Laplace's equation on a square domain with a central point reservoire Could someone please tell me the solution to this problem. I have $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$ on the square domain $-L<x<L, -L<y<L$ with homogeneous Dirichlet boundary conditions: $$u(L,y)=u(-L,y)=u(x,L)=u(x,-L)=0$$ and further condition $u(0,0)=1$. 
I have tried separation of variables but quickly realised that it would only give me the trivial $u=0$. 
This is not for some kind of homework, I am solving the heat equation numerically and would just like to know what the analytical steady state solution is so I may compare. 
Thanks in advance!
 A: There's only one solution to the problem 
$$\begin{cases}
\Delta u =0 & \Omega, \\
u=0 & \partial \Omega
\end{cases}
$$
when $\Omega$ is a domain with Lipschitz continuous boundary, such as the square. This solution is the identically vanishing one, as you can see in various ways; one of them is observing that the energy integral 
$$\int_{\Omega} \lvert \nabla u\rvert^2\, dx$$
must vanish. 
A: If $(0,0)$ is considered a boundary point, with $u(0,0)=1$ as a boundary condition, then there is  no solution of this boundary value problem  among the conditions. Indeed, a slightly generalized maximum principle says that if $u$ is a bounded harmonic function in a bounded domain  $\Omega$ and $u\le 0$ holds on $\partial \Omega$ except for finitely many points, then $u\le 0$ in $\Omega$. 
See also this question where rotational symmetry allows for an elementary proof that an isolated boundary point is not a regular point for the Dirichlet problem.
A: The world of piecewise functions can have innovative possibilities.
You can introduce some jumping points in the normal continuous functions so that the derivatives of the modified functions do not change.
In fact the solution of this PDE can be $u(x,y)=\begin{cases}1&\text{when}~x=y=0\\0&\text{otherwise}\end{cases}$ .
