Abstract and Applied Analysis

Volume 2013, Article ID 256324, 13 pages

http://dx.doi.org/10.1155/2013/256324

## Existence of Nontrivial Solutions and High Energy Solutions for a Class of Quasilinear Schrödinger Equations via the Dual-Perturbation Method

^{1}Department of Mathematics, Honghe University, Mengzi, Yunnan 661100, China^{2}Department of Mathematics, Yunnan Normal University, Kunming, Yunnan 650092, China

Received 23 June 2013; Accepted 12 September 2013

Academic Editor: Mihai Mihǎilescu

Copyright © 2013 Yu Chen and Xian Wu. 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 quasilinear Schrödinger equation of the form , . Under appropriate assumptions on and , existence results of nontrivial solutions and high energy solutions are obtained by the dual-perturbation method.

#### 1. Introduction and Preliminaries

In this paper we consider the quasilinear Schrödinger equation of the form where and . Solutions of (1) are standing waves of the following quasilinear Schrödinger equation: where is a given potential, is a real constant, and and are real functions. The quasilinear Schrödinger equations (2) are derived as models of several physical phenomena; for example, see [1–5]. Several methods can be used to solve (1). For instance, the existence of a positive ground state solution has been proved in [6, 7] by using a constrained minimization argument; the problem is transformed to a semilinear one in [8–11] by a change of variables (dual approach); Nehari method is used to get the existence results of ground state solutions in [12, 13].

Recently, some new methods have been applied to these equations. In [14], the authors prove that the critical points are functions by the Moser’s iteration; then the existence of multibump type solutions is constructed for this class of quasilinear Schrödinger equations. In [15], by analysing the behavior of the solutions for subcritical case, the authors pass to the limit as the exponent approaches to the critical exponent in order to establish the existence of both one-sign and nodal ground state solutions. Another new method which works for these equations is perturbations. In [16] 4-Laplacian perturbations are led into these equations; then high energy solutions are obtained on bounded smooth domain.

In this paper, the perturbation, combined with dual approach, is applied to search the existence of nontrivial solution and sequence of high energy solutions of (1) on the whole space . For simplicity we call this method the dual-perturbation method.

We need the following several notations. Let be the collection of smooth functions with compact support. Let with the inner product and the norm Let the following assumption hold: satisfies and .

Set with the inner product and the norm Then both and are Hilbert spaces.

By the continuity of the for we know that, for each , there exists constant such that where denotes the -norm. In the following, we use or to denote various positive constants. Moreover, we need the following assumptions: there if and if such that uniformly in , there exist and such that for all and , where .

By Lemma 3.4 in [17] we know that, under the assumption , the embedding is compact for each .

Equation (1) is the Euler-Lagrange equation of the energy functional where . Due to the presence of the term , is not well defined in . To overcome this difficulty, a dual approach is used in [9, 10]. Following the idea from these papers, let be defined by on , and on . Then has the following properties: is uniquely defined function and invertible; for all ; for all ;;, ; for all and for all ; for all ; the function is strictly convex; there exists a positive such that there exist positive constants and such that for all ; for all ; for each , there exists such that .

The properties have been proved in [8–11]. It suffices to prove .

Indeed, by , , and , there exist and such that, for , and for , Since there exists a such that (see [10]), we can assume that . For , we have , and hence for , one has , and hence and for , there exist and such that and . Then we have Hence , where .

After the change of variable, can be reduced to From [8, 9, 11] we know that if is a critical point of , that is, for all , then is a weak solution of (1). Particularly, if is a critical point of , then is a classical solution of (1).

A sequence is called a Cerami sequence of if is bounded and in . We say that satisfies the Cerami condition if every Cerami sequence possesses a convergent subsequence.

#### 2. Some Lemmas

Consider the following perturbation functional defined by where . We have the following lemmas.

Lemma 1. * If assumptions , , and hold, then the functional is well defined on and .*

*Proof. * By conditions and , the properties , , , and imply that there exists such that
Hence
for all . By (26) and the continuity of the embedding (),
Hence is well defined in .

Now, we prove that . It suffices to prove that

For any and , by the mean value theorem, (25) and -, we have
The Hölder inequality implies that
Hence, by the Lebesgue theorem, we have
for all . Now, we show that , , are continuous. Indeed, if in , then in for all .

On the space , we define the norm
Then
Moreover, on the space , we define the norm
By (25), we have
where and . Then Theorem A.4 in [18] implies
as . If with and , one has
Hence
and hence
as . Therefore, .

Define
with the norm . On the space , we define the norm
On the space , we define the norm
From in , one has and
as . Since , by the following Lemma 2, we have
If with and , one has
Hence
and hence
as . Therefore, . This completes the proof.

Lemma 2. * Assume that , and
**
Then, for every , , and the operator
**
is continuous. *

*Proof. * Let be a smooth cut-off function such that for and for . Define
We can assume that . Hence
for all . Assume in . Then in and in . As in the proof of Lemma A.1 in [18], there exists a subsequence of and such that and for a.e. . Hence, from (51), one has
a.e. on . It follows from the Lebesgue theorem that in . Consequently, in . Similarly, we can prove in . Since
it follows that in . This completes the proof.

Lemma 3. *Let , , and hold. Then every bounded sequence with possesses a convergent subsequence.*

*Proof. * Since is bounded, then, by the compactness of the embedding (), passing to a subsequence, one has in , in for all , and for a.e. . By (25)
as . Similarly, as . Hence, by the property of , we have
where as . This shows that as . This completes the proof.

The following Lemma 4 has been proved in [10] (see Proposition 2.1(3) in [10]).

Lemma 4. *If a.e. in and , then as .*

#### 3. Main Results

Theorem 5. *Assume conditions , hold. Let be such that . Let be a critical point of with for some constant independent of . Then, up to subsequence, one has in , and is a critical point of . *

*Proof. * By , for , there exists such that
By , for ( is the constant appearing in condition ), we have
where is the constant appearing in condition . Hence
Since , there exists such that
for all . Hence
Since is a critical point of ,
for all . Consequently, taking , by and we have
and hence
for some constant independent of . By the boundedness of , there exists such that
for all . Hence, by the Sobolev embedding theorem, one has

Next, we prove that and , where the positive constant is independent of . Setting , , define , where is a smooth function satisfying for , ; for , and is decreasing in .

This means that , for ; , for ; , for , where .

Let ; then . By (61) . Hence
where
For , . By the properties of and , the mean value theorem implies
Hence
Consequently,
Combining (67) and (68), we have
For any , by and , there exists such that
Combining (66), (72), and (73), one has
By the Hölder inequality and (65),
Moreover,
HenceSince , . Set . Then
Take such that . Since , . Hence, from (65), we have
Since as , taking in (78) with , we have
Set . Then
Inductively, we have
where , and
is convergent as . Let . Then as . Hence
Let ; by (65), we have
Hence, by and (85), we have
By (63) we know that is bounded, and hence is bounded in . Up to subsequence, one has in , in for , and a.e. .

Now, we show that is a critical point of . For any with , by (85), we know that . Take as the test function in (61); we have
By