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Unformatted text preview: CAAM 336 Â· DIFFERENTIAL EQUATIONS Problem Set 9 Â· Solutions Posted Thursday 28 October 2010. Due Wednesday 3 November 2010, 5pm. 1. [40 points: 12 points each for (a) and (b); 6 points for (c); 10 points for (d)] (a) Consider the function u ( x ) = 1 , x âˆˆ [0 , 1 / 3]; , x âˆˆ (1 / 3 , 2 / 3); 1 , x âˆˆ [2 / 3 , 1] . Recall that the eigenvalues of the operator L : C 2 N [0 , 1] â†’ C [0 , 1], Lu = u 00 are Î» n = n 2 Ï€ 2 for n = 0 , 1 ,... with associated (normalized) eigenfunctions Ïˆ ( x ) = 1 and Ïˆ n ( x ) = âˆš 2cos( nÏ€x ) , n = 1 , 2 ,.... We wish to write u ( x ) as a series of the form u ( x ) = âˆž X n =0 a n (0) Ïˆ n ( x ) , where a n (0) = ( u ,Ïˆ n ). Compute these inner products a n (0) = ( u ,Ïˆ n ) by hand and simplify as much as possible. For m = 0 , 2 , 4 , 80, plot the partial sums u ,m ( x ) = m X n =0 a n (0) Ïˆ n ( x ) . (You may superimpose these on one single, welllabeled plot if you like.) (b) Write down a series solution to the homogeneous heat equation u t ( x,t ) = u xx ( x,t ) , < x < 1 , t â‰¥ with Neumann boundary conditions u x (0 ,t ) = u x (1 ,t ) = 0 and initial condition u ( x, 0) = u ( x ). Create a plot showing the solution at times t = 0 , . 002 , . 05 , . 1. You will need to truncate your infinite series to show this plot. Discuss how the number of terms you use in this infinite series affects the accuracy of your plots. (c) Describe the behavior of your solution as t â†’ âˆž . (To do so, write down a formula for the solution in the limit t â†’ âˆž .) (d) How would you expect the solution to the inhomogeneous heat equation u t ( x,t ) = u xx + 1 , < x < 1 , t â‰¥ with Neumann boundary conditions u x (0 ,t ) = u x (1 ,t ) = 0 to behave as t â†’ âˆž ? Solution. (a) To expand u ( x ) in the form u ( x ) = âˆž X n =0 a n (0) Ïˆ n ( x ) , we must compute the coefficients a n (0). For n = 0 we compute a (0) = Z 1 u ( x ) Â· 1d x = Z 1 / 3 1d x + Z 1 2 / 3 1d x = 2 / 3 . For n > 0 we have a n (0) = âˆš 2 Z 1 u ( x )cos( nÏ€x ) dx = âˆš 2 Z 1 / 3 cos( nÏ€x ) dx + Z 1 2 / 3 cos( nÏ€x ) dx = âˆš 2 h sin( nÏ€x ) nÏ€ i 1 / 3 + h sin( nÏ€x ) nÏ€ i 1 2 / 3 = âˆš 2(sin( nÏ€/ 3) sin(2 nÏ€/ 3)) nÏ€ . [GRADERS: this last expression is sufficiently simplified for full credit.] Note that sin(2 nÏ€/ 3) = 2sin( nÏ€/ 3)cos( nÏ€/ 3), and hence sin( nÏ€/ 3) sin(2 nÏ€/ 3) = sin( nÏ€/ 3)(1 2cos( nÏ€/ 3)) . Thus we have a n (0) = 0 in two cases: if n is a multiple of 3, or if cos( nÏ€/ 3) = 1 / 2. The former occurs when n = 3 , 6 , 9 , 12 , 15 ,... , whiles the latter occurs when nÏ€/ 3(mod 2 Ï€ ) = Ï€/ 3 or 5 Ï€/ 3, and hence a n (0) = 0 when n = 1 + 6 p for integers p â‰¥ 0 or n = 1 + 6 p for integers p â‰¥ 1. Together, this implies that for all odd integers n , a n (0) = 0. We end up with the partial sums shown in the following figure. (MATLAB code follows at the end of this solution.) 0.2 0.4 0.6 0.8 10.2 0.2 0.4 0.6 0.8 1 1.2 1.4 x Ïˆ m (x) m=0 m=2 m=4 m=80 (b) We seek a series solution of the form u ( x,t...
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 Fall '09
 Tompson
 Boundary value problem, Boundary conditions, Neumann boundary condition

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