Show that $\frac{\partial}{\partial r}\int_{S_r}u(x,y)ds =0$

Question

Show that for all $0<\rho\leq r$ $\frac{\partial}{\partial r}\int_{S_r}u(x,y)ds =0$.Well, the hypotesis is $u$ is harmonic ($\Delta u=0$) , $0\in \Omega$,and $B_{\rho}$ the ball with radios, $\rho>0$ $B_{\rho}\subset \Omega$ and $\frac{\partial}{\partial r}$ is a unitary radial vector, $S_r$ circle with center in origin. Iam not sure how to star, i need to show that the integral is a constant respect to $r$, i know that the derivate can be introduce under integral sign because $u$ is of class $C^2$ and $S_r$ is compact set(close and bounded in $R^n$) but what can i do? The other side i try to do parts integration but i need to put other function $v$ that let zero on the boundary, but dont go anywhere? Can you help me with a hint please? Thank you

Answer 1

One answer is that this is automatically zero by the spherical mean-value property of harmonic functions. However, this fact is generally used to prove the spherical mean-value property, so let's do it directly.

First for simplicity let's define $\phi(r) = \frac{1}{r^{n - 1}}\int_{S_r} u(x) dS(x)$. Here $x \in \mathbb{R}^n$ and $dS(x)$ represents the surface measure with respect to varying $x$ (the proof is no different in 2D than in $n$D). The first step is realizing that it's hard to take derivatives when the variable is in the limit of integration, so the first thing to do would be to reparametrize the integral via a change of variables to get the $r$ out of the limits and into the integrand. The standard choice is to integrate over the unit sphere. This gives, by changing variables $x \mapsto ry$ for $y \in S_1$, $$ \phi(r) = \int_{S_1}u(ry)dS(y). $$ (Where did the factor of $r^{n - 1}$ go?) Then take a derivative using the chain rule.