Let $u:\mathbb{R}^3 \to \mathbb{R}$ a harmionic function, and let $S_r = \{x^2+y^2+z^2 = r^2 \}$

I want to show that if $0 < r_1 <r_2$, then:

$$ \frac{1}{\mathrm{area}(S_{r_1})} \int_{S_{r_1}} u^2 dS \le \frac{1}{\mathrm{area}(S_{r_2})} \int_{S_{r_2}} u^2 dS$$

I know that: $u^2(0) \le \frac{1}{\mathrm{area}(S_r)} \int_{S_r} u^2 dS$

So I can instead try to show $$\frac{1}{\mathrm{area}(S_{r_1})} \int_{S_{r_1}} u^2 dS \le u^2(0)$$

But I don't see how to do it.

Help would be appreciated.


Theorem(Green's formula). Assume $U$ is a bounded, open subset of $\mathbb R^n$, and $\partial U$ is $C^1$. Let $u,v\in C^2(\bar{U})$. Then $$\int_UDu\cdot Dv\,dx=-\int_U u\Delta v\,dx+\int_{\partial U}\frac{\partial v}{\partial \nu}u\,dS.$$

We can use this to show the $n-$dimensional case.

Set $$\phi(r):=\frac1{\text{area}(S_r)}\int_{S_r}u^2(y)\,dS(y)=\frac1{\text{area}(S_1)}\int_{S_1}u^2(rz)\,dS(z).$$ Then $$\phi'(r)=\frac1{\text{area}(S_1)}\int_{S_1}2u(rz)Du(rz)\cdot z\,dS(z),$$ and consequently, using Green's formula, we compute \begin{align*} \phi'(r)&=2\frac1{\text{area}(S_r)}\int_{S_r}u(y)Du(y)\cdot\frac yr\,dS(y)\\&=2\frac1{\text{area}(S_r)}\int_{S_r}u\frac{\partial u}{\partial \nu}\,dS\\&=2\frac1{\text{area}(S_r)}\left(\int_{B_r}Du\cdot Du\,dx+\int_{B_r}u\Delta u\,dx\right)\\&=2\frac1{\text{area}(S_r)}\int_{B_r}|Du|^2\,dx\geq 0. \end{align*} Hence $\phi(r)$ is increasing.

Addendum: From above we can get $$\frac1{\text{area}(S_r)}\int_{S_r}u^2\,dS=\phi(r)\geq \lim_{r\to 0}\phi(r)=u^2(0),$$ since $\phi$ is continuous at $r=0$, which can be proven using the method in a previous post.

  • 1
    $\begingroup$ @ThomasShelby $D(u^2)=2u Du$? $\endgroup$ – Feng Shao Sep 1 at 14:00
  • $\begingroup$ I am not sure exactly how did you get from the second line to the third line, could you elaborate? In other note, if I declare $\phi (r)$ as you did, then I will have $u^2(0) \le \phi (r)$ and then $0 \le \phi '(r)$ and I think it is enough $\endgroup$ – Gabi G Sep 1 at 14:28
  • $\begingroup$ @GabiG You can just differentiate the integrand: $$\frac{d}{dr}(u^2(rz))=2u(rz)\frac{d}{dr}(u(rz))=2u(rz)Du(rz)\cdot z$$ $\endgroup$ – Feng Shao Sep 1 at 15:05
  • $\begingroup$ @ThomasShelby Yes, it was almost a copy from Evans. Thanks for the link. I will add it in the text tomorrow. $\endgroup$ – Feng Shao Sep 1 at 15:07

For a smooth $f$, $$ \begin{align} \frac{\partial}{\partial r}\frac1{|S_r|}\int_{S_r}f(x)\,\mathrm{d}\sigma(x) &=\frac1{|S_r|}\int_{S_r}n(x)\cdot\nabla f(x)\,\mathrm{d}\sigma(x)\tag1\\ &=\frac1{|S_r|}\int_{B_r}\Delta f(x)\,\mathrm{d}x\tag2\\ \end{align} $$ Explanation:
$(1)$: the derivative of the spherical average is the spherical average of the outward derivative
$(2)$: Divergence Theorem

Furthermore, $$ \Delta f^2=2|\nabla f|^2+2f\Delta f\tag3 $$ Thus, for a harmonic $u$ $$ \begin{align} \frac{\partial}{\partial r}\frac1{|S_r|}\int_{S_r}u(x)^2\,\mathrm{d}\sigma(x) &=\frac1{|S_r|}\int_{B_r}2|\nabla u|^2\,\mathrm{d}x\tag4\\ &\ge0\tag5 \end{align} $$ That is, $\frac1{|S_r|}\int_{S_r}u(x)^2\,\mathrm{d}\sigma(x)$ is a non-decreasing function of $r$.

Thus $u(0)^2\le\frac1{|S_r|}\int_{S_r}u(x)^2\,\mathrm{d}\sigma(x)$ for all $r$, so the suggested attempt probably won't work.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.