# “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?

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 $$u(x,y)\propto R^2-x^2-y^2$$.]
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.