EXISTENCE AND CONCENTRATION OF POSITIVE GROUND STATES FOR SCHR¨ODINGER-POISSON EQUATIONS WITH COMPETING POTENTIAL FUNCTIONS

. This article concerns the Schr¨odinger-Poisson equation


Introduction
We study the Schrödinger-Poisson equation (1.1) This equation has attracted much attention, and still is a stimulating field of research for mathematicians and for physicists.System (1.1) was first introduced in [2] and has been a study object of interest for nonlinear analysis.As a physical model, it describes a charged particle interacting with its own electrostatic field in quantum.And it can be a model to describe semiconductor theory, nonlinear optics and plasma physics.The presence of the nonlinear term f (x, u) simulates the interaction between many particles and external nonlinear perturbations.In fact, it can be described by coupling the nonlinear Schrödinger and Maxwell equations and so it is also known as the Schrödinger-Maxwell system.We refer the readers to [2] and the references therein for the physical aspects of problem (1.1).Especially, the semi-classical state solutions describe the transition from Quantum Mechanics to Newtonian Mechanics from the point of view of physics.
In recent years, (1.1) with V (x) ≡ 1 and ε = 1 has been studied under variant assumptions on f .See for example [1,3,5,7,8,9], and the references therein.In [12], the authors consider Schrödinger-Poisson equation with a non-constant potential and double parameters perturbation: (1.2) They use mountain pass to prove that (1.2) has a ground state solution which possesses the concentrating phenomenon, concentrating around global minimum of the potential V in the semi-classical limit.In [28], the authors studied the existence of positive ground state via the Nehari manifold methods.They multiply the nonlinearity by a potential b(x), that is, As for the concentration of ground state solutions, naturally, there is a competition between the linear potential V (x) and the nonlinear potential b(x), i.e., V (x) wants to attract ground state solutions to its minimum points but b(x) wants to attract ground state solutions to its maximum points.For instance in [4], a potential K(x) before non-local term was added, i.e., under suitable assumptions, for p ∈ (3,5), the authors also obtain a positive ground state solution or positive solution.For other results for this system, see for example [10,13,15,18,21,22,23,30,31,32,33] and the references therein.We should mention that for the equation ) can be written as So in [33], the authors consider the system where 2 < p < 6, the potential V can be sign-changing.Under suitable conditions, they show some concentrations when λ → ∞.See [22] for the generalized extensible beam equations.Motivated by above works, and [29], we study the existence and concentration of ground states to (1.1) with competing potentials.More precisely, we are concerned with the Schrödinger-Poisson equation where 3 < q < p < 5 = 2 * − 1, V , K and P are continuous and bounded positive functions.Q is continuous function and maybe change sign, even be negative.
Using the Nehari manifold and the concentration-compactness, we shall prove that the above problem admits a ground state.Furthermore, we want to prove that these ground states concentrate at a point which locates on the middle ground of the competing potential functions P (x) and Q(x) as ε → 0 + via a concentrationcompactness argument similar to [29].
The previous results of existence and concentration of Schrödinger-Poisson problems (see e.g.[12,Theorem 1.1], [27, Theorems 1.1, 1.2, 1.3], [28, Theorem 1.1]) can not be applied directly to (1.5) when Q = 0 is not a constant potential, especially, when Q is sign-changing or negative.To state our main results, we use the following assumptions: ( Denote the corresponding energy functional by I s and the corresponding least energy by It is well known that C(s) is well defined.Our main result is as follows.
(II) Let the assumptions in (I) be satisfied and let K(x) = 1.Then for ε > 0 small, (1) the positive ground state solution u ε obtained in (I) possesses at most one local (hence global) maximum point x ε in R 3 such that ( Remark 1.2.We point out that the potential K appears both in the first equation and in the second equation of (1.5) which is used in (2.5).In Appendix, we will explain why we let K(x) = 1.
This article is organized as follows.In section 2, we verify the existence of ground states.In section 3, we are devoted to prove the properties of ground states including exponential decay and concentration.

Existence of ground states
Under our assumptions, for the existence of ground states, without loss of generality, we may assume that ε = 1.Then (1.5) becomes (2.1) Let H 1 (R 3 ) denote the usual Sobolev space endowed with the standard scalar product and norm )} is a Hilbert space endowed with the standard scalar product and norm For u ∈ H 1 (R 3 ), we focus on the equation It is well known that there exists a unique φ u ∈ D 1,2 (R 3 ) such that Furthermore, we have Substituting this into (2.1),we can rewrite (2.1) as We formally formulate problem (2.3) in a variational way as For simplicity, define Proof.(1) It is easy to obtain the conclusions by the method in [7] with slight modification (see also [4]).The last splitting property can be obtained by [32,Lemma 2.1].
(2) Noting that K is positive, it is easy to check the conclusions.
(3) It follows form a direct computation, we omit it here.
In view of Lemma 2.1, we can see that (2.5) Thus, for all ϕ ∈ H 1 (R 3 ), we have We define the Nehari manifold of I, as where γ(u) = I (u), u .
(2.8) The next lemma shows that N = ∅.Lemma 2.2.Suppose that u = 0 and 3 < q < p < 2 * − 1.Then there is a unique t = t(u) > 0 such that tu ∈ N and I(ru where Then f (t) = At + Bt 3 − Ct p − Dt q , and hence f (t) > 0 for t small and f (t) < 0 for t large.Hence there is t = t(u) > 0 such that f (t) = 0. Thus which implies tu ∈ N .The uniqueness follows from the fact that the equation The next lemma is crucial for proving our results.
Lemma 2.3.There exists C > 0 such that for any u ∈ N , Proof.Since V, K and P are positive, it follows from γ(u) = 0 that The proof is complete.
According to Lemma 2.3, we can define Then c * ≥ 0. Furthermore, the following lemma shows that c * > 0.
Proof.For any u ∈ N , we have from which we obtain that u > r * > 0, which completes the proof.
Let {u n } ⊂ N be a minimizing sequence of c * , i.e.I(u n ) → c * as n → ∞.In the light of Lemma 2.3, {u n } is bounded in H 1 (R 3 ).Extracting a subsequence if necessary, we have u n u in H 1 (R 3 ) and u n (x) → u(x) a. e. in R 3 .Up to a subsequence, we have the following lemma.
R 3 |u n | p+1 dx has a positive lower bound with respect to n, that is, Proof.Suppose to the contrary R 3 |u n | p+1 dx → 0. Invoking the interpolation inequality, we obtain This is a contradiction.Now, we can assume (extracting a subsequence, if necessary) that We apply the concentration-compactness principle (see [20] or [14]) to . By the concentration-compactness lemma, up to a subsequence, there are three possibilities: 1 (compactness).For any > 0, there is a R > 0 and ρ n dx < .

(dichotomy)
. There exists a β ∈ (0, α), such that for all > 0, there is a R > 0, It is sufficient to show that vanishing and dichotomy do not occur.
Proof.According to concentration-compactness principle,we can suppose that there exists a subsequence of {ρ n }, still denote {ρ n }, β ∈ (0, 1] and {x n } ⊂ R 3 such that for each > 0, there exist r > 0, r < r n , We only need to prove β = 1 to exclude dichotomy.Actually, β = 1 is can be done by using the next lemma.
Let φ n be a cut-off function such that Lemma 2.8.lim n→∞ w n = 0.
Proof.The proof is similar to that in [19] with slight modification.Since it has sign-changing potential Q, here we give the details for completeness.By direct calculations, we obtain ) We have the splitting property In fact, by a direct computation, we obtain Therefore, (2.15) holds.So putting together (2.11)-(2.15),we obtain Let t(v n ) and t(w n ) be the positive values which maximize f (t) := I(tv n ) and I(tw n ).Firstly, we discuss the case t(v n ) ≤ t(w n ) (the other case will be treated later).In this case, (2.17) Our next aim is to find suitable bounds for the sequence {t(v n )}.We claim that there exist 0 < t < 1 < t independent of n such that t(v n ) ∈ (t, t).
In fact, we already know that Since Q is allowed to change sign or be negative, we only have where A is from (2.18) and M is large enough such that t > 1 and moreover Thus, for n large enough, we have (2.21) Case 2. B ≤ 0. Note that are bounded.We can choose M 1 > 0 independent of n, such that And let M 1 be large enough such that t > 1.For n large enough, we have For the case B = 0, it is easy to obtain a similar result.Thus, by (2.16), for all > 0, for n large enough, we obtain Taking into account (2.21), choosing a smaller > 0 if necessary, it holds that For the lower positive bound, we also need to discuss two cases.Case 1. B > 0. Take , where M comes from as in (2.20) and large enough.Note that t < 1, for any t < t, Case 2. B ≤ 0. Note that there exists a L > 0, such that We take . So that it holds Similarly to (2.23), jointly with (2.17), So, by choosing a small > 0, for n large enough, Thus we obtain the lower bound of t(v n ).For all t ∈ (0, t(v n )), noting that t(v n ) ≤ t(w n ), combining with (2.16), we have where {u n } ⊂ N and lim n→∞ I(u n ) = c * are used in the last inequality.Moreover, it is well-known that there exists a D > 0 independent of n such that , and Observing that 0 < t < t < 1 and q < p, one has where In the case t(v n ) > t(w n ), we can argue analogously to conclude that This contradicts (2.9).Evidence now allows us to conclude, that lim n→∞ w n = 0.
Proof.The proof is divided into two cases.
Case (1): {x n } is bounded.In this case, by Lemmas 2.6 and 2.7, u n → u ≡ 0 in L s (R 3 ) for all s ∈ (2, 6).Combining with Lemma 2.2, there is a unique t > 0 such that γ(tu) = 0 and hence ) by following the same method in [19, step 4].By interpolation inequalities, we obtain z n → z ≡ 0 in L s (R 3 ) for all s ∈ [2, 6).Using Lebesgue dominated convergence theorem, one has Similarly, Especially, for N (u n ), Lebesgue dominated convergence theorem can be used, so it also holds lim n→∞ Thus, similarly to case (1), The proof is complete.
In view of Lemma 2.9 and Lemma 2.10, we can define c * * = inf{I(u) : u is a nontrivial solution of (2.1)}.
The next lemma shows that the functional I satisfies the mountain pass geometry.
Proof.Since K is positive and P , Q are bounded, by the Sobolev embedding and result (2) in Lemma 2.1, we have Hence we can choose some α, ρ > 0 such that I(u) ≥ α for all u = ρ.For u ∈ H 1 (R 3 ) \ {0}, we have for t > 0 large enough, where A, B, C, D are similar to Lemma 2.2.Choose e = t 0 u for some suitable t 0 .
As a consequence of the Mountain Pass lemma without (P S) c condition, we define the constant c = inf Using the mountain pass value c and c * * * as the connections, we can prove that the minimizer u is a ground state.
Proof.The original research should be attributed to Rabinowitz (see [17,Proposition 3.11]).For the convenience of readers, we sketch the proof.
Since assume the potential functions V , P and Q are C 1 , according to [29, (i) of Lemma 2.3, and Lemma 2.4], we have the following lemma.
Lemma 3.2.The ground energy function C(s) is locally Lipschitz continuous in s ∈ R 3 .If V , K, P and Q are constant functions, then the least energy depends continuously on them.Lemma 3.3.There exists C > 0 independent with ε such that c ε ≥ C. Furthermore, lim sup In view of our assumptions on V , P and Q, it holds that inf s∈R 3 C(s) can be achieved by some s 0 .Let u 0 be a ground state of Denote the energy functional I s0 .Take a sequence Then there exists a unique θ > 0 such that θw ∈ N ε and θ → 1 as R → ∞, k → ∞ and ε → 0 + .Since Letting R → ∞ and k → ∞ in the above inequality, we have lim sup In view of Lemma 3.3, we obtain the desired conclusion.
Lemma 3.5.There exists ε * > 0 such that, for all ε ∈ (0, ε * ), there exists Proof.We assume, for the sake of contradiction, that there is a sequence By the vanishing lemma, we have combining this with the result (1) of Lemma 3.3, it follows that This is a contradiction.
Proof.We define By following the same methods in [29, Lemma 3.3], we can exclude vanishing and dichotomy.By compactness conditions, for any η > 0, there exists ρ > 0 such that By this and Sobolev embedding theorem, we complete the proof.
Proof.It is similar to that in [29,Lemma 3.5].For readers convenience, we sketch the proof.Using Fatou's lemma and (3.3), we obtain It follows from the above inequalities and Lemma 3.6 that As in [12, Lemma 3.8 and 3.9], the following lemmas hold.
By Lemma 3.9, we have w ε has a unique maximum point p ε .Then v ε has a unique maximum point p ε + y ε and u ε has a unique maximum point x ε = ε(p ε + y ε ).Thus, we have Remark 3.11.In our paper, we assume that our potentials V , P and Q are C 1 functions to ensure that C(s) is continuous.It follows that inf s∈R 3 C(s) can be achieved.Under our assumptions, the result shows that C(s) is Lipschitz continuous (see Lemma 3.2), so we gesture V , P , Q can be weaker than C 1 functions.

1 .
The existence result (I) of Theorem 1.1 follows directly from Section 2. The concentration results (1) and (2) of Theorem 1.1 follow from Lemma 3.8, the result (3) of Theorem 1.1 follows from Lemma 3.10.