Existence and concentration of solutions for nonautomous Schrödinger – Poisson systems with critical growth

In this paper, we study the following Schrödinger–Poisson system { −∆u + u + μφu = λ f (x, u) + u5 in R3, −∆φ = μu2 in R3, where μ, λ > 0 are parameters and f ∈ C(R3 ×R, R). Under certain general assumptions on f (x, u), we prove the existence and concentration of solutions of the above system for each μ > 0 and λ sufficiently large. Our main result can be viewed as an extension of the results by Zhang [Nonlinear Anal. 75(2012), 6391–6401].


Introduction and main results
Consider the following Schrödinger-Poisson system −∆u + u + µφu = λ f (x, u) + u 5  in R 3 , where µ, λ > 0 are parameters and f ∈ C(R 3 × R, R).Equation (1.1) or the more general one arise from several interesting physical fields, such as in quantum electrodynamics, describing the interaction between a charged particle interacting with the electromagnetic field, and also in semiconductor theory and in plasma physics.For more details in physical background we refer to [5,8] and the references therein.

Y. Ye
There are many papers studying the existence of solutions of system (1.2), see [2-4, 7-10, 12-14, 16-22] and their references.A lot of works focus on the study of problem (1.2) with the very special case V = K = 1 and f (x, u) = |u| p−2 u, and existence and multiplicity of positive solutions as well as radial or nonradial symmetric solutions are obtained, see e.g.[2,3,[7][8][9][10]13].The Schrödinger-Poisson system with critical nonlinearity of the form −∆u + u + φu = P(x)|u| 4 u + λQ(x)|u| q−2 u in R 3 , −∆φ = u 2 in R 3 , 2 < q < 6, λ > 0, has been studied in [22].Besides some other conditions, Zhao et al. assume that P ∈ C(R 3 , R), lim |x|→∞ P(x) = P ∞ ∈ (0, +∞) and P(x) ≥ P ∞ and prove the existence of one positive solution for 4 < q < 6 and each λ > 0. It is also proven the existence of one positive solution for q = 4 and λ large enough.Zhang [18] considers the following type of Schrödinger-Poisson system where f ∈ C(R + , R + ) satisfies lim u→+∞ f (u)/u 5 = K > 0 and f (u) ≥ Ku 5 + Du q−1 for some D > 0, which exhibits a critical growth.Applying a combined technique consisting in a truncation argument and a monotonicity trick, he proves that for µ > 0 small, problem (1.3) admits a positive solution for q ∈ (2, 4] with D sufficiently large or q ∈ (4, 6).In [20], the same author studies problem (1.1) when V = 1 and f (x, u) = a(x)|u| p−2 u + λb(x)|u| q−2 u + u 5 , where p, q ∈ (4, 6), λ > 0 is a parameter.Under certain decay rate conditions on K(x), a(x) and b(x), he proves the existence of ground state solution and two nontrivial solutions for λ > 0 small.Recently, the Schödinger-Poisson system with nonconstant coefficient of the following version −∆u has been discussed in Mao et al. [12].Assuming that V is coercive, i.e.V(x) → ∞ as |x| → ∞ and f is local subcritical and 4-superlinear at the origin, the authors prove the existence of nontrivial solution and its asymptotic behavior depending on ε and λ.Motivated by the works described above, in this paper, we try to prove the existence of solutions of problem (1.1) with a much more general nonlinearity in critical growth.Precisely, we make the following hypotheses.
Theorem 1.1 can be viewed as an extension of the main results in [18].Note that, in [18], the existence of solution is obtained by using the radially symmetric Sobolev space H 1 r (R 3 ), where the embedding However, in our case since f is nonradially symmetric, we have to deal with (1.1) in H 1 (R 3 ) and the Sobolev embedding is not compact any more.Moreover, the critical exponential growth makes the problem more complicated.To overcome these difficulties, we use a truncation argument (see [11]) together with careful analysis of the (PS) c λ sequence and prove the (PS) c λ condition holds for a suitable range of c λ indirectly.

Notations
) is the usual Hilbert space endowed with the norm • S denotes the best Sobolev constant • For every 2 ≤ q < 6, denote • C and C i (i = 1, 2, . . . ) denotes various positive constants, which may vary from line to line.

Proof of Theorem 1.1
For simplicity, we assume µ = 1 and denote H = H 1 (R 3 ).We first recall the following well-known facts.
Define the functional associated to problem (1.1) where, for each . For λ sufficiently large, we will find a critical point u λ of I T such that u λ α ≤ T and so we conclude that u λ is also a critical point of I.
Lemma 2.2.The functional I T possesses a mountain pass geometry: (i) there exist constants α, ρ > 0 such that I T (u) ≥ α for all u = ρ; (ii) there exists e ∈ H such that e > ρ and I T (e) < 0.

Lemma 2.3. There is a constant D
Proof.It follows from (2.2) and (2.3) that there exists ε 1 > 0 such that for ε ∈ (0, ε 1 ), (2.4) Since F ≥ 0 for all (x, s), one sees that Thus, using (2.4), there exist t > 0 small and t > 0 large (independent of ε ∈ (0, ε 1 )) such that sup t∈[0,t ]∪[t ,+∞) where ã = c 1 2 q/2 |x|≤1 dx.Choose ε 0 ∈ (0, min 1, ε 1 , r 2 ) such that By the definition of v ε 0 (x), we get and then so that, by ( f 2 ), t q dx = ãε 0 (6−q) Combining this with (2.5) shows that Proof.It follows from (2.1) and ( f 3 ) that which implies that (u n ) n∈N is bounded in H. Thus, going if necessary to a subsequence, we may assume for each bounded domain We claim that u n → u λ in H. Take where A, B, D are nonnegative constants, and define the functionals J T , Ψ T on H by By (2.8), we see that, for any ψ ∈ C ∞ 0 (R 3 ), and where we have used Lebesgue dominated convergent theorem in the last limit.From u n → u λ a.e. in R 3 and 3 .Using the fact ) is bounded.This and the fact Similarly, we deduce that as n → ∞, which implies that J T (u λ ) = 0. Denote v n := u n − u λ .By ( f 1 ) and [23, Lemma 2.2], one obtains that Y. Ye and From the Brezis-Lieb lemma (see [6]), we have (2.17) Furthermore, by [21, Lemma 2.2], we get Hence, using (2.15)-(2.18)and the fact J T (u λ ) = 0, we deduce that and

.21)
Now we estimate the right-hand side of the above inequality.By ( f 1 ) and Young's inequality, we have that for ε > 0 small.Hence, substituting this equality into (2.21) and taking ε = , we deduce that for λ > 0 large

Y. Ye
Thus, by Lemma 2.4, one sees that u n → u λ in H, I T (u λ ) = c λ and I T (u λ ) = 0. Next we show that u λ → 0 as λ → +∞.It follows from the properties of χ and (2.23) that