Startseite Ground state solutions for magnetic Schrödinger equations with polynomial growth
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Ground state solutions for magnetic Schrödinger equations with polynomial growth

  • Yan Wu und Peng Chen EMAIL logo
Veröffentlicht/Copyright: 17. Juni 2024

Abstract

In this article, we investigate the following nonlinear magnetic Schrödinger equations:

( i + A ( x ) ) 2 u + V ( x ) u = f 1 ( x , v 2 ) v , ( i + A ( x ) ) 2 v + V ( x ) v = f 2 ( x , u 2 ) u ,

where V is the electric potential and A is the magnetic potential. Assuming that the nonlinear function f i ( i = 1 , 2 ) satisfies three types of polynomial growth assumptions: super-quadratic, asymptotically quadratic, and local super-quadratic at x , we prove the existence of the Nehari-Pankov type ground state solutions using critical point theory together with the non-Nehari manifold method. The resulting problem engages two major difficulties: the first one is that the associated functional is strongly indefinite, and the second lies in verifying the link geometry and showing the boundedness of Cerami sequences. Our results extend and complement the present ones in the literature.

MSC 2010: 35J20; 35J60; 35Q55

1 Introduction

1.1 Background and motivations

In this article, we are concerned with the following magnetic Schrödinger equations:

(1.1) ( i + A ( x ) ) 2 u + V ( x ) u = f 1 ( x , v 2 ) v , ( i + A ( x ) ) 2 v + V ( x ) v = f 2 ( x , u 2 ) u ,

where i is the imaginary unit and f C ( R N × R , R ) is the reaction. System (1.1) arises from looking for the standing waves solution ( ψ ( x , t ) , φ ( x , t ) ) ( e i E t u ( x ) , e i E t v ( x ) ) ( E R ) of the nonlinear evolution equation:

i ψ t = [ i A ( x ) ] 2 ψ + U ( x ) ψ f ( φ 2 ) φ , x R N , i φ t = [ i A ( x ) ] 2 ψ + U ( x ) ψ f ( ψ 2 ) ψ , x R N ,

where is the Planck constant. U ( x ) is a real electric potential and the nonlinear term and f is a superlinear function. The Schrödinger system, which merges the wave system with the concept of matter wave, is a fundamental postulate in quantum mechanics. The standing wave solution of the Schrödinger equation is applicable in describing optical solutions in light, the movement of superconductors in magnetic fields, and physical phenomena like Bose-Einstein condensates. It plays a vital role in theories of nonlinear optics, electromagnetism, superconductivity, and other related fields. Formally describing the transition from quantum mechanics to classical mechanics involves setting 0 , and therefore, the physical significance lies in the existence of solutions for small values of .

Over the past few decades, significant progress has been made in understanding the existence, multiplicity, and dynamic behavior of the Schrödinger system without magnetic fields (i.e., A 0 ), resulting in numerous excellent results [38,16,22,2426,29]. The introduction of magnetic potential makes the case A 0 more complicated in comparison to case A 0 . For a single equation:

(1.2) ( i + A ) 2 u + V ( x ) u = g ( x , u ) u ,

which has been extensively investigated in the literature based on various assumptions on the potentials A , V and the nonlinearity g ( x , u ) u . To our best knowledge, the first work was studied by Esteban and Lions [13] by assuming that σ ( Δ A + V ) ( 0 , + ) , where Δ A ( i + A ) 2 . They investigated the existence of solutions of (1.2) by solving an appropriate minimization problem for the corresponding energy functional in the case of N = 2 and 3. Arioli and Szulkin [2] considered nontrivial solutions for (1.2) in the cases of the spectrum σ ( Δ A + V ) ( 0 , + ) when g is of subcritical growth and critical growth. Recently, Wen et al. [28] considered the following nonlinear magnetic Schrödinger equation:

( i + A ) 2 u + V ( x ) u = f ( x , u 2 ) u , x R 2

with exponential growth, and they proved the existence of ground state solutions both in the indefinite case with the nonlinear function is subcritical exponential growth and in the definite case with the nonlinear function is critical exponential growth. To overcome the difficulty brought about by the presence of a magnetic field, they used subtle estimates and established a new energy estimate inequality in complex field. For the situation of the Schrödinger equation with a single magnetic field potential, we refer the readers to previous studies [1,17,18] and the references therein.

For the Schrödinger equations, Ding and Liu [11] explored the Schrödinger equations with a magnetic field:

( i ε + A ( x ) ) 2 ω + V ( x ) ω = W ( x ) g ( ω ) ω , ( i ε + A ( x ) ) 2 ω + V ( x ) ω = W ( x ) g ( ω + ω 2 * 2 ) ω .

They studied the existence and concentration phenomena of semiclassical solutions with critical growth of nonlinear terms. Zhang et al. [30] studied the following nonlinear Schrödinger system:

ε i A ( y ) 2 u + V ( y ) u = H u ( x , u , v ) , y R N , ε i A ( y ) 2 v + V ( y ) v = H v ( x , u , v ) , y R N .

They proved the existence of semiclassical solutions for nonlinearity, which is both super-quadratic and subcritical by using generalized linking theorems for strongly indefinite functionals. For more related results about the magnetic Schrödinger system, one can refer to previous studies [14,19,20,31] and the references therein.

Inspired by the aforementioned works [9,10,27,28], we will use non-Nehari approach developed recently by Tang et al. [27] to demonstrate the ground state solutions for the magnetic Schrödinger system (1.1) under three types of polynomial growth conditions: super-quadratic, asymptotically quadratic, and local super-quadratic at infinity. Throughout this article, we assume the following basic hypotheses:

(V) V C ( R N , R + ) , V ( x ) is 1-periodic in x i for i = 1 , , N and

sup [ σ ( Δ A + V ) ( , 0 ) ] < 0 < inf [ σ ( Δ A + V ) ( 0 , ) ] ,

where Δ A ( i + A ) 2 ;

(A) A L loc p ( R N , R N ) with p > 2 , and B = curl A is 1-periodic in x i for i = 1 , 2 , , N ;

(F1) f 1 , f 2 C ( R N , R + ) is 1-periodic in t , 1-periodic in x i for i = 1 , , N , and there exists a constant C > 0 such that

f i ( x , t ) C 1 + t p 2 2 , ( x , t ) R N × R + , i = 1 , 2 ,

where p ( 2 , 2 * ) , and

2 * = 2 N N 2 , if N 3 , + , if N = 1 or 2 ;

(F2) f i ( x , t ) = o ( t ) as t 0 uniformly in x R N , for i = 1 , 2 ;

(F3) f i ( x , t ) is nondecreasing in t on ( 0 , ) , for i = 1 , 2 .

1.2 Main difficulties and ideas

We point out that the strongly indefinite features of the problem (1.1) together with the magnetic potential A ( x ) bring some new difficulties in our analysis, which can be summarized as follows:

  1. the existence and compactness of (PS) sequence become more difficult to demonstrate due to the destruction of the mountain-pass geometry of the energy functional corresponding to the strongly indefinite potential;

  2. due to the presence of magnetic potential A ( x ) that makes (1.1) a complex-valued problem, the existing methods for solving the Schrödinger equation with indefinite real values are invalid for (1.1);

  3. our main obstacle in the case of asymptotically quadratic or locally super-quadratic functions is to achieve the linking geometry and boundedness of Cerami sequences since their arguments become invalid when the nonlinearity is not globally super quadratic;

  4. due to the lack of the strict monotonicity condition, seeking a ground state solution of (1.1) in the Nehari-Pankov manifold is difficult, which needs some new techniques and method.

To circumvent these difficulties, we utilize the spectrum properties of the corresponding operator, which were previously analyzed by Zhao and Ding [32]. This allows us to conveniently decompose the functional space L 2 into a direct sum of two subspaces, E + and E ( E as defined in Section 2).

It is worth noting that in the case of local super-quadratic that we are studying, the local super-quadratic condition allows for nonlinearity to be super-quadratic in some domain and asymptotically quadratic in others, and in this case, we have no global information on the nonlinearity, so the non-Nehari manifold method seems not to work for our problem under the local super-quadratic condition. To prove the existence of ground state solutions in this case about (1.1), motivated by Tang et al. [27], we present a perturbation approach by adding a perturbation term of power function. The idea is to obtain existence of critical points of perturbation functional and to establish suitable estimates for the critical points so that we may pass to the limit to get solutions of the original problem.

1.3 Organization of the paper

Section 2 is dedicated to studying strongly indefinite structure and give some preliminaries. We will recall the abstract linking theorem in the study by Li and Szulkin [15], which will be used to prove the existence of solutions in the three cases we consider. Section 3 focuses on the super quadratic case. To achieve this, we will decompose the space E into appropriate subspaces for the linking structure, utilizing the profile of the spectrum presented by the associated operator. We will then verify the requirements of the abstract result, including compactness, linking geometry, and boundedness of Cerami sequences for the functional associated with problem (1.1). Sections 4 and 5 discuss asymptotic quadratic and local super-quadratic cases, respectively. Our main obstacle in the case of asymptotically quadratic or locally super quadratic functions is to achieve the linking geometry and boundedness of Cerami sequences, since their arguments become invalid when the nonlinearity is not globally super quadratic. Therefore, it is advisable to concentrate on the argument necessary to boundedness of the Cerami sequences, which is more challenging than the super quadratic case.

Notation

  • p denotes the usual norm of L p ( R N , C ) ;

  • B ( x , R ) denotes the ball centered at x with the radius R ;

  • C i and C ˜ i ( i = 0 , 1 , 2 , ) denote positive constants which may be different in different places.

2 The variational setting and preliminaries

Let Δ A + V , then is a self-adjoint operation with domain D ( ) = H 2 ( R N , C ) (see [12]). Let { ( λ ) : λ + } and be the spectral family and the absolute value of , respectively, and denote the square root of by 1 2 . Set = U the polar decomposition of , where U = i d ( 0 ) ( 0 ) commuting with , , and 1 2 . Let E = D ( 1 2 ) and define

E = ( 0 ) E , E 0 = [ ( 0 ) ( 0 ) ] E and E + = [ i d ( 0 ) ] E .

Then for any u E , we have u = u + u 0 + u + , where

u = ( 0 ) u E , u 0 = [ ( 0 ) ( 0 ) ] u E 0 and u + = [ i d ( 0 ) ] u E + .

Under the assumption (V), E 0 = { 0 } = ker { } , then E = E E + is a Hilbert space with the inner product and the corresponding norm defined by

u , v = 1 2 u , 1 2 v 2 , u , v E

and

u = u , u 1 2 , u E ,

respectively. For any u E , it is easy to verify that u = u + u + and

u = ± u , u E ± D ( ) .

To establish the variational structure of problem (1.1), we set

H A 1 ( R N , C ) { u L 2 ( R N , C ) : A u L 2 ( R N , C ) } ,

where A u ( + i A ) u . H A 1 ( R N , C ) is a Hilbert space endowed with the inner product

u , v A Re R N ( A u A v ¯ + u v ¯ ) d x .

where Re ( ω ) denotes the real part of ω C and ω ¯ denotes its conjugate.

In such situation, E = H A 1 ( R N , C ) × H A 1 ( R N , C ) with equivalent norms under ( V ) , E can be embedded continuously in L s ( R N ) for all 2 s 2 * , i.e., there exists an embedding constant γ s > 0 such that

z s γ s z , z E , s [ 2 , 2 * ] .

We define the following inner product on E ,

( z 1 , z 2 ) ( u 1 , u 2 ) + ( v 1 , v 2 ) , z i = ( u i , v i ) E , i = 1 , 2 .

Then E is a Hilbert space, and the associated norm is

z u 2 + v 2 , z = ( u , v ) E .

Under assumptions (V1), (F1), and (F2), the corresponding energy functional of problem (1.1) can be defined as follows:

(2.1) ( z ) = 1 2 ( z + 2 z 2 ) 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x , z = ( u , v ) E .

Set

(2.2) Ψ ( z ) = 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x , z = ( u , v ) E .

Using (F1) and (F2), for any ε > 0 , there exists C ε > 0 such that

(2.3) f i ( x , t ) ε + C ε t p 2 2 and F i ( x , t ) ε t + C ε t p 2 , for i = 1 , 2 , ( x , t ) ( R N , R + ) .

In view of (2.3), is of class C 1 ( E , R ) , and

(2.4) ( z ) , z = ( z + 2 z 2 ) R N [ f 2 ( x , u 2 ) u 2 + f 1 ( x , v 2 ) v 2 ] d x , z = ( u , v ) E .

Moreover, for any z = ( u , v ) , ζ = ( ϕ , ψ ) E ,

(2.5) ( z ) , ζ = z + , ζ + z , ζ Re R N [ f 2 ( x , u 2 ) u ϕ ¯ + f 1 ( x , v 2 ) v ψ ¯ ] d x .

To find a ground state solution of (1.1), we define the following Nehari-Pankov manifold introduced by Pankov [21]:

N { z E \ E : ( z ) , z = ( z ) , ζ = 0 , ζ E } .

N is a natural constraint and contains all nontrivial critical points of the energy functional, and every minimizer z of on the manifold N is a solution, which is called a ground state solution of Nehari-Pankov type.

Lemma 2.1

[15] Let X be a real Hilbert space, X = X X + and X X + , and let φ C 1 ( X , R ) of the form

φ = 1 2 ( z + 2 z 2 ) ψ ( z ) , z = z + z + X X + .

Suppose that the following assumptions are satisfied:

  1. ψ C 1 ( X , R ) is bounded from below and weakly sequentially lower semi-continuous;

  2. ψ is weakly sequentially semi-continuous;

  3. There exist r > ρ > 0 and e X + , e = 1 such that

    κ inf φ ( S ρ + ) > 0 sup φ ( Q ) ,

    where

    S ρ + = { u X + : u = ρ } , Q = { v + s e : v X , s 0 , v + s e r } .

    Then for some c [ κ , sup φ ( Q ) ] , there exists a sequence { u n } X satisfying

    φ ( u n ) c , φ ( u n ) ( 1 + u n ) 0 .

3 Super-quadratic elliptic systems

3.1 Main assumptions and results

(F3’) f i ( x , t ) = g i ( x , t ) + j i ( x , t ) , g i is 1-periodic in x i for i = 1 , , N , g i : R N × R + R + is continuous in t R + for a.e. x R N , g i ( x , t ) = o ( t ) as t 0 , uniformly in x R N ; g i is nondecreasing in t on ( 0 , ) . Furthermore, j i C 1 ( R N , R + ) satisfies that

0 t j i ( x , t ) a ( x ) t + t p 2 ,

where a C ( R N , R ) , lim x a ( x ) = 0 , and 2 < p < 2 * ;

(F4) lim t F i ( x , t ) t = uniformly in x R N , for i = 1 , 2 .

Theorem 3.1

Assume that (V), and (A), (F1)–(F4) are satisfied, then system (1.1) has at least a ground state solution, i.e., it has at least a solution z E such that ( z ) = inf N > 0 .

Theorem 3.2

Assume that (V), (A), (F1), (F2), (F3’), and (F4) are satisfied, then system (1.1) has at least a ground state solution, i.e., it has at least a solution z E such that ( z ) = inf N > 0 .

Example 3.3

F i ( x , t ) = h ( x ) a t ϱ for i = 1 , 2 , where ϱ ( 2 , 2 * ) and h C ( R N , ( 0 , ) ) is 1-periodic in each of x 1 , x 2 , , x N .

Remark 3.4

As far as we know, there have been no scholars who have conducted a systematic study on the existence of ground state solutions for the (1.1) Schrödinger equation system with a magnetic field.

3.2 Proof of the main results

As a direct result of [28, Lemma 3.1], we have the following lemma.

Lemma 3.5

Assume that (F1)–(F3) are satisfied, for any z 1 , z 2 C and k 0 , there holds

(3.1) 1 k 2 2 f i ( x , z 1 2 ) z 1 2 k f i ( x , z 1 2 ) Re ( z 1 z 2 ¯ ) + 1 2 F i ( x , k z 1 + z 2 2 ) 1 2 F i ( x , z 1 2 ) 0 , i = 1 , 2 .

Lemma 3.6

Assume that (V) and (F1)–(F3) are satisfied, Then Ψ is nonnegative, weakly sequentially lower semi-continuous, and Ψ is weakly sequentially continuous.

Proof

Suppose that z n z as n in E . Let z n = ( u n , v n ) , z = ( u , v ) . Going if necessary to a subsequence, we can assume z n z in L loc p ( R N ) for p [ 2 , 2 * ) and z n ( x ) z ( x ) a.e. on R N . Thus,

Ψ ( z ) = 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x = 1 2 R N lim n [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] d x 1 2 liminf n R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] d x = liminf n Ψ ( z n ) .

Next, we show that Ψ is weakly sequentially continuous.

Ψ ( z ) , ζ = Re R N [ f 2 ( x , u 2 ) u ϕ ¯ + f 1 ( x , v 2 ) v ψ ¯ ] d x , z = ( u , v ) , ζ = ( ϕ , ψ ) E .

Firstly, we show that Re R N f 1 ( x , v 2 ) v ψ ¯ d x , v , ψ H A 1 ( R N , C ) is weakly sequentially continuous. Let v n v in H A 1 ( R N , C ) , then v n C 1 for some constants C 1 > 0 . For any ε > 0 , the decay of integral implies that there exists R ε > 0 such that

R N \ B ( 0 , R ε ) ψ 2 < ε 2 .

From (F1) and the Hölder inequality, one has

(3.2) Re R N \ B ( 0 , R ε ) f 1 ( x , v n 2 ) v n ψ ¯ d x R N \ B ( 0 , R ε ) f 1 ( x , v n 2 ) v n ψ d x R N \ B ( 0 , R ε ) C ( 1 + v n p 2 ) v n ψ d x C v n 2 + 2 v n p p 2 + v n 2 p 2 p 1 R N \ B ( 0 , R ε ) ψ 2 d x 1 2 C 1 ε .

Similarly, we have

(3.3) Re R N \ B ( 0 , R ε ) f 1 ( x , v 2 ) v ψ ¯ d x C 1 ε .

Moreover, the absolute continuity of integrals implies that there exists δ > 0 such that

A ψ 2 < ε 2

for any A B ( 0 , R ε ) with meas ( A ) < δ . Since { u n } is bounded in H A 1 , there exists M ε > 0 such that

meas ( Ω n [ M ε , ) ) δ and meas ( Ω [ M ε , ) ) δ ,

where Ω n [ M ε , ) { x B ( 0 , R ε ) : u n ( x ) M ε } and Ω [ M ε , ) { x B ( 0 , R ε ) : u ( x ) M ε } . Set Ω ( M ε ) { x B ( 0 , R ε ) : u ( x ) = M ε } , then similar to the proof of (3.2), one has

(3.4) Re Ω n [ M ε , ) Ω ( M ε ) f 1 ( x , v n 2 ) v n ψ ¯ d x C v n 2 + 2 v n p p 2 + v n 2 p 2 p 1 Ω n [ M ε , ) Ω ( M ε ) ψ 2 d x 1 2 C 2 ε ,

and

(3.5) Re Ω ( M ε ) f 1 ( x , v n 2 ) v n ψ ¯ d x C 2 ε .

Passing to a subsequence, we have u n u in L loc s ( R N ) for 2 s < 2 * and u n u a.e. on R N , then

f 1 ( x , v n 2 ) v n ψ ¯ X v n M ε f 1 ( x , v 2 ) v ψ ¯ X v M ε , a.e. in B ( 0 , R ε ) \ Ω ( M ε ) .

Moreover,

f 1 ( x , v n 2 ) v n ψ ¯ X v n M ε ψ max x B ( 0 , R ε ) , t M ε f ( x , t 2 ) t , x B ( 0 , R ε ) .

Therefore, it is easy to see from the Lebesgue dominated convergence theorem that

(3.6) lim n B ( 0 , R ε ) \ ( Ω n [ M ε , ) Ω ( M ε ) ) f 1 ( x , v n 2 ) v n ψ ¯ d x = lim n B ( 0 , R ε ) \ Ω [ M ε , ) f 1 ( x , v 2 ) v ψ ¯ d x .

Noting that Re ( z 1 z 2 ) z 1 z 2 for any z 1 , z 2 C , one has

(3.7) lim n Re B ( 0 , R ε ) \ ( Ω n [ M ε , ) Ω ( M ε ) ) f 1 ( x , v n 2 ) v n ψ ¯ d x = lim n Re B ( 0 , R ε ) \ Ω [ M ε , ) f 1 ( x , v 2 ) v ψ ¯ d x .

By using (3.2)–(3.7), we have

lim n Re R N f 1 ( x , v n 2 ) v n ψ ¯ d x R N f 1 ( x , v 2 ) v ψ ¯ d x lim n Re R N \ B ( 0 , R ε ) f 1 ( x , v n 2 ) v n ψ ¯ d x R N \ B ( 0 , R ε ) f 1 ( x , v 2 ) v ψ ¯ d x + lim n Re B ( 0 , R ε ) \ ( Ω n [ M ε , ) Ω ( M ε ) ) f 1 ( x , v n 2 ) v n ψ ¯ d x Re B ( 0 , R ε ) \ Ω n [ M ε , ) f 1 ( x , v 2 ) v ψ ¯ d x + lim n Re ( Ω n [ M ε , ) Ω ( M ε ) ) f 1 ( x , v n 2 ) v n ψ ¯ d x Re Ω [ M ε , ) f 1 ( x , v 2 ) v ψ ¯ d x C 3 ε .

Due to the arbitrariness of ε , we have

lim n Re R N f 1 ( x , v n 2 ) v n ψ ¯ d x R N f 1 ( x , v 2 ) v ψ ¯ d x = 0 .

Similarly,

lim n Re R N f 2 ( x , u n 2 ) v n ψ ¯ d x R N f 2 ( x , u 2 ) u ψ ¯ d x = 0 ,

which implies

lim n Ψ ( z n ) , ζ = Ψ ( z ) , ζ .

Hence, Ψ is weakly sequentially continuous.□

Lemma 3.7

Assume that (V) and (F1)–(F3) are satisfied, Then for all k 0 , z = ( u , v ) E , ζ = ( ϕ , ψ ) E ,

( z ) ( k z + ζ ) + 1 2 ζ 2 + 1 k 2 2 ( z ) , z k ( z ) , ζ .

Proof

By (2.1) and (3.1), we have

( z ) ( k z + ζ ) = 1 k 2 2 z + 2 1 k 2 2 z 2 + k z , ζ 1 2 ζ 2 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x + 1 2 R N [ F 2 ( x , k u + ϕ 2 ) + F 1 ( x , k v + ψ 2 ) ] d x = 1 2 ζ 2 + 1 k 2 2 ( z ) , z k ( z ) , ζ 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x + 1 2 R N [ F 2 ( x , k u + ϕ 2 ) + F 2 ( x , k v + ψ 2 ) ] d x + 1 k 2 2 R N [ f 2 ( x , u 2 ) u 2 + f 1 ( x , v 2 ) v 2 ] d x k Re R N [ f 2 ( x , u 2 ) u ϕ ¯ + f 1 ( x , v 2 ) v ψ ¯ ] d x 1 2 ζ 2 + 1 k 2 2 ( z ) , z k ( z ) , ζ .

By using Lemma 3.7, some important corollaries are given as follows, and the proof process will be omitted.

Corollary 3.8

Assume that (V) and (F1)–(F3) are satisfied. Then for all z N ,

( z ) ( k z + ζ ) + 1 2 ζ 2 , k 0 , ζ E .

Corollary 3.9

Assume that (V) and (F1)–(F3) are satisfied. Then for all z E , k 0 ,

( z ) ( k z + ) + k 2 2 z 2 + 1 k 2 2 ( z ) , z + k 2 ( z ) , z .

Lemma 3.10

Assume that (V), (F1)–(F3) are satisfied. Then

  1. There exists ρ > 0 such that

    m inf N Λ inf { ( z ) : z E + , z = ρ } > 0 ;

  2. z + max { z , 2 m } for all z N .

Proof

(i) By (F1) and (F2), there exist p ( 2 , 2 * ) and C 1 > 0 , such that

(3.8) F i ( x , t ) 1 2 γ 2 2 t + C 1 t p 2 , for i = 1 , 2 ( x , t ) R N × R .

By (3.8), we have

(3.9) R N F 2 ( x , u 2 ) d x R N 1 2 γ 2 2 u 2 + C 1 u p d x = 1 2 γ 2 2 u 2 2 + C 1 u p p

and

(3.10) R N F 1 ( x , v 2 ) d x 1 2 γ 2 2 v 2 2 + C 1 v p p .

From (3.9), (3.10), and Corollary 3.8, one has

( z ) = 1 2 z 2 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x 1 2 z 2 1 4 γ 2 2 z 2 2 C 1 2 z p p 1 4 z 2 C 2 z p , z = ( u , v ) E + .

This shows that there exists a ρ > 0 such that (i) holds.

(ii) By Lemma 3.6, Ψ ( z ) > 0 for all z E , so we have for z N

m 1 2 z + 2 1 2 z 2 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x 1 2 z + 2 1 2 z 2 1 2 z + 2 ,

which implies that z + max { z , 2 m } .□

Lemma 3.11

Assume that (V), (F1), (F2), and (F4) are satisfied. Then for every e E + , sup ( E R + e ) < and there exists R e > 0 such that

( z ) 0 , z E R + e , z R e .

Proof

Arguing indirectly, assume that for some sequence { ω n + s n e } E R + e with ω n + s n e such that ( ω n + s n e ) 0 for all n N . Set z n = ( u n , v n ) = ω n + s n e and ξ n = z n z n = ξ n + τ n e , then ξ n + τ n e = 1 . Passing to a subsequence, we may assume that ξ n ξ in E, then ξ n ξ a.e. on R N , ξ n ξ in E, τ n τ , and

(3.11) 0 ( z n ) z n 2 = τ n 2 2 e 2 1 2 ξ n 2 1 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] z n 2 d x .

Case (1) τ = 0 , then it follows from (3.11) that

0 1 2 ξ n 2 + 1 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] z n 2 d x τ n 2 2 e 2 0 .

This implies ξ n 2 0 and 1 = ξ n + τ n e 2 0 , which is a contradiction.

Case (2) τ 0 , then it follows from (3.11) and (F4) that

0 limsup n τ n 2 2 e 2 1 2 ξ n 2 1 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] z n 2 d x τ n 2 2 e 2 1 2 liminf n R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] u n 2 + v n 2 ξ n + τ n e 2 d x τ n 2 2 e 2 1 4 liminf n R N F 2 ( x , u n 2 ) max { u n 2 , v n 2 } + F 1 ( x , v n 2 ) max { u n 2 , v n 2 } ξ n + τ n e 2 d x = ,

which is a contradiction.□

Corollary 3.12

Assume that (V), (F1), (F2), and (F4) are satisfied. Let e E + with e = 1 , Then there exists r 0 > ρ such that sup ( Q ) 0 as r r 0 , where

(3.12) Q = { ζ + s e : ζ E , s 0 , ζ + s e r } .

Lemma 3.13

Assume that (V), (F1), (F2), (F3), and (F4) are satisfied. Then for any z E \ E , N ( E R + z ) , there exist η ( z ) > 0 , ζ ( z ) E such that η ( z ) z + ζ ( z ) N .

Lemma 3.14

Assume that (V), (F1), (F2), and (F4) are satisfied. Then there exists a constant c [ Λ , sup ( Q ) ] and a sequence { z n } E satisfying

( z n ) c , ( z n ) ( 1 + z n ) 0 ,

where Q is defined in (3.12).

Proof

Combining with Lemmas 2.1, 3.6, 3.10, and Corollary 3.12, it is easy to verify Lemma 3.14. The proof will be omitted.□

Lemma 3.15

Assume that (V), (F1), (F2), (F3), and (F4) are satisfied. Then there exists a constant c * [ Λ , m ] and a sequence { z n } E satisfying

(3.13) ( z n ) c * , ( z n ) ( 1 + z n ) 0 .

Lemma 3.16

Assume that (V), (A), (F1), (F2), (F3), and (F4) are satisfied. Then for any { z n } E , satisfying (3.13) is bounded in E.

Proof

To prove the boundedness of { z n } , arguing by contradiction, suppose that z n . Let ω n = z n z n , then ω n = 1 . By the Sobolev embedding theorem, there exists a constant C 3 > 0 such that ω n 2 C 3 .

δ lim n sup y R N B 1 ( y ) ω n + 2 d x = 0 .

Then the Lions’ concentration compactness lemma implies ω n + 0 in L p ( p ( 2 , 2 * ) ) . For fix s > 2 ( c * + 1 ) , combining (F1) with (F2), we see that there exists a constant C ε > 0 , such that

(3.14) F i ( x , t ) ε t + C ε t p 2 , i = 1 , 2 .

which, together with (3.14) and (2.2), yields that

Ψ ( z ) = 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x 1 2 R N [ ε u 2 + C ε u p + ε v 2 + C ε v p ] d x = 1 2 ( ε u 2 2 + ε v 2 2 + C ε u p p + C ε v p p ) = 1 2 ( ε z 2 2 + C ε z p p ) ,

for ε = 1 2 s 2 C 3 2 > 0 , where ( x , t ) ( R N , R + ) , and hence, we have

limsup n Ψ ( s ω n + 2 ) d x ε s 2 2 ω n + 2 2 + C 4 ω n + p p 1 4 + o ( 1 ) .

Let η n = s z n . Combining Corollary 3.9 with Lemma 3.14, we have

c * + o ( 1 ) = ( z n ) ( η n z n + ) + η n 2 2 z n 2 + 1 η n 2 2 ( z n ) , z n + η n 2 ( z n ) , z n = η n 2 2 z n + 2 Ψ ( x , η n z + ) + η n 2 2 z n 2 = η n 2 2 z n 2 Ψ ( x , s ω n + ) > c * + 3 4 + o ( 1 ) .

This leads to a contradiction, so δ > 0 . Hence, there exists { y n } R N such that

(3.15) B 1 ( y n ) ω n 2 d x > δ 2 .

In view of ( A ), for all z Z N , there holds

B ( x + z ) B ( x ) = curl A ( x + z ) curl A ( x ) = 0 ,

which means A ( x + z ) A ( x ) = ϕ z ( x ) , for some ϕ z H loc 1 ( R N , R ) . Define translation ϒ : E × Z N E by letting ϒ z ω ( x ) = ω ( x + z ) e i ϕ z ( x ) . Although a direct computation, we have

R N A [ ϒ z ω ( x ) ] d x = R N [ ω ( x + z ) e i ϕ z ( x ) ] + i A ( x ) ω ( x + z ) e i ϕ z ( x ) d x = R N e i ϕ z ( x ) ω ( x + z ) + i ω ( x + z ) ϕ z ( x ) e i ϕ z ( x ) + i A ( x ) ω ( x + z ) e i ϕ z ( x ) d x = R N [ ω ( x + z ) + i A ( x + z ) ω ( x + z ) ] e i ϕ z ( x ) d x = R N A ω ( x ) d x ,

and

R N ϒ z ω ( x ) q d x = R N ω ( x + z ) e i ϕ z ( x ) q d x = R N ω ( x ) q d x , q [ 2 , ) ,

where we note that e i ϕ z ( x ) = 1 for any x R N due to ϕ z H loc 1 ( R N , R ) . Consequently, for each z Z N , ϒ z is well defined and isometry. Furthermore, let ξ n = ϒ [ y n ] ω n , where [ x ] denotes the largest integer not exceeding x , then (3.15) implies

(3.16) B 1 + N ( 0 ) ξ n 2 d x > δ 2 .

Put z ˜ n ( x ) = z n ( x + k n ) , z ˜ n z n = ξ n , then ξ n = 1 . Passing to a subsequence, we may assume that ξ n ξ on E , and ξ n ξ on L loc 2 a.e. on R N . It is evident that (3.16) implies that ξ 0 . Thus, by virtue of (F4) and Fatou’s lemma, we see that

0 = lim n ( z n ) z n 2 = lim n 1 2 ( ω n + 2 ω n 2 ) 1 2 z n 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] d x 1 2 1 2 liminf n R N F 2 ( x , u n 2 ) 2 max { u n 2 , v n 2 } + F 1 ( x , v n 2 ) 2 max { u n 2 , v n 2 } ξ n 2 d x = .

which is a contradiction. Hence, the statement of Lemma 3.16 is proved.□

Proof of Theorem 3.1

In light of Lemma 3.16, there exists a bounded sequence { z n } E , with z n = ( u n , v n ) . satisfying Lemma 3.15. Hence, there exists a constant C 5 > 0 such that z n 2 C 5 .

If

δ limsup n sup y R N B 1 ( y ) z n 2 d x = 0 ,

then z n = ( u n , v n ) 0 in L p , where p ( 2 , 2 * ) . On the other hand, by virtue of (F1) and (F2), for ε = c * 4 C 5 2 > 0 , there exists a constant C ε > 0 , such that

f i ( x , t ) ε + C ε t p 2 2 , ( x , t ) R N × R + .

On the basis of the aforementioned discussion, we have

limsup n R N [ f 2 ( x , u n 2 ) u n 2 + f 1 ( x , v n 2 ) v n 2 ] d x ε C 5 2 + C ε lim n z n p p = c * 4 .

Thus,

c * = ( z n ) 1 2 ( z n ) , z n + o ( 1 ) = 1 2 R N [ f 2 ( x , u n 2 ) u n 2 + f 1 ( x , v n 2 ) v n 2 ] d x 1 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] d x + o ( 1 ) c * 8 + o ( 1 ) ,

which is a contradiction. Then δ > 0 .

Passing to the subsequence, through a similar argument of the proof of Lemma 3.16, we may assume that there exists { y n } R N , such that

B 1 + N ( y n ) z n 2 d x > δ 2 .

Set ξ n ( x ) = ϒ [ y n ] z n , then

(3.17) B 1 + N ( 0 ) ξ n 2 d x > δ 2 .

Due to the periodic assumption of V ( x ) and f i ( x , t ) , for i = 1 , 2 , , it follows that ξ n = z n and

( ξ n ) c * , ( ξ n ) ( 1 + ξ n ) 0 .

Thus, passing to the subsequence, suppose that ξ n ξ in E , ξ n ξ in L loc 2 , ξ n ( x ) ξ ( x ) a.e on R N . In light of (3.17), we see that ξ 0 . For every ϕ C 0 ( R N ) × C 0 ( R N ) , we have ( ξ ) , ϕ = lim n ( ξ n ) , ϕ = 0 . Hence, ( ξ ) = 0 , which implies that ξ N . Then, ( ξ ) m . On the other way, it follows from (F1) to (F4), Lemmas 3.6, 3.10, and Fatou’s lemma that

m c * = lim n ( z n ) 1 2 ( z n ) , z n = lim n 1 2 R N [ f 2 ( x , u n 2 ) u n 2 + f 1 ( x , v n 2 ) v n 2 F 2 ( x , u n 2 ) F 1 ( x , v n ) 2 ] d x = 1 2 R N [ f 2 ( x , u 2 ) u 2 + f 1 ( x , v 2 ) v 2 F 2 ( x , u 2 ) F 1 ( x , v 2 ) ] d x = ( z ) 1 2 ( z ) , z = ( z ) ,

which implies ( z ) m . So ( z ) = m = inf N > 0 . The proof is completed. Next, we claim that z 0 . By Lemma 3.14 and the concentration compactness arguments, it is easy to prove that has a nontrivial ground state solution z N such that ( z ) = m = inf z N .□

Proof of Theorem 3.2

By applying Lemmas 3.14 and 3.16, we deduce that there exists a bounded sequence z n = ( u n , v n ) E satisfying (3.13). Passing to a subsequence, we have z n z in E . Next, we prove z = ( u , v ) 0 . Arguing by contradiction, suppose that z = 0 , i.e., z n 0 in E , and so z n 0 in L loc s ( R N ) , 2 s < 2 * and z n 0 a.e. on R N . By (F3’), it is easy to show that

lim n R N J 2 ( x , u n 2 ) d x = 0 , lim n R N J 1 ( x , v n 2 ) d x = 0 .

(3.18) Re lim n R N j 2 ( x , u n 2 ) u n ϕ ¯ d x = 0 , Re lim n R N j 1 ( x , v n 2 ) v n ψ ¯ d x = 0 .

Note that

(3.19) ( z ) = G ( z ) + 1 2 [ J 2 ( x , u 2 ) + J 1 ( x , v 2 ) ] d x

and

(3.20) ( z ) , ξ = G , ξ + Re lim n R N [ j 2 ( x , u 2 ) u ϕ ¯ + j 1 ( x , v 2 ) v ψ ¯ ] d x ,

where

G ( z ) = 1 2 ( z + 2 z 2 ) 1 2 R N [ G 2 ( x , u 2 ) + G 1 ( x , v 2 ) ] d x , z = ( u , v ) E .

G ( z ) , ζ = z + , ζ + z , ζ Re R N [ g 2 ( x , u 2 ) u ϕ ¯ + g 1 ( x , v 2 ) v ψ ¯ ] d x , z = ( u , v ) , ζ = ( ϕ , ψ ) E .

From (3.13), (3.18), (3.19), and (3.20), one can obtain that

G ( z n ) c , G ( z n ) ( 1 + z n ) 0 .

Analogous to the proof of Theorem 3.1, through a similar argument of the proof of Lemma 3.16, we can prove that there exists { y n } R N such that

B 1 + N ( y n ) z n 2 d x > δ 2 .

Set ξ n ( x ) = ϒ [ y n ] z n , then

(3.21) B 1 + N ( 0 ) ξ n 2 d x > δ 2 .

Passing to a subsequence, we have ξ n ξ in E , ξ n ξ in L loc s ( R N ) , 2 s < 2 * , and ξ n ξ a.e.on R N . Obviously, (3.21) implies that ξ 0 , we have

G ( ξ n ) c , G ( ξ n ) ( 1 + ξ n ) 0 .

In the same way as the last part of the proof of Theorem 3.1, we can prove that G ( ξ ) = 0 and G ( ξ ) c .

It is easy to show that ξ + 0 . By Lemma 3.15, there exist κ 0 = κ ( ξ ) > 0 and ω 0 = ω ( ξ ) E such that κ 0 ξ + ω 0 N , and so ( κ 0 ξ + ω 0 ) m . Hence, we have

m c G ( ξ ) = G ( κ 0 z + ω 0 ) + 1 2 ω 0 2 + 1 κ 0 2 2 G ( z ) , z κ 0 G ( z ) , ω 0 1 2 R N [ G 2 ( x , u 2 ) + G 1 ( x , v 2 ) ] d x + 1 2 R N [ G 2 ( x , κ 0 u + ϕ 2 ) + G 1 ( x , κ 0 v + ψ 2 ) ] d x + 1 κ 0 2 2 R N [ g 2 ( x , u 2 ) u 2 + g 1 ( x , v 2 ) v 2 ] d x κ 0 R N [ g 2 ( x , u 2 ) u ϕ ¯ + g 1 ( x , v 2 ) v ψ ¯ ] d x G ( κ 0 z + ω 0 ) + 1 2 ω 0 2 = 1 2 ω 0 2 + ( κ 0 z + ω 0 ) + 1 2 R N [ J 2 ( x , κ 0 u + ϕ 2 ) + J 1 ( x , κ 0 v + ψ 2 ) ] d x > ( κ 0 z + ω 0 ) m .

This contradiction implies that z 0 . In the same way as the last part of the proof of Theorem 3.1, we can certify that ( z ) = 0 and ( z ) = m = inf N . This shows that z E is a solution to (1.1) with ( z ) = inf N > 0 . The proof is completed.□

4 Asymptotically quadratic elliptic systems

4.1 Main assumptions and results

(F5) f i ( x , t ) = V ( x ) + h i ( x , t ) for i = 1 , 2 . where V ( x ) C ( R N , R ) is 1-periodic in x i for i = 1 , , N and inf V ( x ) > 0 , and there exists z 0 E + \ 0 such that

z 0 2 ω 2 R N V ( x ) z 0 + ω 2 d x < 0 , ω E ,

h i ( x , t ) 0 , f i ( x , t ) h i ( x , t ) < 0 , H i ( x , t ) = o ( t ) as t , where H i ( x , t ) = 0 t h i ( x , s ) d s for i = 1 , 2 , uniformly in x R N .

(F5’) f i ( x , t ) = V ( x ) + h i ( x , t ) , for i = 1 , 2 , where V ( x ) C ( R N , R ) is 1-periodic in x i for i = 1 , , N and inf V ( x ) > Λ ¯ inf [ σ ( A ) ( 0 , + ) ] , h i ( x , t ) 0 and H i ( x , t ) = o ( t ) as t , where H i ( x , t ) = 0 t h i ( x , s ) d s for i = 1 , 2 uniformly in x R N .

Theorem 4.1

Assume that (V), (A), (F1)–(F3), and (F5) are satisfied, and then system (1.1) has at least a ground state solution, i.e., it has at least a solution z E such that ( z ) = inf N > 0 .

Theorem 4.2

Assume that (V), (A), (F1)–(F3), and (F5’) are satisfied, and then system (1.1) has at least a ground state solution, i.e., it has at least a solution z E such that ( z ) = inf N > 0 .

Example 4.3

f i ( x , t ) = V min { t ϱ , 1 } , where ϱ > 0 and V C ( R N ) is 1-periodic in each of x 1 . x 2 , , x N and inf V > Λ ¯ .

Example 4.4

f i ( x , t ) = V [ 1 ( 1 ln ( e + t ) ) ] for i = 1 , 2 , where V C ( R N ) is 1-periodic in each of x 1 . x 2 , , x N and inf V > Λ ¯ .

Remark 4.5

We point out that, as a consequence of Theorem 3.1, the minimax characterization of the least energy value m is provided by

m = ( z 0 ) = inf v E 0 + \ { 0 } max z E R + v ( z ) ,

where E 0 + is defined by (4.2). Note that this minimax principle is much simpler than the usual characterizations related to the concept of linking.

4.2 Proof of the main results

Employing a standard argument, one can easily check the following lemma.

Lemma 4.6

Assume that (V) and (F1)–(F3) are satisfied, Then Ψ is nonnegative, weakly sequentially lower semi-continuous, and Ψ is weakly sequentially continuous.

Lemma 4.7

Assume that (V) and (F1)–(F3) are satisfied. Then for all θ 0 , z = ( u , v ) E , ζ = ( ϕ , ψ ) E ,

( z ) ( θ z + ζ ) + 1 2 ζ 2 + 1 θ 2 2 ( z ) , z θ ( z ) , ζ .

Corollary 4.8

Assume that (V) and (F1)–(F3) are satisfied. Then for all z N ,

( z ) ( k z + ζ ) + 1 2 ζ 2 , k 0 , ζ E .

Corollary 4.9

Assume that (V) and (F1)–(F3) are satisfied. Then for all z E , k 0 ,

( z ) ( k z + ) + k 2 2 z 2 + 1 k 2 2 ( z ) , z + k 2 ( z ) , z .

Lemma 4.10

Assume that (V), (F1), (F2), and (F5) hold. Then, for any z = ( u , v ) E , τ R , ζ = ( ϕ , ψ ) E , we have

(4.1) τ ( z ) , τ z + 2 ζ τ 2 z + 2 τ z + ζ 2 + ζ 2 R N V ( x ) τ z + ζ 2 d x τ 2 R N u 2 h 2 ( x , u 2 ) f 2 ( x , u 2 ) V ( x ) + v 2 h 1 ( x , v 2 ) f 1 ( x , v 2 ) V ( x ) d x .

Proof

In view of (2.1) and (2.4), we have

τ ( z ) , τ z + 2 ζ = τ 2 z + 2 τ 2 z 2 2 τ z , ζ τ Re R N [ f 2 ( x , u 2 ) u τ u + 2 ϕ ¯ + f 1 ( x , v 2 ) v τ v + 2 ψ ¯ ] d x = τ 2 z + 2 τ 2 z 2 2 τ z , ζ R N [ f 2 ( x , u 2 ) [ τ 2 u 2 + 2 τ Re ( u ϕ ¯ ) ] + f 1 ( x , v 2 ) [ τ 2 v 2 + 2 τ Re ( v ψ ¯ ) ] ] d x = τ 2 z + 2 τ z + ζ 2 + ζ 2 R N V τ u + ϕ 2 d x R N V τ v + ψ 2 d x + R N × { V τ u + ϕ 2 f 2 ( x , u 2 ) [ τ 2 u 2 + 2 τ Re ( u ϕ ¯ ) ] + V τ v + ψ 2 d x f 1 ( x , v 2 ) [ τ 2 v 2 + Re ( v ψ ¯ ) ] } d x τ 2 z + 2 τ z + ζ 2 + ζ 2 R N V τ u + ϕ 2 d x R N V τ v + ψ 2 d x + R N × [ V ( x ) ϕ 2 2 τ u ϕ h 2 ( x , u 2 ) τ 2 u 2 h 2 ( x , u 2 ) ] d x + R N [ V ( x ) ψ 2 2 τ v ψ h 1 ( x , v 2 ) τ 2 v 2 h 1 ( x , v 2 ) ] d x = τ 2 z + 2 τ z + ζ 2 + ζ 2 R N V τ u + ϕ 2 d x R N V τ v + ψ 2 d x + R N [ V ( x ) ϕ τ u h 2 ( x , u 2 ) ] 2 [ τ u h 2 ( x , u 2 ) ] 2 τ 2 u 2 h 2 ( x , u 2 ) V ( x ) V ( x ) d x + R N [ V ( x ) ψ τ v h 1 ( x , v 2 ) ] 2 [ τ v h 1 ( x , v 2 ) ] 2 τ 2 v 2 h 1 ( x , v 2 ) V ( x ) V ( x ) d x τ 2 z + 2 τ z + ζ 2 + ζ 2 R N V ( x ) τ z + ζ 2 d x τ 2 R N u 2 h 2 ( x , u 2 ) f 2 ( x , u 2 ) V ( x ) + v 2 h 1 ( x , v 2 ) f 1 ( x , v 2 ) V ( x ) d x , × z = ( u , v ) E , τ R , ζ E ,

which shows that (4.1) holds. From Lemma 4.10, we have the following corollary immediately.

Lemma 4.11

Assume that (V), (F1), (F2), and (F5) hold. Then, for any z = ( u , v ) N , τ R , ζ E , we have

τ 2 z + 2 τ z + ζ 2 R N V ( x ) τ z + ζ 2 d x ζ 2 + τ 2 R N u 2 h 2 ( x , u 2 ) f 2 ( x , u 2 ) V ( x ) + v 2 h 1 ( x , v 2 ) f 1 ( x , v 2 ) V ( x ) d x .

Similar to the proof of Lemma 3.10, we have the following lemma.

Lemma 4.12

Assume that (V), (F1)–(F3), and (F5) are satisfied. Then

  1. there exists ρ > 0 such that

    m inf N κ inf { ( z ) : z E + , z = ρ } > 0 ;

  2. z + max { z , 2 m } for all z N .

Define a set E 0 + as follows:

(4.2) E 0 + = z E + \ { 0 } : z + 2 ζ 2 R N V ( x ) z + ζ 2 d x < 0 , ζ E .

Lemma 4.13

Suppose that (V), (F1), (F2), and (F5) hold, then for any e E 0 + , sup ( E R + e ) < , there exists R e > 0 such that

( z ) 0 , z E R + e , z R e .

Proof

It is sufficient to show that ( ω + t e ) 0 for t 0 , ω E , and ω + t e R e for large R e > 0 . Arguing indirectly, assume that for some sequence { ω n + s n e } E R + e with ω n + s n e such that ( ω n + s n e ) 0 for all n N . Set z n = ω n + s n e = ( u n , v n ) and ζ n = z n z n = ζ n + τ n e , then ζ n + τ n e = 1 . Passing to a subsequence, we may assume that ζ n ζ in E , then ζ n ζ a.e. on R N , ζ n ζ in E , τ n τ , and

(4.3) 0 ( z n ) z n 2 = τ n 2 2 e 2 1 2 ζ n 2 1 2 R N F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) z n 2 d x .

Clearly, from (4.3), together with ζ n = 1 , we can verify that τ > 0 . Since e E 0 + , there exists a bounded domain Ω R N such that

(4.4) τ 2 e 2 ζ 2 Ω V ( x ) τ e + ζ 2 d x < 0 .

Let H i ( x , t ) = 0 t h i ( x , s ) d s , then F i ( x , t ) = V ( x ) t + H i ( x , t ) . It follow from (4.3) that

0 τ n 2 2 e 2 1 2 ζ n 2 1 2 Ω F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) z n 2 d x = τ n 2 2 e 2 1 2 ζ n 2 1 2 Ω V ( x ) u n 2 + H 2 ( x , u n 2 ) + V ( x ) v n 2 + H 1 ( x , v n 2 ) z n 2 d x = τ n 2 2 e 2 1 2 ζ n 2 1 2 Ω V ( x ) z n 2 z n 2 d x 1 2 Ω H 2 ( x , u n 2 ) + H 1 ( x , v n 2 ) z n 2 d x = τ n 2 2 e 2 1 2 ζ n 2 1 2 Ω V ( x ) τ n e + ζ n 2 d x 1 2 Ω H 2 ( x , u n 2 ) + H 1 ( x , v n 2 ) z n 2 d x .

Clearly, H 2 ( x , u n 2 ) + H 1 ( x , v n 2 ) v n 2 + u n 2 0 as z n . Since ζ n ζ in E , then ζ n ζ in L 2 ( Ω ) , and it is easy to see from the Lebesgue dominated convergence theorem that

Ω H 2 ( x , u n 2 ) + H 1 ( x , v n 2 ) z n 2 d x = Ω H 2 ( x , u n 2 ) + H 1 ( x , v n 2 ) z n 2 ζ n 2 d x = Ω H 2 ( x , u n 2 ) u n 2 + v n 2 + H 1 ( x , v n 2 ) u n 2 + v n 2 ζ n 2 d x = o ( 1 ) .

Hence, 0 τ 2 e 2 ζ 2 Ω V ( x ) τ e + ζ 2 d x , a contradiction to (4.4).□

Lemma 4.14

Suppose that (V), (F1), (F2), and (F5) hold, let e E 0 + with e = 1 . Then there exists r 0 > ρ such that sup ( Q ) 0 for r r 0 , where

(4.5) Q = { ζ + s e : ζ E + , s 0 , ζ + s e r } .

Similar to Lemma 3.13, we have the following lemma.

Lemma 4.15

Assume that (V), (F1)–(F3), and (F5) are satisfied. Then for any z E \ E , N ( E R + z ) , there exist η ( z ) > 0 , ζ ( z ) E such that η ( z ) z + ζ ( z ) N .

Lemma 4.16

Assume that (V), (F1), (F2), and (F4) are satisfied. Then there exists a constant c [ Λ , m ] and a sequence { z n } E satisfying

( z n ) c , ( z n ) ( 1 + z n ) 0 .

where Q is defined in (4.5).

Proof

Choose ζ k N such that

m ( ζ k ) < m + 1 k , k N .

By Lemma 4.12, we have ζ k + 2 m > 0 , since ζ k = ( u k , v k ) E \ { 0 } , it follows from (F1), (F3), and (F5) that

R n u k 2 f 2 ( x , u k 2 ) h 2 ( x , u k 2 ) + v k 2 f 1 ( x , v k 2 ) h 1 ( x , v k 2 ) V ( x ) d x < 0 .

Let e k = ζ k + ζ k + , then e k E + and e k = 1 , By virtue of Corollary 4.11,

τ 2 e k 2 ω 2 R N V ( x ) τ e k + ω 2 d x = τ 2 ζ k + 2 ζ k + 2 ω 2 R N V ( x ) τ ζ k + ζ k + ω τ ζ k + ζ k 2 d x ω τ ζ k + ζ k 2 + τ 2 ζ k + 2 R N u k 2 f 2 ( x , u k 2 ) h 2 ( x , u k 2 ) + v k 2 f 1 ( x , v k 2 ) h 1 ( x , v k 2 ) V ( x ) d x 0 , ω E ,

which yields that e k E 0 + . Then there exists a constant r k > max { ρ , ξ k } such that sup ( Q k ) 0 , where

(4.6) Q k = { ζ + s e k , ζ E , s 0 , ζ + s e k r k } , k N .

Then, by Lemma 3.9, there exist a constant c k [ Λ , sup ( Q k ) ] and a sequence { z k , n } n N E

( z k , n ) c k , ( z k , n ) ( 1 + z k , n ) 0 , k N .

In virtue of Corollary 4.8, we obtain

(4.7) ( ξ k ) ( η ξ k + ζ ) , η 0 , ζ E .

Since ξ k Q k , then by (4.6) and (4.7), we have

( z k , n ) c k < m + 1 k , ( z k , n ) 1 + ( z k , n ) 0 , k N .

We can choose { n k } N such that

m 1 k < ( z k , n k ) < m + 1 k , ( z k , n k ) 1 + ( z k , n k ) < 1 k , k N .

Set z k = z k , n k , k N , then we have

( z n ) c [ k , m ] , ( z n ) 1 + ( z n ) 0 .□

Lemma 4.17

Assume that (V), (A), (F1)–(F3), and (F5) are satisfied. Then there exists a constant c 8 [ k . m ] and a sequence { z n } E satisfying

(4.8) ( z n ) c * , ( z n ) ( 1 + ( z n ) ) 0 ,

is bounded in E.

Proof

To prove the boundedness of { z n } . arguing by contradiction, suppose that z n = ( u n , v n ) such that z n as n . Let ω n = z n z n , then ω n = 1 . By the sobolev embedding theorem, there exists a constant C 0 , such that ω n 2 C 0 . If

δ limsup n sup y R N B 1 ( y ) ω n + 2 d x = 0 ,

then by Lions’ concentration compactness principle, ω n + 0 in L p ( R N ) for 2 < p < 2 * . Fix s > 2 ( c * + 1 ) . combining (F1) and (F5), and we see that there exists a constant C ε > 0 such that

F i ( x , t ) ε t + C ε t p 2 , for i = 1 , 2 .

Thus,

Ψ ( z ) = 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x ε 2 R N [ u 2 + v 2 ] d x + C ε 2 R N [ u p + v p ] d x = ε 2 z 2 2 + C ε 2 z p p ,

for ε = 1 2 s 2 C 0 2 , where ( x , z ) R N × R 2 , and hence, we have

limsup n Ψ ( x , s ω n + ) ε s 2 2 ω n + 2 2 + s p C ε 2 lim n ω n + p p 1 4 C 0 2 ω n + 2 2 1 4 .

Set η n = s z n , by combing Corollary 4.9 with Lemma 4.10, we have

c * + o ( 1 ) = ( z n ) ( η n z n + ) + η n 2 z n 2 + 1 η n 2 2 , ( z n ) , z n + η n 2 ( z n ) , z n = η n 2 2 z n 2 Ψ ( x , η n z n + ) + 1 η n 2 2 ( z n ) , z n + η n 2 ( z n ) , z n = s 2 2 ω n 2 Ψ ( x , s ω n + ) + 1 2 s 2 2 z n 2 ( z n ) , z n + s 2 z n 2 ( z n ) , z n = s 2 2 Ψ ( x , s ω n + ) + o ( 1 ) s 2 2 1 4 + o ( 1 ) = c * + 3 4 + o ( 1 ) ,

which implies that δ > 0 .

Going if necessary to a subsequence, through a similar argument of the proof of Lemma 3.16, we may assume that there exists { y n } R N such that

B 1 + N ( y n ) ω n + 2 d x > δ 2 .

Set ω ˜ n ( x ) = ϒ [ y n ] ω n , then

(4.9) B 1 + N ( 0 ) ω ˜ n + 2 d x > δ 2 .

Passing to a subsequence if necessary, we have ω ˜ n ω ˜ in E , ω ˜ n ω ˜ in L loc p ( R N ) ( p ( 2 , 2 * ) ) and ω ˜ n ω ˜ a.e. on R N . Then (4.9) implies that ω ˜ + 0 , and so ω ˜ 0 .

Now we define z ˜ n ( x ) = ϒ [ y n ] z n . Then z ˜ n z n = w ˜ n w ˜ a.e. on R N and ω ˜ 0 . For x Ω { y R N : ω ˜ ( y ) 0 } . We have lim n z ˜ n ( x ) = . For any ϕ C 0 ( R N ) , setting ϕ = ϒ [ y n ] ϕ n ( x ) , ϕ n ( x ) = ( μ n , ν n ) ,

( z n ) , ϕ n = ( z n + z n , ϕ n ) Re R N [ f 2 ( x , u n 2 ) u n μ ¯ n + f 1 ( x , v n 2 ) v n ν ¯ n ] d x = ( z n + z n , ϕ n ) Re R N { [ V + h 2 ( x , u n 2 ) ] u n μ ¯ n + [ V + h 1 ( x , v n 2 ) ] v n ν ¯ n } d x = ( z n + z n , ϕ n ) Re R N V u n μ ¯ n d x Re R N V v n ν ¯ n d x Re R N [ h 2 ( x , u n 2 ) u n μ ¯ n + h 1 ( x , v n 2 ) v n ν ¯ n ] d x = ( z n + z n , ϕ n ) ( V z n , ϕ n ) L 2 Re R N [ h 2 ( x , u n 2 ) u n μ ¯ n + h 1 ( x , v n 2 ) v n ν ¯ n ] d x = z n { ( ω n + ω n , ϕ n ) ( V ω n , ϕ n ) L 2 Re R N h 2 ( x , u n 2 ) z n ω n u n μ ¯ n + h 1 ( x , v n 2 ) z n ω n v n ν ¯ n d x = z n { ( ω ˜ n + ω ˜ n , ϕ ) ( V ω ˜ n , ϕ ) L 2 Re R N h 2 ( x , u n 2 ) z ˜ n ω ˜ n u n μ ¯ + h 1 ( x , v n 2 ) z ˜ n ω ˜ n v n ν ¯ d x ,

which, together with (4.8), yields

( ω ˜ n + ω ˜ n , ϕ ) ( V ω ˜ n , ϕ ) L 2 Re R N h 2 ( x , u n 2 ) z n ω ˜ n u n μ ¯ + h 1 ( x , v n 2 ) z n ω ˜ n v n ν ¯ d x = o ( 1 ) .

Note that

Re R N h 2 ( x , u n 2 ) z n ω ˜ n u n μ ¯ + h 1 ( x , v n 2 ) z n ω ˜ n v n ν ¯ d x R N [ h 2 ( x , u n 2 ) ω ˜ n μ + h 1 ( x , v n 2 ) ω ˜ n ν ] d x R N [ h 2 ( x , u n 2 ) ω ˜ n ω ˜ μ + h 1 ( x , v n 2 ) ω ˜ n ω ˜ ν ] d x + R N [ h 2 ( x , u n 2 ) ω ˜ μ + h 1 ( x , v n 2 ) ω ˜ ν ] d x C 1 supp ϕ ω ˜ n ω ˜ [ μ + ν ] d x + Ω ω ˜ [ h 2 ( x , u n 2 ) + h 1 ( x , v n 2 ) ] [ μ + ν ] d x = o ( 1 ) .

Hence,

( ω ˜ n + ω ˜ n , ϕ ) ( V ω ˜ n , ϕ ) L 2 = 0 .

Thus, ω ˜ is an eigenfunction of the operator Δ A + ( V V ) , contradicting the fact that has only a continuous spectrum. This contradiction shows that { z n } is bounded.□

Proof of Theorem 4.1

By applying Lemmas 4.16 and 4.17, we deduce that there exists a bounded sequence { z n } E satisfying (4.8). A standard argument shows that { z n } is a nonvanishing sequence. Going if necessary to a subsequence, for some δ > 0 , through a similar argument of the proof of Lemma 3.16, we may assume that there exists { y n } R N such that

B 1 + N ( y n ) z n 2 d x > δ 2 .

Set ω n ( x ) = ϒ [ y n ] z n ( x ) , then

(4.10) B 1 + N ( 0 ) ω n 2 d x > δ 2 .

We have ω n = z n and

(4.11) ( ω n ) c * , ( ω n ) ( 1 + ω n ) 0 .

Passing to a subsequence if necessary, we have ω n ω in E , ω n ω in L loc p ( R N ) ( p ( 2 , 2 * ) ) and ω n ω a.e. on R N . Obviously, (4.10) and (4.11) imply that ω 0 and ( ω ) = 0 . This shows that ω N and so ( ω ) m . On the other hand, by using (4.11) and Fatou’s lemma,

m c * = lim n ( z n ) 1 2 ( z n ) , z n = lim n 1 2 R N [ f 2 ( x , u n 2 ) u n 2 + f 1 ( x , v n 2 ) v n 2 F 2 ( x , u n 2 ) F 1 ( x , v n 2 ) ] d x = 1 2 R N [ f 2 ( x , u 2 ) u 2 + f 1 ( x , v 2 ) v 2 F 2 ( x , u 2 ) F 1 ( x , v 2 ) ] d x = ( z ) 1 2 ( z ) , z = ( z ) .

This shows that ( z ) m and so ( z ) = m = inf N > 0 .□

Similar to the proof of Theorem 4.1, we can prove Theorem 4.2, so we omit its proof process.

5 Local super-quadratic elliptic systems

5.1 Main assumptions and results

With the purpose of studying the local super-linear quadratic case, the following assumptions on the nonlinear f 1 and f 2 are required.

(F6) There exists a domain G R N such that lim t F i ( x , t ) t = for i = 1 , 2 , a.e. x G .

(F7) i ( x , t ) f i ( x , t ) t F i ( x , t ) 0 , for i = 1 , 2 , and there exist C 0 > 0 , R 0 > 0 , and σ ( 0 , 1 ) such that

f i ( x , t 2 ) t t σ 2 N 2 N ( 1 + σ ) ( N 2 ) C 0 i ( x , t ) , for i = 1 , 2 , for t R 0 , if N 3

and for some κ ( 1 , 2 ( 1 σ ) ] such that

f i ( x , t 2 ) t σ κ C 0 i ( x , t ) , for i = 1 , 2 , for t R 0 , if N = 1 , 2 ;

(F7’) i ( x , t ) f i ( x , t ) t F i ( x , t ) 0 , for i = 1 , 2 , and there exists C 0 > 0 , δ 0 ( 0 , Λ ¯ ) and σ ( 0 , 1 ) such that

f i ( x , t 2 ) t Λ ¯ δ 0 f i ( x , t 2 ) t σ 2 N 2 N ( 1 + σ ) ( N 2 ) C 0 i ( x , t ) , for i = 1 , 2 . if N 3 ,

and there exists some κ ( 1 , 2 ( 1 σ ) ] such that

f i ( x , t 2 ) t 2 Λ ¯ δ 0 f i ( x , t 2 ) t t σ κ C 0 i ( x , t ) , for i = 1 , 2 . if N = 1 , 2 .

Theorem 5.1

Assume that (V), (A), (F1)–(F3), (F6), and (F7) are satisfied, and then system (1.1) has at least a ground state solution, i.e., it has at least a solution z E such that ( z ) = inf N > 0 , where N { z E \ E : ( z ) , z = ( z ) , ω = 0 , ω E } .

Theorem 5.2

Assume that (V), (A), (F1), (F2), (F6), and (F7’) are satisfied, and then system (1.1) has a solution z E \ { 0 } such that ( z ) = inf K > 0 , where K { z E \ { 0 } : ( z ) = 0 } .

Remark 5.3

Assume that (F1), (F2), (F3), and (F7) hold. Then (F7’) holds also [27].

Example 5.4

Let N 3 and F i ( x , t ) = sin ( 2 π x 1 ) 2 t ln ( 1 + t ) for i = 1 , 2 . Then

f i ( x , t ) = sin ( 2 π x 1 ) 2 ln ( 1 + t ) + t 1 + t ,

i ( x , t ) = sin ( 2 π x 1 ) 2 t 2 1 + t 0 . for t > 0 .

It is easy to see that f satisfies (F1)–(F3), (F5), (F7), and (F7’) with σ ( 0 , 1 ) and G = ( 1 8 , 3 8 ) × R N 1 .

5.2 Proof of the main results

Employing a standard argument, one can easily check the following lemma.

Lemma 5.5

Assume that (V) and (F1)–(F3) are satisfied, Then Ψ is nonnegative, weakly sequentially lower semi-continuous, and Ψ is weakly sequentially continuous.

Lemma 5.6

Assume that (V) and (F1)–(F3) are satisfied, Then there exists ρ > 0 such that

(5.1) κ ¯ inf { ( z ) : z E + , z = ρ } > 0 .

We omit the proof here since it is standard. Suppose that satisfying G R N is a bounded domain. We can choose e ˜ C 0 ( R N , R + ) C 0 ( G , R + ) satisfying

e ˜ + 2 e ˜ 2 = ( e ˜ , e ˜ ) 2 1 ,

then e ˜ + 0 .

Lemma 5.7

Assume that (V), (F1), (F2), and (F6) are satisfied, Then sup ( E R + e + ) < , and there is R e > 0 such that

( z ) 0 , z E R + e + , z R e .

Proof

Arguing indirectly, assume that for some sequence { ω n + s n e + } E R + e + such that ( ω n + s n e + ) > 0 for all n N and with ω n + s n e + as n . Set z n = ( u n , v n ) = ω n + s n e + and ξ n = z n z n = ξ n + τ n e + , then ξ n + τ n e + = 1 . Passing to a subsequence, we may assume that ξ n ξ in E , then ξ n ξ a.e. on R N , ξ n ξ in E , τ n τ , and

(5.2) 0 ( z n ) z n 2 = τ n 2 2 e + 2 1 2 ξ n 2 1 2 R N [ F 2 ( x , u n ) 2 + F 1 ( x , v n 2 ) ] z n 2 d x τ n 2 2 e 2 1 2 ξ n 2 1 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] z n 2 d x .

Case (1) τ = 0 , it follows from (5.2) that

0 1 2 ξ n 2 + 1 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] z n 2 d x τ n 2 2 e 2 0 .

This implies ξ n 2 0 and 1 = ξ n + τ n e 2 0 , which is a contradiction.

Case (2) τ 0 , we prove that

(5.3) ( ξ + τ e + ) G 0 .

If (5.3) is not true, one has

( ξ + τ e + ) G = 0 .

Since supp e G , we have

0 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] z n 2 d x τ n 2 e + 2 ξ n 2 = ( ( τ n e + + ξ n ) , τ n e + + ξ n ) L 2 ( R N ) = ( ( τ n e + + ξ n ) , τ n e + + ξ n ) L 2 ( G ) + ( ( τ n e + + ξ n ) , τ n e + + ξ n ) L 2 ( R N G ) = ( ( τ n e + + ξ n ) , τ n e + + ξ n ) L 2 ( G ) + ( ( ξ n τ n e ) , ξ n τ n e ) L 2 ( R N G ) = ( ( τ n e + + ξ n ) , τ n e + + ξ n ) L 2 ( G ) ( ( ξ n τ n e ) , ξ n τ n e ) L 2 ( G ) + ( ( ξ τ e ) , ξ τ e ) L 2 ( R N ) + o ( 1 ) = ( ( ξ τ e ) , ξ n τ e ) L 2 ( G ) ξ τ e 2 + o ( 1 ) = ( ( τ e + + τ e ) , τ e + + τ e ) L 2 ( G ) ξ τ e 2 + o ( 1 ) = τ 2 ( e , e ) L 2 ( G ) ξ τ e 2 + o ( 1 ) τ 2 + o ( 1 ) .

This contradiction shows (5.3) holds. Now it follows from (F1), (F6), and Fatou’s lemma that

0 limsup n τ n 2 2 e + 1 2 ξ n 2 1 2 R N [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] z n 2 d x τ n 2 2 e + 1 2 liminf n G [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] u n 2 + v n 2 ξ n + τ n e + 2 d x τ n 2 2 e + 1 2 G liminf n [ F 2 ( x , u n 2 ) + F 1 ( x , v n 2 ) ] 2 max { u n 2 , v n 2 } ξ n + τ n e + 2 d x = ,

which is a contradiction.□

Lemma 5.8

Assume that (V), (F1), (F2), and (F7) hold. Then N .

Proof

By Lemma 5.7, there exists R e ¯ > 0 such that ( z ) 0 for z ( E R + e ¯ + ) \ B R e ¯ ( 0 ) . Moreover, Lemma 5.6 implies that ( s e ¯ + ) > 0 for small s > 0 . Therefore, 0 < sup ( E R + e ¯ + ) < . It is easy to see that is weakly upper semi-continuous on E R + e ¯ + , and hence, ( z 0 ) = sup ( E R + e ¯ + ) for some z 0 E R + e ¯ + . This z 0 is a critical point of E R + e ¯ + , i.e., ( z 0 ) , z 0 = ( z 0 ) , ω = 0 for all ω E R + e ¯ + . Consequently, z 0 N ( E R + e ¯ + ) .□

To prove Theorems 5.1 and 5.2, we define ε ( z ) for any ε 0 as follows:

(5.4) ε ( z ) = ( z ) ε R N z p d x .

Let

(5.5) N ε { z E \ E : ε ( z ) , z = ε ( z ) , ω = 0 , ω E } .

Similar to Lemma 5.8, for ε 0 , we have N ε . Then we define m ε inf N ε ε .

Lemma 5.9

For any z , ξ C and t 0 , there holds

(5.6) 1 t 2 2 p z p z p + t z + ξ p t p z p 2 Re ( z ξ ¯ ) 0 .

Proof

Define a function g ( t ) : R R as follows:

g ( t ) = 1 t 2 2 p z p z p + t z + ξ p t p z p 2 Re ( z ξ ¯ ) .

If z = 0 , then g ( t ) = ξ p 0 . So in the following part, we assume z 0 . Then

g ( 0 ) = 1 2 p z p z p + ξ p > 0 and g ( t ) + as t .

Assume that g ( t ) achieves its minimum at t 0 [ 0 , ) , then we have g ( t 0 ) = 0 , i.e.,

(5.7) p t 0 z p + p z p 2 Re ( z ξ ¯ ) p t 0 z + ξ p 2 Re ( z t 0 z + ξ ¯ ) = 0 .

Set ω = t 0 z + ξ . We discuss in two cases: Re ( z ω ¯ ) = 0 and Re ( z ω ¯ ) 0 .

Case (1) Re ( z ω ¯ ) = 0 , which leads to

g ( t 0 ) = 1 t 0 2 2 p z p z p + t 0 z + ξ p t 0 p z p 2 Re ( z ξ ¯ ) = 1 t 0 2 2 p z p z p + ω p t 0 p z p 2 Re ( z ω t 0 z ¯ ) = 1 t 0 2 2 p z p z p + ω p + t 0 2 p z p = p 2 1 + t 0 2 2 p z p + ω p > 0 .

Case (2) Re ( z ω ¯ ) 0 , which means z 0 and ω 0 . We consider the two possibilities: (i) z = ω , (ii) z ω , Let z = x 1 + i y 1 and ω = x 2 + i y 2 , where x i , y i ( i = 1 , 2 ) denote the real part and imaginary part of z and ω , for z = ω , then

Re ( z ω ¯ ) = x 1 x 2 + y 1 y 2 z ω = z 2 ,

which implies

g ( t 0 ) = 1 t 0 2 2 p z p z p + ω p t 0 p z p 2 Re ( z ω t 0 z ¯ ) 1 t 0 2 2 p z p z p + z p t 0 p z p 2 [ z 2 t 0 z 2 ] = 1 2 p z p [ 1 2 t 0 + t 0 2 ] = 1 2 p z p ( 1 t 0 ) 2 0 .

z ω , without loss of generality, From (5.7), we have

0 = p t 0 z p + p z p 2 Re ( z ω t 0 z ¯ ) p ω p 2 Re ( z ω ¯ ) = z p 2 ω p 2 .

This implies z = ω , which is a contradiction. Hence, we have (5.7) hold.□

Lemma 5.10

Assume that (V), (F1), (F2), and (F3) hold. Then

(5.8) ε ( z ) ε ( k z + ξ ) + 1 2 ξ 2 + 1 k 2 2 ε ( z ) , z k ε ( z ) , ξ , k 0 , z E , ξ E .

Proof

From (2.1), (2.5), (3.1), (5.4), and (5.7), we have

(5.9) ε ( z ) ε ( k z + ξ ) = 1 k 2 2 z + 2 1 k 2 2 z 2 + k z , ξ 1 2 ξ 2 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x + 1 2 R N [ F 2 ( x , k u + ϕ 2 ) + F 1 ( x , k v + ψ 2 ) ] d x ε R N z 2 p d x + ε R N k z + ξ 2 p d x = 1 2 ξ 2 + 1 k 2 2 ε ( z ) , z k ε ( z ) , ξ 1 2 R N [ F 2 ( x , u 2 ) + F 1 ( x , v 2 ) ] d x + 1 2 R N [ F 2 ( x , k u + ϕ 2 ) + F 1 ( x , k v + ψ 2 ) ] d x + 1 k 2 2 R N [ f 2 ( x , u 2 ) u 2 + f 1 ( x , v 2 ) v 2 ] d x k Re R N [ f 2 ( x , u 2 ) u ϕ ¯ + f 1 ( x , v 2 ) v ψ ¯ ] d x ε R N z p d x + ε R N t z + ξ p d x + 1 t 2 2 p ε R N z p d x p t ε R N z p 2 Re ( z ξ ¯ ) d x 1 2 ξ 2 + 1 k 2 2 ε ( z ) , z k ε ( z ) , ξ .

The proof is completed.□

By using Lemma 5.10, some important corollaries are given as follows, and the proof process will be omitted.

Corollary 5.11

Assume that (V) and (F1)–(F3) are satisfied. Then for all z N ε ,

(5.10) ( z ) ( k z + ξ ) + 1 2 ξ 2 , k 0 , ξ E .

Corollary 5.12

Assume that (V), and (F1)–(F3) are satisfied. Then for all z E , k 0 ,

( z ) ( k z + ) + k 2 2 z 2 + 1 k 2 2 ( z ) , z + k 2 ( z ) , z .

Lemma 5.13

Assume that (V), (F1), (F2), and (F3) are satisfied. Then, for ε [ 0 , 1 ] ,

  1. There exists κ ˆ , which does not depend on ε [ 0 , 1 ] such that

    (5.11) ε ( z ) m ε κ ˆ , u N ε .

  2. z + max { z , 2 m ε } for all z N ε .

Proof

By (F1) and (F2), there exists a constant C ε > 0 such that

F i ( x , t ) 1 2 γ 2 2 t + C ε t p 2 , x R N , t R + .

Then

(5.12) F 2 ( x , u 2 ) + F 1 ( x , v 2 ) 1 2 γ 2 2 u 2 + 1 2 γ 2 2 v 2 + C ε u p + C ε v p , x R N , z = ( u , v ) E .

In virtue of (5.10) and (5.12), for any ε [ 0 , 1 ] , one has

ε ( z ) ε ( t z + ) = t 2 2 z + 2 1 2 R N [ F 2 ( x , t u + 2 ) + F 1 ( x , t v + 2 ) ] d x ε R N t z + p d x t 2 2 z + 2 1 2 t 2 2 γ 2 2 u + 2 2 + t 2 2 γ 2 2 v + 2 2 + C ε t p u + p p + C ε t p v + p p ε t p z + p p t 2 4 z + 2 t p C 1 γ p p z + p .

Choose t = t z 1 [ 2 C 1 γ p p p ] 1 p 2 z + , then it follows from above inequality that

ε ( z ) t 2 4 z + 2 C 1 t p γ p p z + p = p 2 4 p [ 4 C 1 γ p p p ] 2 p 2 κ ˆ > 0 , x R N , z = ( u , v ) E , ε [ 0 , 1 ] .

(ii) Similar to Lemma 3.10 (ii), we can prove Lemma 5.13 (ii) hold.□

Lemma 5.14

Assume that (V), (A), (F1), (F2), and (F3) hold. Then for any ε ( 0 , 1 ] , there exists z ε N ε such that

(5.13) ε ( z ε ) = m ε , ε ( z ε ) = 0 .

Proof

By virtue of [23], there exists a bounded sequence { z ε n } E such that

ε ( z ε n ) c , ε ( z ε n ) ( 1 + z ε n ) 0 , n ,

where c [ κ ˆ , m ε ] . Hence, there exists a constant C ˜ 1 > 0 such that z ε n 2 C ˜ 1 . If

δ limsup n sup y R N B 1 ( y ) z ε n 2 d x = 0 .

Then the Lions’ concentration compactness lemma implies z ε n 0 in L p ( p ( 2 , 2 * ) ) . By (F1) and (F2), for ε = c 4 C ˜ 1 2 > 0 , there exists C ε ˜ > 0 such that

f i ( x , t ) = ε + C ε t p 2 2 , ( x , t ) R N × R + , i = 1 , 2 .

Thus,

(5.14) limsup n R N 1 2 [ f 2 ( x , u ε n 2 ) u ε n 2 + f 1 ( x , v ε n 2 ) v ε n 2 ] + p 2 1 ε n z ε n p d x limsup n ε 2 z ε n 2 2 + C ˜ 2 z ε n p p ε 2 C ˜ 1 2 = c 8 .

From (5.4) and (5.14),

c = ε n ( z ε n ) 1 2 ε n ( z ε n ) , z ε n + o ( 1 ) = 1 2 R N [ f 2 ( x , u ε n 2 ) u ε n 2 + f 1 ( x , v ε n 2 ) v ε n 2 ] d x + p 2 ε n R N z ε n p d x 1 2 R N [ F 2 ( x , u ε n 2 ) + F 1 ( x , v ε n 2 ) ] d x ε n R N z ε n p d x + o ( 1 ) 1 2 R N [ f 2 ( x , u ε n 2 ) u ε n 2 + f 1 ( x , v ε n 2 ) v ε n 2 ] d x + p 2 1 ε n R N z ε n p d x + o ( 1 ) c 8 + o ( 1 ) .

That is a contradiction, so we have δ > 0 .

Going if necessary to a subsequence, through a similar argument of the proof of Lemma 3.16, we may assume that there exists { y n } R N such that

B 1 + N ( y n ) z n 2 d x > δ 2 .

Set ω n ( x ) = ϒ [ y n ] z n ( x ) , then

(5.15) B 1 + N ( 0 ) ω n 2 d x > δ 2 .

Hence, ω n = z n , and

(5.16) ε n ( ω n ) c , ε n ( ω n ) ( 1 + ω n ) 0 , n .

Going if necessary to a subsequence, we have ω n ω in E , ω n ω in L loc p ( R N ) ( p ( 2 , 2 * ) ) and ω n ω a.e. on R N , set ω = ( ϕ , ψ ) . Obviously, (5.15) implies that ω 0 . By a standard argument, we have ε n ( ω ) = 0 . Then ω N and ε n ( ω n ) m ε . Moreover, from (5.16), (F3), and Fatou’s lemma, one has

m ε c = lim n ε n ( ω n ) 1 2 ε n ( ω n ) , ω n = lim n R N 1 2 [ 2 ( x , ϕ n 2 ) + 1 ( x , ψ n 2 ) ] + p 2 1 ε n ω n p d x R N lim n 1 2 [ 2 ( x , ϕ n 2 ) + 1 ( x , ψ n 2 ) ] + p 2 1 ε n ω n p d x = R N 1 2 [ 2 ( x , ϕ 2 ) + 1 ( x , ψ 2 ) ] + p 2 1 ε n ω p d x = ε n ( ω ) 1 2 ε n ( ω ) , ω = ε n ( ω ) .

This shows that ε n ( ω ) m ε , and then ε n ( ω ) = m ε .□

Lemma 5.15

Assume that (V), (F1), (F2), and (F3) are satisfied. Then for any ε ( 0 , 1 ] and z E \ E , there exist t ε ( z ) > 0 and ζ ε ( z ) E such that t ε ( z ) + ζ ε ( z ) N ε .

We can easily prove this lemma in a similar way as Lemma 3.13, so we omit it.

Proof of Theorem 5.1

Consider the case N 3 . Since the case where N = 1 , 2 can be dealt with similarly. By Lemma 5.14, there exists z ε N ε such that (5.13) holds, where ε ( 0 , 1 ] .

By Lemma 5.8, N , Then, for z 0 N and ξ E , ( z 0 ) c ˜ 0 and 0 ( z 0 ) , z 0 = 0 ( z 0 ) , ω = 0 hold. In virtue of Lemma 5.15, there exist t ε > 0 and ξ ε E , such that t ε z 0 + ξ ε N ε . By Corollary 5.11 and Lemma 5.13, one has

(5.17) c ˜ = 0 ( z 0 ) 0 ( t ε z 0 + ξ ε ) ε ( t ε z 0 + ξ ε ) m ε k ˆ , ε ( 0 , 1 ] .

Choose a sequence { ε n } ( 0 , 1 ] satisfy ε n 0 as n , and

(5.18) z ε n N ε n , ε n ( z ε n ) = m ε n m ˜ [ κ ˆ , c ˜ ] , ε n ( z ε n ) = 0 .

There are there steps to prove Theorem 5.1.

Step 1: We prove that { z ε n } is bounded in E .

Arguing by contradiction, suppose that z ε n , z ε n = ( u ε n , v ε n ) . Set ω n = z ε n z ε n = ( ϕ n , ψ n ) , then ω n = 1 . By sobolev embedding theorem, going if necessary to a subsequence, we have we have ω n ω in E , ω n ω in L loc p ( R N ) ( p ( 2 , 2 * ) ) and ω n ω a.e. on R N . From 5.18, we have

(5.19) c ˜ ε ( z ε n ) 1 2 ε n ( z ε n ) , z ε n = 1 2 R N [ 2 ( x , u ε n 2 ) + 1 ( x , v ε n 2 ) ] d x + ε n p 2 1 R N z ε n p d x .

In view of Sobolev embedding theorem, there exists a constant C ˜ 3 > 0 such that ω n 2 C ˜ 3 . If

δ limsup n sup y R N B 1 ( y ) ω n + 2 d x = 0 ,

then by Lions’ concentration compactness principle, ω n + 0 in L p ( R N ) ( p ( 2 , 2 * ) ) . Let R > [ 2 ( 1 + c ˜ ) ] 1 2 . From (F1) and (F2), choose ε = 1 4 ( R C ˜ 3 ) 2 , there exists C ˜ 4 > 0 such that

(5.20) limsup n R N 1 2 [ F 2 ( x , R ψ n 2 ) + F 1 ( x , R ϕ n 2 ) ] + ε n R ω n + p d x limsup n 1 2 ε R 2 ω n + 2 2 + C ˜ 4 R p ω n + p p 1 2 ε ( R C ˜ 3 ) 2 = 1 8 .

Let t n = R z ε n . From (5.18), (5.20), and Corollary 5.12, one has

c ˜ m ε n = ε n ( z ε n ) t n 2 2 z ε n + 2 R N 1 2 [ F 2 ( x , t n u ε n + 2 ) + F 1 ( x , t n v ε n + 2 ) ] + ε n t n z ε n + p d x = R 2 2 R N 1 2 [ F 2 ( x , R ϕ n + 2 ) + F 1 ( x , R ψ n + 2 ) ] + ε n R ω n + p d x R 2 2 1 8 + o ( 1 ) > c ˜ + 7 8 + o ( 1 ) ,

which is a contradiction, and then δ > 0 .

Going if necessary to a subsequence, through a similar argument of the proof of Lemma 3.16, we may assume that there exists { y n } R N such that

B 1 + N ( y n ) ω n + 2 d x > δ 2 .

set ω ˜ n ( x ) = ϒ [ y n ] ω n ( x ) , then

(5.21) B 1 + N ( 0 ) ω ˜ n + 2 d x > δ 2 .

Hence, ω ˜ n = ω n = 1 . Passing to a subsequence if necessary, we have ω ˜ n ω ˜ in E , ω ˜ n ω ˜ in L loc p ( R N ) ( p ( 2 , 2 * ) ) and ω ˜ n ω ˜ a.e. on R N . Then (5.21) implies that ω ˜ 0 .

Define z ˜ n = ( u ˜ n , v ˜ n ) = ϒ [ y n ] z ε n ( x + k n ) , note that z ε n = ( u ε n , v ε n ) . Hence, z ˜ n z ε n = ω ˜ n ω ˜ a.e. on R N and ω ˜ 0 , and here, ω ˜ n = ( ϕ ˜ n , ψ ˜ n ) . For any ξ = ( μ , ν ) C 0 ( R N ) . Let ξ = ϒ [ y n ] ξ n and ξ n = ( μ n , ν n ) . From (5.4) and (5.18), we have

0 = ε n ( z ε n ) , z ε n ξ n = z ˜ n + , ξ n + z ˜ n , ξ n Re R N [ f 2 ( x , u ˜ n 2 ) u ˜ n μ ¯ n + f 1 ( x , v ˜ n 2 ) v ˜ n ν ¯ n ] d x p ε n Re R N z ˜ n p 2 z ˜ n ξ ¯ n d x = z ε n [ ω ˜ n + , ξ n + ω ˜ n , ξ n ] Re R N [ f 2 ( x , u ˜ n 2 ) u ˜ n μ ¯ + f 1 ( x , v ˜ n 2 ) v ˜ n ν ¯ ] d x p ε n Re R N z ˜ n p 2 z ˜ n ξ ¯ d x ,

which implies

(5.22) ω ˜ n + , ξ n + ω ˜ n , ξ n = 1 z ε n Re R N [ f 2 ( x , u ˜ n 2 ) u ˜ n μ ¯ + f 1 ( x , v ˜ n 2 ) v ˜ n ν ¯ + p ε n z ˜ n p 2 z ˜ n ξ ¯ ] d x .

By virtue of (F1), (F2), (F7), (5.19), and the Hölder inequality, one can obtain that

(5.23) 1 z ε n Re R N [ f 2 ( x , u ˜ n 2 ) u ˜ n μ ¯ + f 1 ( x , v ˜ n 2 ) v ˜ n ν ¯ + p ε n z ˜ n p 2 z ˜ n ξ ¯ ] d x 1 z ε n 1 σ z ˜ n 0 f 2 ( x , u ˜ n 2 ) u ˜ n μ + f 1 ( x , v ˜ n 2 ) v ˜ n ν + p ε n z ˜ n p 1 ξ z n σ ω ˜ n σ d x = 1 z ε n 1 σ 0 < z ˜ n < R 0 f 2 ( x , u ˜ n 2 ) u ˜ n μ + f 1 ( x , v ˜ n 2 ) v ˜ n ν + p ε n z ˜ n p 1 ξ z n σ ω ˜ n σ d x + 1 z ε n 1 σ z ˜ n R 0 f 2 ( x , u ˜ n 2 ) u ˜ n μ + f 1 ( x , v ˜ n 2 ) v ˜ n ν + p ε n z ˜ n p 1 ξ z n σ ω ˜ n σ d x C 2 ω ˜ n 2 σ ξ 2 2 σ z ε n 1 σ + 1 z ε n 1 σ z ˜ n R 0 f 2 ( x , u ˜ n 2 ) u ˜ n u ˜ n σ μ ω ˜ n σ d x + z ˜ n R 0 f 1 ( x , v ˜ n 2 ) v ˜ n v ˜ n σ ν ω ˜ n σ d x + z ˜ n R 0 p ε n z ˜ n p σ 1 ξ ω ˜ n σ d x C 2 ω ˜ n 2 σ ξ 2 2 σ z ε n 1 σ + ω ˜ n 2 * σ z ε n 1 σ z ˜ n R 0 f 2 ( x , u ˜ n 2 ) u ˜ n u ˜ n σ 2 * 2 * σ 1 d x 2 * σ 1 2 * + z ˜ n R 0 f 1 ( x , v ˜ n 2 ) v ˜ n v ˜ n σ 2 * 2 * σ 1 d x 2 * σ 1 2 * [ μ 2 * + ν 2 * ] + ω ˜ n 2 * σ ξ 2 * z ε n 1 σ z ˜ n R 0 [ p ε n z ˜ n p δ 1 ] 2 * 2 * σ 1 d x 2 * σ 1 2 * C 2 ω ˜ n 2 σ ξ 2 2 σ z ε n 1 σ + C 3 ω ˜ n 2 * σ ξ 2 * z ε n 1 σ z ˜ n R 0 [ 2 ( x , u ˜ n 2 ) + 1 ( x , v ˜ n 2 ) + ε n ( p 2 ) z ε n p ] d x 2 * σ 1 2 * C 4 z ε n 1 σ ξ 2 2 σ + ξ 2 * = o ( 1 ) .

It follows from (5.22) and (5.23) that

ω ˜ n + , ξ + ω ˜ n , ξ = o ( 1 ) . ξ C 0 ( R N ) .

In view of ω ˜ n ω ˜ , one has

( ω ˜ , ξ ) 2 = ω ˜ + , ξ + ω ˜ , ξ = o ( 1 ) .

This implies that ω ˜ = 0 and ω is an eigenfunction of the operator . Note that has only a continuous spectrum. That is a contradiction. Hence, { z ε n } is bounded.

Step 2: We prove that there exists z ˜ E . such that ( z ˜ ) = 0 and ( z ˜ ) m 0 inf N 0 0 = inf N .

By using Lions’ concentration compactness principle, we can prove that there exist a constant δ 1 > 0 , a sequence y n Z N and a subsequence of { z ε n } , which is still denoted by { z ε n } , such that

(5.24) B 1 ( y n ) z ε n 2 d x > δ 1 .

Let z ˆ n = z ε n ( x + y n ) , since E + and E are Z N -translation invariance, we have z ˆ n = z ε n and

(5.25) z ˆ n N ε n , ε n ( z ˆ n ) = m ε n m ˜ [ κ ˆ , c * ] , ( z ˆ n ) = 0 .

Thus there exists z ˜ E \ { 0 } such that, passing to a subsequence if necessary, z ˆ n z ˜ in E , z ˆ n z ˜ in L loc p ( R N ) , p [ 2 , 2 * ) , z ˆ n z ˜ a.e. on R N . Noting that z ˆ = ( u ˆ , v ˆ ) , ξ = ( μ , ν ) . By virtue of (2.1), we have

( z ˜ ) , ξ = z ˜ + , ξ + z ˜ , ξ Re R N [ f 2 ( x , u ˜ 2 ) u ˜ μ ¯ + f 1 ( x , v ˜ 2 ) v ˜ ν ¯ ] d x = lim n z ˆ n + , ξ + z ˆ n , ξ Re R N { f 2 ( x , u ˆ n 2 ) u ˆ n μ ¯ + f 1 ( x , v ˆ n 2 ) v ˆ n ν ¯ + ε n p z ˆ n p 2 z ˆ n ξ ¯ } d x ] = ε n ( z ˆ n ) , ξ = 0 , ξ C 0 ( Ω ) .

This implies that ( z ˜ ) = 0 . Then, z ˜ N , ( z ˜ ) m 0 .

Step 3: we prove that ( z ˜ ) = m 0 .

In view of (5.4), (5.25), and Fatou’s lemma, we have

(5.26) m ˜ = lim n m ε n = lim n ε n ( z ˆ n ) 1 2 ε n ( z ˆ n ) , z ˆ n = lim n 1 2 R N [ 2 ( x , u ˆ n 2 ) + 1 ( x , v ˆ n 2 ) + ε n ( p 2 ) z ˆ n p ] d x lim n 1 2 R N [ 2 ( x , u ˜ n 2 ) + 1 ( x , v ˜ n 2 ) ] d x = ( z ˜ ) 1 2 ( z ˜ ) , z ˜ m 0 .

Let ε > 0 . Then there exists ω ε N such that ( ω ε ) < m 0 + ε . By Lemma 5.15, there exist t n > 0 and ξ n E such that t n ω ε + ξ n N ε n . One has

m 0 + ε > ( ω ε ) = 0 ( ω ε ) 0 ( t n ω ε + ξ n ) ε n ( t n ω ε + ξ n ) m ε n .

Thus,

m ˜ = lim n m ε n m 0 + ε .

Since ε can be any positive number, we have m ˜ m 0 . In view of (5.26), we can obtain that m ˜ = m 0 = ( z ˜ ) .

And the case N = 1 , 2 can be dealt with similarly, we omit it. The proof is completed.□

Lemma 5.16

Assume that (V), (F1), (F2), (F6), and (F7’) hold, Then

  1. ϑ inf { z : z K } > 0 ;

  2. ϱ inf { ( z ) : z K } > 0 .

Proof

We only consider the case where N 3 , since N = 1 , 2 can be dealt with similarly.

(i) Similar to [23], we have K , Let z n = ( u n , v n ) , { z n } K such that z n ϑ . Then

(5.27) z n 2 = Re R N [ f 2 ( x , u n 2 ) u n u n + u n ¯ + f 1 ( x , v n 2 ) v n v n + v n ¯ ] d x .

From (F1), (F2), and (5.27), one has

z n 2 = Re R N [ f 2 ( x , u n 2 ) u n u n + u n ¯ + f 1 ( x , v n 2 ) v n v n + v n ¯ ] d x R N [ f 2 ( x , u n 2 ) u n u n + u n + f 1 ( x , v n 2 ) v n v n + v n ] d x R N ( ε + C ε u n p 2 ) [ u n + 2 u n 2 ] d x + R N ( ε + C ε v n p 2 ) [ v n + 2 v n 2 ] d x R N [ ε u n + 2 + ε v n + 2 ] d x + R N [ C ε u n p 2 u n + 2 + C ε v n p 2 v n + 2 ] d x ε z n + 2 2 + C ε [ u n p p 2 u n + p 2 + v n p p 2 v n + p 2 ] ε z n 2 2 + C ε z n p p ε γ 2 2 z n 2 + C ε γ p p z n p .

Taking ε = 1 2 γ 2 2 , then

z n 2 1 2 z n 2 + C 5 z n p .

Then

(5.28) ϑ + o ( 1 ) = z n ( 2 C 5 ) 1 p 2 > 0 .

This implies that (i) holds.

(ii) Let { z n } K such that ( z n ) ϱ . Then ( z n ) , z ˜ = 0 for any z ˜ E , We have

(5.29) ϱ + o ( 1 ) = ( z n ) 1 2 ( z n ) , z n = 1 2 R N [ 2 ( x , u n 2 ) + 1 ( x , v n 2 ) ] d x .

Let ω n = z n z n = ( ϕ n , ψ n ) , then ω 2 = 1 . Set

(5.30) Ω n { x R N : f i ( x , t n 2 ) Λ ¯ δ 0 } , for i = 1 , 2 .

By using Λ ¯ ω n + 2 2 ω n + 2 , we have

(5.31) Ω n f 2 ( x , u n 2 ) u n u n ϕ n + 2 + f 1 ( x , v n 2 ) v n v n ψ n + 2 ( Λ ¯ δ 0 ) [ ϕ n + 2 2 + ψ n + 2 2 ] = ( Λ ¯ δ 0 ) ω n + 2 2 1 δ 0 Λ ¯ .

Moreover, by virtue of (F7’), (5.28), (5.29), and the Hölder inequality, one can obtain

(5.32) 1 z n 1 σ R N \ Ω n f 2 ( x , u n 2 ) u n u n σ ϕ n σ ϕ n + ϕ n + f 1 ( x , v n 2 ) v n v n σ ψ n σ ψ n + ψ n d x 1 z n 1 σ R N \ Ω n f 2 ( x , u n 2 ) u n u n σ 2 * 2 * 1 σ d x 2 * 1 σ 2 * ϕ 2 * σ ϕ n + ϕ n 2 * + R N \ Ω n f 1 ( x , v n 2 ) v n v n σ 2 * 2 * 1 σ d x 2 * 1 σ 2 * ψ 2 * σ ψ n + ψ n 2 * 1 z n 1 σ R N \ Ω n f 2 ( x , u n 2 ) u n u n σ 2 * 2 * 1 σ d x 2 * 1 σ 2 * + R N \ Ω n f 1 ( x , v n 2 ) v n v n σ 2 * 2 * 1 σ d x 2 * 1 σ 2 * ( ϕ n 2 * σ + ψ n 2 * σ ) ( ϕ n + ϕ n 2 * + ψ n + ψ n 2 * ) 4 ω n 2 * σ ( ω n + ω n 2 * ) z n 1 σ R N \ Ω n [ C 0 2 ( x , u n 2 ) ] 2 * 2 * 1 σ d x 2 * 1 σ 2 * + R N \ Ω n [ C 0 1 ( x , v n 2 ) ] 2 * 2 * 1 σ d x 2 * 1 σ 2 * C 6 z n 1 σ R N \ Ω n [ 2 ( x , u n 2 ) + 1 ( x , v n 2 ) ] d x 2 * 2 * 1 σ C 7 [ ϱ + o ( 1 ) ] 2 * 2 * 1 σ .

From (5.31) and (5.32), we have

1 = z n 2 ( z n ) , z n + z n z n 2 = Re R N { f 2 ( x , u n 2 ) u n u n + u n ¯ + f 1 ( x , v n 2 ) v n v n + v n ¯ } d x z n 2 R N { f 2 ( x , u n 2 ) u n u n + u n + f 1 ( x , v n 2 ) v n v n + v n d x } z n 2 = 1 z n Ω n [ f 2 ( x , u n 2 ) u n ϕ n + ϕ n + f 1 ( x , v n 2 ) v n ψ n + ψ n ] d x + 1 z n 1 σ R N \ Ω n f 2 ( x , u n 2 ) u n u n σ ϕ n σ ϕ n + ϕ n + f 1 ( x , v n 2 ) v n v n σ ψ n σ ψ n + ψ n d x Ω n f 2 ( x , u n 2 ) u n u n ϕ n + 2 + f 1 ( x , v n 2 ) v n v n ψ n + 2 d x + C 7 [ ϱ + o ( 1 ) ] 2 * 2 * 1 σ ( Λ ¯ δ 0 ) ω n + 2 2 1 δ 0 Λ ¯ + C 7 [ ϱ + o ( 1 ) ] 2 * 2 * 1 σ .

Then, we can obtain that ϱ > 0 .□

Proof of Theorem 5.2

Let { z n } K such that ( z n ) ϱ . As in the study by Qin et al. [24], we can easily prove the boundedness of { z n } in E , so we omit it. Then, similar to the proof of Theorem 5.1, we can obtain that there exists z ˜ E \ { 0 } such that ( z ˜ ) = 0 and ( z ˜ ) = ϱ > 0 .□

  1. Funding information: This work is supported by Natural Science Foundation of China (11301297), Natural Science Foundation of Hubei Province (No. 2024AFB730) and Yichang City (No. A-24-3-008), and Open Research Fund of Key Laboratory of Nonlinear Analysis & Applications (Central China Normal University), Ministry of Education, P. R. China.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript. The first author prepared the manuscript, and the second author prepared and checked it.

  3. Conflict of interest: The authors declare that they have no conflict of interest.

  4. Data availability statement: Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.

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Received: 2023-09-03
Revised: 2024-01-01
Accepted: 2024-04-10
Published Online: 2024-06-17

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  74. Temporal periodic solutions of non-isentropic compressible Euler equations with geometric effects
  75. Nodal solutions for a zero-mass Chern-Simons-Schrödinger equation
  76. The modified quasi-boundary-value method for an ill-posed generalized elliptic problem
  77. Nonlocal heat equations with generalized fractional Laplacian
  78. Choquard equations with recurrent potentials
  79. Local and global well-posedness of the Maxwell-Bloch system of equations with inhomogeneous broadening
  80. The Riemann problem for two-layer shallow water equations with bottom topography
  81. Global stability and asymptotic profiles of a partially degenerate reaction diffusion Cholera model with asymptomatic individuals
  82. On Kirchhoff-Schrödinger-Poisson-type systems with singular and critical nonlinearity
  83. Energy-variational solutions for viscoelastic fluid models
  84. Stability on 3D Boussinesq system with mixed partial dissipation
  85. Existence and multiplicity results for non-autonomous second-order Hamiltonian systems
  86. Erratum
  87. Erratum to “Normalized solutions for the Kirchhoff equation with combined nonlinearities in ℝ4
Heruntergeladen am 7.9.2025 von https://www.degruyterbrill.com/document/doi/10.1515/anona-2024-0011/html
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