Radial symmetry of solution for fractional p− Laplacian system

Nonlinear Analysis 196 (2020) 111801

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Radial symmetry of solution for fractional p−Laplacian system✩ Lihong Zhang a , Bashir Ahmad b ,∗, Guotao Wang a,b , Xueyan Ren a a

School of Mathematics and Computer Science, Shanxi Normal University, Linfen, Shanxi 041004, People’s Republic of China b Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia

article

info

Article history: Received 1 October 2019 Accepted 31 January 2020 Communicated by Vicentiu D. Radulescu MSC: 35A16 35R11

abstract In this paper, we investigate the method of moving planes for fractional p-Laplacian system. We firstly discuss the key ingredients for the method of moving planes such as maximum principle for anti-symmetric functions, decay at infinity and boundary estimate. Then we apply the method of moving planes to establish the radial symmetry and the monotonicity of the positive solutions for fractional p−Laplacian system in a unit ball or in the whole space. © 2020 Elsevier Ltd. All rights reserved.

Keywords: Maximum principle for anti-symmetric function Decay at infinity Boundary estimate Radial symmetry and monotonicity Fractional p−Laplacian system

1. Introduction In this paper, we consider the following fractional p−Laplacian system: { (−∆)sp1 u(x) + a1 (x)v(x) = G (v(x)), (−∆)sp2 v(x) + a2 (x)u(x) = F (u(x)),

(1.1)

where (−∆)sp represents fractional p−Laplacian operator defined by ∫ p−2 |u(x) − u(z)| [u(x) − u(z)] (−∆)sp u(x) =CN ,sp lim dz N +sp ϵ→0 RN \B (x) |x − z| ϵ ∫ p−2 |u(x) − u(z)| [u(x) − u(z)] =CN ,sp P V dz, N +sp |x − z| RN ✩ This work is supported by NSFC, China (No.11501342), NSF of Shanxi, China (No.201701D221007) and the Scientific and Technological Innovation Programs of Higher Education Institutions in Shanxi, China (Nos.201802068 and 201802069). All authors equally contributed this manuscript. ∗ Corresponding author. E-mail addresses: [email protected] (L. Zhang), [email protected] (B. Ahmad), [email protected] (G. Wang), [email protected] (X. Ren).

https://doi.org/10.1016/j.na.2020.111801 0362-546X/© 2020 Elsevier Ltd. All rights reserved.

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L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

1,1 P V represents the Cauchy principal value, u(x) ∈ Ls,p ∩ Cloc , with } { ∫ p−1 |1 + u(x)| p−1 dx < ∞ . Ls,p = u ∈ Lloc | N +sp RN 1 + |x|

Observe that the fractional p−Laplacian operator (−∆)sp in RN is a nonlocal differential operator and becomes the fractional Laplacian (−∆)s when p = 2. We can refer to [3,17–19,22,23,31] for the details of fractional Laplacian operators. It is not easy to handle the problems involving a nonlocal fractional Laplacian operator due to the nonlocal characteristic of these operators. The first method to deal with such problems is known as the extended method, which was introduced by Caffarelli and Silvestre [2]. In this method, a nonlocal problem involving the fractional Laplacian is transformed into a higher dimensional local problem. For details and applications of this method, see [1,14,16] and the references therein. The second method is the integral equations method, such as method of moving planes in integral forms and regularity lifting to study their equivalent corresponding integral problems. One can find the details of this method in the articles [6,7,13,15,21,24–26,28,33]. However, in order to apply the extended method or the integral equations method, we have to impose certain additional constraints in some cases. For example, the study of the solution f : RN → R of the equation involving fractional Laplacian needs its extension F : RN × [0, ∞) → R given by { div(y 1−α ∇G) = 0 (x, y) ∈ RN × [0, ∞), G(x, 0) = g(x). Then (−∆)γ/2 g(x) = −CN ,γ lim y 1−γ y→0+

∂G ∂y

x ∈ RN .

When one chooses the integral equations method to study the well-known nonlinear partial differential equation: (−∆)γ/2 g = g (N +γ)(N −γ) , we need an equivalent integral form: ∫ g(x) = RN

1 N −γ

|x − y|

g(y)(N +γ)(N −γ) dy.

In [11], the authors point out that the corresponding extended method or equivalent integral method is workable for equations involving fully nonlinear nonlocal operator like ∫ I(u(x) − u(η)) Hβ (u(x)) = CN ,β P V dη. N +β N |x − η| R For the details of such operators, for instance, see [4]. Notice that Hβ becomes the fractional p−Laplacian p−2 operator (−∆)sp when I(t) = |t| t and β = sp. Therefore, it is essential to explore a method which is directly applicable on this kind of nonlocal operators. In [10], the authors developed the method of moving planes for the fractional Laplacian and applied it to obtain the qualitative properties of solutions for nonlocal equations involving fractional Laplacian operator on different domains, such as radial symmetry, monotonicity and nonexistence of solutions. The method of moving planes is also applicable to an equation involving fully nonlinear nonlocal operators and provides certain information about the solution for the given equation [11]. Since the method employed in [11] or [10] relies on the idea of on non-degeneracy, it cannot be applied to degenerate p−Laplacian problem. In a recent work [9], the authors introduced the concept of boundary estimate to replace the narrow region in the second step of the moving plane method. In [12], the method of moving planes was applied to the fractional Laplacian system to obtain the radial symmetry for its solution. For more works on the topic, we refer the reader to the

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

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papers [5,8,20,27,29,30,32]. Inspired by aforementioned work, we generalize the method of moving planes to extend its scope to the fractional p−Laplacian systems in this paper. As far as authors know, it is the first attempt to apply this method on fractional p−Laplacian systems. We first prove the maximum principle, decay at infinity and boundary estimate for the fractional p−Laplacian system. Then we construct some examples to illustrate the application of the method of moving planes to prove the symmetry of solutions for fractional p−Laplacian systems in a unit ball or in the whole space RN . Notations. We can specify any direction as the x1 −direction. As usual, let Pκ = {x ∈ RN | x1 = κ, for some κ ∈ R} denote the moving plane, with the region to its left given by Σκ = {x ∈ RN | x1 < κ}. We denote by xκ = (2κ−x1 , x2 , . . . , xn ) the reflection point of x about Pκ . Moreover, uκ (x) = u(xκ ), vκ (x) = v(xκ ), x ∈ Σκ , where u(x) and v(x) form a pair of solutions for fractional p−Laplacian system. Our strategy will be based on establishing the following two steps. Step 1 When κ approaches to negative infinity, we get uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0

x ∈ Σκ .

(1.2)

This is regarded as the starting point to move the plane. Once (1.2) is established, we can move Pκ along the x1 −direction to the right. Step 2 We move Pκ to the right until its limiting position as long as (1.2) holds. More precisely, it is assumed that { } κ0 = sup κ | uι (x) − u(x) ≥ 0, vι (x) − v(x) ≥ 0, x ∈ Σι ; ι ≤ κ . Then we prove that u(x) and v(x) are symmetric about Pκ0 . Equivalently, we have uκ0 (x) − u(x) ≡ 0, vκ0 (x) − v(x) ≡ 0, ∀ x ∈ Σκ0 . 2. Maximum principle Theorem 2.1 (Maximum Principle for Anti-symmetric Functions). Let B 1,1 1,1 u(x) ∈ Ls,p1 ∩ Cloc (B), v(x) ∈ Ls,p2 ∩ Cloc (B) are lower semi-continuous on [ ] ⎧ s s ⎪ ⎪ (−∆)p1 uκ (x) − (−∆)p1 u(x) + ϑ1 (x)[ vκ (x) − v(x)] ≥ 0 ⎪ ⎪ ⎪ ⎪ (−∆)sp2 vκ (x) − (−∆)sp2 v(x) + ϑ2 (x) uκ (x) − u(x) ≥ 0 ⎨ uκ (x) − u(x) ≥ 0,

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

vκ (x) − v(x) ≥ 0 [ ] κ κ uκ (x ) − u(x ) = − uκ (x) − u(x) [ ] vκ (xκ ) − v(xκ ) = − vκ (x) − v(x)

be a bounded domain in Σκ , ¯ If B. in B, in B, in Σκ \ B,

(2.1)

in Σκ , in Σκ ,

with ϑ1 (x) · ϑ2 (x) ≤ 1 and ϑi (x) < 0, i = 1, 2, then uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0

in B.

Furthermore, if uκ (x) − u(x) = 0 and vκ (x) − v(x) = 0 at some point in B, then uκ (x) − u(x) = 0, vκ (x) − v(x) = 0

almost everywhere in RN .

Moreover, the conclusions remain valid for an unbounded region B provided that [ ] lim|x|→∞ uκ (x) − u(x) ≥ 0

[ ] and lim|x|→∞ vκ (x) − v(x) ≥ 0.

(2.2)

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Proof . On the contrary, suppose that (2.2) is not valid. Then there exists ξ ∈ B¯ such that [ ] uκ (ξ) − u(ξ) = min uκ (x) − u(x) < 0. B¯

It then follows that (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) ∫ p −2 p −2 |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] − |u(ξ) − u(z)| 1 [u(ξ) − u(z)] dz = CN ,sp1 P V N +sp1 |ξ − z| RN ∫ p −2 p −2 |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] − |u(ξ) − u(z)| 1 [u(ξ) − u(z)] = CN ,sp1 P V dz N +sp1 |ξ − z| Σκ ∫ p −2 p −2 |uκ (ξ) − u(z)| 1 [uκ (ξ) − u(z)] − |u(ξ) − uκ (z)| 1 [u(ξ) − uκ (z)] + CN ,sp1 P V dz N +sp1 |ξ − z κ | Σκ ∫ [ ][ 1 1 p −2 − |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] ≤ CN ,sp1 N +sp1 N +sp1 |ξ − z κ | Σκ |ξ − z| ] p −2 − |u(ξ) − u(z)| 1 [u(ξ) − u(z)] dz ∫ p −2 p −2 |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] − |u(ξ) − u(z)| 1 [u(ξ) − u(z)] + CN ,sp1 N +sp1 |ξ − z κ | Σκ p1 −2

+

|uκ (ξ) − u(z)|

[uκ (ξ) − u(z)] − |u(ξ) − uκ (z)| |ξ −

p1 −2

[u(ξ) − uκ (z)]

N +sp1 zκ|

dz

= CN ,sp1 {A1 + A2 }.

(2.3)

Next we estimate A1 and A2 , respectively. For the first part of A1 , it is easy to have 1 N +sp1

|ξ − z|

>

1 N +sp1

|ξ − z κ |

∀x, z ∈ Σκ .

For the other part of A1 , we can get p1 −2

|uκ (ξ) − uκ (z)| p1 −2

as f (t) = |t|

[uκ (ξ) − uκ (z)] − |u(ξ) − u(z)|

p1 −2

[u(ξ) − u(z)] ≤ 0,

t is a strictly increasing function, and hence the following comparison holds: [uκ (ξ) − uκ (z)] − [u(ξ) − u(z)] = [uκ (ξ) − u(ξ)] − [uκ (z) − u(z)] ≤ 0.

In consequence, we have A1 < 0.

(2.4)

For A2 , we use the mean value theorem to derive ∫ p −2 p −2 |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] − |u(ξ) − uκ (z)| 1 [u(ξ) − uκ (z)] A2 = N +sp 1 |ξ − z κ | Σκ p1 −2

+ ∫ = Σκ

|uκ (ξ) − u(z)|

[uκ (ξ) − u(z)] − |u(ξ) − u(z)| |ξ −

p1 −2

(p1 − 1)|φ(z)|

|ξ −

[u(ξ) − u(z)]

N +sp1 zκ| p1 −2

+ (p1 − 1)|ψ(z)|

N +sp1 zκ|

p1 −2

[ ] dz uκ (ξ) − u(ξ) ,

where uκ (ξ) − uκ (z) < φ(z) < u(ξ) − uκ (z), uκ (ξ) − u(z) < ψ(z) < u(ξ) − u(z).

dz

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

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For each fixed κ, there exists a c > 0 and B|ξ| (x∗ ) ⊂ Σ κ = RN \ Σκ with x∗ = (3|ξ| + ξ1 , ξ2 , ξ3 , . . .), such that ∫ ∫ 1 1 dz ≥ dz N +sp N +sp1 1 B|ξ| (x∗ ) |ξ − z| Σκ |ξ − z κ | ∫ 1 dz ≥ N +sp1 N +sp ∗ 1 |ξ| B|ξ| (x ) 4 ≥c. Applying Lemma 3.1 [7] and generalized first mean value theorem of integrals, there exist positive constants c1 and C such that ∫ p −2 p −2 [ ] (p1 − 1)|φ(z)| 1 + (p1 − 1)|ψ(z)| 1 A2 = dz uκ (ξ) − u(ξ) N +sp 1 |ξ − z κ | Σκ ∫ p −2 [ ] (2.5) c1 |u(ξ) − u(z)| 1 ≤ dz uκ (ξ) − u(ξ) N +sp1 κ |ξ − z | Σκ [ ] ≤C uκ (ξ) − u(ξ) . Combining (2.3)–(2.5), we obtain 0 ≤ (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) + ϑ1 (ξ)[vκ (ξ) − v(ξ)] [ ] [ ] < C uκ (ξ) − u(ξ) + ϑ1 (ξ) vκ (ξ) − v(ξ) . Therefore, we have vκ (ξ) − v(ξ) < 0, which implies that there exists ζ ∈ B¯ such that [ ] vκ (ζ) − v(ζ) = min vκ (x) − v(x) < 0. B¯

In a similar manner, we can find that [ ] (−∆)sp2 vκ (ζ) − (−∆)sp2 v(ζ) < C ′ vκ (ζ) − v(ζ) . From the following contradiction [ ] 0 ≤(−∆)sp2 vκ (ζ) − (−∆)sp2 v(ζ) + ϑ2 (ζ) uκ (ζ) − u(ζ) [ ] [ ]
and vκ (ζ) − v(ζ) = 0,

then ξ and ζ are the minimum of uκ (ξ) − u(ξ) and vκ (ζ) − v(ζ), respectively. Hence [ ] 0 ≤ (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) + ϑ1 (x) vκ (ξ) − v(ξ) [ ] ≤ (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) + ϑ1 (x) vκ (ζ) − v(ζ) ∫ [ ][ 1 1 p −2 − |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] ≤ CN ,sp1 N +sp1 N +sp1 |ξ − z κ | Σκ |ξ − z|

(2.6)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

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] p −2 − |u(ξ) − u(z)| 1 [u(ξ) − u(z)] dz ∫ p −2 p −2 [ ] (p1 − 1)|φ(z)| 1 + (p1 − 1)|ψ(z)| 1 + CN ,sp1 dz uκ (ξ) − u(ξ) N +sp 1 |ξ − z κ | Σκ = CN ,sp1 {B1 + B2 }. Obviously, B2 = 0 when uκ (ξ) − u(ξ) = 0, which implies that B1 ≥ 0. Then p1 −2

|uκ (ξ) − uκ (z)| Since f (t) = |t|

p−2

p1 −2

[uκ (ξ) − uκ (z)] − |u(ξ) − u(z)|

[u(ξ) − u(z)] ≥ 0.

t is an increasing function, therefore [uκ (ξ) − uκ (z)] − [u(ξ) − u(z)] = −[uκ (z) − u(z)] ≥ 0.

Thus, we have no choice but uκ (z) − u(z) = 0 almost everywhere in Σκ , and via the anti-symmetry of uκ (x) − u(x), we have uκ (z) − u(z) = 0

almost everywhere in RN .

Arguing in the same way about vκ (ζ) − v(ζ), we conclude that vκ (z) − v(z) = 0 almost everywhere in RN . When B is an unbounded region such that [ ] lim|x|→∞ uκ (x) − u(x) ≥ 0

and

[ ] lim|x|→∞ vκ (x) − v(x) ≥ 0,

then uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0

x ∈ Σκ

are not satisfied. This can be established by applying the foregoing arguments. The proof is complete. □ 1,1 1,1 Corollary 2.1. Assume that u(x) ∈ Ls,p1 ∩ Cloc (B), v(x) ∈ Ls,p2 ∩ Cloc (B) are lower semi-continuous on ¯ B and B ⊂ Σκ be a bounded domain. If ⎧ in B, (−∆)sp1 uκ (x) − (−∆)sp1 u(x) ≥ 0 ⎪ ⎪ ⎪ ⎪ s s ⎪ in B, ⎪ ⎨ (−∆)p2 vκ (x) − (−∆)p2 v(x) ≥ 0

uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0 ⎪ [ ] ⎪ ⎪ uκ (xκ ) − u(xκ ) = − uκ (x) − u(x) ⎪ ⎪ ⎪ [ ] ⎩ vκ (xκ ) − v(xκ ) = − vκ (x) − v(x)

in Σκ \ B, in Σκ , in Σκ ,

then uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0

in B.

Furthermore, if uκ (x) − u(x) = 0 and vκ (x) − v(x) = 0 at some point in B, then uκ (x) − u(x) = 0, vκ (x) − v(x) = 0 almost everywhere in RN . Moreover, these conclusions hold on an unbounded region B provided we assume that [ ] [ ] lim|x|→∞ uκ (x) − u(x) ≥ 0 and lim|x|→∞ vκ (x) − v(x) ≥ 0. Corollary 2.1 can be obtained directly from Theorem 2.2 in [9].

(2.7)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

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1,1 Theorem 2.2 (Decay at Infinity). Let B ⊂ Σκ be a unbounded domain and u(x) ∈ Ls,p1 ∩ Cloc (B), 1,1 ¯ v(x) ∈ Ls,p2 ∩ Cloc (B) are lower semi-continuous on B. If [ ] ⎧ (−∆)sp1 uκ (x) − (−∆)sp1 u(x) + ϑ1 (x) vκ (x) − v(x) ≥ 0 in B, ⎪ ⎪ ⎪ [ ] ⎪ s s ⎪ ⎪ ⎨ (−∆)p2 vκ (x) − (−∆)p2 v(x) + ϑ2 (x) uκ (x) − u(x) ≥ 0 in B,

uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0 ⎪ [ ] ⎪ κ κ ⎪ uκ (x ) − u(x ) = − uκ (x) − u(x) ⎪ ⎪ ⎪ [ ] ⎩ vκ (xκ ) − v(xκ ) = − vκ (x) − v(x)

with ϑ1 (x) ∼ o(

1 sp ), |x| 1

ϑ2 (x) ∼ o(

in Σκ \ B,

(2.8)

in Σκ , in Σκ ,

1 sp ) for |x| large enough, |x| 2

and ϑi (x) < 0,

i = 1, 2.

If [ ] uκ (ξ) − u(ξ) = min uκ (x) − u(x) < 0

(2.9)

B¯

and [ ] vκ (ζ) − v(ζ) = min vκ (x) − v(x) < 0, B¯

then there exists a constant R > 0 (depending on ci (x), and independent of uκ (x) − u(x), vκ (x) − v(x)) such that at least one of ξ and ζ satisfies |x| < R. (2.10) Proof . It follows from (2.8) and (2.9) that (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) ∫ [ ][ 1 1 p −2 − |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] ≤CN ,sp1 N +sp1 N +sp1 |ξ − z κ | Σκ |ξ − z| ] p −2 − |u(ξ) − u(z)| 1 [u(ξ) − u(z)] dz ∫ p −2 p −2 [ ] (p1 − 1)|φ(z)| 1 + (p1 − 1)|ψ(z)| 1 dz uκ (ξ) − u(ξ) + CN ,sp1 N +sp1 κ |ξ − z | Σκ ∫ p −2 [ ] c1 |u(ξ) − u(z)| 1
(2.11)

where ϱ ∈ Σκ . Whenever κ is fixed, there exists a constant c2 such that for sufficiently large |ξ| ∈ Σκ , ∫ ∫ 1 1 dz ≥ dz N +sp1 N +sp1 κ Σκ |ξ − z | B3|ξ| (ξ)\B2|ξ| (ξ)∩Σ κ |ξ − z| c2 ∽ sp1 . |ξ| Therefore, there exists a C1 such that (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) ≤

] C1 [ sp1 uκ (ξ) − u(ξ) < 0. |ξ|

Repeating the above discussion about vκ (ζ) − v(ζ), we can get (−∆)sp2 vκ (ζ) − (−∆)sp2 v(ζ) ≤

] C2 [ sp2 vκ (ζ) − v(ζ) < 0. |ζ|

(2.12)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

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When κ is negative enough, we have [ ] (−∆)sp2 vκ (ζ) − (−∆)sp2 v(ζ) + ϑ2 (ζ) uκ (ζ) − u(ζ) ] [ ] C2 [ ≤ sp2 vκ (ζ) − v(ζ) + ϑ2 (ζ) uκ (ζ) − u(ζ) |ζ| ] [ ] C2 [ ≤ sp2 vκ (ζ) − v(ζ) + ϑ2 (ζ) uκ (ξ) − u(ξ) |ζ| ] ] 1 [ sp [ ≤ C2 ( sp2 vκ (ζ) − v(ζ) − ϑ1 (ξ)ϑ2 (ζ)|ξ| 1 vκ (ξ) − v(ξ) ) |ζ| ] ] 1 [ sp [ ≤ C2 ( sp2 vκ (ζ) − v(ζ) − ϑ1 (ξ)ϑ2 (ζ)|ξ| 1 vκ (ζ) − v(ζ) ) |ζ| [ ] C2 vκ (ζ) − v(ζ) sp sp ≤ (1 − ϑ1 (ξ)ϑ2 (ζ)|ξ| 1 |ζ| 2 ) sp |ζ| 2 < 0, which is a contradiction. Hence the theorem is proved.

□

Theorem 2.3 (A Boundary Estimate). Let uκ0 (x) − u(x) > 0, vκ0 (x) − v(x) > 0 for x ∈ Σκ0 and κℓ ↘ κ0 , ∗ xℓ , xℓ ∈ Σκℓ such that and

xℓ → x0 ∈ ∂Σκ0 ,

[ ] ∗ ∗ vκℓ (xℓ ) − v(xℓ ) = min vκℓ (x) − v(x) ≤ 0 and

xℓ → x0 ∈ ∂Σκ0 .

[ ] uκℓ (xℓ ) − u(xℓ ) = min uκℓ (x) − u(x) ≤ 0 Σ κℓ

Σ κℓ

Let ρℓ = dist(xℓ , ∂Σκℓ ) ≡ |κℓ − xℓ1 |, and ∗

∗

ρℓ∗ = dist(xℓ , ∂Σκℓ ) ≡ |κℓ − xℓ1 |. If ϑi (x) < 0, i = 1, 2, then } [ ] 1{ lim (−∆)sp1 uκℓ (xℓ ) − (−∆)sp1 u(xℓ ) + ϑ1 (xℓ ) vκℓ (xℓ ) − v(xℓ ) < 0, ρℓ →0 ρℓ and

} ∗ ∗ [ ∗ ∗ ] ∗ 1 { (−∆)sp2 vκℓ (xℓ ) − (−∆)sp2 v(xℓ ) + ϑ2 (xℓ ) uκℓ (xℓ ) − u(xℓ ) < 0. ρℓ∗ →0 ρℓ∗ lim

(2.13)

(2.14)

Proof . Following the steps used to obtain (2.11), we have } [ ] 1{ (−∆)sp1 uκℓ (xℓ ) − (−∆)sp1 u(xℓ ) + ϑ1 (xℓ ) vκℓ (xℓ ) − v(xℓ ) ρℓ ∫ [ ][ CN ,sp1 1 1 p1 −2 ≤ − |uκℓ (xℓ ) − uκℓ (z)| [uκℓ (xℓ ) − uκℓ (z)] N +sp N +sp 1 1 ℓ ℓ κ ρℓ ℓ |x − z | Σκℓ |x − z| ] p1 −2 ℓ ℓ − |u(x ) − u(z)| [u(x ) − u(z)] dz [ ]∫ p −2 p −2 CN ,sp1 uκℓ (xℓ ) − u(xℓ ) (p1 − 1)|φ(z)| 1 + (p1 − 1)|ψ(z)| 1 + dz N +sp 1 ρℓ |xℓ − z κℓ | Σκℓ [ ] + ϑ1 (xℓ ) vκℓ (xℓ ) − v(xℓ ) { } = CN ,sp1 A1ℓ + A2ℓ + A3ℓ .

(2.15)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

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Next, we analyze A1ℓ , A2ℓ and A3ℓ respectively. First, for a part of A1ℓ , assume that h(t) = t−(N +sp)/2 with 2 t = | · · · | . Then we have ] 1 1[ 1 − N +sp N +sp 1 1 ρℓ |xℓ − z| |xℓ − z κℓ | 2(N + sp1 )(κℓ − z1 ) 2(N + sp1 )(κ0 − z1 ) = → N +sp1 +2 N +sp1 +2 |εℓ (z)| |ε0 (z)|

as ℓ → ∞,

where |xℓ − z| ≤ |εℓ (z)| ≤ |xℓ − z κℓ |, and |x0 − z| ≤ |ε0 (z)| ≤ |x0 − z κ0 |. Obviously 2(N + sp1 )(κ0 − z1 ) N +sp1 +2

|ε0 (z)|

> 0.

(2.16)

For the other part of A1ℓ , as [ ] [ ] [ ] [ ] uκ0 (x0 ) − uκ0 (z) − u(x0 ) − u(z) = uκ0 (x0 ) − u(x0 ) − uκ0 (z) − u(z) < 0, and f (t) = |t|

p−2

t is an increasing function when ℓ → ∞ and z ∈ Σκ0 , we get

|uκℓ (xℓ ) − uκℓ (z)|

p1 −2

p1 −2

[uκℓ (xℓ ) − uκℓ (z)] − |u(xℓ ) − u(z)|

[u(xℓ ) − u(z)]

p1 −2

p1 −2

→ |uκ0 (x0 ) − uκ0 (z)|

[uκ0 (x0 ) − uκ0 (z)] − |u(x0 ) − u(z)|

[u(x0 ) − u(z)]

(2.17)

< 0. Combining (2.16) and (2.17) leads to A1ℓ < 0.

(2.18)

A2ℓ ≤ 0.

(2.19)

Similarly, it can be shown that

In consequence, we get [ ] [ ∗ ∗ ] ϑ1 (xℓ ) vκℓ (xℓ ) − v(xℓ ) < ϑ1 (xℓ ) vκℓ (xℓ ) − v(xℓ ) [ ] → ϑ1 (x0 ) vκ0 (x0 ) − v(x0 ) < 0

as ℓ∗ → ∞.

(2.20)

From (2.15), (2.18), (2.19) and (2.20), it follows that (2.13) is valid. Likewise, it can be established that (2.14) is valid. □ Corollary 2.2.

When both ϑ1 (x) and ϑ2 (x) are equal to zero, we have } 1{ (−∆)sp1 uκℓ (xℓ ) − (−∆)sp1 u(xℓ ) < 0, ρℓ →0 ρℓ

(2.21)

} ∗ ∗ 1 { (−∆)sp2 vκℓ (xℓ ) − (−∆)sp2 v(xℓ ) < 0. ρℓ∗ →0 ρℓ∗

(2.22)

lim

and

lim

The proof is similar to that of Theorem 2.3 with A3ℓ equal to zero.

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

10

3. Method of moving planes and its applications 3.1. Symmetry of solutions in a unit ball Consider

⎧ ⎨ (−∆)sp1 u(x) = v q1 (x), in B1 (0), (−∆)sp2 v(x) = uq2 (x), in B1 (0), ⎩ u(x), v(x) = 0, in RN \ B1 (0).

(3.1)

1,1 Theorem 3.1. Let u(x), v(x) ∈ Cloc (B1 (0)) ∩ C(B1 (0)) be a pair of positive solutions for fractional p−Laplacian system (3.1) with q1 ≥ p2 − 1 and q2 ≥ p1 − 1. Then u, v must be radially symmetric and monotone decreasing about the origin.

Proof . Let Bκ = Σκ ∩ B1 (0). Then, for x ∈ Bκ , we have [ ] (−∆)sp1 uκ (x) − (−∆)sp1 u(x) = q1 µqκ1 −1 (x) vκ (x) − v(x) , where µκ (x) lies between v(x) and vκ (x), and hence [ ] (−∆)sp1 uκ (x) − (−∆)sp1 u(x) ≥ q1 v q1 −1 (x) vκ (x) − v(x) .

(3.2)

Analogously, we have [ ] (−∆)sp2 vκ (x) − (−∆)sp2 v(x) = q2 νκq2 −1 (x) uκ (x) − u(x) , where νκ (x) lies between u(x) and uκ (x), and so [ ] (−∆)sp2 vκ (x) − (−∆)sp2 v(x) ≥ q2 uq2 −1 (x) uκ (x) − u(x) .

(3.3)

Step 1 We pick any ray emanating from the origin as the positive direction of x1 . First, we prove that when κ > −1 and κ is close enough to −1, we have uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0,

x ∈ Bκ .

Suppose that (3.4) is not valid. Then there exists a point ξ ∈ Bκ such that [ ] [ ] uκ (ξ) − u(ξ) = min uκ (x) − u(x) = min uκ (x) − u(x) < 0. Bκ

Σκ

Using the earlier arguments, we obtain (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) ∫ [ ][ 1 1 p −2 ≤CN,sp1 − |uκ (ξ) − uκ (z)| 1 [uκ (ξ) − uκ (z)] N +sp1 N +sp 1 κ |ξ − z | Σκ |ξ − z| ] p1 −2 − |u(ξ) − u(z)| [u(ξ) − u(z)] dz ∫ p −2 p −2 [ ] (p1 − 1)|φ(z)| 1 + (p1 − 1)|ψ(z)| 1 + CN,sp1 dz uκ (ξ) − u(ξ) N +sp 1 |ξ − z κ | Σκ ∫ p1 −2 p −2 [ ] (p1 − 1)|φ(z)| + (p1 − 1)|ψ(z)| 1
(3.4)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

11

Let H = Σκ \Bκ with u(z) = 0, ∀z ∈ H. By Lemma 3.1 in [7], there exist positive constants c3 and c4 such that ∫ p −2 p −2 (p1 − 1)|φ(z)| 1 + (p1 − 1)|ψ(z)| 1 dz N +sp 1 |ξ − z κ | Σκ ∫ ∫ p −2 p −2 c3 |u(ξ) − u(z)| 1 c3 |u(ξ)| 1 c4 up1 −2 (ξ) ≥ dz = dz ≥ , N +sp1 N +sp1 ρsp1 |ξ − z κ | Σκ H |ξ − z κ | where ρ = κ + 1 is the width of Bκ in the x1 direction. Observe that (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) ≤ CN,sp1

] c4 up1 −2 (ξ) [ uκ (ξ) − u(ξ) . sp 1 ρ

On the other hand, [ ] (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) ≥ q1 v q1 −1 (ξ) vκ (ξ) − v(ξ) and so vκ (ξ) − v(ξ) < 0. Then there exists a point ζ ∈ Bκ such that [ ] [ ] vκ (ζ) − v(ζ) = min vκ (x) − v(x) = min vκ (x) − v(x) < 0. Bκ

Σκ

It follows from (3.3) that [ ] 0 ≤ (−∆)sp2 vκ (ζ) − (−∆)sp2 v(ζ) − q2 uq2 −1 (ζ) uκ (ζ) − u(ζ) ] [ ] CN,sp2 c4 v p2 −2 (ζ) [ ≤ vκ (ζ) − v(ζ) − q2 uq2 −1 (ζ) uκ (ζ) − u(ζ) ρsp2 ] ] CN,sp2 c4 v p2 −2 (ζ) [ q1 v q1 −1 (ξ)ρsp1 [ ≤ vκ (ζ) − v(ζ) − q2 uq2 −1 (ζ) vκ (ξ) − v(ξ) sp p −2 ρ 2 CN,sp1 c4 u 1 (ξ) [C p2 −2 q2 −1 q1 −1 sp1 ][ ] c v (ζ) q u (ζ)q v (ξ)ρ N,sp2 4 2 1 vκ (ζ) − v(ζ) ≤ − sp p −2 2 1 ρ CN,sp1 c4 u (ξ) <0 when ρ is sufficiently small and q1 ≥ p2 − 1, and q2 ≥ p1 − 1. Therefore, (3.4) is valid when κ is close enough to −1. Step 2 Step 1 provides a starting point to move the plane Pκ . As long as (3.4) holds, we can move the plane Pκ to its limiting position. Equivalently, let κ0 = sup{κ|uι (x) − u(x) ≥ 0, vι (x) − v(x) ≥ 0, x ∈ Σι , ι ≤ κ}, then we will show that κ0 = 0. Letting κ0 < 0, both uκ0 (x) − u(x) and vκ0 (x) − v(x) are constant different from 0. Then it follows from Corollary 2.1 of Maximum principle for anti-symmetric functions that uκ0 (x) − u(x) > 0, vκ0 (x) − v(x) > 0

∀x ∈ Bκ0 .

∗

By the definition of κ0 , we have κℓ ↘ κ0 , and xℓ , xℓ ∈ Bκℓ such that [ ] uκℓ (xℓ ) − u(xℓ ) = min uκℓ (x) − u(x) < 0 Σ κℓ

[ ] ∗ ∗ vκℓ (xℓ ) − v(xℓ ) = min vκℓ (x) − v(x) < 0 Σ κℓ

[ ] and ∇ uκℓ (xℓ ) − u(xℓ ) = 0, [ ∗ ∗ ] and ∇ vκℓ (xℓ ) − v(xℓ ) = 0.

(3.5) (3.6)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

12

∗

∗

Since {xℓ } and {xℓ } converge to x0 and x0 respectively, by the continuity of uκ (x)−u(x) and its derivative with respect to both x and κ, we get [ ] uκ0 (x0 ) − u(x0 ) ≤ 0, x0 ∈ ∂Σκ0 and ∇ uκ0 (x0 ) − u(x0 ) = 0, (3.7) ∗

∗

vκ0 (x0 ) − v(x0 ) ≤ 0,

[ ∗ ∗ ∗ ] x0 ∈ ∂Σκ0 and ∇ vκ0 (x0 ) − v(x0 ) = 0.

(3.8)

Then we have } [ ] 1 1{ (−∆)sp1 uκℓ (xℓ ) − (−∆)sp1 u(xℓ ) = q1 µqκ1ℓ−1 (xℓ ) vκ (xℓ ) − v(xℓ ) ρℓ ρℓ [ ∗ ∗ ] 1 ≥ q1 µqκ1ℓ−1 (xℓ ) vκ (xℓ ) − v(xℓ ) . ρℓ In view of (3.8), we get

[ ∗ ∗ ] 1 q1 µqκ1 −1 (xℓ ) vκ (xℓ ) − v(xℓ ) = 0, ρℓ which contradicts Corollary 2.2 of boundary estimate. Thus we must have lim

ℓ→∞

κ0 = 0 and

u0 (x) − u(x) = 0, v0 (x) − v(x) = 0

∀x ∈ B0 .

As the choice of x1 −direction is arbitrary, we can conclude that u and v are radially symmetric. Also, the monotonicity easily follows from the foregoing argument. This completes the proof. □ 3.2. Symmetry of solutions in RN Consider

{

(−∆)sp1 u(x) + a1 (x)v(x) = G (v(x)), (−∆)sp2 v(x) + a2 (x)u(x) = F (u(x)),

in RN , in RN .

(3.9)

1,1 1,1 Theorem 3.2. Let u(x) ∈ Ls,p1 ∩ Cloc , v(x) ∈ Ls,p2 ∩ Cloc be a pair of positive solutions of Eq. (3.9), with

a1 (x) ∼ o(

1 sp ), |x| 1

a2 (x) ∼ o(

1 sp ) for |x| large, |x| 2

and a1 (x) · a2 (x) ≤ 1, ai (x) < 0,

i = 1, 2.

Assume that for Z > 0, F , G are continuous functions satisfying: (1) : F (Z) and G (Z) are non-negative and non-decreasing about Z; F (Z) N +sp1 N +sp2 (2) : GZ(Z) δ1 and Z δ2 are bounded near Z = 0 and non-increasing with δ1 = N −sp2 , δ2 = N −sp1 . Then u, v must be radially symmetric and monotonically decreasing about some points in RN . Proof . Assume that a(δ1 − 1) ≥ sp1 and b(δ2 − 1) ≥ sp2 . Then we have u(x) ∼ Step 1

1 a, |x|

v(x) ∼

1 |x|

for |x| large.

b

When κ is negative enough, we show that uκ (x) − u(x) ≥ 0, vκ (x) − v(x) ≥ 0, x ∈ Σκ .

Whenever κ is fixed, due to u(x) ∼

1 |x|a , v(x)

∼

1 , |x|b

we have

uκ (x), u(x) → 0,

|x| → ∞.

(3.10)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

13

Thus uκ (x) − u(x) → 0,

|x| → ∞,

vκ (x) − v(x) → 0,

|x| → ∞.

and Assuming that (3.10) is invalid, without losing its generality, let us assume that there exists some ξ ∈ Σκ such that [ ] uκ (ξ) − u(ξ) = min uκ (x) − u(x) < 0. Σκ

From (2.12) in the theorem of decay at infinity, we have (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) ≤

] C1 [ sp1 uκ (ξ) − u(ξ) < 0. |ξ|

By the mean value theorem, we get [ ][ ] (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) = G ′ (θ(ξ)) − a1 (ξ) vκ (ξ) − v(ξ) < 0, where θ(ξ) lies between vκ (ξ) and v(ξ). In view of the assumption on G (Z), we find that vκ (ξ) − v(ξ) < 0. Accordingly, there exists some ζ ∈ Σκ such that [ ] vκ (ζ) − v(ζ) = min vκ (x) − v(x) < 0. Σκ

In consequence, we obtain ( ) ( ) [ ] (−∆)sp1 uκ (ξ) − (−∆)sp1 u(ξ) = G vκ (ξ) − G v(ξ) − a1 (ξ) vκ (ξ) − v(ξ) ( )[ ] G (vκ (ξ))vκδ1 (ξ) G (v(ξ))v δ1 (ξ) = − − a1 ξ vκ (ξ) − v(ξ) δ1 δ1 (ξ) v vκ (ξ) ( )[ ] G (v(ξ)) δ1 G (v(ξ)) δ1 ≥ δ v (ξ) − δ v (ξ) − a1 ξ vκ (ξ) − v(ξ) v 1 (ξ) κ v 1 (ξ) ] ( )[ ] G (v(ξ)) [ δ1 vκ (ξ) − v δ1 (ξ) − a1 ξ vκ (ξ) − v(ξ) = δ v 1 (ξ) [ ] ( )[ ] G (v(ξ)) = δ δ1 η1δ1 −1 (ξ) vκ (ξ) − v(ξ) − a1 ξ vκ (ξ) − v(ξ) 1 v (ξ) [ G (v(ξ)) ( )][ ] δ1 −1 = δ η (ξ) − a ξ vκ (ξ) − v(ξ) , 1 1 1 v δ1 (ξ) where η1 (ξ) lies between vκ (ξ) and v(ξ). Similar to the foregoing arguments, we can obtain uκ (ζ) − u(ζ) < 0, and (−∆)sp2 vκ (ζ) − (−∆)sp2 v(ζ) ≥

[ F (u(ζ)) uδ2 (ζ)

( )] [ ] δ2 η2δ2 −1 (ζ) − a2 ζ uκ (ζ) − u(ζ) ,

where η2 (ζ) lies between uκ (ζ) and u(ζ). Let [ G (v(x)) ( )] c1 (x) = − δ δ1 η1δ1 −1 (x) − a1 x , v 1 (x) and

[ F (u(x)) ( )] δ2 −1 c2 (x) = − δ η (x) − a 2 2 2 x . uδ2 (x)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

14

Since

G (Z) Z δ1

is bounded near Z = 0 and a1 (x) ∼ o(

1 sp ), |x| 1

a2 (x) ∼ o(

therefore,

1 sp ) |x| 2

for |x| large,

c1 (ξ) ∼ o(

1 sp ) |ξ| 1

for |ξ| large,

c2 (ζ) ∼ o(

1 sp ) |ζ| 2

for |ζ| large.

and

When κ is adequately negative, for x ∈ Σκ and |x| → ∞, according to the theorem of decay at infinity, we have uκ (x) − u(x) ≥ 0, or vκ (x) − v(x) ≥ 0. In general, subject to the assumption: uκ (x) − u(x) ≥ 0

x ∈ Σκ ,

we prove that vκ (x) − v(x) ≥ 0. Supposing that there exists ζ ∈ Σκ such that [ ] vκ (ζ) − v(ζ) = min vκ (x) − v(x) < 0, Σκ

we get the contradiction: ] C2 [ sp2 vκ (ζ) − v(ζ) |ζ| ≥ (−∆)sp2 vκ (ζ) − (−∆)sp2 v(ζ) [ F (u(ζ)) ( )][ ] δ2 −1 δ η (ζ) − a ζ uκ (ζ) − u(ζ) ≥ 2 2 δ 2 u (ζ) > 0.

0>

The proof of the Step 1 is complete. Step 2 The Step 1 provides a starting point to move the plane Pκ . As far as (3.10) holds, we can move the plane Pκ to its limiting position. Equivalently, if κ0 = sup{κ|uι (x) − u(x) ≥ 0, vι (x) − v(x) ≥ 0, x ∈ Σι , ι ≤ κ}, then both u and v are symmetric about the limiting plane Pκ0 , or uκ0 (x) − u(x) ≡ 0, vκ0 (x) − v(x) ≡ 0

x ∈ Σκ0 .

(3.11)

Let us assuming that (3.11) is false. Then, based on the Maximum principle for anti-symmetric functions, we obtain uκ0 (x) − u(x) > 0, vκ0 (x) − v(x) > 0 x ∈ Σκ0 . (3.12) ∗

By the definition of κ0 , there is a series κℓ ↘ κ0 , and xℓ , xℓ ∈ Σκℓ such that [ ] [ ] uκℓ (xℓ ) − u(xℓ ) = min uκℓ (x) − u(x) < 0 and ∇ uκℓ (xℓ ) − u(xℓ ) = 0, Σκℓ

∗

∗

[ ] vκℓ (xℓ ) − v(xℓ ) = min vκℓ (x) − v(x) < 0 Σ κℓ

[ ∗ ∗ ] and ∇ vκℓ (xℓ ) − v(xℓ ) = 0.

(3.13)

L. Zhang, B. Ahmad, G. Wang et al. / Nonlinear Analysis 196 (2020) 111801

∗

15

∗

This shows that {xℓ } and {xℓ } converge to x0 , x0 respectively. In fact, by the decay at infinity, at least ∗ ∗ one of {xℓ } and {xℓ } converges. Without losing generality, assume that {xℓ } converges to x0 . If |xℓ | is sufficiently large, then we have ] C1 [ ℓ ℓ sp1 uκℓ (x ) − u(x ) ℓ |x | ≥(−∆)sp1 uκℓ (xℓ ) − (−∆)sp1 u(xℓ ) [ G (u(xℓ )) ( )][ ] ≥ δ2 η1δ1 −1 (xℓ ) − a1 xℓ vκℓ (xℓ ) − v(xℓ ) δ ℓ u 1 (x ) [ G (u(xℓ )) ( ℓ )][ ∗ ∗ ] δ1 −1 ℓ δ η ≥ (x ) − a vκℓ (xℓ ) − v(xℓ ) , 2 1 1 x δ ℓ u 1 (x )

0>

and lim

ℓ→∞

[ G (u(xℓ )) uδ1 (xℓ )

( )] [ ∗ ∗ ] δ2 η1δ1 −1 (xℓ ) − a1 xℓ vκℓ (xℓ ) − v(xℓ ) > 0,

∗

which lead to a contradiction. So {xℓ } must be bounded. From (3.13), by the continuity of vκ (x) − v(x) and its derivative with respect to both x and κ, we obtain [ ] vκ0 (x0 ) − v(x0 ) ≤ 0, x0 ∈ ∂Σκ0 and ∇ vκ0 (x0 ) − v(x0 ) = 0. (3.14) The application of the mean value theorem to (3.9) yields } [ ] 1{ s ℓ s ℓ (−∆) u (x ) − (−∆) u(x ) + a1 (xℓ ) vκℓ (xℓ ) − v(xℓ ) κ p p ℓ 1 1 ρℓ [ ] 1 = ℓ G ′ (η(xℓ )) vκℓ (xℓ ) − v(xℓ ) ρ [ ∗ ∗ ] 1 > ℓ G ′ (η(xℓ )) vκℓ (xℓ ) − v(xℓ ) , ρ

(3.15)

where η(xℓ ) lies between vκℓ (xℓ ) and v(xℓ ). By (3.14), we deduce that (3.15) must tend to zero, which is a contradiction to Theorem 2.3, where the formula is less than zero. Therefore, we get uκ0 (x) − u(x) ≡ 0, vκ0 (x) − v(x) ≡ 0

x ∈ Σκ0 .

(3.16)

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