Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
What do you mean by the Laplacian of a singular function? – Qiaochu Yuan Nov 29 '12 at 22:46
I was just wondering whether $\Phi$ may have discontinuities or not, if $\phi$ is continous – molkee Nov 29 '12 at 23:45
up vote 2 down vote accepted

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 $f \in C^\infty$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.