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R E S E A R C H

Open Access

Infinitely many geometrically distinct

solutions for periodic Schrödinger–Poisson

systems

Jing Chen

1*

and Ning Zhang

2

*Correspondence:

[email protected]

1School of Mathematics and

Computing Sciences, Hunan University of Science and Technology, Xiangtan, P.R. China Full list of author information is available at the end of the article

Abstract

This paper is dedicated to studying the following Schrödinger–Poisson system:

u+V(x)u+K(x)

φ

(x)u=f(x,u), x∈R3,

φ

=K(x)u2, xR3,

whereV(x),K(x), andf(x,u) are periodic inx. By using the non-Nehari manifold method, we establish the existence of ground state solutions for the above problem under some weak assumptions. Moreover, whenfis odd inu, we prove that the above problem admits infinitely many geometrically distinct solutions. Our results improve and complement some related literature.

MSC: 35J10; 35J20

Keywords: Schrödinger–Poisson system; Nehari manifold; Ground state; Geometrically distinct solutions

1 Introduction

In this paper we are concerned with the nonlinear Schrödinger–Poisson system: ⎧

⎨ ⎩

u+V(x)u+K(x)φ(x)u=f(x,u), x∈R3,

φ=K(x)u2, xR3, (1.1)

whereV,K:R3Randf :R3×RRsatisfy the following basic assumptions: (V0) V,KC(R3, (0,)),V(x), andK(x)are 1-periodic inx

1,x2, andx3; (F0) f(x,t)is 1-periodic inx1,x2, andx3;

(F1) fC(R3×R,R),f(x,t) =o(|t|)uniformly inxast0, and there exist constants

C0> 0andp∈(2, 6)such that

f(x,t)C0

1 +|t|p–1, ∀(x,t)∈R3×R.

Schrödinger–Poisson system (also called Schrödinger–Maxwell system) appears in the quantum mechanics model or Hartree–Fock model, which is related to the study of the

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interaction of a charged non-relativistic quantum mechanical particle with the electro-magnetic field. System (1.1) can be described by coupling a nonlinear Schrödinger and a Poisson equation, from physical point of view, the unknown termsuandφare the fields associated to the particle and the electric potential, respectively, the functionsV andK are, respectively, an external potential and nonnegative density charge, the nonlinear term f simulates the interaction effect between particles or external nonlinear perturbations, and the coupled termφ(x)uconcerns the interaction with the field. For more details on the physical aspects, we refer the readers to [5–8,20,23].

Note that when φ≡0, (1.1) reduces to the well-known Schrödinger equation, which has been the object of various investigations; see, for example, [26,33–35,39,40] and the references therein.

Under assumption (V0), the set

E= uH1R3:

R3

|∇u|2+V(x)u2dx< +∞

is a Hilbert space equipped with the norm

u=

R3

|∇u|2+V(x)u2dx 1/2

.

It is well known that the Poisson equation is solved by using the Lax–Milgram theorem. Indeed, as we shall see in Sect.2, for everyuE, uniqueφuD1,2(R3) is obtained, such

that –φ=K(x)u2and so (1.1) can be reduced to a single equation with a nonlocal term

u+V(x)u+K(x)φu(x)u=f(x,u). (1.2)

Moreover, (1.2) is variational and its solutions are the critical points of the functionalΦ

defined onEby

Φ(u) = 1 2

R3

|∇u|2+V(x)u2dx+1 4

R3K(x)

φu(x)u2dx–

R3F(x,u) dx, (1.3)

whereF(x,t) =0tf(x,s) ds. Define

N :=uE:Φ(u),u= 0,u= 0, (1.4)

which is the Nehari manifold ofΦ. Let∗denote the action ofZ3onH1(R3) given by

(k∗u)(x) =u(xk), k∈Z3.

We note that ifu0is a solution of (1.1), then so areku0for allk∈Z3. Set

O(u) :=ku0,k∈Z3

,

which is called the orbit ofu0 with respect to the action ofZ3. Two solutionsu1andu2

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In recent years, there have been rich results for Schrödinger–Poisson systems like (1.1) on the existence of nontrivial solutions, positive solutions, ground states, semi-classical states, and multiple solutions; we refer to [1–3,6,9,14–17,25,29] for the casef(x,u)∼ |u|q–2uwithq∈(4, 6); [4,28,31,36,41] for the casef(x,u)∼ |u|q–2uwithq∈(3, 4]; [13] for the convolution nonlinearity;[18,38] for the critical growth nonlinearity. [19] deals with the multiplicity of solutions for the fractional Schrödinger–Poisson systems. In this paper, we focus on the existence of ground state solutions and infinitely many geometrically dis-tinct solutions for (1.1) in a periodic setting. Let us recall some previous results that led us to the present research.

When the potential and nonlinearity are periodic, that is (V0) and (F0) are satisfied, Zhao and Zhao [41] proved that (1.1) withK(x) = 1 has a ground solution and infinitely many geometrically distinct solutions by using the Nehari manifold approach, wherefand∂f/∂u are continuous and satisfy suitable conditions. Based on the generalized Nehari manifold approach developed by Szulkin and Weth [32], Sun and Ma [31] obtained similar results as those in [41], wheref satisfies (F0), (F1) and the following assumptions:

(SC) lim|t|→∞F|(tx|,4t)= +∞uniformly inx∈R3;

(MT) f(x,t)/|t|3is increasing intonR\ {0}for everyxR3.

Later, Chen and Tang relaxed (SC) and (MT) to the following weaker conditions:

(F2) lim|t|→∞F|(tx|,3t)=∞uniformly inx∈R3; (F3) there existsθ∈(0, 1)such that

f(x,τ)

τ3 –

f(x,) (tτ)3

sign(1 –t) +θV(x)|1 –t

2|

(tτ)2 ≥0, ∀x∈R

3,t> 0,τ = 0,

and established the existence of ground state solutions for (1.1) by means of the non-Nehari manifold method developed by Tang [33–35].

To the best of our knowledge, except for [31,41], there seems to be no result about the existence of infinitely many geometrically distinct solutions for (1.1). Motivated by the work of [9,10,31,33], in the present paper, we shall establish the existence of ground state solutions and infinitely many geometrically distinct solutions for (1.1) under weaker assumptions than previous works.

Before presenting our theorems, in addition to (V0), (F0), (F1), and (F2), we introduce the following assumptions:

(F3) f(x,t)–|t|V3(x)t is nondecreasing inton both(–∞, 0)and(0,∞)for everyx∈R3; (F4) there exists a constantθ∈[0, 1)such that

f(x,t)t– 4F(x,t) +θV(x)t2≥0, ∀(x,t)∈R3×R;

(F5) f(x, –t) = –f(x,t),∀(x,t)∈R3×R.

Theorem 1.1 Assume that V,K,and f satisfy(V0), (F0),and(F1)–(F3).Then Problem (1.1)has a solution u0∈E such thatΦ(u0) =inf> 0.

Theorem 1.2 Assume that V,K,and f satisfy(V0), (F0),and(F1)–(F5).Then Problem (1.1)admits infinitely many pairs of geometrically distinct solutions.

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in [31,41] isf(x,u) =b(x)u3|u|3/2u+|u|ufor all (x,u)R3×R, whereb(x) is 1-periodic

inx1,x2, andx3andinfR3b> 0. In this sense, our results improve and complement those

of [10,31,41].

To prove Theorem1.1, following the idea of [10], we apply the non-Nehari manifold method. Unlike the Nehari manifold approach, the key point of this method lies in finding a minimizing Cerami sequence forΦoutsideNby using a diagonal method. However, the fact that (F3) is weaker than (F3) used in [10] would require our extra efforts. To prove Theorem1.2, inspired by [31,32], we use deformation type arguments and Lusternik– Schnirelman theory. However, sincetf|(tx|,3t)is not increasing, the generalized Nehari

ap-proach developed by [32] does not work. To circumvent this obstacle, we borrow the idea of [11,12] in which Kirchhoff–type problems and Klein–Gordon–Maxwell systems were considered respectively. However, the competing effect of the nonlocal termR3φuu2dx

and the nonlinear termR3F(u) dxin the expression ofΦmakes our problem more

com-plicated.

The paper is organized as follows. In Sect.2, we introduce some notation and prelimi-naries. We complete the proofs of Theorems1.1and1.2in Sects.3and4, respectively.

Throughout this paper, we denote the norm ofLs(R3) byu

s= (

R3|u|sdx)1/sfors≥2,

Br(x) ={y∈R3:|yx|<r}, and positive constants possibly different in different places by

C1,C2, . . . .

2 Notation and preliminaries

Hereafter,H1(R3) is the usual Sobolev space with the standard scalar product and norm

(u,v)H1=

R3(∇uv+uv) dx, u 2

H1=

R3

|∇u|2+u2dx,

and

D1,2R3=uL6R3:∇uL2R3

equipped with the norm defined by

u2D1,2=

R3|∇u| 2dx.

It is easy to show that (1.1) can be reduced to a single equation with a nonlocal term. Namely, for anyKu2L1

loc(R3) such that

R3

R3

u2(x)u2(y)

|xy| dxdy<∞,

the distributional solution

φu(x) =

R3

K(y)u2(y)

|xy| dy= 1 |x|∗Ku

2 (2.1)

of the Poisson equation

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belongs toD1,2(R3) and is the unique weak solution inD1,2(R3) (see, e.g., [29] for more

details), and

R3∇

φuvdx=

R3

K(x)u2vdx, vH1R3, (2.2)

R3

R3

K(x)K(y) |xy| u

2(x)u2(y) dxdy=

R3K(x)

φu(x)u2dx. (2.3)

Moreover,φu(x) > 0 whenu= 0, becauseKdoes (see (V0)). By using Hardy–Littlewood–

Sobolev inequality (see [21] or [22, p. 98]), we have the following inequality:

R3

R3

|u(x)v(y)| |xy| dxdy≤

8√32

3√3πu6/5v6/5, u,vL

6/5R3. (2.4)

Formally, the solutions of (1.1) are then the critical points of the reduced functional (1.3). Indeed, (V0), (F0), and (2.4) imply thatΦis a well-defined functional of classC1and that

Φ(u),v=

R3

uv+V(x)uvdx+

R3

K(x)φu(x)u–f(x,u)

vdx. (2.5)

Hence ifuEis a critical point ofΦ, then the pair (u,φu), withφuas in (2.1), is a solution

of (1.1).

Lemma 2.1 Under assumptions(V0), (F1),and(F2),

Φ(u)≥Φ(tu) +1 –t

4

4

Φ(u),u+(1 –t

2)2

4 ∇u

2

2, ∀uE,t≥0. (2.6)

Proof For anyx∈R3,t0,τ= 0, (F2) yields

1 –t4

4 τf(x,τ) +F(x,tτ) –F(x,τ) + V(x)

4

1 –t22τ2

= 1

t

f(x,τ)

τ3 –

f(x,) (sτ)3 +V(x)

(1 –s2) (sτ)2

s3τ4ds≥0. (2.7)

Note that

Φ(u) = 1 2u

2+1

4

R3

K(x)φu(x)u2dx–

R3

F(x,u) dx (2.8)

and

Φ(u),u=u2+

R3

K(x)φu(x)u2dx–

R3

f(x,u)udx. (2.9)

Thus, by (2.7), (2.8), and (2.9), one has

Φ(u) –Φ(tu) = 1 –t

2

2 u

2+1 –t4

4

R3K(x)

φu(x)u2dx+

R3

F(x,tu) –F(x,u)dx

= 1 –t

4

4

Φ(u),u+(1 –t

2)2

4 ∇u

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+

R3

1 –t4

4 f(x,u)u+F(x,tu) –F(x,u) + V(x)

4

1 –t22u2

dx

≥ 1 –t4 4

Φ(u),u+(1 –t

2)2

4 ∇u

2

2, t≥0.

This shows that (2.6) holds.

Corollary 2.2 Under assumptions(V0), (F1),and(F2),for uN,

Φ(u) =max

t≥0 Φ(tu). (2.10)

Unlike the super-cubic case, to showN =∅in our situation, we have to overcome the competing effect of the nonlocal term. Inspired by Chen and Tang [10], we define a setΛ

as follows:

Λ= uE:

R3

V(x)u2+K(x)φuu2–f(x,u)u

dx< 0

.

Lemma 2.3 Under assumptions(V0)and(F1)–(F3),Λ=∅andNΛ.Then,for any uΛ,there exists unique t(u) > 0such that t(u)uN.

Proof First, we show thatΛ=∅. From (2.4) and Sobolev imbedding theorem, there exists C1> 0 such that

R3φuu2dx≤C1u4 for alluE. For any fixeduEwithu= 0, set

ut(x) =u(tx) fort> 0. By (V0), one has

R3

V(x)(tut)2+K(x)φ(tut)(tut)2–f(x,tut)tut

dx

=t–1

R3V

t–1xu2dx+t–1

R3K

t–1xφuu2dx–

R3

f(t–1x,tu)tu

t3 dx

Vt–1u22+C1Kt–1u4–

R3

f(t–1x,tu)tu

t3 dx, (2.11)

where V∞ =supx∈R3V(x) and K∞ =supx∈R3K(x). Note that, for u(x)= 0, F(t–1x,tu)/

|tu|3+ast+uniformly inxR3by (F3), and (2.7) witht= 0 yields

1

4f(x,τ)τF(x,τ) + V(x)

4 τ

20, xR3,τR, (2.12)

then we have

f(t–1x,tu)tu

|tu|3 →+∞, ast→+∞uniformly inx∈R

3. (2.13)

Thus, it follows from (V0), (2.11), and (2.13) that

R3

V(x)(tut)2+K(x)φ(tut)(tut)2–f(x,tut)tut

dx→–∞, ast→+∞.

Thus, takingv=TuTforTlarge, we havevΛ. Hence,Λ=∅. From (2.5), it is easy to see

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Next, we prove the last part of the lemma. Let uΛbe fixed and define a function g(t) :=Φ(tu),tuon [0,∞). By (F2), one has

f(x,)tτf(x,τ)τt4–V(x)t2– 1(tτ)2, ∀x∈R3,t≥1,τ∈R, (2.14)

which yields

R3

V(x)(tτ)2+K(x)φtτ(tτ)2–f(x,)tτ

dx

t4

R3

V(x)τ2+K(x)φττ2–f(x,τ)τ

dx, ∀t≥1,τ∈R. (2.15)

From (2.5) and (2.15) it follows that

g(t)t2u2+t4

R3

V(x)u2+K(x)φuu2–f(x,u)u

dx

t2

R3V(x)u

2dx, t1. (2.16)

Using (F0), (2.5), and (2.16), it is easy to verify thatg(0) = 0,g(t) > 0 fort> 0 small and g(t) < 0 fortlarge due touΛ. Therefore, there existst0=t(u) > 0 so thatg(t0) = 0 and

t(u)uN. We claim thatt(u) is unique for anyuΛ. In fact, for any given uΛ, let t1,t2> 0 such thatg(t1) =g(t2) = 0. Jointly with (2.6), we have

Φ(t1u)Φ(t2u) +

t41t24 4t14

Φ(t1u),t1u

+(t

2 1–t22)2

4t21

u22

=Φ(t2u) +

(t2 1–t22)2

4t2 1

u22 (2.17)

and

Φ(t2u)Φ(t1u) +

t4 2–t14

4t4 2

Φ(t2u),t2u

+(t

2 2–t12)2

4t2 2

u22

=Φ(t1u) +

(t22–t12)2 4t22u

2

2. (2.18)

(2.17) and (2.18) implyt1=t2. Hence,t(u) > 0 is unique for anyuΛ.

Lemma 2.4 Under assumptions(V0)and(F1)–(F3),then

inf

u∈NΦ(u) :=c=u∈infΛ,u=0maxt≥0 Φ(tu) > 0.

Proof Both Corollary2.2 and Lemma2.3 imply thatc=infuΛ,u=0maxt≥0Φ(tu). Using

Lemma2.1, it is easy to see thatc> 0.

Lemma 2.5 Under assumptions(V0)and(F1)–(F3),there exist a constant c∗∈(0,c]and a sequence{un} ⊂E satisfying

Φ(un)→c∗, Φ(un)1 +un

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Proof By (F1) and (1.3), we know that there existδ0> 0 andρ0> 0 such that

Φ(u)≥ρ0, u=δ0. (2.20)

In view of Lemmas2.3and2.4, we may choosevkNΛsuch that

c–1

k<Φ(vk) <c+ 1

k, k∈N. (2.21)

Using Lemma2.1and (2.20), it is easy to check thatΦ(tvk)≥ρ0for smallt> 0 andΦ(tvk) <

0 for larget> 0 due tovkΛ. SinceΦ(0) = 0, then the mountain pass lemma implies that

there exists a sequence{uk,n}n∈N⊂Esatisfying

Φ(uk,n)→ck, Φ(uk,n)1 +uk,n

→0, k∈N, (2.22)

whereck∈[ρ0,supt≥0Φ(tvk)]. By virtue of Corollary2.2, one hasΦ(vk) =supt≥0Φ(tvk).

Hence, by (2.21) and (2.22), one has

Φ(uk,n)→ck

ρ0,c+

1 k

, Φ(uk,n)1 +uk,n

→0, k∈N. (2.23)

Now, we can choose a sequence{nk} ⊂Nsuch that

Φ(uk,nk)∈

ρ0,c+

1 k

, Φ(uk,nk)1 +uk,nk

<1

k, k∈N. (2.24)

Letuk=uk,nk,k∈N. Then, going if necessary to a subsequence, we have

Φ(un)→c∗∈[ρ0,c], Φ(un)1 +un

→0.

Lemma 2.6 Under assumptions(V0) and (F1)–(F3),any sequence {un} ⊂ E satisfying

(2.19)is bounded in E.

Proof By (2.6) witht= 0, one has

c∗+o(1) =Φ(un) –

1 4

Φ(un),un

≥1

4∇un

2

2. (2.25)

By (V0), it is easy to see that there existsV0> 0 such thatV(x)≥V0for allx∈R3. Since

Φ(un),un=o(1), it follows from (V0), (F1), and the Sobolev embedding inequality that

V0un22≤ un2+

R3

K(x)φunu

2

ndx

=

R3f(un)undx+o(1)

≤1 2V0un

2

2+C2un66+o(1)

≤1 2V0un

2

2+C2S–3∇un62+o(1), (2.26)

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3 Ground state solutions

In this section, we give the proof of Theorem1.1.

Proof of Theorem1.1 Lemma2.5implies the existence of a sequence{un} ⊂Esatisfying

(2.19), then

Φ(un)→c∗> 0,

Φ(un),un

→0. (3.1)

By Lemma2.6,{un}is bounded inE. If

δ:=lim sup

n→∞

sup

y∈R3

B1(y)

|un|2dx= 0,

then by Lion’s concentration compactness principle [24] or [37, Lemma 1.21],un→0

inLs(R3) for 2 <s< 6. Moreover, there exists C

3> 0 such thatun2≤C3. By (F0), for ε=c∗/2C23, there exists> 0 such that

lim sup

n→∞

R3

12f(x,un)unF(x,un)

dx≤3

2εC

2

3+ lim

n→∞un p p=

3c∗

4 . (3.2)

By (V0), (2.3), and (2.4), we have

lim sup

n→∞

R3K(x) φun(x)u

2

ndx=lim sup

n→∞

R3

R3

K(x)K(y) |xy| u

2

n(x)u2n(y) dxdy

K2 lim sup

n→∞

R3

R3

u2

n(x)u2n(y)

|xy| dxdy ≤C1K∞2 lim sup

n→∞

un412/5= 0, (3.3)

where, and in the sequel,C1= 83

2/3√3π. From (1.3), (2.5), (3.1), (3.2), and (3.3), one has

c∗ =Φ(un) –

1 2

Φ(un),un

+o(1)

= –1 4

R3K(x)

φun(x)u2ndx+

R3

1

2f(x,un)unF(x,un)

dx+o(1)

≤3c∗ 4 +o(1).

This contradiction showsδ> 0.

Going if necessary to a subsequence, we may assume the existence ofkn∈Z3such that

B2(kn)

|un|2dx>

δ

2. (3.4)

Letvn(x) =un(x+kn). Then

B2(0)

|vn|2dx>

δ

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SinceV(x),K(x), andf(x,u) are periodic onx, we have

Φ(vn)→c∗∈(0,c], Φ(vn)1 +vn

→0. (3.6)

Passing to a subsequence, we havevnv¯inE,vn→ ¯vinLsloc(R3), 2≤s< 6, andvn(x)→

¯

v(x) a.e. onR3. Thus, (3.5) implies that¯v= 0. For every φC0∞(R3), we have

Φv),φ= lim

n→∞

Φ(vn),φ

= 0.

Hence Φv) = 0. This shows thatv¯ ∈N is a nontrivial solution of Problem (1.1) and

Φ(v)¯ ≥c. It follows from (F2), (3.6), and Fatou’s lemma that

cc∗= lim

n→∞

Φ(vn) –

1 4

Φ(vn),vn

= lim

n→∞

1 4∇vn

2 2+

R3

1

4f(x,vn)vnF(x,vn) + V(x)

4 v

2

n

dx

≥1

4lim infn→∞ ∇vn

2

2+lim infn→∞

R3

1

4f(x,vn)vnF(x,vn) + V(x)

4 v

2

n

dx

≥1 4¯v

2+

R3

1

4f(x,¯v)v¯–F(x,v)¯

dx

=Φ(v) –¯ 1 4

Φv),v¯=Φ(v).¯

This shows thatΦ(v)¯ ≤c, and soΦ(v) =¯ c=infNΦ> 0.

4 Infinitely many geometrically distinct solutions

To prove Theorem1.2, we need some notations. Ford2≥d1> –∞andc∈R, we put

Φd2:=uE:Φ(u)d 2

, Φd1:=

uE:Φ(u)≥d1

, Φd2

d1 :=Φd1∩Φ

d2;

K:=uE\ {0}:Φ(u) = 0, Kc:=

uK:Φ(u) =c.

In view of Theorem1.1, under (V0), (F0), and (F1)–(F3), (1.1) has a nontrivial solution ¯

uH1(RN) satisfyingΦ(u) =¯ c

0:=inf> 0. Therefore,KKc0=∅. Following the

strat-egy of [32], we choose a subsetFofKsuch thatF= –Fand each orbitO(w)Khas a unique representative inF. It suffices to show that the setFis infinite. So from now on we assume by contradiction that

Fis a finite set. (4.1)

Lemma 4.1 κ:=inf{uv:u,vK,u=v}> 0.

Proof Choose{un},{vn} ⊂K such thatunvnκ. Then there existw1,w2∈F and

kn,ln∈Z3 such thatun=w1(·–kn) andvn=w2(·–ln). Putmn=knln. There are two

possible cases.

Case (1).{|mn|}is bounded. Passing to a subsequence,mn=m∈Z3, one has

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Case (2).{|mn|}is unbounded. Passing to a subsequence,|mn| → ∞, one has

κ+o(1) =unvn=w1(·–kn) –w2(·–ln)

=w1–w2(·–mn)=w1

2

+w221/2+o(1) > 0.

Both Cases (1) and (2) show thatκ> 0.

Lemma 4.2 (Discreteness of (PS)-sequences) Let cc0. If {un1},{u2n} ⊂ Φc are two

(PS)-sequences for Φ,then either u1

nu2n →0 as n→ ∞orlim infn→∞u1nu2n

min{κ,√2c0}> 0.

Proof First, we prove the boundedness of Palais–Smale sequences forΦ. Let{un}be such

that

Φ(un)→c> 0, Φ(un)→0. (4.2)

Then it follows from (F4) and (4.2) that, for largen∈N,

c+ 1 +unΦ(un) –

1 4

Φ(un),un

= 1 4un

2+

R3

1

4f(x,un)unF(x,un)

dx

≥1 –θ 4 un

2,

which implies that {un}is bounded in E. Thus, {un1},{u2n} ⊂Φc are two bounded

(PS)-sequences forΦ. Next, we fixpas in (F1), and we distinguish two cases. Case (1).u1

nu2np→0. In this case, we can prove thatlimn→∞u1nu2n= 0 in a similar

fashion as [32, Lemma 2.14]. Case (2).u1

nu2np0. Then again by [37, Lemma 1.21], there existε0> 0 andkn∈Z3

such that, after passing to a subsequence,

B1+N(kn)

u1nu2npdx=max

k∈Z3

B1+N(k)

u1nu2npdx≥ε0.

Using thatΦ is equivariant with respect to translations of the form uu(·–k) with k∈Z3, we may assume that{k

n}is bounded inZ3. We may pass to a subsequence such

that

u1nu1, u2nu2, u1= u2, Φu1=Φu2= 0.

We first consider the case whereu1= 0 andu2= 0, so thatu1,u2K. By Lemma4.1, one

haslim infn→∞u1

nu2nu1–u2 ≥κ. It remains to consider the case where either

u2= 0 oru1= 0. In this case, it is easy to see thatlim inf

n→∞u1nu2nu1–u2 ≥

2c0.

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Esuch that

W(u) ≤2Φ(u), Φ(u),W(u) ≥ Φ(u)2.

(4.3)

Now we consider the Cauchy problem: ⎧

⎨ ⎩

dt = –W(η),

η(0,u) =u. (4.4)

The basic existence-uniqueness theorem for ordinary differential equations implies that, for each uE, (4.4) has a unique solutionη(t,u) defined for t in a maximal interval (T–(u),T+(u)), andη(t,u) is odd with respect touE.

Lemma 4.3 Let uE\(K∪ {0}).Ifinft∈[0,T+(u))Φ(η(t,u)) > –∞, thenlimtT+(u)η(t,u)

exists and is a critical point ofΦ.

Proof From (4.3) and (4.4), we have

d dtΦ

η(t,u)= –Φη(t,u),(t,u)

≤–Φη(t,u)2< 0, ∀t∈0,T+(u). (4.5)

This shows that Φ(η(t,u)) is strictly decreasing on t ∈ [0,T+(u)), and so τ :=

limtT+(u)Φ(η(t,u)) exists.

Case (1).T+(u) < +∞. For 0≤t1<t2<T+(u), from (4.3) and (4.4), we have

η(t2,u) –η(t1,u)

t2

t1

(t,u)dt

≤2 t2

t1

Φη(t,u),(t,u)dt

≤2√t2–t1

t2

t1

Φη(t,u),(t,u)dt 1/2

= 2√t2–t1

Φη(t1,u)

Φη(t2,u)

1/2

≤2√t2–t1

Φ(u) –τ1/2.

SinceT+(u) < +, this implies thatlim

tT+(u)η(t,u) exists and then it must be a critical

point ofΦ(otherwise the trajectorytη(t,u) could be continued beyondT+(u)).

Case (2).T+(u) = +. To prove thatlim

tT+(u)η(t,u) exists, it suffices to show that

for everyε> 0, there exists> 0 such thatη(tε,u) –η(t,u)<ε, ∀t. (4.6)

We suppose by contradiction that (4.6) is false. Then there exist 0 <ε0< 12min{κ,

√ 2c0}

and a sequence{tn} ⊂[0, +∞) such that

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Choose the smallestt1

n∈(tn,tn+1) ands1n∈[tn,t1n] such that

ηt1n,uη(tn,u)=

ε0

3 and Φ ηs1

n,u= min t∈[tn,t1n]

Φη(t,u). (4.8)

Then it follows from (4.3), (4.4), and (4.8) that

ε0

3 =η

tn1,uη(tn,u)

t1

n

tn

(t,u)dt

≤2 t1n

tn

Φη(t,u)dt

2

Φ(η(s1

n,u))

tn1

tn

Φη(t,u)2dt

≤ 2

Φ(η(s1

n,u))

tn1

tn

Φη(t,u),(t,u)dt

= 2

Φ(η(s1

n,u))

Φη(tn,u)

Φηt1n,u.

Since Φ(η(tn,u)) –Φ(η(tn1,u))→0 as n→ ∞, the above implies that Φ(η(s1n,u))

0 as n→ ∞. Similarly we find the largest t2

n∈ (tn,tn+1) and s2n ∈[tn2,tn+1] such that

η(tn+1,u) –η(t2n,u)=

ε0

3 andΦ(η(s 2

n,u))→0 as n→ ∞. Letu1n:=η(s1n,u) andu2n:=

η(s2n,u). Then{u1n}and{u2n}are two Palais–Smale sequences ofΦ such that ε0 3 ≤ u

1

n

u2

n ≤2ε0<min{κ,

2c0}. This, however, contradicts Lemma4.2, hence (4.6) is true. So

limtT+(u)η(t,u) exists, and it must be a critical point ofΦ.

In the following, for a subsetAEandδ> 0, we put(A) :={vE:dist(v,A) <δ}.

Lemma 4.4 Let cc0.Then,for everyδ> 0,there existsε=ε(δ) > 0such that (a) Φc+ε

cεK=Kc;

(b) limtT+(u)Φ(η(t,u)) <cεforuΦc+ε\Uδ(Kc).

Proof In view of (4.1) and theZ3-translation invariance forΦ,Φ(K) :={Φ(w) :wK}is

a finite set. Therefore, there existsεˆ> 0 such that (a) is satisfied forε∈(0,εˆ). Without loss of generality, we may assume (Kc) ⊂ Φc+1 and δ <min{κ,

√ 2c0}.

Next we find ε∈(0,εˆ) such that (b) holds. Let uΦc+εˆ \U

δ(Kc). SinceΦ(η(t,u)) is

strictly decreasing on t∈[0,T+(u)), if Φ(η(t

0,u))cεˆ for somet0∈[0,T+(u)), then

limtT+(u)Φ(η(t,u)) <cεˆ. Thus, we only consider the case whereΦ(η(t,u)) >cεˆfor all

t∈[0,T+(u)). In this case, it follows from Lemma4.3thatlim

tT+(u)η(t,u) exists and is a

critical point ofΦ. Set

α:=infΦ(w):w(Kc)\/2(Kc)

. (4.9)

We claim thatα> 0. Indeed, suppose by contradiction that there exists a sequence{u1

n} ⊂

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condition (4.1) and theZ3-invariance ofΦ, we may assume{u1

n} ⊂(w0)\/2(w0) for

somew0∈Kc. Letu2n=w0. Then

δ

2≤lim supn→∞

u1nu2nδ<min{κ,√2c0},

which contradicts Lemma4.2. Henceα> 0.

Letβ:=sup{Φ(w):w(Kc)\/2(Kc)}and 0 <ε<α

2δ

4β. By Lemma4.3and (a), the

only way (b) can fail is thatη(t,u)→ ˜wKcastT+(u) for someuΦc+ε\(Kc). In

this case we let

t1:=sup

t∈0,T+(u):η(t,u) /(w)˜

and

t2∈

t∈0,T+(u):η(t,u)

2(w)˜

.

(4.10)

Then from (4.3), (4.4), and (4.10), we have

δ

2≤η(t2,u) –η(t1,u)t2

t1

(t,u)dt≤2 t2

t1

Φη(t,u)dt≤2β(t2–t1)

and

Φη(t2,u)

Φη(t1,u)

≤–

t2

t1

Φη(t,u)2dt≤–α2(t2–t1)≤– α2δ

4β.

Hencelimt2→T+(u)Φ(η(t2,u))c+ε

α2δ

4β <c, contrary to our assumption.

Proof of Theorem1.2 Forj∈N, we consider the familyΣjof all closed and symmetric

subsetsAE\ {0}(i.e., A= –A=A) with¯ γ(A)≥j, whereγ denotes the usual Kras-noselskii genus (see, e.g., [27,30]). Moreover, we consider the nondecreasing sequence of Lusternik–Schnirelman values forΦ defined byck:={c∈R:γ(Φc)≥k}fork∈N. We

claim:

Kck=∅ and ck<ck+1, k∈N. (4.11)

To prove this, letk∈Nandc=ck. In view of Lemma4.1,γ(Kc) = 0 ifKc=∅orγ(Kc) = 1

ifKc=∅. By the continuity property of the genus, there existsδ> 0 such thatγ(Uδ(Kc)) =

γ(Kc). Chooseε=ε(δ) > 0 such that the properties of Lemma4.4hold. Then, for every

uΦc+ε\U

δ(Kc),tu∈[0,T+(u)), wheretu:=inf{t∈[0,T+(u)) :Φ(η(t,u))cε}. Since

η(t,u) is odd with respect touandΦis even, it implies thattu=tu. Define a map

h:Φc+ε\(Kc)→Φcε, h(u) =η(tu,u).

Thenhis odd and continuous. Henceγ(Φc+ε\(Kc))≤γ(Φcε)≤k– 1 and therefore

γΦc+εγUδ(Kc)

+k– 1 =γ(Kc) +k– 1.

The definition ofc=ck and ofck+1 implies thatγ(Kc)≥1 ifck+1>ck andγ(Kc) > 1 if

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It follows now from (4.11) that there is an infinite sequence{uk}of pairs of geometrically

distinct critical points ofΦwithΦ(uk) =ck, contrary to (4.1). The proof is finished.

Acknowledgements

The authors would like to thank the referees for their useful suggestions.

Funding

The authors are supported financially by the National Natural Science Foundation of China (No: 11501190) and Hunan Provincial Natural Science Foundation (No: 2019JJ50146).

Availability of data and materials

Not applicable.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The research was carried out in collaboration. All authors read and assured the final manuscript.

Author details

1School of Mathematics and Computing Sciences, Hunan University of Science and Technology, Xiangtan, P.R. China. 2School of Mathematics and Statistics, Central South University, Changsha, P.R. China.

Publisher’s Note

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

Received: 18 January 2019 Accepted: 22 March 2019

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