# Ground State Solution for an Autonomous Nonlinear Schrödinger System

Authors: Min Liu; Jiu Liu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study the following autonomous nonlinear Schrödinger system (discussed in the paper), whereλ,μ, and ν are positive parameters; 2∗=2N/N−2 is the critical Sobolev exponent; and f satisfies general subcritical growth conditions. With the help of the Pohožaev manifold, a ground state solution is obtained.


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## Body

## 1. Introduction and Main Result

In this paper, we consider the following autonomous nonlinear Schrödinger system:(1)−Δu+μu=μfu+λv,x∈ℝN,−Δv+νv=v2∗−2v+λu,x∈ℝN,u,v∈H1ℝN,N≥3,where μ,ν, and λ are positive parameters satisfying 0<λ<μν; 2∗=2N/N−2 is the critical Sobolev exponent; and f satisfies the following conditions:(f1)
f∈Cℝ,ℝ is an odd function.(f2)
lims⟶0+fs/s=0.(f3)
lims⟶+∞fs/s2∗−1=0.(f4)
There existsζ>0 such that Fζ>ζ2/2, where Fζ=∫0ζftdt.Systems of above type arise in nonlinear optics (cf. [1]). It is well known that a solution u,v∈H1ℝN×H1ℝN of system (1) is called a ground state solution if u,v≠0,0 and its energy is minimal among the energy of all the nontrivial solutions.The following nonlinear Schrödinger system(2)−Δu+μu=up−1u+λv,x∈ℝN,−Δv+νv=vq−1v+λu,x∈ℝN,u,v∈H1ℝN,has been studied by many authors. When N≤3, μ=ν=1, p=q=3, and λ>0 small enough, Ambrosetti et al. [2] proved that (2) has multibump solitons. When N≥2, μ=ν=1, 1<p,q<2∗−1, 0<λ<1, and up−1u and vq−1v are replaced by 1+axup−1u and 1+bxvq−1v, Ambrosetti et al. [3] proved that system (2) has a positive ground state solution. When u,v∈D1,2ℝ4, μ=V1x, and ν=V2x satisfy the integral conditions and up−1u,λv,vq−1v and λu are replaced by μ1u3,βuv2,μ2v3, and βu2v, respectively, Liu and Liu [4] proved that (2) has a positive solution. When u,v∈H01Ω, Ω is a smooth bounded domain in ℝ3, p=q=3, and λv, λu are replaced by −βuv2,−βvu2, respectively, Noris and Ramos [5] proved that (2) admits an unbounded sequence of solutions u,v with u>0,v>0, and u≠v for sufficiently large β>0. When N≥3, 1<p<2∗−1,q=2∗−1, and μ,ν>0, 0<λ<μν, Chen and Zou [6] proved that (2) has a positive ground state solution under λ,μ,ν which satisfied certain conditions. When N≥3, 1<p<2∗−1,q=2∗−1, and μ=ax,ν=bx,λ=λx, Li and Tang [7] proved that (2) has a nontrivial solution.Inspired by the above literatures, especially [6], we investigate the existence of ground state solution of system (1). When μfu=up−1u with 1<p<2∗−1, by using the Nehari manifold, Chen and Zou [6] obtained the existence of ground state solution of system (1). But in our paper, without the assumption of the monotonicity of u↦fu/u, we have to adopt a new method to replace the Nehari manifold.The following single Schrödinger equation(3)−Δu+u=fu,u∈H1ℝN,N≥3,has been widely studied by many researchers, and relevant results can been referred to [8–10] and the references therein. By [9], we know that if f satisfies (f1)-(f4); then, equation (3) has a ground state solution. Define
(4)a=infu∈Γ12∫ℝN∇u2+u2dx−∫ℝNFudx,where Γ=u∈H1ℝN:uisanontrivialsolutionofequation3 and define
(5)S=infu∈D1,2ℝN\0∫ℝN∇u2dx∫ℝNu2∗dx2/2∗,where S is the optimal constant of the Sobolev embedding D1,2ℝN⟶L2∗ℝN.The main result of this paper is the following.Theorem 1.
Assume thatμ,ν, and λ are positive parameters satisfying μ>S−N/N−2aN2/N−2 and 0<λ<μν. Suppose that f satisfies (f1)-(f4). Then, system (1) has a ground state solution.Remark 2.
There are some examples of functions that satisfy the assumptions (f1)-(f4), for example, fs=sp−2s with 2<p<2∗ and fs=sp−2s/1+s2 with 4<p<2∗+2.Remark 3.
It is obvious that system (1) has no semitrivial solutions. Indeed, if u,0 is a solution of system (1), then u=0 and if 0,v is a solution of system (1), then v=0.Remark 4.
There are some recent studies on the ground state solutions for other types of Schrödinger equations or systems, for example, [6, 11]. Moreover, in the bounded domain, the existence and the regularity of solutions to differential problems have been widely investigated by using tools of harmonic and real analysis and variational methods, for example, [12–14].


## 2. Preliminaries

In order to make a precise explanation of the results in this paper, we will give some notations.C,Ci denote various positive constants.LpℝN is the usual Lebesgue space endowed with the norm
(6)up=∫ℝNupdx1/p.D1,2ℝN=u∈L2∗ℝN∣∂u/∂xi∈L2ℝN,i=1,2,⋯,N endowed with the norm
(7)uD1,2=∫ℝN∇u2dx1/2.H1ℝN=u∈L2ℝN∣∂u/∂xi∈L2ℝN,i=1,2,⋯,N endowed with the norm
(8)u=∫ℝN∇u2+u2dx1/2.For anyu,v∈H≔H1ℝN×H1ℝN, we set
(9)u,vH=∫ℝN∇u2+μu2+∇v2+νv2dx1/2.For anyu∈H1ℝN, we denote ut=u·/t for all t>0.The weak solutions of (1) correspond to critical points of the functional
(10)Iu,v=12u,vH2−μ∫ℝNFudx−12∗∫ℝNv2∗dx−λ∫ℝNuvdx.Obviously,I∈C1H,ℝ and for all u,v∈H and φ,ψ∈H, we have
(11)I′u,v,φ,ψ=∫ℝN∇u·∇φ+μuφ+∇v·∇ψ+νvψdx−μ∫ℝNfuφdx−∫ℝNv2∗−2vψdx−λ∫ℝNφv+uψdx.Similar to [15, 16], in order to obtain a ground state solution, we define the Pohožaev manifold
(12)P=u,v∈H\0,0:Ju,v=0and consider the constraint minimization problem
(13)m=infu,v∈PIu,v,where J:H⟶ℝ is defined as
(14)Ju,v=N−22∫ℝN∇u2+∇v2dx+N2∫ℝNμu2+νv2dx−μN∫ℝNFudx−N2∗∫ℝNv2∗dx−λN∫ℝNuvdx.We also require the following subcritical system of system (1):
(15)−Δu+μu=μfu+λv,x∈ℝN,−Δv+νv=vq−2v+λu,x∈ℝN,u,v∈H1ℝN,N≥3,where 2<q<2∗, μ,ν, and λ are positive parameters satisfying 0<λ<μν and f satisfies (f1)-(f4). The energy functional of system (15) is
(16)Iqu,v=12u,vH2−μ∫ℝNFudx−1q∫ℝNvqdx−λ∫ℝNuvdx.Define(17)Pq=u,v∈H\0,0:Jqu,v=0andmq=infu,v∈PqIqu,v,where
(18)Jqu,v=N−22∫ℝN∇u2+∇v2dx+N2∫ℝNμu2+νv2dx−μN∫ℝNFudx−Nq∫ℝNvqdx−λN∫ℝNuvdx.


## 3. Proof of Theorem1

The following two lemmas will be used in proof.Lemma 5 (compactness lemma of Strauss, see [9, 10]).
LetP,Q:ℝ⟶ℝ be two continuous functions satisfying
(19)PsQs⟶0ass⟶+∞.
Letun be a sequence of measurable functions: ℝN⟶ℝ such that
(20)supn∫ℝNQunxdx<+∞and Punx⟶υx a.e. in ℝN, as n⟶∞. Then, for any bounded Borel set B, one has
(21)∫BPunx−υxdx⟶0asn⟶+∞.
If one further assumes that(22)PsQs⟶0ass⟶0and unx⟶0 as x⟶+∞, uniformly with respect to n, then Pun converges to υ in L1ℝN as n⟶+∞.Lemma 6 (Strauss inequality, see [17]).
IfN≥2, there exists CN>0 such that, for every ux=ux∈H1ℝN,
(23)ux≤CNu21/2∇u21/2x1−N/2a.e. on ℝN.
Before proving Theorem1, we need to prove a series of lemmas.Lemma 7.
Suppose that (f1)-(f4) hold. Then, the Pohožaev manifold P is not empty.Proof.
From [17], we know that for any ε>0,
(24)uε=NN−2N−2/4εN−2/2ε+x2N−2/4is a positive solution of the following equation:
(25)−Δu=u2∗−2u,x∈ℝN,N≥3.
Define a cut-off functionϕ∈C0∞ℝN,0,1 as
(26)ϕ=1,x∈Bρ,0,x∈ℝN\B2ρ,where ϱ>0 and Bϱ=x∈ℝN,x<ϱ. Let Wε=ϕuε and define Vε=Wε/∫ℝNWε2∗dx1/2∗. By [16], we have
(27)∫ℝNVε2∗dx1/2∗=1,∫ℝNVε2dx=oε1/2,N=3,oεlnε,N=4,oε,N=5.
Takeε>0 small enough such that ∫ℝN1/2∗Vε2∗−ν/2Vε2dx>0. Let U∈H1ℝN be a positive ground state solution of equation (3). Then, we have the following Pohožaev equality:
(28)N−22∫ℝN∇U2dx+N2∫ℝNU2dx=N∫ℝNFUdx.
Then,∫ℝNFU−1/2U2dx>0. Thus, we have
(29)τt≔IUt,Vεt=tN−22∫ℝN∇U2+∇Vε2dx−μtN∫ℝNFU−12U2dx−tN∫ℝN12∗Vε2∗−ν2Vε2dx−λtN∫ℝNUVεdx.
Defines=tN; we have
(30)ηs≔τs1/N=sN−2/N2∫ℝN∇U2+∇Vε2dx−μs∫ℝNFU−12U2dx−s∫ℝN12∗Vε2∗−ν2Vε2dx−λs∫ℝNUVεdx.
We can easily know thatηs>0 for s small enough and ηs<0 for large s. Since d2ηs/ds2<0, ηs is a concave function. Then, there exists a unique s0>0 such that η′s0=0. Hence, there exists a unique t0=s01/N>0 such that τ′t0=0. Then, we have t0τ′t0=JUx/t0,Vεx/t0=0. Then, Ut0,Vεt0∈P.Lemma 8.
Suppose that (f1)-(f4) hold. Then, m=infu,v∈PIu,v>0.Proof.
Since0<λ<μν, there exists 0<θ<1 such that 0<λ<μ1−θν. For any u,v∈P, we have Ju,v=0. By using Young’s inequality, we have
(31)N−22∫ℝN∇u2+∇v2dx+N2∫ℝNμu2+νv2dx=μN∫ℝNFudx+N2∗∫ℝNv2∗dx+λN∫ℝNuvdx≤μNθ2∫ℝNu2dx+NC∫ℝNu2∗dx+N2∗∫ℝNv2∗dx+λN∫ℝNuvdx≤μNθ2∫ℝNu2dx+NC∫ℝNu2∗dx+N2∗∫ℝNv2∗dx+Nμ1−θ2∫ℝNu2dx+Nν2∫ℝNv2dx.
Therefore, we have(32)N−22∫ℝN∇u2+∇v2dx≤NC∫ℝNu2∗dx+N2∗∫ℝNv2∗dx.
By using Sobolev’s inequality, we have(33)N−22∫ℝN∇u2+∇v2dx≤C1∫ℝN∇u2dx2∗/2+∫ℝN∇v2dx2∗/2≤C2∫ℝN∇u2+∇v2dx2∗/2,which implies ∫ℝN∇u2+∇v2dx≥N−2/2C2N−2/2>0. Therefore, we conclude that for any u,v∈P, we have
(34)Iu,v=Iu,v−1NJu,v=12−12∗∫ℝN∇u2dx+12−12∗∫ℝN∇v2dx≥1NN−22C2N−2/2.
Therefore, we havem>0.Lemma 9.
Suppose that (f1)-(f4) hold. Then, m<1/NSN/2.Proof.
LetU∈H1ℝN be a positive ground state solution of equation (3). Then, (28) holds and
(35)a=12∫ℝN∇U2+U2dx−∫ℝNFUdx=12∫ℝN∇U2+U2dx−∫ℝNFUdx−1NN−22∫ℝN∇U2dx+N2∫ℝNU2dx−N∫ℝNFUdx=∫ℝN∇U2dx.
Moreover, we have alsoUμx which is a solution of equation
(36)−Δu+μu=μfu,u∈H1ℝN,N≥3.
Then,Uμx,0∈P. Since μ>S−N/N−2aN2/N−2, we have
(37)m≤IUμx,0=IUμx,0−1NJUμx,0=1N∫ℝN∇Uμx2dx=aμ2−N/2<1NSN/2.Lemma 10.
Suppose that (f1)-(f4) hold. For any un,vn⊂P, if Iun,vn≤C, then un,vn is bounded in H.Proof.
SinceIun,vn≤C, we have
(38)C≥Iun,vn=Iun,vn−1NJun,vn=1N∫ℝN∇un2+∇vn2dx.
Because0<λ<μν, there exists 0<θ<1/2 and α>0 such that 0<λ<μ1−2θν−α. Therefore, we have
(39)N−22∫ℝN∇un2+∇vn2dx+N2∫ℝNμun2+νvn2dx=μN∫ℝNFundx+N2∗∫ℝNvn2∗dx+λN∫ℝNunvndx≤μNθ2∫ℝNun2dx+NC∫ℝNun2∗dx+N2∗∫ℝNvn2∗dx+λN∫ℝNunvndx≤μNθ2∫ℝNun2dx+NC∫ℝNun2∗dx+N2∗∫ℝNvn2∗dx+Nμ1−2θ2∫ℝNun2dx+Nν−α2∫ℝNvn2dx.
Then, we have(40)Nμθ2∫ℝNun2dx+Nα2∫ℝNvn2dx≤CN∫ℝNun2∗dx+N2∗∫ℝNvn2∗dx≤C3∫ℝN∇u2+∇v2dx2∗/2≤C4.
Hence,un,vnis bounded in H.Lemma 11.
Suppose that (f1)-(f4) hold. Then, limq⟶2∗−supmq≤m.Proof.
For anyε∈0,1/2, there exists u,v∈P such that Iu,v<m+ε. Since Ju,v=0, for any t>0, we have
(41)Iut,vt=Iut,vt−tNNJu,v=tN−22−N−22NtN∫ℝN∇u2+∇v2dx.
Defineht=tN−2/2−N−2/2NtN. Through simple calculations, we have h′t=N−2/2tN−3−tN−1. We can easily see that h is increasing for t∈0,1 and h is decreasing for t>1. Then, we have maxt>0Iut,vt=Iu,v and Iut,vt<Iu,v for any t≠1. By calculation, we have Iut,vt<0 for t>N/N−2. Take large T such that
(42)IuT,vT=TN−22∫ℝN∇u2+∇v2dx+TN2∫ℝNμu2+νv2dx−μTN∫ℝNFudx−TN2∗∫ℝNv2∗dx−λTN∫ℝNuvdx≤−1.
Then, there existsσ∈0,2∗ such that
(43)Iqut,vt−Iut,vt=tN2∗∫ℝNv2∗dx−tNq∫ℝNvqdx<ε,for all 2∗−σ<q<2∗ and 0≤t≤T. Then, we have IquT,vT≤−1/2 for all 2∗−σ<q<2∗. Since
(44)Iqut,vt=tN−22∫ℝN∇u2+∇v2dx+tN2∫ℝNμu2+νv2dx−μtN∫ℝNFudx−tNq∫ℝNvqdx−λtN∫ℝNuvdx,Iqut,vt>0 for t small enough. Then, there exists tq∈0,T such that d/dtIqut,vtt=tq=0. So, utq,vtq∈Pq. Hence, we have
(45)mq≤Iqutq,vtq≤Iutq,vtq+ε≤Iu,v+ε<m+2ε,for all 2∗−σ<q<2∗.
From [18, 19], we know that system (15) has a positive and radial ground state solution. Then, for any qn∈2,2∗ and qn⟶2∗−, there exists a positive and radial sequence un,vn⊂H such that
(46)Iqnun,vn=mqn,Iqn′un,vn=0,Jqnun,vn=0.
By Lemmas10 and 11, we know that un,vn is bounded in H.Lemma 12.
Suppose that (f1)-(f4) and (46) hold. Then, liminfn⟶∞mqn>0.Proof.
Similar to the proof of Lemma8, we have
(47)N−22∫ℝN∇un2+∇vn2dx≤NC∫ℝNun2∗dx+Nqn∫ℝNvnqndx.
Using Young’s inequality implies(48)Nqn∫ℝNvnqndx=Nqn∫ℝNvn22∗−qn/2∗−2vn2∗qn−2/2∗−2dx≤Nqn2∗−qn2∗−2∫ℝNvn2dx+Nqnqn−22∗−2∫ℝNvn2∗dx=N2∗∫ℝNvn2∗dx+o1.
Then,(49)N−22∫ℝN∇un2+∇vn2dx≤NC∫ℝNun2∗dx+N2∗∫ℝNvn2∗dx+o1≤C5∫ℝN∇un2+∇vn2dx2∗/2+o1.
So there existsϖ>0 such that up to a subsequence, ∫ℝN∇un2+∇vn2dx+o1≥ϖ. On the other hand,
(50)mqn=Iqnun,vn=Iqnun,vn−1NJqnun,vn=1N∫ℝN∇un2+∇vn2dx.
Then,liminfn⟶∞mqn>0.Proof of Theorem 1.
Because (46) holds, there exists u,v∈H such that unu,vnv in H1ℝN, un⟶u,vn⟶v in LpℝN,2<p<2∗, and unx⟶ux,vnx⟶vx a.e. in ℝN. For any φ,ψ∈H, we have
(51)0=Iqn′un,vn,φ,ψ⟶I′u,v,φ,ψ,i.e., u,v is a solution of system (1). Suppose that u=0. Set Ps=fss and Qs=s2+s2∗. Through Lemma 5 and Lemma 6, we have ∫ℝNPundx⟶0 as n⟶+∞. Since Iqn′un,vn,un,vn=0, by using Young’s inequality, we have
(52)un,vnH=μ∫ℝNfunundx+∫ℝNvnqndx+2λ∫ℝNunvndx≤μ∫ℝNPundx+∫ℝNvn22∗−qn/2∗−2vn2∗qn−2/2∗−2dx+∫ℝNμun2+νvn2dx≤μ∫ℝNPundx+2∗−qn2∗−2∫ℝNvn2dx+qn−22∗−2∫ℝNvn2∗dx+∫ℝNμun2+νvn2dx=∫ℝNvn2∗dx+∫ℝNμun2+νvn2dx+o1.
One has(53)∫ℝN∇vn2dx≤∫ℝN∇un2+∇vn2dx≤∫ℝNvn2∗dx+o1≤∫ℝN∇vn2dxS2∗/2+o1.
So we have (i)∫ℝN∇vn2dx⟶0 or (ii) limsupn⟶∞∫ℝN∇vn2dx≥SN/2. If (i) holds, then we have
(54)mqn=Iqnun,vn−1NJqnun,vn=1N∫ℝN∇un2+∇vn2dx⟶0,which contradicts with Lemma 12. If (ii) holds, then we have
(55)m≥limsupn⟶∞mqn=limsupn⟶∞Iqnun,vn−1NJqnun,vn=limsupn⟶∞1N∫ℝN∇un2+∇vn2dx≥limsupn⟶∞1N∫ℝN∇vn2dx≥1NSN/2.
This is a contradiction. Sou≠0 and through Remark 3, we know that v≠0. Applying the weak lower-semicontinuity of the norm, we have
(56)m≤Iu,v=Iu,v−1NJu,v=1N∫ℝN∇u2+∇v2dx≤liminfn⟶∞1N∫ℝN∇un2+∇vn2dx=liminfn⟶∞Iqnun,vn−1NJqnun,vn=liminfmqnn⟶∞≤m.
This impliesIu,v=m. We complete the proof.



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