The conormal derivative of a function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:21:04Z http://mathoverflow.net/feeds/question/109757 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109757/the-conormal-derivative-of-a-function The conormal derivative of a function Mike 2012-10-15T20:09:24Z 2012-10-17T08:40:34Z <p>Hi!</p> <p>I was wondering about the definition of the conormal derivative of a function $u$ which is given on a domain $\Omega$. It is known that if $-\Delta u = f$, considered as functionals on $H^1_0(\Omega)$, this does not provide enough information to define a conormal derivative of $u$. However, a lot of textbooks, for example "Strongly elliptic systems..." by W. McLean, state that if $-\Delta u = f$ as functionals on $H^1(\Omega)$, then the conormal derivative of $u$ can be defined (just by enforcing Green's formula). I'm not very convinced of this, because saying that $-\Delta u = f$ as functionals on $H^1(\Omega)$ means that i need to have a conormal derivative of $u$ defined already! Or, i can put it like that: either,</p> <ul> <li>the definition of a weak solution $u$ and its conormal derivative has to be done simultaneously. However, i don't see how to do that.</li> <li>Or, the definition of the conormal derivative depends on both, $u$ and $f$. But does this make any sense?</li> </ul> <p>Maybe someone has thought about this and can share his ideas with me.</p> http://mathoverflow.net/questions/109757/the-conormal-derivative-of-a-function/109765#109765 Answer by Delio Mugnolo for The conormal derivative of a function Delio Mugnolo 2012-10-15T21:54:51Z 2012-10-17T08:40:34Z <p>You may define a conormal derivative of $u$ in a very weak sense, just as a distribution. Or you can indeed take a normal derivative in a stronger recalling that each $u\in H^\frac{3}{2}(\Omega)$ s.t. $\Delta u\in L^2(\Omega)$ has a weak normal derivative in $L^2(\Omega)$, see e.g. the classical book of Lions-Magenes if you allow for a smooth boundary of $\Omega$; things are more delicate if $\Omega$ is rougher, say, merely Lipschitz, but can still be dealt with, cf. e.g. <a href="http://dx.doi.org/10.1016/j.jde.2011.06.017" rel="nofollow">this article</a>.</p> <p>Here a weak normal derivative is defined as follows: If $u\in H^1(\Omega)$, then $g\in L^2(\partial \Omega)$ is called its weak normal derivative if the Gauss-Green formula $$\int_\Omega \nabla u \nabla \phi +\int_\Omega \Delta u \phi = \int_{\partial \Omega} g \phi$$ holds for all $\phi \in H^1(\Omega)$. ($g$ need not exist for general $u\in H^1(\Omega)$; but if it exists, it is clearly unique).</p>