# Solutions for a Class of Nonperiodic Superquadratic Hamiltonian Elliptic Systems Involving Gradient Terms

Authors: Shengzhong Duan; Xian Wu
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

In the present paper, we consider the following Hamiltonian elliptic systemHES: -Δu+bx·∇u+Vxu=Hvx,u,v,x∈RN, -Δv-bx·∇v+Vxv=Hux,u,v,x∈RN. A new existence result of nontrivial solutions is obtained for the system HES via variational methods for strongly indefinite problems, which generalizes some known results in the literatures.


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## Body

## 1. Introduction and Main Results

The goal of this paper is to study the existence of solutions for the nonperiodic elliptic systems in Hamiltonian formHES-Δu+bx·∇u+Vxu=Hvx,u,v,x∈RN,-Δv-bx·∇v+Vxv=Hux,u,v,x∈RN,where z=u,v:RN→R1×R1, b=(b1,…,bN)∈C1(RN,RN), V(x)∈C(RN,R), and H∈C1(RN×R2,R). Such a system arises when one is looking for stationary solutions to certain systems of optimal control (Lions [1]) or systems of diffusion equations (Itô [2] and Nagasawa [3]).In recent years, the systems like or similar toHES in the whole space RN were studied by a number of authors. Most of these works focused on the case b(x)=0. An usual way to overcome this difficulty is to consider the corresponding functional in the space of radially symmetric functions. In this way, De Figueiredo and Yang [4] considered the system (1)-Δu+u=Hvx,u,vinRN,-Δv+v=Hux,u,vinRN,where Hx,u,v=F(x,u)+G(x,v), F(x,u)=∫0uf(x,t)dt, and Gx,v=∫0vgx,tdt. They proved that system (1) has a radial solution pair under the assumptions that f(x,t) and g(x,t) are superlinear in t and radially symmetric with respect to x, fx,t≤c(1+tp-1) and gx,t≤c1+tq-1 with 2≤p,q<2N/N-2,N>2. This result was later generalized by Sirakov [5] to the system (2)-Δu+bxu=Hvx,u,vinRN,-Δv+bxv=Hux,u,vinRN.In [6], Zhao et al. considered periodic asymptotically linear elliptic systems (3)-Δu+Vxu=Hvx,u,vinRN,-Δv+Vxv=Hux,u,vinRN,ux⟶0,vx⟶0asx⟶∞,where the potential V is periodic and has a positive bound from below and f(x,t) and g(x,t) are periodic in x, asymptotically linear in t as t→∞. By using critical point theory of strongly indefinite functionals, they obtained the existence of a positive ground state solution as well as infinitely many geometrically distinct solution for systems (3) under the assumptions that f(x,t) and g(x,t) are odd in t. For other results, we refer readers to [7–17].Without assumption of periodicity or radially symmetric about nonlinearities, the problem is quite different in nature and there has not been much work done up to now. In a recent paper [18], Wang et al. considered the following nonperiodic elliptic systems in Hamiltonian form: (4)-Δu+Vxu=Hvx,u,vinRN,-Δv+Vxv=Hux,u,vinRN,ux⟶0,vx⟶0asx⟶∞and obtained the following theorem.Theorem A (see [18, Theorem 1.1]).
Suppose that the following conditions are satisfied:(V0)
V(x)=1.(H0)
There isμ>2 such that 0<μH(x,z)≤Hz(x,z)z if z≠0.(H1)
H(x,z)∈C1(RN×R2,R),Hz(x,z)=hx,zz, where h:RN×R+→R with h(x,s)≥0,h(x,s)=o(1) as s→0 uniformly in x, and there exist p∈[μ,2∗), c1>0 such that h(x,s)≤c1(1+sp-2), where 2∗=2N/N-2.(H2)
There ish∞∈C1(R+,R+) with h∞′(s)>0 for s>0, and h(x,s)→h∞(s) as x→+∞ uniformly in bounded set of s, h∞(s)≤h(x,s) and H(x,z)-H∞(z)≥b(x)zμ for all x∈RN, where b(x)≠0,b(x)≥0 and H∞(z)=∫0zh∞ssds.
Then system (4) has one solution.In the present paper, we are interested in the existence of solutions for Hamiltonian-elliptic systemsHES involving gradient terms and nonperiodic superquadratic nonlinearities. The class of problems treated here has several difficulties. First, the problem is set on RN; a main difficulty when dealing with this problem is the lack of compactness of the Sobolev embedding theorem. Second, the variational functional is strongly indefinite. Therefore, the classic critical point theorem cannot be applied directly. Third, the nonlinearities H(x,u,v) are nonperiodic in variable x and superquadratic at infinity; the method in [6] cannot be applied to obtain the existence of solutions. Finally, the appearance of the gradient terms in the systems also brings us some difficulties; in this case, the variational framework in [18] cannot work any longer. Inspired by recent works of Zhao and Ding [19], we are going to investigate the existence of solutions for the Hamiltonian elliptic systems HES. By using the critical point theory of strongly indefinite functional which was developed recently by Bartsch and Ding [10, 20] and the reduction methods which was developed in [21, 22], we obtain an existence result of problem HES, which generalizes Theorem A.Our fundamental assumptions are as follows:(V1)
V∈C(RN,R) is 1-periodic in xi for i=1,…,N and V0≔minx∈RN⁡V(x)>0.(B0)
b∈C1RN,RN, and div⁡(b)=0.(B1)
b is 1-periodic in xi for i=1,…,N.(H3)
H(x,z)∈C1(RN×R2,[0,+∞)),Hz(x,z)=h(x,z)z, where h:RN×R+→R with h(x,s)≥0,h(x,s)=o(1) as s→0 uniformly in x, and there exist p∈2,2∗, c1>0 such that h(x,s)≤c11+sp-2, where 2∗=2N/N-2.(H4)
H(x,z)/z2→+∞ as |z|→+∞ uniformly in x.(H5)
infx∈RN⁡H~(x,z)>0 whenever z≠0, where H~(x,z)=1/2Hz(x,z)z-H(x,z).(H6)
Hzx,zτ≤c2H~(x,z)zτ for some c2>0,τ>max⁡1,N/2 and all (x,z) with z large enough.(H7)
There ish∞∈C1(R+,R+) with h∞′(s)>0 for s>0, and h(x,s)→h∞(s) as |x|→+∞ uniformly in bounded set of s, h∞(s)≤h(x,s) for all (x,s)∈RN×R+ and h∞(s)<h(x,s) whenever s>0.Now we can state our main result.Theorem 1.
Let(V1), (B0), (B1), and (H3)–(H7) be satisfied. Then system HES has at least one nontrivial solution.Remark 2.
Theorem1 extends and improves Theorem A. First, we only need to assume that the potential V(x) is periodic and has a positive bound from below. Second, the conditions (H4) and (H5) can be obtained by (H0) and (H3). In fact, by (H3), we know that (5)Hzx,z≤c1z+zp-1,∀x,z∈RN×R2.Consequently, by the conditions (H0) and (5), it is easy to see that (H4) and (H5) hold. Furthermore, similar to the proof of Lemma  2.2(i) in [23], the condition (H6) can be obtained by (H0) and (5). Indeed, since p∈(2,2∗), we can obtain that p/p-2>max⁡{1,N/2}. For some τ∈(N/2,p/p-2), τ>1. If z≥1, then there exists c′>0 such that (6)Hzx,z≤c′zp-1.Choose r≥1 so large that (7)1μ≤12-c′τ-1zp-p-2τ,whenever z≥r. Then, by (H0), (6), and (7), we obtain (8)Hx,z≤1μHzx,zz≤12-c′τ-1zp-p-2τHzx,zz≤12-Hzx,zτ-1zτ+1Hzx,zz. It follows that (9)Hzx,zτzτ≤12Hzx,zz-Hx,z=H~x,z.Third, the condition (H3) is weaker than the condition (H1) and the condition (H7) is weaker than the condition (H2). Finally, summing up the above discussion, Theorem A is the special case of Theorem 1 corresponding to b(x)=0.
Throughout this paper, we always assume thatC denote any positive constant and may be different in different places. For 0≠a=(a1,…,aN)∈ZN, we define a∗z(x)≔zx1+a1,…,xN+aN, where z=(u,v):RN→R1×R1.


## 2. Variational Setting

In this section, we will establish variational framework for the systemHES. For the convenience of notation, let ·p denote the usual Lp-norm and ·,·2 be the usual L2-inner product. Let X and Y be two Banach spaces with norms ·X and ·Y; we always choose the equivalent norm (x,y)X×Y=xX2+yY21/2 on the product space X×Y. In particular, if X and Y are two Hilbert spaces with inner products ·,·X and ·,·Y, we choose the inner product x,y,w,zX×Y=x,wX+y,zY on the product space X×Y. In order to continue the discussion, we need the following notations. Set (10)J=0-110,J0=0110, and ∇J=0-∇∇0 is a matrix operator. Let S=-Δ+V denote the Schrödinger operator. Denote A0≔SJ0 and (11)A≔A0+b·∇J=0-Δ-b·∇+V-Δ+b·∇+V0. Then HES can be rewritten as (12)Az=Hzx,z,z∈H1RN,R2.Denote byσ(A) and σess(A) the spectrum and the essential spectrum of the operator A, respectively. Set μ_≔inf⁡σA∩0,∞; then we have the following lemmas.Lemma 3 (see [19, Lemma 2.1]).
Suppose that(B0) and (V1) are satisfied. Then the operator A is a self-adjoint operator on L2RN,R2 with domain D(A)=H2(RN,R2).Lemma 4 (see [19, Lemma 2.3]).
Let(B0), (B1), and (V1) be satisfied. Then(1)
σ(A) = σess(A); that is, A has only essential spectrum;(2)
σ(A)⊂R∖(-V0,V0) and σ(A) is symmetric with respect to origin;(3)
V0≤μ_≤max⁡V.
It follows from Lemmas3 and 4 that the space L2(RN,R2) possesses the orthogonal decomposition (13)L2=L-⊕L+,z=z-+z+, such that A is negative definite (resp., positive definite) in L- (resp. L+). Let A denote the absolute value of A and A1/2 be the square root of A. Let E=DA1/2 be the Hilbert space with the inner product (14)z,w=A1/2z,A1/2wL2, and norm z=(z,z)1/2. E possesses an induced decomposition (15)E=E-⊕E+,E±=E∩L±, which are orthogonal with respect to the inner products ·,·2 and ·,· (the above results can be found in [19]).Lemma 5 (see [19, Lemma 2.4]).
· and ·H1(RN,R2) are equivalent norms. Therefore, E embeds continuously into Lp(RN,R2) for any p∈2,2∗ and compactly into LlocpRN,R2 for p∈[2,2∗), and there exists constant ap such that (16)zp≤apz,∀z∈E,p∈2,2∗.
OnE we define the following functional: (17)Φz=12z+2-z-2-Ψz, where Ψ(z)=∫RNH(x,z)dx. It follows from (H3) that, for any ε>0, there is C(ε)>0 such that (18)Hzx,z≤εz+Cεzp-1,(19)Hx,z≤ε2z2+Cεpzp,for all (x,z). Thus, Lemma 5 implies that Φ is well defined on E. Lemma 4 implies that Φ is strongly indefinite; such type functional appeared extensively when one considers differential equations via critical point theory; see, for example, [24–27] and the references therein. Our hypotheses imply that Φ∈C1(E,R) (see Lemma  3.10 in [27]) and a standard argument shows that the critical points of Φ are solutions of HES.


## 3. The Abstract Critical Point Theorem

In order to study the critical points ofΦ, we now recall a abstract critical point theorem developed recently in [10, 20]. Let E be a Banach space with direct sum E=X⊕Y and corresponding projections PX,PY onto X,Y. We assume that the Banach space X is separable and reflexive. Let S⊂X∗ be a dense subset; for each s∈S there is a seminorm on E defined by (20)ps:E⟶R,psz=sx+yforz=x+y∈E. Denote by TS and Tw the topology induced by seminorm family ps and the weak-topology on E, respectively. Tw∗ denotes the weak∗-topology on E∗. Now, some notations and definitions are needed.For a functionalΦ∈C1(E,R), we write Φa=z∈E∣Φ(z)≥a, Φb=z∈E∣Φ(z)≤b, and Φab=Φa∩Φb.Suppose(Φ0)
for anyc∈R, Φc is TS-closed, and Φ′:Φc,TS→(E∗,Tw∗) is continuous;(Φ1)
for anyc>0, there exists ξ>0 such that z<ξPYz for all z∈Φc;(Φ2)
there existsρ>0 such that κ≔inf⁡Φ(Sρ∩Y)>0, where Sρ≔z∈E:z=ρ.Theorem 6 (see [10] or [20]).
Let(Φ0)–(Φ2) be satisfied and suppose there are R>ρ>0 and e∈Y with e=1 such that supΦ(∂Q)≤κ where Q≔{z=x+te:x∈X,t≥0,z≤R}. Then, Φ has a (Cc)-sequence with κ≤c≤c¯≔sup⁡Φ(Q). Moreover, if Φ satisfies the (C)c-condition for all c≤c¯ then Φ has a critical point z with κ≤Φ(z)≤c¯.Lemma 7 (see [10] or [20]).
Let(21)Φz=12y2-x2-Ψz,forz=x+y∈E=X⊕Y. Suppose(1)
Ψ∈C1(E,R) is bounded from below;(2)
Ψ:(E,Tw)→R is sequentially lower semicontinuous; that is, zn⇀z in E implies Ψ(z)≤liminfn→∞⁡Ψ(zn);(3)
Ψ′:E,Tw→E∗,Tw∗ is sequentially continuous;(4)
ν:E→R,ν(z)=z2 is C1 and ν′:(E,Tw)→(E∗,Tw∗) is sequentially continuous.ThenΦ satisfies Φ0.


## 4. The Limit Equation

In this section, we study the following limit equation related toHES, HES∞Az=h∞zz=H∞′z, where h∞ is given in (H2), H∞(z)=∫0zh∞ssds, and H∞′(z)=∇zH∞(z)=(∇uH∞(z),∇vH∞(z)). By virtue of (H3)–(H7), we have firstly the following lemma.Lemma 8.
h
∞ and H∞ possess the following properties.(i)
h∞z=o(1) as z→0 and h∞z≤c11+zp-2 for some p∈(2,2∗).(ii)
H∞(z)/z2→+∞ as |z|→+∞.(iii)
H~∞(z)>0 whenever z≠0, where H~∞(z)=1/2H∞′(z)z-H∞(z).(iv)
H∞′zτ≤c2H~∞(z)zτ whenever |z| large enough.(v)
H~∞(z)→+∞ as z→+∞.Proof.
(i) It is clear by(H3) and (H7).
(ii) By(H4), for any M>0, there is RM>0 such that H(x,z)/|z|2≥M whenever |z|≥RM. Hence (22)H∞zz2=H∞z-Hx,zz2+Hx,zz2≥H∞z-Hx,zz2+M,
for allz≥RM. Observe that (23)0≤Hx,z-H∞z=∫01Hzx,θz-H∞′θzzdθ;it follows from (H7) that H(x,z)→H∞(z) as |x|→+∞. Letting |x|→+∞ we get H∞(z)/z2≥M for all |z|≥RM.
(iii) SinceH(x,z)→H∞(z) as |x|→+∞, it follows from (H5) that (24)H~∞z=limx→+∞⁡H~x,z>0,ifz≠0.
(iv) By(H6), for |z| large enough (25)H∞′zz≤H∞′z-Hzx,zz+Hzx,zz≤H∞′z-Hzx,zz+c21/τH~x,z1/τ.Letting |x|→+∞, we obtain (26)H∞′zz≤c21/τH~∞z1/τ.
(v) It follows from (ii)–(iv) that(27)c2H~∞z≥H∞′zzτ≥2H∞zz2τ⟶+∞,as |z|→+∞.
Now, we set(28)Ψ∞z=∫RNH∞zdx and define the functional (29)Fz=12z+2-z-2-Ψ∞z,∀z=z-+z+∈E. By Lemma 8, for any ε>0, there is C(ε)>0 such that (30)H∞′z≤εz+Cεzp-1,∀x,z∈RN×R2,(31)H∞z≤ε2z2+Cεpzp,∀x,z∈RN×R2.It follows from (30) and (31) that F∈C1(E,R) is well defined and its critical points are solutions of HES∞.Lemma 9.
F possesses the following properties:(1)
Ψ∞ is weakly sequentially lower semicontinuous and F′ is weak sequentially continuous.(2)
There isρ>0 such that κ:=inf⁡F(∂Bρ∩E+)>0, where ∂Bρ≔z∈E:z=ρ.Proof.
(1) Supposezn⇀z in E. Going if necessary to a subsequence, we can assume zn→z in Llocp(RN,R2) for p∈[2,2∗) and zn(x)→z(x) a.e. in RN. Hence H∞(zn(x))→H∞(z(x)) a.e. in RN. Thus (32)Ψ∞z=∫RNH∞zdx=∫RNlimn→∞⁡H∞zndx≤liminfn→∞⁡∫RNH∞zndx=liminfn→∞⁡Ψ∞zn. Next it is sufficient to show that Ψ∞′ is weak sequentially continuous. Indeed, by (30) and zn→z in LlocpRN,R2, 2≤p<2∗, it follows from Theorem  A.2 in [27] that (33)H∞′zn⟶H∞′z,inLlocp/p-1RN,R2. Furthermore, for each fixed φ∈E, one has that, for any ε1>0, there exists r0>0 such that (34)∫RN∖Br00φxpdx1/p<ε1.Hence, for large n, it follows from (30), (33), (34), and Hölder inequality that (35)∫RNH∞′zn-H∞′zφdx≤∫RNH∞′zn-H∞′zφdx≤∫Br00H∞′zn-H∞′zp/p-1dxp-1/p∫Br00φpdx1/p+∫RN∖Br00εzn+z+Cεznp-1+zp-1φdx≤∫Br00H∞′zn-H∞′zp/p-1dxp-1/p∫Br00φpdx1/p+ε∫RN∖Br00φx2dx1/2zn2+z2+ε1Cεznpp-1+zpp-1. Therefore,(36)Ψ∞′zn,φ-Ψ∞′z,φ=∫RNH∞′zn-H∞′zφdx⟶0,asn⟶∞.
(2) For anyz+∈E+, it follows from (31) that (37)Fz+=12z+2-∫RNH∞z+dx≥12z+2-∫RNε2z+2+Cεpz+pdx≥12-Cεz+2-Cz+p. The conclusion follows because p>2.
Now, we choose a numberγ>0 such that maxx∈RN⁡V(x)<γ. From Lemma 8(ii), there is r>0 such that H∞(z)≥γz2 whenever |z|≥r. Let Eλλ∈R be the spectrum family of the operator A. It follows from Lemma 4 that Eγ-F0L2 is a infinite dimension subspace of E+ and (38)V0z22≤μ_z22≤z2≤γz22,∀z∈Eγ-E0L2.We have the following result.Lemma 10.
For any finite dimensional subspaceEW of Eγ-E0L2, (39)Fz⟶-∞,asz⟶∞,z∈E-⊕EW.Proof.
If not, then there areM>0 and zn⊂E-⊕EW with zn→∞ such that F(zn)≥-M for all n. Denote yn≔zn/zn, passing to a subsequence if necessary; we can assume that yn⇀y, yn-⇀y-, and yn+→y+. Then (40)12yn+2-yn-2≥12yn+2-yn-2-∫RNH∞znzn2dx=Fznzn2≥-Mzn2, which yields that (41)12yn-2≤12yn+2+Mzn2.We claim that y+≠0. Indeed, if not then it follows from (41) that yn-→0. Thus yn→0, which contradicts with yn=1. By (38), we get (42)y+2-y-2-2γy22≤γy+22-y-2-2γy+22-2γy-22≤-γy+22-y-2-2γy-22<0. Hence, there exists R>0 such that (43)y+2-y-2-2γ∫BRy2dx<0, where BR≔x∈RN:x≤R. Note that (44)Fznzn2≤12yn+2-yn-2-∫BRH∞znzn2dx=12yn+2-yn-2-2γ∫BRyn2dx-∫BRH∞zn-γzn2zn2dx≤12yn+2-yn-2-2γ∫BRyn2dx-∫BR∩x∈RN:zn≤rH∞zn-γzn2zn2dx≤12yn+2-yn-2-2γ∫BRyn2dx+γr2BRzn2. Hence(45)0≤limn→∞⁡12yn+2-yn-2-∫BRH∞znzn2dx≤12y+2-y-2-2γ∫BRy2dx<0. Now the desired conclusion follows from this contradiction.As a consequence, we have the following.Lemma 11.
Letκ>0 be given by Lemma 9. Then, letting e∈EW with e=1, there is R1>0 such that F|∂Q≤κ, where Q≔{z=z-+se:z-∈E-,s≥0,z≤R1}.Lemma 12.
Letzn be any (C)c-sequence for F. That is, zn⊂E is such that (46)Fzn⟶c,1+znF′zn⟶0,as n→∞. Then it is bounded and c≥0. Moreover, there is a subsequence still denoted by zn satisfying zn⇀z,F(zn-z)→c-F(z) and F′(zn-z)→0, as n→∞.Proof.
Let{zn}⊂E be such that (47)Fzn⟶c,1+znF′zn⟶0. Then, for large n, one has (48)c+o1≥Fzn-12F′zn,zn=∫RN12h∞znzn2-H∞zndx=∫RNH~∞zndx≥0, which implies c≥0. If zn is unbounded in E, up to a subsequence if necessary, we can assume that zn→+∞. Set wn=zn/zn. Then wn=1 and wnp≤apwn=ap for each 2≤p≤2∗. Note that (49)F′zn,zn+-zn-zn2=1-∫RNh∞znznzn+-zn-zn2dx. Hence, one has (50)limn→∞⁡∫RNh∞znznzn+-zn-zn2dx=limn→∞∫RNh∞znwnwn+-wn-dx=1.On the other hand, for r≥0 and 0≤α<β≤+∞, set (51)gr≔inf⁡H~∞z:z∈R2,z≥r,Onα,β≔x∈RN:α≤znx<β,Sαβ≔inf⁡H~∞zz2:z∈R2,α≤z<β.Then, by Lemma 8, we have g(r)>0 for all r>0 and g(r)→+∞ as r→+∞. For large α>0, one has Sαβ>0 and (52)H~∞znx≥Sαβznx2,∀x∈Onα,β. Consequently, for large n and α, whenever α<β one has (53)c+o1≥∫On0,αH~∞zndx+∫Onα,βH~∞zndx+∫Onβ,+∞H~∞zndx≥∫On0,αH~∞zndx+Sαβ∫Onα,βzn2dx+gβOnβ,+∞. Since H~∞(zn)≥0,limβ→+∞⁡g(β)=+∞ implies (54)limβ→+∞⁡Onβ,+∞=0,uniformly  inn. For any s∈(2,2∗), we choose r∈(s,2∗). Using the Hölder inequality we have (55)∫Onβ,+∞wnsdx≤Onβ,+∞r-s/r∫RNwnrdxs/r≤arsOnβ,+∞r-s/r⟶0, as β→+∞ uniformly in n and (56)∫Onα,βwn2dx≤1zn2∫Onα,βzn2dx≤CSαβzn2⟶0, as n→∞. Therefore,(57)∫Onα,βwnsdx≤∫Onα,βwnrdxs-2/r-2∫Onα,βwn2dxr-s/r-2≤C∫Onα,βwn2dxr-s/r-2⟶0, as n→∞. Let 0<ε<1/3 be given. It follows from Lemma 8 that there is α(ε) such that h∞z<ε/a22 for all |z|≤α(ε). Consequently, we have (58)∫On0,αεh∞znwnwn+-wn-dx≤εa22wn22≤ε,for all n. By Lemma 8, we define μ=2τ/τ-1 and μ′=μ/2. By (55), we can take β(ε) so large that (59)∫Onβε,+∞h∞znwnwn+-wn-dx≤∫Onβε,+∞h∞znτdx1/τ∫Onβε,+∞wn+-wn-wnμ′dx1/μ′≤∫Onβε,+∞c2H~∞zndx1/τ∫RNwn+-wn-μdx1/μ∫Onβε,+∞wnμdx1/μ≤ε, for all n. For fixed 0<α(ε)<β(ε), it follows from (57) that there is n0 such that (60)∫Onαε,βεh∞znwnwn+-wn-dx≤C∫Onαε,βεwn2dx≤ε,∀n≥n0.Now the combination of (58)–(60) implies that for n≥n0(61)∫RNh∞znwnwn+-wn-dx≤3ε<1. This contradicts with (50). Hence, zn is bounded. Passing to a subsequence if necessary, we can assume that zn⇀z in E, zn→z in Llocp(RN,R2) for p∈[2,2∗), and zn(x)→z(x) a.e. in RN. Since HES∞ is an autonomous system, by (V1) and Lemma 8, we know that all of the conditions of Lemma  5.7 in [20] are satisfied. Hence we have F(zn-z)→c-F(z) and F′(zn-z)→0, as n→∞.Lemma 13.
LetK^≔z∈E:F′z=0 be the critical set of F. Then K^∖{0}≠∅ and C^≔inf⁡Fz:z∈K^∖0>0 is attained.Proof.
SetX=E- and Y=E+. Then E=X⊕Y. For any c>0 and z∈Fc, using the fact that Ψ∞≥0 one has (62)0<c≤12z+2-z-2. This yields z-<z+, and hence z≤2z+. F satisfies (Φ1). By virtue of Lemma 7 and conclusion (1) of Lemma 9, it follows that (Φ0) holds. From conclusion  (2) of Lemma 9, we know that (Φ2) holds. Thus, combining with Lemma 11, we know that all of the conditions of Theorem 6 hold. Then there is (C)c sequence zn for F with κ≤c≤sup⁡F(Q). By Lemma 12, (C)c sequence zn is bounded in E. For the concentration functions zn2, there are only two cases needed to be considered: vanishing and nonvanishing. If vanishing occurs, by the vanishing lemma(63)∫RNh∞znzn2dx=o1,∫RNH∞zndx=o1. Hence (64)o1=F′zn,zn=zn+2-zn-2-∫RNh∞znzn2dx, which implies that zn+2-zn-2=o(1). Hence F(zn)→0 as n→∞. This contradicts with F(zn)→c≥κ>0 as n→∞. So nonvanishing occurs; that is, there exist α>0, R<∞ and {an}⊂RN such that (65)liminfn→∞⁡∫Ban,Rzn2dx≥α. Choose gn∈ZN such that gn-an=min⁡g-an:g∈ZN. Thus gn-an≤1/2N. Setting z~n(x)=gn∗zn, by the invariance under translation of F, {z~n} is a (C)c-sequence of F and z~n⇀z~. From (66)liminfn→∞∫B0,R+1/2Nz~n2dx≥α>0, we see that z~≠0, and hence z~ is a nontrivial critical point of F. Therefore (67)K^∖0≠∅.If z∈K^, one has (68)Fz=Fz-12F′z,z=∫RN12h∞zz2-H∞zdx=∫RNH~∞zdx≥0, which implies C^≥0. If C^=0, let zn∈K^∖{0} be such that F(zn)→0 as n→∞. Then {zn} is (C)0-sequence. By Lemma 12, we can assume that zn⇀z∈E. Then (69)Fzn=Fzn-12F′zn,zn=∫RN12h∞znzn2-H∞zndx=∫RNH~∞zndx⟶0. By Lemma 8(i), we see that, for any ε>0, there is 1>r=r(ε)>0 such that (70)h∞z≤ε,wheneverz<r. It follows from Lemma 8(iv) that there exists constant R>r>0 such that (71)h∞z≤c21/τH~∞z1/τ,wheneverz>R. By the continuity of H~∞(z), there exists M0=M0(r,R)>0 such that (72)H~∞z≥M0,wheneverr≤z≤R. Note that h∞(|z|)≤c1(1+zp-2) for some p∈(2,2∗). Hence, there exists M1=M1(R) such that for all r≤z≤R(73)h∞z≤M1≤1H~∞z1/τM1H~∞z1/τ≤M1M0-1/τH~∞z1/τ. Denote Cε=c21/τ+M1M0-1/τ; we obtain(74)h∞z≤ε+CεH~∞z1/τ,∀z∈R2.By Hölder inequality (1/τ+1/τ′=1), we have (75)zn2=∫RNh∞znznzn+-zn-dx≤εzn22+Cε∫RNzn+-zn-znH~∞zn1/τdx≤εzn22+Cε∫RNH~∞zndx1/τ∫RNzn2τ′dx1/τ′≤εa22zn2+a2τ′2Cε∫RNH~∞zndx1/τzn2. Hence 1≤εa22+o(1), a contradiction. Hence C^>0. Finally, we show that there is z∈K^ with F(z)=C^>0. Let zn∈K^∖{0} be such that F(zn)→C^. Then, as before, zn is bounded, and one may assume zn⇀z∈K^∖0. Now (76)C^=limn→∞⁡Fzn=limn→∞⁡∫RNH~∞zndx≥∫RNH~∞zdx=Fz≥C^; that is, F(z)=C^. This completes the proof.Definition 14 (see Ackermann [28]).
A mappingf from Banach space X to another Banach space Y is called BL-split, if for every weakly convergent sequence xn⊂X with xn⇀x it holds that f(xn)-f(xn-x)→f(x) in Y.
In what follows, we use the idea of [21, 28]. For fixed w∈E+, we introduce the functional ϕw:E-→R by (77)ϕwy≔Fw+y=12w2-y2-Ψ∞w+y. Hence one has (78)ϕw′′yz-,z-=-z-2-Ψ∞′′w+yz-,z-=-z-2-∫RNh∞′w+yw+yw+yz-2dx-∫RNh∞w+yz-2dx, for all y,z-∈E-, which implies that ϕw(·) is strictly concave. Moreover (79)ϕwy≤12w2-y2, which implies that ϕw(y)→-∞ as y→∞. Now, it follows from Lemma 9 that ϕw is weakly sequentially upper semicontinuous. Hence, there is a unique strict maximum point ξ(w) for ϕw(·), which is also the only critical point of ϕw on E- and satisfies (80)y≠ξw⟺Fw+y<Fw+ξw,(81)ϕw′ξw,z-=0,i.e.,ξw,z-+Ψ∞′w+ξw,z-=0,for all w∈E+, y, z-∈E-. Now, we define the reduced functional R:E+→R by (82)Rw≔Fw+ξw=12w2-12ξw2-Ψ∞w+ξw. We have the following lemma.Lemma 15.
w
∈
E
+ is a critical point of R if and only if w+ξ(w) is a critical point of F. Moreover, the following conclusions hold:(1)
ξ∈C1E+,E- and ξ(0)=0.(2)
ξ is a bounded map.(3)
ξ, ξ(w)2, ξw22 are all BL-splits.(4)
ξ(wn)⇀ξ(w) in E- if wn⇀w in E+.(5)
ξ is ZN-invariant, i.e., ξ(a∗w)=ξ(w) for all a∈ZN.Proof.
It follows from(H7) that H∞∈C2RN×R2,0,∞ and (83)H∞′′zz1·z1=h∞′zzz1+h∞zz1·z1≥0,∀z,z1∈R2. Hence, combining with (V1), (B0), and (B1), we know that all of the conditions of Lemma  2.6 in [19] hold. So, by Lemma  2.6 in [19], the desired conclusions can be obtained.
By Lemma15, the critical points of R and F are in one-to-one correspondence via the injective map w→w+ξ(w) from E+ into E. Consequently, let (84)K^+≔w∈E+:R′w=0. Then (85)K^=w+ξw:w∈K^+. In particular(86)C^=inf⁡Rw:w∈K^+.Lemma 16.
The sequence{wn}⊂E+ is a bounded (C)c-sequence of R if and only if wn+ξwn⊂E is a bounded (C)c-sequence of F.Proof.
Let{wn}⊂E+ be a (C)c sequence of R; that is, R(wn)→c and 1+wnR′(wn)→0. Since H∞≥0, we have (87)ξwn2=wn2-2∫RNH∞wn+ξwndx-2c+o1≤wn+C2. By (81) and Lemma 15, for all z=z++z-∈E, we have ξ′(wn)z+∈E- and (88)R′wn,z+=wn,z+-ξwn,ξ′wnz+-∫RNH∞′wn+ξwnz++ξ′wnz+dx=wn,z+-∫RNH∞′wn+ξwnz+dx=wn,z+-ξwn,z--∫RNH∞′wn+ξwnz++z-dx=F′wn+ξwn,z. Hence, one has (89)R′wn=supz+≤1⁡R′wn,z+=supz+≤1⁡F′wn+ξwn,z+≤supz+≤1⁡F′wn+ξwn·z+≤F′wn+ξwn. Similarly(90)F′wn+ξwn=supz≤1⁡F′wn+ξwn,z≤supz+≤1⁡R′wn,z+≤R′wn. Consequently(91)R′wn=F′wn+ξwn. Hence (92)ξwnF′wn+ξwn≤wn+CF′wn+ξwn=wn+CR′wn⟶0,1+wn+ξwnF′wn+ξwn≤1+wn+ξwnF′wn+ξwn⟶0.That is, wn+ξwn is (C)c sequence of F. The inverse is obvious. By Lemma 15, the boundedness of ξ yields the equivalence between the boundedness of wn and wn+ξwn. This completes the proof.Next, we discuss the mountain pass geometry of the reduce functionalR. One has the following Lemma.Lemma 17.
R possesses the mountain pass geometry:(1)
There isρ>0 such that inf⁡R(E+∩∂Bρ)>0, where ∂Bρ=z∈E:z=ρ.(2)
There is somee∈E+∖B¯ρ(0) such that R(e)≤0.Proof.
(
1
) By (31) and (80), for any w∈E+, we have (93)Rw=12w2-ξw2-Ψ∞w+ξw=12w2+Fw+ξw-Fw-Ψ∞w≥12w2-Ψ∞w≥12w2-ε2w22-Cεpwpp≥12-Cεw2-Cwp.Hence, for small ρ>0, conclusion (1) holds.
(
2
) Similar to the proof of Lemma 10, we can obtain that (94)Rw⟶-∞,asw⟶∞,w∈EW.Hence conclusion (2) holds.Remark 18.
Lemma17 implies that 0 is an isolated critical point of R. Therefore there is a ν>0 such that w≥ν for all w∈KR, where KR≔w∈E+∣w≠0andR′(w)=0.Remark 19.
If we set(95)N+≔w∈E+∖0:R′w,w=0, then, by Lemma  4.7 in [18], we have that,. for each w∈E+∖{0}, there is a unique sw>0 such that sww∈N+ and Rsww=sups≥0⁡R(sw).Lemma 20.
Letw∈K^+ be such that R(w)=C^ and set Ew=E-⊕R+w. Then (96)supz∈Ew⁡Fz≤Rw=C^.Proof.
For anyz=z-+sw∈Ew, by (80), we obtain (97)Fz≤Fsw+ξsw=Rsw. Since w∈K^+, we have w∈N+. Hence, combining with Remark 19, we obtain R(w)=sups≥0⁡R(sw). Consequently, one has (98)supz∈Ew⁡Fz≤sups≥0⁡Rsw=Rw=C^.


## 5. Proof of the Main Result

In this section we give the proof of Theorem1. Let K≔z∈E:Φ′z=0 be the set of critical points for Φ. First, we study the linking structure for the functional Φ. Similar to the proof of Lemma 9, we have the following three lemmas.Lemma 21.
Ψ is nonnegative and weakly sequentially lower semicontinuous.Lemma 22.
Φ
′ is weakly sequentially continuous.Lemma 23.
There existsr>0 such that κ1≔inf⁡Φ(Sr∩E+)>0, where Sr≔z∈E:z=r.Lemma 24.
There isρ>r>0 such that for any e∈E+ with e=1 and Ee=E-⊕R+e(99)Φz<0,∀z∈Ee∖Bρ.Proof.
For anyz∈Ee, by (H7), we have(100)Φz=12z+2-z-2-∫RNHx,zdx≤12z+2-z-2-∫RNH∞zdx=Fz. Thus, the conclusion follows easily from Lemma 10.
In particular, setz0+∈K^+ with R(z0+)=C^, e=z0+, and Q=z=z-+sz0+:z-∈E-,s≥0,z≤ρ, where ρ is given in Lemma 24. Then we have the following lemma.Lemma 25.
d
=
sup
⁡
Φ
z
:
z
∈
E
e
=
sup
⁡
Φ
(
Q
)
<
C
^.Proof.
By Lemmas23 and 24, we have d=sup⁡Φ(Q)≥κ1. Since Φ is weakly upper semicontinuous on Ee, there is some z1∈Q with z1≠0 such that sup⁡Φ(Q)=Φ(z1). From (H7) and Lemma 20, we obtain (101)d=sup⁡Φz:z∈Ee=sup⁡ΦQ=Φz1<Fz1≤sup⁡FQ≤Rz0+=C^.In what follows, we discuss(C)c sequences of Φ. Firstly we have the following.Lemma 26.
Any(C)c-sequence for Φ is bounded.Proof.
The proof is similar to Lemma12. We omit the details here.Lemma 27 (see [23, TheoremA.2]).
LetΩ be an open set in RN and f∈C(Ω×R,R) be a function such that fx,u≤a(ur+us) for some a>0 and 1≤r<s<∞. Suppose s≤p<∞,r≤q<∞,q>1, un is a bounded sequence in Lp(Ω)∩Lq(Ω), and un→ua.e.inΩ and in LpΩ∩Bρ∩LqΩ∩Bρ for all ρ>0. Then, passing to a subsequence, there exists a sequence vn such that vn→u in Lp(Ω)∩Lq(Ω) and (102)fx,un-fx,un-vn-fx,u⟶0inLq/rΩ+Lp/sΩ, where vn(x)=χ(2x/rn)u(x) and χ∈C∞(R,[0,1]) is such that χ(t)=1 for t≤1, χ(t)=0 for t≥2, rn>0 is a sequence of constants with rn→∞ as n→∞, the space Lp(Ω)∩Lq(Ω) with the norm (103)up∧q≔up+uq, and the space Lp(Ω)+Lq(Ω) with the norm (104)up∨q≔inf⁡vp+wq:v∈LpΩ,w∈LqΩ,u=v+w.Lemma 28.
Let{zn}⊂E be (C)c-sequence of Φ. Then, passing to a subsequence if necessary, we can assume that zn⇀z0 in E. Furthermore, either (i)
zn→z0∈E or(ii)
c≥C^ and there exist a positive integer l≤[c/C^], y1,…,yl∈K^∖{0} and sequences {gni}⊂ZN, i=1,2,…,l, such that, after extraction of a subsequence of zn(105)zn-z~n-∑i=1lgni∗yi⟶0,Φz0+∑i=1lFyi=c,where z~n(x)=η(2|x|/n)z0(x) and η∈C∞(R,[0,1]) is a cut-off function such that η(t)=1 for t≤1, η(t)=0 for t≥2.Proof.
First, Lemma26 implies that any (C)c sequence of Φ is bounded; hence it is a bounded (PS)c sequence. Passing to a subsequence if necessary, we can assume that zn⇀z0∈E with Φ′(z0)=0, zn(x)→z0(x) a.e. in RN and zn→z0 in Lloc(RN,R2). Let η∈C∞(R,[0,1]) be a cut-off function such that η(t)=1 for t≤1, η(t)=0 for t≥2 and define z~n(x)=η(2|x|/n)z0(x); then z~n→z0 in E. Indeed, by Lemma 5, z0∈E implies that for any ε>0 there is a corresponding ρ=ρ(ε)>0 such that (106)∫RN∖Bρ0z02dx≤ε,∫RN∖Bρ0∇z02dx≤ε,(107)z~n-z02≤Cz~n-z0H1RN,R22≤C∫RN∇z~n-z02dx+C∫RNz~n-z02dx=C∫RN∇η2xnz0-z02dx+C∫RNη2xnz0-z02dx≤C∫RNη2xn-12∇z02dx+4Cn2∫RNη′2xn2z02dx+C∫RNη2xn-12z02dx≤C∫Bρ0η2xn-12∇z02dx+4Cn2∫RNη′2xn2z02dx+C∫Bρ0η2xn-12z02dx+Cε.Hence, using the Lebesgue dominated convergence theorem, we obtain (108)z~n-z0⟶0,asn⟶∞. Now we set zn1=zn-z~n. Then, by zn⇀z0, one has zn1⇀0. Moreover, we claim that zn1 is a bounded (PS)c1-sequence for F with c1=c-Φ(z0). In fact, by (H7), for any ε>0, there is a ρ1>0 such that (109)hx,s-h∞s≤ε,wheneverx>ρ1,uniformly in bounded set of s. On the one hand, by (19), taking r=2, s=p, and q=2 in Lemma 27, we know that (110)∫RNHx,zndx=∫RNHx,zn1dx+∫RNHx,z0dx+o1.On the other hand, by (18), taking r=1, s=p-1 and q=2. It follows from Lemma 27 that (111)gn⟶0,inL2RN,R2+Lp/p-1RN,R2, where (112)gnx≔Hzx,znx-Hzx,zn1x-Hzx,z0x. Hence, for each φ∈E with φ=1, one has (113)∫RNHzx,zn-Hzx,zn1-Hzx,z0φdx≤gn2∨p′φ2∧p≤Cgn2∨p′, where p′=p/p-1 and the constant C>0 is independent of φ. Consequently(114)supφ=1⁡∫RNHzx,zn-Hzx,zn1-Hzx,z0φdx=o1.For any z∈E, observe that (115)Φz=12Az,z2-∫RNHx,zdx,Azn=Azn1+Az0+o1.Hence, it follows from (110) and (114) that (116)Φzn=Φzn1+Φz0+o1,Φ′zn=Φ′zn1+Φ′z0+o1, and then (117)Φzn1⟶c-Φz0,Φ′zn1⟶0,as n→∞. Consequently, by (19), (31), and (117), for large n, we obtain (118)c-Φz0+o1=Φzn1≤Fzn1=Φzn1+∫RNHx,zn1dx-∫RNH∞zn1dx≤c-Φz0+o1+∫x≤ρ1Hx,zn1-H∞zn1dx+∫x>ρ1Hx,zn1-H∞zn1dx≤c-Φz0+o1+∫x≤ρ1Hx,zn1-H∞zn1dx+∫x>ρ1∫01hx,θzn1-h∞θzn1zn12dθdx≤c-Φz0+o1+∫x≤ρ1Hx,zn1-H∞zn1dx+ε∫x>ρ1zn12dx≤c-Φz0+o1+C∫x≤ρ1zn12+zn1pdx+ε∫x>ρ1zn12dx. Consequently(119)Fzn1⟶c-Φz0. In what follows, we claim that (120)Ψ′zn1-Ψ∞′zn1⟶0,asn⟶∞.Indeed, since zn1→0 in Llocp(RN) for 2≤p<2∗, by (18), (30), and Theorem  A.2 in [27], we have (121)Hzx,zn1⟶0,inLlocp/p-1RN,R2,H∞′zn1⟶0,inLlocp/p-1RN,R2.Hence, for any w∈E with w=1, it follows from (109) and (121) that (122)Ψ′zn1-Ψ∞′zn1,w=∫RNHzx,zn1-H∞′x,zn1wdx=∫x≤ρ1Hzx,zn1-H∞′zn1wdx+∫x>ρ1Hzx,zn1-H∞′zn1wdx≤∫x≤ρ1Hzx,zn1p/p-1dxp-1/p+∫x≤ρ1H∞′zn1p/p-1dxp-1/p∫x≤ρ1wpdx1/p+∫x>ρ1hzx,zn1-h∞zn1zn1wdx≤∫x≤ρ1Hzx,zn1p/p-1dxp-1/p+∫x≤ρ1H∞′zn1p/p-1dxp-1/p∫x≤ρ1wpdx1/p+εzn1wL2RN,R2wL2RN,R2. Hence, we obtain (123)F′zn1=Φ′zn1+Ψ′zn1-Ψ∞′zn1⟶0. In addition, we have (124)c1+o1=Fzn1-12F′zn1,zn1=∫RNH~∞zn1dx≥0, and c1=0 if and only if zn1→0 in E. In fact, if c1=0, then (125)∫RNH~∞zn1dx⟶0. Hence, using (74) and Hölder inequality (1/τ+1/τ′=1), we obtain (126)zn12=∫RNh∞zn1zn1zn1+-zn1-dx+o1≤εzn122+Cε∫RNzn12H~∞zn11/τdx+o1≤εzn122+Cε∫RNH~∞zn1dx1/τ∫RNzn12τ′dx1/τ′+o1≤εa22zn12+a2τ′2Cε∫RNH~∞zn1dx1/τzn12+o1. This implies that zn1→0 in E. The inverse is obvious. Now, assume that conclusion (i) is false. Then c1>0. For the concentration functions |zn1|2, there are only two cases needed to be considered: vanishing and nonvanishing. If vanishing occurs, by the vanishing lemma and (30)-(31), one has (127)∫RNh∞zn1zn12dx=o1,∫RNH∞zn1dx=o1. Hence (128)c1=limn→∞⁡∫RNH~∞zn1dx=limn→∞⁡∫RN12h∞zn1zn12-H∞zn1dx=0. This is a contradiction. Thus, nonvanishing occurs; that is, there is a sequence an1⊂RN and constants r1, δ1>0 such that (129)zn1L2Ban1,r1≥δ12, for large n. We may choose kn1∈ZN and z¯n1≔kn1∗zn1 such that, passing to a subsequence,(130)z¯n1L2B0,r1+1/2N≥δ12, for all n. Note that F(z¯n1)=F(zn1), ∇F(z¯n1)=∇F(zn1), and z¯n1=zn1. We know that z¯n1 is bounded PSc1-sequence of F. Hence, by Lemma 12, passing to a subsequence if necessary, we can assume that z¯n1⇀z¯1∈K^∖{0},(131)Fwn1⟶c-Φz0-Fz¯1,F′wn1⟶0,as n→∞, where wn1=z¯n1-z¯1. Note that (132)c-Φz0-Fz¯1+o1=Fwn1-12F′wn1,wn1=∫RNH~∞wn1dx≥0. Hence (133)c-Φz0-Fz¯1≥0.Since Φ(z0)≥0 and F(z¯1)≥C^, we obtain that c≥C^ and C^≤F(z¯1)≤c-Φ(z0). There are now two possibilities to consider: F(z¯1)<c-Φ(z0) or F(z¯1)=c-Φ(z0).
Setc1=c-Φ(z0). If c2:=c1-F(z¯1)=0, repeat the above arguments but replace {zn1} and c1 by {wn1} and c2, respectively. We obtain that c2=0 if and only if wn1→0. Consequently, we obtain that wn1→0. Hence the lemma holds with l=1, y1=k1∗z¯1 (where k1∈ZN is chosen to ensure that y1∈K^∖{0}) and gn1=-kn1-k1.
Ifc2>0, then we argue again as above with {zn1} and c1 replaced by {wn} and c2, respectively, and we obtain z¯2∈K^∖{0} with C^≤F(z¯2)≤c2. After at most [c/C^] steps, we obtain the desired conclusion.As a straight consequence of Lemma28, we have the following.Lemma 29.
Φ satisfies Cc condition for all c<C^.We are now in a position to complete the proof of Theorem1.Lemma 30.
Φ satisfies (Φ1).Proof.
For anyc>0 and z∈Φc, using the definition of Φ and H(x,z)≥0, one has (134)0<c≤12z+2-z-2. This yields z-<z+, and hence z<2z+. Φ satisfies (Φ1).Proof of Theorem1.
By Lemma30, Φ satisfies (Φ1). It follows from Lemmas 7, 21, and 22 that (Φ0) holds. By Lemma 23, we know that (Φ2) holds. Lemma 24 shows that the linking condition of Theorem 6 holds. These, together with Lemma 25, yield (C)c-sequence {zn} with κ1≤c≤sup⁡Φ(Q)<C^ for Φ. By virtue of Lemma 29, we can assume that zn→z as n→∞. Furthermore, we have Φ′(z)=0 and Φ(z)≥κ1. This implies that K∖{0}≠∅. The proof is completed.


## 6. Summary

The purpose of this paper is to investigate the existence of nontrivial solutions for a class of Hamiltonian elliptic system inRN in the case that the nonlinearity may not satisfy the standard Ambrosetti-Rabinowitz condition (H0). We obtain rather general conditions for the existence of nontrivial solutions, which extend and improve some recent results in the literature. In this paper, since we have no assumption of periodic about nonlinearity, the problem is more difficult and interesting. We will point out that if the following condition is satisfied:(HP)
H(x,z)is1-periodic  inxifori=1,…,N,then for any 0≠a=(a1,…,aN)∈ZN, we have Φ(a∗z)=Φ(z), ∇Φ(a∗z)=∇Φ(z), and a∗z=z. Hence, combining with Lemmas 7, 21, 22, 23, and 30, similar to the proof of Lemmas 11 and 13, we have the following.Theorem 31.
K
∖
{
0
}
≠
∅ and inf⁡Φz:z∈K∖0>0 is attained; that is, the problem HES has a ground state solution.



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*Source: 9125486-2017-03-28.xml*