Startseite A global compactness result with applications to a Hardy-Sobolev critical elliptic system involving coupled perturbation terms
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A global compactness result with applications to a Hardy-Sobolev critical elliptic system involving coupled perturbation terms

  • Lu Shun Wang , Tao Yang EMAIL logo und Xiao Long Yang
Veröffentlicht/Copyright: 16. November 2022

Abstract

In this article, we study a Hardy-Sobolev critical elliptic system involving coupled perturbation terms:

(0.1) Δ u + V 1 ( x ) u = η 1 η 1 + η 2 u η 1 2 u v η 2 x + α α + β Q ( x ) u α 2 u v β , Δ v + V 2 ( x ) v = η 2 η 1 + η 2 v η 2 2 v u η 1 x + β α + β Q ( x ) v β 2 v u α ,

where n 3 , 2 m < n , x ( x , x ) R m × R n m , η 1 , η 2 > 1 , and η 1 + η 2 = 2 ( n 1 ) n 2 , α , β > 1 and α + β < 2 n n 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) C ( R n ) . Observing that (0.1) is doubly coupled, we first develop two efficient tools (i.e., a refined Sobolev inequality and a variant of the “Vanishing” lemma). On the previous tools, we will establish a global compactness result (i.e., a complete description for the Palais-Smale sequences of the corresponding energy functional) and some existence result for (0.1) via variational method. Our strategy turns out to be very concise because we avoid the use of Levy concentration functions and truncation techniques.

MSC 2010: 35A23; 35B33; 35J50

1 Introduction and main result

In this article, we consider a Hardy-Sobolev critical elliptic system involving coupled perturbation terms:

(1.1) Δ u + V 1 ( x ) u = η 1 2 u η 1 2 u v η 2 x + α α + β Q ( x ) u α 2 u v β , Δ v + V 2 ( x ) v = η 2 2 v η 2 2 v u η 1 x + β α + β Q ( x ) v β 2 v u α ,

where n 3 , 2 m < n , x ( x , x ) R m × R n m , η 1 , η 2 > 1 , and η 1 + η 2 = 2 2 ( n 1 ) n 2 , α , β > 1 and α + β < 2 2 n n 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) C ( R n ) .

Systems of type (1.1) originate from searching standing wave solutions of the following time-dependent Schrödinger systems:

(1.2) i t Ψ 1 = Δ Ψ 1 a 1 ( x ) Ψ 1 + g 1 ( Ψ 1 ) Ψ 1 + 1 G ( Ψ 1 2 , Ψ 2 2 , x ) Ψ 1 , i t Ψ 2 = Δ Ψ 2 a 2 ( x ) Ψ 2 + g 2 ( Ψ 2 ) Ψ 2 + 2 G ( Ψ 1 2 , Ψ 2 2 , x ) Ψ 2 , Ψ j = Ψ j ( x , t ) C , x R n , t > 0 , j = 1 , 2 , Ψ j ( x , t ) 0 , as x + , t > 0 , j = 1 , 2 ,

where i = 1 , a 1 ( x ) , and a 2 ( x ) are potential functions, and g 1 , g 2 , and G are some nonlinearities. System (1.2) has deep realistic backgrounds in various physical problems such as nonlinear optics and Bose-Einstein condensates theory (see [1,20,26]). A standing wave solution of (1.2) is a solution having the following form:

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

where ( u , v ) satisfies the following elliptic system:

(1.3) Δ u + ( a 1 ( x ) + λ 1 ) u = f 1 ( u ) + u F ( u , v , x ) , x R n , Δ v + ( a 2 ( x ) + λ 2 ) v = f 2 ( v ) + v F ( u , v , x ) , x R n ,

with ( λ 1 , λ 2 ) R 2 , f 1 ( u ) = g 1 ( u ) u , f 2 ( v ) = g 2 ( v ) v , and F ( u , v , x ) = 1 2 G ( u 2 , v 2 , x ) . Clearly, system (1.1) is a special case of (1.3) with

V i ( x ) = a i ( x ) + λ i , f i 0 ( i = 1 , 2 ) , F ( u , v , x ) = u ( x ) η 1 v ( x ) η 2 2 x + Q ( x ) u ( x ) α v ( x ) β α + β .

Another motivation on the study of system (1.1) is that it is related to a model describing the dynamics of elliptic galaxies in astrophysics (see [8,9,14,23,25]).

Liu in [21] studied (1.3) with

V i ( x ) = a i ( x ) + λ i , f i 0 ( i = 1 , 2 ) , F ( u , v ) = 2 u ( x ) α v ( x ) β α + β .

In this case, (1.3) became into the elliptic system of two equations:

(1.4) Δ u + V 1 ( x ) u = 2 α α + β u α 2 u v β , x R n , Δ v + V 2 ( x ) v = 2 β α + β u α v β 2 v , x R n ,

where n 3 , α , β > 1 , α + β < 2 n n 2 , V i ( x ) C 1 ( R n ) , and liminf x V i ( x ) = V i , > 0 for i = 1 , 2 . By means of the critical point theory, she found infinitely many approximate solutions in bounded balls. Then she obtained infinitely many solutions to (1.4) by analyzing the structures of the approximate solutions and passing to a limit.

Later on, Liu and Liu in [22] considered the Hardy-Sobolev critical elliptic system:

(1.5) Δ u γ u x 2 = 2 η 1 η 1 + η 2 u η 1 2 u v η 2 x α + λ 1 u , in Ω , Δ v γ v x 2 = 2 η 2 η 1 + η 2 u η 1 v η 2 2 v x α + λ 2 v , in Ω , u = v = 0 , on Ω ,

where Ω is an open bounded domain in R n ( n 3 ) , 0 < α < 2 , γ ( 0 , ( n 2 ) 2 4 ) , λ 1 , λ 2 > 0 , η 1 , η 2 > 1 , and η 1 + η 2 = 2 ( n α ) n 2 . Using the Moser iteration, they proved the exact behavior of positive solutions for (1.5) at the origin. In addition, they established a positive solution to (1.5) via the Mountain Pass lemma (see [2]).

In [29], Tian and Zhang investigated a doubly coupled system:

(1.6) Δ u + u = μ 1 u 2 u + β u v 2 κ v , x R n , Δ v + v = μ 2 v 2 v + β u 2 v κ u , x R n , u 0 , v 0 and u , v H 1 ( R n ) ,

where n = 2 , 3 , μ 2 μ 1 > 0 , κ , and β are linear and nonlinear coupling parameters. By using critical point theory and Liouville type theorem, the authors proved some existence and nonexistence results on the positive solutions of (1.6). Armed with the positive and nondegenerate solution of the scalar equation Δ ω + ω = ω 3 , ω H r 1 ( R n ) , they constructed a synchronized solution branch and proved that for β in certain range and fixed, there existed a series of bifurcations in product space R × H r 1 ( R n ) × H r 1 ( R n ) with parameter κ .

In [6], Benmouloud et al. studied a doubly critical elliptic system:

(1.7) Δ u = 2 α 2 u α 2 u v β + u 2 ( s ) 2 u x s + a 1 ( x ) u , in Ω , Δ v = 2 β 2 u α v β 2 v + v 2 ( s ) 2 v x s + a 2 ( x ) v , in Ω , u = v = 0 , on Ω ,

where n 3 , 0 < s < 2 , 2 ( s ) 2 ( n s ) n 2 , α , β > 1 , and α + β = 2 2 n n 2 , 0 Ω and Ω is an open-bounded domain in R n with C 3 boundary, a 1 ( x ) , a 2 ( x ) C 1 ( Ω ¯ ) and Ω satisfies some extra geometric condition. The authors extended the result of Yan and Yang [31] on equations to an elliptic system involving Sobolev and Hardy-Sobolev critical exponents in Ω . It was worth pointing out that they weakened the conditions on the dimension n and on the potential a ( x ) set in [31]. Actually, by a variational global compactness argument, they relaxed the condition on n and derived infinitely many solutions to (1.7).

In [32], Yang obtained a nontrivial weak solution ( u , v ) D 1 , p ( R n ) × D 1 , p ( R n ) to

(1.8) Δ p u γ 1 u p 2 u x p = u p ( β ) 2 u x β + η 1 p ( α ) u η 1 2 u v η 2 x α , Δ p v γ 2 v p 2 v x p = v p ( β ) 2 v x β + η 2 p ( α ) v η 2 2 v u η 1 x α ,

where n 2 , p ( 1 , n ) , 0 < α , β < p < m < n , γ 1 , γ 2 < ( m p p ) p , x = ( x , x ) R m × R n m , η 1 + η 2 = p ( α ) p ( n α ) n p , and 1 < η 1 η 2 < η 1 + α . The main difficulty is that system (1.8) is invariant under the transformation:

( u ( x ) , v ( x ) ) λ n p p u ( λ x , λ ( x y ) ) , λ n p p v ( λ x , λ ( x y ) ) , x = ( x , x ) R m × R n m ,

where λ > 0 and y R n m . For this reason, the author established an improved Sobolev inequality involving partially weighted Morrey norm, based on which he obtained a Mountain Pass-type solution of (1.8) via variational methods. Furthermore, he studied a fractional Laplace system in [32]. Bhakta et al. in [4] also considered the fractional Laplace system involving small perturbation terms in the dual norms.

The works we have just mentioned earlier is far from being complete. For more information about global compactness results on scalar equations, please refer to [5,10,12,15,16,17, 18,28] and so on. When it comes to systems, one can refer to [4,6,13,21,22,32] and the references therein.

The main purpose of this article is to establish a global compactness result (i.e., a complete description for the Palais-Smale sequences of the corresponding energy functional) and some existence result for the doubly coupled system (1.1). Observing that (1.1) contains both the Hardy-Sobolev critical exponent and the nonlinear coupling exponent, it is very different from the well-studied systems (1.4)–(1.8). To our best knowledge, (1.1) has not been studied before.

Throughout this article, we assume that V 1 ( x ) , V 2 ( x ) , and Q ( x ) satisfy the following conditions:

(V) V i ( x ) C ( R n ) , lim x V i ( x ) = V > 0 , inf x R n V i ( x ) = V i , 0 > 0 , where i = 1 , 2 ;

(Q) Q ( x ) C ( R n ) , lim x Q ( x ) = Q > 0 , inf x R n Q ( x ) = Q 0 > 0 .

For fixed ( u , v ) H 1 ( R n , R 2 ) , the energy functional of (1.1) is defined as follows:

(1.9) I ( u , v ) = 1 2 R n ( u 2 + V 1 ( x ) u 2 + v 2 + V 2 ( x ) v 2 ) d x 1 2 R n ( u + ) η 1 ( v + ) η 2 x d x 1 α + β R n Q ( x ) ( u + ( x ) ) α ( v + ( x ) ) β d x .

We say ( u , v ) H 1 ( R n , R 2 ) is a weak solution to (1.1) if

I ( u , v ) , ( φ , ψ ) = 0 , ( φ , ψ ) H 1 ( R n , R 2 ) .

Clearly, a critical point of I in H 1 ( R n , R 2 ) is a weak solution to (1.1). We say a weak solution ( u , v ) H 1 ( R n , R 2 ) of (1.1) is a semi-trivial solution if u 0 or v 0 . A weak solution ( u , v ) H 1 ( R n , R 2 ) of (1.1) is called a nontrivial solution if

u 0 and v 0 .

If u 0 and v 0 , we say that ( u , v ) is nonnegative.

The limit system of (1.1) involving Sobolev subcritical terms is

(1.10) Δ u + V u = α α + β Q u α 2 u v β , Δ v + V v = β α + β Q v β 2 v u α ,

and the associated energy functional is

(1.11) I ( u , v ) = 1 2 R n ( u 2 + V u 2 + v 2 + V v 2 ) d x Q α + β R n ( u + ( x ) ) α ( v + ( x ) ) β d x , ( u , v ) H 1 ( R n , R 2 ) .

Define

(1.12) D inf ( u , v ) N I ( u , v ) ,

where N { ( u , v ) H 1 ( R n , R 2 ) { ( 0 , 0 ) } : ( I ) ( u , v ) , ( u , v ) = 0 } is the Nehari manifold of (1.10). Following an argument similar to Theorem 5 of [3] or Lemma 2.2 of [4], we see that any ground state solution of system (1.10) is of the following form:

(1.13) W ( x ) 1 + β α α β β 2 V Q 1 α + β 2 , 1 + α β β α α 2 V Q 1 α + β 2 W ( V ( x x 0 ) )

for x 0 R n , where W ( x ) is the unique positive solution of (see [19])

(1.14) Δ u + u = u α + β 2 u , u ( x ) > 0 , in R n , u ( 0 ) = max x R n u ( x ) , u H 1 ( R n ) .

In particular, D is attained by W ( x ) (see (1.13)) and

D = I ( W ( x ) ) = 1 2 1 α + β ( α + β ) α + β α α β β 1 α + β 2 V α + β α + β 2 n 2 Q 2 α + β 2 W α + β α + β .

The limit system of (1.1) containing Hardy-Sobolev critical terms is expressed as follows:

(1.15) Δ u = η 1 2 u η 1 2 u v η 2 x , Δ v = η 2 2 v η 2 2 v u η 1 x ,

and the associated energy functional is expressed as follows:

(1.16) I ( u , v ) = 1 2 R n ( u 2 + v 2 ) d x 1 2 R n u η 1 v η 2 x d x , ( u , v ) D 1 , 2 ( R n , R 2 ) .

Solutions of system (1.15) are closely related to the scalar equation:

(1.17) Δ u = u 2 2 u x , x ( x , x ) R m × R n m , u D 1 , 2 ( R n ) .

In [23], Mancinia et al. proved that all the positive solutions of (1.17) are of the form

(1.18) U ε , ξ ( x , x ) ε 2 n 2 U x ε , x ξ ε for U ( x ) { ( n 2 ) ( m 1 ) } n 2 2 { ( 1 + x ) 2 + x 2 } n 2 2 ,

where ε > 0 and ξ R n m , which minimize the quotient:

S ( n , m ) inf u D 1 , 2 ( R n ) { 0 } R n u 2 d x R n u 2 x d x 2 2 .

In the same way as Theorem 5 of [3] or Lemma 2.2 of [4], we can show that any ground-state solution of system (1.15) is of the following form:

(1.19) U ( x ) 2 η 1 η 1 η 2 η 2 2 1 / ( 2 2 ) , 2 η 2 η 2 η 1 η 1 2 1 / ( 2 2 ) U ε , ξ ( x , x )

for U ε , ξ defined by (1.18). Furthermore, U ( x ) minimizes

(1.20) S ( η 1 , η 2 ) inf ( u , v ) D 1 , 2 ( R n , R 2 ) , u v 0 R n u 2 + v 2 d x R n u η 1 v η 2 x d x 2 2

and

(1.21) I ( U ( x ) ) = [ ( η 1 / η 2 ) η 2 / 2 + ( η 1 / η 2 ) η 1 / 2 ] n 1 2 ( n 1 ) S ( n , m ) n 1 = [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) .

Let X be a Banach space and X 1 be its dual space, Φ C 1 ( X , R ) and c R , we say that { ( u k , v k ) } k = 1 X is a Palais-Smale ((PS) in short) sequence of Φ if

Φ ( u k , v k ) c , Φ ( u k , v k ) 0 in X 1 as k .

Our first main result is the refined Sobolev inequality involving partly weighted Morrey norm (see (1.26) below), which is an efficient tool for further use in the study of (1.1).

Theorem 1.1

Let n 2 , 2 p < n , 0 < α < p < m < n , η 1 + η 2 = p ( α ) = p ( n α ) n p , and 1 < η 1 η 2 < η 1 + α . Then there exists C = C ( n , m , p , α , η 1 , η 2 ) > 0 such that for any ( u , v ) D 1 , p ( R n ) × D 1 , p ( R n ) and for any θ θ ¯ , 2 η 1 p ( α ) , it holds that

(1.22) R n u ( y ) η 1 v ( y ) η 2 y α d y 1 p ( α ) C u D 1 , p ( R n ) θ 2 v D 1 , p ( R n ) θ 2 + η 2 η 1 p ( α ) ( u v ) L p / 2 , n p + r ( R n , y r ) η 1 p ( α ) θ 2 ,

where y = ( y , y ) R m × R n m , θ ¯ = max p p ( α ) , 2 η 1 p ( α ) [ α ( η 2 η 1 ) ] 1 p m p ( α ) p α m , and r = p α p ( α ) .

Remark 1

Inequality (1.22) is a nontrivial extension of the inequality obtained by Palatucci and Pisante, see Theorem 2 of [27].

Benefiting from Theorem 1.1, we then obtain a global compactness result for (1.1).

Theorem 1.2

Let n > 4 , 2 < m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. If { ( u k , v k ) } k = 1 + is a nonnegative Palais-Smale sequence of I at level d 0 , then there exist three sequences { λ k ( i ) } R + ( 1 i l 1 ) , { ζ k ( i ) } R n m ( 1 i l 1 ) , and { ξ k ( j ) ( ξ k ( j ) , ξ k ( j ) ) } R n ( 1 j l 2 ) , 0 u , v H 1 ( R n ) , ( U ( i ) , V ( i ) ) D 1 , 2 ( R n , R 2 ) ( 1 i l 1 ) , ( u ( j ) , v ( j ) ) H 1 ( R n , R 2 ) ( 1 j l 2 ) , ( l 1 , l 2 N ) such that up to a subsequence:

(1.23) ( u k , v k ) ( u , v ) i = 1 l 1 ( λ k ( i ) ) 2 n 2 U ( i ) x λ k ( i ) , x λ k ( i ) + ζ k ( i ) , V ( i ) x λ k ( i ) , x λ k ( i ) + ζ k ( i ) j = 1 l 2 ( u ( j ) ( x ξ k ( j ) ) , v ( j ) ( x ξ k ( j ) ) ) H 1 ( R n , R 2 ) 0

and

(1.24) λ k ( i ) 0 ( 1 i l 1 ) , ξ k ( j ) and ξ k ( j ) ( 1 j l 2 )

as k , where ( u , v ) , ( U ( i ) , V ( i ) ) ( 1 i l 1 ) and ( u ( j ) , v ( j ) ) ( 1 j l 2 ) satisfy

(1.25) U ( i ) 0 ( 0 ) , V ( i ) 0 ( 0 ) , u ( j ) > 0 , v ( j ) > 0 , d = I ( u , v ) + i = 1 l 1 I ( U ( i ) , V ( i ) ) + j = 1 l 2 I ( u ( j ) , v ( j ) ) , I ( u , v ) = 0 , I ( U ( i ) , V ( i ) ) = 0 , ( I ) ( u ( j ) , v ( j ) ) = 0 .

In particular, if u 0 and v 0 , then ( u , v ) is a nonnegative weak solution of (1.1). Note that the corresponding sum in (1.23) will be treated as zero if l i = 0 ( i = 1 , 2 ) .

Remark 2

  1. Because of the lower order terms V 1 ( x ) u and V 2 ( x ) v in (1.1), we must work in H 1 ( R n , R 2 ) rather than D 1 , 2 ( R n , R 2 ) to ensure that the functional I ( u , v ) is well defined. Noting that ( u , v ) L 2 ( R n , R 2 ) and R n u ( x ) α v ( x ) β d x satisfy the translation invariance, this leads to the limit system (1.10). Moreover, R n u η 1 v η 2 x d x satisfies both the scaling invariance and the translation invariance with respect to x R n m , so we obtain another limit system (1.15). These two limit systems create two kinds of bubbles for the blowing up (PS) sequences of (1.1). For any nonnegative (PS) sequence { ( u k , v k ) } of I , ruling out these bubbles yields a nontrivial weak solution to (1.1).

  2. Usually, the scaling rates and the translation elements generated by { u k } and { v k } in (1.23) may not be the same because { u k } and { v k } are distinct. System (1.1) involves two nonlinear coupling parameters, which permit us to obtain a unified sequences of scaling rates and translation elements for both { u k } and { v k } . It is the L 2 -boundedness of the (PS) sequence that leads to λ k ( i ) 0 in (1.24). Note also that n > 4 forces ( U ( i ) , V ( i ) ) L 2 ( R n , R 2 ) .

  3. Similar to Corollary 3.3 of [28], we can prove that any nonnegative (PS) sequence for I at a level which is not of the form i = 1 l 1 I ( U ( i ) , V ( i ) ) + j = 1 l 2 I ( u ( j ) , v ( j ) ) , gives rise to a nontrivial weak solution for (1.1).

  4. When m = n , (1.1) reduces to the purely singular case, and similar results can be proved by modifying (1.22). In this case, the translations ζ k ( i ) in (1.23) have vanished, and we only obtain that ξ k ( j ) as k .

Armed with Theorem 1.2, we finally derive the following existence result for (1.1).

Theorem 1.3

Let n > 4 , 2 < m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. If

V 1 ( x ) V , V 2 ( x ) V , Q ( x ) Q , Q ( x ) Q ,

then (1.1) admits a nontrivial solution ( u , v ) H 1 ( R n , R 2 ) such that

u 0 ( 0 ) , v 0 ( 0 ) , 0 < I ( u , v ) < min [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) , D .

Moreover, we have u v provided

V 1 ( x ) V 2 ( x ) ( V 2 ( x ) ) , η 1 η 2 , α β

or

V 1 ( x ) V 2 ( x ) ( V 2 ( x ) ) , η 1 η 2 , α β .

Remark 3

Theorem 1.3 partly extends the results of [15, 18] on scalar equations to a doubly coupled system. We also observe that (1.6) contains both the linear and nonlinear coupling exponents, while (1.1) involves two nonlinear coupling parameters. Therefore, the study of (1.1) is more complicated than that of (1.6).

The proof of Theorem 1.1 relies on a norm inequality for Riesz potentials on weighted Lebesgue space, see Lemma 3.1. This idea originates from [27]. However, the coupled term R n u ( y ) η 1 v ( y ) η 2 y α d y makes the choice of the weight functions very subtle. In fact, for fixed ( u , v ) D 1 , p ( R n , R 2 ) , we must find a suitable weight function such that, the Riesz potentials corresponding to u and v stay in the same partly weighted L p space, while the product u v belongs to some partly weighted Morrey space. In addition, a precise control on some constants in terms of the partly weighted Morrey norm is more involved when m n and p 2 .

The main difficulties in proving Theorem 1.2 are that the energy functional of (1.1) contains both the Hardy-Sobolev critical term R n u ( x ) η 1 v ( x ) η 2 x d x and the nonlinear coupled term R n Q ( x ) u ( x ) α v ( x ) β d x . This is the reason why we establish the refined Sobolev inequality (1.22) and a variant of the “Vanishing” lemma (see Lemma 2.3). Armed with (1.22), we successfully exclude the blowing up bubbles originated from the Hardy-Sobolev critical term (see Proposition 3.3). After ruling out the blowing up bubbles caused by the unbounded domains (see Proposition 3.4), we then obtain a global compactness result for (1.1). In this way, we avoid the use of Levy concentration functions and truncation techniques, so the arguments become more transparent.

Theorem 1.3 is proved by using the Mountain Pass lemma (see [2]) and the global compactness result (see Theorem 1.2). We can check that the energy functional I satisfies the Mountain Pass geometry, so we obtain a Palais-Smale sequence { ( u k , v k ) } H 1 ( R n , R 2 ) for I at energy level c > 0 . Furthermore, two extra energy estimates indicate that c ( 0 , c ) for some threshold value c (see Proposition 4.2). In view of this, the energy level c is not of the form i = 1 l 1 I ( U ( i ) , V ( i ) ) + j = 1 l 2 I ( u ( j ) , v ( j ) ) , then Theorem 1.2 gives the strong convergence of { ( u k , v k ) } in H 1 ( R n , R 2 ) . Thus, we obtain a weak solution ( u , v ) for (1.1). Clearly, ( u , v ) is nontrivial because (1.1) is doubly coupled and c > 0 .

The rest of the article is organized as follows: in Section 2, we present some preliminary results. In Section 3, we prove Theorems 1.11.2. In Section 4, we prove Theorem 1.3.

Notations: Letting y = ( y , y ) R m × R n m , we denote the norm of L p ( R n , y λ ) by

u L p ( R n , y λ ) R n u ( y ) p y λ d y 1 p ,

where p 1 , 0 m n and 0 λ < n . The partly weighted Morrey space is defined by

L p , γ + λ ( R n , y λ ) { u L loc p ( R n , y λ ) , u L p , γ + λ ( R n , y λ ) < + } ,

where p [ 1 , + ) , γ , λ > 0 , γ + λ ( 0 , n ) , and the corresponding norm is

(1.26) u L p , γ + λ ( R n , y λ ) = sup R > 0 , x R n R γ + λ n B R ( x ) u ( y ) p y λ d y 1 p .

If λ = 0 , we obtain the standard Morrey space L p , γ ( R n ) , and see [27,32]. The space D 1 , p ( R n ) is the completion of C 0 ( R n ) under the norm u D 1 , p ( R n ) p R n u p d x . Denote H 1 ( R n ) = { u L 2 ( R n ) : u L 2 ( R n ) } with the norm ( u 2 2 + u 2 2 ) 1 2 .

For simplicity, (PS) sequence denotes “Palais-Smale” sequence in short. We use “ ” and “ ” to denote the strong and weak convergence in the related function spaces, respectively. , denotes the dual pair for any Banach space X and its dual space X 1 . N = { 1 , 2 , } is the set of natural numbers. R is the set of real numbers. o k ( 1 ) and O k ( 1 ) mean that o k ( 1 ) 0 and O k ( 1 ) C as k + , respectively. Denote u ± max { ± u , 0 } . C and C i will denote positive constants.

2 Preliminaries

In this section, we present some preliminary results.

First, we recall the following Hardy-Sobolev inequalities:

Lemma 2.1

([8], Theorem 2.1) Let n 2 , 1 < p < n , 0 < α p , α < m n , and p ( α ) = p ( n α ) n p . Then there exists a positive constant C = C ( α , p , n , m ) such that

R n u p ( α ) x α d x 1 p ( α ) C R n u p d x 1 p , u D 1 , p ( R n ) .

We also need two important lemmas: the concentration-compactness principle and a variant of the “Vanishing” lemma.

Lemma 2.2

([33], Lemma 2.1) Let { f k } k = 1 be a sequence in L 1 ( R n ) satisfying

f k 0 o n R n , lim k R n f k ( x ) d x = ε > 0 .

Then, up to a subsequence, one of the following two possibilities must occur:

  1. (Vanishing): for all 0 < R < + ,

    lim k sup y R n B R ( y ) f k ( x ) d x = 0 ;

  2. (Nonvanishing): there exist ε 0 > 0 , 0 < R 0 < + , and { y k } R n such that

    liminf k + B R 0 ( y k ) f k ( x ) d x ε 0 > 0 .

Lemma 2.3

Let n 3 . If { ( u k , v k ) } is bounded in H 1 ( R n , R 2 ) , and for some R > 0 , we have

sup y R n B R ( y ) u k ( x ) v k ( x ) d x 0 as k ,

then R n u k ( x ) α v k ( x ) β d x 0 for α , β > 1 satisfying α + β 2 2 ( n 1 ) n 2 .

Proof

The proof is motivated by Lemma 1.21 in [30]. Letting p = 2 ( α + β ) 2 2 α 1 > 1 and q = 2 ( α + β ) 2 2 β 1 > 1 , then we have 1 p + 1 q = 1 . By using Hölder’s inequality with 1 2 + 1 2 p + 1 2 q = 1 , we obtain

(2.1) Ω i u k ( x ) α v k ( x ) β d x = Ω i u k ( x ) v k ( x ) 1 2 u k ( x ) α 1 2 v k ( x ) β 1 2 d x Ω i u k ( x ) v k ( x ) d x 1 2 Ω i u k ( x ) p ( 2 α 1 ) d x 1 2 p Ω i v k ( x ) q ( 2 β 1 ) d x 1 2 q ,

where Ω i is any translation of the set { x ( x 1 , , x n ) R n : x j ( 0 , 1 ] , j = 1 , , n } . As α , β > 1 and α + β 2 , we see that p ( 2 α 1 ) = q ( 2 β 1 ) = 2 ( α + β ) 2 2 , 2 n n 2 . Therefore, we learn from (2.1) and sup y R n B R ( y ) u k ( x ) v k ( x ) d x 0 that

(2.2) Ω i u k ( x ) α v k ( x ) β d x o k ( 1 ) Ω i u k ( x ) 2 + u k ( x ) 2 d x α 2 1 4 Ω i v k ( x ) 2 + v k ( x ) 2 d x β 2 1 4 .

Covering R n by countable many Ω i , we obtain

R n u k ( x ) α v k ( x ) β d x o k ( 1 ) R n u k ( x ) 2 + u k ( x ) 2 d x α 2 1 4 R n v k ( x ) 2 + v k ( x ) 2 d x β 2 1 4 = o k ( 1 ) .

Next, we prove several useful facts about the functionals I and I .

Lemma 2.4

Let n 3 , 2 m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. If { ( u k , v k ) } H 1 ( R n , R 2 ) is a Palais-Smale sequence of I at level d R , then d 0 and { ( u k , v k ) } is bounded in H 1 ( R n , R 2 ) . Moreover, every Palais-Smale sequence for I at a level zero converges strongly to zero.

Proof

From I ( u k , v k ) d and I ( u k , v k ) 0 , we deduce that

d + o k ( 1 ) ( u k , v k ) H 1 ( R n , R 2 ) + o k ( 1 ) = I ( u k , v k ) 1 α + β I ( u k , v k ) , ( u k , v k ) = 1 2 1 α + β R n ( u k 2 + V 1 ( x ) u k 2 + v k 2 + V 2 ( x ) v k 2 ) d x + 1 α + β 1 2 R n ( u k + ) η 1 ( v k + ) η 2 x d x 1 2 1 α + β min { 1 , V 1 , 0 , V 2 , 0 } ( u k 2 2 + u k 2 2 + v k 2 2 + v k 2 2 ) ,

where we use condition (V) in the last inequality. Consequently, we have

d + o k ( 1 ) ( u k , v k ) H 1 ( R n , R 2 ) + o k ( 1 ) C ( u k 2 2 + u k 2 2 + v k 2 2 + v k 2 2 ) 0 .

It follows immediately that d 0 and { ( u k , v k ) } is bounded in H 1 ( R n , R 2 ) . Moreover, we have lim k ( u k , v k ) H 1 ( R n , R 2 ) = 0 if d = 0 .□

Lemma 2.5

Let n 3 , 2 m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. If { ( u k , v k ) } H 1 ( R n , R 2 ) is a Palais-Smale sequence of I at level d R , then { ( u k + , v k + ) } is also a Palais-Smale sequence of I at level d, where u k + max { u k , 0 } and v k + max { v k , 0 } .

Proof

Since { ( u k , v k ) } H 1 ( R n , R 2 ) is a Palais-Smale sequence of I at level d , we have

d + o k ( 1 ) = I ( u k , v k ) = 1 2 R n ( u k 2 + V 1 ( x ) u k 2 + v k 2 + V 2 ( x ) v k 2 ) d x 1 2 R n ( u k + ) η 1 ( v k + ) η 2 x d x 1 α + β R n Q ( x ) ( u k + ( x ) ) α ( v k + ( x ) ) β d x

and

(2.3) I ( u k , v k ) , ( φ , ψ ) = R n ( u k φ + V 1 ( x ) u k φ + v k ψ + V 2 ( x ) v k ψ ) d x η 1 2 R n ( u k + ) η 1 1 ( v k + ) η 2 φ x d x η 2 2 R n ( v k + ) η 2 1 ( u k + ) η 1 ψ x d x α α + β R n Q ( x ) ( u k + ) α 1 ( v k + ) β φ d x β α + β R n Q ( x ) ( u k + ) α ( v k + ) β 1 ψ d x = o k ( 1 ) ( φ , ψ ) H 1 ( R n , R 2 ) , ( φ , ψ ) H 1 ( R n , R 2 ) .

Observe that u k ( x ) = u k + ( x ) u k ( x ) and u k + ( x ) u k ( x ) = 0 , where u k ± max { ± u k , 0 } . Testing (2.3) with ( φ , ψ ) = ( u k , 0 ) , then we deduce from condition (V) that

(2.4) o k ( 1 ) u k H 1 ( R n ) = I ( u k , v k ) , ( u k , 0 ) = R n ( u k 2 + V 1 ( x ) u k 2 ) d x + η 1 2 R n ( u k + ) η 1 1 ( v k + ) η 2 u k x d x + α α + β R n Q ( x ) ( u k + ) α 1 ( v k + ) β u k d x C ( u k 2 2 + u k 2 2 ) .

Obviously, { ( u k , v k ) } is bounded in H 1 ( R n , R 2 ) . Therefore, we have u k H 1 ( R n ) u k H 1 ( R n ) C . Recalling (2.4) and testing (2.3) with ( φ , ψ ) = ( 0 , v k ) , we obtain

(2.5) lim k ( u k , v k ) H 1 ( R n , R 2 ) = 0 ,

which gives

lim k I ( u k + , v k + ) = lim k I ( u k , v k ) = d

and

I ( u k + , v k + ) , ( φ , ψ ) = I ( u k , v k ) , ( φ , ψ ) + o k ( 1 ) ( φ , ψ ) H 1 ( R n , R 2 ) .

Therefore, we have

I ( u k + , v k + ) , ( φ , ψ ) = o k ( 1 ) ( φ , ψ ) H 1 ( R n , R 2 ) , ( φ , ψ ) H 1 ( R n , R 2 ) .

The proof is ended.□

Lemma 2.6

Let ( u , v ) D 1 , 2 ( R n , R 2 ) be a nontrivial critical point of I , then ( u , v ) is a positive solution of (1.15) and there exists C > 0 such that

(2.6) 0 u ( x ) , v ( x ) C 1 + x n 2 , x R n .

Proof

Let ( u , v ) D 1 , 2 ( R n , R 2 ) be a nontrivial critical point of I , then

I ( u , v ) , ( φ , ψ ) = 0 , ( φ , ψ ) D 1 , 2 ( R n , R 2 ) .

Taking ( φ , ψ ) = ( u , 0 ) and ( φ , ψ ) = ( 0 , v ) , respectively, we obtain ( u , v ) D 1 , 2 ( R n , R 2 ) = 0 . Therefore, we have

I ( u + , v + ) , ( φ , ψ ) = 0 , ( φ , ψ ) D 1 , 2 ( R n , R 2 ) .

This implies that ( u , v ) weakly solves

Δ u = η 1 2 ( u + ( x ) ) η 1 1 ( v + ( x ) ) η 2 x 0 , Δ v = η 2 2 ( v + ( x ) ) η 2 1 ( u + ( x ) ) η 1 x 0 .

By standard elliptic regularity theory, we have u , v C 2 ( R n ) when x 0 (see [7]). Hence, u , v 0 and the strong maximum principle gives u > 0 and v > 0 if x 0 . Finally, (2.6) can be proved by modifying the arguments of Proposition 6.4 in [11].□

Lemma 2.7

Let n 3 , 2 m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. If { ( u k , v k ) } H 1 ( R n , R 2 ) is a bounded Palais-Smale sequence of I at level d R and ( u k , v k ) ( u , v ) in H 1 ( R n , R 2 ) , then ( u k u , v k v ) is a Palais-Smale sequence of I at level d I ( u , v ) .

Proof

It is obvious that ( u k u , v k v ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , so we have

(2.7) ( u k u , v k v ) ( 0 , 0 ) in L loc r ( R n , R 2 ) ( 2 r < 2 ) , ( u k u , v k v ) ( 0 , 0 ) a.e. on R n × R n .

For φ , ψ C 0 ( R n ) , there exists a ball B R ( 0 ) such that supp φ , supp ψ B R ( 0 ) . We derive from Hölder’s inequality with α 1 α + β + β α + β + 1 α + β = 1 that

(2.8) R n Q ( x ) ( ( u k u ) + ) α 1 ( ( v k v ) + ) β φ d x C B R ( 0 ) ( ( u k u ) + ) α 1 ( ( v k v ) + ) β φ d x C ( u k u ) + L α + β ( B R ( 0 ) ) α 1 ( v k v ) + L α + β ( B R ( 0 ) ) β φ L α + β ( B R ( 0 ) ) = o k ( 1 ) .

In parallel with (2.8), we have

(2.9) R n Q ( x ) ( ( u k u ) + ) α ( ( v k v ) + ) β 1 ψ d x = o k ( 1 ) .

From η 1 , η 2 > 1 , and η 1 + η 2 = 2 2 ( n 1 ) n 2 , we obtain

η 2 2 1 = 2 η 1 2 1 ( 0 , 1 ) , η 1 2 1 = 2 η 2 2 1 ( 0 , 1 ) .

In view of the aforementioned facts, we can use Hölder’s inequality with η 1 1 2 1 + η 2 2 1 = 1 to obtain

(2.10) R n ( ( u k u ) + ) η 1 1 ( ( v k v ) + ) η 2 x 1 1 2 2 2 1 d x = R n ( ( u k u ) + ) 2 ( η 1 1 ) 2 1 x η 1 1 2 1 ( ( v k v ) + ) 2 η 2 2 1 x η 2 2 1 d x R n ( ( u k u ) + ) 2 x d x η 1 1 2 1 R n ( ( v k v ) + ) 2 x d x η 2 2 1 C .

Consequently, (2.7) and (2.10) indicate that ( ( u k u ) + ) η 1 1 ( ( v k v ) + ) η 2 x 1 1 2 0 in L 2 2 1 ( R n ) . Since φ ( x ) x 1 2 L 2 ( R n ) , we obtain

(2.11) R n ( ( u k u ) + ) η 1 1 ( ( v k v ) + ) η 2 φ x d x R n ( ( u k u ) + ) η 1 1 ( ( v k v ) + ) η 2 x 1 1 2 φ x 1 2 d x = o k ( 1 ) .

Similarly, we also deduce that

(2.12) R n ( ( u k u ) + ) η 1 ( ( v k v ) + ) η 2 1 ψ x d x = o k ( 1 ) .

By using (2.8), (2.9), (2.11), and (2.12), we have

I ( u k u , v k v ) , ( φ , ψ ) = o k ( 1 ) .

In the same way as (2.4), we have

( ( u k u ) , ( v k v ) ) H 1 ( R n , R 2 ) 0 , ( u , v ) H 1 ( R n , R 2 ) = 0 .

From the Brézis-Lieb type Lemma (see [13], Lemma 2.3), we have

R n ( ( u k u ) + ) η 1 ( ( v k v ) + ) η 2 x d x = R n ( u k + ) η 1 ( v k + ) η 2 x d x R n ( u + ) η 1 ( v + ) η 2 x d x + o k ( 1 ) , R n Q ( x ) ( ( u k u ) + ) α ( ( v k v ) + ) β d x = R n Q ( x ) ( u k + ) α ( v k + ) β d x R n Q ( x ) ( u + ) α ( v + ) β d x + o k ( 1 ) .

We end the proof by observing that

I ( u k u , v k v ) = I ( u k , v k ) I ( u , v ) + o k ( 1 ) = d I ( u , v ) + o k ( 1 ) .

3 The proofs of Theorems 1.1–1.2

In this section, we first prove the improved Sobolev inequality (1.22), and see Theorem 1.1. Then, we adopt (1.22) to derive a global compactness result for (1.1), and see Theorem 1.2.

We need a lemma, which is vital for the proof of Theorem 1.1.

Lemma 3.1

([24], Theorem D) Suppose that 0 < s ˜ < n , 1 < p ˜ q ˜ < + , p ˜ = p ˜ p ˜ 1 and that V and W are nonnegative measurable functions on R n , n 1 . If for some σ > 1

(3.1) Q s ˜ n + 1 q ˜ 1 p ˜ 1 Q Q V σ d y 1 q ˜ σ 1 Q Q W ( 1 p ˜ ) σ d y 1 p ˜ σ C σ

for all cubes Q R n , then for any function f L p ˜ ( R n , W ( y ) ) , we have

(3.2) R n s ˜ f ( y ) q ˜ V ( y ) d y 1 q ˜ C C σ R n f ( y ) p ˜ W ( y ) d y 1 p ˜ ,

where C = C ( p ˜ , q ˜ , n ) , and s ˜ f ( y ) = R n f ( z ) y z n s ˜ d z is the Riesz potential of order s ˜ .

In [32], Yang only gives an outline on the proof of (1.22). Actually, it is particularly difficult to deal with the general case p 2 . We provide a detailed proof in the following for the convenience of the readers.

Proof of Theorem 1.1

For any u C 0 ( R n ) , we have

u ( x ) = Δ 1 Δ u ( x ) = 1 ( n 2 ) ω n R n Δ u ( y ) x y n 2 d y = 1 ω n R n ( x y ) u ( y ) x y n d y , n 3 ; u ( x ) = Δ 1 Δ u ( x ) = 1 ω n R n log x y Δ u ( y ) d y = 1 ω n R n ( x y ) u ( y ) x y 2 d y , n = 2 .

Thus, the following inequality holds

(3.3) u ( x ) 1 ω n R n u ( y ) x y n 1 d y 1 ω n 1 ( u ) ( x ) , n 2 ,

where ω n = 2 π n / 2 Γ ( n / 2 ) = B 1 ( 0 ) denotes the surface of unit sphere in R n (see [27]). By density of C 0 ( R n ) in D 1 , p ( R n ) , it is also true for any u D 1 , p ( R n ) ( n 2 ) .

Case (i): η 1 < η 2 . First, let 2 p < m < n , η 1 + η 2 = p ( α ) with 1 < η 1 < η 2 < η 1 + α and denote

t 1 η 2 η 1 α ( 0 , 1 ) .

Take s ˜ = 1 , p ˜ = p , σ = p α > 1 , max p , 2 η 1 t α , 2 η 1 t α 1 p m p ( α ) p α m p ( α ) , 2 η 1 t p ( α ) < q ˜ < 2 η 1 , W ( y ) 1 and V ( y ) = u ( y ) η 1 q ˜ 2 v ( y ) η 2 q ˜ 2 y α in Lemma 3.1, where y = ( y , y ) R m × R n m . Then (3.1) becomes

(3.4) Q 1 n + 1 q ˜ 1 p 1 Q Q V σ d y 1 q ˜ σ C σ .

For any cubes Q R n centered at x , there exists R > 0 such that B R n ( x ) Q B R ( x ) . We deduce 1 n + 1 q ˜ 1 p > 0 by q ˜ < 2 η 1 < p ( α ) , and it results that

(3.5) Q 1 n + 1 q ˜ 1 p 1 Q Q V σ d y 1 q ˜ σ B R ( x ) 1 n + 1 q ˜ 1 p 1 B R n ( x ) B R ( x ) V σ d y 1 q ˜ σ C R 1 + n q ˜ n p R n B R ( x ) V σ d y 1 q ˜ σ .

Second, we verify condition (3.4). From (3.5), it is sufficient to verify

R 1 + n q ˜ n p R n B R ( x ) V σ d y 1 q ˜ σ C σ .

Since 2 η 1 t p ( α ) < q ˜ < 2 η 1 , we can define

ρ 1 2 η 1 q ˜ p ( α ) 1 t ( 0 , 1 ) .

By using Hölder’s inequality and Hardy’s inequality (see Lemma 2.1), we have

(3.6) R n B R ( x ) V σ d y = R n B R ( x ) u η 1 q ˜ 2 σ v ( η 2 q ˜ 2 ) σ y σ α d y = R n B R ( x ) u v η 1 q ˜ 2 σ y t p v ( η 2 η 1 ) σ y ( 1 t ) p d y R n B R ( x ) u v η 1 q ˜ 2 σ t y p d y t B R ( x ) v ( η 2 η 1 ) σ 1 t y p d y 1 t = R n B R ( x ) u v η 1 q ˜ 2 σ t y p d y t B R ( x ) v p y p d y 1 t R n B R ( x ) u v η 1 q ˜ 2 σ t y p d y t [ C v D 1 , p ( R n ) p ] 1 t .

From 2 η 1 t α < q ˜ < 2 η 1 , we have 0 < η 1 q ˜ 2 2 σ p t < 1 . Moreover, we have ρ p 1 η 1 q ˜ 2 2 σ p t < m by q ˜ > 2 η 1 t α 1 p m p ( α ) p α m p ( α ) . Recall that ρ ( 0 , 1 ) , it results to

(3.7) B R ( x ) u v η 1 q ˜ 2 σ t y p d y = B R ( x ) 1 y ρ p u v η 1 q ˜ 2 σ t y ( 1 ρ ) p d y B R ( x ) 1 y ρ p 1 η 1 q ˜ 2 2 σ p t d y 1 η 1 q ˜ 2 2 σ p t B R ( x ) u v p 2 y ( 1 ρ ) p η 1 q ˜ 2 2 σ p t d y η 1 q ˜ 2 2 σ p t B R ( 0 ) 1 y ρ p 1 η 1 q ˜ 2 2 σ p t d y 1 η 1 q ˜ 2 2 σ p t B R ( x ) u v p 2 y r d y η 1 q ˜ 2 2 σ p t = C R n ρ p 1 η 1 q ˜ 2 2 σ p t 1 η 1 q ˜ 2 2 σ p t B R ( x ) u v p 2 y r d y η 1 q ˜ 2 2 σ p t C R ρ p + n n η 1 q ˜ 2 2 σ p t B R ( x ) u v p 2 y r d y η 1 q ˜ 2 2 σ p t ,

where r ( 1 ρ ) p η 1 q ˜ 2 2 σ p t = p α p ( α ) . Substituting (3.7) into (3.6), we obtain

R n B R ( x ) V σ d y C R ρ p t n + t n 1 η 1 q ˜ 2 2 σ p t B R ( x ) u v p 2 y r d y η 1 q ˜ 2 2 σ p v D 1 , p ( R n ) p ( 1 t ) .

Therefore,

R 1 + n q ˜ n p R n B R ( x ) V σ d y 1 q ˜ σ R 1 + n q ˜ n p C R ρ p t n + t n [ 1 η 1 q ˜ 2 2 σ p t ] B R ( x ) u v p 2 y r d y η 1 q ˜ 2 2 σ p v D 1 , p ( R n ) p ( 1 t ) 1 q ˜ σ = C R q ˜ σ ( 1 + n q ˜ n p ) ρ p t n + t n [ 1 η 1 q ˜ 2 2 σ p t ] B R ( x ) u v p 2 y r d y η 1 q ˜ 2 2 σ p v D 1 , p ( R n ) p ( 1 t ) 1 q ˜ σ = C R q ˜ σ ( 1 + n q ˜ n p ) ρ p t n + t n [ 1 η 1 q ˜ 2 2 σ p t ] B R ( x ) u v p 2 y r d y η 1 q ˜ 2 2 σ p 1 q ˜ σ v D 1 , p ( R n ) p ( 1 t ) q ˜ σ = C R q ˜ σ ( 1 + n q ˜ n p ) ρ p t n + t n [ 1 η 1 q ˜ 2 2 σ p t ] η 1 q ˜ 2 2 σ p B R ( x ) u v p 2 y r d y 2 η 1 q ˜ p q ˜ v D 1 , p ( R n ) p ( 1 t ) q ˜ σ = C R n p + r R n B R ( x ) u v p 2 y r d y 2 p η 1 q ˜ 2 q ˜ v D 1 , p ( R n ) p ( 1 t ) q ˜ σ C ( u v ) L p / 2 , n p + r ( R n , y r ) η 1 q ˜ 2 q ˜ v D 1 , p ( R n ) p ( 1 t ) q ˜ σ C σ .

For any ( u , v ) D 1 , p ( R n , R 2 ) , (3.3) indicates that u ( y ) C 1 ( u ) ( y ) and v ( y ) C 1 ( v ) ( y ) . Then, we deduce from Lemma 3.1 that

R n u ( y ) η 1 v ( y ) η 2 y α d y = R n V ( y ) u ( y ) q ˜ 2 v ( y ) q ˜ 2 d y R n V ( y ) u ( y ) q ˜ d y 1 2 R n V ( y ) v ( y ) q ˜ d y 1 2 C R n V ( y ) 1 ( u ) ( y ) q ˜ d y 1 2 R n V ( y ) 1 ( v ) ( y ) q ˜ d y 1 2 ( C C σ ) q ˜ u L p ( R n ) q ˜ 2 v L p ( R n ) q ˜ 2 C ( u v ) L p / 2 , n p + r ( R n , y r ) η 1 q ˜ 2 u D 1 , p ( R n ) q ˜ 2 v D 1 , p ( R n ) q ˜ 2 + p ( 1 t ) σ = C u D 1 , p ( R n ) q ˜ 2 v D 1 , p ( R n ) q ˜ 2 + ( η 2 η 1 ) ( u v ) L p / 2 , n p + r ( R n , y r ) η 1 q ˜ 2 .

Then, for any θ = q ˜ p ( α ) satisfying θ ¯ < θ < 2 η 1 p ( α ) < 1 , we have

R n u ( y ) η 1 v ( y ) η 2 y α d y 1 p ( α ) C u D 1 , p ( R n ) θ 2 v D 1 , p ( R n ) θ 2 + η 2 η 1 p ( α ) ( u v ) L p / 2 , n p + r ( R n , y r ) η 1 p ( α ) θ 2 ,

where r = p α p ( α ) , θ ¯ = max p p ( α ) , 2 η 1 t α p ( α ) , 2 η 1 p ( α ) t α 1 p m p ( α ) p α m , 2 η 1 p ( α ) t and t = 1 η 2 η 1 α . We also observe that θ ¯ can be reduced to θ ¯ = max p p ( α ) , 2 η 1 p ( α ) t α 1 p m p ( α ) p α m .

Case (ii): η 1 = η 2 . Let 2 p < m < n and take s ˜ = 1 , p ˜ = p , max p , p ( α ) α , p ( α ) α p ( α ) p α m p ( α ) < q ˜ < p ( α ) , W ( y ) 1 , V ( y ) = ( u v ) ( y ) p ( α ) q ˜ 2 y α , and σ = p α > 1 in Lemma 3.1, where y = ( y , y ) R m × R n m . The rest is similar to Case (i) by resetting t q ˜ p ( α ) , r p ( 1 t ) α p ( α ) q ˜ = p α p ( α ) , θ = q ˜ p ( α ) ( θ ¯ , 1 ) , and θ ¯ = max p p ( α ) , 1 α 1 p m p ( α ) p α m . Actually, we have

R n ( u v ) ( y ) p ( α ) 2 y α d y 1 p ( α ) C u D 1 , p ( R n ) θ 2 v D 1 , p ( R n ) θ 2 ( u v ) L p / 2 , n p + r ( R n , y r ) 1 θ 2 .

As a consequence of Theorem 1.1, we obtain a useful proposition.

Proposition 3.2

Let n 3 , 2 < m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 . If { ( u k , v k ) } k = 1 is a nonnegative bounded sequence in D 1 , 2 ( R n , R 2 ) with

inf k N R n ( u k + ( y ) ) η 1 ( v k + ( y ) ) η 2 y d y C > 0

for some constant C, then there exist, up to a subsequence, a family of positive numbers { λ k } , a sequence of { x k } R n m , and a pair of functions ( u ˜ , v ˜ ) D 1 , 2 ( R n , R 2 ) such that

λ k n 2 2 u k ( λ k x , λ k ( x x k ) ) , λ k n 2 2 v k ( λ k x , λ k ( x x k ) ) ( u ˜ , v ˜ ) in D 1 , 2 ( R n , R 2 )

with u ˜ 0 and v ˜ 0 .

Proof

Taking p = 2 , α = 1 , and ( u , v ) = ( u k + , v k + ) in Theorem 1.1, then there exists C > 0 such that

0 < C u k + v k + L 1 , n 2 + r ( R n , y r ) u k + L 2 , n 2 + r ( R n , y r ) v k + L 2 , n 2 + r ( R n , y r ) C 1 ,

where r = n 2 n 1 < 2 . In the last inequality, we use the fact that D 1 , 2 ( R n ) L 2 , n 2 + r ( R n , y r ) . For any k K large, we may find λ k > 0 and x k = ( x k , x k ) R m × R n m such that

λ k 2 + r B λ k ( x k ) ( u k + v k + ) ( y ) y r d y > ( u k + v k + ) L 1 , n 2 + r ( R n , y r ) C 2 k C 1 > 0 ,

where y = ( y , y ) R m × R n m . Let x ˜ k x k λ k = ( x ˜ k , x ˜ k ) R m × R n m and

( u ˜ k ( x ) , v ˜ k ( x ) ) λ k n 2 2 u k + ( λ k x , λ k ( x x k ) ) , λ k n 2 2 v k + ( λ k x , λ k ( x x k ) ) ,

then

(3.8) B 1 ( ( x ˜ k , 0 ) ) u ˜ k ( x ) v ˜ k ( x ) x r d x C 1 > 0 ,

where x = ( x , x ) R m × R n m and 0 = ( 0 , , 0 ) R n m . We claim that { x ˜ k } is bounded. Otherwise, x ˜ k + , then for any x B 1 ( ( x ˜ k , 0 ) ) , x x ˜ k 1 for k large. Therefore,

B 1 ( ( x ˜ k , 0 ) ) u ˜ k ( x ) v ˜ k ( x ) x r d x 1 ( x ˜ k 1 ) r B 1 ( ( x ˜ k , 0 ) ) u ˜ k ( x ) v ˜ k ( x ) d x B 1 ( ( x ˜ k , 0 ) ) d x 1 2 2 ( x ˜ k 1 ) r B 1 ( ( x ˜ k , 0 ) ) u ˜ k ( x ) 2 d x 1 2 B 1 ( ( x ˜ k , 0 ) ) v ˜ k ( x ) 2 d x 1 2 C ( x ˜ k 1 ) r u ˜ k D 1 , 2 ( R n ) v ˜ k D 1 , 2 ( R n ) C ˜ ( x ˜ k 1 ) r 0

as k + , which contradicts to (3.8). It must be that x ˜ k C , so there exists R > 0 such that B R ( 0 ) u ˜ k ( x ) v ˜ k ( x ) x r d x C 1 > 0 . By using r < 2 and Hardy’s inequality (see Lemma 2.1), we have

(3.9) 0 < C 1 B R ( 0 ) u ˜ k ( x ) v ˜ k ( x ) x r d x B R ( 0 ) u ˜ k ( x ) 2 x r d x 1 2 B R ( 0 ) v ˜ k ( x ) 2 x r d x 1 2 = B R ( 0 ) u ˜ k ( x ) r x r u ˜ k ( x ) 2 r d x 1 2 B R ( 0 ) v ˜ k ( x ) r x r u ˜ k ( x ) 2 r d x 1 2 B R ( 0 ) u ˜ k ( x ) 2 x 2 d x r 4 B R ( 0 ) u ˜ k ( x ) 2 d x 2 r 4 B R ( 0 ) v ˜ k ( x ) 2 x 2 d x r 4 B R ( 0 ) v ˜ k ( x ) 2 d x 2 r 4 C B R ( 0 ) u ˜ k ( x ) 2 d x 2 r 4 B R ( 0 ) v ˜ k ( x ) 2 d x 2 r 4 .

Since ( u ˜ k , v ˜ k ) D 1 , 2 ( R n , R 2 ) = ( u k , v k ) D 1 , 2 ( R n , R 2 ) C , there exists ( u ˜ , v ˜ ) D 1 , 2 ( R n , R 2 ) such that

( u ˜ k , v ˜ k ) ( u ˜ , v ˜ ) in D 1 , 2 ( R n , R 2 ) , ( u ˜ k , v ˜ k ) ( u ˜ , v ˜ ) in L loc 2 ( R n , R 2 ) .

Finally, (3.9) gives

(3.10) B R ( 0 ) u ˜ ( x ) 2 d x B R ( 0 ) v ˜ ( x ) 2 d x = lim k B R ( 0 ) u ˜ k ( x ) 2 d x B R ( 0 ) v ˜ k ( x ) 2 d x C 1 > 0 ,

so we have u ˜ 0 and v ˜ 0 .□

Benefiting from Proposition 3.2, we can exclude the blowing up bubbles originated from the Hardy-Sobolev critical terms. Moreover, we see that any nonnegative Palais-Smale sequence of I which converges weakly but not strongly to zero in H 1 ( R n , R 2 ) must concentrate on some balls. Actually, we obtain the following Proposition 3.3.

Proposition 3.3

Let n > 4 , 2 < m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. If { ( u k , v k ) } k = 1 H 1 ( R n , R 2 ) is a nonnegative bounded Palais-Smale sequence of I at level d 1 0 such that ( u k , v k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) and

inf k N R n ( u k + ( x ) ) η 1 ( v k + ( x ) ) η 2 x d x C > 0 ,

then there exist, up to a subsequence, a family of positive numbers { λ k } with λ k 0 , a sequence of { x k } R n m and a pair of functions ( u 0 , v 0 ) D 1 , 2 ( R n , R 2 ) such that

λ k n 2 2 u k ( λ k x , λ k ( x x k ) ) , λ k n 2 2 v k ( λ k x , λ k ( x x k ) ) ( u 0 ( x ) , v 0 ( x ) ) in D 1 , 2 ( R n , R 2 )

with u 0 0 and v 0 0 . In addition, ( u 0 , v 0 ) solves (1.15) and the sequence

( z k ( x ) , ν k ( x ) ) u k ( x ) λ k 2 n 2 u 0 x λ k , x λ k + x k , v k ( x ) λ k 2 n 2 v 0 x λ k , x λ k + x k

is a Palais-Smale sequence of I at level d 1 I ( u 0 , v 0 ) .

Proof

(i) We prove the existence of { λ k } such that λ k 0 . It is easy to check that { ( u k , v k ) } satisfy all the assumptions of Proposition 3.2, so there exist, up to a subsequence, a family of positive numbers { λ k } , a sequence of { x k } R n m and a pair of functions ( u 0 , v 0 ) D 1 , 2 ( R n , R 2 ) such that

( u ˜ k ( x ) , v ˜ k ( x ) ) λ k n 2 2 ( u k ( λ k x , λ k ( x x k ) ) , v k ( λ k x , λ k ( x x k ) ) ) ( u 0 ( x ) , v 0 ( x ) )

in D 1 , 2 ( R n , R 2 ) with u 0 0 and v 0 0 . Also, we have

(3.11) ( u ˜ k , v ˜ k ) ( u 0 , v 0 ) in L loc r ( R n , R 2 ) ( 2 r < 2 ) , ( u ˜ k , v ˜ k ) ( u 0 , v 0 ) a.e. on R n × R n , ( u k , v k ) ( 0 , 0 ) in L loc r ( R n , R 2 ) ( 2 r < 2 ) , ( u k , v k ) ( 0 , 0 ) a.e. on R n × R n .

Similar to (3.10), we can prove that there exists R > 0 such that

B R ( 0 ) u 0 ( x ) 2 d x B R ( 0 ) v 0 ( x ) 2 d x = lim k B R ( 0 ) u ˜ k ( x ) 2 d x B R ( 0 ) v ˜ k ( x ) 2 d x C > 0 .

We claim that λ k 0 . Otherwise, λ k + or λ k λ 0 > 0 . If λ k + , we have

0 < C B R ( 0 ) u 0 ( x ) 2 d x B R ( 0 ) v 0 ( x ) 2 d x = lim k B R ( 0 ) u ˜ k ( x ) 2 d x B R ( 0 ) v ˜ k ( x ) 2 d x = lim k λ k 4 B λ k R ( ( 0 , λ k x k ) ) u k ( x ) 2 d x B λ k R ( ( 0 , λ k x k ) ) v k ( x ) 2 d x = 0 .

If λ k λ 0 > 0 and { x k } is bounded, then there exists C 1 > 0 such that

0 < C B R ( 0 ) u 0 ( x ) 2 d x B R ( 0 ) v 0 ( x ) 2 d x = lim k λ k 4 B λ k R ( ( 0 , λ k x k ) ) u k ( x ) 2 d x B λ k R ( ( 0 , λ k x k ) ) v k ( x ) 2 d x lim k λ 0 4 B λ 0 R C 1 ( 0 ) u k ( x ) 2 d x B λ 0 R C 1 ( 0 ) v k ( x ) 2 d x = 0 .

If λ k λ 0 > 0 and x k + , taking a radially decreasing function ϕ ( x ) C 0 ( R n ) such that 0 ϕ ( x ) 1 , ϕ ( x ) 1 on B R ( 0 ) , ϕ ( x ) 0 on R n B 2 R ( 0 ) , then we have

0 < C B R ( 0 ) u 0 ( x ) 2 d x B R ( 0 ) v 0 ( x ) 2 d x lim k R n u ˜ k ( x ) 2 ϕ ( x ) 2 d x R n v ˜ k ( x ) 2 ϕ ( x ) 2 d x = lim k λ k 4 R n u k ( x ) 2 ϕ x λ k , x λ k + x k 2 d x R n v k ( x ) 2 ϕ x λ k , x λ k + x k 2 d x = 0 .

This proves λ k 0 .

(ii) We prove that ( u 0 , v 0 ) is a critical point of I and ( u 0 , v 0 ) solves (1.15). Let ( φ , ψ ) C 0 ( R n , R 2 ) and

( φ k ( x ) , ψ k ( x ) ) λ k 2 n 2 φ x λ k , x λ k + x k , λ k 2 n 2 ψ x λ k , x λ k + x k ,

then we have

(3.12) φ k H 1 ( R n ) 2 = φ k L 2 ( R n ) 2 + φ k L 2 ( R n ) 2 = φ L 2 ( R n ) 2 + λ k 2 φ L 2 ( R n ) 2 φ L 2 ( R n ) 2 , ψ k H 1 ( R n ) 2 = ψ k L 2 ( R n ) 2 + ψ k L 2 ( R n ) 2 = ψ L 2 ( R n ) 2 + λ k 2 ψ L 2 ( R n ) 2 ψ L 2 ( R n ) 2 .

Furthermore, we deduce from λ k 0 and α + β < 2 that

(3.13) R n Q ( x ) ( u k + ( x ) ) α 1 ( v k + ( x ) ) β φ k ( x ) d x = λ k n ( α + β ) ( n 2 ) 2 R n Q ( λ k y , λ k ( y x k ) ) ( u ˜ k ( y ) ) α 1 ( v ˜ k ( y ) ) β φ ( y ) d y = o k ( 1 ) , R n V 1 ( x ) u k ( x ) φ k ( x ) d x = λ k 2 R n V 1 ( λ k y , λ k ( y x k ) ) u ˜ k ( y ) φ ( y ) d y = o k ( 1 ) .

In the same way, we also have

(3.14) R n Q ( x ) ( u k + ( x ) ) α ( v k + ( x ) ) β 1 ψ k ( x ) d x = λ k n ( α + β ) ( n 2 ) 2 R n Q ( λ k y , λ k ( y x k ) ) ( u ˜ k ( y ) ) α ( v ˜ k ( y ) ) β 1 ψ ( y ) d y = o k ( 1 ) , R n V 2 ( x ) v k ( x ) ψ k ( x ) d x = λ k 2 R n V 2 ( λ k y , λ k ( y x k ) ) v ˜ k ( y ) φ ( y ) d y = o k ( 1 ) .

By using (3.12), we have

(3.15) I ( u k , v k ) , ( φ k , ψ k ) = R n ( u k φ k + V 1 ( x ) u k φ k + v k ψ k + V 2 ( x ) v k ψ k ) d x η 1 2 R n ( u k + ) η 1 1 ( v k + ) η 2 φ k x d x η 2 2 R n ( v k + ) η 2 1 ( u k + ) η 1 ψ k x d x α α + β R n Q ( x ) ( u k + ) α 1 ( v k + ) β φ k d x β α + β R n Q ( x ) ( u k + ) α ( v k + ) β 1 ψ k d x = o k ( 1 ) ( φ k , ψ k ) H 1 ( R n , R 2 ) = o k ( 1 ) .

Then, we learn from (3.13), (3.14), and (3.15) that

(3.16) o k ( 1 ) = I ( u k , v k ) , ( φ k , ψ k ) = R n ( u k φ k + v k ψ k ) d x η 1 2 R n ( u k + ) η 1 1 ( v k + ) η 2 φ k x d x η 2 2 R n ( v k + ) η 2 1 ( u k + ) η 1 ψ k x d x + o k ( 1 ) .

A change of variable in (3.16) implies that

o k ( 1 ) = I ( u k , v k ) , ( φ k , ψ k ) = R n ( u ˜ k φ + v ˜ k ψ ) d x η 1 2 R n ( u ˜ k ) η 1 1 ( v ˜ k ) η 2 φ x d x η 2 2 R n ( v ˜ k ) η 2 1 ( u ˜ k ) η 1 ψ x d x + o k ( 1 ) = I ( u ˜ k , v ˜ k ) , ( φ , ψ ) + o k ( 1 ) = I ( u 0 , v 0 ) , ( φ , ψ ) + o k ( 1 ) .

Consequently, ( u 0 , v 0 ) is a critical point of I , and ( u 0 , v 0 ) solves (1.15).

(iii) We claim that { ( z k , ν k ) } H 1 ( R n , R 2 ) and ( z k , ν k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , where

( z k ( x ) , ν k ( x ) ) ( u k ( x ) u ¯ k ( x ) , v k ( x ) v ¯ k ( x ) ) = u k ( x ) λ k 2 n 2 u 0 x λ k , x λ k + x k , v k ( x ) λ k 2 n 2 v 0 x λ k , x λ k + x k .

Observing that { u k } H 1 ( R n ) , u 0 D 1 , 2 ( R n ) , u ¯ k D 1 , 2 ( R n ) 2 = u 0 D 1 , 2 ( R n ) 2 , and

z k H 1 ( R n ) 2 = ( u k u ¯ k ) L 2 ( R n ) 2 + ( u k u ¯ k ) L 2 ( R n ) 2 2 { u k L 2 ( R n ) 2 + u ¯ k L 2 ( R n ) 2 + u k L 2 ( R n ) 2 + u ¯ k L 2 ( R n ) 2 } ,

and it is sufficient to prove u ¯ k L 2 ( R n ) . From (ii) and Lemma 2.6, we know that ( u 0 , v 0 ) is a positive solution of (1.15). Therefore, (2.6) in Lemma 2.6 and n > 4 indicate that

u 0 L r ( R n ) r C R n 1 ( 1 + x n 2 ) r d x C ˜ , r 2 .

In view of this, we have

u ¯ k L r ( R n ) r = λ k n r ( n 2 ) 2 u 0 L r ( R n ) r = o k ( 1 ) , r [ 2 , 2 ) .

In particular, we see that u ¯ k L 2 ( R n ) . Letting φ C 0 ( R n ) , we have R n z k φ d x 0 and R n z k φ d x 0 , so z k 0 in H 1 ( R n ) . Similarly, we have ν k 0 in H 1 ( R n ) .

(iv) Conclusion. By the Brézis-Lieb type Lemma (see [13], Lemma 2.3) and the weak convergence of ( z k , ν k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , similar to Lemma 2.7, we have

I ( z k , ν k ) = I ( u k , v k ) I ( u 0 , v 0 ) + o k ( 1 ) , I ( z k , ν k ) , ( φ , ψ ) = o k ( 1 ) ( φ , ψ ) H 1 ( R n , R 2 ) , ( φ , ψ ) H 1 ( R n , R 2 ) .

Next, we consider the blowing up bubbles caused by the unbounded domains. We have the following.

Proposition 3.4

Let n 3 , 2 m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. If { ( u k , v k ) } k = 1 H 1 ( R n , R 2 ) is a nonnegative bounded Palais-Smale sequence of I at level d 1 0 such that ( u k , v k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) and

R n ( u k + ( x ) ) α ( v k + ( x ) ) β d x C > 0 , R n ( u k + ( x ) ) η 1 ( v k + ( x ) ) η 2 x d x = o k ( 1 ) ,

then there exist, up to a subsequence, a family of { ζ k ( ζ k , ζ k ) } R n × R n m and a pair of functions ( u 0 , v 0 ) H 1 ( R n , R 2 ) such that u 0 ( x ) 0 , v 0 ( x ) 0 , and

( u k ( x + ζ k ) , v k ( x + ζ k ) ) ( u 0 ( x ) , v 0 ( x ) ) in H 1 ( R n , R 2 ) , ζ k + , ζ k + .

In addition, ( u 0 , v 0 ) solves (1.10) and the sequence

( z k ( x ) , ν k ( x ) ) ( u k ( x ) u 0 ( x ζ k ) , v k ( x ) v 0 ( x ζ k ) )

is a Palais-Smale sequence of I at level d 1 I ( u 0 , v 0 ) .

Proof

(i) We prove the existence of { ζ k } such that ζ k + and ζ k + . Since ( u k , v k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , we have

( u k , v k ) ( 0 , 0 ) in L loc r ( R n , R 2 ) ( 2 r < 2 ) , ( u k , v k ) ( 0 , 0 ) a.e. on R n × R n .

In a similar fashion to (2.1), we deduce from α , β > 1 and α + β 2 that

(3.17) 0 < C R n u k ( x ) α v k ( x ) β d x C ˜ R n u k ( x ) v k ( x ) d x 1 2 .

Consequently, there exists δ > 0 such that

R n u k ( x ) v k ( x ) d x δ > 0 .

By using Lemma 2.2, up to a subsequence, two possible cases must occur:

  1. Vanishing: for all 0 < R < + ,

    sup y R n B R ( y ) u k ( x ) v k ( x ) d x 0 as k + .

  2. Nonvanishing: there exist δ ˜ > 0 , 0 < R ¯ < + , { ζ k ( ζ k , ζ k ) } R m × R n m , such that

    (3.18) liminf k + B R ¯ ( ζ k ) u k ( x ) v k ( x ) d x δ ˜ > 0 .

If case (a) occurs, then we learn from Lemma 2.3 that R n u k ( x ) α v k ( x ) β d x 0 , and this contradicts to (3.17). Therefore, case (b) must hold. We claim that ζ k + as k + . Otherwise, there exists C > 0 such that ζ k C , and then we can choose C ¯ R ¯ > 0 such that

B R ¯ ( ζ k ) u k ( x ) v k ( x ) d x B C ¯ ( 0 ) u k ( x ) v k ( x ) d x u k L 2 ( B C ¯ ( 0 ) ) v k L 2 ( B C ¯ ( 0 ) ) 0

as k + , which contradicts to (3.18). So it must be that ζ k + . Denote

( u ˜ k ( x ) , v ˜ k ( x ) ) ( u k ( x + ζ k ) , v k ( x + ζ k ) ) ,

then we derive from ( u ˜ k , v ˜ k ) H 1 ( R n , R 2 ) = ( u k , v k ) H 1 ( R n , , R 2 ) C that there exists a pair of functions ( u 0 , v 0 ) H 1 ( R n , R 2 ) such that ( u ˜ k , v ˜ k ) ( u 0 , v 0 ) in H 1 ( R n , R 2 ) and

( u ˜ k , v ˜ k ) ( u 0 , v 0 ) in L loc r ( R n , R 2 ) ( 2 r < 2 ) , ( u ˜ k , v ˜ k ) ( u 0 , v 0 ) a.e. on R n × R n .

From (3.18), we obtain

0 < δ ˜ liminf k + B R ¯ ( ζ k ) u k ( x ) v k ( x ) d x = liminf k + B R ¯ ( 0 ) u ˜ k ( x ) v ˜ k ( x ) d x liminf k + u ˜ k L 2 ( B R ¯ ( 0 ) ) v ˜ k L 2 ( B R ¯ ( 0 ) ) = u 0 L 2 ( B R ¯ ( 0 ) ) v 0 L 2 ( B R ¯ ( 0 ) ) .

This implies that u 0 0 and v 0 0 . We claim a stronger conclusion ζ k + . Otherwise, there exists M > 0 such that ζ k M . But then we obtain a contradiction that:

o k ( 1 ) = R n ( u k + ( y ) ) η 1 ( v k + ( y ) ) η 2 y d y R n ( u k + ( y ) ) η 1 ( v k + ( y ) ) η 2 y ζ k + ζ k d y = R n ( u ˜ k ( x ) ) η 1 ( v ˜ k ( x ) ) η 2 x + ζ k d x B R ¯ ( 0 ) ( u ˜ k ( x ) ) η 1 ( v ˜ k ( x ) ) η 2 x + M d x 1 R ¯ + M B R ¯ ( 0 ) ( u ˜ k ( x ) ) η 1 ( v ˜ k ( x ) ) η 2 d x C > 0 .

In the last inequality, we use the fact that

0 < δ ˜ B R ¯ ( 0 ) u ˜ k ( x ) v ˜ k ( x ) d x B R ¯ ( 0 ) u ˜ k ( x ) η 1 v ˜ k ( x ) η 2 d x 1 η 2 B R ¯ ( 0 ) u ˜ k ( x ) η 2 η 1 η 2 1 d x 1 1 η 2 .

Consequently, we have η k + .

(ii) We prove that ( u 0 , v 0 ) is a critical point of I and ( u 0 , v 0 ) solves (1.10). Take φ , ψ C 0 ( R n ) and

( φ k ( x ) , ψ k ( x ) ) ( φ ( x ζ k ) , ψ ( x ζ k ) ) .

By using ζ k + and Lemma 3.13 of [7], we obtain

(3.19) R n u k η 1 2 u k v k η 2 φ k ( x ) x d x = R n u ˜ k η 1 2 u ˜ k v ˜ k η 2 φ ( x ) x + ζ k d x = o k ( 1 ) , R n u ˜ k η 1 v ˜ k η 2 x + ζ k d x = o k ( 1 ) , R n v k η 2 2 v k u k η 1 ψ k ( x ) x d x = R n v ˜ k η 2 2 v ˜ k u ˜ k η 1 ψ ( x ) x + ζ k d x = o k ( 1 ) .

Recall that ( u k , v k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) and lim k V 1 ( x + ζ k ) = V , so we have

o k ( 1 ) = R n V 1 ( x ) u k ( x ) φ k ( x ) d x = R n V u ˜ k ( x ) φ ( x ) d x + R n [ V 1 ( x + ζ k ) V ] u ˜ k ( x ) φ ( x ) d x

and

R n [ V 1 ( x + ζ k ) V ] u ˜ k ( x ) φ ( x ) d x C R n V 1 ( x + ζ k ) V 2 φ 2 ( x ) d x 1 2 = o k ( 1 ) .

That is,

(3.20) R n V u ˜ k ( x ) φ ( x ) d x = o k ( 1 ) = R n V 1 ( x ) u k ( x ) φ k ( x ) d x .

In the same way, we can prove that

(3.21) R n V v ˜ k ( x ) ψ ( x ) d x = o k ( 1 ) = R n V 2 ( x ) v k ( x ) ψ k ( x ) d x , R n Q u ˜ k α 1 v ˜ k β φ ( x ) d x = o k ( 1 ) = R n Q ( x ) u k α 1 v k β φ k ( x ) d x , R n Q v ˜ k β 1 u ˜ k α ψ ( x ) d x = o k ( 1 ) = R n Q ( x ) v k β 1 u k α ψ k ( x ) d x .

Observing that { ( u k , v k ) } is a Palais-Smale sequence of I , by using (3.19)–(3.21), we obtain

o k ( 1 ) = I ( u k , v k ) , ( φ k , ψ k ) = I ( u ˜ k , v ˜ k ) , ( φ , ψ ) + o k ( 1 ) = I ( u 0 , v 0 ) , ( φ , ψ ) + o k ( 1 ) .

Hence, ( u 0 , v 0 ) is a critical point of I and ( u 0 , v 0 ) weakly solves (1.10). In addition, from (2.5) of Lemma 2.5, we have u 0 0 and v 0 0 . The strong maximum principle implies that u 0 > 0 and v 0 > 0 .

(iii) We claim that { ( z k , ν k ) } H 1 ( R n , R 2 ) and ( z k , ν k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , where

( z k ( x ) , ν k ( x ) ) ( u k ( x ) u ¯ k ( x ) , v k ( x ) v ¯ k ( x ) ) = ( u k ( x ) u 0 ( x ζ k ) , v k ( x ) v 0 ( x ζ k ) ) .

Observing that { u k } H 1 ( R n ) , u 0 H 1 ( R n ) , u ¯ k H 1 ( R n ) 2 = u 0 H 1 ( R n ) 2 and

z k H 1 ( R n ) 2 = ( u k u ¯ k ) L 2 ( R n ) 2 + ( u k u ¯ k ) L 2 ( R n ) 2 2 { u k L 2 ( R n ) 2 + u ¯ k L 2 ( R n ) 2 + u k L 2 ( R n ) 2 + u ¯ k L 2 ( R n ) 2 } < + ,

we obtain { z k } H 1 ( R n ) . Letting φ C 0 ( R n ) , we have R n z k φ d x 0 and R n z k φ d x 0 , so z k 0 in H 1 ( R n ) . Similarly, we have ν k 0 in H 1 ( R n ) .

(iv) Conclusion. From (3.19), we have

I ( u k , v k ) = 1 2 R n ( u k ( x ) 2 + V 1 ( x ) u k ( x ) 2 + v k ( x ) 2 + V 2 ( x ) v k ( x ) 2 ) d x 1 2 R n u k η 1 v k η 2 x d x 1 α + β R n Q ( x ) u k α v k β d x = 1 2 R n ( u ˜ k ( x ) 2 + V 1 ( x + ζ k ) u ˜ k ( x ) 2 + v ˜ k ( x ) 2 + V 2 ( x + ζ k ) v ˜ k ( x ) 2 ) d x 1 2 R n u ˜ k η 1 v ˜ k η 2 x + ζ k d x 1 α + β R n Q ( x + ζ k ) u ˜ k α v ˜ k β d x = 1 2 R n ( u ˜ k ( x ) 2 + V u ˜ k ( x ) 2 + v ˜ k ( x ) 2 + V v ˜ k ( x ) 2 ) d x 1 α + β R n Q u ˜ k α v ˜ k β d x + o k ( 1 ) .

This implies that I ( u k , v k ) = I ( u ˜ k , v ˜ k ) + o k ( 1 ) . By the Brézis-Lieb type Lemma (see [13], Lemma 2.3) and the weak convergence of ( z k , ν k ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , similar to Lemma 2.7, we have

I ( z k , ν k ) = I ( u ˜ k , v ˜ k ) I ( u 0 , v 0 ) + o k ( 1 ) = I ( u k , v k ) I ( u 0 , v 0 ) + o k ( 1 ) , I ( z k , ν k ) , ( φ , ψ ) = o k ( 1 ) ( φ , ψ ) H 1 ( R n , R 2 ) , ( φ , ψ ) H 1 ( R n , R 2 ) .

At the end of this section, we use Propositions 3.33.4 to prove Theorem 1.2.

Proof of Theorem 1.2

Let { ( u k , v k ) } k = 1 H 1 ( R n , R 2 ) be a Palais-Smale sequence of I at level d 0 , by using Lemma 2.5, we can assume that { ( u k , v k ) } are nonnegative. Then, Lemma 2.4 indicates that { ( u k , v k ) } are bounded. So there exists ( u , v ) H 1 ( R n , R 2 ) such that, up to a subsequence,

(3.22) ( u k , v k ) ( u , v ) in H 1 ( R n , R 2 ) , ( u k , v k ) ( u , v ) in L loc r ( R n , R 2 ) ( 2 r < 2 ) , ( u k , v k ) ( u , v ) a.e. on R n × R n .

Denote ( u k ( 1 ) , v k ( 1 ) ) ( u k u , v k v ) , then

(3.23) ( u k ( 1 ) , v k ( 1 ) ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , ( u k ( 1 ) , v k ( 1 ) ) ( 0 , 0 ) in L loc r ( R n , R 2 ) ( 2 r < 2 ) , ( u k ( 1 ) , v k ( 1 ) ) ( 0 , 0 ) a.e. on R n × R n .

We have ( u k ( 1 ) , v k ( 1 ) ) H 1 ( R n , R 2 ) 2 = ( u k , v k ) H 1 ( R n , R 2 ) 2 ( u , v ) H 1 ( R n , R 2 ) 2 + o k ( 1 ) via the weak convergence. From Lemma 2.7, we see that { ( u k ( 1 ) , v k ( 1 ) ) } is a Palais-Smale sequence of I at level d I ( u , v ) and

(3.24) I ( u k ( 1 ) , v k ( 1 ) ) = I ( u k , v k ) I ( u , v ) + o k ( 1 ) , I ( u k ( 1 ) , v k ( 1 ) ) , ( φ , ψ ) = o k ( 1 ) ( φ , ψ ) H 1 ( R n , R 2 ) , ( φ , ψ ) H 1 ( R n , R 2 ) .

Since ( u k ( 1 ) , v k ( 1 ) ) H 1 ( R n , R 2 ) 2 ( u k , v k ) H 1 ( R n , R 2 ) 2 + o k ( 1 ) C , we can assume that

( u k ( 1 ) , v k ( 1 ) ) H 1 ( R n , R 2 ) 2 0 as k + .

If = 0 , then Theorem 1.2 is true for l 1 = 0 and l 2 = 0 . So we focus on the case > 0 .

Step 1. We exclude the blowing up bubbles originated from the Hardy-Sobolev critical terms. In a similar fashion as Lemma 2.5, we can assume that { ( u k ( 1 ) , v k ( 1 ) ) } k = 1 are nonnegative. Suppose there exists 0 < C < such that

inf k N R n ( u k ( 1 ) ( x ) ) η 1 ( v k ( 1 ) ( x ) ) η 2 x d x C > 0 ,

then { ( u k ( 1 ) , v k ( 1 ) ) } satisfy the assumptions of Proposition 3.3 with d 1 d I ( u , v ) 0 . Consequently, there exist, up to a subsequence, a family of positive numbers { λ k ( 1 ) } with λ k ( 1 ) 0 , a sequence of { ζ k ( 1 ) } R n m and a pair of functions ( U ( 1 ) , V ( 1 ) ) D 1 , 2 ( R n , R 2 ) such that

( λ k ( 1 ) ) n 2 2 ( u k ( 1 ) ( λ k ( 1 ) x , λ k ( 1 ) ( x ζ k ( 1 ) ) ) , v k ( 1 ) ( λ k ( 1 ) x , λ k ( 1 ) ( x ζ k ( 1 ) ) ) ) ( U ( 1 ) ( x ) , V ( 1 ) ( x ) )

in D 1 , 2 ( R n , R 2 ) with U ( 1 ) 0 and V ( 1 ) 0 . In addition, ( U ( 1 ) , V ( 1 ) ) solves (1.15), and the sequence

( z k ( 1 ) ( x ) , ν k ( 1 ) ( x ) ) u k ( 1 ) ( x ) ( λ k ( 1 ) ) 2 n 2 U ( 1 ) x λ k ( 1 ) , x λ k ( 1 ) + ζ k ( 1 ) , v k ( 1 ) ( x ) ( λ k ( 1 ) ) 2 n 2 V ( 1 ) x λ k ( 1 ) , x λ k ( 1 ) + ζ k ( 1 )

is a Palais-Smale sequence of I at level d I ( u , v ) I ( U ( 1 ) , V ( 1 ) ) , i.e.,

(3.25) I ( z k ( 1 ) , ν k ( 1 ) ) = d I ( u , v ) I ( U ( 1 ) , V ( 1 ) ) + o k ( 1 ) , I ( z k ( 1 ) , ν k ( 1 ) ) , ( φ , ψ ) = o k ( 1 ) ( φ , ψ ) H 1 ( R n , R 2 ) , ( φ , ψ ) H 1 ( R n , R 2 ) .

If there still exists 0 < C < + such that

inf k N R n ( u k + ( x ) ) η 1 ( v k + ( x ) ) η 2 x d x inf k N R n ( u k ( 1 ) ( x ) ) η 1 ( v k ( 1 ) ( x ) ) η 2 x d x C > 0 ,

then we repeat the previous argument. Since

inf k N R n ( u k + ( x ) ) η 1 ( v k + ( x ) ) η 2 x d x R n ( u k + ( x ) ) 2 x d x η 1 2 R n ( v k + ( x ) ) 2 x d x η 2 2 C ,

the iterations must stop after finite times. Hence, there exist a positive constant l 1 and a new Palais-Smale sequence { ( u k ( 2 ) , v k ( 2 ) ) } k = 1 of I such that

(3.26) ( u k ( 2 ) , v k ( 2 ) ) ( u k , v k ) ( u , v ) i = 1 l 1 ( λ k ( i ) ) 2 n 2 U ( i ) x λ k ( i ) , x λ k ( i ) + ζ k ( i ) , V ( i ) x λ k ( i ) , x λ k ( i ) + ζ k ( i ) , ( u k ( 2 ) , v k ( 2 ) ) ( 0 , 0 ) in H 1 ( R n , R 2 ) , R n ( u k ( 2 ) ( x ) ) η 1 ( v k ( 2 ) ( x ) ) η 2 x d x = o k ( 1 ) , d = I ( u k ( 2 ) , v k ( 2 ) ) + I ( u , v ) + i = 1 l 1 I ( U ( i ) , V ( i ) ) + o k ( 1 ) ,

with λ k ( i ) 0 and { ζ k ( i ) } R n m .

Step 2. We exclude the blowing up bubbles caused by the unbounded domains. Suppose there exists 0 < C < + such that

R n ( u k ( 2 ) ( x ) ) α ( v k ( 2 ) ( x ) ) β d x C > 0 , R n ( u k ( 2 ) ( x ) ) η 1 ( v k ( 2 ) ( x ) ) η 2 x d x = o k ( 1 ) ,

then we deduce from Proposition 3.4 that, there exist, up to a subsequence, a family of { ξ k ( 1 ) ( ξ k ( 1 ) , ξ k ( 1 ) ) } R n × R n m and a pair of functions ( u ( 1 ) , v ( 1 ) ) H 1 ( R n , R 2 ) such that

( u k ( 2 ) ( x + ξ k ( 1 ) ) , v k ( 2 ) ( x + ξ k ( 1 ) ) ) ( u ( 1 ) , v ( 1 ) ) in H 1 ( R n , R 2 ) , ξ k ( 1 ) + , ξ k ( 1 ) +

with u ( 1 ) > 0 and v ( 1 ) > 0 . In addition, ( u ( 1 ) , v ( 1 ) ) solves (1.10) and the sequence

( u k ( 3 ) ( x ) , v k ( 3 ) ( x ) ) ( u k ( 2 ) ( x ) u ( 1 ) ( x ξ k ( 1 ) ) , v k ( 2 ) ( x ) v ( 1 ) ( x ξ k ( 1 ) ) )

is a Palais-Smale sequence of I at level

d I ( u , v ) i = 1 l 1 I ( U ( i ) , V ( i ) ) I ( u ( 1 ) , v ( 1 ) ) .

If there still exist 0 < C < + such that

R n ( u k ( 3 ) ( x ) ) α ( v k ( 3 ) ( x ) ) β d x C > 0 , R n ( u k ( 3 ) ( x ) ) η 1 ( v k ( 3 ) ( x ) ) η 2 x d x = o k ( 1 ) ,

then one can repeat Step 2 for finite times ( l 2 times) because R n ( u k ( x ) ) α ( v k ( x ) ) β d x C . From (3.26) and Proposition 3.4, we obtain a new Palais-Smale sequence { ( u k ( l 1 + l 2 ) , v k ( l 1 + l 2 ) ) } k = 1 of I such that

(3.27) ( u k ( l 1 + l 2 ) , v k ( l 1 + l 2 ) ) ( u k , v k ) i = 1 l 1 ( λ k ( i ) ) 2 n 2 U ( i ) x λ k ( i ) , x λ k ( i ) + ζ k ( i ) , V ( i ) x λ k ( i ) , x λ k ( i ) + ζ k ( i ) ( u , v ) j = 1 l 2 ( u ( j ) ( x ξ k ( j ) ) , v ( j ) ( x ξ k ( j ) ) ) ,

(3.28) R n ( u k ( l 1 + l 2 ) ( x ) ) α ( v k ( l 1 + l 2 ) ( x ) ) β d x = o k ( 1 ) , R n ( u k ( l 1 + l 2 ) ( x ) ) η 1 ( v k ( l 1 + l 2 ) ( x ) ) η 2 x d x = o k ( 1 ) ,

with

(3.29) λ k ( i ) 0 ( 1 i l 1 ) , ξ k ( j ) and ξ k ( j ) ( 1 j l 2 )

as k , where ( u , v ) , ( U ( i ) , V ( i ) ) ( 1 i l 1 ) , and ( u ( j ) , v ( j ) ) ( 1 j l 2 ) satisfy

(3.30) U ( i ) 0 ( 0 ) , V ( i ) 0 ( 0 ) , u ( j ) > 0 , v ( j ) > 0 , d = I ( u k ( l 1 + l 2 ) , v k ( l 1 + l 2 ) ) + I ( u , v ) + i = 1 l 1 I ( U ( i ) , V ( i ) ) + j = 1 l 2 I ( u ( j ) , v ( j ) ) + o k ( 1 ) , I ( u , v ) = 0 , I ( U ( i ) , V ( i ) ) = 0 , ( I ) ( u ( j ) , v ( j ) ) = 0 .

The fact I ( u k ( l 1 + l 2 ) , v k ( l 1 + l 2 ) ) , ( u k ( l 1 + l 2 ) , v k ( l 1 + l 2 ) ) = o k ( 1 ) , and (3.28) indicate that

C ( u k ( l 1 + l 2 ) , v k ( l 1 + l 2 ) ) H 1 ( R n , R 2 ) 2 R n ( u k ( l 1 + l 2 ) 2 + V 1 ( x ) u k ( l 1 + l 2 ) 2 + v k ( l 1 + l 2 ) 2 + V 2 ( x ) v k ( l 1 + l 2 ) 2 ) d x = R n ( u k ( l 1 + l 2 ) ( x ) ) η 1 ( v k ( l 1 + l 2 ) ( x ) ) η 2 x d x + R n Q ( x ) ( u k ( l 1 + l 2 ) ( x ) ) α ( v k ( l 1 + l 2 ) ( x ) ) β d x R n ( u k ( l 1 + l 2 ) ( x ) ) η 1 ( v k ( l 1 + l 2 ) ( x ) ) η 2 x d x + max y R n Q ( y ) R n ( u k ( l 1 + l 2 ) ( x ) ) α ( v k ( l 1 + l 2 ) ( x ) ) β d x = o k ( 1 ) .

So we have

(3.31) I ( u k ( l 1 + l 2 ) , v k ( l 1 + l 2 ) ) = o k ( 1 ) .

We end the proof by combining (3.27) and (3.30) with (3.31).□

4 The proof of Theorem 1.3

In this section, we shall use Theorem 1.2 to prove Theorem 1.3, i.e., the existence of a nontrivial weak solution to equation (1.1). This relies on the following lemma.

Lemma 4.1

(Mountain pass lemma, [2]) Let ( E , ) be a Banach space and I C 1 ( E , R ) satisfying the following conditions:

  1. I ( 0 ) = 0 ,

  2. There exist ρ , r > 0 such that I ( u ) ρ for all u E with u = r ,

  3. There exists a v 0 E such that lim t + sup I ( t v 0 ) < 0 .

Let t 0 > 0 be such that t 0 v 0 > r and I ( t 0 v 0 ) < 0 , and define

c inf g Γ sup t [ 0 , 1 ] I ( g ( t ) ) ,

where

Γ { g C 0 ( [ 0 , 1 ] , E ) : g ( 0 ) = 0 , g ( 1 ) = t 0 v 0 } .

Then, c ρ > 0 and there exists a (PS) sequence { u k } E for I at level c , i.e.,

lim k + I ( u k ) = c and lim k + I ( u k ) = 0 strongly in E 1 .

We now use Lemma 4.1 to prove the following proposition.

Proposition 4.2

Let n > 4 , 2 < m < n , η 1 , η 2 > 1 , and η 1 + η 2 = 2 , α , β > 1 , and α + β 2 , and V 1 ( x ) , V 2 ( x ) , Q ( x ) satisfy conditions (V)–(Q), respectively. Consider the functional I defined in (1.9), then there exists a (PS) sequence { ( u k , v k ) } H 1 ( R n , R 2 ) for I at level c ( 0 , c ) , i.e.,

(4.1) lim k + I ( u k , v k ) = c and lim k + I ( u k , v k ) = 0 strongly in H 1 ( R n , R 2 ) ,

where

c min [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) , D .

Proof

We now verify the conditions of Lemma 4.1. Similar to Lemma 2.1 of [4], we have

(4.2) R n u α v β d x 1 / ( α + β ) C ( u , v ) H 1 ( R n , R 2 ) , ( u , v ) H 1 ( R n , R 2 ) .

By using (1.20) and (4.2), we have

(4.3) I ( u , v ) = 1 2 R n ( u 2 + V 1 ( x ) u 2 + v 2 + V 2 ( x ) v 2 ) d x 1 2 R n ( u + ) η 1 ( v + ) η 2 x d x 1 α + β R n Q ( x ) ( u + ( x ) ) α ( v + ( x ) ) β d x C [ ( u , v ) H 1 ( R n , R 2 ) 2 ( u , v ) H 1 ( R n , R 2 ) 2 u H 1 ( R n , R 2 ) α + β ] , ( u , v ) H 1 ( R n , R 2 ) .

Since 2 > 2 and α + β > 2 , we learn from (4.3) that there exists r > 0 small enough such that

inf ( u , v ) H 1 ( R n , R 2 ) = r I ( u , v ) > 0 = I ( 0 , 0 ) ,

so ( 1 ) and ( 2 ) of Lemma 4.1 are satisfied. From

I ( t u , t v ) = t 2 2 R n ( u 2 + V 1 ( x ) u 2 + v 2 + V 2 ( x ) v 2 ) d x t 2 2 R n ( u + ) η 1 ( v + ) η 2 x d x t α + β α + β R n Q ( x ) ( u + ( x ) ) α ( v + ( x ) ) β d x ,

we derive that lim t + I ( t u , t v ) = for any ( u , v ) H 1 ( R n , R 2 ) . Consequently, for any fixed ( u 0 , v 0 ) H 1 ( R n , R 2 ) , there exists t u 0 , v 0 > 0 such that t u 0 , v 0 ( u 0 , v 0 ) H 1 ( R n , R 2 ) > r and I ( t u 0 , v 0 ( u 0 , v 0 ) ) < 0 . So ( 3 ) of Lemma 4.1 is satisfied. That is to say, I satisfies the mountain pass geometry.

Let W ( x ) and U ( x ) be defined by (1.13) and (1.19), respectively, then there exist

( u 0 ( x ) , v 0 ( x ) ) U ( x ) , if [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) D ; W ( x ) , if [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) > D

and t u 0 , v 0 > 0 such that t u 0 , v 0 ( u 0 , v 0 ) H 1 ( R n , R 2 ) > r and I ( t u 0 , v 0 ( u 0 , v 0 ) ) < 0 . We can define

c inf g Γ sup t [ 0 , 1 ] I ( g ( t ) ) ,

where

Γ { g C 0 ( [ 0 , 1 ] , H 1 ( R n , R 2 ) ) : g ( 0 ) = ( 0 , 0 ) , g ( 1 ) = t u 0 , v 0 ( u 0 , v 0 ) } .

Clearly, we have c > 0 .

For the case of ( u 0 ( x ) , v 0 ( x ) ) = U ( x ) , we claim that

(4.4) 0 < c < [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) .

From (1.19), we rewrite U ( x ) as follows:

U ( x ) = ( U 1 , U 2 ) = ( τ η 1 , η 2 , τ η 2 , η 1 ) U ε , ξ ( x , x ) = 2 η 1 η 1 η 2 η 2 2 1 / ( 2 2 ) , 2 η 2 η 2 η 1 η 1 2 1 / ( 2 2 ) U ε , ξ ( x , x )

for U ε , ξ defined by (1.18). Direct calculation implies that

(4.5) τ η 1 , η 2 2 + τ η 2 , η 1 2 = τ η 1 , η 2 η 1 τ η 2 , η 1 η 2 = [ ( η 1 / η 2 ) η 2 / 2 + ( η 1 / η 2 ) η 1 / 2 ] n 1 , U ε , ξ 2 2 = R n U ε , ξ 2 x d x = S ( n , m ) n 1 , R n U ε , ξ α + β d x = O ε n ( n 2 ) ( α + β ) 2 , R n U ε , ξ ( x ) 2 d x ε 2 R n C ( 1 + x 2 ) n 2 d x = O ( ε 2 ) for n > 4 .

Letting

h 1 ( t ) I ( t U 1 , t U 2 ) = t 2 2 R n ( U 1 2 + V 1 ( x ) U 1 2 + U 2 2 + V 2 ( x ) U 2 2 ) d x t 2 2 R n U 1 η 1 U 2 η 2 x d x t α + β α + β R n Q ( x ) U 1 α U 2 β d x , t 0 ,

then h 1 ( t ) attains its maximum at some point t ε > 0 because lim t + h 1 ( t ) = and h 1 ( t ) > 0 when t > 0 is sufficiently small. By using h 1 ( t ε ) = 0 , we have

t ε 2 2 R n U 1 η 1 U 2 η 2 x d x + t ε α + β 2 R n Q ( x ) U 1 α U 2 β d x = R n ( U 1 2 + V 1 ( x ) U 1 2 + U 2 2 + V 2 ( x ) U 2 2 ) d x .

Substituting ( U 1 , U 2 ) = ( τ η 1 , η 2 , τ η 2 , η 1 ) U ε , ξ ( x , x ) into the aforementioned equality, we obtain

t ε 2 2 τ η 1 , η 2 η 1 τ η 2 , η 1 η 2 R n U ε , ξ 2 x d x + t ε α + β 2 τ η 1 , η 2 α τ η 2 , η 1 β R n Q ( x ) U ε , ξ α + β d x = ( τ η 1 , η 2 2 + τ η 2 , η 1 2 ) U ε , ξ 2 2 + τ η 1 , η 2 2 R n V 1 ( x ) U ε , ξ 2 d x + τ η 2 , η 1 2 R n V 2 ( x ) U ε , ξ 2 d x .

From (4.5), we see that

(4.6) t ε 2 2 + t ε α + β 2 τ η 1 , η 2 α η 1 τ η 2 , η 1 β η 2 S ( n , m ) n 1 R n Q ( x ) U ε , ξ α + β d x = 1 + τ η 1 , η 2 2 η 1 τ η 2 , η 1 η 2 S ( n , m ) n 1 R n V 1 ( x ) U ε , ξ 2 d x + τ η 2 , η 1 2 η 2 τ η 1 , η 2 η 1 S ( n , m ) n 1 R n V 2 ( x ) U ε , ξ 2 d x .

On the one hand, we learn from (4.5), (4.6), and Q ( x ) > 0 that, for ε > 0 sufficiently small,

(4.7) t ε 2 2 1 + O ( ε 2 ) < 2 .

On the other hand, (4.5), (4.6), and (4.7) indicate that

(4.8) t ε 2 2 1 + τ η 1 , η 2 2 η 1 τ η 2 , η 1 η 2 S ( n , m ) n 1 R n V 1 ( x ) U ε , ξ 2 d x + τ η 2 , η 1 2 η 2 τ η 1 , η 2 η 1 S ( n , m ) n 1 R n V 2 ( x ) U ε , ξ 2 d x 2 α + β 2 2 2 τ η 1 , η 2 α η 1 τ η 2 , η 1 β η 2 S ( n , m ) n 1 R n Q ( x ) U ε , ξ α + β d x 1 + O ( ε 2 ) O ε n ( n 2 ) ( α + β ) 2 > 1 2 .

By using (4.7) and (4.8), it holds that

1 2 < t ε 2 2 < 2

for ε > 0 sufficiently small. In view of the aforementioned facts and n ( n 2 ) ( α + β ) 2 < 2 , we have

c max t > 0 I ( t U ) = I ( t ε U ) max t > 0 t 2 2 R n ( U 1 2 + U 2 2 ) d x t 2 2 R n U 1 η 1 U 2 η 2 x d x + O ( ε 2 ) O ε n ( n 2 ) ( α + β ) 2 = 1 2 1 2 R n U 1 2 + U 2 2 d x R n U 1 η 1 U 2 η 2 / x d x 2 / 2 2 2 2 + O ( ε 2 ) O ε n ( n 2 ) ( α + β ) 2 = [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) + O ( ε 2 ) O ε n ( n 2 ) ( α + β ) 2 < [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 )

for ε > 0 sufficiently small. This proves (4.4).

For the case of ( u 0 ( x ) , v 0 ( x ) ) = W ( x ) , we claim that

(4.9) 0 < c < D .

Notice that W ( x ) = ( W 1 ( x ) , W 2 ( x ) ) is a minimizer of D , I ( W ) = D and

R n ( W 1 2 + V W 1 2 + W 2 2 + V W 2 2 ) d x = Q R n W 1 α W 2 β d x .

Letting

h 2 ( t ) I ( t W ) = t 2 2 R n W 1 2 + V W 1 2 + W 2 2 + V W 2 2 d x Q t α + β α + β R n W 1 α W 2 β d x ,

then we have

h 2 ( t ) = t R n W 1 2 + V W 1 2 + W 2 2 + V W 2 2 d x t α + β 1 R n Q W 1 α W 2 β d x .

Direct calculation indicates that

h 2 ( t ) 0 if t ( 0 , 1 ) , h 2 ( t ) 0 if t ( 1 , + ) .

Then

(4.10) h 2 ( 1 ) = I ( W ) = max t 0 I ( t W ) .

Observing that 2 > 2 and α + β > 2 , so there exists a t W > 0 such that

sup t 0 I ( t W ) = I ( t W W ) .

From (4.10) and the assumptions that V 1 ( x ) , V 2 ( x ) V , Q ( x ) Q , and Q ( x ) Q , we have

c sup t 0 I ( t W ) = I ( t W W ) < I ( t W W ) I ( W ) = D .

This proves (4.9).

From (4.4) and (4.9), we have 0 < c < c min [ S ( η 1 , η 2 ) ] n 1 2 ( n 1 ) , D . By using Lemma 4.1, there exists a (PS) sequence { ( u k , v k ) } H 1 ( R n , R 2 ) for I at level c ( 0 , c ) , i.e.,

lim k + I ( u k , v k ) = c and lim k + I ( u k , v k ) = 0 strongly in H 1 ( R n , R 2 ) .

The proof is ended.□

Proof of Theorem 1.3

Let { ( u k , v k ) } H 1 ( R n , R 2 ) be a (PS) sequence as in Proposition 4.2, i.e.,

I ( u k , v k ) c ( 0 , c ) , I ( u k , v k ) 0 strongly in H 1 ( R n , R 2 ) as k + .

Lemma 2.4 implies that { ( u k , v k ) } are bounded in H 1 ( R n , R 2 ) . From Lemma 2.5, we can assume that { ( u k , v k ) } are nonnegative. Then, Theorem 1.2 is applicable to { ( u k , v k ) } . We claim that l 1 = 0 and l 2 = 0 in Theorem 1.2. Otherwise, if l 1 1 or l 2 1 , we obtain a contradiction

c > c = I ( u , v ) + i = 1 l 1 I ( U ( i ) , V ( i ) ) + j = 1 l 2 I ( u ( j ) , v ( j ) ) c .

Consequently, it must be that l 1 = 0 and l 2 = 0 in Theorem 1.2. Hence, we have

( u k , v k ) ( u , v ) H 1 ( R n , R 2 ) 0 as k , I ( u , v ) = c , I ( u , v ) = 0 .

Since I ( u , v ) = c ( 0 , c ) , it must be that u 0 and v 0 . Furthermore, if

V 1 ( x ) V 2 ( x ) ( V 2 ( x ) ) , η 1 η 2 , α β

or

V 1 ( x ) V 2 ( x ) ( V 2 ( x ) ) , η 1 η 2 , α β ,

and we can prove that u v . In fact, testing (1.1) with ( u , v ) , we obtain

R n ( u 2 + V 1 ( x ) u 2 ) d x = η 1 2 R n ( u + ) η 1 ( v + ) η 2 x d x + α α + β R n Q ( x ) ( u + ( x ) ) α ( v + ( x ) ) β d x , R n ( v 2 + V 2 ( x ) v 2 ) d x = η 2 2 R n ( u + ) η 1 ( v + ) η 2 x d x + β α + β R n Q ( x ) ( u + ( x ) ) α ( v + ( x ) ) β d x .

If u v , we obtain a contradiction by observing the sign of the following equality:

R n ( V 1 ( x ) V 2 ( x ) ) u 2 d x = η 1 η 2 2 R n ( u + ) 2 x d x + α β α + β R n Q ( x ) ( u + ( x ) ) α + β d x .

Acknowledgment

The authors would like to thank Professor Gongbao Li for helpful suggestions on the present article.

  1. Funding information: This work was supported by National Natural Science Foundation of China (Grant No. 11901531) and the excellent doctorial dissertation cultivation grant (Grant No. 30106220328) from Central China Normal University.

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

  3. 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: 2022-02-10
Revised: 2022-04-20
Accepted: 2022-08-14
Published Online: 2022-11-16

© 2023 Lu Shun Wang et al., published by De Gruyter

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

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