Existence of Positive Solutions for a Class of Delay Fractional Differential Equations with Generalization to N-Term

Abstract

We established the existence of a positive solution of nonlinear fractional differential equations 𝔏(𝐷)[𝑥(𝑡)−𝑥(0)]=𝑓(𝑡,𝑥_𝑡), 𝑡∈(0,𝑏] with finite delay 𝑥(𝑡)=𝜔(𝑡), 𝑡∈[−𝜏,0], where lim_𝑡→0 𝑓(𝑡,𝑥_𝑡)=+∞, that is, 𝑓 is singular at 𝑡=0 and 𝑥_𝑡∈𝐶([−𝜏,0],ℝ^≥0). The operator of 𝔏(𝐷) involves the Riemann-Liouville fractional derivatives. In this problem, the initial conditions with fractional order and some relations among them were considered. The analysis rely on the alternative of the Leray-Schauder fixed point theorem, the Banach fixed point theorem, and the Arzela-Ascoli theorem in a cone.

1. Introduction

Fractional differential equations have gained considerable importance due to their varied applications in viscoelasticity, electroanalytical chemistry, and many other physical problems [1–7]. So far there have been several fundamental works done on the fractional derivative and fractional differential equations [1–4, 6]. These works are in the introduction of the theory of the fractional derivative and fractional differential equations and provide a systematic understanding of the fractional calculus such as the existence and the uniqueness of some analytic methods for solving fractional differential equations, namely, the Green’s function method, the Mellin transform method, and the power series.

The existence of positive solutions for fractional differential equations with delay has been carried out by various researchers [8–17]. In [8] the authors have investigated the following type of fractional differential equations:ℒ(𝐷)[𝑥(𝑡)−𝑥(0)]=𝑓𝑡,𝑥_𝑡(],𝑥(𝑡)(𝑡)[],,𝑡∈0,𝑇=𝜙≥0,𝑡∈−𝑤,0(1.1) whereℒ(𝐷)=𝐷^𝛼𝑛−^𝑛−1_𝑗=1𝑝_𝑗(𝑡)𝐷^{𝛼𝑛−𝑗},0<𝛼₁<⋯<𝛼_𝑛<1,𝑝_𝑗(𝑡)=^𝑁𝑗_𝑘=0𝑎_𝑗𝑘𝑡^𝑘,𝑝_𝑗^(2𝑚)(𝑡)≥0,𝑝_𝑗^(2𝑚+1)(𝑡)𝑁≤0,𝑚=0,1,…,^𝑗2,𝑗=1,2,…,𝑛−1,(1.2) and 𝐷^𝛼𝑗 is the standard Riemann-Liouville fractional derivative, 𝑇>0, 𝑤>0, 𝜙∈𝐶=𝐶([−𝑤,0],ℝ⁺) and 𝑓∶𝐼×𝐶→ℝ⁺ is a given continuous function, 𝐼=[0,𝑇].

As a pursuit of this in this paper, we discuss the existence of positive solutions for initial nonlinear fractional differential equations with finite delay,ℒ(𝐷)[𝑥(𝑡)−𝑥(0)]=𝑓𝑡,𝑥_𝑡(],𝑥(𝑡)(𝑡)[],,𝑡∈0,𝑏=𝜔,𝑡∈−𝜏,0(1.3) with initial conditions 𝑥(0)=𝑥₀, [𝐷^{𝛼−𝑛+1}𝑥(𝑡)]_𝑡=0=𝑥_𝑛−1, and [𝐷^{𝛼−𝑛+𝑗}𝑥(𝑡)]_𝑡=0=𝑥_𝑛−𝑗, where ℒ(𝐷)=𝐷^𝛼−∑^𝑛−1_𝑗=1𝑎_𝑗𝐷^𝛼−𝑗, 𝑎_𝑗>0, for all 𝑗,𝑗=1,2,…,𝑛−1, 𝑛−1<𝛼≤𝑛, 𝑛∈ℕ, 𝑓∶(0,𝑏]×𝐶→[0,+∞) is a given continuous function so that lim_𝑡→0𝑓(𝑡,𝑥_𝑡)=+∞ (i.e., 𝑓 is singular at 𝑡=0), where 𝐶 is the space of continuous functions from [−𝜏,0] to ℝ^≥0 and 𝑥_𝑡∈𝐶 defined by 𝑥_𝑡(𝑠)=𝑥(𝑡+𝑠) for each 𝑠∈[−𝜏,0). In the initial conditions of (1.1), we also assume that(i)𝑥_𝑛−1≥0,(ii)𝑥_𝑛−𝑗≥^𝑗−1_𝑘=1𝑎_𝑘𝑥_{𝑘+𝑛−𝑗},∀𝑗=2,3,…,𝑛−1.(1.4) The paper is organized as follows.

In Section 2, we provide some definitions about the fractional derivatives and the fractional integrals as well as we list their basic properties. Required topics of functional analysis such as Banach fixed point theorem and Leray-Schauder Theorem were also introduced. Section 3 deals with existence of a positive solution theorem and it gives an explainable example. The unique positive solution theorem with an explainable example has been discussed in Section 4.

2. Preliminaries

Preliminaries from fractional calculus [1, 4, 6] and functional analysis [17] are outlined below.

Let 𝐸 be a real Banach space with a cone 𝐾.  𝐾 introduces a partial order ≤ in 𝐸 as 𝑥≤𝑦 if 𝑦−𝑥∈𝐾.

Definition 2.1. For 𝑥,𝑦∈𝐸, the order interval ⟨𝑥,𝑦⟩ is defined as 𝑥,𝑦=𝑧∈𝐸∶𝑥≤𝑧≤𝑦.(2.1)

Theorem 2.2 (Leray-Schauder Theorem [17]). Let 𝐸 be a Banach space with 𝐶⊆𝐸 closed and convex. Assume 𝑈 is relatively open subset of 𝐶 with 0∈𝑈 and 𝐹∶𝑈→𝐶 is a continuous, compact map. Then, either (i)𝐹 has fixed point in 𝑈 or (ii) there exists 𝑢∈𝜕𝑈 and 𝛾∈(0,1) with 𝑢=𝛾𝐹𝑢.

Theorem 2.3 (Banach fixed point theorem [17]). Let 𝐾 be a closed subspace of a Banach space Ω. Let 𝐹 be a contraction mapping with Lipschitz constant 𝑙(<1) from 𝐾 to itself. Then, 𝐹 has a unique fixed point 𝑥^∗ in 𝐾. Moreover, if 𝑥₀ is an arbitrary point in 𝐾 and {𝑥_𝑛} is defined by 𝑥_𝑛+1=𝐹𝑥_𝑛,(𝑛=0,1,2,…), then lim_𝑛→∞𝑥_𝑛=𝑥^∗∈𝐾 and 𝑑(𝑥_𝑛,𝑥^∗)≤(𝑘^𝑛/(1−𝑘))𝑑(𝑥₁,𝑥₀).

The complete Gamma function Γ(𝛼) plays an important role in the theory of fractional integral and derivatives. A comprehensive definition of Γ(𝛼) is provided by the Euler limit as Γ(𝛼)=lim_𝑛→∞𝑛!𝑛^𝛼𝛼()⋯().𝛼+1𝛼+𝑛(2.2)

If 𝛼>0, then Γ(𝛼) has the following familiar integral representation:Γ(𝛼)=₀^+∞𝑡^𝛼−1𝑒^−𝑡𝑑𝑡.(2.3) In this paper, the Beta function 𝐵(𝛼,𝛽) is used. We notice that 𝐵(𝛼,𝛽) is closely related to the Gamma function. If 𝛼,𝛽>0, then it has the integral representation 𝐵𝛼,𝛽=¹₀𝑡^𝛼−1(1−𝑡)^𝛽−1Γ(𝛼)Γ𝛽𝑑𝑡=Γ𝛼+𝛽.(2.4)

The definitions of Riemann-Liouville fractional derivative/integral and their properties are given bellow.

Definition 2.4. Let 𝑥∶[𝑎,𝑏]→ℝ and 𝑥∈𝐿¹[𝑎,𝑏]; then the expression 𝐼^𝛼_𝑎+𝑥(𝑡)=1Γ(𝛼)^𝑡_𝑎(𝑡−𝑠)^𝛼−1𝑥(𝑡)𝑑𝑡,𝑥>𝑎,(2.5) is called a left-sided factional integrals of order 𝛼.

Definition 2.5. Let 𝑛 be a positive integer number and 𝛼∈(𝑛−1,𝑛]. Then the left-sided fractional derivative of a function 𝑥∶[𝑎,𝑏]→ℝ is defined as 𝐷^𝛼_𝑎+𝑥(𝑡)=𝐷^𝑛𝐼^𝑛−𝛼𝑥(𝑡),𝑡∈[𝑎,𝑏].(2.6) We denote 𝐷^𝛼_𝑎+𝑥(𝑡) as 𝐷^𝛼_𝑎𝑥(𝑡) and 𝐼^𝛼_𝑎+𝑥(𝑡) as 𝐼^𝛼_𝑎𝑥(𝑡). Further, 𝐷^𝛼𝑥(𝑡) and 𝐼^𝛼𝑥(𝑡) are referred as 𝐷^𝛼𝑥(𝑡) and 𝐼^𝛼𝑥(𝑡), respectively.
If the fractional derivative 𝐷^𝛽_𝑎𝑥(𝑡) is integrable; then 𝐼^𝛼_𝑎𝐷^𝛽_𝑎𝑥(𝑡)=𝐼_𝑎^𝛼−𝛽𝑥(𝑡)−𝐼_𝑎^𝑛−𝛽𝑥(𝑡)_𝑡=𝑎(𝑡−𝑎)^𝛼−𝑛Γ()−𝛼−𝑛+1_{𝑛−1𝑗=1}𝐷_𝑎^𝛽−𝑗𝑥(𝑡)_𝑡=𝑎(𝑡−𝑎)^𝛼−𝑗Γ,𝛼−𝑗+1(2.7) where 𝑛−1≤𝛽<𝑛,𝛽≤𝛼, and 𝑛∈ℕ [9]. Further, if 𝑥∈𝐶[𝑎,𝑏], then {𝐼_𝑎^𝑛−𝛽𝑥(𝑡)}_𝑡=𝑎=0 and (1.4) reduces to 𝐼^𝛼_𝑎𝐷^𝛽_𝑎𝑥(𝑡)=𝐼_𝑎^𝛼−𝛽𝑥(𝑡)−^𝑛−1_𝑗=1𝐷_𝑎^𝛽−𝑗𝑥(𝑡)_𝑡=𝑎(𝑡−𝑎)^𝛼−𝑗Γ.𝛼−𝑗+1(2.8)

3. Existence Theorem

In this section, we show that the initial value problem (1.1) under the conditions among the initial value (i.e., (1.3)) has a positive solution.

Lemma 3.1. Let 𝑔∶(0,𝑏]→ℝ be a continuous function and lim_𝑡→0⁺𝑔(𝑡)=+∞. If there exits 𝜎∈(0,1) such that 𝜎<𝛼 and by letting 𝑡^𝜎𝑔(𝑡) be a continuous function on [0,𝑏], then 𝐻(𝑡)=𝐼^𝛼𝑡^𝜎𝑔(𝑡) is continuous [0,𝑏], where 𝑛−1<𝛼<𝑛, 𝑛∈ℕ.

Proof. Let us consider 𝐿=max𝑡^𝜎𝑔(𝑡), 𝑡∈[0,𝑏]. For all 𝑡∈[0,𝑏) and for given all 𝜖>0, |𝐻()(𝑡=1𝑡+𝜖−𝐻)|Γ(𝛼)₀^𝑡+𝜖(𝑡+𝜖−𝑠)^𝛼−1𝑑𝑠−^𝑡₀(𝑡−𝑠)^𝛼−1=1𝑑𝑠Γ(𝛼)₀^𝑡+𝜖(𝑡+𝜖−𝑠)^𝛼−1𝑠^−𝜎𝑠^𝜎𝑔(𝑠)𝑑𝑠−^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎𝑠^𝜎𝑔(𝑠)≤𝐿𝑑𝑠Γ(𝛼)₀^𝑡+𝜖(𝑡+𝜖−𝑠)^𝛼−1𝑠^−𝜎𝑑𝑠−^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎=𝐿𝑑𝑠Γ(𝛼)𝐵(1−𝜎,𝛼)(𝑡+𝜖)^𝛼−𝜎−𝑡^𝛼−𝜎.(3.1) Hence we conclude that lim_𝜖→0|𝐻(𝑡+𝜖)−𝐻(𝑡)|=0. We notice that a similar result is conclusion for |𝐻(𝑏−𝜖)−𝐻(𝑏)|.

In the following, we want to show that (1.1) is equivalent to an integral equation.

Theorem 3.2. Suppose that 𝑓∶(0,𝑏]×𝐶→[0,+∞) is a given continuous function so that lim_𝑡→0𝑓(𝑡,𝑥_𝑡)=+∞ (i.e., 𝑓 is singular at 𝑡=0), where 𝐶 is the space of continuous functions from [−𝜏,0] to ℝ^≥0 and 𝑥_𝑡∈𝐶 defined by 𝑥_𝑡(𝑠)=𝑥(𝑡+𝑠) for each 𝑠∈[−𝜏,0]. If there exists 𝜎∈(0,1) such that 0<𝜎<𝛼∈(𝑛−1,𝑛) and 𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is a continuous function on [0,𝑏], then the fractional differential equation 𝐷^𝛼−^𝑛−1_𝑗=1𝑎_𝑗𝐷^𝛼−𝑗[𝑥(𝑡)−𝑥(0)]=𝑓𝑡,𝑥_𝑡,𝑡∈(0,𝑏],(3.2) is equivalent to the integral equation 𝑥(𝑡)=𝑥(0)+𝜆(𝑡)+ℒ(𝐼)[𝑥(𝑡)−𝑥(0)]+𝐼^𝛼𝑓𝑡,𝑥_𝑡,𝑡∈[0,𝑏],(3.3) where 𝜆(𝑡)=^𝑛−1_𝑗=1𝜆_𝑗𝑡^𝛼−𝑗,𝜆_𝑗=1Γ𝑥𝛼−𝑗+1^𝑗−^{𝑛−𝑗−1}_𝑘=1𝑎_𝑘𝑥_𝑘+𝑗𝜆,𝑗=1,…,𝑛−2,_𝑛−1=1Γ()(𝐼)=𝛼−𝑛+1,ℒ^𝑛−1_𝑗=1𝑎_𝑗𝐼^𝑗𝑥(𝑡).(3.4)

Proof. From (3.2), we have 𝐼^𝛼𝐷^𝛼[𝑥(𝑡)−𝑥(0)]−^𝑛−1_𝑗=1𝑎_𝑗𝐼^𝛼𝐷^𝛼−𝑗[𝑥(𝑡)−𝑥(0)]=𝐼^𝛼𝑓𝑡,𝑥_𝑡,𝑡∈(0,𝑏].(3.5) By using (2.7), we conclude that 𝐼^𝛼𝐷^𝛼[𝑥(𝑡)−𝑥(0)]=[𝑥(𝑡)−𝑥(0)]−^𝑛−1_𝑗=1𝑥_𝑗𝑡^𝛼−𝑗Γ,𝛼−𝑗+1(3.6) and for 𝑘=1,2,…,𝑛−1, 𝐼^𝛼𝐷^𝛼−𝑘[𝑥(𝑡)−𝑥(0)]=𝐼^𝑘[𝑥(𝑡)−𝑥(0)]−^{𝑛−𝑘−1}_𝑗=1𝑥_𝑗+𝑘𝑡^𝛼−𝑗Γ.𝛼−𝑗+1(3.7) Note that, by Lemma 3.1, 𝐼^𝛼𝑓(𝑡,𝑥_𝑡)=𝐼^𝛼(𝑡^𝜎𝑡^−𝜎𝑓(𝑡,𝑥_𝑡) exists and 𝐷^𝛼𝐼^𝛼𝑓(𝑡,𝑥_𝑡)=𝑓(𝑡,𝑥_𝑡), as 𝑡^−𝜎𝑓(𝑡,𝑥_𝑡) is continuous and 𝐼^𝛼𝑓(𝑡,𝑥_𝑡)∈𝐶[0,𝑏]. In view of (3.5), (3.6), and (3.7), (3.2) is equivalent to the following integral equation: 𝑥(𝑡)=𝑥(0)+𝜆(𝑡)+ℒ(𝐼)[𝑥(𝑡)−𝑥(0)]+𝐼^𝛼𝑓𝑡,𝑥_𝑡,𝑡∈(0,𝑏],(3.8) where 𝜆(𝑡)=^𝑛−1_𝑗=1𝜆_𝑗𝑡^𝛼−𝑗,𝜆_𝑗=1Γ𝑥𝛼−𝑗+1^𝑗−^{𝑛−𝑗−1}_𝑘=1𝑎_𝑘𝑥_𝑘+𝑗𝜆,𝑗=1,…,𝑛−2,_𝑛−1=1Γ()(𝐼)=𝛼−𝑛+1,ℒ^𝑛−1_𝑗=1𝑎_𝑗𝐼^𝑗𝑥(𝑡).(3.9) The proof is completed. Therefore, by Theorem 3.2, the other expression of (1.1) is given as follows: 𝑥(𝑡)=𝑥(0)+𝜆(𝑡)+ℒ(𝐼)[𝑥(𝑡)−𝑥(0)]+𝐼^𝛼𝑓𝑡,𝑥_𝑡(],𝑥(𝑡)(𝑡)[],,𝑡∈0,𝑏=𝜔,𝑡∈−𝜏,0(3.10) where 𝜆(𝑡) and ℒ(𝐼) are mentioned in above.
Let 𝑦∶[−𝜏,𝑏]→[0,+∞) be function defined by 𝑦(𝑡)=𝜔(0)[],𝜔(𝑡)[],,𝑡∈0,𝑏≥0,𝑡∈−𝜏,0(3.11) for each 𝑧∈𝐶([0,𝑏],ℝ) with 𝑧(0)=0; we denote by 𝑧 the function define by 𝑧(𝑡)=𝑧(𝑡)[],[].,𝑡∈0,𝑏0,𝑡∈−𝜏,0(3.12) We can decompose 𝑥(⋅) as 𝑥(𝑡)=𝑧(𝑡)+𝑦(𝑡),𝑡∈[−𝜏,𝑏], which implies that 𝑥_𝑡=𝑧_𝑡+𝑦_𝑡, 𝑡∈[0,𝑏]. Hence by Theorem 3.2, (1.1) is equivalent to the following integral equation: 𝑧(𝑡)=𝜆(𝑡)+ℒ(𝐼)𝑧(𝑡)+𝐼^𝛼𝑓𝑡,𝑧_𝑡+𝑦_𝑡,𝑡∈[0,𝑏].(3.13) Set Ω={𝑧∈𝐶([0,𝑏],ℝ),𝑧(0)=0}, and for each 𝑧∈Ω, let ‖𝑧‖_𝑏 be the seminorm in Ω defined by ‖𝑧‖_𝑏=‖𝑧(0)‖+‖𝑧‖=‖𝑧‖=sup{|𝑧(𝑡)|∶𝑡∈[0,𝑏]}.(3.14)Ω is a Banach space with norm ‖⋅‖_𝑏. Let 𝐾 be a cone of Ω. 𝐾={𝑧∈Ω∶𝑧(𝑡)≥0,𝑡∈[0,𝑏]} and 𝐾^∗=𝑥∈𝐶[−𝜏,𝑏],ℝ^≥0∶𝑥(𝑡)=𝜔(𝑡)≥0,𝑡∈[−𝜏,0].(3.15) Define the operator 𝐹∶𝐾→𝐾 by 𝐹𝑧(𝑡)=𝜆(𝑡)+ℒ(𝐼)𝑧(𝑡)+𝐼^𝛼𝑓𝑡,𝑧_𝑡+𝑦_𝑡,𝑡∈[0,𝑏].(3.16)

Theorem 3.3. Suppose that 𝑓(𝑡,𝑥_𝑡), 𝑡∈(0,𝑏], 𝑥_𝑡∈𝐶, is a continuous function and lim_𝑡→0⁺𝑓(𝑡,⋅)=+∞. If there exits 𝜎∈(0,1) such that 0<𝜎<𝛼∈(𝑛−1,𝑛) and 𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is a continuous function on [0,𝑏], then the operator 𝐹, defined as (3.12), maps bounded set into bounded sets in 𝐾, continuous and completely continuous.

Proof. For all 𝑢∈𝐾, since 𝐹𝑢(𝑡)=𝜆(𝑡)+ℒ(𝐼)𝑢(𝑡)+𝐼^𝛼𝑓(𝑡,𝑢_𝑡+𝑦_𝑡) by Lemma 3.1 and the nonnegativeness of 𝑓, we obtain 𝐹∶𝐾→𝐾.
Since 𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is continuous on [0,𝑏]×[0,+∞), there exists a positive constant 𝑁 such that 𝑡^𝜎𝑓(𝑡,𝑥_𝑡)≤𝑁. Hence 𝐼^𝛼𝑓𝑡,𝑧_𝑡+𝑦_𝑡=𝐼^𝛼𝑡^−𝜎𝑡^𝜎𝑓𝑡,𝑥_𝑡≤𝑁𝐼^𝛼𝑡^−𝜎=𝑁Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎=𝑑𝑠𝑁𝑡^𝛼−𝜎Γ(𝛼)𝐵(1−𝜎,𝛼)≤𝑁𝑏^𝛼−𝜎Γ(𝛼)𝐵(1−𝜎,𝛼).(3.17) Let 𝐺⊂𝐾 be bounded, that is, there exists a positive constant 𝐿 such that ‖𝑧‖_𝑏≤𝐿, for all 𝑧∈𝐶. In view of (3.13), for each 𝑧∈𝐺, we obtain ‖𝐹𝑧(𝑡)‖≤|𝜆(𝑡)|+|ℒ(𝐼)𝑧(𝑡)|+𝐼^𝛼𝑓𝑡,𝑧_𝑡+𝑦_𝑡≤‖𝜆(𝑡)‖+‖𝑧‖_𝑏^𝑛_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝑗+1𝑁𝑏^𝛼−𝜎Γ(𝛼)𝐵()(𝑡)1−𝜎,𝛼≤‖𝜆‖+𝐿^𝑛_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝑗+1𝑁𝑏^𝛼−𝜎Γ(𝛼)𝐵(1−𝜎,𝛼).(3.18) Thus, 𝐹(𝐾) is bounded. In the following text, we would like to show that 𝐹 is continuous. Let 𝑣₀∈𝐾 and ‖𝑣₀‖=𝑐₀, if 𝑣∈𝐾 and ‖𝑣−𝑣₀‖<𝑙, then ‖𝑣‖<𝑙+𝑐₀; by the continuity of 𝑡^𝜎𝑓(𝑡,𝑧_𝑡+𝑦_𝑡), we know that 𝑡^𝜎𝑓(𝑡,𝑧_𝑡+𝑦_𝑡) is uniformly continuous on [0,𝑏]×[0,𝑐]. Thus, for all 𝜖>0 there exists a 𝛿>0 such that 𝑡^𝜎𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝑡^𝜎𝑓𝑡,𝑣_𝑡+𝑦_𝑡<𝜖,(3.19) for all 𝑡∈[0,𝑏] and 𝑢,𝑣∈[0,𝑐] with |𝑢−𝑣|<𝛿. Obviously, if ‖𝑢−𝑣‖<𝛿, then 𝑣₀(𝑡),𝑣(𝑡)∈[0,𝑐] and |𝑣(𝑡)−𝑣₀(𝑡)|<𝛿 for each 𝑡∈[0,𝑏]. Hence, we have 𝑡^𝜎𝑓𝑡,𝑣_𝑡+𝑦_𝑡−𝑡^𝜎𝑓𝑡,𝑣_0𝑡+𝑦_𝑡<𝜖,(3.20) for all 𝑡∈[0,𝑏] and 𝑣∈𝐾 with ‖𝑣−𝑣₀‖<𝛿. For all 𝑡∈[0,𝑏], let 𝑢,𝑣∈𝐾, and |𝑢(𝑡)−𝑣(𝑡)|<𝛿, we choose 𝛿≤𝜖(∑^𝑛−1_𝑗=1(𝑥_𝑗𝑏^𝑗/Γ(𝑗+1)))⁻¹. By using (3.20), we get |𝐹𝑢(𝑡)−𝐹𝑣(𝑡)|≤‖𝑢−𝑣‖_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝐼𝑗+1^𝛼𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝑓𝑡,𝑣_𝑡+𝑦_𝑡=‖𝑢−𝑣‖_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝐼𝑗+1^𝛼𝑡^−𝜎𝑡^𝜎𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝑡^−𝜎𝑡^𝜎𝑓𝑡,𝑣_𝑡+𝑦_𝑡≤‖𝑢−𝑣‖_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+1𝑗+1Γ(𝛼)max_{0≤𝑡≤𝑏}^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎𝑠^𝜎𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝑠^𝜎𝑓𝑡,𝑣_𝑡+𝑦_𝑡≤‖‖𝑑𝑠𝑢−𝑣_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝜖𝑗+1Γ(𝛼)max_{0≤𝑡≤𝑏}^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎=‖‖𝑑𝑠𝑢−𝑣_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝜖𝑗+1Γ(𝛼)max_{0≤𝑡≤𝑏}𝑡^𝛼−𝜎𝐵()≤1−𝜎,𝛼1+𝑏^𝛼−𝜎𝐵(1−𝜎,𝛼)𝜖.(3.21) Finally, we want to prove that the operator 𝐹 is equicontinuous. Let 𝐺⊆𝐾 be bounded, that is, three exits a positive constant 𝑙 such that ‖𝑢‖≤𝑙, for all 𝑢∈𝐺. Suppose, 𝑢∈𝐾,𝑡,𝑟∈[0,𝑏] and 𝑡<𝑟. For a given 𝜖>0, there exists 𝛿>0, so that if |𝑡−𝑟|<𝛿, then |(𝑡)(𝑟≤|𝜆(𝑡)(𝑟+|ℒ(𝐼)𝑢(𝑡)(𝐼)𝑢(𝑟+𝐼𝐹𝑢−𝐹𝑢)|−𝜆)|−ℒ)|^𝛼𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝐼^𝛼𝑓𝑟,𝑢_𝑟+𝑦_𝑟≤^𝑛−1_𝑗=1𝜆_𝑗𝑡^𝛼−𝑗−𝑟^𝛼−𝑗+^𝑛−1_𝑗=1𝑙𝑎_𝑗Γ𝑗^𝑡₀(𝑡−𝑠)^𝑗−1−(𝑟−𝑠)^𝑗−1+𝑑𝑠_{𝑛−1𝑗=1}𝑙𝑎_𝑗Γ𝑗^𝑟_𝑡(𝑟−𝑠)^𝑗−1+1𝑑𝑠Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎𝑠^𝜎𝑓𝑠,𝑢_𝑠+𝑦_𝑠𝑑𝑠−^𝑟₀(𝑟−𝑠)^𝛼−1𝑠^−𝜎𝑠^𝜎𝑓𝑠,𝑢_𝑠+𝑦_𝑠𝑑𝑠≤2^𝑛−1_𝑗=1𝜆_𝑗𝑏^𝛼−𝑗+𝑙^𝑛−1_𝑗=1𝑎_𝑗𝑡^𝑗Γ𝑗+1+𝑙^𝑛−1_𝑗=1𝑎_𝑗Γ|𝑗+1|𝑟−𝑡^𝑗+𝑟^𝑗−𝑡^𝑗,+𝑁𝐵(1−𝜎,𝛼)Γ(𝛼)𝑡^𝛼−𝜎−𝑟^𝛼−𝜎≤2^𝑛−1_𝑗=1𝜆_𝑗𝑏^𝛼−𝑗+𝑙^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ𝑗+1+2𝑙^𝑛−1_𝑗=1𝑎_𝑗Γ𝑗+1+𝑙^𝑛−1_𝑗=1𝑎_𝑗Γ||𝑗+1𝑟−𝑡^𝑗+2𝑁𝐵(1−𝜎,𝛼)Γ(𝛼)𝑏^𝛼−𝜎.(3.22) Set Δ=max{3Δ₁/𝜖,3Δ₂/𝜖,Δ₃,Δ₄,…,Δ_𝑛+1}, where Δ₁=^𝑛−1_𝑗=12𝜆_𝑗𝑏^𝛼−𝑗+𝑙𝑎_𝑗𝑏^𝑗Γ+𝑗+12𝑙𝑎_𝑗Γ,Δ𝑗+1₂=2𝑁Γ(𝛼)𝐵(1−𝜎,𝛼)𝑏^𝛼−𝜎,Δ_𝑗+2=𝑙^𝑛−1_𝑗=1𝑎_𝑗Γ𝑗+1,𝑗=1,2,…,𝑛−1.(3.23)
Case 1. If 0<|𝑟−𝑡|<1<𝑏, we choose 𝛿=𝜖/(3(𝑛−1)). Hence, |(𝑡)(𝑟≤𝜖𝐹𝑢−𝐹𝑢)|3𝜖Δ+3Δ+Δ^𝑛−1_𝑗=1|𝑟−𝑡|^𝑗≤𝜖3𝜖Δ+3Δ+(𝑛−1)Δ𝛿<Δ𝜖.(3.24)Case 2. If 0<1≤|𝑟−𝑡|<𝑏, we choose 𝛿={𝜖/(3(𝑛−1))}^1/(𝑛−1). Hence, |(𝑡)(𝑟≤𝜖𝐹𝑢−𝐹𝑢)|3𝜖Δ+3Δ+Δ^𝑛−1_𝑗=1|𝑟−𝑡|^𝑗≤𝜖3𝜖Δ+3Δ+(𝑛−1)Δ𝛿^𝑛−1<Δ𝜖.(3.25) Therefore, 𝐹(𝐺) is equicontinuous. The Arzela-Ascoli Theorem implies that 𝐹(𝐺) is compact. Thus, 𝐹∶𝐾→𝐾 is completely continuous.

Theorem 3.4. With all the details of (1.1)-(1.3), further, if 𝜎 exists in (0,1) such that 0<𝜎<𝛼∈(𝑛−1,𝑛), 𝑛∈ℕ, and 𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is a continuous function on [0,𝑏], then (1.1) has at least one positive solution 𝑥^∗∈𝐾^∗, satisfying ‖𝑥^∗‖≤max{‖𝜔‖,ℎ}, where ℎ=2Λ/(1−Λ)+1 and Λ is a positive constant which is observed in the proof of the theorem.

Proof. We know that the operator 𝐹∶𝐾→𝐾 is continuous and it is completely continuous by Theorem 3.3. Here we show that there exists an open set 𝑈⊆𝐾, with 𝑧≠𝛾𝐹(𝑧) for 𝛾∈(0,1) and 𝑧∈𝜕𝑈. Let 𝑧∈𝐾 be any solution of 𝑧=𝛾𝐹(𝑧), 𝛾∈(0,1). According to the Theorem 3.3, since 𝐹∶𝐾→𝐾 is continuous and it is completely continuous, we have 𝑧(𝑡)=𝛾𝐹𝑧(𝑡)=𝛾𝜆(𝑡)+ℒ(𝐼)𝑧(𝑡)+𝐼^𝛼𝑓𝑡,𝑧_𝑡+𝑦_𝑡[]≤,𝑡∈0,𝑏_{𝑛−1𝑗=1}𝜆_𝑗𝑏^𝑗+^𝑛−1_𝑗=1𝑎_𝑗𝐼^𝑗𝑧(𝑡)+𝐼^𝛼𝑓𝑡,𝑧_𝑡+𝑦_𝑡≤^𝑛−1_𝑗=1𝜆_𝑗𝑏^𝑗+‖𝑧‖^𝑛−1_𝑗=1𝑎_𝑗Γ𝑏𝑗+1^𝑗+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎𝑠^𝜎𝑓𝑠,𝑧_𝑠+𝑦_𝑠𝑑𝑠.(3.26) Since there exists a positive constant 𝑁 such that ‖𝑠^𝜎𝑓(𝑠,𝑧_𝑠+𝑦_𝑠)‖≤𝑁, as 𝑠^𝜎𝑓(𝑠,𝑧_𝑠+𝑦_𝑠) is continuous on [0,𝑏], therefore 𝑧(𝑡)≤^𝑛−1_𝑗=1𝜆_𝑗𝑏^𝑗+‖𝑧‖^𝑛−1_𝑗=1𝑎_𝑗Γ𝑏𝑗+1^𝑗+𝑁Γ(𝛼)𝐵(1−𝜎,𝛼).(3.27) Set Λ=max{Λ₁,Λ₂,Λ₃}, where Λ₁=^𝑛−1_𝑗=1𝜆_𝑗𝑏^𝑗,Λ₂=^𝑛−1_𝑗=1𝑎_𝑗Γ𝑏𝑗+1^𝑗,Λ₃=𝑁Γ(𝛼)𝐵(1−𝜎,𝛼).(3.28) Equation (3.20) implies that ‖𝑧‖≤Λ₁+‖𝑧‖Λ₂+Λ₃≤2Λ+‖𝑧‖Λ.(3.29) Therefore, we conclude that ‖𝑧‖(1−Λ)≤2Λ. Then, as a result, any solution 𝑧=𝛾𝐹(𝑧) satisfies ‖𝑧‖≠ℎ. Let 𝑈={𝑧∈𝐾∶‖𝑧‖<ℎ}. Theorem 2.2 guarantees that 𝐹 has a fixed point 𝑧∈𝑈. By using Theorem 3.2, (1.1) under the conditions of (1.3) has a positive solution 𝑥^∗∈𝐾 satisfying ‖𝑥^∗‖≤max{‖𝜔‖,ℎ}.

Example 3.5. Consider the following nonlinear fractional differential equation: ℒ(𝐷)[𝑥(𝑡)−𝑥(0)]=𝑡^−1/2𝑥_𝑡(],𝑥(𝑡)[],,𝑡∈0,1=−sin𝑡≥0,𝑡∈−1,0(3.30) where ℒ(𝐷)=𝐷^5/2−𝐷^3/2−𝐷^1/2. Here, 𝑎₁=𝑎₂=1, 𝛼=5/2∈(2,3), 𝑛=3 and 𝑓(𝑡,𝑥_𝑡)=𝑡^−1/2𝑥_𝑡 such that lim_𝑡→0⁺𝑓(𝑡,⋅)=+∞. We select 𝜎=11/12, 0<𝜎<𝛼. Hence 𝑡^𝜎𝑓(𝑡,⋅)=𝑡^5/12𝑥_𝑡 is continuous on [0,1]. We consider the initial conditions 𝑥(0)=0,𝑥₁=𝐷^3/2𝑥(𝑡)_𝑡=0=2,𝑥₂=𝐷^5/2𝑥(𝑡)_𝑡=0=1,𝑥₁≥𝑥₂>0.(3.31) By using Theorem 3.2, the nonlinear fractional differential equation (3.20) is equivalent to an integral equation given below 𝑥(𝑡)=𝜆(𝑡)+ℒ(𝐼)𝑥(𝑡)=𝐼^5/2𝑡^−1/2𝑥_𝑡(],𝑥(𝑡)[],,𝑡∈0,1=−sin𝜋𝑡,𝑡∈−1,0(3.32) where in the view of the procedure proving Theorem 3.2, we have 𝜆(𝑡)=²_𝑗=1𝜆_𝑗𝑡^5/2−𝑗=1Γ()𝑡5/2^3/2+1Γ()𝑡1/2^1/2,ℒ(𝐼)𝑥(𝑡)=²_𝑗=1𝑎_𝑗𝐼^𝑗𝑥(𝑡)=^𝑡₀(1+(𝑡−𝑠))𝑥(𝑠)𝑑𝑠.(3.33) In addition of that we have that Λ=max{Λ₁,Λ₂,Λ₃}, where Λ₁=²_𝑗=1𝜆_𝑗𝑏^𝑗=𝜆₁+𝜆₂=1Γ()+11/2Γ()=75/23√𝜋,Λ₂=²_𝑗=1𝑎_𝑗Γ1𝑗+1=1+2=32,Λ₃=𝑁𝐵Γ521−11,5122=4𝑁3√𝜋𝐵1,5122,𝑁=‖𝑡^11/12𝑥_𝑡‖.(3.34) Therefore, by using the Theorem 3.4, (3.30) has at least one positive solution 𝑥^∗∈𝐾^∗ satisfying ‖𝑥^∗‖≤max{‖𝜔‖,ℎ} where ‖𝜔‖=max_{−1≤𝑡≤0}|−sin𝜋𝑡|=1 and ℎ=Λ/(1−Λ)+1,Λ=max{Λ₁,Λ₂,Λ₃}.

4. Unique Existence of Solution

In this section, we give conditions on 𝑓 and 𝑎₁,𝑎₂,…,𝑎_𝑛, which render a unique positive solution to (1.1).

Theorem 4.1. Let 𝑓∶(0,1)×[0,∞)→[0,∞) be continuous and lim_𝑡→0⁺𝑓(𝑡,⋅)=+∞. Suppose that there exists 𝜎∈(0,1) so that 0<𝜎<𝛼∈(𝑛−1,𝑛), 𝑛∈ℕ, and 𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is a continuous function on [0,𝑏]. If further, the following conditions are satisfied (H1)𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is Lipschitz with respect to the second variable with Lipschitz constant 𝜇, that is, 𝑡^𝜎𝑓𝑡,𝑥_𝑡−𝑡^𝜎𝑓𝑡,𝑧_𝑡≤𝜇‖𝑢−𝑣‖,∀𝑢,𝑣∈𝐾,𝑡∈(0,𝑏],(4.1)(H2)0<^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝑗+1𝜇𝑏^𝛼−𝜎Γ(𝛼)𝐵(1−𝜎,𝛼)<1,(4.2) where 𝑥(𝑡)=𝑢(𝑡)+𝑦(𝑡) and 𝑧(𝑡)=𝑣(𝑡)+𝑦(𝑡); then (1.1) under the conditions of (1.3) has a unique positive solution.

Proof. As it was pointed out in the previous section, (1.1) is equivalent to (3.13). Thus, for 𝑢,𝑣∈𝐾 we obtain |𝐹𝑢(𝑡)−𝐹𝑣(𝑡)|=|ℒ(𝐼)𝑢(𝑡)−ℒ(𝐼)𝑣(𝑟)|+𝐼^𝛼𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝐼^𝛼𝑓𝑟,𝑣_𝑟+𝑦_𝑟≤‖𝑢−𝑣‖_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝐼𝑗+1^𝛼𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝑓𝑡,𝑣_𝑡+𝑦_𝑡=‖𝑢−𝑣‖_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝐼𝑗+1^𝛼𝑡^−𝜎𝑡^𝜎𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝑡^−𝜎𝑡^𝜎𝑓𝑡,𝑣_𝑡+𝑦_𝑡≤‖𝑢−𝑣‖_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+1𝑗+1Γ(𝛼)max_{0≤𝑡≤𝑏}^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎𝑠^𝜎𝑓𝑡,𝑢_𝑡+𝑦_𝑡−𝑠^𝜎𝑓𝑡,𝑣_𝑡+𝑦_𝑡≤‖‖𝑑𝑠𝑢−𝑣_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝜇𝑗+1Γ(𝛼)max_{0≤𝑡≤𝑏}^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎≤‖‖𝑑𝑠𝑢−𝑣_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝜇‖‖𝑗+1𝑢−𝑣_𝑏Γ(𝛼)max_{0≤𝑡≤𝑏}𝑡^𝛼−𝜎𝐵()≤‖‖1−𝜎,𝛼𝑢−𝑣_𝑏^𝑛−1_𝑗=1𝑎_𝑗𝑏^𝑗Γ+𝑗+1𝜇𝑏^𝛼−𝜎Γ(𝛼)𝐵(1−𝜎,𝛼),(4.3) where 𝐹 is given in (3.16). Hence, in view of the Theorem 2.3, 𝐹 will have a unique fixed point in 𝐾, which is the unique positive solution of (1.1).

Example 4.2. Consider the nonlinear fractional differential equation given below ℒ(𝐷𝑥(𝑡)(0=𝜂)[−𝑥)]√𝑡^−t−𝜋1+𝑥_𝑡(𝑠)1+𝑥²_𝑡(𝑠)(]𝑥(𝑡)[],𝑑𝑠,𝑡∈0,𝜋,𝜂>0,=−sin𝑡≥0,𝑡∈−𝜋,0(4.4) where ℒ(𝐷)=𝐷^5/2−𝑎₁𝐷^3/2−𝑎₂𝐷^1/2 with initial conditions 𝑥(0)=0, 𝑥₁=𝐷^3/2𝑥(𝑡)|_𝑡=0=2, 𝑥₂=𝐷^5/2𝑥(𝑡)|_𝑡=0=1, 𝑥₁≥𝑥₂>0, 0<𝑎₁≤1/(4𝜋), 0<𝑎₂≤8/𝜋², and 𝜂=3𝜋^−3/7)/16𝐵(1/12,5/12). Here lim_𝑡→0⁺𝑓(𝑡,𝑥_𝑡)=+∞. We select 𝜎=11/12; then 0<𝜎<𝛼 and 𝑡^𝜎𝑓(𝑡,𝑥_𝑡)=𝑡^5/12𝑓(𝑡,𝑥_𝑡) is continuous on [0,𝜋]. Now, we review the Lipschitz condition with respect to the second variable. Note that from equations (3.11) and (3.12), for 𝑥_𝑡,𝑧_𝑡∈𝐶([−𝜋,0],ℝ^≥0), we have 𝑥_𝑡=𝑢_𝑡+𝑦_𝑡,𝑧_𝑡=𝑣_𝑡+𝑦_𝑡,𝑡∈[−𝜋,𝜋] and 𝑥_𝑡=𝑢(𝑡)+𝑦(𝑡), 𝑧_𝑡=𝑣(𝑡)+𝑦(𝑡), 𝑡∈(0,𝜋]) where 𝑢,𝑣∈𝐾. Without loss of generality, we suppose that 𝑥_𝑡(𝑠)≥𝑦_𝑡(𝑠), for all 𝑠∈[−𝜋,0]. Then, for each 𝑡∈(0,𝜋], 𝑡^11/12𝑓𝑡,𝑥_𝑡−𝑡^11/12𝑓𝑡,𝑧_𝑡=𝜂𝑡^11/12−𝑡_−𝜋1+𝑥(𝑡+𝑠)1+𝑥²()−()𝑡+𝑠1+𝑧𝑡+𝑠1+𝑧²(𝑡+𝑠)𝑑𝑠=𝜂𝑡^11/120_𝑡−𝜋1+𝑥(𝑠)1+𝑥²(𝑠)−1+𝑧(𝑠)1+𝑧²(𝑠)𝑑𝑠≤𝜂𝑡^11/120_𝑡−𝜋1+𝑥(𝑠)1+𝑥²(𝑠)−1+𝑧(𝑠)1+𝑥²(𝑠)𝑑𝑠≤𝜂max_{0≤𝑡≤𝜋}𝑡^11/120_𝑡−𝜋|𝑥(𝑠)−𝑧(𝑠)|1+𝑥²(𝑠)(𝑡)(𝑡)𝑑𝑠≤𝜂‖𝑥−𝑧‖max_{0≤𝑡≤𝜋}𝑡^11/120_𝑡−𝜋𝑑𝑠=𝜂‖𝑢−𝑣‖max_{0≤𝑡≤𝜋}𝑡^11/12(𝜋−𝑡)=𝜂𝜋^23/12‖𝑢−𝑣‖.(4.5) Hence, the condition of (H1) holds with 𝜇=𝜋^23/12𝜂. On other hand, we have 𝑎₁𝑏Γ(2)+𝑎₂𝑏²Γ(3)+𝜇𝜋^11/12Γ()𝐵15/2,512∈().120,1(4.6) Thus, (4.4) satisfies the conditions required by the Theorem 4.1. This Theorem implies that the nonlinear equation (4.4) has a unique solution in 𝐾. By using the Theorem 3.2, (4.4) is equivalent to the following integral equation: 𝑥(𝑡)=1Γ()15/22−𝑡4𝜋^3/2+1Γ()𝑡3/2^1/2+𝑎₁𝐼𝑥(𝑡)+𝑎₂𝐼²𝑥(𝑡)+𝐼^5/2𝑓𝑡,𝑥_𝑡.(4.7) The solution of (4.4) is 𝑥(𝑡)=lim_𝑛→+∞𝑥_𝑛(t), where 𝑥_𝑛+1(𝑡)=𝐹𝑥_𝑛(𝑡). Therefore, the iterated sequence is given by 𝑥₁(𝑡)=𝑎₁𝑡^3/2+𝑎₂𝑡^1/2+𝜂𝑒^−𝑡√𝑡𝐼^5/2−𝑡_−𝜋1+𝑥(𝑠)1+𝑥²(𝑠)𝑑𝑠=𝑎₁𝑡^3/2+𝑎₂𝑡^1/2+𝜂𝑒^−𝑡√𝑡𝐼^5/212ln1+𝑡²1+𝜋²−tan⁻¹𝑡,𝑥₂(𝑡)=𝑎₁𝐼+𝑎₂𝐼²+𝐼^5/2𝑥₁(𝑡)+𝑥₁(𝑡),⋮𝑥_𝑛+1(𝑡)=^𝑛_𝑖=0𝑎₁𝐼+𝑎₂𝐼²+𝐼^{5/2𝑛−𝑖}𝑥₁(𝑡),𝑛=1,2,….(4.8)