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High-energy solutions for coupled Schrödinger systems with critical growth and lack of compactness

  • Wen Guan , Da-Bin Wang EMAIL logo and Huafei Xie
Published/Copyright: August 20, 2025

Abstract

This article deals with the existence of high-energy positive solutions for the following coupled Schrödinger system with critical exponent:

Δ u + V 1 ( x ) u = μ 1 u 3 + β u v 2 , x Ω , Δ v + V 2 ( x ) v = β u 2 v + μ 2 v 3 , x Ω , u , v D 0 1 , 2 ( Ω )

where Ω R 4 is an unbounded exterior domain, Ω , R 4 \ Ω is bounded, V 1 , V 2 L 2 ( Ω ) L loc ( Ω ) are non-negative functions, and μ 1 , μ 2 , and β are the positive constants. Combining variational methods and topological degree theory, we explore the existence of high-energy positive solutions to this system in the case of small perturbations of the source term, i.e., if V 1 2 + V 2 2 and R 4 \ Ω are small enough in a prescribed sense. It is worth noting that our result still holds in the case Ω = R 4 ; hence, this article can be viewed as an extension of recent results for the Benci-Cerami problem in the framework of coupled Schrödinger systems with critical exponent in R 4 .

MSC 2010: 35B33; 35J47; 35J57; 35Q55; 47H11

1 Introduction and main results

The main purpose of this article is to study high-energy positive solutions for the following coupled Schrödinger equations:

(1.1) Δ u + V 1 ( x ) u = μ 1 u 3 + β u v 2 , x Ω , Δ v + V 2 ( x ) v = β u 2 v + μ 2 v 3 , x Ω , u , v D 0 1 , 2 ( Ω ) ,

where Ω R 4 is an unbounded exterior domain, Ω , R 4 \ Ω is bounded, V 1 , V 2 L 2 ( Ω ) L loc ( Ω ) are non-negative functions, and μ 1 , μ 2 , and β are the positive constants.

On the one hand, the motivation to studying the coupled equations (1.1) is inspired from solitary wave solutions of the following nonlinear Schrödinger system:

(1.2) i t Ψ 1 = Δ Ψ 1 μ 1 Ψ 1 2 Ψ 1 β Ψ 2 2 Ψ 1 , x Ω , t > 0 , i t Ψ 2 = Δ Ψ 2 β Ψ 1 2 Ψ 2 μ 2 Ψ 2 2 Ψ 2 , x Ω , t > 0 , Ψ j = Ψ j ( x , t ) C , Ψ j ( x , t ) = 0 , x Ω , t > 0 , j = 1 , 2 ,

where Ω R N ( N 4 ) is a smooth domain and i is the imaginary unit. Systems like (1.2), also known as Gross-Pitaevskii equations, have a very important physical background when N 3 . For instance, in condensed matter physics, system (1.2) arises in the Hartree-Fock theory for a double condensate, i.e., a binary mixture of Bose-Einstein condensates in two different hyperfine states [23]. System (1.2) also appears in nonlinear optics, when problem (1.2) models two optical waves of different frequencies co-propagating in a medium and interacting nonlinearly through the medium, or two polarization components of a wave interacting nonlinearly at some central frequency [1].

In order to find solitary wave solutions, i.e., solutions like

(1.3) ( Ψ 1 ( x , t ) , Ψ 1 ( x , t ) ) = ( e i λ 1 t u ( x ) , e i λ 2 t v ( x ) ) ,

system (1.2) is reduced to the following elliptic system:

(1.4) Δ u + λ 1 u = μ 1 u 3 + β u v 2 , in Ω , Δ v + λ 2 v = β u 2 v + μ 2 v 3 , in Ω , u = v = 0 , on Ω .

We note that a solution ( u , v ) of system (1.4) is said to be nontrivial if both u 0 and v 0 ; we call a solution ( u , v ) semi-trivial if ( u , v ) is of type ( u , 0 ) with u 0 or ( 0 , v ) with v 0 ; we call a solution ( u , v ) positive if u > 0 and v > 0 . From a physical viewpoint, only nontrivial solutions have physical meaning and are worth further investigation. In [1], various analytical and numerical results for system (1.4) were investigated by authors in the case N = 1 . As far as we know, Lin and Wei [28] first studied system (1.4) with general dimension ( N 3 ). We remark that if N 3 , the nonlinearity and the coupling terms in (1.4) are of subcritical growth with respect to Sobolev critical exponent. Since then, many authors have been interested in studying existence, multiplicity, and qualitative properties of solutions for system like (1.4). There are many interesting results in the case N 3 ; we just refer the readers to [2,3,68,14,19,21,22,28,29,3340,45,46,4952] and references therein.

When N = 4 , system (1.4) is a problem with Sobolev critical exponent, which can be seen as a critically coupled perturbed Brezis-Nirenberg problem in dimension 4. Different from N 3 , the discussion on system (1.4) with N = 4 becomes quite delicate and requires some new techniques and ideas due to the lack of compactness. So, compared with the case N 3 , the study of system (1.4) with N = 4 is not frequent in the literature. We refer to Chen and Zou [13] who proved that the critical system (1.4) has a positive ground-state solution for negative β , positive small β , and positive large β , respectively. If λ 1 = λ 2 , the authors obtained the uniqueness of positive ground-state solutions. Furthermore, they also studied the limit behavior of the ground-state solutions as β , and phase separation is expected. To the best of our knowledge, the results obtained in [13] are the first contribution to Schrödinger systems in the critical case. Soon afterward, it turned out that results for the critical system in the higher dimensions ( N 5 , see [15]) are quite different from those obtained if N = 4 . Chen and Zou [16] investigated the existence and symmetry properties of positive ground-state solutions for coupled nonlinear Schrödinger system with critical exponent and a Hardy potential on the whole R N . Note that the positive solutions studied in [13,15,16] are ground-state solutions. Some authors are interested in higher-energy positive solutions to systems like (1.4), cf. [17,25,31,41,43,44]. Some of these articles considered positive solutions to coupled Schrödinger equations with critical exponent when the domain is bounded and has nontrivial topology [41,43,44]. The corresponding results extended the celebrated work for the Coron and Bahri-Coron problems [5,20] to coupled Schrödinger equations. For systems of two coupled equations with critical exponent in a domain Ω R N ( N 3 ) , by proving uniqueness of ground-state solutions and non-degeneracy of a manifold of synchronized type of positive solutions when Ω = R N , Peng et al. [41] succeeded to obtain a positive solution when Ω is bounded and has nontrivial topology. For systems of m -coupled equations with critical exponent in a bounded domain Ω ε R N ( N = 3,4 ) , having k distinct small shrinking holes as the parameter ε 0 , by a perturbation approach based on the Lyapunov-Schmidt finite-dimensional reduction, Pistoia and Soave [43] proved the existence of positive solutions of two different types: either each density concentrates around a different hole, or they have groups of components such that all the components within a single group concentrate around the same point, and different groups concentrate around different points. For systems of m -coupled equations with critical exponent in a bounded symmetric domain Ω ε R 4 with a small shrinking hole B ε ( ξ 0 ) , using Lyapunov-Schmidt finite-dimensional reduction, Pistoia et al. [44] constructed non-synchronized solution that looks like a fountain of positive bubbles concentrating around the same point ξ 0 as ε 0 . In [17,25,31], higher-energy positive solutions to coupled system are considered on whole space R N . Specifically, Clapp and Pistoia [17] obtained that a weakly coupled elliptic system has infinitely many fully nontrivial solutions, which are higher-energy solutions and invariant with respect to some subgroup of O ( N + 1 ) . By proving a global compactness result for coupled Schrödinger system, Liu and Liu [31] studied positive solutions for the following system:

(1.5) Δ u + W 1 ( x ) u = μ 1 u 3 + β u v 2 , x R 4 , Δ v + W 2 ( x ) v = β u 2 v + μ 2 v 3 , x R 4 , u 0 , v 0 , x R 4 , u , v D 1 , 2 ( R 4 ) ,

where W 1 , W 2 L 2 ( R 4 ) L loc ( R 4 ) are non-negative functions and μ 1 , μ 2 , and β are the positive constants. If β > max { μ 1 , μ 2 } and W 1 L 4 ( R 4 ) + W 2 L 4 ( R 4 ) > 0 , the authors obtained the existence of positive solutions to system (1.5) when W 1 L 4 ( R 4 ) and W 2 L 4 ( R 4 ) are suitably small. In fact, the result obtained in [31] generalized the well-known result in [10] due to Benci and Cerami on semilinear Schrödinger equations to the case of coupled nonlinear Schrödinger systems. Very recently, Guo et al. [25] considered system (1.5) with W j ( x ) = λ + W j 0 ( x ) and λ > 0 , W j 0 L 2 ( R 4 ) L loc ( R 4 ) , j = 1 , 2 , and obtained the multiplicity of positive solutions to system (1.5) under W j 0 , j = 1 , 2 and λ are suitable small. We remark that some authors are also interested in sign-changing solutions for elliptic systems with critical exponent (see [12,32,42] for example).

On the other hand, the topic of system (1.1) is closely related to the following classical problems in exterior domains:

(1.6) Δ u + λ u = u p 2 u , x D , u = 0 , x D ,

where D R N ( N 3 ) is an unbounded domain, D is bounded, 2 < p < 2 N N 2 . In the classical study of Benci and Cerami [9], the authors showed that problem (1.6) does not have any ground-state solution. So, they have to search for positive solutions to (1.6) at high-energy levels. To achieve their goal, the authors first analyzed the behavior of Palais-Smale sequences and proved a precise estimate of the energy levels where the Palais-Smale condition fails, then using variational method and Brouwer degree theory succeeded to obtain that the problem (1.6) has at least one positive solution for λ sufficiently small or for R N \ D small enough. We point out that there are very few results for systems in exterior domain. Yu [54] investigated the relationship between the topology of the domain and multiplicity properties of solutions to the following elliptic system:

(1.7) Δ u + λ u = u p 1 u + λ μ v , in Ω , Δ v + λ v = v p 1 v + λ μ u , in Ω , u = v = 0 , on Ω ,

where Ω R N ( N 3 ) is an exterior domain with bounded boundary, λ and μ are the real parameters with 0 < μ < 1 , Ω , R N \ Ω are bounded, and 1 < p < N + 2 N 2 . To be more precise, the author obtained that there exists μ > 0 , such that for every 0 < μ < μ , there is λ ( μ ) > 0 such that for λ > λ ( μ ) , system (1.7) has at least 3 cat Ω ¯ [ Ω ¯ , R N \ B ρ ¯ ] pairs of nontrivial solutions, where ρ ¯ = inf { ρ : R N \ Ω B ρ ( 0 ) } > 0 and cat Ω ¯ [ Ω ¯ , R N \ B ρ ¯ ] denotes the relative category of Ω ¯ with respect to R N \ B ρ ¯ in Ω ¯ . Recently, Liu and Liu [30] studied the existence and multiplicity of positive solutions of the elliptic system

(1.8) Δ u + V 1 ( x ) u = μ 1 ( x ) u 3 + β ( x ) u v 2 , x Ω ε , Δ v + V 2 ( x ) v = β ( x ) u 2 v + μ 2 ( x ) v 3 , x Ω ε , u = v = 0 , x Ω ε ,

where Ω ε is an exterior domain in R N ( 1 N 3 ) such that R N \ Ω ε is far away from the origin and contains a sufficiently large ball, V j , μ j , β C ( R N ) , j = 1 , 2 all tend to positive constants at infinity. Specifically, by considering the interaction of potentials and topology of the domain, the authors proved that system (1.8) has at least three positive solutions for ε > 0 small enough. The main results extended work by Cerami and Molle [11] from single elliptic equations to the coupled nonlinear Schrödinger system. Among other critical systems on R N or bounded domain with C 2 -boundary in R N , Clapp and Szulkin [18] considered the following subcritical m -coupled system:

(1.9) Δ u i + κ i u = μ i u i p 2 u i + j i λ i j β i j u j α i j u i β i j 2 u i , u i H 0 1 ( Ω ) , i , j = 1 , m ,

where Ω is an exterior domain in R N ( N 3 ) , κ i , μ i > 0 , λ i j = λ j i < 0 , α i j , β i j > 1 , α i j = β j i , and α i j + β i j = p ( 2 , 2 N N 2 ) . When the exterior domain Ω is invariant under the action of a closed subgroup G of the group O ( N ) of linear isometries of R N , by a simple variational setting, the authors succeeded to obtain an unbounded sequence of nontrivial solutions whose components are G -invariant.

To the best of our knowledge, there seems to be no existence result for elliptic system with critical exponent in exterior domains in the literature. We remark that, very recently, by establishing global compact lemma, combining variational method with Brouwer degree, Jia et al. [27] obtained a positive solution for a class of Kirchhoff-type nonlocal problem with critical exponent and nonconstant potential function in exterior domains when the hole suitable small. Motivated by the aforementioned works, we are concerned in this article with the existence of high-energy positive solutions to coupled Schrödinger equations (1.1) with critical exponent in exterior domains.

The main result, in the case of small perturbations from infinity of the indefinite potentials, establishes the following existence property of high-energy positive solutions.

Theorem 1.1

Suppose that β > max { μ 1 , μ 2 } and V j , j = 1 , 2 satisfy the following conditions:

( V 1 ) V j L 2 ( Ω ) L loc ( Ω ) , V j ( x ) 0 and V 1 ( x ) + V 2 ( x ) 0 , x Ω ;

( V 2 )

0 < β μ 2 β 2 μ 1 μ 2 V 1 2 + β μ 1 β 2 μ 1 μ 2 V 2 2 < min β 2 μ 1 μ 2 μ 1 ( 2 β μ 1 μ 2 ) , β 2 μ 1 μ 2 μ 2 ( 2 β μ 1 μ 2 ) , 2 S S ,

where S is the best Sobolev constant for the embedding D 1 , 2 ( R 4 ) L 4 ( R 4 ) . Then, there is a small ρ > 0 such that if R N \ Ω B ρ ( 0 ) , the coupled system (1.1) has at least one high-energy positive solution.

Remark 1.1

According to the strong maximum principle, the requirement of assumptions V j L loc ( Ω ) for j = 1 , 2 , is used just to obtain that a nontrivial non-negative solution of system (1.1) is indeed a positive solution of system (1.1). In fact, as discussed in [25], if the assumptions V j L loc ( Ω ) are replaced by other regularity conditions on V j , then the nontrivial non-negative solutions of system (1.1) are actually positive solutions (see Proposition 1.2 in [25]).

Remark 1.2

Throughout the analysis of previous related articles, as far as we know, when discussing critical problems in exterior domain, the critical term is basically used as a small critical perturbation, excepting [27]. Inspired partially by [9,10,27], in the case of small perturbations from infinity of the indefinite potential but nonsmall critical perturbations, we discuss the existence of high-energy positive solutions to coupled Schrödinger equations (1.1) in exterior domains. However, the methods developed in [9,10,27] for the single equation are not directly applicable to system (1.1), some additional difficulties arise due to the competition between the critical term and the coupling terms and the fact how distinguish the solutions from the semi-trivial ones. Besides, for critical elliptic systems in exterior domains, we have to overcome the loss of compactness from both the unboundedness of exterior domain and the critical nonlinearity simultaneously.

Remark 1.3

To our best knowledge, Theorem 1.1 seems to be the first existence result for critical elliptic systems in exterior domains. Furthermore, it is particularly worth noting that our result still holds in the case Ω = R 4 ; hence, it can be viewed as an extension of recent results for the Benci-Cerami problem for coupled Schrödinger systems with critical exponent in R 4 [31].

Our article is organized as follows. In Section 2, we give some preliminary results and obtain that system does not have any ground-state solution. In Section 3, we establish a global compactness result. In Section 4, we provide some estimates that are required to prove Theorem 1.1. By using some auxiliary results obtained in Section 4, we give the proof of Theorem 1.1 in Section 5 of this article. We will use C , C i , i N , to denote positive constants that can differ from line to line.

2 Preliminary results

Throughout this article, without any loss of generality, we may assume that 0 R 4 \ Ω . The norm of u in L r ( Ω ) and L r ( R 4 ) are denoted by u r and u r , R 4 , 1 r < . Define

D 0 1 , 2 ( Ω ) = { u L 4 ( Ω ) : u L 2 ( Ω ) } ,

with the inner product and the norm

( u , v ) Ω = Ω u v d x , u 2 = ( u , u ) Ω .

Let 0 D 0 1 , 2 ( Ω ) × D 0 1 , 2 ( Ω ) with the norm

( u , v ) = Ω u 2 + v 2 d x 1 2 .

Define

D 1 , 2 ( R 4 ) = { u L 4 ( R 4 ) : u L 2 ( R 4 ) } ,

with the inner product and the norm

( u , v ) R 4 = R 4 u v d x , u R 4 2 = ( u , u ) R 4 .

Let 1 D 1 , 2 ( R 4 ) × D 1 , 2 ( R 4 ) with the norm

( u , v ) R 4 = R 4 u 2 + v 2 d x 1 2 .

For considering the positive solution to (1.1), we study the modified system

(2.1) Δ u + V 1 ( x ) u = μ 1 ( u + ) 3 + β u + ( v + ) 2 , x Ω , Δ v + V 1 ( x ) v = β ( u + ) 2 v + + μ 2 ( v + ) 3 , x Ω , u , v D 0 1 , 2 ( Ω ) ,

where u ± = max { ± u , 0 } and v ± = max { ± v , 0 } such that u = u + u and v = v + v . Since V 1 , V 2 L loc ( Ω ) are the nonnegative functions and μ 1 , μ 2 , and β are the positive constants, it follows from the strong maximum principle (see Theorem 8.19 in [24]) that if ( u , v ) is a nontrivial solution of system (2.1), then u ( x ) > 0 , v ( x ) > 0 for x Ω , which shows that ( u , v ) is a positive solution of system (1.1). Therefore, to obtain a positive solution of system (1.1), we only need to prove the existence of a nontrivial solution of system (2.1).

The energy functional associated with equation (2.1) is defined by

Φ ( u , v ) = 1 2 Ω u 2 + v 2 d x + 1 2 Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x 1 4 Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x = 1 2 ( u , v ) 2 + 1 2 Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x 1 4 Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x ,

for ( u , v ) 0 . It is easy to see that solutions to system (2.1) correspond to critical points of the functional Φ ( u , v ) . Moreover, we have

Φ ( u , v ) , ( φ , ψ ) = ( u , φ ) Ω + ( v , ψ ) Ω + Ω ( V 1 ( x ) u φ + V 2 ( x ) v ψ ) d x Ω ( μ 1 ( u + ) 3 φ + 1 2 β u + φ ( v + ) 2 + 1 2 β ( u + ) 2 v + ψ + μ 2 ( v + ) 3 ψ ) d x ,

for ( u , v ) , ( φ , ψ ) 0 .

Define the Nehari manifold as

N { ( u , v ) 0 ( u , v ) ( 0 , 0 ) , G ( u , v ) = 0 } , where G ( u , v ) Φ ( u , v ) , ( u , v ) ,

and

c inf ( u , v ) N Φ ( u , v ) .

Furthermore, we have the following results about N .

Lemma 2.1

Suppose that ( V 1 ) holds, then we have that

  1. N is a C 1 regular manifold diffeomorphic to the

    S + = { ( u , v ) 0 ( u , v ) = 1 , u + 0 or v + 0 } ;

  2. Φ has a positive bound from below on N ;

  3. ( u , v ) is a critical point of Φ if and only if ( u , v ) is a critical point of Φ constrained on N .

Proof

For ( u , v ) S + , defined G ( u , v ) ( t ) by

G ( u , v ) ( t ) = Φ ( t u , t v ) = t 2 2 ( u , v ) 2 + t 2 2 Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x t 4 4 Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x , t > 0 .

Obviously, G ( u , v ) ( 0 ) and G ( u , v ) ( 0 ) = 0 . Thanks to V j ( x ) 0 , j = 1 , 2, we have that

G ( u , v ) ( t ) > 0 , for t > 0 small enough, and G ( u , v ) ( t ) < 0 , for t > 0 large enough .

Hence, there exists t ( u , v ) > 0 such that G ( u , v ) ( t ( u , v ) ) = max t 0 G ( u , v ) ( t ) with G ( u , v ) ( t ( u , v ) ) = 0 and ( t ( u , v ) u , t ( u , v ) v ) N . It is easy to see that t ( u , v ) is unique. Furthermore, we have G ( u , v ) ( t ) > 0 for 0 < t < t ( u , v ) , G ( u , v ) ( t ) < 0 for t > t ( u , v ) .

Since Φ C 2 ( 0 , R ) , G is a C 1 functional. Then, for any ( u , v ) N , we have

(2.2) G ( u , v ) , ( u , v ) = 2 ( u , v ) 2 + 2 Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x 4 Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x = 2 Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x < 0 ,

which implies that (i) holds.

For any ( u , v ) N , since V j ( x ) 0 , j = 1 , 2, and μ 1 , μ 2 , β > 0 , using the Sobolev inequality, there are C 1 , C 2 > 0 such that

0 = ( u , v ) 2 + Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x u 2 + v 2 C 1 u 4 C 2 u 4 .

Then, there is C 3 > 0 such that

(2.3) ( u , v ) C 3 , for any ( u , v ) N .

Consequently, by (2.3), we conclude that

Φ ( u , v ) = 1 2 ( u , v ) 2 + 1 2 Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x 1 4 Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x = 1 4 ( u , v ) 2 + 1 4 Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x > 1 4 C 3 , for any ( u , v ) N ,

which shows that (ii) is satisfied.

If ( u , v ) is a critical point of Φ with ( u , v ) ( 0 , 0 ) , then Φ ( u , v ) = 0 , and thus, G ( u , v ) = 0 . So ( u , v ) is a critical point of Φ constrained on N . If ( u , v ) is a critical point of Φ constrained on N , then there is ς R satisfying Φ ( u , v ) = ς G ( u , v ) . By ( u , v ) N , we have

ς G ( u , v ) , ( u , v ) = Φ ( u , v ) , ( u , v ) = 0 .

Hence, due to (2.2), we obtain ς = 0 , i.e., Φ ( u , v ) = 0 .□

The limit problem of system (2.1) is the following system:

(2.4) Δ u = μ 1 ( u + ) 3 + β u + ( v + ) 2 , x R 4 , Δ v = β ( u + ) 2 v + + μ 2 ( v + ) 3 , x R 4 , u , v D 1 , 2 ( R 4 ) ,

of which the associated functional by

Φ ( u , v ) = 1 2 ( u , v ) R 4 2 1 4 R 4 ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x , ( u , v ) 1 .

Let

N { ( u , v ) 1 ( u , v ) ( 0 , 0 ) , Φ ( u , v ) , ( u , v ) = 0 } , c inf ( u , v ) N Φ ( u , v ) .

To avoid semitrivial solutions in searching for nontrivial solutions of system (2.1), we need to study the following single equation:

(2.5) Δ u + V j ( x ) u = μ j ( u + ) 3 , in Ω , u D 0 1 , 2 ( Ω ) , j = 1 , 2 ,

and the associated energy functional is given by

I j ( u ) = 1 2 u 2 + 1 2 Ω V j ( x ) u 2 d x 1 4 Ω μ j ( u + ) 4 d x .

Let

c j = inf u j I j ( u ) , where j { u D 0 1 , 2 ( Ω ) u 0 , I j ( u ) , u = 0 } .

The limit equations of (2.5) are the following equations:

(2.6) Δ u = μ j ( u + ) 3 in R 4 , u D 1 , 2 ( R 4 ) , j = 1 , 2 ,

and associated energy functional is given by

I j ( u ) = 1 2 u R 4 2 1 4 R 4 μ j ( u + ) 4 d x .

Let

c j = inf u j I j ( u ) , where j { u D 1 , 2 ( R 4 ) u 0 , I j ( u ) , u = 0 } .

For δ > 0 , z R 4 , let

(2.7) Ψ δ , z ( x ) = 2 2 δ δ + x z 2 .

According to results in [4,48], the set { Ψ δ , z δ > 0 , z R 4 } contains all positive solutions of

Δ u = u 3 , in R 4 .

Moreover, we have Ψ δ , z R 4 2 = Ψ δ , z 4 , R 4 4 = S 2 , where S is the best Sobolev constant for the embedding D 1 , 2 ( R 4 ) L 4 ( R 4 ) , i.e.,

S = inf u D 1 , 2 ( R 4 ) \ { 0 } u R 4 2 u 4 , R 4 2 .

It is well-known that c j = I j ( μ j 1 2 Ψ δ , z ) = 1 4 μ j 1 S 2 .

Lemma 2.2

[53] If a L 2 ( R 4 ) , ψ : D 1 , 2 ( R 4 ) R , u R 4 a ( x ) u 2 d x is weakly continuous.

Lemma 2.3

If V j is a nonnegative function in L 2 ( Ω ) , then c j = c j = 1 4 μ j 1 S 2 .

Proof

For any u D 0 1 , 2 ( Ω ) , let u 0 outside Ω , then it can be extended to D 1 , 2 ( R 4 ) . For any u j , there is t u > 0 such that t u u j and

c j I j ( t u u ) = 1 2 t u u R 4 2 1 4 R 4 μ j ( t u u + ) 4 d x = 1 2 t u u 2 1 4 Ω μ j ( t u u + ) 4 d x 1 2 t u u 2 + 1 2 Ω V j ( x ) ( t u u ) 2 d x 1 4 Ω μ j ( t u u + ) 4 d x = I j ( t u u ) I j ( u ) ,

which shows that c j c j .

Next, we prove that c j c j . Let u ˜ n D 0 1 , 2 ( Ω ) be defined by u ˜ n ζ ( x ) μ j 1 2 Ψ n , where Ψ n ( ) = Ψ ( z n ) and Ψ = Ψ 1,0 D 1 , 2 ( R 4 ) , { z n } Ω with z n + as n + , ζ : R 4 [ 0 , 1 ] being defined by

ζ ( x ) = ξ x ρ , ρ inf { τ : R 4 \ Ω B τ ( 0 ) ¯ } ,

where B τ ( x 0 ) { x R 4 : x x 0 < τ } and ξ ( t ) : R + { 0 } [ 0 , 1 ] is a non-decreasing function satisfying

ξ ( t ) = 0 , t 1 and ξ ( t ) = 1 , t 2 .

First, we show that

(2.8) I j ( u ˜ n ) c j and I j ( u ˜ n ) , u ˜ n 0 , as n + .

Obviously, Ψ n 0 in D 1 , 2 ( R 4 ) as n + . Hence, by Lemma 2.2, we have that

Ω V j ( x ) ( u ˜ n ) 2 d x 0 , as n + .

Furthermore, for n large enough, we deduce that

(2.9) Ω ( u ˜ n ) 4 d x R 4 μ j 2 Ψ n 4 d x C 1 B 2 λ ( 0 ) Ψ ( x z n ) 4 d x C 2 B 2 λ ( 0 ) 1 x z n 4 d x = o 1 z n

and

(2.10) u ˜ n μ j 1 2 Ψ n R 4 2 C 3 B 2 λ ( 0 ) Ψ ( x z n ) 2 d x + C 4 B 2 λ ( 0 ) Ψ ( x z n ) 2 d x = o 1 z n .

Thanks to Ψ n R 4 = Ψ R 4 , it follows from (2.9) and (2.10) that (2.8) is satisfied.

For u ˜ n D 0 1 , 2 ( Ω ) , arguing as in Lemma 2.1, there is unique t n > 0 satisfying t n u ˜ n j and then I ( t n u ˜ n ) , t n u ˜ n = 0 .

We prove that t n 1 as n + . According to the definition of u ˜ n , it is easy to obtain that

C 5 u ˜ n C 6 , C 5 u ˜ n 4 C 6 ,

where C 5 , C 6 > 0 are the constants. So, we easily obtain that there exists C > 0 such that t n C . If { t n } is not bounded from above, we can suppose that t n + . It follows from t n u ˜ n j that

u ˜ n 2 = t n 2 μ j u ˜ n 4 4 + o n ( 1 ) + , as n + .

Obviously, it is absurd., i.e., { t n } is bounded from above. Then, by I ( u ˜ n ) , u ˜ n 0 as n + , we easily obtain that t n 1 as n + . So, it follows from (2.8) that I j ( t n u ˜ n ) c j . Thanks to t n u ˜ n j , we succeed to obtain that c j c j . Therefore, the proof is complete.□

The next two lemmas are essentially proved in [31], and here, we omit them.

Lemma 2.4

[31] If β > max { μ 1 , μ 2 } , then c = 1 4 ( ν 1 + ν 2 ) S 2 , where ν 1 = β μ 2 β 2 μ 1 μ 2 and ν 2 = β μ 1 β 2 μ 1 μ 2 .

Lemma 2.5

[31] If β > max { μ 1 , μ 2 } , then any nontrivial solution ( u , v ) of (2.4) must be of the form

( u , v ) = ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) ,

for some δ > 0 and z R 4 . Furthermore, each nontrivial solution of (2.4) is a ground-state solution.

The following proposition indicates that there is no ground-state solution to system (1.1).

Proposition 2.1

If β > max { μ 1 , μ 2 } and V 1 , V 2 L 2 ( Ω ) are nonnegative functions and

V 1 2 + V 2 2 > 0 ,

then c = c and c is not achieved.

Proof

Step 1: Arguing as in the proof of Lemma 2.3, we prove c = c . For any ( u , v ) 0 , let u 0 , v 0 outside Ω , then it can be extended to 1 . For any ( u , v ) N , there is t ( u , v ) > 0 satisfying ( t ( u , v ) u , t ( u , v ) v ) N . So it follows from V j 0 , j = 1 , 2 that

t ( u , v ) 2 = ( u , v ) R 4 2 R 4 ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x = ( u , v ) 2 Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x ( u , v ) 2 + Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x Ω ( μ 1 ( u + ) 4 + 2 β ( u + ) 2 ( v + ) 2 + μ 2 ( v + ) 4 ) d x = 1 .

So, we have

c Φ ( t ( u , v ) u , t ( u , v ) v ) = 1 2 ( t ( u , v ) u , t ( u , v ) v ) R 4 2 1 4 R 4 ( μ 1 ( t ( u , v ) u + ) 4 + 2 β ( t ( u , v ) u + ) 2 ( t ( u , v ) v + ) 2 + μ 2 ( t ( u , v ) v + ) 4 ) d x = t ( u , v ) 2 4 ( u , v ) 2 1 4 ( u , v ) 2 + 1 4 Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x = Φ ( u , v ) ,

from which we conclude that c c .

In the following, we prove c c . Let { ϕ n } , { φ n } D 0 1 , 2 ( Ω ) be defined by

ϕ n ζ ( x ) ν 1 Ψ n , φ n ζ ( x ) ν 2 Ψ n ,

where Ψ n and ζ are defined as in Lemma 2.3. Obviously, ϕ n , φ n 0 . We claim that

(2.11) Φ ( ϕ n , φ n ) c and Φ ( ϕ n , φ n ) , ( ϕ n , φ n ) 0 , as n + .

Thanks to Ψ n R 4 = Ψ R 4 and Ψ n 0 in D 1 , 2 ( R 4 ) as n + , it follows from Lemma 2.2 that

(2.12) Ω V 1 ( x ) ϕ n 2 d x 0 , Ω V 2 ( x ) φ n 2 d x 0 , as n + .

By definition of Ψ , for n large enough,

(2.13) Ω ϕ n 4 d x R 4 ν 1 2 Ψ n 4 d x = B 2 λ ( 0 ) ( ζ ( x ) 1 ) ν 1 Ψ ( x z n ) 4 d x C 1 B 2 λ ( 0 ) 1 x z n 4 d x = o 1 z n

and

(2.14) ϕ n ν 1 Ψ n R 4 2 C 2 B 2 λ ( 0 ) Ψ ( x z n ) 2 d x + C 3 B 2 λ ( 0 ) Ψ ( x z n ) 2 d x = o 1 z n .

Similarly, we can obtain that

(2.15) Ω φ n 4 d x R 4 ν 2 2 Ψ n 4 d x = o 1 z n , Ω ϕ n 2 φ n 2 d x R 4 ν 1 ν 2 Ψ n 4 d x = o 1 z n ,

(2.16) φ n ν 2 Ψ n R 4 2 = o 1 z n ,

which combining with (2.12)–(2.14) show that

Φ ( ϕ n , φ n ) Φ ( ν 1 Ψ 1,0 , ν 2 Ψ 1,0 ) = 1 4 ( ν 1 + ν 2 ) S 2

and

Φ ( ϕ n , φ n ) , ( ϕ n , φ n ) 0 , as n + .

For ( ϕ n , φ n ) 0 , arguing as in Lemma 2.1, there exists unique t n t ( ϕ n , φ n ) > 0 such that ( t n ϕ n , t n φ n ) N and then Φ ( t n ϕ n , t n φ n ) , ( t n ϕ n , t n φ n ) = 0 .

We claim that t n 1 as n + . Indeed, according to the definition of ϕ n and φ n , it is easy to see that

(2.17) C 1 ϕ n , φ n , ϕ n 4 , φ n 4 C 2 ,

where C 1 and C 2 are the positive constants. Hence, we can obtain that there is C > 0 such that t n C . If { t n } is not bounded from above, suppose that t n + . Then, thanks to ( t n ϕ n , t n φ n ) N , we have that

( ϕ n , φ n ) 2 = t n 2 Ω ( μ 1 ( ϕ n + ) 4 + 2 β ( ϕ n + ) 2 ( φ n + ) 2 + μ 2 ( φ n + ) 4 ) d x + o n ( 1 ) + , as n + ,

which is a contradiction due to (2.17). Thus, we obtain that { t n } is bounded from above. So, according to the fact Φ ( ϕ n , φ n ) , ( ϕ n , φ n ) 0 as n + , we can easily obtain t n 1 as n + . Then, it follows from (2.11) that Φ ( t n ϕ n , t n φ n ) c . By ( t n ϕ n , t n φ n ) N , we can conclude that c c .

Consequently, from the aforementioned arguments, we conclude that c = c .

Step 2: To prove that c is not achieved. Suppose that, by contradiction, there exists ( u , v ) N such that Φ ( u , v ) = c = c and Φ ( u , v ) = 0 . Let t ( u , v ) > 0 be such that ( t ( u , v ) u , t ( u , v ) v ) N , then we have

c Φ ( t ( u , v ) u , t ( u , v ) v ) = 1 2 ( t ( u , v ) u , t ( u , v ) v ) R N 2 1 4 R 4 ( μ 1 ( t ( u , v ) u + ) 4 + 2 β ( t ( u , v ) u + ) 2 ( t ( u , v ) v + ) 2 + μ 2 ( t ( u , v ) v + ) 4 ) d x 1 2 ( t ( u , v ) u , t ( u , v ) v ) 2 + 1 2 Ω ( V 1 ( x ) ( t ( u , v ) u ) 2 + V 2 ( x ) ( t ( u , v ) v ) 2 ) d x 1 4 Ω ( μ 1 ( t ( u , v ) u + ) 4 + 2 β ( t ( u , v ) u + ) 2 ( t ( u , v ) v + ) 2 + μ 2 ( t ( u , v ) v + ) 4 ) d x = Φ ( t ( u , v ) u , t ( u , v ) v ) Φ ( u , v ) = c .

Hence, we deduce that

t ( u , v ) = 1 , Ω ( V 1 ( x ) u 2 + V 2 ( x ) v 2 ) d x = 0 ,

which implies that ( u , v ) is a minimizer of c . Since β > max { μ 1 , μ 2 } ,

c = 1 4 ( ν 1 + ν 2 ) S 2 < min 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 = min { c 1 , c 2 } ,

which implies that u 0 and v 0 . Therefore, by the maximum principle, we obtain that u > 0 and v > 0 in R 4 , which is impossible since u = 0 and v = 0 in R 4 \ Ω .□

3 Global compactness result

We first recall some useful properties for the proof of the global compactness result.

Lemma 3.1

[31] (a) If { u n } is a bounded sequence in L 4 ( R 4 ) such that u n u almost everywhere in R 4 as n , then

lim n R 4 ( ( u n + ) 4 ( ( u n u ) + ) 4 ( u + ) 4 ) d x = 0

and

lim n R 4 ( u n + ) 3 ( ( u n u ) + ) 3 ( u + ) 3 4 3 d x = 0 .

(b) If { ( u n , v n ) } is a bounded sequence in L 4 ( R 4 ) × L 4 ( R 4 ) such that ( u n , v n ) ( u , v ) almost everywhere in R 4 as n , then

lim n R 4 ( ( u n + ) 2 ( v n + ) 2 ( ( u n u ) + ) 2 ( ( v n v ) + ) 2 ( u + ) 2 ( v + ) 2 ) d x = 0

and

lim n R 4 u n + ( v n + ) 2 ( u n u ) + ( ( v n v ) + ) 2 u + ( v + ) 2 d x = 0 .

Lemma 3.2

Assume that a L 2 ( R 4 ) . If { u n } is bounded in L 4 ( R 4 ) and u n 0 almost everywhere in R 4 as n , then

lim n R 4 a ( x ) u n 4 3 d x = 0 .

Lemma 3.3

[31] Suppose that { ( u n , v n ) } 1 is a ( P . S . ) d sequence for Φ such that ( u n , v n ) ( 0 , 0 ) in 1 and u n 0 in 1 . Then, there exist a sequence of points { z n } R 4 and a sequence of positive numbers { σ n } such that

(3.1) ( u n 1 ( x ) , v n 1 ( x ) ) = ( σ u n ( σ n x + z n ) , σ v n ( σ n x + z n ) )

converges weakly in 1 to a nontrivial solution ( u , v ) of problem (2.4).

According to Proposition 2.1, we deduce that system (1.1) does not have any ground-state solution. So, for system (1.1), we only need to search for a high-energy solution. For this purpose, the following global compactness is an essential tool.

Theorem 3.1

Assume that β > max { μ 1 , μ 2 } and V 1 , V 2 L 2 ( Ω ) are nonnegative functions. Let { ( u n , v n ) } 0 be a ( P . S . ) d sequence for Φ , i.e.,

(3.2) Φ ( u n , v n ) d and Φ ( u n , v n ) 0 as n + .

Then, there exist a number l N { 0 , 1 , 2 , } , l sequences of numbers { σ n j } R + , points { z n j } R 4 , 1 j l , l + 1 sequences of functions { ( u n ( 0 ) , v n ( 0 ) ) } , { ( u n ( j ) , v n ( j ) ) } 0 , 1 j l , such that for some subsequence, still denoted by { ( u n , v n ) } ,

(3.3) ( a ) ( u n , v n ) = ( u n ( 0 ) , v n ( 0 ) ) + j = 1 l ( σ n j ) 1 u n ( j ) z n j σ n j , ( σ n j ) 1 v n ( j ) z n j σ n j ; ( b ) ( u n ( 0 ) , v n ( 0 ) ) ( u ( 0 ) , u ( 0 ) ) , in 0 , as n + ; ( c ) ( u n ( j ) , u n ( j ) ) ( u ( j ) , v ( j ) ) ( 0 , 0 ) , in 1 , as n + , 1 j l ;

where ( u ( 0 ) , u ( 0 ) ) , ( u ( j ) , v ( j ) ) ( 1 j l ) satisfy

(3.4) Δ u ( 0 ) + V 1 ( x ) u ( 0 ) = μ 1 ( ( u ( 0 ) ) + ) 3 + β ( u ( 0 ) ) + ( ( v ( 0 ) ) + ) 2 , x Ω , Δ v ( 0 ) + V 2 ( x ) v ( 0 ) = β ( ( u ( 0 ) ) + ) 2 ( v ( 0 ) ) + + μ 2 ( ( v ( 0 ) ) + ) 3 , x Ω , u ( 0 ) , v ( 0 ) D 0 1 , 2 ( Ω ) ,

and

(3.5) Δ u ( j ) = μ 1 ( ( u ( j ) ) + ) 3 + β ( u ( j ) ) + ( ( v ( j ) ) + ) 2 , x R 4 , Δ v ( j ) = β ( ( u ( j ) ) + ) 2 ( v ( j ) ) + + μ 2 ( ( v ( j ) ) + ) 3 , x R 4 , u ( j ) , v ( j ) D 1 , 2 ( R 4 ) ,

respectively. Moreover, we have

(3.6) ( u n , v n ) ( u ( 0 ) , v ( 0 ) ) j = 1 l ( σ n j ) 1 u n ( j ) z n j σ n j , ( σ n j ) 1 v n ( j ) z n j σ n j 2 0 as n + ,

and

(3.7) Φ ( u n , v n ) Φ ( u ( 0 ) , v ( 0 ) ) + j = 1 l Φ ( u ( j ) , v ( j ) ) as n + .

Proof

First, we prove that { ( u n , v n ) } is bounded in 0 . It follows from (3.2) that

d + o n ( 1 ) + o n ( 1 ) ( u n , v n ) = Φ ( u n , v n ) 1 4 Φ ( u n , v n ) , ( u n , v n ) = 1 4 ( u n , v n ) 2 + 1 4 Ω ( V 1 ( x ) u n 2 + V 2 ( x ) v n 2 ) d x ,

which combined with V 1 ( x ) , V 2 ( x ) 0 shows that { ( u n , v n ) } is bounded in 0 . So there is ( u ( 0 ) , v ( 0 ) ) 0 such that, up to a subsequence, still denoted by { ( u n , v n ) } ,

(3.8) ( u n , v n ) ( u ( 0 ) , v ( 0 ) ) , in 0 ; ( u n , v n ) ( u ( 0 ) , v ( 0 ) ) a.e. in Ω .

For any φ , ψ C 0 ( Ω ) , by Lemmas 3.1 and 3.2, we have that

Φ ( u n , v n ) , ( φ , ψ ) = ( u n , φ ) Ω + ( v n , ψ ) Ω + Ω ( V 1 ( x ) u n φ + V 2 ( x ) v n ψ ) d x Ω ( μ 1 ( u n + ) 3 φ + 1 2 β u n + φ ( v n + ) 2 + 1 2 β ( u n + ) 2 v n + ψ + μ 2 ( v n + ) 3 ) ψ d x = ( u ( 0 ) , φ ) Ω + ( v ( 0 ) , ψ ) Ω + Ω ( V 1 ( x ) u ( 0 ) φ + V 2 ( x ) v ( 0 ) ψ ) d x Ω ( μ 1 ( ( u ( 0 ) ) + ) 3 φ + 1 2 β ( u ( 0 ) ) + φ ( ( v ( 0 ) ) + ) 2 + 1 2 β ( ( u ( 0 ) ) + ) 2 ( v ( 0 ) ) + ψ + μ 2 ( ( v ( 0 ) ) + ) 3 ) ψ d x = Φ ( u ( 0 ) , v ( 0 ) ) , ( φ , ψ ) + o n ( 1 ) ,

which shows that Φ ( u ( 0 ) , v ( 0 ) ) , ( φ , ψ ) = 0 . It follows that ( u ( 0 ) , v ( 0 ) ) satisfies (3.4).

Let

(3.9) u ^ n ( 1 ) ( x ) ( u n u ( 0 ) ) ( x ) , x Ω , 0 , x R 4 \ Ω , v ^ n ( 1 ) ( x ) = ( v n v ( 0 ) ) ( x ) , x Ω , 0 , x R 4 \ Ω .

Then, it follows from (3.8) that ( u ^ n ( 1 ) , v ^ n ( 1 ) ) 0 in 1 as n + . Thanks to Lemma 2.2, we have

(3.10) Ω ( V 1 ( x ) ( u ^ n ( 1 ) ) 2 + V 2 ( x ) ( v ^ n ( 1 ) ) 2 ) d x ( 0 , 0 ) , as n + .

Furthermore, we have that

(3.11) ( u ^ n ( 1 ) , v ^ n ( 1 ) ) R 4 2 = ( u ^ n ( 1 ) , v ^ n ( 1 ) ) 2 = ( u n , v n ) 2 ( u ( 0 ) , v ( 0 ) ) 2 + o n ( 1 ) , Φ ( ( u ^ n ( 1 ) , v ^ n ( 1 ) ) ) = 1 2 ( u ^ n ( 1 ) , v ^ n ( 1 ) ) R 4 2 + 1 2 Ω ( V 1 ( x ) ( u ^ n ( 1 ) ) 2 + V 2 ( x ) ( v ^ n ( 1 ) ) 2 ) d x 1 4 R 4 ( μ 1 ( ( u ^ n ( 1 ) ) + ) 4 + 2 β ( ( u ^ n ( 1 ) ) + ) 2 ( ( v ^ n ( 1 ) ) + ) 2 + μ 2 ( ( v ^ n ( 1 ) ) + ) 4 ) d x + o n ( 1 ) = 1 2 ( u n , v n ) 2 + 1 2 Ω ( V 1 ( x ) u n 2 + V 2 ( x ) v n 2 ) d x 1 4 Ω ( μ 1 ( u n + ) 4 + 2 β ( u n + ) 2 ( v n + ) 2 + μ 2 ( v n + ) 4 ) d x 1 2 ( u ( 0 ) , v ( 0 ) ) 2 1 2 Ω ( V 1 ( x ) ( u ( 0 ) ) 2 + V 2 ( x ) ( v ( 0 ) ) ) d x + 1 4 Ω ( μ 1 ( ( u 0 ) + ) 4 + 2 β ( ( u 0 ) + ) 2 ( ( v 0 ) + ) 2 + μ 2 ( ( v 0 ) + ) 4 ) d x + o n ( 1 ) = Φ ( u n , v n ) Φ ( u ( 0 ) , v ( 0 ) ) + o n ( 1 ) .

If ( u ^ n ( 1 ) , v ^ n ( 1 ) ) ( 0 , 0 ) in 1 , the theorem is proved with l = 0 .

If ( u ^ n ( 1 ) , v ^ n ( 1 ) ) ( 0 , 0 ) in 1 . For any φ , ψ C 0 ( R 4 ) , thanks to Lemma 3.1 and (3.2), we have that

(3.12) Φ ( u ^ n ( 1 ) , v ^ n ( 1 ) ) , ( φ , ψ ) = ( u ^ n ( 1 ) , φ ) R 4 + ( v ^ n ( 1 ) , ψ ) R 4 + Ω ( V 1 ( x ) u ^ n ( 1 ) φ + V 2 ( x ) v ^ n ( 1 ) ψ ) d x + o n ( 1 ) φ R 4 + o n ( 1 ) ψ R 4 R 4 ( μ 1 ( ( u ^ n ( 1 ) ) + ) 3 φ + 1 2 β ( u ^ n ( 1 ) ) + φ ( ( v ^ n ( 1 ) ) + ) 2 + 1 2 β ( ( u ^ n ( 1 ) ) + ) 2 v n + ψ + μ 2 ( ( v ^ n ( 1 ) ) + ) 3 ) ψ d x = Φ ( u ^ n ( 1 ) , v ^ n ( 1 ) ) , ( φ , ψ ) + o n ( 1 ) φ R 4 + o n ( 1 ) ψ R 4 = Φ ( u n , v n ) , ( φ , ψ ) Φ ( u ( 0 ) , v ( 0 ) ) , ( φ , ψ ) + o n ( 1 ) = o n ( 1 ) ,

which shows that Φ ( u ^ n ( 1 ) , v ^ n ( 1 ) ) ( 0 , 0 ) as n + . Hence, { ( u ^ n ( 1 ) , v ^ n ( 1 ) ) } is a Palais-Smale sequence for Φ and satisfies

( u ^ n ( 1 ) , v ^ n ( 1 ) ) ( 0 , 0 ) , in 1 ; ( u ^ n ( 1 ) , v ^ n ( 1 ) ) ( 0 , 0 ) , in 1 .

Then, by Lemma 3.3, there exist { z n 1 } R 4 , { σ n 1 } R + and ( u ( 1 ) , v ( 1 ) ) 1 such that

( u n ( 1 ) , v n ( 1 ) ) ( σ n 1 u ^ n ( 1 ) ( σ n 1 + z n 1 ) , σ n 1 v ^ n ( 1 ) ( σ n 1 + z n 1 ) ) , ( u n ( 1 ) , v n ( 1 ) ) ( u ( 1 ) , v ( 1 ) ) , in 1 , Φ ( u ( 1 ) , v ( 1 ) ) = 0 , ( u ( 1 ) , v ( 1 ) ) ( 0 , 0 ) .

Thus, ( u ( 1 ) , v ( 1 ) ) is a nonzero solution of system (3.5). Moreover, we have

( u ^ n ( 1 ) , v ^ n ( 1 ) ) R 4 2 = ( u n ( 1 ) , v n ( 1 ) ) R 4 2 = ( u ( 1 ) , v ( 1 ) ) R 4 2 + ( u n ( 1 ) , v n ( 1 ) ) ( u ( 1 ) , v ( 1 ) ) R 4 2 + o n ( 1 ) , Φ ( u ^ n ( 1 ) , v ^ n ( 1 ) ) = Φ ( u n ( 1 ) , v n ( 1 ) ) = Φ ( u ( 1 ) , v ( 1 ) ) + Φ ( u n ( 1 ) u ( 1 ) , v n ( 1 ) v ( 1 ) ) + o n ( 1 ) .

Combining the aforementioned equality and (3.11), we conclude that

(3.13) ( u n , v n ) 2 = ( u ^ n ( 1 ) , v ^ n ( 1 ) ) R 4 2 + ( u ( 0 ) , v ( 0 ) ) 2 + o n ( 1 ) = ( u ( 0 ) , v ( 0 ) ) 2 + ( u ( 1 ) , v ( 1 ) ) R 4 2 + ( u n ( 1 ) u ( 1 ) , v n ( 1 ) v ( 1 ) ) R 4 2 + o n ( 1 ) Φ ( u n , v n ) = Φ ( u ( 0 ) , v ( 0 ) ) + Φ ( u ^ n ( 1 ) , v ^ n ( 1 ) ) + o n ( 1 ) = Φ ( u ( 0 ) , v ( 0 ) ) + Φ ( u ( 1 ) , v ( 1 ) ) + Φ ( u n ( 1 ) u ( 1 ) , v n ( 1 ) v ( 1 ) ) + o n ( 1 ) .

Let ( u ^ n ( 2 ) , v ^ n ( 2 ) ) ( u n ( 1 ) u ( 1 ) , v n ( 1 ) v ( 1 ) ) , if ( u ^ n ( 2 ) , v ^ n ( 2 ) ) ( 0 , 0 ) in 1 ; Theorem is proved with l = 1 .

If ( u ^ n ( 2 ) , v ^ n ( 2 ) ) ( 0 , 0 ) in 1 , we can conclude that { ( u ^ n ( 2 ) , v ^ n ( 2 ) ) } is a Palais-Smale sequence for Φ and satisfies

( u ^ n ( 2 ) , v ^ n ( 2 ) ) ( 0 , 0 ) in 1 ; ( u ^ n ( 2 ) , v ^ n ( 2 ) ) ( 0 , 0 ) , in 1 .

Then, it follows from Lemma 3.3 that there are { z n 2 } R 4 , { σ n 2 } R + and ( u ( 2 ) , v ( 2 ) ) 1 so that

( u n ( 2 ) , u n ( 2 ) ) ( σ n 2 u ^ n ( 2 ) ( σ n 2 + z n 2 ) , σ n 2 v ^ n ( 2 ) ( σ n 2 + z n 2 ) ) , ( u n ( 2 ) , v n ( 2 ) ) ( u ( 2 ) , u ( 2 ) ) in 1 , Φ ( u ( 2 ) , v ( 2 ) ) = 0 , ( u ( 2 ) , v ( 2 ) ) ( 0 , 0 ) ,

which implies that ( u ( 2 ) , v ( 2 ) ) is a nonzero solution of system (3.5). Moreover, one has

( u ^ n ( 2 ) , v ^ n ( 2 ) ) R 4 2 = ( u n ( 2 ) , v n ( 2 ) ) R 4 2 = ( u ( 2 ) , v ( 2 ) ) R 4 2 + ( u n ( 2 ) u ( 2 ) , v n ( 2 ) v ( 2 ) ) R 4 2 + o n ( 1 ) , Φ ( u ^ n ( 2 ) , v ^ n ( 2 ) ) = Φ ( u n ( 2 ) , v n ( 2 ) ) = Φ ( u ( 2 ) , v ( 2 ) ) + Φ ( u n ( 2 ) u ( 2 ) , v n ( 2 ) v ( 2 ) ) + o n ( 1 ) ,

togethers, with (3.13), we conclude that

(3.14) ( u n , v n ) 2 = ( u ( 0 ) , v ( 0 ) ) 2 + ( u ( 1 ) , v ( 1 ) ) R 4 2 + ( u n ( 1 ) u ( 1 ) , v n ( 1 ) v ( 1 ) ) R 4 2 + o n ( 1 ) = ( u ( 0 ) , v ( 0 ) ) 2 + ( u ( 1 ) , v ( 1 ) ) R 4 2 + ( u ^ n ( 2 ) , v ^ n ( 2 ) ) R 4 2 + o n ( 1 ) = ( u ( 0 ) , v ( 0 ) ) 2 + ( u ( 1 ) , v ( 1 ) ) R 4 2 + ( u ( 2 ) , v ( 2 ) ) R 4 2 + ( u n ( 2 ) u ( 2 ) , v n ( 2 ) v ( 2 ) ) R 4 2 + o n ( 1 ) Φ ( u n , v n ) = Φ ( u ( 0 ) , v ( 0 ) ) + Φ ( u ( 1 ) , v ( 1 ) ) + Φ ( u n ( 1 ) u ( 1 ) , v n ( 1 ) v ( 1 ) ) + o n ( 1 ) = Φ ( u ( 0 ) , v ( 0 ) ) + Φ ( u ( 1 ) , v ( 1 ) ) + Φ ( u ^ n ( 2 ) , v ^ n ( 2 ) ) + o n ( 1 ) = Φ ( u ( 0 ) , v ( 0 ) ) + Φ ( u ( 1 ) , v ( 1 ) ) + Φ ( u ( 2 ) , v ( 2 ) ) + Φ ( u n ( 2 ) u ( 2 ) , v n ( 2 ) v ( 2 ) ) + o n ( 1 ) .

Iterating the aforementioned procedures, we can obtain sequences { ( u n ( k 1 ) , u n ( k 1 ) ) } in this way. Let ( u ^ n ( k ) , v ^ n ( k ) ) ( u n ( k 1 ) u ( k 1 ) , v n ( k 1 ) v ( k 1 ) ) , if ( u ^ n ( k ) , v ^ n ( k ) ) 0 in 1 , then theorem is proved with l = k .

If ( u ^ n ( k ) , v ^ n ( k ) ) 0 in 1 , arguing as earlier, { ( u ^ n ( k ) , v ^ n ( k ) ) } is a Palais-Smale sequence for Φ such that

( u ^ n ( k ) , v ^ n ( k ) ) ( 0 , 0 ) , in 1 ; ( u ^ n ( k ) , v ^ n ( k ) ) ( 0 , 0 ) , in 1 .

Then, according to Lemma 3.3, there exist { z n k } R 4 , { σ n k } R + and ( u ( k ) , v ( k ) ) 1 satisfying

( u n ( k ) , v n ( k ) ) ( σ n k u ^ n ( k ) ( σ n k + z n k ) , σ n k v ^ n ( k ) ( σ n k + z n k ) ) , ( u n ( k ) , v n ( k ) ) ( u ( k ) , v ( k ) ) , in 1 , Φ ( u ( k ) , v ( k ) ) = 0 , ( u ( k ) , v ( k ) ) ( 0 , 0 ) .

Thus, ( u ( k ) , v ( k ) ) is a nonzero solution of system (3.5). Furthermore,

(3.15) ( u n , v n ) = ( u ( 0 ) , v ( 0 ) ) 2 + Σ j = 1 k ( u ( j ) , v ( j ) ) R 4 2 + ( u n ( k ) u ( k ) , v n ( k ) v ( k ) ) R 4 2 + o n ( 1 ) , Φ ( u n , v n ) = Φ ( u ( 0 ) , v ( 0 ) ) + Σ j = 1 k Φ ( u ( j ) , v ( j ) ) + Φ ( u n ( k ) u ( k ) , v n ( k ) v ( k ) ) + o n ( 1 ) .

Thanks to ( u ( j ) , v ( j ) ) is solution (2.4) with ( u ( j ) , v ( j ) ) ( 0 , 0 ) , j = 1 , 2 , , k , and there holds

Φ ( u ( j ) , v ( j ) ) c = 1 4 ( ν 1 + ν 2 ) S 2 , j = 1 , 2 , , k .

Then, we conclude that the iteration must terminate at a finite index l 1 , i.e., ( u ^ n ( l + 1 ) , v ^ n ( l + 1 ) ) ( u n ( l ) u ( l ) , v n ( l ) v ( l ) ) ( 0 , 0 ) in 1 . Then, we have

( u n , v n ) = ( u ( 0 ) , v ( 0 ) ) 2 + Σ j = 1 l ( u ( j ) , v ( j ) ) R 4 2 + o n ( 1 ) , Φ ( u n , v n ) = Φ ( u ( 0 ) , v ( 0 ) ) + Σ j = 1 l Φ ( u ( j ) , v ( j ) ) + o n ( 1 ) .

Moreover, it is easy to obtain from the aforementioned discussion that

u n = ( u ( 0 ) + o n ( 1 ) ) + ( σ n 1 ) 1 ( u ( 1 ) + o n ( 1 ) ) x z n 1 σ n 1 + ( σ n 1 σ n 2 ) 1 ( u ( 2 ) + o n ( 1 ) ) x z n 1 σ n 1 z n 2 σ n 1 σ n 2 + ( σ n 1 σ n 2 σ n 3 ) 1 ( u ( 3 ) + o n ( 1 ) ) x z n 1 σ n 1 z n 2 σ n 1 σ n 2 z n 3 σ n 1 σ n 2 σ n 3 + + ( σ n 1 σ n 2 σ n 3 σ n l ) 1 ( u ( l ) + o n ( 1 ) ) x z n 1 σ n 1 z n 2 σ n 1 σ n 2 z n 3 σ n 1 σ n 2 σ n 3 σ n l 1 z n l σ n 1 σ n 2 σ n 3 σ n l

and

v n = ( v ( 0 ) + o n ( 1 ) ) + ( σ n 1 ) 1 ( v ( 1 ) + o n ( 1 ) ) x z n 1 σ n 1 + ( σ n 1 σ n 2 ) 1 ( v ( 2 ) + o n ( 1 ) ) x z n 1 σ n 1 z n 2 σ n 1 σ n 2 + ( σ n 1 σ n 2 σ n 3 ) 1 ( v ( 3 ) + o n ( 1 ) ) x z n 1 σ n 1 z n 2 σ n 1 σ n 2 z n 3 σ n 1 σ n 2 σ n 3 + + ( σ n 1 σ n 2 σ n 3 σ n l ) 1 ( v ( l ) + o n ( 1 ) ) x z n 1 σ n 1 z n 2 σ n 1 σ n 2 z n 3 σ n 1 σ n 2 σ n 3 σ n l 1 z n l σ n 1 σ n 2 σ n 3 σ n l .

So, it follows from rewriting the notations that (3.3)–(3.5) are satisfied.□

Corollary 3.1

Assume that β > max { μ 1 , μ 2 } and V 1 , V 2 L 2 ( Ω ) are nonnegative functions, let { ( u n , v n ) } 0 be a sequence such that

(3.16) Φ ( u n , v n ) c , Φ ( u n , v n ) , ( u n , v n ) = 0 , as n + ,

then we have

(3.17) ( u n , v n ) = ( w n 1 + ν 1 Ψ δ n , z n , w n 2 + ν 2 Ψ δ n , z n ) ,

where { ( w n 1 , w n 2 ) } 1 such that ( w n 1 , w n 2 ) 0 in 1 , and Ψ δ n , z n defined in (2.7) is the positive function such that ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) realizing c .

Proof

It follows from (3.16) that { ( u n , v n ) } is a minimizing sequence for Φ N . Then, according to the Ekeland variational principle [53], there is a sequence { ( u ˜ n , v ˜ n ) } N satisfying

(3.18) Φ ( u ˜ n , v ˜ n ) c , Φ ( u ˜ n , v ˜ n ) ς n G ( u ˜ n , v ˜ n ) 0 , ( u n , v n ) ( u ˜ n , v ˜ n ) 0 , as n + ,

where ς n R . We may assume u ˜ n , v ˜ n 0 .

Next, we prove that Φ ( u ˜ n , v ˜ n ) 0 as n + . From (3.18), we obtain

Φ ( u ˜ n , v ˜ n ) , ( u ˜ n , v ˜ n ) ς n G ( u ˜ n , v ˜ n ) , ( u ˜ n , v ˜ n ) = o n ( 1 ) ( u ˜ n , v ˜ n ) .

It follows from (2.2), (2.3), and the fact { ( u ˜ n , v ˜ n ) } N that G ( u ˜ n , v ˜ n ) , ( u ˜ n , v ˜ n ) < C ˜ < 0 . Therefore, thanks to { ( u ˜ n , v ˜ n ) } N and Φ ( u ˜ n , v ˜ n ) c , it is easy to see that { ( u ˜ n , v ˜ n ) } is bounded. So, ς n 0 as n + . For any ( φ , ψ ) 0 , according to the boundedness of V 1 2 and V 2 2 , using the Hölder inequality, we deduce that

G ( u ˜ n , v ˜ n ) , ( φ , ψ ) = 2 ( u ˜ n , φ ) Ω + 2 ( v ˜ n , ψ ) Ω + 2 Ω ( V 1 ( x ) u ˜ n φ + V 2 ( x ) v ˜ n ψ ) d x 4 Ω ( μ 1 ( u ˜ n + ) 3 φ + 1 2 β u ˜ n + φ ( v ˜ n + ) 2 + 1 2 β ( u ˜ n + ) 2 v ˜ n + ψ + μ 2 ( v ˜ n + ) 3 ) ψ d x C 1 ( u ˜ n , v ˜ n ) ( φ , ψ ) + C 2 ( u ˜ n , v ˜ n ) 3 ( φ , ψ ) + ( C 3 u ˜ n v ˜ n 2 + C 4 u ˜ n 2 v ˜ n ) ( φ , ψ ) ,

which combines with boundedness of { ( u ˜ n , v ˜ n ) } shows that G ( u ˜ n , v ˜ n ) is bounded. Then, we can easily obtain that Φ ( u ˜ n , v ˜ n ) 0 as n + .

Now, for any ( φ , ψ ) 0 , we could easily to obtain that

Φ ( u n , v n ) Φ ( u ˜ n , v ˜ n ) , ( φ , ψ ) 0 , as n + ,

from which we conclude that Φ ( u n , v n ) 0 as n + .

By Theorem 3.1, there exist a number l N and a subsequence of { ( u n , v n ) } , still denoted by { ( u n , v n ) } , such that (3.3)–(3.7) hold.

Case 1: ( u ( 0 ) , v ( 0 ) ) ( 0 , 0 ) and l 1 . Thanks to (3.3)–(3.5), we have

Φ ( u ( 0 ) , v ( 0 ) ) , ( u ( 0 ) , v ( 0 ) ) = 0 , Φ ( u ( j ) , v ( j ) ) , ( u ( j ) , v ( j ) ) = 0 , 1 j l ,

which shows that ( u ( 0 ) , v ( 0 ) ) N , ( u ( j ) , v ( j ) ) N , 1 j l . Hence, we have that Φ ( u ( 0 ) , v ( 0 ) ) c = c , Φ ( u ( j ) , v ( j ) ) c , 1 j l . We deduce that

c = Φ ( u ( 0 ) , v ( 0 ) ) + Σ j = 1 l Φ ( u ( j ) , v ( j ) ) ( l + 1 ) c 2 c ,

which contradicts with the fact c = c .

Case 2: ( u ( 0 ) , v ( 0 ) ) ( 0 , 0 ) and l 2 . Similar to the aforementioned discussion, we obtain a contradiction

c = Σ j = 1 l Φ ( u ( j ) , v ( j ) ) l c 2 c .

Case 3: ( u ( 0 ) , v ( 0 ) ) ( 0 , 0 ) and l = 0 . This case shows that ( u ( 0 ) , v ( 0 ) ) is a ground-state solution of (2.1), which contradicts with Proposition 2.1.

Case 4: ( u ( 0 ) , v ( 0 ) ) ( 0 , 0 ) and l = 0 . If this case holds, we also obtain a contradiction due to the fact Φ ( u ( 0 ) , and v ( 0 ) ) = c .

According to the aforementioned arguments, we must have that ( u ( 0 ) , v ( 0 ) ) ( 0 , 0 ) and l = 1 , i.e., ( u ( 1 ) , v ( 1 ) ) satisfies (3.5), and then, Φ ( u ( 1 ) , v ( 1 ) ) = c = c . This fact shows that ( u ( 1 ) , v ( 1 ) ) is a ground-state solution of (2.4). Consequently, it follows from Lemma 2.5 and the aforementioned discussion, (3.17) holds.□

Corollary 3.2

Assume that β > max { μ 1 , μ 2 } and V 1 , V 2 L 2 ( Ω ) are nonnegative functions, let { ( u n , v n ) } 0 be a sequence of ( P . S . ) d sequence for Φ , i.e.,

(3.19) Φ ( u n , v n ) d and Φ ( u n , v n ) 0 , as n + ,

if d ( c , min { 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c } ) , then the functional Φ satisfying the ( P . S . ) d condition.

Proof

According to Theorem 3.1, there are l N and a subsequence of { ( u n , v n ) } , still denoted by { ( u n , v n ) } , such that (3.3)–(3.7) hold. If l = 0 , we are done. If not, suppose that l > 0 , then thanks to d < 2 c , we deduce that l = 1 and d = Φ ( u ( 0 ) , v ( 0 ) ) + Φ ( u ( 1 ) , v ( 1 ) ) . If ( u ( 0 ) , v ( 0 ) ) ( 0 , 0 ) , according to Proposition 2.1, we have that

d = Φ ( u ( 0 ) , v ( 0 ) ) + Φ ( u ( 1 ) , v ( 1 ) ) > c + c = 2 c ,

which is a contradiction.

If ( u ( 0 ) , v ( 0 ) ) = ( 0 , 0 ) , thanks to u ( 1 ) 0 , v ( 1 ) 0 , it follows from Lemma 2.5 and the fact that the uniqueness of positive solutions for the scalar equation (2.6) that ( u ( 1 ) , v ( 1 ) ) must be, up to translation and dilation, one of the three solutions

( ν 1 Ψ 1,0 , ν 2 Ψ 1,0 ) , 1 μ 1 Ψ 1,0 , 0 , 0 , 1 μ 2 Ψ 1,0 ,

where ν 1 and ν 2 are defined as in Lemma 2.4. So, we obtain that d = c or d = 1 4 μ 1 1 S 2 or d = 1 4 μ 2 1 S 2 . Anyway, we have a contradiction.

Therefore, it follows from the aforementioned discussion that l = 0 , i.e., ( u n , v n ) ( u ( 0 ) , u ( 0 ) ) in 0 .□

4 Basic estimates

Since we assume that 0 R 4 \ Ω , ρ = inf { τ : R 4 \ Ω B τ ( 0 ) ¯ } > 0 . Let

( u ρ , v ρ ) ( ζ ( x ) ν 1 Ψ δ , z , ζ ( x ) ν 2 Ψ δ , z ) = ξ x ρ ν 1 Ψ δ , z , ξ x ρ ν 2 Ψ δ , z ,

where ζ and ξ are defined as in Lemma 2.3. It is easy to see that u ρ , v ρ 0 . Defined J ρ : R 4 × R + 1 by

J ρ ( z , δ ) = ( t ( u ρ , v ρ ) u ρ , t ( u ρ , v ρ ) v ρ ) ,

where t ( u ρ , v ρ ) > 0 satisfies Φ ( J ρ ( z , δ ) ) , J ρ ( z , δ ) = 0 . According to definitions, ( u ρ , v ρ ) and J ρ ( z , δ ) can be seen as elements in 0 and u ρ , v ρ L 4 ( Ω ) . Moreover, we have that

J ρ ( z , δ ) = J ρ ( z , δ ) R 4 , ( u ρ , v ρ ) = ( u ρ , v ρ ) R 4 ,

u ρ 4 = u ρ 4 , R 4 , v ρ 4 = v ρ 4 , R 4 .

Lemma 4.1

Assume that β > max { μ 1 , μ 2 } and V 1 , V 2 L 2 ( Ω ) are the nonnegative functions with V 1 2 + V 1 2 0 , then J ρ ( z , δ ) and ( u ρ , v ρ ) satisfy

  1. J ρ ( z , δ ) is continuous in ( z , δ ) for every ρ ;

  2. Φ ( u ρ , v ρ ) c , Φ ( u ρ , v ρ ) , ( u ρ , v ρ ) 0 , and t ( u ρ , v ρ ) 1 as z + , uniformly for every bounded ρ , and bounded δ away from 0;

  3. as ρ 0 , Φ ( u ρ , v ρ ) c , Φ ( u ρ , v ρ ) , ( u ρ , v ρ ) 0 , and t ( u ρ , v ρ ) 1 as δ 0 or δ + , uniformly in z R 4 .

Proof

It follows from definitions of ( u ρ , v ρ ) and (i) in Lemma 2.1 that (a) hold. By the similar arguments as in the proof of Proposition 2.1, we easily obtain (b).

We now prove (c). We first observe that

u ρ ν 1 Ψ δ , z R 4 2 = R 4 ( u ρ ν 1 Ψ δ , z ) 2 d x C 1 B 2 ρ ( 0 ) Ψ δ , z 2 d x + C 2 ρ x 2 ρ Ψ δ , z ζ ( x ) 2 d x C 1 B 2 ρ ( 0 ) Ψ δ , z 2 d x + C 3 ρ ρ x 2 ρ 1 d x 1 2 ρ x 2 ρ Ψ δ , z 4 d x 1 2 0 as ρ 0 , for any ( δ , z ) R 4 × R + .

It follows that

u ρ 2 = u ρ R 4 2 ν 1 Ψ δ , z R 4 2 , as ρ 0 , for any ( δ , z ) R 4 × R + .

Similarly, we obtain

v ρ 2 = v ρ R 4 2 ν 2 Ψ δ , z R 4 2 , as ρ 0 , for any ( δ , z ) R 4 × R + .

We conclude that

(4.1) ( u ρ , v ρ ) 2 = ( u ρ , v ρ ) R 4 2 ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) R 4 2 , as ρ 0 , for any ( δ , z ) R 4 × R + .

Moreover, as ρ 0 , we have

R 4 u ρ 4 d x R 4 ν 1 Ψ δ , z 4 d x B 2 ρ ( 0 ) u ρ 4 ν 1 Ψ δ , z 4 d x C 4 B 2 ρ ( 0 ) Ψ δ , z 4 d x 0 , for any ( δ , z ) R 4 × R + ,

which yields

(4.2) u ρ 4 2 = u ρ 4 , R 4 2 ν 1 Ψ δ , z 4 , R 4 2 , as ρ 0 , for any ( δ , z ) R 4 × R + .

By similar arguments, we can also obtain that

(4.3) v ρ 4 2 = v ρ 4 , R 4 2 ν 2 Ψ δ , z 4 , R 4 2 , as ρ 0 , for any ( δ , z ) R 4 × R + , R 4 u ρ 2 v ρ 2 d x R 4 ( ν 1 Ψ δ , z ) 2 ( ν 2 Ψ δ , z ) 2 d x as ρ 0 , for any ( δ , z ) R 4 × R + .

Since V j L 2 ( Ω ) for j = 1 , 2 we have

(4.4) R 4 V 1 ( x ) u ρ 2 d x R 4 V 1 ( x ) ν 1 Ψ δ , z 2 d x , as ρ 0 , ( z , δ ) R 4 × R + ,

and

(4.5) R 4 V 2 ( x ) v ρ 2 d x R 4 V 2 ( x ) ν 2 Ψ δ , z 2 d x , as ρ 0 , ( z , δ ) R 4 × R + .

Thanks to V j L 2 ( Ω ) , j = 1 , 2, arguing as in the proof of Lemma 4.2 in [31], for any ε > 0 , there exist δ 1 = δ 1 ( ε ) and δ 2 = δ 2 ( ε ) such that

R 4 V 1 ( x ) ν 1 Ψ δ , z 2 d x = ν 1 R 4 V 1 ( x ) Ψ δ , z 2 d x < ε ,

and

R 4 V 2 ( x ) ν 2 Ψ δ , z 2 d x = ν 2 R 4 V 2 ( x ) Ψ δ , z 2 d x < ε ,

for z R 4 and δ ( 0 , δ 1 ] [ δ 2 , + ) . And then, together with (4.4) and (4.5), we conclude that, as ρ 0 ,

(4.6) R 4 V 1 ( x ) u ρ 2 d x 0 as δ 0 or δ + , z R 4 ,

and

(4.7) R 4 V 2 ( x ) v ρ 2 d x 0 as δ 0 or δ + , z R 4 .

So, it follows from (4.1), (4.2), (4.3), (4.6), (4.7), Φ ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) = c , and ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) N that, as ρ 0 ,

Φ ( u ρ , v ρ ) c and Φ ( u ρ , v ρ ) , ( u ρ , v ρ ) 0 as , δ 0 or δ + for z R 4 .

Then, arguing as in Proposition 2.1, we deduce that as ρ 0 , t ( u ρ , v ρ ) 1 , as δ 0 or δ + for z R 4 . So, the proof of (c) is complete.□

Lemma 4.2

Let ( V 1 ) ( V 2 ) hold, then there is ρ ( 0 , 1 8 ) such that for any ρ < ρ ,

sup ( z , δ ) R 4 × R + Φ ( J ρ ( z , δ ) ) < min 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c .

Proof

Thanks to

Φ ( J ρ ( z , δ ) ) , J ρ ( z , δ ) = Φ ( t ( u ρ , v ρ ) u ρ , t ( u ρ , v ρ ) v ρ ) , ( t ( u ρ , v ρ ) u ρ , t ( u ρ , v ρ ) v ρ ) = 0 ,

combining with (4.1)–(4.5), we conclude that

t ( u ρ , v ρ ) 2 = ( u ρ , v ρ ) 2 + Ω ( V 1 ( x ) u ρ 2 + V 2 ( x ) v ρ 2 ) d x Ω ( μ 1 u ρ 4 + 2 β u ρ 2 v ρ 2 + μ 2 v ρ 4 ) d x ( u ρ , v ρ ) 2 + V 1 2 u ρ 4 2 + V 2 2 v ρ 4 2 Ω ( μ 1 u ρ 4 + 2 β u ρ 2 v ρ 2 + μ 2 v ρ 4 ) d x ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) R 4 2 + V 1 2 ν 1 Ψ δ , z 4 , R 4 2 + V 2 2 ν 2 Ψ δ , z 4 , R 4 2 R 4 ( μ 1 ν 1 2 Ψ δ , z 4 + 2 β ν 1 ν 2 Ψ δ , z 4 + μ 2 ν 2 2 Ψ δ , z 4 ) d x = ( ν 1 + ν 2 ) Ψ δ , z R 4 2 + ( V 1 2 ν 1 + V 2 2 ν 2 ) Ψ δ , z 4 , R 4 2 ( ν 1 + ν 2 ) Ψ δ , z 4 , R 4 4 ,

as ρ 0 , for any ( z , δ ) R 4 × R + . By using Φ ( J ρ ( z , δ ) ) , J ρ ( z , δ ) = 0 again, we have

Φ ( J ρ ( z , δ ) ) = t ( u ρ , v ρ ) 4 4 Ω ( μ 1 u ρ 4 + 2 β u ρ 2 v ρ 2 + μ 2 v ρ 4 ) d x .

Hence, by ( V 1 ) and ( V 2 ) we derive that

lim ρ 0 Φ ( J ρ ( z , δ ) ) = lim ρ 0 t v ρ 4 4 Ω ( μ 1 u ρ 4 + 2 β u ρ 2 v ρ 2 + μ 2 v ρ 4 ) d x 1 4 ( ν 1 + ν 2 ) Ψ δ , z R 4 2 + ( V 1 2 ν 1 + V 2 2 ν 2 ) Ψ δ , z 4 , R 4 2 ( ν 1 + ν 2 ) Ψ δ , z 4 , R 4 4 2 ( ν 1 + ν 2 ) Ψ δ , z 4 , R 4 4 = 1 4 ( ( ν 1 + ν 2 ) Ψ δ , z R 4 2 + ( V 1 2 ν 1 + V 2 2 ν 2 ) Ψ δ , z 4 , R 4 2 ) 2 ( ν 1 + ν 2 ) Ψ δ , z 4 , R 4 4 = 1 4 ( ( ν 1 + ν 2 ) S 2 + ( V 1 2 ν 1 + V 2 2 ν 2 ) S 2 ) 2 ( ν 1 + ν 2 ) S 2 < min 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 1 2 ( ν 1 + ν 2 ) S 2 = min 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c .

Consequently, we obtain that there exists constant ρ ( 0 , 1 8 ) such that for any ρ < ρ ,

sup ( z , δ ) R 4 × R + Φ ( J ρ ( z , δ ) ) < min { 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c } .

The proof is now complete.□

From now on, we always assume Ω fixed with diam ( R 4 \ Ω ) sup { x y : x , y R 4 \ Ω } < ρ , where ρ ( 0 , 1 8 ) is the constant obtained in Lemma 4.2. Hence, for any x 0 R 4 \ Ω , we have R 4 \ Ω B ρ ( x 0 ) . Thus,

ρ = inf { τ : R 4 \ Ω B τ ( 0 ) ¯ } < ρ < 1 8 , R 4 \ Ω B 1 8 ( 0 ) .

Defined χ i : R + R , i = 1 , 2 by

χ 1 ( t ) = 4 , t 1 4 , 1 t , t > 1 4 , and χ 2 ( t ) = 0 , t < 1 , 1 , t 1 .

Define a barycenter-type map α : 1 R 4 and a functional β : 1 R as

α ( u , v ) = 1 ( ν 1 + ν 2 ) S 2 R 4 χ 1 ( x ) x ( u 2 + v 2 ) d x , β ( u , v ) = 1 ( ν 1 + ν 2 ) S 2 R 4 χ 2 ( x ) ( u 2 + v 2 ) d x .

Let

( u , v ) N : ( α ( u , v ) , β ( u , v ) ) = 0 , 0 , 0 , 0 , 1 2 0 .

Lemma 4.3

If z 1 2 , then we have

α ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) = z z + o ( 1 ) , as δ 0 .

Proof

Fix z 1 2 , we have B 1 4 ( 0 ) B ε ( z ) = for any ε > 0 small enough. Then, it follows from the definition of Ψ δ , z that

(4.8) 1 ( ν 1 + ν 2 ) S 2 B 1 4 ( 0 ) 4 x Ψ δ , z 2 ( ν 1 + ν 2 ) d x 1 S 2 B 1 4 ( 0 ) 4 x Ψ δ , z 2 d x C B 1 4 ( 0 ) Ψ δ , z 2 d x C δ z 2 0 , as δ 0 ,

and

(4.9) 1 ( ν 1 + ν 2 ) S 2 A x x Ψ δ , z 2 ( ν 1 + ν 2 ) d x 1 S 2 R 4 \ B 1 4 ( 0 ) Ψ δ , z 2 d x C 1 R 4 \ B 1 4 ( 0 ) Ψ δ , z 2 d x C 2 δ 2 0 , as δ 0 ,

where A = R 4 \ ( B ε ( z ) B 1 4 ( 0 ) ) .

For any x B ε ( z ) , thanks to z 1 2 and ε > 0 small enough, one has

x x z z < C 2 ε .

Therefore, we deduce that

(4.10) z z 1 ( ν 1 + ν 2 ) S 2 B ε ( z ) x x Ψ δ , z 2 ( ν 1 + ν 2 ) d x = 1 S 2 B ε ( z ) z z x x Ψ δ , z 2 d x + 1 S 2 R 4 \ B ε ( z ) z z Ψ δ , z 2 d x C 2 ε S 2 R 4 Ψ δ , z 2 d x + 1 S 2 R 4 \ B ε ( z ) Ψ δ , z 2 d x 0 as δ 0 .

Hence, it follows from (4.8), (4.9), and (4.10) that

α ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) = z z + o ( 1 ) , as δ 0 .

This completes the proof.□

Lemma 4.4

Assume that β > max { μ 1 , μ 2 } and V 1 , V 2 L 2 ( Ω ) are the nonnegative functions with V 1 2 + V 1 2 0 , then we have

(4.11) c 0 inf ( u , v ) Φ ( u , v ) > c .

Furthermore, as ρ 0 , there exist T > 1 2 , 0 < δ 1 < 1 2 , δ 2 > 1 2 such that

(4.12) ( 1 ) β ( J ρ ( z , δ ) ) < 1 2 , z < 1 2 and δ δ 1 ; ( 2 ) α ( J ρ ( z , δ ) ) z z < 1 4 , z 1 2 and δ δ 1 ; ( 3 ) β ( J ρ ( z , δ ) ) > 1 2 , z R 4 and δ δ 2 ; ( 4 ) Φ ( J ρ ( z , δ ) ) < c 0 + c 2 , z R 4 and δ = δ 1 or δ = δ 2 ; ( 5 ) Φ ( J ρ ( z , δ ) ) ( c , c 0 + c 2 ) , z T and δ [ δ 1 , δ 2 ] ; ( 6 ) ( α ( J ρ ( z , δ ) ) , z ) R 4 > 0 , z = T and δ [ δ 1 , δ 2 ] .

Proof

Step 1: To prove (4.11).

It is easy to see that c 0 c . So, to prove (4.11), we suppose c 0 = c by contradiction. Hence, there is { ( u n , v n ) } such that

(4.13) lim n Φ ( u n , v n ) = c , Φ ( u n , v n ) , ( u n , v n ) = 0 , α ( u n , v n ) = ( 0,0,0,0 ) , β ( u n , v n ) = 1 2 .

According to Proposition 2.1, { ( u n , v n ) } is not relatively compact. Then, it follows from Corollary 3.1 that

( u n , v n ) = ( w n 1 + ν 1 Ψ δ n , z n , w n 2 + ν 2 Ψ δ n , z n ) , x R 4 ,

where { z n } R 4 , { δ n } R + , { ( w n 1 , w n 2 ) } 1 with ( w n 1 , w n 2 ) ( 0 , 0 ) in 1 , and ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) is a positive ground-state solution of (2.4) realizing c . In subsequence sense, for ( δ n , z n ) , one of the following conditions holds:

(4.14) ( a ) δ n + , as n + ; ( b ) δ n δ 0 , as n + ; ( c ) δ n 0 and z n z , with z < 1 2 , as n + ; ( d ) δ n 0 , as n + , and z n 1 2 , for n large .

By definitions of α ( u , v ) and β ( u , v ) , thanks to (4.13), we have that

(4.15) α ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) ( 0,0,0,0 ) , β ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) 1 2 , as n + .

If δ n + as n + , we obtain that

β ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) = 1 ( ν 1 + ν 2 ) S 2 R 4 \ B 1 ( 0 ) Ψ δ n , z n 2 ( ν 1 + ν 2 ) d x = 1 1 S 2 B 1 ( 0 ) Ψ δ n , z n 2 d x = 1 + o n ( 1 ) , as n + .

Thanks to (4.15), we conclude a contradiction.

If δ n δ 0 as n + , it follows from Proposition 2.1 that z n + as n + . Then, we have

β ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) = β ( ν 1 Ψ δ , z n , ν 2 Ψ δ , z n ) + o n ( 1 ) = 1 S 2 R 4 \ B 1 ( z n ) Ψ δ , 0 2 d x + o n ( 1 ) = 1 1 S 2 B 1 ( z n ) Ψ δ , 0 2 d x + o n ( 1 ) = 1 + o n ( 1 ) , as n + ,

which is a contradiction.

If δ n 0 and z n z , with z < 1 2 as n + , then

β ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) = β ( ν 1 Ψ δ n , z , ν 2 Ψ δ n , z ) + o n ( 1 ) = 1 S 2 R 4 χ 2 ( x + z ) Ψ δ n , 0 2 d x + o n ( 1 ) = 1 S 2 R 4 \ B 1 ( z ) Ψ δ n , 0 2 d x + o n ( 1 ) = o n ( 1 ) , as n + .

Obviously, it is impossible due to (4.15).

If δ n 0 as n + and z n 1 2 , for n large , according to Lemma 3.3, we obtain

α ( ν 1 Ψ δ n , z n , ν 2 Ψ δ n , z n ) = z n z n + o n ( 1 ) , as n + ,

which contradicts with (4.15).

Consequently, it follows from the aforementioned discussions that c 0 > c .

Step 2: To prove (4.12).

By (c) of Lemma 4.1, as ρ 0 , one has

Φ ( J ρ ( z , δ ) ) = Φ ( t ( u ρ , v ρ ) u ρ , t ( u ρ , v ρ ) v ρ ) c , as δ 0 , uniformly in z R 4 .

Then, it follows from J ρ ( z , δ ) N and Corollary 3.1 that, as ρ 0 ,

(4.16) β ( J ρ ( z , δ ) ) β ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) , as δ 0 , uniformly in z R 4 .

Hence, we conclude that

β ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) = 1 S 2 R 4 \ B 1 ( 0 ) Ψ δ , z 2 d x = 1 S 2 R 4 \ B 1 ( z ) Ψ δ , 0 2 d x = o ( 1 ) , as δ 0 , for z < 1 2 ,

which shows that, as ρ 0 , there is 0 < δ 1 < 1 2 such that (1) holds.

By arguing as earlier, it follows from (c) of Lemma 4.1 and Corollary 3.1 that, as ρ 0 ,

α ( J ρ ( z , δ ) ) α ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) , as δ 0 , uniformly in z R 4 .

Then, according to Lemma 4.3, as ρ 0 , there is 0 < δ 1 < 1 2 so that (2) holds.

By (c) of Lemma 4.1 and Corollary 3.1, as ρ 0 , we also obtain that

(4.17) β ( J ρ ( z , δ ) ) β ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) as δ 0 , uniformly in z R 4 .

Considering

B 1 ( 0 ) Ψ δ , z 2 d x 0 , as δ + ,

one has

β ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) = 1 1 S 2 B 1 ( 0 ) Ψ δ , z 2 d x = 1 + o n ( 1 ) , as δ + .

Therefore, as ρ 0 , according to (4.17), there is δ 2 > 1 2 such that (3) holds.

According to (c) of Lemma 4.1, as ρ 0 , one has

Φ ( J ρ ( z , δ ) ) = Φ ( t ( u ρ , v ρ ) u ρ , t ( u ρ , v ρ ) v ρ ) c as δ 0 , or δ + uniformly in z R 4 .

Thanks to (4.11), as ρ 0 , there exist 0 < δ 1 < 1 2 < δ 2 such that (4) holds.

By J ρ ( z , δ ) N , we have Φ ( J ρ ( z , δ ) ) c . Hence, it follows from Proposition 2.1 that

Φ ( J ρ ( z , δ ) ) > c = c , z R 4 , δ > 0 .

Thanks to (b) of Lemma 4.1, one has

Φ ( J ρ ( z , δ ) ) = Φ ( t ( u ρ , v ρ ) u ρ , t ( u ρ , v ρ ) v ρ ) c , as z + ,

uniformly for every bounded ρ , and bounded δ away from 0. According to (4.11), there is T 1 > 1 2 satisfying

Φ ( J ρ ( z , δ ) ) < c 0 + c 2 , z T 1 , δ 1 δ δ 2 , ρ < 1 ,

which implies that (5) holds.

It follows from (b) of Lemma 4.1, Corollary 3.1, and the fact J ρ ( z , δ ) N that

(4.18) ( α ( J ρ ( z , δ ) ) , z ) R 4 ( α ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) , z ) R 4 , as z + ,

uniformly in δ 1 δ δ 2 , ρ < 1 .

Let

( R 4 ) z + { x R 4 : ( x , z ) R 4 > 0 } and ( R 4 ) z { x R 4 : ( x , z ) R 4 0 } .

Since δ [ δ 1 , δ 2 ] , there exist a large T 2 > 0 and r ( 0 , 1 4 ) such that if z T 2 , the ball B r ( z ˜ ) ( R 4 ) z + with z ˜ satisfying z z ˜ = 1 2 , and by the definition of Ψ δ , z , one has

Ψ δ , z 2 C 0 > 0 , x B r ( z ˜ )

and

Ψ δ , z 2 C 1 x z 6 , x ( R 4 ) z .

Hence, for any z T 2 , δ [ δ 1 , δ 2 ] , one has

( α ( ν 1 Ψ δ , z , ν 2 Ψ δ , z ) , z ) R 4 = 1 S 2 ( R 4 ) z + Ψ δ , z 2 χ 1 ( x ) ( x , z ) R N d x + 1 S 2 ( R 4 ) z Ψ δ , z 2 χ 1 ( x ) ( x , z ) R N d x C 2 1 x x z d x 1 x z 6 z d x C 4 z + C 5 > 0 ,

which shows that (6) holds with T = max { T 1 , T 2 } .□

Let constants δ 1 , δ 2 and T be defined as in Lemma 4.4, define

z , δ { ( δ , z ) R 4 × R + : z < T , δ [ δ 1 , δ 1 ] } .

Denoted

z , δ 1 2 3 4 ,

where

1 = ( z , δ ) R 4 × R + : z < 1 2 , δ = δ 1 ; 2 = ( z , δ ) R 4 × R + : 1 2 z < T , δ = δ 1 ; 3 = { ( z , δ ) R 4 × R + : z T , δ = δ 2 } ; 4 = { ( z , δ ) R 4 × R + : z = T , δ [ δ 1 , δ 1 ] } .

Defined Θ 0 by

Θ { J ρ ( z , δ ) : ( z , δ ) z , δ } .

Obviously, Θ N . Set

γ : γ C ( N , N ) , γ ( u , v ) = ( u , v ) for any ( u , v ) with Φ ( u , v ) < c 0 + c 2 , { D N : D = γ ( Θ ) , γ } .

Lemma 4.5

Assume that β > max { μ 1 , μ 2 } and V 1 , V 2 L 2 ( Ω ) are the nonnegative functions with V 1 2 + V 1 2 0 . Let D , then as ρ 0 , we have

D .

Proof

To achieve our goal, we just prove that for any γ , as ρ 0 , there is ( z , δ ) z , δ so that

(4.19) ( ( α , β ) γ J ρ ) ( z , δ ) = 0 , 0 , 0 , 0 , 1 2 .

γ : R 4 × R + R 4 × R + and : z , δ R 4 × R + be defined by

γ = ( α , β ) γ J ρ , = ( α , β ) J ρ .

First, we prove that, as ρ 0 ,

(4.20) deg γ , z , δ , 0 , 0 , 0 , 0 , 1 2 = deg , z , δ , 0 , 0 , 0 , 0 , 1 2 .

It follows from (4) and (5) of (4.12) that, as ρ 0 ,

Φ ( J ρ ( z , δ ) ) < c 0 + c 2 , ( z , δ ) z , δ .

Thus, as ρ 0 , one has

γ ( J ρ ( z , δ ) ) = J ρ ( z , δ ) , ( z , δ ) z , δ .

Then, we conclude that

γ ( z , δ ) = ( ( α , β ) J ρ ) ( z , δ ) = ( z , δ ) , ( z , δ ) z , δ ,

which implies that (4.20) holds.

Second, we need to prove that as ρ 0 ,

(4.21) deg , z , δ , 0 , 0 , 0 , 0 , 1 2 = 1 .

Define

ϒ ( z , δ , t ) = t ( z , δ ) + ( 1 t ) ( z , δ ) , t [ 0 , 1 ] .

It follows from the homotopy invariance property of the topological degree and the fact

deg ( I d z , δ , δ , z , 0 , 0 , 0 , 0 , 1 2 ) = 1 ,

that, for obtaining (4.21), we just have to prove that as ρ 0 ,

(4.22) ϒ ( z , δ , t ) 0 , 1 2 , for ( z , δ ) z , δ and t [ 0 , 1 ] .

If ( z , δ ) 1 , according to δ 1 < 1 2 and (1) of (4.12), we have

(4.23) t ( β J ρ ) ( z , δ 2 ) + ( 1 t ) δ 1 < 1 2 , t [ 0 , 1 ] .

If ( z , δ ) 2 , considering δ 1 < 1 2 , it follows from (2) of (4.12) that

α ( J ρ ( z , δ ) ) z z < 1 4 ,

which ensures that

(4.24) ( 1 t ) z + t ( α J ρ ) ( z , δ 1 ) ( 1 t ) z + t z z t ( α J ρ ) ( z , δ 1 ) t z z ( 1 t ) z + t t 4 1 2 + t 4 > 0 , t [ 0 , 1 ] .

If ( z , δ ) 3 , by δ 2 > 1 2 and (3) of (4.12), we deduce that

(4.25) t ( β J ρ ) ( z , δ 2 ) + ( 1 t ) δ 2 > ( 1 t ) 1 2 + t 2 = 1 2 , t [ 0 , 1 ] .

If ( z , δ ) 4 , by (6) of (4.12), we have

(4.26) ( t ( α J ρ ) ( z , δ ) + ( 1 t ) z , z ) R 4 = ( 1 t ) z 2 + t ( ( α J ρ ) ( z , δ ) , z ) R 4 > 0 , t [ 0 , 1 ] .

Combining (4.23)-(4.26), we obtain (4.22), and then, (4.21) holds. So, it follows from (4.20) and (4.21) that D .□

5 Proof of the main result

Let

c inf D sup ( u , v ) D Φ ( u , v ) , K c { ( u , v ) N : Φ ( u , v ) = c , Φ ( u , v ) = 0 } , Φ r { ( u , v ) N : Φ ( u , v ) r } , r R .

Let ρ be small enough such that Lemmas 4.2 and 4.5 hold for any ρ < ρ . Fix ρ with ρ < ρ . To prove Theorem 1.1, we need to establish that K c . Suppose by contradiction that K c = . By Lemmas 4.4 and 4.5,

c inf ( u , v ) Φ ( u , v ) = c 0 > c .

In view of Θ , it follows from Lemma 4.2 that c < min { 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c } . Hence, one has

c < c < min 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c .

According to Corollary 3.2, we know that Φ satisfies the Palais-Smale condition in

N ( u , v ) 0 : c < Φ ( u , v ) < min 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c .

So, it follows from a variant due to Hofer [26] of the classical deformation lemma (see [47,53]) that there is a continuous map

ψ : [ 0 , 1 ] × N N

and ε 0 > 0 satisfying

  1. Φ c + ε 0 \ Φ c ε 0 Φ min { 1 4 μ 1 1 S 2 , 1 4 μ 2 1 S 2 , 2 c } \ Φ c 0 + c 2 ;

  2. ψ ( 0 , ( u , v ) ) = ( u , v ) ;

  3. ψ ( t , ( u , v ) ) = ( u , v ) , ( u , v ) Φ c ε 0 ( N \ Φ c + ε 0 ) , t [ 0 , 1 ] ;

  4. ψ ( 1 , Φ c + ε 0 2 ) Φ c ε 0 2 .

According to the definition of c , there exists D such that

c sup ( u , v ) D Φ ( u , v ) < c + ε 0 2 ;

hence, ψ ( 1 , D ) , and by (d), one has that

c sup ( u , v ) ψ ( 1 , D ) Φ ( u , v ) < c ε 0 2 ,

which is impossible. Hence, K c , and the proof of Theorem 1.1 is complete.

Acknowledgments

The authors are very grateful to anonymous referees for their very valuable comments and suggestions helping to the improvement of the manuscript.

  1. Funding information: The research of Da-Bin Wang was supported in part by Natural Science Foundation of Guangdong (2025A1515012082) and the STU Scientific Research Initiation Grant (NTF22031). The research of Huafei Xie was supported in part by National Natural Science Foundation of China (Grant No. 12201232).

  2. Author contributions: All authors contributed equally to the writing of this article. All authors read and approved the final manuscript.

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

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Received: 2025-01-05
Revised: 2025-04-02
Accepted: 2025-06-25
Published Online: 2025-08-20

© 2025 the author(s), published by De Gruyter

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

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