YU Chun, WAN You-yan

(Department of Mathematics, Jianghan University, Wuhan 430056, China)

Abstract: We study the nonlinear Chern-Simons-Schrdinger system with superlinear nonlinearities.By the concentration compactness principle combined with the Nehari manifold, we prove the existence of positive ground state to this problem.Moreover, we obtain that the ground state has exponential decay at infinity.

Keywords: the ground state; the Chern-Simons-Schrdinger system; the variational method; the Nehari manifold; the concentration compactness principle

1 Introduction and Main Result

We are interested in the existence of ground stateu∈H1(R2) to the following Chern-Simons-Schrdinger system (CSS system)

The CSS system comes from the study of the standing wave of Chern-Simons-Schrdinger system which describes the dynamics of large number of particles in a electromagnetic field and the physical background of the high-temperature superconductor, fractional quantum Hall effect and Aharovnov-Bohm scattering.For the more physical background of CSS system, we refer to the references that we will mention below and references therein.Recently,many scholars paid much attention to the CSS system proposed in [1], [2], and [3].Berge,De Bouard,Saut[4]studied the blowing up time-dependent solution and Liu,Smith,Tataru[5]considered the local wellposedness.Byeon, Huh, Seok [6, 7], Huh [8, 9]investigated the existence and non-existence of standing wave solutions by variational methods.Authors in[10–13]obtained a series of existence results of solitary waves.Wan and Tan proved the existence,non-existence,and multiplicity of standing waves to the nonlinear CSS system with an external potentialV(x) without the Ambrosetti-Rabinowitz condition in [14], the existence of nontrivial solutions to Chern-Simons-Schrdinger systems (1.1) withV(x) be a constant and the argument of global compactness withp>4,V(x)∈C(R2)and 06,V(x) satisfying the same condition as [15]in [16].

Inspired by [6, 14–15, 17–18], the purpose of the present paper is to study the existence of ground state for system (1.1) wherep> 6.We consider this problem on the standard Nehari manifold,which is different from the idea in[15]where they use a constraint manifold of Pohozaev-Nehari type and[16]whereV(x)cannot be constants.The componentsAj,j=0,1,2 of the gauge field yieldp>6.The main characteristic of system (1.1) is that the nonlocal termAj,j=0,1,2 depend onuand there is a lack of compactness in R2.By using the variational method joined with the Nehari manifold and concentration compactness principle[17], we can obtain the following result.

Theorem 1.1Ifp> 6, then (1.1) has a positive ground state which has exponential decay at infinity.

The paper is organized as follows.In Section 2, we introduce the framework as well as show some important propositions ofAj,j=0,1,2 and some technical lemmas.In Section 3, we prove the compactness result and Theorem 1.1.

2 Mathematical Framework

LetH1(R2) denote the usual Sobolev space with

Define the functional

From (1.1), integrating by parts, we have

We obtain the derivative ofJinH1(R2) as follow

Notice that

which give

As [15], we have thatJis well defined inH1(R2),J∈C1(H1(R2)), and the weak solution of (1.1) is the critical point of the functionalJby the following two properties.

Proposition 2.1(see [15]) Let 1

(i) Then there is a constantCdepending only onsandqsuch that

where the integral operatorTis given by

(ii) Ifu∈H1(R2), then we have that forj=1,2,

and

Proposition 2.2(see[15]) Suppose thatunconverges toua.e.in R2andunconverges weakly touinH1(R2).LetAj,n:=Aj(un(x)),j=0,1,2.Then

(i)Aj,nconverges toAj(u(x)) a.e.in R2.

Next, we define the Nehari manifold related to the functionals above and discuss the property of the least energy of the critical points.SetX:=H1(R2) and

Lemma 2.3Ifp≥6, then Σ is a smooth manifold.

ProofSet

Then

Byu∈Σ, we have

Therefore, ifp≥6 we have

By the implicit function theorem, Σ is a smooth manifold.

Now we can define critical values of the functional on the Nehari manifold.Let

where Γ :={γ∈C([0,1],X) :γ(0)=0,J(γ(1)) < 0}.These critical values have the following property which is similar to Lemma 2.4 in [16].For the reader’s convenience, we show its proof.

Lemma 2.4

ProofFirst, we showc=c??.Indeed, this will follow if we can prove that for anyu∈X{0}, the rayRt={tu:t≥0} intersects the solution manifold Σ once and only once atθu(θ>0), whereJ(θu),θ≥0, achieves its maximum.

Let

where

We claim that there exists a uniquet0∈(0, +∞)such thath(t0)=0.In fact,by computing,we have that

Next, we showc?=c??.It is easy to see thatc??≥c?.Let us provec??≤c?.Foru∈X{0} fixed, letbe the unique point such thatThen, we can write

with

Letγ∈Γ be a path.If for allthen the inequality is obtained.If there existsγ∈Γ such thatfor allt∈[0,1], then we have

Ifp>6, then

which contradictsJ(γ(1))<0.Hencec?=c??.

3 The Proof of Theorem 1.1

First, we should obtain the following compactness result to prove Theorem 1.1.

Lemma 3.1Let {n} be a minimizing sequence ofc.Ifp>6, then

(i) there exists {un}?Σ such thatn→∞;

(ii) there exists {ξn}?R2such that {vn} is precompact, wherevn(·):=un(·+ξn).

Proof(i) It is a direct consequence of the Ekeland’s variational principle (see [18]).

(ii) We are going to use the concentration compactness principle given in [17].SinceJ(un)→casn→∞,un∈Σ, andp>6 fornlarge, we have

whereA1,n:=A1(un) andA2,n:=A2(un).Then, {un} is bounded inX.For anyn∈N, we consider the following measure

From the concentration compactness lemma in [19], there is a subsequence of {μn}, which we will always denote by {μn}, satisfying one of the three following possibilities.

VanishingAssume that there exists a subsequence of {μn}, such that for allρ>0,

DichotomySuppose there exist a constantwith 0 <

CompactnessThere is a sequence {ξn}?RNsuch that for anyδ>0 there exists a radiusρ>0 such that

Step 1We show vanishing is impossible.

If{μn}is vanishing,then{un}is also vanishing.Namely,there is a subsequence of{un}such that for allρ>0,

By the Lion’s lemma [17],un→0, inLs(R2),s>2.Because

Consequently

which is impossible.

Step 2We prove dichotomy is impossible.

Define

We know thatu2,nconverges to 0 a.e.in R2, andAj(u2,n)→0 a.e.in R2.

and

where

Then

namely,

Then we obtain

Thus we have

We note that

then

By (3.3), (3.4), (3.5), and (3.6), we have

Similarly, we get

Thus, by (3.7), (3.9), and (3.10), we have

Then

Hence

Let

By (3.7)–(3.10), andun∈Σ, we obtain

By Lemma 2.4, for anyn≥1,?θn>0 such thatθnu1,n∈Σ, and then

Case 1Up to a subsequence,g(u1,n)≤0.By (3.13) andp>6, we obtain

which yields thatθn≤1.Hence, for alln≥1,

which is impossible.

Case 2Up to a subsequence,g(u2,n)≤0.We can have the argument as in Case 1.

Case 3Up to a subsequence,g(u1,n) > 0 andg(u2,n) > 0.By (3.12), we have thatg(u1,n)=on(1) andg(u2,n)=on(1).Ifθn≤1+on(1), we can repeat the arguments of Case 1.Assume thatwe get

Hence,u1,n→0 asn→∞inX.Then, we obtain a contradiction with (3.11).

Step 3We show the strong convergence.By the proof above,we have that there exists a subsequence of {μn} such that it is compact, that is, there is a sequence {ξn}?RNsuch that for anyδ>0, there exists a radiusρ>0 such that

We define the new sequence of functionsvn(·)=un(·?ξn)∈X.We have thatAj(vn(·))=Aj(un(·?ξn)),j=1,2 and thusvn∈Σ.Furthermore, from (3.1), we have that for anyδ>0, there is a radiusρ>0 such that

By {un} is bounded inX, {vn} is also bounded inX.Then, there exist a subsequence of{vn} andu∈Xsuch that

asn→∞.According to (3.14), (3.15), and (3.16), we obtain that, takens∈(2, +∞), for anyδ>0, there existsρ>0 such that for anyn≥1 sufficiently large

whereC>0 is the constant of the embeddingwe have

Thus

We have

From (3.15), (3.17), (3.18), and Proposition 2.2, we havewhich yieldsvnstrongly converges touinX.

The Proof of Theorem 1.1From the proof of (ii) in Lemma 3.1, we getu∈Σ andJ(u)=c.Thus,uis a ground state solution of (1.1).Next, we show thatu∈H1(R2) does not change sign.Letu+=max{u,0} andu?=max{?u,0}, thenu=u+?u?.We have that

We obtain

which impliesu?≡0 oru+≡0.We suppose thatu≥0.Combining the Sobolev theorem with the Moser iteration to weak solutionu∈H1(R2)of(1.1),we know thatuis bounded inL∞(R2).Hence, for allq∈[2,∞), there existsC1such thatFurthermore,we obtain thatu∈Cα(R2) for someα∈(0,1).Then, we haveby the standard bootstrap argument.Consequently, we obtainu∈C1,α(R2) for someα∈(0,1) by the classical elliptic estimate.From the maximum principle,we achieve thatu≥0.Last,we show the ground stateuhas exponential decay at infinity.Let Ψ(x)=Me?θ(|x|?L), whereM=max{|u(x)|: |x|=L} and 0<θ<1.Then we haveDefine

withb1>0.From (2.3), choosingη=φR, we obtain

LetR>0 such that |u|p?2≤1 ?θ2for |x|>R.Then

Consequently,φR≡0.It implies exponential decay at infinity.