In this section, by using the method of moving planes in integral forms, we derive the nonexistence of positive solutions to integral system (1.2) and obtain a new Liouville-type theorem on a half-space. To prove the theorems, we need several lemmas.
Let ,
Set
Lemma 3.1 Let be any pair of positive solutions of (1.2). For any , we have
Proof
Obviously, we have
Now, by properties (2.2) and (2.3) of the function and the pair of positive solutions of (1.2), we have
Similarly, we could derive the second inequality in the lemma. This completes the proof of Lemma 3.1. □
Proof of Theorem 1.4 To prove Theorem 1.4, we compare and on . The proof consists of two steps.
In the first step, we start from the very low end of our region , i.e., . We will show that for λ sufficiently small,
(3.3)
In the second step, we will move our plane toward the positive direction of -axis as long as inequality (3.3) holds.
Step 1. Define
and
We show that for sufficiently small positive λ, and must both be measure zero. In fact, by Lemma 3.1, it is easy to verify that
where is valued between and . Therefore, on we have
It follows from the Hardy-Littlewood-Sobolev inequality that
(3.4)
Then by the Hölder inequality,
(3.5)
Similarly, one can show that
(3.6)
Combining (3.5) and (3.6), we arrive at
(3.7)
By the conditions that and , we can choose sufficiently small positive λ such that
Now, inequality (3.7) implies , and therefore must be measure zero. Similarly, one can show that is measure zero. Therefore, (3.3) holds. This completes Step 1.
Step 2. (Move the plane to the limiting position to derive symmetry and monotonicity.)
Inequality (3.3) provides a starting point to move the plane . Now, we start from the neighborhood of and move the plane up as long as (3.3) holds to the limiting position. We will show that the solution must be symmetric about the limiting plane and be strictly monotonically increasing with respect to the variable . More precisely, define
Suppose that for such a , we will show that both and must be symmetric about the plane by using a contradiction argument. Assume that on , we have
We show that the plane can be moved further up. More precisely, there exists an depending on n, α, and the solution such that
(3.8)
In the case
by Lemma 3.1, we have in fact in the interior of . Let
Then, obviously, has measure zero and . The same is true for that of v. From (3.5) and (3.6), we deduce
(3.9)
Again, the conditions that and ensure that one can choose ϵ sufficiently small, so that for all λ in ,
Now, by (3.9), we have , therefore must be measure zero. Similarly, must also be measure zero. This verifies (3.8), therefore both and are symmetric about the plane . Also, the monotonicity easily follows from the argument. This completes the proof of Theorem 1.4. □
Proof of Theorem 1.5 To prove the theorem, firstly we will show that the plane cannot stop at for some , that is, we will prove that .
Suppose that , the process of Theorem 1.4 shows that the plane is the symmetric points of the boundary with respect to the plane , and we derive that and when x is on the plane . This contradicts the pair of positive solutions of (1.2), thus .
Besides, we know that both and of positive solutions of (1.2) are strictly monotonically increasing in the positive direction of -axis, but and , so we come to the conclusion that the pair of positive solutions of (1.2) does not exist.
This completes the proof of Theorem 1.5. □