Laplace's equation in Polar coordinate, an example? Consider Laplace's equation in polar coordinates $$ \frac {1}{r} \frac {\partial} {\partial r}  (r \frac {\partial U} {\partial r}) +  \frac {1} {r^2}  \frac {\partial^2 U} {\partial \theta^2} = 0$$ with $U(r,\theta)$ as the solution, subject to the boundary conditions:

$$U(a,\theta)=\begin{cases}
2\theta && 0 < \theta <\pi\\
0 &&\pi< \theta < 2\pi\\
\end{cases}$$

How can we calculate $\,U(0, \theta)$ ?
 A: The simplest approach to evaluating the value at $r=0$ (and due to the way that Laplace's equation works, the value is independent of $\theta$) is to start by simplifying down the equation system. First, we integrate along $\theta$. This gives
$$
\frac1r\frac{d}{dr}\left(r \frac{d}{dr}\int Ud\theta\right)+\frac1{r^2}\left[\frac{\partial U}{\partial\theta}\right]_b^{b+2\pi} =0 
$$
Again, through the properties of Laplace's equation, we have that $\frac{\partial U}{\partial\theta}|_{\theta=b}=\frac{\partial U}{\partial\theta}|_{\theta=b+2\pi}$ except at the surface, where this relationship may not hold depending on the boundary conditions and the choice of $b$. If we choose $b=0$ or $b=\pi$, then the gradient is different on the two sides, and this could cause a problem (it actually doesn't, but it's better to be rigorous). Choosing any other value within the domain for $b$, we can see that the second term above is zero, and thus
$$
\frac1r\frac{d}{dr}\left(r\frac{d}{dr}\int U d\theta\right) = 0
$$
This can be solved easily to obtain
$$
\int U d\theta = A + \frac{B}r
$$
for some integration constants $A$ and $B$. By the properties of Laplace's equation again, we know that the solution will not go infinite, and thus $B=0$. This gives us that $\int Ud\theta$ must remain constant for all $r$.
In particular, we have
$$
\int_0^{2\pi} U(a,\theta)d\theta = \pi^2
$$
and thus
$$
\int_0^{2\pi} U(0,\theta)d\theta = \pi^2
$$
but $U(0,\theta)=U(0,0)$ is a constant, and thus we have
$$
U(0,\theta) = \frac\pi2
$$
A: If we use separation of variables, we can write: $u(r,\theta) = P(r) Q(\theta)$. The boundary conditions for $u$ yield $P(a) = U(a,\theta)$ and, because of periodicity, $Q(0) = Q(2\pi)$ and $Q'(0) = Q'(2\pi)$. Plugging the new definition for $u$ into the PDE results in:
$$r^2 \left( \frac{1}{r P} \frac{\mathrm{d}}{\mathrm{d}r } \, r \frac{\mathrm{d}P}{\mathrm{d}r}  \right) = -\frac{Q''}{Q} = \lambda \in \mathbb{R}, $$ where the primes denote differentiation with respect the independent variable. The problem for $Q(\theta)$ has as solution: 
\begin{align}
\lambda = 0: \quad & Q = A \theta + B, \\
\lambda > 0: \quad & Q = A \cos{k \theta} + B \sin{k \theta}, \\
\lambda < 0: \quad & Q = A \cosh{\sqrt{|\lambda|} \theta} + B \sinh{\sqrt{|\lambda|} \theta},
\end{align} $k^2 = \lambda$, where the periodicity condition remains to be satisfied. For $\lambda = 0$ we find $A = 0$, while for $\lambda < 0$, $A = B = 0$. For the case $\lambda > 0$, the only nontrivial solution is that corresponding to $k = n = 1,2, \ldots$ (i.e., $k$ is an integer) and therefore:
$$ Q_n(\theta) = A_n \cos{n \theta} + B_n \sin{n \theta}$$  
The problem for $P(r)$ reduces to solving an Euler equation, which upon trying solutions of the type $P(r) = r^s$ yields $s^2-n^2= 0$ and therefore $s = \pm n$. The bounded solution (that for which $|P(0)|<\infty$) is therefore given by:
$$ P_n(r) = C_n r^n $$
When you impose the boundary condition at $r=a$ you can find the constant of integration $C_n$.
I hope you find this useful. 
If somebody finds any typo or any other type of fail in my discussing of the problem, please let me know. Cheers!

Note that we can let $n = 0$ as a solution for the case $\lambda = 0$ since $Q_0(\theta) = A_0$ (which plays the role of the constant $B$)
A: There is a much simpler approach for a question like this, since the function $U$ is harmonic it satisfies the mean value property (my assumption by the way is that $0\le r\le a$), and so $$U(0,\theta)=U(0,0)=\frac{1}{|B_a|}\int_{B_a(0)}U(r,\theta)\,dr\,d\theta=\frac{1}{|B_1(0)|a}\int_{\partial B_a(0)}U(a,\theta)\,d\theta$$
$$=\frac{1}{a\pi}\int_0^{2\pi}U(a,\theta)\,d\theta=\frac{1}{a\pi}\int_0^{\pi}2\theta\,d\theta=\frac{\pi^2}{a\pi}=\frac{\pi}{a}.$$
Note that $U(0,\theta)=U(0,0)$, because rotating a radius of zero does nothing.
