# On the Existence of Ground State Solutions of the Periodic Discrete Coupled Nonlinear Schrödinger Lattice

Authors: Meihua Huang; Zhan Zhou
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We study the existence of ground state solutions of the periodic discrete coupled nonlinear Schrödinger lattice by using the Nehari manifold approach combined with periodic approximations. We show that both of the components of the ground state solutions are not zero.


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## Body

## 1. Introduction

In this paper, we consider the coupled discrete Schrödinger system(1)idundt=-(𝒜u)n+b1nun-a1|un|2un-a3|vn|2un,idvndt=-(𝒜v)n+b2nvn-a2|vn|2vn-a3|un|2vn,
where ai is a positive constant, {bjn} is a real valued T-periodic sequence, i=1,2,3, and j=1,2. 𝒜 is the discrete Laplacian operator defined as (𝒜u)n=un+1+un-1-2un.System (1) could be viewed as the discretization of the two-component system of time-dependent nonlinear Gross-Pitaevskii system (see [1] for more detail) as
(2)iℏ∂tu=-ℏ22mΔu+b1(x)u-a1|u|2u-a3|v|2u,iℏ∂tv=-ℏ22mΔv+b2(x)v-a2|v|2v-a3|u|2v.It is well known that coupled nonlinear Schrödinger equations arise quite naturally in nonlinear optics [2] and Bose-Einstein condensates. Bose-Einstein condensation for a mixture of different interaction atomic species with the same mass was realized in 1997 (see [3]), which stimulated various analytical and numerical results on the ground state solutions of system (2). The discrete nonlinear Schrödinger equations (DNLS) have a crucial role in the modeling of a great variety of phenomena, ranging from solid-state and condensed-matter physics to biology. During the last years, there has been a growing interest in approaches to the existence problem for ground states. We refer to the continuation methods in [4, 5], which have been proved to be powerful for both theoretical considerations and numerical computations (see [6]), to [7], which exploits spatial dynamics and centre manifold reduction, to the variational methods in [8–15], which rely on critical point techniques (linking theorems, the Nehari manifold), and to the Krasnoselskii fixed point theorem together with a suitable compactness criterion [16].The aim of this paper is to study the discrete solitons of (1), that is, solutions of the form
(3)un=exp(-iω1t)ϕn,vn=exp(-iω2t)ψn,n∈ℤ,(4)lim|n|→∞un=0,lim|n|→∞vn=0,
where the amplitudes ϕn and ψn are supposed to be real. Inserting the ansatz of the discrete solitons (3) into (1), we obtain the following equivalent algebraic equations: (5)-(𝒜ϕ)n-ω1ϕn+b1nϕn-a1|ϕn|2ϕn-a3|ψn|2ϕn=0,-(𝒜ψ)n-ω2ψn+b2nψn-a2|ψn|2ψn-a3|ϕn|2ψn=0,
and (4) becomes
(6)lim|n|→∞ϕn=0,lim|n|→∞ψn=0.In fact, we consider the following more general equations:(7)(L1ϕ)n-ω1ϕn-a1|ϕn|2ϕn-a3|ψn|2ϕn=0,(L2ψ)n-ω2ψn-a2|ψn|2ψn-a3|ϕn|2ψn=0,
where L1 and L2 are the second-order difference operators defined by
(8)L1ϕn=αnϕn+1+αn-1ϕn-1+βnϕn,L2ψn=γnψn+1+γn-1ψn-1+δnψn,
where {αn}, {βn}, {γn}, and {δn} are real valued T-periodic sequences, and a1, a2, and a3 are positive numbers. Obviously, (5) is a special case of (7) with αn≡-1, βn≡2+b1n, γn≡-1, and δn≡2+b2n.Since, fori=1,2, the operator Li is a bounded and self-adjoint operator in l2, its spectrum σ(Li) has a band structure; that is, σ(Li) is a union of a finite number of closed intervals [17]. The complement ℝ-σ(Li) consists of a finite number of open intervals called spectral gaps and two of them are semi-infinite which are denoted by (-∞,θ1i) and (θ2i,∞), respectively.In this paper, we consider two types of solutions to (7) as follows: (i) 2kT-periodic, that is, ϕn+2kT=ϕn, ψn+2kT=ψn, and (ii) discrete solitons. Actually, in case (ii), we look for solutions in the space l2×l2=l2(ℤ)×l2(ℤ); then (6) holds naturally. System (7) has a trivial solution ϕn≡0, ψn≡0. We are looking for nontrivial solutions.The main idea in this paper is as follows. First, we consider (7) in a finite 2kT-periodic sequence space, and ωi is not a spectrum of the corresponding operator Li, i=1,2. By using the Nehari manifold approach, we obtain the existence of 2kT-periodic solutions. Then we show that these solutions have upper and lower bounds. Finally, by an approximation technique, we prove that the limit of these solutions exists and is the solution of (7) in l2×l2. Compared with the existence of ground state solutions of the DNLS, the difficulty is that we need to show that both of the components of the ground state solutions are not zero.The remaining of this paper is organized as follows. First, in Section2, we establish the variational framework associated with (7) and introduce the Nehari manifolds. Then, in Section 3, we present a sufficient condition on the existence of 2kT-periodic solutions and nontrivial solutions in l2×l2 of (7).


## 2. Preliminaries and the Nehari Manifold

In this section, we first establish the variational framework associated with (7).LetS be the set of the following form:
(9)S={ϕ={ϕn}:ϕn∈ℝ,n∈ℤ}.For any fixed positive integerk, we define the subspace Xk of S as
(10)Xk={ϕ∈S:ϕn+2kT=ϕn,n∈ℤ}.
Obviously, Xk is isomorphic to R2kT and hence Xk can be equipped with the inner product (·,·)k and norm ∥·∥k as
(11)(ϕ,ψ)k=∑n=-kTkT-1‍ϕnψn,ϕ,ψ∈Xk,(12)∥ϕ∥k=(∑n=-kTkT-1‍ϕn2)1/2,ϕ∈Xk,
respectively. Sometimes, we will consider lp norm on Xk as
(13)∥ϕ∥kp=(∑n=-kTkT-1‍|ϕn|p)1/p,1≤p<∞.
We also define a norm ∥·∥k∞ on Xk by
(14)∥ϕ∥k∞=max{|ϕn|:-kT≤n≤kT-1},ϕ∈Xk.
The symbol X stands for the space X=l2 with the norm
(15)∥ϕ∥=(∑n∈ℤ‍ϕn2)1/2,ϕ∈X,
and the inner product
(16)(ϕ,ψ)=∑n∈ℤ‍ϕnψn,ϕ,ψ∈X.
We also consider lp norm on X as
(17)∥ϕ∥p=(∑n∈ℤ‍|ϕn|p)1/p,1≤p<∞,∥ϕ∥∞=max{|ϕn|:n∈ℤ}.
We mention that
(18)∥ϕ∥kp≤∥ϕ∥kq,∥ϕ∥p≤∥ϕ∥q,
where 1≤q≤p≤∞.Consider the functionalsJk on Xk×Xk and J on X×X defined by
(19)Jk(ϕ,ψ)=12((L1-ω1)ϕ,ϕ)k+12((L2-ω2)ψ,ψ)k-14∑n=-kTkT-1‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2),(20)J(ϕ,ψ)=12((L1-ω1)ϕ,ϕ)+12((L2-ω2)ψ,ψ)-14∑n∈ℤ‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2),
respectively. Then Jk(ϕ,ψ)∈C1(Xk×Xk,ℝ) and its derivative is given by
(21)(Jk′(ϕ,ψ),(ϕ,ψ))=((L1-ω1)ϕ,ϕ)k+((L2-ω2)ψ,ψ)k-∑n=-kTkT-1‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2),
and J(ϕ,ψ)∈C1(X×X,ℝ) and its derivative is given by
(22)(J′(ϕ,ψ),(ϕ,ψ))=((L1-ω1)ϕ,ϕ)+((L2-ω2)ψ,ψ)-∑n∈ℤ‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2).Letζi be the distance from ωi to the spectrum σ(Li); that is,
(23)ζi=θ1i-ωi,i=1,2.
Let ζ=min{ζ1,ζ2}. Then, we have
(24)((Li-ωi)ϕ,ϕ)k≥ζ∥ϕ∥k2,ϕ∈Xk,i=1,2,(25)((Li-ωi)ϕ,ϕ)≥ζ∥ϕ∥2,ϕ∈X,i=1,2.
Furthermore, we let
(26)λ1*=minn∈ℤ{|βn-ω1|-|αn-1|-|αn|},λ1*=maxn∈ℤ{|βn-ω1|+|αn-1|+|αn|},λ2*=minn∈ℤ{|δn-ω2|-|γn-1|-|γn|},λ2*=maxn∈ℤ{|δn-ω2|+|γn-1|+|γn|}.
Then, obviously, λ1*>0 and λ2*>0.Denote(27)λ=maxn∈ℤ{|αn-γn|+|αn-1-γn-1|+|βn-δn|}.
Then
(28)∥(L1-L2)ϕ∥≤λ∥ϕ∥,ϕ∈X.Next, we study the main properties of the Nehari manifolds with the functionalsJk and J.Let(29)Ik(ϕ,ψ)=(Jk′(ϕ,ψ),(ϕ,ψ)),(ϕ,ψ)∈Xk×Xk,I(ϕ,ψ)=(J′(ϕ,ψ),(ϕ,ψ)),(ϕ,ψ)∈X×X.
Then Ik and I are C1 functionals and their derivatives are given by
(30)(Ik′(ϕ,ψ),(ϕ,ψ))=2((L1-ω1)ϕ,ϕ)k+2((L2-ω2)ψ,ψ)k-4∑n=-kTkT-1‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2),(31)(I′(ϕ,ψ),(ϕ,ψ))=2((L1-ω1)ϕ,ϕ)+2((L2-ω2)ψ,ψ)-4∑n∈ℤ‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2),
respectively. The Nehari manifolds are defined as follows:
(32)Nk={(ϕ,ψ)∈Xk×Xk:Ik(ϕ,ψ)=0,(ϕ,ψ)≠0},N={(ϕ,ψ)∈X×X:I(ϕ,ψ)=0,(ϕ,ψ)≠0}.
Note that N contains all critical points of J in X×X.To prove the main results, we need some lemmas on the Nehari manifolds.Lemma 1.
Assume thatω1<θ11 and ω2<θ12 hold. Then the sets Nk and N are nonempty closed C1 submanifolds in Xk×Xk and X×X, respectively. The derivatives Ik′ and I′ are nonzero on the corresponding Nehari manifolds. Moreover, there exists η>0 such that ∥ϕ∥k2+∥ψ∥k2≥η, (ϕ,ψ)∈Nk and ∥ϕ∥2+∥ψ∥2≥η, (ϕ,ψ)∈N.Proof.
The proofs for both cases are similar. We only provide the proof for the case ofNk as an illustration.
First, we show thatNk≠∅.
Let(ϕ,ψ)∈Xk×Xk-{(0,0)}. Then by (21)
(33)Ik(tϕ,tψ)=t2(∑n=-kTkT-1‍((L1-ω1)ϕ,ϕ)k+((L2-ω2)ψ,ψ)k-t2∑n=-kTkT-1‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2)).
Noticing that ((L1-ω1)ϕ,ϕ)k>0 and ((L2-ω2)ψ,ψ)k>0, by (33), we see that Ik(tϕ,tψ)>0 for t>0 small enough and Ik(tϕ,tψ)<0 for t>0 large enough. As a consequence, there exists t0>0 such that Ik(t0ϕ,t0ψ)=0; that is, (t0ϕ,t0ψ)∈Nk.
Let(ϕ,ψ)∈Nk. By (21), (30), and the definition of Nk, we have
(34)(Ik′(ϕ,ψ),(ϕ,ψ))=(Ik′(ϕ,ψ),(ϕ,ψ))-2Ik(ϕ,ψ)=-2∑n=-kTkT-1‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2)<0.
Hence, Ik′≠0, and the implicit function theorem implies that Nk is a C1 submonifold in Xk×Xk.
Now let us prove the last statement of the lemma. Let(ϕ,ψ)∈Nk. By the assumption of Lemma 1 and the definition of Nk, we have
(35)((L1-ω1)ϕ,ϕ)k+((L2-ω2)ψ,ψ)k=∑n=-kTkT-1‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2)≤a*(∥ϕ∥k42+∥ψ∥k42)2≤a*(∥ϕ∥k2+∥ψ∥k2)2,
where a*=max{a1,a2,a3}. Let η=ζ/a*>0. Then, by (24) and (35), it is easy to see that
(36)ζ(∥ϕ∥k2+∥ψ∥k2)≤a*(∥ϕ∥k2+∥ψ∥k2)2,
which implies that
(37)∥ϕ∥k2+∥ψ∥k2≥ζa*=η.
Closedness ofNk is obvious. The proof is completed.Lemma 2.
Assume thatω1<θ11 and ω2<θ12 hold. Then there exists ρ>0 such that Jk(ϕ,ψ)≥ρ for all (ϕ,ψ)∈Nk.Proof.
For any(ϕ,ψ)∈Nk, we have
(38)Jk(ϕ,ψ)=Jk(ϕ,ψ)-14Ik(ϕ,ψ)=14(((L1-ω1)ϕ,ϕ)k+((L2-ω2)ψ,ψ)k)≥ζ4(∥ϕ∥k2+∥ψ∥k2).
By Lemma 1, we know that ∥ϕ∥k2+∥ψ∥k2≥η. Hence, let ρ=(ζη)/4>0. Then (38) implies that Jk(ϕ,ψ)≥ρ. The proof is completed.Lemma 3.
For(ϕ,ψ)∈Nk, the function Jk(tϕ,tψ), t>0, has a unique critical point at t=1, which is, actually, a global maximum. The same statement holds for N and J.Proof.
LetF(t)=Jk(tϕ,tψ), t>0, (ϕ,ψ)∈Nk. Computing the derivative of F, we have
(39)F′(t)=t(1-t2)∑n=-kTkT-1‍(a1ϕn4+a2ψn4+2a3ϕn2ψn2).
This shows that t=1 is a unique maximum point. The proof is completed.Lemma 4.
Let(ϕ-,ψ-) be a minimizer of the functional Jk(ϕ,ψ) constrained on the Nehari manifold Nk; that is,
(40)Jk(ϕ-,ψ-)=mk=inf(ϕ,ψ)∈NJk(ϕ,ψ),
and then (ϕ-,ψ-) is a nontrivial 2kT-periodic solution to (7), which is called a nontrivial periodic ground state solution to (7).Proof.
According to Lagrange multiplier method,(ϕ-,ψ-) is the critical point of the functional Jk(ϕ,ψ)+ΛIk(ϕ,ψ). Thus Ik(ϕ-,ψ-)=0, and for arbitrary (ϕ,ψ)∈Xk×Xk,
(41)(Jk′(ϕ-,ψ-)+ΛIk′(ϕ-,ψ-),(ϕ,ψ))=0.
After taking (ϕ,ψ)=(ϕ-,ψ-), we obtain
(42)Λ(Ik′(ϕ-,ψ-),(ϕ-,ψ-))=0.
But
(43)(Ik′(ϕ-,ψ-),(ϕ-,ψ-))=2((L1-ω1)ϕ-,ϕ-)k+2((L2-ω2)ψ-,ψ-)k-4∑n=-kTkT-1‍(a1(ϕ-n)4+a2(ψ-n)4+2a3(ϕ-n)2(ψ-n)2)=-2∑n=-kTkT-1‍(a1(ϕ-n)4+a2(ψ-n)4+2a3(ϕ-n)2(ψ-n)2)<0.
Thus, Λ=0 and
(44)(Jk′(ϕ-,ψ-),(ϕ,ψ))=0,
for any (ϕ,ψ)∈Xk×Xk. Take (ϕ,ψ)=(e(j),0) and (ϕ,ψ)=(0,e(j)) in (44) for j∈ℤ∩[-kT,kT-1], where
(45)en(j)={1,n=j,0,n≠j.
We see that Jk′(ϕ-,ψ-)=0. Thus, (ϕ-,ψ-) is a nontrivial 2kT-periodic ground state solution to (7). The proof is completed.Through a similar argument to the proof of Lemma4, we get the following lemma.Lemma 5.
Let(ϕ-,ψ-) be a minimizer of the functional J(ϕ,ψ) constrained on the Nehari manifold N; that is,
(46)J(ϕ-,ψ-)=m=inf(ϕ,ψ)∈NJk(ϕ,ψ),
and then (ϕ-,ψ-) is a nontrivial discrete soliton to (7), which is called a nontrivial ground state solution to (7).


## 3. Main Results

In this section, we will establish some sufficient conditions on the existence of2kT-periodic solutions and nontrivial solutions in X×X of (7).We start with the following.Lemma 6.
Assume thatω1<θ11 and ω2<θ12 hold. Then the minimum value in (40) is attained.Proof.
Let{(ϕ(j),ψ(j))}⊂Nk be a minimizing sequence for Jk; that is, Jk(ϕ(j),ψ(j))→mk as j→+∞. The fact that
(47)Jk(ϕ(j),ψ(j))=Jk(ϕ(j),ψ(j))-12Ik(ϕ(j),ψ(j))=14∑n=-kTkT-1‍(a1(ϕn(j))4+a2(ψn(j))4+2a3(ϕn(j))2(ψn(j))2)
shows that ∥(ϕ(j),ψ(j))∥k∞ is bounded. Since the space Xk×Xk is finite dimensional, so the norm ∥(·,·)∥k∞ is equivalent to the Euclidean norm on Xk×Xk, and the sequence {(ϕ(j),ψ(j))} is bounded. Passing to a subsequence, we can assume that (ϕ(j),ψ(j)) converges to (ϕ,ψ)∈Xk×Xk. Since the set Nk is closed and the functional Jk is continuous, we obtain that (ϕ,ψ)∈Nk and Jk(ϕ,ψ)=mk. The proof is completed.To obtain a nontrivial solutions inX×X of (7), we need the following lemma.Lemma 7.
Assume thatω1<θ11 and ω2<θ12 hold. Let (ϕ(k),ψ(k)) be a 2kT-periodic ground state solution, that is, a solution of (40). Then the sequences {mk} and {∥ϕ(k)∥k2+∥ψ(k)∥k2} are bounded above and below away from zero.Proof.
By Lemma2, {mk} is obviously bounded below away from zero. Let
(48)(ϕ~n(k),ψ~n(k))={(1,0),n=0,(0,0),n∈ℤ∩[-kT,kT-1]-{0},
so (ϕ~(k),ψ~(k))∈Xk×Xk-{(0,0)}, as in the beginning of the proof of Lemma 1; there exists tk=(β0-ω1)/a1 such that Ik(tkϕ~(k),tkψ~(k))=0; that is, (tkϕ~(k),tkψ~(k))∈Nk. Therefore, (β0-ω1)2/(4a1)=Jk(tkϕ~(k),tkψ~(k))≥mk. And mk=Jk(ϕ(k),ψ(k)) is bounded above.
Next, we prove that{∥ϕ(k)∥k2+∥ψ(k)∥k2} are bounded above and below away from zero. By the previous proof, we see that mk=Jk(ϕ(k),ψ(k)) is bounded above and below away from zero; that is, there exist ρ>0 and ϱ>0 such that ρ≤Jk(ϕ(k),ψ(k))≤ϱ. Then
(49)Jk(ϕ(k),ψ(k))=Jk(ϕ(k),ψ(k))-14Ik(ϕ(k),ψ(k))=14((L1-ω1)ϕ(k),ϕ(k))k+14((L2-ω2)ψ(k),ψ(k))k≥ζ4(∥ϕ(k)∥k2+∥ψ(k)∥k2).
This implies that
(50)∥ϕ(k)∥k2+∥ψ(k)∥k2≤4ϱζ.
The proof is completed.Lemma 8.
Assume thatλ1*>0 and λ2*>0 hold. Let (ϕ(k),ψ(k)) be a 2kT-periodic ground state solution, that is, a solution of (40). Then there exist positive constants μ and ν such that
(51)μ≤max{∥ϕ(k)∥k∞,∥ψ(k)∥k∞}≤ν.Proof.
By Lemmas4 and 6, we know that (ϕ(k),ψ(k)) is a nontrivial critical point of Jk. Therefore, we have
(52)(L1ϕ(k))n-ω1ϕn(k)=a1|ϕn(k)|2ϕn(k)+a3|ψn(k)|2ϕn(k),(L2ψ(k))n-ω2ψn(k)=a2|ψn(k)|2ψn(k)+a3|ϕn(k)|2ψn(k).
Let mi∈ℤ with -kT≤mi≤kT-1, i=1,2 such that ∥ϕ(k)∥k∞=|ϕm1(k)| and ∥ψ(k)∥k∞=|ψm2(k)|. By the fact that
(53)(L1ϕ(k))m1-ω1ϕm1(k)=a1|ϕm1(k)|2ϕm1(k)+a3|ψm1(k)|2ϕm1(k),(L2ψ(k))m2-ω2ψm2(k)=a2|ψm2(k)|2ψm2(k)+a3|ϕm2(k)|2ψm2(k),
we get
(54)|(L1-ω1)ϕm1(k)|=a1∥ϕ(k)∥k∞3+a3|ψm1(k)|2∥ϕ(k)∥k∞,(55)|(L2-ω2)ψm2(k)|=a2∥ψ(k)∥k∞3+a3|ϕm2(k)|2∥ψ(k)∥k∞.
If one of the components of (ϕ(k),ψ(k)), say ψ(k), is equal to 0, then ϕ(k)≠0. Thus, by (54), we obtain
(56)λ1*≤a1∥ϕ(k)∥k∞2,λ1*≥a1∥ϕ(k)∥k∞2.
By (56), we get
(57)λ1*a1≤∥ϕ(k)∥k∞≤λ1*a1.
Similarly, if ψ(k)≠0, then we have
(58)λ2*a2≤∥ψ(k)∥k∞≤λ2*a2.
Ifϕ(k)≠0 and ψ(k)≠0, then, by (54) and (55), we obtain
(59)λ1*≤a1∥ϕ(k)∥k∞2+a3∥ψ(k)∥k∞2,λ1*≥a1∥ϕ(k)∥k∞2,λ2*≤a3∥ϕ(k)∥k∞2+a2∥ψ(k)∥k∞2,λ2*≥a2∥ψ(k)∥k∞2.
By (59), we get
(60)max{λ1*a1+a3,λ2*a2+a3}≤max{∥ϕ(k)∥k∞,∥ψ(k)∥k∞}≤max{λ1*a1,λ2*a2}.
Let(61)μ=min{λ1*a1,λ2*a2,max{λ1*a1+a3,λ2*a2+a3}},ν=max{λ1*a1,λ2*a2}.
Then, by (57), (58), and (60), we get (51). The proof is completed.Now we are ready to state our main results.Theorem 9.
Assume thatω1<θ11 and ω2<θ12 hold. Then for every positive integer k, (7) possesses a nontrivial 2kT-periodic ground state solution (ϕ(k),ψ(k)) in Xk×Xk. Moreover, there are other three nontrivial 2kT-periodic ground state solutions (-ϕ(k),ψ(k)), (ϕ(k),-ψ(k)), and (-ϕ(k),-ψ(k)) to (7).Proof.
The proof follows easily from Lemmas4 and 6.Theorem 10.
Assume thatωi<θ1i, λi*>0, i=1,2, and
(62)a3>max{λ+θ11-ω1θ11-ω1a1,λ+θ11-ω2θ11-ω1a1,λ+θ12-ω1θ12-ω2a2,λ+θ12-ω2θ12-ω2a2}
holds. Then system (7) has a nontrivial ground state solution (ϕ*,ψ*) in X×X with ϕ*≠0 and ψ*≠0. Moreover, there are other three nontrivial ground state solutions (-ϕ*,ψ*), and (ϕ*,-ψ*), (-ϕ*,-ψ*) to (7).Proof.
Consider the sequence{(ϕ(k),ψ(k))} of 2kT-periodic solutions found in Lemma 6. By Lemma 8, without loss of generality, we can assume that the subsequence {(ϕ(ki),ψ(ki))} of {(ϕ(k),ψ(k))} satisfies μ≤∥ϕ(ki)∥k∞≤ν. Thus, there exists nki∈ℤ such that
(63)|ϕnki(ki)|≥μ.
By the periodicity of the coefficients in (7), we see that {(ϕn+T(k),ψn+T(k))} is also a solution to (7). Making some shifts if necessary, without loss of generality, we can assume that 0≤nki≤T-1 in (63). Moreover, passing to a subsequence {ϕ(kij)} of {ϕ(ki)}, we can also assume that nki=n* and 0≤n*≤T-1. It follows from (37), (50), (51), and (63) that we can choose a subsequence, still denoted by {ϕ(k)} and {ψ(k)}, such that ϕn(k)→ϕn* and ψn(k)→ψn* for all n∈ℤ. Notice that |ϕn**|≥μ. Then (ϕ*,ψ*)∈X×X with (ϕ*,ψ*)≠(0,0). Furthermore, (7) possesses pointwise limits and hence (ϕ*,ψ*) is a nontrivial solution to (7). By the way, (-ϕ*,ψ*), (ϕ*,-ψ*), and (-ϕ*,-ψ*) are nontrivial solutions to (7). Since J(ϕ*,ψ*)=J(-ϕ*,ψ*)=J(ϕ*,-ψ*)=J(-ϕ*,-ψ*), if (ϕ*,ψ*) is a ground state solution, then (-ϕ*,ψ*), (ϕ*,-ψ*), and (-ϕ*,-ψ*) are the ground state solutions.
Now we prove that the solution(ϕ*,ψ*) is a ground state solution. By Lemma 5, we have to show that J(ϕ*,ψ*)=m. Actually, we have proven that, for every sequence kj→∞, passing to a subsequence still denoted by kj and making appropriate shifts, we can suppose that ϕn(kj)→ϕn* and ψn(kj)→ψn* pointwise, where (ϕ*,ψ*)∈X×X is a nontrivial solution. For any positive integer K, we have
(64)liminfj→∞Jkj(ϕ(kj),ψ(kj))=liminfj→∞14∑n=-kjTkjT-1‍(a1(ϕn(kj))4+a2(ψn(kj))4+2a3(ϕn(kj))2(ψn(kj))2)≥liminfj→∞14∑n=-KTKT-1‍(a1(ϕn(kj))4+a2(ψn(kj))4+2a3(ϕn(kj))2(ψn(kj))2)=14∑n=-KTKT-1‍(a1(ϕn*)4+a2(ψn*)4+2a3(ϕn*)2(ψn*)2).
Let K→∞; we obtain that
(65)liminfj→∞Jkj(ϕ(kj),ψ(kj))≥J(ϕ(*),ψ(*))≥m,
and, hence,
(66)liminfk→∞mk≥m.
Now we prove that
(67)limsupk→∞mk≤m.
By Lemma 7, we know that the sequence {mk} is bounded above and below away from zero. We extract a subsequence, still denoted by {mk}, and hence we prove that
(68)limk→∞mk≤m.
Given τ>0, let (ϕ^,ψ^)∈N be such that
(69)J(ϕ^,ψ^)=14∑n∈ℤ‍(a1(ϕ^n)4+a2(ψ^n)4+2a3(ϕ^n)2(ψ^n)2)<m+τ.
Choose t1>1 sufficiently close to 1 such that
(70)14∑n∈ℤ‍(a1(t1ϕ^n)4+a2(t1ψ^n)4+2a3(t1ϕ^n)2(t1ψ^n)2)<m+τ.
We also have that I(t1ϕ^,t1ψ^)<0. By density argument, we can find a finitely supported sequence (ϕ~,ψ~) sufficiently close to (t1ϕ^,t1ψ^) in X such that I(ϕ~,ψ~)<0 and
(71)14∑n∈ℤ‍(a1(ϕ~n)4+a2(ψ~n)4+2a3(ϕ~n)2(ψ~n)2)<m+τ.
Then there exists t2∈(0,1) such that I(t2ϕ~,t2ψ~)=0 and
(72)J(t2ϕ~,t2ψ~)=14∑n∈ℤ‍(a1(t2ϕ~n)4+a2(t2ψ~n)4+2a3(t2ϕ~n)2(t2ψ~n)2)<14∑n∈ℤ‍(a1(ϕ~n)4+a2(ψ~n)4+2a3(ϕ~n)2(ψ~n)2)<m+τ.
Let (ϕ~(k),ψ~(k))∈Xk×Xk be such that ϕ~n(k)=ϕ~n and ψ~n(k)=ψ~n if n∈ℤ∩[-kT,kT-1]. If k is large enough, then Ik(t2ϕ~(k),t2ψ~(k))=I(t2ϕ~,t2ψ~)=0 and
(73)Jk(t2ϕ~(k),t2ψ~(k))=J(t2ϕ~,t2ψ~)<m+τ.
This implies (68). Hence, by (66) and (68), we have J(ϕ*,ψ*)=m.
Finally, we will show thatϕ*≠0 and ψ*≠0. From the above arguments, system (7) has a nontrivial ground state solution (ϕ*,ψ*) in X×X. If one of the components of (ϕ*,ψ*), say ψ*, is equal to 0, then ϕ*≠0 (the proof for the other case is similar). For ϵ small enough, we consider (ϕ*,ϵϕ*)∈(X-{0})×(X-{0}). By a similar argument to the proof of Lemma 1, we know that there exists t* such that I(t*ϕ*,t*ϵϕ*)=0; that is (t*ϕ*,t*ϵϕ*)∈N. By I(t*ϕ*,t*ϵϕ*)=0, we have (t*)2=H1(ϕ*,ϵϕ*)/H2(ϕ*,ϵϕ*), where
(74)H1(ϕ,ψ)=((L1-ω1)ϕ,ϕ)+((L2-ω2)ψ,ψ),H2(ϕ,ψ)=∑n∈ℤ‍(a1(ϕn)4+a2(ψn)4+2a3(ϕn)2(ψn)2).
Moreover, J(t*ϕ*,t*ϵϕ*)=H12(ϕ*,ϵϕ*)/(4H2(ϕ*,ϵϕ*)). Notice that J(ϕ*,0)=inf(ϕ,ψ)∈NJ(ϕ,ψ), J(ϕ*,0)=(a1/4)∑n∈ℤ‍(ϕn*)4, and
(75)H2(ϕ*,0)=H1(ϕ*,0)=((L1-ω1)ϕ*,ϕ*)≥(θ11-ω1)∥ϕ*∥2.
For the sake of simplicity, we let
(76)B=∑n∈ℤ‍(ϕn*)4,D=((L2-ω2)ϕ*,ϕ*).
If ω1≤ω2, then, by (28) and (75),
(77)D=((L2-ω2)ϕ*,ϕ*)=((L1-ω1)ϕ*,ϕ*)+((L2-L1)ϕ*,ϕ*)+((ω1-ω2)ϕ*,ϕ*)≤a1B+λ∥ϕ*∥2≤λ+θ11-ω1θ11-ω1a1B.
This, combined with a3>((λ+θ11-ω1)/(θ11-ω1))a1 and (77) yields a1BD<a1a3B2. If ω2<ω1, then, by (28) and (75),
(78)D=((L2-ω2)ϕ*,ϕ*)=((L1-ω1)ϕ*,ϕ*)+((L2-L1)ϕ*,ϕ*)+((ω1-ω2)ϕ*,ϕ*)=a1B+λ∥ϕ*∥2+(ω1-ω2)∥ϕ*∥2≤a1B+λ+ω1-ω2θ11-ω1a1B=λ+θ11-ω2θ11-ω1a1B.
This, together with a3>((λ+θ11-ω2)/(θ11-ω1))a1 and (78) gives a1BD<a1a3B2. Thus, if a3>max{((λ+θ11-ω1)/(θ11-ω1))a1,((λ+θ11-ω2)/(θ11-ω1))a1}, then we have a1BD<a1a3B2. For ϵ small enough, we have
(79)H12(ϕ*,ϵϕ*)=(((L1-ω1)ϕ*,ϕ*)+((L2-ω2)ϵϕ*,ϵϕ*))2=(a1B+ϵ2D)2=a12B2+2a1BDϵ2+D2ϵ4<a12B2+2a1a3B2ϵ2+a1a2B2ϵ4=a1B(a1B+a2Bϵ4+2a3Bϵ2)=H2(ϕ*,0)H2(ϕ*,ϵϕ*).
Hence, by (79), we have
(80)J(t*ϕ*,t*ϵϕ*)=H12(ϕ*,ϵϕ*)4H2(ϕ*,ϵϕ*)<14H2(ϕ*,0)=J(ϕ*,0)=inf(ϕ,ψ)∈NJ(ϕ,ψ).
This is a contradiction. So, ψ*≠0. The proof is completed.



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*Source: 404369-2013-10-02.xml*