An interior Dirichlet problem for a circle

[fahraynk]

Homework Statement

$$
\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. [/B]

Homework Equations

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.
The books answer is :

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

Now, I can't seem to get hte Fourier expansion right I guess, because I don't 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...

 

[Orodruin]

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.

 

[fahraynk]

Thanks!

 

[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 don't know where their sin term comes from because the Bn integral is 0...

 

[fahraynk]

NVM i figured it out can't delete last post for some reason.