Solve using method of separation of variables: $$\nabla^2u = 0, \ \ \ \ \ a<r<b, 0 \leq \theta < 2\pi \\ u(a,\theta) = 0 \\ u(b,\theta) = 2\sin^2(\theta)$$
So the way I started it was seeing it's in a rectangular domain and since it's polar coord. it is. So have the condition due to periodicity:
$$u(r,0) = u(r, 2\pi) , \frac{\partial u}{\partial \theta}(r,0) = \frac{\partial u}{\partial \theta}(r,2\pi)$$
We also use the laplacian in polar coord. : $$\frac{\partial^2 u}{\partial r^2}+ \frac{1}{r} \frac{\partial^2 u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0 $$
Do we also have the new stuff of :
$u(a,0) = u(a,2\pi) = 0$ and then the same periodicity for $u(b,\theta)$ ?
I'm just confused because in the examples we've seen in lectures it was always with one condition given in the question which was straightforward, but this has two which has got me confused. If someone could help walk me through this it would be really appreciated.
Thank you!
$\textbf{Edit}$: Now that I think about it, I can see how this is more not a circle problem but an annulus problem. I will attempt it using this fact. If you have any hints/tips please do feel free to comment - it would be really helpful