Let $u\in \mathcal C^2(\bar\Omega )$ the solution of $$\begin{cases}\Delta u(x)=0&x\in \Omega\\ u(x)+\frac{\partial u}{\partial \nu}(x)=0&x\in \partial \Omega \end{cases}$$ where $\Omega $ is a bounded domain of $\mathbb R^3$ and $\nabla=\nabla (x) $ is the exterior normal in $x\in\partial \Omega $. Show that $u=0$ is the unique solution to the PDE below. I recall that $\frac{\partial u}{\partial \nu}=\nabla u\cdot \nu$

My attempts

using divergence theorem (on $u\nabla u$), we have that $$\int_{\partial \Omega }u(\nabla u\cdot \nu)ds=\int_\Omega div(u\nabla u)=\int_\Omega \|\nabla u\|^2.$$ Using the PDE, we have $\nabla u\cdot \nu=-u$ and thus$$-\int_{\partial \Omega }u^2(x)ds=\int_\Omega \|\nabla u\|^2.$$ How can I conclude that $\nabla u=0$ ?

  • 1
    $\begingroup$ Even if we understand, the notation $\int_{\partial \Omega }u^2(x)\mathrm ds$ looks strange... At my opinion, it's better to write simply $\int_{\partial \Omega }u^2$ $\endgroup$ – Surb Dec 29 '16 at 10:11


1) What happen if $x^2\leq 0$ ?

2) What happen if $\int_\Omega |f|=0$ ?


If $a$ and $b$ are nonnegative numbers and $a+b=0$, then $a=b=0$ (otherwise we would have $a+b>0)$.

Now, note that $$-\int_{\partial \Omega }u^2(x)ds=\int_\Omega \|\nabla u\|^2\quad \Longrightarrow \quad \int_\Omega \|\nabla u\|^2+\int_{\partial \Omega }u^2(x)ds=0$$ and remember that the integral of a nonnegative function is nonnegative.


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