Schwarz Reflection Principle for Harmonic Functions

Question

Given $\Omega \subset \mathbb{R}^n$ define $\Omega^+ = \Omega \cap \{x_n>0\}$ and $\Omega^0$, $\Omega^-$ analogously let $u \in C^2(\bar{\Omega}^+)$ be harmonic and such that $\frac{\partial u}{\partial x_n}=0$. Then we want to show that the reflection of $u$, i.e. $\tilde{u}=u(x,-x_n)$ for $(x,x_n) \in \Omega^-$, is harmonic on $\Omega^+ \cup \Omega^0 \cup \Omega^-$.

The mean value property seems like the easiest way to go about showing this since it easily gives us that the reflection is harmonic on $\Omega^-$.

How do I use $\frac{\partial u}{\partial x_n}=0$ to conclude that the reflection must be harmonic on the whole of $\Omega^+ \cup \Omega^0 \cup \Omega^-$ though?

Answer 1

The condition $\frac{\partial u}{\partial x_n}=0$ is put in place just to ensure that the extended function is smooth along the hyperplane $x_n=0$. Simply consider the easy example where $n=1$ and $u(x)=kx$ for $x>0$ where $k$ is some constant. In order to have a differentiable extension obtained by reflection you must have $k=0$.

Answer 2

I came across this answer later, and believe that it missed showing that $\tilde{u}$ is harmonic on $\Omega^0$. In case anyone is curious, this can be shown by noting that if $u\in C^2(\Omega^+)\cap C^1(\bar{\Omega^+})$ is harmonic with Neumann boundary conditions, then it obeys the following half-ball mean value property: if $x\in \Omega^0$ and $S_{r,+}(x):=B_r(x)\cap (\mathbb{R}^n)^+$ is the half-sphere centered at $x$, then $$ u(x)=\frac{1}{Area(S_{r,+}(x))}\int_{S_r(x)} u(y)dS(y) $$ The proof of this fact is similar to the proof of the standard mean value property (Evans Thm 2.2.2), with the additional detail that, for a fixed $r$, by the Neumann conditions and Stoke's Theorem, if $B_{r,0}=\bar{B_r}(x)\cap \Omega^0$, $$ \int_{S_{r,+}}\frac{\partial u}{\partial \nu} dS(y)=\int_{S_{r,+}}\frac{\partial u}{\partial \nu} dS(y)-\int_{B_{r,0}(x)}\frac{\partial u}{\partial \nu} dS(y)=\int_{B_{r}(x)\cap \Omega^+}\Delta u dx=0. $$