If $u$ is harmonic then $\frac{\partial}{\partial r}\int_{S_{r}}u(x,y)ds=0$

Question

Let $u:\Omega\rightarrow \mathbb{R}$ be an harmonic function (this is a smooth function) such that $$\Delta u =0 \quad \mathrm{ in }\quad \Omega,$$ where $\Omega\subseteq \mathbb{R}^{2}$ is an open set. Suppose that $0\in \Omega$ and $\rho>0$ such that $\mathcal{B}_{\rho}(0)\subset \Omega$, where $\mathcal{B}_{\rho}(\mathbf{0})$ is the ball of radius $\rho$ centered at $\mathbf{0}=(0,0)$.

Let $\frac{\partial}{\partial r}$ be the radial unitary vector. Show that for all $r>0$ such that $r\leq \rho$ $$\frac{\partial}{\partial r}\int_{S_{r}}u(x,y)\,ds=0,$$ where $S_{r}$ is a circle of radius $r$ centered at $(0,0)$.

Answer 1

This is a result of the divergence theorem applied to $F= \nabla u$ on $V=\mathcal{B}_r(\mathbf{0})$ while $\partial V=S_r$.

Answer 2

Set $$\phi(r) := \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(y) \, dS(y) = \frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} u(x+rz) \, dS(z).$$ Then $$\phi'(r)=\frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} \nabla u(x+rz) \cdot z \, dS(z)$$ and consequently, using [Green's formulas][1] we obatin

\begin{align}\phi'(r) &= \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \nabla u(y) \cdot \frac{u-x}r dS(y) \\ &= \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \frac{\partial u}{\partial \nu} dS(y) \\ &= \frac rn \frac{1}{|B(x,r)|}\int_{B(x,r)} \Delta u(y) \, dy = 0.\end{align}