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Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

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NUMERICAL METHODSThese can then be treated in the way indicated in the previous paragraph. Theextension to more than one dependent variable is straight<strong>for</strong>ward.In practical problems it often happens that boundary conditions applicableto a higher-order equation consist not of the values of the function <strong>and</strong> all itsderivatives at one particular point but of, say, the values of the function at twoseparate end-points. In these cases a solution cannot be found using an explicitstep-by-step ‘marching’ scheme, in which the solutions at successive values of theindependent variable are calculated using solution values previously found. Othermethods have to be tried.One obvious method is to treat the problem as a ‘marching one’, but to use anumber of (intelligently guessed) initial values <strong>for</strong> the derivatives at the startingpoint. The aim is then to find, by interpolation or some other <strong>for</strong>m of iteration,those starting values <strong>for</strong> the derivatives that will produce the given value of thefunction at the finishing point.In some cases the problem can be reduced by a differencing scheme to a matrixequation. Such a case is that of a second-order equation <strong>for</strong> y(x) with constantcoefficients <strong>and</strong> given values of y at the two end-points. Consider the second-orderequationwith the boundary conditionsy ′′ +2ky ′ + µy = f(x), (27.85)y(0) = A, y(1) = B.If (27.85) is replaced by a central difference equation,y i+1 − 2y i + y i−1h 2we obtain from it the recurrence relation+2k y i+1 − y i−12h+ µy i = f(x i ),(1 + kh)y i+1 +(µh 2 − 2)y i +(1− kh)y i−1 = h 2 f(x i ).For h =1/(N − 1) this is in exactly the <strong>for</strong>m of the N × N tridiagonal matrixequation (27.30), withb 1 = b N =1, c 1 = a N =0,a i =1− kh, b i = µh 2 − 2, c i =1+kh, i =2, 3,...,N− 1,<strong>and</strong> y 1 replaced by A, y N by B <strong>and</strong> y i by h 2 f(x i )<strong>for</strong>i =2, 3,...,N − 1. Thesolutions can be obtained as in (27.31) <strong>and</strong> (27.32).27.8 Partial differential equationsThe extension of previous methods to partial differential equations, thus involvingtwo or more independent variables, proceeds in a more or less obvious way. Rather1030

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