Abstract and Applied Analysis

Volume 2014 (2014), Article ID 539639, 10 pages

http://dx.doi.org/10.1155/2014/539639

## Existence of Nontrivial Solutions for Periodic Schrödinger Equations with New Nonlinearities

School of Mathematical Sciences, Huaqiao University, Quanzhou 362021, China

Received 13 April 2014; Accepted 29 May 2014; Published 15 June 2014

Academic Editor: Mihai Mihǎilescu

Copyright © 2014 Shaowei Chen and Dawei Zhang. 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 study the Schrödinger equation: , where is -periodic and is -periodic in the -variables; is in a gap of the spectrum of the operator . We prove that, under some new assumptions for , this equation has a nontrivial solution. Our assumptions for the nonlinearity are very weak and greatly different from the known assumptions in the literature.

#### 1. Introduction and Statement of Results

In this paper, we consider the following Schrödinger equation:
where . For and , we assume the following. (**v**) is 1-periodic in for , 0 is in a spectral gap of , and and lie in the essential spectrum of . Denote
() is 1-periodic in for . And there exist constants and such that
where
() The limit holds uniformly for . And there exists such that
where , . () For any , , where
() There exist and such that, for every with ,
and, for every with ,

*Remark 1. *By the definitions of and , it is easy to verify that, for all ,
Together with as and , this implies that

*Remark 2. *There are many functions satisfying . We give several examples here.

*Example 1. *, , , and

*Example 2. *, , , and

*Example 3. *, , and .

A solution of (1) is called nontrivial if . Our main results are as follows.

Theorem 3. *Suppose and are satisfied. Then (1) has a nontrivial solution.*

Note that the limits and hold uniformly for .

Implying , we have the following corollary.

Corollary 4. *Suppose , , , , and are satisfied. Then (1) has a nontrivial solution.*

It is easy to verify that the condition , for every . And the assumption that as uniformly for imply and . Therefore, we have the following corollary.

Corollary 5. *Suppose , , , and are satisfied. Then (1) has a nontrivial solution.*

Semilinear Schrödinger equations with periodic coefficients have attracted much attention in recent years due to its numerous applications. One can see [1–24] and the references therein. In [2], the authors used the dual variational method to obtain a nontrivial solution of (1) with , where is an asymptotically periodic function. In [20], Troestler and Willem firstly obtained nontrivial solutions for (1) with being a function satisfying the Ambrosetti-Rabinowitz condition:(AR)there exists such that, for every , , where , , and with . Then, in [9], Kryszewski and Szulkin developed some infinite-dimensional linking theorems. Using these theorems, they improved Troestler and Willem’s results and obtained nontrivial solutions for (1) with only satisfying and the (AR) condition. These generalized linking theorems were also used by Li and Szulkin to obtain nontrivial solution for (1) under some asymptotically linear assumptions for (see [11]). In [13] (see also [14]), existence of nontrivial solutions for (1) under and the (AR) condition was also obtained by Pankov and Pflüger through approximating (1) by a sequence of equations defined in bounded domains. In the celebrated paper [17], Schechter and Zou combined a generalized linking theorem with the monotonicity methods of Jeanjean (see [8]). They obtained a nontrivial solution of (1) when exhibits the critical growth. A similar approach was applied by Szulkin and Zou to obtain homoclinic orbits of asymptotically linear Hamiltonian systems (see [19]). Moreover, in [5] (see also [6]), Ding and Lee obtained nontrivial solutions for (1) under some new superlinear assumptions on different from the classical (AR) conditions.

Our assumptions on are very weak and greatly different from the assumptions mentioned above. In fact, our assumptions do not involve the properties of at infinity. It may be asymptotically linear growth at infinity, that is, , or superlinear growth at infinity as well, that is, . Moreover, the assumptions allow in a neighborhood of (see Remark 2).

In this paper, we use the generalized linking theorem for a class of parameter-dependent functionals (see [17, Theorem 2.1] or Proposition 8 in the present paper) to obtain a sequence of approximate solutions for (1). Then, we prove that these approximate solutions are bounded in and (see Lemmas 13 and 14). Finally, using the concentration-compactness principle, we obtain a nontrivial solution of (1).

*Notation*. denotes the open ball of radius and center . For a Banach space , we denote the dual space of by and denote strong and weak convergence in by and , respectively. For , we denote the Fréchet derivative of at by . The Gateaux derivative of is denoted by , . denotes the standard space , and denotes the standard Sobolev space with norm . We use , to mean , .

#### 2. Existence of Approximate Solutions for (1)

Under the assumptions , , and , the functional is of class on , and the critical points of are weak solutions of (1).

There is a standard variational setting for the quadratic form . For the reader’s convenience, we state it here. One can consult [5] or [6] for more details.

Assume that holds, and let be the self-adjoint operator acting on with domain . By virtue of , we have the orthogonal decomposition such that is negative (resp., positive) in (resp., in ). As in [5, Section 2] (see also [6, Chapter 6.2]), let be equipped with the inner product and norm , where denotes the inner product of . From , with equivalent norms. Therefore, continuously embeds in for all if and for all if . In addition, we have the decomposition where is orthogonal with respect to both and . Therefore, for every , there is a unique decomposition with and Moreover, The functional defined by (14) can be rewritten as where

The above variational setting for the functional (14) is standard. One can consult [5] or [6] for more details.

Let be the total orthonormal sequence in . Let , be the orthogonal projections. We define on . The topology generated by is denoted by , and all topological notation related to it will include this symbol.

Lemma 6. *Suppose that holds. Then *(**a**)*, where is defined in ;*(**b**)*for any , there exists with such that .*

*Proof. *(**a**) We apply an indirect argument, and assume by contradiction that
From assumption , is in the essential spectrum of the operator (with domain ):
Then, by Weyl’s criterion (see, e.g., [25, Theorem VII.12] or [26, Theorem 7.2]), there exists a sequence with the properties that , and .

Since , we deduce that for all . Together with the facts that is a continuous periodic function and , this implies
It follows that there exists a constant such that
Note that
Together with (29) and the fact that and , this implies . It contradicts , . Therefore, .

(**b**) It suffices to prove that
From (21), we deduce that . From assumption , is in the essential spectrum of . By Weyl’s criterion, there exists such that and . Multiplying by and then integrating it into , by (20) and (22), we get that
It follows that . Multiplying by and then integrating it into , we get that
It implies that . This together with implies .

*Let and
From assumption (5), we have . Together with the result of Lemma 6, this implies that . Choose
Then by the result of Lemma 6, there exists with such that
Set
Then, is a submanifold of with boundary
*

*Definition 7. *Let . A sequence is called a Palais-Smale sequence at level (-sequence for short) for , if and as .

*The following proposition is proved in [17] (see [17, Theorem 2.1]).*

*Proposition 8. Let . The family of -functional has the form
Assume(a), ;(b) as ;(c) for all , is -sequentially upper semicontinuous; that is, if , then and is weakly sequentially continuous. Moreover, maps bounded sets to bounded sets;(d) there exist with and such that, for all , Then there exists such that the Lebesgue measure of is zero and, for every , there exist and a bounded -sequence for , where satisfies
*

*For and , let
Then
and it is easy to verify that a critical point of is a weak solution of
where
*

*Lemma 9. Suppose that and hold. Then, there exist and such that the Lebesgue measure of is zero and, for every , there exist and a bounded -sequence for , where satisfies
*

*Proof. *For , let
Then, and satisfy assumptions (a) and (b) in Proposition 8, and, by (43), .

From (43) and (20), for any and , we have
Let and be such that . It follows that , , and . In addition, up to a subsequence, we can assume that a.e. in . Then, we have
By Remark 1, for all and . This together with Fatou’s lemma implies
Then, by (49), we obtain
This implies that is -sequentially upper semicontinuous.

If in , then, for any fixed , as ,
This implies that is weakly sequentially continuous. Moreover, it is easy to see that maps bounded sets to bounded sets. Therefore, satisfies assumption (c) in Proposition 8.

Finally, we will verify assumption (d) in Proposition 8 for .

From assumption and as uniformly for , we deduce that, for any , there exists such that
From (49) and (55), we have, for ,
Then by the Sobolev inequality and (by (21) and (22)), we deduce that there exists a constant such that
Choose and such that and . Then, for every , we have
Let be such that and . Then, from (58), we deduce that, for ,

We will prove that as and . Arguing indirectly, assume that, for some sequences and with , there is such that for all . Then, setting , we have , and, up to a subsequence, , and .

First, we consider the case . Dividing both sides of (49) by , we get that

From (5) and the result of Lemma 6, we deduce that
where is defined by (34). Note that, for , we have . This implies that, when is large enough,
By (10), we have, when is large enough,
Combining the above two inequalities yields
We used the inequalities
in the second inequality of (64).

Since for some , by (36), we get that
Note that, by the choice of (see (35)), we have . Then by (64) and the fact that , we have that
It contradicts (60), since as .

Second, we consider the case . In this case, . It follows that
since and . Therefore, the right hand side of (60) is less than when is large enough. However, as , the left hand side of (60) converges to zero. It induces a contradiction.

Therefore, there exists such that
This implies that satisfies assumption (d) in Proposition 8 if . Finally, it is easy to see that
Then, the results of this lemma follow immediately from Proposition 8.

*Lemma 10. Suppose that and are satisfied. Let be fixed, where is the constant in Lemma 9. If is a bounded -sequence for with , then, for every , there exists such that, up to a subsequence, satisfies
*

*Proof. *The proof of this lemma is inspired by the proof of Lemma 3.7 in [19]. Because is a bounded sequence in , up to a subsequence, either (a) or(b)there exist and such that .

If (a) occurs, using the Lions lemma (see, e.g., [21, Lemma 1.21]), a similar argument as for the proof of [19, Lemma 3.6] shows that
It follows that
On the other hand, as is a -sequence of , we have and . It follows that
This contradicts (73). Therefore, case (a) cannot occur.

If case (b) occurs, let . For every ,
Because and are -periodic in every , is still bounded in ,
Up to a subsequence, we assume that in as . Since in , it follows from (75) that . Recall that is weakly sequentially continuous. Therefore, and, by (76), .

Finally, by and Fatou’s lemma

*Lemma 11. There exist and such that, for any , if satisfies , then .*

*Proof. *We adapt the arguments of Yang [23, p. 2626] and Liu [12, Lemma 2.2]. Note that, by and , for any , there exists such that
Let be a critical point of . Then is a solution of
Multiplying both sides of this equation by , respectively, and then integrating into , we get that
It follows that
where and are positive constants related to the Sobolev inequalities and . From the above two inequalities, we obtain
Because , this implies that for some if and are small enough and . The desired result follows.

*Let , where and are the constants that appeared in Lemmas 9 and 11, respectively. Combining Lemmas 9–11, we obtain the following lemma.*

*Lemma 12. Suppose and are satisfied. Then, there exist , , and such that ,
*

*3. A Priori Bound of Approximate Solutions and Proof of the Main Theorem*

*3. A Priori Bound of Approximate Solutions and Proof of the Main Theorem**In this section, we give a priori bound for the sequence of approximate solutions obtained in Lemma 12. We then give the proofs of Theorem 3.*

*Lemma 13. Suppose and are satisfied. Let be the sequence obtained in Lemma 12. Then, and
*

*Proof. *From , we deduce that is a weak solution of (45) with ; that is,
By assumption and the bootstrap argument of elliptic equations, we deduce that .

Multiplying both sides of (85) by and integrating into , we get that
Recall that and . Then by (5), we get that
This together with (86) yields . It follows that on .

Similarly, multiplying both sides of (85) by and integrating into , we can get that on . Therefore, for all .

*Lemma 14. Suppose that , , , , and are satisfied. Let be the sequence obtained in Lemma 12. Then
*

*Proof. *As and , Lemma 11 implies that .

To prove , we apply an indirect argument and assume by contradiction that .

Since , by (81), we get that
It follows that
Set . Then, by (90),
Then, by as , we have that

From Lemma 12,
Then, by , we obtain
From , we have
where is the constant in . As the continuous function is -periodic in every variable, we deduce from (8) that there exists a constant such that
Combining (95) and (96) leads to
Dividing both sides of this inequality by and sending , we obtain

From (7), (21), and (22), we have that
where is the constant defined in .

Since and , we deduce that there exists such that, for every with ,
This together with (98) gives
Combining (99) and (101) yields
This contradicts (92). Therefore, is bounded in .

*Proof of Theorem 3. *Let be the sequence obtained in Lemma 12. From Lemma 14, is bounded in . Therefore, up to a subsequence, either (a) or(b)there exist and such that . According to (72), if case (a) occurs,
Then, by (81) and , we have
This contradicts (see (88)). Therefore, case (a) cannot occur. As case (b) therefore occurs, satisfies . From (14) and (43), we have that
It follows that
By (see Lemma 12), we have . From (106), we have that, for any ,
Together with and , this yields
Finally, by and the weakly sequential continuity of , we have that . Therefore, is a nontrivial solution of (1). This completes the proof.

*Conflict of Interests*

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

*Acknowledgments*

*Acknowledgments**The authors would like to thank the anonymous referees for their comments and suggestions on the paper. Shaowei Chen was supported by Science Foundation of Huaqiao University and Promotion Program for Young and Middle-Aged Teacher in Science and Technology Research of Huaqiao University (ZQN-PY119).*

*References*

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