1. Apr 1, 2013

### Karnage1993

1. The problem statement, all variables and given/known data
Find the steady state temperature $U(r, \theta)$ in one-eighth of a circular ring shown below:

2. Relevant equations

3. The attempt at a solution
I start by assuming a solution of the form $u(r,\theta) = R(r)\Theta(\theta)$. I also note that $u(r,\theta)$ satisfies the equation $u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 0$ where $a \le r \le b$ and $0 \le \theta \le \frac{\pi}{4}$. I know that $r$ is bounded, but I am not sure if the temperature is periodic, ie, if $\Theta(\theta + 2\pi) = \Theta(\theta)$. Where I'm stuck is I do not know how to incorporate the other boundary conditions into what I have, ie, what do I do with the pieces where $u = 0$ and $u = 100$?

2. Apr 2, 2013

### haruspex

Before worrying about boundary conditions, can you develop the general solution?

3. Apr 2, 2013

### Karnage1993

Yes, I believe so. (That is for $0 \le r \le a, 0 \le \theta \le 2\pi$, for some finite $a$, right?)

The process to get to it is quite lengthy, but I can do it.

4. Apr 2, 2013

### HallsofIvy

Staff Emeritus
Well, then, once you have the general solution use the fact that $u(r, 0)= u(r, \pi/4)= 0$, for all r between a and b, to solve for two of the constants.