# On the Ground State to Hamiltonian Elliptic System with Choquard’s Nonlinear Term

Authors: Wenbo Wang; Jianwen Zhou; Yongkun Li
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

In the present paper, we consider the following Hamiltonian elliptic system with Choquard’s nonlinear term−Δu+Vxu=∫ΩGvy/x−yβdygvinΩ,−Δv+Vxv=∫ΩFuy/x−yαdyfuinΩ,u=0,v=0on∂Ω,where Ω⊂ℝN is a bounded domain with a smooth boundary, 0<α<N, 0<β<N, and F is the primitive of f, similarly for G. By establishing a strongly indefinite variational setting, we prove that the above problem has a ground state solution.


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## Body

## 1. Introduction and Main Results

In this paper, we deal with the existence of ground state solutions for the following Hamiltonian elliptic system with Choquard’s nonlinear term:(1)−Δu+Vxu=∫ΩGvyx−yβdygvinΩ,−Δv+Vxv=∫ΩFuyx−yαdyfuinΩ,u=0,v=0on∂Ω,where Ω⊂ℝN is a bounded domain with a smooth boundary, N≥3, 0<α<N, 0<β<N, and F is the primitive of f, similarly for G. For a single equation in whole space, ℝN is closely related to the Choquard-Pekar equation:
(2)−Δu+Vxu=∫ℝNFuyx−yμdyfu,inℝN,0<μ<N.WhenN=3, μ=1, Vx≡1, and fu=u, equation (2) has appeared in several contexts of quantum physics. In 1954, Pekar used equation (2) to describe a polaron at rest in quantum theory. In 1976, to model an electron trapped in its own hole, P. Choquard considered (2) as a certain approximation to Hartree-Fock’s theory of one component plasma (see [1, 2]). In some particular cases, (2) is also known as the Schrödinger-Newton equation which was introduced by Penrose in [3] to describe the self-gravitational collapse of a quantum mechanical wave function.ForN=3, μ=1, Vx1, and fu=u, the existence of ground states of (2) was obtained in [4] by variational methods. Later, Moroz and Van Schaftingen [2] investigated the regularity, radial symmetry, and asymptotic behavior at infinity of positive solutions for a generalized Choquard equation. Other results involving existence, multiplicity, and concentration of Choquard’s problems can be found in [5–12] and the references therein.We also observed that Bhattarai [13] considered the coupled Choquard-type fractional Schrodinger systems:
(3)−Δαu+ω1u=λ1∫ℝNuyp1x−yβdyup1−2u+∫ℝNvyqx−yβdyuq−2u,inℝN,−Δαv+ω2v=λ2∫ℝNvyp2x−yβdyvp2−2v+∫ℝNuyqx−yβdyvq−2v,inℝN,where ω1 and ω2 appear as the Lagrange multipliers, N≥2, 0<α<1, 0<β<N, 2≤p1, p2, and q<N+2α+β/N. Bhattarai looked for minimizers of the L2-constrained minimization problem via concentration compactness techniques. Recently, Giacomoni et al. [14] were also concerned with the coupled Hardy-Littlewood-Sobolev critical nonlinearity fractional Schrӧdinger system in a smooth bounded domain Ω⊂ℝN:
(4)−Δθu=λuq−2u+∫Ωvy2μ∗x−yμdyu2μ∗−2uinΩ,−Δθv=δvq−2v+∫Ωuy2μ∗x−yμdyv2μ∗−2vinΩ,where 2μ∗=2N−μ/N−2θ, 0<q<2. By minimizing over a suitable subset of Nehari’s manifold, they proved the existence of at least two nontrivial solutions for a suitable range of λ and δ. For problems involving Hardy-Littlewood-Sobolev’s critical exponent problems, please see [15–17] and the references therein. Among them, a strongly indefinite Choquard equation was studied in [17].Motivated by the papers mentioned above, in particular, the papers [13, 14, 17], the purpose of the present paper is to investigate the ground state solution of problem (1). To the best of our knowledge, there is no work concerning the existence of ground state solutions to the Choquard-type Hamiltonian elliptic system. For the Hamiltonian elliptic system, we refer the readers to the papers [18–22] and the references therein.Throughout this paper, we will always assumeN≥3 and we suppose that V, g, and f satisfy the following assumptions.
(V)
Vx∈L∞Ω,ℝ, 0 lies in a gap of σ−Δ+Vx, where σ−Δ+Vx is the spectrum of the operator −Δ+Vx.
(H1)
f,g∈Cℝ,ℝ.
(H2)
fu=ou as u⟶0, gv=ov as v⟶0.
(H3)
limsupu→∞fu/up<∞, limsupv→∞gv/vq<∞, where p, q>1 satisfy
(5)N−α/2p+1N+N−β/2q+1N>N−2N.
(H4)
There existsθ>2, such that u,v∈ℝ:
(6)0≤θFu≤fuu,0≤θGv≤gvv.
(H5)
Fu>0 for u≠0 and Gv>0 for v≠0.Before stating our main result, we review the definition ofs,t ground state solution about (1). We call u,v≠0,0 as a s,t ground state solution of (1) in work space E (see Section 2); if u~,v~ is another s,t weak solution (see Definition 2), then the corresponding energy functional Iu,v≤Iu~,v~, where I will be defined in equation (27) or equation (32).Theorem 1.
Suppose that H1, H2, H3, H4, and H5 are satisfied. Then (1) has a s,t ground state solution.To prove Theorem1, here we use a minimizing argument based on the ideas developed in [23]. We are concerned with system (1) in whole space ℝN involving the Hamiltonian elliptic system with Choquard’s nonlinear term; it is a nonlocal problem, which brings about two obstacles. One is to check the linking structure, and the other one is to prove the boundedness of the corresponding (PS) sequences. To avoid these obstacles, we shall deal with our problem in bounded domain. Indeed, the difficulty is still there for the problem on ℝN. It is worth pointing out that the monotonicity condition like [23] is not required on the nonlinear terms f and g; it prevents us from using the standard way (see, e.g., [23, 24]) to check that the minimizer is a critical point. Via the basic leitmotiv from Proposition 3.2 of [25], we shall use the deformation lemma to prove it (see Lemma 13.. We end this section by giving our arrangements of this paper. In Section 2, we establish the variational settings about (1). In Section 3, we provide some lemmas and then prove Theorem 1.


## 2. Variational Settings

For the system(7)−Δu+Vxu=∫ΩGvyx−yβdygv,inΩ,−Δv+Vxv=∫ΩFuyx−yαdyfu,inΩ,if Fu=uθ+1, by Hardy-Littlewood-Sobolev’s inequality (see Theorem 4.3 in [26]), then the term
(8)∫Ω∫ΩFuyFuxx−yαdydx,is well defined if Fu∈LrΩ for t>1 such that 2/r+α/N=2. In view of Sobolev’s embedding theorem, it requires that 1≤θ+1r≤2N/N−2, which leads us to assume that
(9)1−α2N≤θ−1≤2N−αN−2.Since we restrictp≥1, q≥1 and
(10)N−α/2p+1N+N−β/2q+1N>N−2N.So it holds that1−α/2N≤p+1 and 1−β/2N≤q+1 but p or q could be supercritical in the sense that
(11)p+1>2N−αN−2orq+1>2N−βN−2.We remark thatp≥1 is used in (38) and similarly for q≥1. Since p or q could be supercritical, we need the fractional Sobolev spaces (see, e.g., [19, 22, 27]). According to H4, we can choose s, t>0 with s+t=2, such that
(12)p+1<NN−2s2−αN,q+1<NN−2t2−βN.DenoteS=−Δ+Vx. Since the effective domain DS=DS∗ and S is symmetric, so S is self-adjoined on L2Ω. Since the self-adjoint operator is closed, according to the polar decomposition theorem (Theorem VIII.32 in [28] and jointly with Theorem 3.2 and 3.3, Ch IV in [29]), it holds that there is a positive self-adjoint operator S (in fact S=S∗S), with DS=DS and a partial isometry U such that S=US, S and U are uniquely determined and
(13)SU=US,SU=US.It is well known thatU and S are both self-adjoint operators on L2Ω. In view of Theorem 3.35 in Chapter V of [30] and Corollary 5.5.6 in [31], there is a unique square root operator Q such that Q2=S, furthermore, QU=UQ. Denote S1/2≔Q. Let
(14)As≔Ss/2,At≔St/2.According to [19, 27], we consider a basis of L2Ω constituted by eigenfunctions ϕn of
(15)−Δu+Vxu=λuinΩ,u=0on∂Ω,with associated eigenvalues λn. If for u∈L2Ω, we write u=∑n=1∞anϕn, then the effective domain of As and At are
(16)Es≔DAs=u∈L2Ω:∑n=1∞λnsan2<∞,Et≔DAt=u∈L2Ω:∑n=1∞λntan2<∞.They are two Hilbert spaces endowed with the following inner product, respectively,(17)u,vs=Asu,AsvL2,φ,ψt=Atφ,AtψL2,where
(18)As:Es⟶L2Ω,u↦∑n=1∞λns/2anϕn,is an isometric isomorphism. So, As has the inverse As−1, and we denote A−s=As−1, similarly for At. Set E=Es×Et, then E is a Hilbert space with the inner product and norm
(19)z,η=u,φs+v,ψt,z2=us+vt,for z=u,v,η=φ,ψ∈E. We recall the embedding theorem (see [32]); for r>0, the embedding Er⟶LkℝN is continuous for 1≤k≤2N/N−2r and is compact for 1≤k<2N/N−2r. We also consider the bounded self-adjoint operator L:E⟶E defined as follows, for z=u,v,η=φ,ψ:
(20)Lz,η=AsUu,AtψL2+Atv,AsUφL2.A natural question will be asked whether the operatorL:E⟶E is well defined. Here, we only check if AsUu,AtψL2 is well defined. In fact, noting that As and U are self-adjoint operators, we infer that
(21)AsUu,AtψL2=Uu,A2ψL2=u,USψL2=u,SψL2.By the complex interpolation theory (see Chapter 1.15 of [32]), we get
(22)Es=L2Ω,H2Ω1−s/2.SoAsUu,AtψL2<∞.Next, we haveL that has only two eigenvalues −1 and 1, whose corresponding eigenspaces are
(23)E+=u,A−tAsUu:u∈Es,E−=u,−A−tAsUu:u∈Es.Clearly,E=E+⊕E−. Indeed, 0 lies in a gap of σ?+Vx, and using Theorem 3.3 in Chapter IV of [29], we get
(24)L2Ω=L+⊕L−,where L±=u∈L2Ω:Uu=±u.Similar to Proposition 1.1 in [19], for z=u,v, we can easily get
(25)Lz=A−sAtUv,A−tAsUu.We consider the eigenvalue problemLz=λz in E, which yields that
(26)λ2v=A−tAsUA−sAtUv=UUv=Uv++v−=v.Here, we have used the fact thatAsU=UAs and AtU=UAt (see [22]).We consider the corresponding energy functional for (7):
(27)Iu,v=∫ΩAsUuAtvdx−12∫Ω∫ΩFuyFuxx−yαdydx−12∫Ω∫ΩGvyGvxx−yβdydx,u,v∈E.Similarly ([27], p. 61), under our assumptions, using the fact that AsU and At are linear, jointly with Lebesgue’s dominated convergence theorem, the Gateaux derivative of I in the direction η=φ,ψ at z=u,v is defined as
(28)DIz,η=liml→0Iz+lη−Izl=AsUu,AtψL2+AsUϕ,AtvL2−liml→012l∫Ω∫ΩFuy+lφyFux+lφy−FuyFuxx−yαdydx−liml→012l∫Ω∫ΩGvy+lψyGvx+lψy−GvyGvxx−yβdydx=AsUu,AtψL2+AsUϕ,AtvL2−∫Ω∫ΩFuyfuxφxx−yαdydx−∫Ω∫ΩGvygvxψxx−yαdydx.Clearly,DIz,η is linear bounded about η and continuous about z. Thus, I∈C1E,ℝ. And its Fréchet derivative is given by
(29)I′z,η=AsUu,AtψL2+AsUϕ,AtvL2−∫Ω∫ΩFuyfuxφxx−yαdydx−∫Ω∫ΩGvygvxψxx−yαdydxDefinition 2.
We say thatz=u,v∈E\0 is a s,t weak solution of (7), if for each η=φ,ψ∈E, there holds
(30)AsUu,AsψL2+AsUφ,AtvL2=∫Ω∫ΩFuyfuxφxx−yαdydx−∫Ω∫ΩGvygvxψxx−yβdydx.It is easy to check that the critical pointu,v of I is a s,t weak solution of (7). Moreover, for z=u,v∈E, z=z++z−, in view of Lemma 2.1 in [19], it holds that
(31)z±=u±A−sAtUv2,v±A−tAsUu2.Therefore,(32)Iz=12z+2−z−2−12∫Ω∫ΩFuyFuxx−yαdydx−12∫Ω∫ΩGvyGvxx−yβdydx.Remark 3.
Ifinfx∈ΩVx>0, then U=I.


## 3. Existence of Ground States

Following [23] (in page 3804), we introduce the generalized Nehari manifold
(33)M=z∈E\E−:I′z,z=0andI′z,η=0forallη∈E−.We need to prove thatM≠∅.Forz∈E\E−, define γ∈C1ℝ+×F,ℝ given by
(34)γzt,η=Itz++η.It is easy to check the following lemma (see, e.g., Lemma 3.1 in [25]).Lemma 4.
t,η∈ℝ+×E− is a critical point of γz if and only if tz++η∈M.
Fore∈E\E−, set E^e≔E−⊕ℝ+e, where ℝ+=0,+∞.Lemma 5.
For eachz∈E\E−, there exists tzz++ηz∈E^z, such that
(35)Itzz++ηz=maxt≥0,η∈FItz++η.Moreover,tzz++ηz∈M.Proof.
By Lemma4, it suffices to prove that the maximum exists. This follows from the next two lemmas. Indeed, using Lemma 7 and Lemma 8, combining with I0=0, it is easy to check that the maximum point exists.Remark 6.
Even thoughf and g have satisfied the condition S5 in [23], we cannot have the uniqueness of tzz++ηz. So, we cannot get the result similar to Proposition 2.3 in [23].Lemma 7.
There arer>0 and ρ>0 such that infz∈Sρ+Iz≥r, where Sρ+=z∈E+:z=ρ.Proof.
For a givenε>0, by H2, H3, and H4, there is a Cε>0 such that
(36)Fu≤εu2+Cεup+1.
By Hardy-Littlewood-Sobolev’s inequality, one has(37)∫Ω∫ΩFuyFuxx−yαdydx≤CFuLrΩ2≤C∫Ωεu2+Cεup+1rdx2/r≤C1εz4+C2Cεz2p+1,where 2/r+α/N=2. Here we use the fact that
(38)2<2r≤rp+1<2NN−2s.
Similarly, we have(39)∫Ω∫ΩGvyGvxx−yαdydx≤C1εz4+C2Cεz2p+1.
Thus, forz=u,A−tAsUu∈E+, it holds that
(40)Iz≥12z2−εz4−C1εzp+1−C2εzq+1.
Hence, we can choose somer, ρ>0 such that Iz≥r for all z=ρ.Lemma 8.
For everye∈E+\0, there is a Re>0 such that I<0 on E^e\BRe0.Proof.
If not, there existswn⊂E^e such that wn⟶∞ and Iwn≥0.
Without loss of generality, we may assume thate∈E+, e=1. By doing wn,1,wn,2=wn=tne+wn− where tn≥0. Set
(41)w¯n=wnwn=tnwne+wn−wn≔sn2e+ηn−.
Obviously,1=w¯n2=sn2+ηn−2. Jointly with
(42)0≤Iwnwn2=12sn2−ηn−2−12∫Ω∫ΩFwn,1yFwn,1xx−yαwn2dydx−12∫Ω∫ΩGwn,2yGwn,2xx−yβwn2dydx≤12sn2−ηn−2,one has 1/2≤sn≤1. So, for a subsequence, sn⟶s>0, w¯n⇀w¯, w¯nx⟶w¯x, a.e., in Ω. Hence, w¯=se+w¯−x≠0. It means that for
(43)A≔x∈Ω:w¯x≠0,where A>0. Thus, for x∈A, wnx⟶∞.
It follows from H2, H3, and H5 that there exists C1, C2>0 such that
(44)Fu≥C1uθ−C2,Gv≥C1vθ−C2,and
(45)limu→∞Fuu2=∞,limv→∞Gvv2=∞.
Thus,wn=wn,1,wn,2, it follows from Fatou’s lemma that
(46)∫Ω∫ΩFwn,1yFwn,1xx−yαwn2dydx+∫Ω∫ΩGwn,2yFwn,2xx−yβwn2dydx≥C∫AC1wnθdx−C2A∫AFwn,1+Gwn,2wn2wn2wn2dx≥C3∫AFwn,1+Gwn,2wn2wn2wn2dx⟶∞,which would conflict with (38). Here, we use the fact that Ω is bounded.Lemma 9.
M contains all nontrivial critical points.Proof.
Ifz=u,v∈E, z≠0, and I′z=0, by H5 and H6, one has
(47)Iz=Iz−12I′z,z=12∫Ω∫ΩFuyfuxux−Fuxx−yαdydx+12∫Ω∫ΩGvygvxvx−Gvxx−yβdydx≥C1∫ΩFudx∫Ωfuu−Fudx+C2∫ΩGvdx∫Ωgvv−Gvdx>0.
Forz∈E−\0, it follows that
(48)Iz=−12z−2−∫Ω∫ΩFuyFuxx−yαdydx−∫Ω∫ΩGuyGuxx−yβdydx≤0.
Define(49)m≔infz∈MIz.We shall prove thatm is achieved on M and the minimizer is a critical point of I.Lemma 10.
(1)
There existsρ>0 such that
(50)m=infz∈MIz≥infz∈SρIz>0,where Sρ≔z∈E+:z=ρ.
(2)
z+≥maxz−,2m for every z∈M.Proof.
First, let us prove assertion (1). For everyz∈M, with Lemma 7 in hand, we can choose some δ>0, small ρ>0 such that
(51)Iz≥Iρz+z+≥δ.
Second,Iz≤1/2z+2−z−2 for every z∈M, which concludes the proof of assertion (2).Lemma 11.
I is coercive on M.Proof.
If not, there exists a sequencezn⊂M such that zn⟶∞ and Izn≤d for some d∈c,∞. Let wn=wn,1,wn,2≔zn/zn. Up to a subsequence, wn⇀w and wnx⟶wx, a.e., in ℝN.
Ifw=0, one has wn+⟶LpΩ×LqΩ for 1≤p<2N/N−2s, 1≤q<2N/N−2t. So it holds that
(52)limn→∞∫Ω∫ΩFwn,1+yFwn,1+xx−yαdydx=0,limn→∞∫Ω∫ΩGwn,2+yGwn,2+xx−yβdydx=0.
Jointly with Lemma5, we obtain
(53)d≥Izn≥Iswn+≥s24+on1.
It is absurd ifs>4d. So w≠0.
Similar to the proof of Lemma7, it holds that
(54)∫Ω∫ΩFzn,1yFzn,1xx−yαzn2+Gzn,2yFzn,2xx−yβzn2wn2dydx⟶n→∞∞.
Therefore,(55)0≤Iznzn2=12wn+2−wn−2−∫Ω∫ΩFzn,1yFzn,1xx−yαzn2+Gzn,2yFzn,2xx−yβzn2wn2dydx⟶−∞,which is a clear contradiction.Lemma 12.
c=infz∈MIz is achieved.Proof.
The proof is similar to Lemma 3.6 in [33], but for readers’ convenience, we review the details of the proof.
Letzn⊂M be a minimizing sequence. In the light of Lemma 10, zn is bounded in E. Extracting a subsequence if necessary, we have zn⇀z in E. By the continuity of the projections, zn+⇀z+.
Now let us prove thatz+≠0. If not, by Lemma 5 and the result (2) of Lemma 9, we get
(56)m+on1=Izn≥Itzn+=12t2zn+2+on1.
It is impossible ift is large.
By Lemma5, there exists tz>0,ηz∈E− such that
(57)tzz++ηz=tzz++ηz1,tzz++ηz2∈M.
With Lebesgue’s dominated theorem, it follows that(58)m≤Itzz++ηz=12tz2z+2−12ηz−2−∫Ω∫ΩFtzz++ηz1yFtzz++ηz1xx−yαdydx−∫Ω∫ΩGtzz++ηz2yGtzz++ηz2xx−yβdydx=liminfn→∞Itzzn++ηz≤liminfn→∞Izn=c.Lemma 13.
Ifz0∈M is such that
(59)m=infz∈MIz=Iz0,then I′z0=0.Proof.
Assume by contradiction thatI′z0≠0, then δ>0 and θ>0 exist such that
(60)I′z0E∗≥θ,forz∈B2δz0∩B,where B≔z∈E^z0:Iz<m−ε (if necessary, we can choose another z0 and ε small enough such that B2δz0∩B≠∅).
Define(61)T:E^z0⟶ℝ+×E−,given by Ttz0++η=t,η.
DenoteBδ/4≔Bδ/4z0∩B, T~≔TBδ/4, and D≔T~Bδ/4. Define
(62)π≔T~−1:D⟶Bδ/4,given by πt,η=tz0++η.
It is clear that(63)maxt,η∈∂DIπt,η≤m−ε<m.
For(64)0<ε<minε2,θδ8,S≔B2δz0∩B,it yields a deformation Γ (see Lemma 2.3 of [34] or A.4 of [35]) such that
(a)
Γ1,z=z if z∈I−1m−2ε,m+2ε∩S2δ,
(b)
Γ1,Im+ε∩S⊂Im−ε, where S2δ≔z:distz,S≤2δ.
Let us define ah:D⟶E and for every w∈E−, Ψ0w,Ψ1w:D⟶ℝ2 by
(65)ht,η==Γ1,πt,η,Ψ0wt,η=I′πt,η,πt,η,I′πt,η,w,Ψ1wt,η=I′ht,η,ht,η,I′ht,η,w.
Lemma5 and the degree theory now yield that
(66)degΨ0w,D,0=1.
Sincet,η∈T~∂Bδ/4, combining with (53), it therefore holds that
(67)Iπt,η≤m−ε<m−2ε.
It follows from (a) that(68)Ψ0w=Ψ1won∂D.
Consequently, we obtain(69)degΨ1w,D,0=1.
Then, we know that there existst0,η0∈D such that ht0,η0∈M. This implies that
(70)m≤Iht0,η0=IΓ1,πt0,η0.
Noting thatIπt0,η0<m+ε and also πD=Bδ/4⊂S, in view of (b), one has m≤IΓ1,πt0,η0≤m−ε, which is a clear contradiction.Proof of Theorem 14.
The existence ofs,t ground state solution of (1) follows directly from Lemma 12 to Lemma 13.



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