Rellich Integral Identity for Normal Derivatives

Question

Consider the differential operator on $u \in C^2(\overline \Omega)$ with $u \bigr|_{\partial \Omega}=0$ and $\Omega \subseteq \mathbb{R}^n$ $$ Au:= x \cdot \nabla u $$ Then I have seen the following "Rellich Identity" $$ \int_{\partial \Omega} Au \dfrac{\partial u}{\partial \nu} = \int_{\partial \Omega} (\nu \cdot x) \left( \dfrac{\partial u}{\partial \nu} \right)^2 $$ but I cannot show that this holds. It is definitely not true in general that $$ (\nu \cdot x)(\nabla u \cdot \nu) = (x \cdot \nabla u)$$ so it must be some integration by parts trick, but I can't get it.

Answer 1

Fix $p \in \partial \Omega$. Since $u = 0$ on $\partial \Omega$, we have $v \cdot \nabla u = 0$ for any tangent vector $v \in T_p \partial \Omega$. Therefore $\nabla u$ is proportional to the normal vector $\nu$, i.e. $\nabla u = \alpha \nu$ for some $\alpha \in \mathbb{R}$ (which of course depends on $p$).

Taking into account $|\nu|=1$, we see that the integrands are equal: $$ Au \dfrac{\partial u}{\partial \nu} = (x \cdot \alpha \nu) \cdot (\nu \cdot \alpha \nu) = (x \cdot \nu) \cdot \alpha^2 = (\nu \cdot x) \left( \dfrac{\partial u}{\partial \nu} \right)^2. $$ No integration by parts is involved.