let $u$ be harmonic in the domain $U \subset \mathbb{R}^n$ and $B_R(0) \subset U$ and $u(0)=0, u\neq 0$. Let $0<r<R$. Define $a(r):= \frac{1}{r^{n-1}} \int_{\partial B_r(0)} u^2dS, b(r):= \frac{1}{r^{n-2}} \int_{B_r(0)} |\nabla u|^2dx$. The monotony-formula for $u$ harmonic is $b'(r)= \frac{2}{r^{n-2}} \int_{\partial B_r(0)} u_r^2dS$. Show that $a'= \frac{2}{r^{n-1}} \int_{\partial B_r(0)} u u_rdS=\frac{2}{r}b$.

I was able to show the second equality, but with the first one I have a little trouble. I used the transformation formula twice with the diffeomorphism $\Phi : B_1(0) \rightarrow B_r(0), y \mapsto ry=:x$ and $\Phi^{-1}$ with $|det(D\Phi(y))|=r$ to get: $a'= (\frac{r}{r^{n-1}} \int_{\partial B_1(0)} u(ry)^2dS(y))' \\ = \frac{-(n-2)r^{n-3}}{(r^{n-2})^2} \int_{\partial B_1(0)} u(ry)^2dS + \frac{1}{r^{n-2}}\int_{\partial B_1(0)} 2u(ry) u_r(ry)dS \mathrm{\;(with \; product \; rule)}\\ = \frac{-(n-2)}{r^{n-1}} \int_{\partial B_r(0)} u^2 \frac{1}{r}dS + \frac{1}{r^{n-2}}\int_{\partial B_1(0)} 2u u_r \frac{1}{r}dS$.

The last term is what I want to get but the first term should vanish. However I can't see why the first term should be zero, so I'm thinking that maybe my calculations are wrong but I don't know where. Maybe someone could tell me what is wrong in my conclusions.


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