## Monotone formula of harmonic functions

Suppose $B_1\in \mathbb{R}^n$, and $u$ is a harmonic function is $B_1$

$\Delta u=0$

For $r\in (0,1)$, we can define

$\displaystyle E(r)=\int_{B_r}|\nabla u|^2dx$,
$\displaystyle V(r)=\int_{\partial B_r}u^2ds$
$\displaystyle F(r)=\frac{rE(r)}{V(r)}$

In general $F(r)$ is called the frequency of $u$. If $u$ is a homogeneous harmonic polynomial of degree $k$, then $F(r)=k$
Suppose $u=r^k\phi(\theta)$, where $\phi$ is a spherical harmonic polynomial functions on $\mathbb{S}^{n-1}$, then $u_n=kr^{k-1}\phi(\theta)$,

$\displaystyle F(r)=\frac{rE(r)}{V(r)}=k$.

$\mathbf{Thm:}$ $F(r)$ is a non-decreasing function in $r\in(0,1)$

$\mathbf{Proof:}$
Since $\Delta u=0$, then $\Delta u^2=2|\nabla u|^2$. From the Green formula
$\displaystyle E(r)=\frac 12\int_{B_r}\Delta u^2=\int_{\partial B_r}u u_n$
$\displaystyle F'(r)=\frac{E(r)}{V(r)}+\frac{rE'(r)}{V(r)}-\frac{rV'(r)E(r)}{V(r)}=F(r)\left(\frac 1r +\frac{E'(r)}{E(r)}-\frac{V'(r)}{V(r)}\right)$
We will prove that $F'(r)\geq 0$ for $\displaystyle r\in (0,1)$.
$\displaystyle V(r)=\int_{\partial B_r}u^2ds=\int_{\partial B_1}u^2(rw)r^{n-1}dw$
$\displaystyle V'(r)=\int_{\partial B_1}(n-1)r^{n-2}u^2(rw)dw+\int_{\partial B_1}2u(rw)\frac{\partial u}{\partial r}(rw)r^{n-1}dw=\frac{n-1}{r}H(r)+2\int_{\partial B_1}u u_nds$

$\displaystyle \frac{V'(r)}{V(r)}=\frac{n-1}{r}+\frac{2\int_{\partial B_r}u u_n}{\int_{\partial B}u^2}\quad (1)$
$\displaystyle E'(r)=\int_{\partial B_r}|\nabla u|^2ds$

By the Green formula

$\displaystyle \int_{\partial B_r}u_i^2\cdot x^2_jds=r\int_{\partial B_r}u_i^2x_j\nu_jds=r\int_{B_r}2u_iu_{ij}x_j dx+\int_{B_r}u^2_idx\quad (2)$

where $\displaystyle \nu_j=\frac{x_j}{r}$ is the j-th component of outer normal vector of $\displaystyle \partial B_r$.
Again apply green formula

$\displaystyle \int_{ B_r}u_i u_{ij}x_jdx+\int_{B_r}u_{ii}u_jx_jdx+\int_{B_r}u_iu_j\delta_{ij}=\int_{\partial B_r}u_iu_jx_j\nu_i \quad (3)$

Plug (3) into (2), we get
$\displaystyle \int_{\partial B_r}u^2_i x_j^2=r\int_{B_r}u_i^2dx+2\int_{\partial B_r}u_iu_jx_ix_j-2r\int_{B_r}u_{ii}u_jx_j-2r\int_{B_r}u_iu_j\delta_{ij}$
Sum over j and i, noticing $u$ is haromic,

$\displaystyle r^2E(r)=rnD(r)+2r^2\int_{\partial B_r}u_n^2-2rD(r)$

So

$\displaystyle \frac{E'(r)}{E(r)}=\frac{n-2}{r}+\frac{2\int_{\partial B_r}u_n^2}{\int_{\partial B_r}uu_n}\quad (4)$

Plugging (1) and (4) to $F'(r)$, we get

$\displaystyle F'(r)=F(r)\left(\frac{\int_{\partial B_r}u_n^2}{\int_{\partial B_r}uu_n}-\frac{\int_{\partial B_r }uu_n}{\int_{\partial B_r}u^2}\right)\geq 0$

since the cauchy inequality.