#### Abstract

In this paper, we apply the method of the Nehari manifold to study the fractional differential equation , a.e. and where are the left and right Riemann-Liouville fractional integrals of order , respectively. We prove the existence of a ground state solution of the boundary value problem.

#### 1. Introduction

Fractional differential equations have played an important role in many fields such as engineering, science, electrical circuits, diffusion, and applied mathematics (see [1–4]). In recent years, some authors have studied the fractional differential equation by using different methods, such as fixed point theorem, coincidence degree theory, and critical point theory (see [5–22]).

By using the Mountain Pass theorem, Jiao and Zhou [17] studied the existence of solutions for the following boundary value problem: where , and are the left and right Riemann-Liouville fractional integrals of order , respectively, , and is the gradient of with respect to .

By using a critical-points theorem established by G. Bonanno, Bai [19] investigated the following fractional boundary value problem: where , and are the left and right Riemann-Liouville fractional integrals of order , and and are the left and right Caputo fractional derivatives of order .

The authors in [18, 20–22] further studied the existence and multiplicity of solutions for the related problems by critical point theory.

We find that the method of Nehari manifold is seldom used in the above boundary value problem. Inspired by the results in [16–22], we would like to investigate the ground state solution for the following fractional boundary value problem: where , and are the left and right Riemann-Liouville fractional integrals of order , respectively. The technical tool is the method of Nehari manifold. (see [23, 24]).

This paper is organized as follows. In Section 2, some preliminaries on the fractional calculus are presented. In Section 3, we set up the variational framework of problem (3) and give some necessary lemmas. Finally, Section 4 presents the main result and its proof.

#### 2. Preliminaries on the Fractional Calculus

In this section, we will introduce some notations, definitions, and preliminary facts on fractional calculus which are used throughout this paper.

*Definition 1 (left and right Riemann-Liouville fractional integrals). *Let be a function defined on . The left and right Riemann-Liouville fractional integrals of order for function denoted by and function, respectively, are defined by
provided that the right-hand side integral is pointwise defined on .

*Definition 2 (left and right Riemann-Liouville fractional derivatives). *Let be a function defined by . The left and right Riemann-Liouville fractional derivatives of order for function denoted by and function, respectively, are defined by
provided that the right-hand side integral is pointwise defined on .

*Definition 3 (left and right Caputo fractional derivatives). *If and , then the left and right Caputo fractional derivatives of order for function denoted by and function, respectively, are defined by
respectively, where .

Lemma 4 (see [18]). *The left and right Riemann-Liouville fractional integral operators have the property of a semigroup; that is,
**
provided that , and , , or , , .*

Lemma 5 (see [18]). *Assume that and . Then,
**
for .*

Lemma 6 (see [18]). *Assume that . Then,
*

#### 3. A Variational Setting

To apply critical point theory for the existence of solutions for boundary value problem (3), we shall state some basic notations and results [18], which will be used in the proof of our main results.

Throughout this paper, we denote and assume that the following conditions are satisfied.. for every .There are constants and such that for every and .There are constants such that for all and . Here, The map is increasing on , for every and .

Now we construct appropriate function spaces. Denote by the set of all functions with . The fractional derivative space is defined by the closure of with respect to the norm where is the -order left Caputo fractional derivative.

*Remark 7. *If , we define , with respect to the norm
The set is a reflexive and separable Hilbert space.

*Remark 8. *For any , noting the fact , we have .

Lemma 9 (see [17]). *Let and . For all , one has
**
Moreover, if and , then
*

According to (15), we can consider with respect to the equivalent norm

Lemma 10 (see [17]). *Let and . Assume that and the sequence converges weakly to in ; that is, . Then in ; that is, as .*

Similar to the proof of [17, Proposition 4.1], we have the following property.

Lemma 11. *If , for any , one has
*

To obtain a weak solution of boundary value problem (3), we assume that is a sufficiently smooth solution of (3). Multiplying (3) by an arbitrary , we have

Observe that As , we have Then (19) is equivalent to Since (22) is well defined for , the weak solution of (3) can be defined as follows.

*Definition 12. *A weak solution of (3) is a function such that
for every .

We consider the functional , defined by From Theorem 4.1 of [17], we can get that if , then the functional is continuously differentiable on . Since is continuously differentiable on , then for . Hence, a critical point of is a weak solution of problem (3).

#### 4. Main Result

In order to study the solvability of boundary value problem (3), we use the so-called Nehari method. Define where .

There is one-to-one correspondence between the critical points of and weak solutions of boundary value problem (3). Now, we define Then we know any nonzero critical point of must be on . Define

Lemma 13. *Assume the hypotheses ()–() hold. If is a critical point of , then .*

*Proof. *For , together with ,
If is a critical point of , there exists a Lagrange multiplier , such that . Then we have
By (29), , and we have . So we can get that . The proof is complete.

Lemma 14. *Assume the hypotheses ()–() hold. For any , there is a unique such that and one has .*

*Proof. *First, we claim that there exist constants , such that for all and for all . That is, is a strict local minimizer of . In fact, by we can get that
Then together with Lemmas 9 and 11, we have
Choose such that ; then
Choose , such that . Then we have . Let ; then we get that there exist constants , such that for all and for all .

Next, we claim that , as . In fact, by , there exists a constant such that for . On the other hand, we can easily get that there exists a constant such that for . Then together with Lemma 11, we have
Since , we can get that , as .

Let for . From what we have proved, there has at least one such that
We prove next has a unique critical point for . Consider a critical point
Then, together with , we have
So we know that if is a critical point of , then it must be a strict local maximum. This implies the uniqueness.

Finally, from
we see is a critical point if . Define . Then we can get that . The proof is complete.

Lemma 15. *Assume the hypotheses ()–() hold and . Then there exists such that .*

*Proof. *We claim that both and are weakly lower semicontinuous. In fact, according to Lemma 10, if in , then in . Therefore, a.e. . By the Lebesgue dominated convergence theorem, we have , which means that the functional is weakly continuous on . Similarly is weakly continuous on . Since is Hilbet space, together with (17) and Lemma 11, we can easily get that is weakly lower semicontinuous on . Then we get that both and are weakly lower semicontinuous.

Since is continuous for and , there exists , such that
Together with , we get
Let be a minimizing sequence; that is, , as . Then,
By and , we get that is bounded in . Since is a reflexive space, going to a subsequence if necessary, we may assume that in . Then from Lemma 10, in . Since is weakly lower semicontinuous and , we first have
Then we have . In fact, if , then in . By , we get . This is a contradiction with .

Then from Lemma 14, there exists a unique such that . Together with which is weakly lower semicontinuous, we have
Then we get that is achieved at . The proof is complete.

Theorem 16. *Assuming the hypotheses ()–() hold, boundary value problem (3) has a weak solution such that ; that is, boundary value problem (3) has a ground state solution.*

*Proof. *By the Lemmas 14 and 15, we can get that there exists such that . Then the is a critical point of . From Lemma 13 we have . So boundary value problem (3) has a weak solution such that . The proof is complete.

#### 5. Example

In this section, we give an example to illustrate our results.

*Example 1. *Consider the following BVP:
It is easy to verify all the conditions in Theorem 16, so BVP (44) has a ground state solution.

#### Conflict of Interests

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

#### Acknowledgment

This research was supported by the Fundamental Research Funds for the Central Universities (2013XK03).