Dirichlet problem on half circle I have a problem originated from Dirichlet circle problem. It's on half circle meaning the domain of $\theta$ is defined as $0\le\theta\le\pi$ not like the original which is on $0\le\theta\le2\pi$. 
I encountered this on a book which (I think to be mis-typed). cause i can't find any way to solve this problem using Fourier series path which the book suggested in the chapter. I need serious help.
Full problem is:
$\nabla^2u(r,\theta) = u_{rr}+{\frac{1}{r}}u_r+{\frac{1}{r^2}}u_{\theta\theta} = 0, 0\lt r\lt a , 
 0\le\theta\le\pi$
$u(a,\theta)=u_0 (constant)$
$u(r,0)=0$
$u(r,\pi)=0$
And by Fourier path I mean:
$u(r,\theta) = {\frac{a_0}{2}} + \sum_{n=1}^\infty({\frac{r}{a}})^n(a_ncos(n\theta)+b_nsin(n\theta))$
where 
$a_n = {\frac{1}{\pi}}\int_{-\pi}^\pi f(\theta)cos(n\theta)d\theta$
$b_n = {\frac{1}{\pi}}\int_{-\pi}^\pi f(\theta)sin(n\theta)d\theta$
Is it OK to just make all the $\pi$s to ${\frac{\pi}{2}}$ ? looks unacceptable to me.
 A: when $\theta=0$ or $\pi$ all sines are zero, and what is left is a power series in $r$ with $\pm 1$ from cosines and the coefficients $a_n/a^n$.  Since this power series is identically zero, it follows that $a_0=a_1=\dots = 0$.
So you're looking for a solution of the form 
$u(r,\theta) = \sum_{n=1}^\infty (r/a)^n  b_n \sin (n \theta)$ 
Observe that the RHS is also defined for $-\theta$, and since $\sin (-x) = -\sin (x)$, it follows that $u$ extends to $\theta \in (-\pi,0)$ and satisfies $u(r,-\theta) = -u(r,\theta)$. Let's try to solve the problem on the disk, with the condition $u(a,\theta)=1$ for $\theta \in (0,\pi)$ and $u(a,\theta)=-1$ on $\theta \in (-\pi,0)$. 
With that in mind, and using the formulas you provided, we have 
$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \mbox{sgn}(\theta)  \sin (n\theta)d\theta = \frac{2}{\pi} \int_0^\pi \sin (n\theta) d\theta = -\left .\frac{2}{\pi} \frac{\cos(n \theta)}{n}\right|_0^{\pi}=\frac{2}{n \pi}(1- (-1)^n )=\frac{4}{n\pi} \begin{cases} 1 & n \mbox{ odd }\\ 0 & n \mbox{ even} \end{cases}.$$ 
Summarizing, 
$$ u(r,\theta) = \frac{4}{\pi} \sum_{k=0}^\infty (\frac{r}{a})^{2k+1}\frac{\sin ((2k+1) \theta)}{2k+1}.$$ 
