Asma Benhamida
email: asma.benhamida2019@gmail.com

Rejeb Hadiji
email: rejeb.hadiji@u-pec.fr

A system with weights and with critical Sobolev exponent

Keywords: 2010 AMS subject classifications: 35A01, 35A15, 35J57, 35J62 Critical Sobolev exponent, Minimization problem, Non-linear effects variational problem

In this paper, we investigate the minimization problem :

where q = 2N N -2 , N ≥ 4, a and b are two continuous positive weight functions. We show the existence of solutions of the previous minimizing problem under some conditions on a, b, the dimension of the space and the parameter λ.

Introduction

Let Ω ⊂ R N be a given smooth bounded domain, with N ≥ 4. Throughout this paper, we are concerned with the following nonlinear minimization problem :

Q λ = inf (u, v) ∈ (H 1 0 (Ω)) 2 \ {(0, 0)} E λ (u, v), (1)
1 with

E λ (u, v) = 1 2 u 2 q Ω a(x)|∇u(x)| 2 dx+ 1 2 v(x) 2 q Ω b(x)|∇v(x)| 2 dx- λ u q v q Ω u(x)v(x)dx.
where a and b are positive continuous functions on Ω, λ is a real constant and q = 2N N -2 is the critical exponent for the Sobolev embedding

H 1 0 (Ω) ֒→ L q (Ω). (2)
Note that positive minimizers u and v for (1) are non trivial solutions of the Euler-Lagrange equation associated to [START_REF] Alves | On systems of elliptic equations involving subcritical or critical Sobolev exponents[END_REF] namely :

         -div(a(x)∇u) -λv = Λ 1 u 2 * -1 in Ω -div(b(x)∇v) -λu = Λ 2 v 2 * -1 in Ω u ≥ 0, v ≥ 0 in Ω u = v = 0 on ∂Ω, (3)
where Λ 1 , Λ 2 ∈ R are the Lagrange multipliers associated to [START_REF] Alves | On systems of elliptic equations involving subcritical or critical Sobolev exponents[END_REF] .

In the well-known article [START_REF] Brezis | L Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents[END_REF], Brezis and Nirenberg treated the problem (4) in the special case where the weights a and b are positive constant function. They proved that the problem has at least one positive solution for 0 < λ < λ 1 when N ≥ 4 and for λ * < λ < λ 1 when N = 3, where λ 1 denotes the first eigenvalue of -∆ with homogeneous Dirichlet conditions and λ * is a positive constant. In the presence of a non-constant positive and bounded weight h, the scalar problem has been addressed in [START_REF] Hadiji | H Problem with critical Sobolev exponent and with weight[END_REF]. See also [START_REF] Hadiji | H Localization of solutions for nonlinear elliptic problems with weight[END_REF] where the authors consider the existence of minimizers solutions for the equation -div(h(x)∇u) = λu + |u| q-2 u in Ω, u > 0 in Ω, u = 0 ∂Ω.

They proved that the existence of a solution depends on the first eigenvalue λ h 1 of -div(h(x)∇•) in H 1 0 (Ω), on the behavior of the function h in the vicinity of its minima, and on the geometry of the domain Ω In [START_REF] Furtado | Positive and nodal solutions for an elliptic equation with critical growth[END_REF], Furtado and Souza generalized the problem (4), considering a non-homogeneous term |u| q-2 u. We refer also to [START_REF] Bae | H A nonlinear existence result for a quasi-linear elliptic PDE[END_REF][START_REF] Hadiji | R A nonlinear problem with a weight and a nonvanishing boundary datum[END_REF][START_REF] Hadiji | F Existence of solutions of a non-linear eigenvalue problem with a variable weight[END_REF] for more general weights which depend on u and x. There are other results in the scalar case for this kind of problem, see the references [START_REF] Hadiji | R Solutions positives de l'équation -∆u = u p + µu q dans un domaine à trou[END_REF][START_REF] Hadiji | R A nonlinear problem with a weight and a nonvanishing boundary datum[END_REF][START_REF] Hadiji | F Existence of solutions of a non-linear eigenvalue problem with a variable weight[END_REF]. In [START_REF] Alves | On systems of elliptic equations involving subcritical or critical Sobolev exponents[END_REF], the authors consider a critical and subcritical system with critical non-linearity term u α-1 v β , where α > 1 and β > 1. The main difficulty faced when dealing with this problem is the lack of compactness of the embedding [START_REF] Aubin | T Equations différentielles non linéaires et problème de Yamabe concernant la courbure scalaire[END_REF]. We overcome this problem using the presence of linear perturbation terms λu and λv. In [START_REF] Clapp | A Yamabe systems and optimal partitions on manifolds with symmetries[END_REF], the authors considered an arbitrary system with l components, l ≥ 2, for the Yamabe equation on a closed Riemannian manifold; see also [START_REF] Conti | G Nehari's problem and competing species systems[END_REF].

Note that the shape of the domain can have a strong influence on the type of results one can expect. See for example the seminal work of J.M. Coron [START_REF] Coron | M Topologie et cas limite des injections de Sobolev[END_REF].

Geometric and physical motivations, in particular in relation to the Yamabe problem, can be found for example in [START_REF] Aubin | T Equations différentielles non linéaires et problème de Yamabe concernant la courbure scalaire[END_REF], [START_REF] Cordero-Erausquin | C A mass-transportation approach to sharp Sobolev and Gagliardo-Nirenberg inequalities[END_REF] [18], [START_REF] Kholodenko | Towards physically motivated proofs of the Poincaré and geometrization conjectures[END_REF], [START_REF] Lions | L The concentration compactness principle in the calculus of variations. The limit case[END_REF], [START_REF] Lee | H The Yamabe problem[END_REF].

In [START_REF] Kholodenko | Towards physically motivated proofs of the Poincaré and geometrization conjectures[END_REF], the authors connect the Ginzburg-Landau functional used in the physics literature with the Yamabe functional known in mathematics. It is shown that, if properly interpreted, both functionals upon minimization produce the same Ginzburg-Landau-type equations used on critical phenomena.

In this paper, we consider the case where a and b are non constant distinct weights. We prove the existence of positive solutions (u, v) which depend, among others, on the behaviors of the weights a(•) and b(•) near their minima and the dimension of the space.

In order to state the problem and to announce our main results, we introduce some preliminaries. Let us assume the existence of x 0 , in Ω such that, in a neighborhood of x 0 , a and b behave like

a(x) = a(x 0 ) + A k |x -x 0 | k + |x -x 0 | k θ a (x), as x → x 0 , (5) b
(x) = b(x 0 ) + B l |x -x 0 | l + |x -x 0 | l θ b (x), as x → x 0 , (6)
where k > 0, l > 0 and A k , B l are positive constants, θ a (x) and θ b (x) tend to 0 when x tends to x 0 . The parameters k and l will play a critical role in the study of our problem. Indeed, if N ≥ 4 the case (k, l) with k > 2 and l > 2 is treated through a classical procedure.

If 0 < k ≤ 2, 0 < l ≤ 2 the problem is more delicate, we have to assume that the functions a and b satisfy the following additional conditions :

kA k ≤ ã(x) |x -x 0 | k a.e. x ∈ Ω, (7)
and

lB l ≤ b(x) |x -x 0 | l a.e. x ∈ Ω, (8)
where ã(x) := ∇a(x) • (xx 0 ) and b(x) := ∇b(x) • (xx 0 ).

In order to highlight the difficulty, we consider the blow-up of u, v ∈ H 1 0 (Ω) around x = x 0 , see [START_REF] Hadiji | R A nonlinear problem with a weight and a nonvanishing boundary datum[END_REF][START_REF] Hadiji | F Existence of solutions of a non-linear eigenvalue problem with a variable weight[END_REF]. Depending on the parameters k, l and λ different situations occur in the blow-up scale around the point where the weights reach their minimum. More precisely, we consider the function w ε and z ε defined by :

∀ ε > 0, u(x) = ε -(N-2) 2 w ε x-x 0 ε , v(x) = ε -(N-2) 2 z ε x-x 0 ε . (9)
One has

w ε , z ε ∈ H 1 0 (Ω ε ) with Ω ε = {ε -1 y, y ∈ Ω} and w ε L q (Ωε) = z ε L q (Ωε) = u L q (Ω) = v L q (Ω) , thus E λ (u, v) = 1 2 Ωε a(εy + x 0 )|∇w ε (y)| 2 dy+ 1 2 Ωε b(εy + x 0 )|∇z ε (y)| 2 dy-λε 2 Ωε w ε (y)z ε (y)dy.
Consequently, the blow-up around x = x 0 gives

E λ (u, v) ∼ ε→0 1 2 Ωε a(x 0 )|∇w ε (y)| 2 dy+ 1 2 Ωε b(x 0 )|∇z ε (y)| 2 dy-λε 2 Ωε w ε (y)z ε (y)dy+ A k 2 Ωε |εy + x 0 | k |∇w ε (y)| 2 dy + B l 2 Ωε |εy + x 0 | l |∇z ε (y)| 2 dy -λε 2 Ωε w ε (y)z ε (y)dy.
Then, as we can see the three last terms have different weights and the exponents k = 2 and l = 2 are critical for our problem.

Let us finally point out that the lower dimension N = 3 and the exponents k < 2, l < 2 could also be interesting for this problem, but is not yet fully understood. For dimension 3 in the scalar case we refer to [START_REF] Brezis | L Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents[END_REF], [START_REF] Crouau | R Critical Sobolev exponent and the dimension three[END_REF].

The rest of the paper is organized in the following way : in Section 2 we state our main result. In Section 3 we give a sufficient condition for the existence of minimizers. We give precise estimates of the energy in Section 4. In Section 5 we discuss the sign of minimizers. Section 6 contains some non-existence results.

Main Results

Let us suppose that there exists x 0 ∈ Ω such that

a(x 0 ) = min a(x), x ∈ Ω . b(x 0 ) = min b(x), x ∈ Ω .
In this section, we will prove the existence of a solution to the problem [START_REF] Alves | On systems of elliptic equations involving subcritical or critical Sobolev exponents[END_REF]. For this we define: S = inf

u∈H 1 0 (Ω), u L q =1 ∇u 2 2 , (10)
which corresponds to the best constant for the Sobolev embedding H 1 0 (Ω) into L q (Ω). In this paper, we assume that a(x 0 ) = b(x 0 ), and we denote by γ 0 this value. We are now ready to state the main result of this paper Theorem 1. Assume that a, b ∈ H 1 (Ω) ∩ C( Ω) satisfy (5) and (6), respectively. Let λ a 1 , respectively λ b 1 , be the first eigenvalue of -div(a(x)∇•), respectively -div(b(x)∇•) on Ω with zero Dirichlet boundary condition. Define λ1 = min{λ a 1 , λ b 1 }. We have :

(1) If N ≥ 4, k > 2 and l > 2, then Q λ is achieved for every λ ∈ (0, λ a 1 ) ∩ 0, λ b 1 . (2) If N ≥ 5, k = 2 and l = 2, then there exists a constant γ(N) = (N -2)N (N +2) 8(N -1) (A 2 + B 2 ) such that Q λ is achieved for every λ ∈ γ(N), λ1 . (3) If N ≥ 5, k = 2 and l > 2, respectively k > 2 and l = 2, then there exists a constant γ(N) = N (N -2)(N +2) 8(N -1) such that Q λ is achieved for every λ ∈ γ(N)A 2 , λ1 , respectively λ ∈ γ(N)B 2 , λ1 . (4) If N = 4, k = 2
and l > 2, respectively k > 2 and l = 2, then there exists a minimizing solution of Q λ for every λ ∈ Ã2 , λ1 respectively for every λ ∈ B2 , λ1

where Ã2 and B2 are some constants.

We now proceed with the proof of Theorem 1 as follows : First we show that 0 ≤ Q λ < γ 0 S and then we prove that this implies that the infimum Q λ is achieved.

Sufficient Conditions for the Existence of Minimizers

We first prove the existence of Q λ , which is guaranteed by the following result :

Proposition 3.1. Let ϕ a 1 , respectively ϕ b 1 , be the first eigenfunction of -div(a(x)∇•), respectively -div(b(x)∇•), associated to the first eigenvalue λ a 1 , respectively λ b 1 . We have (i) Assume that 0 < λ < λ1 , then Q λ ≥ 0. (ii) For λ ≥ ϕ a 1 L q ϕ b 1 L q Ω ϕ a 1 ϕ b 1 dx |Ω| 1-2 q max(λ a 1 , λ b 1 ), one has Q λ ≤ 0, Proof (i). Let 0 < λ < λ1 (Ω)
, and let u and v be such that u L q = v L q = 1. By the definitions of λ a 1 , λ b 1 and λ1 (Ω) one has

E λ (u, v) ≥ λ1 2 u 2 L 2 + λ1 2 v 2 L 2 -λ Ω uv dx.
By applying the Cauchy-Schwarz inequality , we find

E λ (u, v) ≥ λ1 2 u 2 L 2 + v 2 L 2 -2 u L 2 v L 2 .
Thus

E λ (u, v) ≥ λ1 2 ( u L 2 -v L 2 ) 2 ≥ 0. Consequently, Q λ ≥ 0. (ii) We have Q λ ≤ E λ ( ϕ a 1 ϕ a 1 L q , ϕ b 1 ϕ b 1 L q ),
which implies that,

Q λ ≤ 1 2 ϕ a 1 2 L q Ω a(x)|∇ϕ a 1 | 2 dx+ 1 2 ϕ b 1 2 L q Ω b(x)|∇ϕ b 1 | 2 dx- λ ϕ a 1 L q ϕ b 1 L q Ω ϕ a 1 ϕ b 1 dx.
By the definitions of ϕ a 1 and ϕ b 1 , one has

Q λ ≤ 1 2 ϕ a 1 2 L q λ a 1 Ω |ϕ a 1 | 2 dx + 1 2 ϕ b 1 2 L q λ b 1 Ω |ϕ b 1 | 2 dx - λ ϕ a 1 L q ϕ b 1 L q Ω ϕ a 1 ϕ b 1 dx.
Using the embedding of L q into L 2 , there exists a positive constant

C 1 = |Ω| 1 2 -1 q such that Q λ ≤ 1 2 ϕ a 1 2 L q λ a 1 C 2 1 ϕ a 1 2 L q + 1 2 ϕ b 1 2 L q λ b 1 C 2 1 ϕ b 1 2 L q - λ ϕ a 1 L q ϕ b 1 L q Ω ϕ a 1 ϕ b 1 dx. Thus Q λ ≤ |Ω| 1-2 q max(λ a 1 , λ b 1 ) - λ ϕ a 1 L q ϕ b 1 L q Ω ϕ a 1 ϕ b 1 dx.
Therefore, Q λ ≤ 0, completing the proof of the proposition.

Lemma 3.1. Let 0 < λ < λ1 . If Q λ < γ 0 S, then the infimum in (1) is achieved. Proof Let {U n } ⊂ (H 1 0 (Ω))
2 be a minimizing sequence for (1) that is,

U n L q = 1 (which means u n q = 1, v n q = 1). ( 11
) 1 2 Ω a(x)|∇u n | 2 dx + 1 2 Ω b(x)|∇v n | 2 dx -λ Ω u n v n dx = Q λ + o(1) as n -→ ∞. (12) The sequence {U n } is bounded in (H 1 0 (Ω))
2 . Indeed, from (12), we have

1 2 Ω a(x)|∇u n | 2 dx + 1 2 Ω b(x)|∇v n | 2 dx = λ Ω u n v n dx + Q λ + o(1).
Using Hölder's inequality we have

Ω |u n ||v n |dx ≤ u n L q q-1 v n L q , then 1 2 Ω a(x)|∇u n | 2 dx + Ω b(x)|∇v n | 2 dx ≤ λ u n L q q-1 v n L q + Q λ + o(1)
.

Consequently, we have (u n , v n ) (H 1 0 (Ω)) 2 ≤ C.
Then there is U = (u, v) , up to a subsequence, still denoted by

U n = (u n , v n ) , such that (u n , v n ) ⇀ (u, v) weakly in H 1 0 (Ω) 2 , (u n , v n ) -→ (u, v) strongly in L 2 (Ω) 2 , (u n , v n ) -→ (u, v) a.e. on Ω. with u L q ≤ 1 and v L q ≤ 1. Set w n = u n -u and z n = v n -v, so that (w n , z n ) ⇀ (0, 0) weakly in H 1 0 (Ω) 2 , (w n , z n ) -→ (0, 0) strongly in L 2 (Ω) 2 ,
(w n , z n ) -→ (0, 0) a.e. on Ω.

By [START_REF] Cordero-Erausquin | C A mass-transportation approach to sharp Sobolev and Gagliardo-Nirenberg inequalities[END_REF] and using the definition of S and the fact that min Ω a(x) = γ 0 > 0 and min Ω b(x) = γ 0 > 0, we have

1 2 Ω a(x)|∇u n | 2 dx + Ω b(x)|∇v n | 2 dx ≥ 1 2 2γ 0 S.
By [START_REF] Furtado | Positive and nodal solutions for an elliptic equation with critical growth[END_REF], we have

1 2 Ω a(x)|∇u n | 2 dx + Ω b(x)|∇v n | 2 dx -λ Ω u n v n dx = Q λ + o(1). Which implies that 1 2 Ω a(x)|∇u n | 2 dx + Ω b(x)|∇v n | 2 dx = Q λ + λ Ω u n v n dx + o(1).
Which gives

λ Ω uvdx ≥ γ 0 S -Q λ > 0
and so u ≡ 0 and v ≡ 0. Again using [START_REF] Furtado | Positive and nodal solutions for an elliptic equation with critical growth[END_REF] we obtain

1 2 Ω a(x)|∇u| 2 dx + Ω a(x)|∇w n | 2 dx + Ω b(x)|∇v| 2 dx + Ω b(x)|∇z n | 2 dx -λ Ω u n v n dx = Q λ + o(1). (13)
On the other hand, as {w n } and {z n } are bounded in L q (Ω) and w n -→ 0 for a.e. x in Ω, z n -→ 0 for a.e. x in Ω, then we can use the Brezis-Lieb lemma, see [START_REF] Brezis | E A relation between pointwise convergence of functions and convergence of functionals[END_REF]

u + w n q L q = u q L q + w n q L q + o(1), v + z n q L q = v q L q + z n q L q + o(1). Using (11) we have 1 = u q L q + w n q L q + o(1), 1 = v q L q + z n q L q + o(1). And so, 1 ≤ u 2 L q + w n 2 L q + o(1), ( 14
) 1 ≤ v 2 L q + z n 2 L q + o(1). (15)
We sum the equations ( 14) and ( 15) we obtain

2 ≤ u 2 L q + v 2 L q + 1 γ 0 S Ω a(x)|∇w n | 2 dx + Ω b(x)|∇z n | 2 dx . ( 16
)
Since Q λ > 0, this implies

Q λ ≤ Q λ 2 u 2 L q + Q λ 2 v 2 L q + Q λ 2γ 0 S Ω a(x)|∇w n | 2 dx + Ω b(x)|∇z n | 2 dx . ( 17
)
Adding ( 13) and ( 17) we obtain

1 2 Ω a(x)|∇u| 2 dx + Ω a(x)|∇w n | 2 dx + Ω b(x)|∇v| 2 dx + Ω b(x)|∇z n | 2 dx -λ Ω uvdx ≤ Q λ 2 u 2 L q + Q λ 2 v 2 L q + Q λ 2γ 0 S Ω a(x)|∇w n | 2 dx + Ω b(x)|∇z n | 2 dx . Thus 1 2 Ω a(x)|∇u| 2 dx + Ω b(x)|∇v| 2 dx -λ Ω uvdx ≤ Q λ 2 u 2 L q + Q λ 2 v 2 L q + Q λ 2γ 0 S - 1 2 Ω a(x)|∇w n | 2 dx + Ω b(x)|∇z n | 2 dx . Hence 2E λ (u, v) ≤ Q λ u 2 L q + Q λ v 2 L q + Q λ γ 0 S -1 Ω a(x)|∇w n | 2 dx + Ω b(x)|∇z n | 2 dx + o(1),
and since Q λ < γ 0 S, we know that u = 0 and v = 0. we deduce

2E λ u u L q , v v L q ≤ 1 u 2 L q -1 Ω a(x)|∇u| 2 dx + 1 v 2 L q -1 Ω b(x)|∇v| 2 dx +Q λ u 2 L q + Q λ u 2 L q + λ Ω uvdx -λ Ω uv u L q v L q dx. Then 2E λ ( u u L q , v v L q ) ≤ u 2 L q -1 Q λ - Ω a(x) |∇u| 2 u 2 L q dx -2λ Ω uv u L q v L q dx + v 2 L q -1 Q λ - Ω b(x) |∇v| 2 v 2 L q dx -2λ Ω uv u L q v L q dx +λ Ω uvdx -λ Ω uv u L q v L q dx + 2λ u 2 q -1 Ω uv u L q v L q dx +2λ v 2 L q -1 Ω uv u L q v L q dx + 2Q λ .
On one hand, we have u 2 L q -1 ≤ 0 and v 2 L q -1 ≤ 0, we obtain

2E λ u u L q , v v L q ≤ λ ( u L q v L q -1) + 2 ( u 2 L q -1) + 2 ( v 2 L q -1) u L q v L q Ω uvdx + 2Q λ .
On the other hand, we have u

L q v L q -1 ≤ 0, then 2E λ u u L q , v v L q ≤ 2Q λ .
This means that (u, v) is a minimum of Q λ .

4 Precise estimates of the energy

Q λ < γ 0 S, for all λ > A 2 + B 2 .
(c) For N ≥ 5, and k = 2, l = 2 we have

Q λ < γ 0 S, for all λ > m N (A 2 + B 2 ) .
(d) For N ≥ 5, and k = 2, l > 2 (respectively k > 2, l = 2) we have

Q λ < γ 0 S, for all λ > m N A 2 (respectively λ > m N B 2 ).
(e) For N = 4, and k = 2, l > 2, respectively k > 2, l = 2 we have

Q λ < γ 0 S, for all λ > A 2 , respectively for all λ > B 2 .
where A 2 , B 2 are defined by [START_REF] Clapp | A Yamabe systems and optimal partitions on manifolds with symmetries[END_REF] and (8),

K 3 = R N 1 1 + |x| 2 N -2 dx and m N = N (N -2)(N +2) 8(N -1)
.

Proof We shall estimate the ratio E λ (u, v) defined in [START_REF] Alves | On systems of elliptic equations involving subcritical or critical Sobolev exponents[END_REF], with u = u x 0 ,ε = ζU ε (x -

x 0 ), where for x ∈ R N , U ε (x) = ε N-2 4 ε + |x| 2 N-2 2 and ζ ∈ C ∞ c (Ω) with ζ ≥ 0 and
ϕ ≡ 1 on a neighborhood of x 0 ; for more details see [5] [23]. We recall from [START_REF] Brezis | L Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents[END_REF] that

Ω |∇u x 0 ,ε (x)| 2 dx = K 1 + O(ε N- 2
2 ), ( 18)

Ω |u x 0 ,ε (x)| q dx 2 q = K 2 + O(ε N-2 2 ), (19)
Ω |u x 0 ,ε (x)| 2 dx = K 3 ε + O(ε N-2 2 ), if N 5, ω 4 2 ε| log ε| + o(ε| log ε|), if N = 4, (20)
where K 1 and K 2 are positive constants with K 1 K 2 = S, w 4 is a the area of S 3 and

K 3 = R N 1 1 + |x| 2 N -2 dx.
We know by [START_REF] Hadiji | H Problem with critical Sobolev exponent and with weight[END_REF], the following estimates :

ε N-2 2 2 Ω a(x)|∇u x 0 ,ε (x)| 2 dx ≤            a(x 0 )K 1 2 + O(ε N-2 2 ) if N ≥ 4 and N -2 < k, a(x 0 )K 1 2 + C k 2 ε k 2 + o(ε k 2 ) if N ≥ 4 and N -2 > k, a(x 0 )K 1 2 + (N -2) 2 ω N (A N-2 +M ) 2 ε N-2 4 | log ε| + o(ε N-2 2 | log ε|) if N > 4 and k = N -2, a(x 0 )K 1 2 + A 2 ω 4 ε| log ε| + o(ε| log ε|) if N = 4 and k = 2 (21) and ε N-2 2 Ω b(x)|∇u x 0 ,ε (x)| 2 dx ≤            b(x 0 )K 1 2 + O(ε N-2 2 ) if N ≥ 4 and N -2 < l, b(x 0 )K 1 2 + D l 2 ε l 2 + o(ε l 2 ) if N ≥ 4 and N -2 > l, b(x 0 )K 1 2 + (N -2) 2 ω N (B N-2 +M ) 4 ε N-2 2 | log ε| + o(ε N-2 2 | log ε|) if N > 4 and l = N -2, b(x 0 )K 1 2 + B 2 ω 4 ε| log ε| + o(ε| log ε|) if N = 4 and l = 2 (22)
Where

K 1 = (N -2) 2 R N |y| 2 1 + |y| 2 N dy, K 2 = R N dy 1 + |y| 2 N N-2 N , C k = (N -2) 2 A k R N |y| k+2 1 + |y| 2 N dy, D k = (N -2) 2 B k R N |y| l+2
1 + |y| 2 N dy, M and M ′ are positive constants. We claim that, as ε → 0, by adding ( 19), ( 20), ( 21) and ( 22) we obtain E(u

x 0 ,ε ) ≤                                                                                            a(x 0 )K 1 +o(ε) 2K 2 + b(x 0 )K 1 +o(ε) 2K 2 -λ K 3 K 2 ε if N ≥ 5 k > 2, l > 2, a(x 0 )K 1 +C 2 ε+o(ε) 2K 2 + b(x 0 )K 1 +D 2 ε+o(ε) 2K 2 -λ K 3 K 2 ε if N ≥ 5 k = 2, l = 2, a(x 0 )K 1 +o(ε) 2K 2 + b(x 0 )K 1 +o(ε) 2K 2 -λε ω 4 2 | log ε| K 2 + o(ε| log ε|) if N = 4 k > 2, l > 2, a(x 0 )K 1 +2ω 4 A 2 ε| log ε|+o(| log ε|) 2K 2 + b(x 0 )K 1 +2ω 4 B 2 ε| log ε|+o(ε| log ε|) 2K 2 -λε ω 4 2K 2 | log ε| if N = 4 k = 2, l = 2, a(x 0 )K 1 +C 2 ε+o(ε) 2K 2 + b(x 0 )K 1 +o(ε) K 2 -λ K 3 K 2 ε + O(1) if N ≥ 5 k = 2, l > 2, a(x 0 )K 1 +o(ε) 2K 2 + b(x 0 )K 1 +D 2 ε+o(ε) 2K 2 -λ K 3 K 2 ε + O(1) if N ≥ 5 k > 2, l = 2, a(x 0 )K 1 +o(ε) 2K 2 + b(x 0 )K 1 +2ω 4 B 2 ε| log ε|+o(ε| log ε|) 2K 2 -λε ω 4 2K 2 | log ε| + o(ε| log ε|) if N = 4 k > 2, l = 2, a(x 0 )K 1 +2ω 4 A 2 ε| log ε|+o(| log ε|) 2K 2 + b(x 0 )K 1 +o(ε) 2K 2 -λε ω 4 2K 2 | log ε| + o(ε| log ε|) if N = 4 k = 2, l > 2, a(x 0 )K 1 +C k ε k 2 +o(ε k 2 ) 2K 2 + b(x 0 )K 1 +2ω 4 B 2 ε| log ε|+o(ε| log ε|) 2K 2 -λε ω 4 2K 2 | log ε| + o(ε| log ε|) if N = 4 k < 2, l = 2, a(x 0 )K 1 +2ω 4 A 2 ε| log ε|+o(ε| log ε|) 2K 2 + b(x 0 )K 1 +D k ε l 2 +o(ε l 2 ) 2K 2 -λε ω 4 2 | log ε| + o(ε| log ε|) if N = 4 k = 2, l < 2, Thus (i) If Q λ is achieved in (u, v) then uv ≥ 0.
(ii) There exists a solutions u and v of the minimization problem (1) such that u ≥ 0 and v ≥ 0.

Proof (i) Set F (u, v) = 1 2 Ω a(x)|∇u| 2 dx + 1 2 Ω b(x)|∇v| 2 dx -λ Ω u(x)v(x)dx.
The inequality

F (u, v) ≤ F (|u|, |v|)
gives that -λ

Ω uvdx ≤ -λ Ω |uv|dx, then Ω (u(x)v(x) -|u(x)v(x)|)dx ≥ 0. ( 23
)
We have always

u(x)v(x) -|u(x)v(x)| ≤ 0 ∀x ∈ Ω. Hence Ω (u(x)v(x) -|u(x)v(x)|)dx ≤ 0. ( 24
)
Combining ( 23) and ( 24) we obtain that uv -|uv| = 0 and hence |uv| = uv. Finally for all x ∈ Ω, u(x)v(x) ≥ 0.

(ii) From Lemma 3.1 we know there exists a minimum (u, v). By Proposition 5 (i) we know that |uv| = uv. On the other hand, we have

F (|u n |, |v n |) = 1 2 Ω a(x)|∇u n | 2 dx + 1 2 Ω b(x)|∇v n | 2 dx -λ Ω |u n v n |dx = F (u n , v n ).
Then if we have (u n , v n ) are solutions of ( 1) then (|u n |, |v n |) are also minimizing solutions. Thus, when dealing with (1), one can assume without loss of generality that u ≥ 0 and v ≥ 0.

               -div(a(x)∇u) -λv = Λ 1 u 2 * -1 in Ω -div(b(x)∇v) -λu = Λ 2 v 2 * -1 in Ω ||u|| 2 * = ||u|| 2 * = 1 u ≥ 0, v ≥ 0 in Ω u = v = 0 on ∂Ω, ( 25
)
such that Λ 1 +Λ 1 2 = Q λ ≥ 0 according to Proposition 3.1.

Non-existence Results

In this section, we assume that a, b satisfy ( 5) and ( 6) with k > 2, l > 2 and ( 7), ( 8) if k ≤ 2, l ≤ 2. we obtain in these cases a few non-existence results. We define

ω(a, b) := inf (u,v)∈(H 1 0 (Ω)) 2 \{0} φ a,b (u, v)
where

φ a,b (u, v) := 1 4 Ω ã(x)|∇u| 2 + b(x)|∇v| 2 dx Ω u(x)v(x)dx . We see that ω(a, b) ∈ [-∞, +∞[ .
The main goal of this section is the non-existence results.

Theorem 2. Assume that λ ≤ ω(a, b) and Ω is a strictly star-shaped domain with respect to x 0 . Then (1) has no minimizers.

The proof of this result follows from the following Pohozaev identity; see [START_REF] Pohožaev | I On the eigenfunctions of the equation ∆u + λf (u) = 0[END_REF].

Proposition 6.1. If (u, v) is a solution of (25) then (u, v) satisfies the following identity :

2λ Ω u(x)v(x)dx- 1 2 Ω ∇b(x)•(x-x 0 )|∇v(x)| 2 dx- 1 2 Ω [∇a(x) • (x -x 0 )] |∇u(x)| 2 dx = 1 2 ∂Ω a(x) [(x -x 0 ) • n] ∂u ∂ν 2 dσ(x) + 1 2 ∂Ω b(x) [(x -x 0 ) • n] ∂v ∂ν 2 dσ(x).
where n denotes the outward normal to ∂Ω.

Proof Suppose that (u, v) is a solution of Problem (25). We multiply the first equation in the system by ∇u(x) • (xx 0 ) and we integrate by parts, leading to

Λ 1 Ω |u(x)| q-2 u(x)∇u(x) • (x -x 0 )dx = - N q Λ 1 , (26) λ
Ω v(x)∇u(x)•(x-x 0 )dx = -λ Ω u(x)∇v(x)•(x-x 0 )dx-Nλ Ω v(x)u(x)dx (27)
and

Ω -div(a(x)∇u(x))∇u(x) • (x -x 0 )dx = - N -2 2 Ω a(x)|∇u(x)| 2 dx - 1 2 Ω ∇a(x) • (x -x 0 )|∇u(x)| 2 dx - 1 2 ∂Ω a(x) ∂u ∂ν 2 ((x -x 0 ) • n) dσ(x), (28)
where n denotes the outward normal to ∂Ω. Combining (26), ( 27)and (28) we get

- N -2 2 Ω a(x)|∇u(x)| 2 dx- 1 2 Ω ∇a(x)•(x-x 0 )|∇u(x)| 2 dx- 1 2 ∂Ω a(x) ∂u ∂ν 2 ((x -x 0 ) • n) dσ(x) = λ Ω v(x)∇u(x) • (x -x 0 )dx - N q Λ 1 . (29)
Similarly, we multiply the second equation of (25) by ∇v(x)•(x-x 0 ) and we integrate by parts, we get

- N -2 2 Ω b(x)|∇v(x)| 2 dx- 1 2 Ω ∇b(x)•(x-x 0 )|∇v(x)| 2 dx- 1 2 ∂Ω b(x) ∂v ∂ν 2 ((x -x 0 ) • n) dσ(x) = λ Ω u(x)∇v(x) • (x -x 0 )dx - N q Λ 2 . (30)
On the other hand, multiplying the equations in (25) by N -2 2 u and N -2 2 v respectively, and integrating and summing the obtained results, we get

N -2 2 Ω a(x)|∇u(x)| 2 dx = N -2 2 λ Ω v(x)u(x)dx + Λ 1 N -2 2 (31) and N -2 2 Ω b(x)|∇v(x)| 2 dx = N -2 2 λ Ω u(x)v(x)dx + Λ 2 N -2 2 . (32)
Combining (29), and (31) we obtain

- N -2 2 λ Ω u(x)v(x)dx -λ Ω v(x) [∇u(x) • (x -x 0 )] dx = 1 2 Ω |∇u| 2 [∇a(x) • (x -x 0 )] dx + 1 2 ∂Ω a(x)| ∂u ∂ν |
On the other hand, combining (30), and (32) we get

- N -2 2 λ Ω u(x)v(x)dx -λ Ω u(x) [∇v(x) • (x -x 0 )] dx = 1 2 Ω |∇v(x)| 2 [∇b(x) • (x -x 0 )] dx + 1 2 ∂Ω b(x)| ∂v ∂ν | 2 [(x -x 0 ) • n] dσ(x). ( 34
)
Adding (34) in (33) and using ( 27) we find 2λ

Ω u(x)v(x)dx- 1 2 Ω ∇b(x)•(x-x 0 )|∇v(x)| 2 dx- 1 2 Ω [∇a(x) • (x -x 0 )] |∇u| 2 dx = 1 2 ∂Ω a(x)| ∂u ∂ν | 2 [(x -x 0 ) • n] dσ(x) + 1 2 ∂Ω b(x)| ∂v ∂ν | 2 [(x -x 0 ) • n] dσ(x).
Finally we get the Proposition. Now let us prove Theorem 2. Assume that λ ≤ ω(a, b) and let Ω be strictly starshaped domain with respect to x 0 then (xx 0 ) • n > 0, for all x ∈ ∂Ω.

Suppose that (u, v) is a solution of (25). By the Proposition 6.1 we get 2λ

Ω u(x)v(x)dx- 1 2 Ω ∇b(x)•(x-x 0 )|∇v(x)| 2 dx- 1 2 Ω [∇a(x) • (x -x 0 )] |∇u(x)| 2 dx > 0.
It follows that

λ > 1 4 inf (u,v)∈(H 1 0 (Ω)) 2 \{0} Ω ∇a(x) • (x -x 0 )|∇u(x)| 2 + ∇b(x) • (x -x 0 )|∇v(x)| 2 dx Ω u(x)v(x)dx
which is a contradiction.

Then the problem (25) does not admit solutions. Consequently we obtain the desired result.

Estimates of ω(a, b)

We end this section, using the techniques in [START_REF] Hadiji | H Problem with critical Sobolev exponent and with weight[END_REF], we give some estimates of ω(a, b). Proposition 6.2.

(1) We assume that a, b ∈ C 1 (Ω) and there exists z 0 ∈ Ω such that ã(z 0 ) + b(z 0 ) < 0, then ω(a, b) = -∞.

(2) We assume that a respectively b ∈ H 1 (Ω) ∩ C( Ω) satisfying ( 5) respectively (6) and we have

N 2 16 min A 2 , lB l (diam Ω) l-2 ≤ ω(a, b) ≤ A 2 2 λ 1 (diam Ω) 2 (35)
and Proof Let ϕ ∈ C c (R N ) such that 0 ≤ ϕ ≤ 1 on R N , ϕ ≡ 1 on B(0, 2r) and ϕ ≡ 0 on R N \ B(0, 2r), where 0 < r < 1. Set ϕ j (x) = ϕ(j(xz 0 )) for j ∈ N * , we have

N 2 16 min kA k (diam Ω) k-2 , B 2 ≤ ω(a, b) ≤ B 2 2 λ 1 (diam Ω) 2 (36)
(2.iii) If 0 < k ≤ 2, 0 < l ≤ 2,
ω(a, b) ≤ 1 4 Ω ã(x) + b(x) |∇ϕ j (x)| 2 dx Ω ϕ 2 j (x)dx ≤ 1 4 B(b, 2r j ) ã(x) + b(x) |∇ϕ j (x)| 2 dx B(b, 2r j ) ϕ 2 j (x)dx.
Using the change of variable y = j(xz 0 ), we get

ω(a, b) ≤ j 2 4 B(0,2r) ã( y j + z 0 ) + b( y j + z 0 ) |∇ϕ(y)| 2 dy B(0,2r) ϕ 2 (y)dy .
Applying the Dominated Convergence Theorem, we obtain the desired result when j goes to infinity. Now we will prove (2.i). Using ( 5), ( 6) and since a, b ∈ C 1 (Ω) in a neighborhood V of x 1 , we write

a(x) = a(x 1 ) + A k |x -x 1 | k + θ a (x) (37) b(x) = b(x 1 ) + B l |x -x 1 | l + θ b (x), (38)
where θ a (x) and

θ b (x) ∈ C 1 (V ) are such that lim x→x 1 |θ a (x)| |x -x 1 | k = 0 and lim x→x 2 |θ b (x)| |x -x 1 | l = 0. (39)
From (39), we get the existence of r, such that 0 < r < 1 and

|θ a (x)| ≤ |x -x 0 | k and |θ b (x)| ≤ |x -x 0 | l , for all x ∈ B(x 0 , 2r) ⊂ V. (40)
Let ϕ j (x) = ϕ(j(xx 0 )) define as in the proof of (1); we have

0 ≤ ω(a, b) ≤ 1 4 Ω ã(x) + b(x) |∇ϕ j (x)| 2 dx Ω ϕ 2 j (x)dx .
Using (37) and (38), we obtain

0 ≤ ω(a, b) ≤ 1 4 B(x 0 , 2r j ) kA k |x -x 0 | k + lB l |x -x 0 | l |∇ϕ j (x)| 2 dx B(x 0 , 2r j ) ϕ 2 j (x)dx + 1 4 B(x 0 , 2r j ) (∇θ a (x) • (x -x 0 ) + ∇θ b (x) • (x -x 0 )) |∇ϕ j (x)| 2 dx B(x 0 , 2r j ) ϕ 2 j (x)dx.
By a simple change of variable y = j(xx 0 ) and integrating by parts, we obtain Therefore, for k > 2 and l > 2, when j tends to ∞ we obtain ω ( p, q) = 0. To prove (2.ii), we start by the case k = 2 and l > 2. We show the left-hand inequality in (35). Since a and b satisfy ( 7) and ( 8), respectively, for all pairs (u, v) ∈ E \ {0}, we have

0 ≤ ω(a, b) ≤ kA k 4j k-2 B(0,
φ a,b (u, v) ≥ 1 2 A 2 Ω |(x -x 0 ) • ∇u(x)| 2 dx Ω u(x)v(x)dx + l 4 B l Ω |x -x 0 | l-2 |(x -x 0 ) • ∇v(x)| 2 dx Ω u(x)v(x)dx

21

Let us now prove (2.iii). Since a and b satisfy ( 7) and ( 8), respectively, for all (u, v) ∈ E \ {0}, we have

φ a,b (u, v) ≥ k 4 A k Ω |x -x 0 | k-2 |(x -x 0 ) • ∇u| 2 dx Ω uvdx + l 4 B l Ω |x -x 0 | l-2 |(x -x 0 ) • ∇v| 2 dx Ω uvdx ≥ k 4 A k (diam Ω) k-2 Ω |(x -x 0 ) • ∇u| 2 dx Ω uvdx + l 4 B l (diam Ω) l-2 Ω |(x -x 0 ) • ∇v| 2 dx Ω uvdx .
By applying Lemma 2.1 in [START_REF] Hadiji | H Problem with critical Sobolev exponent and with weight[END_REF] for t = 0, we find

φ a,b (u, v) ≥ k 4 A k (diam Ω) k-2 N 2 2 Ω |u| 2 dx Ω uvdx + l 4 B l (diam Ω) l-2 N 2 2 Ω |v| 2 dx Ω uvdx .
This implies that ω(a, b) ≥ N 2 16 min(kA k (diam Ω) k-2 , lB l (diam Ω) l-2 ).

Proposition 4. 1 .

 1 (a) For N ≥ 4, and k > 2, l > 2 we have Q λ < γ 0 S, for all λ > 0. (b) For N = 4, and k = 2, l = 2 we have

Remark 5 . 1 .

 51 Since u and v are minimizers for (1) we obtain Lagrange multipliers Λ 1 , Λ 2 ∈ R such that the Euler-Lagrange equation formula associated to (1) is

( 2 .

 2 i) If k > 2, l > 2 and a, b ∈ C 1 (Ω), then ω(a, b) = 0. (2.ii) If k = 2, l > 2 or k > 2, l = 2,and a, b satisfying moreover (7) and (8) with ã(x) ≥ 0, b(x) ≥ 0 a.e. x ∈ Ω, then

  a, b satisfy the conditions (7) and (8) respectively, then N 2 16 min kA k (diam Ω) k-2 , lB l (diam Ω) l-2 ≤ ω(a, b).

lB l 4j l- 2 B 4 Bϕ 2

 242 2r) |y| k |∇ϕ(y)| 2 dy B(0,2r) (0,2r) |y| l |∇ϕ(y)| 2 dy B(0,2r) ϕ(y) 2 dy -j (0,2r) θ b ( y j + x 0 ) • div y|∇ϕ(y)| 2 dy B(0,2r) ϕ(y) 2 dy . (y)dy , where C = max y∈B(0,2r) | div(y|∇ϕ(y)| 2 )|.

2 , 4 Ω 4 x 0 + 1 j Ω ψ 2 j 2 Ω |y| 2 |∇ϕ 1 ( 2 Ω 1 Ω 2 Ω |y| 2

 2442212122 x 0 ) • ∇u(x)| 2 dx Ω u(x)v(x)dx + l 4 B l (diam Ω) l-2 Ω |(xx 0 ) • ∇v(x)| 2 dx Ω u(x)v(x)dx.By applying Lemma 2.1 in[START_REF] Hadiji | H Problem with critical Sobolev exponent and with weight[END_REF] for t = 0, we findφ a,b (u, v) lB l (diam Ω) l-2 ).Similarly, we deduce in the case k > 2 and l = 2, thatω(a, b) ≥ N 2 16 min(kA k (diam Ω) k-2 , B 2 ).Now we prove the right-hand inequality in (35) and (36). Let ψ j (x) = ϕ 1 (j(xx 0 )) for j ∈ N large enough, where ϕ 1 is a positive eigenfunction corresponding to the first eigenvalue λ 1 of the operator -∆ in H 1 0 (Ω). We have0 ≤ ω(a, b) ≤ 1 (ã(x) + b(x))|ψ j (x)| 2 dx Ω ψ 2 j (x)dx.Using (4) and (5), we obtain0 ≤ ω(a, b) ≤ x 0 + 1 j Ω 2A 2 |xx 0 | 2 + lB l |xx 0 | l |∇ψ j (x)| 2 dx 4 x 0 , 1 j Ω ψ 2 j (x)dx + x 0 + 1 j Ω (∇θ a (x) • (xx 0 ) + ∇θ b (x) • (xx 0 )) |∇ψ j (x)| 2 dx (x)dx.By a simple change of variable y = j(xx 0 ) and integrating by parts, we have by (39)0 ≤ ω(a, b) ≤ A 2 |y| l |∇ϕ 1 (y)| 2 dy Ω ϕ 1 (y) 2 dy + C j l-|y| l dy Ω ϕ 1 (y) 2 dy , where C = max y∈Ω | div(y∇ϕ 1 (y)| 2 )|. Letting j → ∞ we get 0 ≤ ω(a, b) ≤ A 2 |∇ϕ 1 (y)| 2 dy Ω ϕ 1 (y) 2 dy , therfore 0 ≤ ω(a, b) ≤ A 2 2 λ 1 (diam Ω) 2 .Similarly, we deduce in the case k > 2 and l = 2, that 0 ≤ ω(a, b) ≤ B 2 2 λ 1 (diam Ω) 2 .

[(xx 0 ) • n] dσ(x). (33)

where

A 2 and

B 2 . From these estimates we get the desired result.

Combining Lemma 3.1 and Proposition 4.1 we conclude that the infimum in (1) is achieved. This leads to the conclusion of Theorem 1.

Remark 4.1. We do not know if we can go strictly below a(x 0 )+b(x 0 ) 2 S in the case where N ≥ 4, k < 2 and l < 2.

The Sign of the Minimizers

Now, we will discuss the positivity of the solutions.