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

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 =2N/(N2) for N3.

MSC:35J20, 35J25, 35J65.

1 Introduction

For N3, α>1, β>1 and 2<q<p=α+β< 2 =2N/(N2), 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)=1for 1ik,

and f <1.

(A3) g,h L m ( R N ) L ( R N ) where m=(α+β)/(α+βq), and g,h0.

In [1], if Ω is a smooth and bounded domain in R N (N3), 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=εz,u(z)= u ¯ (εz)andv(z)= v ¯ (εz),

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(εz)|u | α |v | β dz R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) dz.

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(εz)|u | α |v | β dz, 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 dz 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 )dz 1 p R N |u | p dz 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 1ik.

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 dz= lim n Ω | u n | α | v n | β dz Ω |u | α |v | β dz.

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(εz)|u | α |v | β dz>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 (tu,tv)=. 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)
=(2q) ( u , v ) H 2 +(qp) R N f(εz)|u | α |v | β dz<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 (tu,tv).

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(εz)|u | α |v | β dz t q 2 R N ( λ g ( ε z ) | u | q + μ h ( ε z ) | v | q ) dz,

then R (t)=tr(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 τR 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 γR, 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(εz)| u n u | q dz0as n
(3.3)

and

R N h(εz)| v n v | q dz0as n.
(3.4)

Since g L m ( R N ), where m=p/(pq), then for any σ>0, there exists r>0 such that [ B r N ( 0 ) ] c g ( ε z ) p p q dz<σ. By the Hölder inequality and the Sobolev embedding theorem, we get

Similarly, R N h(εz)| v n v | q dz0 as n. By (A1) and u n u, v n v strongly in L loc p ( R N ), we have that

R N f(εz)| u n u | α | v n v | β dz= R N f | u n u | α | v n v | β dz= 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(εz)| u n u | α | v n v | β dz=γ J ε , λ , μ (u,v)+ o n (1).
(3.6)

We may assume that

p n H 2 land R N f(εz)| u n u | α | v n v | β dzlas 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 1ik, 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 1ik, then there exists t 0 >0 such that for 0t< 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 1ik. 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 ij and 1i,jk,

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 |1ik} 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 1ik, 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 1ik. 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 1ik.

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 1ik.

 □

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 dz 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 dzl 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 nN. It is easy to check that {( u n , v n )} is bounded in H. Suppose that R N | u n | α | v n | β dz0 as n. Since

( u n , v n ) H 2 = R N f( ε n z)| u n | α | v n | β dzfor each nN,

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 dz o n (1),

which is a contradiction. Thus, R N | u n | α | v n | β dz0 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 dz c 0 >0,
(4.3)

where 2<l<p=α+β< 2 and p=l(1t)+ 2 t for some t((N2)/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(εz)|u | α |v | β dz 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 /2for any 0<ε< ε 0  and 0<λ+μ< Λ .
(4.9)

For each 1ik, 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 1ik, 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 1ik. 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(εz)|u | α |v | β dz and R N (λg(εz)|u | q +μh(εz)|v | q )dz of the functional J ε , λ , μ by R N f(εz) u + α v + β dz and R N (λg(εz) u + q +μh(εz) v + q )dz, 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. □

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Lin, Hl. Multiple positive solutions for semilinear elliptic systems involving subcritical nonlinearities in R N . Bound Value Probl 2012, 118 (2012). https://doi.org/10.1186/1687-2770-2012-118

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Keywords

  • semilinear elliptic systems
  • subcritical exponents
  • Nehari manifold