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Existence of normalized peak solutions for a coupled nonlinear Schrödinger system

  • Jing Yang EMAIL logo
Published/Copyright: January 27, 2024

Abstract

In this article, we study the following nonlinear Schrödinger system

Δ u 1 + V 1 ( x ) u 1 = α u 1 u 2 + μ u 1 , x R 4 , Δ u 2 + V 2 ( x ) u 2 = α 2 u 1 2 + β u 2 2 + μ u 2 , x R 4 ,

with the constraint R 4 ( u 1 2 + u 2 2 ) d x = 1 , where α > 0 and α > β , μ R , V 1 ( x ) , and V 2 ( x ) are bounded functions. Under some mild assumptions on V 1 ( x ) and V 2 ( x ) , we prove the existence of normalized peak solutions by using the finite dimensional reduction method, combined with the local Pohozaev identities. Because of the interspecies interaction between the components, we aim to obtain some new technical estimates.

MSC 2010: 35A10; 35B99; 35J60

1 Introduction and main result

In this article, we consider the following Schrödinger system with coupled quadratic nonlinearities

(1.1) Δ u 1 + V 1 ( x ) u 1 = α u 1 u 2 + μ u 1 , x R 4 , Δ u 2 + V 2 ( x ) u 2 = α 2 u 1 2 + β u 2 2 + μ u 2 , x R 4 ,

under the constraint

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

where α > 0 and α > β , μ R is a chemical potential, and V 1 ( x ) and V 2 ( x ) are bounded functions.

Such type of systems like (1.1) with quadratic interaction have wide applications in physics, such as Bose-Einstein condensates, plasma physics, and nonlinear optics. For example, the following coupled nonlinear Schrödinger equations

(1.3) i ω k t + Δ ω k + j = 1 K μ j , k ω j 2 ω k = 0

for some μ j , k R and j , k = 1 , 2 , , K , can describe the propagation of solitons with χ ( 3 ) nonlinear fiber couplers in nonlinear optic theory, where the complex function ω k denotes the k th component of the light beam, and j = 1 K ω j 2 is the change in refractive index profile created by all the incoherent components in the light beam. This system has gotten a lot of attentions experimentally and theoretically. If we consider the standing wave solutions for (1.3) of the form ω k ( x , t ) = e i λ k t u k ( x ) with u k a function independent of time, then u k satisfies

(1.4) Δ u k λ k u k + j = 1 K μ j , k u j 2 u k = 0

for k = 1 , 2 , , K . The existence and multiplicity of standing wave solutions to (1.4) and its relates problem have been explored by many authors in recent years, see previous studies by [24,6,9,10,13,22,24,26,30] and the references therein. However, when nonlinear optical effects such as second harmonic generation are investigated in the optical material that has a χ ( 2 ) nonlinear response, it is led to the system (1.1) (c.f. [1,7]). For the existence, multiplicity and asymptotic behavior of solutions for (1.1), we can refer to [2729,31].

We note that the following Gross-Pitaevskii (GP) equations proposed by Gross [15] and Pitaevskii [25] in the 1960s

(1.5) i ω ( x , t ) t = Δ ω ( x , t ) + V ( x ) ω ( x , t ) a ω ( x , t ) 2 ω ( x , t ) , x R N ,

with the constraint

R N ω ( x , t ) 2 d x = 1 ,

has been investigated extensively due to some new experimental advances in quantum phenomena (c.f. [5,11,14]). Here, N 2 , V ( x ) 0 is a real-valued potential and a R is treated as an arbitrary dimensionless parameter. If we want to find a solution for (1.5) of the form ω ( x , t ) = e i μ t u ( x ) , where μ represents the chemical potential of the condensate and u ( x ) is a function independent of time, then the unknown pair ( μ , u ) satisfies the following nonlinear eigenvalue equation

(1.6) Δ u + V ( x ) u = a u 3 + μ u , in R N ,

with the constraint

(1.7) R N u 2 d x = 1 .

For ground states of equation (1.6), one can refer to the studies by Guo et al. [1820] and the references therein, where the main tools are the energy comparison under various assumptions of trapping potential V ( x ) , which can be described equivalently by positive L 2 minimizers of the following functional:

I a = 1 2 R N ( u 2 + V ( x ) u 2 ) d x a 4 R N u 4 d x .

Very recently the existence and the local uniqueness of excites states for a class of degenerated trapping potential with nonisolated critical points were given in the study by Luo et al. [23], where we call any eigenfunction of equations (1.6) and (1.7) whose energy is larger than that of the ground state the excited states in the physics literatures (c.f. [8]). As for two-component Bose-Einstein condensates with the mass constraint, the studies by Guo and Yang [16] showed the existence of the general exited states by using the reduction argument combined with the local Pohozaev identities, which generalize the existence of the ground state with trapping potentials given in the study by Guo et al. [17] to some extent.

To our best knowledge, there seems to be no results on the existence of peak solutions to (1.1) with the L 2 constraint (1.2). So in this article, we want to investigate this problem and suppose the following conditions hold.

( K 1 ): V 1 ( x ) and V 2 ( x ) have m common critical points a 1 , , a m for m 1 with a 1 , , a m the nondegenerated critical points of Γ 1 2 V 1 ( x ) + Γ 2 2 V 2 ( x ) , where Γ 1 and Γ 2 are given below.

( K 2 ): V j ( x ) < V j ( a 1 ) for x B d ( a 1 ) \ { a 1 } and V j ( x ) C θ ( B ¯ d ( a 1 ) ) for some θ ( 0 , 1 ) with some small d > 0 for j = 1 , 2 .

To state our results, we first introduce some notations. Let U be the unique positive solution of the following problem:

Δ u + u = u 2 , in R 4 , u ( 0 ) = max x R 4 u ( x ) , u ( x ) H 1 ( R 4 ) ,

and we denote b * = R 4 U 2 d x . From [21], we know that U ( x ) = U ( x ) is strictly decreasing and satisfies for s 1

D s U ( x ) e x x 3 2 C .

Also ( U 1 , U 2 ) ( Γ 1 U , Γ 2 U ) is the ground state of

(1.8) Δ u 1 + u 1 = α u 1 u 2 in R 4 , Δ u 2 + u 2 = α 2 u 1 2 + β u 2 2 , in R 4 ,

provided that α > β with

Γ 1 = 1 α 2 ( α β ) α , Γ 2 = 1 α .

For any t R + and y R 4 , we define for i = 1 , 2 , U i , t , y t U i ( t ( x y ) ) , and for any δ > 0 , ( u , v ) δ 2 u 1 , δ 2 + v 2 , δ 2 with

u i , δ 2 1 δ R 4 ( u 2 + ( δ + V i ( x ) ) u 2 ) d x .

The first result of our article is as follows.

Theorem 1.1

Assume that ( K 1 ) holds and α > β and α > 0 , then there exists a small constant ε > 0 such that for any ( α , β ) satisfying 3 α 2 β α 3 1 m b * ε , problems (1.1) and (1.2) have a peak solution ( u 1 , μ , u 2 , μ , μ ) depending on α and β with the form

( u 1 , μ ( x ) , u 2 , μ ( x ) ) = j = 1 m U 1 , μ , y μ , j ( x ) + φ 1 , μ ( x ) , j = 1 m U 2 , μ , y μ , j ( x ) + φ 2 , μ ( x ) .

Also as 3 α 2 β α 3 1 m b * , there hold μ and

y μ , j a j = o 1 μ a n d ( φ 1 , μ , φ 2 , μ ) μ = O 1 μ .

Theorem 1.1 tells that we construct a synchronized solution for problems (1.1) and (1.2), where the synchronized solution means that the two components of the solutions for system (1.1) concentrate at the same set of concentrated points. Otherwise, we call it a segregated solution if the components concentrate at two different set of points.

Remark 1.2

If V 1 ( x ) and V 2 ( x ) have no common nondegenerate critical points, that is, V 1 ( x ) has m critical points a 1 , , a m , and while V 2 ( x ) has m ¯ critical points a ¯ 1 , , a ¯ m ¯ with m , m ¯ 1 , then we also can construct a segregated peak solution ( u 1 , μ , u 2 , μ , μ ) with u 1 , μ and u 2 , μ concentrating at these points, respectively.

Now we want to consider the clustering peak solutions to problems (1.1) and (1.2), and the another result is as follows:

Theorem 1.3

Assume that ( K 2 ) holds and α > β and α > 0 , then there exists a small constant ε > 0 such that for any ( α , β ) satisfying 3 α 2 β α 3 1 m b * ε , problems (1.1) and (1.2) have a peak solution ( u 1 , μ , u 2 , μ , μ ) depending on α and β with the form

( u 1 , μ ( x ) , u 2 , μ ( x ) ) = j = 1 m U 1 , μ , y μ , j ( x ) + φ 1 , μ ( x ) , j = 1 m U 2 , μ , y μ , j ( x ) + φ 2 , μ ( x ) .

Also as 3 α 2 β α 3 1 m b * , there hold μ , ( φ 1 , μ , φ 2 , μ ) μ = O 1 μ and

y μ , j a 1 a n d μ y μ , j y μ , l , f o r l j .

Remark 1.4

To our best knowledge, this seems to be the first time to consider the existence of the normalized peak (or bubbling) solutions for problems (1.1) and (1.2). Also we note that only one coefficient a in (1.6) need to be discussed with the constraint mass, while there are two parameters α and β in system (1.1) to be considered, corresponding to the interactions within and between the components, respectively, which makes the problems (1.1) and (1.2) more complicated than the single equation, and we have to analyze the mutual influence of the parameters and the constraint condition very carefully.

We will mainly use the finite dimensional reduction to prove our results, which is an effective way to construct solutions for perturbed elliptic problems. Since the singularly perturbed problem has a small parameter naturally, the approximate solutions can be constructed by the standard steps of the reduction argument. However, the constraint problems (1.1) and (1.2) itself do not provide any natural limiting process explicitly. So to apply the reduction method, we have to instigate the relationship between the constraint conditions and the parameters appeared in the system. This process involve some various Pohozaev identities, which play an important role and need some more accurate estimates.

The structure of this article is organized as follows. We carry out the finite dimensional reduction to study the corresponding problem without constraint in Section 2. In Section 3, we prove Theorems 1.1 and 1.3 is proved in Section 4.

2 The finite dimensional reduction

In this section, we consider the following problem without constraint

(2.1) Δ u 1 + ( λ + V 1 ( x ) ) u 1 = α u 1 u 2 in R 4 , Δ u 2 + ( λ + V 2 ( x ) ) u 2 = α 2 u 1 2 + β u 2 2 in R 4 ,

where λ > 0 is a large parameter.

We want to find a peak solution of equation (2.1) with the form

(2.2) u i , λ ( x ) = j = 1 m U i , λ , y λ , j ( x ) + φ i , λ ( x ) , i = 1 , 2 ,

where U i , λ , y λ U i ( λ ( x y ) ) for some y R 4 . To this aim, we define L λ be the bounded linear operator from H 1 ( R 4 ) × H 1 ( R 4 ) to itself as follows:

L λ ( φ 1 , φ 2 ) , ( ψ 1 , ψ 2 ) λ R 4 φ 1 ψ 1 + ( λ + V 1 ( x ) ) φ 1 ψ 1 α j = 1 m U 1 , λ , y λ , j φ 2 ψ 1 α j = 1 m U 2 , λ , y λ , j φ 1 ψ 1 + R 4 φ 2 ψ 2 + ( λ + V 2 ( x ) ) φ 2 ψ 2 α j = 1 m U 1 , λ , y λ , j φ 1 ψ 2 2 β j = 1 m U 2 , λ , y λ , j φ 2 ψ 2

for any ( ψ 1 , ψ 2 ) , ( φ 1 , φ 2 ) H 1 ( R 4 ) × H 1 ( R 4 ) .

Then to obtain the solution ( u 1 , λ , u 2 , λ ) of equation (2.1) with (2.2) is to solve the following problem:

L λ ( φ 1 , λ , φ 2 , λ ) = l λ + R λ ( φ 1 , λ , φ 2 , λ ) ,

where l λ ( l 1 , λ , l 2 , λ ) with

l 1 , λ = j = 1 m V 1 ( x ) U 1 , λ , y λ , j + α j = 1 m U 1 , λ , y λ , j j = 1 m U 2 , λ , y λ , j j = 1 m U 1 , λ , y λ , j U 2 , λ , y λ , j , l 2 , λ = j = 1 m V 2 ( x ) U 2 , λ , y λ , j + α 2 j = 1 m U 1 , λ , y λ , j 2 j = 1 m U 1 , λ , y λ , j 2 + β j = 1 m U 2 , λ , y λ , j 2 j = 1 m U 2 , λ , y λ , j 2 ,

and R λ ( φ 1 , λ , φ 2 , λ ) ( R 1 , λ ( φ 1 , λ , φ 2 , λ ) , R 2 , λ ( φ 1 , λ , φ 2 , λ ) ) with

R 1 , λ ( φ 1 , λ , φ 2 , λ ) = α φ 1 , λ φ 2 , λ , R 2 , λ ( φ 1 , λ , φ 2 , λ ) = α 2 φ 1 , λ 2 + β φ 2 , λ 2 .

To carry out the reduction argument, we first introduce the following result, which has been proved in [29].

Proposition 2.1

For any α > 0 and α > β , ( U 1 , U 2 ) is nondegenerate for the system (1.8) in H 1 ( R 4 ) × H 1 ( R 4 ) in the sense that the kernel is given by

η ( α , β ) U x i , U x i i = 1 , 2 , 3 , 4

in H 1 ( R 4 ) × H 1 ( R 4 ) , where η ( α , β ) 0 .

Now we define the set

E λ = ( u 1 , u 2 ) H 1 ( R 4 ) × H 1 ( R 4 ) : ( u 1 , u 2 ) , U 1 , λ , y λ , j x k , U 2 , λ , y λ , j x k λ = 0 , k = 1 , , 4 , j = 1 , , m ,

and the norm ( u , v ) λ 2 u 1 , λ 2 + v 2 , λ 2 with

u i , λ 2 1 λ R 4 ( u 2 + ( λ + V i ( x ) ) u 2 ) d x .

Also we set the projection F λ from H 1 ( R 4 ) × H 1 ( R 4 ) to E λ as follows:

F λ ( u 1 , u 2 ) = ( u 1 , u 2 ) i = 1 4 j = 1 m Γ λ , i , j U 1 , λ , y λ , j x i , U 2 , λ , y λ , j x i .

Proposition 2.2

There exist λ 0 , ε 0 , ρ > 0 independent of a j ( j = 1 , , m ) , such that for any λ [ λ 0 , + ) and y λ , j B ε 0 ( a j ) , F λ L λ is bijective in E λ . Also, for any ( u 1 , u 2 ) E λ , it holds

F λ L λ ( u 1 , u 2 ) λ ρ λ ( u 1 , u 2 ) λ .

Proof

By contradiction, suppose that there exist λ n + , y λ n , j a j , ( u 1 , n , u 2 , n ) E λ n such that

(2.3) F λ n L λ n ( u 1 , n , u 2 , n ) λ n λ n n ( u 1 , n , u 2 , n ) λ n ,

with ( u 1 , n , u 2 , n ) λ n 2 = 1 λ n . By (2.3), for any ( ϕ 1 , ϕ 2 ) E λ n , we have

L λ n ( ( u 1 , n , u 2 , n ) , ( ϕ 1 , ϕ 2 ) ) λ n = o ( λ n ) ( u 1 , n , u 2 , n ) λ n ( ϕ 1 , ϕ 2 ) λ n = o ( λ n 1 2 ) ( ϕ 1 , ϕ 2 ) λ n ,

and from this, it holds

L λ n ( ( u 1 , n , u 2 , n ) , ( u 1 , n , u 2 , n ) ) λ n = o ( 1 ) .

Now choosing R sufficiently large such that

( α + 2 β ) j = 1 m U 2 , λ n , y λ n , j < 1 4 ( λ n + V i ( x ) ) , in R 4 \ j = 1 m B R λ n ( y λ n , j ) ,

we find

L λ n ( ( u 1 , n , u 2 , n ) , ( u 1 , n , u 2 , n ) ) λ n λ n 2 ( u 1 , n , u 2 , n ) λ n 2 C λ n j = 1 m B R λ n ( y λ n , j ) ( u 1 , n 2 + u 2 , n 2 ) ,

which gives that

j = 1 m B R λ n ( y λ n , j ) ( u 1 , n 2 + u 2 , n 2 ) C λ n .

To obtain a contradiction, next we want to prove that

(2.4) j = 1 m B R λ n ( y λ n , j ) ( u 1 , n 2 + u 2 , n 2 ) = o 1 λ n .

To this end, we define u ¯ k , n , j ( x ) = 1 λ n u k , n ( x λ n + y λ n , j ) . Then

k = 1 2 R 4 ( u ¯ k , n , j 2 + u ¯ k , n , j 2 ) = O ( λ n ) ( u 1 , n , u 2 , n ) λ n 2 C .

Thus, we can assume that as n + , u ¯ k , n , j u ¯ k , j weakly in H 1 ( R 4 ) × H 1 ( R 4 ) , and u ¯ k , n , j u ¯ k , j strongly in L l o c 2 ( R 4 ) × L l o c 2 ( R 4 ) for k = 1 , 2 and j = 1 , , m . Now we just need to prove that

(2.5) u ¯ k , j = 0 , k = 1 , 2 and j = 1 , , k .

For any ( ϕ 1 , ϕ 2 ) H 1 ( R 2 ) × H 1 ( R 2 ) , we can decompose ( ϕ 1 , ϕ 2 ) as follows:

( ϕ 1 , ϕ 2 ) = F λ n ( ϕ 1 , ϕ 2 ) + i = 1 4 j = 1 m Γ λ n , i , j U 1 , λ n , y λ n , j x i , U 2 , λ n , y λ n , j x i ,

with Γ λ n , i , j = b λ n , i , j ( ϕ 1 , ϕ 2 ) , U 1 , λ n , y λ n , j x i , U 2 , λ n , y λ n , j x i λ n for some constants b λ n , i , j . Then it holds

L λ n ( u 1 , n , u 2 , n ) , ( ϕ 1 , ϕ 2 ) λ n = L λ n ( u 1 , n , u 2 , n ) , F λ n ( ϕ 1 , ϕ 2 ) λ n + i = 1 4 j = 1 m Γ λ n , i , j κ λ n , i , j ,

where κ λ n , i , j = L λ n ( u 1 , n , u 2 , n ) , U 1 , λ n , y λ n , j x i , U 2 , λ n , y λ n , j x i λ n . Also, by (2.3),

L λ n ( u 1 , n , u 2 , n ) , F λ n ( ϕ 1 , ϕ 2 ) λ n = F λ n L λ n ( u 1 , n , u 2 , n ) , F λ n ( ϕ 1 , ϕ 2 ) λ n = o λ n 1 2 ( ϕ 1 , ϕ 2 ) λ n ,

which gives that

(2.6) L λ n ( u 1 , n , u 2 , n ) , ( ϕ 1 , ϕ 2 ) λ n = o λ n 1 2 ( ϕ 1 , ϕ 2 ) λ n + i = 1 4 j = 1 m ϑ λ n , i , j ( ϕ 1 , ϕ 2 ) , U 1 , λ n , y λ n , j x i , U 2 , λ n , y λ n , j x i λ n

for some constants ϑ λ n , i , j . From this, we estimate ϑ λ n , i , j as follows:

i = 1 4 j = 1 m ϑ λ n , i , j U 1 , λ n , y λ n , j x h , U 2 , λ n , y λ n , j x h , U 1 , λ n , y λ n , j x i , U 2 , λ n , y λ n , j x i λ n = L λ n ( u 1 , n , u 2 , n ) , U 1 , λ n , y λ n , j x h , U 2 , λ n , y λ n , j x h λ n + o ( λ n ) = λ n ( u 1 , n , u 2 , n ) , U 1 , λ n , y λ n , j x h , U 2 , λ n , y λ n , j x h λ n + Q n ( u 1 , n , u 2 , n ) + o ( λ n ) = Q n ( u 1 , n , u 2 , n ) + o ( λ n 1 + γ ) ,

where γ > 0 is a small constant and

Q n ( u 1 , n , u 2 , n ) = α R 4 U 1 , λ n , y λ n , j u 2 , n U 1 , λ n , y λ n , j x h α R 4 U 2 , λ n , y λ n , j u 1 , n U 1 , λ n , y λ n , j x h α R 4 U 1 , λ n , y λ n , j u 1 , n U 2 , λ n , y λ n , j x h 2 β R 4 U 2 , λ n , y λ n , j u 2 , n U 2 , λ n , y λ n , j x h .

Noting that ( U 1 , λ n , y λ n , j , U 2 , λ n , y λ n , j ) satisfies that

Δ U 1 , λ n , y λ n , j + λ n U 1 , λ n , y λ n , j = α U 1 , λ n , y λ n , j U 2 , λ n , y λ n , j in R 4 , Δ U 2 , λ n , y λ n , j + λ n U 2 , λ n , y λ n , j = α 2 U 1 , λ n , y λ n , j 2 + β U 2 , λ n , y λ n , j 2 in R 4 ,

we find

Δ U 1 , λ n , y λ n , j x h + λ n U 1 , λ n , y λ n , j x h = α U 2 , λ n , y λ n , j U 1 , λ n , y λ n , j x h + α U 1 , λ n , y λ n , j U 2 , λ n , y λ n , j x h in R 4 , Δ U 2 , λ n , y λ n , j x h + λ n U 2 , λ n , y λ n , j x h = α U 1 , λ n , j U 1 , λ n , y λ n , j x h + 2 β U 2 , λ n , y λ n , j U 2 , λ n , y λ n , j x h in R 4 .

It follows from the definition of E λ n that

Q n ( u 1 , n , u 2 , n ) = o ( λ n 1 + γ ) .

Hence, we find

i = 1 4 j = 1 m ϑ λ n , i , j U 1 , λ n , y λ n , j x i , U 2 , λ n , y λ n , j x i , U 1 , λ n , y λ n , j x h , U 2 , λ n , y λ n , j x h λ n = o ( λ n 1 + γ ) ,

which gives ϑ λ n , i , j = o ( 1 ) . So (2.6) gives that

(2.7) L λ n ( u 1 , n , u 2 , n ) , ( ϕ 1 , ϕ 2 ) λ n = o λ n 1 2 ( ϕ 1 , ϕ 2 ) λ n .

Now setting ϕ ¯ k , j ( x ) = λ n ϕ k ( λ n ( x y λ n , j ) ) and using (2.7), we obtain

L λ n ( u 1 , n , u 2 , n ) , ( ϕ ¯ 1 , j , ϕ ¯ 2 , j ) λ n = i = 1 2 R 4 u ¯ i , n , j ϕ i + 1 + 1 λ n V i z λ n + y λ n , j u ¯ i , n , j ϕ i 2 β R 4 U 2 u ¯ 2 , n , j ϕ 2 α R 4 ( U 1 u ¯ 2 , n , j ϕ 1 + U 1 u ¯ 1 , n , j ϕ 2 + U 2 u ¯ 1 , n , j ϕ 1 ) = o λ n 1 2 ( ϕ ¯ 1 , j , ϕ ¯ 2 , j ) λ n = o ( 1 ) .

Then, ( u ¯ 1 , j , u ¯ 2 , j ) solves

Δ u ¯ 1 , j + u ¯ 1 , j = α u ¯ 1 , j U 2 + α u ¯ 2 , j U 1 in R 4 , Δ u ¯ 2 , j + u ¯ 2 , j = α u ¯ 1 , j U 1 + 2 β u ¯ 2 , j U 2 in R 4 .

From this and the fact that ( U 1 , U 2 ) is nondegenerate, we have

(2.8) ( u ¯ 1 , j , u ¯ 2 , j ) = k = 1 4 τ k , j U 1 x k , U 2 x k .

On the other hand, from ( u 1 , n , u 2 , n ) E λ n and (2.8), it holds

0 = ( u 1 , n , u 2 , n ) , U 1 , λ n , y λ n , j x k , U 2 , λ n , y λ n , j x k λ n = R 4 1 λ n u 1 , n U 1 , λ n , y λ n , j x k + u 2 , n U 2 , λ n , y λ n , j x k + R 4 1 + 1 λ n V 1 ( x ) u 1 , n U 1 , λ n , y λ n , j x k + 1 + 1 λ n V 2 ( x ) u 2 , n U 2 , λ n , y λ n , j x k = R 4 u ¯ 1 , n , j U 1 x k + u ¯ 2 , n , j U 2 x k + R 4 1 + 1 λ n V 1 ( x ) u ¯ 1 , n , j U 1 x k + 1 + 1 λ n V 2 ( x ) u ¯ 2 , n , j U 2 x k = R 4 u ¯ 1 , j U 1 x k + u ¯ 2 , j U 2 x k + R 4 u ¯ 1 , j U 1 x k + u ¯ 2 , j U 2 x k + o ( 1 ) = α R 4 U 2 u ¯ 1 , j U 1 x k + U 1 u ¯ 2 , j U 1 x k + U 1 u ¯ 1 , j U 2 x k + 2 β R 4 U 2 u ¯ 2 , j U 2 x k + o ( 1 ) .

By the property of ( U 1 , U 2 ) , all the constants τ k , j = 0 in (2.8). This gives that (2.5) holds and then (2.4) follows.□

Similar to Proposition 2.2, we have

Proposition 2.3

There exist λ 1 , ε 1 , ρ 1 > 0 independent of a 1 , such that for any λ [ λ 1 , + ) and y λ , k close to a 1 with

(2.9) y λ , k y λ , j ε 1 ln λ λ , f o r k j ,

F λ L λ is bijective in E λ . Moreover, it holds

F λ L λ ( u 1 , u 2 ) λ ρ 1 λ ( u 1 , u 2 ) λ .

Lemma 2.4

We have

(2.10) l λ λ = O ( 1 ) .

Proof

Recall that l λ = ( l 1 , λ , l 2 , λ ) and

( l 1 , λ , l 2 , λ ) , ( ϕ 1 , ϕ 2 ) λ = l 1 , λ , ϕ 1 λ + l 2 , λ , ϕ 2 λ .

Using Hölder inequality, we find

(2.11) l 1 , λ , ϕ 1 λ = R 4 j = 1 m V 1 ( x ) U 1 , λ , y λ , j ϕ 1 + α R 4 j = 1 m U 1 , λ , y λ , j j = 1 m U 2 , λ , y λ , j j = 1 m U 1 , λ , y λ , j U 2 , λ , y λ , j ϕ 1 = O ( ϕ 1 1 , λ ) ,

and

(2.12) l 2 , λ , ϕ 2 λ = R 4 j = 1 m V 2 ( x ) U 2 , λ , y λ , j ϕ 2 + α 2 R 4 j = 1 m U 1 , λ , y λ , j 2 j = 1 m U 1 , λ , y λ , j 2 ϕ 2 + β R 4 j = 1 m U 2 , λ , y λ , j 2 j = 1 m U 2 , λ , y λ , j 2 ϕ 2 = O ( ϕ 2 2 , λ ) .

Hence, (2.12) and (2.11) give (2.10).□

Lemma 2.5

It holds

(2.13) R λ ( φ 1 , λ , φ 2 , λ ) λ = O λ 1 2 ( φ 1 , λ , φ 2 , λ ) λ 2 .

Proof

Note that R λ ( φ 1 , λ , φ 2 , λ ) = ( R 1 , λ ( φ 1 , λ , φ 2 , λ ) , R 2 , λ ( φ 1 , λ , φ 2 , λ ) ) . By the direct computations, we have

(2.14) R 1 , λ ( φ 1 , λ , φ 2 , λ ) , ϕ 1 λ = α R 4 φ 1 , λ φ 2 , λ ϕ 1 C φ 1 , λ L 4 φ 2 , λ L 4 ϕ 1 L 2 = O λ 1 2 φ 1 , λ λ φ 2 , λ λ ϕ 1 λ ,

where we use the following classical Gagliardo-Nirenberg inequality:

(2.15) R 2 ϕ 4 1 4 = O ϕ L 2 1 2 ϕ L 2 1 2 = O λ 1 4 ϕ i , λ , for any ϕ H 1 ( R 4 ) .

With the same argument, it holds

(2.16) R 2 , λ ( φ 1 , λ , φ 2 , λ ) , ϕ 2 λ = O λ 1 2 φ 1 , λ λ φ 2 , λ λ ϕ 2 λ .

Hence, (2.13) follows by (2.16) and (2.14).□

Proposition 2.6

Let λ 0 be as in Proposition 2.2. Then for any λ [ λ 0 , + ) , and y λ , k close to a k for k = 1 , , m , there exists unique φ i , λ E λ , such that

(2.17) R 4 ( u 1 , λ ϕ 1 + u 2 , λ ϕ 2 + ( λ + V 1 ( x ) ) u 1 , λ ϕ 1 + ( λ + V 2 ( x ) ) u 2 , λ ϕ 2 ) = R 4 α u 1 , λ u 2 , λ ϕ 1 + α 2 u 1 , λ 2 ϕ 2 + β u 2 , λ 2 ϕ 2

for any ( ϕ 1 , ϕ 2 ) E λ , where u i , λ ( x ) = k = 1 m U i , λ , y λ , k ( x ) + φ i , λ ( x ) . Moreover, it holds

(2.18) ( φ 1 , λ , φ 2 , λ ) λ C λ .

Proof

To obtain (2.17), it is equivalent to consider the following problem:

(2.19) F λ L λ ( φ 1 , φ 2 ) = F λ l λ + F λ R λ ( φ 1 , φ 2 ) , for ( φ 1 , φ 2 ) E λ .

By Proposition 2.2, we can rewrite (2.19) as follows:

(2.20) ( φ 1 , φ 2 ) = B ( φ 1 , φ 2 ) ( F λ L λ ) 1 F λ l λ + ( F λ L λ ) 1 F λ R λ ( φ 1 , φ 2 ) .

Now we define

M ( φ 1 , φ 2 ) E λ : ( φ 1 , φ 2 ) λ 1 λ 1 γ ,

where γ > 0 is a fixed small constant.

Next we will prove B is a contraction map from M to M . First, from Proposition 2.2 and Lemmas 2.4 and 2.5, we have

B ( φ 1 , φ 2 ) λ C 1 λ l λ λ + 1 λ R λ ( φ 1 , φ 2 ) λ 1 λ 1 γ .

On the other hand, for ( φ 1 , φ 2 ) , ( φ ¯ 1 , φ ¯ 2 ) E λ , it holds

B ( φ 1 , φ 2 ) B ( φ ¯ 1 , φ ¯ 2 ) λ C λ ( R λ ( φ 1 , φ 2 ) λ + R λ ( φ ¯ 1 , φ ¯ 2 ) λ ) 1 λ 1 γ .

By applying the contraction mapping theorem, we find that for any λ [ λ 0 , + ) , y λ , k close to a k for k = 1 , , m , there exists unique ( φ 1 , λ , φ 2 , λ ) E λ depending on a k and λ solving (2.20) and then (2.17) follows. Finally, (2.18) follows from (2.20).□

Similar to Proposition 2.6, using Proposition 2.3 and Lemmas 2.4 and 2.5, we have the following result.

Proposition 2.7

Let λ 1 be as in Proposition 2.3. Then for any λ [ λ 1 , + ) , and y λ , k close to a 1 for k = 1 , , m , satisfying (2.9), (2.17) and (2.18) hold with u i , λ = k = 1 m U i , λ , y λ , k + φ i , λ ( x ) .

3 Proof of Theorem 1.1

From Proposition 2.6, for any λ [ λ 0 , + ) , y λ , j a j for j = 1 , , m , there exist u i , λ and Γ λ , i , j such that

Δ u 1 , λ + ( λ + V 1 ( x ) ) u 1 , λ = α u 1 , λ u 2 , λ + i = 1 4 j = 1 m Γ λ , i , j U 1 , λ , y λ , j x i , in R 4 , Δ u 2 , λ + ( λ + V 2 ( x ) ) u 2 , λ = α 2 u 1 , λ 2 + β u 2 , λ 2 + i = 1 4 j = 1 m Γ λ , i , j U 2 , λ , y λ , j x i , in R 4 ,

where u i , λ ( x ) = j = 1 m U i , λ , y λ , j ( x ) + φ i , λ ( x ) .

Now to obtain a true solution for (2.1), we just need to choose suitable y λ , j such that

(3.1) Γ λ , i , j = 0 , for i = 1 , 2 , 3 , 4 and j = 1 , , m .

To this end, we want to choose ( y λ , 1 , , y λ , m ) solving for any d > 0 ,

(3.2) B d ( y λ , j ) Δ u 1 , λ u 1 , λ x i + ( λ + V 1 ( x ) ) u 1 , λ u 1 , λ x i α u 1 , λ u 2 , λ u 1 , λ x i + B d ( y λ , j ) Δ u 2 , λ u 2 , λ x i + ( λ + V 2 ( x ) ) u 2 , λ u 2 , λ x i α 2 u 1 , λ 2 u 2 , λ x i β u 2 , λ 2 u 2 , λ x i = 0

with i = 1 , 2 , 3 , 4 and j = 1 , , m .

Lemma 3.1

Assume that ( K 1 ) holds. Then there exists some large λ 0 such that for any λ [ λ 0 , + ) , problem (2.1) has a peak solution ( u 1 , λ , u 2 , λ ) with the form

(3.3) u i , λ ( x ) = j = 1 m U i , λ , y λ , j ( x ) + φ i , λ ( x ) ,

satisfying for j = 1 , , m ,

(3.4) y λ , j a j = o 1 λ a n d ( φ 1 , λ , φ 2 , λ ) λ = O 1 λ .

Proof

From the aforementioned discussions, we have to choose suitable y λ , j R 4 for j = 1 , , m satisfying (3.2). Integrating on B d ( y λ , j ) , we find that (3.2) is equivalent to

B d ( y λ , j ) V 1 ( x ) x i u 1 , λ 2 + V 2 ( x ) x i u 2 , λ 2 = B d ( y λ , j ) u 1 , λ 2 ν i 2 B d ( y λ , j ) u 1 , λ x i u 1 , λ ν α B d ( y λ , j ) u 2 , λ u 1 , λ 2 x i + B d ( y λ , j ) ( λ + V 1 ( x ) ) u 1 , λ 2 ν i + B d ( y λ , j ) u 2 , λ 2 ν i 2 B d ( y λ , j ) u 2 , λ x i u 1 , λ ν + B d ( y λ , j ) ( λ + V 2 ( x ) ) u 2 , λ 2 ν i α B d ( y λ , j ) u 1 , λ 2 u 2 , λ x i 2 3 β B d ( y λ , j ) u 2 , λ 3 ν i ,

where ν = ( ν 1 , , ν 4 ) is the outward unit normal of B d ( y λ , j ) . Also it holds

B d ( y λ , j ) u 1 , λ 2 u 2 , λ x i + u 2 , λ u 1 , λ 2 x i = B d ( y λ , j ) u 1 , λ 2 u 2 , λ ν i = O ( e θ λ )

for some θ > 0 . Then we have

(3.5) B d ( y λ , j ) V 1 ( x ) x i u 1 , λ 2 + V 2 ( x ) x i u 2 , λ 2 = O ( e θ λ ) .

On the other hand,

(3.6) B d ( y λ , j ) V 1 ( x ) x i u 1 , λ 2 = B d ( y λ , j ) l = 1 4 2 V 1 ( a j ) x i x l ( x l a j , l ) u 1 , λ 2 + O B d ( y λ , j ) x a j 2 u 1 , λ 2 = m l = 1 4 2 V 1 ( a j ) x i x l ( y λ , j , l a j , l ) R 4 U 1 2 ( x ) + O ( e θ λ + φ 1 , λ 1 , λ 2 ) + o ( y λ , j a j ) = m l = 1 4 2 V 1 ( a j ) x i x l ( y λ , j , l a j , l ) R 4 U 1 2 ( x ) + o 1 λ + y λ , j a j .

Similarly, we have

(3.7) B d ( y λ , j ) V 2 ( x ) x i u 2 , λ 2 = m l = 1 4 2 V 1 ( a j ) x i x l ( y λ , j , l a j , l ) R 4 U 2 2 ( x ) + o 1 λ + y λ , j a j .

Then (3.5), (3.6), and (3.7) give us that

(3.8) l = 1 4 2 V 1 ( a j ) x i x l Γ 1 2 + 2 V 2 ( a j ) x i x l Γ 2 2 ( y λ , j , l a j , l ) = o 1 λ + y λ , j a j .

By the assumption ( K 1 ) , we find

y λ , j a j = o 1 λ for j = 1 , , m .

Lemma 3.2

It holds

(3.9) R 4 U 3 = 3 R 4 U 2 = 3 2 R 4 U 2 .

Proof

Note that U satisfies

(3.10) Δ U + U = U 2 , in R 4 .

By direct computations, we have

R 4 ( U 2 + U 2 ) = R 4 U 3 .

Also multiplying ( x U ) on both sides of equation (3.10) and integrating on R 4 , it holds

R 4 U 2 + 2 R 4 U 2 = 4 3 R 4 U 3 .

So the aforementioned two equalities give (3.9).□

Proposition 3.3

If ( u 1 , λ , u 2 , λ ) is a solution of (2.1) with the form (3.3) satisfying (3.4), then it holds

(3.11) λ m b * 3 2 α Γ 2 Γ 1 2 + β Γ 2 3 R 4 ( u 1 , λ 2 + u 2 , λ 2 ) = b * i = 1 2 j = 1 m Γ i 2 V i ( a j ) + O 1 λ .

Proof

Let ( u 1 , λ , u 2 , λ ) be a solution of equation (2.1), multiplying x y λ , j , u i , λ on both sides of the i th equation of equation (2.1) and integrating on B d ( y λ , j ) , we find

(3.12) B d ( y λ , j ) i = 1 2 ( ( 2 V i ( x ) + V i ( x ) , x y λ , j ) u i , λ 2 + 2 λ u i , λ 2 ) α u 2 , λ u 1 , λ 2 2 3 β u 2 , λ 3 = B d ( y λ , j ) W ( x ) d σ ,

where

W ( x ) i = 1 2 2 u i , λ ν x y λ , j , u i , λ + x y λ , j , ν [ ( λ + V i ( x ) ) u i , λ 2 ] x y λ , j , ν α u 2 , λ u 1 , λ 2 + 2 3 β u 2 , λ 3 .

Now, being u i , λ ( x ) = j = 1 m U i , λ , y λ , j ( x ) + φ i , λ ( x ) with ( φ 1 , λ , φ 2 , λ ) λ = O 1 λ , it holds

B d ( y λ , j ) ( 2 V i ( x ) + V i ( x ) , x y λ , j ) u i , λ 2 = B d ( y λ , j ) ( V i ( x ) , x x λ , j + 2 ( V i ( x ) V i ( a j ) ) ) U i , λ , y λ , j 2 ( x ) + 2 V i ( a j ) B d ( y λ , j ) u i , λ 2 + O B d ( y λ , j ) x a j 2 U i , λ , y λ , j φ i , λ + φ i , λ λ 2 = 2 V i ( a j ) B d ( y λ , j ) u i , λ 2 + O 1 λ = 2 Γ i 2 b * V i ( a j ) + O 1 λ ,

where b * = R 4 U 2 . On the other hand, using (3.9), we find

B d ( y λ , j ) α u 2 , λ u 1 , λ 2 + 2 3 β u 2 , λ 3 = λ R 4 α U 1 2 U 2 + 2 3 β U 2 3 + O 1 λ = 2 λ 3 2 α Γ 2 Γ 1 2 + β Γ 2 3 b * + O 1 λ .

So summing (3.12) from j = 1 to j = m , we find (3.11).□

Now we are in a position to prove Theorem 1.1.

Proof of Theorem 1.1

Now we set b λ = R 4 ( u 1 , λ 2 + u 2 , λ 2 ) and v i , λ = u i , λ b λ , then ( v 1 , λ , v 2 , λ ) satisfies that

Δ v 1 , λ + ( λ + V 1 ( x ) ) v 1 , λ = α b λ 1 2 v 1 , λ v 2 , λ , in R 4 , Δ v 2 , λ + ( λ + V 2 ( x ) ) v 2 , λ = α 2 b λ 1 2 v 1 , λ 2 + β b λ 1 2 v 2 , λ 2 , in R 4 ,

with

R 4 ( v 1 , λ 2 + v 2 , λ 2 ) = 1 .

Therefore, from Proposition 3.3, we obtain that

(3.13) λ ( m D ( α , β ) b * b λ ) = b * i = 1 2 j = 1 m Γ i 2 V i ( a j ) + o ( 1 ) ,

where D ( α , β ) = 3 2 α Γ 2 Γ 1 2 + β Γ 2 3 = 3 α 2 β α 3 . Take ( α * , β * ) is a root of D ( α , β ) = 1 m b * .

If D ( α , β ) 1 m b * as ( α , β ) ( α * , β * ) , letting N 1 R such that N 1 + V i ( a j ) > 0 and λ ¯ = λ N 1 , we have

Δ u 1 , λ + ( λ ¯ + ( N 1 + V 1 ( x ) ) ) u 1 , λ = α u 1 , λ u 2 , λ , in R 4 , Δ u 2 , λ + ( λ ¯ + ( N 1 + V 2 ( x ) ) ) u 2 , λ = α 2 u 1 , λ 2 + β u 2 , λ 2 , in R 4 .

Similar to (3.13), it holds

(3.14) λ ¯ ( m D ( α , β ) b * b λ ¯ ) = b * i = 1 2 j = 1 m Γ i 2 ( N 1 + V i ( a j ) ) + o ( 1 ) > 0 ,

which gives that there exists some λ ¯ 0 large enough such that b λ ¯ 0 1 . In fact, if b λ ¯ 0 > 1 , then from the fact that D ( α , β ) 1 m b * as ( α , β ) ( α * , β * ) , we can take ( α , β ) such that b λ ¯ 0 > m D ( α , β ) b * , which contradicts to (3.14).

Now we define T ( λ ) ( 1 b λ ) λ 2 . If b λ ¯ 0 = 1 , then we have got the solution. Now we consider b λ ¯ 0 < 1 and T ( λ ¯ 0 ) > 0 for ( α , β ) ( α * , β * ) . On the other hand, by (3.13),

lim λ + T ( λ ) = lim λ + ( ( m D ( α , β ) b * b λ ) λ + ( 1 m D ( α , β ) b * ) λ ) λ = lim λ + ( 1 m D ( α , β ) b * + o ( 1 ) ) λ 2 = .

Thus, there exists some λ α , β [ λ ¯ 0 , + ) such that T ( λ α , β ) = 0 , which means that b λ α , β = 1 . So we have derived the existence for (1.1) with μ = λ α , β and R 4 ( u 1 2 + u 2 2 ) = 1 if D ( α , β ) ( 1 m b * , 1 m b * + ε ] for small ε .

When D ( α , β ) 1 m b * as ( α , β ) ( α * , β * ) , letting N 2 R such that N 2 + V i ( a j ) < 0 and λ ˜ = λ N 2 , we have

Δ u 1 , λ + ( λ ˜ + ( N 2 + V 1 ( x ) ) ) u 1 , λ = α u 1 , λ u 2 , λ , in R 4 , Δ u 2 , λ + ( λ ˜ + ( N 2 + V 2 ( x ) ) ) u 2 , λ = α 2 u 1 , λ 2 + β u 2 , λ 2 , in R 4 .

With the same argument as mentioned earlier, we can also obtain the existence of a concentrated solution to (1.1)–(1.2) if D ( α , β ) [ 1 m b * ε , 1 m b * ) .□

4 Proof of Theorem 1.3

In this section, we come to prove Theorem 1.3. First from Proposition 2.7, for any λ [ λ 1 , + ) , y λ , j a 1 , λ y λ , j y λ , l , there exist u i , λ and Γ λ , i , j such that

Δ u 1 , λ + ( λ + V 1 ( x ) ) u 1 , λ = α u 1 , λ u 2 , λ + i = 1 4 j = 1 m Γ λ , i , j U 1 , λ , y λ , j x i , in R 4 , Δ u 2 , λ + ( λ + V 2 ( x ) ) u 2 , λ = α 2 u 1 , λ 2 + β u 2 , λ 2 + i = 1 4 j = 1 m Γ λ , i , j U 2 , λ , y λ , j x i , in R 4

with u i , λ ( x ) = j = 1 m U i , λ , y λ , j ( x ) + φ i , λ ( x ) satisfying (2.18).

Now we define

I λ ( u 1 , u 2 ) = 1 2 i = 1 2 R 4 ( u i 2 + ( λ + V 1 ( x ) ) u i 2 ) α 2 R 4 u 1 2 u 2 β 3 R 4 u 2 3 ,

and Q ( y ) I λ j = 1 m U 1 , λ , y j + φ 1 , λ , j = 1 m U 2 , λ , y j + φ 2 , λ with y = ( y 1 , , y m ) R 4 m .

Then to obtain a true solution of (2.1), we need to choose suitable y λ , j such that

(4.1) Γ λ , i , j = 0 , for i = 1 , 2 , 3 , 4 and j = 1 , , m .

To this aim, similar to Lemma 2.2.13 in [12], we will find the critical points of Q ( y ) because that if y λ = ( y λ , 1 , , y λ , m ) is the critical point of Q ( y ) , then (4.1) holds.

Lemma 4.1

It holds

(4.2) I λ j = 1 m U 1 , λ , y λ , j , j = 1 m U 2 , λ , y λ , j = m b * λ 2 i = 1 2 Γ i 2 + b * 2 i = 1 2 j = 1 m Γ i 2 V i ( y λ , j ) l j ( Γ 0 + o ( 1 ) ) λ 2 e λ y λ , j y λ , l 1 λ y λ , j y λ , l 3 2 + O 1 λ θ 2 + λ l j e 3 2 λ y λ , j y λ , l 1 λ y λ , j y λ , l 3 2 ,

for some constant Γ 0 > 0 .

Proof

First, by the definition of I λ , we have

I λ j = 1 m U 1 , λ , y λ , j , j = 1 m U 2 , λ , y λ , j = λ 2 2 i = 1 2 Γ i 2 R 4 j = 1 m U ( λ ( x y λ , j ) ) 2 + λ 2 2 i = 1 2 Γ i 2 R 4 ( λ + V i ( x ) ) j = 1 m U ( λ ( x y λ , j ) ) 2 λ 3 3 i = 1 2 Γ i 2 R 4 j = 1 m U ( λ ( x y λ , j ) ) 3 ,

where we use the fact that 3 α 2 β α 3 = Γ 1 2 + Γ 2 2 . Now, we find

R 4 j = 1 m U ( λ ( x y λ , j ) ) 2 = 2 m λ b * + l j R 4 U ( λ ( x y λ , j ) ) U ( λ ( x y λ , l ) ) A 1 .

And from the assumption ( K 2 ) and y λ , j y λ , l 0 , we compute that

R 4 ( λ + V i ( x ) ) j = 1 m U ( λ ( x y λ , j ) ) 2 = j = 1 m ( λ + V i ( y λ , j ) ) R 4 U 2 ( λ ( x y λ , j ) ) + j = 1 m R 4 ( V i ( x ) V i ( y λ , j ) ) U 2 ( λ ( x y λ , j ) ) + l j m R 4 ( λ + V i ( x ) ) U ( λ ( x y λ , j ) ) U ( λ ( x y λ , l ) ) = 1 λ m b * + 1 λ 2 j = 1 m V i ( y λ , j ) b * + O 1 λ 2 + 1 2 θ + l j m R 4 λ U ( λ ( x y λ , j ) ) U ( λ ( x y λ , l ) ) A 2 + o 1 λ e λ y λ , j y λ , l 1 λ y λ , j y λ , l 3 2 .

On the other hand, it holds

R 4 j = 1 m U ( λ ( x y λ , j ) ) 3 = 3 k b * λ 2 + 3 R 4 l j U 2 ( λ ( x y λ , j ) ) U ( λ ( x y λ , l ) ) A 3 + O l j R 4 U 3 2 ( λ ( x y λ , j ) ) U 3 2 ( λ ( x y λ , l ) ) = 3 k b * λ 2 + 3 A 3 + O 1 λ 2 e 3 2 λ y λ , j y λ , l 1 λ y λ , j y λ , l 3 2 .

Finally, by using the equation satisfied by U , A 1 + A 2 λ A 3 = 0 . So the aforementioned estimates give (4.2).□

Now we have the following energy expansion.

Lemma 4.2

It holds

(4.3) I λ ( u 1 , λ , u 2 , λ ) = I λ j = 1 m U 1 , λ , y λ , j , j = 1 m U 2 , λ , y λ , j + O 1 λ .

Proof

First, by definition, we have

(4.4) I λ j = 1 m U 1 , λ , y λ , j ( x ) + φ 1 , λ , j = 1 m U 2 , λ , y λ , j ( x ) + φ 2 , λ = I λ j = 1 m U 1 , λ , y λ , j ( x ) , j = 1 m U 2 , λ , y λ , j ( x ) + j = 1 m i = 1 2 R 4 ( U i , λ , y λ , j φ i , λ + ( λ + V i ( x ) ) U i , λ , y λ , j φ i , λ ) α 2 R 4 j = 1 m U 1 , λ , y λ , j 2 φ 2 , λ + 2 j = 1 m U 1 , λ , y λ , j j = 1 m U 1 , λ , y λ , j φ 1 , λ + 2 j = 1 m U 1 , λ , y λ , j φ 1 , λ φ 2 , λ + j = 1 m ( U 2 , λ , y λ , j ) φ 1 , λ 2 + φ 1 , λ 2 φ 2 , λ β 3 R 4 3 j = 1 m U 2 , λ , y λ , j 2 φ 2 , λ + 3 j = 1 m U 2 , λ , y λ , j φ 2 , λ + φ 2 , λ 3 + O ( λ ( φ 1 , λ , φ 2 , λ ) λ 2 ) .

By using the classical Gagliardo-Nirenberg inequality (2.15), similar to (2.14), we have

R 4 j = 1 m U 1 , λ , y λ , j φ 1 , λ φ 2 , λ + j = 1 m ( U 2 , λ , y λ , j ) φ 1 , λ 2 + φ 1 , λ 2 φ 2 , λ + φ 2 , λ 3 = O λ 1 2 ( φ 1 , λ , φ 2 , λ ) λ 2 .

On the other hand, from ( φ 1 , λ , φ 2 , λ ) E λ ,

R 4 ( U i , λ , y λ , j φ i , λ + ( λ + V i ( x ) ) U i , λ , y λ , j φ i , λ ) = 0 .

So, (4.4) gives that

(4.5) I λ j = 1 m U 1 , λ , y λ , j ( x ) + φ 1 , λ , j = 1 m U 2 , λ , y λ , j ( x ) + φ 2 , λ = I λ j = 1 m U 1 , λ , y λ , j ( x ) , j = 1 m U 2 , λ , y λ , j ( x ) λ 2 R 4 α 2 Γ 1 2 + β Γ 2 2 j = 1 m U ( λ ( x y λ , j ) ) 2 φ 2 , λ + α Γ 1 Γ 2 j = 1 m U ( λ ( x y λ , j ) ) 2 φ 1 , λ + O ( λ ( φ 1 , λ , φ 2 , λ ) λ 2 ) .

Now using the classical Gagliardo-Nirenberg inequality (2.15) again, we have

R 4 j = 1 m U ( λ ( x y λ , j ) ) 2 φ i , λ = j = 1 m R 4 U 2 ( λ ( x y λ , j ) ) φ i , λ + l j R 4 U ( λ ( x y λ , j ) ) U ( λ ( x y λ , l ) ) φ i , λ = O 1 λ 5 4 ( φ 1 , λ , φ 2 , λ ) λ .

Then (4.3) follows by the aforementioned estimates.□

Proposition 4.3

Assume that ( K 2 ) holds. Then there exist some large λ 1 such that for any λ [ λ 1 , + ) , problem (2.1) has a peak solution ( u 1 , λ , u 2 , λ ) with the form

(4.6) u i , λ ( x ) = j = 1 m U i , λ , y λ , j ( x ) + φ i , λ

satisfying

(4.7) y λ , j a 1 , λ y λ , j y λ , l , for l j and ( φ 1 , λ , φ 2 , λ ) λ = O 1 λ .

Proof

First, it follows from Lemmas 4.1 and 4.2 that

(4.8) Q ( y ) = m b * λ 2 i = 1 2 Γ i 2 + b * 2 i = 1 2 j = 1 m Γ i 2 V i ( y λ , j ) l j ( Γ 0 + o ( 1 ) ) λ 2 e λ y λ , j y λ , l 1 λ y λ , j y λ , l 3 2 + O 1 λ m i n { θ 2 , 1 } + λ l j e 3 2 λ y λ , j y λ , l 1 λ y λ , j y λ , l 3 2 .

Now we define

= y = ( y 1 , , y m ) : y j B ε ¯ ( a 1 ) , y l y j ε ln λ λ , j = 1 , , m , l j ,

where ε is a small fixed constant. Consider the following problem

(4.9) max y Q ( y ) .

Suppose that it is achieved by y λ . To prove this y λ is a critical point of Q ( y ) , we just need to prove that y λ is the interior point of .

Let y ¯ λ , j , j = 1 , , m , satisfy

y ¯ λ , j a 1 1 λ γ and y ¯ λ , j y ¯ λ , l 1 λ 1 γ if l j ,

where 0 < ε < γ < < 1 is a small fixed constant. Then for y ¯ λ = ( y ¯ λ , 1 , , y ¯ λ , m ) , it holds

Q ( y ¯ λ ) = λ m b * 2 i = 1 2 Γ i 2 + m b * 2 i = 1 2 Γ i 2 V i ( a 1 ) + O 1 λ γ θ 2 .

Suppose that there exists y λ , j 0 such that y λ , j 0 a 1 = ε . Then by (4.8) and the assumption ( K 2 ) , we have

Q ( y λ ) λ m b * 2 i = 1 2 Γ i 2 + ( m 1 ) b * 2 i = 1 2 Γ i 2 V i ( a 1 ) + b * 2 i = 1 2 Γ i 2 V i ( y λ , j 0 ) + o ( 1 ) < Q ( y ¯ λ ) ,

which contradicts the fact that y λ is a maximum point of Q ( y ) in .

Now suppose that there exist y λ , j 0 and y λ , m 0 such that y λ , j 0 y λ , m 0 = ε ln λ λ for m 0 j 0 . Then by using (4.8), we find

Q ( y λ ) λ m b * 2 i = 1 2 Γ i 2 κ ˜ 0 λ 1 ε ( ln λ ) 3 2 < Q ( y ¯ λ )

for some κ ˜ 0 > 0 , which gives a contradiction again. Thus, we have proved that y λ is the interior point of , and thus, it is a critical point of Q ( y ) .□

Finally, we conclude our second main result.

Proof of Theorem 1.3

Now we set b λ = R 4 ( u 1 , λ 2 + u 2 , λ 2 ) and v i , λ = u i , λ b λ , then ( v 1 , λ , v 2 , λ ) satisfies that

Δ v 1 , λ + ( λ + V 1 ( x ) ) v 1 , λ = α b λ 1 2 v 1 , λ v 2 , λ , in R 4 , Δ v 2 , λ + ( λ + V 2 ( x ) ) v 2 , λ = α 2 b λ 1 2 v 1 , λ 2 + β b λ 1 2 v 2 , λ 2 , in R 4 ,

with

R 4 ( v 1 , λ 2 + v 2 , λ 2 ) = 1 .

Therefore, similar to Proposition 3.3, we obtain that

λ ( m D ( α , β ) b * b λ ) = b * i = 1 2 j = 1 m Γ i 2 V i ( a 1 ) + o ( 1 ) .

Next, similar to the proofs of Theorem 1.1, we can finish the proof.□

  1. Funding information: This study was partially supported by NSFC (No. 12226324, 11961043).

  2. Conflict of interest: The authors state no conflict of interest.

  3. 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: 2022-12-26
Accepted: 2023-10-11
Published Online: 2024-01-27

© 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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