Abstract

We consider the following fractional p-Laplacian equation: , where , , , is the fractional -Laplacian, and and for a.e. . has the subcritical growth but higher than ; however, the nonlinearity may change sign. If is coercive, we investigate the existence of ground state solutions for p-Laplacian equation.

1. Introduction

Consider the following nonlinear Schrödinger equation with fractional -Laplacian:where , , , and is the fractional -Laplacian. , , and satisfy the following assumptions:(), there exists a constant such that where meas denotes the Lebesgue measure in ;() for a.e. ;() is measurable, continuous in for a.e. and there are and such that where ;() as uniformly in ;() uniformly in as , where ;() is nondecreasing on .

When , (1) arises in the study of the nonlinear Fractional Schrödinger equation Problems with this type have occurred in many different fields, such as continuum mechanics, phase transition phenomena, population dynamics, and game theory, as they are the typical outcome of stochastically stabilization of Lévy processes; see [1–4].

When and , (4) reduces to be the nonlinear Schrödinger equation Using the Nehari-type monotonicity condition, Szulkin and Weth [5] obtained the existence of ground state solutions for (5). But in this paper, the Nehari manifold is usually not smooth and the Nehari-type monotonicity condition for the nonlinearity is not satisfied; then the Nehari manifold method is invalid. In this paper, we are aimed to obtain ground state solutions for (1) by the so-called non-Nehari manifold method, which is established by Tang [6, 7]. Unlike the Nehari manifold method, the main idea of our approach lies on finding a minimizing sequence for the energy functional outside the Nehari manifold by using the diagonal method.

Now, we are ready to state the main result of this paper.

Theorem 1. Suppose that and hold. Then (1) has a nontrivial ground state solution.

2. Preliminaries

In the paper, we will denote by the infinitesimal as . For the sake of simplicity, the norm of the space will be denoted by , and integrals over the whole will be written .

We define the Gagliardo seminorm by where is a measurable function. Then fractional Sobolev space is given by endowed with the norm For the basic properties of fractional Sobolev spaces, we refer the interested reader to [8]. By condition , we define the fractional Sobolev space with potential by endowed with the norm The energy functional defined by Under our hypotheses, is well defined on . It is well known that , and its derivative is given by for . It is standard to verify that the weak solutions of (1) correspond to the critical points of . Now, we review the main embedding result for the space .

Lemma 2 ([9, Lemma 1]). Under assumption , the embedding is compact for any .

In the following lemma, we will show that has Mountain Pass geometric structure.

Lemma 3. Suppose that , and hold.
(i) There is such that .
(ii) For any , there exists such that .

Proof. (i) By and , we haveBy for and (13), we have By the arbitrariness of and , we get the conclusion.
(ii) Fix ; by , we have as . Thus, there exists such that .

Now, we define the Nehari manifold by It is easy to prove that is not empty. And we have the following lemma.

Lemma 4. Suppose that and hold. Let and ; then .

Proof. By , we have ThenLet ; then . By a simple calculation, we have and for all . Thus,Let , it follows from (18) and (19) that

Lemma 5. Suppose that and hold; let ; then there exist satisfying

Proof. By (i) of Lemma 3, there exist and such that Choose such thatSince for large , Mountain Pass Lemma implies that there exists satisfyingwhere . By Lemma 4, we have Hence, by (23) and (24), we have Now, we can choose a sequence such that Let . Then, going if necessary to a subsequence, we have

3. Proof of Theorem 1

Proof of Theorem 1. In view of Lemma 5, we find a Cerami sequence satisfying (21). By (18), we haveCombining (21) and (28), for big enough, we have It follows that is bounded. Passing to a subsequence, we have in . By Lemma 2, we have in for . Then, by (13) and the Hölder inequality, we haveandIt follows from (30), (31) and Simon inequality () thatOn the other hand, by and , we haveCombining (32) and (33), we have in . Then, by , we have . By (28), Lemma 5, and Fatou’s lemma, we have This shows that and so .