Trace space and Neumann boundary condition - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:20:43Z http://mathoverflow.net/feeds/question/41427 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41427/trace-space-and-neumann-boundary-condition Trace space and Neumann boundary condition Mircea 2010-10-07T16:14:04Z 2010-10-08T06:52:27Z <p>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 $B^3\subset\mathbb R^3$? </p> <p>For example would a $\phi\in L^p(\partial B^3)$, $1&lt; p&lt;2$ make sense? </p> <p>In other words, is $W^{1,p}$ really the right trace space, or else, which is? </p> <p>Where can I find this kind of results?</p> <p>Thanks!</p> http://mathoverflow.net/questions/41427/trace-space-and-neumann-boundary-condition/41432#41432 Answer by Andrey Rekalo for Trace space and Neumann boundary condition Andrey Rekalo 2010-10-07T17:05:31Z 2010-10-07T17:05:31Z <p>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.</p> <p>As for the Neumann problem, the following result is true</p> <blockquote> <p><strong>Theorem.</strong> Let $s>1+1/p$ where $1&lt; p&lt; \infty$. Then the Neumann problem $$\begin{cases} \triangle u=f &amp; \mbox{in }\Omega,\\ \partial_{\nu} u=\phi &amp; \mbox{on }\partial\Omega\end{cases}$$ has a unique solution $u$ in the space $H^{s,p}(\Omega)$ for any $f\in H^{s-2,p}(\Omega)$ and any $\phi\in B^{s-1-1/p,p}(\Omega)$.</p> </blockquote> <p>Have a look at the very accessible <a href="http://books.google.co.uk/books?id=e3D0-onLFd0C&amp;printsec=frontcover&amp;dq=Analytic+Semigroups+and+Semilinear+Initial+Boundary+Value+Problems&amp;source=bl&amp;ots=15BPHCo8LC&amp;sig=pygm9idqBG7ylZgQDAAXCDbcXjg&amp;hl=en&amp;ei=R_2tTIDTGISdOoWI2b8F&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CBUQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow">exposition</a> by Kazuaki Taira.</p> http://mathoverflow.net/questions/41427/trace-space-and-neumann-boundary-condition/41487#41487 Answer by Denis Serre for Trace space and Neumann boundary condition Denis Serre 2010-10-08T06:52:27Z 2010-10-08T06:52:27Z <p>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.</p>