#### 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.