this question is taken from 2014 exam on CE Entrance Exam, Question $32$ on the end of page $6$.
Consider the Laplace equation of following polar coordination,
$$\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial u}{\partial r})+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2} =0 $$
and the boundary condition:
$$u(a,\theta)=\begin{cases}2\theta & 0<\theta<\pi\\0 & \pi<\theta<2\pi\end{cases}$$
Calculate the values of $U(0, \theta)$?
Update: this comes from my idea:
$$ U(0, \theta)= \frac{1}{\pi} \int_{0}^{\pi} 2 \theta d\theta + \frac{1}{\pi} \int 0 \times d \theta = \frac{\pi^2}{\pi} = \pi$$
is there any expert describe in shot that how this value $U(0, \theta)=2\pi$ is find? or maybe this is wrong answer? how we can calculate this value in brief?
The solution to the problem $\Delta u = 0$ for $x \in B(0,a)$ subject to $u=\begin{cases} 2 \theta & 0<\theta<\pi \\ 0 & \pi<\theta<2 \pi \end{cases}$ for $x \in \partial B(0,a)$ satisfies $u(0)=\frac{1}{2\pi} \int_0^{2\pi} u(a,\theta) d \theta = \frac{1}{2 \pi} \int_0^\pi 2 \theta d \theta = \frac{\pi}{2}$. This follows from the mean value property of the Laplace equation.
Note that for a discontinuous BC one must be a bit careful about the exact sense of this solution. Still, for the "right" sense of solution, the mean value property holds. One (unnecessarily elaborate) way to view the solution which gives this result is $u(x)=E_x[f(W_\tau)]$ where $f$ is the solution on the boundary, $W$ is a Wiener process, $E_x$ is expectation over Wiener paths started at $x$, and $\tau$ is the first time for a Wiener path to hit $\partial B(0,a)$. Then for $x=0$, $W_\tau$ is uniformly distributed on the circle, giving the result.