# Integrating Laplace's equation over a sphere

if the Laplacian of $$u$$ is integrated over any volume that encloses the source point, $$\iiint_V \nabla \cdot \nabla u \, d^3V =-1.$$

I can't seem to find an explanation for this elsewhere, can someone explain why this is the case? And would this still apply if the sphere is in thermal equilibrium?

• On the wikipedia page for what? This question is impossible to answer without context. – Bob Knighton Mar 3 at 21:57
• Page for Laplace's equation. – Rob Mar 3 at 21:58
• This is a pure mathematical result. It has nothing to do with thermal equilibrium. – G. Smith Mar 3 at 22:08

To add context missing from the question, we consider a fundamental solution to the Laplace equation

$$\Delta u=-\delta^{(3)}(\textbf{x}-\textbf{x}_0).$$

The answer to the question then becomes very simple. Integrate both sides over a sphere containing the source point $$\textbf{x}_0$$ and recalling $$\Delta u=\boldsymbol{\nabla}\cdot\boldsymbol{\nabla}u$$, we have

$$\oint_V\boldsymbol{\nabla}\cdot\boldsymbol{\nabla}u\,\mathrm{d}V=-1.$$