Existence and Uniqueness Results for Hadamard-Type Fractional Differential Equations with Nonlocal Fractional Integral Boundary Conditions

Abstract

We study the existence and uniqueness of solutions for a fractional boundary value problem involving Hadamard-type fractional differential equations and nonlocal fractional integral boundary conditions. Our results are based on some classical fixed point theorems. Some illustrative examples are also included.

1. Introduction

In this paper, we investigate the following Hadamard boundary value problem: 𝐷^𝑞𝑥(𝑡)=𝑓(𝑡,𝑥(𝑡)),1<𝑞≤2,𝑡∈(1,𝑒),𝑥(1)=0,^𝑚_𝑖=1𝜆_𝑖𝐽^𝛼𝑖𝑥𝜂_𝑖=^𝑛_𝑗=1𝜇_𝑗𝐽^𝛽𝑗𝑥(𝑒)−𝐽^𝛽𝑗𝑥𝜉_𝑗, (1)(2) where 𝐷^𝑞 denotes the Hadamard fractional derivative of order 𝑞,𝑓∶[1,𝑒]×ℝ→ℝ is a continuous function, 𝜂_𝑖,𝜉_𝑗∈(1,𝑒),  𝜆_𝑖,𝜇_𝑗∈ℝ, for all 𝑖=1,2,…,𝑚,  𝑗=1,2,…,𝑛,  𝜂₁<𝜂₂<⋯<𝜂_𝑚,  𝜉₁<𝜉₂<⋯<𝜉_𝑛, and 𝐽^𝜙 is the Hadamard fractional integral of order 𝜙>0(𝜙=𝛼_𝑖,𝛽_𝑗,𝑖=1,2,…,𝑚,𝑗=1,2,…,𝑛).

We mention that integral boundary conditions are encountered in various applications such as population dynamics, blood flow models, chemical engineering, cellular systems, heat transmission, plasma physics, and thermoelasticity.

Condition (2) is a general form of the integral boundary conditions considered in [1] and covers many special cases. For example, if 𝛼_𝑖=𝛽_𝑗=1, for all 𝑖=1,2,…,𝑚 and  𝑗=1,2,…,𝑛, then condition (2) reduces to 𝑥(1)𝜆=0,_1𝜂11𝑥(𝑠)𝑑𝑠𝑠+⋯+𝜆_𝑚^𝜂𝑚₁𝑥(𝑠)𝑑𝑠𝑠=𝜇₁^𝑒_𝜉1𝑥(𝑠)𝑑𝑠𝑠+⋯+𝜇_𝑛^𝑒_𝜉𝑛𝑥(𝑠)𝑑𝑠𝑠. (3)

Fractional differential equations provide appropriate models for describing real world problems, which cannot be described using classical integer order differential equations. The theory of fractional differential equations has received much attention over the past years and has become an important field of investigation due to its extensive applications in numerous branches of physics, economics, and engineering sciences [2–5]. Some recent contributions to the subject can be seen in [1, 6–20] and references cited therein.

It has been noticed that most of the work on this topic is based on Riemann-Liouville and Caputo type fractional differential equations. Another kind of fractional derivatives that appears side by side to Riemann-Liouville and Caputo derivatives in the literature is the fractional derivative due to Hadamard introduced in 1892 [21], which differs from the preceding ones in the sense that the kernel of the integral (in the definition of Hadamard derivative) contains logarithmic function of arbitrary exponent. Details and properties of Hadamard fractional derivative and integral can be found in [2, 22–26]. For some recent results on Hadamard boundary value problem we refer to [27, 28].

We establish a variety of results for the problem (1)-(2) by using classical fixed point theorems. The first result, Theorem 4, relies on Banach contraction mapping principle and concerns an existence and uniqueness result for the solutions of the problem (1)-(2). A second existence and uniqueness result is proved in Theorem 7, via nonlinear contractions and a fixed point theorem due to Boyd and Wong. Existence results are proved in the third result, Theorem 9, by using Krasnoselskii fixed point theorem, and in the fourth result, Theorem 12, by using nonlinear alternative of Leray-Schauder type.

The paper is organized as follows. In Section 2, we recall some preliminary concepts that we need in the sequel and prove a preliminary lemma. Section 3 contains the main results for the problem (1)-(2). In Section 4, some illustrative examples are discussed.

2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus [2] and present preliminary results needed in our proofs later.

Definition 1. The Hadamard derivative of fractional order 𝑞 for a function 𝑓:[1,∞)→ℝ is defined as 𝐷^𝑞𝑓(𝑡)=1Γ𝑡𝑑𝑛−𝑞𝑑𝑡^𝑛𝑡₁𝑡log𝑠^{𝑛−𝑞−1}𝑓(𝑠)𝑠𝑞𝑑𝑠,𝑛−1<𝑞<𝑛,𝑛=+1, (4) where [𝑞] denotes the integer part of the real number 𝑞, log(⋅)=log_𝑒(⋅), and Γ is the Gamma function.

Definition 2. The Hadamard fractional integral of order 𝑞 for a function 𝑓:[1,∞)→ℝ is defined by 𝐽^𝑞𝑓(𝑡)=1Γ𝑞^𝑡₁𝑡log𝑠^𝑞−1𝑓(𝑠)𝑠𝑑𝑠,𝑞>0, (5) provided the integral exists.

For convenience, we set Λ=^𝑚_𝑖=1𝜆_𝑖Γ𝑞Γ𝑞+𝛼_𝑖log𝜂_𝑖^{𝑞+𝛼𝑖−1}−^𝑛_𝑗=1𝜇_𝑗Γ𝑞Γ𝑞+𝛽_𝑗1−log𝜉_𝑗^{𝑞+𝛽𝑗−1}. (6)

Lemma 3. Let Λ≠0, 1<𝑞≤2, 𝛼_𝑖,𝛽_𝑗>0, and 𝜂_𝑖,𝜉_𝑗∈(1,𝑒) for 𝑖=1,2,…,𝑚, 𝑗=1,2,…,𝑛, and ℎ∈𝐶([1,𝑒],ℝ). The unique solution of the following fractional differential equation, 𝐷^𝑞𝑥(𝑡)=ℎ(𝑡),𝑡∈(1,𝑒), (7) subject to the boundary condition, 𝑥(1)=0,^𝑚_𝑖=1𝜆_𝑖𝐽^𝛼𝑖𝑥𝜂_𝑖=^𝑛_𝑗=1𝜇_𝑗𝐽^𝛽𝑗𝑥(𝑒)−𝐽^𝛽𝑗𝑥𝜉_𝑗, (8) is given by the integral equation 𝑥(𝑡)=log𝑡^𝑞−1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝑞+𝛽𝑗}ℎ(𝑒)−𝐽^{𝑞+𝛽𝑗}ℎ𝜉_𝑗−log𝑡^𝑞−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝑞+𝛼𝑖}ℎ𝜂_𝑖+𝐽^𝑞ℎ(𝑡). (9)

Proof. Applying the Hadamard fractional integral of order 𝑞 to both sides of (7), we have 𝑥(𝑡)=𝑧₁log𝑡^𝑞−1+𝑧₂log𝑡^𝑞−2+𝐽^𝑞ℎ(𝑡), (10) where 𝑧₁,𝑧₂∈ℝ.
The condition of 𝑥(1)=0 implies 𝑧₂=0. Therefore, 𝑥(𝑡)=𝑧₁log𝑡^𝑞−1+𝐽^𝑞ℎ(𝑡). (11) For any 𝑝>0, by Definition 2, it follows that 𝐽^𝑝𝑥(𝑡)=𝑧₁Γ𝑞Γ𝑞+𝑝log𝑡^{𝑞+𝑝−1}+𝐽^𝑞+𝑝ℎ(𝑡). (12) The second condition of (8) with (12) leads to 𝑧₁=1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝑞+𝛽𝑗}ℎ(𝑒)−𝐽^{𝑞+𝛽𝑗}ℎ𝜉_𝑗−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝑞+𝛼𝑖}ℎ𝜂_𝑖. (13) Substituting the value of a constant 𝑧₁ into (11), we obtain (9) as required. The proof is completed.

3. Main Results

Let 𝒞=𝐶([1,𝑒],ℝ) denote the Banach space of all continuous functions from [1,𝑒] to ℝ endowed with the norm defined by ‖𝑥‖=sup_{𝑡∈[1,𝑒]}|𝑥(𝑡)|. As in Lemma 3, we define an operator ℱ:𝒞→𝒞 by (ℱ𝑥)(𝑡)=𝐽^𝑞𝑓((𝑠𝑡)−𝑠,𝑥))(log𝑡^𝑞−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖+log𝑡^𝑞−1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)−𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗, (14) with Λ≠0. It should be noticed that problem (1)-(2) has solutions if and only if the operator ℱ has fixed points.

For the sake of convenience, we put 1Φ=Γ+1𝑞+1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗.+𝑞+1 (15)

The first existence and uniqueness result is based on the Banach contraction mapping principle.

Theorem 4. Let 𝑓∶[1,𝑒]×ℝ→ℝ be a continuous function satisfying the assumption that (H₁)there exists a constant 𝐿>0 such that |𝑓(𝑡,𝑥)−𝑓(𝑡,𝑦)|≤𝐿|𝑥−𝑦|, for each 𝑡∈[1,𝑒] and 𝑥,𝑦∈ℝ. If 𝐿Φ<1, (16) where Φ is given by (15), then the boundary value problem (1)-(2) has a unique solution on [1,𝑒].

Proof. We transform the problem (1)-(2) into a fixed point problem,  𝑥=ℱ𝑥, where the operator ℱ is defined by (14). By using Banach’s contraction mapping principle, we will show that ℱ has a fixed point which is a unique solution of problem (1)-(2).
We set sup_{𝑡∈[1,𝑒]}|𝑓(𝑡,0)|=𝑀<∞ and choose 𝑟≥𝑀Φ.1−𝐿Φ (17)
Now, we show that ℱ𝐵_𝑟⊂𝐵_𝑟, where 𝐵_𝑟={𝑥∈𝒞∶‖𝑥‖≤𝑟}. For any 𝑥∈𝐵_𝑟, we have ‖ℱ𝑥‖≤sup_{[]𝑡∈1,𝑒}𝐽^𝑞𝑓((𝑠𝑡)+𝑠,𝑥))(log𝑡^𝑞−1|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖+log𝑡^𝑞−1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)+𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗≤𝐽^𝑞𝑓((𝑠(+𝑓(𝑒)+1𝑠,𝑥))−𝑓𝑠,0)𝑠,0)(|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))−𝑓(𝑠,0)+𝑓(𝑠,0)𝜂_𝑖+1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))−𝑓(𝑠,0)+𝑓(𝑠,0)(𝑒)+𝐽^{𝛽𝑗+𝑞}𝑓((𝑠(+𝑓(𝜉𝑠,𝑥))−𝑓𝑠,0)𝑠,0)_𝑗≤()1𝐿𝑟+𝑀Γ+1𝑞+1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗=()+𝑞+1𝐿𝑟+𝑀Φ≤𝑟. (18) It follows that ℱ𝐵_𝑟⊂𝐵_𝑟.
For 𝑥,𝑦∈𝒞 and for each 𝑡∈[1,𝑒], we have ℱ𝑥(𝑡)−ℱ𝑦(𝑡)≤𝐽^𝑞𝑓((𝑠(𝑠)(𝑡)+𝑠,𝑥))−𝑓𝑠,𝑦log𝑡^𝑞−1|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)𝜂_𝑖+log𝑡^𝑞−1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)(𝑒)+𝐽^{𝛽𝑗+𝑞}𝑓((𝑠(𝑠)×𝜉𝑠,𝑥))−𝑓𝑠,𝑦_𝑗1≤𝐿𝑥−𝑦Γ+1𝑞+1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖×1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗.+𝑞+1=𝐿Φ𝑥−𝑦 (19) The above result implies that ‖ℱ𝑥−ℱ𝑦‖≤𝐿Φ‖𝑥−𝑦‖. As 𝐿Φ<1, therefore ℱ is a contraction. Hence, by the Banach contraction mapping principle, we deduce that ℱ has a fixed point which is the unique solution of the problem (1)-(2).

Next, we give the second existence and uniqueness result by using nonlinear contractions.

Definition 5. Let 𝐸 be a Banach space and let 𝐹∶𝐸→𝐸 be a mapping. 𝐹 is said to be a nonlinear contraction if there exists a continuous nondecreasing function Ψ∶ℝ⁺→ℝ⁺ such that Ψ(0)=0 and Ψ(𝜃)<𝜃 for all 𝜃>0 with the property 𝐹𝑥−𝐹𝑦≤Ψ𝑥−𝑦,∀𝑥,𝑦∈𝐸. (20)

Lemma 6 (see [29]). Let 𝐸 be a Banach space and let 𝐹∶𝐸→𝐸 be a nonlinear contraction. Then 𝐹 has a unique fixed point in 𝐸.

Theorem 7. Let 𝑓:[1,𝑒]×ℝ→ℝ be a continuous function satisfying the assumption (H₂)|𝑓(𝑡,𝑥)−𝑓(𝑡,𝑦)|≤ℎ(𝑡)(|𝑥−𝑦|/(𝐻^∗+|𝑥−𝑦|)),  𝑡∈[1,𝑒],  𝑥,𝑦≥0,  where ℎ∶[1,𝑒]→ℝ⁺  is continuous and a constant 𝐻^∗ is defined by 𝐻^∗=𝐽^𝑞ℎ(𝑒)+1|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}ℎ𝜂_𝑖+1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}ℎ(𝑒)+𝐽^{𝛽𝑗+𝑞}ℎ𝜉_𝑗. (21)Then the boundary value problem (1)-(2) has a unique solution.

Proof. We define the operator ℱ∶𝒞→𝒞 as (14) and a continuous nondecreasing function Ψ∶ℝ⁺→ℝ⁺ by Ψ(𝜃)=𝐻^∗𝜃𝐻^∗+𝜃,∀𝜃≥0. (22) Note that the function Ψ satisfies Ψ(0)=0 and Ψ(𝜃)<𝜃 for all 𝜃>0.
For any 𝑥,𝑦∈𝒞 and for each 𝑡∈[1,𝑒], we have ℱ𝑥(𝑡)−ℱ𝑦(𝑡)≤𝐽^𝑞𝑓((𝑠(𝑠)(𝑡)+𝑠,𝑥))−𝑓𝑠,𝑦log𝑡^𝑞−1|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)𝜂_𝑖+log𝑡^𝑞−1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)(𝑒)+𝐽^{𝛽𝑗+𝑞}𝑓((𝑠(𝑠)×𝜉𝑠,𝑥))−𝑓𝑠,𝑦_𝑗≤𝐽^𝑞ℎ(𝑠)𝑥(𝑠)−𝑦(𝑠)𝐻^∗+𝑥(𝑠)(𝑠(𝑒)+1−𝑦)|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}ℎ(𝑠)𝑥(𝑠)−𝑦(𝑠)𝐻^∗+𝑥(𝑠)(𝑠𝜂−𝑦)^𝑖+1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}ℎ(𝑠)𝑥(𝑠)−𝑦(𝑠)𝐻^∗+𝑥(𝑠)(𝑠(𝑒)−𝑦)+𝐽^{𝛽𝑗+𝑞}ℎ(𝑠)𝑥(𝑠)−𝑦(𝑠)𝐻^∗+𝑥(𝑠)(𝑠𝜉−𝑦)^𝑗≤Ψ𝑥−𝑦𝐻^∗𝐽^𝑞ℎ(𝑒)+1|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}ℎ𝜂_𝑖+1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}ℎ(𝑒)+𝐽^{𝛽𝑗+𝑞}ℎ𝜉_𝑗=Ψ𝑥−𝑦. (23) This implies that ‖ℱ𝑥−ℱ𝑦‖≤Ψ(‖𝑥−𝑦‖). Therefore ℱ is a nonlinear contraction. Hence, by Lemma 6 the operator ℱ has a fixed point which is the unique solution of the problem (1)-(2).

Next, we give an existence result by using Krasnoselskii’s fixed point theorem.

Lemma 8 (Krasnoselskii’s fixed point theorem [30]). Let M be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A, B be the operators such that (a) 𝐴𝑥+𝐵𝑦∈𝑀, whenever 𝑥,𝑦∈𝑀; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists 𝑧∈𝑀 such that z = Az + Bz.

Theorem 9. Assume that 𝑓:[1,𝑒]×ℝ→ℝ is a continuous function satisfying assumption (𝐻₁). In addition we suppose that (H₃)|𝑓(𝑡,𝑥)|≤𝜅(𝑡), ∀(𝑡,𝑥)∈[1,𝑒]×ℝ and 𝜅∈𝐶([1,𝑒],ℝ⁺). If 𝐿Γ𝑞+1<1, (24) then the boundary value problem (1)-(2) has at least one solution on [1,𝑒].

Proof. We define sup_{𝑡∈[1,𝑒]}|𝜅(𝑡)|=‖𝜅‖ and choose a suitable constant 𝑟 as 𝑟≥‖𝜅‖Φ, (25) where Φ is defined by (15). Furthermore, we define the operators 𝒫 and 𝒬 on 𝐵_𝑟={𝑥∈𝒞∶‖𝑥‖≤𝑟} as (𝒫𝑥)(𝑡)=log𝑡^𝑞−1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)−𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗−log𝑡^𝑞−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖,[],(𝑡)𝑡∈1,𝑒𝒬𝑥)(=𝐽^𝑞𝑓(𝑠,𝑥(𝑠))(𝑡),𝑡∈[1,𝑒]. (26) For 𝑥,𝑦∈𝐵_𝑟, we have ≤‖𝜅‖1𝒫𝑥+𝒬𝑦Γ+1𝑞+1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗=‖𝜅‖+𝑞+1Φ≤𝑟. (27) This shows that 𝒫𝑥+𝒬𝑦∈𝐵_𝑟. It follows from assumption (𝐻₁) together with (24) that 𝒬 is a contraction mapping. Since the function 𝑓 is continuous, we have that the operator 𝒫 is continuous. It is easy to verify that ‖‖≤‖𝜅‖1𝒫𝑥|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗.+𝑞+1 (28) Therefore, 𝒫 is uniformly bounded on 𝐵_𝑟.
Next, we prove the compactness of the operator 𝒫. Let us set sup_{(𝑡,𝑥)∈[1,𝑒]×𝐵𝑟}|𝑓(𝑡,𝑥)|=𝑓<∞; consequently we get (𝒫𝑥)𝑡₁−(𝒫𝑥)𝑡₂=log𝑡₁^𝑞−1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)−𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗−log𝑡₁^𝑞−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖−log𝑡₂^𝑞−1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)−𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗+log𝑡₂^𝑞−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖≤𝑓log𝑡₂^𝑞−1−log𝑡₁^𝑞−1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖++𝑞+1𝑓log𝑡₂^𝑞−1−log𝑡₁^𝑞−1|Λ|^𝑛_𝑗=1𝜇_𝑗1−log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗,+𝑞+1 (29) which is independent of 𝑥 and tends to zero as 𝑡₂→𝑡₁. Thus, 𝒫 is equicontinuous. So 𝒫 is relatively compact on 𝐵_𝑟. Hence, by the Arzelá-Ascoli theorem, 𝒫 is compact on 𝐵_𝑟. Thus, all the assumptions of Lemma 8 are satisfied. So the boundary value problem (1)-(2) has at least one solution on [1,𝑒]. The proof is completed.

Remark 10. In the above theorem we can interchange the roles of the operators 𝒫 and 𝒬 to obtain a second result replacing (24) by the following condition: 𝐿|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+𝐿+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗+𝑞+1<1. (30)

Now, our last existence result is based on Leray-Schauder’s nonlinear alternative.

Theorem 11 (nonlinear alternative for single-valued maps [31]). 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 (i.e., 𝐹(𝑈) is a relatively compact subset of C) map. Then either (i)F has 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 12. Assume that 𝑓∶[1,𝑒]×ℝ→ℝ is a continuous function. In addition we suppose that(H₄)there exists a continuous nondecreasing function 𝜓∶[0,∞)→(0,∞) and a function 𝑝∈𝐶([1,𝑒],ℝ⁺) such that 𝑓(𝑡,𝑥)≤𝑝(𝑡)𝜓(|𝑥|)𝑓𝑜𝑟𝑒𝑎𝑐ℎ(𝑡,𝑥)∈[1,𝑒]×ℝ; (31)(H₅)there exists a constant 𝑁>0 such that 𝑁𝑝𝜓(𝑁)Φ>1, (32) where Φ is defined by (15). Then the boundary value problem (1)-(2) has at least one solution on [1,𝑒].

Proof. Firstly, we will show that the operator ℱ, defined by (14), maps bounded sets (balls) into bounded sets in 𝒞. For a positive number 𝑅, let 𝐵_𝑅={𝑥∈𝒞∶‖𝑥‖≤𝑅} be a bounded ball in 𝒞. Then for 𝑡∈[1,𝑒], we have |ℱ𝑥(𝑡)|≤𝐽^𝑞𝑓((𝑠𝑒)+1𝑠,𝑥))(|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖+1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)+𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗≤𝑝𝜓𝑥1(‖‖)Γ+𝑝𝜓𝑥1𝑞+1(‖‖)|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+𝑝𝜓𝑥1+𝑞+1(‖‖)|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗≤𝑝𝜓(𝑅)1+𝑞+1Γ+𝑝𝜓(𝑅)1𝑞+1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+𝑝𝜓(𝑅)1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗+𝑞+1∶=𝐾. (33) Therefore, we conclude that ‖ℱ𝑥‖≤𝐾.
Secondly, we show that ℱ maps bounded sets into equicontinuous sets of 𝒞. Let sup_{(𝑡,𝑥)∈[1,𝑒]×𝐵𝑅}|𝑓(𝑡,𝑥)|=𝑓^∗<∞, 𝜈₁,𝜈₂∈[1,𝑒] with 𝜈₁<𝜈₂ and 𝑥∈𝐵_𝑅. Then we have (ℱ𝑥)𝜈₂−(ℱ𝑥)𝜈₁=𝐽^𝑞𝑓(𝑠,𝑥(𝑠))𝜈₂−log𝜈₂^𝑞−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖+log𝜈₂^𝑞−1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)−𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗−𝐽^𝑞𝑓(𝑠,𝑥(𝑠))𝜈₁+log𝜈₁^𝑞−1Λ^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}𝑓(𝑠,𝑥(𝑠))𝜂_𝑖−log𝜈₁^𝑞−1Λ^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))(𝑒)−𝐽^{𝛽𝑗+𝑞}𝑓(𝑠,𝑥(𝑠))𝜉_𝑗≤𝑓^∗log𝜈₂^𝑞−log𝜈₁^𝑞Γ𝑞+1+𝑓^∗log𝜈₂^𝑞−1−log𝜈₁^𝑞−1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+𝑞+1+𝑓^∗log𝜈₂^𝑞−1−log𝜈₁^𝑞−1|Λ|^𝑛_𝑗=1𝜇_𝑗1−log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗.+𝑞+1 (34) Obviously, the right hand side of the above inequality tends to zero independently of 𝑥∈𝐵_𝑅 as 𝜈₂→𝜈₁. Therefore it follows from the Arzelá-Ascoli theorem that ℱ∶𝒞→𝒞 is completely continuous.
Let 𝑥 be a solution. Then, for 𝑡∈[1,𝑒], following the similar computations as in the first step, we have ‖𝑥‖≤𝑝𝜓𝑥1(‖‖)Γ+𝑝𝜓𝑥1𝑞+1(‖‖)|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+𝑝𝜓𝑥1+𝑞+1(‖‖)|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗=𝑝𝜓𝑥+𝑞+1(‖‖)Φ. (35) Consequently, we have ‖𝑥‖𝑝𝜓(‖𝑥‖)Φ≤1. (36) In view of (H₅), there exists 𝑁 such that ‖𝑥‖≠𝑁. Let us set 𝑈={𝑥∈𝒞∶‖𝑥‖<𝑁}. (37) Note that the operator ℱ∶𝑈→𝒞 is continuous and completely continuous. From the choice of 𝑈, there is no 𝑥∈𝜕𝑈 such that 𝑥=𝜃ℱ𝑥 for some 𝜃∈(0,1). Consequently, by nonlinear alternative of Leray-Schauder type (Theorem 11) we deduce that ℱ has a fixed point in 𝑈, which is a solution of the boundary value problem (1)-(2). This completes the proof.

4. Examples

Example 1. Consider the following boundary value problem for Hadamard fractional differential equation: 𝐷^3/2𝑥(𝑡)=log𝑡⁵𝑒^𝑡(𝑡+2)²|𝑥(𝑡)|(|𝑥(𝑡[],𝑥(1)3+)|),𝑡∈𝐽=1,𝑒=0,2𝐽^1/4𝑥54+15𝐽^3/2𝑥95+3𝐽²𝑥157=𝐽^2/3𝑥(𝑒)−𝐽^2/3𝑥107+5𝐽^9/7𝑥(𝑒)−𝐽^9/7𝑥(2)−2𝐽^11/4𝑥(𝑒)−𝐽^11/4𝑥94. (38)

Here  𝑞=3/2,  𝜆₁=2,  𝜆₂=1/5,  𝜆₃=3,  𝛼₁=1/4,  𝛼₂=3/2,  𝛼₃=2,  𝜂₁=5/4,  𝜂₂=9/5,  𝜂₃=15/7,  𝜇₁=1,  𝜇₂=5,  𝜇₃=−2,  𝛽₁=2/3,  𝛽₂=9/7,  𝛽₃=11/4,  𝜉₁=10/7,  𝜉₂=2,  𝜉₃=9/4,  and  𝑓(𝑡,𝑥)=(log𝑡⁵|𝑥|)/(𝑒^𝑡(𝑡+2)²(3+|𝑥|)). Since 𝑓()≤5𝑡,𝑥−𝑓𝑡,𝑦,27𝑒𝑥−𝑦 (39) then (H₁) is satisfied with 𝐿=5/27𝑒. We can show that Λ=^𝑚_𝑖=1𝜆_𝑖Γ𝑞Γ𝑞+𝛼_𝑖log𝜂_𝑖^{𝑞+𝛼𝑖−1}−^𝑛_𝑗=1𝜇_𝑗Γ𝑞Γ𝑞+𝛽_𝑗1−log𝜉_𝑗^{𝑞+𝛽𝑗−1}1≈−0.6895040549,Φ=Γ+1𝑞+1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗5+𝑞+1≈3.975680952,𝐿Φ=()27𝑒3.975680952≈0.2708465347<1. (40) Hence, by Theorem 4, the boundary value problem (38) has a unique solution on [1,𝑒].

Example 2. Consider the following boundary value problem for Hadamard fractional differential equation: 𝐷^7/4𝑥(𝑡)=𝑒^𝑡(𝑡+1)²|𝑥(𝑡)|(|𝑥(𝑡[],𝑥(1)12+)|),𝑡∈𝐽=1,𝑒=0,4𝐽^6/7𝑥73−23𝐽³𝑥75−2𝐽^5/2𝑥(2)=4𝐽⁵𝑥(𝑒)−𝐽⁵𝑥115+114𝐽^3/4𝑥(𝑒)−𝐽^3/4𝑥16.13 (41)

Here 𝑞=7/4,  𝜆₁=1/4,  𝜆₂=−2/3,  𝜆₃=−2,  𝛼₁=6/7,  𝛼₂=3,  𝛼₃=5/2,  𝜂₁=7/3,  𝜂₂=7/5,  𝜂₃=2,  𝜇₁=4,  𝜇₂=11/4,  𝛽₁=5,  𝛽₂=3/4,  𝜉₁=11/5,  𝜉₂=16/13,  and  𝑓(𝑡,𝑥)=(𝑒^𝑡|𝑥|)/((𝑡+1)²(2+|𝑥|)). We choose ℎ(𝑡)=𝑒^𝑡/4 and that Λ=^𝑚_𝑖=1𝜆_𝑖Γ𝑞Γ𝑞+𝛼_𝑖log𝜂_𝑖^{𝑞+𝛼𝑖−1}−^𝑛_𝑗=1𝜇_𝑗Γ𝑞Γ𝑞+𝛽_𝑗1−log𝜉_𝑗^{𝑞+𝛽𝑗−1}𝐻≈−1.672972140,_∗=𝐽^𝑞ℎ(𝑒)+1|Λ|^𝑚_𝑖=1𝜆_𝑖𝐽^{𝛼𝑖+𝑞}ℎ𝜂_𝑖+1|Λ|^𝑛_𝑗=1𝜇_𝑗𝐽^{𝛽𝑗+𝑞}ℎ(𝑒)+𝐽^{𝛽𝑗+𝑞}ℎ𝜉_𝑗≈1.295076743. (42) Clearly, 𝑓()=𝑒𝑡,𝑥−𝑓𝑡,𝑦^𝑡(1+𝑡)²2|𝑥|−2𝑦|𝑥|𝑦+|𝑥𝑦≤𝑒4+2+2|_𝑡4𝑥−𝑦.1.295076743+𝑥−𝑦 (43) Hence, by Theorem 7, the boundary value problem (41) has a unique solution on [1,𝑒].

Example 3. Consider the following boundary value problem for Hadamard fractional differential equation: 𝐷^6/5𝑥(𝑡)=2sin(𝑥/4)5𝜋+(𝑒^𝑥+1)²+2+cos(𝜋𝑡)[],𝑥(1)𝐽10𝜋+3,𝑡∈𝐽=1,𝑒=0,₄𝑥32−3𝐽^9/4𝑥(2)−10𝐽^1/5𝑥74+6𝐽^7/2𝑥52+143𝐽⁵𝑥119=3𝐽^3/2𝑥(𝑒)−𝐽^3/2𝑥117−7𝐽³𝑥(𝑒)−𝐽³𝑥17+4133𝐽^5/3𝑥(𝑒)−𝐽^5/3𝑥(2). (44)

Here 𝑞=6/5,  𝜆₁=1,  𝜆₂=−3,  𝜆₃=−10,  𝜆₄=6,  𝜆₅=14/3,  𝛼₁=4,  𝛼₂=9/4,  𝛼₃=1/5,  𝛼₄=7/2,  𝛼₅=5,  𝜂₁=3/2,  𝜂₂=2,  𝜂₃=7/4,  𝜂₄=5/2,  𝜂₅=11/9,  𝜇₁=3,  𝜇₂=−7,  𝜇₃=4/3,  𝛽₁=3/2,  𝛽₂=3,  𝛽₃=5/3,  𝜉₁=11/7,  𝜉₂=17/13,  𝜉₃=2,  and  𝑓(𝑡,𝑥)=(2sin(𝑥/4))/(5𝜋+(𝑒^𝑥+1)²)+(2+cos(𝜋𝑡))/(10𝜋+3). Clearly, 𝑓(=()𝑡,𝑥)2sin𝑥/45𝜋+(𝑒^𝑥+1)²+2+cos(𝜋𝑡)≤((|𝑥|10𝜋+32+cos𝜋𝑡))+1.10𝜋 (45) Choosing 𝑝(𝑡)=2+cos(𝜋𝑡) and 𝜓(|𝑥|)=(|𝑥|+1)/(10𝜋), we can show that Λ=^𝑚_𝑖=1𝜆_𝑖Γ𝑞Γ𝑞+𝛼_𝑖log𝜂_𝑖^{𝑞+𝛼𝑖−1}−^𝑛_𝑗=1𝜇_𝑗Γ𝑞Γ𝑞+𝛽_𝑗1−log𝜉_𝑗^{𝑞+𝛽𝑗−1}1≈−9.148087406,Φ=Γ+1𝑞+1|Λ|^𝑚_𝑖=1𝜆_𝑖log𝜂_𝑖^{𝛼𝑖+𝑞}Γ𝛼_𝑖+1+𝑞+1|Λ|^𝑛_𝑗=1𝜇_𝑗1+log𝜉_𝑗^{𝛽𝑗+𝑞}Γ𝛽_𝑗𝑁+𝑞+1≈1.462649525,(3)((𝑁+1)/10𝜋)(1.462649525)>1, (46) which implies that 𝑁>0.1623483851. Hence, by Theorem 12, the boundary value problem (44) has at least one solution on [1,𝑒].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The research of Phollakrit Thiramanus is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand. The research of Jessada Tariboon is supported by King Mongkut’s University of Technology North Bangkok, Thailand. Sotiris K. Ntouyas is a member of Nonlinear Analysis and Applied Mathematics (NAAM) Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.