A PDE exercise from Gilbarg Trudinger. problem 2.2

Prove that if $\Delta u=0$ in $\Omega\subset \mathbb{R}^n$ and $u=\partial u/\partial\nu=0$ on an open smooth portion of $\partial \Omega$, then $u$ is identically zero.

I found a proof here, but I'm not sure if it's the correct proof. Does anyone have any idea? Any hint would be appreciated.

Solution 1:

A very quick (and insightful) proof can be given by applying John's general Holmgren theorem, that is discussed e.g. in the PDE textbooks by Jeffrey Rauch and Fritz John.

I have another, more direct approach, but before going further, I strongly advise you to read the answer below piece by piece, and make an attempt to complete the solution at each break. (I remember having spent a lot of time on this problem.)

The initial idea is similar to that of the blog post you linked to, but it proceeds differently. We work in a small open set $\omega$ that is divided into two parts by the smooth portion of the boundary, and we extend $u$ by zero outside the domain. We want to show that $u$ is harmonic in $\omega$, which would then imply that $u\equiv0$ by analyticity. The boundary conditions ensure that $u$ is a $C^1$ function in $\omega$. It is known that $u\in C^1(\omega)$ is harmonic in $\omega$ iff $$ \int_{\partial B}\partial_\nu u = 0, $$ for all sufficiently small closed balls $B$ in $\omega$, where $\partial_\nu$ is the normal derivative (It is kind of like the mean value property, but it works with derivatives of $u$). This property is satisfied for all small balls that are not centred at $\omega\cap\partial\Omega$. So we assume $B\subset\omega$ is a ball whose centre is on $\omega\cap\partial\Omega$. In this case, we have $$ \int_{\partial B}\partial_\nu u = \int_{\partial B\cap\Omega}\partial_\nu u, $$ because $u=0$ outside $\Omega$. Now we recall that for a bounded domain $U$ with piecewise smooth boundary, and for a function $f$ that is $C^2$ in $U$ and $C^1$ in a neighbourhood of $\overline U$, $$ \int_{U} \Delta f = \int_{\partial U} \partial_\nu \,f. $$ We apply this with $U=B\cap\Omega$ and $f=u$ to conclude that $$ \int_{\partial B\cap\Omega}\partial_\nu u = \int_{\partial( B\cap\Omega)}\partial_\nu u = 0. $$