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.

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