# Laplace's equation inside a circular annulus

1. Oct 3, 2013

### temaire

1. The problem statement, all variables and given/known data

2. Relevant equations

General solution:

Fourier series:

where $r_{1}=a$, $r_{2}=b$, $f_{1}(\theta)=sin(\theta)$, and $f_{2}(\theta)=2sin(\theta)cos(\theta)$.

3. The attempt at a solution

By evaluating the Fourier series shown above, I determined that $a_{o}=b_{o}=a_{n}=b_{n}=c_{n}=d_{n}=0$.

Is this correct?

Last edited: Oct 3, 2013
2. Oct 3, 2013

### brmath

Well, no. You could guess that for yourself because that would mean the answer is identically 0, which is not awfully likely for the diff eq is it?

I looked at $\int_0^{2\pi}sin\theta cos{n \theta}d\theta$ . The best way to evaluate it is to express both the sin and cos in exponential form: $sinx = \frac{e^x-e^{-x}}{2i}$ and similarly for the cosnx. Multiply it all out and regroup your terms. You'll have a sin and a cosh. The former is periodic in $\pi$ the latter is not.

Do try again.

3. Oct 4, 2013

### temaire

I evaluated the integral you mentioned after expressing both the sin and cos in exponential form as follows:

Isn't this still periodic in $\pi$?

4. Oct 4, 2013

### haruspex

How about $2 \sin \theta \cos n \theta = \sin ((n+1)\theta) - \sin ((n-1) \theta)$?

5. Oct 4, 2013

### temaire

Isn't that still always going to be zero at all n, where n is an integer and can't equal +-1, after integrating?

6. Oct 4, 2013

### brmath

****
Here's what I have:

$sinx \cdot cos(nx)$= $\frac{1}{4i}[(e^{ix}-e^{-ix}) \cdot (e^{inx}+e^{-inx})]$ = $\frac{1}{4i}[(e^{i(n+1)x} - e^{-i(n+1)x}) + (e^{(n+1)x} - e^{-(n+1)x})]$

Integrating up the last expression and evaluating at 0 and $2\pi$ gives

$\frac{1}{4(n+1)i}(e^{2(n+1)\pi} + e^{-2(n+1)\pi})$

This is not zero.

Try to be very careful when you do your integration.. Also, note that the product of two periodic functions is periodic, but not necessarily in the period of either.

7. Oct 5, 2013

### haruspex

Quite so, so maybe something else is wrong. But it's much easier than writing out all those exponentials.

8. Oct 5, 2013

### haruspex

Not right. You should get some n-1 terms.

9. Oct 5, 2013

### brmath

You are right, and I'm the one who should be careful about integrals. I do not know by what slip of the mind I lost the i's in the exponents. The other integral is 0 also.

So temaire is right, they are all 0, unless $r_1 = r_2$ (which isn't really an annulus,but whatever).