Abstract

We consider a hybrid fractional sum-difference initial value problem and a hybrid fractional sequential sum-difference initial value problem. The existence results of these two problems are proved by using the hybrid fixed point theorem for three operators in a Banach algebra and the generalized Krasnoselskii’s fixed point theorem, respectively.

1. Introduction

As recognized that fractional difference calculus is a powerful tool used to describe many real world phenomena problems such as physics, chemistry, mechanics, control systems, flow in porous media, and electrical networks [1, 2], this is an impact on a researcher’s motivation to develop the research works in this area. Basic definitions and properties of fractional difference calculus were proposed by Goodrich and Peterson [3]. The developments of the theory related to discrete fractional boundary value problems were studied by many authors (see [448]). In particular Sitthiwirattham [41, 42] studied three-point Caputo fractional difference-fractional sum boundary value problem for sequential Caputo fractional difference equation of the formand three-point fractional sum boundary value problem for sequential Riemann-Liouville fractional difference equation of the formwhere , , , , , is a constant, is a continuous function, , and is the -Laplacian operator.

Calculus which deals with derivatives and integrals of arbitrary orders is known as hybrid differential equations (i.e., quadratic perturbation of a nonlinear differential equation). Hybrid fractional differential equations are initialized to be used to model successfully several physical phenomena (see [4951]). Apparently, this issue has found numerous miscellaneous applications connected with real world problems as they appear in many fields of engineering and science, including biology, chemistry, diffusion, control theory, electromagnetic theory, fluid flow, signal and image processing, fractals theory, fitting of experimental data, potential theory, and viscoelasticity. For some recent developments on the topic, see [5257]. Recently, there are several research works related to boundary value problems for hybrid differential equations (see [5865]). For example, Sun et al. [60] studied the existence of solutions for the boundary value problem of fractional hybrid differential equationswhere denotes the Riemann-Liouville fractional derivative of order , .

Sitho et al. [64] studied existence results for the following initial value problem for hybrid fractional integrodifferential equations,and the initial value problem for hybrid fractional sequential integrodifferential equations,where denotes the Riemann-Liouville fractional derivative of order , is the Riemann-Liouville fractional integral of order , , , , and functions are , , , and with , .

While the boundary value problem for hybrid fractional difference equations has not been studied, to fill this gap, we study a hybrid fractional difference initial value problem of the form,and a hybrid fractional sequential sum-difference initial value problem of the formwhere , , are given constants, , , with , with , , and .

The article is organized as follows. In Section 2, we recall some definitions and basic lemmas used in this work. Then, we present the solutions of (6) and (7) by converting the problem to an equivalent summation equation. In Sections 3 and 4, we prove existence results of problems (6) and (7) by employing the hybrid fixed point theorem for three operators in a Banach algebra and the generalized Krasnoselskii’s fixed point theorem, respectively. We end with some examples to illustrate our results in the last section.

2. Preliminaries

In what follows are the notations, definitions, and lemmas which are used in the main results.

Definition 1 (see [6]). Define the generalized falling function by , for any and for which the right-hand side is defined. If is a pole of the Gamma function and is not a pole, then .

Lemma 2 (see [4]). Assume that the factorial functions are well defined. If , then for any .

Definition 3 (see [6]). For and defined on , the -order fractional sum of is defined bywhere and .

Definition 4 (see [6]). For and defined on , the -order Riemann-Liouville fractional difference of is defined bywhere and is satisfied with .

Lemma 5 (see [5]). For any real number and any positive integer , the following equality holds:or is defined on .

Lemma 6 (see [4]). Let Then,for some , with .

We provide the following lemma dealing with linear variant of the boundary value problems (6) and (7) and give a representation of the solution.

Lemma 7. Let , with , , and . Then, for , the problemhas the unique solutionfor .

Proof. Using the fractional sum of order for (12) and Lemmas 5 and 6, we obtainfor .
Since , , and , it follows thatThus, (13) holds. The proof is completed.

Lemma 8. Let , , with , , and be given. Then, for , the problemhas the unique solutionfor .

Proof. Using the fractional sum of order for (16) and Lemmas 5 and 6, we obtainfor .
Since , , and , we haveThat can be arranged in the formUsing the fractional sum of order for (20), we obtainfor .
Thus (17) holds. Our proof is completed.

Lemma 9 (Arzelá-Ascoli theorem [66]). A set of functions in with the sup norm is relatively compact if and only if it is uniformly bounded and equicontinuous on .

Lemma 10 (see [66]). If a set is closed and relatively compact, then it is compact.

3. Hybrid Fractional Sum-Difference Initial Value Problem (6)

In this section, we aim to show the existence results for problem (6). To accomplish this, we let be a space of all functions and defined a norm and a multiplication in byIn addition, we define operator byClearly, problem (6) has solutions if and only if operator has fixed points. The first shows the existence and uniqueness of a solution to problem (6) by using the Banach contraction principle.

Theorem 11. Assume that , with , and . In addition, suppose thatthere exist constants such that for each and there exists a positive function with bound such that for each and If , then problem (6) has a unique solution.

Proof. We shall show that is a contraction. For any and for each , we haveThus, we have .
Consequently, is a contraction. Therefore, by the Banach fixed point theorem, we get that has a fixed point which is a unique solution of problem (6).

In the second result, we deduce the existence of at least one solution of the initial value problem (6) by using the hybrid fixed point theorem for three operators in a Banach algebra. Clearly, is a Banach algebra with respect to the above norm and multiplication in it.

Theorem 12 (hybrid fixed point theorem for three operators in a Banach algebra [67]). Let be a nonempty, closed convex, and bounded subset of the Banach algebra , and let and be three operators such that(i) and are Lipschitzian with Lipschitz constants and , respectively,(ii) is completely continuous,(iii) for all ,(iv), where . Then, the operator equation has a solution.

Theorem 13. Assume that , with , and . In addition, suppose thatthere exist two positive functions and with bound and , respectively, such that for each and there exists a function and a continuous nondecreasing function such that for each there exists a numberwhere , , andThen problem (6) has a unique solution on .

Proof. Define subset of asWe see that is closed, convex, and bounded subset of the Banach space . By Lemma 7, we define three operators , , and byNote that problem (6) has solutions if and only if the operator has fixed points.
To show that all operators satisfy all the conditions of Theorem 12, we proceed with the following steps.
Step  1. Prove that and are Lipschitzian on .
For any and for each , then by , we haveThis implies that, for all ,Therefore, and are Lipschitzian on with Lipschitz constants and .
Step  2. We prove that is completely continuous on .
Since is continuous, the operator is continuous on . Next, we will prove that the set is uniformly bounded in . For any , we find thatfor all Therefore, , which shows that is uniformly bounded on .
Next we show that is an equicontinuous set in . For any , there exists a positive constant such that, for ,Then we obtainThis implies that the set is an equicontinuous set. From the Arzelá-Ascoli theorem, we find that is completely continuous.
Step  3. for all .
Let and be arbitrary elements such that . Then,We find thatTherefore, .
Step  4. We prove that . Sinceand by , we havewith and .
We see that all the conditions of Theorem 12 are satisfied. Hence, the operator equation has a solution in . In consequence, problem (6) has a solution on . This completes the proof.

4. Hybrid Fractional Sequential Sum-Difference Initial Value Problem (7)

In this section, we prove existence results of problem (7). To accomplish this, we denote that and define the space of all functions with the norm byIn addition, we define the operator byClearly, problem (7) has solutions if and only if the operator has fixed points. The first shows the existence and uniqueness of a solution to problem (7) by using the Banach contraction principle.

Theorem 14. Assume that , with , and . In addition, suppose thatthere exist constants such that for each and there exists a positive function with bound such that for each and If then problem (7) has a unique solution, where

Proof. We shall show that is a contraction. For any and for each , we haveThus, we have .
Consequently, is a contraction. Therefore, by the Banach fixed point theorem, we get that has a fixed point which is a unique solution of problem (7).

In the second result, we deduce the existence of at least one solution of the initial value problem (7) by using the generalized Krasnoselskii’s fixed point theorem.

Theorem 15 (generalized Krasnoselskii’s fixed point theorem [68]). Let be a nonempty, closed convex, and bounded subset of the Banach space . Let and be two operators such that(i) is a contraction,(ii) is completely continuous,(iii) for all Then, the operator equation has a solution.

Theorem 16. Assume that , with , and . In addition, suppose thatthere exist two positive functions and with bound and , respectively, such that, for each and ,there exist functions such thatIfthen problem (7) has a unique solution on .

Proof. Let , , and , and choose a constantWe consider . Define four operators , , and byfor , andProblem (7) has solutions if and only if the operator has fixed points.
The proof is divided into three steps as follows.
Step  1. Verify map bounded sets into bounded sets in .
For each , we obtainThus, . This implies that is uniformly bounded.
Hence, condition (iii) of Theorem 15 holds.
Step  2. Check that is contraction mapping.
For any and for each , by , we haveBy (50), is a contraction mapping. Hence, condition (i) of Theorem 15 holds.
Step  3. Check that is completely continuous on .
The operator is obviously continuous on . Furthermore, is uniformly bounded on sinceFor any , there exists a positive constant such that for Then,This implies that the set is an equicontinuous set. Therefore, by the Arzelá-Ascoli theorem, we find that is completely continuous. Hence, condition (ii) of Theorem 15 holds.
We see that all the assumptions of Theorem 15 are satisfied. Therefore, we can conclude that problem (7) has at least one solution. The proof is completed.

5. Examples

In this section, we provide some examples to illustrate our results.

Example 1. Consider the following fractional difference initial value problem:whereWe set , , and .
Noticing that - hold, for each , we haveThus,Finally, we find thatSo, holds with a number .
Hence, by Theorem 13, problem (59) has a unique solution on .

Example 2. Consider the following fractional difference boundary value problem:for , whereWe let , , , , and
Noticing that - hold, for each , we obtainThus, we obtainFinally, we find thatHence, by Theorem 16, problem (64) has a unique solution on .

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This research was funded by King Mongkut’s University of Technology North Bangkok, Contract no. KMUTNB-GOV-60-71.