An interior Dirichlet problem for a circle

1. Dec 1, 2016

fahraynk

1. The problem statement, all variables and given/known data
$$\bigtriangledown^2=0 for : 0<r<1 \\ BC : u(1,\Theta)= sin(\Theta), 0<\Theta<\pi \\ u(1,\Theta)= 0, pi<\Theta<2\pi \\$$
Basically its an interior dirichlet problem for a circle.

2. Relevant equations

3. The attempt at a solution

The answer is supposed to be $$U(r,\Theta) = \Sigma r^n[a_n cos(n\Theta) + b_n sin(n\Theta)$$
and the a_n, b_n is basically a fourier expansion of the boundary conditions.

$$\frac{r}{2}sin\Theta + \frac{2}{\pi}(\frac{1}{2} - \frac{r^2}{3}cos(2\Theta)-\frac{r^4}{15}cos(4\Theta)-\frac{r^6}{35}cos(6\Theta)...)$$

Now, I cant seem to get hte fourier expansion right I guess, because I dont get this answer.

Heres my attempt ....

$$a_n = \frac{2}{\pi}\int_{0}^{\pi}sin(n\Theta)cos(n\Theta)d\Theta = 0 \\ b_n = \frac{2}{\pi}\int_{0}^{\pi}sin^2(n\Theta) = \frac{1}{\pi} \\ a_0 = \frac{1}{\pi}\int_{0}^{\pi}sin\Theta d\Theta = 0$$

Cant think of how to make this fourier expansion into the books answer...

2. Dec 2, 2016

Orodruin

Staff Emeritus
Why are you computing the integral with $\sin(n\Theta)$? The boundary condition is $\sin(\Theta)$ ...

Also, the integral of sine over half a period is not zero.

3. Dec 2, 2016

fahraynk

Thanks!

4. Dec 12, 2016

fahraynk

'okay I am getting 2/pi + summation [4/pi * [0.5 - r^n/(1-n^2)]cos(nx)]
The Bn term goes to 0 (per wolfram). A0 = 2/pi, An = 2/pi * cos(n*pi +1)/(1-n^2)

Is the books answer wrong? I dont know where their sin term comes from because the Bn integral is 0...

5. Dec 13, 2016

fahraynk

NVM i figured it out cant delete last post for some reason.