MULTIPLICITY RESULTS FOR FRACTIONAL SYSTEMS CROSSING HIGH EIGENVALUES

. We investigate the existence of solutions for a system of nonlocal equations involving the fractional Laplacian operator and with nonlinearities reaching the subcritical growth and interacting, in some sense, with the spectrum of the operator.

With the above decomposition, in order to state and compare our results to the scalar case, it is convenient to rewrite system (1.1) as ∇ is the gradient operator, F (U ) = 2 α+β u + α v + β + ξ 1 u + α+β + ξ 2 v + α+β , T = t r and F 1 = f 1 g 1 .
Let µ 1 , µ 2 be real eigenvalues of the symmetric matrix A, which will assume µ 1 ≤ µ 2 . The interaction of these eigenvalues with the spectrum of the (−∆) s will play an important role in the study of the existence of solutions for the nonlocal gradient system (1.1).
The purpose of this work is to prove the existence of solutions of this class of nonlocal gradient systems of elliptic equations on the hypothesis of an interaction of the eigenvalues µ 1 , µ 2 of the matrix A with eigenvalues of the fractional Laplacian operator (−∆) s . When µ 2 < λ 1,s , this system belongs to the class of the so called Ambrosetti-Prodi type problems [2] which have been studied by several authors in the last decades with different approaches. We also note that the fractional operators appears in phenomena in physics ( [10,28]), in stochastic processes ( [22,35]), in fluid dynamics, dynamical systems, elasticity, obstacle problems and others (see [8,9,12,14,32,33,34] and references therein).
Problem (1.1) is an extension to systems involving fractional Laplacian operator of the equation considered in [27], in which (1.1) was studied in the local operators case (s = 1) and with the particular matrix A = λ 0 0 λ ∈ M 2×2 (R). In [27], the authors established a multiplicity result for the problem −∆u = λu + u p + + f (x) in Ω, u = 0 on ∂Ω provided that the non-homogeneous term f has the form f (x) = h(x)+tϕ 1 (x) (h ∈ L r (Ω) with r > N ), λ is not an eigenvalue of (−∆, H 1 0 (Ω)) and t > T, for some sufficiently large number T = T (h). Still in the local scalar case, but with nonlinearity in the critical growth (p = 2 * − 1), the problem above mentioned has been studied in [16], the authors proved the existence of two solutions when N ≥ 7. This result was extended in [11] using the technique developed in [20]. Works related to this subject in the local scalar case, we recommend [4] and the references therein. For the critical system in the local operators situation, problem (1.1) was studied, for instance, in [17] and [25] when µ 2 < λ 1 and by [23] in the uncoupled case. In this paper, we complement the results achieved in [27], proving that the system (1.1) (or (1.2)) has at least two solutions for sufficiently large values of parameters (t, r), the first solution is negative and obtained explicitly depending on the non-homogeneous terms f and g. The second solution is obtained via the Mountain Pass Theorem when µ 2 < λ 1,s , or applying the Linking Theorem in the case λ k,s < µ 1 ≤ µ 2 < λ k+1,s if k ≥ 1. The resonant case λ k,s = µ 1 for k > 1 is also treated here.
To show the existence of solution difficulties arise when we consider fractional operators. Due to the lack of regularity, we impose that s ≥ 1/2 and we develop similar techniques to these known for the Laplacian operator. In addition to these obstacles, further complications arise due to the presence of the mathematical term α+β that includes either an uncoupled or a coupled nonlinearity. In the case λ k,s ≤ µ 1 ≤ µ 2 < λ k+1,s , it is necessary to require the hypothesis that the constants ξ j are strictly positive. It is important to point out that, with the aid of [18], our results are still valid for the general case ∇F (u, v) when F is a (α + β)−homogeneous nonlinearity, which includes a larger class of functions.
Our main results are Then there exist η, ϑ 0 such that system (1.2) has a solution (u T , v T ) (with u T < 0 and v T < 0 in Ω) for every T ∈ R.
Remark 1. Suppose that det(λ 1,s I − A) > 0 and then the set R is a region between lines satisfying: On the other hand, if det(λ 1,s I − A) > 0 and then the set R satisfies: Note that, since det(λ 1,s I − A) = 0, the lines that define the region R are not parallel. Moreover, if det(λ 1,s I − A) < 0 a similar result can be obtained as in the Remark 1.

2.
Notations and preliminary stuff. For any measurable function u : R N → R the Gagliardo seminorm is defined by The second equality follows by [19,Proposition 3.6] when the above integrals are finite. Then, we consider the fractional Sobolev space , which is a Hilbert space. We use the closed subspace By Theorems 6.5 and 7.1 in [19], the imbedding X(Ω) → L r (Ω) is continuous for r ∈ [1, 2 * s ] and compact for r ∈ [1, 2 * s ). Due to the fractional Sobolev inequality, X(Ω) is a Hilbert space with inner product which induces the norm · X = [ · ] s . Observe that by Proposition 3.6 in [19], we have the following identity Then it is proved that for u, v ∈ X(Ω), in particular, (−∆) s is self-adjoint in X(Ω). Now, we consider the Hilbert space given by the product space equipped with the inner product where µ 1 ≤ µ 2 are the eigenvalues of the symmetric matrix A. In this paper, we consider the following notation for product space S × S := S 2 and for positive and negative part of a function w. Consequently we get w = w + − w − . Definition 2.1. By a solution of (1.1) we mean a weak solution, that is, a pair of

2.1.
A regularity result. Consider the following system where G u , G v : R × R → R are Carathéodory functions which satisfy respectively, the following growth conditions The next result can be found in [15, Lemma 3.1].
(i) problem (2.6) admits an eigenvalue λ 1,s = min σ((−∆) s ) > 0 that can be characterized as follows (2.7) (ii) there exists a non-negative function ϕ 1,s ∈ X(Ω), which is an eigenfunction corresponding to λ 1,s , attaining the minimum in (2.7); (iii) all λ 1,s -eigenfunctions are proportional, and if u is a λ 1,s -eigenfunction, then either u(x) > 0 a.e. in Ω or u(x) < 0 a.e. in Ω; (iv) the set of the eigenvalues of problem (2.6) consists of a sequence {λ k,s } satisfying which is characterized by (vi) Denote by ϕ k,s the eigenfunction associated to the eigenvalue λ k,s , for each k ∈ N. The sequence {ϕ k,s } is an orthonormal basis either of L 2 (Ω) or of X(Ω).
has a solution U = 0 in Y (Ω) if, and only if, the matrix λ k,s I − A is singular for some λ k,s .
Using that the eigenfunctions ϕ k,s constitute an orthonormal basis of X(Ω), we can write the Fourier developments To prove Theorem 1.1, we need the following lemma.
We are ready to prove the existence of a negative solution for the system (1.2).
Proof of Theorem 1.1. We will prove the theorem when the conditions (1.3) and (1.5) hold (other cases (1.3) and (1.4) or (1.8)) are analogous to this and left to the reader).
By Lemma 3.2 and Remark 6, the system Besides, (w, z) = Consequently, if Clearly if u T and v T are negative in Ω, we deduce also that U T is a solution of (1.2). Therefore, to conclude the proof under the conditions (1.3) and (1.5) (see Remark 1), it suffices to show the existence of an unbounded region R ⊂ R 2 where u T and v T are negative in Ω for every T = (t, r) ∈ R.
Indeed, since ϕ 1,s ∈ C 1 (Ω) is strictly positive in Ω, then U = U + U T is a (second) solution of the system (1.2) with U + U T = U T . Therefore, to prove Theorems 1.2 and 1.3 we only have to show that under the hypotheses of these theorems, the system (4.1) has a nonzero solution for every T ∈ R.
Observe that the weak solutions of (4.1) are the critical points of the functional I s : Y (Ω) −→ R given by and that U = 0 is a critical point for I s with I s (0) = 0.
Remark 7. Notice that every nonzero critical point U of I s has positive part U + = 0. Indeed, if U = (u, v) is critical point with u + = v + = 0, then u, v ≤ 0 and consequently U satisfies the system (3.1). Since det(λ j,s I − A) = 0 for all j , by Lemma 3.1, we have that U ≡ 0.

FÁBIO R. PEREIRA
(iii) There exists K > 0 such that for some constant K > 0.
4.1. The Palais-Smale condition for the functional I s . In this subsection we discuss a compactness property for the functional I s , given by the Palais-Smale condition.
Lemma 4.1. If µ 2 < λ 1,s , then the functional I s satisfies the (P S) condition.
Proof. The Fréchet derivative of the functional I s is given by for every (u, v), (ϕ, ψ) ∈ Y (Ω). Let (U n ) ⊂ Y (Ω) be a (PS)-sequence, i.e. satisfying |I s (U n )| ≤ M for some positive real constant M and I s (U n ) → 0 in the dual space Y (Ω) . We are going to prove that there is a convergent subsequence of U n in two steps: Step 1: U n is bounded. Indeed, observe that:

Now, by Remark 8 (i),
with ε n = I s (U n ) Y → 0. Therefore, since u T and v T are negative, ξ 1 , ξ 2 > 0 and α, β > 1, we have that the left hand side of the above inequality is positive, and consequently and On the other hand, by Remark 8 (i) again, Since (U n ) is a (P S)-sequence, by (4.2) and (4.3), we get and consequently, using that µ 2 < λ 1,s , we conclude that (U n ) = (u n , v n ) is bounded in Y (Ω).
Step 2: Conclusion: Passing to a subsequence, we can assume that On the other hand, we have the convergence to zero of I s (U n )(U n − U ); i.e.
as n → ∞. Claim: Indeed, by Remark 8 (iii), there exists a constant K > 0 such that and consequently Since u T < 0 and v T < 0 are negative, we have that (u n + u T ) α+β−1 Therefore, by (4.5), Besides, (∇F (U n + U T ), U n − U ) R 2 → 0 a.e in Ω, so the use of the Dominated Convergence Theorem concludes the claim. Now, by (4.4) and the claim, it follows that U n 2 Y → U 2 Y and consequently U n − U Y → 0.
Proof. We follow the notations of the previous proof.
Therefore, possibly up to a subsequence, W n /β n → 0 a.e. in Ω and strongly in L q (Ω) × L q (Ω), 1 ≤ q < 2 * s ; Y n → Y 0 ∈ Z k a.e. in Ω and strongly in Y (Ω) and L q (Ω) × L q (Ω), 1 ≤ q < 2 * s . Taking β n Y n ∈ Z k as test function, we get Since (U n ) is a (P S)-sequence and On the other hand, since U n = W n + β n Y n , we have that Un βn → Y 0 in L q (Ω) × L q (Ω) for all 1 ≤ q < 2 * s and a.e in Ω. So, by the Dominated Convergence Theorem and by (4.11), it follows that Now, from remark 8 (i), we concluded that Ω F (Y 0 )dx = 0. Finally, using the notation Y 0 = (y 0 1 , y 0 2 ), it follows that (y 0 1 ) + = 0 = (y 0 2 ) + , contradicting Y 0 Y = 1 and Y 0 ∈ Z k with k > 1, which ensures that at least one of the functions is not null and changes sign. Thus (U n ) is bounded and as in the proof of Lemma 4.1, we have that (U n ) admits a convergent subsequence.

5.
Geometry of the functional I s . In this section, we show that the hypothesis µ 2 < λ 1,s ensures that the functional I s satisfies the geometrical conditions of the Mountain Pass Theorem, and has the geometric structure required by the Linking Theorem when λ k,s ≤ µ 1 ≤ µ 2 < λ k+1,s , for some k ≥ 1.

FÁBIO R. PEREIRA
For U = U 1 + RE ∈ Γ 2 , we have Since λ k,s < µ 1 , Now, to estimate the last integral, note that, if U 1 = (u 1 , u 2 ), for R ≥ R 0 , and by (II) each integral on the right can be estimated as follows for i = 1, 2 and w = u, v. Therefore, by (5.3) and by above estimates, Since α + β > 2, for R ≥ R 0 , we have I s (U ) < 0 for all U ∈ Γ 2 . iii) Estimates on Γ 3 : For U ∈ Γ 3 , it follows the estimate Therefore, for all R ≥ R 0 > 0, follows that I s (U ) ≤ 0 for all U ∈ ∂Q, concluding the desired result.
In this case, it is sufficient to estimate the functional on Γ 1 , since the estimates on Γ 2 and Γ 3 , and the proof of a) are made as in the previous proposition.
Therefore, c n 1 y n i → 0 and c n 2 e k i → e k i and consequently max i=1,2 max Ω e k i (x) = 0.
for R sufficiently large. Then On the other hand, if |c 1 | ≥ δ > 0, by the choose of E, we get Proof of Theorems 1.2 and 1.3. If µ 2 < λ 1,s , by Lemma 4.1 and Proposition 5.1, the functional I s satisfies the hypothesis of Mountain Pass Theorem [3], and when λ k,s ≤ µ 1 ≤ µ 2 < λ k+1,s , Lemma 4.2 and Proposition 5.2 (or Lemma 4.3 and Proposition 5.3) ensure the application of Linking Theorem [24]. Thus, in both cases, there exists a non-null solution U for the problem (4.1). By Remark 7, it follows that U + = 0 and therefore, U T and U T + U are distinct solutions for the problem (1.2).