In this section we will use the moving plane method to discuss the symmetry of solutions. The following result was first proved by Gidas, Ni and Nirenberg.

Theorem. Suppose is a positive solution ofin and on

where is locally Lipschitz in . Then is radialy symmetric in and for .

The original proof requires that solutions be up to the boundary. Here we give a method which does not depend on the smoothness of domains nor the smoothness of solutions up to the boundary.

Statement. Suppose that is a bounded domain which is convex in direction and symmetric with respect to the plane . Suppose is a positive solution ofin and on

where is locally Lipschitz in . Then is radialy symmetric in and for .

**Idea of proof**. Write for . We will prove that

for any and with .

Then by letting , we get for any . Then by changing the direction we get the symmetry.

**Proof**. Let for . For , define

the reflection of with the respect to

for .

In we define

for .

Then we have by Mean Value Theorem

in

and on .

where is a bounded function in .

We need to show in for any . This implies in particular that assumes along its maximum in . By Hopf Lemma we have for any such

.

For any close to , we have by the maximum principle for narrow domain. Let be the largest interval of values of such that in . We want to show that . If , by continuity, in and on . Then the Strong Maximum Principle implies in . We will show that for any small

in .

Fix (to be determined). Let be a closed subset in such that . The fact that in implies

for any .

By continuity we have

in .

For small, . We choose in such a way that we may apply the Maximum Principle for domain with small volume to in . Hence we get

in

and then

in .

Therefore we obtain for any small

in .

This contradicts the choice of .

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