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Synchronized vector solutions for the nonlinear Hartree system with nonlocal interaction

  • Fashun Gao , Minbo Yang and Shunneng Zhao EMAIL logo
Published/Copyright: January 3, 2025

Abstract

We are concerned with the following nonlinear Hartree system

Δ u + P 1 ( | x | ) u = α 1 | x | 1 u 2 u + β | x | 1 v 2 u in R 3 , Δ v + P 2 ( | x | ) v = α 2 | x | 1 v 2 v + β | x | 1 u 2 v in R 3 ,

where P 1(r) and P 2(r) are positive radial potentials, α 1 > 0, α 2 > 0 and β R is a coupling constant. We first study nondegeneracy of ground states for the limit system of the above problem. As applications, we show that the nonlinear Hartree system has infinitely many non-radial positive synchronized solutions, whose energy can be arbitrarily large.

2020 Mathematics Subject Classification: Primary 35J47, 35J60; Secondly 35B40, 35B45

1 Introduction and main results

The two-component system with Hartree type nonlinearities

(1.1) i t Φ 1 = Δ Φ 1 + V 1 ( x ) Φ 1 W ( x ) | Φ 1 | 2 Φ 1 β W ( x ) | Φ 2 | 2 Φ 1 , ( t , x ) R + × R N , i t Φ 2 = Δ Φ 2 + V 2 ( x ) Φ 2 W ( x ) | Φ 2 | 2 Φ 2 β W ( x ) | Φ 1 | 2 Φ 2 , ( t , x ) R + × R N

describes the dynamics of boson stars in mean-field theory [1], [2] and attracts much attention in recent years. Here, Φ i : R + × R N C , β > 0 is the scattering length describing the attractive interaction, V i are the external potentials, and W is the response function which possesses information on the mutual interaction between the particles. It is well known that the nonlocal nonlinearities play important roles in the theory of Bose–Einstein condensation (cf. [1]).

It is interesting to study the standing wave solutions of (1.1) with a response function of Riesz potential, i.e., V(x) = |x|μ . Clearly, Φ 1 ( t , x ) = e i E 1 t u ( x ) and Φ 2 ( t , x ) = e i E 2 t v ( x ) solve (1.1) if and only if (u(x), v(x)) is a solution of the Hartree system

(1.2) Δ u + P 1 ( x ) u = | x | μ u 2 u + β | x | μ v 2 u in R N , Δ v + P 2 ( x ) v = | x | μ v 2 v + β | x | μ u 2 v in R N ,

where P i (x) = V i (x) − E i for i = 1, 2. In [3], Wang and Shi studied (1.2) with positive constant potentials and proved the existence and nonexistence of positive ground state solutions. In [4], the authors proved the existence of ground state solution for a singular perturbed problem related to (1.2) when the coupling constant β is large. Later, Gao et al. [5] studied the existence of positive solutions of high energy for the critical coupled Hartree system (1.2) with N ≥ 5, μ = 4, β > max{α 1, α 2} ≥ min{α 1, α 2} > 0, and P 1 , P 2 L N / 2 ( R N ) L loc ( R N ) . Recently, Ye et al. [6] considered the couple critical interaction system (1.2) with axisymmetric potentials and constructed infinitely many solutions by employing the Lyapunov-Schmidt reduction argument and Hatree-type local Pohožaev identities.

Our another motivation comes from the study of existence of positive solution to the single nonlinear Hartree equation. In precise, let u = 0 or v = 0 in (1.2), it goes back to the single Hartree type equation

(1.3) Δ u + P ( x ) u = | x | μ u 2 u in R N .

The existence and uniqueness of ground state solutions to (1.3) has been studied in [7]–[12]. Ackermann [13] proved the existence of infinitely many geometrically distinct weak solutions. Moroz and Van Schaftingen [14] proved the existence, regularity and positivity of radial symmetric ground state.

The nondegeneracy of the ground state solution to the nonlinear Hartree equation has been obtained in a series of papers [7], [12], [15]–[17]. It is a key ingredient in the Lyapunov–Schmidt reduction method of constructing semiclassical or blow-up solutions to the partial differential equation. In [17], Wei and Winter studied the equivalent local Schrödinger–Newton system instead and applied the expansion of the equations by spherical harmonics. Then they use the Liapunov-reduction scheme to construct multibump solutions of the nonlocal partial differential equation with strongly interacting bumps. In [7], Lenzmann found a method that relied on spectral analysis of the linearized operators at the ground states. Later, Chen [15] generalized the nondegeneracy result to higher dimensions and constructed multiple semiclassical solutions to Hartree equation with an external potential by using Lyapunov-Schmidt reduction argument. For other related results, we refer the readers to [13], [18], [19] [13], [18]–[20] for some progress on the topic of Hartree equation. In particular, the authors in [20] proved the existence of infinitely many non-radial solutions for nonlinear Hartree equation (1.3) under some proper assumptions on the function P. Therefore, a natural question to ask is in the following: For the following coupled Hartree system, are there infinitely many non-radial positive synchronized solutions? If yes, can the energy of these solutions be arbitrarily large?

(1.4) Δ u + P 1 ( | x | ) u = α 1 | x | 1 u 2 u + β | x | 1 v 2 u in R 3 , Δ v + P 2 ( | x | ) v = α 2 | x | 1 v 2 v + β | x | 1 u 2 v in R 3 ,

where P 1(|x|) and P 2(|x|) are positive radial potentials, α 1 > 0, α 2 > 0 and β R is a coupling constant. In this paper we answer these questions. We rigorously prove that Hartree system (1.4) has an unbounded sequence of non-radial synchronized vector solutions by employing the Lyapunov-Schmidt reduction argument.

To state our result, we need to introduce potential function P 1(|x|) and P 2(|x|) and satisfy the following condition:

(P 1) There are constants a > 0, m ≥ 3 and θ > 0 such that as r → +,

P 1 ( r ) = 1 + a r m + O 1 r m + θ .

(P 2) There are constants b > 0, n ≥ 3 and ϑ > 0 such that as r → +,

P 2 ( r ) = 1 + b r n + O 1 r n + ϑ .

The main result of this paper reads as follows:

Theorem 1.1.

Suppose that P 1(r) satisfies (P 1) and P 2(r) satisfies (P 2). Then there exists a decreasing sequence { β k } ( α 1 α 2 , 0 ) with β k α 1 α 2 as k such that for β ( α 1 α 2 , 0 ) ( 0 , min { α 1 , α 2 } ) ( max { α 1 , α 2 } , ) and ββ k for any k, (1.4) has infinitely many non-radial positive synchronized solutions, whose energy can be arbitrarily large, provided one of the following two conditions holds:

  1. m < n, a > 0 and b R ; or m > n, a R and b > 0;

  2. m = n, aA + bB > 0, where A and B are defined by

    A = h 0 2 R 3 W 2 d x , B = l 0 2 R 3 W 2 d x with h 0 β α 2 β 2 α 1 α 2 , l 0 β α 1 β 2 α 1 α 2 .

Remark 1.2.

The similar to proof of Theorem 1.1, we can construct an unbounded sequence of non-radial positive vector solutions of segregated type. There exists a β* > 0 such that, for β < β*, (1.4) has infinitely many non-radial positive segregated solutions, whose energy can be arbitrarily large.

We use W to denote the solution of the following problem

(1.5) Δ u + u = | x | 1 | u | 2 u in R 3 .

In [14] the authors showed radial symmetry of solution W yields that W is either positive or negative, and there exist x 0 R 3 and a monotone function vC (0, ) such that for every x R 3 , W(x) = v(|xx 0|). Without loss of generality, we can suppose W > 0 and x 0 = 0, that is, W(x) = W(|x). Chen [15] applied the expansion of the nonlocal term by spherical harmonics to obtain the nondegeneracy of positive bubble solutions W for nonlinear Hartree equation (1.5). More precisely, we summarize this result as follows:

Proposition 1.3.

Assume n = 3. Let L be the linearized operator at W given by

L ( v ) = Δ v + v 2 | x | 1 ( W v ) W | x | 1 W 2 v , v H 2 ( R 3 ) .

Then

kel  L = span W x 1 , W x 2 , W x 3 .

Note that the limit system for (1.4) is

(1.6) Δ u + u = α 1 | x | 1 u 2 u + β | x | 1 v 2 u in R 3 , Δ v + v = α 2 | x | 1 v 2 v + β | x | 1 u 2 v in R 3 ,

and that

( U , V ) = ( h 0 W , l 0 W ) ,

solves (1.6), provided α 1 α 2 < β < min { α 1 , α 2 } or β > max{α 1, α 2}, where h 0 = β α 2 β 2 α 1 α 2 , l 0 = β α 1 β 2 α 1 α 2 .

Since P 1(r) → 1 as r and P 2(r) → 1 as r, we will use (U, V) to build up the approximate solutions for (1.4) with large number of bumps near the infinity. As in [21], let

(1.7) z j = r cos 2 ( j 1 ) π k , r sin 2 ( j 1 ) π k , 0 , j = 1 , , k ,

and

r S k min { m + 1 , n + 1 } 2 π δ k ln k , min { m + 1 , n + 1 } 2 π + δ k ln k ,

where m is the constant in the expansion for P 1, n is the constant in the expansion for P 2, and δ > 0 is a small constant.

The Sobolev space H P i 1 ( R 3 ) is endowed with the standard norm

u P i 2 = R 3 ( | u | 2 + P i ( x ) u 2 ) d x , i = 1,2 .

Define H to be the product space H P 1 1 ( R 3 ) × H P 2 1 ( R 3 ) with the norm

( u , v ) = u P 1 + v P 2 .

Set x = (x′, x″), x R 2 and x R . Define

H P i , s = u H P i 1 ( R 3 ) , u  is even in  x h , h = 2,3 , u ( r cos θ , r sin θ , x ) = u r cos θ + 2 j π k , r sin θ + 2 j π k , x , j = 1 , , k .

where i = 1, 2.

We denote

U r ( x ) j = 1 k U z j ( x ) , V r ( x ) j = 1 k V z j ( x ) ,

where U z j ( x ) = U ( x z j ) and V z j ( x ) = V ( x z j ) . It is easy to check that (U r , V r ) is in H P 1 1 ( R 3 ) × H P 2 1 ( R 3 ) .

Theorem 1.4.

Under the assumptions of Theorem 1.1, there is an integer k 0 > 0, such that for any integer kk 0, (1.4) has a solution (u k , v k ) of the form

( u k , v k ) = ( U r k ( x ) + φ k , V r k ( x ) + ϕ k ) ,

where ( φ k , ϕ k ) H P 1 , s × H P 2 , s , r k ∈ [r 0 k ln k, r 1 k ln k] and as k,

( φ k , ϕ k ) 0 .

Before concluding this introduction, we would like to mention some related results to our problem. Wei and Yan in [21] use a Lyapunov–Schmidt reduction argument to consider the nonlinear Schrödinger equation

(1.8) Δ u + V ( | x | ) u = u p , u > 0 in R N ,

with V(r) → V 0 > 0 as r and they proved the existence of infinitely many positive non-radial solutions for such equation. This technique has also been applied successfully for the study of different problems, for example, Schrödinger-Poisson type system (see [22]), nonlinear Schrödinger system (see [23]), nonlinear Hartree equation (see [20], [24]–[26]). Results on the existence of infinitely many solutions for other elliptic problems can also be found in [27]–[30] and the references therein. In contrast with the single nonlinear Schrödinger equations (1.3) or (1.8), since we deal with a system, we face additional difficulties. Moreover, more involved technical difficulties is needed to deal with the nonlocal term in (1.4) and a more careful analysis of the interaction between the bumps is required

The paper is organized as follows. In Section 2, we give a nondegeneracy result for the limit system (1.6). In Section 3, we carry out the reduction. Section 4, we study the reduced finite dimensional problem and prove Theorem 1.4.

2 Nondegeneracy

Let us consider the following eigenvalue problem

(2.1) Δ v + v + | x | 1 W 2 v = λ | x | 1 ( W v ) W + | x | 1 W 2 v , v H 1 ( R 3 ) .

By the Hardy-Littlewood-Sobolev inequality, Hölder inequality and Sobolev inequality, we can obtain for all v H 1 ( R 3 ) , it holds that

R 3 | x | 1 ( W v ) W v d x C v H 1 2 ,

and

R 3 | x | 1 W 2 v 2 d x C v H 1 2 .

Then we may consider the eigenvalues of problem (2.1) as the following:

Definition 2.1.

The Rayleigh quotient characterization of first eigenvalue of (2.1) can be stated as

(2.2) λ 1 inf v H 1 ( R 3 ) \ { 0 } R 3 ( | v | 2 + v 2 ) d x + R 3 ( | x | 1 W 2 ) v 2 d x R 3 ( | x | 1 ( W v ) ) W v d x + R 3 ( | x | 1 W 2 ) v 2 d x .

Moreover, for any k N + the (k+1)th-eigenvalues can be characterized as follows:

λ k + 1 inf v P k + 1 \ { 0 } R 3 ( | v | 2 + v 2 ) d x + R 3 ( | x | 1 W 2 ) v 2 d x R 3 ( | x | 1 ( W v ) ) W v d x + R 3 ( | x | 1 W 2 ) v 2 d x ,

where

P k + 1 v H 1 ( R 3 ) : R 3 ( v e j + v e j ) d x = 0 , for all j = 1 , , k ,

and e j is the corresponding eigenfunction to λ j .

Choosing v = W in (2.2), since W is the solution of equation (1.5) then we have

λ 1 R 3 ( | W | 2 + W 2 ) d x + R 3 ( | x | 1 W 2 ) W 2 d x R 3 ( | x | 1 W 2 ) W 2 d x + R 3 ( | x | 1 W 2 ) W 2 d x = 1 .

Then from Proposition 1.3 regarding the nondegeneracy of W, we have

Proposition 2.2.

Let λ i , i = 1, 2, …, denote the eigenvalues of (2.1) in increasing order as in Definition 2.1. Then λ 1 ≤ 1 is simple and the corresponding eigenfunction is e 1 = ζW with ζ R , and there exists τ 0 N such that λ τ 0 + 2 = λ τ 0 + 3 = λ τ 0 + 4 = 2 with the corresponding 3-dimensional eigenfunction space spanned by W x 1 , W x 2 , W x 3 . Furthermore, λ τ 0 + 5 > λ τ 0 + 2 = 2 .

For convenience, we denote

e τ 0 + 2 = W x 1 , e τ 0 + 3 = W x 2 , e τ 0 + 4 = W x 3 .

Proposition 2.3.

There exists a decreasing sequence { β k } ( α 1 α 2 , 0 ) with β k α 1 α 2 as k such that for β ( α 1 α 2 , 0 ) ( 0 , min { α 1 , α 2 } ) ( max { α 1 , α 2 } , ) and ββ k for any k, (U, V) is non-degenerate for the system (1.6) in H 1 ( R 3 ) × H 1 ( R 3 ) in the sense that the kernel is given by span { ( θ ( β ) W x j , W x j ) : j = 1,2,3 } , where θ(β) ≠ 0.

Proof. We follow the arguments in [31] or [23]. Now for α 1 α 2 < β < 0 or 0 < β < min{α 1, α 2}, linearization of equations (1.6) at (U, V) gives us

Δ ϕ + ϕ = ( α 1 h 0 + β l 0 ) | x | 1 | W | 2 ϕ + 2 | x | 1 ( W ( α 1 h 0 ϕ + β h 0 l 0 ψ ) ) W in R 3 , Δ ψ + ψ = ( α 2 l 0 + β h 0 ) | x | 1 | W | 2 ψ + 2 | x | 1 ( W ( α 2 l 0 ψ + β h 0 l 0 ϕ ) ) W in R 3 .

We see at once that α 1 h 0 + βl 0 = α 2 l 0 + βh 0 = 1 then

(2.3) Δ ϕ + ϕ = | x | 1 | W | 2 ϕ + 2 | x | 1 ( W ( d ϕ + c ψ ) ) W in R 3 , Δ ψ + ψ = | x | 1 | W | 2 ψ + 2 | x | 1 ( W ( e ψ + c ϕ ) ) W in R 3 ,

where

d = α 1 h 0 , c = β h 0 l 0 , e = α 2 l 0 .

For simplicity, we rewrite system (2.3) as follow

Δ ϕ + ϕ + | x | 1 | W | 2 ϕ = 2 | x | 1 | W | 2 ϕ + 2 | x | 1 ( W ( d ϕ + c ψ ) ) W in R 3 , Δ ψ + ψ + | x | 1 | W | 2 ψ = 2 | x | 1 | U | 2 ψ + 2 | x | 1 ( W ( e ψ + c ϕ ) ) W in R 3 .

Let δ ± = e d 2 c ± 1 2 c ( e d ) 2 + 4 c 2 . For β < 0, a direct computation shows that

Δ ( ϕ δ + ψ ) + ( ϕ δ + ψ ) + | x | 1 | W | 2 ( ϕ δ + ψ ) = 2 | x | 1 | W | 2 ( ϕ δ + ψ ) + 2 | x | 1 ( W ( ϕ δ + ψ ) ) W .

Then we have

ϕ δ + ψ = j = 2 4 c j e τ 0 + j .

Returning to the equation for ψ we get that

Δ ψ + ψ = | x | 1 | W | 2 ψ + 2 | x | 1 W ( e + c δ + ) ψ + c j = 1 3 c j φ j W .

Set ψ = j = 1 γ j e j and f(β) = e + +. Assume that f(β) ≠ λ k for any k, where λ k is defined in Definition 2.1. Using orthogonality we see easily that γ j = 0 for jτ 0 + 2, τ 0 + 3, τ 0 + 4, and γ j = 2 c c τ 0 + j 1 2 f ( β ) for j = τ 0 + 2, τ 0 + 3, τ 0 + 4. Thus, the kernel at (U, V) is a three dimensional space and it can be given by

span 2 δ + c 1 2 f ( β ) + 1 e τ 0 + j , 2 c 1 2 f ( β ) e τ 0 + j : j = 2,3,4 .

We may take θ ( β ) = δ + + 1 2 f ( β ) 2 c . Since c ≠ 0 and 1 − 2e ≠ 0, we have θ(β) ≠ 0. If f(β) = λ j for some j, then similarly γ j = 0 for jτ 0 + 2, τ 0 + 3, τ 0 + 4. It is easy to check that e k is in the kernel when λ k = λ j . Thus the kernel is generated by

span 2 δ + c 1 2 f ( β ) + 1 e τ 0 + j , 2 c 1 2 f ( β ) e τ 0 + j : j = 2,3,4

and span{e k : λ k = λ j }. For min{α 1, α 2} > β > 0 we use δ instead of δ + to get the same result. From [31], f ( β ) = 1 2 if and only if β = 0 and f(β) is monotone decreasing. Also, from [31] there exists a decreasing sequence β k in ( α 1 α 2 , 0 ) such that f(β) = λ k if and only if β = β k . Thus, for β∉{β k }, f(β) = λ j for any j. The same arguments can be used to treat the case β > max{α 1, α 2}, so we omit it. □

3 The reduction argument

Let

Ω j = x = ( x , x ) R 2 × R : x | x | , z j | z j | cos π k , j = 1 , , k ,

where z j is given in (1.7). Then, we have the following basic estimates:

Lemma 3.1.

(Lemma 2.1, [20]) For any x ∈ Ω1 and η ∈ (0, 1), there is a constant C > 0, such that

(3.1) i = 2 k U z i ( x ) C e π η r k e ( 1 η ) | x z 1 | ,

and

(3.2) i = 2 k V z i ( x ) C e π η r k e ( 1 η ) | x z 1 | .

Lemma 3.2.

(Lemma 2.2, [20]) For any q ≥ 1, there is a constant σ > 0, such that

(3.3) U r q ( x ) = U z i q ( x ) + O 1 r σ e 1 2 | x z i | , x Ω i ,

and

(3.4) V r q ( x ) = V z i q ( x ) + O 1 r σ e 1 2 | x z i | , x Ω i .

Denote

T j = U z j r , Y j = V z j r , j = 1 , , k ,

and

E = ( u , v ) H P 1 , s × H P 2 , s : R 3 R 3 | U z j ( x ) | 2 T j ( y ) u ( y ) | x y | d x d y + 2 R 3 R 3 U z j ( x ) u ( x ) U z j ( y ) T j ( y ) | x y | d x d y = 0 , R 3 R 3 | V z j ( x ) | 2 Y j ( y ) v ( y ) | x y | d x d y + 2 R 3 R 3 V z j ( x ) v ( x ) V z j ( y ) Y j ( y ) | x y | d x d y = 0 , j = 1 , , k .

By the energy functional J : H R associated with equation (1.4),

J ( u , v ) = 1 2 R 3 | u | 2 + | v | 2 + P 1 ( x ) u 2 + P 2 ( x ) v 2 d x 1 4 R 3 R 3 α 1 | u ( x ) | 2 | u ( y ) | 2 + α 2 | v ( x ) | 2 | v ( y ) | 2 + 2 β | u ( x ) | 2 | v ( y ) | 2 | x y | d x d y ,

we denote

I ( φ , ϕ ) = J ( U r + φ , V r + ϕ ) , ( φ , ϕ ) E .

Expand I(φ, ϕ) as follows:

I ( φ , ϕ ) = I ( 0,0 ) + l ( φ , ϕ ) + 1 2 L ( φ , ϕ ) + R ( φ , ϕ ) , ( φ , ϕ ) E ,

where

l ( φ , ϕ ) = j = 1 k R 3 ( P 1 ( | x | ) 1 ) U z j φ d x + j = 1 k R 3 ( P 2 ( | x | ) 1 ) V z j φ d x α 1 R 3 R 3 | U r ( x ) | 2 U r ( y ) φ ( y ) | x y | d x d y + α 1 j = 1 k R 3 R 3 | U z j ( x ) | 2 U z j ( y ) φ ( y ) | x y | d x d y α 2 R 3 R 3 | V r ( x ) | 2 V r ( y ) ϕ ( y ) | x y | d x d y + α 2 j = 1 k R 3 R 3 | V z j ( x ) | 2 V z j ( y ) ϕ ( y ) | x y | d x d y β R 3 R 3 | U r ( x ) | 2 V r ( y ) ϕ ( y ) | x y | d x d y + β j = 1 k R 3 R 3 | U z j ( x ) | 2 V z j ( y ) ϕ ( y ) | x y | d x d y β R 3 R 3 | V r ( x ) | 2 U r ( y ) φ ( y ) | x y | d x d y + β j = 1 k R 3 R 3 | V z j ( x ) | 2 U z j ( y ) φ ( y ) | x y | d x d y ,

L ( φ , ϕ ) = R 3 | φ | 2 + P 1 ( x ) φ 2 d x R 3 R 3 α 1 | U r ( x ) | 2 | φ ( y ) | 2 | x y | d x d y + R 3 | ϕ | 2 + P 2 ( x ) ϕ 2 d x R 3 R 3 α 2 | V r ( x ) | 2 | ϕ ( y ) | 2 | x y | d x d y β R 3 R 3 | U r ( x ) | 2 | ϕ ( y ) | 2 + | V r ( x ) | 2 | φ ( y ) | 2 + 4 U r ( x ) φ ( x ) V r ( y ) ϕ ( y ) | x y | d x d y

and

R ( φ , ϕ ) = α 1 R 3 R 3 U r ( x ) φ ( x ) φ 2 ( y ) | x y | d x d y α 1 4 R 3 R 3 φ 2 ( x ) φ 2 ( y ) | x y | d x d y α 2 R 3 R 3 V r ( x ) ϕ ( x ) ϕ 2 ( y ) | x y | d x d y α 2 4 R 3 R 3 ϕ 2 ( x ) ϕ 2 ( y ) | x y | d x d y β R 3 R 3 U r ( x ) φ ( x ) ϕ 2 ( y ) + V r ( x ) ϕ ( x ) φ 2 ( y ) | x y | d x d y β 2 R 3 R 3 φ 2 ( x ) ϕ 2 ( y ) | x y | d x d y .

It is easy to check that there is a bounded linear operator L from E to E, such that

L ( u , v ) , ( φ , ϕ ) = R 3 u φ + P 1 ( x ) u φ d x α 1 R 3 R 3 | U r ( x ) | 2 u ( y ) φ ( y ) | x y | d x d y + R 3 v ϕ + P 2 ( x ) v ϕ d x α 2 R 3 R 3 | V r ( x ) | 2 v ( y ) ϕ ( y ) | x y | d x d y β R 3 R 3 | U r ( x ) | 2 v ( y ) ϕ ( y ) + | V r ( x ) | 2 u ( y ) φ ( y ) + 2 U r ( x ) V r ( y ) ( u ( x ) ϕ ( y ) + φ ( x ) v ( y ) ) | x y | d x d y .

Therefore, we have the following estimates.

Lemma 3.3.

There is a constant C′ > 0, independent of k, such that for any rS k ,

L ( u , v ) C ( u , v ) , ( u , v ) E .

Next, we show that L is invertible in E.

Lemma 3.4.

There is a constant C″ > 0, independent of k, such that for any rS k ,

L ( u , v ) C ( u , v ) , ( u , v ) E .

Proof. Suppose to the contrary that there are k → +, r k S k , and (u k , v k ) ∈ E, with

L ( u k , v k ) = o ( 1 ) ( u k , v k ) .

Then

(3.5) L ( u k , v k ) , ( φ , ϕ ) = o ( 1 ) ( u k , v k ) ( φ , ϕ ) , ( φ , ϕ ) E .

We may assume that ‖(u k , v k )‖2 = k.

By symmetry, we see from (3.5),

(3.6) Ω 1 u k φ + P 1 ( x ) u k φ d x Ω 1 R 3 α 1 | U r ( x ) | 2 u k ( y ) φ ( y ) | x y | d x d y + Ω 1 v k ϕ + P 2 ( x ) v k ϕ d x Ω 1 R 3 α 2 | V r ( x ) | 2 v k ( y ) ϕ ( y ) | x y | d x d y β Ω 1 R 3 | U r ( x ) | 2 v k ( y ) ϕ ( y ) + | V r ( x ) | 2 u k ( y ) φ ( y ) + 2 U r ( x ) V r ( y ) ( u k ( x ) ϕ ( y ) + φ ( x ) v k ( y ) ) | x y | d x d y = 1 k L ( u k , v k ) , ( φ , ϕ ) = o 1 k ( φ , ϕ ) , ( φ , ϕ ) E .

In particular,

Ω 1 | u k | 2 + P 1 ( x ) u k 2 d x Ω 1 R 3 α 1 | U r ( x ) | 2 | u k ( y ) | 2 | x y | d x d y + Ω 1 | v k | 2 + P 2 ( x ) v k 2 d x Ω 1 R 3 α 2 | V r ( x ) | 2 | v k ( y ) | 2 | x y | d x d y β Ω 1 R 3 | U r ( x ) | 2 | v k ( y ) | 2 + | V r ( x ) | 2 | u k ( y ) | 2 + 4 U r ( x ) u k ( x ) V r ( y ) v k ( y ) | x y | d x d y = o ( 1 )

and

(3.7) Ω 1 | u k | 2 + P 1 ( x ) u k 2 + | v k | 2 + P 2 ( x ) v k 2 d x = 1 .

Let u ̄ k ( x ) = u k ( x z 1 ) and v ̄ k ( x ) = v k ( x z 1 ) . Then for any R > 0, since | z 2 z 1 | = r sin π k C ln k , we see that B R (z 1) ⊂ Ω1. As a result, from (3.7), we find that for any R > 0,

B R ( z 1 ) | u ̄ k | 2 + P 1 ( x ) u ̄ k 2 + | v ̄ k | 2 + P 2 ( x ) v ̄ k 2 d x 1 .

So, we may assume that there is a ( u , v ) H 1 ( R 3 ) × H 1 ( R 3 ) , such that as k → +,

u ̄ k u , v ̄ k v , weakly in  H loc 1 ( R 3 ) ,

and

u ̄ k u , v ̄ k v , strongly in  L loc p ( R 3 ) , 2 p < 6 .

Since v ̄ k and u ̄ k is even in x h , h = 2, 3, it is easy to see that u and v are even in x h , h = 2, 3 and satisfy

(3.8) R 3 R 3 | W ( x ) | 2 U ( y ) y 1 ( y ) u ( y ) | x y | d x d y + 2 R 3 R 3 W ( x ) u ( x ) W ( y ) U ( y ) y 1 ( y ) | x y | d x d y + R 3 R 3 | W ( x ) | 2 V ( y ) y 1 ( y ) v ( y ) | x y | d x d y + 2 R 3 R 3 W ( x ) v ( x ) W ( y ) V ( y ) y 1 ( y ) | x y | d x d y = 0 .

Now, we claim that (u, v) satisfies

(3.9) Δ u + u = α 1 R 3 | U ( y ) | 2 | x y | d y u + 2 α 1 R 3 U ( y ) u ( y ) | x y | d y U + β R 3 | V ( y ) | 2 | x y | d y u + 2 β R 3 V ( y ) v ( y ) | x y | d y U in R 3 , Δ v + v = α 2 R 3 | V ( y ) | 2 | x y | d y v + 2 α 2 R 3 V ( y ) v ( y ) | x y | d y V + β R 3 | U ( y ) | 2 | x y | d y v + 2 β R 3 U ( y ) u ( y ) | x y | d y V in R 3 .

Define

E ̃ = ( φ , ϕ ) H 1 ( R 3 ) × H 1 ( R 3 ) : R 3 R 3 | W ( x ) | 2 U ( y ) y 1 ( y ) φ ( y ) | x y | d x d y + 2 R 3 R 3 W ( x ) φ ( x ) W ( y ) U ( y ) y 1 ( y ) | x y | d x d y + R 3 R 3 | W ( x ) | 2 V ( y ) y 1 ( y ) ϕ ( y ) | x y | d x d y + 2 R 3 R 3 W ( x ) ϕ ( x ) W ( y ) V ( y ) y 1 ( y ) | x y | d x d y = 0 .

For any R > 0, let ( φ , ϕ ) C 0 ( B R ( 0 ) ) × C 0 ( B R ( 0 ) ) E ̃ be any function, satisfying that φ and ϕ are even in x h , h = 2, 3. Then φ k φ ( x z 1 ) , ϕ k ϕ ( x z 1 ) C 0 ( B R ( 0 ) ) . With the argument in [21] we find

Ω 1 ( u k φ k + P 1 ( | x | ) u k φ k + v k ψ k + P 2 ( | x | ) v k ψ k ) d x R 3 ( u φ + u φ + v ϕ + v ϕ ) d x .

By Lemma 3.2, we know

Ω 1 R 3 | U r ( x ) | 2 u k ( y ) φ k ( y ) | x y | d x d y = Ω 1 Ω 1 U z 1 2 ( x ) + O 1 r σ e 1 2 β | x z 1 | u k ( y ) φ k ( y ) | x y | d x d y R 3 R 3 | U ( x ) | 2 u ( y ) φ ( y ) | x y | d x d y .

Similarly, we have

Ω 1 R 3 | V r ( x ) | 2 v k ( y ) ϕ k ( y ) | x y | d x d y R 3 R 3 | V ( x ) | 2 v ( y ) ϕ ( y ) | x y | d x d y , Ω 1 R 3 U r ( x ) u k ( y ) U r ( x ) φ k ( y ) | x y | d x d y R 3 R 3 U ( x ) u ( y ) U ( y ) φ ( y ) | x y | d x d y , Ω 1 R 3 V r ( x ) v k ( x ) V r ( y ) ϕ k ( y ) | x y | d x d y R 3 R 3 V ( x ) v ( x ) V ( y ) ϕ ( y ) | x y | d x d y ,

and

Ω 1 R 3 | U r ( x ) | 2 v k ( y ) ϕ k ( y ) + | V r ( x ) | 2 u k ( y ) φ k ( y ) + 2 U r ( x ) V r ( y ) ( u k ( x ) ϕ k ( y ) + φ k ( x ) v k ( y ) ) | x y | d x d y R 3 R 3 | U ( x ) | 2 v ( y ) ϕ ( y ) + | V ( x ) | 2 u ( y ) φ ( y ) + 2 U ( x ) V ( y ) ( u ( x ) ϕ ( y ) + φ ( x ) v ( y ) ) | x y | d x d y .

As a result,

(3.10) R 3 ( u φ + u φ + v ϕ + v ϕ ) d x = α 1 R 3 R 3 | U ( x ) | 2 u ( y ) φ ( y ) | x y | d x d y + 2 α 1 R 3 R 3 U ( x ) u ( x ) U ( y ) φ ( y ) | x y | d x d y + α 2 R 3 R 3 | V ( x ) | 2 v ( y ) ϕ ( y ) | x y | d x d y + 2 α 2 R 3 R 3 V ( x ) v ( x ) V ( y ) ϕ ( y ) | x y | d x d y + β R 3 R 3 | U ( x ) | 2 v ( y ) ϕ ( y ) + | V ( x ) | 2 u ( y ) φ ( y ) + 2 U ( x ) V ( y ) ( u ( x ) ϕ ( y ) + φ ( x ) v ( y ) ) | x y | d x d y .

On the other hand, since u and v are even in x h , h = 2, 3, (3.10) holds for any function ( φ , ϕ ) C 0 ( B R ( 0 ) ) × C 0 ( B R ( 0 ) ) , which is odd in x h , h = 2, 3. Therefore, (3.10) holds for any ( φ , ϕ ) C 0 ( B R ( 0 ) ) × C 0 ( B R ( 0 ) ) E ̃ . By the density of C 0 ( R 3 ) × C 0 ( R 3 ) in H 1 ( R 3 ) × H 1 ( R 3 ) , it is easy to show that (3.10) holds for any function ( φ , ϕ ) E ̃ . But (3.10) holds for φ , ϕ = U x 1 , V x 1 . Thus (3.10) is true for any ( φ , ϕ ) H 1 ( R 3 ) × H 1 ( R 3 ) . So, we have proved (3.9).

From Proposition 2.3, (U, V) is nondegenerate. Since we work in the space of functions which are even in x 2 and x 3, the kernel of (U, V) is given by the one dimensional θ ( β ) W x 1 , W x 1 . Hence we easily check that ( u , v ) = c U x 1 , V x 1 for some c, which implies that (u, v) = (0, 0) since (u, v) satisfies (3.8).

As a consequence, we eventually have

B R ( z 1 ) u k p + v k p d x = o ( 1 ) , R > 0 , 2 p < 6 .

On the other hand, it follows from Lemma 3.1 that for any small η > 0, there is a constant C > 0, such that

(3.11) U r k ( x ) C e ( 1 η ) | x z 1 | , V r k ( x ) C e ( 1 η ) | x z 1 | , x Ω 1 .

Thus, we obtain

Ω 1 | u k | 2 + P 1 ( x ) u k 2 d x Ω 1 R 3 α 1 | U r ( x ) | 2 | u k ( y ) | 2 | x y | d x d y = Ω 1 ( | u k | 2 + P 1 ( | x | ) | u k | 2 ) d x + o ( 1 ) + O ( e ( 1 η ) R ) B R C ( z 1 ) | u k | 12 5 d x 5 6 1 2 Ω 1 ( | u k | 2 + P 1 ( | x | ) | u k | 2 ) d x + o ( 1 ) .

Similar computations give that

Ω 1 | v k | 2 + P 2 ( x ) v k 2 d x Ω 1 R 3 α 2 | V r ( x ) | 2 | v k ( y ) | 2 | x y | d x d y 1 2 Ω 1 ( | v k | 2 + P 1 ( | x | ) | v k | 2 ) d x + o ( 1 ) ,

Ω 1 R 3 | U r ( x ) | 2 | v k ( y ) | 2 + | V r ( x ) | 2 | u k ( y ) | 2 + 4 U r ( x ) u k ( x ) V r ( y ) v k ( y ) | x y | d x d y = O e R 2 Ω 1 u k 12 5 d x 5 6 + Ω 1 v k 12 5 d x 5 6 + o ( 1 ) ,

and

Ω 1 R 3 U r ( x ) u k ( x ) V r ( y ) v k ( y ) | x y | d x d y = O e R 2 Ω 1 u k 12 5 d x 5 6 + Ω 1 v k 12 5 d x 5 6 + o ( 1 ) .

Then, we find

o ( 1 ) = Ω 1 | u k | 2 + P 1 ( x ) u k 2 d x α 1 Ω 1 R 3 | U r ( x ) | 2 | u k ( y ) | 2 | x y | d x d y + Ω 1 | v k | 2 + P 2 ( x ) v k 2 d x α 2 Ω 1 R 3 | V r ( x ) | 2 | v k ( y ) | 2 | x y | d x d y β Ω 1 R 3 | U r ( x ) | 2 | v k ( y ) | 2 + | V r ( x ) | 2 | u k ( y ) | 2 + 4 U r ( x ) u k ( x ) V r ( y ) v k ( y ) | x y | d x d y 1 2 + O e R 2 ,

which is impossible for large k and large R.

This is a contradiction to (3.7). □

Lemma 3.5.

There is a constant C > 0, independent of k, such that

(3.12) | R ( φ , ϕ ) | C ( ( φ , ϕ ) 3 + ( φ , ϕ ) 4 ) ,

(3.13) R ( φ , ϕ ) C ( ( φ , ϕ ) 2 + ( φ , ϕ ) 3 )

and

(3.14) R ( φ , ϕ ) C ( ( φ , ϕ ) + ( φ , ϕ ) 2 ) .

Proof. By a direct computation, we have that for any (u, v), (w, ψ) ∈ E

| R ( φ , ϕ ) | = α 1 R 3 R 3 U r ( x ) φ ( x ) φ 2 ( y ) | x y | d x d y α 1 4 R 3 R 3 φ 2 ( x ) φ 2 ( y ) | x y | d x d y α 2 R 3 R 3 V r ( x ) ϕ ( x ) ϕ 2 ( y ) | x y | d x d y α 2 4 R 3 R 3 ϕ 2 ( x ) ϕ 2 ( y ) | x y | d x d y β R 3 R 3 U r ( x ) φ ( x ) ϕ 2 ( y ) | x y | d x d y β R 3 R 3 V r ( x ) ϕ ( x ) φ 2 ( y ) | x y | d x d y β 2 R 3 R 3 φ 2 ( x ) ϕ 2 ( y ) | x y | d x d y C φ P 1 3 + φ P 1 4 + ϕ P 2 3 + ϕ P 2 4 + φ P 1 ϕ P 2 2 + φ P 1 2 ϕ P 2 + φ P 1 2 ϕ P 2 2 C ( ( φ , ϕ ) 3 + ( φ , ϕ ) 4 ) ,

| R ( φ , ϕ ) , ( u , v ) | = α 1 R 3 R 3 U r ( x ) u ( x ) φ 2 ( y ) | x y | d x d y + 2 α 1 R 3 R 3 U r ( x ) φ ( x ) φ ( y ) u ( y ) | x y | d x d y + α 1 R 3 R 3 φ 2 ( x ) φ ( y ) u ( y ) | x y | d x d y + α 2 R 3 R 3 V r ( x ) v ( x ) ϕ 2 ( y ) | x y | d x d y + 2 α 2 R 3 R 3 V r ( x ) ϕ ( x ) ϕ ( y ) v ( y ) | x y | d x d y + α 2 R 3 R 3 ϕ 2 ( x ) ϕ ( y ) v ( y ) | x y | d x d y + β R 3 R 3 U r ( x ) u ( x ) ϕ 2 ( y ) | x y | d x d y + 2 β R 3 R 3 U r ( x ) φ ( x ) ϕ ( y ) v ( y ) | x y | d x d y + β R 3 R 3 V r ( x ) v ( x ) φ 2 ( y ) | x y | d x d y + 2 β R 3 R 3 V r ( x ) ϕ ( x ) φ ( y ) u ( y ) | x y | d x d y + β R 3 R 3 φ ( x ) u ( x ) ϕ 2 ( y ) | x y | d x d y + β R 3 R 3 φ 2 ( x ) ϕ ( y ) v ( y ) | x y | d x d y C ( ( φ , ϕ ) 2 + ( φ , ϕ ) 3 ) ( u , v ) ,

and similarly,

| R ( φ , ϕ ) ( u , v ) , ( w , ψ ) | C ( ( φ , ϕ ) + ( φ , ϕ ) 2 ) ( u , v ) ( w , ψ ) .

So, (3.13) and (3.14) follow. □

Lemma 3.6.

There is a small σ > 0, such that

(3.15) l k C k m 3 2 + σ + C k n 3 2 + σ .

Proof. By Lemma 3.4 in [20], we can get

j = 1 k R 3 ( P 1 ( | x | ) 1 ) U z j φ d x + R 3 ( P 2 ( | x | ) 1 ) V z j ϕ d x C k m 1 2 + σ + C k n 1 2 + σ ( φ , ϕ ) ,

and

j = 1 k R 3 R 3 | U z j ( x ) | 2 U z j ( y ) φ ( y ) | x y | d x d y R 3 R 3 | U r ( x ) | 2 U r ( y ) φ ( y ) | x y | d x d y + j = 1 k R 3 R 3 | V z j ( x ) | 2 V z j ( y ) ϕ ( y ) | x y | d x d y R 3 R 3 | V r ( x ) | 2 V r ( y ) ϕ ( y ) | x y | d x d y C k 2 e r π k ( φ , ϕ ) .

Similarly, we have

R 3 R 3 | U r ( x ) | 2 V r ( y ) ϕ ( y ) | x y | d x d y j = 1 k R 3 R 3 | U z j ( x ) | 2 V z j ( y ) ϕ ( y ) | x y | d x d y + R 3 R 3 | V r ( x ) | 2 U r ( y ) φ ( y ) | x y | d x d y j = 1 k R 3 R 3 | V z j ( x ) | 2 U z j ( y ) φ ( y ) | x y | d x d y C k 2 e r π k ( φ , ϕ )

by U = h 0 l 0 V . Furthermore, we find

k 2 e π r k k 2 e π m + 1 π δ ln k + e π n + 1 π δ ln k C k m 3 2 + σ + C k n 3 2 + σ

by m ≥ 3 and n ≥ 3. Thus, we have

l k C k m 3 2 + σ + C k n 3 2 + σ .

Proposition 3.7.

There is an integer k 0 > 0, such that for each kk 0, there is a C 1 map from S k to H P 1 , s × H P 2 , s : ( φ , ϕ ) = ( φ ( r ) , ϕ ( r ) ) , r = |x 1| satisfying (φ, ϕ) ∈ E, and

I ( φ , ϕ ) ( φ , ϕ ) , ( g , h ) = 0 , ( g , h ) E .

Moreover, there is a small σ > 0, such that

(3.16) ( φ , ϕ ) C k m 3 2 + σ + C k n 3 2 + σ .

Proof. Since l(φ, ϕ) is a bounded linear functional in E, we know that there is an l k E, such that

l ( φ , ϕ ) = l k , ( φ , ϕ ) .

Thus, finding a critical point for I(φ, ϕ) is equivalent to solving

(3.17) l k + L ( φ , ϕ ) + R ( φ , ϕ ) = 0 .

By Lemma 3.4, L is invertible. Thus, (3.17) can be rewritten as

( φ , ϕ ) = A ( φ , ϕ ) L 1 l k L 1 R ( φ , ϕ ) .

Let

S = ( φ , ϕ ) : ( φ , ϕ ) E , ( φ , ϕ ) 1 k m 3 2 + 1 k n 3 2 .

So, from Lemma 3.6,

A ( φ , ϕ ) C l k + C ( ( φ , ϕ ) 2 + ( φ , ϕ ) 3 ) C k m 3 2 + σ + C k m 3 + C k 3 ( m 3 ) 2 + C k n 3 2 + σ + C k n 3 + C k 3 ( n 3 ) 2 1 k m 3 2 + 1 k n 3 2 .

Thus, A maps S into S.

By (3.14), we have,

A ( φ 1 , ϕ 1 ) A ( φ 2 , ϕ 2 ) = L 1 R ( φ 1 , ϕ 1 ) L 1 R ( φ 2 , ϕ 2 ) C ( ( φ 1 , ϕ 1 ) + ( φ 1 , ϕ 1 ) 2 + ( φ 2 , ϕ 2 ) + ( φ 2 , ϕ 2 ) 2 ) φ 1 φ 2 1 2 ( φ 1 , ϕ 1 ) ( φ 2 , ϕ 2 ) .

Therefore, A is a contraction map from S to S. Combining this inequality with the contraction mapping theorem yields the conclusion. □

4 Proof of Theorem 1.4

Lemma 4.1.

There is a small constant σ > 0, such that

J ( U r , V r ) = k a A r m + b B r n + D k G k K e 2 r π k + O 1 k m + σ + O 1 k n + σ ,

where

A = h 0 2 R 3 W 2 d x , B = l 0 2 R 3 W 2 d x , D = h 0 2 + l 0 2 R 3 R 3 W 2 ( x ) W 2 ( y ) | x y | d x d y , G = 2 β α 1 α 2 4 β 2 α 1 α 2 R 3 R 3 W 2 ( x ) W 2 ( y ) | x y | d x d y ,

and K > 0 is a positive constant.

Proof. Using the symmetry,

J ( U r , V r ) = 1 2 R 3 | U r | 2 + | V r | 2 + P 1 ( x ) U r 2 + P 2 ( x ) V r 2 d x 1 4 R 3 R 3 α 1 | U r ( x ) | 2 | U r ( y ) | 2 + α 2 | V r ( x ) | 2 | V r ( y ) | 2 + 2 β | U r ( x ) | 2 | V r ( y ) | 2 | x y | d x d y = 1 2 R 3 ( P 1 ( | x | ) 1 ) U r 2 d x + 1 2 R 3 ( P 2 ( | x | ) 1 ) V r 2 d x + j = 1 k i = 1 k R 3 R 3 | U z j ( x ) | 2 U z j ( y ) U z i ( y ) | x y | d x d y + j = 1 k i = 1 k R 3 R 3 | V z j ( x ) | 2 V z j ( y ) V z i ( y ) | x y | d x d y 1 4 R 3 R 3 α 1 | U r ( x ) | 2 | U r ( y ) | 2 + α 2 | V r ( x ) | 2 | V r ( y ) | 2 + 2 β | U r ( x ) | 2 | V r ( y ) | 2 | x y | d x d y = 1 2 R 3 ( P 1 ( | x | ) 1 ) U r 2 d x + 1 2 R 3 ( P 2 ( | x | ) 1 ) V r 2 d x + k R 3 R 3 | U ( x ) | 2 | U ( y ) | 2 | x y | d x d y + k i = 2 k R 3 R 3 U z 1 2 ( x ) U z 1 ( y ) U z i ( y ) | x y | d x d y + k R 3 R 3 | V ( x ) | 2 | V ( y ) | 2 | x y | d x d y + k i = 2 k R 3 R 3 V z 1 2 ( x ) V z 1 ( y ) V z i ( y ) | x y | d x d y 1 4 R 3 R 3 α 1 | U r ( x ) | 2 | U r ( y ) | 2 + α 2 | V r ( x ) | 2 | V r ( y ) | 2 + 2 β | U r ( x ) | 2 | V r ( y ) | 2 | x y | d x d y .

It follows from Lemma 3.1 that

R 3 ( P 1 ( | x | ) 1 ) U r 2 d x = k Ω 1 ( P 1 ( | x | ) 1 ) U r 2 d x = k Ω 1 ( P 1 ( | x | ) 1 ) U z 1 + O e 1 2 | z 1 | π k e 1 2 | x z 1 | 2 d x = k Ω 1 ( P 1 ( | x | ) 1 ) U z 1 2 d x + k O Ω 1 | P 1 ( | x | ) 1 | e | z 1 | π k e | x z 1 | d x = k a r m R 3 U 2 d x + O 1 k m + θ .

Similarly, we have

R 3 ( P 2 ( | x | ) 1 ) V r 2 d x = k b r n R 3 V 2 d x + O 1 k n + θ .

Using Lemma 3.1 and the fact that

i = 2 k e | z i z 1 | C e 2 π r k C k m + 1 τ ,  for any  r S k ,

we obtain that if τ > 0 is small enough, for any x ∈ Ω1,

| U r ( x ) | 2 = U z 1 2 ( x ) + 2 U z 1 ( x ) i = 2 k U z i ( x ) + i = 2 k U z i ( x ) 2 = U z 1 2 ( x ) + 2 U z 1 ( x ) i = 2 k U z i ( x ) + O 1 k 2 m + σ .

Thus, we have that

R 3 R 3 | U r ( x ) | 2 | U r ( y ) | 2 | x y | d x d y = k 2 Ω 1 Ω 1 | U r ( x ) | 2 | U r ( y ) | 2 | x y | d x d y = k 2 R 3 R 3 U 2 ( x ) U 2 ( y ) | x y | d x d y + 4 k 2 i = 2 k R 3 R 3 U z 1 2 ( x ) U z 1 ( y ) U z i ( y ) | x y | d x d y + 4 k 2 i = 2 k j = 2 k R 3 R 3 U z 1 ( x ) U z i ( x ) U z 1 ( y ) U z j ( y ) | x y | d x d y + k 2 O 1 k 2 m + σ .

On the other hand, we have

i = 2 k j = 2 k R 3 R 3 U z 1 ( x ) U z i ( x ) U z 1 ( y ) U z j ( y ) | x y | d x d y = C 1 i = 2 k e | z i z 1 | j = 2 k e | z j z 1 | = C 2 e 4 π r k

and

i = 2 k R 3 R 3 U z 1 2 ( x ) U z 1 ( y ) U z i ( y ) | x y | d x d y = C 3 i = 2 k e | z i z 1 | = C 4 e 2 r π k ,

where C 3 > 0 and C 4 > 0 are two positive constants.

So,

R 3 R 3 | U r ( x ) | 2 | U r ( y ) | 2 | x y | d x d y = k 2 R 3 R 3 U 2 ( x ) U 2 ( y ) | x y | d x d y + k 2 C 5 e 2 r π k + O 1 k 2 m 2 + σ ,

where C 5 > 0 is a positive constant. Similarly,

R 3 R 3 | V r ( x ) | 2 | V r ( y ) | 2 | x y | d x d y = k 2 R 3 R 3 V 2 ( x ) V 2 ( y ) | x y | d x d y + k 2 C 6 e 2 r π k + O 1 k 2 n 2 + σ ,

and

R 3 R 3 | U r ( x ) | 2 | V r ( y ) | 2 | x y | d x d y = k 2 R 3 R 3 U 2 ( x ) V 2 ( y ) | x y | d x d y + k 2 C 7 e 2 r π k + O 1 k m + n 2 + σ ,

where C 6 > 0 and C 7 > 0 are positive constant.

As a consequence, we eventually get that

J ( U r , V r ) = k a A r m + b B r n + D k G k K e 2 r π k + O 1 k m + σ + O 1 k n + σ ,

where

A = h 0 2 R 3 W 2 d x , B = l 0 2 R 3 W 2 d x , D = h 0 2 + l 0 2 R 3 R 3 W 2 ( x ) W 2 ( y ) | x y | d x d y G = 2 β α 1 α 2 4 β 2 α 1 α 2 R 3 R 3 W 2 ( x ) W 2 ( y ) | x y | d x d y

and K > 0 is a positive constant and the proof now is complete. □

We are ready to prove Theorem 1.4. Let (φ r , ϕ r ) = (φ(r), ϕ r ) be the map obtained in Proposition 3.7. Define

F ( r ) = J ( U r + φ r , V r + ϕ r ) , r S k .

With the same argument in [21], [23], we can easily check that for k sufficiently large, if r is a critical point of F(r), then (U r + φ r , V r + ϕ r ) is a solution of (1.4).

Proof of Theorem 1.4. It follows from Lemmas 3.3 and 3.5 that

L ( φ r , ϕ r ) C ( φ r , ϕ r ) , | R ( φ r , ϕ r ) | C ( ( φ r , ϕ r ) 3 + ( φ r , ϕ r ) 4 ) .

So, Proposition 3.7 and Lemma 4.1 give

F ( r ) = J ( U r , V r ) + l ( φ r , ϕ r ) + 1 2 L ( φ r , ϕ r ) , ( φ r , ϕ r ) + R ( φ r , ϕ r ) = J ( U r , V r ) + O ( l k ( φ r , ϕ r ) + ( φ r , ϕ r ) 2 + ( φ r , ϕ r ) 3 + ( φ r , ϕ r ) 4 ) = J ( U r , V r ) + O 1 k m 3 + σ + O 1 k n 3 + σ .

We prove the theorem only for the case m = n, since the other case is similar. If m = n, then

F ( r ) = k a A + b B r m + D k G k K e 2 r π k + O 1 k m + σ .

Note that ββ k with {β k } is given in Proposition 2.3. Consider the following maximization problem

(4.1) max { F ( r ) : r S k } ,

where S k is defined in Section 1. Since the function

a A + b B r m k K e 2 r π k

has a maximum point

r ̄ k = m + 1 2 π + o ( 1 ) k ln k ,

which is an interior point of S k , it is easy to check that (4.1) is achieved by some r k , which is in the interior of S k . Thus, r k is a critical point of F(r). As a result

( U r k + φ r k , V r k + ϕ r k )

is a solution of (1.4) and the conclusion follows. □


Corresponding author: Shunneng Zhao, School of Mathematical Sciences, Zhejiang Normal University, Jinhua 321004, People’s Republic of China, E-mail:

Fashun Gao is partially supported by Natural Science Foundation of Henan (242300421382) and NSFC (11901155). Minbo Yang is partially supported by National Key Research and Development Program of China (2022YFA1005700), NSFC (12471114) and ZJNSF (LZ22A010001). Shunneng Zhao is the corresponding author who is partially supported by NSFC (12401146, 12261107) and ZJNSF (LZ22A010001).


Acknowledgments

The authors would like to thank the anonymous referee for his/her useful comments and suggestions which help to improve the presentation of the paper greatly.

  1. Research ethics: Not applicable.

  2. Informed consent: Not applicable.

  3. Author contributions: The authors are listed in alphabetical order of their surnames, and all authors have made equal contributions.

  4. Use of Large Language Models, AI and Machine Learning Tools: None declared.

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

  6. Research funding: The research of first author is partially supported by Natural Science Foundation of Henan (242300421382) and NSFC (11901155). The research of second author is partially supported by National Key Research and Development Program of China (2022YFA1005700), NSFC (12471114) and ZJNSF (LZ22A010001). The research of third author is partially supported by NSFC (12401146, 12261107) and ZJNSF (LZ22A010001).

  7. Data availability: Not applicable.

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Received: 2024-03-01
Accepted: 2024-10-29
Published Online: 2025-01-03

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

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

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