2
$\begingroup$

Suppose for some function $\Phi$ we have:

$ \nabla^2 \Phi(\mathbf{r})=\phi(\mathbf{r}) $

where $\phi(\mathbf{r})$ is some well-behaved smooth function, which is finite everywhere.

Does this mean that $\Phi(\mathbf{r})$ itself doesn't have any singularities?

Could you please point me out any useful theorems?

  • 0
    I was just wondering whether $\Phi$ may have discontinuities or not, if $\phi$ is continous2012-11-29

1 Answers 1

2

The Laplace operator is hypoelliptic (since it's elliptic with smooth coefficients) and any hypoelliptic operator $L$ has the property that if $Lf \in C^\infty$, then $f \in C^\infty$.