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?
$u(r,\theta)=\frac{\log{r}}{\log{a}}$ 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).
For a rectangle, the uniqueness result still holds, but the solution is instead made out of Fourier series (one can construct solutions that are nonzero only on one side of the rectangle and sum them, for example).