Multiple Positive Solutions for a Fractional Boundary Value Problem with Fractional Integral Deviating Argument

Abstract

This work is devoted to the existence of positive solutions for a fractional boundary value problem with fractional integral deviating argument. The proofs of the main results are based on Guo-Krasnoselskii fixed point theorem and Avery and Peterson fixed point theorem. Two examples are given to illustrate the obtained results, ending the paper.

1. Introduction

The study of differential equations with deviating arguments has known a rapid development and still attracts most attention; it is a vast domain constantly enriched and fruitful. Differential equations with deviating arguments appear in many areas of sciences and technology such as in the study of problems related with combustion in rocket motion, in the theory of automatic control, in economics, and biological systems.

There exists a vast amount of literature devoted to the investigation of boundary value problems with deviating arguments (see [1–4]); one can cite the paper of Haloi et al. [5], where the authors studied an abstract initial value problem with deviating arguments by the help of Sobolevskii and Tanabe theory and fixed point theorems. However, few papers exist concerning fractional boundary value problems with deviating arguments, see [3, 4, 6, 7]. We can cite the work of Jankowski [8], where the author studied the positivity of solutions for fourth-order differential equations with deviating arguments and integral boundary conditions, by Avery and Peterson theorem.

Fractional differential equations have been proved to be valuable tools in the modelling of many phenomena in various fields of engineering, such as rheology, fluid flows, electrical networks, viscoelasticity, chemical physics, biosciences, signal processing, systems control theory, electrochemistry, mechanics and diffusion processes; for more details on this subject we refer the reader to [9–12].

In this work, we study a fractional boundary value problem with fractional deviating argument (P): ^𝑐𝐷^𝑞₀₊𝑢(𝑡)+𝑎(𝑡)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑡)𝑢=0,0<𝑡<1,_′(0)=𝑢^′′(0)=0,𝑢(0)=𝛼𝑢(1), (P) where 2<𝑞<3, 0<𝜎<1, ^𝑐𝐷^𝑞₀₊ denotes the Caputo fractional derivative, 𝑓∈𝐶(ℝ,ℝ) is a given function, 𝜃∈𝐶([0,1],[0,1]), 𝑎∈𝐶([0,1],ℝ). We assume that the deviating argument 𝜃 satisfies the delay property 0≤𝜃(𝑡)≤𝑡 for any 𝑡∈[0,1]. Let us remark that when we apply the fractional integral 𝐼^𝜎₀⁺ to the deviating argument 𝜃, then the problem (P) becomes more delayed in time. The existence of a deviation delay in time is necessary in some situations, to avoid the unstable combustion in liquid rocket engine and to contribute in the conversion of the fuel mixture into the product of combustion. No contributions exist, as far as we know, concerning the positivity of solutions for the fractional differential equation with deviating argument (P).

The literature on fractional boundary value problems with deviating argument is very reduced; we can cite Ntouyas et al. [6] work, where the authors investigated the existence of positive solutions to fractional differential equations with advanced arguments by Guo-Krasnoselskii fixed point theorem. Other interesting results on fractional boundary value problems can be found in [13–21].

The organization of this paper is as follows. In Section 2, we provide some necessary background materials and definitions. In Section 3, we present some lemmas which are useful to obtain our main results. In Section 4, we discuss the existence of at least one positive solution of problem (P) by using Guo-Krasnoselskii fixed point theorem on cone; then, under some sufficient conditions on the nonlinear source term, we apply Avery-Peterson theorem to prove the existence of at least three positive solutions. At the end of this section, we give two examples to verify the obtained results.

2. Background Materials and Definitions

In this section, we introduce definitions from fractional calculus theory and preliminary facts which are used throughout this paper.

Definition 2.1 (see [10]). The integral 𝐼^𝛼_𝑎+𝑔(𝑡)=1Γ(𝛼)^𝑡_𝑎𝑔(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠 (2.1) is called Riemann-Liouville fractional integral of 𝑔 of order 𝛼>0 when the right side exists. Here Γ is the gamma function.

Definition 2.2 (see [10]). For a function 𝑔 given on the interval [𝑎,𝑏], the Caputo fractional derivative of 𝑔 of order 𝛼 is defined by ^𝑐𝐷^𝛼_𝑎⁺𝑔(𝑡)=1Γ(𝑛−𝛼)^𝑡_𝑎𝑔^(𝑛)(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠, (2.2) where 𝑛=[𝛼]+1 and [𝛼] denotes the integer part of 𝛼.

Lemma 2.3 (see [10]). Letting 𝛼,𝛽>0 and letting 𝑛=[𝛼]+1, then the following relations hold: ^𝑐𝐷^𝛼₀⁺𝑡^𝛽−1=(Γ(𝛽)/Γ(𝛽−𝛼))𝑡^{𝛽−𝛼−1}, 𝛽>𝑛 and ^𝑐𝐷^𝛼₀⁺𝑡^𝑘=0, 𝑘=0,1,2,…,𝑛−1.

Lemma 2.4 (see [10]). For 𝛼>0, 𝑔(𝑡)∈𝐶(0,1), the homogenous fractional differential equation ^𝑐𝐷^𝛼_𝑎⁺𝑔(𝑡)=0 (2.3) has a solution 𝑔(𝑡)=𝑐₁+𝑐₂𝑡+𝑐₃𝑡²+⋯+𝑐_𝑛𝑡^𝑛−1, (2.4) where 𝑐_𝑖∈𝑅, 𝑖=0,…,𝑛, and 𝑛=[𝛼]+1.

Denote by 𝐿¹([0,1],ℝ) the Banach space of Lebesgue integrable functions from [0,1] into ℝ with the norm ‖𝑦‖_𝐿1=∫¹₀|𝑦(𝑡)|𝑑𝑡. Define 𝐸=𝐶[0,1], equipped with the norm ‖𝑢‖=max_𝑡∈[0,1]|𝑢(𝑡)|.

The following lemmas give some properties of Riemann-Liouville fractional integrals and Caputo fractional derivative.

Lemma 2.5 (see [10]). Let 𝑝,𝑞≥0, 𝑓∈𝐿₁[𝑎,𝑏]. Then 𝐼^𝑝₀₊𝐼^𝑞₀₊𝑓(𝑡)=𝐼₀^𝑝+𝑞₊𝑓(𝑡)=𝐼^𝑞₀₊𝐼^𝑝₀₊𝑓(𝑡) and ^𝑐𝐷^𝑞₀₊𝐼^𝑞₀₊𝑓(𝑡)=𝑓(𝑡), for all 𝑡∈[𝑎,𝑏].

Lemma 2.6 (see [10]). Let 𝛽>𝛼>0. Then the formula ^𝑐𝐷^𝛼₀⁺𝐼^𝛽₀₊𝑓(𝑡)=𝐼₀^𝛽−𝛼₊𝑓(𝑡) holds almost everywhere on 𝑡∈[𝑎,𝑏], for 𝑓∈𝐿₁[𝑎,𝑏], and it is valid at any point 𝑡∈[𝑎,𝑏] if 𝑓∈𝐶[𝑎,𝑏].

Now we present the necessary definitions from the theory of cone in Banach spaces.

Definition 2.7. A nonempty subset 𝑃 of a Banach space 𝐸 is called a cone if 𝑃 is convex, closed and satisfies the conditions(i)𝛼𝑥∈𝑃 for all 𝑥∈𝑃 and any 𝛼∈ℝ₊,(ii)𝑥,−𝑥∈𝑃 imply 𝑥=0.

Definition 2.8. A map 𝐹 is said to be a nonnegative continuous convex functional on a cone 𝑃 of a real Banach space 𝐸 if 𝐹∶𝑃→[0,∞[ is continuous and 𝐹𝑡𝑥+(1−𝑡)𝑦≤𝑡𝐹(𝑥)+(1−𝑡)𝐹𝑦,∀𝑥,𝑦∈𝑃,𝑡∈[0,1]. (2.5) Similarly 𝐹 is a nonnegative continuous concave functional on a cone 𝑃 of a real Banach space 𝐸 if 𝐹𝑡𝑥+(1−𝑡)𝑦≥𝑡𝐹(𝑥)+(1−𝑡)𝐹𝑦,∀𝑥,𝑦∈𝑃,𝑡∈[0,1]. (2.6)

Definition 2.9. A mapping is called completely continuous if it is continuous and maps bounded sets into relatively compact sets.
In the following, we state some fixed point theorems.

Theorem 2.10 (Guo-Krasnoselskii fixed point theorem on cone [22]). Let 𝐸 be a Banach space, and let 𝐾⊂𝐸 be a cone. Assume Ω₁ and Ω₂ are open subsets of 𝐸 with 0∈Ω₁, Ω₁⊂Ω₂ and let 𝒜∶𝐾∩Ω₂⧵Ω₁⟶𝐾 (2.7) be a completely continuous operator such that(i)‖𝒜𝑢‖≤‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₁, and ‖𝒜𝑢‖≥‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₂,(ii)‖𝒜𝑢‖≥‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₁, and ‖𝒜𝑢‖≤‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₂.
Then 𝒜 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

Theorem 2.11 (Avery and Peterson fixed point theorem [23]). Let 𝑃 be a cone in a real Banach space 𝐸. Let 𝜑 and Φ be continuous, nonnegative, and convex functionals on 𝑃, let Λ be a continuous nonnegative, and concave functional on 𝑃, and let Ψ be continuous and nonnegative functional on 𝑃 satisfying Ψ(𝑘𝑢)≤𝑘‖𝑢‖ for 0≤𝑘≤1. Define the sets, 𝑃(𝜑,𝑑)={𝑢∈𝑃,𝜑(𝑢)<𝑑}, 𝑃(𝜑,Λ,𝑏,𝑑)={𝑢∈𝑃,𝑏≤Λ(𝑢),𝜑(𝑢)≤𝑑}, 𝑃(𝜑,Φ,Λ,𝑏,𝑐,𝑑)={𝑢∈𝑃,𝑏≤Λ(𝑢),Φ(𝑢)≤𝑐,𝜑(𝑢)≤𝑑}, 𝑅(𝜑,Ψ,𝑎,𝑑)={𝑢∈𝑃,𝑎≤Ψ(𝑢),𝜑(𝑢)≤𝑑}. For 𝑀 and 𝑑 positive numbers we have Λ(𝑢)≤Ψ(𝑢) and ‖𝑢‖≤𝑀𝜑(𝑢) for any 𝑢∈𝑃(𝜑,𝑑). Assume 𝑇∶𝑃(𝜑,𝑑)→𝑃(𝜑,𝑑) is completely continuous and there exists positive numbers 𝑎, 𝑏, and 𝑐 with 𝑎<𝑏 such that(S1){𝑢∈𝑃(𝜑,Φ,Λ,𝑏,𝑐,𝑑),Λ(𝑢)>𝑏}≠∅ and Λ(𝑇𝑢)>𝑏 for 𝑢∈𝑃(𝜑,Φ,Λ,𝑏,𝑐,𝑑),(S2)Λ(𝑇𝑢)>𝑏 for 𝑢∈𝑃(𝜑,Λ,𝑏,𝑑) with Φ(𝑇𝑢)>𝑐,(S3)0∉𝑅(𝜑,Ψ,𝑎,𝑑) and Ψ(𝑇𝑢)<𝑎 for 𝑢∈𝑅(𝜑,Ψ,𝑎,𝑑) with Ψ(𝑢)=𝑎.
Then 𝑇 has at least three positive fixed points 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝜑,𝑑) such that 𝜑(𝑢_𝑖)≤𝑑, for 𝑖=1,2,3. 𝑏<Λ(𝑢₁), 𝑎<Ψ(𝑢₂), with Λ(𝑢₂)<𝑏 and Ψ(𝑢₃)<𝑎.

3. Some Lemmas

We start by solving an auxiliary problem which allows us to get the expression of the solution. Let us consider the following problem: ^𝑐𝐷^𝑞₀₊𝑢(𝑡)(𝑡)𝑢+𝑦=0,0<𝑡<1,_′(0)=𝑢^′′(0)=0,𝑢(0)=𝛼𝑢(1). (𝑃1)

Lemma 3.1. Assuming that 𝛼≠1 and 𝑦∈𝐶([0,1],ℝ), then the problem (𝑃1) has a unique solution given by 𝑢(𝑡)=1Γ𝑞¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (3.1) where 𝐺()=−()𝑡,𝑠𝑡−𝑠^𝑞−1+𝛼()𝛼−11−𝑠^𝑞−1𝛼,0≤𝑠≤𝑡()𝛼−11−𝑠^𝑞−1,𝑡≤𝑠≤1. (3.2)

Proof. Using Lemmas 2.3 and 2.4, we get 𝑢(𝑡)=−𝐼^𝑞₀₊𝑦(𝑡)+𝑎+𝑏𝑡+𝑐𝑡². (3.3) The boundary conditions 𝑢^′(0)=𝑢^′′(0)=0 give 𝑏=𝑐=0, and the condition 𝑢(0)=𝛼𝑢(1) implies 𝑎=(𝛼/(𝛼−1))𝐼^𝑞₀₊𝑦(1). Substituting 𝑎,𝑏, and 𝑐 by their values in (3.3) we get the desired result.

Some useful estimates of the Green function 𝐺 are given hereafter.

Lemma 3.2. If 𝛼>1, then for all 𝑠,𝑡∈[0,1], the Green function 𝐺(𝑡,𝑠) is nonnegative, continuous and satisfies(i)max_{𝑠,𝑡∈[0,1]}𝐺(𝑡,𝑠)=𝛼/(𝛼−1),(ii)min_{𝑠,𝑡∈[0,𝜏]}𝐺(𝑡,𝑠)=(1−𝜏)^𝑞−1/(𝛼−1) where 0<𝜏<1.

Proof. The proof is easy; then we omit it.

Lemma 3.3. Assuming that 𝛼>1 and 𝑦∈𝐶([0,1],ℝ₊), then the unique solution of problem (𝑃1) satisfies 𝑢(𝑡)≥0 on [0,1]. Moreover we have min_t∈[]0,𝜏𝑢(𝑡)≥1−𝜏²𝛼‖𝑢‖. (3.4)

Proof. From 𝑢^′′(𝑡)=−𝐼₀^𝑞−2₊𝑦(𝑡) and hypothesis on the function 𝑦 we deduce that 𝑢^′′≤0 consequently the function 𝑢 is concave. Moreover from the condition 𝑢(0)=𝛼𝑢(1) we can say that the values 𝑢(0) and 𝑢(1) have the same sign. Since 𝑢(0)=(𝛼/(𝛼−1))𝐼^𝑞₀₊𝑦(1)≥0, then 𝑢(𝑡)≥0 on [0,1]. Let us analyze the boundary conditions; from 𝑢^′′≤0 and the condition 𝑢^′(0)=0 we conclude that 𝑢^′ is also negative; consequently the function 𝑢 is decreasing. From the above discussion we conclude that max_𝑡∈[0,1]𝑢(𝑡)=𝑢(0) and min_{𝑡∈[0,𝜏]}𝑢(𝑡)=𝑢(𝜏). The concavity of 𝑢 allows to write 𝑢(𝜏)−𝑢(0)𝜏≥𝑢(1)−𝑢(𝜏).1−𝜏 (3.5) Using the condition 𝑢(0)=𝛼𝑢(1) it yields 𝑢(𝜏)(0)≥𝑢(0)−𝑢𝛼𝜏(𝜏)𝜏1−𝜏−𝑢(),1−𝜏 (3.6) then 𝑢(𝜏)𝜏1+(0)𝜏1−𝜏≥𝑢1+𝛼().1−𝜏 (3.7) Since (1/(1−𝜏))>1, we have 𝑢(𝜏)1(0)𝜏1−𝜏≥𝑢1+𝛼()(0)𝜏1−𝜏≥𝑢1+𝛼, (3.8) so, 𝑢(𝜏)≥𝑢(0)(1−𝜏)𝛼+𝜏𝛼. (3.9) Using the assumption 𝛼>1, we obtain 𝑢(𝜏)≥1−𝜏²𝛼𝑢(0), (3.10) consequently min_{[]𝑡∈0,𝜏}𝑢(𝑡)≥1−𝜏²𝛼‖𝑢‖. (3.11) This achieves the proof.

4. Existence of Positive Solutions

We first state the assumptions that will be used to prove our main results:(H1)𝜃∈𝐶([0,1],[0,1]) and satisfies the delay property 0≤𝜃(𝑡)≤𝑡, for all 𝑡∈[0,1],(H2)𝑎∈𝐶([0,1],ℝ₊) and ∫^𝜏₀𝑎(𝑠)𝑑𝑠≠0,(H3)𝑓∈𝐶(ℝ₊,ℝ₊) and 𝛼>1.

Define the operator 𝑇∶𝐸→𝐸 by (𝑡)=1𝑇𝑢Γ𝑞¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)𝑑𝑠. (4.1)

Definition 4.1. A function 𝑢 is called positive solution of problem (P) if 𝑢(𝑡)≥0, for all 𝑡∈[0,1] and it satisfies the boundary conditions in (P).
Let us introduce the following notations 𝐴₀=lim_𝑢→0⁺(𝑓(𝑢)/𝑢), 𝐴_∞=lim_𝑢→∞(𝑓(𝑢)/𝑢). The case 𝐴₀=0 and 𝐴_∞=∞ is called superlinear case, and the case 𝐴₀=∞ and 𝐴_∞=0 is called sublinear case.

Theorem 4.2. Assuming that (H1)–(H3) hold, then the problem (P) has at least one positive solution in both cases, superlinear as well as sublinear.

Proof. We apply Guo-Krasnoselskii fixed point theorem on cone. Define the cone 𝑃={𝑢∈𝐸,𝑢nonnegativeon[0,1]andmin_{𝑡∈[0,𝜏]}𝑢(𝑡)≥((1−𝜏²)/𝛼)‖𝑢‖}. It is easy to check that 𝑃 is a nonempty closed and convex subset of 𝐸, so it is a cone. One can check that 𝑇(𝑃)⊂𝑃. It is obvious that 𝑇 is continuous since 𝐺,𝑎 and 𝑓 are continuous. Let us prove that 𝑇 is completely continuous mapping on 𝑃.
Claim 1. 𝑇(𝐵_𝑟) is uniformly bounded, where 𝐵_𝑟={𝑢∈𝑃,‖𝑢‖≤𝑟}. Let us remark that if 𝑢∈𝑃, then min_{[]𝑡∈0,𝜏}𝑢𝐼^𝜎₀⁺𝜃(𝑡)≥1−𝜏²𝛼‖𝑢‖. (4.2) In fact from the delay property, the properties of Riemann fractional integrals and 0<𝜎<1, we have 0≤𝐼^𝜎₀⁺𝜃(𝑡)≤𝑡^𝜎+1Γ(𝜎+2)≤1. (4.3) In view of the concavity of 𝑢 it yields 𝑢𝐼^𝜎₀⁺𝜃(𝑡)≥min_{[]𝑡∈0,𝜏}𝑢(𝑡)≥1−𝜏²𝛼‖𝑢‖. (4.4) Therefore min_{[]𝑡∈0,𝜏}𝑢𝐼^𝜎₀⁺𝜃(𝑡)≥min_{[]𝑡∈0,𝜏}𝑢(𝑡)≥1−𝜏²𝛼‖𝑢‖≥1−𝜏²𝛼𝑢𝐼^𝜎₀⁺𝜃. (4.5) Since the functions 𝑎 and 𝑓 are continuous, then there exists a constant 𝑘 such that max_𝑡∈[0,1]|𝑎(𝑡)𝑓(𝑢(𝐼^𝜎₀⁺𝜃(𝑡)))|=𝑘 for any 𝑢∈𝐵_𝑟. By virtue of Lemma 3.2 we obtain |𝑇𝑢(𝑡)|≤𝑘𝛼()Γ𝑞.𝛼−1 (4.6) Hence 𝑇 is uniformly bounded.
Claim 2. 𝑇 is equicontinuous. We have for any 𝑢∈𝐵_𝑟𝑇^′𝑢(𝑡)=𝑞−1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−2𝑎(𝑠)𝑓𝑢(𝑠),𝑢𝐼^𝜎𝜃(𝑠)≤𝑘𝑑𝑠Γ𝑞−1¹₀(1−𝑠)^𝑞−2𝑘𝑑𝑠=Γ𝑞. (4.7) Therefore 𝑇𝑢𝑡₂−𝑇𝑢𝑡₁=^𝑡2_𝑡1𝑇^′𝑢(𝑡)≤𝑘𝑡𝑑𝑡²−𝑡₁Γ𝑞. (4.8) Thus 𝑇 is equicontinuous; from Ascoli-Arzela Theorem, we deduce that 𝑇 is completely continuous.
Now, we prove the sublinear case. Since 𝐴₀=∞, then for any 𝐴>0 there exists 𝛿>0, such that for any 𝑢, 0<𝑢≤𝛿 then 𝑓(𝑢)≥A𝑢. Set Ω₁={𝑢∈𝐸∶‖𝑢‖<𝛿}. Letting 𝑢∈𝑃∩𝜕Ω₁, we can see that 𝑢(𝐼^𝜎₀⁺𝜃)∈𝑃∩𝜕Ω₁; then from (4.5) we have ‖𝑇𝑢‖=max_[]𝑡∈0,1|(𝑡=𝛼𝑇𝑢)|(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)≥𝑑𝑠𝐴𝛼(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑢𝐼^𝜎₀⁺𝜃(𝑠)𝑑𝑠. (4.9) With the help of (4.4) we get ‖‖≥𝐴𝑇𝑢1−𝜏²‖𝑢‖Γ𝑞¹₀𝑎(𝑠)𝑑𝑠. (4.10) In view of assumption (H2) we can choose 𝐴=Γ(𝑞)/(1−𝜏²)∫¹₀𝑎(𝑠)𝑑𝑠; therefore (4.10) becomes ‖𝑇𝑢‖≥‖𝑢‖.
On the other hand since 𝐴_∞=0, we deduce that for any 𝜀>0 there exists 𝛾>0, such that for any 𝑢, 𝑢≥𝛾 then 𝑓(𝑢)≤𝜀𝑢. Setting 𝑅=max{2𝛿,((𝛼−1)/𝛼)𝛾} and Ω₂={𝑢∈𝐸∶‖𝑢‖<𝑅}, then Ω₁⊂Ω₂ and for 𝑢∈𝑃∩𝜕Ω₂ we have min_{𝑡∈[0,𝜏]}𝑢(𝑡)≥(𝛼/(𝛼−1))‖𝑢‖=𝛼𝑅/(𝛼−1)≥𝛾; consequently 𝑢(𝐼^𝜎₀⁺𝜃(𝑡))≥𝛾. Taking Lemma 3.2 and (4.5) into account it yields (𝑡)≤𝛼𝑇𝑢(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)≤𝑑𝑠𝛼𝜀(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑢𝐼^𝜎₀⁺𝜃(𝑠)≤𝑢𝐼𝑑𝑠𝛼𝜀^𝜎₀⁺𝜃(𝛼−1)Γ𝑞¹₀𝑎(𝑠)‖𝑢‖𝑑𝑠≤𝛼𝜀(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑑𝑠. (4.11) Then if we choose 𝜀=(𝛼−1)Γ(𝑞)/𝛼∫¹₀𝑎(𝑠)𝑑𝑠, we get ‖𝑇𝑢‖≤‖𝑢‖ for all 𝑢∈𝑃∩𝜕Ω₂. The second part of Theorem 2.10 implies that 𝑇 has a fixed point in 𝑃∩(Ω₂⧵Ω₁); that means that 𝑇 has at least a positive solution in 𝑃∩(Ω₂⧵Ω₁). Arguing as above, we prove the superlinear case. The proof is complete.

Let us introduce the following functionals. Defining on 𝑃, the nonnegative, continuous, and concave functional Λ by Λ(𝑢)=min_{𝑡∈[0,𝜏]}|𝑢(𝑡)|, then Λ(𝑢)≤‖𝑢‖. Defining the nonnegative, continuous, and convex functionals 𝜑 and Φ on 𝑃 by 𝜑(𝑢)=Φ(𝑢)=‖𝑢‖ and the nonnegative continuous functional Ψ on 𝑃 by Ψ(𝑢)=‖𝑢‖, then Ψ(𝑘𝑢)≤𝑘‖𝑢‖ for 0≤𝑘≤1.

Theorem 4.3. Let assumptions (H1)–(H3) hold, and assume that there exist positive constants 𝑎,𝑏,𝑐,𝑑,𝜇,𝐿 such that 𝑎<𝑏, 𝜇>(𝛼/(𝛼−1)Γ(𝑞))∫¹₀𝑎(𝑠)𝑑𝑠, 𝐿<(𝛼(1−𝜏)^𝑞−1/(𝛼−1)Γ(𝑞))∫^𝜏₀𝑎(𝑠)𝑑𝑠 and(i)𝑓(𝑢)≤𝑑/𝜇 for 𝑢∈[0,𝑑],(ii)𝑓(𝑢)≥𝑏/𝐿 for 𝑢∈[𝑏,𝑏/𝜉],(iii)𝑓(𝑢)≤𝑎/𝜇 for 𝑢∈[0,𝑎].
Then problem (P) has at least three positive solutions 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝜑,𝑏) such that 𝜑(𝑢_𝑖)≤𝑑, for 𝑖=1,2,3. 𝑏<Λ(𝑢₁), 𝑎<Ψ(𝑢₂), with Λ(𝑢₂)<𝑏 and Ψ(𝑢₃)<𝑎.

Proof. To prove the existence of three positive solutions, we apply Theorem 2.11. Proceeding analogously as in the proof of Theorem 4.2, we prove that the mapping 𝑇 is completely continuous on 𝑃(𝜑,𝑑).
Claim 1. 𝑇(𝑃(𝜑,𝑑))⊂𝑃(𝜑,𝑑).
Letting 𝑢∈𝑃(𝜑,𝑑), then ‖𝑢‖≤𝑑; since 0<𝜃(𝑡)<𝑡, for all 𝑡∈[0,1], then in view of (4.5) we have 𝑢(𝐼^𝜎₀⁺𝜃(𝑡))≤‖𝑢(𝐼^𝜎₀⁺𝜃)‖≤‖𝑢‖≤𝑑. Thus with the help of assumption (i) it yields 𝜑(𝑇𝑢)=‖𝑇𝑢‖=max_[]𝑡∈0,11Γ𝑞¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)≤𝛼𝑑𝑠(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)≤𝑑𝑑𝑠𝜇𝛼(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑑𝑠<𝑑, (4.12) and hence 𝑇𝑢∈𝑃(𝜑,𝑑).
Claim 2. (S1) holds; that is, {𝑢∈𝑃(𝜑,Φ,Λ,𝑏,𝑏/𝜉,𝑑),Λ(𝑢)>𝑏}≠∅ and Λ(𝑇𝑢)>𝑏 for 𝑢∈𝑃(𝜑,Φ,Λ,𝑏,𝑏/𝜉,𝑑). Let 𝑥(𝑡)=(𝑏/2)(1+1/𝜉); since 𝛼>1 and 𝜉<1, then Φ(𝑥)(𝑥)=‖𝑥‖=𝑏=𝜑211+𝜉=𝑏()<𝑏2𝜉𝜉+1𝜉. (4.13)
Moreover we have Λ(𝑥)=min_{[]𝑡∈0,𝜏}𝑥(𝑡)=𝑏211+𝜉>𝑏>𝜉‖𝑥‖. (4.14) Thus 𝑥∈𝑃(𝜑,Φ,Λ,𝑏,𝑏/𝜉,𝑑), so {𝑢∈𝑃(𝜑,Φ,Λ,𝑏,𝑏/𝜉,𝑑),Λ(𝑢)>𝑏}≠∅.
Letting 𝑢∈𝑃(𝜑,Φ,Λ,𝑏,𝑏/𝜉,𝑑), then 𝑏≤𝑢(𝑡)≤𝑏/𝜉 and 𝑏≤𝑢(𝐼^𝜎₀⁺𝜃(𝑡))≤𝑏/𝜉; thus by virtue of Lemma 3.2 and assumption (ii), we obtain Λ(𝑇𝑢)=min_{[]𝑡∈0,𝜏}|(𝑡≥()𝑇𝑢)|1−𝜏^𝑞−1(𝛼−1)Γ𝑞^𝜏₀𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)≥()𝑑𝑠1−𝜏^𝑞−1𝑏(𝛼−1)Γ𝑞𝐿^𝜏₀𝑎(𝑠)𝑑𝑠>𝑏. (4.15)
So condition (S1) is satisfied.
Claim 3. (S2) holds. Letting 𝑢∈𝑃(𝜑,Λ,𝑏,𝑑) such Φ(𝑇𝑢)=‖𝑇𝑢‖>𝑏/𝜉, then Λ(𝑇𝑢)=min_{[]𝑡∈0,𝜏}|𝑇𝑢(𝑡)|≥𝜉‖𝑇𝑢‖>𝑏. (4.16) This implies that claim (S2) holds true.
Claim 4. (S3) holds. Letting 𝑢∈𝑅(𝜑,Ψ,𝑎,𝑑), then 0<𝑎≤‖𝑢‖≤𝑑, and then 0∉𝑅(𝜑,Ψ,𝑎,𝑑). Let 𝑢∈𝑅(𝜑,Ψ,𝑎,𝑑) with Ψ(𝑢)=‖𝑢‖=𝑎; using Lemma 3.2 and assumption (iii) it yields Ψ(𝑇𝑢)=max_[]𝑡∈0,11Γ𝑞¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)=𝛼𝑑𝑠(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑠)≤𝑎𝑑𝑠𝜇𝛼(𝛼−1)Γ𝑞¹₀𝑎(𝑠)𝑑𝑠<𝑎. (4.17) Then (S3) is satisfied.
Finally we conclude by Theorem 2.11 that there exist at least three positive solutions 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝜑,𝑑) such that 𝜑(𝑢_𝑖)≤𝑑, for 𝑖=1,2,3. 𝑏<Λ(𝑢₁), 𝑎<Ψ(𝑢₂), with Λ(𝑢₂)<𝑏 and Ψ(𝑢₃)<𝑎. The proof of Theorem 4.2 is complete.

Remark 4.4. Contrary to paper [8], we cannot consider problem (P) with 0≤𝑡≤𝜃(𝑡)≤1; in fact 𝐼^𝜎₀⁺𝜃(𝑡) can be greater than 1 for 0<𝜎<1.
Now we give two examples to illustrate Theorems 4.2 and 4.3.

Example 4.5. Let us consider the following fractional boundary value problem with fractional deviating argument: ^𝑐𝐷₀^7/3+𝑢(𝑡)+𝑎(𝑡)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑡)𝑢=0,0<𝑡<1,_′(0)=𝑢^′′(0)(0)=5=0,𝑢2𝑢(1), (4.18) where 𝑞=7/3, 𝜎=3/5, 𝛼=5/2, 𝑓(𝑢)=exp(−𝑢), 𝜃(𝑡)=𝑡(1−𝑡)≤𝑡, 𝑎(𝑡)=√1+𝑡, 𝜏=9/10. By simple calculus we obtain 𝐼₀^3/5₊𝜃(𝑡)=(1/Γ(3/5))(𝑡)^3/5((25/24)𝑡−(125/156)𝑡²), ∫₀^9/10𝑎(𝑠)𝑑𝑠=1.0793≠0. It is easy to check that the assumptions (H1)–(H3) hold and that 𝐴₀=∞, 𝐴_∞=0. Applying Theorem 4.2 we deduce that there exists at least one positive solution.

Example 4.6. Let us consider the fractional boundary value problem with fractional deviating argument: ^𝑐𝐷₀^5/2+𝑢(𝑡)+𝑎(𝑡)𝑓𝑢𝐼^𝜎₀⁺𝜃(𝑡)𝑢=0,0<𝑡<1,_′(0)=𝑢^′′(0)=0,𝑢(0)=2𝑢(1), (4.19) where 5𝑞=21,𝜎=2,𝛼=2,𝜃(𝑡)=𝑡²(𝑡)𝑓(𝑢)=𝑢≤𝑡,𝑎=1+𝑡,₂5,0≤𝑢≤2,3015𝑢−5985,2≤𝑢≤3,61,𝑢≥3. (4.20)
It is easy to see that 𝐼₀^1/2₊𝜃(𝑡)=(16/15𝜋)𝑡^5/2≤1 and that hypotheses (H1)–(H3) are satisfied. Let us check the assumptions of Theorem 4.2 for 𝜏=4/5: 2𝜇>Γ(5/2)¹₀()41+𝑠𝑑𝑠=√𝜋𝛼()=2.2568,𝐿<1−𝜏^𝑞−1(𝛼−1)Γ𝑞^𝜏₀𝑎(𝑠)𝑑𝑠=0.15072. (4.21)
If we choose 𝜇=2.5, 𝐿=0.05, 𝑎=2, 𝑏=3, 𝑑≥305/2, 𝜉=50/3, 𝑐=9/50, then the assumptions of Theorem 4.2 are satisfied; consequently, there exist at least three positive solutions 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝜑,𝑑) such that ‖𝑢_𝑖‖≤𝑑=153, 3<min_{𝑡∈[0,4/5]}𝑢₁(𝑡), 2<‖𝑢₂‖, with min_{𝑡∈[0,4/5]}𝑢₂(𝑡)<3 and ‖𝑢₃‖<2.

Acknowledgment

The authors would like to express their thanks to the referee for his helpful comments and suggestions.