solution2 - APPM 4350/5350: Fourier Series and Boundary...

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Unformatted text preview: APPM 4350/5350: Fourier Series and Boundary Value Problems Homework 2 Solutions 1. We are asked to solve u t = k 2 u t 2 (1) subject to u (0 ,t ) = 0 and u ( L,t ) = 0 Assuming a separable solution to the heat equation of the form u ( x,t ) = T ( t ) X ( x ) and substituting into (1) leads to the following equation T kT = X 00 X where the prime indicates differentiation with respect to the appropriate independent variable. Setting each side equal to a constant, call it- , yields the following ODEs T + kT = 0 X 00 + X = 0 To determine the sign on consider the three cases: (a) If =- s < 0, then the solution to the spatial ODE looks like X ( x ) = c 1 cosh sx + c 2 sinh sx Using the boundary condition u (0 ,t ) = 0 implies X (0) = 0 implies c 1 = 0. Similarly, the BC u ( L,t ) = 0 implies X ( L ) = 0 implies c 2 = 0, and thus u ( x,t ) is the trivial solution. (b) If = 0, then the solution to the spatial ODE looks like X ( x ) = c 1 + c 2 x The BC u (0 ,t ) = 0 implies X (0) = 0 implies c 1 = 0. The BC u ( L,t ) = 0 implies X ( L ) = 0 implies c 2 = 0 and again we get u ( x,t ) = 0. (c) If > 0, then the solution to the spatial ODE looks like X ( x ) = c 1 cos x + c 2 sin x The BC u (0 ,t ) = 0 implies X (0) = 0 implies c 1 = 0. The BC u ( L,t ) = 0 implies X ( L ) = 0 implies = ( n L ) 2 for n N and thus X ( x ) = X n ( x ) = b n sin n L x for some constant b n . Because the only nontrivial solution occurs when is positive, well fix > 0. This implies the temporal solution looks like T ( t ) = T n ( t ) = c n e- k ( n L ) 2 t thus u n ( x,t ) = b n sin n L x e- k ( n L ) 2 t where c n has been absorbed into b n . Using the principle of superposition, we can write the general solution to (1) as u ( x,t ) = X n =1 b n sin n L x e- k ( n L ) 2 t (2) The constant b n will depend upon the initial condition u ( x, 0) = f ( x ) and is determined by using the orthogonality of the sine and cosine functions. Observe u ( x, 0) = f ( x ) = X n =1 b n sin n L x so b n is just the Fourier sine series coefficient for f ( x ) on the interval [0 ,L ]. That is b n = 2 L Z L f ( x ) sin n L xdx Using this information we obtain (a) u ( x,t ) = 6 sin ( 9 x L ) e- k 9 L 2 t which is equivalent to the statement that u ( x,t ) has the solution given by (2) where b n = 6 if n = 9 else (b) u ( x,t ) = 3 sin ( x L ) e- k ( L ) 2 t- sin ( 3 L x ) e- k 3 L 2 t which is equivalent to the statement that u ( x,t ) has the solution given by (2) where b n = 3 if n = 1- 1 if n = 3 else (c) Here, u ( x,t ) is given by (2) where b n = 2 L Z L f ( x ) sin n L xdx = 2 L " Z L/ 2 sin n L xdx + Z L L/ 2 2 sin n L xdx # = 2 L 2 L n sin 2 n 4 + 2 L n cos n 2- cos n = 2 n 1 + cos n 2- 2 cos n = 2 n 1 + cos n 2 + 2(- 1) n +1 = 6 n if n is odd b n if n is even where b n = if n = 4 k k N- 4 n if n = 4 k- 2 k N 2. We are asked to solve (1) subject to2....
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solution2 - APPM 4350/5350: Fourier Series and Boundary...

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