# A distributional normal derivative for functions in $H^1(\Omega)$

Let $\Omega$ be a smooth bounded domain with $\partial\Omega = \Gamma$. I have read this.

For all $u \in H^1(\Omega)$ such that $-\Delta u = g \in L^2(\Omega)$ in distribution, we can define the normal derivative $u_\nu \in H^{-\frac 12}(\Gamma)$ such that $$\langle u_\nu, \psi \rangle_{H^{-\frac 12}(\Gamma), H^{\frac 12}(\Gamma)} = -\int_\Omega gD\psi + \int_\Omega \nabla u \nabla D\psi$$ where $D\psi \in H^1(\Omega)$ is an extension of $\psi \in H^{\frac 12}(\Gamma)$.

(eg. here) In other words, we have this notion of a weak normal derivative for $H^1$ functions (usually we need $u \in H^2$).

Question: why do we need to ask for $\Delta u \in L^2(\Omega)$? Why not just define the normal derivative like so: $$\langle u_\nu, \psi \rangle_{H^{-\frac 12}(\Gamma), H^{\frac 12}(\Gamma)} = \langle \Delta u, D\psi\rangle_{H^{1}(\Omega)^*, H^1(\Omega)} + \int_\Omega \nabla u \nabla D\psi?$$

I have not seen something like this in any text except these lecture notes. Can anyone tell me why?

• It might be just a matter of convenience: it is easier to work with an $L^2$ function than a more general distribution. Sep 27, 2014 at 14:35
• $H^1(\Omega)^*$ is not a space of distributions in $\Omega$. For instance, the functional on the left hand side of your equation is in $H^1(\Omega)^*$. This defeats the objective of separating the inhomogeneous term in the PDE from the boundary condition. Sep 28, 2014 at 2:26

It is not possible to define a normal derivative for all $$u \in H^1(\Omega)$$ which depends continuously on $$u$$.
The reason is that all $$C_c^\infty(\Omega)$$ is dense in $$H^1_0(\Omega)$$, but all $$u \in C_c^\infty(\Omega)$$ have zero normal derivative. If we would have a normal derivative depending continuously on $$u \in H^1(\Omega)$$, this implies that it is zero for all $$u \in H^1_0(\Omega)$$ and this is not correct.
Your definition fails since $$\Delta u \in H^1(\Omega)^*$$ is meaningless for $$u \in H^1(\Omega)$$ (how do you define it?).
• I do agree that is not possible to define the normal derivative for all $H^1$ functions. Well $\Delta u \in H^1(\Omega)^*$ can be defined $\langle \Delta u, v \rangle = \int_\Omega \nabla u \nabla v$. Also $C_c^\infty$ is not dense in $H^1$.