Quiz3 - Quiz 3 We consider the problem for u(x t u 2u Q(t x...

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Quiz 3 We consider the problem for u ( x,t ): ∂u ∂t = 2 u ∂x 2 + Q ( t,x ) , 0 < x < π, t > 0 u (0 ,t ) = 0 , u ( π,t ) = 0 t > 0 u ( x, 0) = 0 0 < x < π. (1) The function u ( x,t ) is extended by oddness on ( - π,π ). We recall that ∂u ∂x ( x,t ) is an even function in x and that 2 u ∂x 2 ( x,t ) is again odd in x , so that 2 u ∂x 2 ( x,t ) = X n =1 γ n ( t ) sin nx, with γ n ( t ) = 2 π Z π 0 2 u ∂x 2 ( y,t ) sin nydy. 1. By integrations by parts, show that γ n ( t ) = - n ( t ), where β n ( t ) = 2 π Z π 0 ∂u ∂x ( y,t ) cos nydy. 2. Show that β n ( t ) = nb n ( t ) , where b n ( t ) = 2 π Z π 0 u ( y,t ) sin nydy. 3. Let us define α n ( t ) = 2 π Z π 0 Q ( y,t ) sin nydy . Using (1), show that b n ( t ) satisfies b 0 n ( t ) + n 2 b n ( t ) = α n ( t ) . 4. Deduce b n (0) from (1) and solve for b n ( t ). We recall that the solution of ˙ y + αy = h ( t ) is y ( t ) = e - αt y (0) + Z t 0 e - α ( t - s ) h ( s ) ds. 5.
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Quiz3 - Quiz 3 We consider the problem for u(x t u 2u Q(t x...

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