MathOverflow is a question and answer site for professional mathematicians. 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

In which sense is it possible to solve $\Delta u=0$, $\partial_\nu u=\phi$, for $\int\phi=0$ on a closed domain, say a ball $B^3\subset\mathbb R^3$?

For example would a $\phi\in L^p(\partial B^3)$, $1< p<2$ make sense?

In other words, is $W^{1,p}$ really the right trace space, or else, which is?

Where can I find this kind of results?


share|cite|improve this question
up vote 5 down vote accepted

You can solve the problem with even less regularity than in Rekalo's answer. If $u\in W^{1,p}(\Omega)$, it does not have a normal trace in general. But if you assume in addition that $\Delta u\in L^p(\Omega)$, then the normal trace is well-defined and belongs to $W^{-1/p',p}(\partial\Omega)$, where $p'$ is the conjugate exponent. This space is of negative order, thus is not contained in any $L^q$. It is defined as the dual space of $W^{1/p',p'}(\partial\Omega)$. Since the latter contains the function ${\bf 1}$, it makes sense to say that the integral of $\phi$ is zero: just write $\langle\phi,{\bf 1}\rangle=0$. Under this condition, the Neumann boundary value problem admits a solution, unique up to an additive constant.

share|cite|improve this answer

If you are interested in $L^p$-theory, you are probably looking for solutions belonging to a Sobolev class $H^{s,p}(\Omega)$ with some $s>0$ and $p>1$. In this case, the Besov space $B^{s-1-1/p,p}(\partial\Omega)$ is the "right" trace space. In particular, the restriction map $$\rho: H^{s,p}(\Omega)\to B^{s-1-1/p,p}(\partial\Omega)$$ $$u\mapsto \partial_{\nu} u$$ is well defined for all $s>1+1/p$ and is surjective.

As for the Neumann problem, the following result is true

Theorem. Let $s>1+1/p$ where $1< p< \infty$. Then the Neumann problem $$\begin{cases} \triangle u=f & \mbox{in }\Omega,\\\ \partial_{\nu} u=\phi & \mbox{on }\partial\Omega\end{cases}$$ has a unique solution $u$ in the space $H^{s,p}(\Omega)$ for any $f\in H^{s-2,p}(\Omega)$ and any $\phi\in B^{s-1-1/p,p}(\Omega)$.

Have a look at the very accessible exposition by Kazuaki Taira.

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.