# Namely u s b constant to show this we integrate the

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Is this the steady-state temperature we calculated in part (b)? Namely, u s = B (constant). To show this we integrate the PDE over the boundary, 0 L ¶∂ u ¶∂ t x = 0 L ¶∂ 2 u ¶∂ x 2 x = ¶∂ u ¶∂ x L - ¶∂ u ¶∂ x 0 = 0 Thus we find d dt 0 L u H x , t L x = 0, or 0 L u H x , t L x = constant. We can evaluate the constant from the initial condition. At t = 0, u H x , t L = f H x L , thus 0 L f H x L x = constant But when t Ø¶ , then u H x , t L Ø u s H x L = B . Thus 0 L u H x , L x = 0 L u s H x L x = 0 L B x = BL = 0 L f H x L x from which we conclude that B = 1 L 0 L f H x L x = A 0 2 Problem2.3.7.nb
• Summer '14
• Applied Mathematics, Trigraph, general solution, Hanoi University of Mining and Geology, GHtL

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