# Ngô Quốc Anh

## November 20, 2010

### The Harnack quantity and the local gradient estimate

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 1:27

We first recall the Bochner formula for a gradient vector field over a Riemannian manifold $(M,g)$ which is an important tool in geometric analysis

$\displaystyle\frac{1}{2}\Delta (|\nabla u{|^2}) = |{\nabla ^2}u{|^2} + \langle \nabla \Delta u,\nabla u\rangle + {\rm Ric}(\nabla u,\nabla u)$

which has already discussed [here] where $u$ a smooth function. Over $\mathbb R^n$ with standard Euclidean metric, the Bochner formula reads as follows

$\displaystyle\frac{1}{2}\Delta (|\nabla u{|^2}) = |{\nabla ^2}u{|^2} + \langle \nabla \Delta u,\nabla u\rangle$.

Assume $R_2>R>0$. Let $\phi$ be a cut-off function in $B_{R_2}(0)$ with $\phi=1$ on $B_R(0)$.

Definition. The following quantity

$P=\phi|\nabla w|^2$

is called the Harnack quantity where $w$ is some function.

We shall apply this terminology to study the local gradient estimate for positive solution $u$ to the following general elliptic equation

$\Delta u = f(u)$

in $\mathbb R^n$. We start with setting $w=\log u$ and $F(w)=\frac{f(u)}{u}$. At the maximum point of $P$, we have the first order condition

$\nabla P=0$

which implies that

$\displaystyle\nabla|\nabla w|^2=-\frac{1}{\phi^2}\nabla\phi P$

and the second order condition

$0 \geqslant \Delta P=P_0(\phi)P+\phi \Delta|\nabla w|^2$

where

$\displaystyle P_0(\phi)=\Delta \phi - \frac{2}{\phi^2}|\nabla \phi|^2$.

Using the Bochner identity, we have

$\phi\Delta|\nabla w|^2=2\phi|\nabla^2 w|^2+2\phi\langle \nabla \Delta w,\nabla w\rangle$.

Note that

$\displaystyle \phi|\nabla^2 w|^2 \geqslant \frac{2\phi}{n}|\Delta w|^2$

and

$\displaystyle 2\phi\langle \nabla \Delta w,\nabla w\rangle \geqslant 2F'P-2\phi\langle \nabla |\nabla w|^2,\nabla w\rangle$.

Then for any $\mu>0$, we have

$\displaystyle 2\phi\langle \nabla \Delta w,\nabla w\rangle \geqslant 2F'P-\frac{2}{\mu\phi^2}|\nabla \phi|^2P-\frac{\mu}{\phi}P^2$.

Choose $\mu=\frac{1}{4n}$. Then

$\displaystyle 2\phi\langle \nabla \Delta w,\nabla w\rangle \geqslant 2F'P-\frac{4n}{\phi^2}|\nabla \phi|^2P-\frac{1}{4n\phi}P^2$.

Then

$\displaystyle A(\phi, F')P \geqslant \frac{2}{n}(-P+\phi F)^2-\frac{1}{4n}P^2$

where

$A(\phi, F')=\frac{4n}{\phi}|\nabla \phi|^2-2\phi F'-\phi P_0(\phi)$.

If $P \leqslant 2\phi F$, then we have

$|\nabla w|^2 \leqslant 2F$.

We remark that in this case, we have

$|\nabla u|^2 \leqslant 2u^2 F=2uf(u)$.

Otherwise, we have

$\displaystyle -P+\phi F \leqslant -\frac{P}{2}\leqslant 0$

and

$\displaystyle \frac{2}{n}(-P+\phi F)^2-\frac{1}{4n}P^2 \geqslant \frac{1}{4n}P^2$.

Hence we have in this case

$P\leqslant 4nA(\phi, F')$.

In conclusion, we have on $B_R(0)$

$|\nabla w|^2 \leqslant \max\{ 4nA(\phi, F'), 2F\}$.

This is the first result of a recent paper due to L. Ma and J.C. Wei published in Journal of Functional Analysis [here]. We refer the reader to this paper for further studying.