laplacian of a function, relation to the function 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?  

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 $f \in C^\infty$.
