Let us now consider the following equation
where is a real number satisfying .
Lieb [here] proved among other things that there exist maximizing functions for the Hardy-Littlewoord-Sobolev inequality on
When and , the Euler-Langrange equation for a maximizing is nothing but our integral equation.
Having discussion of fractional Laplacian [here] one can easily see that this integral equation is also closely related to the following family of semilinear PDEs
.
The classification of solution to the integral equation was done by Chen, Li and Ou [here] published in Comm. Pure App. Math. around 2006 via the integral form of the method of moving planes. Our goal is to derive a new approach based on the integral form of the method of moving spheres. This result was due to Zhang and Hao [here] published in J. Math. Anal. Appl. around 2008.
Theorem. Let be a positive solution to the integral equation. Then is radially symmetric and has the form
for some constants and .
Outline of the proof. Let be a positive function on , for and we define
where
.
Set
.
Lemma 1. For any solution of the integral equation, we have
.
This lemma has the same form of the lemma considered in this entry. In fact, the proof is straightforward.
Lemma 2. For , there exists such that
for any and .
This lemma tells us that we can run the method of moving spheres. The idea is as follows: it follows from the form of our solution that our solution is indeed monotone decreasing. Thus starting from a point it must be possible to find such a so that outside a sphere centered at with suitable radius, the attitude of function is lower that that at . The proof is similar to the proof of step 1 in this entry.
Next for each , we define
.
Lemma 3. If for some then
in .
The proof of this lemma is similar to the proof of step 2 in this entry. We do it by contradiction argument. Having all discussion above, we are able to complete the proof of theorem. In fact, w shall prove that is finite for all .
Proof of theorem. If there exists some such that , then by Lemma 3,
.
By the definition of ,
.
Multiply the above by and let we get
.
Thus
.
It now follows from the Lemma 3 that
for any . Thus gives
by the second fundamental lemma [here]. If for any then
.
It now follows from the first fundamental lemma [here] that is constant which contradiction with the integral equation.