Abstract and Applied Analysis

Volume 2014 (2014), Article ID 607078, 7 pages

http://dx.doi.org/10.1155/2014/607078

## Nehari-Type Ground State Positive Solutions for Superlinear Asymptotically Periodic Schrödinger Equations

^{1}Department of Mathematics, Huaihua College, Huaihua, Hunan 418008, China^{2}School of Mathematics and Statistics, Central South University, Changsha, Hunan 410083, China

Received 3 January 2014; Revised 23 May 2014; Accepted 11 June 2014; Published 29 June 2014

Academic Editor: Marco Squassina

Copyright © 2014 Xiaoyan Lin and X. H. Tang. 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.

#### Abstract

We deal with the existence of Nehari-type ground state positive solutions for the nonlinear Schrödinger equation . Under a weaker Nehari condition, we establish some existence criteria to guarantee that the above problem has Nehari-type ground state solutions by using a more direct method in two cases: the periodic case and the asymptotically periodic case.

#### 1. Introduction

Consider the following semilinear Schrödinger equation: where and .

The Schrödinger equation has found a great deal of interest last years because not only it is important in applications but also it provides a good model for developing mathematical methods. Many authors have studied the existence of entire solutions of Schrödinger equations under various stipulations (cf., e.g., [1–28] and the references quoted in them).

When and is periodic, Li et al. [12] made use of a combination of the techniques in [13, 14] with applications of Lions’ concentration compactness principle [26, 29, 30] to establish the following theorem.

Theorem 1 (see [12]). *Assume that and satisfy the following assumptions:*(V0)* and ;*(V1)* is 1-periodic in each of ;*(S0)*, is a Caratheodory function, and there exists a constant such that
*(S1)* is 1-periodic in each of ;*(S2)*, as , uniformly in ;*(S3)*, uniformly in ;*(S4)* is strictly increasing in on for every .**Then problem (1) has a solution such that , where
**
with , and
*

The set is the Nehari manifold, which contains infinitely many elements of . In fact, for any , there exists such that ; see Lemma 9. Since is a solution at which has least “energy” in set , we will call it a Nehari-type ground state solution.

We must point out that “the least energy solution” (which is sometimes also called the ground state solution in some references) is in fact a nontrivial solution which satisfies , where is a very small subset of ; it may contain only one element. In general, it is much more difficult to find a solution for problem (1) with a constraint condition than with one .

To establish the existence of Nehari-type ground state solutions, the so-called Nehari-type condition (S4) seems to be always necessary in the proof of the existence of ground states solutions for problem (1).

In recent paper [15, 20], Theorem 1 has been extended to the case where is in the gap of the spectrum , but an additional assumption on the nonlinearity is assumed in [15].

Motivated by [12, 17, 20], in the present paper, we will develop a more direct method to generalize Theorem 1 by relaxing assumptions (V0), (S3), and (S4) in two cases: the periodic case and the asymptotically periodic case.

In the periodic case, we establish the following two theorems.

Theorem 2. *Assume that and satisfy (V0), (V1), (S1), (S2), and the following assumptions:*()*, and there exists a constant such that
*()*, .;*()* is nondecreasing in on for every .**Then problem (1) has a solution such that .*

Theorem 3. *Assume that and satisfy (), (), (), (), (), (), and (). Then problem (1) has a positive solution such that , where
**
with and , and
*

In the asymptotically periodic case, is allowed to be negative in some bounded domain in . Precisely, we use the following condition instead of (V0).(), for all and there exists a constant such that ()Was first introduced by Deng et al. [6]; it is satisfied if the following assumption holds (see Lemma 7):(), and there exist two constants and a bounded measurable set such that , and where .

In this case, we establish the following two theorems.

Theorem 4. *Assume that and satisfy (), (), (), (), (), and the following assumptions:*(V2)*, , is 1-periodic in each of , for , and for , ;*()*; satisfies (S1), (S2), and (); satisfies that
* *where and with .**Then problem (1) has a solution such that .*

Theorem 5. *Assume that and satisfy (), (), (), (), (), (), and (). Then problem (1) has a positive solution such that .*

In our theorems, we give a new condition () which weakens Nehari-type condition (S3) considerably.

#### 2. Preliminaries

Lemma 6 (see [6, Lemma 2.3]). *Assume that satisfies (). Then there exist two positive constants such that
**
where is the usual norm of *

By Lemma 6, we define an inner product associated with the norm Then is a Hilbert space with this inner product. Moreover, under assumptions () and (), the functional defined by (3) is of class .

Lemma 7. *If () holds, then () does.*

*Proof. *By virtue of (), the Hölder inequality, and the Sobolev inequality, we have
This shows that () holds with .

Lemma 8. *Let be a Banach space. Let be a closed subspace of the metric space and . Define
**
If satisfies
**
then there exists a sequence satisfying
*

*Proof. *For any , define set in and the collection . Let ,
For any and , let and . Then and . Hence,
that is, . Therefore,
These show that the collection is a minimax system for . Since (19) implies
it follows from Theorem 2.4 in [18] that the result is true.

Lemma 9. *Under assumptions (), (), (), and (), for any , there exists such that .*

*Proof. *Let be fixed and define the function on . Clearly we have
It is easy to verify, using (S2) and (), that , for small and for large. Therefore is achieved at a so that and .

Lemma 10. *Under assumptions (), (), and (),
*

*Proof. *For , () yields
It follows that
Note that
Thus, by (3), (28), and (29), one has
This shows that (26) holds.

Corollary 11. *Under assumptions (), (), and (), for ,
*

We define where

Lemma 12. *Under assumptions (), (), (), (), and (), one has that and there exists a sequence satisfying
*

*Proof. *(1) Both Lemma 9 and Corollary 11 imply that . Next, we prove that . By the definition of , we choose a sequence such that
Since for and large, there exist and such that
Let for . Then , and it follows from (35) and (36) that
which implies that . On the other hand, the manifold separates into two components and . By (), one has
It follows that for . By () and (S2), contains a small ball around the origin. Thus every has to cross , because and , and so .

(2) In order to prove the second part of Lemma 12, we apply Lemma 8 with , , and
By () and (S2), there exists such that
Hence we obtain
These show that all assumptions of Lemma 8 are satisfied. Therefore there exists a sequence satisfying (34).

Lemma 13. *Under assumptions (), (), (), (), and (), any sequence satisfying (34) is bounded in .*

*Proof. *To prove the boundedness of , arguing by contradiction, suppose that . Let . Then . By Sobolev embedding theorem, there exists a constant such that
Passing to a subsequence, we may assume that in , in , , and a.e. on .

If , then by Lions’ concentration compactness principle [23, Lemma 1.21], in for . Fix and . By () and (S2), for there exists such that
It follows that
Hence, by using (34), (44), and Lemma 10, one has
which is a contradiction. Thus, .

Going if necessary to a subsequence, we may assume the existence of such that . Let . Then it follows that
Now we define , then , and for some . Passing to a subsequence, we have in , in , , and a.e. on . Thus, (46) implies that . Hence, it follows from (34), (S3), and Fatou’s lemma that
which is a contradiction. Thus is bounded in .

*Remark 14. *In the proof of Lemma 13, (S3) is used only in (47). Hence, it can be weakened to () if is 1-periodic in each of .

#### 3. The Proofs of Theorems

*Proof of Theorem 2. *Lemma 12 implies the existence of a sequence satisfying (34), by a standard argument; we can prove Theorem 2.

*Proof of Theorem 3. *In view of the proofs of Lemmas 12 and 13, we can show that the conclusions of Lemmas 12 and 13 still hold if and are replaced by and , respectively. Hence, there exists a bounded sequence satisfying
where . The rest of the proof is standard, so we omit it.

To prove Theorems 4 and 5, we define functional and as follows: where . Then (), (), and () imply that and

*Proof of Theorem 4. *Lemma 12 implies the existence of a sequence satisfying (34). By Lemma 13, is bounded in . Passing to a subsequence, we have in . Next, we prove .

Arguing by contradiction, suppose that ; that is, in , and so in , , and a.e. on . For any , it follows from (V2) that there exists such that for . Hence,
Since is arbitrary, we have
Similarly, by (), one has
Note that
From (34) and (53)–(55), one has

By a standard argument, we may prove that there exists , going if necessary to a subsequence, such that
Let . Then , and
Since and are periodic, we have
Since is bounded in , passing to a subsequence, we have in , in , , and a.e. on . Obviously, (58) implies that for . By a standard argument, we can prove that and by using (59).

Since , it follows from Lemma 9 that there exists such that , and so . On the other hand, from (49), (51), (V2), (), and (), we have
This contradiction implies that . By a standard argument, we can prove that and . This shows that is a solution for problem (1) with .

*Proof of Theorem 5. *Similar to the proof of Theorem 3, there exists a bounded sequence satisfying (48). Passing to an appropriate subsequence, we have that in . Next, we prove .

Arguing by contradiction, suppose that ; that is, in . Then, in , , and a.e. on . Analogous to the proof of Theorem 4, we can demonstrate that there exists a with for such that and . By a standard argument, we can show that .

Since , it follows from Lemma 9 that there exists such that , and so . On the other hand, from (49), (51), (V2), (), and (), we have
This contradiction shows that . In the same way as the last part of the proof of Theorem 1, we can deduce that and . By a standard argument, we can demonstrate that . Therefore, is a positive solution for problem (1) with .

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

This work is partially supported by the NNSF (no. 11171351) of China.

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