Research article

Schrödinger-Poisson system without growth and the Ambrosetti-Rabinowitz conditions

  • Received: 02 November 2019 Accepted: 08 January 2020 Published: 20 January 2020
  • MSC : 35B45, 35J20, 35J50

  • We consider the following Schrödinger-Poisson system {Δu+V(x)u+ϕu=λf(u)inR3,Δϕ=u2,lim|x|+ϕ=0,inR3. Unlike most other papers on this problem, the Schrödinger-Poisson system without any growth and Ambrosetti-Rabinowitz condition is considered in this paper. Firstly, by Jeanjean's monotonicity trick and the mountain pass theorem, we prove that the problem possesses a positive solution for large value of λ. Secondly, we establish the multiplicity of solutions via the symmetric mountain pass theorem.

    Citation: Chen Huang, Gao Jia. Schrödinger-Poisson system without growth and the Ambrosetti-Rabinowitz conditions[J]. AIMS Mathematics, 2020, 5(2): 1319-1332. doi: 10.3934/math.2020090

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  • We consider the following Schrödinger-Poisson system {Δu+V(x)u+ϕu=λf(u)inR3,Δϕ=u2,lim|x|+ϕ=0,inR3. Unlike most other papers on this problem, the Schrödinger-Poisson system without any growth and Ambrosetti-Rabinowitz condition is considered in this paper. Firstly, by Jeanjean's monotonicity trick and the mountain pass theorem, we prove that the problem possesses a positive solution for large value of λ. Secondly, we establish the multiplicity of solutions via the symmetric mountain pass theorem.


    In this paper, we are interested in finding nontrivial solutions to the following one-parameter family of Schrödinger-Poisson system:

    {Δu+V(x)u+ϕu=λf(u),in R3,Δϕ=u2, lim|x|+ϕ(x)=0,in R3. (1.1)

    In recent years, system (1.1) has been studied widely due to the fact that it arises in several physical phenomena (see [3,4,5,9,18]). From the viewpoint of quantum mechanics, this system describes a charged wave interacting with its own electrostatic field in the case that magnetic effects could be ignored. The terms u and ϕ describe the wave functions associated to the particle and electric potential. The term ϕu is nonlocal and concerns the interactions with electric field. The nonlinearity models the interaction between the particles and external nonlinear perturbations.

    There has been a lot of contributions about the following Schrödinger-Poisson system

    {Δu+V(x)u+ϕu=f(x,u),in R3,Δϕ=u2, lim|x|+ϕ(x)=0,in R3, (1.2)

    which was first introduced in [4]. The case V(x)=1 and f(x,u)=|u|p2u, 2<p<6, has been studied in [7], where Ruiz gave existence and nonexistence results. The existence of a ground state solution of (1.2) with f(x,u)=|u|p2u and 3<p<6 was proved by Azzollini [1]. For the general nonlinearity f and the potential V(x), in [6,12,13,15,20,21], the authors studied the existence and multiplicity of nontrivial solutions for the Schrödinger-Poisson system with superlinear and subcritical growth condition. The following global Ambrosetti-Rabinowitz type condition plays a crucial role in the above mentioned papers:

    0<F(u):=u0f(s)ds1γuf(u), (AR)

    where γ>4. Since the nonlocal term R3ϕuu2 in the energy functional of (1.2) is homogeneous of degree 4, if γ>4 from (A-R) then Ambrosetti-Rabinowitz condition guarantees boundedness of Palais-Smale sequences as well as existence of a mountain pass geometry.

    It is very natural for us to pose the question: Can we replace (A-R) with a weaker condition? When V(x) is periodic or asymptotically periodic and f(u) does not satisfy the Ambrosetti-Rabinowitz condition, Alves, Souto and Soares [2] established the existence of positive ground state solutions by using the mountain pass theorem. In [14] Mao et al. studied the following Schrödinger-Poisson system of the form

    {Δu+V(x)u+εϕu=λf(u),in R3,Δϕ=u2, lim|x|+ϕ(x)=0,in R3,u>0,in R3,

    where f satisfies 0<4F(s)sf(s), for s>0 is small. Under the conditions that ε is small and λ is large, the authors proved the existence of a positive solution. Differently from the above-mentioned results, the purpose of this paper is to present some existence and multiplicity results of solutions of problem (1.1) under the nonlinearity f(t) which possesses only conditions in a neighborhood of the origin. More importantly, we consider the case that f satisfies 0<γF(t)tf(t) where γ(3,4], for t>0 is small. To the best of our knowledge, there are less results in the literatures on the case γ(3,4).

    Firstly, we study problem (1.1) under the following conditions:

    (V1): VC(R3,R), 0<VLV(x) for all xR3 and V(x) is coercive, i.e., lim|x|V(x)=;

    (V2): VC1(R3,R) and 2V(x)+V(x)x0 for a.e. xR3 and V(x)xLr(R3) for some r[32,].

    (f0): f(t)=0 for t0;

    (f1): there exists α(4,2) such that lim supt0+f(t)ttα<+, where 2=6;

    (f2): there exists β(4,2) such that lim inft0+F(t)tβ>0;

    (f3): there exists 3<γ4 such that 0<γF(t)tf(t), for t>0 is small;

    (f1): there exists α(4,2) such that lim supt0f(t)t|t|α<+;

    (f2): there exists β(4,2) such that lim inft0F(t)|t|β>0, where F(t)=t0f(s)ds;

    (f3): there exists ˜γ>4 such that 0<˜γF(t)tf(t), for |t| small and t0;

    (f4): f(t)=f(t), for |t| small.

    Next, we give our main results.

    Theorem 1.1. Assume (V1), (V2) and (f0)(f3) hold. Then the problem (1.1) has a positive solution uX (X is defined in Sect.2) for all sufficiently large λ.

    Remark 1.1. (V2) is used to obtain a special bounded Palais-Smale sequence with Jeanjean's monotonicity trick. That V(x)xLr(R3) for some r[32,] plays an important role in deriving the Pohozaev identity for the weak solutions of (2.2).

    To prove Theorem 1.1, we are faced with several difficulties. On one hand, due to the nonlinearity f without any growth condition at infinity, the natural variational functional associated to (1.1) may be not well defined. Inspired by work of Costa [8] and Huang [10], we modify f(t) to a new well-defined nonlinearity. Furthermore by Moser iteration, we shall show that for large λ, the solutions of the modified problem are the solutions of the original problem.

    On the other hand, different from [8] and [14], since we don't assume the global Ambrosetti-Rabinowitz condition about f(t), the boundedness of Palais-Smale sequence seems hard to verify. We use an argument developed by Jeanjean [11] to overcome this difficulty. Then Pohozaev type identity [19] and the condition (V2) are used to construct a special bounded Palais-Smale sequence for the modified functional Jλ (will be defined in Section 2).

    Theorem 1.2. Assume (V1), (f1)(f3) and (f4) hold. Then for any given positive integer k1 the problem (1.1) has k pairs of solutions ±uiX(i=1,2,...,k) for all sufficiently large λ.

    The key to prove Theorem 1.2 is a priori estimate of the weak solution for the modified problem. Firstly, we modify f to a new nonlinearity h which is odd and satisfies the Ambrosetti-Rabinowitz condition (see Lemma 2.1). By symmetric mountain pass theorem, the modified problem has a sequence of weak solutions. Secondly, it will be shown that the solutions converge to zero in L-norm as λ. Therefore, for λ large, they are solutions of the original problem.

    Remark 1.2. It is evident that the following function satisfies hypotheses (f1)(f3) and (f4):

    f(t)=C1|t|α2t+C2|t|q2t,

    where 4<α<2<q< and C1, C2 are positive constants.

    This paper is organized as follows. In Section 2, we describe the related mathematical tools and give the proof of Theorem 1.1. Theorem 1.2 is proved in Section 3.

    In what follows, C and Ci will denote positive generic constants.

    As usual, the norm of Ls(RN) (s1) is denoted by ||s. Define

    X={uD1,2(R3): R3V(x)u2dx<}

    endowed with the following norm

    u=(RN(|u|2+V(x)u2)dx)1/2.

    By (V1), it well known that XLp(R3) continuously for p[2,6], compactly for p[2,6).

    By the conditions (f0), (f1) and (f2), there exist positive constants δ(0,12), C3 and C4 such that

    F(t)C3tα and F(t)C4tβ for 0t2δ. (2.1)

    For the fixed δ>0, we now consider d(t)C1(R,R) is a cut-off function satisfying

    d(t)={1, if tδ,0, if t2δ,

    |d(t)|2δ and 0d(t)1 for t[δ,2δ]. Define G(t)=d(t)F(t)+(1d(t))F(t), where

    F(t)={C3|t|α, if t>0,0, if t0.

    Set g(t)=G(t). We observe that the conditions (f0)(f3) imply some properties of g(t).

    Lemma 2.1. (1) gC(R,R), g(t)=0, for all t0 and g(t)=o(1) as t0+;

    (2) limt+g(t)t3=+;

    (3) there exists C5>0 such that g(t)C5tα1, for all t0;

    (4) for any T>0, there exists a constant C(T)>0 such that G(t)C(T)tβ for all t[0,T];

    (5) for all t>0, we have 0<γG(t)tg(t).

    By [17], for every uH1(R3), there exists a unique ϕuD1,2(R3) such that

    Δϕu=u2

    and

    R3ϕuvdx=R3u2vdx, for all vD1,2(R3).

    It has the following properties:

    Lemma 2.2. For any uXH1(R3), we have

    (1) ϕu0;

    (2) ϕtu=t2ϕu;

    (3) ϕu2D1,2=R3ϕuu2dxC6|u|412/5C7u4, where C6, C7 are constants.

    We now consider the modified equation of (1.1) given by

    {Δu+V(x)u+ϕu=λg(u),in R3,Δϕ=u2, lim|x|+ϕ(x)=0,in R3. (2.2)

    By definition of G and Lemma 2.1, for uX, the functional associated to (2.2) given by

    Jλ(u)=12R3(|u|2+V(x)u2)dx+14R3ϕu(x)u2dxλR3G(u)dx (2.3)

    is well-defined.

    Noticing that we can not ensure that the modified nonlinearity g satisfies the classical Ambrosetti-Rabinowitz condition, the boundedness of Palais-Smale sequence seems hard to prove. The following abstract result [11] is used to construct a special Palais-Smale sequence.

    Proposition 1. Let X be a Banach space equipped with a norm X and let JR+ be an interval. {Φμ}μJ are C1-functionals on X of the form

    Φμ(u)=A(u)μB(u), for all μJ,

    where B(u)0 for all uX and either A(u)+ or B(u)+ as uX+. Suppose that there exist two points u1, u2X such that

    cμ=infγΓmaxt[0,1]Φμ(γ(t))>max{Φμ(u1),Φμ(u2)}, for all μJ,

    where Γ={γC([0,1],X): γ(0)=u1, γ(1)=u2}. Then, for almost every μJ, there exists a sequence {un(μ)}X such that

    (1) {un(μ)} is bounded in X,

    (2) Φμ(un(μ))cμ,

    (3) Φμ(un(μ))0, in X, where X is dual space of X.

    Furthermore, the map μcμ is continuous from the left.

    Consider a family of functionals

    Jμ,λ(u):=12R3(|u|2+V(x)u2)dx+14R3ϕu(x)u2dxμR3λG(u)dx,uX. (2.4)

    Denote A(u)=12R3(|u|2+V(x)u2)dx+14R3ϕu(x)u2dx, B(u)=R3λG(u)dx and J=[12,1]. Then Jμ,λ(u)=A(u)μB(u). The next lemma ensures that Jμ,λ satisfies all assumptions of Proposition 1.

    Lemma 2.3. Assume that (V1) and (f0)(f2) hold. For all uX, then

    (1) B(u)0;

    (2) A(u) as u;

    (3) there exists u0X, independent of μ, such that Jμ,λ(u0)<0 for all μ[12,1];

    (4) for all μ[12,1], it holds

    cμ=infγΓmaxt[0,1]Jμ,λ(γ(t))>max{Jμ,λ(γ(0)),Jμ,λ(γ(1))},

    where Γ={γC([0,1],X): γ(0)=0, γ(1)=u0}.

    Proof. From Lemma 2.1-(1) and (V1), (1) and (2) are proved directly. To prove (3), let us fix some nonnegative radially symmetric function e(x)C0(R3){0}. Then, for t>0, we have

    J1/2,λ(te)=t22R3|e|2dx+t22R3V(x)e2dx+t44R3ϕe(x)e2dx12R3λG(te)dxt42(1t2e2+12R3ϕe(x)e2dxR3λG(te)t4dx). (2.5)

    By Lemma 2.1-(2), it is easy to see that J1/2,λ(te)<0 for t large.

    It remains to prove (4). By Lemma 2.1-(3) and the Sobolev embedding theorem, we have

    Jμ,λ(u)12R3(|u|2+V(x)u2)dxμR3λG(u)dx12u2C8uα.

    From this, we get cμ>0 and complete the proof.

    Remark 2.1. By Lemma 2.3 and Proposition 1, then for almost every μ[12,1], there exists a sequence {un}X satisfying

    {un} is bounded in X, Jμ,λ(un)cμ and Jμ,λ(un)0 in X. (2.6)

    Lemma 2.4. The sequence {un} given in (2.6), up to subsequence, converges to a positive critical point uμ of Jμ,λ with Jμ,λ(uμ)=cμ.

    Proof. Since {un} is bounded in X, we have

    unuμ in X,  unuμ in Lα(R3),  unuμ a.e. in RN,

    for some uμX. For all φC0(R3), using Lebesgue's Theorem, we have that

    Jμ,λ(un)Jμ,λ(uμ),φ0,

    where we used unuμ in X, Lemma 2.1-(3) and unuμ in Lα(R3). Thus recalling that Jμ,λ(un)0 we indeed have Jμ,λ(uμ)=0.

    Next we prove unuμ in X. Using Lemma 2.1-(3) and the fact unuμ in Lα(R3), we get

    limn+R3(g(un)g(uμ))(unuμ)dx=0.

    Hence

    on(1)=Jμ,λ(un)Jμ,λ(uμ),unuμ=R3(unuμ)2dx+R3V(x)(unuμ)2dx+R3(ϕununϕuμuμ)(unuμ)dxμR3λ(g(un)g(uμ))(unuμ)dx=unuμ2+on(1), (2.7)

    where we used the elementary inequalities

    |R3(ϕununϕuμuμ)(unuμ)dx||R3ϕun(unuμ)2dx|+|R3(ϕunϕuμ)uμ(unuμ)dx||ϕun|6|unuμ|212/5+|ϕunϕuμ|6|unuμ|12/5|u|12/5. (2.8)

    Therefore, unuμ in X. The positivity of uμ follows by a standard argument (see [19]).

    To show the above results are true when μ=1, we need the following remark and lemmas.

    Remark 2.2. Assume that (V1) and (f0)(f2) hold. Then there exist {μn}[12,1] and {uμn}X{0} such that limn+μn=1, uμn>0, Jμn,λ(uμn)=cμnc12 and Jμn,λ(uμn)=0.

    Lemma 2.5. (See [12]) If uX is a critical point of Jμ,λ and (V1) holds, then

    12R3|u|2dx+32R3V(x)u2dx+54R3ϕuu2dx+12R3V(x)xu2dx3μR3λG(u)dx=0. (2.9)

    Lemma 2.6. Assume that (V2) and (f0)(f3) hold. Then the sequence {uμn} obtained in Remark 2.2 is bounded with respect to μn in X.

    Proof. Using the fact Jμn,λ(uμn)c12, Jμn,λ(uμn),uμn=0 and Lemma 2.5, we have

    (3γ2)c12(3γ2)Jμn,λ(uμn)Jμn,λ(uμn),uμn+(γ21)(2.9)L

    and

    (3γ2)c12(γ412)R3(2V(x)+V(x)x)u2μndx+(γ232)R3ϕuμnu2μndx+λR3(uμng(uμn)γG(uμn))dx(γ232)R3ϕuμnu2μndx. (2.10)

    Using (V2), Lemma 2.1 and the fact that 3<γ4, it implies that {R3ϕuμnu2μndx} is bounded.

    Next we prove that uμn is bounded. By Jμn,λ(uμn),uμn=0, we obtain

    γc12(γ21)R3(|uμn|2+V(x)uμn)dx+(γ41)R3ϕuμnu2μndx+R3(uμng(uμn)γG(uμn))dx(γ21)R3(|uμn|2+V(x)uμn)dx(4γ)(6γ)4γ12c12,

    then we complete the proof.

    Lemma 2.7. Assume that (V1), (V2) and (f0)(f3) hold. Then problem (2.2) has at least one positive solution.

    Proof. Using Remark 2.2 and Lemma 2.6, there exist {μn}[12,1] and a bounded sequence {uμn}X{0} such that

    limn+μn=1,  Jμn,λ(uμn)=cμn,  Jμn,λ(uμn)=0.

    Furthermore

    limnJλ(uμn)=limn(Jμn,λ(uμn)+(μn1)R3λG(uμn)dx)=limncμn=c1,

    where we used the fact that μcμ is continuous from the left. By the similar argument, we get

    Jλ(uμn)0 in X.

    Thus {uμn} is a bounded Palais-Smale sequence for Jλ and limnJλ(uμn)=c1. By the argument of Lemma 2.4 again, we complete the proof.

    Indeed, the critical points of Jλ with L-norm not more than δ are also the weak solutions of problem (1.1). So next we shall study the L-estimates for solutions of (2.2), which is essentially contained in the work of Brezis-Kato.

    Lemma 2.8. Assume (V1), (V2), (f0)(f3) hold and uX is a weak solution of problem (2.2). Then uL(R3). Moreover,

    |u|C9λ12αu222α, (2.11)

    where C9>0 only depends on α.

    Proof. Let uX be a weak solution of

    {Δu+V(x)u+ϕu=λg(u),in R3,Δϕ=u2, lim|x|+ϕ(x)=0,in R3, (2.12)

    which is equivalent to

    R3uφdx+R3V(x)uφdx+R3ϕuuφdxλR3g(u)φdx=0, for all φX. (2.13)

    From the above Lemma 2.7, we know that u>0. Let T>0, and define

    uT={u, if 0<uT,T, if uT.

    Choosing φ=u2(η1)Tu in (2.13), where η>1, we get

    R3|u|2u2(η1)Tdx+2(η1){x|u(x)<T}u2(η1)1Tu|u|2dx+R3ϕu(x)u2u2(η1)Tdx+R3V(x)u2u2(η1)Tdx=λR3g(u)u2(η1)Tudx.

    Combining Lemma 2.1-(3) and the nonnegativity of the second, the third and the fourth terms in the left side of the above equation, we obtain

    R3|u|2u2(η1)TdxλR3g(u)u2(η1)TudxλC5R3uαu2(η1)Tdx. (2.14)

    On the other hand, by the Sobolev inequality, we obtain

    (R3(uuη1T)2dx)22C10R3|(uuη1T)|2dxC11R3|u|2u2(η1)Tdx+C10(η1)2R3|u|2u2(η1)TdxC12η2R3|u|2u2(η1)Tdx,

    where we used the fact that (a+b)22(a2+b2).

    By (2.14), the Hölder inequality and the Sobolev embedding theorem, we have

    (R3(uuη1T)2dx)22λC13η2R3uα2u2u2(η1)TdxλC13η2(R3u2dx)α22(R3(uuη1T)222α+2dx)2α+22λC14η2uα2(R3uη222α+2dx)2α+22,

    where we used the fact that 0uTu.

    In what follows, taking ζ=222α+2, we get

    (R3(uuη1T)2dx)22λC14η2uα2|u|2ηηζ.

    Using the Fatou's lemma, letting T+, it follows that

    |u|η2(λC14η2uα2)12η|u|ηζ. (2.15)

    Define ηn+1ζ=2ηn, where n=0,1,2,... and η0=2+2α2. By (2.15) we have

    |u|η12(λC14η21uα2)12η1|u|2η0(λC14uα2)12η1+12η0η1η00η1η11|u|2.

    By iteration we have

    |u|ηn2(λC14uα2)12η0ni=0(ζ2)i(η0)1η0ni=0(ζ2)i(2ζ)1η0ni=0i(ζ2)i|u|2.

    Thus, we obtain |u|C9λ12αu222α.

    By the similar argument in Lemma 2.6, we can obtian the following lemma.

    Lemma 2.9. Let λ>VL2 and uλ be a critical point of Jλ with Jλ(uλ)=c1. Then there exists C16>0 (independent of λ) such that

    uλ2C16c1. (2.16)

    Proof of Theorem 1.1. Let u0C0(RN)X{0} be a nonnegative function such that Jλ(u0)<0. Then, the functional Jλ has the mountain pass geometry and we can define

    dλ=infγΓmaxt[0,1]Jλ(γ(t))>0,

    where Γ={γC([0,1],X): γ(0)=0, γ(1)=u0}. By Lemma 2.7, there exists a positive critical point uλ of Jλ with Jλ(uλ)=dλ. And more remarkable dλ=c1.

    Furthermore, taking T=|u0|, from Lemma 2.1-(4), we obtain

    dλmaxt[0,1]Jλ(tu0)maxt[0,1](t22R3(|u0|2+V(x)u20)dx+t44R3ϕu0u20dxλR3G(tu0)dx)maxt[0,1](t22R3(|u0|2+V(x)u20+12ϕu0u20)dxC17λtβR3uβ0dx)C18λ2β2. (2.17)

    By (2.11), (2.16) and (2.17), we see that

    |uλ|C19λβ22α.

    Then there exists λ0>0 such that for all λ>λ0, we get

    |uλ|δ,

    where δ is fixed in (2.1). Thus, the uλ is a positive solution of the original problem (1.1).

    We start by finding that the conditions (f1) and (f2) imply the existence of positive constants C20,C21 such that

    F(t)C20|t|α (3.1)

    and

    F(t)C21|t|β (3.2)

    with |t| small. Consider ρ(t)C1(R,R) an even cut-off function satisfying:

    ρ(t)={1, if |t|δ,0, if |t|2δ,

    0ρ(t)1, tρ(t)0 and |tρ(t)|2δ, where 0<δ<12 is chosen such that (3.1), (3.2) hold for |t|2δ and (f4) holds for |t|δ. Define

    H(t)=ρ(t)F(t)+(1ρ(t))˜F(t)andh(t)=H(t),

    where ˜F(t)=C22|t|α. From (f1) and the definitions of ρ(t) and h(t), for uX we have

    |h(u)|C23|u|α1. (3.3)

    Lemma 3.1. (See [8]) If f satisfies (f1)(f3), then for all t0, we have

    0<θH(t)th(t),

    where θ=min{α,˜γ}.

    We now consider another modified equation of (1.1) given by

    {Δu+V(x)u+ϕu=λh(u),in R3,Δϕ=u2, lim|x|+ϕ(x)=0,in R3. (3.4)

    By the definition of H(t) and (3.3), the functional associated to (3.4) stated by

    Iλ(u)=12R3(|u|2+V(x)u2)dx+14R3ϕuu2dxλR3H(u)dx, uX (3.5)

    is well-defined. It is well known that its critical points are the weak solutions of (3.4).

    The goal of this section is to prove Theorem 1.2. To this end, we use the Lemma 3.1 to get the boundedness of Palais-Smale sequence. Moreover, by the similar argument in Lemma 2.4, it is easy to show that Iλ satisfies Palais-Smale condition. These are standard results which can be found in textbooks and no proof is given here.

    Lemma 3.2. Assume that (f1)(f3) are satisfied. If uX is a critical point of Iλ, then

    u2C24Iλ(u), (3.6)

    where C24 depends on θ.

    The proof of the above result is quite similar to the one used in Lemma 2.6 and so is omitted.

    Since X is a real, reflexive, and separable Banach space, there exists {ej}jNX such that

    X=¯span{ej:j=1,2,...}.

    We denote

    Yk=span{e1,...,ek},       Zk=¯span{ek+1,...}.

    Lemma 3.3. Set θk,λ=supuYkIλ(u). If λ>1, then

    θk,λC25λ2β2, (3.7)

    where C25 depends on α, β and k.

    Proof. We notice that δ>0 was chosen in Section 3 such that

    H(t)=˜F(t)=C22|t|α for |t|2δ, (3.8)
    H(t)F(t)C26|t|β for |t|2δ. (3.9)

    For uYk, denote Ω1={xR3: |u(x)|2δ}, Ω2={xR3: |u(x)|<2δ}, and let u1=u|Ω1, u2=u|Ω2. Since all norms in Yk are equivalent, from (3.8) and (3.9) we obtain

    R3H(u1)dxˆC(k)|u1|α2

    and

    R3H(u2)dxˆC(k)|u2|β2,

    where ˆC(k) is a positive constant. By the same reason, we can define

    γk=sup{u:uYk, |u|2=1}<.

    Then for uYk, it follows that

    Iλ(u)γk2|u|22+C74u4λˆC(k)|u1|α2λˆC(k)|u2|β2γk2|u1|22+C7γk4|u1|42λˆC(k)|u1|α2+γk2|u2|22+C7γk4|u2|42λˆC(k)|u2|β2¯C1(k)λ2α2+¯C2(k)λ2β2.

    For λ>1, by 4<αβ<2, we have

    θk,λC25λ2β2,

    where C25:=¯C1(k)+¯C2(k).

    Lemma 3.4. Assume that uX is a weak solution of problem (3.4). Then uL(RN). Moreover,

    |u|C27λ12αu222α, (3.10)

    where C27>0 only depends on α.

    The proof of Lemma 3.4 is quite similar to Lemma 2.8 and so is omitted.

    To prove Theorem 1.2, we will apply the following symmetric mountain pass theorem due to Rabinowitz [16].

    Proposition 2. Let X be an infinite dimensional Banach space, JC1(X,R) be even, satisfy (PS) condition and J(0)=0. If X=YZ with dimY<+, and J satisfies

    (1) there are constants ρ,α>0 such that J|BρZα,

    (2) for any finite dimensional subspace WX, there is an R=R(W) such that J0 on WBR(W), then J has a sequence of critical values.

    Remark 3.1. We point out that Iλ satisfies all assumptions of Proposition 2. By Proposition 2, then Iλ possesses a sequence of critical points.

    Proof of Theorem 1.2. Fix an integer k. Choose R>0 such that Iλ(u)0 for all uYk with uR, and for all λ1. For BR={uX: u<R}, let D=BRYk. Define

    Γ:={γC(D,X): γ is odd,  γ(u)=u,if u=R}.

    Let i(A) be the genus of symmetric subset A. For jk, we denote

    Θj={γ(¯DB): γΓ, i(B)kj}

    and

    cj,λ=infAΘjsupuAIλ(u).

    From Proposition 2 and Remark 3.1, and under our conditions on h, we get

    0<c1,λc2,λ...ck,λ.

    Moreover, they are also critical values of Iλ and there exist at least 2k critical points {±uj,λ}kj=1 at these critical values. Since IdΓ, the definition of Θj and Lemma 3.3, we obtain

    cj,λθj,λC(j)λ2β2. (3.11)

    By Lemmas 3.2–3.4, we have

    |uj,λ|C28λβ2(β2)(2α).

    Since 4<β<2, there exists λ1>0 such that |uj,λ|δ, for all λ>λ1. Thus, ±uj,λ(j=1,2,,k) are weak solutions of problem (1.1).

    This work was supported by the National Natural Science Foundation of China (11171220). The authors wish to thank the referees and the editor for their valuable comments and suggestions.

    All authors declare no conflicts of interest in this paper.



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