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


\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


\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.


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


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


\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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Create a free website or blog at

%d bloggers like this: