# Homework Help: Laplace electric potential

1. Oct 8, 2012

### Dustinsfl

Solve Laplace's equation on a circular disk of radius $a$ subject to the piecewise boundary condition
$$u(a,\theta) = \begin{cases} 1, & \frac{\pi}{2} - \epsilon < \theta < \frac{\pi}{2} + \epsilon\\ 0, & \text{otherwise} \end{cases}$$
where $\epsilon \ll 1$.
Physically, this would reflect the electric potential distribution on a conducting disk whose edge is almost completely grounded except a small portion of angular extent $\Delta\theta = 2\epsilon$ around the location $\theta = \frac{\pi}{2}$.
Obtain the solution to this problem and plot the solution for the case of $a = 1$ and $\epsilon = 0.05$.

By separation of variables, we have
$$\begin{cases} \Theta(\theta) = A\cos\lambda\theta + B\sin\lambda\theta\\ R(r) = r^{\pm\lambda} \end{cases}$$
So how do I use the conditions now?

2. Oct 8, 2012

### LCKurtz

The cosines and sines should have square roots over the lambdas, no? And can't you eliminate the $r^{-\lambda}$ by assuming the solution is bounded at the origin? And I'm guessing you have some unstated condition that will give you the eigenvalues $\lambda$. Once you have that you should be able to finish the problem with a Fourier Series to fit the boundary condition at $r=a$.

3. Oct 8, 2012

### Dustinsfl

That is everything. I have $\lambda$ because I set the ODEs to $-\lambda^2$
I am still not sure on how to proceed with the conditions.
$$\sum_{n=1}^{\infty}r^{\lambda_n}(A\cos\lambda_n \theta + B\sin\lambda_n\theta)$$

4. Oct 8, 2012

### LCKurtz

Then $R(r) = r^{\lambda_n}$. If you had the $\lambda_n$ you would just set$$u(a,\theta)=\sum_{n=1}^{\infty}a^{\lambda_n}(A_n \cos \lambda_n \theta + B_n\sin\lambda_n\theta)$$and solve for the $A_n$ and $B_n$ with Fourier series. But first you need to figure out the distinct eigenvalues $\lambda_n=\mu_n^2$ from some explicit or implicit condition in the problem, which I don't know.

Last edited: Oct 8, 2012
5. Oct 8, 2012

### Dustinsfl

All the information is there. Typo I didn't mean -\lambda^2
$$\frac{r^2R''+rR'}{R} = \lambda^2\iff r^2R''+rR'-\lambda^2R$$
The problem is the Cauchy-Euler type.
$$r^n[n(n-1)+n-\lambda^2] = 0\Rightarrow n^2 = \lambda^2\Rightarrow n = \lambda$$

6. Oct 8, 2012

### LCKurtz

OK, I agree about the non-squared powers on the $\lambda_n$ and edited my post accordingly. But I'm afraid I can't help you with regard to determining the distinct values of the $\lambda's$. There is something we are missing which is likely implicit in the polar coordinate function, but I don't know what it is.

7. Oct 8, 2012

### Dustinsfl

We have to have $u(r,\pi) = u(r,-\pi)$ as well as the derivatives equal. I am glad you said that. I felt I missing some BC.
$$u(r,\theta) = \alpha_0 + \beta_0\ln r + \sum_{n=1}^{\infty}r^{\lambda_n}(A_n\cos\lambda_n \theta + B_n\sin\lambda_n\theta)$$
So now how do I use the piecewise condition?

Last edited: Oct 8, 2012
8. Oct 8, 2012

### LCKurtz

Where did that $\alpha_0 + \beta_0\ln r$ suddenly come from? I know the answer to that, but do you have something you need to show me?

So use your new boundary conditions on $\Theta(\theta)$ to figure out the eigenvalues $\lambda_n$. What do you get? You need to do that before you are ready to write the form of the solution for $u(r,\theta)$ and address the boundary condition.