"Overdetermined" Poisson equation Consider the PDE $-\Delta u = f$ on a bounded domain $\Omega \subset \mathbb{R}^n$, where $f \in C^\infty(\bar{\Omega})$. I wish to consider both the boundary conditions $u = 0$ and $\frac{\partial u}{\partial n} = 0$. My question is, are there reasonably well-known "compatibility" conditions under which such equations admit a solution?
 A: Here is one study Overdetermined elliptic problems in physics.
It is proven in $\mathbb{R}^2$ that the Poisson equation $\Delta u=-{\rm constant}$ in $\Omega$, with boundary conditions $u=0$, $\partial u/\partial n={\rm constant}$ on $\delta\Omega$, only has a solution for a circular domain. [That solution is $u(x,y)\propto R^2-x^2-y^2$.]
A: We claim the following: a solution $u$ exists if and only if $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$.

Suppose that $u$ is a solution. Since $u = 0$ on the boundary, we have $$u(x) = \int_\Omega G_\Omega(x, y) f(y) dy,$$ where $G_\Omega(x, y)$ is the Green function. Assuming $\Omega$ is sufficiently regular, one can differentiate under the integral sign and write $$0 = \partial_n u(x) = \int_\Omega \partial_n G_\Omega(x, y) f(y) dy,$$ where $\partial_n$ denotes the derivative in $x$ in the direction normal to the boundary; here of course $x \in \partial \Omega$. The normal derivative of the Green function defines a Poisson kernel. Thus, $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$.
Conversely, if $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$, then $u(x) = \int_\Omega G_\Omega(x, y) f(y) dy$ has all desired properties. Therefore, these two conditions are equivalent, as desired.

Alternatlively, one could argue as follows: if $u$ is a solution and $h$ is a harmonic function in $\Omega$ which is $C^1$ in $\overline{\Omega}$, then $$\int_\Omega h(x) f(x) dx = \int_\Omega h(x) \Delta u(x) = \int_\Omega \Delta h(x) u(x) dx = 0$$ by Green's first identity and the boundary conditions imposed on $u$. Therefore, $f$ is necessarily orthogonal to all harmonic functions.
