Boundary Value Problems for a Class of Sequential Integrodifferential Equations of Fractional Order

Abstract

We investigate the existence of solutions for a sequential integrodifferential equation of fractional order with some boundary conditions. The existence results are established by means of some standard tools of fixed point theory. An illustrative example is also presented.

1. Introduction

Nonlinear boundary value problems of fractional differential equations have received a considerable attention in the last few decades. One can easily find a variety of results ranging from theoretical analysis to asymptotic behavior and numerical methods for fractional equations in the literature on the topic. The interest in the subject has been mainly due to the extensive applications of fractional calculus in the mathematical modeling of several real-world phenomena occurring in physical and technical sciences; see, for example, [1–4]. An important feature of a fractional order differential operator, distinguishing it from an integer-order differential operator, is that it is nonlocal in nature. It means that the future state of a dynamical system or process based on a fractional operator depends on its current state as well as its past states. Thus, differential equations of arbitrary order are capable of describing memory and hereditary properties of some important and useful materials and processes. This feature has fascinated many researchers, and they have shifted their focus to fractional order models from the classical integer-order models. For some recent work on the topic, we refer, for instance, to [5–9]. Recently, in [10], the authors studied sequential fractional differential equations with three-point boundary conditions.

In this paper, we consider a nonlinear Dirichlet boundary value problem of sequential fractional integrodifferential equations given by ^𝑐𝐷^𝛼+𝑘^𝑐𝐷^𝛼−1𝑢(𝑡)=𝑝𝑓(𝑡,𝑢(𝑡))+𝑞𝐼^𝛽𝑔((𝑡,𝑢(0)(1)𝑡,𝑢))0<𝑡<1,=0,𝑢=0, (1)(2) where ^𝑐𝐷^𝛼 denotes the Caputo fractional derivative of order 𝛼,1<𝛼≤2, 𝐼^𝛽(⋅) denotes Riemann-Liouville integral with 0<𝛽<1,  𝑓,𝑔 are given continuous functions, 𝑘≠0, and 𝑝,𝑞 are real constants. We also study the fractional integro-differential equation (1) subject to the following boundary conditions: 𝑢^′(0)(0)(1)𝑢(0)+𝑘𝑢=𝑎,𝑢=𝑏,𝑎,𝑏∈ℝ,=𝑎,𝑢^′(0)=𝑢^′(1),𝑎∈ℝ. (3)(4)

2. Linear Fractional Differential Equations

For 𝛼∈(1,2], we consider the following linear fractional differential equation: ^𝑐𝐷^𝛼+𝑘^𝑐𝐷^𝛼−1𝑢(𝑡)=ℎ(𝑡), (5) where ^𝑐𝐷^𝛼 denotes the Caputo fractional derivative of order 𝛼. Rewriting (1) as ^𝑐𝐷^𝛼(𝑢(𝑡)+𝑘^𝑐𝐷⁻¹𝑢(𝑡))=ℎ(𝑡), we can write its solution as 𝑢(𝑡)+𝑘^𝑐𝐷⁻¹𝑢(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1ℎ(𝑠)𝑑𝑠+𝑐₀+𝑐₁𝑡, (6) where 𝑐₀,𝑐₁ are arbitrary constants. Now, (6) can be expressed as 𝑢(𝑡)=−𝑘^𝑡₀𝑢(𝑠)1𝑑𝑠+Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1ℎ(𝑠)𝑑𝑠+𝑐₀+𝑐₁𝑡. (7) Differentiating (7), we obtain 𝑢^′(𝑡)(𝑡)+1=−𝑘𝑢Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2ℎ(𝑠)𝑑𝑠+𝑐₁, (8) which can alternatively be written as 𝑢(𝑡)𝑒^𝑘𝑡′=𝑒^𝑘𝑡1Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2ℎ(𝑠)𝑑𝑠+𝑐₁. (9)

Integrating from 0 to 𝑡, we have 𝑢(𝑡)=𝐴𝑒^−𝑘𝑡+^𝑡₀𝑒^{−𝑘(𝑡−𝑠)}𝐼^𝛼−1ℎ(𝑠)𝑑𝑠+𝐵, (10) where 𝐴 and 𝐵 are arbitrary constants, and 𝐼^𝛼−1ℎ(𝑡)=^𝑡₀(𝑡−𝑥)^𝛼−2Γ()ℎ(𝑥)𝛼−1𝑑𝑥. (11)

Lemma 1. The unique solution of the linear equation (5) subject to the Dirichlet boundary conditions (2) is given by 𝑢(𝑡)=1−𝑒^−𝑘𝑡𝑒^−𝑘−1¹₀𝑒^{−𝑘(1−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()ℎ(𝑥)+𝛼−1𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()ℎ(𝑥)𝛼−1𝑑𝑥𝑑𝑠. (12)

Proof. Observe that the general solution of (5) is given by (10). Using the given boundary conditions in (10), we find that 1𝐴=−𝐵=1−𝑒^−𝑘1₀𝑒^{−𝑘(1−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()ℎ(𝑥)𝛼−1𝑑𝑥𝑑𝑠. (13) Substituting the values of 𝐴 and 𝐵 in (10) yields the solution (12). This completes the proof.

In the next two lemmas, we present the unique solutions of (5) with different kinds of boundary conditions. We do not provide the proofs for these lemmas as they are similar to that of Lemma 1.

Lemma 2. The unique solution of the problem (5)–(3) is given by 𝑢(𝑡)=𝑒^{𝑘(1−𝑡)}×(𝑏𝑘−𝑎)𝑘−¹₀𝑒^{−𝑘(1−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()ℎ(𝑥)+𝛼−1𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()ℎ(𝑥)𝑎𝛼−1𝑑𝑥𝑑𝑠+𝑘. (14)

Lemma 3. The unique solution of (5) with the boundary conditions (4) is 𝑢(𝑡)=−1−𝑒^−𝑘𝑡𝑘1−𝑒^−𝑘×𝑘¹₀𝑒^{−𝑘(1−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()ℎ(𝑥)−𝛼−1𝑑𝑥𝑑𝑠₁₀(1−𝑠)^𝛼−2Γ()ℎ(𝑠)+𝛼−1𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()ℎ(𝑥)𝛼−1𝑑𝑥𝑑𝑠+𝑎. (15)

3. Existence Results for the Nonlinear Problems

Let 𝒫=𝐶([0,1],ℝ) denote the Banach space of all continuous functions from [0,1] into ℝ endowed with the usual norm defined by ‖𝑥‖=sup{|𝑥(𝑡)|,𝑡∈[0,1]}.

In view of Lemma 1, we transform problem (1)-(2) to an equivalent fixed point problem as 𝑢=𝒱𝑢, (16) where 𝒱∶𝒫→𝒫 is defined by (𝑡)=𝒱𝑢)(1−𝑒^−𝑘𝑡𝑒^−𝑘−1¹₀𝑒^{−𝑘(1−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥+𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠. (17)

In a similar manner, we can define a fixed point operator 𝒱₁∶𝒫→𝒫 for the nonlinear problem (1)–(3) as follows: 𝒱₁𝑢(𝑡)=𝑒^{𝑘(1−𝑡)}(𝑏𝑘−𝑎)𝑘−¹₀𝑒^{−𝑘(1−𝑠)}×𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥+𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥𝑎𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠+𝑘. (18) A fixed point operator 𝒱₂∶𝒫→𝒫 for the nonlinear problem (1)–(4) is defined by 𝒱₂𝑢(𝑡)=−1−𝑒^−𝑘𝑡𝑘1−𝑒^−𝑘×𝑘¹₀𝑒^{−𝑘(1−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()((𝑥𝛼−1×𝑓𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠−𝑝¹₀(1−𝑠)^𝛼−2Γ()𝑓((𝑠𝛼−1𝑠,𝑢))𝑑𝑠−𝑞¹₀(1−𝑠)^{𝛼+𝛽−2}Γ𝑔((𝑠+𝛼+𝛽−1𝑠,𝑢))𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠+𝑎. (19)

We only present the existence results for the problem (1)-(2). Observe that problem (1)-(2) has solutions if the operator equation (16) has fixed points.

For computational convenience, we introduce the following constant: 2𝑄=1−𝑒^−𝑘𝑝Γ𝛼+𝛽+𝑞Γ(𝛼)|𝑘|ΓΓ(𝛼).𝛼+𝛽 (20)

Theorem 4. Assume that 𝑓,𝑔∶[0,1]×ℝ→ℝ are continuous functions satisfying the following condition: 𝐴₁𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)≤𝐿₁||,𝑔()(𝑢−𝑣𝑡,𝑢−𝑔𝑡,𝑣)≤𝐿₂||[],𝐿𝑢−𝑣,∀𝑡∈0,1₁,𝐿₂>0,𝑢,𝑣∈ℝ. (21) Then, the boundary value problem (1)-(2) has a unique solution if 𝐿<1/𝑄, where 𝐿=max{𝐿₁,𝐿₂} and 𝑄 is given by (20).

Proof. Let us define 𝑀=max{𝑀₁,𝑀₂}, where 𝑀₁,𝑀₂ are finite numbers given by sup_𝑡∈[0,1]|𝑓(𝑡,0)|=𝑀₁,sup_𝑡∈[0,1]|𝑔(𝑡,0)|=𝑀₂. Selecting 𝑟≥(𝑄𝑀)/(1−𝐿𝑄), we show that 𝒱𝐵_𝑟⊂𝐵_𝑟, where 𝐵_𝑟={𝑢∈𝒫∶‖𝑢‖≤𝑟}. For 𝑢∈𝐵_𝑟, we have ‖(𝒱𝑢)‖≤sup_[]𝑡∈0,11−𝑒^−𝑘𝑡1−𝑒^−𝑘×¹₀𝑒^{−𝑘(1−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()×𝑓((𝑥+𝑞𝛼−1𝑥,𝑢))𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ×𝑔((𝑥+𝛼+𝛽−1𝑥,𝑢))𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}×𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()×𝑓((𝑥+𝑞𝛼−1𝑥,𝑢))𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ×𝑔((𝑥𝛼+𝛽−1𝑥,𝑢))𝑑𝑥𝑑𝑠≤sup_[]𝑡∈0,11−𝑒^−𝑘𝑡1−𝑒^−𝑘1₀𝑒^{−𝑘(1−𝑠)}×𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()×𝑓((𝑥(+𝑓(+𝑞𝛼−1𝑥,𝑢))−𝑓𝑥,0)𝑥,0)𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ×𝑔((𝑥(+𝑔(+𝛼+𝛽−1𝑥,𝑢))−𝑔𝑥,0)𝑥,0)𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}×𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()×𝑓((𝑥(+𝑓(+𝑞𝛼−1𝑥,𝑢))−𝑓𝑥,0)𝑥,0)𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ×𝑔((𝑥(+𝑔(≤𝑝𝐿𝛼+𝛽−1𝑥,𝑢))−𝑔𝑥,0)𝑥,0)𝑑𝑥𝑑𝑠₁𝑟+𝑀₁×sup_[]𝑡∈0,11−𝑒^−𝑘𝑡1−𝑒^−𝑘×¹₀𝑒^{−𝑘(1−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()+𝛼−1𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()+𝑞𝐿𝛼−1𝑑𝑥𝑑𝑠₂𝑟+𝑀₂×sup_[]𝑡∈0,11−𝑒^−𝑘𝑡1−𝑒^−𝑘×¹₀𝑒^{−𝑘(1−𝑠)𝑠}₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ+𝛼+𝛽−1𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}×^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ≤()2𝛼+𝛽−1𝑑𝑥𝑑𝑠𝐿𝑟+𝑀1−𝑒^−𝑘𝑝Γ𝛼+𝛽+𝑞Γ(𝛼)|𝑘|ΓΓ(𝛼)=()𝛼+𝛽𝐿𝑟+𝑀𝑄≤𝑟, (22) which means that 𝒱𝐵_𝑟⊂𝐵_𝑟.
Now, for 𝑢,𝑣∈𝒫, we obtain ‖𝒱𝑢−𝒱𝑣‖≤sup_[]𝑡∈0,11−𝑒^−𝑘𝑡1−𝑒^−𝑘×¹₀𝑒^{−𝑘(1−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()×𝑓((𝑠((𝑠+𝑞𝛼−1𝑠,𝑢))−𝑓𝑠,𝑣))𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ×𝑔((𝑠((𝑠+𝛼+𝛽−1𝑠,𝑢))−𝑔𝑠,𝑣))𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}×𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()×𝑓((𝑠((𝑠+𝑞𝛼−1𝑠,𝑢))−𝑓𝑠,𝑣))𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ×𝑔((𝑠((𝑠2𝛼+𝛽−1𝑠,𝑢))−𝑔𝑠,𝑣))𝑑𝑥𝑑𝑠≤𝐿1−𝑒^−𝑘𝑝Γ𝛼+𝛽+𝑞Γ(𝛼)|𝑘|ΓΓ(𝛼)‖‖‖‖.𝛼+𝛽𝑢−𝑣=𝐿𝑄𝑢−𝑣 (23) By the given assumption, 𝐿<1/𝑄,  𝒱 is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem).

Our next existence result relies on Krasnoselskii's fixed point theorem.

Lemma 5 (Krasnoselskii, see [11]). Let 𝑀 be a closed, convex, bounded, and nonempty subset of a Banach space 𝑋. Let 𝐴,𝐵 be the operators such that (i) 𝐴𝑥+𝐵𝑦∈𝑀 whenever 𝑥,𝑦∈𝑀, (ii) 𝐴 is compact, and continuous, and (iii) 𝐵 is a contraction mapping. Then, there exists 𝑧∈𝑀 such that 𝑧=𝐴𝑧+𝐵𝑧.

Theorem 6. Let 𝑓,𝑔∶[0,1]×ℝ→ℝ be continuous functions satisfying assumption (𝐴₁), and 𝐴₂𝑓(𝑡,𝑢)≤𝜇₁(𝑡),𝑔(𝑡,𝑢)≤𝜇₂(𝑡),∀(𝑡,𝑢)∈[0,1]×ℝ,𝜇₁,𝜇₂∈𝐶[0,1],ℝ⁺. (24) Then, the problem (1)-(2) has at least one solution on [0,1] provided that 1−𝑒^−𝑘𝑝𝜇₁Γ𝛼+𝛽+𝑞𝜇₂Γ(𝛼)|𝑘|Γ𝛼+𝛽Γ(𝛼)<1, (25) where sup_𝑡∈[0,1]|𝜇_𝑖(𝑡)|=‖𝜇_𝑖‖,𝑖=1,2.

Proof. Let us fix 2𝑟≥1−𝑒^−𝑘𝑝𝜇₁Γ𝛼+𝛽+𝑞𝜇₂Γ(𝛼)|𝑘|Γ𝛼+𝛽Γ(𝛼) (26) and consider 𝐵_𝑟={𝑢∈𝒫∶‖𝑢‖≤𝑟}. We define the operators 𝜓₁ and 𝜓₂ on 𝐵_𝑟 as 𝜓₁𝑢(𝑡)=^𝑡₀𝑒^{−𝑘(𝑡−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥[],𝜓𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠,𝑡∈0,1₂𝑢(𝑡)=1−𝑒^−𝑘𝑡𝑒^−𝑘×−1₁₀𝑒^{−𝑘(1−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥[].𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠,𝑡∈0,1 (27) For 𝑢,𝑣∈𝐵_𝑟, we find that 𝜓₁𝑢+𝜓₂𝑣≤21−𝑒^−𝑘𝑝𝜇₁Γ𝛼+𝛽+𝑞𝜇₂Γ(𝛼)|𝑘|ΓΓ(𝛼)≤𝛼+𝛽𝑟. (28)
Thus, 𝜓₁𝑢+𝜓₂𝑣∈𝐵_𝑟. It follows from assumption (𝐴₁) together with (25) that 𝜓₂ is a contraction mapping. Continuities of 𝑓 and 𝑔 imply that the operator 𝜓₁ is continuous. Also, 𝜓₁ is uniformly bounded on 𝐵_𝑟  as 𝜓₁𝑢≤1−𝑒^−𝑘𝑝𝜇₁Γ𝛼+𝛽+𝑞𝜇₂Γ(𝛼)|𝑘|ΓΓ(𝛼).𝛼+𝛽 (29) Now, we prove the compactness of the operator 𝜓₁. In view of (𝐴₁), we define sup_{(𝑡,𝑢)∈[0,1]×𝐵𝑟}𝑓(𝑡,𝑢)=𝑓,sup_{()∈[]𝑡,𝑢0,1×𝐵𝑟}𝑔(𝑡,𝑢)=𝑔. (30) Consequently, we have 𝜓₁𝑢𝑡₂−𝜓₁𝑢𝑡₁≤𝑒^−𝑘𝑡2−𝑒^{−𝑘𝑡1𝑡1}₀𝑒^𝑘𝑠𝑝𝑓^𝑠₀(𝑠−𝑥)^𝛼−2Γ()+𝑞𝛼−1𝑑𝑥𝑔^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ𝛼+𝛽−1𝑑𝑥𝑑𝑠≤^𝑡2_𝑡1𝑒^{−𝑘(𝑡2−𝑠)}×𝑝𝑓^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑞𝛼−1𝑑𝑥+𝑔^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ𝛼+𝛽−1𝑑𝑥𝑑𝑠 (31) which is independent of 𝑢 and tends to zero as 𝑡₂→𝑡₁. Thus, 𝜓₁ is relatively compact on 𝐵_𝑟. Hence, by the Arzelá-Ascoli theorem, 𝜓₁ is compact on 𝐵_𝑟. Thus, all the assumptions of Lemma 5 are satisfied. So, by the conclusion of Lemma 5, problem (1)-(2) has at least one solution on [0,1].

Now, we show the existence of solutions for the problem (1)-(2) via Leray-Schauder alternative.

Lemma 7 (nonlinear alternative for single valued maps, see [12]). Let 𝐸 be a Banach space, 𝐶 a closed, convex subset of 𝐸,𝑈 an open subset of 𝐶, and 0∈𝑈. Suppose that 𝐹∶𝑈→𝐶 is a continuous, compact (that is, 𝐹(𝑈) is a relatively compact subset of 𝐶) map. Then, either(i)𝐹has a fixed point in 𝑈, or (ii)there is a 𝑢∈𝜕𝑈 (the boundary of 𝑈 in 𝐶) and 𝜆∈(0,1) with 𝑢=𝜆𝐹(𝑢).

Theorem 8. Let 𝑓,𝑔∶[0,1]×ℝ→ℝ be continuous functions and the following assumptions hold.(𝐴₃) There exist functions 𝜎₁,𝜎₂∈𝐶([0,1],ℝ⁺), and nondecreasing functions 𝜓₁,𝜓₂∶ℝ⁺→ℝ⁺ such that |𝑓(𝑡,𝑢)|≤𝜎₁(𝑡)𝜓₁(‖𝑢‖),|𝑔(𝑡,𝑢)|≤𝜎₂(𝑡)𝜓₂(‖𝑢‖),  for  all  (𝑡,𝑢)∈[0,1]×ℝ.(𝐴₄) There exists a constant 𝑀>0 such that 𝑀×21−𝑒^−𝑘𝑝𝜓₁(‖𝑢‖)Γ𝛼+𝛽𝜎₁+𝑞𝜓₂(‖𝑢‖)𝜎₂Γ(𝛼×|𝑘|ΓΓ(𝛼))𝛼+𝛽⁻¹⁻¹>1. (32)Then, the boundary value problem (1)-(2) has at least one solution on [0,1].

Proof. Consider the operator 𝒱∶𝒫→𝒫 with 𝑢=𝒱𝑢, where (𝒱𝑢)(𝑡)=1−𝑒^−𝑘𝑡𝑒^−𝑘×−1₁₀𝑒^{−𝑘(1−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()((𝑥𝛼−1×𝑓𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥+𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥𝛼−1𝑥,𝑢))𝑑𝑥+𝑞^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ((𝑥𝛼+𝛽−1×𝑔𝑥,𝑢))𝑑𝑥𝑑𝑠. (33) We show that 𝒱 maps bounded sets into bounded sets in 𝐶([0,1],ℝ). For a positive number 𝑟, let 𝐵_𝑟={𝑢∈𝐶([0,1],ℝ)∶‖𝑢‖≤𝑟} be a bounded set in 𝐶([0,1],ℝ). Then, 𝑡≤|(𝒱𝑢)()|1−𝑒^−𝑘𝑡1−𝑒^−𝑘1₀𝑒^{−𝑘(1−𝑠)}×𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥+𝑞𝛼−1𝑥,𝑢))𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ𝑔((𝑥+𝛼+𝛽−1𝑥,𝑢))𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝑓((𝑥+𝑞𝛼−1𝑥,𝑢))𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ×𝑔((𝑥≤𝛼+𝛽−1𝑥,𝑢))𝑑𝑥𝑑𝑠1−𝑒^−𝑘𝑡1−𝑒^−𝑘×¹₀𝑒^{−𝑘(1−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝜎𝛼−1¹(𝑥)𝜓₁𝑢+𝑞(‖‖)𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ𝜎𝛼+𝛽−1²(𝑥)𝜓₂(‖𝑢‖)𝑑𝑥𝑑𝑠+^𝑡₀𝑒^{−𝑘(𝑡−𝑠)}𝑝^𝑠₀(𝑠−𝑥)^𝛼−2Γ()𝜎𝛼−1¹(𝑥)𝜓₁𝑢+𝑞(‖‖)𝑑𝑥_𝑠0(𝑠−𝑥)^{𝛼+𝛽−2}Γ𝛼+𝛽−1×𝜎₂(𝑥)𝜓₂𝑢≤𝑝𝜓(‖‖)𝑑𝑥𝑑𝑠₁(𝑟)𝜎₁×1−𝑒^−𝑘𝑡1−𝑒^−𝑘∫¹₀𝑒^{−𝑘(1−𝑠)}∫^𝑠₀(𝑠−𝑥)^𝛼−2Γ()+𝛼−1𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)𝑠}₀(𝑠−𝑥)^𝛼−2Γ()+𝑞𝜎𝛼−1𝑑𝑥𝑑𝑠₂𝜓₂(𝑟)1−𝑒^−𝑘𝑡1−𝑒^−𝑘×¹₀𝑒^{−𝑘(1−𝑠)𝑠}₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ+𝛼+𝛽−1𝑑𝑥𝑑𝑠_𝑡0𝑒^{−𝑘(𝑡−𝑠)𝑠}₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ≤2𝛼+𝛽−1𝑑𝑥𝑑𝑠1−𝑒^−𝑘𝑝𝜓₁(‖𝑢‖)Γ𝛼+𝛽𝜎₁+𝑞𝜓₂(‖𝑢‖)𝜎₂Γ(𝛼×|𝑘|ΓΓ(𝛼))𝛼+𝛽⁻¹. (34) Consequently, ‖𝒱𝑥‖≤21−𝑒^𝑘×𝑝𝜓₁(‖𝑢‖)Γ𝛼+𝛽𝜎₁+𝑞𝜓₂(‖𝑢‖)𝜎₂Γ(𝛼×|𝑘|ΓΓ(𝛼)𝛼+𝛽)⁻¹. (35)
Next, we show that 𝒱 maps bounded sets into equicontinuous sets of 𝐶([0,1],ℝ). Let 𝑡₁,𝑡₂∈[0,1] with 𝑡₁<𝑡₂ and 𝑢∈𝐵_𝑟, where 𝐵_𝑟 is a bounded set of 𝐶([0,1],ℝ). Then, we obtain(𝒱𝑢)𝑡₂−(𝒱𝑢)𝑡₁≤−𝑒^−𝑘𝑡2+𝑒^−𝑘𝑡11−𝑒^−𝑘×¹₀𝑒^{−𝑘(1−𝑠)}𝑝𝜓₁(𝑟)𝜎₁×^𝑠₀(𝑠−𝑥)^𝛼−2Γ()+𝑞𝜓𝛼−1𝑑𝑥₂(𝑟)𝜎₂×^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ+𝑒𝛼+𝛽−1𝑑𝑥𝑑𝑠_−𝑘𝑡2−𝑒^−𝑘𝑡1×^𝑡1₀𝑒^𝑘𝑠𝑝𝜓₁(𝑟)𝜎₁×^𝑠₀(𝑠−𝑥)^𝛼−2Γ()+𝑞𝜓𝛼−1𝑑𝑥₂(𝑟)𝜎₂×^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ𝛼+𝛽−1𝑑𝑥𝑑𝑠+^𝑡2_𝑡1𝑒^{−𝑘(𝑡2−𝑠)}𝑝𝜓₁(𝑟)𝜎₁^𝑠₀(𝑠−𝑥)^𝛼−2Γ()+𝑞𝜓𝛼−1𝑑𝑥₂(𝑟)𝜎₂×^𝑠₀(𝑠−𝑥)^{𝛼+𝛽−2}Γ𝛼+𝛽−1𝑑𝑥𝑑𝑠. (36) Obviously, the right hand side of the previous inequality tends to zero independently of 𝑢∈𝐵_𝑟 as 𝑡₂−𝑡₁→0. As 𝒱 satisfies the previous assumptions, therefore it follows by the Arzelá-Ascoli theorem that 𝒱∶𝐶([0,1],ℝ)→𝐶([0,1],ℝ) is completely continuous.
The proof will be complete by the application of the Leray-Schauder nonlinear alternative (Lemma 7) once we establish the boundedness of the set of all solutions to equations 𝑢=𝜆𝒱𝑢 for 𝜆∈(0,1).
Let 𝑢 be a solution. Then, for 𝑡∈[0,1], and using the computations in proving that 𝒱 is bounded, we have |𝑢(𝑡=|𝜆(𝑡≤2)|𝒱𝑢)()|1−𝑒^−𝑘×𝑝𝜓₁(‖𝑢‖)Γ𝛼+𝛽𝜎₁+𝑞𝜓₂(‖𝑢‖)𝜎₂Γ(𝛼×|𝑘|ΓΓ(𝛼))𝛼+𝛽⁻¹. (37) Consequently, we have ‖𝑢‖×21−𝑒^−𝑘𝑝𝜓₁(‖𝑢‖)Γ𝛼+𝛽𝜎₁+𝑞𝜓₂(‖𝑢‖)𝜎₂Γ(𝛼×)|𝑘|Γ(𝛼+𝛽)Γ(𝛼)⁻¹⁻¹≤1. (38) In view of (𝐴₄), there exists 𝑀 such that ‖𝑢‖≠𝑀. Let us set 𝑈={𝑢∈𝐶([0,1],ℝ)∶‖𝑢‖<𝑀}. (39) Note that the operator 𝒱∶𝑈→𝐶([0,1],ℝ) is continuous and completely continuous. From the choice of 𝑈, there is no 𝑢∈𝜕𝑈 such that 𝑢=𝜆𝒱(𝑢) for some 𝜆∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 7), we deduce that 𝒱 has a fixed point 𝑢∈𝑈 which is a solution of the problem (1)-(2). This completes the proof.

Example 9. Consider a boundary value problem of integro-differential equations of fractional order given by ^𝑐𝐷^3/2+2^𝑐𝐷^1/2𝑢(𝑡)=12𝑓(𝑡,𝑢(𝑡))+𝐼^1/2𝑔((𝑡𝑢(0)(1)𝑡,𝑢)),0<𝑡<1,=0,𝑢=0, (40) where 𝛼=3/2,  𝑘=2,  𝑝=1/2,  𝑞=1,  𝛽=1/2,  𝑓(𝑡,𝑢)=(|𝑢|(2+|𝑢|))/(3(1+|𝑢|))+4𝑡,  𝑔(𝑡,𝑢)=(1/4)tan⁻¹𝑢+cos²𝑡+𝑡³+5. With the given data, it is found that 𝐿₁=2/3,𝐿₂=1/4 as |𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)|≤(2/3)|𝑢−𝑣|,  |𝑔(𝑡,𝑢)−𝑔(𝑡,𝑣)|≤(1/4)|𝑢−𝑣|, and 2𝑄=1−𝑒^−𝑘𝑝Γ𝛼+𝛽+𝑞Γ(𝛼)|𝑘|Γ𝛼+𝛽Γ(𝛼)≃1.3525. (41) Clearly, 𝐿=max{𝐿₁,𝐿₂}=2/3 and 𝐿<1/𝑄. Thus, all the assumptions of Theorem 4 are satisfied. Hence, by the conclusion of Theorem 4, the problem (40) has a unique solution.

Acknowledgment

The authors thank the anonymous referees for their valuable comments. The research of J. J. Nieto has been partially supported by Ministerio de Economia y Competitividad (Spain), project MTM2010-15314, and co-financed by the European Community fund FEDER.