# Ground State Solutions to a Critical Nonlocal Integrodifferential System

Authors: Min Liu; Zhijing Wang; Zhenyu Guo
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

Consider the following nonlocal integrodifferential system:LKu+λ1u+μ1u2⁎-2u+Gu(x,u,v)=0inΩ,LKv+λ2v+μ2v2⁎-2v+Gv(x,u,v)=0inΩ,u=0,v=0inRN∖Ω, where LK is a general nonlocal integrodifferential operator, λ1,λ2,μ1,μ2>0, 2⁎≔2N/N-2s is a fractional Sobolev critical exponent, 0<s<1, N>2s, G(x,u,v) is a lower order perturbation of the critical coupling term, and Ω is an open bounded domain in RN with Lipschitz boundary. Under proper conditions, we establish an existence result of the ground state solution to the nonlocal integrodifferential system.


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## Body

## 1. Introduction

Recently, fractional Sobolev spaces and the corresponding nonlocal equations are applied to many subjects, such as, anomalous diffusion, elliptic problems with measure data, gradient potential theory, minimal surfaces, nonuniformly elliptic problems, optimization, phase transitions, quasigeostrophic flows, singular set of minima of variational functionals, and water waves (see [1] and the references therein). For more details, please refer the book [2]. In [3], the authors considered the following equation:(1)LKu+λu+u2∗-2u+fx,u=0inΩ,u=0inRN∖Ω,where LK is a general nonlocal integrodifferential operator of order s which will be defined later and f is a lower order perturbation of the critical power |u|2∗-2u, and, under proper conditions, an existence result of its solutions was obtained.In present paper, we use some ideas and techniques in [3–5] to extend the results in [3] to the system case.Consider the following nonlocal integrodifferential system:(2)LKu+λ1u+μ1u2∗-2u+Gux,u,v=0inΩ,LKv+λ2v+μ2v2∗-2v+Gvx,u,v=0inΩ,u=0,v=0inRN∖Ω,where λ1,λ2,μ1,μ2>0, 2∗≔2N/N-2s is a fractional Sobolev critical exponent, 0<s<1, N>2s, G(x,u,v) is a lower order perturbation of the critical coupling term, Ω is an open bounded domain in RN with Lipschitz boundary, and LK is a general nonlocal integrodifferential operator defined as (3)LKux=∫RNux+y+ux-y-2uxKydy,x∈RN.Here, K:RN∖{0}→(0,+∞) is a function satisfying(4)mK∈L1RN,wheremx=min⁡1,x2,(5)∃θ>0suchthatKx≥θxN+2s,∀x∈RN∖0,(6)K-x=Kxforanyx∈RN∖0.Define the Hilbert spaceDK(Ω) as the completion of Cc∞(Ω) with respect to the norm ·DK induced by the scalar product 〈·,·〉DK given by (7)u,vDK≔∫R2Nux-uyvx-vyKx-ydxdy.If Ω is an open bounded Lipschitz domain, then DK(Ω) coincides with the Sobolev space(8)X0≔f∈X:f=0a.e.inΩc,where X is a linear space of Lebesgue measurable functions from RN to R such that the restriction to Ω of any function f in X belongs to L2(Ω) and the map (x,y)↦(f(x)-f(y))K(x-y) is in L2R2N∖(Ωc×Ωc),dxdy, and Ωc is the complement of Ω in RN.The solutions of (2) coincide with the critical points of the following energy functional EK:DK→R:(9)EKu,v=12u,vDK2-12λ1u2,Ω2+λ2v2,Ω2-12∗μ1u2∗,Ω2∗+μ2v2∗,Ω2∗-∫ΩGx,ux,vxdx,where |u|p,Ω=∫Ω|u|pdx1/p, 1≤p<+∞, DK(Ω)≔DK(Ω)×DK(Ω), endowed with norm (u,v)DK2≔uDK2+vDK2, and G(x,s,t) is a nonnegative Carathéodory function from Ω×R×R to R; namely,(10)∀δ>0,sup⁡Gx,s,ts+t:a.e.x∈Ω,s,t≔s2+t2≤δ<+∞,(11)lims,t→0,0Gx,s,ts2+t2=0uniformlyinx∈Ω,(12)lims,t→+∞Gx,s,ts2∗+t2∗=0uniformlyinx∈Ω.Consider the following Nehari manifold: (13)MK≔u,v∈DKΩ∖0,0:u,vDK2=λ1u2,Ω2+λ2v2,Ω2+μ1u2∗,Ω2∗+μ2v2∗,Ω2∗+∫ΩGux,ux,vxux+Gvx,ux,vxvxdx,define the ground state energy of (2) by(14)BK≔infu,v∈MKEKu,v,and call a solution (u,v) by a ground state solution if EK(u,v)=BK. Define(15)S¯K≔infu,v∈DKRNu,v≠0,0⁡u,vDK2μ1u2∗,RN2∗+μ2v2∗,RN2∗2/2∗,S¯K,λ1,λ2≔infu,v∈DKΩu,v≠0,0⁡u,vDK2-λ1u2,Ω2-λ2v2,Ω2μ1u2∗,Ω2∗+μ2v2∗,Ω2∗2/2∗.Then S¯K,λ1,λ2≤min⁡{S¯K,SK,λ1,SK,λ2}, where (16)SK,λ1≔infu∈DKΩu≠0uDK2-λ1u2,Ω2μ1u2∗,Ω2∗2/2∗,SK,λ2≔infv∈DKΩv≠0vDK2-λ2v2,Ω2μ2v2∗,Ω2∗2/2∗.Theorem 1.
Assume thatλ1,λ2∈(0,λK(Ω)), where λK(Ω) is the first eigenvalue of the nonlocal operator -LK with homogeneous Dirichlet boundary conditions. If there exists (u0,v0)∈DK(Ω)∖{(0,0)} with u0,v0≥0 a.e. in RN, such that(17)supt>0EKtu0,tv0<sNS¯K,λ1,λ2N/2s,then (2) has a nontrivial ground state solution.


## 2. Preliminaries

We need the following results, which have been proved in [6].Lemma 2.
There exists a positive constantCs such that, for any u∈DK(Ω), (18)Cs∫R2Nux-uy2x-yN+2sdxdy≥u2∗,RN2=u2∗,Ω2,where Cs is only depending on N and s.


## 3. Proof of Theorem1

Lemma 3.
Assume thatG:Ω×R×R→R is a Carathéodory function; that is, (10)–(12) hold. Then, for any ε>0, there exist a(ε)>0 and b(ε)>0 such that for a.e. x∈Ω and, for any s,t∈R,(19)Gx,s,t≤εs2∗+t2∗+aεs+t,(20)Gx,s,t≤εs2+t2+bεs2∗+t2∗.Proof.
For anyε>0, by (12), there exists δ=δ(ε)>0 such that, for a.e. x∈Ω and for any (s,t)∈R2 with (s,t)>δ,(21)Gx,s,t≤εs2∗+t2∗. It follows from (10) that there exists a(ε)=aδ(ε)>0 satisfying, for a.e. x∈Ω and for any (s,t)∈R×R with |(s,t)|≤δ,(22)Gx,s,t≤aεs+t.Combining (21) and (22), we see that (19) holds.
By (11), for any ε>0, there exists σ=σ(ε)>0 such that, for a.e. x∈Ω and for any (s,t)∈R2 with (s,t)<σ,(23)Gx,s,t≤εs2+t2. It is easy to see from (19) that (24)Gx,s,t≤s2∗+t2∗+a1s+t=s2∗+t2∗1+a1s+ts2∗+t2∗.Therefore, there exists b(ε)=bσ(ε)>0 satisfying, for a.e. x∈Ω and for any (s,t)∈R2 with (s,t)≥σ,(25)Gx,s,t≤bεs2∗+t2∗.Then, (20) follows from (23) and (25).Lemma 4.
Assume thatλ1,λ2∈(0,λK(Ω)). Then, there exist ρ,τ>0 and e=(e1,e2)∈DK(Ω) such that (i)
EK(u,v)≥τ for any (u,v)∈DK(Ω) with (u,v)DK=ρ;(ii)
e1,e2≥0 a.e. in RN, (e1,e2)DK>ρ, and EK(e1,e2)<τ.
In particular, if (17) holds, then we can construct e by e=t0(u0,v0), where (u0,v0) is given by (17) and t0>0 is large enough.Proof.
For(u,v)∈DK(Ω) and any ε>0, by (20), Lemma 2, Hs(RN)↪L2∗(RN), and (5), we deduce that (26)EKu,v≥12u,vDK2-12λ1u2,Ω2+λ2v2,Ω2-12∗μ1u2∗,Ω2∗+μ2v2∗,Ω2∗-∫Ωεu2+v2+bεu2∗+v2∗dx≥121-λ1λKΩuDK2+121-λ2λKΩvDK2-εΩ2∗-2/2∗u2∗,Ω2+v2∗,Ω2-μ12∗+bεu2∗,Ω2∗-μ22∗+bεv2∗,Ω2∗≥121-λ1λKΩuDK2+121-λ2λKΩvDK2-εΩ2∗-2/2∗Cs∫R2Nux-uy2x-yN+2sdxdy+∫R2Nvx-vy2x-yN+2sdxdy-μ12∗+bεCs2∗/2∫R2Nux-uy2x-yN+2sdxdy2∗/2-μ22∗+bεCs2∗/2∫R2Nvx-vy2x-yN+2sdxdy2∗/2≥121-λ1λKΩ-εCsΩ2∗-2/2∗θuDK2+121-λ2λKΩ-εCsΩ2∗-2/2∗θvDK2-μ12∗+bεCsθ2∗/2uDK2∗-μ22∗+bεCsθ2∗/2vDK2∗.For ε<θ(λK(Ω)-max⁡{λ1,λ2})/2CsλK(Ω)Ω2∗-2/2∗, we have (27)EKu,v≥au,vDK21-bu,vDK2∗-2,where a and b are suitable positive constants. Let (u,v)∈DK satisfy u,vDK=ρ>0, where ρ is small enough such that 1-bρ2∗-2>0. Therefore, (28)infu,v∈DKΩ∥u,v∥DK=ρEKu,v≥aρ21-bρ2∗-2≔τ>0,which proves (i).
Fix(u,v)∈DK(Ω) with (u,v)DK=1 and u≥0,v≥0 a.e. in RN. Choosing ε=min⁡{μ1,μ2}/2·2∗ in (19), we get that(29)Gx,u,v≤min⁡μ1,μ22·2∗u2∗+v2∗+amin⁡μ1,μ22·2∗u+v.For t>0, by (29), we have (30)EKtu,tv=t22u,vDK2-t22λ1u2,Ω2+λ2v2,Ω2-t2∗2∗μ1u2∗,Ω2∗+μ2v2∗,Ω2∗-∫ΩGx,tu,tvdx≤t22-t2∗min⁡μ1,μ22∗u2∗,Ω2∗+v2∗,Ω2∗+t2∗min⁡μ1,μ22·2∗∫Ωu2∗+v2∗dx+tamin⁡μ1,μ22·2∗∫Ωu+vdx≤t22-min⁡μ1,μ22·2∗u2∗,Ω2∗+v2∗,Ω2∗t2∗+amin⁡μ1,μ22·2∗u2,Ω2+v2,Ω2t,which implies that EK(tu,tv)→-∞ as t→+∞. Then, (ii) follows by taking e=(e1,e2)=(tu,tv) with t sufficiently large.
In particular, if (17) holds, then we may fix (u,v) by (u,v)=(u0,v0)/(u0,v0)DK. Thus, we can construct e by e=(tu0,tv0)/u0,v0DK=t0(u0,v0), where t0 is large enough.Define(31)BK≔infh∈Γsup0<t<1EKht,where (32)Γ≔h∈C0,1,DKΩ:h0=0,h1=e=e1,e2with e=t0(u0,v0) given in Lemma 4. It is standard to see that(33)BK=BK=BK,where BK=inf(u,v)∈DK(Ω)∖{(0,0)}supt>0EK(tu,tv) and BK is given in (14).Lemma 5.
τ ≤ B K < s / N S ¯ K , λ 1 , λ 2 N / 2 s, where τ is given in Lemma 4 and S¯K,λ1,λ2 is defined in (15).Proof.
Noting that, for anyh∈Γ, the function t↦h(t)DK is continuous in [0,1]; we get that h(0)DK=(0,0)DK=0<ρ and h(1)DK=eDK>ρ, where ρ is given by Lemma 4. Thus, we may find th∈(0,1) such that h(th)DK=ρ. Therefore, (34)sup0<t<1EKht≥EKhth≥infu,v∈DKΩu,vDK=ρEKu,v≥τ,which implies that BK≥τ. Recalling that e=t0(u0,v0) and the special pass [0,1]∋t↦tt0(u0,v0) belongs to Γ, by (17), we deduce that (35)BK=infh∈Γsup0<t<1EKht≤supt>0EKtu0,tv0<sNSK,λ1,λ2N/2s.Lemma 6.
Every sequence{(un,vn)} with EK(un,vn)→c and EK′(un,vn)→0 in DK′(Ω) as n→∞ ((PS)c sequence of EK) is bounded in DK(Ω).Proof.
From(36)c+o1=EKun,vn=12un,vnDK2-12λ1un2,Ω2+λ2vn2,Ω2-12∗μ1un2∗,Ω2∗+μ2vn2∗,Ω2∗-∫ΩGx,un,vndxand (37)o1=EK′un,vnun,vn=un,vnDK2-λ1un2,Ω2+λ2vn2,Ω2-μ1un2∗,Ω2∗+μ2vn2∗,Ω2∗-∫ΩGux,un,vnun+Gvx,un,vnvndx,it can be seen that(38)c+o1=EKun,vn-12EK′un,vnun,vn=12-12∗μ1un2∗,Ω2∗+μ2vn2∗,Ω2∗-∫ΩGx,un,vndx+12∫ΩGux,un,vnun+Gvx,un,vnvndx≥sNμ1un2∗,Ω2∗+μ2vn2∗,Ω2∗-k∫ΩGx,un,vndxfor some suitable positive constant k. For u∈DK(u), it follows from Lemma 2 and (5) that(39)∫Ωudx≤Ω2∗-1/2∗u2∗,Ω≤Ω2∗-1/2∗Cs∫R2Nux-uy2x-yN+2sdxdy1/2≤ΘuDKfor some positive constant Θ. By (19) and (39), we have(40)∫ΩGx,un,vndx≤εun2∗,Ω2∗+vn2∗,Ω2∗+aεΘunDK+vnDK≤εun2∗,Ω2∗+vn2∗,Ω2∗+Cεun,vnDK,where C(ε) is a positive constant relevant to ε and Ω. Choosing ε small enough such that s/Nmin⁡{μ1,μ2}-kε>0, then (38) and (40) yield that(41)un2∗,Ω2∗+vn2∗,Ω2∗≤C1+un,vnDK. Letting ε=1 in (40) and noting 0<λ1,λ2<λK(Ω), by (41), we infer that (42)c+o1=EKun,vn≥121-λ1λKΩunDK2+121-λ2λKΩvnDK2-12∗μ1un2∗,Ω2∗+μ2vn2∗,Ω2∗-∫ΩGx,un,vndx≥121-max⁡λ1,λ2λKΩun,vnDK2-max⁡μ1,μ22∗+1un2∗,Ω2∗+vn2∗,Ω2∗-C1un,vnDK≥C1un,vnDK2-C21+un,vnDKfor suitable constants C1 and C2, which implies that {(un,vn)} is bounded in DK(Ω).Proof of Theorem1.
It follows fromG(x,0,0)=0 that EK(0,0)=0<τ, where τ is given in Lemma 4. By Lemma 4, we see that EK has a Mountain Pass Structure (e.g., [7]). By (33) as well as Lemma 5 and Theorem 2.2 in [8], there is (PS)BK sequence {(un,vn)} for EK. By Lemma 6, we get that {(un,vn)} is bounded in DK(Ω), and then, going if necessary to a subsequence, still denoted by {(un,vn)}, there exists (u∞,v∞)∈DK(Ω) such that (un,vn)⇀(u∞,v∞) weakly in DK(Ω); i.e.,(43)∫R2Nunx-unyϕx-ϕyKx-ydxdy+∫R2Nvnx-vnyψx-ψyKx-ydxdy→∫R2Nu∞x-u∞yϕx-ϕyKx-ydxdy+∫R2Nv∞x-v∞yψx-ψyKx-ydxdy for any (ϕ,ψ)∈DK(Ω). It follows from (41) and the boundedness of {(un,vn)} in DK(Ω) that(44)unandvnareboundedinL2∗Ω.Then, it is standard to see that(45)un⇀u∞,vn⇀v∞weaklyinL2∗Ω,un→u∞,vn→v∞stronglyinLpΩ,un→u∞,vn→v∞a.e.  inΩ,un2∗-2un⇀u∞2∗-2u∞,vn2∗-2vn⇀v∞2∗-2v∞weaklyinL2∗/2∗-1Ω,up to a subsequence, where p∈[1,2∗). By (19), (44), and the boundedness of Ω, we see that(46)Gu·,un·,vn·,Gv·,un·,vn·areboundedinL2∗/2∗-1Ω. Since G(x,u,v) is superlinear with respect to u and v, we get that the mappings t↦Gu(x,t,v) and t↦Gv(x,u,t) are continuous in R. Therefore,(47)Gu·,un·,vn·→Gu·,u∞·,v∞·,Gv·,un·,vn·→Gv·,u∞·,v∞·a.e.  inΩ. Then, by (46) and (47), we have (48)Gu·,un·,vn·⇀Gu·,u∞·,v∞·,Gv·,un·,vn·⇀Gv·,u∞·,v∞·weaklyinL2∗/2∗-1Ω.Noting that L2∗(Ω) is the dual space of L2∗/2∗-1(Ω), we deduce that (49)∫ΩGux,un,vnϕdx→∫ΩGux,u∞,v∞ϕdx,∫ΩGvx,un,vnψdx→∫ΩGvx,u∞,v∞ψdxforanyϕ,ψ∈L2∗Ω,and then, in particular,(50)∫ΩGux,un,vnϕdx→∫ΩGux,u∞,v∞ϕdx,∫ΩGvx,un,vnψdx→∫ΩGvx,u∞,v∞ψdxforanyϕ,ψ∈DKΩ. For any (ϕ,ψ)∈DK(Ω), we have (51)0←EK′un,vnϕ,ψ=∫R2Nunx-unyϕx-ϕyKx-ydxdy+∫R2Nvnx-vnyψx-ψyKx-ydxdy-∫Ωλ1unϕ+λ2vnψdx-∫Ωμ1un2∗-2unϕ+μ2vn2∗-2vnψdx-∫ΩGux,un,vnϕ+Gvx,un,vnψdx.Then, it follows from (43), (45), and (50) that (52)∫R2Nu∞x-u∞yϕx-ϕyKx-ydxdy+∫R2Nv∞x-v∞yψx-ψyKx-ydxdy-∫Ωλ1u∞ϕ+λ2v∞ψdx-∫Ωμ1u∞2∗-2u∞ϕ+μ2v∞2∗-2v∞ψdx-∫ΩGux,u∞,v∞ϕ+Gvx,u∞,v∞ψdx=0,which means that (u∞,v∞) is solution of (2). It remains to show that (u∞,v∞) is nontrivial. Suppose by contradiction that (u∞,v∞)=(0,0) in Ω. Then, by (19) and boundedness of {un} and {vn} in L2∗(Ω), we have (53)∫ΩGx,un,vndx≤ε∫Ωun2∗+vn2∗dx+aε∫Ωun+vndx≤Cε+aεun1,Ω+vn1,Ω,and similarly (54)∫ΩGux,un,vnundx≤Cε+aεun1,Ω+vn1,Ω,∫ΩGvx,un,vnvndx≤Cε+aεun1,Ω+vn1,Ω.Noting that un→0,vn→0 in Lp(Ω) for any p∈[1,2∗) and letting n→∞, we get that (55)limsupn→∞∫ΩGx,un,vndx≤Cε,limsupn→∞∫ΩGux,un,vnundx≤Cε,limsupn→∞∫ΩGvx,un,vnvndx≤Cε.Taking ε→0+, we have (56)∫ΩGx,un,vndx,∫ΩGux,un,vnundx,∫ΩGvx,un,vnvndx↓0.Therefore, (57)un,vnDK2-μ1un2∗,Ω2∗-μ2vn2∗,Ω2∗=EK′un,vnun,vn+λ1un2,Ω2+λ2vn2,Ω2+∫ΩGux,un,vnundx+∫ΩGvx,un,vnvndx→0.Assume, up to a subsequence, that (un,vn)DK2→b. Then, μ1un2∗,Ω2∗+μ2vn2∗,Ω2∗→b, and (58)BK+o1=EKun,vn=b2-b2∗+o1,which means that BK=s/Nb. By Lemma 5, we see that b>0. Noting (15), we have (59)un,vnDK2≥S¯K,λ1,λ2μ1un2∗,Ω2∗+μ2vn2∗,Ω2∗2/2∗,which implies that b≥SK,λ1,λ2b2/2∗. Therefore, BK≥s/NSK,λ1,λ22∗/2∗-2=s/NSK,λ1,λ2N/2s, contradicting with Lemma 5. Hence, (u∞,v∞) can not be (0,0). Besides (45), up to a subsequence, we may assume that (60)Gx,un,vn→Gx,u∞,v∞stronglyinL1Ω,Gux,un,vn→Gux,u∞,v∞,Gvx,un,vn→Gvx,u∞,v∞stronglyinLq/q-1Ω,where q∈(2,2∗) is the power sum of G with respect to u and v. From (61)un,ϕDK+vn,ψDK-λ1∫Ωunϕdx-λ2∫Ωvnψdx-μ1∫Ωun2∗-2unϕdx-μ2∫Ωvn2∗-2vnψdx-∫ΩGux,un,vnϕ+Gvx,un,vnψdx=o1,we get that (62)u∞,ϕDK+v∞,ψDK-λ1∫Ωu∞ϕdx-λ2∫Ωv∞ψdx-μ1∫Ωu∞2∗-2u∞ϕdx-μ2∫Ωv∞2∗-2v∞ψdx-∫ΩGux,u∞,v∞ϕ+Gvx,u∞,v∞ψdx=0,where (φ,ψ)∈DK(Ω); i.e., EK′(u,v)=0 in DK′(Ω). Setting ωn=un-u and σn=vn-v, then ωn⇀0,σn⇀0 weakly in DK(Ω). Thus, ωn→0,σn→0 strongly in L2(Ω), ωn→0,σn→0 a.e. in Ω, Gu(x,ωn,σn)ωn→0, Gv(x,ωn,σn)σn→0, G(x,ωn,σn)→0 in L1(Ω), up to a subsequence. Brezis-Lieb Lemma guarantees that(63)un2∗,Ω2∗=u∞2∗,Ω2∗+ωn2∗,Ω2∗+o1,vn2∗,Ω2∗=v∞2∗,Ω2∗+σn2∗,Ω2∗+o1,∫ΩGx,un,vndx=∫ΩGx,u∞,v∞dx+∫ΩGx,ωn,σndx+o1. Since EK′(un,vn)(un,0)=o(1),EK′(un,vn)(0,vn)=o(1) and EK′(u∞,v∞)(u∞,v∞)=0, by (63), we deduce that(64)ωnDK2-μ1ωn2∗,Ω2∗=o1,σnDK2-μ2σn2∗,Ω2∗=o1,and(65)IKωn,σn=EKun,vn-EKu∞,v∞+o1,where (66)IKu,v=12u,vDK2-12∗μ1u2∗,RN2∗+μ2v2∗,RN2∗Assume that (67)limn→∞μ1ωn2∗,Ω2∗=a1,limn→∞μ2σn2∗,Ω2∗=a2,up to a subsequence. By (64), we see that (68)IKωn,σn=12-12∗a1+a2+o1.Letting n→∞, we have(69)BK=limn→∞EKun,vn=12-12∗a1+a2+EKu∞,v∞≥0.
Suppose by contradiction that(u∞,v∞)=(u∞,0), where u∞≢0. Then, u∞ is a nontrivial solution of LKu+λ1u+μ1u2∗-2u=0. Hence, by (69), we deduce that BK≥EK(u∞,0)≥s/NSK,λ1N/2s≥s/NS¯K,λ1,λ2N/2s, contradiction with Lemma 5. Therefore, (u∞,v∞) can not be (u∞,0), where u∞≢0. Similarly, (u∞,v∞) can not be (0,v∞), where v∞≢0. Thus, (2) has a nontrivial ground state solution, which completes the proof.



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