Divergence theorem in $H^1(\Omega)$.

Question

Let $u,v\in H^1(\Omega)$, where $\Omega$ is a Lipschitz domain in $\Bbb R^n$. It is my understanding that the divergence theorem tells us $$\int_\Omega\nabla u\cdot\nabla v\,dx=\int_{\partial\Omega}\frac{\partial u}{\partial n}v\,dS(x)-\int_\Omega v\Delta u \,dx,$$ where $n$ is the unit outward normal to $\partial\Omega$.

My issue with this theorem (which I believe to be true) is that for $u\in H^1(\Omega)$ we do not necessarily have that $\frac{\partial u}{\partial n}\in H^{-\frac{1}{2}}(\partial\Omega)$ or $\Delta u\in H^1(\Omega)^*$ (the dual of $H^1(\Omega)$), so what exactly is meant by the integration on the right hand side? Is it in some sense a duality pairing of the form $$\langle\frac{\partial u}{\partial n}-\Delta u|\cdot\rangle_{H^1(\Omega)^*}:H^1(\Omega)\to\Bbb R,$$ i.e., we are able to deduce that the sum of the two are in the dual space but not the separate entities?

The reason this is of interest to me is that through the divergence theorem I believe one can easily prove that $\frac{\partial u}{\partial n}\in H^{-\frac{1}{2}}(\partial\Omega)$ if and only if $\Delta u\in H^1(\Omega)^*$.

Answer 1

Here is the most general divergence theorem that I am aware of:

Theorem. Let $\Omega$ be an open subset of $\mathbb{R}^N$ of class $C^1$ with a bounded frontier $\Gamma$ and $v \in W^{1,1}(\Omega,\mathbb{R}^N)$. Then $$ \int_\Omega \operatorname{div} v\, dx = \int_\Gamma \gamma_0(v)\cdot \nu \, d\gamma $$ where $\gamma_0$ is the trace operator.

This is Theorem 6.3.4 of Michel Willem's book Functional analysis. It seems to me that if you want to take $v = u \nabla w$ some strong assumption on the regularity of $u$ and $w$ should be made.

Answer 2

I am a bit rusty with all the spaces, but from my point everything should be well defined. If $u\in H^1(\Omega) $, then I can write $u = u_0 + v$ where $u_0 \in H^1_0(\Omega)$ and $v$ is the solution of $$\begin{cases} \Delta v = 0 &\text{ in } \Omega \\ v = u &\text{ on } \partial \Omega. \end{cases}$$ Then $\Delta u = \Delta u_0 \in H_0^{-1}(\Omega) = (H^1(\Omega))^*$.

I am even more rusty with the trace operator, but we can extend the normal $n$ as a Lipschitz function $\tilde{n}$ some distance into the interior and thus define some function $w=\frac{\partial u}{\partial \tilde{n}}$ which is in $L^2$ and thus should have a trace equal to $\frac{\partial u}{\partial n}$ in $H^{-1/2}$, although one would definitely need to look this up somewhere to be on the safe side.

Although in a way, this looks a bit backwards to me, since the easiest way to define operators like $\Delta$ on $H_0^1$ is to say that the result should formally satisfy this theorem for any $v$, since $H^{-1}$ is not actually a space of functions but rather one of distributions.

Answer 3

You should use the divergence theorem:

Divergence Theorem:(see "Finite element methods for Maxwell's equations" by Peter Monk, Theorem 3.24)

Let $\Omega\subset \mathbb R^3$ be a bounded Lipschitz domain with a unit outward normal $\nu$. Then

1) the mapping $$\gamma_\nu: \left[ C^\infty(\overline\Omega)\right]^3\rightarrow H^{-1/2}(\partial \Omega)$$ defined by $$\langle \gamma_\nu(\vec v),\varphi\rangle_{H^{-1/2}(\partial \Omega)\times H^{1/2}(\partial \Omega)}=\int\limits_{\partial\Omega}{\vec v\cdot \vec \nu \varphi ds}=\int\limits_{\Omega}{\vec v\cdot \nabla \varphi dx}+\int\limits_{\Omega}{\nabla\cdot\vec v \,\varphi dx}$$ (this is just Green's theorem, valid for functions $\vec v\in\left[ C^\infty(\overline\Omega)\right]^3$ and $\varphi\in H^1(\Omega)$ which you prove by the density of $C^\infty(\overline\Omega)$ in $H^1(\Omega)$)

can be extended by continuity to a continuous linear map $\gamma_\nu\,\,$ from $H(\text{div};\Omega)$ onto $H^{-1/2}(\partial\Omega)$ and moreover

2) the following Green's theorem holds for functions $\vec v\in H(\text{div};\Omega)$ and $\varphi\in H^1(\Omega)$ $$\int\limits_{\Omega}{\vec v\cdot \nabla \varphi dx}+\int\limits_{\Omega}{\nabla\cdot\vec v \,\varphi dx}=\langle \gamma_\nu(\vec v),\varphi\rangle_{H^{-1/2}(\partial \Omega)\times H^{1/2}(\partial \Omega)}\quad\quad (*)$$

Note 1: The extension by continuity is possible, because $\left[ C^\infty(\overline\Omega)\right]^3$ is dense in $H(\text{div};\Omega)$ which is a normed linear space (also a Hilbert space) and $H^{-1/2}(\partial \Omega)$ is a Banach space.

Note 2: In $(*)$ it is best to use the notation $\langle \gamma_\nu(\vec v),\varphi\rangle_{H^{-1/2}(\partial \Omega)\times H^{1/2}(\partial \Omega)}$ instead of $\int\limits_{\partial\Omega}{\vec v\cdot \vec \nu \varphi ds}$, because for $\vec v\in H(\text{div};\Omega)\setminus \left[C^\infty(\overline\Omega)\right]^3$ the usual trace $\gamma_0(\vec v)$ might not exist as a function in $L^2(\partial\Omega)$ such that $\langle \gamma_\nu(\vec v),\varphi\rangle_{H^{-1/2}(\partial \Omega)\times H^{1/2}(\partial \Omega)}=\int\limits_{\partial\Omega}{\gamma_0(\vec v)\cdot \vec \nu \varphi ds}$

Note 3: If in addition $\vec v\in \left[H^s(\Omega)\right]^3$ with $s>1/2$, then $\gamma_0(\vec v)=\vec v_{|\partial\Omega}\in L^2(\partial\Omega)$ and you can use the notation with the surface integral.

Now for your question:

Let $u,v \in H^1(\Omega)$. Because $\Delta u=\nabla\cdot\nabla u$, then according to the Divergence Theorem above, if in addition $\nabla u\in H(\text{div};\Omega)$, you can write it in the form (using the duality product notation) $$\int\limits_{\Omega}{\nabla u\cdot \nabla v dx}=-\int\limits_{\Omega}{v \Delta u dx}+\langle \gamma_\nu(\nabla u), v\rangle_{H^{-1/2}(\partial \Omega)\times H^{1/2}(\partial \Omega)}$$. Now according to Note 3 if $u\in H^s(\Omega)$ for some $s>3/2$, then you can write $$\int\limits_{\Omega}{\nabla u\cdot \nabla v dx}=-\int\limits_{\Omega}{v \Delta u dx}+\int\limits_{\partial\Omega}{\frac{\partial u}{\partial \nu} v ds}$$