#### Abstract

This paper is concerned with the existence of solutions to a discrete three-point boundary value problem at resonance involving the Riemann-Liouville fractional difference of order . Under certain suitable nonlinear growth conditions imposed on the nonlinear term, the existence result is established by using the coincidence degree continuation theorem. Additionally, a representative example is presented to illustrate the effectiveness of the main result.

#### 1. Introduction

For any number , we denote and , for any with , throughout this paper. It is also worth noting that, in what follows, we appeal to the conventions and for a given function defined on and with .

In this paper, we will consider the existence of solutions for the following discrete fractional three-point boundary value problem: where is a real number, denotes the Riemann-Liouville fractional difference of order , , , , and which implies that the problem (1) is at resonance. We note that the problem (1) happens to be at resonance in the sense that the associated linear homogeneous boundary value problem has , , as a nontrivial solution.

The continuous fractional calculus has received increasing attention within the last ten years or so and the theory of fractional differential equations has been a new important mathematical branch due to its demonstrated applications in various fields of science and engineering. For more details, see  and references therein. Significantly less is known, however, about the discrete fractional calculus, but in recent several years, a lot of papers have appeared; see . For example, in , Atıcı and Eloe explored a discrete fractional conjugate boundary value problem with the Riemann-Liouville fractional difference. To the best of our knowledge, this is a pioneering work on discussing boundary value problems in discrete fractional calculus. After that, Goodrich studied discrete fractional boundary value problems involving the Riemann-Liouville fractional difference intensively and obtained a series of excellent results; see . Bastos et al. in [28, 29] considered the discrete fractional calculus of variations and established the necessary conditions for fractional difference variational problems. Abdeljawad introduced the Caputo fractional difference and developed some useful properties of it in . Ferreira  investigated the existence and uniqueness of solutions for some discrete fractional nonresonance boundary value problems of order less than one by using the Banach fixed point theorem.

Although the solvability of fractional boundary value problems has been studied extensively, there are few papers dealing with it under resonance conditions, besides . Additionally, as far as we know, the existence of solutions to discrete fractional boundary value problems at resonance has not been studied.

Motivated by the aforementioned results, we will investigate the discrete fractional boundary value problem (1) at resonance and establish some sufficient conditions for the existence of solutions to it by using the coincidence degree theory.

For the sake of convenience, we will always assume that the following conditions hold in this paper:, , and ; is continuous.

The remainder of this paper is organized as follows. Section 2 preliminarily provides some necessary basic knowledge for the theory of discrete fractional calculus and the coincidence degree continuation theorem. In Section 3, the existence result of solutions for problem (1) will be established with the help of the coincidence degree theory. Finally in Section 4, a concrete example is provided to illustrate the possible application of the established analytical result.

#### 2. Preliminaries

Since the theory of discrete fractional calculus is in its infancy to some extent, in order to make this paper self-contained, we begin by presenting here some necessary basic definitions and lemmas about it. For more details, see [15, 16, 19, 34].

Definition 1 (see ). For any and , the falling factorial function is defined as provided that the right-hand side is well defined. We appeal to the convention that if is a pole of the Gamma function and is not a pole, then .

Definition 2 (see ). The th discrete fractional sum of a function , for , is defined by Also, we define the trivial sum , .

Definition 3 (see ). The th discrete Riemann-Liouville fractional difference of a function , for , is defined by where is the smallest integer greater than or equal to and is the th order forward difference operator. If , then .

Remark 4. From the Definitions 2 and 3, it is easy to see that maps functions defined on to functions defined on and maps functions defined on to functions defined on , where is the smallest integer greater than or equal to . Also, it is worth reminding the reader that the in (or ) represents an input for the function (or ) and not for the function . For ease of notation, we throughout this paper omit the subscript in and when it does not to lead to domains confusion and general ambiguity.

Lemma 5 (see ). Let . Then , where , , and is the smallest integer greater than or equal to .

Lemma 6 (see ). Let and such that is well defined. Then

Next, we will briefly recall some notations in the frame of Mawhin’s coincidence degree continuous theorem. For more details, see .

Let and be two real Banach spaces. Consider an operator equation , where is a linear operator and is a nonlinear operator. The operator will be called a Fredholm operator of index zero if and is closed in . If is a Fredholm operator of index zero, then there exist continuous projectors and such that , and , . It follows that is invertible and its inverse is denoted by .

If is an open bounded subset of and , the operator will be called -compact on if is bounded and is compact.

Now, we present the coincidence degree continuation theorem as follows, which will be used in the sequel to establish the existence of solutions for problem (1).

Theorem 7 (see ). Let be a Fredholm operator of index zero and let be -compact on . Assume that the following conditions are satisfied:(i) for every ;(ii) for every ;(iii), where is a projection such that .Then the operator equation has at least one solution in .

Finally, we wish to fix our framework for the study of problem (1). First of all, we denote and , and it is clear that and are two Banach spaces when equipped with the usual maximum norm; that is, for any and , and . Next, we define the linear operator by with and the nonlinear operator as Then the problem (1) is equivalent to an operator equation

#### 3. Main Results

In this section, we will establish the existence of at least one solution for the problem (1). To accomplish this, we firstly present here several lemmas which will be used in the sequel.

For convenience, we define the operator by and by Lemma 6, we can find that .

Lemma 8. If holds, then where and are defined by (8) and (12), respectively.

Proof . At first, in view of Lemma 5 and , we can easily verify that (13) holds. Next, we prove that (14) also holds.
For any , then there exists a function such that . Based on Lemma 5, we have where , . From conditions and , we can easily obtain that .
Conversely, for any with , if we set , then , and it is easy to verify that . Moreover, by the relation , we have , which lead to . So we get that (14) holds. The proof is complete.

Lemma 9. If holds, then defined by (8) is a Fredholm operator of index zero.

Proof . In order to show that is a Fredholm operator of index zero, firstly, we consider the following operator defined by where is defined by (12). Evidently, and is a continuous linear projector. In fact, for any , we have that is to say, is idempotent. Hence, is a projector.
From the definition of and (14), it is easy to see that leads to , and if , we can get , which implies that . So, we derive .
For any , set . Since and , we have . Moreover, take . Then can be written as , , for . On the other hand, since , by (14), we can get , which implies that . So, we have and .
Now, since and is closed in , is a Fredholm operator of index zero. The proof is complete.

Let be defined by It is clear that is a linear continuous projector and Also, proceeding as the proof of Lemma 9, we can show that

Define operator by where .

From the definitions of and , it is easy to see that the inverse of is . In fact, if , then we have Also, if , by Lemma 5, we have where . Then it follows from and that which implies that . Consequently, we have for . So, .

Lemma 10. Suppose that holds. If is an open bounded subset and , then is -compact on .

Proof. By the continuity of , we can verify that and are bounded. So we get that is compact. Therefore is -compact on . The proof is complete.

To establish the main result, we need the following conditions.There exist two nonnegative functions with such that There exists a constant such that for each satisfying , .There exists a constant such that for any , , if , then either or

Theorem 11. If hold, then the problem (1) has at least one solution in .

Proof. This proof will be divided into four main steps. Now let us prove the steps one by one.
Step 1. Set and prove that is bounded in .
For , then and , so . By (14), we have . From , there exists a constant such that . Since , by Lemma 5, we have Considering , we get Consequently, by (29), (30), Lemma 6, , and the monotonicity of functions and on , we have So, by the fact that in and (31), we can derive that is bounded.
Step 2. Set and prove that is bounded in .
For any , there exists a constant such that , , and . So it follows from (14) that . By virtue of and the fact that is decreasing for on , we can derive that , which implies that is bounded in .
Step 3. Set and prove that is bounded in , where is a linear isomorphism defined by For any , there exists such that If , then . Hence . By , we get . If , then . For , if , then, from , we can obtain that Therefore, we have which is a contradiction. So, is bounded in .
Step 4. Let be a bounded open set such that and prove that It follows from Lemma 10 that is -compact on . Then by Steps 1 and 2, we have (i) for every ;(ii) for every .
At last, we prove that condition (iii) of Theorem 7 is satisfied. Let According to the arguments in Step 3, we have and therefore, via the homotopy property of degree, we get that which implies that condition (iii) of Theorem 7 is satisfied. Then by Theorem 7, we can conclude that has at least one solution in ; that is, (1) has at least one solution in . The proof is completed.

#### 4. An Illustrative Example

In this section, we will illustrate the possible application of the above established analytical result with a concrete example.

Example 1. Consider the following discrete fractional boundary value problem: where It is obvious that is continuous. Corresponding to problem (1), there exist , , , , , and Therefore, the problem (40) is at resonance.

Choosing , , , then we have Hence, the condition in Theorem 11 holds.

Let . If holds for any , then we can easily verify that which implies that condition of Theorem 11 holds.

Furthermore, we can choose , to show that the condition of Theorem 11 holds. In fact, for any satisfying , we can get that So, by the fact that for and (45), we can derive that, for any satisfying , which implies that (27) in of Theorem 11 holds. Therefore, all conditions of Theorem 11 hold. Hence, we can conclude that problem (40) has at least one solution.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The author sincerely thanks the reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original paper. The research is supported by the Longdong University Grant XYZK-1402.