2 v, we may write the derivatives of v asv ′ = e −x2 (y ′ − 2xy),v ′′ = e −x2 (y ′′ − 4xy ′ +4x 2 y − 2y).Substituting these expressions into (18.131), and dividing through by e −x2 , finally yieldsHermite’s equation,y ′′ − 2xy +2ny =0,thus demonstrating that y =(−1) n e x2 d n (e −x2 )/dx n is indeed a solution. Moreover, sincethis solution is clearly a polynomial of order n, it must be some multiple of H n (x). Thenormalisation is easily checked by noting that, from (18.130), the highest-order term is(2x) n , which agrees with the expression (18.128). ◭Mutual orthogonalityWe saw in section 17.4 that Hermite’s equation could be cast in Sturm–Liouvilleform with p = e −x2 , q =0,λ =2n and ρ = e −x2 , and its natural interval is thus[−∞, ∞]. Since the Hermite polynomials H n (x) are solutions of the equation andare regular at the end-points, they must be mutually orthogonal over this intervalwith respect to the weight function ρ = e −x2 ,i.e.∫ ∞−∞dnH n (x)H k (x)e −x2 dx =0 ifn ≠ k.This result may also be proved directly using the Rodrigues’ formula (18.130).Indeed, the normalisation, when k = n, is most easily found in this way.626

18.9 HERMITE FUNCTIONS◮Show thatI ≡∫ ∞−∞H n (x)H n (x)e −x2 dx =2 n n! √ π. (18.132)Using the Rodrigues’ formula (18.130), we may write∫ ∞∫ ∞I =(−1) n H n (x) dnd n H n0 dx n (e−x2 ) dx = e −x2 dx,−∞ dx nwhere, in the second equality, we have integrated by parts n times and used the fact thatthe boundary terms all vanish. From (18.128) we see that d n H n /dx n =2 n n!. Thus we haveI =2 n n!∫ ∞−∞e −x2 dx =2 n n! √ π,where, in the second equality, we use the standard result for the area under a Gaussian(see section 6.4.2). ◭The above orthogonality and normalisation conditions allow any (reasonable)function in the interval −∞ ≤ x

SPECIAL FUNCTIONSDifferentiating this form k times with respect to h gives∞∑ H n(n − k)! hn−k = ∂k G ∂k= ex2∂hk ∂h k e−(x−h)2 =(−1) k e x2 ∂k∂x k e−(x−h)2 .n=kRelabelling the summation on the LHS using the new index m = n − k, weobtain∞∑ H m+km! hm =(−1) k x2∂ke∂x k e−(x−h)2 .m=0Setting h = 0 in this equation, we findH k (x) =(−1) k e x2dx k (e−x2 ),which is the Rodrigues’ formula (18.130) for the Hermite polynomials. ◭The generating function (18.133) is also useful for determining special valuesof the Hermite polynomials. In particular, it is straightforward to show thatH 2n (0) = (−1) n (2n)!/n! andH 2n+1 (0) = 0.Recurrence relationsThe two most useful recurrence relations satisfied by the Hermite polynomialsare given bydkH n+1 (x) =2xH n (x) − 2nH n−1 (x), (18.134)H n(x) ′ =2nH n−1 (x). (18.135)The first relation provides a simple iterative way of evaluating the nth Hermitepolynomials at some point x = x 0 , given the values of H 0 (x) andH 1 (x) atthatpoint. For proofs of these recurrence relations, see exercise 18.5.18.10 Hypergeometric functionsThe hypergeometric equation has the formx(1 − x)y ′′ +[c − (a + b +1)x]y ′ − aby =0, (18.136)and has three regular singular points, at x =0, 1, ∞, but no essential singularities.The parameters a, b and c are given real numbers.In our discussions of Legendre functions, associated Legendre functions andChebyshev functions in sections 18.1, 18.2 and 18.4, respectively, it was noted thatin each case the corresponding second-order differential equation had three regularsingular points, at x = −1, 1, ∞, and no essential singularities. The hypergeometricequation can, in fact, be considered as the ‘canonical form’ for second-orderdifferential equations with this number of singularities. It may be shown § that,§ See, for example, J. Mathews and R. L. Walker, Mathematical Methods of Physics, 2nd edn (ReadingMA: Addision–Wesley, 1971).628

18.10 HYPERGEOMETRIC FUNCTIONSby making appropriate changes of the independent and dependent variables,any second-order differential equation with three regular singularities and anordinary point at infinity can be transformed into the hypergeometric equation(18.136) with the singularities at = −1, 1 and ∞. As we discuss below, this allowsLegendre functions, associated Legendre functions and Chebyshev functions, forexample, to be written as particular cases of hypergeometric functions, which arethe solutions to (18.136).Since the point x = 0 is a regular singularity of (18.136), we may find at leastone solution in the form of a Frobenius series (see section 16.3):y(x) =∞∑a n x n+σ . (18.137)Substituting this series into (18.136) and dividing through by x σ−1 ,weobtain∞∑n=0n=0{(1 − x)(n + σ)(n + σ − 1) + [c − (a + b +1)x](n + σ) − abx} a n x n =0.(18.138)Setting x = 0, so that only the n = 0 term remains, we obtain the indicial equationσ(σ − 1) + cσ = 0, which has the roots σ = 0 and σ =1− c. Thus, providedc is not an integer, one can obtain two linearly independent solutions of thehypergeometric equation in the form (18.137).For σ = 0 the corresponding solution is a simple power series. Substitutingσ = 0 into (18.138) and demanding that the coefficient of x n vanishes, we find therecurrence relationn[(n − 1) + c]a n − [(n − 1)(a + b + n − 1) + ab]a n−1 =0,(18.139)which, on simplifying and replacing n by n + 1, yields the recurrence relation(a + n)(b + n)a n+1 =(n +1)(c + n) a n. (18.140)It is conventional to make the simple choice a 0 = 1. Thus, provided c is not anegative integer or zero, we may write the solution as follows:F(a, b, c; x) =1+ ab x a(a +1)b(b +1)+ + ···c 1! c(c +1) 2!(18.141)= Γ(c) ∞∑ Γ(a + n)Γ(b + n) x nΓ(a)Γ(b) Γ(c + n) n! , (18.142)n=0where F(a, b, c; x) is known as the hypergeometric function or hypergeometricseries, and in the second equality we have used the property (18.154) of the629x 2

SPECIAL FUNCTIONSgamma function. § It is straightforward to show that the hypergeometric seriesconverges in the range |x| < 1. It also converges at x =1ifc>a+ b andat x = −1 ifc>a+ b − 1. We also note that F(a, b, c; x) is symmetric in theparameters a and b, i.e.F(a, b, c; x) =F(b, a, c; x).The hypergeometric function y(x) = F(a, b, c; x) is clearly not the generalsolution to the hypergeometric equation (18.136), since we must also consider thesecond root of the indicial equation. Substituting σ =1− c into (18.138) anddemanding that the coefficient of x n vanishes, we find that we must haven(n +1− c)a n − [(n − c)(a + b + n − c)+ab]a n−1 =0,which, on comparing with (18.139) and replacing n by n + 1, yields the recurrencerelation(a − c +1+n)(b − c +1+n)a n+1 = a n .(n + 1)(2 − c + n)We see that this recurrence relation has the same form as (18.140) if one makesthe replacements a → a − c +1,b → b − c + 1 and c → 2 − c. Thus, provided c,a−b and c−a−b are all non-integers, the general solution to the hypergeometricequation, valid for |x| < 1, may be written asy(x) =AF(a, b, c; x)+Bx 1−c F(a − c +1,b− c +1, 2 − c; x),(18.143)where A and B are arbitrary constants to be fixed by the boundary conditions onthe solution. If the solution is to be regular at x =0,onerequiresB =0.18.10.1 Properties of hypergeometric functionsSince the hypergeometric equation is so general in nature, it is not feasible topresent a comprehensive account of the hypergeometric functions. Nevertheless,we outline here some of their most important properties.Special casesAs mentioned above, the general nature of the hypergeometric equation allows usto write a large number of elementary functions in terms of the hypergeometricfunctions F(a, b, c; x). Such identifications can be made from the series expansion(18.142) directly, or by transformation of the hypergeometric equation into a morefamiliar equation, the solutions to which are already known. Some particularexamples of well known special cases of the hypergeometric function are asfollows:§ We note that it is also common to denote the hypergeometric function by 2 F 1 (a, b, c; x). Thisslightly odd-looking notation is meant to signify that, in the coefficient of each power of x, thereare two parameters (a and b) in the numerator and one parameter (c) in the denominator.630

18.10 HYPERGEOMETRIC FUNCTIONSF(a, b, b; x) =(1− x) −a , F( 1 2 , 1 2 , 3 2 ; x2 )=x −1 sin −1 x,F(1, 1, 2; −x) =x −1 ln(1 + x), F( 1 2 , 1, 3 2 ; −x2 )=x −1 tan −1 x,lim m→∞ =ex , F( 1 2 , 1, 3 2 ; x2 )= 1 2 x−1 ln[(1 + x)/(1 − x)],F( 1 2 , − 1 2 , 1 2 ;sin2 x)=cosx, F(m +1, −m, 1; (1 − x)/2) = P m (x),F( 1 2 ,p,p;sin2 x)=secx, F(m, −m, 1 2 ;(1− x)/2) = T m(x),where m is an integer, P m (x) isthemth Legendre polynomial and T m (x) isthemth Chebyshev polynomial of the first kind. Some of these results are proved inexercise 18.11.◮Show that F(m, −m, 1 2 ;(1− x)/2) = T m(x).Let us prove this result by transforming the hypergeometric equation. The form of theresult suggests that we should make the substitution x =(1− z)/2 into (18.136), in whichcase d/dx = −2d/dz. Thus, letting u(z) =y(x) and setting a = m, b = −m and c =1/2,(18.136) becomes(1 − z)2(1 + z)(−2) 2 d2 u2On simplifying, we obtaindz 2 + [12]− z− (m − m +1)1 (−2) du − (m)(−m)u =0.2 dz(1 − z 2 ) d2 udz − z du2 dz + m2 u =0,which has the form of Chebyshev’s equation, (18.54). This equation has u(z) =T m (z) asits power series solution, and so F(m, −m, 1 ;(1− z)/2) and T 2 m(z) are equal to within anormalisation factor. On comparing the expressions (18.141) and (18.56) at x =0,i.e.atz = 1, we see that they both have value 1. Hence, the normalisations already agree andwe obtain the required result. ◭Integral representationOne of the most useful representations for the hypergeometric functions is interms of an integral, which may be derived using the properties of the gammaand beta functions discussed in section 18.12. The integral representation readsF(a, b, c; x) =Γ(c)Γ(b)Γ(c − b)∫ 1and requires c>b>0 for the integral to converge.◮Prove the result (18.144).0t b−1 (1 − t) c−b−1 (1 − tx) −a dt,From the series expansion (18.142), we haveF(a, b, c; x) =Γ(c) ∞∑ Γ(a + n)Γ(b + n) x nΓ(a)Γ(b) Γ(c + n) n!=n=0Γ(c)Γ(a)Γ(b)Γ(c − b)631∞∑n=0Γ(a + n)B(b + n, c − b) xnn! ,(18.144)

SPECIAL FUNCTIONSwhere in the second equality we have used the expression (18.165) relating the gamma andbeta functions. Using the definition (18.162) of the beta function, we then findΓ(c)∞∑F(a, b, c; x) =Γ(a + n) xnΓ(a)Γ(b)Γ(c − b)n!=Γ(c)Γ(b)Γ(c − b)∫ 10n=0dt t b−1 (1 − t) c−b−1∫ 10t b+n−1 (1 − t) c−b−1 dt∞∑n=0Γ(a + n) (tx) nΓ(a) n! ,where in the second equality we have rearranged the expression and reversed the orderof integration and summation. Finally, one recognises the sum over n as being equal to(1 − tx) −a , and so we obtain the final result (18.144). ◭The integral representation may be used to prove a wide variety of properties ofthe hypergeometric functions. As a simple example, on setting x = 1 in (18.144),and using properties of the beta function discussed in section 18.12.2, one quicklyfinds that, provided c is not a negative integer or zero and c>a+ b,F(a, b, c;1)=Γ(c)Γ(c − a − b)Γ(c − a)Γ(c − b) .Relationships between hypergeometric functionsThere exist a great many relationships between hypergeometric functions withdifferent arguments. These are most easily derived by making use of the integralrepresentation (18.144) or the series form (18.141). It is not feasible to list all therelationships here, so we simply note two useful examples, which readF(a, b, c; x) =(1− x) c−a−b F(c − a, c − b, c; x), (18.145)F ′ (a, b, c; x) = ab F(a +1,b+1,c+1;x), (18.146)cwhere the prime in the second relation denotes d/dx. The first result followsstraightforwardly from the integral representation using the substitution t =(1 − u)/(1 − ux), whereas the second result may be proved more easily from theseries expansion.In addition to the above results, one may also derive relationships betweenF(a, b, c; x) and any two of the six ‘contiguous functions’ F(a ± 1,b,c; x), F(a, b ±1,c; x) andF(a, b, c ± 1; x). These ‘contiguous relations’ serve as the recurrencerelations for the hypergeometric functions. An example of such a relationship is(c − a)F(a − 1,b,c; x)+(2a − c − ax + bx)F(a, b, c; x)+a(x − 1)F(a +1,b,c; x)=0.Repeated application of such relationships allows one to express F(a+l,b+m, c+n; x), where l,m,n are integers (with c+n not equalling a negative integer or zero),as a linear combination of F(a, b, c; x) and one of its contiguous functions.632

18.11 CONFLUENT HYPERGEOMETRIC FUNCTIONS18.11 Confluent hypergeometric functionsThe confluent hypergeometric equation has the formxy ′′ +(c − x)y ′ − ay = 0; (18.147)it has a regular singularity at x = 0 and an essential singularity at x = ∞.This equation can be obtained by merging two of the singularities of the ordinaryhypergeometric equation (18.136). The parameters a and c are given real numbers.◮Show that setting x = z/b in the hypergeometric equation, and letting b →∞, yields theconfluent hypergeometric equation.Substituting x = z/b into (18.136), with d/dx = bd/dz, and letting u(z) =y(x), we obtain(bz 1 − z ) d 2 udu+[bc − (a + b +1)z] − abu =0,b dz2 dzwhich clearly has regular singular points at z =0,b and ∞. Ifwenowmergethelasttwosingularities by letting b →∞,weobtainzu ′′ +(c − z)u ′ − au =0,where the primes denote d/dz. Hence u(z) must satisfy the confluent hypergeometricequation. ◭In our discussion of Bessel, Laguerre and associated Laguerre functions, it wasnoted that the corresponding second-order differential equation in each case had asingle regular singular point at x = 0 and an essential singularity at x = ∞. Fromtable 16.1, we see that this is also true for the confluent hypergeometric equation.Indeed, this equation can be considered as the ‘canonical form’ for secondorderdifferential equations with this pattern of singularities. Consequently, as wemention below, the Bessel, Laguerre and associated Laguerre functions can all bewritten in terms of the confluent hypergeometric functions, which are the solutionsof (18.147).The solutions of the confluent hypergeometric equation are obtained fromthose of the ordinary hypergeometric equation by again letting x → x/b andcarrying out the limiting process b →∞. Thus, from (18.141) and (18.143), twolinearly independent solutions of (18.147) are (when c is not an integer)y 1 (x) =1+ a x a(a +1) z 2+ + ···≡ M(a, c; x), (18.148)c 1! c(c +1) 2!y 2 (x) =x 1−c M(a − c +1, 2 − c; x), (18.149)where M(a, c; x) is called the confluent hypergeometric function (or Kummerfunction). § It is worth noting, however, that y 1 (x) is singular when c =0, −1, −2,...and y 2 (x) is singular when c =2, 3, 4,... . Thus, it is conventional to take the§ We note that an alternative notation for the confluent hypergeometric function is 1 F 1 (a, c; x).633

SPECIAL FUNCTIONSsecond solution to (18.147) as a linear combination of (18.148) and (18.149) givenbyU(a, c; x) ≡π []M(a, c; x) M(a − c +1, 2 − c; x)− x1−c .sin πc Γ(a − c +1)Γ(c) Γ(a)Γ(2 − c)This has a well behaved limit as c approaches an integer.18.11.1 Properties of confluent hypergeometric functionsThe properties of confluent hypergeometric functions can be derived from thoseof ordinary hypergeometric functions by letting x → x/b and taking the limitb →∞, in the same way as both the equation and its solution were derived. Ageneral procedure of this sort is called a confluence process.Special casesThe general nature of the confluent hypergeometric equation allows one to writea large number of elementary functions in terms of the confluent hypergeometricfunctions M(a, c; x). Once again, such identifications can be made from the seriesexpansion (18.148) directly, or by transformation of the confluent hypergeometricequation into a more familiar equation for which the solutions are alreadyknown. Some particular examples of well known special cases of the confluenthypergeometric function are as follows:M(a, a; x) =e x ,M(1, 2; 2x) = ex sinh x,xM(−n, 1; x) =L n (x), M(−n, m +1;x) = n!m!(n + m)! Lm n (x),M(−n, 1 2 ; x2 )= (−1)n n!(2n)! H 2n(x), M(−n, 3 2 ; x2 ), = (−1)n n!2(2n +1)!M(ν + 1 2 , 2ν +1;2ix) =ν!eix ( x 2 )−ν J ν (x), M( 1 2 , 3 2 ; −x2 )=√ π2x erf(x),H 2n+1 (x),xwhere n and m are integers, L m n (x) is an associated Legendre polynomial, H n (x) isa Hermite polynomial, J ν (x) is a Bessel function and erf(x) is the error functiondiscussed in section 18.12.4.Integral representationUsing the integral representation (18.144) of the ordinary hypergeometric function,exchanging a and b and carrying out the process of confluence givesM(a, c, x) =Γ(c)Γ(a)Γ(c − a)634∫ 10e tx t a−1 (1 − t) c−a−1 dt,(18.150)

18.12 THE GAMMA FUNCTION AND RELATED FUNCTIONSwhich converges provided c>a>0.◮Prove the result (18.150).Since F(a, b, c; x) is unchanged by swapping a and b, we may write its integral representation(18.144) asΓ(c)F(a, b, c; x) =t a−1 (1 − t) c−a−1 (1 − tx) −b dt.Γ(a)Γ(c − a) 0Setting x = z/b and taking the limit b →∞,weobtainM(a, c; z) =Γ(c)Γ(a)Γ(c − a)∫ 10∫ 1t a−1 (1 − t) c−a−1 limb→∞(1 − tz bSince the limit is equal to e tz , we obtain result (18.150). ◭) −bdt.Relationships between confluent hypergeometric functionsA large number of relationships exist between confluent hypergeometric functionswith different arguments. These are straightforwardly derived using the integralrepresentation (18.150) or the series form (18.148). Here, we simply note twouseful examples, which readM(a, c; x) =e x M(c − a, c; −x), (18.151)M ′ (a, c; x) = a M(a +1,c+1;x), (18.152)cwhere the prime in the second relation denotes d/dx. The first result followsstraightforwardly from the integral representation, and the second result may beproved from the series expansion (see exercise 18.19).In an analogous manner to that used for the ordinary hypergeometric functions,one may also derive relationships between M(a, c; x) and any two of thefour ‘contiguous functions’ M(a ± 1,c; x) andM(a, c ± 1; x). These serve as therecurrence relations for the confluent hypergeometric functions. An example ofsuch a relationship is(c − a)M(a − 1,c; x)+(2a − c + x)M(a, c; x) − aM(a +1,c; x) =0.18.12 The gamma function and related functionsMany times in this chapter, and often throughout the rest of the book, we havemade mention of the gamma function and related functions such as the beta anderror functions. Although not derived as the solutions of important second-orderODEs, these convenient functions appear in a number of contexts, and so herewe gather together some of their properties. This final section should be regardedmerely as a reference containing some useful relations obeyed by these functions;a minimum of formal proofs is given.635

SPECIAL FUNCTIONSThe gamma function Γ(n) is defined by18.12.1 The gamma functionΓ(n) =∫ ∞0x n−1 e −x dx, (18.153)which converges for n>0, where in general n is a real number. Replacing n byn + 1 in (18.153) and integrating the RHS by parts, we findΓ(n +1)=∫ ∞0x n e −x dx= [ −x n e −x] ∫ ∞∞0 + nx n−1 e −x dx= n∫ ∞00x n−1 e −x dx,from which we obtain the important resultΓ(n +1)=nΓ(n). (18.154)From (18.153), we see that Γ(1) = 1, and so, if n is a positive integer,Γ(n +1)=n!. (18.155)In fact, equation (18.155) serves as a definition of the factorial function even fornon-integer n. For negative n the factorial function is defined by(n + m)!n! =(n + m)(n + m − 1) ···(n +1) , (18.156)where m is any positive integer that makes n + m>0. Different choices of m(> −n) do not lead to different values for n!. A plot of the gamma function isgiven in figure 18.9, where it can be seen that the function is infinite for negativeinteger values of n, in accordance with (18.156). For an extension of the factorialfunction to complex arguments, see exercise 18.15.By letting x = y 2 in (18.153), we immediately obtain another useful representationof the gamma function given byΓ(n) =2∫ ∞0y 2n−1 e −y2 dy. (18.157)Setting n = 1 2we find the resultΓ ( ∫) ∞ ∫ ∞12 =2 e −y2 dy = e −y2 dy = √ π,0−∞where have used the standard integral discussed in section 6.4.2. From this result,Γ(n) for half-integral n can be found using (18.154). Some immediately derivablefactorial values of half integers are(−32)!=−2√ π,(−12)!=√ π,( 12636) (!=12√ π, 3)2 !=34√ π.

18.12 THE GAMMA FUNCTION AND RELATED FUNCTIONSΓ( n)642−4−3−2−11 23 4n−2−4−6Figure 18.9The gamma function Γ(n).Moreover, it may be shown for non-integral n that the gamma function satisfiesthe important identityΓ(n)Γ(1 − n) =πsin nπ . (18.158)This is proved for a restricted range of n in the next section, once the betafunction has been introduced.It can also be shown that the gamma function is given byΓ(n +1)= √ (2πn n n e −n 1+ 112n + 1288n 2 − 139 )51 840n 3 + ... = n!,(18.159)which is known as Stirling’s asymptotic series. For large n the first term dominates,and son! ≈ √ 2πn n n e −n ; (18.160)this is known as Stirling’s approximation. This approximation is particularly usefulin statistical thermodynamics, when arrangements of a large number of particlesaretobeconsidered.◮Prove Stirling’s approximation n! ≈ √ 2πn n n e −n for large n.From (18.153), the extended definition of the factorial function (which is valid for n>−1)is given byn! =∫ ∞0x n e −x dx =637∫ ∞0e n ln x−x dx. (18.161)

SPECIAL FUNCTIONSIf we let x = n + y, then(ln x =lnn +ln 1+ y )n=lnn + y n − y22n 2 + y33n 3 − ··· .Substituting this result into (18.161), we obtain∫ ∞[n! = exp n(ln n + y ···) ]−nn − y22n + − n − y dy.2Thus, when n is sufficiently large, we may approximate n! by∫ ∞n! ≈ e n ln n−n e −y2 /(2n) dy = e n ln n−n√ 2πn = √ 2πn n n e −n ,−∞which is Stirling’s approximation (18.160). ◭The beta function is defined by18.12.2 The beta functionB(m, n) =∫ 10x m−1 (1 − x) n−1 dx, (18.162)which converges for m>0, n>0, where m and n are, in general, real numbers.By letting x =1− y in (18.162) it is easy to show that B(m, n) =B(n, m). Otheruseful representations of the beta function may be obtained by suitable changesof variable. For example, putting x =(1+y) −1 in (18.162), we find that∫ ∞y n−1 dyB(m, n) =. (18.163)0 (1 + y)m+nAlternatively, if we let x =sin 2 θ in (18.162), we obtain immediatelyB(m, n) =2∫ π/20sin 2m−1 θ cos 2n−1 θdθ. (18.164)The beta function may also be written in terms of the gamma function asB(m, n) = Γ(m)Γ(n)Γ(m + n) . (18.165)◮Prove the result (18.165).Using (18.157), we haveΓ(n)Γ(m) =4=4∫ ∞0∫ ∞ ∫ ∞0 0x 2n−1 e −x2 dx∫ ∞0y 2m−1 e −y2 dyx 2n−1 y 2m−1 e −(x2 +y 2) dx dy.638

18.12 THE GAMMA FUNCTION AND RELATED FUNCTIONSChanging variables to plane polar coordinates (ρ, φ) givenbyx = ρ cos φ, y = ρ sin φ, weobtainΓ(n)Γ(m) =4=4∫ π/2 ∫ ∞0 0∫ π/20ρ 2(m+n−1) e −ρ2 sin 2m−1 φ cos 2n−1 φ ρ dρ dφsin 2m−1 φ cos 2n−1 φdφ∫ ∞0ρ 2(m+n)−1 e −ρ2 dρ= B(m, n)Γ(m + n),where in the last line we have used the results (18.157) and (18.164). ◭The result (18.165) is useful in proving the identity (18.158) satisfied by thegamma function, since∫ ∞y n−1 dyΓ(n)Γ(1 − n) =B(1 − n, n) =0 1+y ,where, in the second equality, we have used the integral representation (18.163).For 0

SPECIAL FUNCTIONSwhich is the required result. ◭We note that is it conventional to define, in addition, the functionsγ(a, x)Γ(a, x)P (a, x) ≡ , Q(a, x) ≡Γ(a) Γ(a) ,which are also often called incomplete gamma functions; it is clear that Q(a, x) =1 − P (a, x).18.12.4 The error functionFinally, we mention the error function, which is encountered in probability theoryand in the solutions of some partial differential equations. The error function isrelated to the incomplete gamma function by erf(x) =γ( 1 2 ,x2 )/ √ π and is thusgiven byerf(x) = √ 2 ∫ xe −u2 du =1− 2 ∫ ∞√ e −u2 du. (18.167)π πFrom this definition we can easily see that0erf(0) = 0, erf(∞) =1, erf(−x) =−erf(x).By making the substitution y = √ 2u in (18.167), we find√ ∫ √22xerf(x) = e −y2 /2 dy.π 0The cumulative probability function Φ(x) for the standard Gaussian distribution(discussed in section 30.9.1) may be written in terms of the error function asfollows:Φ(x) = √ 1 ∫ xe −y2 /2 dy2π −∞= 1 2 + √ 1 ∫ xe −y2 /2 dy2π 0= 1 2 + 1 ( ) x√22 erf .It is also sometimes useful to define the complementary error functionerfc(x) =1− erf(x) = √ 2 ∫ ∞e −u2 du = Γ( 1 2 ,x2 )√ . (18.168)π πxx18.1 Use the explicit expressions18.13 Exercises640

18.13 EXERCISES√√Y0 0 = 1, Y 04π 1 = 3cos θ,4π√√Y ±1301= ∓ sin θ exp(±iφ), Y8π 2 = 5(3 16π cos2 θ − 1),√√Y ±1±22= ∓ sin θ cos θ exp(±iφ), Y2=158πto verify for l =0, 1, 2that1532π sin2 θ exp(±2iφ),l∑|Yl m (θ, φ)| 2 2l +1=4π ,m=−land so is independent of the values of θ and φ. Thisistrueforanyl, buta general proof is more involved. This result helps to reconcile intuition withthe apparently arbitrary choice of polar axis in a general quantum mechanicalsystem.18.2 Express the functionf(θ, φ) =sinθ[sin 2 (θ/2) cos φ + i cos 2 (θ/2) sin φ]+sin 2 (θ/2)as a sum of spherical harmonics.18.3 Use the generating function for the Legendre polynomials P n (x) to show that∫ 1P 2n+1 (x) dx =(−1) n (2n)!02 2n+1 n!(n +1)!and that, except for the case n =0,∫ 10P 2n (x) dx =0.18.4 Carry through the following procedure as a proof of the result∫ 1I n = P n (z)P n (z) dz = 2−12n +1 .(a) Square both sides of the generating-function definition of the Legendrepolynomials,∞∑(1 − 2zh + h 2 ) −1/2 = P n (z)h n .(b) Express the RHS as a sum of powers of h, obtaining expressions for thecoefficients.(c) Integrate the RHS from −1 to 1 and use the orthogonality property of theLegendre polynomials.(d) Similarly integrate the LHS and expand the result in powers of h.(e) Compare coefficients.18.5 The Hermite polynomials H n (x) may be defined byShow thatΦ(x, h) =exp(2xh − h 2 )=n=0∞∑n=0∂ 2 Φ ∂Φ ∂Φ− 2x +2h∂x2 ∂x ∂h =0,6411n! H n(x)h n .

SPECIAL FUNCTIONSand hence that the H n (x) satisfy the Hermite equationy ′′ − 2xy ′ +2ny =0,where n is an integer ≥ 0.Use Φ to prove that(a) H n(x) ′ =2nH n−1 (x),(b) H n+1 (x) − 2xH n (x)+2nH n−1 (x) =0.18.6 A charge +2q is situated at the origin and charges of −q are situated at distances±a from it along the polar axis. By relating it to the generating function for theLegendre polynomials, show that the electrostatic potential Φ at a point (r, θ, φ)with r>ais given byΦ(r, θ, φ) =2q ∞∑ ( a) 2sP2s (cos θ).4πɛ 0 r rs=118.7 For the associated Laguerre polynomials, carry through the following exercises.(a) Prove the Rodrigues’ formulaL m n (x) = ex x −m d nn! dx n (xn+m e −x ),taking the polynomials to be defined byn∑L m n (x) = (−1) k (n + m)!k!(n − k)!(k + m)! xk .k=0(b) Prove the recurrence relations(n +1)L m n+1(x) =(2n + m +1− x)L m n (x) − (n + m)L m n−1(x),x(L m n ) ′ (x) =nL m n (x) − (n + m)L m n−1(x),but this time taking the polynomial as defined byL m n (x) =(−1) m dmdx L n+m(x)mor the generating function.18.8 The quantum mechanical wavefunction for a one-dimensional simple harmonicoscillator in its nth energy level is of the formψ(x) =exp(−x 2 /2)H n (x),where H n (x) isthenth Hermite polynomial. The generating function for thepolynomials is∞∑ HG(x, h) =e 2hx−h2 n (x)= h n .n!n=0(a) Find H i (x) fori =1, 2, 3, 4.(b) Evaluate by direct calculation∫ ∞−∞e −x2 H p (x)H q (x) dx,(i) for p =2,q = 3; (ii) for p =2,q = 4; (iii) for p = q = 3. Check youranswers against the expected values 2 p p! √ πδ pq .642

18.13 EXERCISES[ You will find it convenient to use∫ ∞x 2n e −x2 dx = (2n)!√ π−∞2 2n n!for integer n ≥ 0. ]18.9 By initially writing y(x) asx 1/2 f(x) and then making subsequent changes ofvariable, reduce Stokes’ equation,d 2 y+ λxy =0,dx2 to Bessel’s equation. √Hence show that a solution that is finite at x =0isamultiple of x 1/2 J 1/3 ( 2 λx3 ).318.10 By choosing a suitable form for h in their generating function,[ ( zG(z,h) =exp h − 1 )] ∞∑= J n (z)h n ,2 hn=−∞show that integral repesentations of the Bessel functions of the first kind aregiven, for integral m, by∫ 2πJ 2m (z) = (−1)m cos(z cos θ) cos2mθ dθ m ≥ 1,π 0∫ 2πJ 2m+1 (z) = (−1)m+1 cos(z cos θ) sin(2m +1)θdθ m≥ 0.π 018.11 Identify the series for the following hypergeometric functions, writing them interms of better known functions:(a) F(a, b, b; z),(b) F(1, 1, 2; −x),(c) F( 1 , 1, 3 ; 2 2 −x2 ),(d) F( 1 , 1 , 3 ; 2 2 2 x2 ),(e) F(−a, a, 1 2 ;sin2 x); this is a much more difficult exercise.18.12 By making the substitution z =(1− x)/2 and suitable choices for a, b and c,convert the hypergeometric equation,z(1 − z) d2 udu+[c − (a + b +1)z ] − abu =0,dz2 dzinto the Legendre equation,(1 − x 2 ) d2 y dy− 2x + l(l +1)y =0.dx2 dxHence, using the hypergeometric series, generate the Legendre polynomials P l (x)for the integer values l =0, 1, 2, 3. Comment on their normalisations.18.13 Find a change of variable that will allow the integral∫ ∞√u − 1I =1 (u +1) du 2to be expressed in terms of the beta function, and so evaluate it.18.14 Prove that, if m and n are both greater than −1, then∫ ∞u mI =0 (au 2 + b) du = Γ[ 1 (m 2 +1)]Γ[1 (n +1)]2 (m+n+2)/2 2a (m+1)/2 b (n+1)/2 Γ[ 1 (m + n +2)].2643

SPECIAL FUNCTIONSDeduce the value of∫ ∞(u +2) 2J =du.0 (u 2 +4)5/218.15 The complex function z! is defined byz! =∫ ∞0u z e −u dufor Re z>−1.For Re z ≤−1 it is defined by(z + n)!z! =(z + n)(z + n − 1) ···(z +1) ,where n is any (positive) integer > −Re z. Being the ratio of two polynomials, z!is analytic everywhere in the finite complex plane except at the poles that occurwhen z is a negative integer.(a) Show that the definition of z! forRez ≤−1 is independent of the value ofn chosen.(b) Prove that the residue of z! at the pole z = −m, wherem is an integer > 0,is (−1) m−1 /(m − 1)!.18.16 For −1 < Re z0, occurs in some statistical mechanics problems. By first consideringthe integral∫ ∞J = e iu(k+ia) du,and a suitable variation of it, show that I =(π/a) exp(a 2 )erfc(a), where erfc(x)is the complementary error function.18.18 Consider two series expansions of the error function as follows.(a) Obtain a series expansion of the error function erf(x) in ascending powersof x. How many terms are needed to give a value correct to four significantfigures for erf(1)?(b) Obtain an asymptotic expansion that can be used to estimate erfc(x) forlarge x (> 0) in the form of a seriesaerfc(x) =R(x) =e −x2 nx . n n=0Consider what bounds can be put on the estimate and at what point theinfinite series should be terminated in a practical estimate. In particular,estimate erfc(1) and test the answer for compatibility with that in part (a).18.19 For the functions M(a, c; z) that are the solutions of the confluent hypergeometricequation,0644∞∑

18.13 EXERCISES(a) use their series representation to prove thatb d M(a, c; z) =aM(a +1,c+1;z);dz(b) use an integral representation to prove thatM(a, c; z) =e z M(c − a, c; −z).18.20 The Bessel function J ν (z) can be considered as a special case of the solutionM(a, c; z) of the confluent hypergeometric equation, the connection beingM(a, ν +1;−z/a)lim= z −ν/2 J ν (2 √ z).a→∞ Γ(ν +1)Prove this equality by writing each side in terms of an infinite series and showingthat the series are the same.18.21 Find the differential equation satisfied by the function y(x) defined by∫ xy(x) =Ax −n e −t t n−1 dt ≡ Ax −n γ(n, x),0and, by comparing it with the confluent hypergeometric function, express y asa multiple of the solution M(a, c; z) of that equation. Determine the value of Athat makes y equal to M.18.22 Show, from its definition, that the Bessel function of the second kind, and ofintegral order ν, can be written asY ν (z) = 1 [ ∂Jµ (z)− (−1) ν ∂J ]−µ(z).π ∂µ∂µµ=νUsing the explicit series expression for J µ (z), show that ∂J µ (z)/∂µ can be writtenas( z)J ν (z)ln + g(ν,z),2and deduce that Y ν (z) can be expressed asY ν (z) = 2 ( z)π J ν(z)ln + h(ν,z),2where h(ν,z), like g(ν,z), is a power series in z.18.23 Prove two of the properties of the incomplete gamma function P (a, x 2 ) as follows.(a) By considering its form for a suitable value of a, show that the error functioncan be expressed as a particular case of the incomplete gamma function.(b) The Fresnel integrals, of importance in the study of the diffraction of light,are given by∫ x ( π)∫ x ( π)C(x) = cos2 t2 dt, S(x) = sin2 t2 dt.0Show that they can be expressed in terms of the error function by[ √ ] πC(x)+iS(x) =A erf2 (1 − i)x ,where A is a (complex) constant, which you should determine. Hence expressC(x)+iS(x) in terms of the incomplete gamma function.0645

SPECIAL FUNCTIONS18.24 The solutions y(x, a) of the equationd 2 ydx − ( 1 2 4 x2 + a)y =0 (∗)are known as parabolic cylinder functions.(a) If y(x, a) is a solution of (∗), determine which of the following are alsosolutions: (i) y(a, −x), (ii) y(−a, x), (iii) y(a, ix) and(iv)y(−a, ix).(b) Show that one solution of (∗), even in x, isy 1 (x, a) =e −x2 /4 M( 1 a + 1 , 1 , 1 2 4 2 2 x2 ),where M(α, c, z) is the confluent hypergeometric function satisfyingz d2 M+(c − z) dM − αM =0.dz 2 dzYou may assume (or prove) that a second solution, odd in x, isgivenbyy 2 (x, a) =xe −x2 /4 M( 1 a + 3 , 3 , 1 2 4 2 2 x2 ).(c) Find, as an infinite series, an explicit expression for e x2 /4 y 1 (x, a).(d) Using the results from part (a), show that y 1 (x, a) can also be written asy 1 (x, a) =e x2 /4 M(− 1 a + 1 , 1 , − 1 2 4 2 2 x2 ).(e) By making a suitable choice for a deduce that()∞∑ b n x 2n∞∑1+(2n)! = ex2 /2 (−1) n b n x 2n1+,(2n)!n=1n=1where b n = ∏ nr=1 (2r − 3 ). 218.14 Hints and answers18.1 Note that taking the square of the modulus eliminates all mention of φ.18.3 Integrate both sides of the generating function definition from x =0tox =1,and then expand the resulting term, (1+h 2 ) 1/2 , using a binomial expansion. Showthat 1/2 C m can be written as [ (−1) m−1 (2m − 2)! ]/[2 2m−1 m!(m − 1)! ].18.5 Prove the stated equation using the explicit closed form of the generating function.Then substitute the series and require the coefficient of each power of h to vanish.(b) Differentiate result (a) and then use (a) again to replace the derivatives.18.7 (a) Write the result of using Leibnitz’ theorem on the product of x n+m and e −x as afinite sum, evaluate the separated derivatives, and then re-index the summation.(b) For the first recurrence relation, differentiate the generating function withrespect to h and then use the generating function again to replace the exponential.Equating coefficients of h n then yields the result. For the second, differentiate thecorresponding relationship for the ordinary Laguerre polynomials m times.18.9 x 2 f ′′ + xf ′ +(λx 3 − 1 4 )f = 0. Then, in turn, set x3/2 = u, and 2 3 λ1/2 u = v; thenvsatisfies Bessel’s equation with ν = 1 3 .18.11 (a) (1 − z) −a .(b)x −1 ln(1 + x). (c) Compare the calculated coefficients with thoseof tan −1 x. F( 1 2 , 1, 3 2 ; −x2 )=x −1 tan −1 x. (d) x −1 sin −1 x. (e) Note that a termcontaining x 2n can only arise from the first n + 1 terms of an expansion in powersof sin 2 x; make a few trials. F(−a, a, 1 2 ;sin2 x)=cos2ax.18.13 Looking for f(x) =u such that u + 1 is an inverse power of x with f(0) = ∞ andf(1) = 1 leads to f(x) =2x −1 − 1. I = B( 1 2 , 3 2 )/√ 2=π/(2 √ 2).646

18.14 HINTS AND ANSWERS18.15 (a) Show that the ratio of two definitions based on m and n, withm>n>−Re z,is unity, independent of the actual values of m and n.(b) Consider the limit as z →−m of (z + m)z!, with the definition of z! basedonn where n>m.18.17 Express the integrand in partial fractions and use J, as given, and J ′ =∫ ∞0exp[ −iu(k −ia)]du to express I as the sum of two double integral expressions.Reduce them using the standard Gaussian integral, and then make a change ofvariable 2v = u +2a.18.19 (b) Using the representationΓ(b)M(a, b; z) =e zt t a−1 (1 − t) b−a−1 dtΓ(b − a)Γ(a) 0allows the equality to be established, without actual integration, by changing theintegration variable to s =1− t.18.21 Calculate y ′ (x) andy ′′ (x) and then eliminate x −1 e −x to obtain xy ′′ +(n+1+x)y ′ +ny =0;M(n, n +1;−x). Comparing the expansion of the hypergeometric serieswith the result of term by term integration of the expansion of the integrandshows that A = n.18.23 (a) If the dummy variable in the incomplete gamma function is t, make the changeof variable y =+ √ t.Nowchoosea so that 2(a − 1) + 1 = 0; erf(x) =P ( 1 2 ,x2 ).(b) Change the integration √ variable u in the standard representation of the RHSto s, givenbyu = 1 2 π(1 − i)s, and note that (1 − i) 2 = −2i. A =(1+i)/2. Frompart (a), C(x)+iS(x) = 1 (1 + i)P ( 1 , − 1 πi 2 2 2 x2 ).∫ 1647

19Quantum operatorsAlthough the previous chapter was principally concerned with the use of linearoperators and their eigenfunctions in connection with the solution of givendifferential equations, it is of interest to study the properties of the operatorsthemselves and determine which of them follow purely from the nature of theoperators, without reference to specific forms of eigenfunctions.19.1 Operator formalismThe results we will obtain in this chapter have most of their applications in thefield of quantum mechanics and our descriptions of the methods will reflect this.In particular, when we discuss a function ψ that depends upon variables such asspace coordinates and time, and possibly also on some non-classical variables, ψwill usually be a quantum-mechanical wavefunction that is being used to describethe state of a physical system. For example, the value of |ψ| 2 for a particularset of values of the variables is interpreted in quantum mechanics as being theprobability that the system’s variables have that set of values.To this end, we will be no more specific about the functions involved thanattaching just enough labels to them that a particular function, or a particularset of functions, is identified. A convenient notation for this kind of approachis that already hinted at, but not specifically stated, in subsection 17.1, wherethe definition of an inner product is given. This notation, often called the Diracnotation, denotes a state whose wavefunction is ψ by | ψ〉; sinceψ belongs to avector space of functions, | ψ〉 is known as a ket vector. Ket vectors, or simply kets,must not be thought of as completely analogous to physical vectors. Quantummechanics associates the same physical state with ke iθ | ψ〉 as it does with | ψ〉for all real k and θ and so there is no loss of generality in taking k as 1 and θas 0. On the other hand, the combination c 1 | ψ 1 〉 + c 2 | ψ 2 〉,where| ψ 1 〉 and | ψ 2 〉648

19.1 OPERATOR FORMALISMrepresent different states, is a ket that represents a continuum of different statesas the complex numbers c 1 and c 2 are varied.If we need to specify a state more closely – say we know that it correspondsto a plane wave with a wave number whose magnitude is k – then we indicatethis with a label; the corresponding ket vector would be written as | k〉. If we alsoknew the direction of the wave then | k〉 would be the appropriate form. Clearly,in general, the more labels we include, the more precisely the corresponding stateis specified.The Dirac notation for the Hermitian conjugate (dual vector) of the ket vector| ψ〉 is written as 〈ψ| and is known as a bra vector; the wavefunction describing thisstate is ψ ∗ , the complex conjugate of ψ. The inner product of two wavefunctions∫ψ ∗ φdvis then denoted by 〈ψ| φ〉 or, more generally if a non-unit weight functionρ is involved, by〈ψ| ρ| φ〉,evaluated as∫ψ ∗ (r)φ(r)ρ(r) dr. (19.1)Given the (contrived) names for the two sorts of vectors, an inner product like〈ψ| φ〉 becomes a particular type of ‘bra(c)ket’. Despite its somewhat whimsicalconstruction, this type of quantity has a fundamental role to play in the interpretationof quantum theory, because expectation values, probabilities and transitionrates are all expressed in terms of them. For physical states the inner product ofthe corresponding ket with itself, with or without an explicit weight function, isnon-zero, and it is usual to take〈ψ|ψ〉 =1.Although multiplying a ket vector by a constant does not change the statedescribed by the vector, acting upon it with a more general linear operator Aresults (in general) in a ket describing a different state. For example, if ψ is a statethat is described in one-dimensional x-space by the wavefunction ψ(x) =exp(−x 2 )and A is the differential operator ∂/∂x, then| ψ 1 〉 = A| ψ〉 ≡|Aψ〉is the ket associated with the state whose wavefunction is ψ 1 (x) =−2x exp(−x 2 ),clearly a different state. This allows us to attach a meaning to an expression suchas 〈φ|A| ψ〉 through the equation〈φ|A| ψ〉 = 〈φ| ψ 1 〉, (19.2)i.e. it is the inner product of | ψ 1 〉 and | φ〉. We have already used this notation inequation (19.1), but there the effect of the operator A was merely multiplicationby a weight function.If it should happen that the effect of an operator acting upon a particular ket649

QUANTUM OPERATORSis to produce a scalar multiple of that ket, i.e.A| ψ〉 = λ| ψ〉, (19.3)then, just as for matrices and differential equations, | ψ〉 is called an eigenket or,more usually, an eigenstate of A, with corresponding eigenvalue λ; tomarkthisspecial property the state will normally be denoted by | λ〉, rather than by themore general | ψ〉. Taking the Hermitian conjugate of this ket vector eigenequationgives a bra vector equation,〈ψ|A † = λ ∗ 〈ψ|. (19.4)It should be noted that the complex conjugate of the eigenvalue appears in thisequation. Should the action of A on |ψ〉 produce an unphysical state (usuallyone whose wavefunction is identically zero, and is therefore unacceptable as aquantum-mechanical wavefunction because of the required probability interpretation)we denote the result either by 0 or by the ket vector |∅〉 according tocontext. Formally, |∅〉 can be considered as an eigenket of any operator, but onefor which the eigenvalue is always zero.If an operator A is Hermitian (A † = A) then its eigenvalues are real andthe eigenstates can be chosen to be orthogonal; this can be shown in the sameway as in chapter 17 (but using a different notation). As indicated there, thereality of their eigenvalues is one reason why Hermitian operators form thebasis of measurement in quantum mechanics; in that formulation of physics, theeigenvalues of an operator are the only possible values that can be obtained whena measurement of the physical quantity corresponding to the operator is made.Actual individual measurements must always result in real values, even if theyare combined in a complex form (x + iy or re iθ ) for final presentation or analysis,and using only Hermitian operators ensures this. The proof of the reality of theeigenvalues using the Dirac notation is given below in a worked example.In the same notation the Hermitian property of an operator A is representedby the double equality〈Aφ|ψ〉 = 〈φ|A| ψ〉 = 〈φ|Aψ〉.It should be remembered that the definition of an Hermitian operator involvesspecifying boundary conditions that the wavefunctions considered must satisfy.Typically, they are that the wavefunctions vanish for large values of the spatialvariables upon which they depend; this deals with most physical systems sincethey are nearly all formally infinite in extent. Some model systems require thewavefunction to be periodic or to vanish at finite values of a spatial variable.Depending on the nature of the physical system, the eigenvalues of a particularlinear operator may be discrete, part of a continuum, or a mixture of both. Forexample, the energy levels of the bound proton–electron system (the hydrogenatom) are discrete, but if the atom is ionised and the electron is free, the energy650

19.1 OPERATOR FORMALISMspectrum of the system is continuous. This system has discrete negative andcontinuous positive eigenvalues for the operator corresponding to the total energy(the Hamiltonian).◮Using the Dirac notation, show that the eigenvalues of an Hermitian operator are real.Let | a〉 be an eigenstate of Hermitian operator A corresponding to eigenvalue a, thenHence,⇒⇒A| a〉 = a| a〉,〈a|A| a〉 = 〈a|a| a〉 = a〈a| a〉,and〈a|A † = a ∗ 〈a|,〈a|A † | a〉 = a ∗ 〈a| a〉,〈a|A| a〉 = a ∗ 〈a| a〉, since A is Hermitian.(a − a ∗ )〈a| a〉 = 0,⇒ a = a ∗ , since 〈a| a〉 ≠0.Thus a is real. ◭It is not our intention to describe the complete axiomatic basis of quantummechanics, but rather to show what can be learned about linear operators, andin particular about their eigenvalues, without recourse to explicit wavefunctionson which the operators act.Before we proceed to do that, we close this subsection with a number of results,expressed in Dirac notation, that the reader should verify by inspection or byfollowing the lines of argument sketched in the statements. Where a sum overa complete set of eigenvalues is shown, it should be replaced by an integral forthose parts of the eigenvalue spectrum that are continuous. With the notationthat |a n 〉 is an eigenstate of Hermitian operator A with non-degenerate eigenvaluea n (or, if a n is k-fold degenerate, then a set of k mutually orthogonal eigenstateshas been constructed and the states relabelled), we have the following results.A| a n 〉 = a n | a n 〉,〈a m |a n 〉 = δ mn (orthonormality of eigenstates), (19.5)A(c n | a n 〉 + c m | a m 〉)=c n a n | a n 〉 + c m a m | a m 〉 (linearity). (19.6)The definitions of the sum and product of two operators are(A + B)| ψ〉 ≡A| ψ〉 + B| ψ〉, (19.7)AB| ψ〉 ≡A(B| ψ〉) (≠ BA| ψ〉 in general), (19.8)⇒ A p | a n 〉 = a p n| a n 〉. (19.9)651

QUANTUM OPERATORSIf A| a n 〉 = a| a n 〉 for all N 1 ≤ n ≤ N 2 ,then∑N 2| ψ〉 = d n | a n 〉 satisfies A| ψ〉 = a| ψ〉 for any set of d i .n=N 1For a general state | ψ〉,| ψ〉 =∞∑c n | a n 〉,wherec n = 〈a n |ψ〉. (19.10)n=0This can also be expressed as the operator identity,∞∑1= | a n 〉〈a n |, (19.11)in the sense thatn=0n=0∞∑∞∑| ψ〉 =1| ψ〉 = | a n 〉〈a n |ψ〉 = c n | a n 〉.It also follows that( ∞)(∑∞)∑∞∑∞∑1=〈ψ|ψ〉 = c ∗ m〈a m | c n |a n 〉 = c ∗ mc n δ mn = |c n | 2 .m=0n=0m,nn=0 (19.12)Similarly, the expectation value of the physical variable corresponding to A is∞∑∞∑〈ψ|A| ψ〉 = c ∗ m〈a m | A| a n 〉c n = c ∗ m〈a m | a n | a n 〉c nm,nm,nm,nn=0∞∑∞∑= c ∗ mc n a n δ mn = |c n | 2 a n . (19.13)n=019.1.1 Commutation and commutatorsAs has been noted above, the product AB of two linear operators may or maynot be equal to the product BA. ThatisAB| ψ〉 is not necessarily equal to BA| ψ〉.If A and B are both purely multiplicative operators, multiplication by f(r)and g(r) say, then clearly the order of the operations is immaterial, the result| f(r)g(r)ψ〉 being obtained in both cases. However, consider a case in which Ais the differential operator ∂/∂x and B is the operator ‘multiply by x’. Then thewavefunction describing AB| ψ〉 is∂(xψ(x)) = ψ(x)+x∂ψ∂x ∂x ,652

19.1 OPERATOR FORMALISMwhilstthatforBA| ψ〉 is simplywhich is not the same.If the resultx ∂ψ∂x ,AB| ψ〉 = BA| ψ〉is true for all ket vectors | ψ〉, thenA and B are said to commute; otherwise theyare non-commuting operators.A convenient way to express the commutation properties of two linear operatorsis to define their commutator, [ A, B ], by[ A, B ] | ψ〉 ≡AB| ψ〉−BA| ψ〉. (19.14)Clearly two operators that commute have a zero commutator. But, for the examplegiven above we have that[ ∂∂x ,x ]ψ(x) =(ψ(x)+x ∂ψ ) (− x ∂ψ )= ψ(x) =1× ψ∂x ∂xor, more simply, that[ ∂∂x ,x ]= 1; (19.15)in words, the commutator of the differential operator ∂/∂x and the multiplicativeoperator x is the multiplicative operator 1. It should be noted that the order ofthe linear operators is important and that[ A, B ] = − [ B,A] . (19.16)Clearly any linear operator commutes with itself and some other obvious zerocommutators (when operating on wavefunctions with ‘reasonable’ properties) are:[ A, I ] , where I is the identity operator;[ A n ,A m ] , for any positive integers n and m;[ A, p(A) ] , where p(x) is any polynomial in x;[ A, c ] , where A is any linear operator and c is any constant;[ f(x),g(x) ] , where the functions are mutiplicative;[ A(x),B(y) ] , where the operators act on different variables, with[ ] ∂∂x , ∂as a specific example.∂y653

QUANTUM OPERATORSSimple identities amongst commutators include the following:[ A, B + C ] = [ A, B ] + [ A, C ] , (19.17)[ A + B,C ] = [ A, C ] + [ B,C ] , (19.18)[ A, BC ] = ABC − BCA + BAC − BAC=(AB − BA)C + B(AC − CA)= [ A, B ] C + B [ A, C ] , (19.19)[ AB, C ] = A [ B,C ] + [ A, C ] B. (19.20)◮If A and B are two linear operators that both commute with their commutator, prove that[ A, B n ] = nB n−1 [ A, B ] and that [ A n ,B] = nA n−1 [ A, B ].Define C n by C n = [ A, B n ]. We aim to find a reduction formula for C n :C n = [ A, B B ] n−1= [ A, B ] B n−1 + B [ A, B ] n−1 , using (19.19),= B n−1 [ A, B ] + B [ A, B ] n−1 ,since[[A, B ] ,B] =0,= B n−1 [ A, B ] + BC n−1 , the required reduction formula,= B n−1 [ A, B ] + B{B n−2 [ A, B ] + BC n−2 }, applying the formula,=2B n−1 [ A, B ] + B 2 C n−2= ···= nB n−1 [ A, B ] + B n C 0 .However, C 0 = [ A, I ] =0andsoC n = nB n−1 [ A, B ].Using equation (19.16) and interchanging A and B in the result just obtained, we findas stated in the question. ◭[ A n ,B] = − [ B,A n ] = −nA n−1 [ B,A] = nA n−1 [ A, B ] ,As the power of a linear operator can be defined, so can its exponential;this situation parallels that for matrices, which are of course a particular set ofoperators that act upon state functions represented by vectors. The definitionfollows that for the exponential of a scalar or matrix, namelyexp A =∞∑n=0A nn! . (19.21)Related functions of A, suchassinA and cos A, can be defined in a similar way.Since any linear operator commutes with itself, when two functions of itare combined in some way, the result takes a form similar to that for thecorresponding functions of scalar quantities. Consider, for example, the functionf(A) defined by f(A) =2sinA cos A. Expressing sin A and cos A in terms of their654

19.1 OPERATOR FORMALISMdefining series, we havef(A) =2∞∑m=0(−1) m A 2m+1(2m +1)!∞∑n=0(−1) n A 2n.(2n)!Writing m + n as r and replacing n by s, we havewherec r =f(A) =2∞∑( r∑A 2r+1r=0 s=0∞∑=2 (−1) r c r A 2r+1 ,r∑s=0r=0(−1) r−s(2r − 2s +1)!1(2r − 2s + 1)! (2s)! = 1(2r +1)!)(−1) s(2s)!r∑2r+1 C 2s .By adding the binomial expansions of 2 2r+1 =(1+1) 2r+1 and 0 = (1 − 1) 2r+1 ,itcan easily be shown thatIt then follows that2 2r+1 =2s=0s=0r∑2r+1 2 2rC 2s ⇒ c r =(2r +1)! .2sinA cos A =2∞∑r=0(−1) r A 2r+1 2 2r(2r +1)!=∞∑r=0(−1) r (2A) 2r+1(2r +1)!= sin 2A,a not unexpected result.However, if two (or more) linear operators that do not commute are involved,combining functions of them is more complicated and the results less intuitivelyobvious. We take as a particular case the product of two exponential functionsand, even then, take the simplified case in which each linear operator commuteswith their commutator (so that we may use the results from the previous workedexample).◮If A and B are two linear operators that both commute with their commutator, show thatexp(A) exp(B) =exp(A + B + 1 [ A, B ] ).2We first find the commutator of A and exp λB, whereλ is a scalar quantity introduced for655

QUANTUM OPERATORSlater algebraic convenience:[ ] [∞∑A, eλB= A,==∞∑n=0∞∑n=1n=0(λB) nn!]∞∑=n=0λ nn! [ A, Bn ]λ nn! nBn−1 [ A, B ] , using the earlier result,λ nn! nBn−1 [ A, B ]∞∑ λ m B m= λ [ A, B ] ,writingm = n − 1,m!m=0= λe λB [ A, B ] .Now consider the derivative with respect to λ of the functionf(λ) =e λA e λB e −λ(A+B) .In the following calculation we use the fact that the derivative of e λC is Ce λC ; this is thesame as e λC C, since any two functions of the same operator commute. Differentiating thethree-factor product givesdfdλ = eλA Ae λB e −λ(A+B) + e λA e λB Be −λ(A+B) + e λA e λB (−A − B)e −λ(A+B)= e λA (e λB A + λe λB [ A, B ] )e −λ(A+B) + e λA e λB Be −λ(A+B)− e λA e λB Ae −λ(A+B) − e λA e λB Be −λ(A+B)= e λA λe λB [ A, B ] e −λ(A+B)= λ [ A, B ] f(λ).In the second line we have used the result obtained above to replace Ae λB ,andinthelastline have used the fact that [ A, B ] commutes with each of A and B, and hence with anyfunction of them.Integrating this scalar differential equation with respect to λ and noting that f(0) = 1,we obtainln f = 1 2 λ2 [ A, B ] ⇒ e λA e λB e −λ(A+B) = f(λ) =e 1 2 λ2 [ A,B ] .Finally, post-multiplying both sides of the equation by e λ(A+B) and setting λ =1yieldse A e B = e 1 2 [ A,B ]+A+B . ◭19.2 Physical examples of operatorsWe now turn to considering some of the specific linear operators that play apart in the description of physical systems. In particular, we will examine theproperties of some of those that appear in the quantum-mechanical descriptionof the physical world.As stated earlier, the operators corresponding to physical observables are restrictedto Hermitian operators (which have real eigenvalues) as this ensures thereality of predicted values for experimentally measured quantities. The two basic656

19.2 PHYSICAL EXAMPLES OF OPERATORSquantum-mechanical operators are those corresponding to position r and momentump. One prescription for making the transition from classical to quantummechanics is to express classical quantities in terms of these two variables inCartesian coordinates and then make the component by component substitutionsr → multiplicative operator r and p → differential operator − i∇.(19.22)This generates the quantum operators corresponding to the classical quantities.For the sake of completeness, we should add that if the classical quantity containsa product of factors whose corresponding operators A and B do not commute,then the operator 1 2(AB + BA) is to be substituted for the product.The substitutions (19.22) invoke operators that are closely connected with thetwo that we considered at the start of the previous subsection, namely x and∂/∂x. One,x, corresponds exactly to the x-component of the prescribed quantumposition operator; the other, however, has been multiplied by the imaginaryconstant −i, where is the Planck constant divided by 2π. This has the (subtle)effect of converting the differential operator into the x-component of an Hermitianoperator; this is easily verified using integration by parts to show that it satisfiesequation (17.16). Without the extra imaginary factor (which changes sign undercomplex conjugation) the two sides of the equation differ by a minus sign.Making the differential operator Hermitian does not change in any essentialway its commutation properties, and the commutation relation of the two basicquantum operators reads[[ p x ,x] = −i ∂ ]∂x ,x = −i. (19.23)Corresponding results hold when x is replaced, in both operators, by y or z.However, it should be noted that if different Cartesian coordinates appear in thetwo operators then the operators commute, i.e.[ p x ,y] = [ p x ,z] = [ p y ,x ] = [ p y ,z ] = [ p z ,x] = [ p z ,y] =0.(19.24)As an illustration of the substitution rules, we now construct the Hamiltonian(the quantum-mechanical energy operator) H for a particle of mass m movingin a potential V (x, y, z) when it has one of its allowed energy values, i.e itsenergy is E n ,whereH|ψ n 〉 = E n |ψ n 〉. This latter equation when expressed in aparticular coordinate system is the Schrödinger equation for the particle. In termsof position and momentum, the total classical energy of the particle is given byE = p22m + V (x, y, z) =p2 x + p 2 y + p 2 z+ V (x, y, z).2mSubstituting −i∂/∂x for p x (and similarly for p y and p z ) in the first term on the657

QUANTUM OPERATORSRHS gives(−i) 2 ∂ ∂2m ∂x ∂x + (−i)2 ∂ ∂2m ∂y ∂y + (−i)2 ∂ ∂2m ∂z ∂z .The potential energy V , being a function of position only, becomes a purelymultiplicative operator, thus creating the full expression for the Hamiltonian,( ∂2H = − 22m∂x 2 + ∂2∂y 2 + ∂2∂z 2 )+ V (x, y, z),and giving the corresponding Schrödinger equation as(Hψ n = − 2 ∂ 2 )ψ n2m ∂x 2 + ∂2 ψ n∂y 2 + ∂2 ψ n∂z 2 + V (x, y, z)ψ n = E n ψ n .We are not so much concerned in this section with solving such differentialequations, but with the commutation properties of the operators from which theyare constructed. To this end, we now turn our attention to the topic of angularmomentum, the operators for which can be constructed in a straightforwardmanner from the two basic sets.19.2.1 Angular momentum operatorsAs required by the substitution rules, we start by expressing angular momentumin terms of the classical quantities r and p, namely L = r × p with CartesiancomponentsL z = xp y − yp x , L x = yp z − zp y , L y = zp x − xp z .Making the substitutions (19.22) yields as the corresponding quantum-mechanicaloperators(L z = −i x ∂∂y − y ∂ ),∂x(L x = −i y ∂ ∂z − z ∂ ), (19.25)∂y(L y = −i z ∂∂x − x ∂ ).∂zIt should be noted that for xp y , say, x and ∂/∂y commute, and there is noambiguity about the way it is to be carried into its quantum form. Further, sincethe operators corresponding to each of its factors commute and are Hermitian,the operator corresponding to the product is Hermitian. This was shown directlyfor matrices in exercise 8.7, and can be verified using equation (17.16).The first question that arises is whether or not these three operators commute.658

19.2 PHYSICAL EXAMPLES OF OPERATORSConsider firstNow consider(L x L y = − 2 y ∂ ∂z − z ∂ )(z ∂∂y ∂x − x ∂ )∂z(= − 2 y ∂ ∂2 ∂2+ yz − yx∂x ∂z∂x ∂z 2 − ∂ 2z2 ∂2+ zx∂y∂x ∂y∂z(L y L x = − 2 z ∂∂x − x ∂ )(y ∂ ∂z ∂z − z ∂ )∂y(= − 2 zy ∂2∂x∂z − ∂ 2z2 ∂2− xy∂x∂y ∂z 2 + x ∂ ∂2+ xz∂y ∂z∂yThese two expressions are not the same. The difference between them, i.e. thecommutator of L x and L y , is given by[ ]Lx ,L y = Lx L y − L y L x = (x 2 ∂∂y − y ∂ )= iL z . (19.26)∂xThis, and two similar results obtained by permutting x, y and z cyclically,summarise the commutation relationships between the quantum operators correspondingto the three Cartesian components of angular momentum:[ ]Lx ,L y = iLz ,[ ]Ly ,L z = iLx , (19.27)[ L z ,L x ] = iL y .As well as its separate components of angular momentum, the total angularmomentum associated with a particular state |ψ〉 is a physical quantity of interest.This is measured by the operator corresponding to the sum of squares of itscomponents,).).L 2 = L 2 x + L 2 y + L 2 z. (19.28)This is an Hermitian operator, as each term in it is the product of two Hermitianoperators that (trivially) commute. It might seem natural to want to ‘take thesquare root’ of this operator, but such a process is undefined and we will notpursue the matter.We next show that, although no two of its components commute, the totalangular momentum operator does commute with each of its components. In theproof we use some of the properties (19.17) to (19.20) and result (19.27). We begin659

QUANTUM OPERATORSwith[L 2 ,L z]=[L2x + L 2 y + L 2 z,L z]= L x [ L x ,L z ] + [ L x ,L z ] L x[ ] [ ]+ L y Ly ,L z + Ly ,L z Ly + [ L 2 ]z,L z= L x (−i)L y +(−i)L y L x + L y (i)L x +(i)L x L y +0=0.Thus operators L 2 and L z commute and, continuing in the same way, it can beshown that[L 2 ,L x]=[L 2 ,L y]=[L 2 ,L z]=0. (19.29)Eigenvalues of the angular momentum operatorsWe will now use the commutation relations for L 2 and its components to findthe eigenvalues of L 2 and L z , without reference to any specific wavefunction. Inother words, the eigenvalues of the operators follow from the structure of theircommutators. There is nothing particular about L z ,andL x or L y could equallywell have been chosen, though, in general, it is not possible to find states that aresimultaneously eigenstates of two or more of L x , L y and L z .To help with the calculation, it is convenient to define the two operatorsU ≡ L x + iL y and D ≡ L x − iL y .These operators are not Hermitian; they are in fact Hermitian conjugates, in thatU † = D and D † = U, but they do not represent measurable physical quantities.We first note their multiplication and commutation properties:UD =(L x + iL y )(L x − iL y )=L 2 x + L 2 y + i [ ]L y ,L x= L 2 − L 2 z + L z , (19.30)DU =(L x − iL y )(L x + iL y )=L 2 x + L2 y − i [ ]L y ,L x= L 2 − L 2 z − L z , (19.31)[ L z ,U] = [ L z ,L x ] + i [ ]L z ,L y = iLy + L x = U, (19.32)[ L z ,D] = [ L z ,L x ] − i [ ]L z ,L y = iLy − L x = −D. (19.33)In the same way as was shown for matrices, it can be demonstrated that if twooperators commute they have a common set of eigenstates. Since L 2 and L zcommute they possess such a set; let one of the set be |ψ〉 withL 2 |ψ〉 = a|ψ〉 and L z |ψ〉 = b|ψ〉.Now consider the state |ψ ′ 〉 = U|ψ〉 and the actions of L 2 and L z upon it.660

19.2 PHYSICAL EXAMPLES OF OPERATORSConsider first L 2 |ψ ′ 〉, recalling that L 2 commutes with both L x and L y and hencewith U:L 2 |ψ ′ 〉 = L 2 U|ψ〉 = UL 2 |ψ〉 = Ua|ψ〉 = aU|ψ〉 = a|ψ ′ 〉.Thus, |ψ ′ 〉 is also an eigenstate of L 2 , corresponding to the same eigenvalue as|ψ〉. Now consider the action of L z :L z |ψ ′ 〉 = L z U|ψ〉=(UL z + U)|ψ〉, using[ L z ,U] = U,= Ub|ψ〉 + U|ψ〉=(b + )U|ψ〉=(b + )|ψ ′ 〉.Thus, |ψ ′ 〉 is also an eigenstate of L z , but with eigenvalue b + .In summary, the effect of U acting upon |ψ〉 is to produce a new state thathas the same eigenvalue for L 2 and is still an eigenstate of L z , though with thateigenvalue increased by . An exactly analogous calculation shows that the effectof D acting upon |ψ〉 is to produce another new state, one that also has the sameeigenvalue for L 2 and is also still an eigenstate of L z , though with the eigenvaluedecreased by in this case. For these reasons, U and D are usually known asladder operators.It is clear that, by starting from any arbitrary eigenstate and repeatedly applyingeither U or D, we could generate a series of eigenstates, all of which have theeigenvalue a for L 2 , but increment in their L z eigenvalues by ±. However, wealso have the physical requirement that, for real values of the z-component, itssquare cannot exceed the square of the total angular momentum, i.e. b 2 ≤ a. Thusb has a maximum value c that satisfiesc 2 ≤ a but (c + ) 2 >a;let the corresponding eigenstate be |ψ u 〉 with L z |ψ u 〉 = c|ψ u 〉. Now it is still truethatL z U|ψ u 〉 =(c + )U|ψ u 〉,and, to make this compatible with the physical constraint, we must have thatU|ψ u 〉 is the zero ket vector |∅〉. Now, using result (19.31), we haveDU|ψ u 〉 =(L 2 − L 2 z − L z )|ψ u 〉,⇒ 0|∅〉= D|∅〉=(a 2 − c 2 − c)|ψ u 〉,⇒ a = c(c + ).This gives the relationship between a and c. We now establish the possible formsfor c.If we start with eigenstate |ψ u 〉, which has the highest eigenvalue c for L z ,and661

QUANTUM OPERATORSoperate repeatedly on it with the (down) ladder operator D, we will generate astate |ψ d 〉 which, whilst still an eigenstate of L 2 with eigenvalue a, hasthelowestphysically possible value, d say, for the eigenvalue of L z . If this happens after noperations we will have that d = c − n andL z |ψ d 〉 =(c − n)|ψ d 〉.Arguing in the same way as previously that D|ψ d 〉 must be an unphysical ketvector, we conclude that0|∅〉= U|∅〉= UD|ψ d 〉=(L 2 − L 2 z + L z )|ψ d 〉, using (19.30),=[a − (c − n) 2 + (c − n)]|ψ d 〉⇒ a =(c − n) 2 − (c − n).Equating the two results for a givesc 2 + c = c 2 − 2cn + n 2 2 − c + n 2 ,2c(n +1)=n(n +1),c = 1 2 n.Since n is necessarily integral, c is an integer multiple of 1 2. This result is validirrespective of which eigenstate |ψ〉 we started with, though the actual value ofthe integer n depends on |ψ u 〉 and hence upon |ψ〉.Denoting 1 2n by l we can say that the possible eigenvalues of the operatorL z , and hence the possible results of a measurement of the z-component of theangular momentum of a system, are given byl, (l − 1), (l − 2), ... , −l.The value of a for all 2l + 1 of the corresponding states,|ψ u 〉, D|ψ u 〉, D 2 |ψ u 〉, ... , D 2l |ψ u 〉,is l(l +1) 2 .The similarity of form between this eigenvalue and that appearing in Legendre’sequation is not an accident. It is intimately connected with the facts (i) that L 2is a measure of the rotational kinetic energy of a particle in a system centredon the origin, and (ii) that in spherical polar coordinates L 2 has the same formas the angle-dependent part of ∇ 2 , which, as we have seen, is itself proportionalto the quantum-mechanical kinetic energy operator. Legendre’s equation andthe associated Legendre equation arise naturally when ∇ 2 ψ = f(r) issolvedinspherical polar coordinates using the method of separation of variables discussedin chapter 21.The derivation of the eigenvalues l(l+1) 2 and m, with −l ≤ m ≤ l, dependsonly on the commutation relationships between the corresponding operators. Any662

19.2 PHYSICAL EXAMPLES OF OPERATORSother set of four operators with the same commutation structure would resultin the same eigenvalue spectrum. In fact, quantum mechanically, orbital angularmomentum is restricted to cases in which n is even and so l is an integer; thisis in accord with the requirement placed on l if solutions to ∇ 2 ψ = f(r) thatarefinite on the polar axis are to be obtained. The non-classical notion of internalangular momentum (spin) for a particle provides a set of operators that are ableto take both integral and half-integral multiples of as their eigenvalues.We have already seen that, for a state |l, m〉 that has a z-component ofangular momentum m, the state U|l, m〉 is one with its z-component of angularmomentum equal to (m +1). But the new state ket vector so produced is notnecessarily normalised so as to make 〈l, m +1| l, m +1〉 = 1. We will concludethis discussion of angular momentum by calculating the coefficients µ m and ν m inthe equationsU|l, m〉 = µ m |l, m +1〉 and D|l, m〉 = ν m |l, m − 1〉on the basis that 〈l, r | l, r〉 = 1 for all l and r.To do so, we consider the inner product I = 〈l, m |DU| l, m〉, evaluated in twodifferent ways. We have already noted that U and D are Hermitian conjugatesand so I canbewrittenasI = 〈l, m |U † U| l, m〉 = µ ∗ m〈l, m | l, m〉µ m = |µ m | 2 .But, using equation (19.31), it can also be expressed asThus we are required to haveI = 〈l, m |L 2 − L 2 z − L z | l, m〉= 〈l, m |l(l +1) 2 − m 2 2 − m 2 | l, m〉=[l(l +1) 2 − m 2 2 − m 2 ] 〈l, m | l, m〉=[l(l +1)− m(m +1)] 2 .|µ m | 2 =[l(l +1)− m(m +1)] 2 ,but can choose that all µ m are real and non-negative (recall that |m| ≤l). Asimilar calculation can be used to calculate ν m . The results are summarised in theequationsU| l, m〉 = √ l(l +1)− m(m +1)| l, m +1〉, (19.34)D| l, m〉 = √ l(l +1)− m(m − 1) | l, m − 1〉. (19.35)It can easily be checked that U|l, l〉 = |∅〉= D|l, −l〉.663

QUANTUM OPERATORS19.2.2 Uncertainty principlesThe next topic we explore is the quantitative consequences of a non-zero commutatorfor two quantum (Hermitian) operators that correspond to physicalvariables.As previously noted, the expectation value in a state |ψ〉 of the physical quantityA corresponding to the operator A is E[A] =〈ψ| A |ψ〉. Any one measurement ofA can only yield one of the eigenvalues of A. But if repeated measurements couldbe made on a large number of identical systems, a discrete or continuous rangeof values would be obtained. It is a natural extension of normal data analysisto measure the uncertainty in the value of A by the observed variance in themeasured values of A, denoted by (∆A) 2 and calculated as the average value of(A − E[A]) 2 . The expected value of this variance for the state |ψ〉 is given by〈ψ |(A − E[A]) 2 |ψ〉.We now give a mathematical proof that there is a theoretical lower limit forthe product of the uncertainties in any two physical quantities, and we start byproving a result similar to the Schwarz inequality. Let |u〉 and |v〉 be any two statevectors and let λ be any real scalar. Then consider the vector |w〉 = |u〉 + λ|v〉 and,in particular, note that0 ≤〈w | w〉 = 〈u | u〉 + λ(〈u | v〉 + 〈v | u〉)+λ 2 〈v | v〉.This is a quadratic inequality in λ and therefore the quadratic equation formedby equating the RHS to zero must have no real roots. The coefficient of λ is(〈u | v〉 + 〈v | u〉) =2Re〈u | v〉 and its square is thus ≥ 0. The condition for no realroots of the quadratic is therefore0 ≤ (〈u | v〉 + 〈v | u〉) 2 ≤ 4〈u | u〉〈v | v〉. (19.36)This result will now be applied to state vectors constructed from |ψ〉, the statevector of the particular system for which we wish to establish a relationship betweenthe uncertainties in the two physical variables corresponding to (Hermitian)operators A and B. WetakeThenFurther,|u〉 =(A − E[A]) | ψ〉 and |v〉 = i(B − E[B]) | ψ〉. (19.37)〈u | u〉 = 〈ψ |(A − E[A]) 2 |ψ〉 =(∆A) 2 ,〈v | v〉 = 〈ψ |(B − E[B]) 2 |ψ〉 =(∆B) 2 .〈u | v〉 = 〈ψ | (A − E[A])i(B − E[B]) | ψ〉= i〈ψ | AB | ψ〉−iE[A]〈ψ | B | ψ〉−iE[B]〈ψ | A | ψ〉 + iE[A]E[B]〈ψ | ψ〉= i〈ψ | AB | ψ〉−iE[A]E[B].664

19.2 PHYSICAL EXAMPLES OF OPERATORSIn the second line, we have moved expectation values, which are purely numbers,out of the inner products and used the normalisation condition 〈ψ|ψ〉 = 1.Similarly〈v | u〉 = −i〈ψ | BA| ψ〉 + iE[A]E[B].Adding these two results givesand substitution into (19.36) yields〈u | v〉 + 〈v | u〉 = i〈ψ | AB − BA|ψ〉,0 ≤ (i〈ψ | AB − BA|ψ〉) 2 ≤ 4(∆A) 2 (∆B) 2At first sight, the middle term of this inequality might appear to be negative, butthis is not so. Since A and B are Hermitian, AB −BA is anti-Hermitian, as is easilydemonstrated. Since i is also anti-Hermitian, the quantity in the parentheses inthe middle term is real and its square non-negative. Rearranging the equation andexpressing it in terms of the commutator of A and B gives the generalised formof the Uncertainty Principle. For any particular state |ψ〉 of a system, this providesthe quantitative relationship between the minimum value that the product of theuncertainties in A and B can have and the expectation value, in that state, oftheir commutator,Immediate observations include the following:(∆A) 2 (∆B) 2 ≥ 1 4 |〈ψ | [ A, B ] | ψ〉|2 . (19.38)(i) If A and B commute there is no absolute restriction on the accuracy withwhich the corresponding physical quantities may be known. That is not tosay that ∆A and ∆B will always be zero, only that they may be.(ii) If the commutator of A and B is a constant, k ≠ 0, then the RHS ofequation (19.38) is necessarily equal to 1 4 |k|2 , whatever the form of |ψ〉,and it is not possible to have ∆A =∆B =0.(iii) Since the RHS depends upon |ψ〉, it is possible, even for two operatorsthat do not commute, for the lower limit of (∆A) 2 (∆B) 2 to be zero. Thiswill occur if the commutator [ A, B ] is itself an operator whose expectationvalue in the particular state |ψ〉 happens to be zero.To illustrate the third of these, we might consider the components of angularmomentum discussed in the previous subsection. There, in equation (19.27), wefound that the commutator of the operators corresponding to the x- and y-components of angular momentum is non-zero; in fact, it has the value iL z .This means that if the state |ψ〉 of a system happened to be such that 〈ψ|L z |ψ〉 =0,as it would if, for example, it were the eigenstate of L z , |ψ〉 = |l, 0〉, then therewould be no fundamental reason why the physical values of both L x and L yshould not be known exactly. Indeed, if the state were spherically symmetric, and665

QUANTUM OPERATORShence formally an eigenstate of L 2 with l = 0, all three components of angularmomentum could be (and are) known to be zero.◮Working in one dimension, show that the minimum value of the product ∆p x × ∆x foraparticleis 1 . Find the form of the wavefunction that attains this minimum value for a2particle whose expectation values for position and momentum are ¯x and ¯p, respectively.We have already seen, in (19.23) that the commutator of p x and x is −i, aconstant.Therefore, irrespective of the actual form of |ψ〉, the RHS of (19.38) is 1 4 2 (see observation(ii) above). Thus, since all quantities are positive, taking the square roots of both sides ofthe equation shows directly that∆p x × ∆x ≥ 1 . 2Returning to the derivation of the Uncertainty Principle, we see that the inequality becomesan equality only when(〈u | v〉 + 〈v | u〉) 2 =4〈u | u〉〈v | v〉.The RHS of this equality has the value 4||u|| 2 ||v|| 2 and so, by virtue of Schwarz’s inequality,we have4‖u‖ 2 ‖v‖ 2 =(〈u | v〉 + 〈v | u〉) 2≤ (|〈u | v〉| + |〈v | u〉|) 2≤ (‖u‖‖v|| + ‖v‖‖u‖) 2=4‖u‖ 2 ‖v‖ 2 .Since the LHS is less than or equal to something that has the same value as itself, all ofthe inequalities are, in fact, equalities. Thus 〈u|v〉 = ‖u‖‖v‖, showing that |u〉 and |v〉 areparallel vectors, i.e. |u〉 = µ|v〉 for some scalar µ.We now transform this condition into a constraint that the wavefunction ψ = ψ(x)must satisfy. Recalling the definitions (19.37) of |u〉 and |v〉 in terms of |ψ〉, we have(−i d )dx − ¯p ψ = µi(x − ¯x)ψ,dψ[ µ(x − ¯x) − i¯p ]ψ =0.dx + 1 [ µ(x − ¯x)2The IF for this equation is exp− i¯px ], giving2 { [d µ(x − ¯x)2ψ expdx2− i¯px ]}=0,which, in turn, leads to[ ] ( )µ(x − ¯x)2 i¯pxψ(x) =A exp − exp .2From this it is apparent that the minimum uncertainty product ∆p x × ∆x is obtained whenthe probability density |ψ(x)| 2 has the form of a Gaussian distribution centred on ¯x. Thevalue of µ is not fixed by this consideration and it could be anything (positive); a largevalue for µ would yield a small value for ∆x but a correspondingly large one for ∆p x . ◭666

19.2 PHYSICAL EXAMPLES OF OPERATORS19.2.3 Annihilation and creation operatorsAs a final illustration of the use of operator methods in physics we consider theirapplication to the quantum mechanics of a simple harmonic oscillator (s.h.o.).Although we will start with the conventional description of a one-dimensionaloscillator, using its position and momentum, we will recast the description interms of two operators and their commutator and show that many importantconclusions can be reached from studying these alone.The Hamiltonian for a particle of mass m with momentum p moving in aone-dimensional parabolic potential V (x) = 1 2 kx2 isH = p22m + 1 2 kx2 = p22m + 1 2 mω2 x 2 ,where its classical frequency of oscillation ω is given by ω 2 = k/m. We recall thatthe corresponding operators, p and x, do not commute and that [ p, x ] = −i.In analogy with the ladder operators used when discussing angular momentum,we define two new operators:√ mω2 x + ip √2mωand A † ≡√ mω2 x − ip √2mω.A ≡(19.39)Since both x and p are Hermitian, A and A † are Hermitian conjugates, thoughneither is Hermitian and they do not represent physical quantities that can bemeasured.Now consider the two products A † A and AA † :A † A = mω 2x2 − ipx2 + ixp2 + p22mω = H ω − i 2 [ p, x ] = H ω − 2 ,AA † = mω 2x2 + ipx2 − ixp2 + p22mω = H ω + i 2 [ p, x ] = H ω + 2 .From these it follows thatand thatFurther,Similarly,H = 1 2 ω(A† A + AA † ) (19.40)[A, A† ] = . (19.41)[ H,A] = [ 12 ω(A† A + AA † ),A ]= 1 2 ω ( A † 0+ [ A † ,A ] A + A [ A † ,A ] +0A †)= 1 2ω(−A − A) =−ωA. (19.42)[H,A† ] = ωA † (19.43).Before we apply these relationships to the question of the energy spectrum ofthe s.h.o., we need to prove one further result. This is that if B is an Hermitianoperator then 〈ψ | B 2 | ψ〉 ≥0forany |ψ〉. The proof, which involves introducing667

QUANTUM OPERATORSan arbitrary complete set of orthonormal base states |φ i 〉 and using equation(19.11), is as follows:〈ψ | B 2 | ψ〉 = 〈ψ | B × 1 × B | ψ 〉= ∑ i= ∑ i= ∑ i= ∑ i= ∑ i〈ψ | B |φ i 〉〈φ i | B |ψ〉〈ψ | B |φ i 〉 ( 〈φ i | B |ψ〉 ∗) ∗〈ψ | B |φ i 〉 ( 〈ψ | B † |φ i 〉 ) ∗〈ψ | B |φ i 〉〈ψ | B |φ i 〉 ∗ , since B is Hermitian,|〈ψ | B |φ i 〉| 2 ≥ 0.We note, for future reference, that the Hamiltonian H for the s.h.o. is the sum oftwo terms each of this form and therefore conclude that 〈ψ|H|ψ〉 ≥0 for all |ψ〉.The energy spectrum of the simple harmonic oscillatorLet the normalised ket vector |n〉 (or |E n 〉) denote the nth energy state of the s.h.o.with energy E n . Then it must be an eigenstate of the (Hermitian) Hamiltonian Hand satisfyH|n〉 = E n |n〉 with 〈m|n〉 = δ mn .Now consider the state A|n〉 and the effect of H upon it:HA|n〉 = AH|n〉−ωA|n〉, using (19.42),= AE n |n〉−ωA|n〉=(E n − ω)A|n〉.Thus A|n〉 is an eigenstate of H corresponding to energy E n − ω and must besome multiple of the normalised ket vector |E n − ω〉, i.e.A| E n 〉≡A|n〉 = c n |E n − ω〉,where c n is not necessarily of unit modulus. Clearly, A is an operator thatgenerates a new state that is lower in energy by ω; it can thus be compared tothe operator D, which has a similar effect in the context of the z-component ofangular momentum. Because it possesses the property of reducing the energy ofthe state by ω, which, as we will see, is one quantum of excitation energy for theoscillator, the operator A is called an annihilation operator. Repeated applicationof A, m times say, will produce a state whose energy is mω lower than that ofthe original:A m |E n 〉 = c n c n−1 ···c n−m+1 |E n − mω〉. (19.44)668

19.2 PHYSICAL EXAMPLES OF OPERATORSIn a similar way it can be shown that A † parallels the operator U of our angularmomentum discussion and creates an additional quantum of energy each time itis applied:(A † ) m |E n 〉 = d n d n+1 ···d n+m−1 |E n + mω〉. (19.45)It is therefore known as a creation operator.As noted earlier, the expectation value of the oscillator’s energy operator〈ψ|H|ψ〉 must be non-negative, and therefore it must have a lowest value. Letthis be E 0 , with corresponding eigenstate |0〉. Since the energy-lowering propertyof A applies to any eigenstate of H, in order to avoid a contradiction we musthave that A|0〉 = |∅〉. It then follows from (19.40) thatH|0〉 = 1 2 ω(A† A + AA † )|0〉= 1 2 ωA† A|0〉 + 1 2 ω(A† A + )|0〉, using (19.41),=0+0+ 1 2ω|0〉. (19.46)This shows that the commutator structure of the operators and the form of theHamiltonian imply that the lowest energy (its ground-state energy) is 1 2ω; thisis a result that has been derived without explicit reference to the correspondingwavefunction. This non-zero lowest value for the energy, known as the zero-pointenergy of the oscillator, and the discrete values for the allowed energy statesare quantum-mechanical in origin; classically such an oscillator could have anynon-negative energy, including zero.Working back from this result, we see that the energy levels of the s.h.o. are12 ω, 32 ω, 52 ω,... ,(m + 1 2)ω,..., and that the corresponding (unnormalised)ket vectors can be written as|0〉, A † |0〉, (A † ) 2 |0〉, ... , (A † ) m |0〉, ... .This notation, and elaborations of it, are often used in the quantum treatment ofclassical fields such as the electromagnetic field. Thus, as the reader should verify,A(A † ) 3 A 2 A † A(A † ) 4 |0〉 is a state with energy 9 2 ω, whilst A(A† ) 3 A 5 A † A(A † ) 4 |0〉 isnot a physical state at all.The normalisation of the eigenstatesIn order to make quantitative calculations using the previous results we need toestablish the values of the c n and d n that appear in equations (19.44) and (19.45).To do this, we first establish the operator recurrence relationA m (A † ) m = A m−1 (A † ) m A + mA m−1 (A † ) m−1 . (19.47)669

QUANTUM OPERATORSThe proof, which makes repeated use of [ A, A † ] = , is as follows:A m (A † ) m = A m−1 AA † (A † ) m−1= A m−1 (A † A + )(A † ) m−1= A m−1 A † A(A † ) m−1 + A m−1 (A † ) m−1= A m−1 A † (A † A + )(A † ) m−2 + A m−1 (A † ) m−1= A m−1 (A † ) 2 A(A † ) m−2 + A m−1 A † (A † ) m−2 + A m−1 (A † ) m−1= A m−1 (A † ) 2 (A † A + )(A † ) m−3 +2A m−1 (A † ) m−1..= A m−1 (A † ) m A + mA m−1 (A † ) m−1 .Now we take the expectation values in the ground state |0〉 of both sides ofthis operator equation and note that the first term on the RHS is zero since itcontains the term A|0〉. The non-vanishing terms are〈0 | A m (A † ) m | 0〉 = m〈0 | A m−1 (A † ) m−1 | 0〉.The LHS is the square of the norm of (A † ) m |0〉, and, from equation (19.45), it isequal to|d 0 | 2 |d 1 | 2 ···|d m−1 | 2 〈0 | 0〉.Similarly, the RHS is equal tom |d 0 | 2 |d 1 | 2 ···|d m−2 | 2 〈0 | 0〉.It follows that |d m−1 | 2 = m and, taking all coefficients as real, d m = √ (m +1).Thus the correctly normalised state of energy (n + 1 2), obtained by repeatedapplication of A † to the ground state, is given by|n〉 =(A† ) n|0〉. (19.48)(n! n )1/2To evaluate the c n , we note that, from the commutator of A and A † ,[A, A† ] |n〉 = AA † |n〉−A † A|n〉 |n〉 = √ (n +1) A |n +1〉−c n A † |n − 1〉= √ √(n +1) c n+1 |n〉−c n n |n〉, = √ √(n +1) c n+1 − c n n,which has the obvious solution c n = √ n. To summarise:c n = √ n and d n = √ (n +1). (19.49)We end this chapter with another worked example. This one illustrates howthe operator formalism that we have developed can be used to obtain results670

19.3 EXERCISESthat would involve a number of non-trivial integrals if tackled using explicitwavefunctions.◮Given that the first-order change in the ground-state energy of a quantum system whenit is perturbed by a small additional term H ′ in the Hamiltonian is 〈0|H ′ |0〉, find the firstorderchange in the energy of a simple harmonic oscillator in the presence of an additionalpotential V ′ (x) =λx 3 + µx 4 .From the definitions of A and A † , equation (19.39), we can writex = √ 1 (A + A † ) ⇒ H ′ λ=2mω (2mω) (A + 3/2 A† ) 3 µ+(2mω) (A + 2 A† ) 4 .We now compute successive values of (A + A † ) n |0〉 for n =1, 2, 3, 4, remembering thatA |n〉 = √ n |n − 1〉 and A † |n〉 = √ (n +1) |n +1〉 :(A + A † ) |0〉 =0+ 1/2 |1〉,(A + A † ) 2 |0〉 = |0〉 + √ 2 |2〉,(A + A † ) 3 |0〉 =0+ 3/2 |1〉 +2 3/2 |1〉 + √ 6 3/2 |3〉=3 3/2 |1〉 + √ 6 3/2 |3〉,(A + A † ) 4 |0〉 =3 2 |0〉 + √ 18 2 |2〉 + √ 18 2 |2〉 + √ 24 2 |4〉.To find the energy shift we need to form the inner product of each of these state vectorswith |0〉. But|0〉 is orthogonal to all |n〉 if n ≠ 0. Consequently, the term 〈0| (A + A † ) 3 |0〉in the expectation value is zero, and in the expression for 〈0| (A + A † ) 4 |0〉 only the firstterm is non-zero; its value is 3 2 . The perturbation energy is thus given by〈0 | H ′ | 0〉 = 3µ2(2mω) . 2It could have been anticipated on symmetry grounds that the expectation of λx 3 , anodd function of x, would be zero, but the calculation gives this result automatically. Thecontribution of the quadratic term in the perturbation would have been much harder toanticipate! ◭19.3 Exercises19.1 Show that the commutator of two operators that correspond to two physicalobservables cannot itself correspond to another physical observable.19.2 By expressing the operator L z , corresponding to the z-component of angularmomentum, in spherical polar coordinates (r, θ, φ), show that the angular momentumof a particle about the polar axis cannot be known at the same time asits azimuthal position around that axis.19.3 In quantum mechanics, the time dependence of the state function |ψ〉 of a systemis given, as a further postulate, by the equationi ∂ |ψ〉 = H|ψ〉,∂twhere H is the Hamiltonian of the system. Use this to find the time dependenceof the expectation value 〈A〉 of an operator A that itself has no explicit timedependence. Hence show that operators that commute with the Hamiltoniancorrespond to the classical ‘constants of the motion’.671

QUANTUM OPERATORSFor a particle of mass m moving in a one-dimensional potential V (x), proveEhrenfest’s theorem:〈 〉d〈p x 〉 dVd〈x〉= −and = 〈p x〉dt dxdt m .19.4 Show that the Pauli matricesS x = 1 2 (0 11 0), S y = 1 2 (0 −ii 0)( )1 0, S z = 1 ,2 0 −1which are used as the operators corresponding to intrinsic spin of 1 in nonrelativisticquantum mechanics, satisfy S 2 x = S 2 y = S 2 z = 1 4 2 I, and have the2same commutation properties as the components of orbital angular momentum.Deduce that any state |ψ〉 represented by the column vector (a, b) T is an eigenstateof S 2 with eigenvalue 3 2 /4.19.5 Find closed-form expressions for cos C and sin C, whereC is the matrixC =(1 11 −1Demonstrate that the ‘expected’ relationshipscos 2 C +sin 2 C = I and sin 2C =2sinC cos Care valid.19.6 Operators A and B anticommute. Evaluate (A + B) 2n for a few values of n andhence propose an expression for c nr in the expansionn∑(A + B) 2n = c nr A 2n−2r B 2r .r=0Prove your proposed formula for general values of n, using the method ofinduction.Show that∞∑ n∑cos(A + B) = d nr A 2n−2r B 2r ,n=0 r=0where the d nr are constants whose values ( you should ) determine.0 1By taking as A the matrix A =, confirm that your answer is1 0consistent with that obtained in exercise 19.5.19.7 Expressed in terms of the annihilation and creation operators A and A † discussedin the text, a system has an unperturbed Hamiltonian H 0 = ωA † A. The systemis disturbed by the addition of a perturbing Hamiltonian H 1 = gω(A + A † ),where g is real. Show that the effect of the perturbation is to move the wholeenergy spectrum of the system down by g 2 ω.19.8 For a system of N electrons in their ground state |0〉, the Hamiltonian isN∑ p 2 xH =n+ p 2 y n+ p 2 N∑z n+ V (x n ,y n ,z n ).2mn=1n=1Show that [ ]p 2 x n,x n = −2ipxn , and hence that the expectation value of thedouble commutator [[x, H ] ,x], wherex = ∑ Nn=1 x n,isgivenby〈0 | [[x, H ] ,x] | 0〉 = N2m .672).

19.3 EXERCISESNow evaluate the expectation value using the eigenvalue properties of H, namelyH|r〉 = E r |r〉, and deduce the sum rule for oscillation strengths,∞∑r=0(E r − E 0 )|〈r | x | 0〉| 2 = N22m .19.9 By considering the functionF(λ) =exp(λA)B exp(−λA),where A and B are linear operators and λ is a parameter, and finding itsderivatives with respect to λ, prove thate A Be −A = B + [ A, B ] + 1 2! [ A, [ A, B ]]+ 1 [ A, [ A, [ A, B ]]]+ ··· .3!Use this result to express( ) ( )iLx θ−iLx θexp L y expas a linear combination of the angular momentum operators L x , L y and L z .19.10 For a system containing more than one particle, the total angular momentum Jand its components are represented by operators that have completely analogouscommutation relations to those for the operators for a single particle, i.e. J 2 haseigenvalue j(j +1) 2 and J z has eigenvalue m j for the state |j,m j 〉. The usualorthonormality relationship 〈j ′ ,m ′ j | j,m j〉 = δ j ′ j δ m ′jm jis also valid.A system consists of two (distinguishable) particles A and B. ParticleA is inan l = 3 state and can have state functions of the form |A, 3,m A 〉, whilst B isin an l = 2 state with possible state functions |B,2,m B 〉. The range of possiblevalues for j is |3 − 2| ≤j ≤|3+2|, i.e.1≤ j ≤ 5, and the overall state functioncanbewrittenas|j,m j 〉 =∑m A m B| A, 3,m A 〉|B,2,m B 〉.m A +m B =m jC jm jThe numerical coefficients C jm jm A m Bare known as Clebsch–Gordon coefficients.Assume (as can be shown) that the ladder operators U(AB) andD(AB) forthe system can be written as U(A) +U(B) andD(A) +D(B), respectively, andthat they lead to relationships equivalent to (19.34) and (19.35) with l replacedby j and m by m j .(a) Apply the operators to the (obvious) relationship|AB, 5, 5〉 = |A, 3, 3〉|B,2, 2〉to show that√√6|AB, 5, 4〉 = |A, 3, 2〉|B,2, 2〉 + 4|A, 3, 3〉|B,2, 1〉.10 10(b) Find, to within an overall sign, the real coefficients c and d in the expansion|AB, 4, 4〉 = c|A, 3, 2〉|B,2, 2〉 + d|A, 3, 3〉|B,2, 1〉by requiring it to be orthogonal to |AB, 5, 4〉. Check your answer by consideringU(AB)|AB, 4, 4〉.(c) Find, to within an overall sign, and as efficiently as possible, an expressionfor |AB, 4, −3〉 as a sum of products of the form |A, 3,m A 〉|B,2,m B 〉.673

QUANTUM OPERATORS19.4 Hints and answers19.1 Show that the commutator is anti-Hermitian.19.3 Use the Hermitian conjugate of the given equation to obtain the time dependenceof 〈ψ|. The rate of change of 〈ψ|A|ψ〉 is i〈ψ| [ H,A] |ψ〉. Note that [ H,p x ] =[ V,p x ] and [ H,x] = [ p 2 x,x ] /2m.19.5 Show that C 2 =2I.cos C =cos √ ( )1 02, sin C = sin √ ( )2 1 1√ .0 12 1 −119.7 Express the total Hamiltonian in terms of B = A + gI and determine the valueof [ B,B ] † .19.9 Show that, if F (n) is the nth derivative of F(λ), then F (n+1) = [ A, F ] (n) . Use a Taylorseries in λ to evaluate F(1), using derivatives evaluated at λ = 0. Successivelyreduce the level of nesting of each multiple commutator by using the result ofevaluating the previous term. The given expression reduces to cos θL y − sin θL z .674

20Partial differential equations:general and particular solutionsIn this chapter and the next the solution of differential equations of typestypically encountered in the physical sciences and engineering is extended tosituations involving more than one independent variable. A partial differentialequation (PDE) is an equation relating an unknown function (the dependentvariable) of two or more variables to its partial derivatives with respect tothose variables. The most commonly occurring independent variables are thosedescribing position and time, and so we will couch our discussion and examplesin notation appropriate to them.As in other chapters we will focus our attention on the equations that arisemost often in physical situations. We will restrict our discussion, therefore, tolinear PDEs, i.e. those of first degree in the dependent variable. Furthermore, wewill discuss primarily second-order equations. The solution of first-order PDEswill necessarily be involved in treating these, and some of the methods discussedcan be extended without difficulty to third- and higher-order equations. We shallalso see that many ideas developed for ordinary differential equations (ODEs)can be carried over directly into the study of PDEs.In this chapter we will concentrate on general solutions of PDEs in termsof arbitrary functions and the particular solutions that may be derived fromthem in the presence of boundary conditions. We also discuss the existence anduniqueness of the solutions to PDEs under given boundary conditions.In the next chapter the methods most commonly used in practice for obtainingsolutions to PDEs subject to given boundary conditions will be considered. Thesemethods include the separation of variables, integral transforms and Green’sfunctions. This division of material is rather arbitrary and has been made onlyto emphasise the general usefulness of the latter methods. In particular, it willbe readily apparent that some of the results of the present chapter are infact solutions in the form of separated variables, but arrived at by a differentapproach.675

PDES: GENERAL AND PARTICULAR SOLUTIONS20.1 Important partial differential equationsMost of the important PDEs of physics are second-order and linear. In order togain familiarity with their general form, some of the more important ones willnow be briefly discussed. These equations apply to a wide variety of differentphysical systems.Since, in general, the PDEs listed below describe three-dimensional situations,the independent variables are r and t, wherer is the position vector and t istime. The actual variables used to specify the position vector r are dictated by thecoordinate system in use. For example, in Cartesian coordinates the independentvariables of position are x, y and z, whereas in spherical polar coordinates theyare r, θ and φ. The equations may be written in a coordinate-independent manner,however, by the use of the Laplacian operator ∇ 2 .The wave equation20.1.1 The wave equation∇ 2 u = 1 ∂ 2 uc 2 ∂t 2 (20.1)describes as a function of position and time the displacement from equilibrium,u(r,t), of a vibrating string or membrane or a vibrating solid, gas or liquid. Theequation also occurs in electromagnetism, where u may be a component of theelectric or magnetic field in an elecromagnetic wave or the current or voltagealong a transmission line. The quantity c is the speed of propagation of the waves.◮ Find the equation satisfied by small transverse displacements u(x, t) of a uniform string ofmass per unit length ρ held under a uniform tension T , assuming that the string is initiallylocated along the x-axis in a Cartesian coordinate system.Figure 20.1 shows the forces acting on an elemental length ∆s of the string. If the tensionT in the string is uniform along its length then the net upward vertical force on theelement is∆F = T sin θ 2 − T sin θ 1 .Assuming that the angles θ 1 and θ 2 are both small, we may make the approximationsin θ ≈ tan θ. Since at any point on the string the slope tan θ = ∂u/∂x, theforcecanbewritten[ ∂u(x +∆x, t)∆F = T−∂x]∂u(x, t)∂x≈ T ∂2 u(x, t)∆x,∂x 2where we have used the definition of the partial derivative to simplify the RHS.This upward force may be equated, by Newton’s second law, to the product of themass of the element and its upward acceleration. The element has a mass ρ ∆s, whichisapproximately equal to ρ ∆x if the vibrations of the string are small, and so we haveρ ∆x ∂2 u(x, t)= T ∂2 u(x, t)∆x.∂t 2 ∂x 2676

20.1 IMPORTANT PARTIAL DIFFERENTIAL EQUATIONSuTθ 2∆sθ 1Txx +∆xxFigure 20.1tension T .The forces acting on an element of a string under uniformDividing both sides by ∆x we obtain, for the vibrations of the string, the one-dimensionalwave equation∂ 2 u∂x = 1 ∂ 2 u2 c 2 ∂t , 2where c 2 = T/ρ. ◭The longitudinal vibrations of an elastic rod obey a very similar equation tothat derived in the above example, namely∂ 2 u∂x 2 = ρ ∂ 2 uE ∂t 2 ;here ρ is the mass per unit volume and E is Young’s modulus.The wave equation can be generalised slightly. For example, in the case of thevibrating string, there could also be an external upward vertical force f(x, t) perunit length acting on the string at time t. The transverse vibrations would thensatisfy the equationT ∂2 u∂x 2 + f(x, t) u=ρ∂2 ∂t 2 ,which is clearly of the form ‘upward force per unit length = mass per unit length× upward acceleration’.Similar examples, but involving two or three spatial dimensions rather than one,are provided by the equation governing the transverse vibrations of a stretchedmembrane subject to an external vertical force density f(x, y, t),( ∂ 2 )uT∂x 2 + ∂2 u∂y 2 + f(x, y, t) =ρ(x, y) ∂2 u∂t 2 ,where ρ is the mass per unit area of the membrane and T is the tension.677

PDES: GENERAL AND PARTICULAR SOLUTIONSThe diffusion equation20.1.2 The diffusion equationκ∇ 2 u = ∂u(20.2)∂tdescribes the temperature u in a region containing no heat sources or sinks; italso applies to the diffusion of a chemical that has a concentration u(r,t). Theconstant κ is called the diffusivity. The equation is clearly second order in thethree spatial variables, but first order in time.◮Derive the equation satisfied by the temperature u(r,t) at time t for a material of uniformthermal conductivity k, specific heat capacity s and density ρ. Express the equation inCartesian coordinates.Let us consider an arbitrary volume V lying within the solid and bounded by a surface S(this may coincide with the surface of the solid if so desired). At any point in the solidthe rate of heat flow per unit area in any given direction ˆr is proportional to minus thecomponent of the temperature gradient in that direction and so is given by (−k∇u) · ˆr. Thetotal flux of heat out of the volume V per unit time is given by− dQdt∫∫S= (−k∇u) · ˆn dS∫∫∫= ∇ · (−k∇u) dV , (20.3)Vwhere Q is the total heat energy in V at time t and ˆn is the outward-pointing unit normalto S; note that we have used the divergence theorem to convert the surface integral intoa volume integral.We can also express Q as a volume integral over V ,∫∫∫Q = sρu dV ,and its rate of change is then given byVdQdt = ∫∫∫Vsρ ∂u dV , (20.4)∂twhere we have taken the derivative with respect to time inside the integral (see section 5.12).Comparing (20.3) and (20.4), and remembering that the volume V is arbitrary, we obtainthe three-dimensional diffusion equationκ∇ 2 u = ∂u∂t ,where the diffusion coefficient κ = k/(sρ). To express this equation in Cartesian coordinates,we simply write ∇ 2 in terms of x, y and z to obtainκ( )∂ 2 u∂x + ∂2 u2 ∂y + ∂2 u= ∂u2 ∂z 2 ∂t . ◭The diffusion equation just derived can be generalised tok∇ 2 u + f(r,t)=sρ ∂u∂t .678

20.1 IMPORTANT PARTIAL DIFFERENTIAL EQUATIONSThe second term, f(r,t), represents a varying density of heat sources throughoutthe material but is often not required in physical applications. In the most generalcase, k, s and ρ may depend on position r, in which case the first term becomes∇ · (k∇u). However, in the simplest application the heat flow is one-dimensionalwith no heat sources, and the equation becomes (in Cartesian coordinates)∂ 2 u∂x 2 = sρ ∂uk ∂t .Laplace’s equation,20.1.3 Laplace’s equation∇ 2 u =0, (20.5)may be obtained by setting ∂u/∂t = 0 in the diffusion equation (20.2), anddescribes (for example) the steady-state temperature distribution in a solid inwhich there are no heat sources – i.e. the temperature distribution after a longtime has elapsed.Laplace’s equation also describes the gravitational potential in a region containingno matter or the electrostatic potential in a charge-free region. Further, itapplies to the flow of an incompressible fluid with no sources, sinks or vortices;in this case u is the velocity potential, from which the velocity is given by v = ∇u.Poisson’s equation,20.1.4 Poisson’s equation∇ 2 u = ρ(r), (20.6)describes the same physical situations as Laplace’s equation, but in regionscontaining matter, charges or sources of heat or fluid. The function ρ(r) iscalled the source density and in physical applications usually contains somemultiplicative physical constants. For example, if u is the electrostatic potentialin some region of space, in which case ρ is the density of electric charge, then∇ 2 u = −ρ(r)/ɛ 0 ,whereɛ 0 is the permittivity of free space. Alternatively, u mightrepresent the gravitational potential in some region where the matter density isgiven by ρ; then∇ 2 u =4πGρ(r), where G is the gravitational constant.The Schrödinger equation20.1.5 Schr-odinger’s equation− 22m ∇2 u + V (r)u = i ∂u∂t , (20.7)679

PDES: GENERAL AND PARTICULAR SOLUTIONSdescribes the quantum mechanical wavefunction u(r,t) of a non-relativistic particleof mass m; is Planck’s constant divided by 2π. Like the diffusion equation it issecond order in the three spatial variables and first order in time.20.2 General form of solutionBefore turning to the methods by which we may hope to solve PDEs such asthose listed in the previous section, it is instructive, as for ODEs in chapter 14, tostudy how PDEs may be formed from a set of possible solutions. Such a studycan provide an indication of how equations obtained not from possible solutionsbut from physical arguments might be solved.For definiteness let us suppose we have a set of functions involving twoindependent variables x and y. Without further specification this is of course avery wide set of functions, and we could not expect to find a useful equation thatthey all satisfy. However, let us consider a type of function u i (x, y) inwhichx andy appear in a particular way, such that u i can be written as a function (howevercomplicated) of a single variable p, itself a simple function of x and y.Let us illustrate this by considering the three functionsu 1 (x, y) =x 4 +4(x 2 y + y 2 +1),u 2 (x, y) =sinx 2 cos 2y +cosx 2 sin 2y,u 3 (x, y) = x2 +2y +23x 2 +6y +5 .These are all fairly complicated functions of x and y and a single differentialequation of which each one is a solution is not obvious. However, if we observethat in fact each can be expressed as a function of the variable p = x 2 +2y alone(withnootherx or y involved) then a great simplification takes place. Writtenin terms of p the above equations becomeu 1 (x, y) =(x 2 +2y) 2 +4=p 2 +4=f 1 (p),u 2 (x, y) =sin(x 2 +2y) =sinp = f 2 (p),u 3 (x, y) = (x2 +2y)+23(x 2 +2y)+5 = p +23p +5 = f 3(p).Let us now form, for each u i , the partial derivatives ∂u i /∂x and ∂u i /∂y. Ineachcase these are (writing both the form for general p and the one appropriate toour particular case, p = x 2 +2y)∂u i∂x = df i(p) ∂pdp ∂x =2xf′ i,∂u i∂y = df i(p) ∂pdp ∂y =2f′ i,for i = 1, 2, 3. All reference to the form of f i can be eliminated from these680

20.3 GENERAL AND PARTICULAR SOLUTIONSequations by cross-multiplication, obtaining∂p ∂u i∂yor, for our specific form, p = x 2 +2y,∂x = ∂p∂x∂u i∂y ,∂u i∂x = x∂u i∂y . (20.8)It is thus apparent that not only are the three functions u 1 , u 2 u 3 solutions of thePDE (20.8) but so also is any arbitrary function f(p) of which the argument p hasthe form x 2 +2y.20.3 General and particular solutionsIn the last section we found that the first-order PDE (20.8) has as a solution anyfunction of the variable x 2 +2y. This points the way for the solution of PDEsof other orders, as follows. It is not generally true that an nth-order PDE canalways be considered as resulting from the elimination of n arbitrary functionsfrom its solution (as opposed to the elimination of n arbitrary constants for annth-order ODE, see section 14.1). However, given specific PDEs we can try tosolve them by seeking combinations of variables in terms of which the solutionsmay be expressed as arbitrary functions. Where this is possible we may expect ncombinations to be involved in the solution.Naturally, the exact functional form of the solution for any particular situationmust be determined by some set of boundary conditions. For instance, if the PDEcontains two independent variables x and y then for complete determination ofits solution the boundary conditions will take a form equivalent to specifyingu(x, y) along a suitable continuum of points in the xy-plane (usually along a line).We now discuss the general and particular solutions of first- and secondorderPDEs. In order to simplify the algebra, we will restrict our discussionto equations containing just two independent variables x and y. Nevertheless,the method presented below may be extended to equations containing severalindependent variables.20.3.1 First-order equationsAlthough most of the PDEs encountered in physical contexts are second order(i.e. they contain ∂ 2 u/∂x 2 or ∂ 2 u/∂x∂y, etc.), we now discuss first-order equationsto illustrate the general considerations involved in the form of the solution andin satisfying any boundary conditions on the solution.The most general first-order linear PDE (containing two independent variables)681

PDES: GENERAL AND PARTICULAR SOLUTIONSis of the formA(x, y) ∂u ∂u+ B(x, y) + C(x, y)u = R(x, y), (20.9)∂x ∂ywhere A(x, y), B(x, y), C(x, y) andR(x, y) are given functions. Clearly, if eitherA(x, y) orB(x, y) is zero then the PDE may be solved straightforwardly as afirst-order linear ODE (as discussed in chapter 14), the only modification beingthat the arbitrary constant of integration becomes an arbitrary function of x or yrespectively.◮Find the general solution u(x, y) ofx ∂u∂x +3u = x2 .Dividing through by x we obtain∂u∂x + 3ux = x,which is a linear equation with integrating factor (see subsection 14.2.4)(∫ ) 3expx dx =exp(3lnx) =x 3 .Multiplying through by this factor we find∂∂x (x3 u)=x 4 ,which, on integrating with respect to x, givesx 3 u = x55 + f(y),where f(y) isanarbitrary function of y. Finally, dividing through by x 3 , we obtain thesolutionu(x, y) = x25 + f(y)x . ◭ 3When the PDE contains partial derivatives with respect to both independentvariables then, of course, we cannot employ the above procedure but must seekan alternative method. Let us for the moment restrict our attention to the specialcase in which C(x, y) =R(x, y) = 0 and, following the discussion of the previoussection, look for solutions of the form u(x, y) =f(p) wherep is some, at presentunknown, combination of x and y. We then have∂u∂x = df(p) ∂pdp ∂x ,∂u∂y = df(p) ∂pdp ∂y ,682

20.3 GENERAL AND PARTICULAR SOLUTIONSwhich, when substituted into the PDE (20.9), give[A(x, y) ∂p]∂p df(p)+ B(x, y) =0.∂x ∂y dpThis removes all reference to the actual form of the function f(p) since fornon-trivial p we must haveA(x, y) ∂p ∂p+ B(x, y) =0. (20.10)∂x ∂yLet us now consider the necessary condition for f(p) to remain constant as xand y vary; this is that p itself remains constant. Thus for f to remain constantimplies that x and y must vary in such a way thatdp = ∂p ∂pdx + dy =0. (20.11)∂x ∂yThe forms of (20.10) and (20.11) are very alike and become the same if werequire thatdxA(x, y) = dyB(x, y) . (20.12)By integrating this expression the form of p can be found.◮Forx ∂u ∂u− 2y =0, (20.13)∂x ∂yfind (i) the solution that takes the value 2y +1 on the line x =1, and (ii) a solution thathas the value 4 at the point (1, 1).If we seek a solution of the form u(x, y) =f(p), we deduce from (20.12) that u(x, y) willbe constant along lines of (x, y) thatsatisfydxx = dy−2y ,which on integrating gives x = cy −1/2 . Identifying the constant of integration c with p 1/2(to avoid fractional powers), we conclude that p = x 2 y. Thus the general solution of thePDE (20.13) isu(x, y) =f(x 2 y),where f is an arbitrary function.We must now find the particular solutions that obey each of the imposed boundaryconditions. For boundary condition (i) a little thought shows that the particular solutionrequired isu(x, y) =2(x 2 y)+1=2x 2 y +1. (20.14)For boundary condition (ii) some obviously acceptable solutions areu(x, y) =x 2 y +3,u(x, y) =4x 2 y,u(x, y) =4.683

PDES: GENERAL AND PARTICULAR SOLUTIONSEach is a valid solution (the freedom of choice of form arises from the fact that uis specified at only one point (1, 1), and not along a continuum (say), as in boundarycondition (i)). All three are particular examples of the general solution, which may bewritten, for example, asu(x, y) =x 2 y +3+g(x 2 y),where g = g(x 2 y)=g(p) is an arbitrary function subject only to g(1) = 0. For thisexample, the forms of g corresponding to the particular solutions listed above are g(p) =0,g(p) =3p − 3, g(p) =1− p. ◭As mentioned above, in order to find a solution of the form u(x, y) =f(p) werequire that the original PDE contains no term in u, but only terms containingits partial derivatives. If a term in u is present, so that C(x, y) ≠ 0 in (20.9),then the procedure needs some modification, since we cannot simply divide outthe dependence on f(p) to obtain (20.10). In such cases we look instead fora solution of the form u(x, y) =h(x, y)f(p). We illustrate this method in thefollowing example.◮Find the general solution ofx ∂u∂x +2∂u − 2u =0. (20.15)∂yWe seek a solution of the form u(x, y) =h(x, y)f(p), with the consequence that∂u∂x = ∂h df(p) ∂pf(p)+h∂x dp ∂x ,∂u∂y = ∂h df(p) ∂pf(p)+h∂y dp ∂y .Substituting these expressions into the PDE (20.15) and rearranging, we obtain(x ∂h) (∂x +2∂h ∂y − 2h f(p)+ x ∂p )∂x +2∂p h df(p) =0.∂y dpThe first factor in parentheses is just the original PDE with u replaced by h. Therefore, ifh is any solution of the PDE, however simple, this term will vanish, to leave(x ∂p )∂x +2∂p h df(p) =0,∂y dpfrom which, as in the previous case, we obtainx ∂p∂x +2∂p ∂y =0.From (20.11) and (20.12) we see that u(x, y) will be constant along lines of (x, y) thatsatisfydxx = dy 2 ,which integrates to give x = c exp(y/2). Identifying the constant of integration c with p wefind p = x exp(−y/2). Thus the general solution of (20.15) isu(x, y) =h(x, y)f(x exp(− 1 y)), 2where f(p) is any arbitrary function of p and h(x, y) is any solution of (20.15).684

20.3 GENERAL AND PARTICULAR SOLUTIONSIf we take, for example, h(x, y) =expy, which clearly satisfies (20.15), then the generalsolution isu(x, y) =(expy)f(x exp(− 1 y)). 2Alternatively, h(x, y) =x 2 also satisfies (20.15) and so the general solution to the equationcan also be writtenu(x, y) =x 2 g(x exp(− 1 y)), 2where g is an arbitrary function of p; clearly g(p) =f(p)/p 2 . ◭20.3.2 Inhomogeneous equations and problemsLet us discuss in a more general form the particular solutions of (20.13) foundin the second example of the previous subsection. It is clear that, so far as thisequation is concerned, if u(x, y) is a solution then so is any multiple of u(x, y) orany linear sum of separate solutions u 1 (x, y)+u 2 (x, y). However, when it comesto fitting the boundary conditions this is not so.For example, although u(x, y) in (20.14) satisfies the PDE and the boundarycondition u(1,y)=2y + 1, the function u 1 (x, y) =4u(x, y) =8xy + 4, whilstsatisfying the PDE, takes the value 8y+4 on the line x = 1 and so does not satisfythe required boundary condition. Likewise the function u 2 (x, y) =u(x, y)+f 1 (x 2 y),for arbitrary f 1 , satisfies (20.13) but takes the value u 2 (1,y)=2y +1+f 1 (y) onthe line x = 1, and so is not of the required form unless f 1 is identically zero.Thus we see that when treating the superposition of solutions of PDEs twoconsiderations arise, one concerning the equation itself and the other connectedto the boundary conditions. The equation is said to be homogeneous if the factthat u(x, y) is a solution implies that λu(x, y), for any constant λ, is also a solution.However, the problem is said to be homogeneous if, in addition, the boundaryconditions are such that if they are satisfied by u(x, y) then they are also satisfiedby λu(x, y). The last requirement itself is referred to as that of homogeneousboundary conditions.For example, the PDE (20.13) is homogeneous but the general first-orderequation (20.9) would not be homogeneous unless R(x, y) = 0. Furthermore,the boundary condition (i) imposed on the solution of (20.13) in the previoussubsection is not homogeneous though, in this case, the boundary conditionu(x, y) = 0on the line y =4x −2would be, since u(x, y) =λ(x 2 y − 4) satisfies this condition for any λ and, being afunction of x 2 y, satisfies (20.13).The reason for discussing the homogeneity of PDEs and their boundary conditionsis that in linear PDEs there is a close parallel to the complementary-functionand particular-integral property of ODEs. The general solution of an inhomogeneousproblem can be written as the sum of any particular solution of theproblem and the general solution of the corresponding homogeneous problem (as685

PDES: GENERAL AND PARTICULAR SOLUTIONSfor ODEs, we require that the particular solution is not already contained in thegeneral solution of the homogeneous problem). Thus, for example, the generalsolution of∂u∂x − x ∂u + au = f(x, y), (20.16)∂ysubject to, say, the boundary condition u(0,y)=g(y), is given byu(x, y) =v(x, y)+w(x, y),where v(x, y) is any solution (however simple) of (20.16) such that v(0,y)=g(y)and w(x, y) is the general solution of∂w∂x − x∂w + aw =0, (20.17)∂ywith w(0,y) = 0. If the boundary conditions are sufficiently specified then the onlypossible solution of (20.17) will be w(x, y) ≡ 0andv(x, y) will be the completesolution by itself.Alternatively, we may begin by finding the general solution of the inhomogeneousequation (20.16) without regard for any boundary conditions; it is just thesum of the general solution to the homogeneous equation and a particular integralof (20.16), both without reference to the boundary conditions. The boundaryconditions can then be used to find the appropriate particular solution from thegeneral solution.We will not discuss at length general methods of obtaining particular integralsof PDEs but merely note that some of those methods available for ordinarydifferential equations can be suitably extended. §◮Find the general solution ofy ∂u∂x − x ∂u =3x. (20.18)∂yHence find the most general particular solution (i) which satisfies u(x, 0) = x 2 and (ii) whichhas the value u(x, y) =2at the point (1, 0).This equation is inhomogeneous, and so let us first find the general solution of (20.18)without regard for any boundary conditions. We begin by looking for the solution of thecorresponding homogeneous equation ((20.18) but with the RHS equal to zero) of theform u(x, y) =f(p). Following the same procedure as that used in the solution of (20.13)we find that u(x, y) will be constant along lines of (x, y) thatsatisfydxy = dy−x⇒x22 + y22 = c.Identifying the constant of integration c with p/2, we find that the general solution of the§ See for example H. T. H. Piaggio, An Elementary Treatise on Differential Equations and theirApplications (London: G. Bell and Sons, Ltd, 1954), pp. 175 ff.686

20.3 GENERAL AND PARTICULAR SOLUTIONShomogeneous equation is u(x, y) =f(x 2 + y 2 ) for arbitrary function f. Now by inspectiona particular integral of (20.18) is u(x, y) =−3y, and so the general solution to (20.18) isu(x, y) =f(x 2 + y 2 ) − 3y.Boundary condition (i) requires u(x, 0) = f(x 2 )=x 2 ,i.e.f(z) =z, and so the particularsolution in this case isu(x, y) =x 2 + y 2 − 3y.Similarly, boundary condition (ii) requires u(1, 0) = f(1) = 2. One possibility is f(z) =2z,and if we make this choice, then one way of writing the most general particular solutionisu(x, y) =2x 2 +2y 2 − 3y + g(x 2 + y 2 ),where g is any arbitrary function for which g(1) = 0. Alternatively, a simpler choice wouldbe f(z) = 2, leading tou(x, y) =2− 3y + g(x 2 + y 2 ). ◭Although we have discussed the solution of inhomogeneous problems onlyfor first-order equations, the general considerations hold true for linear PDEs ofhigher order.20.3.3 Second-order equationsAs noted in section 20.1, second-order linear PDEs are of great importance indescribing the behaviour of many physical systems. As in our discussion of firstorderequations, for the moment we shall restrict our discussion to equations withjust two independent variables; extensions to a greater number of independentvariables are straightforward.The most general second-order linear PDE (containing two independent variables)has the formA ∂2 u∂x 2 + B ∂2 u∂x∂y + C ∂2 u∂y 2 + D ∂u∂x + E ∂u + Fu = R(x, y), (20.19)∂ywhere A,B,...,F and R(x, y) are given functions of x and y. Because of the natureof the solutions to such equations, they are usually divided into three classes, adivision of which we will make further use in subsection 20.6.2. The equation(20.19) is called hyperbolic if B 2 > 4AC, parabolic if B 2 =4AC and elliptic ifB 2 < 4AC. Clearly, if A, B and C are functions of x and y (rather than justconstants) then the equation might be of different types in different parts of thexy-plane.Equation (20.19) obviously represents a very large class of PDEs, and it isusually impossible to find closed-form solutions to most of these equations.Therefore, for the moment we shall consider only homogeneous equations, withR(x, y) = 0, and make the further (greatly simplifying) restriction that, throughoutthe remainder of this section, A,B,...,F are not functions of x and y but merelyconstants.687

PDES: GENERAL AND PARTICULAR SOLUTIONSWe now tackle the problem of solving some types of second-order PDE withconstant coefficients by seeking solutions that are arbitrary functions of particularcombinations of independent variables, just as we did for first-order equations.Following the discussion of the previous section, we can hope to find suchsolutions only if all the terms of the equation involve the same total numberof differentiations, i.e. all terms are of the same order, although the numberof differentiations with respect to the individual independent variables may bedifferent. This means that in (20.19) we require the constants D, E and F to beidentically zero (we have, of course, already assumed that R(x, y) iszero),sothatwe are now considering only equations of the formA ∂2 u∂x 2 + B ∂2 u∂x∂y + C ∂2 u=0, (20.20)∂y2 where A, B and C are constants. We note that both the one-dimensional waveequation,∂ 2 u∂x 2 − 1 ∂ 2 uc 2 ∂t 2 =0,and the two-dimensional Laplace equation,∂ 2 u∂x 2 + ∂2 u∂y 2 =0,are of this form, but that the diffusion equation,κ ∂2 u∂x 2 − ∂u∂t =0,is not, since it contains a first-order derivative.Since all the terms in (20.20) involve two differentiations, by assuming a solutionof the form u(x, y) =f(p), where p is some unknown function of x and y (or t),we may be able to obtain a common factor d 2 f(p)/dp 2 as the only appearance off on the LHS. Then, because of the zero RHS, all reference to the form of f canbe cancelled out.We can gain some guidance on suitable forms for the combination p = p(x, y)by considering ∂u/∂x when u is given by u(x, y) =f(p), for then∂u∂x = df(p) ∂pdp ∂x .Clearly differentiation of this equation with respect to x (or y) will not lead to asingle term on the RHS, containing f only as d 2 f(p)/dp 2 , unless the factor ∂p/∂xis a constant so that ∂ 2 p/∂x 2 and ∂ 2 p/∂x∂y are necessarily zero. This shows thatp must be a linear function of x. In an exactly similar way p must also be a linearfunction of y, i.e.p = ax + by.If we assume a solution of (20.20) of the form u(x, y) =f(ax+by), and evaluate688

20.3 GENERAL AND PARTICULAR SOLUTIONSthe terms ready for substitution into (20.20), we obtain∂ 2 u∂x 2 = a2 d2 f(p)dp 2 ,which on substitution give∂u∂x = adf(p) dp ,∂u∂y = bdf(p) dp ,∂ 2 u∂x∂y = f(p)abd2 dp 2 ,∂ 2 u∂y 2 = b2 d2 f(p)dp 2 ,(Aa 2 + Bab + Cb 2) d 2 f(p)dp 2 =0. (20.21)This is the form we have been seeking, since now a solution independent ofthe form of f can be obtained if we require that a and b satisfyAa 2 + Bab + Cb 2 =0.From this quadratic, two values for the ratio of the two constants a and b areobtained,b/a =[−B ± (B 2 − 4AC) 1/2 ]/2C.If we denote these two ratios by λ 1 and λ 2 then any functions of the two variablesp 1 = x + λ 1 y,p 2 = x + λ 2 ywill be solutions of the original equation (20.20). The omission of the constantfactor a from p 1 and p 2 is of no consequence since this can always be absorbedinto the particular form of any chosen function; only the relative weighting of xand y in p is important.Since p 1 and p 2 are in general different, we can thus write the general solutionof (20.20) asu(x, y) =f(x + λ 1 y)+g(x + λ 2 y), (20.22)where f and g are arbitrary functions.Finally, we note that the alternative solution d 2 f(p)/dp 2 = 0 to (20.21) leadsonly to the trivial solution u(x, y) =kx + ly + m, for which all second derivativesare individually zero.◮ Find the general solution of the one-dimensional wave equation∂ 2 u∂x − 1 ∂ 2 u2 c 2 ∂t =0. 2This equation is (20.20) with A =1,B =0andC = −1/c 2 , and so the values of λ 1 and λ 2are the solutions of1 − λ2c =0, 2namely λ 1 = −c and λ 2 = c. This means that arbitrary functions of the quantitiesp 1 = x − ct, p 2 = x + ct689

PDES: GENERAL AND PARTICULAR SOLUTIONSwill be satisfactory solutions of the equation and that the general solution will beu(x, t) =f(x − ct)+g(x + ct), (20.23)where f and g are arbitrary functions. This solution is discussed further in section 20.4. ◭The method used to obtain the general solution of the wave equation may alsobe applied straightforwardly to Laplace’s equation.◮ Find the general solution of the two-dimensional Laplace equation∂ 2 u∂x + ∂2 u=0. (20.24)2 ∂y2 Following the established procedure, we look for a solution that is a function f(p) ofp = x + λy, where from (20.24) λ satisfies1+λ 2 =0.This requires that λ = ±i, and satisfactory variables p are p = x ± iy. The general solutionrequired is therefore, in terms of arbitrary functions f and g,u(x, y) =f(x + iy)+g(x − iy). ◭It will be apparent from the last two examples that the nature of the appropriatelinear combination of x and y depends upon whether B 2 > 4AC or B 2 < 4AC.This is exactly the same criterion as determines whether the PDE is hyperbolicor elliptic. Hence as a general result, hyperbolic and elliptic equations of theform (20.20), given the restriction that the constants A, B and C are real, have assolutions functions whose arguments have the form x+αy and x+iβy respectively,where α and β themselves are real.The one case not covered by this result is that in which B 2 =4AC, i.e.aparabolic equation. In this case λ 1 and λ 2 are not different and only one suitablecombination of x and y results, namelyu(x, y) =f(x − (B/2C)y).To find the second part of the general solution we try, in analogy with thecorresponding situation for ordinary differential equations, a solution of the formu(x, y) =h(x, y)g(x − (B/2C)y).Substituting this into (20.20) and using A = B 2 /4C results in()A ∂2 h∂x 2 + B ∂2 h∂x∂y + C ∂2 h∂y 2 g =0.Therefore we require h(x, y) to be any solution of the original PDE. There areseveral simple solutions of this equation, but as only one is required we take thesimplest non-trivial one, h(x, y) =x, to give the general solution of the parabolicequationu(x, y) =f(x − (B/2C)y)+xg(x − (B/2C)y). (20.25)690

20.3 GENERAL AND PARTICULAR SOLUTIONSWe could, of course, have taken h(x, y) =y, but this only leads to a solution thatis already contained in (20.25).◮Solve∂ 2 u∂x +2 ∂2 u2 ∂x∂y + ∂2 u∂y =0, 2subject to the boundary conditions u(0,y)=0and u(x, 1) = x 2 .From our general result, functions of p = x + λy will be solutions provided1+2λ + λ 2 =0,i.e. λ = −1 and the equation is parabolic. The general solution is thereforeu(x, y) =f(x − y)+xg(x − y).The boundary condition u(0,y) = 0 implies f(p) ≡ 0, whilst u(x, 1) = x 2 yieldsxg(x − 1) = x 2 ,which gives g(p) =p + 1, Therefore the particular solution required isu(x, y) =x(p +1)=x(x − y +1). ◭To reinforce the material discussed above we will now give alternative derivationsof the general solutions (20.22) and (20.25) by expressing the original PDEin terms of new variables before solving it. The actual solution will then becomealmost trivial; but, of course, it will be recognised that suitable new variablescould hardly have been guessed if it were not for the work already done. Thisdoes not detract from the validity of the derivation to be described, only fromthe likelihood that it would be discovered by inspection.We start again with (20.20) and change to new variablesζ = x + λ 1 y, η = x + λ 2 y.With this change of variables, we have from the chain rule thatUsing these and the fact that∂∂x = ∂∂ζ + ∂∂η ,∂∂y = λ ∂1∂ζ + λ ∂2∂η .equation (20.20) becomesA + Bλ i + Cλ 2 i =0 fori =1, 2,[2A + B(λ 1 + λ 2 )+2Cλ 1 λ 2 ] ∂2 u∂ζ∂η =0.691

PDES: GENERAL AND PARTICULAR SOLUTIONSThen, providing the factor in brackets does not vanish, for which the requiredcondition is easily shown to be B 2 ≠4AC, weobtainwhich has the successive integrals∂ 2 u∂ζ∂η =0,∂u∂η = F(η),u(ζ,η)=f(η)+g(ζ).This solution is just the same as (20.22),u(x, y) =f(x + λ 2 y)+g(x + λ 1 y).If the equation is parabolic (i.e. B 2 =4AC), we instead use the new variablesζ = x + λy, η = x,and recalling that λ = −(B/2C) we can reduce (20.20) toA ∂2 u∂η 2 =0.Two straightforward integrations give as the general solutionu(ζ,η)=ηg(ζ)+f(ζ),whichintermsofx and y has exactly the form of (20.25),u(x, y) =xg(x + λy)+f(x + λy).Finally, as hinted at in subsection 20.3.2 with reference to first-order linearPDEs, some of the methods used to find particular integrals of linear ODEscan be suitably modified to find particular integrals of PDEs of higher order. Insimple cases, however, an appropriate solution may often be found by inspection.◮Find the general solution of∂ 2 u∂x + ∂2 u=6(x + y).2 ∂y2 Following our previous methods and results, the complementary function isu(x, y) =f(x + iy)+g(x − iy),and only a particular integral remains to be found. By inspection a particular integral ofthe equation is u(x, y) =x 3 + y 3 , and so the general solution can be writtenu(x, y) =f(x + iy)+g(x − iy)+x 3 + y 3 . ◭692

20.4 THE WAVE EQUATION20.4 The wave equationWe have already found that the general solution of the one-dimensional waveequation isu(x, t) =f(x − ct)+g(x + ct), (20.26)where f and g are arbitrary functions. However, the equation is of such generalimportance that further discussion will not be out of place.Let us imagine that u(x, t) =f(x − ct) represents the displacement of a string attime t and position x. It is clear that all positions x and times t for which x − ct =constant will have the same instantaneous displacement. But x − ct = constantis exactly the relation between the time and position of an observer travellingwith speed c along the positive x-direction. Consequently this moving observersees a constant displacement of the string, whereas to a stationary observer, theinitial profile u(x, 0) moves with speed c along the x-axis as if it were a rigidsystem. Thus f(x − ct) represents a wave form of constant shape travelling alongthe positive x-axis with speed c, the actual form of the wave depending uponthe function f. Similarly, the term g(x + ct) is a constant wave form travellingwith speed c in the negative x-direction. The general solution (20.23) representsa superposition of these.If the functions f and g are the same then the complete solution (20.23)represents identical progressive waves going in opposite directions. This mayresult in a wave pattern whose profile does not progress, described as a standingwave. As a simple example, suppose both f(p) andg(p) have the form §Then (20.23) can be written asf(p) =g(p) =A cos(kp + ɛ).u(x, t) =A[cos(kx − kct + ɛ)+cos(kx + kct + ɛ)]=2A cos(kct)cos(kx + ɛ).The important thing to notice is that the shape of the wave pattern, given by thefactor in x, is the same at all times but that its amplitude 2A cos(kct) dependsupon time. At some points x that satisfycos(kx + ɛ) =0there is no displacement at any time; such points are called nodes.So far we have not imposed any boundary conditions on the solution (20.26).The problem of finding a solution to the wave equation that satisfies given boundaryconditions is normally treated using the method of separation of variables§ In the usual notation, k is the wave number (= 2π/wavelength) and kc = ω, the angular frequencyof the wave.693

PDES: GENERAL AND PARTICULAR SOLUTIONSdiscussed in the next chapter. Nevertheless, we now consider D’Alembert’s solutionu(x, t) of the wave equation subject to initial conditions (boundary conditions) inthe following general form:∂u(x, 0)initial displacement, u(x, 0) = φ(x); initial velocity, = ψ(x).∂tThe functions φ(x) andψ(x) are given and describe the displacement and velocityof each part of the string at the (arbitrary) time t =0.It is clear that what we need are the particular forms of the functions f and gin (20.26) that lead to the required values at t = 0. This means thatφ(x) =u(x, 0) = f(x − 0) + g(x +0), (20.27)∂u(x, 0)ψ(x) = = −cf ′ (x − 0) + cg ′ (x +0), (20.28)∂twhere it should be noted that f ′ (x − 0) stands for df(p)/dp evaluated, after thedifferentiation, at p = x − c × 0; likewise for g ′ (x +0).Looking on the above two left-hand sides as functions of p = x ± ct, buteverywhere evaluated at t = 0, we may integrate (20.28) between an arbitrary(and irrelevant) lower limit p 0 and an indefinite upper limit p to obtain∫1 pψ(q) dq + K = −f(p)+g(p),c p 0the constant of integration K depending on p 0 . Comparing this equation with(20.27), with x replaced by p, we can establish the forms of the functions f andg asf(p) = φ(p) − 1 ∫ pψ(q) dq − K 2 2c p 02 , (20.29)g(p) = φ(p) + 1 ∫ pψ(q) dq + K 2 2c p 02 . (20.30)Adding (20.29) with p = x − ct to (20.30) with p = x + ct gives as the solution tothe original problem∫ x+ctu(x, t) = 1 2 [φ(x − ct)+φ(x + ct)] + 1 ψ(q) dq, (20.31)2c x−ctin which we notice that all dependence on p 0 has disappeared.Each of the terms in (20.31) has a fairly straightforward physical interpretation.In each case the factor 1/2 represents the fact that only half a displacement profilethat starts at any particular point on the string travels towards any other positionx, the other half travelling away from it. The first term 1 2φ(x − ct) arises fromthe initial displacement at a distance ct to the left of x; this travels forwardarriving at x at time t. Similarly, the second contribution is due to the initialdisplacement at a distance ct to the right of x. The interpretation of the final694

20.5 THE DIFFUSION EQUATIONterm is a little less obvious. It can be viewed as representing the accumulatedtransverse displacement at position x due to the passage past x of all parts ofthe initial motion whose effects can reach x within a time t, both backward andforward travelling.The extension to the three-dimensional wave equation of solutions of the typewe have so far encountered presents no serious difficulty. In Cartesian coordinatesthe three-dimensional wave equation is∂ 2 u∂x 2 + ∂2 u∂y 2 + ∂2 u∂z 2 − 1 ∂ 2 uc 2 =0. (20.32)∂t2 In close analogy with the one-dimensional case we try solutions that are functionsof linear combinations of all four variables,p = lx + my + nz + µt.It is clear that a solution u(x, y, z, t) =f(p) will be acceptable provided that) (l 2 + m 2 + n 2 − µ2 d 2 f(p)c 2 dp 2 =0.Thus, as in the one-dimensional case, f can be arbitrary provided thatl 2 + m 2 + n 2 = µ 2 /c 2 .Using an obvious normalisation, we take µ = ±c and l, m, n as three numberssuch thatl 2 + m 2 + n 2 =1.In other words (l,m,n) are the Cartesian components of a unit vector ˆn thatpoints along the direction of propagation of the wave. The quantity p can bewritten in terms of vectors as the scalar expression p = ˆn · r ± ct, and the generalsolution of (20.32) is thenu(x, y, z, t) =u(r,t)=f(ˆn · r − ct)+g(ˆn · r + ct), (20.33)where ˆn is any unit vector. It would perhaps be more transparent to write ˆnexplicitly as one of the arguments of u.20.5 The diffusion equationOne important class of second-order PDEs, which we have not yet consideredin detail, is that in which the second derivative with respect to one variableappears, but only the first derivative with respect to another (usually time). Thisis exemplified by the one-dimensional diffusion equationκ ∂2 u(x, t)∂x 2695= ∂u∂t , (20.34)

PDES: GENERAL AND PARTICULAR SOLUTIONSin which κ is a constant with the dimensions length 2 × time −1 . The physicalconstants that go to make up κ in a particular case depend upon the nature ofthe process (e.g. solute diffusion, heat flow, etc.) and the material being described.With (20.34) we cannot hope to repeat successfully the method of subsection20.3.3, since now u(x, t) is differentiated a different number of times on the twosides of the equation; any attempted solution in the form u(x, t) =f(p) withp = ax + bt will lead only to an equation in which the form of f cannot becancelled out. Clearly we must try other methods.Solutions may be obtained by using the standard method of separation ofvariables discussed in the next chapter. Alternatively, a simple solution is alsogiven if both sides of (20.34), as it stands, are separately set equal to a constantα (say), so that∂ 2 u∂x 2 = α κ ,These equations have the general solutions∂u∂t = α.u(x, t) = α2κ x2 + xg(t)+h(t) and u(x, t) =αt + m(x)respectively and may be made compatible with each other if g(t) is taken asconstant, g(t) =g (where g could be zero), h(t) =αt and m(x) =(α/2κ)x 2 + gx.An acceptable solution is thusu(x, t) = α2κ x2 + gx + αt + constant. (20.35)Let us now return to seeking solutions of equations by combining the independentvariables in particular ways. Having seen that a linear combination ofx and t will be of no value, we must search for other possible combinations. Ithas been noted already that κ has the dimensions length 2 × time −1 and so thecombination of variablesη = x2κtwill be dimensionless. Let us see if we can satisfy (20.34) with a solution of theform u(x, t) =f(η). Evaluating the necessary derivatives we have∂u∂x = df(η) ∂ηdη ∂x = 2x df(η)κt dη ,∂ 2 u∂x 2 = 2 (df(η) 2xκt dη+ κt) 2d 2 f(η)dη 2 ,∂u∂t = − x2 df(η)κt 2 dη .Substituting these expressions into (20.34) we find that the new equation can be696

20.5 THE DIFFUSION EQUATIONwritten entirely in terms of η,4η d2 f(η)dη 2+(2+η) df(η)dηThis is a straightforward ODE, which can be solved as follows. Writing f ′ (η) =df(η)/dη, etc., we havef ′′ (η)f ′ (η) = − 12η − 1 4⇒ ln[η 1/2 f ′ (η)] = − η 4 + c⇒ f ′ (η) = A ( −η)η exp 1/2 4∫ η ( −µ)⇒ f(η) =A µ −1/2 exp dµ.η 04If we now write this in terms of a slightly different variableζ = η1/22 = x2(κt) , 1/2then dζ = 1 4 η−1/2 dη, and the solution to (20.34) is given by∫ ζu(x, t) =f(η) =g(ζ) =B exp(−ν 2 ) dν. (20.36)ζ 0Here B is a constant and it should be noticed that x and t appear on the RHSonly in the indefinite upper limit ζ, and then only in the combination xt −1/2 .Ifζ 0 is chosen as zero then u(x, t) is, to within a constant factor, § the error functionerf[x/2(κt) 1/2 ], which is tabulated in many reference books. Only non-negativevalues of x and t are to be considered here, so that ζ ≥ ζ 0 .Let us try to determine what kind of (say) temperature distribution and flowthis represents. For definiteness we take ζ 0 = 0. Firstly, since u(x, t) in (20.36)depends only upon the product xt −1/2 , it is clear that all points x at times t suchthat xt −1/2 has the same value have the same temperature. Put another way, atany specific time t the region having a particular temperature has moved alongthe positive x-axis a distance proportional to the square root of t. Thisisatypicaldiffusion process.Notice that, on the one hand, at t = 0 the variable ζ →∞and u becomesquite independent of x (except perhaps at x = 0); the solution then represents auniform spatial temperature distribution. On the other hand, at x = 0 we havethat u(x, t) is identically zero for all t.=0.§ Take B =2π −1/2 to give the usual error function normalised in such a way that erf(∞) =1.Seethe Appendix.697

PDES: GENERAL AND PARTICULAR SOLUTIONS◮An infrared laser delivers a pulse of (heat) energy E to a point P on a large insulatedsheet of thickness b, thermal conductivity k, specific heat s and density ρ. The sheet isinitially at a uniform temperature. If u(r, t) is the excess temperature a time t later, at apoint that is a distance r (≫ b) from P , then show that a suitable expression for u isu(r, t) = α t exp (− r22βt), (20.37)where α and β are constants. (Note that we use r instead of ρ to denote the radial coordinatein plane polars so as to avoid confusion with the density.)Further, (i) show that β =2k/(sρ); (ii) demonstrate that the excess heat energy in thesheet is independent of t, and hence evaluate α; and (iii) prove that the total heat flow pastany circle of radius r is E.The equation to be solved is the heat diffusion equationk∇ 2 u(r,t)=sρ ∂u(r,t) .∂tSince we only require the solution for r ≫ b we can treat the problem as two-dimensionalwith obvious circular symmetry. Thus only the r-derivative term in the expression for ∇ 2 uis non-zero, giving(k ∂r ∂u )= sρ ∂ur ∂r ∂r ∂t , (20.38)where now u(r,t)=u(r, t).(i) Substituting the given expression (20.37) into (20.38) we obtain( ))2kα r2βt 2 2βt − 1 exp(− r2= sρα ( ) r22βt t 2 2βt − 1 exp(− r22βtfrom which we find that (20.37) is a solution, provided β =2k/(sρ).(ii) The excess heat in the system at any time t is∫ ∞∫ ∞( )rbρs u(r, t)2πr dr =2πbρsα00 t exp − r2dr2βt=2πbρsαβ.The excess heat is therefore independent of t and so must be equal to the total heatinput E, implying thatEα =2πbρsβ = E4πbk .(iii) The total heat flow past a circle of radius r is−2πrbk∫ ∞0∂u(r, t)dt = −2πrbk∂r[= E exp∫ ∞0(− r2E4πbkt),( ))−rexp(− r2dtβt 2βt∞= E for all r.2βt)]0As we would expect, all the heat energy E deposited by the laser will eventually flow pasta circle of any given radius r. ◭698

20.6 CHARACTERISTICS AND THE EXISTENCE OF SOLUTIONS20.6 Characteristics and the existence of solutionsSo far in this chapter we have discussed how to find general solutions to varioustypes of first- and second-order linear PDE. Moreover, given a set of boundaryconditions we have shown how to find the particular solution (or class of solutions)that satisfies them. For first-order equations, for example, we found that if thevalue of u(x, y) is specified along some curve in the xy-plane then the solution tothe PDE is in general unique, but that if u(x, y) is specified at only a single pointthen the solution is not unique: there exists a class of particular solutions all ofwhich satisfy the boundary condition. In this section and the next we make morerigorous the notion of the respective types of boundary condition that cause aPDE to have a unique solution, a class of solutions, or no solution at all.20.6.1 First-order equationsLet us consider the general first-order PDE (20.9) but now write it asA(x, y) ∂u ∂u+ B(x, y) = F(x, y, u). (20.39)∂x ∂ySuppose we wish to solve this PDE subject to the boundary condition thatu(x, y) =φ(s) is specified along some curve C in the xy-plane that is describedparametrically by the equations x = x(s) andy = y(s), where s is the arc lengthalong C. The variation of u along C is therefore given byduds = ∂u dx∂x ds + ∂u dy∂y ds = dφds . (20.40)We may then solve the two (inhomogeneous) simultaneous linear equations(20.39) and (20.40) for ∂u/∂x and ∂u/∂y, unless the determinant of the coefficientsvanishes (see section 8.18), i.e. unless∣ dx/ds dy/dsA B ∣ =0.At each point in the xy-plane this equation determines a set of curves calledcharacteristic curves (or just characteristics), which thus satisfyB dxds − Ady ds =0,or, multiplying through by ds/dx and dividing through by A,dy B(x, y)=dx A(x, y) . (20.41)However, we have already met (20.41) in subsection 20.3.1 on first-order PDEs,where solutions of the form u(x, y) =f(p), where p is some combination of x and y,699

PDES: GENERAL AND PARTICULAR SOLUTIONSwere discussed. Comparing (20.41) with (20.12) we see that the characteristics aremerely those curves along which p is constant.Since the partial derivatives ∂u/∂x and ∂u/∂y may be evaluated provided theboundary curve C does not lie along a characteristic, defining u(x, y) =φ(s)along C is sufficient to specify the solution to the original problem (equationplus boundary conditions) near the curve C, in terms of a Taylor expansionabout C. Therefore the characteristics can be considered as the curves alongwhich information about the solution u(x, y) ‘propagates’. This is best understoodby using an example.◮Find the general solution ofx ∂u ∂u− 2y∂x ∂y = 0 (20.42)that takes the value 2y +1 on the line x =1betweeny =0andy =1.We solved this problem in subsection 20.3.1 for the case where u(x, y) takes the value2y +1 along the entire line x = 1. We found then that the general solution to the equation(ignoring boundary conditions) is of the formu(x, y) =f(p) =f(x 2 y),for some arbitrary function f. Hence the characteristics of (20.42) are given by x 2 y = cwhere c is a constant; some of these curves are plotted in figure 20.2 for various values ofc. Furthermore, we found that the particular solution for which u(1,y)=2y +1for all ywas given byu(x, y) =2x 2 y +1.In the present case the value of x 2 y is fixed by the boundary conditions only betweeny =0andy = 1. However, since the characteristics are curves along which x 2 y, and hencef(x 2 y), remains constant, the solution is determined everywhere along any characteristicthat intersects the line segment denoting the boundary conditions. Thus u(x, y) =2x 2 y +1is the particular solution that holds in the shaded region in figure 20.2 (corresponding to0 ≤ c ≤ 1).Outside this region, however, the solution is not precisely specified, and any function ofthe formu(x, y) =2x 2 y +1+g(x 2 y)will satisfy both the equation and the boundary condition, provided g(p) = 0 for0 ≤ p ≤ 1. ◭In the above example the boundary curve was not itself a characteristic andfurthermore it crossed each characteristic once only. For a general boundary curveC this may not be the case. Firstly, if C is itself a characteristic (or is just a singlepoint) then information about the solution cannot ‘propagate’ away from C, andso the solution remains unspecified everywhere except on C.The second possibility is that C (although not a characteristic itself) crossessome characteristics more than once, as in figure 20.3. In this case specifying thevalue of u(x, y) along the curve PQdetermines the solution along all the characteristicsthat intersect it. Therefore, also specifying u(x, y) along QR can overdeterminethe problem solution and generally results in there being no solution.700

20.6 CHARACTERISTICS AND THE EXISTENCE OF SOLUTIONSy2c =11−11xy = c/x 2x =1Figure 20.2 The characteristics of equation (20.42). The shaded region showswhere the solution to the equation is defined, given the imposed boundarycondition at x = 1 between y =0andy = 1, shown as a bold vertical line.yRPCQxFigure 20.3A boundary curve C that crosses characteristics more than once.20.6.2 Second-order equationsThe concept of characteristics can be extended naturally to second- (and higher-)order equations. In this case let us write the general second-order linear PDE(20.19) as(A(x, y) ∂2 u∂x 2 + B(x, y) ∂2 u∂x∂y + C(x, y) ∂2 u∂y 2 = F x, y, u, ∂u∂x , ∂u ). (20.43)∂y701

PDES: GENERAL AND PARTICULAR SOLUTIONSyCdrdydxˆn dsxFigure 20.4point.A boundary curve C and its tangent and unit normal at a givenFor second-order equations we might expect that relevant boundary conditionswould involve specifying u, or some of its first derivatives, or both, along asuitable set of boundaries bordering or enclosing the region over which a solutionis sought. Three common types of boundary condition occur and are associatedwith the names of Dirichlet, Neumann and Cauchy. They are as follows.(i) Dirichlet: The value of u is specified at each point of the boundary.(ii) Neumann: The value of ∂u/∂n, thenormal derivative of u, isspecifiedateach point of the boundary. Note that ∂u/∂n = ∇u · ˆn, whereˆn is thenormal to the boundary at each point.(iii) Cauchy: Bothu and ∂u/∂n are specified at each point of the boundary.Let us consider for the moment the solution of (20.43) subject to the Cauchyboundary conditions, i.e. u and ∂u/∂n are specified along some boundary curveC in the xy-plane defined by the parametric equations x = x(s), y = y(s), s beingthe arc length along C (see figure 20.4). Let us suppose that along C we haveu(x, y) =φ(s) and∂u/∂n = ψ(s). At any point on C the vector dr = dx i + dy j isa tangent to the curve and ˆn ds = dy i − dx j is a vector normal to the curve. Thuson C we have∂u dr≡∇u ·∂s ds = ∂u dx∂x ds + ∂u dy∂y ds = dφ(s) ,ds∂u∂n≡∇u · ˆn =∂u∂xdyds − ∂u∂ydxds = ψ(s).These two equations may then be solved straightforwardly for the first partialderivatives ∂u/∂x and ∂u/∂y along C. Using the chain rule to writedds = dx ∂ds ∂x + dy ∂ds ∂y ,702

20.6 CHARACTERISTICS AND THE EXISTENCE OF SOLUTIONSwe may differentiate the two first derivatives ∂u/∂x and ∂u/∂y along the boundaryto obtain the pair of equationsddsdds( ) ∂u∂x( ) ∂u∂y= dx ∂ 2 uds= dxds∂x 2 + dy ∂ 2 uds ∂x∂y ,∂ 2 u∂x∂y + dy ∂ 2 uds ∂y 2 .We may now solve these two equations, together with the original PDE (20.43),for the second partial derivatives of u, except where the determinant of theircoefficients equals zero,A B Cdx dy0∣ ds ds∣ =0.∣Expanding out the determinant,A( ) 2 dy− Bds0( dxdsdxdsMultiplying through by (ds/dx) 2 we obtainAdyds∣)( ) dy+ Cds( ) 2 dx=0.ds( ) dy 2− B dy + C =0, (20.44)dx dxwhich is the ODE for the curves in the xy-plane along which the second partialderivatives of u cannot be found.As for the first-order case, the curves satisfying (20.44) are called characteristicsof the original PDE. These characteristics have tangents at each point given by(when A ≠0)dydx = B ± √ B 2 − 4AC. (20.45)2AClearly, when the original PDE is hyperbolic (B 2 > 4AC), equation (20.45)defines two families of real curves in the xy-plane; when the equation is parabolic(B 2 =4AC) it defines one family of real curves; and when the equation is elliptic(B 2 < 4AC) it defines two families of complex curves. Furthermore, when A,B and C are constants, rather than functions of x and y, the equations of thecharacteristics will be of the form x + λy = constant, which is reminiscent of theform of solution discussed in subsection 20.3.3.703

PDES: GENERAL AND PARTICULAR SOLUTIONSctx − ct =constant0Lxx + ct =constantFigure 20.5 The characteristics for the one-dimensional wave equation. Theshaded region indicates the region over which the solution is determined byspecifying Cauchy boundary conditions at t = 0 on the line segment x =0tox = L.◮Find the characteristics of the one-dimensional wave equation∂ 2 u∂x − 1 ∂ 2 u2 c 2 ∂t =0. 2This is a hyperbolic equation with A =1,B =0andC = −1/c 2 . Therefore from (20.44)the characteristics are given by( ) 2 dx= c 2 ,dtand so the characteristics are the straight lines x − ct =constantandx + ct =constant.◭The characteristics of second-order PDEs can be considered as the curves alongwhich partial information about the solution u(x, y) ‘propagates’. Consider a pointin the space that has the independent variables as its coordinates; unless bothof the two characteristics that pass through the point intersect the curve alongwhich the boundary conditions are specified, the solution will not be determinedat that point. In particular, if the equation is hyperbolic, so that we obtain twofamilies of real characteristics in the xy-plane, then Cauchy boundary conditionspropagate partial information concerning the solution along the characteristics,belonging to each family, that intersect the boundary curve C. Thesolutionuis then specified in the region common to these two families of characteristics.For instance, the characteristics of the hyperbolic one-dimensional wave equationin the last example are shown in figure 20.5. By specifying Cauchy boundary704

20.7 UNIQUENESS OF SOLUTIONSEquation type Boundary Conditionshyperbolic open Cauchyparabolic open Dirichlet or Neumannelliptic closed Dirichlet or NeumannTable 20.1 The appropriate boundary conditions for different types of partialdifferential equation.conditions u and ∂u/∂t on the line segment t =0,x =0toL, the solution isspecified in the shaded region.As in the case of first-order PDEs, however, problems can arise. For example,if for a hyperbolic equation the boundary curve intersects any characteristicmore than once then Cauchy conditions along C can overdetermine the problem,resulting in there being no solution. In this case either the boundary curve Cmust be altered, or the boundary conditions on the offending parts of C must berelaxed to Dirichlet or Neumann conditions.The general considerations involved in deciding which boundary conditions areappropriate for a particular problem are complex, and we do not discuss themany further here. § We merely note that whether the various types of boundarycondition are appropriate (in that they give a solution that is unique, sometimesto within a constant, and is well defined) depends upon the type of second-orderequation under consideration and on whether the region of solution is boundedby a closed or an open curve (or a surface if there are more than two independentvariables). Note that part of a closed boundary may be at infinity if conditionsare imposed on u or ∂u/∂n there.It may be shown that the appropriate boundary-condition and equation-typepairings are as given in table 20.1.For example, Laplace’s equation ∇ 2 u = 0 is elliptic and thus requires eitherDirichlet or Neumann boundary conditions on a closed boundary which, as wehave already noted, may be at infinity if the behaviour of u is specified there(most often u or ∂u/∂n → 0 at infinity).20.7 Uniqueness of solutionsAlthough we have merely stated the appropriate boundary types and conditionsfor which, in the general case, a PDE has a unique, well-defined solution, sometimesto within an additive constant, it is often important to be able to provethat a unique solution is obtained.§ For a discussion the reader is referred, for example, to P. M. Morse and H. Feshbach, Methods ofTheoretical Physics, Part I (New York: McGraw-Hill, 1953), chap. 6.705

PDES: GENERAL AND PARTICULAR SOLUTIONSAs an important example let us consider Poisson’s equation in three dimensions,∇ 2 u(r) =ρ(r), (20.46)with either Dirichlet or Neumann conditions on a closed boundary appropriateto such an elliptic equation; for brevity, in (20.46), we have absorbed any physicalconstants into ρ. We aim to show that, to within an unimportant constant, thesolution of (20.46) is unique if either the potential u or its normal derivative∂u/∂n is specified on all surfaces bounding a given region of space (including, ifnecessary, a hypothetical spherical surface of indefinitely large radius on which uor ∂u/∂n is prescribed to have an arbitrarily small value). Stated more formallythis is as follows.Uniqueness theorem. If u is real and its first and second partial derivatives arecontinuous in a region V and on its boundary S, and ∇ 2 u = ρ in V and eitheru = f or ∂u/∂n = g on S, whereρ, f and g are prescribed functions, then u isunique (at least to within an additive constant).◮Prove the uniqueness theorem for Poisson’s equation.Let us suppose on the contrary that two solutions u 1 (r)andu 2 (r) both satisfy the conditionsgiven above, and denote their difference by the function w = u 1 − u 2 . We then have∇ 2 w = ∇ 2 u 1 −∇ 2 u 2 = ρ − ρ =0,so that w satisfies Laplace’s equation in V . Furthermore, since either u 1 = f = u 2 or∂u 1 /∂n = g = ∂u 2 /∂n on S, we must have either w =0or∂w/∂n =0onS.If we now use Green’s first theorem, (11.19), for the case where both scalar functionsare taken as w we have ∫[w∇ 2 w +(∇w) · (∇w) ] ∫dV = w ∂wVS ∂n dS.However, either condition, w =0or∂w/∂n = 0, makes the RHS vanish whilst the firstterm on the LHS vanishes since ∇ 2 w =0inV . Thus we are left with∫|∇w| 2 dV =0.VSince |∇w| 2 can never be negative, this can only be satisfied if∇w = 0,i.e. if w, and hence u 1 − u 2 , is a constant in V .If Dirichlet conditions are given then u 1 ≡ u 2 on (some part of) S and hence u 1 = u 2everywhere in V . For Neumann conditions, however, u 1 and u 2 can differ throughout Vby an arbitrary (but unimportant) constant. ◭The importance of this uniqueness theorem lies in the fact that if a solution toPoisson’s (or Laplace’s) equation that fits the given set of Dirichlet or Neumannconditions can be found by any means whatever, then that solution is the correctone, since only one exists. This result is the mathematical justification for themethod of images, which is discussed more fully in the next chapter.706

20.8 EXERCISESWe also note that often the same general method, used in the above examplefor proving the uniqueness theorem for Poisson’s equation, can be employed toprove the uniqueness (or otherwise) of solutions to other equations and boundaryconditions.20.8 Exercises20.1 Determine whether the following can be written as functions of p = x 2 +2y only,and hence whether they are solutions of (20.8):(a) x 2 (x 2 − 4) + 4y(x 2 − 2) + 4(y 2 − 1);(b) x 4 +2x 2 y + y 2 ;(c) [x 4 +4x 2 y +4y 2 +4]/[2x 4 + x 2 (8y +1)+8y 2 +2y].20.2 Find partial differential equations satisfied by the following functions u(x, y) forall arbitrary functions f and all arbitrary constants a and b:(a) u(x, y) =f(x 2 − y 2 );(b) u(x, y) =(x − a) 2 +(y − b) 2 ;(c) u(x, y) =y n f(y/x);(d) u(x, y) =f(x + ay).20.3 Solve the following partial differential equations for u(x, y) with the boundaryconditions given:(a) x ∂u + xy = u, u =2y on the line x =1;∂x(b) 1 + x ∂u = xu, u(x, 0) = x.∂y20.4 Find the most general solutions u(x, y) of the following equations, consistent withthe boundary conditions stated:(a) y ∂u∂x − x ∂u∂y =0,u(x, 0)=1+sinx;(b) i ∂u∂x =3∂u ∂y , u =(4+3i)x2 on the line x = y;(c)sin x sin y ∂u∂u+cosx cos y∂x ∂y =0,u =cos2y on x + y = π/2;∂u ∂u(d) +2x∂x ∂y =0, u = 2 on the parabola y = x2 .20.5 Find solutions of1 ∂ux ∂x + 1 ∂uy ∂y =0for which (a) u(0,y)=y and (b) u(1, 1) = 1.20.6 Find the most general solutions u(x, y) of the following equations consistent withthe boundary conditions stated:(a) y ∂u∂x − x ∂u∂y =3x, u = x2 on the line y =0;707

PDES: GENERAL AND PARTICULAR SOLUTIONS(b) y ∂u∂x − x ∂u =3x, u(1, 0) = 2;∂y(c) y 2 ∂u ∂u+ x2∂x ∂y = x2 y 2 (x 3 + y 3 ), no boundary conditions.20.7 Solvesin x ∂u ∂u+cosx∂x ∂y =cosxsubject to (a) u(π/2,y)=0and(b)u(π/2,y)=y(y +1).20.8 A function u(x, y) satisfies2 ∂u∂x +3∂u ∂y =10,and takes the value 3 on the line y =4x. Evaluate u(2, 4).20.9 If u(x, y) satisfies∂ 2 u∂x − 3 ∂2 u u2 ∂x∂y +2∂2 ∂y =0 2and u = −x 2 and ∂u/∂y =0fory =0andallx, find the value of u(0, 1).20.10 Consider the partial differential equation∂ 2 u∂x − 3 ∂2 u u2 ∂x∂y +2∂2 ∂y =0. (∗)2(a) Find the function u(x, y) that satisfies (∗) and the boundary condition u =∂u/∂y =1wheny =0forallx. Evaluate u(0, 1).(b) In which region of the xy-plane would u be determined if the boundarycondition were u = ∂u/∂y =1wheny =0forallx>0?20.11 In those cases in which it is possible to do so, evaluate u(2, 2), where u(x, y) isthe solution of2y ∂u∂x − x ∂u∂y = xy(2y2 − x 2 )that satisfies the (separate) boundary conditions given below.(a) u(x, 1) = x 2 for all x.(b) u(x, 1) = x 2 for x ≥ 0.(c) u(x, 1) = x 2 for 0 ≤ x ≤ 3.(d) u(x, 0) = x for x ≥ 0.(e) u(x, 0) = x for all x.(f) u(1, √ 10) = 5.(g) u( √ 10, 1) = 5.20.12 Solve6 ∂2 u∂x − 5 ∂2 u2 ∂x∂y + ∂2 u∂y =14, 2subject to u =2x +1and∂u/∂y =4− 6x, both on the line y =0.20.13 By changing the independent variables in the previous exercise toξ = x +2y and η = x +3y,show that it must be possible to write 14(x 2 +5xy +6y 2 )intheformf 1 (x +2y)+f 2 (x +3y) − (x 2 + y 2 ),and determine the forms of f 1 (z) andf 2 (z).708

20.8 EXERCISES20.14 Solve∂ 2 u u∂x∂y +3∂2 = x(2y +3x).∂y2 20.15 Find the most general solution of ∂ 2 u/∂x 2 + ∂ 2 u/∂y 2 = x 2 y 2 .20.16 An infinitely long string on which waves travel at speed c has an initial displacement{sin(πx/a), −a ≤ x ≤ a,y(x) =0, |x| >a.It is released from rest at time t = 0, and its subsequent displacement is describedby y(x, t).By expressing the initial displacement as one explicit function incorporatingHeaviside step functions, find an expression for y(x, t) atageneraltimet>0. Inparticular, determine the displacement as a function of time (a) at x =0,(b)atx = a, and(c)atx = a/2.20.17 The non-relativistic Schrödinger equation (20.7) is similar to the diffusion equationin having different orders of derivatives in its various terms; this precludessolutions that are arbitrary functions of particular linear combinations of variables.However, since exponential functions do not change their forms underdifferentiation, solutions in the form of exponential functions of combinations ofthe variables may still be possible.Consider the Schrödinger equation for the case of a constant potential, i.e. fora free particle, and show that it has solutions of the form A exp(lx+my +nz +λt),where the only requirement is that− 2 (l 2 + m 2 + n 2) = iλ.2mIn particular, identify the equation and wavefunction obtained by taking λ as−iE/, andl, m and n as ip x /,ip y / and ip z /, respectively, where E is theenergy and p the momentum of the particle; these identifications are essentiallythe content of the de Broglie and Einstein relationships.20.18 Like the Schrödinger equation of the previous exercise, the equation describingthe transverse vibrations of a rod,a 4 ∂4 u∂x + ∂2 u4 ∂t =0, 2has different orders of derivatives in its various terms. Show, however, that ithas solutions of exponential form, u(x, t) =A exp(λx + iωt), provided that therelation a 4 λ 4 = ω 2 is satisfied.Use a linear combination of such allowed solutions, expressed as the sum ofsinusoids and hyperbolic sinusoids of λx, to describe the transverse vibrations ofa rod of length L clamped at both ends. At a clamped point both u and ∂u/∂xmust vanish; show that this implies that cos(λL)cosh(λL) = 1, thus determiningthe frequencies ω at which the rod can vibrate.20.19 An incompressible fluid of density ρ and negligible viscosity flows with velocity valong a thin, straight, perfectly light and flexible tube, of cross-section A which isheld under tension T . Assume that small transverse displacements u of the tubeare governed by∂ 2 u∂t +2v ∂2 u(v 2 ∂x∂t + 2 − T ) ∂ 2 uρA ∂x =0. 2(a) Show that the general solution consists of a superposition of two waveformstravelling with different speeds.709

PDES: GENERAL AND PARTICULAR SOLUTIONS(b) The tube initially has a small transverse displacement u = a cos kx and issuddenly released from rest. Find its subsequent motion.20.20 A sheet of material of thickness w, specific heat capacity c and thermal conductivityk is isolated in a vacuum, but its two sides are exposed to fluxes ofradiant heat of strengths J 1 and J 2 . Ignoring short-term transients, show that thetemperature difference between its two surfaces is steady at (J 2 − J 1 )w/2k, whilsttheir average temperature increases at a rate (J 2 + J 1 )/cw.20.21 In an electrical cable of resistance R and capacitance C, each per unit length,voltage signals obey the equation ∂ 2 V/∂x 2 = RC∂V/∂t. This has solutions of theform given in (20.36) and also of the form V = Ax + D.(a) Find a combination of these that represents the situation after a steadyvoltage V 0 is applied at x =0attimet =0.(b) Obtain a solution describing the propagation of the voltage signal resultingfrom the application of the signal V = V 0 for 0

20.9 HINTS AND ANSWERS20.25 The Klein–Gordon equation (which is satisfied by the quantum-mechanical wavefunctionΦ(r) of a relativistic spinless particle of non-zero mass m) is∇ 2 Φ − m 2 Φ=0.Show that the solution for the scalar field Φ(r) inanyvolumeV bounded byasurfaceS is unique if either Dirichlet or Neumann boundary conditions arespecified on S.20.9 Hints and answers20.1 (a) Yes, p 2 − 4p − 4; (b) no, (p − y) 2 ;(c)yes,(p 2 +4)/(2p 2 + p).20.3 Each equation is effectively an ordinary differential equation, but with a functionof the non-integrated variable as the constant of integration;(a) u = xy(2 − ln x); (b) u = x −1 (1 − e y )+xe y .20.5 (a) (y 2 − x 2 ) 1/2 ;(b)1+f(y 2 − x 2 ), where f(0) = 0.20.7 u = y + f(y − ln(sin x)); (a) u =ln(sinx); (b) u = y +[y − ln(sin x)] 2 .20.9 General solution is u(x, y) =f(x + y) +g(x + y/2). Show that 2p = −g ′ (p)/2,and hence g(p) =k − 2p 2 , whilst f(p) =p 2 − k, leading to u(x, y) =−x 2 + y 2 /2;u(0, 1) = 1/2.20.11 p = x 2 +2y 2 ; u(x, y) =f(p)+x 2 y 2 /2.(a) u(x, y) =(x 2 +2y 2 + x 2 y 2 − 2)/2; u(2, 2) = 13. The line y = 1 cuts eachcharacteristic in zero or two distinct points, but this causes no difficulty withthe given boundary conditions.(b) As in (a).(c) The solution is defined over the space between the ellipses p =2andp = 11;(2, 2) lies on p = 12, and so u(2, 2) is undetermined.(d) u(x, y) =(x 2 +2y 2 ) 1/2 + x 2 y 2 /2; u(2, 2) = 8 + √ 12.(e) The line y = 0 cuts each characteristic in two distinct points. No differentiableform of f(p) givesf(±a) =±a respectively, and so there is no solution.(f) The solution is only specified on p = 21, and so u(2, 2) is undetermined.(g) The solution is specified on p = 12, and so u(2, 2) = 5 + 1 (4)(4) = 13.220.13 The equation becomes ∂ 2 f/∂ξ∂η = −14, with solution f(ξ,η) =f(ξ)+g(η)−14ξη,which can be compared with the answer from the previous question; f 1 (z) =10z 2and f 2 (z) =5z 2 .20.15 u(x, y) =f(x + iy)+g(x − iy)+(1/12)x 4 (y 2 − (1/15)x 2 ). In the last term, x andy may be interchanged. There are (infinitely) many other possibilities for thespecific PI, e.g. [ 15x 2 y 2 (x 2 + y 2 ) − (x 6 + y 6 )]/360.20.17 E = p 2 /(2m), the relationship between energy and momentum for a nonrelativisticparticle; u(r,t)=A exp[i(p · r − Et)/], a plane wave of wave numberk = p/ and angular frequency ω = E/ travelling in the direction p/p.20.19 (a) c = v ± α where α 2 = T/ρA;(b) u(x,[t) =a cos[k(x − vt)] cos(kαt) − (va/α)sin[k(x − vt)] sin(kαt).20.21 (a) V 0 1 − (2/ √ π) ∫ ]12 x(CR/t)1/2 exp(−ν 2 ) dν ;(b) consider the input as equivalent to V 0 applied at t = 0 and continued and−V 0 applied at t = T and continued;V (x, t) = 2V ∫ 10 2 x[CR/(t−T )] 1/2√ exp ( −ν 2) dν;π 12 x(CR/t) 1/2(c) For t ≫ T , maximum at x =[2t/(CR)] 1/2 with value V 0T exp(− 1 ) 2.(2π) 1/2 t711

PDES: GENERAL AND PARTICULAR SOLUTIONS20.23 (a) Parabolic, open, Dirichlet u(x, 0) given, Neumann ∂u/∂x =0atx = ±L/2for all t;(b) elliptic, closed, Dirichlet;(c) elliptic, closed, Neumann ∂u/∂n = σ/ɛ 0 ;(d) hyperbolic, open, Cauchy.20.25 Follow an argument similar to that in section 20.7 and argue that the additionalterm ∫ m 2 |w| 2 dV must be zero, and hence that w = 0 everywhere.712

21Partial differential equations:separation of variables and othermethodsIn the previous chapter we demonstrated the methods by which general solutionsof some partial differential equations (PDEs) may be obtained in terms ofarbitrary functions. In particular, solutions containing the independent variablesin definite combinations were sought, thus reducing the effective number of them.In the present chapter we begin by taking the opposite approach, namely thatof trying to keep the independent variables as separate as possible, using themethod of separation of variables. We then consider integral transform methodsby which one of the independent variables may be eliminated, at least fromdifferential coefficients. Finally, we discuss the use of Green’s functions in solvinginhomogeneous problems.21.1 Separation of variables: the general methodSuppose we seek a solution u(x, y, z, t) to some PDE (expressed in Cartesiancoordinates). Let us attempt to obtain one that has the product form §u(x, y, z, t) =X(x)Y (y)Z(z)T (t). (21.1)A solution that has this form is said to be separable in x, y, z and t, and seekingsolutions of this form is called the method of separation of variables.As simple examples we may observe that, of the functions(i) xyz 2 sin bt, (ii) xy + zt, (iii) (x 2 + y 2 )z cos ωt,(i) is completely separable, (ii) is inseparable in that no single variable can beseparated out from it and written as a multiplicative factor, whilst (iii) is separablein z and t but not in x and y.§ It should be noted that the conventional use here of upper-case (capital) letters to denote thefunctions of the corresponding lower-case variable is intended to enable an easy correspondencebetween a function and its argument to be made.713

PDES: SEPARATION OF VARIABLES AND OTHER METHODSWhen seeking PDE solutions of the form (21.1), we are requiring not thatthere is no connection at all between the functions X, Y , Z and T (for example,certain parameters may appear in two or more of them), but only that X doesnot depend upon y, z, t, thatY does not depend on x, z, t, andsoon.For a general PDE it is likely that a separable solution is impossible, butcertainly some common and important equations do have useful solutions ofthis form, and we will illustrate the method of solution by studying the threedimensionalwave equation∇ 2 u(r) = 1 ∂ 2 u(r)c 2 ∂t 2 . (21.2)We will work in Cartesian coordinates for the present and assume a solutionof the form (21.1); the solutions in alternative coordinate systems, e.g. sphericalor cylindrical polars, are considered in section 21.3. Expressed in Cartesiancoordinates (21.2) takes the formsubstituting (21.1) gives∂ 2 u∂x 2 + ∂2 u∂y 2 + ∂2 u∂z 2 = 1 c 2 ∂ 2 u∂t 2 ; (21.3)d 2 Xdx 2 YZT + X d2 Ydy 2 ZT + XY d2 Zdz 2 T = 1 c 2 XY Z d2 Tdt 2 ,which can also be written asX ′′ YZT + XY ′′ ZT + XY Z ′′ T = 1 c 2 XY ZT′′ , (21.4)whereineachcasetheprimesrefertotheordinary derivative with respect to theindependent variable upon which the function depends. This emphasises the factthat each of the functions X, Y , Z and T has only one independent variable andthus its only derivative is its total derivative. For the same reason, in each termin (21.4) three of the four functions are unaltered by the partial differentiationand behave exactly as constant multipliers.If we now divide (21.4) throughout by u = XY ZT we obtainX ′′X + Y ′′Y + Z ′′Z = 1 T ′′c 2 T . (21.5)This form shows the particular characteristic that is the basis of the method ofseparation of variables, namely that of the four terms the first is a function of xonly, the second of y only, the third of z only and the RHS a function of t onlyand yet there is an equation connecting them. This can only be so for all x, y, zand t if each of the terms does not in fact, despite appearances, depend upon thecorresponding independent variable but is equal to a constant, the four constantsbeing such that (21.5) is satisfied.714

21.1 SEPARATION OF VARIABLES: THE GENERAL METHODSince there is only one equation to be satisfied and four constants involved,there is considerable freedom in the values they may take. For the purposes ofour illustrative example let us make the choice of −l 2 , −m 2 , −n 2 ,forthefirstthree constants. The constant associated with c −2 T ′′ /T must then have the value−µ 2 = −(l 2 + m 2 + n 2 ).Having recognised that each term of (21.5) is individually equal to a constant(or parameter), we can now replace (21.5) by four separate ordinary differentialequations (ODEs):X ′′X = Y ′′−l2 ,Y = Z ′′−m2 ,Z = 1 T ′′−n2 ,c 2 T = −µ2 . (21.6)The important point to notice is not the simplicity of the equations (21.6) (thecorresponding ones for a general PDE are usually far from simple) but that, bythe device of assuming a separable solution, a partial differential equation (21.3),containing derivatives with respect to the four independent variables all in oneequation, has been reduced to four separate ordinary differential equations (21.6).The ordinary equations are connected through four constant parameters thatsatisfy an algebraic relation. These constants are called separation constants.The general solutions of the equations (21.6) can be deduced straightforwardlyand areX(x) =A exp(ilx)+B exp(−ilx),Y (y) =C exp(imy)+D exp(−imy),Z(z) =E exp(inz)+F exp(−inz),T (t) =G exp(icµt)+H exp(−icµt),(21.7)where A, B,...,H are constants, which may be determined if boundary condtionsare imposed on the solution. Depending on the geometry of the problem andany boundary conditions, it is sometimes more appropriate to write the solutions(21.7) in the alternative formX(x) =A ′ cos lx + B ′ sin lx,Y (y) =C ′ cos my + D ′ sin my,Z(z) =E ′ cos nz + F ′ sin nz,T (t) =G ′ cos(cµt)+H ′ sin(cµt),(21.8)for some different set of constants A ′ ,B ′ ,...,H ′ . Clearly the choice of how bestto represent the solution depends on the problem being considered.As an example, suppose that we take as particular solutions the four functionsX(x) =exp(ilx),Z(z) =exp(inz),Y(y) =exp(imy),T(t) =exp(−icµt).715

PDES: SEPARATION OF VARIABLES AND OTHER METHODSThis gives a particular solution of the original PDE (21.3)u(x, y, z, t) =exp(ilx) exp(imy) exp(inz) exp(−icµt)= exp[i(lx + my + nz − cµt)],which is a special case of the solution (20.33) obtained in the previous chapterand represents a plane wave of unit amplitude propagating in a direction givenby the vector with components l,m,n in a Cartesian coordinate system. In theconventional notation of wave theory, l, m and n are the components of thewave-number vector k, whose magnitude is given by k =2π/λ, whereλ is thewavelength of the wave; cµ is the angular frequency ω of the wave. This givesthe equation in the formu(x, y, z, t) =exp[i(k x x + k y y + k z z − ωt)]= exp[i(k · r − ωt)],and makes the exponent dimensionless.The method of separation of variables can be applied to many commonlyoccurring PDEs encountered in physical applications.◮Use the method of separation of variables to obtain for the one-dimensional diffusionequationκ ∂2 u∂x = ∂u2 ∂t , (21.9)a solution that tends to zero as t →∞for all x.Here we have only two independent variables x and t and we therefore assume a solutionof the formu(x, t) =X(x)T (t).Substituting this expression into (21.9) and dividing through by u = XT (and also by κ)we obtainX ′′X = T ′κT .Now, arguing exactly as above that the LHS is a function of x onlyandtheRHSisafunction of t only, we conclude that each side must equal a constant, which, anticipatingthe result and noting the imposed boundary condition, we will take as −λ 2 . This gives ustwo ordinary equations,X ′′ + λ 2 X =0, (21.10)T ′ + λ 2 κT =0, (21.11)which have the solutionsX(x) =A cos λx + B sin λx,T (t) =C exp(−λ 2 κt).Combining these to give the assumed solution u = XT yields (absorbing the constant Cinto A and B)u(x, t) =(A cos λx + B sin λx)exp(−λ 2 κt). (21.12)716

21.2 SUPERPOSITION OF SEPARATED SOLUTIONSIn order to satisfy the boundary condition u → 0ast →∞, λ 2 κ must be > 0. Since κis real and > 0, this implies that λ is a real non-zero number and that the solution issinusoidal in x and is not a disguised hyperbolic function; this was our reason for choosingthe separation constant as −λ 2 . ◭As a final example we consider Laplace’s equation in Cartesian coordinates;this may be treated in a similar manner.◮Use the method of separation of variables to obtain a solution for the two-dimensionalLaplace equation,∂ 2 u∂x + ∂2 u=0. (21.13)2 ∂y2 If we assume a solution of the form u(x, y) =X(x)Y (y) then, following the above method,and taking the separation constant as λ 2 , we findX ′′ = λ 2 X, Y ′′ = −λ 2 Y.Taking λ 2 as > 0, the general solution becomesu(x, y) =(A cosh λx + B sinh λx)(C cos λy + D sin λy). (21.14)An alternative form, in which the exponentials are written explicitly, may be useful forother geometries or boundary conditions:u(x, y) =[A exp λx + B exp(−λx)](C cos λy + D sin λy), (21.15)with different constants A and B.If λ 2 < 0 then the roles of x and y interchange. The particular combination of sinusoidaland hyperbolic functions and the values of λ allowed will be determined by the geometricalproperties of any specific problem, together with any prescribed or necessary boundaryconditions. ◭We note here that a particular case of the solution (21.14) links up with the‘combination’ result u(x, y) =f(x + iy) of the previous chapter (equations (20.24)and following), namely that if A = B and D = iC then the solution is the sameas f(p) =AC exp λp with p = x + iy.21.2 Superposition of separated solutionsIt will be noticed in the previous two examples that there is considerable freedomin the values of the separation constant λ, the only essential requirement beingthat λ has the same value in both parts of the solution, i.e. the part dependingon x and the part depending on y (or t). This is a general feature for solutionsin separated form, which, if the original PDE has n independent variables, willcontain n − 1 separation constants. All that is required in general is that weassociate the correct function of one independent variable with the appropriatefunctions of the others, the correct function being the one with the same valuesof the separation constants.If the original PDE is linear (as are the Laplace, Schrödinger, diffusion andwave equations) then mathematically acceptable solutions can be formed by717

PDES: SEPARATION OF VARIABLES AND OTHER METHODSsuperposing solutions corresponding to different allowed values of the separationconstants. To take a two-variable example: ifu λ1 (x, y) =X λ1 (x)Y λ1 (y)is a solution of a linear PDE obtained by giving the separation constant the valueλ 1 , then the superpositionu(x, y) =a 1 X λ1 (x)Y λ1 (y)+a 2 X λ2 (x)Y λ2 (y)+···= ∑ a i X λi (x)Y λi (y)i(21.16)is also a solution for any constants a i , provided that the λ i are the allowed valuesof the separation constant λ given the imposed boundary conditions. Note thatif the boundary conditions allow any of the separation constants to be zero thenthe form of the general solution is normally different and must be deduced byreturning to the separated ordinary differential equations. We will encounter thisbehaviour in section 21.3.The value of the superposition approach is that a boundary condition, say thatu(x, y) takes a particular form f(x) wheny = 0, might be met by choosing theconstants a i such thatf(x) = ∑ a i X λi (x)Y λi (0).iIn general, this will be possible provided that the functions X λi (x) form a completeset – as do the sinusoidal functions of Fourier series or the spherical harmonicsdiscussed in subsection 18.3.◮A semi-infinite rectangular metal plate occupies the region 0 ≤ x ≤∞and 0 ≤ y ≤ b inthe xy-plane. The temperature at the far end of the plate and along its two long sides isfixed at 0 ◦ C. If the temperature of the plate at x =0is also fixed and is given by f(y), findthe steady-state temperature distribution u(x,y) of the plate. Hence find the temperaturedistribution if f(y) =u 0 ,whereu 0 is a constant.The physical situation is illustrated in figure 21.1. With the notation we have used severaltimes before, the two-dimensional heat diffusion equation satisfied by the temperatureu(x, y, t) is( )∂ 2 uκ∂x + ∂2 u= ∂u2 ∂y 2 ∂t ,with κ = k/(sρ). In this case, however, we are asked to find the steady-state temperature,which corresponds to ∂u/∂t = 0, and so we are led to consider the (two-dimensional)Laplace equation∂ 2 u∂x + ∂2 u2 ∂y =0. 2We saw that assuming a separable solution of the form u(x, y) =X(x)Y (y) led tosolutions such as (21.14) or (21.15), or equivalent forms with x and y interchanged. Inthe current problem we have to satisfy the boundary conditions u(x, 0)=0=u(x, b) andso a solution that is sinusoidal in y seems appropriate. Furthermore, since we requireu(∞,y) = 0 it is best to write the x-dependence of the solution explicitly in terms of718

21.2 SUPERPOSITION OF SEPARATED SOLUTIONSybu =0u = f(y)u → 00 u =0xA semi-infinite metal plate whose edges are kept at fixed tem-Figure 21.1peratures.exponentials rather than of hyperbolic functions. We therefore write the separable solutionin the form (21.15) asu(x, y) =[A exp λx + B exp(−λx)](C cos λy + D sin λy).Applying the boundary conditions, we see firstly that u(∞,y) = 0 implies A =0ifwetake λ>0. Secondly, since u(x, 0) = 0 we may set C = 0, which, if we absorb the constantD into B, leaves us withu(x, y) =B exp(−λx)sinλy.But, using the condition u(x, b) = 0, we require sin λb =0andsoλ must be equal to nπ/b,where n is any positive integer.Using the principle of superposition (21.16), the general solution satisfying the givenboundary conditions can therefore be written∞∑u(x, y) = B n exp(−nπx/b) sin(nπy/b), (21.17)n=1for some constants B n . Notice that in the sum in (21.17) we have omitted negative values ofn since they would lead to exponential terms that diverge as x →∞.Then = 0 term is alsoomitted since it is identically zero. Using the remaining boundary condition u(0,y)=f(y)we see that the constants B n must satisfy∞∑f(y) = B n sin(nπy/b). (21.18)n=1This is clearly a Fourier sine series expansion of f(y) (see chapter 12). For (21.18) tohold, however, the continuation of f(y) outside the region 0 ≤ y ≤ b must be an oddperiodic function with period 2b (see figure 21.2). We also see from figure 21.2 that ifthe original function f(y) does not equal zero at either of y =0andy = b then itscontinuation has a discontinuity at the corresponding point(s); nevertheless, as discussedin chapter 12, the Fourier series will converge to the mid-points of these jumps and hencetend to zero in this case. If, however, the top and bottom edges of the plate were held notat 0 ◦ C but at some other non-zero temperature, then, in general, the final solution wouldpossess discontinuities at the corners x =0,y =0andx =0,y = b.Bearing in mind these technicalities, the coefficients B n in (21.18) are given byB n = 2 ∫ b ( nπy)f(y)sin dy. (21.19)bb0719

PDES: SEPARATION OF VARIABLES AND OTHER METHODSf(y)−b 0 b yFigure 21.2The continuation of f(y) for a Fourier sine series.Therefore, if f(y) =u 0 (i.e. the temperature of the side at x = 0 is constant along itslength), (21.19) becomes∫ bB n = 2 b 0[= − 2u 0bTherefore the required solution isu 0 sin( nπy)dyb) ] bb( nπynπ cos b= − 2u 0nπ [(−1)n − 1] =u(x, y) = ∑ n odd4u(0nπ exp0{4u0 /nπ for n odd,0 for n even.− nπxb)sin( nπy). ◭bIn the above example the boundary conditions meant that one term in eachpart of the separable solution could be immediately discarded, making the problemmuch easier to solve. Sometimes, however, a little ingenuity is required inwriting the separable solution in such a way that certain parts can be neglectedimmediately.◮Suppose that the semi-infinite rectangular metal plate in the previous example is replacedby one that in the x-direction has finite length a. The temperature of the right-hand edgeis fixed at 0 ◦ C and all other boundary conditions remain as before. Find the steady-statetemperature in the plate.As in the previous example, the boundary conditions u(x, 0) = 0 = u(x, b) suggest a solutionthat is sinusoidal in y. In this case, however, we require u =0onx = a (rather than atinfinity) and so a solution in which the x-dependence is written in terms of hyperbolicfunctions, such as (21.14), rather than exponentials is more appropriate. Moreover, sincethe constants in front of the hyperbolic functions are, at this stage, arbitrary, we maywrite the separable solution in the most convenient way that ensures that the conditionu(a, y) = 0 is straightforwardly satisfied. We therefore writeu(x, y) =[A cosh λ(a − x)+B sinh λ(a − x)](C cos λy + D sin λy).Now the condition u(a, y) = 0 is easily satisfied by setting A = 0. As before theconditions u(x, 0)=0=u(x, b) implyC =0andλ = nπ/b for integer n. Superposing the720

21.2 SUPERPOSITION OF SEPARATED SOLUTIONSsolutions for different n we then obtainu(x, y) =∞∑B n sinh[nπ(a − x)/b] sin(nπy/b), (21.20)n=1for some constants B n . We have omitted negative values of n in the sum (21.20) since therelevant terms are already included in those obtained for positive n. Again the n =0termis identically zero. Using the final boundary condition u(0,y)=f(y) as above we find thatthe constants B n must satisfyf(y) =∞∑B n sinh(nπa/b) sin(nπy/b),n=1and, remembering the caveats discussed in the previous example, the B n are therefore givenby∫ b2B n =f(y)sin(nπy/b) dy. (21.21)b sinh(nπa/b) 0For the case where f(y) =u 0 , following the working of the previous example gives(21.21) asB n =4u 0nπ sinh(nπa/b)for n odd, B n =0 forn even. (21.22)The required solution is thusu(x, y) = ∑ 4u 0nπ sinh(nπa/b) sinh[nπ(a − x)/b]sin( nπy/b ) .n oddWe note that, as required, in the limit a →∞this solution tends to the solution of theprevious example. ◭Often the principle of superposition can be used to write the solution toproblems with more complicated boundary conditions as the sum of solutions toproblems that each satisfy only some part of the boundary condition but whenadded togther satisfy all the conditions.◮Find the steady-state temperature in the (finite) rectangular plate of the previous example,subject to the boundary conditions u(x, b) =0, u(a, y) =0and u(0,y)=f(y) as before, butnow, in addition, u(x, 0) = g(x).Figure 21.3(c) shows the imposed boundary conditions for the metal plate. Although wecould find a solution to this problem using the methods presented above, we can arrive atthe answer almost immediately by using the principle of superposition and the result ofthe previous example.Let us suppose the required solution u(x, y) is made up of two parts:u(x, y) =v(x, y)+w(x, y),where v(x, y) is the solution satisfying the boundary conditions shown in figure 21.3(a),721

PDES: SEPARATION OF VARIABLES AND OTHER METHODSyyb0b0f(y) 0000axg(x)ax(a)(b)yb0f(y)0g(x)ax(c)Figure 21.3Superposition of boundary conditions for a metal plate.whilst w(x, y) is the solution satisfying the boundary conditions in figure 21.3(b). It is clearthat v(x, y) is simply given by the solution to the previous example,v(x, y) = ∑ [ nπ(a − x)B n sinhbn odd]sin( nπy),bwhere B n is given by (21.21). Moreover, by symmetry, w(x, y) must be of the same form asv(x, y) but with x and a interchanged with y and b, respectively, and with f(y) in (21.21)replaced by g(x). Therefore the required solution can be written down immediately withoutfurther calculation asu(x, y) = ∑ [ nπ(a − x)n sinhn oddBb]sinthe B n being given by (21.21) and the C n byC n =2a sinh(nπb/a)( nπy)+ ∑ n sinhbn oddC∫ a0g(x)sin(nπx/a) dx.[ nπ(b − y)a] ( nπx)sin ,aClearly, this method may be extended to cases in which three or four sides of the platehave non-zero boundary conditions. ◭As a final example of the usefulness of the principle of superposition we nowconsider a problem that illustrates how to deal with inhomogeneous boundaryconditions by a suitable change of variables.722

21.2 SUPERPOSITION OF SEPARATED SOLUTIONS◮A bar of length L is initially at a temperature of 0 ◦ C. One end of the bar (x =0)is heldat 0 ◦ C and the other is supplied with heat at a constant rate per unit area of H. Findthetemperature distribution within the bar after a time t.With our usual notation, the heat diffusion equation satisfied by the temperature u(x, t) isκ ∂2 u∂x = ∂u2 ∂t ,with κ = k/(sρ), where k is the thermal conductivity of the bar, s is its specific heatcapacity and ρ is its density.The boundary conditions can be written as∂u(L, t)u(x, 0) = 0, u(0,t)=0,= H ∂x k ,the last of which is inhomogeneous. In general, inhomogeneous boundary conditions cancause difficulties and it is usual to attempt a transformation of the problem into anequivalent homogeneous one. To this end, let us assume that the solution to our problemtakes the formu(x, t) =v(x, t)+w(x),where the function w(x) is to be suitably determined. In terms of v and w the problembecomes( )∂ 2 vκ∂x + d2 w= ∂v2 dx 2 ∂t ,v(x, 0) + w(x) =0,v(0,t)+w(0) = 0,∂v(L, t)∂x+ dw(L)dx= H k .There are several ways of choosing w(x) so as to make the new problem straightforward.Using some physical insight, however, it is clear that ultimately (at t = ∞), when alltransients have died away, the end x = L will attain a temperature u 0 such that ku 0 /L = Hand there will be a constant temperature gradient u(x, ∞) =u 0 x/L. We therefore choosew(x) = Hxk .Since the second derivative of w(x) is zero, v satisfies the diffusion equation and theboundary conditions on v are now given byv(x, 0) = − Hxk , v(0,t)=0, ∂v(L, t)=0,∂xwhich are homogeneous in x.From (21.12) a separated solution for the one-dimensional diffusion equation isv(x, t) =(A cos λx + B sin λx)exp(−λ 2 κt),corresponding to a separation constant −λ 2 .Ifwerestrictλ to be real then all thesesolutions are transient ones decaying to zero as t →∞. These are just what is required toadd to w(x) to give the correct solution as t →∞. In order to satisfy v(0,t) = 0, however,we require A = 0. Furthermore, since∂v∂x = B exp(−λ2 κt)λ cos λx,723

PDES: SEPARATION OF VARIABLES AND OTHER METHODSf(x)−L0Lx−HL/kFigure 21.4 The appropriate continuation for a Fourier series containingonly sine terms.in order to satisfy ∂v(L, t)/∂x =0werequirecosλL =0,andsoλ is restricted to thevaluesλ = nπ2L ,where n is an odd non-negative integer, i.e. n =1, 3, 5,... .Thus, to satisfy the boundary condition v(x, 0) = −Hx/k, we must have∑ ( nπx)B n sin = − Hx2L k ,n oddin the range x =0tox = L. In this case we must be more careful about the continuationof the function −Hx/k, for which the Fourier sine series is required. We want a series thatis odd in x (sine terms only) and continuous as x =0andx = L (no discontinuities, sincethe series must converge at the end-points). This leads to a continuation of the functionas shown in figure 21.4, with a period of L ′ =4L. Following the discussion of section 12.3,since this continuation is odd about x = 0 and even about x = L ′ /4=L it can indeed beexpressed as a Fourier sine series containing only odd-numbered terms.The corresponding Fourier series coefficients are found to beB n = −8HL (−1) (n−1)/2kπ 2 n 2and thus the final formula for u(x, t) isu(x, t) = Hxk− 8HLkπ 2∑n odd(−1) (n−1)/2n 2sinfor n odd,( nπx) ( )exp − kn2 π 2 t,2L 4L 2 sρgiving the temperature for all positions 0 ≤ x ≤ L and for all times t ≥ 0. ◭We note that in all the above examples the boundary conditions restricted theseparation constant(s) to an infinite number of discrete values, usually integers.If, however, the boundary conditions allow the separation constant(s) λ to takea continuum of values then the summation in (21.16) is replaced by an integralover λ. This is discussed further in connection with integral transform methodsin section 21.4.724

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES21.3 Separation of variables in polar coordinatesSo far we have considered the solution of PDEs only in Cartesian coordinates,but many systems in two and three dimensions are more naturally expressedin some form of polar coordinates, in which full advantage can be taken ofany inherent symmetries. For example, the potential associated with an isolatedpoint charge has a very simple expression, q/(4πɛ 0 r), when polar coordinates areused, but involves all three coordinates and square roots when Cartesians areemployed. For these reasons we now turn to the separation of variables in planepolar, cylindrical polar and spherical polar coordinates.Most of the PDEs we have considered so far have involved the operator ∇ 2 ,e.g.the wave equation, the diffusion equation, Schrödinger’s equation and Poisson’sequation (and of course Laplace’s equation). It is therefore appropriate that werecall the expressions for ∇ 2 when expressed in polar coordinate systems. Fromchapter 10, in plane polars, cylindrical polars and spherical polars, respectively,we have∇ 2 = 1 (∂ρ ∂ )+ 1 ∂ 2ρ ∂ρ ∂ρ ρ 2 ∂φ 2 , (21.23)∇ 2 = 1 (∂ρ ∂ )+ 1 ∂ 2ρ ∂ρ ∂ρ ρ 2 ∂φ 2 + ∂2∂z 2 , (21.24)∇ 2 = 1 (∂r 2 r 2 ∂ )+ 1 (∂∂r ∂r r 2 sin θ ∂ )1 ∂ 2+sin θ ∂θ ∂θ r 2 sin 2 θ ∂φ 2 . (21.25)Of course the first of these may be obtained from the second by taking z to beidentically zero.21.3.1 Laplace’s equation in polar coordinatesThe simplest of the equations containing ∇ 2 is Laplace’s equation,∇ 2 u(r) =0. (21.26)Since it contains most of the essential features of the other more complicatedequations, we will consider its solution first.Laplace’s equation in plane polarsSuppose that we need to find a solution of (21.26) that has a prescribed behaviouron the circle ρ = a (e.g. if we are finding the shape taken up by a circular drumskinwhen its rim is slightly deformed from being planar). Then we may seek solutionsof (21.26) that are separable in ρ and φ (measured from some arbitrary radiusas φ = 0) and hope to accommodate the boundary condition by examining thesolution for ρ = a.725

PDES: SEPARATION OF VARIABLES AND OTHER METHODSThus, writing u(ρ, φ) =P (ρ)Φ(φ) and using the expression (21.23), Laplace’sequation (21.26) becomes(Φ ∂ρ ∂P )+ P ∂ 2 Φρ ∂ρ ∂ρ ρ 2 ∂φ 2 =0.Now, employing the same device as previously, that of dividing through byu = P Φ and multiplying through by ρ 2 , results in the separated equation)ρP∂∂ρ(ρ ∂P∂ρ+ 1 ∂ 2 ΦΦ ∂φ 2 =0.Following our earlier argument, since the first term on the RHS is a function ofρ only, whilst the second term depends only on φ, we obtain the two ordinaryequationsρP(dρ dP )= n 2 , (21.27)dρ dρ1 d 2 ΦΦ dφ 2 = −n2 , (21.28)where we have taken the separation constant to have the form n 2 for laterconvenience; for the present, n is a general (complex) number.Let us first consider the case in which n ≠ 0. The second equation, (21.28), thenhas the general solutionΦ(φ) =A exp(inφ)+B exp(−inφ). (21.29)Equation (21.27), on the other hand, is the homogeneous equationρ 2 P ′′ + ρP ′ − n 2 P =0,which must be solved either by trying a power solution in ρ or by making thesubstitution ρ =expt as described in subsection 15.2.1 and so reducing it to anequation with constant coefficients. Carrying out this procedure we findP (ρ) =Cρ n + Dρ −n . (21.30)Returning to the solution (21.29) of the azimuthal equation (21.28), we cansee that if Φ, and hence u, is to be single-valued and so not change when φincreases by 2π then n must be an integer. Mathematically, other values of n arepermissible, but for the description of real physical situations it is clear that thislimitation must be imposed. Having thus restricted the possible values of n inone part of the solution, the same limitations must be carried over into the radialpart, (21.30). Thus we may write a particular solution of the two-dimensionalLaplace equation asu(ρ, φ) =(A cos nφ + B sin nφ)(Cρ n + Dρ −n ),726

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESwhere A, B, C, D are arbitrary constants and n is any integer.We have not yet, however, considered the solution when n =0.Inthiscase,the solutions of the separated ordinary equations (21.28) and (21.27), respectively,are easily shown to beΦ(φ) =Aφ + B,P (ρ) =C ln ρ + D.But, in order that u = P Φ is single-valued, we require A = 0, and so the solutionfor n = 0 is simply (absorbing B into C and D)u(ρ, φ) =C ln ρ + D.Superposing the solutions for the different allowed values of n, we can writethe general solution to Laplace’s equation in plane polars asu(ρ, φ) =(C 0 ln ρ + D 0 )+∞∑n=1(A n cos nφ + B n sin nφ)(C n ρ n + D n ρ −n ),(21.31)where n can take only integer values. Negative values of n have been omittedfrom the sum since they are already included in the terms obtained for positiven. We note that, since ln ρ is singular at ρ = 0, whenever we solve Laplace’sequation in a region containing the origin, C 0 must be identically zero.◮A circular drumskin has a supporting rim at ρ = a. If the rim is twisted so that itis displaced vertically by a small amount ɛ(sin φ +2sin2φ), whereφ is the azimuthalangle with respect to a given radius, find the resulting displacement u(ρ, φ) over the entiredrumskin.The transverse displacement of a circular drumskin is usually described by the twodimensionalwave equation. In this case, however, there is no time dependence and sou(ρ, φ) solves the two-dimensional Laplace equation, subject to the imposed boundarycondition.Referring to (21.31), since we wish to find a solution that is finite everywhere insideρ = a, werequireC 0 =0andD n =0foralln>0. Now the boundary condition at therim requiresu(a, φ) =D 0 +∞∑C n a n (A n cos nφ + B n sin nφ) =ɛ(sin φ +2sin2φ).n=1Firstly we see that we require D 0 =0andA n =0foralln. Furthermore, we musthave C 1 B 1 a = ɛ, C 2 B 2 a 2 =2ɛ and B n =0forn>2. Hence the appropriate shape for thedrumskin (valid over the whole skin, not just the rim) isu(ρ, φ) = ɛρ asin φ +2ɛρ2a 2sin 2φ = ɛρ a(sin φ + 2ρ a sin 2φ ). ◭727

PDES: SEPARATION OF VARIABLES AND OTHER METHODSLaplace’s equation in cylindrical polarsPassing to three dimensions, we now consider the solution of Laplace’s equationin cylindrical polar coordinates,(1 ∂ρ ∂u )+ 1 ∂ 2 uρ ∂ρ ∂ρ ρ 2 ∂φ 2 + ∂2 u=0. (21.32)∂z2 We note here that, even when considering a cylindrical physical system, if thereis no dependence of the physical variables on z (i.e. along the length of thecylinder) then the problem may be treated using two-dimensional plane polars,as discussed above.For the more general case, however, we proceed as previously by trying asolution of the formu(ρ, φ, z) =P (ρ)Φ(φ)Z(z),which, on substitution into (21.32) and division through by u = P ΦZ, gives(1 dρ dP )+ 1 d 2 ΦPρdρdρ Φρ 2 dφ 2 + 1 d 2 ZZ dz 2 =0.The last term depends only on z, and the first and second (taken together) dependonly on ρ and φ. Taking the separation constant to be k 2 , we find1 d 2 ZZ dz 2 = k2 ,(1 dρ dP )+ 1 d 2 ΦPρdρdρ Φρ 2 dφ 2 + k2 =0.The first of these equations has the straightforward solutionZ(z) =E exp(−kz)+F exp kz.Multiplying the second equation through by ρ 2 ,weobtain(ρ dρ dP )+ 1 d 2 ΦP dρ dρ Φ dφ 2 + k2 ρ 2 =0,in which the second term depends only on Φ and the other terms depend onlyon ρ. Taking the second separation constant to be m 2 , we find1 d 2 ΦΦ dφ 2 = −m2 , (21.33)ρ d (ρ dP )+(k 2 ρ 2 − m 2 )P =0. (21.34)dρ dρThe equation in the azimuthal angle φ has the very familiar solutionΦ(φ) =C cos mφ + D sin mφ.728

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESAs in the two-dimensional case, single-valuedness of u requires that m is aninteger. However, in the particular case m = 0 the solution isΦ(φ) =Cφ + D.This form is appropriate to a solution with axial symmetry (C = 0) or one that ismultivalued, but manageably so, such as the magnetic scalar potential associatedwith a current I (in which case C = I/(2π) andD is arbitrary).Finally, the ρ-equation (21.34) may be transformed into Bessel’s equation oforder m by writing µ = kρ. This has the solutionP (ρ) =AJ m (kρ)+BY m (kρ).The properties of these functions were investigated in chapter 16 and will notbe pursued here. We merely note that Y m (kρ) is singular at ρ = 0, and so, whenseeking solutions to Laplace’s equation in cylindrical coordinates within someregion containing the ρ = 0 axis, we require B =0.The complete separated-variable solution in cylindrical polars of Laplace’sequation ∇ 2 u = 0 is thus given byu(ρ, φ, z) =[AJ m (kρ)+BY m (kρ)][C cos mφ + D sin mφ][E exp(−kz)+F exp kz].(21.35)Of course we may use the principle of superposition to build up more generalsolutions by adding together solutions of the form (21.35) for all allowed valuesof the separation constants k and m.◮A semi-infinite solid cylinder of radius a has its curved surface held at 0 ◦ C and its baseheld at a temperature T 0 . Find the steady-state temperature distribution in the cylinder.The physical situation is shown in figure 21.5. The steady-state temperature distributionu(ρ, φ, z) must satisfy Laplace’s equation subject to the imposed boundary conditions. Letus take the cylinder to have its base in the z = 0 plane and to extend along the positivez-axis. From (21.35), in order that u is finite everywhere in the cylinder we immediatelyrequire B =0andF = 0. Furthermore, since the boundary conditions, and hence thetemperature distribution, are axially symmetric, we require m =0,andsothegeneralsolution must be a superposition of solutions of the form J 0 (kρ)exp(−kz) for all allowedvalues of the separation constant k.The boundary condition u(a, φ, z) = 0 restricts the allowed values of k, sincewemusthave J 0 (ka) = 0. The zeros of Bessel functions are given in most books of mathematicaltables, and we find that, to two decimal places,J 0 (x) =0 forx =2.40, 5.52, 8.65, ....Writing the allowed values of k as k n for n =1, 2, 3,... (so, for example, k 1 =2.40/a), therequired solution takes the formu(ρ, φ, z) =∞∑A n J 0 (k n ρ)exp(−k n z).n=1729

PDES: SEPARATION OF VARIABLES AND OTHER METHODSzu =0 u =0ayxu = T 0Figure 21.5 A uniform metal cylinder whose curved surface is kept at 0 ◦ Cand whose base is held at a temperature T 0 .By imposing the remaining boundary condition u(ρ, φ, 0) = T 0 , the coefficients A n can befound in a similar way to Fourier coefficients but this time by exploiting the orthogonalityof the Bessel functions, as discussed in chapter 16. From this boundary condition werequire∞∑u(ρ, φ, 0) = A n J 0 (k n ρ)=T 0 .n=1If we multiply this expression by ρJ 0 (k r ρ) and integrate from ρ =0toρ = a, and use theorthogonality of the Bessel functions J 0 (k n ρ), then the coefficients are given by (18.91) asA n = 2T 0a 2 J1 2(k J 0 (k n ρ)ρdρ. (21.36)na) 0The integral on the RHS can be evaluated using the recurrence relation (18.92) ofchapter 16,ddz [zJ 1(z)] = zJ 0 (z),whichonsettingz = k n ρ yields1 dk n dρ [k nρJ 1 (k n ρ)] = k n ρJ 0 (k n ρ).Therefore the integral in (21.36) is given by∫ a[ 1J 0 (k n ρ)ρdρ= ρJ 1 (k n ρ) = 1 aJ 1 (k n a),k n k n0∫ a730] a0

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESand the coefficients A n may be expressed asA n = 2T 0a 2 J 2 1 (k na)[ ] aJ1 (k n a)=k n2T 0k n aJ 1 (k n a) .The steady-state temperature in the cylinder is then given by∞∑ 2T 0u(ρ, φ, z) =k n aJ 1 (k n a) J 0(k n ρ)exp(−k n z). ◭n=1We note that if, in the above example, the base of the cylinder were not kept ata uniform temperature T 0 , but instead had some fixed temperature distributionT (ρ, φ), then the solution of the problem would become more complicated. Insuch a case, the required temperature distribution u(ρ, φ, z)isingeneralnot axiallysymmetric, and so the separation constant m is not restricted to be zero but maytake any integer value. The solution will then take the form∞∑ ∞∑u(ρ, φ, z) = J m (k nm ρ)(C nm cos mφ + D nm sin mφ) exp(−k nm z),m=0 n=1where the separation constants k nm are such that J m (k nm a) = 0, i.e. k nm a is the nthzero of the mth-order Bessel function. At the base of the cylinder we would thenrequire∞∑ ∞∑u(ρ, φ, 0) = J m (k nm ρ)(C nm cos mφ + D nm sin mφ) =T (ρ, φ).m=0 n=1(21.37)The coefficients C nm could be found by multiplying (21.37) by J q (k rq ρ)cosqφ,integrating with respect to ρ and φ over the base of the cylinder and exploitingthe orthogonality of the Bessel functions and of the trigonometric functions. TheD nm could be found in a similar way by multiplying (21.37) by J q (k rq ρ)sinqφ.Laplace’s equation in spherical polarsWe now come to an equation that is very widely applicable in physical science,namely ∇ 2 u = 0 in spherical polar coordinates:(1 ∂r 2 r 2 ∂u )+ 1 (∂∂r ∂r r 2 sin θ ∂u )1 ∂ 2 u+sin θ ∂θ ∂θ r 2 sin 2 =0. (21.38)θ ∂φ2 Our method of procedure will be as before; we try a solution of the formu(r, θ, φ) =R(r)Θ(θ)Φ(φ).Substituting this in (21.38), dividing through by u = RΘΦ and multiplying by r 2 ,we obtain(1 dr 2 dR )+ 1 (dsin θ dΘ )1 d 2 Φ+R dr dr Θsinθ dθ dθ Φsin 2 =0. (21.39)θ dφ2 731

PDES: SEPARATION OF VARIABLES AND OTHER METHODSThe first term depends only on r and the second and third terms (taken together)depend only on θ and φ. Thus (21.39) is equivalent to the two equations(1 dr 2 dR )= λ, (21.40)R dr dr(1 dsin θ dΘ )1 d 2 Φ+Θsinθ dθ dθ Φsin 2 = −λ. (21.41)θ dφ2 Equation (21.40) is a homogeneous equation,r 2 d2 Rdr 2dR+2r − λR =0,drwhich can be reduced, by the substitution r =expt (and writing R(r) =S(t)), tod 2 Sdt 2This has the straightforward solution+ dS − λS =0.dtS(t) =A exp λ 1 t + B exp λ 2 t,and so the solution to the radial equation isR(r) =Ar λ1 + Br λ2 ,where λ 1 + λ 2 = −1 andλ 1 λ 2 = −λ. We can thus take λ 1 and λ 2 as given by land −(l +1);λ then has the form l(l + 1). (It should be noted that at this stagenothing has been either assumed or proved about whether l is an integer.)Hence we have obtained some information about the first factor in theseparated-variable solution, which will now have the formu(r, θ, φ) = [ Ar l + Br −(l+1)] Θ(θ)Φ(φ), (21.42)where Θ and Φ must satisfy (21.41) with λ = l(l +1).The next step is to take (21.41) further. Multiplying through by sin 2 θ andsubstituting for λ, it too takes a separated form:[ sin θΘddθ(sin θ dΘdθ)+ l(l +1)sin 2 θ]+ 1 d 2 Φ=0. (21.43)Φ dφ2 Taking the separation constant as m 2 , the equation in the azimuthal angle φhas the same solution as in cylindrical polars, namelyΦ(φ) =C cos mφ + D sin mφ.As before, single-valuedness of u requires that m is an integer; for m = 0 we againhave Φ(φ) =Cφ + D.732

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESHaving settled the form of Φ(φ), we are left only with the equation satisfied byΘ(θ), which is(sin θ dsin θ dΘ )+ l(l +1)sin 2 θ = m 2 . (21.44)Θ dθ dθA change of independent variable from θ to µ =cosθ will reduce this to aform for which solutions are known, and of which some study has been made inchapter 16. Puttingµ =cosθ,dµd= − sin θ,dθdθ = −(1 − µ2 ) 1/2 ddµ ,the equation for M(µ) ≡ Θ(θ) reads[d(1 − µ 2 ) dM ]]+[l(l +1)− m2dµ dµ1 − µ 2 M =0. (21.45)This equation is the associated Legendre equation, which was mentioned in subsection18.2 in the context of Sturm–Liouville equations.We recall that for the case m = 0, (21.45) reduces to Legendre’s equation, whichwas studied at length in chapter 16, and has the solutionM(µ) =EP l (µ)+FQ l (µ). (21.46)We have not solved (21.45) explicitly for general m, but the solutions were givenin subsection 18.2 and are the associated Legendre functions P m l (µ) andQm l (µ),wherePl m (µ) =(1− µ 2 |m|/2 d|m|)dµ P l(µ), (21.47)|m|and similarly for Q m l (µ). We then haveM(µ) =EP m l (µ)+FQ m l (µ); (21.48)here m must be an integer, 0 ≤|m| ≤l. We note that if we require solutions toLaplace’s equation that are finite when µ =cosθ = ±1 (i.e. on the polar axiswhere θ =0,π), then we must have F = 0 in (21.46) and (21.48) since Q m l (µ)diverges at µ = ±1.It will be remembered that one of the important conditions for obtainingfinite polynomial solutions of Legendre’s equation is that l is an integer ≥ 0.This condition therefore applies also to the solutions (21.46) and (21.48) and isreflected back into the radial part of the general solution given in (21.42).Now that the solutions of each of the three ordinary differential equationsgoverning R, Θ and Φ have been obtained, we may assemble a complete separated-733

PDES: SEPARATION OF VARIABLES AND OTHER METHODSvariable solution of Laplace’s equation in spherical polars. It isu(r, θ, φ) =(Ar l + Br −(l+1) )(C cos mφ + D sin mφ)[EP m l (cos θ)+FQ m l (cos θ)],(21.49)where the three bracketted factors are connected only through the integer parametersl and m, 0≤|m| ≤l. As before, a general solution may be obtainedby superposing solutions of this form for the allowed values of the separationconstants l and m. As mentioned above, if the solution is required to be finite onthe polar axis then F = 0 for all l and m.◮An uncharged conducting sphere of radius a is placed at the origin in an initially uniformelectrostatic field E. Show that it behaves as an electric dipole.The uniform field, taken in the direction of the polar axis, has an electrostatic potentialu = −Ez = −Ercos θ,where u is arbitrarily taken as zero at z = 0. This satisfies Laplace’s equation ∇ 2 u =0,asmust the potential v when the sphere is present; for large r the asymptotic form of v muststill be −Er cos θ.Since the problem is clearly axially symmetric, we have immediately that m =0,andsincewerequirev to be finite on the polar axis we must have F = 0 in (21.49). Thereforethe solution must be of the form∞∑v(r, θ, φ) = (A l r l + B l r −(l+1) )P l (cos θ).l=0Now the cos θ-dependence of v for large r indicates that the (θ, φ)-dependence of v(r, θ, φ)is given by P1 0 (cos θ) =cosθ. Thus the r-dependence of v must also correspond to anl = 1 solution, and the most general such solution (outside the sphere, i.e. for r ≥ a) isv(r, θ, φ) =(A 1 r + B 1 r −2 )P 1 (cos θ).Theasymptoticformofv for large r immediately gives A 1 = −E and so yields the solution(v(r, θ, φ) = −Er + B )1cos θ.r 2Since the sphere is conducting, it is an equipotential region and so v must not depend onθ for r = a. This can only be the case if B 1 /a 2 = Ea, thus fixing B 1 . The final solution istherefore)v(r, θ, φ) =−Er(1 − a3cos θ.r 3Since a dipole of moment p gives rise to a potential p/(4πɛ 0 r 2 ), this result shows that thesphere behaves as a dipole of moment 4πɛ 0 a 3 E, because of the charge distribution inducedon its surface; see figure 21.6. ◭Often the boundary conditions are not so easily met, and it is necessary touse the mutual orthogonality of the associated Legendre functions (and thetrigonometric functions) to obtain the coefficients in the general solution.734

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES− +++−− − ++− − θ +− a +− +−+−+− + − +Figure 21.6 Induced charge and field lines associated with a conductingsphere placed in an initially uniform electrostatic field.◮A hollow split conducting sphere of radius a is placed at the origin. If one half of itssurface is charged to a potential v 0 and the other half is kept at zero potential, find thepotential v inside and outside the sphere.Let us choose the top hemisphere to be charged to v 0 and the bottom hemisphere to beat zero potential, with the plane in which the two hemispheres meet perpendicular to thepolar axis; this is shown in figure 21.7. The boundary condition then becomes{v0 for 0

PDES: SEPARATION OF VARIABLES AND OTHER METHODSzav = v 0θrφyxv =0−aFigure 21.7 A hollow split conducting sphere with its top half charged to apotential v 0 and its bottom half at zero potential.where in the last line we have used (21.50). The integrals of the Legendre polynomials areeasily evaluated (see exercise 17.3) and we findA 0 = v 02 , A 1 = 3v 04a , A 2 =0, A 3 = − 7v 016a 3 , ··· ,so that the required solution inside the sphere isv(r, θ, φ) = v 02[1+ 3r2a P 1(cos θ) − 7r38a 3 P 3(cos θ)+···Outside the sphere (for r>a) we require the solution to be bounded as r tends toinfinity and so in (21.51) we must have A l =0foralll. In this case, by imposing theboundary condition at r = a we require∞∑v(a, θ, φ) = B l a −(l+1) P l (cos θ),l=0where v(a, θ, φ) is given by (21.50). Following the above argument the coefficients in theexpansion are given byB l a −(l+1) =2l +12so that the required solution outside the sphere isv(r, θ, φ) = v 0a2r∫ 1v 0 P l (µ)dµ,[1+ 3a2r P 1(cos θ) − 7a38r 3 P 3(cos θ)+···0].]. ◭736

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESIn the above example, on the equator of the sphere (i.e. at r = a and θ = π/2)the potential is given byv(a, π/2,φ)=v 0 /2,i.e. mid-way between the potentials of the top and bottom hemispheres. This isso because a Legendre polynomial expansion of a function behaves in the sameway as a Fourier series expansion, in that it converges to the average of the twovalues at any discontinuities present in the original function.If the potential on the surface of the sphere had been given as a function of θand φ, then we would have had to consider a double series summed over l andm (for −l ≤ m ≤ l), since, in general, the solution would not have been axiallysymmetric.Finally, we note in general that, when obtaining solutions of Laplace’s equationin spherical polar coordinates, one finds that, for solutions that are finite on thepolar axis, the angular part of the solution is given byΘ(θ)Φ(φ) =P m l (cos θ)(C cos mφ + D sin mφ),where l and m are integers with −l ≤ m ≤ l. This general form is sufficientlycommon that particular functions of θ and φ called spherical harmonics aredefined and tabulated (see section 18.3).21.3.2 Other equations in polar coordinatesThe development of the solutions of ∇ 2 u = 0 carried out in the previous subsectioncan be employed to solve other equations in which the ∇ 2 operator appears. Sincewe have discussed the general method in some depth already, only an outline ofthe solutions will be given here.Let us first consider the wave equation∇ 2 u = 1 ∂ 2 uc 2 ∂t 2 , (21.52)and look for a separated solution of the form u = F(r)T (t), so that initially weare separating only the spatial and time dependences. Substituting this form into(21.52) and taking the separation constant as k 2 we obtain∇ 2 F + k 2 d 2 TF =0,dt 2 + k2 c 2 T =0. (21.53)The second equation has the simple solutionT (t) =A exp(iωt)+B exp(−iωt), (21.54)where ω = kc; this may also be expressed in terms of sines and cosines, of course.The first equation in (21.53) is referred to as Helmholtz’s equation; we discuss itbelow.737

PDES: SEPARATION OF VARIABLES AND OTHER METHODSWe may treat the diffusion equationκ∇ 2 u = ∂u∂tin a similar way. Separating the spatial and time dependences by assuming asolution of the form u = F(r)T (t), and taking the separation constant as k 2 ,wefind∇ 2 F + k 2 dTF =0,dt + k2 κT =0.Just as in the case of the wave equation, the spatial part of the solution satisfiesHelmholtz’s equation. It only remains to consider the time dependence, whichhas the simple solutionT (t) =A exp(−k 2 κt).Helmholtz’s equation is clearly of central importance in the solutions of thewave and diffusion equations. It can be solved in polar coordinates in much thesame way as Laplace’s equation, and indeed reduces to Laplace’s equation whenk = 0. Therefore, we will merely sketch the method of its solution in each of thethree polar coordinate systems.Helmholtz’s equation in plane polarsIn two-dimensional plane polar coordinates, Helmholtz’s equation takes the form(1 ∂ρ ∂F )+ 1 ∂ 2 Fρ ∂ρ ∂ρ ρ 2 ∂φ 2 + k2 F =0.If we try a separated solution of the form F(r) = P (ρ)Φ(φ), and take theseparation constant as m 2 , we findd 2 Φdφ 2 + m2 φ =0,d 2 Pdρ 2 + 1 ( )dPρ dρ + k 2 − m2ρ 2 P =0.As for Laplace’s equation, the angular part has the familiar solution (if m ≠0)Φ(φ) =A cos mφ + B sin mφ,or an equivalent form in terms of complex exponentials. The radial equationdiffers from that found in the solution of Laplace’s equation, but by making thesubstitution µ = kρ it is easily transformed into Bessel’s equation of order m(discussed in chapter 16), and has the solutionP (ρ) =CJ m (kρ)+DY m (kρ),where Y m is a Bessel function of the second kind, which is infinite at the origin738

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESand is not to be confused with a spherical harmonic (these are written with asuperscript as well as a subscript).Putting the two parts of the solution together we haveF(ρ, φ) =[A cos mφ + B sin mφ][CJ m (kρ)+DY m (kρ)]. (21.55)Clearly, for solutions of Helmholtz’s equation that are required to be finite at theorigin, we must set D =0.◮Find the four lowest frequency modes of oscillation of a circular drumskin of radius awhose circumference is held fixed in a plane.The transverse displacement u(r,t) of the drumskin satisfies the two-dimensional waveequation∇ 2 u = 1 ∂ 2 uc 2 ∂t , 2with c 2 = T/σ,whereT is the tension of the drumskin and σ is its mass per unit area.From (21.54) and (21.55) a separated solution of this equation, in plane polar coordinates,that is finite at the origin isu(ρ, φ, t) =J m (kρ)(A cos mφ + B sin mφ)exp(±iωt),where ω = kc. Since we require the solution to be single-valued we must have m as aninteger. Furthermore, if the drumskin is clamped at its outer edge ρ = a then we alsorequire u(a, φ, t) = 0. Thus we needJ m (ka) =0,which in turn restricts the allowed values of k. The zeros of Bessel functions can beobtained from most books of tables; the first few areJ 0 (x) =0 forx ≈ 2.40, 5.52, 8.65,...,J 1 (x) =0 forx ≈ 3.83, 7.02, 10.17,...,J 2 (x) =0 forx ≈ 5.14, 8.42, 11.62 ... .The smallest value of x for which any of the Bessel functions is zero is x ≈ 2.40, whichoccurs for J 0 (x). Thus the lowest-frequency mode has k =2.40/a and angular frequencyω =2.40c/a. Sincem = 0 for this mode, the shape of the drumskin is(u ∝ J 0 2.40 ρ );athis is illustrated in figure 21.8.Continuing in the same way, the next three modes are given byω =3.83 c (a , u ∝ J 1 3.83 ρ ) (cos φ, J 1 3.83 ρ )sin φ;aaω =5.14 c (a , u ∝ J 2 5.14 ρ ) (cos 2φ, J 2 5.14 ρ )sin 2φ;aaω =5.52 c (a , u ∝ J 0 5.52 ρ ).aThese modes are also shown in figure 21.8. We note that the second and third frequencieshave two corresponding modes of oscillation; these frequencies are therefore two-folddegenerate. ◭739

PDES: SEPARATION OF VARIABLES AND OTHER METHODSaω =2.40c/aω =3.83c/aω =5.14c/aω =5.52c/aFigure 21.8 The modes of oscillation with the four lowest frequencies for acircular drumskin of radius a. The dashed lines indicate the nodes, where thedisplacement of the drumskin is always zero.Helmholtz’s equation in cylindrical polarsGeneralising the above method to three-dimensional cylindrical polars is straightforward,and following a similar procedure to that used for Laplace’s equationwe find the separated solution of Helmholtz’s equation takes the form[ (√ ) (√ )]F(ρ, φ, z) = AJ m k2 − α 2 ρ + BY m k2 − α 2 ρ× (C cos mφ + D sin mφ)[E exp(iαz)+F exp(−iαz)],where α and m are separation constants. We note that the angular part of thesolution is the same as for Laplace’s equation in cylindrical polars.Helmholtz’s equation in spherical polarsIn spherical polars, we find again that the angular parts of the solution Θ(θ)Φ(φ)are identical to those of Laplace’s equation in this coordinate system, i.e. they arethe spherical harmonics Yl m (θ, φ), and so we shall not discuss them further.The radial equation in this case is given byr 2 R ′′ +2rR ′ +[k 2 r 2 − l(l +1)]R =0, (21.56)which has an additional term k 2 r 2 R compared with the radial equation for theLaplace solution. The equation (21.56) looks very much like Bessel’s equation.In fact, by writing R(r) =r −1/2 S(r) and making the change of variable µ = kr,it can be reduced to Bessel’s equation of order l + 1 2, which has as its solutionsS(µ) = J l+1/2 (µ) and Y l+1/2 (µ) (see section 18.6). The separated solution to740

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESHelmholtz’s equation in spherical polars is thusF(r, θ, φ) =r −1/2 [AJ l+1/2 (kr)+BY l+1/2 (kr)](C cos mφ + D sin mφ)×[EPl m (cos θ)+FQ m l (cos θ)]. (21.57)For solutions that are finite at the origin we require B = 0, and for solutionsthat are finite on the polar axis we require F = 0. It is worth mentioning thatthe solutions proportional to r −1/2 J l+1/2 (kr) andr −1/2 Y l+1/2 (kr), when suitablynormalised, are called spherical Bessel functions of the first and second kind,respectively, and are denoted by j l (kr) andn l (µ) (see section 18.6).As mentioned at the beginning of this subsection, the separated solution ofthe wave equation in spherical polars is the product of a time-dependent part(21.54) and a spatial part (21.57). It will be noticed that, although this solutioncorresponds to a solution of definite frequency ω = kc, the zeros of the radialfunction j l (kr) are not equally spaced in r, except for the case l = 0 involvingj 0 (kr), and so there is no precise wavelength associated with the solution.To conclude this subsection, let us mention briefly the Schrödinger equationfor the electron in a hydrogen atom, the nucleus of which is taken at the originand is assumed massive compared with the electron. Under these circumstancesthe Schrödinger equation is− 22m ∇2 u −e2 u4πɛ 0 r = i∂u ∂t .For a ‘stationary-state’ solution, for which the energy is a constant E and the timedependentfactor T in u is given by T (t) =A exp(−iEt/), the above equation issimilar to, but not quite the same as, the Helmholtz equation. § However, as withthe wave equation, the angular parts of the solution are identical to those forLaplace’s equation and are expressed in terms of spherical harmonics.The important point to note is that for any equation involving ∇ 2 , provided θand φ do not appear in the equation other than as part of ∇ 2 , a separated-variablesolution in spherical polars will always lead to spherical harmonic solutions. Thisis the case for the Schrödinger equation describing an atomic electron in a centralpotential V (r).21.3.3 Solution by expansionIt is sometimes possible to use the uniqueness theorem discussed in the previouschapter, together with the results of the last few subsections, in which Laplace’sequation (and other equations) were considered in polar coordinates, to obtainsolutions of such equations appropriate to particular physical situations.§ For the solution by series of the r-equation in this case the reader may consult, for example, L.Schiff, Quantum Mechanics (New York: McGraw-Hill, 1955), p. 82.741

PDES: SEPARATION OF VARIABLES AND OTHER METHODSPzθr−aOayxFigure 21.9 The polar axis Oz is taken as normal to the plane of the ring ofmatter and passing through its centre.We will illustrate the method for Laplace’s equation in spherical polars and firstassume that the required solution of ∇ 2 u = 0 can be written as a superpositionin the normal way:u(r, θ, φ) =∞∑l=0 m=−ll∑(Ar l + Br −(l+1) )Pl m (cos θ)(C cos mφ + D sin mφ).(21.58)Here, all the constants A, B, C, D may depend upon l and m, and we haveassumed that the required solution is finite on the polar axis. As usual, boundaryconditions of a physical nature will then fix or eliminate some of the constants;for example, u finite at the origin implies all B = 0, or axial symmetry impliesthat only m = 0 terms are present.The essence of the method is then to find the remaining constants by determiningu at values of r, θ, φ for which it can be evaluated by other means, e.g. by directcalculation on an axis of symmetry. Once the remaining constants have been fixedby these special considerations to have particular values, the uniqueness theoremcan be invoked to establish that they must have these values in general.◮Calculate the gravitational potential at a general point in space due to a uniform ring ofmatter of radius a and total mass M.Everywhere except on the ring the potential u(r) satisfies the Laplace equation, and so ifwe use polar coordinates with the normal to the ring as polar axis, as in figure 21.9, asolution of the form (21.58) can be assumed.We expect the potential u(r, θ, φ) totendtozeroasr →∞, and also to be finite at r =0.At first sight this might seem to imply that all A and B, and hence u, must be identicallyzero, an unacceptable result. In fact, what it means is that different expressions must applyto different regions of space. On the ring itself we no longer have ∇ 2 u = 0 and so it is not742

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESsurprising that the form of the expression for u changes there. Let us therefore take twoseparate regions.In the region r>a(i) we must have u → 0asr →∞, implying that all A =0,and(ii) the system is axially symmetric and so only m = 0 terms appear.With these restrictions we can write as a trial form∞∑u(r, θ, φ) = B l r −(l+1) Pl 0 (cos θ). (21.59)l=0The constants B l are still to be determined; this we do by calculating directly the potentialwhere this can be done simply – in this case, on the polar axis.Considering a point P on the polar axis at a distance z (> a) from the plane of the ring(taken as θ = π/2), all parts of the ring are at a distance (z 2 + a 2 ) 1/2 from it. The potentialat P is thus straightforwardlyGMu(z,0,φ)=− , (21.60)(z 2 + a 2 )1/2where G is the gravitational constant. This must be the same as (21.59) for the particularvalues r = z, θ =0,andφ undefined. Since Pl 0(cos θ) =P l(cos θ) withP l (1) = 1, puttingr = z in (21.59) gives∞∑ B lu(z,0,φ)= . (21.61)zl+1 l=0However, expanding (21.60) for z>a(as it applies to this region of space) we obtainu(z,0,φ)=− GM [1 − 1 ( a) 2 3( a) 4+ − ···],z 2 z 8 zwhich on comparison with (21.61) gives §B 0 = −GM,B 2l = − GMa2l (−1) l (2l − 1)!!for l ≥ 1, (21.62)2 l l!B 2l+1 =0.We now conclude the argument by saying that if a solution for a general point (r, θ, φ)exists at all, which of course we very much expect on physical grounds, then it must be(21.59) with the B l given by (21.62). This is so because thus defined it is a function withno arbitrary constants and which satisfies all the boundary conditions, and the uniquenesstheorem states that there is only one such function. The expression for the potential in theregion r>ais therefore[]u(r, θ, φ) =− GM ∞∑ (−1) l (2l − 1)!!( a) 2l1+P2l (cos θ) .r2 l l! rl=1The expression for r

PDES: SEPARATION OF VARIABLES AND OTHER METHODSComparing this expression for r = z, θ = 0 with the z

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESwhere the coefficients R l (r) andF l (r) in the Legendre polynomial expansionsare functions of r. Since in any particular problem ρ is given, we can find thecoefficients F l (r) in the expansion in the usual way (see subsection 18.1.2). It thenonly remains to find the coefficients R l (r) in the expansion of the solution u.Writing ∇ 2 in spherical polars and substituting (21.64) and (21.65) into (21.63)we obtain∞∑[Pl (cos θ)l=0r 2ddr(r 2 dR )l+ R ldrdr 2 sin θ dθ(sin θ dP )]l(cos θ)=dθ∞∑F l (r)P l (cos θ).l=0(21.66)However, if, in equation (21.44) of our discussion of the angular part of thesolution to Laplace’s equation, we set m = 0 we conclude that(1 dsin θ dP )l(cos θ)= −l(l +1)P l (cos θ).sin θ dθ dθSubstituting this into (21.66), we find that the LHS is greatly simplified and weobtain∞∑[ ( 1 dr 2 r 2 dR )l− l(l +1)R ]∞∑ldr dr r 2 P l (cos θ) = F l (r)P l (cos θ).l=0This relation is most easily satisfied by equating terms on both sides for eachvalue of l separately, so that for l =0, 1, 2,... we have(1 dr 2 r 2 dR )l− l(l +1)R ldr dr r 2 = F l (r). (21.67)This is an ODE in which F l (r) is given, and it can therefore be solved forR l (r). The solution to Poisson’s equation, u, is then obtained by making thesuperposition (21.64).◮In a certain system, the electric charge density ρ is distributed as follows:{Ar cos θ for 0 ≤ r

PDES: SEPARATION OF VARIABLES AND OTHER METHODSThis can be rearranged to giver 2 R 1 ′′ +2rR 1 ′ − 2R 1 = − Ar3 ,ɛ 0where the prime denotes differentiation with respect to r. The LHS is homogeneous andthe equation can be reduced by the substitution r =expt, and writing R 1 (r) =S(t), to¨S + Ṡ − 2S = − A ɛ 0exp 3t, (21.68)where the dots indicate differentiation with respect to t.This is an inhomogeneous second-order ODE with constant coefficients and can bestraightforwardly solved by the methods of subsection 15.2.1 to giveRecalling that r =expt we findS(t) =c 1 exp t + c 2 exp(−2t) −A10ɛ 0exp 3t.R 1 (r) =c 1 r + c 2 r −2 −A r 3 .10ɛ 0Since we are interested in the region r

21.4 INTEGRAL TRANSFORM METHODS21.4 Integral transform methodsIn the method of separation of variables our aim was to keep the independentvariables in a PDE as separate as possible. We now discuss the use of integraltransforms in solving PDEs, a method by which one of the independent variablescan be eliminated from the differential coefficients. It will be assumed that thereader is familiar with Laplace and Fourier transforms and their properties, asdiscussed in chapter 13.The method consists simply of transforming the PDE into one containingderivatives with respect to a smaller number of variables. Thus, if the originalequation has just two independent variables, it may be possible to reduce thePDE into a soluble ODE. The solution obtained can then (where possible) betransformed back to give the solution of the original PDE. As we shall see,boundary conditions can usually be incorporated in a natural way.Which sort of transform to use, and the choice of the variable(s) with respectto which the transform is to be taken, is a matter of experience; we illustrate thisin the example below. In practice, transforms can be taken with respect to eachvariable in turn, and the transformation that affords the greatest simplificationcan be pursued further.◮A semi-infinite tube of constant cross-section contains initially pure water. At time t =0,one end of the tube is put into contact with a salt solution and maintained at a concentrationu 0 . Find the total amount of salt that has diffused into the tube after time t, if the diffusionconstant is κ.The concentration u(x, t) attimet and distance x from the end of the tube satisfies thediffusion equationκ ∂2 u∂x = ∂u2 ∂t , (21.71)which has to be solved subject to the boundary conditions u(0,t)=u 0 for all t andu(x, 0) = 0 for all x>0.Since we are interested only in t>0, the use of the Laplace transform is suggested.Furthermore, it will be recalled from chapter 13 that one of the major virtues of Laplacetransformations is the possibility they afford of replacing derivatives of functions by simplemultiplication by a scalar. If the derivative with respect to time were so removed, equation(21.71) would contain only differentiation with respect to a single variable. Let us thereforetake the Laplace transform of (21.71) with respect to t:∫ ∞∫κ ∂2 u∞0 ∂x exp(−st) dt = ∂uexp(−st) dt.2 0 ∂tOn the LHS the (double) differentiation is with respect to x, whereas the integration iswith respect to the independent variable t. Therefore the derivative can be taken outsidethe integral. Denoting the Laplace transform of u(x, t) byū(x, s) and using result (13.57)to rewrite the transform of the derivative on the RHS (or by integrating directly by parts),we obtainκ ∂2 ū= sū(x, s) − u(x, 0).∂x2 But from the boundary condition u(x, 0) = 0 the last term on the RHS vanishes, and the747

PDES: SEPARATION OF VARIABLES AND OTHER METHODSsolution is immediate:ū(x, s) =A exp(√ ) ( √ )s sκ x + B exp −κ x ,where the constants A and B may depend on s.We require u(x, t) → 0asx →∞and so we must also have ū(∞,s) = 0; consequentlywe require that A = 0. The value of B is determined by the need for u(0,t)=u 0 and hencethat∫ ∞ū(0,s)= u 0 exp(−st) dt = u 00s .We thus conclude that the appropriate expression for the Laplace transform of u(x, t) isū(x, s) = u ( √ )0 ss exp −κ x . (21.72)To obtain u(x, t) from this result requires the inversion of this transform – a task that isgenerally difficult and requires a contour integration. This is discussed in chapter 24, butfor completeness we note that the solution is( )] xu(x, t) =u 0[1 − erf √ ,4κtwhere erf(x) is the error function discussed in the Appendix. (The more complete sets ofmathematical tables list this inverse Laplace transform.)In the present problem, however, an alternative method is available. Let w(t) betheamount of salt that has diffused into the tube in time t; thenand its transform is given by¯w(s) ===w(t) =∫ ∞0∫ ∞0∫ ∞0∫ ∞0dt exp(−st)dx∫ ∞0ū(x, s) dx.u(x, t) dx,∫ ∞0u(x, t) dxu(x, t)exp(−st) dtSubstituting for ū(x, s) from (21.72) into the last integral and integrating, we obtain¯w(s) =u 0 κ 1/2 s −3/2 .This expression is much simpler to invert, and referring to the table of standard Laplacetransforms (table 13.1) we findw(t) =2(κ/π) 1/2 u 0 t 1/2 ,which is thus the required expression for the amount of diffused salt at time t. ◭The above example shows that in some circumstances the use of a Laplacetransformation can greatly simplify the solution of a PDE. However, it will havebeen observed that (as with ODEs) the easy elimination of some derivatives isusually paid for by the introduction of a difficult inverse transformation. Thisproblem, although still present, is less severe for Fourier transformations.748

21.4 INTEGRAL TRANSFORM METHODS◮An infinite metal bar has an initial temperature distribution f(x) along its length. Findthe temperature distribution at a later time t.We are interested in values of x from −∞ to ∞, which suggests Fourier transformationwith respect to x. Assuming that the solution obeys the boundary conditions u(x, t) → 0and ∂u/∂x → 0as|x| →∞, we may Fourier-transform the one-dimensional diffusionequation (21.71) to obtain∫κ ∞√2π−∞∂ 2 u(x, t)∂x 2 exp(−ikx) dx = 1 √2π∂∂t∫ ∞−∞u(x, t)exp(−ikx) dx,where on the RHS we have taken the partial derivative with respect to t outside theintegral. Denoting the Fourier transform of u(x, t) byũ(k, t), and using equation (13.28) torewrite the Fourier transform of the second derivative on the LHS, we then have−κk 2 ∂ũ(k, t)ũ(k, t) = .∂tThis first-order equation has the simple solutionũ(k, t) =ũ(k, 0) exp(−κk 2 t),where the initial conditions giveũ(k, 0) = √ 1 ∫ ∞u(x, 0) exp(−ikx) dx2π −∞= √ 1 ∫ ∞f(x)exp(−ikx) dx = ˜f(k).2πThus we may write the Fourier transform of the solution as−∞ũ(k, t) =˜f(k)exp(−κk 2 t)= √ 2π ˜f(k)˜G(k, t), (21.73)where we have defined the function ˜G(k, t) =( √ 2π) −1 exp(−κk 2 t). Since ũ(k, t) canbewritten as the product of two Fourier transforms, we can use the convolution theorem,subsection 13.1.7, to write the solution asu(x, t) =∫ ∞−∞G(x − x ′ ,t)f(x ′ ) dx ′ ,where G(x, t) is the Green’s function for this problem (see subsection 15.2.5). This functionis the inverse Fourier transform of ˜G(k, t) and is thus given byG(x, t) = 12π= 12π∫ ∞−∞∫ ∞−∞Completing the square in the integrand we findG(x, t) = 1 ( ) ∫ ∞2π exp − x24κt= 1 (2π exp − x24κt= 1 √4πκtexpexp(−κk 2 t)exp(ikx) dk[exp −κt) ∫ ∞(− x24κtexp−∞exp−∞),(k 2 − ixκt k )]dk.[−κt(k − ix ) ] 2dk2κt(−κtk ′2) dk ′where in the second line we have made the substitution k ′ = k − ix/(2κt), and in the last749

PDES: SEPARATION OF VARIABLES AND OTHER METHODSut 1t 2t 3x = axFigure 21.10 Diffusion of heat from a point source in a metal bar: the curvesshow the temperature u at position x for various times t 1

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSthe infinite range in x, rather than of the transform method itself. In fact themethod of separation of variables would yield the same solutions, since in theinfinite-range case the separation constant is not restricted to take on an infiniteset of discrete values but may have any real value, with the result that the sumover λ becomes an integral, as mentioned at the end of section 21.2.◮An infinite metal bar has an initial temperature distribution f(x) along its length. Findthe temperature distribution at a later time t using the method of separation of variables.This is the same problem as in the previous example, but we now seek a solution byseparating variables. From (21.12) a separated solution for the one-dimensional diffusionequation is given byu(x, t) =[A exp(iλx)+B exp(−iλx)] exp(−κλ 2 t),where −λ 2 is the separation constant. Since the bar is infinite we do not require thesolution to take a given form at any finite value of x (for instance at x =0)andsothereis no restriction on λ other than its being real. Therefore instead of the superposition ofsuch solutions in the form of a sum over allowed values of λ we have an integral overall λ,u(x, t) = √ 1 ∫ ∞A(λ)exp(−κλ 2 t)exp(iλx) dλ, (21.75)2π−∞where in taking λ from −∞ to ∞ we need include only one of the complex exponentials;we have taken a factor 1/ √ 2π out of A(λ) for convenience. We can see from (21.75)that the expression for u(x, t) has the form of an inverse Fourier transform (where λ isthe transform variable). Therefore, Fourier-transforming both sides and using the Fourierinversion theorem, we findũ(λ, t) =A(λ)exp(−κλ 2 t).Now, the initial boundary condition requiresu(x, 0) = √ 1 ∫ ∞A(λ)exp(iλx) dλ = f(x),2π−∞from which, using the Fourier inversion theorem once more, we see that A(λ) =˜f(λ).Therefore we haveũ(λ, t) =˜f(λ)exp(−κλ 2 t),which is identical to (21.73) in the previous example (but with k replaced by λ), and henceleads to the same result. ◭21.5 Inhomogeneous problems – Green’s functionsIn chapters 15 and 17 we encountered Green’s functions and found them a usefultool for solving inhomogeneous linear ODEs. We now discuss their usefulness insolving inhomogeneous linear PDEs.For the sake of brevity we shall again denote a linear PDE byLu(r) =ρ(r), (21.76)where L is a linear partial differential operator. For example, in Laplace’s equation751

PDES: SEPARATION OF VARIABLES AND OTHER METHODSwe have L = ∇ 2 , whereas for Helmholtz’s equation L = ∇ 2 +k 2 . Note that we havenot specified the dimensionality of the problem, and (21.76) may, for example,represent Poisson’s equation in two or three (or more) dimensions. The readerwill also notice that for the sake of simplicity we have not included any timedependence in (21.76). Nevertheless, the following discussion can be generalisedto include it.As we discussed in subsection 20.3.2, a problem is inhomogeneous if the factthat u(r) is a solution does not imply that any constant multiple λu(r) isalsoasolution. This inhomogeneity may derive from either the PDE itself or from theboundary conditions imposed on the solution.In our discussion of Green’s function solutions of inhomogeneous ODEs (seesubsection 15.2.5) we dealt with inhomogeneous boundary conditions by making asuitable change of variable such that in the new variable the boundary conditionswere homogeneous. In an analogous way, as illustrated in the final exampleof section 21.2, it is usually possible to make a change of variables in PDEs totransform between inhomogeneity of the boundary conditions and inhomogeneityof the equation. Therefore let us assume for the moment that the boundaryconditions imposed on the solution u(r) of (21.76) are homogeneous. This mostcommonly means that if we seek a solution to (21.76) in some region V thenon the surface S that bounds V the solution obeys the conditions u(r) =0or∂u/∂n =0,where∂u/∂n is the normal derivative of u at the surface S.We shall discuss the extension of the Green’s function method to the direct solutionof problems with inhomogeneous boundary conditions in subsection 21.5.2,but we first highlight how the Green’s function approach to solving ODEs canbe simply extended to PDEs for homogeneous boundary conditions.21.5.1 Similarities to Green’s functions for ODEsAs in the discussion of ODEs in chapter 15, we may consider the Green’sfunction for a system described by a PDE as the response of the system to a ‘unitimpulse’ or ‘point source’. Thus if we seek a solution to (21.76) that satisfies somehomogeneous boundary conditions on u(r) then the Green’s function G(r, r 0 )forthe problem is a solution ofLG(r, r 0 )=δ(r − r 0 ), (21.77)where r 0 lies in V . The Green’s function G(r, r 0 ) must also satisfy the imposed(homogeneous) boundary conditions.It is understood that in (21.77) the L operator expresses differentiation withrespect to r as opposed to r 0 . Also, δ(r − r 0 ) is the Dirac delta function (seechapter 13) of dimension appropriate to the problem; it may be thought of asrepresenting a unit-strength point source at r = r 0 .Following an analogous argument to that given in subsection 15.2.5 for ODEs,752

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSif the boundary conditions on u(r) are homogeneous then a solution to (21.76)that satisfies the imposed boundary conditions is given by∫u(r) = G(r, r 0 )ρ(r 0 ) dV (r 0 ), (21.78)where the integral on r 0 is over some appropriate ‘volume’. In two or moredimensions, however, the task of finding directly a solution to (21.77) that satisfiesthe imposed boundary conditions on S can be a difficult one, and we return tothis in the next subsection.An alternative approach is to follow a similar argument to that presented inchapter 17 for ODEs and so to construct the Green’s function for (21.76) as asuperposition of eigenfunctions of the operator L, provided L is Hermitian. Byanalogy with an ordinary differential operator, a partial differential operator isHermitian if it satisfies∫[∫] ∗v ∗ (r)Lw(r) dV = w ∗ (r)Lv(r) dV ,VVwhere the asterisk denotes complex conjugation and v and w are arbitrary functionsobeying the imposed (homogeneous) boundary condition on the solution ofLu(r) =0.The eigenfunctions u n (r), n =0, 1, 2,...,ofL satisfyLu n (r) =λ n u n (r),where λ n are the corresponding eigenvalues, which are all real for an Hermitianoperator L. Furthermore, each eigenfunction must obey any imposed (homogeneous)boundary conditions. Using an argument analogous to that given inchapter 17, the Green’s function for the problem is given by∞∑ u n (r)u ∗G(r, r 0 )=n(r 0 ). (21.79)λ nn=0From (21.79) we see immediately that the Green’s function (irrespective of howit is found) enjoys the propertyG(r, r 0 )=G ∗ (r 0 , r).Thus, if the Green’s function is real then it is symmetric in its two arguments.Once the Green’s function has been obtained, the solution to (21.76) is againgiven by (21.78). For PDEs this approach can become very cumbersome, however,and so we shall not pursue it further here.21.5.2 General boundary-value problemsAs mentioned above, often inhomogeneous boundary conditions can be dealtwith by making an appropriate change of variables, such that the boundary753

PDES: SEPARATION OF VARIABLES AND OTHER METHODSconditions in the new variables are homogeneous although the equation itself isgenerally inhomogeneous. In this section, however, we extend the use of Green’sfunctions to problems with inhomogeneous boundary conditions (and equations).This provides a more consistent and intuitive approach to the solution of suchboundary-value problems.For definiteness we shall consider Poisson’s equation∇ 2 u(r) =ρ(r), (21.80)but the material of this section may be extended to other linear PDEs of the form(21.76). Clearly, Poisson’s equation reduces to Laplace’s equation for ρ(r) =0andso our discussion is equally applicable to this case.We wish to solve (21.80) in some region V bounded by a surface S, which mayconsist of several disconnected parts. As stated above, we shall allow the possibilitythat the boundary conditions on the solution u(r) may be inhomogeneous on S,although as we shall see this method reduces to those discussed above in thespecial case that the boundary conditions are in fact homogeneous.The two common types of inhomogeneous boundary condition for Poisson’sequation are (as discussed in subsection 20.6.2):(i) Dirichlet conditions, in which u(r) is specified on S, and(ii) Neumann conditions, in which ∂u/∂n is specified on S.In general, specifying both Dirichlet and Neumann conditions on S overdeterminesthe problem and leads to there being no solution.The specification of the surface S requires some further comment, since Smay have several disconnected parts. If we wish to solve Poisson’s equationinside some closed surface S then the situation is straightforward and is shownin figure 21.11(a). If, however, we wish to solve Poisson’s equation in the gapbetween two closed surfaces (for example in the gap between two concentricconducting cylinders) then the volume V is bounded by a surface S that hastwo disconnected parts S 1 and S 2 , as shown in figure 21.11(b); the direction ofthe normal to the surface is always taken as pointing out of the volume V .Asimilar situation arises when we wish to solve Poisson’s equation outside someclosed surface S 1 . In this case the volume V is infinite but is treated formallyby taking the surface S 2 as a large sphere of radius R and letting R tend toinfinity.In order to solve (21.80) subject to either Dirichlet or Neumann boundaryconditions on S, we will remind ourselves of Green’s second theorem, equation(11.20), which states that, for two scalar functions φ(r) andψ(r) defined in somevolume V bounded by a surface S,∫∫(φ∇ 2 ψ − ψ∇ 2 φ) dV = (φ∇ψ − ψ∇φ) · ˆn dS, (21.81)V754S

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSVVSˆnS 1S 2ˆnˆn(a)Figure 21.11regions V .Surfaces used for solving Poisson’s equation in different(b)where on the RHS it is common to write, for example, ∇ψ · ˆn dS as (∂ψ/∂n) dS.The expression ∂ψ/∂n stands for ∇ψ · ˆn, the rate of change of ψ in the directionof the unit outward normal ˆn to the surface S.The Green’s function for Poisson’s equation (21.80) must satisfy∇ 2 G(r, r 0 )=δ(r − r 0 ), (21.82)where r 0 lies in V . (As mentioned above, we may think of G(r, r 0 )asthesolutionto Poisson’s equation for a unit-strength point source located at r = r 0 .) Let usfor the moment impose no boundary conditions on G(r, r 0 ).If we now let φ = u(r) andψ = G(r, r 0 ) in Green’s theorem (21.81) then weobtain∫[u(r)∇ 2 G(r, r 0 ) − G(r, r 0 ) ∇ 2 u(r) ] dV (r)V∫ [= u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r),S ∂n∂nwhere we have made explicit that the volume and surface integrals are withrespect to r. Using (21.80) and (21.82) the LHS can be simplified to give∫[u(r)δ(r − r 0 ) − G(r, r 0 )ρ(r)] dV (r)V∫ [= u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r). (21.83)S ∂n∂nSince r 0 lies within the volume V ,∫u(r)δ(r − r 0 ) dV (r) =u(r 0 ),Vand thus on rearranging (21.83) the solution to Poisson’s equation (21.80) can be755

PDES: SEPARATION OF VARIABLES AND OTHER METHODSwritten as∫u(r 0 )=V∫ [G(r, r 0 )ρ(r) dV (r)+ u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r).S ∂n∂n(21.84)Clearly, we can interchange the roles of r and r 0 in (21.84) if we wish. (Rememberalso that, for a real Green’s function, G(r, r 0 )=G(r 0 , r).)Equation (21.84) is central to the extension of the Green’s function methodto problems with inhomogeneous boundary conditions, and we next discuss itsapplication to both Dirichlet and Neumann boundary-value problems. But, beforedoing so, we also note that if the boundary condition on S is in fact homogeneous,so that u(r) =0or∂u(r)/∂n =0onS, then demanding that the Green’s functionG(r, r 0 ) also obeys the same boundary condition causes the surface integral in(21.84) to vanish, and we are left with the familiar form of solution given in(21.78). The extension of (21.84) to a PDE other than Poisson’s equation isdiscussed in exercise 21.28.21.5.3 Dirichlet problemsIn a Dirichlet problem we require the solution u(r) of Poisson’s equation (21.80)to take specific values on some surface S that bounds V ,i.e.werequirethatu(r) =f(r) onS where f is a given function.If we seek a Green’s function G(r, r 0 ) for this problem it must clearly satisfy(21.82), but we are free to choose the boundary conditions satisfied by G(r, r 0 )insuch a way as to make the solution (21.84) as simple as possible. From (21.84),we see that by choosingG(r, r 0 )=0 forr on S (21.85)the second term in the surface integral vanishes. Since u(r) =f(r) onS, (21.84)then becomes∫∫u(r 0 )= G(r, r 0 )ρ(r) dV (r)+ f(r) ∂G(r, r 0)dS(r). (21.86)VS ∂nThus we wish to find the Dirichlet Green’s function that(i) satisfies (21.82) and hence is singular at r = r 0 ,and(ii) obeys the boundary condition G(r, r 0 )=0forr on S.In general, it is difficult to obtain this function directly, and so it is useful toseparate these two requirements. We therefore look for a solution of the formG(r, r 0 )=F(r, r 0 )+H(r, r 0 ),where F(r, r 0 ) satisfies (21.82) and has the required singular character at r = r 0 butdoes not necessarily obey the boundary condition on S, whilst H(r, r 0 ) satisfies756

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSthe corresponding homogeneous equation (i.e. Laplace’s equation) inside V butis adjusted in such a way that the sum G(r, r 0 ) equals zero on S. The Green’sfunction G(r, r 0 ) is still a solution of (21.82) since∇ 2 G(r, r 0 )=∇ 2 F(r, r 0 )+∇ 2 H(r, r 0 )=∇ 2 F(r, r 0 )+0=δ(r − r 0 ).The function F(r, r 0 ) is called the fundamental solution and will clearly takedifferent forms depending on the dimensionality of the problem. Let us firstconsider the fundamental solution to (21.82) in three dimensions.◮Find the fundamental solution to Poisson’s equation in three dimensions that tends to zeroas |r| →∞.We wish to solve∇ 2 F(r, r 0 )=δ(r − r 0 ) (21.87)in three dimensions, subject to the boundary condition F(r, r 0 ) → 0as|r| →∞.Sincetheproblem is spherically symmetric about r 0 , let us consider a large sphere S of radius Rcentred on r 0 , and integrate (21.87) over the enclosed volume V . We then obtain∫∫∇ 2 F(r, r 0 ) dV = δ(r − r 0 ) dV =1, (21.88)Vsince V encloses the point r 0 . However, using the divergence theorem,∫∫∇ 2 F(r, r 0 ) dV = ∇F(r, r 0 ) · ˆn dS, (21.89)Vwhere ˆn is the unit normal to the large sphere S at any point.Since the problem is spherically symmetric about r 0 , we expect thatF(r, r 0 )=F(|r − r 0 |)=F(r),i.e. that F has the same value everywhere on S. Thus, evaluating the surface integral in(21.89) and equating it to unity from (21.88), we have §4πr 2 dFdr ∣ =1.r=RIntegrating this expression we obtainF(r) =− 14πr +constant,but, since we require F(r, r 0 ) → 0as|r| →∞, the constant must be zero. The fundamentalsolution in three dimensions is consequently given by1F(r, r 0 )=−4π|r − r 0 | . (21.90)This is clearly also the full Green’s function for Poisson’s equation subject to the boundarycondition u(r) → 0as|r| →∞. ◭VUsing (21.90) we can write down the solution of Poisson’s equation to find,S§ A vertical bar to the right of an expression is a common alternative to enclosing the expression insquare brackets; as usual, the subscript shows the value of the variable at which the expression isto be evaluated.757

PDES: SEPARATION OF VARIABLES AND OTHER METHODSfor example, the electrostatic potential u(r) due to some distribution of electriccharge ρ(r). The electrostatic potential satisfies∇ 2 u(r) =− ρ ɛ 0,where u(r) → 0as|r| →∞. Since the boundary condition on the surface atinfinity is homogeneous the surface integral in (21.86) vanishes, and using (21.90)we recover the familiar solution∫u(r 0 )=ρ(r)dV (r), (21.91)4πɛ 0 |r − r 0 |where the volume integral is over all space.We can develop an analogous theory in two dimensions. As before the fundamentalsolution satisfies∇ 2 F(r, r 0 )=δ(r − r 0 ), (21.92)where δ(r−r 0 ) is now the two-dimensional delta function. Following an analogousmethod to that used in the previous example, we find the fundamental solutionin two dimensions to be given byF(r, r 0 )= 12π ln |r − r 0| + constant. (21.93)From the form of the solution we see that in two dimensions we cannot applythe condition F(r, r 0 ) → 0as|r| →∞, and in this case the constant does notnecessarily vanish.We now return to the task of constructing the full Dirichlet Green’s function. Todo so we wish to add to the fundamental solution a solution of the homogeneousequation (in this case Laplace’s equation) such that G(r, r 0 )=0onS, asrequiredby (21.86) and its attendant conditions. The appropriate Green’s function isconstructed by adding to the fundamental solution ‘copies’ of itself that represent‘image’ sources at different locations outside V . Hence this approach is called themethod of images.In summary, if we wish to solve Poisson’s equation in some region V subject toDirichlet boundary conditions on its surface S then the procedure and argumentare as follows.(i) To the single source δ(r − r 0 ) inside V add image sources outside VN∑q n δ(r − r n ) with r n outside V,n=1where the positions r n and the strengths q n of the image sources are to bedetermined as described in step (iii) below.758

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS(ii) Since all the image sources lie outside V , the fundamental solution correspondingto each source satisfies Laplace’s equation inside V . Thus wemay add the fundamental solutions F(r, r n ) corresponding to each imagesource to that corresponding to the single source inside V , obtaining theGreen’s functionN∑G(r, r 0 )=F(r, r 0 )+ q n F(r, r n ).(iii) Now adjust the positions r n and strengths q n of the image sources sothat the required boundary conditions are satisfied on S. For a DirichletGreen’s function we require G(r, r 0 )=0forr on S.(iv) The solution to Poisson’s equation subject to the Dirichlet boundarycondition u(r) =f(r) onS is then given by (21.86).In general it is very difficult to find the correct positions and strengths for theimages, i.e. to make them such that the boundary conditions on S are satisfied.Nevertheless, it is possible to do so for certain problems that have simple geometry.In particular, for problems in which the boundary S consists of straight lines (intwo dimensions) or planes (in three dimensions), positions of the image pointscan be deduced simply by imagining the boundary lines or planes to be mirrorsin which the single source in V (at r 0 ) is reflected.◮Solve Laplace’s equation ∇ 2 u =0in three dimensions in the half-space z>0, given thatu(r) =f(r) on the plane z =0.The surface S bounding V consists of the xy-plane and the surface at infinity. Therefore,the Dirichlet Green’s function for this problem must satisfy G(r, r 0 )=0onz =0andG(r, r 0 ) → 0as|r| →∞. Thus it is clear in this case that we require one image source at aposition r 1 that is the reflection of r 0 in the plane z = 0, as shown in figure 21.12 (so thatr 1 lies in z

PDES: SEPARATION OF VARIABLES AND OTHER METHODSzV+r 0yx−r 1Figure 21.12 The arrangement of images for solving Laplace’s equation inthe half-space z>0.We may evaluate this normal derivative by writing the Green’s function (21.94) explicitlyin terms of x, y and z (and x 0 , y 0 and z 0 ) and calculating the partial derivative with respectto z directly. It is usually quicker, however, to use the fact that §∇|r − r 0 | = r − r 0|r − r 0 | ; (21.96)thus∇G(r, r 0 )= r − r 04π|r − r 0 | − r − r 13 4π|r − r 1 | . 3Since r 0 =(x 0 ,y 0 ,z 0 )andr 1 =(x 0 ,y 0 , −z 0 ) the normal derivative is given by− ∂G(r, r 0)= −k · ∇G(r, r 0 )∂z= − z − z 04π|r − r 0 | + z + z 03 4π|r − r 1 | . 3Therefore on the surface z = 0, and writing out the dependence on x, y and z explicitly,we have− ∂G(r, r 0)2z 0∂z ∣ =z=04π[(x − x 0 ) 2 +(y − y 0 ) 2 + z .0 2]3/2 Inserting this expression into (21.95) we obtain the solutionu(x 0 ,y 0 ,z 0 )= z ∫ ∞ ∫ ∞0f(x, y)dx dy. ◭2π −∞ −∞ [(x − x 0 ) 2 +(y − y 0 ) 2 + z0 2 ]3/2An analogous procedure may be applied in two-dimensional problems. For§ Since |r − r 0 | 2 =(r − r 0 ) · (r − r 0 ) we have ∇|r − r 0 | 2 =2(r − r 0 ), from which we obtain∇(|r − r 0 | 2 ) 1/2 = 1 2(r − r 0 )2 (|r − r 0 | 2 ) 1/2 = r − r 0|r − r 0 | .Note that this result holds in two and three dimensions.760

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSexample, in solving Poisson’s equation in two dimensions in the half-space x>0we again require just one image charge, of strength q 1 = −1, at a position r 1 thatis the reflection of r 0 in the line x = 0. Since we require G(r, r 0 )=0whenr lieson x = 0, the constant in (21.93) must equal zero, and so the Dirichlet Green’sfunction isG(r, r 0 )= 1 (ln |r − r0 |−ln |r − r 1 | ) .2πClearly G(r, r 0 ) tends to zero as |r| →∞. If, however, we wish to solve the twodimensionalPoisson equation in the quarter space x>0, y>0, then more imagepoints are required.◮A line charge in the z-direction of charge density λ is placed at some position r 0 in thequarter-space x>0, y>0. Calculate the force per unit length on the line charge due tothe presence of thin earthed plates along x =0and y =0.Here we wish to solve Poisson’s equation,∇ 2 u = − λ ɛ 0δ(r − r 0 ),in the quarter space x>0, y>0. It is clear that we require three image line chargeswith positions and strengths as shown in figure 21.13 (all of which lie outside the regionin which we seek a solution). The boundary condition that the electrostatic potential u iszero on x =0andy = 0 (shown as the ‘curve’ C in figure 21.13) is then automaticallysatisfied, and so this system of image charges is directly equivalent to the original situationof a single line charge in the presence of the earthed plates along x =0andy = 0. Thusthe electrostatic potential is simply equal to the Dirichlet Green’s functionu(r) =G(r, r 0 )=−λ (ln |r − r0 |−ln |r − r 1 | +ln|r − r 2 |−ln |r − r 3 | ) ,2πɛ 0which equals zero on C and on the ‘surface’ at infinity.The force on the line charge at r 0 , therefore, is simply that due to the three line chargesat r 1 , r 2 and r 3 . The elecrostatic potential due to a line charge at r i , i = 1, 2 or 3, is givenby the fundamental solutionu i (r) =∓λ ln |r − r i | + c,2πɛ 0the upper or lower sign being taken according to whether the line charge is positive ornegative, respectively. Therefore the force per unit length on the line charge at r 0 , due tothe one at r i ,isgivenby−λ∇u i (r)∣ = ± λ2 r 0 − r ir=r02πɛ 0 |r 0 − r i | . 2Adding the contributions from the three image charges shown in figure 21.13, the totalforce experienced by the line charge at r 0 is given by(F =λ2− r 0 − r 12πɛ 0 |r 0 − r 1 | + r 0 − r 22 |r 0 − r 2 | − r )0 − r 3,2 |r 0 − r 3 | 2where, from the figure, r 0 − r 1 =2y 0 j, r 0 − r 2 =2x 0 i +2y 0 j and r 0 − r 3 =2x 0 i. Thus, in761

PDES: SEPARATION OF VARIABLES AND OTHER METHODSy−λ+λr 0r 3C xx 0y 0V+λr 2−λr 1Figure 21.13 The arrangement of images for finding the force on a linecharge situated in the (two-dimensional) quarter-space x>0, y>0, when theplanes x =0andy = 0 are earthed.terms of x 0 and y 0 , the total force on the line charge due to the charge induced on theplates is given by(F =λ2− 1 j + 2x 0 i +2y 0 j− 1 )i2πɛ 0 2y 0 4x 2 0 +4y2 02x 0( )λ 2 y2= −04πɛ 0 (x 2 0 + y2 0 ) i + x2 0j . ◭x 0 y 0Further generalisations are possible. For instance, solving Poisson’s equation inthe two-dimensional strip −∞

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSVza+1 r 0A−a|r 0 | r 1xyB−aFigure 21.14 The arrangement of images for solving Poisson’s equationoutside a sphere of radius a centred at the origin. For a charge +1 at r 0 ,theimage point r 1 is given by (a/|r 0 |) 2 r 0 and the strength of the image charge is−a/|r 0 |.By symmetry we expect the image point r 1 to lie on the same radial line as the originalsource, r 0 , as shown in figure 21.14, and so r 1 = kr 0 where k

PDES: SEPARATION OF VARIABLES AND OTHER METHODSTherefore, in order that G =0at|r| = a, the strength of the image charge must be−a/|r 0 |. Consequently, the Dirichlet Green’s function for the exterior of the sphere is1G(r, r 0 )=−4π|r − r 0 | + a/|r 0 |4π |r − (a 2 /|r 0 | 2 )r 0 | .For a less formal treatment of the same problem see exercise 21.22. ◭If we seek solutions to Poisson’s equation in the interior of a sphere then theabove analysis still holds, but r and r 0 are now inside the sphere and the imager 1 lies outside it.For two-dimensional Dirichlet problems outside the circle |r| = a, we are ledby arguments similar to those employed previously to use the same image pointas in the three-dimensional case, namelyr 1 =a2|r 0 | 2 r 0. (21.100)As illustrated below, however, it is usually necessary to take the image strengthas −1 in two-dimensional problems.◮Solve Laplace’s equation in the two-dimensional region |r| ≤a, subject to the boundarycondition u = f(φ) on |r| = a.In this case we wish to find the Dirichlet Green’s function in the interior of a disc ofradius a, so the image charge must lie outside the disc. Taking the strength of the imageto be −1, we haveG(r, r 0 )= 12π ln |r − r 0|− 12π ln |r − r 1| + c,where r 1 =(a 2 /|r 0 | 2 )r 0 lies outside the disc, and c is a constant that includes the strengthof the image charge and does not necessarily equal zero.Since we require G(r, r 0 )=0when|r| = a, the value of the constant c is determined,and the Dirichlet Green’s function for this problem is given byG(r, r 0 )= 1 (ln |r − r 0 |−ln2π∣ r −a2|r 0 | r 2 0∣ − ln |r )0|. (21.101)aUsing plane polar coordinates, the solution to the boundary-value problem can be writtenas a line integral around the circle ρ = a:∫u(r 0 )= f(r) ∂G(r, r 0)dlC ∂n∫ 2π= f(r) ∂G(r, r 0)0 ∂ρ ∣ adφ. (21.102)ρ=aThe normal derivative of the Green’s function (21.101) is given by∂G(r, r 0 )∂ρ= r|r| · ∇G(r, r 0)= r2π|r| ·( r − r0|r − r 0 | − r − r )1. (21.103)2 |r − r 1 | 2764

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSUsing the fact that r 1 =(a 2 /|r 0 | 2 )r 0 and the geometrical result (21.99), we find that∂G(r, r 0 )∂ρ ∣ = a2 −|r 0 | 2ρ=a2πa|r − r 0 | . 2In plane polar coordinates, r = ρ cos φ i + ρ sin φ j and r 0 = ρ 0 cos φ 0 i + ρ 0 sin φ 0 j,and so( )∂G(r, r 0 )1a 2 − ρ 2 0∂ρ ∣ =ρ=a2πa a 2 + ρ 2 0 − 2aρ 0 cos(φ − φ 0 ) .On substituting into (21.102), we obtain∫ 2πu(ρ 0 ,φ 0 )= 12π 0which is the solution to the problem. ◭(a 2 − ρ 2 0 )f(φ) dφa 2 + ρ 2 0 − 2aρ 0 cos(φ − φ 0 ) , (21.104)21.5.4 Neumann problemsIn a Neumann problem we require the normal derivative of the solution ofPoisson’s equation to take on specific values on some surface S that bounds V ,i.e. we require ∂u(r)/∂n = f(r) onS, wheref is a given function. As we shall see,much of our discussion of Dirichlet problems can be immediately taken over intothe solution of Neumann problems.As we proved in section 20.7 of the previous chapter, specifying Neumannboundary conditions determines the relevant solution of Poisson’s equation towithin an (unimportant) additive constant. Unlike Dirichlet conditions, Neumannconditions impose a self-consistency requirement. In order for a solution u to exist,it is necessary that the following consistency condition holds:∫ ∫∫∫fdS = ∇u · ˆn dS = ∇ 2 udV = ρdV, (21.105)SSwhere we have used the divergence theorem to convert the surface integral intoa volume integral. As a physical example, the integral of the normal componentof an electric field over a surface bounding a given volume cannot be chosenarbitrarily when the charge inside the volume has already been specified (Gauss’stheorem).Let us again consider (21.84), which is central to our discussion of Green’sfunctions in inhomogeneous problems. It reads∫∫ [u(r 0 )= G(r, r 0 )ρ(r) dV (r)+ u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r).VS ∂n∂nAs always, the Green’s function must obeyV∇ 2 G(r, r 0 )=δ(r − r 0 ),where r 0 lies in V . In the solution of Dirichlet problems in the previous subsection,we chose the Green’s function to obey the boundary condition G(r, r 0 )=0onS765V

PDES: SEPARATION OF VARIABLES AND OTHER METHODSand, in a similar way, we might wish to choose ∂G(r, r 0 )/∂n = 0 in the solution ofNeumann problems. However, in general this is not permitted since the Green’sfunction must obey the consistency condition∫∫∂G(r, r 0 )dS = ∇G(r, r 0 ) · ˆn dS = ∇ 2 G(r, r 0 ) dV =1.S ∂nSVThe simplest permitted boundary condition is therefore∂G(r, r 0 )= 1 for r on S,∂n Awhere A is the area of the surface S; this defines a Neumann Green’s function.If we require ∂u(r)/∂n = f(r) onS, the solution to Poisson’s equation is given by∫u(r 0 )= G(r, r 0 )ρ(r) dV (r)+ 1 ∫∫u(r) dS(r) − G(r, r 0 )f(r) dS(r)V A SS∫∫= G(r, r 0 )ρ(r) dV (r)+〈u(r)〉 S − G(r, r 0 )f(r) dS(r), (21.106)Vwhere 〈u(r)〉 S is the average of u over the surface S and is a freely specifiable constant.For Neumann problems in which the volume V is bounded by a surface S atinfinity, we do not need the 〈u(r)〉 S term. For example, if we wish to solve a Neumannproblem outside the unit sphere centred at the origin then r>ais the regionV throughout which we require the solution; this region may be considered as beingbounded by two disconnected surfaces, the surface of the sphere and a surfaceat infinity. By requiring that u(r) → 0as|r| →∞,theterm〈u(r)〉 S becomes zero.As mentioned above, much of our discussion of Dirichlet problems can betaken over into the solution of Neumann problems. In particular, we may use themethod of images to find the appropriate Neumann Green’s function.◮Solve Laplace’s equation in the two-dimensional region |r| ≤a subject to the boundarycondition ∂u/∂n = f(φ) on |r| = a, with ∫ 2πf(φ) dφ =0as required by the consistency0condition (21.105).Let us assume, as in Dirichlet problems with this geometry, that a single image charge isplaced outside the circle atr 1 =a2|r 0 | r 0,2where r 0 is the position of the source inside the circle (see equation (21.100)). Then, from(21.99), we have the useful geometrical result|r − r 1 | = a|r 0 | |r − r 0| for |r| = a. (21.107)Leaving the strength q of the image as a parameter, the Green’s function has the formG(r, r 0 )= 1 (ln |r − r0 | + q ln |r − r 1 | + c ) . (21.108)2π766S∫

21.6 EXERCISESUsing plane polar coordinates, the radial (i.e. normal) derivative of this function is givenby∂G(r, r 0 )∂ρ= r|r| · ∇G(r, r 0)= r2π|r| ·[ r − r0|r − r 0 | + q(r − r ]1).2 |r − r 1 | 2Using (21.107), on the perimeter of the circle ρ = a the radial derivative takes the form∂G(r, r 0 )∂ρ ∣ = 1 [ ]|r| 2 − r · r 0+ q|r|2 − q(a 2 /|r 0 | 2 )r · r 0ρ=a2π|r| |r − r 0 | 2 (a 2 /|r 0 | 2 )|r − r 0 | 2= 1 1 [ ]|r| 2 + q|r2πa |r − r 0 | 2 0 | 2 − (1 + q)r · r 0 ,where we have set |r| 2 = a 2 in the second term on the RHS, but not in the first. If we takeq = 1, the radial derivative simplifies to∂G(r, r 0 )∂ρ ∣ = 1ρ=a2πa ,or 1/L, whereL is the circumference, and so (21.108) with q = 1 is the required NeumannGreen’s function.Since ρ(r) = 0, the solution to our boundary-value problem is now given by (21.106) as∫u(r 0 )=〈u(r)〉 C − G(r, r 0 )f(r) dl(r),where the integral is around the circumference of the circle C. In plane polar coordinatesr = ρ cos φ i + ρ sin φ j and r 0 = ρ 0 cos φ 0 i + ρ 0 sin φ 0 j, and again using (21.107) we findthat on C the Green’s function is given byG(r, r 0 )| ρ=a= 12π= 12π= 12πC[( ) ]aln |r − r 0 | +ln|r 0 | |r − r 0| + c(ln |r − r 0 | 2 +ln a )|r 0 | + c {ln [ a 2 + ρ 2 0 − 2aρ 0 cos(φ − φ 0 ) ] +ln a }+ c . (21.109)ρ 0Since dl = adφ on C, the solution to the problem is given byu(ρ 0 ,φ 0 )=〈u〉 C − a ∫ 2πf(φ)ln[a 2 + ρ 2 0 − 2aρ 0 cos(φ − φ 0 )] dφ.2π 0The contributions of the final two terms terms in the Green’s function (21.109) vanishbecause ∫ 2πf(φ) dφ = 0. The average value of u around the circumference, 〈u〉0 C , is a freelyspecifiable constant as we would expect for a Neumann problem. This result should becompared with the result (21.104) for the corresponding Dirichlet problem, but it shouldbe remembered that in the one case f(φ) is a potential, and in the other the gradient of apotential. ◭21.6 Exercises21.1 Solve the following first-order partial differential equations by separating thevariables:(a) ∂u∂x − x ∂u∂y =0;∂u ∂u(b)x − 2y∂x ∂y =0.767

PDES: SEPARATION OF VARIABLES AND OTHER METHODS21.2 A cube, made of material whose conductivity is k, has as its six faces the planesx = ±a, y = ±a and z = ±a, and contains no internal heat sources. Verify thatthe temperature distributionu(x, y, z, t) =A cos πx ( )πzsina a exp − 2κπ2 ta 2obeys the appropriate diffusion equation. Across which faces is there heat flow?What is the direction and rate of heat flow at the point (3a/4,a/4,a)attimet = a 2 /(κπ 2 )?21.3 The wave equation describing the transverse vibrations of a stretched membraneunder tension T and having a uniform surface density ρ is( )∂ 2 uT∂x + ∂2 u= ρ ∂2 u2 ∂y 2 ∂t . 2Find a separable solution appropriate to a membrane stretched on a frame oflength a and width b, showing that the natural angular frequencies of such amembrane are given by( )ω 2 = π2 T n2ρ a + m2,2 b 2where n and m are any positive integers.21.4 Schrödinger’s equation for a non-relativistic particle in a constant potential regioncan be taken as( )− 2 ∂ 2 u2m ∂x + ∂2 u2 ∂y + ∂2 u= i ∂u2 ∂z 2 ∂t .(a) Find a solution, separable in the four independent variables, that can bewritten in the form of a plane wave,ψ(x, y, z, t) =A exp[i(k · r − ωt)].Using the relationships associated with de Broglie (p = k) and Einstein(E = ω), show that the separation constants must be such thatp 2 x + p 2 y + p 2 z =2mE.(b) Obtain a different separable solution describing a particle confined to a boxof side a (ψ must vanish at the walls of the box). Show that the energy ofthe particle can only take the quantised valuesE = 2 π 22ma 2 (n2 x + n 2 y + n 2 z),where n x , n y and n z are integers.21.5 Denoting the three terms of ∇ 2 in spherical polars by ∇ 2 r, ∇ 2 θ , ∇2 φin an obviousway, evaluate ∇ 2 r u, etc. for the two functions given below and verify that, in eachcase, although the individual terms are not necessarily zero their sum ∇ 2 u is zero.Identify the corresponding values of l and m.((a) u(r, θ, φ) = Ar 2 + B ) 3cos 2 θ − 1.r 3 2(b) u(r, θ, φ) =(Ar + B )sin θ exp iφ.r 221.6 Prove that the expression given in equation (21.47) for the associated Legendrefunction Pl m (µ) satisfies the appropriate equation, (21.45), as follows.768

21.6 EXERCISES(a) Evaluate dPl m(µ)/dµ and d2 Pl m(µ)/dµ2 , using the forms given in (21.47), andsubstitute them into (21.45).(b) Differentiate Legendre’s equation m times using Leibnitz’ theorem.(c) Show that the equations obtained in (a) and (b) are multiples of each other,and hence that the validity of (b) implies that of (a).21.7 Continue the analysis of exercise 10.20, concerned with the flow of a very viscousfluid past a sphere, to find the full expression for the stream function ψ(r, θ). Atthe surface of the sphere r = a, the velocity field u = 0, whilst far from the sphereψ ≃ (Ur 2 sin 2 θ)/2.Show that f(r) can be expressed as a superposition of powers of r, anddetermine which powers give acceptable solutions. Hence show thatψ(r, θ) = U )(2r 2 − 3ar + a3sin 2 θ.4r21.8 The motion of a very viscous fluid in the two-dimensional (wedge) region −α

PDES: SEPARATION OF VARIABLES AND OTHER METHODSx = 0, and free at the other, x = L, show that the angular frequency of vibrationω satisfies( ) ( )ω 1/2 Lω 1/2 Lcosh = − sec .aa[ At a clamped end both u and ∂u/∂x vanish, whilst at a free end, where there isno bending moment, ∂ 2 u/∂x 2 and ∂ 3 u/∂x 3 are both zero. ]21.12 A membrane is stretched between two concentric rings of radii a and b (b >a).If the smaller ring is transversely distorted from the planar configuration by anamount c|φ|, −π ≤ φ ≤ π, show that the membrane then has a shape given byu(ρ, φ) = cπ ln(b/ρ)2 ln(b/a) − 4c ∑( )a m b2mπ m 2 (b 2m − a 2m ) ρ − m ρm cos mφ.m odd21.13 A string of length L, fixed at its two ends, is plucked at its mid-point by anamount A and then released. Prove that the subsequent displacement is given by∞∑[ ] [ ]8A (−1) n (2n +1)πx (2n +1)πctu(x, t) =π 2 (2n +1) sin cos,2 LLn=0where, in the usual notation, c 2 = T/ρ.Find the total kinetic energy of the string when it passes through its unpluckedposition, by calculating it in each mode (each n) and summing, using the result∞∑ 1(2n +1) = π22 8 .0Confirm that the total energy is equal to the work done in plucking the stringinitially.21.14 Prove that the potential for ρ0andcosφ

21.6 EXERCISES21.18 A sphere of radius a and thermal conductivity k 1 is surrounded by an infinitemedium of conductivity k 2 in which far away the temperature tends to T ∞ .A distribution of heat sources q(θ) embedded in the sphere’s surface establishsteady temperature fields T 1 (r, θ) inside the sphere and T 2 (r, θ) outside it. It canbe shown, by considering the heat flow through a small volume that includespart of the sphere’s surface, that∂T 1k 1∂r − k ∂T 22 = q(θ) on r = a.∂rGiven thatq(θ) = 1 ∞∑q n P n (cos θ),an=0find complete expressions for T 1 (r, θ) andT 2 (r, θ). What is the temperature atthe centre of the sphere?21.19 Using result (21.74) from the worked example in the text, find the generalexpression for the temperature u(x, t) in the bar, given that the temperaturedistribution at time t =0isu(x, 0) = exp(−x 2 /a 2 ).21.20 Working in spherical polar coordinates r =(r, θ, φ), but for a system that hasazimuthal symmetry around the polar axis, consider the following gravitationalproblem.(a) Show that the gravitational potential due to a uniform disc of radius a andmass M, centred at the origin, is given for raby[GM1 − 1 ( a) 2P2 (cos θ)+ 1 ( a) 4P4 (cos θ) − ···],r 4 r8 rwhere the polar axis is normal to the plane of the disc.(b) Reconcile the presence of a term P 1 (cos θ), which is odd under θ → π − θ,with the symmetry with respect to the plane of the disc of the physicalsystem.(c) Deduce that the gravitational field near an infinite sheet of matter of constantdensity ρ per unit area is 2πGρ.21.21 In the region −∞

PDES: SEPARATION OF VARIABLES AND OTHER METHODS21.22 Point charges q and −qa/b (with a0 for the solution of∇ 2 Φ = 0 with Φ specified in cylindrical polar coordinates (ρ, φ, z) on the planez =0by{1 for ρ ≤ 1,Φ(ρ, φ, z) =1/ρ for ρ>1.Determine the variation of Φ(0, 0,z)alongthez-axis.21.24 Electrostatic charge is distributed in a sphere of radius R centred on the origin.Determine the form of the resultant potential φ(r) at distances much greater thanR, as follows.(a) Express in the form of an integral over all space the solution of∇ 2 φ = − ρ(r) .ɛ 0(b) Show that, for r ≫ r ′ ,|r − r ′ | = r − r · r′r( ) 1+O .r(c) Use results (a) and (b) to show that φ(r) hastheformφ(r) = M r + d · r ( ) 1+O .r 3 r 3Find expressions for M and d, and identify them physically.21.25 Find, in the form of an infinite series, the Green’s function of the ∇ 2 operator forthe Dirichlet problem in the region −∞

21.7 HINTS AND ANSWERSin V and takes the specified form φ = f on S, the boundary of V .TheGreen’s function, G(r, r ′ ), to be used satisfies∇ 2 G − m 2 G = δ(r − r ′ )and vanishes when r is on S.(b) When V is all space, G(r, r ′ ) can be written as G(t) =g(t)/t, wheret = |r − r ′ |and g(t) is bounded as t →∞. Find the form of G(t).(c) Find φ(r) in the half-space x>0ifρ(r) =δ(r − r 1 )andφ = 0 both on x =0and as r →∞.21.28 Consider the PDE Lu(r) =ρ(r), for which the differential operator L is given byL = ∇ · [p(r)∇]+q(r),where p(r) andq(r) are functions of position. By proving the generalised form ofGreen’s theorem,∫ ∮(φLψ − ψLφ) dV = p(φ∇ψ − ψ∇φ) · ˆn dS,VSshow that the solution of the PDE is given by∫∮ [u(r 0 )= G(r, r 0 )ρ(r) dV (r)+ p(r) u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r),VS∂n∂nwhere G(r, r 0 ) is the Green’s function satisfying LG(r, r 0 )=δ(r − r 0 ).21.7 Hints and answers21.1 (a) C exp[λ(x 2 +2y)]; (b) C(x 2 y) λ .21.3 u(x, y, t) = sin(nπx/a) sin(mπy/b)(A sin ωt + B cos ωt).21.5 (a) 6u/r 2 , −6u/r 2 ,0,l =2(or−3), m =0;(b) 2u/r 2 ,(cot 2 θ − 1)u/r 2 ; −u/(r 2 sin 2 θ), l =1(or−2), m = ±1.21.7 Solutions of the form r l give l as −1, 1, 2, 4. Because of the asymptotic form ofψ, anr 4 term cannot be present. The coefficients of the three remaining terms aredetermined by the two boundary conditions u = 0 on the sphere and the form ofψ for large r.21.9 Express cos 2 φ in terms of cos 2φ; T (ρ, φ) =A + B/2+(Bρ 2 /2a 2 )cos2φ.21.11 (A cos mx + B sin mx + C cosh mx + D sinh mx)cos(ωt + ɛ), with m 4 a 4 = ω 2 .21.13 E n =16ρA 2 c 2 /[(2n +1) 2 π 2 L]; E =2ρc 2 A 2 /L = ∫ A[2Tv/( 1 L)] dv.0 221.15 Note that the boundary value function is a square wave that is symmetric in φ.21.17 Since there is no heat flow at x = ±a, use a series of period 4a, u(x, 0) = 100 for0

PDES: SEPARATION OF VARIABLES AND OTHER METHODS21.25 The terms in G(r, r 0 ) that are additional to the fundamental solution are1∞∑ { [(x(−1) n − x0 ) 2 +(y − y 0 ) 2 +(z +(−1) n z 0 − nc) 2] −1/24πn=2+ [ (x − x 0 ) 2 +(y − y 0 ) 2 +(z +(−1) n z 0 + nc) 2] } −1/2.21.27 (a) As given in equation (21.86), but with r 0 replaced by r ′ .(b) Move the origin to r ′ and integrate the defining Green’s equation to obtain4πt 2 dG ∫ tdt − m2 G(t ′ )4πt ′2 dt ′ =1,0leading to G(t) =[−1/(4πt)]e −mt .(c) φ(r) =[−1/(4π)](p −1 e −mp − q −1 e −mq ), where p = |r − r 1 | and q = |r − r 2 | withr 1 =(x 1 ,y 1 ,z 1 )andr 2 =(−x 1 ,y 1 ,z 1 ).774

22Calculus of variationsIn chapters 2 and 5 we discussed how to find stationary values of functions of asingle variable f(x), of several variables f(x,y,...) and of constrained variables,where x,y,... are subject to the n constraints g i (x,y,...)=0,i =1, 2,...,n.Inallthese cases the forms of the functions f and g i were known, and the problem wasone of finding the appropriate values of the variables x, y etc.We now turn to a different kind of problem in which we are interested inbringing about a particular condition for a given expression (usually maximisingor minimising it) by varying the functions on which the expression depends. Forinstance, we might want to know in what shape a fixed length of rope shouldbe arranged so as to enclose the largest possible area, or in what shape it willhang when suspended under gravity from two fixed points. In each case we areconcerned with a general maximisation or minimisation criterion by which thefunction y(x) that satisfies the given problem may be found.The calculus of variations provides a method for finding the function y(x).The problem must first be expressed in a mathematical form, and the formmost commonly applicable to such problems is an integral. Ineachoftheabovequestions, the quantity that has to be maximised or minimised by an appropriatechoice of the function y(x) may be expressed as an integral involving y(x) andthe variables describing the geometry of the situation.In our example of the rope hanging from two fixed points, we need to findthe shape function y(x) that minimises the gravitational potential energy of therope. Each elementary piece of the rope has a gravitational potential energyproportional both to its vertical height above an arbitrary zero level and to thelength of the piece. Therefore the total potential energy is given by an integralfor the whole rope of such elementary contributions. The particular function y(x)for which the value of this integral is a minimum will give the shape assumed bythe hanging rope.775

CALCULUS OF VARIATIONSyabxFigure 22.1 Possible paths for the integral (22.1). The solid line is the curvealong which the integral is assumed stationary. The broken curves representsmall variations from this path.So in general we are led by this type of question to study the value of anintegral whose integrand has a specified form in terms of a certain functionand its derivatives, and to study how that value changes when the form ofthe function is varied. Specifically, we aim to find the function that makes theintegral stationary, i.e. the function that makes the value of the integral a localmaximum or minimum. Note that, unless stated otherwise, y ′ is used to denotedy/dx throughout this chapter. We also assume that all the functions we need todeal with are sufficiently smooth and differentiable.Let us consider the integral22.1 The Euler–Lagrange equationI =∫ baF(y, y ′ ,x) dx, (22.1)where a, b and the form of the function F are fixed by given considerations,e.g. the physics of the problem, but the curve y(x) is to be chosen so as tomake stationary the value of I, which is clearly a function, or more accurately afunctional, of this curve, i.e. I = I[y(x)]. Referring to figure 22.1, we wish to findthe function y(x) (given, say, by the solid line) such that first-order small changesin it (for example the two broken lines) will make only second-order changes inthe value of I.Writing this in a more mathematical form, let us suppose that y(x) is thefunction required to make I stationary and consider making the replacementy(x) → y(x)+αη(x), (22.2)where the parameter α is small and η(x) is an arbitrary function with sufficientlyamenable mathematical properties. For the value of I to be stationary with respect776

22.2 SPECIAL CASESto these variations, we requiredIdα∣ = 0α=0for all η(x). (22.3)Substituting (22.2) into (22.1) and expanding as a Taylor series in α we obtainI(y, α) ==∫ ba∫ baF(y + αη, y ′ + αη ′ ,x) dxF(y, y ′ ,x) dx +∫ ba( )∂F ∂Fαη +∂y ∂y ′ αη′ dx +O(α 2 ).With this form for I(y, α) the condition (22.3) implies that for all η(x) werequire∫ b( ∂FδI =a ∂y η + ∂F )∂y ′ η′ dx =0,where δI denotes the first-order variation in the value of I due to the variation(22.2) in the function y(x). Integrating the second term by parts this becomes[η ∂F ] b ∫ b[ ∂F∂y ′ +a a ∂y − d ( )] ∂Fdx ∂y ′ η(x) dx =0. (22.4)In order to simplify the result we will assume, for the moment, that the end-pointsare fixed, i.e. not only a and b are given but also y(a) andy(b). This restrictionmeans that we require η(a) =η(b) = 0, in which case the first term on the LHS of(22.4) equals zero at both end-points. Since (22.4) must be satisfied for arbitraryη(x), it is easy to see that we require∂F∂y = ddx( ∂F∂y ′ ). (22.5)This is known as the Euler–Lagrange (EL) equation, and is a differential equationfor y(x), since the function F is known.22.2 Special casesIn certain special cases a first integral of the EL equation can be obtained for ageneral form of F.22.2.1 F does not contain y explicitlyIn this case ∂F/∂y = 0, and (22.5) can be integrated immediately giving∂F∂y ′ = constant. (22.6)777

CALCULUS OF VARIATIONSB(b, y(b))dsdxdyA(a, y(a))Figure 22.2An arbitrary path between two fixed points.◮Show that the shortest curve joining two points is a straight line.Let the two points be labelled A and B and have coordinates (a, y(a)) and (b, y(b))respectively (see figure 22.2). Whatever the shape of the curve joining A to B, the lengthof an element of path ds is given byds = [ (dx) 2 +(dy) 2] 1/2=(1+y ′2 ) 1/2 dx,and hence the total path length along the curve is given byL =∫ ba(1 + y ′2 ) 1/2 dx. (22.7)We must now apply the results of the previous section to determine that path which makesL stationary (clearly a minimum in this case). Since the integral does not contain y (orindeed x) explicitly, we may use (22.6) to obtaink = ∂F y ′=∂y ′ (1 + y ′2 ) . 1/2where k is a constant. This is easily rearranged and integrated to giveky = x + c,(1 − k 2 )1/2which, as expected, is the equation of a straight line in the form y = mx + c, withm = k/(1 − k 2 ) 1/2 . The value of m (or k) can be found by demanding that the straight linepasses through the points A and B and is given by m =[y(b) − y(a)]/(b − a). Substitutingthe equation of the straight line into (22.7) we find that, again as expected, the total pathlength is given byL 2 =[y(b) − y(a)] 2 +(b − a) 2 . ◭778

22.2 SPECIAL CASESydsdydxxFigure 22.3A convex closed curve that is symmetrical about the x-axis.22.2.2 F does not contain x explicitlyIn this case, multiplying the EL equation (22.5) by y ′ and using(dy ′ ∂F ) ( )dx ∂y ′ = y ′ d ∂F ′′ ∂Fdx ∂y ′ + y∂y ′we obtainy ′ ∂F ∂F+ y′′∂y ∂y ′ = d (y ′ ∂F )dx ∂y ′ .But since F is a function of y and y ′ only, and not explicitly of x, theLHSofthis equation is just the total derivative of F, namely dF/dx. Hence, integratingwe obtainF − y ′ ∂F∂y ′ = constant. (22.8)◮Find the closed convex curve of length l that encloses the greatest possible area.Without any loss of generality we can assume that the curve passes through the originand can further suppose that it is symmetric with respect to the x-axis; this assumptionis not essential. Using the distance s along the curve, measured from the origin, as theindependent variable and y as the dependent one, we have the boundary conditionsy(0) = y(l/2) = 0. The element of area shown in figure 22.3 is then given byand the total area bydA = ydx= y [ (ds) 2 − (dy) 2] 1/2,A =2∫ l/20y(1 − y ′2 ) 1/2 ds; (22.9)here y ′ stands for dy/ds rather than dy/dx. Since the integrand does not contain s explicitly,779

CALCULUS OF VARIATIONSwe can use (22.8) to obtain a first integral of the EL equation for y, namelyy(1 − y ′2 ) 1/2 + yy ′2 (1 − y ′2 ) −1/2 = k,where k is a constant. On rearranging this givesky ′ = ±(k 2 − y 2 ) 1/2 ,which, using y(0) = 0, integrates toy/k = sin(s/k). (22.10)The other end-point, y(l/2) = 0, fixes the value of k as l/(2π) to yieldy =l2π2πssin .lFrom this we obtain dy =cos(2πs/l) ds and since (ds) 2 =(dx) 2 +(dy) 2 we find also thatdx = ± sin(2πs/l) ds. This in turn can be integrated and, using x(0) = 0, gives x in termsof s asx − l2π = − l2π2πscos .lWe thus obtain the expected result that x and y lie on the circle of radius l/(2π) givenby(x − l ) 2+ y 2 = l22π 4π . 2Substituting the solution (22.10) into the expression for the total area (22.9), it is easilyverified that A = l 2 /(4π). A much quicker derivation of this result is possible using planepolar coordinates. ◭The previous two examples have been carried out in some detail, even thoughthe answers are more easily obtained in other ways, expressly so that the methodis transparent and the way in which it works can be filled in mentally at almostevery step. The next example, however, does not have such an intuitively obvioussolution.◮Two rings, each of radius a, are placed parallel with their centres 2b apart and on acommon normal. An open-ended axially symmetric soap film is formed between them (seefigure 22.4). Find the shape assumed by the film.Creating the soap film requires an energy γ per unit area (numerically equal to the surfacetension of the soap solution). So the stable shape of the soap film, i.e. the one thatminimises the energy, will also be the one that minimises the surface area (neglectinggravitational effects).It is obvious that any convex surface, shaped such as that shown as the broken line infigure 22.4(a), cannot be a minimum but it is not clear whether some shape intermediatebetween the cylinder shown by solid lines in (a), with area 4πab (or twice this for thedouble surface of the film), and the form shown in (b), with area approximately 2πa 2 , willproduce a lower total area than both of these extremes. If there is such a shape (e.g. that infigure 22.4(c)), then it will be that which is the best compromise between two requirements,the need to minimise the ring-to-ring distance measured on the film surface (a) and theneed to minimise the average waist measurement of the surface (b).We take cylindrical polar coordinates as in figure 22.4(c) and let the radius of the soapfilm at height z be ρ(z) withρ(±b) =a. Counting only one side of the film, the element of780

22.3 SOME EXTENSIONSbzρ−ba(a) (b) (c)Figure 22.4Possible soap films between two parallel circular rings.surface area between z and z + dz isso the total surface area is given bydS =2πρ [ (dz) 2 +(dρ) 2] 1/2,S =2π∫ b−bρ(1 + ρ ′2 ) 1/2 dz. (22.11)Since the integrand does not contain z explicitly, we can use (22.8) to obtain an equationfor ρ that minimises S, i.e.ρ(1 + ρ ′2 ) 1/2 − ρρ ′2 (1 + ρ ′2 ) −1/2 = k,where k is a constant. Multiplying through by (1 + ρ ′2 ) 1/2 , rearranging to find an explicitexpression for ρ ′ and integrating we findcosh −1 ρk = z k + c.where c is the constant of integration. Using the boundary conditions ρ(±b) =a, werequire c =0andk such that a/k =coshb/k (if b/a istoolarge,nosuchk can be found).Thus the curve that minimises the surface area isρ/k =cosh(z/k),and in profile the soap film is a catenary (see section 22.4) with the minimum distancefrom the axis equal to k. ◭22.3 Some extensionsIt is quite possible to relax many of the restrictions we have imposed so far. Forexample, we can allow end-points that are constrained to lie on given curves ratherthan being fixed, or we can consider problems with several dependent and/orindependent variables or higher-order derivatives of the dependent variable. Eachof these extensions is now discussed.781

CALCULUS OF VARIATIONS22.3.1 Several dependent variablesHere we have F = F(y 1 ,y 1 ′ ,y 2,y 2 ′ ,...,y n,y n,x)whereeachy ′ i = y i (x). The analysisin this case proceeds as before, leading to n separate but simultaneous equationsfor the y i (x),∂F∂y i= ddx( ∂F∂y ′ i), i =1, 2,...,n. (22.12)22.3.2 Several independent variablesWith n independent variables, we need to extremise multiple integrals of the form∫ ∫ ∫ ()∂y ∂y ∂yI = ··· F y, , ,..., ,x 1 ,x 2 ,...,x n dx 1 dx 2 ···dx n .∂x 1 ∂x 2 ∂x nUsing the same kind of analysis as before, we find that the extremising functiony = y(x 1 ,x 2 ,...,x n ) must satisfy∂Fn∑( )∂y = ∂ ∂F, (22.13)∂x i ∂y xiwhere y xi stands for ∂y/∂x i .i=122.3.3 Higher-order derivativesIf in (22.1) F = F(y, y ′ ,y ′′ ,...,y (n) ,x) then using the same method as beforeand performing repeated integration by parts, it can be shown that the requiredextremising function y(x) satisfies∂F∂y − ddx( ∂F∂y ′ )+ d2dx 2 ( ∂F∂y ′′ )− ···+(−1) n dndx n ( ∂F∂y (n) )=0, (22.14)provided that y = y ′ = ··· = y (n−1) = 0 at both end-points. If y, or any of itsderivatives, is not zero at the end-points then a corresponding contribution orcontributions will appear on the RHS of (22.14).22.3.4 Variable end-pointsWe now discuss the very important generalisation to variable end-points. Suppose,as before, we wish to find the function y(x) that extremises the integralI =∫ baF(y, y ′ ,x) dx,but this time we demand only that the lower end-point is fixed, while we allowy(b) to be arbitrary. Repeating the analysis of section 22.1, we find from (22.4)782

22.3 SOME EXTENSIONSy(x)+η(x)∆yy(x)h(x, y) =0∆xbFigure 22.5 Variation of the end-point b along the curve h(x, y) =0.that we require[η ∂F ] b∂y ′ +a∫ ba[ ∂F∂y − d ( )] ∂Fdx ∂y ′ η(x) dx =0. (22.15)Obviously the EL equation (22.5) must still hold for the second term on the LHSto vanish. Also, since the lower end-point is fixed, i.e. η(a) = 0, the first term onthe LHS automatically vanishes at the lower limit. However, in order that it alsovanishes at the upper limit, we require in addition that∣∂F ∣∣∣x=b∂y ′ =0. (22.16)Clearly if both end-points may vary then ∂F/∂y ′ must vanish at both ends.An interesting and more general case is where the lower end-point is againfixed at x = a, but the upper end-point is free to lie anywhere on the curveh(x, y) = 0. Now in this case, the variation in the value of I due to the arbitraryvariation (22.2) is given to first order by[ ] ∂F b ∫ b( ∂FδI =∂y ′ η +a a ∂y − d )∂Fdx ∂y ′ ηdx+ F(b)∆x, (22.17)where ∆x is the displacement in the x-direction of the upper end-point, asindicated in figure 22.5, and F(b) is the value of F at x = b. In order for (22.17)to be valid, we of course require the displacement ∆x to be small.From the figure we see that ∆y = η(b) +y ′ (b)∆x. Since the upper end-pointmust lie on h(x, y) = 0 we also require that, at x = b,∂h ∂h∆x + ∆y =0,∂x ∂ywhich on substituting our expression for ∆y and rearranging becomes( )∂h ∂h+ y′ ∆x + ∂h η =0. (22.18)∂x ∂y ∂y783

CALCULUS OF VARIATIONSAx = x 0xyBFigure 22.6 A frictionless wire along which a small bead slides. We seek theshape of the wire that allows the bead to travel from the origin O to the linex = x 0 in the least possible time.Now, from (22.17) the condition δI = 0 requires, besides the EL equation, thatat x = b, the other two contributions cancel, i.e.F∆x + ∂F η =0. (22.19)∂y′Eliminating ∆x and η between (22.18) and (22.19) leads to the condition that atthe end-point(F − y ′ ∂F ) ∂h∂y ′ ∂y − ∂F ∂h∂y ′ =0. (22.20)∂xIn the special case where the end-point is free to lie anywhere on the vertical linex = b, we have ∂h/∂x = 1 and ∂h/∂y = 0. Substituting these values into (22.20),we recover the end-point condition given in (22.16).◮A frictionless wire in a vertical plane connects two points A and B, A being higher than B.Let the position of A be fixed at the origin of an xy-coordinate system, but allow B to lieanywhere on the vertical line x = x 0 (see figure 22.6). Find the shape of the wire such thatabeadplacedonitatA will slide under gravity to B in the shortest possible time.This is a variant of the famous brachistochrone (shortest time) problem, which is oftenused to illustrate the calculus of variations. Conservation of energy tells us that the particlespeed is given byv = dsdt = √ 2gy,where s is the path length along the wire and g is the acceleration due to gravity. Sincethe element of path length is ds =(1+y ′2 ) 1/2 dx, the total time taken to travel to the linex = x 0 is given by√∫ x=x0dst =x=0 v = √ 1 ∫ x01+y ′2dx.2g 0 yBecause the integrand does not contain x explicitly, we can use (22.8) with the specificform F = √ 1+y ′2 / √ y to find a first integral; on simplification this yields[1/2y(1 + y )] ′2 = k,784

22.4 CONSTRAINED VARIATIONwhere k is a constant. Letting a = k 2 and solving for y ′ we findy ′ = dy √ a − ydx = ,ywhich on substituting y = a sin 2 θ integrates to givex = a (2θ − sin 2θ)+c.2Thus the parametric equations of the curve are given byx = b(φ − sin φ)+c, y = b(1 − cos φ),where b = a/2 andφ =2θ; they define a cycloid, the curve traced out by a point onthe rim of a wheel of radius b rolling along the x-axis. We must now use the end-pointconditions to determine the constants b and c. Since the curve passes through the origin,we see immediately that c =0.Nowsincey(x 0 ) is arbitrary, i.e. the upper end-point canlie anywhere on the curve x = x 0 , the condition (22.20) reduces to (22.16), so that we alsorequire∣ ∂F ∣∣∣x=x0 y ′= √ =0,∂y ′ y(1 + y ′2 ) ∣x=x0which implies that y ′ =0atx = x 0 . In words, the tangent to the cycloid at B must beparallel to the x-axis; this requires πb = x 0 . ◭22.4 Constrained variationJust as the problem of finding thestationary values of a function f(x, y) subject tothe constraint g(x, y) = constant is solved by means of Lagrange’s undeterminedmultipliers (see chapter 5), so the corresponding problem in the calculus ofvariations is solved by an analogous method.Suppose that we wish to find the stationary values ofI =∫ bsubject to the constraint that the value ofJ =a∫ baF(y, y ′ ,x) dx,G(y, y ′ ,x) dxis held constant. Following the method of Lagrange undetermined multipliers letus define a new functionalK = I + λJ =∫ ba(F + λG) dx,and find its unconstrained stationary values. Repeating the analysis of section 22.1we find that we require∂F∂y − d ( ) [ ∂F ∂Gdx ∂y ′ + λ∂y − d ( )] ∂Gdx ∂y ′ =0,785

CALCULUS OF VARIATIONS−ayOaxFigure 22.7A uniform rope with fixed end-points suspended under gravity.which, together with the original constraint J = constant, will yield the requiredsolution y(x).This method is easily generalised to cases with more than one constraint by theintroduction of more Lagrange multipliers. If we wish to find the stationary valuesof an integral I subject to the multiple constraints that the values of the integralsJ i be held constant for i =1, 2,...,n, then we simply find the unconstrainedstationary values of the new integraln∑K = I + λ i J i .1◮Find the shape assumed by a uniform rope when suspended by its ends from two pointsat equal heights.We will solve this problem using x (see figure 22.7) as the independent variable. Lettheropeoflength2L be suspended between the points x = ±a, y =0(L>a)andhave uniform linear density ρ. We then need to find the stationary value of the rope’sgravitational potential energy,∫∫ aI = −ρg yds= −ρg y(1 + y ′2 ) 1/2 dx,with respect to small changes in the form of the rope but subject to the constraint thatthe total length of the rope remains constant, i.e.∫ ∫ aJ = ds = (1 + y ′2 ) 1/2 dx =2L.We thus define a new integral (omitting the factor −1 fromI for brevity)K = I + λJ =−a∫ a−a−a(ρgy + λ)(1 + y ′2 ) 1/2 dxand find its stationary values. Since the integrand does not contain the independentvariable x explicitly, we can use (22.8) to find the first integral:(ρgy + λ)(1+y ′2) 1/2 (− (ρgy + λ) 1+y ′2) −1/2y ′2 = k,786

22.5 PHYSICAL VARIATIONAL PRINCIPLESwhere k is a constant; this reduces to( ) 2 ρgy + λy ′2 =− 1.kMaking the substitution ρgy + λ = k cosh z, this can be integrated easily to give( )k ρgy + λρg cosh−1 = x + c,kwhere c is the constant of integration.We now have three unknowns, λ, k and c, that must be evaluated using the two endconditions y(±a) = 0 and the constraint J =2L. The end conditions giveρg(a + c)cosh = λ k k+ c)=coshρg(−a ,kand since a ≠ 0, these imply c =0andλ/k =cosh(ρga/k). Putting c =0intotheconstraint, in which y ′ = sinh(ρgx/k), we obtain∫ a[ ( ρgx)] 1/22L = 1+sinh 2 dx−ak= 2k ( ρga)ρg sinh .kCollecting together the values for the constants, the form adopted by the rope is thereforey(x) = kρg[cosh( ρgxk)− cosh( ρgakwhere k is the solution of sinh(ρga/k) =ρgL/k. This curve is known as a catenary. ◭)],22.5 Physical variational principlesMany results in both classical and quantum physics can be expressed as variationalprinciples, and it is often when expressed in this form that their physicalmeaning is most clearly understood. Moreover, once a physical phenomenon hasbeen written as a variational principle, we can use all the results derived in thischapter to investigate its behaviour. It is usually possible to identify conservedquantities, or symmetries of the system of interest, that otherwise might be foundonly with considerable effort. From the wide range of physical variational principleswe will select two examples from familiar areas of classical physics, namelygeometric optics and mechanics.22.5.1 Fermat’s principle in opticsFermat’s principle in geometrical optics states that a ray of light travelling in aregion of variable refractive index follows a path such that the total optical pathlength (physical length × refractive index) is stationary.787

CALCULUS OF VARIATIONSyBn 2xθ 1θ 2n 1AFigure 22.8 Path of a light ray at the plane interface between media withrefractive indices n 1 and n 2 ,wheren 2

22.5 PHYSICAL VARIATIONAL PRINCIPLESyOdx l xFigure 22.9 Transverse displacement on a taut string that is fixed at twopoints a distance l apart.states that in moving from one configuration at time t 0 to another at time t 1 themotion of such a system is such as to make∫ t1L = L(q 1 ,q 2 ...,q n , ˙q 1 , ˙q 2 ,...,˙q n ,t) dt (22.21)t 0stationary. The Lagrangian L is defined, in terms of the kinetic energy T andthe potential energy V (with respect to some reference situation), by L = T − V .Here V is a function of the q i only, not of the ˙q i . Applying the EL equation to Lwe obtain Lagrange’s equations,∂L= d ( ) ∂L, i =1, 2,...,n.∂q i dt ∂˙q i◮Using Hamilton’s principle derive the wave equation for small transverse oscillations of ataut string.In this example we are in fact considering a generalisation of (22.21) to a case involvingone isolated independent coordinate t, togetherwithacontinuum in which the q i becomethe continuous variable x. The expressions for T and V therefore become integrals over xrather than sums over the label i.If ρ and τ are the local density and tension of the string, both of which may depend onx, then, referring to figure 22.9, the kinetic and potential energies of the string are givenbyand (22.21) becomes∫ lT =0L = 1 2ρ2∫ t1( ) 2 ∫ ∂ylτdx, V =∂t0 2t 0dt∫ l0[ρ( ) 2 ∂y− τ∂t789( ) 2 ∂ydx∂x( ) ] 2 ∂ydx.∂x

CALCULUS OF VARIATIONSUsing (22.13) and the fact that y does not appear explicitly, we obtain(∂ρ ∂y )− ∂ (τ ∂y )=0.∂t ∂t ∂x ∂xIf, in addition, ρ and τ do not depend on x or t then∂ 2 y∂x = 1 ∂ 2 y2 c 2 ∂t , 2where c 2 = τ/ρ. This is the wave equation for small transverse oscillations of a tautuniform string. ◭22.6 General eigenvalue problemsWe have seen in this chapter that the problem of finding a curve that makes thevalue of a given integral stationary when the integral is taken along the curveresults, in each case, in a differential equation for the curve. It is not a greatextension to ask whether this may be used to solve differential equations, bysetting up a suitable variational problem and then seeking ways other than theEuler equation of finding or estimating stationary solutions.We shall be concerned with differential equations of the form Ly = λρ(x)y,where the differential operator L is self-adjoint, so that L = L † (with appropriateboundary conditions on the solution y) andρ(x) is some weight function, asdiscussed in chapter 17. In particular, we will concentrate on the Sturm–Liouvilleequation as an explicit example, but much of what follows can be applied toother equations of this type.We have already discussed the solution of equations of the Sturm–Liouvilletype in chapter 17 and the same notation will be used here. In this section,however, we will adopt a variational approach to estimating the eigenvalues ofsuch equations.Suppose we search for stationary values of the integral∫ b []I = p(x)y ′2 (x) − q(x)y 2 (x) dx, (22.22)awith y(a) =y(b) =0andp and q any sufficiently smooth and differentiablefunctions of x. However, in addition we impose a normalisation conditionJ =∫ baρ(x)y 2 (x) dx = constant. (22.23)Here ρ(x) is a positive weight function defined in the interval a ≤ x ≤ b, butwhich may in particular cases be a constant.Then, as in section 22.4, we use undetermined Lagrange multipliers, § and§ We use −λ, rather than λ, so that the final equation (22.24) appears in the conventional Sturm–Liouville form.790

22.6 GENERAL EIGENVALUE PROBLEMSconsider K = I − λJ given byK =∫ ba[py ′2 − (q + λρ)y 2] dx.On application of the EL equation (22.5) this yields(dp dy )+ qy + λρy =0, (22.24)dx dxwhich is exactly the Sturm–Liouville equation (17.34), with eigenvalue λ. Now,since both I and J are quadratic in y and its derivative, finding stationary valuesof K is equivalent to finding stationary values of I/J. This may also be shownby considering the functional Λ = I/J, for whichδΛ =(δI/J) − (I/J 2 ) δJ=(δI − ΛδJ)/J= δK/J.Hence, extremising Λ is equivalent to extremising K. Thus we have the importantresult that finding functions y that make I/J stationary is equivalent to findingfunctions y that are solutions of the Sturm–Liouville equation; the resulting valueof I/J equals the corresponding eigenvalue of the equation.Of course this does not tell us how to find such a function y and, naturally, tohave to do this by solving (22.24) directly defeats the purpose of the exercise. Wewill see in the next section how some progress can be made. It is worth recallingthat the functions p(x), q(x) andρ(x) can have many different forms, and so(22.24) represents quite a wide variety of equations.We now recall some properties of the solutions of the Sturm–Liouville equation.The eigenvalues λ i of (22.24) are real and will be assumed non-degenerate (forsimplicity). We also assume that the corresponding eigenfunctions have been madereal, so that normalised eigenfunctions y i (x) satisfy the orthogonality relation (asin (17.24))∫ bay i y j ρdx= δ ij . (22.25)Further, we take the boundary condition in the form[ ] x=by i py j′ = 0; (22.26)x=athis can be satisfied by y(a) =y(b) = 0, but also by many other sets of boundaryconditions.791

CALCULUS OF VARIATIONS◮Show that∫ b(y′j py ′ i − y j qy i)dx = λi δ ij . (22.27)aLet y i be an eigenfunction of (22.24), corresponding to a particular eigenvalue λ i ,sothat( )py′ ′i +(q + λi ρ)y i =0.Multiplying this through by y j and integrating from a to b (the first term by parts) weobtain( )[y ] b ∫ b∫ bj py′i y j(py ′ i) ′ dx + y j (q + λ i ρ)y i dx =0. (22.28)−aaThe first term vanishes by virtue of (22.26), and on rearranging the other terms and using(22.25), we find the result (22.27). ◭We see at once that, if the function y(x) minimises I/J, i.e. satisfies the Sturm–Liouville equation, then putting y i = y j = y in (22.25) and (22.27) yields J andI respectively on the left-hand sides; thus, as mentioned above, the minimisedvalue of I/J is just the eigenvalue λ, introduced originally as the undeterminedmultiplier.a◮For a function y satisfying the Sturm–Liouville equation verify that, provided (22.26) issatisfied, λ = I/J.Firstly, we multiply (22.24) through by y to givey(py ′ ) ′ + qy 2 + λρy 2 =0.Now integrating this expression by parts we have[ ] b ∫ b) ∫ bypy ′ −(py ′2 − qy 2 dx + λ ρy 2 dx =0.aaThe first term on the LHS is zero, the second is simply −I and the third is λJ. Thusλ = I/J. ◭a22.7 Estimation of eigenvalues and eigenfunctionsSince the eigenvalues λ i of the Sturm–Liouville equation are the stationary valuesof I/J (see above), it follows that any evaluation of I/J must yield a value that liesbetween the lowest and highest eigenvalues of the corresponding Sturm–Liouvilleequation, i.e.λ min ≤ I J ≤ λ max,where, depending on the equation under consideration, either λ min = −∞ and792

22.7 ESTIMATION OF EIGENVALUES AND EIGENFUNCTIONSλ max is finite, or λ max = ∞ and λ min is finite. Notice that here we have departedfrom direct consideration of the minimising problem and made a statement abouta calculation in which no actual minimisation is necessary.Thus, as an example, for an equation with a finite lowest eigenvalue λ 0 anyevaluation of I/J provides an upper bound on λ 0 . Further, we will now show thatthe estimate λ obtained is a better estimate of λ 0 than the estimated (guessed)function y is of y 0 , the true eigenfunction corresponding to λ 0 . The sense in which‘better’ is used here will be clear from the final result.Firstly, we expand the estimated or trial function y in terms of the completeset y i :y = y 0 + c 1 y 1 + c 2 y 2 + ··· ,where, if a good trial function has been guessed, the c i will be small. Using (22.25)we have immediately that J =1+ ∑ i |c i| 2 . The other required integral isI =∫ ba[ (p y 0 ′ + ∑ i2 (c i y i) ′ − q y 0 + ∑ i) ] 2c i y i dx.On multiplying out the squared terms, all the cross terms vanish because of(22.27) to leaveλ = I J= λ 0 + ∑ i |c i| 2 λ i1+ ∑ j |c j| 2= λ 0 + ∑ i|c i | 2 (λ i − λ 0 )+O(c 4 ).Hence λ differs from λ 0 by a term second order in the c i , even though y differedfrom y 0 by a term first order in the c i ; this is what we aimed to show. We noticeincidentally that, since λ 0

CALCULUS OF VARIATIONSy(x)1(c)0.80.6(b)(a)0.4(d)0.20.20.40.60.8x1Figure 22.10 Trial solutions used to estimate the lowest eigenvalue λ of−y ′′ = λy with y(0) = y ′ (1) = 0. They are: (a) y = sin(πx/2), the exact result;(b) y =2x − x 2 ;(c)y = x 3 − 3x 2 +3x; (d)y =sin 2 (πx/2).◮Estimate the lowest eigenvalue of the equationwith boundary conditions− d2 y= λy, 0 ≤ x ≤ 1, (22.29)dx2 y(0) = 0, y ′ (1) = 0. (22.30)We need to find the lowest value λ 0 of λ for which (22.29) has a solution y(x) that satisfies(22.30). The exact answer is of course y = A sin(xπ/2) and λ 0 = π 2 /4 ≈ 2.47.Firstly we note that the Sturm–Liouville equation reduces to (22.29) if we take p(x) =1,q(x) =0andρ(x) = 1 and that the boundary conditions satisfy (22.26). Thus we are ableto apply the previous theory.We will use three trial functions so that the effect on the estimate of λ 0 of making betteror worse ‘guesses’ can be seen. One further preliminary remark is relevant, namely that theestimate is independent of any constant multiplicative factor in the function used. Thisis easily verified by looking at the form of I/J. We normalise each trial function so thaty(1) = 1, purely in order to facilitate comparison of the various function shapes.Figure 22.10 illustrates the trial functions used, curve (a) being the exact solutiony =sin(πx/2). The other curves are (b) y(x) =2x − x 2 ,(c)y(x) =x 3 − 3x 2 +3x, and(d)y(x) =sin 2 (πx/2). The choice of trial function is governed by the following considerations:(i) the boundary conditions (22.30) must be satisfied.(ii) a ‘good’ trial function ought to mimic the correct solution as far as possible, butit may not be easy to guess even the general shape of the correct solution in somecases.(iii) the evaluation of I/J should be as simple as possible.794

22.8 ADJUSTMENT OF PARAMETERSIt is easily verified that functions (b), (c) and (d) all satisfy (22.30) but, so far as mimickingthe correct solution is concerned, we would expect from the figure that (b) would besuperior to the other two. The three evaluations are straightforward, using (22.22) and(22.23):∫ 10λ b =(2 − 2x)2 dx∫ 1(2x − 0 x2 ) 2 dx = 4/38/15 =2.50∫ 10λ c =(3x2 − 6x +3) 2 dx∫ 10 (x3 − 3x 2 +3x) 2 dx = 9/59/14 =2.80∫ 10λ d =(π2 /4) sin 2 (πx) dx∫ 1= π2 /80 sin4 (πx/2) dx 3/8 =3.29.We expected all evaluations to yield estimates greater than the lowest eigenvalue, 2.47,and this is indeed so. From these trials alone we are able to say (only) that λ 0 ≤ 2.50.As expected, the best approximation (b) to the true eigenfunction yields the lowest, andtherefore the best, upper bound on λ 0 . ◭We may generalise the work of this section to other differential equations ofthe form Ly = λρy, whereL = L † . In particular, one findswhere I and J are now given byI =∫ baλ min ≤ I J ≤ λ max,y ∗ (Ly) dx and J =∫ baρy ∗ ydx. (22.31)It is straightforward to show that, for the special case of the Sturm–Liouvilleequation, for whichLy = −(py ′ ) ′ − qy,the expression for I in (22.31) leads to (22.22).22.8 Adjustment of parametersInstead of trying to estimate λ 0 by selecting a large number of different trialfunctions, we may also use trial functions that include one or more parameterswhich themselves may be adjusted to give the lowest value to λ = I/J andhence the best estimate of λ 0 . The justification for this method comes from theknowledge that no matter what form of function is chosen, nor what values areassigned to the parameters, provided the boundary conditions are satisfied λ cannever be less than the required λ 0 .To illustrate this method an example from quantum mechanics will be used.The time-independent Schrödinger equation is formally written as the eigenvalueequation Hψ = Eψ, whereH is a linear operator, ψ the wavefunction describinga quantum mechanical system and E the energy of the system. The energy795

CALCULUS OF VARIATIONSoperator H is called the Hamiltonian and for a particle of mass m moving in aone-dimensional harmonic oscillator potential is given byH = − 22m dx 2 + kx22 , (22.32)where is Planck’s constant divided by 2π.◮Estimate the ground-state energy of a quantum harmonic oscillator.Using (22.32) in Hψ = Eψ, the Schrödinger equation is− 2 d 2 ψ2m dx + kx2 ψ = Eψ, −∞

22.9 EXERCISES22.9 Exercises22.1 A surface of revolution, whose equation in cylindrical polar coordinates is ρ =ρ(z), is bounded by the circles ρ = a, z = ±c (a >c). Show that the functionthat makes the surface integral I = ∫ ρ −1/2 dS stationary with respect to smallvariationsisgivenbyρ(z) =k + z 2 /(4k), where k =[a ± (a 2 − c 2 ) 1/2 ]/2.22.2 Show that the lowest value of the integral∫ B(1 + y ′2 ) 1/2dx,A ywhere A is (−1, 1) and B is (1, 1), is 2 ln(1+ √ 2). Assume that the Euler–Lagrangeequation gives a minimising curve.22.3 The refractive index n of a medium is a function only of the distance r from afixed point O. Prove that the equation of a light ray, assumed to lie in a planethrough O, travelling in the medium satisfies (in plane polar coordinates)( ) 21 dr= r2 n 2 (r)r 2 dφ a 2 n 2 (a) − 1,where a is the distance of the ray from O at the point at which dr/dφ =0.If n = [1+(α 2 /r 2 )] 1/2 and the ray starts and ends far from O, find its deviation(the angle through which the ray is turned), if its minimum distance from O is a.22.4 The Lagrangian for a π-meson is given byL(x,t)= 1 (˙φ 2 −|∇φ| 2 − µ 2 φ 2 ),2where µ is the meson mass and φ(x,t) is its wavefunction. Assuming Hamilton’sprinciple, find the wave equation satisfied by φ.22.5 Prove the following results about general systems.(a) For a system described in terms of coordinates q i and t, show that if t doesnot appear explicitly in the expressions for x, y and z (x = x(q i ,t), etc.) thenthe kinetic energy T is a homogeneous quadratic function of the ˙q i (it mayalso involve the q i ). Deduce that ∑ i ˙q i(∂T/∂˙q i )=2T .(b) Assuming that the forces acting on the system are derivable from a potentialV , show, by expressing dT /dt in terms of q i and ˙q i ,thatd(T + V )/dt =0.22.6 For a system specified by the coordinates q and t, show that the equation ofmotion is unchanged if the Lagrangian L(q, ˙q, t) is replaced byL 1 = L + dφ(q,t) ,dtwhere φ is an arbitrary function. Deduce that the equation of motion of a particlethat moves in one dimension subject to a force −dV (x)/dx (x being measuredfrom a point O) is unchanged if O is forced to move with a constant velocity v(x still being measured from O).22.7 In cylindrical polar coordinates, the curve (ρ(θ),θ,αρ(θ)) lies on the surface ofthe cone z = αρ. Show that geodesics (curves of minimum length joining twopoints) on the cone satisfyρ 4 = c 2 [β 2 ρ ′2 + ρ 2 ],where c is an arbitrary constant, but β has to have a particular value. Determinethe form of ρ(θ) and hence find the equation of the shortest path on the conebetween the points (R,−θ 0 ,αR)and(R,θ 0 ,αR).[ You will find it useful to determine the form of the derivative of cos −1 (u −1 ). ]797

CALCULUS OF VARIATIONS22.8 Derive the differential equations for the plane-polar coordinates, r and φ, ofaparticle of unit mass moving in a field of potential V (r). Find the form of V ifthe path of the particle is given by r = a sin φ.22.9 You are provided with a line of length πa/2 and negligible mass and some leadshot of total mass M. Use a variational method to determine how the lead shotmust be distributed along the line if the loaded line is to hang in a circular arcof radius a when its ends are attached to two points at the same height. Measurethe distance s along the line from its centre.22.10 Extend the result of subsection 22.2.2 to the case of several dependent variablesy i (x), showing that, if x does not appear explicitly in the integrand, then a firstintegral of the Euler–Lagrange equations isn∑F − y ′ ∂Fi =constant.∂y ′ i=1 i22.11 A general result is that light travels through a variable medium by a path whichminimises the travel time (this is an alternative formulation of Fermat’s principle).With respect to a particular cylindrical polar coordinate system (ρ, φ, z), the speedof light v(ρ, φ) is independent of z. If the path of the light is parameterised asρ = ρ(z),φ= φ(z), use the result of the previous exercise to show thatv 2 (ρ ′2 + ρ 2 φ ′2 +1)is constant along the path.For the particular case when v = v(ρ) =b(a 2 +ρ 2 ) 1/2 , show that the two Euler–Lagrange equations have a common solution in which the light travels along ahelical path given by φ = Az + B, ρ = C, provided that A has a particular value.22.12 Light travels in the vertical xz-plane through a slab of material which lies betweenthe planes z = z 0 and z =2z 0 , and in which the speed of light v(z) =c 0 z/z 0 .Using the alternative formulation of Fermat’s principle, given in the previousquestion, show that the ray paths are arcs of circles.Deduce that, if a ray enters the material at (0,z 0 ) at an angle to the vertical,π/2 − θ, ofmorethan30 ◦ , then it does not reach the far side of the slab.22.13 A dam of capacity V (less than πb 2 h/2) is to be constructed on level ground nextto a long straight wall which runs from (−b, 0) to (b, 0).Thisistobeachievedbyjoining the ends of a new wall, of height h, to those of the existing wall. Showthat, in order to minimise the length L of new wall to be built, it should formpart of a circle, and that L is then given by∫ bdx−b (1 − λ 2 x 2 ) , 1/2where λ is found fromVhb = sin−1 µ− (1 − µ2 ) 1/22 µ 2 µand µ = λb.22.14 In the brachistochrone problem of subsection 22.3.4 show that if the upper endpointcan lie anywhere on the curve h(x, y) = 0, then the curve of quickest descenty(x) meetsh(x, y) = 0 at right angles.22.15 The Schwarzchild metric for the static field of a non-rotating spherically symmetricblack hole of mass M is given by((ds) 2 = c 2 1 − 2GM )(dt) 2 −c 2 r(dr) 21 − 2GM/(c 2 r) − r2 (dθ) 2 − r 2 sin 2 θ (dφ) 2 .Considering only motion confined to the plane θ = π/2, and assuming that the798

22.9 EXERCISESpath of a small test particle is such as to make ∫ ds stationary, find two firstintegrals of the equations of motion. From their Newtonian limits, in whichGM/r, ṙ 2 and r 2 ˙φ 2 are all ≪ c 2 , identify the constants of integration.22.16 Use result (22.27) to evaluateJ =∫ 1−1(1 − x 2 )P ′ m(x)P ′ n(x) dx,where P m (x) is a Legendre polynomial of order m.22.17 Determine the minimum value that the integralJ =∫ 10[x 4 (y ′′ ) 2 +4x 2 (y ′ ) 2 ] dxcan have, given that y is not singular at x =0andthaty(1) = y ′ (1) = 1. Assumethat the Euler–Lagrange equation gives the lower limit, and verify retrospectivelythat your solution makes the first term on the LHS of equation (22.15) vanish.22.18 Show that y ′′ − xy + λx 2 y = 0 has a solution for which y(0) = y(1) = 0 andλ ≤ 147/4.22.19 Find an appropriate, but simple, trial function and use it to estimate the lowesteigenvalue λ 0 of Stokes’ equation,d 2 y+ λxy =0, with y(0) = y(π) =0.dx2 Explain why your estimate must be strictly greater than λ 0 .22.20 Estimate the lowest eigenvalue, λ 0 , of the equationd 2 ydx − 2 x2 y + λy =0, y(−1) = y(1) = 0,using a quadratic trial function.22.21 A drumskin is stretched across a fixed circular rim of radius a. Small transversevibrations of the skin have an amplitude z(ρ, φ, t) thatsatisfies∇ 2 z = 1 ∂ 2 zc 2 ∂t 2in plane polar coordinates. For a normal mode independent of azimuth, z =Z(ρ)cosωt, find the differential equation satisfied by Z(ρ). By using a trialfunction of the form a ν − ρ ν , with adjustable parameter ν, obtain an estimate forthe lowest normal mode frequency.[ The exact answer is (5.78) 1/2 c/a.]22.22 Consider the problem of finding the lowest eigenvalue, λ 0 , of the equation(1 + x 2 ) d2 y dy+2x + λy =0, y(±1) = 0.dx2 dx(a) Recast the problem in variational form, and derive an approximation λ 1 toλ 0 by using the trial function y 1 (x) =1− x 2 .(b) Show that an improved estimate λ 2 is obtained by using y 2 (x) =cos(πx/2).(c) Prove that the estimate λ(γ) obtained by taking y 1 (x) +γy 2 (x) asthetrialfunction isλ(γ) = 64/15 + 64γ/π − 384γ/π3 +(π 2 /3+1/2)γ 2.16/15 + 64γ/π 3 + γ 2Investigate λ(γ) numerically as γ is varied, or, more simply, show thatλ(−1.80) = 3.668, an improvement on both λ 1 and λ 2 .799

CALCULUS OF VARIATIONS22.23 For the boundary conditions given below, obtain a functional Λ(y) whose stationaryvalues give the eigenvalues of the equation(1 + x) d2 y dy+(2+x)dx2 dx + λy =0, y(0) = 0, y′ (2) = 0.Derive an approximation to the lowest eigenvalue λ 0 using the trial functiony(x) =xe −x/2 . For what value(s) of γ wouldy(x) =xe −x/2 + β sin γxbe a suitable trial function for attempting to obtain an improved estimate of λ 0 ?22.24 This is an alternative approach to the example in section 22.8. Using the notation∫of that section, the expectation value of the energy of the state ψ is given byψ ∗ Hψdv. Denote the eigenfunctions of H by ψ i ,sothatHψ i = E i ψ i , and, sinceH is self-adjoint (Hermitian), ∫ ψj ∗ψ i dv = δ ij .(a) By writing any function ψ as ∑ c j ψ j and following an argument similar tothat in section 22.7, show that∫ψ ∗ HψdvE = ∫ ≥ E 0 ,ψ∗ ψdvthe energy of the lowest state. This is the Rayleigh–Ritz principle.(b) Using the same trial function as in section 22.8, ψ =exp(−αx 2 ), show thatthe same result is obtained.22.25 This is an extension to section 22.8 and the previous question. With the groundstate(i.e. the lowest-energy) wavefunction as exp(−αx 2 ), take as a trial functionthe orthogonal wave function x 2n+1 exp(−αx 2 ), using the integer n as a variableparameter. Use either Sturm–Liouville theory or the Rayleigh–Ritz principle toshow that the energy of the second lowest state of a quantum harmonic oscillatoris ≤ 3ω/2.22.26 The Hamiltonian H for the hydrogen atom is− 22m ∇2 −q24πɛ 0 r .For a spherically symmetric state, as may be assumed for the ground state, theonly relevant part of ∇ 2 is that involving differentiation with respect to r.(a) Define the integrals J n byJ n =∫ ∞0r n e −2βr dr∫and show that, for a trial wavefunction of the form exp(−βr) withβ>0,ψ ∗ Hψdv and ∫ ψ ∗ ψdv(see exercise 22.24(a)) can be expressed as aJ 1 − bJ 2and cJ 2 respectively, where a, b and c are factors which you should determine.(b) Show that the estimate of E is minimised when β = mq 2 /(4πɛ 0 2 ).(c) Hence find an upper limit for the ground-state energy of the hydrogen atom.In fact, exp(−βr) is the correct form for the wavefunction and the limit givesthe actual value.22.27 The upper and lower surfaces of a film of liquid, which has surface energy perunit area (surface tension) γ and density ρ, have equations z = p(x) andz = q(x),respectively. The film has a given volume V (per unit depth in the y-direction)and lies in the region −L

22.10 HINTS AND ANSWERStotal energy (per unit depth) of the film consists of its surface energy and itsgravitational energy, and is expressed by∫ L∫ L]E = ρg (p 2 − q 2 ) dx + γ[(1 + p ′2 ) 1/2 +(1+q ′2 ) 1/2 dx.2−L(a) Express V in terms of p and q.(b) Show that, if the total energy is minimised, p and q must satisfyp ′2(1 + p ′2 ) − q ′2=constant.1/2 (1 + q ′2 )1/2(c) As an approximate solution, consider the equationsp = a(L −|x|), q = b(L −|x|),where a and b are sufficiently small that a 3 and b 3 can be neglected comparedwith unity. Find the values of a and b that minimise E.22.28 A particle of mass m moves in a one-dimensional potential well of the formV (x) =−µ 2 α 2m sech 2 αx,where µ and α are positive constants. As in exercise 22.26, the expectation value〈E〉 of the energy of the system is ∫ ψ ∗ Hψdx, where the self-adjoint operatorH is given by −( 2 /2m)d 2 /dx 2 + V (x). Using trial wavefunctions of the formy = A sech βx, show the following:(a) for µ = 1, there is an exact eigenfunction of H, with a corresponding 〈E〉 ofhalf of the maximum depth of the well;(b) for µ = 6, the ‘binding energy’ of the ground state is at least 10 2 α 2 /(3m).[ You will find it useful to note that for u, v ≥ 0, sech u sech v ≥ sech(u + v). ]22.29 The Sturm–Liouville equation can be extended to two independent variables, xand z, with little modification. In equation (22.22), y ′2 is replaced by (∇y) 2 andthe integrals of the various functions of y(x, z) become two-dimensional, i.e. theinfinitesimal is dx dz.The vibrations of a trampoline 4 units long and 1 unit wide satisfy the equation∇ 2 y + k 2 y =0.By taking the simplest possible permissible polynomial as a trial function, showthat the lowest mode of vibration has k 2 ≤ 10.63 and, by direct solution, that theactual value is 10.49.−L22.10 Hints and answers22.1 Note that the integrand, 2πρ 1/2 (1 + ρ ′2 ) 1/2 , does not contain z explicitly.22.3 I = ∫ n(r)[r 2 +(dr/dφ) 2 ] 1/2 dφ. Take axes such that φ = 0 when r = ∞.If β =(π − deviation angle)/2 thenβ = φ at r = a, and the equation reduces to∫β∞(a 2 + α 2 ) = dr1/2 −∞ r(r 2 − a 2 ) , 1/2which can be evaluated by putting r = a(y + y −1 )/2, or successively r = a cosh ψ,y =expψ to yield a deviation of π[(a 2 + α 2 ) 1/2 − a]/a.801

CALCULUS OF VARIATIONS22.5 (a) ∂x/∂t =0andsoẋ = ∑ i ˙q i∂x/∂q i ; (b) use∑( )d ∂T˙q i = d dt ∂˙q i dt (2T ) − ∑ ii22.7 Use result (22.8); β 2 =1+α 2 .¨q i∂T∂˙q i.Put ρ = uc to obtain dθ/du = β/[u(u 2 − 1) 1/2 ]. Remember that cos −1 is amultivalued function; ρ(θ) =[R cos(θ 0 /β)]/[cos(θ/β)].22.9 −λy ′ (1 − y ′2 ) −1/2 = 2gP(s), y = y(s), P (s) = ∫ s0 ρ(s′ ) ds ′ . The solution, y =−a cos(s/a), and 2P (πa/4) = M together give λ = −gM. Therequiredρ(s) isgiven by [M/(2a)] sec 2 (s/a).22.11 Note that the φ E–L equation is automatically satisfied if v ≠ v(φ). A =1/a.22.13 Circle is λ 2 x 2 +[λy +(1− λ 2 b 2 ) 1/2 ] 2 =1.Usethefactthat ∫ ydx = V/h todetermine the condition on λ.22.15 Denoting (ds) 2 /(dt) 2 by f 2 , the Euler–Lagrange equation for φ gives r 2 ˙φ = Af,where A corresponds to the angular momentum of the particle. Use the resultof exercise 22.10 to obtain c 2 − (2GM/r) =Bf, where, to first order in smallquantities,cB = c 2 − GM + 1 r 2 (ṙ2 + r 2 ˙φ2 ),which reads ‘total energy = rest mass + gravitational energy + radial andazimuthal kinetic energy’.22.17 Convert the equation to the usual form, by writing y ′ (x) =u(x), and obtainx 2 u ′′ +4xu ′ − 4u = 0 with general solution Ax −4 + Bx. Integrating a second timeand using the boundary conditions gives y(x) =(1+x 2 )/2 andJ =1;η(1) = 0,since y ′ (1) is fixed, and ∂F/∂u ′ =2x 4 u ′ =0atx =0.22.19 Using y =sinx as a trial function shows that λ 0 ≤ 2/π. The estimate must be>λ 0 since the trial function does not satisfy the original equation.22.21 Z ′′ + ρ −1 Z ′ +(ω/c) 2 Z =0,withZ(a) =0andZ ′ (0) = 0; this is an SL equationwith p = ρ, q = 0 and weight function ρ/c 2 .Estimateofω 2 =[c 2 ν/(2a 2 )][0.5 −2(ν +2) −1 +(2ν +2) −1 ] −1 , which minimises to c 2 (2 + √ 2) 2 /(2a 2 )=5.83c 2 /a 2 whenν = √ 2.22.23 Note that the original equation is not self-adjoint; it needs an integrating factorof e x .Λ(y) =[ ∫ 2(1 + 0 x)ex y ′2 dx]/[ ∫ 20 ex y 2 dx; λ 0 ≤ 3/8. Since y ′ (2) must equal 0,γ =(π/2)(n + 1 ) for some integer n.222.25 E 1 ≤ (ω/2)(8n 2 +12n +3)/(4n + 1), which has a minimum value 3ω/2 wheninteger n =0.22.27 (a) V = ∫ L(p − q) dx. (c)UseV =(a − −L b)L2 to eliminate b from the expressionfor E; now the minimisation is with respect to a alone. The values for a and bare ±V/(2L 2 ) − Vρg/(6γ).22.29 The SL equation has p =1,q =0,andρ =1.Use u(x, z) =x(4 − x)z(1 − z) as a trial function; numerator = 1088/90, denominator= 512/450. Direct solution k 2 =17π 2 /16.802

23Integral equationsIt is not unusual in the analysis of a physical system to encounter an equationin which an unknown but required function y(x), say, appears under an integralsign. Such an equation is called an integral equation, and in this chapter we discussseveral methods for solving the more straightforward examples of such equations.Before embarking on our discussion of methods for solving various integralequations, we begin with a warning that many of the integral equations met inpractice cannot be solved by the elementary methods presented here but mustinstead be solved numerically, usually on a computer. Nevertheless, the regularoccurrence of several simple types of integral equation that may be solvedanalytically is sufficient reason to explore these equations more fully.We shall begin this chapter by discussing how a differential equation can betransformed into an integral equation and by considering the most commontypes of linear integral equation. After introducing the operator notation andconsidering the existence of solutions for various types of equation, we go onto discuss elementary methods of obtaining closed-form solutions of simpleintegral equations. We then consider the solution of integral equations in terms ofinfinite series and conclude by discussing the properties of integral equations withHermitian kernels, i.e. those in which the integrands have particular symmetryproperties.23.1 Obtaining an integral equation from a differential equationIntegral equations occur in many situations, partly because we may always rewritea differential equation as an integral equation. It is sometimes advantageous tomake this transformation, since questions concerning the existence of a solutionare more easily answered for integral equations (see section 23.3), and,furthermore, an integral equation can incorporate automatically any boundaryconditions on the solution.803

INTEGRAL EQUATIONSWe shall illustrate the principles involved by considering the differential equationy ′′ (x) =f(x, y), (23.1)where f(x, y) can be any function of x and y but not of y ′ (x). Equation (23.1)thus represents a large class of linear and non-linear second-order differentialequations.We can convert (23.1) into the corresponding integral equation by first integratingwith respect to x to obtainy ′ (x) =Integrating once more, we findy(x) =∫ x0du∫ x0∫ u0f(z,y(z)) dz + c 1 .f(z,y(z)) dz + c 1 x + c 2 .Provided we do not change the region in the uz-plane over which the doubleintegral is taken, we can reverse the order of the two integrations. Changing theintegration limits appropriately, we findy(x) ==∫ x0∫ x0f(z,y(z)) dz∫ xzdu + c 1 x + c 2 (23.2)(x − z)f(z,y(z)) dz + c 1 x + c 2 ; (23.3)this is a non-linear (for general f(x, y)) Volterra integral equation.It is straightforward to incorporate any boundary conditions on the solutiony(x) by fixing the constants c 1 and c 2 in (23.3). For example, we might have theone-point boundary condition y(0) = a and y ′ (0) = b, for which it is clear thatwe must set c 1 = b and c 2 = a.23.2 Types of integral equationFrom (23.3), we can see that even a relatively simple differential equation suchas (23.1) can lead to a corresponding integral equation that is non-linear. In thischapter, however, we will restrict our attention to linear integral equations, whichhave the general formg(x)y(x) =f(x)+λ∫ baK(x, z)y(z) dz. (23.4)In (23.4), y(x) is the unknown function, while the functions f(x), g(x) andK(x, z)are assumed known. K(x, z) is called the kernel of the integral equation. Theintegration limits a and b are also assumed known, and may be constants orfunctions of x, andλ is a known constant or parameter.804

23.3 OPERATOR NOTATION AND THE EXISTENCE OF SOLUTIONSIn fact, we shall be concerned with various special cases of (23.4), which areknown by particular names. Firstly, if g(x) = 0 then the unknown function y(x)appears only under the integral sign, and (23.4) is called a linear integral equationof the first kind. Alternatively, if g(x) = 1, so that y(x) appears twice, once insidethe integral and once outside, then (23.4) is called a linear integral equation ofthe second kind. In either case, if f(x) = 0 the equation is called homogeneous,otherwise inhomogeneous.We can distinguish further between different types of integral equation by theform of the integration limits a and b. If these limits are fixed constants then theequation is called a Fredholm equation. If, however, the upper limit b = x (i.e. itis variable) then the equation is called a Volterra equation; such an equation isanalogous to one with fixed limits but for which the kernel K(x, z) =0forz>x.Finally, we note that any equation for which either (or both) of the integrationlimits is infinite, or for which K(x, z) becomes infinite in the range of integration,is called a singular integral equation.23.3 Operator notation and the existence of solutionsThere is a close correspondence between linear integral equations and the matrixequations discussed in chapter 8. However, the former involve linear, integral relationsbetween functions in an infinite-dimensional function space (see chapter 17),whereas the latter specify linear relations among vectors in a finite-dimensionalvector space.Since we are restricting our attention to linear integral equations, it will beconvenient to introduce the linear integral operator K, whose action on anarbitrary function y is given byKy =∫ baK(x, z)y(z) dz. (23.5)This is analogous to the introduction in chapters 16 and 17 of the notation L todescribe a linear differential operator. Furthermore, we may define the Hermitianconjugate K † byK † y =∫ baK ∗ (z,x)y(z) dz,where the asterisk denotes complex conjugation and we have reversed the orderof the arguments in the kernel.It is clear from (23.5) that K is indeed linear. Moreover, since K operates onthe infinite-dimensional space of (reasonable) functions, we may make an obviousanalogy with matrix equations and consider the action of K on a function f asthat of a matrix on a column vector (both of infinite dimension).When written in operator form, the integral equations discussed in the previoussection resemble equations familiar from linear algebra. For example, the805

INTEGRAL EQUATIONSinhomogeneous Fredholm equation of the first kind may be written as0=f + λKy,which has the unique solution y = −K −1 f/λ, provided that f ≠ 0 and the inverseoperator K −1 exists.Similarly, we may write the corresponding Fredholm equation of the secondkind asy = f + λKy. (23.6)In the homogeneous case, where f = 0, this reduces to y = λKy, which isreminiscent of an eigenvalue problem in linear algebra (except that λ appears onthe other side of the equation) and, similarly, only has solutions for at most acountably infinite set of eigenvalues λ i . The corresponding solutions y i are calledthe eigenfunctions.In the inhomogeneous case (f ≠ 0), the solution to (23.6) can be writtensymbolically asy =(1− λK) −1 f,again provided that the inverse operator exists. It may be shown that, in general,(23.6) does possess a unique solution if λ ≠ λ i ,i.e.whenλ does not equal one ofthe eigenvalues of the corresponding homogeneous equation.When λ does equal one of these eigenvalues, (23.6) may have either manysolutions or no solution at all, depending on the form of f. If the function f isorthogonal to every eigenfunction of the equationthat belongs to the eigenvalue λ ∗ ,i.e.〈g|f〉 =g = λ ∗ K † g (23.7)∫ bag ∗ (x)f(x) dx =0for every function g obeying (23.7), then it can be shown that (23.6) has manysolutions. Otherwise the equation has no solution. These statements are discussedfurther in section 23.7, for the special case of integral equations with Hermitiankernels, i.e. those for which K = K † .23.4 Closed-form solutionsIn certain very special cases, it may be possible to obtain a closed-form solutionof an integral equation. The reader should realise, however, when faced with anintegral equation, that in general it will not be soluble by the simple methodspresented in this section but must instead be solved using (numerical) iterativemethods, such as those outlined in section 23.5.806

23.4 CLOSED-FORM SOLUTIONS23.4.1 Separable kernelsThe most straightforward integral equations to solve are Fredholm equationswith separable (or degenerate) kernels. A kernel is separable if it has the formn∑K(x, z) = φ i (x)ψ i (z), (23.8)i=1where φ i (x) areψ i (z) are respectively functions of x only and of z only and thenumber of terms in the sum, n, is finite.Let us consider the solution of the (inhomogeneous) Fredholm equation of thesecond kind,y(x) =f(x)+λ∫ baK(x, z)y(z) dz, (23.9)which has a separable kernel of the form (23.8). Writing the kernel in its separatedform, the functions φ i (x) may be taken outside the integral over z to obtainn∑∫ by(x) =f(x)+λ φ i (x) ψ i (z)y(z) dz.i=1Since the integration limits a and b are constant for a Fredholm equation, theintegral over z in each term of the sum is just a constant. Denoting these constantsbyc i =the solution to (23.9) is found to be∫ bay(x) =f(x)+λaψ i (z)y(z) dz, (23.10)n∑c i φ i (x), (23.11)where the constants c i can be evalutated by substituting (23.11) into (23.10).◮Solve the integral equationy(x) =x + λ∫ 10i=1(xz + z 2 )y(z) dz. (23.12)The kernel for this equation is K(x, z) =xz + z 2 , which is clearly separable, and using thenotation in (23.8) we have φ 1 (x) =x, φ 2 (x) =1,ψ 1 (z) =z and ψ 2 (z) =z 2 . From (23.11)the solution to (23.12) has the formy(x) =x + λ(c 1 x + c 2 ),where the constants c 1 and c 2 are given by (23.10) asc 1 =c 2 =∫ 10∫ 10z[z + λ(c 1 z + c 2 )] dz = 1 3 + 1 3 λc 1 + 1 2 λc 2,z 2 [z + λ(c 1 z + c 2 )] dz = 1 4 + 1 4 λc 1 + 1 3 λc 2.807

INTEGRAL EQUATIONSThese two simultaneous linear equations may be straightforwardly solved for c 1 and c 2 togive24 + λ18c 1 =and c72 − 48λ − λ 2 2 =72 − 48λ − λ , 2so that the solution to (23.12) isy(x) =(72 − 24λ)x +18λ72 − 48λ − λ 2 . ◭In the above example, we see that (23.12) has a (finite) unique solution providedthat λ is not equal to either root of the quadratic in the denominator of y(x).The roots of this quadratic are in fact the eigenvalues of the correspondinghomogeneous equation, as mentioned in the previous section. In general, if theseparable kernel contains n terms, as in (23.8), there will be n such eigenvalues,although they may not all be different.Kernels consisting of trigonometric (or hyperbolic) functions of sums or differencesof x and z are also often separable.◮Find the eigenvalues and corresponding eigenfunctions of the homogeneous Fredholmequationy(x) =λ∫ π0sin(x + z) y(z) dz. (23.13)The kernel of this integral equation can be written in separated form asK(x, z) =sin(x + z) =sinx cos z +cosx sin z,so, comparing with (23.8), we have φ 1 (x) = sinx, φ 2 (x) = cosx, ψ 1 (z) = cosz andψ 2 (z) =sinz.Thus, from (23.11), the solution to (23.13) has the formwhere the constants c 1 and c 2 are given byc 1 = λc 2 = λ∫ π0∫ π0y(x) =λ(c 1 sin x + c 2 cos x),cos z (c 1 sin z + c 2 cos z) dz = λπ 2 c 2, (23.14)sin z (c 1 sin z + c 2 cos z) dz = λπ 2 c 1. (23.15)Combining these two equations we find c 1 =(λπ/2) 2 c 1 , and, assuming that c 1 ≠0,thisgives λ = ±2/π, the two eigenvalues of the integral equation (23.13).By substituting each of the eigenvalues back into (23.14) and (23.15), we find thatthe eigenfunctions corresponding to the eigenvalues λ 1 =2/π and λ 2 = −2/π are givenrespectively byy 1 (x) =A(sin x +cosx) and y 2 (x) =B(sin x − cos x), (23.16)where A and B are arbitrary constants. ◭808

23.4 CLOSED-FORM SOLUTIONS23.4.2 Integral transform methodsIf the kernel of an integral equation can be written as a function of the differencex − z of its two arguments, then it is called a displacement kernel. An integralequation having such a kernel, and which also has the integration limits −∞ to∞, may be solved by the use of Fourier transforms (chapter 13).If we consider the following integral equation with a displacement kernel,y(x) =f(x)+λ∫ ∞−∞K(x − z)y(z) dz, (23.17)the integral over z clearly takes the form of a convolution (see chapter 13).Therefore, Fourier-transforming (23.17) and using the convolution theorem, weobtainỹ(k) =˜f(k)+ √ 2πλ ˜K(k)ỹ(k),which may be rearranged to give˜f(k)ỹ(k) =1 − √ 2πλ ˜K(k) . (23.18)Taking the inverse Fourier transform, the solution to (23.17) is given byy(x) = √ 1 ∫ ∞ ˜f(k) exp(ikx)2π −∞ 1 − √ 2πλ ˜K(k) dk.If we can perform this inverse Fourier transformation then the solution can befound explicitly; otherwise it must be left in the form of an integral.◮Find the Fourier transform of the function{1 if |x| ≤a,g(x) =0 if |x| >a.Hence find an explicit expression for the solution of the integral equation∫ ∞sin(x − z)y(x) =f(x)+λy(z) dz. (23.19)−∞ x − zFind the solution for the special case f(x) =(sinx)/x.The Fourier transform of g(x) is given directly by˜g(k) = √ 1 ∫ a[ ] a√1 exp(−ikx) 2 sin kaexp(−ikx) dx = √ = .2π −a2π (−ik)−aπ k(23.20)The kernel of the integral equation (23.19) is K(x − z) = [sin(x − z)]/(x − z). Using(23.20), it is straightforward to show that the Fourier transform of the kernel is˜K(k) ={√π/2 if |k| ≤1,0 if |k| > 1.(23.21)809

INTEGRAL EQUATIONSThus, using (23.18), we find the Fourier transform of the solution to beỹ(k) ={˜f(k)/(1 − πλ) if |k| ≤1,˜f(k) if |k| > 1.(23.22)Inverse Fourier-transforming, and writing the result in a slightly more convenient form,the solution to (23.19) is given by( ) ∫ 1 1 1y(x) =f(x)+1 − πλ − 1 √ ˜f(k)exp(ikx) dk2π −1= f(x)+ πλ ∫1 1√ ˜f(k)exp(ikx) dk. (23.23)1 − πλ 2π −1It is clear from (23.22) that when λ = 1/π, which is the only eigenvalue of thecorresponding homogeneous equation to (23.19), the solution becomes infinite, as wewould expect.For the special case f(x) =(sinx)/x, the Fourier transform ˜f(k) is identical to that in(23.21), and the solution (23.23) becomesy(x) = sin x ( ) ∫ πλ 1 1√ πx + √ exp(ikx) dk1 − πλ 2π −1 2= sin x ( [ ] k=1πλ 1 exp(ikx)x + 1 − πλ)2 ixk=−1= sin x ( ) ( ) πλ sin x 1 sin xx + 1 − πλ x = 1 − πλ x. ◭If, instead, the integral equation (23.17) had integration limits 0 and x (somaking it a Volterra equation) then its solution could be found, in a similar way,by using the convolution theorem for Laplace transforms (see chapter 13). Wewould find¯f(s)ȳ(s) =1 − λ ¯K(s) ,where s is the Laplace transform variable. Often one may use the dictionary ofLaplace transforms given in table 13.1 to invert this equation and find the solutiony(x). In general, however, the evaluation of inverse Laplace transform integralsis difficult, since (in principle) it requires a contour integration; see chapter 24.As a final example of the use of Fourier transforms in solving integral equations,we mention equations that have integration limits −∞ and ∞ and a kernel of theformK(x, z) =exp(−ixz).Consider, for example, the inhomogeneous Fredholm equationy(x) =f(x)+λ∫ ∞−∞exp(−ixz) y(z) dz. (23.24)The integral over z is clearly just (a multiple of) the Fourier transform of y(z),810

23.4 CLOSED-FORM SOLUTIONSso we can writey(x) =f(x)+ √ 2πλỹ(x). (23.25)If we now take the Fourier transform of (23.25) but continue to denote theindependent variable by x (i.e. rather than k, for example), we obtainỹ(x) =˜f(x)+ √ 2πλy(−x). (23.26)Substituting (23.26) into (23.25) we findy(x) =f(x)+ √ √ ]2πλ[˜f(x)+ 2πλy(−x) ,but on making the change x →−x and substituting back in for y(−x), this givesy(x) =f(x)+ √ [ 2πλ˜f(x)+2πλ 2 f(−x)+ √ ]2πλ˜f(−x)+2πλ 2 y(x) .Thus the solution to (23.24) is given byy(x) =1[1 − (2π) 2 λ 4 f(x)+(2π) 1/2 λ˜f(x)+2πλ 2 f(−x)+(2π) 3/2 λ 3˜f(−x) ].(23.27)Clearly, (23.24) possesses a unique solution provided λ ≠ ±1/ √ 2π or ±i/ √ 2π;these are easily shown to be the eigenvalues of the corresponding homogeneousequation (for which f(x) ≡ 0).◮Solve the integral equation) ∫ ∞y(x) =exp(− x2+ λ exp(−ixz) y(z) dz, (23.28)2−∞where λ is a real constant. Show that the solution is unique unless λ has one of two particularvalues. Does a solution exist for either of these two values of λ?Following the argument given above, the solution to (23.28) is given by (23.27) withf(x) =exp(−x 2 /2). In order to write the solution explicitly, however, we must calculatethe Fourier transform of f(x). Using equation (13.7), we find ˜f(k) =exp(−k 2 /2), fromwhich we note that f(x) has the special property that its functional form is identical tothat of its Fourier transform. Thus, the solution to (23.28) is given byy(x) =11 − (2π) 2 λ 4 [1+(2π) 1/2 λ +2πλ 2 +(2π) 3/2 λ 3] exp(− x22).(23.29)Since λ is restricted to be real, the solution to (23.28) will be unique unless λ = ±1/ √ 2π,at which points (23.29) becomes infinite. In order to find whether solutions exist for eitherof these values of λ we must return to equations (23.25) and (23.26).Let us first consider the case λ =+1/ √ 2π. Putting this value into (23.25) and (23.26),we obtainy(x) =f(x)+ỹ(x), (23.30)ỹ(x) =˜f(x)+y(−x). (23.31)811

INTEGRAL EQUATIONSSubstituting (23.31) into (23.30) we findy(x) =f(x)+˜f(x)+y(−x),but on changing x to −x and substituting back in for y(−x), this givesy(x) =f(x)+˜f(x)+f(−x)+˜f(−x)+y(x).Thus, in order for a solution to exist, we require that the function f(x) obeysf(x)+˜f(x)+f(−x)+˜f(−x) =0.This is satisfied if f(x) =−˜f(x), i.e. if the functional form of f(x) is minus the form of itsFourier transform. We may repeat this analysis for the case λ = −1/ √ 2π, and, in a similarway, we find that this time we require f(x) =˜f(x).In our case f(x) = exp(−x 2 /2), for which, as we mentioned above, f(x) = ˜f(x).Therefore, (23.28) possesses no solution when λ =+1/ √ 2π but has many solutions whenλ = −1/ √ 2π. ◭A similar approach to the above may be taken to solve equations with kernelsof the form K(x, y) =cosxy or sin xy, either by considering the integral over y ineach case as the real or imaginary part of the corresponding Fourier transformor by using Fourier cosine or sine transforms directly.23.4.3 DifferentiationA closed-form solution to a Volterra equation may sometimes be obtained bydifferentiating the equation to obtain the corresponding differential equation,which may be easier to solve.◮Solve the integral equationy(x) =x −∫ x0xz 2 y(z) dz. (23.32)Dividing through by x, weobtain∫y(x)xx=1− z 2 y(z) dz,0which may be differentiated with respect to x to give[ ][ ]d y(x)y(x)= −x 2 y(x) =−x 3 .dx xxThis equation may be integrated straightforwardly, and we find[ ] y(x)ln = − x4x 4 + c,where c is a constant of integration. Thus the solution to (23.32) has the form)y(x) =Ax exp(− x4, (23.33)4where A is an arbitrary constant.Since the original integral equation (23.32) contains no arbitrary constants, neithershould its solution. We may calculate the value of the constant, A, by substituting thesolution (23.33) back into (23.32), from which we find A =1.◭812

23.5 NEUMANN SERIES23.5 Neumann seriesAs mentioned above, most integral equations met in practice will not be of thesimple forms discussed in the last section and so, in general, it is not possible tofind closed-form solutions. In such cases, we might try to obtain a solution in theform of an infinite series, as we did for differential equations (see chapter 16).Let us consider the equationy(x) =f(x)+λ∫ baK(x, z)y(z) dz, (23.34)where either both integration limits are constants (for a Fredholm equation) orthe upper limit is variable (for a Volterra equation). Clearly, if λ were small thena crude (but reasonable) approximation to the solution would bey(x) ≈ y 0 (x) =f(x),where y 0 (x) stands for our ‘zeroth-order’ approximation to the solution (and isnot to be confused with an eigenfunction).Substituting this crude guess under the integral sign in the original equation,we obtain what should be a better approximation:y 1 (x) =f(x)+λ∫ baK(x, z)y 0 (z) dz = f(x)+λ∫ baK(x, z)f(z) dz,which is first order in λ. Repeating the procedure once more results in thesecond-order approximationy 2 (x) =f(x)+λ= f(x)+λ∫ ba∫ baK(x, z)y 1 (z) dz∫ b ∫ bK(x, z 1 )f(z 1 ) dz 1 + λ 2 dz 1 K(x, z 1 )K(z 1 ,z 2 )f(z 2 ) dz 2 .It is clear that we may continue this process to obtain progressively higher-orderapproximations to the solution. Introducing the functionsK 1 (x, z) =K(x, z),K 2 (x, z) =K 3 (x, z) =∫ ba∫ bK(x, z 1 )K(z 1 ,z) dz 1 ,and so on, which obey the recurrence relationaK n (x, z) =a∫ bdz 1 K(x, z 1 )K(z 1 ,z 2 )K(z 2 ,z) dz 2 ,a∫ baK(x, z 1 )K n−1 (z 1 ,z) dz 1 ,813a

INTEGRAL EQUATIONSwe may write the nth-order approximation asn∑∫ by n (x) =f(x)+ λ m K m (x, z)f(z) dz. (23.35)m=1The solution to the original integral equation is then given by y(x) =lim n→∞ y n (x), provided the infinite series converges. Using (23.35), this solutionmay be written asy(x) =f(x)+λ∫ bwhere the resolvent kernel R(x, z; λ) is given byR(x, z; λ) =aaR(x, z; λ)f(z) dz, (23.36)∞∑λ m K m+1 (x, z). (23.37)m=0Clearly, the resolvent kernel, and hence the series solution, will convergeprovided λ is sufficiently small. In fact, it may be shown that the series convergesin some domain of |λ| provided the original kernel K(x, z) is bounded in such away that∫ b ∫ b|λ| 2 dx |K(x, z)| 2 dz < 1. (23.38)aa◮Use the Neumann series method to solve the integral equationy(x) =x + λ∫ 10xzy(z) dz. (23.39)Following the method outlined above, we begin with the crude approximation y(x) ≈y 0 (x) =x. Substituting this under the integral sign in (23.39), we obtain the next approximationy 1 (x) =x + λ∫ 10xzy 0 (z) dz = x + λRepeating the procedure once more, we obtainy 2 (x) =x + λ= x + λ∫ 10∫ 10xzy 1 (z) dz∫ 1(xz z + λz )dz = x +30xz 2 dz = x + λx 3 ,( ) λ3 + λ2x.9For this simple example, it is easy to see that by continuing this process the solution to(23.39) is obtained as[ ( ) 2 ( ) 3λ λ λy(x) =x +3 + + + ···]x.3 3Clearly the expression in brackets is an infinite geometric series with first term λ/3 and814

23.6 FREDHOLM THEORYcommon ratio λ/3. Thus, provided |λ| < 3, this infinite series converges to the valueλ/(3 − λ), and the solution to (23.39) isy(x) =x +λx3 − λ = 3x3 − λ . (23.40)Finally, we note that the requirement that |λ| < 3 may also be derived very easily fromthe condition (23.38). ◭23.6 Fredholm theoryIn the previous section, we found that a solution to the integral equation (23.34)can be obtained as a Neumann series of the form (23.36), where the resolventkernel R(x, z; λ) is written as an infinite power series in λ. This solution is validprovided the infinite series converges.A related, but more elegant, approach to the solution of integral equationsusing infinite series was found by Fredholm. We will not reproduce Fredholm’sanalysis here, but merely state the results we need. Essentially, Fredholm theoryprovides a formula for the resolvent kernel R(x, z; λ) in (23.36) in terms of theratio of two infinite series:D(x, z; λ)R(x, z; λ) = . (23.41)d(λ)The numerator and denominator in (23.41) are given byD(x, z; λ) =d(λ) =∞∑n=0∞∑n=0(−1) nD n (x, z)λ n , (23.42)n!(−1) nd n λ n , (23.43)n!where the functions D n (x, z) and the constants d n are found from recurrencerelations as follows. We start withD 0 (x, z) =K(x, z) and d 0 =1, (23.44)where K(x, z) is the kernel of the original integral equation (23.34). The higherordercoefficients of λ in (23.43) and (23.42) are then obtained from the tworecurrence relationsd n =∫ bD n (x, z) =K(x, z)d n − naD n−1 (x, x) dx, (23.45)∫ baK(x, z 1 )D n−1 (z 1 ,z) dz 1 . (23.46)Although the formulae for the resolvent kernel appear complicated, they areoften simple to apply. Moreover, for the Fredholm solution the power series(23.42) and (23.43) are both guaranteed to converge for all values of λ, unlike815

INTEGRAL EQUATIONSNeumann series, which converge only if the condition (23.38) is satisfied. Thus theFredholm method leads to a unique, non-singular solution, provided that d(λ) ≠0.In fact, as we might suspect, the solutions of d(λ) = 0 give the eigenvalues of thehomogeneous equation corresponding to (23.34), i.e. with f(x) ≡ 0.◮Use Fredholm theory to solve the integral equation (23.39).Using (23.36) and (23.41), the solution to (23.39) can be written in the form∫ 1D(x, z; λ)y(x) =x + λ R(x, z; λ)zdz= x + λzdz. (23.47)00 d(λ)In order to find the form of the resolvent kernel R(x, z; λ), we begin by settingD 0 (x, z) =K(x, z) =xz and d 0 =1and use the recurrence relations (23.45) and (23.46) to obtaind 1 =∫ 10D 0 (x, x) dx =∫ 10∫ 1x 2 dx = 1 3 ,D 1 (x, z) = xz ∫ 13 − xz1zdz 2 1 = xz [ z303 − xz 1=0.3Applying the recurrence relations again we find that d n =0andD n (x, z) =0forn>1.Thus, from (23.42) and (23.43), the numerator and denominator of the resolvent respectivelyare given byD(x, z; λ) =xz and d(λ) =1− λ 3 .Substituting these expressions into (23.47), we find that the solution to (23.39) is givenby∫ 1xz 2y(x) =x + λ1 − λ/3 dz[= x + λ0x1 − λ/3z 33] 10] 10= x + λx3 − λ = 3x3 − λ ,which, as expected, is the same as the solution (23.40) found by constructing a Neumannseries. ◭23.7 Schmidt–Hilbert theoryThe Schmidt–Hilbert (SH) theory of integral equations may be considered asanalogous to the Sturm–Liouville (SL) theory of differential equations, discussedin chapter 17, and is concerned with the properties of integral equations withHermitian kernels. An Hermitian kernel enjoys the propertyK(x, z) =K ∗ (z,x), (23.48)and it is clear that a special case of (23.48) occurs for a real kernel that is alsosymmetric with respect to its two arguments.816

23.7 SCHMIDT–HILBERT THEORYLet us begin by considering the homogeneous integral equationy = λKy,where the integral operator K has an Hermitian kernel. As discussed in section23.3, in general this equation will have solutions only for λ = λ i ,wheretheλ iare the eigenvalues of the integral equation, the corresponding solutions y i beingthe eigenfunctions of the equation.By following similar arguments to those presented in chapter 17 for SL theory,it may be shown that the eigenvalues λ i of an Hermitian kernel are real andthat the corresponding eigenfunctions y i belonging to different eigenvalues areorthogonal and form a complete set. If the eigenfunctions are suitably normalised,we have〈y i |y j 〉 =∫ bay ∗ i (x)y j(x) dx = δ ij . (23.49)If an eigenvalue is degenerate then the eigenfunctions corresponding to thateigenvalue can be made orthogonal by the Gram–Schmidt procedure, in a similarway to that discussed in chapter 17 in the context of SL theory.Like SL theory, SH theory does not provide a method of obtaining the eigenvaluesand eigenfunctions of any particular homogeneous integral equation withan Hermitian kernel; for this we have to turn to the methods discussed in theprevious sections of this chapter. Rather, SH theory is concerned with the generalproperties of the solutions to such equations. Where SH theory becomesapplicable, however, is in the solution of inhomogeneous integral equations withHermitian kernels for which the eigenvalues and eigenfunctions of the correspondinghomogeneous equation are already known.Let us consider the inhomogeneous equationy = f + λKy, (23.50)where K = K † and for which we know the eigenvalues λ i and normalisedeigenfunctions y i of the corresponding homogeneous problem. The function fmay or may not be expressible solely in terms of the eigenfunctions y i ,andtoaccommodate this situation we write the unknown solution y as y = f + ∑ i a iy i ,where the a i are expansion coefficients to be determined.Substituting this into (23.50), we obtainf + ∑ ia i y i = f + λ ∑ ia i y iλ i+ λKf, (23.51)wherewehaveusedthefactthaty i = λ i Ky i . Forming the inner product of both817

INTEGRAL EQUATIONSsides of (23.51) with y j , we find∑a i 〈y j |y i 〉 = λ ∑ a i〈y j |y i 〉 + λ〈y j |Kf〉. (23.52)λii iSince the eigenfunctions are orthonormal and K is an Hermitian operator,we have that both 〈y j |y i 〉 = δ ij and 〈y j |Kf〉 = 〈Ky j |f〉 = λ −1j 〈y j |f〉. Thus thecoefficients a j are given bya j = λλ−1 j 〈y j |f〉1 − λλ −1 = λ〈y j|f〉jλ j − λ , (23.53)and the solution isy = f + ∑ a i y i = f + λ ∑ 〈y i |f〉λii i − λ y i. (23.54)This also shows, incidentally, that a formal representation for the resolvent kernelisR(x, z; λ) = ∑ iy i (x)y ∗ i (z)λ i − λ . (23.55)If f can be expressed as a linear superposition of the y i ,i.e.f = ∑ i b iy i ,thenb i = 〈y i |f〉 and the solution can be written more briefly asy = ∑ 1 − λλ −1 y i . (23.56)iiWe see from (23.54) that the inhomogeneous equation (23.50) has a uniquesolution provided λ ≠ λ i ,i.e.whenλ is not equal to one of the eigenvalues ofthe corresponding homogeneous equation. However, if λ does equal one of theeigenvalues λ j then, in general, the coefficients a j become singular and no (finite)solution exists.Returning to (23.53), we notice that even if λ = λ j a non-singular solution tothe integral equation is still possible provided that the function f is orthogonalto every eigenfunction corresponding to the eigenvalue λ j ,i.e.〈y j |f〉 =∫ bab iy ∗ j (x)f(x) dx =0.The following worked example illustrates the case in which f can be expressed interms of the y i . One in which it cannot is considered in exercise 23.14.◮Use Schmidt–Hilbert theory to solve the integral equationy(x) = sin(x + α)+λ∫ π0sin(x + z)y(z) dz. (23.57)It is clear that the kernel K(x, z) =sin(x + z) is real and symmetric in x and z and is818

23.8 EXERCISESthus Hermitian. In order to solve this inhomogeneous equation using SH theory, however,we must first find the eigenvalues and eigenfunctions of the corresponding homogeneousequation.In fact, we have considered the solution of the corresponding homogeneous equation(23.13) already, in subsection 23.4.1, where we found that it has two eigenvalues λ 1 =2/πand λ 2 = −2/π, with eigenfunctions given by (23.16). The normalised eigenfunctions arey 1 (x) = √ 1 (sin x +cosx) and y 2 (x) = √ 1 (sin x − cos x) (23.58)π πand are easily shown to obey the orthonormality condition (23.49).Using (23.54), the solution to the inhomogeneous equation (23.57) has the formy(x) =a 1 y 1 (x)+a 2 y 2 (x), (23.59)where the coefficients a 1 and a 2 are given by (23.53) with f(x) =sin(x + α). Therefore,using (23.58),1a 1 =1 − πλ/21a 2 =1+πλ/2∫ π0∫ π01√ π(sin z +cosz) sin(z + α) dz =1√ π(sin z − cos z) sin(z + α) dz =√ π(cos α +sinα),2 − πλ√ π(cos α − sin α).2+πλSubstituting these expressions for a 1 and a 2 into (23.59) and simplifying, we find thatthe solution to (23.57) is given byy(x) =11 − (πλ/2) 2 [sin(x + α)+(πλ/2) cos(x − α)]. ◭23.1 Solve the integral equation∫ ∞023.8 Exercisescos(xv)y(v) dv =exp(−x 2 /2)for the function y = y(x) forx>0. Note that for x

INTEGRAL EQUATIONS23.5 Solve for φ(x) the integral equationφ(x) =f(x)+λ∫ 10[( xy) n ( y) ] n+ φ(y) dy,xwhere f(x) is bounded for 0

23.8 EXERCISES(b) Obtain the eigenvalues and eigenfunctions over the interval [0, 2π] if∞∑ 1K(x, y) = cos nx cos ny.nn=123.8 By taking its Laplace transform, and that of x n e −ax , obtain the explicit solutionof∫ x]f(x) =e[x −x + (x − u)e u f(u) du .Verify your answer by substitution.23.9 For f(t) =exp(−t 2 /2), use the relationships of the Fourier transforms of f ′ (t) andtf(t) tothatoff(t) itself to find a simple differential equation satisfied by ˜f(ω),the Fourier transform of f(t), and hence determine ˜f(ω) to within a constant.Use this result to solve the integral equationfor h(t).23.10 Show that the equation∫ ∞−∞f(x) =x −1/3 + λ0e −t(t−2x)/2 h(t) dt = e 3x2 /8∫ ∞0f(y)exp(−xy) dyhas a solution of the form Ax α + Bx β . Determine the values of α and β, andshowthat those of A and B are11 − λ 2 Γ( 1 3 )Γ( 2 3 ) andλΓ( 2 3 )1 − λ 2 Γ( 1 3 )Γ( 2 3 ) ,where Γ(z) is the gamma function.23.11 At an international ‘peace’ conference a large number of delegates are seatedaround a circular table with each delegation sitting near its allies and diametricallyopposite the delegation most bitterly opposed to it. The position of a delegate isdenoted by θ, with0≤ θ ≤ 2π. Thefuryf(θ) felt by the delegate at θ is the sumof his own natural hostility h(θ) and the influences on him of each of the otherdelegates; a delegate at position φ contributes an amount K(θ − φ)f(φ). Thusf(θ) =h(θ)+∫ 2π0K(θ − φ)f(φ) dφ.Show that if K(ψ) takestheformK(ψ) =k 0 + k 1 cos ψ thenf(θ) =h(θ)+p + q cos θ + r sin θand evaluate p, q and r. A positive value for k 1 implies that delegates tend toplacate their opponents but upset their allies, whilst negative values imply thatthey calm their allies but infuriate their opponents. A walkout will occur if f(θ)exceeds a certain threshold value for some θ. Isthismorelikelytohappenforpositive or for negative values of k 1 ?23.12 By considering functions of the form h(x) = ∫ x(x − y)f(y) dy, show that the0solution f(x) oftheintegralequationf(x) =x + 1 2satisfies the equation f ′′ (x) =f(x).821∫ 10|x − y|f(y) dy

INTEGRAL EQUATIONSBy examining the special cases x =0andx = 1, show that2f(x) =(e +3)(e +1) [(e +2)ex − ee −x ].23.13 The operator M is defined byMf(x) ≡∫ ∞−∞K(x, y)f(y) dy,where K(x, y) = 1 inside the square |x|

23.9 HINTS AND ANSWERS23.9 Hints and answers23.1 Define y(−x) =y(x) and use the cosine Fourier transform inversion theorem;y(x) =(2/π) 1/2 exp(−x 2 /2).23.3 f ′′ (x) − f(x) =expx; α =3/4, β =1/2, γ =1/4.23.5 (a) φ(x) =f(x) − (1+2n)F n x n − (1 − 2n)F −n x −n . (b) There are no solutions forλ =[1± (1 − 4n 2 ) −1/2 ] −1 unless F ±n =0orF n /F −n = ∓[(1 − 2n)/(1 + 2n)] 1/2 .23.7 (a) a (i)n= ∫ ba h n(x)ψ (i) (x) dx; (b) use (1/ √ π)cosnx and (1/ √ π)sinnx; M is diagonal;eigenvalues λ k = k/π with eigenfunctions ψ (k) (x) =(1/ √ π)coskx.23.9 d˜f/dω = −ω˜f, leading to ˜f(ω) =Ae −ω2 /2 . Rearrange the integral as a convolutionand deduce that ˜h(ω) = Be −3ω2 /2 ; h(t) = Ce −t2 /6 , where resubstitution andGaussian normalisation show that C = √ 2/(3π).23.11 p = k 0 H/(1 − 2πk 0 ), q = k 1 H c /(1 − πk 1 )andr = k 1 H s /(1 − πk 1 ),where H = ∫ 2πh(z) dz, H0 c = ∫ 2πh(z)coszdz,andH0 s = ∫ 2πh(z)sinzdz. Positive0values of k 1 (≈ π −1 ) are most likely to cause a conference breakdown.23.13 For eigenvalue 0 : f(x) =0for|x|

24Complex variablesThroughout this book references have been made to results derived from the theoryof complex variables. This theory thus becomes an integral part of the mathematicsappropriate to physical applications. Indeed, so numerous and widespreadare these applications that the whole of the next chapter is devoted to a systematicpresentation of some of the more important ones. This current chapter developsthe general theory on which these applications are based. The difficulty with it,from the point of view of a book such as the present one, is that the underlyingbasis has a distinctly pure mathematics flavour.Thus, to adopt a comprehensive rigorous approach would involve a largeamount of groundwork in analysis, for example formulating precise definitionsof continuity and differentiability, developing the theory of sets and making adetailed study of boundedness. Instead, we will be selective and pursue only thoseparts of the formal theory that are needed to establish the results used in the nextchapter and elsewhere in this book.In this spirit, the proofs that have been adopted for some of the standardresults of complex variable theory have been chosen with an eye to simplicityrather than sophistication. This means that in some cases the imposed conditionsare more stringent than would be strictly necessary if more sophisticated proofswere used; where this happens the less restrictive results are usually stated aswell. The reader who is interested in a fuller treatment should consult one of themany excellent textbooks on this fascinating subject. §One further concession to ‘hand-waving’ has been made in the interests ofkeeping the treatment to a moderate length. In several places phrases such as ‘canbe made as small as we like’ are used, rather than a careful treatment in termsof ‘given ɛ>0, there exists a δ>0 such that’. In the authors’ experience, some§ For example, K. Knopp, Theory of Functions, Part I (New York: Dover, 1945); E. G. Phillips,Functions of a Complex Variable with Applications 7th edn (Edinburgh: Oliver and Boyd, 1951); E.C. Titchmarsh, The Theory of Functions (Oxford: Oxford University Press, 1952).824

24.1 FUNCTIONS OF A COMPLEX VARIABLEstudents are more at ease with the former type of statement, despite its lack ofprecision, whilst others, those who would contemplate only the latter, are usuallywell able to supply it for themselves.24.1 Functions of a complex variableThe quantity f(z) is said to be a function of the complex variable z if to every valueof z in a certain domain R (a region of the Argand diagram) there correspondsone or more values of f(z). Stated like this f(z) could be any function consistingof a real and an imaginary part, each of which is, in general, itself a function of xand y. If we denote the real and imaginary parts of f(z) byu and v, respectively,thenf(z) =u(x, y)+iv(x, y).In this chapter, however, we will be primarily concerned with functions that aresingle-valued, so that to each value of z there corresponds just one value of f(z),and are differentiable in a particular sense, which we now discuss.A function f(z) that is single-valued in some domain R is differentiable at thepoint z in R if the derivative[ ]f(z +∆z) − f(z)f ′ (z) = lim(24.1)∆z→0 ∆zexists and is unique, in that its value does not depend upon the direction in theArgand diagram from which ∆z tends to zero.◮Show that the function f(z) =x 2 − y 2 + i2xy is differentiable for all values of z.Considering the definition (24.1), and taking ∆z =∆x + i∆y, we havef(z +∆z) − f(z)∆z= (x +∆x)2 − (y +∆y) 2 +2i(x +∆x)(y +∆y) − x 2 + y 2 − 2ixy∆x + i∆y= 2x∆x +(∆x)2 − 2y∆y − (∆y) 2 +2i(x∆y + y∆x +∆x∆y)∆x + i∆y=2x + i2y + (∆x)2 − (∆y) 2 +2i∆x∆y.∆x + i∆yNow, in whatever way ∆x and ∆y are allowed to tend to zero (e.g. taking ∆y =0andletting ∆x → 0 or vice versa), the last term on the RHS will tend to zero and the uniquelimit 2x + i2y will be obtained. Since z was arbitrary, f(z) withu = x 2 − y 2 and v =2xyis differentiable at all points in the (finite) complex plane. ◭We note that the above working can be considerably reduced by recognisingthat, since z = x + iy, we can write f(z) asf(z) =x 2 − y 2 +2ixy =(x + iy) 2 = z 2 .825

COMPLEX VARIABLESWe then find that[ (z +∆z)f ′ 2 − z 2 ] [ (∆z) 2 ]+2z∆z(z) = lim= lim∆z→0 ∆z∆z→0 ∆z( )= lim ∆z +2z =2z,∆z→0from which we see immediately that the limit both exists and is independent ofthe way in which ∆z → 0. Thus we have verified that f(z) =z 2 is differentiablefor all (finite) z. We also note that the derivative is analogous to that found forreal variables.Although the definition of a differentiable function clearly includes a wideclass of functions, the concept of differentiability is restrictive and, indeed, somefunctions are not differentiable at any point in the complex plane.◮Show that the function f(z) =2y +ix is not differentiable anywhere in the complex plane.In this case f(z) cannot be written simply in terms of z, and so we must consider thelimit (24.1) in terms of x and y explicitly. Following the same procedure as in the previousexample we findf(z +∆z) − f(z)∆z2y +2∆y + ix + i∆x − 2y − ix=∆x + i∆y2∆y + i∆x=∆x + i∆y .In this case the limit will clearly depend on the direction from which ∆z → 0. Suppose∆z → 0 along a line through z of slope m, sothat∆y = m∆x, thenlim∆z→0[ f(z +∆z) − f(z)∆z] [ ]2∆y + i∆x= lim= 2m + i∆x, ∆y→0 ∆x + i∆y 1+im .This limit is dependent on m and hence on the direction from which ∆z → 0. Since thisconclusion is independent of the value of z, and hence true for all z, f(z) =2y + ix isnowhere differentiable. ◭A function that is single-valued and differentiable at all points of a domain Ris said to be analytic (or regular) inR. A function may be analytic in a domainexcept at a finite number of points (or an infinite number if the domain isinfinite); in this case it is said to be analytic except at these points, which arecalled the singularities of f(z). In our treatment we will not consider cases inwhich an infinite number of singularities occur in a finite domain.826

24.2 THE CAUCHY–RIEMANN RELATIONS◮Show that the function f(z) =1/(1 − z) is analytic everywhere except at z =1.Since f(z) is given explicitly as a function of z, evaluation of the limit (24.1) is somewhateasier. We find[ ]f(z +∆z) − f(z)f ′ (z) = lim∆z→0 ∆z[ ( 1 1= lim∆z→0 ∆z 1 − z − ∆z − 1 )]1 − z= lim∆z→0[1(1 − z − ∆z)(1 − z)]=1(1 − z) 2 ,independently of the way in which ∆z → 0, provided z ≠ 1. Hence f(z) is analyticeverywhere except at the singularity z =1.◭24.2 The Cauchy–Riemann relationsFrom examining the previous examples, it is apparent that for a function f(z)to be differentiable and hence analytic there must be some particular connectionbetween its real and imaginary parts u and v.By considering a general function we next establish what this connection mustbe. If the limit[ ]f(z +∆z) − f(z)L = lim(24.2)∆z→0 ∆zis to exist and be unique, in the way required for differentiability, then any twospecific ways of letting ∆z → 0 must produce the same limit. In particular, movingparallel to the real axis and moving parallel to the imaginary axis must do so.This is certainly a necessary condition, although it may not be sufficient.If we let f(z) =u(x, y)+iv(x, y) and∆z =∆x + i∆y then we havef(z +∆z) =u(x +∆x, y +∆y)+iv(x +∆x, y +∆y),and the limit (24.2) is given by[ u(x +∆x, y +∆y)+iv(x +∆x, y +∆y) − u(x, y) − iv(x, y)L = lim∆x, ∆y→0∆x + i∆yIf we first suppose that ∆z is purely real, so that ∆y = 0, we obtain[ u(x +∆x, y) − u(x, y)L = lim+ i∆x→0 ∆xv(x +∆x, y) − v(x, y)∆x].]= ∂u∂x + i ∂v∂x , (24.3)provided each limit exists at the point z. Similarly, if ∆z is taken as purelyimaginary, so that ∆x = 0, we find[ ]u(x, y +∆y) − u(x, y) v(x, y +∆y) − v(x, y)L = lim+ i = 1 ∂u∆y→0 i∆yi∆yi ∂y + ∂v∂y . (24.4)827

COMPLEX VARIABLESFor f to be differentiable at the point z, expressions (24.3) and (24.4) mustbe identical. It follows from equating real and imaginary parts that necessaryconditions for this are∂u∂x = ∂v∂yand∂v∂x = − ∂u∂y . (24.5)These two equations are known as the Cauchy–Riemann relations.We can now see why for the earlier examples (i) f(z) =x 2 − y 2 + i2xy mightbe differentiable and (ii) f(z) =2y + ix could not be.(i) u = x 2 − y 2 , v =2xy:∂u ∂v=2x = and∂x ∂y(ii) u =2y, v = x:∂u ∂v=0= but∂x ∂y∂v ∂u=2y = −∂x ∂y ,∂v∂u=1≠ −2 =−∂x ∂y .It is apparent that for f(z) to be analytic something more than the existenceof the partial derivatives of u and v with respect to x and y is required; thissomething is that they satisfy the Cauchy–Riemann relations.We may enquire also as to the sufficient conditions for f(z) to be analytic inR. It can be shown § that a sufficient condition is that the four partial derivativesexist, are continuous and satisfy the Cauchy–Riemann relations. It is the additionalrequirement of continuity that makes the difference between the necessaryconditions and the sufficient conditions.◮In which domain(s) of the complex plane is f(z) =|x|−i|y| an analytic function?Writing f = u + iv it is clear that both ∂u/∂y and ∂v/∂x arezeroinallfourquadrantsand hence that the second Cauchy–Riemann relation in (24.5) is satisfied everywhere.Turning to the first Cauchy–Riemann relation, in the first quadrant (x >0, y>0) wehave f(z) =x − iy so that∂u∂x =1,∂v∂y = −1,which clearly violates the first relation in (24.5). Thus f(z) is not analytic in the firstquadrant.Following a similiar argument for the other quadrants, we find∂u= −1∂xor +1 for x0, respectively,∂v= −1∂yor +1 for y>0andy

24.2 THE CAUCHY–RIEMANN RELATIONSSince x and y are related to z and its complex conjugate z ∗ byx = 1 2 (z + z∗ ) and y = 1 2i (z − z∗ ), (24.6)we may formally regard any function f = u + iv as a function of z and z ∗ , ratherthan x and y. If we do this and examine ∂f/∂z ∗ we obtain∂f∂z ∗ = ∂f ∂x∂x ∂z ∗ + ∂f ∂y∂y ∂z ∗( ∂u=∂x ∂x)( + i ∂v ) 12= 1 ( ∂u2 ∂x − ∂v )+ i ∂y 2( ∂u+∂y + i ∂v∂y( ∂v∂x + ∂u∂y)(− 1 )2i). (24.7)Now, if f is analytic then the Cauchy–Riemann relations (24.5) must be satisfied,and these immediately give that ∂f/∂z ∗ is identically zero. Thus we conclude thatif f is analytic then f cannot be a function of z ∗ and any expression representingan analytic function of z can contain x and y only in the combination x + iy, notin the combination x − iy.We conclude this section by discussing some properties of analytic functionsthat are of great practical importance in theoretical physics. These can be obtainedsimply from the requirement that the Cauchy–Riemann relations must be satisfiedby the real and imaginary parts of an analytic function.The most important of these results can be obtained by differentiating thefirst Cauchy–Riemann relation with respect to one independent variable, and thesecond with respect to the other independent variable, to obtain the two chainsof equalities(∂ ∂u∂x ∂x(∂ ∂v∂x ∂x)= ∂∂x)= − ∂∂x( ∂v∂y)= ∂∂y( ∂u∂y)( ∂v∂x= − ∂∂y)= − ∂∂y( ) ∂u∂x( ∂u∂y= − ∂∂y),( ∂v∂yThus both u and v are separately solutions of Laplace’s equation in two dimensions,i.e.∂ 2 u∂x 2 + ∂2 u∂y 2 = 0 and ∂ 2 v∂x 2 + ∂2 v=0. (24.8)∂y2 We will make significant use of this result in the next chapter.A further useful result concerns the two families of curves u(x, y) = constantand v(x, y) = constant, where u and v are the real and imaginary parts of anyanalytic function f = u + iv. As discussed in chapter 10, the vector normal to thecurve u(x, y) = constant is given by∇u = ∂u∂x i + ∂u j, (24.9)∂y829).

COMPLEX VARIABLESwhere i and j are the unit vectors along the x- andy-axes, respectively. A similarexpression exists for ∇v, the normal to the curve v(x, y) = constant. Taking thescalar product of these two normal vectors, we obtain∇u · ∇v = ∂u ∂v∂x ∂x + ∂u ∂v∂y ∂y= − ∂u ∂u∂x ∂y + ∂u ∂u∂y ∂x =0,where in the last line we have used the Cauchy–Riemann relations to rewritethe partial derivatives of v as partial derivatives of u. Since the scalar productof the normal vectors is zero, they must be orthogonal, and the curves u(x, y) =constant and v(x, y) = constant must therefore intersect at right angles.◮Use the Cauchy–Riemann relations to show that, for any analytic function f = u + iv, therelation |∇u| = |∇v| must hold.From (24.9) we have( ) 2 ( ) 2 ∂u ∂u|∇u| 2 = ∇u · ∇u = + .∂x ∂yUsing the Cauchy–Riemann relations to write the partial derivatives of u in terms of thoseof v, weobtain( ) 2 ( ) 2 ∂v ∂v|∇u| 2 = + = |∇v| 2 ,∂y ∂xfrom which the result |∇u| = |∇v| follows immediately. ◭24.3 Power series in a complex variableThe theory of power series in a real variable was considered in chapter 4, whichalso contained a brief discussion of the natural extension of this theory to a seriessuch as∞∑f(z) = a n z n , (24.10)n=0where z is a complex variable and the a n are, in general, complex. We nowconsider complex power series in more detail.Expression (24.10) is a power series about the origin and may be used forgeneral discussion, since a power series about any other point z 0 can be obtainedby a change of variable from z to z − z 0 .Ifz were written in its modulus andargument form, z = r exp iθ, expression (24.10) would becomef(z) =∞∑a n r n exp(inθ). (24.11)n=0830

24.3 POWER SERIES IN A COMPLEX VARIABLEThis series is absolutely convergent if∞∑|a n |r n , (24.12)n=0which is a series of positive real terms, is convergent. Thus tests for the absoluteconvergence of real series can be used in the present context, and of these themost appropriate form is based on the Cauchy root test. With the radius ofconvergence R defined by1R = lim |a n| 1/n , (24.13)n→∞the series (24.10) is absolutely convergent if |z| R.If |z| = R then no particular conclusion may be drawn, and this case must beconsidered separately, as discussed in subsection 4.5.1.A circle of radius R centred on the origin is called the circle of convergenceof the series ∑ a n z n .ThecasesR = 0 and R = ∞ correspond, respectively, toconvergence at the origin only and convergence everywhere. For R finite theconvergence occurs in a restricted part of the z-plane (the Argand diagram). Fora power series about a general point z 0 , the circle of convergence is, of course,centred on that point.◮Find the parts of the z-plane for which the following series are convergent:∞∑ z n∞(i)n! , (ii) ∑∞∑n!z n z n, (iii)n .n=0n=0n=1(i) Since (n!) 1/n behaves like n as n →∞we find lim(1/n!) 1/n =0.HenceR = ∞ and theseries is convergent for all z. (ii) Correspondingly, lim(n!) 1/n = ∞. Thus R =0andtheseries converges only at z = 0. (iii) As n →∞,(n) 1/n has a lower limit of 1 and hencelim(1/n) 1/n =1/1 = 1. Thus the series is absolutely convergent if the condition |z| < 1issatisfied. ◭Case (iii) in the above example provides a good illustration of the fact thaton its circle of convergence a power series may or may not converge. For thisparticular series, the circle of convergence is |z| = 1, so let us consider theconvergence of the series at two different points on this circle. Taking z =1,theseries becomes∞∑ 1n =1+1 2 + 1 3 + 1 4 + ··· ,n=1which is easily shown to diverge (by, for example, grouping terms, as discussed insubsection 4.3.2). Taking z = −1, however, the series is given by∞∑ (−1) n= −1+ 1 n2 − 1 3 + 1 4 − ··· ,n=1831

COMPLEX VARIABLESwhich is an alternating series whose terms decrease in magnitude and whichtherefore converges.The ratio test discussed in subsection 4.3.2 may also be employed to investigatethe absolute convergence of a complex power series. A series is absolutelyconvergent if|a n+1 ||z| n+1 |a n+1 ||z|limn→∞ |a n ||z| n = lim < 1 (24.14)n→∞ |a n |and hence the radius of convergence R of the series is given by1R = limn→∞|a n+1 |.|a n |For instance, in case (i) of the previous example, we have1R = lim n!n→∞ (n +1)! = lim 1n→∞ n +1 =0.Thus the series is absolutely convergent for all (finite) z, confirming the previousresult.Before turning to particular power series, we conclude this section by statingthe important result § that the power series ∑ ∞0 a nz n has a sum that is an analyticfunction of z inside its circle of convergence.As a corollary to the above theorem, it may further be shown that if f(z) =∑an z n then, inside the circle of convergence of the series,f ′ (z) =∞∑na n z n−1 .n=0Repeated application of this result demonstrates that any power series can bedifferentiated any number of times inside its circle of convergence.24.4 Some elementary functionsIn the example at the end of the previous section it was shown that the functionexp z defined by∞∑ z nexp z =(24.15)n!n=0is convergent for all z of finite modulus and is thus, by the discussion ofthe previous section, an analytic function over the whole z-plane. Like its§ For a proof see, for example, K. F. Riley, Mathematical Methods for the Physical Sciences (Cambridge:Cambridge University Press, 1974), p. 446. Functions that are analytic in the whole z-plane are usually called integral or entire functions.832

24.4 SOME ELEMENTARY FUNCTIONSreal-variable counterpart it is called the exponential function; also like its realcounterpart it is equal to its own derivative.The multiplication of two exponential functions results in a further exponentialfunction, in accordance with the corresponding result for real variables.◮Show that exp z 1 exp z 2 =exp(z 1 + z 2 ).From the series expansion (24.15) of exp z 1 and a similar expansion for exp z 2 ,itisclearthat the coefficient of z1 rzs 2 in the corresponding series expansion of exp z 1 exp z 2 is simply1/(r!s!).But, from (24.15) we also have∞∑ (z 1 + z 2 ) nexp(z 1 + z 2 )=.n!n=0In order to find the coefficient of z1 rzs 2 in this expansion, we clearly have to consider theterm in which n = r + s, namely(z 1 + z 2 ) r+s 1 (=r+s)C 0 z r+s1+ ···+ r+s C s z r(r + s)! (r + s)!1z2 s + ···+ r+s C r+s z r+s2 .The coefficient of z1 rzs 2 in this is given byr+s 1 (r + s)! 1C s =(r + s)! s!r! (r + s)! = 1r!s! .Thus, since the corresponding coefficients on the two sides are equal, and all the seriesinvolved are absolutely convergent for all z, we can conclude that exp z 1 exp z 2 =exp(z 1 +z 2 ). ◭As an extension of (24.15) we may also define the complex exponent of a realnumber a>0 by the equationa z = exp(z ln a), (24.16)where ln a is the natural logarithm of a. The particular case a = e and the factthat ln e = 1 enable us to write exp z interchangeably with e z .Ifz is real then thedefinition agrees with the familiar one.The result for z = iy,exp iy =cosy + i sin y, (24.17)has been met already in equation (3.23). Its immediate extension isexp z =(expx)(cos y + i sin y). (24.18)As z varies over the complex plane, the modulus of exp z takes all real positivevalues, except that of 0. However, two values of z that differ by 2πki, for anyinteger k, produce the same value of exp z, as given by (24.18), and so exp z isperiodic with period 2πi. If we denote exp z by t, then the strip −π

COMPLEX VARIABLESderived from them (e.g. tan and tanh), the identities they satisfy and theirderivative properties are also just as for real variables. In view of this we will notgive them further attention here.The inverse function of exp z is given by w, the solution ofexp w = z. (24.19)This inverse function was discussed in chapter 3, but we mention it again herefor completeness. By virtue of the discussion following (24.18), w is not uniquelydefined and is indeterminate to the extent of any integer multiple of 2πi. Ifweexpress z asz = r exp iθ,where r is the (real) modulus of z and θ is its argument (−π

24.5 MULTIVALUED FUNCTIONS AND BRANCH CUTSOn the RHS let us write t as follows:t = r exp[i(θ +2kπ)],where k is an integer. We then obtain[ ]1t 1/n (θ +2kπ)=exp ln r + in n[ ]= r 1/n (θ +2kπ)exp i ,nwhere k =0, 1,...,n− 1; for other values of k we simply recover the roots already found.Thus t has n distinct nth roots. ◭24.5 Multivalued functions and branch cutsIn the definition of an analytic function, one of the conditions imposed wasthat the function is single-valued. However, as shown in the previous section, thelogarithmic function, a complex power and a complex root are all multivalued.Nevertheless, it happens that the properties of analytic functions can still beapplied to these and other multivalued functions of a complex variable providedthat suitable care is taken. This care amounts to identifying the branch points ofthe multivalued function f(z) in question. If z is varied in such a way that itspath in the Argand diagram forms a closed curve that encloses a branch point,then, in general, f(z) will not return to its original value.For definiteness let us consider the multivalued function f(z) =z 1/2 and expressz as z = r exp iθ. From figure 24.1(a), it is clear that, as the point z traverses anyclosed contour C that does not enclose the origin, θ will return to its originalvalue after one complete circuit. However, for any closed contour C ′ that doesenclose the origin, after one circuit θ → θ +2π (see figure 24.1(b)). Thus, for thefunction f(z) =z 1/2 , after one circuitr 1/2 exp(iθ/2) → r 1/2 exp[i(θ +2π)/2] = −r 1/2 exp(iθ/2).In other words, the value of f(z) changes around any closed loop enclosing theorigin; in this case f(z) →−f(z). Thus z = 0 is a branch point of the functionf(z) =z 1/2 .We note in this case that if any closed contour enclosing the origin is traversedtwice then f(z) =z 1/2 returns to its original value. The number of loops arounda branch point required for any given function f(z) to return to its original valuedepends on the function in question, and for some functions (e.g. Ln z, which alsohas a branch point at the origin) the original value is never recovered.In order that f(z) may be treated as single-valued, we may define a branch cutin the Argand diagram. A branch cut is a line (or curve) in the complex planeand may be regarded as an artificial barrier that we must not cross. Branch cutsare positioned in such a way that we are prevented from making a complete835

COMPLEX VARIABLESy Cy yrθxrθxxC ′(a) (b) (c)Figure 24.1 (a) A closed contour not enclosing the origin; (b) a closedcontour enclosing the origin; (c) a possible branch cut for f(z) =z 1/2 .circuit around any one branch point, and so the function in question remainssingle-valued.For the function f(z) =z 1/2 , we may take as a branch cut any curve startingat the origin z = 0 and extending out to |z| = ∞ in any direction, since allsuch curves would equally well prevent us from making a closed loop aroundthe branch point at the origin. It is usual, however, to take the cut along eitherthe real or the imaginary axis. For example, in figure 24.1(c), we take the cut asthe positive real axis. By agreeing not to cross this cut, we restrict θ to lie in therange 0 ≤ θ

24.6 SINGULARITIES AND ZEROS OF COMPLEX FUNCTIONSyzyyir 1θ 1r 2iixxx−iθ 2−i−i(a) (b) (c)Figure 24.2 (a) Coordinates used in the analysis of the branch points off(z) =(z 2 +1) 1/2 ; (b) one possible arrangement of branch cuts; (c) anotherpossible branch cut, which is finite.(iii) z = −i but not z = i, thenθ 1 → θ 1 , θ 2 → θ 2 +2π and so f(z) →−f(z);(iv) both branch points, then θ 1 → θ 1 +2π, θ 2 → θ 2 +2π and so f(z) → f(z).Thus, as expected, f(z) changes value around loops containing either z = i or z = −i (butnot both). We must therefore choose branch cuts that prevent us from making a completeloop around either branch point; one suitable choice is shown in figure 24.2(b).For this f(z), however, we have noted that after traversing a loop containing both branchpoints the function returns to its original value. Thus we may choose an alternative, finite,branch cut that allows this possibility but still prevents us from making a complete looparound just one of the points. A suitable cut is shown in figure 24.2(c). ◭24.6 Singularities and zeros of complex functionsA singular point of a complex function f(z) is any point in the Argand diagramat which f(z) fails to be analytic. We have already met one sort of singularity,the branch point, and in this section we will consider other types of singularityas well as discuss the zeros of complex functions.If f(z) has a singular point at z = z 0 but is analytic at all points in someneighbourhood containing z 0 but no other singularities, then z = z 0 is called anisolated singularity. (Clearly, branch points are not isolated singularities.)The most important type of isolated singularity is the pole. Iff(z) has the formf(z) =g(z)(z − z 0 ) n , (24.23)where n is a positive integer, g(z) is analytic at all points in some neighbourhoodcontaining z = z 0 and g(z 0 ) ≠0,thenf(z) has a pole of order n at z = z 0 .Analternative (though equivalent) definition is thatlim [(z − z 0 ) n f(z)] = a, (24.24)z→z 0837

COMPLEX VARIABLESwhere a is a finite, non-zero complex number. We note that if the above limit isequal to zero, then z = z 0 is a pole of order less than n, orf(z) is analytic there;if the limit is infinite then the pole is of an order greater than n. It may also beshown that if f(z) has a pole at z = z 0 ,then|f(z)| →∞as z → z 0 from anydirection in the Argand diagram. § If no finite value of n can be found such that(24.24) is satisfied, then z = z 0 is called an essential singularity.◮Find the singularities of the functions(i) f(z) = 11 − z − 1 , (ii) f(z) =tanhz.1+z(i) If we write f(z) asf(z) = 11 − z − 11+z = 2z(1 − z)(1 + z) ,we see immediately from either (24.23) or (24.24) that f(z) has poles of order 1 (or simplepoles) atz =1andz = −1.(ii) In this case we writef(z) =tanhz = sinh z exp z − exp(−z)=cosh z exp z +exp(−z) .Thus f(z) has a singularity when exp z = − exp(−z) or, equivalently, whenexp z =exp[i(2n +1)π]exp(−z),where n is any integer. Equating the arguments of the exponentials we find z =(n + 1 )πi, 2for integer n.Furthermore, using l’Hôpital’s rule (see chapter 4) we havelimz→(n+ 1 2 )πi{[z − (n +12 )πi]sinhzcosh z}= limz→(n+ 1 2 )πi{[z − (n +12Therefore, from (24.24), each singularity is a simple pole. ◭})πi]coshz +sinhz=1.sinh zAnother type of singularity exists at points for which the value of f(z) takesan indeterminate form such as 0/0 but lim z→z0 f(z) exists and is independentof the direction from which z 0 is approached. Such points are called removablesingularities.◮Show that f(z) =(sinz)/z has a removable singularity at z =0.It is clear that f(z) takes the indeterminate form 0/0 atz = 0. However, by expandingsin z as a power series in z, we findf(z) = 1 (z ···)− z3z 3! + z55! − =1− z23! + z45! − ··· .§ Although perhaps intuitively obvious, this result really requires formal demonstration by analysis.838

24.7 CONFORMAL TRANSFORMATIONSThus lim z→0 f(z) = 1 independently of the way in which z → 0, and so f(z) has a removablesingularity at z =0.◭An expression common in mathematics, but which we have so far avoidedusing explicitly in this chapter, is ‘z tends to infinity’. For a real variable suchas |z| or R, ‘tending to infinity’ has a reasonably well defined meaning. For acomplex variable needing a two-dimensional plane to represent it, the meaning isnot intrinsically well defined. However, it is convenient to have a unique meaningand this is provided by the following definition: the behaviour of f(z) at infinityis given by that of f(1/ξ) atξ =0,whereξ =1/z.◮Find the behaviour at infinity of (i) f(z) = a + bz −2 , (ii) f(z) = z(1 + z 2 ) and (iii)f(z) =expz.(i) f(z) =a + bz −2 : on putting z =1/ξ, f(1/ξ) =a + bξ 2 , which is analytic at ξ =0;thus f is analytic at z = ∞.(ii) f(z) =z(1 + z 2 ): f(1/ξ) =1/ξ +1/ξ 3 ; thus f has a pole of order 3 at z = ∞.(iii) f(z) =expz : f(1/ξ) = ∑ ∞0 (n!)−1 ξ −n ; thus f has an essential singularity at z = ∞. ◭We conclude this section by briefly mentioning the zeros of a complex function.As the name suggests, if f(z 0 )=0thenz = z 0 is called a zero of the functionf(z). Zeros are classified in a similar way to poles, in that iff(z) =(z − z 0 ) n g(z),where n is a positive integer and g(z 0 ) ≠0,thenz = z 0 is called a zero of ordern of f(z). If n =1thenz = z 0 is called a simple zero. It may further be shownthat if z = z 0 is a zero of order n of f(z) then it is also a pole of order n of thefunction 1/f(z).We will return in section 24.11 to the classification of zeros and poles in termsof their series expansions.24.7 Conformal transformationsWe now turn our attention to the subject of transformations, by which we meana change of coordinates from the complex variable z = x + iy to another, sayw = r + is, by means of a prescribed formula:w = g(z) =r(x, y)+is(x, y).Under such a transformation, or mapping, the Argand diagram for the z-variableis transformed into one for the w-variable, although the complete z-plane mightbe mapped onto only a part of the w-plane, or onto the whole of the w-plane, oronto some or all of the w-plane covered more than once.We shall consider only those mappings for which w and z are related by afunction w = g(z) and its inverse z = h(w) with both functions analytic, exceptpossibly at a few isolated points; such mappings are called conformal. Their839

COMPLEX VARIABLESyz 1z 2sC ′ 1w 1w 2C 1C 2z 0θ 2 θ 1xw = g(z)φ 2w 0φ 1rC ′ 2Figure 24.3 Two curves C 1 and C 2 in the z-plane, which are mapped ontoC 1 ′ and C′ 2 in the w-plane.important properties are that, except at points at which g ′ (z), and hence h ′ (z), iszero or infinite:(i) continuous lines in the z-plane transform into continuous lines in thew-plane;(ii) the angle between two intersecting curves in the z-plane equals the anglebetween the corresponding curves in the w-plane;(iii) the magnification, as between the z-plane and the w-plane, of a small lineelement in the neighbourhood of any particular point is independent ofthe direction of the element;(iv) any analytic function of z transforms to an analytic function of w andvice versa.Result (i) is immediate, and results (ii) and (iii) can be justified by the followingargument. Let two curves C 1 and C 2 pass through the point z 0 in the z-planeand let z 1 and z 2 be two points on their respective tangents at z 0 , each a distanceρ from z 0 . The same prescription with w replacing z describes the transformedsituation; however, the transformed tangents may not be straight lines and thedistances of w 1 and w 2 from w 0 have not yet been shown to be equal. Thissituation is illustrated in figure 24.3.In the z-plane z 1 and z 2 are given byz 1 − z 0 = ρ exp iθ 1 and z 2 − z 0 = ρ exp iθ 2 .The corresponding descriptions in the w-plane arew 1 − w 0 = ρ 1 exp iφ 1 and w 2 − w 0 = ρ 2 exp iφ 2 .The angles θ i and φ i are clear from figure 24.3. The transformed angles φ i arethose made with the r-axis by the tangents to the transformed curves at their840

24.7 CONFORMAL TRANSFORMATIONSpoint of intersection. Since any finite-length tangent may be curved, w i is morestrictly given by w i − w 0 = ρ i exp i(φ i + δφ i ), where δφ i → 0asρ i → 0, i.e. asρ → 0.Now since w = g(z), where g is analytic, we have( ) ( )w1 − w 0w2 − w 0lim= lim= dgz 1→z 0 z 1 − z 0 z2→z 0 z 2 − z 0 dz ∣ ,z=z0which may be written as{ρ1limρ→0 ρ exp[i(φ 1 + δφ 1 − θ 1 )]}{ }ρ2= limρ→0 ρ exp[i(φ 2 + δφ 2 − θ 2 )] = g ′ (z 0 ).(24.25)Comparing magnitudes and phases (i.e. arguments) in the equalities (24.25) givesthe stated results (ii) and (iii) and adds quantitative information to them, namelythat for small line elementsρ 1ρ ≈ ρ 2ρ ≈|g′ (z 0 )|, (24.26)φ 1 − θ 1 ≈ φ 2 − θ 2 ≈ arg g ′ (z 0 ). (24.27)For strict comparison with result (ii), (24.27) must be written as θ 1 −θ 2 = φ 1 −φ 2 ,with an ordinary equality sign, since the angles are only defined in the limitρ → 0 when (24.27) becomes a true identity. We also see from (24.26) thatthe linear magnification factor is |g ′ (z 0 )|; similarly, small areas are magnified by|g ′ (z 0 )| 2 .Since in the neighbourhoods of corresponding points in a transformation anglesare preserved and magnifications are independent of direction, it follows that smallplane figures are transformed into figures of the same shape, but, in general, onesthat are magnified and rotated (though not distorted). However, we also notethat at points where g ′ (z) = 0, the angle arg g ′ (z) through which line elements arerotated is undefined; these are called critical points of the transformation.The final result (iv) is perhaps the most important property of conformaltransformations. If f(z) is an analytic function of z and z = h(w) is also analytic,then F(w) =f(h(w)) is analytic in w. Its importance lies in the further conclusionsit allows us to draw from the fact that, since f is analytic, the real and imaginaryparts of f = φ + iψ are necessarily solutions of∂ 2 φ∂x 2 + ∂2 φ∂y 2 = 0 and ∂ 2 ψ∂x 2 + ∂2 ψ=0. (24.28)∂y2 Since the transformation property ensures that F =Φ+iΨ is also analytic, wecan conclude that its real and imaginary parts must themselves satisfy Laplace’sequation in the w-plane:∂ 2 Φ∂r 2+ ∂2 Φ∂s 2 = 0 and ∂ 2 Ψ∂r 2841+ ∂2 Ψ=0. (24.29)∂s2

COMPLEX VARIABLESysi Pw = g(z)Q R S TxR ′ P ′ Q ′T ′S ′rFigure 24.4 Transforming the upper half of the z-plane into the interior ofthe unit circle in the w-plane, in such a way that z = i is mapped onto w =0and the points x = ±∞ are mapped onto w =1.Further, suppose that (say) Re f(z) =φ is constant over a boundary C in thez-plane; then Re F(w) = Φ is constant over C in the z-plane. But this is the sameas saying that Re F(w) is constant over the boundary C ′ in the w-plane, C ′ beingthe curve into which C is transformed by the conformal transformation w = g(z).This result is exploited extensively in the next chapter to solve Laplace’s equationfor a variety of two-dimensional geometries.Examples of useful conformal transformations are numerous. For instance,w = z + b, w =(expiφ)z and w = az correspond, respectively, to a translation byb, a rotation through an angle φ and a stretching (or contraction) in the radialdirection (for a real). These three examples can be combined into the generallinear transformation w = az +b, where,ingeneral,a and b are complex. Anotherexample is the inversion mapping w =1/z, which maps the interior of the unitcircle to the exterior and vice versa. Other, more complicated, examples also exist.◮Show that if the point z 0 lies in the upper half of the z-plane then the transformationw =(expiφ) z − z 0z − z0∗maps the upper half of the z-plane into the interior of the unit circle in the w-plane. Hencefind a similar transformation that maps the point z = i onto w =0and the points x = ±∞onto w =1.Taking the modulus of w, we have|w| =∣ (exp iφ) z − z ∣ 0∣∣∣ z − z0∗ ∣ = z − z 0z − z0∗ ∣ .However, since the complex conjugate z0 ∗ is the reflection of z 0 in the real axis, if z and z 0both lie in the upper half of the z-plane then |z − z 0 |≤|z − z0 ∗ |; thus |w| ≤1, as required.We also note that (i) the equality holds only when z lies on the real axis, and so this axisis mapped onto the boundary of the unit circle in the w-plane; (ii) the point z 0 is mappedonto w = 0, the origin of the w-plane.By fixing the images of two points in the z-plane, the constants z 0 and φ can also befixed. Since we require the point z = i to be mapped onto w = 0, we have immediately842

24.7 CONFORMAL TRANSFORMATIONSysw 5w 1w 4w = g(z) φφ 41x 1 x 2 x 3 x 4 x 5xrφ 2 φ 3w 2w 3φ 5Figure 24.5 Transforming the upper half of the z-plane into the interiorof a polygon in the w-plane, in such a way that the points x 1 ,x 2 ,...,x n aremapped onto the vertices w 1 ,w 2 ,...,w n of the polygon with interior anglesφ 1 ,φ 2 ,...,φ n .z 0 = i. By further requiring z = ±∞ to be mapped onto w = 1, we find 1 = w =expiφand so φ = 0. The required transformation is thereforew = z − iz + i ,and is illustrated in figure 24.4. ◭We conclude this section by mentioning the rather curious Schwarz–Christoffeltransformation. § Suppose, as shown in figure 24.5, that we are interested in a(finite) number of points x 1 ,x 2 ,...,x n on the real axis in the z-plane. Then bymeans of the transformation{ ∫ z}w = A (ξ − x 1 ) (φ1/π)−1 (ξ − x 2 ) (φ2/π)−1 ···(ξ − x n ) (φn/π)−1 dξ + B, (24.30)0we may map the upper half of the z-plane onto the interior of a closed polygon inthe w-plane having n vertices w 1 ,w 2 ,...,w n (which are the images of x 1 ,x 2 ,...,x n )with corresponding interior angles φ 1 ,φ 2 ,...,φ n , as shown in figure 24.5. Thereal axis in the z-plane is transformed into the boundary of the polygon itself.The constants A and B are complex in general and determine the position,size and orientation of the polygon. It is clear from (24.30) that dw/dz =0atx = x 1 ,x 2 ,...,x n , and so the transformation is not conformal at these points.There are various subtleties associated with the use of the Schwarz–Christoffeltransformation. For example, if one of the points on the real axis in the z-plane(usually x n ) is taken at infinity, then the corresponding factor in (24.30) (i.e. theone involving x n ) is not present. In this case, the point(s) x = ±∞ are consideredas one point, since they transform to a single vertex of the polygon in the w-plane.§ Strictly speaking, the use of this transformation requires an understanding of complex integrals,which are discussed in section 24.8.843

COMPLEX VARIABLESysibw 3w = g(z)φ 3x 1 x 2−1 1xw 1φ 1 φ 2 w 2−aarFigure 24.6 Transforming the upper half of the z-plane into the interior ofa triangle in the w-plane.We can also map the upper half of the z-plane into an infinite open polygonby considering it as the limiting case of some closed polygon.◮Find a transformation that maps the upper half of the z-plane into the triangular regionshown in figure 24.6 in such a way that the points x 1 = −1 and x 2 = 1 are mappedinto the points w = −a and w = a, respectively, and the point x 3 = ±∞ is mapped intow = ib. Hence find a transformation that maps the upper half of the z-plane into the region−a

24.8 COMPLEX INTEGRALSyw 3 s w 3w = g(z)x1 x 2−1 1xφ 1 φ 2w 1 w 2−aarFigure 24.7 Transforming the upper half of the z-plane into the interior ofthe region −a

COMPLEX VARIABLESyBC 2C 1xC 3AFigure 24.8A and B.Some alternative paths for the integral of a function f(z) betweenThe question of when such an integral exists will not be pursued, except to statethat a sufficient condition is that dx/dt and dy/dt are continuous.◮Evaluate the complex integral of f(z) =z −1 along the circle |z| = R, starting and finishingat z = R.The path C 1 is parameterised as follows (figure 24.9(a)):z(t) =R cos t + iR sin t, 0 ≤ t ≤ 2π,whilst f(z) isgivenbyf(z) = 1x + iy = x − iyx 2 + y . 2Thus the real and imaginary parts of f(z) areu =xx 2 + y = R cos t and v = −y2 R 2 x 2 + y = − R sin t .2 R 2Hence, using expression (24.34),∫ ∫12π∫C 1z dz = cos t2π( − sin t0 R(−R sin t) dt − 0 R∫ 2π∫cos t2π( − sin t+ i0 RR cos tdt+ i 0 R=0+0+iπ + iπ =2πi. ◭)R cos tdt)(−R sin t) dt (24.35)With a bit of experience, the reader may be able to evaluate integrals likethe LHS of (24.35) directly without having to write them as four separate realintegrals. In the present case,∫C 1∫dz 2πz = −R sin t + iR cos tdt =0 R cos t + iR sin t846∫ 2π0idt=2πi. (24.36)

24.8 COMPLEX INTEGRALSyRtyyC 1 C 2Rt s =1t =0x −RR x −RR xC 3biRC3a(a) (b) (c)Figure 24.9details.Different paths for an integral of f(z) =z −1 . See the text forThis very important result will be used many times later, and the following shouldbe carefully noted: (i) its value, (ii) that this value is independent of R.In the above example the contour was closed, and so it began and ended atthe same point in the Argand diagram. We can evaluate complex integrals alongopen paths in a similar way.◮Evaluate the complex integral of f(z) =z −1 along the following paths (see figure 24.9):(i) the contour C 2 consisting of the semicircle |z| = R in the half-plane y ≥ 0,(ii) the contour C 3 made up of the two straight lines C 3a and C 3b .(i) This is just as in the previous example, except that now 0 ≤ t ≤ π. With this change,we have from (24.35) or (24.36) that∫dz= πi. (24.37)C 2z(ii) The straight lines that make up the countour C 3 may be parameterised as follows:C 3a , z =(1− t)R + itR for 0 ≤ t ≤ 1;C 3b , z = −sR + i(1 − s)R for 0 ≤ s ≤ 1.With these parameterisations the required integrals may be written∫ ∫dz 1∫C 3z = −R + iR1R + t(−R + iR) dt + −R − iRds. (24.38)iR + s(−R − iR)0If we could take over from real-variable theory that, for real t, ∫ (a+bt) −1 dt = b −1 ln(a+bt)even if a and b are complex, then these integrals could be evaluated immediately. However,to do this would be presuming to some extent what we wish to show, and so the evaluation8470

COMPLEX VARIABLESmust be made in terms of entirely real integrals. For example, the first is given by∫ 10∫−R + iR1R(1 − t)+itR dt =0∫ 1(−1+i)(1 − t − it)(1 − t) 2 + t 2 dt∫2t − 11=0 1 − 2t +2t dt + i 12 0 1 − 2t +2t dt 2= 1 [] 1[ (ln(1 − 2t +2t 2 ) + i t −12tan −1 212022[ π(2 − − π )]= πi2 2 .=0+ i 2The second integral on the RHS of (24.38) can also be shown to have the value πi/2. Thus∫dzC 3z = πi. ◭Considering the results of the preceding two examples, which have commonintegrands and limits, some interesting observations are possible. Firstly, the twointegrals from z = R to z = −R, along C 2 and C 3 , respectively, have the samevalue, even though the paths taken are different. It also follows that if we took aclosed path C 4 , given by C 2 from R to −R and C 3 traversed backwards from −Rto R, then the integral round C 4 of z −1 would be zero (both parts contributingequal and opposite amounts). This is to be compared with result (24.36), in whichclosed path C 1 , beginning and ending at the same place as C 4 , yields a value 2πi.It is not true, however, that the integrals along the paths C 2 and C 3 are equalfor any function f(z), or, indeed, that their values are independent of R in general.◮Evaluate the complex integral of f(z) =Rez along the paths C 1 , C 2 and C 3 shown infigure 24.9.(i) If we take f(z) =Rez and the contour C 1 then∫∫ 2πRe zdz= R cos t(−R sin t + iR cos t) dt = iπR 2 .C 1 0(ii) Using C 2 as the contour,∫∫ πRe zdz= R cos t(−R sin t + iR cos t) dt = 1 2 iπR2 .C 2 0(iii) Finally the integral along C 3 = C 3a + C 3b is given by∫∫ 1∫ 1Re zdz= (1 − t)R(−R + iR) dt + (−sR)(−R − iR) dsC 3 00= 1 2 R2 (−1+i)+ 1 2 R2 (1 + i) =iR 2 . ◭)] 10The results of this section demonstrate that the value of an integral between thesame two points may depend upon the path that is taken between them but, atthe same time, suggest that, under some circumstances, the value is independentof the path. The general situation is summarised in the result of the next section,848

24.9 CAUCHY’S THEOREMnamely Cauchy’s theorem, which is the cornerstone of the integral calculus ofcomplex variables.Before discussing Cauchy’s theorem, however, we note an important resultconcerning complex integrals that will be of some use later. Let us consider theintegral of a function f(z) along some path C. IfM is an upper bound on thevalue of |f(z)| on the path, i.e. |f(z)| ≤M on C, andL is the length of the path C,then∫∫∫∣ f(z) dz∣ ≤ |f(z)||dz| ≤M dl = ML. (24.39)CcCIt is straightforward to verify that this result does indeed hold for the complexintegrals considered earlier in this section.24.9 Cauchy’s theoremCauchy’s theorem states that if f(z) is an analytic function, and f ′ (z) is continuousat each point within and on a closed contour C, then∮f(z) dz =0. (24.40)CIn this statement and from now on we denote an integral around a closed contourby ∮ C .To prove this theorem we will need the two-dimensional form of the divergencetheorem, known as Green’s theorem in a plane (see section 11.3). This says thatif p and q are two functions with continuous first derivatives within and on aclosed contour C (bounding a domain R) inthexy-plane, then∫∫R( ∂p∂x + ∂q∂y)dxdy =∮C(p dy− qdx). (24.41)With f(z) =u + iv and dz = dx + idy, this can be applied to∮∮∮I = f(z) dz = (u dx− vdy)+i (v dx+ udy)to give∫∫ [ ∂(−u)I =∂yRCC+ ∂(−v) ] ∫∫ [ ∂(−v)dx dy + i+ ∂u ]dx dy. (24.42)∂xR ∂y ∂xNow, recalling that f(z) is analytic and therefore that the Cauchy–Riemannrelations (24.5) apply, we see that each integrand is identically zero and thus I isalso zero; this proves Cauchy’s theorem.In fact, the conditions of the above proof are more stringent than they needbe. The continuity of f ′ (z) is not necessary for the proof of Cauchy’s theorem,849C

COMPLEX VARIABLESyBC 1RC 2AxFigure 24.10 Two paths C 1 and C 2 enclosing a region R.analyticity of f(z) within and on C being sufficient. However, the proof thenbecomes more complicated and is too long to be given here. §The connection between Cauchy’s theorem and the zero value of the integralof z −1 around the composite path C 4 discussed towards the end of the previoussection is apparent: the function z −1 is analytic in the two regions of the z-planeenclosed by contours (C 2 and C 3a )and(C 2 and C 3b ).◮Suppose two points A and B in the complex plane are joined by two different paths C 1and C 2 . Show that if f(z) is an analytic function on each path and in the region enclosedby the two paths, then the integral of f(z) is the same along C 1 and C 2 .The situation is shown in figure 24.10. Since f(z) isanalyticinR, it follows from Cauchy’stheorem that we have∫ ∫∮f(z) dz − f(z) dz = f(z) dz =0,C 1 C 2 C 1 −C 2since C 1 − C 2 forms a closed contour enclosing R. Thus we immediately obtain∫∫f(z) dz = f(z) dz,C 1 C 2and so the values of the integrals along C 1 and C 2 are equal. ◭An important application of Cauchy’s theorem is in proving that, in somecases, it is possible to deform a closed contour C into another contour γ in sucha way that the integrals of a function f(z) around each of the contours have thesame value.§ The reader may refer to almost any book that is devoted to complex variables and the theory offunctions.850

24.10 CAUCHY’S INTEGRAL FORMULAyCC 1γC 2Figure 24.11 The contour used to prove the result (24.43).x◮Consider two closed contours C and γ in the Argand diagram, γ being sufficiently smallthat it lies completely within C. Show that if the function f(z) is analytic in the regionbetween the two contours then∮∮f(z) dz = f(z) dz. (24.43)CγTo prove this result we consider a contour as shown in figure 24.11. The two close parallellines C 1 and C 2 join γ and C, which are ‘cut’ to accommodate them. The new contour Γso formed consists of C, C 1 , γ and C 2 .Within the area bounded by Γ, the function f(z) is analytic, and therefore, by Cauchy’stheorem (24.40),∮f(z) dz =0. (24.44)ΓNow the parts C 1 and C 2 of Γ are traversed in opposite directions, and in the limit lie ontop of each other, and so their contributions to (24.44) cancel. Thus∮ ∮f(z) dz + f(z) dz =0. (24.45)CThe sense of the integral round γ is opposite to the conventional (anticlockwise) one, andso by traversing γ in the usual sense, we establish the result (24.43). ◭A sort of converse of Cauchy’s theorem is known as Morera’s theorem, whichstates that if f(z) is a continuous function of z in a closed domain R bounded byacurveC and, further, ∮ Cf(z) dz =0,thenf(z) is analytic in R.γ24.10 Cauchy’s integral formulaAnother very important theorem in the theory of complex variables is Cauchy’sintegral formula, which states that if f(z) is analytic within and on a closed851

COMPLEX VARIABLEScontour C and z 0 is a point within C thenf(z 0 )= 1 ∮f(z)dz. (24.46)2πi C z − z 0This formula is saying that the value of an analytic function anywhere inside aclosed contour is uniquely determined by its values on the contour § and that thespecific expression (24.46) can be given for the value at the interior point.We may prove Cauchy’s integral formula by using (24.43) and taking γ to be acircle centred on the point z = z 0 , of small enough radius ρ that it all lies insideC. Then, since f(z) is analytic inside C, the integrand f(z)/(z − z 0 ) is analyticin the space between C and γ. Thus, from (24.43), the integral around γ has thesame value as that around C.We then use the fact that any point z on γ is given by z = z 0 + ρ exp iθ (andso dz = iρ exp iθ dθ). Thus the value of the integral around γ is given by∮I =γf(z)z − z 0dz == i∫ 2π0∫ 2π0f(z 0 + ρ exp iθ)iρ exp iθ dθρ exp iθf(z 0 + ρ exp iθ) dθ.If the radius of the circle γ is now shrunk to zero, i.e. ρ → 0, then I → 2πif(z 0 ),thus establishing the result (24.46).An extension to Cauchy’s integral formula can be made, yielding an integralexpression for f ′ (z 0 ):f ′ (z 0 )= 1 ∫2πi Cunder the same conditions as previously stated.◮Prove Cauchy’s integral formula for f ′ (z 0 ) given in (24.47).f(z)dz, (24.47)(z − z 0 )2To show this, we use the definition of a derivative and (24.46) itself to evaluatef ′ f(z 0 + h) − f(z 0 )(z 0 ) = limh→0 h[ ∮ (1 f(z) 1= limh→0 2πi C h z − z 0 − h − 1 ) ]dzz − z 0[ ∮]1f(z)= limh→0 2πi (z − z 0 − h)(z − z 0 ) dz= 12πiwhich establishes result (24.47). ◭∮CCf(z)(z − z 0 ) 2 dz,§ The similarity between this and the uniqueness theorem for the Laplace equation with Dirichletboundary conditions (see chapter 20) is apparent.852

24.11 TAYLOR AND LAURENT SERIESFurther, it may be proved by induction that the nth derivative of f(z) is alsogiven by a Cauchy integral,f (n) (z 0 )= n! ∮f(z) dz. (24.48)2πi C (z − z 0 )n+1Thus, if the value of the analytic function is known on C then not only may thevalue of the function at any interior point be calculated, but also the values ofall its derivatives.The observant reader will notice that (24.48) may also be obtained by theformal device of differentiating under the integral sign with respect to z 0 inCauchy’s integral formula (24.46):f (n) (z 0 )= 1 ∮∂ n [ ] f(z)dz2πi (z − z 0 )= n!2πi∮C ∂z0nCf(z) dz(z − z 0 ) n+1 .◮Suppose that f(z) is analytic inside and on a circle C of radius R centred on the pointz = z 0 .If|f(z)| ≤M on the circle, where M is some constant, show that|f (n) (z 0 )|≤ Mn!R . (24.49)nFrom (24.48) we have|f (n) (z 0 )| = n!∮2π ∣and on using (24.39) this becomes|f (n) (z 0 )|≤ n! M2π RThis result is known as Cauchy’s inequality. ◭Cf(z) dz(z − z 0 ) n+1 ∣ ∣∣∣,Mn!2πR =n+1R . nWe may use Cauchy’s inequality to prove Liouville’s theorem, which states thatif f(z) is analytic and bounded for all z then f is a constant. Setting n =1in(24.49) and letting R →∞, we find |f ′ (z 0 )| =0andhencef ′ (z 0 ) = 0. Since f(z) isanalytic for all z, wemaytakez 0 as any point in the z-plane and thus f ′ (z) =0for all z; this implies f(z) = constant. Liouville’s theorem may be used in turn toprove the fundamental theorem of algebra (see exercise 24.9).24.11 Taylor and Laurent seriesFollowing on from (24.48), we may establish Taylor’s theorem forfunctionsofacomplex variable. If f(z) is analytic inside and on a circle C of radius R centredon the point z = z 0 ,andz is a point inside C, then∞∑f(z) = a n (z − z 0 ) n , (24.50)n=0853

COMPLEX VARIABLESwhere a n is given by f (n) (z 0 )/n!. The Taylor expansion is valid inside the regionof analyticity and, for any particular z 0 , can be shown to be unique.To prove Taylor’s theorem (24.50), we note that, since f(z) is analytic insideand on C, we may use Cauchy’s formula to write f(z) asf(z) = 12πi∮Cn=0f(ξ)dξ, (24.51)ξ − zwhere ξ lies on C. Now we may expand the factor (ξ − z) −1 as a geometric seriesin (z − z 0 )/(ξ − z 0 ),1ξ − z = 1 ∑∞ ( ) n z − z0,ξ − z 0 ξ − z 0so (24.51) becomesf(z) = 1 ∮f(ξ)2πi C ξ − z 0= 1 ∞∑2πi= 12πin=0∞ ∑n=0(z − z 0 ) n ∮( z − z0ξ − z 0) nCdξf(ξ)dξ(ξ − z 0 )n+1∞∑(z − z 0 ) n 2πif(n) (z 0 ), (24.52)n!n=0where we have used Cauchy’s integral formula (24.48) for the derivatives off(z). Cancelling the factors of 2πi, we thus establish the result (24.50) witha n = f (n) (z 0 )/n!.◮Show that if f(z) and g(z) are analytic in some region R, and f(z) =g(z) within somesubregion S of R, thenf(z) =g(z) throughout R.It is simpler to consider the (analytic) function h(z) =f(z)−g(z), and to show that becauseh(z) =0inS it follows that h(z) = 0 throughout R.If we choose a point z = z 0 in S, then we can expand h(z) in a Taylor series about z 0 ,h(z) =h(z 0 )+h ′ (z 0 )(z − z 0 )+ 1 2 h′′ (z 0 )(z − z 0 ) 2 + ··· ,which will converge inside some circle C that extends at least as far as the nearest part ofthe boundary of R, sinceh(z) isanalyticinR. But since z 0 lies in S, we haveh(z 0 )=h ′ (z 0 )=h ′′ (z 0 )=···=0,and so h(z) =0insideC. We may now expand about a new point, which can lie anywherewithin C, and repeat the process. By continuing this procedure we may show that h(z) =0throughout R.This result is called the identity theorem and, in fact, the equality of f(z) andg(z)throughout R follows from their equality along any curve of non-zero length in R, orevenat a countably infinite number of points in R. ◭So far we have assumed that f(z) is analytic inside and on the (circular)contour C. If,however,f(z) has a singularity inside C at the point z = z 0 ,thenitcannot be expanded in a Taylor series. Nevertheless, suppose that f(z) has a pole854

24.11 TAYLOR AND LAURENT SERIESof order p at z = z 0 but is analytic at every other point inside and on C. Thenthe function g(z) =(z − z 0 ) p f(z) is analytic at z = z 0 , and so may be expandedas a Taylor series about z = z 0 :∞∑g(z) = b n (z − z 0 ) n . (24.53)n=0Thus, for all z inside C, f(z) will have a power series representation of the formf(z) =a −p(z − z 0 ) p + ···+ a −1+ a 0 + a 1 (z − z 0 )+a 2 (z − z 0 ) 2 + ··· ,z − z 0(24.54)with a −p ≠ 0. Such a series, which is an extension of the Taylor expansion, iscalled a Laurent series. By comparing the coefficients in (24.53) and (24.54), wesee that a n = b n+p . Now, the coefficients b n in the Taylor expansion of g(z) areseen from (24.52) to be given byb n = g(n) (z 0 )= 1 ∮g(z)dz,n! 2πi (z − z 0 )n+1and so for the coefficients a n in (24.54) we havea n = 12πi∮g(z)1dz =(z − z 0 )n+1+p2πi∮f(z)dz,(z − z 0 )n+1an expression that is valid for both positive and negative n.The terms in the Laurent series with n ≥ 0 are collectively called the analyticpart, whilst the remainder of the series, consisting of terms in inverse powers ofz − z 0 , is called the principal part. Depending on the nature of the point z = z 0 ,the principal part may contain an infinite number of terms, so thatf(z) =+∞∑n=−∞a n (z − z 0 ) n . (24.55)In this case we would expect the principal part to converge only for |(z − z 0 ) −1 |less than some constant, i.e. outside some circle centred on z 0 . However, theanalytic part will converge inside some (different) circle also centred on z 0 .Ifthelatter circle has the greater radius then the Laurent series will converge in theregion R between the two circles (see figure 24.12); otherwise it does not convergeat all.In fact, it may be shown that any function f(z) that is analytic in a regionR between two such circles C 1 and C 2 centred on z = z 0 can be expressed asa Laurent series about z 0 that converges in R. We note that, depending on thenature of the point z = z 0 , the inner circle may be a point (when the principalpart contains only a finite number of terms) and the outer circle may have aninfinite radius.We may use the Laurent series of a function f(z) about any point z = z 0 to855

COMPLEX VARIABLESyRz 0C 1C 2xFigure 24.12 The region of convergence R for a Laurent series of f(z) abouta point z = z 0 where f(z) has a singularity.classify the nature of that point. If f(z) is actually analytic at z = z 0 ,thenin(24.55) all a n for n0;(ii) it is not possible to find such a lowest value of −p.In case (i), f(z) is of the form (24.54) and is described as having a pole of orderp at z = z 0 ; the value of a −1 (not a −p ) is called the residue of f(z) at the polez = z 0 , and will play an important part in later applications.For case (ii), in which the negatively decreasing powers of z − z 0 do notterminate, f(z) is said to have an essential singularity. These definitions should becompared with those given in section 24.6.◮Find the Laurent series off(z) =1z(z − 2) 3about the singularities z =0and z =2(separately). Hence verify that z =0is a pole oforder 1 and z =2is a pole of order 3, and find the residue of f(z) at each pole.To obtain the Laurent series about z = 0, we make the factor in parentheses in the856

24.11 TAYLOR AND LAURENT SERIESdenominator take the form (1 − αz), where α is some constant, and thus obtain1f(z)=−8z(1 − z/2) 3=− 1 [ (1+(−3) − z )+ (−3)(−4) (− z ) 2 (−3)(−4)(−5)(+ − z ) 3+ ···]8z2 2! 2 3! 2= − 1 8z − 316 − 3z16 − 5z232 − ··· .Since the lowest power of z is −1, the point z = 0 is a pole of order 1. The residue of f(z)at z = 0 is simply the coefficient of z −1 in the Laurent expansion about that point and isequal to −1/8.The Laurent series about z = 2 is most easily found by letting z =2+ξ (or z − 2=ξ)and substituting into the expression for f(z) toobtainf(z) =1(2 + ξ)ξ 3 = 1= 12ξ 3 [1 −2ξ 3 (1 + ξ/2)( ) 2 ( ξ ξ−2 2( ξ2)+) 3 ( ) 4 ξ+ − ···]2= 12ξ − 13 4ξ + 12 8ξ − 1 16 + ξ 32 − ···1=2(z − 2) − 13 4(z − 2) + 12 8(z − 2) − 116 + z − 232 − ··· .From this series we see that z = 2 is a pole of order 3 and that the residue of f(z) atz =2is 1/8. ◭As we shall see in the next few sections, finding the residue of a functionat a singularity is of crucial importance in the evaluation of complex integrals.Specifically, formulae exist for calculating the residue of a function at a particular(singular) point z = z 0 without having to expand the function explicitly as aLaurent series about z 0 and identify the coefficient of (z − z 0 ) −1 . The type offormula generally depends on the nature of the singularity at which the residueis required.◮Suppose that f(z) has a pole of order m at the point z = z 0 . By considering the Laurentseries of f(z) about z 0 , derive a general expression for the residue R(z 0 ) of f(z) at z = z 0 .Hence evaluate the residue of the functionat the point z = i.f(z) =exp iz(z 2 +1) 2If f(z) has a pole of order m at z = z 0 , then its Laurent series about this point has theformf(z) =a −m(z − z 0 ) + ···+ a −1m (z − z 0 ) + a 0 + a 1 (z − z 0 )+a 2 (z − z 0 ) 2 + ··· ,which, on multiplying both sides of the equation by (z − z 0 ) m ,gives(z − z 0 ) m f(z) =a −m + a −m+1 (z − z 0 )+···+ a −1 (z − z 0 ) m−1 + ··· .857

COMPLEX VARIABLESDifferentiating both sides m − 1times,weobtaind m−1∞∑dz [(z − z 0) m f(z)] = (m − 1)! a m−1 −1 + b n (z − z 0 ) n ,n=1for some coefficients b n . In the limit z → z 0 , however, the terms in the sum disappear, andafter rearranging we obtain the formula{}1 d m−1R(z 0 )=a −1 = limz→z0 (m − 1)! dz [(z − z 0) m f(z)] , (24.56)m−1which gives the value of the residue of f(z) at the point z = z 0 .If we now consider the functionexp izf(z) =(z 2 +1) = exp iz2 (z + i) 2 (z − i) , 2we see immediately that it has poles of order 2 (double poles) at z = i and z = −i. Tocalculate the residue at (for example) z = i, we may apply the formula (24.56) with m =2.Performing the required differentiation, we obtainddz [(z − i)2 f(z)] = d [ ] exp izdz (z + i) 21=(z + i) [(z + 4 i)2 i exp iz − 2(exp iz)(z + i)].Setting z = i, we find the residue is given byR(i) = 1 1 (−4ie −1 − 4ie −1) = − i1! 162e . ◭An important special case of (24.56) occurs when f(z) has a simple pole (a poleof order 1) at z = z 0 . Then the residue at z 0 is given byR(z 0 ) = lim [(z − z 0 )f(z)] . (24.57)z→z0If f(z) has a simple pole at z = z 0 and, as is often the case, has the formg(z)/h(z), where g(z) is analytic and non-zero at z 0 and h(z 0 ) = 0, then (24.57)becomes(z − z 0 )g(z)(z − z 0 )R(z 0 ) = lim= g(z 0 ) limz→z0 h(z)z→z0 h(z)1= g(z 0 ) limz→z0 h ′ (z) = g(z 0)h ′ (z 0 ) , (24.58)where we have used l’Hôpital’s rule. This result often provides the simplest wayof determining the residue at a simple pole.24.12 Residue theoremHaving seen from Cauchy’s theorem that the value of an integral round a closedcontour C is zero if the integrand is analytic inside the contour, it is natural toask what value it takes when the integrand is not analytic inside C. Theanswerto this is contained in the residue theorem, which we now discuss.858

24.12 RESIDUE THEOREMSuppose the function f(z) has a pole of order m at the point z = z 0 ,andsocan be written as a Laurent series about z 0 of the form∞∑f(z) = a n (z − z 0 ) n . (24.59)n=−mNow consider the integral I of f(z) around a closed contour C that enclosesz = z 0 , but no other singular points. Using Cauchy’s theorem, this integral hasthe same value as the integral around a circle γ of radius ρ centred on z = z 0 ,since f(z) is analytic in the region between C and γ. On the circle we havez = z 0 + ρ exp iθ (and dz = iρ exp iθ dθ), and so∮I = f(z) dz==γ∞∑∮a n (z − z 0 ) n dzn=−m∞∑∫ 2πa n iρ n+1 exp[i(n +1)θ] dθ.n=−mFor every term in the series with n ≠ −1, we have∫ 2π[ iρiρ n+1 n+1 exp[i(n +1)θ]exp[i(n +1)θ] dθ =0i(n +1)but for the n = −1 termweobtain0∫ 2π0idθ=2πi.] 2π0=0,Therefore only the term in (z − z 0 ) −1 contributes to the value of the integralaround γ (and therefore C), and I takes the value∮I = f(z) dz =2πia −1 . (24.60)CThus the integral around any closed contour containing a single pole of generalorder m (or, by extension, an essential singularity) is equal to 2πi times the residueof f(z) atz = z 0 .If we extend the above argument to the case where f(z) is continuous withinand on a closed contour C and analytic, except for a finite number of poles,within C, then we arrive at the residue theorem∮f(z) dz =2πi ∑ R j , (24.61)Cjwhere ∑ j R j is the sum of the residues of f(z) at its poles within C.The method of proof is indicated by figure 24.13, in which (a) shows the originalcontour C referred to in (24.61) and (b) shows a contour C ′ giving the same value859

COMPLEX VARIABLESCC ′C(a)(b)Figure 24.13 The contours used to prove the residue theorem: (a) the originalcontour; (b) the contracted contour encircling each of the poles.to the integral, because f is analytic between C and C ′ . Now the contribution tothe C ′ integral from the polygon (a triangle for the case illustrated) joining thesmall circles is zero, since f is also analytic inside C ′ . Hence the whole value ofthe integral comes from the circles and, by result (24.60), each of these contributes2πi times the residue at the pole it encloses. All the circles are traversed in theirpositive sense if C is thus traversed and so the residue theorem follows. Formally,Cauchy’s theorem (24.40) is a particular case of (24.61) in which C encloses nopoles.Finally we prove another important result, for later use. Suppose that f(z) hasa simple pole at z = z 0 and so may be expanded as the Laurent seriesf(z) =φ(z)+a −1 (z − z 0 ) −1 ,where φ(z) is analytic within some neighbourhood surrounding z 0 . We wish tofind an expression for the integral I of f(z) along an open contour C, whichisthe arc of a circle of radius ρ centred on z = z 0 given by|z − z 0 | = ρ, θ 1 ≤ arg(z − z 0 ) ≤ θ 2 , (24.62)where ρ is chosen small enough that no singularity of f, other than z = z 0 , lieswithin the circle. Then I is given by∫∫∫I = f(z) dz = φ(z) dz + a −1 (z − z 0 ) −1 dz.CCIf the radius of the arc C is now allowed to tend to zero, then the first integraltends to zero, since the path becomes of zero length and φ is analytic andtherefore continuous along it. On C, z = ρe iθ and hence the required expressionfor I is( ∫ θ2)1I = lim f(z) dz = lim a −1 ρ→0∫Cρ→0 θ 1ρe iθ iρeiθ dθ = ia −1 (θ 2 − θ 1 ). (24.63)860C

24.13 DEFINITE INTEGRALS USING CONTOUR INTEGRATIONWe note that result (24.60) is a special case of (24.63) in which θ 2 is equal toθ 1 +2π.24.13 Definite integrals using contour integrationThe remainder of this chapter is devoted to methods of applying contour integrationand the residue theorem to various types of definite integral. However, threeapplications of contour integration, in which obtaining a value for the integral isnot the prime purpose of the exercise, have been postponed until chapter 25. Theyare the location of the zeros of a complex polynomial, the evaluation of the sumsof certain infinite series and the determination of inverse Laplace transforms.For the integral evalations considered here, not much preamble is given since,for this material, the simplest explanation is felt to be via a series of workedexamples that can be used as models.Suppose that an integral of the form24.13.1 Integrals of sinusoidal functions∫ 2π0F(cos θ, sin θ) dθ (24.64)is to be evaluated. It can be made into a contour integral around the unit circleC by writing z =expiθ, and hencecos θ = 1 2 (z + z−1 ), sin θ = − 1 2 i(z − z−1 ), dθ = −iz −1 dz. (24.65)This contour integral can then be evaluated using the residue theorem, providedthe transformed integrand has only a finite number of poles inside the unit circleandnoneonit.◮Evaluate∫ 2πcos 2θI =dθ, b > a > 0. (24.66)0 a 2 + b 2 − 2ab cos θBy de Moivre’s theorem (section 3.4),cos nθ = 1 2 (zn + z −n ). (24.67)Using n = 2 in (24.67) and straightforward substitution for the other functions of θ in(24.66) givesI =i ∮z 4 +12ab C z 2 (z − a/b)(z − b/a) dz.Thus there are two poles inside C, a double pole at z = 0 and a simple pole at z = a/b(recall that b>a).We could find the residue of the integrand at z = 0 by expanding the integrand as aLaurent series in z and identifying the coefficient of z −1 . Alternatively, we may use the861

COMPLEX VARIABLESformula (24.56) with m = 2. Choosing the latter method and denoting the integrand byf(z), we haveddz [z2 f(z)] = d []z 4 +1dz (z − a/b)(z − b/a)= (z − a/b)(z − b/a)4z3 − (z 4 +1)[(z − a/b)+(z − b/a)].(z − a/b) 2 (z − b/a) 2Now setting z = 0 and applying (24.56), we findR(0) = a b + b a .For the simple pole at z = a/b, equation (24.57) gives the residue as[ ] (a/b) 4 +1R(a/b) = lim (z − a/b)f(z) =z→(a/b)(a/b) 2 (a/b − b/a)= − a4 + b 4ab(b 2 − a 2 ) .Therefore by the residue theoremI =2πi ×i [ a 2 + b 22ab ab]− a4 + b 4=ab(b 2 − a 2 )2πa 2b 2 (b 2 − a 2 ) . ◭24.13.2 Some infinite integralsWe next consider the evaluation of an integral of the form∫ ∞−∞where f(z) has the following properties:f(x) dx,(i) f(z) is analytic in the upper half-plane, Im z ≥ 0, except for a finite numberof poles, none of which is on the real axis;(ii) on a semicircle Γ of radius R (figure 24.14), R times the maximum of |f|on Γ tends to zero as R →∞(a sufficient condition is that zf(z) → 0as|z| →∞);(iii) ∫ 0−∞ f(x) dx and ∫ ∞0f(x) dx both exist.Since∫∣ f(z) dz∣ ≤ 2πR × (maximum of |f| on Γ),Γcondition (ii) ensures that the integral along Γ tends to zero as R →∞, afterwhich it is obvious from the residue theorem that the required integral is givenby∫ ∞−∞f(x) dx =2πi × (sum of the residues at poles with Im z>0).(24.68)862

24.13 DEFINITE INTEGRALS USING CONTOUR INTEGRATIONyΓ−RORxFigure 24.14A semicircular contour in the upper half-plane.◮Evaluate∫ ∞dxI =, where a is real.0 (x 2 + a 2 )4The complex function (z 2 + a 2 ) −4 has poles of order 4 at z = ±ai, ofwhichonlyz = aiis in the upper half-plane. Conditions (ii) and (iii) are clearly satisfied. For higher-orderpoles, formula (24.56) for evaluating residues can be tiresome to apply. So, instead, we putz = ai + ξ and expand for small ξ to obtain §The coefficient of ξ −1 is given by(1(z 2 + a 2 ) = 14 (2aiξ + ξ 2 ) = 11 − iξ ) −4.4 (2aiξ) 4 2a1 (−4)(−5)(−6)(2a) 4 3!( ) 3 −i= −5i2a 32a , 7and hence by the residue theorem∫ ∞dx−∞ (x 2 + a 2 ) = 10π4 32a , 7and so I =5π/(32a 7 ). ◭Condition (i) of the previous method required there to be no poles of theintegrand on the real axis, but in fact simple poles on the real axis can beaccommodated by indenting the contour as shown in figure 24.15. The indentationat the pole z = z 0 is in the form of a semicircle γ of radius ρ in the upper halfplane,thus excluding the pole from the interior of the contour.§ This illustrates another useful technique for determining residues.863

COMPLEX VARIABLESyΓγ−RORxFigure 24.15 An indented contour used when the integrand has a simplepole on the real axis.What is then obtained from a contour integration, apart from the contributionsfor Γ and γ, is called the principal value of the integral, defined as ρ → 0byP∫ R−Rf(x) dx ≡∫ z0−ρ−Rf(x) dx +∫ Rz 0+ρf(x) dx.The remainder of the calculation goes through as before, but the contributionfrom the semicircle, γ, must be included. Result (24.63) of section 24.12 showsthat since only a simple pole is involved its contribution is−ia −1 π, (24.69)where a −1 is the residue at the pole and the minus sign arises because γ istraversed in the clockwise (negative) sense.We defer giving an example of an indented contour until we have establishedJordan’s lemma; we will then work through an example illustrating both. Jordan’slemma enables infinite integrals involving sinusoidal functions to be evaluated.For a function f(z) of a complex variable z, if(i) f(z) is analytic in the upper half-plane except for a finite number of poles in Im z>0,(ii) the maximum of |f(z)| →0 as |z| →∞in the upper half-plane,(iii) m>0,then∫I Γ = e imz f(z) dz → 0 as R →∞, (24.70)Γwhere Γ is the same semicircular contour as in figure 24.14.Note that this condition (ii) is less stringent than the earlier condition (ii) (see thestart of this section), since we now only require M(R) → 0 and not RM(R) → 0,where M is the maximum § of |f(z)| on |z| = R.§ More strictly, the least upper bound.864

24.13 DEFINITE INTEGRALS USING CONTOUR INTEGRATIONThe proof of the lemma is straightforward once it has been observed that, for0 ≤ θ ≤ π/2,1 ≥ sin θ ≥ 2 θ π . (24.71)Then, since on Γ we have | exp(imz)| = | exp(−mR sin θ)|,∫∫ π∫ π/2I Γ ≤ |e imz f(z)| |dz| ≤MR e −mR sin θ dθ =2MR e −mR sin θ dθ.ΓThus, using (24.71),∫ π/2I Γ ≤ 2MR e −mR(2θ/π) dθ = πM (1 − e−mR ) < πM0mm ;hence, as R →∞, I Γ tends to zero since M tends to zero.◮Find the principal value of∫ ∞−∞cos mxx − a0dx, for a real, m>0.Consider the function (z − a) −1 exp(imz); although it has no poles in the upper half-planeit does have a simple pole at z = a, and further |(z − a) −1 |→0as|z| →∞. We will use acontour like that shown in figure 24.15 and apply the residue theorem. Symbolically,∫ a−ρ ∫ ∫ R ∫+ + + =0. (24.72)−RγNow as R →∞and ρ → 0 we have ∫ → 0, by Jordan’s lemma, and from (24.68) andΓ(24.69) we obtain∫ ∞e imxP−∞ x − a dx − iπa −1 =0, (24.73)where a −1 is the residue of (z − a) −1 exp(imz) atz = a, which is exp(ima). Then taking thereal and imaginary parts of (24.73) givesPP∫ ∞−∞∫ ∞−∞a+ρcos mxdx = −π sin ma,x − asin mxdx = π cos ma,x − aΓ0as required,as a bonus. ◭24.13.3 Integrals of multivalued functionsWe have discussed briefly some of the properties and difficulties associated withcertain multivalued functions such as z 1/2 or Ln z. It was mentioned that onemethod of managing such functions is by means of a ‘cut plane’. A similartechnique can be used with advantage to evaluate some kinds of infinite integralinvolving real functions for which the corresponding complex functions are multivalued.A typical contour employed for functions with a single branch point865

COMPLEX VARIABLESyΓγABCDxFigure 24.16 A typical cut-plane contour for use with multivalued functionsthat have a single branch point located at the origin.located at the origin is shown in figure 24.16. Here Γ is a large circle of radius Rand γ is a small one of radius ρ, both centred on the origin. Eventually we willlet R →∞and ρ → 0.The success of the method is due to the fact that because the integrand ismultivalued, its values along the two lines AB and CD joining z = ρ to z = Rare not equal and opposite although both are related to the corresponding realintegral. Again an example provides the best explanation.◮Evaluate∫ ∞dxI =, a > 0.0 (x + a) 3 x1/2 We consider the integrand f(z) =(z + a) −3 z −1/2 and note that |zf(z)| →0onthetwocircles as ρ → 0andR →∞. Thus the two circles make no contribution to the contourintegral.The only pole of the integrand inside the contour is at z = −a (and is of order 3).To determine its residue we put z = −a + ξ and expand (noting that (−a) 1/2 equalsa 1/2 exp(iπ/2) = ia 1/2 ):1(z + a) 3 z = 11/2 ξ 3 ia 1/2 (1 − ξ/a) 1/2= 1 (1+ 1 ξiξ 3 a 1/2 2 a + 3 ξ ···)28 a + .2The residue is thus −3i/(8a 5/2 ).The residue theorem (24.61) now gives∫ ∫ ∫+ +ABΓDC∫ ( ) −3i+ =2πi .γ 8a 5/2866

24.14 EXERCISESWe have seen that ∫ and ∫ vanish, and if we denote z by x along the line AB then itΓ γhas the value z = x exp 2πi along the line DC (note that exp 2πi must not be set equal to1 until after the substitution for z has been made in ∫ ). Substituting these expressions,DCThusand∫ ∞0∫dx0(x + a) 3 x + 1/2∞dx[x exp 2πi + a] 3 x 1/2 exp( 1 2(1 − 1 ) ∫ ∞dx 3π=exp πi 0 (x + a) 3 x1/2 4a 5/2I = 1 2 ×3π4a 5/2 . ◭2πi)=3π4a 5/2 .Several other examples of integrals of multivalued functions around a varietyof contours are included in the exercises that follow.24.14 Exercises24.1 Find an analytic function of z = x + iy whose imaginary part is(y cos y + x sin y)expx.24.2 Find a function f(z), analytic in a suitable part of the Argand diagram, for whichsin 2xRe f =cosh 2y − cos 2x .Wherearethesingularitiesoff(z)?24.3 Find the radii of convergence of the following Taylor series:∞∑ z n(a)ln n , (b) ∑ ∞n!z nn , n n=2n=1∞∑∞∑( ) n 2n + p(c) z n n ln n , (d)z n , with p real.nn=1n=124.4 Find the Taylor series expansion about the origin of the function f(z) defined by∞∑ ( pz)f(z) = (−1) r+1 sin ,rr=1where p is a constant. Hence verify that f(z) is a convergent series for all z.24.5 Determine the types of singularities (if any) possessed by the following functionsat z =0andz = ∞:(a) (z − 2) −1 , (b) (1 + z 3 )/z 2 , (c) sinh(1/z),(d) e z /z 3 , (e) z 1/2 /(1 + z 2 ) 1/2 .24.6 Identify the zeros, poles and essential singularities of the following functions:(a) tan z, (b) [(z − 2)/z 2 ] sin[1/(1 − z)], (c) exp(1/z),(d) tan(1/z), (e) z 2/3 .867

COMPLEX VARIABLES24.7 Find the real and imaginary parts of the functions (i) z 2 , (ii) e z , and (iii) cosh πz.By considering the values taken by these parts on the boundaries of the region0 ≤ x, y ≤ 1, determine the solution of Laplace’s equation in that region thatsatisfies the boundary conditionsφ(x, 0) = 0, φ(0,y)=0,φ(x, 1) = x, φ(1,y)=y +sinπy.24.8 Show that the transformation∫ z1w =dζ0 (ζ 3 − ζ)1/2transforms the upper half-plane into the interior of a square that has one cornerat the origin of the w-plane and sides of length L, whereL =∫ π/20cosec 1/2 θdθ.24.9 The fundamental theorem of algebra states that, for a complex polynomial p n (z)of degree n, the equation p n (z) = 0 has precisely n complex roots. By applyingLiouville’s theorem (see the end of section 24.10) to f(z) =1/p n (z), prove thatp n (z) = 0 has at least one complex root. Factor out that root to obtain p n−1 (z)and, by repeating the process, prove the above theorem.24.10 Show that, if a is a positive real constant, the function exp(iaz 2 ) is analytic and→ 0as|z| →∞for 0 < arg z ≤ π/4. By applying Cauchy’s theorem to a suitablecontour prove that24.11 The function∫ ∞0cos(ax 2 ) dx =√ π8a .f(z) =(1− z 2 ) 1/2of the complex variable z is defined to be real and positive on the real axis inthe range −1

24.14 EXERCISES24.14 Prove that, for α>0, the integral∫ ∞t sin αt0 1+t dt 2has the value (π/2) exp(−α).24.15 Prove that∫ ∞cos mx0 4x 4 +5x 2 +1 dx = π (4e −m/2 − e −m) for m>0.624.16 Show that the principal value of the integral∫ ∞cos(x/a)dx−∞ x 2 − a 2is −(π/a)sin1.24.17 The following is an alternative (and roundabout!) way of evaluating the Gaussianintegral.(a) Prove that the integral of [exp(iπz 2 )]cosec πz around the parallelogram withcorners ±1/2 ± R exp(iπ/4) has the value 2i.(b) Show that the parts of the contour parallel to the real axis do not contributewhen R →∞.(c) Evaluate the integrals along the other two sides by putting z ′ = r exp(iπ/4)and working in terms of z ′ + 1 and 2z′ − 1 . Hence, by letting R →∞show2that∫ ∞−∞e −πr2 dr =1.24.18 By applying the residue theorem around a wedge-shaped contour of angle 2π/n,with one side along the real axis, prove that the integral∫ ∞dx0 1+x , nwhere n is real and ≥ 2, has the value (π/n)cosec (π/n).24.19 Using a suitable cut plane, prove that if α is real and 0

COMPLEX VARIABLES24.22 The equation of an ellipse in plane polar coordinates r, θ, with one of its foci atthe origin, isl=1− ɛ cos θ,rwhere l is a length (that of the latus rectum) and ɛ (0

25Applications of complex variablesIn chapter 24, we developed the basic theory of the functions of a complexvariable, z = x + iy, studied their analyticity (differentiability) properties andderived a number of results concerned with values of contour integrals in thecomplex plane. In this current chapter we will show how some of those resultsand properties can be exploited to tackle problems arising directly from physicalsituations or from apparently unrelated parts of mathematics.In the former category will be the use of the differential properties of the realand imaginary parts of a function of a complex variable to solve problems involvingLaplace’s equation in two dimensions, whilst an example of the latter mightbe the summation of certain types of infinite series. Other applications, such asthe Bromwich inversion formula for Laplace transforms, appear as mathematicalproblems that have their origins in physical applications; the Bromwich inversionenables us to extract the spatial or temporal response of a system to an initialinput from the representation of that response in ‘frequency space’ – or, morecorrectly, imaginary frequency space.Other topics that will be considered are the location of the (complex) zeros ofa polynomial, the approximate evaluation of certain types of contour integralsusing the methods of steepest descent and stationary phase, and the so-called‘phase-integral’ solutions to some differential equations. For each of these a briefintroduction is given at the start of the relevant section and to repeat them herewould be pointless. We will therefore move on to our first topic of complexpotentials.25.1 Complex potentialsTowards the end of section 24.2 of the previous chapter it was shown that the realand the imaginary parts of an analytic function of z are separately solutions ofLaplace’s equation in two dimensions. Analytic functions thus offer a possible way871

APPLICATIONS OF COMPLEX VARIABLESyxFigure 25.1 The equipotentials (dashed circles) and field lines (solid lines)for a line charge perpendicular to the z-plane.of solving some two-dimensional physical problems describable by a potentialsatisfying ∇ 2 φ = 0. The general method is known as that of complex potentials.We also found that if f = u + iv is an analytic function of z then any curveu = constant intersects any curve v = constant at right angles. In the context ofsolutions of Laplace’s equation, this result implies that the real and imaginaryparts of f(z) have an additional connection between them, for if the set ofcontours on which one of them is a constant represents the equipotentials of asystem then the contours on which the other is constant, being orthogonal toeach of the first set, must represent the corresponding field lines or stream lines,depending on the context. The analytic function f is the complex potential. Itis conventional to use φ and ψ (rather than u and v) to denote the real andimaginary parts of a complex potential, so that f = φ + iψ.As an example, consider the functionf(z) = −q ln z (25.1)2πɛ 0in connection with the physical situation of a line charge of strength q per unitlength passing through the origin, perpendicular to the z-plane (figure 25.1). Itsreal and imaginary parts areφ = −q ln |z|, ψ = −q arg z. (25.2)2πɛ 0 2πɛ 0The contours in the z-plane of φ = constant are concentric circles and those ofψ = constant are radial lines. As expected these are orthogonal sets, but in additionthey are, respectively, the equipotentials and electric field lines appropriate to872

25.1 COMPLEX POTENTIALSthe field produced by the line charge. The minus sign is needed in (25.1) becausethe value of φ must decrease with increasing distance from the origin.Suppose we make the choice that the real part φ of the analytic function fgives the conventional potential function; ψ could equally well be selected. Thenwe may consider how the direction and magnitude of the field are related to f.◮Show that for any complex (electrostatic) potential f(z) the strength of the electric fieldis given by E = |f ′ (z)| and that its direction makes an angle of π − arg[f ′ (z)] with thex-axis.Because φ = constant is an equipotential, the field has componentsE x = − ∂φ∂xandE y = − ∂φ∂y . (25.3)Since f is analytic, (i) we may use the Cauchy–Riemann relations (24.5) to change thesecond of these, obtainingE x = − ∂φ∂xandE y = ∂ψ∂x ; (25.4)(ii) the direction of differentiation at a point is immaterial and sodfdz = ∂f∂x = ∂φ∂x + i ∂ψ∂x = −E x + iE y . (25.5)From these it can be seen that the field at a point is given in magnitude by E = |f ′ (z)|and that it makes an angle with the x-axis given by π − arg[f ′ (z)]. ◭It will be apparent from the above that much of physical interest can becalculated by working directly in terms of f and z. In particular, the electric fieldvector E may be represented, using (25.5) above, by the quantityE = E x + iE y = −[f ′ (z)] ∗ .Complex potentials can be used in two-dimensional fluid mechanics problemsin a similar way. If the flow is stationary (i.e. the velocity of the fluid does notdepend on time) and irrotational, and the fluid is both incompressible and nonviscous,then the velocity of the fluid can be described by V = ∇φ, whereφ is thevelocity potential and satisfies ∇ 2 φ = 0. If, for a complex potential f = φ + iψ,the real part φ is taken to represent the velocity potential then the curves ψ =constant will be the streamlines of the flow. In a direct parallel with the electricfield, the velocity may be represented in terms of the complex potential byV = V x + iV y =[f ′ (z)] ∗ ,the difference of a minus sign reflecting the same difference between the definitionsof E and V. The speed of the flow is equal to |f ′ (z)|. Points where f ′ (z) = 0, andthus the velocity is zero, are called stagnation points of the flow.Analogously to the electrostatic case, a line source of fluid at z = z 0 , perpendicularto the z-plane (i.e. a point from which fluid is emerging at a constant rate),873

APPLICATIONS OF COMPLEX VARIABLESis described by the complex potentialf(z) =k ln(z − z 0 ),where k is the strength of the source. A sink is similarly represented, but with kreplaced by −k. Other simple examples are as follows.(i) The flow of a fluid at a constant speed V 0 and at an angle α to the x-axisis described by f(z) =V 0 (exp iα)z.(ii) Vortex flow, in which fluid flows azimuthally in an anticlockwise directionaround some point z 0 , the speed of the flow being inversely proportionalto the distance from z 0 , is described by f(z) =−ik ln(z − z 0 ), where k isthe strength of the vortex. For a clockwise vortex k is replaced by −k.◮ Verify that the complex potential( )f(z) =V 0 z + a2zis appropriate to a circular cylinder of radius a placed so that it is perpendicular to auniform fluid flow of speed V 0 parallel to the x-axis.Firstly, since f(z) isanalyticexceptatz = 0, both its real and imaginary parts satisfyLaplace’s equation in the region exterior to the cylinder. Also f(z) → V 0 z as z →∞,sothat Re f(z) → V 0 x, which is appropriate to a uniform flow of speed V 0 in the x-directionfar from the cylinder.Writing z = r exp iθ and using de Moivre’s theorem we have[]f(z) =V 0 r exp iθ + a2r exp(−iθ) ( ) )= V 0 r + a2cos θ + iV 0(r − a2sin θ.rrThus we see that the streamlines of the flow described by f(z) are given by( )ψ = V 0 r − a2sin θ =constant.rIn particular, ψ =0onr = a, independently of the value of θ, andsor = a must be astreamline. Since there can be no flow of fluid across streamlines, r = a must correspondto a boundary along which the fluid flows tangentially. Thus f(z) is a solution of Laplace’sequation that satisfies all the physical boundary conditions of the problem, and so, by theuniqueness theorem, it is the appropriate complex potential. ◭By a similar argument, the complex potential f(z) =−E(z − a 2 /z) (note theminus signs) is appropriate to a conducting circular cylinder of radius a placedperpendicular to a uniform electric field E in the x-direction.The real and imaginary parts of a complex potential f = φ + iψ have anotherinteresting relationship in the context of Laplace’s equation in electrostatics orfluid mechanics. Let us choose φ as the conventional potential, so that ψ representsthe stream function (or electric field, depending on the application), and consider874

25.1 COMPLEX POTENTIALSyQxPˆnFigure 25.2 A curve joining the points P and Q. Also shown is ˆn, the unitvector normal to the curve.the difference in the values of ψ at any two points P and Q connected by somepath C, as shown in figure 25.2. This difference is given by∫ Q ∫ Q( )∂ψ ∂ψψ(Q) − ψ(P )= dψ = dx +PP ∂x ∂y dy ,which, on using the Cauchy–Riemann relations, becomes∫ Q(ψ(Q) − ψ(P )= − ∂φ)∂φdx +P ∂y ∂x dy∫ Q∫ Q∂φ= ∇φ · ˆn ds =PP ∂n ds,where ˆn is the vector unit normal to the path C and s is the arc length along thepath; the last equality is written in terms of the normal derivative ∂φ/∂n ≡∇φ· ˆn.Now suppose that in an electrostatics application, the path C is the surface ofa conductor; then∂φ∂n = − σ ,ɛ 0where σ is the surface charge density per unit length normal to the xy-plane.Therefore −ɛ 0 [ψ(Q) − ψ(P )] is equal to the charge per unit length normal to thexy-plane on the surface of the conductor between the points P and Q. Similarly,in fluid mechanics applications, if the density of the fluid is ρ and its velocity isV then∫ Q∫ Qρ[ψ(Q) − ψ(P )] = ρ ∇φ · ˆn ds = ρ V · ˆn dsPis equal to the mass flux between P and Q per unit length perpendicular to thexy-plane.875P

APPLICATIONS OF COMPLEX VARIABLES◮ A conducting circular cylinder of radius a is placed with its centre line passing throughthe origin and perpendicular to a uniform electric field E in the x-direction. Find the chargeper unit length induced on the half of the cylinder that lies in the region x

25.2 APPLICATIONS OF CONFORMAL TRANSFORMATIONSyssxrr(a) z-plane (b) w-plane (c) w-planeFigure 25.3 The equipotential lines (broken) and field lines (solid) (a) for aninfinite charged conducting plane at y =0,wherez = x + iy, and after thetransformations (b) w = z 2 and (c) w = z 1/2 of the situation shown in (a).◮ Find the complex electrostatic potential associated with an infinite charged conductingplate y =0, and thus obtain those associated with(i) a semi-infinite charged conducting plate (r >0, s =0);(ii) the inside of a right-angled charged conducting wedge (r > 0, s = 0 andr =0, s>0).Figure 25.3(a) shows the equipotentials (broken lines) and field lines (solid lines) for theinfinite charged conducting plane y = 0. Suppose that we elect to make the real part ofthe complex potential coincide with the conventional electrostatic potential. If the plate ischarged to a potential V then clearlyφ(x, y) =V − ky, (25.6)where k is related to the charge density σ by k = σ/ɛ 0 , since physically the electric field Ehas components (0,σ/ɛ 0 )andE = −∇φ.Thus what is needed is an analytic function of z, ofwhichtherealpartisV − ky. Thiscan be obtained by inspection, but we may proceed formally and use the Cauchy–Riemannrelations to obtain the imaginary part ψ(x, y) as follows:∂ψ∂y = ∂φ∂x =0 and ∂ψ∂x = − ∂φ∂y = k.Hence ψ = kx + c and, absorbing c into V , the required complex potential isf(z) =V − ky + ikx = V + ikz. (25.7)(i) Now consider the transformationw = g(z) =z 2 . (25.8)This satisfies the criteria for a conformal mapping (except at z = 0) and carries the upperhalf of the z-plane into the entire w-plane; the equipotential plane y =0goesintothehalf-plane r>0, s =0.By the general results proved, f(z), when expressed in terms of r and s, will give acomplex potential whose real part will be constant on the half-plane in question; we877

APPLICATIONS OF COMPLEX VARIABLESdeduce thatF(w) =f(z) =V + ikz = V + ikw 1/2 (25.9)is the required potential. Expressed in terms of r, s and ρ =(r 2 + s 2 ) 1/2 , w 1/2 is given by[ ( ) 1/2 ( ) ] 1/2 ρ + r ρ − rw 1/2 = ρ 1/2 + i, (25.10)2ρ2ρand, in particular, the electrostatic potential is given byΦ(r, s) =ReF(w) =V −k √2[(r 2 + s 2 ) 1/2 − r ] 1/2. (25.11)The corresponding equipotentials and field lines are shown in figure 25.3(b). Using results(25.3)–(25.5), the magnitude of the electric field is|E| = |F ′ (w)| = | 1 2 ikw−1/2 | = 1 2 k(r2 + s 2 ) −1/4 .(ii) A transformation ‘converse’ to that used in (i),w = g(z) =z 1/2 ,has the effect of mapping the upper half of the z-plane into the first quadrant of thew-plane and the conducting plane y =0intothewedger>0, s =0andr =0,s>0.The complex potential now becomesF(w) =V + ikw 2= V + ik[(r 2 − s 2 )+2irs], (25.12)showing that the electrostatic potential is V −2krs and that the electric field has componentsE =(2ks,2kr). (25.13)Figure 25.3(c) indicates the approximate equipotentials and field lines. (Note that, in bothtransformations, g ′ (z) iseither0or∞ at the origin, and so neither transformation isconformal there. Consequently there is no violation of result (ii), given at the start ofsection 24.7, concerning the angles between intersecting lines.) ◭The method of images, discussed in section 21.5, can be used in conjunction withconformal transformations to solve some problems involving Laplace’s equationin two dimensions.◮A wedge of angle π/α with its vertex at z =0is formed by two semi-infinite conductingplates, as shown in figure 25.4(a). A line charge of strength q per unit length is positionedat z = z 0 , perpendicular to the z-plane. By considering the transformation w = z α ,findthecomplex electrostatic potential for this situation.Let us consider the action of the transformation w = z α on the lines defining the positionsof the conducting plates. The plate that lies along the positive x-axis is mapped onto thepositive r-axis in the w-plane, whereas the plate that lies along the direction exp(iπ/α) ismapped into the negative r-axis, as shown in figure 25.4(b). Similarly the line charge at z 0is mapped onto the point w 0 = z0 α.From figure 25.4(b), we see that in the w-plane the problem can be solved by introducinga second line charge of opposite sign at the point w0 ∗ , so that the potential Φ = 0 alongthe r-axis. The complex potential for such an arrangement is simplyF(w) =−q ln(w − w 0 )+q ln(w − w2πɛ 0 2πɛ0).∗0878

25.3 LOCATION OF ZEROSφ =0yπ/αz 0w = z α sw 0φ =0xΦ=0Φ=0r(a)(b)w ∗ 0Figure 25.4 (a) An infinite conducting wedge with interior angle π/α and aline charge at z = z 0 ; (b) after the transformation w = z α , with an additionalimagechargeplacedatw = w0 ∗.Substituting w = z α into the above shows that the required complex potential in theoriginal z-plane isf(z) =q ( ) z α − z0∗αln. ◭2πɛ 0 z α − z0αIt should be noted that the appearance of a complex conjugate in the finalexpression is not in conflict with the general requirement that the complexpotential be analytic. It is z ∗ that must not appear; here, z0 ∗α is no more than aparameter of the problem.25.3 Location of zerosThe residue theorem, relating the value of a closed contour integral to the sum ofthe residues at the poles enclosed by the contour, was discussed in the previouschapter. One important practical use of an extension to the theorem is that oflocating the zeros of functions of a complex variable. The location of such zeroshas a particular application in electrical network and general oscillation theory,since the complex zeros of certain functions (usually polynomials) give the systemparameters (usually frequencies) at which system instabilities occur. As the basisof a method for locating these zeros we next prove three important theorems.(i) If f(z) has poles as its only singularities inside a closed contour C and isnot zero at any point on C then∮f ′ (z)C f(z) dz =2πi ∑ (N j − P j ). (25.14)jHere N j is the order of the jth zero of f(z) enclosed by C. Similarly P j is theorder of the jth pole of f(z) inside C.To prove this we note that, at each position z j , f(z) can be written asf(z) =(z − z j ) mj φ(z), (25.15)879

APPLICATIONS OF COMPLEX VARIABLESwhere φ(z) is analytic and non-zero at z = z j and m j is positive for a zero andnegative for a pole. Then the integrand f ′ (z)/f(z) takes the formf ′ (z)f(z) = m j+ φ′ (z)z − z j φ(z) . (25.16)Since φ(z j ) ≠ 0, the second term on the RHS is analytic; thus the integrand has asimple pole at z = z j , with residue m j . For zeros m j = N j and for poles m j = −P j ,and thus (25.14) follows from the residue theorem.(ii) If f(z) is analytic inside C and not zero at any point on it then2π ∑ jN j =∆ C [arg f(z)], (25.17)where ∆ C [x] denotes the variation in x around the contour C.Since f is analytic, there are no P j ; further, sincef ′ (z)f(z) = d [Ln f(z)], (25.18)dzequation (25.14) can be writtenHowever,2πi ∑ ∮N j =Cf ′ (z)f(z) dz =∆ C[Ln f(z)]. (25.19)∆ C [Ln f(z)] = ∆ C [ln |f(z)|]+i∆ C [arg f(z)], (25.20)and, since C is a closed contour, ln |f(z)| must return to its original value; so thereal term on the RHS is zero. Comparison of (25.19) and (25.20) then establishes(25.17), which is known as the principle of the argument.(iii) If f(z) and g(z) are analytic within and on a closed contour C and|g(z)| < |f(z)| on C then f(z) andf(z) +g(z) have the same number of zerosinside C; thisisRouché’s theorem.With the conditions given, neither f(z) norf(z)+g(z) can have a zero on C.So, applying theorem (ii) with an obvious notation,2π ∑ j N j(f + g) =∆ C [arg(f + g)]=∆ C [arg f]+∆ C [arg(1 + g/f)]=2π ∑ k N k(f)+∆ C [arg(1 + g/f)]. (25.21)Further, since |g| < |f| on C, 1+g/f always lies within a unit circle centredon z = 1; thus its argument always lies in the range −π/2 < arg(1 + g/f)

25.3 LOCATION OF ZEROSpolynomials, only the behaviour of a single term in the function need be consideredif the contour is chosen appropriately. For example, for a polynomial,f(z)+g(z) = ∑ N0 b iz i , only the properties of its largest power, taken as f(z), needbe investigated if a circular contour is chosen with radius R sufficiently large that,on the contour, the magnitude of the largest power term, |b N R N |, is greater thanthe sum of the magnitudes of all other terms. It is obvious that f(z) =b N z N hasN zeros inside |z| = R (all at the origin); consequently, f + g also has N zerosinside the same circle.The corresponding situation, in which only the properties of the polynomial’ssmallest power, again taken as f(z), need be investigated is a circular contourwith a radius R chosen sufficiently small that, on the contour, the magnitude ofthe smallest power term (usually the constant term in a polynomial) is greaterthan the sum of the magnitudes of all other terms. Then, a similar argument tothat given above shows that, since f(z) =b 0 has no zeros inside |z| = R, neitherdoes f + g.Aweakformofthemaximum-modulus theorem may also be deduced. Thisstates that if f(z) is analytic within and on a simple closed contour C then |f(z)|attains its maximum value on the boundary of C. The proof is as follows.Let |f(z)| ≤M on C with equality at at least one point of C. Now supposethat there is a point z = a inside C such that |f(a)| >M. Then the functionh(z) ≡ f(a) is such that |h(z)| > |−f(z)| on C, and thus, by Rouché’s theorem,h(z) andh(z) − f(z) have the same number of zeros inside C. Buth(z) (≡ f(a))has no zeros inside C, and, again by Rouché’s theorem, this would imply thatf(a) − f(z) has no zeros in C. However, f(a) − f(z) clearly has a zero at z = a,and so we have a contradiction; the assumption of the existence of a point z = ainside C such that |f(a)| >Mmust be invalid. This establishes the theorem.The stronger form of the maximum-modulus theorem, which we do not prove,states, in addition, that the maximum value of f(z) is not attained at any interiorpoint except for the case where f(z) is a constant.◮ Show that the four zeros of h(z) =z 4 + z +1 occur one in each quadrant of the Arganddiagram and that all four lie between the circles |z| =2/3 and |z| =3/2.Putting z = x and z = iy shows that no zeros occur on the real or imaginary axes. Theymust therefore occur in conjugate pairs, as can be shown by taking the complex conjugateof h(z) =0.Now take C as the contour OXY O shown in figure 25.5 and consider the changes∆[arg h] in the argument of h(z) asz traverses C.(i) OX: argh is everywhere zero, since h is real, and thus ∆ OX [arg h] =0.(ii) XY : z = R exp iθ andsoargh changes by an amount∆ XY [arg h] =∆ XY [arg z 4 ]+∆ XY [arg(1 + z −3 + z −4 )]{=∆ XY [arg R 4 e 4iθ ]+∆ XY arg[1+O(R −3 )] }=2π +O(R −3 ). (25.22)881

APPLICATIONS OF COMPLEX VARIABLESyYROXxFigure 25.5 A contour for locating the zeros of a polynomial that occur inthe first quadrant of the Argand diagram.(iii) YO: z = iy and so arg h =tan −1 y/(y 4 + 1), which starts at O(R −3 ) and finishes at0asy goes from large R to 0. It never reaches π/2 because y 4 +1=0hasnorealpositive root. Thus ∆ YO [arg h] =0.Hence for the complete contour ∆ C [arg h] =0+2π +0+O(R −3 ), and, if R is allowedto tend to infinity, we deduce from (25.17) that h(z) has one zero in the first quadrant.Furthermore, since the roots occur in conjugate pairs, a second root must lie in the fourthquadrant, and the other pair must lie in the second and third quadrants.To show that the zeros lie within the given annulus in the z-plane we apply Rouché’stheorem, as follows.(i) With C as |z| =3/2, f = z 4 , g = z +1.Now|f| =81/16 on C and |g| ≤1+|z|

25.4 SUMMATION OF SERIES◮By considering∮π cot πzC (a + z) dz, 2where a is not an integer and C is a circle of large radius, evaluate∞∑ 1(a + n) . 2n=−∞The integrand has (i) simple poles at z = integer n, for−∞

APPLICATIONS OF COMPLEX VARIABLESIm sLRe sλFigure 25.6 The integration path of the inverse Laplace transform is alongthe infinite line L. Thequantityλ must be positive and large enough for allpoles of the integrand to lie to the left of L.25.5 Inverse Laplace transformAs a further example of the use of contour integration we now discuss a methodwhereby the process of Laplace transformation, discussed in chapter 13, can beinverted.It will be recalled that the Laplace transform ¯f(s) ofafunctionf(x), x ≥ 0, isgiven by¯f(s) =∫ ∞0e −sx f(x) dx, Re s>s 0 . (25.24)In chapter 13, functions f(x) were deduced from the transforms by means of aprepared dictionary. However, an explicit formula for an unknown inverse maybe written in the form of an integral. It is known as the Bromwich integral and isgiven byf(x) = 12πi∫ λ+i∞λ−i∞e sx¯f(s) ds, λ > 0, (25.25)where s is treated as a complex variable and the integration is along the line Lindicated in figure 25.6. The position of the line is dictated by the requirementsthat λ is positive and that all singularities of ¯f(s) lie to the left of the line.That (25.25) really is the unique inverse of (25.24) is difficult to show for generalfunctions and transforms, but the following verification should at least make it884

25.5 INVERSE LAPLACE TRANSFORMΓRΓRΓRLLL(a) (b) (c)Figure 25.7 Some contour completions for the integration path L of theinverse Laplace transform. For details of when each is appropriate see themain text.plausible:∫ λ+i∞f(x) = 1∫ ∞ds e sx e −su f(u) du, Re(s) > 0, i.e. λ>0,2πi λ−i∞0= 1 ∫ ∞ ∫ λ+i∞du f(u) e s(x−u) ds2πi 0λ−i∞= 1 ∫ ∞ ∫ ∞du f(u) e λ(x−u) e ip(x−u) i dp, putting s = λ + ip,2πi 0−∞= 1 ∫ ∞f(u)e λ(x−u) 2πδ(x − u) du2π 0{ f(x) x ≥ 0,=(25.26)0 x

APPLICATIONS OF COMPLEX VARIABLESIdeally, we would like the contribution to the integral from the circular arc Γ totend to zero as its radius R →∞. Using a modified version of Jordan’s lemma,it may be shown that this is indeed the case if there exist constants M>0andα>0 such that on Γ|¯f(s)| ≤ M R α .Moreover, this condition always holds when ¯f(s) hastheform¯f(s) = P (s)Q(s) ,where P (s) andQ(s) are polynomials and the degree of Q(s) is greater than thatof P (s).When the contribution from the part-circle Γ tends to zero as R →∞,wehave from the residue theorem that the inverse Laplace transform (25.25) is givensimply byf(t) = ∑ (residues of ¯f(s)e sx at all poles ) . (25.27)◮Find the function f(x) whose Laplace transform iss¯f(s) =s 2 − k , 2where k is a constant.It is clear that ¯f(s) is of the form required for the integral over the circular arc Γ to tendto zero as R →∞, and so we may use the result (25.27) directly. Now¯f(s)e sx se sx=(s − k)(s + k) ,and thus has simple poles at s = k and s = −k. Using (24.57) the residues at each pole canbe easily calculated asR(k) = kekxand R(−k) = ke−kx2k2k .Thus the inverse Laplace transform is given by(f(x) = 1 2 e kx + e −kx) =coshkx.This result may be checked by computing the forward transform of cosh kx. ◭Sometimes a little more care is required when deciding in which half-plane toclose the contour C.◮Find the function f(x) whose Laplace transform is¯f(s) = 1 s (e−as − e −bs ),where a and b are fixed and positive, with b>a.From (25.25) we have the integralf(x) = 12πi∫ λ+i∞λ−i∞e (x−a)s − e (x−b)ssds. (25.28)886

25.5 INVERSE LAPLACE TRANSFORMf(x)1abxFigure 25.8b>a.The result of the Laplace inversion of ¯f(s) =s −1 (e −as −e −bs )withNow, despite appearances to the contrary, the integrand has no poles, as may be confirmedby expanding the exponentials as Taylor series about s = 0. Depending on the value of x,several cases arise.(i) For xb. (25.30)(iii) For a

APPLICATIONS OF COMPLEX VARIABLES25.6 Stokes’ equation and Airy integralsMuch of the analysis of situations occurring in physics and engineering is concernedwith what happens at a boundary within or surrounding a physical system.Sometimes the existence of a boundary imposes conditions on the behaviour ofvariables describing the state of the system; obvious examples include the zerodisplacement at its end-points of an anchored vibrating string and the zeropotential contour that must coincide with a grounded electrical conductor.More subtle are the effects at internal boundaries, where the same non-vanishingvariable has to describe the situation on either side of the boundary but itsbehaviour is quantitatively, or even qualitatively, different in the two regions. Inthis section we will study an equation, Stokes’ equation, whose solutions havethis latter property; as well as solutions written as series in the usual way, we willfind others expressed as complex integrals.The Stokes’ equation can be written in several forms, e.g.d 2 yd+ λxy =0;dx2 2 yd+ xy =0;dx2 2 ydx 2 = xy.We will adopt the last of these, but write it asd 2 y= zy (25.32)dz2 to emphasis that its complex solutions are valid for a complex independentvariable z, though this also means that particular care has to be exercised whenexamining their behaviour in different parts of the complex z-plane. The otherforms of Stokes’ equation can all be reduced to that of (25.32) by suitable(complex) scaling of the independent variable.25.6.1 The solutions of Stokes’ equationIt will be immediately apparent that, even for z restricted to be real and denotedby x, the behaviour of the solutions to (25.32) will change markedly as x passesthrough x = 0. For positive x they will have similar characteristics to the solutionsof y ′′ = k 2 y,wherek is real; these have monotonic exponential forms, eitherincreasing or decreasing. On the other hand, when x is negative the solutionswill be similar to those of y ′′ + k 2 y = 0, i.e. oscillatory functions of x. Thisisjust the sort of behaviour shown by the wavefunction describing light diffractedby a sharp edge or by the quantum wavefunction describing a particle near tothe boundary of a region which it is classically forbidden to enter on energygrounds. Other examples could be taken from the propagation of electromagneticradiation in an ion plasma or wave-guide.Let us examine in a bit more detail the behaviour of plots of possible solutionsy(z) of Stokes’ equation in the region near z = 0 and, in particular, what may888

25.6 STOKES’ EQUATION AND AIRY INTEGRALS(b)y(c)(a)(a)(c)z(b)Figure 25.9 Behaviour of the solutions y(z) ofStokes’equationnearz =0for various values of λ = −y ′ (0). (a) with λ small, (b) with λ large and (c) withλ appropriate to the Airy function Ai(z).happen in the region z>0. For definiteness and ease of illustration (see figure25.9), let us suppose that both y and z, and hence the derivatives of y, are real andthat y(0) is positive; if it were negative, our conclusions would not be changedsince equation (25.32) is invariant under y(z) →−y(z). The only difference wouldbe that all plots of y(z) would be reflected in the z-axis.We first note that d 2 y/dx 2 , and hence also the curvature of the plot, has thesame sign as z, i.e. it has positive curvature when z>0, for so long as y(z)remains positive there. What will happen to the plot for z>0 therefore dependscrucially on the value of y ′ (0). If this slope is positive or only slightly negativethe positive curvature will carry the plot, either immediately or ultimately, furtheraway from the z-axis. On the other hand, if y ′ (0) is negative but sufficiently largein magnitude, the plot will cross the y = 0 line; if this happens the sign of thecurvature reverses and again the plot will be carried ever further from the z-axis,only this time towards large negative values.Between these two extremes it seems at least plausible that there is a particularnegative value of y ′ (0) that leads to a plot that approaches the z-axis asymptotically,never crosses it (and so always has positive curvature), and has a slopethat, whilst always negative, tends to zero in magnitude. There is such a solution,known as Ai(z), whose properties we will examine further in the followingsubsections. The three cases are illustrated in figure 25.9.The behaviour of the solutions of (25.32) in the region z

APPLICATIONS OF COMPLEX VARIABLESforward, in that, whatever the sign of y at any particular point z, the curvaturealways has the opposite sign. Consequently the curve always bends towards thez-axis, crosses it, and then bends towards the axis again. Thus the curve exhibitsoscillatory behaviour. Furthermore, as −z increases, the curvature for any given|y| gets larger; as a consequence, the oscillations become increasingly more rapidand their amplitude decreases.25.6.2 Series solution of Stokes’ equationObtaining a series solution of Stokes’ equation presents no particular difficultywhen the methods of chapter 16 are used. The equation, written in the formd 2 y− zy =0,dz2 has no singular points except at z = ∞. Every other point in the z-plane isan ordinary point and so two linearly independent series expansions about it(formally with indicial values σ = 0 and σ = 1) can be found. Those about z =0take the forms ∑ ∞0 a nz n and ∑ ∞0 b nz n+1 . The corresponding recurrence relationsare(n +3)(n +2)a n+3 = a n and (n +4)(n +3)b n+3 = b n ,andthetwoseries(witha 0 = b 0 = 1) take the formsy 1 (z) =1+z3(3)(2) + z 6(6)(5)(3)(2) + ··· ,y 2 (z) =z +z4(4)(3) + z 7(7)(6)(4)(3) + ··· .The ratios of successive terms for the two series are thusa n+3 z n+3 z 3a n z n =(n +3)(n +2)andb n+3 z n+4 z 3b n z n+1 =(n +4)(n +3) .It follows from the ratio test that both series are absolutely convergent for all z.A similar argument shows that the series for their derivatives are also absolutelyconvergent for all z. Any solution of the Stokes’ equation is representable as asuperposition of the two series and so is analytic for all finite z; itisthereforeanintegral function with its only singularity at infinity.25.6.3 Contour integral solutionsWe now move on to another form of solution of the Stokes’ equation (25.32), onethat takes the form of a contour integral in which z appears as a parameter in890

25.6 STOKES’ EQUATION AND AIRY INTEGRALSthe integrand. Consider the contour integraly(z) =∫ baf(t) exp(zt) dt, (25.33)in which a, b and f(t) are all yet to be chosen. Note that the contour is in thecomplex t-plane and that the path from a to b can be distorted as required solong as no poles of the integrand are trapped between an original path and itsdistortion.Substitution of (25.33) into (25.32) yields∫ bat 2 f(t) exp(zt) dt =∫ bazf(t) exp(zt) dt= [ f(t) exp(zt) ] b a −df(t)exp(zt) dt.a dtIf we could choose the limits a and b so that the end-point contributions vanishthen Stokes’ equation would be satisfied by (25.33), provided f(t) satisfiesdf(t)+ t 2 f(t) =0dt⇒ f(t) =A exp(− 1 3 t3 ), (25.34)where A is any constant.To make the end-point contributions vanish we must choose a and b such thatexp(− 1 3 t3 + zt) = 0 for both values of t. This can only happen if |a| →∞and|b| →∞and, even then, only if Re (t 3 ) is positive. This condition is satisfied if2nπ − 1 2 π

APPLICATIONS OF COMPLEX VARIABLESIm tC 1C 2C 3Re tFigure 25.10 The contours used in the complex t-plane to define the functionsAi(z) andBi(z).though sometimes the sense of traversal is reversed. Consequently there arerelationships connecting Ai and Bi when the rotated variables are used as theirarguments. As two examples,Ai(z)+ΩAi(Ωz)+Ω 2 Ai(Ω 2 z)=0, (25.37)Bi(z) =i[Ω 2 Ai(Ω 2 z) − ΩAi(Ωz)]=e −πi/6 Ai(ze −2πi/3 )+e πi/6 Ai(ze 2πi/3 ).(25.38)Since the only requirements for the integral paths is that they start and end inthe correct sectors, we can distort path C 1 so that it lies on the imaginary axisfor virtually its whole length and just to the left of the axis at its two ends. Thisenables us to obtain an alternative expression for Ai(z), as follows.Setting t = is, wheres is real and −∞

25.6 STOKES’ EQUATION AND AIRY INTEGRALScontour integral (25.35) with the particular solution of Stokes’ equation thatdecays monotonically to zero for real z>0as|z| →∞. As discussed in subsection25.6.1, all solutions except the one called Ai(z) tendto±∞ as z (real) takes onincreasingly large positive values and so their asymptotic forms reflect this. In aworked example in subsection 25.8.2 we use the method of steepest descents (asaddle-point method) to show that the function defined by (25.39) has exactlythe characteristic asymptotic property expected of Ai(z) (see page 911). It followsthat it is the same function as Ai(z), up to a real multiplicative constant.The choice of definition (25.36) as the other named solution Bi(z) ofStokes’equation is a less obvious one. However, it is made on the basis of its behaviour fornegative real values of z. As discussed earlier, Ai(z) oscillates almost sinusoidallyin this region, except for a relatively slow increase in frequency and an evenslower decrease in amplitude as −z increases. The solution Bi(z) is chosen to bethe particular function that exhibits the same behaviour as Ai(z) exceptthatitisin quadrature with Ai, i.e. it is π/2 out of phase with it. Specifically, as x →−∞,Ai(x) ∼Bi(x) ∼(1 2|x|3/2√ sin + π ), (25.40)2πx1/4 3 4(1 2|x|3/2√ cos + π2πx1/4 3 4). (25.41)There is a close parallel between this choice and that of taking sine and cosinefunctions as the basic independent solutions of the simple harmonic oscillatorequation. Plots of Ai(z) andBi(z) forrealz are shown in figure 25.11.◮By choosing a suitable contour for C 1 in (25.35), express Ai(0) in terms of the gammafunction.With z set equal to zero, (25.35) takes the formAi(0) = 1 ∫exp(− 1 32πit3 ) dt.C 1We again use the freedom to choose the specific line of the contour so as to make theactual integration as simple as possible.HereweconsiderC 1 as made up of two straight-line segments: one along the linearg t =4π/3, starting at infinity in the correct sector and ending at the origin; the otherstarting at the origin and going to infinity along the line arg t =2π/3, thus ending in thecorrect final sector. On each, we set 1 3 t3 = s, wheres is real and positive on both lines.Then dt = e 4πi/3 (3s) −2/3 ds on the first segment and dt = e 2πi/3 (3s) −2/3 ds on the second.893

APPLICATIONS OF COMPLEX VARIABLES10.5−10−5z−0.5Figure 25.11z.The functions Ai(z) (full line) and Bi(z) (broken line) for realThen we haveAi(0) = 12πi∫ 0∞∫ ∞= 3−2/3e −s e 4πi/3 (3s) −2/3 ds + 12πi∫ ∞e −s (−e 4πi/3 + e 2πi/3 )s −2/3 ds2πi 0√3i 3−2/3∫ ∞=e −s s −2/3 ds2πi= 3−1/62π Γ( 1 3 ),00e −s e 2πi/3 (3s) −2/3 dswhere we have used the standard integral defining the gamma function in the last line. ◭Finally in this subsection we should mention that the Airy functions and theirderivatives are closely related to Bessel functions of orders ± 1 3 and ± 2 3and that894

25.7 WKB METHODSthere exist many representations, both as linear combinations and as indefiniteintegrals, of one in terms of the other. §25.7 WKB methodsThroughout this book we have had many occasions on which it has been necessaryto solve the equationd 2 ydx 2 + k2 0f(x)y = 0 (25.42)when the notionally general function f(x) has been, in fact, a constant, usuallythe unit function f(x) = 1. Then the solutions have been elementary and of theform A sin k 0 x or A cos k 0 x with arbitrary but constant amplitude A.Explicit solutions of (25.42) for a non-constant f(x) are only possible in alimited number of cases, but, as we will show, some progress can be made if f(x)is a slowly varying function of x, in the sense that it does not change much in arange of x of the order of k0 −1.We will also see that it is possible to handle situations in which f(x) iscomplex; this enables us to deal with, for example, the passage of waves throughan absorbing medium. Developing such solutions will involve us in finding theintegrals of some complex quantities, integrals that will behave differently in thevarious parts of the complex plane – hence their inclusion in this chapter.25.7.1 Phase memoryBefore moving on to the formal development of WKB methods we discuss theconcept of phase memory which is the underlying idea behind them.Let us first suppose that f(x) is real, positive and essentially constant overarangeofx and define n(x) as the positive square root of f(x); n(x) isthenalso real, positive and essentially constant over the same range of x. We adoptthis notation so that the connection can be made with the description of anelectromagnetic wave travelling through a medium of dielectric constant f(x)and, consequently, refractive index n(x). The quantity y(x) would be the electricor magnetic field of the wave. For this simplified case, in which we can omit the§ These relationships and many other properties of the Airy functions can be found in, for example,M. Abramowitz and I. A. Stegun (eds), Handbook of Mathematical Functions (New York: Dover,1965) pp. 446–50. So called because they were used, independently, by Wentzel, Kramers and Brillouin to tacklecertain wave-mechanical problems in 1926, though they had earlier been studied in some depth byJeffreys and used as far back as the first half of the nineteenth century by Green.895

APPLICATIONS OF COMPLEX VARIABLESx-dependence of n(x), the solution would be (as usual)y(x) =A exp(−ik 0 nx), (25.43)with both A and n constant. The quantity k 0 nx would be real and would be calledthe ‘phase’ of the wave; it increases linearly with x.As a first variation on this simple picture, we may allow f(x) to be complex,though, for the moment, still constant. Then n(x) is still a constant, albeit acomplex one:n = µ + iν.The solution is formally the same as before; however, whilst it still exhibitsoscillatory behaviour, the amplitude of the oscillations either grows or declines,depending upon the sign of ν:y(x) =A exp(−ik 0 nx) =A exp[ −ik 0 (µ + iν)x ]=A exp(k 0 νx) exp(−ik 0 µx).This solution with ν negative is the appropriate description for a wave travellingin a uniform absorbing medium. The quantity k 0 (µ + iν)x is usually called thecomplex phase of the wave.We now allow f(x), and hence n(x), to be both complex and varying withposition, though, as we have noted earlier, there will be restrictions on howrapidly f(x) may vary if valid solutions are to be obtained. The obvious extensionof solution (25.43) to the present case would bey(x) =A exp[ −ik 0 n(x)x ], (25.44)but direct substitution of this into equation (25.42) givesy ′′ + k 2 0n 2 y = −k 2 0(n ′2 x 2 +2nn ′ x)y − ik 0 (n ′′ x +2n ′ )y.Clearly the RHS can only be zero, as is required by the equation, if n ′ and n ′′ areboth very small, or if some unlikely relationship exists between them.To try to improve on this situation, we consider how the phase φ of the solutionchanges as the wave passes through an infinitesimal thickness dx of the medium.The infinitesimal (complex) phase change dφ for this is clearly k 0 n(x) dx, andtherefore will be∫ x∆φ = k 0 n(u) dufor a finite thickness x of the medium. This suggests that an improvement on(25.44) might be∫ x)y(x) =A exp(−ik 0 n(u) du . (25.45)0This is still not an exact solution as nowy ′′ (x)+k 2 0n 2 (x)y(x) =−ik 0 n ′ (x)y(x).0896

25.7 WKB METHODSThis still requires k 0 n ′ (x) to be small (compared with, say, k0 2n2 (x)), but is someimprovement (not least in complexity!) on (25.44) and gives some measure of theconditions under which the solution might be a suitable approximation.The integral in equation (25.45) embodies what is sometimes referred to as thephase memory approach; it expresses the notion that the phase of the wave-likesolution is the cumulative effect of changes it undergoes as it passes through themedium. If the medium were uniform the overall change would be proportionalto nx, as in (25.43); the extent to which it is not uniform is reflected in the amountby which the integral differs from nx.The condition for solution (25.45) to be a reasonable approximation can bewritten as n ′ k0 −1 ≪ n 2 or, in words, the change in n over an x-range of k0 −1 shouldbe small compared with n 2 . For light in an optical medium, this means that therefractive index n, which is of the order of unity, must change very little over adistance of a few wavelengths.For some purposes the above approximation is adequate, but for others furtherrefinement is needed. This comes from considering solutions that are still wavelikebut have amplitudes, as well as phases, that vary with position. These are theWKB solutions developed and studied in the next three subsections.25.7.2 Constructing the WKB solutionsHaving formulated the notion of phase memory, we now construct the WKBsolutions of the general equation (25.42), in which f(x) can now be both positiondependentand complex. As we have already seen, it is the possibility of a complexphase that permits the existence of wave-like solutions with varying amplitudes.Since n(x) is calculated as the square root of f(x), there is an ambiguity in itsoverall sign. In physical applications this is normally resolved unambiguously byconsiderations such as the inevitable increase in entropy of the system, but, so faras dealing with purely mathematical questions is concerned, the ambiguity mustbe borne in mind.The process we adopt is an iterative one based on the assumption that thesecond derivative of the complex phase with respect to x is very small and canbe approximated at each stage of the iteration. So we start with equation (25.42)and look for a solution of the formy(x) =A exp[ iφ(x)], (25.46)where A is a constant. When this is substituted into (25.42) the equation becomes[ ( ) ]2 dφ− + i d2 φdx dx 2 + k2 0n 2 (x) y(x) =0. (25.47)Setting the quantity in square brackets to zero produces a non-linear equation for897

APPLICATIONS OF COMPLEX VARIABLESwhich there is no obvious solution for a general n(x). However, on the assumptionthat d 2 φ/dx 2 is small, an iterative solution can be found.As a first approximation φ ′′ is ignored, and the solutiondφdx ≈±k 0n(x)is obtained. From this, differentiation gives an approximate value ford 2 φdx 2 ≈±k dn0dx ,which can be substituted into equation (25.47) to give, as a second approximationfor dφ/dx, the expression[] 1/2dφdx ≈± k0n 2 2 dn(x) ± ik 0dx(= ± k 0 n 1 ± i dn···)2k 0 n 2 dx +≈±k 0 n + i dn2n dx .This can now be integrated to give an approximate expression for φ(x) as follows:∫ xφ(x) =± k 0 n(u) du + i ln[ n(x)], (25.48)x 02where the constant of integration has been formally incorporated into the lowerlimit x 0 of the integral. Now, noting that exp(i 1 2 i ln n) =n−1/2 , substitution of(25.48) into equation (25.46) givesy ± (x) =A [ ∫ x]n exp ± ik 1/2 0 n(u) du(25.49)x 0as two independent WKB solutions of the original equation (25.42). This result isessentially the same as that in (25.45) except that the amplitude has been dividedby √ n(x), i.e. by [ f(x)] 1/4 .Sincef(x) may be complex, this may introduce anadditional x-dependent phase into the solution as well as the more obvious changein amplitude.◮Find two independent WKB solutions of Stokes’ equation in the formd 2 y+ λxy =0, with λ real and > 0.dx2 The form of the equation is the same as that in (25.42) with f(x) =x, andthereforen(x) =x 1/2 . The WKB solutions can be read off immediately using (25.49), so long aswe remember that although f(x) is real, it has four fourth roots and that therefore theconstant appearing in a solution can be complex. Two independent WKB solutions arey ± (x) = A ±|x| 1/4 exp [± i √ λ∫ x√ udu]= A ±|x| 1/4 exp [± i 2√ λ3 x3/2 ].898(25.50)

25.7 WKB METHODSThe precise combination of these two solutions that is required for any particular problemhas to be determined from the problem. ◭When Stokes’ equation is applied more generally to functions of a complexvariable, i.e. the real variable x is replaced by the complex variable z, ithassolutions whose type of behaviour depends upon where z lies in the complexplane. For the particular case λ = −1, when Stokes’ equation takes the formd 2 ydz 2 = zyand the two WKB solutions (with the inverse fourth root written explicitly) arey 1,2 (z) = A [1,2z exp ∓ 2 ]1/4 3 z3/2 , (25.51)one of the solutions, Ai(z) (see section 25.6), has the property that it is realwhenever z is real, whether positive or negative. For negative real z it hassinusoidal behaviour, but it becomes an evanescent wave for real positive z.Since the function z 3/2 has a branch point at z = 0 and therefore has an abrupt(complex) change in its argument there, it is clear that neither of the two functionsin (25.51), nor any fixed combination of them, can be equal to Ai(z) for all valuesof z. More explicitly, for z real and positive, Ai(z) is proportional to y 1 (z), whichis real and has the form of a decaying exponential function, whilst for z real andnegative, when z 3/2 is purely imaginary and y 1 (z) andy 2 (z) are both oscillatory,it is clear that Ai(z) must contain both y 1 and y 2 with equal amplitudes.The actual combinations of y 1 (z) andy 2 (z) needed to coincide with these twoasymptotic forms of Ai(z) are as follows.[1For z real and > 0, c 1 y 1 (z) =2 √ πz exp − 2 ]1/4 3 z3/2 . (25.52)For z real and < 0, c 2 [ y 1 (z)e iπ/4 − y 2 (z)e −iπ/4 ]=1√ π(−z)1/4 sin [ 23 (−z)3/2 + π 4]. (25.53)Therefore it must be the case that the constants used to form Ai(z) fromthe solutions (25.51) change as z moves from one part of the complex plane toanother. In fact, the changes occur for particular values of the argument of z;these boundaries are therefore radial lines in the complex plane and are knownas Stokes lines. For Stokes’ equation they occur when arg z is equal to 0, 2π/3 or4π/3.The general occurrence of a change in the arbitrary constants used to makeup a solution, as its argument crosses certain boundaries in the complex plane, isknown as the Stokes phenomenon and is discussed further in subsection 25.7.4.899

APPLICATIONS OF COMPLEX VARIABLES◮Apply the WKB method to the problem of finding the quantum energy levels E of aparticle of mass m bound in a symmetrical one-dimensional potential well V (x) that hasonly a single minimum. The relevant Schrödinger equation is− 2 d 2 ψ+ V (x)ψ = Eψ.2m dx2 Relate the problem close to each of the classical ‘turning points’, x = ± a at which E −V (x) =0, to Stokes’ equation and assume that it is appropriate to use the solution Ai(x)given in equations (25.52) and (25.53) at x = a. Show that if the general WKB solution inthe ‘classically allowed’ region −a

25.7 WKB METHODSfor some constant A. This form is only valid for negative z close to z = 0 and is notappropriate within the well as a whole, where the approximation (25.55) leading to Stokes’equation is not valid. However, it does allow us to determine the correct combination ofthe WKB solutions found earlier for the proper continuation inside the well of the solutionfound for z>0. This isψ 1 (x) = √ A (∫ asin k(u) du + π ).k(x) x 4A similar argument gives the continuation inside the well of the evanescent solutionrequired in the region x 0 in the range −a

APPLICATIONS OF COMPLEX VARIABLESThe result just proved gives∫ an√ 2mV0−a n(a 2sn − x 2s ) 1/2 dx =(n + 1 )π. 2Writing x = va n shows that the integral is proportional to a s+1n I s ,whereI s is the integralbetween −1 and +1 of (1 − v 2s ) 1/2 and does not depend upon n. Thus E n ∝ a 2sn anda s+1n ∝ (n + 1 ), implying that E 2 n ∝ (n + 1 2 )2s/s+1 .Although not asked for, we note that the above result indicates that, for a simpleharmonic oscillator, for which s = 1, the energy levels [ E n ∼ (n + 1 ) ] are equally spaced,2whilst for very large s, corresponding to a square well, the energy levels vary as n 2 .Bothof these results agree with what is found from detailed analyses of the individual cases. ◭25.7.3 Accuracy of the WKB solutionsWe may also ask when we can expect the WKB solutions to the Stokes’ equationto be reasonable approximations. Although our final form for the WKB solutionsis not exactly that used when the condition |n ′ k0 −1|≪|n2| was derived, it shouldgive the same order of magnitude restriction as a more careful analysis. For thederivation of (25.51), k0 2 = −1, n(z) =[f(z)]1/2 = z 1/2 , and the criterion becomes12 |z−1/2 |≪|z|, or, in round terms, |z| 3 ≫ 1.For the more general equation, typified by (25.42), the condition for the validityof the WKB solutions can usually be satisfied by making some quantity, often |z|,sufficiently large. Alternatively, a parameter such as k 0 can be made large enoughthat the validity criterion is satisfied to any pre-specified level. However, from apractical point of view, natural physical parameters cannot be varied at will, andrequiring z to be large may well reduce the value of the method to virtually zero.It is normally more useful to try to obtain an improvement on a WKB solutionby multiplying it by a series whose terms contain increasing inverse powers ofthe variable, so that the result can be applied successfully for moderate, and notjust excessively large, values of the variable.We do not have the space to discuss the properties and pitfalls of suchasymptotic expansions in any detail, but exercise 25.18 will provide the readerwith a model of the general procedure. A few particular points that should benoted are given as follows.(i) If the multiplier is analytic as z →∞, then it will be represented by aseries that is convergent for |z| greater than some radius of convergenceR.(ii) If the multiplier is not analytic as z →∞, as is usually the case, then themultiplier series eventually diverges and there is a z-dependent optimalnumber of terms that the series should contain in order to give the bestaccuracy.(iii) For a fixed value of arg z, the asymptotic expansion of the multiplier isunique. However, the same asymptotic expansion can represent more than902

25.7 WKB METHODSone function and the same function may need different expansions fordifferent values of arg z.Finally in this subsection we note that, although the form of equation (25.42) mayappear rather restrictive, in that it contains no term in y ′ , the results obtained sofar can be applied to an equation such asd 2 ydz 2 + P (z)dy + Q(z)y =0. (25.56)dzTo make this possible, a change of either the dependent or the independentvariable is made. For the former we write( ∫ 1 zY (z) =y(z)exp P (u) du)⇒d2 Y(Q2dz 2 + − 1 4 P 2 − 1 )dPY =0,2 dzwhilst for the latter we introduce a new independent variable ζ defined by( ∫dζz)( ) 2dz =exp − P (u) du ⇒d2 y dzdζ 2 + Q y =0.dζIn either case, equation (25.56) is reduced to the form of (25.42), though itwill be clear that the two sets of WKB solutions (which are, of course, onlyapproximations) will not be the same.25.7.4 The Stokes phenomenonAs we saw in subsection 25.7.2, the combination of WKB solutions of a differentialequation required to reproduce the asymptotic form of the accurate solution y(z)of the same equation, varies according to the region of the z-plane in which z lies.We now consider this behaviour, known as the Stokes phenomenon, in a littlemore detail.Let y 1 (z) andy 2 (z) be the two WKB solutions of a second-order differentialequation. Then any solution Y (z) of the same equation can be written asymptoticallyasY (z) ∼ A 1 y 1 (z)+A 2 y 2 (z), (25.57)where, although we will be considering (abrupt) changes in them, we will continueto refer to A 1 and A 2 as constants, as they are within any one region. In order toproduce the required change in the linear combination, as we pass over a Stokesline from one region of the z-plane to another, one of the constants must change(relative to the other) as the border between the regions is crossed.At first sight, this may seem impossible without causing a discernible discontinuityin the representation of Y (z). However, we must recall that the WKBsolutions are approximations, and that, as they contain a phase integral, forcertain values of arg z the phase φ(z) will be purely imaginary and the factors903

APPLICATIONS OF COMPLEX VARIABLESexp[± iφ(z) ] will be purely real. What is more, one such factor, known as thedominant term, will be exponentially large, whilst the other (the subdominant term)will be exponentially small. A Stokes line is precisely where this happens.We can now see how the change takes place without an observable discontinuityoccurring. Suppose that y 1 (z) is very large and y 2 (z) is very small on a Stokes line.Then a finite change in A 2 will have a negligible effect on Y (z); in fact, Stokesshowed, for some particular cases, that the change is less than the uncertaintyin y 1 (z) arising from the approximations made in deriving it. Since the solutionwith any particular asymptotic form is determined in a region bounded by twoStokes lines to within an overall multiplicative constant and the original equationis linear, the change in A 2 when one of the Stokes lines is crossed must beproportional to A 1 ,i.e.A 2 changes to A 2 + SA 1 ,whereS is a constant (the Stokesconstant) characteristic of the particular line but independent of A 1 and A 2 .Itshould be emphasised that, at a Stokes line, if the dominant term is not presentin a solution, then the multiplicative constant in the subdominant term cannotchange as the line is crossed.As an example, consider the Bessel function J 0 (z) ofzeroorder.Itissinglevalued,differentiable everywhere, and can be written as a series in powers of z 2 .Itis therefore an integral even function of z. However, its asymptotic approximationsfor two regions of the z-plane, Re z>0andz real and negative, are given byJ 0 (z) ∼ √ 1 1 √z(e iz e −iπ/4 + e −iz e iπ/4) , | arg(z)| < 12π2 π, | arg(z−1/2 )| < 1 4 π,J 0 (z) ∼ 1 √2π1 √z(e iz e 3iπ/4 + e −iz e iπ/4) , arg(z) =π, arg(z −1/2 )=− 1 2 π.We note in passing that neither of these expressions is naturally single-valued,and a prescription for taking the square root has to be given. Equally, neither isan even function of z. For our present purpose the important point to note isthat, for both expressions, on the line arg z = π/2 both z-dependent exponentsbecome real. For large |z| the second term in each expression is large; this is thedominant term, and its multiplying constant e iπ/4 is the same in both expressions.Contrarywise, the first term in each expression is small, and its multiplyingconstant does change, from e −iπ/4 to e 3iπ/4 ,asargz passes through π/2 whilstincreasing from 0 to π. It is straightforward to calculate the Stokes constant forthis Stokes line as follows:S = A 2(new) − A 2 (old)A 1= e3iπ/4 − e −iπ/4e iπ/4= e iπ/2 − e −iπ/2 =2i.If we had moved (in the negative sense) from arg z = 0 to arg z = −π, the relevantStokes line would have been arg z = −π/2. There the first term in each expressionis dominant, and it would have been the constant e iπ/4 in the second term thatwould have changed. The final argument of z −1/2 would have been +π/2.904

25.8 APPROXIMATIONS TO INTEGRALSFinally, we should mention that the lines in the z-plane on which the exponentsin the WKB solutions are purely imaginary, and the two solutions haveequal amplitudes, are usually called the anti-Stokes lines. For the general Bessel’sequation they are the real positive and real negative axes.25.8 Approximations to integralsIn this section we will investigate a method of finding approximations to thevalues or forms of certain types of infinite integrals. The class of integrals tobe considered is that containing integrands that are, or can be, represented byexponential functions of the general form g(z) exp[ f(z) ]. The exponents f(z) maybe complex, and so integrals of sinusoids can be handled as well as those withmore obvious exponential properties. We will be using the analyticity propertiesof the functions of a complex variable to move the integration path to a partof the complex plane where a general integrand can be approximated well by astandard form; the standard form is then integrated explicitly.The particular standard form to be employed is that of a Gausssian function ofa real variable, for which the integral between infinite limits is well known. Thisform will be generated by expressing f(z) as a Taylor series expansion about apoint z 0 , at which the linear term in the expansion vanishes, i.e. where f ′ (z) =0.Then, apart from a constant multiplier, the exponential function will behave likeexp[ 1 2 f′′ (z 0 )(z − z 0 ) 2 ] and, by choosing an appropriate direction for the contourto take as it passes through the point, this can be made into a normal Gaussianfunction of a real variable and its integral may then be found.25.8.1 Level lines and saddle pointsBefore we can discuss the method outlined above in more detail, a number ofobservations about functions of a complex variable and, in particular, about theproperties of the exponential function need to be made. For a general analyticfunction,f(z) =φ(x, y)+iψ(x, y), (25.58)of the complex variable z = x + iy, we recall that, not only do both φ and ψsatisfy Laplace’s equation, but ∇φ and ∇ψ are orthogonal. This means that thelines on which one of φ and ψ is constant are exactly the lines on which the otheris changing most rapidly.Let us apply these observations to the functionh(z) ≡ exp[ f(z) ] = exp(φ) exp(iψ), (25.59)recalling that the functions φ and ψ are themselves real. The magnitude of h(z),given by exp(φ), is constant on the lines of constant φ, which are known as the905

APPLICATIONS OF COMPLEX VARIABLESFigure 25.12 A greyscale plot with associated contours of the value of |h(z)|,where h(z) =exp[i(z 3 +6z 2 − 15z + 8)], in the neighbourhood of one of itssaddle points; darker shading corresponds to larger magnitudes. The plot alsoshows the two level lines (thick solid lines) through the saddle and part of theline of steepest descents (dashed line) passing over it. At the saddle point, theangle between the line of steepest descents and a level line is π/4.level lines of the function. It follows that the direction in which the magnitudeof h(z) changes most rapidly at any point z is in a direction perpendicular tothe level line passing through that point. This is therefore the line through z onwhich the phase of h(z), namely ψ(z), is constant. Lines of constant phase aretherefore sometimes referred to as lines of steepest descent (or steepest ascent).We further note that |h(z)| can never be negative and that neither φ nor ψcan have a finite maximum at any point at which f(z) is analytic. This latterobservation follows from the fact that at a maximum of, say, φ(x, y), both ∂ 2 φ/∂x 2and ∂ 2 φ/∂y 2 would have to be negative; if this were so, Laplace’s equation couldnot be satisfied, leading to a contradiction. A similar argument shows that aminimum of either φ or ψ is not possible wherever f(z) is analytic. A morepositive conclusion is that, since the two unmixed second partial derivatives∂ 2 φ/∂x 2 and ∂ 2 φ/∂y 2 must have opposite signs, the only possible conclusionabout a point at which ∇φ is defined and equal to zero is that the point is asaddle point of h(z). An example of a saddle point is shown as a greyscale plotin figure 25.12 and, more pictorially, in figure 5.2.906

25.8 APPROXIMATIONS TO INTEGRALSFrom the observations contained in the two previous paragraphs, we deducethat a path that follows the lines of steepest descent (or ascent) can never forma closed loop. On such a path, φ, and hence |h(z)|, must continue to decrease(increase) until the path meets a singularity of f(z). It also follows that if a levelline of h(z) forms a closed loop in the complex plane, then the loop must enclosea singularity of f(z). This may (if φ →∞) or may not (if φ →−∞) produce asingularity in h(z).We now turn to the study of the behaviour of h(z) at a saddle point and howthis enables us to find an approximation to the integral of h(z) alongacontourthat can be deformed to pass through the saddle point. At a saddle point z 0 ,atwhich f ′ (z 0 ) = 0, both ∇φ and ∇ψ are zero, and consequently the magnitude andphase of h(z) are both stationary. The Taylor expansion of f(z) at such a pointtakes the formf(z) =f(z 0 )+0+ 1 2! f′′ (z 0 )(z − z 0 ) 2 +O(z − z 0 ) 3 . (25.60)We assume that f ′′ (z 0 ) ≠ 0 and write it explicitly as f ′′ (z 0 ) ≡ Ae iα , thus definingthe real quantities A and α. Ifithappensthatf ′′ (z 0 ) = 0, then two or moresaddle points coalesce and the Taylor expansion must be continued until the firstnon-vanishing term is reached; we will not consider this case further, though thegeneral method of proceeding will be apparent from what follows. If we alsoabbreviate the (in general) complex quantity f(z 0 )tof 0 , then (25.60) takes theformf(z) =f 0 + 1 2 Aeiα (z − z 0 ) 2 +O(z − z 0 ) 3 . (25.61)To study the implications of this approximation for h(z), we write z − z 0 asρe iθ with ρ and θ both real. Then|h(z)| = | exp(f 0 )| exp[ 1 2 Aρ2 cos(2θ + α)+O(ρ 3 )]. (25.62)This shows that there are four values of θ for which |h(z)| is independent of ρ (tosecond order). These therefore correspond to two crossing level lines given by(θ = 1 2 ±12 π − α) (and θ = 1 2 ±32 π − α) . (25.63)The two level lines cross at right angles to each other. It should be noted thatthe continuations of the two level lines away from the saddle are not straightin general. At the saddle they have to satisfy (25.63), but away from it the linesmust take whatever directions are needed to make ∇φ = 0. In figure 25.12 one ofthe level lines (|h| = 1) has a continuation (y = 0) that is straight; the other doesnot and bends away from its initial direction x =1.So far as the phase of h(z) is concerned, we havearg[ h(z)]=arg(f 0 )+ 1 2 Aρ2 sin(2θ + α)+O(ρ 3 ),which shows that there are four other directions (two lines crossing at right907

APPLICATIONS OF COMPLEX VARIABLESangles) in which the phase of h(z) is independent of ρ. They make angles of π/4with the level lines through z 0 and are given byθ = − 1 2 α, θ = 1 2 (±π − α), θ = π − 1 2 α.From our previous discussion it follows that these four directions will be thelines of steepest descent (or ascent) on moving away from the saddle point. Inparticular, the two directions for which the term cos(2θ + α) in (25.62) is negativewill be the directions in which |h(z)| decreases most rapidly from its value at thesaddle point. These two directions are antiparallel, and a steepest descents pathfollowing them is a smooth locally straight line passing the saddle point. It isknown as the line of steepest descents (l.s.d.) through the saddle point. Note that‘descents’ is plural as on this line the value of |h(z)| decreases on both sides of thesaddle. This is the line which we will make the path of the contour integral ofh(z) follow. Part of a typical l.s.d. is indicated by the dashed line in figure 25.12.25.8.2 Steepest descents methodTo help understand how an integral along the line of steepest descents can behandled in a mechanical way, it is instructive to consider the case where thefunction f(z) =−βz 2 and h(z) =exp(−βz 2 ). The saddle point is situated atz = z 0 = 0, with f 0 = f(z 0 )=1andf ′′ (z 0 )=−2β, implying that A =2|β| andα = ±π +argβ, withthe± sign chosen to put α in the range 0 ≤ α

25.8 APPROXIMATIONS TO INTEGRALSany of the continuations to infinity of the level lines that pass through the saddle.In practical applications the start- and end-points of the path are nearly alwaysat singularities of f(z) with Re f(z) →−∞and |h(z)| →0.We now set out the complete procedure for the simplest form of integralevaluation that uses a method of steepest descents. Extensions, such as includinghigher terms in the Taylor expansion or having to pass through more than onesaddle point in order to have appropriate termination points for the contour, canbe incorporated, but the resulting calculations tend to be long and complicated,and we do not have space to pursue them in a book such as this one.As our general integrand we take a function of the form g(z)h(z), where, asbefore, h(z) =exp[f(z) ]. The function g(z) should neither vary rapidly nor havezeros or singularities close to any saddle point used to evaluate the integral.Rapidly varying factors should be incorporated in the exponent, usually in theform of a logarithm. Provided g(z) satisfies these criteria, it is sufficient to treatit as a constant multiplier when integrating, assigning to it its value at the saddlepoint, g(z 0 ).Incorporating this and retaining only the first two non-vanishing terms inequation (25.61) gives the integrand asg(z 0 ) exp(f 0 ) exp[ 1 2 Aeiα (z − z 0 ) 2 ]. (25.64)From the way in which it was defined, it follows that on the l.s.d. the imaginarypart of f(z) isconstant(=Imf 0 ) and that the final exponent in (25.64) is eitherzero (at z 0 ) or negative. We can therefore write it as −s 2 ,wheres is real. Further,since exp[ f(z)] → 0 at the start- and end-points of the contour, we must havethat s runs from −∞ to +∞, the sense of s being chosen so that it is negativeapproaching the saddle and positive when leaving it.Making this change of variable,12 Aeiα (z − z 0 ) 2 = −s 2 , with dz = ±√2A exp[ 1 2 i(π − α)]ds, (25.65)allows us to express the contribution to the integral from the neighbourhood ofthe saddle point as√ ∫ 2 ∞±g(z 0 ) exp(f 0 )A exp[ 1 2 i(π − α)] exp(−s 2 ) ds.−∞The simple saddle-point approximation assumes that this is the only contribution,and gives as the value of the contour integral∫√2πg(z) exp[ f(z)]dz = ±CA g(z 0) exp(f 0 ) exp[ 1 2 i(π − α)], (25.66)where we have used the standard result that ∫ ∞−∞ exp(−s2 ) ds = √ π. The overall ±909

APPLICATIONS OF COMPLEX VARIABLESsign is determined by the direction θ in the complex plane in which the distortedcontour passes through the saddle point. If − 1 2 π

25.8 APPROXIMATIONS TO INTEGRALSto the l.s.d., i.e. on the imaginary t-axis, provides some reassurance. Whether µ ispositive or negative,h(5 + iµ) = exp(50 + 10iµ − 25 − 10iµ + µ 2 )=exp(25+µ 2 ).This is greater than h(5) for all µ and increases as |µ| increases, showing thatthe integration path really does lie at a minimum of h(t) for a traversal in thisdirection.We now give a fully worked solution to a problem that could not be easilytackled by elementary means.◮ Apply the saddle-point method to the function defined by∫ ∞F(x) = 1 cos( 1 3πs3 + xs) ds0to show that its form for large positive real x is one that tends asymptotically to zero, henceenabling F(x) to be identified with the Airy function, Ai(x).We first express the integral as an exponential function and then make the change ofvariable s = x 1/2 t to bring it into the canonical form ∫ g(t)exp[f(t)]dt as follows:∫ ∞F(x) = 1 cos( 1 3πs3 + xs) ds0= 1 ∫ ∞exp[ i( 1 32πs3 + xs)]ds−∞= 1 ∫ ∞x 1/2 exp[ ix 3/2 ( 1 32πt3 + t)]dt.−∞We now seek to find an approximate expression for this contour integral by deforming itspath along the real t-axis into one passing over a saddle point of the integrand. Consideredas a function of t, the multiplying factor x 1/2 /2π is a constant, and any effects due to theproximity of its zeros and singularities to any saddle point do not arise.The saddle points are situated where0=f ′ (t) =ix 3/2 (t 2 +1) ⇒ t = ±i.For reasons discussed later, we choose to use the saddle point at t = t 0 = i . At this point,f(i) =ix 3/2 (− 1 i + i) =− 2 3 3 x3/2 and Ae iα ≡ f ′′ (i) =ix 3/2 (2i) =−2x 3/2 ,and so A =2x 3/2 and α = π.Now, expanding f(t) around t = i by setting t = i + ρe iθ , we havef(t) =f(i)+0+ 1 2! f′′ (i)(t − i) 2 +O[(t − i) 3 ]= − 2 3 x3/2 + 1 2 2x3/2 e iπ ρ 2 e 2iθ +O(ρ 3 ).For the l.s.d. contour that crosses the saddle point we need the second term in this lastline to decrease as ρ increases. This happens if π +2θ = ±π, i.e.ifθ =0orθ = −π (or+π); thus, the l.s.d. through the saddle is orientated parallel to the real t-axis. Given theinitial contour direction, the deformed contour should approach the saddle point from thedirection θ = −π and leave it along the line θ =0.Since−π/2 < 0 ≤ π/2, the overall signof the ‘omnibus’ approximation formula is determined as positive.911

APPLICATIONS OF COMPLEX VARIABLESFinally, putting the various values into the formula yields( ) 1/2 2πF(x) ∼ + g(i)exp[f(i)]exp[ 1 i(π − α)]2A( ) 1/2 (2π x 1/2=+2x 3/2 2π exp − 2 )3 x3/2 exp[ 1 i(π − π)]2(1=2 √ πx exp − 2 )1/4 3 x3/2 .This is the leading term in the asymptotic expansion of F(x),which,asshowninequation(25.39), is a particular contour integral solution of Stokes’ equation. The fact that it tendsto zero in a monotonic way as x → +∞ allows it to be identified with the Airy function,Ai(x).We may ask why the saddle point at t = −i was not used. The answer to this is asfollows. Of course, any path that starts and ends in the right sectors will suffice, butif another saddle point exists close to the one used, then the Taylor expansion actuallyemployed is likely to be less effective than if there were no other saddle points or if therewere only distant ones.An investigation of the same form as that used at t =+i shows that the saddle at t = −iis higher by a factor of exp( 4 3 x3/2 ) and that its l.s.d. is orientated parallel to the imaginaryt-axis. Thus a path that went through it would need to go via a region of largish negativeimaginary t, over the saddle at t = −i, and then, when it reached the col at t =+i, bendsharply and follow part of the same l.s.d. as considered earlier. Thus the contribution fromthe t = −i saddle would be incomplete and roughly half of that from the t =+i saddlewould still have to be included. The more serious error would come from the first of these,as, clearly, the part of the path that lies in the plane Re t = 0 is not symmetric and is farfrom Guassian-like on the side nearer the origin. The Gaussian-path approximation usedwill therefore not be a good one, and, what is more, the resulting error will be magnified bya factor exp( 4 3 x3/2 ) compared with the best estimate. So, both on the grounds of simplicityand because the effect of the other (neglected) saddle point is likely to be less severe, wechoose to use the one at t =+i. ◭25.8.3 Stationary phase methodIn the previous subsection we showed how to use the saddle points of anexponential function of a complex variable to evaluate approximately a contourintegral of that function. This was done by following the lines of steepest descentthat passed through the saddle point; these are lines on which the phase of theexponential is constant but its amplitude is varying at the maximum possiblerate for that function. We now introduce an alternative method, one that entirelyreverses the roles of amplitude and phase. To see how such an alternative approachmight work, it is useful to study how the integral of an exponential function ofa complex variablecan be represented as the sum of infinitesimal vectors in thecomplex plane.We start by studying the familiar integralI 0 =∫ ∞−∞exp(−z 2 ) dz, (25.67)912

25.8 APPROXIMATIONS TO INTEGRALSwhich we already know has the value √ π when z is real. This choice of demonstrationmodel is not accidental, but is motivated by the fact that, as we havealready shown, in the neighbourhood of a saddle point all exponential integrandscan be approximated by a Gaussian function of this form.The same integral can also be thought of as an integral in the complex plane,in which the integration contour happens to be along the real axis. Since theintegrand is analytic, the contour could be distorted into any other that had thesame end-points, z = −∞ and z =+∞, both on the real axis.As a particular possibility, we consider an arc of a circle of radius R centredon z =0.Itiseasilyshownthatcos2θ ≥ 1+4θ/π for −π/4

APPLICATIONS OF COMPLEX VARIABLESfrom which it follows that C(∞) =S(∞) = 1 2 . Clearly, C(−∞) =S(−∞) =− 1 2 .We are now in a position to examine these two equivalent ways of evaluating I 0in∫terms of sums of infinitesmal vectors in the complex plane. When the integral∞−∞ exp(−z2 ) dz is evaluated as a real integral, or a complex one along the realz-axis, each element dz generates a vector of length exp(−z 2 ) dz in an Arganddiagram, usually called the amplitude–phase diagram for the integral. For thisintegration, whilst all vector contributions lie along the real axis, they do differ inmagnitude, starting vanishingly small, growing to a maximum length of 1 × dz,and then reducing until they are again vanishingly small. At any stage, theirvector sum (in this case, the same as their algebraic sum) is a measure of theindefinite integral∫ xI(x) = exp(−z 2 ) dz. (25.70)−∞The total length of the vector sum when x →∞is, of course, √ π, and it shouldnot be overlooked that the sum is a vector parallel to (actually coinciding with)the real axis in the amplitude–phase diagram. Formally this indicates that theintegral is real. This ‘ordinary’ view of evaluating the integral generates the sameamplitude–phase diagram as does the method of steepest descents. This is becausefor this particular integrand the l.s.d. never leaves the real axis.Now consider the same integral evaluated using the form of equation (25.69).Here, each contribution, as the integration variable goes from u to u + du, isofthe formg(u) du =cos( 1 2 πu2 ) du + i sin( 1 2 πu2 ) du.As infinitesimal vectors in the amplitude–phase diagram, all g(u) du have the samemagnitude du, but their directions change continuously. Near u =0,whereu 2is small, the change is slow and each vector element is approximately equal to√2π exp(−iπ/4) du; these contributions are all in phase and add up to a significantvector contribution in the direction θ = −π/4. This is illustrated by the centralpart of the curve in part (b) of figure 25.13, in which the amplitude–phase diagramfor the ‘ordinary’ integration, discussed above, is drawn as part (a).Part (b) of the figure also shows that the vector representing the indefiniteintegral (25.70) initially (s large and negative) spirals out, in a clockwise sense,from around the point 0 + i0 in the amplitude–phase diagram and ultimately (slarge and positive) spirals in, in an anticlockwise direction, to the point √ π + i0.The total curve is called a Cornu spiral. In physical applications, such as thediffraction of light at a straight edge, the relevant limits of integration aretypically −∞ and some finite value x. Then, as can be seen, the resulting vectorsum is complex in general, with its magnitude (the distance from 0 + i0 tothepoint on the spiral corresponding to z = x) growing steadily for x0.914

25.8 APPROXIMATIONS TO INTEGRALS(a)(b)π/4β(c)√ π∫ ∞Figure 25.13 Amplitude–phase diagrams for the integral−∞ exp(−z2 ) dzusing different contours in the complex z-plane. (a) Using the real axis, as inthe steepest descents method. (b) Using the level line z = u exp(− 1 iπ) that4passes through the saddle point, as in the stationary phase method. (c) Usinga path that makes a positive angle β (< π/4) with the z-axis.The final curve, 25.13(c), shows the amplitude-phase diagram corresponding toan integration path that is along a line making a positive angle β (0

APPLICATIONS OF COMPLEX VARIABLESstationary, the magnitude of any factor, g(z), multiplying the exponential function,exp[ f(z)]∼ exp[ Ae iα (z − z 0 ) 2 ], is at least comparable to its magnitude elsewhere,then this result can be used to obtain an approximation to the value of theintegral of h(z) =g(z) exp[ f(z) ]. This is the basis of the method of stationaryphase.Returning to the behaviour of a function exp[ f(z) ] at one of its saddle points,we can now see how the considerations of the previous paragraphs can be appliedthere. We already know, from equation (25.62) and the discussion immediatelyfollowing it, that in the equationh(z) ≈ g(z 0 ) exp(f 0 )exp{ 1 2 Aρ2 [cos(2θ + α)+i sin(2θ + α)]}(25.71)the second exponent is purely imaginary on a level line, and equal to zero atthe saddle point itself. What is more, since ∇ψ = 0 at the saddle, the phase isstationary there; on one level line it is a maximum and on the other it is aminimum. As there are two level lines through a saddle point, a path on whichthe amplitude of the integrand is constant could go straight on at the saddlepoint or it could turn through a right angle. For the moment we assume that itruns continuously through the saddle.On the level line for which the phase at the saddle point is a minimum, we canwrite the phase of h(z) as approximatelyarg g(z 0 )+Imf 0 + v 2 ,where v is real, iv 2 = 1 2 Aeiα (z − z 0 ) 2 and, as previously, Ae iα = f ′′ (z 0 ). Thene iπ/4 dv = ±√A2 eiα/2 dz, (25.72)leading to an approximation to the integral of∫∫ ∞ Ah(z) dz ≈±g(z 0 ) exp(f 0 ) exp(iv )√ 2 2 exp[ i( 1 4 π − 1 2 α)]dv−∞= ± g(z 0 ) exp(f 0 ) √ π exp(iπ/4)√A2 exp[ i( 1 4 π − 1 2 α)]√2π= ±A g(z 0) exp(f 0 ) exp[ 1 2i(π − α)]. (25.73)Result (25.68) was used to obtain the second line above. The ± ambiguity is againresolved by the direction θ of the contour; it is positive if −3π/4

25.8 APPROXIMATIONS TO INTEGRALSare self-cancelling, as discussed previously. However, the ends of the contour mustbe in regions where the integrand is vanishingly small, and so at each end ofthe level line we need to add a further section of path that makes the contourterminate correctly.Fortunately, this can be done without adding to the value of the integral. Thisis because, as noted in the second paragraph following equation (25.67), far fromthe saddle the level line will be at a finite height up a ‘precipitous cliff’ thatseparates the region where the integrand grows without limit from the one whereit tends to zero. To move down the cliff-face into the zero-level valley requiresan ever smaller step the further we move away from the saddle; as the integrandis finite, the contribution to the integral is vanishingly small. In figure 25.12, thisadditional piece of path length might, for example, correspond to the infinitesimalmove from a point on the large positive x-axis (where h(z) has value 1) to a pointjust above it (where h(z) ≈ 0).Now that formula (25.73) has been justified, we may note that it is exactlythe same as that for the method of steepest descents, equation (25.66). A similarcalculation using the level line on which the phase is a maximum also reproducesthe steepest-descents formula. It would appear that ‘all roads lead to Rome’.However, as we explain later, some roads are more difficult than others. Wherea problem involves using more than one saddle point, if the steepest-descentsapproach is tractable, it will usually be the more straight forward to apply.Typical amplitude-phase diagrams for an integration along a level line thatgoes straight through the saddle are shown in parts (a) and (b) of figure 25.14.The value of the integral is given, in both magnitude and phase, by the vectorv joining the initial to the final winding points and, of course, is the same inboth cases. Part (a) corresponds to the case of the phase being a minimum at thesaddle; the vector path crosses v at an angle of −π/4. When a path on which thephase at the saddle is a maximum is used, the Cornu spiral is as in part (b) ofthe figure; then the vector path crosses v at an angle of +π/4. As can be seen,the two spirals are mirror images of each other.Clearly a straight-through level line path will start and end in different zerolevelvalleys. For one that turns through a right angle at the saddle point, theend-point could be in a different valley (for a function such as exp(−z 2 ), there isonly one other) or in the same one. In the latter case the integral will give a zerovalue, unless a singularity of h(z) happens to have been enclosed by the contour.Parts (c) and (d) of figure 25.14 illustrate the phase–amplitude diagrams for thesetwo cases. In (c) the path turns through a right angle (+π/2, as it happens) at thesaddle point, but finishes up in a different valley from that in which it started. In(d) it also turns through a right angle but returns to the same valley, albeit closeto the other precipice from that near its starting point. This makes no differenceand the result is zero, the two half spirals in the diagram producing resultantsthat cancel.917

APPLICATIONS OF COMPLEX VARIABLESvv(a)(b)v(c)(d)Figure 25.14 Amplitude–phase diagrams for stationary phase integration. (a)Using a straight-through path on which the phase is a minimum. (b) Usinga straight-through path on which the phase is a maximum. (c) Using a levelline that turns through +π/2 at the saddle point but starts and finishes indifferent valleys. (d) Using a level line that turns through a right angle butfinishes in the same valley as it started. In cases (a), (b) and (c) the integralvalue is represented by v (see text). In case (d) the integral has value zero.We do not have the space to consider cases with two or more saddle points,but even more care is needed with the stationary phase approach than whenusing the steepest-descents method. At a saddle point there is only one l.s.d. butthere are two level lines. If more than one saddle point is required to reach theappropriate end-point of an integration, or an intermediate zero-level valley hasto be used, then care is needed in linking the corresponding level lines in such away that the links do not make a significant, but unknown, contribution to theintegral. Yet more complications can arise if a level line through one saddle pointcrosses a line of steepest ascent through a second saddle.We conclude this section with a worked example that has direct links to thetwo preceding sections of this chapter.918

25.8 APPROXIMATIONS TO INTEGRALS◮In the worked example in subsection 25.8.2 the function∫ ∞F(x) = 1 cos( 1 3πs3 + xs) ds(∗)0was shown to have the properties associated with the Airy function, Ai(x), whenx>0. Usethe stationary phase method to show that, for x

APPLICATIONS OF COMPLEX VARIABLESAV 0 e iωt˜LCDRI RECLFigure 25.15 The inductor–capacitor–resistor network for exercise 25.1.Bwhich can also be simplified, and gives[ (12+2 √ π(−x) exp i1/4 3 (−x)3/2 − π )].4Adding the two contributions and taking the real part of the sum, though this is notnecessary here because the sum is real anyway, we obtain(22F(x) =2 √ π(−x) cos 1/4 3 (−x)3/2 − π )4(1 2= √ sin π(−x)1/43 (−x)3/2 + π ),4in agreement with the asymptotic form given in (25.53). ◭25.9 Exercises25.1 In the method of complex impedances for a.c. circuits, an inductance L isrepresented by a complex impedance Z L = iωL and a capacitance C by Z C =1/(iωC). Kirchhoff’s circuit laws,∑I i = 0 at a node and ∑ Z i I i = ∑ V j around any closed loop,iijare then applied as if the circuit were a d.c. one.Apply this method to the a.c. bridge connected as in figure 25.15 to showthat if the resistance R is chosen as R =(L/C) 1/2 then the amplitude of thecurrent, I R , through it is independent of the angular frequency ω of the applieda.c. voltage V 0 e iωt .Determine how the phase of I R , relative to that of the voltage source, varieswith the angular frequency ω.25.2 A long straight fence made of conducting wire mesh separates two fields andstands one metre high. Sometimes, on fine days, there is a vertical electric fieldover flat open countryside. Well away from the fence the strength of the field isE 0 . By considering the effect of the transformation w =(1−z 2 ) 1/2 on the real and920

25.9 EXERCISESimaginary z-axes, find the strengths of the field (a) at a point one metre directlyabove the fence, (b) at ground level one metre to the side of the fence, and (c)at a point that is level with the top of the fence but one metre to the side of it.What is the direction of the field in case (c)?25.3 For the function( ) z + cf(z) =ln ,z − cwhere c is real, show that the real part u of f is constant on a circle of radiusc cosech u centred on the point z = c coth u. Use this result to show that theelectrical capacitance per unit length of two parallel cylinders of radii a, placedwith their axes 2d apart, is proportional to [cosh −1 (d/a)] −1 .25.4 Find a complex potential in the z-plane appropriate to a physical situation inwhich the half-plane x>0, y = 0 has zero potential and the half-plane xa, s =0andthehalf-plane r

APPLICATIONS OF COMPLEX VARIABLESThen, by the principle of the argument, the number of zeros inside C is given bythe integer (2π) −1 ∆ C [arg f(z)].It can be shown that the zeros of z 4 + z + 1 lie one in each quadrant. Use theabove method to show that the zeros in the second and third quadrants have|z| < 1.25.9 Prove that∞∑ 1n 2 + 3 n + =4π.1−∞ 4 8Carry out the summation numerically, say between −4 and 4, and note howmuch of the sum comes from values near the poles of the contour integration.25.10 This exercise illustrates a method of summing some infinite series.(a) Determine the residues at all the poles of the functionπ cot πzf(z) =a 2 + z , 2where a is a positive real constant.(b) By evaluating, in two different ways, the integral I of f(z) along the straightline joining −∞ − ia/2 and+∞−ia/2, show that∞∑ 1 π coth πa= − 1a 2 + n2 2a 2a . 2 n=1(c) Deduce the value of ∑ ∞1 n−2 .25.11 By considering the integral of( ) 2 sin αz παz sin πz , α < π 2 ,around a circle of large radius, prove that∞∑(−1) m−1 sin2 mα= 1 (mα) 2 2 .m=125.12 Use the Bromwich inversion, and contours similar to that shown in figure 25.7(a),to find the functions of which the following are the Laplace transforms:(a) s(s 2 + b 2 ) −1 ;(b) n!(s − a) −(n+1) ,withn a positive integer and s>a;(c) a(s 2 − a 2 ) −1 ,withs>|a|.Compare your answers with those given in a table of standard Laplace transforms.25.13 Find the function f(t) whose Laplace transform is¯f(s) = e−s − 1+s.s 225.14 A function f(t) has the Laplace transformF(s) = 1 ( ) s + i2i ln ,s − ithe complex logarithm being defined by a finite branch cut running along theimaginary axis from −i to i.(a) Convince yourself that, for t>0, f(t) can be expressed as a closed contourintegral that encloses only the branch cut.922

25.9 EXERCISES(b) Calculate F(s) on either side of the branch cut, evaluate the integral andhence determine f(t).(c) Confirm that the derivative with respect to s of the Laplace transformintegral of your answer is the same as that given by dF/ds.25.15 Use the contour in figure 25.7(c) to show that the function with Laplace transforms −1/2 is (πx) −1/2 .[ For an integrand of the form r −1/2 exp(−rx) change variable to t = r 1/2 .]25.16 Transverse vibrations of angular frequency ω on a string stretched with constanttension T are described by u(x, t) =y(x) e −iωt ,whered 2 ydx + ω2 m(x)y(x) =0.2 THere, m(x) =m 0 f(x) is the mass per unit length of the string and, in the generalcase, is a function of x. Find the first-order W.K.B. solution for y(x).Due to imperfections in its manufacturing process, a particular string hasa small periodic variation in its linear density of the form m(x) = m 0 [1 +ɛ sin(2πx/L)], where ɛ ≪ 1. A progressive wave (i.e. one in which no energy islost) travels in the positive x-direction along the string. Show that its amplitudefluctuates by ± 1 ɛ of its value A 4 0 at x = 0 and that, to first order in ɛ, the phaseof the wave is√ɛωL m0 πx2π Tsin2 Lahead of what it would be if the string were uniform, with m(x) =m 0 .25.17 The equation(d 2 ydz + ν + 1 2 2 − 1 )4 z2 y =0,sometimes called the Weber–Hermite equation, has solutions known as paraboliccylinder functions. Find, to within (possibly complex) multiplicative constants,the two W.K.B. solutions of this equation that are valid for large |z|. In each case,determine the leading term and show that the multiplicative correction factor isoftheform1+O(ν 2 /z 2 ).Identify the Stokes and anti-Stokes lines for the equation. On which of theStokes lines is the W.K.B. solution that tends to zero for z large, real and negative,the dominant solution?25.18 A W.K.B. solution of Bessel’s equation of order zero,d 2 ydz + 1 dy2 z dz + y =0, (∗)valid for large |z| and −π/2 < arg z < 3π/2, is y(z) =Az −1/2 e iz .Obtainanimprovement on this by finding a multiplier of y(z) in the form of an asymptoticexpansion in inverse powers of z as follows.(a) Substitute for y(z) in(∗) and show that the equation is satisfied to O(z −5/2 ).(b) Now replace the constant A by A(z) and find the equation that must besatisfied by A(z). Look for a solution of the form A(z) =z ∑ σ ∞n=0 a nz −n ,where a 0 = 1. Show that σ = 0 is the only acceptable solution to the indicialequation and obtain a recurrence relation for the a n .(c) To within a (complex) constant, the expression y(z) =A(z)z −1/2 e iz is theasymptotic expansion of the Hankel function H (1)0(z). Show that it is adivergent expansion for all values of z and estimate, in terms of z, the valueof N such that ∑ Nn=0 a nz −n−1/2 e iz gives the best estimate of H (1)0 (z).923

APPLICATIONS OF COMPLEX VARIABLES25.19 The function h(z) ofthecomplexvariablez is defined by the integralh(z) =∫ i∞−i∞exp(t 2 − 2zt) dt.(a) Make a change of integration variable, t = iu, and evaluate h(z) usingastandard integral. Is your answer valid for all finite z?(b) Evaluate the integral using the method of steepest descents, considering inparticular the cases (i) z is real and positive, (ii) z is real and negative and(iii) z is purely imaginary and equal to iβ, whereβ is real. In each case sketchthe corresponding contour in the complex t-plane.(c) Evaluate the integral for the same three cases as specified in part (b) usingthe method of stationary phases. To determine an appropriate contour thatpasses through a saddle point t = t 0 ,writet = t 0 +(u + iv) and apply thecriterion for determining a level line. Sketch the relevant contour in eachcase, indicating what freedom there is to distort it.Comment on the accuracy of the results obtained using the approximate methodsadopted in (b) and (c).25.20 Use the method of steepest descents to show that an approximate value for theintegralF(z) =∫ ∞−∞exp[ iz( 1 5 t5 + t)]dt,where z is real and positive, is( ) 1/2 2πexp(−βz) cos(βz − 1 8zπ),where β =4/(5 √ 2).25.21 The stationary phase approximation to an integral of the formF(ν) =∫ bag(t)e iνf(t) dt, |ν| ≫1,where f(t) is a real function of t and g(t) is a slowly varying function (whencompared with the argument of the exponential), can be written as( ) 1/2 2π ∑ Ng(t n ){ [F(ν) ∼√ exp i νf(t n )+ π ( )]}|ν| An 4 sgn νf ′′ (t n ) ,n=1where the t n are the N stationary points of f(t) that lie in a

25.10 HINTS AND ANSWERSt = −i and then approaches the origin in the fourth quadrant in a curve thatis ultimately antiparallel to the positive real axis. The other contour, C 1 ,isthemirror image of this in the real axis; it is confined to the upper half-plane, passesthrough t = i and is antiparallel to the real t-axis at both of its extremities. Thecontribution to J ν (z) from the curve C k is 1 2 H(k) ν , the function H ν(k) being knownas a Hankel function.Using the method of steepest descents, establish the leading term in an asymptoticexpansion for H ν(1) for z real, large and positive. Deduce, without detailedcalculation, the corresponding result for H ν(2) . Hence establish the asymptoticform of J ν (z) for the same range of z.25.23 Use the method of steepest descents to find an asymptotic approximation, validfor z large, real and positive, to the function defined by∫F ν (z) = exp(−iz sin t + iνt) dt,Cwhere ν is real and non-negative and C is a contour that starts at t = −π + i∞and ends at t = −i∞.25.10 Hints and answers25.1 Apply Kirchhoff’s laws to three independent loops, say ADBA, ADEA andDBED. Eliminate other currents from the equations to obtain I R = ω 0 CV 0 [(ω0 2 −ω 2 − 2iωω 0 )/(ω0 2 + ω2 )], where ω0 2 =(LC)−1 ; |I R | = ω 0 CV 0 ; the phase of I R istan −1 [(−2ωω 0 )/(ω0 2 − ω2 )].25.3 Set c coth u 1 = −d, c coth u 2 =+d, |c cosech u| = a and note that the capacitanceis proportional to (u 2 − u 1 ) −1 .25.5 ξ = constant, ellipses x 2 (a+1) −2 +y 2 (a−1) −2 = c 2 /(4a 2 ); η = constant, hyperbolaex 2 (cos α) −2 − y 2 (sin α) −2 = c 2 . The curves are the cuts −c ≤ x ≤ c, y =0and|x| ≥c, y =0.Thecurvesforη =2π are the same as those for η =0.25.7 (a) For a quarter-circular contour enclosing the first quadrant, the change in theargument of the function is 0 + 8(π/2) + 0 (since y 8 + 5 = 0 has no real roots);(b) one negative real zero; a conjugate pair in the second and third quadrants,− 3 , −1 ± i.225.9 Evaluate∫π cot πz( 1 + z)( dz1+ z)2 4around a large circle centred on the origin; residue at z = −1/2 is 0; residue atz = −1/4 is4π cot(−π/4).25.11 The behaviour of the integrand for large |z| is |z| −2 exp [ (2α − π)|z| ]. The residueat z = ±m, foreachintegerm, issin 2 (mα)(−1) m /(mα) 2 . The contour contributesnothing.Required summation = [ total sum − (m = 0 term) ]/2.25.13 Note that ¯f(s) has no pole at s =0.Fort1 in the left half-plane. For 0

APPLICATIONS OF COMPLEX VARIABLES25.17 Use the binomial theorem to expand, in inverse powers of z, both the squareroot in the exponent and the fourth root in the multiplier, working to O(z −2 ).The leading terms are y 1 (z) =Ce −z2 /4 z ν and y 2 (z) =De z2 /4 z −(ν+1) . Stokes lines:arg z =0,π/2,π,3π/2; anti-Stokes lines: arg z =(2n +1)π/4 forn =0, 1, 2, 3. y 1is dominant on arg z = π/2 or3π/2.25.19 (a) i √ πe −z2 , valid for all z, including i √ π exp(β 2 ) in case (iii).(b) The same values as in (a). The (only) saddle point, at t 0 = z, is traversed inthe direction θ =+ 1 π in all cases, though the path in the complex t-plane varies2with each case.(c) The same values as in (a). The level lines are v = ±u. In cases (i) and (ii) thecontour turns through a right angle at the saddle point.All three methods give exact answers in this case of a quadratic exponent.25.21 Saddle points at t 1 = −z and t 2 =2z with f 1 ′′ = −18z and f′′ 2 =18z.Approximation is[( π) 1/2 cos(7νz 3 − 1 π) 4+ cos(20νz3 − 1 π)]4.9zν 1+z 2 1+4z 225.23 Saddle point at t 0 =cos −1 (ν/z) is traversed in the direction θ = − 1 4 π. F ν(z) ≈(2π/z) 1/2 exp [ i(z − 1 2 νπ − 1 4 π)]. 926

26TensorsIt may seem obvious that the quantitative description of physical processes cannotdepend on the coordinate system in which they are represented. However, we mayturn this argument around: since physical results must indeed be independent ofthe choice of coordinate system, what does this imply about the nature of thequantities involved in the description of physical processes? The study of theseimplications and of the classification of physical quantities by means of themforms the content of the present chapter.Although the concepts presented here may be applied, with little modification,to more abstract spaces (most notably the four-dimensional space–time ofspecial or general relativity), we shall restrict our attention to our familiar threedimensionalEuclidean space. This removes the need to discuss the properties ofdifferentiable manifolds and their tangent and dual spaces. The reader who isinterested in these more technical aspects of tensor calculus in general spaces,and in particular their application to general relativity, should consult one of themany excellent textbooks on the subject. §Before the presentation of the main development of the subject, we begin byintroducing the summation convention, which will prove very useful in writingtensor equations in a more compact form. We then review the effects of a changeof basis in a vector space; such spaces were discussed in chapter 8. This isfollowed by an investigation of the rotation of Cartesian coordinate systems, andfinally we broaden our discussion to include more general coordinate systems andtransformations.§ For example, R. D’Inverno, Introducing Einstein’s Relativity (Oxford: Oxford University Press,1992); J. Foster and J. D. Nightingale, A Short Course in General Relativity (New York: Springer,2006); B. F. Schutz, A First Course in General Relativity (Cambridge; Cambridge University Press1985).927

TENSORS26.1 Some notationBefore proceeding further, we introduce the summation convention for subscripts,since its use looms large in the work of this chapter. The convention is thatany lower-case alphabetic subscript that appears exactly twice in any term of anexpression is understood to be summed over all the values that a subscript inthat position can take (unless the contrary is specifically stated). The subscriptedquantities may appear in the numerator and/or the denominator of a term in anexpression. This naturally implies that any such pair of repeated subscripts mustoccur only in subscript positions that have the same range of values. Sometimesthe ranges of values have to be specified but usually they are apparent from thecontext.The following simple examples illustrate what is meant (in the three-dimensionalcase):(i) a i x i stands for a 1 x 1 + a 2 x 2 + a 3 x 3 ;(ii) a ij b jk stands for a i1 b 1k + a i2 b 2k + a i3 b 3k ;(iii) a ij b jk c k stands for ∑ 3 ∑ 3j=1 k=1 a ijb jk c k ;(iv)(v)∂v i∂x istands for ∂v 1∂x 1+ ∂v 2∂x 2+ ∂v 3∂x 3;∂ 2 φstands for ∂2 φ∂x i ∂x i ∂x 2 + ∂2 φ1∂x 2 2+ ∂2 φ∂x 2 .3Subscripts that are summed over are called dummy subscripts and the othersfree subscripts. It is worth remarking that when introducing a dummy subscriptinto an expression, care should be taken not to use one that is already present,either as a free or as a dummy subscript. For example, a ij b jk c kl cannot, and mustnot, be replaced by a ij b jj c jl or by a il b lk c kl , but could be replaced by a im b mk c klor by a im b mn c nl . Naturally, free subscripts must not be changed at all unless theworking calls for it.Furthermore, as we have done throughout this book, we will make frequentuse of the Kronecker delta δ ij , which is defined by{1 if i = j,δ ij =0 otherwise.When the summation convention has been adopted, the main use of δ ij is toreplace one subscript by another in certain expressions. Examples might includeandb j δ ij = b i ,a ij δ jk = a ij δ kj = a ik . (26.1)928

26.2 CHANGE OF BASISIn the second of these the dummy index shared by both terms on the left-handside (namely j) has been replaced by the free index carried by the Kronecker delta(namely k), and the delta symbol has disappeared. In matrix language, (26.1) canbe written as AI = A, whereA is the matrix with elements a ij and I is the unitmatrix having the same dimensions as A.In some expressions we may use the Kronecker delta to replace indices in anumber of different ways, e.g.a ij b jk δ ki = a ij b ji or a kj b jk ,where the two expressions on the RHS are totally equivalent to one another.26.2 Change of basisIn chapter 8 some attention was given to the subject of changing the basis set (orcoordinate system) in a vector space and it was shown that, under such a change,different types of quantity behave in different ways. These results are given insection 8.15, but are summarised below for convenience, using the summationconvention. Although throughout this section we will remind the reader that weare using this convention, it will simply be assumed in the remainder of thechapter.If we introduce a set of basis vectors e 1 , e 2 , e 3 into our familiar three-dimensional(vector) space, then we can describe any vector x in terms of its componentsx 1 ,x 2 ,x 3 with respect to this basis:x = x 1 e 1 + x 2 e 2 + x 3 e 3 = x i e i ,where we have used the summation convention to write the sum in a morecompact form. If we now introduce a new basis e ′ 1 , e′ 2 , e′ 3 related to the old one bye ′ j = S ij e i (sum over i), (26.2)where the coefficient S ij is the ith component of the vector e ′ j with respect to theunprimed basis, then we may write x with respect to the new basis asx = x ′ 1e ′ 1 + x ′ 2e ′ 2 + x ′ 3e ′ 3 = x ′ ie ′ i(sum over i).If we denote the matrix with elements S ij by S, then the components x ′ i and x iin the two bases are related byx ′ i =(S −1 ) ij x j(sum over j),where, using the summation convention, there is an implicit sum over j fromj =1toj = 3. In the special case where the transformation is a rotation of thecoordinate axes, the transformation matrix S is orthogonal and we havex ′ i =(S T ) ij x j = S ji x j (sum over j). (26.3)929

TENSORSScalars behave differently under transformations, however, since they remainunchanged. For example, the value of the scalar product of two vectors x · y(which is just a number) is unaffected by the transformation from the unprimedto the primed basis. Different again is the behaviour of linear operators. If alinear operator A is represented by some matrix A in a given coordinate systemthen in the new (primed) coordinate system it is represented by a new matrix,A ′ = S −1 AS.In this chapter we develop a general formulation to describe and classify thesedifferent types of behaviour under a change of basis (or coordinate transformation).In the development, the generic name tensor is introduced, and certainscalars, vectors and linear operators are described respectively as tensors of zeroth,first and second order (the order –orrank – corresponds to the number ofsubscripts needed to specify a particular element of the tensor). Tensors of thirdand fourth order will also occupy some of our attention.26.3 Cartesian tensorsWe begin our discussion of tensors by considering a particular class of coordinatetransformation – namely rotations – and we shall confine our attention strictlyto the rotation of Cartesian coordinate systems. Our object is to study the propertiesof various types of mathematical quantities, and their associated physicalinterpretations, when they are described in terms of Cartesian coordinates andthe axes of the coordinate system are rigidly rotated from a basis e 1 , e 2 , e 3 (lyingalong the Ox 1 , Ox 2 and Ox 3 axes) to a new one e ′ 1 , e′ 2 , e′ 3 (lying along the Ox′ 1 ,Ox ′ 2 and Ox′ 3 axes).Since we shall be more interested in how the components of a vector or linearoperator are changed by a rotation of the axes than in the relationship betweenthe two sets of basis vectors e i and e ′ i , let us define the transformation matrix Las the inverse of the matrix S in (26.2). Thus, from (26.2), the components of aposition vector x, in the old and new bases respectively, are related byx ′ i = L ij x j . (26.4)Because we are considering only rigid rotations of the coordinate axes, thetransformation matrix L will be orthogonal, i.e. such that L −1 = L T .Thereforethe inverse transformation is given byx i = L ji x ′ j. (26.5)The orthogonality of L also implies relations among the elements of L thatexpress the fact that LL T = L T L = I. In subscript notation they are given byL ik L jk = δ ij and L ki L kj = δ ij . (26.6)Furthermore, in terms of the basis vectors of the primed and unprimed Cartesian930

26.3 CARTESIAN TENSORSx 2x ′ 1x ′ 2θθθO x 1Figure 26.1 Rotation of Cartesian axes by an angle θ about the x 3 -axis. Thethree angles marked θ and the parallels (broken lines) to the primed axesshow how the first two equations of (26.7) are constructed.coordinate systems, the transformation matrix is given byL ij = e ′ i · e j .We note that the product of two rotations is also a rotation. For example,suppose that x ′ i = L ijx j and x ′′i = M ij x ′ j ; then the composite rotation is describedbyx ′′i = M ij x ′ j = M ij L jk x k =(ML) ik x k ,corresponding to the matrix ML.◮Find the transformation matrix L corresponding to a rotation of the coordinate axesthrough an angle θ about the e 3 -axis (or x 3 -axis), as shown in figure 26.1.Taking x as a position vector – the most obvious choice – we see from the figure thatthe components of x with respect to the new (primed) basis are given in terms of thecomponents in the old (unprimed) basis byx ′ 1 = x 1 cos θ + x 2 sin θ,x ′ 2 = −x 1 sin θ + x 2 cos θ, (26.7)x ′ 3 = x 3 .The (orthogonal) transformation matrix is thusThe inverse equations arein line with (26.5). ◭⎛L = ⎝cos θ sin θ 0− sin θ cos θ 00 0 1⎞⎠ .x 1 = x ′ 1 cos θ − x ′ 2 sin θ,x 2 = x ′ 1 sin θ + x ′ 2 cos θ, (26.8)x 3 = x ′ 3,931

TENSORS26.4 First- and zero-order Cartesian tensorsUsing the above example as a guide, we may consider any set of three quantitiesv i , which are directly or indirectly functions of the coordinates x i and possiblyinvolve some constants, and ask how their values are changed by any rotation ofthe Cartesian axes. The specific question to be answered is whether the specificforms v ′ i in the new variables can be obtained from the old ones v i using (26.4),v ′ i = L ij v j . (26.9)If so, the v i are said to form the components of a vector or first-order Cartesiantensor v. By definition, the position coordinates are themselves the componentsof such a tensor.The first-order tensor v does not change under rotation of thecoordinate axes; nevertheless, since the basis set does change, from e 1 , e 2 , e 3 toe ′ 1 , e′ 2 , v′ 3 , the components of v must also change. The changes must be such thatv = v i e i = v ie ′ ′ i (26.10)is unchanged.Since the transformation (26.9) is orthogonal, the components of any suchfirst-order Cartesian tensor also obey a relation that is the inverse of (26.9),v i = L ji v ′ j. (26.11)We now consider explicit examples. In order to keep the equations to reasonableproportions, the examples will be restricted to the x 1 x 2 -plane, i.e. there areno components in the x 3 -direction. Three-dimensional cases are no different inprinciple – but much longer to write out.◮Which of the following pairs (v 1 ,v 2 ) form the components of a first-order Cartesian tensorin two dimensions?:(i) (x 2 , −x 1 ), (ii) (x 2 ,x 1 ), (iii) (x 2 1,x 2 2).We shall consider the rotation discussed in the previous example, and to save space wedenote cos θ by c and sin θ by s.(i) Here v 1 = x 2 and v 2 = −x 1 , referred to the old axes. In terms of the new coordinatesthey will be v ′ 1 = x′ 2 and v′ 2 = −x′ 1 ,i.e.v ′ 1 = x′ 2 = −sx 1 + cx 2v ′ 2 = −x′ 1 = −cx 1 − sx 2 .(26.12)Now if we start again and evaluate v 1 ′ and v′ 2 as given by (26.9) we find thatv 1 ′ = L 11v 1 + L 12 v 2 = cx 2 + s(−x 1 )v 2 ′ = L (26.13)21v 1 + L 22 v 2 = −s(x 2 )+c(−x 1 ).The expressions for v 1 ′ and v′ 2 in (26.12) and (26.13) are the same whatever the valuesof θ (i.e. for all rotations) and thus by definition (26.9) the pair (x 2 , −x 1 ) is a first-orderCartesian tensor.932

26.4 FIRST- AND ZERO-ORDER CARTESIAN TENSORS(ii) Here v 1 = x 2 and v 2 = x 1 . Following the same procedure,v ′ 1 = x ′ 2 = −sx 1 + cx 2v ′ 2 = x ′ 1 = cx 1 + sx 2 .But, by (26.9), for a Cartesian tensor we must havev ′ 1 = cv 1 + sv 2 = cx 2 + sx 1v ′ 2 =(−s)v 1 + cv 2 = −sx 2 + cx 1 .These two sets of expressions do not agree and thus the pair (x 2 ,x 1 ) is not a first-orderCartesian tensor.(iii) v 1 = x 2 1 and v 2 = x 2 2 . As in (ii) above, considering the first component alone issufficient to show that this pair is also not a first-order tensor. Evaluating v 1 ′ directly giveswhilst (26.9) requires thatwhich is quite different. ◭v ′ 1 = x ′ 12 = c 2 x 2 1 +2csx 1 x 2 + s 2 x 2 2,v ′ 1 = cv 1 + sv 2 = cx 2 1 + sx 2 2,There are many physical examples of first-order tensors (i.e. vectors) that will befamiliar to the reader. As a straightforward one, we may take the set of Cartesiancomponents of the momentum of a particle of mass m, (mẋ 1 ,mẋ 2 ,mẋ 3 ). This settransforms in all essentials as (x 1 ,x 2 ,x 3 ), since the other operations involved,multiplication by a number and differentiation with respect to time, are quiteunaffected by any orthogonal transformation of the axes. Similarly, accelerationand force are represented by the components of first-order tensors.Other more complicated vectors involving the position coordinates more thanonce, such as the angular momentum of a particle of mass m, namely J =x × p = m(x × ẋ), are also first-order tensors. That this is so is less obvious incomponent form than for the earlier examples, but may be verified by writingout the components of J explicitly or by appealing to the quotient law to bediscussed in section 26.7 and using the Cartesian tensor ɛ ijk from section 26.8.Having considered the effects of rotations on vector-like sets of quantities wemay consider quantities that are unchanged by a rotation of axes. In our previousnomenclature these have been called scalars but we may also describe them astensors of zero order. They contain only one element (formally, the number ofsubscripts needed to identify a particular element is zero); the most obvious nontrivialexample associated with a rotation of axes is the square of the distance ofa point from the origin, r 2 = x 2 1 + x2 2 + x2 3 . In the new coordinate system it willhave the form r ′2 = x ′ 12 + x ′ 22 + x ′ 32 , which for any rotation has the same value asx 2 1 + x2 2 + x2 3 . 933

TENSORSIn fact any scalar product of two first-order tensors (vectors) is a zero-ordertensor (scalar), as might be expected since it can be written in a coordinate-freeway as u · v.◮By considering the components of the vectors u and v with respect to two Cartesiancoordinate systems (related by a rotation), show that the scalar product u · v is invariantunder rotation.In the original (unprimed) system the scalar product is given in terms of components byu i v i (summed over i), and in the rotated (primed) system byu ′ iv ′ i = L ij u j L ik v k = L ij L ik u j v k = δ jk u j v k = u j v j ,where we have used the orthogonality relation (26.6). Since the resulting expression in therotated system is the same as that in the original system, the scalar product is indeedinvariant under rotations. ◭The above result leads directly to the identification of many physically importantquantities as zero-order tensors. Perhaps the most immediate of these isenergy, either as potential energy or as an energy density (e.g. F · dr, eE · dr, D · E,B · H, µ · B), but others, such as the angle between two directed quantities, areimportant.As mentioned in the first paragraph of this chapter, in most analyses of physicalsituations it is a scalar quantity (such as energy) that is to be determined. Suchquantities are invariant under a rotation of axes and so it is possible to work withthe most convenient set of axes and still have confidence in the results.Complementing the way in which a zero-order tensor was obtained from twofirst-order tensors, so a first-order tensor can be obtained from a zero-order tensor(i.e. a scalar). We show this by taking a specific example, that of the electric fieldE = −∇φ; this is derived from a scalar, the electrostatic potential φ, and hascomponentsE i = − ∂φ∂x i. (26.14)Clearly, E is a first-order tensor, but we may prove this more formally byconsidering the behaviour of its components (26.14) under a rotation of thecoordinate axes, since the components of the electric field E ′ i are then given by(E i ′ = − ∂φ ) ′= − ∂φ′∂x i ∂x ′ = − ∂x j ∂φi∂x ′ = L ij E j , (26.15)i∂x jwhere (26.5) has been used to evaluate ∂x j /∂x ′ i . Now (26.15) is in the form(26.9), thus confirming that the components of the electric field do behave as thecomponents of a first-order tensor.934

26.5 SECOND- AND HIGHER-ORDER CARTESIAN TENSORS◮If v i are the components of a first-order tensor, show that ∇ · v = ∂v i /∂x i is a zero-ordertensor.In the rotated coordinate system ∇ · v is given by( ) ′ ∂vi= ∂v′ i= ∂x j ∂∂v k(L∂x i ∂x ′ i∂x ′ ik v k )=L ij L ik ,i∂x j ∂x jsince the elements L ij are not functions of position. Using the orthogonality relation (26.6)we then find∂v i′∂x ′ i∂v k ∂v k= L ij L ik = δ jk = ∂v j.∂x j ∂x j ∂x jHence ∂v i /∂x i is invariant under rotation of the axes and is thus a zero-order tensor; thiswas to be expected since it can be written in a coordinate-free way as ∇ · v. ◭26.5 Second- and higher-order Cartesian tensorsFollowing on from scalars with no subscripts and vectors with one subscript,we turn to sets of quantities that require two subscripts to identify a particularelement of the set. Let these quantities by denoted by T ij .Taking (26.9) as a guide we define a second-order Cartesian tensor as follows:the T ij form the components of such a tensor if, under the same conditions asfor (26.9),andT ′ij = L ik L jl T kl (26.16)T ij = L ki L lj T ′ kl. (26.17)At the same time we may define a Cartesian tensor of general order as follows.The set of expressions T ij···k form the components of a Cartesian tensor if, for allrotations of the axes of coordinates given by (26.4) and (26.5), subject to (26.6),the expressions using the new coordinates, Tij···k ′ are given byT ij···k ′ = L ip L jq ···L kr T pq···r (26.18)andT ij···k = L pi L qj ···L rk T pq···r. ′ (26.19)It is apparent that in three dimensions, an Nth-order Cartesian tensor has 3 Ncomponents.Since a second-order tensor has two subscripts, it is natural to display itscomponents in matrix form. The notation [T ij ] is used, as well as T, to denotethe matrix having T ij as the element in the ith row and jth column. §We may think of a second-order tensor T as a geometrical entity in a similarway to that in which we viewed linear operators (which transform one vector into§ We can also denote the column matrix containing the elements v i of a vector by [v i ].935

TENSORSanother, without reference to any coordinate system) and consider the matrixcontaining its components as a representation of the tensor with respect to aparticular coordinate system. Moreover, the matrix T =[T ij ], containing thecomponents of a second-order tensor, behaves in the same way under orthogonaltransformations T ′ = LTL T as a linear operator.However, not all linear operators are second-order tensors. More specifically,the two subscripts in a second-order tensor must refer to the same coordinatesystem. In particular, this means that any linear operator that transforms a vectorinto a vector in a different vector space cannot be a second-order tensor. Thus,although the elements L ij of the transformation matrix are written with twosubscripts, they cannot be the components of a tensor since the two subscriptseach refer to a different coordinate system.As examples of sets of quantities that are readily shown to be second-ordertensors we consider the following.(i) The outer product of two vectors. Letu i and v i , i =1, 2, 3, be the componentsof two vectors u and v, and consider the set of quantities T ij defined byT ij = u i v j . (26.20)The set T ij are called the components of the the outer product of u and v. Underrotations the components T ij becomeT ij ′ = u ′ iv j ′ = L ik u k L jl v l = L ik L jl u k v l = L ik L jl T kl , (26.21)which shows that they do transform as the components of a second-order tensor.Use has been made in (26.21) of the fact that u i and v i are the components offirst-order tensors.The outer product of two vectors is often denoted, without reference to anycoordinate system, asT = u ⊗ v. (26.22)(This is not to be confused with the vector product of two vectors, which is itselfa vector and is discussed in chapter 7.) The expression (26.22) gives the basis towhich the components T ij of the second-order tensor refer: since u = u i e i andv = v i e i , we may write the tensor T asT = u i e i ⊗ v j e j = u i v j e i ⊗ e j = T ij e i ⊗ e j . (26.23)Moreover, as for the case of first-order tensors (see equation (26.10)) we notethat the quantities T ij ′ are the components of the same tensor T, but referred toa different coordinate system, i.e.T = T ij e i ⊗ e j = T ije ′ ′ i ⊗ e ′ j.These concepts can be extended to higher-order tensors.936

26.5 SECOND- AND HIGHER-ORDER CARTESIAN TENSORS(ii) The gradient of a vector. Suppose v i represents the components of a vector;let us consider the quantities generated by forming the derivatives of each v i ,i =1, 2, 3, with respect to each x j , j =1, 2, 3, i.e.T ij = ∂v i.∂x jThese nine quantities form the components of a second-order tensor, as can beseen from the fact thatT ′ij = ∂v′ i∂x ′ j= ∂(L ikv k ) ∂x l ∂v k∂x l ∂x ′ = L ik L jl = L ik L jl T kl .j∂x lIn coordinate-free language the tensor T may be written as T = ∇v and hencegives meaning to the concept of the gradient of a vector, a quantity that was notdiscussed in the chapter on vector calculus (chapter 10).A test of whether any given set of quantities forms the components of a secondordertensor can always be made by direct substitution of the x ′ i in terms of thex i , followed by comparison with the right-hand side of (26.16). This procedure isextremely laborious, however, and it is almost always better to try to recognisethe set as being expressible in one of the forms just considered, or to makealternative tests based on the quotient law of section 26.7 below.◮Show that the T ij given by(x2T =[T ij ]= 2−x 1 x 2−x 1 x 2 x 2 1are the components of a second-order tensor.)(26.24)Again we consider a rotation θ about the e 3 -axis. Carrying out the direct evaluation firstwe obtain, using (26.7),T 11 ′ = x ′ 2 2 = s 2 x 2 1 − 2scx 1 x 2 + c 2 x 2 2,T 12 ′ = −x ′ 1x ′ 2 = scx 2 1 +(s 2 − c 2 )x 1 x 2 − scx 2 2,T 21 ′ = −x ′ 1x ′ 2 = scx 2 1 +(s 2 − c 2 )x 1 x 2 − scx 2 2,T 22 ′ = x ′ 1 2 = c 2 x 2 1 +2scx 1 x 2 + s 2 x 2 2.Now, evaluating the right-hand side of (26.16),T 11 ′ = ccx 2 2 + cs(−x 1 x 2 )+sc(−x 1 x 2 )+ssx 2 1,T 12 ′ = c(−s)x 2 2 + cc(−x 1 x 2 )+s(−s)(−x 1 x 2 )+scx 2 1,T 21 ′ =(−s)cx 2 2 +(−s)s(−x 1 x 2 )+cc(−x 1 x 2 )+csx 2 1,T 22 ′ =(−s)(−s)x 2 2 +(−s)c(−x 1 x 2 )+c(−s)(−x 1 x 2 )+ccx 2 1.After reorganisation, the corresponding expressions are seen to be the same, showing, asrequired, that the T ij are the components of a second-order tensor.The same result could be inferred much more easily, however, by noting that the T ijare in fact the components of the outer product of the vector (x 2 , −x 1 ) with itself. That(x 2 , −x 1 ) is indeed a vector was established by (26.12) and (26.13). ◭937

TENSORSPhysical examples involving second-order tensors will be discussed in the latersections of this chapter, but we might note here that, for example, magneticsusceptibility and electrical conductivity are described by second-order tensors.26.6 The algebra of tensorsBecause of the similarity of first- and second-order tensors to column vectors andmatrices, it would be expected that similar types of algebraic operation can becarried out with them and so provide ways of constructing new tensors from oldones. In the remainder of this chapter, instead of referring to the T ij (say) as thecomponents of a second-order tensor T, we may sometimes simply refer to T ijas the tensor. It should always be remembered, however, that the T ij are in factjust the components of T in a given coordinate system and that T ij ′ refers to thecomponents of the same tensor T in a different coordinate system.The addition and subtraction of tensors follows an obvious definition; namelythat if V ij···k and W ij···k are (the components of) tensors of the same order, thentheir sum and difference, S ij···k and D ij···k respectively, are given byS ij···k = V ij···k + W ij···k ,D ij···k = V ij···k − W ij···k ,for each set of values i,j,...,k.ThatS ij···k and D ij···k are the components oftensors follows immediately from the linearity of a rotation of coordinates.It is equally straightforward to show that if the T ij···k are the components ofa tensor, then so is the set of quantities formed by interchanging the order of (apair of) indices, e.g. T ji···k .If T ji···k is found to be identical with T ij···k then T ij···k is said to be symmetricwith respect to its first two subscripts (or simply ‘symmetric’, for second-ordertensors). If, however, T ji···k = −T ij···k for every element then it is an antisymmetrictensor. An arbitrary tensor is neither symmetric nor antisymmetric but can alwaysbe written as the sum of a symmetric tensor S ij···k and an antisymmetric tensorA ij···k :T ij···k = 1 2 (T ij···k + T ji···k )+ 1 2 (T ij···k − T ji···k )= S ij···k + A ij···k .Of course these properties are valid for any pair of subscripts.In (26.20) in the previous section we had an example of a kind of ‘multiplication’of two tensors, thereby producing a tensor of higher order – in that case twofirst-order tensors were multiplied to give a second-order tensor. Inspection of(26.21) shows that there is nothing particular about the orders of the tensorsinvolved and it follows as a general result that the outer product of an Nth-ordertensor with an Mth-order tensor will produce an (M + N)th-order tensor.938

26.7 THE QUOTIENT LAWAn operation that produces the opposite effect – namely, generates a tensorof smaller rather than larger order – is known as contraction and consists ofmaking two of the subscripts equal and summing over all values of the equalisedsubscripts.◮Show that the process of contraction of an Nth-order tensor produces another tensor, oforder N − 2.Let T ij···l···m···k be the components of an Nth-order tensor, thenij···l···m···k = L ip L jq ···L lr ···L ms ···L kn T pq···r···s···n .} {{ }N factorsT ′Thus if, for example, we make the two subscripts l and m equal and sum over all valuesof these subscripts, we obtainT ij···l···l···k ′ = L ip L jq ···L lr ···L ls ···L kn T pq···r···s···n= L ip L jq ···δ rs ···L kn T pq···r···s···n= L ip L jq ···L kn} {{ }(N − 2) factorsT pq···r···r···n ,showing that T ij···l···l···k are the components of a (different) Cartesian tensor of orderN − 2. ◭For a second-rank tensor, the process of contraction is the same as taking thetrace of the corresponding matrix. The trace T ii itself is thus a zero-order tensor(or scalar) and hence invariant under rotations, as was noted in chapter 8.The process of taking the scalar product of two vectors can be recast into tensorlanguage as forming the outer product T ij = u i v j of two first-order tensors u andv and then contracting the second-order tensor T so formed, to give T ii = u i v i ,ascalar (invariant under a rotation of axes).As yet another example of a familiar operation that is a particular case of acontraction, we may note that the multiplication of a column vector [u i ]byamatrix [B ij ] to produce another column vector [v i ],B ij u j = v i ,can be looked upon as the contraction T ijj of the third-order tensor T ijk formedfrom the outer product of B ij and u k .26.7 The quotient lawThe previous paragraph appears to give a heavy-handed way of describing afamiliar operation, but it leads us to ask whether it has a converse. To put thequestion in more general terms: if we know that B and C are tensors and alsothatA pq···k···m B ij···k···n = C pq···mij···n , (26.25)939

TENSORSdoes this imply that the A pq···k···m also form the components of a tensor A? HereA, B and C are respectively of Mth, Nth and (M +N −2)th order and it should benoted that the subscript k that has been contracted may be any of the subscriptsin A and B independently.The quotient law for tensors states that if (26.25) holds in all rotated coordinateframes then the A pq···k···m do indeed form the components of a tensor A. Toproveit for general M and N is no more difficult regarding the ideas involved than toshow it for specific M and N, but this does involve the introduction of a largenumber of subscript symbols. We will therefore take the case M = N = 2, butit will be readily apparent that the principle of the proof holds for general Mand N.We thus start with (say)A pk B ik = C pi , (26.26)where B ik and C pi are arbitrary second-order tensors. Under a rotation of coordinatesthe set A pk (tensor or not) transforms into a new set of quantities thatwe will denote by A ′ pk. We thus obtain in succession the following steps, using(26.16), (26.17) and (26.6):A ′ pk B′ ik = C′ pi (transforming (26.26)),= L pq L ij C qj (since C is a tensor),= L pq L ij A ql B jl (from (26.26)),= L pq L ij A ql L mj L nl B mn ′ (since B is a tensor),= L pq L nl A ql B in ′ (since L ij L mj = δ im ).Now k on the left and n on the right are dummy subscripts and thus we maywrite(A ′ pk − L pq L kl A ql )B ′ ik =0. (26.27)Since B ik , and hence Bik ′ , is an arbitrary tensor, we must haveA ′ pk = L pq L kl A ql ,showing that the A ′ pkare given by the general formula (26.18) and hence thatthe A pk are the components of a second-order tensor. By following an analogousargument, the same result (26.27) and deduction could be obtained if (26.26) werereplaced byA pk B ki = C pi ,i.e. the contraction being now with respect to a different pair of indices.Use of the quotient law to test whether a given set of quantities is a tensor isgenerally much more convenient than making a direct substitution. A particularway in which it is applied is by contracting the given set of quantities, having940

26.8 THE TENSORS δ ij AND ɛ ijkN subscripts, with an arbitrary Nth-order tensor (i.e. one having independentlyvariable components) and determining whether the result is a scalar.◮Use the quotient law to show that the elements of T, equation (26.24), are the componentsof a second-order tensor.The outer product x i x j is a second-order tensor. Contracting this with the T ij given in(26.24) we obtainT ij x i x j = x 2 2x 2 1 − x 1 x 2 x 1 x 2 − x 1 x 2 x 2 x 1 + x 2 1x 2 2 =0,which is clearly invariant (a zeroth-order tensor). Hence by the quotient theorem T ij mustalso be a tensor. ◭26.8 The tensors δ ij and ɛ ijkIn many places throughout this book we have encountered and used the twosubscriptquantity δ ij defined by{1 if i = j,δ ij =0 otherwise.Let us now also introduce the three-subscript Levi–Civita symbol ɛ ijk , the valueof which is given by⎧⎪⎨ +1 if i, j, k is an even permutation of 1, 2, 3,ɛ ijk = −1 if i, j, k is an odd permutation of 1, 2, 3,⎪⎩ 0 otherwise.We will now show that δ ij and ɛ ijk are respectively the components of a secondanda third-order Cartesian tensor. Notice that the coordinates x i do not appearexplicitly in the components of these tensors, their components consisting entirelyof 0 and 1.In passing, we also note that ɛ ijk is totally antisymmetric, i.e. it changes signunder the interchange of any pair of subscripts. In fact ɛ ijk , or any scalar multipleof it, is the only three-subscript quantity with this property.Treating δ ij first, the proof that it is a second-order tensor is straightforwardsince if, from (26.16), we consider the equationδ ′ kl = L ki L lj δ ij = L ki L li = δ kl ,we see that the transformation of δ ij generates the same expression (a patternof 0’s and 1’s) as does the definition of δ ij ′ in the transformed coordinates. Thusδ ij transforms according to the appropriate tensor transformation law and istherefore a second-order tensor.Turning now to ɛ ijk , we have to consider the quantityɛ ′ lmn = L li L mj L nk ɛ ijk . (26.28)941

TENSORSLet us begin, however, by noting that we may use the Levi–Civita symbol towrite an expression for the determinant of a 3 × 3 matrix A,|A|ɛ lmn = A li A mj A nk ɛ ijk , (26.29)which may be shown to be equivalent to the Laplace expansion (see chapter 8). §Indeed many of the properties of determinants discussed in chapter 8 can beproved very efficiently using this expression (see exercise 26.9).◮Evaluate the determinant of the matrix⎛A = ⎝ 2 1 −33 4 01 −2 1⎞⎠ .Setting l =1,m =2andn = 3 in (26.29) we find|A| = ɛ ijk A 1i A 2j A 3k= (2)(4)(1) − (2)(0)(−2) − (1)(3)(1) + (−3)(3)(−2)+ (1)(0)(1) − (−3)(4)(1) = 35,which may be verified using the Laplace expansion method. ◭We can now show that the ɛ ijk are in fact the components of a third-order tensor.Using (26.29) with the general matrix A replaced by the specific transformationmatrix L, we can rewrite the RHS of (26.28) in terms of |L|ɛ ′ lmn = L li L mj L nk ɛ ijk = |L|ɛ lmn .Since L is orthogonal its determinant has the value unity, and so ɛ ′ lmn = ɛ lmn.Thus we see that ɛ ′ lmn has exactly the properties of ɛ ijk but with i, j, k replaced byl,m,n, i.e. it is the same as the expression ɛ ijk written using the new coordinates.This shows that ɛ ijk is a third-order Cartesian tensor.In addition to providing a convenient notation for the determinant of a matrix,δ ij and ɛ ijk can be used to write many of the familiar expressions of vectoralgebra and calculus as contracted tensors. For example, provided we are usingright-handed Cartesian coordinates, the vector product a = b × c has as itsith component a i = ɛ ijk b j c k ; this should be contrasted with the outer productT = b ⊗ c, which is a second-order tensor having the components T ij = b i c j .§ This may be readily extended to an N × N matrix A, i.e.|A|ɛ i1 i 2···i N= A i1 j 1A i2 j 2 ···A iN j Nɛ j1 j 2···j N,where ɛ i1 i 2···i Nequals 1 if i 1 i 2 ···i N is an even permutation of 1, 2,...,N and equals −1 ifitisanodd permutation; otherwise it equals zero.942

26.8 THE TENSORS δ ij AND ɛ ijk◮Write the following as contracted Cartesian tensors: a·b, ∇ 2 φ, ∇×v, ∇(∇·v), ∇×(∇×v),(a × b) · c.The corresponding (contracted) tensor expressions are readily seen to be as follows:a · b = a i b i = δ ij a i b j ,∇ 2 φ =∂2 φ ∂ 2 φ= δ ij ,∂x i ∂x i ∂x i ∂x j∂v k(∇×v) i = ɛ ijk ,∂x j[∇(∇ · v)] i =∂ ( ) ∂vj ∂ 2 v j= δ jk ,∂x i ∂x j ∂x i ∂x k( )∂ ∂v m∂ 2 v m[∇×(∇×v)] i = ɛ ijk ɛ klm = ɛ ijk ɛ klm ,∂x j ∂x l ∂x j ∂x l(a × b) · c = δ ij c i ɛ jkl a k b l = ɛ ikl c i a k b l . ◭An important relationship between the ɛ- andδ- tensors is expressed by theidentityɛ ijk ɛ klm = δ il δ jm − δ im δ jl . (26.30)To establish the validity of this identity between two fourth-order tensors (theLHS is a once-contracted sixth-order tensor) we consider the various possiblecases.The RHS of (26.30) has the values+1 if i = l and j = m ≠ i, (26.31)−1 if i = m and j = l ≠ i, (26.32)0 for any other set of subscript values i, j, l, m. (26.33)In each product on the LHS k has the same value in both factors and for anon-zero contribution none of i, l, j, m can have the same value as k. Since thereare only three values, 1, 2 and 3, that any of the subscripts may take, the onlynon-zero possibilities are i = l and j = m or vice versa but not all four subscriptsequal (since then each ɛ factor is zero, as it would be if i = j or l = m). Thisreproduces (26.33) for the LHS of (26.30) and also the conditions (26.31) and(26.32). The values in (26.31) and (26.32) are also reproduced in the LHS of(26.30) since(i) if i = l and j = m, ɛ ijk = ɛ lmk = ɛ klm and, whether ɛ ijk is +1 or −1, theproduct of the two factors is +1; and(ii) if i = m and j = l, ɛ ijk = ɛ mlk = −ɛ klm and thus the product ɛ ijk ɛ klm (nosummation) has the value −1.This concludes the establishment of identity (26.30).943

TENSORSA useful application of (26.30) is in obtaining alternative expressions for vectorquantities that arise from the vector product of a vector product.◮Obtain an alternative expression for ∇×(∇×v).As shown in the previous example, ∇×(∇×v) can be expressed in tensor form as∂ 2 v m[∇×(∇×v)] i = ɛ ijk ɛ klm∂x j ∂x l=(δ il δ jm − δ im δ jl ) ∂2 v m∂x j ∂x l= ∂ ( ) ∂vj−∂2 v i∂x i ∂x j ∂x j ∂x j=[∇(∇ · v)] i −∇ 2 v i ,where in the second line we have used the identity (26.30). This result has alreadybeen mentioned in chapter 10 and the reader is referred there for a discussion of itsapplicability. ◭By examining the various possibilities, it is straightforward to verify that, moregenerally,∣ δ ip δ iq δ ir ∣∣∣∣∣ɛ ijk ɛ pqr =δ jp δ jq δ jr(26.34)∣ δ kp δ kq δ krand it is easily seen that (26.30) is a special case of this result. From (26.34) wecan derive alternative forms of (26.30), for example,ɛ ijk ɛ ilm = δ jl δ km − δ jm δ kl . (26.35)The pattern of subscripts in these identities is most easily remembered by notingthat the subscripts on the first δ on the RHS are those that immediately follow(cyclically, if necessary) the common subscript, here i, ineachɛ-term on the LHS;the remaining combinations of j,k,l,m as subscripts in the other δ-terms on theRHS can then be filled in automatically.Contracting (26.35) by setting j = l (say) we obtain, since δ kk = 3 when usingthe summation convention,ɛ ijk ɛ ijm =3δ km − δ km =2δ km ,and by contracting once more, setting k = m, we further find thatɛ ijk ɛ ijk =6. (26.36)26.9 Isotropic tensorsIt will have been noticed that, unlike most of the tensors discussed (except forscalars), δ ij and ɛ ijk have the property that all their components have valuesthat are the same whatever rotation of axes is made, i.e. the component values944

26.9 ISOTROPIC TENSORSare independent of the transformation L ij . Specifically, δ 11 has the value 1 inall coordinate frames, whereas for a general second-order tensor T all we knowis that if T 11 = f 11 (x 1 ,x 2 ,x 3 )thenT 11 ′ = f 11(x ′ 1 ,x′ 2 ,x′ 3 ). Tensors with the formerproperty are called isotropic (or invariant) tensors.It is important to know the most general form that an isotropic tensor can take,since the description of the physical properties, e.g. the conductivity, magneticsusceptibility or tensile strength, of an isotropic medium (i.e. a medium havingthe same properties whichever way it is orientated) involves an isotropic tensor.In the previous section it was shown that δ ij and ɛ ijk are second- and third-orderisotropic tensors; we will now show that, to within a scalar multiple, they are theonly such isotropic tensors.Let us begin with isotropic second-order tensors. Suppose T ij is an isotropictensor; then, by definition, for any rotation of the axes we must have thatT ij = T ij ′ = L ik L jl T kl (26.37)for each of the nine components.First consider a rotation of the axes by 2π/3 about the (1, 1, 1) direction; thistakes Ox 1 , Ox 2 , Ox 3 into Ox ′ 2 , Ox′ 3 , Ox′ 1 respectively. For this rotation L 13 =1,L 21 =1,L 32 = 1 and all other L ij = 0. This requires that T 11 = T 11 ′ = T 33.Similarly T 12 = T 12 ′ = T 31. Continuing in this way, we find:(a) T 11 = T 22 = T 33 ;(b) T 12 = T 23 = T 31 ;(c) T 21 = T 32 = T 13 .Next, consider a rotation of the axes (from their original position) by π/2about the Ox 3 -axis. In this case L 12 = −1, L 21 =1,L 33 = 1 and all other L ij =0.Amongst other relationships, we must have from (26.37) that:T 13 =(−1) × 1 × T 23 ;T 23 =1× 1 × T 13 .Hence T 13 = T 23 = 0 and therefore, by parts (b) and (c) above, each element T ij =0 except for T 11 , T 22 and T 33 , which are all the same. This shows that T ij = λδ ij .◮Show that λɛ ijk is the only isotropic third-order Cartesian tensor.The general line of attack is as above and so only a minimum of explanation will be given.T ijk = T ijk ′ = L il L jm L kn T lmn (in all, there are 27 elements).Rotate about the (1, 1, 1) direction: this is equivalent to making subscript permutations1 → 2 → 3 → 1. We find(a) T 111 = T 222 = T 333 ,(b) T 112 = T 223 = T 331 (and two similar sets),(c) T 123 = T 231 = T 312 (and a set involving odd permutations of 1, 2, 3).945

TENSORSRotate by π/2 about the Ox 3 -axis: L 12 = −1, L 21 =1,L 33 = 1, the other L ij =0.(d) T 111 =(−1) × (−1) × (−1) × T 222 = −T 222 ,(e) T 112 =(−1) × (−1) × 1 × T 221 ,(f) T 221 =1× 1 × (−1) × T 112 ,(g) T 123 =(−1) × 1 × 1 × T 213 .Relations (a) and (d) show that elements with all subscripts the same are zero. Relations(e), (f) and (b) show that all elements with repeated subscripts are zero. Relations (g) and(c) show that T 123 = T 231 = T 312 = −T 213 = −T 321 = −T 132 .In total, T ijk differs from ɛ ijk by at most a scalar factor, but since ɛ ijk (and hence λɛ ijk )has already been shown to be an isotropic tensor, T ijk must be the most general third-orderisotropic Cartesian tensor. ◭Using exactly the same procedures as those employed for δ ij and ɛ ijk ,itmaybeshown that the only isotropic first-order tensor is the trivial one with all elementszero.26.10 Improper rotations and pseudotensorsSo far we have considered rigid rotations of the coordinate axes described byan orthogonal matrix L with |L| = +1, (26.4). Strictly speaking such transformationsare called proper rotations. We now broaden our discussion to includetransformations that are still described by an orthogonal matrix L but for which|L| = −1; these are called improper rotations.This kind of transformation can always be considered as an inversion of thecoordinate axes through the origin represented by the equationx ′ i = −x i , (26.38)combined with a proper rotation. The transformation may be looked uponalternatively as one that changes an initially right-handed coordinate system intoa left-handed one; any prior or subsequent proper rotation will not change thisstate of affairs. The most obvious example of a transformation with |L| = −1 isthe matrix corresponding to (26.38) itself; in this case L ij = −δ ij .As we have emphasised in earlier chapters, any real physical vector v may beconsidered as a geometrical object (i.e. an arrow in space), which can be referredto independently of any coordinate system and whose direction and magnitudecannot be altered merely by describing it in terms of a different coordinate system.Thus the components of v transform as v ′ i = L ijv j under all rotations (proper andimproper).We can define another type of object, however, whose components may alsobe labelled by a single subscript but which transforms as v ′ i = L ijv j under properrotations and as v ′ i = −L ij v j (note the minus sign) under improper rotations. Inthis case, the v i are not strictly the components of a true first-order Cartesiantensor but instead are said to form the components of a first-order Cartesianpseudotensor or pseudovector.946

26.10 IMPROPER ROTATIONS AND PSEUDOTENSORSx 3x ′ 1vpv ′Ox 2Ox 1x ′ 2p ′x ′ 3Figure 26.2 The behaviour of a vector v and a pseudovector p under areflection through the origin of the coordinate system x 1 ,x 2 ,x 3 giving the newsystem x ′ 1 ,x′ 2 ,x′ 3 .It is important to realise that a pseudovector (as its name suggests) is not ageometrical object in the usual sense. In particular, it should not be consideredas a real physical arrow in space, since its direction is reversed by an impropertransformation of the coordinate axes (such as an inversion through the origin).This is illustrated in figure 26.2, in which the pseudovector p is shown as a brokenline to indicate that it is not a real physical vector.Corresponding to vectors and pseudovectors, zeroth-order objects may bedivided into scalars and pseudoscalars – the latter being invariant under rotationbut changing sign on reflection.We may also extend the notion of scalars and pseudoscalars, vectors and pseudovectors,to objects with two or more subscripts. For two subcripts, as definedpreviously, any quantity with components that transform as T ij ′ = L ikL jl T kl underall rotations (proper and improper) is called a second-order Cartesian tensor.If, however, T ij ′ = L ikL jl T kl under proper rotations but T ij ′ = −L ikL jl T kl underimproper ones (which include reflections), then the T ij are the components ofa second-order Cartesian pseudotensor. In general the components of Cartesianpseudotensors of arbitary order transform asT ′ij···k = |L|L il L jm ···L kn T lm···n , (26.39)where |L| is the determinant of the transformation matrix.For example, from (26.29) we have that|L|ɛ ijk = L il L jm L kn ɛ lmn ,947

TENSORSbut since |L| = ±1 we may rewrite this asɛ ijk = |L|L il L jm L kn ɛ lmn .From this expression, we see that although ɛ ijk behaves as a tensor under properrotations, as discussed in section 26.8, it should properly be regarded as a thirdorderCartesian pseudotensor.◮If b j and c k are the components of vectors, show that the quantities a i = ɛ ijk b j c k formthe components of a pseudovector.In a new coordinate system we havea ′ i = ɛ ′ ijkb ′ jc ′ k= |L|L il L jm L kn ɛ lmn L jp b p L kq c q= |L|L il ɛ lmn δ mp δ nq b p c q= |L|L il ɛ lmn b m c n= |L|L il a l ,from which we see immediately that the quantities a i form the components of a pseudovector.◭The above example is worth some further comment. If we denote the vectorswith components b j and c k by b and c respectively then, as mentioned insection 26.8, the quantities a i = ɛ ijk b j c k are the components of the real vectora = b × c, provided that we are using a right-handed Cartesian coordinate system.However, in a coordinate system that is left-handed the quantitites a ′ i = ɛ′ ijk b′ j c′ kare not the components of the physical vector a = b × c, which has, instead, thecomponents −a ′ i . It is therefore important to note the handedness of a coordinatesystem before attempting to write in component form the vector relation a = b×c(which is true without reference to any coordinate system).It is worth noting that, although pseudotensors can be useful mathematicalobjects, the description of the real physical world must usually be in terms oftensors (i.e. scalars, vectors, etc.). § For example, the temperature or density of agas must be a scalar quantity (rather than a pseudoscalar), since its value doesnot change when the coordinate system used to describe it is inverted throughthe origin. Similarly, velocity, magnetic field strength or angular momentum canonly be described by a vector, and not by a pseudovector.At this point, it may be useful to make a brief comment on the distinctionbetween active and passive transformations of a physical system, as this differenceoften causes confusion. In this chapter, we are concerned solely with passive trans-§ In fact the quantum-mechanical description of elementary particles, such as electrons, protons andneutrons, requires the introduction of a new kind of mathematical object called a spinor, whichisnot a scalar, vector, or more general tensor. The study of spinors, however, falls beyond the scopeof this book.948

26.11 DUAL TENSORSformations, for which the physical system of interest is left unaltered, and onlythe coordinate system used to describe it is changed. In an active transformation,however, the system itself is altered.As an example, let us consider a particle of mass m that is located at a positionx relative to the origin O and hence has velocity ẋ. The angular momentum ofthe particle about O is thus J = m(x × ẋ). If we merely invert the Cartesiancoordinates used to describe this system through O, neither the magnitude nordirection of any these vectors will be changed, since they may be consideredsimply as arrows in space that are independent of the coordinates used to describethem. If, however, we perform the analogous active transformation onthe system, by inverting the position vector of the particle through O, thenitis clear that the direction of particle’s velocity will also be reversed, since itis simply the time derivative of the position vector, but that the direction ofits angular momentum vector remains unaltered. This suggests that vectors canbe divided into two categories, as follows: polar vectors (such as position andvelocity), which reverse direction under an active inversion of the physical systemthrough the origin, and axial vectors (such as angular momentum), whichremain unchanged. It should be emphasised that at no point in this discussionhave we used the concept of a pseudovector to describe a real physicalquantity. §26.11 Dual tensorsAlthough pseudotensors are not themselves appropriate for the description ofphysical phenomena, they are sometimes needed; for example, we may use thepseudotensor ɛ ijk to associate with every antisymmetric second-order tensor A ij(in three dimensions) a pseudovector p i given byp i = 1 2 ɛ ijkA jk ; (26.40)p i is called the dual of A ij . Thus if we denote the antisymmetric tensor A by thematrix⎛0 A 12 −A 31⎞A =[A ij ]= ⎝ −A 12 0 A 23⎠A 31 −A 23 0then the components of its dual pseudovector are (p 1 ,p 2 ,p 3 )=(A 23 ,A 31 ,A 12 ).§ The scalar product of a polar vector and an axial vector is a pseudoscalar. It was the experimentaldetection of the dependence of the angular distribution of electrons of (polar vector) momentump e emitted by polarised nuclei of (axial vector) spin J N upon the pseudoscalar quantity J N · p e thatestablished the existence of the non-conservation of parity in β-decay.949

TENSORS◮Using (26.40), show that A ij = ɛ ijk p k .By contracting both sides of (26.40) with ɛ ijk , we findɛ ijk p k = 1 ɛ 2 ijkɛ klm A lm .Using the identity (26.30) then givesɛ ijk p k = 1 (δ 2 ilδ jm − δ im δ jl )A lm= 1 (A 2 ij − A ji )= 1 (A 2 ij + A ij )=A ij ,whereinthelastlineweusethefactthatA ij = −A ji . ◭By a simple extension, we may associate a dual pseudoscalar s with everytotally antisymmetric third-rank tensor A ijk , i.e. one that is antisymmetric withrespect to the interchange of every possible pair of subscripts; s is given bys = 1 3! ɛ ijkA ijk . (26.41)Since A ijk is a totally antisymmetric three-subscript quantity, we expect it toequal some multiple of ɛ ijk (since this is the only such quantity). In fact A ijk = sɛ ijk ,as can be proved by substituting this expression into (26.41) and using (26.36).26.12 Physical applications of tensorsIn this section some physical applications of tensors will be given. First-ordertensors are familiar as vectors and so we will concentrate on second-order tensors,starting with an example taken from mechanics.Consider a collection of rigidly connected point particles of which the αth,which has mass m (α) and is positioned at r (α) with respect to an origin O, istypical. Suppose that the rigid assembly is rotating about an axis through O withangular velocity ω.The angular momentum J about O of the assembly is given byJ = ∑ (r (α) × p (α)) .αBut p (α) = m (α) ṙ (α) and ṙ (α) = ω × r (α) , for any α, and so in subscript form thecomponents of J are given byJ i = ∑ αm (α) ɛ ijk x (α)jẋ (α)k= ∑ α= ∑ α= ∑ αm (α) ɛ ijk x (α)jɛ klm ω l x m(α)m (α) (δ il δ jm − δ im δ jl )x (α)j x m (α) ω lm (α) [ (r(α) ) 2δil − x (α)ix (α)l]ω l ≡ I il ω l , (26.42)where I il is a symmetric second-order Cartesian tensor (by the quotient rule, see950

26.12 PHYSICAL APPLICATIONS OF TENSORSsection 26.7, since J and ω are vectors). The tensor is called the inertia tensor at Oof the assembly and depends only on the distribution of masses in the assemblyand not upon the direction or magnitude of ω.A more realistic situation obtains if a continuous rigid body is considered. Inthis case, m (α) must be replaced everywhere by ρ(r) dx dy dz and all summationsby integrations over the volume of the body. Written out in full in Cartesians,the inertia tensor for a continuous body would have the form⎛ ∫(y 2 + z 2 )ρdV − ∫ xyρ dV − ∫ ⎞xzρ dVI =[I ij ]= ⎝ − ∫ ∫xyρ dV (z 2 + x 2 )ρdV − ∫ yzρ dV− ∫ xzρ dV − ∫ ⎠ ,∫yzρ dV (x 2 + y 2 )ρdVwhere ρ = ρ(x, y, z) is the mass distribution and dV stands for dx dy dz; theintegrals are to be taken over the whole body. The diagonal elements of thistensor are called the moments of inertia and the off-diagonal elements without theminus signs are known as the products of inertia.◮Show that the kinetic energy of the rotating system is given by T = 1 2 I jlω j ω l .By an argument parallel to that already made for J, the kinetic energy is given by∑T = 1 m ( (α) ṙ (α) · ṙ (α))2= 1 2= 1 2= 1 2α∑αm (α) ɛ ijk ω j x (α)kɛ ilmω l x (α)m∑m (α) (δ jl δ km − δ jm δ kl )x (α)kx(α) m ω j ω lα∑ [ (m (α) δ ) ]jl r(α) 2− x(α)x (α) ω j ω lα= 1 I 2 jlω j ω l .Alternatively, since J j = I jl ω l we may write the kinetic energy of the rotating system asT = 1 J 2 jω j . ◭The above example shows that the kinetic energy of the rotating body can beexpressed as a scalar obtained by twice contracting ω with the inertia tensor. Italso shows that the moment of inertia of the body about a line given by the unitvector ˆn is I jlˆn j ˆn l (or ˆn T Iˆn in matrix form).Since I (≡ I jl ) is a real symmetric second-order tensor, it has associated with itthree mutually perpendicular directions that are its principal axes and have thefollowing properties (proved in chapter 8):(i) with each axis is associated a principal moment of inertia λ µ , µ =1, 2, 3;(ii) when the rotation of the body is about one of these axes, the angularvelocity and the angular momentum are parallel and given byJ = Iω = λ µ ω,i.e. ω is an eigenvector of I with eigenvalue λ µ ;951jl

TENSORS(iii) referred to these axes as coordinate axes, the inertia tensor is diagonalwith diagonal entries λ 1 ,λ 2 ,λ 3 .Two further examples of physical quantities represented by second-order tensorsare magnetic susceptibility and electrical conductivity. In the first case we have(in standard notation)M i = χ ij H j , (26.43)and in the second casej i = σ ij E j . (26.44)Here M is the magnetic moment per unit volume and j the current density(current per unit perpendicular area). In both cases we have on the left-hand sidea vector and on the right-hand side the contraction of a set of quantities withanother vector. Each set of quantities must therefore form the components of asecond-order tensor.For isotropic media M ∝ H and j ∝ E, but for anisotropic materials such ascrystals the susceptibility and conductivity may be different along different crystalaxes, making χ ij and σ ij general second-order tensors, although they are usuallysymmetric.◮The electrical conductivity σ in a crystal is measured by an observer to have componentsas shown:⎛[σ ij ]= ⎝ 1 √ ⎞√ 2 02 3 1 ⎠ . (26.45)0 1 1Show that there is one direction in the crystal along which no current can flow. Does thecurrent flow equally easily in the two perpendicular directions?The current density in the crystal is given by j i = σ ij E j ,whereσ ij ,relativetotheobserver’scoordinate system, is given by (26.45). Since [σ ij ] is a symmetric matrix, it possessesthree mutually perpendicular eigenvectors (or principal axes) with respect to which theconductivity tensor is diagonal, with diagonal entries λ 1 ,λ 2 ,λ 3 , the eigenvalues of [σ ij ].As discussed in chapter 8, the eigenvalues of [σ ij ] are given by |σ − λI| = 0. Thus werequire∣ √ ∣∣∣∣∣ 1√− λ 2 02 3− λ 10 1 1− λ ∣ =0,from which we find(1 − λ)[(3 − λ)(1 − λ) − 1] − 2(1 − λ) =0.This simplifies to give λ =0, 1, 4 so that, with respect to its principal axes, the conductivitytensor has components σ ij ′ given by ⎛[σ ij] ′ = ⎝ 4 0 0⎞0 1 0 ⎠ .0 0 0Since j ′ i = σ ′ ijE ′ j, we see immediately that along one of the principal axes there is no currentflow and along the two perpendicular directions the current flows are not equal. ◭952

26.12 PHYSICAL APPLICATIONS OF TENSORSWe can extend the idea of a second-order tensor that relates two vectors to asituation where two physical second-order tensors are related by a fourth-ordertensor. The most common occurrence of such relationships is in the theory ofelasticity. This is not the place to give a detailed account of elasticity theory,but suffice it to say that the local deformation of an elastic body at any interiorpoint P can be described by a second-order symmetric tensor e ij called the straintensor. It is given bye ij = 1 ( ∂ui+ ∂u )j,2 ∂x j ∂x iwhere u is the displacement vector describing the strain of a small volume elementwhose unstrained position relative to the origin is x. Similarly we can describethe stress in the body at P by the second-order symmetric stress tensor p ij ;thequantity p ij is the x j -component of the stress vector acting across a plane throughP whose normal lies in the x i -direction. A generalisation of Hooke’s law thenrelates the stress and strain tensors bywhere c ijkl is a fourth-order Cartesian tensor.p ij = c ijkl e kl (26.46)◮Assuming that the most general fourth-order isotropic tensor isc ijkl = λδ ij δ kl + ηδ ik δ jl + νδ il δ jk , (26.47)find the form of (26.46) for an isotropic medium having Young’s modulus E and Poisson’sratio σ.For an isotropic medium we must have an isotropic tensor for c ijkl , and so we assume theform (26.47). Substituting this into (26.46) yieldsp ij = λδ ij e kk + ηe ij + νe ji .But e ij is symmetric, and if we write η + ν =2µ, then this takes the formp ij = λe kk δ ij +2µe ij ,in which λ and µ are known as Lamé constants. It will be noted that if e ij =0fori ≠ jthen the same is true of p ij , i.e. the principal axes of the stress and strain tensors coincide.Now consider a simple tension in the x 1 -direction, i.e. p 11 = S but all other p ij =0.Then denoting e kk (summed over k) byθ we have, in addition to e ij =0fori ≠ j, thethreeequationsS = λθ +2µe 11 ,0=λθ +2µe 22 ,0=λθ +2µe 33 .Adding them givesS = θ(3λ +2µ).Substituting for θ from this into the first of the three, and recalling that Young’s modulusis defined by S = Ee 11 ,givesE asE =µ(3λ +2µ). (26.48)λ + µ953

TENSORSFurther, Poisson’s ratio is defined as σ = −e 22 /e 11 (or −e 33 /e 11 ) and is thus( ) ( )( ) 1 λθ 1 λσ =e 11 2µ = Ee11e 11 2µ 3λ +2µ = λ2(λ + µ) . (26.49)Solving (26.48) and (26.49) for λ and µ gives finallyp ij =σE(1 + σ)(1 − 2σ) e kkδ ij + E(1 + σ) e ij. ◭26.13 Integral theorems for tensorsIn chapter 11, we discussed various integral theorems involving vector and scalarfields. Most notably, we considered the divergence theorem, which states that, forany vector field a, ∫∮∇ · a dV = a · ˆn dS, (26.50)VSwhere S is the surface enclosing the volume V and ˆn is the outward-pointing unitnormal to S at each point.Writing (26.50) in subscript notation, we have∫∂a kdV = a k ˆn k dS. (26.51)V ∂x k SAlthough we shall not prove it rigorously, (26.51) can be extended in an obviousmanner to relate integrals of tensor fields, rather than just vector fields, overvolumes and surfaces, with the result∫∂T ij···k···mdV = T ij···k···m ˆn k dS.V ∂x k SThis form of the divergence theorem for general tensors can be very useful invector calculus manipulations.◮A vector field a satisfies ∇ · a =0inside some volume V and a · ˆn =0on the boundary∫ surface S. By considering the divergence theorem applied to T ij = x i a j , show thata dV = 0.VApplying the divergence theorem to T ij = x i a j we find∫∫∮∂T ij ∂(x i a j )dV = dV = x i a j ˆn j dS =0,V ∂x j V ∂x j Ssince a j ˆn j = 0. By expanding the volume integral we obtain∫V∫∫∂(x i a j ) ∂x idV = a j dV +∂x j V ∂x j∫= δ ij a j dVV∫= a i dV =0,V∮∮Vx i∂a j∂x jdVwhere in going from the first to the second line we used ∂x i /∂x j = δ ij and ∂a j /∂x j =0.◭954

26.14 NON-CARTESIAN COORDINATESThe other integral theorems discussed in chapter 11 can be extended in asimilar way. For example, written in tensor notation Stokes’ theorem states that,for a vector field a i ,∫∮∂a kɛ ijk ˆn i dS = a k dx k .S ∂x j CFor a general tensor field this has the straightforward extension∫Sɛ ijk∂T lm···k···n∂x jˆn i dS =∮CT lm···k···n dx k .26.14 Non-Cartesian coordinatesSo far we have restricted our attention to the study of tensors when they aredescribed in terms of Cartesian coordinates and the axes of coordinates are rigidlyrotated, sometimes together with an inversion of axes through the origin. In theremainder of this chapter we shall extend the concepts discussed in the previoussections by considering arbitrary coordinate transformations from one generalcoordinate system to another. Although this generalisation brings with it severalcomplications, we shall find that many of the properties of Cartesian tensorsare still valid for more general tensors. Before considering general coordinatetransformations, however, we begin by reminding ourselves of some properties ofgeneral curvilinear coordinates, as discussed in chapter 10.The position of an arbitrary point P in space may be expressed in terms of thethree curvilinear coordinates u 1 ,u 2 ,u 3 . We saw in chapter 10 that if r(u 1 ,u 2 ,u 3 )isthe position vector of the point P then at P there exist two sets of basis vectorse i = ∂r and ɛ i = ∇u i , (26.52)∂u iwhere i =1, 2, 3. In general, the vectors in each set neither are of unit length norform an orthogonal basis. However, the sets e i and ɛ i are reciprocal systems ofvectors and soe i · ɛ j = δ ij . (26.53)In the context of general tensor analysis, it is more usual to denote the secondset of vectors ɛ i in (26.52) by e i , the index being placed as a superscript todistinguish it from the (different) vector e i , which is a member of the first set in(26.52). Although this positioning of the index may seem odd (not least becauseof the possibility of confusion with powers) it forms part of a slight modificationto the summation convention that we will adopt for the remainder of this chapter.This is as follows: any lower-case alphabetic index that appears exactly twice inany term of an expression, once as a subscript and once as a superscript, istobesummed over all the values that an index in that position can take (unless the955

TENSORScontrary is specifically stated). All other aspects of the summation conventionremain unchanged.With the introduction of superscripts, the reciprocity relation (26.53) should berewritten so that both sides of (26.54) have one subscript and one superscript, i.e.ase i · e j = δ j i . (26.54)The alternative form of the Kronecker delta is defined in a similar way topreviously, i.e. it equals unity if i = j and is zero otherwise.For similar reasons it is usual to denote the curvilinear coordinates themselvesby u 1 ,u 2 ,u 3 , with the index raised, so thate i = ∂r∂u i and e i = ∇u i . (26.55)From the first equality we see that we may consider a superscript that appears inthe denominator of a partial derivative as a subscript.Given the two bases e i and e i , we may write a general vector a equally well interms of either basis as follows:a = a 1 e 1 + a 2 e 2 + a 3 e 3 = a i e i ;a = a 1 e 1 + a 2 e 2 + a 3 e 3 = a i e i .The a i are called the contravariant components of the vector a and the a ithe covariant components, the position of the index (either as a subscript orsuperscript) serving to distinguish between them. Similarly, we may call the e i thecovariant basis vectors and the e i the contravariant ones.◮Show that the contravariant and covariant components of a vector a are given by a i = a·e iand a i = a · e i respectively.For the contravariant components, we finda · e i = a j e j · e i = a j δj i = a i ,where we have used the reciprocity relation (26.54). Similarly, for the covariant components,a · e i = a j e j · e i = a j δ j i= a i . ◭The reason that the notion of contravariant and covariant components ofa vector (and the resulting superscript notation) was not introduced earlier isthat for Cartesian coordinate systems the two sets of basis vectors e i and e i areidentical and, hence, so are the components of a vector with respect to eitherbasis. Thus, for Cartesian coordinates, we may speak simply of the componentsof the vector and there is no need to differentiate between contravariance andcovariance, or to introduce superscripts to make a distinction between them.If we consider the components of higher-order tensors in non-Cartesian coordinates,there are even more possibilities. As an example, let us consider a956

26.15 THE METRIC TENSORsecond-order tensor T. Using the outer product notation in (26.23), we may writeT in three different ways:T = T ij e i ⊗ e j = T i je i ⊗ e j = T ij e i ⊗ e j ,where T ij , T i j and T ij are called the contravariant, mixed and covariant componentsof T respectively. It is important to remember that these three sets ofquantities form the components of the same tensor T but refer to different (tensor)bases made up from the basis vectors of the coordinate system. Again, if we areusing Cartesian coordinates then all three sets of components are identical.We may generalise the above equation to higher-order tensors. Componentscarrying only superscripts or only subscripts are referred to as the contravariantand covariant components respectively; all others are called mixed components.26.15 The metric tensorAny particular curvilinear coordinate system is completely characterised at eachpoint in space by the nine quantitiesg ij = e i · e j , (26.56)which, as we will show, are the covariant components of a symmetric second-ordertensor g called the metric tensor.Since an infinitesimal vector displacement can be written as dr = du i e i , we findthat the square of the infinitesimal arc length (ds) 2 can be written in terms of themetric tensor as(ds) 2 = dr · dr = du i e i · du j e j = g ij du i du j . (26.57)It may further be shown that the volume element dV is given bydV = √ gdu 1 du 2 du 3 , (26.58)where g is the determinant of the matrix [ g ij ], which has the covariant componentsof the metric tensor as its elements.If we compare equations (26.57) and (26.58) with the analogous ones in section10.10 then we see that in the special case where the coordinate system is orthogonal(so that e i · e j =0fori ≠ j) the metric tensor can be written in terms of thecoordinate-system scale factors h i , i =1, 2, 3as{h2g ij = i i = j,0 i ≠ j.Its determinant is then given by g = h 2 1 h2 2 h2 3 .957

TENSORS◮Calculate the elements g ij of the metric tensor for cylindrical polar coordinates. Hencefind the square of the infinitesimal arc length (ds) 2 and the volume dV for this coordinatesystem.As discussed in section 10.9, in cylindrical polar coordinates (u 1 ,u 2 ,u 3 )=(ρ, φ, z) andsothe position vector r of any point P may be writtenr = ρ cos φ i + ρ sin φ j + z k.From this we obtain the (covariant) basis vectors:e 1 = ∂r = cosφ i +sinφ j;∂ρe 2 = ∂r = −ρ sin φ i + ρ cos φ j;∂φe 3 = ∂r = k. (26.59)∂zThus the components of the metric tensor [g ij ]=[e i · e j ] are found to be⎛G =[g ij ]= ⎝ 1 0 0⎞0 ρ 2 0 ⎠ , (26.60)0 0 1from which we see that, as expected for an orthogonal coordinate system, the metric tensoris diagonal, the diagonal elements being equal to the squares of the scale factors of thecoordinate system.From (26.57), the square of the infinitesimal arc length in this coordinate system is givenby(ds) 2 = g ij du i du j =(dρ) 2 + ρ 2 (dφ) 2 +(dz) 2 ,and, using (26.58), the volume element is found to bedV = √ gdu 1 du 2 du 3 = ρdρdφdz.These expressions are identical to those derived in section 10.9. ◭We may also express the scalar product of two vectors in terms of the metrictensor:a · b = a i e i · b j e j = g ij a i b j , (26.61)where we have used the contravariant components of the two vectors. Similarly,using the covariant components, we can write the same scalar product asa · b = a i e i · b j e j = g ij a i b j , (26.62)where we have defined the nine quantities g ij = e i ·e j . As we shall show, they formthe contravariant components of the metric tensor g and are, in general, differentfrom the quantities g ij . Finally, we could express the scalar product in terms ofthe contravariant components of one vector and the covariant components of theother,a · b = a i e i · b j e j = a i b j δ i j = a i b i , (26.63)958

26.15 THE METRIC TENSORwhere we have used the reciprocity relation (26.54). Similarly, we could writea · b = a i e i · b j e j = a i b j δ j i = a i b i . (26.64)By comparing the four alternative expressions (26.61)–(26.64) for the scalarproduct of two vectors we can deduce one of the most useful properties ofthe quantities g ij and g ij . Since g ij a i b j = a i b i holds for any arbitrary vectorcomponents a i , it follows thatg ij b j = b i ,which illustrates the fact that the covariant components g ij ofthemetrictensorcan be used to lower an index. In other words, it provides a means of obtainingthe covariant components of a vector from its contravariant components. By asimilar argument, we haveg ij b j = b i ,so that the contravariant components g ij can be used to perform the reverseoperation of raising an index.It is straightforward to show that the contravariant and covariant basis vectors,e i and e i respectively, are related in the same way as other vectors, i.e. bye i = g ij e j and e i = g ij e j .We also note that, since e i and e i are reciprocal systems of vectors in threedimensionalspace (see chapter 7), we may writee i =e j × e ke i · (e j × e k ) ,for the combination of subscripts i, j, k =1, 2, 3 and its cyclic permutations. Asimilar expression holds for e i in terms of the e i -basis. Moreover, it may be shownthat |e 1 · (e 2 × e 3 )| = √ g.◮Show that the matrix [g ij ] is the inverse of the matrix [g ij ]. Hence calculate the contravariantcomponents g ij of the metric tensor in cylindrical polar coordinates.Using the index-lowering and index-raising properties of g ij and g ij on an arbitrary vectora, we findδka i k = a i = g ij a j = g ij g jk a k .But, since a is arbitrary, we must haveg ij g jk = δk. i (26.65)Denoting the matrix [g ij ]byG and [g ij ]byĜ, equation (26.65) can be written in matrixform as ĜG = I, whereI is the unit matrix. Hence G and Ĝ are inverse matrices of eachother.959

TENSORSThus, by inverting the matrix G in (26.60), we find that the elements g ij are given incylindrical polar coordinates byĜ =[g ij ]=⎛⎝ 1 0 0 1/ρ 00⎞⎠ . ◭0 0 1So far we have not considered the components of the metric tensor gj i with onesubscript and one superscript. By analogy with (26.56), these mixed componentsare given bygj i = e i · e j = δ j i ,and so the components of gji are identical to those of δj i.Wemaythereforeconsider the δj i to be the mixed components of the metric tensor g.26.16 General coordinate transformations and tensorsWe now discuss the concept of general transformations from one coordinatesystem, u 1 ,u 2 ,u 3 , to another, u ′1 ,u ′2 ,u ′3 . We can describe the coordinate transformusing the three equationsu ′i = u ′i (u 1 ,u 2 ,u 3 ),for i =1, 2, 3, in which the new coordinates u ′i can be arbitrary functions of the oldones u i rather than just represent linear orthogonal transformations (rotations)of the coordinate axes. We shall assume also that the transformation can beinverted, so that we can write the old coordinates in terms of the new ones asu i = u i (u ′1 ,u ′2 ,u ′3 ),As an example, we may consider the transformation from spherical polar toCartesian coordinates, given byx = r sin θ cos φ,y = r sin θ sin φ,z = r cos θ,which is clearly not a linear transformation.The two sets of basis vectors in the new coordinate system, u ′1 ,u ′2 ,u ′3 , are givenas in (26.55) bye ′ i = ∂r and e ′i = ∇u ′i . (26.66)∂u ′iConsidering the first set, we have from the chain rule that∂r∂u j = ∂u′ i∂r∂u j ∂u ′ i ,960

26.16 GENERAL COORDINATE TRANSFORMATIONS AND TENSORSso that the basis vectors in the old and new coordinate systems are related bye j = ∂u′ i∂u j e′ i. (26.67)Now, since we can write any arbitrary vector a in terms of either basis asa = a ′i e ′ i = a j e j = a j ∂u′i∂u j e′ i,it follows that the contravariant components of a vector must transform asa ′i = ∂u′ i∂u j a j . (26.68)In fact, we use this relation as the defining property for a set of quantities a i toform the contravariant components of a vector.◮Find an expression analogous to (26.67) relating the basis vectors e i and e ′i in the twocoordinate systems. Hence deduce the way in which the covariant components of a vectorchange under a coordinate transformation.If we consider the second set of basis vectors in (26.66), e ′i = ∇u ′i , we have from the chainrule that∂u j∂x = ∂u j ∂u ′i∂u ′i ∂xand similarly for ∂u j /∂y and ∂u j /∂z. So the basis vectors in the old and new coordinatesystems are related bye j = ∂u j∂u ′i e′i . (26.69)For any arbitrary vector a,a = a ′ ie ′i = a j e j = a j∂u je′i∂u′iand so the covariant components of a vector must transform asa ′ i = ∂u j∂u a j. (26.70)′iAnalogously to the contravariant case (26.68), we take this result as the defining propertyof the covariant components of a vector. ◭We may compare the transformation laws (26.68) and (26.70) with those fora first-order Cartesian tensor under a rigid rotation of axes. Let us considera rotation of Cartesian axes x i throughanangleθ about the 3-axis to a newset x ′i , i =1, 2, 3, as given by (26.7) and the inverse transformation (26.8). It isstraightforward to show that∂x j= ∂x′ i∂x ′ i∂x j = L ij,961

TENSORSwhere the elements L ij are given by⎛L = ⎝cos θ sin θ⎞0− sin θ cos θ 0 ⎠ .0 0 1Thus (26.68) and (26.70) agree with our earlier definition in the special case of arigid rotation of Cartesian axes.Following on from (26.68) and (26.70), we proceed in a similar way to definegeneral tensors of higher rank. For example, the contravariant, mixed andcovariant components, respectively, of a second-order tensor must transform asfollows:contravariant components,mixed components,covariant components,T ′ ij = ∂u′ i∂u k ∂u ′ j∂u l T kl ;T ′ ij = ∂u′ i∂u k ∂u l∂u ′ j T k l ;T ′ ij = ∂uk∂u ′ i∂u l∂u ′ j T kl.It is important to remember that these quantities form the components of thesame tensor T but refer to different tensor bases made up from the basis vectorsof the different coordinate systems. For example, in terms of the contravariantcomponents we may writeT = T ij e i ⊗ e j = T ′ ij e ′ i ⊗ e ′ j.We can clearly go on to define tensors of higher order, with arbitrary numbersof covariant (subscript) and contravariant (superscript) indices, by demandingthat their components transform as follows:T ′ ij···klm···n = ∂u′ i∂u ′ j···∂u′ k∂u d ∂u e ∂uf∂u a ∂u b ∂u c ···∂u ′ l∂u′m∂u ′n T ab···c de···f. (26.71)Using the revised summation convention described in section 26.14, the algebraof general tensors is completely analogous to that of the Cartesian tensorsdiscussed earlier. For example, as with Cartesian coordinates, the Kroneckerdelta is a tensor provided it is written as the mixed tensor δj i sinceδ ′i j = ∂u′i ∂u l∂u k ∂u ′j δk l = ∂u′i ∂u k ∂u′i∂u k =∂u′j∂u = ′j δi j,where we have used the chain rule to justify the third equality. This also showsthat δj i is isotropic. As discussed at the end of section 26.15, the δj i can beconsidered as the mixed components of the metric tensor g.962

26.17 RELATIVE TENSORS◮Show that the quantities g ij = e i · e jtensor.form the covariant components of a second-orderIn the new (primed) coordinate system we haveg ij ′ = e ′ i · e ′ j,but using (26.67) for the inverse transformation, we havee ′ i = ∂uk∂u e k,′iand similarly for e ′ j .Thuswemaywriteg ij ′ = ∂uk ∂u le∂u ′i ∂u ′j k · e l = ∂uk ∂u l∂u ′i ∂u g kl,′jwhich shows that the g ij are indeed the covariant components of a second-order tensor(themetrictensorg). ◭A similar argument to that used in the above example shows that the quantitiesg ij form the contravariant components of a second-order tensor which transformsaccording tog ′ ij = ∂u′ i∂u ′ j∂u k ∂u l gkl .In the previous section we discussed the use of the components g ij and g ij inthe raising and lowering of indices in contravariant and covariant vectors. Thiscan be extended to tensors of arbitrary rank. In general, contraction of a tensorwith g ij will convert the contracted index from being contravariant (superscript)to covariant (subscript), i.e. it is lowered. This can be repeated for as many indicesare required. For example,Similarly contraction with g ij raises an index, i.e.T ij = g ik T k j = g ik g jl T kl . (26.72)T ij = g ik T jk = gik g jl T kl . (26.73)That (26.72) and (26.73) are mutually consistent may be shown by using the factthat g ik g kj = δ i j .26.17 Relative tensorsIn section 26.10 we introduced the concept of pseudotensors in the context of therotation (proper or improper) of a set of Cartesian axes. Generalising to arbitrarycoordinate transformations leads to the notion of a relative tensor.For an arbitrary coordinate transformation from one general coordinate system963

TENSORSu i to another u ′i , we may define the Jacobian of the transformation (see chapter 6)as the determinant of the transformation matrix [∂u ′i /∂u j ]: this is usually denotedbyJ =∂u ′∣ ∂u ∣ .Alternatively, we may interchange the primed and unprimed coordinates toobtain |∂u/∂u ′ | =1/J: unfortunately this also is often called the Jacobian of thetransformation.Using the Jacobian J, we define a relative tensor of weight w as one whosecomponents transform as follows:T ′ ij···klm···n = ∂u′ i∂u ′ j···∂u′ k∣∂u d ∂u e ∂uf∂u a ∂u b ∂u c ∂u ′ l∂u ′ m ···∂u ′ n T ab···c ∂u ∣∣∣wde···f ∣∂u ′ .(26.74)Comparing this expression with (26.71), we see that a true (or absolute) generaltensor may be considered as a relative tensor of weight w =0.Ifw = −1, on theother hand, the relative tensor is known as a general pseudotensor and if w =1as a tensor density.It is worth comparing (26.74) with the definition (26.39) of a Cartesian pseudotensor.For the latter, we are concerned only with its behaviour under a rotation(proper or improper) of Cartesian axes, for which the Jacobian J = ±1. Thus,general relative tensors of weight w = −1 andw = 1 would both satisfy thedefinition (26.39) of a Cartesian pseudotensor.◮If the g ij are the covariant components of the metric tensor, show that the determinant gof the matrix [g ij ] is a relative scalar of weight w =2.The components g ij transform asg ij ′ = ∂uk ∂u l∂u ′i ∂u g kl.′jDefining the matrices U =[∂u i /∂u ′j ], G =[g ij ]andG ′ =[g ij ′ ], we may write this expressionasG ′ = U T GU.Taking the determinant of both sides, we obtain∣ g ′ = |U| 2 g =∂u ∣∣∣2∣ g,∂u ′which shows that g is a relative scalar of weight w =2.◭From the discussion in section 26.8, it can be seen that ɛ ijk is a covariantrelative tensor of weight −1. We may also define the contravariant tensor ɛ ijk ,which is numerically equal to ɛ ijk but is a relative tensor of weight +1.If two relative tensors have weights w 1 and w 2 respectively then, from (26.74),964

26.18 DERIVATIVES OF BASIS VECTORS AND CHRISTOFFEL SYMBOLSthe outer product of the two tensors, or any contraction of them, is a relativetensor of weight w 1 + w 2 . As a special case, we may use ɛ ijk and ɛ ijk to constructpseudovectors from antisymmetric tensors and vice versa, in an analogous wayto that discussed in section 26.11.For example, if the A ij are the contravariant components of an antisymmetrictensor (w =0)thenp i = 1 2 ɛ ijkA jkare the covariant components of a pseudovector (w = −1), since ɛ ijk has weightw = −1. Similarly, we may show thatA ij = ɛ ijk p k .26.18 Derivatives of basis vectors and Christoffel symbolsIn Cartesian coordinates, the basis vectors e i are constant and so their derivativeswith respect to the coordinates vanish. In a general coordinate system, however,the basis vectors e i and e i are functions of the coordinates. Therefore, in orderthat we may differentiate general tensors we must consider the derivatives of thebasis vectors.First consider the derivative ∂e i /∂u j . Since this is itself a vector, it can bewritten as a linear combination of the basis vectors e k , k =1, 2, 3. If we introducethe symbol Γ k ijto denote the coefficients in this combination, we have∂e i∂u j =Γk ije k . (26.75)The coefficient Γ k ij is the kth component of the vector ∂e i/∂u j .Usingthereciprocityrelation e i · e j = δj i , these 27 numbers are given (at each point in space)byΓ k ij = e k · ∂e i∂u j . (26.76)Furthermore, by differentiating the reciprocity relation e i · e j = δj i with respectto the coordinates, and using (26.76), it is straightforward to show that thederivatives of the contravariant basis vectors are given by∂e i∂u j = −Γi kje k . (26.77)The symbol Γ k ijis called a Christoffel symbol (of the second kind), but, despiteappearances to the contrary, these quantities do not form the components of athird-order tensor. It is clear from (26.76) that in Cartesian coordinates Γ k ij =0for all values of the indices i, j and k.965

TENSORS◮Using (26.76), deduce the way in which the quantities Γ k ijtransform under a generalcoordinate transformation, and hence show that they do not form the components of athird-order tensor.In a new coordinate systemΓ ′k ij = e ′k · ∂e′ i∂u , ′jbut from (26.69) and (26.67) respectively we have, on reversing primed and unprimedvariables,e ′k = ∂u′k∂u n en and e ′ i = ∂ul∂u e l.′iTherefore in the new coordinate system the quantities Γ ′k ij are given byΓ ′k ij = ∂u′k∂u n en ·= ∂u′k∂u n en ·( )∂ ∂ul∂u ′j ∂u e ′i l( )∂ 2 u l∂u ′j ∂u e ′i l + ∂ul ∂e l∂u ′i ∂u ′j= ∂u′k ∂ 2 u le n · e∂u n ∂u ′j ∂u ′i l + ∂u′k ∂u l ∂u me n · ∂e l∂u n ∂u ′i ∂u ′j ∂u m= ∂u′k ∂ 2 u l+ ∂u′k ∂u l ∂u m∂u l ∂u ′j ∂u ′i ∂u n ∂u ′i ∂u ′j Γn lm, (26.78)where in the last line we have used (26.76) and the reciprocity relation e n · e l = δl n.From(26.78), because of the presence of the first term on the right-hand side, we concludeimmediately that the Γ k ijdo not form the components of a third-order tensor. ◭In a given coordinate system, in principle we may calculate the Γ k ijusing(26.76). In practice, however, it is often quicker to use an alternative expression,which we now derive, for the Christoffel symbol in terms of the metric tensor g ijand its derivatives with respect to the coordinates.Firstly we note that the Christoffel symbol Γ k ijis symmetric with respect tothe interchange of its two subscripts i and j. This is easily shown: since∂e i∂u j =∂2 r∂u j ∂u i =∂2 r∂u i ∂u j = ∂e j∂u i ,it follows from (26.75) that Γ k ij e k =Γ k ji e k. Taking the scalar product with e l andusing the reciprocity relation e k · e l = δk l gives immediately thatΓ l ij =Γ l ji.To obtain an expression for Γ k ijwe then use g ij = e i · e j and consider thederivative∂g ij∂u k = ∂e i∂u k · e j + e i · ∂e j∂u k=Γ l ike l · e j + e i · Γ l jke l=Γ l ikg lj +Γ l jkg il , (26.79)966

26.18 DERIVATIVES OF BASIS VECTORS AND CHRISTOFFEL SYMBOLSwhere we have used the definition (26.75). By cyclically permuting the free indicesi, j, k in (26.79), we obtain two further equivalent relations,and∂g jk∂u i =Γ l jig lk +Γ l kig jl (26.80)∂g ki∂u j =Γ l kjg li +Γ l ijg kl . (26.81)If we now add (26.80) and (26.81) together and subtract (26.79) from the result,we find∂g jk∂u i+ ∂g ki∂u j − ∂g ij∂u k =Γl jig lk +Γ l kig jl +Γ l kjg li +Γ l ijg kl − Γ l ikg lj − Γ l jkg il=2Γ l ijg kl ,where we have used the symmetry properties of both Γ l ij and g ij. Contractingboth sides with g mk leads to the required expression for the Christoffel symbol interms of the metric tensor and its derivatives, namelyΓ m ij = 1 2 gmk (∂gjk∂u i+ ∂g ki∂u j − ∂g )ij∂u k . (26.82)◮Calculate the Christoffel symbols Γ m ij for cylindrical polar coordinates.We may use either (26.75) or (26.82) to calculate the Γ m ij for this simple coordinate system.In cylindrical polar coordinates (u 1 ,u 2 ,u 3 )=(ρ, φ, z), the basis vectors e i are given by(26.59). It is straightforward to show that the only derivatives of these vectors with respectto the coordinates that are non-zero are∂e ρ∂φ = 1 ρ e φ,∂e φ∂ρ = 1 ρ e φ,∂e φ∂φ = −ρe ρ.Thus, from (26.75), we have immediately thatΓ 2 12 =Γ 2 21 = 1 ρand Γ 1 22 = −ρ. (26.83)Alternatively, using (26.82) and the fact that g 11 =1,g 22 = ρ 2 , g 33 = 1 and the othercomponents are zero, we see that the only three non-zero Christoffel symbols are indeedΓ 2 12 =Γ2 21 and Γ1 22 .ThesearegivenbyΓ 2 12 =Γ 2 21 = 1 ∂g 222g 22 ∂u = 1 ∂1 2ρ 2 ∂ρ (ρ2 )= 1 ρ ,Γ 1 22 = − 1 ∂g 222g 11 ∂u = − 1 ∂1 2 ∂ρ (ρ2 )=−ρ,which agree with the expressions found directly from (26.75) and given in (26.83 ). ◭967

TENSORS26.19 Covariant differentiationFor Cartesian tensors we noted that the derivative of a scalar is a (covariant)vector.Thisisalsotrueforgeneral tensors, as may be shown by considering thedifferential of a scalardφ = ∂φ∂u i dui .Since the du i are the components of a contravariant vector and dφ is a scalar,we have by the quotient law, discussed in section 26.7, that the quantities ∂φ/∂u imust form the components of a covariant vector. As a second example, if thecontravariant components in Cartesian coordinates of a vector v are v i , then thequantities ∂v i /∂x j form the components of a second-order tensor.However, it is straightforward to show that in non-Cartesian coordinates differentiationof the components of a general tensor, other than a scalar, with respectto the coordinates does not in general result in the components of another tensor.◮Show that, in general coordinates, the quantities ∂v i /∂u j do not form the components ofa tensor.We may show this directly by considering( ) ∂vi ′= ∂v′i= ∂uk ∂v ′i∂u j ∂u ′j ∂u ′j ∂u k( )= ∂uk ∂ ∂u ′i∂u ′j ∂u k ∂u l vl= ∂uk ∂u ′i ∂v l∂u ′j ∂u l ∂u + ∂uk ∂ 2 u ′ik ∂u ′j ∂u k ∂u l vl . (26.84)The presence of the second term on the right-hand side of (26.84) shows that the ∂v i /∂x jdo not form the components of a second-order tensor. This term arises because the‘transformation matrix’ [∂u ′i /∂u j ] changes as the position in space at which it is evaluatedis changed. This is not true in Cartesian coordinates, for which the second term vanishesand ∂v i /∂x j is a second-order tensor. ◭We may, however, use the Christoffel symbols discussed in the previous sectionto define a new covariant derivative of the components of a tensor that doesresult in the components of another tensor.Let us first consider the derivative of a vector v with respect to the coordinates.Writing the vector in terms of its contravariant components v = v i e i , we find∂v∂u j = ∂vi∂u j e i + v i ∂e i∂u j , (26.85)where the second term arises because, in general, the basis vectors e i are not968

26.19 COVARIANT DIFFERENTIATIONconstant (this term vanishes in Cartesian coordinates). Using (26.75) we write∂v∂u j = ∂vi∂u j e i + v i Γ k ije k .Since i and k are dummy indices in the last term on the right-hand side, we mayinterchange them to obtain( )∂v∂u j = ∂vi∂v∂u j e i + v k Γ i ikje i =∂u j + vk Γ i kj e i . (26.86)The reason for the interchanging the dummy indices, as shown in (26.86), is thatwe may now factor out e i . The quantity in parentheses is called the covariantderivative, for which the standard notation isv i ; j ≡ ∂vi∂u j +Γi kjv k , (26.87)the semicolon subscript denoting covariant differentiation. A similar short-handnotation also exists for the partial derivatives, a comma being used for theseinstead of a semicolon; for example, ∂v i /∂u j is denoted by v i ,j. In Cartesiancoordinates all the Γ i kjare zero, and so the covariant derivative reduces to thesimple partial derivative ∂v i /∂u j .Using the short-hand semicolon notation, the derivative of a vector may bewritten in the very compact form∂v∂u j = vi ; je iand, by the quotient rule (section 26.7), it is clear that the v i ; j are the (mixed)components of a second-order tensor. This may also be verified directly, usingthe transformation properties of ∂v i /∂u j and Γ i kjgiven in (26.84) and (26.78)respectively.In general, we may regard the v i ; j as the mixed components of a secondordertensor called the covariant derivative of v and denoted by ∇v. In Cartesiancoordinates, the components of this tensor are just ∂v i /∂x j .◮Calculate v i ; i in cylindrical polar coordinates.Contracting (26.87) we obtainNow from (26.83) we havev i ; i = ∂vi∂u i +Γ i kiv k .Γ i 1i =Γ 1 11 +Γ 2 12 +Γ 3 13 =1/ρ,Γ i 2i =Γ 1 21 +Γ 2 22 +Γ 3 23 =0,Γ i 3i =Γ 1 31 +Γ 2 32 +Γ 3 33 =0,969

TENSORSand sov i ; i = ∂vρ∂ρ + ∂vφ∂φ + ∂vz∂z + 1 ρ vρ= 1 ∂ρ ∂ρ (ρvρ )+ ∂vφ∂φ + ∂vz∂z .This result is identical to the expression for the divergence of a vector field in cylindricalpolar coordinates given in section 10.9. This is discussed further in section 26.20. ◭So far we have considered only the covariant derivative of the contravariantcomponents v i of a vector. The corresponding result for the covariant componentsv i may be found in a similar way, by considering the derivative of v = v i e i andusing (26.77) to obtainv i; j = ∂v i∂u j − Γk ijv k . (26.88)Comparing the expressions (26.87) and (26.88) for the covariant derivative ofthe contravariant and covariant components of a vector respectively, we see thatthere are some similarities and some differences. It may help to remember thatthe index with respect to which the covariant derivative is taken (j in this case), isalso the last subscript on the Christoffel symbol; the remaining indices can thenbe arranged in only one way without raising or lowering them. It only remains tonote that for a covariant index (subscript) the Christoffel symbol carries a minussign, whereas for a contravariant index (superscript) the sign is positive.Following a similar procedure to that which led to equation (26.87), we mayobtain expressions for the covariant derivatives of higher-order tensors.◮By considering the derivative of the second-order tensor T with respect to the coordinateu k , find an expression for the covariant derivative T ij ; kof its contravariant components.Expressing T in terms of its contravariant components, we have∂T∂u = ∂k ∂u (T ij e k i ⊗ e j )ij∂T=∂u e k i ⊗ e j + T ij ∂e i∂u ⊗ e k j + T ij e i ⊗ ∂e j∂u . kUsing (26.75), we can rewrite the derivatives of the basis vectors in terms of Christoffelsymbols to obtainij∂T ∂T=∂uk ∂u e k i ⊗ e j + T ij Γ l ike l ⊗ e j + T ij e i ⊗ Γ l jke l .Interchanging the dummy indices i and l in the second term and j and l in the third termon the right-hand side, this becomes( )∂T ∂Tij∂u = +Γ i k ∂ulkT lj +Γ j k lk T il e i ⊗ e j ,970

26.20 VECTOR OPERATORS IN TENSOR FORMwhere the expression in parentheses is the required covariant derivativeT ij ij∂T; k= +Γ i∂ulkT lj +Γ j k lk T il . (26.89)Using (26.89), the derivative of the tensor T with respect to u k cannowbewrittenintermsof its contravariant components as∂T∂u = T ij k ; k e i ⊗ e j . ◭Results similar to (26.89) may be obtained for the the covariant derivatives ofthe mixed and covariant components of a second-order tensor. Collecting theseresults together, we haveT ij ; k = T ij ,k +Γi lkT lj +Γ j lk T il ,T i j; k = T i j,k +Γ i lkT l j − Γ l jkT i l,T ij; k = T ij, k − Γ l ikT lj − Γ l jkT il ,where we have used the comma notation for partial derivatives. The position ofthe indices in these expressions is very systematic: for each contravariant index(superscript) on the LHS we add a term on the RHS containing a Christoffelsymbol with a plus sign, and for every covariant index (subscript) we add acorresponding term with a minus sign. This is extended straightforwardly totensors with an arbitrary number of contravariant and covariant indices.We note that the quantities T ij ; k , T i j; k and T ij; k are the components of thesame third-order tensor ∇T with respect to different tensor bases, i.e.∇T = T ij ; k e i ⊗ e j ⊗ e k = T i j; ke i ⊗ e j ⊗ e k = T ij; k e i ⊗ e j ⊗ e k .We conclude this section by considering briefly the covariant derivative of ascalar. The covariant derivative differs from the simple partial derivative withrespect to the coordinates only because the basis vectors of the coordinatesystem change with position in space (hence for Cartesian coordinates there is nodifference). However, a scalar φ does not depend on the basis vectors at all andso its covariant derivative must be the same as its partial derivative, i.e.φ ; j = ∂φ∂u j = φ ,j. (26.90)26.20 Vector operators in tensor formIn section 10.10 we used vector calculus methods to find expressions for vectordifferential operators, such as grad, div, curl and the Laplacian, in general orthogonalcurvilinear coordinates, taking cylindrical and spherical polars as particularexamples. In this section we use the framework of general tensors that we havedeveloped to obtain, in tensor form, expressions for these operators that are validin all coordinate systems, whether orthogonal or not.971

TENSORSIn order to compare the results obtained here with those given in section10.10 for orthogonal coordinates, it is necessary to remember that here we areworking with the (in general) non-unit basis vectors e i = ∂r/∂u i or e i = ∇u i .Thus the components of a vector v = v i e i are not the same as the components ˆv iappropriate to the corresponding unit basis ê i . In fact, if the scale factors of thecoordinate system are h i , i =1, 2, 3, then v i = ˆv i /h i (no summation over i).As mentioned in section 26.15, for an orthogonal coordinate system with scalefactors h i we have{h2g ij = i if i = j,0 otherwise{1/hand g ij 2= i if i = j,0 otherwise,and so the determinant g of the matrix [g ij ] is given by g = h 2 1 h2 2 h2 3 .GradientThe gradient of a scalar φ is given by∇φ = φ ; i e i = ∂φ∂u i ei , (26.91)since the covariant derivative of a scalar is the same as its partial derivative.DivergenceReplacing the partial derivatives that occur in Cartesian coordinates with covariantderivatives, the divergence of a vector field v in a general coordinate systemis given by∇ · v = v i ; i = ∂vi∂u i +Γ i kiv k .Using the expression (26.82) for the Christoffel symbol in terms of the metrictensor, we find(Γ i ki = 1 ∂gil2 gil ∂u k + ∂g kl∂u i − ∂g )ki∂u l = 1 ∂g 2 gil il∂u k . (26.92)The last two terms have cancelled becauseg il ∂g kl∂u i = g li ∂g ki∂u l = g il ∂g ki∂u l ,where in the first equality we have interchanged the dummy indices i and l, andin the second equality have used the symmetry of the metric tensor.We may simplify (26.92) still further by using a result concerning the derivativeof the determinant of a matrix whose elements are functions of the coordinates.972

26.20 VECTOR OPERATORS IN TENSOR FORM◮Suppose A =[a ij ], B =[b ij ] and that B = A −1 . By considering the determinant a = |A|,show that∂a∂u = ab ji ∂a ijk ∂u . kIf we denote the cofactor of the element a ij by ∆ ij then the elements of the inverse matrixare given by (see chapter 8)However, the determinant of A is given byb ij = 1 a ∆ji . (26.93)a = ∑ ja ij ∆ ij ,in which we have fixed i and written the sum over j explicitly, for clarity. Partiallydifferentiating both sides with respect to a ij , we then obtain∂a=∆ ij , (26.94)∂a ijsince a ij does not occur in any of the cofactors ∆ ij .Now, if the a ij depend on the coordinates then so will the determinant a and, by thechain rule, we have∂a∂u = ∂a ∂a ij ∂ak ∂a ij ∂u k =∆ij ij∂u = ab ji ∂a ijk ∂u , (26.95)kin which we have used (26.93) and (26.94). ◭Applying the result (26.95) to the determinant g of the metric tensor, andremembering both that g ik g kj = δ i j and that gij is symmetric, we obtain∂g∂u k = ∂g ggij ij∂u k . (26.96)Substituting (26.96) into (26.92) we find that the expression for the Christoffelsymbol can be much simplified to giveΓ i ki = 1 ∂g2g ∂u k = √ 1 ∂ √ gg ∂u k .Thus finally we obtain the expression for the divergence of a vector field in ageneral coordinate system as∇ · v = v i ; i = √ 1 ∂g ∂u j (√ gv j ). (26.97)LaplacianIf we replace v by ∇φ in ∇ · v then we obtain the Laplacian ∇ 2 φ. From (26.91),we havev i e i = v = ∇φ = ∂φ∂u i ei ,973

TENSORSand so the covariant components of v are given by v i = ∂φ/∂u i . In (26.97),however, we require the contravariant components v i . These may be obtained byraising the index using the metric tensor, to givev j = g jk jk ∂φv k = g∂u k .Substituting this into (26.97) we obtain∇ 2 φ = √ 1 (∂ √gg jk ∂φ )g ∂u j ∂u k . (26.98)◮Use (26.98) to find the expression for ∇ 2 φ in an orthogonal coordinate system with scalefactors h i , i =1, 2, 3.For an orthogonal coordinate system √ g = h 1 h 2 h 3 ;further,g ij =1/h 2 i if i = j and g ij =0otherwise. Therefore, from (26.98) we have( )∇ 2 φ = 1 ∂ h 1 h 2 h 3 ∂φ,h 1 h 2 h 3 ∂u j h 2 j∂u jwhich agrees with the results of section 10.10. ◭CurlThe special vector form of the curl of a vector field exists only in three dimensions.We therefore consider a more general form valid in higher-dimensional spaces aswell. In a general space the operation curl v is defined by(curl v) ij = v i; j − v j; i ,which is an antisymmetric covariant tensor.In fact the difference of derivatives can be simplified, sincev i; j − v j; i = ∂v i∂u j − Γl ijv l − ∂v j∂u i +Γ l jiv l= ∂v i∂u j − ∂v j∂u i ,where the Christoffel symbols have cancelled because of their symmetry properties.Thus curl v can be written in terms of partial derivatives as(curl v) ij = ∂v i∂u j − ∂v j∂u i .Generalising slightly the discussion of section 26.17, in three dimensions we mayassociate with this antisymmetric second-order tensor a vector with contravariantcomponents,(∇×v) i = − 12 √ g ɛijk (curl v) jk= − 1 (2 √ ∂vjg ɛijk ∂u k − ∂v )k∂u j = √ 1 ɛ ijk ∂v kg ∂u j ;974

26.21 ABSOLUTE DERIVATIVES ALONG CURVESthis is the analogue of the expression in Cartesian coordinates discussed insection 26.8.26.21 Absolute derivatives along curvesIn section 26.19 we discussed how to differentiate a general tensor with respectto the coordinates and introduced the covariant derivative. In this section weconsider the slightly different problem of calculating the derivative of a tensoralong a curve r(t) that is parameterised by some variable t.Let us begin by considering the derivative of a vector v along the curve. If weintroduce an arbitrary coordinate system u i with basis vectors e i , i =1, 2, 3, thenwe may write v = v i e i and so obtaindvdt = dvidt e i + v i de idt= dvidt e i + v i ∂e i du k∂u k dt ;here the chain rule has been used to rewrite the last term on the right-hand side.Using (26.75) to write the derivatives of the basis vectors in terms of Christoffelsymbols, we obtaindvdt = dvidt e i +Γ j dukikvi dt e j.Interchanging the dummy indices i and j in the last term, we may factor out thebasis vector and find( )dv dvidt = dt +Γi jkv j duk e i .dtThe term in parentheses is called the absolute (or intrinsic) derivative of thecomponents v i along the curve r(t)and is usually denoted byδv iδt ≡ dvidt +Γi jkv j duk = v i du k; kdt dt .With this notation, we may writedvdt = δviδt e i = v i du k; kdt e i. (26.99)Using the same method, the absolute derivative of the covariant componentsv i of a vector is given byδv iδt ≡ v du ki; kdt .Similarly, the absolute derivatives of the contravariant, mixed and covariant975

TENSORScomponents of a second-order tensor T areδT ij≡ T ij du k; kδt dt ,δT i j≡ T i du kj; kδt dt ,δT ij du k≡ T ij; kδt dt .The derivative of T along the curve r(t) may then be written in terms of, forexample, its contravariant components asdTdtijδT=δt e i ⊗ e j = T ij du k; kdt e i ⊗ e j .26.22 GeodesicsAs an example of the use of the absolute derivative, we conclude this chapterwith a brief discussion of geodesics. A geodesic in real three-dimensional spaceis a straight line, which has two equivalent defining properties. Firstly, it is thecurve of shortest length between two points and, secondly, it is the curve whosetangent vector always points in the same direction (along the line). Althoughin this chapter we have considered explicitly only our familiar three-dimensionalspace, much of the mathematical formalism developed can be generalised to moreabstract spaces of higher dimensionality in which the familiar ideas of Euclideangeometry are no longer valid. It is often of interest to find geodesic curves insuch spaces by using the defining properties of straight lines in Euclidean space.We shall not consider these more complicated spaces explicitly but will determinethe equation that a geodesic in Euclidean three-dimensional space (i.e.a straight line) must satisfy, deriving it in a sufficiently general way that ourmethod may be applied with little modification to finding the equations satisfiedby geodesics in more abstract spaces.Let us consider a curve r(s), parameterised by the arc length s from some pointon the curve, and choose as our defining property for a geodesic that its tangentvector t = dr/ds always points in the same direction everywhere on the curve, i.e.dt= 0. (26.100)dsAlternatively, we could exploit the property that the distance between two pointsis a minimum along a geodesic and use the calculus of variations (see chapter 22);this would lead to the same final result (26.101).If we now introduce an arbitrary coordinate system u i with basis vectors e i ,i =1, 2, 3, then we may write t = t i e i , and from (26.99) we finddtds = du kti ; kds e i = 0.976

26.23 EXERCISESWriting out the covariant derivative, we obtain( )dtids +Γi jkt j duk e i = 0.dsBut, since t j = du j /ds, it follows that the equation satisfied by a geodesic isd 2 u ids 2du j du k+Γi jk =0. (26.101)ds ds◮Find the equations satisfied by a geodesic (straight line) in cylindrical polar coordinates.From (26.83), the only non-zero Christoffel symbols are Γ 1 22 = −ρ and Γ2 12 =Γ2 21 =1/ρ.Thus the required geodesic equations are( )d 2 u 1 du 2 du 2ds 2 +Γ1 22ds ds =0 ⇒ d 2 2ρ dφds − ρ =0,2 dsd 2 u 2ds 2 +2Γ2 12du 1 du 2ds ds =0 ⇒ d 2 φds + 2 2 ρd 2 u 3ds 2 =0 ⇒d 2 zds =0. ◭2dρ dφds ds =0,26.23 Exercises26.1 Use the basic definition of a Cartesian tensor to show the following.(a) That for any general, but fixed, φ,(u 1 ,u 2 )=(x 1 cos φ − x 2 sin φ, x 1 sin φ + x 2 cos φ)are the components of a first-order tensor in two dimensions.(b) That( )x22 x 1 x 2x 1 x 2 x 2 1is not a tensor of order 2. To establish that a single element does nottransform correctly is sufficient.26.2 The components of two vectors, A and B, and a second-order tensor, T, are givenin one coordinate system by⎛A = ⎝ 1 ⎞ ⎛0 ⎠ , B = ⎝ 0 ⎞ ⎛1 ⎠ , T = ⎝ 2 √ ⎞√ 3 03 4 0 ⎠ .000 0 2In a second coordinate system, obtained from the first by rotation, the componentsof A and B are⎛A ′ = 1 ⎝2√301⎞ ⎛⎠ , B ′ = 1 ⎝2⎞−1√0 ⎠ .3Find the components of T in this new coordinate system and hence evaluate,with a minimum of calculation,T ij T ji , T ki T jk T ij , T ik T mn T ni T km .977

TENSORS26.3 In section 26.3 the transformation matrix for a rotation of the coordinate axeswas derived, and this approach is used in the rest of the chapter. An alternativeview is that of taking the coordinate axes as fixed and rotating the componentsof the system; this is equivalent to reversing the signs of all rotation angles.Using this alternative view, determine the matrices representing (a) a positiverotation of π/4 about the x-axis and (b) a rotation of −π/4 about the y-axis.Determine the initial vector r which, when subjected to (a) followed by (b),finishes at (3, 2, 1).26.4 Show how to decompose the Cartesian tensor T ij into three tensors,T ij = U ij + V ij + S ij ,where U ij is symmetric and has zero trace, V ij is isotropic and S ij has only threeindependent components.26.5 Use the quotient law discussed in section 26.7 to show that the array⎛⎞⎝ y2 + z 2 − x 2 −2xy −2xz−2yx x 2 + z 2 − y 2 −2yz ⎠−2zx −2zy x 2 + y 2 − z 2forms a second-order tensor.26.6 Use tensor methods to establish the following vector identities:(a) (u × v) × w =(u · w)v − (v · w)u;(b) curl (φu) =φ curl u + (grad φ) × u;(c) div (u × v) =v · curl u − u · curl v;(d) curl (u × v) =(v · grad)u − (u · grad)v + u div v − v div u;(e) grad 1 (u · u) =u × curl u +(u · grad)u.226.7 Use result (e) of the previous question and the general divergence theorem fortensors to show that, for a vector field A,∫S[A(A · dS) −12 A2 dS ] =∫V[A divA − A × curl A] dV ,where S is the surface enclosing volume V .26.8 A column matrix a has components a x ,a y ,a z and A is the matrix with elementsA ij = −ɛ ijk a k .(a) What is the relationship between column matrices b and c if Ab = c?(b) Find the eigenvalues of A and show that a is one of its eigenvectors. Explainwhy this must be so.26.9 Equation (26.29),|A|ɛ lmn = A li A mj A nk ɛ ijk ,is a more general form of the expression (8.47) for the determinant of a 3 × 3matrix A. The latter could have been written as|A| = ɛ ijk A i1 A j2 A k3 ,whilst the former removes the explicit mention of 1, 2, 3 at the expense of anadditional Levi–Civita symbol. As stated in the footnote on p. 942, (26.29) canbereadilyextendedtocoverageneralN × N matrix.Use the form given in (26.29) to prove properties (i), (iii), (v), (vi) and (vii)of determinants stated in subsection 8.9.1. Property (iv) is obvious by inspection.For definiteness take N = 3, but convince yourself that your methods of proofwould be valid for any positive integer N.978

26.23 EXERCISES26.10 A symmetric second-order Cartesian tensor is defined byT ij = δ ij − 3x i x j .Evaluate the following surface integrals, each taken over the surface of the unitsphere:∫∫∫(a) T ij dS; (b) T ik T kj dS; (c) x i T jk dS.26.11 Given a non-zero vector v, find the value that should be assigned to α to makeP ij = αv i v j and Q ij = δ ij − αv i v jinto parallel and orthogonal projection tensors, respectively, i.e. tensors thatsatisfy, respectively, P ij v j = v i , P ij u j =0andQ ij v j =0,Q ij u j = u i ,foranyvectoru that is orthogonal to v.Show, in particular, that Q ij is unique, i.e. that if another tensor T ij has thesame properties as Q ij then (Q ij − T ij )w j =0forany vector w.26.12 In four dimensions, define second-order antisymmetric tensors, F ij and Q ij ,anda first-order tensor, S i , as follows:(a) F 23 = H 1 , Q 23 = B 1 and their cyclic permutations;(b) F i4 = −D i , Q i4 = E i for i =1, 2, 3;(c) S 4 = ρ, S i = J i for i =1, 2, 3.Then, taking x 4 as t and the other symbols to have their usual meanings inelectromagnetic theory, show that the equations ∑ j ∂F ij/∂x j = S i and ∂Q jk /∂x i +∂Q ki /∂x j + ∂Q ij /∂x k = 0 reproduce Maxwell’s equations. In the latter i, j, k is anyset of three subscripts selected from 1, 2, 3, 4, but chosen in such a way that theyare all different.26.13 In a certain crystal the unit cell can be taken as six identical atoms lying at thecorners of a regular octahedron. Convince yourself that these atoms can also beconsidered as lying at the centres of the faces of a cube and hence that the crystalhas cubic symmetry. Use this result to prove that the conductivity tensor for thecrystal, σ ij ,mustbeisotropic.26.14 Assuming that the current density j and the electric field E appearing in equation(26.44) are first-order Cartesian tensors, show explicitly that the electrical conductivitytensor σ ij transforms according to the law appropriate to a second-ordertensor.The rate W at which energy is dissipated per unit volume, as a result of thecurrent flow, is given by E · j. Determine the limits between which W must lie fora given value of |E| as the direction of E is varied.26.15 In a certain system of units, the electromagnetic stress tensor M ij is given byM ij = E i E j + B i B j − 1 δ 2 ij(E k E k + B k B k ),where the electric and magnetic fields, E and B, are first-order tensors. Show thatM ij is a second-order tensor.Consider a situation in which |E| = |B|, but the directions of E and B arenot parallel. Show that E ± B are principal axes of the stress tensor and findthe corresponding principal values. Determine the third principal axis and itscorresponding principal value.26.16 A rigid body consists of four particles of masses m, 2m, 3m, 4m, respectivelysituated at the points (a, a, a), (a, −a, −a), (−a, a, −a), (−a, −a, a) and connectedtogether by a light framework.(a) Find the inertia tensor at the origin and show that the principal moments ofinertia are 20ma 2 and (20 ± 2 √ 5)ma 2 .979

TENSORS(b) Find the principal axes and verify that they are orthogonal.26.17 A rigid body consists of eight particles, each of mass m, held together by lightrods. In a certain coordinate frame the particles are at positions±a(3, 1, −1), ±a(1, −1, 3), ±a(1, 3, −1), ±a(−1, 1, 3).Show that, when the body rotates about an axis through the origin, if the angularvelocity and angular momentum vectors are parallel then their ratio must be40ma 2 ,64ma 2 or 72ma 2 .26.18 The paramagnetic tensor χ ij of a body placed in a magnetic field, in which itsenergy density is − 1 µ 2 0M · H with M i = ∑ j χ ijH j ,is⎛⎝ 2k 0 0 3k 0k⎞⎠ .0 k 3kAssuming depolarizing effects are negligible, find how the body will orientateitself if the field is horizontal, in the following circumstances:(a) the body can rotate freely;(b) the body is suspended with the (1, 0, 0) axis vertical;(c) the body is suspended with the (0, 1, 0) axis vertical.26.19 A block of wood contains a number of thin soft-iron nails (of constant permeability).A unit magnetic field directed eastwards induces a magnetic moment inthe block having components (3, 1, −2), and similar fields directed northwardsand vertically upwards induce moments (1, 3, −2) and (−2, −2, 2) respectively.Show that all the nails lie in parallel planes.26.20 For tin, the conductivity tensor is diagonal, with entries a, a, and b when referredto its crystal axes. A single crystal is grown in the shape of a long wire of length Land radius r, the axis of the wire making polar angle θ with respect to the crystal’s3-axis. Show that the resistance of the wire is L(πr 2 ab) ( −1 a cos 2 θ + b sin 2 θ ) .26.21 By considering an isotropic body subjected to a uniform hydrostatic pressure(no shearing stress), show that the bulk modulus k, defined by the ratio of thepressure to the fractional decrease in volume, is given by k = E/[3(1−2σ)] whereE is Young’s modulus and σ is Poisson’s ratio.26.22 For an isotropic elastic medium under dynamic stress, at time t the displacementu i and the stress tensor p ij satisfy( ∂ukp ij = c ijkl + ∂u )l∂x l ∂x kand∂p ij= ρ ∂2 u i∂x j ∂t , 2where c ijkl is the isotropic tensor given in equation (26.47) and ρ is a constant.Show that both ∇ · u and ∇×u satisfy wave equations and find the correspondingwave speeds.980

26.23 EXERCISES26.23 A fourth-order tensor T ijkl has the propertiesT jikl = −T ijkl , T ijlk = −T ijkl .Prove that for any such tensor there exists a second-order tensor K mn such thatT ijkl = ɛ ijm ɛ kln K mnand give an explicit expression for K mn . Consider two (separate) special cases, asfollows.(a) Given that T ijkl is isotropic and T ijji = 1, show that T ijkl is uniquely determinedand express it in terms of Kronecker deltas.(b) If now T ijkl has the additional propertyT klij = −T ijkl ,show that T ijkl has only three linearly independent components and find anexpression for T ijkl in terms of the vectorV i = − 1 ɛ 4 jklT ijkl .26.24 Working in cylindrical polar coordinates ρ, φ, z, parameterise the straight line(geodesic) joining (1, 0, 0) to (1,π/2, 1) in terms of s, the distance along the line.Show by substitution that the geodesic equations, derived at the end of section26.22, are satisfied.26.25 In a general coordinate system u i , i =1, 2, 3, in three-dimensional Euclideanspace, a volume element is given bydV = |e 1 du 1 · (e 2 du 2 × e 3 du 3 )|.Show that an alternative form for this expression, written in terms of the determinantg of the metric tensor, is given bydV = √ gdu 1 du 2 du 3 .Show that, under a general coordinate transformation to a new coordinatesystem u ′i , the volume element dV remains unchanged, i.e. show that it is a scalarquantity.26.26 By writing down the expression for the square of the infinitesimal arc length (ds) 2in spherical polar coordinates, find the components g ij ofthemetrictensorinthiscoordinate system. Hence, using (26.97), find the expression for the divergenceof a vector field v in spherical polars. Calculate the Christoffel symbols (of thesecond kind) Γ i jkin this coordinate system.26.27 Find an expression for the second covariant derivative v i; jk ≡ (v i; j ) ; k of a vectorv i (see (26.88)). By interchanging the order of differentiation and then subtractingthe two expressions, we define the components R l ijkof the Riemann tensor asv i; jk − v i; kj ≡ R l ijkv l .Show that in a general coordinate system u i these components are given byR l ijk = ∂Γl ik∂u j− ∂Γl ij∂u k+Γ m ikΓ l mj − Γ m ijΓ l mk.By first considering Cartesian coordinates, show that all the components R l ijk ≡ 0for any coordinate system in three-dimensional Euclidean space.In such a space, therefore, we may change the order of the covariant derivativeswithout changing the resulting expression.981

TENSORS26.28 A curve r(t) is parameterised by a scalar variable t. Show that the length of thecurve between two points, A and B, isgivenby∫ √ BduL = g i du jijA dt dt dt.Using the calculus of variations (see chapter 22), show that the curve r(t) thatminimises L satisfies the equationd 2 u idt 2du j du k+Γi jk = ¨ṡ du idt dt s dt ,where s is the arc length along the curve, ṡ = ds/dt and ¨s = d 2 s/dt 2 . Hence, showthat if the parameter t is of the form t = as + b, wherea and b are constants,then we recover the equation for a geodesic (26.101).[ A parameter which, like t, is the sum of a linear transformation of s and atranslation is called an affine parameter. ]26.29 We may define Christoffel symbols of the first kind byΓ ijk = g il Γ l jk.Show that these are given byΓ kij = 1 2By permuting indices, verify that(∂gik∂u j+ ∂g jk∂u i− ∂g ij∂u k ).∂g ij∂u =Γ k ijk +Γ jik .Using the fact that Γ l jk =Γl kj, show thatg ij; k ≡ 0,i.e. that the covariant derivative of the metric tensor is identically zero in allcoordinate systems.26.24 Hints and answers26.1 (a) u ′ 1 = x 1 cos(φ − θ) − x 2 sin(φ − θ), etc.;(b) u ′ 11 = s2 x 2 1 − 2scx 1x 2 + c 2 x 2 2 ≠ c2 x 2 2 + csx 1x 2 + scx 1 x 2 + s 2 x 2 1 .26.3 (a) (1/ √ 2)( √ 2, 0, 0; 0, 1, −1; 0, 1, 1). (b) (1/ √ 2)(1, 0, −1; 0, √ 2, 0; 1, 0, 1).r =(2 √ 2, −1+ √ 2, −1 − √ 2) T .26.5 Twice contract the array with the outer product of (x, y, z) with itself, thusobtaining the expression −(x 2 + y 2 + z 2 ) 2 , which is an invariant and therefore ascalar.26.7 Write A j (∂A i /∂x j )as∂(A i A j )/∂x j − A i (∂A j /∂x j ).26.9 (i) Write out the expression for |A T |, contract both sides of the equation with ɛ lmnand pick out the expression for |A| ontheRHS.Notethatɛ lmn ɛ lmn is a numericalscalar.(iii) Each non-zero term on the RHS contains any particular row index once andonly once. The same can be said for the Levi–Civita symbol on the LHS. Thusinterchanging two rows is equivalent to interchanging two of the subscripts ofɛ lmn , and thereby reversing its sign. Consequently, the magnitude of |A| remainsthe same but its sign is changed.(v) If, say, A pi = λA pj , for some particular pair of values i and j and all p then,982

26.24 HINTS AND ANSWERSin the (multiple) summation on the RHS, each A nk appears multiplied by (withno summation over i and j)ɛ ijk A li A mj + ɛ jik A lj A mi = ɛ ijk λA lj A mj + ɛ jik A lj λA mj =0,since ɛ ijk = −ɛ jik . Consequently, grouped in this way all terms are zero and|A| =0.(vi) Replace A mj by A mj + λA lj and note that λA li A lj A nk ɛ ijk = 0 by virtue ofresult (v).(vii) If C = AB,|C|ɛ lmn = A lx B xi A my B yj A nz B zk ɛ ijk .Contract this with ɛ lmn and show that the RHS is equal to ɛ xyz |A T |ɛ xyz |B|. Itthenfollows from result (i) that |C| = |A||B|.26.11 α = |v| −2 . Note that the most general vector has components w i = λv i +µu (1)i+νu (2)i,where both u (1) and u (2) are orthogonal to v.26.13 Construct the orthogonal transformation matrix S for the symmetry operationof (say) a rotation of 2π/3 about a body diagonal and, setting L = S −1 = S T ,construct σ ′ = LσL T and require σ ′ = σ. Repeat the procedure for (say) a rotationof π/2 about the x 3 -axis. These together show that σ 11 = σ 22 = σ 33 and thatall other σ ij = 0. Further symmetry requirements do not provide any additionalconstraints.26.15 The transformation of δ ij has to be included; the principal values are ±E · B.The third axis is in the direction ±B × E with principal value −|E| 2 .26.17 The principal moments give the required ratios.26.19 The principal permeability, in direction (1, 1, 2), has value 0. Thus all the nails liein planes to which this is the normal.26.21 Take p 11 = p 22 = p 33 = −p, and p ij = e ij =0fori ≠ j, leading to −p =(λ+2µ/3)e ii . The fractional volume change is e ii ; λ and µ are as defined in (26.46)and the worked example that follows it.26.23 Consider Q pq = ɛ pij ɛ qkl T ijkl and show that K mn = Q mn /4 has the required property.(a) Argue from the isotropy of T ijkl and ɛ ijk for that of K mn and hence that itmust be a multiple of δ mn . Show that the multiplier is uniquely determined andthat T ijkl =(δ il δ jk − δ ik δ jl )/6.(b) By relabelling dummy subscripts and using the stated antisymmetry property,show that K nm = −K mn . Show that −2V i = ɛ min K mn and hence that K mn = ɛ imn V i .26.25T ijkl = ɛ kli V j − ɛ klj V i .Use |e 1 · (e 2 × e 3 )| = √ g.Recall that √ g ′ = |∂u/∂u ′ | √ g and du ′1 du ′2 du ′3 = |∂u ′ /∂u| du 1 du 2 du 3 .26.27 (v i; j ) ; k =(v i; j ) ,k − Γ l ik v l; j − Γ l jk v i; l and v i; j = v i, j − Γ m ijv m . If all components of atensor equal zero in one coordinate system, then they are zero in all coordinatesystems.26.29 Use g il g ln = δi n and g ij = g ji . Show that( )∂gijg ij;k =∂u − Γ k jik − Γ ijk e i ⊗ e jandthenusetheearlierresult.983

27Numerical methodsIt happens frequently that the end product of a calculation or piece of analysisis one or more algebraic or differential equations, or an integral that cannot beevaluated in closed form or in terms of tabulated or pre-programmed functions.From the point of view of the physical scientist or engineer, who needs numericalvalues for prediction or comparison with experiment, the calculation or analysisis thus incomplete.With the ready availability of standard packages on powerful computers forthe numerical solution of equations, both algebraic and differential, and for theevaluation of integrals, in principle there is no need for the investigator to doanything other than turn to them. However, it should be a part of every engineer’sor scientist’s repertoire to have some understanding of the kinds of procedure thatare being put into practice within those packages. The present chapter indicates(at a simple level) some of the ways in which analytically intractable problemscan be tackled using numerical methods.In the restricted space available in a book of this nature, it is clearly notpossible to give anything like a full discussion, even of the elementary points thatwill be made in this chapter. The limited objective adopted is that of explainingand illustrating by simple examples some of the basic principles involved. Inmany cases, the examples used can be solved in closed form anyway, but this‘obviousness’ of the answers should not detract from their illustrative usefulness,and it is hoped that their transparency will help the reader to appreciate some ofthe inner workings of the methods described.The student who proposes to study complicated sets of equations or makerepeated use of the same procedures by, for example, writing computer programsto carry out the computations, will find it essential to acquire a good understandingof topics hardly mentioned here. Amongst these are the sensitivity ofthe adopted procedures to errors introduced by the limited accuracy with whicha numerical value can be stored in a computer (rounding errors) and to the984

27.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONSerrors introduced as a result of approximations made in setting up the numericalprocedures (truncation errors). For this scale of application, books specificallydevoted to numerical analysis, data analysis and computer programming shouldbe consulted.So far as is possible, the method of presentation here is that of indicatingand discussing in a qualitative way the main steps in the procedure, and thenof following this with an elementary worked example. The examples have beenrestricted in complexity to a level at which they can be carried out with a pocketcalculator. Naturally it will not be possible for the student to check all thenumerical values presented, unless he or she has a programmable calculator orcomputer readily available, and even then it might be tedious to do so. However,it is advisable to check the initial step and at least one step in the middle ofeach repetitive calculation given in the text, so that how the symbolic equationsare used with actual numbers is understood. Clearly the intermediate step shouldbe chosen to be at a point in the calculation at which the changes are stillsufficiently large that they can be detected by whatever calculating device isused.Where alternative methods for solving the same type of problem are discussed,for example in finding the roots of a polynomial equation, we have usuallytaken the same example to illustrate each method. This could give the mistakenimpression that the methods are very restricted in applicability, but it is felt bythe authors that using the same examples repeatedly has sufficient advantages, interms of illustrating the relative characteristics of competing methods, to justifydoing so. Once the principles are clear, little is to be gained by using newexamples each time, and, in fact, having some prior knowledge of the ‘correctanswer’ should allow the reader to judge the efficiency and dangers of particularmethods as the successive steps are followed through.One other point remains to be mentioned. Here, in contrast with every otherchapter of this book, the value of a large selection of exercises is not clear cut.The reader with sufficient computing resources to tackle them can easily devisealgebraic or differential equations to be solved, or functions to be integrated(which perhaps have arisen in other contexts). Further, the solutions of theseproblems will be self-checking, for the most part. Consequently, although anumber of exercises are included, no attempt has been made to test the full rangeof ideas treated in this chapter.27.1 Algebraic and transcendental equationsThe problem of finding the real roots of an equation of the form f(x) =0,wheref(x) is an algebraic or transcendental function of x, is one that can sometimesbe treated numerically, even if explicit solutions in closed form are not feasible.985

NUMERICAL METHODSf(x)141210 f(x) =x 5 − 2x 2 − 3864200.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6x1.8−2−4Figure 27.10 ≤ x ≤ 1.9.A graph of the function f(x) =x 5 − 2x 2 − 3forx in the rangeExamples of the types of equation mentioned are the quartic equation,and the transcendental equation,ax 4 + bx + c =0,x − 3 tanh x =0.The latter type is characterised by the fact that it contains, in effect, a polynomialof infinite order on the LHS.We will discuss four methods that, in various circumstances, can be used toobtain the real roots of equations of the above types. In all cases we will take asthe specific equation to be solved the fifth-order polynomial equationf(x) ≡ x 5 − 2x 2 − 3=0. (27.1)The reasons for using the same equation each time were discussed in the introductionto this chapter.For future reference, and so that the reader may follow some of the calculationsleading to the evaluation of the real root of (27.1), a graph of f(x) in the range0 ≤ x ≤ 1.9 is shown in figure 27.1.Equation (27.1) is one for which no solution can be found in closed form, thatis in the form x = a, wherea does not explicitly contain x. The general scheme tobe employed will be an iterative one in which successive approximations to a realroot of (27.1) will be obtained, each approximation, it is to be hoped, being betterthan the preceding one; certainly, we require that the approximations convergeand that they have as their limit the sought-for root. Let us denote the required986

27.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONSroot by ξ and the values of successive approximations by x 1 , x 2 , ..., x n , ... . Then,for any particular method to be successful,lim x n = ξ, where f(ξ) =0. (27.2)n→∞However, success as defined here is not the only criterion. Since, in practice,only a finite number of iterations will be possible, it is important that the valuesof x n be close to that of ξ for all n>N,whereN is a relatively low number;exactly how low it is naturally depends on the computing resources available andthe accuracy required in the final answer.So that the reader may assess the progress of the calculations that follow, werecord that to nine significant figures the real root of equation (27.1) has thevalueξ =1.495 106 40. (27.3)We now consider in turn four methods for determining the value of this root.27.1.1 Rearrangement of the equationIf equation (27.1), f(x) = 0, can be recast into the formx = φ(x), (27.4)where φ(x) isaslowly varying function of x, then an iteration schemex n+1 = φ(x n ) (27.5)will often produce a fair approximation to the root ξ after a few iterations,as follows. Clearly, ξ = φ(ξ), since f(ξ) = 0; thus, when x n is close to ξ,the next approximation, x n+1 , will differ little from x n , the actual size of thedifference giving an order-of-magnitude indication of the inaccuracy in x n+1(when compared with ξ).In the present case, the equation can be writtenx =(2x 2 +3) 1/5 . (27.6)Because of the presence of the one-fifth power, the RHS is rather insensitiveto the value of x used to compute it, and so the form (27.6) fits the generalrequirements for the method to work satisfactorily. It remains only to choose astarting approximation. It is easy to see from figure 27.1 that the value x =1.5would be a good starting point, but, so that the behaviour of the procedure atvalues some way from the actual root can be studied, we will make a poorerchoice, x 1 =1.7.With this starting value and the general recurrence relationshipx n+1 =(2x 2 n +3) 1/5 , (27.7)987

NUMERICAL METHODSn x n f(x n )1 1.7 5.422 1.544 18 1.013 1.506 86 2.28 × 10 −14 1.497 92 5.37 × 10 −25 1.495 78 1.28 × 10 −26 1.495 27 3.11 × 10 −37 1.495 14 7.34 × 10 −48 1.495 12 1.76 × 10 −4Table 27.1 Successive approximations to the root of (27.1) using the methodof rearrangement.n A n f(A n ) B n f(B n ) x n f(x n )1 1.0 −4.0000 1.7 5.4186 1.2973 −2.69162 1.2973 −2.6916 1.7 5.4186 1.4310 −1.09573 1.4310 −1.0957 1.7 5.4186 1.4762 −0.34824 1.4762 −0.3482 1.7 5.4186 1.4897 −0.10165 1.4897 −0.1016 1.7 5.4186 1.4936 −0.02896 1.4936 −0.0289 1.7 5.4186 1.4947 −0.0082Table 27.2 Successive approximations to the root of (27.1) using linearinterpolation.successive values can be found. These are recorded in table 27.1. Although notstrictly necessary, the value of f(x n ) ≡ x 5 n − 2x 2 n − 3 is also shown at each stage.It will be seen that x 7 and all later x n agree with the precise answer (27.3) towithin one part in 10 4 . However, f(x n )andx n − ξ are both reduced by a factorof only about 4 for each iteration; thus a large number of iterations would beneeded to produce a very accurate answer. The factor 4 is, of course, specificto this particular problem and would be different for a different equation. Thesuccessive values of x n are shown in graph (a) of figure 27.2.27.1.2 Linear interpolationIn this approach two values, A 1 and B 1 ,ofx are chosen with A 1

27.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS6(a)6(b)44x 1x 22x 22−2−4642−2−41.0 1.2 1.4 1.6ξ(c)x 3x 3x 14x 3 x 42ξ1.0 1.2 1.4x1.61.0 1.2 1.4 1.62x2−2ξx 1−41.0−2−4(d)61.2x 11.4x 3ξ1.6Figure 27.2 Graphical illustrations of the iteration methods discussed inthe text: (a) rearrangement; (b) linear interpolation; (c) binary chopping;(d) Newton–Raphson.with n =1.Next,f(x 1 ) is evaluated and the process repeated after replacingeither A 1 or B 1 by x 1 , according to whether f(x 1 ) has the same sign as f(A 1 )orf(B 1 ), respectively. In figure 27.2(b), A 1 is the one replaced.As can be seen in the particular example that we are considering, with thismethod there is a tendency, if the curvature of f(x) is of constant sign nearthe root, for one of the two ends of the successive chords to remain unchanged.Starting with the initial values A 1 = 1 and B 1 =1.7, the results of the firstfive iterations using (27.8) are given in table 27.2 and indicated in graph (b) offigure 27.2. As with the rearrangement method, the improvement in accuracy,as measured by f(x n )andx n − ξ, is a fairly constant factor at each iteration(approximately 3 in this case), and for our particular example there is little tochoose between the two. Both tend to their limiting value of ξ monotonically,from either higher or lower values, and this makes it difficult to estimate limitswithin which ξ can safely be presumed to lie. The next method to be describedgives at any stage a range of values within which ξ is known to lie.989

NUMERICAL METHODSn A n f(A n ) B n f(B n ) x n f(x n )1 1.0000 −4.0000 1.7000 5.4186 1.3500 −2.16102 1.3500 −2.1610 1.7000 5.4186 1.5250 0.59683 1.3500 −2.1610 1.5250 0.5968 1.4375 −0.99464 1.4375 −0.9946 1.5250 0.5968 1.4813 −0.25735 1.4813 −0.2573 1.5250 0.5968 1.5031 0.15446 1.4813 −0.2573 1.5031 0.1544 1.4922 −0.05527 1.4922 −0.0552 1.5031 0.1544 1.4977 0.04878 1.4922 −0.0552 1.4977 0.0487 1.4949 −0.0085Table 27.3chopping.Successive approximations to the root of (27.1) using binary27.1.3 Binary choppingAgain two values of x, A 1 and B 1 , that straddle the root are chosen, such thatA 1

27.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONSn x n f(x n )1 1.7 5.422 1.545 01 1.033 1.498 87 7.20 × 10 −24 1.495 13 4.49 × 10 −45 1.495 106 40 2.6 × 10 −86 1.495 106 40 —Table 27.4 Successive approximations to the root of (27.1) using the Newton–Raphson method.of (notionally) constructing the chord between two points on the curve of f(x)against x, the tangent to the curve is notionally constructed at each successivevalue of x n , and the next value, x n+1 , is taken as the point at which the tangentcuts the axis f(x) = 0. This is illustrated in graph (d) of figure 27.2.If the nth value is x n , the tangent to the curve of f(x) at that point has slopef ′ (x n ) and passes through the point x = x n , y = f(x n ). Its equation is thusy(x) =(x − x n )f ′ (x n )+f(x n ). (27.10)The value of x at which y = 0 is then taken as x n+1 ; thus the condition y(x n+1 )=0yields, from (27.10), the iteration schemex n+1 = x n − f(x n)f ′ (x n ) . (27.11)This is the Newton–Raphson iteration formula. Clearly, if x n is close to ξ then x n+1is close to x n , as it should be. It is also apparent that if any of the x n comes closeto a stationary point of f, sothatf ′ (x n ) is close to zero, the scheme is not goingto work well.For our standard example, (27.11) becomesx n+1 = x n − x5 n − 2x 2 n − 35x 4 = 4x5 n − 2x 2 n +3n − 4x n 5x 4 . (27.12)n − 4x nAgain taking a starting value of x 1 =1.7, we obtain in succession the entriesin table 27.4. The different values are given to an increasing number of decimalplaces as the calculation proceeds; f(x n ) is also recorded.It is apparent that this method is unlike the previous ones in that the increasein accuracy of the answer is not constant throughout the iterations but improvesdramatically as the required root is approached. Away from the root the behaviourof the series is less satisfactory, and from its geometrical interpretation it can beseen that if, for example, there were a maximum or minimum near the root thenthe series could oscillate between values on either side of it (instead of ‘homingin’ on the root). The reason for the good convergence near the root is discussedin the next section.991

NUMERICAL METHODSOf the four methods mentioned, no single one is ideal, and, in practice, somemixture of them is usually to be preferred. The particular combination of methodsselected will depend a great deal on how easily the progress of the calculationmay be monitored, but some combination of the first three methods mentioned,followed by the NR scheme if great accuracy were required, would be suitablefor most situations.27.2 Convergence of iteration schemesFor iteration schemes in which x n+1 can be expressed as a differentiable functionof x n , for example the rearrangement or NR methods of the previous section, apartial analysis of the conditions necessary for a successful scheme can be madeas follows.Suppose the general iteration formula is expressed asx n+1 = F(x n ) (27.13)((27.7) and (27.12) are examples). Then the sequence of values x 1 ,x 2 ,...,x n ,... isrequired to converge to the value ξ that satisfies bothf(ξ) =0 and ξ = F(ξ). (27.14)If the error in the solution at the nth stage is ɛ n ,i.e.x n = ξ + ɛ n ,thenξ + ɛ n+1 = x n+1 = F(x n )=F(ξ + ɛ n ). (27.15)For the iteration process to converge, a decreasing error is required, i.e. |ɛ n+1 |

27.2 CONVERGENCE OF ITERATION SCHEMESy = xyx n x n+1 x n+2y = F(x)ξxFigure 27.3 Illustration of the convergence of the iteration scheme x n+1 =F(x n )when0

NUMERICAL METHODSn x n+1 ɛ n1 8.5 4.52 5.191 1.193 4.137 1.4 × 10 −14 4.002 257 2.3 × 10 −35 4.000 000 637 6.4 × 10 −76 4 —Table 27.5(27.20).Successive approximations to √ 16 using the iteration scheme◮The following is an iteration scheme for finding the square root of X:x n+1 = 1 (x n + X ). (27.20)2 x nShow that it has second-order convergence and illustrate its efficiency by finding, say, √ 16starting with a very poor guess, √ 16 = 1.If this scheme does converge to ξ then ξ will satisfyξ = 1 (ξ + X )⇒ ξ 2 = X,2 ξas required. The iteration function F is given byF(x) = 1 2(x + X x),and so, since ξ 2 = X,F ′ (ξ) = 1 2(1 − X x 2 )x=ξ=0,whilst( ) XF ′′ (ξ) = = 1 x 3 x=ξξ ≠0.Thus the procedure has second-order, but not third-order, convergence.We now show the procedure in action. Table 27.5 gives successive values of x n and ofɛ n , the difference between x n and the true value, 4. As we can see, the scheme is crudeinitially, but once x n gets close to ξ, it homes in on the true value extremely rapidly. ◭27.3 Simultaneous linear equationsAs we saw in chapter 8, many situations in physical science can be describedapproximately or exactly by a set of N simultaneous linear equations in N994

27.3 SIMULTANEOUS LINEAR EQUATIONSvariables (unknowns), x i , i =1, 2,...,N. The equations take the general formA 11 x 1 + A 12 x 2 + ···+ A 1N x N = b 1 ,A 21 x 1 + A 22 x 2 + ···+ A 2N x N = b 2 , (27.21)..A N1 x 1 + A N2 x 2 + ···+ A NN x N = b N ,where the A ij are constants and form the elements of a square matrix A. Theb iare given and form a column matrix b. IfA is non-singular then (27.21) can besolved for the x i using the inverse of A, according to the formulax = A −1 b.This approach was discussed at length in chapter 8 and will not be consideredfurther here.27.3.1 Gaussian eliminationWe follow instead a continuation of one of the earliest techniques acquired by astudent of algebra, namely the solving of simultaneous equations (initially onlytwo in number) by the successive elimination of all the variables but one. This(known as Gaussian elimination) is achieved by using, at each stage, one of theequations to obtain an explicit expression for one of the remaining x i in termsof the others and then substituting for that x i in all other remaining equations.Eventually a single linear equation in just one of the unknowns is obtained. Thisis then solved and the result is resubstituted in previously derived equations (inreverse order) to establish values for all the x i .This method is probably very familiar to the reader, and so a specific exampleto illustrate this alone seems unnecessary. Instead, we will show how a calculationalong such lines might be arranged so that the errors due to the inherent lack ofprecision in any calculating equipment do not become excessive. This can happenif the value of N is large and particularly (and we will merely state this) if theelements A 11 ,A 22 ,...,A NN on the leading diagonal of the matrix in (27.21) aresmall compared with the off-diagonal elements.The process to be described is known as Gaussian elimination with interchange.The only, but essential, difference from straightforward elimination is that beforeeach variable x i is eliminated, the equations are reordered to put the largest (inmodulus) remaining coefficient of x i on the leading diagonal.We will take as an illustration a straightforward three-variable example, whichcan in fact be solved perfectly well without any interchange since, with simplenumbers and only two eliminations to perform, rounding errors do not havea chance to build up. However, the important thing is that the reader should995

NUMERICAL METHODSappreciate how this would apply in (say) a computer program for a 1000-variablecase, perhaps with unforseeable zeros or very small numbers appearing on theleading diagonal.◮Solve the simultaneous equations(a) x 1 +6x 2 −4x 3 = 8,(b) 3x 1 −20x 2 +x 3 = 12,(c) −x 1 +3x 2 +5x 3 = 3.(27.22)Firstly, we interchange rows (a) and (b) to bring the term 3x 1 onto the leading diagonal. Inthe following, we label the important equations (I), (II), (III), and the others alphabetically.A general (i.e. variable) label will be denoted by j.(I) 3x 1 −20x 2 +x 3 = 12,(d) x 1 +6x 2 −4x 3 = 8,(e) −x 1 +3x 2 +5x 3 = 3.For (j) = (d) and (e), replace row (j) byrow (j) − a j1× row (I),3where a j1 is the coefficient of x 1 in row (j), to give the two equations(II)( )6+20 x23+ ( )−4 − 1 x33= 8− 12 3(f)( )3 −20 x23+ ( )5+ 1 x33= 3+ 12 3Now |6+ 2020| > |3 − | and so no interchange is required before the next elimination. To3 3eliminate x 2 ,replacerow(f)by( )−113row (f) − × row (II).This gives[(III) 16] (−13) x3 =7+ 11 3 38 338Collecting together and tidying up the final equations, we have(I) 3x 1 −20x 2 +x 3 = 12,(II) 38x 2 −13x 3 = 12,(III) x 3 = 2.Starting with (III) and working backwards, it is now a simple matter to obtainx 1 =10, x 2 =1, x 3 =2. ◭38327.3.2 Gauss–Seidel iterationIn the example considered in the previous subsection an explicit way of solvinga set of simultaneous equations was given, the accuracy obtainable being limitedonly by the rounding errors in the calculating facilities available, and the calculationwas planned to minimise these. However, in some situations it may be thatonly an approximate solution is needed. If, for a large number of variables, this is996

27.3 SIMULTANEOUS LINEAR EQUATIONSthe case then an iterative method may produce a satisfactory degree of precisionwith less calculation. Such a method, known as Gauss–Seidel iteration, isbasedupon the following analysis.The problem is again that of finding the components of the column matrix xthat satisfiesAx = b (27.23)when A and b are a given matrix and column matrix, respectively.The steps of the Gauss–Seidel scheme are as follows.(i) Rearrange the equations (usually by simple division on both sides of eachequation) so that all diagonal elements of the new matrix C are unity, i.e.(27.23) becomesCx = d, (27.24)where C = I − F, andF has zeros as its diagonal elements.(ii) Step (i) producesand this forms the basis of an iteration scheme,Fx + d = Ix = x, (27.25)x n+1 = Fx n + d, (27.26)where x n is the nth approximation to the required solution vector ξ.(iii) To improve the convergence, the matrix F, which has zeros on its leadingdiagonal, can be written as the sum of two matrices L and U that havenon-zero elements only below and above the leading diagonal, respectively:L ij =U ij ={Fij if i>j,0 otherwise,{Fij if i

NUMERICAL METHODSn x 1 x 2 x 31 2 2 22 4 0.1 1.343 12.76 1.381 2.3234 9.008 0.867 1.8815 10.321 1.042 2.0396 9.902 0.987 1.9887 10.029 1.004 2.004Table 27.6 Successive approximations to the solution of simultaneous equations(27.29) using the Gauss–Seidel iteration method.◮Obtain an approximate solution to the simultaneous equationsx 1 +6x 2 −4x 3 = 8,3x 1 −20x 2 +x 3 = 12,−x 1 +3x 2 +5x 3 = 3.These are the same equations as were solved in subsection 27.3.1.(27.29)Divide the equations by 1, −20 and 5, respectively, to givex 1 +6x 2 − 4x 3 =8,−0.15x 1 + x 2 − 0.05x 3 = −0.6,−0.2x 1 +0.6x 2 + x 3 =0.6.Thus, set out in matrix form, (27.28) is, in this case, given by⎛⎝ x ⎞ ⎛1x 2⎠ = ⎝ 0 0 0⎞ ⎛0.15 0 0 ⎠ ⎝ x ⎞1x 2⎠x 30.2 −0.6 0 x 3n+1n+1⎛+ ⎝ 0 −6 4⎞ ⎛0 0 0.05 ⎠ ⎝ x ⎞ ⎛1x 2⎠ + ⎝ 8⎞−0.6 ⎠ .0 0 0 x 3 0.6nSuppose initially (n = 1) we guess each component to have the value 2. Then the successivesets of values of the three quantities generated by this scheme are as shown in table 27.6.Even with the rather poor initial guess, a close approximation to the exact result, x 1 = 10,x 2 =1,x 3 = 2, is obtained in only a few iterations. ◭27.3.3 Tridiagonal matricesAlthough for the solution of most matrix equations Ax = b the number ofoperations required increases rapidly with the size N × N of the matrix (roughlyas N 3 ), for one particularly simple kind of matrix the computing required increasesonly linearly with N. This type often occurs in physical situations in which objectsin an ordered set interact only with their nearest neighbours and is one in whichonly the leading diagonal and the diagonals immediately above and below it998

. . .. . .27.3 SIMULTANEOUS LINEAR EQUATIONScontain non-zero entries. Such matrices are known as tridiagonal matrices. Theymay also be used in numerical approximations to the solutions of certain typesof differential equation.A typical matrix equation involving a tridiagonal matrix is as follows:b 1c 10a 2c 3b 2c 2a 3b 3a N–1b N–1c N–10a Nb N. . .. . .. . .x 1y 1x 2y 2x 3y 3=x Nx N–1y N–1y N(27.30)So as to keep the entries in the matrix as free from subscripts as possible, wehave used a, b and c to indicate subdiagonal, leading diagonal and superdiagonalelements, respectively. As a consequence, we have had to change the notation forthe column matrix on the RHS from b to (say) y.In such an equation the first and last rows involve x 1 and x N , respectively, andso the solution could be found by letting x 1 be unknown and then solving in turneach row of the equation in terms of x 1 , and finally determining x 1 by requiringthe next-to-last line to generate for x N an equation compatible with that givenby the last line. However, if the matrix is large this becomes a very cumbersomeoperation, and a simpler method is to assume a form of solutionSince the ith line of the matrix equation isx i−1 = θ i−1 x i + φ i−1 . (27.31)a i x i−1 + b i x i + c i x i+1 = y i ,we must have, by substituting for x i−1 ,that(a i θ i−1 + b i )x i + c i x i+1 = y i − a i φ i−1 .This is also in the form of (27.31), but with i replaced by i+1. Thus the recurrenceformulae for θ i and φ i areθ i =−c ia i θ i−1 + b i,φ i = y i − a i φ i−1a i θ i−1 + b i, (27.32)provided the denominator does not vanish for any i. From the first of the matrixequations it follows that θ 1 = −c 1 /b 1 and φ 1 = y 1 /b 1 . The equations may nowbe solved for the x i in two stages without carrying through an unknown quantity.First, all the θ i and φ i are generated using (27.32) and the values of θ 1 and φ 1 ,then, as a second stage, (27.31) is used to evaluate the x i , starting with x N (= φ N )and working backwards.999

NUMERICAL METHODS◮Solve the following tridiagonal matrix equation, in which only non-zero elements areshown:⎛⎞ ⎛ ⎞ ⎛ ⎞1 2x 1 4−1 2 1x 232 −1 2x ⎜ 3 1 13−3⎟ ⎜ x =4 ⎟ ⎜ 10. (27.33)⎟⎝3 4 2 ⎠ ⎝ x 5⎠ ⎝ 7 ⎠−2 2 x 6 −2The solution is set out in table 27.7, in which the arrows indicate the general flow of thecalculation. First, the columns of a i , b i , c i and y i are filled in from the original equation(27.33) and then the recurrence relations (27.32) are used to fill in the successive rowsstarting at the top; on each row we work from left to right as far as and including the φ icolumn. Finally, the bottom entry in the the x i column is set equal to the bottom entry inthe completed φ i column and the rest of the x i column is completed by using (27.31) andworking up from the bottom. Thus the solution is x 1 =2;x 2 =1;x 3 =3;x 4 = −1; x 5 =a i b i c i a i θ i−1 + b i θ i y i a i φ i−1 φ i x i↓ 0 1 2 → 1 −2 4 0 4 2 ↑↓ −1 2 1 → 4 −1/4 3 −4 7/4 1 ↑↓ 2 −1 2 → −3/2 4/3 −3 7/2 13/3 3 ↑↓ 3 1 1 → 5 −1/5 10 13 −3/5 −1 ↑↓ 3 4 2 → 17/5 −10/17 7 −9/5 44/17 2 ↑↓ −2 2 0 → 54/17 0 −2 −88/17 1 → 1 ↑Table 27.7 The solution of tridiagonal matrix equation (27.33). The arrowsindicate the general flow of the calculation, as described in the text.2; x 6 =1.◭27.4 Numerical integrationAs noted at the start of this chapter, with modern computers and computerpackages – some of which will present solutions in algebraic form, where thatis possible – the inability to find a closed-form expression for an integral nolonger presents a problem. But, just as for the solution of algebraic equations, itis extremely important that scientists and engineers should have some idea of theprocedures on which such packages are based. In this section we discuss some ofthe more elementary methods used to evaluate integrals numerically and at thesame time indicate the basis of more sophisticated procedures.The standard integral evaluation has the formI =∫ baf(x) dx, (27.34)where the integrand f(x) may be given in analytic or tabulated form, but for thecases under consideration no closed-form expression for I can be obtained. All1000

27.4 NUMERICAL INTEGRATION(a) (b) (c)f(x)f(x)f i+1f if i+1f i+1hhhhf if i−1x i+1x i x i+1/2 x i+1 x i+1x iFigure 27.4 (a) Definition of nomenclature. (b) The approximation in usingthe trapezium rule; f(x) is indicated by the broken curve. (c) Simpson’s ruleapproximation; f(x) is indicated by the broken curve. The solid curve is partof the approximating parabola.x i−1x inumerical evaluations of I are based on regarding I as the area under the curveof f(x) between the limits x = a and x = b and attempting to estimate that area.The simplest methods of doing this involve dividing up the interval a ≤ x ≤ binto N equal sections, each of length h =(b − a)/N. The dividing points arelabelled x i , with x 0 = a, x N = b, i running from 0 to N. The point x i is a distanceih from a. The central value of x in a strip (x = x i + h/2) is denoted for brevityby x i+1/2 , and for the same reason f(x i ) is written as f i . This nomenclature isindicated graphically in figure 27.4(a).So that we may compare later estimates of the area under the curve with thetrue value, we next obtain an exact expression for I, even though we cannotevaluate it. To do this we need to consider only one strip, say that between x iand x i+1 . For this strip the area is, using Taylor’s expansion,∫ h/2−h/2∫ h/2f(x i+1/2 + y) dy =−h/2∞∑==n=0∞∑∞∑f (n) (x i+1/2 ) ynn! dyn=0∫ h/2f (n)i+1/2f (n)i+1/2n even−h/2y nn! dy) n+12 h. (27.35)(n +1)!(2It should be noted that, in this exact expression, only the even derivatives off survive the integration and all derivatives are evaluated at x i+1/2 . Clearly1001

NUMERICAL METHODSother exact expressions are possible, e.g. the integral of f(x i + y) over the range0 ≤ y ≤ h, but we will find (27.35) the most useful for our purposes.Although the preceding discussion has implicitly assumed that both of thelimits a and b are finite, with the consequence that N is finite, the general methodcan be adapted to treat some cases in which one of the limits is infinite. It issufficient to consider one infinite limit, as an integral with limits −∞ and ∞ canbe considered as the sum of two integrals, each with one infinite limit.Consider the integralI =∫ ∞af(x) dx,where a is chosen large enough that the integrand is monotonically decreasingfor x>aand falls off more quickly than x −2 . The change of variable t =1/xconverts this integral into∫ 1/a( )1 1I =0 t 2 f dt.tIt is now an integral over a finite range and the methods indicated earlier can beapplied to it. The value of the integrand at the lower end of the t-range is zero.In a similar vein, integrals with an upper limit of ∞ and an integrand that isknown to behave asymptotically as g(x)e −αx ,whereg(x) is a smooth function, canbe converted into an integral over a finite range by setting x = −α −1 ln αt. Again,the lower limit, a, for this part of the integral should be positive and chosenbeyond the last turning point of g(x). The part of the integral for x

27.4 NUMERICAL INTEGRATIONThis provides a very simple expression for estimating integral (27.34); its accuracyis limited only by the extent to which h can be made very small (and hence Nvery large) without making the calculation excessively long. Clearly the estimateprovided is exact only if f(x) is a linear function of x.The error made in calculating the area of the strip when the trapezium rule isused may be estimated as follows. The values used are f i and f i+1 , as in (27.36).These can be expressed accurately in terms of f i+1/2 and its derivatives by theTaylor seriesf i+1/2±1/2 = f i+1/2 ± h 2 f′ i+1/2 + 1 ( ) 2 hf i+1/2 ′′2! 2± 1 ( ) 3 hf (3)3! 2i+1/2 + ··· .ThusA i (estim.) = 1 2 h(f i + f i+1 )[= h f i+1/2 + 1 ( h2! 2whilst, from the first few terms of the exact result (27.35),A i (exact) = hf i+1/2 + 2 ( h3! 2Thus the error ∆A i = A i (estim.) − A i (exact) is given by∆A i = ( 18 − ) 124 h 3 f i+1/2 ′′ +O(h5 )) 2f ′′i+1/2 +O(h4 )) 3f ′′i+1/2 +O(h5 ).≈ 1 12 h3 f i+1/2 ′′ .The total error in I(estim.) is thus given approximately by],∆I(estim.) ≈ 1 12 nh3 〈f ′′ 〉 = 1 12 (b − a)h2 〈f ′′ 〉, (27.38)where 〈f ′′ 〉 represents an average value for the second derivative of f over theinterval a to b.◮Use the trapezium rule with h =0.5 to evaluateI =∫ 20(x 2 − 3x +4)dx,and, by evaluating the integral exactly, examine how well (27.38) estimates the error.With h =0.5, we will need five values of f(x) =x 2 − 3x + 4 for use in formula (27.37).They are f(0) = 4, f(0.5) = 2.75, f(1) = 2, f(1.5) = 1.75 and f(2) = 2. Putting these into(27.37) givesI(estim.) = 0.5 (4 + 2 × 2.75 + 2 × 2 + 2× 1.75 + 2) = 4.75.2The exact value is[ ] x32I(exact) =3 − 3x22 +4x =4 2 . 301003

NUMERICAL METHODSThe difference between the estimate of the integral and the exact answer is 1/12. Equation(27.38) estimates this error as 2 × 0.25 ×〈f ′′ 〉/12. Our (deliberately chosen!) integrand isone for which 〈f ′′ 〉 can be evaluated trivially. Because f(x) is a quadratic function of x,its second derivative is constant, and equal to 2 in this case. Thus 〈f ′′ 〉 has value 2 and(27.38) estimates the error as 1/12; that the estimate is exactly right should be no surprisesince the Taylor expansion for a quadratic polynomial about any point always terminatesafter three terms and so no higher-order terms in h have been ignored in (27.38). ◭27.4.2 Simpson’s ruleWhereas the trapezium rule makes a linear interpolation of f, Simpson’s ruleeffectively mimics the local variation of f(x) using parabolas. The strips aretreated two at a time (figure 27.4(c)) and therefore their number, N, should bemade even.In the neighbourhood of x i ,fori odd, it is supposed that f(x) can be adequatelyrepresented by a quadratic form,f(x i + y) =f i + ay + by 2 . (27.39)In particular, applying this to y = ±h yields two expressions involving bf i+1 = f(x i + h) =f i + ah + bh 2 ,f i−1 = f(x i − h) =f i − ah + bh 2 ;thusbh 2 = 1 2 (f i+1 + f i−1 − 2f i ).Now, in the representation (27.39), the area of the double strip from x i−1 tox i+1 is given byA i (estim.) =∫ h−h(f i + ay + by 2 ) dy =2hf i + 2 3 bh3 .Substituting for bh 2 then yields, for the estimated area,A i (estim.) = 2hf i + 2 3 h × 1 2 (f i+1 + f i−1 − 2f i )= 1 3 h(4f i + f i+1 + f i−1 ),an expression involving only given quantities. It should be noted that the valuesof neither b nor a need be calculated.For the full integral,(I(estim.) = 1 3 h f 0 + f N +4 ∑f m +2 ∑ )f m . (27.40)m oddm evenIt can be shown, by following the same procedure as in the trapezium rule case,that the error in the estimated area is approximately∆I(estim.) ≈(b − a)180 h4 〈f (4) 〉.1004

27.4 NUMERICAL INTEGRATION27.4.3 Gaussian integrationIn the cases considered in the previous two subsections, the function f wasmimicked by linear and quadratic functions. These yield exact answers if fitself is a linear or quadratic function (respectively) of x. This process couldbe continued by increasing the order of the polynomial mimicking-function soas to increase the accuracy with which more complicated functions f could benumerically integrated. However, the same effect can be achieved with less effortby not insisting upon equally spaced points x i .The detailed analysis of such methods of numerical integration, in which theintegration points are not equally spaced and the weightings given to the values ateach point do not fall into a few simple groups, is too long to be given in full here.Suffice it to say that the methods are based upon mimicking the given functionwith a weighted sum of mutually orthogonal polynomials. The polynomials, F n (x),are chosen to be orthogonal with respect to a particular weight function w(x), i.e.∫ baF n (x)F m (x)w(x) dx = k n δ nm ,where k n is some constant that may depend upon n. Often the weight function isunity and the polynomials are mutually orthogonal in the most straightforwardsense; this is the case for Gauss–Legendre integration for which the appropriatepolynomials are the Legendre polynomials, P n (x). This particular scheme isdiscussed in more detail below.Other schemes cover cases in which one or both of the integral limits a and bare not finite. For example, if the limits are 0 and ∞ and the integrand containsa negative exponential function e −αx , a simple change of variable can cast itinto a form for which Gauss–Laguerre integration would be particularly wellsuited. This form of quadrature is based upon the Laguerre polynomials, forwhich the appropriate weight function is w(x) =e −x . Advantage is taken of this,and the handling of the exponential factor in the integrand is effectively carriedout analytically. If the other factors in the integrand can be well mimicked bylow-order polynomials, then a Gauss–Laguerre integration using only a modestnumber of points gives accurate results.If we also add that the integral over the range −∞ to ∞ of an integrandcontaining an explicit factor exp(−βx 2 ) may be conveniently calculated using ascheme based on the Hermite polynomials, the reader will appreciate the closeconnection between the various Gaussian quadrature schemes and the sets ofeigenfunctions discussed in chapter 18. As noted above, the Gauss–Legendrescheme, which we discuss next, is just such a scheme, though its weight function,being unity throughout the range, is not explicitly displayed in the integrand.Gauss–Legendre quadrature can be applied to integrals over any finite rangethough the Legendre polynomials P l (x) on which it is based are only defined1005

NUMERICAL METHODSand orthogonal over the interval −1 ≤ x ≤ 1, as discussed in subsection 18.1.2.Therefore, in order to use their properties, the integral between limits a and b in(27.34) has to be changed to one between the limits −1 and +1. This is easilydone with a change of variable from x to z given byz = 2x − b − a ,b − aso that I becomesI = b − a ∫ 1g(z) dz, (27.41)2 −1in which g(z) ≡ f(x).The n integration points x i for an n-point Gauss–Legendre integration aregiven by the zeros of P n (x), i.e. the x i are such that P n (x i ) = 0. The integrand g(x)is mimicked by the (n − 1)th-degree polynomialG(x) =n∑i=1P n (x)(x − x i )P ′ n(x i ) g(x i),which coincides with g(x) at each of the points x i , i =1, 2,...,n. To see this itshould be noted thatP n (x)limx→x k (x − x i )P n(x ′ i ) = δ ik.It then follows, to the extent that g(x) is well reproduced by G(x), that∫ 1−1g(x) dx ≈n∑i=1g(x i )P ′ n(x i )∫ 1−1P n (x)x − x idx. (27.42)The expressionw(x i ) ≡ 1 ∫ 1P n (x)P n(x ′ dxi ) −1 x − x ican be shown, using the properties of Legendre polynomials, to be equal to2w i =(1 − x 2 i )|P n(x ′ i )| , 2which is thus the weighting to be attached to the factor g(x i ) in the sum (27.42).The latter then becomes∫ 1n∑g(x) dx ≈ w i g(x i ). (27.43)−1In fact, because of the particular properties of Legendre polynomials, it can beshown that (27.43) integrates exactly any polynomial of degree up to 2n − 1. Theerror in the approximate equality is of the order of the 2nth derivative of g, and1006i=1

27.4 NUMERICAL INTEGRATIONso, provided g(x) is a reasonably smooth function, the approximation is a goodone.Taking 3-point integration as an example, the three x i are the zeros of P 3 (x) =12 (5x3 − 3x), namely 0 and ±0.774 60, and the corresponding weights are21 × ( − 2) 3 2= 8 2and9(1 − 0.6)× ( )6 2= 5 9 .2Table 27.8 gives the integration points (in the range −1 ≤ x i ≤ 1) and thecorresponding weights w i for a selection of n-point Gauss–Legendre schemes.◮Using a 3-point formula in each case, evaluate the integral∫ 11I =0 1+x dx, 2(i) using the trapezium rule, (ii) using Simpson’s rule, (iii) using Gaussian integration. Alsoevaluate the integral analytically and compare the results.(i) Using the trapezium rule, we obtainI = 1 × [ ( 12 2 f(0) + 2f 1) ]2 + f(1)[= 1 4 1+8+ ] 15 2 =0.7750.(ii) Using Simpson’s rule, we obtainI = 1 × [ ( 13 2 f(0) + 4f 1) ]2 + f(1)[= 1 6 1+16+ ] 15 2 =0.7833.(iii) Using Gaussian integration, we obtainI = 1 − 0 ∫ 1dz2 −1 1+ 1 (z +1)24{}= 1 0.555 56 [f(−0.774 60) + f(0.774 60)] +0.888 89f(0)2{}= 1 0.555 56 [0.987 458 + 0.559 503] +0.888 89 × 0.82=0.785 27.(iv) Exact evaluation gives∫ 1dxI =0 1+x = [ tan −1 x ] 1= π =0.785 40.2 04In practice, a compromise has to be struck between the accuracy of the result achievedand the calculational labour that goes into obtaining it. ◭Further Gaussian quadrature procedures, ones that utilise the properties of theChebyshev polynomials, are available for integrals over finite ranges when theintegrands involve factors of the form (1 − x 2 ) ±1/2 . In the same way as decreasinglinear and quadratic exponentials are handled through the weight functions inGauss–Laguerre and Gauss–Hermite quadrature, respectively, the square root1007

NUMERICAL METHODSGauss–Legendre integration∫ 1f(x) dx =−1i=1n∑w i f(x i )±x i w i ±x i w in =2 n =90.57735 02692 1.00000 00000 0.00000 00000 0.33023 935500.32425 34234 0.31234 70770n =3 0.61337 14327 0.26061 069640.00000 00000 0.88888 88889 0.83603 11073 0.18064 816070.77459 66692 0.55555 55556 0.96816 02395 0.08127 43884n =4 n =100.33998 10436 0.65214 51549 0.14887 43390 0.29552 422470.86113 63116 0.34785 48451 0.43339 53941 0.26926 671930.67940 95683 0.21908 63625n =5 0.86506 33667 0.14945 134920.00000 00000 0.56888 88889 0.97390 65285 0.06667 134430.53846 93101 0.47862 867050.90617 98459 0.23692 68851 n =120.12523 34085 0.24914 70458n =6 0.36783 14990 0.23349 253650.23861 91861 0.46791 39346 0.58731 79543 0.20316 742670.66120 93865 0.36076 15730 0.76990 26742 0.16007 832850.93246 95142 0.17132 44924 0.90411 72564 0.10693 932600.98156 06342 0.04717 53364n =70.00000 00000 0.41795 91837 n =200.40584 51514 0.38183 00505 0.07652 65211 0.15275 338710.74153 11856 0.27970 53915 0.22778 58511 0.14917 298650.94910 79123 0.12948 49662 0.37370 60887 0.14209 610930.51086 70020 0.13168 86384n =8 0.63605 36807 0.11819 453200.18343 46425 0.36268 37834 0.74633 19065 0.10193 011980.52553 24099 0.31370 66459 0.83911 69718 0.08327 674160.79666 64774 0.22238 10345 0.91223 44283 0.06267 204830.96028 98565 0.10122 85363 0.96397 19272 0.04060 142980.99312 85992 0.01761 40071Table 27.8 The integration points and weights for a number of n-pointGauss–Legendre integration formulae. The points are given as ±x i and thecontributions from both +x i and −x i must be included. However, the contributionfrom any point x i = 0 must be counted only once.1008

27.4 NUMERICAL INTEGRATIONfactor is treated accurately in Gauss–Chebyshev integration. Thus∫ 1f(x)n∑√ dx ≈ w i f(x i ), (27.44)1 − x2−1where the integration points x i are the zeros of the Chebyshev polynomials ofthe first kind T n (x) andw i are the corresponding weights. Fortunately, both setsare analytic and can be written compactly for all n asx i =cos (i − 1 2 )π , w i = π for i =1,... ,n. (27.45)nnNote that, for any given n, all points are weighted equally and that no specialaction is required to deal with the integrable singularities at x = ±1; they aredealt with automatically through the weight function.For integrals involving factors of the form (1−x 2 ) 1/2 , the corresponding formula,based on Chebyshev polynomials of the second kind U n (x), is∫ 1f(x) √ n∑1 − x 2 dx ≈ w i f(x i ), (27.46)−1with integration points and weights given, for i =1,... ,n,byiπx i =cosn +1 , w i = π iπn +1 sin2 n +1 . (27.47)For discussions of the many other schemes available, as well as their relativemerits, the reader is referred to books devoted specifically to the theory ofnumerical analysis. There, details of integration points and weights, as well asquantitative estimates of the error involved in replacing an integral by a finitesum, will be found. Table 27.9 gives the points and weights for a selection ofGauss–Laguerre and Gauss–Hermite schemes. §i=1i=127.4.4 Monte Carlo methodsSurprising as it may at first seem, random numbers may be used to carry outnumerical integration. The random element comes in principally when selectingthe points at which the integrand is evaluated, and naturally does not extend tothe actual values of the integrand!For the most part we will continue to use as our model one-dimensionalintegrals between finite limits, as typified by equation (27.34). Extensions to coverinfinite or multidimensional integrals will be indicated briefly at the end of thesection. It should be noted here, however, that Monte Carlo methods – the name§ They, and those presented in table 27.8 for Gauss–Legendre integration, are taken from the muchmore comprehensive sets to be found in M. Abramowitz and I. A. Stegun (eds), Handbook ofMathematical Functions (New York: Dover, 1965).1009

NUMERICAL METHODSGauss–Laguerre and Gauss–Hermite integration∫ ∞e −x f(x) dx =0i=1n∑w i f(x i )∫ ∞e −x2 f(x) dx =−∞i=1n∑w i f(x i )x i w i ±x i w in =2 n =20.58578 64376 0.85355 33906 0.70710 67812 0.88622 692553.41421 35624 0.14644 66094n =3n =3 0.00000 00000 1.18163 590060.41577 45568 0.71109 30099 1.22474 48714 0.29540 897522.29428 03603 0.27851 773366.28994 50829 0.01038 92565 n =40.52464 76233 0.80491 40900n =4 1.65068 01239 0.08131 283540.32254 76896 0.60315 410431.74576 11012 0.35741 86924 n =54.53662 02969 0.03888 79085 0.00000 00000 0.94530 872059.39507 09123 0.00053 92947 0.95857 24646 0.39361 932322.02018 28705 0.01995 32421n =50.26356 03197 0.52175 56106 n =61.41340 30591 0.39866 68111 0.43607 74119 0.72462 959523.59642 57710 0.07594 24497 1.33584 90740 0.15706 732037.08581 00059 0.00361 17587 2.35060 49737 0.00453 0009912.6408 00844 0.00002 33700n =7n =6 0.00000 00000 0.81026 461760.22284 66042 0.45896 46740 0.81628 78829 0.42560 725261.18893 21017 0.41700 08308 1.67355 16288 0.05451 558282.99273 63261 0.11337 33821 2.65196 13568 0.00097 178125.77514 35691 0.01039 919759.83746 74184 0.00026 10172 n =815.9828 73981 0.00000 08985 0.38118 69902 0.66114 701261.15719 37124 0.20780 23258n =7 1.98165 67567 0.01707 798300.19304 36766 0.40931 89517 2.93063 74203 0.00019 960411.02666 48953 0.42183 127792.56787 67450 0.14712 63487 n =94.90035 30845 0.02063 35145 0.00000 00000 0.72023 521568.18215 34446 0.00107 40101 0.72355 10188 0.43265 1559012.7341 80292 0.00001 58655 1.46855 32892 0.08847 4527419.3957 27862 0.00000 00317 2.26658 05845 0.00494 362433.19099 32018 0.00003 96070Table 27.9 The integration points and weights for a number of n-pointGauss–Laguerre and Gauss–Hermite integration formulae. Where the pointsare given as ±x i , the contributions from both +x i and −x i must be included.However, the contribution from any point x i = 0 must be counted only once.1010

27.4 NUMERICAL INTEGRATIONhas become attached to methods based on randomly generated numbers – inmany ways come into their own when used on multidimensional integrals overregions with complicated boundaries.It goes without saying that in order to use random numbers for calculationalpurposes a supply of them must be available. There was a time when they wereprovided in book form as a two-dimensional array of random digits in the range0 to 9. The user could generate the successive digits of a random number ofany desired length by selecting their positions in the table in any predeterminedand systematic way. Nowadays all computers and nearly all pocket calculatorsoffer a function which supplies a sequence of decimal numbers, ξ, that,forallpractical purposes, are randomly and uniformly chosen in the range 0 ≤ ξ

NUMERICAL METHODSaveraged:t = 1 nn∑f(ξ i ). (27.49)i=1Stratified samplingHere the range of x is broken up into k subranges,0=α 0

27.4 NUMERICAL INTEGRATIONwill have a very small variance. Further, any error in inverting the relationshipbetween η and ξ will not be important since f(η)/g(η) will be largely independentof the value of η.As an example, consider the function f(x) =[tan −1 (x)] 1/2 , which is not analyticallyintegrable over the range (0, 1) but is well mimicked by the easily integratedfunction g(x) =x 1/2 (1 − x 2 /6). The ratio of the two varies from 1.00 to 1.06 as xvaries from 0 to 1. The integral of g over this range is 0.619 048, and so it has tobe renormalised by the factor 1.615 38. The value of the integral of f(x) from0to 1 can then be estimated by averaging the value of[tan −1 (η)] 1/2(1.615 38) η 1/2 (1 − 1 6 η2 )for random variables η which are such that G(η) is uniformly distributed on(0, 1). Using batches of as few as ten random numbers gave a value 0.630 for θ,with standard deviation 0.003. The corresponding result for crude Monte Carlo,using the same random numbers, was 0.634 ± 0.065. The increase in precision isobvious, though the additional labour involved would not be justified for a singleapplication.Control variatesThe control-variate method is similar to, but not the same as, importance sampling.Again, an analytically integrable function that mimics f(x) in shape hasto be found. The function, known as the control variate, is first scaled so as tomatch f as closely as possible in magnitude and then its integral is found inclosed form. If we denote the scaled control variate by h(x), then the estimate ofθ is computed ast =∫ 10[f(x) − h(x)] dx +∫ 10h(x) dx. (27.51)The first integral in (27.51) is evaluated using (crude) Monte Carlo, whilst thesecond is known analytically. Although the first integral should have been renderedsmall by the choice of h(x), it is its variance that matters. The method relieson the following result (see equation (30.136)):V [t − t ′ ]=V [t]+V [t ′ ] − 2Cov[t, t ′ ],and on the fact that if t estimates θ whilst t ′ estimates θ ′ using the same randomnumbers, then the covariance of t and t ′ can be larger than the variance of t ′ ,and indeed will be so if the integrands producing θ and θ ′ are highly correlated.To evaluate the same integral as was estimated previously using importancesampling, we take as h(x) the function g(x) used there, before it was renormalised.Again using batches of ten random numbers, the estimated value for θ was foundto be 0.629 ± 0.004, a result almost identical to that obtained using importance1013

NUMERICAL METHODSsampling, in both value and precision. Since we knew already that f(x) andg(x)diverge monotonically by about 6% as x varies over the range (0, 1), we couldhave made a small improvement to our control variate by scaling it by 1.03 beforeusing it in equation (27.51).Antithetic variatesAs a final example of a method that improves on crude Monte Carlo, and one thatis particularly useful when monotonic functions are to be integrated, we mentionthe use of antithetic variates. This method relies on finding two estimates t andt ′ of θ that are strongly anticorrelated (i.e. Cov[t, t ′ ] is large and negative) andusing the resultV [ 1 2 (t + t′ )] = 1 4 V [t]+ 1 4 V [t′ ]+ 1 2 Cov[t, t′ ].For example, the use of 1 2[f(ξ) +f(1 − ξ)] instead of f(ξ) involves only twiceas many evaluations of f, and no more random variables, but generally givesan improvement in precision significantly greater than this. For the integral off(x) =[tan −1 (x)] 1/2 , using as previously a batch of ten random variables, anestimate of 0.623 ± 0.018 was found. This is to be compared with the crudeMonte Carlo result, 0.634 ± 0.065, obtained using the same number of randomvariables.For a fuller discussion of these methods, and of theoretical estimates of theirefficiencies, the reader is referred to more specialist treatments. For practical implementationschemes, a book dedicated to scientific computing should be consulted. §Hit or miss methodWe now come to the approach that, in spirit, is closest to the activities that gaveMonte Carlo methods their name. In this approach, one or more straightforwardyes/no decisions are made on the basis of numbers drawn at random – the endresult of each trial is either a hit or a miss! In this section we are concernedwith numerical integration, but the general Monte Carlo approach, in whichone estimates a physical quantity that is hard or impossible to calculate directlyby simulating the physical processes that determine it, is widespread in modernscience. For example, the calculation of the efficiencies of detector arrays inexperiments to study elementary particle interactions are nearly always carriedout in this way. Indeed, in a normal experiment, far more simulated interactionsare generated in computers than ever actually occur when the experiment istaking real data.As was noted in chapter 2, the process of evaluating a one-dimensional integralf(x)dx can be regarded as that of finding the area between the curve y = f(x)∫ ba§ e.g. W. H. Press, S. A. Teukolsky, W. T. Vetterling and B. P. Flannery, Numerical Recipes in C: TheArtofScientificComputing, 2nd edn (Cambridge: Cambridge University Press, 1992).1014

27.4 NUMERICAL INTEGRATIONy = f(x)y = cx = ax = bxFigure 27.5 A simple rectangular figure enclosing the area (shown shaded)which is equal to ∫ bf(x) dx.aand the x-axis in the range a ≤ x ≤ b. It may not be possible to do thisanalytically, but if, as shown in figure 27.5, we can enclose the curve in a simplefigure whose area can be found trivially then the ratio of the required (shaded)area to that of the bounding figure, c(b − a), is the same as the probability that arandomly selected point inside the boundary will lie below the line.In order to accommodate cases in which f(x) can be negative in part of thex-range, we treat a slightly more general case. Suppose that, for a ≤ x ≤ b, f(x)is bounded and known to lie in the range A ≤ f(x) ≤ B; then the transformationz = x − ab − awill reduce the integral ∫ baf(x) dx to the formA(b − a)+(B − A)(b − a)∫ 10h(z) dz, (27.52)whereh(z) = 1 [f ((b − a)z + a) − A] .B − AIn this form z lies in the range 0 ≤ z ≤ 1andh(z) lies in the range 0 ≤ h(z) ≤ 1,i.e. both are suitable for simulation using the standard random-number generator.It should be noted that, for an efficient estimation, the bounds A and B shouldbe drawn as tightly as possible –preferably, but not necessarily, they should beequal to the minimum and maximum values of f in the range. The reason forthis is that random numbers corresponding to values which f(x) cannot reachadd nothing to the estimation but do increase its variance.It only remains to estimate the final integral on the RHS of equation (27.52).This we do by selecting pairs of random numbers, ξ 1 and ξ 2 , and testing whether1015

NUMERICAL METHODSh(ξ 1 ) >ξ 2 . The fraction of times that this inequality is satisfied estimates the valueof the integral (without the scaling factors (B − A)(b − a)) since the expectationvalue of this fraction is the ratio of the area below the curve y = h(z) totheareaof a unit square.To illustrate the evaluation of multiple integrals using Monte Carlo techniques,consider the relatively elementary problem of finding the volume of an irregularsolid bounded by planes, say an octahedron. In order to keep the descriptionbrief, but at the same time illustrate the general principles involved, let us supposethat the octahedron has two vertices on each of the three Cartesian axes, one oneither side of the origin for each axis. Denote those on the x-axis by x 1 (< 0) andx 2 (> 0), and similarly for the y- andz-axes. Then the whole of the octahedroncan be enclosed by the rectangular parallelepipedx 1 ≤ x ≤ x 2 , y 1 ≤ y ≤ y 2 , z 1 ≤ z ≤ z 2 .Any point in the octahedron lies inside or on the parallelepiped, but any pointin the parallelepiped may or may not lie inside the octahedron.The equation of the plane containing the three vertex points (x i , 0, 0), (0,y j , 0)and (0, 0,z k )isx+ y + z =1 fori, j, k =1, 2, (27.53)x i y j z kand the condition that any general point (x, y, z) lies on the same side of theplane as the origin is thatx+ y + z − 1 ≤ 0. (27.54)x i y j z kFor the point to be inside or on the octahedron, equation (27.54) must thereforebe satisfied for all eight of the sets of i, j and k given in (27.53).Thus an estimate of the volume of the octahedron can be made by generatingrandom numbers ξ from the usual uniform distribution and then using them insets of three, according to the following scheme.With integer m labelling the mth set of three random numbers, calculatex = x 1 + ξ 3m−2 (x 2 − x 1 ),y = y 1 + ξ 3m−1 (y 2 − y 1 ),z = z 1 + ξ 3m (z 2 − z 1 ).Define a variable n m as 1 if (27.54) is satisfied for all eight combinations of i, j, kvalues and as 0 otherwise. The volume V can then be estimated using 3M randomnumbers from the formulaV(x 2 − x 1 )(y 2 − y 1 )(z 2 − z 1 ) = 1 M1016M∑n m .m=1

27.4 NUMERICAL INTEGRATIONIt will be seen that, by replacing each n m in the summation by f(x, y, z)n m ,this procedure could be extended to estimate the integral of the function f overthe volume of the solid. The method has special valueif f is too complicated tohave analytic integrals with respect to x, y and z or if the limits of any of theseintegrals are determined by anything other than the simplest combinations of theother variables. If large values of f are known to be concentrated in particularregions of the integration volume, then some form of stratified sampling shouldbe used.It will be apparent that this general method can be extended to integralsof general functions, bounded but not necessarily continuous, over volumes withcomplicated bounding surfaces and, if appropriate, in more than three dimensions.Random number generationEarlier in this subsection we showed how to evaluate integrals using sequences ofnumbers that we took to be distributed uniformly on the interval 0 ≤ ξ

NUMERICAL METHODSon (0, 1) and then take as the random number y the value of F −1 (ξ). We nowillustrate this with a worked example.◮Find an explicit formula that will generate a random number y distributed on (−∞, ∞)according to the Cauchy distribution( a)dyf(y) dy =π a 2 + y , 2given a random number ξ uniformly distributed on (0, 1).The first task is to determine the indefinite integral:∫ y ( a)dtF(y) =−∞ π a 2 + t = 1 y 2 π tan−1 a + 1 2 .Now, if y is distributed as we wish then F(y) is uniformly distributed on (0, 1). This followsfrom the fact that the derivative of F(y) isf(y). We therefore set F(y) equaltoξ andobtainξ = 1 y π tan−1 a + 1 2 ,yieldingy = a tan[π(ξ − 1 )]. 2This explicit formula shows how to change a random number ξ drawn from a populationuniformly distributed on (0, 1) into a random number y distributed according to theCauchy distribution. ◭Look-up tables operate as described below for cumulative distributions F(y)that are non-invertible, i.e. F −1 (y) cannot be expressed in closed form. Theyare especially useful if many random numbers are needed but great samplingaccuracy is not essential. The method for an N-entry table can be summarised asfollows. Define w m by F(w m )=m/N for m =1, 2,...,N, and store a table ofy(m) = 1 2 (w m + w m−1 ).As each random number y is needed, calculate k as the integral part of Nξ andtake y as given by y(k).Normally, such a look-up table would have to be used for generating randomnumbers with a Gaussian distribution, as the cumulative integral of a Gaussian isnon-invertible. It would be, in essence, table 30.3, with the roles of argument andvalue interchanged. In this particular case, an alternative, based on the centrallimit theorem, can be considered.With ξ i generated in the usual way, i.e. uniformly distributed on the interval0 ≤ ξ

27.5 FINITE DIFFERENCESmany values of ξ i for each value of y and is a very poor approximation if thewings of the Gaussian distribution have to be sampled accurately. For nearly allpractical purposes a Gaussian look-up table is to be preferred.27.5 Finite differencesIt will have been noticed that earlier sections included several equations linkingsequential values of f i and the derivatives of f evaluated at one of the x i .Inthis section, by way of preparation for the numerical treatment of differentialequations, we establish these relationships in a more systematic way.Again we consider a set of values f i of a function f(x) evaluated at equallyspaced points x i , their separation being h. As before, the basis for our discussionwill be a Taylor series expansion, but on this occasion about the point x i :f i±1 = f i ± hf i ′ + h22! f′′ i ± h33! f(3) i + ··· . (27.56)In this section, and subsequently, we denote the nth derivative evaluated at x iby f (n)i .From (27.56), three different expressions that approximate f (1)ican be derived.The first of these, obtained by subtracting the ± equations, is( )f (1) dfi≡ = f i+1 − f i−1dx 2hx i− h23! f(3) i− ··· . (27.57)The quantity (f i+1 − f i−1 )/(2h) is known as the central difference approximationto f (1)iand can be seen from (27.57) to be in error by approximately (h 2 /6)f (3)i.An alternative approximation, obtained from (27.56+) alone, is given by( )f (1) dfi ≡ = f i+1 − f i− h dxx ih 2! f(2) i − ··· . (27.58)The forward difference approximation, (f i+1 − f i )/h, is clearly a poorer approximation,since it is in error by approximately (h/2)f (2)i as compared with (h 2 /6)f (3)i .Similarly, the backward difference (f i − f i−1 )/h obtained from (27.56−) is not asgood as the central difference; the sign of the error is reversed in this case.This type of differencing approximation can be continued to the higher derivativesof f in an obvious manner. By adding the two equations (27.56±), a centraldifference approximation to f (2)i can be obtained:(f (2) d 2 )fi≡dx 2 ≈ f i+1 − 2f i + f i−1h 2 . (27.59)The error in this approximation (also known as the second difference of f) iseasilyshowntobeabout(h 2 /12)f (4)i.Of course, if the function f(x) is a sufficiently simple polynomial in x, all1019

NUMERICAL METHODSderivatives beyond a particular one will vanish and there is no error in takingthe differences to obtain the derivatives.◮The following is copied from the tabulation of a second-degree polynomial f(x) at valuesof x from 1 to 12 inclusive:2, 2, ?, 8, 14, 22, 32, 46, ?, 74, 92, 112.The entries marked ? were illegible and in addition one error was made in transcription.Complete and correct the table. Would your procedure have worked if the copying errorhad been in f(6)?Write out the entries again in row (a) below, and where possible calculate first differencesin row (b) and second differences in row (c). Denote the jth entry in row (n) by(n) j .(a) 2 2 ? 8 14 22 32 46 ? 74 92 112(b) 0 ? ? 6 8 10 14 ? ? 18 20(c) ? ? ? 2 2 4 ? ? ? 2Because the polynomial is second-degree, the second differences (c) j , which are proportionalto d 2 f/dx 2 , should be constant, and clearly the constant should be 2. That is, (c) 6 shouldequal 2 and (b) 7 should equal 12 (not 14). Since all the (c) j = 2, we can conclude that(b) 2 =2,(b) 3 =4,(b) 8 = 14, and (b) 9 = 16. Working these changes back to row (a) showsthat (a) 3 =4,(a) 8 = 44 (not 46), and (a) 9 = 58.The entries therefore should read(a) 2, 2, 4, 8, 14, 22, 32, 44, 58, 74, 92, 112,where the amended entries are shown in bold type.It is easily verified that if the error were in f(6) no two computable entries in row (c)would be equal, and it would not be clear what the correct common entry should be.Nevertheless, trial and error might arrive at a self-consistent scheme. ◭27.6 Differential equationsFor the remaining sections of this chapter our attention will be on the solutionof differential equations by numerical methods. Some of the general difficultiesof applying numerical methods to differential equations will be all too apparent.Initially we consider only the simplest kind of equation – one of first order,typically represented bydy= f(x, y), (27.60)dxwhere y is taken as the dependent variable and x the independent one. If thisequation can be solved analytically then that is the best course to adopt. Butsometimes it is not possible to do so and a numerical approach becomes theonly one available. In fact, most of the examples that we will use can be solvedeasily by an explicit integration, but, for the purposes of illustration, this is anadvantage rather than the reverse since useful comparisons can then be madebetween the numerically derived solution and the exact one.1020

27.6 DIFFERENTIAL EQUATIONSx h y(exact)0.01 0.1 0.5 1.0 1.5 2 30 (1) (1) (1) (1) (1) (1) (1) (1)0.5 0.605 0.590 0.500 0 −0.500 −1 −2 0.6071.0 0.366 0.349 0.250 0 0.250 1 4 0.3681.5 0.221 0.206 0.125 0 −0.125 −1 −8 0.2232.0 0.134 0.122 0.063 0 0.063 1 16 0.1352.5 0.081 0.072 0.032 0 −0.032 −1 −32 0.0823.0 0.049 0.042 0.016 0 0.016 1 64 0.050Table 27.10 The solution y of differential equation (27.61) using the Eulerforward difference method for various values of h. The exact solution is alsoshown.27.6.1 Difference equationsConsider the differential equationdy= −y, y(0) = 1, (27.61)dxand the possibility of solving it numerically by approximating dy/dx by a finitedifference along the lines indicated in section 27.5. We start with the forwarddifference( ) dy≈ y i+1 − y i, (27.62)dxx ihwhere we use the notation of section 27.5 but with f replaced by y. Inthisparticular case, it leads to the recurrence relation( ) dyy i+1 = y i + h = y i − hy i =(1− h)y i . (27.63)dxiThus, since y 0 = y(0) = 1 is given, y 1 = y(0 + h) =y(h) can be calculated, andso on (this is the Euler method). Table 27.10 shows the values of y(x) obtainedif this is done using various values of h and for selected values of x. The exactsolution, y(x) =exp(−x), is also shown.It is clear that to maintain anything like a reasonable accuracy only very smallsteps, h, can be used. Indeed, if h is taken to be too large, not only is the accuracybad but, as can be seen, for h>1 the calculated solution oscillates (when itshould be monotonic), and for h>2 it diverges. Equation (27.63) is of the formy i+1 = λy i , and a necessary condition for non-divergence is |λ| < 1, i.e. 0

NUMERICAL METHODSx y(estim.) y(exact)−0.5 (1.648) —0 (1.000) (1.000)0.5 0.648 0.6071.0 0.352 0.3681.5 0.296 0.2232.0 0.056 0.1352.5 0.240 0.0823.0 −0.184 0.050Table 27.11 The solution of differential equation (27.61) using the Milnecentral difference method with h =0.5 and accurate starting values.more accurate, but of course still approximate, central difference. A more accuratemethod based on central differences (Milne’s method) gives the recurrence relation( ) dyy i+1 = y i−1 +2h(27.64)dxiin general and, in this particular case,y i+1 = y i−1 − 2hy i . (27.65)An additional difficulty now arises, since two initial values of y are needed.The second must be estimated by other means (e.g. by using a Taylor series,as discussed later), but for illustration purposes we will take the accurate value,y(−h) =exph, as the value of y −1 .Ifh is taken as, say, 0.5 and (27.65) is appliedrepeatedly, then the results shown in table 27.11 are obtained.Although some improvement in the early values of the calculated y(x) isnoticeable, as compared with the corresponding (h =0.5) column of table 27.10,this scheme soon runs into difficulties, as is obvious from the last two rows of thetable.Some part of this poor performance is not really attributable to the approximationsmade in estimating dy/dx but to the form of the equation itself andhence of its solution. Any rounding error occurring in the evaluation effectivelyintroduces into y some contamination by the solution ofdydx =+y.This equation has the solution y(x) =expx and so grows without limit; ultimatelyit will dominate the sought-for solution and thus render the calculations totallyinaccurate.We have only illustrated, rather than analysed, some of the difficulties associatedwith simple finite-difference iteration schemes for first-order differential equations,1022

27.6 DIFFERENTIAL EQUATIONSbut they may be summarised as (i) insufficiently precise approximations to thederivatives and (ii) inherent instability due to rounding errors.27.6.2 Taylor series solutionsSince a Taylor series expansion is exact if all its terms are included, and the limitsof convergence are not exceeded, we may seek to use one to evaluate y 1 , y 2 ,etc.for an equationdy= f(x, y), (27.66)dxwhen the initial value y(x 0 )=y 0 is given.The Taylor series isy(x + h) =y(x)+hy ′ (x)+ h22! y′′ (x)+ h33! y(3) (x)+··· . (27.67)In the present notation, at the point x = x i this is writteny i+1 = y i + hy (1)i + h22! y(2) i+ h33! y(3) i + ··· . (27.68)But, for the required solution y(x), we know that( )y (1) dyi ≡ = f(x i ,y i ), (27.69)dxx iand the value of the second derivative at x = x i , y = y i can be obtained from it:y (2)i= ∂f∂x + ∂f dy∂y dx = ∂f∂x + f ∂f∂y . (27.70)This process can be continued for the third and higher derivatives, all of whichare to be evaluated at (x i ,y i ).Having obtained expressions for the derivatives y (n)iin (27.67), two alternativeways of proceeding are open to us:(i) equation (27.68) is used to evaluate y i+1 , the whole process is repeated toobtain y i+2 ,andsoon;(ii) equation (27.68) is applied several times but using a different value of heach time, and so the corresponding values of y(x + h) are obtained.It is clear that, on the one hand, approach (i) does not require so many terms of(27.67) to be kept, but, on the other hand, the y i (n) have to be recalculated ateach step. With approach (ii), fairly accurate results for y may be obtained forvalues of x close to the given starting value, but for large values of h a largenumber of terms of (27.67) must be kept. As an example of approach (ii) wesolve the following problem.1023

NUMERICAL METHODSx y(estim.) y(exact)0 1.0000 1.00000.1 1.2346 1.23460.2 1.5619 1.56250.3 2.0331 2.04080.4 2.7254 2.77780.5 3.7500 4.0000Table 27.12The solution of differential equation (27.71) using a Taylor series.◮Find the numerical solution of the equationdydx =2y3/2 , y(0) = 1, (27.71)for x =0.1 to 0.5 in steps of 0.1. Compare it with the exact solution obtained analytically.Since the right-hand side of the equation does not contain x explicitly, (27.70) is greatlysimplified and the calculation becomes a repeated application ofy (n+1)i= ∂y(n) dy∂y dx = f ∂y(n)∂y .The necessary derivatives and their values at x =0,wherey = 1, are given below:y(0) = 1 1y ′ =2y 3/2 2y ′′ =(3/2)(2y 1/2 )(2y 3/2 )=6y 2 6y (3) =(12y)2y 3/2 =24y 5/2 24y (4) =(60y 3/2 )2y 3/2 = 120y 3 120y (5) = (360y 2 )2y 3/2 = 720y 7/2 720Thus the Taylor expansion of the solution about the origin (in fact a Maclaurin series) isy(x)=1+2x + 6 2! x2 + 243! x3 + 1204! x4 + 7205! x5 + ··· .Hence, y(estim.) = 1 + 2x +3x 2 +4x 3 +5x 4 +6x 5 . Values calculated from this are givenin table 27.12. Comparison with the exact values shows that using the first six terms givesa value that is correct to one part in 100, up to x =0.3. ◭27.6.3 Prediction and correctionAn improvement in the accuracy obtainable using difference methods is possibleif steps are taken, sometimes retrospectively, to allow for inaccuracies in approximatingderivatives by differences. We will describe only the simplest schemes ofthis kind and begin with a prediction method, usually called the Adams method.1024

27.6 DIFFERENTIAL EQUATIONSThe forward difference estimate of y i+1 , namely( ) dyy i+1 = y i + h = y i + hf(x i ,y i ), (27.72)dxiwould give exact results if y were a linear function of x in the range x i ≤ x ≤ x i +h.The idea behind the Adams method is to allow some relaxation of this andsuppose that y can be adequately approximated by a parabola over the intervalx i−1 ≤ x ≤ x i+1 . In the same interval, dy/dx can then be approximated by a linearfunction:f(x, y) = dydx ≈ a + b(x − x i) for x i − h ≤ x ≤ x i + h.The values of a and b are fixed by the calculated values of f at x i−1 and x i ,whichwe may denote by f i−1 and f i :Thuswhich yieldsa = f i ,∫ xi+h[y i+1 − y i ≈x ib = f i − f i−1.hf i + (f i − f i−1 )h](x − x i ) dx,y i+1 = y i + hf i + 1 2 h(f i − f i−1 ). (27.73)The last term of this expression is seen to be a correction to result (27.72). Thatit is, in some sense, the second-order correction,12 h2 y (2)i−1/2 ,to a first-order formula is apparent.Such a procedure requires, in addition to a value for y 0 , a value for either y 1 ory −1 ,sothatf 1 or f −1 can be used to initiate the iteration. This has to be obtainedby other methods, e.g. a Taylor series expansion.Improvements to simple difference formulae can also be obtained by usingcorrection methods. In these, a rough prediction of the value y i+1 is made first,and then this is used in a better formula, not originally usable since it, in turn,requires a value of y i+1 for its evaluation. The value of y i+1 is then recalculated,using this better formula.Such a scheme based on the forward difference formula might be as follows:(i) predict y i+1 using y i+1 = y i + hf i ;(ii) calculate f i+1 using this value;(iii) recalculate y i+1 using y i+1 = y i + h(f i + f i+1 )/2. Here (f i + f i+1 )/2 hasreplaced the f i used in (i), since it better represents the average value ofdy/dx in the interval x i ≤ x ≤ x i + h.1025

NUMERICAL METHODSSteps (ii) and (iii) can be iterated to improve further the approximation to theaverage value of dy/dx, but this will not compensate for the omission of higherorderderivatives in the forward difference formula.Many more complex schemes of prediction and correction, in most casescombining the two in the same process, have been devised, but the reader isreferred to more specialist texts for discussions of them. However, because itoffers some clear advantages, one group of methods will be set out explicitly inthe next subsection. This is the general class of schemes known as Runge–Kuttamethods.27.6.4 Runge–Kutta methodsThe Runge–Kutta method of integratingdy= f(x, y) (27.74)dxis a step-by-step process of obtaining an approximation for y i+1 by starting fromthe value of y i . Among its advantages are that no functions other than f are used,no subsidiary differentiation is needed and no additional starting values need becalculated.To be set against these advantages is the fact that f is evaluated using somewhatcomplicated arguments and that this has to be done several times for each increasein the value of i. However, once a procedure has been established, for exampleon a computer, the method usually gives good results.The basis of the method is to simulate the (accurate) Taylor series for y(x i + h),not by calculating all the higher derivatives of y at the point x i but by takinga particular combination of the values of the first derivative of y evaluated ata number of carefully chosen points. Equation (27.74) is used to evaluate thesederivatives. The accuracy can be made to be up to whatever power of h is desired,but, naturally, the greater the accuracy, the more complex the calculation, and,in any case, rounding errors cannot ultimately be avoided.The setting up of the calculational scheme may be illustrated by consideringthe particular case in which second-order accuracy in h is required. To secondorder, the Taylor expansion is( )y i+1 = y i + hf i + h2 df, (27.75)2 dxx iwhere( ) ( df ∂f=dxx i∂x + f ∂f )≡ ∂f i∂yx i∂x + f ∂f ii∂y ,the last step being merely the definition of an abbreviated notation.1026

27.6 DIFFERENTIAL EQUATIONSWe assume that this can be simulated by a formy i+1 = y i + α 1 hf i + α 2 hf(x i + β 1 h, y i + β 2 hf i ), (27.76)which in effect uses a weighted mean of the value of dy/dx at x i and its value atsome point yet to be determined. The object is to choose values of α 1 , α 2 , β 1 andβ 2 such that (27.76) coincides with (27.75) up to the coefficient of h 2 .Expanding the function f in the last term of (27.76) in a Taylor series of itsown, we obtainf(x i + β 1 h, y i + β 2 hf i )=f(x i ,y i )+β 1 h ∂f i∂x + β 2hf i∂f i∂y +O(h2 ).Putting this result into (27.76) and rearranging in powers of h, we obtain()y i+1 = y i +(α 1 + α 2 )hf i + α 2 h 2 ∂f iβ 1∂x + β ∂f i2f i . (27.77)∂yComparing this with (27.75) shows that there is, in fact, some freedom remainingin the choice of the α’s and β’s. In terms of an arbitrary α 1 (≠1),α 2 =1− α 1 , β 1 = β 2 =12(1 − α 1 ) .One possible choice is α 1 =0.5, giving α 2 =0.5, β 1 = β 2 = 1. In this case theprocedure (equation (27.76)) can be summarised bywherey i+1 = y i + 1 2 (a 1 + a 2 ), (27.78)a 1 = hf(x i ,y i ),a 2 = hf(x i + h, y i + a 1 ).Similar schemes giving higher-order accuracy in h can be devised. Two suchschemes, given without derivation, are as follows.(i) To order h 3 ,wherey i+1 = y i + 1 6 (b 1 +4b 2 + b 3 ), (27.79)b 1 = hf(x i ,y i ),b 2 = hf(x i + 1 2 h, y i + 1 2 b 1),b 3 = hf(x i + h, y i +2b 2 − b 1 ).1027

NUMERICAL METHODS(ii) To order h 4 ,wherey i+1 = y i + 1 6 (c 1 +2c 2 +2c 3 + c 4 ), (27.80)c 1 = hf(x i ,y i ),c 2 = hf(x i + 1 2 h, y i + 1 2 c 1),c 3 = hf(x i + 1 2 h, y i + 1 2 c 2),c 4 = hf(x i + h, y i + c 3 ).27.6.5 IsoclinesThe final method to be described for first-order differential equations is not somuch numerical as graphical, but since it is sometimes useful it is included here.The method, known as that of isoclines, involves sketching for a number ofvalues of a parameter c those curves (the isoclines) in the xy-plane along whichf(x, y) =c, i.e. those curves along which dy/dx is a constant of known value. Itshould be noted that isoclines are not generally straight lines. Since a straightline of slope dy/dx at and through any particular point is a tangent to the curvey = y(x) at that point, small elements of straight lines, with slopes appropriateto the isoclines they cut, effectively form the curve y = y(x).Figure 27.6 illustrates in outline the method as applied to the solution ofdy= −2xy. (27.81)dxThe thinner curves (rectangular hyperbolae) are a selection of the isoclines alongwhich −2xy is constant and equal to the corresponding value of c. The smallcross lines on each curve show the slopes (= c) that solutions of (27.81) musthave if they cross the curve. The thick line is the solution for which y =1atx = 0; it takes the slope dictated by the value of c on each isocline it crosses. Theanalytic solution with these properties is y(x) =exp(−x 2 ).27.7 Higher-order equationsSo far the discussion of numerical solutions of differential equations has beenin terms of one dependent and one independent variable related by a first-orderequation. It is straightforward to carry out an extension to the case of severaldependent variables y [r] governed by R first-order equations:dy [r]dx = f [r](x, y [1] ,y [2] ,...,y [R] ), r =1, 2,...,R.We have enclosed the label r in brackets so that there is no confusion between,say, the second dependent variable y [2] and the value y 2 of a variable y at the1028

27.7 HIGHER-ORDER EQUATIONSy1.00.80.6y0.40.20.20.40.60.8c−1.0−0.8−0.6−0.4−0.2−0.11.0 xFigure 27.6 The isocline method. The cross lines on each isocline show theslopes that solutions of dy/dx = −2xy must have at the points where theycross the isoclines. The heavy line is the solution with y(0) = 1, namelyexp(−x 2 ).second calculational point x 2 . The integration of these equations by the methodsdiscussed in the previous section presents no particular difficulty, provided thatall the equations are advanced through each particular step before any of themis taken through the following step.Higher-order equations in one dependent and one independent variable can bereduced to a set of simultaneous equations, provided that they can be written inthe formd R ydx R = f(x, y, y′ ,...,y (R−1) ), (27.82)where R is the order of the equation. To do this, a new set of variables p [r] isdefined byp [r] = dr y, r =1, 2,...,R− 1. (27.83)dxr Equation (27.82) is then equivalent to the following set of simultaneous first-orderequations:dydx = p [1],dp [r]dx = p [r+1], r =1, 2,...,R− 2, (27.84)dp [R−1]= f(x, y, p [1] ,...,p [R−1] ).dx1029

NUMERICAL METHODSThese can then be treated in the way indicated in the previous paragraph. Theextension to more than one dependent variable is straightforward.In practical problems it often happens that boundary conditions applicableto a higher-order equation consist not of the values of the function and all itsderivatives at one particular point but of, say, the values of the function at twoseparate end-points. In these cases a solution cannot be found using an explicitstep-by-step ‘marching’ scheme, in which the solutions at successive values of theindependent variable are calculated using solution values previously found. Othermethods have to be tried.One obvious method is to treat the problem as a ‘marching one’, but to use anumber of (intelligently guessed) initial values for the derivatives at the startingpoint. The aim is then to find, by interpolation or some other form of iteration,those starting values for the derivatives that will produce the given value of thefunction at the finishing point.In some cases the problem can be reduced by a differencing scheme to a matrixequation. Such a case is that of a second-order equation for y(x) with constantcoefficients and given values of y at the two end-points. Consider the second-orderequationwith the boundary conditionsy ′′ +2ky ′ + µy = f(x), (27.85)y(0) = A, y(1) = B.If (27.85) is replaced by a central difference equation,y i+1 − 2y i + y i−1h 2we obtain from it the recurrence relation+2k y i+1 − y i−12h+ µy i = f(x i ),(1 + kh)y i+1 +(µh 2 − 2)y i +(1− kh)y i−1 = h 2 f(x i ).For h =1/(N − 1) this is in exactly the form of the N × N tridiagonal matrixequation (27.30), withb 1 = b N =1, c 1 = a N =0,a i =1− kh, b i = µh 2 − 2, c i =1+kh, i =2, 3,...,N− 1,and y 1 replaced by A, y N by B and y i by h 2 f(x i )fori =2, 3,...,N − 1. Thesolutions can be obtained as in (27.31) and (27.32).27.8 Partial differential equationsThe extension of previous methods to partial differential equations, thus involvingtwo or more independent variables, proceeds in a more or less obvious way. Rather1030

27.8 PARTIAL DIFFERENTIAL EQUATIONSthan an interval divided into equal steps by the points at which solutions to theequations are to be found, a mesh of points in two or more dimensions has to beset up and all the variables given an increased number of subscripts.Considerations of the stability, accuracy and feasibility of particular calculationalschemes are the same as for the one-dimensional case in principle, but inpractice are too complicated to be discussed here.Rather than note generalities that we are unable to pursue in any quantitativeway, we will conclude this chapter by indicating in outline how two familiarpartial differential equations of physical science can be set up for numericalsolution. The first of these is Laplace’s equation in two dimensions,∂ 2 φ∂x 2 + ∂2 φ=0, (27.86)∂y2 the value of φ being given on the perimeter of a closed domain.A grid with spacings ∆x and ∆y in the two directions is first chosen, so that,for example, x i stands for the point x 0 + i∆x and φ i,j for the value φ(x i ,y j ). Next,using a second central difference formula, (27.86) is turned intoφ i+1,j − 2φ i,j + φ i−1,j(∆x) 2 + φ i,j+1 − 2φ i,j + φ i,j−1(∆y) 2 =0,(27.87)for i =0, 1,...,N and j =0, 1,...,M.If(∆x) 2 = λ(∆y) 2 then this becomes therecurrence relationshipφ i+1,j + φ i−1,j + λ(φ i,j+1 + φ i,j−1 )=2(1+λ)φ i,j . (27.88)The boundary conditions in their simplest form (i.e. for a rectangular domain)mean thatφ 0,j , φ N,j , φ i,0 , φ i,M (27.89)have predetermined values. Non-rectangular boundaries can be accommodated,either by more complex boundary-value prescriptions or by using non-Cartesiancoordinates.To find a set of values satisfying (27.88), an initial guess of a complete set ofvalues for the φ i,j is made, subject to the requirement that the quantities listedin (27.89) have the given fixed values; those values that are not on the boundaryare then adjusted iteratively in order to try to bring about condition (27.88)everywhere. Clearly one scheme is to set λ = 1 and recalculate each φ i,j as themean of the four current values at neighbouring grid-points, using (27.88) directly,and then to iterate this recalculation until no value of φ changes significantlyafter a complete cycle through all values of i and j. This procedure is the simplestof such ‘relaxation’ methods; for a slightly more sophisticated scheme see exercise27.26 at the end of this chapter. The reader is referred to specialist books forfuller accounts of how this approach can be made faster and more accurate.1031

NUMERICAL METHODSOur final example is based upon the one-dimensional diffusion equation forthe temperature φ of a system:∂φ∂t = φκ∂2 ∂x 2 . (27.90)If φ i,j stands for φ(x 0 + i∆x, t 0 + j∆t) then a forward difference representationof the time derivative and a central difference representation for the spatialderivative lead to the following relationship:φ i,j+1 − φ i,j∆t= κ φ i+1,j − 2φ i,j + φ i−1,j(∆x) 2 . (27.91)This allows the construction of an explicit scheme for generating the temperaturedistribution at later times, given that it is known at some earlier time:φ i,j+1 = α(φ i+1,j + φ i−1,j )+(1− 2α)φ i,j , (27.92)where α = κ∆t/(∆x) 2 .Although this scheme is explicit, it is not a good one because of the asymmetricway in which the differences are formed. However, the effect of this can beminimised if we study and correct for the errors introduced in the following way.Taylor’s series for the time variable givesφ i,j+1 = φ i,j +∆t ∂φ i,j+ (∆t)2 ∂ 2 φ i,j∂t 2! ∂t 2 + ··· , (27.93)using the same notation as previously. Thus the first correction term to the LHSof (27.91) is− ∆t ∂ 2 φ i,j2 ∂t 2 . (27.94)The first term omitted on the RHS of the same equation is, by a similar argument,−κ 2(∆x)2 ∂ 4 φ i,j4! ∂x 4 . (27.95)But, using the fact that φ satisfies (27.90), we obtain∂ 2 φ∂t 2= ∂ ∂t( ) ( )κ ∂2 φ∂x 2 = κ ∂2 ∂φ∂x 2 = κ 2 ∂4 φ∂t ∂x 4 , (27.96)and so, to this accuracy, the two errors (27.94) and (27.95) can be made to cancelif α is chosen such that− κ2 ∆t2= − 2κ(∆x)2 , i.e. α = 1 4!6 .1032

27.9 EXERCISES27.9 Exercises27.1 Use an iteration procedure to find the root of the equation 40x =expx to foursignificant figures.27.2 Using the Newton–Raphson procedure find, correct to three decimal places, theroot nearest to 7 of the equation 4x 3 +2x 2 − 200x − 50 = 0.27.3 Show the following results about rearrangement schemes for polynomial equations.(a) That if a polynomial equation g(x) ≡ x m − f(x) = 0, where f(x) is apolynomial of degree less than m and for which f(0) ≠ 0, is solved usinga rearrangement iteration scheme x n+1 =[f(x n )] 1/m , then, in general, thescheme will have only first-order convergence.(b) By considering the cubic equationx 3 − ax 2 +2abx − (b 3 + ab 2 )=0for arbitrary non-zero values of a and b, demonstrate that, in special cases, thesame rearrangement scheme can give second- (or higher-) order convergence.27.4 The square root of a number N is to be determined by means of the iterationscheme[ ( ) ]x n+1 = x n 1 − N − x2n f(N) .Determine how to choose f(N) so that the process has second-order convergence.Given that √ 7 ≈ 2.65, calculate √ 7 as accurately as a single application of theformula will allow.27.5 Solve the following set of simultaneous equations using Gaussian elimination(including interchange where it is formally desirable):x 1 +3x 2 +4x 3 +2x 4 =0,2x 1 +10x 2 − 5x 3 + x 4 =6,4x 2 +3x 3 +3x 4 =20,−3x 1 +6x 2 +12x 3 − 4x 4 =16.27.6 The following table of values of a polynomial p(x) of low degree contains anerror. Identify and correct the erroneous value and extend the table up to x =1.2.x p(x) x p(x)0.0 0.000 0.5 0.1650.1 0.011 0.6 0.2160.2 0.040 0.7 0.2450.3 0.081 0.8 0.2560.4 0.128 0.9 0.24327.7 Simultaneous linear equations that result in tridiagonal matrices can sometimesbe treated as three-term recurrence relations, and their solution may be foundin a similar manner to that described in chapter 15. Consider the tridiagonalsimultaneous equationsx i−1 +4x i + x i+1 =3(δ i+1,0 − δ i−1,0 ), i =0, ±1, ±2,... .Prove that, for i>0, the equations have a general solution of the form x i =αp i + βq i ,wherep and q are the roots of a certain quadratic equation. Show thata similar result holds for i

NUMERICAL METHODS27.8 A possible rule for obtaining an approximation to an integral is the mid-pointrule, givenby∫ x0 +∆xf(x) dx =∆xf(x 0 + 1 2 ∆x)+O(∆x3 ).x 0Writing h for ∆x, and evaluating all derivates at the mid-point of the interval(x, x +∆x), use a Taylor series expansion to find, up to O(h 5 ), the coefficients ofthe higher-order errors in both the trapezium and mid-point rules. Hence find alinear combination of these two rules that gives O(h 5 ) accuracy for each step ∆x.27.9 Although it can easily be shown, by direct calculation, that∫ ∞e −x cos(kx) dx = 101+k , 2the form of the integrand is appropriate for Gauss–Laguerre numerical integration.Using a 5-point formula, investigate the range of values of k for which theformula gives accurate results. At about what value of k do the results becomeinaccurate at the 1% level?27.10 Using the points and weights given in table 27.9, answer the following questions.(a) A table of unnormalised Hermite polynomials H n (x) has been spattered withink blots and gives H 5 (x) as32x 5 −?x 3 + 120x and H 4 (x) as?x 4 −?x 2 + 12,where the coefficients marked ? cannot be read. What should they read?(b) What is the value of the integral∫ ∞e −2x2I =−∞ 4x 2 +3x +1 dx,as given by a 7-point integration routine?27.11 Consider the integrals I p defined byI p =∫ 1−1x 2p√1 − x2 dx.(a) By setting x =sinθ and using the results given in exercise 2.42, show that I phas the valueI p =2 2p − 1 2p − 32p 2p − 2 ··· 1 π2 2 .(b) Evaluate I p for p =1, 2,... ,6 using 5- and 6-point Gauss–Chebyshev integration(convenientlyrunonaspreadsheetsuchasExcel)and compare theresults with those in (a). In particular, show that, as expected, the 5-pointscheme first fails to be accurate when the order of the polynomial numerator(2p) exceeds (2×5)−1 = 9. Likewise, verify that the 6-point scheme evaluatesI 5 accurately but is in error for I 6 .27.12 In normal use only a single application of n-point Gaussian quadrature is made,using a value of n that is estimated from experience to be ‘safe’. However, it isinstructive to examine what happens when n is changed in a controlled way.(a) Evaluate the integral∫ 5 √I n = 7x − x2 − 10 dx2using n-point Gauss–Legendre formulae for n =2, 3,... ,6. Estimate (to 4s.f.) the value I ∞ you would obtain for very large n and compare it with theresult I obtained by exact integration. Explain why the variation of I n withn is monotonically decreasing.1034

27.9 EXERCISES(b) Try to repeat the processes described in (a) for the integrals∫ 51J n = √7x − x2 − 10 dx.2WhyisitverydifficulttoestimateJ ∞ ?27.13 Given a random number η uniformly distributed on (0, 1), determine the functionξ = ξ(η) that would generate a random number ξ distributed as(a) 2ξ√ on 0 ≤ ξ

NUMERICAL METHODS(b) Substitute them into the predictor equation and, by making that expressionfor ȳ n+1 coincide with the true Taylor series for y n+1 up to order h 3 , establishsimultaneous equations that determine the values of a 1 ,a 2 and a 3 .(c) Find the Taylor series for f n+1 and substitute it and that for f n−1 into thecorrector equation. Make the corrected prediction for y n+1 coincide with thetrue Taylor series by choosing the weights b 1 ,b 2 and b 3 appropriately.(d) The values of the numerical solution of the differential equationdy 2(1 + x)y + x3/2=dx 2x(1 + x)at three values of x are given in the following table:x 0.1 0.2 0.3y(x) 0.030 628 0.084 107 0.150 328Use the above predictor–corrector scheme to find the value of y(0.4) andcompare your answer with the accurate value, 0.225 577.27.18 If dy/dx = f(x, y) then show thatd 2 fdx = ∂2 f2 ∂x +2f ∂2 f2 ∂x∂y + ∂2 ff2∂y + ∂f ( ) 2∂f ∂f2 ∂x ∂y + f .∂yHence verify, by substitution and the subsequent expansion of arguments inTaylor series of their own, that the scheme given in (27.79) coincides with theTaylor expansion (27.68), i.e.y i+1 = y i + hy (1)i+ h2iup to terms in h 3 .27.19 To solve the ordinary differential equation2! y(2)+ h33! y(3) i+ ··· ,du= f(u, t)dtfor f = f(t), the explicit two-step finite difference schemeu n+1 = αu n + βu n−1 + h(µf n + νf n−1 )may be used. Here, in the usual notation, h is the time step, t n = nh, u n = u(t n )and f n = f(u n ,t n ); α, β, µ, andν are constants.(a) A particular scheme has α =1,β =0,µ=3/2 andν = −1/2. By consideringTaylor expansions about t = t n for both u n+j and f n+j , show that this schemegives errors of order h 3 .(b) Find the values of α, β, µ and ν that will give the greatest accuracy.27.20 Set up a finite difference scheme to solve the ordinary differential equationx d2 φdx + dφ2 dx =0in the range 1 ≤ x ≤ 4, subject to the boundary conditions φ(1) = 2 anddφ/dx =2atx =4.UsingN equal increments, ∆x, inx, obtain the generaldifference equation and state how the boundary conditions are incorporatedinto the scheme. Setting ∆x equal to the (crude) value 1, obtain the relevantsimultaneous equations and so obtain rough estimates for φ(2),φ(3) and φ(4).Finally, solve the original equation analytically and compare your numericalestimates with the accurate values.1036

27.9 EXERCISES27.21 Write a computer program that would solve, for a range of values of λ, thedifferential equationdydx = 1√ , y(0) = 1,x2 + λy2 using a third-order Runge–Kutta scheme. Consider the difficulties that mightarise when λ

NUMERICAL METHODS27.25 Laplace’s equation,∂ 2 V∂x + ∂2 V2 ∂y =0, 2is to be solved for the region and boundary conditions shown in figure 27.7.V =80−∞40404040404040∞202020V =0Figure 27.7Region, boundary values and initial guessed solution values.Starting from the given initial guess for the potential values V , and using thesimplest possible form of relaxation, obtain a better approximation to the actualsolution. Do not aim to be more accurate than ± 0.5 units, and so terminate theprocess when subsequent changes would be no greater than this.27.26 Consider the solution, φ(x, y), of Laplace’s equation in two dimensions using arelaxation method on a square grid with common spacing h. Asinthemaintext,denote φ(x 0 + ih, y 0 + jh) byφ i,j . Further, define φ m,ni,jbyφ m,ni,j≡ ∂m+n φ∂x m ∂y nevaluated at (x 0 + ih, y 0 + jh).(a) Show thatφ 4,0i,j+2φ 2,2i,j+ φ 0,4i,j=0.(b) Working up to terms of order h 5 , find Taylor series expansions, expressed interms of the φ m,ni,j ,for S ±,0 = φ i+1,j + φ i−1,j ,S 0,± = φ i,j+1 + φ i,j−1 .(c) Find a corresponding expansion, to the same order of accuracy, for φ i±1,j+1 +φ i±1,j−1 and hence show thatS ±,± = φ i+1,j+1 + φ i+1,j−1 + φ i−1,j+1 + φ i−1,j−1has the form)+h44φ 0,0i,j+2h 2 (φ 2,0i,j+ φ 0,2i,j6 (φ4,0 i,j+6φ 2,2i,j+ φ 0,4i,j ).(d) Evaluate the expression 4(S ±,0 +S 0,± )+S ±,± and hence deduce that a possiblerelaxation scheme, good to the fifth order in h, is to recalculate each φ i,j asthe weighted mean of the current values of its four nearest neighbours (eachwith weight 1 1) and its four next-nearest neighbours (each with weight ).5 201038

27.10 HINTS AND ANSWERS27.27 The Schrödinger equation for a quantum mechanical particle of mass m movingin a one-dimensional harmonic oscillator potential V (x) =kx 2 /2is− 2 d 2 ψ2m dx + kx2 ψ= Eψ.2 2For physically acceptable solutions, the wavefunction ψ(x) must be finite at x =0,tend to zero as x →±∞andbenormalised,sothat ∫ |ψ| 2 dx =1.Inpractice,these constraints mean that only certain (quantised) values of E, the energy ofthe particle, are allowed. The allowed values fall into two groups: those for whichψ(0) = 0 and those for which ψ(0) ≠0.Show that if the unit of length is taken as [ 2 /(mk)] 1/4 and the unit of energyis taken as (k/m) 1/2 , then the Schrödinger equation takes the formd 2 ψdy 2 +(2E′ − y 2 )ψ =0.Devise an outline computerised scheme, using Runge–Kutta integration, that willenable you to:(a) determine the three lowest allowed values of E;(b) tabulate the normalised wavefunction corresponding to the lowest allowedenergy.You should consider explicitly:(i) the variables to use in the numerical integration;(ii) how starting values near y = 0 are to be chosen;(iii) how the condition on ψ as y →±∞is to be implemented;(iv) how the required values of E are to be extracted from the results of theintegration;(v) how the normalisation is to be carried out.27.10 Hints and answers27.1 5.370.27.3 (a) ξ ≠0andf ′ (ξ) ≠ 0 in general; (b) ξ = b, but f ′ (b) = 0 whilst f(b) ≠0.27.5 Interchange is formally needed for the first two steps, though in this case no errorwill result if it is not carried out; x 1 = −12, x 2 =2,x 3 = −1, x 4 =5.27.7 The quadratic equation is z 2 +4z +1=0;α + β − 3=x 0 = α ′ + β ′ +3.With p = −2+ √ 3andq = −2 − √ 3, β must be zero for i>0andα ′ must bezero for i0, x i =0fori =0,x i = −3(−2 − √ 3) ifor i0; ξ =ln2η for 0

NUMERICAL METHODSV =80−∞40.541.8 46.7 48.4 46.741.840.5∞16.8 20.4 16.8V =0Figure 27.8 The solution to exercise 27.25.27.15 1 − x 2 /2+x 4 /8 − x 6 /48; 1.0000, 0.9950, 0.9802, 0.9560, 0.9231, 0.8825; exactsolution y =exp(−x 2 /2).27.17 (b) a 1 =23/12, a 2 = −4/3, a 3 =5/12.(c) b 1 =5/12, b 2 =2/3, b 3 = −1/12.(d) ȳ(0.4) = 0.224 582, y(0.4) = 0.225 527 after correction.27.19 (a) The error is 5h 3 u (3)n /12 + O(h 4 ).(b) α = −4, β =5,µ=4andν =227.21 For λ positive the solutions are (boringly) monotonic functions of x. Withy(0)given, there are no real solutions at all for any negative λ!27.23 (a) Setting A∆t = c∆x gives, for example, u(0, 2) = (1 − c) 2 , u(1, 2) = 2c(1 − c),u(2, 2) = c 2 . For stability, 0

28Group theoryFor systems that have some degree of symmetry, full exploitation of that symmetryis desirable. Significant physical results can sometimes be deduced simply by astudy of the symmetry properties of the system under investigation. Consequentlyit becomes important, for such a system, to identify all those operations (rotations,reflections, inversions) that carry the system into a physically indistinguishablecopy of itself.The study of the properties of the complete set of such operations formsone application of group theory. Though this is the aspect of most interest tothe physical scientist, group theory itself is a much larger subject and of greatimportance in its own right. Consequently we leave until the next chapter anydirect applications of group theoretical results and concentrate on building upthe general mathematical properties of groups.28.1 GroupsAs an example of symmetry properties, let us consider the sets of operations,such as rotations, reflections, and inversions, that transform physical objects, forexample molecules, into physically indistinguishable copies of themselves, so thatonly the labelling of identical components of the system (the atoms) changes inthe process. For differently shaped molecules there are different sets of operations,but in each case it is a well-defined set, and with a little practice all members ofeach set can be identified.As simple examples, consider (a) the hydrogen molecule, and (b) the ammoniamolecule illustrated in figure 28.1. The hydrogen molecule consists of two atomsH of hydrogen and is carried into itself by any of the following operations:(i) any rotation about its long axis;(ii) rotation through π about an axis perpendicular to the long axis andpassing through the point M that lies midway between the atoms;1041

GROUP THEORYNHMHHH(a)(b)HFigure 28.1(a) The hydrogen molecule, and (b) the ammonia molecule.(iii) inversion through the point M;(iv) reflection in the plane that passes through M and has its normal parallelto the long axis.These operations collectively form the set of symmetry operations for the hydrogenmolecule.The somewhat more complex ammonia molecule consists of a tetrahedron withan equilateral triangular base at the three corners of which lie hydrogen atomsH, whilst a nitrogen atom N is sited at the fourth vertex of the tetrahedron. Theset of symmetry operations on this molecule is limited to rotations of π/3 and2π/3 about the axis joining the centroid of the equilateral triangle to the nitrogenatom, and reflections in the three planes containing that axis and each of thehydrogen atoms in turn. However, if the nitrogen atom could be replaced by afourth hydrogen atom, and all interatomic distances equalised in the process, thenumber of symmetry operations would be greatly increased.Once all the possible operations in any particular set have been identified, itmust follow that the result of applying two such operations in succession will beidentical to that obtained by the sole application of some third (usually different)operation in the set – for if it were not, a new member of the set would havebeen found, contradicting the assumption that all members have been identified.Such observations introduce two of the main considerations relevant to decidingwhether a set of objects, here the rotation, reflection and inversion operations,qualifies as a group in the mathematically tightly defined sense. These two considerationsare (i) whether there is some law for combining two members of the set,and (ii) whether the result of the combination is also a member of the set. Theobvious rule of combination has to be that the second operation is carried outon the system that results from application of the first operation, and we havealready seen that the second requirement is satisfied by the inclusion of all suchoperations in the set. However, for a set to qualify as a group, more than thesetwo conditions have to be satisfied, as will now be made clear.1042

28.1 GROUPS28.1.1 Definition of a groupAgroupG is a set of elements {X,Y,...}, together with a rule for combiningthem that associates with each ordered pair X, Y a ‘product’ or combination lawX • Y for which the following conditions must be satisfied.(i) For every pair of elements X, Y that belongs to G, the product X • Y alsobelongs to G. (This is known as the closure property of the group.)(ii) For all triples X, Y , Z the associative law holds; in symbols,X • (Y • Z) =(X • Y ) • Z. (28.1)(iii) There exists a unique element I, belonging to G, with the property thatI • X = X = X • I (28.2)for all X belonging to G. This element I is known as the identity elementof the group.(iv) For every element X of G, there exists an element X −1 , also belonging toG, such thatX −1 is called the inverse of X.X −1 • X = I = X • X −1 . (28.3)An alternative notation in common use is to write the elements of a group Gas the set {G 1 ,G 2 ,...} or, more briefly, as {G i }, a typical element being denotedby G i .It should be noticed that, as given, the nature of the operation • is not stated. Itshould also be noticed that the more general term element, rather than operation,has been used in this definition. We will see that the general definition of agroup allows as elements not only sets of operations on an object but also sets ofnumbers, of functions and of other objects, provided that the interpretation of •is appropriately defined.In one of the simplest examples of a group, namely the group of all integersunder addition, the operation • is taken to be ordinary addition. In this group therole of the identity I is played by the integer 0, and the inverse of an integer X is−X. That requirements (i) and (ii) are satisfied by the integers under addition istrivially obvious. A second simple group, under ordinary multiplication, is formedby the two numbers 1 and −1; in this group, closure is obvious, 1 is the identityelement, and each element is its own inverse.It will be apparent from these two examples that the number of elements in agroup can be either finite or infinite. In the former case the group is called a finitegroup and the number of elements it contains is called the order of the group,which we will denote by g; an alternative notation is |G| but has obvious dangers1043

GROUP THEORYif matrices are involved. In the notation in which G = {G 1 ,G 2 ,...,G n } the orderof the group is clearly n.As we have noted, for the integers under addition zero is the identity. Forthe group of rotations and reflections, the operation of doing nothing, i.e. thenull operation, plays this role. This latter identification may seem artificial, butit is an operation, albeit trivial, which does leave the system in a physicallyindistinguishable state, and needs to be included. One might add that without itthe set of operations would not form a group and none of the powerful resultswe will derive later in this and the next chapter could be justifiably applied togive deductions of physical significance.In the examples of rotations and reflections mentioned earlier, • has been takento mean that the left-hand operation is carried out on the system that resultsfrom application of the right-hand operation. ThusZ = X • Y (28.4)means that the effect on the system of carrying out Z isthesameaswouldbe obtained by first carrying out Y and then carrying out X. The order of theoperations should be noted; it is arbitrary in the first instance but, once chosen,must be adhered to. The choice we have made is dictated by the fact that mostof our applications involve the effect of rotations and reflections on functions ofspace coordinates, and it is usual, and our practice in the rest of this book, towrite operators acting on functions to the left of the functions.It will be apparent that for the above-mentioned group, integers under ordinaryaddition, it is true thatY • X = X • Y (28.5)for all pairs of integers X, Y . If any two particular elements of a group satisfy(28.5), they are said to commute under the operation •; if all pairs of elements ina group satisfy (28.5), then the group is said to be Abelian. Thesetofallintegersforms an infinite Abelian group under (ordinary) addition.As we show below, requirements (iii) and (iv) of the definition of a groupare over-demanding (but self-consistent), since in each of equations (28.2) and(28.3) the second equality can be deduced from the first by using the associativityrequired by (28.1). The mathematical steps in the following arguments are allvery simple, but care has to be taken to make sure that nothing that has notyet been proved is used to justify a step. For this reason, and to act as a modelin logical deduction, a reference in Roman numerals to the previous result,or to the group definition used, is given over each equality sign. Such explicitdetailed referencing soon becomes tiresome, but it should always be available ifneeded.1044

28.1 GROUPS◮Using only the first equalities in (28.2) and (28.3), deduce the second ones.Consider the expression X −1 • (X • X −1 );X −1 • (X • X −1 ) (ii) =(X −1 −1 (iv)• X) • X = I • X −1(iii)= X −1 . (28.6)But X −1 belongs to G, and so from (iv) there is an element U in G such thatU • X −1 = I.(v)Form the product of U with the first and last expressions in (28.6) to giveU • (X −1 • (X • X −1 −1 (v))) = U • X = I. (28.7)Transforming the left-hand side of this equation givesU • (X −1 • (X • X −1 )) (ii) =(U • X −1 ) • (X • X −1 )(v)= I • (X • X −1 )(iii)= X • X −1 . (28.8)Comparing (28.7), (28.8) shows thatX • X −1 = I, (iv) ′i.e. the second equality in group definition (iv). SimilarlyX • I (iv) = X • (X −1 • X) (ii) =(X • X −1 ) • X(iv) ′= I • X(iii)= X. (iii ′ )i.e. the second equality in group definition (iii). ◭The uniqueness of the identity element I canalsobedemonstratedratherthanassumed. Suppose that I ′ , belonging to G, also has the propertyTake X as I, thenFurther, from (iii ′ ),I ′ • X = X = X • I ′ for all X belonging to G.I ′ • I = I. (28.9)X = X • I for all X belonging to G,1045

GROUP THEORYand setting X = I ′ givesI ′ = I ′ • I. (28.10)It then follows from (28.9), (28.10) that I = I ′ , showing that in any particulargroup the identity element is unique.In a similar way it can be shown that the inverse of any particular elementis unique. If U and V are two postulated inverses of an element X of G, byconsidering the productU • (X • V )=(U • X) • V,it can be shown that U = V . The proof is left to the reader.Given the uniqueness of the inverse of any particular group element, it followsthat(U • V • ···• Y • Z) • (Z −1 • Y −1 • ···• V −1 • U −1 )=(U • V • ···• Y ) • (Z • Z −1 ) • (Y −1 • ···• V −1 • U −1 )=(U • V • ···• Y ) • (Y −1 • ···• V −1 • U −1 )..= I,where use has been made of the associativity and of the two equations Z • Z −1 = Iand I • X = X. Thus the inverse of a product is the product of the inverses inreverse order, i.e.(U • V • ···• Y • Z) −1 =(Z −1 • Y −1 • ···• V −1 • U −1 ). (28.11)Further elementary results that can be obtained by arguments similar to thoseabove are as follows.(i) Given any pair of elements X, Y belonging to G, there exist uniqueelements U, V, also belonging to G, such thatX • U = Y and V • X = Y.Clearly U = X −1 • Y ,andV = Y • X −1 , and they can be shown to beunique. This result is sometimes called the division axiom.(ii) The cancellation law can be stated as follows. IfX • Y = X • Zfor some X belonging to G, thenY = Z. Similarly,implies the same conclusion.Y • X = Z • X1046

28.1 GROUPSLMKFigure 28.2 Reflections in the three perpendicular bisectors of the sides ofan equilateral triangle take the triangle into itself.(iii) Forming the product of each element of G with a fixed element X of Gsimply permutes the elements of G; this is often written symbolically asG • X = G. If this were not so, and X • Y and X • Z were not differenteven though Y and Z were, application of the cancellation law would leadto a contradiction. This result is called the permutation law.In any finite group of order g, any element X when combined with itself toform successively X 2 = X • X, X 3 = X • X 2 , ... will, after at most g − 1suchcombinations, produce the group identity I. OfcourseX 2 , X 3 , ... are some ofthe original elements of the group, and not new ones. If the actual number ofcombinations needed is m−1, i.e. X m = I, thenm is called the order of the elementX in G. The order of the identity of a group is always 1, and that of any otherelement of a group that is its own inverse is always 2.◮Determine the order of the group of (two-dimensional) rotations and reflections that takea plane equilateral triangle into itself and the order of each of the elements. The group isusually known as 3m (to physicists and crystallographers) or C 3v (to chemists).There are two (clockwise) rotations, by 2π/3 and4π/3, about an axis perpendicular tothe plane of the triangle. In addition, reflections in the perpendicular bisectors of the threesides (see figure 28.2) have the defining property. To these must be added the identityoperation. Thus in total there are six distinct operations and so g = 6 for this group.To reproduce the identity operation either of the rotations has to be applied three times,whilst any of the reflections has to be applied just twice in order to recover the originalsituation. Thus each rotation element of the group has order 3, and each reflection elementhas order 2. ◭A so-called cyclic group is one for which all members of the group can begenerated from just one element X (say). Thus a cyclic group of order g can bewritten asG = { I,X,X 2 ,X 3 ,...,X g−1} .1047

GROUP THEORYIt is clear that cyclic groups are always Abelian and that each element, apartfrom the identity, has order g, the order of the group itself.28.1.2 Further examples of groupsIn this section we consider some sets of objects, each set together with a law ofcombination, and investigate whether they qualify as groups and, if not, why not.We have already seen that the integers form a group under ordinary addition,but it is immediately apparent that (even if zero is excluded) they do not doso under ordinary multiplication. Unity must be the identity of the set, but therequisite inverse of any integer n, namely1/n, does not belong to the set ofintegers for any n other than unity.Other infinite sets of quantities that do form groups are the sets of all realnumbers, or of all complex numbers, under addition, and of the same two setsexcluding 0 under multiplication. All these groups are Abelian.Although subtraction and division are normally considered the obvious counterpartsof the operations of (ordinary) addition and multiplication, they are notacceptable operations for use within groups since the associative law, (28.1), doesnot hold. Explicitly,X − (Y − Z) ≠(X − Y ) − Z,X ÷ (Y ÷ Z) ≠(X ÷ Y ) ÷ Z.From within the field of all non-zero complex numbers we can select just thosethat have unit modulus, i.e. are of the form e iθ where 0 ≤ θ

28.2 FINITE GROUPS28.2 Finite groupsWhilst many properties of physical systems (e.g. angular momentum) are relatedto the properties of infinite, and, in particular, continuous groups, the symmetryproperties of crystals and molecules are more intimately connected with those offinite groups. We therefore concentrate in this section on finite sets of objects thatcan be combined in a way satisfying the group postulates.Although it is clear that the set of all integers does not form a group underordinary multiplication, restricted sets can do so if the operation involved is multiplication(mod N) for suitable values of N; this operation will be explained below.As a simple example of a group with only four members, consider the set Sdefined as follows:S = {1, 3, 5, 7} under multiplication (mod 8).To find the product (mod 8) of any two elements, we multiply them together inthe ordinary way, and then divide the answer by 8, treating the remainder afterdoing so as the product of the two elements. For example, 5 × 7 = 35, which ondividing by 8 gives a remainder of 3. Clearly, since Y × Z = Z × Y , the full setof different products is1 × 1=1, 1 × 3=3, 1 × 5=5, 1 × 7=7,3 × 3=1, 3 × 5=7, 3 × 7=5,5 × 5=1, 5 × 7=3,7 × 7=1.The first thing to notice is that each multiplication produces a member of theoriginal set, i.e. the set is closed. Obviously the element 1 takes the role of theidentity, i.e. 1 × Y = Y for all members Y of the set. Further, for each element Yof the set there is an element Z (equal to Y , as it happens, in this case) such thatY × Z = 1, i.e. each element has an inverse. These observations, together with theassociativity of multiplication (mod 8), show that the set S is an Abelian groupof order 4.It is convenient to present the results of combining any two elements of agroup in the form of multiplication tables – akin to those which used to appear inelementary arithmetic books before electronic calculators were invented! Writtenin this much more compact form the above example is expressed by table 28.1.Although the order of the two elements being combined does not matter herebecause the group is Abelian, we adopt the convention that if the product in ageneral multiplication table is written X • Y then X is taken from the left-handcolumn and Y is taken from the top row. Thus the bold ‘7’ in the table is theresult of 3 × 5, rather than of 5 × 3.Whilst it would make no difference to the basic information content in a tableto present the rows and columns with their headings in random orders, it is1049

GROUP THEORY1 3 5 71 1 3 5 73 3 1 7 55 5 7 1 37 7 5 3 1Table 28.1 The table of products for the elements of the group S = {1, 3, 5, 7}under multiplication (mod 8).usual to list the elements in the same order in both the vertical and horizontalheadings in any one table. The actual order of the elements in the common list,whilst arbitrary, is normally chosen to make the table have as much symmetry aspossible. This is initially a matter of convenience, but, as we shall see later, someof the more subtle properties of groups are revealed by putting next to each otherelements of the group that are alike in certain ways.Some simple general properties of group multiplication tables can be deducedimmediately from the fact that each row or column constitutes the elements ofthe group.(i) Each element appears once and only once in each row or column of thetable; this must be so since G • X = G (the permutation law) holds.(ii) The inverse of any element Y can be found by looking along the rowin which Y appears in the left-hand column (the Y th row), and notingthe element Z at the head of the column (the Zth column) in whichthe identity appears as the table entry. An immediate corollary is thatwhenever the identity appears on the leading diagonal, it indicates thatthe corresponding header element is of order 2 (unless it happens to bethe identity itself).(iii) For any Abelian group the multiplication table is symmetric about theleading diagonal.To get used to the ideas involved in using group multiplication tables, we nowconsider two more sets of integers under multiplication (mod N):S ′ = {1, 5, 7, 11} under multiplication (mod 24), andS ′′ = {1, 2, 3, 4} under multiplication (mod 5).These have group multiplication tables 28.2(a) and (b) respectively, as the readershould verify.If tables 28.1 and 28.2(a) for the groups S and S ′ are compared, it will be seenthat they have essentially the same structure, i.e if the elements are written as{I,A,B,C} in both cases, then the two tables are each equivalent to table 28.3.For S, I =1,A =3,B =5,C = 7 and the law of combination is multiplication(mod 8), whilst for S ′ , I =1,A =5,B =7,C = 11 and the law of combination1050

28.2 FINITE GROUPS(a)1 5 7 111 1 5 7 115 5 1 11 77 7 11 1 511 11 7 5 1(b)1 2 3 41 1 2 3 42 2 4 1 33 3 1 4 24 4 3 2 1Table 28.2 (a) The multiplication table for the group S ′ = {1, 5, 7, 11} undermultiplication (mod 24). (b) The multiplication table for the group S ′′ ={1, 2, 3, 4} under multiplication (mod 5).I A B CI I A B CA A I C BB B C I AC C B A ITable 28.3The common structure exemplified by tables 28.1 and 28.2(a).1 i −1 −i1 1 i −1 −ii i −1 −i 1−1 −1 −i 1 i−i −i 1 i −1Table 28.4 The group table for the set {1,i,−1, −i} under ordinary multiplicationof complex numbers.is multiplication (mod 24). However, the really important point is that the twogroups S and S ′ have equivalent group multiplication tables – they are said tobe isomorphic, a matter to which we will return more formally in section 28.5.◮Determine the behaviour of the set of four elements{1,i,−1, −i}under the ordinary multiplication of complex numbers. Show that they form a group anddetermine whether the group is isomorphic to either of the groups S (itself isomorphic toS ′ ) and S ′′ defined above.That the elements form a group under the associative operation of complex multiplicationis immediate; there is an identity (1), each possible product generates a member of the setand each element has an inverse (1, −i, −1, i, respectively). The group table has the formshown in table 28.4.We now ask whether this table can be made to look like table 28.3, which is thestandardised form of the tables for S and S ′ . Since the identity element of the group(1) will have to be represented by I, and ‘1’ only appears on the leading diagonal twicewhereas I appears on the leading diagonal four times in table 28.3, it is clear that no1051

GROUP THEORY1 i −1 −i1 1 i −1 −ii i −1 −i 1−1 −1 −i 1 i−i −i 1 i −11 2 4 31 1 2 4 32 2 4 3 14 4 3 1 23 3 1 2 4Table 28.5 A comparison between tables 28.4 and 28.2(b), the latter with itscolumns reordered.I A B CI I A B CA A B C IB B C I AC C I A BTable 28.6 The common structure exemplified by tables 28.4 and 28.2(b), thelatter with its columns reordered.amount of relabelling (or, equivalently, no allocation of the symbols A, B, C, amongsti, −1, −i) can bring table 28.4 into the form of table 28.3. We conclude that the group{1,i,−1, −i} is not isomorphic to S or S ′ . An alternative way of stating the observation isto say that the group contains only one element of order 2 whilst a group correspondingto table 28.3 contains three such elements.However, if the rows and columns of table 28.2(b) – in which the identity does appeartwice on the diagonal and which therefore has the potential to be equivalent to table 28.4 –are rearranged by making the heading order 1, 2, 4, 3 then the two tables can be comparedin the forms shown in table 28.5. They can thus be seen to have the same structure, namelythat shown in table 28.6.We therefore conclude that the group of four elements {1,i,−1, −i} under ordinary multiplicationof complex numbers is isomorphic to the group {1, 2, 3, 4} under multiplication(mod 5). ◭What we have done does not prove it, but the two tables 28.3 and 28.6 are infact the only possible tables for a group of order 4, i.e. a group containing exactlyfour elements.28.3 Non-Abelian groupsSo far, all the groups for which we have constructed multiplication tables havebeen based on some form of arithmetic multiplication, a commutative operation,with the result that the groups have been Abelian and the tables symmetricabout the leading diagonal. We now turn to examples of groups in which somenon-commutation occurs. It should be noted, in passing, that non-commutationcannot occur throughout a group, as the identity always commutes with anyelement in its group.1052

28.3 NON-ABELIAN GROUPSAs a first example we consider again as elements of a group the two-dimensionaloperations which transform an equilateral triangle into itself (see the end ofsubsection 28.1.1). It has already been shown that there are six such operations:the null operation, two rotations (by 2π/3 and4π/3 about an axis perpendicularto the plane of the triangle) and three reflections in the perpendicular bisectorsof the three sides. To abbreviate we will denote these operations by symbols asfollows.(i) I is the null operation.(ii) R and R ′ are (clockwise) rotations by 2π/3 and4π/3 respectively.(iii) K, L, M are reflections in the three lines indicated in figure 28.2.Some products of the operations of the form X • Y (where it will be recalledthat the symbol • means that the second operation X is carried out on the systemresulting from the application of the first operation Y ) are easily calculated:R • R = R ′ , R ′ • R ′ = R, R • R ′ = I = R ′ • RK • K = L • L = M • M = I.(28.12)Others, such as K • M, are more difficult, but can be found by a little thought,or by making a model triangle or drawing a sequence of diagrams such as thosefollowing.xxK • M = K == R ′xxshowing that K • M = R ′ . In the same way,M • K = M == Rx x xxshows that M • K = R, andR • L = R == Kx x x xshows that R • L = K.Proceeding in this way we can build up the complete multiplication table(table 28.7). In fact, it is not necessary to draw any more diagrams, as allremaining products can be deduced algebraically from the three found above and1053

GROUP THEORYI R R ′ K L MI I R R ′ K L MR R R ′ I M K LR ′ R ′ I R L M KK K L M I R R ′L L M K R ′ I RM M K L R R ′ ITable 28.7 The group table for the two-dimensional symmetry operations onan equilateral triangle.the more self-evident results given in (28.12). A number of things may be noticedabout this table.(i) It is not symmetric about the leading diagonal, indicating that some pairsof elements in the group do not commute.(ii) There is some symmetry within the 3×3 blocks that form the four quartersof the table. This occurs because we have elected to put similar operationsclose to each other when choosing the order of table headings – the tworotations (or three if I is viewed as a rotation by 0π/3) are next to eachother, and the three reflections also occupy adjacent columns and rows.We will return to this later.That two groups of the same order may be isomorphic carries over to non-Abelian groups. The next two examples are each concerned with sets of sixobjects; they will be shown to form groups that, although very different in naturefrom the rotation–reflection group just considered, are isomorphic to it.We consider first the set M of six orthogonal 2 × 2 matrices given by)) ( )−1I =( 1 00 1( ) −1 0C =0 1A =D =(−12√32− √ 32− 1 2( 12− √ 32− √ 32− 1 2)B =E =− √ 32√232− 1 2( √1 32 2√32− 1 2) (28.13)the combination law being that of ordinary matrix multiplication. Here we useitalic, rather than the sans serif used for matrices elsewhere, to emphasise thatthe matrices are group elements.Although it is tedious to do so, it can be checked that the product of anytwo of these matrices, in either order, is also in the set. However, the result isgenerally different in the two cases, as matrix multiplication is non-commutative.The matrix I clearly acts as the identity element of the set, and during the checkingfor closure it is found that the inverse of each matrix is contained in the set, I,C, D and E being their own inverses. The group table is shown in table 28.8.1054

28.3 NON-ABELIAN GROUPSI A B C D EI I A B C D EA A B I E C DB B I A D E CC C D E I A BD D E C B I AE E C D A B ITable 28.8 The group table, under matrix multiplication, for the set M ofsix orthogonal 2 × 2 matrices given by (28.13).The similarity to table 28.7 is striking. If {R,R ′ ,K,L,M} of that table arereplaced by {A, B, C, D, E} respectively, the two tables are identical, without eventhe need to reshuffle the rows and columns. The two groups, one of reflectionsand rotations of an equilateral triangle, the other of matrices, are isomorphic.Our second example of a group isomorphic to the same rotation–reflectiongroup is provided by a set of functions of an undetermined variable x. Thefunctions are as follows:f 1 (x) =x, f 2 (x) =1/(1 − x), f 3 (x) =(x − 1)/x,f 4 (x) =1/x, f 5 (x) =1− x, f 6 (x) =x/(x − 1),and the law of combination isf i (x) • f j (x) =f i (f j (x)),i.e. the function on the right acts as the argument of the function on the left toproduce a new function of x. It should be emphasised that it is the functionsthat are the elements of the group. The variable x is the ‘system’ on which theyact, and plays much the same role as the triangle does in our first example of anon-Abelian group.To show an explicit example, we calculate the product f 6 • f 3 . The productwill be the function of x obtained by evaluating y/(y − 1), when y is set equal to(x − 1)/x. Explicitly(x − 1)/xf 6 (f 3 )=(x − 1)/x − 1 =1− x = f 5(x).Thus f 6 • f 3 = f 5 . Further examples aref 2 • f 2 =11 − 1/(1 − x) = x − 1 = f 3 ,xandf 6 • f 6 =x/(x − 1)x/(x − 1) − 1 = x = f 1. (28.14)1055

GROUP THEORYThe multiplication table for this set of six functions has all the necessary propertiesto show that they form a group. Further, if the symbols f 1 ,f 2 ,f 3 ,f 4 ,f 5 ,f 6 arereplaced by I,A,B,C,D,E respectively the table becomes identical to table 28.8.This justifies our earlier claim that this group of functions, with argument substitutionas the law of combination, is isomorphic to the group of reflections androtations of an equilateral triangle.28.4 Permutation groupsThe operation of rearranging n distinct objects amongst themselves is called apermutation of degree n, and since many symmetry operations on physical systemscan be viewed in that light, the properties of permutations are of interest. Forexample, the symmetry operations on an equilateral triangle, to which we havealready given much attention, can be considered as the six possible rearrangementsof the marked corners of the triangle amongst three fixed points in space, muchas in the diagrams used to compute table 28.7. In the same way, the symmetryoperations on a cube can be viewed as a rearrangement of its corners amongsteight points in space, albeit with many constraints, or, with fewer complications,as a rearrangement of its body diagonals in space. The details will be left untilwe review the possible finite groups more systematically.The notations and conventions used in the literature to describe permutationsare very varied and can easily lead to confusion. We will try to avoid this by usingletters a,b,c,... (rather than numbers) for the objects that are rearranged by apermutation and by adopting, before long, a ‘cycle notation’ for the permutationsthemselves. It is worth emphasising that it is the permutations, i.e.theactsofrearranging, and not the objects themselves (represented by letters) that formthe elements of permutation groups. The complete group of all permutations ofdegree n is usually denoted by S n or Σ n . The number of possible permutations ofdegree n is n!, and so this is the order of S n .Suppose the ordered set of six distinct objects {a bcdef} is rearranged bysome process into {b efadc}; then we can represent this mathematically asθ{a bcdef} = {b efadc},where θ is a permutation of degree 6. The permutation θ can be denoted by[256143],sincethefirstobject,a, is replaced by the second, b, the secondobject, b, is replaced by the fifth, e, the third by the sixth, f, etc.Theequationcan then be written more explicitly asθ{a bcdef} =[256143]{a bcdef} = {b efadc}.If φ is a second permutation, also of degree 6, then the obvious interpretation ofthe product φ • θ of the two permutations isφ • θ{a bcdef} = φ(θ{a bcdef}).1056

28.4 PERMUTATION GROUPSSuppose that φ is the permutation [4 5 3 6 2 1]; thenφ • θ{a bcdef} =[453621][256143]{a bcdef}=[453621]{b efadc}= {a dfceb}=[146352]{a bcdef}.Written in terms of the permutation notation this result is[453621][256143]=[146352].A concept that is very useful for working with permutations is that of decompositioninto cycles. The cycle notation is most easily explained by example. Forthe permutation θ given above:the 1st object, a, has been replaced by the 2nd, b;the 2nd object, b, has been replaced by the 5th, e;the 5th object, e, has been replaced by the 4th, d;the 4th object, d, has been replaced by the 1st, a.This brings us back to the beginning of a closed cycle, which is convenientlyrepresented by the notation (1 2 5 4), in which the successive replacementpositionsareenclosed,insequence,inparentheses.Thus(1254)means2nd→ 1st, 5th → 2nd, 4th → 5th, 1st → 4th. It should be noted that the objectinitially in the first listed position replaces that in the final position indicated inthe bracket – here ‘a’ is put into the fourth position by the permutation. Clearlythe cycle (5 4 1 2), or any other that involved the same numbers in the samerelative order, would have exactly the same meaning and effect. The remainingtwo objects, c and f, are interchanged by θ or, more formally, are rearrangedaccording to a cycle of length 2, a transposition, represented by (3 6). Thus thecomplete representation (specification) of θ isθ =(1254)(36).The positions of objects that are unaltered by a permutation are either placed bythemselves in a pair of parentheses or omitted altogether. The former is recommendedas it helps to indicate how many objects are involved – important whenthe object in the last position is unchanged, or the permutation is the identity,which leaves all objects unaltered in position! Thus the identity permutation ofdegree 6 isI = (1)(2)(3)(4)(5)(6),though in practice it is often shortened to (1).It will be clear that the cycle representation is unique, to within the internalabsolute ordering of the numbers in each bracket as already noted, and that1057

GROUP THEORYeach number appears once and only once in the representation of any particularpermutation.The order of any permutation of degree n within the group S n canbereadofffrom the cyclic representation and is given by the lowest common multiple (LCM)of the lengths of the cycles. Thus I has order 1, as it must, and the permutationθ discussed above has order 4 (the LCM of 4 and 2).Expressed in cycle notation our second permutation φ is (3)(1 4 6)(2 5), andthe product φ • θ is calculated as(3)(1 4 6)(2 5) • (1 2 5 4)(3 6){a bcdef} = (3)(1 4 6)(2 5){b efadc}= {a dfceb}= (1)(5)(2 4 3 6){a bcdef}.i.e. expressed as a relationship amongst the elements of the group of permutationsof degree 6 (not yet proved as a group, but reasonably anticipated), this resultreads(3)(1 4 6)(2 5) • (1 2 5 4)(3 6) = (1)(5)(2 4 3 6).We note, for practice, that φ has order 6 (the LCM of 1, 3, and 2) and that theproduct φ • θ has order 4.The number of elements in the group S n of all permutations of degree n isn! and clearly increases very rapidly as n increases. Fortunately, to illustrate theessential features of permutation groups it is sufficient to consider the case n =3,which involves only six elements. They are as follows (with labelling which thereader will by now recognise as anticipatory):I = (1)(2)(3) A =(123) B =(132)C = (1)(2 3) D = (3)(1 2) E = (2)(1 3)It will be noted that A and B have order 3, whilst C, D and E have order 2. Asperhaps anticipated, their combination products are exactly those correspondingto table 28.8, I, C, D and E being their own inverses. For example, putting in allsteps explicitly,D • C{a bc} = (3)(1 2) • (1)(2 3){a bc}= (3)(12){a cb}= {c ab}=(321){a bc}=(132){a bc}= B{a bc}.In brief, the six permutations belonging to S 3 form yet another non-Abelian groupisomorphic to the rotation–reflection symmetry group of an equilateral triangle.1058

28.5 MAPPINGS BETWEEN GROUPS28.5 Mappings between groupsNow that we have available a range of groups that can be used as examples,we return to the study of more general group properties. From here on, whenthere is no ambiguity we will write the product of two elements, X • Y , simplyas XY , omitting the explicit combination symbol. We will also continue to use‘multiplication’ as a loose generic name for the combination process betweenelements of a group.If G and G ′ are two groups, we can study the effect of a mappingΦ:G → G ′of G onto G ′ .IfX is an element of G we denote its image in G ′ under the mappingΦbyX ′ =Φ(X).A technical term that we have already used is isomorphic. We will now defineit formally. Two groups G = {X,Y,...} and G ′ = {X ′ ,Y ′ ,...} are said to beisomorphic if there is a one-to-one correspondencebetween their elements such thatX ↔ X ′ ,Y↔ Y ′ , ···XY = Z implies X ′ Y ′ = Z ′and vice versa.In other words, isomorphic groups have the same (multiplication) structure,although they may differ in the nature of their elements, combination law andnotation. Clearly if groups G and G ′ are isomorphic, and G and G ′′ are isomorphic,then it follows that G ′ and G ′′ are isomorphic. We have already seen an example offour groups (of functions of x, of orthogonal matrices, of permutations and of thesymmetries of an equilateral triangle) that are isomorphic, all having table 28.8as their multiplication table.Although our main interest is in isomorphic relationships between groups, thewider question of mappings of one set of elements onto another is of someimportance, and we start with the more general notion of a homomorphism.Let G and G ′ be two groups and Φ a mapping of G → G ′ . If for every pair ofelements X and Y in G(XY ) ′ = X ′ Y ′then Φ is called a homomorphism, and G ′ is said to be a homomorphic image of G.The essential defining relationship, expressed by (XY ) ′ = X ′ Y ′ , is that thesame result is obtained whether the product of two elements is formed first andthe image then taken or the images are taken first and the product then formed.1059

GROUP THEORYThree immediate consequences of the above definition are proved as follows.(i) If I is the identity of G then IX = X for all X in G. ConsequentlyX ′ =(IX) ′ = I ′ X ′ ,for all X ′ in G ′ . Thus I ′ is the identity in G ′ . In words, the identity elementof G maps into the identity element of G ′ .(ii) Further,I ′ =(XX −1 ) ′ = X ′ (X −1 ) ′ .That is, (X −1 ) ′ =(X ′ ) −1 . In words, the image of an inverse is the sameelement in G ′ as the inverse of the image.(iii) If element X in G is of order m, i.e.I = X m ,thenI ′ =(X m ) ′ =(XX m−1 ) ′ = X ′ (X m−1 ) ′ = ···= X} ′ X ′ {{ ···X}′ .m factorsIn words, the image of an element has the same order as the element.What distinguishes an isomorphism from the more general homomorphism arethe requirements that in an isomorphism:(I) different elements in G must map into different elements in G ′ (whereas ina homomorphism several elements in G may have the same image in G ′ ),that is, x ′ = y ′ must imply x = y;(II) any element in G ′ must be the image of some element in G.An immediate consequence of (I) and result (iii) for homomorphisms is thatisomorphic groups each have the same number of elements of any given order.For a general homomorphism, the set of elements of G whose image in G ′is I ′ is called the kernel of the homomorphism; this is discussed further in thenext section. In an isomorphism the kernel consists of the identity I alone. Toillustrate both this point and the general notion of a homomorphism, considera mapping between the additive group of real numbers R and the multiplicativegroup of complex numbers with unit modulus, U(1). Suppose that the mappingR→U(1) isΦ:x → e ix ;then this is a homomorphism since(x + y) ′ → e i(x+y) = e ix e iy = x ′ y ′ .However, it is not an isomorphism because many (an infinite number) of theelements of R have the same image in U(1). For example, π, 3π, 5π,... in R allhave the image −1 inU(1) and, furthermore, all elements of R of the form 2πn,where n is an integer, map onto the identity element in U(1). The latter set formsthe kernel of the homomorphism.1060

28.6 SUBGROUPS(a)I A B C D EI I A B C D EA A B I E C DB B I A D E CC C D E I A BD D E C B I AE E C D A B I(b)I A B CI I A B CA A I C BB B C I AC C B A ITable 28.9 Reproduction of (a) table 28.8 and (b) table 28.3 with the relevantsubgroups shown in bold.For the sake of completeness, we add that a homomorphism for which (I) aboveholds is said to be a monomorphism (or an isomorphism into), whilst a homomorphismfor which (II) holds is called an epimorphism (or an isomorphism onto). If,in either case, the other requirement is met as well then the monomorphism orepimorphism is also an isomorphism.Finally, if the initial and final groups are the same, G = G ′ , then the isomorphismG → G ′ is termed an automorphism.28.6 SubgroupsMore detailed inspection of tables 28.8 and 28.3 shows that not only do thecomplete tables have the properties associated with a group multiplication table(see section 28.2) but so do the upper left corners of each table taken on theirown. The relevant parts are shown in bold in the tables 28.9(a) and (b).This observation immediately prompts the notion of a subgroup. A subgroupof a group G can be formally defined as any non-empty subset H = {H i } ofG, the elements of which themselves behave as a group under the same rule ofcombination as applies in G itself. As for all groups, the order of the subgroup isequal to the number of elements it contains; we will denote it by h or |H|.Any group G contains two trivial subgroups:(i) G itself;(ii) the set I consisting of the identity element alone.All other subgroups of G are termed proper subgroups. In a group with multiplicationtable 28.8 the elements {I,A,B} form a proper subgroup, as do {I,A} in agroup with table 28.3 as its group table.Some groups have no proper subgroups. For example, the so-called cyclicgroups, mentioned at the end of subsection 28.1.1, have no subgroups otherthan the whole group or the identity alone. Tables 28.10(a) and (b) show themultiplication tables for two of these groups. Table 28.6 is also the group tablefor a cyclic group, that of order 4.1061

GROUP THEORY(a)I A BI I A BA A B IB B I A(b)I A B C DI I A B C DA A B C D IB B C D I AC C D I A BD D I A B CTable 28.10 The group tables of two cyclic groups, of orders 3 and 5. Theyhave no proper subgroups.It will be clear that for a cyclic group G repeated combination of any elementwith itself generates all other elements of G, before finally reproducing itself. So,for example, in table 28.10(b), starting with (say) D, repeated combination withitself produces, in turn, C, B, A, I and finally D again. As noted earlier, in anycyclic group G every element, apart from the identity, is of order g, the order ofthe group itself.The two tables shown are for groups of orders 3 and 5. It will be proved insubsection 28.7.2 that the order of any group is a multiple of the order of any ofits subgroups (Lagrange’s theorem), i.e. in our general notation, g is a multipleof h. It thus follows that a group of order p, wherep is any prime, must be cyclicand cannot have any proper subgroups. The groups for which tables 28.10(a) and(b) are the group tables are two such examples. Groups of non-prime order may(table 28.3) or may not (table 28.6) have proper subgroups.As we have seen, repeated multiplication of an element X (not the identity)by itself will generate a subgroup {X,X 2 ,X 3 ,...}. The subgroup will clearly beAbelian, and if X is of order m, i.e.X m = I, the subgroup will have m distinctmembers. If m is less than g – though, in view of Lagrange’s theorem, m mustbe a factor of g – the subgroup will be a proper subgroup. We can deduce, inpassing, that the order of any element of a group is an exact divisor of the orderof the group.Some obvious properties of the subgroups of a group G, which can be listedwithout formal proof, are as follows.(i) The identity element of G belongs to every subgroup H.(ii) If element X belongs to a subgroup H, so does X −1 .(iii) The set of elements in G that belong to every subgroup of G themselvesform a subgroup, though this may consist of the identity alone.Properties of subgroups that need more explicit proof are given in the followingsections, though some need the development of new concepts before theycan be established. However, we can begin with a theorem, applicable to allhomomorphisms, not just isomorphisms, that requires no new concepts.Let Φ : G → G ′ be a homomorphism of G into G ′ ;then1062

28.7 SUBDIVIDING A GROUP(i) the set of elements H ′ in G ′ that are images of the elements of G forms asubgroup of G ′ ;(ii) the set of elements K in G that are mapped onto the identity I ′ in G ′ formsa subgroup of G.As indicated in the previous section, the subgroup K is called the kernel of thehomomorphism.To prove (i), suppose Z and W belong to H ′ , with Z = X ′ and W = Y ′ ,whereX and Y belong to G. Thenand therefore belongs to H ′ ,andZW = X ′ Y ′ =(XY ) ′Z −1 =(X ′ ) −1 =(X −1 ) ′and therefore belongs to H ′ . These two results, together with the fact that I ′belongs to H ′ , are enough to establish result (i).To prove (ii), suppose X and Y belong to K; then(XY ) ′ = X ′ Y ′ = I ′ I ′ = I ′(closure),I ′ =(XX −1 ) ′ = X ′ (X −1 ) ′ = I ′ (X −1 ) ′ =(X −1 ) ′and therefore X −1 belongs to K. These two results, together with the fact that Ibelongs to K, are enough to establish (ii). An illustration of this result is providedby the mapping Φ of R→U(1) considered in the previous section. Its kernelconsists of the set of real numbers of the form 2πn, wheren is an integer; it formsa subgroup of R, the additive group of real numbers.In fact the kernel K of a homomorphism is a normal subgroup of G. Thedefining property of such a subgroup is that for every element X in G and everyelement Y in the subgroup, XY X −1 belongs to the subgroup. This property iseasily verified for the kernel K, since(XY X −1 ) ′ = X ′ Y ′ (X −1 ) ′ = X ′ I ′ (X −1 ) ′ = X ′ (X −1 ) ′ = I ′ .Anticipating the discussion of subsection 28.7.2, the cosets of a normal subgroupthemselves form a group (see exercise 28.16).28.7 Subdividing a groupWe have already noted, when looking at the (arbitrary) order of headings in agroup table, that some choices appear to make the table more orderly than doothers. In the following subsections we will identify ways in which the elementsof a group can be divided up into sets with the property that the members of anyone set are more like the other members of the set, in some particular regard,1063

GROUP THEORYthan they are like any element that does not belong to the set. We will find thatthese divisions will be such that the group is partitioned, i.e. the elements will bedivided into sets in such a way that each element of the group belongs to one,and only one, such set.We note in passing that the subgroups of a group do not form such a partition,not least because the identity element is in every subgroup, rather than being inprecisely one. In other words, despite the nomenclature, a group is not simply theaggregate of its proper subgroups.28.7.1 Equivalence relations and classesWe now specify in a more mathematical manner what it means for two elementsof a group to be ‘more like’ one another than like a third element, as mentionedin section 28.2. Our introduction will apply to any set, whether a group or not,but our main interest will ultimately be in two particular applications to groups.We start with the formal definition of an equivalence relation.An equivalence relation on a set S is a relationship X ∼ Y , between twoelements X and Y belonging to S, in which the definition of the symbol ∼ mustsatisfy the requirements of(i) reflexivity, X ∼ X;(ii) symmetry, X ∼ Y implies Y ∼ X;(iii) transitivity, X ∼ Y and Y ∼ Z imply X ∼ Z.Any particular two elements either satisfy or do not satisfy the relationship.The general notion of an equivalence relation is very straightforward, andthe requirements on the symbol ∼ seem undemanding; but not all relationshipsqualify. As an example within the topic of groups, if it meant ‘has the sameorder as’ then clearly all the requirements would be satisfied. However, if it meant‘commutes with’ then it would not be an equivalence relation, since although Acommutes with I, andI commutes with C, this does not necessarily imply that Acommutes with C, as is obvious from table 28.8.It may be shown that an equivalence relation on S divides up S into classes C isuch that:(i) X and Y belong to the same class if, and only if, X ∼ Y ;(ii) every element W of S belongs to exactly one class.This may be shown as follows. Let X belong to S, and define the subset S X ofS to be the set of all elements U of S such that X ∼ U. Clearly by reflexivityX belongs to S X . Suppose first that X ∼ Y , and let Z be any element of S Y .Then Y ∼ Z, and hence by transitivity X ∼ Z, which means that Z belongs toS X . Conversely, since the symmetry law gives Y ∼ X, ifZ belongs to S X then1064

28.7 SUBDIVIDING A GROUPthis implies that Z belongs to S Y . These two results together mean that the twosubsets S X and S Y have the same members and hence are equal.Now suppose that S X equals S Y .SinceY belongs to S Y it also belongs to S Xand hence X ∼ Y . This completes the proof of (i), once the distinct subsets oftype S X are identified as the classes C i . Statement (ii) is an immediate corollary,the class in question being identified as S W .The most important property of an equivalence relation is as follows.Two different subsets S X and S Y can have no element in common, and the collectionof all the classes C i is a ‘partition’ of S, i.e. every element in S belongs to one, andonly one, of the classes.To prove this, suppose S X and S Y have an element Z in common; then X ∼ Zand Y ∼ Z and so by the symmetry and transitivity laws X ∼ Y .Bytheabovetheorem this implies S X equals S Y . But this contradicts the fact that S X and S Yare different subsets. Hence S X and S Y can have no element in common.Finally, if the elements of S are used in turn to define subsets and hence classesin S, every element U is in the subset S U that is either a class already found orconstitutes a new one. It follows that the classes exhaust S, i.e.everyelementisin some class.Having established the general properties of equivalence relations, we now turnto two specific examples of such relationships, in which the general set S has themore specialised properties of a group G and the equivalence relation ∼ is chosenin such a way that the relatively transparent general results for equivalencerelations can be used to derive powerful, but less obvious, results about theproperties of groups.28.7.2 Congruence and cosetsAs the first application of equivalence relations we now prove Lagrange’s theoremwhich is stated as follows.Lagrange’s theorem. If G is a finite group of order g and H is a subgroup of G oforder h then g is a multiple of h.We take as the definition of ∼ that, given X and Y belonging to G, X ∼ Y ifX −1 Y belongs to H. This is the same as saying that Y = XH i for some elementH i belonging to H; technically X and Y are said to be left-congruent with respectto H.This defines an equivalence relation, since it has the following properties.(i) Reflexivity: X ∼ X, sinceX −1 X = I and I belongs to any subgroup.(ii) Symmetry: X ∼ Y implies that X −1 Y belongs to H and so, therefore, doesits inverse, since H is a group. But (X −1 Y ) −1 = Y −1 X and, as this belongsto H, it follows that Y ∼ X.1065

GROUP THEORY(iii) Transitivity: X ∼ Y and Y ∼ Z imply that X −1 Y and Y −1 Z belong to Hand so, therefore, does their product (X −1 Y )(Y −1 Z)=X −1 Z,fromwhichit follows that X ∼ Z.With ∼ proved as an equivalence relation, we can immediately deduce that itdivides G into disjoint (non-overlapping) classes. For this particular equivalencerelation the classes are called the left cosets of H. Thus each element of G is inone and only one left coset of H. The left coset containing any particular X isusually written XH, and denotes the set of elements of the form XH i (one ofwhich is X itself since H contains the identity element); it must contain h differentelements, since if it did not, and two elements were equal,XH i = XH j ,we could deduce that H i = H j and that H contained fewer than h elements.From our general results about equivalence relations it now follows that theleft cosets of H are a ‘partition’ of G into a number of sets each containing hmembers. Since there are g members of G and each must be in just one of thesets, it follows that g is a multiple of h. This concludes the proof of Lagrange’stheorem.The number of left cosets of H in G is known as the index of H in G and iswritten [G : H]; numerically the index = g/h. For the record we note that, forthe trivial subgroup I, which contains only the identity element, [G : I] =g andthat, for a subgroup J of subgroup H, [G : H][H : J ]=[G : J ].The validity of Lagrange’s theorem was established above using the far-reachingproperties of equivalence relations. However, for this specific purpose there is amore direct and self-contained proof, which we now give.Let X be some particular element of a finite group G of order g, andH be asubgroup of G of order h, with typical element Y i . Consider the set of elementsXH ≡{XY 1 ,XY 2 ,...,XY h }.This set contains h distinct elements, since if any two were equal, i.e. XY i = XY jwith i ≠ j, this would contradict the cancellation law. As we have already seen,the set is called a left coset of H.We now prove three simple results.• Two cosets are either disjoint or identical. Suppose cosets X 1 H and X 2 H havean element in common, i.e. X 1 Y 1 = X 2 Y 2 for some Y 1 ,Y 2 in H. ThenX 1 =X 2 Y 2 Y1 −1 , and since Y 1 and Y 2 both belong to H so does Y 2 Y1 −1 ; thus X 1belongs to the left coset X 2 H. Similarly X 2 belongs to the left coset X 1 H.Consequently, either the two cosets are identical or it was wrong to assumethat they have an element in common.1066

28.7 SUBDIVIDING A GROUP• Two cosets X 1 H and X 2 H are identical if, and only if, X2 −1X1 belongs to H. IfX2 −1X1 belongs to H then X 1 = X 2 Y i for some i, andX 1 H = X 2 Y i H = X 2 H,since by the permutation law Y i H = H. Thus the two cosets are identical.Conversely, suppose X 1 H = X 2 H.ThenX2 −1X1H = H. But one element ofH (on the left of the equation) is I; thus X2 −1X1 must also be an element of H(on the right). This proves the stated result.• Every element of G is in some left coset XH. This follows trivially since Hcontains I, and so the element X i is in the coset X i H.The final step in establishing Lagrange’s theorem is, as previously, to note thateach coset contains h elements, that the cosets are disjoint and that every one ofthe g elements in G appears in one and only one distinct coset. It follows thatg = kh for some integer k.As noted earlier, Lagrange’s theorem justifies our statement that any group oforder p, wherep is prime, must be cyclic and cannot have any proper subgroups:since any subgroup must have an order that divides p, this can only be 1 or p,corresponding to the two trivial subgroups I and the whole group.It may be helpful to see an example worked through explicitly, and we againuse the same six-element group.◮Find the left cosets of the proper subgroup H of the group G that has table 28.8 as itsmultiplication table.The subgroup consists of the set of elements H = {I,A,B}. We note in passing that it hasorder 3, which, as required by Lagrange’s theorem, is a divisor of 6, the order of G. Asinall cases, H itself provides the first (left) coset, formally the cosetIH = {II,IA,IB} = {I,A,B}.We continue by choosing an element not already selected, C say, and formCH = {CI,CA,CB} = {C,D,E}.These two cosets of H exhaust G, and are therefore the only cosets, the index of H in Gbeing equal to 2.This completes the example, but it is useful to demonstrate that it would not havemattered if we had taken D, say, instead of I to form a first cosetDH = {DI, DA, DB} = {D, E, C},and then, from previously unselected elements, picked B, say:BH = {BI,BA,BB} = {B,I,A}.The same two cosets would have resulted. ◭It will be noticed that the cosets are the same groupings of the elementsof G which we earlier noted as being the choice of adjacent column and rowheadings that give the multiplication table its ‘neatest’ appearance. Furthermore,1067

GROUP THEORYif H is a normal subgroup of G then its (left) cosets themselves form a group (seeexercise 28.16).28.7.3 Conjugates and classesOur second example of an equivalence relation is concerned with those elementsX and Y of a group G that can be connected by a transformation of the formY = G −1i XG i ,whereG i is an (appropriate) element of G. Thus X ∼ Y if thereexists an element G i of G such that Y = G −1i XG i . Different pairs of elements Xand Y will, in general, require different group elements G i . Elements connectedin this way are said to be conjugates.We first need to establish that this does indeed define an equivalence relation,as follows.(i) Reflexivity: X ∼ X, sinceX = I −1 XI and I belongs to the group.(ii) Symmetry: X ∼ Y implies Y = G −1i XG i and therefore X =(G −1i ) −1 YG −1i .Since G i belongs to G so does G −1i , and it follows that Y ∼ X.(iii) Transitivity: X ∼ Y and Y ∼ Z imply Y = G −1i XG i and Z = G −1j YG jand therefore Z = G −1j G −1i XG i G j =(G i G j ) −1 X(G i G j ). Since G i and G jbelong to G so does G i G j , from which it follows that X ∼ Z.These results establish conjugacy as an equivalence relation and hence showthat it divides G into classes, two elements being in the same class if, and only if,they are conjugate.Immediate corollaries are:(i) If Z is in the class containing I thenZ = G −1i IG i = G −1i G i = I.Thus, since any conjugate of I canbeshowntobeI, the identity must bein a class by itself.(ii) If X is in a class by itself thenY = G −1i XG imust imply that Y = X. ButX = G i G −1i XG i G −1ifor any G i ,andsoX = G i (G −1i XG i )G −1i = G i YG −1i = G i XG −1i ,i.e. XG i = G i X for all G i .Thus commutation with all elements of the group is a necessary (andsufficient) condition for any particular group element to be in a class byitself. In an Abelian group each element is in a class by itself.1068

28.7 SUBDIVIDING A GROUP(iii) In any group G the set S of elements in classes by themselves is an Abeliansubgroup (known as the centre of G). We have shown that I belongs to S,and so if, further, XG i = G i X and YG i = G i Y for all G i belonging to Gthen:(a) (XY )G i = XG i Y = G i (XY ), i.e. the closure of S, and(b) XG i = G i X implies X −1 G i = G i X −1 , i.e. the inverse of X belongsto S.Hence S is a group, and clearly Abelian.Yet again for illustration purposes, we use the six-element group that hastable 28.8 as its group table.◮Find the conjugacy classes of the group G having table 28.8 as its multiplication table.As always, I is in a class by itself, and we need consider it no further.Consider next the results of forming X −1 AX, asX runs through the elements of G.I −1 AI A −1 AA B −1 AB C −1 AC D −1 AD E −1 AE= IA = IA = AI = CE = DC = ED= A = A = A = B = B = BOnly A and B are generated. It is clear that {A, B} is one of the conjugacy classes of G.This can be verified by forming all elements X −1 BX; again only A and B appear.We now need to pick an element not in the two classes already found. Suppose wepick C. JustasforA, we compute X −1 CX, asX runs through the elements of G. Thecalculations can be done directly using the table and give the following:X : I A B C D EX −1 CX : C E D C E DThus C, D and E belong to the same class. The group is now exhausted, and so the threeconjugacy classes are{I}, {A, B}, {C,D,E}. ◭In the case of this small and simple, but non-Abelian, group, only the identityis in a class by itself (i.e. only I commutes with all other elements). It is also theonly member of the centre of the group.Other areas from which examples of conjugacy classes can be taken includepermutations and rotations. Two permutations can only be (but are not necessarily)in the same class if their cycle specifications have the same structure.For example, in S 5 the permutations (1 3 5)(2)(4) and (2 5 3)(1)(4) could be inthe same class as each other but not in the class that contains (1 5)(2 4)(3). Anexample of permutations with the same cycle structure yet in different conjugacyclasses is given in exercise 29. 10.In the case of the continuous rotation group, rotations by the same angle θabout any two axes labelled i and j are in the same class, because the groupcontains a rotation that takes the first axis into the second. Without going into1069

GROUP THEORYmathematical details, a rotation about axis i can be represented by the operatorR i (θ), and the two rotations are connected by a relationship of the formR j (θ) =φ −1ij R i (θ)φ ij ,in which φ ij is the member of the full continuous rotation group that takes axisi into axis j.28.8 Exercises28.1 For each of the following sets, determine whether they form a group under the operationindicated (where it is relevant you may assume that matrix multiplicationis associative):(a) the integers (mod 10) under addition;(b) the integers (mod 10) under multiplication;(c) the integers 1, 2, 3, 4, 5, 6 under multiplication (mod 7);(d) the integers 1, 2, 3, 4, 5 under multiplication (mod 6);(e)all matrices of the form(a a− b0 bwhere a and b are integers (mod 5) and a ≠0≠ b, under matrix multiplication;(f) those elements of the set in (e) that are of order 1 or 2 (taken together);(g) all matrices of the form⎛⎝ 1 a 0 1 00⎞⎠ ,b c 1where a, b, c are integers, under matrix multiplication.28.2 Which of the following relationships between X and Y are equivalence relations?Give a proof of your conclusions in each case:(a) X and Y are integers and X − Y is odd;(b) X and Y are integers and X − Y is even;(c) X and Y are people and have the same postcode;(d) X and Y are people and have a parent in common;(e) X and Y are people and have the same mother;(f) X and Y are n×n matrices satisfying Y = PXQ,whereP and Q are elementsof a group G of n × n matrices.28.3 Define a binary operation • on the set of real numbers byx • y = x + y + rxy,where r is a non-zero real number. Show that the operation • is associative.Prove that x • y = −r −1 if, and only if, x = −r −1 or y = −r −1 . Hence provethat the set of all real numbers excluding −r −1 forms a group under the operation•.1070),

28.8 EXERCISES28.4 Prove that the relationship X ∼ Y , defined by X ∼ Y if Y can be expressed inthe formY = aX + bcX + d ,with a, b, c and d as integers, is an equivalence relation on the set of real numbersR. Identify the class that contains the real number 1.28.5 The following is a ‘proof’ that reflexivity is an unnecessary axiom for an equivalencerelation.Because of symmetry X ∼ Y implies Y ∼ X. Then by transitivity X ∼ Y andY ∼ X imply X ∼ X. Thus symmetry and transitivity imply reflexivity, whichtherefore need not be separately required.Demonstrate the flaw in this proof using the set consisting of all real numbers plusthe number i. Show by investigating the following specific cases that, whether ornot reflexivity actually holds, it cannot be deduced from symmetry and transitivityalone.(a) X ∼ Y if X + Y is real.(b) X ∼ Y if XY is real.28.6 Prove that the set M of matricesA =(a b0 cwhere a, b, c are integers (mod 5) and a ≠0≠ c, form a non-Abelian groupunder matrix multiplication.Show that the subset containing elements of M that are of order 1 or 2 donot form a proper subgroup of M,(a) using Lagrange’s theorem,(b) by direct demonstration that the set is not closed.28.7 S is the set of all 2 × 2 matrices of the form( )w xA =, where wz − xy =1.y zShow that S is a group under matrix multiplication. Which element(s) have order2? Prove that an element A has order 3 if w + z +1=0.28.8 Show that, under matrix multiplication, matrices of the form( )a0 + aM(a 0 , a) =1 i −a 2 + a 3 i,a 2 + a 3 i a 0 − a 1 iwhere a 0 and the components of column matrix a =(a 1 a 2 a 3 ) T are realnumbers satisfying a 2 0 + |a|2 = 1, constitute a group. Deduce that, under thetransformation z → Mz, wherez is any column matrix, |z| 2 is invariant.28.9 If A is a group in which every element other than the identity, I, hasorder2,prove that A is Abelian. Hence show that if X and Y are distinct elements of A,neither being equal to the identity, then the set {I,X,Y ,XY } forms a subgroupof A.Deduce that if B is a group of order 2p, withp a prime greater than 2, then Bmust contain an element of order p.28.10 The group of rotations (excluding reflections and inversions) in three dimensionsthat take a cube into itself is known as the group 432 (or O in the usual chemicalnotation). Show by each of the following methods that this group has 24 elements.1071),

GROUP THEORY(a) Identify the distinct relevant axes and count the number of qualifying rotationsabout each.(b) The orientation of the cube is determined if the directions of two of its bodydiagonals are given. Consider the number of distinct ways in which one bodydiagonal can be chosen to be ‘vertical’, say, and a second diagonal made tolie along a particular direction.28.11 Identify the eight symmetry operations on a square. Show that they form agroup D 4 (known to crystallographers as 4mm andtochemistsasC 4v ) having oneelement of order 1, five of order 2 and two of order 4. Find its proper subgroupsand the corresponding cosets.28.12 If A and B are two groups, then their direct product, A × B, isdefinedtobethe set of ordered pairs (X,Y ), with X an element of A, Y an element of Band multiplication given by (X,Y )(X ′ ,Y ′ )=(XX ′ ,YY ′ ). Prove that A × B is agroup.Denote the cyclic group of order n by C n and the symmetry group of a regularn-sided figure (an n-gon) by D n – thus D 3 is the symmetry group of an equilateraltriangle, as discussed in the text.(a) By considering the orders of each of their elements, show (i) that C 2 × C 3 isisomorphic to C 6 , and (ii) that C 2 × D 3 is isomorphic to D 6 .(b) Are any of D 4 , C 8 , C 2 × C 4 , C 2 × C 2 × C 2 isomorphic?28.13 Find the group G generated under matrix multiplication by the matrices( ) ( )0 10 iA =, B =.1 0i 0Determine its proper subgroups, and verify for each of them that its cosetsexhaust G.28.14 Show that if p is prime then the set of rational number pairs (a, b), excluding(0, 0), with multiplication defined by(a, b) • (c, d) =(e, f), where (a + b √ p)(c + d √ p)=e + f √ p,forms an Abelian group. Show further that the mapping (a, b) → (a, −b) isanautomorphism.28.15 Consider the following mappings between a permutation group and a cyclicgroup.(a) Denote by A n the subset of the permutation group S n that contains all theeven permutations. Show that A n is a subgroup of S n .(b) List the elements of S 3 in cycle notation and identify the subgroup A 3 .(c) For each element X of S 3 ,letp(X) =1ifX belongs to A 3 and p(X) =−1 ifitdoes not. Denote by C 2 the multiplicative cyclic group of order 2. Determinethe images of each of the elements of S 3 for the following four mappings:Φ 1 : S 3 → C 2 X → p(X),Φ 2 : S 3 → C 2 X →−p(X),Φ 3 : S 3 → A 3 X → X 2 ,Φ 4 : S 3 → S 3 X → X 3 .(d) For each mapping, determine whether the kernel K is a subgroup of S 3 and,if so, whether the mapping is a homomorphism.28.16 For the group G with multiplication table 28.8 and proper subgroup H = {I,A,B},denote the coset {I,A,B} by C 1 and the coset {C,D,E} by C 2 .Formthesetofall possible products of a member of C 1 with itself, and denote this by C 1 C 1 .1072

28.8 EXERCISESSimilarly compute C 2 C 2 , C 1 C 2 and C 2 C 1 . Show that each product coset is equal toC 1 or to C 2 ,andthata2× 2 multiplication table can be formed, demonstratingthat C 1 and C 2 are themselves the elements of a group of order 2. A subgrouplike H whose cosets themselves form a group is a normal subgroup.28.17 The group of all non-singular n × n matrices is known as the general lineargroup GL(n) and that with only real elements as GL(n, R). If R ∗ denotes themultiplicative group of non-zero real numbers, prove that the mapping Φ :GL(n, R) → R ∗ , defined by Φ(M) =detM, is a homomorphism.Show that the kernel K of Φ is a subgroup of GL(n, R). Determine its cosetsand show that they themselves form a group.28.18 The group of reflection–rotation symmetries of a square is known as D 4 ;letX be one of its elements. Consider a mapping Φ : D 4 → S 4 , the permutationgroup on four objects, defined by Φ(X) = the permutation induced by X onthe set {x, y, d, d ′ }, where x and y are the two principal axes, and d and d ′the two principal diagonals, of the square. For example, if R is a rotation byπ/2, Φ(R) = (12)(34). Show that D 4 is mapped onto a subgroup of S 4 and, byconstructing the multiplication tables for D 4 and the subgroup, prove that themapping is a homomorphism.28.19 Given that matrix M is a member of the multiplicative group GL(3, R), determine,for each of the following additional constraints on M (applied separately), whetherthe subset satisfying the constraint is a subgroup of GL(3, R):(a) M T = M;(b) M T M = I;(c) |M| =1;(d) M ij =0forj>iand M ii ≠0.28.20 The elements of the quaternion group, Q, aretheset{1, −1,i,−i, j, −j,k,−k},with i 2 = j 2 = k 2 = −1, ij = k and its cyclic permutations, and ji = −k andits cyclic permutations. Find the proper subgroups of Q and the correspondingcosets. Show that the subgroup of order 2 is a normal subgroup, but that theother subgroups are not. Show that Q cannot be isomorphic to the group 4mm(C 4v ) considered in exercise 28.11.28.21 Show that D 4 , the group of symmetries of a square, has two isomorphic subgroupsof order 4. Show further that there exists a two-to-one homomorphism from thequaternion group Q, of exercise 28.20, onto one (and hence either) of these twosubgroups, and determine its kernel.28.22 Show that the matrices⎛M(θ, x, y) = ⎝cos θ − sin θ xsin θ cos θ y0 0 1where 0 ≤ θ

GROUP THEORYm 1 (π)m 2 (π)m 3 (π)m 4 (π)Figure 28.3 The notation for exercise 28.11.28.9 Hints and answers28.1 § (a) Yes, (b) no, there is no inverse for 2, (c) yes, (d) no, 2 × 3 is not in the set,(e) yes, (f) yes, they form a subgroup of order 4, [1, 0; 0, 1] [4, 0; 0, 4] [1, 2; 0, 4][4, 3; 0, 1], (g) yes.28.3 x • (y • z) =x + y + z + r(xy + xz + yz)+r 2 xyz =(x • y) • z. Show that assumingx • y = −r −1 leads to (rx +1)(ry +1)=0. The inverse of x is x −1 = −x/(1 + rx);show that this is not equal to −r −1 .28.5 (a) Consider both X = i and X ≠ i. Here, i ≁ i. (b) In this case i ∼ i, but theconclusion cannot be deduced from the other axioms. In both cases i is in a classby itself and no Y , as used in the false proof, can be found.28.7 † Use |AB| = |A||B| =1×1 = 1 to prove closure. The inverse has w ↔ z, x ↔−x,y ↔−y, giving |A −1 | = 1, i.e. it is in the set. The only element of order 2 is −I;A 2 can be simplified to [−(w +1), −x; −y, −(z + 1)].28.9 If XY = Z, show that Y = XZ and X = ZY,thenformYX. Note that theelements of B can only have orders 1, 2 or p. Suppose they all have order 1 or2; then using the earlier result, whilst noting that 4 does not divide 2p, leadstoa contradiction.28.11 Using the notation indicated in figure 28.3, R being a rotation of π/2 about anaxis perpendicular to the square, we have: I has order 1; R 2 , m 1 , m 2 , m 3 , m 4 haveorder 2; R, R 3 have order 4.subgroup {I,R,R 2 ,R 3 } has cosets {I,R,R 2 ,R 3 }, {m 1 ,m 2 ,m 3 ,m 4 };subgroup {I,R 2 ,m 1 ,m 2 } has cosets {I,R 2 ,m 1 ,m 2 }, {R,R 3 ,m 3 ,m 4 };subgroup {I,R 2 ,m 3 ,m 4 } has cosets {I,R 2 ,m 3 ,m 4 }, {R,R 3 ,m 1 ,m 2 };subgroup {I,R 2 } has cosets {I,R 2 }, {R,R 3 }, {m 1 ,m 2 }, {m 3 ,m 4 };subgroup {I,m 1 } has cosets {I,m 1 }, {R,m 3 }, {R 2 ,m 2 }, {R 3 ,m 4 };subgroup {I,m 2 } has cosets {I,m 2 }, {R,m 4 }, {R 2 ,m 1 }, {R 3 ,m 3 };subgroup {I,m 3 } has cosets {I,m 3 }, {R,m 2 }, {R 2 ,m 4 }, {R 3 ,m 1 };subgroup {I,m 4 } has cosets {I,m 4 }, {R,m 1 }, {R 2 ,m 3 }, {R 3 ,m 2 }.28.13 G = {I, A, B, B 2 , B 3 , AB, AB 2 , AB 3 }. The proper subgroups are as follows:{I, A}, {I, B 2 }, {I, AB 2 }, {I, B, B 2 , B 3 }, {I, B 2 , AB, AB 3 }.28.15 (b) A 3 = {(1), (123), (132)}.(d) For Φ 1 , K = {(1), (123), (132)} is a subgroup.For Φ 2 , K = {(23), (13), (12)} is not a subgroup because it has no identity element.For Φ 3 , K = {(1), (23), (13), (12)} is not a subgroup because it is not closed.§ Where matrix elements are given as a list, the convention used is [row 1; row 2; ...], individualentries in each row being separated by commas.1074

28.9 HINTS AND ANSWERS≠For Φ 4 , K = {(1), (123), (132)} is a subgroup.Only Φ 1 is a homomorphism; Φ 4 fails because, for example, [(23)(13)] ′(23) ′ (13) ′ .28.17 Recall that, for any pair of matrices P and Q, |PQ| = |P||Q|. K is the set of allmatrices with unit determinant. The cosets of K are the sets of matrices whosedeterminants are equal; K itself is the identity in the group of cosets.28.19 (a) No, because the set is not closed, (b) yes, (c) yes, (d) yes.28.21 Each subgroup contains the identity, a rotation by π, andtworeflections.Thehomomorphism is ±1 → I, ±i → R 2 , ±j → m x , ±k → m y with kernel {1, −1}.28.23 There are 10 elements in all: I, rotations R i (i =1, 4) and reflections m j (j =1, 5).(a) There are five proper subgroups of order 2, {I,m j } and one proper subgroupof order 5, {I,R,R 2 ,R 3 ,R 4 }.(b) Four conjugacy classes, {I}, {R,R 4 }, {R 2 ,R 3 }, {m 1 ,m 2 ,m 3 ,m 4 ,m 5 }.1075

29Representation theoryAs indicated at the start of the previous chapter, significant conclusions canoften be drawn about a physical system simply from the study of its symmetryproperties. That chapter was devoted to setting up a formal mathematical basis,group theory, with which to describe and classify such properties; the currentchapter shows how to implement the consequences of the resulting classificationsand obtain concrete physical conclusions about the system under study. Theconnection between the two chapters is akin to that between working withcoordinate-free vectors, each denoted by a single symbol, and working with acoordinate system in which the same vectors are expressed in terms of components.The ‘coordinate systems’ that we will choose will be ones that are expressed interms of matrices; it will be clear that ordinary numbers would not be sufficient,as they make no provision for any non-commutation amongst the elementsof a group. Thus, in this chapter the group elements will be represented bymatrices that have the same commutation relations as the members of the group,whatever the group’s original nature (symmetry operations, functional forms,matrices, permutations, etc.). For some abstract groups it is difficult to give awritten description of the elements and their properties without recourse to suchrepresentations. Most of our applications will be concerned with representationsof the groups that consist of the symmetry operations on molecules containingtwo or more identical atoms.Firstly, in section 29.1, we use an elementary example to demonstrate the kindof conclusions that can be reached by arguing purely on symmetry grounds. Thenin sections 29.2–29.10 we develop the formal side of representation theory andestablish general procedures and results. Finally, these are used in section 29.11to tackle a variety of problems drawn from across the physical sciences.1076

29.1 DIPOLE MOMENTS OF MOLECULESABA ′ B ′(a) HCl (b) CO 2 (c) O 3Figure 29.1 Three molecules, (a) hydrogen chloride, (b) carbon dioxide and(c) ozone, for which symmetry considerations impose varying degrees ofconstraint on their possible electric dipole moments.29.1 Dipole moments of moleculesSome simple consequences of symmetry can be demonstrated by consideringwhether a permanent electric dipole moment can exist in any particular molecule;three simple molecules, hydrogen chloride, carbon dioxide and ozone, are illustratedin figure 29.1. Even if a molecule is electrically neutral, an electric dipolemoment will exist in it if the centres of gravity of the positive charges (due toprotons in the atomic nuclei) and of the negative charges (due to the electrons)do not coincide.For hydrogen chloride there is no reason why they should coincide; indeed, thenormal picture of the binding mechanism in this molecule is that the electron fromthe hydrogen atom moves its average position from that of its proton nucleus tosomewhere between the hydrogen and chlorine nuclei. There is no compensatingmovement of positive charge, and a net dipole moment is to be expected – andis found experimentally.For the linear molecule carbon dioxide it seems obvious that it cannot havea dipole moment, because of its symmetry. Putting this rather more rigorously,we note that any rotation about the long axis of the molecule leaves it totallyunchanged; consequently, any component of a permanent electric dipole perpendicularto that axis must be zero (a non-zero component would rotate althoughno physical change had taken place in the molecule). That only leaves the possibilityof a component parallel to the axis. However, a rotation of π radiansabout the axis AA ′ shown in figure 29.1(b) carries the molecule into itself, asdoes a reflection in a plane through the carbon atom and perpendicular to themolecular axis (i.e. one with its normal parallel to the axis). In both cases the twooxygen atoms change places but, as they are identical, the molecule is indistinguishablefrom the original. Either ‘symmetry operation’ would reverse the signof any dipole component directed parallel to the molecular axis; this can only becompatible with the indistinguishability of the original and final systems if theparallel component is zero. Thus on symmetry grounds carbon dioxide cannothave a permanent electric dipole moment.1077

REPRESENTATION THEORYFinally, for ozone, which is angular rather than linear, symmetry does notplace such tight constraints. A dipole-moment component parallel to the axisBB ′ (figure 29.1(c)) is possible, since there is no symmetry operation that reversesthe component in that direction and at the same time carries the molecule intoan indistinguishable copy of itself. However, a dipole moment perpendicular toBB ′ is not possible, since a rotation of π about BB ′ would both reverse anysuch component and carry the ozone molecule into itself – two contradictoryconclusions unless the component is zero.In summary, symmetry requirements appear in the form that some or allcomponents of permanent electric dipoles in molecules are forbidden; they donot show that the other components do exist, only that they may. The greaterthe symmetry of the molecule, the tighter the restrictions on potentially non-zerocomponents of its dipole moment.In section 23.11 other, more complicated, physical situations will be analysedusing results derived from representation theory. In anticipation of these results,and since it may help the reader to understand where the developments in thenext nine sections are leading, we make here a broad, powerful, but rather formal,statement as follows.If a physical system is such that after the application of particular rotations orreflections (or a combination of the two) the final system is indistinguishable fromthe original system then its behaviour, and hence the functions that describe itsbehaviour, must have the corresponding property of invariance when subjected tothe same rotations and reflections.29.2 Choosing an appropriate formalismAs mentioned in the introduction to this chapter, the elements of a finite groupG can be represented by matrices; this is done in the following way. A suitablecolumn matrix u, known as a basis vector, § is chosen and is written in terms ofits components u i ,thebasis functions, asu =(u 1 u 2 ··· u n ) T .Theu i may be ofa variety of natures, e.g. numbers, coordinates, functions or even a set of labels,though for any one basis vector they will all be of the same kind.Once chosen, the basis vector can be used to generate an n-dimensional representationof the group as follows. An element X of the group is selected andits effect on each basis function u i is determined. If the action of X on u 1 is toproduce u ′ 1 , etc. then the set of equationsu ′ i = Xu i (29.1)§ This usage of the term basis vector is not exactly the same as that introduced in subsection 8.1.1.1078

29.2 CHOOSING AN APPROPRIATE FORMALISMgenerates a new column matrix u ′ =(u ′ 1 u′ 2 ··· u′ n) T . Having established u and u ′we can determine the n × n matrix, M(X) say, that connects them byu ′ = M(X)u. (29.2)It may seem natural to use the matrix M(X) so generated as the representativematrix of the element X; in fact, because we have already chosen the conventionwhereby Z = XY implies that the effect of applying element Z is the same as thatof first applying Y and then applying X to the result, one further step has to betaken. So that the representative matrices D(X) may follow the same convention,i.e.D(Z) =D(X)D(Y ),and at the same time respect the normal rules of matrix multiplication, it isnecessary to take the transpose of M(X) as the representative matrix D(X).Explicitly,and (29.2) becomesD(X) =M T (X) (29.3)u ′ = D T (X)u. (29.4)Thus the procedure for determining the matrix D(X) that represents the groupelement X in a representation based on basis vector u is summarised by equations(29.1)–(29.4). §This procedure is then repeated for each element X of the group, and theresulting set of n × n matrices D = {D(X)} is said to be the n-dimensionalrepresentation of G having u as its basis. The need to take the transpose of eachmatrix M(X) is not of any fundamental significance, since the only thing thatreally matters is whether the matrices D(X) have the appropriate multiplicationproperties – and, as defined, they do.In cases in which the basis functions are labels, the actions of the groupelements are such as to cause rearrangements of the labels. Correspondingly thematrices D(X) contain only ‘1’s and ‘0’s as entries; each row and each columncontains a single ‘1’.§ An alternative procedure in which a row vector is used as the basis vector is possible. Definingequations of the form u T X = u T D(X) are used, and no additional transpositions are needed todefine the representative matrices. However, row-matrix equations are cumbersome to write outand in all other parts of this book we have adopted the convention of writing operators (here thegroup element) to the left of the object on which they operate (here the basis vector).1079

REPRESENTATION THEORY◮For the group S 3 of permutations on three objects, which has group multiplication table28.8 on p. 1055, with (in cycle notation)I = (1)(2)(3), A =(123), B =(132C = (1)(2 3), D = (3)(1 2), E = (2)(1 3),use as the components of a basis vector the ordered letter tripletsu 1 = {PQR}, u 2 = {QRP}, u 3 = {RPQ},u 4 = {PRQ}, u 5 = {QPR}, u 6 = {RQP}.Generate a six-dimensional representation D = {D(X)} of the group and confirm that therepresentative matrices multiply according to table 28.8, e.g.D(C)D(B) =D(E).It is immediate that the identity permutation I = (1)(2)(3) leaves all u i unchanged, i.e.u ′ i = u i for all i. The representative matrix D(I) is thus I 6 ,the6× 6 unit matrix.We next take X as the permutation A = (1 2 3) and, using (29.1), let it act on each ofthe components of the basis vector:u ′ 1 = Au 1 =(123){PQR} = {QRP} = u 2u ′ 2 = Au 2 =(123){QRP} = {RPQ} = u 3..u ′ 6 = Au 6 =(123){RQP} = {QPR} = u 5 .The matrix M(A) has to be such that u ′ = M(A)u (here dots replace zeros to aid readability):⎛u ′ =⎜⎝u 2u 3u 1u 6u 4u 5⎞ ⎛=⎟ ⎜⎠ ⎝· 1 · · · ·· · 1 · · ·1 · · · · ·· · · · · 1· · · 1 · ·· · · · 1 ·⎞ ⎛⎟ ⎜⎠ ⎝u 1u 2u 3u 4u 5u 6⎞≡ M(A)u.⎟⎠D(A) isthenequaltoM T (A).The other D(X) are determined in a similar way. In general, ifXu i = u j ,then [M(X)] ij = 1, leading to [D(X)] ji =1and[D(X)] jk =0fork ≠ i. For example,Cu 3 = (1)(23){RPQ} = {RQP} = u 6implies that [D(C)] 63 =1and[D(C)] 6k =0fork =1, 2, 4, 5, 6. When calculated in full⎛⎞⎛⎞· · · 1 · ·· 1 · · · ·· · · · 1 ·· · 1 · · ·· · · · · 11 · · · · ·D(C) =⎜ 1 · · · · ·, D(B) =⎟⎜ · · · · · 1,⎟⎝ · 1 · · · · ⎠⎝ · · · 1 · · ⎠· · 1 · · ·· · · · 1 ·⎛⎞· · · · · 1· · · 1 · ·· · · · 1 ·D(E) =⎜ · 1 · · · ·,⎟⎝ · · 1 · · · ⎠1 · · · · ·1080

29.2 CHOOSING AN APPROPRIATE FORMALISMP1P3P3R322Q11 2RQ R(a) (b) (c)QFigure 29.2 Diagram (a) shows the definition of the basis vector, (b) showsthe effect of applying a clockwise rotation of 2π/3 and (c) shows the effect ofapplying a reflection in the mirror axis through Q.from which it can be verified that D(C)D(B) =D(E). ◭Whilst a representation obtained in this way necessarily has the same dimensionas the order of the group it represents, there are, in general, square matrices ofboth smaller and larger dimensions that can be used to represent the group,though their existence may be less obvious.One possibility that arises when the group elements are symmetry operationson an object whose position and orientation can be referred to a spacecoordinate system is called the natural representation. In it the representativematrices D(X) describe, in terms of a fixed coordinate system, what happensto a coordinate system that moves with the object when X is applied. Thereis usually some redundancy of the coordinates used in this type of representation,since interparticle distances are fixed and fewer than 3N coordinates,where N is the number of identical particles, are needed to specify uniquelythe object’s position and orientation. Subsection 29.11.1 gives an example thatillustrates both the advantages and disadvantages of the natural representation.We continue here with an example of a natural representation that has no suchredundancy.◮Use the fact that the group considered in the previous worked example is isomorphic tothe group of two-dimensional symmetry operations on an equilateral triangle to generate athree-dimensional representation of the group.Label the triangle’s corners as 1, 2, 3 and three fixed points in space as P, Q, R, so thatinitially corner 1 lies at point P, 2 lies at point Q, and 3 at point R. We take P, Q, R asthe components of the basis vector.In figure 29.2, (a) shows the initial configuration and also, formally, the result of applyingthe identity I to the triangle; it is therefore described by the basis vector, (P Q R) T .Diagram (b) shows the the effect of a clockwise rotation by 2π/3, corresponding toelement A in the previous example; the new column matrix is (Q R P) T .Diagram (c) shows the effect of a typical mirror reflection – the one that leaves thecorner at point Q unchanged (element D in table 28.8 and the previous example); the newcolumn matrix is now (R Q P) T .In similar fashion it can be concluded that the column matrix corresponding to elementB, rotationby4π/3, is (R P Q) T , and that the other two reflections C and E result in1081

REPRESENTATION THEORYcolumn matrices (P R Q) T and (Q P R) T respectively. The forms of the representativematrices M nat (X), (29.2), are now determined by equations such as, for element E,⎛⎝ Q ⎞ ⎛P ⎠ = ⎝ 0 1 0⎞ ⎛1 0 0 ⎠ ⎝ P ⎞Q ⎠R 0 0 1 Rimplying that⎛D nat (E) = ⎝ 0 1 0⎞T⎛1 0 0 ⎠ = ⎝ 0 1 0⎞1 0 0 ⎠ .0 0 1 0 0 1In this way the complete representation is obtained as⎛D nat (I) = ⎝ 1 0 0⎞⎛0 1 0 ⎠ , D nat (A) = ⎝ 0 0 1⎞⎛1 0 0 ⎠ , D nat (B) = ⎝ 0 1 0⎞0 0 1 ⎠ ,0 0 10 1 01 0 0⎛D nat (C) = ⎝ 1 0 0⎞⎛0 0 1 ⎠ , D nat (D) = ⎝ 0 0 1⎞⎛0 1 0 ⎠ , D nat (E) = ⎝ 0 1 0⎞1 0 0 ⎠ .0 1 01 0 00 0 1It should be emphasised that although the group contains six elements this representationis three-dimensional. ◭We will concentrate on matrix representations of finite groups, particularlyrotation and reflection groups (the so-called crystal point groups). The generalideas carry over to infinite groups, such as the continuous rotation groups, but ina book such as this, which aims to cover many areas of applicable mathematics,some topics can only be mentioned and not explored. We now give the formaldefinition of a representation.Definition. A representation D = {D(X)} of a group G is an assignment of a nonsingularsquare n × n matrix D(X) to each element X belonging to G, such that(i) D(I) =I n , the unit n × n matrix,(ii) D(X)D(Y )=D(XY ) for any two elements X and Y belonging to G, i.e.thematrices multiply in the same way as the group elements they represent.As mentioned previously, a representation by n × n matrices is said to be ann-dimensional representation of G. The dimension n is not to be confused withg, the order of the group, which gives the number of matrices needed in therepresentation, though they might not all be different.A consequence of the two defining conditions for a representation is that thematrix associated with the inverse of X is the inverse of the matrix associatedwith X. This follows immediately from setting Y = X −1 in (ii):D(X)D(X −1 )=D(XX −1 )=D(I) =I n ;henceD(X −1 )=[D(X)] −1 .1082

29.2 CHOOSING AN APPROPRIATE FORMALISMAs an example, the four-element Abelian group that consists of the set {1,i,−1, −i}under ordinary multiplication has a two-dimensional representation based on thecolumn matrix (1 i) T :( )( )1 00 −1D(1) =, D(i) =,D(−1) =0 1( −1 00 −1), D(−i) =1 0( 0 1−1 0The reader should check that D(i)D(−i) =D(1), D(i)D(i) =D(−1) etc., i.e. thatthe matrices do have exactly the same multiplication properties as the elementsof the group. Having done so, the reader may also wonder why anybody wouldbother with the representative matrices, when the original elements are so muchsimpler to handle! As we will see later, once some general properties of matrixrepresentations have been established, the analysis of large groups, both Abelianand non-Abelian, can be reduced to routine, almost cookbook, procedures.An n-dimensional representation of G is a homomorphism of G into the set ofinvertible n × n matrices (i.e. n × n matrices that have inverses or, equivalently,have non-zero determinants); this set is usually known as the general linear groupand denoted by GL(n). In general the same matrix may represent more than oneelement of G; if, however, all the matrices representing the elements of G aredifferent then the representation is said to be faithful, and the homomorphismbecomes an isomorphism onto a subgroup of GL(n).A trivial but important representation is D(X) =I n for all elements X of G.Clearly both of the defining relationships are satisfied, and there is no restrictionon the value of n. However, such a representation is not a faithful one.To sum up, in the context of a rotation–reflection group, the transposes ofthe set of n × n matrices D(X) thatmakeuparepresentationD may be thoughtof as describing what happens to an n-component basis vector of coordinates,(x y ···) T , or of functions, (Ψ 1 Ψ 2 ···) T ,theΨ i themselves being functionsof coordinates, when the group operation X is carried out on each of thecoordinates or functions. For example, to return to the symmetry operationson an equilateral triangle, the clockwise rotation by 2π/3, R, carries the threedimensionalbasis vector (x y z) T into the column matrix⎛⎜⎝− 1 2 x + √ 32 y− √ 32 x − 1 2 yzwhilst the two-dimensional basis vector of functions (r 2 3z 2 − r 2 ) T is unaltered,as neither r nor z is changed by the rotation. The fact that z is unchanged byany of the operations of the group shows that the components x, y, z actuallydivide (i.e. are ‘reducible’, to anticipate a more formal description) into two sets:1083⎞⎟⎠).

REPRESENTATION THEORYone comprises z, which is unchanged by any of the operations, and the othercomprises x, y, which change as a pair into linear combinations of themselves.This is an important observation to which we return in section 29.4.29.3 Equivalent representationsIf D is an n-dimensional representation of a group G, andQ is any fixed invertiblen × n matrix (|Q| ≠ 0), then the set of matrices defined by the similaritytransformationD Q (X) =Q −1 D(X)Q (29.5)also forms a representation D Q of G, said to be equivalent to D. Wecanseefromacomparison with the definition in section 29.2 that they do form a representation:(i) D Q (I) =Q −1 D(I)Q = Q −1 I n Q = I n ,(ii) D Q (X)D Q (Y )=Q −1 D(X)QQ −1 D(Y )Q = Q −1 D(X)D(Y )Q= Q −1 D(XY )Q = D Q (XY ).Since we can always transform between equivalent representations using a nonsingularmatrix Q, we will consider such representations to be one and the same.Despite the similarity of words and manipulations to those of subsection 28.7.1,that two representations are equivalent does not constitute an ‘equivalence relation’– for example, the reflexive property does not hold for a general fixedmatrix Q. However, if Q were not fixed, but simply restricted to belonging toa set of matrices that themselves form a group, then (29.5) would constitute anequivalence relation.The general invertible matrix Q that appears in the definition (29.5) of equivalentmatrices describes changes arising from a change in the coordinate system(i.e. in the set of basis functions). As before, suppose that the effect of an operationX on the basis functions is expressed by the action of M(X) (whichisequalto D T (X)) on the corresponding basis vector:u ′ = M(X)u = D T (X)u. (29.6)A change of basis would be given by u Q = Qu and u ′ Q = Qu′ , and we may writeThis is of the same form as (29.6), i.e.u ′ Q = Qu ′ = QM(X)u = QD T (X)Q −1 u Q . (29.7)u ′ Q = D T Q T(X)u Q, (29.8)where D Q T(X) =(Q T ) −1 D(X)Q T is related to D(X) by a similarity transformation.Thus D Q T(X) represents the same linear transformation as D(X), but with1084

29.3 EQUIVALENT REPRESENTATIONSrespect to a new basis vector u Q ; this supports our contention that representationsconnected by similarity transformations should be considered as the samerepresentation.◮For the four-element Abelian group consisting of the set {1,i,−1, −i} under ordinarymultiplication, discussed near the end of section 29.2, change the basis vector from u =(1 i) T to u Q =(3− i 2i − 5) T . Find the real transformation matrix Q. Show that thetransformed representative matrix for element i, D Q T(i), isgivenby( )17 −29D Q T(i) =10 −17and verify that D T Q T (i)u Q = iu Q .Firstly, we solve the matrix equation( )3 − i=2i − 5( )(a b 1c d i),with a, b, c, d real. This gives Q and hence Q −1 as( )( )3 −12 1Q =, Q−5 2−1 =.5 3Following (29.7) we now find the transpose of D Q T(i) as( )( )( ) ( )3 −1 0 1 2 1 17 10QD T (i)Q −1 ==−5 2 −1 0 5 3 −29 −17and hence D Q T(i) is as stated. Finally,(17 10D T Q T(i)u Q =−29 −17= i(3 − i2i − 5)( ) ( )3 − i 1+3i=2i − 5 −2 − 5i)= iu Q ,as required. ◭Although we will not prove it, it can be shown that any finite representationof a finite group of linear transformations that preserve spatial length (or, inquantum mechanics, preserve the magnitude of a wavefunction) is equivalent to1085

REPRESENTATION THEORYa representation in which all the matrices are unitary (see chapter 8) and so fromnow on we will consider only unitary representations.29.4 Reducibility of a representationWe have seen already that it is possible to have more than one representationof any particular group. For example, the group {1,i,−1, −i} under ordinarymultiplication has been shown to have a set of 2 × 2 matrices, and a set of fourunit n × n matrices I n , as two of its possible representations.Consider two or more representations, D (1) , D (2) , ...,D (N) , which may beof different dimensions, of a group G. Now combine the matrices D (1) (X),D (2) (X), ...,D (N) (X) that correspond to element X of G into a larger blockdiagonalmatrix:D (1) (X )0D(X ) =D (2) (X )(29.9). . .0D (N) (X )Then D = {D(X)} is the matrix representation of the group obtained by combiningthe basis vectors of D (1) , D (2) , ...,D (N) into one larger basis vector. If, knowingly orunknowingly, we had started with this larger basis vector and found the matricesof the representation D to have the form shown in (29.9), or to have a formthat can be transformed into this by a similarity transformation (29.5) (using,of course, the same matrix Q for each of the matrices D(X)) then we would saythat D is reducible and that each matrix D(X) can be written as the direct sum ofsmaller representations:D(X) =D (1) (X) ⊕ D (2) (X) ⊕ ···⊕ D (N) (X).It may be that some or all of the matrices D (1) (X), D (2) (X), ...,D (N) themselvescan be further reduced – i.e. written in block diagonal form. For example,suppose that the representation D (1) , say, has a basis vector (x y z) T ; then, forthe symmetry group of an equilateral triangle, whilst x and y are mixed togetherfor at least one of the operations X, z is never changed. In this case the 3 × 3representative matrix D (1) (X) can itself be written in block diagonal form as a1086

29.4 REDUCIBILITY OF A REPRESENTATION2 × 2matrixanda1× 1 matrix. The direct-sum matrix D(X) can now be writtena bc d10D(X ) =D (2) (X )(29.10). . .0D (N) (X )but the first two blocks can be reduced no further.When all the other representations D (2) (X), ... have been similarly treated,what remains is said to be irreducible and has the characteristic of being blockdiagonal, with blocks that individually cannot be reduced further. The blocks areknown as the irreducible representations of G, often abbreviated to the irreps ofG, and we denote them by ˆD (i) . They form the building blocks of representationtheory, and it is their properties that are used to analyse any given physicalsituation which is invariant under the operations that form the elements of G.Any representation can be written as a linear combination of irreps.If, however, the initial choice u of basis vector for the representation D isarbitrary, as it is in general, then it is unlikely that the matrices D(X) willassume obviously block diagonal forms (it should be noted, though, that sincethe matrices are square, even a matrix with non-zero entries only in the extremetop right and bottom left positions is technically block diagonal). In general, itwill be possible to reduce them to block diagonal matrices with more than oneblock; this reduction corresponds to a transformation Q to a new basis vectoru Q , as described in section 29.3.In any particular representation D, each constituent irrep ˆD (i) may appear anynumber of times, or not at all, subject to the obvious restriction that the sum ofall the irrep dimensions must add up to the dimension of D itself. Let us say thatˆD (i) appears m i times. The general expansion of D is then writtenD = m 1 ˆD (1) ⊕ m 2 ˆD (2) ⊕ ···⊕ m N ˆD (N) , (29.11)where if G is finite so is N.This is such an important result that we shall now restate the situation insomewhat different language. When the set of matrices that forms a representation1087

REPRESENTATION THEORYof a particular group of symmetry operations has been brought to irreducibleform, the implications are as follows.(i) Those components of the basis vector that correspond to rows in therepresentation matrices with a single-entry block, i.e. a 1 × 1block,areunchanged by the operations of the group. Such a coordinate or functionis said to transform according to a one-dimensional irrep of G. Intheexample given in (29.10), that the entry on the third row forms a 1 × 1block implies that the third entry in the basis vector (x y z ···) T ,namely z, is invariant under the two-dimensional symmetry operations onan equilateral triangle in the xy-plane.(ii) If, in any of the g matrices of the representation, the largest-sized blocklocated on the row or column corresponding to a particular coordinate(or function) in the basis vector is n × n, then that coordinate (or function)is mixed by the symmetry operations with n − 1 others and is said totransform according to an n-dimensional irrep of G. Thus in the matrix(29.10), x is the first entry in the complete basis vector; the first row ofthe matrix contains two non-zero entries, as does the first column, and sox is part of a two-component basis vector whose components are mixedby the symmetry operations of G. The other component is y.The result (29.11) may also be formulated in terms of the more abstract notionof vector spaces (chapter 8). The set of g matrices that forms an n-dimensionalrepresentation D of the group G can be thought of as acting on column matricescorresponding to vectors in an n-dimensional vector space V spanned by the basisfunctions of the representation. If there exists a proper subspace W of V ,suchthat if a vector whose column matrix is w belongs to W then the vector whosecolumn matrix is D(X)w also belongs to W , for all X belonging to G, thenitfollows that D is reducible. We say that the subspace W is invariant under theactions of the elements of G. With D unitary, the orthogonal complement W ⊥ ofW ,i.e.thevectorspaceV remaining when the subspace W has been removed, isalso invariant, and all the matrices D(X) split into two blocks acting separatelyon W and W ⊥ .BothW and W ⊥ may contain further invariant subspaces, inwhich case the matrices will be split still further.As a concrete example of this approach, consider in plane polar coordinatesρ, φ the effect of rotations about the polar axis on the infinite-dimensional vectorspace V of all functions of φ that satisfy the Dirichlet conditions for expansionas a Fourier series (see section 12.1). We take as our basis functions the set{sin mφ, cos mφ} for integer values m =0, 1, 2,...; this is an infinite-dimensionalrepresentation (n = ∞) and, since a rotation about the polar axis can be throughany angle α (0 ≤ α

29.4 REDUCIBILITY OF A REPRESENTATIONNow, for some k, consider a vector w in the space W k spanned by {sin kφ, cos kφ},say w = a sin kφ + b cos kφ. Under a rotation by α about the polar axis, a sin kφbecomes a sin k(φ + α), which can be written as a cos kα sin kφ + a sin kα cos kφ, i.eas a linear combination of sin kφ and cos kφ; similarly cos kφ becomes anotherlinear combination of the same two functions. The newly generated vector w ′ ,whose column matrix w ′ is given by w ′ = D(α)w, therefore belongs to W k forany α and we can conclude that W k is an invariant irreducible two-dimensionalsubspace of V . It follows that D(α) is reducible and that, since the result holdsfor every k, in its reduced form D(α) has an infinite series of identical 2 × 2 blockson its leading diagonal; each block will have the form( )cos α − sin α.sin α cos αWe note that the particular case k = 0 is special, in that then sin kφ = 0 andcos kφ = 1, for all φ; consequently the first 2×2 block in D(α) is reducible furtherand becomes two single-entry blocks.A second illustration of the connection between the behaviour of vector spacesunder the actions of the elements of a group and the form of the matrix representationof the group is provided by the vector space spanned by the sphericalharmonics Y lm (θ, φ). This contains subspaces, corresponding to the differentvalues of l, that are invariant under the actions of the elements of the full threedimensionalrotation group; the corresponding matrices are block-diagonal, andthose entries that correspond to the part of the basis containing Y lm (θ, φ) forma(2l +1)× (2l + 1) block.To illustrate further the irreps of a group, we return again to the group G oftwo-dimensional rotation and reflection symmetries of an equilateral triangle, orequivalently the permutation group S 3 ; this may be shown, using the methods ofsection 29.7 below, to have three irreps. Firstly, we have already seen that the setM of six orthogonal 2 × 2 matrices given in section (28.3), equation (28.13), isisomorphic to G. These matrices therefore form not only a representation of G,but a faithful one. It should be noticed that, although G contains six elements,the matrices are only 2 × 2. However, they contain no invariant 1 × 1 sub-block(which for 2 × 2 matrices would require them all to be diagonal) and neither canall the matrices be made block-diagonal by the same similarity transformation;they therefore form a two-dimensional irrep of G.Secondly, as previously noted, every group has one (unfaithful) irrep in whichevery element is represented by the 1 × 1 matrix I 1 , or, more simply, 1.Thirdly an (unfaithful) irrep of G is given by assignment of the one-dimensionalset of six ‘matrices’ {1, 1, 1, −1, −1, −1} to the symmetry operations {I,R,R ′ ,K,L, M} respectively, or to the group elements {I,A,B,C,D,E} respectively; seesection 28.3. In terms of the permutation group S 3 , 1 corresponds to evenpermutations and −1 to odd permutations, ‘odd’ or ‘even’ referring to the number1089

REPRESENTATION THEORYof simple pair interchanges to which a permutation is equivalent. That theseassignments are in accord with the group multiplication table 28.8 should bechecked.Thus the three irreps of the group G (i.e. the group 3m or C 3v or S 3 ), are, usingthe conventional notation A 1 ,A 2 , E (see section 29.8), as follows:where( )1 0M I =, M A =0 1(−1)0M C =0 1, M D =ElementI A B C D EA 1 1 1 1 1 1 1(29.12)E M I M A M B M C M D M EIrrep A 2 1 1 1 −1 −1 −1(−12√32− √ 32− 1 2(12− √ 32− √ 32− 1 2)(−12− √ )32, M B = √3,2− 1 2) ( √ )1 32 2, M E = .√32− 1 229.5 The orthogonality theorem for irreducible representationsWe come now to the central theorem of representation theory, a theorem thatjustifies the relatively routine application of certain procedures to determinethe restrictions that are inherent in physical systems that have some degree ofrotational or reflection symmetry. The development of the theorem is long andquite complex when presented in its entirety, and the reader will have to referelsewhere for the proof. §The theorem states that, in a certain sense, the irreps of a group G are asorthogonal as possible, as follows. If, for each irrep, the elements in any onepositionineachoftheg matricesareusedtomakeupg-component columnmatrices then(i) any two such column matrices coming from different irreps are orthogonal;(ii) any two such column matrices coming from different positions in thematrices of the same irrep are orthogonal.This orthogonality is in addition to the irreps’ being in the form of orthogonal(unitary) matrices and thus each comprising mutually orthogonal rows andcolumns.§ See, e.g., H. F. Jones, Groups, Representations and Physics (Bristol: Institute of Physics, 1998); J.F. Cornwell, Group Theory in Physics, vol 2 (London: Academic Press, 1984); J-P. Serre, LinearRepresentations of Finite Groups (New York: Springer, 1977).1090

29.5 THE ORTHOGONALITY THEOREM FOR IRREDUCIBLE REPRESENTATIONSMore mathematically, if we denote the entry in the ith row and jth column of amatrix D(X) by[D(X)] ij ,and ˆD (λ) and ˆD (µ) are two irreps of G having dimensionsn λ and n µ respectively, then∑ [ ]ˆD (λ) ∗ [(X) ˆD (X)](µ) = g δ ik δ jl δ λµ . (29.13)ijkl nXλThis rather forbidding-looking equation needs some further explanation.Firstly, the asterisk indicates that the complex conjugate should be taken ifnecessary, though all our representations so far have involved only real matrixelements. Each Kronecker delta function on the right-hand side has the value 1if its two subscripts are equal and has the value 0 otherwise. Thus the right-handside is only non-zero if i = k, j = l and λ = µ, all at the same time.Secondly, the summation over the group elements X means that g contributionshave to be added together, each contribution being a product of entries drawnfrom the representative matrices in the two irreps ˆD (λ) = { ˆD (λ) (X)} and ˆD (µ) ={ ˆD (µ) (X)}. Theg contributions arise as X runs over the g elements of G.Thus, putting these remarks together, the summation will produce zero if either(i) the matrix elements are not taken from exactly the same position in everymatrix, including cases in which it is not possible to do so because theirreps ˆD (λ) and ˆD (µ) have different dimensions, or(ii) even if ˆD (λ) and ˆD (µ) do have the same dimensions and the matrix elementsare from the same positions in every matrix, they are different irreps, i.e.λ ≠ µ.Some numerical illustrations based on the irreps A 1 ,A 2 and E of the group 3m(or C 3v or S 3 ) will probably provide the clearest explanation (see (29.12)).(a) Take i = j = k = l = 1, with ˆD (λ) =A 1 and ˆD (µ) =A 2 . Equation (29.13)then reads1(1) + 1(1) + 1(1) + 1(−1) + 1(−1) + 1(−1) = 0,as expected, since λ ≠ µ.(b) Take (i, j) as(1, 2) and (k, l) as(2, 2), corresponding to different matrixpositions within the same irrep ˆD (λ) = ˆD (µ) = E. Substituting in (29.13)gives(0(1) +− √ 32) (−12)+( √32) (− ) (12 +0(1)+− √ 32) (−12)+( √32) (−12)=0.(c) Take (i, j) as(1, 2), and (k, l) as(1, 2), corresponding to the same matrixpositions within the same irrep ˆD (λ) = ˆD (µ) = E. Substituting in (29.13)gives(0(0)+− √ 32)(− √ 32)+( √32)( √32) (+0(0)+− √ 32)(− √ 32)+( √32)( √32)= 6 2 .1091

REPRESENTATION THEORY(d) No explicit calculation is needed to see that if i = j = k = l = 1, withˆD (λ) = ˆD (µ) =A 1 (or A 2 ), then each term in the sum is either 1 2 or (−1) 2and the total is 6, as predicted by the right-hand side of (29.13) since g =6and n λ =1.29.6 CharactersThe actual matrices of general representations and irreps are cumbersome towork with, and they are not unique since there is always the freedom to changethe coordinate system, i.e. the components of the basis vector (see section 29.3),and hence the entries in the matrices. However, one thing that does not changefor a matrix under such an equivalence (similarity) transformation – i.e. undera change of basis – is the trace of the matrix. This was shown in chapter 8,but is repeated here. The trace of a matrix A is the sum of its diagonal elements,n∑Tr A =or, using the summation convention (section 26.1), simply A ii . Under a similaritytransformation, again using the summation convention,i=1A ii[D Q (X)] ii =[Q −1 ] ij [D(X)] jk [Q] ki=[D(X)] jk [Q] ki [Q −1 ] ij=[D(X)] jk [I] kj=[D(X)] jj ,showing that the traces of equivalent matrices are equal.This fact can be used to greatly simplify work with representations, though withsome partial loss of the information content of the full matrices. For example,using trace values alone it is not possible to distinguish between the two groupsknown as 4mm and ¯42m, orasC 4v and D 2d respectively, even though the twogroups are not isomorphic. To make use of these simplifications we now definethe characters of a representation.Definition. The characters χ(D) of a representation D of a group G are defined asthe traces of the matrices D(X), one for each element X of G.At this stage there will be g characters, but, as we noted in subsection 28.7.3,elements A, B of G in the same conjugacy class are connected by equations ofthe form B = X −1 AX. It follows that their matrix representations are connectedby corresponding equations of the form D(B) =D(X −1 )D(A)D(X),andsobytheargument just given their representations will have equal traces and hence equalcharacters. Thus elements in the same conjugacy class have the same characters,1092

29.6 CHARACTERS3m I A, B C, D, EA 1 1 1 1 z; z 2 ; x 2 + y 2A 2 1 1 −1 R zE 2 −1 0 (x, y); (xz, yz); (R x ,R y ); (x 2 − y 2 , 2xy)Table 29.1 The character table for the irreps of group 3m (C 3v or S 3 ). Theright-hand column lists some common functions that transform according tothe irrep against which each is shown (see text).though, in general, these will vary from one representation to another. However,it might also happen that two or more conjugacy classes have the same charactersin a representation – indeed, in the trivial irrep A 1 , see (29.12), every elementinevitably has the character 1.For the irrep A 2 of the group 3m, the classes {I}, {A, B} and {C,D,E} havecharacters 1, 1 and −1, respectively, whilst they have characters 2, −1 and0respectively in irrep E.We are thus able to draw up a character table for the group 3m as shownin table 29.1. This table holds in compact form most of the important informationon the behaviour of functions under the two-dimensional rotational andreflection symmetries of an equilateral triangle, i.e. under the elements of group3m. The entry under I for any irrep gives the dimension of the irrep, since itis equal to the trace of the unit matrix whose dimension is equal to that ofthe irrep. In other words, for the λth irrep χ (λ) (I) =n λ ,wheren λ is its dimension.In the extreme right-hand column we list some common functions of Cartesiancoordinates that transform, under the group 3m, according to the irrep on whoseline they are listed. Thus, as we have seen, z, z 2 ,andx 2 + y 2 are all unchangedby the group operations (though x and y individually are affected) and so arelisted against the one-dimensional irrep A 1 . Each of the pairs (x, y), (xz, yz), and(x 2 − y 2 , 2xy), however, is mixed as a pair by some of the operations, and so thesepairs are listed against the two-dimensional irrep E: each pair forms a basis setfor this irrep.The quantities R x , R y and R z refer to rotations about the indicated axes;they transform in the same way as the corresponding components of angularmomentum J, and their behaviour can be established by examining how thecomponents of J = r × p transform under the operations of the group. To dothis explicitly is beyond the scope of this book. However, it can be noted thatR z , being listed opposite the one-dimensional A 2 , is unchanged by I and by therotations A and B but changes sign under the mirror reflections C, D, andE, aswould be expected.1093

REPRESENTATION THEORY29.6.1 Orthogonality property of charactersSome of the most important properties of characters can be deduced from theorthogonality theorem (29.13),∑ [ ]ˆD (λ) ∗ [(X) ˆD (X)](µ) = g δ ik δ jl δ λµ .ijkl n λXIf we set j = i and l = k, so that both factors in any particular term in thesummation refer to diagonal elements of the representative matrices, and thensum both sides over i and k, we obtain∑ ∑n λ n∑ µ [ ]ˆD (λ) ∗ [(X) ˆD (µ)ii(X)]kk = g ∑n λn λXi=1 k=1n µ∑δ ik δ ik δ λµ .i=1 k=1Expressed in term of characters, this reads∑ [χ (λ) (X) ] ∗χ (µ) (X) = g ∑n λδniiδ 2 λµ = g ∑n λ1 × δ λµ = gδ λµ .λ n λXi=1i=1(29.14)In words, the (g-component) ‘vectors’ formed from the characters of the variousirreps of a group are mutually orthogonal, but each one has a squared magnitude(the sum of the squares of its components) equal to the order of the group.Since, as noted in the previous subsection, group elements in the same classhave the same characters, (29.14) can be written as a sum over classes rather thanelements. If c i denotes the number of elements in class C i and X i any element ofC i ,then∑ [c i χ (λ) (X i ) ] ∗χ (µ) (X i )=gδ λµ . (29.15)iAlthough we do not prove it here, there also exists a ‘completeness’ relation forcharacters. It makes a statement about the products of characters for a fixed pairof group elements, X 1 and X 2 , when the products are summed over all possibleirreps of the group. This is the converse of the summation process defined by(29.14). The completeness relation states that∑ [χ (λ) (X 1 ) ] ∗χ (λ) (X 2 )= g δ C1Cc2, (29.16)1λwhere element X 1 belongs to conjugacy class C 1 and X 2 belongs to C 2 . Thus thesum is zero unless X 1 and X 2 belong to the same class. For table 29.1 we canverify that these results are valid.(i) For ˆD (λ) = ˆD (µ) =A 1 or A 2 , (29.15) reads1(1) + 2(1) + 3(1) = 6,1094

29.7 COUNTING IRREPS USING CHARACTERSwhilst for ˆD (λ) = ˆD (µ) = E, it gives1(2 2 ) + 2(1) + 3(0) = 6.(ii) For ˆD (λ) =A 2 and ˆD (µ) = E, say, (29.15) reads1(1)(2) + 2(1)(−1) + 3(−1)(0) = 0.(iii) For X 1 = A and X 2 = D, say, (29.16) reads1(1) + 1(−1) + (−1)(0) = 0,whilst for X 1 = C and X 2 = E, both of which belong to class C 3 for whichc 3 =3,1(1) + (−1)(−1) + (0)(0) = 2 = 6 3 .29.7 Counting irreps using charactersThe expression of a general representation D = {D(X)} in terms of irreps, asgiven in (29.11), can be simplified by going from the full matrix form to that ofcharacters. ThusD(X) =m 1 ˆD (1) (X) ⊕ m 2 ˆD (2) (X) ⊕ ···⊕ m N ˆD (N) (X)becomes, on taking the trace of both sides,χ(X) =N∑m λ χ (λ) (X). (29.17)λ=1Given the characters of the irreps of the group G to which the elements X belong,and the characters of the representation D = {D(X)}, theg equations (29.17)can be solved as simultaneous equations in the m λ , either by inspection or bymultiplying both sides by [ χ (µ) (X) ] ∗and summing over X, making use of (29.14)and (29.15), to obtainm µ = 1 ∑ [χ (µ) (X) ] ∗ 1 ∑ [χ(X) = c i χ (µ) (X i ) ] ∗χ(Xi ). (29.18)ggXiThat an unambiguous formula can be given for each m λ , once the characterset (the set of characters of each of the group elements or, equivalently, ofeach of the conjugacy classes) of D is known, shows that, for any particulargroup, two representations with the same characters are equivalent. This stronglysuggests something that can be shown, namely, the number of irreps = the numberof conjugacy classes. The argument is as follows. Equation (29.17) is a set ofsimultaneous equations for N unknowns, the m λ , some of which may be zero. Thevalue of N is equal to the number of irreps of G. There are g different values ofX, but the number of different equations is only equal to the number of distinct1095

REPRESENTATION THEORYconjugacy classes, since any two elements of G in the same class have the samecharacter set and therefore generate the same equation. For a unique solutionto simultaneous equations in N unknowns, exactly N independent equations areneeded. Thus N is also the number of classes, establishing the stated result.◮Determine the irreps contained in the representation of the group 3m in the vector spacespanned by the functions x 2 , y 2 , xy.We first note that although these functions are not orthogonal they form a basis set for arepresentation, since they are linearly independent quadratic forms in x and y and any otherquadratic form can be written (uniquely) in terms of them. We must establish how theytransform under the symmetry operations of group 3m. Weneedtodosoonlyforarepresentativeelement of each conjugacy class, and naturally we take the simplest in each case.ThefirstclasscontainsonlyI (as always) and clearly D(I) isthe3× 3 unit matrix.The second class contains the rotations, A and B, and we choose to find D(A). Since,under A,x → − 1 √ √332 x + 2 y and y → − 2 x − 1 2 y,it follows thatx 2 → 1 4 x2 − √ 3xy + 32 4 y2 , y 2 → 3 4 x2 + √ 3xy + 12 4 y2 (29.19)andxy → √ 34 x2 − 1 xy − √ 32 4 y2 . (29.20)Hence D(A) can be deduced and is given below.The third and final class contains the reflections, C, D and E; oftheseC is much theeasiest to deal with. Under C, x →−x and y → y, causingxy to change sign but leavingx 2 and y 2 unaltered. The three matrices needed are thus⎛⎛D(I) =I 3 , D(C) = ⎝ 1 0 0⎞0 1 0 ⎠ , D(A) = ⎜⎝0 0 −114343− √ 34 2√1 34 2√34− √ 34− 1 2their traces are respectively 3, 1 and 0.It should be noticed that much more work has been done here than is necessary, sincethe traces can be computed immediately from the effects of the symmetry operations on thebasis functions. All that is needed is the weight of each basis function in the transformedexpression for that function; these are clearly 1, 1, 1 for I, and 1 , 1 , − 1 for A, from (29.19)4 4 2and (29.20), and 1, 1, −1 forC, from the observations made just above the displayedmatrices. The traces are then the sums of these weights. The off-diagonal elements of thematrices need not be found, nor need the matrices be written out.From (29.17) we now need to find a superposition of the characters of the irreps thatgives representation D in the bottom line of table 29.2.By inspection it is obvious that D =A 1 ⊕ E, but we can use (29.18) formally:m A1 = 1 [1(1)(3) + 2(1)(0) + 3(1)(1)] = 1,6m A2 = 1 [1(1)(3) + 2(1)(0) + 3(−1)(1)] = 0,6m E = 1 [1(2)(3) + 2(−1)(0) + 3(0)(1)] = 1.6Thus A 1 and E appear once each in the reduction of D, andA 2 not at all. Table 29.1gives the further information, not needed here, that it is the combination x 2 + y 2 thattransforms as a one-dimensional irrep and the pair (x 2 − y 2 , 2xy) thatformsabasisofthe two-dimensional irrep, E. ◭1096⎞⎟⎠ ;

29.7 COUNTING IRREPS USING CHARACTERSClassesIrrep I AB CDEA 1 1 1 1A 2 1 1 −1E 2 −1 0D 3 0 1Table 29.2 The characters of the irreps of the group 3m and of the representationD, which must be a superposition of some of them.29.7.1 Summation rules for irrepsThe first summation rule for irreps is a simple restatement of (29.14), with µ setequal to λ; itthenreads∑ [χ (λ) (X) ] ∗χ (λ) (X) =g.XIn words, the sum of the squares (modulus squared if necessary) of the charactersof an irrep taken over all elements of the group adds up to the order of thegroup. For group 3m (table 29.1), this takes the following explicit forms:for A 1 , 1(1 2 )+2(1 2 )+3(1 2 )=6;for A 2 , 1(1 2 )+2(1 2 )+3(−1) 2 =6;for E, 1(2 2 )+2(−1) 2 +3(0 2 )=6.We next prove a theorem that is concerned not with a summation within an irrepbut with a summation over irreps.Theorem. If n µ is the dimension of the µth irrep of a group G then∑n 2 µ = g,where g is the order of the group.µProof. Define a representation of the group in the following way. Rearrangethe rows of the multiplication table of the group so that whilst the elements ina particular order head the columns, their inverses in the same order head therows. In this arrangement of the g × g table, the leading diagonal is entirelyoccupied by the identity element. Then, for each element X of the group, take asrepresentative matrix the multiplication-table array obtained by replacing X by1 and all other element symbols by 0. The matrices D reg (X) so obtained form theregular representation of G; theyareeachg × g, have a single non-zero entry ‘1’in each row and column and (as will be verified by a little experimentation) have1097

REPRESENTATION THEORY(a)I A BI I A BA A B IB B I A(b)I A BI I A BB B I AA A B ITable 29.3 (a) The multiplication table of the cyclic group of order 3, and(b) its reordering used to generate the regular representation of the group.the same multiplication structure as the group G itself, i.e. they form a faithfulrepresentation of G.Although not part of the proof, a simple example may help to make theseideas more transparent. Consider the cyclic group of order 3. Its multiplicationtable is shown in table 29.3(a) (a repeat of table 28.10(a) of the previous chapter),whilst table 29.3(b) shows the same table reordered so that the columns arestill labelled in the order I, A, B but the rows are now labelled in the orderI −1 = I, A −1 = B, B −1 = A. The three matrices of the regular representation arethen⎛D reg (I) = ⎝1 0⎞00 1 00 0 1⎛⎠ , D reg (A) = ⎝0 1⎞00 0 11 0 0⎛⎠ , D reg (B) = ⎝0 0 1⎞1 0 0 ⎠ .0 1 0An alternative, more mathematical, definition of the regular representation of agroup is[D reg (G k ) ] {1 ifij = Gk G j = G i ,0 otherwise.We now return to the proof. With the construction given, the regular representationhascharactersasfollows:χ reg (I) =g, χ reg (X) =0 ifX ≠ I.We now apply (29.18) to D reg to obtain for the number m µ of times that the irrepˆD (µ) appears in D reg (see 29.11))m µ = 1 ∑ [χ (µ) (X) ] ∗χ reg (X) = 1 [χ (µ) (I) ] ∗χ reg (I) = 1 ggg n µg = n µ .XThus an irrep ˆD (µ) of dimension n µ appears n µ times in D reg , and so by countingthe total number of basis functions, or by considering χ reg (I), we can conclude1098

29.7 COUNTING IRREPS USING CHARACTERSthat∑n 2 µ = g. (29.21)µThis completes the proof.As before, our standard demonstration group 3m provides an illustration. Inthis case we have seen already that there are two one-dimensional irreps and onetwo-dimensional irrep. This is in accord with (29.21) since1 2 +1 2 +2 2 =6, which is the order g of the group.Another straightforward application of the relation (29.21), to the group withmultiplication table 29.3(a), yields immediate results. Since g =3,noneofitsirreps can have dimension 2 or more, as 2 2 = 4 is too large for (29.21) to besatisfied. Thus all irreps must be one-dimensional and there must be three ofthem (consistent with the fact that each element is in a class of its own, and thatthere are therefore three classes). The three irreps are the sets of 1 × 1 matrices(numbers)A 1 = {1, 1, 1} A 2 = {1,ω,ω 2 } A ∗ 2 = {1,ω 2 ,ω},where ω =exp(2πi/3); since the matrices are 1 × 1, the same set of nine numberswould be, of course, the entries in the character table for the irreps of the group.The fact that the numbers in each irrep are all cube roots of unity is discussedbelow. As will be noticed, two of these irreps are complex – an unusual occurrencein most applications – and form a complex conjugate pair of one-dimensionalirreps. In practice, they function much as a two-dimensional irrep, but this is tobe ignored for formal purposes such as theorems.A further property of characters can be derived from the fact that all elementsin a conjugacy class have the same order. Suppose that the element X has orderm, i.e.X m = I. This implies for a representation D of dimension n that[D(X)] m = I n . (29.22)Representations equivalent to D are generated as before by using similaritytransformations of the formD Q (X) =Q −1 D(X)Q.In particular, if we choose the columns of Q to be the eigenvectors of D(X) then,as discussed in chapter 8,⎛⎞λ 1 0 ··· 0D Q (X) =0 λ. 2 ⎜ .⎝. ⎟. .. 0 ⎠0 ··· 0 λ n1099

REPRESENTATION THEORYwhere the λ i are the eigenvalues of D(X). Therefore, from (29.22), we have that⎛λ m ⎞ ⎛⎞1 0 ··· 0 1 0 ··· 0..0 λ m 2 .⎜ .⎝. ⎟. .. 0 ⎠ = 0 1 .⎜ .⎝. ⎟. .. 0 ⎠ .0 ··· 0 λ m n 0 ··· 0 1Hence all the eigenvalues λ i are mth roots of unity, and so χ(X), the trace ofD(X), is the sum of n of these. In view of the implications of Lagrange’s theorem(section 28.6 and subsection 28.7.2), the only values of m allowed are the divisorsof the order g of the group.29.8 Construction of a character tableIn order to decompose representations into irreps on a routine basis usingcharacters, it is necessary to have available a character table for the group inquestion. Such a table gives, for each irrep µ of the group, the character χ (µ) (X)of the class to which group element X belongs. To construct such a table thefollowing properties of a group, established earlier in this chapter, may be used:(i) the number of classes equals the number of irreps;(ii) the ‘vector’ formed by the characters from a given irrep is orthogonal tothe ‘vector’ formed by the characters from a different irrep;(iii) ∑ µ n2 µ = g, wheren µ is the dimension of the µth irrep and g is the orderof the group;(iv) the identity irrep (one-dimensional with all characters equal to 1) is presentfor every group;(v) ∑ ∣X∣χ (µ) (X) ∣ 2 = g.(vi) χ (µ) (X) isthesumofn µ mth roots of unity, where m is the order of X.◮Construct the character table for the group 4mm (or C 4v ) using the properties of classes,irreps and characters so far established.The group 4mm is the group of two-dimensional symmetries of a square, namely rotationsof 0, π/2, π and 3π/2 and reflections in the mirror planes parallel to the coordinate axesand along the main diagonals. These are illustrated in figure 29.3. For this group there areeight elements:• the identity, I;• rotations by π/2 and3π/2, R and R ′ ;• arotationbyπ, Q ;• four mirror reflections m x , m y , m d and m d ′.Requirements (i) to (iv) at the start of this section put tight constraints on the possiblecharacter sets, as the following argument shows.The group is non-Abelian (clearly Rm x ≠ m x R), and so there are fewer than eightclasses, and hence fewer than eight irreps. But requirement (iii), with g = 8, then implies1100

29.8 CONSTRUCTION OF A CHARACTER TABLEm d ′m xm dm yFigure 29.3 The mirror planes associated with 4mm, the group of twodimensionalsymmetries of a square.that at least one irrep has dimension 2 or greater. However, there can be no irrep withdimension 3 or greater, since 3 2 > 8, nor can there be more than one two-dimensionalirrep, since 2 2 +2 2 = 8 would rule out a contribution to the sum in (iii) of 1 2 from theidentity irrep, and this must be present. Thus the only possibility is one two-dimensionalirrep and, to make the sum in (iii) correct, four one-dimensional irreps.Therefore using (i) we can now deduce that there are five classes. This same conclusioncan be reached by evaluating X −1 YX for every pair of elements in G, as in the descriptionof conjugacy classes given in the previous chapter. However, it is tedious to do so andcertainly much longer than the above. The five classes are I, Q, {R, R ′ }, {m x , m y }, {m d , m d ′}.It is straightforward to show that only I and Q commute with every element of thegroup, so they are the only elements in classes of their own. Each other class must haveat least 2 members, but, as there are three classes to accommodate 8 − 2 = 6 elements,there must be exactly 2 in each class. This does not pair up the remaining 6 elements, butdoes say that the five classes have 1, 1, 2, 2, and 2 elements. Of course, if we had startedby dividing the group into classes, we would know the number of elements in each classdirectly.We cannot entirely ignore the group structure (though it sometimes happens that theresults are independent of the group structure – for example, all non-Abelian groups oforder 8 have the same character table!); thus we need to note in the present case thatm 2 i = I for i = x, y, d or d ′ and, as can be proved directly, Rm i = m i R ′ for the same fourvalues of label i. We also recall that for any pair of elements X and Y , D(XY )=D(X)D(Y ).We may conclude the following for the one-dimensional irreps.(a) In view of result (vi), χ(m i )=D(m i )=±1.(b) Since R 4 = I, result (vi) requires that χ(R) is one of 1, i, −1, −i. But, sinceD(R)D(m i )=D(m i )D(R ′ ), and the D(m i ) are just numbers, D(R) =D(R ′ ). FurtherD(R)D(R) =D(R)D(R ′ )=D(RR ′ )=D(I) =1,and so D(R) =±1 =D(R ′ ).(c) D(Q) =D(RR) =D(R)D(R) =1.If we add this to the fact that the characters of the identity irrep A 1 are all unity then wecan fill in those entries in character table 29.4 shown in bold.Suppose now that the three missing entries in a one-dimensional irrep are p, q and r,where each can only be ±1. Then, allowing for the numbers in each class, orthogonality1101

REPRESENTATION THEORY4mm I Q R, R ′ m x , m y m d , m d ′A 1 1 1 1 1 1A 2 1 1 1 −1 −1B 1 1 1 −1 1 −1B 2 1 1 −1 −1 1E 2 −2 0 0 0Table 29.4 The character table deduced for the group 4mm. For an explanationof the entries in bold see the text.with the characters of A 1 requires that1(1)(1) + 1(1)(1) + 2(1)(p) + 2(1)(q) + 2(1)(r) =0.The only possibility is that two of p, q, andr equal −1 and the other equals +1. Thiscan be achieved in three different ways, corresponding to the need to find three furtherdifferent one-dimensional irreps. Thus the first four lines of entries in character table 29.4can be completed. The final line can be completed by requiring it to be orthogonal to theother four. Property (v) has not been used here though it could have replaced part of theargument given. ◭29.9 Group nomenclatureThe nomenclature of published character tables, as we have said before, is erraticand sometimes unfortunate; for example, often E is used to represent, not onlya two-dimensional irrep, but also the identity operation, where we have used I.Thus the symbol E might appear in both the column and row headings of atable, though with quite different meanings in the two cases. In this book we useroman capitals to denote irreps.One-dimensional irreps are regularly denoted by A and B, B being used if arotation about the principal axis of 2π/n has character −1. Here n is the highestinteger such that a rotation of 2π/n is a symmetry operation of the system, andthe principal axis is the one about which this occurs. For the group of operationson a square, n = 4, the axis is the perpendicular to the square and the rotationin question is R. The names for the group, 4mm and C 4v , derive from the factthat here n is equal to 4. Similarly, for the operations on an equilateral triangle,n = 3 and the group names are 3m and C 3v , but because the rotation by 2π/3 hascharacter +1 in all its one-dimensional irreps (see table 29.1), only A appears inthe irrep list.Two-dimensional irreps are denoted by E, as we have already noted, and threedimensionalirreps by T, although in many cases the symbols are modified byprimes and other alphabetic labels to denote variations in behaviour from oneirrep to another in respect of mirror reflections and parity inversions. In the studyof molecules, alternative names based on molecular angular momentum propertiesare common. It is beyond the scope of this book to list all these variations, or to1102

29.10 PRODUCT REPRESENTATIONSgive a large selection of character tables; our aim is to demonstrate and justifythe use of those found in the literature specifically dedicated to crystal physics ormolecular chemistry.Variations in notation are not restricted to the naming of groups and theirirreps, but extend to the symbols used to identify a typical element, and henceall members, of a conjugacy class in a group. In physics these are usually of thetypes n z , ¯n z or m x . The first of these denotes a rotation of 2π/n about the z-axis,and the second the same thing followed by parity inversion (all vectors r go to−r), whilst the third indicates a mirror reflection in a plane, in this case the planex =0.Typical chemistry symbols for classes are NC n , NC 2 n, NC x n , NS n , σ v , σ xy .Herethe first symbol N, where it appears, shows that there are N elements in theclass (a useful feature). The subscript n has the same meaning as in the physicsnotation, but σ rather than m is used for a mirror reflection, subscripts v, d or h orsuperscripts xy, xz or yz denoting the various orientations of the relevant mirrorplanes. Symmetries involving parity inversions are denoted by S; thus S n is thechemistry analogue of ¯n. None of what is said in this and the previous paragraphshould be taken as definitive, but merely as a warning of common variations innomenclature and as an initial guide to corresponding entities. Before using anyset of group character tables, the reader should ensure that he or she understandsthe precise notation being employed.29.10 Product representationsIn quantum mechanical investigations we are often faced with the calculation ofwhat are called matrix elements. These normally take the form of integrals over allspace of the product of two or more functions whose analytic forms depend on themicroscopic properties (usually angular momentum and its components) of theelectrons or nuclei involved. For ‘bonding’ calculations involving ‘overlap integrals’there are usually two functions involved, whilst for transition probabilities a thirdfunction, giving the spatial variation of the interaction Hamiltonian, also appearsunder the integral sign.If the environment of the microscopic system under investigation has somesymmetry properties, then sometimes these can be used to establish, withoutdetailed evaluation, that the multiple integral must have zero value. We nowexpress the essential content of these ideas in group theoretical language.Suppose we are given an integral of the form∫∫J = Ψφdτ or J = Ψξφdτto be evaluated over all space in a situation in which the physical system isinvariant under a particular group G of symmetry operations. For the integral to1103

REPRESENTATION THEORYbe non-zero the integrand must be invariant under each of these operations. Ingroup theoretical language, the integrand must transform as the identity, the onedimensionalrepresentation A 1 of G; more accurately, some non-vanishing part ofthe integrand must do so.An alternative way of saying this is that if under the symmetry operationsof G the integrand transforms according to a representation D and D does notcontain A 1 amongst its irreps then the integral J is necessarily zero. It should benoted that the converse is not true; J maybezeroevenifA 1 is present, since theintegral, whilst showing the required invariance, may still have the value zero.It is evident that we need to establish how to find the irreps that go to makeup a representation of a double or triple product when we already know theirreps according to which the factors in the product transform. The method isestablished by the following theorem.Theorem. For each element of a group the character in a product representation isthe product of the corresponding characters in the separate representations.Proof. Suppose that {u i } and {v j } are two sets of basis functions, that transformunder the operations of a group G according to representations D (λ) and D (µ)respectively. Denote by u and v the corresponding basis vectors and let X be anelement of the group. Then the functions generated from u i and v j by the actionof X are calculated as follows, using (29.1) and (29.4):[ (Du ′ i = Xu i =(λ) (X) ) ]Tu = [ D (λ) (X) ]iii u i + ∑ [ (D (λ) (X) ) ] T u l,ill≠i[ (Dv j ′ = Xv j =(µ) (X) ) ]Tv = [ D (µ) (X) ]jjj v j + ∑ [ (D (µ) (X) ) ] T v m.jmm≠jHere [D(X)] ij is just a single element of the matrix D(X) and[D(X)] kk =[D T (X)] kkis simply a diagonal element from the matrix – the repeated subscript does notindicate summation. Now, if we take as basis functions for a product representationD prod (X) the products w k = u i v j (where the n λ n µ various possible pairs ofvalues i, j are labelled by k), we have also thatw k ′ = Xw k = Xu i v j =(Xu i )(Xv j )= [ D (λ) (X) ] [ii D (µ) (X) ] jj u iv j + terms not involving the product u i v j .This is to be compared with[ (Dw k ′ = Xw k =prod (X) ) ]Tw = [ D prod (X) ]kkk w k + ∑ [ (D prod (X) ) ] T w n,knn≠kwhere D prod (X) is the product representation matrix for element X of the group.The comparison shows that[D prod (X) ] kk = [ D (λ) (X) ] [ii D (µ) (X) ] jj .1104

29.11 PHYSICAL APPLICATIONS OF GROUP THEORYIt follows thatn λn µ∑χ prod [(X) = D prod (X) ] kkk=1n λ n µ∑ ∑ [= D (λ) (X) ] [ii D (µ) (X) ] jj=i=1 j=1{nλ∑ [D (λ) (X) ] iii=1}{ nµ∑ [D (µ) (X) ] }jjj=1= χ (λ) (X) χ (µ) (X). (29.23)This proves the theorem, and a similar argument leads to the corresponding resultfor integrands in the form of a product of three or more factors.An immediate corollary is that an integral whose integrand is the product oftwo functions transforming according to two different irreps is necessarily zero. Tosee this, we use (29.18) to determine whether irrep A 1 appears in the productcharacter set χ prod (X):m A1= 1 ∑ [χ(A 1) (X) ] ∗χ prod (X) = 1 ∑χ prod (X) = 1 ∑χ (λ) (X)χ (µ) (X).gg gXXXWe have used the fact that χ (A1) (X) =1forallX but now note that, by virtue of(29.14), the expression on the right of this equation is equal to zero unless λ = µ.Any complications due to non-real characters have been ignored – in practice,they are handled automatically as it is usually Ψ ∗ φ, rather than Ψφ, that appearsin integrands, though many functions are real in any case, and nearly all charactersare.Equation (29.23) is a general result for integrands but, specifically in the contextof chemical bonding, it implies that for the possibility of bonding to exist, thetwo quantum wavefunctions must transform according to the same irrep. This isdiscussed further in the next section.29.11 Physical applications of group theoryAs we indicated at the start of chapter 28 and discussed in a little more detail atthe beginning of the present chapter, some physical systems possess symmetriesthat allow the results of the present chapter to be used in their analysis. Weconsider now some of the more common sorts of problem in which these resultsfind ready application.1105

REPRESENTATION THEORY1y42x3Figure 29.4manganese.A molecule consisting of four atoms of iodine and one of29.11.1 Bonding in moleculesWe have just seen that whether chemical bonding can take place in a moleculeis strongly dependent upon whether the wavefunctions of the two atoms forminga bond transform according to the same irrep. Thus it is sometimes useful to beable to find a wavefunction that does transform according to a particular irrepof a group of transformations. This can be done if the characters of the irrep areknown and a sensible starting point can be guessed. We state without proof thatstarting from any n-dimensional basis vector Ψ ≡ (Ψ 1 Ψ 2 ··· Ψ n ) T ,where{Ψ i }is a set of wavefunctions, the new vector Ψ (λ) ≡ (Ψ (λ)1Ψ (λ)2··· Ψ n (λ) ) T generatedbyΨ (λ)i= ∑ Xχ (λ)∗ (X)XΨ i (29.24)will transform according to the λth irrep. If the randomly chosen Ψ happens notto contain any component that transforms in the desired way then the Ψ (λ) sogenerated is found to be a zero vector and it is necessary to select a new startingvector. An illustration of the use of this ‘projection operator’ is given in the nextexample.◮Consider a molecule made up of four iodine atoms lying at the corners of a square in thexy-plane, with a manganese atom at its centre, as shown in figure 29.4. Investigate whetherthe molecular orbital given by the superposition of p-state (angular momentum l = 1)atomic orbitalsΨ 1 =Ψ y (r − R 1 )+Ψ x (r − R 2 ) − Ψ y (r − R 3 ) − Ψ x (r − R 4 )can bond to the d-state atomic orbitals of the manganese atom described by either (i) φ 1 =(3z 2 −r 2 )f(r) or (ii) φ 2 =(x 2 −y 2 )f(r), wheref(r) is a function of r and so is unchanged byany of the symmetry operations of the molecule. Such linear combinations of atomic orbitalsare known as ring orbitals.We have eight basis functions, the atomic orbitals Ψ x (N) andΨ y (N), where N =1, 2, 3, 4and indicates the position of an iodine atom. Since the wavefunctions are those of p-statesthey have the forms xf(r) oryf(r) and lie in the directions of the x- andy-axes shown inthe figure. Since r is not changed by any of the symmetry operations, f(r) can be treated asa constant. The symmetry group of the system is 4mm, whose character table is table 29.4.1106

29.11 PHYSICAL APPLICATIONS OF GROUP THEORYCase (i). The manganese atomic orbital φ 1 =(3z 2 − r 2 )f(r), lying at the centre of themolecule, is not affected by any of the symmetry operations since z and r are unchangedby them. It clearly transforms according to the identity irrep A 1 . We therefore need toknow which combination of the iodine orbitals Ψ x (N) andΨ y (N), if any, also transformsaccording to A 1 .We use the projection operator (29.24). If we choose Ψ x (1) as the arbitrary onedimensionalstarting vector, we unfortunately obtain zero (as the reader may wish toverify), but Ψ y (1) is found to generate a new non-zero one-dimensional vector transformingaccording to A 1 . The results of acting on Ψ y (1) with the various symmetry elements Xcan be written down by inspection (see the discussion in section 29.2). So, for example, theΨ y (1) orbital centred on iodine atom 1 and aligned along the positive y-axisischangedby the anticlockwise rotation of π/2 produced by R ′ into an orbital centred on atom 4and aligned along the negative x-axis; thus R ′ Ψ y (1) = −Ψ x (4). The complete set of groupactions on Ψ y (1) is:I, Ψ y (1); Q, −Ψ y (3); R, Ψ x (2); R ′ , −Ψ x (4);m x , Ψ y (1); m y , −Ψ y (3); m d , Ψ x (2); m d ′, −Ψ x (4).Now χ (A1) (X) =1forallX, so (29.24) states that the sum of the above results for XΨ y (1),all with weight 1, gives a vector (here, since the irrep is one-dimensional, just a wavefunction)that transforms according to A 1 and is therefore capable of forming a chemicalbond with the manganese wavefunction φ 1 .ItisΨ (A1) =2[Ψ y (1) − Ψ y (3) + Ψ x (2) − Ψ x (4)],though, of course, the factor 2 is irrelevant. This is precisely the ring orbital Ψ 1 given inthe problem, but here it is generated rather than guessed beforehand.Case (ii). The atomic orbital φ 2 =(x 2 − y 2 )f(r) behaves as follows under the action oftypical conjugacy class members:I, φ 2 ; Q, φ 2 ; R, (y 2 − x 2 )f(r) =−φ 2 ; m x ,φ 2 ; m d , −φ 2 .From this we see that φ 2 transforms as a one-dimensional irrep, but, from table 29.4, thatirrep is B 1 not A 1 (the irrep according to which Ψ 1 transforms, as already shown). Thusφ 2 and Ψ 1 cannot form a bond. ◭The original question did not ask for the the ring orbital to which φ 2 maybond, but it can be generated easily by using the values of XΨ y (1) calculated incase (i) and now weighting them according to the characters of B 1 :Ψ (B1) =Ψ y (1) − Ψ y (3) + (−1)Ψ x (2) − (−1)Ψ x (4)+Ψ y (1) − Ψ y (3) + (−1)Ψ x (2) − (−1)Ψ x (4)=2[Ψ y (1) − Ψ x (2) − Ψ y (3) + Ψ x (4)].Now we will find the other irreps of 4mm present in the space spanned bythe basis functions Ψ x (N) andΨ y (N); at the same time this will illustrate theimportant point that since we are working with characters we are only interestedin the diagonal elements of the representative matrices. This means (section 29.2)that if we work in the natural representation D nat we need consider only thosefunctions that transform, wholly or partially, into themselves. Since we have noneed to write out the matrices explicitly, their size (8 × 8) is no drawback. All theirreps spanned by the basis functions Ψ x (N) andΨ y (N) can be determined byconsidering the actions of the group elements upon them, as follows.1107

REPRESENTATION THEORY(i) Under I all eight basis functions are unchanged, and χ(I) =8.(ii) The rotations R, R ′ and Q change the value of N in every case and soall diagonal elements of the natural representation are zero and χ(R) =χ(Q) =0.(iii) m x takes x into −x and y into y and, for N = 1 and 3, leaves N unchanged,with the consequences (remember the forms of Ψ x (N) andΨ y (N)) thatΨ x (1) →−Ψ x (1), Ψ x (3) →−Ψ x (3),Ψ y (1) → Ψ y (1), Ψ y (3) → Ψ y (3).Thus χ(m x ) has four non-zero contributions, −1, −1, 1 and 1, togetherwith four zero contributions. The total is thus zero.(iv) m d and m d ′ leave no atom unchanged and so χ(m d )=0.The character set of the natural representation is thus 8, 0, 0, 0, 0, which, eitherby inspection or by applying formula (29.18), shows thatD nat =A 1 ⊕ A 2 ⊕ B 1 ⊕ B 2 ⊕ 2E,i.e. that all possible irreps are present. We have constructed previously thecombinations of Ψ x (N) and Ψ y (N) that transform according to A 1 and B 1 .The others can be found in the same way.29.11.2 Matrix elements in quantum mechanicsIn section 29.10 we outlined the procedure for determining whether a matrixelement that involves the product of three factors as an integrand is necessarilyzero. We now illustrate this with a specific worked example.◮Determine whether a ‘dipole’ matrix element of the form∫J = Ψ d1 xΨ d2 dτ,where Ψ d1 and Ψ d2 are d-state wavefunctions of the forms xyf(r) and (x 2 − y 2 )g(r) respectively,can be non-zero (i) inamoleculewithsymmetryC 3v (or 3m), such as ammonia, and(ii) in a molecule with symmetry C 4v (or 4mm), such as the MnI 4 molecule considered inthe previous example.We will need to make reference to the character tables of the two groups. The table forC 3v is table 29.1 (section 29.6); that for C 4v is reproduced as table 29.5 from table 29.4 butwith the addition of another column showing how some common functions transform.We make use of (29.23), extended to the product of three functions. No attention needbe paid to f(r) andg(r) as they are unaffected by the group operations.Case (i). From the character table 29.1 for C 3v , we see that each of xy, x and x 2 − y 2forms part of a basis set transforming according to the two-dimensional irrep E. Thus wemay fill in the array of characters (using chemical notation for the classes, except thatwe continue to use I rather than E) as shown in table 29.6. The last line is obtained by1108

29.11 PHYSICAL APPLICATIONS OF GROUP THEORY4mm I Q R, R ′ m x , m y m d , m d ′A 1 1 1 1 1 1 z; z 2 ; x 2 + y 2A 2 1 1 1 −1 −1 R zB 1 1 1 −1 1 −1 x 2 − y 2B 2 1 1 −1 −1 1 xyE 2 −2 0 0 0 (x, y); (xz, yz); (R x ,R y )Table 29.5 The character table for the irreps of group 4mm (or C 4v ). Theright-hand column lists some common functions, or, for the two-dimensionalirrep E, pairs of functions, that transform according to the irrep against whichthey are shown.Function Irrep ClassesI 2C 3 3σ vxy E 2 −1 0x E 2 −1 0x 2 − y 2 E 2 −1 0product 8 −1 0Table 29.6 The character sets, for the group C 3v (or 3mm), of three functionsand of their product x 2 y(x 2 − y 2 ).Function Irrep ClassesI C 2 2C 6 2σ v 2σ dxy B 2 1 1 −1 −1 1x E 2 −2 0 0 0x 2 − y 2 B 1 1 1 −1 1 −1product 2 −2 0 0 0Table 29.7 The character sets, for the group C 4v (or 4mm), of three functions,and of their product x 2 y(x 2 − y 2 ).multiplying together the corresponding characters for each of the three elements. Now, byinspection, or by applying (29.18), i.e.m A1 = 1 [1(1)(8) + 2(1)(−1) + 3(1)(0)] = 1,6we see that irrep A 1 does appear in the reduced representation of the product, and so Jis not necessarily zero.Case (ii). From table 29.5 we find that, under the group C 4v , xy and x 2 − y 2 transformas irreps B 2 and B 1 respectively and that x is part of a basis set transforming as E. Thusthe calculation table takes the form of table 29.7 (again, chemical notation for the classeshas been used).Here inspection is sufficient, as the product is exactly that of irrep E and irrep A 1 iscertainly not present. Thus J is necessarily zero and the dipole matrix element vanishes. ◭1109

REPRESENTATION THEORYy 3x 3y 1y2x 1x 2Figure 29.5An equilateral array of masses and springs.29.11.3 Degeneracy of normal modesAs our final area for illustrating the usefulness of group theoretical results weconsider the normal modes of a vibrating system (see chapter 9). This analysishas far-reaching applications in physics, chemistry and engineering. For a givensystem, normal modes that are related by some symmetry operation have the samefrequency of vibration; the modes are said to be degenerate. It can be shown thatsuch modes span a vector space that transforms according to some irrep of thegroup G of symmetry operations of the system. Moreover, the degeneracy ofthe modes equals the dimension of the irrep. As an illustration, we consider thefollowing example.◮Investigate the possible vibrational modes of the equilateral triangular arrangement ofequal masses and springs shown in figure 29.5. Demonstrate that two are degenerate.Clearly the symmetry group is that of the symmetry operations on an equilateral triangle,namely 3m (or C 3v ), whose character table is table 29.1. As on a previous occasion, it ismost convenient to use the natural representation D nat of this group (it almost alwayssaves having to write out matrices explicitly) acting on the six-dimensional vector space(x 1 , y 1 , x 2 , y 2 , x 3 , y 3 ). In this example the natural and regular representations coincide, butthis is not usually the case.We note that in table 29.1 the second class contains the rotations A (by π/3) and B (by2π/3), also known as R and R ′ . This class is known as 3 z in crystallographic notation, orC 3 in chemical notation, as explained in section 29.9. The third class contains C, D, E, thethree mirror reflections.Clearly χ(I) = 6. Since all position labels are changed by a rotation, χ(3 z )=0.Forthemirror reflections the simplest representative class member to choose is the reflection m y inthe plane containing the y 3 -axis, since then only label 3 is unchanged; under m y , x 3 →−x 3and y 3 → y 3 , leading to the conclusion that χ(m y ) = 0. Thus the character set is 6, 0, 0.Using (29.18) and the character table 29.1 shows thatD nat =A 1 ⊕ A 2 ⊕ 2E.1110

29.11 PHYSICAL APPLICATIONS OF GROUP THEORYHowever, we have so far allowed x i , y i to be completely general, and we must now identifyand remove those irreps that do not correspond to vibrations. These will be the irrepscorresponding to bodily translations of the triangle and to its rotation without relativemotion of the three masses.Bodily translations are linear motions of the centre of mass, which has coordinatesx =(x 1 + x 2 + x 3 )/3 and y =(y 1 + y 2 + y 3 )/3).Table 29.1 shows that such a coordinate pair (x, y) transforms according to the twodimensionalirrep E; this accounts for one of the two such irreps found in the naturalrepresentation.It can be shown that, as stated in table 29.1, planar bodily rotations of the triangle –rotations about the z-axis, denoted by R z – transform as irrep A 2 . Thus, when the linearmotions of the centre of mass, and pure rotation about it, are removed from our reducedrepresentation, we are left with E⊕A 1 . So, E and A 1 must be the irreps corresponding to theinternal vibrations of the triangle – one doubly degenerate mode and one non-degeneratemode.The physical interpretation of this is that two of the normal modes of the system havethe same frequency and one normal mode has a different frequency (barring accidentalcoincidences for other reasons). It may be noted that in quantum mechanics the energyquantum of a normal mode is proportional to its frequency. ◭In general, group theory does not tell us what the frequencies are, since it isentirely concerned with the symmetry of the system and not with the values ofmasses and spring constants. However, using this type of reasoning, the resultsfrom representation theory can be used to predict the degeneracies of atomicenergy levels and, given a perturbation whose Hamiltonian (energy operator) hassome degree of symmetry, the extent to which the perturbation will resolve thedegeneracy. Some of these ideas are explored a little further in the next sectionand in the exercises.29.11.4 Breaking of degeneraciesIf a physical system has a high degree of symmetry, invariant under a group G ofreflections and rotations, say, then, as implied above, it will normally be the casethat some of its eigenvalues (of energy, frequency, angular momentum etc.) aredegenerate. However, if a perturbation that is invariant only under the operationsof the elements of a smaller symmetry group (a subgroup of G) is added, some ofthe original degeneracies may be broken. The results derived from representationtheory can be used to decide the extent of the degeneracy-breaking.The normal procedure is to use an N-dimensional basis vector, consisting of theN degenerate eigenfunctions, to generate an N-dimensional representation of thesymmetry group of the perturbation. This representation is then decomposed intoirreps. In general, eigenfunctions that transform according to different irreps nolonger share the same frequency of vibration. We illustrate this with the followingexample.1111

REPRESENTATION THEORYMMMFigure 29.6masses.A circular drumskin loaded with three symmetrically placed◮A circular drumskin has three equal masses placed on it at the vertices of an equilateraltriangle, as shown in figure 29.6. Determine which degenerate normal modes of the drumskincan be split in frequency by this perturbation.When no masses are present the normal modes of the drum-skin are either non-degenerateor two-fold degenerate (see chapter 21). The degenerate eigenfunctions Ψ of the nth normalmode have the formsJ n (kr)(cos nθ)e ±iωt or J n (kr)(sin nθ)e ±iωt .Therefore, as explained above, we need to consider the two-dimensional vector spacespanned by Ψ 1 =sinnθ and Ψ 2 =cosnθ. This will generate a two-dimensional representationof the group 3m (or C 3v ), the symmetry group of the perturbation. Taking the easiestelement from each of the three classes (identity, rotations, and reflections) of group 3m,we haveIΨ 1 =Ψ 1 , IΨ 2 =Ψ 2 ,AΨ 1 =sin [ n ( θ − 2 π)] = ( cos 2 nπ) Ψ3 3 1 − ( sin 2 nπ) Ψ3 2 ,AΨ 2 =cos [ n ( θ − 2 π)] = ( cos 2 nπ) Ψ3 3 2 + ( sin 2 nπ) Ψ3 1 ,CΨ 1 = sin[n(π − θ)] = −(cos nπ)Ψ 1 ,CΨ 2 =cos[n(π − θ)] = (cos nπ)Ψ 2 .The three representative matrices are therefore(cos23D(I) =I 2 , D(A) =nπ − sin 2 nπ)( )3 − cos nπ 0sin 2 nπ cos 2 nπ , D(C) =.0 cosnπ3 3The characters of this representation are χ(I) =2,χ(A) =2cos(2nπ/3) and χ(C) =0.Using (29.18) and table 29.1, we find that(m A1 = 1 6 2+4cos2nπ) = m3 A2(m E = 1 6 4 − 4cos2nπ) .3Thus{A 1 ⊕ A 2 if n =3, 6, 9, ...,D =E otherwise.Hence the normal modes n =3, 6, 9, ... each transform under the operations of 3m1112

29.12 EXERCISESas the sum of two one-dimensional irreps and, using the reasoning given in the previousexample, are therefore split in frequency by the perturbation. For other values of n therepresentation is irreducible and so the degeneracy cannot be split. ◭29.12 Exercises29.1 A group G hasfourelementsI,X,Y and Z, which satisfy X 2 = Y 2 = Z 2 =XY Z = I. Show that G is Abelian and hence deduce the form of its charactertable.Show that the matrices(1 0D(I) =0 1D(Y )=(−1 −p0 1)(−1 0, D(X) =0 −1), D(Z) =(1 p0 −1where p is a real number, form a representation D of G. Find its characters anddecompose it into irreps.29.2 Using a square whose corners lie at coordinates (±1, ±1), form a natural representationof the dihedral group D 4 . Find the characters of the representation,and, using the information (and class order) in table 29.4 (p. 1102), express therepresentation in terms of irreps.Now form a representation in terms of eight 2 × 2 orthogonal matrices, byconsidering the effect of each of the elements of D 4 on a general vector (x, y).Confirm that this representation is one of the irreps found using the naturalrepresentation.29.3 The quaternion group Q (see exercise 28.20) has eight elements {±1, ±i, ±j,±k}obeying the relationsi 2 = j 2 = k 2 = −1, ij = k = −ji.Determine the conjugacy classes of Q and deduce the dimensions of its irreps.Show that Q is homomorphic to the four-element group V, which is generated bytwo distinct elements a and b with a 2 = b 2 =(ab) 2 = I. Find the one-dimensionalirreps of V and use these to help determine the full character table for Q.29.4 Construct the character table for the irreps of the permutation group S 4 asfollows.(a) By considering the possible forms of its cycle notation, determine the numberof elements in each conjugacy class of the permutation group S 4 ,andshowthat S 4 has five irreps. Give the logical reasoning that shows they must consistof two three-dimensional, one two-dimensional, and two one-dimensionalirreps.(b) By considering the odd and even permutations in the group S 4 , establish thecharacters for one of the one-dimensional irreps.(c) Form a natural matrix representation of 4 × 4 matrices based on a set ofobjects {a, b, c, d}, which may or may not be equal to each other, and, byselecting one example from each conjugacy class, show that this naturalrepresentation has characters 4, 2, 1, 0, 0. In the four-dimensional vectorspace in which each of the four coordinates takes on one of the four valuesa, b, c or d, the one-dimensional subspace consisting of the four points withcoordinates of the form {a, a, a, a} is invariant under the permutation groupand hence transforms according to the invariant irrep A 1 . The remainingthree-dimensional subspace is irreducible; use this and the characters deducedabove to establish the characters for one of the three-dimensional irreps, T 1 .1113),),

REPRESENTATION THEORY(d) Complete the character table using orthogonality properties, and check thesummation rule for each irrep. You should obtain table 29.8.Typical element and class sizeIrrep (1) (12) (123) (1234) (12)(34)1 6 8 6 3A 1 1 1 1 1 1A 2 1 −1 1 −1 1E 2 0 −1 0 2T 1 3 1 0 −1 −1T 2 3 −1 0 1 −1Table 29.8 The character table for the permutation group S 4 .29.5 In exercise 28.10, the group of pure rotations taking a cube into itself was foundto have 24 elements. The group is isomorphic to the permutation group S 4 ,considered in the previous question, and hence has the same character table, oncecorresponding classes have been established. By counting the number of elementsin each class, make the correspondences below (the final two cannot be decidedpurely by counting, and should be taken as given).Permutation Symbol Actionclass type (physics)(1) I none(123) 3 rotations about a body diagonal(12)(34) 2 z rotation of π about the normal to a face(1234) 4 z rotations of ±π/2 about the normal to a face(12) 2 d rotation of π about an axis through thecentres of opposite edgesReformulate the character table 29.8 in terms of the elements of the rotationsymmetry group (432 or O) of a cube and use it when answering exercises 29.7and 29.8.29.6 Consider a regular hexagon orientated so that two of its vertices lie on the x-axis.Find matrix representations of a rotation R through 2π/6 and a reflection m y inthe y-axis by determining their effects on vectors lying in the xy-plane . Showthat a reflection m x in the x-axis can be written as m x = m y R 3 , and that the 12elements of the symmetry group of the hexagon are given by R n or R n m y .Using the representations of R and m y as generators, find a two-dimensionalrepresentation of the symmetry group, C 6 , of the regular hexagon. Is it a faithfulrepresentation?29.7 In a certain crystalline compound, a thorium atom lies at the centre of a regularoctahedron of six sulphur atoms at positions (±a, 0, 0), (0, ±a, 0), (0, 0, ±a). Thesecan be considered as being positioned at the centres of the faces of a cube ofside 2a. The sulphur atoms produce at the site of the thorium atom an electricfield that has the same symmetry group as a cube (432 or O).The five degenerate d-electron orbitals of the thorium atom can be expressed,relative to any arbitrary polar axis, as(3 cos 2 θ − 1)f(r), e ±iφ sin θ cos θf(r), e ±2iφ sin 2 θf(r).A rotation about that polar axis by an angle φ ′ effectively changes φ to φ − φ ′ .1114

29.12 EXERCISESUse this to show that the character of the rotation in a representation based onthe orbital wavefunctions is given by1+2cosφ ′ +2cos2φ ′and hence that the characters of the representation, in the order of the symbolsgiven in exercise 29.5, is 5, −1, 1, −1, 1. Deduce that the five-fold degeneratelevel is split into two levels, a doublet and a triplet.29.8 Sulphur hexafluoride is a molecule with the same structure as the crystallinecompound in exercise 29.7, except that a sulphur atom is now the central atom.The following are the forms of some of the electronic orbitals of the sulphuratom, together with the irreps according to which they transform under thesymmetry group 432 (or O).Ψ s = f(r) A 1Ψ p1 = zf(r) T 1Ψ d1 =(3z 2 − r 2 )f(r) EΨ d2 =(x 2 − y 2 )f(r) EΨ d3 = xyf(r) T 2The function x transforms according to the irrep T 1 . Use the above data todetermine whether dipole matrix elements of the form J = ∫ φ 1 xφ 2 dτ can benon-zero for the following pairs of orbitals φ 1 ,φ 2 in a sulphur hexafluoridemolecule: (a) Ψ d1 , Ψ s ;(b)Ψ d1 , Ψ p1 ;(c)Ψ d2 , Ψ d1 ;(d)Ψ s , Ψ d3 ;(e)Ψ p1 , Ψ s .29.9 The hydrogen atoms in a methane molecule CH 4 form a perfect tetrahedronwith the carbon atom at its centre. The molecule is most conveniently describedmathematically by placing the hydrogen atoms at the points (1, 1, 1), (1, −1, −1),(−1, 1, −1) and (−1, −1, 1). The symmetry group to which it belongs, the tetrahedralgroup (¯43m or T d ),hasclassestypifiedbyI, 3,2 z , m d and ¯4 z , where the firstthree are as in exercise 29.5, m d is a reflection in the mirror plane x − y =0and¯4 z is a rotation of π/2 about the z-axis followed by an inversion in the origin. Areflection in a mirror plane can be considered as a rotation of π about an axisperpendicular to the plane, followed by an inversion in the origin.Thecharactertableforthegroup¯43m is very similar to that for the group432, and has the form shown in table 29.9.Typical element and class size Functions transformingIrreps I 3 2 z¯4z m d according to irrep1 8 3 6 6A 1 1 1 1 1 1 x 2 + y 2 + z 2A 2 1 1 1 −1 −1E 2 −1 2 0 0 (x 2 − y 2 , 3z 2 − r 2 )T 1 3 0 −1 1 −1 (R x ,R y ,R z )T 2 3 0 −1 −1 1 (x, y, z); (xy, yz, zx)Table 29.9The character table for group ¯43m.By following the steps given below, determine how many different internal vibrationfrequencies the CH 4 molecule has.(a) Consider a representation based on the twelve coordinates x i ,y i ,z i fori =1, 2, 3, 4. For those hydrogen atoms that transform into themselves, arotation through an angle θ about an axis parallel to one of the coordinateaxes gives rise in the natural representation to the diagonal elements 1 for1115

REPRESENTATION THEORYthe corresponding coordinate and 2 cos θ for the two orthogonal coordinates.If the rotation is followed by an inversion then these entries are multipliedby −1. Atoms not transforming into themselves give a zero diagonal contribution.Show that the characters of the natural representation are 12, 0, 0,0, 2 and hence that its expression in terms of irreps isA 1 ⊕ E ⊕ T 1 ⊕ 2T 2 .(b) The irreps of the bodily translational and rotational motions are included inthis expression and need to be identified and removed. Show that when thisis done it can be concluded that there are three different internal vibrationfrequencies in the CH 4 molecule. State their degeneracies and check thatthey are consistent with the expected number of normal coordinates neededto describe the internal motions of the molecule.29.10 Investigate the properties of an alternating group and construct its charactertable as follows.(a) The set of even permutations of four objects (a proper subgroup of S 4 )is known as the alternating group A 4 . List its twelve members using cyclenotation.(b) Assume that all permutations with the same cycle structure belong to thesame conjugacy class. Show that this leads to a contradiction, and hencedemonstrates that, even if two permutations have the same cycle structure,they do not necessarily belong to the same class.(c) By evaluating the productsp 1 = (123)(4) • (12)(34) • (132)(4) and p 2 = (132)(4) • (12)(34) • (123)(4)deduce that the three elements of A 4 with structure of the form (12)(34)belong to the same class.(d) By evaluating products of the form (1α)(βγ) • (123)(4) • (1α)(βγ), where α, β, γare various combinations of 2, 3, 4, show that the class to which (123)(4)belongs contains at least four members. Show the same for (124)(3).(e) By combining results (b), (c) and (d) deduce that A 4 has exactly four classes,and determine the dimensions of its irreps.(f) Using the orthogonality properties of characters and noting that elements ofthe form (124)(3) have order 3, find the character table for A 4 .29.11 Use the results of exercise 28.23 to find the character table for the dihedral groupD 5 , the symmetry group of a regular pentagon.29.12 Demonstrate that equation (29.24) does, indeed, generate a set of vectors transformingaccording to an irrep λ, by sketching and superposing drawings of anequilateral triangle of springs and masses, based on that shown in figure 29.5.CCCA B AB 30 ◦ AB30 ◦ (a) (b) (c)Figure 29.7 The three normal vibration modes of the equilateral array. Mode(a) is known as the ‘breathing mode’. Modes (b) and (c) transform accordingto irrep E and have equal vibrational frequencies.1116

29.13 HINTS AND ANSWERS(a) Make an initial sketch showing an arbitrary small mass displacement from,say, vertex C. Draw the results of operating on this initial sketch with eachof the symmetry elements of the group 3m (C 3v ).(b) Superimpose the results, weighting them according to the characters of irrepA 1 (table 29.1 in section 29.6) and verify that the resultant is a symmetricalarrangement in which all three masses move symmetrically towards (or awayfrom) the centroid of the triangle. The mode is illustrated in figure 29.7(a).(c)Start again, this time considering a displacement δ of C parallel to the x-axis.Form a similar superposition of sketches weighted according to the charactersof irrep E (note that the reflections are not needed). The resultant containssome bodily displacement of the triangle, since this also transforms accordingto E. Show that the displacement of the centre of mass is ¯x = δ, ȳ =0.Subtract this out, and verify that the remainder is of the form shown infigure 29.7(c).(d) Using an initial displacement parallel to the y-axis, and an analogous procedure,generate the remaining normal mode, degenerate with that in (c) andshown in figure 29.7(b).29.13 Further investigation of the crystalline compound considered in exercise 29.7shows that the octahedron is not quite perfect but is elongated along the (1, 1, 1)direction with the sulphur atoms at positions ±(a+δ,δ,δ), ±(δ,a+δ,δ), ±(δ,δ,a+δ), where δ ≪ a. This structure is invariant under the (crystallographic) symmetrygroup 32 with three two-fold axes along directions typified by (1, −1, 0). Thelatter axes, which are perpendicular to the (1, 1, 1) direction, are axes of twofoldsymmetry for the perfect octahedron. The group 32 is really the threedimensionalversion of the group 3m and has the same character table as table 29.1(section 29.6). Use this to show that, when the distortion of the octahedron isincluded, the doublet found in exercise 29.7 is unsplit but the triplet breaks upinto a singlet and a doublet.29.13 Hints and answers29.1 There are four classes and hence four one-dimensional irreps, which must haveentriesasfollows:1,1,1,1; 1,1,−1, −1; 1, −1, 1, −1; 1, −1, −1, 1. Thecharacters of D are 2, −2, 0, 0 and so the irreps present are the last two of these.29.3 There are five classes {1}, {−1}, {±i}, {±j}, {±k}; there are four one-dimensionalirreps and one two-dimensional irrep. Show that ab = ba. The homomorphismis ±1 → I, ±i → a, ±j → b, ±k → ab. V is Abelian and hence has fourone-dimensional irreps.In the class order given above, the characters for Q are as follows:ˆD (1) , 1, 1, 1, 1, 1; ˆD (2) , 1, 1, 1, −1, −1; ˆD (3) , 1, 1, −1, 1, −1;ˆD (4) , 1, 1, −1, −1, 1; ˆD (5) , 2, −2, 0, 0, 0.29.5 Note that the fourth and fifth classes each have 6 members.29.7 The five basis functions of the representation are multiplied by 1, e −iφ′ , e +iφ′ ,e −2iφ′ , e +2iφ′ as a result of the rotation. The character is the sum of these forrotations of 0, 2π/3, π, π/2, π; D rep =E+T 2 .29.9 (b) The bodily translation has irrep T 2 and the rotation has irrep T 1 . The irrepsof the internal vibrations are A 1 ,E,T 2 , with respective degeneracies 1, 2, 3,making six internal coordinates (12 in total, minus three translational, minusthree rotational).29.11 There are four classes and hence four irreps, which can only be the identityirrep, one other one-dimensional irrep, and two two-dimensional irreps. In theclass order {I}, {R,R 4 }, {R 2 ,R 3 }, {m i } the second one-dimensional irrep must1117

REPRESENTATION THEORY(because of orthogonality) have characters 1, 1, 1, −1. The summation rulesand orthogonality require the other two character sets to be 2, (−1 + √ 5)/2,(−1 − √ 5)/2, 0 and 2, (−1 − √ 5)/2, (−1+ √ 5)/2, 0. Note that R has order 5 andthat, e.g., (−1+ √ 5)/2 =exp(2πi/5) + exp(8πi/5).29.13 The doublet irrep E (characters 2, −1, 0) appears in both 432 and 32 and sois unsplit. The triplet T 1 (characters 3, 0, 1) splits under 32 into doublet E(characters 2, −1, 0) and singlet A 1 (characters 1, 1, 1).1118

30ProbabilityAll scientists will know the importance of experiment and observation and,equally, be aware that the results of some experiments depend to a degree onchance. For example, in an experiment to measure the heights of a random sampleof people, we would not be in the least surprised if all the heights were found tobe different; but, if the experiment were repeated often enough, we would expectto find some sort of regularity in the results. Statistics, which is the subject of thenext chapter, is concerned with the analysis of real experimental data of this sort.First, however, we discuss probability. To a pure mathematician, probability is anentirely theoretical subject based on axioms. Although this axiomatic approach isimportant, and we discuss it briefly, an approach to probability more in keepingwith its eventual applications in statistics is adopted here.We first discuss the terminology required, with particular reference to theconvenient graphical representation of experimental results as Venn diagrams.The concepts of random variables and distributions of random variables are thenintroduced. It is here that the connection with statistics is made; we assert thatthe results of many experiments are random variables and that those results havesome sort of regularity, which is represented by a distribution. Precise definitionsof a random variable and a distribution are then given, as are the definingequations for some important distributions. We also derive some useful quantitiesassociated with these distributions.30.1 Venn diagramsWe call a single performance of an experiment a trial and each possible resultan outcome. Thesample space S of the experiment is then the set of all possibleoutcomes of an individual trial. For example, if we throw a six-sided die then thereare six possible outcomes that together form the sample space of the experiment.At this stage we are not concerned with how likely a particular outcome might1119

PROBABILITYAiiiiiiBSivFigure 30.1A Venn diagram.be (we will return to the probability of an outcome in due course) but ratherwill concentrate on the classification of possible outcomes. It is clear that somesample spaces are finite (e.g. the outcomes of throwing a die) whilst others areinfinite (e.g. the outcomes of measuring people’s heights). Most often, one is notinterested in individual outcomes but in whether an outcome belongs to a givensubset A (say) of the sample space S; these subsets are called events. For example,we might be interested in whether a person is taller or shorter than 180 cm, inwhich case we divide the sample space into just two events: namely, that theoutcome (height measured) is (i) greater than 180 cm or (ii) less than 180 cm.A common graphical representation of the outcomes of an experiment is theVenn diagram. A Venn diagram usually consists of a rectangle, the interior ofwhich represents the sample space, together with one or more closed curves insideit. The interior of each closed curve then represents an event. Figure 30.1 showsa typical Venn diagram representing a sample space S and two events A andB. Every possible outcome is assigned to an appropriate region; in this examplethere are four regions to consider (marked i to iv in figure 30.1):(i) outcomes that belong to event A but not to event B;(ii) outcomes that belong to event B but not to event A;(iii) outcomes that belong to both event A and event B;(iv) outcomes that belong to neither event A nor event B.◮A six-sided die is thrown. Let event A be ‘the number obtained is divisible by 2’ and eventB be ‘the number obtained is divisible by 3’. Draw a Venn diagram to represent these events.It is clear that the outcomes 2, 4, 6 belong to event A and that the outcomes 3, 6 belongto event B. Of these, 6 belongs to both A and B. The remaining outcomes, 1, 5, belong toneither A nor B. The appropriate Venn diagram is shown in figure 30.2. ◭In the above example, one outcome, 6, is divisible by both 2 and 3 and sobelongs to both A and B. This outcome is placed in region iii of figure 30.1, whichis called the intersection of A and B and is denoted by A ∩ B (see figure 30.3(a)).If no events lie in the region of intersection then A and B are said to be mutuallyexclusive or disjoint. In this case, often the Venn diagram is drawn so that theclosed curves representing the events A and B do not overlap, so as to make1120

30.1 VENN DIAGRAMSA42 6 3BS15Figure 30.2 The Venn diagram for the outcomes of the die-throwing trialsdescribed in the worked example.ABABS(a)S(b)ABAĀS(c)S(d)Figure 30.3 Venn diagrams: the shaded regions show (a) A ∩ B, theintersectionof two events A and B, (b)A ∪ B, the union of events A and B, (c)the complement Ā of an event A, (d)A − B, those outcomes in A that do notbelong to B.graphically explicit the fact that A and B are disjoint. It is not necessary, however,to draw the diagram in this way, since we may simply assign zero outcomes tothe shaded region in figure 30.3(a). An event that contains no outcomes is calledthe empty event and denoted by ∅. The event comprising all the elements thatbelong to either A or B, or to both, is called the union of A and B and is denotedby A ∪ B (see figure 30.3(b)). In the previous example, A ∪ B = {2, 3, 4, 6}.It is sometimes convenient to talk about those outcomes that do not belong toa particular event. The set of outcomes that do not belong to A is called thecomplement of A and is denoted by Ā (see figure 30.3(c)); this can also be writtenas Ā = S − A. It is clear that A ∪ Ā = S and A ∩ Ā = ∅.The above notation can be extended in an obvious way, so that A − B denotesthe outcomes in A that do not belong to B. It is clear from figure 30.3(d) thatA − B canalsobewrittenasA ∩ ¯B. Finally, when all the outcomes in event B(say) also belong to event A, but A may contain, in addition, outcomes that do1121

PROBABILITYA1B24756 3C8SFigure 30.4regions.The general Venn diagram for three events is divided into eightnot belong to B, thenB is called a subset of A, a situation that is denoted byB ⊂ A; alternatively, one may write A ⊃ B, which states that A contains B. Inthiscase, the closed curve representing the event B is often drawn lying completelywithin the closed curve representing the event A.The operations ∪ and ∩ are extended straightforwardly to more than twoevents. If there exist n events A 1 ,A 2 ,...,A n , in some sample space S, then theevent consisting of all those outcomes that belong to one or more of the A i is theunion of A 1 ,A 2 ,...,A n and is denoted byA 1 ∪ A 2 ∪ ···∪ A n . (30.1)Similarly, the event consisting of all the outcomes that belong to every one of theA i is called the intersection of A 1 ,A 2 ,...,A n and is denoted byIf, for any pair of values i, j with i ≠ j,A 1 ∩ A 2 ∩ ···∩ A n . (30.2)A i ∩ A j = ∅ (30.3)then the events A i and A j are said to be mutually exclusive or disjoint.Consider three events A, B and C with a Venn diagram such as is shown infigure 30.4. It will be clear that, in general, the diagram will be divided into eightregions and they will be of four different types. Three regions correspond to asingle event; three regions are each the intersection of exactly two events; oneregion is the three-fold intersection of all three events; and finally one regioncorresponds to none of the events. Let us now consider the numbers of differentregions in a general n-event Venn diagram.For one-event Venn diagrams there are two regions, for the two-event casethere are four regions and, as we have just seen, for the three-event case there areeight. In the general n-event case there are 2 n regions, as is clear from the factthat any particular region R lies either inside or outside the closed curve of anyparticular event. With two choices (inside or outside) for each of n closed curves,there are 2 n different possible combinations with which to characterise R. Oncen1122

30.1 VENN DIAGRAMSgets beyond three it becomes impossible to draw a simple two-dimensional Venndiagram, but this does not change the results.The 2 n regions will break down into n + 1 types, with the numbers of each typeas follows § no events, n C 0 =1;one event but no intersections, n C 1 = n;two-fold intersections, n C 2 = 1 2n(n − 1);three-fold intersections, n C 3 = 1 3!n(n − 1)(n − 2);.an n-fold intersection, n C n =1.That this makes a total of 2 n can be checked by considering the binomialexpansion2 n =(1+1) n =1+n + 1 2n(n − 1) + ···+1.Using Venn diagrams, it is straightforward to show that the operations ∩ and∪ obey the following algebraic laws:commutativity, A ∩ B = B ∩ A, A ∪ B = B ∪ A;associativity, (A ∩ B) ∩ C = A ∩ (B ∩ C), (A ∪ B) ∪ C = A ∪ (B ∪ C);distributivity, A ∩ (B ∪ C) =(A ∩ B) ∪ (A ∩ C),A ∪ (B ∩ C) =(A ∪ B) ∩ (A ∪ C);idempotency, A ∩ A = A, A ∪ A = A.◮Show that (i) A ∪ (A ∩ B) =A ∩ (A ∪ B) =A, (ii) (A − B) ∪ (A ∩ B) =A.(i) Using the distributivity and idempotency laws above, we see thatA ∪ (A ∩ B) =(A ∪ A) ∩ (A ∪ B) =A ∩ (A ∪ B).By sketching a Venn diagram it is immediately clear that both expressions are equal toA. Nevertheless, we here proceed in a more formal manner in order to deduce this resultalgebraically. Let us begin by writingX = A ∪ (A ∩ B) =A ∩ (A ∪ B), (30.4)from which we want to deduce a simpler expression for the event X. Using the first equalityin (30.4) and the algebraic laws for ∩ and ∪, wemaywriteA ∩ X = A ∩ [A ∪ (A ∩ B)]=(A ∩ A) ∪ [A ∩ (A ∩ B)]= A ∪ (A ∩ B) =X.§ The symbols n C i ,fori =0, 1, 2,...,n, are a convenient notation for combinations; they and theirproperties are discussed in chapter 1.1123

PROBABILITYSince A ∩ X = X we must have X ⊂ A. Now, using the second equality in (30.4) in asimilar way, we findA ∪ X = A ∪ [A ∩ (A ∪ B)]=(A ∪ A) ∩ [A ∪ (A ∪ B)]= A ∩ (A ∪ B) =X,from which we deduce that A ⊂ X. Thus,sinceX ⊂ A and A ⊂ X, we must conclude thatX = A.(ii) Since we do not know how to deal with compound expressions containing a minussign, we begin by writing A − B = A ∩ ¯B as mentioned above. Then, using the distributivitylaw, we obtain(A − B) ∪ (A ∩ B) =(A ∩ ¯B) ∪ (A ∩ B)= A ∩ ( ¯B ∪ B)= A ∩ S = A.In fact, this result, like the first one, can be proved trivially by drawing a Venn diagram. ◭Further useful results may be derived from Venn diagrams. In particular, it issimple to show that the following rules hold:(i) if A ⊂ B then Ā ⊃ ¯B;(ii) A ∪ B = Ā ∩ ¯B;(iii) A ∩ B = Ā ∪ ¯B.Statements (ii) and (iii) are known jointly as de Morgan’s laws and are sometimesuseful in simplifying logical expressions.◮There exist two events A and B such that(X ∪ A) ∪ (X ∪ Ā) =B.Find an expression for the event X in terms of A and B.We begin by taking the complement of both sides of the above expression: applying deMorgan’s laws we obtain¯B =(X ∪ A) ∩ (X ∪ Ā).We may then use the algebraic laws obeyed by ∩ and ∪ to yield¯B = X ∪ (A ∩ Ā) =X ∪ ∅ = X.Thus, we find that X = ¯B. ◭30.2 ProbabilityIn the previous section we discussed Venn diagrams, which are graphical representationsof the possible outcomes of experiments. We did not, however, giveany indication of how likely each outcome or event might be when any particularexperiment is performed. Most experiments show some regularity. By this wemean that the relative frequency of an event is approximately the same on eachoccasion that a set of trials is performed. For example, if we throw a die N1124

30.2 PROBABILITYtimes then we expect that a six will occur approximately N/6 times (assuming,of course, that the die is not biased). The regularity of outcomes allows us todefine the probability, Pr(A), as the expected relative frequency of event A in alarge number of trials. More quantitatively, if an experiment has a total of n Soutcomes in the sample space S, andn A of these outcomes correspond to theevent A, then the probability that event A will occur isPr(A) = n A. (30.5)n S30.2.1 Axioms and theoremsFrom (30.5) we may deduce the following properties of the probability Pr(A).(i) For any event A in a sample space S,0 ≤ Pr(A) ≤ 1. (30.6)If Pr(A) =1thenA is a certainty; if Pr(A) =0thenA is an impossibility.(ii) For the entire sample space S we havePr(S) = n S=1, (30.7)n Swhich simply states that we are certain to obtain one of the possibleoutcomes.(iii) If A and B are two events in S then, from the Venn diagrams in figure 30.3,we see thatn A∪B = n A + n B − n A∩B , (30.8)the final subtraction arising because the outcomes in the intersection ofA and B are counted twice when the outcomes of A are added to thoseof B. Dividing both sides of (30.8) by n S , we obtain the addition rule forprobabilitiesPr(A ∪ B) =Pr(A)+Pr(B) − Pr(A ∩ B). (30.9)However, if A and B are mutually exclusive events (A ∩ B = ∅) thenPr(A ∩ B) = 0 and we obtain the special casePr(A ∪ B) =Pr(A)+Pr(B). (30.10)(iv) If Ā is the complement of A then Ā and A are mutually exclusive events.Thus, from (30.7) and (30.10) we have1=Pr(S) =Pr(A ∪ Ā) =Pr(A)+Pr(Ā),from which we obtain the complement lawPr(Ā) =1− Pr(A). (30.11)1125

PROBABILITYThis is particularly useful for problems in which evaluating the probabilityof the complement is easier than evaluating the probability of the eventitself.◮Calculate the probability of drawing an ace or a spade from a pack of cards.Let A be the event that an ace is drawn and B the event that a spade is drawn. Itimmediately follows that Pr(A) = 4 = 113and Pr(B) = = 1 . The intersection of A and52 13 52 4B consists of only the ace of spades and so Pr(A ∩ B) = 1 . Thus, from (30.9)52Pr(A ∪ B) = 1 + 1 − 1 = 4 .13 4 52 13In this case it is just as simple to recognise that there are 16 cards in the pack that satisfythe required condition (13 spades plus three other aces) and so the probability is 16 . ◭ 52The above theorems can easily be extended to a greater number of events. Forexample, if A 1 ,A 2 ,...,A n are mutually exclusive events then (30.10) becomesPr(A 1 ∪ A 2 ∪ ···∪ A n )=Pr(A 1 )+Pr(A 2 )+···+Pr(A n ). (30.12)Furthermore, if A 1 ,A 2 ,...,A n (whether mutually exclusive or not) exhaust S, i.e.are such that A 1 ∪ A 2 ∪ ···∪ A n = S, thenPr(A 1 ∪ A 2 ∪ ···∪ A n )=Pr(S) =1. (30.13)◮A biased six-sided die has probabilities 1 p, p, p, p, p, 2p of showing 1, 2, 3, 4, 5, 62respectively. Calculate p.Given that the individual events are mutually exclusive, (30.12) can be applied to givePr(1 ∪ 2 ∪ 3 ∪ 4 ∪ 5 ∪ 6) = 1 13p + p + p + p + p +2p = p.2 2The union of all possible outcomes on the LHS of this equation is clearly the samplespace, S, andsoPr(S) = 13 p. 2Now using (30.7),132p =Pr(S) =1 ⇒ p = . ◭2 13When the possible outcomes of a trial correspond to more than two events,and those events are not mutually exclusive, the calculation of the probability ofthe union of a number of events is more complicated, and the generalisation ofthe addition law (30.9) requires further work. Let us begin by considering theunion of three events A 1 , A 2 and A 3 , which need not be mutually exclusive. Wefirst define the event B = A 2 ∪ A 3 and, using the addition law (30.9), we obtainPr(A 1 ∪ A 2 ∪ A 3 )=Pr(A 1 ∪ B) =Pr(A 1 )+Pr(B) − Pr(A 1 ∩ B).(30.14)1126

30.2 PROBABILITYHowever, we may write Pr(A 1 ∩ B) asPr(A 1 ∩ B) =Pr[A 1 ∩ (A 2 ∪ A 3 )]=Pr[(A 1 ∩ A 2 ) ∪ (A 1 ∩ A 3 )]=Pr(A 1 ∩ A 2 )+Pr(A 1 ∩ A 3 ) − Pr(A 1 ∩ A 2 ∩ A 3 ).Substituting this expression, and that for Pr(B) obtained from (30.9), into (30.14)we obtain the probability addition law for three general events,Pr(A 1 ∪ A 2 ∪ A 3 )=Pr(A 1 )+Pr(A 2 )+Pr(A 3 ) − Pr(A 2 ∩ A 3 ) − Pr(A 1 ∩ A 3 )− Pr(A 1 ∩ A 2 )+Pr(A 1 ∩ A 2 ∩ A 3 ). (30.15)◮Calculate the probability of drawing from a pack of cards one that is an ace or is a spadeor shows an even number (2,4,6,8,10).If, as previously, A is the event that an ace is drawn, Pr(A) = 4 . Similarly the event B,52that a spade is drawn, has Pr(B) = 13 . The further possibility C, that the card is even (but52not a picture card) has Pr(C) = 20 . The two-fold intersections have probabilities52Pr(A ∩ B) = 1 52 , Pr(A ∩ C) =0, Pr(B ∩ C) = 5 52 .There is no three-fold intersection as events A and C are mutually exclusive. HencePr(A ∪ B ∪ C) = 1 31[(4+13+20)− (1+0+5)+(0)] =52 52 .The reader should identify the 31 cards involved. ◭When the probabilities are combined to calculate the probability for the unionof the n general events, the result, which may be proved by induction upon n (seethe answer to exercise 30.4), isPr(A 1 ∪ A 2 ∪ ···∪ A n )= ∑ iPr(A i ) − ∑ i,jPr(A i ∩ A j )+ ∑ i,j,kPr(A i ∩ A j ∩ A k )− ···+(−1) n+1 Pr(A 1 ∩ A 2 ∩ ···∩ A n ). (30.16)Each summation runs over all possible sets of subscripts, except those in whichany two subscripts in a set are the same. The number of terms in the summationof probabilities of m-fold intersections of the n events is given by n C m (as discussedin section 30.1). Equation (30.9) is a special case of (30.16) in which n = 2 andonly the first two terms on the RHS survive. We now illustrate this result with aworked example that has n = 4 and includes a four-fold intersection.1127

PROBABILITY◮Find the probability of drawing from a pack a card that has at least one of the followingproperties:A, itisanace;B, it is a spade;C, it is a black honour card (ace, king, queen, jack or 10);D, it is a black ace.Measuring all probabilities in units of 1 , the single-event probabilities are52Pr(A) =4, Pr(B) =13, Pr(C) =10, Pr(D) =2.The two-fold intersection probabilities, measured in the same units, arePr(A ∩ B) =1, Pr(A ∩ C) =2, Pr(A ∩ D) =2,Pr(B ∩ C) =5, Pr(B ∩ D) =1, Pr(C ∩ D) =2.The three-fold intersections have probabilitiesPr(A ∩ B ∩ C) =1, Pr(A ∩ B ∩ D) =1, Pr(A ∩ C ∩ D) =2, Pr(B ∩ C ∩ D) =1.Finally, the four-fold intersection, requiring all four conditions to hold, is satisfied only bythe ace of spades, and hence (again in units of 1 52Pr(A ∩ B ∩ C ∩ D) =1.Substituting in (30.16) givesP = 1 20[(4+13+10+2)− (1+2+2+5+1+2)+(1+1+2+1)− (1)] =52 52 . ◭We conclude this section on basic theorems by deriving a useful generalexpression for the probability Pr(A ∩ B) thattwoeventsA and B both occur inthe case where A (say) is the union of a set of n mutually exclusive events A i .Inthis caseA ∩ B =(A 1 ∩ B) ∪ ···∪ (A n ∩ B),where the events A i ∩ B are also mutually exclusive. Thus, from the addition law(30.12) for mutually exclusive events, we findPr(A ∩ B) = ∑ Pr(A i ∩ B). (30.17)iMoreover, in the special case where the events A i exhaust the sample space S, wehave A ∩ B = S ∩ B = B, and we obtain the total probability lawPr(B) = ∑ iPr(A i ∩ B). (30.18)30.2.2 Conditional probabilitySo far we have defined only probabilities of the form ‘what is the probability thatevent A happens?’. In this section we turn to conditional probability, the probabilitythat a particular event occurs given the occurrence of another, possibly related,event. For example, we may wish to know the probability of event B, drawingan1128

30.2 PROBABILITYace from a pack of cards from which one has already been removed, given thatevent A, the card already removed was itself an ace, has occurred.We denote this probability by Pr(B|A) and may obtain a formula for it byconsidering the total probability Pr(A ∩ B) =Pr(B ∩ A) that both A and B willoccur. This may be written in two ways, i.e.From this we obtainPr(A ∩ B) =Pr(A)Pr(B|A)=Pr(B)Pr(A|B).Pr(A|B) =Pr(A ∩ B)Pr(B)(30.19)andPr(B ∩ A)Pr(B|A) = . (30.20)Pr(A)In terms of Venn diagrams, we may think of Pr(B|A) as the probability of B inthe reduced sample space defined by A. Thus, if two events A and B are mutuallyexclusive thenPr(A|B) =0=Pr(B|A). (30.21)When an experiment consists of drawing objects at random from a given setof objects, it is termed sampling a population. We need to distinguish betweentwo different ways in which such a sampling experiment may be performed. Afteran object has been drawn at random from the set it may either be put asideor returned to the set before the next object is randomly drawn. The former istermed ‘sampling without replacement’, the latter ‘sampling with replacement’.◮Find the probability of drawing two aces at random from a pack of cards (i) when thefirst card drawn is replaced at random into the pack before the second card is drawn, and(ii) when the first card is put aside after being drawn.Let A be the event that the first card is an ace, and B the event that the second card is anace. NowPr(A ∩ B) =Pr(A)Pr(B|A),and for both (i) and (ii) we know that Pr(A) = 4 = 1 . 52 13(i) If the first card is replaced in the pack before the next is drawn then Pr(B|A) =Pr(B) = 4 = 1 ,sinceA and B are independent events. We then have52 13Pr(A ∩ B) =Pr(A)Pr(B) = 1 13 × 113 = 1169 .(ii) If the first card is put aside and the second then drawn, A and B are not independentand Pr(B|A) = 3 , with the result that51Pr(A ∩ B) =Pr(A)Pr(B|A) = 1 13 × 351 = 1221 . ◭1129

PROBABILITYTwo events A and B are statistically independent if Pr(A|B) =Pr(A) (or equivalentlyif Pr(B|A) =Pr(B)). In words, the probability of A given B is then the sameas the probability of A regardless of whether B occurs. For example, if we throwa coin and a die at the same time, we would normally expect that the probabilityof throwing a six was independent of whether a head was thrown. If A and B arestatistically independent then it follows thatPr(A ∩ B) =Pr(A)Pr(B). (30.22)In fact, on the basis of intuition and experience, (30.22) may be regarded as thedefinition of the statistical independence of two events.The idea of statistical independence is easily extended to an arbitrary numberof events A 1 ,A 2 ,...,A n . The events are said to be (mutually) independent ifPr(A i ∩ A j )=Pr(A i )Pr(A j ),Pr(A i ∩ A j ∩ A k )=Pr(A i )Pr(A j )Pr(A k ),..Pr(A 1 ∩ A 2 ∩ ···∩ A n )=Pr(A 1 )Pr(A 2 ) ···Pr(A n ),for all combinations of indices i, j and k for which no two indices are the same.Even if all n events are not mutually independent, any two events for whichPr(A i ∩ A j )=Pr(A i )Pr(A j ) are said to be pairwise independent.We now derive two results that often prove useful when working with conditionalprobabilities. Let us suppose that an event A is the union of n mutuallyexclusive events A i .IfB is some other event then from (30.17) we havePr(A ∩ B) = ∑ Pr(A i ∩ B).iDividing both sides of this equation by Pr(B), and using (30.19), we obtainPr(A|B) = ∑ iPr(A i |B), (30.23)which is the addition law for conditional probabilities.Furthermore, if the set of mutually exclusive events A i exhausts the samplespace S then, from the total probability law (30.18), the probability Pr(B) of someevent B in S can be written asPr(B) = ∑ iPr(A i )Pr(B|A i ). (30.24)◮A collection of traffic islands connected by a system of one-way roads is shown in figure30.5. At any given island a car driver chooses a direction at random from those available.What is the probability that a driver starting at O will arrive at B?In order to leave O the driver must pass through one of A 1 , A 2 , A 3 or A 4 , which thusform a complete set of mutually exclusive events. Since at each island (including O) thedriver chooses a direction at random from those available, we have that Pr(A i )= 1 for 41130

30.2 PROBABILITYA 4A 3OA 1A 2BFigure 30.5A collection of traffic islands connected by one-way roads.i =1, 2, 3, 4. From figure 30.5, we see also thatPr(B|A 1 )= 1 , Pr(B|A 3 2)= 1 , Pr(B|A 3 3)=0, Pr(B|A 4 )= 2 = 1 . 4 2Thus, using the total probability law (30.24), we find that the probability of arriving at Bis given byPr(B) = ∑ (Pr(A i )Pr(B|A i )= 1 1 + 1 +0+ ) 14 3 3 2 =7. ◭ 24iFinally, we note that the concept of conditional probability may be straightforwardlyextended to several compound events. For example, in the case of threeevents A, B, C, we may write Pr(A ∩ B ∩ C) in several ways, e.g.Pr(A ∩ B ∩ C) =Pr(C)Pr(A ∩ B|C)=Pr(B ∩ C)Pr(A|B ∩ C)=Pr(C)Pr(B|C)Pr(A|B ∩ C).◮Suppose {A i } is a set of mutually exclusive events that exhausts the sample space S. IfBand C are two other events in S, show thatPr(B|C) = ∑ iPr(A i |C)Pr(B|A i ∩ C).Using (30.19) and (30.17), we may writePr(C)Pr(B|C) =Pr(B ∩ C) = ∑ iPr(A i ∩ B ∩ C). (30.25)Each term in the sum on the RHS can be expanded as an appropriate product ofconditional probabilities,Pr(A i ∩ B ∩ C) =Pr(C)Pr(A i |C)Pr(B|A i ∩ C).Substituting this form into (30.25) and dividing through by Pr(C) gives the requiredresult. ◭1131

PROBABILITY30.2.3 Bayes’ theoremIn the previous section we saw that the probability that both an event A and arelated event B will occur can be written either as Pr(A)Pr(B|A) orPr(B)Pr(A|B).Hencefrom which we obtain Bayes’ theorem,Pr(A)Pr(B|A) =Pr(B)Pr(A|B),Pr(A|B) = Pr(A) Pr(B|A). (30.26)Pr(B)This theorem clearly shows that Pr(B|A) ≠Pr(A|B), unless Pr(A) =Pr(B). It issometimes useful to rewrite Pr(B), if it is not known directly, asso that Bayes’ theorem becomesPr(B) =Pr(A)Pr(B|A)+Pr(Ā)Pr(B|Ā)Pr(A|B) =Pr(A)Pr(B|A)Pr(A)Pr(B|A) +Pr(Ā)Pr(B|Ā) . (30.27)◮Suppose that the blood test for some disease is reliable in the following sense: for peoplewho are infected with the disease the test produces a positive result in 99.99% of cases; forpeople not infected a positive test result is obtained in only 0.02% of cases. Furthermore,assume that in the general population one person in 10 000 people is infected. A person isselected at random and found to test positive for the disease. What is the probability thatthe individual is actually infected?Let A be the event that the individual is infected and B be the event that the individualtests positive for the disease. Using Bayes’ theorem the probability that a person who testspositive is actually infected isPr(A|B) =Pr(A)Pr(B|A)Pr(A)Pr(B|A) +Pr(Ā)Pr(B|Ā) .Now Pr(A) =1/10000 = 1 − Pr(Ā), and we are told that Pr(B|A) = 9999/10000 andPr(B|Ā) =2/10000. Thus we obtain1/10000 × 9999/10000Pr(A|B) =(1/10000 × 9999/10000) + (9999/10000 × 2/10000) = 1 3 .Thus, there is only a one in three chance that a person chosen at random, who testspositive for the disease, is actually infected.At a first glance, this answer may seem a little surprising, but the reason for the counterintuitiveresult is that the probability that a randomly selected person is not infected is9999/10000, which is very high. Thus, the 0.02% chance of a positive test for an uninfectedperson becomes significant. ◭1132

30.3 PERMUTATIONS AND COMBINATIONSWe note that (30.27) may be written in a more general form if S is not simplydivided into A and Ā but, rather, into any set of mutually exclusive events A i thatexhaust S. Using the total probability law (30.24), we may then writePr(B) = ∑ iPr(A i )Pr(B|A i ),so that Bayes’ theorem takes the formPr(A|B) =Pr(A)Pr(B|A)∑i Pr(A i)Pr(B|A i ) , (30.28)where the event A need not coincide with any of the A i .As a final point, we comment that sometimes we are concerned only with therelative probabilities of two events A and C (say), given the occurrence of someother event B. From (30.26) we then obtain a different form of Bayes’ theorem,Pr(A|B)Pr(C|B) = Pr(A)Pr(B|A)Pr(C)Pr(B|C) , (30.29)which does not contain Pr(B) at all.30.3 Permutations and combinationsIn equation (30.5) we defined the probability of an event A in a sample space S asPr(A) = n A,n Swhere n A is the number of outcomes belonging to event A and n S is the totalnumber of possible outcomes. It is therefore necessary to be able to count thenumber of possible outcomes in various common situations.30.3.1 PermutationsLet us first consider a set of n objects that are all different. We may ask inhow many ways these n objects may be arranged, i.e. how many permutations ofthese objects exist. This is straightforward to deduce, as follows: the object in thefirst position may be chosen in n different ways, that in the second position inn − 1 ways, and so on until the final object is positioned. The number of possiblearrangements is thereforen(n − 1)(n − 2) ···(1) = n! (30.30)Generalising (30.30) slightly, let us suppose we choose only k (< n) objectsfrom n. The number of possible permutations of these k objects selected from nis given byn!n(n − 1)(n − 2) ···(n − k +1) =} {{ } (n − k)! ≡ n P k . (30.31)k factors1133

PROBABILITYIn calculating the number of permutations of the various objects we have sofar assumed that the objects are sampled without replacement – i.e. once an objecthas been drawn from the set it is put aside. As mentioned previously, however,we may instead replace each object before the next is chosen. The number ofpermutations of k objects from n with replacement may be calculated very easilysince the first object can be chosen in n different ways, as can the second, thethird, etc. Therefore the number of permutations is simply n k . This may also beviewed as the number of permutations of k objects from n where repetitions areallowed, i.e. each object may be used as often as one likes.◮Find the probability that in a group of k people at least two have the same birthday(ignoring 29 February).It is simplest to begin by calculating the probability that no two people share a birthday,as follows. Firstly, we imagine each of the k people in turn pointing to their birthday ona year planner. Thus, we are sampling the 365 days of the year ‘with replacement’ and sothe total number of possible outcomes is (365) k . Now (for the moment) we assume that notwo people share a birthday and imagine the process being repeated, except that as eachperson points out their birthday it is crossed off the planner. In this case, we are samplingthe days of the year ‘without replacement’, and so the possible number of outcomes forwhich all the birthdays are different is365 365!P k =(365 − k)! .Hence the probability that all the birthdays are different is365!p =(365 − k)! 365 . kNow using the complement rule (30.11), the probability q that two or more people havethe same birthday is simply365!q =1− p =1−(365 − k)! 365 . kThis expression may be conveniently evaluated using Stirling’s approximation for n! whenn is large, namelyn! ∼ √ ( n) n2πn ,eto give( ) 365−k+0.5 365q ≈ 1 − e −k .365 − kIt is interesting to note that if k = 23 the probability is a little greater than a half thatat least two people have the same birthday, and if k = 50 the probability rises to 0.970.This can prove a good bet at a party of non-mathematicians! ◭So far we have assumed that all n objects are different (or distinguishable). Letus now consider n objects of which n 1 are identical and of type 1, n 2 are identicaland of type 2, ...,n m are identical and of type m (clearly n = n 1 + n 2 + ···+ n m ).From (30.30) the number of permutations of these n objects is again n!. However,1134

30.3 PERMUTATIONS AND COMBINATIONSthe number of distinguishable permutations is onlyn!n 1 !n 2 ! ···n m ! , (30.32)since the ith group of identical objects can be rearranged in n i ! ways withoutchanging the distinguishable permutation.◮A set of snooker balls consists of a white, a yellow, a green, a brown, a blue, a pink, ablack and 15 reds. How many distinguishable permutations of the balls are there?In total there are 22 balls, the 15 reds being indistinguishable. Thus from (30.32) thenumber of distinguishable permutations is22!(1!)(1!)(1!)(1!)(1!)(1!)(1!)(15!) = 22! = 859 541 760. ◭15!30.3.2 CombinationsWe now consider the number of combinations of various objects when their orderis immaterial. Assuming all the objects to be distinguishable, from (30.31) we seethat the number of permutations of k objects chosen from n is n P k = n!/(n − k)!.Now, since we are no longer concerned with the order of the chosen objects, whichcan be internally arranged in k! different ways, the number of combinations of kobjects from n is( )n!n(n − k)!k! ≡ n C k ≡ for 0 ≤ k ≤ n, (30.33)kwhere, as noted in chapter 1, n C k is called the binomial coefficient since it alsoappears in the binomial expansion for positive integer n, namelyn∑(a + b) n =n C k a k b n−k . (30.34)k=0◮A hand of 13 playing cards is dealt from a well-shuffled pack of 52. What is the probabilitythat the hand contains two aces?Since the order of the cards in the hand is immaterial, the total number of distinct handsis simply equal to the number of combinations of 13 objects drawn from 52, i.e. 52 C 13 .However, the number of hands containing two aces is equal to the number of ways, 4 C 2 ,in which the two aces can be drawn from the four available, multiplied by the number ofways, 48 C 11 , in which the remaining 11 cards in the hand can be drawn from the 48 cardsthat are not aces. Thus the required probability is given by4 C 48 2 C 11= 4! 48! 13!39!52C 13 2!2! 11!37! 52!= (3)(4) (12)(13)(38)(39)2 (49)(50)(51)(52) =0.213 ◭1135

PROBABILITYAnother useful result that may be derived using the binomial coefficients is thenumber of ways in which n distinguishable objects can be divided into m piles,with n i objects in the ith pile, i =1, 2,...,m (the ordering of objects within eachpile being unimportant). This may be straightforwardly calculated as follows. Wemay choose the n 1 objects in the first pile from the original n objects in n C n1 ways.The n 2 objects in the second pile can then be chosen from the n − n 1 remainingobjects in n−n1 C n2 ways, etc. We may continue in this fashion until we reach the(m − 1)th pile, which may be formed in n−n1−···−nm−2 C nm−1 ways. The remainingobjects then form the mth pile and so can only be ‘chosen’ in one way. Thus thetotal number of ways of dividing the original n objects into m piles is given bythe productN = n n−nC 1n1 C ···n−n1−···−nm−2 n2 C nm−1n! (n − n 1 )!=n 1 !(n − n 1 )! n 2 !(n − n 1 − n 2 )! ··· (n − n 1 − n 2 − ···− n m−2 )!n m−1 !(n − n 1 − n 2 − ···− n m−2 − n m−1 )!n! (n − n 1 )!=···(n − n 1 − n 2 − ···− n m−2 )!n 1 !(n − n 1 )! n 2 !(n − n 1 − n 2 )! n m−1 !n m !n!=n 1 !n 2 ! ···n m ! . (30.35)These numbers are called multinomial coefficients since (30.35) is the coefficient ofx n11 xn2 2 ···xnm m in the multinomial expansion of (x 1 + x 2 + ···+ x m ) n , i.e. for positiveinteger n∑(x 1 + x 2 + ···+ x m ) n n!=n 1 !n 2 ! ···n m ! xn1 1 xn2 2 ···xnm m .n 1 ,n 2 ,... ,nmn 1 +n 2 +···+nm=nFor the case m =2,n 1 = k, n 2 = n − k, (30.35) reduces to the binomial coefficientn C k . Furthermore, we note that the multinomial coefficient (30.35) is identical tothe expression (30.32) for the number of distinguishable permutations of n objects,n i of which are identical and of type i (for i =1, 2,...,mand n 1 +n 2 +···+n m = n).A few moments’ thought should convince the reader that the two expressions(30.35) and (30.32) must be identical.◮In the card game of bridge, each of four players is dealt 13 cards from a full pack of 52.What is the probability that each player is dealt an ace?From (30.35), the total number of distinct bridge dealings is 52!/(13!13!13!13!). However,the number of ways in which the four aces can be distributed with one in each hand is4!/(1!1!1!1!) = 4!; the remaining 48 cards can then be dealt out in 48!/(12!12!12!12!)ways. Thus the probability that each player receives an ace is4! 48! (13!) 4 24(13) 4=(12!) 4 52! (49)(50)(51)(52) =0.105. ◭As in the case of permutations we might ask how many combinations of kobjects can be chosen from n with replacement (repetition). To calculate this, we1136

30.3 PERMUTATIONS AND COMBINATIONSmay imagine the n (distinguishable) objects set out on a table. Each combinationof k objects can then be made by pointing to k of the n objectsinturn(withrepetitions allowed). These k equivalent selections distributed amongst n differentbut re-choosable objects are strictly analogous to the placing of k indistinguishable‘balls’ in n different boxes with no restriction on the number of balls in each box.A particular selection in the case k =7,n = 5 may be symbolised asxxx||x|xx|x.This denotes three balls in the first box, none in the second, one in the third, twoin the fourth and one in the fifth. We therefore need only consider the number of(distinguishable) ways in which k crosses and n − 1 vertical lines can be arranged,i.e. the number of permutations of k + n − 1 objects of which k are identicalcrosses and n − 1 are identical lines. This is given by (30.33) as(k + n − 1)!= n+k−1 C k . (30.36)k!(n − 1)!We note that this expression also occurs in the binomial expansion for negativeinteger powers. If n is a positive integer, it is straightforward to show that (seechapter 1)∞∑(a + b) −n = (−1) k n+k−1 C k a −n−k b k ,k=0where a istakentobelargerthanb in magnitude.◮A system contains a number N of (non-interacting) particles, each of which can be inany of the quantum states of the system. The structure of the set of quantum states is suchthat there exist R energy levels with corresponding energies E i and degeneracies g i (i.e. theith energy level contains g i quantum states). Find the numbers of distinct ways in whichthe particles can be distributed among the quantum states of the system such that the ithenergy level contains n i particles, for i =1, 2,...,R, in the cases where the particles are(i) distinguishable with no restriction on the number in each state;(ii) indistinguishable with no restriction on the number in each state;(iii) indistinguishable with a maximum of one particle in each state;(iv) distinguishable with a maximum of one particle in each state.It is easiest to solve this problem in two stages. Let us first consider distributing the Nparticles among the R energy levels, without regard for the individual degenerate quantumstates that comprise each level. If the particles are distinguishable then the number ofdistinct arrangements with n i particles in the ith level, i =1, 2,...,R, is given by (30.35) asN!n 1 !n 2 ! ···n R ! .If, however, the particles are indistinguishable then clearly there exists only one distinctarrangement having n i particles in the ith level, i =1, 2,...,R . If we suppose that thereexist w i ways in which the n i particles in the ith energy level can be distributed amongthe g i degenerate states, then it follows that the number of distinct ways in which the N1137

PROBABILITYparticles can be distributed among all R quantum states of the system, with n i particles inthe ith level, is given by⎧N!R∏⎪⎨w i for distinguishable particles,n 1 !n 2 ! ···n R !i=1W {n i } = R∏⎪⎩ w ifor indistinguishable particles.(30.37)i=1It therefore remains only for us to find the appropriate expression for w i in each of thecases (i)–(iv) above.Case (i). If there is no restriction on the number of particles in each quantum state,then in the ith energy level each particle can reside in any of the g i degenerate quantumstates. Thus, if the particles are distinguishable then the number of distinct arrangementsis simply w i = g n ii. Thus, from (30.37),W {n i } =N!n 1 !n 2 ! ···n R !R∏i=1g n ii = N!R∏i=1g n iin i ! .Such a system of particles (for example atoms or molecules in a classical gas) is said toobey Maxwell–Boltzmann statistics.Case (ii). If the particles are indistinguishable and there is no restriction on the numberin each state then, from (30.36), the number of distinct arrangements of the n i particlesamong the g i states in the ith energy level isw i = (n i + g i − 1)!n i !(g i − 1)! .Substituting this expression in (30.37), we obtainR∏ (n i + g i − 1)!W {n i } =n i !(g i − 1)! .i=1Such a system of particles (for example a gas of photons) is said to obey Bose–Einsteinstatistics.Case (iii). If a maximum of one particle can reside in each of the g i degenerate quantumstates in the ith energy level then the number of particles in each state is either 0 or 1.Since the particles are indistinguishable, w i is equal to the number of distinct arrangementsin which n i states are occupied and g i − n i states are unoccupied; this is given byThus, from (30.37), we havew i = g iC ni =g i !n i !(g i − n i )! .R∏ g i !W {n i } =n i !(g i − n i )! .i=1Such a system is said to obey Fermi–Dirac statistics, and an example is provided by anelectron gas.Case (iv). Again, the number of particles in each state is either 0 or 1. If the particlesare distinguishable, however, each arrangement identified in case (iii) can be reordered inn i ! different ways, so thatw i = g g i !iP ni =(g i − n i )! .1138

30.4 RANDOM VARIABLES AND DISTRIBUTIONSSubstituting this expression into (30.37) givesR∏ g i !W {n i } = N!n i !(g i − n i )! .i=1Such a system of particles has the names of no famous scientists attached to it, since itappears that it never occurs in nature. ◭30.4 Random variables and distributionsSuppose an experiment has an outcome sample space S. A real variable X thatis defined for all possible outcomes in S (so that a real number – not necessarilyunique – is assigned to each possible outcome) is called a random variable (RV).The outcome of the experiment may already be a real number and hence a randomvariable, e.g. the number of heads obtained in 10 throws of a coin, or the sum ofthe values if two dice are thrown. However, more arbitrary assignments are possible,e.g. the assignment of a ‘quality’ rating to each successive item produced by amanufacturing process. Furthermore, assuming that a probability can be assignedto all possible outcomes in a sample space S, it is possible to assign a probabilitydistribution to any random variable. Random variables may be divided into twoclasses, discrete and continuous, and we now examine each of these in turn.30.4.1 Discrete random variablesA random variable X that takes only discrete values x 1 , x 2 , ...,x n , with probabilitiesp 1 , p 2 , ...,p n , is called a discrete random variable. The number of valuesn for which X has a non-zero probability is finite or at most countably infinite.As mentioned above, an example of a discrete random variable is the number ofheads obtained in 10 throws of a coin. If X is a discrete random variable, we candefine a probability function (PF) f(x) that assigns probabilities to all the distinctvalues that X can take, such that{pi if x = x i ,f(x) =Pr(X = x) =(30.38)0 otherwise.A typical PF (see figure 30.6) thus consists of spikes, at valid values of X, whoseheight at x corresponds to the probability that X = x. Since the probabilitiesmust sum to unity, we requiren∑f(x i )=1. (30.39)i=1We may also define the cumulative probability function (CPF) of X, F(x), whosevalue gives the probability that X ≤ x, sothatF(x) =Pr(X ≤ x) = ∑ f(x i ). (30.40)x i≤x1139

PROBABILITYf(x)2pF(x)1p12 p 12 3 4 5 6x1 2 3 4 5 6(a)(b)Figure 30.6 (a) A typical probability function for a discrete distribution, thatfor the biased die discussed earlier. Since the probabilities must sum to unitywe require p =2/13. (b) The cumulative probability function for the samediscrete distribution. (Note that a different scale has been used for (b).)Hence F(x) is a step function that has upward jumps of p i at x = x i , i =1, 2,...,n, and is constant between possible values of X. We may also calculatethe probability that X lies between two limits, l 1 and l 2 (l 1

30.4 RANDOM VARIABLES AND DISTRIBUTIONSf(x)l 1 a b l 2xFigure 30.7 The probability density function for a continuous random variableX that can take values only between the limits l 1 and l 2 . The shaded areaunder the curve gives Pr(a

PROBABILITY◮A random variable X has a PDF f(x) given by Ae −x in the interval 0

30.5 PROPERTIES OF DISTRIBUTIONSIn many circumstances, however, random variables do not depend on oneanother, i.e. they are independent. As an example, for a person drawn at randomfrom a population, we might expect height and IQ to be independent randomvariables. Let us suppose that X and Y are two random variables with probabilitydensity functions g(x) andh(y) respectively. In mathematical terms, X and Y areindependent RVs if their joint probability density function is given by f(x, y) =g(x)h(y). Thus, for independent RVs, if X and Y are both discrete thenPr(X = x i , Y = y j )=g(x i )h(y j )or, if X and Y are both continuous, thenPr(x

PROBABILITYthe series is absolutely convergent or that the integral exists, as the case may be.From its definition it is straightforward to show that the expectation value hasthe following properties:(i) if a is a constant then E[a] =a;(ii) if a is a constant then E[ag(X)] = aE[g(X)];(iii) if g(X) =s(X)+t(X) thenE[ g(X)] = E[ s(X)] + E[t(X)].It should be noted that the expectation value is not a function of X but isinstead a number that depends on the form of the probability density functionf(x) and the function g(x). Most of the standard quantities used to characterisef(x) are simply the expectation values of various functions of the random variableX. We now consider these standard quantities.30.5.1 MeanThe property most commonly used to characterise a probability distribution isits mean, which is defined simply as the expectation value E[X] of the variable Xitself. Thus, the mean is given by{∑iE[X] =x if(x i ) for a discrete distribution,∫ (30.46)xf(x) dx for a continuous distribution.The alternative notations µ and 〈x〉 are also commonly used to denote the mean.If in (30.46) the series is not absolutely convergent, or the integral does not exist,we say that the distribution does not have a mean, but this is very rare in physicalapplications.◮The probability of finding a 1s electron in a hydrogen atom in a given infinitesimal volumedV is ψ ∗ ψdV, where the quantum mechanical wavefunction ψ is given byψ = Ae −r/a 0.Find the value of the real constant A and thereby deduce the mean distance of the electronfrom the origin.Let us consider the random variable R = ‘distance of the electron from the origin’. Sincethe 1s orbital has no θ- orφ-dependence (it is spherically symmetric), we may considerthe infinitesimal volume element dV as the spherical shell with inner radius r and outerradius r + dr. Thus, dV =4πr 2 dr and the PDF of R is simplyPr(r

30.5 PROPERTIES OF DISTRIBUTIONSIntegrating by parts we find A =1/(πa 3 0 )1/2 . Now, using the definition of the mean (30.46),we find∫ ∞E[R] = rf(r) dr = 4 r 3 e −2r/a 0dr.0a 3 0 0The integral on the RHS may be integrated by parts and takes the value 3a 4 0 /8; consequentlywe find that E[R] =3a 0 /2. ◭∫ ∞30.5.2 Mode and medianAlthough the mean discussed in the last section is the most common measureof the ‘average’ of a distribution, two other measures, which do not rely on theconcept of expectation values, are frequently encountered.The mode of a distribution is the value of the random variable X at which theprobability (density) function f(x) has its greatest value. If there is more than onevalue of X for which this is true then each value may equally be called the modeof the distribution.The median M of a distribution is the value of the random variable X at whichthe cumulative probability function F(x) takes the value 1 2 ,i.e.F(M) = 1 2 . Relatedto the median are the lower and upper quartiles Q l and Q u of the PDF, whichare defined such thatF(Q l )= 1 4 , F(Q u)= 3 4 .Thus the median and lower and upper quartiles divide the PDF into four regionseach containing one quarter of the probability. Smaller subdivisions are alsopossible, e.g. the nth percentile, P n , of a PDF is defined by F(P n )=n/100.◮Find the mode of the PDF for the distance from the origin of the electron whose wavefunctionwas given in the previous example.We found in the previous example that the PDF for the electron’s distance from the originwas given byf(r) = 4r2 e −2r/a 0. (30.47)a 3 0Differentiating f(r) with respect to r, weobtaindfdr = 8ra 3 0(1 − r a 0)e −2r/a 0.Thus f(r) has turning points at r =0andr = a 0 ,wheredf/dr = 0. It is straightforwardto show that r = 0 is a minimum and r = a 0 is a maximum. Moreover, it is also clear thatr = a 0 is a global maximum (as opposed to just a local one). Thus the mode of f(r) occursat r = a 0 . ◭1145

PROBABILITY30.5.3 Variance and standard deviationThe variance of a distribution, V [X], also written σ 2 , is defined byV [X] =E [ {∑(X − µ) 2] j=(x j − µ) 2 f(x j )∫(x − µ) 2 f(x) dxfor a discrete distribution,for a continuous distribution.(30.48)Here µ has been written for the expectation value E[X] ofX. As in the case ofthe mean, unless the series and the integral in (30.48) converge the distributiondoes not have a variance. From the definition (30.48) we may easily derive thefollowing useful properties of V [X]. If a and b are constants then(i) V [a] =0,(ii) V [aX + b] =a 2 V [X].The variance of a distribution is always positive; its positive square root isknown as the standard deviation of the distribution and is often denoted by σ.Roughly speaking, σ measures the spread (about x = µ) of the values that X canassume.◮Find the standard deviation of the PDF for the distance from the origin of the electronwhose wavefunction was discussed in the previous two examples.Inserting the expression (30.47) for the PDF f(r) into (30.48), the variance of the randomvariable R is given by∫ ∞V [R] = (r − µ) 2 4r2 e −2r/a 0dr = 4 (r 4 − 2r 3 µ + r 2 µ 2 )e −2r/a 0dr,0 a 3 0a 3 0 0where the mean µ = E[R] =3a 0 /2. Integrating each term in the integrand by parts weobtain∫ ∞V [R] =3a 2 0 − 3µa 0 + µ 2 = 3a2 04 .Thus the standard deviation of the distribution is σ = √ 3a 0 /2. ◭We may also use the definition (30.48) to derive the Bienaymé–Chebyshevinequality, which provides a useful upper limit on the probability that randomvariable X takes values outside a given range centred on the mean. Let us considerthe case of a continuous random variable, for which∫Pr(|X − µ| ≥c) = f(x) dx,|x−µ|≥cwhere the integral on the RHS extends over all values of x satisfying the inequality1146

30.5 PROPERTIES OF DISTRIBUTIONS|x − µ| ≥c. From (30.48), we find that∫∫σ 2 ≥ (x − µ) 2 f(x) dx ≥ c 2|x−µ|≥c|x−µ|≥cf(x) dx. (30.49)The first inequality holds because both (x − µ) 2 and f(x) are non-negative forall x, and the second inequality holds because (x − µ) 2 ≥ c 2 over the range ofintegration. However, the RHS of (30.49) is simply equal to c 2 Pr(|X − µ| ≥c),and thus we obtain the required inequalityPr(|X − µ| ≥c) ≤ σ2c 2 .A similar derivation may be carried through for the case of a discrete randomvariable. Thus, for any distribution f(x) that possesses a variance we have, forexample,Pr(|X − µ| ≥2σ) ≤ 1 4and Pr(|X − µ| ≥3σ) ≤ 1 9 .30.5.4 MomentsThe mean (or expectation) of X is sometimes called the first moment of X, sinceit is defined as the sum or integral of the probability density function multipliedby the first power of x. By a simple extension the kth moment of a distributionis defined by{∑µ k ≡ E[X k j]=xk j f(x j) for a discrete distribution,∫x k f(x) dx for a continuous distribution.(30.50)For notational convenience, we have introduced the symbol µ k to denote E[X k ],the kth moment of the distribution. Clearly, the mean of the distribution is thendenoted by µ 1 , often abbreviated simply to µ, as in the previous subsection, asthis rarely causes confusion.A useful result that relates the second moment, the mean and the variance ofa distribution is proved using the properties of the expectation operator:V [X] =E [ (X − µ) 2]= E [ X 2 − 2µX + µ 2]= E [ X 2] − 2µE[X]+µ 2= E [ X 2] − 2µ 2 + µ 2In alternative notations, this result can be written= E [ X 2] − µ 2 . (30.51)〈(x − µ) 2 〉 = 〈x 2 〉−〈x〉 2 or σ 2 = µ 2 − µ 2 1.1147

PROBABILITY◮A biased die has probabilities p/2, p, p, p, p, 2p of showing 1, 2, 3, 4, 5, 6 respectively. Find(i) the mean, (ii) the second moment and (iii) the variance of this probability distribution.By demanding that the sum of the probabilities equals unity we require p =2/13. Now,using the definition of the mean (30.46) for a discrete distribution,E[X] = ∑ jx j f(x j )=1× 1 p + 2× p + 3× p + 4× p + 5× p + 6× 2p2= 53 2 p = 532 × 213 = 5313 .Similarly, using the definition of the second moment (30.50),E[X 2 ]= ∑ x 2 j f(x j )=1 2 × 1 p 2 +22 p +3 2 p +4 2 p +5 2 p +6 2 × 2pj= 2532 p = 25313 .Finally, using the definition of the variance (30.48), with µ =53/13, we obtainV [X] = ∑ (x j − µ) 2 f(x j )j=(1− µ) 2 1 p +(2− 2 µ)2 p +(3− µ) 2 p +(4− µ) 2 p +(5− µ) 2 p +(6− µ) 2 2p( ) 3120= p = 480169 169 .It is easy to verify that V [X] =E [ X 2] − (E[X]) 2 . ◭In practice, to calculate the moments of a distribution it is often simpler to usethe moment generating function discussed in subsection 30.7.2. This is particularlytrue for higher-order moments, where direct evaluation of the sum or integral in(30.50) can be somewhat laborious.30.5.5 Central momentsThe variance V [X] is sometimes called the second central moment of the distribution,since it is defined as the sum or integral of the probability density functionmultiplied by the second power of x − µ. The origin of the term ‘central’ is that bysubtracting µ from x before squaring we are considering the moment about themean of the distribution, rather than about x = 0. Thus the kth central momentof a distribution is defined asν k ≡ E [ {∑(X − µ) k] j=(x j − µ) k f(x j ) for a discrete distribution,∫ (30.52)(x − µ) k f(x) dx for a continuous distribution.It is convenient to introduce the notation ν k for the kth central moment. ThusV [X] ≡ ν 2 and we may write (30.51) as ν 2 = µ 2 − µ 2 1 . Clearly, the first centralmoment of a distribution is always zero since, for example in the continuous case,∫∫∫ν 1 = (x − µ)f(x) dx = xf(x) dx − µ f(x) dx = µ − (µ × 1) = 0.1148

30.5 PROPERTIES OF DISTRIBUTIONSWe note that the notation µ k and ν k for the moments and central momentsrespectively is not universal. Indeed, in some books their meanings are reversed.We can write the kth central moment of a distribution in terms of its kth andlower-order moments by expanding (X − µ) k in powers of X. We have alreadynoted that ν 2 = µ 2 − µ 2 1 , and similar expressions may be obtained for higher-ordercentral moments. For example,In general, it is straightforward to show thatν 3 = E [ (X − µ 1 ) 3]= E [ X 3 − 3µ 1 X 2 +3µ 2 1X − µ 3 ]1= µ 3 − 3µ 1 µ 2 +3µ 2 1µ 1 − µ 3 1= µ 3 − 3µ 1 µ 2 +2µ 3 1. (30.53)ν k = µ k − k C 1 µ k−1 µ 1 + ···+(−1) r k C r µ k−r µ r 1 + ···+(−1) k−1 ( k C k−1 − 1)µ k 1.(30.54)Once again, direct evaluation of the sum or integral in (30.52) can be rathertedious for higher moments, and it is usually quicker to use the moment generatingfunction (see subsection 30.7.2), from which the central moments can be easilyevaluated as well.◮The PDF for a Gaussian distribution (see subsection 30.9.1) with mean µ and varianceσ 2 is given byf(x) = 1 [ ]σ √ 2π exp (x − µ)2− .2σ 2Obtain an expression for the kth central moment of this distribution.As an illustration, we will perform this calculation by evaluating the integral in (30.52)directly. Thus, the kth central moment of f(x) isgivenbyν k =∫ ∞−∞= 1σ √ 2π= 1σ √ 2π(x − µ) k f(x) dx∫ ∞−∞∫ ∞−∞](x − µ) k (x − µ)2exp[−2σ 2dx)y k exp(− y2dy, (30.55)2σ 2where in the last line we have made the substitution y = x − µ. It is clear that if k isodd then the integrand is an odd function of y and hence the integral equals zero. Thus,ν k =0ifk is odd. When k is even, we could calculate ν k by integrating by parts to obtaina reduction formula, but it is more elegant to consider instead the standard integral (seesubsection 6.4.2)I =∫ ∞−∞exp(−αy 2 ) dy = π 1/2 α −1/2 ,1149

PROBABILITYand differentiate it repeatedly with respect to α (see section 5.12). Thus, we obtain∫dI∞dα = − y 2 exp(−αy 2 ) dy = − 1 2 π1/2 α −3/2−∞d 2 ∫I ∞dα = y 4 exp(−αy 2 ) dy =( 1 )( 3 2 2 2 )π1/2 α −5/2−∞.d n ∫I∞dα n =(−1)n y 2n exp(−αy 2 ) dy =(−1) n ( 1 )( 3 ) ···( 1 (2n − 2 2 2 1))π1/2 α −(2n+1)/2 .−∞Setting α =1/(2σ 2 ) and substituting the above result into (30.55), we find (for k even)ν k =( 1 )( 3 ) ···( 1 (k − 2 2 2 1))(2σ2 ) k/2 = (1)(3) ···(k − 1)σ k . ◭One may also characterise a probability distribution f(x) using the closelyrelated normalised and dimensionless central momentsγ k ≡ν k= ν kν k/2 σ k .2From this set, γ 3 and γ 4 are more commonly called, respectively, the skewnessand kurtosis of the distribution. The skewness γ 3 of a distribution is zero if it issymmetrical about its mean. If the distribution is skewed to values of x smallerthan the mean then γ 3 < 0. Similarly γ 3 > 0 if the distribution is skewed to highervalues of x.From the above example, we see that the kurtosis of the Gaussian distribution(subsection 30.9.1) is given byγ 4 = ν 4ν22 = 3σ4σ 4 =3.It is therefore common practice to define the excess kurtosis of a distributionas γ 4 − 3. A positive value of the excess kurtosis implies a relatively narrowerpeak and wider wings than the Gaussian distribution with the same mean andvariance. A negative excess kurtosis implies a wider peak and shorter wings.Finally, we note here that one can also describe a probability density functionf(x) in terms of its cumulants, which are again related to the central moments.However, we defer the discussion of cumulants until subsection 30.7.4, since theirdefinition is most easily understood in terms of generating functions.30.6 Functions of random variablesSuppose X is some random variable for which the probability density functionf(x) is known. In many cases, we are more interested in a related random variableY = Y (X), where Y (X) is some function of X. What is the probability density1150

30.6 FUNCTIONS OF RANDOM VARIABLESfunction g(y) for the new random variable Y ? We now discuss how to obtainthis function.30.6.1 Discrete random variablesIf X is a discrete RV that takes only the values x i , i =1, 2,...,n,thenY mustalso be discrete and takes the values y i = Y (x i ), although some of these valuesmay be identical. The probability function for Y is given by{∑jg(y) =f(x j) if y = y i ,(30.56)0 otherwise,where the sum extends over those values of j for which y i = Y (x j ). The simplestcase arises when the function Y (X) possesses a single-valued inverse X(Y ). In thiscase, only one x-value corresponds to each y-value, and we obtain a closed-formexpression for g(y) given by{f(x(yi )) if y = y i ,g(y) =0 otherwise.If Y (X) does not possess a single-valued inverse then the situation is morecomplicated and it may not be possible to obtain a closed-form expression forg(y). Nevertheless, whatever the form of Y (X), one can always use (30.56) toobtain the numerical values of the probability function g(y) aty = y i .30.6.2 Continuous random variablesIf X is a continuous RV, then so too is the new random variable Y = Y (X). Theprobability that Y lies in the range y to y + dy is given by∫g(y) dy = f(x) dx, (30.57)where dS corresponds to all values of x for which Y lies in the range y to y + dy.Once again the simplest case occurs when Y (X) possesses a single-valued inverseX(Y ). In this case, we may writeg(y) dy =∣∫ x(y+dy)x(y)dS∣ ∫ ∣∣∣ x(y)+| dxf(x ′ ) dx ′ =x(y)dy | dyf(x ′ ) dx ′ ,from which we obtaing(y) =f(x(y))dx∣ dy ∣ . (30.58)1151

PROBABILITYθlighthouseLbeamOcoastlineyFigure 30.8The illumination of a coastline by the beam from a lighthouse.◮A lighthouse is situated at a distance L from a straight coastline, opposite a point O, andsends out a narrow continuous beam of light simultaneously in opposite directions. The beamrotates with constant angular velocity. If the random variable Y is the distance along thecoastline, measured from O, of the spot that the light beam illuminates, find its probabilitydensity function.The situation is illustrated in figure 30.8. Since the light beam rotates at a constant angularvelocity, θ is distributed uniformly between −π/2 and π/2, and so f(θ) =1/π. Nowy = L tan θ, which possesses the single-valued inverse θ =tan −1 (y/L), provided that θ liesbetween −π/2 andπ/2. Since dy/dθ = L sec 2 θ = L(1 + tan 2 θ)=L[1 + (y/L) 2 ], from(30.58) we findg(y) = 1 dθπ ∣ dy ∣ = 1for −∞

30.6 FUNCTIONS OF RANDOM VARIABLESYy + dyydx 1 dx 2XFigure 30.9 Illustration of a function Y (X) whose inverse X(Y ) is a doublevaluedfunction of Y . The range y to y + dy corresponds to X being either inthe range x 1 to x 1 + dx 1 or in the range x 2 to x 2 + dx 2 .This result may be generalised straightforwardly to the case where the range y toy + dy corresponds to more than two x-intervals.◮The random variable X is Gaussian distributed (see subsection 30.9.1) with mean µ andvariance σ 2 . Find the PDF of the new variable Y =(X − µ) 2 /σ 2 .It is clear that X(Y ) is a double-valued function of Y . However, in this case, it isstraightforward to obtain single-valued functions giving the two values of x that correspondto a given value of y; thesearex 1 = µ − σ √ y and x 2 = µ + σ √ y,where √ y is taken tomean the positive square root.The PDF of X is given byf(x) = 1 [ ]σ √ 2π exp (x − µ)2− .2σ 2Since dx 1 /dy = −σ/(2 √ y)anddx 2 /dy = σ/(2 √ y), from (30.59) we obtaing(y) = 1∣ ∣∣∣σ √ 2π exp(− 1 y) −σ22 √ y ∣ + 1∣ ∣∣∣σ √ 2π exp(− 1 y) σ22 √ y ∣= 12 √ π ( 1 2 y)−1/2 exp(− 1 y). 2As we shall see in subsection 30.9.3, this is the gamma distribution γ( 1 , 1 ). ◭ 2 230.6.3 Functions of several random variablesWe may extend our discussion further, to the case in which the new randomvariable is a function of several other random variables. For definiteness, let usconsider the random variable Z = Z(X,Y ), which is a function of two otherRVs X and Y . Given that these variables are described by the joint probabilitydensity function f(x, y), we wish to find the probability density function p(z) ofthe variable Z.1153

PROBABILITYIf X and Y are both discrete RVs thenp(z) = ∑ i,jf(x i ,y j ), (30.60)where the sum extends over all values of i and j for which Z(x i ,y j )=z. Similarly,if X and Y are both continuous RVs then p(z) is found by requiring that∫∫p(z) dz = f(x, y) dx dy, (30.61)dSwhere dS is the infinitesimal area in the xy-plane lying between the curvesZ(x, y) =z and Z(x, y) =z + dz.◮Suppose X and Y are independent continuous random variables in the range −∞ to ∞,with PDFs g(x) and h(y) respectively. Obtain expressions for the PDFs of Z = X + Y andW = XY .Since X and Y are independent RVs, their joint PDF is simply f(x, y) =g(x)h(y). Thus,from (30.61), the PDF of the sum Z = X + Y is given by∫ ∞∫ z+dz−xp(z) dz = dx g(x) dy h(y)−∞z−x(∫ ∞)= g(x)h(z − x) dx dz.−∞Thus p(z) istheconvolution of the PDFs of g and h (i.e. p = g ∗ h, see subsection 13.1.7).In a similar way, the PDF of the product W = XY is given by∫ ∞∫ (w+dw)/|x|q(w) dw = dx g(x) dy h(y)−∞w/|x|(∫ ∞= g(x)h(w/x) dx )dw ◭−∞|x|The prescription (30.61) is readily generalised to functions of n random variablesZ = Z(X 1 ,X 2 ,...,X n ), in which case the infinitesimal ‘volume’ element dS is theregion in x 1 x 2 ···x n -space between the (hyper)surfaces Z(x 1 ,x 2 ,...,x n )=z andZ(x 1 ,x 2 ,...,x n )=z + dz. In practice, however, the integral is difficult to evaluate,since one is faced with the complicated geometrical problem of determining thelimits of integration. Fortunately, an alternative (and powerful) technique existsfor evaluating integrals of this kind. One eliminates the geometrical problem byintegrating over all values of the variables x i without restriction, while shiftingthe constraint on the variables to the integrand. This is readily achieved bymultiplying the integrand by a function that equals unity in the infinitesimalregion dS and zero elsewhere. From the discussion of the Dirac delta function insubsection 13.1.3, we see that δ(Z(x 1 ,x 2 ,...,x n )−z) dz satisfies these requirements,and so in the most general case we have∫∫ ∫p(z) = ··· f(x 1 ,x 2 ,...,x n )δ(Z(x 1 ,x 2 ,...,x n ) − z) dx 1 dx 2 ...dx n ,(30.62)1154

30.6 FUNCTIONS OF RANDOM VARIABLESwhere the range of integration is over all possible values of the variables x i .Thisintegral is most readily evaluated by substituting in (30.62) the Fourier integralrepresentation of the Dirac delta function discussed in subsection 13.1.4, namelyδ(Z(x 1 ,x 2 ,...,x n ) − z) = 1 ∫ ∞e ik(Z(x1,x2,...,xn)−z) dk. (30.63)2π −∞This is best illustrated by considering a specific example.◮A general one-dimensional random walk consists of n independent steps, each of whichcan be of a different length and in either direction along the x-axis. If g(x) is the PDF forthe (positive or negative) displacement X along the x-axis achieved in a single step, obtainan expression for the PDF of the total displacement S after n steps.The total displacement S is simply the algebraic sum of the displacements X i achieved ineach of the n steps, so thatS = X 1 + X 2 + ···+ X n .Since the random variables X i are independent and have the same PDF g(x), their jointPDF is simply g(x 1 )g(x 2 ) ···g(x n ). Substituting this into (30.62), together with (30.63), weobtain∫ ∞ ∫ ∞ ∫ ∞p(s) = ··· g(x 1 )g(x 2 ) ···g(x n ) 1 ∫ ∞e ik[(x 1+x 2 +···+x n)−s] dk dx 1 dx 2 ···dx n2π= 12π−∞ −∞∫ ∞−∞−∞−∞(∫ ∞ndk e −iks g(x)e dx) ikx . (30.64)−∞It is convenient to define the characteristic function C(k) of the variable X asC(k) =∫ ∞−∞g(x)e ikx dx,which is simply related to the Fourier transform of g(x). Then (30.64) may be written asp(s) = 1 ∫ ∞e −iks [C(k)] n dk.2π −∞Thus p(s) can be found by evaluating two Fourier integrals. Characteristic functions willbe discussed in more detail in subsection 30.7.3. ◭30.6.4 Expectation values and variancesIn some cases, one is interested only in the expectation value or the varianceof the new variable Z rather than in its full probability density function. Fordefiniteness, let us consider the random variable Z = Z(X,Y ), which is a functionof two RVs X and Y with a known joint distribution f(x, y); the results we willobtain are readily generalised to more (or fewer) variables.It is clear that E[Z] andV [Z] can be obtained, in principle, by first using themethods discussed above to obtain p(z) and then evaluating the appropriate sumsor integrals. The intermediate step of calculating p(z) is not necessary, however,since it is straightforward to obtain expressions for E[Z] andV [Z] in terms of1155

PROBABILITYthe variables X and Y . For example, if X and Y are continuous RVs then theexpectation value of Z is given by∫∫∫E[Z] = zp(z) dz = Z(x, y)f(x, y) dx dy. (30.65)An analogous result exists for discrete random variables.Integrals of the form (30.65) are often difficult to evaluate. Nevertheless, wemay use (30.65) to derive an important general result concerning expectationvalues. If X and Y are any two random variables and a and b are arbitraryconstants then by letting Z = aX + bY we findE[aX + bY ]=aE[X]+bE[Y ].Furthermore, we may use this result to obtain an approximate expression for theexpectation value E[ Z(X,Y )] of any arbitrary function of X and Y . Letting µ X =E[X] andµ Y = E[Y ], and provided Z(X,Y ) can be reasonably approximated bythe linear terms of its Taylor expansion about the point (µ X ,µ Y ), we haveZ(X,Y ) ≈ Z(µ X ,µ Y )+( ∂Z∂X)(X − µ X )+( ∂Z∂Y)(Y − µ Y ),(30.66)where the partial derivatives are evaluated at X = µ X and Y = µ Y . Taking theexpectation values of both sides, we find( )( )∂Z∂ZE[ Z(X,Y )] ≈ Z(µ X ,µ Y )+ (E[X]−µ X )+ (E[Y ]−µ Y )=Z(µ X ,µ Y ),∂X∂Ywhich gives the approximate result E[ Z(X,Y )] ≈ Z(µ X ,µ Y ).By analogy with (30.65), the variance of Z = Z(X,Y ) is given by∫∫∫V [Z] = (z − µ Z ) 2 p(z) dz = [Z(x, y) − µ Z ] 2 f(x, y) dx dy,(30.67)where µ Z = E[Z]. We may use this expression to derive a second useful result. IfX and Y are two independent random variables, so that f(x, y) =g(x)h(y), anda, b and c are constants then by setting Z = aX + bY + c in (30.67) we obtainV [aX + bY + c] =a 2 V [X]+b 2 V [Y ]. (30.68)From (30.68) we also obtain the important special caseV [X + Y ]=V [X − Y ]=V [X]+V [Y ].Provided X and Y are indeed independent random variables, we may obtainan approximate expression for V [ Z(X,Y )], for any arbitrary function Z(X,Y ),in a similar manner to that used in approximating E[ Z(X,Y )] above. Taking the1156

30.7 GENERATING FUNCTIONSvariance of both sides of (30.66), and using (30.68), we find( ) 2 ( ) 2 ∂Z∂ZV [ Z(X,Y )] ≈ V [X]+ V [Y ], (30.69)∂X∂Ythe partial derivatives being evaluated at X = µ X and Y = µ Y .30.7 Generating functionsAs we saw in chapter 16, when dealing with particular sets of functions f n ,each member of the set being characterised by a different non-negative integern, it is sometimes possible to summarise the whole set by a single function of adummy variable (say t), called a generating function. The relationship betweenthe generating function and the nth member f n of the set is that if the generatingfunction is expanded as a power series in t then f n is the coefficient of t n .Forexample, in the expansion of the generating function G(z,t) =(1− 2zt + t 2 ) −1/2 ,the coefficient of t n is the nth Legendre polynomial P n (z), i.e.∞∑G(z,t) =(1− 2zt + t 2 ) −1/2 = P n (z)t n .We found that many useful properties of, and relationships between, the membersof a set of functions could be established using the generating function and otherfunctions obtained from it, e.g. its derivatives.Similar ideas can be used in the area of probability theory, and two types ofgenerating function can be usefully defined, one more generally applicable thanthe other. The more restricted of the two, applicable only to discrete integraldistributions, is called a probability generating function; this is discussed in thenext section. The second type, a moment generating function, can be used withboth discrete and continuous distributions and is considered in subsection 30.7.2.From the moment generating function, we may also construct the closely relatedcharacteristic and cumulant generating functions; these are discussed insubsections 30.7.3 and 30.7.4 respectively.n=030.7.1 Probability generating functionsAs already indicated, probability generating functions are restricted in applicabilityto integer distributions, of which the most common (the binomial, the Poissonand the geometric) are considered in this and later subsections. In such distributionsa random variable may take only non-negative integer values. The actualpossible values may be finite or infinite in number, but, for formal purposes,all integers, 0, 1, 2,... are considered possible. If only a finite number of integervalues can occur in any particular case then those that cannot occur are includedbut are assigned zero probability.1157

PROBABILITYIf, as previously, the probability that the random variable X takes the value x nis f(x n ), then∑f(x n )=1.nIn the present case, however, only non-negative integer values of x n are possible,and we can, without ambiguity, write the probability that X takes the value n asf n , with∞∑f n =1. (30.70)We may now define the probability generating function Φ X (t) byn=0Φ X (t) ≡∞∑f n t n . (30.71)It is immediately apparent that Φ X (t) =E[t X ] and that, by virtue of (30.70),Φ X (1) = 1.Probably the simplest example of a probability generating function (PGF) isprovided by the random variable X defined by{1 if the outcome of a single trial is a ‘success’,X =0 if the trial ends in ‘failure’.If the probability of success is p and that of failure q (= 1 − p) thenn=0Φ X (t) =qt 0 + pt 1 +0+0+···= q + pt. (30.72)This type of random variable is discussed much more fully in subsection 30.8.1.In a similar but slightly more complicated way, a Poisson-distributed integervariable with mean λ (see subsection 30.8.4) has a PGFΦ X (t) =∞∑n=0e −λ λ nt n = e −λ e λt . (30.73)n!We note that, as required, Φ X (1) = 1 in both cases.Useful results will be obtained from this kind of approach only if the summation(30.71) can be carried out explicitly in particular cases and the functions derivedfrom Φ X (t) can be shown to be related to meaningful parameters. Two suchrelationships can be obtained by differentiating (30.71) with respect to t. Takingthe first derivative we finddΦ X (t)dt∞∑∞∑= nf n t n−1 ⇒ Φ ′ X(1) = nf n = E[X], (30.74)n=01158n=0

30.7 GENERATING FUNCTIONSand differentiating once more we obtaind 2 Φ X (t)∞∑dt 2 = n(n − 1)f n t n−2 ⇒ Φ ′′ X (1) = ∑∞ n(n − 1)f n = E[X(X − 1)].n=0n=0(30.75)Equation (30.74) shows that Φ ′ X (1) gives the mean of X. Using both (30.75) and(30.51) allows us to writeΦ ′′ X(1) + Φ ′ X(1) − [ Φ ′ X(1) ] 2= E[X(X − 1)] + E[X] − (E[X])2= E [ X 2] − E[X]+E[X] − (E[X]) 2= E [ X 2] − (E[X]) 2= V [X], (30.76)and so express the variance of X in terms of the derivatives of its probabilitygenerating function.◮A random variable X is given by the number of trials needed to obtain a first successwhen the chance of success at each trial is constant and equal to p. Find the probabilitygenerating function for X and use it to determine the mean and variance of X.Clearly, at least one trial is needed, and so f 0 =0.Ifn (≥ 1) trials are needed for the firstsuccess, the first n − 1 trials must have resulted in failure. ThusPr(X = n) =q n−1 p, n ≥ 1, (30.77)where q =1− p is the probability of failure in each individual trial.The corresponding probability generating function is thus∞∑∞∑Φ X (t) = f n t n = (q n−1 p)t nn=0n=1= p ∞∑(qt) n = p qq × qt1 − qt = pt1 − qt , (30.78)n=1where we have used the result for the sum of a geometric series, given in chapter 4, toobtain a closed-form expression for Φ X (t). Again, as must be the case, Φ X (1) = 1.To find the mean and variance of X we need to evaluate Φ ′ X (1) and Φ′′ X (1). Differentiating(30.78) givesΦ ′ pX(t) =⇒ Φ ′(1 − qt)X(1) = p 2 p = 1 2 p ,Φ ′′ X(t) =2pq ⇒ Φ ′′(1 − qt)X(1) = 2pq = 2q3 p 3 p . 2Thus, using (30.74) and (30.76),E[X] =Φ ′ X(1) = 1 p ,V [X] =Φ ′′ X(1) + Φ ′ X(1) − [Φ ′ X(1)] 2= 2qp 2 + 1 p − 1 p 2 = q p 2 .A distribution with probabilities of the general form (30.77) is known as a geometricdistribution and is discussed in subsection 30.8.2. This form of distribution is common in‘waiting time’ problems (subsection 30.9.3). ◭1159

PROBABILITYnr = nrFigure 30.10The pairs of values of n and r used in the evaluation of Φ X+Y (t).Sums of random variablesWe now turn to considering the sum of two or more independent randomvariables, say X and Y , and denote by S 2 the random variableS 2 = X + Y.If Φ S2 (t) isthePGFforS 2 , the coefficient of t n in its expansion is given by theprobability that X + Y = n and is thus equal to the sum of the probabilities thatX = r and Y = n − r for all values of r in 0 ≤ r ≤ n. Since such outcomes fordifferent values of r are mutually exclusive, we havePr(X + Y = n) =∞∑Pr(X = r)Pr(Y = n − r). (30.79)r=0Multiplying both sides of (30.79) by t n and summing over all values of n enablesus to express this relationship in terms of probability generating functions asfollows:∞∑∞∑ n∑Φ X+Y (t) = Pr(X + Y = n)t n = Pr(X = r)t r Pr(Y = n − r)t n−rn=0=n=0 r=0∞∑r=0 n=r∞∑Pr(X = r)t r Pr(Y = n − r)t n−r .The change in summation order is justified by reference to figure 30.10, whichillustrates that the summations are over exactly the same pairs of values of n andr, but with the first (inner) summation over the points in a column rather thanover the points in a row. Now, setting n = r + s gives the final result,Φ X+Y (t) =∞∑Pr(X = r)t rr=0∞ ∑s=0Pr(Y = s)t s=Φ X (t)Φ Y (t), (30.80)1160

30.7 GENERATING FUNCTIONSi.e. the PGF of the sum of two independent random variables is equal to theproduct of their individual PGFs. The same result can be deduced in a less formalway by noting that if X and Y are independent thenE [ t X+Y ] = E [ t X] E [ t Y ] .Clearly result (30.80) can be extended to more than two random variables bywriting S 3 = S 2 + Z etc., to given∏Φ ∑(n Xi)(t) = Φ Xi (t), (30.81)i=1and, further, if all the X i have the same probability distribution,i=1Φ (∑ ni=1 Xi)(t) =[Φ X(t)] n . (30.82)This latter result has immediate application in the deduction of the PGF for thebinomial distribution from that for a single trial, equation (30.72).Variable-length sums of random variablesAs a final result in the theory of probability generating functions we show how tocalculate the PGF for a sum of N random variables, all with the same probabilitydistribution, when the value of N is itself a random variable but one with aknown probability distribution. In symbols, we wish to find the distributionofS N = X 1 + X 2 + ···+ X N , (30.83)where N is a random variable with Pr(N = n) = h n and PGF χ N (t) =∑hn t n .The probability ξ k that S N = k is given by a sum of conditional probabilities,namely § ∞∑ξ k = Pr(N = n)Pr(X 0 + X 1 + X 2 + ···+ X n = k)n=0∞∑= h n × coefficient of t k in [Φ X (t)] n .n=0Multiplying both sides of this equation by t k and summing over all k, weobtain§ Formally X 0 = 0 has to be included, since Pr(N = 0) may be non-zero.1161

PROBABILITYan expression for the PGF Ξ S (t) ofS N :∞∑∞∑ ∑∞Ξ S (t) = ξ k t k = t k h n × coefficient of t k in [Φ X (t)] nk=0==k=0∞∑n=0h nn=0∑∞k=0∞∑h n [Φ X (t)] nn=0t k × coefficient of t k in [Φ X (t)] n= χ N (Φ X (t)). (30.84)In words, the PGF of the sum S N is given by the compound function χ N (Φ X (t))obtained by substituting Φ X (t) fort in the PGF for the number of terms N inthe sum. We illustrate this with the following example.◮The probability distribution for the number of eggs in a clutch is Poisson distributed withmean λ, and the probability that each egg will hatch is p (and is independent of the size ofthe clutch). Use the results stated in (30.72) and (30.73) to show that the PGF (and hencethe probability distribution) for the number of chicks that hatch corresponds to a Poissondistribution having mean λp.The number of chicks that hatch is given by a sum of the form (30.83) in which X i =1ifthe ith chick hatches and X i = 0 if it does not. As given by (30.72), Φ X (t) is thus (1−p)+pt.The value of N is given by a Poisson distribution with mean λ; thus, from (30.73), in theterminology of our previous discussion,χ N (t) =e −λ e λt .We now substitute these forms into (30.84) to obtainΞ S (t) =exp(−λ)exp[λΦ X (t)]=exp(−λ)exp{λ[(1 − p)+pt]}=exp(−λp)exp(λpt).But this is exactly the PGF of a Poisson distribution with mean λp.That this implies that the probability is Poisson distributed is intuitively obvious since,in the expansion of the PGF as a power series in t, every coefficient will be preciselythat implied by such a distribution. A solution of the same problem by direct calculationappears in the answer to exercise 30.29. ◭30.7.2 Moment generating functionsAs we saw in section 30.5 a probability function is often expressed in terms ofits moments. This leads naturally to the second type of generating function, amoment generating function. For a random variable X, and a real number t, themoment generating function (MGF) is defined byM X (t) =E [ {∑e tX] if(x=etxi i ) for a discrete distribution,∫e tx f(x) dx for a continuous distribution.(30.85)1162

30.7 GENERATING FUNCTIONSThe MGF will exist for all values of t provided that X is bounded and alwaysexists at the point t =0whereM(0) = E(1) = 1.It will be apparent that the PGF and the MGF for a random variable Xare closely related. The former is the expectation of t X whilst the latter is theexpectation of e tX :Φ X (t) =E [ t X] , M X (t) =E [ e tX] .The MGF can thus be obtained from the PGF by replacing t by e t , and viceversa. The MGF has more general applicability, however, since it can be usedwith both continuous and discrete distributions whilst the PGF is restricted tonon-negative integer distributions.As its name suggests, the MGF is particularly useful for obtaining the momentsof a distribution, as is easily seen by noting thatE [ e tX] = E[1+tX + t2 X 2 ]+ ···2!=1+E[X]t + E [ X 2] t 22! + ··· .Assuming that the MGF exists for all t around the point t = 0, we can deducethat the moments of a distribution are given in terms of its MGF by∣E[X n ]= dn M X (t) ∣∣∣t=0dt n . (30.86)Similarly, by substitution in (30.51), the variance of the distribution is given byV [X] =M ′′ X(0) − [ M ′ X(0) ] 2, (30.87)where the prime denotes differentiation with respect to t.◮The MGF for the Gaussian distribution (see the end of subsection 30.9.1) is given byM X (t) =exp ( µt + 1 2 σ2 t 2) .Find the expectation and variance of this distribution.Using (30.86),M X(t) ′ = ( µ + σ 2 t ) exp ( µt + 1 2 σ2 t 2) ⇒ E[X] =M X(0) ′ = µ,M X(t) ′′ = [ σ 2 +(µ + σ 2 t) 2] exp ( µt + 1 2 σ2 t 2) ⇒ M X(0) ′′ = σ 2 + µ 2 .Thus, using (30.87),V [X] =σ 2 + µ 2 − µ 2 = σ 2 .That the mean is found to be µ and the variance σ 2 justifies the use of these symbols inthe Gaussian distribution. ◭The moment generating function has several useful properties that follow fromits definition and can be employed in simplifying calculations.1163

PROBABILITYScaling and shiftingIf Y = aX + b, wherea and b are arbitrary constants, thenM Y (t) =E [ e tY ] = E [ e t(aX+b)] = e bt E [ e atX] = e bt M X (at). (30.88)This result is often useful for obtaining the central moments of a distribution. If theMFG of X is M X (t) then the variable Y = X−µ has the MGF M Y (t) =e −µt M X (t),which clearly generates the central moments of X, i.e.( )E[(X − µ) n ]=E[Y n ]=M (n) dnY (0) = dt n [e−µt M X (t)] .t=0Sums of random variablesIf X 1 ,X 2 ,...,X N are independent random variables and S N = X 1 + X 2 + ···+ X NthenM SN (t) =E [ e ] tSN = E [ [ N]e t(X1+X2+···+XN)] ∏= E e tXi .Since the X i are independent,N∏M SN (t) = E [ e tXi] N∏= M Xi (t). (30.89)i=1In words, the MGF of the sum of N independent random variables is the productof their individual MGFs. By combining (30.89) with (30.88), we obtain the moregeneral result that the MGF of S N = c 1 X 1 + c 2 X 2 + ···+ c N X N (where the c i areconstants) is given byM SN (t) =i=1i=1N∏M Xi (c i t). (30.90)i=1Variable-length sums of random variablesLet us consider the sum of N independent random variables X i (i =1, 2,...,N), allwith the same probability distribution, and let us suppose that N is itself a randomvariable with a known distribution. Following the notation of section 30.7.1,S N = X 1 + X 2 + ···+ X N ,where N is a random variable with Pr(N = n) =h n and probability generatingfunction χ N (t) = ∑ h n t n . For definiteness, let us assume that the X i are continuousRVs (an analogous discussion can be given in the discrete case). Thus, the1164

30.7 GENERATING FUNCTIONSprobability that value of S N lies in the interval s to s + ds is given by §Pr(s

PROBABILITYso that C X (t) =M X (it), where M X (t) istheMGFofX. Clearly, the characteristicfunction and the MGF are very closely related and can be used interchangeably.Because of the formal similarity between the definitions of C X (t) andM X (t), thecharacteristic function possesses analogous properties to those listed in the previoussection for the MGF, with only minor modifications. Indeed, by substituting itfor t in any of the relations obeyed by the MGF and noting that C X (t) =M X (it),we obtain the corresponding relationship for the characteristic function. Thus, forexample, the moments of X are given in terms of the derivatives of C X (t) byE[X n ]=(−i) n C (n)X (0).Similarly, if Y = aX + b then C Y (t) =e ibt C X (at).Whether to describe a random variable by its characteristic function or by itsMGF is partly a matter of personal preference. However, the use of the CF doeshave some advantages. Most importantly, the replacement of the exponential e tXin the definition of the MGF by the complex oscillatory function e itX in the CFmeans that in the latter we avoid any difficulties associated with convergence ofthe relevant sum or integral. Furthermore, when X is a continous RV, we seefrom (30.91) that C X (t) is related to the Fourier transform of the PDF f(x). Asa consequence of Fourier’s inversion theorem, we may obtain f(x) fromC X (t) byperforming the inverse transformf(x) = 1 ∫ ∞C X (t)e −itx dt.2π−∞30.7.4 Cumulant generating functionAs mentioned at the end of subsection 30.5.5, we may also describe a probabilitydensity function f(x) in terms of its cumulants. These quantities may be expressedin terms of the moments of the distribution and are important in sampling theory,which we discuss in the next chapter. The cumulants of a distribution are bestdefined in terms of its cumulant generating function (CGF), given by K X (t) =ln M X (t) whereM X (t) is the MGF of the distribution. If K X (t) isexpandedasapower series in t then the kth cumulant κ k of f(x) is the coefficient of t k /k!:t 2K X (t) =lnM X (t) ≡ κ 1 t + κ 22! + κ t 33 + ··· . (30.92)3!Since M X (0) = 1, K X (t) contains no constant term.◮Find all the cumulants of the Gaussian distribution discussed in the previous example.The moment generating function for the Gaussian distribution is M X (t) =exp ( µt + 1 2 σ2 t 2) .Thus, the cumulant generating function has the simple formK X (t) =lnM X (t) =µt + 1 2 σ2 t 2 .1166

30.7 GENERATING FUNCTIONSComparing this expression with (30.92), we find that κ 1 = µ, κ 2 = σ 2 and all othercumulants are equal to zero. ◭We may obtain expressions for the cumulants of a distribution in terms of itsmoments by differentiating (30.92) with respect to t to givedK X= 1 dM X.dt M X dtExpanding each term as power series in t and cross-multiplying, we obtaint(κ 2···)(1 + κ 2 t + κ 32! + t 2···)1+µ 1 t + µ 22! + t=(µ 2···)1 + µ 2 t + µ 32! + ,and, on equating coefficients of like powers of t on each side, we findµ 1 = κ 1 ,µ 2 = κ 2 + κ 1 µ 1 ,µ 3 = κ 3 +2κ 2 µ 1 + κ 1 µ 2 ,µ 4 = κ 4 +3κ 3 µ 1 +3κ 2 µ 2 + κ 1 µ 3 ,.µ k = κ k + k−1 C 1 κ k−1 µ 1 + ···+ k−1 C r κ k−r µ r + ···+ κ 1 µ k−1 .Solving these equations for the κ k , we obtain (for the first four cumulants)κ 1 = µ 1 ,κ 2 = µ 2 − µ 2 1 = ν 2 ,κ 3 = µ 3 − 3µ 2 µ 1 +2µ 3 1 = ν 3 ,κ 4 = µ 4 − 4µ 3 µ 1 +12µ 2 µ 2 1 − 3µ2 2 − 6µ4 1 = ν 4 − 3ν2 2 . (30.93)Higher-order cumulants may be calculated in the same way but become increasinglylengthy to write out in full.The principal property of cumulants is their additivity, which may be provedby combining (30.92) with (30.90). If X 1 , X 2 , ..., X N are independent randomvariables and K Xi (t) for i =1, 2,...,N is the CGF for X i then the CGF ofS N = c 1 X 1 + c 2 X 2 + ···+ c N X N (where the c i are constants) is given byK SN (t) =N∑K Xi (c i t).i=1Cumulants also have the useful property that, under a change of origin X →X + a the first cumulant undergoes the change κ 1 → κ 1 + a but all higher-ordercumulants remain unchanged. Under a change of scale X → bX, cumulant κ rundergoes the change κ r → b r κ r .1167

PROBABILITYDistribution Probability law f(x) MGF E[X] V [X]binomial n C x p x q n−x (pe t + q) n np npq( ) r pnegative binomial r+x−1 C x p r q x rq1 − qe t pgeometric q x−1 pe t 1p1 − qe t phypergeometricPoisson(Np)!(Nq)!n!(N−n)!x!(Np−x)!(n−x)!(Nq−n+x)!N!λ xx! e−λ e λ(et −1)npλrqp 2qp 2N − nN − 1 npqλTable 30.1Some important discrete probability distributions.30.8 Important discrete distributionsHaving discussed some general properties of distributions, we now consider themore important discrete distributions encountered in physical applications. Theseare discussed in detail below, and summarised for convenience in table 30.1; werefer the reader to the relevant section below for an explanation of the symbolsused.30.8.1 The binomial distributionPerhaps the most important discrete probability distribution is the binomial distribution.This distribution describes processes that consist of a number of independentidentical trials with two possible outcomes, A and B = Ā. We may callthese outcomes ‘success’ and ‘failure’ respectively. If the probability of a successis Pr(A) =p then the probability of a failure is Pr(B) =q =1− p. Ifweperformn trials then the discrete random variableX = number of times A occurscan take the values 0, 1, 2,...,n; its distribution amongst these values is describedby the binomial distribution.We now calculate the probability that in n trials we obtain x successes (and son − x failures). One way of obtaining such a result is to have x successes followedby n−x failures. Since the trials are assumed independent, the probability of this ispp ···p ×} {{ }qq ···q} {{ }= p x q n−x .x times n − x timesThis is, however, just one permutation of x successes and n − x failures. The total1168

30.8 IMPORTANT DISCRETE DISTRIBUTIONSf(x)f(x)n =5,p =0.6 n =5,p =0.1670.40.40.30.30.20.20.1 0.100 1 2 3 4 5 x00 1 2 3 4 5 xf(x)f(x)n = 10, p =0.6 n = 10, p =0.1670.4 0.40.3 0.30.2 0.20.10.10 0 1 2 3 4 5 6 7 8 9 10 x0 0 1 2 3 4 5 6 7 8 910 xFigure 30.11 Some typical binomial distributions with various combinationsof parameters n and p.number of permutations of n objects, of which x are identical and of type 1 andn − x are identical and of type 2, is given by (30.33) asn!x!(n − x)! ≡ n C x .Therefore, the total probability of obtaining x successes from n trials isf(x) =Pr(X = x) = n C x p x q n−x = n C x p x (1 − p) n−x , (30.94)which is the binomial probability distribution formula. When a random variableX follows the binomial distribution for n trials, with a probability of success p,we write X ∼ Bin(n, p). Then the random variable X is often referred to as abinomial variate. Some typical binomial distributions are shown in figure 30.11.◮If a single six-sided die is rolled five times, what is the probability that a six is thrownexactly three times?Here the number of ‘trials’ n = 5, and we are interested in the random variableX = number of sixes thrown.Since the probability of a ‘success’ is p = 1 , the probability of obtaining exactly three sixes6in five throws is given by (30.94) as( ) 3 ( ) (5−3)5! 1 5Pr(X =3)==0.032. ◭3!(5 − 3)! 6 61169

PROBABILITYFor evaluating binomial probabilities a useful result is the binomial recurrenceformulaPr(X = x +1)= p ( ) n − xPr(X = x), (30.95)q x +1which enables successive probabilities Pr(X = x + k), k =1, 2,..., to be calculatedonce Pr(X = x) is known; it is often quicker to use than (30.94).◮The random variable X is distributed as X ∼ Bin(3, 1 ). Evaluate the probability function2f(x) using the binomial recurrence formula.The probability Pr(X = 0) may be calculated using (30.94) and isThe ratio p/q = 1 / 1 2 2(30.95), we findPr(X =0)= 3 C 0( 12) 0 ( 12) 3=18 .= 1 in this case and so, using the binomial recurrence formulaPr(X =1)=1× 3 − 00+1 × 1 8 = 3 8 ,Pr(X =2)=1× 3 − 11+1 × 3 8 = 3 8 ,Pr(X =3)=1× 3 − 22+1 × 3 8 = 1 8 ,results which may be verified by direct application of (30.94). ◭We note that, as required, the binomial distribution satifiesn∑f(x) =x=0n∑n C x p x q n−x =(p + q) n =1.x=0Furthermore, from the definitions of E[X] andV [X] for a discrete distribution,we may show that for the binomial distribution E[X] =np and V [X] =npq. Thedirect summations involved are, however, rather cumbersome and these resultsare obtained much more simply using the moment generating function.The moment generating function for the binomial distributionTo find the MGF for the binomial distribution we consider the binomial randomvariable X to be the sum of the random variables X i , i =1, 2,...,n, which aredefined by{1 if a ‘success’ occurs on the ith trial,X i =0 if a ‘failure’ occurs on the ith trial.1170

30.8 IMPORTANT DISCRETE DISTRIBUTIONSThusM i (t) =E [ e tXi] = e 0t × Pr(X i =0)+e 1t × Pr(X i =1)=1× q + e t × p= pe t + q.From (30.89), it follows that the MGF for the binomial distribution is given byn∏M(t) = M i (t) =(pe t + q) n . (30.96)i=1We can now use the moment generating function to derive the mean andvariance of the binomial distribution. From (30.96)M ′ (t) =npe t (pe t + q) n−1 ,and from (30.86)E[X] =M ′ (0) = np(p + q) n−1 = np,where the last equality follows from p + q =1.Differentiating with respect to t once more givesM ′′ (t) =e t (n − 1)np 2 (pe t + q) n−2 + e t np(pe t + q) n−1 ,and from (30.86)E[X 2 ]=M ′′ (0) = n 2 p 2 − np 2 + np.Thus, using (30.87)V [X] =M ′′ (0) − [ M ′ (0) ] 2= n 2 p 2 − np 2 + np − n 2 p 2 = np(1 − p) =npq.Multiple binomial distributionsSuppose X and Y are two independent random variables, both of which aredescribed by binomial distributions with a common probability of success p, butwith (in general) different numbers of trials n 1 and n 2 ,sothatX ∼ Bin(n 1 ,p)and Y ∼ Bin(n 2 ,p). Now consider the random variable Z = X + Y .Wecouldcalculate the probability distribution of Z directly using (30.60), but it is mucheasier to use the MGF (30.96).Since X and Y are independent random variables, the MGF M Z (t) of the newvariable Z = X + Y is given simply by the product of the individual MGFsM X (t) andM Y (t). Thus, we obtainM Z (t) =M X (t)M Y (t) =(pe t + q) n1 (pe t + q) n1 =(pe t + q) n1+n2 ,which we recognise as the MGF of Z ∼ Bin(n 1 + n 2 ,p). Hence Z is also describedby a binomial distribution.This result may be extended to any number of binomial distributions. If X i ,1171

PROBABILITYi =1, 2,...,N, is distributed as X i ∼ Bin(n i ,p)thenZ = X 1 + X 2 + ···+ X N isdistributed as Z ∼ Bin(n 1 + n 2 + ···+ n N ,p), as would be expected since the resultof ∑ i n i trials cannot depend on how they are split up. A similar proof is alsopossible using either the probability or cumulant generating functions.Unfortunately, no equivalent simple result exists for the probability distributionof the difference Z = X − Y of two binomially distributed variables.30.8.2 The geometric and negative binomial distributionsA special case of the binomial distribution occurs when instead of the number ofsuccesses we consider the discrete random variableX = number of trials required to obtain the first success.The probability that x trials are required in order to obtain the first success, issimply the probability of obtaining x − 1 failures followed by one success. If theprobability of a success on each trial is p, thenforx>0f(x) =Pr(X = x) =(1− p) x−1 p = q x−1 p,where q =1− p. This distribution is sometimes called the geometric distribution.The probability generating function for this distribution is given in (30.78). Byreplacing t by e t in (30.78) we immediately obtain the MGF of the geometricdistributionM(t) =pet1 − qe t ,from which its mean and variance are found to beE[X] = 1 p , V[X] = q p 2 .Another distribution closely related to the binomial is the negative binomialdistribution. This describes the probability distribution of the random variableX = number of failures before the rth success.One way of obtaining x failures before the rth success is to have r − 1 successesfollowed by x failures followed by the rth success, for which the probability ispp ···p} {{ }r − 1times× qq ···q × p = p r q x .} {{ }x timesHowever, the first r + x − 1 factors constitute just one permutation of r − 1successes and x failures. The total number of permutations of these r + x − 1objects, of which r − 1 are identical and of type 1 and x are identical and of type1172

30.8 IMPORTANT DISCRETE DISTRIBUTIONS2, is r+x−1 C x . Therefore, the total probability of obtaining x failures before therth success isf(x) =Pr(X = x) = r+x−1 C x p r q x ,which is called the negative binomial distribution (see the related discussion onp. 1137). It is straightforward to show that the MGF of this distribution is( ) p rM(t) =1 − qe t ,and that its mean and variance are given byE[X] = rq and V [X] = rqpp 2 .30.8.3 The hypergeometric distributionIn subsection 30.8.1 we saw that the probability of obtaining x successes in nindependent trials was given by the binomial distribution. Suppose that these n‘trials’ actually consist of drawing at random n balls, from a set of N such ballsof which M are red and the rest white. Let us consider the random variableX = number of red balls drawn.On the one hand, if the balls are drawn with replacement then the trials areindependent and the probability of drawing a red ball is p = M/N each time.Therefore, the probability of drawing x red balls in n trials is given by thebinomial distribution asn!Pr(X = x) =x!(n − x)! px (1 − p) n−x .On the other hand, if the balls are drawn without replacement the trials are notindependent and the probability of drawing a red ball depends on how many redballs have already been drawn. We can, however, still derive a general formulafor the probability of drawing x red balls in n trials, as follows.The number of ways of drawing x red balls from M is M C x , and the numberof ways of drawing n − x white balls from N − M is N−M C n−x . Therefore, thetotal number of ways to obtain x red balls in n trials is M C N−M x C n−x . However,the total number of ways of drawing n objects from N is simply N C n . Hence theprobability of obtaining x red balls in n trials isM C N−M x C n−xPr(X = x) =NC nM!(N − M)!=x!(M − x)! (n − x)!(N − M − n + x)!=n!(N − n)!, (30.97)N!(Np)!(Nq)! n!(N − n)!x!(Np − x)!(n − x)!(Nq − n + x)! N! , (30.98)1173

PROBABILITYwhere in the last line p = M/N and q =1− p. This is called the hypergeometricdistribution.By performing the relevant summations directly, it may be shown that thehypergeometric distribution has meanE[X] =n M N = npand varianceV [X] =nM(N − M)(N − n)N 2 (N − 1)= N − nN − 1 npq.◮In the UK National Lottery each participant chooses six different numbers between 1and 49. In each weekly draw six numbered winning balls are subsequently drawn. Find theprobabilities that a participant chooses 0, 1, 2, 3, 4, 5, 6 winning numbers correctly.The probabilities are given by a hypergeometric distribution with N (the total number ofballs) = 49, M (the number of winning balls drawn) = 6, and n (the number of numberschosen by each participant) = 6. Thus, substituting in (30.97), we find6 C 43 0 C 6Pr(0) = = 16 49C 6 2.29 , Pr(1) = C 43 1 C 5= 149C 6 2.42 ,6 C 43 2 C 4Pr(2) = = 16 49C 6 7.55 , Pr(3) = C 43 3 C 3= 149C 6 56.6 ,6 C 43 4 C 2Pr(4) = = 16 49C 6 1032 , Pr(5) = C 43 5 C 1= 149C 6 54 200 ,It can easily be seen thatPr(6) =6 C 43 6 C 0 1=49C 6 13.98 × 10 . 6as expected. ◭6∑Pr(i) =0.44 + 0.41 + 0.13 + 0.02 + O(10 −3 )=1,i=0Note that if the number of trials (balls drawn) is small compared with N, Mand N − M then not replacing the balls is of little consequence, and we mayapproximate the hypergeometric distribution by the binomial distribution (withp = M/N); this is much easier to evaluate.30.8.4 The Poisson distributionWe have seen that the binomial distribution describes the number of successfuloutcomes in a certain number of trials n. The Poisson distribution also describesthe probability of obtaining a given number of successes but for situationsin which the number of ‘trials’ cannot be enumerated; rather it describes thesituation in which discrete events occur in a continuum. Typical examples of1174

30.8 IMPORTANT DISCRETE DISTRIBUTIONSdiscrete random variables X described by a Poisson distribution are the numberof telephone calls received by a switchboard in a given interval, or the numberof stars above a certain brightness in a particular area of the sky. Given a meanrate of occurrence λ of these events in the relevant interval or area, the Poissondistribution gives the probability Pr(X = x) that exactly x events will occur.We may derive the form of the Poisson distribution as the limit of the binomialdistribution when the number of trials n →∞and the probability of ‘success’p → 0, in such a way that np = λ remains finite. Thus, in our example of atelephone switchboard, suppose we wish to find the probability that exactly xcalls are received during some time interval, given that the mean number of callsin such an interval is λ. Let us begin by dividing the time interval into a largenumber, n, of equal shorter intervals, in each of which the probability of receivinga call is p. Asweletn →∞then p → 0, but since we require the mean numberof calls in the interval to equal λ, we must have np = λ. The probability of xsuccesses in n trials is given by the binomial formula asn!Pr(X = x) =x!(n − x)! px (1 − p) n−x . (30.99)Now as n →∞, with x finite, the ratio of the n-dependent factorials in (30.99)behaves asymptotically as a power of n, i.e.n!limn→∞ (n − x)! = lim n(n − 1)(n − 2) ···(n − x +1)∼ n→∞ nx .Alsolim lim(1− (1 − p) λ/pn→∞ p→0 p)n−x = limp→0 (1 − p) x = e−λ1 .Thus, using λ = np, (30.99) tends to the Poisson distributionf(x) =Pr(X = x) = e−λ λ x, (30.100)x!which gives the probability of obtaining exactly x calls in the given time interval.As we shall show below, λ is the mean of the distribution. Events following aPoisson distribution are usually said to occur randomly in time.Alternatively we may derive the Poisson distribution directly, without consideringa limit of the binomial distribution. Let us again consider our exampleof a telephone switchboard. Suppose that the probability that x calls have beenreceived in a time interval t is P x (t). If the average number of calls received in aunit time is λ then in a further small time interval ∆t the probability of receivinga call is λ∆t, provided ∆t is short enough that the probability of receiving two ormore calls in this small interval is negligible. Similarly the probability of receivingno call during the same small interval is simply 1 − λ∆t.Thus, for x>0, the probability of receiving exactly x calls in the total interval1175

PROBABILITYt +∆t is given byP x (t +∆t) =P x (t)(1 − λ∆t)+P x−1 (t)λ∆t.Rearranging the equation, dividing through by ∆t and letting ∆t → 0, we obtainthe differential recurrence equationdP x (t)= λP x−1 (t) − λP x (t). (30.101)dtFor x = 0 (i.e. no calls received), however, (30.101) simplifies todP 0 (t)= −λP 0 (t),dtwhich may be integrated to give P 0 (t) =P 0 (0)e −λt . But since the probability P 0 (0)of receiving no calls in a zero time interval must equal unity, we have P 0 (t) =e −λt .This expression for P 0 (t) may then be substituted back into (30.101) with x =1to obtain a differential equation for P 1 (t) that has the solution P 1 (t) =λte −λt .We may repeat this process to obtain expressions for P 2 (t),P 3 (t),...,P x (t), and wefindP x (t) = (λt)xx! e−λt . (30.102)By setting t = 1 in (30.102), we again obtain the Poisson distribution (30.100) forobtaining exactly x calls in a unit time interval.If a discrete random variable is described by a Poisson distribution of mean λthen we write X ∼ Po(λ). As it must be, the sum of the probabilities is unity:∞∑Pr(X = x) =e −λx=0∞ ∑x=0λ xx! = e−λ e λ =1.From (30.100) we may also derive the Poisson recurrence formula,Pr(X = x +1)=λx +1 Pr(X = x) for x =0, 1, 2,..., (30.103)which enables successive probabilities to be calculated easily once one is known.◮A person receives on average one e-mail message per half-hour interval. Assuming thatthe e-mails are received randomly in time, find the probabilities that in any particular hour0, 1, 2, 3, 4, 5 messages are received.Let X = number of e-mails received per hour. Clearly the mean number of e-mails perhour is two, and so X follows a Poisson distribution with λ =2,i.e.Pr(X = x) = 2xx! e−2 .Thus Pr(X =0)=e −2 =0.135, Pr(X =1)=2e −2 =0.271, Pr(X =2)=2 2 e −2 /2! = 0.271,Pr(X =3)=2 3 e −2 /3! = 0.180, Pr(X =4)=2 4 e −2 /4! = 0.090, Pr(X =5)=2 5 e −2 /5! =0.036. These results may also be calculated using the recurrence formula (30.103). ◭1176

30.8 IMPORTANT DISCRETE DISTRIBUTIONS0.3f(x)f(x)λ =1 λ =20.30.20.20.1 0.10001 2 3 4 5 x0 1 2 3 4 5 67xf(x)0.3λ =50.20.1001234 567 8 9 10 11xFigure 30.12 Three Poisson distributions for different values of the parameterλ.The above example illustrates the point that a Poisson distribution typicallyrises and then falls. It either has a maximum when x is equal to the integer partof λ or, if λ happens to be an integer, has equal maximal values at x = λ − 1andx = λ. The Poisson distribution always has a long ‘tail’ towards higher values of Xbut the higher the value of the mean the more symmetric the distribution becomes.Typical Poisson distributions are shown in figure 30.12. Using the definitions ofmean and variance, we may show that, for the Poisson distribution, E[X] =λ andV [X] =λ. Nevertheless, as in the case of the binomial distribution, performingthe relevant summations directly is rather tiresome, and these results are muchmore easily proved using the MGF.The moment generating function for the Poisson distributionThe MGF of the Poisson distribution is given byM X (t) =E [ e tX] =∞∑ e tx e −λ λ x ∑∞= e −λx!x=01177x=0(λe t ) xx!= e −λ e λet = e λ(et −1)(30.104)

PROBABILITYfrom which we obtainM X(t) ′ =λe t e λ(et−1) ,M X(t) ′′ =(λ 2 e 2t + λe t )e λ(et−1) .Thus, the mean and variance of the Poisson distribution are given byE[X] =M ′ X(0) = λ and V [X] =M ′′ X(0) − [M ′ X(0)] 2 = λ.The Poisson approximation to the binomial distributionEarlier we derived the Poisson distribution as the limit of the binomial distributionwhen n →∞and p → 0 in such a way that np = λ remains finite, where λ is themean of the Poisson distribution. It is not surprising, therefore, that the Poissondistribution is a very good approximation to the binomial distribution for largen (≥ 50, say) and small p (≤ 0.1, say). Moreover, it is easier to calculate as itinvolves fewer factorials.◮In a large batch of light bulbs, the probability that a bulb is defective is 0.5%. Forasample of 200 bulbs taken at random, find the approximate probabilities that 0, 1 and 2 ofthe bulbs respectively are defective.Let the random variable X = number of defective bulbs in a sample. This is distributedas X ∼ Bin(200, 0.005), implying that λ = np =1.0. Since n is large and p small, we mayapproximate the distribution as X ∼ Po(1), giving−1 1xPr(X = x) ≈ ex! ,from which we find Pr(X =0)≈ 0.37, Pr(X =1)≈ 0.37, Pr(X =2)≈ 0.18. For comparison,it may be noted that the exact values calculated from the binomial distribution are identicalto those found here to two decimal places. ◭Multiple Poisson distributionsMirroring our discussion of multiple binomial distributions in subsection 30.8.1,let us suppose X and Y are two independent random variables, both of whichare described by Poisson distributions with (in general) different means, so thatX ∼ Po(λ 1 )andY ∼ Po(λ 2 ). Now consider the random variable Z = X + Y .Wemay calculate the probability distribution of Z directly using (30.60), but we mayderive the result much more easily by using the moment generating function (orindeed the probability or cumulant generating functions).Since X and Y are independent RVs, the MGF for Z is simply the product ofthe individual MGFs for X and Y . Thus, from (30.104),M Z (t) =M X (t)M Y (t) =e λ1(et −1) e λ2(et −1) = e (λ1+λ2)(et −1) ,which we recognise as the MGF of Z ∼ Po(λ 1 + λ 2 ). Hence Z is also Poissondistributed and has mean λ 1 + λ 2 . Unfortunately, no such simple result holds forthe difference Z = X − Y of two independent Poisson variates. A closed-form1178

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONSexpression for the PDF of this Z does exist, but it is a rather complicatedcombination of exponentials and a modified Bessel function. §◮Two types of e-mail arrive independently and at random: external e-mails at a mean rateof one every five minutes and internal e-mails at a rate of two every five minutes. Calculatethe probability of receiving two or more e-mails in any two-minute interval.LetX = number of external e-mails per two-minute interval,Y = number of internal e-mails per two-minute interval.Since we expect on average one external e-mail and two internal e-mails every five minuteswe have X ∼ Po(0.4) and Y ∼ Po(0.8). Letting Z = X + Y we have Z ∼ Po(0.4+0.8) =Po(1.2). NowPr(Z ≥ 2) = 1 − Pr(Z

PROBABILITYDistribution Probability law f(x) MGF E[X] V [X][ ]1Gaussianσ √ 2π exp (x − µ)2− exp(µt + 12σ 2 2 σ2 t 2 ) µ σ 2( ) λexponential λe −λx 1 1λ − tλ λ( ) 2rλλgammaΓ(r) (λx)r−1 e −λx r rλ − tλ λ 2( ) n/211chi-squared2 n/2 Γ(n/2) x(n/2)−1 e −x/2 n 2n1 − 2t1e bt − e at a + b (b − a) 2uniformb − a(b − a)t2 12Table 30.2Some important continuous probability distributions.The probability density function for a Gaussian distribution of a randomvariable X, with mean E[X] =µ and variance V [X] =σ 2 ,takestheformf(x) = 1σ √ 2π exp [− 1 2( x − µ) 2]. (30.105)σThe factor 1/ √ 2π arises from the normalisation of the distribution,∫ ∞−∞f(x)dx =1;the evaluation of this integral is discussed in subsection 6.4.2. The Gaussiandistribution is symmetric about the point x = µ and has the characteristic ‘bell’shape shown in figure 30.13. The width of the curve is described by the standarddeviation σ: ifσ is large then the curve is broad, and if σ is small then the curveis narrow (see the figure). At x = µ ± σ, f(x) falls to e −1/2 ≈ 0.61 of its peakvalue; these points are points of inflection, where d 2 f/dx 2 = 0. When a randomvariable X follows a Gaussian distribution with mean µ and variance σ 2 ,wewriteX ∼ N(µ, σ 2 ).The effects of changing µ and σ are only to shift the curve along the x-axis orto broaden or narrow it, respectively. Thus all Gaussians are equivalent in thata change of origin and scale can reduce them to a standard form. We thereforeconsider the random variable Z =(X − µ)/σ, for which the PDF takes the formφ(z) = 1 √2πexp(− z22), (30.106)which is called the standard Gaussian distribution and has mean µ = 0 andvariance σ 2 = 1. The random variable Z is called the standard variable.From (30.105) we can define the cumulative probability function for a Gaussian1180

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONS0.40.30.2µ =3σ =1σ =20.1σ =3−6 −4 −2 2 3 4 6 8 10 12Figure 30.13 The Gaussian or normal distribution for mean µ =3andvarious values of the standard deviation σ.Φ(a)0.40.30.20.1φ(z)Φ(a)Φ(z)1z−2 0 a−4 a 2 4 −2 −1 y 2Figure 30.14 On the left, the standard Gaussian distribution φ(z); the shadedarea gives Pr(Z

PROBABILITYIt is usual only to tabulate Φ(z) forz>0, since it can be seen easily, fromfigure 30.14 and the symmetry of the Gaussian distribution, that Φ(−z) =1−Φ(z);see table 30.3. Using such a table it is then straightforward to evaluate theprobability that Z lies in a given range of z-values. For example, for a and bconstant,Pr(Z a)=1− Φ(a),Pr(a

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONSΦ(z) .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .53590.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .57530.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .61410.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .65170.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .68790.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .72240.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .75490.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .78520.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .81330.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .83891.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .86211.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .88301.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .90151.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .91771.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .93191.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .94411.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .95451.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .96331.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .97061.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .97672.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .98172.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .98572.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .98902.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .99162.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .99362.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .99522.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .99642.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .99742.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .99812.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .99863.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .99903.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .99933.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .99953.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .99973.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998Table 30.3 The cumulative probability function Φ(z) for the standard Gaussiandistribution, as given by (30.108). The units and the first decimal placeof z are specified in the column under Φ(z) and the second decimal place isspecified by the column headings. Thus, for example, Φ(1.23) = 0.8907.1183

PROBABILITY◮Sawmill A produces boards whose lengths are Gaussian distributed with mean 209.4 cmand standard deviation 5.0 cm. A board is accepted if it is longer than 200 cm but isrejected otherwise. Show that 3% of boards are rejected.Sawmill B produces boards of the same standard deviation but of mean length 210.1 cm.Find the proportion of boards rejected if they are drawn at random from the outputs of Aand B in the ratio 3:1.Let X = length of boards from A, sothatX ∼ N(209.4, (5.0) 2 )and( ) ( )200 − µ 200 − 209.4Pr(X

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONSNow using Φ(−z) =1− Φ(z) gives( ) µ − 140Φ=1− 0.030 = 0.970.σUsing table 30.3 again, we findµ − 140=1.88. (30.113)σSolving the simultaneous equations (30.112) and (30.113) gives µ = 173.5, σ =17.8. ◭The moment generating function for the Gaussian distributionUsing the definition of the MGF (30.85),M X (t) =E [ e tX] ∫ ∞[]1=−∞ σ √ 2π exp (x − µ)2tx −2σ 2 dx= c exp ( µt + 1 2 σ2 t 2) ,where the final equality is established by completing the square in the argumentof the exponential and writing∫ ∞{1c =−∞ σ √ 2π exp − [x − (µ + σ2 t)] 2 }2σ 2 dx.However, the final integral is simply the normalisation integral for the Gaussiandistribution, and so c = 1 and the MGF is given byM X (t) =exp ( µt + 1 2 σ2 t 2) . (30.114)We showed in subsection 30.7.2 that this MGF leads to E[X] =µ and V [X] =σ 2 ,as required.Gaussian approximation to the binomial distributionWe may consider the Gaussian distribution as the limit of the binomial distributionwhen the number of trials n →∞but the probability of a success p remainsfinite, so that np →∞also. (This contrasts with the Poisson distribution, whichcorresponds to the limit n →∞and p → 0 with np = λ remaining finite.) Inother words, a Gaussian distribution results when an experiment with a finiteprobability of success is repeated a large number of times. We now show howthis Gaussian limit arises.The binomial probability function gives the probability of x successes in n trialsasn!f(x) =x!(n − x)! px (1 − p) n−x .Taking the limit as n →∞(and x →∞) we may approximate the factorials byStirling’s approximationn! ∼ √ ( n) n2πne1185

PROBABILITYx f(x) (binomial) f(x) (Gaussian)0 0.0001 0.00011 0.0016 0.00142 0.0106 0.00923 0.0425 0.03954 0.1115 0.11195 0.2007 0.20916 0.2508 0.25757 0.2150 0.20918 0.1209 0.11199 0.0403 0.039510 0.0060 0.0092Table 30.4 Comparison of the binomial distribution for n =10andp =0.6with its Gaussian approximation.to obtainf(x) ≈ √ 1 ( x) −x−1/2 ( n − x2πn nn= √ 1 [exp − ( )x + 1 x2πn2 ln) −n+x−1/2p x (1 − p) n−xn − ( n − x + 1 2]+ x ln p +(n − x)ln(1− p) .)lnn − xnBy expanding the argument of the exponential in terms of y = x − np, where1 ≪ y ≪ np and keeping only the dominant terms, it can be shown thatf(x) ≈ √ 1[1√ exp − 1 (x − np) 2 ],2πn p(1 − p) 2 np(1 − p)which is of Gaussian form with µ = np and σ = √ np(1 − p).Thus we see that the value of the Gaussian probability density function f(x) isa good approximation to the probability of obtaining x successes in n trials. Thisapproximation is actually very good even for relatively small n. For example, ifn =10andp =0.6 then the Gaussian approximation to the binomial distributionis (30.105) with µ =10× 0.6 =6andσ = √ 10 × 0.6(1 − 0.6) = 1.549. Theprobability functions f(x) for the binomial and associated Gaussian distributionsfor these parameters are given in table 30.4, and it can be seen that the Gaussianapproximation is a good one.Strictly speaking, however, since the Gaussian distribution is continuous andthe binomial distribution is discrete, we should use the integral of f(x) fortheGaussian distribution in the calculation of approximate binomial probabilities.More specifically, we should apply a continuity correction so that the discreteinteger x in the binomial distribution becomes the interval [x − 0.5, x+0.5] in1186

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONSthe Gaussian distribution. Explicitly,Pr(X = x) ≈ 1 ∫ x+0.5[σ √ exp − 1 ( u − µ) 2]du.2π x−0.5 2 σThe Gaussian approximation is particularly useful for estimating the binomialprobability that X lies between the (integer) values x 1 and x 2 ,Pr(x 1 15.5) = Pr Z> √ =Pr(Z>−1.06)18=Pr(Z

PROBABILITYStirling’s approximation for large x givesx! ≈ √ ( x) x2πxeimplying thatln x! ≈ ln √ 2πx + x ln x − x,which, on substituting into (30.115), yieldsln f(x) ≈−λ + x ln λ − (x ln x − x) − ln √ 2πx.Since we expect the Poisson distribution to peak around x = λ, we substituteɛ = x − λ to obtain{ [ (ln f(x) ≈−λ +(λ + ɛ) ln λ − ln λ 1+ ɛ )]}+(λ + ɛ) − ln √ 2π(λ + ɛ).λUsing the expansion ln(1 + z) =z − z 2 /2+···, we find( ) ɛln f(x) ≈ ɛ − (λ + ɛ)λ − ɛ22λ 2 − ln √ ( ) ɛ2πλ −λ − ɛ22λ 2≈− ɛ22λ − ln √ 2πλ,when only the dominant terms are retained, after using the fact that ɛ is of theorder of the standard deviation of x, i.e.oforderλ 1/2 . On exponentiating thisresult we obtainf(x) ≈ √ 1 exp[−2πλ](x − λ)2,2λwhich is the Gaussian distribution with µ = λ and σ 2 = λ.The larger the value of λ, the better is the Gaussian approximation to thePoisson distribution; the approximation is reasonable even for λ = 5, but λ ≥ 10is safer. As in the case of the Gaussian approximation to the binomial distribution,a continuity correction is necessary since the Poisson distribution is discrete.◮E-mail messages are received by an author at an average rate of one per hour. Find theprobability that in a day the author receives 24 messages or more.We first define the random variableX = number of messages received in a day.Thus E[X] =1× 24 = 24, and so X ∼ Po(24). Since λ>10 we may approximate thePoisson distribution by X ∼ N(24, 24). Now the standard variable isZ = X √ − 24 ,24and, using the continuity correction, we find()23.5 − 24Pr(X >23.5) = Pr Z> √24=Pr(Z>−0.102) = Pr(Z

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONSIn fact, almost all probability distributions tend towards a Gaussian when thenumbers involved become large – that this should happen is required by thecentral limit theorem, which we discuss in section 30.10.Multiple Gaussian distributionsSuppose X and Y are independent Gaussian-distributed random variables, sothat X ∼ N(µ 1 ,σ1 2)andY ∼ N(µ 2,σ2 2 ). Let us now consider the random variableZ = X + Y . The PDF for this random variable may be found directly using(30.61), but it is easier to use the MGF. From (30.114), the MGFs of X and YareM X (t) =exp ( µ 1 t + 1 2 σ2 1t 2) , M Y (t) =exp ( µ 2 t + 1 2 σ2 2t 2) .Using (30.89), since X and Y are independent RVs, the MGF of Z = X + Y issimply the product of M X (t) andM Y (t). Thus, we haveM Z (t) =M X (t)M Y (t) =exp ( µ 1 t + 1 2 σ2 1t 2) exp ( µ 2 t + 1 2 σ2 2t 2)=exp [ (µ 1 + µ 2 )t + 1 2 (σ2 1 + σ2)t 2 2] ,which we recognise as the MGF for a Gaussian with mean µ 1 + µ 2 and varianceσ 2 1 + σ2 2 . Thus, Z is also Gaussian distributed: Z ∼ N(µ 1 + µ 2 , σ 2 1 + σ2 2 ).A similar calculation may be performed to calculate the PDF of the randomvariable W = X − Y . If we introduce the variable Ỹ = −Y then W = X + Ỹ ,where Ỹ ∼ N(−µ 1 , σ 2 1 ). Thus, using the result above, we find W ∼ N(µ 1 −µ 2 , σ 2 1 + σ2 2 ).◮An executive travels home from her office every evening. Her journey consists of a trainride, followed by a bicycle ride. The time spent on the train is Gaussian distributed withmean 52 minutes and standard deviation 1.8 minutes, while the time for the bicycle journeyis Gaussian distributed with mean 8 minutes and standard deviation 2.6 minutes. Assumingthese two factors are independent, estimate the percentage of occasions on which the wholejourney takes more than 65 minutes.We first define the random variablesX =timespentontrain, Y = time spent on bicycle,so that X ∼ N(52, (1.8) 2 )andY ∼ N(8, (2.6) 2 ). Since X and Y are independent, the totaljourney time T = X + Y is distributed asT ∼ N(52 + 8, (1.8) 2 +(2.6) 2 )=N(60, (3.16) 2 ).The standard variable is thusZ = T − 603.16 ,and the required probability is given by()65 − 60Pr(T >65) = Pr Z> =Pr(Z>1.58) = 1 − 0.943 = 0.057.3.16Thus the total journey time exceeds 65 minutes on 5.7% of occasions. ◭1189

PROBABILITYThe above results may be extended. For example, if the random variablesX i , i =1, 2,...,n, are distributed as X i ∼ N(µ i ,σi 2 ) then the random variableZ = ∑ i c iX i (where the c i are constants) is distributed as Z ∼ N( ∑ i c iµ i , ∑ i c2 i σ2 i ).30.9.2 The log-normal distributionIf the random variable X follows a Gaussian distribution then the variableY = e X is described by a log-normal distribution. Clearly, if X can take valuesin the range −∞ to ∞, thenY will lie between 0 and ∞. The probability densityfunction for Y is found using the result (30.58). It isg(y) =f(x(y))dx∣ dy ∣ = 1 []σ √ 12π y exp (ln y − µ)2−2σ 2 .We note that µ and σ 2 are not the mean and variance of the log-normaldistribution, but rather the parameters of the corresponding Gaussian distributionfor X. The mean and variance of Y , however, can be found straightforwardlyusing the MGF of X, which reads M X (t) =E[e tX ]=exp(µt + 1 2 σ2 t 2 ). Thus, themean of Y is given byand the variance of Y readsE[Y ]=E[e X ]=M X (1) = exp(µ + 1 2 σ2 ),V [Y ]=E[Y 2 ] − (E[Y ]) 2 = E[e 2X ] − (E[e X ]) 2= M X (2) − [M X (1)] 2 =exp(2µ + σ 2 )[exp(σ 2 ) − 1].In figure 30.15, we plot some examples of the log-normal distribution for variousvalues of the parameters µ and σ 2 .30.9.3 The exponential and gamma distributionsThe exponential distribution with positive parameter λ is given by{λe−λxfor x>0,f(x) =(30.116)0 for x ≤ 0and satisfies ∫ ∞−∞f(x) dx = 1 as required. The exponential distribution occurs naturallyif we consider the distribution of the length of intervals between successiveevents in a Poisson process or, equivalently, the distribution of the interval (i.e.the waiting time) before the first event. If the average number of events per unitinterval is λ then on average there are λx events in interval x, so that from thePoisson distribution the probability that there will be no events in this interval isgiven byPr(no events in interval x) =e −λx .1190

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONSg(y)10.80.6µ =0,σ =0µ =0,σ =0.5µ =0,σ =1.5µ =1,σ =10.40.20012 3 4yFigure 30.15 The PDF g(y) for the log-normal distribution for various valuesof the parameters µ and σ.The probability that an event occurs in the next infinitestimal interval [x, x + dx]is given by λdx,sothatPr(the first event occurs in interval [x, x + dx]) = e −λx λdx.Hence the required probability density function is given byf(x) =λe −λx .The expectation and variance of the exponential distribution can be evaluated as1/λ and (1/λ) 2 respectively. The MGF is given byM(t) =λλ − t . (30.117)We may generalise the above discussion to obtain the PDF for the intervalbetween every rth event in a Poisson process or, equivalently, the interval (waitingtime) before the rth event. We begin by using the Poisson distribution to give−λx (λx)r−1Pr(r − 1 events occur in interval x) =e(r − 1)! ,from which we obtain−λx (λx)r−1Pr(rth event occurs in the interval [x, x + dx]) = e(r − 1)! λdx.Thus the required PDF isλf(x) =(r − 1)! (λx)r−1 e −λx , (30.118)which is known as the gamma distribution of order r with parameter λ. Althoughour derivation applies only when r is a positive integer, the gamma distribution is1191

PROBABILITYf(x)10.80.6r =10.40.2r =2r =5r =1000 2 4 6 8 10 12 14 16 18 20xFigure 30.16 The PDF f(x) for the gamma distributions γ(λ, r) withλ =1and r =1, 2, 5, 10.defined for all positive r by replacing (r −1)! by Γ(r) in (30.118); see the appendixfor a discussion of the gamma function Γ(x). If a random variable X is describedby a gamma distribution of order r with parameter λ, wewriteX ∼ γ(λ, r);we note that the exponential distribution is the special case γ(λ, 1). The gammadistribution γ(λ, r) is plotted in figure 30.16 for λ = 1 and r =1, 2, 5, 10. Forlarge r, the gamma distribution tends to the Gaussian distribution whose meanand variance are specified by (30.120) below.The MGF for the gamma distribution is obtained from that for the exponentialdistribution, by noting that we may consider the interval between every rth eventin a Poisson process as the sum of r intervals between successive events. Thus therth-order gamma variate is the sum of r independent exponentially distributedrandom variables. From (30.117) and (30.90), the MGF of the gamma distributionis therefore given by( ) r λM(t) = , (30.119)λ − tfrom which the mean and variance are found to beE[X] = r λ , V[X] = r λ 2 . (30.120)We may also use the above MGF to prove another useful theorem regardingmultiple gamma distributions. If X i ∼ γ(λ, r i ), i =1, 2,...,n, are independentgamma variates then the random variable Y = X 1 + X 2 + ···+ X n has MGFn∏( ) λri( ) λ r1+r 2+···+r nM(t) ==. (30.121)λ − t λ − ti=1Thus Y is also a gamma variate, distributed as Y ∼ γ(λ, r 1 + r 2 + ···+ r n ).1192

30.9 IMPORTANT CONTINUOUS DISTRIBUTIONS30.9.4 The chi-squared distributionIn subsection 30.6.2, we showed that if X is Gaussian distributed with mean µ andvariance σ 2 , such that X ∼ N(µ, σ 2 ), then the random variable Y =(x − µ) 2 /σ 2is distributed as the gamma distribution Y ∼ γ( 1 2 , 1 2). Let us now consider nindependent Gaussian random variables X i ∼ N(µ i ,σi 2 ), i =1, 2,...,n, and definethe new variableχ 2 n =n∑ (X i − µ i ) 2. (30.122)i=1Using the result (30.121) for multiple gamma distributions, χ 2 n must be distributedas the gamma variate χ 2 n ∼ γ( 1 2 , 1 2n), which from (30.118) has the PDF1f(χ 2 2n)=Γ( 1 2 n)(1 2 χ2 n) (n/2)−1 exp(− 1 2 χ2 n)1=2 n/2 Γ( 1 n) (n/2)−1 exp(− 12 n)(χ2 2 χ2 n). (30.123)This is known as the chi-squared distribution of order n and has numerousapplications in statistics (see chapter 31). Setting λ = 1 2 and r = 1 2n in (30.120),we find thatE[χ 2 n]=n,σ 2 iV [χ 2 n]=2n.An important generalisation occurs when the n Gaussian variables X i are notlinearly independent but are instead required to satisfy a linear constraint of theformc 1 X 1 + c 2 X 2 + ···+ c n X n =0, (30.124)in which the constants c i are not all zero. In this case, it may be shown (seeexercise 30.40) that the variable χ 2 n defined in (30.122) is still described by a chisquareddistribution, but one of order n − 1. Indeed, this result may be triviallyextended to show that if the n Gaussian variables X i satisfy m linear constraintsof the form (30.124) then the variable χ 2 n defined in (30.122) is described by achi-squared distribution of order n − m.30.9.5 The Cauchy and Breit–Wigner distributionsA random variable X (in the range −∞ to ∞) that obeys the Cauchy distributionis described by the PDFf(x) = 1 π119311+x 2 .

PROBABILITYf(x)0.80.6x 0 =0,x 0 =2,Γ=1 Γ=10.40.20x 0 =0,Γ=3−4 −2 0 2 4xFigure 30.17 The PDF f(x) for the Breit–Wigner distribution for differentvalues of the parameters x 0 and Γ.This is a special case of the Breit–Wigner distributionf(x) = 1 π12 Γ14 Γ2 +(x − x 0 ) , 2which is encountered in the study of nuclear and particle physics. In figure 30.17,we plot some examples of the Breit–Wigner distribution for several values of theparameters x 0 and Γ.We see from the figure that the peak (or mode) of the distribution occursat x = x 0 . It is also straightforward to show that the parameter Γ is equal tothe width of the peak at half the maximum height. Although the Breit–Wignerdistribution is symmetric about its peak, it does not formally possess a mean sincethe integrals ∫ 0−∞ xf(x) dx and ∫ ∞0xf(x) dx both diverge. Similar divergences occurfor all higher moments of the distribution.30.9.6 The uniform distributionFinally we mention the very simple, but common, uniform distribution, whichdescribes a continuous random variable that has a constant PDF over its allowedrange of values. If the limits on X are a and b then{1/(b − a) for a ≤ x ≤ b,f(x) =0 otherwise.The MGF of the uniform distribution is found to beM(t) = ebt − e at(b − a)t ,1194

30.10 THE CENTRAL LIMIT THEOREMand its mean and variance are given byE[X] = a + b2− a)2, V[X] =(b .1230.10 The central limit theoremIn subsection 30.9.1 we discussed approximating the binomial and Poisson distributionsby the Gaussian distribution when the number of trials is large. We nowdiscuss why the Gaussian distribution is so common and therefore so important.The central limit theorem may be stated as follows.Central limit theorem. Suppose that X i , i =1, 2,...,n,areindependent randomvariables, each of which is described by a probability density function f i (x) (thesemay all be different) with a mean µ i and a variance σi 2 . The random variable Z =(∑i X i)/n, i.e. the ‘mean’ of the Xi , has the following properties:(i) its expectation value is given by E[Z] = (∑ i µ i)/n;(ii) its variance is given by V [Z] = (∑ )i σ2 i /n 2 ;(iii) as n →∞the probability function of Z tends to a Gaussian with correspondingmean and variance.We note that for the theorem to hold, the probability density functions f i (x)must possess formal means and variances. Thus, for example, if any of the X iwere described by a Cauchy distribution then the theorem would not apply.Properties (i) and (ii) of the theorem are easily proved, as follows. FirstlyE[Z] = 1 n (E[X 1]+E[X 2 ]+···+ E[X n ]) = 1 ∑n (µ i1 + µ 2 + ···+ µ n )=µ in ,a result which does not require that the X i are independent random variables. Ifµ i = µ for all i then this becomesE[Z] = nµ n = µ.Secondly, if the X i are independent, it follows from an obvious extension of(30.68) that[ 1V [Z] =Vn (X 1 + X 2 + ···+ X n )]= 1 ∑n (V 2 [X i1]+V [X 2 ]+···+ V [X n ]) =σ2 in 2 .Let us now consider property (iii), which is the reason for the ubiquity ofthe Gaussian distribution and is most easily proved by considering the momentgenerating function M Z (t) ofZ. From (30.90), this MGF is given byn∏( ) tM Z (t) = M Xi ,ni=11195

PROBABILITYwhere M Xi (t) istheMGFoff i (x). Now( ) tM Xi =1+ t n n E[X i]+ 1 t 22n 2 E[X2 i ]+···t=1+µ in + 1 2 (σ2 i + µ 2 i ) t2n 2 + ··· ,and as n becomes large( ) ( t µi tM Xi ≈ expnn + 1 t 2 )2 σ2 in 2 ,as may be verified by expanding the exponential up to terms including (t/n) 2 .Thereforen∏(µi tM Z (t) ≈ expn + 1 t 2 ) (∑2 σ2 iin 2 =expµ ∑ )in t + 1 i σ2 i2n 2 t 2 .i=1Comparing this with the form of the MGF for a Gaussian distribution, (30.114),we can see that the probability density function g(z) ofZ tends to a Gaussian distributionwith mean ∑ i µ i/n and variance ∑ i σ2 i /n2 . In particular, if we considerZ to be the mean of n independent measurements of the same random variable X(so that X i = X for i =1, 2,...,n) then, as n →∞, Z has a Gaussian distributionwith mean µ and variance σ 2 /n.We may use the central limit theorem to derive an analogous result to (iii)above for the product W = X 1 X 2 ···X n of the n independent random variablesX i . Provided the X i only take values between zero and infinity, we may writeln W =lnX 1 +lnX 2 + ···+lnX n ,which is simply the sum of n new random variables ln X i . Thus, provided thesenew variables each possess a formal mean and variance, the PDF of ln W willtend to a Gaussian in the limit n →∞, and so the product W will be describedby a log-normal distribution (see subsection 30.9.2).30.11 Joint distributionsAs mentioned briefly in subsection 30.4.3, it is common in the physical sciences toconsider simultaneously two or more random variables that are not independent,in general, and are thus described by joint probability density functions. We willreturn to the subject of the interdependence of random variables after firstpresenting some of the general ways of characterising joint distributions. Wewill concentrate mainly on bivariate distributions, i.e. distributions of only tworandom variables, though the results may be extended readily to multivariatedistributions. The subject of multivariate distributions is large and a detailedstudy is beyond the scope of this book; the interested reader should therefore1196

30.11 JOINT DISTRIBUTIONSconsult one of the many specialised texts. However, we do discuss the multinomialand multivariate Gaussian distributions, in section 30.15.The first thing to note when dealing with bivariate distributions is that thedistinction between discrete and continuous distributions may not be as clear asfor the single variable case; the random variables can both be discrete, or bothcontinuous, or one discrete and the other continuous. In general, for the randomvariables X and Y , the joint distribution will take an infinite number of valuesunless both X and Y have only a finite number of values. In this chapter wewill consider only the cases where X and Y are either both discrete or bothcontinuous random variables.30.11.1 Discrete bivariate distributionsIn direct analogy with the one-variable (univariate) case, if X is a discrete randomvariable that takes the values {x i } and Y one that takes the values {y j } then theprobability function of the joint distribution is defined as{Pr(X = xi , Y = y j ) for x = x i , y = y j ,f(x, y) =0 otherwise.We may therefore think of f(x, y) as a set of spikes at valid points in the xy-plane,whose height at (x i ,y i ) represents the probability of obtaining X = x i and Y = y j .The normalisation of f(x, y) implies∑ ∑f(x i ,y j )=1, (30.125)ijwhere the sums over i and j take all valid pairs of values. We can also define thecumulative probability functionF(x, y) = ∑ ∑f(x i ,y j ), (30.126)x i≤x y j≤yfrom which it follows that the probability that X lies in the range [a 1 ,a 2 ]andYlies in the range [b 1 ,b 2 ] is given byPr(a 1

PROBABILITY30.11.2 Continuous bivariate distributionsIn the case where both X and Y are continuous random variables, the PDF ofthe joint distribution is defined byf(x, y) dx dy =Pr(x

30.12 PROPERTIES OF JOINT DISTRIBUTIONS30.11.3 Marginal and conditional distributionsGiven a bivariate distribution f(x, y), we may be interested only in the probabilityfunction for X irrespective of the value of Y (or vice versa). This marginaldistribution of X is obtained by summing or integrating, as appropriate, thejoint probability distribution over all allowed values of Y . Thus, the marginaldistribution of X (for example) is given by{∑jf X (x) =f(x, y j) for a discrete distribution,∫ (30.130)f(x, y) dy for a continuous distribution.It is clear that an analogous definition exists for the marginal distribution of Y .Alternatively, one might be interested in the probability function of X giventhat Y takes some specific value of Y = y 0 ,i.e.Pr(X = x|Y = y 0 ). This conditionaldistribution of X is given byg(x) = f(x, y 0)f Y (y 0 ) ,where f Y (y) is the marginal distribution of Y . The division by f Y (y 0 ) is necessaryin order that g(x) is properly normalised.30.12 Properties of joint distributionsThe probability density function f(x, y) contains all the information on the jointprobability distribution of two random variables X and Y . In a similar mannerto that presented for univariate distributions, however, it is conventional tocharacterise f(x, y) by certain of its properties, which we now discuss. Onceagain, most of these properties are based on the concept of expectation values,which are defined for joint distributions in an analogous way to those for singlevariabledistributions (30.46). Thus, the expectation value of any function g(X,Y )of the random variables X and Y is given by{∑ ∑i jE[g(X,Y )] =g(x i,y j )f(x i ,y j ) for the discrete case,g(x, y)f(x, y) dx dy for the continuous case.∫ ∞ ∫ ∞−∞ −∞30.12.1 MeansThe means of X and Y are defined respectively as the expectation values of thevariables X and Y . Thus, the mean of X is given by{∑ ∑i jE[X] =µ X =x if(x i ,y j ) for the discrete case,∫ ∞ ∫ ∞−∞ −∞ xf(x, y) dx dy for the continuous case. (30.131)E[Y ] is obtained in a similar manner.1199

PROBABILITY◮Show that if X and Y are independent random variables then E[XY ]=E[X]E[Y ].LetusconsiderthecasewhereX and Y are continuous random variables. Since X andY are independent f(x, y) =f X (x)f Y (y), so thatE[XY ]=∫ ∞ ∫ ∞xyf X (x)f Y (y) dx dy =∫ ∞xf X (x) dx∫ ∞−∞ −∞−∞−∞An analogous proof exists for the discrete case. ◭yf Y (y) dy = E[X]E[Y ].30.12.2 VariancesThe definitions of the variances of X and Y are analogous to those for thesingle-variable case (30.48), i.e. the variance of X is given by{∑ ∑V [X] =σX 2 = i j (x i − µ X ) 2 f(x i ,y j ) for the discrete case,∫ ∞ ∫ ∞−∞ −∞ (x − µ X) 2 f(x, y) dx dy for the continuous case. (30.132)Equivalent definitions exist for the variance of Y .30.12.3 Covariance and correlationMeans and variances of joint distributions provide useful information abouttheir marginal distributions, but we have not yet given any indication of how tomeasure the relationship between the two random variables. Of course, it maybe that the two random variables are independent, but often this is not so. Forexample, if we measure the heights and weights of a sample of people we wouldnot be surprised to find a tendency for tall people to be heavier than short peopleand vice versa. We will show in this section that two functions, the covarianceand the correlation, can be defined for a bivariate distribution and that these areuseful in characterising the relationship between the two random variables.The covariance of two random variables X and Y is defined byCov[X,Y ]=E[(X − µ X )(Y − µ Y )], (30.133)where µ X and µ Y are the expectation values of X and Y respectively. Clearlyrelated to the covariance is the correlation of the two random variables, definedbyCorr[X,Y ]= Cov[X,Y ] , (30.134)σ X σ Ywhere σ X and σ Y are the standard deviations of X and Y respectively. It can beshown that the correlation function lies between −1 and +1. If the value assumedis negative, X and Y are said to be negatively correlated, if it is positive they aresaid to be positively correlated andifitiszerotheyaresaidtobeuncorrelated.We will now justify the use of these terms.1200

30.12 PROPERTIES OF JOINT DISTRIBUTIONSOne particularly useful consequence of its definition is that the covarianceof two independent variables, X and Y , is zero. It immediately follows from(30.134) that their correlation is also zero, and this justifies the use of the term‘uncorrelated’ for two such variables. To show this extremely important propertywe first note thatCov[X,Y ]=E[(X − µ X )(Y − µ Y )]= E[XY − µ X Y − µ Y X + µ X µ Y ]= E[XY ] − µ X E[Y ] − µ Y E[X]+µ X µ Y= E[XY ] − µ X µ Y . (30.135)Now, if X and Y are independent then E[XY ]=E[X]E[Y ]=µ X µ Y and soCov[X,Y ] = 0. It is important to note that the converse of this result is notnecessarily true; two variables dependent on each other can still be uncorrelated.In other words, it is possible (and not uncommon) for two variables X and Yto be described by a joint distribution f(x, y) thatcannot be factorised into aproduct of the form g(x)h(y), but for which Corr[X,Y ] = 0. Indeed, from thedefinition (30.133), we see that for any joint distribution f(x, y) that is symmetricin x about µ X (or similarly in y) we have Corr[X,Y ]=0.We have already asserted that if the correlation of two random variables ispositive (negative) they are said to be positively (negatively) correlated. We havealso stated that the correlation lies between −1 and +1. The terminology suggeststhat if the two RVs are identical (i.e. X = Y ) then they are completely correlatedand that their correlation should be +1. Likewise, if X = −Y then the functionsare completely anticorrelated and their correlation should be −1. Values of thecorrelation function between these extremes show the existence of some degreeof correlation. In fact it is not necessary that X = Y for Corr[X,Y ] = 1; it issufficient that Y is a linear function of X, i.e.Y = aX + b (with a positive). If ais negative then Corr[X,Y ]=−1. To show this we first note that µ Y = aµ X + b.NowY = aX + b = aX + µ Y − aµ X ⇒ Y − µ Y = a(X − µ X ),and so using the definition of the covariance (30.133)Cov[X,Y ]=aE[(X − µ X ) 2 ]=aσX.2It follows from the properties of the variance (subsection 30.5.3) that σ Y = |a|σ Xand so, using the definition (30.134) of the correlation,Corr[X,Y ]= aσ2 X|a|σX2 = a|a| ,which is the stated result.It should be noted that, even if the possibilities of X and Y being non-zero aremutually exclusive, Corr[X,Y ] need not have value ±1.1201

PROBABILITY◮A biased die gives probabilities 1 p, p, p, p, p, 2p of throwing 1, 2, 3, 4, 5, 6 respectively.2If the random variable X is the number shown on the die and the random variable Y isdefined as X 2 , calculate the covariance and correlation of X and Y .We have already calculated in subsections 30.2.1 and 30.5.4 thatp = 253, E[X] =13Using (30.135), we obtainNow E[X 3 ]isgivenby13 , E[ X 2] = 25313, V[X] =480169 .Cov[X,Y ]=Cov[X,X 2 ]=E[X 3 ] − E[X]E[X 2 ].E[X 3 ]=1 3 × 1 p 2 +(23 +3 3 +4 3 +5 3 )p +6 3 × 2p= 13132 p = 101,and the covariance of X and Y is given byCov[X,Y ] = 101 − 5313 × 25313 = 3660169 .The correlation is defined by Corr[X,Y ]=Cov[X,Y ]/σ X σ Y . The standard deviation ofY may be calculated from the definition of the variance. Letting µ Y = E[X 2 ]= 253 gives13σY 2 = p ( )1 2 2 ( )− µ Y + p 2 2 2 ( )− µ Y + p 3 2 2 ( )− µ Y + p 4 2 2− µ Y2+ p ( )5 2 2 ( )− µ Y +2p 6 2 2− µ Y187 356 28 824= p =169 169 .We deduce thatCorr[X,Y ]= 3660√ √169 169169 28 824 480 ≈ 0.984.Thus the random variables X and Y display a strong degree of positive correlation, as wewould expect. ◭We note that the covariance of X and Y occurs in various expressions. Forexample, if X and Y are not independent thenV [X + Y ]=E [ (X + Y ) 2] − (E[X + Y ]) 2= E [ X 2] +2E[XY ]+E [ Y 2] −{(E[X]) 2 +2E[X]E[Y ]+(E[Y ]) 2 }= V [X]+V [Y ]+2(E[XY ] − E[X]E[Y ])= V [X]+V [Y ]+2Cov[X,Y ].1202

30.12 PROPERTIES OF JOINT DISTRIBUTIONSMore generally, we find (for a, b and c constant)V [aX + bY + c] =a 2 V [X]+b 2 V [Y ]+2ab Cov[X,Y ].(30.136)Note that if X and Y are in fact independent then Cov[X,Y ] = 0 and we recoverthe expression (30.68) in subsection 30.6.4.We may use (30.136) to obtain an approximate expression for V [ f(X,Y )]for any arbitrary function f, even when the random variables X and Y arecorrelated. Approximating f(X,Y ) by the linear terms of its Taylor expansionabout the point (µ X ,µ Y ), we have( )( )∂f∂ff(X,Y ) ≈ f(µ X ,µ Y )+ (X − µ X )+ (Y − µ Y ),∂X∂Y(30.137)where the partial derivatives are evaluated at X = µ X and Y = µ Y . Taking thevariance of both sides, and using (30.136), we find( ) 2 ( ) 2 ( )( )∂f∂f∂f ∂fV [ f(X,Y )] ≈ V [X]+ V [Y ]+2Cov[X,Y ].∂X∂Y∂X ∂Y(30.138)Clearly, if Cov[X,Y ] = 0, we recover the result (30.69) derived in subsection 30.6.4.We note that (30.138) is exact if f(X,Y ) is linear in X and Y .For several variables X i , i =1, 2,...,n, we can define the symmetric (positivedefinite) covariance matrix whose elements areand the symmetric (positive definite) correlation matrixV ij =Cov[X i ,X j ], (30.139)ρ ij =Corr[X i ,X j ].The diagonal elements of the covariance matrix are the variances of the variables,whilst those of the correlation matrix are unity. For several variables, (30.138)generalises toV [f(X 1 ,X 2 ,...,X n )] ≈ ∑ i( ∂f∂X i) 2V [X i ]+ ∑ i∑j≠i( ∂f∂X i)( ∂f∂X j)Cov[X i ,X j ],where the partial derivatives are evaluated at X i = µ Xi .1203

PROBABILITY◮A card is drawn at random from a normal 52-card pack and its identity noted. The cardis replaced, the pack shuffled and the process repeated. Random variables W, X, Y, Z aredefined as follows:W =2 if the drawn card is a heart; W =0otherwise.X =4 if the drawn card is an ace, king, or queen; X =2if the card isajackorten;X =0otherwise.Y =1 if the drawn card is red; Y =0otherwise.Z =2 if the drawn card is black and an ace, king or queen; Z =0otherwise.Establish the correlation matrix for W, X, Y, Z.The means of the variables are given byµ W =2× 1 = 1 , µ 4 2 X = ( ) ( )4 × 313 + 2 ×213 =16, 13µ Y =1× 1 = 1 , µ 2 2 Z =2× 6 = 3 . 52 13The variances, calculated from σU 2 = V [U] =E [ U 2] − (E[U]) 2 ,whereU = W , X, Y orZ, areσW 2 = ( ) (4 × 1 4 − 1) 22 =3, 4 σ2 X = ( 16 × 313σ 2 Y = ( 1 × 1 2)−( 12) 2=14 , σ2 Z = ( 4 × 652) ( ) (+ 4 ×2 − 16)−( 31313) 2=69169 .13) 2=472169 ,The covariances are found by first calculating E[WX] etc. and then forming E[WX]−µ W µ Xetc.E[WX]=2(4) ( ) (352 +2(2) 2)52 =8, Cov[W,X]= 8 − ( 1 16)13 13 2 13 =0,E[WY] = 2(1) ( )14 =1, Cov[W,Y]= 1 − ( 1 1)2 2 2 2 =1, 4(E[WZ]=0, Cov[W,Z]=0− 1 3)2 13 = −3, 26E[XY ] = 4(1) ( ) (652 +2(1) 4)52 =88, Cov[X,Y ]= − ( 16 1)13 13 13 2 =0,E[XZ] = 4(2) ( 652)=1212, Cov[X,Z]= − 1613 13 13( 3)13 =108, 169(E[YZ]=0, Cov[Y,Z]=0− 1 3)2 13 = −3. 26The correlations Corr[W,X] and Corr[X,Y ] are clearly zero; the remainder are given by(Corr[W,Y]= 1 3 × ) 1 −1/24 4 4 = 0.577,(Corr[W,Z]=− 3 3 × ) 69 −1/226 4 169 = −0.209,Corr[X,Z]= 108169Corr[Y,Z]=− 326( 472169 × 69) −1/2169 = 0.598,( 1 × ) 69 −1/24 169 = −0.361.Finally, then, we can write down the correlation matrix:⎛⎞1 0 0.58 −0.21⎜ 0 1 0 0.60 ⎟ρ = ⎝0.58 0 1 −0.36⎠ .−0.21 0.60 −0.36 11204

30.13 GENERATING FUNCTIONS FOR JOINT DISTRIBUTIONSAs would be expected, X is uncorrelated with either W or Y , colour and face-value beingtwo independent characteristics. Positive correlations are to be expected between W andY and between X and Z; both correlations are fairly strong. Moderate anticorrelationsexist between Z and both W and Y , reflecting the fact that it is impossible for W and Yto be positive if Z is positive. ◭Finally, let us suppose that the random variables X i , i =1, 2,...,n, are relatedto a second set of random variables Y k = Y k (X 1 ,X 2 ,...,X n ), k =1, 2,...,m.Byexpanding each Y k as a Taylor series as in (30.137) and inserting the resultingexpressions into the definition of the covariance (30.133), we find that the elementsof the covariance matrix for the Y k variables are given byCov[Y k ,Y l ] ≈ ∑ ∑( )( )∂Yk ∂YlCov[X i ,X j ].∂Xi j i ∂X j(30.140)It is straightforward to show that this relation is exact if the Y k are linearcombinations of the X i . Equation (30.140) can then be written in matrix form asV Y = SV X S T , (30.141)where V Y and V X are the covariance matrices of the Y k and X i variables respectivelyand S is the rectangular m × n matrix with elements S ki = ∂Y k /∂X i .30.13 Generating functions for joint distributionsIt is straightforward to generalise the discussion of generating function in section30.7 to joint distributions. For a multivariate distribution f(X 1 ,X 2 ,...,X n ) ofnon-negative integer random variables X i , i =1, 2,...,n, we define the probabilitygenerating function to beΦ(t 1 ,t 2 ,...,t n )=E[t X11 tX2 2 ···tXn n ].As in the single-variable case, we may also define the closely related momentgenerating function, which has wider applicability since it is not restricted tonon-negative integer random variables but can be used with any set of discreteor continuous random variables X i (i =1, 2,...,n). The MGF of the multivariatedistribution f(X 1 ,X 2 ,...,X n ) is defined asM(t 1 ,t 2 ,...,t n )=E[e t1X1 e t2X2 ···e tnXn ]=E[e t1X1+t2X2+···+tnXn ](30.142)and may be used to evaluate (joint) moments of f(X 1 ,X 2 ,...,X n ). By performinga derivation analogous to that presented for the single-variable case in subsection30.7.2, it can be shown that∂m1+m2+···+mn M(0, 0,...,0)E[X m11 Xm2 2···Xn mn]= . (30.143)∂t m11 ∂tm2 2 ···∂tmn n1205

PROBABILITYFinally we note that, by analogy with the single-variable case, the characteristicfunction and the cumulant generating function of a multivariate distribution aredefined respectively asC(t 1 ,t 2 ,...,t n )=M(it 1 ,it 2 ,...,it n ) and K(t 1 ,t 2 ,...,t n )=lnM(t 1 ,t 2 ,...,t n ).◮Suppose that the random variables X i , i =1, 2,...,n,aredescribedbythePDFf(x) =f(x 1 ,x 2 ,...,x n )=N exp(− 1 2 xT Ax),where the column vector x =(x 1 x 2 ··· x n ) T , A is an n × n symmetric matrix and Nis a normalisation constant such that∫∫ ∞ ∫ ∞ ∫ ∞f(x) d n x ≡ ··· f(x 1 ,x 2 ,...,x n ) dx 1 dx 2 ···dx n =1.∞Find the MGF of f(x).−∞−∞−∞From (30.142), the MGF is given by∫M(t 1 ,t 2 ,...,t n )=N exp(− 1 2 xT Ax + t T x) d n x, (30.144)∞where the column vector t =(t 1 t 2 ··· t n ) T . In order to evaluate this multiple integral,we begin by noting thatx T Ax − 2t T x =(x − A −1 t) T A(x − A −1 t) − t T A −1 t,which is the matrix equivalent of ‘completing the square’. Using this expression in (30.144)and making the substitution y = x − A −1 t,weobtainM(t 1 ,t 2 ,...,t n )=c exp( 1 2 tT A −1 t), (30.145)where the constant c is given by∫c = N exp(− 1 2 yT Ay) d n y.∞From the normalisation condition for N, weseethatc = 1, as indeed it must be in orderthat M(0, 0,...,0) = 1. ◭30.14 Transformation of variables in joint distributionsSuppose the random variables X i , i =1, 2,...,n, are described by the multivariatePDF f(x 1 ,x 2 ...,x n ). If we wish to consider random variables Y j , j =1, 2,...,m,related to the X i by Y j = Y j (X 1 ,X 2 ,...,X m ) then we may calculate g(y 1 ,y 2 ,...,y m ),the PDF for the Y j , in a similar way to that in the univariate case by demandingthat|f(x 1 ,x 2 ...,x n ) dx 1 dx 2 ···dx n | = |g(y 1 ,y 2 ,...,y m ) dy 1 dy 2 ···dy m |.From the discussion of changing the variables in multiple integrals given inchapter 6 it follows that, in the special case where n = m,g(y 1 ,y 2 ,...,y m )=f(x 1 ,x 2 ...,x n )|J|,1206

30.15 IMPORTANT JOINT DISTRIBUTIONSwhere∣ ∣∣∣∣∣∣∣∣∣ ∂x 1 ∂x∣n ∣∣∣∣∣∣∣∣∣...J ≡ ∂(x ∂y 1 ∂y 11,x 2 ...,x n )∂(y 1 ,y 2 ,...,y n ) = .. .. . ,∂x 1 ∂x n...∂y n ∂y nis the Jacobian of the x i with respect to the y j .◮Suppose that the random variables X i , i =1, 2,...,n, are independent and Gaussian distributedwith means µ i and variances σi2 respectively. Find the PDF for the new variablesZ i =(X i − µ i )/σ i , i =1, 2,...,n. By considering an elemental spherical shell in Z-space,find the PDF of the chi-squared random variable χ 2 n = ∑ ni=1 Z i 2 .Since the X i are independent random variables,[]1n∑ (x i − µ i ) 2f(x 1 ,x 2 ,...,x n )=f(x 1 )f(x 2 ) ···f(x n )=exp −.(2π) n/2 σ 1 σ 2 ···σ n 2σ 2 i=1 iTo derive the PDF for the variables Z i ,werequire|f(x 1 ,x 2 ,...,x n ) dx 1 dx 2 ···dx n | = |g(z 1 ,z 2 ,...,z n ) dz 1 dz 2 ···dz n |,and, noting that dz i = dx i /σ i ,weobtain)1g(z 1 ,z 2 ,...,z n )=(−(2π) exp 1 n∑z n/2 i2 .2i=1Let us now consider the random variable χ 2 n = ∑ ni=1 Z i 2 , which we may regard as thesquare of the distance from the origin in the n-dimensional Z-space. We now require thatg(z 1 ,z 2 ,...,z n ) dz 1 dz 2 ···dz n = h(χ 2 n)dχ 2 n.If we consider the infinitesimal volume dV = dz 1 dz 2 ···dz n to be that enclosed by then-dimensional spherical shell of radius χ n and thickness dχ n then we may write dV =dχ n , for some constant A. We thus obtainAχ n−1nh(χ 2 n)dχ 2 n ∝ exp(− 1 2 χ2 n)χ n−1n dχ n ∝ exp(− 1 2 χ2 n)χn n−2 dχ 2 n,where we have used the fact that dχ 2 n =2χ n dχ n . Thus we see that the PDF for χ 2 n is givenbyh(χ 2 n)=B exp(− 1 2 χ2 n)χ n−2n ,for some constant B. This constant may be determined from the normalisation condition∫ ∞h(χ 2 n) dχ 2 n =10and is found to be B = [2 n/2 Γ( 1 2 n)]−1 . This is the nth-order chi-squared distributiondiscussed in subsection 30.9.4. ◭30.15 Important joint distributionsIn this section we will examine two important multivariate distributions, themultinomial distribution, which is an extension of the binomial distribution, andthe multivariate Gaussian distribution.1207

PROBABILITY30.15.1 The multinomial distributionThe binomial distribution describes the probability of obtaining x ‘successes’ fromn independent trials, where each trial has only two possible outcomes. This maybe generalised to the case where each trial has k possible outcomes with respectiveprobabilities p 1 , p 2 , ..., p k . If we consider the random variables X i , i =1, 2,...,n,to be the number of outcomes of type i in n trials then we may calculate theirjoint probability functionf(x 1 ,x 2 ,...,x k )=Pr(X 1 = x 1 ,X 2 = x 2 , ..., X k = x k ),wherewemusthave ∑ ki=1 x i = n. Inn trials the probability of obtaining x 1outcomes of type 1, followed by x 2 outcomes of type 2 etc. is given byp x11 px2 2 ···pxk k .However, the number of distinguishable permutations of this result isn!x 1 !x 2 ! ···x k ! ,and thusn!f(x 1 ,x 2 ,...,x k )=x 1 !x 2 ! ···x k ! px1 1 px2 2 ···pxk k . (30.146)This is the multinomial probability distribution.If k = 2 then the multinomial distribution reduces to the familiar binomialdistribution. Although in this form the binomial distribution appears to be afunction of two random variables, it must be remembered that, in fact, sincep 2 =1− p 1 and x 2 = n − x 1 , the distribution of X 1 is entirely determined by theparameters p and n. ThatX 1 has a binomial distribution is shown by rememberingthat it represents the number of objects of a particular type obtained fromsampling with replacement, which led to the original definition of the binomialdistribution. In fact, any of the random variables X i has a binomial distribution,i.e. the marginal distribution of each X i is binomial with parameters n and p i .Itimmediately follows thatE[X i ]=np i and V [X i ] 2 = np i (1 − p i ). (30.147)◮At a village fête patrons were invited, for a 10 p entry fee, to pick without looking sixtickets from a drum containing equal large numbers of red, blue and green tickets. If fiveor more of the tickets were of the same colour a prize of 100 p was awarded. A consolationaward of 40 p was made if two tickets of each colour were picked. Was a good time had byall?In this case, all types of outcome (red, blue and green) have the same probabilities. Theprobability of obtaining any given combination of tickets is given by the multinomialdistribution with n =6,k =3andp i = 1 , i =1, 2, 3.31208

30.15 IMPORTANT JOINT DISTRIBUTIONS(i) The probability of picking six tickets of the same colour is given byPr (six of the same colour) = 3 × 6! ( ) 6 ( ) 0 ( ) 0 1 1 1= 16!0!0! 3 3 3 243 .The factor of 3 is present because there are three different colours.(ii) The probability of picking five tickets of one colour and one ticket of anothercolour isPr(five of one colour; one of another) = 3 × 2 × 6! ( ) 5 ( ) 1 ( ) 0 1 1 1= 4 5!1!0! 3 3 3 81 .The factors of 3 and 2 are included because there are three ways to choose thecolour of the five matching tickets, and then two ways to choose the colour of theremaining ticket.(iii) Finally, the probability of picking two tickets of each colour isPr (two of each colour) = 6! ( ) 2 ( ) 2 ( ) 2 1 1 1= 102!2!2! 3 3 3 81 .Thus the expected return to any patron was, in pence,( 1100243 + 4 ) (+ 40 × 10 )=10.29.8181A good time was had by all but the stallholder! ◭30.15.2 The multivariate Gaussian distributionA particularly interesting multivariate distribution is provided by the generalisationof the Gaussian distribution to multiple random variables X i , i =1, 2,...,n.If the expectation value of X i is E(X i )=µ i then the general form of the PDF isgiven by[f(x 1 ,x 2 ,...,x n )=N exp − 1 2∑ ∑ij]a ij (x i − µ i )(x j − µ j ) ,where a ij = a ji and N is a normalisation constant that we give below. If we writethe column vectors x =(x 1 x 2 ··· x n ) T and µ =(µ 1 µ 2 ··· µ n ) T ,anddenote the matrix with elements a ij by A thenf(x) =f(x 1 ,x 2 ,...,x n )=N exp [ − 1 2 (x − µ)T A(x − µ) ] ,where A is symmetric. Using the same method as that used to derive (30.145) itis straightforward to show that the MGF of f(x) is given byM(t 1 ,t 2 ,...,t n )=exp ( µ T t + 1 2 tT A −1 t ) ,where the column matrix t =(t 1 t 2 ··· t n ) T . From the MGF, we find thatE[X i X j ]= ∂2 M(0, 0,...,0)= µ i µ j +(A −1 ) ij ,∂t i ∂t j1209

PROBABILITYand thus, using (30.135), we obtainCov[X i ,X j ]=E[(X i − µ i )(X j − µ j )] = (A −1 ) ij .Hence A is equal to the inverse of the covariance matrix V of the X i , see (30.139).Thus, with the correct normalisation, f(x) is given byf(x) =1(2π) n/2 (det V) 1/2 exp [ − 1 2 (x − µ)T V −1 (x − µ) ] .(30.148)◮Evaluate the integral∫I = exp [ − 1 (x − 2 µ)T V −1 (x − µ) ] d n x,∞where V is a symmetric matrix, and hence verify the normalisation in (30.148).We begin by making the substitution y = x − µ to obtain∫I = exp(− 1 2 yT V −1 y) d n y.∞Since V is a symmetric matrix, it may be diagonalised by an orthogonal transformation tothe new set of variables y ′ = S T y,whereS is the orthogonal matrix with the normalisedeigenvectors of V as its columns (see section 8.16). In this new basis, the matrix V becomesV ′ = S T VS =diag(λ 1 ,λ 2 ,...,λ n ),where the λ i are the eigenvalues of V. Also, since S is orthogonal, det S = ±1, and sod n y = |det S| d n y ′ = d n y ′ .Thus we can write I as∫ ∞ ∫ ∞ ∫ ( )∞n∑ y ′ 2iI = ··· exp − dy 1 ′ dy 2 ′ ···dy n′−∞ −∞ −∞2λ ii=1n∏∫ ( )∞= exp − y′ 2idy i ′ =(2π) n/2 (λ 1 λ 2 ···λ n ) 1/2 , (30.149)i=1−∞ 2λ iwhere we have used the standard integral ∫ ∞−∞ exp(−αy2 ) dy =(π/α) 1/2 (see subsection6.4.2). From section 8.16, however, we note that the product of eigenvalues in (30.149) isequal to det V. Thus we finally obtainI =(2π) n/2 (det V) 1/2 ,and hence the normalisation in (30.148) ensures that f(x) integrates to unity. ◭The above example illustrates some importants points concerning the multivariateGaussian distribution. In particular, we note that the Y i′ are independentGaussian variables with mean zero and variance λ i . Thus, given a general set ofn Gaussian variables x with means µ and covariance matrix V, one can alwaysperform the above transformation to obtain a new set of variables y ′ , which arelinear combinations of the old ones and are distributed as independent Gaussianswith zero mean and variances λ i .This result is extremely useful in proving many of the properties of the mul-1210

30.16 EXERCISEStivariate Gaussian. For example, let us consider the quadratic form (multipliedby 2) appearing in the exponent of (30.148) and write it as χ 2 n,i.e.χ 2 n =(x − µ) T V −1 (x − µ). (30.150)From (30.149), we see that we may also write it asn∑χ 2 y ′ 2in = ,λ ii=1which is the sum of n independent Gaussian variables with mean zero and unitvariance. Thus, as our notation implies, the quantity χ 2 n is distributed as a chisquaredvariable of order n. As illustrated in exercise 30.40, if the variables X i arerequired to satisfy m linear constraints of the form ∑ ni=1 c iX i =0thenχ 2 n definedin (30.150) is distributed as a chi-squared variable of order n − m.30.16 Exercises30.1 By shading or numbering Venn diagrams, determine which of the following arevalid relationships between events. For those that are, prove the relationshipusing de Morgan’s laws.(a) ( ¯X ∪ Y )=X ∩ Ȳ .(b) ¯X ∪ Ȳ = (X ∪ Y ).(c) (X ∪ Y ) ∩ Z =(X ∪ Z) ∩ Y .(d) X ∪ (Y ∩ Z) =(X ∪ Ȳ ) ∩ ¯Z.(e) X ∪ (Y ∩ Z) =(X ∪ Ȳ ) ∪ ¯Z.30.2 Given that events X,Y and Z satisfy(X ∩ Y ) ∪ (Z ∩ X) ∪ ( ¯X ∪ Ȳ )=(Z ∪ Ȳ ) ∪{[( ¯Z ∪ ¯X) ∪ ( ¯X ∩ Z)] ∩ Y },prove that X ⊃ Y , and that either X ∩ Z = ∅ or Y ⊃ Z.30.3 A and B each have two unbiased four-faced dice, the four faces being numbered1, 2, 3, 4. Without looking, B tries to guess the sum x of the numbers on thebottom faces of A’s two dice after they have been thrown onto a table. If theguess is correct B receives x 2 euros, but if not he loses x euros.Determine B’s expected gain per throw of A’s dice when he adopts each of thefollowing strategies:(a) he selects x at random in the range 2 ≤ x ≤ 8;(b) he throws his own two dice and guesses x to be whatever they indicate;(c) he takes your advice and always chooses the same value for x. Which numberwould you advise?30.4 Use the method of induction to prove equation (30.16), the probability additionlaw for the union of n general events.30.5 Two duellists, A and B, take alternate shots at each other, and the duel is overwhen a shot (fatal or otherwise!) hits its target. Each shot fired by A has aprobability α of hitting B, and each shot fired by B has a probability β of hittingA. Calculate the probabilities P 1 and P 2 , defined as follows, that A will win sucha duel: P 1 , A fires the first shot; P 2 , B fires the first shot.If they agree to fire simultaneously, rather than alternately, what is the probabilityP 3 that A will win, i.e. hit B without being hit himself?1211

PROBABILITY30.6 X 1 ,X 2 ,...,X n are independent, identically distributed, random variables drawnfrom a uniform distribution on [0, 1]. The random variables A and B are definedbyA =min(X 1 ,X 2 ,...,X n ), B = max(X 1 ,X 2 ,...,X n ).For any fixed k such that 0 ≤ k ≤ 1 , find the probability, p 2 n, that bothA ≤ k and B ≥ 1 − k.Check your general formula by considering directly the cases (a) k =0,(b)k = 1 , 2(c) n =1and(d)n =2.30.7 A tennis tournament is arranged on a straight knockout basis for 2 n players,and for each round, except the final, opponents for those still in the competitionare drawn at random. The quality of the field is so even that in any match it isequally likely that either player will win. Two of the players have surnames thatbegin with ‘Q’. Find the probabilities that they play each other(a) in the final,(b) at some stage in the tournament.30.8 This exercise shows that the odds are hardly ever ‘evens’ when it comes to dicerolling.(a) Gamblers A and B each roll a fair six-faced die, and B wins if his score isstrictly greater than A’s. Show that the odds are 7 to 5 in A’s favour.(b) Calculate the probabilities of scoring a total T from two rolls of a fair diefor T =2, 3,...,12. Gamblers C and D each roll a fair die twice and scorerespective totals T C and T D , D winning if T D >T C . Realising that the oddsare not equal, D insists that C should increase her stake for each game. Cagrees to stake £1.10 per game, as compared to D’s £1.00 stake. Who willshow a profit?30.9 An electronics assembly firm buys its microchips from three different suppliers;half of them are bought from firm X, whilst firms Y and Z supply 30% and20%, respectively. The suppliers use different quality-control procedures and thepercentages of defective chips are 2%, 4% and 4% for X, Y and Z, respectively.The probabilities that a defective chip will fail two or more assembly-line tests are40%, 60% and 80%, respectively, whilst all defective chips have a 10% chanceof escaping detection. An assembler finds a chip that fails only one test. What isthe probability that it came from supplier X?30.10 As every student of probability theory will know, Bayesylvania is awash withnatives, not all of whom can be trusted to tell the truth, and lost, and apparentlysomewhat deaf, travellers who ask the same question several times in an attemptto get directions to the nearest village.One such traveller finds himself at a T-junction in an area populated by theAsciis and Bisciis in the ratio 11 to 5. As is well known, the Biscii always lie, butthe Ascii tell the truth three quarters of the time, giving independent answers toall questions, even to immediately repeated ones.(a) The traveller asks one particular native twice whether he should go to theleft or to the right to reach the local village. Each time he is told ‘left’. Shouldhe take this advice, and, if he does, what are his chances of reaching thevillage?(b) The traveller then asks the same native the same question a third time, andfor a third time receives the answer ‘left’. What should the traveller do now?Have his chances of finding the village been altered by asking the thirdquestion?1212

30.16 EXERCISES30.11 A boy is selected at random from amongst the children belonging to families withn children. It is known that he has at least two sisters. Show that the probabilitythat he has k − 1brothersis(n − 1)!(2 n−1 − n)(k − 1)!(n − k)! ,for 1 ≤ k ≤ n − 2 and zero for other values of k. Assume that boys and girls areequally likely.30.12 Villages A, B, C and D are connected by overhead telephone lines joining AB,AC, BC, BD and CD. As a result of severe gales, there is a probability p (thesame for each link) that any particular link is broken.(a) Show that the probability that a call can be made from A to B is1 − p 2 − 2p 3 +3p 4 − p 5 .(b) Show that the probability that a call can be made from D to A is1 − 2p 2 − 2p 3 +5p 4 − 2p 5 .30.13 A set of 2N + 1 rods consists of one of each integer length 1, 2,...,2N,2N +1.Three, of lengths a, b and c, are selected, of which a is the longest. By consideringthe possible values of b and c, determine the number of ways in which a nondegeneratetriangle (i.e. one of non-zero area) can be formed (i) if a is even,and (ii) if a is odd. Combine these results appropriately to determine the totalnumber of non-degenerate triangles that can be formed with the 2N + 1 rods,and hence show that the probability that such a triangle can be formed from arandom selection (without replacement) of three rods is(N − 1)(4N +1).2(4N 2 − 1)30.14 A certain marksman never misses his target, which consists of a disc of unit radiuswith centre O. The probability that any given shot will hit the target within adistance t of O is t 2 ,for0≤ t ≤ 1. The marksman fires n independendent shotsat the target, and the random variable Y is the radius of the smallest circle withcentre O that encloses all the shots. Determine the PDF for Y and hence findthe expected area of the circle.The shot that is furthest from O is now rejected and the corresponding circledetermined for the remaining n − 1 shots. Show that its expected area isn − 1n +1 π.30.15 The duration (in minutes) of a telephone call made from a public call-box is arandom variable T . The probability density function of T is⎧⎪⎨ 0 t

PROBABILITYfighting, kittens are removed at random, one at a time, until peace is restored.Show, by induction, that the expected number of kittens finally remaining isN(x, y) =xy +1 + yx +1 .30.17 If the scores in a cup football match are equal at the end of the normal period ofplay, a ‘penalty shoot-out’ is held in which each side takes up to five shots (fromthe penalty spot) alternately, the shoot-out being stopped if one side acquires anunassailable lead (i.e. has a lead greater than its opponents have shots remaining).If the scores are still level after the shoot-out a ‘sudden death’ competition takesplace.In sudden death each side takes one shot and the competition is over if oneside scores and the other does not; if both score, or both fail to score, a furthershot is taken by each side, and so on. Team 1, which takes the first penalty, hasa probability p 1 , which is independent of the player involved, of scoring and aprobability q 1 (= 1 − p 1 ) of missing; p 2 and q 2 are defined likewise.Define Pr(i : x, y) as the probability that team i has scored x goals after yattempts, and let f(M) be the probability that the shoot-out terminates after atotal of M shots.(a) Prove that the probability that ‘sudden death’ will be needed isf(11+) =5∑( 5 C r ) 2 (p 1 p 2 ) r (q 1 q 2 ) 5−r .r=0(b) Give reasoned arguments (preferably without first looking at the expressionsinvolved) which show thatf(M =2N) =for N =3, 4, 5and2N−6∑r=0f(M =2N +1)={ }p2 Pr(1 : r, N)Pr(2:5− N + r, N − 1)+ q 2 Pr(1 : 6 − N + r, N)Pr(2:r, N − 1)2N−5∑r=0{ }p1 Pr(1 : 5 − N + r, N)Pr(2:r, N)+ q 1 Pr(1 : r, N)Pr(2:5− N + r, N)for N =3, 4.(c) Give an explicit expression for Pr(i : x, y) and hence show that if the teamsare so well matched that p 1 = p 2 =1/2 thenf(2N) =f(2N +1)=2N−6∑r=02N−5∑r=0( ) 1N!(N − 1)!62 2N r!(N − r)!(6 − N + r)!(2N − 6 − r)! ,( ) 1(N!) 22 2N r!(N − r)!(5 − N + r)!(2N − 5 − r)! .(d) Evaluate these expressions to show that, expressing f(M) in units of 2 −8 ,wehaveM 6 7 8 9 10 11+f(M) 8 24 42 56 63 63Give a simple explanation of why f(10) = f(11+).1214

30.16 EXERCISES30.18 A particle is confined to the one-dimensional space 0 ≤ x ≤ a, and classicallyitcanbeinanysmallintervaldx with equal probability. However, quantummechanics gives the result that the probability distribution is proportional tosin 2 (nπx/a), where n is an integer. Find the variance in the particle’s positionin both the classical and quantum-mechanical pictures, and show that, althoughthey differ, the latter tends to the former in the limit of large n, in agreementwith the correspondence principle of physics.30.19 A continuous random variable X has a probability density function f(x); thecorresponding cumulative probability function is F(x). Show that the randomvariable Y = F(X) is uniformly distributed between 0 and 1.30.20 For a non-negative integer random variable X, in addition to the probabilitygenerating function Φ X (t) defined in equation (30.71), it is possible to define theprobability generating function∞∑Ψ X (t) = g n t n ,where g n is the probability that X>n.(a) Prove that Φ X and Ψ X are related byΨ X (t) = 1 − Φ X(t).1 − t(b) Show that E[X] is given by Ψ X (1) and that the variance of X can beexpressed as 2Ψ ′ X (1) + Ψ X(1) − [Ψ X (1)] 2 .(c) For a particular random variable X, the probability that X>nis equal toα n+1 ,with0

PROBABILITY30.23 A point P is chosen at random on the circle x 2 + y 2 = 1. The random variableX denotes the distance of P from (1, 0). Find the mean and variance of X andthe probability that X is greater than its mean.30.24 As assistant to a celebrated and imperious newspaper proprietor, you are giventhe job of running a lottery, in which each of his five million readers will have anequal independent chance, p, of winning a million pounds; you have the job ofchoosing p. However, if nobody wins it will be bad for publicity, whilst if morethan two readers do so, the prize cost will more than offset the profit from extracirculation – in either case you will be sacked! Show that, however you choosep, there is more than a 40% chance you will soon be clearing your desk.30.25 The number of errors needing correction on each page of a set of proofs followsa Poisson distribution of mean µ. The cost of the first correction on any page isα and that of each subsequent correction on the same page is β. Prove that theaverage cost of correcting a page isα + β(µ − 1) − (α − β)e −µ .30.26 In the game of Blackball, at each turn Muggins draws a ball at random from abag containing five white balls, three red balls and two black balls; after beingrecorded, the ball is replaced in the bag. A white ball earns him $1, whilst ared ball gets him $2; in either case, he also has the option of leaving with hiscurrent winnings or of taking a further turn on the same basis. If he draws ablack ball the game ends and he loses all he may have gained previously. Findan expression for Muggins’ expected return if he adopts the strategy of drawingup to n balls, provided he has not been eliminated by then.Show that, as the entry fee to play is $3, Muggins should be dissuaded fromplaying Blackball, but, if that cannot be done, what value of n would you advisehim to adopt?30.27 Show that, for large r, the value at the maximum of the PDF for the gammadistribution of order r with parameter λ is approximately λ/ √ 2π(r − 1).30.28 A husband and wife decide that their family will be complete when it includestwo boys and two girls – but that this would then be enough! The probabilitythat a new baby will be a girl is p. Ignoring the possibility of identical twins,show that the expected size of their family is( )12pq − 1 − pq ,where q =1− p.30.29 The probability distribution for the number of eggs in a clutch is Po(λ), and theprobability that each egg will hatch is p (independently of the size of the clutch).Show by direct calculation that the probability distribution for the number ofchicks that hatch is Po(λp) and so justify the assumptions made in the workedexample at the end of subsection 30.7.1.30.30 A shopper buys 36 items at random in a supermarket, where, because of thesales tax imposed, the final digit (the number of pence) in the price is uniformlyand randomly distributed from 0 to 9. Instead of adding up the bill exactly,she rounds each item to the nearest 10 pence, rounding up or down with equalprobability if the price ends in a ‘5’. Should she suspect a mistake if the cashierasks her for 23 pence more than she estimated?30.31 Under EU legislation on harmonisation, all kippers are to weigh 0.2000 kg, andvendors who sell underweight kippers must be fined by their government. Theweight of a kipper is normally distributed, with a mean of 0.2000 kg and astandard deviation of 0.0100 kg. They are packed in cartons of 100 and largequantities of them are sold.Every day, a carton is to be selected at random from each vendor and tested1216

30.16 EXERCISESaccording to one of the following schemes, which have been approved for thepurpose.(a) The entire carton is weighed, and the vendor is fined 2500 euros if theaverage weight of a kipper is less than 0.1975 kg.(b) Twenty-five kippers are selected at random from the carton; the vendor isfined 100 euros if the average weight of a kipper is less than 0.1980 kg.(c) Kippers are removed one at a time, at random, until one has been foundthat weighs more than 0.2000 kg; the vendor is fined 4n(n − 1) euros, wheren is the number of kippers removed.Which scheme should the Chancellor of the Exchequer be urging his governmentto adopt?30.32 In a certain parliament, the government consists of 75 New Socialites andthe opposition consists of 25 Preservatives. Preservatives never change theirmind, always voting against government policy without a second thought; NewSocialites vote randomly, but with probability p that they will vote for their partyleader’s policies.Following a decision by the New Socialites’ leader to drop certain manifestocommitments, N of his party decide to vote consistently with the opposition. Theleader’s advisors reluctantly admit that an election must be called if N is suchthat, at any vote on government policy, the chance of a simple majority in favourwould be less than 80%. Given that p =0.8, estimate the lowest value of N thatwould precipitate an election.30.33 A practical-class demonstrator sends his twelve students to the storeroom tocollect apparatus for an experiment, but forgets to tell each which type ofcomponent to bring. There are three types, A, B and C, held in the stores (inlarge numbers) in the proportions 20%, 30% and 50%, respectively, and eachstudent picks a component at random. In order to set up one experiment, oneunit each of A and B and two units of C are needed. Let Pr(N) be the probabilitythat at least N experiments can be set up.(a) Evaluate Pr(3).(b) Find an expression for Pr(N) intermsofk 1 and k 2 , the numbers of componentsof types A and B respectively selected by the students. Show that Pr(2)canbewrittenintheform6∑ ∑8−iPr(2) = (0.5) 12 12 C i (0.4) i 12−i C j (0.6) j .i=2(c) By considering the conditions under which no experiments can be set up,show that Pr(1) = 0.9145.30.34 The random variables X and Y take integer values, x and y, both ≥ 1, and suchthat 2x + y ≤ 2a, wherea is an integer greater than 1. The joint probabilitywithin this region is given byPr(X = x, Y = y) =c(2x + y),where c is a constant, and it is zero elsewhere.Show that the marginal probability Pr(X = x) is6(a − x)(2x +2a +1)Pr(X = x) = ,a(a − 1)(8a +5)and obtain expressions for Pr(Y = y), (a) when y is even and (b) when y is odd.Show further thatE[Y ]= 6a2 +4a +1.8a +51217j=2

PROBABILITY[ You will need the results about series involving the natural numbers given insubsection 4.2.5. ]30.35 The continuous random variables X and Y have a joint PDF proportional toxy(x − y) 2 with 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Find the marginal distributionsfor X and Y and show that they are negatively correlated with correlationcoefficient − 2 . 330.36 A discrete random variable X takes integer values n =0, 1,... ,N with probabilitiesp n . A second random variable Y is defined as Y =(X − µ) 2 ,whereµ is theexpectation value of X. Prove that the covariance of X and Y is given byN∑N∑Cov[X,Y ]= n 3 p n − 3µ n 2 p n +2µ 3 .n=0Now suppose that X takes all of its possible values with equal probability, andhence demonstrate that two random variables can be uncorrelated, even thoughone is defined in terms of the other.30.37 Two continuous random variables X and Y have a joint probability distributionf(x, y) =A(x 2 + y 2 ),where A is a constant and 0 ≤ x ≤ a,0≤ y ≤ a. Show that X and Y are negativelycorrelated with correlation coefficient −15/73. By sketching a rough contourmap of f(x, y) and marking off the regions of positive and negative correlation,convince yourself that this (perhaps counter-intuitive) result is plausible.30.38 A continuous random variable X is uniformly distributed over the interval [−c, c].Asampleof2n + 1 values of X is selected at random and the random variableZ is defined as the median of that sample. Show that Z is distributed over [−c, c]with probability density function(2n +1)!f n (z) =(n!) 2 (2c) 2n+1 (c2 − z 2 ) n .Find the variance of Z.30.39 Show that, as the number of trials n becomes large but np i = λ i , i =1, 2,...,k− 1,remains finite, the multinomial probability distribution (30.146),n!M n (x 1 ,x 2 ,...,x k )=x 1 !x 2 ! ···x k ! px 11 px 22 ···px kk ,can be approximated by a multiple Poisson distribution with k − 1 factors:n=0∏k−1M n(x ′ 1 ,x 2 ,...,x k−1 )=i=1e −λ iλ x ii.x i !(Write ∑ k−1ip i = δ and express all terms involving subscript k in terms of n andδ, either exactly or approximately. You will need to use n! ≈ n ɛ [(n − ɛ)!] and(1 − a/n) n ≈ e −a for large n.)(a) Verify that the terms of M n ′ when summed over all values of x 1 ,x 2 ,...,x k−1adduptounity.(b) If k =7andλ i =9foralli =1, 2,...,6, estimate, using the appropriateGaussian approximation, the chance that at least three of x 1 ,x 2 ,...,x 6 willbe 15 or greater.30.40 The variables X i , i =1, 2,...,n, are distributed as a multivariate Gaussian, withmeans µ i and a covariance matrix V. IftheX i are required to satisfy the linear1218

30.17 HINTS AND ANSWERSconstraint ∑ ni=1 c iX i =0,wherethec i are constants (and not all equal to zero),show that the variableχ 2 n =(x − µ) T V −1 (x − µ)follows a chi-squared distribution of order n − 1.30.17 Hints and answers30.1 (a) Yes, (b) no, (c) no, (d) no, (e) yes.30.3 Show that, if p x /16 is the probability that the total will be x, then the corrspondinggain is [p x (x 2 + x) − 16x]/16. (a) A loss of 0.36 euros; (b) a gain of 27/64 euros;(c) a gain of 2.5 euros, provided he takes your advice and guesses ‘5’ each time.30.5 P 1 = α(α + β − αβ) −1 ; P 2 = α(1 − β)(α + β − αβ) −1 ; P 3 = P 2 .30.7 If p r is the probability that before the rth round both players are still in thetournament (and therefore have not met each other), show thatp r+1 = 1 (2 n+1−r − 214 2 n+1−r − 1 p r and hence that p r =2) r−12 n+1−r − 12 n − 1 .(a) The probability that they meet in the final is p n =2 −(n−1) (2 n − 1) −1 .(b) The probability that they meet at some stage in the tournament is given bythe sum ∑ nr=1 p r(2 n+1−r − 1) −1 =2 −(n−1) .30.9 The relative probabilities are X : Y : Z = 50 : 36 : 8 (in units of 10 −4 ); 25/47.30.11 Take A j as the event that a family consists of j boys and n − j girls, and B asthe event that the boy has at least two sisters. Apply Bayes’ theorem.30.13 (i) For a even, the number of ways is 1 + 3 + 5 + ···+(a − 3), and (ii) for a oddit is 2+4+6+···+(a − 3). Combine the results for a =2m and a =2m +1,with m running from 2 to N, to show that the total number of non-degeneratetriangles is given by N(4N +1)(N − 1)/6. The number of possible selections of aset of three rods is (2N + 1)(2N)(2N − 1)/6.30.15 Show that k = e 2 and that the average duration of a call is 1 minute. Let p nbe the probability that the call ends during the interval 0.5(n − 1) ≤ t

PROBABILITY30.23 Mean = 4/π. Variance = 2 − (16/π 2 ). Probability that X exceeds its mean=1− (2/π)sin −1 (2/π) =0.561.30.25 Consider, separately, 0, 1 and ≥ 2errorsonapage.30.27 Show that the maximum occurs at x =(r − 1)/λ, and then use Stirling’s approximationto find the maximum value.30.29 Pr(k chicks hatching) = ∑ ∞n=kPo(n, λ) Bin(n, p).30.31 There is not much to choose between the schemes. In (a) the critical value ofthe standard variable is −2.5 and the average fine would be 15.5 euros. For(b) the corresponding figures are −1.0 and 15.9 euros. Scheme (c) is governedby a geometric distribution with p = q = 1 , and leads to an expected fine2of ∑ ∞n=1 4n(n − 1)( 1 2 )n . The sum can be evaluated by differentiating the result∑ ∞n=1 pn = p/(1 − p) with respect to p, and gives the expected fine as 16 euros.30.33 (a) [12!(0.5) 6 (0.3) 3 (0.2) 3 ]/(6!3!3!)=0.0624.30.35 You will need to establish the normalisation constant for the distribution (36),the common mean value (3/5) and the common standard deviation (3/10). Themarginal distributions are f(x) =3x(6x 2 − 8x + 3), and the same function of y.The covariance has the value −3/50, yielding a correlation of −2/3.30.37 A = 3/(24a 4 ); µ X = µ Y = 5a/8; σX 2 = σ2 Y = 73a2 /960; E[XY ] = 3a 2 /8;Cov[X,Y ]=−a 2 /64.30.39 (b) With the continuity correction Pr(x i ≥ 15) = 0.0334. The probability that atleast three are 15 or greater is 7.5 × 10 −4 .1220

31StatisticsIn this chapter, we turn to the study of statistics, which is concerned withthe analysis of experimental data. In a book of this nature we cannot hopeto do justice to such a large subject; indeed, many would argue that statisticsbelongs to the realm of experimental science rather than in a mathematicstextbook. Nevertheless, physical scientists and engineers are regularly called uponto perform a statistical analysis of their data and to present their results in astatistical context. Therefore, we will concentrate on this aspect of a much moreextensive subject. § 31.1 Experiments, samples and populationsWe may regard the product of any experiment as a set of N measurements of somequantity x or set of quantities x,y,...,z. This set of measurements constitutes thedata. Each measurement (or data item) consists accordingly of a single number x ior a set of numbers (x i ,y i ,...,,z i ), where i =1,...,,N. For the moment, we willassume that each data item is a single number, although our discussion can beextended to the more general case.As a result of inaccuracies in the measurement process, or because of intrinsicvariability in the quantity x being measured, one would expect the N measuredvalues x 1 ,x 2 ,...,x N to be different each time the experiment is performed. We may§ There are, in fact, two separate schools of thought concerning statistics: the frequentist approachand the Bayesian approach. Indeed, which of these approaches is the more fundamental is still amatter of heated debate. Here we shall concentrate primarily on the more traditional frequentistapproach (despite the preference of some of the authors for the Bayesian viewpoint!). For a fullerdiscussion of the frequentist approach one could refer to, for example, A. Stuart and K. Ord,Kendall’s Advanced Theory of Statistics, vol. 1 (London: Edward Arnold, 1994) or J. F. Kenneyand E. S. Keeping, Mathematics of Statistics (New York: Van Nostrand, 1954). For a discussionof the Bayesian approach one might consult, for example, D. S. Sivia, Data Analysis: A BayesianTutorial (Oxford: Oxford University Press, 1996).1221

STATISTICStherefore consider the x i as a set of N random variables. In the most general case,these random variables will be described by some N-dimensional joint probabilitydensity function P (x 1 ,x 2 ,...,x N ). § In other words, an experiment consisting of Nmeasurements is considered as a single random sample from the joint distribution(or population) P (x), where x denotes a point in the N-dimensional data spacehaving coordinates (x 1 ,x 2 ,...,x N ).The situation is simplified considerably if the sample values x i are independent.In this case, the N-dimensional joint distribution P (x) factorises into the productof N one-dimensional distributions,P (x) =P (x 1 )P (x 2 ) ···P (x N ). (31.1)In the general case, each of the one-dimensional distributions P (x i ) may bedifferent. A typical example of this occurs when N independent measurementsare made of some quantity x but the accuracy of the measuring procedure variesbetween measurements.It is often the case, however, that each sample value x i is drawn independentlyfrom the same population. In this case, P (x) is of the form (31.1), but, in addition,P (x i ) has the same form for each value of i. The measurements x 1 ,x 2 ,...,x Nare then said to form a random sample of size N from the one-dimensionalpopulation P (x). This is the most common situation met in practice and, unlessstated otherwise, we will assume from now on that this is the case.31.2 Sample statisticsSuppose we have a set of N measurements x 1 ,x 2 ,...,x N . Any function of thesemeasurements (that contains no unknown parameters) is called a sample statistic,or often simply a statistic. Sample statistics provide a means of characterising thedata. Although the resulting characterisation is inevitably incomplete, it is usefulto be able to describe a set of data in terms of a few pertinent numbers. We nowdiscuss the most commonly used sample statistics.§ In this chapter, we will adopt the common convention that P (x) denotes the particular probabilitydensity function that applies to its argument, x. This obviates the need to use a different letterfor the PDF of each new variable. For example, if X and Y are random variables with differentPDFs, then properly one should denote these distributions by f(x) andg(y), say. In our shorthandnotation, these PDFs are denoted by P (x) andP (y), where it is understood that the functionalform of the PDF may be different in each case.1222

31.2 SAMPLE STATISTICS188.7 204.7 193.2 169.0168.1 189.8 166.3 200.0Table 31.1 Experimental data giving eight measurements of the round triptime in milliseconds for a computer ‘packet’ to travel from Cambridge UK toCambridge MA.31.2.1 AveragesThe simplest number used to characterise a sample is the mean, whichforNvalues x i , i =1, 2,...,N, is defined by¯x = 1 NN∑x i . (31.2)In words, the sample mean is the sum of the sample values divided by the numberof values in the sample.◮Table 31.1 gives eight values for the round trip time in milliseconds for a computer ‘packet’to travel from Cambridge UK to Cambridge MA. Find the sample mean.Using (31.2) the sample mean in milliseconds is given by¯x = 1 (188.7 + 204.7 + 193.2 + 169.0 + 168.1 + 189.8 + 166.3 + 200.0)8= 1479.8 = 184.975.8Since the sample values in table 31.1 are quoted to an accuracy of one decimal place, it isusual to quote the mean to the same accuracy, i.e. as ¯x = 185.0. ◭Strictly speaking the mean given by (31.2) is the arithmetic mean and this is byfar the most common definition used for a mean. Other definitions of the meanare possible, though less common, and include(i) the geometric mean,i=1(ii) the harmonic mean,( N) 1/N∏¯x g = x i , (31.3)¯x h =i=1N∑ Ni=1 1/x , (31.4)i(iii) the root mean square,¯x rms =( ∑Ni=1 x2 iN) 1/2. (31.5)1223

STATISTICSIt should be noted that, ¯x, ¯x h and ¯x rms would remain well defined even if somesample values were negative, but the value of ¯x g could then become complex.The geometric mean should not be used in such cases.◮Calculate ¯x g , ¯x h and ¯x rms for the sample given in table 31.1.The geometric mean is given by (31.3) to be¯x g = (188.7 × 204.7 × ···× 200.0) 1/8 = 184.4.The harmonic mean is given by (31.4) to be8¯x h =(1/188.7) + (1/204.7) + ···+(1/200.0) = 183.9.Finally, the root mean square is given by (31.5) to be¯x rms = [ 18 (188.72 + 204.7 2 + ···+ 200.0 2 ) ] 1/2= 185.5. ◭Two other measures of the ‘average’ of a sample are its mode and median. Themode is simply the most commonly occurring value in the sample. A sample maypossess several modes, however, and thus it can be misleading in such cases touse the mode as a measure of the average of the sample. The median of a sampleis the halfway point when the sample values x i (i =1, 2,...,N) are arranged inascending (or descending) order. Clearly, this depends on whether the size ofthe sample, N, is odd or even. If N is odd then the median is simply equal tox (N+1)/2 , whereas if N is even the median of the sample is usually taken to be12 (x N/2 + x (N/2)+1 ).◮Find the mode and median of the sample given in table 31.1.From the table we see that each sample value occurs exactly once, and so any value maybe called the mode of the sample.To find the sample median, we first arrange the sample values in ascending order andobtain166.3, 168.1, 169.0, 188.7, 189.8, 193.2, 200.0, 204.7.Since the number of sample values N = 8, which is even, the median of the sample is1(x 2 4 + x 5 )= 1 (188.7 + 189.8) = 189.25. ◭231.2.2 Variance and standard deviationThe variance and standard deviation both give a measure of the spread of valuesin a sample about the sample mean ¯x. Thesample variance is defined bys 2 = 1 NN∑(x i − ¯x) 2 , (31.6)i=11224

31.2 SAMPLE STATISTICSand the sample standard deviation is the positive square root of the samplevariance, i.e.s = √ 1 N∑(x i − ¯x)N2 . (31.7)i=1◮Find the sample variance and sample standard deviation of the data given in table 31.1.We have already found that the sample mean is 185.0 to one decimal place. However,when the mean is to be used in the subsequent calculation of the sample variance it isbetter to use the most accurate value available. In this case the exact value is 184.975, andso using (31.6),s 2 = 1 [(188.7 − 184.975) 2 + ···+ (200.0 − 184.975) 2]8= 1608.368= 201.0,where once again we have quoted the result to one decimal place. The sample standarddeviation is then given by s = √ 201.0 =14.2. As it happens, in this case the differencebetween the true mean and the rounded value is very small compared with the variationof the individual readings about the mean and using the rounded value has a negligibleeffect; however, this would not be so if the difference were comparable to the samplestandard deviation. ◭Using the definition (31.7), it is clear that in order to calculate the standarddeviation of a sample we must first calculate the sample mean. This requirementcan be avoided, however, by using an alternative form for s 2 . From (31.6), we seethats 2 = 1 N= 1 NN∑(x i − ¯x) 2i=1N∑x 2 i − 1 Ni=1N∑2x i¯x + 1 Ni=1= x 2 − 2¯x 2 + ¯x 2 = x 2 − ¯x 2N∑¯x 2i=1We may therefore write the sample variance s 2 as( ) 2s 2 = x 2 − ¯x 2 = 1 N∑x 2 1N∑i − x i , (31.8)NNi=1from which the sample standard deviation is found by taking the positive squareroot. Thus, by evaluating the quantities ∑ Ni=1 x i and ∑ Ni=1 x2 i for our sample, wecan calculate the sample mean and sample standard deviation at the same time.1225i=1

STATISTICS◮Calculate ∑ Ni=1 x i and ∑ Ni=1 x2 i for the data given in table 31.1 and hence find the meanand standard deviation of the sample.From table 31.1, we obtainN∑x i = 188.7 + 204.7+···+ 200.0 = 1479.8,i=1N∑x 2 i = (188.7) 2 + (204.7) 2 + ···+ (200.0) 2 = 275 334.36.i=1Since N = 8, we find as before (quoting the final results to one decimal place)¯x = 1479.88√( ) 2275 334.36 1479.8= 185.0, s =−=14.2. ◭8831.2.3 Moments and central momentsBy analogy with our discussion of probability distributions in section 30.5, thesample mean and variance may also be described respectively as the first momentand second central moment of the sample. In general, for a sample x i , i =1, 2,...,N, we define the rth moment m r and rth central moment n r asm r = 1 Nn r = 1 NN∑x r i , (31.9)i=1N∑(x i − m 1 ) r . (31.10)i=1Thus the sample mean ¯x and variance s 2 may also be written as m 1 and n 2respectively. As is common practice, we have introduced a notation in whicha sample statistic is denoted by the Roman letter corresponding to whicheverGreek letter is used to describe the corresponding population statistic. Thus, weuse m r and n r to denote the rth moment and central moment of a sample, sincein section 30.5 we denoted the rth moment and central moment of a populationby µ r and ν r respectively.This notation is particularly useful, since the rth central moment of a sample,m r , may be expressed in terms of the rth- and lower-order sample moments n r in away exactly analogous to that derived in subsection 30.5.5 for the correspondingpopulation statistics. As discussed in the previous section, the sample variance isgiven by s 2 = x 2 − ¯x 2 but this may also be written as n 2 = m 2 − m 2 1 ,whichistobecompared with the corresponding relation ν 2 = µ 2 −µ 2 1 derived in subsection 30.5.3for population statistics. This correspondence also holds for higher-order central1226

31.2 SAMPLE STATISTICSmoments of the sample. For example,n 3 = 1 N= 1 NN∑(x i − m 1 ) 3i=1N∑(x 3 i − 3m 1 x 2 i +3m 2 1x i − m 3 1)i=1= m 3 − 3m 1 m 2 +3m 2 1m 1 − m 3 1= m 3 − 3m 1 m 2 +2m 3 1, (31.11)which may be compared with equation (30.53) in the previous chapter.Mirroring our discussion of the normalised central moments γ r of a populationin subsection 30.5.5, we can also describe a sample in terms of the dimensionlessquantitiesg k = n k= n kn k/2 s k ;2g 3 and g 4 are called the sample skewness and kurtosis. Likewise, it is common todefine the excess kurtosis of a sample by g 4 − 3.31.2.4 Covariance and correlationSo far we have assumed that each data item of the sample consists of a singlenumber. Now let us suppose that each item of data consists of a pair of numbers,so that the sample is given by (x i ,y i ), i =1, 2,...,N.We may calculate the sample means, ¯x and ȳ, and sample variances, s 2 x ands 2 y,ofthex i and y i values individually but these statistics do not provide anymeasure of the relationship between the x i and y i . By analogy with our discussionin subsection 30.12.3 we measure any interdependence between the x i and y i interms of the sample covariance, which is given byV xy = 1 N∑(x i − ¯x)(y i − ȳ)Ni=1= (x − ¯x)(y − ȳ)= xy − ¯xȳ. (31.12)Writing out the last expression in full, we obtain the form most useful forcalculations, which reads(V xy = 1 N) (∑x i y i − 1 N)(∑ N)∑NN 2 x i y i .i=11227i=1i=1

STATISTICSr xy =0.0 r xy =0.1 r xy =0.5yxr xy = −0.7 r xy = −0.9 r xy =0.99Figure 31.1 Scatter plots for two-dimensional data samples of size N = 1000,with various values of the correlation r. No scales are plotted, since the valueof r is unaffected by shifts of origin or changes of scale in x and y.We may also define the closely related sample correlation byr xy = V xy,s x s ywhich can take values between −1 and +1. If the x i and y i are independent thenV xy =0=r xy , and from (31.12) we see that xy = ¯xȳ. It should also be notedthat the value of r xy is not altered by shifts in the origin or by changes in thescale of the x i or y i . In other words, if x ′ = ax + b and y ′ = cy + d, wherea,b, c, d are constants, then r x′ y ′ = r xy. Figure 31.1 shows scatter plots for severaltwo-dimensional random samples x i ,y i of size N = 1000, each with a differentvalue of r xy .◮Ten UK citizens are selected at random and their heights and weights are found to be asfollows (to the nearest cm or kg respectively):Person A B C D E F G H I JHeight (cm) 194 168 177 180 171 190 151 169 175 182Weight (kg) 75 53 72 80 75 75 57 67 46 68Calculate the sample correlation between the heights and weights.In order to find the sample correlation, we begin by calculating the following sums (wherex i are the heights and y i are the weights)∑∑x i = 1757, y i = 668,i1228i

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS∑x 2 i = 310 041,i∑yi 2 = 45 746,i∑x i y i = 118 029.ThesampleconsistsofN = 10 pairs of numbers, so the means of the x i and of the y i aregiven by ¯x = 175.7 andȳ =66.8. Also, xy = 11 802.9. Similarly, the standard deviationsof the x i and y i are calculated, using (31.8), as√( ) 2310 041 1757s x = − =11.6,s y =10√45 746−1010( ) 2 668=10.6.10Thus the sample correlation is given byxy − ¯xȳ 11 802.9 − (175.7)(66.8)r xy = = =0.54.s x s y (11.6)(10.6)Thus there is a moderate positive correlation between the heights and weights of thepeople measured. ◭It is straightforward to generalise the above discussion to data samples ofarbitrary dimension, the only complication being one of notation. We chooseto denote the i th data item from an n-dimensional sample as (x (1)i,x (2)i,...,x (n)i),where the bracketted superscript runs from 1 to n and labels the elements withina given data item whereas the subscript i runs from 1 to N and labels the dataitems within the sample. In this n-dimensional case, we can define the samplecovariance matrix whose elements areV kl = x (k) x (l) − x (k) x (l)and the sample correlation matrix with elementsr kl = V kl.s k s lBoth these matrices are clearly symmetric but are not necessarily positive definite.i31.3 Estimators and sampling distributionsIn general, the population P (x) from which a sample x 1 ,x 2 ,...,x N is drawnis unknown. Thecentral aim of statistics is to use the sample values x i to infercertain properties of the unknown population P (x), such as its mean, variance andhigher moments. To keep our discussion in general terms, let us denote the variousparameters of the population by a 1 ,a 2 ,..., or collectively by a. Moreover, we makethe dependence of the population on the values of these quantities explicit bywriting the population as P (x|a). For the moment, we are assuming that thesample values x i are independent and drawn from the same (one-dimensional)population P (x|a), in which caseP (x|a) =P (x 1 |a)P (x 2 |a) ···P (x N |a).1229

STATISTICSSuppose, we wish to estimate the value of one of the quantities a 1 ,a 2 ,...,whichwe will denote simply by a. Since the sample values x i provide our only source ofinformation, any estimate of a must be some function of the x i , i.e. some samplestatistic. Such a statistic is called an estimator of a and is usually denoted by â(x),where x denotes the sample elements x 1 ,x 2 ,...,x N .Since an estimator â is a function of the sample values of the random variablesx 1 ,x 2 ,...,x N , it too must be a random variable. In other words, if a number ofrandom samples, each of the same size N, are taken from the (one-dimensional)population P (x|a) then the value of the estimator â will vary from one sample tothe next and in general will not be equal to the true value a. This variation in theestimator is described by its sampling distribution P (â|a). From section 30.14, thisis given byP (â|a) dâ = P (x|a) d N x,where d N x is the infinitesimal ‘volume’ in x-space lying between the ‘surfaces’â(x) =â and â(x) =â + dâ. The form of the sampling distribution generallydepends upon the estimator under consideration and upon the form of thepopulation from which the sample was drawn, including, as indicated, the truevalues of the quantities a. It is also usually dependent on the sample size N.◮The sample values x 1 ,x 2 ,...,x N are drawn independently from a Gaussian distributionwith mean µ and variance σ. Suppose that we choose the sample mean ¯x as our estimatorˆµ of the population mean. Find the sampling distributions of this estimator.The sample mean ¯x is given by¯x = 1 N (x 1 + x 2 + ···+ x N ),where the x i are independent random variables distributed as x i ∼ N(µ, σ 2 ). From ourdiscussion of multiple Gaussian distributions on page 1189, we see immediately that ¯x willalso be Gaussian distributed as N(µ, σ 2 /N). In other words, the sampling distribution of¯x is given byP (¯x|µ, σ) =[1√2πσ2 /N exp −Note that the variance of this distribution is σ 2 /N. ◭](¯x − µ)2. (31.13)2σ 2 /N31.3.1 Consistency, bias and efficiency of estimatorsFor any particular quantity a, we may in fact define any number of differentestimators, each of which will have its own sampling distribution. The qualityof a given estimator â may be assessed by investigating certain properties of itssampling distribution P (â|a). In particular, an estimator â is usually judged onthe three criteria of consistency, bias and efficiency, each of which we now discuss.1230

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONSConsistencyAn estimator â is consistent if its value tends to the true value a in the large-samplelimit, i.e.lim â = a.N→∞Consistency is usually a minimum requirement for a useful estimator. An equivalentstatement of consistency is that in the limit of large N the samplingdistribution P (â|a) of the estimator must satisfylim P (â|a) → δ(â − a).N→∞BiasThe expectation value of an estimator â is given by∫∫E[â] = âP (â|a) dâ = â(x)P (x|a) d N x, (31.14)where the second integral extends over all possible values that can be taken bythe sample elements x 1 ,x 2 ,...,x N . This expression gives the expected mean valueof â from an infinite number of samples, each of size N. Thebias of an estimatorâ is then defined asb(a) =E[â] − a. (31.15)We note that the bias b does not depend on the measured sample valuesx 1 ,x 2 ,...,x N . In general, though, it will depend on the sample size N, the functionalform of the estimator â and, as indicated, on the true properties a ofthe population, including the true value of a itself. If b =0thenâ is called anunbiased estimator of a.◮An estimator â is biased in such a way that E[â] =a + b(a), wherethebiasb(a) is givenby (b 1 − 1)a + b 2 and b 1 and b 2 are known constants. Construct an unbiased estimator of a.Let us first write E[â] is the clearer formE[â] =a +(b 1 − 1)a + b 2 = b 1 a + b 2 .The task of constructing an unbiased estimator is now trivial, and an appropriate choiceis â ′ =(â − b 2 )/b 1 , which (as required) has the expectation valueE[â ′ ]= E[â] − b 2b 1= a. ◭EfficiencyThe variance of an estimator is given by∫∫V [â] = (â − E[â]) 2 P (â|a) dâ = (â(x) − E[â]) 2 P (x|a) d N x(31.16)1231

STATISTICSand describes the spread of values â about E[â] that would result from a largenumber of samples, each of size N. An estimator with a smaller variance is saidto be more efficient than one with a larger variance. As we show in the nextsection, for any given quantity a of the population there exists a theoretical lowerlimit on the variance of any estimator â. This result is known as Fisher’s inequality(or the Cramér–Rao inequality) and readsV [â] ≥(1+ ∂b∂a) 2 /E[ ]− ∂2 ln P∂a 2 , (31.17)where P stands for the population P (x|a) andb is the bias of the estimator.Denoting the quantity on the RHS of (31.17) by V min ,theefficiency e of anestimator is defined ase = V min /V [â].An estimator for which e = 1 is called a minimum-variance or efficient estimator.Otherwise, if e

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONSand, on differentiating twice with respect to µ, we find∂ 2 ln P= − N ∂µ 2 σ . 2This is independent of the x i and so its expectation value is also equal to −N/σ 2 .Withbset equal to zero in (31.17), Fisher’s inequality thus states that, for any unbiased estimatorˆµ of the population mean,V [ˆµ] ≥ σ2N .Since V [¯x] =σ 2 /N, the sample mean ¯x is a minimum-variance estimator of µ. ◭31.3.2 Fisher’s inequalityAs mentioned above, Fisher’s inequality provides a lower limit on the variance ofany estimator â of the quantity a; itreads(V [â] ≥ 1+ ∂b ) 2 / [ ]E − ∂2 ln P∂a∂a 2 , (31.18)where P stands for the population P (x|a) andb is the bias of the estimator.We now present a proof of this inequality. Since the derivation is somewhatcomplicated, and many of the details are unimportant, this section can be omittedon a first reading. Nevertheless, some aspects of the proof will be useful whenthe efficiency of maximum-likelihood estimators is discussed in section 31.5.◮Prove Fisher’s inequality (31.18).The normalisation of P (x|a) isgivenby∫P (x|a) d N x =1, (31.19)where d N x = dx 1 dx 2 ···dx N and the integral extends over all the allowed values of thesample items x i . Differentiating (31.19) with respect to the parameter a, weobtain∫∫∂P ∂ ln P∂a dN x =∂a PdN x =0. (31.20)We note that the second integral is simply the expectation value of ∂ ln P/∂a,wheretheaverage is taken over all possible samples x i , i =1, 2,...,N.Further,byequatingthetwoexpressions for ∂E[â]/∂a obtained by differentiating (31.15) and (31.14) with respect to awe obtain, dropping the functional dependencies, a second relationship,1+ ∂b ∫∂a = â ∂P ∫∂a dN x =â ∂ ln P∂a PdN x. (31.21)Now, multiplying (31.20) by α(a), where α(a) isany function of a, and subtracting theresult from (31.21), we obtain∫[â − α(a)] ∂ ln P∂aPdN x =1+ ∂b∂a .At this point we must invoke the Schwarz inequality proved in subsection 8.1.3. The proof1233

STATISTICSis trivially extended to multiple integrals and shows that for two real functions, g(x) andh(x),(∫ )(∫ ) (∫2g 2 (x) d N x h 2 (x) d N x ≥ g(x)h(x) d x) N . (31.22)If we now let g =[â − α(a)] √ P and h =(∂ ln P/∂a) √ P , we find{∫} [ ∫ ( ) 2 (∂ ln P[â − α(a)] 2 Pd N xPd x]N ≥ 1+ ∂b ) 2.∂a∂aOn the LHS, the factor in braces represents the expected spread of â-values around thepoint α(a). The minimum value that this integral may take occurs when α(a) =E[â].Making this substitution, we recognise the integral as the variance V [â],andsoobtaintheresult(V [â] ≥ 1+ ∂b ) [ 2 ∫ ( ) ] 2 −1∂ ln PPd N x . (31.23)∂a ∂aWe note that the factor in brackets is the expectation value of (∂ ln P/∂a) 2 .Fisher’s inequality is, in fact, often quoted in the form (31.23). We may recover the form(31.18) by noting that on differentiating (31.20) with respect to a we obtain∫ ( ∂ 2 ln PP + ∂ ln P )∂Pd N x =0.∂a 2 ∂a ∂aWriting ∂P/∂a as (∂ ln P/∂a)P and rearranging we find that∫ ( ) 2 ∫∂ ln P∂Pd N 2 ln Px = − Pd N x.∂a∂a 2Substituting this result in (31.23) gives(V [â] ≥− 1+ ∂b ) 2 [∫ ∂ 2 ln P∂a ∂a 2] −1Pd N x .Since the factor in brackets is the expectation value of ∂ 2 ln P/∂a 2 , we have recoveredresult (31.18). ◭31.3.3 Standard errors on estimatorsFor a given sample x 1 ,x 2 ,...,x N , we may calculate the value of an estimator â(x)for the quantity a. It is also necessary, however, to give some measure of thestatistical uncertainty in this estimate. One way of characterising this uncertaintyis with the standard deviation of the sampling distribution P (â|a), which is givensimply byσâ =(V [â]) 1/2 . (31.24)If the estimator â(x) were calculated for a large number of samples, each of sizeN, then the standard deviation of the resulting â values would be given by (31.24).Consequently, σâ is called the standard error on our estimate.In general, however, the standard error σâ depends on the true values of some1234

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONSor all of the quantities a and they may be unknown. When this occurs, one mustsubstitute estimated values of any unknown quantities into the expression for σâin order to obtain an estimated standard error ˆσâ. One then quotes the result asa = â ± ˆσâ.◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with standard deviation σ =1. The sample values are as follows (to two decimalplaces):2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Estimate the population mean µ, quoting the standard error on your result.We have shown in the final worked example of subsection 31.3.1 that, in this case, ¯x isa consistent, unbiased, minimum-variance estimator of µ and has variance V [¯x] =σ 2 /N.Thus, our estimate of the population mean with its associated standard error isˆµ = ¯x ± √ σ =1.11 ± 0.32.NIf the true value of σ had not been known, we would have needed to use an estimatedvalue ˆσ in the expression for the standard error. Useful basic estimators of σ are discussedin subsection 31.4.2. ◭It should be noted that the above approach is most meaningful for unbiasedestimators. In this case, E[â] =a and so σâ describes the spread of â-values aboutthe true value a. For a biased estimator, however, the spread about the true valuea is given by the root mean square error ɛâ, which is defined byɛ 2 â = E[(â − a)2 ]= E[(â − E[â]) 2 ]+(E[â] − a) 2= V [â]+b(a) 2 .We see that ɛ 2 âis the sum of the variance of â and the square of the bias and socan be interpreted as the sum of squares of statistical and systematic errors. Fora biased estimator, it is often more appropriate to quote the result asa = â ± ɛâ.As above, it may be necessary to use estimated values â in the expression for theroot mean square error and thus to quote only an estimate ˆɛâ of the error.31.3.4 Confidence limits on estimatorsAn alternative (and often equivalent) way of quoting a statistical error is with aconfidence interval. Let us assume that, other than the quantity of interest a, thequantities a have known fixed values. Thus we denote the sampling distribution1235

STATISTICSP (â|a)αβâ α (a)â β (a)âFigure 31.2 The sampling distribution P (â|a) ofsomeestimatorâ for a givenvalue of a. The shaded regions indicate the two probabilities Pr(â â β (a)) = β.of â by P (â|a). For any particular value of a, one can determine the two valuesâ α (a) andâ β (a) such thatPr(â â β (a)) =∫ âα(a)−∞∫ ∞â β(a)P (â|a) dâ = α, (31.25)P (â|a) dâ = β. (31.26)This is illustrated in figure 31.2. Thus, for any particular value of a, the probabilitythat the estimator â lies within the limits â α (a) andâ β (a) is given byPr(â α (a) < â

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONSP (â|a − ) P (â|a + )αâ obsβâFigure 31.3 An illustration of how the observed value of the estimator, â obs ,and the given values α and β determine the two confidence limits a − and a + ,which are such that â α (a + )=â obs = â β (a − ).each of size N, were analysed then the interval [a − ,a + ] would contain the truevalue a on a fraction 1 − α − β of the occasions.The interval [a − ,a + ] is called a confidence interval on a at the confidencelevel 1 − α − β. The values a − and a + themselves are called respectively thelower confidence limit and the upper confidence limit at this confidence level. Inpractice, the confidence level is often quoted as a percentage. A convenient wayof presenting our results is∫ âobs−∞∫ ∞P (â|a + ) dâ = α, (31.28)P (â|a − ) dâ = β. (31.29)â obsThe confidence limits may then be found by solving these equations for a − anda + either analytically or numerically. The situation is illustrated graphically infigure 31.3.Occasionally one might not combine the results (31.28) and (31.29) but useeither one or the other to provide a one-sided confidence interval on a. Wheneverthe results are combined to provide a two-sided confidence interval, however, theinterval is not specified uniquely by the confidence level 1 − α − β. In other words,there are generally an infinite number of intervals [a − ,a + ] for which (31.27) holds.To specify a unique interval, one often chooses α = β, resulting in the centralconfidence interval on a. All cases can be covered by calculating the quantitiesc = â − a − and d = a + − â and quoting the result of an estimate asa = â +d−c.So far we have assumed that the quantities a other than the quantity of interesta are known in advance. If this is not the case then the construction of confidencelimits is considerably more complicated. This is discussed in subsection 31.3.6.1237

STATISTICS31.3.5 Confidence limits for a Gaussian sampling distributionAn important special case occurs when the sampling distribution is Gaussian; ifthe mean is a and the standard deviation is σâ then]1 (â − a)2P (â|a, σâ) = √ exp[−2πσâ2 2σâ2 . (31.30)For almost any (consistent) estimator â, the sampling distribution will tend tothis form in the large-sample limit N →∞, as a consequence of the central limittheorem. For a sampling distribution of the form (31.30), the above procedurefor determining confidence intervals becomes straightforward. Suppose, from oursample, we obtain the value â obs for our estimator. In this case, equations (31.28)and (31.29) become)(âobs − a +Φ= α,σâ)(âobs − a −1 − Φ= β,σâwhere Φ(z) is the cumulative probability function for the standard Gaussian distribution,discussed in subsection 30.9.1. Solving these equations for a − and a + givesa − = â obs − σâΦ −1 (1 − β), (31.31)a + = â obs + σâΦ −1 (1 − α); (31.32)we have used the fact that Φ −1 (α) =−Φ −1 (1−α) to make the equations symmetric.The value of the inverse function Φ −1 (z) can be read off directly from table 30.3,given in subsection 30.9.1. For the normally used central confidence interval onehas α = β. In this case, we see that quoting a result using the standard error, asa = â ± σâ, (31.33)is equivalent to taking Φ −1 (1 − α) = 1. From table 30.3, we find α =1− 0.8413 =0.1587, and so this corresponds to a confidence level of 1 − 2(0.1587) ≈ 0.683.Thus, the standard error limits give the 68.3% central confidence interval.◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with standard deviation σ =1. The sample values are as follows (to two decimalplaces):2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Find the 90% central confidence interval on the population mean µ.Our estimator ˆµ is the sample mean ¯x. As shown towards the end of section 31.3, thesampling distribution of ¯x is Gaussian with mean E[¯x] and variance V [¯x] =σ 2 /N. Sinceσ = 1 in this case, the standard error is given by σˆx = σ/ √ N =0.32. Moreover, insubsection 31.3.3, we found the mean of the above sample to be ¯x =1.11.1238

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONSFor the 90% central confidence interval, we require α = β =0.05. From table 30.3, wefindΦ −1 (1 − α) =Φ −1 (0.95) = 1.65,and using (31.31) and (31.32) we obtaina − = ¯x − 1.65σ¯x =1.11 − (1.65)(0.32) = 0.58,a + = ¯x +1.65σ¯x =1.11 + (1.65)(0.32) = 1.64.Thus, the 90% central confidence interval on µ is [0.58, 1.64]. For comparison, the truevalue used to create the sample was µ =1.◭In the case where the standard error σâ in (31.33) is not known in advance,one must use a value ˆσâ estimated from the sample. In principle, this complicatessomewhat the construction of confidence intervals, since properly one shouldconsider the two-dimensional joint sampling distribution P (â, ˆσâ|a). Nevertheless,in practice, provided ˆσâ is a fairly good estimate of σâ the above procedure maybe applied with reasonable accuracy. In the special case where the sample valuesx i are drawn from a Gaussian distribution with unknown µ and σ, itisinfactpossible to obtain exact confidence intervals on the mean µ, for a sample of anysize N, using Student’s t-distribution. This is discussed in subsection 31.7.5.31.3.6 Estimation of several quantities simultaneouslySuppose one uses a sample x 1 ,x 2 ,...,x N to calculate the values of several estimatorsâ 1 , â 2 ,...,â M (collectively denoted by â) of the quantities a 1 ,a 2 ,...,a M(collectively denoted by a) that describe the population from which the sample wasdrawn. The joint sampling distribution of these estimators is an M-dimensionalPDF P (â|a) given byP (â|a) d M â = P (x|a) d N x.◮Sample values x 1 ,x 2 ,...,x N are drawn independently from a Gaussian distribution withmean µ and standard deviation σ. Suppose we choose the sample mean ¯x and sample standarddeviation s respectively as estimators ˆµ and ˆσ. Find the joint sampling distribution ofthese estimators.Since each data value x i in the sample is assumed to be independent of the others, thejoint probability distribution of sample values is given by[ ∑iP (x|µ, σ) =(2πσ 2 ) −N/2 exp −(x ]i − µ) 2.2σ 2We may rewrite the sum in the exponent as follows:∑(x i − µ) 2 = ∑ (x i − ¯x + ¯x − µ) 2ii= ∑ i(x i − ¯x) 2 +2(¯x − µ) ∑ i= Ns 2 + N(¯x − µ) 2 ,1239(x i − ¯x)+ ∑ i(¯x − µ) 2

STATISTICSwhere in the last line we have used the fact that ∑ i (x i − ¯x) = 0. Hence, for given valuesof µ and σ, the sampling distribution is in fact a function only of the sample mean ¯x andthe standard deviation s. Thus the sampling distribution of ¯x and s must satisfy{P (¯x, s|µ, σ) d¯xds=(2πσ 2 ) −N/2 exp − N[(¯x − }µ)2 + s 2 ]dV , (31.34)2σ 2where dV = dx 1 dx 2 ···dx N is an element of volume in the sample space which yieldssimultaneously values of ¯x and s that lie within the region bounded by [¯x, ¯x + d¯x] and[s, s + ds]. Thus our only remaining task is to express dV in terms of ¯x and s and theirdifferentials.Let S be the point in sample space representing the sample (x 1 ,x 2 ,...,x N ). For givenvalues of ¯x and s, we require the sample values to satisfy both the condition∑x i = N¯x,iwhich defines an (N − 1)-dimensional hyperplane in the sample space, and the condition∑(x i − ¯x) 2 = Ns 2 ,iwhich defines an (N − 1)-dimensional hypersphere. Thus S is constrained to lie in theintersection of these two hypersurfaces, which is itself an (N −2)-dimensional hypersphere.Now, the volume of an (N − 2)-dimensional hypersphere is proportional to s N−1 . It follows√that the volume √ dV between two concentric (N − 2)-dimensional hyperspheres of radiusNs and N(s + ds) andtwo(N − 1)-dimensional hyperplanes corresponding to ¯x and¯x + d¯x isdV = As N−2 ds d¯x,where A is some constant. Thus, substituting this expression for dV into (31.34), we find])N(¯x − µ)2P (¯x, s|µ, σ) =C 1 exp[− C2σ 2 2 s N−2 exp(− Ns2 = P (¯x|µ, σ)P (s|σ),2σ 2 (31.35)where C 1 and C 2 are constants. We have written P (¯x, s|µ, σ) in this form to show that itseparates naturally into two parts, one depending only on ¯x and the other only on s. Thus,¯x and s are independent variables. Separate normalisations of the two factors in (31.35)require( ) 1/2 ( ) (N−1)/2 NN1C 1 =and C2πσ 2 2 =22σ 2 Γ ( ,1(N − 1))2where the calculation of C 2 requires the use of the gamma function, discussed in theAppendix. ◭The marginal sampling distribution of any one of the estimators â i is givensimply by∫ ∫P (â i |a) = ··· P (â|a) dâ 1 ···dâ i−1 dâ i+1 ···dâ M ,and the expectation value E[â i ] and variance V [â i ]ofâ i are again given by (31.14)and (31.16) respectively. By analogy with the one-dimensional case, the standarderror σâi on the estimator â i is given by the positive square root of V [â i ]. With1240

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONSseveral estimators, however, it is usual to quote their full covariance matrix. ThisM × M matrix has elements∫V ij =Cov[â i , â j ]= (â i − E[â i ])(â j − E[â j ])P (â|a) d M â∫= (â i − E[â i ])(â j − E[â j ])P (x|a) d N x.Fisher’s inequality can be generalised to the multi-dimensional case. Adaptingthe proof given in subsection 31.3.2, one may show that, in the case where theestimators are efficient and have zero bias, the elements of the inverse of thecovariance matrix are given by[ ](V −1 ) ij = E − ∂2 ln P, (31.36)∂a i ∂a jwhere P denotes the population P (x|a) from which the sample is drawn. Thequantity on the RHS of (31.36) is the element F ij of the so-called Fisher matrixF of the estimators.◮Calculate the covariance matrix of the estimators ¯x and s in the previous example.As shown in (31.35), the joint sampling distribution P (¯x, s|µ, σ) factorises, and so theestimators ¯x and s are independent. Thus, we conclude immediately thatCov[¯x, s] =0.Since we have already shown in the worked example at the end of subsection 31.3.1 thatV [¯x] =σ 2 /N, it only remains to calculate V [s]. From (31.35), we find∫ ∞) ( ) r/2E[s r ]=C 2 s N−2+r exp(− Ns2 2 Γ ( 1(N − 1+r))2ds =02σ 2 N Γ ( 1(N − 1)) σr ,2where we have evaluated the integral using the definition of the gamma function given inthe Appendix. Thus, the expectation value of the sample standard deviation is( ) 1/2(2 Γ1E[s] =N) 2N Γ ( σ, (31.37)1(N − 1))2and its variance is given by⎧ [⎨⎩ N − 1 − 2 Γ ( 1N) 2V [s] =E[s 2 ] − (E[s]) 2 = σ2NΓ ( 12 (N − 1)) ] 2⎫⎬⎭We note, in passing, that (31.37) shows that s is a biased estimator of σ. ◭The idea of a confidence interval can also be extended to the case where severalquantities are estimated simultaneously but then the practical construction of aninterval is considerably more complicated. The general approach is to constructan M-dimensional confidence region R in a-space. By analogy with the onedimensionalcase, for a given confidence level of (say) 1 − α, one first constructs1241

STATISTICSa region ˆR in â-space, such that∫∫P (â|a) d M â =1− α.ˆRA common choice for such a region is that bounded by the ‘surface’ P (â|a) =constant. By considering all possible values a and the values of â lying withinthe region ˆR, one can construct a 2M-dimensional region in the combined space(â, a). Suppose now that, from our sample x, the values of the estimators areâ i,obs , i =1, 2,...,M. The intersection of the M ‘hyperplanes’ â i = â i,obs withthe 2M-dimensional region will determine an M-dimensional region which, whenprojected onto a-space, will determine a confidence limit R at the confidencelevel 1 − α. It is usually the case that this confidence region has to be evaluatednumerically.The above procedure is clearly rather complicated in general and a simplerapproximate method that uses the likelihood function is discussed in subsection31.5.5. As a consequence of the central limit theorem, however, in thelarge-sample limit, N →∞, the joint sampling distribution P (â|a) will tend, ingeneral, towards the multivariate GaussianP (â|a) =1(2π) M/2 |V| 1/2 exp [ − 1 2 Q(â, a)] , (31.38)where V is the covariance matrix of the estimators and the quadratic form Q isgiven byQ(â, a) =(â − a) T V −1 (â − a).Moreover, in the limit of large N, the inverse covariance matrix tends to theFisher matrix F given in (31.36), i.e. V −1 → F.For the Gaussian sampling distribution (31.38), the process of obtaining confidenceintervals is greatly simplified. The surfaces of constant P (â|a) correspondto surfaces of constant Q(â, a), which have the shape of M-dimensional ellipsoidsin â-space, centred on the true values a. In particular, let us suppose that theellipsoid Q(â, a) =c (where c is some constant) contains a fraction 1 − α of thetotal probability. Now suppose that, from our sample x, we obtain the values â obsfor our estimators. Because of the obvious symmetry of the quadratic form Qwith respect to a and â, it is clear that the ellipsoid Q(a, â obs )=c in a-space thatis centred on â obs should contain the true values a with probability 1 − α. ThusQ(a, â obs )=c defines our required confidence region R at this confidence level.This is illustrated in figure 31.4 for the two-dimensional case.It remains only to determine the constant c corresponding to the confidencelevel 1 − α. As discussed in subsection 30.15.2, the quantity Q(â, a) is distributedas a χ 2 variable of order M. Thus, the confidence region corresponding to the1242

31.4 SOME BASIC ESTIMATORSâ 2a 2(a)(b)a truea trueâ obsâ obsâ 1a 1Figure 31.4 (a) The ellipse Q(â, a) =c in â-space. (b) The ellipse Q(a, â obs )=cin a-space that corresponds to a confidence region R at the level 1 − α, whenc satisfies (31.39).confidence level 1 − α is given by Q(a, â obs )=c, wheretheconstantc satisfies∫ c0P (χ 2 M) d(χ 2 M)=1− α, (31.39)and P (χ 2 M ) is the chi-squared PDF of order M, discussed in subsection 30.9.4. Thisintegral may be evaluated numerically to determine the constant c. Alternatively,some reference books tabulate the values of c corresponding to given confidencelevels and various values of M. A representative selection of values of c is givenin table 31.2; there the number of degrees of freedom is denoted by the moreusual n, rather than M.31.4 Some basic estimatorsIn many cases, one does not know the functional form of the population fromwhich a sample is drawn. Nevertheless, in a case where the sample valuesx 1 ,x 2 ,...,x N are each drawn independently from a one-dimensional populationP (x), it is possible to construct some basic estimators for the moments and centralmoments of P (x). In this section, we investigate the estimating properties of thecommon sample statistics presented in section 31.2. In fact, expectation valuesand variances of these sample statistics can be calculated without prior knowledgeof the functional form of the population; they depend only on the sample size Nand certain moments and central moments of P (x).31.4.1 Population mean µLet us suppose that the parent population P (x) has mean µ and variance σ 2 .Anobvious estimator ˆµ of the population mean is the sample mean ¯x. Provided µand σ 2 are both finite, we may apply the central limit theorem directly to obtain1243

STATISTICS% 99 95 10 5 2.5 1 0.5 0.1n =1 1.57 10 −4 3.93 10 −3 2.71 3.84 5.02 6.63 7.88 10.832 2.01 10 −2 0.103 4.61 5.99 7.38 9.21 10.60 13.813 0.115 0.352 6.25 7.81 9.35 11.34 12.84 16.274 0.297 0.711 7.78 9.49 11.14 13.28 14.86 18.475 0.554 1.15 9.24 11.07 12.83 15.09 16.75 20.526 0.872 1.64 10.64 12.59 14.45 16.81 18.55 22.467 1.24 2.17 12.02 14.07 16.01 18.48 20.28 24.328 1.65 2.73 13.36 15.51 17.53 20.09 21.95 26.129 2.09 3.33 14.68 16.92 19.02 21.67 23.59 27.8810 2.56 3.94 15.99 18.31 20.48 23.21 25.19 29.5911 3.05 4.57 17.28 19.68 21.92 24.73 26.76 31.2612 3.57 5.23 18.55 21.03 23.34 26.22 28.30 32.9113 4.11 5.89 19.81 22.36 24.74 27.69 29.82 34.5314 4.66 6.57 21.06 23.68 26.12 29.14 31.32 36.1215 5.23 7.26 22.31 25.00 27.49 30.58 32.80 37.7016 5.81 7.96 23.54 26.30 28.85 32.00 34.27 39.2517 6.41 8.67 24.77 27.59 30.19 33.41 35.72 40.7918 7.01 9.39 25.99 28.87 31.53 34.81 37.16 42.3119 7.63 10.12 27.20 30.14 32.85 36.19 38.58 43.8220 8.26 10.85 28.41 31.41 34.17 37.57 40.00 45.3121 8.90 11.59 29.62 32.67 35.48 38.93 41.40 46.8022 9.54 12.34 30.81 33.92 36.78 40.29 42.80 48.2723 10.20 13.09 32.01 35.17 38.08 41.64 44.18 49.7324 10.86 13.85 33.20 36.42 39.36 42.98 45.56 51.1825 11.52 14.61 34.38 37.65 40.65 44.31 46.93 52.6230 14.95 18.49 40.26 43.77 46.98 50.89 53.67 59.7040 22.16 26.51 51.81 55.76 59.34 63.69 66.77 73.4050 29.71 34.76 63.17 67.50 71.42 76.15 79.49 86.6660 37.48 43.19 74.40 79.08 83.30 88.38 91.95 99.6170 45.44 51.74 85.53 90.53 95.02 100.4 104.2 112.380 53.54 60.39 96.58 101.9 106.6 112.3 116.3 124.890 61.75 69.13 107.6 113.1 118.1 124.1 128.3 137.2100 70.06 77.93 118.5 124.3 129.6 135.8 140.2 149.4Table 31.2 The tabulated values are those which a variable distributed as χ 2with n degrees of freedom exceeds with the given percentage probability. Forexample, a variable having a χ 2 distribution with 14 degrees of freedom takesvalues in excess of 21.06 on 10% of occasions.1244

31.4 SOME BASIC ESTIMATORSexact expressions, valid for samples of any size N, for the expectation value andvariance of ¯x. From parts (i) and (ii) of the central limit theorem, discussed insection 30.10, we immediately obtainE[¯x] =µ, V [¯x] = σ2N . (31.40)Thus we see that ¯x is an unbiased estimator of µ. Moreover, we note that thestandard error in ¯x is σ/ √ N, and so the sampling distribution of ¯x becomes moretightly centred around µ as the sample size N increases. Indeed, since V [¯x] → 0as N →∞, ¯x is also a consistent estimator of µ.In the limit of large N, we may in fact obtain an approximate form for thefull sampling distribution of ¯x. Part (iii) of the central limit theorem (see section30.10) tells us immediately that, for large N, the sampling distribution of ¯x isgiven approximately by the Gaussian form[ ]1P (¯x|µ, σ) ≈ √2πσ2 /N exp (¯x − µ)2−2σ 2 ./NNote that this does not depend on the form of the original parent population.If, however, the parent population is in fact Gaussian then this result is exactfor samples of any size N (as is immediately apparent from our discussion ofmultiple Gaussian distributions in subsection 30.9.1).31.4.2 Population variance σ 2An estimator for the population variance σ 2 is not so straightforward to defineas one for the mean. Complications arise because, in many cases, the true meanof the population µ is not known. Nevertheless, let us begin by considering thecase where in fact µ is known. In this event, a useful estimator is( )̂σ 2 = 1 N∑(x i − µ) 2 1N∑= x 2 i − µ 2 . (31.41)NNi=1i=1◮Show that ̂σ 2 is an unbiased and consistent estimator of the population variance σ 2 .The expectation value of ̂σ 2 is given by[E[ ̂σ 2 ]= 1 N]N E ∑− µ 2 = E[x 2 i ] − µ 2 = µ 2 − µ 2 = σ 2 ,i=1x 2 ifrom which we see that the estimator is unbiased. The variance of the estimator is[V [ ̂σ 2 ]=1 N]N V ∑+ V [µ 2 ]= 1 2 N V [x2 i ]= 1 N (µ 4 − µ 2 2),i=1x 2 iin which we have used that fact that V [µ 2 ]=0andV [x 2 i ]=E[x4 i ] − (E[x2 i ])2 = µ 4 − µ 2 2 ,1245

STATISTICSwhere µ r is the rth population moment. Since ̂σ 2 is unbiased and V [ ̂σ 2 ] → 0asN →∞,showing that it is also a consistent estimator of σ 2 , the result is established. ◭If the true mean of the population is unknown, however, a natural alternativeis to replace µ by ¯x in (31.41), so that our estimator is simply the sample variances 2 given by( ) 2s 2 = 1 N∑x 2 1N∑i − x i .NNi=1In order to determine the properties of this estimator, we must calculate E[s 2 ]and V [s 2 ]. This task is straightforward but lengthy. However, for the investigationof the properties of a central moment of the sample, there exists a useful trickthat simplifies the calculation. We can assume, with no loss of generality, thatthe mean µ 1 of the population from which the sample is drawn is equal to zero.With this assumption, the population central moments, ν r ,areidenticaltothecorresponding moments µ r , and we may perform our calculation in terms of thelatter. At the end, however, we replace µ r by ν r in the final result and so obtaina general expression that is valid even in cases where µ 1 ≠0.◮Calculate E[s 2 ] and V [s 2 ] for a sample of size N.The expectation value of the sample variance s 2 for a sample of size N is given by[ ] ⎡( ) ⎤E[s 2 ]= 1 ∑N E x 2 i − 1 2∑N E ⎣ x 2 i⎦ii⎡= 1 N NE[x2 i ] − 1 N E ⎢∑2 ⎣ii=1x 2 i + ∑ i,jj≠ix i x j⎤⎥⎦ . (31.42)The number of terms in the double summation in (31.42) is N(N − 1), so we findE[s 2 ]=E[x 2 i ] − 1 N 2 (NE[x2 i ]+N(N − 1)E[x i x j ]).Now, since the sample elements x i and x j are independent, E[x i x j ]=E[x i ]E[x j ]=0,assuming the mean µ 1 of the parent population to be zero. Denoting the rth moment ofthe population by µ r , we thus obtainE[s 2 ]=µ 2 − µ 2N = N − 1N µ 2 = N − 1N σ2 , (31.43)where in the last line we have used the fact that the population mean is zero, and soµ 2 = ν 2 = σ 2 . However, the final result is also valid in the case where µ 1 ≠0.Using the above method, we can also find the variance of s 2 , although the algebra israther heavy going. The variance of s 2 is given byV [s 2 ]=E[s 4 ] − (E[s 2 ]) 2 , (31.44)where E[s 2 ] is given by (31.43). We therefore need only consider how to calculate E[s 4 ],1246

31.4 SOME BASIC ESTIMATORSwhere s 4 is given bys 4 =[ ∑i x2 iN( ∑− i x ) ] 2 2iN= (∑ i x2 i )2− 2 (∑ i x2 i )(∑ i x i) 2+ (∑ i x i) 4. (31.45)N 2 N 3 N 4We will consider in turn each of the three terms on the RHS. In the first term, the sum( ∑ i x2 i )2 can be written as( ) 2∑x 2 i = ∑ x 4 i + ∑ x 2 i x 2 j ,iii,jj≠iwhere the first sum contains N terms and the second contains N(N − 1) terms. Since thesample elements x i and x j are assumed independent, we have E[x 2 i x2 j ]=E[x2 i ]E[x2 j ]=µ2 2 ,and so⎡( ) ⎤ 2∑E ⎣ x 2 ⎦i = Nµ 4 + N(N − 1)µ 2 2.iTurning to the second term on the RHS of (31.45),( )( ) 2∑ ∑x 2 i x i = ∑ x 4 i + ∑ x 3 i x j + ∑ x 2 i x 2 j + ∑ x 2 i x j x k .iiii,ji,ji,j,kj≠ij≠ik≠j≠iSince the mean of the population has been assumed to equal zero, the expectation valuesof the second and fourth sums on the RHS vanish. The first and third sums contain Nand N(N − 1) terms respectively, and so⎡( )( ) ⎤ 2∑ ∑E ⎣ x 2 i x i⎦ = Nµ 4 + N(N − 1)µ 2 2.iiFinally, we consider the third term on the RHS of (31.45), and write( ) 4∑x i = ∑ x 4 i + ∑ x 3 i x j + ∑ x 2 i x 2 j + ∑ x 2 i x j x k + ∑ x i x j x k x l .iii,ji,ji,j,ki,j,k,lj≠ij≠ik≠j≠il≠k≠j≠iThe expectation values of the second, fourth and fifth sums are zero, and the first and thirdsums contain N and 3N(N − 1) terms respectively (for the third sum, there are N(N − 1)/2ways of choosing i and j, and the multinomial coefficient of x 2 i x 2 j is 4!/(2!2!) = 6). Thus⎡( ) ⎤ 4∑E ⎣ x i⎦ = Nµ 4 +3N(N − 1)µ 2 2.iCollecting together terms, we therefore obtainE[s 4 (N − 1)2]= µN 3 4 + (N − 1)(N2 − 2N +3)µ 2N2, (31.46)3which, together with the result (31.43), may be substituted into (31.44) to obtain finallyV [s 2 (N − 1)2 (N − 1)(N − 3)]= µN 3 4 − µ 2N 3 2= N − 1 [(N − 1)νN 3 4 − (N − 3)ν2], 2 (31.47)1247

STATISTICSwhere in the last line we have used again the fact that, since the population mean is zero,µ r = ν r . However, result (31.47) holds even when the population mean is not zero. ◭From (31.43), we see that s 2 is a biased estimator of σ 2 , although the biasbecomes negligible for large N. However, it immediately follows that an unbiasedestimator of σ 2 is given simply bŷσ 2 =NN − 1 s2 , (31.48)where the multiplicative factor N/(N − 1) is often called Bessel’s correction. Thusin terms of the sample values x i , i =1, 2,...,N, an unbiased estimator of thepopulation variance σ 2 is given bŷσ 2 = 1N − 1N∑(x i − ¯x) 2 . (31.49)Using (31.47), we find that the variance of the estimator ̂σ 2 is( 2V [ ̂σ N2 ]=V [sN − 1) 2 ]= 1 (ν 4 − N − 3 )N N − 1 ν2 2 ,where ν r is the rth central moment of the parent population. We note that,since E[ ̂σ 2 ]=σ 2 and V [ ̂σ 2 ] → 0asN →∞, the statistic ̂σ 2 is also a consistentestimator of the population variance.i=131.4.3 Population standard deviation σThe standard deviation σ of a population is defined as the positive square root ofthe population variance σ 2 (as, indeed, our notation suggests). Thus, it is commonpractice to take the positive square root of the variance estimator as our estimatorfor σ. Thus, we takeˆσ =( ̂σ2) 1/2, (31.50)where ̂σ 2 is given by either (31.41) or (31.48), depending on whether the populationmean µ is known or unknown. Because of the square root in the definition ofˆσ, it is not possible in either case to obtain an exact expression for E[ ˆσ] andV [ ˆσ]. Indeed, although in each case the estimator is the positive square root ofan unbiased estimator of σ 2 ,itisnot itself an unbiased estimator of σ. However,the bias does becomes negligible for large N.◮Obtain approximate expressions for E[ ˆσ] and V [ ˆσ] for a sample of size N in the casewhere the population mean µ is unknown.As the population mean is unknown, we use (31.50) and (31.48) to write our estimator in1248

31.4 SOME BASIC ESTIMATORSthe form( ) 1/2 Nˆσ =s,N − 1where s is the sample standard deviation. The expectation value of this estimator is givenby( ) 1/2 ( ) 1/2 NNE[ ˆσ] =E[(s 2 ) 1/2 ] ≈(E[s 2 ]) 1/2 = σ.N − 1N − 1An approximate expression for the variance of ˆσ may be found using (31.47) and is givenbyV [ ˆσ] =NN − 1 V [(s2 ) 1/2 ] ≈NN − 1[ d≈N [ 1N − 1] 2d(s 2 ) (s2 ) 1/2 V [s 2 ]s 2 =E[s 2 ]V [s4s]s 2 ].2 2 =E[s 2 ]Using the expressions (31.43) and (31.47) for E[s 2 ]andV [s 2 ] respectively, we obtainV [ ˆσ] ≈ 1 (ν 4 − N − 3 )4Nν 2 N − 1 ν2 2 . ◭31.4.4 Population moments µ rWe may straightforwardly generalise our discussion of estimation of the populationmean µ (= µ 1 ) in subsection 31.4.1 to the estimation of the rth populationmoment µ r . An obvious choice of estimator is the rth sample moment m r .Theexpectation value of m r is given byE[m r ]= 1 NN∑i=1E[x r i ]= Nµ rN = µ r,and so it is an unbiased estimator of µ r .The variance of m r may be found in a similar manner, although the calculationis a little more complicated. We find thatV [m r ]=E[(m r − µ r ) 2 ]⎡( ) ⎤ 2= 1 ∑N 2 E ⎣ x r i − Nµ r⎦i⎡⎤= 1 N 2 E ⎣ ∑ x 2ri + ∑ ∑∑x r i x r j − 2Nµ r x r i + N 2 µ 2 ⎦rii j≠ii= 1 N µ 2r − µ 2 r + 1 ∑ ∑N 2 E[x r i x r j]. (31.51)ij≠i1249

STATISTICSHowever, since the sample values x i are assumed to be independent, we haveE[x r i x r j]=E[x r i ]E[x r j]=µ 2 r . (31.52)The number of terms in the sum on the RHS of (31.51) is N(N −1), and so we findV [m r ]= 1 N µ 2r − µ 2 r + N − 1Nµ2 r = µ 2r − µ 2 rN . (31.53)Since E[m r ]=µ r and V [m r ] → 0asN →∞,therth sample moment m r is alsoa consistent estimator of µ r .◮Find the covariance of the sample moments m r and m s for a sample of size N.We obtain the covariance of the sample moments m r and m s in a similar manner to thatused above to obtain the variance of m r . From the definition of covariance, we haveCov[m r ,m s ]=E[(m r − µ r )(m s − µ s )][( )( )]= 1 ∑ ∑N E x r 2 i − Nµ r x s j − Nµ sij= 1 N 2 E ⎡⎣ ∑ ix r+si+ ∑ i⎤∑∑ ∑x r i x s j − Nµ r x s j − Nµ s x r i + N 2 µ r µ s⎦Assuming the x i to be independent, we may again use result (31.52) to obtainj≠iCov[m r ,m s ]= 1 N [Nµ 2 r+s + N(N − 1)µ r µ s − N 2 µ r µ s − N 2 µ s µ r + N 2 µ r µ s ]= 1 N µ r+s + N − 1Nµ rµ s − µ r µ s= µ r+s − µ r µ s.NWe note that by setting r = s, we recover the expression (31.53) for V [m r ]. ◭ji31.4.5 Population central moments ν rWe may generalise the discussion of estimators for the second central moment ν 2(or equivalently σ 2 ) given in subsection 31.4.2 to the estimation of the rth centralmoment ν r . In particular, we saw in that subsection that our choice of estimatorfor ν 2 depended on whether the population mean µ 1 is known; the same is truefor the estimation of ν r .Let us first consider the case in which µ 1 is known. From (30.54), we may writeν r asν r = µ r − r C 1 µ r−1 µ 1 + ···+(−1) k r C k µ r−k µ k 1 + ···+(−1) r−1 ( r C r−1 − 1)µ r 1.If µ 1 is known, a suitable estimator is obviouslyˆν r = m r − r C 1 m r−1 µ 1 + ···+(−1) k r C k m r−k µ k 1 + ···+(−1) r−1 ( r C r−1 − 1)µ r 1,where m r is the rth sample moment. Since µ 1 and the binomial coefficients are1250

31.4 SOME BASIC ESTIMATORS(known) constants, it is immediately clear that E[ˆν r ]=ν r ,andsoˆν r is an unbiasedestimator of ν r . It is also possible to obtain an expression for V [ˆν r ], though thecalculation is somewhat lengthy.In the case where the population mean µ 1 is not known, the situation is morecomplicated. We saw in subsection 31.4.2 that the second sample moment n 2 (ors 2 )isnot an unbiased estimator of ν 2 (or σ 2 ). Similarly, the rth central moment ofa sample, n r , is not an unbiased estimator of the rth population central momentν r . However, in all cases the bias becomes negligible in the limit of large N.As we also found in the same subsection, there are complications in calculatingthe expectation and variance of n 2 ; these complications increase considerably forgeneral r. Nevertheless, we have derived already in this chapter exact expressionsfor the expectation value of the first few sample central moments, which are validfor samples of any size N. From (31.40), (31.43) and (31.46), we findE[n 1 ]=0,E[n 2 ]= N − 1N ν 2, (31.54)E[n 2 2]= N − 1N 3 [(N − 1)ν 4 +(N 2 − 2N +3)ν2].2By similar arguments it can be shown that(N − 1)(N − 2)E[n 3 ]=N 2 ν 3 , (31.55)E[n 4 ]= N − 1N 3 [(N 2 − 3N +3)ν 4 +3(2N − 3)ν2]. 2 (31.56)From (31.54) and (31.55), we see that unbiased estimators of ν 2 and ν 3 areˆν 2 =NN − 1 n 2, (31.57)ˆν 3 =N 2(N − 1)(N − 2) n 3, (31.58)where (31.57) simply re-establishes our earlier result that ̂σ 2 = Ns 2 /(N − 1) is anunbiased estimator of σ 2 .Unfortunately, the pattern that appears to be emerging in (31.57) and (31.58)is not continued for higher r, as is seen immediately from (31.56). Nevertheless,in the limit of large N, the bias becomes negligible, and often one simply takesˆν r = n r . For large N, it may be shown thatE[n r ] ≈ ν rV [n r ] ≈ 1 N (ν 2r − ν 2 r + r 2 ν 2 ν 2 r−1 − 2rν r−1 ν r+1 )Cov[n r ,n s ] ≈ 1 N (ν r+s − ν r ν s + rsν 2 ν r−1 ν s−1 − rν r−1 ν s+1 − sν s−1 ν r+1 )1251

STATISTICS31.4.6 Population covariance Cov[x, y] and correlation Corr[x, y]So far we have assumed that each of our N independent samples consists ofa single number x i . Let us now extend our discussion to a situation in whicheach sample consists of two numbers x i ,y i , which we may consider as beingdrawn randomly from a two-dimensional population P (x, y). In particular, wenow consider estimators for the population covariance Cov[x, y] and for thecorrelation Corr[x, y].When µ x and µ y are known, an appropriate estimator of the population covarianceisĈov[x, y] =xy − µ x µ y =This estimator is unbiased since(1N)N∑x i y i − µ x µ y . (31.59)[]E[Ĉov[x, y] = 1 N]N E ∑x i y i − µ x µ y = E[x i y i ] − µ x µ y =Cov[x, y].i=1Alternatively, if µ x and µ y are unknown, it is natural to replace µ x and µ y in(31.59) by the sample means ¯x and ȳ respectively, in which case we recover thesample covariance V xy = xy − ¯xȳ discussed in subsection 31.2.4. This estimatoris biased but an unbiased estimator of the population covariance is obtained byformingĈov[x, y] =i=1NN − 1 V xy. (31.60)◮Calculate the expectation value of the sample covariance V xy for a sample of size N.The sample covariance is given by( ) ( )( )1 ∑ 1 ∑ 1 ∑V xy = x i y i − x i y j .NN NiijThus its expectation value is given by[ ] [( )(E[V xy ]= 1 ∑N E x i y i − 1 ∑ ∑N E x 2 iiij⎡⎤= E[x i y i ] − 1 N E ⎢∑2 ⎣ x i y i + ∑ ⎥x i y j ⎦ii,jj≠ix j)]1252

31.4 SOME BASIC ESTIMATORSSince the number of terms in the double sum on the RHS is N(N − 1), we haveE[V xy ]=E[x i y i ] − 1 N 2 (NE[x iy i ]+N(N − 1)E[x i y j ])= E[x i y i ] − 1 N (NE[x iy 2 i ]+N(N − 1)E[x i ]E[y j ])= E[x i y i ] − 1 ( ) N − 1E[xi y i ]+(N − 1)µ x µ y = Cov[x, y],NNwhere we have used the fact that, since the samples are independent, E[x i y j ]=E[x i ]E[y j ]. ◭It is possible to obtain expressions for the variances of the estimators (31.59)and (31.60) but these quantities depend upon higher moments of the populationP (x, y) and are extremely lengthy to calculate.Whether the means µ x and µ y are known or unknown, an estimator of thepopulation correlation Corr[x, y] is given byy]Ĉorr[x, y] =Ĉov[x, , (31.61)ˆσ x ˆσ ywhere Ĉov[x, y], ˆσ x and ˆσ y are the appropriate estimators of the population covarianceand standard deviations. Although this estimator is only asymptoticallyunbiased, i.e. for large N, it is widely used because of its simplicity. Once againthe variance of the estimator depends on the higher moments of P (x, y) andisdifficult to calculate.In the case in which the means µ x and µ y are unknown, a suitable (but biased)estimator isV xyĈorr[x, y] =N =NN − 1 s x s y N − 1 r xy, (31.62)where s x and s y are the sample standard deviations of the x i and y i respectivelyand r xy is the sample correlation. In the special case when the parent populationP (x, y) is Gaussian, it may be shown that, if ρ = Corr[x, y],E[r xy ]=ρ − ρ(1 − ρ2 )+O(N −2 ),2N(31.63)V [r xy ]= 1 N (1 − ρ2 ) 2 +O(N −2 ), (31.64)from which the expectation value and variance of the estimator Ĉorr[x, y] maybe found immediately.We note finally that our discussion may be extended, without significant alteration,to the general case in which each data item consists of n numbersx i ,y i ,...,z i .1253

STATISTICS31.4.7 A worked exampleTo conclude our discussion of basic estimators, we reconsider the set of experimentaldata given in subsection 31.2.4. We carry the analysis as far as calculatingthe standard errors in the estimated population parameters, including the populationcorrelation.◮Ten UK citizens are selected at random and their heights and weights are found to be asfollows (to the nearest cm or kg respectively):Person A B C D E F G H I JHeight (cm) 194 168 177 180 171 190 151 169 175 182Weight (kg) 75 53 72 80 75 75 57 67 46 68Estimate the means, µ x and µ y , and standard deviations, σ x and σ y , of the two-dimensionaljoint population from which the sample was drawn, quoting the standard error on the estimatein each case. Estimate also the correlation Corr[x, y] of the population, and quote thestandard error on the estimate under the assumption that the population is a multivariateGaussian.In subsection 31.2.4, we calculated various sample statistics for these data. In particular,we found that for our sample of size N = 10,¯x = 175.7, ȳ =66.8,s x =11.6, s y =10.6, r xy =0.54.Let us begin by estimating the means µ x and µ y . As discussed in subsection 31.4.1, thesample mean is an unbiased, consistent estimator of the population mean. Moreover, thestandard error on ¯x (say) is σ x / √ N. In this case, however, we do not know the true valueof σ x and we must estimate it using ̂σ x = √ N/(N − 1)s x . Thus, our estimates of µ x andµ y , with associated standard errors, areˆµ x = ¯x ±ˆµ y = ȳ ±s x√N − 1= 175.7 ± 3.9,s y√N − 1=66.8 ± 3.5.We now turn to estimating σ x and σ y . As just mentioned, our estimate of σ x (say)is ̂σ x = √ N/(N − 1)s x . Its variance (see the final line of subsection 31.4.3) is givenapproximately byV [ ˆσ] ≈ 1 (ν 4 − N − 3 )4Nν 2 N − 1 ν2 2 .Since we do not know the true values of the population central moments ν 2 and ν 4 ,wemust use their estimated values in this expression. We may take ˆν 2 = ̂σ x 2 =(ˆσ) 2 ,whichwehave already calculated. It still remains, however, to estimate ν 4 .Asimpliedneartheendof subsection 31.4.5, it is acceptable to take ˆν 4 = n 4 . Thus for the x i and y i values, we have(ˆν 4 ) x = 1 N(ˆν 4 ) y = 1 NN∑(x i − ¯x) 4 = 53 411.6i=1N∑(y i − ȳ) 4 = 27 732.5i=11254

31.5 MAXIMUM-LIKELIHOOD METHODSubstituting these values into (31.50), we obtain( ) 1/2 Nˆσ x =s x ± ( ˆV [ ˆσ x ]) 1/2 =12.2 ± 6.7, (31.65)N − 1( ) 1/2 Nˆσ y =s y ± ( ˆV [ ˆσ y ]) 1/2 =11.2 ± 3.6. (31.66)N − 1Finally, we estimate the population correlation Corr[x, y], which we shall denote by ρ.From (31.62), we haveˆρ =NN − 1 r xy =0.60.Under the assumption that the sample was drawn from a two-dimensional Gaussianpopulation P (x, y), the variance of our estimator is given by (31.64). Since we do not knowthe true value of ρ, we must use our estimate ˆρ. Thus, we find that the standard error ∆ρin our estimate is given approximately by∆ρ ≈ 10 ( ) 1[1 − (0.60) 2 ] 2 =0.05. ◭9 1031.5 Maximum-likelihood methodThe population from which the sample x 1 ,x 2 ,...,x N is drawn is, in general,unknown. In the previous section, we assumed that the sample values were independentand drawn from a one-dimensional population P (x), and we consideredbasic estimators of the moments and central moments of P (x). We did not, however,assume a particular functional form for P (x). We now discuss the processof data modelling, in which a specific form is assumed for the population.In the most general case, it will not be known whether the sample values areindependent, and so let us consider the full joint population P (x), where x is thepoint in the N-dimensional data space with coordinates x 1 ,x 2 ,...,x N .Wethenadopt the hypothesis H that the probability distribution of the sample values hassome particular functional form L(x; a), dependent on the values of some set ofparameters a i , i =1, 2,...,m. Thus, we haveP (x|a,H)=L(x; a),where we make explicit the conditioning on both the assumed functional form andon the parameter values. L(x; a) is called the likelihood function. Hypotheses of thistype form the basis of data modelling and parameter estimation. One proposes aparticular model for the underlying population and then attempts to estimate fromthe sample values x 1 ,x 2 ,...,x N the values of the parameters a defining this model.◮A company measures the duration (in minutes) of the N intervals x i , i = 1, 2,...,Nbetween successive telephone calls received by its switchboard. Suppose that the samplevalues x i are drawn independently from the distribution P (x|τ) =(1/τ)exp(−x/τ), whereτis the mean interval between calls. Calculate the likelihood function L(x; τ).Since the sample values are independent and drawn from the stated distribution, the1255

STATISTICSL(x; τ)1N =5L(x; τ)1N =100.50.500τ2 4 6 8 10 12 14 16 18 2000τ2 4 6 8 10 12 14 16 18 20L(x; τ)1N =20L(x; τ)1N =500.50.500τ2 4 6 8 10 12 14 16 18 2000τ2 4 6 8 10 12 14 16 18 20Figure 31.5 Examples of the likelihood function (31.67) for samples of differentsize N. In each case, the true value of the parameter is τ =4andthesample values x i are indicated by the short vertical lines. For the purposesof illustration, in each case the likelihood function is normalised so that itsmaximum value is unity.likelihood is given byL(x; τ) =P (x i |τ)P (x 2 |τ) ···P (x N |τ)= 1 (τ exp − x )1 1(τ τ exp − x )2··· 1 (τ τ exp − x )Nτ= 1 [τ exp − 1 ]N τ (x 1 + x 2 + ···+ x N ) . (31.67)which is to be considered as a function of τ, given that the sample values x i are fixed. ◭The likelihood function (31.67) depends on just a single parameter τ. Plotsofthe likelihood function, considered as a function of τ, are shown in figure 31.5 forsamples of different size N. The true value of the parameter τ used to generate thesample values was 4. In each case, the sample values x i are indicated by the shortvertical lines. For the purposes of illustration, the likelihood function in eachcase has been scaled so that its maximum value is unity (this is, in fact, commonpractice). We see that when the sample size is small, the likelihood function isvery broad. As N increases, however, the function becomes narrower (its width isinversely proportional to √ N) and tends to a Gaussian-like shape, with its peakcentred on 4, the true value of τ. We discuss these properties of the likelihoodfunction in more detail in subsection 31.5.6.1256

31.5 MAXIMUM-LIKELIHOOD METHODL(x; a)(a)L(x; a)(b)âaâaL(x; a)(c)L(x; a)(d)âaâaFigure 31.6 Typical shapes of one-dimensional likelihood functions L(x; a)encountered in practice, when, for illustration purposes, it is assumed that theparameter a is restricted to the range zero to infinity. The ML estimator inthe various cases occurs at: (a) the only stationary point; (b) one of severalstationary points; (c) an end-point of the allowed parameter range that is nota stationary point (although stationary points do exist); (d) an end-point ofthe allowed parameter range in which no stationary point exists.31.5.1 The maximum-likelihood estimatorSince the likelihood function L(x; a) gives the probability density associated withany particular set of values of the parameters a, our best estimate â of theseparameters is given by the values of a for which L(x; a) is a maximum. This iscalled the maximum-likelihood estimator (or ML estimator).In general, the likelihood function can have a complicated shape when consideredas a function of a, particularly when the dimensionality of the space ofparameters a 1 ,a 2 ,...,a M is large. It may be that the values of some parametersare either known or assumed in advance, in which case the effective dimensionalityof the likelihood function is reduced accordingly. However, even when thelikelihood depends on just a single parameter a (either intrinsically or as theresult of assuming particular values for the remaining parameters), its form maybe complicated when the sample size N is small. Frequently occurring shapes ofone-dimensional likelihood functions are illustrated in figure 31.6, where we haveassumed, for definiteness, that the allowed range of the parameter a is zero toinfinity. In each case, the ML estimate â is also indicated. Of course, the ‘shape’ ofhigher-dimensional likelihood functions may be considerably more complicated.In many simple cases, however, the likelihood function L(x; a) has a single1257

STATISTICSmaximum that occurs at a stationary point (the likelihood function is then termedunimodal). In this case, the ML estimators of the parameters a i , i =1, 2,...,M,may be found without evaluating the full likelihood function L(x; a). Instead, onesimply solves the M simultaneous equations∂L∂a i∣∣∣∣a=â=0 fori =1, 2,...,M. (31.68)Since ln z is a monotonically increasing function of z (and therefore has thesame stationary points), it is often more convenient, in fact, to maximise thelog-likelihood function, lnL(x; a), with respect to the a i . Thus, one may, as analternative, solve the equations∂ ln L∂a i∣∣∣∣a=â=0 fori =1, 2,...,M. (31.69)Clearly, (31.68) and (31.69) will lead to the same ML estimates â of the parameters.In either case, it is, of course, prudent to check that the point a = â is a localmaximum.◮Find the ML estimate of the parameter τ in the previous example, in terms of the measuredvalues x i , i =1, 2,...,N.From (31.67), the log-likelihood function in this case is given byln L(x; τ) =N∑i=1( 1lni/τ)τ e−x = −N∑ (i=1ln τ + x iτ). (31.70)Differentiating with respect to the parameter τ and setting the result equal to zero, we find∂ ln L∂τ= −N∑( 1τ − x )i=0.τ 2Thus the ML estimate of the parameter τ is given byi=1ˆτ = 1 NN∑x i , (31.71)i=1which is simply the sample mean of the N measured intervals. ◭In the previous example we assumed that the sample values x i were drawnindependently from the same parent distribution. The ML method is more flexiblethan this restriction might seem to imply, and it can equally well be applied tothe common case in which the samples x i are independent but each is drawnfrom a different distribution.1258

31.5 MAXIMUM-LIKELIHOOD METHOD◮In an experiment, N independent measurements x i of some quantity are made. Supposethat the random measurement error on the i th sample value is Gaussian distributed withmean zero and known standard deviation σ i . Calculate the ML estimate of the true valueµ of the quantity being measured.As the measurements are independent, the likelihood factorises:L(x; µ, {σ k })=N∏P (x i |µ, σ i ),where {σ k } denotes collectively the set of known standard deviations σ 1 ,σ 2 ,...,σ N .Theindividual distributions are given by1P (x i |µ, σ i )= √ exp[− (x ]i − µ) 2.2πσ2i2σi2and so the full log-likelihood function is given byln L(x; µ, {σ k })=− 1 N∑[ln(2πσi 2 )+ (x ]i − µ) 2.2σ 2 i=1iDifferentiating this expression with respect to µ and setting the result equal to zero, wefindN∂ ln L∂µ = ∑ x i − µ=0,σi2from which we obtain the ML estimatorˆµ =i=1i=1∑ Ni=1 (x i/σ 2 i )∑ Ni=1 (1/σ2 i ) . (31.72)This estimator is commonly used when averaging data with different statistical weightsw i =1/σi 2. We note that when all the variances σ2 i have the same value the estimatorreduces to the sample mean of the data x i . ◭There is, in fact, no requirement in the ML method that the sample valuesbe independent. As an illustration, we shall generalise the above example to acase in which the measurements x i are not all independent. This would occur, forexample, if these measurements were based at least in part on the same data.◮In an experiment N measurements x i of some quantity are made. Suppose that the randommeasurement errors on the samples are drawn from a joint Gaussian distribution with meanzero and known covariance matrix V. Calculate the ML estimate of the true value µ of thequantity being measured.From (30.148), the likelihood in this case is given by1L(x; µ, V) =(2π) N/2 |V| exp [ − 1 (x − 1/2 2 µ1)T V −1 (x − µ1) ] ,where x is the column matrix with components x 1 ,x 2 ,...,x N and 1 is the column matrixwith all components equal to unity. Thus, the log-likelihood function is given by[ln L(x; µ, V) =− 1 2 N ln(2π)+ln|V| +(x − µ1) T V −1 (x − µ1) ] .1259

STATISTICSDifferentiating with respect to µ and setting the result equal to zero gives∂ ln L∂µ = 1T V −1 (x − µ1) =0.Thus, the ML estimator is given by∑ˆµ = 1T V −1 x1 T V −1 1 = i,j (V −1 ) ij x j∑i,j (V .−1 ) ijIn the case of uncorrelated errors in measurement, (V −1 ) ij = δ ij /σi 2 and our estimatorreduces to that given in (31.72). ◭In all the examples considered so far, the likelihood function has been effectivelyone-dimensional, either instrinsically or under the assumption that the values ofall but one of the parameters are known in advance. As the following exampleinvolving two parameters shows, the application of the ML method to theestimation of several parameters simultaneously is straightforward.◮In an experiment N measurements x i of some quantity are made. Suppose the random erroron each sample value is drawn independently from a Gaussian distribution of mean zero butunknown standard deviation σ (which is the same for each measurement). Calculate the MLestimates of the true value µ of the quantity being measured and the standard deviation σof the random errors.In this case the log-likelihood function is given byln L(x; µ, σ) =− 1 N∑[ln(2πσ 2 )+ (x ]i − µ) 2.2σ 2 i=1Taking partial derivatives of ln L with respect to µ and σ and setting the results equal tozero at the joint estimate ˆµ, ˆσ, weobtainN∑ x i − ˆµ=0, (31.73)ˆσ 2 i=1N∑ (x i − ˆµ) 2 N∑ 1− =0. (31.74)ˆσ 3 ˆσi=1i=1In principle, one should solve these two equations simultaneously for ˆµ and ˆσ, but in thiscase we notice that the first is solved immediately byˆµ = 1 N∑x i = ¯x,Ni=1where ¯x is the sample mean. Substituting this result into the second equation, we findˆσ = √ 1 N∑(x i − ¯x)N2 = s,i=1where s is the sample standard deviation. As shown in subsection 31.4.3, s is a biasedestimator of σ. The reason why the ML method may produce a biased estimator isdiscussed in the next subsection. ◭1260

31.5 MAXIMUM-LIKELIHOOD METHOD31.5.2 Transformation invariance and bias of ML estimatorsAn extremely useful property of ML estimators is that they are invariant toparameter transformations. Suppose that, instead of estimating some parametera of the assumed population, we wish to estimate some function α(a) oftheparameter. The ML estimator ˆα(a) is given by the value assumed by the functionα(a) at the maximum point of the likelihood, which is simply equal to α(â). Thus,we have the very convenient propertyˆα(a) =α(â).We do not have to worry about the distinction between estimating a and estimatinga function of a. Thisisnot true, in general, for other estimation procedures.◮A company measures the duration (in minutes) of the N intervals x i , i =1, 2,...,N,between successive telephone calls received by its switchboard. Suppose that the samplevalues x i are drawn independently from the distribution P (x|τ) =(1/τ)exp(−x/τ). FindtheML estimate of the parameter λ =1/τ.This is the same problem as the first one considered in subsection 31.5.1. In terms of thenew parameter λ, the log-likelihood function is given byN∑N∑ln L(x; λ) = ln(λe −λx i)= (ln λ − λx i ).i=1Differentiating with respect to λ and setting the result equal to zero, we have∂ ln LN∑( 1=i)∂λ λ − x =0.i=1Thus, the ML estimator of the parameter λ is given by( ) −11N∑ˆλ = x i = ¯x −1 . (31.75)Ni=1Referring back to (31.71), we see that, as expected, the ML estimators of λ and τ arerelated by ˆλ =1/ˆτ. ◭Although this invariance property is useful it also means that, in general, MLestimators may be biased. In particular, one must be aware of the fact that evenif â is an unbiased ML estimator of a it does not follow that the estimator ˆα(a) isalso unbiased. In the limit of large N, however, the bias of ML estimators alwaystends to zero. As an illustration, it is straightforward to show (see exercise 31.8)that the ML estimators ˆτ and ˆλ in the above example have expectation valuesE[ˆτ] =τ and E[ˆλ] = N λ. (31.76)N − 1In fact, since ˆτ = ¯x and the sample values are independent, the first result followsimmediately from (31.40). Thus, ˆτ is unbiased, but ˆλ =1/ˆτ is biased, albeit thatthe bias tends to zero for large N.1261i=1

STATISTICS31.5.3 Efficiency of ML estimatorsWe showed in subsection 31.3.2 that Fisher’s inequality puts a lower limit on thevariance V [â] of any estimator of the parameter a. Under our hypothesis H onp. 1255, the functional form of the population is given by the likelihood function,i.e. P (x|a,H)=L(x; a). Thus, if this hypothesis is correct, we may replace P byL in Fisher’s inequality (31.18), which then readsV [â] ≥(1+ ∂b∂a) 2 /E[ ]− ∂2 ln L∂a 2 ,where b is the bias in the estimator â. We usually denote the RHS by V min .An important property of ML estimators is that if there exists an efficientestimator â eff ,i.e.oneforwhichV [â eff ]=V min ,thenitmust be the ML estimatoror some function thereof. This is easily shown by replacing P by L in the proofof Fisher’s inequality given in subsection 31.3.2. In particular, we note that theequality in (31.22) holds only if h(x) =cg(x), where c is a constant. Thus, if anefficient estimator â eff exists, this is equivalent to demanding that∂ ln L= c[â eff − α(a)].∂aNow, the ML estimator â ML is given by∂ ln L∂a ∣ =0 ⇒ c[â eff − α(â ML )] = 0,a=âMLwhich, in turn, implies that â eff must be some function of â ML .◮Show that the ML estimator ˆτ given in (31.71) is an efficient estimator of the parameter τ.As shown in (31.70), the log-likelihood function in this case isN∑ (ln L(x; τ) =− ln τ + x )i.τi=1Differentiating twice with respect to τ, we find∂ 2 ln LN∑( 1=∂τ 2 τ − 2x ) ()i= N 1 − 2 N∑x 2 τ 3 τ 2 i , (31.77)τNi=1i=1and so the expectation value of this expression is[ ] ∂ 2 ln LE = N (1 − 2 )∂τ 2 τ 2 τ E[x i] = − N τ , 2where we have used the fact that E[x] =τ. Setting b = 0 in (31.18), we thus find that forany unbiased estimator of τ,V [ˆτ] ≥ τ2N .From (31.76), we see that the ML estimator ˆτ = ∑ i x i/N is unbiased. Moreover, usingthe fact that V [x] =τ 2 , it follows immediately from (31.40) that V [ˆτ] =τ 2 /N. Thus ˆτ is aminimum-variance estimator of τ. ◭1262

31.5 MAXIMUM-LIKELIHOOD METHOD31.5.4 Standard errors and confidence limits on ML estimatorsThe ML method provides a procedure for obtaining a particular set of estimatorsâ ML for the parameters a of the assumed population P (x|a). As for any other setof estimators, the associated standard errors, covariances and confidence intervalscan be found as described in subsections 31.3.3 and 31.3.4.◮A company measures the duration (in minutes) of the 10 intervals x i , i =1, 2,...,10,between successive telephone calls made to its switchboard to be as follows:0.43 0.24 3.03 1.93 1.16 8.65 5.33 6.06 5.62 5.22.Supposing that the sample values are drawn independently from the probability distributionP (x|τ) =(1/τ)exp(−x/τ), find the ML estimate of the mean τ and quote an estimate ofthe standard error on your result.As shown in (31.71) the (unbiased) ML estimator ˆτ in this case is simply the sample mean¯x =3.77. Also, as shown in subsection 31.5.3, ˆτ is a minimum-variance estimator withV [ˆτ] =τ 2 /N. Thus, the standard error in ˆτ is simply√ τ . (31.78)Nσˆτ =Since we do not know the true value of τ, however, we must instead quote an estimate ˆσˆτof the standard error, obtained by substituting our estimate ˆτ for τ in (31.78). Thus, wequote our final result asτ = ˆτ ±ˆτ √N=3.77 ± 1.19. (31.79)For comparison, the true value used to create the sample was τ =4.◭For the particular problem considered in the above example, it is in fact possibleto derive the full sampling distribution of the ML estimator ˆτ using characteristicfunctions, and it is given byP (ˆτ|τ) =NN ˆτ N−1 ((N − 1)! τ N exp − Nˆτ ), (31.80)τwhere N is the size of the sample. This function is plotted in figure 31.7 for thecase τ = 4 and N = 10, which pertains to the above example. Knowledge of theanalytic form of the sampling distribution allows one to place confidence limitson the estimate ˆτ obtained, as discussed in subsection 31.3.4.◮Using the sample values in the above example, obtain the 68% central confidence intervalon the value of τ.For the sample values given, our observed value of the ML estimator is ˆτ obs =3.77. Thus,from (31.28) and (31.29), the 68% central confidence interval [τ − ,τ + ] on the value of τ isfound by solving the equations∫ ˆτobs−∞∫ ∞P (ˆτ|τ + ) dˆτ =0.16,ˆτ obsP (ˆτ|τ − ) dˆτ =0.16,1263

STATISTICSP (ˆτ|τ)0.40.30.20.1002 4 6 8 10 12 14ˆτFigure 31.7 The sampling distribution P (ˆτ|τ) for the estimator ˆτ for the caseτ =4andN = 10.where P (ˆτ|τ) is given by (31.80) with N = 10. The above integrals can be evaluatedanalytically but the calculations are rather cumbersome. It is much simpler to evaluatethem by numerical integration, from which we find [τ − ,τ + ]=[2.86, 5.46]. Alternatively,we could quote the estimate and its 68% confidence interval asτ =3.77 +1.69−0.91 .Thus we see that the 68% central confidence interval is not symmetric about the estimatedvalue, and differs from the standard error calculated above. This is a result of the (non-Gaussian) shape of the sampling distribution P (ˆτ|τ), apparent in figure 31.7. ◭In many problems, however, it is not possible to derive the full samplingdistribution of an ML estimator â in order to obtain its confidence intervals.Indeed, one may not even be able to obtain an analytic formula for its standarderror σâ. This is particularly true when one is estimating several parameter âsimultaneously, since the joint sampling distribution will be, in general, verycomplicated. Nevertheless, as we discuss below, the likelihood function L(x; a)itself can be used very simply to obtain standard errors and confidence intervals.The justification for this has its roots in the Bayesian approach to statistics, asopposed to the more traditional frequentist approach we have adopted here. Wenow give a brief discussion of the Bayesian viewpoint on parameter estimation.31.5.5 The Bayesian interpretation of the likelihood functionAs stated at the beginning of section 31.5, the likelihood function L(x; a) isdefined byP (x|a,H)=L(x; a),1264

31.5 MAXIMUM-LIKELIHOOD METHODwhere H denotes our hypothesis of an assumed functional form. Now, usingBayes’ theorem (see subsection 30.2.3), we may writeP (x|a,H)P (a|H)P (a|x,H)= , (31.81)P (x|H)which provides us with an expression for the probability distribution P (a|x,H)of the parameters a, given the (fixed) data x and our hypothesis H, in terms ofother quantities that we may assign. The various terms in (31.81) have specialformal names, as follows.• The quantity P (a|H) ontheRHSistheprior probability, which represents ourstate of knowledge of the parameter values (given the hypothesis H) before wehave analysed the data.• This probability is modified by the experimental data x through the likelihoodP (x|a,H).• When appropriately normalised by the evidence P (x|H), this yields the posteriorprobability P (a|x,H), which is the quantity of interest.• The posterior encodes all our inferences about the values of the parameters a.Strictly speaking, from a Bayesian viewpoint, this entire function, P (a|x,H), isthe ‘answer’ to a parameter estimation problem.Given a particular hypothesis, the (normalising) evidence factor P (x|H) isunimportant, since it does not depend explicitly upon the parameter values a.Thus, it is often omitted and one considers only the proportionality relationP (a|x,H) ∝ P (x|a,H)P (a|H). (31.82)If necessary, the posterior distribution can be normalised empirically, by requiringthat it integrates to unity, i.e. ∫ P (a|x,H) d m a = 1, where the integral extends overall values of the parameters a 1 ,a 2 ,...,a m .The prior P (a|H) in (31.82) should reflect our entire knowledge concerning thevalues of the parameters a, before the analysis of the current data x. For example,there may be some physical reason to require some or all of the parameters tolie in a given range. If we are largely ignorant of the values of the parameters,we often indicate this by choosing a uniform (or very broad) prior,P (a|H) = constant,in which case the posterior distribution is simply proportional to the likelihood.In this case, we thus haveP (a|x,H) ∝ L(x; a). (31.83)In other words, if we assume a uniform prior then we can identify the posteriordistribution (up to a normalising factor) with L(x; a), considered as a function ofthe parameters a.1265

STATISTICSThus, a Bayesian statistician considers the ML estimates â ML of the parametersto be the values that maximise the posterior P (a|x,H) under the assumption ofa uniform prior. More importantly, however, a Bayesian would not calculate thestandard error or confidence interval on this estimate using the (classical) methodemployed in subsection 31.3.4. Instead, a far more straightforward approach isadopted. Let us assume, for the moment, that one is estimating just a singleparameter a. Using (31.83), we may determine the values a − and a + such thatPr(a a + |x,H)= L(x; a) da = β.a +where∫it is assumed that the likelihood has been normalised in such a way thatL(x; a) da = 1. Combining these equations gives∫ a+Pr(a − ≤ a

31.5 MAXIMUM-LIKELIHOOD METHODL(x; τ)0.40.30.20.1002 4 6 8 10 12 14τFigure 31.8 The likelihood function L(x; τ) (normalised to unit area) for thesample values given in the worked example in subsection 31.5.4 and indicatedhere by short vertical lines.68.3% Bayesian central confidence interval. Even when L(x; a) is not Gaussian,however, (31.85) is often used as a measure of the standard error.◮For the sample data given in subsection 31.5.4, use the likelihood function to estimate thestandard error ˆσˆτ in the ML estimator ˆτ and obtain the Bayesian 68% central confidenceinterval on τ.We showed in (31.67) that the likelihood function in this case is given byL(x; τ) = 1 [τ exp − 1 ]N τ (x 1 + x 2 + ···+ x N ) .where x i , i =1, 2,...,N, denotes the sample value and N = 10. This likelihood function isplotted in figure 31.8, after normalising (numerically) to unit area. The short vertical linesin the figure indicate the sample values. We see that the likelihood function peaks at theML estimate ˆτ =3.77 that we found in subsection 31.5.4. Also, from (31.77), we have( )∂ 2 ln L= N 21 − 2 N∑x∂τ 2i .τ τNi=1Remembering that ˆτ = ∑ i x i/N, our estimate of the standard error in ˆτ is( ∣ ) −1/2ˆσˆτ = − ∂2 ln L= √∂τ∣∣∣τ=ˆτ ˆτ =1.19,2 Nwhich is precisely the estimate of the standard error we obtained in subsection 31.5.4.It should be noted, however, that in general we would not expect the two estimates ofstandard error made by the different methods to be identical.In order to calculate the Bayesian 68% central confidence interval, we must determinethe values a − and a + that satisfy (31.84) with α = β =0.16. In this case, the calculationcan be performed analytically but is somewhat tedious. It is trivial, however, to determinea − and a + numerically and we find the confidence interval to be [3.16, 6.20]. Thus we canquote our result with 68% central confidence limits asτ =3.77 +2.43−0.61 .1267

STATISTICSBy comparing this result with that given towards the end of subsection 31.5.4, we see that,as we might expect, the Bayesian and classical confidence intervals differ somewhat. ◭The above discussion is generalised straightforwardly to the estimation ofseveral parameters a 1 ,a 2 ,...,a M simultaneously. The elements of the inverse ofthe covariance matrix of the ML estimators can be approximated by∣(V −1 ) ij = − ∂2 ln L ∣∣∣a=â. (31.86)∂a i ∂a jFrom (31.36), we see that (at least for unbiased estimators) the expectation valueof (31.86) is equal to the element F ij of the Fisher matrix.The construction of a multi-dimensional Bayesian confidence region is alsostraightforward. For a given confidence level 1 − α (say), it is most commonto construct the confidence region as the M-dimensional region R in a-space,bounded by the ‘surface’ L(x; a) = constant, for which∫L(x; a) d M a =1− α,Rwhere it is assumed that L(x; a) is normalised to unit volume. Moreover, wesee from (31.83) that (assuming a uniform prior probability) we may obtain themarginal posterior distribution for any parameter a i simply by integrating thelikelihood function L(x; a) over the other parameters:∫ ∫P (a i |x,H)= ··· L(x; a) da 1 ···da i−1 da i+1 ···da M .Here the integral extends over all possible values of the parameters, and againis∫it assumed that the likelihood function is normalised in such a way thatL(x; a) d M a = 1. This marginal distribution can then be used as above todetermine Bayesian confidence intervals on each a i separately.◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with unknown mean µ and standard deviation σ. The sample values are asfollows (to two decimal places):2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Find the Bayesian 95% central confidence intervals on µ and σ separately.The likelihood function in this case is[L(x; µ, σ) =(2πσ 2 ) −N/2 exp − 1N]∑(x2σ 2 i − µ) 2 . (31.87)i=1Assuming uniform priors on µ and σ (over their natural ranges of −∞ → ∞ and 0 → ∞respectively), we may identify this likelihood function with the posterior probability, as in(31.83). Thus, the marginal posterior distribution on µ is given by∫ [∞1P (µ|x,H) ∝σ exp − 1N]∑(x N 2σ 2 i − µ) 2 dσ.01268i=1

31.5 MAXIMUM-LIKELIHOOD METHODBy substituting σ =1/u (so that dσ = −du/u 2 ) and integrating by parts either (N − 2)/2or (N − 3)/2 times, we findP (µ|x,H) ∝ [ N(¯x − µ) 2 + Ns 2] −(N−1)/2,where we have used the fact that ∑ i (x i − µ) 2 = N(¯x − µ) 2 + Ns 2 , ¯x being the sample meanand s 2 the sample variance. We may now obtain the 95% central confidence interval byfinding the values µ − and µ + for which∫ µ−P (µ|x,H) dµ =0.025 and P (µ|x,H) dµ =0.025.−∞µ +The normalisation of the posterior distribution and the values µ − and µ + are easilyobtained by numerical integration. Substituting in the appropriate values N = 10, ¯x =1.11and s =1.01, we find the required confidence interval to be [0.29, 1.97].To obtain a confidence interval on σ, we must first obtain the corresponding marginalposterior distribution. From (31.87), again using the fact that ∑ i (x i−µ) 2 = N(¯x−µ) 2 +Ns 2 ,this is given byP (σ|x,H) ∝ 1 ( ) ∫ ∞]σ exp − Ns2N(¯x − µ)2exp[− dµ.N 2σ 2 −∞ 2σ 2Noting that the integral of a one-dimensional Gaussian is proportional to σ, we concludethatP (σ|x,H) ∝ 1 ( )σ exp − Ns2 .N−1 2σ 2The 95% central confidence interval on σ can then be found in an analogous manner tothat on µ, by solving numerically the equations∫ σ−0P (σ|x,H) dσ =0.025andWe find the required interval to be [0.76, 2.16]. ◭∫ ∞∫ ∞σ +P (σ|x,H) dσ =0.025.31.5.6 Behaviour of ML estimators for large NAs mentioned in subsection 31.3.6, in the large-sample limit N →∞, the samplingdistribution of a set of (consistent) estimators â, whether ML or not, will tend,in general, to a multivariate Gaussian centred on the true values a. Thisisadirect consequence of the central limit theorem. Similarly, in the limit N →∞thelikelihood function L(x; a) also tends towards a multivariate Gaussian but onecentred on the ML estimate(s) â. Thus ML estimators are always asymptoticallyconsistent. This limiting process was illustrated for the one-dimensional case byfigure 31.5.Thus, as N becomes large, the likelihood function tends to the formwhere Q denotes the quadratic formL(x; a) =L max exp [ − 1 2 Q(a, â)] ,Q(a, â) =(a − â) T V −1 (a − â)1269

STATISTICSand the matrix V −1 is given by( ) ∣V−1ij = − ∂2 ln L ∣∣∣a=â.∂a i ∂a jMoreover, in the limit of large N, this matrix tends to the Fisher matrix given in(31.36), i.e. V −1 → F. Hence ML estimators are asymptotically minimum-variance.Comparison of the above results with those in subsection 31.3.6 shows thatthe large-sample limit of the likelihood function L(x; a) has the same form as thelarge-sample limit of the joint estimator sampling distribution P (â|a). The onlydifference is that P (â|a) is centred in â-space on the true values â = a whereasL(x; a) is centred in a-space on the ML estimates a = â. From figure 31.4 and itsaccompanying discussion, we therefore conclude that, in the large-sample limit,the Bayesian and classical confidence limits on the parameters coincide.31.5.7 Extended maximum-likelihood methodIt is sometimes the case that the number of data items N in our sample is itself arandom variable. Such experiments are typically those in which data are collectedfor a certain period of time during which events occur at random in some way,as opposed to those in which a prearranged number of data items are collected.In particular, let us consider the case where the sample values x 1 ,x 2 ,...,x N aredrawn independently from some distribution P (x|a) and the sample size N is arandom variable described by a Poisson distribution with mean λ, i.e.N ∼ Po(λ).The likelihood function in this case is given byL(x,N; λ, a) = λN ∏NN! e−λ P (x i |a), (31.88)i=1and is often called the extended likelihood function. The function L(x; λ, a) canbe used as before to estimate parameter values or obtain confidence intervals.Two distinct cases arise in the use of the extended likelihood function, dependingon whether the Poisson parameter λ is a function of the parameters a or is anindependent parameter.Let us first consider the case in which λ is a function of the parameters a. From(31.88), we can write the extended log-likelihood function asN∑N∑ln L = N ln λ(a) − λ(a)+ ln P (x i |a) =−λ(a)+ ln[λ(a)P (x i |a)].i=1where we have ignored terms not depending on a. The ML estimates â of theparameters can then be found in the usual way, and the ML estimate of thePoisson parameter is simply ˆλ = λ(â). The errors on our estimators â will be, ingeneral, smaller than those obtained in the usual likelihood approach, since ourestimate includes information from the value of N as well as the sample values x i .1270i=1

31.6 THE METHOD OF LEAST SQUARESThe other possibility is that λ is an independent parameter and not a functionof the parameters a. In this case, the extended log-likelihood function isln L = N ln λ − λ +N∑ln P (x i |a), (31.89)where we have omitted terms not depending on λ or a. Differentiating withrespect to λ and setting the result equal to zero, we find that the ML estimate ofλ is simplyˆλ = N.By differentiating (31.89) with respect to the parameters a i and setting the resultsequal to zero, we obtain the usual ML estimates â i of their values. In this case,however, the errors in our estimates will be larger, in general, than those in thestandard likelihood approach, since they must include the effect of statisticaluncertainty in the parameter λ.i=131.6 The method of least squaresThe method of least squares is, in fact, just a special case of the method ofmaximum likelihood. Nevertheless, it is so widely used as a method of parameterestimation that it has acquired a special name of its own. At the outset, let ussuppose that a data sample consists of a set of pairs (x i ,y i ), i =1, 2,...,N.Forexample, these data might correspond to the temperature y i measured at variouspoints x i along some metal rod.For the moment, we will suppose that the x i are known exactly, whereas thereexists a measurement error (or noise) n i on each of the values y i . Moreover, letus assume that the true value of y at any position x is given by some functiony = f(x; a) that depends on the M unknown parameters a. Theny i = f(x i ; a)+n i .Our aim is to estimate the values of the parameters a from the data sample.Bearing in mind the central limit theorem, let us suppose that the n i are drawnfrom a Gaussian distribution with no systematic bias and hence zero mean. In themost general case the measurement errors n i might not be independent but bedescribed by an N-dimensional multivariate Gaussian with non-trivial covariancematrix N, whose elements N ij =Cov[n i ,n j ] we assume to be known. Under theseassumptions it follows from (30.148), that the likelihood function isL(x, y; a) =1(2π) N/2 |N| 1/2 exp [ − 1 2 χ2 (a) ] ,1271

STATISTICSwhere the quantity denoted by χ 2 is given by the quadratic formN∑χ 2 (a) = [y i − f(x i ; a)](N −1 ) ij [y j − f(x j ; a)] = (y − f) T N −1 (y − f).i,j=1(31.90)In the last equality, we have rewritten the expression in matrix notation bydefining the column vector f with elements f i = f(x i ; a). We note that in the(common) special case in which the measurement errors n i are independent, theircovariance matrix takes the diagonal form N = diag(σ1 2,σ2 2 ,...,σ2 N ), where σ i isthe standard deviation of the measurement error n i . In this case, the expression(31.90) for χ 2 reduces toN∑[ ] 2χ 2 yi − f(x i ; a)(a) =.σ ii=1The least-squares (LS) estimators â LS of the parameter values are defined asthose that minimise the value of χ 2 (a); they are usually determined by solvingthe M equations∣∂χ 2 ∣∣∣a=âLS=0 fori =1, 2,...,M. (31.91)∂a iClearly, if the measurement errors n i are indeed Gaussian distributed, as assumedabove, then the LS and ML estimators of the parameters a coincide. Becauseof its relative simplicity, the method of least squares is often applied to cases inwhich the n i are not Gaussian distributed. The resulting estimators â LS are not theML estimators, and the best that can be said in justification is that the method isan obviously sensible procedure for parameter estimation that has stood the testof time.Finally, we note that the method of least squares is easily extended to the casein which each measurement y i depends on several variables, which we denoteby x i . For example, y i might represent the temperature measured at the (threedimensional)position x i in a room. In this case, the data is modelled by afunction y = f(x i ; a), and the remainder of the above discussion carries throughunchanged.31.6.1 Linear least squaresWe have so far made no restriction on the form of the function f(x; a). It sohappens, however, that, for a model in which f(x; a) isalinear function of theparameters a 1 ,a 2 ,...,a M , one can always obtain analytic expressions for the LSestimators â LS and their variances. The general form of this kind of model isM∑f(x; a) = a i h i (x), (31.92)i=11272

31.6 THE METHOD OF LEAST SQUARESwhere {h 1 (x),h 2 (x),...,h M (x)} is some set of linearly independent fixed functionsof x, often called the basis functions. Note that the functions h i (x) themselves maybe highly non-linear functions of x. The ‘linear’ nature of the model (31.92) refersonly to its dependence on the parameters a i . Furthermore, in this case, it maybe shown that the LS estimators â i have zero bias and are minimum-variance,irrespective of the probability density function from which the measurement errorsn i are drawn.In order to obtain analytic expressions for the LS estimators â LS ,itisconvenientto write (31.92) in the formf(x; a) =M∑R ij a j , (31.93)where R ij = h j (x i ) is an element of the response matrix R of the experiment. Theexpression for χ 2 given in (31.90) can then be written, in matrix notation, asj=1χ 2 (a) =(y − Ra) T N −1 (y − Ra). (31.94)The LS estimates of the parameters a are now found, as shown in (31.91), bydifferentiating (31.94) with respect to the a i and setting the resulting expressionsequal to zero. Denoting by ∇χ 2 the vector with elements ∂χ 2 /∂a i , we find∇χ 2 = −2R T N −1 (y − Ra). (31.95)This can be verified by writing out the expression (31.94) in component form anddifferentiating directly.◮Verify result (31.95) by formulating the calculation in component form.To make the derivation less cumbersome, let us adopt the summation convention discussedin section 26.1, in which it is understood that any subscript that appears exactly twice inany term of an expression is to be summed over all the values that a subscript in thatposition can take. Thus, writing (31.94) in component form, we haveχ 2 (a) =(y i − R ik a k )(N −1 ) ij (y j − R jl a l ).Differentiating with respect to a p gives∂χ 2= −R ik δ kp (N −1 ) ij (y j − R jl a l )+(y i − R ik a k )(N −1 ) ij (−R jl δ lp )∂a p= −R ip (N −1 ) ij (y j − R jl a l ) − (y i − R ik a k )(N −1 ) ij R jp , (31.96)where δ ij is the Kronecker delta symbol discussed in section 26.1. By swapping the indicesi and j in the second term on the RHS of (31.96) and using the fact that the matrix N −1is symmetric, we obtain∂χ 2= −2R ip (N −1 ) ij (y j − R jk a k )∂a p= −2(R T ) pi (N −1 ) ij (y j − R jk a k ). (31.97)If we denote the vector with components ∂χ 2 /∂a p , p =1, 2,...,M,by∇χ 2 and write theRHS of (31.97) in matrix notation, we recover the result (31.95). ◭1273

STATISTICSSetting the expression (31.95) equal to zero at a = â, we find−2R T N −1 y +2R T N −1 Râ =0.Provided the matrix R T N −1 R is not singular, we may solve this equation for â toobtainâ =(R T N −1 R) −1 R T N −1 y ≡ Sy, (31.98)thus defining the M×N matrix S. It follows that the LS estimates â i , i =1, 2,...,M,are linear functions of the original measurements y j , j =1, 2,...,N.Moreover,using the error propagation formula (30.141) derived in subsection 30.12.3, wefind that the covariance matrix of the estimators â i is given byV ≡ Cov[â i , â j ]=SNS T =(R T N −1 R) −1 . (31.99)The two equations (31.98) and (31.99) contain the complete method of leastsquares. In particular, we note that, if one calculates the LS estimates using(31.98) then one has already obtained their covariance matrix (31.99).◮Prove result (31.99).Using the definition of S given in (31.98), the covariance matrix (31.99) becomesV = SNS T=[(R T N −1 R) −1 R T N −1 ]N[(R T N −1 R) −1 R T N −1 ] T .Using the result (AB ···C) T = C T ···B T A T for the transpose of a product of matrices andnoting that, for any non-singular matrix, (A −1 ) T =(A T ) −1 we findV =(R T N −1 R) −1 R T N −1 N(N T ) −1 R[(R T N −1 R) T ] −1=(R T N −1 R) −1 R T N −1 R(R T N −1 R) −1=(R T N −1 R) −1 ,where we have also used the fact that N is symmetric and so N T = N. ◭It is worth noting that one may also write the elements of the (inverse)covariance matrix as(V −1 ) ij = 1 ( ∂ 2 χ 2,2 ∂a i ∂a j)a=âwhich is the same as the Fisher matrix (31.36) in cases where the measurementerrors are Gaussian distributed (and so the log-likelihood is ln L = −χ 2 /2). Thisproves,atleastforthiscase,ourearlierstatementthattheLSestimatorsareminimum-variance. In fact, since f(x; a) is linear in the parameters a, onecanwrite χ 2 exactly asχ 2 (a) =χ 2 (â)+ 1 M∑( ∂ 2 χ 2(a i − â i )(a j − â j ),2 ∂a i ∂a ji,j=1)a=âwhich is quadratic in the parameters a i . Hence the form of the likelihood function1274

31.6 THE METHOD OF LEAST SQUARESy7654321001234 5xFigure 31.9 A set of data points with error bars indicating the uncertaintyσ =0.5 onthey-values. The straight line is y = ˆmx + ĉ, where ˆm and ĉ arethe least-squares estimates of the slope and intercept.L ∝ exp(−χ 2 /2) is Gaussian. From the discussions of subsections 31.3.6 and31.5.6, it follows that the ‘surfaces’ χ 2 (a) = c, where c is a constant, boundellipsoidal confidence regions for the parameters a i . The relationship between thevalue of the constant c and the confidence level is given by (31.39).◮An experiment produces the following data sample pairs (x i ,y i ):x i : 1.85 2.72 2.81 3.06 3.42 3.76 4.31 4.47 4.64 4.99y i : 2.26 3.10 3.80 4.11 4.74 4.31 5.24 4.03 5.69 6.57where the x i -values are known exactly but each y i -value is measured only to an accuracyof σ =0.5. Assuming the underlying model for the data to be a straight line y = mx + c,find the LS estimates of the slope m and intercept c and quote the standard error on eachestimate.The data are plotted in figure 31.9, together with error bars indicating the uncertainty inthe y i -values. Our model of the data is a straight line, and so we havef(x; c, m) =c + mx.In the language of (31.92), our basis functions are h 1 (x) =1andh 2 (x) =x and our modelparameters are a 1 = c and a 2 = m. From (31.93) the elements of the response matrix areR ij = h j (x i ), so that⎛ ⎞1 x 11 x 2R = ⎜⎝.⎟. ⎠ , (31.100)1 x Nwhere x i are the data values and N = 10 in our case. Further, since the standard deviationon each measurement error is σ, we have N = σ 2 I,whereI is the N × N identity matrix.Because of this simple form for N, the expression (31.98) for the LS estimates reduces toâ = σ 2 (R T R) −1 1 σ 2 RT y =(R T R) −1 R T y. (31.101)Note that we cannot expand the inverse in the last line, since R itself is not square and1275

STATISTICShence does not possess an inverse. Inserting the form for R in (31.100) into the expression(31.101), we find( ) (ĉ=∑i 1 ∑i x ) −1 ( ∑∑iˆmi x ∑∑i y )ii i x2 ii x iy i( )( )1=x2−¯x Nȳ.N(x 2 − ¯x 2 ) −¯x 1 NxyWe thus obtain the LS estimatesxy − ¯x ȳˆm = and ĉ = x2 ȳ − ¯x xy= ȳ − ˆm¯x, (31.102)x 2 − ¯x 2 x 2 − ¯x 2where the last expression for ĉ shows that the best-fit line passes through the ‘centreof mass’ (¯x, ȳ) of the data sample. To find the standard errors on our results, we mustcalculate the covariance matrix of the estimators. This is given by (31.99), which in ourcase reduces toσ 2V = σ 2 (R T R) −1 =N(x 2 − ¯x 2 )( )x2−¯x. (31.103)−¯x 1The standard error on each estimator is simply the positive square root of the correspondingdiagonal element, i.e. σĉ = √ V 11 and σ ˆm = √ V 22 , and the covariance of the estimators ˆmand ĉ is given by Cov[ĉ, ˆm] =V 12 = V 21 . Inserting the data sample averages and momentsinto (31.102) and (31.103), we findc = ĉ ± σĉ =0.40 ± 0.62 and m = ˆm ± σ ˆm =1.11 ± 0.17.The ‘best-fit’ straight line y = ˆmx + ĉ is plotted in figure 31.9. For comparison, the truevalues used to create the data were m =1andc =1.◭The extension of the method to fitting data to a higher-order polynomial, suchas f(x; a) =a 1 + a 2 x + a 3 x 2 , is obvious. However, as the order of the polynomialincreases the matrix inversions become rather complicated. Indeed, even when thematrices are inverted numerically, the inversion is prone to numerical instabilities.A better approach is to replace the basis functions h m (x) =x m , m =1, 2,...,M,with a set of polynomials that are ‘orthogonal over the data’, i.e. such thatN∑h l (x i )h m (x i )=0 forl ≠ m.i=1Such a set of polynomial basis functions can always be found by using the Gram–Schmidt orthogonalisation procedure presented in section 17.1. The details of thisapproach are beyond the scope of our discussion but we note that, in this case,the matrix R T R is diagonal and may be inverted easily.31.6.2 Non-linear least squaresIf the function f(x; a) isnot linear in the parameters a then, in general, it isnot possible to obtain an explicit expression for the LS estimates â. Instead,onemust use an iterative (numerical) procedure, which we now outline. In practice,1276

31.7 HYPOTHESIS TESTINGhowever, such problems are best solved using one of the many commerciallyavailable software packages.One begins by making a first guess a 0 for the values of the parameters. At thispoint in parameter space, the components of the gradient ∇χ 2 will not be equalto zero, in general (unless one makes a very lucky guess!). Thus, for at least somevalues of i, we have∣∂χ 2 ∣∣∣a=a≠0.∂a i 0Our aim is to find a small increment δa in the values of the parameters, such that∣∂χ 2 ∣∣∣a=a0= 0 for all i. (31.104)∂a i +δaIf our first guess a 0 were sufficiently close to the true (local) minimum of χ 2 ,we could find the required increment δa by expanding the LHS of (31.104) as aTaylor series about a = a 0 , keeping only the zeroth-order and first-order terms:∣ ∣∂χ 2 ∣∣∣a=a0≈ ∂χ2 ∣∣∣a=a+∂a i +δa∂a i 0M∑j=1∣∂ 2 χ 2 ∣∣∣a=aδa j . (31.105)∂a i ∂a j 0Setting this expression to zero, we find that the increments δa j may be found bysolving the set of M linear equationsM∑j=1∣ ∣∂ 2 χ 2 ∣∣∣a=aδa j = − ∂χ2 ∣∣∣a=a.∂a i ∂a j 0 ∂a i 0It most cases, however, our first guess a 0 will not be sufficiently close to the trueminimum for (31.105) to be an accurate approximation, and consequently (31.104)will not be satisfied. In this case, a 1 = a 0 + δa is (hopefully) an improved guessat the parameter values; the whole process is then repeated until convergence isachieved.It is worth noting that, when one is estimating several parameters a, thefunction χ 2 (a) maybevery complicated. In particular, it may possess numerouslocal extrema. The procedure outlined above will converge to the local extremum‘nearest’ to the first guess a 0 . Since, in fact, we are interested only in the localminimum that has the absolute lowest value of χ 2 (a), it is clear that a large partof solving the problem is to make a ‘good’ first guess.31.7 Hypothesis testingSo far we have concentrated on using a data sample to obtain a number or a setof numbers. These numbers may be estimated values for the moments or centralmoments of the population from which the sample was drawn or, more generally,the values of some parameters a in an assumed model for the data. Sometimes,1277

STATISTICShowever, one wishes to use the data to give a ‘yes’ or ‘no’ answer to a particularquestion. For example, one might wish to know whether some assumed modeldoes, in fact, provide a good fit to the data, or whether two parameters have thesame value.31.7.1 Simple and composite hypothesesIn order to use data to answer questions of this sort, the question must beposed precisely. This is done by first asserting that some hypothesis is true.The hypothesis under consideration is traditionally called the null hypothesisand is denoted by H 0 . In particular, this usually specifies some form P (x|H 0 )for the probability density function from which the data x are drawn. If thehypothesis determines the PDF uniquely, then it is said to be a simple hypothesis.If, however, the hypothesis determines the functional form of the PDF but not thevalues of certain parameters a on which it depends then it is called a compositehypothesis.One decides whether to accept or reject the null hypothesis H 0 by performingsome statistical test, as described below in subsection 31.7.2. In fact, formallyone uses a statistical test to decide between the null hypothesis H 0 and thealternative hypothesis H 1 . We define the latter to be the complement H 0 of thenull hypothesis within some restricted hypothesis space known (or assumed) inadvance. Hence, rejection of H 0 implies acceptance of H 1 , and vice versa.As an example, let us consider the case in which a sample x is drawn from aGaussian distribution with a known variance σ 2 but with an unknown mean µ.If one adopts the null hypothesis H 0 that µ = 0, which we write as H 0 : µ =0,then the corresponding alternative hypothesis must be H 1 : µ ≠ 0. Note that,in this case, H 0 is a simple hypothesis whereas H 1 is a composite hypothesis.If, however, one adopted the null hypothesis H 0 : µ

31.7 HYPOTHESIS TESTINGP (t|H 0 )αt crittP (t|H 1 )βt crittFigure 31.10 The sampling distributions P (t|H 0 )andP (t|H 1 ) of a test statistict. The shaded areas indicate the (one-tailed) regions for which Pr(t >t crit |H 0 )=α and Pr(t t crit |H 0 )= P (t|H 0 ) dt = α; (31.106)t critthis is indicated by the shaded region in the upper half of figure 31.10. Equally,a (one-tailed) rejection region could consist of values of t less than some valuet crit . Alternatively, one could define a (two-tailed) rejection region by two valuest 1 and t 2 such that Pr(t 1

STATISTICSfalse (in which case H 1 is true). The probability β (say)thatsuchanerrorwilloccur is, in general, difficult to calculate, since the alternative hypothesis H 1 isoften composite. Nevertheless, in the case where H 1 is a simple hypothesis, it isstraightforward (in principle) to calculate β. Denoting the corresponding samplingdistribution of t by P (t|H 1 ), the probability β is the integral of P (t|H 1 )overthecomplement of the rejection region, called the acceptance region. For example, inthe case corresponding to (31.106) this probability is given byβ =Pr(tc, (31.107)P (t|H 1 )where c is some constant determined by the required significance level.In the case where the test statistic t is a simple scalar quantity, the Neyman–Pearson lemma is also useful in deciding which such statistic is the ‘best’ inthe sense of having the maximum power for a given significance level α. From(31.107), we can see that the best statistic is given by the likelihood ratiot(x) = P (x|H 0)P (x|H 1 ) . (31.108)and that the corresponding rejection region for H 0 is given by t

31.7 HYPOTHESIS TESTING◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with standard deviation σ =1. The mean µ of the distribution is known toequal either zero or unity. The sample values are as follows:2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Test the null hypothesis H 0 : µ =0at the 10% significance level.The restricted nature of the hypothesis space means that our null and alternative hypothesesare H 0 : µ =0andH 1 : µ = 1 respectively. Since H 0 and H 1 are both simple hypotheses,the best test statistic is given by the likelihood ratio (31.108). Thus, denoting the meansby µ 0 and µ 1 , we havet(x) = exp [ − 1 2∑i (x i − µ 0 ) 2]exp [ ∑− 1 2 i (x i − µ 1 ) 2] = exp [ − 1 2∑i (x2 i − 2µ 0 x i + µ 2 0 )]exp [ − 1 2∑i (x2 i− 2µ 1 x i + µ 2 1 )]=exp [ (µ 0 − µ 1 ) ∑ i x i − 1 2 N(µ2 0 − µ 2 1) ] .Inserting the values µ 0 =0andµ 1 = 1, yields t =exp(−N¯x + 1 N), where ¯x is the2sample mean. Since − ln t is a monotonically decreasing function of t, however, we mayequivalently use as our test statisticv = − 1 N ln t + 1 2 = ¯x,where we have divided by the sample size N and added 1 for convenience. Thus we2may take the sample mean as our test statistic. From (31.13), we know that the samplingdistribution of the sample mean under our null hypothesis H 0 is the Gaussian distributionN(µ 0 ,σ 2 /N), where µ 0 =0,σ 2 =1andN = 10. Thus ¯x ∼ N(0, 0.1).Since ¯x is a monotonically decreasing function of t, our best rejection region for a givensignificance α is ¯x >¯x crit ,where¯x crit depends on α. Thus, in our case, ¯x crit is given by( )¯xcrit − µ 0α =1− Φ=1− Φ(10¯x crit ),σwhere Φ(z) is the cumulative distribution function for the standard Gaussian. For a 10%significance level we have α =0.1 and, from table 30.3 in subsection 30.9.1, we find¯x crit =0.128. Thus the rejection region on ¯x is¯x >0.128.From the sample, we deduce that ¯x =1.11, and so we can clearly reject the null hypothesisH 0 : µ = 0 at the 10% significance level. It can, in fact, be rejected at a much highersignificance level. As revealed on p. 1239, the data was generated using µ =1.◭31.7.4 The generalised likelihood-ratio testIf the null hypothesis H 0 or the alternative hypothesis H 1 is composite (or bothare composite) then the corresponding distributions P (x|H 0 )andP (x|H 1 )arenot uniquely determined, in general, and so we cannot use the Neyman–Pearsonlemma to obtain the ‘best’ test statistic t. Nevertheless, in many cases, there stillexists a general procedure for constructing a test statistic t which has useful1281

STATISTICSproperties and which reduces to the Neyman–Pearson statistic (31.108) in thespecial case where H 0 and H 1 are both simple hypotheses.Consider the quite general, and commonly occurring, case in which thedata sample x is drawn from a population P (x|a) with a known (or assumed)functional form but depends on the unknown values of some parametersa 1 ,a 2 ,...,a M . Moreover, suppose we wish to test the null hypothesis H 0 thatthe parameter values a lie in some subspace S of the full parameter spaceA. In other words, on the basis of the sample x it is desired to test thenull hypothesis H 0 :(a 1 ,a 2 ,...,a M lies in S) against the alternative hypothesisH 1 :(a 1 ,a 2 ,...,a M lies in S), where S is A − S.Since the functional form of the population is known, we may write down thelikelihood function L(x; a) for the sample. Ordinarily, the likelihood will havea maximum as the parameters a are varied over the entire parameter space A.This is the usual maximum-likelihood estimate of the parameter values, whichwe denote by â. If, however, the parameter values are allowed to vary only overthe subspace S then the likelihood function will be maximised at the point â S ,which may or may not coincide with the global maximum â. Now, let us take asour test statistic the generalised likelihood ratiot(x) = L(x; â S)L(x; â) , (31.109)where L(x; â S ) is the maximum value of the likelihood function in the subspaceS and L(x; â) is its maximum value in the entire parameter space A. Itisclearthat t is a function of the sample values only and must lie between 0 and 1.We will concentrate on the special case where H 0 is the simple hypothesisH 0 : a = a 0 . The subspace S then consists of only the single point a 0 . Thus(31.109) becomest(x) = L(x; a 0)L(x; â) , (31.110)and the sampling distribution P (t|H 0 ) can be determined (in principle). As in theprevious subsection, the best rejection region for a given significance α is simplyt

31.7 HYPOTHESIS TESTING◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with standard deviation σ =1. The sample values are as follows:2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Test the null hypothesis H 0 : µ =0at the 10% significance level.We must test the (simple) null hypothesis H 0 : µ = 0 against the (composite) alternativehypothesis H 1 : µ ≠ 0. Thus, the subspace S is the single point µ =0,whereasA is theentire µ-axis. The likelihood function is1L(x; µ) =(2π) exp [ ∑− 1 N/2 2 i (x i − µ) 2] ,which has its global maximum at µ = ¯x. The test statistic t is then given byt(x) = L(x;0)L(x; ¯x) = exp [ ∑ ]− 1 2 i x2 iexp [ − 1 2∑i (x i − ¯x) 2] =exp( − 1 N¯x2) .2It is in fact more convenient to consider the test statisticv = −2lnt = N¯x 2 .Since −2lnt is a monotonically decreasing function of t, the rejection region now becomesv>v crit ,where∫ ∞P (v|H 0 ) dv = α, (31.111)v critα being the significance level of the test. Thus it only remains to determine the samplingdistribution P (v|H 0 ). Under the null hypothesis H 0 ,weexpect¯x to be Gaussian distributed,with mean zero and variance 1/N. Thus, from subsection 30.9.4, v will follow a chi-squareddistribution of order 1. Substituting the appropriate form for P (v|H 0 ) in (31.111) and settingα =0.1, we find by numerical integration (or from table 31.2) that v crit = N¯x 2 crit =2.71.Since N = 10, the rejection region on ¯x at the 10% significance level is thus¯x 0.52.As noted before, for this sample ¯x =1.11, and so we may reject the null hypothesisH 0 : µ = 0 at the 10% significance level. ◭The above example illustrates the general situation that if the maximumlikelihoodestimates â of the parameters fall in or near the subspace S then thesample will be considered consistent with H 0 and the value of t will be nearunity. If â is distant from S then the sample will not be in accord with H 0 andordinarily t will have a small (positive) value.It is clear that in order to prescribe the rejection region for t, orforarelatedstatistic u or v, it is necessary to know the sampling distribution P (t|H 0 ). If H 0is simple then one can in principle determine P (t|H 0 ), although this may provedifficult in practice. Moreover, if H 0 is composite, then it may not be possibleto obtain P (t|H 0 ), even in principle. Nevertheless, a useful approximate form forP (t|H 0 ) exists in the large-sample limit. Consider the null hypothesisand the a 0 iH 0 :(a 1 = a 0 1,a 2 = a 0 2,...,a R = a 0 R),where R ≤ Mare fixed numbers. (In fact, we may fix the values of any subset1283

STATISTICScontaining R of the M parameters.) If H 0 is true then it follows from ourdiscussion in subsection 31.5.6 (although we shall not prove it) that, when thesample size N is large, the quantity −2lnt follows approximately a chi-squareddistribution of order R.31.7.5 Student’s t-testStudent’s t-test is just a special case of the generalised likelihood ratio test appliedto a sample x 1 ,x 2 ,...,x N drawn independently from a Gaussian distribution forwhich both the mean µ and variance σ 2 are unknown, and for which one wishesto distinguish between the hypothesesH 0 : µ = µ 0 , 0

31.7 HYPOTHESIS TESTINGThe sum of squares in the denominator of (31.112) may be put into the form∑i (x i − µ 0 ) 2 = N(¯x − µ 0 ) 2 + ∑ i (x i − ¯x) 2 .Thus, on dividing the numerator and denominator in (31.112) by ∑ i (x i − ¯x) 2 andrearranging, the generalised likelihood ratio λ can be written) −N/2λ =(1+ t2,N − 1where we have defined the new variablet = ¯x − µ 0s/ √ N − 1 . (31.113)Since t 2 is a monotonically decreasing function of λ, the corresponding rejectionregion is t 2 > c, where c is a positive constant depending on the requiredsignificance level α. It is conventional, however, to use t itself as our test statistic,in which case our rejection region becomes two-tailed and is given bytt crit , (31.114)where t crit is the positive square root of the constant c.The definition (31.113) and the rejection region (31.114) form the basis ofStudent’s t-test. It only remains to determine the sampling distribution P (t|H 0 ).At the outset, it is worth noting that if we write the expression (31.113) for tin terms of the standard estimator ˆσ = √ Ns 2 /(N − 1) of the standard deviationthen we obtaint = ¯x − µ 0ˆσ/ √ N . (31.115)If, in fact, we knew the true value of σ anduseditinthisexpressionfort thenit is clear from our discussion in section 31.3 that t would follow a Gaussiandistribution with mean 0 and variance 1, i.e. t ∼ N(0, 1). When σ is not known,however, we have to use our estimate ˆσ in (31.115), with the result that t isno longer distributed as the standard Gaussian. As one might expect fromthe central limit theorem, however, the distribution of t does tend towards thestandard Gaussian for large values of N.As noted earlier, the exact distribution of t, valid for any value of N, wasfirstdiscovered by William Gossett. From (31.35), if the hypothesis H 0 is true then thejoint sampling distribution of ¯x and s is given by)P (¯x, s|H 0 )=Cs N−2 exp(− Ns22σ 2 exp[− N(¯x − µ 0) 2 ]2σ 2 ,(31.116)where C is a normalisation constant. We can use this result to obtain the jointsampling distribution of s and t by demanding thatP (¯x, s|H 0 ) d¯xds= P (t, s|H 0 ) dt ds.1285

STATISTICSUsing (31.113) to substitute for ¯x − µ 0 in (31.116), and noting that d¯x =(s/ √ N − 1) dt, we find( )]P (¯x, s|H 0 ) d¯xds= As N−1 exp[− Ns22σ 2 1+ t2dt ds,N − 1where A is another normalisation constant. In order to obtain the samplingdistribution of t alone, we must integrate P (t, s|H 0 ) with respect to s over itsallowed range, from 0 to ∞. Thus, the required distribution of t alone is given by∫ ∞∫ ∞( )]P (t|H 0 )= P (t, s|H 0 ) ds = A s N−1 exp[− Ns2002σ 2 1+ t2ds.N − 1(31.117)To carry out this integration, we set y = s{1+[t 2 /(N − 1)]} 1/2 , which on substitutioninto (31.117) yields) −N/2 ∫ ∞)P (t|H 0 )=A(1+ t2y N−1 exp(− Ny2N − 1 02σ 2 dy.Since the integral over y does not depend on t, it is simply a constant. We thusfind that that the sampling distribution of the variable t is1 Γ ( 12P (t|H 0 )= √ N) ((N − 1)π Γ ( 12 (N − 1))1+ t2N − 1) −N/2,(31.118)wherewehaveusedthecondition ∫ ∞−∞ P (t|H 0) dt = 1 to determine the normalisationconstant (see exercise 31.18).The distribution (31.118) is called Student’s t-distribution with N − 1 degrees offreedom. A plot of Student’s t-distribution is shown in figure 31.11 for variousvalues of N. For comparison, we also plot the standard Gaussian distribution,to which the t-distribution tends for large N. As is clear from the figure, thet-distribution is symmetric about t = 0. In table 31.3 we list some critical pointsof the cumulative probability function C n (t) ofthet-distribution, which is definedby∫ tC n (t) = P (t ′ |H 0 ) dt ′ ,−∞where n = N − 1 is the number of degrees of freedom. Clearly, C n (t) is analogousto the cumulative probability function Φ(z) of the Gaussian distribution, discussedin subsection 30.9.1. For comparison purposes, we also list the critical points ofΦ(z), which corresponds to the t-distribution for N = ∞.1286

31.7 HYPOTHESIS TESTINGP (t|H 0 )0.50.40.30.2N =3N =2N =10N =50.10−4 −3 −2 −1 0 1 2 3 4tFigure 31.11 Student’s t-distribution for various values of N. The brokencurve shows the standard Gaussian distribution for comparison.◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with unknown mean µ and unknown standard deviation σ. The sample valuesare as follows:2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Test the null hypothesis H 0 : µ =0at the 10% significance level.For our null hypothesis, µ 0 = 0. Since for this sample ¯x =1.11, s =1.01 and N = 10, itfollows from (31.113) that¯xt =s/ √ N − 1 =3.33.The rejection region for t is given by (31.114) where t crit is such thatC N−1 (t crit )=1− α/2,and α is the required significance of the test. In our case α =0.1 andN = 10, and fromtable 31.3 we find t crit =1.83. Thus our rejection region for H 0 at the 10% significancelevel ist1.83.For our sample t =3.30 and so we can clearly reject the null hypothesis H 0 : µ =0atthislevel. ◭It is worth noting the connection between the t-test and the classical confidenceinterval on the mean µ. The central confidence interval on µ at the confidencelevel 1 − α is the set of values for which−t crit < ¯x − µs/ √ N − 1

STATISTICSC n (t) 0.5 0.6 0.7 0.8 0.9 0.950 0.975 0.990 0.995 0.999n =1 0.00 0.33 0.73 1.38 3.08 6.31 12.7 31.8 63.7 318.32 0.00 0.29 0.62 1.06 1.89 2.92 4.30 6.97 9.93 22.33 0.00 0.28 0.58 0.98 1.64 2.35 3.18 4.54 5.84 10.24 0.00 0.27 0.57 0.94 1.53 2.13 2.78 3.75 4.60 7.175 0.00 0.27 0.56 0.92 1.48 2.02 2.57 3.37 4.03 5.896 0.00 0.27 0.55 0.91 1.44 1.94 2.45 3.14 3.71 5.217 0.00 0.26 0.55 0.90 1.42 1.90 2.37 3.00 3.50 4.798 0.00 0.26 0.55 0.89 1.40 1.86 2.31 2.90 3.36 4.509 0.00 0.26 0.54 0.88 1.38 1.83 2.26 2.82 3.25 4.3010 0.00 0.26 0.54 0.88 1.37 1.81 2.23 2.76 3.17 4.1411 0.00 0.26 0.54 0.88 1.36 1.80 2.20 2.72 3.11 4.0312 0.00 0.26 0.54 0.87 1.36 1.78 2.18 2.68 3.06 3.9313 0.00 0.26 0.54 0.87 1.35 1.77 2.16 2.65 3.01 3.8514 0.00 0.26 0.54 0.87 1.35 1.76 2.15 2.62 2.98 3.7915 0.00 0.26 0.54 0.87 1.34 1.75 2.13 2.60 2.95 3.7316 0.00 0.26 0.54 0.87 1.34 1.75 2.12 2.58 2.92 3.6917 0.00 0.26 0.53 0.86 1.33 1.74 2.11 2.57 2.90 3.6518 0.00 0.26 0.53 0.86 1.33 1.73 2.10 2.55 2.88 3.6119 0.00 0.26 0.53 0.86 1.33 1.73 2.09 2.54 2.86 3.5820 0.00 0.26 0.53 0.86 1.33 1.73 2.09 2.53 2.85 3.5525 0.00 0.26 0.53 0.86 1.32 1.71 2.06 2.49 2.79 3.4630 0.00 0.26 0.53 0.85 1.31 1.70 2.04 2.46 2.75 3.3940 0.00 0.26 0.53 0.85 1.30 1.68 2.02 2.42 2.70 3.3150 0.00 0.26 0.53 0.85 1.30 1.68 2.01 2.40 2.68 3.26100 0.00 0.25 0.53 0.85 1.29 1.66 1.98 2.37 2.63 3.17200 0.00 0.25 0.53 0.84 1.29 1.65 1.97 2.35 2.60 3.13∞ 0.00 0.25 0.52 0.84 1.28 1.65 1.96 2.33 2.58 3.09Table 31.3 The confidence limits t of the cumulative probability functionC n (t) for Student’s t-distribution with n degrees of freedom. For example,C 5 (0.92) = 0.8. The row n = ∞ is also the corresponding result for thestandard Gaussian distribution.where t crit satisfies C N−1 (t crit )=α/2. Thus the required confidence interval is¯x −√ t crits< µ < ¯x + √ t crits.N − 1 N − 1Hence, in the above example, the 90% classical central confidence interval on µis0.49

31.7 HYPOTHESIS TESTINGdistributions. In particular, let us consider the case where we have two independentsamples of sizes N 1 and N 2 , drawn respectively from Gaussian distributions witha common variance σ 2 but with possibly different means µ 1 and µ 2 . On the basisof the samples, one wishes to distinguish between the hypothesesH 0 : µ 1 = µ 2 , 0

STATISTICSwhere α is the required significance level of the test. In our case we set α =0.05, and fromtable 31.3 with n = 16 we find that t crit =2.12. The rejection region is thereforet2.12.Since t = −2.17 for our samples, we can reject the null hypothesis H 0 : µ 1 = µ 2 , althoughonly by a small margin. (Indeed, it is easily shown that one cannot reject H 0 at the 2%significance level). The 95% central confidence interval on ω = µ 1 − µ 2 is given by( ) 1/2 ( ) 1/2 N1 + N 2N1 + N 2¯w − ˆσt crit < ω < ¯w + ˆσt crit ,N 1 N 2 N 1 N 2where t crit is given by (31.120). Thus, we find−26.1

31.7 HYPOTHESIS TESTINGλ(u)0.100.0500λ crita b10 20 30 40uFigure 31.12 The sampling distribution P (u|H 0 )forN = 10; this is a chisquareddistribution for N − 1 degrees of freedom.distribution with unknown µ and σ, and we wish to distinguish between the twohypothesesH 0 : σ 2 = σ 2 0, −∞

STATISTICShowever, it is difficult to determine a and b for a given significance level α, soaslightly different rejection region, which we now describe, is usually adopted.The sampling distribution P (u|H 0 ) may be found straightforwardly from thesampling distribution of s given in (31.35). Let us first determine P (s 2 |H 0 )bydemanding thatfrom which we findP (s 2 |H 0 )= P (s|H 0)2sP (s|H 0 ) ds = P (s 2 |H 0 ) d(s 2 ),( ) (N−1)/2 N (s 2 ) (N−3)/2 ( )=2σ02 Γ ( 12 (N − 1)) exp − Ns22σ02 . (31.122)Thus, the sampling distribution of u = Ns 2 /σ0 2 is given byP (u|H 0 )=12 (N−1)/2 Γ ( 12 (N − 1))u(N−3)/2 exp ( − 1 2 u) .We note, in passing, that the distribution of u is precisely that of an (N − 1) thorderchi-squared variable (see subsection 30.9.4), i.e. u ∼ χ 2 N−1 . Although it doesnot give quite the best test, one then takes the rejection region to be0

31.7 HYPOTHESIS TESTINGWe now turn to Fisher’s F-test. Let us suppose that two independent samplesof sizes N 1 and N 2 are drawn from Gaussian distributions with means andvariances µ 1 ,σ1 2 and µ 2,σ2 2 respectively, and we wish to distinguish between thetwo hypothesesH 0 : σ 2 1 = σ 2 2 and H 1 : σ 2 1 ≠ σ 2 2.In this case, the generalised likelihood ratio is found to beλ = (N [1 + N 2 ) (N1+N2)/2 F(N1 − 1)/(N 2 − 1) ] N 1/2N N1/21N N2/2 [2 1+F(N1 − 1)/(N 2 − 1) ] ,(N 1+N 2)/2where F is given by the variance ratioF = N 1s 2 1 /(N 1 − 1)N 2 s 2 2 /(N 2 − 1) ≡ u2v 2 (31.123)and s 1 and s 2 are the standard deviations of the two samples. On plotting λ as afunction of F, it is apparent that the rejection region λ

STATISTICSdistribution P (v 2 ,F|H 0 ) is obtained by requiring thatP (v 2 ,F|H 0 ) d(v 2 ) dF = P (u 2 |H 0 )P (v 2 |H 0 ) d(u 2 ) d(v 2 ).(31.125)In order to find the distribution of F alone, we now integrate P (v 2 ,F|H 0 ) withrespect to v 2 from 0 to ∞, from which we obtainP (F|H 0 )(N1 − 1=N 2 − 1) (N1−1)/2B ( 1F (N1−3)/2 (2 (N 1 − 1), 1 2 (N 2 − 1) )1+ N ) −(N1+N1 − 12−2)/2N 2 − 1 F ,(31.126)where B ( 12 (N 1 − 1), 1 2 (N 2 − 1) ) is the beta function defined in the Appendix.P (F|H 0 ) is called the F-distribution (or occasionally the Fisher distribution) with(N 1 − 1, N 2 − 1) degrees of freedom.◮Evaluate the integral ∫ ∞0 P (v2 ,F|H 0 ) d(v 2 ) to obtain result (31.126).From (31.125), we haveP (F|H 0 )=AF (N 1−3)/2∫ ∞0(v 2 ) (N 1+N 2 −4)/2 exp{− [(N }1 − 1)F +(N 2 − 1)]v 2d(v 2 ).2σ 2Making the substitution x =[(N 1 − 1)F +(N 2 − 1)]v 2 /(2σ 2 ), we obtain[P (F|H 0 )=A2σ 2(N 1 − 1)F +(N 2 − 1)[2σ 2= A(N 1 − 1)F +(N 2 − 1)] (N1 +N 2 −2)/2F (N 1−3)/2∫ ∞0x (N 1+N 2 −4)/2 e −x dx] (N1 +N 2 −2)/2F (N 1−3)/2 Γ ( 12 (N 1 + N 2 − 2) ) ,where in the last line we have used the definition of the gamma function given in theAppendix. Using the further result (18.165), which expresses the beta function in terms ofthe gamma function, and the expression for A given in (31.124), we see that P (F|H 0 )isindeed given by (31.126). ◭As it does not matter whether the ratio F given in (31.123) is defined as u 2 /v 2or as v 2 /u 2 , it is conventional to put the larger sample variance on the top, sothat F is always greater than or equal to unity. A large value of F indicates thatthe sample variances u 2 and v 2 are very different whereas a value of F close tounity means that they are very similar. Therefore, for a given significance α, itis1294

31.7 HYPOTHESIS TESTINGC n1 ,n 2(F) n 1 = 1 2 3 4 5 6 7 8n 2 =1 161 200 216 225 230 234 237 2392 18.5 19.0 19.2 19.2 19.3 19.3 19.4 19.43 10.1 9.55 9.28 9.12 9.01 8.94 8.89 8.854 7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.045 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.826 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.157 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.738 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.449 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.2310 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.0720 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.4530 4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.2740 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.1850 4.03 3.18 2.79 2.56 2.40 2.29 2.20 2.13100 3.94 3.09 2.70 2.46 2.31 2.19 2.10 2.03∞ 3.84 3.00 2.60 2.37 2.21 2.10 2.01 1.94n 1 = 9 10 20 30 40 50 100 ∞n 2 =1 241 242 248 250 251 252 253 2542 19.4 19.4 19.4 19.5 19.5 19.5 19.5 19.53 8.81 8.79 8.66 8.62 8.59 8.58 8.55 8.534 6.00 5.96 5.80 5.75 5.72 5.70 5.66 5.635 4.77 4.74 4.56 4.50 4.46 4.44 4.41 4.376 4.10 4.06 3.87 3.81 3.77 3.75 3.71 3.677 3.68 3.64 3.44 3.38 3.34 3.32 3.27 3.238 3.39 3.35 3.15 3.08 3.04 3.02 2.97 2.939 3.18 3.14 2.94 2.86 2.83 2.80 2.76 2.7110 3.02 2.98 2.77 2.70 2.66 2.64 2.59 2.5420 2.39 2.35 2.12 2.04 1.99 1.97 1.91 1.8430 2.21 2.16 1.93 2.69 1.79 1.76 1.70 1.6240 2.12 2.08 1.84 1.74 1.69 1.66 1.59 1.5150 2.07 2.03 1.78 1.69 1.63 1.60 1.52 1.44100 1.97 1.93 1.68 1.57 1.52 1.48 1.39 1.28∞ 1.88 1.83 1.57 1.46 1.39 1.35 1.24 1.00Table 31.4 Values of F for which the cumulative probability function C n1 ,n 2(F)of the F-distribution with (n 1 ,n 2 ) degrees of freedom has the value 0.95. Forexample, for n 1 =10andn 2 =6,C n1 ,n 2(4.06) = 0.95.customary to define the rejection region on F as F>F crit ,whereC n1,n 2(F crit )=∫ Fcrit1P (F|H 0 ) dF = α,and n 1 = N 1 − 1 and n 2 = N 2 − 1 are the numbers of degrees of freedom.Table 31.4 lists values of F crit corresponding to the 5% significance level (i.e.α =0.05) for various values of n 1 and n 2 .1295

STATISTICS◮Suppose that two classes of students take the same mathematics examination and thefollowing percentage marks are obtained:Class 1: 66 62 34 55 77 80 55 60 69 47 50Class 2: 64 90 76 56 81 72 70Assuming that the two sets of examinations marks are drawn from Gaussian distributions,test the hypothesis H 0 : σ1 2 = σ2 2at the 5% significance level.The variances of the two samples are s 2 1 =(12.8)2 and s 2 2 =(10.3)2 and the sample sizesare N 1 =11andN 2 = 7. Thus, we haveu 2 = N 1s 2 1N 1 − 1 = 180.2 and v2 = N 2s 2 2N 2 − 1 = 123.8,where we have taken u 2 to be the larger value. Thus, F = u 2 /v 2 =1.46 to two decimalplaces. Since the first sample contains eleven values and the second contains seven values,we take n 1 =10andn 2 = 6. Consulting table 31.4, we see that, at the 5% significancelevel, F crit =4.06. Since our value lies comfortably below this, we conclude that there isno statistical evidence for rejecting the hypothesis that the two samples were drawn fromGaussian distributions with a common variance. ◭It is also common to define the variable z = 1 2ln F, the distribution of whichcan be found straightfowardly from (31.126). This is a useful change of variablesince it can be shown that, for large values of n 1 and n 2 , the variable z isdistributed approximately as a Gaussian with mean 1 2 (n−1 2 − n −11 ) and variance12 (n−1 2 + n −11 ). 31.7.7 Goodness of fit in least-squares problemsWe conclude our discussion of hypothesis testing with an example of a goodnessof-fittest. In section 31.6, we discussed the use of the method of least squares inestimating the best-fit values of a set of parameters a in a given model y = f(x; a)for a data set (x i ,y i ), i =1, 2,...,N. We have not addressed, however, the questionof whether the best-fit model y = f(x; â) does, in fact, provide a good fit to thedata. In other words, we have not considered thus far how to verify that thefunctional form f of our assumed model is indeed correct. In the language ofhypothesis testing, we wish to distinguish between the two hypothesesH 0 : model is correct and H 1 : model is incorrect.Given the vague nature of the alternative hypothesis H 1 , we clearly cannot usethe generalised likelihood-ratio test. Nevertheless, it is still possible to test thenull hypothesis H 0 at a given significance level α.The least-squares estimates of the parameters â 1 , â 2 ,...,â M , as discussed insection 31.6, are those values that minimise the quantityN∑χ 2 (a) = [y i − f(x i ; a)](N −1 ) ij [y j − f(x j ; a)] = (y − f) T N −1 (y − f).i,j=11296

31.7 HYPOTHESIS TESTINGIn the last equality, we rewrote the expression in matrix notation by defining thecolumn vector f with elements f i = f(x i ; a). The value χ 2 (â) at this minimum canbe used as a statistic to test the null hypothesis H 0 , as follows. The N quantitiesy i − f(x i ; a) are Gaussian distributed. However, provided the function f(x j ; a) islinear in the parameters a, the equations (31.98) that determine the least-squaresestimate â constitute a set of M linear constraints on these N quantities. Thus,as discussed in subsection 30.15.2, the sampling distribution of the quantity χ 2 (â)will be a chi-squared distribution with N − M degrees of freedom (d.o.f), which hasthe expectation value and varianceE[χ 2 (â)] = N − M and V [χ 2 (â)] = 2(N − M).Thus we would expect the value of χ 2 (â) to lie typically in the range (N − M) ±√ 2(N − M). A value lying outside this range may suggest that the assumed modelfor the data is incorrect. A very small value of χ 2 (â) is usually an indication thatthe model has too many free parameters and has ‘over-fitted’ the data. Morecommonly, the assumed model is simply incorrect, and this usually results in avalue of χ 2 (â) that is larger than expected.One can choose to perform either a one-tailed or a two-tailed test on thevalue of χ 2 (â). It is usual, for a given significance level α, to define the one-tailedrejection region to be χ 2 (â) >k,wheretheconstantk satisfies∫ ∞kP (χ 2 n ) dχ2 n = α (31.127)and P (χ 2 n) is the PDF of the chi-squared distribution with n = N − M degrees offreedom (see subsection 30.9.4).◮An experiment produces the following data sample pairs (x i ,y i ):x i : 1.85 2.72 2.81 3.06 3.42 3.76 4.31 4.47 4.64 4.99y i : 2.26 3.10 3.80 4.11 4.74 4.31 5.24 4.03 5.69 6.57where the x i -values are known exactly but each y i -value is measured only to an accuracyof σ =0.5. At the one-tailed 5% significance level, test the null hypothesis H 0 that theunderlying model for the data is a straight line y = mx + c.These data are the same as those investigated in section 31.6 and plotted in figure 31.9. Asshown previously, the least squares estimates of the slope m and intercept c are given byˆm =1.11 and ĉ =0.4. (31.128)Since the error on each y i -value is drawn independently from a Gaussian distribution withstandard deviation σ, we haveN∑[ ] 2χ 2 yi − f(x i ; a)N∑ [ yi − mx i − c] 2(a) ==. (31.129)σσi=1i=1Inserting the values (31.128) into (31.129), we obtain χ 2 ( ˆm, ĉ) =11.5. In our case, thenumber of data points is N = 10 and the number of fitted parameters is M = 2. Thus, the1297

STATISTICSnumber of degrees of freedom is n = N − M = 8. Setting n =8andα =0.05 in (31.127)we find from table 31.2 that k =15.51. Hence our rejection region isχ 2 ( ˆm, ĉ) > 15.51.Since above we found χ 2 ( ˆm, ĉ) = 11.5, we cannot reject the null hypothesis that theunderlying model for the data is a straight line y = mx + c. ◭As mentioned above, our analysis is only valid if the function f(x; a) is linearin the parameters a. Nevertheless, it is so convenient that it is sometimes appliedin non-linear cases, provided the non-linearity is not too severe.31.8 Exercises31.1 A group of students uses a pendulum experiment to measure g, the accelerationof free fall, and obtains the following values (in m s −2 ): 9.80, 9.84, 9.72, 9.74,9.87, 9.77, 9.28, 9.86, 9.81, 9.79, 9.82. What would you give as the best value andstandard error for g as measured by the group?31.2 Measurements of a certain quantity gave the following values: 296, 316, 307, 278,312, 317, 314, 307, 313, 306, 320, 309. Within what limits would you say there isa 50% chance that the correct value lies?31.3 The following are the values obtained by a class of 14 students when measuringa physical quantity x: 53.8, 53.1, 56.9, 54.7, 58.2, 54.1, 56.4, 54.8, 57.3, 51.0, 55.1,55.0, 54.2, 56.6.(a) Display these results as a histogram and state what you would give as thebest value for x.(b) Without calculation, estimate how much reliance could be placed upon youranswer to (a).(c) Databooks give the value of x as 53.6 with negligible error. Are the dataobtained by the students in conflict with this?31.4 Two physical quantities x and y are connected by the equationy 1/2 x=ax 1/2 + b ,and measured pairs of values for x and y are as follows:x: 10 12 16 20y: 409 196 114 94Determine the best values for a and b by graphical means, and (either by handor by using a built-in calculator routine) by a least-squares fit to an appropriatestraight line.31.5 Measured quantities x and y are known to be connected by the formulay =axx 2 + b ,where a and b are constants. Pairs of values obtained experimentally arex: 2.0 3.0 4.0 5.0 6.0y: 0.32 0.29 0.25 0.21 0.18Use these data to make best estimates of the values of y that would be obtainedfor (a) x =7.0, and (b) x = −3.5. As measured by fractional error, which estimateis likely to be the more accurate?1298

31.8 EXERCISES31.6 Prove that the sample mean is the best linear unbiased estimator of the populationmean µ as follows.(a) If the real numbers a 1 ,a 2 ,...,a n satisfy the constraint ∑ ni=1 a i = C, whereCis a given constant, show that ∑ ni=1 a2 i is minimised by a i = C/n for all i.(b) Consider the linear estimator ˆµ = ∑ ni=1 a ix i . Impose the conditions (i) that itis unbiased and (ii) that it is as efficient as possible.31.7 A population contains individuals of k types in equal proportions. A quantity Xhas mean µ i amongst individuals of type i and variance σ 2 , which has the samevalue for all types. In order to estimate the mean of X over the whole population,two schemes are considered; each involves a total sample size of nk. In the firstthe sample is drawn randomly from the whole population, whilst in the second(stratified sampling) n individuals are randomly selected from each of the k types.Show that in both cases the estimate has expectationµ = 1 kk∑µ i ,i=1but that the variance of the first scheme exceeds that of the second by an amount1k∑(µk 2 i − µ) 2 .n31.8 Carry through the following proofs of statements made in subsections 31.5.2 and31.5.3 about the ML estimators ˆτ and ˆλ.i=1(a) Find the expectation values of the ML estimators ˆτ and ˆλ given, respectively,in (31.71) and (31.75). Hence verify equations (31.76), which show that, eventhough an ML estimator is unbiased, it does not follow that functions of itare also unbiased.(b) Show that E[ˆτ 2 ]=(N+1)τ 2 /N and hence prove that ˆτ is a minimum-varianceestimator of τ.31.9 Each of a series of experiments consists of a large, but unknown, number n(≫ 1) of trials in each of which the probability of success p is the same, but alsounknown. In the ith experiment, i =1, 2,...,N, the total number of successes isx i (≫ 1). Determine the log-likelihood function.Using Stirling’s approximation to ln(n − x), show thatd ln(n − x)dnand hence evaluate ∂( n C x )/∂n.≈1+ln(n − x),2(n − x)By finding the (coupled) equations determining the ML estimators ˆp and ˆn,show that, to order n −1 , they must satisfy the simultaneous ‘arithmetic’ and‘geometric’ mean constraintsˆnˆp = 1 NN∑N∏ (x i and (1 − ˆp) N =i=1i=11 − x iˆn).1299

STATISTICS31.10 This exercise is intended to illustrate the dangers of applying formalised estimatortechniques to distributions that are not well behaved in a statistical sense.The following are five sets of 10 values, all drawn from the same Cauchydistribution with parameter a.(i) 4.81 −1.24 1.30 −0.23 2.98−1.13 −8.32 2.62 −0.79 −2.85(ii) 0.07 1.54 0.38 −2.76 −8.821.86 −4.75 4.81 1.14 −0.66(iii) 0.72 4.57 0.86 −3.86 0.30−2.00 2.65 −17.44 −2.26 −8.83(iv) −0.15 202.76 −0.21 −0.58 −0.140.36 0.44 3.36 −2.96 5.51(v) 0.24 −3.33 −1.30 3.05 3.991.59 −7.76 0.91 2.80 −6.46Ignoring the fact that the Cauchy distribution does not have a finite variance (oreven a formal mean), show that â, theMLestimatorofa, has to satisfy∑101s(â) ==5.1+x 2 i=1 i/â2(∗)Using a programmable calculator, spreadsheet or computer, find the value ofâ that satisfies (∗) for each of the data sets and compare it with the valuea =1.6 used to generate the data. Form an opinion regarding the variance of theestimator.Show further that if it is assumed that (E[â]) 2 = E[â 2 ], then E[â] =ν 1/22,whereν 2 is the second (central) moment of the distribution, which for the Cauchydistribution is infinite!31.11 According to a particular theory, two dimensionless quantities X and Y haveequal values. Nine measurements of X gave values of 22, 11, 19, 19, 14, 27, 8,24 and 18, whilst seven measured values of Y were 11, 14, 17, 14, 19, 16 and14. Assuming that the measurements of both quantities are Gaussian distributedwith a common variance, are they consistent with the theory? An alternativetheory predicts that Y 2 = π 2 X; are the data consistent with this proposal?31.12 On a certain (testing) steeplechase course there are 12 fences to be jumped, andany horse that falls is not allowed to continue in the race. In a season of racinga total of 500 horses started the course and the following numbers fell at eachfence:Fence: 1 2 3 4 5 6 7 8 9 10 11 12Falls: 62 75 49 29 33 25 30 17 19 11 15 12Use this data to determine the overall probability of a horse’s falling at a fence,and test the hypothesis that it is the same for all horses and fences as follows.(a) Draw up a table of the expected number of falls at each fence on the basisof the hypothesis.(b) Consider for each fence i the standardised variableestimated falls − actual fallsz i =standard deviation of estimated falls ,and use it in an appropriate χ 2 test.(c) Show that the data indicates that the odds against all fences being equallytesting are about 40 to 1. Identify the fences that are significantly easier orharder than the average.1300

31.8 EXERCISES31.13 A similar technique to that employed in exercise 31.12 can be used to testcorrelations between characteristics of sampled data. To illustrate this considerthe following problem.During an investigation into possible links between mathematics and classicalmusic, pupils at a school were asked whether they had preferences (a) betweenmathematics and english, and (b) between classical and pop music. The resultsare given below.Classical None PopMathematics 23 13 14None 17 17 36English 30 10 40By computing tables of expected numbers, based on the assumption that nocorrelations exist, and calculating the relevant values of χ 2 , determine whetherthere is any evidence for(a) a link between academic and musical tastes, and(b) a claim that pupils either had preferences in both areas or had no preference.You will need to consider the appropriate value for the number of degrees offreedom to use when applying the χ 2 test.31.14 Three candidates X, Y and Z were standing for election to a vacant seat ontheir college’s Student Committee. The members of the electorate (current firstyearstudents, consisting of 150 men and 105 women) were each allowed tocross out the name of the candidate they least wished to be elected, the othertwo candidates then being credited with one vote each. The following data areknown.(a) X received 100 votes from men, whilst Y received 65 votes from women.(b) Z received five more votes from men than X received from women.(c) The total votes cast for X and Y were equal.Analyse this data in such a way that a χ 2 test can be used to determine whethervoting was other than random (i) amongst men and (ii) amongst women.31.15 A particle detector consisting of a shielded scintillator is being tested by placing itnear a particle source whose intensity can be controlled by the use of absorbers.It might register counts even in the absence of particles from the source becauseof the cosmic ray background.The number of counts n registered in a fixed time interval as a function of thesource strength s is given in as:source strength s: 0 1 2 3 4 5 6counts n: 6 11 20 42 44 62 61At any given source strength, the number of counts is expected to be Poissondistributed with meann = a + bs,where a and b are constants. Analyse the data for a fit to this relationship andobtain the best values for a and b together with their standard errors.(a) How well is the cosmic ray background determined?(b) What is the value of the correlation coefficient between a and b? Isthisconsistent with what would happen if the cosmic ray background wereimagined to be negligible?(c) Do the data fit the expected relationship well? Is there any evidence that thereported data ‘are too good a fit’?1301

STATISTICS31.16 The function y(x) is known to be a quadratic function of x. The following tablegives the measured values and uncorrelated standard errors of y measured atvarious values of x (in which there is negligible error):x 1 2 3 4 5y(x) 3.5 ± 0.5 2.0 ± 0.5 3.0 ± 0.5 6.5 ± 1.0 10.5 ± 1.0Construct the response matrix R using as basis functions 1, x,x 2 . Calculate thematrix R T N −1 R and show that its inverse, the covariance matrix V, hastheform⎛⎞V = 1 12 592 −9708 1580⎝−9708 8413 −1461⎠ .9184 1580 −1461 269Use this matrix to find the best values, and their uncertainties, for the coefficientsof the quadratic form for y(x).31.17 The following are the values and standard errors of a physical quantity f(θ)measured at various values of θ (in which there is negligible error):θ 0 π/6 π/4 π/3f(θ) 3.72 ± 0.2 1.98 ± 0.1 −0.06 ± 0.1 −2.05 ± 0.1θ π/2 2π/3 3π/4 πf(θ) −2.83 ± 0.2 1.15 ± 0.1 3.99 ± 0.2 9.71 ± 0.4Theory suggests that f should be of the form a 1 + a 2 cos θ + a 3 cos 2θ. Show thatthe normal equations for the coefficients a i are481.3a 1 + 158.4a 2 − 43.8a 3 = 284.7,158.4a 1 + 218.8a 2 +62.1a 3 = −31.1,−43.8a 1 +62.1a 2 + 131.3a 3 = 368.4.(a) If you have matrix inversion routines available on a computer, determine thebest values and variances for the coefficients a i and the correlation betweenthe coefficients a 1 and a 2 .(b) If you have only a calculator available, solve for the values using a Gauss–Seidel iteration and start from the approximate solution a 1 =2,a 2 = −2,a 3 =4.31.18 Prove that the expression given for the Student’s t-distribution in equation (31.118)is correctly normalised.31.19 Verify that the F-distribution P (F) given explicitly in equation (31.126) is symmetricbetween the two data samples, i.e. that it retains the same form but with N 1and N 2 interchanged, if F is replaced by F ′ = F −1 . Symbolically, if P ′ (F ′ )isthedistribution of F ′ and P (F) =η(F,N 1 ,N 2 ), then P ′ (F ′ )=η(F ′ ,N 2 ,N 1 ).31.20 It is claimed that the two following sets of values were obtained (a) by randomlydrawing from a normal distribution that is N(0, 1) and then (b) randomlyassigning each reading to one of two sets A and B:Set A: −0.314 0.603 −0.551 −0.537 −0.160 −1.635 0.7190.610 0.482 −1.757 0.058Set B: −0.691 1.515 −1.642 −1.736 1.224 1.423 1.165Make tests, including t- andF-tests, to establish whether there is any evidencethat either claims is, or both claims are, false.1302

31.9 HINTS AND ANSWERS31.9 Hints and answers31.1 Note that the reading of 9.28 m s −2 is clearly in error, and should not be used inthe calculation; 9.80 ± 0.02 m s −2 .31.3 (a) 55.1. (b) Note that two thirds of the readings lie within ±2 of the mean andthat 14 readings are being used. This gives a standard error in the mean ≈ 0.6.(c) Student’s t has a value of about 2.5 for 13 d.o.f. (degrees of freedom), andtherefore it is likely at the 3% significance level that the data are in conflict withthe accepted value.31.5 Plot or calculate a least-squares fit of either x 2 versus x/y or xy versus y/xto obtain a ≈ 1.19 and b ≈ 3.4. (a) 0.16; (b) −0.27. Estimate (b) is the moreaccurate because, using the fact that y(−x) =−y(x), it is effectively obtained byinterpolation rather than extrapolation.31.7 Recall that, because of the equal proportions of each type, the expected numbersof each type in the first scheme is n. Show that the variance of the estimator forthe second scheme is σ 2 /(kn). When calculating that for the first scheme, recallthat x 2 i= µ 2 i + σ 2 and note that µ 2 i can be written as (µ i − µ + µ) 2 .31.9 The log-likelihood function is()N∑N∑N∑ln L = ln n C xi + x i ln p + Nn − x i ln(1 − p);∂( n C x )∂ni=1≈ ln( nn − x)−i=1x2n(n − x) .Ignore the second term on the RHS of the above to obtainN∑( ) nln + N ln(1 − p) =0.n − x ii=131.11 ¯X =18.0 ± 2.2, Ȳ =15.0 ± 1.1. ˆσ =4.92 giving t =1.21 for 14 d.o.f., andis significant only at the 75% level. Thus there is no significant disagreementbetween the data and the theory. For the second theory, only the mean valuescan be tested as Y 2 will not be Gaussian distributed. The difference in the meansis Ȳ 2 − π 2 ¯X =47± 36 and is only significantly different from zero at the 82%level. Again the data is consistent with the proposed theory.31.13 Consider how many entries may be chosen freely in the table if all row andcolumn totals are to match the observed values. It should be clear that for anm × n table the number of degrees of freedom is (m − 1)(n − 1).(a) In order to make the fractions expressing each preference or lack of preferencecorrect, the expected distribution, if there were no correlation, mustbeClassical None PopMathematics 17.5 10 22.5None 24.5 14 31.5English 28 16 36This gives a χ 2 of 12.3 for four d.o.f., making it less than 2% likely, that nocorrelation exists.(b) The expected distribution, if there were no correlation, isMusic preference No music preferenceAcademic preference 104 26No academic preference 56 141303i=1

STATISTICSThis gives a χ 2 of 1.2 for one d.o.f and no evidence for the claim.31.15 As the distribution at each value of s is Poisson, the best estimate of themeasurement error is the square root of the number of counts, i.e. √ n(s). Linearregression gives a =4.3 ± 2.1 andb =10.06 ± 0.94.(a) The cosmic ray background must be present, since n(0) ≠ 0 but its value ofabout 4 is uncertain to within a factor 2.(b) The correlation coefficient between a and b is −0.63. Yes; if a were reducedtowards zero then b would have to be increased to compensate.(c) Yes, χ 2 =4.9 for five d.o.f., which is almost exactly the ‘expected’ value,neither too good nor too bad.31.17 a 1 =2.02 ± 0.06, a 2 = −2.99 ± 0.09, a 3 =4.90 ± 0.10; r 12 = −0.60.31.19 Note that |dF| = |dF ′ /F ′2 | and write1+ N [ ][1 − 1N1 − 1as1+ (N ]2 − 1)F ′.(N 2 − 1)F ′ (N 2 − 1)F ′ N 1 − 11304

IndexWhere the discussion of a topic runs over two consecutive pages, reference is madeonly to the first of these. For discussions spread over three or more pages the first andlast page numbers are given; these references are usually to the major treatment of thecorresponding topic. Isolated refences to a topic, including those appearing on consecutivepages, are listed individually. Some long topics are split, e.g. ‘Fourier transforms’ and‘Fourier transforms, examples’. The letter ‘n’ after a page number indicates that the topicis discussed in a footnote on the relevant page.A, B, one-dimensional irreps, 1090, 1102, 1109Abelian groups, 1044absolute convergence of series, 124, 831absolute derivative, 975–977acceleration vector, 335Adams method, 1024Adams–Moulton–Bashforth, predictor-correctorscheme, 1035addition rule for probabilities, 1125, 1130addition theorem for spherical harmonicsYl m (θ, φ), 594adjoint, see Hermitian conjugateadjoint operators, 559–564adjustment of parameters, 795Airy integrals, 890–894Ai(z), 889algebra ofcomplex numbers, 85functions in a vector space, 556matrices, 251power series, 134series, 131tensors, 938–941vectors, 213in a vector space, 242in component form, 218algebraic equations, numerical methods for, seenumerical methods for algebraic equationsalternating group, 1116alternating series test, 130ammonia molecule, symmetries of, 1042Ampère’s rule (law), 381, 408amplitude modulation of radio waves, 444amplitude-phase diagram, 914analytic (regular) functions, 826angle between two vectors, 221angular frequency, 693nin Fourier series, 419angular momentum, 933, 949and irreps, 1093of particle system, 950–952of particles, 338of solid body, 396, 951vector representation, 238angular momentum operatorcomponent, 658total, 659angular velocity, vector representation, 223, 238,353annihilation and creation operators, 667anti-Hermitian matrices, 271eigenvalues, 276–278imaginary nature, 277eigenvectors, 276–278orthogonality, 277anti-Stokes line, 905anticommutativity of vector or cross product,222antisymmetric functions, 416and Fourier series, 419and Fourier transforms, 445antisymmetric matrices, 2701305

INDEXgeneral properties, see anti-Hermitianmatricesantisymmetric tensors, 938, 941antithetic variates, in Monte Carlo methods,1014aperture function, 437approximately equal ≈, definition, 132arbitrary parameters for ODE, 469arc length ofplane curves, 73space curves, 341arccosech, arccosh, arccoth, arcsech, arcsinh,arctanh, see hyperbolic functions, inversesArchimedean upthrust, 396, 410area element inCartesian coordinates, 188plane polars, 202area ofcircle, 71ellipse, 71, 207parallelogram, 223region, using multiple integrals, 191–193surfaces, 346as vector, 393–395, 408area, maximal enclosure, 779arg, argument of a complex number, 87Argand diagram, 84, 825argument, principle of the, 880arithmetic series, 117arithmetico-geometric series, 118arrays, see matricesassociated Laguerre equation, 535, 621–624as example of Sturm–Liouville equation, 566,622natural interval, 567, 622associated Laguerre polynomials L m n (x), 621as special case of confluent hypergeometricfunction, 634generating function, 623orthogonality, 622recurrence relations, 624Rodrigues’ formula, 622associated Legendre equation, 535, 587–593, 733,768general solution, 588as example of Sturm–Liouville equation, 566,590, 591general solution, 588natural interval, 567, 590, 591associated Legendre functions, 587–593of first kind Pl m (x), 588, 733, 768generating function, 592normalisation, 590orthogonality, 590, 591recurrence relations, 592Rodrigues’ formula, 588of second kind Q m l (x), 588associative law foradditionin a vector space of finite dimensionality,242in a vector space of infinite dimensionality,556of complex numbers, 86of matrices, 251of vectors, 213convolution, 447, 458group operations, 1043linear operators, 249multiplicationof a matrix by a scalar, 251ofavectorbyascalar,214of complex numbers, 88of matrices, 253multiplication by a scalarin a vector space of finite dimensionality,242in a vector space of infinite dimensionality,556atomic orbitals, 1115d-states, 1106, 1108, 1114p-states, 1106s-states, 1144auto-correlation functions, 450automorphism, 1061auxiliary equation, 493repeated roots, 493average value, see mean valueaxial vectors, 949backward differences, 1019basis functionsfor linear least squares estimation, 1273in a vector space of infinite dimensionality,556of a representation, 1078change in, 1084, 1087, 1092basis vectors, 217, 243, 929, 1078derivatives, 965–968Christoffel symbol Γ k ij , 965for particular irrep, 1106–1108, 1116linear dependence and independence, 217non-orthogonal, 245orthonormal, 244required properties, 217Bayes’ theorem, 1132Bernoulli equation, 477Bessel correction to variance estimate, 1248Bessel equation, 535, 602–607, 614, 615as example of Sturm–Liouville equation, 566natural interval, 608Bessel functions J ν (x), 602–614, 729, 738as special case of confluent hypergeometricfunction, 634generating function, 613graph of, 606integral relationships, 610integral representation, 613–614orthogonality, 608–6111306

INDEXrecurrence relations, 611–612second kind Y ν (x), 607graph of, 607series, 604ν = 0, 606ν = ±1/2, 605spherical j l (x), 615, 741zeros of, 729, 739Bessel inequality, 246, 559best unbiased estimator, 1232beta function, 638bias of estimator, 1231bilinear transformation, general, 110binary chopping, 990binomial coefficient n C k , 27–30, 1135–1137elementary properties, 26identities, 27in Leibnitz’ theorem, 49negative n, 29non-integral n, 29binomial distribution Bin(n, p), 1168–1171and Gaussian distribution, 1185and Poisson distribution, 1174, 1177mean and variance, 1171MGF, 1170recurrence formula, 1169binomial expansion, 25–30, 140binormal to space curves, 342birthdays, different, 1134bivariate distributions, 1196–1207conditional, 1198continuous, 1197correlation, 1200–1207and independence, 1200matrix, 1203–1207positive and negative, 1200uncorrelated, 1200covariance, 1200–1207matrix, 1203expectation (mean), 1199independent, 1197, 1200marginal, 1198variance, 1200Boltzmann distribution, 171bonding in molecules, 1103, 1105–1108Born approximation, 149, 575Bose–Einstein statistics, 1138boundary conditionsand characteristics, 700and Laplace equation, 764, 766for Green’s functions, 512, 514–516inhomogeneous, 515for ODE, 468, 470, 501for PDE, 681, 685–687for Sturm–Liouville equations, 564homogeneous and inhomogeneous, 685, 723,752, 754superposition solutions, 718–724types, 702–705bra vector 〈ψ|, 649brachistochrone problem, 784Bragg formula, 237branch cut, 835branch points, 835Bromwich integral, 884bulk modulus, 980calculus of residues, see zeros of a function of acomplex variable and contour integrationcalculus of variationsconstrained variation, 785–787estimation of ODE eigenvalues, 790Euler–Lagrange equation, 776Fermat’s principle, 787Hamilton’s principle, 788higher-order derivatives, 782several dependent variables, 782several independent variables, 782soap films, 780variable end-points, 782–785calculus, elementary, 41–76cancellation law in a group, 1046canonical form, for second-order ODE, 516card drawing, see probabilitycarrier frequency of radio waves, 445Cartesian coordinates, 217Cartesian tensors, 930–955algebra, 938–941contraction, 939definition, 935first-order, 932–935from scalar, 934general order, 935–954integral theorems, 954isotropic, 944–946physical applications, 934, 939–941, 950–954second-order, 935–954, 968symmetry and antisymmetry, 938tensor fields, 954zero-order, 932–935from vector, 935Cartesian tensors, particularconductivity, 952inertia, 951strain, 953stress, 953susceptibility, 952catenary, 781, 787Cauchyboundary conditions, 702distribution, 1152inequality, 853integrals, 851–853product, 131root test, 129, 831theorem, 849Cauchy–Riemann relations, 827–830, 849, 873,875in terms of z and z ∗ , 829central differences, 10191307

INDEXcentral limit theorem, 1194–1196central moments, see moments, centralcentre of a group, 1069centre of mass, 195of hemisphere, 195of semicircular lamina, 197centroid, 195of plane area, 195of plane curve, 197of triangle, 216CF, see complementary functionchain rule for functions ofone real variable, 46several real variables, 157change of basis, see similarity transformationschange of variablesand coordinate systems, 158–160in multiple integrals, 199–207evaluation of Gaussian integral, 202–204general properties, 206in RVD, 1150–1157character tables, 10933m or 32 or C 3v or S 3 , 1093, 1097, 1108, 1110,11174mm or C 4v or D 4 , 1102, 1106, 1108, 1113A 4 , 1116D 5 , 1116S 4 or 432 or O, 1114, 1115¯43m or T d , 1115construction of, 1100–1102quaternion, 1113characteristic equation, 280normal mode form, 319of recurrence relation, 499characteristic functions, see moment generatingfunctions (MGFs)characteristicsand boundary curves, 700multiple intersections, 700, 705and the existence of solutions, 699–705first-order equations, 699second-order equations, 703and equation type, 703characters, 1092–1096, 1100–1102and conjugacy classes, 1092, 1095counting irreps, 1095definition, 1092of product representation, 1104orthogonality properties, 1094, 1102summation rules, 1097charge (point), Dirac δ-function respresentation,441charged particle in electromagnetic fields, 370Chebyshev equation, 535, 595–602as example of Sturm–Liouville equation, 566,599general solution, 597natural interval, 567, 599polynomial solutions, 552Chebyshev functions, 595–602Chebyshev polynomialsof first kind T n (x), 596as special case of hypergeometric function,631generating function, 601graph of, 597normalisation, 600orthogonality, 599recurrence relations, 601Rodrigues’ formula, 599of second kind U n (x), 597generating function, 601graph of, 598normalisation, 600orthogonality, 599Rodrigues’ formula, 599chi-squared (χ 2 ) distribution, 1192and goodness of fit, 1297and likelihood-ratio test, 1283, 1292and multiple estimators, 1243percentage points tabulation, 1244test for correlation, 1301Cholesky separation, 313Christoffel symbol Γ k ij , 965–968from metric tensor, 966, 973circlearea of, 71equation for, 16circle of convergence, 831Clairaut equation, 483classes and equivalence relations, 1064closure of a group, 1043closure property of eigenfunctions of anHermitian operator, 563cofactor of a matrix element, 259column matrix, 250column vector, 250combinations (probability), 1133–1139common ratio in geometric series, 117commutation law for group elements, 1044commutative law foradditionin a vector space of finite dimensionality,242in a vector space of infinite dimensionality,556of complex numbers, 86of matrices, 251of vectors, 213complex scalar or dot product, 222convolution, 447, 458inner product, 244multiplicationofavectorbyascalar,214of complex numbers, 88scalar or dot product, 220commutatorof two matrices, 309of two operators, 653, 656comparison test, 1251308

INDEXcomplement, 1121probability for, 1125complementary equation, 490complementary error function, 640complementary function (CF), 491for ODE, 492partially known, 506repeated roots of auxiliary equation, 493completeness ofbasis vectors, 243eigenfunctions of an Hermitian operator, 560,563eigenvectors of a normal matrix, 275spherical harmonics Yl m (θ, φ), 594completing the squareas a means of integration, 66for quadratic equations, 35for quadratic forms, 1206to evaluate Gaussian integral, 436, 749complex conjugatez ∗ , of complex number, 89–91, 829of a matrix, 256–258of scalar or dot product, 222properties of, 90complex exponential function, 92, 833complex Fourier series, 424complex integrals, 845–849, see also zeros of afunction of a complex variable and contourintegrationAiry integrals, 890–894Cauchy integrals, 851–853Cauchy’s theorem, 849definition, 845Jordan’s lemma, 864Morera’s theorem, 851of z −1 , 846principal value, 864residue theorem, 858–860WKB methods, 895–905complex logarithms, 99, 834principal value of, 100, 834complex numbers, 83–114addition and subtraction of, 85applications to differentiation and integration,101argument of, 87associativity ofaddition, 86multiplication, 88commutativity ofaddition, 86multiplication, 88complex conjugate of, see complex conjugatecomponents of, 84de Moivre’s theorem, see de Moivre’s theoremdivision of, 91, 94from roots of polynomial equations, 83imaginary part of, 83modulus of, 87multiplication of, 88, 94as rotation in the Argand diagram, 88notation, 84polar representation of, 92–95real part of, 83trigonometric representation of, 93complex potentials, 871–876and fluid flow, 873equipotentials and field lines, 872for circular and elliptic cylinders, 876for parallel cylinders, 921for plates, 877–879, 921for strip, 921for wedges, 878under conformal transformations, 876–879complex power series, 133complex powers, 99complex variables, see functions of a complexvariable and power series in a complexvariable and complex integralscomponentsof a complex number, 84of a vector, 217in a non-orthogonal basis, 234uniqueness, 243conditional (constrained) variation, 785–787conditional convergence, 124conditional distributions, 1198conditional probability, see probability,conditionalconesurface area of, 74volume of, 75confidence interval, 1236confidence region, 1241confluence process, 634confluent hypergeometric equation, 535, 633as example of Sturm–Liouville equation, 566general solution, 633confluent hypergeometric functions, 633contiguous relations, 635integral representation, 634recurrence relations, 635special cases, 634conformal transformations (mappings), 839–879applications, 876–879examples, 842–844properties, 839–842Schwarz–Christoffel transformation, 843congruence, 1065conic sections, 15eccentricity, 17parametric forms, 17standard forms, 16conjugacy classes, 1068–1070element in a class by itself, 1068conjugate roots of polynomial equations, 99connectivity of regions, 383conservative fields, 387–389necessary and sufficient conditions, 387–389potential (function), 3891309

INDEXconsistency, of estimator, 1230constant coefficients in ODE, 492–503auxiliary equation, 493constants of integration, 62, 468constrained variation, 785–787constraints, stationary values under, seeLagrange undetermined multiplerscontinuity correction for discrete RV, 1186continuity equation, 404contour integration, 861–867, 887infinite integrals, 862–867inverse Laplace transforms, 884–887residue theorem, 858–867sinusoidal functions, 861summing series, 882contraction of tensors, 939contradiction, proof by, 32–34contravariantbasis vectors, 961derivative, 965components of tensor, 956definition, 961control variates, in Monte Carlo methods, 1013convergence of infinite series, 831absolute, 124, 831complex power series, 133conditional, 124necessary condition, 125power series, 132under various manipulations, see powerseries, manipulationratio test, 832rearrangement of terms, 124tests for convergence, 125–131alternating series test, 130comparison test, 125grouping terms, 129integral test, 128quotient test, 127ratio comparison test, 127ratio test (D’Alembert), 126, 132root test (Cauchy), 129, 831convergence of numerical iteration schemes,992–994convolutionFourier tranforms, see Fourier transforms,convolutionLaplace tranforms, see Laplace transforms,convolutionconvolution theoremFourier transforms, 448Laplace transforms, 457coordinate geometry, 15–18conic sections, 15straight line, 15coordinate systems, see Cartesian, curvilinear,cylindrical polar, plane polar and sphericalpolar coordinatescoordinate transformationsand integrals, see change of variablesand matrices, see similarity transformationsgeneral, 960–965relative tensors, 963tensor transformations, 962weight, 964orthogonal, 932coplanar vectors, 225Cornu spiral, 914correlation functions, 449–451auto-correlation, 450cross-correlation, 449energy spectrum, 450Parseval’s theorem, 451Wiener–Kinchin theorem, 450correlation matrix, of sample, 1229correlation of bivariate distributions, 1200–1207correlation of sample data, 1229correlation, chi-squared test, 1301correspondence principle in quantum mechanics,1215cosets and congruence, 1065cosh, hyperbolic cosine, 102, 833, see alsohyperbolic functionscosine, cos(x)in terms of exponential functions, 102Maclaurin series for, 140orthogonality relations, 417counting irreps, see characters, counting irrepscoupled pendulums, 329, 331covariance matrixof linear least squares estimators, 1274of sample, 1229covariance of bivariate distributions, 1200–1207covariance of sample data, 1229covariantbasis vector, 961derivative, 965components of tensor, 956definition, 961derivative, 968of scalar, 971semi-colon notation, 969differentiation, 968–971CPF, see probability functions, cumulativeCramér–Rao (Fisher’s) inequality, 1232, 1233Cramer determinant, 299Cramer’s rule, 299cross product, see vector productcross-correlation functions, 449crystal lattice, 148crystal point groups, 1082cube roots of unity, 98cube, rotational symmetries of, 1114cumulants, 1166curl of a vector field, 353as a determinant, 353as integral, 398, 400curl curl, 356in curvilinear coordinates, 368in cylindrical polars, 3601310

INDEXin spherical polars, 362Stoke’s theorem, 406–409tensor form, 974current-carrying wire, magnetic potential, 729curvature, 52–55circle of, 53of a function, 52of space curves, 342radius of, 53curves, see plane curves and space curvescurvilinear coordinates, 364–369basis vectors, 364length and volume elements, 365scale factors, 364surfaces and curves, 364tensors, 955–977vector operators, 367–369cut plane, 865cycle notation for permutations, 1057cyclic groups, 1061, 1098cyclic relation for partial derivatives, 157cycloid, 370, 785cylinders, conducting, 874, 876cylindrical polar coordinates, 357–361area element, 360basis vectors, 358Laplace equation, 728–731length element, 360vector operators, 357–361volume element, 360δ-function (Dirac), see Dirac δ-functionδ ij , δ j i , Kronecker delta, tensor, see Kroneckerdelta, δ ij , δ j i ,tensorD’Alembert’s ratio test, 126, 832in convergence of power series, 132D’Alembert’s solution to wave equation, 694damped harmonic oscillators, 239and Parseval’s theorem, 451data modelling, maximum-likelihood, 1255de Broglie relation, 436, 709, 768de Moivre’s theorem, 95, 861applications, 95–99finding the nth roots of unity, 97solving polynomial equations, 98trigonometric identities, 95–97deconvolution, 449defective matrices, 278, 311degeneracybreaking of, 1111–1113of normal modes, 1110degenerate (separable) kernel, 807degenerate eigenvalues, 275, 282degreeof ODE, 468of polynomial equation, 2del ∇, see gradient operator (grad)del squared ∇ 2 (Laplacian), 352, 676as integral, 400in curvilinear coordinates, 368in cylindrical polar coordinates, 360in polar coordinates, 725in spherical polar coordinates, 362, 741tensor form, 973delta function (Dirac), see Dirac δ-functiondependent random variables, 1196–1205derivative, see also differentiationabsolute, 975–977covariant, 968Fourier transform of, 444Laplace transform of, 455normal, 350of basis vectors, 336of composite vector expressions, 337of function of a complex variable, 825of function of a function, 46of hyperbolic functions, 106–109of products, 44–46, 48–50of quotients, 47of simple functions, 44of vectors, 334ordinary, first, second and nth, 42partial, see partial differentiationtotal, 154derivative method for second series solution ofODE, 545–548determinant formand ɛ ijk , 942for curl, 353determinants, 259–263adding rows or columns, 262and singular matrices, 263as product of eigenvalues, 287evaluationusing ɛ ijk , 942using Laplace expansion, 259identical rows or columns, 262in terms of cofactors, 259interchanging two rows or two columns, 262Jacobian representation, 201, 205, 207notation, 259of Hermitian conjugate matrices, 262of order three, in components, 260of transpose matrices, 261product rule, 262properties, 261–263, 978relationship with rank, 267removing factors, 262secular, 280diagonal matrices, 268diagonalisation of matrices, 285–288normal matrices, 286properties of eigenvalues, 287simultaneous, 331diamond, unit cell, 234die throwing, see probabilitydifference method for summation of series, 119difference schemes for differential equations,1020–1023, 1030–10321311

INDEXdifference, finite, see finite differencesdifferentiablefunction of a complex variable, 825–827function of a real variable, 42differentialdefinition, 43exact and inexact, 155of vector, 338, 344total, 154differential equations, see ordinary differentialequations and partial differential equationsdifferential equations, particularassociated Laguerre, 535, 566, 621–624associated Legendre, 535, 566, 587–593Bernoulli, 477Bessel, 535, 566, 602–607, 614Chebyshev, 535, 566, 595–602Clairaut, 483confluent hypergeometric, 535, 566, 633diffusion, 678, 695–698, 716, 723, 1032Euler, 504Euler–Lagrange, 776Helmholtz, 737–741Hermite, 535, 566, 624–628hypergeometric, 535, 566, 628–632Lagrange, 789Laguerre, 535, 566, 616–621Laplace, 679, 690, 717, 718, 1031Legendre, 534, 535, 566, 577–586Legendre linear, 503–505Poisson, 679, 744–746Schrödinger, 679, 741, 768, 795simple harmonic oscillator, 535, 566Sturm–Liouville, 790wave, 676, 689, 693–695, 714, 737, 790differential operators, see linear differentialoperatordifferentiation, see also derivativeas gradient, 42as rate of change, 41chain rule, 46covariant, 968–971from first principles, 41–44implicit, 47logarithmic, 48notation, 43of Fourier series, 424of integrals, 178of power series, 135partial, see partial differentiationproduct rule, 44–46, 48–50quotient rule, 47theorems, 55–57using complex numbers, 101diffraction, see Fraunhofer diffractiondiffusion equation, 678, 688, 695–698combination of variables, 696–698integral transforms, 747numerical methods, 1032separation of variables, 716simple solution, 696superposition, 723diffusion of solute, 678, 696, 747dihedral group, 1113, 1116dimension of irrep, 1088dimensionality of vector space, 243dipole matrix elements, 208, 1108, 1115dipole moments of molecules, 1077Dirac δ-function, 355, 405, 439–443and convolution, 447and Green’s functions, 511, 512as limit of various distributions, 443as sum of harmonic waves, 442definition, 439Fourier transform of, 443impulses, 441point charges, 441properties, 439reality of, 443relation to Fourier transforms, 442relation to Heaviside (unit step) function, 441three-dimensional, 441, 452Dirac notation, 648direct product, of groups, 1072direct sum ⊕, 1086direction cosines, 221Dirichlet boundary conditions, 702, 852nGreen’s functions, 754, 756–765method of images, 758–765Dirichlet conditions, for Fourier series, 415disc, moment of inertia, 208discontinuous functions and Fourier series,420–422discrete Fourier transforms, 462disjoint events, see mutually exclusive eventsdisplacement kernel, 809distance from aline to a line, 231line to a plane, 232point to a line, 229point to a plane, 230distributive law foraddition of matrix products, 254convolution, 447, 458inner product, 244linear operators, 249multiplicationof a matrix by a scalar, 251of a vector by a complex scalar, 222ofavectorbyascalar,214multiplication by a scalarin a vector space of finite dimensionality,242in a vector space of infinite dimensionality,556scalar or dot product, 220vector or cross product, 222div, divergence of vector fields, 352as integral, 398in curvilinear coordinates, 3671312

INDEXin cylindrical polars, 360in spherical polars, 362tensor form, 972divergence theoremfor tensors, 954for vectors, 401in two dimensions, 384physical applications, 404related theorems, 403division axiom in a group, 1046division of complex numbers, 91dominant term, in Stokes phenomenon, 904dot product, see scalar productdouble integrals, see multiple integralsdrumskin, see membranedual tensors, 949dummy variable, 61ɛ ijk , Levi-Civita symbol, tensor, 941–946and determinant, 942identities, 943isotropic, 945vector products, 942weight, 964e x , see exponential functionE, two-dimensional irrep, 1090, 1102, 1108eccentricity, of conic sections, 17efficiency, of estimator, 1231eigenequation for differential operators, 554more general form, 555, 571–573eigenfrequencies, 319estimation using Rayleigh–Ritz method,327–329eigenfunctionscompleteness for Hermitian operators, 560,563construction of a real set for an Hermitianoperator, 563definition, 555normalisation for Hermitian operators, 562of integral equations, 817of simple harmonic oscillators, 555orthogonality for Hermitian operators,561–563eigenvalues, 272–282, see Hermitian operatorscharacteristic equation, 280continuous and discrete, 650definition, 272degenerate, 282determination, 280–282estimation for ODE, 790estimation using Rayleigh–Ritz method,327–329notation, 273of anti-Hermitian matrices, seeanti-Hermitian matricesof Fredholm equations, 808of general square matrices, 278of Hermitian matrices, see Hermitian matricesof integral equations, 808, 816of linear differential operatorsadjustment of parameters, 795definition, 555error in estimate of, 793estimation, 790–796higher eigenvalues, 793, 800simple harmonic oscillator, 555of linear operators, 272of normal matrices, 273–276of representative matrices, 1100of unitary matrices, 278under similarity transformation, 287eigenvectors, 272–282characteristic equation, 280definition, 272determination, 280–282normalisation condition, 273notation, 273of anti-Hermitian matrices, seeanti-Hermitian matricesof commuting matrices, 278of general square matrices, 278of Hermitian matrices, see Hermitian matricesof linear operators, 272of normal matrices, 273–276of unitary matrices, 278stationary properties for quadratic andHermitian forms, 290Einstein relation, 436, 709, 768elastic deformations, 953electromagnetic fieldsflux, 395Maxwell’s equations, 373, 408, 979electrostatic fields and potentialscharged split sphere, 735conducting cylinder in uniform field, 876conducting sphere in uniform field, 734from charge density, 745, 758from complex potential, 873infinite charged plate, 759, 877infinite wedge with line charge, 878ininite charged wedge, 877of line charges, 761, 872semi-infinite charged plate, 877sphere with point charge, 764ellipsearea of, 71, 207, 385as section of quadratic surface, 292equation for, 16ellipsoid, volume of, 207elliptic PDE, 687, 690empty event ∅, 1121end-points for variationscontributions from, 782fixed, 777variable, 782–785energy levels ofparticle in a box, 768simple harmonic oscillator, 6421313

INDEXenergy spectrum and Fourier transforms, 450,451entire functions, 832nenvelopes, 173–175equations of, 174to a family of curves, 173epimorphism, 1061equilateral triangle, symmetries of, 1047, 1052,1081, 1110equivalence relations, 1064–1066, 1068and classes, 1064congruence, 1065–1067examples, 1070equivalence transformations, see similaritytransformationsequivalent representations, 1084–1086, 1099error function, erf(x), 640, 697, 748as special case of confluent hypergeometricfunction, 634error termsin Fourier series, 430in Taylor series, 139errors, first and second kind, 1280essential singularity, 838, 856estimation of eigenvalueslinear differential operator, 792–795Rayleigh–Ritz method, 327–329estimators (statistics), 1229best unbiased, 1232bias, 1231central confidence interval, 1237confidence interval, 1236confidence limits, 1236confidence region, 1241consistency, 1230efficiency, 1231maximum-likelihood, 1256minimum-variance, 1232standard error, 1234Euler equationdifferential, 504, 522trigonometric, 93Euler method, numerical, 1021Euler–Lagrange equation, 776special cases, 777–781even functions, see symmetric functionsevents, 1120complement of, 1121empty ∅, 1121intersection of ∩, 1120mutually exclusive, 1129statistically independent, 1129union of ∪, 1121exact differentials, 155exact equations, 472, 505condition for, 472non-linear, 519expectation values, see probability distributions,meanexponential distribution, 1190from Poisson, 1190MGF, 1191exponential functionMaclaurin series for, 140of a complex variable, 92, 833relation with hyperbolic functions, 102F-distribution (Fisher), 1290–1296critical points table, 1295logarithmic form, 1296Fabry–Pérot interferometer, 146factorial function, general, 636factorisation, of a polynomial equation, 7faithful representation, 1083, 1098Fermat’s principle, 787, 798Fermi–Dirac statistics, 1138Fibonacci series, 525field lines and complex potentials, 872fieldsconservative, 387–389scalar, 347tensor, 954vector, 347fields, electrostatic, see electrostatic fields andpotentialsfields, gravitational, see gravitational fields andpotentialsfinite differences, 1019central, 1019for differential equations, 1020–1023forward and backward, 1019from Taylor series, 1019, 1026schemes for differential equations, 1030–1032finite groups, 1043first law of thermodynamics, 176first-order differential equations, see ordinarydifferential equationsFisher distribution, see F-distribution (Fisher)Fisher matrix, 1241, 1268Fisher’s inequality, 1232, 1233fluidsArchimedean upthrust, 396, 410complex velocity potential, 873continuity equation, 404cylinder in uniform flow, 874flow, 873flux, 395, 875irrotational flow, 353sources and sinks, 404, 873stagnation points, 873velocity potential, 409, 679vortex flow, 408, 874forward differences, 1019Fourier cosine transforms, 446Fourier series, 415–432and separation of variables, 719–722, 724coefficients, 417–419, 425complex, 424differentiation, 424Dirichlet conditions, 4151314

INDEXdiscontinuous functions, 420–422error term, 430integration, 424non-periodic functions, 422–424orthogonality of terms, 417complex case, 425Parseval’s theorem, 426standard form, 417summation of series, 427symmetry considerations, 419uses, 415Fourier series, examplessquare-wave, 418x, 424, 425x 2 , 422x 3 , 424Fourier sine transforms, 445Fourier transforms, 433–453as generalisation of Fourier series, 433–435convolution, 446–449and the Dirac δ-function, 447associativity, commutativity, distributivity,447definition, 447resolution function, 446convolution theorem, 448correlation functions, 449–451cosine transforms, 446deconvolution, 449definition, 435discrete, 462evaluation using convolution theorem, 448for integral equations, 809–812for PDE, 749–751Fourier-related (conjugate) variables, 436in higher dimensions, 451–453inverse, definition, 435odd and even functions, 445Parseval’s theorem, 450properties: differentiation, exponentialmultiplication, integration, scaling,translation, 444relation to Dirac δ-function, 442sine transforms, 445Fourier transforms, examplesconvolution, 448damped harmonic oscillator, 451Dirac δ-function, 443exponential decay function, 435Gaussian (normal) distribution, 435rectangular distribution, 442spherically symmetric functions, 452two narrow slits, 448two wide slits, 438, 448Fourier’s inversion theorem, 435Fraunhofer diffraction, 437–439diffraction grating, 461two narrow slits, 448two wide slits, 438, 448Fredholm integral equations, 805eigenvalues, 808operator form, 806with separable kernel, 807Fredholm theory, 815Frenet–Serret formulae, 343Fresnel integrals, 913Frobenius series, 539Fuch’s theorem, 539function of a matrix, 255functional, 776functions of a complex variable, 825–839,853–858analyticity, 826behaviour at infinity, 839branch points, 835Cauchy integrals, 851–853Cauchy–Riemann relations, 827–830conformal transformations, 839–844derivative, 825differentiation, 825–830identity theorem, 854Laplace equation, 829, 871Laurent expansion, 855–858multivalued and branch cuts, 835–837, 885particular functions, 832–835poles, 837power series, 830–832real and imaginary parts, 825, 830singularities, 826, 837–839Taylor expansion, 853–855zeros, 839, 879–882functions of one real variabledecomposition into even and odd functions,416differentiation of, 41–50Fourier series, see Fourier seriesintegration of, 59–72limits, see limitsmaxima and minima of, 50–52stationary values of, 50–52Taylor series, see Taylor seriesfunctions of several real variableschain rule, 157differentiation of, 151–179integration of, see multiple integrals,evaluationmaxima and minima, 162–167points of inflection, 162–167rates of change, 153–155saddle points, 162–167stationary values, 162–167Taylor series, 160–162fundamental solution, 757fundamental theorem ofalgebra, 83, 85, 868calculus, 61complex numbers, see de Moivre’s theoremgamma distribution, 1153, 1191gamma function1315

INDEXas general factorial function, 636definition and properties, 636graph of, 637Gauss’s theorem, 765Gauss–Seidel iteration, 996–998Gaussian (normal) distribution N(µ, σ 2 ),1179–1189and binomial distribution, 1185and central limit theorem, 1195and Poisson distribution, 1187continuity correction, 1186CPF, 1018, 1181tabulation, 1182Fourier transform, 435integration with infinite limits, 202–204mean and variance, 1180–1184MGF, 1185, 1188multiple, 1188multivariate, 1209random number generation, 1018sigma limits, 1183standard variable, 1180Gaussian elimination with interchange, 995Gaussian integration, 1005–1009points and weights, 1008, 1010general tensorsalgebra, 938–941contraction, 939contravariant, 961covariant, 961dual, 949metric, 957–960physical applications, 957–960, 976pseudotensors, 964tensor densities, 964generalised likelihood ratio, 1282generating functionsassociated Laguerre polynomials, 623associated Legendre polynomials, 592Bessel functions, 613Chebyshev polynomials, 601Hermite polynomials, 627Laguerre polynomials, 620Legendre polynomials, 584–586generating functions in probability, 1157–1167,see also moment generating functions andprobability generating functionsgeodesics, 797, 976, 982geometric distribution, 1159, 1172geometric series, 117Gibbs’ free energy, 178Gibbs’ phenonmenon, 421gradient of a function ofone variable, 42several real variables, 153–155gradient of scalar, 348–352tensor form, 972gradient of vector, 937, 969gradient operator (grad), 348as integral, 398in curvilinear coordinates, 367in cylindrical polars, 360in spherical polars, 362tensor form, 972Gram–Schmidt orthogonalisation ofeigenfunctions of Hermitian operators, 562eigenvectors ofHermitian matrices, 277normal matrices, 275functions in a Hilbert space, 557gravitational fields and potentialsLaplace equation, 679Newton’s law, 339Poisson equation, 679, 744uniform disc, 771uniform ring, 742Green’s functions, 568–571, 751–767and boundary conditions, 512, 514and Dirac δ-function, 511and partial differential operators, 753and Wronskian, 527diffusion equation, 749Dirichlet problems, 756–765for ODE, 185, 511–516Neumann problems, 765–767particular integrals from, 514Poisson’s equation, 755Green’s theoremsapplications, 706, 754, 849in a plane, 384–387, 407in three dimensions, 402ground-state energyharmonic oscillator, 796hydrogen atom, 800group multiplication tables, 1050order three, 1062order four, 1050, 1052, 1061order five, 1062order six, 1055, 1061grouping terms as a test for convergence, 129groupsAbelian, 1044associative law, 1043cancellation law, 1046centre, 1069closure, 1043cyclic, 1061definition, 1043–1046direct product, 1072division axiom, 1046elements, 1043order, 1047finite, 1043identity element, 1043–1046inverse, 1043, 1046isomorphic, 1051mappings between, 1059–1061homomorphic, 1059–1061image, 1059isomorphic, 10591316

INDEXnomenclature, 1102non-Abelian, 1052–1056order, 1043, 1081, 1082, 1094, 1097, 1100permutation law, 1047subgroups, see subgroupsgroups, examples1and−1 under multiplication, 1043alternating, 1116complex numbers e iθ , 1048functions, 1055general linear, 1073integers under addition, 1043integers under multiplication (mod N),1049–1051matrices, 1054permutations, 1056–1058quaternion, 1073rotation matrices, 1048symmetries of a square, 1100symmetries of an equilateral triangle, 1047H n (x), see Hermite polynomialsHamilton’s principle, 788Hamiltonian, 796Hankel functions H ν (1) (x), H ν (2) (x), 607Hankel transforms, 459harmonic oscillatorsdamped, 239, 451ground-state energy, 796Schrödinger equation, 796simple, see simple harmonic oscillatorheat flowdiffusion equation, 678, 696, 723in bar, 723, 749, 770in thin sheet, 698Heaviside function, 441relation to Dirac δ-function, 441Heisenberg’s uncertainty principle, 435–437Helmholtz equation, 737–741cylindrical polars, 740plane polars, 738spherical polars, 740–741Helmholtz potential, 177hemisphere, centre of mass and centroid, 195Hermite equation, 535, 624–628as example of Sturm–Liouville equation, 566natural interval, 567Hermite polynomials H n (x), 625as special case of confluent hypergeometricfunction, 634generating function, 627graph of, 625normalisation, 626orthogonality, 626recurrence relations, 628Rodrigues’ formula, 626Hermitian conjugate, 256–258and inner product, 258product rule, 257Hermitian forms, 288–292positive definite and semi-definite, 290stationary properties of eigenvectors, 290Hermitian kernel, 816Hermitian matrices, 271eigenvalues, 276–278reality, 276eigenvectors, 276–278orthogonality, 277Hermitian operators, 559–564and physical variables, 650boundary condition for simple harmonicoscillators, 560eigenfunctionscompleteness, 560, 563orthogonality, 561–563eigenvaluesreality, 561Green’s functions, 568–571importance of, 555, 560in Sturm–Liouville equations, 564properties, 561–564superposition methods, 568–571higher-order differential equations, see ordinarydifferential equationsHilbert spaces, 557–559hit or miss, in Monte Carlo methods, 1014homogeneousboundary conditions, see boundaryconditions, homogeneous andinhomogeneousdifferential equations, 490dimensionally consistent, 475, 521simultaneous linear equations, 293homomorphism, 1059–1061kernel of, 1060representation as, 1083Hooke’s law, 953hydrogen atoms-states, 1144electron wavefunction, 208ground-state energy, 800hydrogen molecule, symmetries of, 1041hyperbolaas section of quadratic surface, 292equation for, 16hyperbolic functions, 102–109, 833calculus of, 106–109definitions, 102, 833graphs, 102identities, 104in equations, 105inverses, 105graphs, 106trigonometric analogies, 102–104hyperbolic PDE, 687, 690hypergeometric distribution, 1173mean and variance, 1173hypergeometric equation, 535, 628–632as example of Sturm–Liouville equation, 566,5671317

INDEXgeneral solution, 630natural interval, 567hypergeometric functions, 628–632contiguous relations, 632integral representation, 631recurrence relations, 632special cases, 630hypothesis testing, 1277–1298errors, first and second kind, 1280generalised likelihood ratio, 1282generalised likelihood-ratio test, 1281goodness of fit, 1296Neyman–Pearson test, 1280null, 1278power, 1280rejection region, 1279simple or composite, 1278statistical tests, 1278test statistic, 1278i, j, k (unit vectors), 219i, squarerootof−1, 84identity element of a group, 1043–1046uniqueness, 1043, 1045identity matrices, 254, 255identity operator, 249images, method of, see method of imagesimaginary part or term of a complex number, 83importance sampling, in Monte Carlo methods,1012improperintegrals, 70rotations, 946–948impulses, δ-function respresentation, 441incomplete gamma function, 639independent random variables, 1156, 1200index of a subgroup, 1066indices, of regular singular points, 540indicial equation, 540distinct roots with non-integral difference,540–542repeated roots, 542, 546, 547roots differ by integer, 542, 546induction, proof by, 31inequalitiesamongst integrals, 72Bessel, 246, 559Schwarz, 246, 559triangle, 246, 559inertia, see also moments of inertiamoments and products, 951tensor, 951inexact differentials, 155inexact equation, 473infinite integrals, 70contour integration, 862–867infinite series, see seriesinflectiongeneral points of, 52stationary points of, 50–52inhomogeneousboundary conditions, see boundaryconditions, homogeneous andinhomogeneousdifferential equations, 490simultaneous linear equations, 293inner product in a vector space, see also scalarproductof finite dimensionality, 244and Hermitian conjugate, 258commutativity, 244distributivity over addition, 244of infinite dimensionality, 557integral equationseigenfunctions, 817eigenvalues, 808, 816Fredholm, 805from differential equations, 803homogeneous, 805linear, 804first kind, 805second kind, 805nomenclature, 804singular, 805Volterra, 805integral equations, methods fordifferentiation, 812Fredholm theory, 815integral transforms, 809–812Fourier, 809–812Laplace, 810Neumann series, 813–815Schmidt–Hilbert theory, 816–819separable (degenerate) kernels, 807integral functions, 832nintegral test for convergence of series, 128integral transforms, see also Fourier transformsand Laplace transformsgeneral form, 459Hankel transforms, 459Mellin transforms, 459integrals, see also integrationcomplex, see complex integralsdefinite, 59double, see multiple integralsFourier transform of, 444improper, 70indefinite, 62inequalities, 72, 559infinite, 70Laplace transform of, 456limitscontaining variables, 188fixed, 59variable, 61line, see line integralsmultiple, see multiple integralsnon-zero, 1104of vectors, 339properties, 601318

INDEXtriple, see multiple integralsundefined, 59integrand, 59integrating factor (IF), 506first-order ODE, 473–475integration, see also integralsapplications, 72–76finding the length of a curve, 73mean value of a function, 72surfaces of revolution, 74volumes of revolution, 75as area under a curve, 59as the inverse of differentiation, 61formal definition, 59from first principles, 59in plane polar coordinates, 70logarithmic, 64multiple, see multiple integralsmultivalued functions, 865–867, 885of Fourier series, 424of functions of several real variables, seemultiple integralsof hyperbolic functions, 106–109of power series, 135of simple functions, 62of singular functions, 70of sinusoidal functions, 63, 861integration constant, 62integration, methods forby inspection, 62by parts, 67–69by substitution, 65–67t substitution, 65change of variables, see change of variablescompleting the square, 66contour, see contour integrationGaussian, 1005–1009numerical, 1000–1009partial fractions, 64reduction formulae, 69stationary phase, 912–920steepest descents, 908–912trigonometric expansions, 63using complex numbers, 101intersection ∩, probability for, 1120, 1128intrinsic derivative, see absolute derivativeinvariant tensors, see isotropic tensorsinverse hyperbolic functions, 105inverse integral transformsFourier, 435Laplace, 454, 884–887uniqueness, 454inverse matrices, 263–266elements, 264in solution of simultaneous linear equations,295product rule, 266properties, 265inverse of a linear operator, 249inverse of a product in a group, 1046inverse of element in a groupuniqueness, 1043, 1046inversion theorem, Fourier’s, 435inversions asimproper rotations, 946symmetry operations, 1041irregular singular points, 534irreps, 1087counting, 1095dimension n λ , 1097direct sum ⊕, 1086identity A 1 , 1100, 1104n-dimensional, 1088, 1089, 1102number in a representation, 1087, 1095one-dimensional, 1088, 1089, 1093, 1099, 1102orthogonality theorem, 1090–1092projection operators for, 1107reduction to, 1096summation rules for, 1097–1099irrotational vectors, 353isobaric ODE, 476non-linear, 521isoclines, method of, 1028, 1037isomorphic groups, 1051–1056, 1058, 1059isomorphism (mapping), 1060isotope decay, 484, 525isotropic (invariant) tensors, 944–946, 953iteration schemesconvergence of, 992–994for algebraic equations, 986–994for differential equations, 1025for integral equations, 813–816Gauss–Seidel, 996–998order of convergence, 993J ν (x), see Bessel functionsj, squarerootof−1, 84j l (x), see spherical Bessel functionsJacobiansanalogy with derivatives, 207and change of variables, 206definition intwo dimensions, 201three dimensions, 205general properties, 206in terms of a determinant, 201, 205, 207joint distributions, see bivariate distributionsand multivariate distributionsJordan’s lemma, 864kernel of a homomorphism, 1060, 1063kernel of an integral transform, 459kernel of integral equationsdisplacement, 809Hermitian, 816of form exp(−ixz), 810–812of linear integral equations, 804resolvent, 814, 815separable (degenerate), 807ket vector |ψ〉, 6481319

INDEXkinetic energy of oscillating system, 316Klein–Gordon equation, 711, 772Kronecker delta δ ij and orthogonality, 244Kronecker delta, δ ij , δ j i , tensor, 928, 941–946,956, 962identities, 943isotropic, 945vector products, 942Kummer function, 633kurtosis, 1150, 1227L n (x), see Laguerre polynomialsL m n (x), see associated Laguerre polynomialsL’Hôpital’s rule, 142–144Lagrange equations, 789and energy conservation, 797Lagrange undetermined multipliers, 167–173and ODE eigenvalue estimation, 792application to stationary properties of theeigenvectors of quadratic and Hermitianforms, 290for functions of more than two variables,169–173in deriving the Boltzmann distribution,171–173integral constraints, 785with several constraints, 169–173Lagrange’s identity, 226Lagrange’s theorem, 1065and the order of a subgroup, 1062and the order of an element, 1062Lagrangian, 789, 797Laguerre equation, 535, 616–621as example of Sturm–Liouville equation, 566,619natural interval, 567, 619Laguerre polynomials L n (x), 617as special case of confluent hypergeometricfunction, 634generating function, 620graph of, 617normalisation, 619orthogonality, 619recurrence relations, 620Rodrigues’ formula, 618Lamé constants, 953lamina: mass, centre of mass and centroid,193–195Laplace equation, 679expansion methods, 741–744in two dimensions, 688, 690, 717, 718and analytic functions, 829and conformal transformations, 876–879numerical method for, 1031, 1038plane polars, 725–727separated variables, 717in three dimensionscylindrical polars, 728–731spherical polars, 731–737uniqueness of solution, 741with specified boundary values, 764, 766Laplace expansion, 259Laplace transforms, 453–459, 884convolutionassociativity, commutativity, distributivity,458definition, 457convolution theorem, 457definition, 453for ODE with constant coefficients, 501–503for PDE, 747–748inverse, 454, 884–887uniqueness, 454properties: translation, exponentialmultiplication, etc., 456table for common functions, 455Laplace transforms, examplesconstant, 453derivatives, 455exponential function, 453integrals, 456polynomial, 453Laplacian, see del squared ∇ 2 (Laplacian)Laurent expansion, 855–858analytic and principal parts, 855region of convergence, 855least squares, method of, 1271–1277basis functions, 1273linear, 1272non-linear, 1276response matrix, 1273Legendre equation, 534, 535, 577–586as example of Sturm–Liouville equation, 566,583associated, see associated Legendre equationgeneral solution, 578, 580natural interval, 567, 583Legendre functions P l (x), 577–586associated Legendre functions, 768of second kind Q l (x), 579graph of, 580Legendre linear equation, 503Legendre polynomials P l (x), 578as special case of hypergeometric function,631associated Legendre functions, 733generating function, 584–586graph of, 579in Gaussian integration, 1006normalisation, 578, 582orthogonality, 583, 735recurrence relations, 585, 586Rodrigues’ formula, 581Leibnitz’ rule for differentiation of integrals, 178Leibnitz’ theorem, 48–50length ofa vector, 218plane curves, 73, 341space curves, 341tensor form, 9821320

INDEXlevel lines, 905, 906Levi-Civita symbol, see ɛ ijk , Levi-Civita symbol,tensorlikelihood function, 1255limits, 141–144definition, 141L’Hôpital’s rule, 142–144of functions containing exponents, 142of integrals, 59containing variables, 188of products, 141of quotients, 141–144of sums, 141line charge, electrostatic potential, 872, 878line integralsand Cauchy integrals, 851–853and Stokes’ theorem, 406–409of scalars, 377–387of vectors, 377–389physical examples, 381round closed loop, 386line of steepest descents, 908line, vector equation of, 226linear dependence and independencedefinition in a vector space, 242of basis vectors, 217relationship with rank, 267linear differential operator L, 511, 545, 554adjoint L † , 559eigenfunctions, see eigenfunctionseigenvalues, see eigenvalues, of lineardifferential operatorsfor Sturm-Liouville equation, 564–568Hermitian, 555, 559–564self-adjoint, 559linear equations, differentialfirst-order ODE, 474general ODE, 490–517ODE with constant coefficients, 492–503ODE with variable coefficients, 503–517linear equations, simultaneous, see simultaneouslinear equationslinear independence of functions, 491Wronskian test, 491, 532linear integral operator K, 805and Schmidt–Hilbert theory, 816–818Hermitian conjugate, 805inverse, 806linear interpolation for algebraic equations, 988linear least squares, method of, 1272linear moleculesnormal modes of, 320–322symmetries of, 1077linear operators, 247–249associativity, 249distributivity over addition, 249eigenvalues and eigenvectors, 272in a particular basis, 248inverse, 249non-commutativity, 249particular: identity, null or zero, singular andnon-singular, 249properties, 249linear vector spaces, see vector spaceslines of steepest descent, 906Liouville’s theorem, 853Ln of a complex number, 99, 834ln (natural logarithm)Maclaurin series for, 140of a complex number, 99, 834log-likelihood function, 1258longitudinal vibrations in a rod, 677lottery (UK), and hypergeometric distribution,1174lower triangular matrices, 269Maclaurin series, 138standard expressions, 140Madelung constant, 149magnetic dipole, 220magnitude of a vector, 218in terms of scalar or dot product, 221mappings between groups, see groups, mappingsbetweenmarginal distributions, 1198mass of non-uniform bodies, 193matrices, 241–307as a vector space, 252as arrays of numbers, 249as representation of a linear operator, 249column, 250elements, 249minors and cofactors, 259identity or unit, 254row, 250zero or null, 254matrices, algebra of, 250addition, 251change of basis, 283–285Cholesky separation, 313diagonalisation, see diagonalisation ofmatricesmultiplication, 252–254and common eigenvalues, 278commutator, 309non-commutativity, 254multiplication by a scalar, 251normal modes, see normal modesnumerical methods, see numerical methodsfor simultaneous linear equationssimilarity transformations, see similaritytransformationssimultaneous linear equations, seesimultaneous linear equationssubtraction, 251matrices, derivedadjoint, 256–258complex conjugate, 256–258Hermitian conjugate, 256–258inverse, see inverse matrices1321

INDEXtranspose, 250matrices, properties ofanti- or skew-symmetric, 270anti-Hermitian, see anti-Hermitian matricesdeterminant, see determinantsdiagonal, 268eigenvalues, see eigenvalueseigenvectors, see eigenvectorsHermitian, see Hermitian matricesnormal, see normal matricesnullity, 293order, 249orthogonal, 270rank, 267square, 249symmetric, 270trace or spur, 258triangular, 269tridiagonal, 998–1000, 1030unitary, see unitary matricesmatrix elements in quantum mechanicsas integrals, 1103dipole, 1108, 1115maxima and minima (local) of a function ofconstrained variables, see Lagrangeundetermined multipliersone real variable, 50–52sufficient conditions, 51several real variables, 162–167sufficient conditions, 164, 167maximum modulus theorem, 881maximum-likelihood, method of, 1255–1271and Bayesian approach, 1264bias, 1260data modelling, 1255estimator, 1256extended, 1270log-likelihood function, 1258parameter estimation, 1255transformation invariance, 1260Maxwell’selectromagnetic equations, 373, 408, 979thermodynamic relations, 176–178Maxwell–Boltzmann statistics, 1138mean µfrom MGF, 1163from PGF, 1158of RVD, 1144of sample, 1223of sample: geometric, harmonic, root meansquare, 1223mean value of a function ofone variable, 72several variables, 199mean value theorem, 56median of RVD, 1145membranedeformed rim, 725–727normal modes, 739, 1112transverse vibrations, 677, 739, 768, 1112method of images, 706, 758–765, 878disc (section of cylinder), 764, 766infinite plate, 759intersecting plates in two dimensions, 761sphere, 762–764, 772metric tensor, 957–960, 963and Christoffel symbols, 966and scale factors, 957, 972covariant derivative of, 982determinant, 957, 964derivative of, 973length element, 957raising or lowering index, 959, 963scalar product, 958volume element, 957, 981MGF, see moment generating functionsMilne’s method, 1022minimum-variance estimator, 1232minor of a matrix element, 259mixed, components of tensor, 957, 962, 969ML estimators, 1256bias, 1260confidence limits, 1262efficiency, 1261transformation invariance, 1260mode of RVD, 1145modulo, mod N, multiplication, 1049modulusof a complex number, 87of a vector, see magnitude of a vectormoleculesbonding in, 1103, 1105–1108dipole moments of, 1077symmetries of, 1077moment generating functions (MGFs),1162–1167and central limit theorem, 1195and PGF, 1163mean and variance, 1163particular distributionsbinomial, 1170exponential, 1191Gaussian, 1163, 1185Poisson, 1177properties, 1163moments (of distributions)central, 1148of RVD, 1147moments (of forces), vector representation of,223moments of inertiaand inertia tensor, 951definition, 198of disc, 208of rectangular lamina, 198of right circular cylinder, 209of sphere, 205perpendicular axes theorem, 209momentum as first-order tensor, 933monomorphism, 10611322

INDEXMonte Carlo methods, of integration, 1009–1017antithetic variates, 1014control variates, 1013crude, 1011hit or miss, 1014importance sampling, 1012multiple integrals, 1016random number generation, 1017stratified sampling, 1012Morera’s theorem, 851multinomial distribution, 1208and multiple Poisson distribution, 1218multiple angles, trigonometric formulae, 10multiple integralsapplication in findingarea and volume, 191–193mass, centre of mass and centroid, 193–195mean value of a function of severalvariables, 199moments of inertia, 198change of variablesdouble integrals, 200–204general properties, 206triple integrals, 204definitions ofdouble integrals, 187triple integrals, 190evaluation, 188–190notation, 188, 189, 191order of integration, 188, 191multiplication tables for groups, see groupmultiplication tablesmultiplication theorem, see Parseval’s theoremmultivalued functions, 835–837integration of, 865–867multivariate distributions, 1196, 1207–1211change of variables, 1206Gaussian, 1209multinomial, 1208mutually exclusive events, 1120, 1129n l (x), see spherical Bessel functionsnabla ∇, see gradient operator (grad)natural intervalfor associated Laguerre equation, 567, 622for associated Legendre equation, 567, 590,591for Bessel equation, 608for Chebyshev equation, 567, 599for Hermite equation, 567for Laguerre equation, 567, 619for Legendre equation, 567, 583for simple harmonic oscillator equation, 567for Sturm–Liouville equations, 565, 567natural logarithm, see ln and Lnnatural numbers, in series, 31, 121natural representations, 1081, 1110necessary and sufficient conditions, 34negative binomial distribution, 1172negative function, 556negative vector, 242Neumann boundary conditions, 702Green’s functions, 754, 765–767method of images, 765–767self-consistency, 765Neumann functions Y ν (x), 607Neumann series, 813–815Newton–Raphson (NR) method, 990–992order of convergence, 993Neyman–Pearson test, 1280nodes of oscillation, 693non-Abelian groups, 1052–1056of functions, 1055of matrices, 1054of permutations, 1056–1058of rotations and reflections, 1052non-Cartesian coordinates, see curvilinear,cylindrical polar, plane polar and sphericalpolar coordinatesnon-linear differential equations, see ordinarydifferential equations, non-linearnon-linear least squares, method of, 1276norm offunction, 557vector, 244normalto coordinate surface, 366to plane, 228to surface, 346, 350, 390normal derivative, 350normal distribution, see Gaussian (normal)distributionnormal matrices, 272eigenvectorscompleteness, 275orthogonality, 275eigenvectors and eigenvalues, 273–276normal modes, 316–329characteristic equation, 319coupled pendulums, 329, 331definition, 320degeneracy, 1110–1113frequencies of, 319linear molecular system, 320–322membrane, 739, 1112normal coordinates, 320normal equations, 320rod–string system, 317–320symmetries of, 322normal subgroups, 1063normalisation ofeigenfunctions, 562eigenvectors, 273functions, 557vectors, 219null (zero)matrix, 254, 255operator, 249space, of a matrix, 293vector, 214, 242, 5561323

INDEXnull operation, as identity element of group,1044nullity, of a matrix, 293numerical methods for algebraic equations,985–992binary chopping, 990convergence of iteration schemes, 992–994linear interpolation, 988Newton–Raphson, 990–992rearrangement methods, 987numerical methods for integration, 1000–1009Gaussian integration, 1005–1009mid-point rule, 1034Monte Carlo, 1009nomenclature, 1001Simpson’s rule, 1004trapezium rule, 1002–1004numerical methods for ordinary differentialequations, 1020–1030accuracy and convergence, 1021Adams method, 1024difference schemes, 1021–1023Euler method, 1021first-order equations, 1021–1028higher-order equations, 1028–1030isoclines, 1028Milne’s method, 1022prediction and correction, 1024–1026reduction to matrix form, 1030Runge–Kutta methods, 1026–1028Taylor series methods, 1023numerical methods for partial differentialequations, 1030–1032diffusion equation, 1032Laplace’s equation, 1031minimising error, 1032numerical methods for simultaneous linearequations, 994–1000Gauss–Seidel iteration, 996–998Gaussian elimination with interchange, 995matrix form, 994–1000tridiagonal matrices, 998–1000O(x), order of, 132observables in quantum mechanics, 277, 560odd functions, see antisymmetric functionsODE, see ordinary differential equations (ODEs)operatorsHermitian, see Hermitian operatorslinear, see linear operators and lineardifferential operator and linear integraloperatoroperators (quantum)angular momentum, 656–663annihilation and creation, 667coordinate-free, 648–671eigenvalues and eigenstates, 649physical examplesangular momentum, 658Hamiltonian, 657order ofapproximation in Taylor series, 137nconvergence of iteration schemes, 993group, 1043group element, 1047ODE, 468permutation, 1058recurrence relations (series), 497subgroup, 1061and Lagrange’s theorem, 1065tensor, 930ordinary differential equations (ODE), see alsodifferential equations, particularboundary conditions, 468, 470, 501complementary function, 491degree, 468dimensionally homogeneous, 475exact, 472, 505first-order, 468–484first-order higher-degree, 480–484soluble for p, 480soluble for x, 481soluble for y, 482general form of solution, 468–470higher-order, 490–523homogeneous, 490inexact, 473isobaric, 476, 521linear, 474, 490–517non-linear, 518–523exact, 519isobaric (homogeneous), 521x absent, 518y absent, 518order, 468ordinary point, see ordinary points of ODEparticular integral (solution), 469, 492, 494singular point, see singular points of ODEsingular solution, 469, 481, 482, 484ordinary differential equations, methods forcanonical form for second-order equations,516eigenfunctions, 554–573equations containing linear forms, 478–480equations with constant coefficients, 492–503Green’s functions, 511–516integrating factors, 473–475Laplace transforms, 501–503numerical, 1020–1030partially known CF, 506separable variables, 471series solutions, 531–550, 604undetermined coefficients, 494variation of parameters, 508–510ordinary points of ODE, 533, 535–538indicial equation, 543orthogonal lines, condition for, 12orthogonal matrices, 270, 929, 930general properties, see unitary matricesorthogonal systems of coordinates, 3641324

INDEXorthogonal transformations, 932orthogonalisation (Gram–Schmidt) ofeigenfunctions of an Hermitian operator, 562eigenvectors of a normal matrix, 275functions in a Hilbert space, 557orthogonality ofeigenfunctions of an Hermitian operator,561–563eigenvectors of a normal matrix, 275eigenvectors of an Hermitian matrix, 277functions, 557terms in Fourier series, 417, 425vectors, 219, 244orthogonality properties of characters, 1094,1102orthogonality theorem for irreps, 1090–1092orthonormalbasis functions, 557basis vectors, 244under unitary transformation, 285oscillations, see normal modesoutcome, of trial, 1119outer product of two vectors, 936P l (x), see Legendre polynomialsPl m (x), see associated Legendre functionsPappus’ theorems, 195–197parabola, equation for, 16parabolic PDE, 687, 690parallel axis theorem, 238parallel vectors, 223parallelepiped, volume of, 225parallelogram equality, 247parallelogram, area of, 223, 224parameter estimation (statistics), 1229–1255,1298Bessel correction, 1248error in mean, 1298maximum-likelihood, 1255mean, 1243variance, 1245–1248parameters, variation of, 508–510parametric equationsof conic sections, 17of cycloid, 370, 785of space curves, 340of surfaces, 345parity inversion, 1102Parseval’s theoremconservation of energy, 451for Fourier series, 426for Fourier transforms, 450partial derivative, see partial differentiationpartial differential equations (PDE), 675–707,713–767, see also differential equations,particulararbitrary functions, 680–685boundary conditions, 681, 699–707, 723characteristics, 699–705and equation type, 703equation types, 687, 710first-order, 681–687general solution, 681–692homogeneous, 685inhomogeneous equation and problem,685–687, 744–746, 751–767particular solutions (integrals), 685–692second-order, 687–698partial differential equations (PDE), methods forchange of variables, 691, 696–698constant coefficients, 687general solution, 689integral transform methods, 747–751method of images, see method of imagesnumerical, 1030–1032separation of variables, see separation ofvariablessuperposition methods, 717–724with no undifferentiated term, 684partial differentiation, 151–179as gradient of a function of several realvariables, 151chain rule, 157change of variables, 158–160definitions, 151–153properties, 157cyclic relation, 157reciprocity relation, 157partial fractions, 18–25and degree of numerator, 21as a means of integration, 64complex roots, 22in inverse Laplace transforms, 454, 502repeated roots, 23partial sum, 115particular integrals (PI), 469, see also ordinarydifferential equation, methods for andpartial differential equations, methods forpartition of agroup, 1064set, 1065parts, integration by, 67–69path integrals, see line integralsPDE, see partial differential equationsPDFs, 1140pendulums, coupled, 329, 331periodic function representation, see Fourierseriespermutation groups S n , 1056–1058cycle notation, 1057permutation law in a group, 1047permutations, 1133–1139degree, 1056distinguishable, 1135order of, 1058symbol n P k , 1133perpendicular axes theorem, 209perpendicular vectors, 219, 244PF, see probability functionsPGF, see probability generating functions1325

INDEXphase memory, 895phase, complex, 896PI, see particular integralsplane curves, length of, 73in Cartesian coordinates, 73in plane polar coordinates, 74plane polar coordinates, 70, 336arc length, 74, 361area element, 202, 361basis vectors, 336velocity and acceleration, 337plane waves, 695, 716planesand simultaneous linear equations, 300vector equation of, 227plates, conducting, see also complex potentials,for platesline charge near, 761point charge near, 759point charges, δ-function respresentation, 441point groups, 1082points of inflection of a function ofone real variable, 50–52several real variables, 162–167Poisson distribution Po(λ), 1174–1179and Gaussian distribution, 1187as limit of binomial distribution, 1174, 1177mean and variance, 1176MGF, 1177multiple, 1178recurrence formula, 1176Poisson equation, 575, 679, 744–746fundamental solution, 757Green’s functions, 753–767uniqueness, 705–707Poisson summation formula, 461Poisson’s ratio, 953polar coordinates, see plane polar and cylindricalpolar and spherical polar coordinatespolar representation of complex numbers, 92–95polar vectors, 949pole, of a function of a complex variablecontours containing, 861–867, 884–887order, 837, 856residue, 856–858polynomial equations, 1–10conjugate roots, 99factorisation, 7multiplicities of roots, 4number of roots, 83, 85, 868properties of roots, 9real roots, 1solution of, using de Moivre’s theorem, 98polynomial solutions of ODE, 538, 548–550populations, sampling of, 1222positive (semi-) definite quadratic and Hermitianforms, 290positive semi-definite norm, 244potential energy ofion in a crystal lattice, 148magnetic dipole in a field, 220oscillating system, 317potential functionand conservative fields, 389complex, 871–876electrostatic, see electrostatic fields andpotentialsgravitational, see gravitational fields andpotentialsvector, 389power seriesand differential equations, see series solutionsof differential equationsinterval of convergence, 132Maclaurin, see Maclaurin seriesmanipulation: difference, differentiation,integration, product, substitution, sum, 134Taylor, see Taylor seriespower series in a complex variable, 133, 830–832analyticity, 832circle and radius of convergence, 133, 831convergence tests, 831, 832form, 830power, in hypothesis testing, 1280powers, complex, 99, 833prediction and correction methods, 1024–1026,1035prime, non-existence of largest, 34principal axes ofCartesian tensors, 951–953conductivity tensors, 952inertia tensors, 951quadratic surfaces, 292rotation symmetry, 1102principal normals of space curves, 342principal value ofcomplex integrals, 864complex logarithms, 100, 834principle of the argument, 880probability, 1124–1211axioms, 1125conditional, 1128–1133Bayes’ theorem, 1132combining, 1130definition, 1125for intersection ∩, 1120for union ∪, 1121, 1125–1128probability distributions, 1139, see alsoindividual distributionsbivariate, see bivariate distributionschange of variables, 1150–1157cumulants, 1166generating functions, see moment generatingfunctions and probability generatingfunctionsmean µ, 1144mean of functions, 1145mode, median and quartiles, 1145moments, 1147–1150multivariate, see multivariate distributions1326

INDEXstandard deviation σ, 1146variance σ 2 , 1146probability functions (PFs), 1139cumulative (CPFs), 1139, 1141density functions (PDFs), 1140probability generating functions (PGFs),1157–1162and MGF, 1163binomial, 1161definition, 1158geometric, 1159mean and variance, 1158Poisson, 1158sums of RV, 1161trials, 1158variable sums of RV, 1161product rule for differentiation, 44–46, 48–50products of inertia, 951projection operators for irreps, 1107, 1116projection tensors, 979proper rotations, 946proper subgroups, 1061pseudoscalars, 947, 950pseudotensors, 946–950, 964pseudovectors, 946–950Q l (x), see Legendre polynomialsQ m l (x), see associated Legendre functionsquadratic equationscomplex roots of, 83properties of roots, 10roots of, 2quadratic forms, 288–292completing the square, 1206positive definite and semi-definite, 290quadratic surfaces, 292removing cross terms, 289stationary properties of eigenvectors, 290quantum mechanics, from classical mechanics,657quantum operators, see operators (quantum)quartiles, of RVD, 1145quaternion group, 1073, 1113quotient law for tensors, 939–941quotient rule for differentiation, 47quotient test for series, 127radius of convergence, 133, 831radius of curvatureof plane curves, 53of space curves, 342radius of torsion of space curves, 343random number generation, 1017random numbers, non-uniform distribution,1035random variable distributions, see probabilitydistributionsrandom variables (RV), 1119, 1139–1143continuous, 1140–1143dependent, 1196–1205discrete, 1139independent, 1156, 1200sums of, 1160–1162uncorrelated, 1200range of a matrix, 293rank of matrices, 267and determinants, 267and linear dependence, 267rank of tensors, see order of tensorrate of change of a function ofone real variable, 41several real variables, 153–155ratio comparison test, 127ratio test (D’Alembert), 126, 832in convergence of power series, 132ratio theorem, 215and centroid of a triangle, 216Rayleigh–Ritz method, 327–329, 800real part or term of a complex number, 83real roots, of a polynomial equation, 1rearrangement methods for algebraic equations,987reciprocal vectors, 233, 366, 955, 959reciprocity relation for partial derivatives, 157rectangular distribution, 1194Fourier transform of, 442recurrence relations (functions), 585, 611associated Laguerre polynomials, 624associated Legendre functions, 592Chebyshev polynomials, 601confluent hypergeometric functions, 635Hermite polynomials, 628hypergeometric functions, 632Laguerre polynomials, 620Legendre polynomials, 586recurrence relations (series), 496–501characteristic equation, 499coefficients, 536, 538, 999first-order, 497second-order, 499higher-order, 501reducible representations, 1084, 1086reduction formulae for integrals, 69reflectionsand improper rotations, 946as symmetry operations, 1041reflexivity, and equivalence relations, 1064regular functions, see analytic functionsregular representations, 1097, 1110regular singular points, 534, 538–540relative velocities, 218remainder term in Taylor series, 138repeated roots of auxiliary equation, 493representation, 1076definition, 1082dimension of, 1078, 1082equivalent, 1084–1086faithful, 1083, 1098generation of, 1078–1084, 1112irreducible, see irreps1327

INDEXnatural, 1081, 1110product, 1103–1105reducible, 1084, 1086regular, 1097, 1110counting irreps, 1098unitary, 1086representative matrices, 1079block-diagonal, 1086eigenvalues, 1100inverse, 1083number needed, and order of group, 1082of identity, 1082residueat a pole, 856–858theorem, 858–860resolution function, 446resolvent kernel, 814, 815response matrix, for linear least squares, 1273rhomboid, volume of, 237Riemann tensor, 981Riemann theorem for conditional convergence,124Riemann zeta series, 128, 129right hand screw rule, 222Rodrigues’ formula forassociated Laguerre polynomials, 622associated Legendre functions, 588Chebyshev polynomials, 599Hermite polynomials, 626Laguerre polynomials, 618Legendre polynomials, 581Rolle’s theorem, 55root test (Cauchy), 129, 831rootsof a polynomial equation, 2properties, 9of unity, 97rope, suspended at its ends, 786rotation groups (continuous), invariantsubspaces, 1088rotation matrices as a group, 1048rotation of a vector, see curlrotationsas symmetry operations, 1041axes and orthogonal matrices, 930, 931, 961improper, 946–948invariance under, 934product of, 931proper, 946Rouché’s theorem, 880–882row matrix, 250Runge–Kutta methods, 1026–1028RV, see random variablesRVD (random variable distributions), seeprobability distributionssaddle point method of integration, 908saddle points, 162and integral evaluation, 905sufficient conditions, 164, 167samplingcorrelation, 1227covariance, 1227space, 1119statistics, 1222–1229with or without replacement, 1129scalar fields, 347derivative along a space curve, 349gradient, 348–352line integrals, 377–387rate of change, 349scalar product, 219–222and inner product, 244and metric tensor, 958and perpendicular vectors, 219, 244for vectors with complex components, 221in Cartesian coordinates, 221invariance, 930, 939scalar triple product, 224–226cyclic permutation of, 225in Cartesian coordinates, 225determinant form, 225interchange of dot and cross, 225scalars, 212invariance, 930zero-order tensors, 933scale factors, 359, 362, 364and metric tensor, 957, 972scattering in quantum mechanics, 463Schmidt–Hilbert theory, 816–819Schrödinger equation, 679constant potential, 768hydrogen atom, 741numerical solution, 1039variational approach, 795Schwarz inequality, 246, 559Schwarz–Christoffel transformation, 843second differences, 1019second-order differential equations, see ordinarydifferential equations and partial differentialequationssecular determinant, 280self-adjoint operators, 559–564, see alsoHermitian operatorssemicircle, angle in, 18semicircular lamina, centre of mass, 197separable kernel in integral equations, 807separable variables in ODE, 471separation constants, 715, 717separation of variables, for PDE, 713–746diffusion equation, 716, 722–724, 737, 751expansion methods, 741–744general method, 713–717Helmholtz equation, 737–741inhomogeneous boundary conditions, 722–724inhomogeneous equations, 744–746Laplace equation, 717–722, 725–737, 741polar coordinates, 725–746separation constants, 715, 717superposition methods, 717–7241328

INDEXwave equation, 714–716, 737, 739series, 115–141convergence of, see convergence of infiniteseriesdifferentiation of, 131finite and infinite, 116integration of, 131multiplication by a scalar, 131multiplication of (Cauchy product), 131notation, 116operations, 131summation, see summation of seriesseries, particulararithmetic, 117arithmetico-geometric, 118Fourier, see Fourier seriesgeometric, 117Maclaurin, 138, 140power, see power seriespowers of natural numbers, 121Riemann zeta, 128, 129Taylor, see Taylor seriesseries solutions of differential equations, 531–550about ordinary points, 535–538about regular singular points, 538–540Frobenius series, 539convergence, 536indicial equation, 540linear independence, 540polynomial solutions, 538, 548–550recurrence relation, 536, 538second solution, 537, 544–548derivative method, 545–548Wronskian method, 544, 580shortest path, 778and geodesics, 976, 982similarity transformations, 283–285, 929, 1092properties of matrix under, 284unitary transformations, 285simple harmonic oscillator, 555energy levels of, 642, 902equation, 535, 566operator formalism, 667simple poles, 838Simpson’s rule, 1004simultaneous linear equations, 292–307and intersection of planes, 300homogeneous and inhomogeneous, 293singular value decomposition, 301–307solution usingCramer’s rule, 299inverse matrix, 295numerical methods, see numerical methodsfor simultaneous linear equationssine, sin(x)in terms of exponential functions, 102Maclaurin series for, 140orthogonality relations, 417singular and non-singularintegral equations, 805linear operators, 249matrices, 263singular integrals, see improper integralssingular point (singularity), 826, 837–839essential, 838, 856removable, 838singular points of ODE, 533irregular, 534particular equations, 535regular, 534singular solution of ODE, 469, 481, 482, 484singular value decompositionand simultaneous linear equations, 301–307singular values, 302sinh, hyperbolic sine, 102, 833, see alsohyperbolic functionsskew-symmetric matrices, 270skewness, 1150, 1227Snell’s law, 788soap films, 780solenoidal vectors, 352, 389solid angleas surface integral, 395subtended by rectangle, 411solid: mass, centre of mass and centroid,193–195source density, 679space curves, 340–344arc length, 341binormal, 342curvature, 342Frenet–Serret formulae, 343parametric equations, 340principal normal, 342radius of curvature, 342radius of torsion, 343tangent vector, 342torsion, 342spaces, see vector spacesspan of a set of vectors, 242sphere, vector equation of, 228spherical Bessel functions, 615, 741of first kind j l (z), 615of second kind n l (z), 615spherical harmonics Yl m (θ, φ), 593–595addition theorem, 594spherical polar coordinates, 361–363area element, 362basis vectors, 362length element, 362vector operators, 361–363volume element, 205, 362spur of a matrix, see trace of a matrixsquare matrices, 249square, symmetries of, 1100square-wave, Fourier series for, 418stagnation points of fluid flow, 873standard deviation σ, 1146of sample, 1224standing waves, 6931329

INDEXstationary phase, method of, 912–920stationary valuesof functions ofone real variable, 50–52several real variables, 162–167of integrals, 776under constraints, see Lagrange undeterminedmultipliersstatistical tests, and hypothesis testing , 1278statistics, 1119, 1221–1298describing data, 1222–1229estimating parameters, 1229–1255, 1298steepest descents, method of, 908–912Stirling’sapproximation, 637, 1185asymptotic series, 637Stokes constant, in Stokes phenomenon, 904Stokes line, 899, 903Stokes phenomenon, 903dominant term, 904Stokes constant, 904subdominant term, 904Stokes’ equation, 643, 799, 888–894Airy integrals, 890–894qualitative solutions, 888series solution, 890Stokes’ theorem, 388, 406–409for tensors, 955physical applications, 408related theorems, 407strain tensor, 953stratified sampling, in Monte Carlo methods,1012streamlines and complex potentials, 873stress tensor, 953stress waves, 980stringloaded, 798plucked, 770transverse vibrations of, 676, 789Student’s t-distributionnormalisation, 1286plots, 1287Student’s t-test, 1284–1290comparison of means, 1289critical points table, 1288one- and two-tailed confidence limits, 1288Sturm–Liouville equations, 564boundary conditions, 564examples, 566associated Laguerre, 566, 622associated Legendre, 566, 590, 591Bessel, 566, 608–611Chebyshev, 566, 599confluent hypergeometric, 566Hermite, 566hypergeometric, 566Laguerre, 566, 619Legendre, 566, 583manipulation to self-adjoint form, 565–568natural interval, 565, 567two independent variables, 801variational approach, 790–795weight function, 790zeros of eigenfunctions, 573subdominant term, in Stokes phenomenon, 904subgroups, 1061–1063index, 1066normal, 1063order, 1061Lagrange’s theorem, 1065proper, 1061trivial, 1061submatrices, 267subscripts and superscripts, 928contra- and covariant, 956covariant derivative, 969dummy, 928free, 928partial derivative, 969summation convention, 928, 955substitution, integration by, 65–67summation convention, 928, 955summation of series, 116–124arithmetic, 117arithmetico-geometric, 118contour integration method, 882difference method, 119Fourier series method, 427geometric, 117powers of natural numbers, 121transformation methods, 122–124differentiation, 122integration, 122substitution, 123superposition methodsfor ODE, 554, 568–571for PDE, 717–724surface integralsand divergence theorem, 401Archimedean upthrust, 396, 410of scalars, vectors, 389–396physical examples, 395surfaces, 345–347area of, 346cone, 74solid, and Pappus’ theorem, 195–197sphere, 346coordinate curves, 346normal to, 346, 350of revolution, 74parametric equations, 345quadratic, 292tangent plane, 346SVD, see singular value decompositionsymmetric functions, 416and Fourier series, 419and Fourier transforms, 445symmetric matrices, 270general properties, see Hermitian matrices1330

INDEXsymmetric tensors, 938symmetry in equivalence relations, 1064symmetry operationson molecules, 1041order of application, 1044t-test, see Student’s t-testt substitution, 65tan −1 x, Maclaurin series for, 140tangent planes to surfaces, 346tangent vectors to space curves, 342tanh, hyperbolic tangent, see hyperbolicfunctionsTaylor series, 136–141and finite differences, 1019, 1026and Taylor’s theorem, 136–139, 853approximation errors, 139in numerical methods, 992, 1003as solution of ODE, 1023for functions of a complex variable, 853–855for functions of several real variables, 160–162remainder term, 138required properties, 136standard forms, 136tensors, see Cartesian tensors and Cartesiantensors, particular and general tensorstest statistic, 1278tetrahedral group, 1115tetrahedronmass of, 194volume of, 192thermodynamicsfirst law of, 176Maxwell’s relations, 176–178top-hat function, see rectangular distributiontorque, vector representation of, 223torsion of space curves, 342total derivative, 154total differential, 154trace of a matrix, 258and second-order tensors, 939as sum of eigenvalues, 280, 287invariance under similarity transformations,284, 1092trace formula, 287transcendental equations, 986transformation matrix, 283, 289transformationsactive and passive, 948conformal, 839–844coordinate, see coordinate transformationssimilarity, see similarity transformationstransforms, integral, see integral transforms andFourier transforms and Laplace transformstransients, and the diffusion equation, 723transitivity in equivalence relations, 1064transpose of a matrix, 250, 255product rule, 256transverse vibrationsmembrane, 677, 739, 768rod, 769string, 676trapezium rule, 1002–1004trial functionsfor eigenvalue estimation, 793for particular integrals of ODE, 494trials, 1119triangle inequality, 246, 559triangle, centroid of, 216triangular matrices, 269tridiagonal matrices, 998–1000, 1030, 1033trigonometric identities, 10–15triple integrals, see multiple integralstriple scalar product, see scalar triple producttriple vector product, see vector triple productturning point, 50uncertainty principle (Heisenberg), 435–437from commutator, 664undetermined coefficients, method of, 494undetermined multipliers, see Lagrangeundetermined multipliersuniform distribution, 1194union ∪, probability for, 1121, 1125, 1128uniqueness theoremKlein–Gordon equation, 711Laplace equation, 741Poisson equation, 705–707unit step function, see Heaviside functionunit vectors, 219unitarymatrices, 271eigenvalues and eigenvectors, 278representations, 1086transformations, 285upper triangular matrices, 269variable end-points, 782–785variable, dummy, 61variables, separation of, see separation ofvariablesvariance σ 2 , 1146from MGF, 1163from PGF, 1159of dependent RV, 1203of sample, 1224variation of parameters, 508–510variation, constrained, 785–787variational principles, physical, 787–790Fermat, 787Hamilton, 788variations, calculus of, see calculus of variationsvector operators, 347–369acting on sums and products, 354combinations of, 355–357curl, 353, 368del ∇, 348del squared ∇ 2 , 352divergence (div), 352geometrical definitions, 398–4001331

INDEXgradient operator (grad), 348–352, 367identities, 356, 978Laplacian, 352, 368non-Cartesian, 357–369tensor forms, 971–975curl, 974divergence, 972gradient, 972Laplacian, 973vector product, 222–224anticommutativity, 222definition, 222determinant form, 224in Cartesian coordinates, 224non-associativity, 222vector spaces, 242–247, 1113associativity of addition, 242basis vectors, 243commutativity of addition, 242complex, 242defining properties, 242dimensionality, 243group actions on, 1088inequalities: Bessel, Schwarz, triangle, 246invariant, 1088, 1113matrices as an example, 252of infinite dimensionality, 556–559associativity of addition, 556basis functions, 556commutativity of addition, 556defining properties, 556Hilbert spaces, 557–559inequalities: Bessel, Schwarz, triangle, 559parallelogram equality, 247real, 242span of a set of vectors in, 242vector triple product, 226identities, 226non-associativity, 226vectorsas first-order tensors, 932as geometrical objects, 241base, 336column, 250compared with scalars, 212component form, 217examples of, 212graphical representation of, 212irrotational, 353magnitude of, 218non-Cartesian, 336, 358, 362notation, 212polar and axial, 949solenoidal, 352, 389span of, 242vectors, algebra of, 212–234addition and subtraction, 213in component form, 218angle between, 221associativity of addition and subtraction, 213commutativity of addition and subtraction,213multiplication by a complex scalar, 222multiplication by a scalar, 214multiplication of, see scalar product andvector productouter product, 936vectors, applicationscentroid of a triangle, 216equation of a line, 226equation of a plane, 227equation of a sphere, 228finding distance from aline to a line, 231line to a plane, 232point to a line, 229point to a plane, 230intersection of two planes, 228vectors, calculus of, 334–369differentiation, 334–339, 344integration, 339line integrals, 377–389surface integrals, 389–396volume integrals, 396vectors, derived quantitiescurl, 353derivative, 334differential, 338, 344divergence (div), 352reciprocal, 233, 366, 955, 959vector fields, 347curl, 406divergence, 352flux, 395rate of change, 350vectors, physicalacceleration, 335angular momentum, 238angular velocity, 223, 238, 353area, 393–395, 408area of parallelogram, 223, 224force, 212, 213, 220moment or torque of a force, 223velocity, 335velocity vectors, 335Venn diagrams, 1119–1124vibrationsinternal, see normal modeslongitudinal, in a rod, 677transversemembrane, 677, 739, 768, 799, 801rod, 769string, 676, 789Volterra integral equation, 804, 805differentiation methods, 812Laplace transform methods, 810volume elementscurvilinear coordinates, 365cylindrical polars, 360spherical polars, 205, 3621332

INDEXvolume integrals, 396and divergence theorem, 401volume ofcone, 75ellipsoid, 207parallelepiped, 225rhomboid, 237tetrahedron, 192volumesas surface integrals, 397, 401in many dimensions, 210of regions, using multiple integrals, 191–193volumes of revolution, 75and surface area & centroid, 195–197zero-order tensors, 932–935zeros of a function of a complex variable, 839location of, 879–882, 921order, 839, 856principle of the argument, 880Rouché’s theorem, 880, 882zeros of Sturm-Liouville eigenfunctions, 573zeros, of a polynomial, 2zeta series (Riemann), 128, 129z-plane, see Argand diagramwave equation, 676, 688, 790boundary conditions, 693–695characteristics, 704from Maxwell’s equations, 373in one dimension, 689, 693–695in three dimensions, 695, 714, 737standing waves, 693wave number, 437, 693nwave packet, 436wave vector, k, 437wavefunction of electron in hydrogen atom, 208Weber functions Y ν (x), 607wedge product, see vector productweightof relative tensor, 964of variable, 477weight function, 555, 790Wiener–Kinchin theorem, 450WKB methods, 895–905accuracy, 902general solutions, 897phase memory, 895the Stokes phenomenon, 903work doneby force, 381vector representation, 220Wronskianand Green’s functions, 527for second solution of ODE, 544, 580from ODE, 532test for linear independence, 491, 532X-ray scattering, 237Yl m (θ, φ), see spherical harmonicsY ν (x), Bessel functions of second kind, 607Young’s modulus, 677, 953z, as a complex number, 84z ∗ , as complex conjugate, 89–91zero (null)matrix, 254, 255operator, 249unphysical state |∅〉, 650vector, 214, 242, 5561333