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Discrete Dynamics in Nature and Society
Volume 2018, Article ID 2352789, 8 pages
Research Article

Existence Results for Fractional Differential Equations with Multistrip Riemann–Stieltjes Integral Boundary Conditions

1College of Science, China Agricultural University, Beijing 100083, China
2School of Mathematics and Physics, University of Science and Technology Beijing, Beijing 100083, China

Correspondence should be addressed to Huihui Pang; moc.361@0002hhp

Received 29 December 2017; Revised 28 February 2018; Accepted 20 March 2018; Published 24 April 2018

Academic Editor: Chris Goodrich

Copyright © 2018 Tongling Lv et al. 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.


This paper is concerned with the existence and multiplicity of the positive solutions for a fractional boundary value problem with multistrip Riemann–Stieltjes integral boundary conditions. Our results are based on the Leggett–Williams fixed point theorem. In the end, two examples are worked out to illustrate our main work.

1. Introduction

Nowadays, differential equations with fractional order have gained much attention and importance since they provided valuable tools for their applications in various sciences, such as gas dynamics, nuclear physics, electrodynamics of complex medium, and polymer rheology. With this advantage, fractional order models are regarded as more realistic and practical. For more details about fractional differential equations, we refer the readers to the monographs [14] and papers [59].

Many scholars have studied the existence of nonlinear fractional differential equations with a variety boundary conditions. However, it is better to impose nonlocal conditions because they can accurately describe the actual phenomenon. Some authors studied multipoint boundary value problems; for example, [10] discussed the infinite-point boundary value problems where ,  ,  ,   with . Existence result of at least two positive solutions is given via fixed point theorem in a cone.

Different from [10], some work focuses on the solvability of the fractional differential equations with integral boundary conditions. The details are found in [1116] and the references therein. In [17], Sun and Zhao investigated the following fractional differential equation with integral boundary conditions: where . By using the monotone iteration method and some inequalities technique, the existence result of positive solutions is obtained.

By the same method, Zhang [18] discussed the following fractional differential equation with Riemann–Stieltjes integral boundary conditions: where ,  ,   is a bounded variation, and denotes a Riemann–Stieltjes integral with a signed measure. This includes both the multipoint and a Riemann integral in a single framework.

Motivated by the wide applications of nonlocal boundary value problems and the results mentioned above, we consider the following fractional differential equation with multistrip Riemann–Stieltjes integral boundary conditions:where ,  ,  ,  , and is the standard Riemann-Liouville derivative; the nonlinear term is related to the lower derivative of the function . We emphasize that multistrip integral boundary conditions in (5) state that the value of unknown function at the right end point of the given interval is equal to the linear combination of the Riemann–Stieltjes integral values of the unknown function on the subinterval , for . The consideration of the fractional differential equation together with multistrip Riemann–Stieltjes integral boundary conditions makes problem (4) and (5) new. The proof of our main results is based on the Leggett–Williams fixed point theorem in a cone, which we present now.

Theorem 1 (Leggett–Williams fixed point theorem). Let be a cone in a real Banach space , , be a nonnegative continuous concave functional on such that for all , and Suppose that is completely continuous and there exist constants such that , and for ; for ; for , with Then has at least three fixed points , , and , which satisfy

If there holds , then condition implies condition of Theorem 1. Throughout this paper, we always make the following assumptions: is continuous;,  ,  ;,  ,  , and is an increasing function of bounded variation;, where

2. Preliminaries

In this section, we will present several definitions and lemmas that are necessary for the proof of our main results.

Definition 2 (see [1]). The Riemann-Liouville fractional integral of order of a function is given by provided the right side is pointwise defined on

Definition 3 (see [1]). The Riemann-Liouville fractional derivative of order of a function is given by where ,   denotes the integer part of number , provided the right side is pointwise defined on

From the definitions of Riemann-Liouville’s derivative, we can obtain the following statement.

Lemma 4. Let ; if we assume , then the fractional differential equation has , for some , as a unique solution, where is the smallest integer greater than or equal to

Lemma 5. Let ; if we assume , then for some , where is the smallest integer greater than or equal to

Remark 6. The following properties are useful for our discussion: (1), for ;(2), for

Lemma 7. Suppose that holds. For , the unique solution of is , in which where

Proof. In view of Lemma 5, we reduce problem (11) to an equivalent integral equation where are arbitrary constants. Consequently the general solution of the problem (11) can be written as By , we find . Set in (15), then Together with the boundary condition , we have Hence the unique solution of (11) is Furthermore, Then Hence, the solution of (11) is

Lemma 8. The function defined by (13) satisfies the following inequality:

Proof. For , we have , and then For , since , we have Then the proof is completed.

For convenience, denote is introduced . It is obvious that ,   and for .

The following properties of the Green function play an important role in this paper.

Lemma 9. The Green function defined by (12) satisfies the following properties:(1) is continuous on ;(2) for ;(3) for (4) for ;(5) for

Proof. and hold obviously; we only show that are true.
For any , by (12), (13), and the right inequality of (22), we get For any , by (12), (13), and the left inequality of (22), we get By the definition of and , we have Therefore, From (13), it is evident that It follows from (29) and (30) that The proof of the Lemma is completed.

3. Existence Result

Define the space is endowed with the ordering if , for all , and endowed with the norm , where .

Lemma 10. is a Banach space.

Proof. Let be a Cauchy sequence in the space Then clearly and are Cauchy sequences in the space Therefore, and converge to some and on uniformly and We need to proof that
Note that By the convergence of , we haveuniformly for On the other hand, by Lemma 5 one has , for and some . Further, we can obtain From (33) and (34), we have Taking the -order derivative on both sides of (35) yields In view of Remark 6 and Lemma 4, we find that This completes the proof.

Define the cone by

Let the nonnegative continuous concave functional on the cone be defined by

Lemma 11. Assume conditions hold. For any , define the operator by and then is completely continuous.

Proof. First, we prove that In view of the nonnegativeness and continuity of and ,   is continuous and for . Hence
Next, we show is uniformly bounded. Let be bounded; that is, there exists a positive constant such that , for all Let ; then for , from the Lemma 9, we have Hence, is bounded.
Finally, we show is equicontinuous. Indeed, for any ,  ,  , we have Note that, applying the mean value theorem, we arrive at and , which implies that Moreover, Therefore, (43) and (44) imply that is equicontinuous for all By means of the Arzela-Ascoli theorem, is completely continuous.

For convenience, we denote

Theorem 12. Assume that conditions hold, there exist nonnegative numbers , and satisfies the following conditions: , for ;, for ;, for Then BVP (4) and (5) has at least three positive solutions , , and such that

Proof. We will verity that the conditions of Theorem 1 are satisfied.
Let We first prove that is completely continuous. From Lemma 11, we only need to prove that For any , we have ,   , for all The assumption implies for Consequently, for , Thus, and further to get . Therefore is completely continuous.
Similarly, the conditions of Theorem 1 can be obtained by the assumption Here we do not do more explanation.
Finally, in order to verity , we make It is easy to find that If , we have ,  for Then that is, for all This shows that condition of Theorem 1 is also satisfied.
From the above, BVP (4) and (5) has at least three positive solutions , , and such that The proof is completed.

4. Example

Here we provide two cases to verify the feasibility and breadth of the conclusion, where the strip intervals in boundary condition (5) satisfy intersection relation and inclusion relation in Examples 1 and 2, respectively.

Example 1. Consider the boundary value problem of nonlinear fractional differential equations as follows:where , and satisfy and ,   do not contain each other. Let for and It is easy to see that satisfies condition
Take By a simple calculation, we obtain ,  ,  ,  ,  
Set , , such that , and, in addition, , for ;, for ;