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 ,

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 123 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, 45678.

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 .

Our main results are as follows.

Theorem 1.1 Suppose (H1) 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 1.2 Suppose (H1) 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 1.3 Suppose (H1), (H2) hold. Then the problem (1.1) has a positive solution.

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 ∗ ,

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 ∗ .

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 RN.

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 .

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

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

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

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

In the sequel, we have

By the choice of φ i , we obtain

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

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 ) .

Suppose r > 0 is sufficiently small such that r + l < R and φi 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:

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 ,

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 (u0,ν0)∈H×H is a positive solution to problem (1.1). For all 0≤μi≤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 .

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 } .

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∈Bρ(ξi)∖{ξi},

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∈Bρ(ξi)∖{ξi}. This proves the theorem. □

We first establish a compactness result.

Lemma 4.1 Suppose that (H1) holds. Then J satisfies the (PS)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 111213 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 .

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˜xj=0 or ν˜xj≥S0N2, 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 ξi 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 .

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 ,