# Sign Changing Solutions for Coupled Critical Elliptic Equations

Authors: Xin Wang; Xiaorui Yue
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we consider the coupled elliptic system with a Sobolev critical exponent. We show the existence of a sign changing solution for problemP for the coupling parameter −μ1μ2<β<0. We also construct multiple sign changing solutions for the symmetric case.


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## Body

## 1. Introduction

In this paper, we consider the following coupled elliptic system with a Sobolev critical exponent:(1)P−Δu1+λ1u1=ν1u1p1−2u1+μ1u12∗−2u1+βu12∗/2−2u1u22∗/2,inΩ,−Δu2+λ2u2=ν2u2p2−2u2+μ2u22∗−2u2+βu12∗/2u22∗/2−2u2,inΩ,u1=u2=0on∂Ω,where Ω⊂ℝN is a bounded smooth domain, N≥6, 2<pj<2∗, λj∈−λ1Ω,0,νj,μj>0 for j=1,2, and λ1Ω is the first eigenvalue of −Δ with the Dirichlet boundary condition.In recent years, the following coupled elliptic system has attracted much interest:(2)−Δu1+λ1u1=μ1u13+βu22u1,x∈Ω,−Δu2+λ2u2=μ2u23+βu12u2,x∈Ω,u1=u2=0,x∈∂Ω.Ω
⊂
ℝ
N is a bounded smooth domain or Ω=ℝN, N≥3. System (2) arises when we consider the standing wave solutions of the two coupled Gross–Pitaevskii equations:(3)−i∂∂tΦ1=ΔΦ1+μ1Φ12Φ1+βΦ22Φ1,x∈Ω,t>0,−i∂∂tΦ2=ΔΦ2+μ2Φ22Φ2+βΦ12Φ2,x∈Ω,t>0,Φj=Φjx,t∈ℂ,j=1,2,Φjx,t=0,x∈∂Ω,t>0,j=1,2.The system above has applications in many physical problems (see [1, 2]). It also arises in the Hartree–Fock theory for a double condensate, which is a binary mixture of Bose–Einstein condensates in two different hyperfine states 1 and 2 (see [3] and the references therein). Considering the solitary wave solutions of system (3), we set Φjx,t=eiλjtujx for j=1,2. Then, it is reduced to system (2). For the subcritical case, i.e., N=3, the existence of least energy and other finite energy solutions as well as the existence and multiplicity of positive and sign-changing solutions are studied in [2, 4–23] and the references therein. For the critical case, i.e., N=42∗:=2N/N−2=4, the existence of a positive least energy solution is proved when β is negative, positive small, and positive large in [24]. For the higher dimension N≥5, the authors in [25, 26] also consider the following critical case:(4)−Δu1+λ1u1=μ1u12∗−1+βu12∗/2−1u22∗/2,inΩ,−Δu2+λ2u2=μ2u22∗−1+βu12∗/2u22∗/2−1,inΩ,u1,u2≥0inΩ,u1=u2=0on∂Ω.Note that whenN=4, system (4) is the same as system (2). But interestingly, the authors in [25] find different results for the higher dimension case N≥5 from that of N=4. The authors in [27] also consider the even case:(5)−Δu1+λ1u1=μ1u12∗−2u1+βu12∗/2−2u1u22∗/2,inΩ,−Δu2+λ2u2=μ2u22∗−2u2+βu12∗/2u22∗/2−2u2,inΩ,u1=u2=0on∂Ω.They show that whenN≥6, system (5) has a sign changing solution for any β<0.Recall that whenβ=0, system (4) becomes the famous Brezis–Nirenberg problem:(6)−Δuj+λjuj=μjuj2∗−1,inΩ,uj≥0inΩ,uj=0on∂Ω,j=1,2,which are two different independent scalar equations. In [28], the authors show that when N≥4,−λ1Ω<λj<0, and μj=1, equation (6) has a ground state solution. For the even case,(7)−Δu+λju=μju2∗−2u,inΩ,u=0on∂Ω,j=1,2.The authors in [29] show that when N≥6,−λ1Ω<λj<0, and μj>0, equation (7) has sign changing solutions for j=1,2. In [30, 31], when N≥4,λj<0,andμj>0, the authors obtained nontrivial solutions of (7) for j=1,2.In fact, in the pioneering paper [28] of Brezis and Nirenberg, they also study a more general equation including the following classical case:(8)−Δu+λju=νjupj−1+μju2∗−1,inΩ,u≥0inΩ,u=0on∂Ω,j=1,2,where 2<pj<2∗andνj>0. From [28], we know that when N≥4,νj,μj>0,andλj∈−λ1Ω,0, then problem (8) has positive ground state solutions v1,v2∈C2Ω∩CΩ¯ with the energy:(9)0<Bj:=12∫Ω∇vj2+λjvj2−∫Ωνjpjvjp−12∗∫Ωμjvj2∗<1Nμj−N−2/2SN/2,j=1,2,where S is the sharp imbedding constant from D1,2ℝN into L2∗ℝN.Based on these, firstly, we try to show similar results in [29–31] for the even case of the general equation (8), i.e.,(10)−Δu+λju=νjupj−2u+μju2∗−2u,inΩ,u=0on∂Ω,j=1,2.Then, naturally and interestingly, we guess that the more general system corresponding to system (5) also has similar result of the existence of sign changing solutions. That is, we consider the more general critical elliptic system as follows:(11)−Δu1+λ1u1=ν1u1p1−2u1+μ1u12∗−2u1+βu12∗/2−2u1u22∗/2,inΩ,−Δu2+λ2u2=ν2u2p2−2u2+μ2u22∗−2u2+βu12∗/2u22∗/2−2u2,inΩ,u1=u2=0on∂Ω.Precisely, we get the following results.Theorem 1.
Assume thatN≥6,2<pj<2∗,λj∈−λ1Ω,0,andνj,μj>0, for j=1,2; then, (10) has sign changing solutions for j=1,2 and the energy is the least one among all sign changing solutions.
Then, we can consider the caseλ1=λ2=λandp1=p2=p. The following equation is one case of (10):(12)−Δu+λu=νup−2u+u2∗−2u,inΩ,u=0on∂Ω,where ν>0. Let v and w be the sign changing solutions of (12) when N≥6,λ∈−λ1Ω,0, and N≥4,λ≤−λ1Ω, respectively (the latter case really exists and we shall show it later). By Lemma 1 in [25], the system below has a positive solution:(13)μ1a12∗−2+βa12∗/2−2a22∗/2=1,μ2a22∗−2+βa12∗/2a22∗/2−2=1.
Then, we can construct solutions of (11) by using vandw as in [2, 24, 25].Theorem 2.
Leta1,a2 be a solution of (13). Assume that ν1=ν/a1p−2andν2=ν/a2p−2,μ1,μ2,β>0 and one of the following happens:(1)
N≥6,−λ1Ω<λ1=λ2<0,(2)
N≥4,λ1=λ2≤−λ1Ω,
Thena1v,a2vanda1w,a2w are sign changing solutions of system (11).We also get a least energy semi-sign changing solution for system (11).Theorem 3.
Assume thatN≥6,2<pj<2∗,−λ1Ω<λj<0,νj,μj>0, and −μ1μ2<β<0. Then, system (11) has a semi-sign changing solution with one component changing sign and the other one positive, and the energy is the least one among all these kinds of solutions.Next, we give the proof of Theorem1 in Section 2. And we shall show Theorems 2 and 3 in Sections 3 and 4, respectively.


## 2. Proof of Theorem 1

In this section, we consider scalar case (10). Fixing j=1 or 2, we show the existence result. The proof is similar to that in [29]. Considering we need this result in the following two Theorems 2 and 3, we verify it also for the completeness of the current paper. The working space and some notations shall be given firstly. We assume that all the integrations below are taken over Ω if without special specification. Since λj>−λ1Ω, we can define the equivalent inner product in H01Ω by(14)u,vj=∫∇u,∇v+λjuv,which gives rise to a norm denoted by ⋅j. We also use uq=∫Ωuqdx1/q1≤q<∞ and u0=∫Ω∇u2dx1/2 for convenience. Then, the energy functional of equation (10) is(15)Iju=12uj2−1pj∫νjujpj−12∗∫μjuj2∗.Recallvj is a ground state solution of (8), that is,(16)Bj:=Ijvj=infIju:u∈H01Ω\0,Ij′uu=0.For the sign changing case, we define the manifold as follows:(17)Sj:=u∈H01Ω:u±≡0,Ij′u±u±=0.where u±:=±maxu,0 and the condition in the definition of (17) is that(18)u±j2=νj∫u±pj+μj∫u±2∗.It is easy to check thatSj≠∅. Then, we define(19)ℬj:=infu∈SjIju.We need a conclusion in [32].Lemma 1 (see [32]).
Consider a rectangleR=Π1sai,bi⊂ℝs and a continuous function Φ=Φ1,…,Φs:R⟶ℝs. If Φixi=ai>0>Φixi=bi hold for all i=1,…,s, then Φ has a zero inside R.Similarly as in [29], we set(20)Eju:=0,u=0,νj∫upj+μj∫u2∗uj2,u≠0.Then, (18) is equivalent to Eju+=Eju−=1. Define(21)ℋj:=h∈C0,1×0,1,H01Ω:ht,0=0,h0,s≥0,h1,s≤0,Ijht,1≤0,Ejhs,1≥2,∀t,s∈0,1.Sinceμj>0, it is easy to see that ℋj≠∅. Then, we have the following lemma.Lemma 2.
ℬ
j
=
inf
h
∈
ℋ
j
sup
u
∈
h
0,1
×
0,1
I
j
u
.Proof.
For anyu∈Sj, t,s≥0, we have(22)Ijtu+−su−=12t2u+j2−νjpjtpj∫u+pj−μj2∗t2∗∫u+2∗+12s2u−j2−νjpjspj∫u−pj−μj2∗s2∗∫u−2∗:=f+t+f−s.
Then,(23)supt,s≥0Ijtu+−su−=f+1+f−1=Iju.
Denote the maph∈ℋj satisfying(24)h0,1×0,1⊂tu+−su−:t,s≥0.
By (23), we have(25)ℬj≥infh∈ℋjsupu∈h0,1×0,1Iju.
On the other hand, for anyh∈ℋj, we have(26)Ejht+−Ejht−≥0,t∈0,s:s∈0,1,≤0,t∈1,s:s∈0,1,Ejht++Ejht−−2<0,t∈t,0:t∈0,1,≥0,t∈t,1:t∈0,1.
Then, by Lemma1, we have that there exists t0∈0,1×0,1 such that Ejht0+=Ejht0−=1, i.e., u0:=ht0∈Sj. Consequently,(27)ℬj≤infh∈ℋjsupu∈h0,1×0,1Iju.
This completes the proof.
Now, we have an upper estimate forℬj.Lemma 3.
ℬ
j
<
c
j
∗
:=
B
j
+
1
/
N
μ
j
−
N
−
2
/
2
S
N
/
2
.Proof.
Sincevj∈CΩ¯ and vj=0 on ∂Ω, there exists Bx0,2r0:=x:x−x0≤2r⊂Ω for some x0 near ∂Ω and r>0 small to be decided later. Take a cutoff function ϕ∈C01Bx0,2r with 0≤ϕ≤1 and ϕx≡1 for x−x0≤r. Let(28)Uε,x0x:=NN−2ε2N−2/4ε2+x−x02N−2/2,x∈ℝN.
Then (see [33, 34]), Uε,x0 is the solution of the following equation:(29)−Δu=u2∗−2u,u∈D1,2ℝN.
Setuε:=ϕUε,x0; then (see [35]),(30)∫∇uε2=SN/2+OεN−2,∫uε2∗=SN/2+OεN,∫uε2≥Cε2+OεN−2,∫uε≤CεN−2/2,∫uεpj−1≤CεN−23−pj+4/2,∫uε2∗−1≤CεN−2/2.
Now, we show that(31)supt,s∈ℝIjtvj+suε<cj∗.
In fact, by (30) and N−23−pj+4/2>N−2/2 since pj<2∗, we have(32)Ijtvj+suε≤12t2vjj2−νjpjtpj∫vjpj−μj2∗t2∗∫vj2∗+12s2uεj2−μj2∗s2∗∫uε2∗+C1tsΔvjL∞Bx0,2ruε1+tsvjL∞Bx0,2ruε1+C2vjL∞Bx0,2r∫suεpj−1+∫suε2∗−1≤Bj+1Nμj−N−2/2SN/2+C3vjL∞Bx0,2rεN−2/2−C4ε2<Bj+1Nμ−N−2/2SN/2,forr>0small enough.
Then by Lemma2, the conclusion follows.□
Set(33)G:=u∈H01Ω:Eju±−1<12.
Then, we show thatIj satisfies a local PS condition in the following sense.Lemma 4.
Under the assumptions in Theorem1, for any c<c∗, if un⊂G satisfies that(34)Ijun⟶c,I′un⟶0,asn⟶∞,then we have that un has a convergent subsequence in H01Ω.Proof.
It is easy to check thatun is bounded in H01Ω, so there exists u∈H01Ω such that un⇀u weakly in H01Ω up to a subsequence and un⟶u strongly in LqΩ for 1≤q<2∗. Then, Ij′u=0 and un±⇀u± weakly in H01Ω. By the Sobolev inequality, we have that there exist positive constants C0andC1 independent of n such that un±j≥C0>0 and then(35)Ijun=Ijun−1pjIj′unun+o1=12−1pjunj2+1pj−12∗μj∫un2∗≥C1>0.
Since(36)o1=Ij′un−Ij′uun±−u±=un±−u±02−μj∫un±−u±2∗+o1≥un±−u±021−SN/N−2μjun±−u±j2∗−2+o1,we have either un±⟶u± strongly in H01Ω or un±−u±02≥μjN−2/2SN/2. Considering the latter case, we have(37)Ijun±=Ijun±−u±+Iju±+o1=1Nun±−u±02+Iju±+o1≥1Nμj−N−2/2SN/2+o1.
Therefore,(38)cj∗>limn⟶∞Ijun=limn⟶∞Ijun++Ijun−≥Bj+1NμJ−N−2/2SN/2=cj∗,is a contradiction. That is, we must have un⟶u strongly in H01Ω and u±≠0.
Now we are ready to show Theorem 1 based on a deformation lemma (see [36–38] for instance).Proof of Theorem 1.
Obviously,Ij is coercive on Sj and then there exists a minimizing sequence un⊂Sj. Let hn∈ℋj such that(39)hn0,1×0,1⊂tun+−sun−:t,s≥0.
Then, by Lemma2, we have(40)limn⟶∞supu∈hn0,1×0,1Iju=limn⟶∞Ijun=ℬj.
By the well-known deformation lemma, we have that there exists a sequenceu^n⊂H01Ω such that(41)distu^n,hn0,1×0,1⟶0,(42)Iju^n⟶ℬj,Ij′u^n⟶0,asn⟶∞.
Obviously,u^n is bounded and there exists u^∈H01Ω such that u^n⇀u^ weakly in H01Ω; then, Ij′u^=0 by the definition of weak convergence.
Now, we claim thatu^n⊂G. In fact, by (41), there exists a sequence u˜n with(43)u˜n=tnun+−snun−∈hn0,1×0,1,distu^n,u˜n⟶0.
Then, forn large enough depending on ξ>0,(44)Iju˜n+=Ijtnun+>Ijun+−ξ≥Bj−ξ,(45)Iju˜n−=Ijsnun−>Ijun−−ξ≥Bj−ξ.
(44), (45), and (43) implies that u^n±≠0. Since Ij′u^nu^n±=o1, the claim comes true.
Therefore, by Lemmas3 and 4, we have that u^n⟶u^ strongly in H01Ω with u^±≠0, i.e., u^ is a sign changing solution of equation (10) and the energy Iju^ is the least one among all sign changing solutions of (10).


## 3. Proof of Theorem 2

Firstly, inspired by [30, 31], we show the following theorem about the scalar equation by the minimax method (see [39]), and we would like to just give the sketch for the completeness of the current paper. We also use the same notations as in Section 2 whenever no confusion arises.Denoteφ1,φ2,φ3, … as the L2 normalized eigenfunctions of −Δ, corresponding to positive eigenvalues 0<λ1Ω<λ2Ω≤λ3Ω≤…↑+∞, counted with their multiplicity. We set Hk:=φ1,φ2,…,φk as the k dimensional space in H01Ω. For the fixed λj<0, define(46)λj+:=minλkΩ:−λj<λkΩ.We may suppose thatλj+=λk0Ω for some k0∈ℕ. Especially, we set λ0Ω≡0.Theorem 4.
Assume thatN≥4,2<pj<2∗,λj<0,​andνj,μj>0, for j=1,2, then equation (10) has a nontrivial solution. Moreover, when λj≤−λ1Ω, the solution changes sign.Proof.
It is easy to check thatIj satisfies the following geometry.(a)
Foru∈Hk0−1⊥, there exist r0,α0>0 such that Iju>α0 with u0=r0 and Iju>0 for 0<u0<r0(b)
For any finite dimensional subspaceH˜⊂H01Ω, it holds that I^j0∩H˜ is bounded in H01Ω, where I^j0:=u∈H01Ω:Iju≥0
As in [36], we define the minimax value:(47)c^j:=suph∈Λinfu∈∂B1∩Hk0−1⊥Ijhu,where(48)Λ:=h∈CH01Ω,H01Ω:hIntB1⊂I^j0∪Br0,his an odd homeomorphism ofH01ΩontoH01Ω.
Obviously,h=r0id∈Λ. We also set another minimax value:(49)b^j:=infK∈Λ^supu∈KIju,where(50)Λ^:=K⊂H01Ω:Kis compact,symmetric,andγK∩h∂B1≥k0, for allh∈Λ,and γ is the Krasnoselskii genus. It can be verified that Λ^≠∅. By properties (a) and (b), we can show that(51)0<α0≤c^j≤b^j<c^j∗:=1Nμj−N−2/2SN/2,and Ij satisfies the PSc^j condition. Also by a deformation lemma, there exists a PSc^j sequence un of Ij and then un⟶u in H01Ω. That is, u is a nontrivial solution of equation (10).
Multiply the equations in (10) with the first eigenfunction φ1 and integrate over Ω; we have(52)λ1Ω+λj∫Ωuφ1=∫Ωνjupj−2uφ1+μju2∗−2u2φ1.
Then, it is easy to see that whenλj≤−λ1Ω,νj,μj>0 any solution of (10) must change sign.Proof of Theorem 2.
The proof is by direct computation. Recall thatvandw are sign changing solutions of (12) being well defined. Since a1,a2 satisfy (13), for any φ∈C0∞Ω, we have(53)ν1∫a1vp−2a1vφ+μ1∫a1v2∗−2a1vφ+β∫a1v2∗/2−2a1va2v2∗/2φ=a1ν∫vp−2vϕ+∫v2∗−2v=a1v,φ1,That is, a1v,a2v satisfy the first equation in system (11). Similarly, we can verify that Theorem 2 holds.


## 4. Proof of Theorem 3

Inspired by [27, 29], we now consider the general case of system (11), i.e., we do not assume that λ1=λ2. We always assume that β<0andνj,μj>0 for j=1,2. Firstly, we introduce the product space H:=H01Ω×H01Ω as the working space. Define the inner product as(54)u,v=∑j=12uj,vjj,foru=u1,u2,v=v1,v2∈H,which gives rise to a norm on H denoted by ⋅. Recall that solutions of (11) are critical points of the following energy functional:(55)Iu=12u2−∫∑j=12νjpujpj−12∗∫∑j=12μjuj2∗+2βu12∗/2u22∗/2.Fora,b∈ℝ with a≤b, we denote(56)Ib:=u∈H:Iu≤b,I^a:=u∈H:Iu≥a,I−1a,b:=Ib∩I^a.Similar toSj related to the scalar equation, we define(57)S:=u1,u2∈H:u1±,u2≡0,I′u1±,u2u1±,0=I′u1,u20,u2=0.That is,(58)u1±12=∫ν1u1±p1+μ1u1±2∗+βu1±2∗/2u22∗/2,u222=∫ν2u2p2+μ2u12∗+βu12∗/2u22∗/2.LetU1,U2∈C0∞Ω with U1±,U2≡0 and suppU1∩suppU2=∅, it is easy to see that there exist t1,t2,s>0 such that t1U1+−t2U2−,sU2∈S. Thus, S≠∅. By Sobolev inequality, it is easy to see that for any u∈S, there exists a constant C0>0 such that(59)u1±1,u22,u1±2∗,u22∗≥C0>0.Set(60)3ε0:=min12−1p1C02,12−1p1C02.Then, we define(61)ℬ:=infu∈SIu.We firstly have a lower bound forℬ.Lemma 5.
Assume that−μ1μ2≤β<0; then I is coercive on S and ℬ>0.Proof.
We may assume thatp1≥p2. For u∈S, note that for β≥−μ1μ2, it holds that(62)∫∑jμjuj2∗+2βu12∗/2u22∗/2≥0.
Then, we have(63)Iu=Iu−1p2I′uu1,u2=12−1p2u2+1p2−1p1∫ν1u1p1+1p2−12∗∫∑jμjujj2∗+2βu12∗/2u22∗/2≥12−1p2u2.
(63) implies that I is coercive on S. Then, by (59), we have ℬ>0.□
We shall give an upper bound forℬ later by defining another manifold and the infimum on it. The idea is similar to that in [27], but since the corresponding equation (10) is different and more complex as well as system (11), some new tricks should be used in the current paper, so we shall give the details of the proofs. Let(64)ℳ:=u1,u2∈H:u1,u2≡0,I′u1,u2u1,0=I′u1,u20,u2=0.
Obviously,S⊂ℳ. Then, we define(65)B:=infu∈ℳIu.
Before the estimate for the bound ofℬ, we need some preliminaries.Lemma 6.
Assume that−μ1μ2≤β<0; then, for any u∈S, supt1,t2,s≥0It1u1+−t2u1−,su2 is attained only by t1=t2=s=1. Similarly, for any u∈ℳ, supt,s>0Itu1,su2 is attained only by t=s=1.Proof.
Foru∈S,t1,t2,s≥0; set(66)Ft1,t2,s:=It1u1+−t2u1−,su2=12t12u+12+t22u−12+s2u222−∫ν1p1t1p1u1+p+ν1p1t2p1u1−p1+ν1p2sp2u2p2−12∗∫μ1t12∗u1+2∗+μ1t22∗u1−2∗+μ2s2∗u22∗−2β2∗s2∗/2∫t12∗/2u1+2∗/2+t22∗/2u1−2∗/2u22∗/2.
Then,Ft11,1,1=Ft21,1,1=Fs1,1,1=0 and(67)Ft1t11,1,1=u1+12−p1−1∫ν1u1+p1−2∗−1∫μ1u1+2∗−2∗2−1β∫u1+2∗/2u22∗/2,Ft2t21,1,1=u1−12−p1−1∫ν1u1−p1−2∗−1∫μ1u1−2∗−2∗2−1β∫u1−2∗/2u22∗/2,Fss1,1,1=u222−p2−1∫ν2u2p2−2∗−1∫μ2u22∗−2∗2−1β∫u12∗/2u22∗/2,Ft1t21,1,1=Ft2t11,1,1=0,Ft1s1,1,1=Fst11,1,1=−2∗2β∫u1+2∗/2u22∗/2,Ft2s1,1,1=Fst21,1,1=−2∗2β∫u1−2∗/2u22∗/2.
Set(68)HF:=Ft1t1Ft1t2Ft1sFt2t1Ft2t2Ft2sFst1Fst2Fss1,1,1.
Denote(69)A1±=∫ν1u1±p1,A2=∫ν2u2p1,D1±=∫μ1u1±2∗,D2=∫μ2u22∗,D1±=∫βu1±2∗/2u22∗/2,D=∫βu12∗/2u22∗/2.
Then,(70)a11:=Ft1t11,1,1=−p1−2A1+−2∗−2D1+−2−2∗2D+<0,a22:=Ft2t21,1,1=−p1−2A1−−2∗−2D1−−2−2∗2D−<0,a33:=Fss1,1,1=−p2−2A2−2∗−2D2−2−2∗2D<0.
Note thatD=D++D−,D2≤D1++D1−D2,and2D≤D1++D1−+D2. We can show that HF is negative definite. Thus, 1,1,1 is the unique maximum point of Ft1,t2,s.Lemma 7.
Assume that−μ1μ2<β<0; then, for any u∈H with u1,u2≡0, we have that there exist unique tu,su>0 such that tuu1,suu2∈ℳ. Moreover, tuandsu are continuous with respect to u.Proof.
Proof. For any u∈H with u1,u2≡0, we denote(71)E1=u112,E2=u222,(72)A1=∫ν1u1p1,A2=∫ν2u2p2,(73)D1=∫μ1u12,D2=∫μ2u22∗,(74)D=∫βu12∗/2u22∗/2.
Then,D<0 and D1D2>D2. tu1,su2∈ℳ for some t,s>0 is equivalent to(75)t2E1=tp1A1+t2∗D1+t2∗/2s2∗/2D,(76)s2E2=sp2A2+s2∗D2+t2∗/2s2∗/2D,t,s>0.
By (75), we have(77)s2∗/2=t2−2∗/2E1−tp1A1−t2∗D1D:=GtD:=t2−2∗/2G˜tD,where s>0 implies that G˜t<0. Since G′˜t<0 and for t>0 small enough, G˜t>0, there exists a unique t0>0 such that G˜t0=0; thus, t>t0. Combining (76) and (77), it is left to show that(78)Ft:=GtD4/2∗−1t−2∗/2E2−GtD2p2/2∗−1t−2∗/2A2−GtDt−2∗/2D2−D,has a solution t>t0. Note that 2−2∗/2<p1−2∗/2<2∗/2 and 4/2∗−1<2p2/2∗−1<1; by direct calculation, we can check that(79)limt⟶+∞Ft=D1D2−D2D<0,limt⟶t0Ft0=−D>0.
This implies that (75) and (76) have a positive solution tu,su. The uniqueness of tu,su follows from Lemma 6.
For the continuity oftu,su with respect to u, we take a sequence un with un⟶u strongly in H. We may assume that tu,su=1,1 by replacing un by tunu1n,sunu2n if necessary. That is, u∈ℳ. Using similar denotations to (74) and rewriting tun,sun=tn,sn for convenience, we have(80)tn2E1n=tnp1A1n+tn2∗D1n+tn2∗/2sn2∗/2Dn,(81)sn2E2=snp2A2n+sn2∗D2n+tn2∗/2sn2∗/2Dn,(82)E1=A1+D1+D,E2=A2+D2+D.
Iftn⟶∞, then by (81), we have sn⟶∞. Therefore,(83)D2=limn⟶+∞Dn2=limn⟶+∞tn2−2∗/2E1n−tnp1−2∗/2A1n−tn2∗/2D1nsn2∗/2⋅sn2−2∗/2E2n−snp2−2∗/2A2n−sn2∗/2D2ntn2∗/2=limn⟶+∞tn2−2∗E1n−tnp1−2∗A1n−D1nsn2−2∗E2n−snp2−2∗A2n−D2n=limn⟶+∞D1nD2n=D1D2>D2,is a contradiction. Thus, tnandsn are uniformly bounded and we may assume that tn⟶t^,sn⟶s^ as n⟶+∞. By (80) and (81), we have t^u1,s^u2∈ℳ. Then, t^=s^=1, that is, tun,sun⟶tu,su.
Now, we show the following:Lemma 8.
Assume that−μ1μ2<β<0, then(84)ℬ<minB+1Nμ1−N−2/2SN/2,ℬ1+1Nμ2−N−2/2SN/2.Proof.
(i)
We first show that ℬ<B+1/Nμ1−N−2/2SN/2. In [27], the authors show that (11) has a positive least energy solution; we denote it as U1,U2 with the energy(85)IU1,U2=B<min1Nμ1−N−2/2SN/2+B2,B1+1Nμ2−N−2/2SN/2,andU1,U2∈C2Ω∩CΩ¯. For r>0 small, take xr∈Ω such that distxr,∂Ω=4r and(86)U1x+U2x≤Cr,∀x∈Bxr,3r.Letϕ∈C0∞B0,2 with 0≤ϕ≤1 and ϕ≡1 on B0,1. Set ϕrx:=1−ϕx−xr/r. Then, there exists a positive constant such that ∇ϕrx≤c/r. Define U1r,U2r:=ϕrU1,ϕrU2. Then, for j=1,2, we have(87)∫∇Ujr2≤∫∇Uj2+CrN,∫Ujrq≥∫Ujq−CrN+q,q=2,pj,2∗.By Lemma7, there exist tr,sr>0 such that trU1r,srU2r∈ℳ. That is,(88)tr2U1r12=trpν1∫U1rp1+tr2∗μ1∫U1r2∗+βtr2∗/2sr2∗/2∫U1r2∗/2U2r2∗/2,(89)sr2U2r12=srpν2∫U2rp2+sr2∗μ1∫U2r2∗+βtr2∗/2sr2∗/2∫U1r2∗/2U2r2∗/2.Then similar to the proof of (83), it can be shown that trandsr are uniformly bounded. Thus, up to a subsequence, there exist t0,s0>0 such that tr⟶t0andsr⟶s0. Let r⟶0 in (88) and (89); it implies that t0U1,s0U2∈ℳ. Then, t0=s0=1, and we may assume that 1/2≤trandsr≤2 for r>0 small. By (87), we have(90)ItrU1r,srU2r≤ItrU1,srU2+CrN≤IU1,U2+CrN=B+CrN.LetVjrj=1,2 be the solution of problem (8) when Ω=B0,r; then, trU1r−V1r⋅−xr,srU2r∈S and from [40], it holds that(91)IjVjr≤1Nμj−N−2/2SN/2−C1r2N−4/N−4,j=1,2.
Thus,(92)ℬ≤ItrU1r−V1r⋅−xr,srU2r=ItrU1r,srU2r+I1V1r⋅−xr≤B+CrN+1Nμ1−N−2/2SN/2−C1r2N−4/N−4<B+1Nμ1−N−2/2SN/2,forr>0small.(ii)
Next, we show thatℬ<ℬ1+1/Nμ2−N−2/2SN/2. Let wj be the sign changing solution of (8) with Ijwj=ℬj for j=1,2. Then, wj∈C2Ω∩CΩ¯ and wj+>0:=x∈Ω:wjx>0≠∅. For any R>0, we may take yR∈wj+>0 with distyR,∂wj+>0=4R. Then, Byr,3r⊂wj+>0 and wj+≤CR,∀x∈ByR,3R. Recall ϕ and ϕR in the above step i; we define U1R:=ϕRw1+. Since I1′w1w1+=0, we have(93)w1+12=ν1∫w1+p1+μ1∫w1+2∗.
Then, there existstR>0 such that(94)tRU1R+12=ν1∫tRU1R+p1+μ1∫tRU1R+2∗,and ItRU1R,0=maxt>0ItU1R,0. It is easy to see that tRU1R−w1−,V2R⋅−yR∈S, where V2R is the positive solution of problem (8) in the ball B0,R for j=2. Then by (91), we have(95)ℬ≤ItRU1R−w1−,V2R⋅−yR=ItRU1R,0+I1w1−+I2V2R⋅−yR≤I1w1++CRN+I1w1−+1Nμ2−N−2/2SN/2−C1R2N−4/N−4<ℬ1+1Nμ2−N−2/2SN/2,forR>0small.
Foru1,u2∈S, we define u1ρ:=1−ρu1+−ρu1−,ρ∈0,1. Let uρ:=u1ρ,u2 and recall that tuρ,suρ is defined in Lemma 7; then, we can define a continuous map ηu:0,1⟶ℳ by(96)ηuρ:=tuρu1ρ,suρu2.
Then, we have the following conclusion and recall thatε0 is defined in (60).Lemma 9.
For anyu∈S with Iu<ℬ+ε0, there exists a small ρo∈0,1/4 such that(97)I′ηuρ0tuρ0u1ρ0+,0<0,(98)I′ηu1−ρ0tu1−ρ0u11−ρ0+,0>0,(99)Iηuρ0,Iηu1−ρ0<ℬ−ε0.Proof.
It is easy to see that there exist constantsC1andC2>0 such that(100)C1≤tuρ,suρ≤C2,∀ρ∈0,1.
SinceI′ηuρtuρu1ρ,0=0 and u1ρ−=ρu1−, we have(101)I′ηuρtuρu1ρ+,0=−I′ηuρtuρρu1−,0≤−tuρ2ρ2u1−12+tuρp1ρp1ν1∫u1−p1+tuρ2∗ρ2∗μ1∫u1−2∗≤C12ρ2u1−2+C2p1ρp1ν1∫u1−p1+C22∗ρ2∗μ∫u1−2∗<0,forρ>0small.
Similarly, note thatu11−ρ+=ρu1+; we have(102)I′ηu1−ρtu1−ρu11−ρ+,0≥tu1−ρ2ρ2u1+12−tu1−ρp1ρp1ν1∫u1+p1+tu1−ρ2∗ρ2∗μ1∫u1+2∗≥C12ρ2u1−12−C2p1ρp1ν1∫u1−p1−C22∗ρ2∗μ1∫u1−2∗>0,forρ>0small.
Sinceu∈S, we have u1+,u2,−u1−,u2∈ℳ. Then,(103)Iηu0=Iu1+,u2=Iu1,u2−12u1−12−ν1p1∫u1−p1−μ12∗∫u1−2∗+2β2∗∫u1−2∗/2u22∗/2=Iu1,u2−12−1p1u1−12−1p1−12∗∫μ1u1−2∗+2β2∗−1p1∫u1−2∗/2u22∗/2≤Iu1,u2−12−1p1u1−12<ℬ−2ε0.
Similarly, we have(104)Iηu1=I−u1−,u2=Iu1,u2−12u1+2−ν1p1∫u1+p1−μ12∗∫u1+2∗+2β2∗∫u1+2∗/2u22∗/2≤Iu1,u2−12−1p1u1+2<ℬ−2ε0.
SinceI′andI are continuous with respect to ρ, (101), (102), (103), and (104) imply (97), (98), and (99), respectively.□
Now, we show the existence of aPSℬ sequence of I. Precisely, we have the following important lemma.Lemma 10.
There exist a sequenceun⊂ℳ and a constant C>0 such that(105)Iun⟶ℬ,I′un⟶0,asn⟶+∞;u1n±1,u2n2≥C,∀n≥1.Proof.
By Lemma5 and the Ekeland’s variational principle, there exists a minimizing sequence u˜n=u˜1n,u˜2n⊂S such that(106)Iu˜n<ℬ+ε0,I′u˜n⟶0.
Letηnρ=ηn1ρ,ηn2ρ:=ηu˜nρ; by Lemma 9, there exists ρn∈0,1/4 such that(107)I′ηnρnηn1ρn+,0<0,(108)I′ηn1−ρnηn11−ρn+,0>0,(109)Iηnρn,Iηn1−ρn<ℬ−ε0.
We claim that there exists a sequenceun⊂ℳ such that(110)distun,ηnρn,1−ρn⟶0,(111)I′ℳun⟶0,Iun⟶ℬ.
If not, then we assume that there exists smallε1>0 such that(112)∀u∈Iℳ−1ℬ−ε1,ℬ+ε1∩Sε1:Iℳ′u≥ε1,where(113)Iℳ−1ℬ−ε1,ℬ+ε1:=u∈ℳ:Iu∈ℬ−ε1,ℬ+ε1,S:=ηnρn,1−ρn,and Sε1 is the ε1 neighborhood of S. Then, by the well-known deformation lemma, there exist a continuous map ξ:0,1×ℳ⟶ℳ and an ε2∈0,minε0,ε1/4 such that(1)
ξ0,u=u,∀u∈ℳ(2)
ξt,u=u,∀u∉I−1ℬ−2ε2,ℬ+2ε2,t∈0,1(3)
ξ1,Iℳℬ+ε2∩S⊂Iℳℬ−ε2
By Lemma6, for n large enough, we have that(114)supρ∈ρn,1−ρnIηnρ≤Iu˜n<ℬ+ε2.
Then, by the properties (2) and (3) ofξ and (109), we have(115)supρ∈ρn,1−ρnIξ1,ηnρ≤ℬ−ε2,(116)ξ1,ηnρn=ηnρn,ξ1,ηn1−ρn=ηn1−ρn.
Similarly as in (20), we define(117)Eu1,u2:=0,u1=0,ν1∫u1p1+μ1∫u12∗+β∫u12∗/2u22∗/2u112,u1≠0.
We rewriteηρ:=η1ρ,η2ρ:=ξ1,ηnρ for convenience. ηρ∈ℳ implies that Eη1ρ,η2ρ=1. By (107) and (108), we have that(118)Eη1ρn+,η2ρn>1,Eη11−ρn+,η21−ρn<1.
Then, by the continuity ofE, we can define(119)ρ0:=maxρ∈ρn,1−ρn:Eη1ρ+,η2ρ=1.
By definition ofρ0, there exists δ0∈0,1−ρn−ρ0 such that Eη1ρ+,η2ρ<1 for any ρ∈ρ0,ρ0+δ0. Then, Eη1ρ−,η2ρ>1 for any ρ∈ρ0,ρ0+δ0. Thus, Eη1ρ0−,η2ρ0≥1. Hence, Eη1ρ0±,η2ρ0=1. That is, ηρ0∈S. But by (115), we have Iηρ0<ℬ, a contradiction with the definition of ℬ. Therefore, the claim (111) becomes true.
Now, we can chooseρ^n∈ρn,1−ρn such that un−ηnρ^n⟶0,n⟶∞. Since Iun⟶ℬ, we have Iηnρ^n⟶ℬ. Recall that u˜n∈S and Iu˜n⟶ℬ; then by Lemma 6, we have ηnρ^n−u˜n⟶0. By (59), we have that there exists a constant C>0 independent of n such that u1n±1,u2n2≥C>0.
Set(120)J1u1,u2:=I′u1,u2u1,0,J2u1,u2:=I′u1,u20,u2.
By (111), there exist two Lagrange multipliers L1n,L2n∈ℝ such that(121)Ju1n,u2n:=I′u1n,u2n+L1nJ1′u1n,u2n+L2nJ2′u1n,u2n⟶0.
Then,Ju1n,u2n,u1n,0=o1 and Ju1n,u2n,0,u2n=o1; i.e.,(122)L1np1−2∫ν1u1np1+2∗−2∫μ1u1n2∗+2−2∗2∫βu1n2∗/2u2n2∗/2=L2n2∗2∫βu1n2∗/2u2n2∗/2+o1,(123)L2np2−2∫ν2u2np2+2∗−2∫μ2u2n2∗+2−2∗2∫βu1n2∗/2u2n2∗/2=L1n2∗2∫βu1n2∗/2u2n2∗/2+o1.
Similarly as in (74), we denote(124)A1n=∫ν1u1np1,A2n=∫ν2u2np2,D1n=∫μ1u1n2∗,D2n=∫μ2u2n2∗,Dn=∫βu1n2∗/2u2n2∗/2.
Note thatu1n,u2n⊂ℳ, we have(125)u1n12=A1n+D1n−Dn,u2n22=A2n+D2n−D.
Consequently, by (59), we have(126)A1n+D1n>Dn,A2n+D2n>D.
Set(127)S1n:=p1−2p2−2A1nA2n+p1−22∗−2A1nD2n+p1−22−2∗2A1nDn+p2−22∗−2A2nD1n+p2−22−2∗2A2nDn,S2n:=2∗−22D1nD2n+2∗−22−2∗2D1nDn+2∗−22−2∗2D2nDn−22∗−2Dn2.
Then, by (122) and (123), we have(128)o1=L1nL2nS1n+S2n:=L1nL2nSn.
Note that(129)Dn≤12D1n+D2n,Dn2≤D1nD2n.We have(130)S2n≥2∗−22D1nD2n−Dn2≥0.
Thus,Sn>C>0 for some constant C>0. By (128), we have L1nL2n=o1. Without loss of generality, we may assume that L1n=o1; then by (123) and (59), we have L2n=o1. Then, (121) implies that I′u1n,u2n⟶0. Thus, we complete the proof.
We also need an important lemma which is proved in [25].Lemma 11 (see [25]).
Assume thatu1n,u2n⇀u1,u2 weakly in H as n⟶∞; then, passing to a subsequence, it holds that(131)limn⟶∞∫u1n2∗/2u2n2∗/2−u12∗/2u22∗/2−u1n−u12∗/2u2n−u22∗/2=0.Proof of Theorem 3.
Proof of Theorem 3. Letun⊂ℳ be the sequence obtained in Lemma 10. Then, up to a subsequence, there exists u∈H such that un⇀u weakly in H. Then, I′u=0 and(132)u1n12=∫ν1u1np1+μ1u1n2∗+βu1n2∗/2u2n2∗/2,(133)u2n22=∫ν2u2np2+μ2u1n2∗+βu1n2∗/2u2n2∗/2,(134)u1n±12=∫ν1u1n±p1+μ1u1n±2∗+βu1n±2∗/2u2n2∗/2+o1.
Setwn=w1n,w2n:=un−u. By Lemma 11, (133), and (134), we have(135)∫∇w1n±2=∫μ1w1n±2∗+βw1n±2∗/2w2n2∗/2+o1,(136)∫∇w2n2=∫μ2w2n2∗+βw1n2∗/2w2n2∗/2+o1.
Then,(137)ℬ=limn⟶∞Iun=limn⟶∞1Nun2+∫∑j12∗−1pjνjujnpj=1Nu2+∫∑j12∗−1pjνjujpj+limn⟶∞1N∫∇w1n+2+∇w1n−2+∇w2n2.Case 1.
1. Assume thatu1,u2≡0. Then, ujn⟶0 in L2Ω and LpjΩ for j=1,2. From (135) and (136), we have(138)∫∇w1n±2≤μ1S−2∗/2∫∇w1n±22∗/2,∫∇w2n2≤μ2S−2∗/2∫∇w2n22∗/2.Then,(139)∫∇w1n±2≥μ1−N−2/2SN/2,(140)∫∇w2n2≥μ2−N−2/2SN/2.
Thus, by (137), we have(141)ℬ≥2Nμ1−N−2/2SN/2+1Nμ2−N−2/2SN/2,a contradiction with Lemma 8, (85), and (9). Therefore, Case 1 is impossible.Case 2.
Assume thatu1≡0andu2≡0. Then, (139) still holds and I′0,u2=0 which implies that u2 is a nontrivial solution of (10) with j=2. Thus, I0,u2≥B2. By (137), we have(142)ℬ≥B2+2Nμ1−N−2/2SN/2,a contradiction with Lemma 8 and (85). Therefore, Case 2 is impossible.Case 3.
Assume thatu1≡0andu2≡0. Then, (140) still holds and I′u1,0=0 which implies that u1 is a nontrivial solution of (10) with j=1. Thus, Iu1,0≥B1.(i)
Ifu1n⟶u1 strongly in H01Ω, then u1 is a sign changing solution of (10) with j=1. So Iu1,0≥ℬ1. By (137), we have(143)ℬ≥ℬ1+1Nμ2−N−2/2SN/2,a contradiction with Lemma 8.(ii)
Ifu1n↛u1 strongly in H01Ω, then by (132) and Brezis–Lieb lemma, we can also show that(144)∫∇w1n2≥μ1−N−2/2SN/2.
By (137), we have(145)ℬ≥B1+1Nμ1−N−2/2SN/2+1Nμ2−N−2/2SN/2,a contradiction with Lemma 8 and (85).
Therefore, Case 3 is impossible.
Now, we have shown thatu1,u2≡0 and then u1,u2 is a nontrivial solution of (11). If u1n↛u1 strongly in H01Ω, then (144) holds and(146)ℬ≥B+1Nμ1−N−2/2SN/2,a contradiction with Lemma 8. Thus, u1n⟶u1 strongly in H01Ω and then u1 is sign changing, Iu≥ℬ. On the other hand, by (137), we have(147)Iu≤limn⟶∞Iun=ℬ.
Then, we haveIu=ℬ and un⟶u strongly in H. Then u1,u2∈S and Iu1,u2=ℬ. Similarly to the proof in Lemma 10, we have I′u1,u2=0. By the maximum principle, we have u2>0 in Ω. Consequently, u1,u2 is a least energy sign changing solution of system (11) with u1 sign changing and u2 positive. Thus, we complete the proof.


## 5. Conclusion

In this paper, we study a coupled nonlinear Schrödinger system with critical exponents which arise in many physical problems. By the modified Nehari manifold method and some mathematical skills for estimations of the energy, we show the existence of the least energy sign changing solution for the general system. Besides, we construct multiple solutions of the system for the symmetrical case.



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