Mathematical Problems in Engineering

Volume 2018, Article ID 5438592, 8 pages

https://doi.org/10.1155/2018/5438592

## Resonant Integral Boundary Value Problems for Caputo Fractional Differential Equations

^{1}Department of Applied Mathematics, Shandong University of Science and Technology, Qingdao 266590, China^{2}State Key Laboratory of Mining Disaster Prevention and Control Cofounded by Shandong Province and the Ministry of Science and Technology, Shandong University of Science and Technology, Qingdao 266590, China

Correspondence should be addressed to Yujun Cui; moc.361@102027jyc

Received 3 May 2018; Accepted 30 July 2018; Published 7 August 2018

Academic Editor: Zhen-Lai Han

Copyright © 2018 Wenjie Ma 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.

#### Abstract

This paper deals with the following Caputo fractional differential equations with Riemann-Stieltjes integral boundary conditions , where denotes the standard Caputo derivative, ; denotes the Riemann-Stieltjes integrals of with respect to . By mean of coincidence degree theory, we obtain the existence of solutions for the above fractional BVP at resonance. In the end, according to the main results, we give a typical example.

#### 1. Introduction

Fractional differential equations have various applications in chemical engineering, blood flow phenomena, physics, signal and image processing, fitting of experimental data, control theory, etc. Fractional differential equations also are a fundamental tool for the description of process of natural phenomena. In past years, many authors studied the existence of solutions or positive solutions of various fractional differential equations such as fractional boundary value problem at resonance [1–9], fractional boundary value problem at nonresonance [10–23], Caputo fractional derivative problem [24, 25], impulsive problem [26], conjugate boundary value problems [27, 28], integral boundary value problem [29], variational structure problem [30], fractional -Laplace problem [31], fractional lower and upper solution problem [32, 33], solitons [34, 35], etc.

For example, Alberto et al. [11] considered the following fractional differential equations: where , , denotes the Caputo fractional derivative, and is a continuous function. The existence of solutions is obtained by using the Guo-Krasnoselskii fixed point theorem. Similar arguments for related issues are treated in [12].

In [7], by mean of coincidence degree theory, the existence of solution of integral BVP for nonlinear differential equation is proved:where , is the Riemann-Liouville derivative, , and satisfies Caratheodory conditions. Similar method for different nonlinear differential equations, with either Caputo or Riemann-Liouville derivatives, and using suitable coincidence degree theory in each case, has allowed proving the existence of solutions in many papers; see also [1–5, 24, 25, 33].

In [5], the authors studied nonlocal fractional differential equation: where , , , is the Riemann-Liouville derivative, and is continuous function. They solved the above differential equation by using iterative technique and the fixed point index theory. We can also refer to [16, 17].

In this paper, we are concerned with the existence of solution for the following differential equation with boundary value conditions:where denotes the Caputo fractional derivative, , and is a continuous function. The BVP (4) is a resonance problem, because , if and only if its associated linear homogeneous BVP have nontrivial solution. Here, we assume that the following two conditions hold: , . is a continuous function.

As is known to all, there are few studies on such boundary value problems. Therefore, by mean of coincidence degree theory, we consider the existence of solutions for fractional differential equations (4) with Riemann-Stieltjes integral boundary conditions at resonance.

#### 2. Preliminaries

For the convenience of the reading, we introduce the theorem and lemma of fractional calculus theory.

Lemma 1 (see [36]). *Put , ; the fractional differential equationhas the general solution , where is the smallest integer greater than or equal to .*

Lemma 2 (see [36]). *Given that , then where is the smallest integer greater than or equal to .*

Let be two Banach spaces and be a linear operator. Assume that is a Fredholm operator of index zero; , are two continuous projectors such that Thus is invertible. We denote the inverse of the mapping by (generalized inverse operator of ). If is an open bounded subset of such that , will be called -compact on if is bounded and is compact.

Theorem 3 (see [37]). * is a linear operator, and is a Fredholm operator of index zero, and is L-compact on . The equation has at least one solution in if and only if the following three conditions are satisfied:*(1)* for every .*(2)* for every .*(3)*, where is a projection such that .*

*In this paper, let with the norm in which Let with the norm Operator is defined as with *

*Define operator as follows: And so problem (4) becomes .*

*3. Main Results*

*In this section, we continue to give the following assumptions.*

* There exist functions with such that *

* There exists positive constant such that, for , if for , then *

* There exists constant such that eitherorholds if *

*Based on the above, we will prove the following important theorem.*

*Theorem 4. Suppose that - is satisfied; there exists at least one solution of the fractional differential equation (4) in .*

*For convenience, we first give some important lemmas as follows.*

*Lemma 5. Assume that is satisfied; then is Fredholm operator with index zero, and a linear continuous projector is defined by Furthermore, define a linear operator as follows: such that .*

*Proof. *It follows from Lemma 1 and that According to , we have Moreover, we can obtain that On one hand, suppose ; then there exist :Then we have This together with , , we can get and And so we obtain that On the other hand, if satisfies we let Then we conclude that and That is, , then In conclusion, We give projection operator as We get , , and For any , together with , we have It is easy to obtain that which implies Next operator is defined as follows: Noting that means is a projection operator. And obviously, For any , let , and we have , so Moreover, by simple calculation, we can get . Above all,

To summarize, we can know that is closed subspace of ; ; and so is a Fredholm operator of index zero.

Next we will prove that is the inverse of It is clear thatFor each , we have and This implies that So

*Lemma 6. Assume that is a bounded open subset of , then is -compact on if .*

*Proof. *By the continuity of , we obtain and are bounded, so there exists constant , such that for and . , let , and we have andAs a result of are uniformly continuous on . So we have that are equicontinuous, and, by Ascoli-Arzela theorem, we have that is compact. Thus, is -compact.

*Lemma 7. Set is bounded if conditions , , and are satisfied, and *

*Proof. *Suppose that we have and If , by condition , we have . If , then If , we suppose , then which contradicts with

*Lemma 8. Set is bounded if - are satisfied.*

*Proof. *Take , then . Thus we haveAccording to , there exists constant such that Because ,If , we have , then and therefore .

By , we have Then we obtain thatAccording to the boundary condition , one hasSo we get Thus, from , we obtain that So we have Therefore, is bounded.

*Lemma 9. Set is bounded if , , and hold.*

*Proof. *Let , then , and we can get According to we have , that is to say, is bounded.

*In summary, we can prove Theorem 4.*

*Proof of Theorem 4. *Assume that is bounded open subset of , and, according to Lemma 6, we get that is -compact on . Form Lemmas 8 and 9, we have (1);(2) Set It follows from Lemma 7 that we get , , so we have According to the above proof and Theorem 3, Theorem 4 holds.

*Example 10. *Considering the following fractional differential equation,Obviously, , , then , and thus it is at resonance. Let then where Taking , then we have , and That is, holds. Finally, taking , when , we conclude that then is satisfied. It follows from Theorem 4 that (54) has at least one solution.

*Data Availability*

*The data used to support the findings of this study are included within the article.*

*Conflicts of Interest*

*The authors declare that they have no conflicts of interest.*

*Acknowledgments*

*The research is supported by the National Natural Science Foundation (NNSF) of China (11371221 and 11571207), Shandong Natural Science Foundation (SDNSF) (ZR2018MA011), and the Tai’shan Scholar Engineering Construction Fund of Shandong Province of China.*

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