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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 λ , yields the following ODE’s T + kλT = 0 X 00 + λX = 0 To determine the sign on λ consider the three cases: (a) If λ = 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 λ = 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 λ > 0, then the solution to the spatial ODE looks like X ( x ) = c 1 cos √ λx + c 2 sin √ λx 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 λ = ( nπ L ) 2 for n ∈ N and thus X ( x ) = X n ( x ) = b n sin nπ L x for some constant b n . Because the only nontrivial solution occurs when λ is positive, we’ll fix λ > 0. This implies the temporal solution looks like T ( t ) = T n ( t ) = c n e k ( nπ L ) 2 t thus u n ( x,t ) = b n sin nπ L x e k ( nπ 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 nπ L x e k ( nπ 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 nπ 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 nπ L xdx Using this information we obtain (a) u ( x,t ) = 6 sin ( 9 πx L ) e k “ 9 π 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 ( πx L ) e k ( π L ) 2 t sin ( 3 π L x ) e k “ 3 π 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 nπ L xdx = 2 L " Z L/ 2 sin nπ L xdx + Z L L/ 2 2 sin nπ L xdx # = 2 L 2 L nπ sin 2 nπ 4 + 2 L nπ cos nπ 2 cos nπ = 2 nπ 1 + cos nπ 2 2 cos nπ = 2 nπ 1 + cos nπ 2 + 2( 1) n +1 = 6 nπ if n is odd ˜ b n if n is even where ˜ b n = if n = 4 k k ∈ N 4 nπ if n = 4 k 2 k ∈ N 2. We are asked to solve (1) subject to2....
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This note was uploaded on 11/12/2010 for the course CHEN 3320 at Colorado.
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 FALCONER,J

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