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Unformatted text preview: ACM 100c  Methods of Applied Mathematics Week 2 April 4, 2009 1 Solving the initial boundary value problem We can use the solutions generated by separation of variables to solve boundary value problems of the form T t = k 2 T x 2 < x < L with boundary conditions T (0 ,t ) = 0 , T ( L,t ) = 0 , and with initial conditions T ( x, 0) = f ( x ) . Note we previously got solutions of the form u ( x,t ) = exp( k 2 t ) sin( x ) u ( x,t ) = exp( k 2 t ) cos( x ) . We would like to satisfy the boundary conditions at x = 0 and x = L . The cosine form of the solution cannot do this, but the sine form will at least satisfy the condition at x = 0 . To satisfy the condition at x = L we can pick the value of . In fact this is just the Sturm Liouville eigenvalue problem we encountered last term. By setting = n L , n = 1 , 2 , 3 , , we see that the solutions u n ( x,t ) = A n exp k ( n 2 2 /L 2 ) t sin( nx/L ) satisfy both boundary conditions. However these solutions u n , while satisfying the bound ary conditions, satisfy the initial condition u n ( x, 0) = A n sin( n/L ) . 1 In order to solve the initial value problem with u ( x, 0) = f ( x ) we note that we can actually use any superposition of the solutions u n ( x,t ) : u ( x,t ) = X n =1 A n exp n 2 2 kt L 2 sin( nx/L ) , which corresponds to an initial condition given by u ( x, 0) = X n =1 A n sin( nx/L ) . We recognize here the Fourier sine series over the interval < x < L and we know from ACM 95b that such a Fourier series is complete and can be used to represent any square integrable function over the interval < x < L . So setting f ( x ) = X n =0 A n sin( nx/L ) , we can use the orthogonality of the Fourier sine series to solve for A n : A n = 2 L Z L f ( x ) sin( nx/L ) dx. This series now satisfies the boundary conditions and the initial conditions. We have therefore derived a series solution to the heat equation of the form u ( x,t ) = X n =1 A n exp k ( n 2 2 /L 2 ) t sin( nx/L ) < x < L, t > . One question that immediately arises is whether this series makes any sense. In other words, does it converge? We recall that while the Fourier series coefficients for a generic function f ( x ) exist as long as f ( x ) is integrable, meaning that Z L  f ( x )  dx exists, the series coefficients A n can decay with n very slowly leading to nonuniform convergence of the series. However, we see that even if this is the case for a given f ( x ) , as soon as t > the argument of the exponential terms in the series terms in the series u ( x,t ) = X n =1 A n exp k ( n 2 2 /L 2 ) t sin( nx/L ) becomes nonzero and the series coefficients decay as A n exp k ( n 2 2 /L 2 ) t 2 as n . This implies the resulting series converges uniformly and rapidly and all the derivatives of u also converge uniformly and rapidly. This is actually a generic feature of the heat equation for these boundary conditions. The smoothness of the solution isof the heat equation for these boundary conditions....
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This note was uploaded on 12/12/2009 for the course ACM 95c taught by Professor Nilesa.pierce during the Spring '09 term at Caltech.
 Spring '09
 NilesA.Pierce

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