Let $u: \Omega \subset \mathbb{R}^n \rightarrow \mathbb{R}$ be a harmonic function, i.e. $\Delta u=0$. Let $x_0 \in \Omega$ be a point. Assume $n \geq 2$.
- Simplify the following formula $$ \operatorname{div}\left(\left(\boldsymbol{x}-\boldsymbol{x}_0\right)|\nabla u(\boldsymbol{x})|^2-2 \nabla u\left(\left(\boldsymbol{x}-\boldsymbol{x}_0\right) \cdot \nabla u(\boldsymbol{x})\right)\right) $$
- Use the formula you derive to prove the monotonicity of the function $D(r):=\dfrac{1}{r^{n-2}} \displaystyle\int_{B_r\left(x_0\right)}|\nabla u(x)|^2 d V(x)$ for $r>0$, where $B_r(\boldsymbol{x}_0)$ is the ball $|\boldsymbol{x}-\boldsymbol{x}_0|^2 \leq r^2$. One may just work on the case when $n=3$.
For the first question, I simplify by \begin{align*} \nabla [(\boldsymbol{x}-\boldsymbol{x}_0)|\nabla u(\boldsymbol{x})|^2] &= \nabla \cdot (\boldsymbol{x}-\boldsymbol{x}_0)|\nabla u(\boldsymbol{x})|^2+ (\boldsymbol{x}-\boldsymbol{x}_0)\nabla \cdot |\nabla u(\boldsymbol{x})|^2\\ &= \big(\nabla \cdot (\boldsymbol{x}-\boldsymbol{x}_0)+ (\boldsymbol{x}-\boldsymbol{x}_0)\nabla\big)|\nabla u(\boldsymbol{x})|^2 \\ &=\big(n+ (\boldsymbol{x}-\boldsymbol{x}_0)\nabla\big)|\nabla u(\boldsymbol{x})|^2 \end{align*} and \begin{align*} &\nabla(-2\nabla u((\boldsymbol{x}-\boldsymbol{x_0})\cdot \nabla u(\boldsymbol{x})))\\ =& -2\Delta u((\boldsymbol{x}-\boldsymbol{x_0})\cdot \nabla u(\boldsymbol{x})))\cdot \nabla \big((\boldsymbol{x}-\boldsymbol{x_0})\cdot \nabla u(\boldsymbol{x})\big) =0 \end{align*} since $u$ is harmonic. I use $P$ as a short hand for $\left(\left(\boldsymbol{x}-\boldsymbol{x}_0\right)|\nabla u(\boldsymbol{x})|^2-2 \nabla u\left(\left(\boldsymbol{x}-\boldsymbol{x}_0\right) \cdot \nabla u(\boldsymbol{x})\right)\right)$ The result I get is $$ \operatorname{div}P= \big(n+ (\boldsymbol{x}-\boldsymbol{x}_0)\nabla\big)|\nabla u(\boldsymbol{x})|^2 . $$ Then I am stuck. I don't know how to simplify the item $(\boldsymbol{x}-\boldsymbol{x_0})\nabla$.
For the second question, I start with $n=3$ and I put $\boldsymbol{x}_0 = \boldsymbol{0}$. Now the boundary of $B_r(\boldsymbol{x_0})$ is $\Sigma:x^2 + y^2 + z^2 = r^2$. Hence the normal vector of $B_r(\boldsymbol{0})$ is $\boldsymbol{v} = (2x,2y,2z)$. Then by the Divergence Theorem \begin{align*} \frac{1}{r}\int_{B_r\left(x_0\right)}|\nabla u(x)|^2 d V(x) =\frac{1}{r}\int_{\Sigma}\frac{P \cdot \boldsymbol{v}}{n+(\boldsymbol{x}-\boldsymbol{x}_0)\cdot \nabla}d \boldsymbol{x} \\ \end{align*} I think the equation is wrong since $(\boldsymbol{x}-\boldsymbol{x}_0)\cdot \nabla$ is not a number. What is the problem and how can I continue? Any help will be appreciated.