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Existence result for semilinear elliptic systems involving critical exponents

S Khademloo*, M Farzinejad and O Khazaee kohpar

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Department of Basic Sciences, Babol University of Technology, Babol, 47148-71167, Iran

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Boundary Value Problems 2012, 2012:119  doi:10.1186/1687-2770-2012-119


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/119


Received:14 July 2012
Accepted:4 October 2012
Published:24 October 2012

© 2012 Khademloo et al.; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper we deal with the existence of a positive solution for a class of semilinear systems of multi-singular elliptic equations which involve Sobolev critical exponents. In fact, by the analytic techniques and variational methods, we prove that there exists at least one positive solution for the system.

MSC: 35J60, 35B33.

Keywords:
semilinear elliptic system; nontrivial solution; critical exponent; variational method

1 Introduction

We consider the following elliptic system:

{ L u = σ α 2 u | u | α 2 | ν | β + η u | u | 2 2 + a 1 u + a 2 ν , Ω , L ν = σ β 2 ν | ν | β 2 | u | α + λ ν | ν | 2 2 + a 2 u + a 3 ν , Ω , u = ν = 0 , Ω , (1.1)

where Ω R N ( N 3 ) is a smooth bounded domain such that ξ i Ω , i = 1 , 2 , , k , k 2 , are different points, 0 μ i < μ ¯ : = ( N 2 2 ) 2 , L : = Δ i = 1 k μ i | x ξ i | 2 , η , λ , σ 0 , a 1 , a 2 , a 3 R , 1 < α , β < 2 1 , α + β = 2 .

We work in the product space H × H , where the space H : = H 0 1 ( Ω ) is the completion of C 0 ( Ω ) with respect to the norm ( Ω | | 2 d x ) 1 2 .

In resent years many publications [1-3] concerning semilinear elliptic equations involving singular points and the critical Sobolev exponent have appeared. Particularly in the last decade or so, many authors used the variational method and analytic techniques to study the existence of positive solutions of systems of the form of (1.1) or its variations; see, for example, [4-8].

Before stating the main result, we clarify some terminology. Since our method is variational in nature, we need to define the energy functional of (1.1) on H × H

J ( u , ν ) = 1 2 Ω ( | u | 2 + | ν | 2 i = 1 k μ i ( u 2 + ν 2 ) | x ξ i | 2 ) d x 1 2 Ω ( a 1 u 2 + 2 a 2 u ν + a 3 ν 2 ) d x σ σ 2 Ω | u | α | u | β d x η 2 Ω | u | 2 d x λ 2 Ω | ν | 2 d x .

Then J ( u , ν ) belongs to C 1 ( H × H , R ) . A pair of functions ( u 0 , ν 0 ) H × H is said to be a solution of (1.1) if ( u 0 , ν 0 ) ( 0 , 0 ) , and for all ( φ , ϕ ) H × H , we have

Ω ( u 0 φ + ν 0 ϕ i = 1 k μ i ( u 0 φ + ν 0 ϕ ) | x ξ i | 2 ( a 1 u 0 φ + a 2 φ ν 0 + a 2 ϕ u 0 + a 3 ν 0 ϕ ) σ ( α φ u 0 | u 0 | α 2 | ν 0 | β + β ϕ ν 0 | ν 0 | β 2 | u 0 | α ) η | u 0 | 2 2 u 0 φ λ | ν 0 | 2 2 ν 0 ϕ ) d x = J ( u 0 , ν 0 ) , ( φ , ϕ ) = 0 .

Standard elliptic arguments show that

u , ν C 2 ( Ω { ξ 1 , , ξ k } ) C 1 ( Ω ¯ { ξ 1 , , ξ k } ) .

The following assumptions are needed:

( H 1 ) η + λ + σ > 0 , 0 μ 1 μ 2 μ k < μ ¯ 1 and i = 1 k μ i < μ ¯ , α + β > 1 , α + β = 2 , a 1 , a 2 , a 3 > 0 , a 1 a 3 a 2 2 > 0 ,

( H 2 ) 0 < λ 1 λ 2 < Λ 1 ( μ ) , where Λ 1 ( μ ) is the first eigenvalue of L, λ 1 , λ 2 are the eigenvalues of the matrix A = ( a 1 a 2 a 2 a 3 ) .

The quadratic from Q ( u , ν ) : = ( u , ν ) A ( u , ν ) T = a 1 u 2 + 2 a 2 u ν + a 3 ν 2 is positively defined and satisfies

λ 1 ( u 2 + ν 2 ) a 1 u 2 + 2 a 2 u ν + a 3 ν 2 λ 2 ( u 2 + ν 2 ) u , ν H . (1.2)

Our main results are as follows.

Theorem 1.1Suppose ( H 1 ) holds. Then for any solution ( u , v ) H × H of problem (1.1), there exists a positive constant C 1 such that

max { | u ( x ) | , | ν ( x ) | } C 1 | x ξ i | ( μ ¯ μ ¯ μ i ) , x B ρ 1 ( ξ i ) { ξ i } ,

where ρ 1 > 0 and B ρ 1 ( ξ i ) Ω .

Theorem 1.2Suppose ( H 1 ) holds. Then for any positive solution ( u , v ) H × H of problem (1.1), there exists a positive constant C 2 such that B ρ 2 ( ξ i ) Ω and

min { u ( x ) , ν ( x ) } C 2 | x ξ i | ( μ ¯ μ ¯ μ i ) , x B ρ 2 ( ξ i ) { ξ i } ,

where ρ 2 > 0 .

Theorem 1.3Suppose ( H 1 ), ( H 2 ) hold. Then the problem (1.1) has a positive solution.

2 Preliminaries

On H × H , we use the norm

( u , ν ) H × H = ( Ω ( | u | 2 + | ν | 2 ) d x ) 1 2 .

Using the Young inequality, the following best constant is well defined:

S μ i : = inf u D 1 , 2 ( R N ) { 0 } R N ( | u | 2 μ i u 2 | x ξ i | 2 ) d x ( R N | u | 2 d x ) 2 / 2 , (2.1)

where D 1 , 2 ( R N ) is the completion of C 0 ( R N ) with respect to the norm ( R N | | 2 d x ) 1 2 .

We infer that S μ i is attained in R N by the functions

V μ i , ε ξ i ( x ) = ε 2 N 2 U μ i ( x ξ i ε ) , ε > 0 ,

where

U μ i ( x ξ i ) = ( 2 N ( μ ¯ μ i ) μ ¯ ) μ ¯ / 2 | x ξ i | ( μ ¯ μ ¯ μ i ) ( 1 + | x ξ i | 2 μ ¯ μ i μ ¯ ) μ ¯ .

For all η , λ , σ 0 , η + λ + σ > 0 , α + β > 1 , α + β = 2 , by the Young and Hardy-Sobolev inequalities, the following constant is well defined on D : = ( D 1 , 2 ( R N ) { 0 } ) 2 :

S η , λ , σ ( μ i ) : = inf ( u , ν ) D R N ( | u | 2 + | ν | 2 μ i u 2 + ν 2 | x ξ i | 2 ) d x ( R N ( η | u | 2 + λ | ν | 2 + σ | u | α | ν | β ) d x ) 2 / 2 . (2.2)

Set

u μ , ε ξ ( x ) = ψ ( x ) V μ , ε ξ ( x ) = ε 2 N 2 ψ ( x ) U μ ( x ξ ε ) ,

where ξ Ω , 0 μ < μ ¯ , ψ C 0 ( B ρ ( ξ ) ) satisfies 0 ψ 1 and ψ 1 , x B ρ 2 ( ξ ) , for all ρ > 0 small. Then for any 0 < μ < μ ¯ , by [9] we have the following estimates:

Ω ( | u μ , ε ξ | 2 μ ( u μ , ε ξ ) 2 | x ξ | 2 ) d x = S μ N 2 + o ( ε 2 μ ¯ μ ) , Ω | u μ , ε ξ | 2 d x = S μ N 2 + o ( ε 2 N N 2 μ ¯ μ ) ,

and for any a R N { 0 } ,

Ω | u μ , ε 0 | 2 | x + ξ | 2 d x = { ε 2 | ξ | 2 R N U μ 2 ( x ) d x + o ( ε 2 ) if  μ < μ ¯ 1 , C μ 2 ω N | ξ | 2 ε 2 | log ε | + o ( ε 2 ) if  μ = μ ¯ 1 , Ω | u μ , ε ξ | 2 d x = { o 1 ( ε 2 ) if  0 μ < μ ¯ 1 , o 1 ( ε 2 | log ε | ) if  μ = μ ¯ 1 ,

where C μ = ( 4 N ( μ ¯ μ ) N 2 ) N 2 4 , ω N is the volume of the unit ball in R N .

3 Asymptotic behavior of solutions

Proof of Theorem 1.1 Suppose ( u 0 , ν 0 ) H × H is a nontrivial solution to problem (1.1). For all 0 μ i μ ¯ define

u ( x ) = | x ξ i | γ u 0 ( x ) and ν ( x ) = | x ξ i | γ ν 0 ( x ) , where  γ = ( μ ¯ μ ¯ μ i ) .

It is not difficult to verify that u , ν H 0 1 ( Ω , | x ξ i | 2 γ ) and satisfy

{ div ( | x ξ i | 2 γ u ) = σ α 2 | x ξ i | 2 γ u | u | α 2 | ν | β + η | x ξ i | 2 γ u | u | 2 2 + a 1 | x ξ i | 2 γ u + a 2 | x ξ i | 2 γ ν + j = 1 j i k μ j | x ξ j | 2 | x ξ i | 2 γ u , div ( | x ξ i | 2 γ ν ) = σ β 2 | x ξ i | 2 γ ν | ν | β 2 | u | α + λ | x ξ i | 2 γ ν | ν | 2 2 + a 2 | x ξ i | 2 γ u + a 3 | x ξ i | 2 γ ν + j = 1 j i k μ j | x ξ j | 2 | x ξ i | 2 γ ν . (3.1)

Let R > 0 small enough such that B R ( ξ i ) Ω and ξ i B R ( ξ i ) for j i . Also, let φ i C 0 ( B R ( ξ i ) ) be a cut-off function. Set

u n : = min { | u | , n } ; ν n : = min { | ν | , n } ; ϕ 1 i : = φ i 2 u u n 2 ( s 1 ) , ϕ 2 i : = φ i 2 ν ν n 2 ( s 1 ) ,

where s , n > 1 . Multiplying the first equation of (3.1) by ϕ 1 i and the second one by ϕ 2 i respectively and integrating, we have

Ω | x ξ i | 2 γ u ϕ 1 i = σ α 2 Ω | x ξ i | 2 γ u | u | α 2 | ν | β ϕ 1 i + η Ω | x ξ i | 2 γ × u | u | 2 2 ϕ 1 i + a 1 Ω | x ξ i | 2 γ u ϕ 1 i + a 2 Ω | x ξ i | 2 γ × ν ϕ 1 i + j = 1 j i k μ j | x ξ j | 2 | x ξ i | 2 γ u ϕ 1 i .

Note that ϕ 1 i = 2 φ i u u n 2 ( s 1 ) φ i + φ i 2 u n 2 ( s 1 ) u + 2 ( s 1 ) φ i 2 u n 2 ( s 1 ) u n .

Then

Ω | x ξ i | 2 γ u ϕ 1 i = 2 Ω | x ξ i | 2 γ φ i u u n 2 ( s 1 ) φ i u + Ω | x ξ i | 2 γ φ i 2 × u n 2 ( s 1 ) | u | 2 + 2 ( s 1 ) Ω | x ξ i | 2 γ φ i 2 u n 2 ( s 1 ) | u n | 2 .

By the Cauchy inequality and the Young inequality, we get

(3.2)

The same result holds for Ω | x ξ i | 2 γ ϕ 2 i ν .

By letting ψ 1 ( x ) = φ i u u n ( s 1 ) , ψ 2 ( x ) = φ i ν ν n ( s 1 ) , we have

(3.3)

Using Caffarelli-Kohn-Nirenberg inequality [10], we infer that

(3.4)

Define

ω ( x ) : = max { u ( x ) , ν ( x ) } , ω n ( x ) : = min { ω ( x ) , n } .

Then ω n ( x ) : = max { u n ( x ) , ν n ( x ) } . Now, from the Hölder inequality, we deduce that

(3.5)

(3.6)

In the sequel, we have

(3.7)

By the choice of φ i , we obtain

(3.8)

So, from (3.4) to (3.8) it follows that

(3.9)

Take s = 2 2 and φ i ( x ) to be a constant near the zero. Letting n , we infer that ω L 2 2 2 ( B R ( ξ i ) , | x ξ i | 2 γ ) and so

u , ν L 2 2 2 ( B R ( ξ i ) , | x ξ i | 2 γ ) . (3.10)

Suppose r > 0 is sufficiently small such that r + l < R and φ i is a cut-off function with the properties | φ i | < 1 l and φ i ( x ) = 1 in B r ( ξ i ) .

Set t : = 2 2 2 ( 2 2 ) , δ : = 2 γ t 2 γ ( t 1 ) .

Then we have the following results:

(3.11)

where we used the Hölder inequality. From (3.9) in combination with (3.11), it follows that

( B r + l ( ξ i ) | x ξ i | 2 γ | ω | 2 s ) 1 2 s C 1 2 s l 1 2 s ( B r + l ( ξ i ) | x ξ i | 2 γ | ω | p ¯ 0 s ) 1 p ¯ 0 s , (3.12)

where p ¯ 0 = 2 t t 1 < 2 .

Denote s = χ j , χ = p p ¯ 0 and l = ρ j , j 1 , where χ 1 , 2 γ < N and p ¯ 0 χ j = p χ j 1 . Using (3.12) recursively, we get

( B r ( ξ i ) | ω | 2 χ j ) 1 2 χ j r γ χ j ( B r ( ξ i ) | ω | 2 χ j | x ξ i | 2 γ ) 1 2 χ j r γ χ j C i = 1 j 1 2 χ j ρ i = 1 j i χ i ( B r + ρ ( ξ i ) | x ξ i | 2 γ | ω | 2 ) 1 2 ,

we have χ j as j . Note that the infinite sums on the right-hand side converge, then we obtain that ω L ( B r ( ξ i ) ) , particularly, we have u , ν L ( Ω ) . Thus,

u 0 ( x ) = | x ξ i | γ u ( x ) M 1 | x ξ i | γ for  x B r ( ξ i ) { ξ i } ,

where M 1 = max { u L ( B r ( ξ i ) ) , 1 i k } .

ν 0 ( x ) = | x ξ i | γ ν ( x ) M 2 | x ξ i | γ for  x B r ( ξ i ) { ξ i } ,

where M 1 = max { ν L ( B r ( ξ i ) ) , 1 i k } . The proof is complete. □

Proof of Theorem 1.2 Suppose ( u 0 , ν 0 ) H × H is a positive solution to problem (1.1). For all 0 μ i μ ¯ , set

u ( x ) = | x ξ i | γ u 0 ( x ) and ν ( x ) = | x ξ i | γ ν 0 ( x ) .

Then

{ div ( | x ξ i | 2 γ u ) = σ α 2 | x ξ i | 2 γ u α 1 ν β + η | x ξ i | 2 γ u 2 1 + a 1 | x ξ i | 2 γ u + a 2 | x ξ i | 2 γ ν + j = 1 j i k μ j | x ξ j | 2 | x ξ i | 2 γ u , div ( | x ξ i | 2 γ ν ) = σ β 2 | x ξ i | 2 γ ν β 1 u α + λ | x ξ i | 2 γ ν 2 1 + a 2 | x ξ i | 2 γ u + a 3 | x ξ i | 2 γ ν + j = 1 j i k μ j | x ξ j | 2 | x ξ i | 2 γ ν . (3.13)

Choose 0 < ρ 0 < ρ and define n ( t ) = min | x ξ i | = t u ( x ) for ρ 0 t ρ . Let

n ( ρ 0 ) = A ρ 0 2 μ ¯ μ i + B , n ( ρ ) = A ρ 2 μ ¯ μ i + B , where A = n ( ρ ) n ( ρ 0 ) ρ 2 μ ¯ μ i ρ 0 2 μ ¯ μ i , B = n ( ρ 0 ) ρ 2 μ ¯ μ i n ( ρ ) ρ 0 2 μ ¯ μ i ρ 2 μ ¯ μ i ρ 0 2 μ ¯ μ i .

It is easy to verify that

div ( | x ξ i | 2 ( μ ¯ μ ¯ μ i ) ( A | x ξ i | 2 μ ¯ μ i + B ) ) = 0 x Ω { ξ i } . (3.14)

Combining (3.13) with (3.14), we get

div ( | x ξ i | 2 ( μ ¯ μ ¯ μ i ) ( u A | x ξ i | 2 μ ¯ μ i + B ) ) 0 , x B ρ ( ξ i ) B ρ 0 ( ξ i ) , u ( x ) ( A | x ξ i | 2 μ ¯ μ i + B ) 0 , x ( B ρ ( ξ i ) B ρ 0 ( ξ i ) ) .

Therefore, by the maximum principle in B ρ ( ξ i ) B ρ 0 ( ξ i ) , we obtain

u ( x ) ( A | x ξ i | 2 μ ¯ μ i + B ) 0 , x B ρ ( ξ i ) B ρ 0 ( ξ i ) .

Thus, for all x B ρ ( ξ i ) B ρ 0 ( ξ i ) ,

u ( x ) A | x ξ i | 2 μ ¯ μ i + B = | x ξ i | 2 μ ¯ μ i ρ 2 μ ¯ μ i ρ 0 2 μ ¯ μ i ρ 2 μ ¯ μ i n ( ρ 0 ) + ρ 0 2 μ ¯ μ i | x ξ i | 2 μ ¯ μ i ρ 0 2 μ ¯ μ i ρ 2 μ ¯ μ i n ( ρ ) | x ξ i | 2 μ ¯ μ i ρ 0 2 μ ¯ μ i | x ξ i | 2 μ ¯ μ i ρ 0 2 μ ¯ μ i ρ 2 μ ¯ μ i | x ξ i | 2 μ ¯ μ i n ( ρ ) .

Taking ρ 0 0 , we conclude u ( x ) n ( ρ ) = min | x ξ i | = ρ u ( x ) > 0 for all x B ρ ( ξ i ) { ξ i } .

Similar result also holds for ν ( x ) . Therefore, we have

u 0 ( x ) = | x ξ i | γ u ( x ) | x ξ i | γ min | x ξ i | = ρ u ( x ) = | x ξ i | γ C i | x ξ i | γ min i = 1 , 2 , , k C i = | x ξ i | γ N 1 .

For any x B ρ ( ξ i ) { ξ i } ,

ν 0 ( x ) = | x ξ i | γ ν ( x ) | x ξ i | γ min | x ξ i | = ρ ν ( x ) = | x ξ i | γ C i ´ | x ξ i | γ min i = 1 , 2 , , k C i ´ = | x ξ i | γ N 2 .

For any x B ρ ( ξ i ) { ξ i } . This proves the theorem. □

4 Local ( P S ) c -condition and the existence of positive solutions

We first establish a compactness result.

Lemma 4.1Suppose that ( H 1 ) holds. ThenJsatisfies the ( P S ) c -condition for all

c < c : = 1 N min { S η , λ , σ N 2 ( μ 1 ) , , S η , λ , σ N 2 ( μ k ) , ( S 0 ) N 2 } = 1 N S η , λ , σ N 2 ( μ k ) .

Proof Suppose that { ( u n , ν n ) } H × H satisfies J ( u n , ν n ) c < c and J ( u n , ν n ) 0 . The standard argument shows that { ( u n , ν n ) } is bounded in H × H .

For some ( u , ν ) H × H , we have

( u n , ν n ) ( u , ν ) weakly in  H × H , ( u n , ν n ) ( u , ν ) weakly in  L 2 ( Ω , | x ξ i | 2 ) × L 2 ( Ω , | x ξ i | 2 ) , ( u n , ν n ) ( u , ν ) weakly in  L 2 ( Ω ) × L 2 ( Ω ) , ( u n , ν n ) ( u , ν ) strongly in  L q 1 ( Ω ) × L q 2 ( Ω ) , q 1 , q 2 [ 1 , 2 ) , ( u n , ν n ) ( u , ν ) a.e. in  Ω .

Therefore, ( u , ν ) is a solution to (1.1). Then by the concentration-compactness principle [11-13] and up to a subsequence, there exist an at most countable set J , a set of different points { x j } j J Ω ξ i i = 1 k , nonnegative real numbers τ ˜ x j , ν ˜ x j , j J , and τ ˜ ξ i , ν ˜ ξ i , γ ˜ ξ i ( 1 i k ) such that the following convergence holds in the sense of measures:

| u n | 2 + | ν n | 2 d τ ˜ | u | 2 + | ν | 2 + j J τ ˜ x j δ x j + i = 1 k τ ˜ ξ i δ ξ i , u n 2 + ν n 2 | x ξ i | 2 d γ ˜ = u 2 + ν 2 | x ξ i | 2 + γ ˜ ξ i δ ξ i , η | u n | 2 + λ | ν n | 2 + σ | u n | α | ν n | β d ν ˜ = η | u | 2 + λ | ν | 2 + σ | u | α | ν | β + j J ν ˜ x j δ x j + i = 1 k ν ˜ ξ i δ ξ i .

By the Sobolev inequalities [10], we have

S μ i ν ˜ ξ i 2 2 τ ˜ ξ i μ i γ ˜ ξ i , 1 i k . (4.1)

We claim that J is finite, and for any j J , ν ˜ x j = 0 or ν ˜ x j S 0 N 2 .

In fact, let ε > 0 be small enough for any 1 i k , ξ i B ε ( x j ) and B ε ( x i ) B ε ( x j ) = for i j , i , j J . Let ϕ ε j be a smooth cut-off function centered at x j such that 0 ϕ ε j 1 , ϕ ε j = 1 for | x x j | ε 2 , ϕ ε j = 0 for | x x j | ε and | ϕ ε j | 4 ε . Then

Then we have

0 = lim ε 0 lim n J λ ( u n , ν n ) , ( u n ϕ ε j , ν n ϕ ε j ) τ ˜ x j ν ˜ x j .

By the Sobolev inequality, S 0 ν ˜ x j 2 2 τ ˜ x j ; and then we deduce that ν ˜ x j = 0 or ν ˜ x j S 0 N 2 , which implies that J is finite.

Now, we consider the possibility of concentration at points ξ i ( 1 i k ), for ε > 0 small enough that x j B ε ( ξ i ) for all j J and B ε ( ξ i ) B ε ( ξ j ) = for i j and 1 i , j k . Let φ ε i be a smooth cut-off function centered at ξ i such that 0 φ ε i 1 , φ ε i = 1 for | x ξ i | ε and | φ ε i | 4 ε . Then

Thus, we have

0 = lim ε 0 lim n J λ ( u n , ν n ) , ( u n φ ε i , ν n φ ε i ) τ ˜ ξ i μ i γ ˜ ξ i ν ˜ ξ i . (4.2)

From (4.1) and (4.2) we derive that S μ i ν ˜ ξ i 2 2 ν ˜ ξ i , 1 i k , and then either ν ˜ ξ i = 0 or ν ˜ ξ i S μ i N 2 . On the other hand, from the above arguments, we conclude that

c = lim n ( J ( u n , ν n ) 1 2 J ( u n , ν n ) , ( u n , ν n ) ) = 1 N lim n Ω ( η | u n | 2 + λ | ν n | 2 + σ | u n | α | ν n | β ) d x = 1 N ( Ω ( η | u | 2 + λ | ν | 2 + σ | u | α | ν | β ) d x + j J ν ˜ x j + i = 1 k ν ˜ ξ i ) = 1 N ( j J ν ˜ x j + i = 1 k ν ˜ ξ i ) + J ( u , ν ) .

If ν ˜ ξ i = ν ˜ x j = 0 for all i { 1 , , k } and j J , then c = 0 , which contradicts the assumption that c > 0 . On the other hand, if there exists an i { 1 , , k } such that ν ˜ ξ i 0 or there exists a j J with ν ˜ x j 0 , then we infer that

c 1 N min { ( S η , λ , σ ( 0 ) ) N / 2 , ( S η , λ , σ ( μ 1 ) ) N / 2 , , ( S η , λ , σ ( μ k ) ) N / 2 } = 1 N ( S η , λ , σ ( μ k ) ) N / 2 ,

which contradicts our assumptions. Hence, ( u n , ν n ) ( u , ν ) , as n in H × H . □

First, under the assumptions ( H 1 ), ( H 2 ), we have the following notations:

f η , λ , σ ( τ ) = ( 1 + τ 2 ) S μ k ( η + σ τ β + λ τ 2 ) 2 2 , τ > 0 ; f η , λ , σ ( τ min ) : = min τ > 0 f η , λ , σ ( τ ) > 0 , σ > 0 ,

where τ min > 0 is a minimal point of f η , λ , σ ( τ ) , and therefore a root of the equation

α σ τ β σ β τ β 2 2 λ τ 2 2 + 2 η = 0 , τ > 0 .

Lemma 4.2Suppose that ( H 1 ) holds. Then we have

(i) S η , λ , σ ( μ ) = f η , λ , σ ( τ min )

(ii) S η , λ , σ ( μ ) has the minimizers ( V μ , ε ξ ( x ) , τ min V μ , ε ξ ( x ) ) , ε > 0 , where V μ , ε ξ ( x ) are the extremal functions of S η , λ , σ ( μ ) defined as in (2.2).

Proof The argument is similar to that of [6]. □

Lemma 4.3Under the assumptions of ( H 1 ), we have

sup J ( t u ε , μ k , t ( τ min u ε , μ k ) ) < c = 1 N ( S η , λ , σ ( μ k ) ) N / 2 .

Proof Suppose ( H 1 ) holds. Define the function

g ( t ) : = J ( t u ε , μ k , t ( τ min u ε , μ k ) ) , t 0 .

Note that lim t + g ( t ) = and g ( t ) > 0 as t is close to 0. Thus, sup t 0 g ( t ) is attained at some finite t ε > 0 with g ( t ε ) = 0 . Furthermore, c < t ε < c , where c and c are the positive constants independent of ε. By using (1.2), we have

g ( t ) t 2 2 ( 1 + τ min 2 ) ( Ω ( | u ε , μ k | 2 μ k u ε , μ k 2 | x ξ i | 2 λ 1 u ε , μ k 2 ) d x ) t 2 2 ( σ τ min β + η + λ τ min 2 ) Ω | u ε , μ k | 2 d x .

Note that

max ( t 2 2 B 1 t 2 2 B 2 ) = 1 N ( B 1 B 2 2 / 2 ) N / 2 , B 1 > 0 , B 2 > 0 , (4.3)

and 0 μ μ ¯ 1 and so 2 < 2 μ ¯ μ .

From (4.3), Lemma 4.2 and Lemma 4.3, it follows that

g ( t ε ) 1 N ( ( 1 + τ min 2 ) Ω ( | u ε , μ k | 2 μ k u ε , μ k 2 | x ξ k | 2 λ 1 u ε , μ k 2 ) d x ( ( σ τ min β + η + λ τ min 2 ) Ω | u ε , μ k | 2 d x ) 2 / 2 ) N / 2 1 N ( f ( τ min ) S ( μ k ) × ( S ( μ k ) ) N / 2 + O ( ε 2 μ ¯ μ ) C ε 2 ( S ( μ k ) ) ( N 2 ) / 2 + O ( ε 2 μ ¯ μ ) ) 1 N ( f ( τ min ) ) N / 2 + O ( ε 2 μ ¯ μ ) C ε 2 < 1 N ( S η , λ , σ ( μ k ) ) N / 2 = c ,

so g ( t ε ) < c . Hence, g ( t ε ) < c , t 0 and

sup t 0 g ( t ) = sup t 0 J ( t u ε , μ k , t ( τ min u ε , μ k ) ) < c , if  μ < μ ¯ 1 . (4.4)

 □

Proof of Theorem 1.3 Set c : = inf h Γ max t [ 0 , 1 ] J ( h ( t ) ) , where

Γ = { h C ( [ 0 , 1 ] , H × H ) | h ( 0 ) = ( 0 , 0 ) , J ( h ( 1 ) ) < 0 } .

Suppose that ( H 1 ) holds. For all ( u , ν ) H × H { ( 0 , 0 ) } , from the Young and Hardy-Sobolev inequalities, it follows that

J ( u , ν ) C ( u 2 + ν 2 ) C ( u 2 + ν 2 ) C ( u , ν ) 2 C ( u , ν ) 2 ,

and there exists a constant ρ > 0 small such that

b : = inf ( u , ν ) = ρ J ( u , ν ) > 0 = J ( 0 , 0 ) .

Since J ( t u , t ν ) as t , there exists t 0 > 0 such that ( t 0 u , t 0 ν ) > ρ and J ( t 0 u , t 0 ν ) < 0 . By the mountain-pass theorem [14], there exists a sequence { ( u n , ν n ) } H × H such that J ( u n , ν n ) c and J ( u n , ν n ) 0 , as n .

From Lemma 4.2 it follows that

0 < c sup t [ 0 , 1 ] J ( t t 0 u ε , μ k , t t 0 τ min u ε , μ k ) sup t 0 J ( t u ε , μ k , t τ min u ε , μ k ) < c .

By Lemma 4.1 there exists a subsequence of { ( u n , ν n ) } , still denoted by { ( u n , ν n ) } , such that ( u n , ν n ) ( u , ν ) strongly in H × H . Thus, we get a critical point ( u , ν ) of J satisfying (1.1), and c is a critical value. Set u + = max { u , 0 } .

Replacing respectively u, ν with u + and ν + in terms of the right-hand side of (1.1) and repeating the above process, we can get a nonnegative nontrivial solution ( u , ν ) of (1.1). If u 0 , we get ν 0 by (1.1) and the assumption a 2 > 0 . Similarly, if ν 0 , we also have u 0 . There, u , ν 0 . From the maximum principle, it follows that u , ν > 0 in Ω. □

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Each of the authors, SK, MF and OKK contributed to each part of this work equally and read and approved the final version of the manuscript.

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