Assignment_8_solutions.pdf - ECE 205 Winter 2017 Assignment...

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ECE 205 - Winter 2017 Assignment 8 solutions Due 02-04-2017 Note: Please submit the full solutions to all questions. In Questions 1 and 2, we consider the heat equation in one spatial dimension, for the temperature, U , of a rod of length L : U t = α 2 U xx , where α 2 is the thermal diffusivity. 1: Heat equation ( I ) Consider the initial condition U ( x, 0) = 100 x L . a) ( i ) Solve the PDE with the boundary conditions: U (0 , t ) = U ( L, t ) = 0 , ( t > 0) . ( ii ) What is the equilibrium state for this boundary value problem, i.e. solve the PDE U t = α 2 U xx , U t = 0, U (0 , t ) = U ( L, t ) = 0, t > 0. ( iii ) Is your solution from a ( i ) asymptotic to your solution to a ( ii ) as t → ∞ ? b) ( i ) Solve the PDE with the boundary conditions: U x (0 , t ) = U x ( L, t ) = 0 , ( t > 0) . ( ii ) What is the equilibrium state for this Boundary value problem, i.e. solve the PDE: U t = α 2 U xx , U t = 0, U x (0 , t ) = U x ( L, t ) = 0 . ( t > 0). ( iii ) Is your solution from b ( i ) asymptotic to your solution to b ( ii ) as t → ∞ ? ( II ) a) Consider now the general Boundary value problem: U t = α 2 U xx , U (0 , t ) = U ( L, t ) = 0 , t > 0 . Let us consider the mean square temperature: M ( t ) = 1 L L Z 0 U 2 ( x, t ) dx. Use the Leibniz rule for differentiating under the integral and integration by parts to show that this function is not increasing. b) Consider now the general boundary value problem: U t = α 2 U xx , U x (0 , t ) = U x ( L, t ) = 0 , t > 0 . Let us consider the total energy: E ( t ) = k L Z 0 U ( x, t ) dx, for some positive constant k . Use the Leibniz rule for differentiating under the integral and integration by parts to show that the function is constant, and explain your answer. 1
ECE 205 - Winter 2017 Assignment 8 solutions Due 02-04-2017 Solution ( I ) We try U ( x, t ) = X ( x ) T ( t ) and we get: ˙ T α 2 T = X 00 X = k, where K is a constant. X (0) = X ( L ) = 0 k = - λ 2 for a non trivial solution. Therefore X ( x ) = A sin( λx ) and X ( L ) = 0 implies λ = L for n Z . A solution is X n ( x ) = A n sin ( nπx L ) . Then ˙ T α 2 T = - L 2 T n = c n e - nπα L 2 t . We absorb c n into A n and we get: U n ( x, t ) A n e - β 2 n t sin nπx L , where β n = nπα L . The general solution is a sum of such terms due to the linearity so, U ( x, t ) = n = X n =1 A n e - β 2 n t sin( λ n x ) , where λ n = L and β n = αλ n . Note that we can combine the solutions with negative integer values with those of positive integer values. We want U ( x, 0) = 100 x L , then n = X n =1 A n sin( λ n x ) = 100 x L ⇐⇒ A n = 2 L L Z 0 100 x L sin nπx L dx = - 200 nπL [ L cos( )] = 200 ( - 1) n +1 . The above integration is performed by parts. Thus, U ( x, t ) = n = X n =1 200 ( - 1) n +1 e - β 2 n t sin( λ n x ) . ( ii ) If we solve the PDE with U t = 0 we get, U xx = 0. Therefore, U ( x, t ) = c 1 x + c 2 . With U (0 , t ) = U ( L, t ) = 0, we get c 1 = c 2 = 0. Therefore, the equilibrium solution is U e ( x, t ) = 0 . ( iii ) We have lim t →∞ e - β 2 n t = 0 . Thus for our general solution from ( I ), all terms die away and lim t →∞ U ( x, t ) = 0 = U e ( x, t ) . The general solution tends to the equilibrium solution.
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ECE 205 - Winter 2017 Assignment 8 solutions Due 02-04-2017 b) As before, we have U ( x, t ) = X ( x ) T ( t ) with ˙ T α 2 T = X 00 X = k, where k is a constant. Now we also have X 0 (0) = X 0 ( L ) = 0. Therefore, if k = 0 then X = c 1 is a solution. If k = λ 2 > 0, then X ( x ) = c 2 e λx + c 3 e - λx is a solution. But X 0 (0) = X 0 ( L ) = 0 implies c 2 = c 3 = 0. If k = λ 2 < 0, then we have X ( x ) = c 4 sin( λx ) + c 5 cos( λx ). The condition X 0 (0) = 0 gives c 4 = 0. The condition X 0 ( L ) = 0 implies λ = λ n = L where n Z . Thus X n ( x ) = d n

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