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?