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Abstract and Applied Analysis
Volume 2013, Article ID 398164, 11 pages
Research Article

Existence of Prescribed -Norm Solutions for a Class of Schrödinger-Poisson Equation

1Department of Mathematics, Soochow University, Suzhou, Jiangsu 215006, China
2Cumt College of Sciences, China University of Mining and Technology, Xuzhou, Jiangsu 221116, China

Received 26 May 2013; Accepted 15 July 2013

Academic Editor: Kanishka Perera

Copyright © 2013 Yisheng Huang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


By using the standard scaling arguments, we show that the infimum of the following minimization problem: −   can be achieved for and small, where . Moreover, the properties of and the associated Lagrange multiplier are also given if .

1. Introduction

In this paper, we consider the nonlinear Schrödinger-Poisson type equation: whereis a parameter,, anddenotes the convolution. Problems like (1) have attracted considerable attentions recently since a pair, solution of (1), corresponds to a solitary wave of the formof the evolution equation: which was obtained by approximation of a special case of Hatree-Fock equation with the frequencydescribing a quantum mechanical system of many particles. For more mathematical and physical background of (2), we refer to [14] and the references therein.

In the case that the frequencyis a fixed and assigned parameter, the critical points of the following functional defined in: are the solutions of (1), whereis obviously well defined and is afunctional for each(cf. [5]). Such case has been extensively studied by using variational methods in the past decades including the existence, nonexistence, and multiplicity of solutions; see, for example, [512] and the references therein.

On the other hand, the physicists are often interested in the solutions with prescribed-norm and unknown frequency, such a solution is called a “normalized solution,” which is associated with the existence of stable standing waves. Precisely, by a “normalized solution”of (1), we mean that Clearly, this kind of solutions can be obtained as the constrained critical points of thefunctional on the constraint Thus, the frequencycannot be fixed any longer and it will appear as a Lagrange multiplier associated with the critical pointon. Among all the critical points ofconstrained on, we are interested in the ones with minimal energy since the corresponded standing waves are orbitally stable under the flow of (2) and can provide us some information on the dynamics of (2). Therefore, we are reduced to study the minimization problem for. Here we note that, for each,if, andif(cf. [13, Remark 1.1] or (15) below). When(now), by using a mountain pass argument, it was proved in [14] thathas a critical point constrained onat a strictly positive energy level forsmall, and this critical point is orbitally unstable.

The main difficulty of considering (7) is the lack of compactness for the (bounded) minimizing sequence. We recall that the necessary and sufficient condition due to Lions [15, 16] in order that any minimizing sequence for (7) is relatively compact is the strong subadditivity inequality:

In the range, by using the standard scaling arguments, Bellazzini and Siciliano in [17] proved that (8) holds forlarge. In the range, Bellazzini and Siciliano also showed in [18] that (8) holds forsmall, where they developed a new abstract theorem which guarantees the following conditionforsmall:

(MD) The functionis monotone decreasing.

We remark that their abstract theorem heavily relies on the behavior ofnear zero; that is, to use the abstract theorem, one has to verify some extra conditions, such as these are unnecessary if one can show (8) by using the standard scaling arguments like [17]. However, as mentioned in [18], the authors were not sure whether (8) can be proved or not by using the standard scaling arguments if. Therefore, the first aim of this paper is to show that (8) holds forsmall whenby using the standard scaling arguments. To achieve this aim, we introduce a new subsetof(see details in Section 3), then we consider the minimization problem (7) constrained oninstead of, and we use the standard scaling arguments to prove that (8) holds forsmall. Moreover, we can get a specific estimate onthat allows us to obtain the sign and the behavior of the Lagrange multiplierif; these are not considered in [18].

The other aim of this paper is to study the properties of the Lagrange multiplierand the ratiocorresponding to the solutionof (1) with. It is known thatandare interpreted in physics as the frequency and the ratio between the infimum of the energy of the standing waves with fixed charge and the charge itself, respectively, and the relevance of the energy/charge ratio for the existence of standing waves in field theories has been discussed under a general framework in [19].

Our main results read as follows.

Theorem 1. All the minimizing sequences for (7) are precompact inup to translations provided that one of the following conditions holds (1)and, whereis defined by (12); (2)andfor some. In particular, (1) has a solutionsuch thatand. Moreover, if the above assumption (1) holds andis a solution of (1) withand, then,andas, respectively.

Theorem 2. Letand let . Ifis a solution of (1) with, then we have (i),,asand there exists a positive constant, independent of, such that; (ii)there exists a positive constant, independent of, such that. In particular, if, then.

Remarks. (a) We point out that parts of Theorem 1 are already contained in [18, Theorem 4.1]. In the proof of Theorem 1, within hand, we can obtain some additional information of the Lagrange multiplierand the ratiowhen, and these are not contained in [18, Theorem 4.1]. However, we do not know whetheris optimal or not.

(b) Theorem 2shows that (1) has only the zero solution ifand. In the case of, it was shown in [5, 20] (see also [13, Remark 1.4]) that there existssuch that (1) has only the zero solution for. The nonexistence results of nonzero solutions of (1) were also discussed in [13] for.

(c) As we have anticipated, the existence of minimizers foris related to the existence and stability of the standing wave solutions to (2). For the existence of stable standing wave solutions to (2), we refer to [4, 14, 17, 18, 20, 21] and the references therein.

This paper is organized as follows. In Section 2, we give some preliminaries. Section 3 is devoted to the proof of the main theorems, especially in the proof of Theorem 1, we first define a new subset of and then analyze the properties of minimizing sequences forconstrained on the new subset, and finally, we prove that (8) holds whenand, respectively.

2. Preliminaries

Throughout this paper, all the functions, unless otherwise stated, are complex valued, but for simplicity we will write,anddefined in the following: (i)is the usual Lebesgue space endowed with the norm, where; (ii)is the usual Sobolev space endowed with the norm (iii)is the completion ofwith respect to the norm (iv)is the best Sobolev imbedding constant ofdefined as Moreover, the letterwill denote a suitable positive constant, whose value may change in the same line, and the symboldenotes a quantity which goes to zero. We also useto denote a bounded quantity.

Let, and then, for each,is the unique solution of the Poisson equation and is usually interpreted as the Coulombian potential of the electrostatic field generated by the charge density. Evidently, see, for example [5], For each, letand, and we have, that is,. Let it is clear thatfor allsinceas.

We now recall an abstract result on the constrained minimization problem where,,is assumed, and for some functional.

Lemma 3 (see [17, 18]). Let. Letandbe a minimizing sequence forweakly convergent, up to translations, to a nonzero function. Assume that (8) holds and that Then. In particular it follows thatand.

As pointed out in [18], Lemma 3 is a variant of the concentration-compactness principle of Lions [15, 16]. Assumption (18) shows thatpossesses the Brizis-Lieb splitting property and (19) is the homogeneity of. If, in addition, the condition (8) holds, then one can show that dichotomy does not occur; that is,. Furthermore, if (20) and (21) are also fulfilled, thenstrongly converges toin. Finally we recall the following results obtained in [17, 18].

Lemma 4 (see [18]). If, thenfor all.

Lemma 5 (see [17, Lemma ]). If, then, for every, the functionalis bounded below and coercive on.

Remark 6. For, it follows from Lemmas 4 and 5 that each minimizing sequence foris bounded from below and above by two positive constants inand, up to a subsequence, respectively.

3. Proof of the Main Theorems

Before proving our main theorems, we need some preliminary lemmas. First, we set wherewithandis a functional ondefined as It was shown in [13, Lemma 2.1] that ifis a constrained critical point ofonassociated with the Lagrange multiplierthen, which is nothing but a linear combination of(recall thatis given by (3)) and the following Pohozaev identity for (1) (cf. [5, 9]) The following lemma shows thatis well defined.

Lemma 7. Letand let . For eachwith, there exists a uniquesuch that; moreover,.

Proof. We divide the proof into two cases.
Case   1 . Let, for simplicity, and we will write,and, the derivatives ofon, instead of,and. From (15), we have Noting thatsince, then, by (25),and; thus there existssuch that. If there exists anothersuch that, without loss of generality, we may assume that, and then we get a contradiction. Therefore,is unique and it is clear that. Moreover,because of.
Case  2  . By Lemma 4, we know that the set. Let, iffor all; that is,is strictly increasing, then we obtain thatfor all. However, it is easy to see that; this is a contradiction. On the other hand, we know thatas; hence there is asuch that,and Next, we will show thatis unique. Arguing by contradiction, suppose that there is anothersuch that, without loss of generality, we may assume that, and then we have According to (28), there existssuch that. After a simple calculation, we get Ifthen, by (29),for all, which contradicts. If, then, by (30),for all. Noting that, we have again a contradiction. Therefore,is unique.

Lemma 8. Letand. For eachsuch thatasandfor all, there exists a bounded sequencesuch thatandaswithfor all; that is,is also a minimizing sequence forconstrained on.

Proof. It follows from Lemma 7 that, for each, there existssuch thatand; therefore, we have as, that is,is a minimizing sequence. Next, we will show thatis bounded. Indeed, from Remark 6,andare bounded from below and above by two positive constants inand, respectively. Noting that; therefore,is bounded from below and above by two positive constants.

Remark 9. Thanks to the Lemma 8, we know that, and, in the following, we will consider the minimization problem (7) restricted toinstead of. By Lemmas 4 and 8, for each, ifsatisfyingas, then, up to a subsequence, we may assume that. It follows from Lemma 5 thatis bounded in; by the results of [17, 18], we may assume thatasin.
The following estimates of the elements ofare crucial to proving the strong subadditivity inequality (8).

Lemma 10. Letand. For each, it holds

Proof. Since, Noting that(see (13)), by using the Hölder inequality, we get which implies that On the other hand, we have this concludes the proof of this lemma.

Remark 11. Let. It was shown in [13, Theorem 1.1] thatif and only if, where the positive numberis defined as Therefore, after a simple calculation, we can show that both of Lemmas 7 and 10 hold ifand.
Motivated by [17], we will use the standard scaling arguments to prove that the strong subadditivity inequality (8) holds for. First, we consider the case of.

Lemma 12. For, let Then

Proof. By Lemma 8 and Remark 9, for eachsatisfyingas, we may assume that, for all,, which implies that Noting that(), we have where We calculate the derivative ofon: Letting, we see from (14) that Furthermore,
Now we divide the value of into two cases to discuss.
Case  1 . It follows from Lemma 10, (14), and the Hölder inequality that
Case  2 . Again by Lemma 10, (14), and the Hölder inequality, we have Let Then by (47), (48), and (49), we know that, for each, there holds for all. On the other hand, for each, it follows from (41), (44), (45), (46), and Lemma 10 that, for alland all, This, together with the mean value theorem and (41), yields that for alland all, whereanddepend only on,and. By (42), we have then Clearly,(cf. (50)) is strictly increasing on, and thenis strictly decreasing onsince.
Let For each, let without loss of generality, we may assume that. Choosing, then by (50) we know that. (a)If, then by (54) (b)If, then there existssuch that. Therefore
It follows from (54) that Combining the above cases (a) and (b), we can show that Thus the conclusion of this lemma holds.

Remark 13. For the case of, it has been proved in [4, 17] that the strong subadditivity inequality (8) holds forsmall. By using the result of [17], we can give a specific estimate of lower bound ofsuch that (8) holds; that is, (8) holds for all. However, if we pluginto (49), then we have, which coincides with the one given in [17].

Next, we will show (8) for. We point out that the case ofis quite different from the case ofsince the inequality (48) does not hold anymore. Inspired by [18], we will give some estimates forin Lemmas 14 and 15, and these are crucial for the proof of (8) if.

Lemma 14. Letandbe fixed. If there existssuch thatand then there exist positive constantsanddependent onand, such that

Proof. From the assumptions of the lemma, we see that By (60), (62), and (63), we deduce that Combining (62) and (64), and using Lemma 10, we also obtain For each, let, we have. It follows from (60), (64), and (65) that Set, thensinceandis a fixed positive constant. From the above inequality, we see that for some positive constantsanddepending onand.

Lemma 15. Suppose thatandsatisfyingandfor all. Then there exists a positive constantdependent on, such that

Proof. Following the line of the proof of Lemma 14, we arrive that which, together with (14) and the Höder inequality, implies that and then Combining (62), (69), and (71), we have and this completes the proof.

Lemma 16. If, then there exists a positive constantsuch that

Proof. Suppose thatandsatisfyingas. It follows from Remark 9 that, up to a subsequence,for all. By Lemma 5, it is easy to see thatis bounded in. Noting that, then, by (42), we have whereis given by (43). Obviously, sinceand (34) holds. Moreover, for alland.
We claim that there existssuch that for eachand for eachsatisfyingandas, we have
Indeed, if not, we can findandsuch thatasand for each,as, but. For, there existssuch that, and it can be deduced from Lemma 14 that whereandare positive constants dependent onand. On the other hand, we know that for eachthere existssuch that. Then by Lemma 15, we obtain whereis a positive constant depending only on. Noting that (78) holds for all, by (78) and (79), we deduce that which is a contradiction forlarge sinceimplies Thus we have shown the claim. Now for eachand for allwithandas, using (77), we have By (76), similarly as in the proofs of (45) and (51), we get that Now, we can chooseso small that there exists a positive constantdependent on,and, such that Since, for each,, it follows that, for each, where, namely,for all. Thus we complete the proof of this lemma by using the arguments in the proof of Lemma 12.

Lemma 17. Let. Assume thatis a solution of (1) with. (a)If, then. (b)Ifand, then

Proof. Sinceis a solution of (1), it follows that Thus, from (87) and (88), after a simple calculation, we have which yields thatholds. Moreover, ifand, then (89) implies that Thus we get from (88) that </