A harmonic function is non-decreasing 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.
Thank you advance for your supports
 A: 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.
