We are trying to give a brief proof of harnack inequality of the divergence form

Harnack inequality says that on such that , any weak solution of satisfies

Define for , then

We will prove that

If , then , , only depends on

If , then ,

If , ,

If , ,

If , ,

Choose , then

That is

where . The left hand side is

Let ,

From the embedding , there is a constant

Then we will have

Combing with , we can get

Let us suppose , is cut off function on , on for some and outside , , then means

Denote .Suppose , and choose and , from ,

, for all

because we know .

Let us consider the case to get the other inequality involves .

Choose , , apply , we can get

, for

Remember from the proposition, the case needs , that is .

So by iteration

where , because .

By similar iteration, from , we get

where .

If we can prove that , for some , then combine and ,

We need the following theorem, which is actually the consequence of John-Nirenberg thm and poincare inequality

Let , where is convex, and suppose there exists a constant such that

for any balls

Then there exist positive constants and depending only on such that

There exists two numbers , such that

for any and .

Let , and on and

while

So letting , then

By holder inequality

By the theorem we have stated, there exists such that

Multiply them together

This is equivalent to

Carefully examine the proof of this lemma, not only for some , but also for , we also have

So we can choose large enough such that , then

Hence we complete the proof of Harnack inequality.

consult much to the GT’s book and MA600A: Theory of Partial Differential Equation, Roger Moser.