Generalized Schrodinger-Poisson type systems

In this paper we study some Schrodinger-Poisson type systems on a bounded domain, with Dirichlet boundary condition on both the variables.


Introduction
This paper deals with the following problem    −∆u + εqΦf (u) = η|u| p−1 u in Ω, −∆Φ = 2qF (u) in Ω, where Ω ⊂ R 3 is a bounded domain with smooth boundary ∂Ω, 1 < p < 5, q > 0, ε, η = ±1, f : R → R is a continuous function and F (s) = s 0 f (t) dt. When the function f (t) = t and ε = η = 1, this system represents the well known Schrödinger-Poisson (or Schrödinger-Maxwell) equations, briefly SPE, that have been widely studied in the recent past. In the pioneer paper of Benci and Fortunato [5], the linear version of SPE (where η = 0) has been approached as an eigenvalue problem. In [16] the authors have proved the existence of infinitely many solutions for SPE when p > 4, whereas in [19] an analogous result has been found for almost any q > 0 and p ∈ ]2, 5[. A multiplicity result has been obtained in [20] for any q > 0 and p sufficiently close to the critical exponent 5, by using the abstract Lusternik-Schnirelmann theory. For the sake of completeness we mention also [15] where Neumann condition on Φ is assumed on ∂Ω, [1] and the references within for results on SPE in unbounded domains and [8,9] for the Klein-Gordon-Maxwell system in a bounded domain. If f (t) = t and ε = −1, the system is equivalent to a nonlocal nonlinear problem related with the following well known Choquard equation in the whole space R 3 ∆u + u − 1 |x| * u 2 u = 0.
We refer to [12,13] for more details on the Choquard equation and to [14] for a recent result on a system in R 3 strictly related with ours.
Up to our knowledge, problem (P) has not been investigated when a more general function f appears instead of the identity. Since problem (P) possesses a variational structure, our aim is to find weak assumptions on f in order to apply the usual variational techniques. In particular, the first step in a classical approach to such a type of systems consists in the use of the reduction method. To this end, we need to assume suitable growth conditions on f which allow us to invert the Laplace operator and thus to solve the second equation of the system. Then, after we have reduced the problem to a single equation, we find critical points of a one variable functional, checking geometrical and compactness assumptions of the Mountain Pass Theorem. If on one hand a suitable use of some a priori estimates makes quite immediate to show that geometrical hypotheses are verified (at least for small q), on the other some technical difficulties arise in getting boundedness for the Palais-Smale sequences. If η = 1, we use a suitable truncation argument based on an idea of Berti and Bolle [6] and Jeanjean and Le Coz [10] (see also [3,11]) and we are able to show the existence of a bounded Palais-Smale sequence of the functional taking q sufficiently small. If we had the sufficient compactness, we would conclude by extracting any strongly convergent subsequence from this bounded Palais-Smale sequence. However the growth hypothesis we assume on f does not permit to deduce compactness on the variable Φ. Indeed the exponent 4 turns out to be critical and in this sense we are justified to refer to (f ) as the critical growth condition for the function f. The first result in this paper is the following.
for all s ∈ R. Then, there existsq > 0 such that for all 0 < q q problem has at least a nontrivial solution.
In this situation the contrasting nonlocal and local nonlinear terms perturb the functional in a way which in some sense recalls the typical concave-convex power-like nonlinearity. As a consequence, the functional's geometry depends on the ratio of magnitude between r and p. We get the following result. Theorem 1.2. If p+1 2 < r < 5, problem (P r ) has infinitely many solutions for any q > 0. If r = p+1 2 , then there exists an increasing sequence (q n ) n such that, if q q n , problem (P r ) has at least n couples of solutions. If 1 < r < p+1 2 , then there exists an increasing sequence (q n ) n such that, if q q n , problem (P r ) has at least 2n couples of solutions. Finally, if p < 5 r, then the problem has no nontrivial solution.
The paper is organized as follows: in Section 2 we introduce the functional setting where we study the problem (P f ) and the variational tools we use; in Section 3 we provide the proof of Theorem 1.1; in Section 4 we consider the system (P r ) and prove Theorem 1.2.
We point out the fact that in the sequel we will use the symbols C, C 1 , C 2 , C 3 and so on, to denote positive constants whose value might change from line to line.

Variational tools
Standard arguments can be used to prove that problem (P f ) is variational and the related C 1 functional Since, by (f ), F : In particular, we are allowed to consider the following map which is continuously differentiable by the Implicit Function Theorem applied to ∂ Φ H where, for any Since, for every u ∈ H 1 0 (Ω), then and Moreover, we have the following estimates.
Equation (2) allows us to define on H 1 0 (Ω) the C 1 one variable functional By using standard variational arguments as those in [5], the following result can be easily proved.
, then the following propositions are equivalent: So we are led to look for critical points of I q . To this end, we need to investigate the compactness property of its Palais-Smale sequences. It is easy to see that the standard arguments used to prove boundedness do not work. Indeed, assuming that (u n ) n ∈ (H 1 0 (Ω)) N is a Palais-Smale sequence, namely (I q (u n )) n is bounded and I ′ q (u n ) → 0 in H −1 , we obtain the following inequality Since, by (1), In the classical SPE (ε = 1 and f (t) = t) we should deduce the boundedness of the sequence (u n ) n for p 3. In our general situation we need a different approach. Let T > 0 and χ : [0, +∞[→ [0, 1] be a smooth function such that χ ′ L ∞ 2 and We define a new functional I T q : H 1 0 (Ω) → R as follows for all u ∈ H 1 0 (Ω). We are going to find a critical point u ∈ H 1 0 (Ω) of this new functional such that u H 1 0 T in order to get solutions of our problem.

Proof of Theorem 1.1
We prove that the functional I T q satisfies Mountain Pass geometrical assumptions. More precisely, we have the following result. (ii) there exist constants ρ, α > 0 such that (iii) there exists a functionū ∈ H 1 0 (Ω) with ū H 1 0 > ρ such that I T q (ū) < 0.
If ε = −1, by using (5) and the immersion of H 1 0 (Ω) into L p (Ω) spaces, we get Thus, if q is such that q 2 C 1 < 1 and ρ is small enough, there exists α > 0 such that I T q (u) α for all u ∈ H 1 0 (Ω) with u H 1 0 = ρ. To prove (iii), let us consider u ∈ H 1 0 (Ω), u = 0 and t > 2T Then, we have and so, for t large enough, I T q (tu) is negative.
Thus we can complete the proof of Theorem 1.1.
where Γ = {γ ∈ C([0, 1], H 1 0 (Ω)) | γ(0) = 0, I T q (γ(1)) < 0}. Certainly there exists a Palais-Smale sequence at mountain pass level m T q , that is a sequence (u n ) n in H 1 0 (Ω) such that and As a first step we prove that there existsT > 0 andq > 0 such that for any 0 < q q there exists a Palais-Smale sequence (u n ) n of I q at the level mT q such that, up to a subsequence, u n H 1 0 (Ω) T for any n ∈ N. Let T > 0, q > 0 and (u n ) n in H 1 0 (Ω) be a Palais-Smale sequence of I T q at level m T q . We make some preliminary computations. The first is an estimate on the mountain pass level m T q . Let u ∈ H 1 0 (Ω), u = 0 andt > 0 be such that the pathγ(t) = ttu belongs to Γ. For all t ∈ [0, 1] it is From ( Then, from (10) we get m T q C + q 2 (C 5 T 2 + C 6 T 10 ).
From (9) we deduce that Since the functional is identically equal to 0, we have that On the other hand, multiplying the second equation of (P f ) by Φ ′ u [u] and integrating, we have that We deduce that B n = C n for any n ∈ N. By using (11) and (12) we get where For (f ) and (5) we have also the following estimate max (|A n |, |B n |, |D n |) q 2 (C 1 T 2 + C 2 T 10 ).
We show that, if T is sufficiently large, then lim sup n u n H 1 0 T . By contradiction, we will assume that there exists a subsequence (relabeled (u n ) n ) such that for all n ∈ N we have u n H 1 0 > T . By our contradiction hypothesis, (13) and (14), we obtain, for n large enough, with σ > 0 small. If T 2 − σT > C, we can findq such that for any q q the previous inequality turns out to be a contradiction. The contradiction arises from the assumption that lim sup n u n H 1 0 T . So we have that the sequence (u n ) n possesses a subsequence which is bounded in the H 1 0 (Ω) norm by T and such that, for every n ∈ N, I T q (u n ) coincides with I q (u n ). The last step is to prove that there existsq > 0 such that for any 0 < q q there exists a Palais-Smale sequence (u n ) n of I q which is, up to a subsequence, weakly convergent to a nontrivial critical point of LetT andq be given by the first step, and consider any 0 < q q. We know that there exists a Palais-Smale sequence of the functional I q at the level m q := mT q , such that T , for any n ∈ N. (15) Up to subsequences, there exist u 0 ∈ H 1 0 (Ω) and Φ 0 ∈ H 1 0 (Ω) such that By (f ) and (16) we also have Now we show that Φ 0 = Φ u0 and that (u 0 , Φ 0 ) is a weak nontrivial solution of (P f ). Let us consider a test function ψ ∈ C ∞ 0 (Ω). From the second equation of our problem we obtain Passing to the limit and using (17) and (18), we have that So Φ 0 is a weak solution of −∆Φ = 2qF (u 0 ), and then, by uniqueness, it is Φ 0 = Φ u0 . Since (u n ) n is a Palais-Smale sequence, for any ψ ∈ C ∞ 0 (Ω) we obtain that Passing to the limit, by (16) and (19) we have is a weak solution of (P f ). It remains to prove that u 0 = 0. Assume by contradiction that u 0 = 0. By compactness we obtain that u n → 0 in L p+1 (Ω). On the other hand, since I ′ q (u n ), u n → 0, we deduce that, up to subsequences, Of course l q > 0. Otherwise from (20) we would deduce that u n → 0 in H 1 0 (Ω) and then 0 < m q = lim n I q (u n ) = 0. By (15) we also have that l q T . By using Holder inequality, Sobolev inequalities, (f ), (4) and (7) we have that Passing to the limit, by (20) we deduce that l q q 2 C 1 l 2 q + C 2 l 10 q that is 1 q 2 C 1 l q + C 2 l 9 q q 2 C 1T + C 2T 9 .
We conclude observing that the previous inequality does not hold if we take q q < min q, ( This section deals with the study of the system (P r ). We divide the proof of Theorem 1.2 in two parts, concerning respectively the existence and the non existence of a solution according to the value of r.

The existence result
Here, we suppose that 1 < r < 5. In analogy to problem (P f ), we use a variational approach finding the solutions as critical points of the C 1 functional where, for any u ∈ H 1 0 (Ω), Φ u ∈ H 1 0 (Ω) is the unique positive solution of We have the following estimates.
and for any k > 0 it is Proof. The first part of the lemma is a consequence of the fact that, since 1 < r < 5, then for any u ∈ H 1 0 (Ω) the function |u| r ∈ L 6 5 (Ω) and we can argue as in Section 2 to define the map Φ and to deduce (22). In order to prove (23), we proceed as in [18]: multiplying (21) by |u| and integrating we get

Inequality (23) follows since it is
In the following lemma we establish the compactness of the Palais-Smale sequences of the functional I q,r . Lemma 4.2. If 1 < r < 5 the functional I q,r satisfies the Palais-Smale condition.
Thus we can complete the proof of the Theorem 1.2.
Proof of Theorem 1.2. We first suppose that r is subcritical and we deal with each case separately. case 1: p + 1 2 < r < 5 We show that for any q > 0 and for any finite dimensional subspace of H 1 0 (Ω), the functional I q,r satisfies the assumptions of [4, Theorem 2.23]. By (22) and the Sobolev embedding H 1 0 (Ω) ֒→ L 6 5 r (Ω) it is I q,r (u) 1 2 Ω |∇u| 2 dx − C u 2r then assumption (I 1 ) holds since I q,r (u) α > 0 if u H 1 0 (Ω) is sufficiently small. By Lemma 4.2, also assumption (I 3 ) holds. (I 4 ) can be checked by an easy computation. In order to prove (I 7 ), we show that for any finite dimensional subspace E of H 1 0 (Ω) there exists a ball Bρ such that I q,r | E∩∂Bρ < 0. Let E be a finite dimensional subspace of H 1 0 (Ω). It is easy to see that for ρ > 0 and u ∈ H 1 0 (Ω)