Trace space and Neumann boundary condition - MathOverflow

Trace space and Neumann boundary condition

2010-10-07T16:14:04Z

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?

Thanks!

Answer by Andrey Rekalo for Trace space and Neumann boundary condition

2010-10-07T17:05:31Z

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.

Answer by Denis Serre for Trace space and Neumann boundary condition

2010-10-08T06:52:27Z

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.