Multiplicity Results to a Conformable Fractional Differential Equations Involving Integral Boundary Condition

Meng, Shuman; Cui, Yujun

doi:https://doi.org/10.1155/2019/8402347

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Abstract Introduction Preliminaries Data Availability Conflicts of Interest Acknowledgments References Copyright Related Articles

Research Article | Open Access

Volume 2019 | Article ID 8402347 | https://doi.org/10.1155/2019/8402347

Multiplicity Results to a Conformable Fractional Differential Equations Involving Integral Boundary Condition

Shuman Meng¹and Yujun Cui²

Academic Editor: Mohamed Boutayeb

Received17 Oct 2018

Accepted12 Feb 2019

Published01 Apr 2019

Abstract

In this article, by using topological degree theory couple with the method of lower and upper solutions, we study the existence of at least three solutions to Riemann-Stieltjes integral initial value problem of the type , , where is the standard conformable fractional derivative of order , , and . Simultaneously, the fixed point theorem for set-valued increasing operator is applied when considering the given problem.

1. Introduction

In recent years, fractional differential equations have exerted tremendous influence on some mathematical models of research processes and phenomena in many fields such as electrochemistry, heat conduction, underground water flow, and porous media. A growing number of papers deal with the existence or multiplicity of solutions of initial value problem and boundary value problem for fractional differential equations [1–11]. Recently, the authors [12] give an interesting fractional derivative called the “conformal fractional derivative”, which depends on the limit definition of the function derivative. Moreover, readers can find in [13] the properties of conformable fractional derivatives that are similar to ordinary differential ones. Other related work on conformable fractional differential equation can be found in [12–17] and the references therein.

Whether it is ordinary differential equations or fractional differential equations, the existence of solutions to boundary value problems with integral boundary conditions has been studied in the applied sciences and physics. For more important content of the typical theory of differential equations and integral equations with boundary value problems are obtained [18, 19]. Lots of results have been established for differential equations and differential systems with integral boundary conditions by using upper and lower solution, fixed point theory and fixed point theorem; see [20–28].

Now there are more and more articles to prove that there are multiple solutions for integral boundary [29–35]. For example, [31] introduced the system of fractional differential equations where , , , are the standard the Riemann-Liouville derivatives of orders , . By means of the Guo-Krasnoselskii fixed point theorem and the property of degree to obtain the existence and multiplicity of positive solutions to integral boundary value conditions and where are nondecreasing functions.

Inspired by the above work, we consider the existence and multiple solutions of the following fractional differential equation involving integral boundary condition: where , denotes the Riemann-Stieltjes integral with positive Stieltjes measure. is the standard conformable fractional derivative of order of at , defined by If exists on , then . Obviously, if is differentiable.

In order to study the existence and multiplicity results of the problem (4), this paper is arranged as follows: after this introduction, in Section 2, we briefly show some necessary definitions and lemmas that are used to prove our main results. In Section 3, we shall define a modified bounded function to discuss the existence of solutions for a conformable fractional differential equation with initial value condition. Moreover, in this part we employ two fixed point theorems, which are Schauder’s fixed point theorem and the fixed point theorem for set-valued increasing operator, respectively. In addition, using two pairs of upper and lower solutions and the property of degree theory, we deduce that problem (4) has at least three solutions.

2. Preliminaries

In the following, let , then is a Banach space with the norm .

Definition 1 (see [12]). Let ; the conformable fractional integral starting from a point of a function of order is defined as

Lemma 2 (see [14]). Let be differentiable and . Then, for all we have

Lemma 3 (see [15]). Let , , and the function , be -differentiable on ; then(i) for all constant function ;(ii);(iii);(iv);(v) for all function .

Lemma 4 ([15] (mean value theorem)). Let an interval , and the function , if it satisfies(1) is continuous on ;(2) is -differentiable for some on .Then there exists a constant , such that .

Definition 5. A function is called a lower solution of problem (4), if it satisfies If inequalities (8), (9) are reversed, then is an upper solution of problem (4).

Definition 6. A function is said to be a strict lower solution of problem (4), if the inequality (8), (9) is strict for .

Lemma 7 (see [13]). is a given function that satisfies(i) is continuous on ;(ii) is -differentiable for some .Then we have the following:(1)if for all , then is increasing on ;(2)if for all , then is decreasing on .

The following lemma is a direct consequence of the application of the definition of conformable fractional derivative and Lemma 7.

Lemma 8. Let ; the function is continuous and -differentiable on .(1)If has an extreme value at , then ;(2)if has a maximum (minimum) value at , then (3)if has a maximum (minimum) value at , then

Theorem 9 ([36, 37] (fixed point theorem for set-valued increasing operator)). Let be a partially ordered set, be a nonempty closed set of , and be a set-valued increasing operator. Assume that(i)any totally ordered subset of is a relatively compact,(ii)for all , is a compact set in ,(iii)there exist and such that .Then has a fixed point in ; that is, there exists such that .

Throughout the paper, we list some hypotheses.

() is continuous.

() , is continuous at and is nondecreasing at .

() Assume that is lower and upper solution of problem (4) with .

3. Main Results

Based on the above preparations and the assumptions mentioned, we first consider the following initial value problem (IVP) where . We suppose that holds. Applying Lemma 2, it is easy to prove that (12) is equivalent to the following integral equation

Define integral operator by Then, is a solution of (12) if and only if is a solution of the operator equation , that is, a fixed point of operator .

Theorem 10. Assume that - hold. Then for a fixed number , there exists a solution of problem (12) such that , .

Proof. Firstly we define the following modified function: Obviously is continuous and bounded on , so there exists such that .
Secondly, we consider the modified initial value problem The above initial value problem (16) is equivalent to the following integral equation: For any , we have Choose , and let Obviously, is a bounded convex subset in . From the above argument, the operator defined as follows is well defined, It is easy to verify that is compact. By Schauder’s fixed point theorem, has a fixed point in . Subsequently, is the solution of problem (16). Also, by the definition of and Definition 5, we have and Hence and are lower and upper solutions of (16).
Our final job is to apply the method in [38] to illustrate that if (16) has upper and lower solutions, hence any solution of differential equation (16) must satisfy , and then is a solution of (12).
We now prove that any solution of (16) does satisfy We need to show that , . Analogously we can also show that . If on , let , then has a positive maximum at some . By the initial boundary conditions, we derive , so . Then there will be two cases:
(i) , then by Lemma 8, we have , that is and (ii) , then by Lemma 8, we have and .
For case (i), by the definition of and Definition 5, we can get which contradicts with (24). This implies that has no positive local maximum on .
For case (ii), if , then , . Notice that has no positive local maximum on , and there exists such that and But for every , we have again a contradiction. We complete the proof.

Theorem 11. Assume that - hold, then problem (4) has a solution with , .

Proof. It follows from Theorem 10 that for any fixed , IVP (12) have a solution ; we can also mark it as satisfies , . The following two cases are considered:
(i) . It is easy to obtain that . Using Definition 5, we can conclude that i.e., , then is the solution of problem (4).
(ii) . Define the set-valued operator on as follows: It can be seen from Theorem 10 that for any fixed , , this shows that . Moreover, from the definition of , we can obtain that (4) has a solution with , if has a fixed point. In order to apply the fixed point theorem for set-valued increasing operator, we shall prove the following:
Step 1. The sequence of must have subsequence converging to for .
Assume that converges to . By using Theorem 10, we know that the solution of (12) related to satisfies , that is to say is uniformly bounded. Moreover, by the continuity of , we conclude that is bounded on for all . Let with . Using Lemma 4, we can conclude that Therefore, is equicontinuous. It means that is completely continuous sequences; by Arzela-Ascoli theorems, we get has subsequences and when . Note that satisfy Letting , we obtain that This shows that is a solution of problem (4); hence, .
Step 2. is set-valued increasing operator, that is, , and , such that .
Without loss of generality, we assume that , for such that and , where is the solution of (16) related to . Therefore, we give the modified function for and and consider the following modified initial value problem: It is easy to obtain that , are the lower and upper solutions of initial value problem (34). By the help of Theorem 10, we deduce that problem (34) has existence solution with . Subsequently, . Let , then . So the operator is set-valued increasing operator. The theorem is now proved.

Above we have studied the solution of (4), and then we present the main results of this paper.

Theorem 12. Suppose that , and the following conditions hold.
are two lower solutions and are two upper solutions of (4) such that , , and .
, are strict lower and upper solutions of problem (4).
, where .
Then problem (4) has at least three solutions (), such that , , , and on .

Proof. Similar to Theorem 10, we define the following modified function : Obviously is continuous and bounded on , so there exists such that .
Let , where We consider the modified boundary value problem It is clear that the solution of the above problem satisfies and then is a solution of problem (4).
By Lemma 2, it is easy to know that (36) is equivalent to the following integral equation: Integrating (37) with respect to form to , Combining with (37) and (38), we reduce (36) to the following equivalent integral equation: Now, we prove the existence of at least three solutions of the problem (36). Note that is a bounded convex subset in . Define the following operator Then, is a solution of (36) if and only if is a solution of the equation , that is, a fixed point of . For , we get So and is compact. By the topological degree theory, we have Let and It follows from the fact that then .
Next, by , we conclude that (36) has no solution on . Without loss of generality, we assume that (36) has a solution on . Thus there exists such that and for . It follows from Lemma 8 that when . But we noticed that is a strict lower solution of (36), thus This is a contradiction. If and for , from the boundary condition and , we have This is a contradiction again. So (36) has no solution on .
The additivity of degree implies that If we prove that we can get and there are solutions in , , , respectively.
Now we show that . Firstly, we show that .
Define functions in a similar way, and we consider the modified boundary value problem Equation (51) is equivalent to the operator equation , where Similar to the above argument, we can conclude that any solution of (51) satisfies , which in view of leads to on ; therefore, Moreover, since , we have It follows from (53) and (54) that Taking into account of on we deduce that Similarly, we can show that . Therefore, there exist three solutions , , of problem (4). The proof is finished.

Data Availability

The datasets 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 of China (11371221, 11571207, 51774197), Shandong Natural Science Foundation (ZR2018MA011), and the Tai’shan Scholar Engineering Construction Fund of Shandong Province of China.

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Copyright © 2019 Shuman Meng and Yujun Cui. 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.

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