From PDE Evans (2nd edition), page 515, we are given
$$\sum_{i=1}^n \left(\left(Du \cdot x + \frac{n-p}p u \right)p|Du|^{p-2}u_{x_i}-|Du|^px_i \right)_{x_i}=0. \tag{10}$$
Then the author goes on to say:
Application: monotonicity formulas. Assume that $u$ is a smooth solution of the $p$-Laplacian PDE within some region $U \subset \mathbb{R}^n$ and that the ball $B(0,r)$ lies within $U$. If we integrate the divergence identity $\text{(10)}$ over $B(0,r)$ and simplify using the Gauss-Green Theorem, we discover that $$(n-p)\int_{B(0,r)} |Du|^p \, dx = r \int_{\partial B(0,r)} |Du|^p - p|Du|^{p-2} u_r^2 \, dS,$$ where $u_r:= Du \cdot \frac x{|x|}.$
How can I derive the aforementioned equality?
My work so far:
On the LHS of $\text{(10)}$, I distributed the $p|Du|^{p-2}u_{x_i}$ to the terms contained in the inner set of parentheses. Here's what I mean by this: $$(Du \cdot x)p|Du|^{p-2}u_{x_i}+(n-p)u|Du|^{p-2} u_{x_i} = |Du|^p x_i.$$ Then I algebraically rearranged this to obtain $$(n-p)u|Du|^{p-2} u_{x_i}=|Du|^p x_i-(Du \cdot x)p|Du|^{p-2}u_{x_i}.$$ I got stuck after this though. Should I integrate over $B(0,r)$ to both sides of this now?
Let's rewrite $(10)$ as $$ \sum_{i=1}^n \left((n-p) u |Du|^{p-2}u_{x_i}\right)_{x_i} = - \sum_{i=1}^n \left(\left(Du \cdot x \right)p|Du|^{p-2}u_{x_i}-|Du|^px_i \right)_{x_i} \tag{A} $$ Expanding the left hand side of (A) using the product rule, we get $$ \sum_{i=1}^n \left((n-p) u_{x_i} |Du|^{p-2}u_{x_i}\right) =(n-p) |Du|^p $$ because $\sum_{i=1}^n \left(|Du|^{p-2}u_{x_i}\right)_{x_i}=0$ (the $p$-Laplace equation). Integration over $B(0,r)$ gives the left hand side of the claimed identity.
Next, turn to the right hand side of (A). It represents the divergence of the vector field $$\mathbf F(x) = -p|Du|^{p-2} (Du \cdot x) Du + |Du|^px$$ Integrating the divergence of $\mathbf F$ over $B(0,r)$ gives the flux integral of $\mathbf F$ across $\partial B(0,r)$. The flux is computed by taking the dot product of $\mathbf F$ with the unit normal $x/r$, and integrating that. We have $$\mathbf F(x)\cdot x/r = - p|Du|^{p-2} (Du \cdot x) u_r + |Du|^p |x|$$ Simplify this a bit: $(Du \cdot x) =r u_r$ and $|x|=r$. So, $$\mathbf F(x)\cdot x/r = - p|Du|^{p-2} r u_r^2 + |Du|^p r$$ The integral of this expression over $\partial B(0,r)$ is what you see in the claimed identity.