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Unformatted text preview: CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 6 Solutions Posted Wednesday 29 September 2010. Due Wednesday 6 October 2010, 5pm. General advice: You may compute any integrals you encounter using symbolic mathematics tools such as WolframAlpha, Mathematica, or the Symbolic Math Toolbox in MATLAB. 1. [60 points: 13 points for (a), (b), (c), and (e); 8 points for (d)] Solve the following boundary value problems using the spectral method. For each problem, (i) write down the expansions of the right hand side functions as linear combinations of the eigenfunctions; (ii) write down the sum for the solution u obtained from the spectral method; and (iii) produce a plot showing the sum of the first twenty terms in the series for u . For parts (c)(e) you may use the eigenvalues and eigenfunctions computed in Problem 1(d) of Problem Set 5, and the results of Section 5.2.3 of the text. (a) d 2 u dx 2 ( x ) = e x , u (0) = 1, u (1) = 0. (b) d 2 u dx 2 ( x ) 10 u ( x ) = e x , u (0) = 0, u (1) = 0. (c) d 2 u dx 2 ( x ) = x + sin( x ), u (0) = du dx (1) = 0 . (d) d 2 u dx 2 ( x ) = x + sin( x ), u (0) = du dx (1) = 1 . (e) d 2 u dx 2 ( x ) = f ( x ), du dx (0) = u (1) = 0 , where f ( x ) = 1 , < x < 1 / 2; , 1 / 2 < x < 1 . (This f is not continuous; follow the usual procedure and see if you obtain a sensible answer.) Solution. (a) First we solve the problem with homogeneous Dirichlet boundary conditions using the spectral method, then we will add a correction to satisfy the inhomogeneous boundary conditions. For the operator with homogeneous Dirichlet conditions, we have eigenvalues k = k 2 2 with associated (normalized) eigenfunctions u k ( x ) = 2sin( kx ). We shall call the solution to the problem with homogeneous Dirichlet conditions b u , which is given by the spectral method b u = X k =1 ( f,u k ) k u k . To compute ( e x ,u k ), integrate twice by parts to obtain Z 1 e x 2sin( kx ) dx = 2 e x sin( kx ) 1 2 k Z 1 e x cos( kx ) dx = 2 k Z 1 e x cos( kx ) dx = 2 k e x cos( kx ) 1 + 1 k Z 1 e x sin( kx ) dx = 2 k ( 1 e ( 1) k ) 2 k 2 2 Z 1 e x sin( kx ) dx, from which we discern that 2 1 + 1 k 2 2 Z 1 e x sin( kx ) dx = 2 k (1 ( 1) k e ) . We conclude that ( f,u k ) = Z 1 e x 2sin( kx ) dx = 2 k 1 + k 2 2 (1 ( 1) k e ) . The spectral method thus gives b u = X k =1 ( f,u k ) k u k = X k =1 2(1 ( 1) k e ) k (1 + k 2 2 ) sin( kx ) . Now we need to add some function w to b u that will produce a u = b u + w that satisfies both the differential equation and the inhomogeneous boundary conditions. We want d 2 u dx 2 = d 2 b u dx 2 d 2 w dx 2 = e x , but since we already have d 2 b u dx 2 = e x , we need d 2 w dx 2 = 0 ....
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 Fall '09
 Tompson

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