Let us go back to the following problem that we have discussed before in this topic. Denote by the ball centered at the origin with radius . Here we just consider our problem within a ball. We will take care the case of whole space .
Theorem. Suppose is a positive solution of
in and on
where is locally Lipschitz in . Then is radially symmetric in .
Since we haven’t provided any solutions to this problem, here I just figure out the idea and several steps needed during the proof.
Definition. For each unit vector and each number , define a hyperplane by
The reflection in of any point is
and the reflection in of any function is defined by
for all .
Obviously, is parallel to for any . This is because vector is perpendicular to for any . Since vector is chosen to be a unit vector, it is easy to check that the distance from the origin to exactly equals . The following lemma plays an important role in our argument
Lemma. If has the property that
for each unit vector and for all , then is spherically symmetric (i.e. depends only on ).
Proof. We only need to prove that for any two points and satisfying and we have .
Indeed, if we choose we are able to prove . To check that, we have
The proof of lemma follows.
We are now in a position to derive a preliminary sketch of the method of reflection in hyperplanes. Our aim is to show that solutions to the problem satisfy
for each unit vector and all .
To this end, we fix a unit vector and contemplate all positions of the hyperplane from to . Let
The procedure is to prove that (for fixed but arbitrary )
if whenever .The proof of the above estimate proceeds in two stages. The first is for small positive value , i.e. when is very closed to , say . The key ingredient used here is to apply the maximum principle in the thin domains. Once this stage can be done, we are then able to start moving the plane continuously to the left as long as the above estimate still holds. The second stage is a demonstration that one the estimate holds in an interval it can fail only as . However, this is not the case. Once this whole argument has been established, a continuity argument for shows that
if ; in other words, that
if and . But the same result for the direction implies that
if and .
These two inequalities yield the desired result that for all .
See also: L.E. Fraenkel, An introduction to maximum principles and symmetry in elliptic problems, Cambridge, 2000.