#### Abstract

We discuss the existence of positive solutions to a class of fractional boundary value problem with changing sign nonlinearity and advanced arguments where is the standard Riemann-Liouville derivative, is continuous, ,, and is the advanced argument. Our analysis relies on a nonlinear alternative of Leray-Schauder type. An example is given to illustrate our results.

#### 1. Introduction

Fractional differential equations (FDEs) have been of great interest for the past three decades. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, and so forth (see [1, 2]). Therefore, the theory of FDEs has been developed very quickly. There has been a significant development in fractional differential equations in recent years; see .

In , the author studied existence of positive solutions in case of the nonlinear fractional differential equation as follows: where is the standard Riemann-Liouville fractional derivative, is continuous, and . In , the author applied the Avery-Peterson fixed point theorem to obtain sufficient conditions of the existence of multiple solutions to the following problem: where is continuous and is a nonnegative continuous function defined on .

Motivated by [5, 10], in this paper, we consider the existence of positive solution of the following boundary value problem for nonlinear fractional differential equation with changing sign nonlinearity and advanced arguments: where denotes a linear functional on given by involving a Stieltjes integral with a suitable function of bounded variation. It is important to indicate that we did not assume that is positive to all positive . The measure can be a signed measure.

Put ; let us introduce the following assumptions:(H1) is continuous, and ;(H2), and on ;(H3) may change sign; is not identically zero on any subinterval on ;(H4), where .

#### 2. Basic Definitions and Preliminaries

In this section, we present some preliminaries and lemmas that are useful to the proof of our main results. For convenience, we also present the necessary definitions from fractional calculus theory here. These definitions can be found in the recent literature.

Definition 1. The fractional integral of order of a function is given by provided that the right-hand side is pointwise defined on .

Definition 2. The fractional derivative of order of a continuous function is given by where and denotes the integral part of number , provided that the right-hand side is pointwise defined on .

Lemma 3. Let ; then where being the smallest integer greater than or equal to .

Consider the following boundary value problem:

Lemma 4. Assume that and ; then problem (7) has the unique solution given by the following formula: where

Theorem 5. Let be a Banach space with closed and convex. Assume that is a relatively open subset of with and is a continuous, compact map. Then either(i) has a fixed point in or(ii)there exist and with .

#### 3. Existence of Positive Solutions

Let us denote by the Banach space of all continuous real functions on endowed with the sup norm and let be the cone:

Lemma 6. Let assumptions (H1)–(H4) hold. Moreover, we assume that assumptions (H5)-(H6) hold with(H5),(H6) is continuous, , and there is such that where . Then, for every , there exists a positive number such that, for , the nonlinear fractional differential equation, has a positive solution with as and where

Proof. It is easy to know from (9), (H5), and (H6) that . By Lemma 4, (12) has a unique solution in : For , we define two operators and by where It is easy to show that and are completely continuous. We claim that operators and have the same fixed points in . In fact, let ; then So Let ; then . So , and hence This shows that fixed points of are solutions of (12). We will apply the nonlinear alternative of Leray-Schauder type to prove that has at least one fixed point for small .
Let be such that Suppose that , where ; then Since , there exists a unique such that
Let and be such that . We claim that . In fact, That is, , which implies that . Let . By Theorem 5, has a fixed point . Moreover, combining (21) with the expression of operator , we obtain that Hence (12) has a positive solution . Note that as ; we get that as .

Theorem 7. Suppose that (H1)–(H6) hold. Then there exists a positive number such that (3) has at least one positive solution for .

Proof. Let Then for each . We have . Choose such that . There is such that for ; then Fix , and let be such that where is given by Lemma 6, and for with .
Let . We look for a solution of the form , where is the solution of (12), given by Lemma 6. Thus solves the following equation: where .
Now, we need to prove the existence of . Consider the following equation: where Obviously, (31) is equivalent to the operator equation: It is easy to show that operator is completely continuous. Let and such that . That is, We claim that . Suppose on the contrary that . Then, by (28) and (29), we get From (27), we get Using (34)–(36), for each , we obtain that In particular, which is a contradiction. And so the claim is proved. Let . By Theorem 5, has a fixed point . Consequently, . This proves that there exists this is the solution of (30). Hence satisfies (37) and Lemma 6; then we get that is, is a positive solution of (3). So the proof of Theorem 7 is complete.

#### 4. An Example

In this section, we give an example to illustrate the result of this paper. Consider the following nonlinear fractional differential equation:

Let and . Obviously, all assumptions (H1)–(H3) hold. In the following, we will verify that assumptions (H4)–(H6) hold also.

(i) It is obvious that implies (H4).

(ii) By direct calculation, we have so assumption (H5) holds.

(iii) Finally, we check assumption (H6). It means that there exists such that . Note that We now verify that there exists such that that is, By simple calculation, we get Setting , then inequality (44) holds. Similarly, there exists such that Let . By (44)–(47), we obtain that there exists such that Thus assumption (H6) holds. By applying Theorem 7, we know that there exists a number such that (40) has at least one positive solution for .

#### Conflict of Interests

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

#### Acknowledgments

Zhaocai Hao acknowledges the support from NSFC (11371221) and the Education Department of Shandong Province Science and Technology Plan Project (J13LI01). The authors are grateful to the anonymous referees for their helpful suggestions and comments.