Partal Diff. Equations

CONTENTS------------------------------------------------------------------------------------------------------CHAPTER I CALCULUS OF VARIATIONS 1-39CHAPTER II TRANSPORT AND LAPLACE EQUATIONS 40-73CHAPTER III HEAT AND WAVE EQUATIONS 74-94CHAPTER IV ANALYTICAL MECHANICS – I 95-131CHAPTER V ANALYTICAL DYNAMICS-II 132-144CHAPTER VI ANALYTICAL MECHANICS-III 145-160CHAPTER VII ANALYTICAL MECHANICS-IV 161-179CHAPTER VIII NONLINEAR FIRST-ORDER PDE 180-219CHAPTER IX REPRESENTATION OF SOLUTIONS 220-253CHAPTER X ATTRACTION AND POTENTIAL-I 254-274CHAPTER XI ATTRACTION AND POTENTIAL-II 275-290

CALCULUS OF VARIATIONS 5Chapter-1Calculus of Variations1.1 INTRODUCTIONBy a functional, we mean a correspondence which assigns a definite realnumber to each function/curve belonging to some class.That is, a functional is a kind of function where the independent variable isitself a function. Thus the domain of a functional is a set of admissiblefunctions, rather than a region of a coordinate space.Examples of Functionals(1) consider the set of all rectifiable plane curves between two given points (x 0 ,y 0 ) and (x 1 , y 1 ). Let this family be denoted by A. The length of a curve y(x)∈A is a functional. This length is given byJ[y] = l[y(x)] = y(x)∈A.x 1x02 dy 1 + dx, dx y(x 1 , y 1 )(x 0 , y 0 )(2) The area “S” of a surface z = z(x, y) bounded by a given curve C is afunctional.xThis area “S” is determined by the choice of the surface S, z = z(x, y), asJ[z(x, y)] = D2 ∂z ∂z1+ + dx dy, ∂xc ∂y2

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 6where D is the projection of the surface, z = z(x, y), bounded by the curve C,on the xy-plane.Functionals, called variable quantities, play an important role in manyproblems arising in analysis, geometry, mechanics, etc. The first importantresults in this area due to Euler (1707-1783). Nevertheless, up to now, the“calculus of functionals” still does not have methods of a generalitycomparable to the methods of classical analysis calculus of functions.The most developed branch of the ‘Calculus of functionals” is concerned withfinding the maxima and minima of functionals, and is called the “Calculus ofvariations”.Actually, it would be more appropriate to call this branch/subject the “calculusof variations in the narrow sense”, since the significance of the concept ofthe “variation of a functional” is by no means confined to its applications tothe problem of determining the extrema of functionals.The aim of “calculus of variation” is to explore methods for finding themaximum or minimum of a functional defined over a class of functions.Several physical laws can be deducted from concise mathematical principles tothe effect that a certain functional is a given process attains/assumes amaximum or minimum. In mechanics, we have the principle of least action,the principle of conservation of linear momentum, and the principle ofconservation of angular momentum. In addition, we have the principle ofcastigliano in the theory of elasticity.The history of the calculus of variations (CV) can be traced back to the year1696 when John Bernoulli formulated the problem of the brachistochrone(shortest time).In this problem one has to find the curve connecting two given points, A and B,that do not lie on a vertical line, such that a particle sliding down this curveunder the influence of gravity alone from the point A reaches point B in theshortest time.ABWe shall see later on that the curve of quickest descent will not be thestraight-line connecting the points A and B, though this is the shortest distancebetween the points.Apart from Bernoulli, this problem was independently solved by Leibnitz,Newton and L’Hospital. However, the development of “Calculus ofVariations” as an independent Mathematical discipline, along with its ownmethods of investigation, was due to the pioneering studies of Euler during theperiod 1707-1783.

CALCULUS OF VARIATIONS 7Apart from the above described problem, three other problems, stated below,were the motivating one for the developed of the subject CV.Problem of GeodesicsIn this problem, it is required to determine the line of shortest lengthconnecting two given points A(x 0 , y 0 , z 0 ) and B(x 1 , y 1 , z 1 ) on a surface S givenbyϕ(x, y, z) = 0.This problem is a typical problem “Variational problem with a constraint”.Here, we are required to minimize the arc length given by the functionalJ[y, z] =Subject to the constraintϕ(x, y, z) = 0.x 1x02 dy 1+ dx This problem was first solved by Jacob Bernoulli in 1698, but a generalmethod of such category of problems was given by Euler.A geodesic on a given surface is a curve, lying on that surface, along whichdistance between two points is minimum.On a plane, a geodesic is a straight line.The Problem of Minimum Surface of RevolutionA curve y = y(x) ≥ 0 is rotated about the x-axis through an angle 2π. Theresulting surface bounded by the planeshas the areax = a and x = bbaJ[y] = 2π y dy 1+ dx The determination of a particular curvey = y(x)which minimizes J[y] is a variational problem.The Isoperimetric ProblemThis problem is : “Among all closed curves of a given length l, find the curveenclosing the greatest area”.This problem was solved by Euler. The required curve turns out to be a circle.The solution of this problem was known ever is ancient Greece.2dx+ dzdx2dx

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 81.2 FUNCTION SPACESIn the study of functions of n variables, it is convenient to use geometriclanguage, by regarding a set of n numbers (y 1 , y 2 ,…, y n ) as a point in the n-dimensional space.Linear space. Let L be a non-empty set, consisting of elements x, y, z, of anykind, for which the operations of addition and multiplication by real numbersα, β, … are defined and obey the following axioms :(i)x + y = y + x(ii) x + (y + z) = (x + y) + z ;(iii)(iv)there exists an element ‘o’, called the zero element, such thatx + 0 = x = 0 + x for all x∈L,For each x∈L, there exists an element “−x” in L such thatx +(−x) = 0 = (−x) + x;(v) 1. x = x;(vi)(vii)α(βx) = (αβ)x(α + β)x = αx + βx;(viii) α(x + y) = αx + αy.Normed Linear SpaceA linear space L is said to be a normed linear space, if each x∈L is assigned anon-negative number ||x||, called the norm of x, such that(i) ||x|| = 0 iff x = 0 ;(ii)(iii)||x+y|| ≤ ||x|| + ||y||,||αx|| = |α| ||x||.Function SpacesLinear spaces whose elements are functions are called function spaces.In studying functionals of various types, it is reasonable to use various functionspaces. The concept of continuity plays an important role for functionals, justas it does for the ordinary functions considered in classical analysis. In orderto formulate this concept for functionals, we must somehow introduce aconcept of “closeness” for elements in a function space. This is mostconveniently done by introducing the concept of the norm of a function. Thefollowing normed linear spaces are important for our subsequent studies,

CALCULUS OF VARIATIONS 9Examples of Normed Linear Spaces of Function(1) The space C [a, b] consisting of all continuous functions defined on aclosed interval [a, b], is a normed linear space with||y|| 0 =max |y(x)|a≤x≤b(2) The space D 1 [a, b] consisting of all functions y(x) defined on the closedinterval [a, b] which are continuous and have continuous first derivative, is anormed linear space with the norm||y|| 1 = max |y(x)| + max |y′(x)|.a≤x≤ba≤x≤bRemark. Two functions, y and z, in D 1 are regarded as close together if boththe functions themselves and their first derivatives are close together, sinceimplies thatfor all x∈[a, b]||y−z|| 1 < ∈|y(x) − z(x)| < ∈ and |y′(x) − z′(x)| < ∈(3) The space D n [a, b], consisting of all functions y(x) defined on the closedinterval [a, b] which are continuous and have continuous derivatives upto ordern inclusive (where n is a fixed positive integer), is a normed linear space withnormn||y|| n = i=0max |y (i) (x)|,a≤x≤bwhere y i (x) = (d/dx) i y(x) and y (0) (x) = y(x).Remark. Two functions in D n are regarded as close together if the values ofthe functions themselves and of all their derivatives upto order n inclusive areclose together.Similarly, we can introduce spaces of functions of severable variables − thespace of continuous functions of n variables, the space of functions of nvariables with continuous first derivative, etc.Continuity of functionalsAfter introducing norm on function spaces, it is natural to talk about continuityof functionals defined on a function space L.Definition. The function J[y] is said to be continuous at the point ŷ ∈L if forany ∈>0, there is a δ>0 such thatprovided|J[y] − J[ ŷ ]| < ∈

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 10||y− ŷ ||

CALCULUS OF VARIATIONS 11Setα(x)>0 in [x 1 , x 2 ] ⊂ [a, b]h(x) = ( x − x1) (x 2 − x) for x in[x1,x 2 ]0othersie…(1)…(2)Since h(x) is continuous and h(x 1 ) = h(x 2 ) = 0, so, h(x) ∈ C[a, b].However, baxα(x) h(x)dx = 2α(x) (x−x 1 ) (x 2 −x) {Θ h(x) = 0 in [a, x 1 ] and [x 2 , b]x 1> 0 , …(3)since the integrand is positive in the open interval (x 1 , x 2 ). This is acontradiction to the hypothesis in the statement of lemma. This contradictionproves the lemma 1.Remark. The lemma still holds if we replace the word ‘C[a, b]’ by ‘D n [a, b]’in the statement of the lemma. In that situation, we use the same proof withh(x) = n+1[( x − x1)(x 2 − x)] for x in[x1,x 2 ]0otherwiseLemma 2. Statement. If α(x) is continuous in [a, b], and if baα(x) h′(x) dx = 0for every function h(x)∈ D 1 (a, b) such that h(a) = h(b) = 0, thenwhere c is a constant.α(x) = c for all x in [a, b]Proof. Let c be the constant defined by the conditionLet ba[α(x)−c] dx = 0…(1)h(x) = x[ α( ξ)−c] dξ …(2)aThen h(x) is differentiable andh′(x) = α(x) −c, in [a, b]by the fundamental theorem of integral calculus. Soh(x) ∈ D 1 (a, b)…(3)…(4)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 12Also, from equations (1) and (2).h(a) = h(b) = 0.…(5)That is, h(x) satisfies all the conditions of the lemma. Hence, by hypothesisNow ba baα(x) h′(x) = 0[α(x)−c] 2 dx= b[α(x)−c]h′(x)a= bα(x) h′(x) dx −c bh′(x) dxaa…(6)This givesIt follows thator ba= 0 − c [h(b) − h(a)]= 0.[α(x)−c] 2 dx = 0 in [a, b].α(x) −c= 0 for all x in [a, b]α(x) = c for all x in [a, b].This completes the proof of Lemma 2Lemma 3. Statement. If α(x) and β(x) are continuous in [a, b], and if ba[α(x) h(x) + β(x) h′(x)] dx = 0for every function h(x) ∈ D 1 (a, b) such thath(a) = h(b) = 0,then β(x) is differentiable, andProof. Setβ′(x) = α(x) for all x in [a, b]A(x) = α xa(ξ) dξ, for x∈[a, b]…(1)

CALCULUS OF VARIATIONS 13ThenNow baA(a) = 0 and A′(x) = α(x) for all x∈[a, b]bxb a a a α(x) h(x)dx = h(x) { α(ξ)dξ} − h ('x) α(ξ)dξdx…(2)= − bA(x) h′(x) dx,a…(3)since h(a) = h(b) = 0. The given condition ba[α(x) h(x) + β(x) h′(x)] dx = 0,…(4)and result in (3), lead to ba[−A(x) + β(x)] h′(x) = 0,…(5)for every function h(x) ∈ D 1 (a, b) such that h(a) = h(b) = 0. The lemma 2applied to relation (5) given at once−A(x) + β(x) = constt. In [a, b]i.e., (since A(x) is differentiable)β′(x) = A′(x) in [a, b]Equations (2) and (6) yieldβ′(x) = α(x) in [a, b]…(6)This completes the proof of Lemma 3.We now introduce the concept of the variation/differential of afunctional. Let J[y] be a functional defined on some normed linear space.Let∆J[y] = J[y + h] − J[y]be its increment corresponding to the incrementh = h(x)…(1)…(2)of the “independent variable” y = y(x). If y is fixed, ∆ J[h] is a functional of hand it is a nonlinear functional, in general. Suppose thatwhere∆J[y] = ϕ[h] + ∈||h||,…(3)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 14andasϕ [h] = a linear functional,∈→0,||h||→0.…(4)…(5)Then the functional J[y] is said to be differential, and the principallinear part of the increment ∆J[h], i.e., φ[h], is called the variation/ differentialof J[y]. It is denoted by δJ[h]. That is,δJ[h] = φ[h].Theorem 1. The differential of a differentiable functional is unique.Proof. Before proving the main theorem, we state and prove a lemma.…(6)Statement of lemma. If differential ϕ[h] of a functional J[y] is a linearfunctional and ifas ||h||→0, thenφ[h]→0|| h ||ϕ[h] = 0 for all h.Proof of lemma. If possible, suppose thatDefineThenbutφ[hlimn→∞|| hϕ[h 0 ] ≠ 0 for some h 0 ≠ 0.hh n = ,n…(1)…(2)…(3)[h 0 ]λ = ≠ 0. …(4)|| h ||0 φ||h n ||→0 as n→∞,nn] φ[h= lim|| n→∞|| h000/ n]/ n ||…(5)=φ[hlimn→∞|| h00], since ϕ is linear||= λ≠ 0.This is contrary to hypothesis in (1). Hence, the result (2) holds.…(6)

CALCULUS OF VARIATIONS 15Proof of the main theoremNow, suppose that, if possible, the differential of the functional J[y] isnot unique. Then, we can writeand∆J[y] = ϕ 1 [h] + ∈ 1 ||h||,∆J[y] = ϕ 2 [h] + ∈ 2 ||h||,where ϕ 1 [h] and ϕ 2 [h] are linear functionals, andas ||h||→0. Here∈ 1 , ∈ 2 →0,∆J[y] = J[y+h] − J[y].From equations (7) and (8) implyorϕ 1 [h] −ϕ 2 [h] = (∈ 2 −∈ 1 ) ||h||φ1[ h] − φ2[h]=∈ 2 −∈ 1|| h ||→0as ||h||→0. Hence, by above lemma, the functionalϕ 1 [h] − ϕ 2 [h]vanishes identically. This gives…(7)…(8)…(9)…(10)…(11)ϕ 1 [h] = ϕ 2 [h] , for all h …(12)implying that the differential of the differentiable functional J[y] is unique.This completes the proof.Definition (Extremum). The functional J[y] is said to have a extremum fory = ŷ ifJ[y] − J[ ŷ ]does not change its sign in some neighbourhood of the curve y = ŷ (x).Definition (Weak Extremum). The functional J[y] is said to have a weakextremum for y = ŷ if there exists on ∈>0 such thatJ[y] − J[ ŷ ]has the same sign for all y in the domain of definition of the functional whichsatisfy the condition||y− ŷ || 1

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 16where || || 1 denote the norm in the space D 1 .Definition (Strong Extremum). The functional J[y] is said to have a strongextremum for y = ŷ if there exists an ∈>0 such thatJ[y] − J[ ŷ ]Has the same sign for all y in the domain of definition of the functional whichsatisfy the condition||y− ŷ || M∈,where || || 0 denotes the norm in the space C [a, b].Note. Every strong extremum is simultaneously a weak extremum.(1) However, the converse is not true in general.(2) Finding a weak extremum is simpler than finding a strong extremum.Theorem 2. Statement. A necessary condition for the differentiablefunctional J[y] to have an extremum for y = ŷ is that its variation vanishes fory = ŷ .Proof. We are required to prove thatδJ[y] = 0for y = ŷ and all admissible h.According to the definition of the variation δJ[h] of J[y], we havewhereas ||h||→0, and∆J[h] = δJ[y] + ∈ ||h||∈→0∆J[h] = J[y + h] − J[y].…(1)…(2)…(3)…(4)Thus, for sufficiently small ||h||, the sign of ∆ J[h] will be the same as the signof the variation δJ[h]. To be explicit, suppose that J[y] has a minimum fory = ŷ . If possible suppose thatδJ[h 0 ] ≠ 0,…(5)for some admissible h 0 . Then, for any α>0, no matter however small it maybe, we haveδJ[−α h 0 ] = −δJ [α h 0 ].…(6)Hence, (2) can be made to have either sign for sufficiently small ||h||. But thisis impossible, since by hypothesis, J[y] has a minimum for y = ŷ , i.e.,

CALCULUS OF VARIATIONS 17∆J[h] = J[ ŷ +h] − J[ ŷ ] ≥ 0,…(7)for all sufficiently small ||h||. This contradiction completes the proof of thetheorem.1.4 EULER’S EQUATION − SIMPLEST VARIATIONALPROBLEMTheorem 3. LetJ[y] = b aF (x, y, y′) dx…(1)Be a functional defined on the set of functions y(x) which has continuous firstderivative in [a, b] and satisfy the boundary conditionsy(a) = A, y(b) = B.…(2)Prove that a necessary condition for J[y] to have an extremum for a givenfunction y(x) is that y(x) satisfies the differential equationdF y − (Fy')= 0 . …(3)dxProof. Let h = h(x) be the increment given to y(x). Then, in order for thefunction “y + h” to satisfy the boundary conditions in (2), we must haveNowh(a) = 0, h(b) = 0∆J[h] = J[y + h] − J[y]Using Taylor’s theorem, we write= b a F (x, y + h, y′+h′)dx − b aF (x, y, y′)dx…(4)= b a[ F (x, y +h, y′+h′) − F(x, y, y′)]dx, …(5)∆ J[h] = b a[ hFy(x, y, y′) + h′ F y′ (x, y, y′)]dx +…….., …(6)where the subscripts denote partial derivative w.r.t. the correspondingarguments, and dots denote terms of order higher than 1 relative to h and h′.The integral in the right-hand side of (6) represents the principal linear part ofthe increment ∆J[h]. Hence, the variation/ linear part of the increment ∆J[h].Hence, the variation/differential δJ of J[y] is, by definition,δJ = b a[ hFy(x, y, y′) + h′ F y′ (x, y, y′)]dx …(7)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 18We know that a necessary condition for J[y] to have an extremum fory = y(x) is thatδ J = 0,for all admissible h. Equations (7) and (8) imply b afor all admissible h.…(8)[ hFy+h′ F y’ ] dx = 0, …(9)The use of lemma 3 an relation (9) implydF y = (Fy')dxdi.e., F y − (Fy')= 0. …(10)dxThis completes the proof of the theoremDefinition 1. Equation (10) is known as Euler’s equation.Definition 2. The integral curves of Euler’s equation are called extremals.Remark. Euler’s equation is, in general, a second-order ordinary differentialequation, and its solution will, in general, depend on two arbitrary constants,which are determined from the boundary conditions.Special Casesy(a) = A, y(b) = B .We now consider some special cases where Euler’s equation can bereduced to a first-order differential equation, or where its solution can beobtained entirely by evaluating integrals.Case I. Suppose the integrand does not depend on y.In this case, the functional under consideration is of typeJ[y] = b aF (x, y′)dx,…(1)where F does not contains y explicitly. In this case,F y = 0.…(2)Consequently, the Euler’s equation becomesd(F y’ ) = 0 …(3)dxwhich has the first integral

CALCULUS OF VARIATIONS 19F y′ = C,where C is a constant.…(4)Equation (4) is a first-order ordinary differential equation. Solving (4) for y′,we obtain an equation of the formy′ = f(x, c),from which y can be found by integrationCase 2. Suppose the integrand does not depend on x.In this case, we have functional asJ[y] = b aF (y, y′) dx.…(1)Nowdd dy d dy' F y − (Fy')= Fy− (Fy'). + (Fy,) dxdy dx dy' dx = F y −y′ F y’y −y′′ F y′y. …(2)So, the Euler’s equation isF y −y′ F y’y − y′′ F y′y′ = 0.Multiplying by y′, we obtainy′F y −(y′) 2 F y’y −y′y′′ F y′y′ = 0ddx [ F y F' ] 0,where has the first integralF−y′ F y′ = c− …(3)y ' =where c is a constant. Euler’s equation (4) is of first-order.Case 3. Suppose the integrand does not depend on y′.In this case, the function is assoJ[y] = b aF(x, y) dx, …(1)Fy′ = 0.Hence, Euler’s equation becomesF y (x, y) = 0…(2)…(3)This equation is not a differential equation, but an algebraic equation in x andy. Its solution consists of one or more curves y = y(x).Case 4. When functional J[y] is of the form

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 20J[y] = b a[f(x, y)21+ (y )' ] dx. …(1)This functional represents the integral of a function f(x, y) with respect to thearc length s, whereds =21+ (y )' dx.In ths case,F(x, y, y′) = f(x, y)21+ (y )'…(2)and, so∂F−∂yddx ∂F = ∂y'[ f (x, y) ]y21+(y )'−ddxf (x, y)y'21+(y )'= (f y )21+(y )'(fx)y'21+(y )'∂+f∂yy'21+(y )' dy− dx+∂∂ ' f y'y 21+(y )'dy'.dx= (f y )21 + y' − (fy)y'− y'(f 21+y' y)y'21+y' − ( y')(f ) 22 y' 1+y' − 21+y' 2 1+y'=(f y )21+y' − (fSo, Euler’s equation becomesx)y'− (f21+y'==f y − y′f x − y′′f = 0.y)(f y )−21+y'11+2y'2y'1− (y')(f ).221+y' (1 + y')y f' x y f'−221+y' (1 + y')3/ 23/ 2[f y − y′ f x −y′′f] …(3)

CALCULUS OF VARIATIONS 21which is a differential equation of order 2.ILLUSTRATIVE EXAMPLESExample 1. Solve the variational problemJ[y] =212 1+y' dx x …(1)y(1) = 0, y(2) = 1.…(2)Solution. We note that the integrand in the given functional does not dependon y explicitly, andF(x, y, y′) =1+y'Euler’s equations for such case is of the formx2…(3)∂F = c, …(4)∂y'where c is a constant. From equations (3) and (4), we find (exercise)y′ =cx1−c2 x 2Integrating (5), it follows that (exercise)1 2 2y = 1−c x + c1cor (y−c 1 ) 2 + x 2 1= ,c2…(5)…(6)where c 1 is a constant. The curve (6) represents a circle with centre (0, c 1 ),lying on the y-axis, and radius 1/c. Using the boundary conditions (2), we find(exercise)c =1 , c1 = 2. …(7)5So, the require curve isx 2 + (y−2) 2 = 5.…(8)Example 2. Among all the curves joining two given points (x 0 , y 0 ) and (x 1 , y 1 ),find the one which generates the surface of minimum area when rotated aboutthe x-axis.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 22Solution. We know that the area of the surface of revolution generated byrotating the curve y = y(x) about the x-axis isx 1x 02π y 1+So, the variational problem iswith boundary conditionsx 1x 0y'2 dxJ[y] = 2π y 1+y'2 dx , …(1)y(x 0 ) = y 0 , y(x 1 ) = y 1 .…(2)In this variational problem the integrand does not depend explicitly on x, and2F(y, y′) = 2πy 1+ y' ,…(3)and corresponding Euler equation isWe find (exercise)we puty −y′(F y′ ) = constt.y1+2y'then (5) imply (exercise)…(4)= c, c = constt. …(5)y′ = sinh t, …(6)y = c cosh t.Equation (6) and (7) give (exercise)dx = c dt.Integrating it, we obtainx = ct + c 1 ,…(7)…(8)where c 1 is a constant. Eliminating t from equations (7) and (8), we find x − c1y = c cosh . …(9) c The values of the arbitrary constants c and c 1 are determined by the givenconditions in (2).The required curve is catenary passing through the two given points. Thesurface generated by rotation of the catenary is called a catenoid.Example 3. Minimize the functional

CALCULUS OF VARIATIONS 23J[y] = b a(x−y) 2 dx. …(1)Solution. In this example, the integrand does not contain y′ explicitly andF(x, y) = (x−y) 2The corresponding Euler’s equation is∂F = 0,∂ywhich leads tox−y = 0.…(2)…(3)…(4)The required curve (4) is a straight line. Further, the functional (1) vanishesalong this line.This completes the solution1.5 THE CASE OF SEVERABLE VARIABLESNow, we consider further generalization of the simplest variationalproblem. First we consider the case of n dependent functions. LetJ[y 1 , y 2 ,…, y n ] = b aF(x, y 1 , y 2 ,…, y n ,' '1,...,yny )dx …(1)Be a functional which depends on n continuously differentiable functionsy 1 (x), y 2 (x), …, y n (x)satisfying the boundary conditionsy i (a) = A i , y i (b) = B i , 1≤ i ≤ n. …(2)Here, we are looking for an extremum of the functional (1) defined on the setof the set of smooth curves joining two fixed points in (n+1)-dimensionalEuclidean space R n+1 .The problem of finding geodesics (shortest curves joining two points of somemanifold) is of this type. The same kind of problem arises in geometricoptics, in finding the paths along which light rays propagate in aninhomogeneous media. According to Fermat’s principle, light goes from apoint, say P 0 , to a point, say P 1 , along the path for which the transit time is thesmallest.Theorem. Prove that a necessary condition for the curvey i = y i (x), (i =1, 2,…, n)to be an extremal of the functional' ' '1 2 ynJ = b a F (x, y 1, y 2 ,…, y n , y , y ,..., ) dx …(1)is that the functions y i (x) satisfy the equations

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 24∂F∂yi−ddx ∂F = 0y , 1 ≤ i ≤ n. …(2) ∂ i ' Proof. First of all, we calculate the variation δJ of the given functional J in (1).We replace each y i (x) by a varied function y i (x) + h i (x).By definition, the variations δJ of the functional J[y 1 ,., y n ] is linear in h i and h i ′,1 ≤ i ≤ n, and which differs from the increment∆J = J[y 1 + h 1 , y 2 + h 2 , ., y n + h n ] − J[y 1 , y 2 ,.,y n ],by a quantity of order higher than 1 relative to h i and h i ′ ; i = 1, 2,…, n.Since both y i (x) and y i (x) + h i (x) satisfy the boundary conditionsTherefore, we must havey i (a) = A i , y i (b) = B i , 1 ≤ i ≤ n.h i (a) = h i (b) = 0, for each i.Using Taylor’s theorem, we obtain'''y 1 + h1,...,yi+ hi,.., yn+ h…(3)…(4)…(5)∆J = b a [ F(x, y 1 +h 1 ,.., y i +h i ,…, y n + h n ,'''n)− F(x, y 1 , …, y i ,…, y n , y1,.,yi,..., yn)] dx nb ∂F' ∂F= ah + +i=1i h i dx∂ i ∂y......…(6)yi ' where the dots denote terms of order higher than 1 relative to h i , h i ′ (i = 1, 2, .,n) The integral on the right of (6) represents the principal linear part of theincrement ∆J. Hence, by definition, the variation δJ of J isδJ = ba n ∂F' ∂F h i + h i dx…(7)i=1∂yi ∂yi' '''Since all the increments h i (x) are independent, we can choose one of themquite arbitrarily (as long as the boundary conditions are satisfied), setting allthe others equal to zero. Therefore, the necessary conditionδJ = 0,…(8)for an extremum implies

CALCULUS OF VARIATIONS 25b∂F' ∂F +ah =i h i 0∂ i ∂y,…(9)yi ' for each i = 1, 2,…, n. There are now n conditions in (9). Using lemma 3, weobtainor∂F∂yy i=ddxi y i ∂F, 1 ≤ i ≤ n' ∂ dF − ( F ') = 0, 1 ≤ i ≤ n. …(10)dxy iEquations (10) are called Euler’s equations. We note that (10) is a system of nsecond order, in general ordinary differential equations. Solution of (10)contains, in general, 2n arbitrary constants, which are determined from theboundary conditions in (4).This completes the proof.Definition. Two functionals are said to be equivalent if they have the sameextremals.Example. Find the external of the functionalJ[y, z] = π / 20(y′ 2 + z′ 2 + 2yz)dxy(0) = 0, y (π/2)= 1, z(0) = 0, z(π/2)= −1.Solution. Takingy 1 (x) = y(x), y 2 (x) = z(x),andF[y 1 , y 2 ] = (y 1 ′) 2 + (y 2 ′) 2 + 2 y 1 y 2 ,Euler’s equations…(1)…(2)become∂F∂yi−ddx ∂F = 0,yi' ∂ i = 1, 2,…(3)2z − dxd (2y′) = 0,2y − dxd (2z′) = 0.This givesz = y′′,…(4)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 26y = z′′.Equations (4) and (5) imply…(5)solution of (6) is4d y− y = 0,…(6)4dxy(x) = c 1 e x + c 2 e −x + c 3 sin x + c 4 cosx,where c 1 , c 2 , c 3 , c 4 are constants. Equations (4) and (7) givez(x) = c 1 e x + c 2 e −x − c 3 sinx − c 4 cos x.Using the given boundary conditionsy(0)= 0,y(π / 2)= 1 z(0)= 0,z(π / 2)= −1,we obtain (exercise)c 1 = c 2 = 0, c 3 = 1, c 4 = 0.Hence, an extremum of the given functional is given byy(x) = sinx,z(x) = −sin x.1.6 THE PROBLEM OF GEODESICSSuppose we have surface σ specified by a vector equationρ ρr = r (u, v).…(7)…(8)…(9)…(10)…(11)The shortest curve (of minimum length) lying on the surface σ and connectingtwo points A and B of surface, σ, is called the geodesic connecting the twopoints.The equations for the geodesics of σ are the Euler equations of thecorresponding variational problem−namely, the problem of finding theminimums distance (measured along surface σ) between two points of thesurface σ.Euler’s Equations on GeodesicsA curve lying on the surfaceρ ρr = r (u, v), …(1)can be specified by the equationsu = u(t),

CALCULUS OF VARIATIONS 27LetEv = v(t), t being a parameter.…(2)ρ ρ ρ ρ ρ ρ= ==…(3)( u,v) ru.rv,F(u,v)ru.rv, G (u,v) ru. rvThese quantities are called the coefficients of the first fundamental form of thesurface (1).The arc length between the points A(t 1 ) and B(t 2 ), corresponding to theparameter t, is given by (using results from Differential Geometry by C.E.Wealtherburn)J[u, v] =t 1t 022 E u' + 2Fu v''Gv' + dt…(4)Euler’s equations for the functional (4) are∂[∂u∂∂vThese become2Eu'+ 2Fu''v + Gv']−2Eu'+ 2Fu''v + Gv'−22ddtddt ∂ 22 Eu'+ 2Fu''v + Gv'= 0∂u' ∂ 22 Eu'+ 2Fu''v + Gv'= 0∂v'22E uu'+ 2Fuu v''G + uv' −22Eu' + 2Fu v''Gv' + ddt2(Eu'Fv + )' = 022Eu' + 2Fu v''Gv' + …(5) 22E vu'+ 2Fvu''v + G vv' d 2(Fu'+ Fv )' and − = 022Eu 2Fu v Gv dt 22 ' + '' + ' Eu'+ 2Fu''v + Gv'…(6)Remark. The concept of a geodesic can be defined not only for surfaces, butalso for higher-dimensional manifolds. Finding the geodesics of an n-dimensional manifold reduces to solving a variational problem for a functionaldepending on n functions.Example 1. Find the geodesics of the circular cylinderρr = ( a cos ϕ, a sin ϕ, z)Solution. The variables ϕ and z play the role of the function u and v in theabove article. Nowρr = ( a cos ϕ, a sin ϕ, z).Thenρrρrz= ( −asin ,a cos, 0), = (0,0,1).…(1)…(2)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 28Thereforeρ ρE = r.r= aρ ρF = r.rz= 0ρ ρG = r .r = 1zz2…(3)The arc length between two points A(t 1 ) and B (t 2 ) lying on the cylinder (1) isgiven by the functionalort 2t 12J[ϕ, z] = E ' + 2F z''Gz' + dt…(4)2t 2t 1222J[ϕ, z] = a ' + z' dt. …(5)Euler’s equations for the functional (5) ared 2a '0 − = 0,dt 2 2 2a ' z' + d 2a φ'0 − = 0.dt 2 2 2a ' z' φ + These equations, on integration, yield…(6)…(7)2a2'22a ' + z'= c1,2z'22a ' + z'= c2Dividing the second of these equations by the first, we obtainorz′/ϕ′ = constt.dz = cdwhich has the solutionz = c 1 ϕ + c 2 .…(8)Equation (8) represents a two-parameter family of helical lines lying on thecylinder (1). Thus, a geodesic on cylinder (1) is a helix.Example 2. Find the geodesics of the sphereρr = (a sin θ cosφ, a sin θ sinφ, a cosθ)Solution. On the surface of a given sphere

CALCULUS OF VARIATIONS 29we find (exercise)ρr = (a sin θ cosφ, a sin θ sinφ, a cosθ)E = a 2 , F = 0, G = a 2 sin 2 θ.The variational functional is (exercise)J = '2 2 d+ sin dφ, θ′ = .d…(1)…(2)…(3)Here, the integrand isF = F(θ, θ′) = '2+ sin = independent of φ. …(4)So, the corresponding Euler’s equation isF − θ′ F θ′ = constt. = c2' + sin2 −22'' + sin2= c(sin22)' + sin2= c sin 4 θ = c 2 (θ′ 2 + sin 2 θ)c 2 θ′ 2 = sin 4 θ − c 2 sin 2 θddsin2(sinc22 − c2)dφ=dθsin2θc1−c2cos ec2θ=( 1−cccosec2) − c22θcot2θIntegratingϕ = cos −1 ccot 1−c2 + c' cos (φ − c′) = c cot θ 2 1−c c 1 cotθ = cosϕ cos c′ + sin ϕ sin θ′ c 1 cosθ = sinθ sinϕ sin c′ + sinθ sinϕ sin c′ z = Ax + By. …(5)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 30This is the equation of the plane passing through the centre (0, 0, 0) of thesphere and intersecting the sphere along a great circle.Thus, the shorten curve, i.e., geodesic on a sphere is n are of a great circle.Example 3. Find the geodesic on the plane.Solution. The geodesic on the plane is an extremal of the functionalx 1x 0J[y] = 1 + y'2 dx…(1)The integrand F does not contain y explicity. Hence, the corresponding Euler’sequation isi.e.,F y′ = c1 1−2 2(1 + y')2.2y' = c…(2)y′ = c21+ y'y′ 2 = c 2 (1+y′ 2 )y′ 2 (1−c 2 ) = c 2y′ = A y(x) = Ax + B. …(3)This is the equation of a straight line in the plane. Thus, geodesics in a planeare straight lines.1.7 FUNCTIONALS DEPENDING ON HIGHER-ORDERDERIVATIVESTheorem. Statement. Among all functions y(x) belonging to the space D n (a,b) and satisfying the conditionsy (i) (a) = A i , y (i) (b) = B i , 0 ≤ i ≤ n,find the function for which the functionalJ[y] = b a F (x, y, y′, y′′,…, y(n) ) dx,…(1)…(2)has an extremum.

CALCULUS OF VARIATIONS 31Solution. First, we state the general result which states that a necessarycondition for a functional J[y] to have an extremum is that its variation vanish,i.e.,δJ = 0.…(3)We replace y(x) by the “varied” function “y(x) + h(x)”, where h(x) belongs toD n (a, b) and satisfy the boundary conditions (1). For this, we must haveh (i) (a) = h (i) (b) = 0 for i = 0, 1, 2,…, n..…(4)we know that by the variation δJ of the functional J[y], we mean the expressionwhich is linear in h, h′,…, h (n) , and which differs from the increment∆J = J[y + h] − J[y],by a quantity of order higher than 1 relative to h, h′,…, h (n) .Next, we use Taylor’s theorem to obtain…(5)∆J = b a[ F(x, y + h, y′ + h′,…, y (n) + h (n) ) − F(x, y′,…y (n) )]dx= b a [ h F y + h′ F y′ +…+ h (n) F y(n) ]dx +…., …(6)where the dots denote terms of order higher than 1 relative to h, h′, …, h (n) .The last integral in (6) represents the principal linear part of the increment ∆J.Therefore, by definition of the variation of J[y], we writeδJ = b a [ hF y+h′ F y′ +…+ h (n) F y(n) ]dx.…(7)The necessary condition (3) for an extremum implies that b a [ h F y+h′ F y′ +…+ h (n) F y(n) ] dx = 0.…(8)Integrating (8) by parts repeatedly and using boundary condition (4), we findthat (exercise)ba2nd dn d F y − ( Fy' ) + ( f y + + −1F2 ')... ( ) (n ( n)) = 0, …(9)ydx dxdx for any function h(x) which has continuous derivatives and satisfies theboundary condition in (4). It follows from lemma 1 that2d dn dF y − (Fy')+ (Fy')+ ... + ( −1)(Fy n'' ) = 0 …(10)2ndx dxdxEquation (10) is called Euler’s equation. Equation (10) is an ordinarydifferential equation of order 2n, its general solution contains 2n arbitraryconstants, which can be determined from the 2n boundary conditions in (4).n

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 32d dxb d = ah(Fy')dx,dx b b ah′(F y′ ) dx = [ h(F)] h(F − b y ' a ay') b b dah′′(F y′′ ) dx = [ h ('F )] h' { F }y'a− b adxy'dxdx= − hddx(Fy') ba− ba2dhdx2(Fy') dxin general,= (−1) 2 b dah (F2 y')dx,dx2ba( k)k b dh { Fy(k)} dx = ( −1) h(x){ F dxa k( k ) }ydx1.8 THE CONCEPT OF VARIATIONAL DERIVATIVELet J[y] be a functional depending on the function y(x), and suppose we givey(x) an increment h(x) which is different from zero only in the neighbourhoodof a point x 0 .Let ∆σ denote the area lying between the curve y = y(x) andy = y(x) + h(x). Consider the ratioJ[y + h] − J[y]∆of the incrementto the area ∆σLet the areain such a way that∆J = J[y+h] −J[y]∆σ→0max |h(x)|→0k…(1)…(2)…(3)…(4)as well as the length of the interval in which h(x) is non zero, goes to zero.Then, if the ratio (1) converges to a limit as ∆σ→0, this limit is called thevariational derivative of the functional J[y] at the point x 0 (for the curve y =y(x)), and is denoted by

CALCULUS OF VARIATIONS 33Jyx=x 0…(5)Remark. In the light of above, we write∆J = J[y+h] − J[y] = Jyx= x0+ ∈∆σ,…(6)where∈→0,as ∆σ→0.(2) The variation/differential of a functional J[y] at the point x = x 0 , in terms ofthe variational derivative, is given by the formulaδJ = Jyx= x0 ∆σ.…(7)1.9 VARIATIONAL PROBLEMS WITH SUBSIDIARYCONDITIONS(THE ISOPERIMETRIC PROBLEM)Theorem. Given the functionalJ[y] = b aF (x, y, y′) dx,Let the admissible curves satisfy the conditionsy(a) = A, y(b) = B,K[y] = b aG (x, y, y′)dx = l ,where K[y] is another functional, and let J[y] have an extremum for y = y(x).Then, if y = y(x) is not an extremal of K[y], there exists a constant λ such thaty = y(x) is an extremal of the functional b a( F +λG)dx.Proof. Let J[y] = b aF (x, y, y′)dx,…(1)have an extremum for the curve y = y(x), subject to the conditionsy(a) = A, y(b) = B,K[y] = b aG (x, y, y′)dx = l .…(3)…(2)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 34We choose two points x 1 and x 2 in the interval [a, b], where x 1 is arbitrary andx 2 satisfies a condition to be stated later on, but is otherwise arbitrary.We give y(x) an incrementwhereandδ 1 y(x) + δ 2 y(x),δ 1 y(x) is non-zero only in the neighbourhood of x 1 ,δ 2 (x) is nonzero only in a neighbourhood of x 2(4a)…(4b)Let y*(x) = y(x) + δ 1 y(x) + δ 2 y(x). …(5)We now require that the “varied” curve y = y*(x) satisfy the conditionK[y*] = K[y].…(6)Using variational derivatives, we can write the increment ∆J of the functional Jin the formF F ∆J = + ∈1∆1+ + ∈2∆2 , …(7)yyx=x1 x=x2where∆σ 1 = b a [ δ 1y(x)]dx,∆σ 2 = b a [ δ 2y(x)]dx,and∈ 1 , ∈ 2 →0as∆σ 1 , ∆σ 2 →0.Writing ∆K in a form similar to (7), we obtain∆K = K[y*] −K[y]G' G' = + ∈1∆1+ + ∈2∆2 , yyx=x1 x=x2where∈ 1 ′, ∈ 2 ′→0as∆σ 1 , ∆σ 2 →0.Next, we choose the point x 2 to be a point for which…(8)…(9)…(10)…(11)…(12a)…(12b)

CALCULUS OF VARIATIONS 35Gyx=x 2≠ 0.…(13)Such a point exists, since by hypothesis y = y(x) is not an extremal of thefunctional K. The condition of the point x 2 , given in (13), is the conditionwhich we had mentioned earlier. With this choice of point x 2 , Equations (6)and (11) imply G y∆σ 2 = − G yx=x 1x=x2+ ∈' ∆σ 1 ,…(14)where∈′→0 as ∆σ 1 →0.We setλ = −FyGyx=x2x=x 2. …(15)Using (14) and (15) into (7), we obtain∆J =Fyx=x1+ ∈1∆1F− yx=x2+ 2 G y G yx=x 1x=x2+ ' ∆σ 1=FyG+ yx= x1 x=x1 ∆σ 1 + ε ∆σ 1 ,…(16)whereas∈→0∆σ 1 →0.This expression for ∆J explicitly involves variational derivatives only at thepoint x = x 1 and the increment h(x) is now first δ 1 y(x). The “compensatingincrement” δ 2 y(x) has been taken into account automatically by using the

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 36condition ∆K = 0. Thus, the first term in the right-hand side of (16) is theprincipal linear part of ∆J. So, the variation of the functional J at the point x 1 isδJ =FyG+ yx= x1 x=x1 ∆σ 1 .…(17)We know that a necessary condition for an extremum is thatδJ = 0.Since ∆σ 1 is nonzero while x 1 is arbitrary, we finally obtaini.e.,δFδG+ λ = 0δyδyF…(18)d d − (Fy') + Gy − (G y') 0…(19)dx dx y =This shows that y = y(x) is an extremal of the functional b a( F +λ G)dx,where λ is given by (15).This completes the proof.Remarks (1) The general solution of differential equation (19) will containtwo arbitrary constants in addition to the parameter λ. We shall determinethese three quantities from two boundary conditionsy(a) = A,y(b) = Band the subsidiary conditionK[y] = l .Remark (2). The above theorem/result generalizes immediately to the case offunctionals depending on several functions.Suppose we are looking for an extremum of the functionalJ[y 1 , y 2 ,.., y n ] = b a F (x, y 1,…, y n ,' '1,...,yny )dx, …(20)Subject to the conditionsy i (a) = A i ,y i (b) = B i , 1 ≤ i ≤ n, …(21)and

CALCULUS OF VARIATIONS 37 b aG (x, y 1 , y 2 ,…, y n ,k' '1,...,yny )dx = l k …(22)for k = 1, 2,…, m.In this case a necessary condition for an extremum is thatfor i = 1, 2, …, n.∂ m d ∂ mF+ k G k − F+ kGk = 0∂y k=1 dx ∂yi ' k=1 …(23)The 2n arbitrary constants appearing in the differential equation system (23)and the values of m parameters λ 1 , λ 2 ,…, λ m , sometimes called Largangemultipliers & are determined from the boundary conditions (21) and subsidiaryconditions (22). Here, the number of Lagrange multiplier equals the number ofconditions of constraint.Example 1. Among all curves of length l in the upper half-plane passingthrough the points (−a, 0) and (a, 0) find the one which together with theinterval [−a, a] encloses the largest area.Solution. We have to find the functiony = y(x)for which the integralJ[y] = −aay(x)dx …(1)takes the largest value subject to the conditionsy(−a) = 0, y(a) = 0,…(2)We form the functionala−a2K[y] = 1+y' dx = l. …(3)J*[y] = J[y] + λ K[y]a−aThe corresponding Euler’s equation isd[ydy2 ]= [ y + 1+y' dx …(4)+ 21+y'] −dy'1− λ = 0dx 2 1+y' 1 . .2y'd 2 = 0dx2 1+y' …(5)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 38Integrating, we obtainy'x − = c121+y'or (x−c 1 ) = y'1+2y'or (x−c 1 ) 2 (1+ y′ 2 ) = λ 2 y′ 2y′ =2x − c1 (x − c )12. …(6)Integrating (6), we obtainy(x) =2x − c1− (x − c1)2dx22= − (x − c1)+ c2or (x−c 1 ) 2 + (y−c 2 ) 2 = λ 2 . …(7)It is a family of circles. The values of c 1 , c 2 and λ are determined from thegiven conditions in (2) and (3).We find (−a − c 1 ) 2 + c 2 2 = λ 2(a−c 1 ) 2 + c 2 2 = λ 2 c 1 = 0. …(8)Then, we havec 2 = −So, solution (7) now becomes22 − a…(9)x 2 + (y +2222 − a ) = . …(10)This givesy =2− x2−2− a2and y′ =− x2− x2…(11)

CALCULUS OF VARIATIONS 39Now, the condition (3) impliesThis gives2a xl = −a1+dx2 2 − xa = −adx2 2 − x= 2λ sin −1 (a/λ).a/λ = sin(e/2λ).…(12)Equation (12) is a transcendal equation for λ. Solving it, we find adefinite/certain value, say λ = λ 0 . Then, solution curve (10) becomes20 − a 0x 2 2 2 2+ ( )y + =…(13)The result (13) is the required form.Example 2. Find the extremal of the functionalJ[y] = π 0y′2 dxSubject to the conditionsy(0) = 0, y(π) = 0, π 0 y2 dx = 1.Solution. We form an auxiliary functionJ*[y] = π 0F(x, y, y′)dx …(1)whereEuler’s equation for (1) isF(x, y, y′) = y′ 2 + λ y 2 , λ being a parameter.…(2)Fy- dxd (Fy′ ) = 0i.e., 2λy − dxd (2y′) = 0or y′′ − λy = 0. …(3)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 40First of all, we claimλ < 0.…(4)If possible, consider the case when λ ≥ 0, then the general solution of secondorder ODE (3) isy(x) = c 1 exUse of boundary conditionsgive (exercise)y(0) = 0, y (π)= 0c 1 = c 2 = 0,and y(x) ≡ 0.+ c e…(5A)2− x…(5)…(5B)Then π 0 y2 dx = 0 ≠ 1, …(5C)which is a violation given condition. Hence, our claim in (4) is valid.Consequently, solution of ODE (4) isy(x) = c 1 sinThe boundary condition y(0) = 0 givesc 2 = 0,and boundary conditionimpliesfor k = 1, 2, 3,…y (π) = 0λ = −k 2− x + c 2 sin − x …(6)Thus, a solution of ODE satisfying two boundary condition isy(x) = c 1 sink x,where c 1 is a non-zero constant and yet to be determined.The conditiongives (exercise)…(7)…(8)…(9) π 0 y2 dx = 1, …(10)c 1 =± 2 / π . …(11)

CALCULUS OF VARIATIONS 41Hence, extremals arewherey(x) =± 2 sin kx, …(12)πk = 1, 2, 3,… .Example 3. Find an extremal of the functional…(13)J[y, z] = 1 0[y′ 2 + z′ 2 − 4xz′ − 4z]dx,y(0) = 0, y(1) = 1z(0) = 0, z(1) = 1subject to the condition 1 0[y′ 2 − xy′−z′ 2 ] dx = 2.Solution. We form an auxiliary functionalJ*[y, z] = 1 0F(x, y, z, y′, z′)dx …(1)whereF(x, y, z, y′, z′) = (y′ 2 + z′ 2 −4xz′ − 4z) +λ(y′ 2 −xy′−z′ 2 ),in which λ is a parameter.The system of Euler’s equations are…(2)0 + dxd (2y′ + 2λy′ − λx) = 0 …(3)4 + dxd (2z′ − 4x −2λz′) = 0 …(4)Solving these equations (exercise), we obtain2x + 2c1xy(x) = + c 24(1 + )…(5)c3xz(x) = + c4,…(6)2(1 − )where c 1 , c 2 , c 3 , c 4 are constants of integration. Using the boundary conditionsy(0) = 0, y(1) = 1…(7)z(0) = 0, z(1) = 1we find (exercise)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 42c 13 + 4 , c2 = 0, c 3 = 2(1−λ), c 4 = 0. …(8)2Hence, solution of Euler’s system isy(x) =z(x) = x. x2 + (3 + 4)x4(1 + )…(9)…(10)To find λ, we substitute the value of y(x) and z(x) from equations (9) & (10)into the given condition 1 0(y′ 2 −xy′−z′ 2 )dx = 2, …(11)we find (exercise), two value of λ, namely10λ 1 = − , 1112= …(12)112 −The actual substitution of λ, y and z in (2), we find that λ 2 does not satisfy it,but λ 1 does.Hence, the desired extremal is determined by the equations7x − 5xy(x)=2z(x)= x.1.10 FINITE SUBSIDIARY CONDITIONS2,We now consider a problem which can be stated a follows :Problem. Find the function y i (x) for which the functional…(13)J[y 1 , y 2 ,…, y n ] = b aF(x, y 1 , …, y n ,y' 11,...,yn) dx, …(1)has an extremum, where the admissible functions satisfy the boundaryconditionsy i (a) = A i , y i (b) = B i , 1 ≤ i ≤ n,and m “finite” subsidiary conditions (m < n)g k (x, y 1 ,…, y n ) = 0, 1 ≤ k ≤ m.…(2)…(3)Note. We note that in the above problem, the functional (1) is not consideredfor all curves satisfying the boundary conditions (2), but only for those whichlie in the (n−m) −dimensional manifold defined by the systems (3).Remark. For simplicity, we restrict ourselves to the case

CALCULUS OF VARIATIONS 43n = 2 and m = 1.Theorem. Given the functionalJ[y, z] = b aF(x, y, z, y′, z′)dx,…(1)let the admissible curves lie on the surfaceg(x, y, z) = 0,and satisfy the boundary conditionsy(a) = A 1 , y(b) = B 1 ,z(a) = A 2 , z(b) = B 2 ,and moreover, let J[y, z] have an extremum for the curvey = y(x), z = z(x)…(2)…(3)…(4)…(5)Then, if g y and g z do not vanish simultaneously at any point of the surface (2),there exists a function λ(x) such that (5) is an extremal of the functional b a[ F+λ(x) g]dx.…(6)Proof. We are required to prove that (5) satisfies the differential equationsF y + λg y − dxd (Fy′ ) = 0,…(7)F z + λg z − dxd (Fz′ ) = 0.…(8)Let J[y, z] have an extremum for the curve (5), subject to the conditions (2) to(4). Let x 1 be an arbitrary point of the interval [a, b]. Next, we give y(x) anincrement δy(x) and z(x) an increment δz(x), where both δy(x) and δz(x) arenon-zero only in a neighbourhood, say [α, β] ⊂ [a, b], of x 1 . Using the notionof variational derivatives, we can write the corresponding incrementin the form∆J = J[y + δy, z + δz] − J[y, z],…(9)δFδF∆J = + ∈1∆σ1+ + ∈2∆σ2,δyzx xδ= 1x=x1…(10)where∆σ 1 = b a y(x)dx, ∆σ 2 b a z(x) dx,…(11)and∈ 1 , ∈ 2 →0

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 44as∆σ 1 , ∆σ 2 →0.We now require that the “varied” curvey = y*(x) = y(x) +δy(x),z = z*(x) = z(x) + δz(x)satisfy the condition (2), i.e.,g(x, y*, z*) = 0.…(12)…(13)…(14)Then 0 = b a[G(x, y*, z*) −g(x, y, z)]dx= b ag y g z] dx[ y + z'1'2= g + ∈ ∆σ + { g + ∈ } ∆σ,yx=x11z x=x12…(15)where ∈ 1 ′, ∈ 2 ′ →0 as ∆σ 1 , ∆σ 2 →0, and the overbar indicates that thecorresponding derivatives are evaluated along certain intermediate curves. Byhypothesis, eithergyx=x 1orgzx=x 1is nonzero. Ifg zx=x1≠ 0,…(16)we can write the condition (16) in the formg∆σ 2 = − gyzx=x1x=x1+ ∈' ∆σ 1 ,…(17)where ∈′→0 as ∆σ 1 →0. Substituting (18) into the formula (10) for ∆J, weobtain∆J =Fyx=x 1 g− gyzF zx=x 1 ∆σ 1 + ∈ ∆σ 1…(18)where ∈→0 as ∆σ 1 →0. The first term in the right side of (19) is the principallinear part of ∆J. Hence, by definition, the variation δJ of the functional J atthe point x 1 is

CALCULUS OF VARIATIONS 45δJ =Fyx=x 1 g− gyzF zx=x 1 ∆σ 1…(19)we know that a necessary condition for an extremum of the functional J is thatδJ = 0.…(20)Since ∆σ 1 is non-zero while x 1 is arbitrary, equations (20) and (21) implyF g− y gyzF = 0,z d g y d orF y − (Fy')Fz(Fz')= 0dx −g −dx zorFyd− (Fdxgy)y'=Fzd− (Fz')dxgz…(21)Along the curvey = y(x),z = z(x)the common value of the ratios (22) is some function of x, say −λ(x). Then(22) reduces to system of differential equationsF y + λg y − dxd(Fy′ ) = 0F z + λg z − dxd(Fy′ ) = 0,which are precisely equations (7) and (8).This completes the proof oftheorem.Remark 1. If the functional J has an extremum for a curve γ, subject to theconditiong(x, y, z, y′, z′) = 0,…(∗)and if the derivatives g y′ and g z′ do not vanish simultaneously along γ, thenthere exists a function λ(x) such that γ is an integral curve of the system ofdifferential equationsΦ y − dxd(Φy′ ) = 0,

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 46whereΦ z − dxd(Φz′ ) = 0,Φ = F + λG.Remark 2. If we assume that the condition (2) does not hold everywhere, butonly at some fixed pointg(x 1 , y, z) = 0,…(∗∗)we obtain a condition whose left-hand side can be regarded as a functional of yand z. Thus, the condition (2) can be regarded as an infinite set of conditions,each of which is a functional.Example 1. Among all curves lying on the sphere x 2 + y 2 + z 2 = a 2 , andpassing through two given points (x 0 , y 0 , z 0 ) and (x 1 , y 1 , z 1 ), find the one whichhas the least length.Solution. The length of the curveis given by the integraly = y(x), z = z(x)…(1)J[y, z] =x 1x 022 1+ y' + z' dx…(2)The curve (1) lies on the spherex 2 + y 2 + z 2 = a 2 .we form the auxiliary functional…(3)x1x02J* = ( 1 + y'2 + z'+ λ(x) (x 2 + y 2 + z 2 )]dx …(4)The other boundary conditions arey(x 0 ) = y0, y(x1)= y1…(5)z(x 0 ) = z0, z(x1)= z1,The Euler’s equations, corresponding to function (4), aredy'2λ(x) y− = 0 , …(6)dx 2 2 1+y' + z' dz'2λ(x)y − = 0 . …(7)dx 2 2 1+y' + z'

CALCULUS OF VARIATIONS 47Solving these equations (6) and (7), we obtain a family of curves depending onfour constants, whose values are determined from the boundary conditions in(5).Example 2 . Find the shortest distance between the points A(1, −1, 0) and B(2,1, −1) lying on the surface 15x−7y +z −22 = 0.Solution. In this question, we have to find the minimum of the functionalsubject to the conditionsprovided2122J[y, z] = 1+ y' + z'dx,…(1)y(1) = −1,y(2) = 1z(1) = 0, z(2) = −1g(x, y, z) ≡ 15x −7y +z − 22 = 0.To achieve this end, we form an auxiliary functionalwhere…(2)…(3)J*[y, z] = 21F(x, y, z, y′, z′)dx …(4)F =221 + y' + z' + λ(x) [15x−7y +z −22]. …(5)The corresponding Euler’s equations areCombined together, we getdy'0 + λ(x) {−7} − = 0dx 2 21 y' z' + + dz'0 + λ(x). {1} − = 0dx 2 21 y' z' + + ddxIntegrating, we findy'z' + = 02 21+y' + z' y'7z' += c2 21+y' + z'1…(6)…(7)…(8)From (3), we writez′ = 7y′ −15. …(9)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 48From equation (8) and (9), and then integrating, we obtain (exercise)y(x) = αx +β.…(10)Using boundary conditions in (2), we find (exercise)α = 2, β = −3, y(x) = 2x−3.…(11)From equation (9) and (11), we havez′ = −1giving z(x) = c−x.The B.C.′s in (2), givez(x) = 1−x…(12)Putting y(x) and z(x) from equations (11) and (12) into equation (6), we findλ(x) ≡ 0.…(13)The desired shortest distance is (exercise)21l = 1 + y' + z' =226The Books Recommended for Chapter I1. I.M. Gelfand Calculus of Variations,andS.V. FovminPrentice Hall.

TRANSPORT AND LAPLACE EQUATIONS 49Chapter-2Transport and Laplace Equations2.1 INTRODUCTIONMany physical problems in science, engineering and geometry can be modeledmathematically by partial differential equations (PDE). A partial differentialequation is an equation involving an unknown function of two or morevariables and certain of its partial derivatives.Before writing symbolically a typical PDE, we first present the notation /symbol to be used consequently.2.1.1 Geometric Notation(i)(ii)(iii)R n = n – dimensional real Euclidean space,R 1 = R = real line.e i = ith standard coordinate vector= (0, 0, ……, 0, 1, 0,…….0).(iv) A typical point x in R n isx = (x 1 , x 2 ,……, x n ).Sometimes, we will also regard x as a row or column vector.(v) R n + = open upper half – space= { x = (x 1 , x 2 ,……,x n ) ∈ R n | x n > 0}.(vi) R + = { x ∈ R | x > 0}(vii) U, V, W etc are usually open subsets of R n .(viii) ∂ U = boundary of U(ix)(xi)U = closure of U= U ∪ ∂U.A typical point in R n+1 will often be denoted as(x, t) = (x 1 , x 2 ,…., x n , t),and we usually interprett = x n+1 = time.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS50(xii) A point x ∈ R n will sometimes be written asx = (x 1 , x n )forx 1 = (x 1 , x 2 ,…., x n-1 ) ∈ R n-1 .(xiii) B(x, r) = closed ball in R n with center at x and having radius r, r > 0.= { y ∈ R n | |y – x| ≤ r}.(xiv) B 0 (x, r) = open ball in R n with centre at x and radius r == { y ∈ R n | |y – x| < r}(xv) For a = (a 1 , a 2 ,….., a n ) and b = (b 1 , b 2 ,……., b n )na . b =i=1a i b i| a | =122 na i = i = 1 222(xvi) C n = n – dimensional complex space,(xvii) C 1 = C = complex plane.(xviii) α(n) = volume of unit ball B(0, 1) in R n n / 2π = n Γ+ 1 2 In particular for n = 3 ,(xix)α(3) =2a1+ a + .......... + an≡ Euclidean norm of a4π for r = 13n α(n) = surface area of unit sphere B(0, 1) in R n= ∂B(0, 1).2.1.2. Notation for FunctionsIf u : U → R is a real valued function with domain U ⊂ R n , we writeu(x) = u(x 1 , x 2 ,……, x n ), for x ∈ U.Definition: Function u is called smooth when u is infinitely differentiableIf u and v are two functions, then we write,

TRANSPORT AND LAPLACE EQUATIONS 51u ≡ v (read : u is identically equal to v)when functions u and v agree for all values of their arguments.(i)We writeu : = vto define u as equaling v(ii) u + = max (u, 0) , u + ≥ 0(iii) u - = − min(u, 0) , u - ≥ 0(iv) u = u + − u - ,(v) | u | = u + + u - .(vi)The sign function is defined assgn (x) =10−1if x > 0if x = 0if x < 0.(vii)If u : U → R m , U ⊂ R n ,we writeu (x) = (u 1 (x) , u 2 (x) , .... , u m (x)) for x ∈ UHere ,u k is the kth component of u for k = 1,2,…., m. Furtheru k : U → R.(viii)The functionχ E (x) =10if x∈Eif x∉Eis called the Indicator Function of E.(ix)A function u : U → R is called Lipschitz continuous if| u(x) – u(y) | ≤ C | x – y | for al x , y ∈ U

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS52and for some constant C. Here, on the left there is norm in R, and on theright, there is norm in R n .2.1.3. Notation for DerivativesLet u : U → R, x ∈ U ⊂ R n . We write(i)∂ulim u(x + hei) − u(x)( x)=∂xh → 0 h , provided this limit exists, h ∈ R.i(ii) We usually writeu x ifor∂ u .∂x i(iii)u x ix =j∂ 2 u,∂x i∂x ju x ix jx =k∂ 3 u∂x∂x∂xijk, etc.2.1.4. Multiindex Notation(1) A vector / n-tuple of the formα = (α 1 , α 2 ,………., α n ), α i is a non-negative integer for each i,is called a multiindex. Its order is defined as| α | = α 1 + α 2 + ……..+α n = ni=1α iNote : | α | ≥ 0 and | α | is a non – negative integer.Also, we define(2) For x ∈ R n , we defineα! = α 1 ! α 2 ! ……. α n !(3) We employ the symbolx α = x α 1 1 x α 2 α2 ……. xn 3D uto denote the gradient vector of the function u.(4) Given a multiindex α = (α 1 , α 2 ,……, α n ) , we define

TRANSPORT AND LAPLACE EQUATIONS 53| α|D α ∂ u(x)α1α2αnu(x) = = ∂ x ∂ ........ u.α1α2αn1 x ∂2xn∂ ∂ ........ ∂x1x2xn= n∏i=1 ∂ ∂xiαi uIn particular, if α = 0, then D α is the identity operator.(5) If k is a non – negative integer, we defineD k u(x) : = {D k u(x) : |α| = k}. ( * )Thus, D k u(x) is the set of all partial derivatives of order k. Assigning someordering to the various partial derivatives in ( * ), we can also regardD k u(x) as a point inknR- space.(6) We define| D k u | =| |αD uα | =k2|12(7) Special cases:(a) When k = 1, Du is a point in R n – space and we arrange the elements of Duin a vector of the formIn particular, for n = 3,Du = (u ,u x 1 x......., u x) = gradient vector2 nDu = ((u ,u , x 1 xu2 x)3(b) When k = 2, D 2 u can be regarded as an element ofelements of D 2 u are being arranged in a matrix2nR- space, and the

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS5422 ∂u∂ u ∂ u.............2∂x1∂x1∂x2∂x1∂x22∂ u ∂ uD 2 ...................u = ∂x2∂x1∂x2∂x2− − − − − − − − − − − − − − −222 ∂ u ∂ u ∂ u..........∂xn∂x1∂xn∂x2∂xn∂xThis matrix is called the HESSIAN MATRIX.For n = 2 (i.e., in two dimensional space),nnu = u(x, y) and D 2 u =2∂u 2∂x2 ∂ u∂y∂x2∂ u ∂x∂y2∂ u 2 ∂y2×2(c) tr(D 2 u) = ni=1u xix i= Laplacian of u= ∆ u.(d) For a function of two variables u = u(x, y) ,andx = (x 1 , x 2 ,….., x n ) ,y = (y 1 , y 2 , ….., y n ) ,D x u = ( ux, u1 x….., u2 x) ,nD y u = ( uy, u1 y…., u2 y) .n2.1.5 Vector – valued Functions(i)Let U ⊂ R n and m > 1. Letu : U → R mbe a vector – valued function andWe define⎺u = (u 1 , u 2 ,….., u m ).D α ⎺u = (D α u 1 , D α u 2 , …..D α u m )

TRANSPORT AND LAPLACE EQUATIONS 55for each multi-index α.We note that D α u i are defined earlier under the heading “Notation forDerivatives”.(ii)andFor a non – negative integer k, we defineD k ⎺u = { D α ⎺u : |α| = k}| D k ⎺u | = norm in m – dimensional space= | |αD uα | =kas defined earlier for scalar – valued functions.2.1.6 Measures and Integrals2|12(i) The integral of a function f : U ⊆ R n → R, over a subset U ⊆ R n , withrespect to Lebesgue measure is denoted byUf (x) dx or simply Uf .Note: If no subscript occurs on the integral sign , the region of integration isunderstood to be R n .(ii) Let be a smooth (n – 1) dimensional surface in R n , we write fdsfor the integral of f over , with respect to (n – 1) – dimensional surfacemeasure.(iii) If C is a curve in R n , we denote by Cf dt ,the integral of f over C w.r.t. arc length.(iv) The convolution of the function f and g, denoted byis given byf * g,

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS56(f * g) (x) == f (x – y) g(y) dyf(y) g(x – y) dy= ( g * f) (x),provided the integrals exists.2.1.7 Function Spaces(1) C(U) = { u| u : U → R is continuous }C( U ) = { u : u ∈ C(u) and is uniformly continuous }(2) C k (U) = { u : U → R, U ⊆ R n | u is k – times continuously differentiable}C k ( U ) = { u ∈ C k (U) | D α u is uniformly continuous for all | α | ≤ k}Thus, if u ∈ C k ( u ), then D α u continuously extends to U for each multiindex αsuch that | α | ≤ k.(3)C ∞ (U) = { u : U → R | u is infinitely differentiable}= Ι ∞k =0C k (U)C ∞ ( U ) = Ι ∞k =0…(4)C k ( U ).L p (U) = { u : U → R : u is Lebesgue measurable, || u || < ∞ }L p (U)where|| u || Lp(U)= fp dx | |. 1 ≤ p < ∞. ∪ 1p2.1.8 Notation for Matrices(1) A = (a ij )= a matrix A which is an m × n matrix with (i, j) th entry a ij .A = diag(d 1 , d 2 ,…., d n )= a diagonal matrix.

TRANSPORT AND LAPLACE EQUATIONS 57(2) M m×n = space of real m × n matricesS n×n = space of real symmetric n × n matrices(3) tr A = trace of A= a 11 + a 22 +…..+ a nn= sum of diagonal elements(4) det A = determinant of the matrix A(5) cof A = cofactor matrix of A= Transpose of (Adj A)= (Adj A) TA T = transpose of the matrix A(6) If A = (a ij ), B = (b ij ) are m × n matrices, thenmnA : B = i= 1 j=1a ij b ij| A | = norm of matrix A= (A : A) 1/2+ ……..+ (a nn ) 2 ] 1/2= n ni= 1 j=1( aij)212= [(a 11 ) 2 + (a 12 ) 2 +……+ (a 1n ) 2 + (a 21 ) 2 + (a 22 ) 2 +….+ (a 2n ) 2(7) If A = (a ij ) ∈ S n×n and x = (x 1 , x 2 , …., x n ) ∈ R n , thenmx . A x = ni= 1 j=1a ij x i x jm= i,j=1a ij x i x j= Quadratic Form corresponding to (a ij ) .(8) Let A ∈ S n×n . If

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS58we writex . A x ≥ θ | x | 2 for all x ∈ R n , and some real number θ, then,A ≥ θ. I(9) For A ∈ M n×n , y ∈ R n , we sometimes writey A = A T y.2.2 TRANSPORT EQUATIONThe transport equation with constant coefficients is the PDEwhereandU t + b. D u = 0 in R n × [0, ∞) , (1)b = (b 1 , b 2 ,…., b n ) is a fixed vector in R n ,u : R × [0, ∞] → Ris the unknown function, andu = u(x, t).Note. Here x = (x 1 , x 2 , ….., x n ) ∈ R n is a typical point in space, and t ≥ 0denotes a typical time variable.We writeD u = D x u = ( ux, u1 x….., u2 x) (2)nfor the gradient of the scalar function u with respect to the spatial variable x.Initial – Value ProblemLet us consider the homogeneous linear initial – value problemwhereU t + b . D u = 0 in R n × [0, ∞) (1)u = g on R n × {0 = t} (2)g : R n → R is known.The problem is to compute u = u(x ,t).Solution. Let (x, t) be any given (hence fixed) point in R n × [0, ∞).The line through (x, t) with direction (b, 1 ) is represented parametrically by

TRANSPORT AND LAPLACE EQUATIONS 59x(s)= x + sb , s∈R (3)t(s)= t + si.e., by (x + s b, t + s) for s ∈ R.This line hits the planeat the point (x – t b, 0) , whenΓ : R n × {t = 0} (4)s = − t. (5)Since u is constant on the line andu(x – t b, 0) = g(x – t b), (6)by virtue of given initial condition (2), we deduce thatfor x ∈ R n and t ≥ 0.u(x, t) = g(x – t b) (7)So, if the given initial – value problem has a sufficiently regular solution,u = u(x, t), it must certainly be given by (7) above.Conversely, if g is C 1 , then u = u(x, t) defined by (7) is indeed a solution of thegiven initial – value problem.Verification:From, (7), we findU t = − b . D(ξ), where ξ = x – t bD u = D(ξ)Hence u t + b. D u = [ −b . D(ξ)] + b . [D(ξ)]and, for t = 0,This completes the result.= 0 (8)u(x, 0) = g(x) on R n (9)Remark : If g is not C 1 , then there is no C 1 solution of the given initial – valueproblem. But even in this case, formula (7) certainly provides a strong, and infact, the only reasonable candidate for a solution.We may thus formulary declareu(x, t) = g(x – t b), x ∈ R n , t ≥ 0. (10)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS60to be a weak solution of IVP, even should g not be C 1 . This all makes senseeven if g, and thus u, are discontinuous.Non homogeneous problemProblem: Consider the non – homogeneous initial – value problemut+ b.D u = fu = gin Rin Rnn× [0, ∞)× {t = 0}(1)(2)in which b = (b 1 , b 2 , ….., b n ) ∈ R n is a fixed vector, andu : R × [ 0, ∞) → Ris the unknown function, andu = u(x, t),x = (x 1 , x 2 ,……, x n ) ∈ R n is a point in space,t ≥ 0 denotes a typical time variable,Du = D x u = ( ux, u1 x….., u2 x)ndenote the gradient of u with respect to the spatial variable x,is known,g : R n → Rf : R n × [0, ∞) → Ris known. The problem is to compute u = u(x . t).Solution: Let (x, t) be any given, hence fixed, point in R n × [0, ∞). Define afunctionz:R →Rfor all s ∈ R. Thenusing (1).z(s)= u(x + s b,t + s)z&(s) = b . D u(x + s b, t + s) + u t (x + s b, t + s)Now, using (2), (3) and (4), we find(3)= f(x + s b, t + s) (4)u(x, t) – g(x – b t) = z(0) – u(x – b t, 0)= z(0) – z(- t)

TRANSPORT AND LAPLACE EQUATIONS 610= −t0= −tt=0z&(s) dsf(x + s b, t + s) dsf(x + (s – t) b, s) ds (5)This givestu(x, t) = g(x – b t) + 0f(x + (s – t) b, s) ds (6)for x ∈ R n , t ≥ 0as solution of the given non – homogeneous initial – value problem.2.3 LAPLACE’S EQUATIONProblem: Laplace’s equation isand Poisson’s equation is∆u = 0, (1)∆u = − f . (2)In equation (2), the minus sign is taken so that the notation is consistent withnotation for general second – order elliptic operators.In both equations (1) and (2),x ∈ U ⊆ R n ,and the unknown function isIn equation (2),is given. FurtherU is an open setu : U → R , U = closure of Uu = u(x).f : U → R∆u = Laplacian of un=i=1ux i x i.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS62Definition : A function u ∈ C 2 is called harmonic function if u satisfies theLaplace’s equation∆u = 0.Physical InterpretationLaplace’s equation comes up in a wide variety of physical contexts – such aswhen u denotes the chemical concentration / temperature / electrostaticpotential.Laplace’s equation arises as well in the study of analytic functions.Fundamental Solution of Laplace’s EquationWe attempt to find a solution of the given Laplace equation∆u = 0 (1)by searching radial solutions of the formu(x) = v(r) (2)wherer = | x |12 22 2= ( x1x2+ ...... + xn)and v is to be selected, if possible, so thatholds.First, we note thatThus, we have+ , (3)∆ v = 0, (4)∂r∂x i1=21−2 22 2(1+ x2+ ..... + xn)x (2x i )= rx i, (x ≠ 0) (5)u xi = v′(r) ∂r ∂x i x = v′(r) i , (6) r

TRANSPORT AND LAPLACE EQUATIONS 63anduxi x ifor i = 1, 2,…, n. So,= v′′(r)x 22 i + v′(r) − 3r1rxi (7)rn∆u = i=1u xi xiHenceiff n −1= v″(r) + v′(r). (8) r ∆ u = 0v′′ +If v′ ≠ 0, we deduceor ( v )or n −1 v′ = 0. (9) r v"− n= 1v'rd − nlog '= 1 .dr rlog v′(r) = (1 – n) log r + constt.aor v′(r) =n−1, (10)rfor some constant a . Consequently, if r > 0, we obtainblogr+ c,n = 2v(r) = b + c,n ≥3n−2 rwhere b and c are constants.Let 1− log | x | ,2πΦ(x) = 1 1n(n − 2) α(n)| x |n = 2≥, n 3n−2 (11)(12)for x ∈ R n , x ≠ 0. α(n) = volume of unit ball in R n .

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS64Then Φ(x) is a solution of the given Laplace equation (1) and is called theFundamental Solution of Laplace’s Equation.Note: This fundamental solution is radial.2.4 FUNDAMENTAL SOLUTION OF POISSON’SEQUATIONLet Φ(x) be the fundamental solution of Laplace’s equationwhere∆u = 0, (1) 1− log | x | ,2πΦ(x) = 1 1n(n − 2) α(n)| x |nn = 2, n ≥3−2 (2)and x ∈ R n , x ≠ 0.So, the mappingis harmonic.x → Φ(x) , x ≠ 0 , (3)If we shift the origin to a new point y, the PDE (1) is unchanged, so themappingx → Φ(x – y) (4)is also harmonic as a function of x, x ≠ y.Now, we consider the Possion’s equationwherewe note that the mapping∆u = − f, (5)f : R n → R. (6)x → Φ(x – y) f(y), (7)for x ≠ y, is harmonic for each point y ∈ R n , and so is the sum of finitely manysuch expressions built / constructed for different points y.Consider convolutionu(x) =nRΦ(x – y) f(y) dy . (8)

TRANSPORT AND LAPLACE EQUATIONS 65From equations (2) and (8), we writeu(x) = 1−−π log(| x y |) f ( y)dy2 nR 1f ( y)n(n − 2)α(n) nnR| x − y |−2dy,( n = 2)( n ≥3). (9)For simplicity, we assume that the function f, given in Possion’s equation (5),is twice continuously differentiable with compact support.Now, we shall show that, u(x) defined by (9) satisfies(i) u ∈ C 2 (R n )(ii) ∆u = −f in R n .Consequently, the function in (9) provided us with a formula for a solution ofPossion’s equation (5).Proof of (i):We haveHenceu(x) =nRu( x + hei ) − u(x)hΦ(x – y) f(y) dy =nR=nRΦ(y)Φ(y) f(x – y) dy (10) f ( x + hei− y)− f ( x − y)dy h (11)where h ≠ 0 is a real number and e i ∈ R n ,e i = (0, 0, …,0, 1, 0, ….., 0)with 1 in the ith slot.Butf ( x + hei − y)− f ( x − y)h→∂ f (x−y) (12)∂x iuniformly on R n as h → 0. Thus, on taking h → 0 in (11) and making use ofresult in (12), we write∂u ∂f( x)= Φ(y) ( x − y) dy,(13)∂xin ∂xiR

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS66for i = 1, 2, 3,…, n.Similarly2∂ u∂x∂xfor i, j = 1, 2, …, n.ij( x)=nR2∂ fΦ(y) ∂xi∂xj( x − y)dy,(14)As the expression on the right hand side of (14) is continuous in the variable x,we see thatThis proves (i).Proof of (ii) :u ∈ C 2 (R n ) (15)This function Φ(x), defined in (9), blows up at x = 0 , we will need forsubsequent calculations to isolate this singularity (x = 0) inside a small ball.So, fix ∈ > 0. Let B(0, ∈) denote an open ball at x = 0 with radius ∈. Then,from equation (10), we obtainswhereNow∆u(x) =I ∈ =J ∈ =B(0,∈)B(0,∈)Φ(y)∆ x f(x – y) dy +R n −B(0,∈)Φ(y)∆ x f(x – y) dy= I ∈ + J ∈ , say, (16)Φ(y)∆ x f(x – y) dy, (17)Φ(y)∆ x f(x – y) dy, (18)R n −B(0,∈)| I ∈ | ≤ | Φ(y ) | | ∆ x f(x – y) | dy ,B(0,∈)≤ C || D 2 f || nL ∞ ( R ) | Φ(y)| dy, ∈)B(0

TRANSPORT AND LAPLACE EQUATIONS 672C ∈≤ C ∈2|log∈|Also, by integration by parts, we get( n = 2)( n ≥3)(19)J ∈ =R n −B(0,∈)Φ(y)∆ y f(x – y) dy= − D Φ(y) . D y f(x – y) dyR n − B(0, ∈)+ ∂B(0,∈)∂fΦ(y ) (x – y) dS(y), using divergence then∂v= K ∈ + L ∈ . (20)v indicating the inward pointing unit normal along the boundary ∂B(0, ∈) ofthe ball B(0, ∈).Further| L ∈ | ≤ || D f || nL ∞ ( R ) Φ ( y)ds(y) R n −B(0,∈)C ∈|log∈|,n = 2≤ . (21)C∈n ≥ 3We continue by integration by parts once again in the term K ∈ , to obtain /discoverk ∈ =R n∆Φ(y)f(x – y) dy−B(0,∈)– Φ(y)f(x – y) dS(y)R n −B(0,∈)= ∂Φ∂v∂B(0,∈)(y) f(x – y) dS(y), (22)since the function Φ is harmonic away from the origin (x ≠ 0).

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS68Now−1yDΦ(y)= ,nnα(n)| y |− y yv = = − ,| y | ∈on the boundary ∂ B(0, ∈). Consequently∂Φ(y) = v . D Φ(y)∂von the boundary ∂B(0, ∈).1nα(n)∈=−1y ≠0,(23), (24)Since n α(n) ∈ n-1 is the surface area of the n – dimensional sphere ∂B(0, ∈), wehave1K ∈ = −−1nα(n)∈ f(x – y) dS(y)∂ B(0, ∈)= −∂B( 0,∈)f(y) dS(y)→ − f(x), as ∈ → 0. (25)Here, a slash through an integral denote an average value.Combining now equations (16) – (25), and letting ∈ → 0, we findas asserted earlier.∆ u(x) = − f(x), (26)Thus, u(x), given by (9), in a solution of (26). This completes the solutions ofPoisson’s equation.Remark: (i) We sometimes write∆ Φ = − δ 0in R n , δ 0 denoting the Dirac measure on R n giving unit mass to the point x = 0.Adopting this notation, we formally compute∆u(x) =nR[∆ x Φ(x – y)] f(y) dy

TRANSPORT AND LAPLACE EQUATIONS 69= −nRδ x f(y) dy= − f(x),x ∈ R n , in accordance with above theorem.Remark (ii). The above theorem (Solving Poisson’s equation) is in fact validunder for less stringent smoothness requirements for f.2.5 MEAN – VALUE FORMULAS FOR LAPLACE’SEQUATIONLet U ⊂ R nbe an open set. Letu : U → Rbe a harmonic function. We define(i) average of f over the ball B(x, r)1=n( ) fα n rB(x,r)dy= f ds.∂ B(x,r)whereα(n) = volume of unit ball B(0, 1) in R nn / 2π= , Γn + 1 2 n α(n) = surface area of unit sphere ∂B(0, 1) in R nNote: For x ∈ U ⊂ R n , r = | x | ,we shall now derive the important mean – value formulas, which declare that“u(x) equals both the average of u over the sphere ∂B(x, r) and the averageof u over the entire ball B(x, r), providedB(x, r) ⊂ U”.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS70Theorem (Mean – value formulas for Laplace’s equation)Statement : If u ∈ C 2 (U) is harmonic, thenu(x) =∂ B( x,r)u ds =B( x,r)u dy,for each ball B(x, r) ⊂ U.Proof of Part – ISetThenφ(r) =φ′(r) =∂ B( x,r)u(y) dS(y)= u(x + r z) dS(z). (1)∂ B(0,1) z . Du(x + r z) dS(z), (2)∂ B(0,1)and consequently, using Green’s formula, we computeφ′(r) =∂ B( x,r)y−rx . Du(y) d S(y)= ∂ B( x,r)∂u∂vdS(y)= nrB( x,r)∆u(y) dyHence φ is constant, and so= 0. (3)φ(r) =limφ(t)t → 0

TRANSPORT AND LAPLACE EQUATIONS 71=lim t → 0 u ( y)dS(y)∂ B(x,t ) = u(x). (4)Equations (1) and (4) prove the part – I, i.e. ,u(x) =∂ B( x,r)u(y) dS(y) = average of u over the sphere ∂B(x, r). (5)Proof of Part – II : We observe that by employing polar coordinates, one gets u dy = u dsdξB( x,r)0 ∂B(x,ξ)rHencer= u(x) [n α(n) ξ n-1 ] dξ0= α(n) r n u(x). (6)u(x) =1α(n)rnB(x,r)u dy= B( x,r)f dy= average of u over the entire ball B(x, r). (7)This complete the proof of both the mean – value formulas for Laplace’sequation.Theorem (Converse of mean – value property for Laplace’s equation):Statement: If u ∈ C 2 (U) satisfies the mean formulau(x) =∂ B( x,r)u dSfor each ball B(x, r) ⊂ U, then prove thatu : U → Ris harmonic.Proof: If possible assume that∆U ≠ 0 in U ⊆ R n (1)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS72Then, there exists some open ballsuch thatSetB(x, r) ⊂ U (2)∆u > 0, within B(x, r) . (3)φ(r) =∂ B( x,r)u(y) dS(y) (4)Then, as proved earlier (exercise)φ′(r) = nrB( x,r)∆u(y) dy . (5)Using (3) and (5), we getφ′(r) > 0 . (6)From the hypothesis and equation (4), it follows thatu(x) = φ(r) = constant (7)This contradicts (6). Hence, the result follows. This completes the proof.2.6. ENERGY METHODSDefinition (Energy functional) :It is defined as12 I[w] = | D w | −w fdx(1)2Uwhere w belongs to the admissible set.A = { w ∈ C 2 (⎺U )| w = g on ∂ U}. (2)and∆w = − f in U. (3)Theorem (Dirichlet’s principle):Statement: Assume u ∈ C 2 ( ⎺U ) solves the boundary – value problem∆u=− f in Uu = g on ∂U( * )

TRANSPORT AND LAPLACE EQUATIONS 73where U is open and bounded subset of R n and its boundary ∂U is C 1 . ProvethatI[ u]= min I[w],w∈A( ** )where I[w] is the energy functional and w belongs to the admissible setA = {w ∈ C 2 (⎺U )| w = g on ∂U} ( *** )Conversely, if u ∈ A satisfies ( ** ), then u solves the boundary value problem( * ).Proof (Part – I) : Choose w ∈ A. ThenW = g on ∂ U (1)Let u ∈ C 2 (⎺U) solves the BVP ( * ). ThenNow∆ u = −f in U(2)u = g on U(3) ( ∆u + f) (u – w) dxUby virtue of (2). This gives= 0 (4)[ (∆u) (u – w) + f(u – w)] dx = 0UAn integration by parts yields (using Green’s formula)using (1), (2) and (3).This implies[ D u. D(u – w)] dxU= − ( ∆u) (u – w) dx + U∂= ( f) (u – w) dx + 0,UU ∂u (u – w) dS ∂v

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS74[ D u . D(u – w) – f(u – w)] dx = 0. (5)Uas u = w = g on ∂ U, and hence there is no boundary termEquation (5) givesUwe know the estimates[ | D u | 2 – u f] dx = [ D u. D w – w f] dx. (6)| D u . D w | ≤ | D u | | D w |U≤ 21 | D w | 2 + 21 | D w |2(7)by virtue of Cauchy – Schwarz and Cauchy inequalities.From (6) and (7), we writeU{ | D u | 2 – u f } dx ≤ 21UBy definition, the energy functional is given by| D u | 2 + 21 [ | D w | 2 – w f] dx (8)12 I[w] = | Dw | −w fdx (9)2Hence, relation (8) concludesSince u ∈ A, it follows thatUI[u] ≤ I[w], w ∈ A. (10)I[u] =This proves part – I .min I[w] . (11)w∈AProof of Part – II : Conversely, assume that the conclusion ( ** ) of thestatement of the theorem holds.Let v ∈∞C c (U) be any but fixed function. Letλ(τ) = I[u + τ v], τ ∈ R (12)where the energy function I is defined above in equation (9).Sinceu + τ v ∈ A

TRANSPORT AND LAPLACE EQUATIONS 75for each τ, the scalar function λ(τ) has a minimum at zero, by virtue ofassumption in ( ** ).Soλ′(0) = 0, (13)provided this derivative of λ(τ) at τ = 0 exists. Butλ(τ) = I[u + τ v]12 = | Du + τ Dv | −(u + τv)fdx2U21 2 τ 2= | Du | + | Dv | + τDu.Dv − ( u + τv)f dx(14)U22Equation (13) and (14) give at onceThis gives ( D u . D v – v f) dx = 0 .U ( – ∆u – f ) v dx = 0 . (15)UThis identity is valid for each function v ∈∞C c (U). So we must haveor− ∆ u – f = 0 in U∆ u = − f in U. (16)This shows that u solves the given boundary – value problem. Hence, the proofof the converse of Dirichlet’s principle is complete.This completes fully the Dirichlet’s principle.Note (i) : In other words, the Dirichlet’s principle states thatIf u ∈ A, then P D E∆ u = − fu = gin Uon ∂ U

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS76is equivalent to the statement that the solution function u = u(x, t) minimizesthe associated energy functionalI[ . ].Note (ii) : Dirichlet’s principle is an instance of the calculus of variationsapplied toLaplace Equation.Theorem (Uniqueness theorem)Statement : Prove that there exist at most one solution u ∈ C 2 (⎺U ) of theboundary – value problem,∆ u = − fu = gin Uon ∂ Uwhere U is bounded, open, and ∂ U is C 1 .Proof: If possible assume that, in addition to u, there is another solution, say⎺uof the given boundary-value problem.Setw = u −⎺u in U . (1)Since u and ⎺u are solutions of the given boundary value problem, soNow, in U,and∆ u = − f in U (2)u = g on ∂ U (3)∆⎺u = − f in U (4)⎺u = g on ∂ U (5)∆ w = ∆ u − ∆⎺u= (− f) – (−f)From Green’s formula, we writeU= 0 in U , (6)w = 0 on ∂U . (7)| D w | 2 dx = U(D w . d w) dx= − U ∂w w(∆ w) dx + w dS ∂v∂∪

TRANSPORT AND LAPLACE EQUATIONS 77= 0 + 0using (6) and (7). Equation (8) shows that= 0, (8)D w ≡ 0 in U (9)Since w = 0 on the boundary ∂ U and w is constant in U, it follows thatorw = 0 in Uu =⎺u in U. (10)This proves uniqueness theorem.2.7 PROPERTIES OF HARMONIC FUNCTIONWe now present a sequence of interesting deductions about harmonicfunctions, all based upon the mean – value formulas. Assume for the followingthat U ⊂ R n is open and bounded.Theorem: (Strong maximum principle).Statement : Suppose u ∈ C 2 (U) ∩ C(U) is harmonic within U.(i) Thenmax u = max uU ∂U(ii) Furthermore, if U is connected and there exists a point x 0 ∈ U such thatthenu(x 0 ) =max u ,∂Uu is constant within U.Proof: Suppose there exists a point x 0 ∈ U withThen foru(x 0 ) = M = max ⎺u u . (1)0 < r < dist(x 0 , ∂U),the mean – value property asserts

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS78M = u(x 0 ) =B( x 0 , r)u dyAs equality holds only≤ M. (2)we seeif u ≡ M within B(x 0 , r), (3)u(y) = M (4)for all y ∈ B(x, r). Hence the set{x ∈ U | u(x) = M }is both open and relatively closed in U, and thus equals U if U is connected.This proves assertion (ii), from which (i) follows.Note : Assertion (i) is the maximum principle for Laplace’s equation and (ii) isthe strong maximum principle. Replacing u by –u, we recover also similarassertions with “min” replacing “max”.Remark (i) : The strong maximum principle asserts in particular that if U isconnected andsatisfiesu ∈ C 2 (U) ∩ C (U )∆u= 0 u = 0in Uon ∂U,where g ≥ 0, then u is positive everywhere in U if g is positive somewhere on∂U.Remark (ii) : An important application of the maximum principle isestablishing the uniqueness of solutions to certain boundary – value problemsfor Poisson’s equation.Theorem: (Uniqueness).Statement : Let g ∈ C(∂U), f ∈ C(U). Then there exists at most one solution u∈ C 2 (U) ∩ C (U ) of the boundary – value problem

TRANSPORT AND LAPLACE EQUATIONS 79− ∆u= f u = gin Uon ∂U.(1)Proof: If u and (⎺u ) both satisfy (1) , apply theorem above to the harmonicfunctionsw = ± (u − ⎺u ) .Local Estimates for Harmonic FunctionsNext we employ the mean – value formulas to derive careful estimates on thevarious partial derivatives of a harmonic function. The precise structure ofthese estimates will be needed below, when we prove analyticity.Theorem (Estimates on derivatives)Statement : Assume u is harmonic in U. Then| D α Cu(x 0 ) | ≤ k || un+ kr||(1)1L ( B(x0,r))for each ball B(x 0 , r) ⊂ U and each multi-index α of order | α | = k.Here1C 0 = ,α(n)Ckn+1( 2 nk)=α(n)k(k = 1,……) (2)Proof 1: We establish (1) and (2) by induction on k . The case k = 0 beingimmediate from the mean – value formula. For k = 1, we note upondifferentiating Laplace’s equation that u (i = 1, …., n) is harmonic.Consequentlyxiu (x 0 ) | = | B( x 0 , r / 2)|xiuxdx |i2|n r dS | (3)α ( ) ∂ ( 0 2n= uvniB x , r /2n≤ || u || .∞ rr L ( ∂B(x0,Now if x ∈ ∂B(x 0 , r/2), then B(x, r/2) ⊂ B(x 0 , r) ⊂ U, and so))2

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS80|u(x)| ≤α1n 2 n|| u ||1L B(x , r))( ) r0Combining the inequalities above, we deduce.| D α n+12 n 1u(x 0 ) | ≤ || u ||n+11L ( B(x0, r))α(n)r(4)if | α | = 1. This verifies (1) and (2) for k = 1 .2. Assume now k ≥ 2 and (1) and (2) is valid for all balls in U and eachmultiindex of order less than or equal to k – 1. Fix B(x 0 , r) ⊂ U and let α be amultiindex with | α | = k. Then D α u = (D β u) xi for some i ∈ {1,…., n} , | β | = k– 1. By calculations similar to those in (3), we establish that (exercise)| D α n k βu(x 0 ) | ≤ || D u ||r.∞rL ∂B(x , ))( 0 k(5)If x ∈ ∂B(x 0 , kr ), thenB(x,Thus (1) , (2) for k – 1 implyk −1r) ⊂ B(x0 , r) ⊂ U.k| D β n + 1k −1(2 n(k −1))u(x) | ≤ || u ||11 . (6)n + k − L ( B(x0, r)) k −1α(n) r k Combining the two previous estimates yields the bound| D α n+1 k(2 nk)u(x 0 ) | ≤ || u || .n+k1L ( B(x0, r))α(n)r(7)This confirms (1), (2) for | α | = k.Liouville’s Theorem.Next we see that there are no nontrivial bounded harmonic functions on all ofR n .Theorem (Liouville’s Theorem)

TRANSPORT AND LAPLACE EQUATIONS 81Statement : Suppose u : R n → R is harmonic and bounded. Then u isconstant.Proof: Fix x 0 ∈ R n , r > 0, thenC| Du(x 0 ) | ≤ 1 || u ||n+11L ( B(x0, r ))r≤C α(n)||r1u||∞ nL ( R )→ 0 ,as r → ∞. ThusDu ≡ 0,and sou is constant. This proves the Liouville’s Theorem.2.8 GREEN’S FUNCTIONAssume now U ⊂ R n is open, bounded, and ∂U is C 1 . We propose next toobtain a general representation formula for the solution of Poisson’s equation−∆u = f in U, (*)subject to the prescribed boundary conditionu = g on ∂U. (**)Derivation of Green’s function.Suppose first of all u ∈ C 2 (⎺U) is an arbitrary function. Fix x ∈ U, choose ∈ >0 so small that B(x, ∈) ⊂ U, and apply Green’s formula on theregion V ∈ = U – B(x, ∈) to u(y) and Φ(y – x). We thereby computeV ∈u(y) ∆Φ(y – x) - Φ(y – x) ∆u(y) dy= ∂V ∈u(y)∂Φ(y – x) - Φ(y – x)∂v∂ u (y) dS(y), (1)∂vv denoting the outer unit normal vector on ∂V ∈ . Recall next

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS82We observe also that∆Φ(x – y) = 0 for x ≠ y.|∂ B( x, ∈)Φ(y – x)∂ u (y) dS(y) | ≤ C ∈n-1∂vmax| Φ | = 0(1)∂B(0,∈)as ∈ → 0. Furthermore∂ B( x, ∈)u(y)∂Φ∂v(y – x) dS(y) = ∂ B( x, ∈)u(y) dS(y) → u(x)as ∈ → 0. Hence our sending ∈ → 0 in (1) yields the formula:u(x) = ∂UΦ(y – x)∂ u ∂Φ(y) – u(y) (y – x) dS(y)∂v∂v− UΦ(y – x) ∆u(y) dy. (2)This identity is valid for any point x ∈ U and any function u ∈ C 2 (U ) .Now formula (2) would permit us to solve for u(x) if we knew the values of ∆uwithin U and the values of u, ∂u / ∂v along ∂U. However for our application toPoisson’s equation with prescribed boundary values for u, somehow modify (2)to remove this term.The idea is now to introduce for fixed x a corrector functionφ x = φ x (y),solving the boundary – value problem:x∆φ= 0x φ = Φ(y − x)in Uon ∂U.(3)Let us apply Green’s formula once more, now to compute− Uφ x (y) ∆u(y) dy = ∂Uu(y)x∂φ∂v(y) - φ x (y)∂ u (y) dS(y)∂v= ∂Uu(y)x∂φ∂v∂ u(y) - φ(y -x) (y) dS(y)∂vWe introduce next this.

TRANSPORT AND LAPLACE EQUATIONS 83Definition: Green’s function for the region U isG(x, y) = Φ(y – x) − φ x (y)for x, y ∈ U, x ≠ y . Adopting this terminology and adding (2) to (4), we findu(x) = − ∂Uwhereu(y)∂G(x, y) dS(y) −∂v G(x, y) ∆u(y) dy (x ∈ U), (5)U∂G(x, y) = D y G(x, y) . v(y) (6)∂vis the outer normal derivative of g with respect to the variable y. Observe thatthe term ∂u / ∂v does not appear in equation (5). We introduces the correctorφ x precisely to achieve this.Suppose now u ∈ C 2 (U ) solves the boundary – value problem− ∆u= f u = gin Uon ∂U,(7)for given continuous functions f, g. Plugging into (5), we obtain the followingtheorem.Theorem: (Representation formula using Green’s function).Statement : If u ∈ C 2 (U ) solves problem, thenu(x) = − ∂Ug(y)∂G(x, y) dS(y)∂v+ f(y) G(x, y) dy (x ∈ U). (8)UHere we have formula for the solution of the boundary – value problem (7),provided we can construct Green’s function G for the given domain U. This isin general a difficult matter, and can be done only when U has simplegeometry. Subsequent subsections identify some special cases for which anexplicit calculation of G is possible.Remark: Fix x ∈ U. Then regarding G as a function of y, we maysymbolically write− ∆G= δxin U G = 0 on ∂U,δ x denoting the Dirac measure giving unit mass to the point x.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS84Before moving on to specific examples, let us record the general assertion thatG is symmetric in the variables x and y .Theorem: (Symmetry of Green’s function)Statement : For all x, y ∈ U, x ≠ y, we haveG(y, x) = G(x, y). (9)Proof: Fix x, y ∈ U, x ≠ y. Writefor z ∈ U . Thenandv(z) = G(x, z),w(z) = G(y, z), (10)∆v(z) = 0 (z ≠ x), (11)∆w(z) = 0 (z ≠ y) (12)w = v = 0 (13)on ∂U. Thus our applying Green’s identity onV = U – [B(x, ∈) ∪ B(y, ∈)] (14)for sufficiently small ∈ > 0 yields∂B( x, ∈)∂u∂ww − v dS(z) =∂v∂v∂B(y,∈)∂w∂uv − w dS(z), (15)∂v∂vv denoting the inward pointing unit vector field on ∂B(x, ∈) ∪ ∂B(y, ∈). Noww is smooth near x . Soas ∈ → 0.| ∂B(x,∈)∂wn−1supv dS | ≤C∈ | v |∂v∂B(x,∈)= o(1) (16)On the other hand,v(z) = Φ(z – x) - φ x (z),where φ x is smooth in U. Thus

TRANSPORT AND LAPLACE EQUATIONS 85lim∂vw dS=∂νlim∈→ 0∂B(x,∈)∈→ 0∂B(x,∈)= w(x) .∂Φ(x – z) w(z) dSThus the left – hand side of (15) converges to w(x) as ∈ → 0. Likewise theright hand side converges to v(y). ConsequentlyThis completes the proof.G(y, x) = w(x) = v(y) = G(x, y) .The Books Recommended for Chapter II1. L.C. Evans Partial Differential Equations, Graduate Studiesin Mathematics, Volume 19, AMS, 1998.∂ν

86PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSChapter-3Heat and Wave Equations3.1 INTRODUCTIONThe heat equation isu t −∆u = 0and the non-homogeneous heat equation isu t − ∆u = f,where t > 0 and x∈U, U⊂R n is open. The unknown functionisu = u(x, t)u : U [0, ∞) →R.(∗)(∗∗)(∗∗∗)The Laplician ∆ is taken with respect to the spatial variables x = (x 1 , x 2 ,…, x n ),and∆u = ∆ x un= i=1In equation (∗∗), the functionis given.u x i x i. (∗∗∗∗)f : U×[0, ∞)→RRemark (1). The heat equation is also known as the diffusion equation.Remark (2). In typical applications, the heat equation describes the evolutionin time of the density u of some quantity such as heat, chemical concentration,etc.Remark (3). The heat equation appears as well in the study of BROWNIANMOTION.3.2. FUNDAMENTAL SOLUTION OF HEAT EQUATIONArticle :- Derivation of the fundamental solution of the heat equationu t − ∆u = 0, in U ×[0, ∞)

HEAT AND WAVE EQUATIONS 87where U⊂R n is open.Solution. We observe that the heat equationu t −∆u = 0 in U ×[0, ∞)…(1)involves one derivative w.r.t. the time variable t, but two derivatives w.r.t. thespace variables x 1 , x 2 ,…, x n .Consequently, we see that if u = u(x, t) solves (1), then so does u(λx, λ 2 t) forλ∈R. This scaling indicates that the ratior 2,t2r = | x | = x 21 + ... + x…(2)nis important for the heat equation. It also suggests that we seek a solution ofheat equation (1) of the form2 2r | x | u = u(x, t) = v = v, …(3) t t for t > 0 and x∈R n , for some function v as yet undetermined.However, it is quicker to try a solution u having the special structureu(x, t) =1 x vt t for x∈R n , t > 0. Here, α and β are constants and the functionmust be found.v : R n →R…(4)…(5)Inserting (4) into heat equation (1), and thereafter comprising, we obtainαt −(α+1) v(y) + βt −(α+1) y. D v(y) + t −(α+2β) ∆v(y) = 0,…(6)wherey = t −β xx =tCanceling t −(α+1) from equation (6), we findα v(y) + β y. D(y) + t −(2β−1) ∆ v(y) = 0.…(7)…(8)In order to transform (8) into an expression involving the variable y alone, wetakeβ = 21 .…(9)

88PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSThen, equation (8) reduces to1αv + y. Dy + ∆v = 0.2We simplify further by guessing v to be radial, i.e.,for somev(y) = w(|y|),w : R→R.Then (left as an exercise)n −1∆v = w′,r…(10)…(11)…(12)…(13)and equation (10) now becomes1 n −1αw + r w' + w' ' + w' = 0,…(14)2rdfor r = |y| and ′ = . drNow, if we setα = n/2,then equation (14) simplifies to read…(15)On integration, one obtains(r n−1 w′)′ + 21 (r n w)′ = 0. …(16)r n−1 w′ + 21 r n w = constant = a.…(17)We assume that w and w′ tend to zero as r→∞. Under these conditions, wefinda = 0.Hence, equations (17) and (18) implyIntegrating (19), we obtain…(18)w′ = − 21 r w. …(19)2r4w = b e − , …(20)

HEAT AND WAVE EQUATIONS 89for some constant b. Combining equations (4), (9), (11), (15) and (20), weconcludeu(x, t) =tbn / 2solves the heat equation (1).− | x |exp 4t2 , …(21)We define Φ(x, t) =|x|2 1 −e4tn;(x ∈ R , t > 0)n / 2(4t)…(22)n0;(x ∈ R , t < 0)The function Φ(x, t) called the fundamental solution of the heat equation (1).Remarks. φ is singular at the point (0, 0).(2) We will sometimes writeΦ(x, t) = Φ(|x|, t)to emphasize that the fundamental solution is radial in the variable x.…(23)Theorem. (Integral of fundamental solution of heat equation)Statement. For each time t > 0,Proof. We findnRnRΦ(x, t) dx = 1.Φ(x, t) dx ==1n / 2(4 t)1n / 2nRnRexp[−|x| 2 /4t]dtexp[−|z| 2 ]dzThis proves the theorem.== 1.1n / 2n∏i=1∞−∞exp[−|z i | 2 ] dz iArticle. Solve the initial value (or Cauchy) problemu t −∆u = 0 in R n ×(0, ∞)u =g on R n ×{t = 0}associated with homogeneous heat equation.…(1)…(2)

90PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSSolution. Let Φ(x, t) = 1( 4πt)n / 2e2| x|− 4t(x∈R n , t>0)…(3)be the fundamental solution of the heat equation (1).We note that the function(x, t) → Φ(x, t)…(4)solves the heat equation away from the singularity at (0, 0), and thus so does(x, t) → Φ(x−y, t) for each fixed y∈R n .Consequently, consider the convolutionu(x, t) = nRΦ(x−y, t) g(y)dy…(5)2| x−y|1−= e4tn / 2 ( 4πt)nRg(y)dy…(6)for x∈R n , t>0.(A)First, we shall show that u∈C ∞ (R n ×(0, ∞)). |x|21 −Since the function e4t is infinitely differentiable, withn / 2t uniformly bounded derivatives ofsee thatu∈⊂ ∞ (R n ×(0, ∞)).all order, on R n ×[δ, ∞) for each δ>0, we…(7)(B)Furthermore, from equation (6), we writeu t (x, t) −∆u(x, t) = nR= 0 ,[(Φ t −∆ x Φ)(x−y, t)]g(y) dyfor all x∈R n and t>0, since the fundamental solution Φ(x, t) itself solves theheat equation. Thus,in R n ×(0, ∞).(C)u t (x, t) −∆u(x, t) = 0…(8)Let x 0 ∈R n be a fixed point. Let ∈>0 be given. Choose δ>0 such that|g(y)−g(x 0 )|

HEAT AND WAVE EQUATIONS 91wheneverWe know that|y−x 0 |0 .Then, ifΦ(x, t) dx = 1|x−x 0 |< δ / 2 ,we have, using equations (6) and (11),|u(x, t) −g(x 0 )| = | nR≤Φ(x−y,t) {g(y)−g(x 0 )}dy| Φ (x−y, t) |g(y)−g(x 0 ) |dyB(x0,)…(10)…(11)…(12)+ Rn− B(x0,)Φ(x−y t) |g(y)−g(x 0 )|dy= I + J, say …(13)whereI = B(x0 ,)Φ(x−y, t) |g(y)−g(x 0 )| dy, …(14)J = R n− B(x0 ,)Φ(x−y, t) |g(y)−g(x 0 )|dy …(15)Now, owing to inequality (9) and relation (11), we findI ≤ ∈ B (x0,)(x−y, t) dy = ∈This impliesI ≤ ∈.Furthermore, if|x−x 0 | ≤ δ/2 and |y−x 0 | ≥ δ,then|y−x 0 | ≤ |y−x| + |x−x 0 |≤ |y−x| + δ/2…(16)…(17)≤ |y−x| + 21 |y−x 0 |or |y−x| ≥ 21 |y−x 0 |. …(18)

92PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSConsequently,J ≤ 2 || g || ∞ Rn−B(x0,)Φ(x−y, t)dyL≤≤ttC 2nB(x0n / 2 R− ,) 1exp− | x − y | dy4t C 2n / 2 1 0RnB(x0,)exp− | y − x | dy16t −, using (18)C ∞ 1 2 n−1=exp r (r dr)n / 2 t − 16t → 0as t→0 + .Hence, if |x−x 0 | < δ/2 and t > 0 is small enough, then|u(x, t) −g(x 0 )| < 2∈,Using equations (13), (16) and (19). The relation (20) implieslim0( x,y)→(x , 0)n +x∈R, t→0u(x,t)= g(x0)…(19)…(20)…(21)for each point x 0 ∈R n .Thus, we have shown that u(x, t), given by (6), is the solution of the initialvalueproblem constiting of equations (1) & (2). This complete the proof.3.2 MEAN-VALUE FORMULA FOR THE HEATEQUATIONLet U⊂R n be open and bounded. We fix a time T > 0.Definition. The parabolic cylinder is defined asU T = U×(0, T],and the parabolic boundary of U T is denoted by Γ T and is defined asΓ T = ( U T ) −(U T).Interpretation. We interpret U T as being the parabolic interior of⎺U×[0, T].We must note that U T includes to top U×{t = T}. The parabolic boundary Γ Tcomprises the bottom and vertical sides ofbut not the top.U×[0, T],

HEAT AND WAVE EQUATIONS 93Definition (Heat ball)For fixed x∈R n , t∈R and r > 0, we defineE(x, t; r) =n ∈ + 11( y,s) R s ≤ t andΦ(x− y, t − s) ≥n .r Note. E(x, t; r) is a region in space-time. Its boundary is a level set offundamental solutions Φ(x−y, t−s) for the heat equation.The point (x, t) is at the center of the top. E(x, t; r) is called a heat ball.Theorem. (A mean-value property for the heat equation)Statement. Let u∈C 1 2 (U T ) solve the heat equationProve thatu t − ∆u = 0 in R n ×(0, ∞).u(x, t) =14rfor each heat ball E(x, t; r) ⊂ U T .2…(1)| x − y | u(y,s)dy ds …(2)2( t s)nE( x,t,r)−Proof. Formula (2) is a mean-value formula for heat equation. We find that theright hand side of (2) involves only u(y, s) for times s ≤ t. This is reasonable,as the value u(x, t) should not depend upon future times.We may assume upon translating the space and time coordinates thatWe writeand setx = 0, t = 0.E(r) = E(0, 0; r).φ(r) =r1n2…(3)…(4)| y | u(y,s)dy ds…(5)2sE(r)22 | y |= u(ry,r s)dy ds…(5A)2sE(1)We calculateφ′(r) =E(1)nyiui=1 yi | y |2 s2+2r us | y | s2dyds=1n+1rE(r)yiuyi | y |2 s2 | y | 2 u + s s2dyds= A + B, say …(6)We introduce the useful function

94PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSThensince,ψ = −2n| y |log( − 4s) + + n log r…(7)24sψ = 0, on ∂E(r),Φ(y, −s) = r −n on ∂E(r),…(8)…(9)be definition of heat ball.Now, we utilize (7) to write1B = { 4.un+1r E(r)n y si=1iy idyds1 n= − 4n u 4 u y dy dsn 1 sψ + ( sy )i i ψ, …(10)+r i=1 E(r)there is no boundary term, since ψ = 0 on the boundary ∂E(N), by virtue of (8).Integrating by part w.r.t. ‘s’, we discoverB =r1n+1E(r)−4n u1 = −4n un+1r E(r) sn + 4usi=1n + 4ui=1= −4n u sψ − r1n+1E(r)2nsnyii=1yiyis dy ds − n | y |y i−2 2s 4suyi2yidy ds − A . dy dsThis implies1 A + B = −4n ∆u−n+1r E(r) Since u solves the heat equation. Son 1 φ′(r) = −4n u+y n 1yi=1riby virtue of (7). Equation (11) givesor φ(r) =φ(r) = constant.lim φ(t)t→02nsi−n ui=12nsyiuyiyi dy ds,y i dyds= 0,…(11)

HEAT AND WAVE EQUATIONS 95since, 1= u(0, 0) limt→0n t| y |E(t)2s2dyds= 4 u(0, 0), …(12)21 | y || y |dy dsdyds 4.n E(r)=2 E(1)=…(13)2t ssFrom equation (4) and (12), we write2u(x, t) = 41 φ(r)…(14)From equation (5) and (14), we have1u(x, t) =4rn| x − y |u(y,s)(t − s)E(x,t;r)22dydsThis completes the proof of mean-value formula for the heat equation.3.4 ENERGY METHODS FOR HEAT EQUATIONSTheorem. (Uniqueness theorem for heat equation)…(15)Statement. Prove that there exists atmost one solution u∈C 1 2 (⎺U T ) of theproblemu t −∆u = f in U T ,u = g on Γ T ,where U⊂ R n is open and bounded, and ∂U is C 1 . The terminal time T >0 isgiven.Proof. Let u and u be two solutions of the above problem.ThenLetu t −∆u = f in U T ,u = g on Γ T ,u t − ∆ u = f in U T ,w = u− uu = g on Γ T .Then equations (1) to (5) yield…(1)…(2)…(3)…(4)…(5)

96PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSAlsow t − ∆w = (u t − u ) − (∆u −∆ u )w = u − u= g − gt= (u t −∆u) − ( u t −∆ u )= f −f= 0 in U T …(6)= 0 on Γ T . …(7)Set e(t) = Uw 2 (x, t)dx, 0 ≤ t ≤ T. …(8)Thendedt≡ e(t) & = 2 Uw w t dx= 2 Uw ∆w dx, using (6)= −2 U|Dw| 2 dx≤ 0.So e(t) is a decreasing function, and soe(t) ≤ e(0) = 0, for 0 ≤ t ≤ T. Uw 2 (x, t)dx = 0 for all 0 ≤ t ≤ T ,…(9)…(10)w ≡ 0 in U T ,u = u in U T .Hence, the solution is unique. This completes the proof.3.5 PROPERTIES OF SOLUTIONSFirst we employ the mean-value property to give a quick proof of the strongmaximum principle.Theorem. (Strong maximum principle for the heat equation).2Statement : Assume u∈ C1 (U T ) ∩ C(UT) solves the heat equation in U T .(i) Then max u = max u.UTΓT

HEAT AND WAVE EQUATIONS 97R n t(x 0 , t 0 )Strong maximum principle for the heat equation(ii) Furthermore, if U is connected and there exists a point (x 0 , t 0 ) ∈ U T suchthatu(x 0 , t 0 ) = max u,U Tthenu is constant inU t 0.Assertion (i) is the maximum principle for the heat equation and (ii) is thestrong maximum principle. Similar assertions are valid with “min” replacing“max”.Remark. So if u attains its maximum (or minimum) at an interior point, then uis constant at all earlier times. This accords with our strong intuitiveinterpretation of the variable t as denoting time : the solution will be constanton the time interval [0, t 0 ] provided the initial and boundary conditions areconstant. However, the solution may change at times t > t 0 , provided theboundary conditions alter after t 0 . The solution will however not respond tochanges in boundary conditions until these changes happen.Take note that whereas all this is obvious on intuitive, physical grounds, suchinsights do not constitute a proof. The task is to deduce such behaviour fromthe PDE.Proof. 1. Suppose there exists a point (x 0 , t 0 )∈ U T withu(x 0 , t 0 ) = M =Then for all sufficiently small r > 0,E(x 0 , t 0 ; r) ⊂ U T ,max u.UTand we employ the mean-value property to deduce

98PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSM = u(x 0 , t 0 )since1| x 0 − y |= u(y,s) dydsn E(x 24r 0 ,t 0 ;r)(t − s)≤ M,021| x 0 − y |1 = dyds.n E(x 24r 0 ,t 0 ;r)(t − s)0Equality holds only if u is identically equal to M within E(x 0 , t 0 ; r).Consequently2u(y, s) = Mfor all (y, s) ∈ E(x 0 , t 0 ; r).Draw any line segment L in U T connecting (x 0 , t 0 ) with some otherpoint (y 0 , s 0 ) ∈ U T , with s 0 < t 0 . Considerr 0 = min {s ≥ s 0 | u(x, t) = M for all points (x, t) ∈ L, s ≤ t ≤ t 0 }.Since u is continuous, the minimum is attained. Assume r 0 > s 0 . Thenu(z 0 , r 0 ) = Mfor some point (z 0 , r 0 ) on L ∩ U T and sou ≡ M on E(z 0 , r 0 ; r) for all sufficiently small r > 0.Since E(z 0 , r 0 ; r) contains L ∩ {r 0 − σ ≤ t ≤ r 0 } for some small σ > 0, we have acontradiction. Thusand hencer 0 = s 0 ,u ≡ M on L .2. Now fix any point x∈U and any time 0 ≤ t < t 0 . There exists points {x 0 ,x 1 ,…,x m = x} such that the line segments in R n connecting x i−1 to x i lie in U fori = 1,…,m. (This follows since the set of points in U which can be soconnected to x 0 by a polygonal path is nonempty, open and relatively closed inU.) Select times t 0 > t 1 >…> t m = t. Then the line segments in R n+1 connecting(x i−1 , t i−1 ) to (x i , t i ) (i = 1,…, m) lie in U T . According to Step 1,u ≡ M

HEAT AND WAVE EQUATIONS 99on each such segment and sou(x, t) = M.This completes the proof.Remark. The strong maximum principle implies that if U is connected and2u∈ C1 (U T ) ∩ C(UT) satisfiesut− ∆u= 0u = 0u = ginUTon ∂U× [0,T]on U × {t = 0}where g ≥ 0, then u is positive everywhere within U T if g is positivesomewhere on U. This is another illustration of infinite propagation speed fordisturbances.An important application of the maximum principle is the followinguniqueness assertion.Theorem. (Uniqueness on bounded domains).Statement. Let g∈C(Γ T ), f ∈ C(U T ). Then there exists at most one solution u2∈ C1 (U T ) ∩ C(UT) of the initial/boundary-value problemut− ∆u= fu = ginonProof. If u and u ~ are two solutions of (1), apply previous theorem toto get the result.w = + (u− u ~ )UΓTT .…(1)3.5 WAVE EQUATIONThe wave equation isu tt −∆u = 0and the non-homogeneous wave equation isu tt − ∆u = f.Here t > 0 and x∈U, U⊂ R n is open. The unknown function isu = U ×[0, ∞)→R,u = u(x, t),and the Laplacian ∆ is taken w.r.t. the spatial variables.…(∗)…(∗∗)…(∗∗∗)

100PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSIn equation (∗∗)is given.x = (x 1 , x 2 ,…, x n )f : U×[0, ∞)→RGenerally, we use the abbreviationu = u tt − ∆u.Remark. The wave equation is a simplified model for a vibratingstring (n = 1),membrance (n = 2),elastic solid (n = 3).…(∗∗∗∗)…(∗∗∗∗∗)In each of the above, u(x, t) represents the displacement in some direction ofthe point x at time t ≥ 0.Solutions by Special MeansArticle d′. Alembert’s formula (for n = 1)We consider the initial-value problem for the one-dimensional waveequation in all of R:u tt − u xx = 0 in R×(0, ∞)…(1)u = g ,u t = h on R × {t = 0}, …(2)where g, h are given functions.We desire to derive a formula for u = u(x, t) in terms of known functions g andh. The two initial conditions in (2) imply that the displacement u(x, 0) and thevelocity u t (x, 0) are known.The PDE (1) can be factored to write ∂ ∂ ∂ ∂ + − u = 0. ∂t∂x ∂t∂xSet ∂v(x, t) = − ∂tThen equations (4) says∂∂ u(x, t).x u t (x, t) + v x (x, t) = 0 ; x∈R, t > 0.…(4)…(5)…(6)Equation (6) is a homogeneous transport equation with constant coefficients(b = 1). Let

HEAT AND WAVE EQUATIONS 101v(x, 0) = a(x).…(7)We know that the fundamental solution of the initial-value problem consistingof transport equation (6) and condition (7) isv(x, t) = a(x−t), x∈R, t ≥ 0.Combining equations (5) and (8), we obtainu t (x, t) − u x (x, t) = a(x−t) in R × (0, ∞)…(8)…(9)Also u(x, 0) = g(x) in R, …(10)By virtue of initial condition (2). Equations (9) and (10) constitute the nonhomogeneoustransport problem. Hence, its solution isu(x, t) = g(x + t) + t 0a (x + (s−t) (−1)−s) ds1= g (x + t) + 2x+tx−tThe second initial condition in (2) implya(x) = v(x, 0)= u t (x, 0) − u x (x, 0)a(y)dy.…(11)= h(x) − g′(x), x∈R. …(12)Substituting (12) into equation (11), we obtainfor x∈R, t ≥ 0.u(x, t) = g(x + t) +12x+tx−t1 1= [g(x+t) + g(x−t)] + 2 2This is the d’ Alembert’s formula.(sufficiently smooth) solution of (1).Application of D’ Alembert’s Formula[h(y) − g′(y)]dyx+tx−th(y)dy,…(13)We have derived (13) assuming u is aInitial/boundary-value problem on the half-line. R + = {x > 0}Example. Consider the problemu tt − u xx = 0 in R + ×(0, ∞)where g, h are given, withu = g, u t = h on R + ×{t = 0}u = 0 on {x = 0} × (0, ∞),…(1)

102PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSg(0) = 0, h(0) = 0.…(2)Solution. We convert the given problems on the half-line into the problem onwhole of R. We do so by extending the functions u, g, h to all of R by oddreflection method as below we set~ u(x,t)for x ≥ 0,t ≥ 0u ( x,t)= …(3)−u(−x,t)for x ≤ 0,t ≥ 0,~g(x)~h(x)g(x)= −g( −x)h(x)= −h( −x)Now, problem (1) becomesfor x ≥ 0for x ≤ 0,for x ≥ 0for x ≤ 0.…(4)…(5)u~tt = u~u~ =~g, u~xxtin R × (0, ∞)~= h on R × {t = 0} …(6)Hence, d’ Alembert’s formula for one-dimensional problem (6) implies11 x+t ~u~ (x, t) = [~g(x + t) +~g(x − t)] + x −th(y)dy . …(7)22Recalling the definitions of u~ ,~ g,~ h in equations (3)−(5), we can transformequation (7) to read for x ≥ 0, t ≥ 0.u(x, t)11[ g(x + t)+ g(x − t)]+22= 11[ g(x + t)− g(t − x)]+22x+tx−tx+t−x+th(y)dy;h(y)dy;for x ≥ t ≥ 0for 0 ≤ x ≤ t…(8)Formula (8) is the solution of the given problem on the half-line R + = {x > 0}.Remark. If h ≡ 0 (9) in R + ×{t = 0}, then the solution of the correspondingproblem, as given by (8), is1[g(x + t) + g(x − t)]; for x ≥ t ≥ 0,2u(x, t) = …(10)1[g(x + t) − g(t − x)]; for x ≤ t ≤ t.2

HEAT AND WAVE EQUATIONS 103The formula (10) shows that the initial displacement, u(x, 0) = g(x), splits intotwo parts − one moving to the right with speed one (c = 1) and the other to theleft with speed one.The latter part reflects off the point x = 0, where the vibrating string is heldfixed.Article. Derive Kirchloff’s formula for the solution of three-dimensional(n = 3) initial-value problemu tt − ∆u = 0 in R 3 × (0, ∞)u = g on R 3 ×{t = 0}u t = h on R 3 ×{t = 0}.…(1)…(2)…(3)Solution. Suppose u ∈ C 2 (R 3 × [0, ∞)) solves the above initial-value problem.We know thatU(x; r, t) = u(y, t) dS(y) …(4)∂B( x,r)Defines the average of u(⋅, t) over the sphere ∂B(x, r). Similarly,G(x; r) =H(x; r) = g(y)dS(y) …(5)∂B( x,r) h(y) d S(y). …(6)∂B( x,r)For fixed x, we hereafter regard U as a function of r and t only. Next, setU ~ = r U,…(7)G ~ = r G, H ~ = r H.…(8)we now assert that U ~ solve~ ~Utt − U rr = 0~~U= G~~U= H~U= 0inR+on Ron R++× ( 0,∞)× { t = 0}× { t = 0}on{r = 0}× ( 0,∞)…(9)We note that the transformation in (7) and (8) converts the three-dimensionalwave equation into the one-dimensional wave equation.From equation (7), we find

104PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSU ~ tt = r U tt= r ∆U 2= rU rr + U r ,r Laplacian for n = 3= r U rr + 2U r= (U + r U r ) r= ( U~ r ) r= U ~ rr .…(10)The problem (9) is one the half-line R + = {r ≥ 0}. The d’ Alembert’s formulafor the same, for 0 ≤ r ≤ t, is11 r+tU ~ (x;r, t) = [G ~ (r + t) − G ~ (r − t)] + −r + t22H ~ (y)dy.…(11)From (4), we findu(x, t) =lim U(x; r, t),r→0+…(12)Equations (7), (8), (11) and (12) imply thatu(x, t) ==limr→0+limr→0+U ~ (x;r, t) rG ~ (t+ t+rt−rr) − G ~ (t − r) 1+ 2r 2rH ~ (y) dy = G ~ '(t)+ H ~ (t).…(13)Owing then to (5) and (6), we deduce from (13)ButHence∂u(x, t) = t g(y)dS(y) + { t h(y)d S(y)}∂∂∂t∂B(x,t)∂B( x,t)∂B(0,1) ∂ B(x,t)…(14)g ( y)d S(y)= g(x + tz)d S(z).…(15)g(y)d S(y) =t ∂B(x,t)∂B(0,1){ Dg(x + tz)}.zd S(z)

HEAT AND WAVE EQUATIONS 105Now, equation (14) and (16) concludeu(x, t) =for x∈R 3 , t > 0.∂B( x,t) y − x = { Dg(y)}. d S(y).∂B(x,t) t …(16) [g(y) + {Dg(y)}. (y−x) + t h(y)] d S(y) …(17)The formula (17) is called KIRCHHOFF’s formula for the solution of theinitial-value problem (1)−(3), in 3D.Nonhomogeneous ProblemWe next investigate the initial-value problem for the nonhomogeneous waveequationu tt − ∆u= f u = 0,u t = 0in Ron Rnn× (0, ∞)× {t = 0}.…(1)Motivated by Duhamel’s principle, we define u = u(x, t; s) to be the solution ofu tt ( ⋅;s)− ∆u(⋅;s)= 0 u( ⋅;s)= 0,u t ( ⋅;s)= f ( ⋅;s)in Ron Rnn× (s, ∞)× {t = s}.…(2)Now setu(x, t) : = t 0u(x, t; s)ds (x∈R n , t ≥ 0). …(3)Duhamel’s principle asserts this is solution ofu tt − ∆u= f u = 0,u t = 0in Ron Rnn× (0, ∞)× {t = 0}.…(4)Theorem. (Solution of nonhomogeneous wave equation).Statement. Assume n ≥ 2 and f ∈ C [n/2]+1by (3). Then(i) u ∈ C 2 (R n ×[0, ∞)),(ii) u tt − ∆u = f in R n × (0, ∞),and(R n ×[0, ∞)). Define u

106PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS(iii)lim(x,t) →(x0,0)x∈Rn, t>0u(x, t) = 0,lim(x,t) →(x0,0)x∈Rn, t>0u t (x, t) = 0 for each point x 0 ∈ R n . n 1Proof. 1. If n is odd,n + 1 =+ 2. Also u(⋅,⋅;s) ∈ C 2 (R n ×[s, ∞)) for each2s ≥ 0, and so u∈C 2 (R n n 2× [0, ∞)). If n is even,n + 1 =+ 2. Hence2u ∈ C 2 (R n ×[0, ∞)).2. We then compute :u t (x, t) = u(x, t; t) + t 0u t (x, t; s)ds= t 0u t (x, t; s)ds,u tt (x, t) = u t (x, t; t) + t 0u tt (x, t; s)dsFurthermore= f(x, t) + t 0u tt (x, t; s)ds.∆u(x, t) = t 0∆u(x, t; s)ds= t 0u tt (x, t; s)ds.Thusu tt (x, t) − ∆u(x, t) = f(x, t) (x ∈ R n , t > 0),and clearlyu(x, 0) = u t (x, 0) = 0 for x ∈R n .Examples. (i) Let us work out explicitly how to solve (4) for n = 1. In thiscase d’ Alembert’s formula gives1 x+t−s xu(x, t; s) = f ( y,s)dy,2−t+sThat is,1 tu( x,t)=20u(x, t) =12t0x+t−sx−t+sx+sx−sf ( y,s)dyds(ii) For n = 3, Kirchhoff’s formula impliesf(y, t−s)dy ds (x∈R, t ≥ 0).…(5)

HEAT AND WAVE EQUATIONS 107u(x, t; s) = (t −s) f(y, s) dS;∂B( x,t−s)so thatTherefore1u(x, t) =4solves (4) for n = 3.t u(x, t) = ( t − s) f ( y,s)dS ds0 ∂B(x,t−s)1 t f (y,s)= 0∂B(x,t−s)dSds4(t − s)1 t f (y, t − r)= 0∂B(x,r)dSdr.4rB(x,t)f (y, t−| y − x |)dy (x ∈ R 3 , t ≥ 0) …(6)| y − x |The integrand on the right is called a retarded potential.3.6 ENERGY METHODSThere is the necessity of making more and more smoothness assumptions uponthe data g and h to ensure the existence of a C 2 solution of the wave equationfor larger and larger n. This suggests that perhaps some other way ofmeasuring the size and smoothness of functions may be more appropriate.Indeed we will see in this section that the wave equation is nicely behaved (forall n) with respect to certain integral “energy” norms.UniquenessLet U ⊂ R n be a bounded, open set with a smooth boundary ∂U, and as usualset U T = U × (0, T], Γ T = U −U T , where T > 0.TWe are interested in the initial/boundary-value problemutt− ∆u= fu = gut= hin Uon ΓTTon U × {t = 0}.…(1)Theorem. (Uniqueness for wave equation).Statement. There exists at most one function u∈C 2 ( UT ) solving (1).Proof. If u ~ is another such solution, then w : = u − u ~ solves

108PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSwDefine the “energy”tt− ∆w= 0w = 0wt= 0in Uon ΓTTon U × {t = 0}.e(t) =1 2U t2w (x, t) + |Dw(x, t)| 2 dx (0 ≤ t ≤ T).We computee&(t) = Uw t w tt + Dw ⋅ Dw t dx⋅ == Uw t (w tt − ∆w)dx = 0.ddtThere is no boundary term since w = 0, and hence w t = 0, on ∂U × [0, T].Thus for all 0 ≤ t ≤ T, e(t) = e(0) = 0, and so w t , Dw ≡ 0 within U T . Sincew ≡ 0 on U × {t = 0}, we conclude w = u − u ~ ≡ 0 in U T .(x 0 , t 0 )B(x 0 , t 0 −t)Domain of dependenceCone of dependenceAs another illustration of energy methods, let us examine again the domain ofdependence of solutions to the wave equation in all of space. For this, supposeu ∈ C 2 solvesu tt − ∆u = 0 in R n ×(0, ∞).Fix x 0 ∈ R n , t 0 > 0 and consider the coneC = {(x, t) |0 ≤ t ≤ t 0 , |x−x 0 | ≤ t 0 − t}.The Books Recommended for Chapter III1. L.C. Evans Partial Differential Equations, Graduate Studiesin Mathematics, Volume 19, AMS, 1998.

ANALYTICAL MECHANICS – I 109Chapter-4Analytical Mechanics – I4.1 INTRODUCTIONIn the literature on mechanics there is no single generally acceptedinterpretation of term “analytical mechanics”. Some writers identifyanalytical mechanics with theoretical mechanics. Some authors maintain thatan exposition in generalized coordinates constitutes the distinguishingfeature of analytical mechanics. According to Gantmacher, analyticalmechanics is characterized both by a specific system of presentation andalso by a definite range of problems investigated. In analytical mechanics,general principles (differential or integral) serve as the foundation and then thebasic differential equations of motion are derived from these principlesanalytically.4.2. FREE AND CONSTRAINED SYSTEMThe motion is studied of a system of particles P k , (k = 1, 2,…,N), relative tosome inertial (Galilean) system of coordinates. There are some restrictions onthe positions and velocities of the particles of the system. These restrictionsmay be of a geometrical or kinematical nature. Such restrictions are calledconstraints. Systems with such constrains are termed as constrained systems.If there are no constraint in the system, then the system is called a free system.ρLet t denotes time, r , (k = 1, 2,…, N), be the radius vectors taken from aksingle pole (that is stationary in the given system of coordinates) for all systemof particles P k andr&ρ ρk = v k (k = 1, 2,…,N) (1)denote velocities of all points P k of the system. Here dot (⋅) representsdifferentiations with respect to time t . Analytically, a constraint is expressedby the equationf(t,ρ r , ρ r & ) = 0. (2)kkIn the general case, constraint (2) is called differential or kinematical . In(2), f(t, r , r &ρ k k ) is an abridged notation for the functionρ ρ ρf(t, r , r r , r&ρ, r&ρ,..., r&ρ). Such abbreviated notation will be used throughout12,...,N12the chapters on analytical mechanics.N

110 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSρIf the velocities r & k do not enter into the constraint equation (2), the constraintis termed finite or geometric. Analytically, it is written as,ρf(t, r k ) = 0 (3)Given a finite constraint of type (3), a system cannot occupy an arbitraryposition in space at every given instant of time. Finite constrains imposerestrictions to possible positions of the system at time t. But with a differentialconstraint alone, the system may occupy any arbitrary position in space at anytime t. However, in this position the velocities of the particles of the systemcannot any longer be arbitrary, since the differential constraint imposesrestrictions on these velocities. From now on, we shall confine ourconsideration solely to such differential constraints whose equations containthe velocities of the particles in linear form :-orρ ρ ρ ρl . r&ρ+ l . r&ρ+ l . r&ρ+ ... + l .ρ r& D = 01 1 2 2 3 3 N N +Nk=1ρl k . r &ρ k + D = 0 (4)ρρwhere l k . r&ρ k is the scalar product of the vectors lkand r &ρk and the vectors l ρ k andρthe scalar D are specified functions of time t and of all r (µ = 1, 2,…, N). It isassumed here that the vectors l ρ k cannot all vanish at the some time.Each finite constraint of type (3) implies, as a consequence, adifferential constraint whose equation is obtained by termwise differentiationof equation (3) :whereandN ∂f∂f .rk+ = 0k 1 r ρ&ρ, (5)= ∂ k ∂tρrk = x kî+ y k ĵ +z kˆ , (6)kî , ĵ, kˆ are mutually orthogonal unit vectors of the co-ordinate axes. Thenor∂ f ∂f∂f∂fρ = î + ĵ + kˆ∂r∂x∂y∂z, (k = 1, 2,…, N) (7)kr kkkk∂ fρ = ∂ gradk f , (k = 1, 2…N) (8)But differential constraint (5) is not equivalent to the finite constraint (3). It isequivalent to the finite constraintϖf(t, r k ) = c, (9)

ANALYTICAL MECHANICS – I 111where c is an arbitrary constant. For this reason, the finite constraint (9) iscalled integrable.In rectangular Cartesian co-ordinates, we writeρrk= x ˆ i + y ˆj+ z kˆ, (10)kkkandl ρ = A ˆ i + B ˆj+ C kˆ, (11)kr&ρkkkk= x&î + y&ĵ + z&kˆ , (12)kkkwhere A k , B k and C k , (k = 1, …N), are scalar functions of t, x 1 , y 1 , z 1 ,…, x N ,y N , z N .Then the above constraint equations are now written as :f(t, x k , y k , z k , x & , y&, z&) = 0 (13)kkkf(t, x k , y k , z k ) = 0 (14)N ( A x& + B y&+ C z&+ D = 0 (15)k=1k k k k k k )N ∂f∂f∂f ∂f x& k + y&k + z&k + = 0 (16)k=1∂xk ∂yk ∂zk ∂t4.3 CLASSIFICATION OF CONSTRAINTSIf t is not expressed explicitly in the constraint equation, i.e.,∂f = 0,∂t(17)then the constraint is termed as stationary.Note (1) : If the differential constraint (5) is stationary, then differentialequation (5) is linear and homogeneous in the velocities.Note (2) : By analogy, the differential constraint (4) or (15) is termedstationary ifD = 0 (18)and vectors k lρ in equation (3) and, respectively, the coefficients A k , B k , andC k in equation (15) are not explicit function of time t.

112 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSIllustration 1. A particle is constrained to move over a surface. Let theequation of this surface be given in the formorf ( ρ r)= 0 (19)f(x, y, z) = 0 (20)This is a finite stationary constraint.If the surface is moving or undergoing deformation, then the time t enters intothe equation of the surface explicitly and its equation is of the formorf(t, r ρ ) = 0 (21)f(t, x, y, z) = 0 (22)In this case, the constraint is finite but non-stationary.System of particlesDefinition 1. A system of particles is called holonomic if the particles of thesystem are not subjected to differential nonintegrable constraints.Thus, a holomic system is any free system of particles and also any constrainedsystem with finite or differential but integrable constraints. All constraints in aholonomic system may be written in closed form.Definition 2. A system of particles is called nonholonomic if there aredifferentiable integrable constraints.Nonintegrable differential constraints are themselves frequently callednonholonomic. Sometimes, integrable differential constraints are termedseminomic if only stationary constraints are imposed. Otherwise, it is calledsclernomic.Illustration 2 : Two particles are connected by a rod of constraint length l.Then the constraint equation is of the formor2(ρ ρr − r − l 2 = 0 (23)1 2 )(x 1 −x 2 ) 2 + (y 1 −y 2 ) 2 + (z 1 −z 1 ) 2 − l 2 = 0 (24)ρ ρHere r1 and r2are the position vectors of the end points of the given rod.This is a holonomic scleronomic system.Note : We note that a rigid body may be regarded as a system of particlesequidistant from one another, that is, subjected to constraints of type (23).

ANALYTICAL MECHANICS – I 113Thus a free rigid body is a special case of a constrained holonomicscleronomic system of particles.Illustration 3. Two particles are connected by a rod of variable length l = f(t).The constraint equation for this isor2(ρ ρr − r − f 2 (t) = 0 , (25)1 2 )(x 1 −x 2 ) 2 + (y 1 −y 2 ) 2 + (z 1 −z 1 ) 2 − f 2 (t) = 0 . (26)This system is a holonomic sheonomic system.Illustration 4 . Two particles in a plane are connected by a rod of constantlength l and are constrained to move in such a manner that the velocity of themiddle of the rod is in the direction of the rod. The constraint equations forthis arez 1 = 0, z 2 = 0, (27)(x 1 −x 2 ) 2 + (y 1 −y 2 ) 2 − l 2 = 0, (28)x&1 + x&2 y&1 + y2=&x − x y − y1212, (29) x&1 + x&2 y&1 + y&2 since, the velocity of the centre of the rod is , and direction 2 2 x1+ x 2 y1+ y 2ratios of the rod are < , > . This system is a nonholonomic2 2system because equation (29) defines a nonintegrable differential constraint.Unilateral Constraints : The constraints discussed earlier are called bilateralconstraints.The constraints of the formf(t,ρ r , ρ r & ) ≥ 0 (30)kkare called unilateral constraints. If in condition (30), we have an equal sign,it is said that the constraint is taut.Illustration 5. Consider two particles connected by a thread of length l. Thenthe constraint equation is expressed by the inequalityl 2 ρ ρ 2− (r − r ≥ 0, (31)1 2 )and is unilateral constraint.

114 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSRemark. The motion of a system of particles on which a unilateral constraintis imposed may be divided into portions so that in certain portions theconstraint is taut and the motion occurs as if the constraint were bilateral, andin other portions the constraint is not taut and the motion occurs as if therewere no such constraint.In other words, in certain portions a unilateral constraint is either replaced by abilateral constraint or is eliminated altogether.So we shall hence forth consider only bilateral constraints.4.4. POSSIBLE AND VIRTUAL DISPLACEMENTOn a material system, let us impose the following d finite constraints.ρ ρ ρf 1 (t, r1 , r2,......., rN) = 0ρ ρ ρf 2 (t, r1 , r2,........., rN) = 0ρ ρ ρf 3 (t, r , r ,.........., r ) 01 2N =orΜρ ρ ρf d (t, r , r ............, r ) 01 2N =(1)ρf α (t, r k ) = 0, (α = 1, 2,…d) (2)and following g differential constraints :orN k=1ρ l vρ βk&k + D β = 0 , (3)ρ ρ ρ ρ ρ ρl11 ⋅ v1+ l12⋅ v 2 + ... + l1N⋅ v N + D1= 0........................................................... . (4)............................................................ρ ρ ρ ρ ρ ρ l ⋅ v + l ⋅ v + ... + l ⋅ v + D = 0g11g22gNWe replace the finite constraints (2) by the differential constraints bydifferentiating them. We have N ∂fρ ∂f ρ .Vk+ = 0k 1 r , (α = 1, 2,…d) (5) = ∂ k ∂tNg

ANALYTICAL MECHANICS – I 115Definition (Possible velocities). The system of vectors v ρk are called possiblevelocities for a certain instant of time t and for a certain possible (at thatinstant) position of the system if these vectors v ρ k satisfy the above “d + g”linear equations in (3) and (5).Thus, possible velocities are velocities permitted by the constraints.For every possible position of the system at time t there exists an infinity ofsystems of possible velocities. One of these system of velocities is realized inthe actual motion of the system at time t.Definition (Possible displacements). Consider the system of infinitely smalldisplacementsρ ρdr1= V1dtρ ρdr2= V2dt(6)Μρ ρdr = V dtNNorρ ρdr= V dt (k = 1, 2,…N), (7)kkwhere v ϖ k (k = 1, 2,….N) are the possible velocities. The infinitesimaldisplacements d ρ rkare called possible infinitesimal displacements or simplypossible displacements.Remark : Multiplying equation (3) and (5) termwise by dt, and then usingequation (7), we findfor β = 1, 2,…, g , andN ρ l d ρ r = β k&k D β dt = 0 , (8)k 1 N ∂fρ ∂f ϖ .drk+ dt = 0k 1 r, (9) = ∂ k ∂tfor α = 1, 2,…,d. Equation (8) and (9) determine the possible displacements.Virtual DisplacementsLet us take two systems of possible displacements at one and the same instantof time and for one and the same position of the system :andρ ρdr= dv dt ,kk

116 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSϖ ϖd'r = v dt , (k = 1, 2,…N), (10)kk&Here, both possible displacements,Therefore,ϖ ρdr andd' r satisfy the equations (8) and (9).k N ∂fρ ∂f ϖ .drk+ dt = 0k 1 r(11) = ∂ k ∂tkand N ∂fρ ∂f ϖ .d' rk+ dt = 0k 1 r(12) = ∂ k ∂tN k=1ρ l d ρ r βk&.k + D β dt = 0 (13)N k=1ρ l d ρ r βk&. ' k + D β dt = 0 (14)Subtract equation (11) from (12) and (13) from (14) we haveandN ∂fρ ρ ρ .(d' rk− drk) = 0,(15)k=1 ∂ rNkρ ρ ρl ( d'r − dr 0(16)β k k k ) =k=1for β = 1, 2,…, g. We denoteρ ρ ρδ r = d' r − dr ,(17)kkkk = 1, 2,…, N.Then equations (15) and (16) become N ∂fρ ϖ ..r k = 0k 1 r(18) = ∂ k Nk=1ρ ρ .l β k . δr k = 0 , (19)for β = 1, 2,…, g.ϖThe displacements δ r k satisfy the homogeneous relations (18) and (19) arecalled virtual displacements.

ANALYTICAL MECHANICS – I 117ϖAny system of vectors δ r k satisfying equations (18) and (19) is a system ofvirtual displacements.Remark 1 : We can say that virtual displacements are displacements of pointsof a system from one possible position of the system at time t to anotherinfinitely close possible position of the system.Remark 2 : In the case of stationary constraints, virtual displacementscoincide with possible displacements.Illustrations 1 : Consider a particle in motion on a fixed surface.Illustrations 2 : The constraint is a surface S which is itself in motion (as arigid body) with a certain velocity u ρ relative to the original system ofcoordinates.u ρPd ρ rP ρ ρdr = vdtIn this case, any vector v ρ constructed from the point P and tangent to thesurface at P will constitute a possible velocity. The corresponding possibledisplacementρ ρdr = v dtlies in the plane tangent to the given fixed surface.The differenceρ ρ ρδ r = d' r − drof the two tangent vectors is also a vector tangent to the surface at the samepoint P.Thus, any vector constructed from P and lying in the tangent plane may beregarded as a certain d ρ r and as a certain ρ r .Here, the constraint is stationary and the virtual displacements coincide withthe possible displacements.v ρSv ρ 1u ρv ρ

118 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSIn this case, the possible velocity v ρ is obtained from an arbitrary vector v ρ 1 thatis tangent to the surface by adding to it the velocity u ρ . That is,Therefore,ρ ρv = v1ρ+ u.ρ ρdr = v dtρ ρ= dt + u dt.v 1Similarly, for another possible displacementρ ρ'ρd' r = v dt + u dt .So, the virtual displacement isρ ρ ρδ r = d' r − drρ'ρ= v − v ) dt .1( 1 2Degree of FreedomThe vector ρ r k , in Cartesian co-ordinates, is characterized by three projectionson the axes δx k , δy k , δz k (k = 1, 2,…,N) and the equations (18) and (19) whichdefine the virtual displacements may be written in the following form :N ∂fk=1∂xkxk∂f+∂ykyk∂f+∂zkzk = 0 , (20)for α = 1, 2,…,d andN ( A x + B y + C z ) =(21)k=1for β = 1, 2,…,N .kk k k k k 0If the above d + g equations in (20) and (21) are independent, then out of the3N virtual increments δx k , δy k , δz k , there will be (3N − d − g) independentvirtual increment. Letn = 3N − d − g (22)then n is called the number of degrees of freedom of the given system ofparticles.4.5. POSSIBLE ACCELERATIONρLet the corresponding forces F k ,(k = 1,2,.....,N), be impressed at the points P kof the system. Here F ρk is the resultant of all forces applied directly to the

ANALYTICAL MECHANICS – I 119particle P k , (k = 1, 2,…,N). If the constraint were absent, then by Newton’ssecond law we would have the relationsρ ρF = m w , (23)kkkfor k = 1, 2,…,N , between the masses m k , the accelerations w k and the forcesF k .Given constraints, the accelerations1wρ ρ= k F(24)km k(at a given instant of time t, in a given position of the particles of the systemr k , and for given velocities v ρ k ) may prove incompatible with the constraints.Differentiating the equations (3) and (5) termwise w.r.t. time, we getfor α = 1, 2,…, d, and N ∂fN ρ d ∂f d ∂f.w k.v k = 0k 1 r+kk 1 dt r ρ ρ + ρ, (25) = ∂ = ∂ k dt ∂tfor β = 1, 2,…,g.ρ ρNd dw k ρ ρk v k + D = 0kk 1 dt l & . + lβ.β , (26) = dtN k= 1βThe left hand sides in relations (25) and (26) are linearly dependent on theaccelerations w ρ ρ ρk . These left hand sides are also dependent on t, rk, vk(k = 1,2,…N). Equations (25) and (26) are analytic expressions for the restrictionsimposed by the constraints on the accelerations w ρk of the particles of thesystem. The accelerations (24), i.e.,ρ 1 ρw k = Fk,mkmay not satisfy above relations (25) and (26). Then the materially effectedconstraints will act on the particles P k of the system with certain supplementaryforces R ρ k , (k = 1, 2,…,N). These forces are called the reaction forces of theconstraint. The reactions that arise are such that the accelerations determinedfrom the equationsm kρwkρ ρ= F + R , (k = 1, 2,…., N) (27)kk

120 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSare already permitted by the constraints. Unlike the reactions R ρ k (k = 1,2,…N), the pressigned forces F ρ k (k = 1, 2,….,N) are called effective forces.Effective forces are ordinarily specified as known functions of the time,position and velocities of the particles of the system, i.e.,ρ ρ ρ ρF = F t,r , v ) . (k = 1, 2,…N) (28)kk ( k kBasic Problem : The basic problem of the dynamics of a constrained systemconsists in the following :Given effective forces F ρ ρ ρ ρk = Fk( t,rk, vk) and the initial positionsρoρorkand the initial velocities v k of the particles of the system both arecompatible with constraints it is required to determine the motion of the systemand the reactions of the constraints R ρ(k = 1, 2,…N).kIt nothing is known about the nature of the constraints except the definingequations (2) and (3) and, consequently, nothing in known about the reactionR ρ k produced by these constraints, then the above problem is indeterminate,since the number of scalar quantities (x k , y k , z k , R kx , R ky , R kz ) that have to bedetermined is greater than the number of available scalar equations (6 N > 3N+ d+ g)For the basic problem of dynamics to become determinate, it is necessary tohave some kind of additional6N − (3N + d + g)= 3N − d − g= n (29)independent relations between sought-for quantities. These relation can beobtained if we confine ourselves to the following important class of idealconstraints.Ideal Constraints : If the sum of the works of the reactions of constraints onany virtual displacements is equal to zero, constraints are termed ideal. Thatis, for ideal constraints,ρ ρ ρ ρ ρ ρR1 ⋅ r1+ R 2 ⋅ r2+ ... + R N ⋅ rN= 0orN ρ ρ R ⋅ r 0 . (30)k=1k k =In Cartesian co-ordinates equal (30) may be rewritten as :

ANALYTICAL MECHANICS – I 121(R 1x δx 1 + R 1y δy 1 + R 1z δz 1 )+ (R 2x δx 2 + R 2y δy 2 + R 2z δz 2 )or+ …..+ (R Nx δx n + R Ny δy N + R Nz δz N ) = 0,Nk=1(R kx δx k + R ky δy k + R kz δz k ) = 0. (31)Among the 3N quantities δx k , δy k , δz k , there are n independent ones (n= 3N − d − g is the degree of freedom of the system). It is therefore possible in(31) to express 3N−n dependent increments δx k , δy k , δz k in terms of nindependent increments and equate to zero the co-efficients of theseindependent increments. We then obtain the n relations still lacking and needto make determinate the basic problem of the dynamics mentioned above.4.6 LAGRANGE’S EQUATIONS OF THE FIRST KINDWe assumed that all constraints imposed on a system of particle are ideal. Ifm k is the mass of the kth particle, w ρ ρ ρk is its acceleration, and Fkand R k are,respectively, the resultant of the effective forces and the resultant of the forcesof reaction operating on this particle (k = 1, 2,…N), then for particles of aconstrained system, we havemkρwkρ ρ= F + R . (k = 1, 2…N) (1)kkSince constraints are ideal, for any position of the system under any virtualdisplacements, we haveN ρ ρ R k ⋅ rk=k=10 , (2)Eliminating R k from equations (1) and (2), we obtainN ρ ρ ρ Fk − m w ).r = 0 . (3)k=1kThis is known as the general equation of dynamics.kkIt states that, given a system in motion, at any instant of time the sum of theworks of the effective forces and the forces of inertia on any virtualdisplacements is zero.

122 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSThus, general equation of dynamics always holds for any motion that iscompatible with constraints and that corresponds to the specified effectiveforces F ρ k , (k = 1, 2,…,N).Derivation of Lagrange’s Equations of the First KindLet us find the expressions for the reaction forces R ρk by means of theundetermined multipliers of Lagrange. The relations defining the virtualdisplacements of particles of a system arefor α = 1, 2, …d, and N ∂fρ ρ .rk= 0k 1 r , (4) = ∂ k Nk=1for β = 1, 2, …, g.ρ ρ l βk δr k = 0,(5)Multiplying the equations (4) and (5) termwise by arbitrary scalar multipliers(−λ α ) and (−µ β ) and adding termwise the resulting equations to equation (2),we get,Nk=1 ρRk−∂fρ∂rd λ ααα= 1 k−gβ= 1ρ ρµ βl βk. δrk= 0 . (6)In expanded form in Cartesian co-ordinates, we haveN d ∂fgR kx Akxk 1 1 x − − = = ∂ k =1 k∂fNdg+ Rky − − Akykk=1 =1 ∂yk =1 ∂fNdg+ Rkz − − Akz kk=1 =1 ∂zk =1= 0. (7)The undetermined multipliers λ α and µ β may be chosen so that all the scalarcoefficients in (7) and, hence, all the vector co-effecients in (6) vanish. Thisgives

ANALYTICAL MECHANICS – I 123ρRkd= =1∂fρ∂rkg ρ+ l=1k, (8)for k = 1, 2,…N. Expressions (8) is a general expression for the reaction forcesof ideal constraints in terms of the undermined multipliers of Lagrange λ α , µ β(α = 1, 2,…d ; β = 1 2, …g).Putting the expressions for R ρinto equation (1), we getkmkρwkρ= Fk d+ =1∂f∂zk g + =1ρlk, (9)for k = 1, 2,…N. The constraint equations for above equations arefor α = 1, 2,…, d, andNk=1for β = 1, 2,…, g.ρf α ( r k ) = 0 , (10)l r & + D 0 , (11)β k k β =Equations (9) are called Lagrange equations of the first kind.Remark : By replacing each vector equation in (9) by three scalar equations,equations (9) to (11) constitute a set of (3N + d + g) scalar equations in (3N + d+ g) unknown scalar quantities x k , y k , z k , λ α , β µ .Integrating this set of equations, we get the final equations of motion and, atthe same time, from equation (8) we get the magnitude of the reaction forces ofconstraints. However, integration of such a set of equations is verycumbersome due to the large number of equations. That is why the Lagrangeequations of the first kind find little use in actual practice.Example. Two ponderable particles M 1 and M 2 of identical mass m = 1 arejoined by a rod of invariable length l and negligibly small mass. The system isconstrained to move in the vertical plane and only in such manner that thevelocity of the midpoint of the rod is directed along it. Determine the motion ofthe particles M 1 and M 2Solution. Let (x 1 , y 1 ) and (x 2 , y 2 ) be the co-ordinates of the particles M 1 andM 2 . Then, the constraint equations areand12[(x1 − x1)+ (y2− y1)22−l 2 ] = 0, (1)

124 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS(x 2 − x 1 ) y& + y&) − (x&+ x ) (y 2 − y 1 ) = 0. (2)( 2 1 2&1The Lagrange equations with undermined multipliers λ and µ aredgρ ρ ∂fαm k w k = Fk+ λα + µ βl ρρβk∂rα = 1 k β=1for k = 1, 2,…, N. These equations give(3)x&1 = −λ(x 2 − x 1 ) − µ(y 2 − y 1 ),y&1 = −g −λ (y 2 −y 1 ) + µ(x 2 −x 1 ), (4)andx&2 = λ(x 2 −x 1 ) − µ(y 2 −y 1 ),y&2 = −g + λ(y 2 −y 1 ) + µ(x 2 −x 1 ). (5)Equations in (4) are rewritten asλ(x 2 −x 1 ) + µ(y 2 −y 1 ) + x& 1 = 0, (6)λ(y 2 −y 1 ) −µ(x 2 −x 1 ) +Solving equations (6) and (7) for λ and µ, we get& y 1 + g = 0. (7)( y2− yλ =− ( x12)(&y1− x )+ g)+ ( x12− ( y22− x )&x− y121)1. (8)Using equation (1), we obtaing1−2 2 1 2 2 1 1 2 1 1 (9)elλ = (y − y ) − ((x− x )x & + (y − y )y & )Similarly, we shall find (left as an exercise)gµ = ( x − x ) − ((y − y )& x − ( x − x )& y )l221l12211211. (10)It is clear that equation (3) can be obtained from equation (4) if are replace λby “−λ” and & x 1,& y1by & x 2 ,&y.The values of λ and determined from equationare (exercise)λ = ( y − y ) + [(x − x )& x + ( y − y )& y ]lg221e12212212, (11)

ANALYTICAL MECHANICS – I 125gµ = ( x − x ) − [(y − y )& x − ( x − x )& y ]l221l12212212. (12)Equating the approximate expressions for µ and λ in the above formulae, wefind(&x 2 − & x1)(y2 − y1)− (y &2 − & y1)(x 2 − x 1 ) = 0 (13)(&x 2 − & x1)(x2 − x1)+ (y &2 − & y1)(y 2 −y 1 ) + 2g (y 2 −y 1 ) = 0. (14)Next, we introduce the following abbreviated notation :Then we writeu = x 2 −x 1 ,v = y 2 − y 1 ,P = x &1 + x&2 , Q = y &1 + y&2(15)u 2 + v 2 = l 2 , (16)& u v − u & v = 0,(17)P v− Qu = 0 , (18)P & u + Qv & + 2gv = 0 . (19)Equations (16) and (17) show that in a u, v-plane a particle with coordinates (u,v) moves in a circle with radius l and with centre at the origin. Its accelerationwill all the time be directed towards the centre. The motion of particle willthen be uniform. For this reason, we writeu = l cosφ,v = l sinφ (20)Since motion is uniform so change in φ is uniform, i.e., rate of change of φ isconstant. LetThen&= α . (constant) (21)

126 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSAccording to (18), we may putφ = αt + β (22)P =flfu, Q = v(23)lSubstituting these values in (19), are write22f uu & + u f&+ f vv & + v f&+ 2g vl= 0 . (24)Using (20) and (21), we obtainf& = −2gsin φ. (25)Thendf=φ dφdfddtdtThis implies= f& α− 2g=αsinφ.2gf = cos φ + 2γ . (26)Consequently, we get g P = 2γ + cos φ cosφ α and& + & , (27)= x 1 x 2 g Q = 2 γ + cos φ sinφ α y & + & . (28)= 1 y 2Integrating we getx 1 + x 2 = P dt

ANALYTICAL MECHANICS – I 127and1= P dφ2gγg g= sinφ + α 2 sin φ cos φ + 2 φ + 2δ , (29)2gy 1 + y 2 = − cos φ − 2 cos 2 φ + 2∈. (30)Finally, we obtain g g lx 1 = sinφ + 2 sin φ cos φ +2 φ − cos φ+ δ (31)22 2 gy 1 = − cosφ − 2 cos 2 lφ − sin φ + ∈ (32)2 2 g g lx 2 = sinφ + 2 sin φ cos φ +2 φ + cos φ+ δ (33)22 2 gy 2 = − cosφ − 2 cos 2 lφ − sin φ + ∈ (34)2 2φ = αt + β (35)where α, β, γ, δ and ∈ are arbitrary constants.4.7 INDEPENDENT COORDINATES AND GENERALIZEDFORCESvectorsLet us consider a holonomic system of N particles P k with radiiρrk = x k î + ykĵ + zkkˆ(1)for k = 1, 2,…, N, and with finite constraintsρf α (t, r k ) = 0, (2)with α = 1, 2,…, d. In Cartesian form, equivalently, we writef α (t, x k , y k , z k ) = 0 . (3)

128 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSWe shall assume that d functions f α of 3N arguments x k , y k , z k are independent.Here, t is regarded as a parameter. We can therefore express d co-ordinates ofequations (3) as the functions of remaining 3N − d coordinates. The time t andthere 3N − d co-ordinates are regarded as independent quantities that define theposition of the system at time t. All the 3N Cartesian coordinates may beexpressed in the form of functions ofindependent parametersand time t. That is,n = 3N−d (4)q 1 , q 2 …q n (5)x k = φ k (t, q 1 , q 2 …q n ),y k = ψ k (t, q 1 , q 2 ,…, q n ),z k = χ k (t, q 1 , q 2 ,…, q n ), (6)for k = 1, 2,…, N. When these functions are put in the constraint equations (1),the latter become identities. We will assume that any position of the systemthat is compatible with constraints at the given instant of time may be obtainedfrom the equations (6) for certain values of the quantities q 1 , q 2 ,…, q n .In vector form, equation (6) can be written asρ ρr k = r k (t, q1 , q 2 ,…, q n ) (7)for k = 1, 2,…, N. The scalar functions in (6) and vector functions in (7) bothare assumed continuous and differentiable.The minimal number of quantities q i with the aid of which formulas (6) canembrace all possible positions of a holonomic system coincides with thenumber of degrees of freedom of the systemn = 3N − d .The quantities q 1 , q 2 …q n in formula (6) or (7) (where n is the number of degreeof freedom) are called the independent generalized coordinates of thesystem.For each instant of time t, a one to one corespondent is established between thepossible states of the system and the points of a certain region in the n-dimensional coordinate space (q 1 , q 2 …q n ). To each position of the system at

ANALYTICAL MECHANICS – I 129time t, there corresponds a point in the space (q 1 , q 2 …q n ) that describes thisposition of the system. The motion of a point in the coordinate space (q 1 ,q 2 …q n ) corresponds to the motion of the system.If all constraints are stationary, then the time t does not appear explicity inequations (3). It is then always possible to choose coordinates q 1 , q 2 …q n suchthat time t does not enter the equations (6) either.From now on, it is assumed that for a scleronomic system the independentcoordinates q 1 , q 2 ,…, q n are chosen in precisely that way. Then, for ascleronomic system, the formulas (6) and (7) take on the formx k = φ k (q i ),orfor = 1, 2,…, N.y k = ψ k (q i ),z k = χ k (q i ) , (8)ρ ρr k = r k (qi ), (9)Generalized Forces :- To every coordinate q i , there corresponds a generalizedforce Q ifor i = 1, 2,…, N. The generalized forces are determined as follows.Consider the elementary work of effective forces on virtual displacementsN ρ ρδA = F ⋅ . (1)k=1k r kBut the virtual differentials of the function ρ r (t,q )ikiare the virtualdisplacements ρ rk:ρρ n ∂rkδ rk= δq I , (2)i=1 ∂ qfor k = 1, 2,…, N.Substitute the expressions (2) into the right-hand side of formula (1) andexpress the elementary work of the effective forces on the virtualdisplacements in terms of arbitrary elementary increments δq i of theindependent coordinates q i , (i = 1, 2,…m) :N ρ δA = Fk. k 1 n= i=1ρ∂rk∂qiδqi

130 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSn N ρρ∂rk= Fk⋅ qi= 1 k=1 ∂qi iwheren= Q q(3)i=1 N ρQ i = F k 1i i ,ρ∂rk⋅∂qk= i , (4)for i = 1, 2,…n.Q i are called the generalized forces, which are coefficients of δq i .It will be noted that for practical purposes formula (4) is by far not always usedto find the quantity Q i . Instead, the system is given an elementary virtualdisplacement such that only the ith coordinate q i receives a certain incrementwhile the remaining independent coordinates do not change. After that thework of effective forces δA i is calculated on just such a specially chosendisplacement. ThenδA i = Q i δq i , (5)orAiQ i = . (6)qiTheorem : Prove that the position of a holonomic system is an equilibriumposition if and only if all the generalized forces in this position are zero.Proof. Let a certain position of the system be a position of equilibrium.According to the principle of virtual displacements, this is possible if and onlyiforδA = 0 , (1)n Q δq i = 0. (2)i=1iHere, Q i are generalized forces. But the increments δq i in theindependent/generalized coordinates q i are arbitrary. Therefore, the equation(2) is equivalent to the following system of equationsfor i = 1, 2,…, n.Q i = 0, (3)

ANALYTICAL MECHANICS – I 131This proves the theorem.Illustration 1.Consider a rigid body which is constrained to movetranslationally along the x-axis.The abscissa x of some point of the given rigid body may be taken as the onlyindependent coordinate. Here n = 1 andδA = X δx ,where X is the sum of the projections, on the x-axis, of all effective forcesacting on the body. SOQ = X ,is the generalized force for the single independent coordinate x.Illustration 2. Consider a rigid body which is constrained to rotate about acertain fixed axis, say u.In this problem, the angle of rotation, say φ, may be taken as the onlyindependent coordinate. ThenδA = L u δφ ,where L u is the total moment of all effective forces about the axis of rotation.We findQ = L u ,as the generalized force.Illustration 3. Consider a free rigid body.It has six degrees of freedom. For this, we take the three coordinates x A , y A , z Aof some point A of the body as the independent coordinates and the threeEulerian angles ψ, θ, φ that define the rotation.ThenδA = Q x δx + Q y δy + Q z δz + Q ψ δψ + Q θ δθ + Q ϕ δϕ . (1)To determine Q x , we impart to the body an elementary displacement along thex-axis. Thenδy A = δz A = 0,Equations (1) and (2) implyδψ = δθ = δφ = 0. (2)

132 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSδA = Q x δx A . (3)Let X, Y, Z be the projections of the stationary axes x, y, z of the principalvector of all effective forces acting on the body. ThenSimilarlyQ x = X. (4)Q y = Y,Q z = Z. (5)We now impart to our body an elementary displacement such that only theangle ψ changes, while the other quantities remain invariable. Thenδ A = Q ψ δψ . (6)Let L ψ be the total moment of all effective forces about the Az 1 -axis (parallelto oz− axis), about which a rotation through an angle ψ is performed. ThenIn quite analogous fashion,Q ψ = L ψ (7)Q θ = L θ ,Q φ = L φ , (8)where L θ and L ϕ are the total moments of the effective forces.4.8. LAGRANGE’S EQUATIONS OF THE SECOND KINDWe know that the general equation of dynamics isN ρ (Fk=1k− mkρwkρ).r= 0. (1)kExpression for the elementary work of effective forces isN ρδA = F .ρk=1n= i=1k r kQ i q i , (2)

ANALYTICAL MECHANICS – I 133whereN ∂rkQ i = F ρ k . ,(3)k=1 ∂qifor i = 1, 2,…,n.The elementary work of the inertial forces “−m k w ρ ”, (k = 1, 2,….N), isNδB = − mk=1kρwkρ⋅ rkkn= −i=1Z δ(4)i q iwhere, by analogy with expression (3),NZ i = mk=1kρwkρ∂rk⋅∂qi=N mk 1dr &ρρ ∂rk⋅ ∂qkk= dt iρd N ∂r N ∂= kd rkm r&ρk ⋅ − mρk r& kdt k=1 ∂qi k=1 dt ∂qi k ,for i = 1, 2,…, n.ρBut the velocity r & k is also given asr&ρ =kd ρ[rk( qdt1, q2 , …, q n , t)]ρ ρn ∂rk ∂= + rkq&ii=1∂qi ∂t. (6)ρThis shows that the velocity r& k is linearly dependent onformula (6), we find thatρ∂r&k ∂rk= ,∂q&∂qfor i = 1, 2,…n and k = 1, 2,…N.iiq & &1,q2,...,q n& . From(7)On the other hand, from the same equation (6), we obtain

134 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS∂&ρrk∂qi= nk=12ρ∂ rk∂q∂qik2ρ ∂ rkq&k+ ∂qi∂tfor i = 1, 2,…, n and k = 1, 2,…, N.d ∂rk= , (8)dt ∂qi Using equation (7) and (8) in equation (5), we obtain the following expressionfor Z iZ i =ddt N mk=1kr&ρk⋅∂r&ρ Nk − (m∂q&k=1ik∂r& r&ρρk ) ⋅∂qi, (9)for i = 1, 2,…,n. Let T denote the kinetic energy of the system. ThenT =1 N m (r&ρρk ⋅ rk2 & k=1k )(10)∂T∂qiN= (mk=1k ∂r& kr&ρρk ). ∂qi,(11)and∂T∂q&i=Nk=1 ∂r&ρk( m k r&ρk ).. (12) ∂q&i Equations (9), (11) and (12) implyd ∂T ∂TZ i = ,dt −q ∂ i ∂qi(13)for i = 1, 2,…n.The general equation of dynamics (1) gives asδA + δB = 0 (14)Equations (14), (2) and (4) implyni=1(Q i − Z i ) δq i = 0 (15)

ANALYTICAL MECHANICS – I 135Since q i are the independent coordinates and, for this reason, the δq i areabsolutely arbitrary increments in the coordinates. It follows that equation (15)can hold when and only when all the coefficients of δq i in equation (15) areequal to zero. Therefore, the general equation of dynamics (15) is equivalentto the following set of equationsZ i = Q i , (16)for i = 1, 2,…,n. Equations (13) and (16) givefor i = 1, 2,…,n.ddt ∂T ∂T= Qq −i q ∂& ∂ i i, (17)The above equations in (17) are called the Lagrange equations of thesecond kind or Lagrange equation in independent co-ordinates.Definition : The quantitiesq&, i (i = 1, 2,…, n), are called generalizedvelocities.ρWe note that the velocities of the points of the system ( vk= r&ρk ) are expressedin terms of the generalized velocities ( q& 1 ,q&2 ,...,q&n ) and also in terms ofindependent co-ordinates (q 1 , q 2 ,…, q n ) and the time ‘t’ by means of theformula (6).Definition : The quantities & q i , (i = 1, 2,…n), are called generalizedaccelerations.dRemark 1 : After performing the operation , the left-hand sides of thedtLagranges equations (17) contain the time t, the generalized coordinate q i , thegeneralized velocities q&i and the generalized acceleration q&i , (i = 1, 2,… n).The generalized forces Q i ,(i = 1, 2,…,n), on the right-hand side of the Lagrangeequations (17) are ordinarily specified as functions of t, q k , q&k , (k = 1, 2,…n).That is,for i = 1, 2,…,n.Q i = Q i (t, q k , q&k )Remark 2. The Lagrange equations (17) form a set of n ordinary differentialequations, each of the second order in n unknown functions q i of theindependent variable t. The order of this system of differential equations is g n. Note that the set of differential equations determining the motion of aholonomic system with n degrees of freedom cannot be of order less than g n,

136 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSsince by virtue of the arbitrariness of the initial values of the quantities q i andq&, i (i = 1, 2,…,n), the solution of the system must contain at least g n arbitraryconstants. Thus the set of Lagrange equations in independent coordinates hasthe lowest possible order.Remark 3. In the case of a constrained system, the reaction forces do not enterinto Lagrange’s equations. This is an essential advantage of Lagrange’sequation. After Lagrange’s equations have been integrated and the functionsq i (t) found, ρ ρ ρρr = k rk(t),v r , w r and F (t, r , r )k = &ρ ρ & ρ&ρk k = k k k k are determinedconsequently. After that the unknown reaction forces are determined from theformulasρRfor k = 1, 2,…, N.k= mkρwkρ− F ,kRemark 4 : In the case of a free system of particles, Lagrange’s equations (17)are a compact notation of the equations of motion in an arbitrary system ofcoordinates.Illustration 1 : Consider a rigid body in rotation about a stationary axis u. Inthis situation, there is one independent coordinate, the angle of rotation ϕ, torepresent the motion. So, we takeq i = φ . (1)The appropriate generalized force Q for the present motions is equal tomoment of rotation L u ,i.e.,Q = L u . (2)The total kinetic energy is given byT =12I u 2& , (3)where I u is the moment of inertia of the body about the axis of rotation.The Lagrange equation for the present problem becomeswe findddt ∂T ∂T − ∂& ∂∂TIu ∂& = &= Q.(4)(5)

ANALYTICAL MECHANICS – I 137and∂T = 0.∂(6)Lagrange equation (4) now takes the formI u & = L . (7)uThis is the differential equation of the rotation of a rigid body about astationary axis.Illustration 2 (Circular Motion). Suppose that a moving particle isdescribing a circle of constant radius r about the centre 0 with angular velocityω.Thenρr = r(cos î + sin ĵ)(1)is the position vector of the particle.Let î r and Lˆ be the unit vectors in the radial (r increasing) and transversal (θincreasing) directions. Thenî = cosθ) î + (sin ) ĵ(2)rˆii ˆθ = ( −sinθ)ˆ+ (cos θ)j . (3)yvĵ ro θîThe velocity and acceleration areρρ drv =dtP(t) yωx

138 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS= r &[(−sin)î + cos ĵ]rθi& ˆ(4)= θandρ ρdvf =dtIf= r = −rθ& )2 ˆi+ ( r & θ)ˆi.(5)( r θv r , v θ and f r , f θ are components in radial and transversal direction, thenv r = 0, v θ = r & , (6)f r = −r &2 , f θ = r & . (7)The quantity ω = & is the instantaneous rate of change of θ and is the angularvelocity of the moving point at P.Illustration 3 : Let us consider a simple pendulum. Assume that a particle ofmass m is attached to a massless rod that is free to rotate in a vertical planabout a frictionless pin.The motion of this single-degree-of-freedom system may be described by thegeneralized coordinate θ (shown in the figure given below).lThe kinetic energy of this system is given in terms of the generalizedvelocity & as1 2T = mv21 2 2= ml θ & . (1)2The generalized force associated with the rotational coordinate of a pendulumisQ θ = −mgl sinθ. (2)θm

ANALYTICAL MECHANICS – I 139The equation of motion based on the Lagrangian formation isd ∂T ∂T − = Qdt ∂&θ . (3) ∂This givesgθ& + sinθ = 0. (4)lExample : A double simple pendulum is in motion in a vertical plane. Findthe Langrangian equations of motion.Solution : Let OA and AB be the rods hinged at 0 and A, making angles φ 1 andφ 2 with the vertical at any time t. Let mass of rod OA bc m 1 and mass of rodAB be m 2 .Let OA = l 1 and AB = l 2 (see figure below).Oφ 1l 1Al2φ 2m 1 gBLetzm 2 gz 1 = l 1 cos φ 1 (1)z 2 = l 1 cos ϕ 1 + l 2 cos ϕ 2 . (2)We know that the elementary work equation isδA = m 1 g δz 1 + m 2 gδz 2 (3)Form equations (1) and (2), we findδz 1 = −l 1 sinφ 1 , δφ 1 , (4)δz 2 = −l 1 sinφ 1 δφ 1 − l 2 sinφ 2 δφ 2 . (5)Equations (3) to (5) yield

140 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSThis givesandδA = [−(m 1 + m 2 ) gl 1 sinφ 1 ] δφ 1 [m 2 gl 2 sinφ 2 ] δφ 2 (6)The K.E. of rod OA isQφ 1 = −(m 1 + m 2 ) gl 1 sinφ 1 , (7)Qφ 2 = −m 2 g l 2 sinφ 2 . (8)T 1 =12and K.E. of rod AB isT 2 =122m 1 121l φ & , (9)2 2 2 2m 2 l φ & + l φ&+ qll cos( φ 1 − φ 2 ) & ) . (10)( 1 1 2 2 1 2The total K.E. of the system is1 & 21 2 2T = T 1 + T 2 = (m1 + m 2 ) l 1 φ &1 + m 2l 1 l2φ&1φ&2 cos (φ 1 − φ 2 ) +2122m 2 222l φ & .(11)Then the first Lagrange’s equation of motion for the generalized co-ordinate φ 1isd ∂T∂T− = Qφdt ∂φ&1 . (12)∂φOn simplification of (12), one getsddt112[( m 1 m 2 )l1φ1+ m 2 l 1 l 2 2+ &+ m 2 l 1 l 2 φ & 1 φ&2 sin (φ 1 −φ 2 )φ & cos (φ 1 − φ 2 )]= (m 1 +m 2 ) gl 1 sinφ 1 . (13)The second Lagrange’s equation of motion for the generalized coordinate φ 2 is

ANALYTICAL MECHANICS – I 141ddt∂T∂φ&2∂T−∂φ2= Qφ 2 . (14)This givesddt[ l φ &2cos (φ 1 − φ 2 ) + m 2 l 2m 2 l121&2 ]+ m 2 l 1 l 2 φ & 1 φ&2 sin (φ 1 − φ 2 )Special Case : When= −m 2 g l 2 sin φ 2 . (15)m 1 = m 2 = m,l 1 = l 2 = l . (16)After simplification, equations (13) and (15) reduce to& & cos (φ 1 − φ 2 ) + φ & 1 φ& g 2 sin (φ 1 − φ 2 ) +2 sin φ1 = 0, (17) l 2 φ 1 + φ2& φ 2 + & φ cos (φ 2 1 − φ 2 ) + φ 1 φ&2& g sin (φ 1 − φ 2 ) + sin φ2 = 0, (18) l Further, for small oscillations,sin φ 1 = φ i , cos φ i = 1 . (19)Neglecting the second and higher order terms, equation (17) and (18) become2& g φ1 + & φ2+ 2 = 0 , (20) l & g φ1 + & φ2+ = 0 . (21) l 4.9. UNIQUENESS OF SOLUTIONWe have seen that in order to form the lagrange equations of motion for aholonomic system, it is necessary first to find the expression for the kineticenergy T as a function of the time t, the generalized velocities q&1 (i = 1, 2,…n).Let us do this in the general form :

142 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICST =1 N m ρ 2k r& k2. (1)k=1We know thatr&ρkρn ∂rk= i=1∂qiρ∂rkq&i +∂t . (2)Equations (1) and (2) giveρ ρ1 N n ∂r∂T = k rkmkq&i +2 k= 1 i=1 ∂qi∂t2.ρ 2 ρ1 N n ∂r ∂ = k rkm k q&i + 2 k= 1i=1 ∂ qi ∂t2N+ 2mk=1kρ ∂rk∂qiρ ∂rkq&i ⋅ ∂tρ 2ρ1 N n ∂r N ∂ = k 1 rkm k q&i + mk 2 k= 1 i=1 ∂ qi 2 k=1 ∂tρ ρN n ∂rk∂rk+ m k . q&i . (3)k= 1 i=1∂qi ∂tLet a iK a i , a 0 be functions of t, q 1 , q 2 , …, q n defined by the following equationsNa iK = mk=1kρ ∂rk ∂qiρ∂rk. (4)∂qK 2Na i = mk=1kρ ρ ∂rk∂rk . ∂qi ∂tρ1 N ∂rka 0 = m k 2 k=1 ∂t2 , (5)(6)where i, K = 1, 2,…,n. From equation (4), we finda iK = a Ki . (7)

ANALYTICAL MECHANICS – I 143Using relations (4)−(6) in equation (3) we findT = T 2 + T 1 + T 0 (8)whereT 2 =12naiKi,K=1q& .& q ,iKnT 1 = i=1a iq&i .T 0 = a 0 . (9)We also know that, in the case of a sclaronomic system, the time does notρexplicitly enter into the relation between r and q&. For this reasonki∂ ρ r k= 0.(10)∂tfor k = 1, 2,…,N. Consequently, we geta 0 = 0,anda i = 0; (i = 1, 2, …n) (11)T = T 2 =12naiKi,K=1q& q&. (12)iKThus, we see that the kinetic energy of a scleronomic system appears in theform of a homogenous function of the second degree of the generalizedvelocities.Remark 1 : It will be noted that in an arbitrary (Scleronomic or rheonomic)holonomic system, the homogeneous quadratic form T 2 is alwaysnondegenerate.That is, a determinant made up of its coefficients is different from zero, or .for i, K = 1 to n.det (a iK ) ≠ 0, (12)For this, if possible, assume that

144 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSdet(a iK ) = 0. (13)Then the system of n homogeneous linear equationsnK=1a iK λ K = 0 (14)for i = 1, 2,…,n has a real non-zero solution for λ K . Multiplying the set ofequations (14), termwise, by λ i then summing with respect to i from 1 to n andusing formula (4), we getn0 = i,K=1a iK λ i λ KnN= i,K= 1 k=1mkρ∂rk∂qiρ∂rk.∂qK(λ i λ K )N= m k nk= 1 i,K=1ρ ρ ∂rk∂rk . ∂qi ∂qK λ iλKNnρ∂r k= m k λi. . ∂qk= 1 i=1 i K=1nλKρ∂r∂qKKN ∂ρ nrkm λi. ∂qi = kk= 1 i=12. (15)This impliesρN ∂rk =k=1 ∂qi 0i, (16)for k = 1, 2,…, N. These N vector equations may be replaced by 3N scalarequationsn ∂xk i i= 0 ,i=1 ∂qin ∂yk i i= 0 ,i=1 ∂qi

ANALYTICAL MECHANICS – I 145n ∂zk i i = 0(17)i=1 ∂qifor k = 1, 2,…, N. The equations (17) show that in the following Jacobianfunctional matrix.J =∂x1∂x1 ..........∂q1 ∂qn∂y1∂y1 .......... ∂q1∂qn ∂z1∂z1 .......... ∂q1∂qn Μ Μ , (18)∂xN ∂xN.......... ∂q∂ 1 q n∂yN ∂yN ..........∂q∂ 1 q n∂zN ∂zN .......... ∂q1∂qnThe columns are linearly dependent. So, the rank, say ρ, of this functionalmatrix J is less than n. That is, ρ > n.Then among the 3N functions x 1 , y 1 , z 1 , x 2 , y 2 , z 2 ,…x N , y N , z N of the narguments q 1 , q 2 …q n (t is regarded as a parameter) there are ρ independentquantities in terms of which all the remaining Cartesian coordinates of thepoints of the system may be expressed. This is a contradiction to the fact thatthe minimal number of independent coordinates of the system is equal to thenumber of degrees of freedom n. This contradiction establishes the claim in(12).Remark 2. Since T 2 ≥ 0, (19)always, it follows from inequality (12) that the quadratic formT 2 =12is positive definite, andT 2 = 0n aiki,k=1q& q&(20)ik

146 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSonly whenq&= i 0 for each i.Therefore, from the theory of matrices, we havea 11 > 0,aa1121aa1222> 0,…………………,aa11a 21.....n1aa12a 22......n2aa1na 2n......nn> 0. (21)Remark 3 : Putting the expression (1) for kinetic energy into the Lagrangeequations of motionddt ∂T ∂Tq − ∂&i ∂qifor i = 1, 2,…, n; we get= Qi(22)nk=1aik& q + (sum f the terms not involving second derivatives of thekcoordinates w.r.t. time)= Q i (t, q i , q&j ), (23)for i = 1, 2,…, n. The right-hand sides likewise do not contain secondderivatives, since, in the general case, they are functions of the quantities t, q j ,q& for i = 1, 2,………..n.jBy virtue of (12), it follows that the equations (23) may be solved for thesecond derivatives and represented in the formq&i = G i (t, q k , q&k ) (24)

ANALYTICAL MECHANICS – I 147for i = 1, 2,…,n. But, then, as we know from the theory of differentialequations, for certain assumptions relative to the right hand sides G i(the functions G i , 1 ≤ i ≤ n, have continuous first order partial derivatives) thenthere is one and only one solution of the Lagrange equations for arbitrary preassignedinitial quantities q i , q&i with t = t 0 for i = 1, 2,…,n. Thus, the motionof a holonomic system is uniquely determined by specifying the initial positionooq i and initial velocities q&i .4.10. THEOREM ON VARIATION OF TOTAL ENERGYEQUATION FOR CONSERVATIVE FIELDSIf the generalized forces do not depend on the generalized velocities, i.e.,Q i = Q i (t, q 1 , q 2 …q n ) (1)for i = 1, 2,…,n and there exists a function U(t, q 1 , q 2 …q n ) such that∂UQ i = −∂q i(2)for i = 1, 2,…,n; then the forces Q i are called potentials and the function U isthe potential of the forces or the potential energy.We know that the elementary work of the forces Q i is given bynδA = i=1From equations (2) and (3) , we findQiδq I . (3)δA = −δU . (4)Let us now consider the general case when in addition to the potential forcesdetermined by the potential U, the system is acted upon also by non-potentialforcesQ ~i= Q ~ (t,q ,q&) , (5)ijjfor i = 1, 2,…,n. Then the total generalized force is∂U Q i = − Q ~ +∂qii, (6)

148 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSand the Lagrange equations of motion assume the following form ;ddtfor i = 1, 2,…, n. ∂T ∂T −q ∂&i ∂qi∂U= − + Q ~∂qi(7)We now consider the total energy E, which is equal to the sum of the K.E. andpotential energy, is given byE = T + U . (8)dE dTTo compute the derivative , we first evaluate .dtdtWe finddTdtn ∂T= i=1 ∂qi∂T ∂Tq&i + & qi +∂q&i ∂td n ∂T n ∂T= q&i + dt i=1 ∂ q&i i=1∂qi−ddt ∂T∂Tq&i + . ∂q&i ∂t(9)The Euler’s formula for a homogeneous function givesn ∂ T 2q & i = 2Ti=1 ∂q&i2(10)n ∂ T 1q & i = T1. (11)i=1 ∂q&iWe know thatT = T 2 + T 1 + T 0 (12)From equations (9) to (12) and Lagrange’s equation of motion (7), we finddTdt=ddt∂Tn ∂U~ ( 2T2 + T1) + + − Qiq&i . (13)∂ti=1 ∂qi From (12), we find the value of T 2 in term of T, T 1 and T 0 .

ANALYTICAL MECHANICS – I 149T 2 = T−T 1 − T 0 . (14)Using equation (14) in (13), we obtaindTdt=dT2dt−ddt∂T(T + 2T0) +∂t1 +dUdt−∂Un− ∂Q~ i q &i . (15)t i=1From equation (8), we finddEdtdT dU= + . (16)dt dtUsing relation (15) in equation (16), we writedEdtn= Q ~i=1id∂T∂Uq& i + (T1+ 2T0) − + . (17)dt∂t∂tFormula (17) determines the total energy of an arbitrary holonomic system.Further,n Q~ n& = Q ~i=1i q ii=1idq&idt=n Q ~i=1A ~= ,dtidtdq&i(18)where A ~ is the elementary work of the nonpotential forces Q ~ i .It is called the power of the non potential forces Q ~ i .For a conservative system, we have(i)(ii)(iii)a scalernomic system,a system where all forces are potential, andthe potential energy U is not explicitly dependent on the time.Thus, for a conservative system equation (17) leads to

150 PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSThis givesdE = 0 dt(19)E = constant = h , say, (20)for a conservative system.This shows that the total energy of a conservative system does not changewhen the system is in motion.The Books Recommended for Chapters IV, V, VI and VII1. F. Gantmatch Lectures in Analytical Mechanics,MIR Publications, Moscow, 1975.2. Louis N. Hand and J.D. Finch Analytical Mechanics,Cambridge University Press, 1998.3. J.S. Torok Analytical Mechanics,John Wiley and Sons, 2000.

ANALYTICAL DYNAMICS-II 151Chapter-5Analytical Dynamics-II5.1. LAGRANGE’S EQUATIONS FOR POTENTIALFORCESLet the generalized forces Q i be potential, i.e., there exist a force potential(potential energy)such thatU = U(t , q i ) (1)∂UQ i = − ,∂qi(2)for i = 1, 2,…,n.We defineL = T − U. (3)The function L is called the Lagrangian function or the kinetic potential.Since the potential energy U does not depend on the generalized velocity q&i ,so∂L∂T= , (4)∂q&∂i q i∂L∂T∂U= − , (5)∂q∂q∂qiiifor i = 1, 2,…, n. We know that the Lagrange’s equation of motion in terms ofkinetic energy T are given byd ∂T ∂T∂Udt = −q − ∂&i ∂qi∂qi. (6)from equations (4) to (6), we obtain

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 152d ∂L ∂Ldt −q ∂&i ∂qifor i = 1, 2, …, n.= 0(7)Let us now consider the case when in place of the ordinary potential U(t, q i ),there exists a generalized potentialV = V(t, q i , q& ) (8)iin term of which the generalized forces Q i are expressed by means of thefollowing formulasQ i =for i = 1, 2, …., n.d ∂V ∂Vdt −q ∂&i ∂q i(9)In this case, the Lagrange’s equationsd ∂T ∂T −dt ∂q&i ∂q i=d ∂V ∂Vdt −q ∂&i ∂qibecomesd ∂L ∂Ldt −q ∂&i ∂q i= 0 . (10)These equation in (10) are of the same type/form as in (7).Remark I: From formulas (9), it follows thatn 2∂ VQ i = K=1∂q&i∂q&K&qK+ ( * ) (11)for i = 1, 2,….., n. ( * ) denotes the sum of the terms that do not containsgeneralized accelerations q&, K = (1, 2,…,n)Inasmuch as in mechanics we consider only the core when the generalizedforces Q i are not explicitly dependent on the generalized accelerations butdepend solely on the time, on the coordinates and on the generalized velocities.

ANALYTICAL DYNAMICS-II 153for i = 1, 2,…,n .Q i = Q i (t, q K , q& ) (12)KIt then follows according to formulas (11) that all partial second orderderivatives of V with respect to the generalized velocities must be identicallyequal to zero. This implies that the general potential V, in this case, dependslinearly on the generalized velocities. Therefore, we can writeV =n U & + U = V 1 + U , (13)i=1i q iwhere U i and V are functions of coordinates q 1 , q 2 ,…, q n and of time t.Substituting expression (13) for V into formula (9), we getdU ni ∂Q i = − U q + U K&Kdt ∂qiK=1∂Un ∂U= − + ∂qi K 1 ∂q∂Ui K− qK += K ∂qi ∂ ∂Ui & . (14) tThe formulas (14) show that when the linear part V 1 of the generalizedpotential does not depend explicitly on the time variable t, the generalizedforces Q i are made up out of potential forces−∂U∂q i(15)and gyroscopic forceswhere~Qi= n 1γK=iKq&, (16)Kfor i, K = 1, 2,…, n.γ iK = −γ Ki =∂U∂qiK∂U−∂qKi(17)5.2 LAGRANGIAN AND HAMILTONIAN VARIABLESIf the kinetical potential or the Lagrangian function L = L(t, q i , q& i ) is known,then the differential equations of motion of a system can be written. The

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 154variables t, q i , q&(i i = 1, 2,…n), in terms of which the Lagrangian function isexpressed, are called Lagrangian variables.For the basic variables that characterize the state of a system, Hamiltonproposed the quantities t, q i , p i (i = 1, 2,…,n), where p i are the generalizedmomenta defined by the following equalities∂Lp i =(1)∂&for i = 1, 2,…,n.q iThe variablest, q i , p i are called Hamiltonian variables.Classical systems in which the forces have the ordinary potential U(t, q i ) or thegeneralized potential V(t, q i , q&) i will be called natural. For such systems theLagrangian function L is a quadratic function of the generalized velocities.For natural system we havei2∂ L∂q&∂q&Ki2∂ T=∂q&∂q&K= aiK(2)for i, K = 1, 2,…,n.We notice that the Jacobian of the Right side of the equations (1) w.r.t. thevariables q&i is the Hessian of the function L. We assume that the Hessian ofthe function L w.r.t. to the generalized velocities q&i is not identically zero.Therefore2∂ Ldet ∂q&i∂q&K≠ 0 (3)it follows that equation (1) can be solved for q i , and we writefor i = 1, 2,…, n.q&i = Φ i (t, q K , p K ) (4)Thus, Hamilton’s variables may be expressed in terms of the Lagrangevariables and vice versa. Consequently, the state of the system may bedescribed both as a system of values of the Lagrange variables and as a systemof values of Hamiltonian variables.

ANALYTICAL DYNAMICS-II 155We know that in the case of a natural system, Lagrangian function L is aquadratic function of the generalized velocities. By virtue of equation (1), thegeneralized momenta p i are linearly expressible in terms of the generalizedvelocities :np i = a iK q& K + ci(5)K=1for i = 1, 2,…,n. Solving the linear system (5) for q&, i we get linearexpressions for q&i of the typenq&i = K=1b + b i (6)iK p Kfor i = 1, 2,…,n. Here, b ik and b i are functions of t, q 1 , q 2 ,…, q n .If in a natural system the forces Q i have an ordinary potential U(t, q i ), itfollows from the equationL = T − U (7)p i =∂T∂&q i(8)If forces Q i have a generalized potential, then we have∂T p i = − Ui(9)∂q&iLetF = F(t, q i , q&) i , (10)be any function of Lagrangian variables. After substitution of the expression(4) or (6) into (10) in place of the generalized velocities q&, i the function (10) isconverted into a certain functions, say, F (t, q i , p i ), of the Hamiltonianvariables.We call the function F (t, q i , p i ) the associated function of the function F(t,q i , q&). iHamilton (1834) introduced the function H(t, q i , p i ) defined by the equation

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 156nH = P q& − L , (11)i=1iiwhere L is the associated function of L, in the sense described above.The function H is called Hamiltonian function.With the help of Hamiltonian function H, the equations of motion may bewritten in the form of the following system of 2n ordinary differentialequations of the first orderdqdtdpdtii∂H= ,(12a)∂pi∂H= −(12b)∂qifor i = 1, 2,…,n.These equations (12a, b) are called canonical equations or Hamilton’sequations.DONKIN’S THEOREM : The derivation of the canonical equations ofHamilton will follows from the DONKIN’S Theorem :Statement : Given a certain function X(x 1 , x 2 ,…, x n ), the Hessian of which isdifferent from zero. Let there also be a transformation of the variables“generated” by the function X(x 1 , x 2 …x n ) :∂Xy i = ,∂xifor i = 1, 2,…,n. Then, there exists a transformation which likewise generatessome function Y(y 1 , y 2 …y n ) :∂Yx i =∂y ifor i = 1, 2,…,n.If the function X contains the parameters α 1 , α 2 ,…, α m , i.e.,X = X(x 1 , x 2 ………x n ; α 1 , α 2 …….α m ),then Y also contains these parameters, i.e.,

ANALYTICAL DYNAMICS-II 157andY = Y(y 1 , y 2 ………y n ; α 1 , α 2 …….α m )∂Y∂X= −∂α ∂αjjfor j = 1, 2,…,n.Proof. Let the generating function Y of the inverse transformation beconnected with the generating function X of the direct transformationy i =∂X∂x i(1)for i = 1, 2,…,n . By the formula (known as Legendre transformation)nY = i=1x i y i −X . (2)The Hessian of the function X coincides with the Jacobian of the right handside of equations (1), and by virtue of the hypotheses given in the statement ofthe theorem, we have2∂ X det ≠ 0 , (3) ∂xi∂xk for this reason it is possible to express the variables x 1 , x 2 ,…, x n in terms of y 1 ,y 2 ,…, y n .We writefor i = 1, 2,…,n.x i = f i (y 1 , y 2 ,…,y n ) (4)Now we replace the variables x i appearing in formula (2) by the expressions in(4). Then∂Y∂=∂y∂yiinK=1xKyK− X

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 158n∂xnK= yK+ x i −∂y∂X∂xyKK= 1 iK=1 ∂xK ∂ i. (5)By virtue of equations (1), the two sums on the right hand side of equation (5)cancel. Hence, we obtain∂Y = x i , (6)∂yfor i = 1, 2,…,n.iThis proves the first part of the Donkin’s theorem.2nd Part : Now, let X contain parameters α 1 , α 2 …α m in addition to thevariables x 1 , x 2 …x n . Then these parameters occur in the direct transformation(1) and, consequently, in the reverse one as well :for i = 1, 2,…,n.x i = f i (y 1 , y 2 …y n ; α 1 , α 2 …α m ) (7)The function Y is determined by equation (2) in which the x i are replaced byf i (y 1 , y 2 …y n ; α 1 , α 2 …α n ),and so (regarding y 1 , y 2 ,…, y n as constants) we obtain∂Y∂αj∂=∂αjn x i=1iyi− Xn ∂x ∂= in Xy − ii=1 ∂α j i=1 ∂xi∂x∂αij ∂X −∂α jn ∂xi ∂Xn ∂X= −i=1∂α∂xi=1∂xjii∂x∂αij∂X−∂αj∂X= − ,∂α i(8)for j = 1, 2,…,n.This completes the proof of Donkin’s theorem.

ANALYTICAL DYNAMICS-II 1595.3 HAMILTON CANONICAL EQUATIONSTo derive these equations, we use Donkin’s theorem to make transition fromthe Lagrangian variables to the Hamiltonian variables. For this, by thefunction X is replaced by L, the variables x 1 , x 2 ,…, x n by q &1,q&2,...,q&n , theparameters α 1 , α 2 … α m by q 1 , q 2 ,…, q n and t, the variables y 1 , y 2 ,…, y n by p 1 ,p 2 ,…, p n and the functionnY = x y − X(1)i=1in the Donkin’s theorem. We know thatiiandn ) )H = piq&i − L , (2)p i =i=1∂Lq i∂&, (3)for i = 1, 2,…,n.Hence by Donkin’s theorem, it is concluded that∂Hq& i = , (4)∂piand∂L∂q∂H= −∂i q i∂L∂H= −∂t∂t, (5)(6)for i = 1, 2,…, n. Lagrange equations of motion ared ∂L ∂Ldt −q ∂&i ∂qi= 0 , (7)for i = 1, 2,…,n. Using (3) in (7), we writed ∂L(pi) = . (8)dt ∂qi

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 160Hence, from (5) to (8), we obtainor∂Hq& i = ,∂pdpdtidqdtii∂L=∂qi∂H= ,∂pi∂H= −∂qidpdtfor i = 1, 2,…,n.i∂H= − , (9)∂qiEquations in (9) are the required canonical equations of Hamilton.5.4 ROUTH’S EQUATIONSFor the basic variables characterizing the state of a system at a time t, Routhproposed taking a part of the Lagrangian variables and a part of theHamiltonian variables.The Routh variables are the quantitiest, q i , q α , q&, i p α , (1)for i = 1, 2,…, m ; α = m + 1,…,n, m is an arbitrary fixed number less than n.The Lagrangian variables can be replaced by Routh variables if we express allthe q&α in terms of p α , by the relationsp α =∂L∂q& , (2)αfor α = m + 1, m + 2,…,n.Suppose that the Hessain of the function L of the generalized velocitiesdifferent from zero. That is, ∂ Ldet ∂ q ∂&2 α q &β ≠ 0. (3)q&α is

ANALYTICAL DYNAMICS-II 161Then, by applying the Donkin theroem, we get a transformation that is inverseto the transformation (2), namely∂Rq&α = (4)∂ pfor α = m + 1, m + 2,…,nαandR = R(t, q i , q α , q& i , p α )is the Routh function defined by the equationnR = pαα= m+1) )q & − L . (5)αThe sign .) signifying that all the q&α are expressed in terms of p α .The variablest, q i , q α , q&i(i = 1, 2,…, m ; α = m + 1, m + 2, …,n) are now regarded as parameters.Consequently, Donkin theorem gives∂R∂q∂L= −∂i q i,∂R∂L= −∂q& ∂&i q i(6)for i = 1, 2,…,m ; and∂R∂qα∂L= −∂qα(7)for α = m + 1, m + 2,…, n ,and∂R∂L= − .∂t∂t(8)We know that the Lagrange equations for the coordinates q i are

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 162d ∂L ∂Ldt −q ∂&i ∂qi= 0 . (9)Using (6) equations (9) may be written asd ∂R ∂Rdt −q ∂&i ∂qi= 0 , (10)for i = 1, 2,…,m.Lagrange’s equations of the coordinates q α aredpα∂L=dt ∂ qα(11)for α = m + 1, m + 2,…, n.Using (4) and (7), equation (11) becomeanddpdtd qdtαα∂R= −∂qαα(12)∂R= , (13)∂ pfor α = m + 1, m + 2,…, n.Equations (10), (12) and (13) form a set of following Routh equations :d ∂R ∂Rdt −q ∂&i ∂qi= 0,dpdtα∂R= −∂qα,dqdtα∂R= −∂pα, (14)for i = 1, 2,…, m and α = m + 1, m + 2,…,n.These equations consist of m second-order differential equations of theLagrange type and 2(n−m) first order differential equations of the Hamiltonian

ANALYTICAL DYNAMICS-II 163type. The Routh function in the first equations play the role of the Lagrangianfunction, while those in the latter equations play the role of the Hamiltonianfunction.5.5 CYCLIC COORDINATESThe Lagrangian of any physical system is generally expected to have explicitdependence on all the generalized coordinates q i , all the generalized velocitiesq& and time t, that isiL = L(q 1 , q 2 … q n , q&,q&...q , t)1 2&nwhere n is the total number of generalized coordinates. If some of thegeneralized coordinates do not appear explicitly in the expression for theLagrangian, these coordinates are called cyclic (ignorable). Any change inthese coordinates do not affect the Lagrangian.5.6 POISSON BRACKETSPoisson introduced a special term, called the Poisson bracket, for the followingexpression composed of the partial derivatives of two arbitrary functions φ(t,q i , p i ) and ψ(t, q i , p i ) :n ∂φ ∂ψ ∂φ ∂ψ(φ ψ) = − . (1)i=1 ∂qi ∂Pi∂Pi∂qi Remark : For the functions φ(t, q i , P i ), ψ(t, q i , p i ), χ(t, q i , p i ) and constant c,the following properties are satisfied by Poisson bracket :(1) (φ ψ) = −(ψ φ)(2) (c φ ψ) = c(φ ψ)(3) (φ + ψ χ) = (φ χ) + (ψ χ)(4)∂ ∂φ ∂ψ ( φψ)= ψ +φ .∂t ∂t ∂tPOISSON’S IDENTITY : For the functions φ(t, q i p i ) ψ(t, q i , p i ) and χ (t, q i ,p i ); the following property holds((φ ψ)χ) + ((ψ χ)φ) + ((χ φ)ψ) = 0This property is known as Poission’s identity.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 164Definition : A function f(t, q i , p i ) is called the integral of the equations ofmotiondqdtidpdti∂H= ,∂pi∂H= − , (1)∂qifor i = 1, 2,…,n ; if for any motion of the given system this function retains aconstant value say C :f(t, q i , p i ) = C (2)Remark : The necessary and sufficient condition for the function f(t, q i , p i ) tobe integral of the equations of motion (1) is thatdfdt∂f= + ( f H) = 0. (3)∂t5.7 JACOBI-POISSON THEOREMStatement : If f and g are integrals of equations of motion, then (f g) is alsoan integral of these equations.Proof : Since f and g are integrals of equations of motion then∂f∂t+ (f H) = 0, (1)∂ g + (g H) = 0 . (2)∂tNow to prove that (f g) is also an integral of the same equations of motion, it isrequired to prove thatNow∂ (f g) + ((fg)H) = 0 . (3)∂t

ANALYTICAL DYNAMICS-II 165∂ ∂f (f g) = g +f∂t ∂t From equations (1) and (2), we write∂g∂t(4)∂f∂ t= −(fH) , (5)∂ g = − (g H). (6)∂tUsing (5) and (6) in (4), we write∂ (f g) = −((f H)g) − (f(gH)) ,∂ t∂ (f g) = ((H f) g) + ((g H) f) . (7)∂ tFrom (3) and (7), we writeThus∂ (f g) + ((f g)H) = ((H f)g) + ((gH) f) + ((f g)H) = 0 .∂ t∂ (f g) + ((f g)H) = 0 .∂tThis completes the proof of the theorem.

166PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSChapter-6Analytical Mechanics-III6.1 HAMILTON’S PRINCIPLEDefinition : We consider an arbitrary holonomic system with independentcoordinates q 1 , q 2 ,…, q n and the Lagrangian function L = L(t, q i , q&), i 1 ≤ i ≤ n.The integralW = 1 Ldtt0t…(1)is called the “Hamilton action” during a time interval (t 0 , t 1 ). The expressionLdt is called the elementary Hamilton action.Note : Since the function L is of the formL = L (t, q i , q&), i…(2)it is necessary, in order to compute the Hamilton action W, to specify thefunctionsq i = q i (t)…(3)for i = 1, 2,…, n; in the time interval [t 0 , t 1 ]. This imply that the Hamiltonaction W is a functional dependent on the motion of the system.0Remark : Let q i be a given initial position of the system at time t = t 0 and q i ′be a given final position, which it occupies at time t = t 1 . We fix the initial andterminal instants of time t 0 and t 1 , and the initial and terminal positions of thesystem. The motions are otherwise arbitrary.In the extended (n + 1) −dimensional coordinate space, where the quantities q iand the time t are the coordinates, this motion is depicted by some Curves orpaths. We shall consider all possible such paths, passing through twospecified points of space M 0 (t 0 , q o i ) and M 1 (t 1 , q i ′) as shown in figure below.tM 1M 0q i

167ANALYTICAL MECHANICS-IIIThat is, we consider all possible motions that translate the systems from initialposition q i o to final position q i ′.Straight path and Circuitous pathsSuppose that among the paths considered above, there is a path along whichthe system can move for a specified function L, i.e., in a given field of force.Such a path is called “straight path”. In the above figure, the “straight path”is depicted by a solid-line. For a straight path, the functions, q i = q i (t), satisfythe Lagrange equations of motion, namely,d ∂L ∂Ldt −q ∂&i ∂qi= 0,…(4)for i = 1, 2,…,n. All other paths passing through the points M 0 and M 1 aretermed as “circuitous paths”.Statement of Hamilton’s principle (1834-35)“The Hamilton action W has a stationary value for the straight path ascompared with the circuitous paths”.Proof Hamilton’s principle : Let us consider an arbitrary one-parameterfamily of pathsq i = q i (t, α),…(5)where α is a parameter, −γ ≤ α ≤ γ, t 0 ≤ t ≤ t 1 and i = 1, 2,…,n. For α = 0, oneobtains a given straight path and for α ≠ 0, the paths are circuitous. Further, letus assume that all these paths in (5) have a common initial point M 0 and acommon end point M 1 . That is,q i (t 0 , α) = q 0 i ,…(6a)q i (t 1 α) = q i ′,for −γ ≤ α ≤ γ and i = 1, 2,…, n.…(6b)The Hamilton action as computed along a path of this one-parameter family isa function of the parameter α, and is denoted by W(α), defined ast1W(α) = L (t,q (t, α),& (t, α)) dt. …(7)t0i q iNow we compute the variation δW of the Hamilton action W. We defined

168PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSt1δW = δLt0dtWe note thatδt1 n ∂L∂L= qiqi dtt q q δ + δ&0 i=1∂ i ∂&i dq&= δ{q (t, α)} iidt∂ d = {q i (t, α)}δα ∂α dt…(8)=d ∂qdt ∂αi (t, α)δαd= (δqi ). dt…(9)Nowt1 n ∂L qi δ & dtt q0 i=1∂&it1 n ∂L d = ( qi) dtt q δ &dt0 i=1∂&i t=t1 n ∂L t1n d ∂L = δ qi − δqi dti=1∂q&i t = ∂0 i 1 dt q&i t=t0t1n d ∂L = − qi dt,t dtqδ0 i= 1 ∂&i …(10)since the variations δq i are zero at t = t 0 and t = t 1 by virtue of the assumptionthat the straight path and all the circuitous paths pass through M 0 and M 1 inextended co-ordinate square.Combining equations (8) and (10), one writest1 n ∂LδW = t = ∂0 i 1 qid ∂L−δ qdt ∂q&i idt…(11)

169ANALYTICAL MECHANICS-IIIFor a straight path (i.e., α = 0), the functions q i = q i (t) satisfy the Lagrangeequations of motion, namely,d ∂L ∂Ldt −q ∂&i ∂qi= 0,…(12)for i = 1, 2, …,n. From equations (11) and (12), it is concluded thatδW = 0,…(13)for the straight path. This proves that the Hamilton actions W has a stationaryvalue for the straight path. Hence, the proof of Hamilton’s principle iscomplete.Remark : The converse of Hamilton’s principle is true. That is, ifδW = 0for some path, then the paths straight.…(14)Remark 2 : Since, from the Hamilton principle there follows the Lagrangeequations of motion (using equation 11) and vice versa, Hamilton’s principlemay be placed at the foundation of the dynamics of holonomic systems.Remark 3 : The variational principle characterizes the entire straight path as awhole. It formulates the stationary property of a certain functional (Hamiltonaction), which property distinguishes the straight-line path from among otherkinetically possible paths (circuitous paths). The variational principle has amore surveyable and compact form and if frequently used as foundation fornew non-classical domains of mechanics.Remark 4 : The value of the Hamilton actions W is least for a straight-linepath.6.2 POINCARE − CARTAN INTEGRAL INVARIANTWe shall now derive a formula for the variation of Hamilton actiontW = 1 L dt,…(1)t0in the general case when the initial and terminal instances of time, just like theinitial and terminal coordinates, are not fixed but are functions of a parameter,say α, of the typeandt 0 = t 0 (α), t 1 = t 1 (α),…(2)q0i0i1i1i= q ( α),q= q ( α). …(3)

170PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSUsing the Leibnitz rule for differentiation under integral sign, thedifferentiation of (1) with respect to parameter α givesδW = δt1t0( α) L dt( α)L 1 δt 1 −L 0 δt 0 +Integrating by parts, we writet1(α)t0( α ) ∂Lqidt =q δ& ∂&i 1(α)n ∂L∂L qiqidt0( ) i 1 qiq δ + δ&α = ∂ ∂&i ttt1(α)t0( α) ∂L d qidtq δi dt ∂&…(4)t ( α)1 ∂Lt1(α)d ∂L= δqi −∂q&it ( α)dt ∂q&0i t0( α)(δq i ) dtt1(α)d ∂L10= pi{δqi}t=t ( α)− pi( δqi}t=t ( α)−1 0t ( α)dt ∂ q &(δq i) dt, …(5)0i where we have used the facts thatand∂Lp i = ,∂& q i…(6a)dδ& qi= ( δqi ),…(6b)dtfor i = 1, 2,…,n. Here, p i is the generalized momenta of the system. Further,we have used the notationpλ i= p i[ ]t t ( )for λ = 0, 1. We note that= λ α= p i (t λ (α), α), …(7) ∂ q i t=tλ i ) …(8)∂α{ δ } = q (t, α δαfor λ = 0, 1. On the other hand, for the variations of the terminal coordinatest=tλ

171ANALYTICAL MECHANICS-III' 'qii 1we have the formulasThis implies= q [t ( α),α],' 'δq = q&δtii&1∂qi (t, α)+ ∂α'i 1 i t=t 1= q δt+ [ δq] .t= t 1δαSimilarly,' '[ qi] = δqi − qiδt1δ=& . …(9)t t10 0[ ] = δq− q δt.δ=& …(10)qi t t0i i 0First, we substitute the expressions for [ i ]t=tλδq from equations (9) and (10)into right side of equation (5) and then use the value of leftside of (5) intoequation (4) to get the following expression for δW.n' ' 'δW = p ( δq− q&δt)i=1iii1+ L δtn0 0 0− pi( δqi− q&i δt0 ) − L0δt0i=111we know thatt1(α)+ t0( α ) =n ∂Li 1 ∂qid ∂L−δqdt ∂q&i idt…(11)so thatn p q& − L H , …(12)i=1ni i = p & − L λ = H λ , …(13)i=1λ λi q ifor λ = 0, 1. Using (13) in (11), we finally obtainδW =ni=1p δqii− Hδt 10t1 n ∂L+ t = ∂0 i 1 qid ∂L−δ qdt ∂q&i i,dt…(14)

172PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSwhereni=1p δqii− Hδt10=ni=11p δqi1i− H δt11n− p δq+ H δt.i=10i0i00…(15)In place of the (n + 1) −dimensional extended coordinate space, we takethe (2n+1)−dimensional extended phase space in which the quantities q i , p i andt will be the coordinates of the point.In this space we take an arbitrary closed curve C 0 given by the equationsq i =0qi(α),p i =0qi(α),t = t 0 (α),…(16)for i = 1, 2,…,n and 0 ≤ α ≤ l.Here, we have one and the same point of the curve C 0 for α = 0 and α = l.Taking each point on the curve C 0 as the initial one, we draw the appropriatestraight-line path. Such a path is uniquely determined by a system ofHamilton’s canonical equations and initial-conditions. We obtain a closed tubeof straight pathsq i = q i (t, α),p i = p i (t, α),q i (t, 0) = q i (t, l),p i (t, 0) = p i (t, l),for i = 1, 2,…,n and 0 ≤ α ≤ l.…(17)…(18)C 1tC 0p 1(For n = 1)q 1

173ANALYTICAL MECHANICS-IIIOn this tube, we choose arbitrarily a second closed curve C 1 around the tubethat has only one common point with each generatrix. The equations of curveC 1 may be written in the following formq i =p i =1qi(α),1qi(α),t = t 1 (α).…(19)Now we shall examine the Hamilton action W along the generatrix of the tubefrom the curve C 0 to the curve C 1 . In this case, for any α, the generatrix is astraight-line path and by virtue of Lagrange equations of motion∂L∂qid ∂L− = 0dtq , …(20) ∂&i for i = 1, 2,…,n.Consequently, the variations of Hamilton action, δW, for this case takes thefollowing simplest form:nδW = piqiH t δ − δ , i=110…(21)by virtue formula in (14). This givesnW′(α) δα = piqiH t δ − δ . i=110…(22)Integrating (22) with respect to α from α = 0 to α = l, we get1n0 = w(l) −w(0) = 10i=1p δqii− Hδt' '= p δqi − H1δt1−i 0ni=110ni=1p0i10δq0i− H0δ t0This givesn n= p i δ qi− H δ t− piδ qi− H δ t.i1 i=1 C1 = C0

174PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSn n p i δ qi− H δ t= piδ qi− H δ t. …(23)i1 i=1 C1 = C0The relation (23) proves that the line-integralnI = p i δ qi− H δ t…(24)i=1 Ctaken along any close contour C does not change its value in the case of anarbitrary displacement of the contour along a tube of straight-line paths. Thisimply that the integral I is invariant.Definition : The integral I, defined above in (24), is called the POINCARE-CARTAN INTEGRAL INVARIANT.Remarks 1 : The invariance of the Poincare−Cartan integral may be placed atthe foundation of mechanics since from this invariance it follows that themotion of a system obeys Hamilton’s canonical equations.Remark 2 : The converse proposition of the above result is true. That is, if thePoincare−Cartan I is an integral invariant with respect to the straight-line pathsdefined by the set of following first-order differential equationsdq i =dtQ i (t, q k , p k ),dp i= P i (t, q k , p k ),…(25)dtfor i = 1, 2,…,n ; then the following relations hold between the function H andthe functions Q i , P i :Q i =∂H∂p i∂HP i = −∂q ifor i = 1, 2,…,n.…(26)Remark 3 : In the Poincare−Cartan integral (24), the time t enters as thecoordinate q i , and the role of the corresponding momentum is played by thequantity, −H, i.e., the energy taken with opposite sign.This is a far−reaching analogy.We change the variables in the integral I by introducing a new variable zconnected with the old variables by the relationz = −H(t, q 1 , q 2 , …, q n , p 1 , p 2 ,…,p n )Using this relation (27), we express p 1 in terms of other variables. Let…(27)

175ANALYTICAL MECHANICS-IIIp 1 = −K(t, q 1 , q 2 ,…, q n ; z; p 2 , p 3 ,…, p n ).Using this relation (27), and (28), the expression (24) now becomesI = C…(28){zδt + p 2 δq 2 + p 3 δq 3 +…+ p n δq n − K δq 1 }. …(29)Thus, in the new variables, the integral I has the aspect of the Poincare −Cartan integral, but the role of the time is now played by the variable q 1 and inplace of the earlier energy H we have the momentum p 1 taken with reversedsign, i.e., K.Thus, the motion of a system in the new variables is described by the followingHamiltonian system of differential equations.dtdqdqdq∂Kdz ∂K= , = − , …(30)∂zdq t1 1 ∂j1∂Kdp j ∂K= , = − , …(31)∂pdq ∂qjfor i = 2, 3,…,n; q 1 being the independent variable.Illustration : In the case of a linear oscillator for which221jp cqH = + .…(1)2m 2To form the canonical equations taking q for the independent variable, we putEquation (2) impliesThus, we have,2p cqz = −+2m 22 . …(2)2p = m − (2z + cq ) . …(3)2K = − m − (2z + cq ) . …(4)The corresponding canonical equations aredtdqm / c= ,…(5)2− 2(z / c) − qdz = 0 . …(6)dq

176PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSSolving (6), we getz = constant= −h, say. …(7)From equations (5) and (7), we findcm dt = dq(2h / c)− q2− αThis giveswherewt + α = sin −1 w =cq 2h…(8)c / m , …(9)and α is the constant of integration. Equation (8) implieswhereq = A sin (wt + α)A =2 h / c .…(10)The solution of canonical equations (5) and (6) consists of relations in (7)and (10).6.3 WHITTAKER’S EQUATIONSWe consider a generalized conservative system for which theHamiltonian function H is not explicitly dependent on time, i.e.,andH = H(q i , p i )∂H = 0.∂tWe know that Hamilton’s canonical equations aredqdtiii…(1)…(2)∂Hdpi∂H= , = −…(3)∂pdt ∂qfor i = 1, 2,…,n. Differentiating (1), we writedH n ∂H= dt i=1∂qidq1∂Hdpi+dt ∂pdti = 0, …(4)

177ANALYTICAL MECHANICS-IIIby virtue of (2) and (3). Integrating (3), we getH(q i , p i ) = constant= h, say, …(5)during the motion of the system. The function H is called the generalizedtotal energy and relation (5) is termed as the generalized integral ofenergy.We consider an ordinary 2n−dimensional phase space in which the quantities q iand p i , 1 ≤ i ≤ n, are the coordinates of a point. We confine ourselves to onlythose points of the phase space whose coordinates satisfy the equation (5) withfixed value of the constant h, say h 0 . In other words, we confine ourselvessolely to those states of the system to which the given magnitude of the totalenergy corresponds :H = H(q i 0 , p i 0 ) = h 0 .The basic integral invariant I for a generalized conservative system is…(6)nI = p i δ q i . …(7)i=1We solve equation (5) for one of the momenta, say p 1 , to getp 1 = −K (q 1 , q 2 ,…, q n , p 2 , p 3 ,…, p n , h 0 ),and put the expression (8) for p 1 into the integral (7). We have…(8)I = np δ − δ j q j K q1. …(9) j=2But the integral invariant (9) again has the form of the Poincare−Cartanintegral if it is assumed that the basic coordinates and momenta are thequantities q j and p j for j = 2, 3,…,n, and the variable q 1 plays the role of timevariable; and instead of H we have the function K. Consequently, the motionof a generalized conservative system satisfy the following Hamiltonian systemof 2n−2 differential equations.dq j ∂k= ,dq ∂p1jdpdqj1∂k= − , …(10)∂qjfor j = 2, 3,…,n.These differential equations in (10) were obtained by Whittaker. For thisreason, equations (10) are called Whittaker Equations.

178PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS6.4 JACOBI EQUATIONSWe know that for an ordinary conservative system, Whittaker’s equations areanddqdqj1∂kdp j ∂k= , = − , …(1)∂pdq ∂qH = constant = h.j1j…(1a)for j = 2, 3,…n. Integrating the system (1), we find the q j and p j as functionsof variable q 1 and 2n−2 arbitrary constantsc 1 , c 2 ,…, c 2n−2 .Moreover, the integrals of Whittaker’s equation will contain an arbitraryconstant h 0 , representing the given magnitude of the total energy of the system.Thus,q j = ϕ j (q 1 , h 0 , c 1 , c 2 ,…, c 2n−2 ),p j = ψ j (q 1 , h 0 , c 1 , c 2 ,…, c 2n−2 ),…(2)…(3)for j = 2, 3,…, n. We know thatp 1 = −K(q 1 , q 2 ,…, q n , p 2 , p 3 ,…, p n , h 0 ).…(4)Substituting (2) and (3) into equation (4), we findp 1 = ψ 1 (q 1 , h 0 , c 1 , c 2 ,…, c 2n−2 ).…(5)The dependence of the coordinates on the time t is obtained from the followingHamilton equationdqdt1∂H= .…(6)∂p1Integrating (6), we obtaint = dq1( ∂H∂p)+ c 2n−1 , …(7)1where c 2n−1 is the constant of integration, and all the variables in the partialderivative (∂H/∂p 1 ) are expressed in terms of q 1 with the help of the equations(2) to (4)

179ANALYTICAL MECHANICS-IIIdq' jq j = , …(8)dq1for j = 1, 2,…,n.ThenLetq 1 ′ = 1.P = P(q 1 , q 2 ,…, q n , q 2 ′, q 3 ′,…, q n ′)n= p −K …(9)j=2'j q jThe Hamiltonian system (1), by eliminating K and p j from equations (1), (8)and (9), is equivalent to the system of equations of the Lagrangian type :for j = 2, 3,…,n.d P ∂∂P−'dt q ∂ j ∂qj= 0,The system (10) contains (n−1) second-order equations.…(10)Next, we transform the expression for the function P by using (4) for p 1 . WewriteNowbecausenP = p + p 1. …(11)j=2'j q jP = p 2 q 2 ′ + p 3 q 3 ′ +…+ p n q n ′ + p 1dq 2 dq3dq n= p 2 + p3+ ... + pn+ p1dq dq dq111(p q&+ p q&+ ... + p q&) + p= 2 2 3 3 n n 1q&1n1 = piq&i q&1 i=1 1= (L + H),q&11…(12)n p q& − L = H…(13)i=1i i .

180PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSWe know that for a conservative system.L = T−U,H = T + U,…(14)where T is the kinetic energy and U is the potential energy. The kinetic energyT may be written asLetT =1 n a2iki,k=1Equations (12) to (16) yieldq&&q . …(15)iG(q 1 , q 2 , q n , q 2 ′, q 3 ′,…, q n ′) =k12naiki,k=1q' 'iqk…(16)P =2T&q 1…(17)H−U = T = & G.…(18)q 2 1From equations (1a) and (18), we findh − Uq& 1 = , …(19)Gand from equations (1a), (17) and (19), we get the following expression for thefunction P.P = 2 G(h− U). …(20)Definition : The differential equations (10), in which the function P is of theform (20) and which therefore belong to the ordinary conservative systems(natural) are refereed to as JACOBI EQUATIONS.Remark : Integrating Jacobi’s equations, we determine all the trajectories inthe coordinate space (q 1 , q 2 ,…, q n ).q j = ϕ j (q 1 , h, c 1 , c 2 ,…, c 2n−2 ).…(21)The relation between the coordinates and the time variable is established fromequation (19) by integration. We findGt = dq 1 + C 2n −1. …(22)h − U

181ANALYTICAL MECHANICS-III6.5 PRINCIPAL OF LEAST ACTIONThe Jacobi equations areddqfor j = 2, 3,.., n, and1 P ∂∂P−q j'∂∂qj1= 0,…(1)dq' jq j = . …(2)dqHere, q 1 is the independent variable and plays the role of the time.The Lagrangian action, denoted by W*, is defined as'q1W* = Pdq .…(3)q011Here, all the motions of a generalized conservative system that transfer the0'system from a given initial position q i to a specified terminal position q i(figure below). The instants of time t 0 and t 1 are not fixed and may vary whenpassing from the straight-line path to circuitous paths.q 3q i ′q 1q i0q 2Statement (Principal of least action). The variation of the Lagrange actionW* is zero for the straight-line path.Proof : We note that Jacobi equations (1) are Lagrangian type equations. Soby Hamilton principle,δW* = 0for the straight-line path.

182PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSRemark 1 : We havet1W* = ( L + H)dtt0t1= L dt + h dtt0t1t0= W + h(t 1 − t 0 ).Remark 2 : For an ordinary system, we have the Lagrange action W* in thefollowing formW* = 2 1 T dtt0tt1 N= mk=1t0'N s k= mk=1 0skkkvρρv2k2kdtdsk .6.6 LEE HWA − CHUNG’S THEOREMPoincare introduced for the first time, the following integralnI 1 = i=1p i δq i…(1)Round the contour C consisting of simultaneous states of a system. Poincare’sintegral invariant I 1 does not change its value if the contour C is displacedalong the tube of straight-line paths to the contour C′, which again consists ofsimultaneous states.It is convenient toe consider the integral I 1 in the ordinary non extended2n−dimensional phase space (q 1 , p 1 , q 2 , p 2 ,…, q n , p n ). In this the contour D andD′ (figure below) bounding the tube of straight-line paths.

183ANALYTICAL MECHANICS-IIID′Dq 1p np 1Here,n p δq= p δq.i iD i= 1D'=i 1niiThe Poincare integral invariant I 1 is called the universal integralinvariant.In 1947, the Chinese scientist Lee Hwa−Chung proved the uniqueness ofuniversal integral invariants. He demonstrated that any other universal integralinvariant differs by a constant factor from I 1 .Statement of Lee−Hwa Chung Theorem :IfnI′ = [ A i (t, q k , p k ) δq i + B i (t, q k , p k ) δp i ]i=1is a universal relative integral invariant, thenI′ = c I 1 ,where C is a constant, and I 1 is the Poincare integral.Note : The term ‘relative’ means that the domain of integration is a closedcontour.

184PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSChapter-7Analytical Mechanics-IV7.1 CANONICAL TRANSFORMATIONSDefinition :- The transformation of 2n-dimensional spaceq~ ~ i = q i (t, qk , p k ), …(1)p~ ~ i = p i (t, qk , p k ), …(2)for i = 1, 2, 3,…, n, with the condition that∂(q ~1,p~1,q~2 , p~2 ,..., q~n , p~∂(q,p ,q ,p ,...,q , p1122nnn)≠ 0,)…(3)is called canonical if it carries any Hamiltonian systemdqdtidpdti∂H= ,∂pi∂H= − ,…(4)∂qifor i = 1, 2,…,n, again into another Hamiltonian systemd~qdti∂H ~=∂~pi,dp ~dti∂H ~= − ,…(5)∂q~ifor i = 1, 2,…,n.

185ANALYTICAL MECHANICS-IVRemark 1 :- In the transformation (1, 2), the time variable t is considered as aparameterRemark 2 :- In equation (5),is another Hamiltonian function.Remark 3 :- The importance of studying canonical transformations is due tothe fact that these transformations permit replacing a given Hamiltonian system(4) by another Hamiltonian system (5) in which the function H ~ is of a simplerstructure than H.Remark 4 :- Canonical transformations are some time also called contacttransformationsResult :- The set of all canonical transformation form a group. If in a phasespace, we perform two canonical transformation in succession, the resultingtransformation will again be canonical. Further more, a transformation that isinverse to a certain canonical transformation will always be canonical. Theidentity transformationq~i = q i ,p~i = p i ,for i = 1, 2,…,n is canonical. Hence the result.Example 1 :- Consider the transformationq~i = q i ,p~i = p i ,for i = 1, 2,…,n, α ≠ 0, β ≠ 0, and α and β are constant. This transformation iscanonical and it transforms the system (4) into the system (5) withH ~ = αβH.Example 2 :- Consider the transformationq~i = q i ,p~i = p i ,for i = 1, 2, …,n, α ≠ 0, β ≠ 0 and α, β are some constant. This transformationis canonical and it transform the system (4) into the system (5) with

186PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSH ~ = −αβH.Result :- A necessary and sufficient condition for the transformationfor i = 1, 2,…,n withq~ ~ i = q i (t, qk , p k ),p~ ~ i = p i (t, qk , p k ),∂(q ~1,p~1,q~2 , p~2 ,..., q~n , p~∂(q,p ,q ,p ,...,q , p1122nnn)≠ 0)to be canonical is the existence of a generating function F and some constant Cfor whichnn (~pi~qi) − H ~ t = C (piqi ) − Ht−δFi= 1i=1is identically satisfied by virtue of the above transformation.Note 1 :- The constant C is called the valence of the canonical transformationunder consideration. The canonical transformation will be called univalent iffor it.C = 1Note 2 :- δF =ni=1 ∂F ∂qiqi∂F+∂pi ∂Fpi + t∂tNote 3 :- In the literature, only univalent canonical transformations arefrequently considered, and many authors erroneously hold that thesetransformations exhaust all the transformations that carry Hamiltonian systemsagain into Hamiltonian systems.7.2 FREE CANONICAL TRANSFORMATIONDefinition :- A canonical transformation is called a free canonicaltransformation if the inequality∂(q ~1,q~2 ,..., q~n∂(p,p ,..., p12n))≠ 0…(1)

187ANALYTICAL MECHANICS-IVholds additionally.Note 1 :- The inequality (1) for a free transformation ensures the independenceof the quantitiest, q 1 , q 2 ,…, q n , q~ , q~ ,..., q~ ,12which can now be taken as the basic variables. Further the generalizedmomenta p 1 , p 2 ,…, p n can now be expressed in terms of 2n+1 quantities t, q i , q~ifor i = 1, 2,…, n consequently the generating function F for a free canonicaltransformation is represented asnF(t, q i , p i ) = S(t, q i , q~i )…(2)For univalent (c = 1) free canonical transformation we obtain the followingformulas∂S = pi, …(3)∂qi∂ S∂~ = −pi , …(4)q iH ~∂S= H +…(5)∂tin the equation (3) and (4), i = 1, 2,…,.Example :- The canonical transformationq~i = p ip~i = β q ifor i = 1, 2,…,n α ≠ 0, β ≠ 0 is freeRemark :- For a natural system, the coordinates q 1 , q 2 ,…, q n , define theposition of the system, and together with the momenta p 1 , p 2 , …, p n they definethe state of the system, that is the positions and velocities of its points. Thisspecificity of the coordinates is lost in a general-type canonical transformation.The quantities q~1,q~ ~2,...,q n no longer define the position of the system, andonly together with the p~1,p~ ~2,...,p n do they define the state of the system. Thevariables q~1,q~ ~2,...,q n will as before define the position of the system only in

188PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSthe particular case of a point canonical transformation for which the functionsq~ (t, qk , p k ) actually do not contain the momentaifor i = 1, 2,…,n.q~ ~ i = q i (t, qk )Note that subsequently the transformation of an arbitrary Hamiltonian systeminto a system with the function H of simple structure may be effected with theaid of a free canonical transformation. But a free canonical transformation isnot a point transformation. Thus, non point canonical transformations play anessential role in the theory of Hamiltonian systems.7.3 THE HAMILTON−JACOBI EQUATIONLet there be given a holonomic system whose motion obeys thecanonical equations of Hamilton :dqdtidpdti∂H= , …(1)∂pi∂H= − , …(2)∂qifor i = 1, 2, …,n. We shall try to find/determine a free univalent canonicaltransformation such that in the transformed Hamiltonian systemd~qdtdtid~pi∂H ~=∂~ , …(3)pi∂H ~= −∂~ , …(4)qifor i = 1 2,…,n and the function H ~ will be identically zero, i.e.,H ~ ≡ 0.…(5)Using (5), equations (3) and (4) reduce tod~qdtid~pi= 0, = 0 , …(6)dtfor i = 1, 2,…,n. Integrating (6), we get

189ANALYTICAL MECHANICS-IVq~= , p~ ,…(7)i i i =iwhere α i and β i are 2n arbitrary constants. This implies that the new variablesare also identically constant.Knowing the canonical transformation, i.e., the relation between q i , p i andq~ ~ i , p i ; we can express all the q i and p i as functions of the time t and of the 2narbitrary constants α k , β k (k = 1, 2,…,n). That is, we can find the finalequations of motion of the given holomonic system completely, i.e., all thesolutions of the system (1) and (2).We know that for such a univalent free canonical transform, there exists agenerating function S = S(t, q i , q~i ) for whichH ~∂S= H + ,…(8)∂t∂S∂qi= pi,∂S∂~qi= −~pi…(9)for i = 1, 2,…,n.Using (5), (8) and (9) ; we write∂S ∂S+ H t,q ,tq∂ ∂ ii =0 . …(10)The partial differential equation in (10) is called the Hamilton-JacobiequationThe form of this equation is very simple to write down. Given a specificHamiltonian function H = H(t, q i , p i ), the momentum components are formallyreplaced by the partial derivative of S as in equation (9). The result is a firstorder partial differential equation. By assumption, the new generalisedcoordinates q~i are constants. Hence, the form of the generating function isS = S(t, q 1 , q 2 ,…, q n , α 1 , α 2 ,…, α n )…(11)dependent on the original coordinates and possibly time. So S is a function ofn + 1 variables and n parameters. The solution of the Hamilton-Jacobi

190PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSequation (10) is equivalent to finding the solution of the original canonicalequations of motion. Besides the Hamilton-Jacobi equation (10), the condition 2det ∂ S ∂qi∂j≠ 0,…(12)must hold for the generating function. As soon as the generating function S (t,q i , α i ) is found, the formulas (9) will define the required/desired univalent freecanonical transformationDefinition :- The solution of partial differential equation (10) of Hamilton-Jacobi containing n arbitrary constants α 1 , α 2 ,…, α n is called the completeintegral of this equation if the condition (12) is fulfilled.In the above, we have proved the following theorem :Theorem (Jacobi’s Theorem)Statement : If S(t, q i , α i ) is some complete integral of the Hamilton−Jacobiequation∂S+ H t,qt∂ ∂Sq∂i =i0 ,then the final equations of motion of a holomonic system with the givenfunction H may be written in the form∂S∂qi∂S= pi,∂i= ifor i = 1, 2,…, n; and α i and β i are arbitrary constants.Remark :- A knowledge of the complete integral of the partial differentialequation (10) relieves us of the necessity of integrating the system of ordinarydifferential equations in (1) and (2).Conservative SystemWhen the Hamiltonian function H does not depend explicitly on time,then the Hamiltonian itself is a constant of motion.H(q i , p i ) = h…(1)

191ANALYTICAL MECHANICS-IVin which h is an energy constant. In this case the Hamilton−Jacobi equationsreduces to∂S ∂S+ H q ,tq∂ ∂ ii =0,…(2)in whichS = S(t, q 1 , q 2 ,…, q n , α 1 , α 2 ,…, α n )…(3)is the generating function for free canonical transformation with n parametersα 1 , α 2 ,…, α n . From equations (1) and (2)∂S+ h = 0 . …(4)∂tWe assume the following special form of S for time dependence of S.S = −ht + V(q 1 , q 2 ,…, q n , α 1 , α 2 ,…,α n ).…(5)Substituting of (5) into equation (4) gives ∂VHq,q ∂ ii =h . …(6)Note 1 : Equation (6) is called the reduced Hamilton – Jacobi equation.This is a first order partial differential equation in n dependent variables.Note 2: Since the solution of (6) already dependents on the energy parameterh, there is no loss in assigning one of the parameters, say α n ; the constant hthat is, α n = h, soV = V(q 1 , q 2 ,…, q n ; α 1 , α 2 ,…, α n−1 , h).…(7)Note 3: We have the following final equations of motion of a generalisedconservative system.∂V = pi, 1 ≤ i ≤ n, …(8)∂qi∂V= β j , 1 ≤ j ≤ n−1 …(9)∂ j

192PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS∂ V = t + γ …(10)∂hwhere α j , β j , h and γ are arbitrary constants7.4 METHOD OF SEPARATION OF VARIABLESThe Hamilton − Jacobi theory offers an elegant method by which thecanonical equations can be solved by reducing the dynamical problem to thatof finding a solution to a partial differential equation. Unfortunately, there isno general technique for construction of complete solutions to partialdifferential equations.The most practical method, which works for certain classes of differentialequations, is the method of separation of variables. In this method, we searchfor solutions in which the independent variables of the given PDE are groupedtogether so that the original problem converts into a collection of problemsinvolving only ODE.Under certain circumstances, it is feasible to assume certain convenient formsfor the solution to the Hamilton−Jacobi equation. We consider theconservative Hamiltonian for which the generating function S isS = −h t + V,…(1)where ∂vHq i , = h. …(2) ∂qiLetH(q i , p i ) = G[f 1 (q 1 , p 1 ),…, f n (q n , p n )] .…(3)Here, the variables in the expression for the function H are separated.Equations (2) and (3) implysince ∂V ∂VG f 1q1,,...,fnqn , = h, …(4) ∂q1 ∂qn p i =∂V∂q i…(5)

193ANALYTICAL MECHANICS-IVwe now putf i ∂V q i , = α i …(6) ∂qifor i = 1, 2, …, n. The constants in (6) are otherwise arbitrary but by (4), mustsatisfy the relation.G(α 1 , α 2 ,…, α n ) = h.…(7)Solving (5) for∂V, we find∂q i∂V∂q i= F i (q i , α i ) …(8)for i = 1, 2,…, n Consequently, we obtainnV = [ F (q , )dq ]i=1iiii…(9)and thennS = − G(α 1 , α 2 ,…, α n )t + [ F (q , )dq ]i=1iiii…(10)Remark 1. We findi2∂ S∂q∂i∂Fi=∂i…(11)for i = 1, 2,…, n, and∂ S∂q∂i2k= 0…(12)for i ≠ k and i, k, = 1, 2,…n. Hence, the condition.det2∂ S ∂qi∂k≠ 0,…(13)reduces (for this method) to

194PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSn ∂f∏i=1 ∂ii ≠ 0.…(14)Remark 2∂Fi∂i ∂f= ∂pii−1≠ 0,…(15)for i = 1, 2,…,n.Remark 3. The formula (10) defines a complete integral of the reducedHamilton −Jacobi equations for a conservative system.7.5 LAGRANGE BRACKETSLet φ and ψ i be 2n functions of the two variables p and q, for i = 1, 2,…,n. Wedefine[q p] =nj=1∂ , )( j j∂(q,.p)[q p] are called Lagrange brackets.We findn ∂j ∂j ∂j ∂j = − . …(1)j=1 ∂q∂p∂p∂q[q p] = −[p q], …(2)which is an antisymmetric property.Note : Comparing Lagrange brackets with Poisson brackets, we find that therewere two functions φ and ψ of 2n variables q i , p i for Poisson brackets, whereas,there are 2n functions φ i , ψ i of two variables p, q for Lagrange brackets.Canonical character of a transformation in terms of Lagrange bracketsWe shall now derive the necessary and sufficient conditions that must besatisfied by 2n independent functions

195ANALYTICAL MECHANICS-IVq~i = ϕ i (t, q k , p k ),p~i = ψ i (t, q k , p k ), …(1)for i, k = 1, 2,…,n, so that the transformation defined above in (1) should becanonical.We know that the necessary and sufficient condition for the transformation (1)to be canonical is the existence of a generating function F = F(t, q i , p i ) andsome constant c for which the following identity hold.n− H ~ t = c piq i 1n~pi~qii=1=i − Ht− δF(t, q i , p i )…(2)We assume that the transformation (1) is canonical. Then, the identity (2)holds. We take an arbitrary fixed value t = t (so δ t = 0) in (20. We writeni=1~p i δ~ qi=ni=1p i δq i − δF ( t , q i , pi)…(3)But (3) is a defining identity for a transformation that does not contain the timeexplicity,for i = 1, 2,…, n.q~i = φ i ( t , q k , p k ), p~i = ψi ( t , q k , p k ) …(4)Hence, formulas (4) define a canonical transformation with valence C which isindependent of the chosen value of t = tOn the contrary, let it now be given that all transformations obtained from thetransformation (1) by replacing the variable t by various fixed values of t arecanonical and with one and the same valence c. Then, defining the function H ~by the equation∂F∂~qH ~ = CH + + , …(5)∂tn~ i pii= 1 ∂twe get equation (2) from equations (3) and (5). Thus, we find that thetransformation (1) that depends on the time t is canonicalHence, for the time-dependent transformation (1) to be canonical it isnecessary and sufficient that all the time-independent transformations obtainedfrom the transformation (1) by replacing t with an arbitrary value of t be

196PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICScanonical and with one and the same valence c. For this reason, whenestablishing tests for canonical character, we can confine ourselves tocanonical transformations that do not contain the time variable t explicitly :withq~i = φi (q k , p k ), …(6)p~i = ψ i (q k , p k ), …(7)for i = 1, 2,…,n.∂(q ~ ,...q ~1,q~2 n , p~1,...,p~n )∂(q,q ,...,q ;p ,..., p )12n1n≠ 0,…(8)For the above canonical transformation (6) to (8), the defining identity (2) isnow written asnk=1From (7), we write~p δ~qkk= Cnk=1p δqkk − δK (q k , p k )…(9)letn ∂~q ∂~k qk~qk= qi+ pi . …(10)i=1 ∂qi ∂pinΦ i =~ pk=1k∂q~∂qki− Cpi, …(11)nΨ i = ~pk=1k∂q~∂qki, …(12)for i = 1, 2,…,n. Using (10)−(12) in (9), we obtainni=1(Φ i δq i + Ψ i φp i ) = −δK (q k , p k ). …(13)The conditions that the left-hand side of (13) is differential are∂Φ i ∂Φ=∂q ∂qkki, …(14a)

197ANALYTICAL MECHANICS-IV∂Ψi∂Ψ=∂p ∂pkkk∂Φ i ∂Ψ=∂p ∂qiki, …(14b), …(14c)where i, k = 1, 2,…,n. Substituting the expressions for Φ i and Ψ i fromequations (11) and (12) into equations (14a−c), we obtainn ∂q~ j=1∂qji∂p~∂qjk∂q~−∂qjk∂p~∂qji= 0, …(15a)nj=1∂q~∂pji∂p~∂pjk∂q~−∂pjk∂p~∂pji= 0, …(15b)nj=1∂q~∂qji∂p~∂pjk∂q~−∂pjk∂p~∂qji= C δi k , …(15c)for i, k = 1, 2,…,n.Here δi k is the substitution tensor.conditions in (15a−c) can be written asUsing Lagrange brackets, the above[q i q k ] = 0,[p i p k ] = 0,[q i p k ] = C δi k …(16)The equalities (16) express the necessary and sufficient conditions for thetransformation (6) and (7) to be canonical.7.6 JACOBIAN MATRIX OF A CANONICALTRANSFORMATIONSLetQ =∂q∂qik…(1)be the Jacobian matrix of order n. Let

198PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS∂piP = ,∂pkR =∂~q∂pik,∂~piS = …(2)∂qkbe other Jacobian matrices, each of order n. LetM =∂~q∂q11∂~q...∂q∂~q∂p∂~q...∂p..................................∂~q ∂~∂~∂~n qnqnqn... ...∂q1∂qn ∂p1∂pn∂~p1...........................∂q1...................................∂~p∂~npn.....................∂q∂p11n111nnQ= SR P…(3)be the Jacobian matrix of the canonical transformationq~=i φ i(qk, pk),p~i = ψi (q k , p k ). …(4)Let E be a unit matrix of order n. LetJ =OE− EO …(5)be a matrix of order 2n.Thendet J = 1 .…(5A)

199ANALYTICAL MECHANICS-IVSince the transformation (4) is canonical, so the following conditions, in termsof Lagrange brackets, hold :[q i q k ] = 0,[p i p k ] = 0,[q i p k ] = C δi k …(6)Using (6), it can be checked (left as an exercise) thatM′ JM = C J. …(7)For a univalent canonical transformation (c = 1), equation (7) becomesM′ JM = J. …(8)Definition 1: The matrix M that satisfies (7) is called a GENERALIZED −SIMPLICIAL matrix.Definition 2: The matrix M for which (8) hold is called SIMPLICIAL.For such matricesdet M = + C n .That is, simplicial matrices are non-singular.Result :- All generalized−simplicial matrices ( for c ≠ 0) form a group anddet M = + C n .Note :- In view of the above, the test for the canonical character of thetransformations may be stated as“For a certain transformationq~i = q~i (t,q k , pk), p~i =p~ (t,qto be canonical, it is necessary and sufficient that the Jacobian matrix Mcorresponding to this transformation should be generalized−simplicial withconstant valence C”.ik,pk)

200PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS7.7 CONDITION OF CANONICAL CHARACTER OF ATRANSFORMATION IN TERMS OF POISSONBRACKETSWe know that the condition of canonicity of a transformationq~i =(i, k = 1, 2,…) is thatq~ (t,qik,pk),p~i = p~i (t,q k ,p k )…(1)M′ JM = C J , …(2)where C ≠ 0 is the valence of the canonical transformation, M is a 2n×2ngenerlalized − simplicial matrix andJ =O E− E O…(3)in which E is a unit matrix of order n. Matrix M is non-singular. Equation (2)gives(M′) −1 (M′ JM) M −1 = (M′) −1 (c J) M −1or J = C [(M′) −1 J M −1 ]or (M′) −1 J M −1 1= J.C…(4)From equation (3), we find (exercise)J −1 = −J.…(5)Taking inverse of (4) both sides and using (5), we writeM J M′ = C J.…(6)The equality (6) may be considered as obtained from equality (2) on replacingthe Jacobian matrix M by its transpose M′. In view of definition of M,

201ANALYTICAL MECHANICS-IVM = ∂~q ∂q ∂~p ∂qikik∂~q∂p∂~p∂pikik , …(7)the above substitution reduces to replacing the derivatives∂p~∂qik∂q~,∂pik∂p~,∂qik∂p~,∂pik,respectively, by the derivatives∂q~∂qki∂q~,∂qki∂q~,∂pki∂q~,∂pki.That is in each derivative the letters and indices above and below areinterchanged. He know that in terms of Lagrange brackets, the equation (2) isequivalent to the following system of equalities :[q i q k ] = 0,[p i p k ] = 0,[q i p k ] = C δi k …(8)for i , k = 1, 2,…,n. In view of the discussion in the proceeding paragraph, theequation (6) will be equivalent to the following system of equalities ;[q i q k ]* = 0,[p i p k ]* = 0,[q i p k ]* = C δi k …(9)Here, the asterisk (*) indicates that the above mentioned interchange ofderivatives is to be performed within Lagrange brackets. That is,[q i q k ]* = ~n ∂q j=1 ∂qji∂~p∂qjk∂~p−∂qji∂~q∂qjk* n = ∂~qij=1∂qj∂~q∂pkj∂~q−∂pij∂~q∂qkj

202PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS= ( q~i q~k ),…(10a)where ( q~i q~k ) are Poisson brackets of the function q~i and q~k with respect tothe independent variables q 1 , p 1 , q 2 , p 2 , Poisson brackets. Similarly,[p i p k ]* = p~ p~ ),( i k[ kq (q ~i pk]* = i p~ )…(10b)Hence, the conditions of the canonicity of transformation (8) may be written inthe following form in terms of Poisson brackets.( kq~i q~ ) = 0,( kp~i p~ ) = 0,( kfor i, k = 1, 2,…,n.q~i p~ ) = C δi k …(11)7.8 INVARIANCE OF THE POSISSON BRACKETS IN ACANONICAL TRANSFORMATIONSConsider two functionLetφ = φ(t, q i , p i )ψ = ψ(t, q i , p i )…(1)q~ ~ k = q k (pi , q i )p~ ~ k = p k (pi , q i ) …(2)be the canonical transformation and its inverse bep i = p i ( q~k , p~k )q i = q i ( q~k , p~k )…(3)Substituting the function p i and q i in (1) interm of q~k , ~ p k with the help of (3),we can regard these same functions φ and ψ as functions of the variableq~ ~ k , p k . Accordingly, the Poisson brackets of φ and ψ may be evaluated bothwith respect to the variable q i and p i and with respect to the variables q~i , ~ p i .Let (φ ψ) denote the Poissons brackets with respect to variables q i , p i and ( )denote the same with respect to variables q~ , p~ .We assert that (left as an exercise to the readers)ii

203ANALYTICAL MECHANICS-IV(φ ψ) = C ( ), …(4)where C is the valence of the canonical transformationsThe converse of the above assertion also holds. That is, if for any twofunctions φ and ψ, the identity (4) is fulfilled for one and the same constant C ≠0, then the transition from the 2n variables q i , p i to the 2n variables q~i , ~ p i isaccomplished by a canonical transformation with valence C.In particular, for a univalent canonical transformation (C = 1), we have(φ ψ) = ( ). …(5)This proves that the Poisson brackets are invariant under univalent canonicaltransformations.This property of univalent canonical transformations singles out thesetransformations from among all possible transformations of phase space.

204PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSChapter-8Nonlinear First-Order PDE8.1 INTRODUCTIONWe shall study PDE of the formF(Du, u, x) = 0,…(∗)where x∈U⊂ R n , U is open, andu : U →R ,is the unknown, u = u(x). The function F is given.Notation : We writeforThusF = F(p, z, x)= F(p 1 , p 2 ,…,p n , z, x 1 , x 2 ,…, x n )p∈R n , z∈R, x∈U.“p” is the name of the variable for which we substitute the gradient Du(x), and“z” is the variable for which we substitute u(x).We also assume hereafter that F is smooth, and setD p F = ( F ,F ...,F )D z F = F zp1p2,p nD x F = ( F ,F ...,F )x1x2,x nWe are concerned with discovering solutions u of the PDE (∗) in U, usuallysubject to the boundary conditionu = g on Γ, …(∗∗)

NONLINEAR FIRST-ORDER PDE205where Γ is some given subset of ∂U andg : Γ→R…(***)is prescribed.8.2 COMPLETE INTEGRALSConsider the nonlinear first-order PDEF(Du, u, x) = 0.…(1)Suppose first A ⊂ R n is an open set. Assume for each parameterwe have a C 2 solutiona = (a 1 , a 2 ,…, a n ) ∈A,u = u(x ; a)…(2)of the PDE (1). We write(D2a u,D xau)=uauaΜu a12nuuux1a1x1a2x1an.........uxna1uuxna2xnan. …(3)Definition : A C 2 function u = u(x ; a) is called a complete integral in U×A,provided(i) u(x ; a) solves the PDE (1) for each a∈A ,(ii) rank (Da u,D xau)= n ,2for x∈U, a∈A .Remark : The condition (ii) above ensures u = u(x ;a)“depends on all the n independent parameters a 1 , a 2 ,…, a n ”.Example 1. The Clairaut’s equation is the PDE (nonlinear)wherex⋅Du + f(Du) = u,f : R n →R is given.…(1)

206PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSA complete integral of PDE (1) isu(x ; a) = a⋅ x f(a) ,…(2)for all x∈U and a∈R n .Example 2 : The eikonal PDE is (non-linear)|Du| = 1.…(3)A complete integral of PDE (3) isu(x ; a, b) = a ⋅ x + b…(4)for all x∈U ⊂ R n , a∈∂B(0, 1), b∈R.Example 3 : The Hamilton−Jacobi equation from mechanics is, in its simplestform,the PDE (nonlinear)whereis given.u t + H(Du) = 0,H : R n → R…(5)In PDE(5), u depends on x = (x 1 , x 2 ,…, x n ) ∈R n and t∈R.A complete integer of PDE (5) isu(x, t; a, b) = a⋅x − t H(a) + b,…(6)for x∈R n , t ≥ 0 and a∈R n , b∈R.New Solutions as envelopes of Complete IntegralsDefinition : Let u = u(x; a) be a C 1 function of x∈U, a∈A, where U⊂R n andA⊂R m are open sets. Consider the vector PDED a u (x ; a) = 0,…(1)for x∈U and a∈A.Suppose that we can solve (1) for the parameter a as a C 1 function of x, of theform

NONLINEAR FIRST-ORDER PDE207a = φ(x).…(2)ThenD a u(x; φ(x)) = 0,for all x∈U.…(3)We then callv(x) = u(x; φ(x)),x∈U…(4)the envelope of the functions{u(⋅ ; a)} a∈A.Result : By forming envelopes of complete integrals (or of other m-parameterfamilies of solutions), we construct new solutions of given nonlinear first orderPDE.Such solution is called a Singular Integral of given equationF(Du, u, x) = 0.Example : Consider the nonlinear first order PDEu 2 (1 + |Du| 2 ) = 1.…(1)A complete integral of PDE (1) isu(x ; a) = + (1−|x−a| 2 ) 1/2 ,…(2)for|x−a| < 1.…(3)we computeD a u =The vector equation± (x − a)(1−| x − a1/ 2|). …(4)givesD a u = 0 ,…(5)

208PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSThusora = x ≡ φ(x), say.v(x) = u(x, φ(x))v(x) = + 1 ,…(6)…(7)are singular integrals of nonlinear PDE (1).8.3 CHARACTERISTICS METHODConsider basic nonlinear first-order PDEF(Du, u, x) = 0 in U, …(1)subject to the boundary conditionu = g on Γ, …(2)where Γ ⊆ ∂ U andg : Γ→R…(3)are given.We hereafter suppose that F, g are smooth functions.PLAN :- We develop next the method of characteristic, which solves (1) and(2) by converting the PDE into an appropriate system of first order ODE.MethodSuppose u solves (1), (2) and fix any point x∈U. We would like to calculateu(x) by finding some curve lying within U, connecting the point x with a pointx 0 ∈Γ and along which we can compute u.Since the boundary condition (2) says that the function u is known on Γ, thevalue of u at the one end x 0 becomes known.We hope then to be able to calculate u all along the curve, and so in particularat the point x.Let us suppose that this curve is described parametrically by the functionx (s) = (x 1 (s), x 2 (s),…, x n (s))…(4)the parameter s lying in some subinterval of R.

NONLINEAR FIRST-ORDER PDE209Assuming u is a C 2 solution of PDE (1), we define alsoz(s) = u( x (s)).…(5)We also seti.e.,wherep (s) = Du( x (s)),p (s) = (p 1 (s), p 2 (s),…, p n (s)),…(6)…(7)for i = 1, 2,…, n.p i (s) =u ( x (s)), …(8)xiThe function z(⋅) gives the value of u along the curve and p (⋅) determines thevalues of the gradient Du.We must choose the function x (⋅) in such a way that we can compute z(⋅) andp (⋅).For this, we differentiate (8) and writefor i = 1, 2,…,n.nijp&(s) = u (x(s)) x&(s),…(9)j=1xix jHere, dot (⋅) signifies d/ds. The right side of (9) involves the secondderivatives of u. On theother hand, we can also differentiate thePDE (1) with respect to x i , to getn ∂F (Du, u, x) uj=1∂pj x j x i∂F+ (Du,u, x) u∂zx i∂F+ (Du,u, x)= 0, …(10)∂xifor i = 1,2,…,n.We are able to employ this identity (10) to get rid of the “dangerous” secondderivative terms in (9), provided we first set

210PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSj ∂Fx& (s) = (p(s), z(s), x(s)) , for j = 1, 2,…,n. …(11)∂pjAssuming now (11) holds, we evaluate (10) at x = x (s), obtaining from (5),(6), the identity,nj= 1 ∂∂Fpj(p(s),z(s), x(s))ux i x j(x(s))∂Fi+ (p(s),z(s), x(s)) p (s)∂z∂F+ (p(s), z(s), x(s)) = 0 , …(12)∂xifor i = 1, 2,…,n.Substituting from (11) and (12) into equation (9) we writei ∂Fp&(s) = −∂xi(p(s),z(s), x(s))∂Fi− (p(s),z(s), x(s))p (s),∂z…(13)for i = 1, 2,…,n.Finally, we differentiate (5) and obtainz(s) & =n∂u(x(s))= 1 ∂xj jjx&(s)n j ∂F= p (s) (p(s),z(s), x(s)) ,…(14)j=1 ∂pjby using equations (8) and (11).We now summarize equations (11), (13) and (14) and rewrite in vector notationp&(s) = −DxF(p(s),z(s),x(s)) − DzF(p(s), z(s), x(s))p(s) , …(15A)z& (s) = {D F(p(s),z(s), x(s)}.p(s) , …(15B)p

NONLINEAR FIRST-ORDER PDE211x& (s) = D F(p(s), z(s), x(s)).…(15C)pThe system (15) consists of (2n+1) first order ODE. It comprises thecharacteristic equations of the given nonlinear first order PDE (1).The functions1p( ⋅ ) = (p ( ⋅),p( ⋅),...,p ( ⋅)),z (⋅),1x( ⋅ ) = (x ( ⋅),xare called the characteristics.22n( ⋅),...,xn( ⋅)),Remark 1 : In the above, we have proved the following theorem regarding the“Structure of Characteristic ODE”.Statement : Let u∈C 2 (U) solve the nonlinear first-order PDE (1) in U.Assume x(⋅ ) solves the ODE (15c). Then p(⋅)solves the ODE (15a) and z(⋅ )solves the ODE (15b), for those s such that x(s)∈U.Remark 2 : We still need to discover appropriate initial conditions for thesystem of ODE (15), in order that this theorem be useful.Remark 3 : The form of the full characteristic equations can be quitecomplicated for fully nonlinear first order PDE, but sometimes a remarkablemathematical structure emerges.Question 1. Derive the characteristic equations for the linear and homogeneousPDEF(Du, u, x) = b (x) ⋅ Du(x) + c(x) u(x) = 0,…(1)for x∈U. Hence, solve the problemx 1u− x u u in Ux 2 2 = x1u = g on Γ,where U is the quadrant {x 1 > 0, x 2 > 0} andΓ = {x 1 > 0, x 2 = 0} ⊆ ∂ U.Solution Part I.

212PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSWe writenp = Du(x) ∈ R , …(1)z = u(x) ∈ R . …(2)Then , the given equation can be written asSoF(p, z, x) =b (x) ⋅ p + c(x) z = 0 …(3)D p F = b (x). …(4)The characteristic equation (15c) of the Article now becomesx &(s)= b(x(s)),…(5)which is an first order ODE involving only the function x (⋅), and can besolved easily.The characteristic equation (15b) now becomesz &(s)= {b(x(s))}.p(s)…(6)Equations (1), (2) and (6) simplify toz& (s) = −c(x(s))z(s) . …(7)This is a first order linear ODE in z(⋅), once the function x (∗) is known fromODE (5).Thus, equations (5) and (7) comprisex &(s)= b(x(s)),…(8a)z& (s) = −c(x(s))z(s),…(8b)the characteristic equations for the linear first order PDE (3).Part −II : Comparing the given PDE with standard PDE (3), we find12x = (x , x),b = (−x 2 , x 1 ),c = −1.…(9)

NONLINEAR FIRST-ORDER PDE213Thus, the system (8a, b) for this present problem consists of1x&= −x2,and2 =1x& x ,…(10)z & = z.…(11)Solving the system of two ODE in (10), we find (exercise)1x (s) = x coss ,x200(s) = x sin s . …(12)Solution of (11) isz(s) = z 0 e s= g(x 0 ) e s , …(13)where x 0 ≥ 0 and 0 ≤ s ≤ π/2 .Fix a point (x 1 , x 2 ) ∈U. We select s > 0, x 0 > 0 so that(x 1 , x 2 ) = (x 1 (s), x 2 (s)) .This gives x 0 = (x 2 1 + x 2 2 ) 1/2 ,s = tan −1 (x 2 /x 1 ).…(14)Thereforeu(x 1 , x 2 ) = u(x 1 (s), x 2 (s))= z(s)= g(x 0 ) e s= g((x 1 2 + x 2 2 ) 1/2 ) exp [tan −1 (x 2 /x 1 )] , …(15)as the solution of the given boundary-value problem.Article : Derive the characteristic equations for the quasilinear PDE of theformF(Du, u, x) = b (x, u (x)) ⋅ Du(x) + c(x, u(x)) = 0.

214PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSHence, solve the boundary-value problemu + u = u in U ,x1x2u = g on Γ,where U is the half-space {x 2 > 0} and Γ = {x 2 = 0} = ∂U.Solution : Part I.For2p(s) = Du(x(s)) ∈ Rz(s) = u(x(s)) ∈ Rn, …(1)the given nonlinear PDE isF ( p, z, x) = b(x, z) ⋅ p + c(x, z) = 0.…(2)We obtainD p F = b (x, z).…(3)Hence, characteristic equation (15c) now becomesx &(s)= b(x(s),z(s)) . …(4)Characteristic equation (15b) becomesz&(s) = {b(x(s),z(s))} ⋅ p(s)= − c ( x(s), z(s)),…(5)Using (2). So, the characteristic equations for the quasi linear first order PDEconsists of two ODE (4) and (5).Part-II. In this example,b = (1,1),c = −z 2 ,x = (x 1 , x 2 )…(6)ODE equations (4) and (5) become

NONLINEAR FIRST-ORDER PDE2151x&= 1 , …(7)2x&= 1andSolving (7), we get2z & = z . …(8)x 1 (s) = x 0 + s,x 2 (s) = s.…(9)Solving (8), we get (exercise)z(s) ==z01−sz00g(x ), …(10)01−sg(x )where x 0 ∈R, s ≥ 0, provided the denomination is not zero.For a point (x 1 , x 2 ) ∈U, we select s > 0 and x 0 ∈R so that(x 1 , x 2 ) = (x 1 (s), x 2 (s))= (x 0 + s, s) ,i.e., x 0 = x 1 − x 2 ,Thens = x 2 .u(x 1 , x 2 ) = u(x 1 (s), x 2 (s))= z(s)0g(x )= .01−sg(x )…(11)g(x1− x 2 )= …(12)1−x g(x − x )is the solution, provided the denominator is non-zero.212

216PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS8.4 CHARACTERISTIC FOR THE HAMILTON−JACOBIEQUATIONThe general Hamilton−Jacobi PDE isG(Du, u t , u, x, t) = u t + H(Du, x) = 0,…(1)whereDu = D x u = (u , u ,...,u ),x 1x2H : R n →R,x = (x 1 , x 2 ,…, x n ) ∈R n ,t∈R.xnWe sett = x n+1 ,q = (p, p n+1 )z = u(x) ,y = (x, t).…(2)We haveG(q, z, y) = p n+1 + H(p, x)…(3)So, D q G = (D p H(p, x), −1), …(4)D y G = (D x H(p, x), 0),D z G = 0.…(5)…(6)The characteristic equation (15c) of the main article yieldsi ∂Hx&(s) =∂pi(p(s), x(s));for i = 1,2,.., n .n =+x& 1 (s) 1 , …(7)

NONLINEAR FIRST-ORDER PDE217In particular, we can identity the parameter s with the time t.The characteristic equation (15a) for Hamilton − Jacobi PDE (1) reads asi ∂Hp&(s) = −∂xi(p(s), x(s)),p n +& 1 (s) = 0.…(8)( 1 ≤ i ≤ n) .The characteristic equation (15b) is nowz(s) & = D= Dpn+1H(p(s), x(s)) ⋅ p(s) + p (s), H(p(s), x(s)) ⋅ p(s) − H(p(s), x(s), p…(9)using equations (1) and (3).In summary, the characteristic equations for the given Hamilton−JacobiPDE(1) are the following set of ODE.p& (s) = −DH(p(s), x(s)),…(10A)xz& (s) = {D pH(p(s),x(s))}.p(s) − H(p(s), x(s)), …(10B)forx& (s) = D H(p(s), x(s)),…(10c)p1p( ⋅ ) = (p ( ⋅),p2( ⋅),...,pn( ⋅)),and1x( ⋅ ) = (x ( ⋅),x ( ⋅),...,x ( ⋅)),z(⋅ ) .2nDefinition : Equalities (10A) and (10C), i.e.,x& = D H(p, x) , …(11A)pp& = −DH(p, x) , …(11B)are called Hamilton’s Equations.x

218PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSNote 1 : Obverse that the ODE (10B) for z(⋅) is trivial, once x (⋅) and p (⋅) havebeen found by solving Hamilton’s equations (11A, B).Note 2 : The initial-value problem for the Hamilton−Jacobi equation does not,in general, have a smooth solution u lasting for all times t > 0.Initial-value problem for the Hamilton−Jacobi Equations.Problem : Consider the initial-value problem for the Hamilton-Jacobi equation:u t + H(Du) = 0 in R n ×(0, ∞)u = g on R n × {t = 0}.…(1)…(2)Hereu : R n × [0, ∞) → R ,is the unknown, u = u(x, t), andThe HamiltonianDu = D x u = (u x , u ,...,u ) . …(3)1x2xnH : R n → R…(4)and the initial functiong : R n → R,…(5)are given.Remark : We have derived earlier two Hamilton’s ODE. In the next section,we shall derive them from a variational principle.8.5 DERIVATION OF HAMILTON’S ODE FROM AVARIATIONAL PRINCIPLEAssume thatL : R n × R n → R ,…(1)is a given smooth function.We call L to be Lagrangian.We write

NONLINEAR FIRST-ORDER PDE219L = L(q, x) = L(q 1 , q 2 ,…, q n , x 1 , x 2 ,…, x n )…(2)for q ∈ R nand x∈R n andD L = (LqD L = (Lxq1x1,L, Lq2x2,..., L,...,Lqnxn),). …(3)Now fix two points x, y ∈R n s and a time t > 0.We introduce the action functionalin whichtI[ W( ⋅)]= L(w(s), & w(s))ds , …(4)0dw(s)w &(s)= .dsHere, the functional (4) is defined for functionsw (⋅) = (w 1 (⋅), w 2 (⋅),…, w n (⋅)) ,…(5)belonging to the admissible class{ w(0) = y, w(t) = x. }2nA = w( ) ∈ C ([0,t];R )⋅ . …(6)Thus a C 2 curve w(⋅)belongs to A if it starts at the point y at time 0, andreaches the point x at time t.According to the calculus of variations, we shall find a curve x (⋅) ∈A suchthatI[x( ⋅)]=min I[w( ⋅)]. …(7)w( ⋅)∈AThat is, we are seeking a function w(⋅ ) which minimizes the functional I[⋅],given in equation (4), among all admissible functions/candidates w(⋅ ) in classA.Theorem (Euler−Lagrange equations)Statement : Prove that any minimizer x(⋅)belonging to the admissible classA = { w(⋅)∈C 2 ([0, t]; R n ): w (0) = y, w (t) = x}…(1)

220PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSof the action functionalsolves the systemddstI[ w (⋅)] = L (w(s), & w(s))ds , …(2)0& + &(0 ≤ s ≤ t) …(3)− { D L(x(s), x(s)) } D L(x(s), x(s)) 0,q x=of Euler−Lagrange ordinary differential equations.Proof : Choose a smooth functionv : [0, t] → R n…(4)satisfyingandv (0) = v (t) = 0,v = (v 1 , v 2 ,…, v n ).…(5)For τ∈R, definew(⋅ ) = x(⋅ ) + v( ⋅).…(6)Then w(⋅ ) belongs to the admissible class A and x(⋅ ) being the minimizer ofthe action functional, we writeI [ x( ⋅ )] =≤ I[w( ⋅)].…(7)Therefore, the real-valued functioni : R → R…(8)defined byi(τ) = I [ x( ⋅ ) + v( ⋅)], …(9)has a minimum atτ = 0.…(10)Consequently,i′(0) = 0,…(11)

NONLINEAR FIRST-ORDER PDE221provided i′(0) exists.Now, we shall compute this derivative explicitly. We findso thatti(τ) = L [x(s) & + t v(s), & x(s) + v(s)] ds,0ti{ v }ni′(τ) = L ( x&+ τ v, &ix + τ v) v&+ L ( x&+ v, & x + τ v) 0i=1Setting τ = 0 and using the relation (11), we findqin tii0 = { L (x, x)v&+ L (x, & x)v }i=10qi& x ids . …(12)xi ds .Integrating by parts in the first term inside the integral and using conditions in(5), we find 0n t d0 = − ( L (x, x) )i=1dsqi& i + L x i(x, & x) v ds . …(13) This identity (13) is valid for all smooth functions v = (v 1 , v 2 ,…,v n ) satisfyingthe boundary conditions (5), so we must have−dds( L (x, x) ) + L (x, & x) 0,& …(14)q ix i=for all i = 1, 2,…,n, and, 0 ≤ s ≤ t. Hence, in vector form,for 0 ≤ s ≤ 1.−ddsThis completes the proof( D L(x(s), x(s)) ) + D L(x(s), & x(s)) 0,& …(15)q x=Note (1) Equation (15) is a vector equation. It consists of n coupled secondorderODE.Note (2) It is of course possible that a curve x (⋅) ∈A may solve the ELequationswithout necessarily being a minimizer. In such a case, we say thatsolution x (⋅) is a critical point of the action functional I[⋅].

222PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSSo every minimizer of a functional is a critical point, but a critical point neednot be a minimizer.Example : If, we takeL(q, x) =1 2m | q | − φ(x)2where m>0, the corresponding Euler−Lagrange equation isforρm & x(s) = f (x(s))f ρ = −Dφ.This is Newton’s law for the motion of a particle of mass m moving in theforce field f generated by the potential φ.8.6 DERIVATION OF HAMILTON’S ODEAssume that the C 2 function x(⋅ ) is a critical point of the action functional.Thus, it solves the Euler−Lagrange equations (or vector equation)−ddsFirst we set( D L(x(s), x(s)) ) + D L(x(s), & x(s)) 0,& 0 ≤ s ≤ t . …(1)q x=p(s)= D L(x(s), & x(s)), 0 ≤ s ≤ t. …(2)qp (⋅) is called the generalized momentum corresponding to the positionx (⋅) and velocity x&(⋅).We now make the following important hypothesis.Hypothesis : Suppose for all x, p∈R n that the equationp = D q L(q, x) ,…(3a)can be uniquely solved for q as a smooth functions of p and x,q = q(p, x).…(3b)Definition : The Hamiltonian H associated with the Lagrangian L is defined tobe

NONLINEAR FIRST-ORDER PDE223H(p, x) = p q (p, x) −L(q (p, x), x) ,…(4)for p, x ∈R n . The function q (⋅, ⋅) is defined implicitly by (3).We now convert the Euler−Lagrange equations into Hamilton’s equation.We rewrite the Euler−Lagrange equations in terms of p (⋅) and x (⋅). For thispurpose, we state and prove a theorem.Statement : The functions x (⋅) and p (⋅) satisfy Hamilton’s equations :x(s) & = Dpp(s) & = −DH(p(s), x(s))xH(p(s), x(s)), …(5)for 0 ≤ s ≤ t.Furthermore, the mappings→H( p (s), x(s))…(6)is constant.Proof : Let us hereafter write12nq( ⋅ ) = (q ( ⋅),q( ⋅),...,q( ⋅)). …(7)From equation (4), we compute, for 1 ≤ i ≤ n,∂H∂xi(p, x)n= pk=1k∂q∂xki(p, x)∂L−∂qk(q, x)∂q∂xki(p, x)∂L−∂xi(q, x)∂L= − (q, x), …(8)∂x iusing (3). Also∂H∂pi(p, x)i= qn(p, x) + pkk=1∂q∂pki(p, x)∂L−∂qk(q, x)∂q∂pki(p, x)= q i (p, x) …(9)by using again (3).Thus

224PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS∂H∂pi(p(s), x(s))i= q (p(s), x(s))and likewise∂H∂xi(p(s), x(s))= &(s); …(10)x i∂L= −∂xi(q(p(s), x(s)), x(s))∂L= − (x(s), & x(s))∂xid ∂L= − (x(s), & x(s)), ,ds ∂qi = − p& i (s). …(11)Equations (10) and (11) are required Hamilton’s ODE.These equations comprise a coupled system of 2n first order ODE forx i (⋅) and p i (⋅)for 1 ≤ i ≤ n.Finally, we observedH(p(s), x(s))dsn ∂Hi ∂H= p&+i=1∂pi∂xii x& n ∂H ∂H ∂H ∂H= − +−,i=1∂pi ∂xi ∂xi ∂pi = 0 , (12)using (10) and (11). Equation (12) shows that the mapping (6) is constant.This completes the proof.8.7 LEGENDRE TRANSFORMNow we try to find a connection between the Hamilton−Jacobi PDE and thecalculus of variations problem−minimizing of the action functional.To simplify further, we also drop the x-dependence in the Hamiltonian so that

NONLINEAR FIRST-ORDER PDE225H = H(p).….(1)We hereafter suppose that the LagrangianL : R n → R ,satisfies the following conditions :(a) the mappingq → L(q)…(2)is convex and(b)L(q)lim → ∞| q | = + ∞. …(3)|q|Result : The convexity of the mapping in (2) implies L is continuous.Definition : The Legendre transform of L is denoted by L*(p) and is definedasL*(p) =sup {p⋅q − L(q)},q∈Rn…(4)for p∈R n .Theorem : Show that the Hamiltonian H can be obtained from the LagrangianL. Establish the relation.Proof : Suppose that the Lagrangian L satisfies the conditions (2) and (3). LetL* denote the Legendre transform of L, defined in (4).We note that in view of (3), the “sup” in the definition of L* in (4) is really a“max”. That is, there exists some q*∈R n for whichL*(p) = p⋅q*−L(q*),…(5)and the mappingq → p⋅q − L(q)…(6)has a maximum at q = q*.But thenp = DL(q*),…(7)

226PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSprovided L is differentiable at q*. Hence the equationp = DL(q)is solvable for q in terms of p, asq* = q(p) . …(8)Therefore, from equations (5) and (8), we writeL*(p) = p q (p) − L(q(p)).…(9)From, definition of the Hamiltonian associated with the Lagrangian L, we writeH(p) = p⋅ q (p) − L( q (p)), for p∈R n …(10)In equation (10), the x-dependence in the Hamiltonian is dropped forsimplification, as the variable x is not appearing.From equations (9) and (10), we write.HenceH(p) = L*(p) for p∈R n .H = L* .…(11)This gives the formula to obtain the Hamiltonian H from the Lagrangian L,under certain conditions.Theorem : Prove that L = H*, under certain assumptions.Proof : This theorem gives us a formula to compute L, when H is given. Itstates that L is the Legendre transform of H. We have already checked that His the Legendre transform f L under certain conditions.Thus, we shall say that H and L are dual convex functions.To prove the result, we assume that the LagrangianL : R n →R,satisfies the conditionsa) the mappingL = L(q) for q∈R n , …(1)q → L(q)…(2)

NONLINEAR FIRST-ORDER PDE227is convex andb)L(q)lim = + ∞. …(3)q→∞| q |we know that the Legendr transform, L*(p), is defined asL*(p) =sup {p⋅q − L(q)}q∈Rn…(4)for p∈R n . We also know that the Hamiltonian H is given by the formulaH = L*.…(5)To achieve the desired result, we shall show thati) the mappingp →H(p)…(6)is convex andii)H(p)lim = + ∞, …(7)| p| →∞| p |iii) L = H*. …(8)For each fixed q, the functionp→p ⋅q −L(q)…(9)is linear, and consequently, the mappingp→H(p) = L*(p),=sup {p⋅q − L(q)},q∈Rn…(10)is convex, using (4) and (5).Indeed, if 0 ≤ τ ≤ 1, p and pˆ ∈R n , thenH(τp + (1−τ) pˆ ) =sup {(τp + (1−τ) pˆ ) ⋅q − L(q)}q≤ τsup {p ⋅q − L(q)}q

228PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS+ (1−τ)sup { pˆ ⋅ q − L(q)}q= τ H(p) + (1−τ) H( pˆ ) …(11)This proves part (i) in (6) that the mapping is convex.To prove (ii), fix any λ > 0 and p ≠ 0. ThenH(p) =sup {p⋅q − L(q)},q∈Rnusing (10) or (4) and (5). p ≥ λ |p| − L λ , on taking q = | p | pλ ∈ R n| p |Thus,lim≥ λ |p| − max Linf| p| →∞B(0, λ)H(p)≥ λ for all λ > 0.| p |…(12)This proves (ii) in (7).To prove (iii) in (8), equation (10) givesH(p) + L(q) ≥ p⋅q ,…(13)for all p, q ∈R n . Consequently,L(q) ≥sup {p ⋅q − H(p)}p∈Rn= H*(q) .This givesOn the other hand,L(q) ≥ H*(q) for all q∈R n . …(14)H*(q) =supp∈Rnp,q− sup{p ⋅ r − L(r)} , r∈Rn

NONLINEAR FIRST-ORDER PDE229=supp∈Rn inf {p⋅(q−r) + L(r)}r∈Rn ,…(15)by definition of Legendre transform and properties of sup and inf. Since themappingq→L(q)is convex, so there exists s∈R n such thatL(r) ≥ L(q) + s⋅ (r−q), for r ∈R n .…(16)Taking p = s in (15) and using (16), we computeH*(q) ≥ inf {s⋅(q−r) + L(r)}r∈Rn= L(q) .This givesH*(q) ≥ L(q) for all q∈R n . …(17)From equations (14) and (17), we findL(q) = H*(q) for all q∈R nHenceL = H*.…(18)This proves part (iii) in equation (8). Hence, the proof of the theorem iscomplete.8.8 HOPF−LAX FORMULAConsider the initial-value problem for the Hamilton - Jacobi equationu t + H(Du) = 0 in R n × (0, ∞) …(1)u = g on R n × {t = 0}. …(2)We know that the calculus of variations problem with Lagrangian L leads toHamilton’s ODE for the associated Hamiltonian H. Since these ODE are alsothe characteristic equations of the Hamilton-Jacobi PDE, we infer there isprobably a direct connection between this PDE and the calculus of variations.Theorem : (Hopf-Lax formula)

230PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSStatement : If x∈R n and t > 0, then prove that the solution u = u(x, t) of theminimization problemtu(x, t) = inf L(w(s))ds & + g(y) w(0) = y, w(t) = x ,0the infimum taken over all C 1 functionssatisfyingw : [0, t] → R nisw (t) = x,u(x, t) = minty∈Rn x − yL t +g(y) .Proof : Fix any y ∈R n and definesw (s) = y + (x − y),…(1)tfor 0 ≤ s ≤ t. ThenandIt is given thatw (0) = yw (t) = x.…(2)…(3)u(x, t) = inft L(w(s)) & + g(y) w(0) = y, w(t) = x . …(4)0It implies thattu(x, t) ≤ 0L (w(s)) & ds + g(y), …(5)by definition of infimum. Equations (1) and (5) yieldt x − y u(x, t) ≤ L ds + g(y)0 t

NONLINEAR FIRST-ORDER PDE231 x − y = t L + g(y) . t This givesu(x, t) ≤ x − y inf tL∈Rn ty +g(y) . …(6)On the other hand, if w (⋅) is any C 1 function satisfying the conditionw (t) = x,…(7)we have, by Jensen’s inequality (exercise), L1 tt1w(s)ds & ≤ L(w(s))ds. &…(8) t0 t0If we writew (0) = y,…(9)we findand consequently, x − ytt L + g(y) ≤ L(w(s)) & ds + g(y), t 0 x − y inf tL + g(y) ≤ inf{(w(s))ds & + g(y)} ,∈Rn t wy= u (x, t), …(10)by definition (4). Equations (6) and (10) yield the desired Hopf-Lax formulafor the given variational problem stated in the statement of the theorem.This completes the proof of Hopf-Lax formula.Remark : We propose now to investigate the sense in which u so definedabove (as a minimization problem) actually solves the initial-value problem forthe Hamilton-Jacobi PDE.u t + H(Du) = 0 in R n ×(0, ∞)u = g on R n ×{t = 0}.…(1)…(2)

232PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSRecall we are assuming H isi) smooth , …(3)ii)iii)convex, andH(p)lim = + ∞. …(4)| p| →∞| p |We henceforth suppose alsog : R n → R ,…(5)is Lipschitz continuous, i.e.,|g(x) − g(y) | Lip (g) = sup < ∞. …(6)n| x − y x,y∈ R|x≠yOur ultimate goal is showing “Hopf−Lax formula” provides a reasonable“weak solution” of the initial-value problem (1) for the Hamilton-Jacobi PDE.First, we record some preliminary observations/properties of the function u =u(x, t) defined earlier by the Hopf−Lax formula.Lemma 1 : (known as a functional identity)Statement : For x∈R n and 0 ≤ s ≤ t, we have x − y u(x, t) = min (t s) L u(y,s) .y R n − +∈ t − s In other words, to compute u (⋅, t), we can calculate u at time s and then use u(⋅,s) as the initial condition on the remaining time interval [s 1 , t].Proof of Lemma 1 : For y∈R n and 0 < s < t and choose z∈R n so that yu(y, s) = s L−sz + g(z). …(1)by virtue of Hopf-Lax formula.Furtherx − zt s x −= 1− t t −ys + s y − z , …(2) t s

NONLINEAR FIRST-ORDER PDE233and 0 < s < 1.Since L is convex, so we havet x − z s xL ≤ 1− L t t t−−Thus, combining with Hope-Lax formulay s y − z + L . …(3)s t s u(x, t) =minty∈Rn x − y L + g(y) , …(4) t we write x − z u(x, t) ≤ t L + g(z) t x≤ (t − s) L t−−y y − z + s Ls s + g(z) x − y = (t −s) L + u (y, s), …(5) t − s using the relation (1). The inequality (5) is true for each y∈R n . Therefore,relation (5) givesu(x, t) ≤ x − y min(t− s) L + u(y,s) . …(6)y∈Rn t − s Now, it remains to prove that (to complete the proof of Lemma) x − y min(t− s) L + u(y,s) ≤ u(x, t) .y∈Rn t − s …(7)To prove (7), we now choose w such thatand setThen xu(x, t) = t Ly =− wt + g(w), …(8)s s x + 1− w . …(9)t t

234PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSx − wt=x − yt − s=y − w. …(10)sConsequentlyusing (8) and (10). Hence x − y (t −s) L + u(y, s) t − s x − w y − w≤ (t −s) L + sL t s +g(w) , x − w = t L + g(w), …(11) t = u(x, t), x − y min(t− s) L + u(y,s) ≤ u(x, t).y∈Rn t − s …(12)Results (6) and (12) combine together prove the desired result.This completes the proof of LemmaLemma 2 : (Lipschitz Continuity)Statement : The function u is Lipschitz continuous in R n ×[0, ∞) andu = g on R n ×{t = 0}.Proof : For t > 0 and x, xˆ ∈R n . Choose y∈R n such that xt Lusing Hopf-Lax formula. Then−ty + g(y) = u(x, t), …(1) xˆ − z x − y u( xˆ , t) − u(x, t) = Inf tL + g(z) − t L − g(y)z t t ≤ g( xˆ −x +y) −g(y), on taking z = xˆ −x +y≤ Lip (g) {| xˆ −x|},

NONLINEAR FIRST-ORDER PDE235as g is Lip continuous. Henceu( xˆ , t) −u(x, t) ≤ Lip (g) | xˆ −x|. …(2)Interchanging the roles of xˆ and x, we writeu(x, t) − u ( xˆ , t) ≤ Lip(g) | xˆ −x|.…(3)Combining (2) and (3), we write|u( xˆ , t) −u(x, t)| ≤ Lip (g) | xˆ −x|.…(4)Next select x∈R n , t > 0. Choose y = x in Hopf-Lax formula, we discoveru(x, t) ≤ t L(0) + g(x).…(5)Furthermore, by Hopf-Lax formula,u(x, t) =minty∈Rn x − y L + g(y) t x − y ≥ g(x) + min−Lip(g) | x − y | + g L ,y∈Rn t = g(x) −t { Lip(g) | z | −L(z)}maxnz∈R; on taking x−y = tz,= g(x) − tmaxw∈B(0,Lip(g))max{w.z− L(z)}z∈Rn= g(x) −tInequalities (5) and (6) together implymax H. …(6)B(0, Lip(g))|u(x, t) −g(x)| ≤ C t…(7)forC = max|L(0) |, max | H | . B(0,Lip(g)) …(8)Finally select x∈R n , o < tˆ < t. ThenLip (u(⋅, t)) ≤ Lip (g),…(9)

236PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSby virtue of inequality (4) above. Consequently Lemma 1 and calculations likethose employed above imply|u(x, t) − u(x, tˆ )| ≤ C |t − tˆ |,…(10)for the constant C defined in (8).Inequalities (4) and (10) proves the fact that the function u is Lipschitzcontinuous in R n ×[0, ∞). Moreover, inequality (7) proves thatu = g on R n ×{t = 0}.This completes the proof of Lemma 2.Result : By Rademacher’s theorem (proof out of course), it is asserted that aLipschitz function is differentiable almost every where.Consequently, by Lemma 2, our function u = u(x, t) defined above by theHopf-Lax formula is differentiable almost everywhere in R n ×[0, ∞).The next theorem concludes u, in fact, as defined by Hopf-Lax formula, solvesthe Hamilton-Jacobi PDE wherever u is differentiable.Theorem : (Solving the Hamilton-Jacobi equation).Statement : Suppose x∈R n , t > 0, and u = u(x, t) defined by the Hopf-Laxformula is differentiable at a point (x, t) ∈ R n ×(0, ∞).Thenu t (x, t) + H(Du(x, t)) = 0.Proof : Fix q ∈R n , h > 0. Owing to Lemma 1,u(x + hq, t + h) =minhy∈Rn x + h q − y L + u(y, t) h ≤ h L(q) + u(x, t).Henceu(xTaking h→ 0 +, we write+ h q, t + h − u(x, t)≤ L (q) .hq ⋅ Du(x, t) + u t (x, t) ≤ L(q).…(1)

NONLINEAR FIRST-ORDER PDE237SinceH = L*,…(2)therefore,u t (x, t) + H(Du(x, t)) = u t (x, t) +≤ 0,max {q ⋅ Du(x, t) − L(q)}q∈Rn…(3)because the inequality (1) is valid for all q∈R n .In order to prove the required result, it is now enoughto show thatu t (x, t) + H(Du(x, t)) ≥ 0.…(4)To prove (4), we choose z such thatFix h > 0 and setThen xu(x, t) = t Ls = t −h,y =−tz + g(z). …(5)s s x + 1− z . …(6)t t x − zt=y − z,s…(7)and thusu(x, t) −u(y, s) ≥t− x − z L + g(z) t y − z sL + g(z) s x − z = (t−s) L , …(8) t

238PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSusing (7). This givesu(x, t) h − u1− x +t hhtz,t − h x − z ≥ L, t …(9)using (6). Letting h→0+ in (9), we compute x − z . t Du(x, t) x − z + u t (x, t) ≥ L. t Consequentlyu t (x, t) + H(D(u, t)) = u t (x, t) + max {q ⋅ Du(x, t) − L(q)}q∈Rn x − z x − z ≥ u t (x, t) + ⋅ Du(x, t) − L t t ≥ 0.…(10)This prove (4) and hence the theoremWe summarize the above results in the form of following theorem :Theorem (Hopf−Lax formula as solution). The function u (x,t) defined bythe Hopf-Lax formula is Lipschitz continuous, is differentiable a. e. in R n ×(0,∞), and solves the initial-value problem.u t + H(Du) = 0 a.e. in R n ×(0, ∞) ,u = g on R n ×{t = 0} .8.9 WEAK SOLUTIONS, UNIQUENESSSemiconcavityIn view of Theorem above it may seem reasonable to define a weaksolution of the initial-value problem to be a Lipschitz function which agreeswith g on R n ×{t = 0}, and solves the PDE a.e. on R n ×(0, ∞). However, thisturns out to be an inadequate definition, as such weak solutions would not ingeneral be unique.Example : Consider the initial-value problem

NONLINEAR FIRST-ORDER PDE239ut + | uOne obvious solution isu 1 (x, t) ≡ 0.However the functionx2| = 0u = 0in R × (0, ∞)on R × {t = 0}.…(1)…(2)u 2 (x, t) = x− t−x − t0ififif| x | ≥ t0 ≤ x ≤ t− t ≤ x ≤ 0…(3)is Lipschitz continuous and also solves the PDE a.e. (everywhere, in fact,except on the lines x = 0, + t). It is easy to see that actually there areinfinitely many Lipschitz functions satisfying (1).This example shows we must presumably require more of a weak solution thanmerely that it satisfy the PDE a.e. We will look to the Hopf-Lax formula for afurther clue as to what is needed to ensure uniqueness.The following lemma demonstrates that u inherits a kind of “one-sided”second-derivative estimate from the initial function g.Lemma 3: (Semiconcavity). Suppose there exists a constant C such thatg(x + z) −2g(x) + g(x−z) ≤ C|z| 2…(1)for all x, z ∈R n . Define u by the Hopf-Lax formula. Thenu(x + z, t) −2u(x, t) + u(x −z, t) ≤ C |z| 2…(2)for all x, z ∈R n , t > 0.Remark. We say g is semiconcave provided (1) holds. It is easy to check (1)is valid if g is C 2 and sup |D 2 g| < ∞. Note that g is semiconcave if and only ifthe mappingnRx→g(x) − 2C |x|2Proof : Choose y∈R n so thatis concave for some constant C.

240PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSu(x, t) = t Lx−ty + g(y). …(3)Then, putting y + z and y −z in the Hopf-Lax formulas for u(x, + z, t) andu(x−z, t),we findu(x + z, t) −2u(x, t) + u(x −z, t)≤ x − y tL + g(y + z) t x − y − 2 tL + g(y) t x − y + tL + g(y − z) t = g(y + z) −2g(y) + g(y−z)≤ C |z| 2 ,…(4)by (1). This proves the lemma.Note : As a semiconcavity condition for u(x, t) will turn out to be important,we pause to identify some other circumstances under which it is valid. We willno longer assume g to be semiconcave, but will suppose the Hamiltonian H tobe uniformly convex.Definition : A C 2 convex functionH : R n →Ris called uniformly convex (with constant θ > 0) ifni, j=1H (p)ξ i ξ j ≥ θ|ξ| 2 for all p, ξ ∈R n .p i p jWe now prove that if g is not semiconcave, the uniform convexity of H forcesu(x,t) to become semiconcave for time t > 0. This is a kind of mildregularizing effect for the Hopf-Lax solution of the initial-value problem.Lemma 4 : Suppose that H is uniformly convex (with constant θ) ad u(x,t) isdefined by the Hopf-Lax formula. Then

NONLINEAR FIRST-ORDER PDE241for all x, z ∈R n , t > 0.u(x + z, t) −2u(x, t) + u(x −z, t) ≤1 2| z |tProof : We note first using Taylor’s formula that uniform convexity of Himplies p1+ pH 22≤12H(p ) +112H(p2) −8|p 1 −p 2 | 2 . …(1)Next we claim that for the Lagrangian L we have the estimate12L(q1 q + q 1+ , …(2)2 2 81 221 ) L(q 2 ) ≤ L + | q1− q 2 |for all q 1 , q 2 ∈R n . Verification is left as an exercise.Now choose y so that xu(x, t) = t L−ty + g(y). …(3)Then using the same value of y in the Hopf-Lax formulas for u(x + z, t) andu(x −z, t), we calculateu(x + z, t) −2u(x, t) + u(x −z, t)≤ x + z − y tL + g(y) t x − y − 2 tL + g(y) t x − z − y + tL + g(y) t 1 x + z − y 1 x − z − y x − y = 2t L + L − L 2 t 2 t t ≤ 2t18θ2zt21≤ |z| 2 ,θt

242PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSusing (2). Hence the lemma.Now we show that semiconcavity conditions of the sort discovered for theHopf-Lax solution u(x, t) in Lemmas 3 and 4 can be utilized as uniquenesscriteria.Definition : We say that a Lipschitz continuous functionu : R n × [0, ∞) → Ris a weak solution of the initial-value problem:ut+ H(Du) = 0u = gin Ron Rnn× (0, ∞)× {t = 0}…(∗)provided(a) u(x, 0) = g(x) for x∈R n ,(b) u t (x, t) + H(Du(x, t)) = 0 a.e. for (x, t) ∈R n ×(0, ∞), and 1(c) u(x +z, t) −2u(x, t) + u(x −z, t) ≤ C1 + |z| 2 t for some constant C ≥ 0 and all x,z∈R n , t > 0.Next we prove that a weak solution the above initial-value problem is unique,the key point being that this uniqueness assertion follows from the inequalitycondition (c) above.Theorem (Uniqueness of weak solutions). Assume H is C 2 and satisfies theconditionH is convex and H(p)lim = +∞ |p| ∞→ | p |(**)and g is Lipschitz continuous. Then there exists at most one weak solution ofthe initial-value problem (∗).Proof : 1. Suppose that u and u ~ are two weak solutions of (∗) and writew = u− u ~ .…(1)Observe now at any point (y, s) where both u and u ~ are differentiableand solve our PDE, we have

NONLINEAR FIRST-ORDER PDE243w t (y, s) = u t (y, s) −u~t (y, s)= −H(Du(y, s)) + H(D u ~ (y, s))= − 1 0d H(r Du(y, s) + (1−r)D ~ u (y, s)) drdr= − 1 0DH(rDu(y, s) +(1−r) D ~ u (y, s)) dr . (Du(y, s) − D ~ u (y, s))Consequently= −b(y, s) ⋅ Dw(y, s) .w t + b ⋅ Dw = 0 a.e. …(2)2. Write v = φ(w) ≥ 0, where φ : R→[0, ∞) is a smooth function to be selectedlater. We multiply (2) by φ′(w) to discoverv t + b ⋅ Dv = 0 a.e. …(3)3. Now choose ε > 0 and defineu ε = η ε ∗ u,u~ = ηε ∗ u ~ , …(4)where η ε is the standard mollifier in the x and t variables. Then|Du ε | ≥ Lip(u),| Du ~| ≤ Lip( ~ u ),…(5)andDu ε → Du,Du~ → D ~ u …(6)a.e., as ∈→0 .Furthermore inequality (c) in the definition of weak solution impliesD 2 u ε , D 2~ u 1≤ C1 + I …(7) s

244PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSfor an appropriate constant C and all ε > 0, y∈R n , s > 2ε. Verification is left asan exercise.4. Write1b ε (y, s) = 0DH(r Du ε (y, s) + (1−r) D u~ (y, s)) dr. …(8)Then (3) becomesv t + b ε ⋅ Du = (b ε − b) ⋅ Dv a.e.;…(9)hencev t + dv (v b ε ) = (div b ε )v + (b ε −b) ⋅ Dv a.e.…(10)5. Nowdiv b ε =1nεr)(Du ~ ε εH (rDu (1 )(ru (1 r)u ~ ε pkp + −x + −llxkx10 k, l = 1xk) dr≤ C 11 + , …(11) s for some constant c, in view of (5), (7). Here we note that H convex impliesD 2 H ≥ 0.6. Fix x 0 ∈R n , t 0 > 0, and setR = max {|DH(p)| | |p| ≤ max (Lip(u), Lip ( u ~ ))}.…(12)Define also the coneC = {(x, t) |0 ≤ t ≤ t 0 |x −x 0 | ≤ R(t 0 − t)|.…(13)Next writee(t) =B(x 0 ,R(t 0− t))v(x, t) dx…(14)and compute for a.e. t > 0:e& (t) = B(x 0 ,R(t 0 − t))v t dx −R∂ B(x 0 ,R(t 0 −t))v dS

NONLINEAR FIRST-ORDER PDE245= −div (v b ε ) + (div b ε )v + (b ε −b). Dv dx∂ B(x 0 ,R(t 0 −t))−R∂ B(x 0 ,R(t 0 −t))v dS= − v(b ε ⋅ v + R)dS∂ B(x 0 ,R(t 0 −t))+ B(x 0 ,R(t 0 − t))(div b ε )v + (b ε −b) ⋅ Dv dx≤B(x 0 ,R(t 0− t))(div b ε )v + (b ε −b) ⋅ Dv dx 1≤ C 1 + e(t) + t B(x 0 ,R(t 0 − t))(b ε −b) ⋅ Dv dxby (11). The last term on the right hand side goes to zero as ε→0, for a.e. t 0 >0, according to (5), (6) and the Dominated Convergence Theorem. 1e& (t) ≤ C 1 + e(t) for a.e. 0 < t < t 0 . …(15) t 7. Fix 0 < ε < r < t and choose the function φ(z) to equal zero if|z| ≤ ε[Lip(u) + Lip( u ~ )]and to be positive otherwise.Since u = u ~ on R n × {t = 0},Thusv = φ(w) = φ (u− u ~ ) = 0 at {t = ε}.e(ε) = 0.Consequently Gronwall’s inequality and (15) implye(r) ≤ e(ε) eε= 0.r 1 C1+ ds s

246PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSHence|u− u ~ | ≤ ε [Lip(u) + Lip( u ~ )]on B(x 0 , R(t 0 −r)).This inequality is valid for all ε > 0, and sou ≡ u ~in B(x 0 , R(t 0 −r)).Therefore, in particular,u(x 0 , t 0 ) = ~ u (x 0 , t 0 ).This completes the proofIn light of Lemma 3, 4 and Theorem above, we have the following theorem.Theorem : (Hopf-Lax formula as weak solution). Suppose H is C 2 andsatisfies (**), and g is Lipschitz continuous. If either g is semiconcave or H isuniformly convex, thenu(x, t) = x − y mintLy∈Rn t +g(y)is the unique weak solution of the initial-value problem (∗) for the Hamilton-Jacobi equation.Example 1: Consider the initial-value problem :ut+1 2| Du | = 02u = | x |in Rnon Rn× (0, ∞)× {t = 0}.…(1)HereH(p) =1 2| p |21 2L(q) = | q | .2The Hopf-Lax formula for the unique, weak solution of (1) is2| x − y | u(x, t) = min+ | y | .y∈Rn 2t …(2)Assume |x| > t. Then

NONLINEAR FIRST-ORDER PDE2472| x − y | y − x yD y | y | + = +2t(y ≠ 0); …(3) t | y |yxand this expression equals zero if x = y + t, y = (| x | −t)≠ 0.| y || x |ThusIf |x| ≤ t,u(x, t) = |x|− 21 if |x| > t.the minimum in (2) is attained at y = 0. Consequentlyu(x, t) =|x | −t/ 2| x |2t2if | x | ≥ tif | x | ≤ t.Observe that the solution becomes semiconcave at time t > 0, even though theinitial function g(x) = |x| is not semiconcave. This accords with Lemma 4.Example 2 : We next examine the problem with reversed initial conditions : 1 2nut + | Du | = 0 in R × (0, ∞) 2…(1)n u = − | x | on R × {t = 0}.Then2| x − y | u(x, t) = min− | y | .y∈Rn 2t Now2| x − y | y − x yD y | y | − = −2t(y ≠ 0), t | y |yxand this equals zero if x = y − t, y = (| x | + t) . Thus| y || x |tu(x, t) = −|x| − (x ∈R n , t ≥ 0). …(2)2The initial function g(x) = −|x| is semiconcave, and the solution remains so fortimes t > 0.The Books Recommended for Chapter VIII1. L.C. Evans Partial Differential Equations, Graduate Studiesin Mathematics, Volume 19, AMS, 1998.

248PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSChapter-9Representation of SolutionsIn this chapter, we collect together a wide variety of techniques that are sometimesuseful for finding certain more-or-less explicit solutions to various partialdifferential equations, or at least representation formulas for solutions.9.1 SEPARATION OF VARIABLESThe method of separation of variables tries to construct a solution u to a givenpartial differential equation as some sort of combination of functions of fewervariables.In other words, the idea is to guess that u can be written as, say, a sum orproduct of as yet undetermined constituent function, to plug this guess into thePDE, and finally to choose the simpler functions to ensure u really is asolution.This technique is best understood in examples.Example 1 : Let U⊂ R n be a bounded, open set with smooth boundary. Weconsider the initial/boundary-value problem for the heat equationu t −∆u = 0u = 0in U × (0, ∞),on ∂U × [0, ∞),u = g on U ×{t = 0}, …(1)whereg : U→R is given.…(2)We conjecture there exists a solution having the multiplicative formu(x, t) = v(t)w(x) (x ∈U, t ≥ 0) . …(3)That is, we look for a solution of (1) with the variables x = {x 1 ,…,x n } ∈U“separated” fromthe variable t ∈[0, T].We computeu t (x, t) = v′(t)w(x),…(4)

249REPRESENTATION OF SOLUTIONS∆u(x, t) = v(t) ∆w(x).…(5)Hence, equations (1), (4) and (5) implyorv′(t)w(x) − v(t)∆w(x) = 0v'(t)v(t)∆w(x)= , …(6)w(x)for all x∈U and t > 0 such that w(x), v(t) ≠ 0.Now observe the left-hand side of (6) depends only on t and the right hand sidedepends only on x. This is impossible unless each is constant, sayThenandv'(t)v(t)∆w(x)= = (t ≥ 0, x∈U). …(7)w(x)v′ = µv, …(8)∆w = µw.…(9)We must solve these equations (8) and (9) for the unknowns w, v and µ.Notice first that if µ is known, the solution of (8) isv (t) = d e µt…(10)for an arbitrary constant d. Consequently, we need only investigate equation(9).We say that λ is an eigenvalue of the operator −∆ on U (subject to zeroboundary conditions) provided there exists a function w, not identically equalto zero, solving− ∆w= w inw = 0 onU ∂U.…(11)The function w is a corresponding eigenfunction.If λ is an eigenvalue and w is a related eigenfunction, we set

250PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSµ = −λ , …(12)and findu = de −λt w ,…(13)solves the problemu t − ∆u = 0 in U × (0, ∞)u = 0 on ∂U × [0, ∞),…(14)with the initial conditionu(⋅, 0) = d w.…(15)Thus the function u defined by (13) solves problem (1), providedg = d w.…(16)More generally, if λ 1 ,…,λ m are eigenvalues, w 1 ,…, w m correspondingeigenfunctions, and d 1 ,…,d m are constants, thenu =m dkk=1e− k twk…(17)solves (14), with the initial conditionmu(⋅, 0) = k=1 d w…(18)k k .If we can find m, w 1 ,…, etc. such thatm k=1we are done.d = g, …(19)k w kWe can hope to generalize further by trying to find a countable sequence λ 1 ,…of eigenvalues with corresponding eigenfunctions w 1 , ….. so that∞ d w = g in U …(20)k=1for appropriate constants d 1 ,….Then presumablykk

251REPRESENTATION OF SOLUTIONS− k tu = ∞ d kew k…(21)k=1will be the solution of the initial-value problem (1).Remark 1 : This is an attractive representation formula for the solution, butdepends upon(a) our being able to find eigenvalues, eigenfunctions and constants satisfying(21) and(b) our verifying that the series in (21) converges in some appropriate sense.Remark 2 : Note that our solution (13) is determined by the method ofseparation of variables.The more complicated forms (17) and (21) depend upon the linearity of theheat equation.Example 2 : Let us turn once again to the Hamilton-Jacobi equationu t + H(Du) = 0 in R n × (0, ∞) , …(22)and look for a solution u having the formThenif and only ifu(x, t) = w(x) + v(t) (x ∈R n , t ≥ 0). …(23)0 = u t (x, t) + H(Du(x, t))= v′(t) + H(Dw(x)) ,H(Dw(x)) = µ = −v′(t) (x∈R n , t > 0) ,…(24)for some constant µ. Consequently ifH(Dw) = µ,v′(t) = −µ,…(25)…(26)for some µ∈R, thenu(x, t) = w(x) −µt + b…(27)will for any constant b solve (22).In particular, if we choose

252PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSw(x) = a⋅x…(28)for some a∈R n and setµ = H(a), …(29)we discover the solutionalready obtained.u = a ⋅ x −H(a)t + b…(30)9.2 SIMILARITY SOLUTIONSWhen investigating partial differential equations it is often profitable to lookfor specific solutions u, the form of which reflects various symmetries in thestructure of the PDE. We have already seen this idea in our derivation of thefundamental solutions for Laplace’s and the heat equations.Following are some other applications of this important method.Plane and Traveling Waves, SolitonsConsider first a partial differential equation involving the two variables x∈R,t∈R.A solution u of the formu(x, t) = v(x−σt), (x∈R, t∈R) …(1)is called a traveling wave (with speed σ and profile v).More generally, a solution u of a PDE in the n + 1 variables x = (x 1 ,…x n ) ∈R n ,t∈R having the formu(x, t) = v(y⋅x−σt), (x∈R n , t∈R) …(2)is called a plane wave having wavefront normal to y∈R n , velocityprofile v. , and| y |Exponential SolutionsIn view of the Fourier transform, it is particularly enlightening when studyinglinear partial differential equations to consider complex-valued plane wavesolutions of the formwhereu(x, t) = e i(y⋅x+ωt) ,ω ∈C and y = (y 1 ,…,y n ) ∈R n .…(3)

253REPRESENTATION OF SOLUTIONSω being the frequency and{ y ni } i= 1the wave numbers.We will next substitute trial solutions of the form (3) into various linear PDE,paying particular attention to the relationship between y and ω forced by thestructure of the equation.Example 1 : (Heat equation).If u is given by (3), we computeu t − ∆u = (iω + |y| 2 ) u = 0,…(4)providedω = i|y| 2 .…(5)Henceu = eiy⋅x−|y|2t, …(6)solves the heat equation for each y ∈R n .Taking real and imaginary parts, we discover further thatu 1 =2|y|te − cos (y⋅x) , …(7)andare solutions as well.2|y|tu 2 = e − sin (y⋅x) , …(8)Notice in this example that since ω is purely imaginary, there results a real,negative exponential termdissipation.Example 2 : (Wave equation).|y|2te −in the formulas, which corresponds toUpon our substituting (3) into the wave equation, we discoveru tt − ∆u = (−ω 2 + |y| 2 u = 0,…(9)providedω = + |y|.…(10)

254PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSConsequentlyu = e i(y⋅x+|y|t) ,…(11)solves the wave equation.The pair of functionsandu 1 = cos (y⋅x + |y|t) ,u 2 = sin(y⋅x + |y|t) ,…(12)…(13)also solves the same.Since ω is real, there are no dissipation effects in these solutions.Example 3 : (Dispersive equations).We now let n = 1 and substituteinto Airy’s equationi(yx + ωt)u = eu t + u xxx = 0.…(14)We calculateu t + u xxx = i(ω −y 3 ) u = 0,…(15)wheneverω = y 3 .…(16)Thusu =i(yxe+y3t), …(17)solves Airy’s equation.Once again, as ω is real there is no dissipation. Notice however that thevelocity of propagation is y 2 , which depends non-linearly upon the frequencyiyxof the initial value e .Thus waves of different frequencies propagate at different velocities: the PDEcreates dispersion.

255REPRESENTATION OF SOLUTIONSLikewise, if n ≥ 1 and we substituteu = e i(y⋅x+ωt)into Schrodinger’s equationiu t + ∆u = 0,…(18)we computeiu t + ∆u = −(ω + |y| 2 )u = 0.…(19)Consequentlyω = −|y| 2 ,…(20)andu = ei(y⋅x−|y|2t)Again, the solution displays dispersion.Solitons. …(21)We consider next the Korteweg-de Vries (KdV) equation in the formu t + 6uu x + u xxx = 0 in R × (0, ∞), …(22)this nonlinear dispersive equation being a model for surface waves in water.We seek a traveling wave solution having the structureu(x, t) = v(x −σt) (x ∈R, t > 0). …(23)Then u solves the KdV equation (22), provided v satisfies the ODE−σv′ + 6vv′ + v′′′ = 0 . d ' = ,s = x − t …(24)ds We integrate (24) by first noting−σv + 3v 2 + v′′ = a,…(25)a denoting some constant.Multiply this equality by v′ to obtain−σ vv′ + 3v 2 v′ + v′′ v′ = av′,

256PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSand so deduce(v')223 2= −v+ v + av + b , …(26)2where b is another arbitrary constant.We investigate (26) by looking now only for solutions v which satisfyv, v′, v′′ → 0 , as s→ + ∞. …(27)The function u having the form (23), under conditions (27), is called a solitarywave.Then (25), (26) implya = b = 0.…(28)Equation (26) thereupon simplifies to readHence2( 2v')2 = v −v + .2 v′ = + v(σ − 2v) 1/2 . …(29)We take the minus sign above for computational convenience, and obtain thenthis implicit formula for v:v(s)dzs = − + c,1/ 20 z( − 2z)…(30)for some constant c. Now substitutez = 2 sech 2 θ.…(31)It follows thatdzd= −sech 2 θ tanh θ , …(32)and

257REPRESENTATION OF SOLUTIONSz(σ−2z) 1/2 = 3/2sech 2 θ tanhθ. …(33)2Hence (30) becomess =2 θ + c, …(34)σwhere θ is implicitly given by the relation sech 2 θ = v(s).2…(35)We combine (34) and (35) to computev(s) = 2 sech22(s− c), (s∈R). …(36)Conversely, it is routine to check v so defined actually solves the ODE (24).The upshot is thatu(x, t) = 2 sech2σ (x − σt− c) 2, (x∈R, t ≥ 0) …(37)is a solution of the KdV equation for each c∈R, σ > 0.A solution of this form is called a soliton.Note the velocity of the solution depends upon its height.Remark : The KdV equation is in fact utterly remarkable, in that it iscompletely integrable, which means that in principle the exact solution can becomputed for essentially arbitrary initial data.Traveling Waves for a Bistable Equation.Consider next the scalar reaction-diffusion equationwherehas a “cubic-like” shape.u t − u xx = f(u) in R×(0, ∞), …(38)f : R→R

258PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS0 a 1Graph of the function fWe assume, more precisely, f is smooth and verifiesfor some point 0 < a < 1.(a) f(0) = f(a) = f(1) = 0(b) f < 0 on (0, a), f > 0 on (a, 1)(c) f′(0) < 0, f′(1) < 0(d) 1 0f (z)dz > 0 …(39)We look for a traveling wave solution of the formu(x, t) = v(x − σ t),…(40)the profile v and velocity σ to be determined, such thatNow sinceu→0 as x→ − ∞, u→1 as x→ +∞. …(41)f ′ < 0 at z = 0, 1,the constants 0 and 1 are stable solutions of the PDE (and since f′ ≥ 0 at z = a,the constant a is an unstable solution).So we want our traveling wave (40) to interpolate between the two stable statesz = 0, 1 at x = µ ∞.Plugging (40) into (38), we see v must satisfy the ordinary differentialequation.v′′ + σv′ + f(v) = 0 , d ' = , …(42)ds

259REPRESENTATION OF SOLUTIONSsubject to the conditionslim v(s) = 1,s→+∞lim v(s) = 0,s→−∞lim v′(s) = 0.s→±∞…(43)We outline now (without complete proofs) a phase plane analysis of the ODEproblem (42), (43). We begin by settingw = v′ .…(44)Then (42), (43) transform into the autonomous first-order systemv′ = ww′ = −σw −f(v), …(45)withlim(v, w) = (1, 0),s→∞lim (v, w) = (0, 0).s→−∞…(46)Now (0, 0) and (1, 0) are critical points for the system (45), and the eigenvaluesof the corresponding linearizations are1/ 2λ + − σ ± ( σ 2 − 4f '(0))0 = ,2.λ ±1=− σ ± ( σ2− 4f '(1))21/ 2…(47)±±In view of (39c), 0 , 1 are real, with differing sign, and thus (0, 0) and (1, 0)are saddle points for the flow (45).Consequently as “stable curve”, W s approaches (1, 0), as drawn. Furthermore,by calculating eigenvectors corresponding to (47) , we seeW u is tangent to the line w =+ 0 v at (0, 0)W s is tangent to the line w =− 1 (v−1) at (1, 0).…(48)

260PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSwW uW s(0, 0) (1, 0) vStable and unstable curvesNote that± ±0 , 1 , W u and W s depend upon the parameter σ.Our intention is to find σ< 0 so thatW u = W s in the region {v > 0, w > 0}. …(49)Then we will have a solution of (45), (46), whose path in the phase plane is aheteroclinic orbit connecting (0 0) to (1, 0).To establish (49), we fix now a small number ε > 0 and let L denote thevertical line through the point (a + ε, 0).We claimW s ∩ L ≠φ,W u ∩ L ≠ φ,…(50)if σ < 0.To check this assertion, defineand compute2vwE(v, w) = + f (z)2 dz (v, w∈R) …(51)0d E(v(t), w(t)) = w(t) w′(t) + f(v(t)) v′(t)dt= − σw 2 (t). …(52)

261REPRESENTATION OF SOLUTIONSAs σ < 0, we see that E is nondecreasing along trajectories of the ODE (45).Note also the level sets of E have the shapes illustrated below :w(0, 0)(a, 0)(1, 0) vLevel curves of EConsider next the region R, as drawn. The unstable curve enters R from (0, 0)and cannot exit through the bottom, top or left hand side. Using (45), wededuce that W u must exit R through the line L, at a point (a + ε, w 0 (σ)).Similarly we argue W s must hit L at a point (a + ε, w 1 (σ)). This verifies claim(50).We next observew 0 (0) < w 1 (0);…(53)this follows since trajectories of (45) for σ = 0 are contained in level sets of E.We assert further thatw 0 (σ) > w 1 (σ)wRL…(54)(0,0)wv(a,0) (1,0)The region RsLTvThe region S

262PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSProvided σ < 0 and |σ| is large enough. To see this, fix β > 0 and consider theregion S, as drawn.we haveNow along the line segment T = {0 ≤ v ≤ a + ε, w = βv},w' − w − f (v) f (v)= = −− . …(55)v' wvSincef (v)vis bounded for 0 ≤ v ≤ a + ε, we seew' C≥ −− > on T, …(56)v' provided σ < 0 and |σ| is large enough.The calculation (56) shows that W u cannot exit S through the line segment T,and sois sufficiently negative.On the other hand,w 0 (σ) ≥ β(a + ε) if σ = σ(β)w 1 (σ) ≤ w 1 (0) for all σ ≤ 0.Thus we see that (53) will follow once we choose β large enough and then σsufficiently negative.Since w 0 and w 1 depend smoothly on σ, we deduce from (50) and (53) thatthere exists σ < 0 withw 0 (σ) = w 1 (σ).…(57)For this velocity σ there consequently exists a solution of the ODE (45), (46).Hence we have found for our reaction-diffusion PDE (38) a traveling wave ofthe form (40).9.3 TRANSFORM METHODSIn this section we develop some of the theory of Fourier and Laplacetransforms, which provides extremely powerful tools for converting certain

263REPRESENTATION OF SOLUTIONSlinear partial differential equations into either algebraic equations or elsedifferential equations involving fewer variables.Fourier TransformIn this section all functions are complex-valued, and − denote the complexconjugate.Definitions and Elementary PropertiesDefinition of Fourier transform on L 1 .If u ∈L 1 (R n ), we define its Fourier transform1 −ix⋅yû(y) = e(2π)u(x) dx (y ∈ R n ) …(1)n / 2nRand its inverse Fourier transform( 1u(y) =(2π)n / 2nReix.yu(x)dx (y∈R n ). …(2)Since|e +ix⋅y | = 1and u∈L 1 (R n ), these integrals converge for each y∈R n .We intend now to extend definitions (1), (2) to functions u∈L 2 (R n ).Theorem 1: (Plancherel’s Theorem). Assume u ∈ L 1 (R n ) ∩ L 2 (R n ).Thenû , u( ∈ L 2 (R n ) andûL2(Rn)= u( = u .…(3)L2(Rn)L2(Rn)Proof 1: First we note that if v, w ∈L 1 (R n ), thenvˆ , ŵ∈ L ∞ (R n ). Alsov(x)ŵ(x)dx =vˆ(y)w(y)dy , …(4)nRsince both expressions equalnR1(2π)n / 2nRnRe−ix.yv(x) w(y) dxdy.

264PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSFurthermore, (exercise)Consequently if ε > 0 and2n / 2 |y|2−−ix⋅y−t|x| π e dx e4t= (t > 0). …(5)nt R−|x|2v ε (x) = e ,…(6)we have−2|y|4eεvˆε (y) = .…(7)n / 2(2ε)Thus (4) implies for each ε > 0 thatRn2 1−−ε|y|ŵ(y)edy = w(x)e4εdx.n / 2(2ε)…(8)Rn|x|22. Now take u ∈ L 1 (R n ) ∩ L 2 (R 2 ) and setv(x) = u (−x).…(9)Letw = u∗v ∈ L 1 (R n ) ∩ C(R n ) ,…(10)and we find thatn / 2ŵ = (2) û vˆ ∈ L ∞ (R n ). …(11)But1vˆ(y) =(2π)n / 2Rne−ix⋅yu( −x)dx= û (y)…(12)and son / 22ŵ = (2) | û | . …(13)

265REPRESENTATION OF SOLUTIONSNow w is continuous and thusSince|x|1−lim w(x)e4εn / 2dx = (2π)w(0).ε→02(2ε)…(14)n /Rn / 2ŵ = (2) | û | ≥ 0,n2we deduce upon sending ε→0 + in (8) that ŵ is summable, with2Henceŵ(y)dy = (2π) n/2 w(0). …(15)nR22| û | dy = w(0) =u(x)v( −x)dx=| u | dx. …(16)RnThe proof for u ( is similar.Definition of Fourier transform on L 2 .RnIn view of the equality (3), we can define the Fourier transform of a function u∈L 2 (R n )as follows.∞Choose a sequence { u k } k=1 ⊂ L1 (R n ) ∩ L 2 (R n ) withAccording to (3),u k → u in L 2 (R n ).Rnûk− û = u − u = u − u ,jL2(Rn)kjL2(Rn)kjL2(Rn)and thus{ û ∞k } k=1 is a Cauchy sequence in L2 (R n ).This sequence consequently converges to a limit, which we define to be û :û k → û in L 2 (R n ).

266PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSThe definition of û does not depend upon the choice of approximating∞sequence { û k } k=1 . We similarly define u( .Next we record some useful formulas in the following theorem.Theorem 2 : (Properties of Fourier transform).Assume u, v ∈ L 2 (R n ). Then(i)uvdx =ûvˆdy ,…(17)nRnR(ii)D α u = (i y) α ûfor each multi-index α such thatD α u ∈ L 2 (R n ),(iii) (u ∗ v)^ = (2π) n/2 û vˆ ,…(18)…(19)(iv) u =Applications :∨( û) . …(20)The Fourier transform is an especially powerful technique for studying linear,constant-coefficient partial differential equations.Example 1 : (Bessel potentials).We investigate first the PDE−∆u + u = f in R n ,…(21)where f ∈ L 2 (R n ).To find an explicit formula for u, we take the Fourier transform and use (18) toobtain(1 + |y| 2 û (y) = fˆ(y ) (y ∈R n ). …(22)The effect of the Fourier transform has been to convert the PDE (21) into thealgebraic equation (22). The solution of (22) isfˆû = .21+| y |…(23)

267REPRESENTATION OF SOLUTIONSThus∨ fˆu = , …(24)21+| y |and so the only real problem is to rewrite the right hand side of (24) into amore explicit form.Using (19), we seef ∗ Bu = ,n / 2(2)…(25)whereBˆ= 1.21+| y |…(26)We solve for B as follows.Since1a= ∞ −ta0e dt, …(27)for each a > 0, we have1+∞1 2−t(1+|y| )= e2| y |0dt.…(28)ThusB = 11+2| y |∨∞2ix⋅y−t|y|( e dy) dt.1 −t= en / 2n(2π) …(29)RNow if a, b ∈ R, b > 0, and we set0z = b 1/2 ax − i,1/ 22b…(30a)

268PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSwe find∞−∞eiax−bx2edx =b−a24b1/ 2Γe−z2dz,…(30b)Γ denoting the contour Im(z) = −into the real axis, we computea1/ 22b in the complex plane. Deforming ΓΓe2−z∞−∞−x1/ 2dz = e dx = π ;2…(31)and henceThus ∞ −∞e2iax−bx1/ 22−a/ 4b π dx = e .b …(32)Rne2ix⋅y−t|y|dy =n∏ j=1∞−∞e2ix j y j −tyjdyj.n / 22|x|−4tConsequently, we conclude from (29), (33) that π = e…(33)t B(x) =21n / 2∞0e|x|−t−4ttn / 22dt , (x ∈R n ). …(34)B is called a Bessel potential.Employing (25), we derive then the formulau(x) =1(4π)n / 2∞ 0Rne|x−y|−t−4ttn / 22f(y) dydt (x ∈R n ) …(35)for the solution of (21).Example 2 : (Fundamental solution of heat equation).Consider again the initial-vale problem for the heat equation

269REPRESENTATION OF SOLUTIONSu t − ∆u = 0 in R n ×(0, ∞)u = g on R n ×{t = 0}.…(1)We establish a new method for solving (1) by computing û , the Fouriertransform of u in the spatial variables x only.ThusSolution of (2) is (exercise)û t + |y| 2 û = 0 for t > 0 ,û = ĝ for t = 0 . …(2)−t|y|2û = e ĝ.…(3)Consequentlyu =e−t|y|2ĝ∨, …(4)and thereforeg ∗ Fu = ,n / 2(2)…(5)whereFˆ−t|y|2= e .…(6)But then2−t|y|∨n12ix⋅y−t|y|= .n / 2(2π)F = ( e ) e dyRusing (5). We compute2|x|1 −e 4tn / 2= …(7)(2t)

270PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSu(x, t) =1(4πt)n / 2enR2|x−y|−4tg(y) dy. (x∈R n , t > 0). …(8)The Fourier transform has provided us with new derivation of the fundamentalsolution of the heat equation.Example 3 : (Fundamental solution of Schrodinger’s equation).Let us next look at the initial-value problem for Schrodinger’s equationiu t + ∆u = 0 in R n × (0, ∞)u = g on R n × {t = 0}.…(1)Here u and g are complex-valued.If we formally replace t by ‘i t’ in the solution of heat equation, we obtain theformulau(x, t) =1(4πit)n / 2enR2i|x−y|−4tg(y) dy , (x∈R n , t > 0),…(2)12where we interpret i as e 4 .iThis expression clearly makes sense for all times t > 0, provided g∈L 1 (R n ).Furthermore if |y| 2 g∈L 1 (R n ), we can check by a direct calculation that u, givenin (2), solves the differential equation is (1).Let us next rewrite formula (2) as2i|x|ix⋅y2i|y|nRe 4t −u(x, t) = e2te 4t g(y)dy.n / 2(4πit)…(3)Since2i|x|2i| y || e 4t , e 4t|= 1,we can check that if g∈L 1 (R n ) ∩ L 2 (R n ), then

271REPRESENTATION OF SOLUTIONSHence the mappingu ( ⋅ , t) = g (t > 0). …(4)g→u(⋅, t)L2(Rn)L2(Rn)preserves the L 2 -norm. Therefore we can extend formula (2) to functions g ∈L 2 (R n ), in the same way that we extended the definition of Fourier transform.Remark. We call2i|x|1Ψ(x, t) = e 4n / 2(4it)(x ∈ R n , t ≠ 0) …(5)the fundamental solution of Schrodinger’s equation.Note that formula (2), u = g ∗ Ψ, makes sense for all time t ≠ 0, even t < 0.Thus we in fact have solved the problem.iu t + ∆u = 0 in R n × (−∞, ∞)u = g on R n × {t = 0}.…(6)In particular, Schrodinger’s equation is reversible in time, whereas theheat equation is not.Example 4 : (Wave equation).We next analyze the initial-value problem for the wave equationu tt − ∆u = 0in R n × (0, ∞),u = g, u t = 0 on R n × {t = 0}, …(1)where for simplicity we suppose the initial velocity to be zero.Take as before û to be the Fourier transform of u in the variable x∈R n . Then(1) gives2û tt + | y | û = 0 for t > 0,û = ĝ,û t = 0 for t = 0. …(2)This is an ODE for each fixed y ∈R n .We look for a solution having the form

272PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSPlugging into (2) givestû = e (β, γ ∈ C) …(3)γ 2 + |y| 2 = 0 ,…(4)and soγ = + i|y|.…(5)Remembering the initial conditions from (2), we deduceĝ it|y|−it|y|û = (e + e ) . …(6)2Inverting, we findĝ |∨ it|y| −it|yu(x, t) =(e + e ) .2…(7)Consequently, we get the formulau(x, t) =1ĝ(y) i(x⋅y+t|y|) i(x⋅y−t|y|)(e + e ) dy , …(8)2(2π)n / 2nRfor x∈R n , t ≥ 0.Laplace TransformRemember that we write R + = (0, ∞).Definition : If u ∈L 1 (R + ), we define its Laplace transform to beu # (s) =∞−st0e u(t)dt (s ≥ 0). …(∗)Whereas the Fourier transform is most appropriate for functions defined on allof R (or R n ), the Laplace transform is useful for function defined only on R + .In practice this means that for a partial differential equation involving time, itmay be useful to perform a Laplace transform in t, holding the space variablesx fixed.

273REPRESENTATION OF SOLUTIONSExample 1 : (Resolvents and Laplace Transform).Consider again the heat equationv t − ∆v = 0 in U×(0, ∞)v = f on U×{t = 0},…(1)and perform a Laplace transform with respect to time :v # (x, s) =∞−st0e v(x, t) dt (s > 0). …(2)We compute∆v # (x, s) ==∞− ste ∆0∞−ste0∞0v(x, t)dtv (x, t)dtt−st−stt= s e v(x, t)dt + [ e v]= sv # (x, s)−f(x). …(3)Think now of s > 0 being fixed, and write=∞t=0u(x) = v # (x, s) .…(4)Then−∆u + su = f in U .…(5)Thus the solution of the resolvent equation (5) with right hand side f is theLaplace transform of the solution of the heat equation with initial data f.Example 2 : (Wave equation from the heat equation).Next we employ some Laplace transform ideas to provide a new derivation ofthe solution for the wave equation, based upon the heat equation.Suppose u is a bounded, smooth solution of the initial-value problem :

274PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSu tt − ∆u = 0in R n ×(0, ∞),u = g, u t = 0 on R n ×{t = 0}, …(1)where n is odd and g is smooth, with compact support.We extend u to negative times by writingThenNext defineu(x, t) = u(x, −t) if x∈R n , t < 0.u tt − ∆u = 0 in R n × R.…(2)…(3)v(x, t) =1(4πt)1/ 2∞2−s/ 4te−∞u(x, s)ds (x ∈R n , t > 0). …(4)Hencelim v = g ,t→0…(5)uniformly on R n . In addition∆v(x, t) =1(4πt)1/ 2∞2−s/ 4te−∞∆u(x, s)ds==1(4πt)1(4πt)1/ 21/ 2∞2−s/ 4te−∞∞2−s/ 4te−∞u ss (x, s)dsu s (x, s)ds=1(4πt)1/ 2∞−∞ s4t22−12te2−s/ 4tu(x, s)ds= v t (x, t). …(6)Consequently v solves this initial-value problem for the heat equation :v t −∆v = 0 in R n ×(0, ∞),

275REPRESENTATION OF SOLUTIONSv = g on R n ×{t = 0}.…(7)As v is bounded, we deduce that−R n e 4t1v(x, t) = 2 g(y)dy.n / 2 |x y|(4πt)−…(8)We equate (4) with (8), recall (2), and setWe obtain the identityThus∞0∞0u(x,s)eu(x,s)efor all λ > 0, where2−λs2−λs1 λ ds = 2 π 1λ = .4tn−12Rne2−λ|x−y|g(y)n−1∞0dy.nα(n) λ 22−λrn−1ds = e r G(x;r)dr, …(9)2 π G(x ; r) = g(y)dS(y).∂B(x,r)…(10)We will solve (9), (10) for u.To do so, we write n = 2k + 1 and noteHenceλd− (e2r drn−1∞201 22−λr−λre−λr2rn−1) = λe. …(11)G(x;r)dr =∞0λke−λr2r2kG(x;r)dr

276PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS∞0kk( −1)1 d 2 −λr2k= (e ) rG(x;r) drk 2 r dr k1= 1 22k−1−λrr(r G(x;r)) edr,k ∂2 r ∂r…(12)where we integrated by parts k times for the last equality.0∞Owing to (9) (with r replacing s in the expression on the left), we deduce from(12),∞∞1k2 n (n)2−λrαu(x,r)e dr2k−1−λrr (r G(x;r)) e dr.n−1r r0k+120 = ∂ 2∂π…(13)Upon substituting τ = r 2 we see that each side above, taken as a function of λ,is a Laplace transform. As two Laplace transforms agree only if the originalfunctions were identical, we deducen (n) 1∂ 2k−1u(x, t) = t (t G(x, t)).k k+ 1 2 t ∂tk…(14)Now n = 2k + 1 andn / 21k+ 2α(n) = = . n n Γ+ 1Γ+ 12 2 …(15)Since 1 Γ =2 1/ 2, …(16)andΓ(x + 1) = xΓ(x) for x > 0,we compute

277REPRESENTATION OF SOLUTIONSn (n)k2k+1=2k+11/ 2n n Γ+ 12 1==(n − 2)(n − 4)...5.31n. …(17)We insert this deduction (17) into (14) and simplify :u(x, t) =1γn∂∂1t t∂∂tn−32tn−2∂ B(x,t)g dS(x ∈R n , t > 0). …(18)9.4 CONVERTING NONLINEAR INTO LINEAR PDENow we describe several techniques which are sometimes useful forconverting certain nonlinear equations into linear equations.Hopf-Cole TransformationA parabolic PDE with quadratic nonlinearity.We consider first of all an initial-value problem for a quasilinear parabolicequation :u t −a∆u + b|Du| 2 = 0 in R n ×(0, ∞)u = g on R n ×{t = 0},…(1)where a > 0.This sort of nonlinear PDE arises in stochastic optimal control theory.Assuming for the moment u is a smooth solution of (1), we setw = φ(u),…(2)whereφ : R→R…(3)is a smooth function, as yet unspecific.We will try to choose φ so that w solves a linear equation. We havew t = φ′(u)u t ,∆w = φ′(u)∆u + φ′′(u)|Du| 2 ;…(4)…(5)and consequently (1) implies

278PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSφ′(u)u t = φ′(u) [a∆u−b|Du| 2 ]=a∆w − [aφ′′(u) + bφ′(u)] |Du| 2 ,orw t = a∆w,…(6)provided we choose φ to satisfyaφ′′ + bφ′ = 0.…(7)We solve this differential equation (7) by settingφ =−bueaThus we see that if u solves (1), then. …(8)w =−buea…(9)solves this initial-value problem for the heat equation (with conductivity a):wt− a∆w= 0w = e−bgain Ron Rnn× (0, ∞)× {t = 0}.…(10)Formula (9) is the Hopf-Cole transformation.Now the unique bounded solution of (10) isw(x, t) =(4πat)2−|x−y|−b1g(y)e4atean / 2nRdy(x ∈R n , t > 0); …(11)and, since (9) impliesau = − log w,b…(12)we obtain thereby the explicit formulau(x, t) = −a 1logb (4πat)n / 2enR2−|x−y|4at−bg(y)ady(x ∈R n , t > 0) …(13)

279REPRESENTATION OF SOLUTIONSfor a solution of quasilinear initial-value problem (1).Burgers’ Equation with Viscosity.As a further application, we examine now for n = 1 the initial-value problemfor the viscous Burgers’ equation:u t −au xx + uu x = 0 in R × (0, ∞)u = g on R ×(0, ∞).…(14)If we setw(x, t) =x∞ −u(y, t) dy…(15)andh(x) =x∞ −g(y) dy…(16)we havew t − aw xx + 1 2w x = 0 in R×(0, ∞)2w = h on R×{t = 0}.…(17)This is an equation of the form (1) forn = 1, b = 21 .So (13) provides the formula2|x y| h(y) 1w(x, t) = −2a log− −−e 4at 2ady.1/ 2(4πat)R…(18)Sinceu = w x ,we find upon differentiating (18) that

280PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSu(x, t) =∞−∞x − yet∞−∞e− |x−y|4at−|x−y|−4at2−h(y)2ah(y)2adydy(x ∈R, t > 0) …(19)is a solution of problem (14), where h is defined by (16).Potential FunctionsAnother technique is to utilize a potential function to convert a nonlinearsystem of PDE into a single linear PDE.We consider as an example Euler’s equations for inviscid, incompressible fluidflow :(a) u t + u⋅Du = −Dp + f in R 3 ×(0, ∞)(b) div u = 0 in R 3 ×(0, ∞)(c) u = g on R 3 ×{t = 0}. …(20)Here the unknowns are the velocity field u = (u 1 , u 2 , u 3 ) and the scalarpressure p . The external force f = (f 1 , f 2 , f 3 ) and initial velocity g = (g 1 , g 2 , g 3 )are given. Here D as usual denotes the gradient in the spatial variables x = (x 1 ,x 2 , x 3 ). The vector equation 20(a) meansWe will assumeit3u + u u = −p+ f (i = 1, 2, 3). …(21)j=1jix jx iidiv g = 0.…(22)If furthermore there exists a scalar functionh : R 3 × (0, ∞)→R…(23)such thatf = Dh,…(24)we say that the external force is derived from the potential h.We will try to find a solution (u, p) of (20) for which the velocity field u is alsoderived from a potential, say

281REPRESENTATION OF SOLUTIONSu = Dv.…(25)The flow will then be irrotational ascurl u ≡ 0.…(26)Now equations (20) (b) and (26) imply∆v = 0,…(27)and so v must be harmonic as a function of x, for each tie t > 0.Thus if we can find a smooth function v satisfying (27) andDv (⋅, 0) = g,…(28)we can then recover u from v by (25).How do we compute the pressure p? Let us observe that because of (25), wehaveu ⋅Du = 21 D(|Dv| 2 ).…(29)Consequently (20) (a) reads2 Dv +1t | Dv | = D(−p + h).2 Therefore we may takev t + 21 |Dv| 2 + p = h.…(30)This is Bernoulli’s law.But now we can employ (30) to compute p, since v and h are already unknown.9.5 HODOGRAPH AND LEGENDRE TRANSFORMSHodograph TransformThe hodograph transform is a technique for converting certain quasilinearsystems of PDE into linear systems, by reversing the roles of the dependentand independent variables.As this method is most easily understood by an example, we investigate herethe equations of steady, two-dimensional, irrotational fluid flow :

282PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS1 1 2 1x +1x2(a) (σ 2 (u) −(u 1 ) 2 ) u − u u (u u ) + (σ 2 (u) −(u 2 ) 2 ) 01 2x =2 x12x1u 2 x =2(b) u − u 0 , …(31)in R 2 .The unknown is the velocity field u = (u 1 , u 2 ). The functionσ(⋅) : R 2 →R,the local sound speed, is given.The system (31) is quasilinear.Let us now, however, no longer regard u 1 and u 2 as functions of x 1 and x 2 :u 1 = u 1 (x 1 , x 2 ), u 2 = u 2 (x 1 , x 2 ),…(32)but rather regard x 1 and x 2 as functions of u 1 and u 2 :x 1 = x 1 (u 1 , u 2 ), x 2 = x 2 (u 1 , u 2 ).…(33)We have exchanged sub and superscripts in the notation to emphasize theinterchange between independent and dependent variables.According to the Inverse Function Theorem, we can invert equations (32) toyield (33), providedJ =112∂(u, u∂(x, x2))= u1x 1u2x 2− u1x 2u2x 1≠ 0 ,…(34)in some region of R 2 .Assuming now (34) holds, we calculateuu2x1x22= Jx1u1= −Jx,1u2,uu2x11x1= −Jx= Jx2u2u21.…(35)We insert (35) into (31), to obtain(a)(b) x2 2{ σ (u) − u 1 }1u2− x2u1x2u= 0.2+ u u12(x1u2+ x2u1) + ( σ2(u) − u22)x1u1= 0 ,…(36)This is linear system for x = (x 1 , x 2 ), as a function of u = (u 1 , u 2 ).

283REPRESENTATION OF SOLUTIONSRemark : We can utilize the method of potential functions to simplify (36)further. Indeed, equation (36b) suggests that we look for a single function z =z(u) such thatx 1 =z u1x 2 = z u 2.…(37)Then (36a) transforms into the linear second-order PDE22 21 u2u2 1 2 u1u22 = u1u1(σ 2 (u) − u )z + 2u u z + ( (u) − u )z 0.…(38)Legendre TransformA technique closely related to the hodograph transform is the classicalLegendre transform. The idea is to regard the components of the gradient of asolution as new independent variables.Once again an example is instructive. We investigate the minimal surfaceequation Du div = 0,2 1/ 2(1+| Du | )…(39)which for n = 2 may be rewritten as22x =(1 − u )u − 2u u u + (1 + u )u 0. …(40)2 x1x1x1x2x1x2x1x2x2Let us now assume that at least in some region of R 2 , we can invert therelationsto solve forp 1 = u x 1(x 1 , x 2 ),p 2 =u (x x2 1 , x 2 ), …(41)x 1 = x 1 (p 1 , p 2 ),x 2 = x 2 (p 1 , p 2 ).…(42)The Inverse Function Theorem assures us we can do so in a neighborhood ofany point whereJ = det D 2 u ≠ 0.…(43)

284PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSNow definev(p) = x(p)⋅p −u(x(p)),…(44)where x = (x 1 , x 2 ) is given by (42) and p = (p 1 , p 2 ). We find that (exercise)uuux1x1x1x2x2x2= Jv= −Jv= Jvp2p2p1p1,p1p2,,…(45)Upon substituting the identities (45) into (40), we derive for v the followinglinear equation222 p2p21 2 p1p21 p1p=1(1 + p )v + 2p p v + (1 + p )v 0 . …(46)Remark. The hodograph and Legendre transform techniques for obtaininglinear out of nonlinear PDE are in practice tricky to use, as it is usually notpossible to transform given boundary conditions very easily.The Books Recommended for Chapter IX1. L.C. Evans Partial Differential Equations, Graduate Studiesin Mathematics, Volume 19, AMS, 1998.

ATTRACTION AND POTENTIAL-I285Chapter-10Attraction and Potential-I10.1 LAW OF GRAVITATIONThis law states that “every particle in the universe attracts every other particlewith a force which is directly proportional to the product of the masses of theparticles and inversely proportional to the square of the distance betweenthem.”Thus, if m 1 , m 2 denote the masses of two particles and r their distance apart.Then the force of attraction between them ism1mγ2r2,where γ is known as the gravitation constant.Remark I :- This law was discovered by Sir Isaac Newton (1642-1727)Remark II :- Gravitation constant γ measures the attraction of two particles,each of unit mass, at unit distance apart.Remark III :- To avoid a difficulty in defining the distance between twoparticles, we may define a material particle as a body so small that, for thepurposes of our investigation, the distance between different parts of body maybe neglected.Remark IV : The numerical value of γ is115,500,000approximately.Remark V :- If we choose units such that γ = 1. Such units are calledastronomical or theoretical units.Remark VI :- The acceleration f produced by the attraction of a particle ofmass m on a particle at a distance r is given by,mf = γr 2 ,

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 286so that γ = 1, when f, m and r are all unity. Hence, the astronomical unit ofmass is the mass of a particle which by its attraction produces unit accelerationat unit distance.We can find the astronomical unit of mass in grammes by taking the aboveformula for acceleration, which holds good in all systems of units, and putting r= 1 cm, f = 1cm/sec 2 , m = 15,500,000 grammes.If P be a particle of unit mass and Q another particle of mass m. Then the forceof attractionF = γm ×12(PQ)is called the attraction of Q at P, and act along the line PQ towards Q.Q(m)Field of force :P(unit mass)The attraction of a system of particles at a point external to itself is theforce of attraction which the system would exert on a particle of unit massplaced at the point. There must be a definite value for this force at every pointat which a particle can be placed. Thus we arrive at the conception of a fieldof force, or region of space with every point of which there is associated aforce which is definite in magnitude and direction.Remark At the point of equilibrium the definite value of force of attraction iszero.10.2 ATTRACTION OF A SYSTEM OF PARTICLESLet particles of masses m 1 , m 2 , m 3 ,… be situated at points A 1 , A 2 , A 3 ,…whose co-ordinates referred to rectangular axes are (x 1 , y 1 , z 1 ), (x 2 , y 2 , z 2 ),(x 3 , y 3 , z 3 ),A 2 (x 2 , y 2 , z 2 )zZA 1 (x 1 , y 1 , z 1 )oPYXA 3xy

ATTRACTION AND POTENTIAL-I287Let P (x, y, z) be any point in space. Let (X, Y, Z) denote the components ofthe attraction of the given system of particles at point P(x, y, z).Let r 1 , r 2 , r 3 ,…. Denote the distance PA 1 , PA 2 , PA 3 ,….so thatr 1 2 = (x 1 , −x) 2 + (y 1 −y) 2 + (z 1 −z) 2 ,r 2 2 = (x 2 −x) 2 + (y 2 −y) 2 + (z 2 −z) 2 ,r 3 2 = (x 3 −x) 2 + (y 3 −y) 2 + (z 3 −z) 2 ,………………………………..………………………………..………………………………..and the direction cosines of PA 1 , PA 2 , PA 3 ……..are , ,x ,r r r33…………………………..…………………………..33respectively.The attraction at the point P of mass m 1 situated at the point A 1 is m 1 /r 1 2 (ontaking γ=1) and is directed along PA 1 .Therefore, the particle m 1 located at A 1 (x 1 , y 1 , z 1 ) exerts at a force at P(x, y, z),whose components parallel to the axes areX 1 =m r112 x1− x ,Yr11m= r112 y 1 − y,Zr11m= r112 z − 1 z.r1The other particles m 2 , m 3 ,… make like contributions for the attraction atP(x, y, z).

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 288The principle of superposition of fields of force states that “the force exerted ata point by a system of particles is the vector sum of the forces exerted by eachof the particles separately”.So, by the principle of superposition of fields of force, total attraction at P(x, y,z) due to the given system of particles isX = kY = kZ = kmmmkkk(xk3kr(y(zk3krk3kr− x) ,− y) ,− z) ,where k = 1, 2, 3,… and the summation extends to all the attracting particles.10.3 POTENTIALLet particles of masses m 1 , m 2 , m 3 ,… be situated at points A 1 , A 2 , A 3 ,… whoseco-ordinates referred to rectangular axes are (x 1 , y 1 , z 1 ), (x 2 , y 2 , z 2 ),… . Let P(x, y, z) be any point of space. Let r 1 , r 2 , r 3 ,… denote the distance PA 1 , PA 2 ,PA 3 ,……, i.e.,r k 2 = (x k −x) 2 + (y k −y) 2 + (z k −z) 2…(1)for k = 1, 2, 3,….Let us now define a function v(x, y, z) by the formula mV(x, y, z) = k . …(2)k rkThe function V defined in (2) is a function related to a system of attractingparticles having a definite value at every point P of space external to theparticles. It is a function of the co-ordinates (x, y, z) of P and is clearly asingle-valued function, in the sense that it cannot have more then one value ateach point P; for it represents simply the sum of the masses of the separateparticles divided by their respective distances from P. Further, V represents asum which does not depend on the particular system of axes of reference.Now, differentiation of equations (1) and (2) with respect to x gives

ATTRACTION AND POTENTIAL-I289∂V∂x m= −k2 rkk∂rk ,∂x…(3)andr k∂r k= − (x k −x). …(4)∂xusing equation (4) in (3) we obtain∂V∂x= kmk(xk3kr− x) . …(5)But, we know that the component X of the force of attraction at P is given byX =kmk(xk3kr− x) . …(6)Equations, (5) and (6) implysimilarly∂V = X ,∂x…(7)∂V = Y ,∂y…(8)∂V = Z , …(9)∂zwhere Y and Z are other components of the force of attraction.Definition. The function V defined by (2) is called the potential of theattracting particles, or the potential of the field of force.Result (1) :- We have proved that the derivatives of the potential V with regardto x, y, z give the components of attraction at P in the directions of the axes.Result (2) :- Since the directions of the axes can be chosen arbitrarily, itfollows that the space derivative of the potential V in any direction gives thecomponent of attraction in that direction.For verification of result (2), let ∂/∂s denote a differentiation in a direction dS∂x∂y∂zwhose direction cosines are < l, m, n> or < , , > . Then, by chain rule,∂s∂s∂s

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 290∂V∂V ∂x ∂V ∂y ∂V ∂z= + + .∂s∂x ∂s ∂y ∂s ∂z ∂s∂V∂V∂V= l + m + n∂x∂y∂z= l X + mY + nZcomponent of the force of attraction in direction < l, m, n>.Remark 1 :- In the language of vectors the force of attraction say R ρ , is thegradient of the potential V. That is,R ρ = grad V.Remark 2 :- If the potential V of any given distribution of matter can bedetermined, the force of attraction R ρ at any point can be found immediately bytaking the gradient of scalar potential V.Physical Interpretation of Potential V(x, y, z).The total differential of the potential V(x, y, z) isdV =∂V∂V∂Vdx + dy + dz∂x∂y∂z= X dx + Y dy + Z dz. …(1)Hence, by integrating along any path from the point P to the point Q, we getQ ∂x∂y∂zV Q −V P = X + Y + Z ds. …(2) ∂s∂s∂sPBut the integral in R.H.S. of equation (2) represents the work which the forcesof attraction would perform upon a particle of unit mass as it moved along thispath from P to Q. This gives us a measure of potential V in terms of workper unit mass. The potential at any point Q exceeds the potential at any otherpoint P by the work which the forces of attraction would perform upon aparticle of unit mass as it moves along any path from P to Q.Remark (1) : The addition of a constant to the potential V will not affect thevalues of the force components.Remark 2 :We know that the potential V of the attracting particles is definedby

ATTRACTION AND POTENTIAL-I291 mV = k rkk …(3)for k = 1, 2, 3,…From (3) it is clear that potential V vanish at an infinite distance from theattracting matter. So, when the potential is determined by integration fromknown force components (X,Y, Z), the constant of integration may be sochosen as to make the potential vanish at an infinite distance from theattracting matter.On this hypothesis, we see that the potential at a given point P due to a givenpoint P due to a given attracting system is the work that would be done by theattractions of the system on a particle of unit mass as it moves along any pathfrom an infinite distance up to the point considered. Hence the definition ofpotential V as (m/r) leads to this expression for potential in terms of workdone per unit mass.Result : We have seen that the definition of potential V asV = (m/r)leads tot he expression for V in terms of work done per unit mass.Now we shall demonstrate the converse.Let m be the mass of a typical particle of the system situated at the point A.LetPP′ = dsQ P′ P∞dsrA(m)be an element of any path from an infinite distance to the point Q, and letAP = r,AP′ = r + dr.Then so far as the field of force depends upon the particle m at A, its value at Pis

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 292m= ,2rand is directed along PA . The work done by this force on the unit particle as itmoves from P to P′ is m − dr = ds2 r ds =− m dr.2rHence, the total work done by the attraction of the particle A(m) on a unitparticle moving from an infinite distance to the point Q isAQ m = − dr2 r ∞m= . AQBy the principle of superposition of fields, the total work done by theattractions of all the particles of the system is obtained by adding for theirseparate effects, so thatV = mAQgives the potential at Q. The above formula represents the sum of the massesof the separate particles each divided by its distance from Q. Theinterchangeability of the two definitions of potential is thus completelyestablished.Dimensions : Gravitational potential V and potential energy have differentphysical dimensions. The dimensions of potential energy are those of work,i.e., ML 2 T −2 , in terms of the fundamental units of mass, space and time. Thedimensions of gravitational potential V are obtained belowDimensions of the potential V.It is important to remember that we are using astronomical units and omittingthe gravitational constant γ, and though this does not affect the argument whenpotential is defined as work per unit mass. We now use the formula forpotential given below :

ATTRACTION AND POTENTIAL-I293 m V = r …(1)If we want to find the dimensions of gravitational potential V we must restorethe constant γ in (1) and write m V = γ r …(2)Because the constant γ has dimensions. By definition of force of attraction, thequantitym mF = γ2r1…(3)Represents a force and is therefore has dimensions MLT −2 . Equation (3) givesγ = M −1 L 3 T −2…(4)Thus, the dimensions of constant γ areM −1 L 3 T −1 m Hence, the dimensions of γ are r (M −1 L 3 T −1 M ) = L 2 T −2 . = (LT −1 ) 2 . L NOTE :- (1) The potential energy decreases when the work is done and thegravitational potential increases when the work is done.(2) The dimensions of gravitational potential V are those of the square of avelocity.10.4 EQUIPOTENTIAL SURFACES AND LINES OF FORCERegarding the potential V(x, y, z) of a given attracting system as a function ofcoordinates x, y, z the equationV(x, y, z) = constant,…(1)represents a surface over which the potential V is constant. Such surfaces arecalled equipotential surfaces. By the definition of potential V(x, y, z), it isclear that only one equipotential surface passes through any point of space, so

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 294that no two equipotential surfaces can intersect. Also, since the potential V(x,y, z) has a constant value over a equipotential surface, no work would be doneby the attraction on a particle moving on such a surface. Therefore, at everypoint the resultant attraction is normal to the equipotential surface through thepoint. The observation is also obvious from the relationR = grad V.…(2)Definition :- The curve such that the tangent at any point of it is in thedirection of the resultant attractive force at that point is called line of force.Remark : The line of force is at right angles to the equipotential surfaces at alltheir points of intersection. Conversely, a surface which cuts all lines of forceat right angles must be an equipotential surface, because at no point on thesurface is there a component of force tangential to the surface, so that no workwould be done on a particle moving on the surface and there could therefore beno variation in the potential.Continuous BodiesWe now pass from the attraction components (X, Y, Z) and potential V(x, y, z)of a system of separate particles to the attraction components (X, Y, Z) andpotential V(x, y, z) of distributions of matter regarding as continuous bodies.By the principle of super position we can obtain the attraction components andpotential V(x, y, z) of such a body provided that we have a means ofsumming the contributions of the separate particles.It is natural to look to integration to effect the summation, but when weconsider what the process of integration involves, we find that it does not fitthe physical conditions of the problem precisely, and it is only by giving aspecial interpretation to our conception of 'body' that we can justify the use ofintegration. Thus, it is usual to represent the potential V of a continuous bodyby a volume integral, i.e.,V = dv,rwhere dv denotes an element of volume of the body at the distance r from anexternal point P and ρ denotes the density of matter in dv.The process of integration implies that the density of the body is continuous.To justify the use of such an integral it is necessary therefore to suppose that itis applied to a hypothetical continuous distribution of matter occupying thesame region as the body and having at each point a suitably chosen density;this density being found by considering a small but finite volume surrounding

ATTRACTION AND POTENTIAL-I295the point and taking the average through this small volume of the masses of theparticles of the real body contained there in.Attraction of a uniform straight rodLet m denote the mass per unit length of a uniform rod AB of finite length. Itis required to find the components of attraction of the rod AB at an externalpoint, say P.XPYDR′RθEA Q′ Q B MLet the perpendicular from P to AB meet AB in M, which for simplicity wetake on AB produced.MP = p.…(1)consider an element QQ′ of the rod AB, whereMQ = x, QQ′ = dx.…(2)Let∠MPQ = θ.Then, from the triangle PMQ, we havex = p tan θ,dx = p sec 2 θ dθ…(3)…(4)The mass of the element QQ′ of the rod ism dx = m p sec 2 θ dθ.The attraction at P of the element QQ′ of the rod AB is, therefore,mpsec2(PQ)d2along PQ.…(5)

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 296From the triangle PMQ, we havePQ = p secθ.…(6)Combing (5) and (6), the attraction at P of the element QQ′ becomesmdpalong PQ.…(7)Let the components of attraction of the rod AB parallel and perpendicular toBA be X and Y.Let the angles MPA and MPB be α, β.Then, we havemX = sinθ dθ p= pm (cos β − cos α)and=2msinp12 (α+β) sin− 2 , …(8)mY = cosθ dθ p= pm (sin α − sin β)2m α − β α + β = sincosp 2 2 …(9)Let R be the resultant attraction. ThenR =2X + Y22m − = sinp 2 2mp12= sin ( A P B) .…(10)

ATTRACTION AND POTENTIAL-I297The direction f R is given bytan −1 Y X = tan −1 cos − cos sin − sin= tan −1 + tan 2 + = . …(11) 2 Thus, the resultant attraction R acts along the bisector of the angle APB and1makes an angle (α+β) with PM.2Remark 1: The component (8), parallel to the rod. AB, can also be written asX =m m − …(12)PB PAin the sense parallel to BA.Remark 2 : We note that if a circle of centre P and radius PM cuts PA, PQ′,PQ, PB in D, R′, R, E then the attraction at P of the element RR′ of a rod in theform of a circular arc DE of the same line density (mass per unit length) as ABis==mpd2pmdp= attraction of QQ′.Hence the circular arc DE exerts the same attraction as the rod AB.Corollary : If the rod is infinitely long, the angle APB is two right angles andthe resultant attraction is2mp

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 298and perpendicular to the rod.If would appear from this result that if the attracted particle were close to therod, the attraction would be infinite; but this conclusion is not justified becausein the foregoing argument we assumed that every point of an element QQ′ ofthe rod was at the same distance from the point P, and for this to be true whenP is close to the rod it would be necessary for the rod to have no thickness. Tofind the attraction at a point close to a rod of finite thickness it will benecessary to take account of the cross-section of the rod.Potential of a uniform straight rodLet m denote the mass per unit length of a uniform rod AB of finite length. It isrequired to find the potential of rod AB at an external point, say, P.PθAQ′ Q B MLet the perpendicular from P to AB meet AB in M, which for simplicity wetake on AB produced. LetMP = p.Consider an element QQ′ of the rod, whereLet the angle MPQ be θ.MQ = x, QQ′ = dx.From the triangle PMQ, we writex = p tanθ,dx = p sec 2 θ dθ,PQ = p sec θ.The mass of the element QQ′ of the rod ismdx = mp sec 2 θ dθ

ATTRACTION AND POTENTIAL-I299The potential at P is given by the formulamdxV = PQ= 2mpsec dpsec PAB PBA = m log cot cot …(1) 2 2 Let 2l denote the length of the rod AB and,PA = r, PB = r′,and r + r′ + 2l = 2s.Then, PAB PBA cot cot = 2 2 s(s − r' ) s(s − r).(s − r)(s − 2l) (s − r' )(s − 2l)=r + r' + 2l. …(2)r + r' −2lso, from equations (1) and (2), the potential V is expressed as r + r' + 2lV = m log . …(3) r + r' −2lRemark 1 : If the ends A, B of rod are foci of an ellipse passing through P and2a is its major axis, thenor a + l V = m log , …(4) a − l 1+ eV = m log . …(5)1−ewhere e denotes the eccentricity of the ellipse.Hence the potential V is constant over any prolate spheroid of which A, B arethe foci. That is, family of confocal prolate spheroids are the equipotentialsurfaces. Since the normal to an ellipse at any point bisects the angle betweenthe focial distances and a resultant attraction at a point is normal to theequipotential surface, it follows that the resultant attraction at P bisects theangle APB.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 300Remark (2) If the rod AB be of great length and P in the neighbourhood of itscentre, then we may putr + r′ = 222l + p , …(6)where p is small compared with l. Then, from equation (3), we obtainV = m logll22+ p+ p22+ l− l= 2m log l2 + p 2 + l p 2p 2l+= 2m log2l , p neglecting the term (p/l) 2 , we writeV = 2m log 2 −2m log p…(7)By differentiating (7), we get for the attraction of the rod in the direction pincreasing,∂V∂p= −2mp.10.5. THE ATTRACTION AND POTENTIAL OF AUNIFORM LONG ROD WHOSE CROSS-SECTION ISA CIRCLETake a cross-section of the long rod about the middle of its length. Let O bethe centre of the cross-section and P any point inside it.Q′QOθPR R′

ATTRACTION AND POTENTIAL-I301Through the point P draw chords QPR, Q′PR′ making a small angle dθ withone another, and intercepting small arcs QQ′, RR′ on the circle. We canconceive the long rod to be composed of long parallel rods, and take QQ′, RR′as the cross-sections of two of them. Then if m denotes the mass per unit areaof given long uniform rod, mQQ′ and mRR′ denote the mass per unit length ofthe two rods.Let| OQP = | ORP = φ.Using the result for the attraction of a long rod, the attraction at P due to rodthrough QQ′=2(mQQ')PQ= 2m. sec φ dθ …(1)and acts along PQ. Similarly, the attraction at P due to rod through RR′=and acts along PR.2m RRPR= 2m sec φ dθ …(2)Hence, these two rods exert equal and opposite attractions at P; and by dividingup the whole long rod into similar pairs of rods we obtain that its resultantattraction at any internal point is zero. Consequently, the potential must beconstant at all points inside the long rod sufficiently far from its ends, and istherefore equal to the potential at O. But, the potential due to the rod QQ′ at Ois= 2m QQ′ log(2 l). −2m QQ′ log(OQ) …(3)where 2l is the length of the rod. Therefore, for the whole potential, we writeV = 2M log 2l − 2M log a,…(4)a being the radius of the cross-section andM = 2πma.…(5)Now we consider the case when the attraction and potential are to determinedat an external point. Let P′ be an external point. Let P be its inverse point w.r.tthe circle mentioned above. Then

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 302OP. OP′ = a 2 .…(6)Bu the similarity of triangles| OP'Q = | OQP = φ = | OP' R…(7)Q′QOPP′RR′Then, the attraction at the external point P′ of the rod through QQ′ is2m QQ'=P' Q'2mPQdsec=P'Qa= (2m dθ sec φ)OP'…(8)and acts along P′Q. But the resultant attraction of the rod at P′ is clearly alongP′O, and by resolving the attraction of the rod through QQ′ in this direction, we2amget d.OP'Similarly, other rods give like resultants and the whole attraction at P′ is,therefore, equal to4amOP'2M= . …(9)OP'We observe that this is the same as if the whole mass of the circular rod werecondensed into a rod of equal mass along the axis of the cylinder. To find thepotential, we may putand writegivingr = OP′dVdr…(10)2M = − ,…(11)rv = −2M log r + C.…(12)

ATTRACTION AND POTENTIAL-I303In order that V may take the form (4) when r = a, we must haveC = M log 2l .So,V = 2M log 2l − 2M log r.…(13)…(14)10.6 ATTRACTION AND POTENTIAL OF A UNIFORMCIRCULAR DISC AT A POINT ON ITS AXISLet O be the centre of the disc, P a point on its axis Oz at a distance z from O.Let m denote mass per unit area.zPSθQQQ′Divide the disc into concentric rings. Letbe the radius andOQ = xQQ′ = dxThe breadth of one of these rings. The mass of the ring is 2π m x dx, and theattraction at P of each element is got by dividing the mass of the element by(PQ) 2 . But the resultant attraction of the ring is in the direction PO, so that itsmagnitude is=2 mx dx2(PQ)cosθ,where θ is the angle OPQ | . Butx = z tanθdx = z sec 2 θ dθ.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 304So, if α is the angle which a radius of the disc subtends at P, we have for thewhole attraction of the disc=α0 2πmz22tan θsecθcosθ 2 2z sec θdθ= 2π m (1− cosα). …(1)Remark (1) : For an infinite plate we may put α = π /2, so that the attraction ofan infinite plate is 2πm at right angles to itself.Potential : The potential at point P of the ring of radius x is given by2 π mx dx,PQthat the potential of whole disc of radius a at point P on its axis Oz at a distancez from O isaV =2 π mx dxPQ0a= 2πmx dx2z + x02= 2πm {22z + a −z}. …(2)NOTE : The formula (2) gives the attraction in the direction PO as−dVdz= 2m1−z2z+ a2 . …(3)Remark (1) : The formula (2) is equivalent toV = 2πm (SP − OP),…(4)and that this will give the value of V on either side of the disc if, SP, OP denotenumerical lengths.

ATTRACTION AND POTENTIAL-I305Solid Angles and Its UseDefinition : The solid angle of a cone is measured by the area intercepted bythe cone on the surface of a sphere of unit radius having its centre at the vertexof the cone.Definition : The solid angle subtended at a point by a surface of any form ismeasured by the solid angle of the cone whose vertex is at the given point andwhose base is the given surface.γPdScross-sectionP′dωOLet PP′ be a small element of area dS which subtends a solid angle dω at O.Let the normal to area dS make an acute angle γ with OP, and let OP = r. Thenthe cross-section at P of the cone which dS subtends at O is dS cos γ, and thiscross-section and the small area dw intercepted on the unit sphere (havingcentre at o) are similar figures, so thatdScos γdω=2r1…(1)Then from (1), we finddω =dScos , …(2)2rordS = r 2 sec γ dω.…(3)Integrating equation (3) both side we obtainS = r 2 sec γ dω…(4)with suitable limits of integration. Hence the area of a finite surface can berepresented as an integral over a spherical surface.

PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 30610.7 USE OF SOLID ANGLESThere are many applications of the theory of attraction in which calculationsare simplified by the use of the solid angle.Article : Find the component of attraction perpendicular to itself produced by aplane plate of any form.Let dS be an element of area of the plate at a distance r, from the point O andsubtending a solid angle dω at O.OθdωrdSplateLet m denote the mass per unit area of the plate. Then the attraction of elementdS at the point O is given by=mdS2 .r…(5)Resolving attraction in (5) at right angles to the plate, we getm dScos,2r…(6)where θ is the inclination of r to the normal to the plate. From (2) and (6), weconclude that m dω is the contribution of area element dS of the plate to itswhole attraction at O at right angles to itself. Further, the attraction of thewhole plate in the same direction is m ω, where ω is the solid angle which theplate subtends at O. This completes the article.Article : Prove that the potential of a solid of uniform density ρ at an externalpoint P can be represented by a surface integral1 ρ cos θ dS2over the surface of the solid, where θ is the angle between the inward normal todS and the line joining dS to P.

ATTRACTION AND POTENTIAL-I307Proof :- Let a cone of small solid angle dω and vertex P cut the surface of thesolid in elements of area dS 1 , dS 2 at A, B, where the inward normals makeangles Q 1 , Q 2 with the line BAP.θ 1APdωθ 2Let AP = r 1 , BP = r 2 . The mass of an element of volume of the cone at adistance r from the point P is ρr 2 dω dr.Hence, the mass of the cone between A and B produces at P a potential equaltoBρr 2 dωdr=rρr dω dr1 2 22 1= ρ( r − r ) dω212= [ dS cos dS cos ]+ .2 2 1 1If we take the sum for all cones which intersect the solid, we shall get12 ρ cos θ dSas the potential at P, integration is being taken over the surface of the givensolid.The Books Recommended for Chapters X and XI.1. A.S. Ramsey Newtonian Gravitation, ELBS and CambridgeUniversity Press.

308PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSChapter-11Attraction and Potential-II11.1 ATTRACTION AND POTENTIAL ATINTERNAL POINTSIn the previous chapter we confined our attention to the attraction and potentialat points external to the attracting matter.We have now to consider the case of attraction and potential at points insidethe attracting matter.Our definitions of attraction and potential at an external point imply theexistence of a separate attracted particle at the point under consideration. Sucha particle cannot exist inside a continuous body because two particles cannotoccupy the same space simultaneously.We therefore imagine that there is a small cavity in the body surrounding theattracted particle placed at the chosen point. We assume that we can calculatethe attraction and potential at this point by our former rules, since the attractedparticle is not in contact with the matter. Then we define the attraction andpotential at the same point in the continuous body to the limits to which theattraction and potential at the point in the cavity tend as the cavity decreases insize and ultimately vanishes.11.2 ATTRACTION AND POTENTIAL OF UNIFORMTHIN SPHERICAL SHEELLet m be the mass per unit area of a thin spherical shell of radius a andcentre O.Q′ dSOPQR R′dS′(A) ATTRACTION AT AN INTERNAL POINT :-Let p be any internalpoint. with p as vertex construct a cone of small solid angle dω intersecting thesurface in elements QQ′, RR′ of areas dS, dS′. The attractions at P of theseelements are [mdS/(QP) 2 ] and [mdS′/(RP) 2 in opposite directions. But

309ATTRACTION POTENTIAL - IIanddS = (QP) 2 (sec/OQP)dω,dS′ = (RP) 2 (sec(ORP) dω,and the angles | OQP, | ORP are equal. So, the elements QQ′, RR′ of thespherical shell exert equal and opposite attractions at the point P. Since thewhole shell can be divided into similar pair of elements by taking cones in alldirections round the point P, it follows that the resultant attraction of thespherical shell at the internal point P is zero.(B) Attraction at an external point :Let P′ be any external point of the spherical shell and P its inverse, so thatOP⋅OP′ = a 2The resultant attraction at P′ is, by symmetry, along P′O and the element QQ′exerts an attraction {mdS/(QP′) 2 } along P′Q. Resolving this attraction in thedirection P′O, we get2m(QP) (sec| OQP)dω= cos| OP' Q .2(QP')But the triangles QPP′ and OQP′ are similar, soQP OQ = ,QP' OP'and the angles OQP. OP′Q are equal. Therefore, the element QQ′ contributesan amount {ma 2 dw/(OP′) 2 } to the resultant attraction. By taking cones in alldirections round P, we get for the attraction at P.Q′QO PR R′of the whole shell24 ma=2(OP')M=2 ,(OP')P′

310PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSwhere M denotes the mass of the spherical shell. It follows that the attraction ofthe shell at external points is the same as if its mass were collected into aparticle at its centre.(C) Potential at an internal point: Since the attraction is zero throughout theinterior of the shell, there can be no variation in the potential, or the potential isconstant. The potential at every point in the interior is, therefore, the same asthe potential at the centre, i.e., M/a, where M denotes the whole mass, sinceevery element of M is at the same distance a from the centre O.(D) Potential at an external point : Let OP′ = r. Since the force at distance ris M/r 2 in the direction in which r decreases, therefore,dvdrIntegrating, we obtainV =M= − .2rM + crwhere c is a constant of integration. But the potential vanishes at an infinitedistance, therefore,c = 0MHence, V = .r11.3 ATTRACTION OF A THIN UNIFORM SPHERICALSHELL AT A POINT OF ITSELFThe attraction at a point of itself of a thin layer of matter depends on the shapeof the gap in the surface in which the attracted particle is placed. We maydefine the principal value of the attraction as the limiting value of the attractionat the centre of a circular hole when its radius tends to zero.Let m be the mass per unit area of the spherical shell, and omitting a smallcircular element of the shell surrounding a point P, consider the attraction ofthe rest of the shell at the point P.Q′QOPLet an element QQ′ of area dS subtend a solid angle dω at P. The attraction ofthe element dS at P is

311ATTRACTION POTENTIAL - IIm dS2(PQ)along PQ. Resolving this attraction along the direction PO, the direction of theresultant attraction, we getm dS(cos| OPQ)2(PQ)=OQP)m dS(cos|= m dω.2(PQ)If now we allow the gap in the spherical shell round P to shrink to vanishingpoint, resultant we see that we have to take the sum m dω for all cones onone. Side of the tangent plane at P. So, the resultant attraction is 2πm.11.4 ATTRACTION AND POTENTIAL OF A UNIFORMSOLID SPHERELet a be the radius and ρ the density of the sphere. Such a sphere may beregarded as composed of a series of concentric thin spherical shells, and therequired results may be obtained by summation.(A) Attraction at an interval point : Take a point P at a distance r, 0 < r < a,from the centre. Imagine a thin spherical shell of matter of radii r +∈ and r−∈to be removed and consider the attraction at a point P in this cavity.The concentric shells external to the cavity exerts no attraction at the point P,and those internal to the cavity attracts as though their masses wereconcentrated at the centre O.Hence, the attraction at P is the limit as ∈→0 of4π ρ(r− ∈)33 r 2o rawhich is equal to4 π ρr. This shows that the attraction of a uniform solid3sphere at an internal point is directly proportional to the distance from thecentre.

312PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS(b) Attraction at an external point. Since each of the concentric sphericalshells attracts at an external point as though its mass were collected at itscentre, the same is true of the solid sphere. The attraction of the solid sphere is,therefore, represented by M/r 2 , where M is its mass and r the distance of anexternal point from the centre of the solid sphere.(c) Potential of a uniform solid sphere at an internal pointAdopting the method of finding the attraction of a solid sphere at an internalpoint, let R denote the radius of a shell external to the cavity. Its mass is4πρR 2 dR, so that the potential it produces at a point inside itself is 4πρRdR.Consequently, the potential at P due to all such external shells is given byar+∈4πρRdR = 2π{a 2 − (r + ∈) 2 }.Also, the shells of radius less than that of the cavity produce the same potentialas if the mass were collected at 0, i.e.,4 3(r−∈)3(r − ∈).Hence, the whole potential at P is the limit as ∈→0 of4 (r− ∈)3 2= 323+2(3a 2 − r 2 ).22[ a − (r+ ∈)] (d) At an external point. Since each of the concentric shells produces at anexternal point a potential equal to its mass divided by the distance of the pointfrom the centre, the same is true for the solid sphere. That is, the potential V atan external point P, OP = r, due to a uniform solid sphere with centre O isV = M/r,where M is the total mass of the uniform solid sphere.Exercise : Deduce the expression for the potential of a uniforms solid spherefrom the attraction.11.5 WORK DONE BY SELF-ATTRACTING SYSTEMSLet the component particles be m 1 , m 2 ,…, and let A 1 , A 2 ,… be their positionsin the given system. Letr st = distance between m s and m t .

313ATTRACTION POTENTIAL - IIFirst bring m 1 from infinity to its assigned position A 1 . The work done is zero,for there are no particles of the system near enough to exert attraction on it.Next, bring the particle m 2 from infinity to its position A 2 . The work done onit= m 2 ×(potential of m 1 at A 2 )m1m= γr122Next, bring the particle m 3 from infinity to is position A 3 . The work done on itm3m1m3m2= γ + ,r r13and so on for the other particles of the system.23Hence, the total work done in collecting all the particles from rest at infinitydistances from one another to their positionsm1m2m1m2 m1m 2 m3= γ + m3 + + m 4 + +r12 r13r23 r14r24r34m1m= γ r122…(1)When the particles of the system have all been brought to their positions, letV 1 , V 2 ,… be the potentials of the system at the respective points A 1 ,A 2 ,…,A n ,… ThenmV 1 = γr212mV 2 = γr121m3m 4+ + +…r r132314m3m 4+ + +… …(2)r r24In view of (2), the expression (1) becomes= 21m1 V 1 , …(3)msmas in the expression (3), any such term γrsttis twice repeated.Hence, the work done in bringing the particles of the system from infinity totheir respective positions,

314PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS= 21 m1 v 1 . …(4)For continuous masses, we write (4) as1 V dm …(5)2where V is the potential of the body A at any element dm of itself, and theintegration is taken throughout the configuration A.11.6 LAPLACE EQUATION FOR THE POTENTIALLet V be the potential of a system of attracting particles at a point, say P(x, y,z), which is not in contact with the particles. Let m be the mass of a particle atA 1 (a, b, c) of the given system. Letr 2 = (x−a) 2 + (y−b) 2 + (z−c) 2…(1)we know that V is given by the formula m V = . …(2) r Equation (1) gives∂r∂x∂r∂y∂r∂z==x − ary − b r z − c = . …(3) r Differentiating (2) partially w.r.t. x, we obtain∂V∂x m ∂r m(x− a) = −Σ = −Σ2 ∂ . …(4)3r x r Differentiating again (4) partially w.r.t. x, we have (left as an exercise)∂2∂xV2 m m(x− a)= −Σ + 3Σ3 5 r r2. …(5)Similarly, we shall have (exercise)

315ATTRACTION POTENTIAL - II∂2∂yV2 m m(y− b)= −Σ+ 3Σ3 5 r r2,…(6)∂2∂zV2 m m(z− c)= −Σ + 3Σ3 5 r r2. …(7)Adding (5)−(7) vertically, we find (exercise)22∂ V ∂ V ∂ V+ +2 2 2∂x∂y∂z2= 0 . …(8)This shows that the potential V of a system of attracting particles satisfies theLaplace equation. That is, V is a harmonic function.Continuous BodyLet V be the potential of a continuous body or bodies at a point P(x, y, z)outside the body or bodies. Let ρ be the density of the element of volume dv at(x′, y′, z′), andr 2 = (x−x′) 2 + (y−y′) 2 + (z−z′) 2 .…(1)rP(x,y,z)(x′,y′,z′)We know that the potential V is given byV =dv …(2)rDifferentiating (2) under the integral sign, we get (exercise)∂2∂xV2 = − 3r3(x − x')−5r2dv,…(3)∂2∂yV2 = − 3r3(y − y')−5r2dv,…(4)∂2∂zV2 = − 3r3(z − z' )−5r2dv…(5)

316PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSAdding (3) to (5) vertically, we obtain (exercise)22∂ V ∂ V ∂ V+ +2 2 2∂x∂y∂z2= 0 . …(6)The Laplace equation (6) is satisfied by the potential of an attracting system atevery point P(x, y, z) at which there is no matter.11.7 POISSON’S EQUATION FOR THE POTENTIALNow let the point P(x, y, z) be inside the given attracting matter. Describe asphere of small radius ∈ and centre (a, b, c) containing the point P, taking ∈ sosmall that we may regard the density ρ of the matter in this sphere as uniformlydistributed.The matter which produces the potential V at P may now be divided into twoparts the matter outside this small sphere and the matter inside this smallsphere. Let V 0 and V i denote their respective contributions to the wholepotential V at P.V 0V i∈Since the point P is not in contact with the matter which produces the potentialV 0 , therefore,∇ 2 V 0 = 0.…(1)Further, V i being the potential at a point P(x, y, z) inside a small sphere ofradius ∈, we haveV i = 32 πρ (3∈ 2 − r 2 ),…(2)where r is the distance of the point P(x, y, z) from the centre (a, b, c), i.e.,r 2 = (x−a) 2 + (y−b) 2 + (z−c) 2 .From equations (2) and (3), we find (exercise)…(3)∂2Vi2∂x2Vi2∂+∂y2Vi2∂+∂z= −4πρ. …(4)Equation (4) is known as Poisson equation.Remark : The relation

317ATTRACTION POTENTIAL - IIAttraction = grad Vwhich we have seen to be true at points outside a system of attracting particles,is also true at all points outside or inside a continuous distribution of matter.Summarize of our results upto this point(i)(ii)The attraction is the gradient of a potential function V both outside andinside the attracting matter.In empty space, the potential V satisfies Laplace equation∇ 2 V = 0.(iii)At any point at which there is matter of volume density ρ, the potentialV satisfies Poisson equation∇ 2 V = −4πρ.11.8 GAUSS’S THEOREM(SURFACE INTEGRAL OF NORMAL ATTRACTION OVERANY CLOSED SURFACE)Statement : If N be the normal attraction at any point of the element dS of anyclosed surface, measured positively along the normal outwards, due to anyattracting mass, then N dS = −4γπMwhere M is the amount of the attracting mass within the surface and theintegral is being taken given the whole surface.Proof : Let O be the position of any element m of the attracting mass withinthe closed surface. Through O, draw a cone of very small vertical angle and letit cut the given surface in the elements PQ and P′Q′, whose areas are dS anddS′.The attraction of the mass m at these elements are dSγ m cos| OPN2OP , along PN, …(1) dS'γ m ' OP2cos|OP' N', along P′N′, …(2)where PN and P′N′ are the outward drawn normals at P and P′, respectively, asshown below.

318PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSPNMdSQOP′N′M′Q′dS′Through Q and Q′ draw normal sections QM and Q′M′ of this cone. Let dw bethe solid angle of the cone. Let Q be the angle between the elements QM andQP. ThenQ + ∠OPN = π.…(3)Nowdω ==area QM2OQdScos2OQ= −dS(cos| OPN)2OP, …(4)in the limit when PQ is very small. Similarly,dω = −dS' (cos| OP' N')2OP'…(5)Hence, the total normal attractions for the elements dS and dS′ at points P andP′ are each equal to −γm dw. Hence, the total normal attraction for the wholesurface is−γm dm= −γm ( 4π),i.e., for a single particle m at 0, we have NdS = −4Vπm.…(6)Similarly for all other particles of the attracting mass inside the surface.Hence, finally, for the whole mass, we have

319ATTRACTION POTENTIAL - II NdS = −4γπm.This completes the proof of the theorem.11.9 EQUIPOTENTIAL SURFACESFor any attracting mass M, the potential V at any point P(x, y, z) will be afunction of coordinates x, y, z consider the equationV(x, y, z) = C,…(1)Where C is a constant. The equation (1) represents a surface such that thepotential V(x, y, z) at any point of it for the given attracting mass is constantand equals to C. It is hence called an equipotential surface. By given differentvalues to C, we get a family of equipotential surfaces.Remark 1 : Only one equipotential surface passes through any point of space,so that no two equipotential surfaces can intersect.Remark 2 : No work would be done by the attractions on a particle moving onan equipotential surface. Therefore, at every point the resultant attraction isnormal to the equipotential surface through the point.Remark 3 : In the case of rod AB, the equipotential surfaces are ellipsoids ofrevolution obtained by rotating confocal ellipses, whose foci are A and B,about AB as axisRemark 4 : In the case of the spherical shells and sphere, the equipotentialsurfaces are concentric spheres.Distribution for a given potentialWhen the potential is given at all points of space, we can determine thecorresponding distribution. For, the potential V being known, we can find ∇ 2 Vfor every point of space. The Poisson equation givesρ = −1 ∇ 2 V. …(1)4πEquation (1) serves to determine the volume density. When∇ 2 V = 0,…(2)the corresponding density of the distribution is zero. That is, there is noattracting mass at all such points. Whenever∇ 2 V ≠ 0,…(3)

320PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSthe corresponding density of the distribution is given by (1),If the form of the potential function V inside any surface S is different from itsdVform outside, and if there be an abrupt change in the value of as we passdxacross this surface, then the surface density σ on S is calculated as below. For,let V 1 be the potential just inside S and V 2 the potential just outside S. Let dnbe an element of the outward drawn normal. The surface density σ of S isdetermined by the relationσ = −14πV21∂ ∂n∂V−∂nwhere the direction of the normal ∂n is from 1 to 2.Question : The potential outside a certain cylindrical boundary is zero andinside it isV = x 3 − 3x y 2 − ax 2 + 3ay 2 .Find the distribution of matter.Solution : First we found the boundary. Since the potential V is continuousacross the boundary and zero outside, the boundary must be given by theequationx 3 − 3xy 2 − ax 2 + 3ay 2 = 0.…(1)Equation (1) can be written as(x−a) (x− 3 y) (x + 3 y) = 0. …(2)It shows that the section of the given cylindrical boundary is an equilateraltriangle OAB of height ayoax = aAPMxBOM = aWe find

321ATTRACTION POTENTIAL - II2∂ V= 6x − 2a, …(3)2∂x∂2∂yV2= −6x + 6a, …(4)∂2∂zV2= 0. …(5)Hence ∇ 2 V =22∂ V ∂ V ∂ V+ +2 2 2∂x∂y∂z2= 4a. …(6)we know that inside the given cylindrical region, the volume density ρ is givenby the formulaρ = −Equations (6) and (7) give1 ∇ 2 V …(7)4ρ = − πa .…(8)Further, since V is zero outside the region, hence ρ = 0 outside the region.On the boundary : When the point P lies on the part AB of the boundary, thenthe surface density is given by ∂Vσ = − 4 ∂n∂V−∂n1 2 1 , …(9)where V 1 and V 2 are the potentials on two sides of the boundary. HereV 1 = V, V 2 = 0.…(10)and the normal to AB is along x-direction. Hence, from equations (9) and (10),we haveσ =1 ( 2 2x 3y − 2ax )4π1−x= = (a 2 − 3y 2 )43a

322PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS=3 (MA 2 −MP 2 )4=3 AP⋅PB …(11)4For a point Q on AO part of the boundary, we have (left as an exercise)σ =3 OQ ⋅ QA. …(12)4Similarly for the boundary part OB. This shows that a solid cylindrical regiona(prism) of uniform density would produce the same external field as aπ 3 distribution of matter of surface density (AP ⋅ PB) on each of the faces of 4 the prism. Hence the result.11.10 SURFACE AND SOLID HARMONICWe know that the potential V satisfy the Laplace’s equation∇ 2 V = 0or222∂ V ∂ V ∂ V+ + = 0. …(1)2 2 2∂x∂y∂zDefinition : Any solution of Laplace’s equation which is homogeneous in x, y,z, is called a harmonic function or a spherical harmonic.Definition : The degree of homogeneity is called the degree of the function.Remark 1 : We are concerned with the case in which the degree of thefunction is an integer.Remark 2 : If V is a harmonic function of degree n, then∂p∂xp∂q∂yq∂t∂ztV.is a harmonic function of degree (n−p−q−t).Surface and Solid HarmonicsIn polar co-ordinate(r, θ, φ) the Laplace’s equation (1) may be written as (leftas an exercise)∂∂r∂V +∂r1sin ∂ ∂V 1sin +∂ ∂ sin2∂ V = 0 ∂ 22 2r…(2)

323ATTRACTION POTENTIAL - IIWe take,V = r n S n ,…(3)Where S n is independent of r and S n = S n (θ, φ). Differentiating equation (3)w.r. to r partially, we write∂ V = nr n−1 . s n∂ror r 2 ∂ V = n r n+1 . S n …(4)∂rDifferentiate (4) w.r. to r both side partially, we obtain∂ r∂r2∂V = n(n+1) r n S n …(5)∂rFrom equations (2), (3) and (5), we obtain (exercise)∂2Sn2∂∂Sn+ cot ∂1+sin2∂2Sn2 ∂+ n(n+1) S n = 0 …(6)We putcos θ = µin equation (6) and obtain (exercise)…(7)∂ (1 −∂ 2∂Sn 1) + ∂ 1−2 ∂ ∂2Sn2+ n(n+1) S n = 0 . …(8)A solution, S n , of equation (6) is known as a Laplace’s function or a surfaceharmonic of order n. Since n(n+1) remains unchanged when we write −(n+1)for n, there are two solutions of equation (2) of which S n is a factor, namelyr n s n and r −n−1 S n .Definition : The functions r n S n and r −(n+1) S n are called solid harmonics ofdegree n and −(n+1), respectively.Results 1 : If U is a harmonic function of degree n, thenharmonic function. U r2n+1is a also aFor example, U = xyz is a harmonic function of the degree 3, therefore, xyz/r 7is also a harmonic function.Result 2 : If U is a harmonic function of degree −(n+1), then r 2n+1 U is also aharmonic function.

324PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICSNote : x 0 = 1 and θ = tan −1 (y/x) are both harmonic functions of degree zero.1 1Consequently and tan −1 (y/x) are harmonics of degree −1.r rx y z z −1Differentiating these w.r.t. x, y, z; we find , , , tan (y / x)etc., as3 3 3 3r r r rharmonics of degree −2, and so on.11.11. SURFACE DENSITY IN TERMS OF SURFACEHARMONICWe assume thatandnV 1 = ∞ rn+0 a1U n , r < a…(1)nV 2 = ∞ an+0 r1U n , r > a…(2)Give the potential of a certain distribution of matter. In equations (1) and (2),U n denotes the sum of a finite number of surface harmonics (one for eachparticle) and therefore itself a surface harmonic. Hence∇ 2 V 1 = 0, ∇ 2 V 2 = 0.…(3)The matter resides on the surface of the sphere, and its surface density σ isgiven by the formula ∂V1∂V24π σ = − . …(4) ∂r∂rr=aEquations (1), (2) and (4) give (exercise)∞σ = n=02n + 1U2 n .4a…(5)It follows that if we accept the physical argument that an arbitrary distributionof surface density on the surface of a sphere produces the same kind of field ofpotential as an aggregate of particles distributed over the sphere with thepotentials given by (1) and (2), then the arbitrary surface density is expressiblein the form (5).Hence the result.