Solve the following boundary value problems for the Laplace equation on the semi-annular domain:
$ 1 < x^2 + y^2 < 2, y > 0 $
$u(x, y) = 0, x^2 + y^2 = 1, u(x, y) = 1, x^2 + y^2 = 2, u(x, 0) = 0$
Effort:
We move to polar coordinate and need to solve $u(r,\theta)$ such that:
$$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}=0$$
with
$$ u(1,\theta) = 0, 0 < \theta < \pi, u(\sqrt{2},\theta) = 1, 0 < \theta < \pi,u(r,0) =0,u(r,\pi) =0, 0 \le r \le 2 $$
The general solution as we learned in class is :
$$u(r, \theta) = \frac{a_0}{2}+ \sum_{n=1}^{\infty} \left( a_n r^n \cos(n\theta) + b_n r^n \sin(n\theta) \right)$$
The only boundary condition that yields non trivial solution is $u(\sqrt{2},\theta) = 1, 0 < \theta < \pi $ and we get
$$u(\sqrt{2}, \theta) = \frac{a_0}{2}+ \sum_{n=1}^{\infty} \left( a_n 2^{\frac{n}{2}} \cos(n\theta) + b_n 2^{\frac{n}{2}} \sin(n\theta) \right) =1$$
But now I get a solution of $ u(r,\theta) =1$ which I don't think is correct.