Existence of normalized solutions for the coupled elliptic system with quadratic nonlinearity

Jun Wang; Xuan Wang; Song Wei

doi:10.1515/ans-2022-0010

Article Open Access

Existence of normalized solutions for the coupled elliptic system with quadratic nonlinearity

Jun Wang , Xuan Wang and Song Wei

Published/Copyright: June 6, 2022

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From the journal Advanced Nonlinear Studies Volume 22 Issue 1

Abstract

In the present paper, we study the existence of the normalized solutions for the following coupled elliptic system with quadratic nonlinearity

− Δ u − λ 1 u = μ 1 ∣ u ∣ u + β u v in R N , − Δ v − λ 2 v = μ 2 ∣ v ∣ v + β 2 u 2 in R N ,

where u , v satisfying the additional condition

∫ R N u 2 d x = a 1 , ∫ R N v 2 d x = a 2 .

On the one hand, we prove the existence of minimizer for the system with L 2 -subcritical growth ( N ≤ 3 ). On the other hand, we prove the existence results for different ranges of the coupling parameter β > 0 with L 2 -supercritical growth ( N = 5 ). Our argument is based on the rearrangement techniques and the minimax construction.

Keywords: normalized solutions; quadratic nonlinearity; variational methods

MSC 2010: 35J61; 35J20; 35Q55; 49J40

1 Introduction

In the present paper, we study the existence of solutions ( λ 1 , λ 2 , u ˜ , v ˜ ) to the coupled elliptic equations:

(1.1) − Δ u − λ 1 u = μ 1 ∣ u ∣ u + β u v in R N , − Δ v − λ 2 v = μ 2 ∣ v ∣ v + β 2 u 2 in R N ,

satisfying the constraint

(1.2) ∫ R N u 2 d x = a 1 , ∫ R N v 2 d x = a 2 .

Here, N ≤ 3 or N = 5 , a 1 , a 2 are fixed and β > 0 . Prescribing the L 2 masses from the beginning, we call this type of solutions as normalized solutions. The frequencies λ 1 , λ 2 appearing as Lagrange multipliers are included in the unknown. The mass ‖ ⋅ ‖ L 2 often has a clear physical meaning; on this account, normalized solutions are particularly interesting from a physical point of view.

The system (1.1) has important physic meaning. That is, we consider the following generalized three-coupled system:

(1.3) i ∂ t V 1 = − Δ V 1 − μ 1 ∣ V 1 ∣ p − 2 V 1 − β V 2 ¯ V 3 , i ∂ t V 2 = − Δ V 2 − μ 2 ∣ V 2 ∣ p − 2 V 2 − β V 1 ¯ V 3 , i ∂ t V 3 = − Δ V 3 − μ 3 ∣ V 3 ∣ p − 2 V 3 − β V 1 V 2 ,

where V i ( i = 1 , 2 , 3 ) are complex valued functions of ( t , x ) ∈ R × R N , p > 2 , N ≤ 3 , μ i > 0 ( i = 1 , 2 , 3 ) , and β ∈ R . The system (1.3) is a reduced system studied in [12,13, 14,33] and related to the Raman amplification in a plasma. It is an instability phenomenon taking place when an incident laser field propagates into a plasma. This kind of model was first introduced by the previous paper [35] to describe the Raman scattering in a plasma. In [9], a modified model was derived to describe the nonlinear interaction between a laser beam and a plasma. From the physical point of view, when an incident laser field enters a plasma, it is backscattered by a Raman type process. These two waves interact to create an electronic plasma wave. The three waves combine to create a variation of the density of the ions, which have itself an influence on the three proceedings waves. The system describing this phenomenon is composed of three Schrödinger equations coupled to a wave equation and reads in a suitable dimensionless form. For a complete description of this model as well as a precise description of the physical coefficients, we refer to [9,10, 11,14] and the references therein. Recently, the previous paper [13] studied the case V 1 = V 2 . That is, the system

(1.4) i ∂ t V 1 = − Δ V 1 − μ 1 ∣ V 1 ∣ p − 2 V 1 − β V 2 ¯ V 1 , i ∂ t V 2 = − 2 Δ V 2 − 2 μ 2 ∣ V 2 ∣ p − 2 V 2 − β V 1 2 .

Correspondingly, the stationary system is given by

(1.5) − Δ u 1 + λ 1 u 1 = μ 1 ∣ u 1 ∣ p − 2 u 1 + β u 1 u 2 , − Δ u 2 + λ 2 u 2 = μ 2 ∣ u 2 ∣ p − 2 u 2 + β / 2 u 1 2 .

For the case p = 3 , λ 1 = λ 2 , and μ 2 = 1 , [13] proved the existence of multiple synchronous solutions for (1.5). On the other hand, they prove the stability and instability results for the synchronous solution and ( 0 , e i w t ψ w ) , where ψ w denotes the unique positive solution − Δ u + w u = ∣ u ∣ p − 2 u , u ∈ H 1 ( R N ) . For more general results on this direction, one can refer to the recent papers [38,40,41].

In the present paper, we study the existence of L 2 ( R N ) -normalized solutions of (1.5) with p = 3 . On the other hand, the problem under investigation can be seen from the following generalized nonlinear elliptic system:

(1.6) − Δ u 1 = λ 1 u 1 + f 1 ( u 1 ) + ∂ 1 F 1 ( u 1 , u 2 ) , − Δ u 2 = λ 2 u 2 + f 2 ( u 2 ) + ∂ 2 F 1 ( u 1 , u 2 ) , u 1 , u 2 ∈ H 1 ( R N ) , N ≥ 2 ,

under the condition (1.2). This problem is associated with models for binary mixtures of ultracold quantum gases. In the last decades, many authors investigated the case of homogeneous nonlinearities, i.e., f i ( u i ) = μ i ∣ u i ∣ p i − 2 u i , F ( u 1 , u 2 ) = β ∣ u 1 ∣ r 1 ∣ u 2 ∣ r 2 , with positive constants β , μ i , p i , r i . That is, the existence of L 2 -normalized of the system:

(1.7) − Δ u 1 = λ 1 u 1 + μ 1 ∣ u 1 ∣ p 1 − 2 u 1 + r 1 β ∣ u 1 ∣ r 1 − 2 ∣ u 2 ∣ r 2 u 1 , − Δ u 2 = λ 2 u 2 + μ 2 ∣ u 2 ∣ p 2 − 2 u 2 + r 2 β ∣ u 2 ∣ r 1 ∣ u 2 ∣ r 2 − 2 u 2 .

This kind of the system comes from mean field models for binary mixtures of Bose-Einstein condensates or for binary gases of fermion atoms in degenerate quantum states. For information about the physical background of the problem (1.7), we can refer to [1,2,3,7,8,15,17,24,26,27,29,30, 31,32,36,37,39,42,44,46] and reference therein. Let 2 ∗ = 2 N N − 2 if N ≥ 3 , and 2 ∗ = ∞ if N = 1 , 2 . For fix N ≥ 2 , p 1 , p 2 ∈ ( 2 , 2 ∗ ) and β , μ 1 , μ 2 , r 1 , r 2 , a 1 , a 2 > 0 with 2 ≤ r 1 + r 2 < 2 ∗ , [3] proved the existence of the normalized solution of (1.7). In [17], the authors proved the existence of multiple normalized solutions. Recently, many papers considered the existence of normalized solutions for the following coupled cubic elliptic system:

(1.8) − Δ u − λ 1 u = μ 1 u 3 + β u v 2 in R N , − Δ v − λ 2 v = μ 2 v 3 + β u 2 v in R N , ∫ R N u 2 d x = a 1 2 and ∫ R N v 2 d x = a 2 2 .

For example, see [4,5, 6,28] and the references therein. In 2016, Bartsch et al. [4] proved that there exist β 1 , β 2 > 0 such that if β ∈ ( 0 , β 1 ) ∪ ( β 2 , ∞ ) , then the aforementioned systems have a solution ( λ 1 , λ 2 , u , v ) in N = 3 . Later, [5] proved the existence of the normalized solutions for the focusing-repulsive case μ i > 0 , β < 0 . Nearly, [6] proved the existence of normalized solutions using bifurcation methods.

Motivated by previous works [4,5], in this paper, we consider the existence of normalized solutions for the system (1.1) with quadratic nonlinearity. On the one hand, we prove the existence of minimizer for the system with L 2 -subcritical growth ( N ≤ 3 ). On the other hand, we prove the existence results for different ranges of the coupling parameter β > 0 with L 2 -supercritical growth ( N = 5 ). Compared to the previous cubic elliptic system, we have some difficulties in studying the existence of the normalized solution of (1.1). That is, since the coupled term ∫ R N u 2 v d x in the energy functional is not symmetric, we need make different estimates to dealing with it. Moreover, this term is not positive defined and leads us to add another constraint to get the positive normalized solution (see our forthcoming paper [43]).

We first consider the L 2 -subcritical case N ≤ 3 . Then, the following results hold.

Theorem 1.1

For N ≤ 3 , assume that β > 0 , μ 1 , μ 2 , a 1 , a 2 > 0 are given. Then, there exists a solution ( λ 1 , λ 2 , u ˜ , v ˜ ) to equations (1.1) and (1.2) with λ 1 , λ 2 < 0 , and u ˜ , v ˜ ∈ C 2 ( R N ) are positive, radially symmetric, and decrease with r = ∣ x ∣ .

Next we consider the L 2 -supercritical case N = 5 . Obviously, solutions of (1.1) and (1.2) will be obtained as critical points of the energy functional J : H 1 ( R N ) × H 1 ( R N ) ↦ R defined by

(1.9) J ( u , v ) ≔ 1 2 ∫ R N ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − μ 1 3 ∫ R N ∣ u ∣ 3 d x − μ 2 3 ∫ R N ∣ v ∣ 3 d x − β 2 ∫ R N u 2 v d x

constrained on the L 2 -spheres in H 1 ( R N ) × H 1 ( R N ) . Set

(1.10) S ( a 1 , a 2 ) ≔ ( u , v ) ∈ H 1 ( R N ) × H 1 ( R N ) : ∫ R N u 2 d x = a 1 , ∫ R N v 2 d x = a 2 .

To state our results, we shall introduce the minimization problem:

(1.11) m ( a 1 , a 2 ) ≔ inf ( u , v ) ∈ S ( a 1 , a 2 ) J ( u , v ) .

It is natural that the minimizers of the problem (1.11) are critical points of J ∣ S ( a , b ) . Then, we have the following results.

Theorem 1.2

Let a 1 , a 2 , μ 1 , μ 2 > 0 be fixed, N = 5 and let β 1 > 0 be defined by

(1.12) max 1 a 1 μ 1 4 , 1 a 2 μ 2 4 = 1 a 1 ( μ 1 + β 1 ) 4 + 1 a 2 ( μ 2 + β 1 ) 4 .

If 0 < β < β 1 , then (1.1)–(1.2) has a solution ( λ ˜ 1 , λ ˜ 2 , u ˜ , v ˜ ) such that λ ˜ 1 , λ ˜ 2 < 0 , and u ˜ and v ˜ are both positive and radial.

Finally, we consider the existence normalized solutions for large β > 0 . To state the main results, we introduce a Pohozaev-type constraint as follows:

(1.13) V ≔ { ( u , v ) ∈ H 1 ( R 5 ) × H 1 ( R 5 ) : G ( u , v ) = 0 } ,

where

(1.14) G ( u , v ) ≔ ∫ R 5 ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − 5 6 ∫ R 5 ( μ 1 ∣ u ∣ 3 d x + μ 2 ∣ v ∣ 3 ) d x − 5 4 β ∫ R 5 u 2 v d x .

Then, we have the following results.

Theorem 1.3

For N = 5 , let a 1 , a 2 , μ 1 , μ 2 > 0 be fixed, and let β 2 > 0 be defined by

(1.15) ( a 1 + a 2 ) 5 12 μ 1 a 1 3 2 + 3 2 β 2 a 1 a 2 1 2 + μ 2 a 2 3 2 4 = min 1 a 1 μ 1 4 , 1 a 2 μ 2 4 .

If β > β 2 , then (1.1) and (1.2) has a solution ( λ ¯ 1 , λ ¯ 2 , u ¯ , v ¯ ) such that λ ¯ 1 , λ ¯ 2 < 0 , and u ¯ and v ¯ are both positive and radial. Moreover, ( λ ¯ 1 , λ ¯ 2 , u ¯ , v ¯ ) is a ground state solution in the sense that

(1.16) J ( u ¯ , v ¯ ) = inf { J ( u , v ) : ( u , v ) ∈ V } = inf { J ( u , v ) : ( u , v ) is a solution of ( 1.1 ) – ( 1.2 ) for some λ 1 , λ 2 }

holds.

Remark 1.4

In Theorem 1.3, we cannot use the Rayleigh-type quotient to describe the ground state of (1.1) as in [4]. This is because we cannot find the unique maximum of the functional J if the term ∫ R N u 2 v d x is negative.
It is an interesting issue to prove the existence of normalized solutions of (1.1)–(1.2) when β < 0 .
In Theorems 1.1–1.3, we can only study the existence of the normalized solutions of (1.1) and (1.2) if N ≤ 3 or N = 4 . For the L 2 -critical case N = 4 , the additional assumptions and different arguments are needed. We give the result in the forthcoming paper [43].

This paper is organized as follows. In Section 2, we establish some basic results. The proof of Theorem 1.1 is given in Section 3. Sections 4 and 5 are devoted to giving the proof of Theorems 1.2 and 1.3.

2 Preliminaries

Throughout this paper, we shall use the following notations:

H 1 ( R N ) is the usual Sobolev space endowed with the norm ‖ u ‖ 2 = ∫ R N ( ∣ ∇ u ∣ 2 + ∣ u ∣ 2 ) d x .
H rad 1 ( R N ) = { u ∈ H 1 ( R N ) : u is a radial sysmetry function } .
L p ( R N ) is the usual Lebesgue space with norm ∣ u ∣ p p = ∫ R N ∣ u ∣ p d x .

In this section, we recall some basic conclusions. First, we have the following Gagliardo-Nirenberg inequality (see [3]):

(2.1) ∣ u ∣ p ≤ p 2 ∣ Q ∣ 2 p − 2 1 p ∣ ∇ u ∣ 2 N ( p − 2 ) / ( 2 p ) ∣ u ∣ 2 1 − N ( p − 2 ) / ( 2 p ) ,

where p ∈ [ 2 , 2 ∗ ) , and the function Q is the unique ground state solution to

− N ( p − 2 ) 4 Δ Q + 1 + p − 2 4 ( 2 − N ) Q = ∣ Q ∣ p − 2 Q , x ∈ R N .

Next we give some rearrangement inequalities results for system (1.1). For N ≤ 3 , let u be a measurable function on R N . Let mes ( A ) be the N -dimensional Lebesgue measure of a Lebesgue measurable set A ⊂ R N . It is said to vanish at infinity if mes ( x ∈ R N : ∣ u ( x ) ∣ > t ) < ∞ for every t > 0 . In view of two measurable functions u , v , which vanish at infinity with t > 0 , we define A ∗ ( u , v , t ) ≔ { x ∈ R N : ∣ x ∣ < r } , where r > 0 is chosen such that

mes ( x ∈ R N : ∣ x ∣ < r ) = mes ( x ∈ R N : ∣ u ( x ) ∣ > t ) + mes ( x ∈ R N : ∣ v ( x ) ∣ > t )

and { u , v } ∗ defined by

{ u , v } ∗ ( x ) ≔ ∫ 0 ∞ χ A ∗ ( u , v , t ) ( x ) d t ,

where χ A is a characteristic function of the set A ⊂ R N . Then, we have the following properties for the aforementioned rearrangement. For the details of the proof, one can refer to [18, Lemma A.1].

Lemma 2.1

The function { u , v } ∗ is radially symmetric, nonincreasing, and lower semi-continuous. Moreover, for each t > 0 , there holds { x ⊂ R N : { u , v } ∗ > t } = A ∗ ( u , v , t ) .
Let Φ : [ 0 , ∞ ) → [ 0 , ∞ ) be nondecreasing, lower semi-continuous, continuous at 0 and Φ ( 0 ) = 0 . Then { Φ ( u ) , Φ ( v ) } ∗ = Φ ( { u , v } ) ∗ .
∣ { u , v } ∗ ∣ p p = ∣ u ∣ p p + ∣ v ∣ p p for 1 ≤ p < ∞ .
If u , v ∈ H 1 ( R N ) , then { u , v } ∗ ∈ H 1 ( R N ) and ∣ ∇ { u , v } ∗ ∣ 2 2 ≤ ∣ ∇ u ∣ 2 2 + ∣ ∇ v ∣ 2 2 . In addition, if u , v ∈ ( H 1 ( R N ) ∩ C 1 ( R N ) ) \ { 0 } are radially symmetric, positive, and nonincreasing, then
∫ R N ∣ ∇ { u , v } ∗ ∣ 2 d x < ∫ R N ∣ ∇ u ∣ 2 d x + ∫ R N ∣ ∇ v ∣ 2 d x .
Let u 1 , u 2 , v 1 , v 2 ≥ 0 be Borel measurable functions, which vanish at infinity, then
∫ R N ( u 1 u 2 + v 1 v 2 ) d x ≤ ∫ R N { u 1 , v 1 } ∗ { u 2 , v 2 } ∗ d x .

To this end, we introduce some facts concerning the following scalar equation:

(2.2) − Δ w + w = ∣ w ∣ w , w > 0 , in R 5 , w ( 0 ) = max w and w ∈ H 1 ( R 5 ) .

It is well known that (2.2) has a unique positive solution w 0 . Moreover, w 0 is radial symmetric and decays at infinity. Set

(2.3) C 0 ≔ ∫ R 5 w 0 2 d x and C 1 ≔ ∫ R 5 ∣ w 0 ∣ 3 d x .

For a , μ ∈ R fixed, let us search for ( λ , w ) ∈ R × H 1 ( R 5 ) , with λ < 0 in R , solving

(2.4) − Δ w − λ w = μ ∣ w ∣ w , in R 5 , w ( 0 ) = max w and ∫ R 5 w 2 = a .

Solution w of (2.4) can be found as critical points of I μ : H 1 ( R 5 ) ↦ R , defined by

(2.5) I μ ( w ) ≔ 1 2 ∫ R 5 ∣ ∇ w ∣ 2 d x − μ 3 ∫ R 5 ∣ w ∣ 3 d x ,

constrained on the L 2 -sphere T a = { u ∈ H 1 ( R N ) : ∫ R N u 2 = a } , and λ appears as Lagrange multipliers. It is easy to see that this solution can be obtained by the scaling of solutions of (2.2). Let

(2.6) P ( a , μ ) ≔ w ∈ T a : ∫ R 5 ∣ ∇ w ∣ 2 = 5 μ 6 ∫ R 5 w 3 .

Then, we have the following results for P ( a , μ ) .

Lemma 2.2

If w is a solution of (2.4), then w ∈ P ( a , μ ) . Furthermore, the positive solution w of (2.4) minimizes I μ on P ( a , μ ) .

Proof

The proof is similar to [19, Lemmas 2.7, 2.10] (also see [17]). We omit the details here.□

Proposition 2.3

Problem (2.4) has a unique positive solution ( λ a , μ , w a , μ ) defined by

(2.7) λ a , μ ( x ) ≔ − C 0 2 μ 4 a 2 and w a , μ ( x ) ≔ C 0 2 μ 5 a 2 w 0 C 0 μ 2 a x .

The function w a , μ satisfies

(2.8) ∫ R 5 ∣ ∇ w a , μ ∣ 2 = 5 C 0 C 1 6 μ 4 a

and

(2.9) ∫ R 5 ∣ w a , μ ∣ 3 = C 0 C 1 μ 5 a .

(2.10) l ( a , μ ) ≔ inf u ∈ P ( a , μ ) I μ ( u ) = I μ ( w a , μ ) = C 0 C 1 12 μ 4 a .

The value l ( a , μ ) is called least energy level of problem (2.4).

Proof

Since w 0 is the solution of (2.2), it is straightforward to check that w a , μ is a solution of (2.4) for λ = λ a , μ < 0 (here w a , μ is given in [4] in the same way). By [21], it is the only positive solution. To obtain (2.8) and (2.9), we can use the explicit expression of w a , μ . With a change of variables,

(2.11) ∫ R 5 ∣ ∇ w a , μ ∣ 2 d x = C 0 4 μ 10 a 4 ⋅ C 0 2 μ 4 a 2 ∫ R 5 ∇ w 0 C 0 μ 2 a x 2 d x = C 0 6 μ 14 a 6 ⋅ C 0 μ 2 a − 5 ∫ R 5 ∣ w 0 ( x ) ∣ 3 d x = 5 C 0 C 1 6 μ 4 a ,

(2.12) ∫ R 5 ∣ w a , μ ∣ 3 d x = C 0 6 μ 15 a 6 ⋅ C 0 μ 2 a − 5 ∫ R 5 ∣ w 0 ( x ) ∣ 3 d x = C 0 C 1 μ 5 a ,

where the last equality follows by Lemma 2.2 with a = C 0 and μ = 1 . It is not difficult to check (2.9) and the least energy level of I μ on P ( a , μ ) .□

3 Proof of Theorem 1.1

In this section, we are devoted to giving the proof of Theorem 1.1. To accomplish this, we need the following main results for the least energy level of (1.11).

Lemma 3.1

Assume N ≤ 3 and α > 0 . Then, we have the following results.

For a , b ≥ 0 , one has that m ( a , b ) is well defined, where m ( a , b ) is given in (1.11).
If a b > 0 , then m ( a , b ) < 0 .
If a b = 0 , then m ( a , b ) = 0 .
If ( a n , b n ) → ( a , b ) as n → ∞ with a , b > 0 , then m ( a n , b n ) → m ( a , b ) as n → ∞ .
If 0 ≤ c ≤ a , 0 ≤ d ≤ b , then m ( a , b ) ≤ m ( c , d ) + m ( a − c , b − d ) .

Proof

( i ) For any a , b ≥ 0 and ( u , v ) ∈ S ( a , b ) , we infer from Hölder’s inequality, Young’s inequality, and inequality (2.1) that

(3.1) J ( u , v ) = 1 2 ∫ R N ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − μ 1 3 ∫ R N ∣ u ∣ 3 d x − μ 2 3 ∫ R N ∣ v ∣ 3 d x − β 2 ∫ R N u 2 v d x ≥ 1 2 ∫ R N ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − μ 1 3 ∫ R N ∣ u ∣ 3 d x − μ 2 3 ∫ R N ∣ v ∣ 3 d x − β 2 ∫ R N ∣ u ∣ 3 d x 2 3 ∫ R N ∣ v ∣ 3 d x 1 3 ≥ 1 2 ∫ R N ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − μ 1 3 ∫ R N ∣ u ∣ 3 d x − μ 2 3 ∫ R N ∣ v ∣ 3 d x − β 3 ∫ R N ∣ u ∣ 3 d x − β 6 ∫ R N ∣ v ∣ 3 d x = 1 2 ∫ R N ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − 1 3 ( μ 1 + β ) ∫ R N ∣ u ∣ 3 d x − 1 3 μ 2 + β 2 ∫ R N ∣ v ∣ 3 d x ≥ 1 2 ( ∣ ∇ u ∣ 2 2 + ∣ ∇ v ∣ 2 2 ) − ( μ 1 + β ) a 6 − N 4 2 ∣ Q ∣ 2 ∣ ∇ u ∣ 2 N 2 − μ 2 + β 2 b 6 − N 4 2 ∣ Q ∣ 2 ∣ ∇ v ∣ 2 N 2 .

We infer from N ≤ 3 that J is bounded from below on S ( a , b ) . That is, m ( a , b ) is well defined.

( i i ) Given u , v ∈ S ( a , b ) with u > 0 , v > 0 , set u t ( x ) = t u t 2 N x , v t ( x ) = t v t 2 N x for t > 0 . Then, we have ∣ u t ∣ 2 2 = ∣ u ∣ 2 2 = a and ∣ v t ∣ 2 2 = ∣ v ∣ 2 2 = b . Moreover, a direct computation shows that

(3.2) J ( u t , v t ) = 1 2 ∫ R N ( ∣ ∇ u t ∣ 2 + ∣ ∇ v t ∣ 2 ) d x − μ 1 3 ∫ R N ∣ u t ∣ 3 d x − μ 2 3 ∫ R N ∣ v t ∣ 3 d x − β 2 ∫ R N u t 2 v t d x = t 4 N 2 ∫ R N ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − μ 1 3 t ∫ R N ∣ u ∣ 3 d x − μ 2 3 t ∫ R N ∣ v ∣ 3 d x − β 2 t ∫ R N u 2 v d x .

Then, one sees J ( u t , v t ) < 0 for t > 0 small enough. By the definition of m ( a , b ) , one gets that m ( a , b ) < 0 .

( i i i ) If a = b = 0 , then the fact m ( a , b ) = 0 is trivial. In the following, we only prove the case a > 0 and b = 0 . (The proof of the case a = 0 , b > 0 can be obtained in a similar way.) Given ( u , 0 ) ∈ S ( a , 0 ) , set u t ( x ) = t u t 2 N x for t > 0 , then equality (3.2) implies that

(3.3) 0 ≤ m ( a , 0 ) = t 4 N 2 ∫ R N ∣ ∇ u ∣ 2 d x − μ 1 3 t ∫ R N ∣ u ∣ 3 d x ≤ 0 ,

since t can be arbitrarily small. One has m ( a , 0 ) = 0 .

( i v ) Without the loss of generality, assume that a n , b n > 0 . By definition of m ( a n , b n ) , there exists for any 0 < ε < 1 , ( u n , v n ) ∈ S ( a n , b n ) such that

(3.4) J ( u n , v n ) ≤ m ( a n , b n ) + ε .

Noting that 0 < ε < 1 and m ( a n , b n ) < 0 . We infer from ( i ) , the inequalities (3.1) and (3.4) that J ( u n , v n ) is bounded. Then, there exists C > 0 such that

(3.5) ∣ ∇ u n ∣ 2 ≤ C , ∣ ∇ v n ∣ 2 ≤ C , ∣ u n ∣ 3 ≤ C , ∣ v n ∣ 3 ≤ C , ∫ R N u n 2 v n d x ≤ C .

Set

w n ≔ u n ∣ u n ∣ 2 a 1 2 and z n ≔ v n ∣ v n ∣ 2 b 1 2 .

Then, ( w n , z n ) ∈ S ( a , b ) . We infer from (3.5) that

(3.6) m ( a , b ) ≤ J ( w n , z n ) = 1 2 ∫ R N ( ∣ ∇ w n ∣ 2 + ∣ ∇ z n ∣ 2 ) d x − μ 1 3 ∫ R N ∣ w n ∣ 3 d x − μ 2 3 ∫ R N ∣ z n ∣ 3 d x − β 2 ∫ R N w n 2 z n d x = 1 2 ∫ R N a a n ∣ ∇ u n ∣ 2 + b b n ∣ ∇ v n ∣ 2 d x − μ 1 3 a a n 3 2 ∫ R N ∣ u n ∣ 3 d x − μ 2 3 b b n 3 2 ∫ R N ∣ z n ∣ 3 d x − β 2 a b b n a n b n ∫ R N u n 2 v n d x = J ( u n , v n ) + 1 2 a a n − 1 ∫ R N ∣ ∇ u n ∣ 2 d x + 1 2 b b n − 1 ∫ R N ∇ ∣ v n ∣ 2 d x − μ 1 3 a a n 3 2 − 1 ∫ R N ∣ u n ∣ 3 d x − μ 2 3 b b n 3 2 − 1 ∫ R N ∣ v n ∣ 3 d x − β 2 a b b n a n b n − 1 ∫ R N u n 2 v n d x = J ( u n , v n ) + o ( 1 ) ,

where o ( 1 ) → 0 as n → ∞ . Hence, we deduce from (3.4) and (3.6) that

m ( a , b ) ≤ m ( a n , b n ) + ε + o ( 1 ) .

By using the similar arguments, one obtains

m ( a n , b n ) ≤ m ( a , b ) + ε + o ( 1 ) .

Hence, the desired results follow from the fact that ε > 0 is arbitrary small.

( v ) For any ε > 0 , there exist ( u 1 , v 1 ) ∈ S ( c , d ) and ( u 2 , v 2 ) ∈ S ( a − c , b − d ) such that

(3.7) J ( u 1 , v 1 ) ≤ m ( c , d ) + ε 2 , J ( u 2 , v 2 ) ≤ m ( a − c , b − d ) + ε 2 .

Letting w 1 ≔ { u 1 , u 2 } ∗ and w 2 ≔ { v 1 , v 2 } ∗ . According to Lemma 2.1 ( i i i ) – ( i v ) , one obtains

∣ { u 1 , u 2 } ∗ ∣ 2 2 = ∣ u 1 ∣ 2 2 + ∣ u 2 ∣ 2 2 = c + ( a − c ) = a and ∣ { v 1 , v 2 } ∗ ∣ 2 2 = ∣ v 1 ∣ 2 2 + ∣ v 2 ∣ 2 2 = d + ( b − d ) = b .

That is, ( w 1 , w 2 ) ∈ S ( a , b ) . We infer from the definition of m ( a , b ) and Lemma 2.1 ( i i ) , ( i v ) , and ( v ) that

m ( a , b ) ≤ J ( w 1 , w 2 ) = 1 2 ∫ R N ( ∣ ∇ w 1 ∣ 2 + ∣ ∇ w 2 ∣ 2 ) d x − μ 1 3 ∫ R N ∣ w 1 ∣ 3 d x − μ 2 3 ∫ R N ∣ w 2 ∣ 3 d x − β 2 ∫ R N w 1 2 w 2 d x = 1 2 ∫ R N ( ∣ ∇ { u 1 , u 2 } ∗ ∣ 2 + ∣ ∇ { v 1 , v 2 } ∗ ∣ 2 ) d x − μ 1 3 ∫ R N ∣ { u 1 , u 2 } ∗ ∣ 3 d x − μ 2 3 ∫ R N ∣ { v 1 , v 2 } ∗ ∣ 3 d x − β 2 ∫ R N { u 1 2 , u 2 2 } ∗ { v 1 , v 2 } ∗ d x ≤ 1 2 ∫ R N ( ∣ ∇ u 1 ∣ 2 + ∣ ∇ v 1 ∣ 2 ) d x − μ 1 3 ∫ R N ∣ u 1 ∣ 3 d x − μ 2 3 ∫ R N ∣ v 1 ∣ 3 d x − β 2 ∫ R N u 1 2 v 1 d x + 1 2 ∫ R N ( ∣ ∇ u 2 ∣ 2 + ∣ ∇ v 2 ∣ 2 ) d x − μ 1 3 ∫ R N ∣ u 2 ∣ 3 d x − μ 2 3 ∫ R N ∣ v 2 ∣ 3 d x − β 2 ∫ R N u 2 2 v 2 d x = J ( u 1 , v 1 ) + J ( u 2 , v 2 ) ≤ m ( c , d ) + m ( a − c , b − d ) + ε .

In light of the arbitrariness of ε , we deduce that

m ( a , b ) ≤ m ( c , d ) + m ( a − c , b − d ) .

This finishes the proof.□

The next lemma states the Brézis-Lieb type results. For the proof, one can refer to [22, Lemma 3.2].

Lemma 3.2

Assume ( u n , v n ) ⇀ ( u , v ) in H 1 ( R N ) × H 1 ( R N ) for N ≤ 3 . Then

∫ R N ( u n 2 v n − ( u n − u ) 2 ( v n − v ) ) d x = ∫ R N u 2 v d x + o ( 1 ) ,

where o ( 1 ) → 0 as n → ∞ .

In the next lemma, we study the compactness of the minimizing sequences of (1.11).

Lemma 3.3

Every minimizing sequence for problem (1.11) is, up to translation, strongly convergent in L p ( R N ) × L p ( R N ) with N ≤ 3 and p ∈ ( 2 , 2 ∗ ) .

Proof

Let { ( u n , v n ) } ∈ S ( a , b ) be a minimizing sequence for m ( a , b ) . We infer from inequality (3.1) and m ( a , b ) < 0 that { ( u n , v n ) } is bounded in H 1 ( R N ) × H 1 ( R N ) . Assume that

sup y ∈ R N ∫ B ( y , 1 ) ( ∣ u n ∣ 2 + ∣ v n ∣ 2 ) d x → 0 , n → ∞ .

Then, we have u n → 0 and v n → 0 in L 3 ( R N ) (see [25, Lemma I.1]). From Hölder’s inequality, one sees that

J ( u n , v n ) + μ 1 3 ∫ R N ∣ u n ∣ 3 d x + μ 2 3 ∫ R N ∣ v n ∣ 3 d x + β 2 ∫ R N ∣ u n ∣ 3 d x 2 3 ∫ R N ∣ v n ∣ 3 d x 1 3 = 1 2 ∫ R N ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x − β 2 ∫ R 3 u n 2 v n d x + β 2 ∫ R N ∣ u n ∣ 3 d x 2 3 ∫ R N ∣ v n ∣ 3 d x 1 3 ≥ 1 2 ∫ R N ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x ≥ 0 .

One has m ( a , b ) ≥ 0 as n → ∞ . This contradicts the fact that m ( a , b ) < 0 due to Lemma 3.1 ( i i ) . Consequently, for α > 0 , there exists a sequence { y n } ⊂ R N such that

∫ B ( y n , 1 ) ( ∣ u n ∣ 2 + ∣ v n ∣ 2 ) d x ≥ α > 0 .

Hence, we have ( u , v ) ∈ H 1 ( R N ) × H 1 ( R N ) and ( u , v ) ≠ ( 0 , 0 ) such that ( u n ( ⋅ − y n ) , v n ( ⋅ − y n ) ) ⇀ ( u , v ) in H 1 ( R N ) × H 1 ( R N ) . We define

φ n ( ⋅ ) ≔ u n ( ⋅ ) − u ( ⋅ + y n ) and ψ n ( ⋅ ) ≔ v n ( ⋅ ) − v ( ⋅ + y n ) .

Thus, it suffices to show that φ n ( ⋅ ) → 0 , ψ n ( ⋅ ) → 0 in L p ( R N ) for all p ∈ ( 2 , 2 ∗ ) . Suppose by contradiction that there exists p ∈ ( 2 , 2 ∗ ) such that ( φ n , ψ n ) ↛ ( 0 , 0 ) in L p ( R N ) × L p ( R N ) . By using [25, Lemma I.1] again, we know that there exists a sequence { z n } ∈ R N such that

( φ n ( ⋅ − z n ) , ψ n ( ⋅ − z n ) ) ⇀ ( φ , ψ ) ≠ ( 0 , 0 ) in H 1 ( R N ) × H 1 ( R N ) .

Now, we infer from the translations invariance, the Brézis-Lieb Lemma (see [45, Lemma 1.32]) and Lemma 3.2 that

(3.8) J ( u n , v n ) = J ( u n ( ⋅ − y n ) , v n ( ⋅ − y n ) ) = J ( u n ( ⋅ − y n ) − u , v n ( ⋅ − y n ) − v ) + J ( u , v ) + o ( 1 ) = J ( φ n ( ⋅ − y n ) , ψ n ( ⋅ − y n ) ) + J ( u , v ) + o ( 1 ) = J ( φ n ( ⋅ − z n ) , ψ n ( ⋅ − z n ) ) + J ( u , v ) + o ( 1 ) = J ( φ n ( ⋅ − z n ) − φ , ψ n ( ⋅ − z n ) − ψ ) + J ( φ , ψ ) + J ( u , v ) + o ( 1 )

and

(3.9) ∣ u n ( ⋅ − y n ) ∣ 2 = ∣ φ n ( ⋅ − y n ) ∣ 2 2 + ∣ u ∣ 2 2 + o ( 1 ) = ∣ φ n ( ⋅ − z n ) ∣ 2 2 + ∣ u ∣ 2 2 + o ( 1 ) = ∣ φ n ( ⋅ − y n ) − φ ∣ 2 2 + ∣ φ ∣ 2 2 + ∣ u ∣ 2 2 + o ( 1 )

and

(3.10) ∣ v n ( ⋅ − y n ) ∣ 2 = ∣ v n ( ⋅ − y n ) ∣ 2 2 + ∣ v ∣ 2 2 + o ( 1 ) = ∣ ψ n ( ⋅ − z n ) ∣ 2 2 + ∣ v ∣ 2 2 + o ( 1 ) = ∣ ψ n ( ⋅ − y n ) − ψ ∣ 2 2 + ∣ ψ ∣ 2 2 + ∣ v ∣ 2 2 + o ( 1 ) ,

where and subsequently o ( 1 ) → 0 as n → ∞ . Thus,

(3.11) ∣ φ n ( ⋅ − z n ) − φ ∣ 2 2 = ∣ u n ( ⋅ − y n ) ∣ 2 2 − ∣ φ ∣ 2 2 − ∣ u ∣ 2 2 + o ( 1 ) = a − ∣ φ ∣ 2 2 − ∣ u ∣ 2 2 + o ( 1 ) = c + o ( 1 )

and

(3.12) ∣ ψ n ( ⋅ − z n ) − ψ ∣ 2 2 = ∣ v n ( ⋅ − y n ) ∣ 2 2 − ∣ ψ ∣ 2 2 − ∣ v ∣ 2 2 + o ( 1 ) = b − ∣ ψ ∣ 2 2 − ∣ v ∣ 2 2 + o ( 1 ) = d + o ( 1 ) ,

where c ≔ a − ∣ φ ∣ 2 2 − ∣ u ∣ 2 2 ≥ 0 , d ≔ b − ∣ ψ ∣ 2 2 − ∣ v ∣ 2 2 ≥ 0 . Hence, ( φ n ( ⋅ − z n ) − φ , ψ n ( ⋅ − z n ) − ψ ) ∈ S ( c , d ) . One infers from (3.9)–(3.12) and Lemma 3.1 ( i i i ) that

(3.13) m ( c , d ) ≤ J ( φ n ( ⋅ − z n ) − φ , ψ n ( ⋅ − z n ) − ψ ) .

Recalling that J ( u n , v n ) → m ( a , b ) , one obtains

(3.14) m ( a , b ) ≥ m ( c , d ) + J ( φ , ψ ) + J ( u , v ) .

If J ( φ , ψ ) > m ( ∣ φ ∣ 2 2 , ∣ ψ ∣ 2 2 ) or J ( u , v ) > m ( ∣ u ∣ 2 2 , ∣ v ∣ 2 2 ) , then we infer from the estimate (3.14) and Lemma 3.1 ( v ) that

(3.15) m ( a , b ) ≥ m ( c , d ) + J ( φ , ψ ) + J ( u , v ) > m ( c , d ) + m ( ∣ φ ∣ 2 2 , ∣ ψ ∣ 2 2 ) + m ( ∣ u ∣ 2 2 , ∣ v ∣ 2 2 ) ≥ m ( a , b ) ,

which is incompatible. Hence, we have J ( φ , ψ ) = m ( ∣ φ ∣ 2 2 , ∣ ψ ∣ 2 2 ) and J ( u , v ) = m ( ∣ u ∣ 2 2 , ∣ v ∣ 2 2 ) . Since J ( ∣ u ∣ , ∣ v ∣ ) ≤ J ( u , v ) , without the loss of generality, we may assume that u , v , φ , ψ ≥ 0 . Since ( u , v ) ≠ ( 0 , 0 ) is a solution to system (1.1) with some ( λ 1 , λ 2 ) , by the elliptic regularity theory and the maximum principle, we infer that u , v ∈ C 2 ( R N ) and v > 0 . Similarly, we have φ , ψ ∈ C 2 ( R N ) and ψ > 0 . We denote by u ˜ , v ˜ , φ ˜ , ψ ˜ the classical Schwarz symmetric-decreasing rearrangement of u , v , φ , ψ . By the properties of the Schwarz rearrangement, one has

∫ R N u ˜ 2 d x = ∣ u ∣ 2 2 , ∫ R N v ˜ 2 d x = ∣ v ∣ 2 2 , ∫ R N φ ˜ 2 d x = ∣ φ ∣ 2 2 , ∫ R N ψ ˜ 2 d x = ∣ ψ ∣ 2 2 .

Thus, we know that

m ( ∣ u ∣ 2 2 , ∣ v ∣ 2 2 ) ≤ J ( u ˜ , v ˜ ) ≤ J ( u , v ) , m ( ∣ φ ∣ 2 2 , ∣ ψ ∣ 2 2 ) ≤ J ( φ ˜ , ψ ˜ ) ≤ J ( φ , ψ ) .

This implies that

J ( u ˜ , v ˜ ) = m ( ∣ u ∣ 2 2 , ∣ v ∣ 2 2 ) , J ( φ ˜ , ψ ˜ ) = m ( ∣ φ ∣ 2 2 , ∣ ψ ∣ 2 2 ) .

In view of Lemma 2.1 ( i i ) , ( i v ) , ( v ) and [23, Lemma 3.4] (also see [20]), we have that

∫ R N ( ∣ ∇ { u ˜ , φ ˜ } ∗ ∣ 2 ) d x < ∫ R N ( ∣ ∇ u ˜ ∣ 2 + ∣ ∇ φ ˜ ∣ 2 ) d x ≤ ∫ R N ( ∣ ∇ u ∣ 2 + ∣ ∇ φ ∣ 2 ) d x , ∫ R N ( ∣ ∇ { v ˜ , ψ ˜ } ∗ ∣ 2 ) d x < ∫ R N ( ∣ ∇ v ˜ ∣ 2 + ∣ ∇ ψ ˜ ∣ 2 ) d x ≤ ∫ R N ( ∣ ∇ v ∣ 2 + ∣ ∇ ψ ∣ 2 ) d x

and

∫ R N ( { u ˜ , φ ˜ } ∗ ) 2 { v ˜ , ψ ˜ } ∗ d x = ∫ R N { u ˜ 2 , φ ˜ 2 } ∗ { v ˜ , ψ ˜ } ∗ d x ≥ ∫ R N ( u ˜ 2 v ˜ + φ ˜ 2 ψ ˜ ) d x = ∫ R N ( u 2 ˜ v ˜ + φ 2 ˜ ψ ˜ ) d x ≥ ∫ R N ( u 2 v + φ 2 ψ ) d x .

Thus,

(3.16) J ( u , v ) + J ( φ , ψ ) > J ( { u ˜ , φ ˜ } ∗ , { v ˜ , ψ ˜ } ∗ ) .

Moreover, we infer from Lemma 2.1 ( i i i ) that

(3.17) ∫ R N ( { u ˜ , φ ˜ } ∗ ) 2 d x = ∫ R N ( u ˜ 2 , φ ˜ 2 ) d x = ∫ R N ( u 2 + φ 2 ) d x , ∫ R N ( { v ˜ , ψ ˜ } ∗ ) 2 d x = ∫ R N ( v ˜ 2 , ψ ˜ 2 ) d x = ∫ R N ( v 2 + ψ 2 ) d x .

From (3.15)–(3.17) and Lemma 3.1 ( v ) , one deduces that

(3.18) m ( a , b ) ≥ m ( c , d ) + J ( φ , ψ ) + J ( u , v ) > m ( c , d ) + J ( { u ˜ , φ ˜ } ∗ , { v ˜ , ψ ˜ } ∗ ) ≥ m ( c , d ) + m ( a − c , b − d ) ≥ m ( a , b ) .

This is a contradiction.□

Now we are ready to give the proof of Theorem 1.1.

Proof of Theorem 1.1

Let { ( u n , v n ) } be a minimizing sequence for the functional J on S ( a 1 , a 2 ) . In consideration of inequality (3.1) and Lemma 3.3, we may assume that there exist { y n } ∈ R N and ( u , v ) ∈ H 1 ( R N ) × H 1 ( R N ) such that

(3.19) ( u n ( ⋅ − y n ) , v n ( ⋅ − y n ) ) ⇀ ( u , v ) in H 1 ( R N ) × H 1 ( R N ) , ( u n ( ⋅ − y n ) , v n ( ⋅ − y n ) ) → ( u , v ) in L 3 ( R N ) × L 3 ( R N )

for n → ∞ . For simplify the notation, we denote α n = u n ( ⋅ − y n ) and β n = v n ( ⋅ − y n ) . Then,

(3.20) ∫ R N ( α n 2 β n − u 2 v ) d x = ∫ R N ( α n 2 β n − α n 2 v + α n 2 v − u 2 v ) d x ≤ ∫ R N ( ∣ α n 2 β n − α n 2 v ∣ + ∣ α n 2 v − u 2 v ∣ ) d x = ∫ R N α n 2 ∣ β n − v ∣ d x + ∫ R N ∣ ( α n + u ) ( α n − u ) v ∣ d x ≤ ∫ R N ∣ α n ∣ 3 d x 2 3 ∫ R N ∣ β n − v ∣ 3 d x 1 3 + ∫ R N ∣ ( α n + u ) ( α n − u ) ∣ 3 2 d x 2 3 ∫ R N ∣ v ∣ 3 d x 1 3 ≤ ∣ α n ∣ 3 2 ∣ β n − v ∣ 3 + ∣ α n + u ∣ 3 ∣ α n − u ∣ 3 ∣ v ∣ 3 → 0 , n → ∞ .

Due to the weak lower semi-continuity of the norm, (3.20) induces that

(3.21) J ( u , v ) ≤ liminf n → ∞ J ( u n , v n ) = m ( a 1 , a 2 ) < 0 .

Thus, recalling (3.1), we conclude that u ≠ 0 , v ≠ 0 . Furthermore,

∣ u ∣ 2 2 ≤ liminf n → ∞ ∣ u n ∣ 2 2 = lim n → ∞ ∣ u n ∣ 2 2 = a 1 , ∣ v ∣ 2 2 ≤ liminf n → ∞ ∣ v n ∣ 2 2 = lim n → ∞ ∣ v n ∣ 2 2 = a 2 .

Now we focus on showing ∣ u ∣ 2 2 = a 1 and ∣ v ∣ 2 2 = a 2 . Assume by contradiction that ∣ u ∣ 2 2 ≔ c < a 1 or ∣ v ∣ 2 2 ≔ d < a 2 . By the definition of m ( c , d ) and inequality (3.21), it follows m ( c , d ) ≤ J ( u , v ) ≤ m ( a 1 , a 2 ) . From Lemma 3.1( i i ) and ( v ), we deduce

J ( u , v ) ≤ m ( c , d ) ≤ m ( a 1 , a 2 ) + m ( c − a 1 , d − a 2 ) < m ( a 1 , a 2 ) ≤ J ( u , v ) .

This is contradiction. This leads to the fact that ( u , v ) ∈ S ( a 1 , a 2 ) .

Let u ˜ and v ˜ denote the Schwarz spherical rearrangement of ∣ u ∣ and ∣ v ∣ . Note that u ˜ , v ˜ ∈ H 1 ( R N ) , ∣ u ˜ ∣ 2 2 = a 1 , ∣ v ˜ ∣ 2 2 = a 2 , and J ( u ˜ , v ˜ ) ≤ J ( u , v ) . Thus, ( u ˜ , v ˜ ) is a weak solution of equations (1.1) and (1.2), where the parameters λ 1 , λ 2 ∈ R are determined by the Lagrange’s multiplier rule. As a consequence of the elliptic regularity theory and the maximum principle in [22], we know that u ˜ , v ˜ ∈ C 2 ( R N ) and u ˜ , v ˜ > 0 . According to Lemma 2.3 in [22], λ 1 < 0 , λ 2 < 0 . The proof is completed.□

4 Proof of Theorem 1.2

In this section, we follow the idea of [4] to give the proof of Theorem 1.2. To avoid compactness issue, we work in the standard radial space H rad 1 ( R 5 ) × H rad 1 ( R 5 ) . That is, we look for the critical points of J constrained on H rad 1 ( R 5 ) × H rad 1 ( R 5 ) . For any a > 0 , the set S a is defined by

(4.1) S a ≔ w ∈ H rad 1 ( R 5 ) : ∫ R 5 w 2 d x = a .

For s ∈ R and w ∈ H 1 ( R 5 ) , we define the dilation

(4.2) ( s ⋆ w ) ( x ) ≔ e 5 s 2 w ( e s x ) .

It is easy to check that s ⋆ w ∈ S a for every s ∈ R and w ∈ S a . Moreover, we also have the following conclusions for the functional I μ ( I μ denotes the functional for the scalar equation (2.5)).

Lemma 4.1

For any w ∈ H 1 ( R 5 ) and for every μ > 0 , there holds,

I μ ( s ⋆ w ) = e 2 s 2 ∫ R 5 ∣ ∇ w ∣ 2 d x − e 5 2 s 3 μ ∫ R 5 ∣ w ∣ 3 d x , ∂ ∂ s I μ ( s ⋆ w ) = e 2 s ∫ R 3 ∣ ∇ w ∣ 2 d x − 5 e 5 2 s 6 μ ∫ R 5 ∣ w ∣ 3 d x .

In particular, if w = w a , μ , then

∂ ∂ s I μ ( s ⋆ w a , μ ) is > 0 if s < 0 , = 0 if s = 0 , < 0 if s > 0 .

Since the proof of Lemma 4.1 is similar to [4, Lemma 3.1], we omit the details here. For a 1 , a 2 , μ 1 , μ 2 > 0 , let β 1 = β 1 ( a 1 , a 2 , μ 1 , μ 2 ) > 0 be defined by (1.12). We have the following conclusions for the functionals J and I .

Lemma 4.2

For 0 < β < β 1 , there holds

inf { J ( u , v ) : ( u , v ) ∈ P ( a 1 , μ 1 + β ) × P ( a 2 , μ 2 + β ) } > max { l ( a 1 , μ 1 ) , l ( a 2 , μ 2 ) } ,

where l ( a i , μ i ) is defined by (2.10).

Proof

From the definition of I μ in (2.6) and the virtue of Young’s inequality, we infer that for ( u , v ) ∈ P ( a 1 , μ 1 + β ) × P ( a 2 , μ 2 + β ) ,

J ( u , v ) = I μ 1 ( u ) + I μ 2 ( v ) − β 2 ∫ R 5 u 2 v d x ≥ I μ 1 ( u ) + I μ 2 ( v ) − β 3 ∫ R 5 ∣ u ∣ 3 d x − β 6 ∫ R 5 ∣ v ∣ 3 d x ≥ I μ 1 ( u ) + I μ 2 ( v ) − β 3 ∫ R 5 ∣ u ∣ 3 d x − β 3 ∫ R 5 ∣ v ∣ 3 d x = 1 2 ∫ R 5 ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − μ 1 + β 3 ∫ R 5 ∣ u ∣ 3 d x − μ 2 + β 3 ∫ R 5 ∣ v ∣ 3 d x = I μ 1 + β ( u ) + I μ 2 + β ( v ) ≥ inf u ∈ P ( a 1 , μ 1 + β ) I μ 1 + β ( u ) + inf v ∈ P ( a 2 , μ 2 + β ) I μ 2 + β ( v ) = l ( a 1 , μ 1 + β ) + l ( a 2 , μ 2 + β ) .

The conclusion holds if and only if

max { l ( a 1 , μ 1 ) , l ( a 2 , μ 2 ) } < l ( a 1 , μ 1 + β ) + l ( a 2 , μ 2 + β ) .

By the definition of l ( a 1 , μ 1 ) , l ( a 2 , μ 2 ) , l ( a 1 , μ 1 + β ) , l ( a 2 , μ 2 + β ) in Proposition 2.3, one obtains

(4.3) C 0 C 1 12 a 1 ( μ 1 + β ) 4 + C 0 C 1 12 a 2 ( μ 2 + β ) 4 > max C 0 C 1 12 a 1 μ 1 4 , C 0 C 1 12 a 2 μ 2 4 .

This ends the proof.□

Now we fix 0 < β < β 1 = β 1 ( a 1 , a 2 , μ 1 , μ 2 ) and choose ε > 0 such that

(4.4) inf { J ( u , v ) : ( u , v ) ∈ P ( a 1 , μ 1 + β ) × P ( a 2 , μ 2 + β ) } > max { l ( a 1 , μ 1 ) , l ( a 2 , μ 2 ) } + ε .

We define the following

(4.5) w 1 ≔ w a 1 , μ 1 + β and w 2 ≔ w a 2 , μ 2 + β ,

and for i = 1 , 2 ,

(4.6) φ i ( s ) ≔ I μ i ( s ⋆ w i ) and ψ i ( s ) ≔ ∂ ∂ s I μ i + β ( s ⋆ w i ) .

The next lemma is a major building block for the proof of the linking structure of J .

Lemma 4.3

For i = 1 , 2 , there exists ρ i < 0 and R i > 0 , depending on ε and on β , such that

0 < φ i ( ρ i ) < ε and φ i ( R i ) ≤ 0 ;
ψ i ( s ) > 0 for any s < 0 and ψ i ( s ) < 0 for every s > 0 . In particular, ψ i ( ρ i ) > 0 and ψ i ( R i ) < 0 .

Proof

By using the change of variables as in Lemma 4.1, we deduce that for w a , μ ∈ P ( a , μ )

φ i = I μ ( s ⋆ w a i , μ i + β ) = e 2 s 2 ∫ R 5 ∣ ∇ w a i , μ i + β ∣ 2 d x − e 5 2 s 3 μ i ∫ R 5 ∣ w a i , μ i + β ∣ 3 d x .

We deduce that φ i ( s ) → 0 + as s → − ∞ , and φ i ( s ) → − ∞ as s → + ∞ . Thus, there exist ρ i and R i satisfying ( i ) . Conclusion ( i i ) follows directly from Lemma 4.1.□

Set Q ≔ [ ρ 1 , R 1 ] × [ ρ 2 , R 2 ] . We define

γ 0 ( t 1 , t 2 ) ≔ ( t 1 ⋆ w 1 , t 2 ⋆ w 2 ) ∈ S a 1 × S a 2 , ∀ ( t 1 , t 2 ) ∈ Q ¯

and

Γ ≔ { γ ∈ C ( Q ¯ , S a 1 × S a 2 ) : γ = γ 0 on ∂ Q } .

Then, the minimax structure of the functional J is based on (4.4) and the following two lemmas.

Lemma 4.4

There holds

sup ∂ Q J ( γ 0 ) ≤ max { l ( a 1 , μ 1 ) , l ( a 2 , μ 2 ) } + ε .

Proof

We infer from u , v ≥ 0 and β > 0 that

J ( u , v ) = I μ 1 ( u ) + I μ 2 ( v ) − β 2 ∫ R 5 u 2 v d x ≤ I μ 1 ( u ) + I μ 2 ( v )

for every ( u , v ) ∈ S a 1 × S a 2 . This and Lemma 4.3 imply that

J ( t 1 ⋆ w 1 , ρ 2 ⋆ w 2 ) ≤ I μ 1 ( t 1 ⋆ w 1 ) + I μ 2 ( ρ 2 ⋆ w 2 ) = I μ 1 ( t 1 ⋆ w 1 ) + φ 2 ( ρ 2 ) ≤ I μ 1 ( t 1 ⋆ w 1 ) + ε ≤ sup s ∈ R I μ 1 ( s ⋆ w 1 ) + ε = sup s ∈ R I μ 1 ( s ⋆ w a 1 μ 1 ) + ε .

To get the estimates of the last term, we infer from Proposition 2.3 that

w a i , μ i = s ¯ i ⋆ w i for e s ¯ i 2 ≔ 6 ∫ R 5 ∣ ∇ w i ∣ 2 d x 5 ∫ R 5 μ i w i 3 d x = μ i + β μ i .

By the definition of s ⋆ w , it is easy to check that s 1 ⋆ ( s 2 ⋆ w ) = ( s 1 + s 2 ) ⋆ w for every s 1 , s 2 ∈ R and w ∈ H 1 ( R 5 ) . Thus, we have

sup s ∈ R I μ 1 ( s ∗ w 1 ) = sup s ∈ R I μ 1 ( s ∗ w a 1 , μ 1 ) .

By using Lemma 4.1, the supremum on the right hand side is achieved for s = 0 , and hence,

(4.7) J ( t 1 ⋆ w 1 , ρ 2 ⋆ w 2 ) ≤ l ( a 1 , μ 1 ) + ε ∀ t 1 ∈ [ ρ 1 , R 1 ] .

Similarly, we have

(4.8) J ( ρ 1 ⋆ w 1 , t 2 ⋆ w 2 ) ≤ l ( a 2 , μ 2 ) + ε ∀ t 1 ∈ [ ρ 2 , R 2 ] .

The value of J ( γ 0 ) on the remaining sides of ∂ Q is smaller. Indeed, by Lemma 4.3 and similar calculations to the derivation mentioned earlier, one can obtain

(4.9) J ( t 1 ⋆ w 1 , R 2 ⋆ w 2 ) ≤ I μ 1 ( t 1 ⋆ w 1 ) + I μ 2 ( R 2 ⋆ w 2 ) ≤ sup s ∈ R I μ 1 ( s ⋆ w 1 ) = l ( a 1 , μ 1 )

for every t 1 ∈ [ ρ 1 , R 1 ] . Notice that, in a similar way,

(4.10) J ( R 1 ⋆ w 1 , t 2 ⋆ w 2 ) ≤ l ( a 2 , μ 2 ) , ∀ t 2 ∈ [ ρ 2 , R 2 ] .

This finishes the proof.□

In the following, we show that the class Γ links with P ( a 1 , μ 1 + β ) × P ( a 2 , μ 2 + β ) .

Lemma 4.5

For every γ ∈ Γ , there exists ( t 1 , γ , t 2 , γ ) ∈ Q such that γ ( t 1 , γ , t 2 , γ ) ∈ P ( a 1 , μ 1 + β ) × P ( a 2 , μ 2 + β ) .

Proof

For γ ∈ Γ , we use the notation γ ( t 1 , t 2 ) = ( γ 1 ( t 1 , t 2 ) , γ 2 ( t 1 , t 2 ) ) ∈ S a 1 × S a 2 . Let us consider the map F γ : Q → R 2 defined by

F γ ( t 1 , t 2 ) ≔ ∂ ∂ s I μ 1 + β ( s ⋆ γ 1 ( t 1 , t 2 ) ) s = 0 , ∂ ∂ s I μ 2 + β ( s ⋆ γ 2 ( t 1 , t 2 ) ) s = 0 ,

where

∂ ∂ s I μ i + β ( s ⋆ γ i ( t 1 , t 2 ) ) s = 0 = ∂ ∂ s e 2 s 2 ∫ R 5 ∣ ∇ γ i ( t 1 , t 2 ) ∣ 2 d x − e 5 2 s 3 ( μ i + β ) ∫ R 5 γ i 3 ( t 1 , t 2 ) d x s = 0 = ∫ R 5 ∣ ∇ γ i ( t 1 , t 2 ) ∣ 2 d x − 5 6 ( μ i + β ) ∫ R 5 γ i 3 ( t 1 , t 2 ) d x .

We deduce that

F γ ( t 1 , t 2 ) = ( 0 , 0 ) if and only if γ ( t 1 , t 2 ) ∈ P ( a 1 , μ 1 + β ) × P ( a 2 , μ 2 + β ) .

Now we focus on the solution of F γ ( t 1 , t 2 ) = ( 0 , 0 ) in Q for every γ ∈ Γ . The results can be obtained by the standard degree theory (see [4]). We shall show that the oriented path F γ ( ∂ + Q ) has winding number equal to 1 with respect to the origin of R 2 . In this way, we observe that F γ ( ∂ + Q ) = F γ 0 ( ∂ + Q ) depends only on the choice of γ 0 and not on γ . Recalling the definition of ψ i given in (4.6), it is natural to show that

F γ 0 ( t 1 , t 2 ) = e 2 t 1 ∫ R 5 ∣ ∇ w 1 ∣ 2 d x − 5 e 5 2 t 1 6 ( μ 1 + β ) ∫ R 5 ∣ w 1 ∣ 3 d x , e 2 t 2 ∫ R 5 ∣ ∇ w 2 ∣ 2 d x − 5 e 5 2 t 2 6 ( μ 2 + β ) ∫ R 5 ∣ w 2 ∣ 3 d x = ( ψ 1 ( t 1 ) , ψ 2 ( t 2 ) ) .

Hence, Lemma 4.3 ( i i ) completely describes the restriction of F γ 0 on ∂ Q . Denote ι ( ρ , P ) as the winding number of the curve ρ with respect to the point P . Particularly, by virtue of the topological degree,

deg ( F γ , Q , ( 0 , 0 ) ) = ι ( F γ 0 ( ∂ + Q ) , ( 0 , 0 ) ) = 1 .

Thus, there exists γ ( t 1 , γ , t 2 , γ ) ∈ P such that F γ ( t 1 , γ , t 2 , γ ) = ( 0 , 0 ) . Thus, the desired result is obtained.□

From Lemmas 4.4 and 4.5, we can apply the minimax principle [16, Theorem 3.2] to J on Γ . Then, we obtain the Palais-Smale sequence for the constrained functional J on S a 1 × S a 2 . However, we do not know the boundedness of a Palais-Smale sequence. To accomplish this, we borrow the idea of [19].

Lemma 4.6

There exists a Palais-Smale sequence ( u n , v n ) for J on S a 1 × S a 2 at the level

(4.11) c ≔ inf γ ∈ Γ max ( t 1 , t 2 ) ∈ Q J ( γ ( t 1 , t 2 ) ) > max { l ( a 1 , μ 1 ) , l ( a 2 , μ 2 ) }

satisfying the additional condition

(4.12) ∫ R 5 ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x − 5 6 ∫ R 5 μ 1 ∣ u n ∣ 3 + μ 2 ∣ v n ∣ 3 + 3 2 β u n 2 v n d x = o ( 1 ) ,

where o ( 1 ) → 0 as n → ∞ . Furthermore, u n − , v n − → 0 a.e. in R 5 as n → ∞ .

Proof

We first define the augmented functional J ˜ : R × S a 1 × S a 2 → R by

J ˜ ( s , u , v ) ≔ J ( s ⋆ u , s ⋆ v ) .

Set

γ ˜ 0 ( t 1 , t 2 ) ≔ ( 0 , γ 0 ( t 1 , t 2 ) ) = ( 0 , t 1 ⋆ w 1 , t 2 ⋆ w 2 )

and

Γ ˜ ≔ { γ ˜ ∈ C ( Q , R × S a 1 × S a 2 ) : γ ˜ = γ ˜ 0 on ∂ Q }

by using the minimax principle in [16, Theorem 3.2] to the functional J ˜ with the minimax class Γ ˜ , we find a Palais-Smale sequence for J ˜ at level

c ˜ ≔ inf γ ˜ ∈ Γ ˜ ( t 1 , t 2 ) ∈ Q J ˜ ( γ ˜ ( t 1 , t 2 ) ) .

Since J ˜ ( γ ˜ 0 ) = J ( γ 0 ) on ∂ Q , it follows from Lemmas 4.4 and 4.5 that assumptions of the minimax principle will be satisfied if we prove that c ˜ = c . Next, we use the notation

γ ˜ ( t 1 , t 2 ) = ( s ( t 1 , t 2 ) , γ 1 ( t 1 , t 2 ) , γ 2 ( t 1 , t 2 ) )

for any γ ˜ ⊂ Γ ˜ and ( t 1 , t 2 ) ∈ Q . It follows that

J ˜ ( γ ˜ ( t 1 , t 2 ) ) = J ( s ( t 1 , t 2 ) ⋆ γ 1 ( t 1 , t 2 ) , s ( t 1 , t 2 ) ⋆ γ 2 ( t 1 , t 2 ) )

and ( s ( ⋅ ) ⋆ γ 1 ( ⋅ ) , s ( ⋅ ) ⋆ γ 2 ( ⋅ ) ) ∈ Γ . Thus, c ˜ = c , and the minimax principle is applicable. It is not difficult to directly check that J ˜ ( s , u , v ) = J ˜ ( s , ∣ u ∣ , ∣ v ∣ ) in view of the definition of J ˜ . Moreover,

J ˜ ( γ ˜ ( t 1 , t 2 ) ) = J ( s ( t 1 , t 2 ) ⋆ γ 1 ( t 1 , t 2 ) , s ( t 1 , t 2 ) ⋆ γ 2 ( t 1 , t 2 ) ) = J ˜ ( 0 , s ( t 1 , t 2 ) ⋆ γ 1 ( t 1 , t 2 ) , s ( t 1 , t 2 ) ⋆ γ 2 ( t 1 , t 2 ) ) .

In this way, using the notation of Theorem 3.2 in [16], we can choose the minimizing sequence γ n = ( s n , γ 1 , n , γ 2 , n ) for c ˜ satisfying the additional conditions s n ≡ 0 , γ 1 , n ( t 1 , t 2 ) ≥ 0 a.e. in R 5 for every ( t 1 , t 2 ) ∈ Q , γ 2 , n ( t 1 , t 2 ) ≥ 0 a.e. in R 5 for every ( t 1 , t 2 ) ∈ Q . Thus, we follow [16, Theorem 3.2] to deduce that there exists a Palais-Smale sequence ( s ˜ n , u ˜ n , v ˜ n ) for J ˜ on R × S a 1 × S a 2 at level c , and such that

(4.13) lim n → ∞ ∣ s ˜ n ∣ + dist H 1 ( ( u ˜ n , v ˜ n ) , γ ˜ n ( Q ) ) = 0 .

To obtain a Palais-Smale sequence for J at level c satisfying (4.12), we follow closely the approach of [19, Lemma 2.4] with minor changes. Finally, we infer from (4.13) that u n − , v n − → 0 a.e. in R 5 as n → ∞ . Moreover, the lower estimate for c follows from Lemma 4.4.□

To complete the proof of Theorem 1.2, we want to show that ( u n , v n ) is strongly convergent in H 1 ( R 5 , R 2 ) to a limit ( u , v ) . This can be accomplished by proving the following:

d J ∣ S a 1 × S a 2 ( u n , v n ) → 0 and ( u n , v n ) ∈ S a 1 × S a 2

holds for all n . Now we focus on the following statement to achieve desired results.

Lemma 4.7

The sequence { ( u n , v n ) } is bounded in H 1 ( R 5 , R 2 ) . Furthermore, there exists C ¯ > 0 such that

∫ R 5 ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x ≥ C ¯ for all n .

Proof

We deduce from (4.12) that

J ( u n , v n ) = 1 10 ∫ R 5 ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x + o ( 1 ) ,

where o ( 1 ) → 0 as n → ∞ . Hence, the desired result is a consequence of J ( u n , v n ) → c > 0 .□

By the previous results, we know that there exists a subsequence such that ( u n , v n ) ⇀ ( u ˜ , v ˜ ) weakly in H 1 ( R 5 ) , strongly in L 3 ( R 5 ) , and a.e. in R 5 . Moreover, we have u ˜ , v ˜ ≥ 0 in R 5 . Since the compact embedding H 1 ( R 5 ) ↪ L p ( R 5 ) ( ∀ p ∈ ( 2 , 10 / 3 ) ) , we cannot conclude that ( u ˜ , v ˜ ) ∈ S a 1 × S a 2 directly. Observe that for every ( φ , ψ ) ∈ H 1 ( R 5 , R 2 ) , d J ∣ S a 1 × S a 2 ( u n , v n ) → 0 implies that there exists two sequences of real numbers { λ 1 , n } and { λ 2 , n } such that

(4.14) ∫ R 5 ∇ u n ⋅ ∇ φ + ∇ v n ⋅ ∇ ψ − μ 1 u n 2 φ − μ 2 v n 2 ψ − β u n u n 2 ψ + v n φ d x − ∫ R 5 ( λ 1 , n u n φ + λ 2 , n ψ ) d x = o ( 1 ) ‖ ( φ , ψ ) ‖ H 1 ,

with o ( 1 ) → 0 as n → ∞ .

Lemma 4.8

Both { λ 1 , n } and { λ 2 , n } are bounded sequences, and at least one of them is converging, up to a subsequence, to a strictly negative value.

Proof

The value of the { λ i , n } can be found using ( u n , 0 ) and ( 0 , v n ) as test functions in (4.14):

λ 1 , n a 1 2 = ∫ R 5 ( ∣ ∇ u n ∣ 2 − μ 1 ∣ u n ∣ 3 − β u n 2 v n ) d x − o ( 1 ) , λ 2 , n a 2 2 = ∫ R 5 ∣ ∇ v n ∣ 2 − μ 2 ∣ v n ∣ 3 − β 2 u n 2 v n d x − o ( 1 ) ,

with o ( 1 ) → 0 as n → ∞ . Then, in view of the boundedness of { ( u n , v n ) } in H 1 , the boundedness of { λ i , n } is obtained. Furthermore, equality (4.12) and Lemma 4.7 imply that

λ 1 , n a 1 2 + λ 2 , n a 2 2 = ∫ R 5 ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 − μ 1 ∣ u n ∣ 3 − μ 2 ∣ v n ∣ 3 − 3 2 β u n 2 v n d x − o ( 1 ) = − 1 5 ∫ R 5 ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x + o ( 1 ) ≤ − C ¯ 10

for n sufficiently large. It follows that at least one sequence of { λ i , n } is negative and bounded away from 0.□

Next we assume that λ 1 , n → λ 1 ∈ R and λ 2 , n → λ 2 ∈ R . The sign of the limit values plays a crucial role in our argument, as the next statement will illustrate.

Lemma 4.9

If λ 1 < 0 ( resp . λ 2 < 0 ) , then u n → u ¯ (resp. v n → v ¯ ) strongly in H 1 ( R 5 ) .

Proof

Let us suppose that λ 1 < 0 . By weak convergence in H 1 ( R 5 ) , strong convergence in L 3 ( R 5 ) , and using (4.14), we have that

o ( 1 ) = ( d J ( u n , v n ) − d J ( u ˜ , v ˜ ) ) [ ( u n − u ˜ , 0 ) ] − λ 1 ∫ R 5 ( u n − u ˜ ) 2 d x = d J ( u n , v n ) [ ( u n − u ˜ , 0 ) ] − d J ( u ˜ , v ˜ ) [ ( u n − u ˜ , 0 ) ] − λ 1 ∫ R 5 ( u n − u ˜ ) 2 d x = ∫ R 5 ∇ u n ⋅ ∇ ( u n − u ˜ ) d x − ∫ R 5 ∇ u ˜ ⋅ ∇ ( u n − u ˜ ) d x − ∫ R 5 ( ∣ u n ∣ 3 − ∣ u ˜ ∣ 3 ) ( u n − u ˜ ) d x − ∫ R 5 ( β u n v n ( u n − u ˜ ) − λ 1 ( u n − u ˜ ) 2 ) d x = ∫ R 5 ( ∣ ∇ ( u n − u ˜ ) ∣ 2 − λ 1 ( u n − u ˜ ) 2 ) d x + o ( 1 ) ,

with o ( 1 ) → 0 as n → ∞ . Since λ 1 < 0 , it follows that this is equivalent to the strong convergence in H 1 . One can use the similar arguments to prove the case λ 2 < 0 .□

Remark 4.10

Note that Lemmas 4.7–4.9 do not depend on the value of β . Hence, we can use them in the proof of Theorem 1.3.

Now we are ready to give the proof of Theorem 1.2.

Proof of Theorem 1.2

Let { ( u n , v n ) } ∈ S a 1 × S a 2 be given by Lemma 4.7. By using (4.14), the weak convergence ( u n , v n ) ⇀ ( u ˜ , v ˜ ) , and the convergence of λ 1 , n → λ 1 ∈ R and λ 2 , n → λ 2 ∈ R , we deduce that ( u ˜ , v ˜ ) is a solution of (1.1). It remains to prove that ( u ˜ , v ˜ ) ∈ S a 1 × S a 2 . Without the loss of generality, we may suppose that λ 1 < 0 by Lemma 4.8. Moreover, Lemmas 4.6 and 4.9 imply that u n → u ˜ strongly in H 1 ( R 5 ) and u ˜ ∈ S a 1 . If λ 2 < 0 , we can deduce in the same way that v n → v ˜ strongly in H 1 ( R 5 ) and v ˜ ∈ S a 2 . Then, the proof is completed. We use the contradiction arguments. Suppose that λ 2 ≥ 0 and v n ↛ v ˜ strongly in H 1 ( R 5 ) . Note that any weak solution of (1.1) is smooth by regularity. Moreover, we know that v ˜ > 0 by using maximum principle. Hence, we deduce that

(4.15) − Δ v ˜ = λ 2 v ˜ + μ 2 v ˜ 2 + β 2 u ˜ 2 ≥ c v ˜ 3 2 in R 5 ,

where c > 0 is a constant. Then, it follows from [34, Theorem 8.4] that v ˜ ≡ 0 . Particularly, this implies that u ˜ solves

− Δ u ˜ − λ 1 u ˜ = μ 1 ∣ u ˜ ∣ u ˜ , in R 5 , u ˜ > 0 , in R 5 , ∫ R 5 u ˜ 2 = a 1 .

So that u ˜ ∈ P ( a 1 , μ 1 ) and I μ 1 ( u ˜ ) = l ( a 1 , μ 1 ) . On the other hand, we infer from (4.12) that

(4.16) c = lim n → ∞ J ( u n , v n ) = lim n → ∞ 1 12 ∫ R 5 μ 1 ∣ u n ∣ 3 + 3 2 β u n 2 v n + μ 2 ∣ v n ∣ 3 d x = μ 1 12 ∫ R 5 ∣ u ˜ ∣ 3 d x = I μ 1 ( u ˜ ) = l ( a 1 , μ 1 ) .

This contradicts Lemma 4.6.□

5 Proof of Theorem 1.3

In this section, we shall follow the idea of [4] to give the proof of Theorem 1.3. To accomplish this, we first show the existence of a positive solution ( u ¯ , v ¯ ) . Then, we characterize it as a ground state in the second one, in the sense that

J ( u ¯ , v ¯ ) = inf { J ( u , v ) : ( u , v ) ∈ V } = inf { J ( u , v ) : ( u , v ) is a solution of (1.1)–(1.2) for some λ 1 , λ 2 } .

To give the proof of Theorem 1.3, we shall use a mountain pass lemma arguments, which has been developed in [4]. For ( u , v ) ∈ S a 1 × S a 2 , we define the functional

J ( s ⋆ ( u , v ) ) = e 2 s 2 ∫ R 5 ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x − e 5 2 s 3 ∫ R 5 μ 1 ∣ u ∣ 3 + 3 2 β u 2 v + μ 2 ∣ v ∣ 3 d x ,

where s ⋆ u is defined in (4.2). We use the notation s ⋆ ( u , v ) = ( s ⋆ u , s ⋆ v ) for simplicity. Then, it is easy to check the following conclusions hold.

Lemma 5.1

lim s → − ∞ ∫ R 5 ∣ ∇ ( s ⋆ u ) ∣ 2 + ∣ ∇ ( s ⋆ v ) ∣ 2 = 0 , lim s → + ∞ ∫ R 5 ∣ ∇ ( s ⋆ u ) ∣ 2 + ∣ ∇ ( s ⋆ v ) ∣ 2 = + ∞

and

lim s → − ∞ J ( s ⋆ ( u , v ) ) = 0 + , lim s → + ∞ J ( s ⋆ ( u , v ) ) = − ∞ .

The next lemma states a mountain pass structure of the problem.

Lemma 5.2

There exists K > 0 sufficiently small such that for the sets

A ≔ ( u , v ) ∈ S a 1 × S a 2 : ∫ R 5 ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ≤ K

and

B ≔ ( u , v ) ∈ S a 1 × S a 2 : ∫ R 5 ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 = 2 K ,

and there holds

J ( u , v ) > 0 on A and sup A J < inf B J .

Proof

By using the Gagliardo-Nirenberg inequality (2.1), we have

(5.1) ∫ R 5 μ 1 ∣ u ∣ 3 + 3 2 β u 2 v + μ 2 ∣ v ∣ 3 d x ≤ C ∫ R 5 ( ∣ u ∣ 3 + ∣ v ∣ 3 ) d x ≤ C ∫ R 5 ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 5 4

for every ( u , v ) ∈ S a 1 × S a 2 , provided C > 0 depends on μ 1 , μ 2 , β , a 1 , a 2 > 0 but not on the particular choice of ( u , v ) . Now taking K to be determined, if ( u 1 , v 1 ) ∈ B and ( u 2 , v 2 ) ∈ A , we have

J ( u 1 , v 1 ) − J ( u 2 , v 2 ) ≥ 1 2 ∫ R 5 ∣ ∇ u 1 ∣ 2 + ∣ ∇ v 1 ∣ 2 − ∫ R 5 ∣ ∇ u 2 ∣ 2 + ∣ ∇ v 2 ∣ 2 − 1 3 ∫ R 5 μ 1 ∣ u ∣ 1 3 + 3 2 β u 1 2 v 1 + μ 2 ∣ v ∣ 1 3 ≥ K 2 − C 3 ( 2 K ) 5 4 ≥ K 4 ,

where K > 0 is sufficiently small. Moreover, we take K smaller such that

J ( u 2 , v 2 ) ≥ 1 2 ∫ R 5 ∣ ∇ u 2 ∣ 2 + ∣ ∇ v 2 ∣ 2 d x − C 3 ∫ R 5 ∣ ∇ u 2 ∣ 2 + ∣ ∇ v 2 ∣ 2 5 4 d x > 0

for every ( u 2 , v 2 ) ∈ A .□

Next our aim is to introduce a suitable minimax class. Recalling the definition of w a , μ in Proposition 2.3. Then, we define

(5.2) C ≔ ( u , v ) ∈ S a 1 × S a 2 : ∫ R 5 ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ≥ 3 K and J ( u , v ) ≤ 0 .

It follows from Lemma 5.1 that there exist s 1 < 0 and s 2 > 0 , such that

s 1 ⋆ w a 1 , C 0 a 1 1 2 , w a 2 , C 0 a 2 1 2 ≕ ( u ¯ 1 , v ¯ 1 ) ∈ A and s 2 ⋆ w a 1 , C 0 a 1 1 2 , w a 2 , C 0 a 2 1 2 ≕ ( u ¯ 2 , v ¯ 2 ) ∈ C .

We define

(5.3) Γ ≔ { γ ∈ C ( [ 0 , 1 ] , S a 1 × S a 2 ) : γ ( 0 ) = ( u ¯ 1 , v ¯ 1 ) and γ ( 1 ) = ( u ¯ 2 , v ¯ 2 ) } .

The mountain pass lemma is applicable for J on the minimax class Γ , which comes from Lemma 5.2 and the continuity of the L 2 -norm of the gradient in the topology of H 1 . With similar calculations as in Lemma 4.6, we deduce the following conclusions.

Lemma 5.3

There exists a Palais-Smale sequence ( u n , v n ) for J on S a 1 × S a 2 at the level

d ≔ inf γ ∈ Γ max t ∈ [ 0 , 1 ] J ( γ ( t ) ) ,

satisfying the additional condition (4.12):

∫ R 5 ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x − 5 6 ∫ R 5 μ 1 ∣ u n ∣ 3 + 3 2 β u n 2 v n + μ 2 ∣ v n ∣ 3 d x = o ( 1 ) ,

where o ( 1 ) → 0 as n → ∞ . Furthermore, u n − , v n − → 0 a.e. in R 5 as n → ∞ .

Similar to the previous section, we want to show that ( u n , v n ) → ( u ¯ , v ¯ ) in H 1 ( R 5 ) × H 1 ( R 5 ) . As in the Remark 4.10 and β > 0 small, we know that ( u ¯ , v ¯ ) is a solution of (1.1) and (1.2). By Lemma 4.6, we know that up to a subsequence, ( u n , v n ) ⇀ ( u ¯ , v ¯ ) weakly in H 1 ( R 5 , R 2 ) , strongly in L 3 ( R 5 , R 2 ) , a.e. in R 5 . In addition, we can also suppose that one of them, say λ 1 is strictly negative. Therefore, as Lemma 4.9, we know that u n → u ¯ strongly in H 1 ( R 5 ) . If by contradiction v n ↛ v ¯ strongly in H 1 ( R 5 ) , then λ 2 ≥ 0 . By using the similar argument as in the proof of Theorem 1.2, we infer that v ¯ ≡ 0 . Thus, as (4.16), we know that d = l ( a 1 , μ 1 ) (see Proposition 2.3). In the following, we need find the contradiction. To accomplish this, we recall the definition of β 2 = β 2 ( a 1 , a 2 , μ 1 , μ 2 ) > 0 , see (1.15).

Lemma 5.4

If β > β 2 , then

sup s ∈ R J s ⋆ ( w a 1 , C 0 a 1 1 2 , w a 2 , C 0 a 2 1 2 ) < min { l ( a 1 , μ 1 ) , l ( a 2 , μ 2 ) } .

Proof

From Proposition 2.3 and (2.3), a direct computation shows that

(5.4) ∫ R 5 s ⋆ w a 1 , C 0 a 1 1 2 2 s ⋆ w a 2 , C 0 a 2 1 2 d x = ∫ R 5 e 15 2 s w a 1 , ( C 0 a 1 ) 1 2 ( e s x ) 2 ⋅ w a 2 , ( C 0 a 2 ) 1 2 ( e s x ) d x = e 5 2 s ∫ R 5 C 0 2 C 0 a 1 5 2 a 1 2 2 ⋅ C 0 2 C 0 a 2 5 2 a 2 2 w 0 3 d x = e 5 2 s a 1 a 2 1 2 C 0 3 2 ∫ R 5 w 0 3 d x = e 5 2 s C 1 a 1 a 2 1 2 C 0 3 2 .

By using Proposition 2.3 again, we can explicitly compute the maximum in s of the function

J ( s ⋆ ( w a 1 , C 0 / a 1 , w a 2 , C 0 / a 2 ) ) = 5 e 2 s 12 C 1 a 1 C 0 + C 1 a 2 C 0 − e 5 2 s 3 μ 1 C 1 a 1 3 2 C 0 3 2 + 3 β C 1 a 1 a 2 1 2 2 C 0 3 2 + μ 2 C 1 a 2 3 2 C 0 3 2 .

That is, the maximum is given by

max s ∈ R J ( s ⋆ ( w a 1 , C 0 / a 1 , w a 2 , C 0 / a 2 ) ) = C 1 C 0 ( a 1 + a 2 ) 5 12 μ 1 a 1 3 2 + 3 2 β a 1 a 2 3 2 + μ 2 a 2 3 2 4 .

Recall the definitions of β 2 , l ( a 1 , μ 1 ) and l ( a 2 , μ 2 ) , then the conclusion holds if β > β 2 .□

Next we prove the existence of a positive solution at level d .

Existence of a positive solution at level d

Now we argue by contradiction. Suppose that v n ↛ v ¯ strongly in H 1 ( R 5 ) . Then, we have v ¯ ≡ 0 and d = l ( a 1 , μ 1 ) . We define the path

γ ( t ) ≔ ( ( ( 1 − t ) s 1 + t s 2 ) ⋆ ( w a 1 , μ 1 , w a 2 , μ 2 ) ) .

It is easy to show that γ ∈ Γ , and as a result of Lemma 5.4,

d = inf γ ∈ Γ max t ∈ [ 0 , 1 ] J ( γ ( t ) ) ≤ sup t ∈ [ 0 , 1 ] J ( γ ( t ) ) ≤ sup s ∈ R J ( s ⋆ ( w a 1 , μ 1 , w a 2 , μ 2 ) ) < ℓ ( a 1 , μ 1 ) .

This is a contradiction.

Next we introduce the variational characterization for ( u ¯ , v ¯ ) . That is, we shall prove the following conclusion.

J ( u ¯ , v ¯ ) = inf { J ( u , v ) : ( u , v ) ∈ V } = inf { J ( u , v ) : ( u , v ) is a solution of (1.1)–(1.2) for some λ 1 , λ 2 } ,

where V is given in (1.13). We also recall the definitions of A and C (see Lemma 5.2 and (5.2)). Set

A + ≔ { ( u , v ) ∈ A : u , v ≥ 0 a.e. in R 5 }

and

C + ≔ { ( u , v ) ∈ C : u , v ≥ 0 a.e. in R 5 } .

For any ( u 1 , v 1 ) ∈ A + and ( u 2 , v 2 ) ∈ C + , let

Γ ( u 1 , v 1 , u 2 , v 2 ) ≔ { γ ∈ C ( [ 0 , 1 ] , S a 1 × S a 2 ) : γ ( 0 ) = ( u 1 , v 1 ) and γ ( 1 ) = ( u 2 , v 2 ) } .

The next lemma we give another characterization of d . This can be done by using similar arguments as in [4, Lemma 4.5]. Here, we omit the details.

Lemma 5.5

The sets A + and C + are connected by arcs, so that

(5.5) d ≔ inf γ ∈ Γ ( u 1 , v 1 , u 2 , v 2 ) max t ∈ [ 0 , 1 ] J ( γ ( t ) )

for every ( u 1 , v 1 ) ∈ A + and ( u 2 , v 2 ) ∈ C + .

Let us recall the set

V ≔ { ( u , v ) ∈ T a 1 × T a 2 : G ( u , v ) = 0 } .

Its radial subset is given by

(5.6) V rad ≔ { ( u , v ) ∈ S a 1 × S a 2 : G ( u , v ) = 0 } ,

where

G ( u , v ) = ∫ R 5 ( ∣ ∇ u n ∣ 2 + ∣ ∇ v n ∣ 2 ) d x − 5 6 ∫ R 5 μ 1 ∣ u n ∣ 3 + μ 2 ∣ v n ∣ 3 + 3 2 β u n 2 v n d x .

The next lemma proves some properties for the set V .

Lemma 5.6

If ( u , v ) is a solution of (1.1) and (1.2) for some λ 1 , λ 2 ∈ R , then ( u , v ) ∈ V .

Proof

We use ( u , v ) as test function in (1.1). Then, we obtain

(5.7) ∫ R 5 ∣ ∇ u ∣ 2 d x − λ 1 ∫ R 5 u 2 d x = ∫ R 5 ( μ 1 ∣ u ∣ 3 + β u 2 v ) d x , ∫ R 5 ∣ ∇ v ∣ 2 d x − λ 2 ∫ R 5 v 2 d x = ∫ R 5 β 2 u 2 v + μ 2 ∣ v ∣ 3 d x .

Hence, we infer that

(5.8) λ 1 ∫ R 5 u 2 d x + λ 2 ∫ R 5 v 2 d x = ∫ R 5 ∣ ∇ u ∣ 2 d x + ∫ R 5 ∣ ∇ v ∣ 2 d x − ∫ R 5 μ 1 ∣ u ∣ 3 + 3 2 β u 2 v + μ 2 ∣ v ∣ 3 d x .

Moreover, it is easy to deduce that the Pohozaev identity for system (1.1) is given by

(5.9) 3 ∫ R 5 ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x = 5 ∫ R 5 ( λ 1 u 2 + λ 2 v 2 ) + 10 3 μ 1 ∣ u ∣ 3 + 3 2 β u 2 v + μ 2 ∣ v ∣ 3 d x .

From (5.8) and (5.9), we obtain the desired results.□

Now we define

Ψ ( u , v ) ( s ) ≔ J ( s ⋆ ( u , v ) ) ,

where ( u , v ) ∈ T a 1 × T a 2 . Then, it is easy to check the following results.

Lemma 5.7

For every ( u , v ) ∈ T a 1 × T a 2 and v ≥ 0 , then there exists a unique s ( u , v ) ∈ R such that s ( u , v ) ⋆ ( u , v ) ∈ V . Moreover, s ( u , v ) is the unique critical point of Ψ ( u , v ) , which is a strict maximum.

Lemma 5.8

There holds inf V J = inf V rad J .

Proof

To prove this lemma by contradiction, we suppose there exists ( u , v ) ∈ V such that

(5.10) 0 < J ( u , v ) < inf V rad J .

For ( u , v ) ∈ H 1 ( R 5 ) × H 1 ( R 5 ) , let { u , v } ∗ denotes its Schwarz spherical rearrangement as before. The properties of Schwarz symmetrization leads to J ( { u , v } ∗ ) ≤ J ( u , v ) and G ( { u , v } ∗ ) ≤ G ( u , v ) = 0 . Thus, there exists s 0 ≤ 0 such that G ( s 0 ⋆ ( { u , v } ∗ ) ) = 0 . One implies that

J ( s 0 ⋆ ( { u , v } ∗ ) ) ≤ e 2 s 0 J ( { u , v } ∗ ) .

But in fact, by using that G ( s 0 ⋆ ( { u , v } ∗ ) ) = G ( u , v ) = 0 , we have

(5.11) J ( s 0 ⋆ ( { u , v } ∗ ) ) = e 2 s 0 10 ∫ R 5 ( ∣ ∇ u ∗ ∣ 2 + ∣ ∇ v ∗ ∣ 2 ) d x ≤ e 2 s 0 10 ∫ R 5 ( ∣ ∇ u ∣ 2 + ∣ ∇ v ∣ 2 ) d x = e 2 s 0 J ( u , v ) .

Thus, with the assumption of (5.10) and applying (5.11),

0 < J ( u , v ) < inf V rad J ≤ J ( s 0 ⋆ ( { u , v } ∗ ) ) ≤ e 2 s 0 J ( u , v ) ,

which contradicts s 0 ≤ 0 .□

Finally, we give the proof of Theorem 1.3.

Proof of Theorem 1.3

Recalling that any solution of (1.1) and (1.2) stays in V . If we show

(5.12) J ( u ¯ , v ¯ ) = d ≤ inf { J ( u , v ) : ( u , v ) ∈ V rad } ,

then J ( u ¯ , v ¯ ) = inf V J by using Lemma 5.8. To prove (5.12), we choose an arbitrary ( u , v ) ∈ V rad and show that J ( u , v ) ≥ d . First, we choose an arbitrary ( u , v ) ∈ V rad , since ( ∣ u ∣ , ∣ v ∣ ) ∈ V rad and J ( u , v ) = J ( ∣ u ∣ , ∣ v ∣ ) , it is possible to suppose that u , v ≥ 0 a.e. in R 5 . Now let us consider the function Ψ ( u , v ) . By Lemma 5.1, there exists s 0 ≫ 1 such that ( − s 0 ) ⋆ ( u , v ) ∈ A + and s 0 ⋆ ( u , v ) ∈ C + . Thus, the continuous path

γ ( t ) ≔ ( ( 2 t − 1 ) s 0 ) ⋆ ( u , v ) t ∈ [ 0 , 1 ]

connects A + with C + , and by Lemmas 5.5 and 5.7, we infer that

d ≤ max t ∈ [ 0 , 1 ] J ( γ ( t ) ) = J ( u , v ) .

Consequently, this holds for all the elements in V rad . Thus, inequality (5.12) holds.□

Acknowledgements

The authors thank the anonymous referees for his/her very valuable comments, which led to an improvement of this paper. This work was supported by NNSF of China (Grant No. 11971202), Outstanding Young Foundation of Jiangsu Province No. BK20200042.

Funding information: This work was supported by NNSF of China (Grant No. 11971202), Outstanding Young Foundation of Jiangsu Province No. BK20200042.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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Received: 2021-08-16

Revised: 2022-04-26

Accepted: 2022-04-28

Published Online: 2022-06-06

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https://doi.org/10.1515/ans-2022-0010

Keywords for this article

normalized solutions; quadratic nonlinearity; variational methods

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