How to prove that if $-\nabla^2 \Phi = \rho$ and $\rho$ is bounded then $\Phi$ is bounded.

Question

This problem is obviously inspired by the study of electrostatics, and requires a little more detail than in the title. Precisely, suppose that two functions defined in 3-$d$ space, $\Phi(\mathbf{x})$ and $\rho(\mathbf{x})$, that are related through Poisson's equation: $$-\nabla^2 \Phi = \rho.$$ If it is given that $\rho$ is bounded from above and below ($R_L \le \rho \le R_M$) in some finite volume $V$, is it possible to prove that $\Phi$ is bounded in that same volume?

I can argue that the presence of point charges and line charges produces divergences in $\Phi$, but surface charges do not. The pattern is that charge densities with support on sets with dimension $0$ and $1$ produce divergent values of $\Phi$, but those on sets of dimension $2$ do not, so $3$ must not either. Is that sufficient? Is there a more rigorous argument that can be made based on Green's theorem with Dirichlet boundary conditions? $$\Phi(\mathbf{x}) = \int_V \frac{\rho(\mathbf{x} - \mathbf{x}')}{4\pi|\mathbf{x} - \mathbf{x}'|} \operatorname{d}^3\mathbf{x} + \Phi_0(\mathbf{x}),$$ where $\Phi_0(\mathbf{x})$ is the field produced by the boundary conditions/charges external to $V$ and is bounded in $V$ (it satisfies $\nabla^2 \Phi_0 = 0$ in $V$).

Answer 1

Because the integral kernel $\frac{1}{4\pi \lvert \mathbf{x} \rvert}$ is positive, the analysis is quite straightforward for this: if we have $ R_L \leq \rho \leq R_M $, then positivity of the integral ($\int f>0$ if $f>0$) implies that $$ R_L\int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV \leq \int_V \frac{\rho(\mathbf{x}-\mathbf{x}')}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV \leq R_M\int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV $$ To show that the integral $\int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV$ is bounded, if $V$ is bounded, we can find a large ball $B$ of radius $R$ centred on $\mathbf{x}$ so that $V$ is contained in this ball whenever $\mathbf{x} \in V$. But again since $\frac{1}{4\pi \lvert \mathbf{x} \rvert}>0$, we have $$ \int_V \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV < \int_B \frac{1}{4\pi \lvert \mathbf{x}-\mathbf{x}' \rvert} \, dV = \frac{1}{4\pi}\int_0^R \frac{1}{r} 4\pi r^2 \, dr = \int_0^R r \, dr = \frac{1}{2}R^2, $$ using polar coordinates centred at $\mathbf{x}$.

Answer 2

The elliptic regularity theorem says that if you have an elliptic PDE $Lu=f$ on a bounded domain with nice enough boundary, subject to consistent boundary conditions, the coefficients of $L$ have $2k$ continuous derivatives, and $f$ is square integrable, then $u$ has $2k$ continuous derivatives (and in particular is bounded). Poisson's equation for a bounded forcing has these properties (provided that the sum of the fluxes from the boundary conditions and the forcing is zero). This is a bit laborious to prove; one should first go through the elliptic existence/uniqueness theory of developing Lax-Milgram to get solutions in an appropriate Sobolev space, then proceed with pointwise derivative estimates on the weak solution that you now know exists and is unique. It is carried out in full detail in Evans Chapter 5.