# A harmonic function is non-decreasing [closed]

We consider $$u:U\to\mathbb{R}$$ be a harmonic function. Now, I need to prove that for every $$x_{0}\in U$$, the function $$\left( {0,{\rm{dist}}\left( {{x_0},\partial U} \right)} \right)\ni\rho \mapsto \dfrac{1}{{{\rho ^{n - 2}}}}\int_{{B_\rho }\left( {{x_0}} \right)} {{{\left| {Du} \right|}^2}dx}$$

is non-decreasing.

We know that if $$u$$ is a harmonic function then $$u$$ be also a subharmonic function. But I am referring Example 2 in Section 8.6.2 in Partial Differential Equations Lawrence C.Evan with Monotonicity formula for harmonic functions . But I do not have an idea approaching the problem to solve because I confuse with them.

• The proof in Evan, he proves for the general case. But I am confusing when he differentiate the integral $\frac{d}{{dr}}\left( {\frac{1}{{{r^{n - p}}}}\int_{B\left( {0,r} \right)} {{{\left| {Du} \right|}^p}dx} } \right)$. After calculating we have it is greater than or equal zero and this is a monotonicity formula. – Phạm Trương Hoàng Nhân Jun 6 at 15:16
As is mentioned in the comments, the proof you're looking is precisely the example in Evans that you've been referred to except with $$p=2$$. From what I understand, your issue is in the first line when Evans computes $$\frac d {dr} \big ( \frac 1 {r^{n-2}}\int_{B(0,r)} \vert Du \vert^2 d x\big )$$ and possibly integrating (10) over $$B(0,r)$$. The product rule for derivatives and polar coordinates (see Section C.3 Thm 4 (ii) in Evans) yield \begin{align} \frac d {dr} \bigg ( {r^{2-n}}\int_{B(0,r)} \vert Du \vert^2 d x\bigg ) &= (2-n)r^{1-n}\int_{B(0,r)} \vert Du \vert^2 d x + r^{2-n}\frac d {dr} \int_{B(0,r)} \vert Du \vert^2 d x \\ &= \frac{2-n}{r^{n-1}}\int_{B(0,r)} \vert Du \vert^2 d x + \frac{1}{r^{n-2}} \int_{\partial B(0,r)} \vert Du \vert^2 d S \tag{1} \end{align} which is the second line in Evans' computation. Setting $$p=2$$ in (10) in the same example gives \begin{align*} \sum_{i=1}^n \big ( \vert Du\vert^2 x_i\big)_{x_i} &=2 \sum_{i=1}^n \bigg ( \bigg ( Du\cdot x + \frac{n-2}{2} u\bigg) u_{x_i} \bigg)_{x_i} \end{align*} which can also be written as $$\mathrm{div} (\vert Du\vert^2x) = 2 \mathrm{div}\bigg ( \bigg ( \vert x \vert u_r + \frac{n-2}{2} u\bigg) Du \bigg )$$ since $$u_r = Du \cdot \frac x {\vert x \vert }$$. Hence, the divergence theorem implies \begin{align} r \int_{\partial B(0,r)}\vert Du\vert^2 d S &= \int_{\partial B(0,r)} \vert Du\vert^2(x \cdot \nu )d S \\ &= \int_{B(0,r)} \mathrm{div} (\vert Du\vert^2x) dx \\ &= 2 \int_{B(0,r)}\mathrm{div}\bigg ( \bigg ( \vert x \vert u_r + \frac{n-2}{2} u\bigg) Du \bigg ) dx\\ &= 2 \int_{\partial B(0,r)} \vert x \vert u_r ( Du \cdot \nu ) d S+ (n-2)\int_{\partial B(0,r)} u( Du \cdot \nu )dS \\ &= 2r\int_{\partial B(0,r)} u_r^2dS + (n-2)\int_{\partial B(0,r)} u\frac{\partial u}{\partial \nu}dS \\ &= 2r\int_{\partial B(0,r)} u_r^2dS + (n-2)\int_{ B(0,r)} \vert Du \vert^2 dx \tag{2} \end{align} where the last line follow from the fact $$u$$ is harmonic. The above equality is again given in Evans. Substituting (2) into (1) gives \begin{align*} \frac d {dr} \bigg ( {r^{2-n}}\int_{B(0,r)} \vert Du \vert^2 d x\bigg ) &= \frac{1}{r^{n-1}} \bigg( 2r\int_{\partial B(0,r)} u_r^2dS - r \int_{\partial B(0,r)}\vert Du\vert^2 d S \bigg ) + \frac{1}{r^{n-2}} \int_{\partial B(0,r)} \vert Du \vert^2 d S \\ &= \frac 2 {r^{n-2}}\int_{\partial B(0,r)} u_r^2dS \geqslant 0 \end{align*} as required.