[Faller] 03FALL - HWK4 SOLN

# [Faller] 03FALL - HWK4 SOLN - ECH 159 Fall 2002 Solution...

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Unformatted text preview: ECH 159, Fall 2002, Solution for Homework 3, October 16, 2002 Problem 1 The heat equation in a one-dimensional rod with heat source or sink proportional to temperature reads ∂ t u = k∂ 2 x u- αu. The sign of α decides between two different scenarios: α > 0 The rod loses heat. This corresponds to a wire in lateral thermal contact with a big, cold background. The wire radiates heat out to the side α < 0 The rod picks up heat. Either the same as above with a hot background or a exothermal chemical reaction takes place with a reaction rate propor- tional to temperature. Boundary conditions: u (0 ,t ) = u ( L,t ) = 0 (a) Determine the steady state solution for all three possible cases (including α = 0). (b) Solve the full problem for α > 0 and the initial condition u ( x,t ) = f ( x ). (c) Same as (b) but α < 0, Hint: The two problems are not symmetric. Solution (a) Steady state → ∂ u = 0 → k∂ 2 x u- αu = 0 Case 1: α = 0 ∂ 2 x u = 0 → u = c 1 x + c , Boundary conditions u (0) = 0 → c = 0 ,u ( L ) = 0 → c 1 = 0 → u ( x ) = 0 Case 2: α > k∂ 2 x u = αu → ∂ 2 x u = α k u As α > 0 we come to hyperbolic functions: u = c 1 sinh( p α k x ) + c 2 cosh( p α k x ) Boundary conditions: u (0) = 0 → c 2 = 0, u ( L ) = 0 → c 1 = 0, or √ αk = 0. As α > → c 1 = 0 → u ( x ) = 0 Case 3: α < k∂ 2 x u =-| α | u → ∂ 2 x u =- | α | k u This leads to u = c 1 sin( q | α | k x ) + c 2 cos( q | α | k x ) Boundary conditions u (0) = 0 → c 2 = 0, u ( L ) = 0 → c 1 = 0, or q | α | k L = nπ This means that if the length of the rod is L = nπ q k | α | there exist non-zero solutions. These are the eigenmodes or standing waves of the problem. (b) α > Solve the problem by separation of variables u ( x,t ) = X ( x ) T ( t ) leading to d t T ( t ) T ( t ) = k d 2 x X X- α = ± λ 2 Solving the time equation d t T = ± λ 2 t , we exclude λ = 0 as we dealt with the steady state already, leading to T ( t ) = T exp ± λ 2 t For α > 0 we saw that the steady state is u = 0 and physically we expect that if we have a negative source, i.e. heat is taken out of the system, that we have- λ 2 So we deal with the space problem k d 2 x X- αX =- λ 2 X leading to d 2 x X =- λ 2- α k X , let’s define β 2 = λ 2- α k . For β > 0 we find trigonometric solutions X ( x ) = c 1 sin( βx ) + c 2 cos( βx ) and for β < 0 we find hyperbolic solutions X ( x ) = c 1 sinh( βx ) + c 2 cosh( βx ). In this latter case the boundary conditions enforce X ( x ) = 0. In the trigonometric case we use X (0) = 0 to find c 2 = 0 and X ( L ) = 0 to find βL = nπ . Remembering the definition of β we find n 2 π 2 L 2 = λ 2- α k . So our eigenvalues are λ = q kn 2 π 2 L 2 + α and the general solution is u ( x,t ) = ∞ X n =1 A n sin( nπ L x )exp- ( kn 2 π 2 L 2 + α ) t ....
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[Faller] 03FALL - HWK4 SOLN - ECH 159 Fall 2002 Solution...

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