Existence Results for Fractional Difference Equations with Three-Point Fractional Sum Boundary Conditions

Abstract

We consider a discrete fractional boundary value problem of the form Δ^𝛼𝑢(𝑡)=𝑓(𝑡+𝛼−1,𝑢(𝑡+𝛼−1)),𝑡∈[0,𝑇]_ℕ0∶={0,1,…,𝑇},𝑢(𝛼−2)=0,𝑢(𝛼+𝑇)=Δ^−𝛽𝑢(𝜂+𝛽), where 1<𝛼≤2, 𝛽>0, 𝜂∈[𝛼−2,𝛼+𝑇−1]_ℕ𝛼−2∶={𝛼−2,𝛼−1,…,𝛼+𝑇−1}, and 𝑓∶[𝛼−1,𝛼,…,𝛼+𝑇−1]_ℕ𝛼−1×ℝ→ℝ is a continuous function. The existence of at least one solution is proved by using Krasnoselskii's fixed point theorem and Leray-Schauder's nonlinear alternative. Some illustrative examples are also presented.

1. Introduction

Fractional differential and difference equations have received increasing attention during recent years since fractional derivatives provide an excellent instrument for the description of memory and hereditary properties of various materials and processes. For a reader interested in the systematic development of the topic, a good account of papers dealing with fractional or discrete fractional boundary value problems can be found in [1–25] and references cited therein.

In this paper we consider a nonlinear discrete fractional boundary value problem of the form Δ^𝛼𝑢(𝑡)((,[]=𝑓𝑡+𝛼−1,𝑢𝑡+𝛼−1))𝑡∈0,𝑇_ℕ0{},𝑢()()∶=0,1,…,𝑇𝛼−2=0,𝑢𝛼+𝑇=Δ^−𝛽𝑢𝜂+𝛽, (1) where1<𝛼≤2,𝛽>0,𝜂∈[𝛼−2,𝛼+𝑇−1]_ℕ𝛼−2∶={𝛼−2,𝛼−1,…,𝛼+𝑇−1}, and𝑓∶[𝛼−1,…,𝛼+𝑇−1]_ℕ𝛼−1×ℝ→ℝis a continuous function. Letℕbe the nonnegative integers. Denote[𝑖,𝑗]_ℕ𝑖={𝑘∈ℕ∣𝑖≤𝑘≤𝑗}.

Problem (1) was studied recently by authors in [26] where existence and uniqueness results are obtained by using Banach’s contraction mapping principle, the nonlinear contraction, and Schaefer’s fixed point theorem. Here we continue this study by proving some new existence results by using Krasnoselskii’s fixed point theorem and Leray-Schauder’s nonlinear alternative. Thus the results of this paper complement the results of [26].

The plan of this paper is as follows. In Section 2 we recall some definitions and basic lemmas. Also we give a representation for the solution to (1) by converting the problem to an equivalent summation equation. In Section 3, using this representation, we prove our main results. Some illustrative examples are presented in Section 4.

2. Preliminaries

In this section, we introduce notations, definitions, and lemmas needed in the following.

Definition 1. One defines the generalized falling function by𝑡^𝛼∶=Γ(𝑡+1)/Γ(𝑡+1−𝛼), for any𝑡and𝛼for which the right-hand side is defined. If𝑡+1−𝛼is a pole of the Gamma function and𝑡+1is not a pole, then𝑡^𝛼=0.

Definition 2 (see [17]). The𝛼th fractional sum of a function𝑓, for𝛼>0, is defined by Δ^−𝛼𝑓(𝑡)=Δ^−𝛼𝑓()1𝑡;𝑎∶=Γ(𝛼)^𝑡−𝛼_𝑠=𝑎(𝑡−𝜎(𝑠))^𝛼−1𝑓(𝑠), (2) for𝑡∈{𝑎+𝛼,𝑎+𝛼+1,…}∶=ℕ_𝑎+𝛼and𝜎(𝑠)=𝑠+1. One also defines the𝛼th fractional difference for𝛼>0byΔ^𝛼𝑓(𝑡)∶=Δ^𝑁Δ^𝛼−𝑁𝑓(𝑡), where𝑡∈ℕ_𝑎+𝛼and𝑁∈ℕare chosen so that0≤𝑁−1<𝛼≤𝑁.

Lemma 3. Let𝑡and𝛼be any numbers for which𝑡^𝛼and𝑡^𝛼−1are defined. ThenΔ𝑡^𝛼=𝛼𝑡^𝛼−1.

Lemma 4 (see [17]). Let0≤𝑁−1<𝛼≤𝑁.Then Δ^−𝛼Δ^𝛼𝑦(𝑡)=𝑦(𝑡)+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯+𝐶_𝑁𝑡^𝛼−𝑁, (3) for some𝐶_𝑖∈ℝ, with1≤𝑖≤𝑁.

Theorem 5 (Arzelá-Ascoli theorem, see [27]). Let𝑋be a Banach space and𝑆={𝑠(𝑡)}a function family of continuous mappings𝑠∶[𝑎,𝑏]→𝑋. If S is uniformly bounded and equicontinuous, and, for any𝑡^∗∈[𝑎,𝑏], the set{𝑠(𝑡^∗)}is relatively compact, then there exists a uniformly convergent function sequence{𝑠_𝑛(𝑡)}for𝑛=1,2,…,𝑡∈[𝑎,𝑏]in𝑆.

Lemma 6 (Mazur Lemma, see [28]). If𝑆is a compact subset of Banach space𝑋, then its convex closure conv  𝑆 is compact.

To define the solution of the boundary value problem (1) we need the following lemma, proved in [26] which deals with linear variant of the boundary value problem (1) and gives a representation of the solution.

Lemma 7. Let Λ=^{𝜂−𝛼+2}_𝑠=1𝜂+𝛽−𝑠−𝛼+1^𝛽−1Γ(𝑠+𝛼−1)Γ𝛽Γ(𝑠)−Γ(𝛼+𝑇+1)Γ(𝑇+2)≠0 (4) and let ℎ∶[𝛼−1,…,𝛼+𝑇−1]_ℕ𝛼−1→ℝbe given. Then the problem Δ^𝛼𝑢(𝑡)=ℎ(𝑡+𝛼−1),𝑡∈[0,𝑇]_ℕ0(],𝑢()𝑢(),𝛼∈1,2𝛼−2=0,𝛼+𝑇=Δ^−𝛽𝑢𝜂+𝛽,𝜂∈[𝛼−2,𝛼+𝑇−1]_ℕ𝛼−2,𝛽>0, (5) has a unique solution 𝑢(𝑡)𝑡=−^𝛼−1(𝛼)1ΛΓΓ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1ℎ()−𝜉+𝛼−1_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1ℎ()+1𝑠+𝛼−1Γ(𝛼)^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1ℎ(𝑠+𝛼−1). (6)

3. Main Results

Let𝐶([𝛼−2,𝛼+𝑇]_ℕ𝛼−2,ℝ)be the Banach space of all continuous functions𝑥with the norm‖𝑥‖=max{|𝑥(𝑡)|,𝑡∈[𝛼−2,𝛼+𝑇]_ℕ𝛼−2}.Define an operator𝐹∶𝐶([𝛼−2,𝛼+𝑇]_ℕ𝛼−2,ℝ)→𝐶([𝛼−2,𝛼+𝑇]_ℕ𝛼−2,ℝ)by (𝑡)𝑡𝐹𝑥=−^𝛼−1(𝛼)1ΛΓΓ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1𝑓(𝑥(−𝜉+𝛼−1,𝜉+𝛼−1))_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1((+1×𝑓𝑠+𝛼−1,𝑥𝑠+𝛼−1))Γ(𝛼)^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1×𝑓(𝑠+𝛼−1,𝑥(𝑠+𝛼−1)), (7) for𝑡∈[𝛼−2,𝛼+𝑇]_ℕ𝛼−2, whereΛ≠0is defined by (4). It is easy to see that the problem (1) has solutions if and only if the operator𝐹has fixed points.

In the following, for the sake of convenience, we set a constant Γ()Ω=𝑇+𝛼+1|Λ|Γ(𝛼)Γ()×1𝑇+2Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1+Γ(𝑇+𝛼+1)Γ()Γ().𝑇+1𝛼+1 (8)

Our first result is based on the Krasnoselskii’s fixed point theorem [29].

Theorem 8. Let𝐾be a bounded closed convex and nonempty subset of a Banach space𝑋. Let𝐴,𝐵be operators such that: (i)𝐴𝑥+𝐵𝑦∈𝐾whenever𝑥,𝑦∈𝐾(ii)𝐴is compact and continuous(iii)𝐵is a contraction mapping. Then there exists𝑧∈𝐾such that𝑧=𝐴𝑧+𝐵𝑧.

Theorem 9. Assume that(H₁) there exists a constant𝐿>0such that|𝑓(𝑡,𝑥)−𝑓(𝑡,𝑦)|≤𝐿|𝑥−𝑦|for each𝑡∈[𝛼−2,𝛼+𝑇]_ℕ𝛼−2and𝑥,𝑦∈ℝ;(H₂)|𝑓(𝑡,𝑢)|≤𝜇(𝑡), for all(𝑡,𝑢)∈[𝛼−2,𝑇+𝛼]_ℕ𝛼−2×ℝ, with𝜇∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ⁺);(H₃) the set𝑆₁={(𝑡−𝜎(𝑠))^𝛼−1𝑓(𝑠+𝛼−1,𝑢(𝑠+𝛼−1))∣𝑢∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ),𝑠∈{−1,…,𝑇+1}}is relatively compact for any𝑡∈[0,𝑇]_ℕ0.
If 𝐿Ω<1, (9) whereΩis given by (8), then the boundary value problem (1) has at least one solution on[𝛼−2,𝑇+𝛼]_ℕ𝛼−2.

Proof. By settingmax_{𝑡∈[𝛼−2,𝑇+𝛼]ℕ𝛼−2}|𝜇(𝑡)|=‖𝜇‖and choosing a constant 𝑅≥𝜇Ω, (10) we now consider a suitable ball𝐵_𝑅={𝑢∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2)∶‖𝑢‖≤𝑅}.
In view of Lemma 7, we define the operatorsℱ₁andℱ₂on the ball𝐵_𝑅as ℱ₁𝑢(𝑡)=1Γ(𝛼)^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1((,ℱ×𝑓𝜉+𝛼−1,𝑢𝜉+𝛼−1))₂𝑢(𝑡)𝑡=−^𝛼−1(𝛼)1ΛΓΓ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1((−×𝑓𝜉+𝛼−1,𝑢𝜉+𝛼−1))_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×𝑓(𝑠+𝛼−1,𝑢(𝑠+𝛼−1)). (11)
For𝑢,𝑣∈𝐵_𝑅, we have ℱ₁𝑢(𝑡)+ℱ₂𝑣(𝑡)≤max_{[]𝑡∈𝛼−2,𝑇+𝛼ℕ𝛼−2}𝜇Γ(𝛼)^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1+𝜇𝑡^𝛼−1|Λ|Γ(𝛼)×1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1=𝜇Γ(𝛼)max_{[]𝑡∈𝛼−2,𝛼+𝑇ℕ𝛼−2}^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1+𝜇|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×max_{[]𝑡∈𝛼−2,𝛼+𝑇ℕ𝛼−2}𝑡^𝛼−1=𝜇max_{[]𝑡∈𝛼−2,𝛼+𝑇ℕ𝛼−2}Γ(𝑡+1)Γ()Γ()+𝜇𝑡−𝛼+1𝛼+1|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×max_{[]𝑡∈𝛼−2,𝛼+𝑇ℕ𝛼−2}Γ(𝑡+1)Γ()≤𝜇Γ()𝑡−𝛼+2𝑇+𝛼+1Γ()Γ()+𝜇𝑇+1𝛼+1|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×Γ(𝑇+𝛼+1)Γ()=𝜇𝑇+2Ω≤𝑅. (12)
This implies that‖ℱ₁𝑢+ℱ₂𝑣‖≤𝑅. Therefore,ℱ₁𝑢+ℱ₂𝑣∈𝐵_𝑅. Now, we will prove thatℱ₂is a contraction mapping. For any𝑥,𝑦∈𝐶([𝛼−2,𝛼+𝑇]_ℕ𝛼−2,ℝ), we have ℱ₂𝑥(𝑡)−ℱ₂𝑦(𝑡≤1)|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1×𝑓((()+𝜉+𝛼−1,𝑥𝜉+𝛼−1))−𝑓𝜉+𝛼−1,𝑦𝜉+𝛼−1_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×𝑓((()𝑠+𝛼−1,𝑥𝑠+𝛼−1))−𝑓𝑠+𝛼−1,𝑦𝑠+𝛼−1×max_{[]𝑡∈𝛼−2,𝛼+𝑇ℕ𝛼−2}𝑡^𝛼−1≤𝐿𝑥−𝑦|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×max_{[]𝑡∈𝛼−2,𝛼+𝑇ℕ𝛼−2}Γ(𝑡+1)Γ()≤𝐿𝑡−𝛼+2𝑥−𝑦|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×Γ(𝑇+𝛼+1)Γ().𝑇+2≤𝐿Ω𝑥−𝑦 (13) By (9), we get thatℱ₂is a contraction mapping.
Next, we will show thatℱ₁is compact and continuous. Continuity of𝑓coupled with the assumption(𝐻₂)implies that the operatorℱ₁is continuous and uniformly bounded on𝐵_𝑅as ℱ₁𝑢≤‖𝑢‖Γ(𝑇+𝛼+1)Γ(𝛼)Γ()Γ().𝑇+1𝛼+1 (14)
From(𝐻₂), there exists a constant𝑁>0such thatsup_{(𝑡,𝑢)∈[𝛼−2,𝑇+𝛼]ℕ𝛼−2×𝐵𝑅}|𝑓(𝑡,𝑢)|≤𝑁. For any𝜀>0, there exists a neighborhood𝑈of𝑡,𝑡∈[𝛼−2,𝛼+𝑇]_ℕ𝛼−2, such that, for𝑡₁,𝑡₂∈𝑈, 𝑡₂𝑡₂−1⋯𝑡₁+1𝑡₂−𝛼𝑡₂−𝛼−1⋯𝑡₁<(𝛼)Γ()−𝛼+1−1𝛼Γ𝑇−𝛼+1𝑁Γ(𝑇+1)𝜀. (15)
Then, we have ℱ₁𝑢𝑡₂−ℱ₁𝑢𝑡₁=1Γ(𝛼)^𝑡2−𝛼_𝑠=0𝑡₂−𝜎(𝑠)^𝛼−1((−1×𝑓𝜉+𝛼−1,𝑢𝜉+𝛼−1))Γ(𝛼)^𝑡1−𝛼_𝑠=0𝑡₁−𝜎(𝑠)^𝛼−1((≤1×𝑓𝜉+𝛼−1,𝑢𝜉+𝛼−1))Γ(𝛼)^𝑡1−𝛼_𝑠=0𝑡₂−𝜎(𝑠)^𝛼−1−𝑡₁−𝜎(𝑠)^𝛼−1((+1×𝑓𝑠+𝛼−1,𝑢𝑠+𝛼−1))Γ(𝛼)^𝑡2−𝛼_{𝑠=𝑡1−𝛼+1}𝑡₂−𝜎(𝑠)^𝛼−1((≤𝑁×𝑓𝑠+𝛼−1,𝑢𝑠+𝛼−1))Γ(𝛼)^𝑡1−𝛼_𝑠=0𝑡₂−𝜎(𝑠)^𝛼−1−^𝑡1−𝛼_𝑠=0𝑡₁−𝜎(𝑠)^𝛼−1+^𝑡2−𝛼_{𝑠=𝑡1−𝛼+1}𝑡₂−𝜎(𝑠)^𝛼−1=𝑁(𝛼)Γ𝑡𝛼Γ²+1Γ𝑡₂−Γ𝑡−𝛼+1¹+1Γ𝑡₁=𝑁−𝛼+1(𝛼)Γ𝑡𝛼Γ¹+1Γ𝑡₁×Γ𝑡−𝛼+1₂+1Γ𝑡₁−𝛼+1Γ𝑡₂−𝛼+1Γ𝑡₁=𝑁+1−1(𝛼)Γ()𝛼Γ𝑇+1Γ()×𝑡𝑇−𝛼+1₂𝑡₂−1⋯𝑡₁+1𝑡₂−𝛼𝑡₂−𝛼−1⋯𝑡₁−𝛼+1−1<𝜀. (16) This means that the set ℱ₁(𝐵_𝑅) is an equicontinuous set.
In view of Lemma 6 and the condition that𝑆₁is relatively compact, we have thatconv𝑆₁is compact. For any𝑡^∗∈[0,𝑇]_ℕ0, 𝐹₁𝑢_𝑛𝑡^∗=1Γ(𝛼)^{𝑡∗−𝛼}_𝑠=0𝑡^∗−𝜎(𝑠)^𝛼−1((=1×𝑓𝑠+𝛼−1,𝑢𝑠+𝛼−1))Γ(𝛼)𝜃_𝑛, (17) where 𝜃_𝑛=^{𝑡∗−𝛼}_𝑠=0𝑡^∗−𝜎(𝑠)^𝛼−1𝑓(𝑠+𝛼−1,𝑢(𝑠+𝛼−1)). (18)
Since conv𝑆₁is convex and compact, we have that𝜃_𝑛∈conv𝑆₁. Therefore, for any𝑡^∗∈[0,𝑇]_ℕ0, the set(𝐹₁𝑢_𝑛)(𝑡^∗), for𝑛=1,2,…, is relatively compact. From Theorem 5, every(𝐹₁𝑢_𝑛)(𝑡)contains a uniformly convergent subsequence(𝐹₁𝑢_𝑛𝑘)(𝑡), for𝑘=1,2,…, on∈[0,𝑇]_ℕ0which means that the set is relatively compact. Sinceℱ₁(𝐵_𝑅)is a bounded, equicontinuous, and relatively compact set, we have that𝐹₁is compact on𝐵_𝑅. Therefore, all the assumptions of Theorem 8 are satisfied, and the conclusion of Theorem 8 implies that boundary value problem (1) has at least one solution on[𝛼−2,𝑇+𝛼]_ℕ𝛼−2. This completes the proof.

Corollary 10. Assume that there exists a constant𝐺>0such that|𝑓(𝑡,𝑢)|≤𝐺for any𝑡∈[0,𝑇]_ℕ0and 𝑢∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2×ℝ), and the set 𝑆={(𝑡−𝜎(𝑠))^𝛼−1𝑓(𝑠+𝛼−1,𝑢(𝑠+𝛼−1))∣𝑢∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ), and 𝑠∈{1,…,𝑇+1}} is relatively compact for every𝑡∈[0,𝑇]_ℕ0; then there exists at least one solution of boundary value problem (1) on 𝑡∈[0,𝑇]_ℕ0.

Proof. Letting 𝜇(𝑡)≡𝐺for𝑡∈[𝛼−2,𝑇+𝛼]_ℕ𝛼−2, we get the result by using Theorem 9.

The second result is based on Leray-Schauder nonlinear alternative [30].

Theorem 11. Let𝐸be a Banach space,𝐶a closed, convex subset of𝐸,𝑈an open subset of 𝐶, and0∈𝑈. Suppose that𝐹∶𝑈→𝐶is a continuous, compact (i.e.,𝐹(𝑈)is a relatively compact subset of𝐶) map. Then either(i)𝐹has a fixed point in𝑈 or (ii)there are𝑢∈𝜕𝑈(the boundary of𝑈in𝐶) and𝜆∈(0,1)with𝑢=𝜆𝐹(𝑢).

Theorem 12. Let𝑓∶[𝛼−2,𝑇+𝛼]_ℕ𝛼−2×ℝ→ℝbe a continuous function. Assume that(H₄) there exist a function𝑝∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ⁺) and a continuous nondecreasing functionΨ∶ℝ⁺→ℝ⁺such that|𝑓(𝑡,𝑢)|≤𝑝(𝑡)Ψ(‖𝑢‖), for all(𝑡,𝑢)∈[𝛼−2,𝑇+𝛼]_ℕ𝛼−2×ℝ;(H₅) the set𝑆=𝑆₁∩𝑆₂is relatively compact for any𝑡∈[0,𝑇]_ℕ0, where𝑆₁is defined as in(𝐻₃)and𝑆₂={𝑡^𝛼−1𝑓(𝑠+𝛼−1,𝑢(𝑠+𝛼−1))∣𝑢∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ),𝑠∈{−1,…,𝑇+1}};(H₆) there exists a constant𝑀>0such that 𝑀Γ(𝛼)Ψ(𝑀)𝑝Φ>1, (19)where Γ()Φ=𝑇+𝛼+1|Λ|Γ()×1𝑇+2Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1. (20)
Then the boundary value problem (1) has at least one solution on[𝛼−2,𝑇+𝛼]_ℕ𝛼−2.

Proof. We consider the operator equation𝑢=𝐹(𝑢), where𝐹∶𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2)→𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2)defined by (7).
Now, we will show that𝐹maps bounded sets into bounded sets in𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ). For a positive number𝑟, let𝐵_𝑟={𝑢∈𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ)∣‖𝑢‖≤𝑟}be a bounded set in𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ). Then |(𝑡≤𝑡𝐹𝑢)|_𝛼−1|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1×𝑓((+𝜉+𝛼−1,𝑢𝜉+𝛼−1))_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×𝑓((+1𝑠+𝛼−1,𝑢𝑠+𝛼−1))Γ(𝛼)^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1×𝑓((≤𝑡𝑠+𝛼−1,𝑢𝑠+𝛼−1))_𝛼−1|Λ|Γ(𝛼)1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1𝑝()Ψ𝑢+𝜉+𝛼−1(‖‖)_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1()Ψ𝑢+1×𝑝𝑠+𝛼−1(‖‖)Γ(𝛼)^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1()Ψ𝑢≤Ψ𝑢)×𝑝𝑠+𝛼−1(‖‖)(‖‖)(𝑇+𝛼^𝛼−1|Λ|Γ(𝛼)×1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1()+×𝑝𝜉+𝛼−1_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1𝑝()+Ψ𝑢𝑠+𝛼−1(‖‖)Γ(𝛼)^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1𝑝()=Ψ𝑢𝑠+𝛼−1(‖‖)Γ(𝛼)Γ(𝑇+𝛼+1)|Λ|Γ()×1𝑇+2Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1()+×𝑝𝑠+𝛼−1_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1𝑝()+𝑠+𝛼−1_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1𝑝(𝑠+𝛼−1). (21)
Consequently, ‖‖≤Ψ(𝑟𝑝𝐹𝑢)Γ(𝛼)×Γ(𝑇+𝛼+1)|Λ|Γ()×1𝑇+2Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1. (22)
Next we show that𝐹maps bounded sets into equicontinuous sets of𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ). Let𝑢∈𝐵_𝑟, where𝐵_𝑟is a bounded set of𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ). For any𝜀>0, there exists a neighborhood𝑈of𝑡,𝑡∈[𝛼−2,𝑇+𝛼]_ℕ𝛼−2, such that, for𝑡₁,𝑡₂∈𝑈, 𝑡^𝛼−1₂−𝑡^𝛼−1₁𝑝Ψ(𝑟)|Λ|Γ(𝛼)×1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1+𝑝Ψ(𝑟)Γ(𝛼)Γ(𝑇+1)Γ()×𝑡𝑇−𝛼+1₂𝑡₂−1⋯𝑡₁+1𝑡₂−𝛼𝑡₂−𝛼−1⋯𝑡₁−𝛼+1−1<𝜀. (23)
Then, we have 𝐹𝑢𝑡₂−𝐹𝑢𝑡₁=−𝑡^𝛼−1₂−𝑡^𝛼−1₁(𝛼)×1ΛΓΓ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1((−×𝑓𝜉+𝛼−1,𝑢𝜉+𝛼−1))_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1((+1×𝑓𝑠+𝛼−1,𝑢𝑠+𝛼−1))Γ(𝛼)^𝑡1−𝛼_𝑠=0𝑡₂−𝜎(𝑠)^𝛼−1−𝑡₁−𝜎(𝑠)^𝛼−1((+×𝑓𝑠+𝛼−1,𝑢𝑠+𝛼−1))_{𝑡2−𝛼𝑠=𝑡1−𝛼+1}𝑡₂−𝜎(𝑠)^𝛼−1((≤𝑡×𝑓𝑠+𝛼−1,𝑢𝑠+𝛼−1))_𝛼−12−𝑡^𝛼−1₁𝑝Ψ(𝑟)|Λ|Γ(𝛼)×1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1+𝑝Ψ(𝑟)Γ(𝛼)×^𝑡1−𝛼_𝑠=0𝑡₂−𝜎(𝑠)^𝛼−1−𝑡₁−𝜎(𝑠)^𝛼−1+^𝑡2−𝛼_{𝑠=𝑡1−𝛼+1}𝑡₂−𝜎(𝑠)^𝛼−1=𝑡^𝛼−1₂−𝑡^𝛼−1₁𝑝Ψ(𝑟)|Λ|Γ(𝛼)×1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1+𝑝Ψ(𝑟)Γ(𝛼)Γ(𝑇+1)Γ()×𝑡𝑇−𝛼+1₂𝑡₂−1⋯𝑡₁+1𝑡₂−𝛼𝑡₂−𝛼−1⋯𝑡₁−𝛼+1−1<𝜀. (24)
This means that the set𝐹(𝐵_𝑟)is an equicontinuous set. By(𝐻₅), Lemma 6, and Theorem 5, it follows that𝐹∶𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ)is completely continuous.
Let𝑢=𝐹𝑢. Then, for𝑡∈[𝛼−2,𝑇+𝛼]_ℕ𝛼−2 and using the computations as in the first step, we get |𝑢(𝑡≤𝑡)|^𝛼−1|Λ|Γ(𝛼)×1Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1×𝑓((+𝜉+𝛼−1,𝑥𝜉+𝛼−1))_𝑇𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1×𝑓((+1𝑠+𝛼−1,𝑥𝑠+𝛼−1))Γ(𝛼)^𝑡−𝛼_𝑠=0(𝑡−𝜎(𝑠))^𝛼−1×𝑓((≤Ψ𝑢𝑝𝑠+𝛼−1,𝑥𝑠+𝛼−1))(‖‖)Γ(𝛼)×Γ(𝑇+𝛼+1)|Λ|Γ()×1𝑇+2Γ𝛽^𝜂_𝑠=𝛼^𝑠−𝛼_𝜉=0𝜂+𝛽−𝜎(𝑠)^𝛽−1×(𝑠−𝜎(𝜉))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1+^𝑇_𝑠=0(𝑇+𝛼−𝜎(𝑠))^𝛼−1. (25)
Consequently, ‖𝑢‖Γ(𝛼)Ψ(‖𝑢‖)𝑝Φ≤1. (26) In view of(𝐻₄), there exists𝑀such that‖𝑢‖<𝑀. Let us set 𝑈=𝑢∈𝐶[𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ∣‖𝑢‖<𝑀. (27)
Note that the operator𝐹∶𝑈→𝐶([𝛼−2,𝑇+𝛼]_ℕ𝛼−2,ℝ)is continuous and completely continuous. From the choice of𝑈, there is no𝑢∈𝜕𝑈such that𝑢=𝜆𝐹𝑢for some𝜆∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Theorem 11), we deduce that𝐹has a fixed point𝑢∈𝑈which is a solution of the problem (1). This completes the proof.

4. Some Examples

In this section, in order to illustrate our result, we consider two examples.

Example 1. Consider the following three-point fractional sum boundary value problem: Δ^3/2𝑥(𝑡)=𝑒^−sin2𝑡(𝑡+7/2)²⋅|𝑥||𝑥|1+1,𝑡∈2,92_ℕ1/2,𝑥−12𝑥=0,112=Δ^−4/3𝑥236. (28)

Here𝛼=3/2,𝛽=4/3,𝜂=5/2,𝑇=4, and 𝑓(𝑡,𝑥)=𝑒^−sin2𝑡/(𝑡+3)²⋅|𝑥|/(|𝑥|+1).Also we find Λ=³_𝑠=1(10/3−𝑠)^1/3Γ(𝑠+1/2)()Γ()Γ(𝑠)−Γ()1/31/313/2Γ(6)Γ()=−283≠0,Ω=13/2||Γ()Γ(6)×1−2833/2Γ(4/3)^5/2_𝑠=3/2^𝑠−3/2_𝜉=052+43−𝜎(𝑠)^4/3−1×(𝑠−𝜎(𝜉))^3/2−1+⁴_𝑠=034+2−𝜎(𝑠)^3/2−1+Γ(4+3/2+1)Γ(4+1)Γ(3/2+1)≈10.29. (29)

Since|𝑓(𝑡,𝑥)−𝑓(𝑡,𝑦)|≤(4/49)|𝑥−𝑦|is satisfied with𝛾=4/49, we can show that 4𝛾Ω≈49×10.29=0.84<1. (30)

Hence, by Theorem 9, the boundary value problem (28) has at least one solution on[1/2,9/2]_ℕ1/2.

Example 2. Consider the following three-point fractional sum boundary value problem: Δ^3/2𝑥(𝑡)=𝑡+1/2()+()𝑥100𝜋sin𝜋𝑥𝑡+3/2²1+𝑥²,1𝑡∈2,92_ℕ1/2,𝑥−12𝑥=0,112=Δ^−4/3𝑥236. (31)

Here 𝛼=3/2,𝛽=4/3,𝜂=5/2,𝑇=4,𝑓(𝑡,𝑥)=(𝑡/100𝜋)sin(𝜋𝑥)+(𝑡+1)𝑥²/(1+𝑥²),Λ=−283≠0, and Φ=8.168. It is clear that |𝑓(𝑡,𝑥)|=|(𝑡/100𝜋)sin(𝜋𝑥)+(𝑡+1)𝑥²/(1+𝑥²)|≤(𝑡+1)‖𝑥‖/100+1.

We choose𝑝(𝑡)=𝑡+1,Ψ(‖𝑥‖)=‖𝑥‖/100+1and find that 𝑀Γ(𝛼)𝑝Φ>𝑀100+1 (32) which implies𝑀>103.27. Hence, by Theorem 12, the boundary value problem (31) has at least one solution on [1/2,9/2]_ℕ1/2.

Acknowledgments

The authors would like to thank the reviewers for their valuable comments and suggestions on the paper. This research of T. Sitthiwirattham and J. Tariboon is supported by King Mongkut’s University of Technology North Bangkok, Thailand. Sotiris K. Ntouyas is Member of Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, King Abdulaziz University, Jeddah, Saudi Arabia.