Poisson equation in a ring I would like to solve:
$-\Delta u = 1$ in $\Omega $ where $\Omega = \{ (x,y): 1 \leq x^2+y^2 \leq4 \}$
with $u=0$ in $\{ (x,y): x^2+y^2 = 4 \}$ and $\frac{\partial u}{\partial n}=0$ in $\{ (x,y): x^2+y^2 = 1 \}$.
I dont know how to start neither what method should I use. Can anyone give me some hints?
My first thought is to use separation of variables but it didn't work out.
 A: In plane polars [i.e. $x = r \cos(\theta), y = r\sin(\theta)$], the Laplacian is given by $$\Delta = \frac{\partial^2}{\partial r^2} + \frac{1}{r} \frac{\partial }{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2}.$$ Thus the equation becomes $$-u_{rr} - \frac 1 r u_r - \frac 1 {r^2} u_{\theta \theta} = 1, \,\,\,\,\, \theta \in [0, 2\pi), \,\,\,\,\, r \in (1,2).$$ The normal derivative at $r =1$ is the same as the negative $r$-derivative, so the boundary conditions are $$u_r = 0 \,\,\, \text{ on } \,\,\, r = 1, \,\,\,\,\,\,\,\,\,\,\,\,\,u = 0 \,\,\, \text{ on } \,\,\, r = 2.$$ Now you can separate variables and solve. 
Edit [addressing the comment]:
I'm not sure how you got that. Where did the factors of $\frac 1 r$ and the $\theta$-derivatives go? 
Now that I think about it though, my previous hint may have been misleading because I'm nort sure you can separate variables since it isn't a homogeneous system. Here is the solution:
Since the domain is radial, it makes sense that the solution shouldn't depend on $\theta$. Making this assumption ${u_\theta \theta} = 0$ so you end up with $$u_{rr} + \frac 1 r u_r = -1 \,\,\,\, \implies \,\,\,\, ru_{rr} + u_r = r \,\,\,\, \frac{d}{dr}(ru_{r}) = r.$$ Integrating from $1$ up to some $r \in (1,2)$, we get $$ru_r - 1\cdot u_r(1,\cdot) = \frac{r^2 - 1}{2}.$$ But $u_r(1,\cdot) \equiv 0$, so $$u_r = \frac{r}{2} - \frac{1}{2r}.$$ Now integrate from some $r \in (1,2)$ to $2$ to see $$u(2,\cdot) - u(r,\cdot) = \frac{4-r^2}{4} + \frac{1}2 \log (r) - \frac{1}{2} \log(2).$$ Again, $u(2,\cdot) \equiv 0$, so we wind up with $$u(r,\theta) = \frac{r^2-4}{4} + \frac{\log(2) - \log(r)}{2}.$$ Of course you can turn this back into $x,y$ if you want $(r= \sqrt{x^2+y^2} )$.
