# Determ the stationary temp with a PDE

1. May 18, 2006

I have a square area with the length a. The temperature surrounding the square is T_0 except at the top where it's T_0(1+sin(pi*x/a)). They ask for the stationary temperature in the area. In other words, how can the temperature u(x,y) inside the area be written when the time = infinity.

The first thing I do is that I realize that u(x,y) can be written u(x,y) = sum(X(x)*Y(y)).
I also think that it should be a nice starting point to create v = u-T_0, that gives me that the surrounding temperature is 0 everywhere except at the top where it's T_0*sin(pi*x/a)

experience tell me that X(x) = sin(k*pi*x/a)

But how do I find Y(y)? cant seem to get it right.

2. May 26, 2006

### JohanL

Start with the heat equation in 2dimensions, with du/dt=0 because you want the stationary solution. Let u(x,y)=X(x)Y(y) and put it into the p.d.e.
Then you separate variables.

$$X''/X=-Y''/Y=-\lambda^2$$

with cos/sin and exponential solutions. The homogenous boundary conditions eliminatate some of the constants.

and you end up with something like

$$u(x,y)=\sum_{m=0}^\infty B_m*sinh(m\pi y)*sin(m\pi x)$$

$$u(x,a)=\sum_{m=0}^\infty[B_m*sinh(m\pi a)]*sin(m\pi x)$$

where B_m*sinh(mpia) are the coefficients in a fourier sin-series

Last edited: May 26, 2006