Prove by the divergence theorem that $\int_{\partial \Omega} u\frac{\partial u}{\partial n}dS \ge 0$. For what harmonic function does equality hold?

Let $\Omega\subset R^n$ be a domain with a smooth boundary. Let $u\in C^2(\bar{\Omega})$ be harmonic on $\Omega$.

Prove using the divergence theorem that $\int_{\partial \Omega} u\frac{\partial u}{\partial n}dS \ge 0$ where we recall $\frac{\partial u}{\partial n} = N · \nabla u$ is the outward normal derivative of u and N is the unit outward normal vector along the boundary ∂Ω. For what harmonic function $u$ does the equality hold?

Well the diveregence theorem states that $\int_\Omega \nabla ·F dV = \int_{\partial \Omega} F·n dS$. If we let $F=u$ in the theorem, then $\int_{\partial \Omega} u\frac{\partial u}{\partial n}dS=\int_{\partial \Omega} F(N·\nabla F)dS$. I am not sure where to go from here.

Solution 1:

Since $\Delta u = 0$ on $\Omega$, therefore by Green's identity, we have $$ \int_{\partial\Omega}u\dfrac{\partial u}{\partial\mathbf{n}}dS = \int_{\Omega} u\Delta udx + \int_{\Omega} |\nabla u|^2dx = \int_{\Omega} |\nabla u|^2dx \ge 0 $$ The equality holds if and only if $|\nabla u| = 0$, hence $u$ is a constant function.

NOTE: You can use Divergence Theorem to prove Green's identity $$ \int_{\Omega}v\Delta udx = \int_{\partial\Omega} v\dfrac{\partial u}{\partial\mathbf{n}}dS - \int_{\Omega}\nabla u\cdot\nabla vdx $$