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Math Help - temperature PDE

  1. #1
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    temperature PDE

    Can someone help me with this?

    An infinitely long plate of width L has its two parallel sides maintained at temperature zero and its other side at constant temperature T. What is the rate of change?

    I have that Laplace's equation holds and u(x,0) = T, u(0,y) = 0, u(L,y) = 0 with the lim as y goes to infinity of u(x,y) finite.

    I know how to do the separable equations stuff, but can someone help me with the eigenfunctions (i.e. what are X(x) and Y(y) for u(x,y) = XY). Thanks.
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    Can someone help me with this?

    An infinitely long plate of width L has its two parallel sides maintained at temperature zero and its other side at constant temperature T. What is the rate of change?

    I have that Laplace's equation holds and u(x,0) = T, u(0,y) = 0, u(L,y) = 0 with the lim as y goes to infinity of u(x,y) finite.

    I know how to do the separable equations stuff, but can someone help me with the eigenfunctions (i.e. what are X(x) and Y(y) for u(x,y) = XY). Thanks.
    Let u(x,y) = X(x)Y(y) then u_{xx}+u_{yy} = 0 \implies \frac{X''}{X} = - \frac{Y''}{Y}=k .

    In the first case, k=0, we get X'' = Y'' = 0 and so u(x,y) = (ax+b)(cy+d). Now we use the boundary conditions. Since u(0,y) = 0 \implies b(cy+d) = 0. So either cy+d is identically zero or b=0. In the former case if cy+d is identically zero it would mean u(x,y) is a trivial solution. Thus, b=0 and so u(x,y) = ax(cy+d). But under the condition u(L,y)=0 we can only get the trivial solution. But of course the trivial solution does not solve this Dirichlet problem and so we can reject the trivial solution.

    In the second case, k>0, we get X'' +kX=0 and Y''-kY=0 so u(x,y) = (a\sin (\sqrt{k}x) + b\cos( \sqrt{k}x))(c\sinh(\sqrt{k}y) + d\cosh(\sqrt{k}y)). The condition u(0,y) =0 \implies b(c\sinh(\sqrt{k}y) + d\cosh(\sqrt{k}y)) = 0. To have a non-trivial solution u(x,y) we must have b=0. Therefore, u(x,y) = a\sin( \sqrt{k} x)(c\sinh(\sqrt{k}y) + d\cosh(\sqrt{k}y)). The condition u(L,y)=0\implies a\sin( \sqrt{k}L) (c\sinh(\sqrt{k}y) + d\cosh(\sqrt{k}y)) = 0. To have a non-trivial solution for u we must have a\not = 0 with \sin(\sqrt{k}L) = 0 \implies \sqrt{k}L = \pi n, n\in \mathbb{Z}^+. Thus, k = \tfrac{\pi^2 n^2}{L^2}. At this point we can assume a=1 by letting it get absorbed into the constants. Thus, so far we established the solutions need to have form u(x,y) = \sin \tfrac{\pi n x}{L} ( c\sinh \tfrac{\pi n y}{L} + d\cosh \tfrac{\pi n y}{L}) = 0. These are our eigenfunctions. If c_n,d_n are arbitray numbers for n\geq 1 we see that \sum_{n=1}^N \sin \tfrac{\pi n x}{L} ( c_n\sinh \tfrac{\pi n y}{L} + d_n\cosh \tfrac{\pi n y}{L}) is a solution to this equation (without the last condition yet). We are told that the solution statys bounded. The only way that can happen is if the c_n = -d_n (why is that?). We would like to satisfy the condition u(x,0) = T therefore we want \sum_{n=1}^N d_n\sin \tfrac{\pi n x}{L} = T for 0<x<L. The sad case here is that finite sums do not do, we need to now think of infinite sums of sines. By Fourier analysis (assuming I did not make mistakes) d_n = \tfrac{2T}{\pi n}(1 - (-1)^n). This gives us a solution u(x,y) = \sum_{n=1}^{\infty} \sin \tfrac{\pi n x}{L} ( -\tfrac{2T}{\pi n}(1-(-1)^n)\sinh \tfrac{\pi n y}{L} + \tfrac{2T}{\pi n}(1-(-1)^n) \cosh \tfrac{\pi n y}{L})


    You think about the final case.
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