Prove that $- \Delta(\eta^2 |\nabla u|^2) \le C |\nabla u|^2$ for $u$ harmonic.

Question

Let $B_1 = B(0, 1) \subset \mathbb R^n$, $\eta \in C^\infty_0(B_1)$ such that $\eta = 1$ on $B_{1/2}$ and let $u \in C^2(\overline{B_1})$ be a harmonic function. I want to show that $$- \Delta(\eta^2 |\nabla u|^2) \le C |\nabla u|^2$$ for $C > 0$, where $|\nabla u|^2 = \sum_i (\partial_i u)^2$. We have \begin{align} - \Delta(\eta^2 |\nabla u|^2) &= - \Delta \eta^2 |\nabla u|^2 - \eta^2 \Delta |\nabla u|^2 - 2 \nabla \eta^2 \cdot \nabla|\nabla u|^2 \end{align} where $2 \nabla \eta^2 \cdot \nabla|\nabla u|^2$ is the dot product in $\mathbb R^n$. We easily find that $$\Delta |\nabla u|^2 = \sum_{i,j = 1}^n (\partial_{ij} u )^2 \ge 0,$$ so that, in addition of the fact that $\eta \in C_0^\infty(B_1)$, we get $$- \Delta(\eta^2 |\nabla u|^2) \le C |\nabla u|^2 - 2 \nabla \eta^2 \cdot \nabla|\nabla u|^2.$$ I tried a lot of things to bound $- 2 \nabla \eta^2 \cdot \nabla|\nabla u|^2$ by a constant times $|\nabla u|^2$ but I am not able to get rid of the $\nabla$ before the absolute value in $\nabla|\nabla u|^2$.. Any idea how to do that ?

Answer 1

The trick is that you do not throw away $$- \eta^2 \Delta |\nabla u|^2= -\eta^2 \sum_{i,j} (u_{ij})^2$$

Note

$$\nabla \eta^2 \cdot \nabla|\nabla u|^2 = 2\eta \nabla \eta \cdot \nabla|\nabla u|^2 \le 2|\eta| | \nabla \eta|\ |\nabla|\nabla u|^2 | $$ and \begin{align} |\nabla|\nabla u|^2 |^2 &= \sum_i (\partial_i \langle \nabla u, \nabla u\rangle)^2 \\ &= 4\sum_i (\langle \nabla u, \nabla u_i\rangle)^2 \\ &\le 4 |\nabla u|^2 \sum_i |\nabla u_i|^2 \\ &= 4|\nabla u|^2 \sum_{i,j} u_{ij}^2 \\ \Rightarrow |\nabla |\nabla u|^2| &\le 2 |\nabla u| \sqrt{\sum_{i,j} u_{ij}^2} \end{align} So we have (using $ab\le \frac{1}{2\epsilon} a^2 + \frac{\epsilon}{2} b^2)$)

\begin{align} -2\nabla \eta^2 \cdot \nabla|\nabla u|^2 & \le 4 |\eta| |\nabla \eta| |\nabla u | \sqrt{\sum_{i,j} u_{ij}^2} \\ &= 4\left( |\eta|\sqrt{\sum_{i,j} u_{ij}^2} \right)\bigg( |\nabla \eta| |\nabla u | \bigg) \\ &\le 4 \left( \frac{1}{4} \eta^2 \sum_{i,j} u_{ij}^2 + |\nabla \eta|^2 |\nabla u |^2 \right) \\ &= \eta^2 \sum_{i,j} u_{ij}^2 + 4|\nabla \eta|^2 |\nabla u |^2 . \end{align}

So we have $$ - \Delta(\eta^2 |\nabla u|^2)\le (- \Delta \eta^2 +4 |\nabla \eta|^2 )|\nabla u|^2\le C|\nabla u|^2$$