# Radial Symmetry and Monotonicity of Solutions to a System Involving Fractional p-Laplacian in a Ball

Authors: Linfen Cao; Xiaoshan Wang; Zhaohui Dai
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

In this paper, we study a nonlinear system involving the fractional p-Laplacian in a unit ball and establish the radial symmetry and monotonicity of its positive solutions. By using the direct method of moving planes, we prove the following result. For0<s,t<1,p>0, if u and v satisfy the following nonlinear system -Δpsux=fvx;-Δptvx=gux,x∈B10;ux,vx=0,x∉B10. and f,g are nonnegative continuous functions satisfying the following: (i) f(r) and g(r) are increasing for r>0; (ii) f′(r)/rp-2, g′(r)/rp-2 are bounded near r=0. Then the positive solutions (u,v) must be radially symmetric and monotone decreasing about the origin.


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## Body

## 1. Introduction

Nonlinear equations involving fractional powers of the Laplacian are currently actively studied. The fractional and nonlocal operators of elliptic type due to concrete real world have been applied in finance, thin obstacle problem, optimization, quasi-geostrophic flow, etc. In recent years, there have been tremendous interests in developing the problems related to fractional Laplacian (see in [1–5] and the reference therein). Fractional p-Laplacian is a generalization of the fractional Laplacian, but there are few interesting results about these nonlinear equations involving fractional p-Laplacian. In [6], the authors considered the positive solution to the following problem involving the fractional p-Laplacian:(1)-Δpsux=gux,x∈Rn.where -Δps is defined by(2)-Δpsux=Cn,spPV∫Rnux-uzp-2ux-uzx-zn+spdz,where PV stands for the Cauchy principal value. In order to make sense the integral, we require that u∈Lsp∩Cloc1,1 with(3)Lsp=u∈Lloc1∣∫Rnuxp-11+xn+spdx<∞.They show that if u(x)∈Lsp∩Cloc1,1 is a positive solution to (1) with lim|x|→∞u(x)=0, and g′(s)≤0 for s>0 sufficiently small, then u must be radially symmetric and monotone decreasing about some point in Rn.The fractional p-Laplacian-Δps is a special case of the following fully nonlinear operators Fα:(4)Fαux=Cn,αPV∫RnGux-uyx-yn+αdy,where G is at least local Lipschitz continuous, G(0)=0, and α>0 (see [7] for the introductions of these operators, one also see [8] for systems case). In this case α=sp and G(t)=tp-2t.Whenp=2, the fractional p-Laplacian -Δps becomes the well-known fractional Laplacian operator -Δs. In [2], the authors considered the following system involving fractional Laplacian on the whole space:(5)-Δsux=fvx,-Δtvx=gux,x∈Rn,ux,vx≥0,x∈Rn.Assume that, for r≥0, f,g are nonnegative continuous functions satisfying the following: (a) f(r) and g(r) are nondecreasing about r; (b) f(r)/rp, g(r)/rq are bounded near r=0 and nonincreasing with p=n+2s/n-2t and q=n+2t/n-2s. If u and v are nonnegative solutions for (5), then either u and v are constant or f(v)=C1vn+2s/n-2t and g(u)=C2un+2t/n-2s.Motivated by the ideas of [2, 6], our main concern in this paper is to study the following nonlinear system involving fractional p-Laplacian in a unit ball:(6)-Δpsux=fvx,-Δptvx=gux,x∈B10,ux,vx=0,x∉B10,where 0<s,t<1. We obtain the following main theorem by the direct method of moving planes.Theorem 1.
Assume thatu(x)∈Lsp(B1(0))∩Cloc1,1(B1(0)) and v(x)∈Ltp(B1(0))∩Cloc1,1(B1(0)) are positive solutions of system (6); f,g are nonnegative continuous functions satisfying the following:
(i)f(r) and g(r) are increasing about r>0.
(ii)f′(r)/rp-2, g′(r)/rp-2 are bounded near r=0.
Then the positive solutions(u,v) must be radially symmetric and monotone decreasing about the origin.Remark 2.
In [3, 9], the authors introduced the direct method of moving planes, respectively, to prove symmetry, monotonicity, and nonexistence of solutions to various differential equations on the whole space and on a half space, etc., such as ones involving fractional Laplacian in [2, 3], fractional p-Laplacian in [6], and fully nonlinear operators Fα in [4, 8]. This direct method of moving planes has some advantages, which overcome the necessary of imposing extra assumption on the solutions when using the extension method (see also [10]) or the equivalent integral equation method (see also [11]).The key ingredients of the direct method are maximum principle for antisymmetric functions and key boundary estimate lemma. For convenience we put both of them in the appendix.


## 2. Proof of Theorem1

As usual, we choose any direction to thex1 direction; let(7)Tλ=x=x1,x2,…,xn∈Rn∣x1=λ,λ∈Rbe the moving planes,(8)Σλ=x∈Rn∣x1<λbe the region to the left of the plane, and(9)xλ=2λ-x1,x2,…,xnbe the reflection of x about plane Tλ, and uλ(x)=u(xλ), vλ(x)=v(xλ).Let(10)Uλx=uλx-ux,Vλx=vλx-vxdenote(11)Ωλ=Σλ∩B10.Then in Ωλ, we have(12)-Δpsuλx--Δpsux=fvλx-fvx=f′ξλxVλxand(13)-Δptvλx--Δptvx=guλx-gux=g′ηλxUλx,where ξλ(x) is valued between vλ(x) and v(x); ηλ(x) is valued between uλ(x) and u(x).Step 1.
In this step, we show that, forλ>-1 and sufficiently closed to -1, it holds(14)Uλx,Vλx≥0,∀x∈Ωλ.We prove (14) by a contradiction argument, then there is at least one inequality that does not hold. Without loss of generality, we assume Uλ(x)<0 at some point in Ωλ, then there exists some x~∈Ωλ, such that(15)Uλx~=minΩλ⁡Uλx<0,then we will show that Vλ(x~)<0 and derive a contradiction.Since the fractional p-Laplacian is a special case of the fully nonlinear operatorsFα(.), when(16)α=sp,Gt=tp-2t,we need an analysis lemma about G(t) in [6].Lemma 3.
ForG(t)=tp-2t, it is well-known that, by the mean value theorem, we have(17)Gt2-Gt1=G′ξt2-t1.Then there exists a constant c0>0, such that(18)ξ≥c0max⁡t1,t2.Then we have(19)-Δpsuλx~--Δpsux~=Cn,spPV∫RnGuλx~-uλy-Gux~-uyx~-yn+spdy=Cn,spPV∫ΣλGuλx~-uλy-Gux~-uyx~-yn+spdy+Cn,spPV∫ΣλGuλx~-uy-Gux~-uλyx~-yλn+spdy=Cn,spPV∫ΣλGuλx~-uλy-Gux~-uy1x~-yn+sp-1x~-yλn+spdy+∫ΣλGuλx~-uλy-Gux~-uy+Guλx~-uy-Gux~-uλyx~-yλn+spdy≤Cn,spPV∫ΣλGt1x~,y-Gt2x~,y+Gt3x~,y-Gt4x~,yx~-yλn+spdy≤Cn,spPVUλx~∫ΣλG′ξy+G′ηyx~-yλn+spdy=Cn,spPVUλx~Ix~.Here(20)t1x~,y=uλx~-uλy,t2x~,y=ux~-uλy,t3x~,y=uλx~-uy,t4x~,y=ux~-uy,and(21)t1x~,y<ξx~,y<t2x~,y,t3x~,y<ηx~,y<t4x~,y.We now estimateI(x~) for x~∈Ωλ to derive a contradiction when λ is sufficiently close to -1. To this end, we need to use Lemma 3. Let D=Σλ∖Ωλ. Noticing that u(y)=0 in D, for some positive constant c1 and c2, we have(22)Ix~=∫Σλp-2ξp-2+ηp-2x~-yλn+spdy≥∫Σλp-2c0t2p-2+t4p-2x~-yλn+spdy≥∫Σλc1t4yp-2x~-yλn+spdy≥c1∫Dup-2x~x~-yλn+spdy≥c2up-2x~δsp,where δ=λ+1 is the width of the region Ωλ in the x1-direction.Hence(23)-Δpsuλx~--Δpsux~≤CUλx~up-2x~δsp<0.Together with (12) and f′(·)>0, it is easy to deduce that(24)Vλx~<0,and(25)CUλx~up-2x~δsp≥f′ξλx~Vλx~,and we have(26)Uλx~≥Cf′ξλx~Vλx~δspup-2x~.From (24), there exists x¯∈Ωλ such that(27)Vλx¯=minΩλ⁡Vλx<0.Similar to (23), we can derive(28)-Δptvλx¯--Δptvx¯≤CVλx¯vp-2x¯δtp<0,and, combining (13) and (26), we can deduce that(29)CVλx¯vp-2x¯δtp≥g′ηλx¯Uλx¯≥g′ηλx¯Uλx~≥Cf′ξλx~Vλx~δspg′ηλx¯up-2x~,then(30)Vλx¯≥f′ξλx~g′ηλx¯up-2x~vp-2x¯δs+tpVλx~=AVλx~,where(31)A=f′ξλx~g′ηλx¯up-2x~vp-2x¯δs+tp.When λ is sufficiently close to -1, δ becomes sufficiently small; combining assumptions (i) and (ii) in Theorem 1, we can deduce that A is sufficiently small and we can let A<1. According to (24), inequality (30) becomes(32)Vλx¯≥AVλx~>Vλx~.This is a contradiction with the fact thatx¯ is the negative minimum of Vλ(x). Therefore (14) must be true for λ sufficiently closed to -1.Step 2.
Inequality (14) provides a starting point, from which we move plane Tλ toward the right as long as (14) holds to its limiting position to show that u,v are symmetric about the limiting plane. More precisely, let(33)λ0=λ≤0∣Uμx,Vμx≥0,x∈Ωμ,∀μ≤λ.We will show that(34)λ0=0,and(35)Uλ0x≡0,Vλ0x≡0,x∈Ωλ0.
Suppose thatλ0<0, and Uλ0(x),Vλ0(x)≥0 but Uλ0(x)≢0 and Vλ0(x)≢0, from that nonnegative functions Uλ0(x) or Vλ0(x) are positive at some point in Ωλ0; combining (12), (13), and the fact that f′(·),g′(·)>0, by using the strong maximum principle of Theorem A.1, we obtain(36)Uλ0x>0,Vλ0x>0,x∈Ωλ0.In fact, if there exists somex0∈Ωλ0, such that(37)Uλ0x0=minΩλ0⁡Uλ0=0,we have,(38)-Δpsuλ0x0--Δpsux0=Cn,spPV∫Σλ0Guλ0x0-uλ0y-Gux0-uy1x0-yn+sp-1x0-yλ0n+spdy+∫Σλ0Guλ0x0-uλ0y-Gux0-uy+Guλ0x0-uy-Gux0-uλ0yx0-yλ0n+spdy=Cn,spPV∫Σλ0Guλ0x0-uλ0y-Gux0-uy1x0-yn+sp-1x0-yλ0n+spdy+Uλ0x0∫Σλ0G′ξy+G′ηyx0-yλ0n+spdy=Cn,spPV∫Σλ0Guλ0x0-uλ0y-Gux0-uy1x0-yn+sp-1x0-yλ0n+spdy<0.On the other hand,(39)-Δpsuλ0x0--Δpsux0=fvλ0x-fvx=f′ξλ0xVλ0x≥0.A contradiction with (38), hence (36) must be true.By the definition ofλ0, there exists a sequence 0>λk↘λ0, and xk∈Ωλk, such that(40)Uλkxk=minΣλk⁡Uλkx<0,∇Uλkxk=0,and, combining (23) and (24), there also exists a sequence 0>λk↘λ0 and xl∈Ωλk, such that(41)Vλkxl=minΣλk⁡Vλkx<0,∇Vλkxl=0.There is a subsequence of{xk}and{xl} that converges to some point x0 and x1, respectively. And from (40), (41), and the continuity of Uλ(x),Vλ(x) and its derivative with respect to both x and λ, we have(42)Uλ0x0≤0,hencexo∈∂Σλ0;∇Uλ0x0=0,and(43)Vλ0x1≤0,hencex1∈∂Σλ0;∇Vλ0x1=0.Then, we have(44)limδk→0¯⁡1δk-Δpsuλkxk--Δpsuxk=f′ξλkxkVλkxkδk≥f′ξλkxkVλkxlδk⟶0,and(45)limδk→0¯⁡1δk-Δptvλkxl--Δptvxl=g′ηλkxlUλkxlδk≥g′ηλkxlUλkxkδk⟶0.This will contradict with TheoremA.2. Therefore, we have(46)λ0=0,Uλ0x=Vλ0x≡0,∀x∈Ωλ0. Since x1 direction can be chosen arbitrarily, we conclude that u and v must be radially symmetry and monotone decreasing about the origin.This completes the proof of Theorem1.



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*Source: 1565731-2018-08-01.xml*