Is every smooth function from $\mathbb{R}^n \to \mathbb R$ with compact support the laplacian of some function?

Question

Given a smooth function $g:\mathbb R^n \to \mathbb R $ with compact support, is it true that there exists a function $u$ such that $g=\Delta u$?

Answer 1

Yes. A solution is given by the convolution of g with the fundamental solution (see http://en.wikipedia.org/wiki/Newtonian_potential).

The solution $u$ is not unique because one could always add to $u$ a harmonic function $h$ (satisfying $\Delta h = 0$).

If you specify some sort of boundary data for $u$, then there is a unique solution. For example, suppose the support of $g$ is contained in some open set $\Omega$ (you may need some sort of boundary regularity on $\Omega$). Then there is a unique solution $u$ to $\Delta u = g$ satisfying $u = 0$ on $\partial \Omega$. (Now the solution is unique by the maximum principle.)