On a resonant and superlinear elliptic system

We prove existence of solutions for a class of nonhomogeneous elliptic system with asymmetric nonlinearities that are resonant at −∞ and superlinear at +∞. The proof is based on topological degree arguments. A priori bounds for the solutions are obtained by adapting the method of BrezisTurner.

Moreover, we also assume that the functions f and g are such that f, g ∈ L r with r > N.
In the sequel, we will present our resonance assumption. We begin writing the system (1) in an equivalent matricial form The matrix A has real eigenvalues Denote by λ 1 the first eigenvalue of (−∆, H 1 0 (Ω)) and φ 1 is the positive eigenfunction associated to λ 1 with |φ 1 | L 2 = 1. We will assume that λ 1 = ξ. This assumption is called resonance hypothesis. Note that, in this case λ 1 > a and Φ = (bφ 1 , (λ 1 −a)φ 1 ) is the positive eigenfunction of the following eigenvalue problem associated to the eigenvalue λ 1 (A) = 1. The above eigenvalue problem is studied in the Appendix (see [3] for more details). Our main results is Theorem 1. Assume that (2) and (3) hold. In addition, suppose that λ 1 = ξ and Then system (1) possesses at least one strong solution.
Our proof is based upon ideas found in [6]. In this paper, the authors get a solution of the following superlinear elliptic equation under the assumptions 1 < p < N +1 N −1 , f ∈ L r with r > N and Ω f φ 1 < 0. The proof of the main result in [6] uses the thecnique introduced in [1]. The method consists in getting a priori bounds, using Hardy-Sobolev type inequalities, with topological degree arguments. This approach was also used in [12] to study an Ambrosetti-Prodi type problem. Similar problems, under Dirichelt and Neumann boundary condition, can be found in [4,11,13] (see also [5] for a different method to those in [6]) . We also mention that in [6], the authors consider the Hamiltonian systems , under appropriate conditions. Let H be the Hilbert space H 1 0 (Ω) × H 1 0 (Ω) equipped with the norm The space C 1 0 (Ω) is defined as C 1 0 (Ω) = u ∈ C 1 (Ω); u = 0 on ∂Ω and we equip the Banach space C 1 0 (Ω) × C 1 0 (Ω) with the norm U C 1 0 (Ω) = u C 1 0 (Ω) + v C 1 0 (Ω) .

2.
A priori estimates. Let us first remark the following lemma based on the Hardy-Sobolev inequality. The proof can be found in [6].
Under the assumptions of Theorem 1, there is an increasing continuous function ρ : R + → R + , depending only on p, q and Ω, such that ρ (0) = 0 and , for all U ∈ H solution of (4).
Proof. Let U ∈ H be a weak solution of (4). Using Φ = (bφ 1 , (λ 1 − a) φ 1 ) as a test function in (4), we find By Hölder inequality and using (6), it results Now, the proof will be divided in two parts, according to the sign of t.
Case 1. t ≥ 0. By (7) we obtain a bound for t. We need to find an estimate for U 1 .
Multiplying the equation (4) by U 1 and using the decomposition of U , we obtain The variational characterization of the eigenvalues implies that By Hölder inequality and using the Sobolev embedding theorem, we have Now, we will estimate the integral in the right-hand side of (8):

FABIANA MARIA FERREIRA AND FRANCISCO ODAIR DE PAIVA
Using the fact that (u 1 ) By applying Lemma 1, we obtain with α, α ∈ (0, 1) and δ, δ ∈ (1, 2). Using the inequality (6), By inequality (7), we have Replacing the estimate (9) in (8), we get By Young's inequality, we deduce that Therefore Replacing the values α, α , δ and δ , given by Lemma 1, we conclude A bootstrap argument applied to (1) and the previous bound of U in H-norm will give us the claim of the Theorem.
Case 2. t < 0. By Hopf's Maximum Principle, there is > 0 such that where η denotes the exterior normal derivative at the boundary of Ω. Let us remember that our solution U = (u, v) of problem (1), as well as U 1 = (u 1 , v 1 ), belongs to Let 0 be the supremum of the 's above. We claim that . This is a contradiction with the fact that u + , v + ≡ 0, b > 0, λ 1 > a and t < 0. Hence, , and we conclude that, Thus it is enough to find an a priori bound for U 1 C 1 0 (Ω) . We note that inequality (8) remains true for t < 0. Thus, in this case u + < |u 1 | and v + < |v 1 |. Hence, by Lemma 1, we have And using (6) we get Replacing (12) in (8), we find Since α, α ∈ (0, 1) and δ, δ ∈ (1, 2) by Young's inequality we deduce that We now use that U 1 solves the problem a bootstrap argument to this equation gives us that with constants γ, η ≥ 1. Replacing (13) in (14), result U 1 2,r ≤ ρ( F r ).
3. Some lemmas. In this section we prove two lemmas. The first lemma will be used to prove that all solutions of the system are non-degenerate, since f and g are small enough. The second will be used in the computation of the index of such solutions. In order to do that we need to consider the linearization of problem (1) at some solution (u 0 , v 0 ), which is Lemma 2. There exists > 0 such that for any m, k ∈ L ∞ , such that m, k ≥ 0 a.e., m > 0 or k > 0 in a set of positive measure, and m ∞ , k ∞ < , then the system possesses only the trivial solution w = z = 0.

Lemma 3.
There exists > 0 such that for any m, k ∈ L ∞ satisfying the conditions in the previous lemma, the following eigenvalue problem has only one eigenvalue µ in the interval (0, 1).

Proof of the Theorem 1. Let T
. Notice that, T F is continuous, compact and T F (u, v) = (u, v) if, and only if, U = (u, v) solves (1). Choose F 1 = (f 1 , g 1 ) such that f 1 = −(γbφ 1 ) p and g 1 = − (γ (λ 1 − a) φ 1 ) q with γ > 0. Then, it is easy to see that, U 1 = (γbφ 1 , γ (λ 1 − a) φ 1 ) is a solution of problem (1) with F = F 1 . Moreover, from the a-priori estimates of Theorem 2, taking γ small enough, we can apply Lemma 2 for m(x) = pu p−1 + and k(x) = qv q−1 + , where (u, v) is any arbitrary solution of (1) with F = F 1 . It can be conclude, as a consequence of Lemma 2, that (u, v) is non-degenerate. In addition, for some R (u,v) sufficiently small, we have where β is a number of characteristic values, between 0 and 1, of the eigenvalue problem By Lemma 3, we can assume that that β = 1. We can also conclude that (u, v) is an isolated solution. Moreover, the number of the solutions is finite. Then for R large enough deg(I − T F1 , B C 1 0 (Ω) 2 (0, R), 0) = (−1) = 0.
Then H(τ, U ) = 0 if, and only if, U is a solution of the system By the a priori bounds, Theorem 2, all solutions of the previous system are uniformly bounded in C 1 0 (Ω) × C 1 0 (Ω). Hence for R > 0 large enough, we have H(τ, U ) = 0 for all (τ, U ) ∈ [0, 1] × ∂B C 1 0 (Ω) 2 (0, R). It follows that deg (H(0, U ) Appendix. Let M 2 (Ω) be the set of all symmetric matrices of the form where the functions a, b, c ∈ C(Ω, R) satisfy 1. b(x) ≥ 0 for all x ∈Ω and b = 0; 2. max x∈Ω max{a(x), c(x)} > 0. We can consider the weighted eigenvalue problem and obtain a sequence of eigenvalues where F k varies over all k-dimensional subspaces of H. Moreover, λ k (A) → ∞ when k → ∞. The eigenvalue λ 1 (A) is positive, simple, isolated and the associated eigenfunction can be choose with positive components. Additional, if we set V k = span{Φ A 1 , ..., Φ A k }, with Φ A k the associated eigenfunction, we can decompose H as H = V k ⊕ V ⊥ k . Furthermore, the followings variational inequalities hold and These results can be found in detail in [2,7,10]. Proof. We argue as in [8]. Let j > 0, since the extreme in (20) is achieved, exists F j ∈ H such that 1 λ j (A) = inf{ Ω Az, z ; z ∈ F j e z = 1}.
Pick z ∈ F j with z = 1, in this moment we have two cases: Case 1. y achieves the infimum in (23). Then z is an eigenfunction associated to µ j (A) and so, by the unique continuation property we obtain Thus, in any case, 1 λ j (A) < Ω By, y .