Infinitely many solutions for fractional Schrödinger equation with potential vanishing at infinity

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R E S E A R C H

Open Access

Inﬁnitely many solutions for fractional

Schrödinger equation with potential

vanishing at inﬁnity

Yongzhen Yun

1*

_{, Tianqing An}

1

_{, Jiabin Zuo}

2

_{and Dafang Zhao}

1,3 *_{Correspondence:}

[email protected] 1_{College of Science, Hohai}

University, Nanjing, P.R. China Full list of author information is available at the end of the article

Abstract

The paper investigates the following fractional Schrödinger equation:

(–)s_u₊_V₍_x₎_u₌_K₍_x₎_f₍_u_), _x_∈_RN_,

where 0 <s< 1, 2s<N, (–)s_{is the fractional Laplacian operator of order}_s_._V₍_x_),_K₍_x₎

are nonnegative continuous functions andf(x) is a continuous function satisfying some conditions. The existence of inﬁnitely many solutions for the above equation is presented by using a variant fountain theorem, which improves the related

conclusions on this topic. The interesting result of this paper is the potentialV(x) vanishing at inﬁnity, i.e., lim_|x|→+∞V(x) = 0.

MSC: 35J87; 35J20; 49J40; 47G20

Keywords: Fractional Laplacian; Fountain theorem; Fractional Schrödinger equation; Weak solution

1 Introduction

This paper deals with the following fractional Schrödinger equation:

(–)su+V(x)u=K(x)f(u), x∈RN, (1.1)

where 0 <s< 1, 2s<N,V(x),K(x) are nonnegative continuous functions, which satisfy some conditions, and the functionf is continuous onRN_{. (–}₎s_{is the fractional Laplacian} operator of orders, i.e., (–)s_u₌_F–1₍_|_ξ_|2s_F_u_{), where}_F _{is the usual Fourier transform} inRN_{. Especially, the potential}_V₍_x_{) vanishes at inﬁnity, i.e.,}_lim

|x|→+∞V(x) = 0 (shortly

V(∞) = 0).

Since Laskin in [1,2], for the ﬁrst time, studied the existence of standing wave solutions for fractional Schrödinger equation of the form

ih∂ψ

∂t = (–)

s_ψ₊_V₍_x₎_ψ_–_f₍_x_,_ψ_), ₍_t_,_x₎_∈_R_×_RN_, _(1.2)

the fractional Schrödinger equation has become an important model in fractional quan-tum mechanics. Recently, the fractional Schrödinger equation has attracted a large

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ber of studies for its property, a lot of papers deal with the existence of solutions of it by using critical point theory, see [3–8] and the references therein.

Whens= 1, the fractional Schrödinger equation (1.1) reduces to the following classical nonlinear Schrödinger equation:

–u+V(x)u=K(x)f(u), x∈RN. (1.3)

There have been many works focusing on the existence of solutions of equation (1.3) in the last decades, such as sign-changing solutions, ground state solutions, radially symmetric solutions, multiplicity of standing wave solutions, and so on, see [9,10]. Alves and Souto in [9] studied equation (1.3) with the potential functionV(x) vanishing at inﬁnity. They proposed the following assumptions forV(x),K(x):

(I) For anyx∈RN_,_V₍_x_),_K₍_x_{) > 0}_{holds and}_K₍_x₎_∈_L∞₍_RN₎_. (II) For someR> 0and|An| ≤R, thenlimr→+∞

An∩Bcr(0)K(x)dx= 0, where{An} ⊂R N

is a sequence of Borel sets. (III) K_V(₍x_x)₎∈L∞(RN₎_.

(IV) Forp∈(2, 2∗), then K(x)

V(x)2

∗–p 2∗–2

→0 as|x| →+∞.

Then they obtained a positive ground state solution of equation (1.3) by using the varia-tional method.

In [9], Alves and Souto studied the classical nonlinear Schrödinger equation (1.3), but for fractional Schrödinger equation of the form (1.1), since it is difficult to get compact embedding in fractional Sobolev spaces, especially when the potential functionV(x) van-ishes at infinity. There are few papers dealing with this aspect. In [11], Li et al. obtained a positive solution of the fractional Schrödinger equation with vanishing at infinity by using the variational method. Based on the variational method in [12] of Yang and Zhao, three solutions have been got for a class of fractional Schrödinger equations with vanishing at infinity. Especially, as far as we know, there are no results that study the existence of in-finitely many solutions for problem (1.1). We focus on this topic and obtain some new conclusions. We make the following assumptions for the functionf. Note that we do not need all assumptions to hold simultaneously.

(f1) For allt∈R,f(–t) = –f(t)holds.

(f2) f(t)t≥0holds for allt∈R, there existsv∈(1, 2)such that

f(t)≤c1

1 +|t|v–1 for allt∈Rand somec1> 0.

(f3) lim_|t|→0f(_tt)= 0.

(f4) There exists a constantd> 0such thatlim|t|→∞infF_|(_t_|t)≥d, whereF(t) =

t 0f(s)ds. (f5) f(t)t≥0holds for allt∈R, and there existc2> 0andp∈(2, 2∗s)such that

f(t)≤c2

1 +|t|p–1 for allt∈Rand somec2> 0.

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(f7) There exist two constantsμ> 2andc3> 0such that

f(t)t–μF(t)≥c3

1 +t2 for allt∈R.

In the following, we present the main results of this paper.

Theorem 1.1 If the functions V,K satisfy(I)–(III)and f satisﬁes(f1)–(f4),then(1.1)has inﬁnitely many small energy solutions uk∈H for every k∈N.

Theorem 1.2 Assume that V,K satisfy(I)–(III),f satisﬁes(f1)and(f5)–(f7),then(1.1) admits inﬁnitely many high energy solutions uk_∈_{H for every k}_≥_k

0(k0∈N).

Throughout this paper,s∈(0, 1) is a ﬁxed constant, · pis the usual norm ofLp(RN), Br(x) is an open ball centered atxwith radiusr,ci(i= 1, 2, . . .), andCrepresents diﬀerent positive constants.

We organize this paper as follows. The preliminaries of a fractional Sobolev space and the variant fountain theorems are presented in Sect.2. In Sect.3, we give the proofs of Theorems1.1and1.2.

2 Preliminaries

In this section, we ﬁrstly provide a short review of fractional Sobolev spaces. For more information, we refer to [13,14].

Fors∈(0, 1), the fractional Sobolev spaceHs₍_RN_{) is deﬁned by}

HsRN:=

u∈L2RN:

RN

1 +|ξ|2sFu(ξ)2dξ< +∞

,

and the norm is

uHs=

RNu 2_dx₊

RN|

ξ|2sFu(ξ)2dξ 1

2 .

For problem (1.1), the working space is deﬁned as follows:

H:=

u∈HsRN:

RN|

ξ|2sFu(ξ)2dξ+

RNV(x)u

2_dx_{< +}_∞

.

Apparently,His a Hilbert space with the inner product

(u,v) :=

RN|

ξ|2sFu(ξ)Fv(ξ)dξ+

RNV(x)uv dx

and the norm

u:=

RN|

ξ|2sFu(ξ)2dξ+

RN

V(x)u2dx

1 2 .

Denote byLr

k(RN) the weighted Lebesgue space

Lr_kRN:=

u∈RN:

RN

K(x)|u|rdx< +∞

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equipped with the norm

uLr

k(RN):= _RNK(x)|u| r_dx

1 r .

In the following, we present the deﬁnition of a weak solution of (1.1).

Deﬁnition 2.1 We say thatu∈His a weak solution of (1.1) if

RN|

ξ|2sFu(ξ)Fv(ξ)dξ+

RNV(x)uv dx=

RNK(x)f(u)v dx, ∀v∈H.

Deﬁne the energy functional for problem (1.1) by

I(u) :=1 2u

2_–

RNK(x)F(u)dx.

Under our assumptions,I∈C1₍_H_,_R_{) is well-deﬁned and Gâteaux-diﬀerentiable on}_H_{, that} is,

I(u,v)=

RN

(u(x) –u(y))(v(x) –v(y))

|x–y|N+2s dx dy+

RNV(x)uv dx

–

RNK(x)f(u)v dx for allu,v∈H.

Moreover, the critical points of energy functionalIare solutions to problem (1.1).

Lemma 2.2([13]) Let0 <s< 1,N≥1such that2s<N,then

u_L2∗s(RN₎≤CuHs₍_RN₎, ∀u∈Hs

RN_,

where C=C(N,s) > 0and2_s∗=_N2_–2N_s is the fractional critical exponent.Furthermore,the embedding Hs₍_RN₎_→_Lq₍_RN₎_{is continuous for every}₂_≤_q_≤₂∗

s and is locally compact for all2≤q< 2∗_s.

Lemma 2.3([11]) Assume that(I)–(III)are satisﬁed,then the embedding H→Lq_K(RN)is compact whenever q∈[2, 2∗_s).

Remark2.4 From Lemma2.3, we obtain the following inequality:

uq

Lq_k(RN₎≤d1u

q _{for q}_∈_{2, 2}∗ s

, (2.1)

whered1is the best constant forH→LqK(RN).

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LetXbe a Banach space and the norm is · . Denote by{Xj}a sequence of subspaces ofXanddimXj<∞for eachj∈N; moreover,X=

j∈NXj. We deﬁne

Mk:= k

j=0

Xj, Nk:=

∞

j=k Xj,

and

Yk:=

u∈Mk:u ≤φk

, Zk:=

u∈Nk:u=ψk

,

whereφk>ψk> 0. Forλ∈[1, 2],Iλ:H→Ris a family ofC1-functionals deﬁned by

Iλ(u) :=Φ(u) –λΨ(u),

whereΦ(u) :=1₂u2andΨ(u) :=_RNK(x)F(u)dx.

Lemma 2.5([15]) Let X be a Banach space.If the functional Iλ(u)satisﬁes:

(A1) Forλ∈[1, 2],Iλ(u)maps bounded sets into bounded sets uniformly and

Iλ(–u) =Iλ(u)holds for allu∈X.

(A2) Ψ(u)≥0holds for allu∈X,andΨ(u)→ ∞asu → ∞on any ﬁnite dimensional subspace ofX.

(A3) Forλ∈[1, 2],there existφk>ψk> 0such that

αk(λ) = inf u∈Nk,u=φk

Iλ(u)≥0 >βk(λ) = inf u∈Mk,u=ψk

Iλ(u)

and

γk(λ) = inf u∈Nk,φk≥u

Iλ(u) ask→ ∞.

Then there existλn→1,un(λn)∈Mnsuch that

I_λ_n|Mn

u(λn)

= 0 and Iλn

u(λn)

→ωk∈

γk(2),βk(1)

as n→ ∞.

In particular, if{u(λn)} possesses a convergent subsequence,then I1 has inﬁnitely many nontrivial critical points{uk} ∈X\ {0}satisfying I1(uk)→0–as k→ ∞.

Lemma 2.6([16]) Let X be a Banach space.If the functional Iλ(u)satisﬁes:

(B1) Forλ∈[1, 2],Iλ(u)maps bounded sets into bounded sets uniformly and

Iλ(–u) =Iλ(u)holds for allu∈X.

(B2) Ψ(u)≥0holds for allu∈X,Φ(u)→ ∞orΨ(u)→ ∞whenu → ∞. (B3) There existsφk>ψk> 0such that

ek(λ) = inf u∈Nk,u=ψk

Iλ(u) >gk(λ) = max u∈Mk,u=φk

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Then

ek(λ)≤hk(λ) = inf

η∈Γk

max

u∈YkIλ

γ(u), ∀λ∈[1, 2],

whereΓk={η∈C(Yk,X) :ηis odd,η|∂Yk= id,k≥2}.Furthermore,for a.e.λ∈[1, 2],there exists{uk

n(λ)}∞n=1such that

sup

n

uk_n(λ)<∞, I_λuk_n(λ)→ ∞ and Iλ

uk_n(λ)→gk(λ) as n→ ∞.

3 Proofs of the main results

In this section, by using the variant fountain theorem, we prove Theorem1.1and Theo-rem1.2, respectively.

3.1 Proof of Theorem1.1

To begin with, we give some lemmas.

Lemma 3.1 If p∈(1, 2∗_s),then we have that

ζk= sup u∈Nk,u=1 RN

K(x)|u|pdx

1 p

→0, k→ ∞.

The proof is similar to Lemma 3.8 in [17], here we omit it.

Lemma 3.2 If(f2)and(f4)hold,then for all u∈H,Ψ(u)≥0holds and ifu → ∞,then

Ψ(u)→ ∞on any ﬁnite dimensional subspace of H.

Proof From (f2), it is clear thatΨ(u)≥0 holds for allu∈H. LetH0be any ﬁnite dimen-sional subspace ofH. In the following we will prove thatΨ(u)→ ∞whenu → ∞onH0.

Firstly, we show that, for allu∈H0⊂H, there is

measx∈RN:K(x)u(x)≥θu≥θ, (3.1)

whereθ is a constant withθ > 0. To prove it, arguing by contradiction, we assume that there existsun∈H0\ {0}such that

meas

x∈RN :K(x)un(x)≥ 1 nun

< 1

n, ∀n∈N.

Letvn=un_un, thenvn= 1 and

meas

x∈RN :K(x)vn(x)≥ 1 n

<1

n, ∀n∈N. (3.2)

Sincevn= 1, there existsv∈H0withv= 1,vn→vholds inH. Using Lemma2.3, we have

RNK(x)

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Fromv= 0 andK(x) > 0, there exist two constantsδ0> 0 andδ1> 0 such that

measx∈RN:v(x)≥δ0

≥δ0, (3.4)

and

measx∈RN:K(x)v(x)≥δ1

≥δ1. (3.5)

For anyn∈RN_{, we denote}

Dn:=

x∈RN:K(x)vn(x)< 1 n

,

DC_n:=

x∈RN:K(x)vn(x)≥ 1 n

,

and

D0:=

x∈RN:v(x)≥δ0

.

Fornsuﬃciently large, by (3.2), (3.4), and (3.5), we obtain

meas(Dn∩D0)≥meas(D0) –meas

DC_n≥2δ0 3 .

Therefore,

RNK(x)

vn(x) –v(x) 2

dx

≥

Dn∩D0

K(x)vn(x) –v(x) 2

dx

≥

Dn∩D0

K(x)v(x)2– 2vn(x)v(x)dx

≥δ0 δ1– 2 n

meas{Dn∩D0}

≥2δ0 3 δ1–

2 n

≥2δ0δ1 9 > 0,

which contradicts (3.3), thus, (3.1) holds. By condition (f4), for allx∈RN_and_u_≥_δ

2> 0, one has

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DeﬁneDu:={x∈RN :K(x)|u(x)| ≥θu}foru∈H0\ {0}. By (3.1), for anyu∈H0 with

u ≥δ2

θ,K(x)|u(x)| ≥δ2holds asx∈Du. Consequently, from (3.6) we obtain

Ψ(u)≥

RNK(x)F(u)dx

≥

Du

K(x)F(u)dx

≥d

Du

K(x)u(x)dx

≥dθumeas{Du}

≥dθ2u,

which implies thatΨ(u)→ ∞whenu → ∞onH0⊂H.

Lemma 3.3 If(f2)–(f4)hold,then forλ∈[1, 2],there existsφk>ψk> 0such that

αk(λ) := inf u∈Nk,u=φk

Iλ(u)≥0 >βk(λ) := inf u∈Mk,u=ψk

Iλ(u)

and

γk(λ) := inf u∈Nk,φk≥u

Iλ(u) as k→ ∞.

Proof From Lemma3.1, forp∈(1, 2∗_s), we obtain

RNK(x)|u|

p_dx_≤_ζp

kup. (3.7)

By (f2) and (f3), for anyε> 0, one has

F(u)≤ε|u|2+cε|u|v, (3.8)

wherecε> 0 depending onε. From (2.1) and (3.8), foru∈Nk, we have

Iλ(u)≥

1 2u

2_–_λ

RNK(x)

ε|u|2+cε|u|v

dx

≥1

2u 2_{– 2}_ε

RNK(x)|u|

2_dx_{– 2}_c

ε

RNK(x)|u| v_dx

≥1

2u 2_{– 2}_d

1εu2– 2cεζkvuv.

Letd1ε<₁₂1, then for anyu∈Nkwithu< 1, we know that

Iλ(u)≥

1 3u

2_{– 2}_c

εζkvuv. (3.9)

Chooseφk= (12cεζ_kv)

1

2–v_{, then}_φ_k_{> 0 and}_φ_k→_{0 as}_k→ ∞_{. For any}_λ∈_{[1, 2], we obtain}

αk(λ) = inf u∈Nk,u=φk

Iλ(u)≥

1 6

12cεζ_kv

2 2–v₌1

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From (3.9), for anyu∈Nkwithu ≤φkandλ∈[1, 2], one has

Iλ(u)≥–2cεζ_kvuv≥–2cεζ_kvφ_kv,

which means that

γk(λ) = inf u∈Nk,u≤φk

Iλ(u)≥–2cεζkvφkv→0+ ask→ ∞.

Hence, forφk→0 ask→ ∞, we get

γk(λ) = inf u∈Nk,u≤φk

Iλ(u)→0 ask→ ∞.

By (f2)–(f4), there is

F(u)≥d|u|–ε|u|2–cε|u|v. (3.10)

Hence, ifu∈Mk, by the equivalence of norms on a ﬁnite dimensional space, one has

Iλ(u)≤

1 2u

2_–

RNK(x)F(u)dx

≤1

2u 2_–_d

RNK(x)|u|dx+

ε

RNK(x)|u| 2_dx₊_c

ε

RNK(x)|u| v_dx

≤ u2_–_c

4u+c5uv.

Therefore, we chooseψk> 0 small enough satisfyingφk>ψk> 0 such that

βk(λ) = inf u∈Mk,u=ψk

Iλ(u) < 0, ∀λ∈[1, 2].

Lemma 3.4 If(f1)–(f3)hold and forλ∈[1, 2],k∈N,there existλn→1and u(λn)∈Mn such that

Iλn|Mn

u(λn)

= 0 and Iλn

u(λn)

→ωk∈

γk(2),βk(1)

as n→ ∞,

then{u(λn)}possesses a convergent subsequence in H.

Proof Firstly, we show that{u(λn)}is bounded inH. Using (3.8) and the Hölder inequality, forλn∈[1, 2] withλn→1 asn→ ∞, one has

u(λn) 2

= 2Iλn

u(λn)

+ 2λn

RN

K(x)Fu(λn)

dx

≤2(ωk+ 1) + 2λn

RNK(x)

εu(λn)2+cεu(λn)v

dx

≤2(ωk+ 1) +c6u(λn) 2

+c7u(λn) v

.

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we haveu(λn)→ustrongly inLpk(RN) forp∈[2, 2∗s). LetPn:H→Yndenote the orthog-onal projection operator, then

0 =Iλn|Yn

u(λn)

=PnIλn

u(λn)

.

ThusPnIλn(u(λn)),u(λn) –u= 0. Fromu(λn)uinH, we know thatI

1(u),u(λn) –u= 0 asn→ ∞. According to the property of the orthogonal projection operator, one ﬁnds

u(λn) –u 2

=Pn

u(λn)

–u2

=PnIλn

u(λn)

,u(λn) –u

–I₁(u),u(λn) –u

+λn

RN

K(x)fu(λn)

Pn

u(λn) –u

dx

–

RNK(x)f(u)

u(λn) –u

dx.

From conditions (f1) and (f2), forε> 0 small, we get

f(u)≤ε|u|+cε|u|v–1, (3.11)

wherecε> 0 depending onε. Using the Hölder inequality, we obtain

λn

RNK

(x)fu(λn)

Pn

u(λn) –u

dx

≤λn

RNK(x)

εu(λn)+cεu(λn) v–1

Pn

u(λn) –u

dx

≤2ε

RN

K(x)u(λn)Pn

u(λn) –u

dx+ 2cε

RN

K(x)u(λn) v–1

Pn

u(λn) –u

dx

≤2εK∞

RN

u(λn)Pn

u(λn) –u

dx+ 2cεK∞

RN

u(λn) v–1

Pn

u(λn) –u

dx

≤2εK∞un2Pn

u(λn) –u₂+ 2cεK∞unv2–1Pn

u(λn) –u₂→0

asn→ ∞. From (3.11) and the Hölder inequality, one has

_RNK(x)f(u)

u(λn) –u

dx

≤ RNK(x)

ε|u|+cε|u|v–1

u(λn) –u

dx

≤εK∞

RN|u|

u(λn) –u

dx+cεK∞

RN|u| v–1_u₍_λ

n) –u

dx

≤εK∞un2u(λn) –u₂+cεK∞unv2–1u(λn) –u₂→0 asn→ ∞.

Consequently,un→uinHasn→ ∞.

Proof of Theorem1.1 By (f2)–(f3) andλ∈[1, 2], we know clearly thatIλ(u) maps bounded

sets into bounded sets. Condition (f1) implies thatIλ(–u) =Iλ(u), then (A1) of Lemma2.5

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From Lemmas3.2and3.3, we have (A2) of Lemma2.5holds. Thanks to Lemma2.5, for eachk∈N, there existλn→1,un(λn)∈Mnsuch that

Iλn|Yn

u(λn)

= 0 and Iλn

u(λn)

→ωk∈

γk(2),βk(1)

asn→ ∞.

It follows from Lemma3.4that{u(λn)}has a convergent subsequence for anyk∈N. Using Lemma2.5,I1has inﬁnitely many nontrivial critical points{uk}satisfying

I1(uk) = 1 2uk

2_–

RN

K(x)F(uk)dx→0– ask→ ∞

for anyk∈N. Consequently, problem (1.1) owns inﬁnitely many small energy solutions.

The proof of Theorem1.1is completed.

3.2 Proof of Theorem1.2

Lemma 3.5 Assume that(f3), (f5),and(f6)hold.Then there existsφk>ψk> 0such that

Iλ(u) >gk(λ) = max u∈Mk,u=φk

Iλ(u), ∀λ∈[1, 2].

Proof By (f3) and (f5), for anyε> 0, we get

f(u)≤ε|u|+cε|u|p–1, (3.12)

where the constantcε> 0 depending onε.

By (3.7), foru∈Nkandε> 0 withλε<1₃, one has

Iλ(u)≥

1 2u

2_–λε 2

RNK(x)|u|

2_dx_–λCε

p

RNK(x)|u| p_dx

≥1

3u 2_–_c

8ζkpup.

Chooseγk= (6C8ζ_kpup) 1

2–p_{, then for any}_u∈_N

kwithu=ψk, we can deduce

Iλ(u)≥

1 6

6C8ζkp

2 2–p₌1

6ψ 2

k> 0. (3.13)

Therefore,

Iλ(u) >

1 6ψ

2

k> 0, ∀λ∈[1, 2]. (3.14)

Notice that (3.1) still holds here. Hence, there existsεk> 0 such that

meas(Gu)≥εk, ∀u∈Nk\ {0}, (3.15)

where Gu:={x∈RN :K(x)|u(x)| ≥εku}. By (f5), there exists a constant Rk > 0 with

|u| ≥Rksuch that

F(u)≥ 1

ε_k3|u|

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Consequently, using (3.15), for anyu∈Nkwithu ≥ Rkε_k,|u(x)| ≥Rkholds for allx∈Gu.

Therefore, by (3.15) and (3.16), one has

Iλ(u)≤

1 2u

2_–

RN

K(x)F(u)dx

≤1

2u 2_–

Gu

K(x)F(u)dx

≤1

2u 2_– 1

ε3_k

Gu

K(x)|u|2dx

≤1

2u

2_–Rku

ε_k2 meas{Gu}

≤1

2u

2_–Rku

εk

≤1

2u 2_–_u2

= –1 2u

2_.

Finally, for anyu=φk>max{ψk,Rεkk}, we obtain

gk(λ) = max u∈Wk,u=ρk

Iλ(u)≤–

r_k2

2 < 0, ∀λ∈[1, 2].

Proof of Theorem1.2 From (f6) and (f7), forλ∈[1, 2], we know thatIλ(u) maps bounded

sets into bounded sets. By (f1), for all (λ,u)∈[1, 2]×H, we haveIλ(–u) =Iλ(u). Therefore,

(B1) of Lemma2.6is satisﬁed. From condition (f6)

lim

|t|→∞ F(t)

|t|2 =∞,

we know thatΨ(u)≥0 holds foru∈H. By the deﬁnition ofΦ(u),Φ(u)→ ∞holds when

u → ∞, which means that (B2) of Lemma2.6holds. Using Lemma3.5, we have (B3) of Lemma2.6holds. Therefore, by Lemma2.6, for a.e.λ∈[1, 2], there exists{uk

n(λ)}∞n=1for k≥k0withk∈N, one has

sup

n

uk_n(λ)<∞, Iλ

uk_n(λ)→ ∞ and Iλ

uk_n(λ)→gk(λ) asn→ ∞. (3.17)

Using Lemma2.6, for allλ∈[1, 2], we also have

ek(λ)≤hk(λ) = inf

δ∈Γk

max

u∈YkIλ

γ(u).

Letξk= (6C7ζkpup) 1

2–p_{, then}_ξ

k→ ∞ask→ ∞. Fork≥k0withk∈N, by (3.13), we have hk(λ)≥ek(λ)≥ξk, then

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where ρk:=maxu∈YkIλ(γ(u)),Γk:={δ∈C(Bk,H) :δis odd,δ|∂Yk = id,k≥2}with Yk:=

{u∈Yk:u ≤ξk}.

Chooseλm→1 asm→ ∞forλm∈[1, 2]. Thanks to (3.17), we know that{ukn(λm)}is bounded, which implies{uk

n(λm)}possesses a weakly convergent subsequence. Similar to the proof of Lemma3.4, we know that{uk

n(λm)}has a strong convergent subsequence as n→ ∞. Assume thatlimn→∞ukn(λm) =uk(λm) form∈Nandk≥k0. By (3.17) and (3.18), one has

I_λ_muk(λm)

= 0 and Iλm

uk(λm)

∈[ξk,ρk] fork≥k0. (3.19)

We claim that{uk_n(λm)}is bounded inH. If the claim is not true, we haveuk(λm) → ∞ asm→ ∞. Letvm:= u

k₍_λ m)

uk₍_λ_m₎, thenvm= 1. Without loss of generality, we assumevm→v inH. Thanks to Lemma2.3, we obtainvm→vinLpk(RN) forp∈[2, 2∗s). In the following, we divide two steps to show our assumption is not true.

Step 1. Assumev(x)= 0 inH. Since

1 2–

Iλm(uk(λm))

uk₍_λ m)2

=λm

RN

K(x)F(uk₍_λ m))

uk₍_λ m)2

dx

=λm

RN

|vm(x)|K(x)F(uk(λm))

|uk₍_λ m)|2

dx,

wherevm(x)= 0,m= 1, 2, 3, . . . . By (3.19), (f6), and Fatou’s lemma, one has

1

2=lim infm→∞ λm

RN

|vm(x)|K(x)F(uk(λm))

|uk₍_λ m)|2

dx→0,

which is in contradiction with our claim.

Step 2. Assumev(x) = 0 inH. Using (3.19), we have

μ

2 – 1 =

μIλm(uk(λm)) –Iλm(u k₍_λ

m)),uk(λm)

uk₍_λ m)2

+λm

RN

|vm(x)|2K(x)[μF(uk(λm)) –f(uk(λm))uk(λm)]

|uk₍_λ m)|2

dx.

Therefore,

RN

|uk₍_λ m)|2

dx→μ

2 – 1 (3.20)

asm→ ∞. However, according to (f7), one has

lim sup

m→∞

|uk₍_λ

m)|2 ≤

0. (3.21)

Combining with (3.20) and (3.21), we haveμ₂– 1≤0, i.e.,μ≤2, we deduce a contradiction to our assumption. Therefore,{uk_n(λm)}is bounded inH. Similar to the proof of Lemma3.4, we know that{uk

n(λm)}owns a strongly convergent subsequence withuk∈Hfor allk≥k0 andk∈N. Assume

lim

m→∞u k₍_λ

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Using (3.19) andβk→ ∞ask→ ∞, for allk≥k0andk∈N, we have

I1

uk= 0, I1

uk∈[βk,δk]→ ∞ ask→ ∞,

which implies thatukis a nontrivial critical point ofI1. Hence, fork≥k0,k∈N, inﬁnitely many nontrivial critical pointsuk_of_I

1are obtained. From the above discussion, we know that problem (1.1) has inﬁnitely many nontrivial solutions with high energy, that is,

I1

uk=1 2u

k2 –

RN

K(x)Fukdx→ ∞ ask→ ∞.

The proof of Theorem1.2is completed.

Acknowledgements

The authors thank the anonymous referees for invaluable comments and insightful suggestions which improved the presentation of this manuscript.

Funding

The work is supported by the Fundamental Research Funds for Central Universities (2017B19714 and 2017B07414), Natural Science Foundation of Jiangsu Province (BK20180500), the Postgraduate Research & Practice Innovation Program of Jiangsu Province (SJKY19_0431), and Natural Science Foundation of Jilin Engineering Normal University (XYB201814).

Availability of data and materials

Not applicable.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Each of the authors contributed to each part of this study equally, all authors read and approved the ﬁnal manuscript.

Author details

1_{College of Science, Hohai University, Nanjing, P.R. China.}2_{Faculty of Applied Sciences, Jilin Engineering Normal} University, Changchun, P.R. China.3_{School of Mathematics and Statistic, Hubei Normal University, Huangshi, P.R. China.}

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 6 November 2018 Accepted: 21 March 2019

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