Well, I have the problem: $\Delta u(x,y)=0 \text{ on } \Omega_{a}=\{(x,y)\in\mathbb{R}^{2}:a<\lVert (x,y)\rVert<1\}$

$u(x,y)=1 \text{ on } \lVert (x,y)\rVert=a$

$u(x,y)=0 \text{ on } \lVert (x,y)\rVert=1$

And the question, is there any solution of this that is not radial?

Well, my approach is that the domain is symmetric so as the Laplace operator is invariant by rotation we can obtain something but I don't see this very clear. And if this works, why in the problem on the $B(0,1)$ doesn't admit radial solutions?

• Do you know how to prove uniqueness of solutions to Laplace's equation? Commented Jun 9, 2017 at 19:09
• I think so, studying the uniqueness of the problem with homogeneous boundary conditions, and with the maximum and minimum principle we can conclude that $u$ is 0, not? Commented Jun 9, 2017 at 19:24
• Yes, that's the idea. So find a radial solution and show that it's unique. Commented Jun 9, 2017 at 19:47
• Thank you very much, I think I understand the solution, we can think on radial solutions because the 0 is not include on our domain and adjusting our parametres of the fundamental solution we can obtain the unique solution. Commented Jun 9, 2017 at 19:56
• But one more question, this is possible because the domain is "radial" not? If the domain is a rectangle we can't find radial solutions not? Commented Jun 9, 2017 at 19:57

$u(r,\theta)=\frac{\log{r}}{\log{a}}$ is a radial solution. We now show that a solution to this problem is unique: let $u,v$ be two solutions. Then $w=u-v=0$ on both parts of the boundary and is harmonic. Then if $w \neq 0$ (so $u \neq v$) $$0 < \int_{\Omega_a} \lvert \nabla w \rvert^2 \, dV = \int_{\partial \Omega_a} w \nabla w \cdot d\mathbf{s} - \int_{\Omega_a} w\Delta w \, dV = 0,$$ since $w=0$ on the boundary and $\Delta w=0$ on $\Omega_a$, which is a contradiction unless $w=0$ everywhere, so $u=v$. Hence $u$ is the only solution.
There is a radial solution to Laplace's equation on $B(0,1)$ with the boundary condition $u(1,\theta)=0$, but it's exactly zero, by exactly the same argument (and indeed, this is the limit as $a\to 0$ of the $\Omega_a$ solution).