Symmetry and nonexistence of positive solutions for fractional systems

This paper is devoted to study the nonexistence results of positive solutions for the following fractional H$\acute{e}$non system \begin{eqnarray*}\left\{ \begin{array}{lll}&(-\triangle)^{\alpha/2}u=|x|^av^p,~~~&x\in R^n,&(-\triangle)^{\alpha/2}v=|x|^bu^q,~~~&x\in R^n,&u\geq0, v\geq 0, \end{array} \right. \end{eqnarray*} where $0<\alpha<2$, $0<p,q<\infty$, $a$, $b$ $\geq0$, $n\geq2$. Using a direct method of moving planes, we prove non-existence of positive solution in the subcritical case.

The fractional Laplacian in R n is a nonlocal pseudo-differential operator taking the form (−△) α/2 u(x) = C n,α P V R n u(x) − u(y) |x − y| n+α dy = C n,α lim ε→0 R n \Bε (x) u(x) − u(y) |x − y| n+α dy, (1.2) where C n,α is a normalization constant. This operator is well defined in S, the Schwartz space of rapidly decreasing C ∞ functions in R n . In this space, it can also be equivalently defined in terms of the Fourier transform where Fu is the Fourier transform of u. One can extend this operator to a wider space of distributions: Then in this space, we defined (−△) α/2 u as a distribution by In our paper, we use the direct method of moving planes introduced by [1] to derive the symmetry of positive solutions and then deduce the nonexistence of positive solutions under explain how this paper improved previous result by listing their conditions [8]. The following is our main theorems.
Theorem 1.1. Let (u, v) ∈ L α ∩ L ∞ loc be a pair of nonnegative solution to (1.1) and 1 < p < n+α+a n−α , 1 < q < n+α+b n−α . Then u and v are radially symmetric and decreasing about the origin.
Using the above theorems, we can show the following nonexistence of positive solutions to (1.1) or (1.3).
As a consequence of the proof in Theorem 1.3, we obtain the following nonexistence result immediately.
Remark 1.1. In [8], the authors applied the method of moving planes in integral forms to study the symmetry of positive solutions for the fractional system (1.1), then derived the nonexistence of positive solutions. However, due to technical restrictions to apply the integral method, the authors have to assume that u ∈ L β loc (R n ) and v ∈ L γ loc (R n ) where β = n(q−1) In this paper,we manage to derive the same nonexistence result without imposing extra integrability conditions, by using a direct application of the moving of the moving planes to the differential equations.
In recent years, the fractional Hénon-type problem has received a lot of attention. In [15], the authors considered the integral equation where n−s n−α < p < α * (s) − 1 with α * (s) = 2(n−s) n−α . They proved the nonexistence of positive solutions for the equation by the method of moving planes in integral forms and established the equivalence between the above integral equation and the following partial differential equation In [14], the authors studied the following weighted system of partial differential equations They first established the equivalence between the differential system and an integral system Then, in the critical case n−s p+1 + n−t q+1 = n − α, they showed that every pair of positive solutions (u(x), v(x)) is radially symmetric about the origin. While in the subcritical case, they proved the nonexistence of positive solutions.
In [21], the authors considered the following fractional Laplacian equation where n ≥ 2 and α is any real number between 0 and 2. They proved that the only solution is constant. As an application, they obtained an equivalence between a semi-linear differential equation and the corresponding integral equation For more similar results, please see [3], [6], [9] and the references therein.
The paper is organized as follows. In Section 2, we use the method of moving planes to prove Theorem 1.1. In Section 3, we establish the equivalence between problem (1.1) and integral system (1.3). Finally, in Section 4 we complete the proof of Theorem 1.3 and 1.4.

The Symmetry of Positive Solutions
Without any decay conditions on u and v , we are not able to carry the method of moving planes on u and v directly. To circumvent this difficulty, we make a Kelvin transform. Let . Then In a similar way, we have We first give some basic notations before starting moving the planes. Then we start moving planes on system (2.4).
be the moving plane, be the region to the left of the plane, and Then system (2.4) becomes (2.5) Before starting moving planes, we need some lemmas in [1].
then for sufficiently small δ, we have Furthermore, if ϕ = 0 at some point in Ω, then These conclusions hold for unbounded region Ω if we further assume that

Proof of Theorem 1.1
Now,we start moving planes.
Step.1: we show that when λ sufficiently negative, This is guaranteed by the fact that, for ǫ sufficiently small and λ sufficiently negative, it holds that we will prove it in the Appendices. Without loss of generality, we suppose that U λ (x 0 ) < 0, then On the other hand, By (2.15) and (2.16), we can deduce that which is a contradiction. This proves that V λ (x 0 ) < 0.
In this part, we show that λ 0 = 0 Suppose that λ 0 < 0 we show that the plane T λ can be moved further right. To be more rigorous, there exists some δ > 0, such that for any λ ∈ (λ 0 , λ 0 + δ), we have This is a contradiction with the definition of λ 0 . In fact, when λ 0 < 0 , we have If not, there exists somex such that It follows that On the other hand A contradiction with (2.22), so does V λ 0 . This proves (2.21). We claim that for λ 0 < 0 and ε > 0 sufficiently small, the proof will be given in the appendix. It follows from (2.21) that there exists a constant C 0 > 0, such that From decay at infinity, we have From narrow region principle, To see this, in Lemma 2.1, we let, for any sufficiently small η > 0, H = Σ λ \ B η (0 λ ) and the narrow region Ω = (Σ − λ \Σ λ 0 −δ )\B η (0 λ ), while the lower bound of C(x) can be seen from (2.19).
Similarly, one can move the plane T λ from the +∞ to the left and show that W λ 0 ≤ 0, ∀x ∈ Σ λ . Now we have shown that

Proof of theorem 1.1
Proof. So far, we have proved that u, v is symmetric about the plane T 0 . Since the x 1 direction can be chosen arbitrarily, we have actually shown that u, v is radially symmetric about 0. Let x 1 x 2 be any points centered at 0, i.e., Then, Let Hence, This completes the proof.
3 The equivalence between problem (1.1) and the integral form (1.3) In this section, we prove the equivalence between problem (1. 1) and (1.3).
Proof of Theorem 1.2 Let (u, v) be a pair of positive solution to (1.1), we first show that (3.29) where G R (x, y) is the Green function of fractional Laplacian on B R (0).
It is easy to see that By the Maximum Principle, we derive From Proposition 2 in [21], we have Thus we proved (3.28).
Next, we will show that c 1 = c 2 = 0. Without lose of generality, we may assume that c 2 > 0, then from (3.28), we have But it is impossible, hence c 1 = c 2 = 0. Therefore We complete our proof.

The nonexistence of positive solutions
In this section, we prove Theorem 1.3 and Theorem 1.4. First we need some Lemmas.
Proof. For any R > 0, we need to show Here we only show B R (0) |x| b v q+1 dx < ∞, R → ∞, the other can be proved in a similar way.
Let k = 1, then Multiply both sides of (4.45) by |x| b u q (x) and integrate on R n , we have On the other hand, from the integration by parts formula, it follows From Lemma 4.1, we have Hence when R → ∞, Therefore, Using the similar argument on the second equation of (4.44), we also have Adding (4.46) and (4.47) together, and using the fact |x − y| 2 = x · (x − y) + y · (y − x), we have Then (4.48) becomes That is Because 1 < p < n+α+a n−α and 1 < q < n+α+b n−α , thus n+b q+1 + n+a p+1 = n − α and problem (1.1) admits no positive solutions.
This completes our proof.
Proof of Theorem 1.4 Theorem 1.4 is a a direct consequence of Theorem 1.3.
Together with (5.50), it yields that Through an identical argument, one can show that (5.6) holds for V λ (x) as well.
Lemma 5.2. For λ 0 < 0, if either of U λ 0 V λ 0 is not identically 0, then there exist some constant C and ε > 0 small such that Proof. From Lemma 2.2 in [7], we have the integral equation there exists some x 0 such that V λ 0 (x 0 ) > 0. Thus, for some δ > 0 small, it holds that v p λ 0 (y) − v p (y) ≥ C > 0, y ∈ B δ (x 0 ). Therefore, In a same way, one can show that V λ 0 (x) also satisfies (5.55).