Open Access

Multiple positive solutions for semilinear elliptic systems involving subcritical nonlinearities in R N

Boundary Value Problems20122012:118

https://doi.org/10.1186/1687-2770-2012-118

Received: 29 March 2012

Accepted: 4 October 2012

Published: 24 October 2012

Abstract

In this paper, we investigate the effect of the coefficient f ( x ) of the subcritical nonlinearity. Under some assumptions, for sufficiently small ε , λ , μ > 0 , there are at least k (≥1) positive solutions of the semilinear elliptic systems

{ ε 2 Δ u ¯ + u ¯ = λ g ( x ) | u ¯ | q 2 u ¯ + α α + β f ( x ) | u ¯ | α 2 u ¯ | v ¯ | β in  R N ; ε 2 Δ v ¯ + v ¯ = μ h ( x ) | v ¯ | q 2 v ¯ + β α + β f ( x ) | u ¯ | α | v ¯ | β 2 v ¯ in  R N ; u ¯ , v ¯ H 1 ( R N ) ,

where α > 1 , β > 1 , 2 < q < p = α + β < 2 = 2 N / ( N 2 ) for N 3 .

MSC:35J20, 35J25, 35J65.

Keywords

semilinear elliptic systemssubcritical exponentsNehari manifold

1 Introduction

For N 3 , α > 1 , β > 1 and 2 < q < p = α + β < 2 = 2 N / ( N 2 ) , we consider the semilinear elliptic systems

where ε , λ , μ > 0 .

Let f, g and h satisfy the following conditions:

(A1) f is a positive continuous function in R N and lim | x | f ( x ) = f > 0 .

(A2) there exist k points a 1 , a 2 , , a k in R N such that
f ( a i ) = max x R N f ( x ) = 1 for  1 i k ,

and f < 1 .

(A3) g , h L m ( R N ) L ( R N ) where m = ( α + β ) / ( α + β q ) , and g , h 0 .

In [1], if Ω is a smooth and bounded domain in R N ( N 3 ), they considered the following system:
{ ε 2 Δ u ¯ λ 1 u ¯ = μ 1 u ¯ 3 + β u ¯ v ¯ 2 in  Ω ; ε 2 Δ v ¯ λ 2 v ¯ = μ 2 v ¯ 3 + β u ¯ 2 v ¯ in  Ω ; u ¯ > 0 , v ¯ > 0 ,
and proved the existence of a least energy solution in Ω for sufficiently small ε > 0 and β ( , β 0 ) . Lin and Wei also showed that this system has a least energy solution in R N for ε = 1 and β ( 0 , β 0 ) . In this paper, we study the effect of f ( z ) of ( E ¯ ε , λ , μ ). Recently, many authors [25] considered the elliptic systems with subcritical or critical exponents, and they proved the existence of a least energy positive solution or the existence of at least two positive solutions for these problems. In this paper, we construct the k compact Palais-Smale sequences which are suitably localized in correspondence of k maximum points of f. Then we could show that under some assumptions (A1)-(A3), for sufficiently small ε , λ , μ > 0 , there are at least k (≥1) positive solutions of the elliptic system ( E ε , λ , μ ). By the change of variables
x = ε z , u ( z ) = u ¯ ( ε z ) and v ( z ) = v ¯ ( ε z ) ,
System ( E ¯ ε , λ , μ ) is transformed to
Let H = H 1 ( R N ) × H 1 ( R N ) be the space with the standard norm
( u , v ) H = [ R N ( | u | 2 + u 2 ) d z + R N ( | v | 2 + v 2 ) d z ] 1 / 2 .
Associated with the problem ( E ε , λ , μ ), we consider the C 1 -functional J ε , λ , μ , for ( u , v ) H ,
J ε , λ , μ ( u , v ) = 1 2 ( u , v ) H 2 1 α + β R N f ( ε z ) | u | α | v | β d z 1 q R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z .
Actually, the weak solution ( u , v ) H of ( E ε , λ , μ ) is the critical point of the functional J ε , λ , μ , that is, ( u , v ) H satisfies

for any ( φ 1 , φ 2 ) H .

We consider the Nehari manifold
M ε , λ , μ = { ( u , v ) H { ( 0 , 0 ) } | J ε , λ , μ ( u , v ) , ( u , v ) = 0 } ,
(1.1)
where
J ε , λ , μ ( u , v ) , ( u , v ) = ( u , v ) H 2 R N f ( ε z ) | u | α | v | β d z R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z .

The Nehari manifold M ε , λ , μ contains all nontrivial weak solutions of ( E ε , λ , μ ).

Let
S α , β = inf u , v H 1 ( R N ) { ( 0 ) } ( u , v ) H 2 ( R N | u | α | v | β d z ) 2 / ( α + β ) ,
(1.2)
then by [[2], Theorem 5], we have
S α , β = [ ( α β ) β α + β + ( β α ) α α + β ] S p ,
where p = α + β and S p is the best Sobolev constant defined by
S p = inf u H 1 ( R N ) { 0 } R N ( | u | 2 + u 2 ) d z ( R N | u | p d z ) 2 / p .
For the semilinear elliptic systems ( λ = μ = 0 )
we define the energy functional I ε ( u , v ) = 1 2 ( u , v ) H 2 1 α + β R N f ( ε z ) | u | α | v | β d z , and
N ε = { ( u , v ) H { ( 0 , 0 ) } | I ε ( u , v ) , ( u , v ) = 0 } .
If f max z R N f ( z ) (=1), then we define I max ( u , v ) = 1 2 ( u , v ) H 2 1 α + β R N | u | α | v | β d z and
θ max = inf ( u , v ) N max I max ( u , v ) ,

where N max = { ( u , v ) H { ( 0 , 0 ) } | I max ( u , v ) , ( u , v ) = 0 } .

It is well known that this problem
has the unique, radially symmetric and positive ground state solution w H 1 ( R N ) . Define I ¯ max ( u ) = 1 2 R N ( | u | 2 + u 2 ) d z 1 p R N | u | p d z and θ ¯ max = inf u N ¯ max I ¯ max ( u ) , where
N ¯ max = { u H 1 ( R N ) { 0 } | I ¯ max ( u ) , u = 0 } .
Moreover, we have that
θ ¯ max = p 2 2 p S p p p 2 > 0 . (See Wang [6, Theorems 4.12 and 4.13].)

This paper is organized as follows. First of all, we study the argument of the Nehari manifold M ε , λ , μ . Next, we prove that the existence of a positive solution ( u 0 , v 0 ) M ε , λ , μ of ( E ε , λ , μ ). Finally, in Section 4, we show that the condition (A2) affects the number of positive solutions of ( E ε , λ , μ ); that is, there are at least k critical points ( u i , v i ) M ε , λ , μ of J ε , λ , μ such that J ε , λ , μ ( u i , v i ) = β ε , λ , μ i ((PS)-value) for 1 i k .

Theorem 1.1 ( E ε , λ , μ ) has at least one positive solution ( u 0 , v 0 ) , that is, ( E ¯ ε , λ , μ ) admits at least one positive solution.

Theorem 1.2 There exist two positive numbers ε 0 and Λ such that ( E ε , λ , μ ) has at least k positive solutions for any 0 < ε < ε 0 and 0 < λ + μ < Λ , that is, ( E ¯ ε , λ , μ ) admits at least k positive solutions.

2 Preliminaries

By studying the argument of Han [[7], Lemma 2.1], we obtain the following lemma.

Lemma 2.1 Let Ω R N (possibly unbounded) be a smooth domain. If u n u , v n v weakly in H 0 1 ( Ω ) , and u n u , v n v almost everywhere in Ω, then
lim n Ω | u n u | α | v n v | β d z = lim n Ω | u n | α | v n | β d z Ω | u | α | v | β d z .

Note that J ε , λ , μ is not bounded from below in H. From the following lemma, we have that J ε , λ , μ is bounded from below on M ε , λ , μ .

Lemma 2.2 The energy functional J ε , λ , μ is bounded from below on M ε , λ , μ .

Proof For ( u , v ) M ε , λ , μ , by (1.1), we obtain that
J ε , λ , μ ( u , v ) = ( 1 2 1 q ) ( u , v ) H 2 + ( 1 q 1 p ) R N f ( ε z ) | u | α | v | β d z > 0 ,

where p = α + β . Hence, we have that J ε , λ , μ is bounded from below on M ε , λ , μ . □

We define
θ ε , λ , μ = inf ( u , v ) M ε , λ , μ J ε , λ , μ ( u , v ) .
Lemma 2.3 (i) There exist positive numbers σ and d 0 such that J ε , λ , μ ( u , v ) d 0 for ( u , v ) H = σ ;
  1. (ii)

    There exists ( u ¯ , v ¯ ) H { ( 0 , 0 ) } such that ( u ¯ , v ¯ ) H > σ and J ε , λ , μ ( u ¯ , v ¯ ) < 0 .

     
Proof (i) By (1.2), the Hölder inequality ( p 1 = p p q , p 2 = p q ) and the Sobolev embedding theorem, we have that
J ε , λ , μ ( u , v ) = 1 2 ( u , v ) H 2 1 p R N f ( ε z ) | u | α | v | β d z 1 q R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z 1 2 ( u , v ) H 2 1 p S α , β p / 2 ( u , v ) H p 1 q Max S p q 2 ( λ + μ ) ( u , v ) H q ,
where p = α + β and Max = max { g m , h m } . Hence, there exist positive σ and d 0 such that J ε , λ , μ ( u , v ) d 0 for ( u , v ) H = σ .
  1. (ii)
    For any ( u , v ) H { ( 0 , 0 ) } , since
    J ε , λ , μ ( t u , t v ) = t 2 2 ( u , v ) H 2 t p p R N f ( ε z ) | u | α | v | β d z t q q R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z ,
     

then lim t J ε , λ , μ ( t u , t v ) = . Fix some ( u , v ) H { ( 0 , 0 ) } , there exists t ¯ > 0 such that ( t ¯ u , t ¯ v ) H > σ and J ε , λ , μ ( t ¯ u , t ¯ v ) < 0 . Let ( u ¯ , v ¯ ) = ( t ¯ u , t ¯ v ) . □

Define
ψ ( u , v ) = J ε , λ , μ ( u , v ) , ( u , v ) .
Then for ( u , v ) M ε , λ , μ , we obtain that
ψ ( u , v ) , ( u , v ) = 2 ( u , v ) H 2 p R N f ( ε z ) | u | α | v | β d z q R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z = ( p q ) R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z ( p 2 ) ( u , v ) H 2
(2.1)
= ( 2 q ) ( u , v ) H 2 + ( q p ) R N f ( ε z ) | u | α | v | β d z < 0 .
(2.2)

Lemma 2.4 For each ( u , v ) H { ( 0 , 0 ) } , there exists a unique positive number t u , v such that ( t u , v u , t u , v v ) M ε , λ , μ and J ε , λ , μ ( t u , v u , t u , v v ) = sup t 0 J ε , λ , μ ( t u , t v ) .

Proof Fixed ( u , v ) H { ( 0 , 0 ) } , we consider
R ( t ) = J ε , λ , μ ( t u , t v ) = t 2 2 ( u , v ) H 2 t p p R N f ( ε z ) | u | α | v | β d z t q q R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z .
Since R ( 0 ) = 0 , lim t R ( t ) = , by Lemma 2.3(i), then sup t 0 R ( t ) is achieved at some t u , v > 0 . Moreover, we have that R ( t u , v ) = 0 , that is, ( t u , v u , t u , v v ) M ε , λ , μ . Next, we claim that t u , v is a unique positive number such that R ( t u , v ) = 0 . Consider
r ( t ) = ( u , v ) H 2 t p 2 R N f ( ε z ) | u | α | v | β d z t q 2 R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z ,
then R ( t ) = t r ( t ) . Since r ( 0 ) = ( u , v ) H 2 > 0 ,
r ( t ) = ( p 2 ) t p 3 R N f ( ε z ) | u | α | v | β d z ( q 2 ) t q 3 R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z < 0 ,

there exists a unique positive number t ¯ u , v such that r ( t ¯ u , v ) = 0 . It follows that R ( t ¯ u , v ) = 0 . Hence, t ¯ u , v = t u , v . □

Remark 2.5 By Lemma 2.3(i) and Lemma 2.4, then θ ε , λ , μ d 0 > 0 for some constant d 0 .

Lemma 2.6 Let ( u 0 , v 0 ) M ε , λ , μ satisfy
J ε , λ , μ ( u 0 , v 0 ) = min ( u , v ) M ε , λ , μ J ε , λ , μ ( u , v ) = θ ε , λ , μ ,

then ( u 0 , v 0 ) is a solution of ( E ε , λ , μ ).

Proof By (2.2), ψ ( u , v ) , ( u , v ) < 0 for ( u , v ) M ε , λ , μ . Since J ε , λ , μ ( u 0 , v 0 ) = min ( u , v ) M ε , λ , μ J ε , λ , μ ( u , v ) , by the Lagrange multiplier theorem, there is τ R such that J ε , λ , μ ( u 0 , v 0 ) = τ ψ ( u 0 , v 0 ) in H 1 . Then we have
0 = J ε , λ , μ ( u 0 , v 0 ) , ( u 0 , v 0 ) = τ ψ ( u 0 , v 0 ) , ( u 0 , v 0 ) .

It follows that τ = 0 and J ε , λ , μ ( u 0 , v 0 ) = 0 in H 1 . Therefore, ( u 0 , v 0 ) is a nontrivial solution of ( E ε , λ , μ ) and J ε , λ , μ ( u 0 , v 0 ) = θ ε , λ , μ . □

3 (PS) γ -condition in H for J ε , λ , μ

First of all, we define the Palais-Smale (denoted by (PS)) sequence and (PS)-condition in H for some functional J.

Definition 3.1 (i) For γ R , a sequence { ( u n , v n ) } is a (PS) γ -sequence in H for J if J ( u n , v n ) = γ + o n ( 1 ) and J ( u n , v n ) = o n ( 1 ) strongly in H 1 as n , where H 1 is the dual space of H;
  1. (ii)

    J satisfies the (PS) γ -condition in H if every (PS) γ -sequence in H for J contains a convergent subsequence.

     

Applying Ekeland’s variational principle and using the same argument as in Cao-Zhou [8] or Tarantello [9], we have the following lemma.

Lemma 3.2 (i) There exists a (PS) -sequence { ( u n , v n ) } in M ε , λ , μ for J ε , λ , μ .

In order to prove the existence of positive solutions, we want to prove that J ε , λ , μ satisfies the (PS) γ -condition in H for γ ( 0 , p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) ) .

Lemma 3.3 J ε , λ , μ satisfies the (PS) γ -condition in H for γ ( 0 , p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) ) .

Proof Let { ( u n , v n ) } be a (PS) γ -sequence in H for J ε , λ , μ such that J ε , λ , μ ( u n , v n ) = γ + o n ( 1 ) and J ε , λ , μ ( u n , v n ) = o n ( 1 ) in H 1 . Then
γ + c n + d n ( u n , v n ) H q J ε , λ , μ ( u n , v n ) 1 q J ε , λ , μ ( u n , v n ) , ( u n , v n ) = ( 1 2 1 q ) ( u n , v n ) H 2 + ( 1 q 1 p ) R N f ( ε z ) | u n | α | v n | β d z q 2 2 q ( u n , v n ) H 2 ,
where c n = o n ( 1 ) , d n = o n ( 1 ) as n . It follows that { ( u n , v n ) } is bounded in H. Hence, there exist a subsequence { ( u n , v n ) } and ( u , v ) H such that
Moreover, we have that J ε , λ , μ ( u , v ) = 0 in H 1 . We use the Brézis-Lieb lemma to obtain (3.1) and (3.2) as follows:
(3.1)
(3.2)
Next, we claim that
R N g ( ε z ) | u n u | q d z 0 as  n
(3.3)
and
R N h ( ε z ) | v n v | q d z 0 as  n .
(3.4)
Since g L m ( R N ) , where m = p / ( p q ) , then for any σ > 0 , there exists r > 0 such that [ B r N ( 0 ) ] c g ( ε z ) p p q d z < σ . By the Hölder inequality and the Sobolev embedding theorem, we get
Similarly, R N h ( ε z ) | v n v | q d z 0 as n . By (A1) and u n u , v n v strongly in L loc p ( R N ) , we have that
R N f ( ε z ) | u n u | α | v n v | β d z = R N f | u n u | α | v n v | β d z = o n ( 1 ) .
(3.5)
Let p n = ( u n u , v n v ) . By (3.1)-(3.4) and Lemma 2.1, we deduce that
p n H 2 = ( u n H 2 + v n H 2 ) ( u H 2 + v H 2 ) + o n ( 1 ) = R N f ( ε z ) | u n | α | v n | β d z + R N ( λ g ( ε z ) | u n | q + μ h ( ε z ) | v n | q ) d z R N f ( ε z ) | u | α | v | β d z R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z + o n ( 1 ) = R N f ( ε z ) | u n u | α | v n v | β d z + o n ( 1 ) ,
and
1 2 p n H 2 1 α + β R N f ( ε z ) | u n u | α | v n v | β d z = γ J ε , λ , μ ( u , v ) + o n ( 1 ) .
(3.6)
We may assume that
p n H 2 l and R N f ( ε z ) | u n u | α | v n v | β d z l as  n .
(3.7)
Recall that
S α , β = inf u , v H 1 ( R N ) { ( 0 ) } ( u , v ) H 2 ( R N | u | α | v | β d z ) 2 / p , where  p = α + β .
If l > 0 , by (3.5), then
S α , β l 2 p = S α , β ( R N f ( ε z ) | u n u | α | v n v | β d z ) 2 / p + o n ( 1 ) = S α , β ( R N f | u n u | α | v n v | β d z ) 2 / p + o n ( 1 ) ( f ) 2 / p p n H 2 + o n ( 1 ) = ( f ) 2 / p l .
This implies l ( S α , β ) p / ( p 2 ) / ( f ) 2 / ( p 2 ) . By (3.6) and (3.7), we obtain that
γ = ( 1 2 1 p ) l + J ε , λ , μ ( u , v ) p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) ,

which is a contradiction. Hence, l = 0 , that is, ( u n , v n ) ( u , v ) strongly in H. □

4 Existence of k solutions

Let w H 1 ( R N ) be the unique, radially symmetric and positive ground state solution of equation (E 0) in R N . Recall the facts (or see Bahri-Li [10], Bahri-Lions [11], Gidas-Ni-Nirenberg [12] and Kwong [13]):
  1. (i)

    w L ( R N ) C loc 2 , θ ( R N ) for some 0 < θ < 1 and lim | z | w ( z ) = 0 ;

     
  2. (ii)
    for any ε > 0 , there exist positive numbers C 1 , C 2 ε and C 3 ε such that for all z R N
    C 2 ε exp ( ( 1 + ε ) | z | ) w ( z ) C 1 exp ( | z | )
     
and
| w ( z ) | C 3 ε exp ( ( 1 ε ) | z | ) .
By Lien-Tzeng-Wang [14], then
S p = R N ( | w | 2 + w 2 ) d z ( R N w p d z ) 2 / p .
(4.1)
For 1 i k , we define
w ε i ( z ) = w ( z a i ε ) , where  f ( a i ) = max z R N f ( z ) = 1 .

Clearly, w ε i ( z ) H 1 ( R N ) .

First of all, we want to prove that
lim ε 0 + sup t 0 J ε , λ , μ ( t α w ε i , t β w ε i ) p 2 2 p ( S α , β ) p / ( p 2 ) uniformly in  i .
Lemma 4.1 For λ > 0 and μ > 0 , then
lim ε 0 + sup t 0 J ε , λ , μ ( t α w ε i , t β w ε i ) p 2 2 p ( S α , β ) p / ( p 2 ) uniformly in i .
Moreover, we have that
0 < θ ε , λ , μ p 2 2 p ( S α , β ) p / ( p 2 ) .
Proof Part I: Since J ε , λ , μ is continuous in H, J ε , λ , μ ( 0 , 0 ) = 0 , and { ( α w ε i , β w ε i ) } is uniformly bounded in H for any ε > 0 and 1 i k , then there exists t 0 > 0 such that for 0 t < t 0 and any ε > 0 ,
J ε , λ , μ ( t α w ε i , t β w ε i ) < p 2 2 p ( S α , β ) p / ( p 2 ) uniformly in  i .
From (A1), we have that inf z R N f ( z ) > 0 . Then
J ε , λ , μ ( t α w ε i , t β w ε i ) t 2 2 ( α w , β w ) H 2 t α + β α + β ( inf z R N f ( z ) ) R N | α w | α | β w | β d z as  t .
It follows that there exists t 1 > 0 such that for t > t 1 and any ε > 0 ,
J ε , λ , μ ( t α w ε i , t β w ε i ) < p 2 2 p ( S α , β ) p / ( p 2 ) uniformly in  i .
From now on, we only need to show that
lim ε 0 + sup t 0 t t 1 J ε , λ , μ ( t w ε i ) p 2 2 p ( S α , β ) p / ( p 2 ) uniformly in  i .
Since
sup t 0 ( t 2 2 a t α + β α + β b ) = α + β 2 2 ( α + β ) ( a b 2 α + β ) α + β α + β 2 , where  a , b > 0 ,
and by (4.1), then
(4.2)
For t 0 t t 1 , by (4.2), we have that
J ε , λ , μ ( t α w ε i , t β w ε i ) = t 2 2 ( α w ε i , β w ε i ) H 2 t α + β α + β R N f ( ε z ) | α w ε i | α | β w ε i | β d z t q q R N ( λ g ( ε z ) | α w ε i | q + μ h ( ε z ) | β w ε i | q ) d z p 2 2 p ( S α , β ) p / ( p 2 ) + t 1 p p R N ( 1 f ( ε z ) ) | α w ε i | α | α w ε i | β d z .
Since
then
lim ε 0 + sup t 0 t t 1 J ε , λ , μ ( t α w ε i , t β w ε i ) p 2 2 p ( S α , β ) p / ( p 2 ) ,
that is, for λ > 0 and μ > 0 ,
lim ε 0 + sup t 0 J ε , λ , μ ( t α w ε i , t β w ε i ) p 2 2 p ( S α , β ) p / ( p 2 ) uniformly in  i .
Part II: By Lemma 2.4, there is a number t ε i > 0 such that ( t ε i α w ε i , t ε i β w ε i ) M ε , λ , μ , where 1 i k . Hence, from the result of Part I, we have that for λ > 0 and μ > 0 ,
0 < θ ε , λ , μ lim ε 0 + sup t 0 J ε , λ , μ ( t α w ε i , t β w ε i ) p 2 2 p ( S α , β ) p / ( p 2 ) .

 □

Proof of Theorem 1.1 By Lemma 3.2, there exists a (PS) -sequence { ( u n , v n ) } in M ε , λ , μ for J ε , λ , μ . Since 0 < θ ε , λ , μ p 2 2 p ( S α , β ) p / ( p 2 ) < p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) for λ > 0 and μ > 0 , by Lemma 3.3, there exist a subsequence { ( u n , v n ) } and ( u 0 , v 0 ) H such that ( u n , v n ) ( u 0 , v 0 ) strongly in H. It is easy to check that ( u 0 , v 0 ) is a nontrivial solution of ( E ε , λ , μ ) and J ε , λ , μ ( u 0 , v 0 ) = θ ε , λ , μ . Since J ε , λ , μ ( u 0 , v 0 ) = J λ , μ ( | u 0 | , | v 0 | ) and ( | u 0 | , | v 0 | ) M ε , λ , μ , by Lemma 2.6, we may assume that u 0 0 , v 0 0 . Applying the maximum principle, u 0 > 0 and v 0 > 0 in Ω. □

Choosing 0 < ρ 0 < 1 such that
B ρ 0 N ( a i ) ¯ B ρ 0 N ( a j ) ¯ = for  i j  and  1 i , j k ,
where B ρ 0 N ( a i ) ¯ = { z R N | | z a i | ρ 0 } and f ( a i ) = max z R N f ( z ) = 1 , define K = { a i | 1 i k } and K ρ 0 / 2 = i = 1 k B ρ 0 / 2 N ( a i ) ¯ . Suppose i = 1 k B ρ 0 N ( a i ) ¯ B r 0 N ( 0 ) for some r 0 > 0 . Let Q ε be given by
Q ε ( u , v ) = R N χ ( ε z ) | u | α | v | β d z R N | u | α | v | β d z ,

where χ : R N R N , χ ( z ) = z for | z | r 0 and χ ( z ) = r 0 z / | z | for | z | > r 0 .

For each 1 i k , we define

By Lemma 2.4, there exists t ε i > 0 such that ( t ε i α w ε i , t ε i β w ε i ) M ε , λ , μ for each 1 i k . Then we have the following result.

Lemma 4.2 There exists ε 1 > 0 such that if ε ( 0 , ε 1 ) , then Q ε ( t ε i α w ε i , t ε i β w ε i ) K ρ 0 / 2 for each 1 i k .

Proof Since
Q ε ( t ε i α w ε i , t ε i β w ε i ) = R N χ ( ε z ) | w ( z a i ε ) | p d z R N | w ( z a i ε ) | p d z = R N χ ( ε z + a i ) | w ( z ) | p d z R N | w ( z ) | p d z a i as  ε 0 + ,
there exists ε 1 > 0 such that
Q ε ( t ε i α w ε i , t ε i β w ε i ) K ρ 0 / 2 for any  ε ( 0 , ε 1 )  and each  1 i k .

 □

We need the following lemmas to prove that β λ , μ i < β ˜ λ , μ i for sufficiently small ε, λ, μ.

Lemma 4.3 θ max = p 2 2 p ( S α , β ) p / ( p 2 ) .

Proof From Part I of Lemma 4.1, we obtain sup t 0 I max ( t α w ε i , t β w ε i ) = p 2 2 p ( S α , β ) p / ( p 2 ) uniformly in i. Similarly to Lemma 2.4, there is a sequence { s max i } R + such that ( s max i α w ε i , s max i β w ε i ) N max and
θ max I max ( s max i α u ε i , s max i β u ε i ) = sup t 0 J max ( t α u ε i , t β u ε i ) = p 2 2 p ( S α , β ) p / ( p 2 ) .
Let { ( u n , v n ) } N max be a minimizing sequence of θ max for I max . It follows that ( u n , v n ) H 2 = R N | u n | α | v n | β d z and
θ max = 1 2 ( u n , v n ) H 2 1 p R N | u n | α | v n | β d z + o n ( 1 ) = p 2 2 p ( u n , v n ) H 2 + o n ( 1 ) .

We may assume that ( u n , v n ) H 2 l and R N | u n | α | v n | β d z l as n , where l = 2 p p 2 θ max > 0 . By the definition of S α , β , then S α , β l 2 p l . We can deduce that S α , β l p 2 p = ( 2 p p 2 θ max ) p 2 p , that is, p 2 2 p ( S α , β ) p / ( p 2 ) θ max . □

Lemma 4.4 There exists a number δ 0 > 0 such that if ( u , v ) N ε and I ε ( u , v ) θ max + δ 0 , then Q ε ( u , v ) K ρ 0 / 2 for any 0 < ε < ε 1 .

Proof On the contrary, there exist the sequences { ε n } R + and { ( u n , v n ) } N ε n such that ε n 0 , I ε n ( u n , v n ) = θ max ( > 0 ) + o n ( 1 ) as n and Q ε n ( u n , v n ) K ρ 0 / 2 for all n N . It is easy to check that { ( u n , v n ) } is bounded in H. Suppose that R N | u n | α | v n | β d z 0 as n . Since
( u n , v n ) H 2 = R N f ( ε n z ) | u n | α | v n | β d z for each  n N ,
then
θ max + o n ( 1 ) = I ε n ( u n , v n ) = ( 1 2 1 p ) R N f ( ε n z ) | u n | α | v n | β d z o n ( 1 ) ,
which is a contradiction. Thus, R N | u n | α | v n | β d z 0 as n . Similarly to the concentration-compactness principle (see Lions [15, 16] or Wang [[6], Lemma 2.16]), then there exist a constant c 0 > 0 and a sequence { z n ˜ } R N such that
B N ( z n ˜ ; 1 ) | u n | α l p | v n | β l p d z c 0 > 0 ,
(4.3)
where 2 < l < p = α + β < 2 and p = l ( 1 t ) + 2 t for some t ( ( N 2 ) / N , 1 ) . Let ( u n ˜ ( z ) , v n ˜ ( z ) ) = ( u n ( z + z n ˜ ) , v n ( z + z n ˜ ) ) . Then there are a subsequence { ( u n ˜ , v n ˜ ) } and ( u ˜ , v ˜ ) H such that u n ˜ u ˜ and v n ˜ v ˜ weakly in H 1 ( R N ) . Using the similar computation of Lemma 2.4, there is a sequence { s max n } R + such that ( s max n u n ˜ , s max n v n ˜ ) N max and
0 < θ max I max ( s max n u n ˜ , s max n v n ˜ ) = I max ( s max n u n , s max n v n ) I ε n ( s max n u n , s max n v n ) I ε n ( u n , v n ) = θ max + o n ( 1 ) as  n .
We deduce that a subsequence { s max n } satisfies s max n s 0 > 0 . Then there are a subsequence { ( s max n u n ˜ , s max n v n ˜ ) } and ( s 0 u ˜ , s 0 v ˜ ) H such that s max n u n ˜ s 0 u ˜ and s max n v n ˜ s 0 v ˜ weakly in H 1 ( R N ) . By (4.3), then u ˜ 0 and v ˜ 0 . Applying Ekeland’s variational principle, there exists a (PS) -sequence { ( U n , V n ) } for I max and ( U n s max n u n ˜ , V n s max n v n ˜ ) H = o n ( 1 ) . Similarly to the proof of Lemma 3.3, there exist a subsequence { ( U n , V n ) } and ( U 0 , V 0 ) H such that U n U 0 , V n V 0 strongly in H 1 ( R N ) and I max ( U 0 , V 0 ) = θ max . Now, we want to show that there exists a subsequence { z n } = { ε n z n ˜ } such that z n z 0 K .
  1. (i)
    Claim that the sequence { z n } is bounded in R N . On the contrary, assume that | z n | , then
    θ max = I max ( U 0 , V 0 ) < 1 2 ( U 0 , V 0 ) H 2 1 p R N f | U 0 | α | V 0 | β d z lim inf n [ ( s max n ) 2 2 ( u n ˜ , v n ˜ ) H 2 ( s max n ) p p R N f ( ε n z + z n ) | u n ˜ | α | v n ˜ | β d z ] = lim inf n [ ( s max n ) 2 2 ( u n , v n ) H 2 ( s max n ) p p R N f ( ε n z ) | u n | α | v n | β d z ] = lim inf n I ε n ( s max n u n , s max n v n ) lim inf n I ε n ( u n , v n ) = θ max ,
     
which is a contradiction.
  1. (ii)
    Claim that z 0 K . On the contrary, assume that z 0 K , that is, f ( z 0 ) < 1 = max z R N f ( z ) . Then use the argument of (i) to obtain that
    θ max = I max ( U 0 , V 0 ) I max ( s 0 U 0 , s 0 V 0 ) < ( s 0 ) 2 2 ( U 0 , V 0 ) H 2 ( s 0 ) p p R N f ( z 0 ) | U 0 | α | V 0 | β d z lim inf n [ ( s max n ) 2 2 ( u n ˜ , v n ˜ ) H 2 ( s max n ) p p R N f ( ε n z + z n ) | u n ˜ | α | v n ˜ | β d z ] θ max ,
     

which is a contradiction.

Since ( U n s max n u n ˜ , V n s max n v n ˜ ) H = o n ( 1 ) and U n U 0 , V n V 0 strongly in H 1 ( R N ) , we have that
Q ε n ( u n , v n ) = R N χ ( ε n z ) | u n ˜ ( z z n ˜ ) | α | v n ˜ ( z z n ˜ ) | β d z R N | u n ˜ ( z z n ˜ ) | α | v n ˜ ( z z n ˜ ) | β d z = R N χ ( ε n z + ε n z n ˜ ) | U 0 | α | V 0 | β d z R N | U 0 | α | V 0 | β d z z 0 K ρ 0 / 2 as  n ,

which is a contradiction.

Hence, there exists δ 0 > 0 such that if ( u , v ) N ε and I ε ( u , v ) θ max + δ 0 , then Q ε ( u , v ) K ρ 0 / 2 for any 0 < ε < ε 1 . □

Lemma 4.5 If ( u , v ) M ε , λ , μ and J ε , λ , μ ( u , v ) θ max + δ 0 / 2 , then there exists a number Λ > 0 such that Q ε ( u , v ) K ρ 0 / 2 for any 0 < ε < ε 1 and 0 < λ + μ < Λ .

Proof Using the similar computation in Lemma 2.4, we obtain that there is the unique positive number
s ε = ( ( u , v ) H 2 R N f ( ε z ) | u | α | v | β d z ) 1 / ( p 2 )
such that ( s ε u , s ε v ) N ε . We want to show that there exists Λ 0 > 0 such that if 0 < λ + μ < Λ 0 , then s ε < c for some constant c > 0 (independent of u and v). First, for ( u , v ) M ε , λ , μ ,
0 < d 0 θ ε , λ , μ J ε , λ , μ ( u , v ) θ max + δ 0 / 2 .
Since J ε , λ , μ ( u , v ) , ( u , v ) = 0 , then
θ max + δ 0 / 2 J ε , λ , μ ( u , v ) = ( 1 2 1 q ) ( u , v ) H 2 + ( 1 q 1 p ) R N f ( ε z ) | u | α | v | β d z q 2 2 q ( u , v ) H 2 , that is, ( u , v ) H 2 c 1 = 2 q q 2 ( θ max + δ 0 / 2 ) ,
(4.4)
and
d 0 J ε , λ , μ ( u , v ) = ( 1 2 1 p ) ( u , v ) H 2 ( 1 q 1 p ) Ω ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z p 2 2 p ( u , v ) H 2 , that is, ( u , v ) H 2 c 2 = 2 p p 2 d 0 .
(4.5)
Moreover, we have that
Ω f ( ε z ) | u | α | v | β d z = ( u , v ) H 2 R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z c 2 Max S p q 2 ( λ + μ ) c 1 q / 2 ,
where Max = max { g m , h m } . It follows that there exists Λ 0 > 0 such that for 0 < λ + μ < Λ 0
R N f ( ε z ) | u | α | v | β d z c 2 Max S p q 2 ( λ + μ ) ( c 1 ) q / 2 > 0 .
(4.6)
Hence, by (4.4), (4.5) and (4.6), s ε < c for some constant c > 0 (independent of u and v) for 0 < λ + μ < Λ 0 . Now, we obtain that
θ max + δ 0 / 2 J ε , λ , μ ( u , v ) = sup t 0 J ε , λ , μ ( t u , t v ) J ε , λ , μ ( s ε u , s ε v ) = 1 2 ( s ε u , s ε v ) H 2 1 p R N f ( ε z ) | s ε u | α | s ε v | β d z 1 q R N ( λ g ( ε z ) | s ε u | q + μ h ( ε z ) | s ε v | q ) d z I ε ( s ε u , s ε v ) 1 q R N ( λ g ( ε z ) | s ε u | q + μ h ( ε z ) | s ε v | q ) d z .
From the above inequality, we deduce that for any 0 < ε < ε 1 and 0 < λ + μ < Λ 0 ,
I ε ( s ε u , s ε v ) θ max + δ 0 / 2 + 1 q R N ( λ g ( ε z ) | s ε u | q + μ h ( ε z ) | s ε v | q ) d z θ max + δ 0 / 2 + Max ( λ + μ ) S p q 2 ( s ε u , s ε v ) H q < θ max + δ 0 / 2 + Max S p q 2 ( λ + μ ) c q ( c 1 ) q / 2 .
Hence, there exists Λ ( 0 , Λ 0 ) such that for 0 < λ + μ < Λ ,
I ε ( s ε u , s ε v ) θ max + δ 0 , where  ( s ε u , s ε v ) N ε .
By Lemma 4.4, we obtain
Q ε ( s ε u , s ε v ) = R N χ ( ε z ) | s ε u | α | s ε v | β d z R N | s ε u | α | s ε v | β d z K ρ 0 / 2 ,

or Q ε ( u , v ) K ρ 0 / 2 for any 0 < ε < ε 0 and 0 < λ + μ < Λ . □

Since f < 1 , then by Lemma 4.3,
θ max = p 2 2 p ( S α , β ) p / ( p 2 ) < p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) .
(4.7)
By Lemmas 4.1, 4.2 and (4.7), for any 0 < ε < ε 0 ( < ε 1 ) and 0 < λ + μ < Λ ,
β ε , λ , μ i J ε , λ , μ ( t ε i α w ε i , t ε i β w ε i ) < p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) .
(4.8)
Applying above Lemma 4.5, we get that
β ˜ ε , λ , μ i θ max + δ 0 / 2 for any  0 < ε < ε 0  and  0 < λ + μ < Λ .
(4.9)
For each 1 i k , by (4.8) and (4.9), we obtain that
β ε , λ , μ i < β ˜ ε , λ , μ i for any  0 < ε < ε 0  and  0 < λ + μ < Λ .
It follows that
β ε , λ , μ i = inf ( u , v ) O ε , λ , μ i O ε , λ , μ i J ε , λ , μ ( u , v ) for any  0 < ε < ε 0  and  0 < λ + μ < Λ .

Then applying Ekeland’s variational principle and using the standard computation, we have the following lemma.

Lemma 4.6 For each 1 i k , there is a ( PS ) β ε , λ , μ i -sequence { ( u n , v n ) } O ε , λ , μ i in H for J ε , λ , μ .

Proof See Cao-Zhou [8]. □

Proof of Theorem 1.2 For any 0 < ε < ε 0 and 0 < λ + μ < Λ , by Lemma 4.6, there is a ( PS ) β ε , λ , μ i -sequence { ( u n , v n ) } O ε , λ , μ i for J ε , λ , μ where 1 i k . By (4.8), we obtain that
β ε , λ , μ i < p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) .

Since J ε , λ , μ satisfies the (PS) γ -condition for γ ( , p 2 2 p ( S α , β ) p / ( p 2 ) ( f ) 2 / ( p 2 ) ) , then J ε , λ , μ has at least k critical points in M ε , λ , μ for any 0 < ε < ε 0 and 0 < λ + μ < Λ . Set u + = max { u , 0 } and v + = max { v , 0 } . Replace the terms R N f ( ε z ) | u | α | v | β d z and R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) d z of the functional J ε , λ , μ by R N f ( ε z ) u + α v + β d z and R N ( λ g ( ε z ) u + q + μ h ( ε z ) v + q ) d z , respectively. It follows that ( E ε , λ , μ ) has k nonnegative solutions. Applying the maximum principle, ( E ε , λ , μ ) admits at least k positive solutions. □

Declarations

Acknowledgements

The author was grateful for the referee’s helpful suggestions and comments.

Authors’ Affiliations

(1)
Department of Natural Sciences in the Center for General Education, Chang Gung University, Tao-Yuan, R.O.C

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© Lin; licensee Springer 2012

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