# temperature PDE

• February 9th 2009, 02:44 PM
PvtBillPilgrim
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.
• February 9th 2009, 09:32 PM
ThePerfectHacker
Quote:

Originally Posted by PvtBillPilgrim
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. 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.