Positive Solutions to Nonlinear Higher-Order Nonlocal Boundary Value Problems for Fractional Differential Equations

Abstract

We study existence of positive solutions to nonlinear higher-order nonlocal boundary value problems corresponding to fractional differential equation of the type ^𝑐𝒟^𝛿₀₊𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0, 𝑡∈(0,1), 0<𝑡<1. 𝑢(1)=𝛽𝑢(𝜂)+𝜆₂, 𝑢^′(0)=𝛼𝑢^′(𝜂)−𝜆₁, 𝑢^′′(0)=0, 𝑢^′′′(0)=0⋯𝑢^(𝑛−1)(0)=0, where, 𝑛−1<𝛿<𝑛, 𝑛(≥3)∈ℕ, 0<𝜂,𝛼,𝛽<1, the boundary parameters 𝜆₁,𝜆₂∈ℝ⁺ and ^𝑐𝐷^𝛿₀₊ is the Caputo fractional derivative. We use the classical tools from functional analysis to obtain sufficient conditions for the existence and uniqueness of positive solutions to the boundary value problems. We also obtain conditions for the nonexistence of positive solutions to the problem. We include examples to show the applicability of our results.

1. Introduction

Fractional calculus goes back to the beginning of the theory of differential calculus and is developing since the 17th century through the pioneering work of Leibniz, Euler, Abel, Liouville, Riemann, Letnikov, Weyl, and many others. Fractional calculus is the generalization of ordinary integration and differentiation to an arbitrary order. For almost 300 years, it was seen as interesting but abstract mathematical concept. Nevertheless the applications of fractional calculus just emerged in the last few decades in various areas of physics, chemistry, engineering, biosciences, electrochemistry, and diffusion processes. For details, we refer the readers to [1–5].

The existence and uniqueness of solutions for fractional differential equations is well studied in [6–10] and references therein. It should be noted that most of the papers and books on fractional calculus are devoted to the solvability of initial value problems for fractional differential equations. In contrast, the theory of boundary value problems for nonlinear fractional differential equations has received attention quiet recently, and many aspects of the theory need to be further investigated.

There are some recent development dealing with the existence and multiplicity of positive solutions to nonlinear boundary value problems for fractional differential equations, see, for example, [11–18] and the reference therein. However, few results can be found in the literature concerning the existence of positive solutions to nonlinear three-point boundary value problems for fractional differential equations. For example, Li and coauthors [19] obtained sufficient conditions for the existence and multiplicity results to the following three point fractional boundary value problem 𝒟^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)+𝑓𝑡,𝑢))=0,0<𝑡<1,1<𝛼≤2,=0,𝒟^𝛽₀₊𝑢(1)=𝑎𝒟^𝛽₀₊𝑢(𝜉), (1.1) where 𝒟^𝛼₀₊ is standard Riemann-Liouville fractional order derivative.

Bai [20] studied the existence and uniqueness of positive solutions to the following three-point boundary value problem for fractional differential equations 𝒟^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)𝜂(1),+𝑓𝑡,𝑢))=0,0<𝑡<1,1<𝛼≤2,=0,𝛽𝑢=𝑢 (1.2) where 0<𝛽𝜂^𝛼−1<1, 0<𝜂<1, 𝒟^𝛼₀₊ is standard Riemann-Liouville fractional order derivative. The function 𝑓 is assumed to be continuous on [0,1]×[0,∞).

The purpose of the present work is to investigate sufficient conditions for the existence, uniqueness, and nonexistence of positive solutions to more general boundary value problems for higher-order nonlinear fractional differential equations ^𝑐𝒟^𝛿₀₊𝑢(𝑡)((𝑡()𝑢(1)𝜂+𝑓𝑡,𝑢))=0,𝑡∈0,1,𝑛−1<𝛿<𝑛,=𝛽𝑢+𝜆₂,𝑢^′(0)=𝛼𝑢^′𝜂−𝜆₁,𝑢^′′(0)=0,𝑢^′′′(0)=0⋯𝑢^(𝑛−1)(0)=0, (1.3) where, 𝑛−1<𝛿<𝑛, 𝑛=3,4,5,…; 0<𝜂,𝛼,𝛽<1, the boundary parameters 𝜆₁,𝜆₂∈ℝ⁺ and ^𝑐𝒟^𝛿₀₊, is the Caputo fractional derivative. The function 𝑓 is assumed to be continuous and nonnegative on [0,1]×[0,∞). To the best of our knowledge, existence and uniqueness of positive solution to the boundary value problem (1.3) have never been studied previously.

This paper is organized as follows: in Section 2, we recall some basic definitions and preliminary results which are needed for our main results. In Section 3, we study existence and uniqueness and nonexistence of positive solutions to the boundary value problem (1.3) under certain assumptions on the function 𝑓. Moreover, examples are provided to illustrate the applicability of main results.

2. Background Materials and Lemmas

For the convenience of the readers, in this section, we provide definitions of Riemann-Liouville fractional integral and fractional derivative and some of their basic properties which will be helpful in the forth coming investigations.

Definition 2.1 (see [2, 5]). For a function 𝜙∶(𝑎,∞)→ℝ, the Riemann-Liouville fractional integral of order 𝛼>0 is defined by ℐ^𝛼_𝑎+𝜙(𝑡)=1Γ(𝛼)^𝑡_𝑎(𝑡−𝑠)^𝛼−1𝜙(𝑠)𝑑𝑠, (2.1)
The provided that the integral on the right hand side exists. For 𝛼,𝛽≥0, the fractional integral satisfies the semigroup property ℐ^𝛼₀₊ℐ^𝛽₀₊𝜙(𝑡)=ℐ^𝛼+𝛽₀₊𝜙(𝑡)=ℐ^𝛽₀₊ℐ^𝛼₀₊𝜙(𝑡)almosteverywhereon[0,1]. (2.2) In addition, if 𝜙∈𝐶[0,1] or if 𝛼+𝛽≥1, then the identity is true for every 𝑡∈[0,1].

Definition 2.2 (see [2, 5]). The standard Riemann-Liouville fractional derivative of order 𝛼>0 of a function 𝜙∶(𝑎,∞)→ℝ is given by 𝒟^𝛼_𝑎+𝜙(𝑡)=(𝑑/𝑑𝑡)^𝑛ℐ^𝑛−𝛼_𝑎+𝜙(𝑡), where 𝑎∈ℝ, 𝑛=[𝑠]+1, provided that the right hand side is pointwise defined on (𝑎,∞).

Definition 2.3. For a given function 𝜙∶(𝑎,∞)→ℝ, the Caputo fractional derivative of order 𝛼>0 is defined by ^𝑐𝒟^𝛼_𝑎+𝜙(𝑡)=ℐ^𝑛−𝛼_𝑎+𝜙^(𝑛)(𝑡), where 𝑎∈ℝ, 𝑛=[𝑠]+1.

Lemma 2.4 (see [2]). If 𝛼>𝛽>0, then ^𝑐𝒟^𝛽₀₊ℐ^𝛼₀₊𝜙(𝑡)=ℐ^𝛼−𝛽₀₊𝜙(𝑡). In particular, if 𝑚 is positive integer and 𝛿>𝑚, then (𝑑^𝑚/𝑑𝑡^𝑚)(ℐ^𝛿₀₊𝜙(𝑡))=ℐ^𝛿−𝑚₀₊𝜙(𝑡).

The following two lemmas play a fundamental role to obtain an equivalent integral representation to the boundary value problem (1.3).

Lemma 2.5 (see [2]). Let 𝛼>0, then ℐ^𝛼𝑐₀₊𝒟^𝛼₀₊𝜙(𝑡)=𝜙(𝑡)−^𝑛−1_𝑘=0𝜙^𝑘(0)𝑡𝑘!^𝑘=𝜙(𝑡)+𝑐₁+𝑐₂𝑡+𝑐₃𝑡²+⋯+𝑐_𝑛𝑡^𝑛−1, (2.3) where 𝑐_𝑖=𝜙^𝑘−1(0)/(𝑘−1)!, 𝑖=1,2,…,𝑛 and 𝑛−1<𝛼≤𝑛.

Lemma 2.6 (see [2]). For 𝛼>0, the fractional differential equation ^𝑐𝒟^𝛼₀₊𝜙(𝑡)=0 has a general solution 𝜙(𝑡)=𝑐₁+𝑐₂𝑡+𝑐₃𝑡²+⋯+𝑐_𝑛𝑡^𝑛−1, where 𝑐_𝑖∈ℝ, 𝑖=1,2,…,𝑛 and 𝑛−1<𝛼≤𝑛.

For the existence of positive solutions to the boundary value problem (1.3), we use the following fixed point theorem due to Krasnosel’skii.

Theorem 2.7 (see [21]). Let ℰ be a Banach space, and let 𝒫⊂ℰ be a cone. Assume that Ω₁, Ω₂ are open subsets of ℰ such that 0∈Ω₁⊂Ω₁⊂Ω₂. Let 𝒜∶𝒫∩(Ω₂⧵Ω₁)→ℰ be completely continuous such that either (i)‖𝒜𝑢‖≤‖𝑢‖, for 𝑢∈𝒫∩𝜕Ω₁ and ‖𝒜𝑢‖≥‖𝑢‖, for 𝑢∈𝒫∩𝜕Ω₂, or (ii)‖𝒜𝑢‖≥‖𝑢‖, for 𝑢∈𝒫∩𝜕Ω₁ and ‖𝒜𝑢‖≤‖𝑢‖, for 𝑢∈𝒫∩𝜕Ω₂. Then, 𝒜 has a fixed point in 𝒫∩(Ω₂⧵Ω₁).

3. Main Results

Lemma 3.1. Let ℎ∈𝐶[0,1], then the unique solution of the linear problem ^𝑐𝒟^𝛿₀₊𝑢(𝑡)+ℎ(𝑡)=0,𝑡∈(0,1),𝑛−1<𝛿<𝑛,𝑛(≥3)∈ℕ,𝑢(1)=𝛽𝑢𝜂+𝜆₂,𝑢^′(0)=𝛼𝑢^′𝜂−𝜆₁,𝑢^′′(0)=0,𝑢^′′′(0)=0⋯𝑢^(𝑛−1)(0)=0 (3.1)(3.2) is given by 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠+¹₀𝐻𝑡;𝜂,𝑠ℎ(𝑠)𝑑𝑠+𝜓(𝑡), (3.3) where, 𝐺()=()𝑡,𝑠1−𝑠^𝛿−1−(𝑡−𝑠)^𝛿−1Γ(𝛿)(),𝑠≤𝑡,1−𝑠^𝛿−1Γ(𝛿)𝐻=𝛽),𝑡≤𝑠,𝑡;𝜂,𝑠(1−𝑠^𝛿−1−𝜂−𝑠^𝛿−1Γ(𝛿)+𝛼𝑡1−𝛽1−𝛽𝜂−1−𝛽𝜂−𝑠^𝛿−2()Γ()𝛽()1−𝛼1−𝛽𝛿−1,𝑠≤𝜂,1−𝑠^𝛿1−𝛽Γ(𝛿−1),𝜂≤𝑠, (3.4) and 𝜓(𝑡)=((1−𝛽𝜂−(1−𝛽)𝑡)/(1−𝛼)(1−𝛽))𝜆₁+𝜆₂/(1−𝛽).

Proof. In view of Lemma 2.5, (3.1) is equivalent to the integral equation 𝑢(𝑡)=−𝐼^𝛿₀₊ℎ(𝑡)+𝑐₁+𝑐₂𝑡+𝑐₃𝑡²+⋯+𝑐_𝑛𝑡^𝑛−1. (3.5) Using Lemma 2.4, we obtain 𝑢^′(𝑡)=−𝐼^𝛿−1₀₊ℎ(𝑡)+𝑐₂+2𝑐₃𝑡+⋯+(𝑛−1)𝑐_𝑛𝑡^𝑛−2,𝑢^′′(𝑡)=−𝐼^𝛿−2₀₊ℎ(𝑡)+2𝑐₃+⋯+(𝑛−1)(𝑛−2)𝑐_𝑛𝑡^𝑛−3,⋮𝑢^(𝑛−1)(𝑡)=−𝐼^{𝛿−(𝑛−1)}₀₊ℎ(𝑡)+(𝑛−1)!𝑐_𝑛, (3.6) where 𝑐₁,𝑐₂,…,𝑐_𝑛∈ℝ. The boundary conditions, 𝑢^′′(0)=0, 𝑢^′′′(0)=0,…,𝑢^(𝑛−1)(0)=0, lead to 𝑐₃=0, 𝑐₄=0,…,𝑐_𝑛=0. Using the boundary conditions 𝑢^′(0)=𝛼𝑢^′(𝜂)−𝜆₁, 𝑢(1)=𝛽𝑢(𝜂)+𝜆₂ and the fact that 𝐼^𝛿−1₀₊ℎ(0)=0, we obtain (1−𝛼)𝑐₂=−𝛼𝐼^𝛿−1₀₊ℎ𝜂−𝜆₁,1−𝛽𝑐₁+1−𝛽𝜂𝑐₂=𝐼^𝛿₀₊ℎ(1)−𝛽𝐼^𝛿₀₊ℎ𝜂+𝜆₂, (3.7) which implies that 𝑐₂=−(𝛼/(1−𝛼))𝐼^𝛿−1₀₊ℎ(𝜂)−(𝜆₁/(1−𝛼)) and 𝑐₁=1𝐼1−𝛽^𝛿0+ℎ(1)−𝛽𝐼1−𝛽^𝛿0+ℎ𝜂+𝛼1−𝛽𝜂()𝐼1−𝛼1−𝛽^𝛿−10+ℎ𝜂+1−𝛽𝜂𝜆₁()+𝜆1−𝛼1−𝛽².1−𝛽 (3.8) Hence, the unique solution of the linear fractional boundary value problem (3.1), (3.2) is given by 𝑢(𝑡)=−𝐼^𝛿₀₊ℎ(𝑡)+1𝐼1−𝛽^𝛿0+ℎ(1)−𝛽𝐼1−𝛽^𝛿0+ℎ𝜂+𝛼1−𝛽𝜂−𝛼1−𝛽𝑡(𝐼1−𝛼)1−𝛽^𝛿−10+ℎ𝜂+1−𝛽𝜂(−𝑡1−𝛼)1−𝛽𝜆1−𝛼¹+𝜆₂1−𝛽=−𝐼^𝛿₀₊ℎ(𝑡)+𝐼^𝛿₀₊ℎ(1)+𝛽𝐼1−𝛽^𝛿0+ℎ(1)−𝛽𝐼1−𝛽^𝛿0+ℎ𝜂+𝛼1−𝛽𝜂−𝛼1−𝛽𝑡(𝐼1−𝛼)1−𝛽^𝛿−10+ℎ𝜂+1−𝛽𝜂(−𝑡1−𝛼)1−𝛽𝜆1−𝛼¹+𝜆₂=1−𝛽_𝑡0(1−𝑠)^𝛿−1−(𝑡−𝑠)^𝛿−1Γ(𝛿)ℎ(𝑠)𝑑𝑠+¹_𝑡(1−𝑠)^𝛿−1Γ(𝛿)ℎ(𝑠)+𝑑𝑠_𝜂0𝛽(1−𝑠)^𝛿−1−𝛽𝜂−𝑠^𝛿−1Γ(𝛿)+𝛼𝑡1−𝛽1−𝛽𝜂−𝛼1−𝛽𝜂−𝑠^𝛿−2()Γ()ℎ(𝑠)+1−𝛼1−𝛽𝛿−1𝑑𝑠_1𝜂𝛽(1−𝑠)^𝛿−1Γ(𝛿)ℎ(𝑠)1−𝛽𝑑𝑠+1−𝛽𝜂()−𝑡1−𝛼1−𝛽𝜆1−𝛼¹+𝜆₂=1−𝛽₁₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠+¹₀𝐻𝑡;𝜂,𝑠ℎ(𝑠)𝑑𝑠+𝜓(𝑡). (3.9)

Lemma 3.2. The functions 𝐺(𝑡,𝑠) and 𝐻(𝑡;𝜂,𝑠) satisfy the following properties: (i)𝐺(𝑡,𝑠)≥0, 𝐻(𝑡;𝜂,𝑠)≥0 and 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) for all 0≤𝑠, 𝑡≤1,(ii)For 𝑠∈(0,1), 0<𝜉<𝜏<1, min_{𝜉≤𝑡≤𝜏}𝐺(𝑡,𝑠)≥(1−𝜏^𝛿−1)max_0≤𝑡≤1𝐺(𝑡,𝑠)=(1−𝜏^𝛿−1)𝐺(𝑠,𝑠), (iii)𝛽(1−𝜂^𝛿−1)(1−𝑠)^𝛿−1≤(1−𝛽)Γ(𝛿)𝐻(𝑡;𝜂,𝑠)<2(𝛿−1)(1−𝑠)^𝛿−2, (iv)for 𝑠∈(0,1),0<𝜉<𝜏<1, min_{𝜉≤𝑡≤𝜏}𝐻(𝑡;𝜂,𝑠)≥(1−𝜏)max_0≤𝑡≤1𝐻(𝑡;𝜂,𝑠).

Proof. (i) For 𝛿>1, in view of the expression for 𝐺(𝑡,𝑠), it follows that 𝐺(𝑡,𝑠)≥0 and 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) for all 0≤𝑠, 𝑡≤1.
For 𝑠≤𝜂 and 0<𝛽<1, we have 1−𝛽𝜂−(1−𝛽)𝑡>(1−𝛽𝜂)(1−𝑡). Hence, 𝐻(𝑡,𝜂,𝑠)>0. Also, we note that 𝐻(𝑡,𝜂,𝑠)>0 for 𝑠≥𝜂.
(ii) Now, max_[]𝑡∈0,1𝐺()=()𝑡,𝑠1−𝑠^𝛿−1Γ(𝛿). (3.10) Since 𝐺()=1𝑡,𝑠Γ(𝛿)(1−𝑠)^𝛿−1−(𝑡−𝑠)^𝛿−1=max_[]𝑡∈0,1𝐺()−()𝑡,𝑠𝑡−𝑠^𝛿−1Γ(𝛿)≥max_[]𝑡∈0,1𝐺(𝑡,𝑠)−𝜏^𝛿−1(1−𝑠/𝜏)^𝛿−1Γ(𝛿),𝑡∈𝜉,𝜏, (3.11) which implies that min_{𝑡∈𝜉,𝜏}𝐺(𝑡,𝑠)≥1−𝜏^𝛿−1max_[]𝑡∈0,1𝐺(𝑡,𝑠),for𝑠∈(0,1). (3.12) Now, for 𝑠≤𝜂, we have 𝐻=𝛽)𝑡;𝜂,𝑠(1−𝑠^𝛿−1−𝜂−𝑠^𝛿−1Γ(𝛿)+𝛼𝑡1−𝛽1−𝛽𝜂−1−𝛽𝜂−𝑠^𝛿−2()Γ()≤𝛽()1−𝛼1−𝛽𝛿−11−𝑠^𝛿−1Γ(𝛿)+𝛼1−𝛽𝜂−𝑠^𝛿−2(Γ(𝛿)<2()1−𝛼)1−𝛽𝛿−1)(1−𝑠^𝛿−2(Γ(𝛿).1−𝛼)1−𝛽 (3.13) For 𝑠≥𝜂, obviously, 𝐻(𝜂,𝑠)<2(𝛿−1)(1−𝑠)^𝛿−2/(1−𝛼)(1−𝛽)Γ(𝛿). From the expression of 𝐻(𝜂,𝑠), it clearly follows that 𝐻≥𝛽𝑡;𝜂,𝑠1−𝜂^𝛿−1(1−𝑠)^𝛿−1Γ(𝛿).1−𝛽 (3.14)
(iii) From the definition of 𝐻(𝑡;𝜂,𝑠), we have 𝜕𝐻=−𝜕𝑡𝑡;𝜂,𝑠𝜂−𝑠^𝛿−2(1−𝛼)Γ(𝛿−1)≤0. (3.15) Therefore, 𝐻(𝑡;𝜂,𝑠) is nonincreasing in 𝑡, so its minimum value occurs at 𝑡=𝜏 for 𝑡∈[𝜉,𝜏], and its maximum value occurs at 𝑡=0 for 𝑡∈[0,1]. That is, min_{𝜉≤𝑡≤𝜏}𝐻=𝛽)𝑡;𝜂,𝑠(1−𝑠^𝛿−1−𝜂−𝑠^𝛿−1Γ(𝛿)+𝛼𝜏1−𝛽1−𝛽𝜂−1−𝛽𝜂−𝑠^𝛿−2Γ(𝛿),1−𝛽max_0≤𝑡≤1𝐻=𝛽)𝑡;𝜂,𝑠(1−𝑠^𝛿−1−𝜂−𝑠^𝛿−1Γ(𝛿)+𝛼1−𝛽1−𝛽𝜂𝜂−𝑠^𝛿−2Γ(𝛿).1−𝛽 (3.16)(3.17) Since (1−𝑠)^𝛿−1−(𝜂−𝑠)^𝛿−1≥0 and 1−𝜏<1, therefore (1−𝑠)^𝛿−1−𝜂−𝑠^𝛿−1≥(1−𝜏)(1−𝑠)^𝛿−1−𝜂−𝑠^𝛿−1. (3.18) Also, as 1−𝛽≤1−𝛽𝜂, therefore 1−𝛽𝜂−1−𝛽𝜏≥1−𝛽𝜂−1−𝛽𝜂𝜏=(1−𝜏)1−𝛽𝜂. (3.19) Subletting (3.18) and (3.19) in (3.16), we have min_{𝜉≤𝑡≤𝜏}𝐻≥()𝛽)𝑡;𝜂,𝑠1−𝜏(1−𝑠^𝛿−1−𝜂−𝑠^𝛿−1Γ(𝛿)+𝛼1−𝛽1−𝛽𝜂𝜂−𝑠^𝛿−2Γ(𝛿)=()1−𝛽1−𝜏max_0≤𝑡≤1𝐻𝑡;𝜂,𝑠. (3.20)

Remark 3.3. For 𝜆₁,𝜆₂>0, min_{𝜉≤𝑡≤𝜏}𝜓(𝑡)≥(1−𝜏)max_0≤𝑡≤1𝜓(𝑡).

Proof. As (𝑑/𝑑𝑡)𝜓(𝑡)=−𝜆₁/(1−𝛼)<0, therefore 𝜓(𝑡) is a decreasing function. Hence, max_0≤𝑡≤1𝜓(𝑡)(0)=𝜆=𝜓1−𝛽𝜂₁()+𝜆1−𝛼1−𝛽²,1−𝛽min_{𝜉≤𝑡≤𝜏}𝜓(𝑡)(𝜏)=𝜏𝜆=𝜓1−𝛽𝜂−1−𝛽₁(+𝜆1−𝛼)1−𝛽²≥1−𝛽1−𝜏1−𝛽1−𝛽𝜂𝜆1−𝛼¹+𝜆₂=(1−𝜏)max_0≤𝑡≤1𝜓(𝑡). (3.21) Thus, we have min_{𝜉≤𝑡≤𝜏}𝜓(𝑡)≥(1−𝜏)max_0≤𝑡≤1𝜓(𝑡).
Let ℬ=𝐶[0,1] be endowed with the norm ‖𝑢‖=max_0≤𝑡≤1|𝑢(𝑡)|. Define a cone 𝒫⊂ℬ by 𝒫=𝑢∈ℬ∶min_{𝜉≤𝑡≤𝜏}𝑢(𝑡)≥(1−𝜏)‖𝑢‖, (3.22) and an operator 𝒜∶ℬ→ℬ by 𝒜𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+¹₀𝐻𝑡;𝜂,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝜓(𝑡). (3.23) By Lemma 3.1, the boundary value problem (1.3) has a solution if and only if 𝒜 has a fixed point.

Lemma 3.4. The operator 𝒜∶𝒫→𝒫 is completely continuous.

Proof. Firstly, we show that 𝒜(𝑃)⊂𝑃. From (3.23), Lemma 3.2, and Remark 3.3, we have min_{𝜉≤𝑡≤𝜏}(𝒜𝑢(𝑡))≥1−𝜏^𝛿−11₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+(1−𝜏)¹₀max_[]𝑡∈0,1𝐻𝑓((𝑠+()𝑡;𝜂,𝑠𝑠,𝑢))𝑑𝑠1−𝜏max_[]𝑡∈0,1𝜓(𝑡)≥(1−𝜏)¹₀𝐺(𝑠,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+¹₀max_[]𝑡∈0,1𝐻𝑡;𝜂,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠+max_[]𝑡∈0,1𝜓(𝑡)≥(1−𝜏)‖𝒜𝑢‖. (3.24) Hence, we have 𝒜(𝑃)⊂𝑃.
Next, we show that 𝒜 is uniformly bounded. For fixed ℓ>0, consider a bounded subset ℳ of 𝒫 defined by ℳ=𝑢∈𝒫∶‖𝑢‖≤ℓ,ℓ∈ℝ⁺, (3.25) and define 𝒦=max_{0≤𝑢≤ℓ}𝑓(𝑡,𝑢(𝑡))+1, then for 𝑢∈ℳ, we have |𝒜𝑢(𝑡)|≤¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+¹₀𝐻𝑡;𝜂,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠+max_𝑡∈[0,1]𝜓(𝑡)≤𝒦Γ(𝛿)¹₀(1−𝑠)^𝛿−1()𝑑𝑠+2𝒦𝛿−1(1−𝛼)1−𝛽Γ(𝛿)¹₀(1−𝑠)^𝛿−2𝜆𝑑𝑠+1−𝛽𝜂₁+(1−𝛼)𝜆₂(≤𝒦1−𝛼)1−𝛽()ℐ1−𝛼1−𝛽^𝛿0+(1)+2ℐ^𝛿₀₊(1)+1−𝛽𝜂𝜆₁+(1−𝛼)𝜆₂()=()𝒦1−𝛼1−𝛽3𝛿−1𝛿()()+𝜆1−𝛼1−𝛽𝛿−11−𝛽𝜂₁+(1−𝛼)𝜆₂().1−𝛼1−𝛽 (3.26) Hence, 𝒜(ℳ) is bounded.
Finally, we show that 𝒜 is equicontinuous. Define 𝜎=𝛿(1−𝛼)Γ(𝛿)𝜀/𝒦[𝛿(1−𝛼)+𝛼𝜂^𝛼−1+𝜆₁] and choose 𝑡>𝜏 such that 𝑡−𝜏<𝜎. Then, for all 𝜀>0 and 𝑢∈ℳ, we have |𝒜𝑢(𝑡)−𝒜𝑢(𝜏)|=¹₀(𝐺(𝑡,𝑠)−𝐺(𝜏,𝑠))𝑓(𝑠,𝑢(𝑠))𝑑𝑠+¹₀𝐻𝑓((𝑠−𝜆𝑡;𝜂,𝑠−𝐻𝑡;𝜂,𝑠𝑠,𝑢))𝑑𝑠₁()1−𝛼𝑡−𝜏≤𝒦¹₀𝐺(𝑡,𝑠)−𝐺(𝜏,𝑠)𝑑𝑠+¹₀𝐻𝜆𝑡;𝜂,𝑠−𝐻𝑡;𝜂,𝑠𝑑𝑠+¹()11−𝛼𝑡−𝜏=𝒦Γ(𝛿)^𝑡₀(𝑡−𝑠)^𝛿−1−(𝜏−𝑠)^𝛿−1𝛼()𝑑𝑠+𝑡−𝜏(1−𝛼)Γ(𝛿−1)×^𝜂₀𝜂−𝑠^𝛿−2𝜆𝑑𝑠+¹()=𝒦1−𝛼𝑡−𝜏(𝛿)𝑡𝛿Γ^𝛿−𝜏^𝛿+𝛼𝜂^𝛿−1+𝜆₁(.1−𝛼𝑡−𝜏) (3.27) Using the mean value theorem, we obtain 𝑡^𝛿−𝜏^𝛿≤𝛿(𝑡−𝜏)<𝛿𝜎. Hence, it follows that |(𝑡)(𝜏<𝛿()𝒜𝑢−𝒜𝑢)|𝒦𝜎1−𝛼+𝛼𝜂^𝛼−1+𝜆₁𝛿(1−𝛼)Γ(𝛿)<𝜀. (3.28) By means of Arzela-Ascoli theorem 𝒜∶𝒫→𝒫 is completely continuous operator.

For convenience, we introduce following notations: 𝒦¹_{𝛼,𝛽,𝛿}=3𝐼^𝛿₀₊𝜓₁(1)+6𝐼^𝛿−1₀₊𝜓₁(1)(1−𝛼)1−𝛽⁻¹,𝒦²_{𝜉,𝛽,𝜏,𝛿}=1−𝜏^𝛿−1ℐ1−𝛽^𝛿𝜉+𝜓₂(𝜏)⁻¹,𝑐_𝛼=13(1−𝛼),𝑐_𝛽=131−𝛽,𝛾=1−𝜏^𝛿−1. (3.29)

Theorem 3.5. If there exist constants 𝜌₁,𝜌₂∈ℝ⁺ such that 𝜌₂>𝜌₁ and functions 𝜓₁,𝜓₂∈𝐿[0,1] such that (i)𝑓(𝑡,𝑢)≥𝜌₂𝒦²_{𝜉,𝛽,𝜏,𝛿}𝜓₂(𝑡), for(𝑡,𝑢)∈[0,1]×[𝛾𝜌₂,𝜌₂],(ii)𝑓(𝑡,𝑢)≤𝜌₁𝒦¹_{𝛼,𝛽,𝜂}𝜓₁(𝑡), for(𝑡,𝑢)∈[0,1]×[0,𝜌₁], then the boundary value problem (1.3) has at least one positive solution for 𝜆₁, 𝜆₂ small enough and has no positive solution for 𝜆₁, 𝜆₂ large enough.

Proof. By Lemma 3.4, the operator 𝒜 is completely continuous. The proof is divided into two steps.Step 1. We prove that the boundary value problem (1.3) has at least one positive solution.
Define Ω_𝜌1={𝑢∈ℬ∶‖𝑢‖<𝜌₁} an open subset of ℬ and choose 𝜆₁, 𝜆₂ such that 𝜆₁≤𝑐_𝛼𝜌₁ and 𝜆₂≤𝑐_𝛽𝜌₁. Then, for any 𝑢∈𝒫∩𝜕Ω_𝜌1, we have ‖𝑢‖=𝜌₁ and in view of Lemma 3.2 and (3.23), it follows that |𝒜𝑢(𝑡)|=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+¹₀𝐻𝑓((𝑠𝑡𝑡;𝜂,𝑠𝑠,𝑢))𝑑𝑠+1−𝛽𝜂−1−𝛽(𝜆1−𝛼)1−𝛽¹+𝜆₂1−𝛽≤𝐼^𝛿₀₊𝑓((1+21,𝑢))()𝐼1−𝛼1−𝛽^𝛿−10+𝑓((1+𝜆1,𝑢))1−𝛽𝜂₁+(1−𝛼)𝜆₂(1−𝛼)1−𝛽≤𝒦¹_{𝛼,𝛽,𝛿}𝜌₁𝐼^𝛿₀₊𝜓₁(1)+2𝐼^𝛿−1₀₊𝜓₁(1)()+𝜆1−𝛼1−𝛽¹+𝜆1−𝛼²≤𝜌1−𝛽₁3+𝜌₁3+𝜌₁3=‖𝑢‖, (3.30) which implies that ‖𝒜𝑢‖≤‖𝑢‖, for 𝑢∈𝒫∩𝜕Ω_𝜌1.
Define Ω_𝜌2={𝑢∈𝒫∶‖𝑢‖<𝜌₂}. For any 𝑡∈[𝜉,𝜏] and 𝑢∈𝒫∩𝜕Ω_𝜌2, using Lemma 3.2, we have min_{𝜉≤𝑡<𝜏}𝑢(𝑡)≥1−𝜏^𝛿−1‖𝑢‖. (3.31) Therefore, by (3.3) and Remark 3.3, we have the following estimate: |𝐴𝑢(𝑡)|=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+¹₀𝐻𝑓((𝑠𝑡𝑡;𝜂,𝑠𝑠,𝑢))𝑑𝑠+1−𝛽𝜂−1−𝛽(𝜆1−𝛼)1−𝛽¹+𝜆₂≥1−𝛽1−𝜏^𝛿−1+𝛽1−𝜂^𝛿−1𝐼1−𝛽^𝛿𝜉+𝑓((𝜏+𝑡𝜏,𝑢))1−𝛽𝜂−1−𝛽()𝜆1−𝛼1−𝛽¹+𝜆₂1−𝛽≥𝜌₂𝒦²_{𝜉,𝛽,𝜏,𝛿}1−𝜏^𝛿−1+𝛽1−𝜏^𝛿−1𝐼1−𝛽^𝛿𝜉+𝜓₂(𝜏)=𝜌₂𝒦²_{𝜉,𝛽,𝜏,𝛿}1−𝜏^𝛿−1𝐼1−𝛽^𝛿𝜉+𝜓₂(𝜏)=𝜌₂=‖𝑢‖, (3.32) which implies that ‖𝒜𝑢‖≥‖𝑢‖ for 𝑢∈𝒫∩𝜕Ω_𝜌2
Hence, by Theorem 2.7, it follows that 𝒜 has a fixed point 𝑢 in 𝒫∩(Ω_𝜌2⧵Ω_𝜌1).Step 2. Now, we prove that for large values of 𝜆₁, 𝜆₂, the boundary value problem (1.3) has no positive solution. Otherwise, for 𝑖=1,2, there exists 0<𝜆_𝑖1<𝜆_𝑖2⋯<𝜆_𝑖𝑛<⋯, with lim_𝑛→∞𝜆_𝑖𝑛=+∞, such that for any positive integer 𝑛, the boundary value problem ^𝑐𝒟^𝛿₀₊𝑢(𝑡)((𝑡()𝑢(1)𝜂+𝑓𝑡,𝑢))=0,𝑡∈0,1,0<𝑡<1,=𝛽𝑢+𝜆_2𝑛,𝑢^′(0)=𝛼𝑢^′𝜂−𝜆_1𝑛,𝑢^′′(0)=0,𝑢^′′′(0)=0⋯𝑢^(𝑛−1)(0)=0 (3.33) has a positive solution given by 𝑢_𝑛(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝑢_𝑛(𝑠)𝑑𝑠+¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝑢_𝑛(𝑠)𝑡𝑑𝑠+1−𝛽𝜂−1−𝛽(𝜆1−𝛼)1−𝛽^1𝑛+𝜆_2𝑛.1−𝛽 (3.34) If 𝑡>𝑛, then 𝑢_𝑛(𝑡)≥(1−𝑡)𝜆_1𝑛/(1−𝛼)(1−𝛽)+𝜆_2𝑛/(1−𝛽) and if 𝑡≤𝑛, then 𝑢_𝑛(𝑡)≥(1−𝜂)𝜆_1𝑛/(1−𝛼)(1−𝛽)+𝜆_2𝑛/(1−𝛽). Therefore, 𝑢_𝑛→∞ as 𝑛→∞.
Let 𝜌₁=min((1/2)𝜌₂,𝜆₁/𝑐_𝛼,𝜆₂/𝑐_𝛽) and define Ω𝜌₁={𝑢∈𝒫∶𝜌₁<‖𝑢_𝑛‖<𝜌₁}. For 𝑢∈Ω𝜌₁, using (3.3) and Remark 3.3, we have 𝑢_𝑛(𝑡)≥¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝑢_𝑛(𝑠)𝑑𝑠+¹₀𝐻𝑡;𝜂,𝑠𝑓𝑠,𝑢_𝑛(𝑠)+𝑡𝑑𝑠1−𝛽𝜂−1−𝛽()𝜆1−𝛼1−𝛽^1𝑛+𝜆_2𝑛≥11−𝛽21−𝜏^𝛿−1+𝛽1−𝜂^𝛿−1ℐ1−𝛽^𝛿𝜉+𝑓1,𝑢_𝑛(1≥𝜌₂)𝒦²_{𝛼,𝛽,𝜂}1−𝛽1−𝛽1−𝜏^𝛿−1+𝛽1−𝜂^𝛿−1ℐ^𝛿_𝜉+𝜓₂(1)>2𝜌₁>2𝑢_𝑛, (3.35) which is a contradiction. Hence, the boundary value problem (1.3) have no positive solution for 𝜆₁, 𝜆₂ large enough.

Theorem 3.6. Assume that (𝐴) is satisfied and there exists real-valued function 𝜙(𝑡)∈𝐿[0,1] such that 𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)≤𝜙(𝑡)|𝑢−𝑣|,for[][),ℐ𝑡∈0,1,𝑢,𝑣∈0,∞_𝛿0+𝜙(1)+2(ℐ1−𝛼)1−𝛽^𝛿−10+𝜙(1)<1, (3.36) then the boundary value problem (1.3) has a unique positive solution.

Proof. For 𝑢,𝑣∈𝒫, using (3.23) and Lemma 3.2, we obtain |𝒜𝑢(𝑡)−𝒜𝑣(𝑡)|≤¹₀𝐺(𝑓((𝑠((𝑠+𝑡,𝑠)𝑠,𝑢))−𝑓𝑠,𝑣))𝑑𝑠₁₀𝐻𝑓((𝑠((𝑠<‖𝑡;𝜂,𝑠𝑠,𝑢))−𝑓𝑠,𝑣))𝑑𝑠𝑢−𝑣‖¹₀(1−𝑠)^𝛿−1Γ(𝛿)𝜙(𝑠)𝑑𝑠+¹₀2(𝛿−1)(1−𝑠)^𝛿−2()Γ(𝛿)𝜙(𝑠)=‖ℐ1−𝛼1−𝛽𝑑𝑠𝑢−𝑣‖^𝛿₀₊𝜙(1)+2()ℐ1−𝛼1−𝛽^𝛿−10+𝜙(1)<‖𝑢−𝑣‖, (3.37) where we have taken into account (1−𝑠)^𝛿−1≤(1−𝑠)^𝛿−2, we obtain the relation ℐ^𝛿₀₊𝜙(1)≤(1/(𝛿−1))ℐ^𝛿−1₀₊𝜙(1). Hence, it follows by Banach contraction principle that the boundary value problem (1.3) has a unique positive solution.

Example 3.7. Consider the boundary value problem ^𝑐𝒟^𝛿₀₊𝑢(𝑡)=𝑡²𝑒^(3/2)𝑡+181+𝑡²+𝑡²𝑢²(1+sin𝑡)(),𝑢(1)𝜂16𝜋,𝑡∈0,1=𝛽𝑢+𝜆₂,𝑢^′(0)=𝛼𝑢^′𝜂−𝜆₁,𝑢^′′(0)=0,𝑢^′′′(0)=0, (3.38) where, 𝛿=7/2, 𝜂=1/2, 𝛼=2/3, and 𝛽=1/3. Let 𝑓(𝑡,𝑢)=(𝑡²𝑒^(3/2)𝑡+1)/8(1+𝑡²)+𝑡²𝑢²(1+sin𝑡)/16𝜋, (𝑡,𝑢)∈[0,1]×[0,∞]. For 𝜌₁=1,𝜌₂=550, we observe that 𝑓()≤1𝑡,𝑢81𝑒+𝜋𝑡²,for()∈[]×[],𝑓()≥1𝑡,𝑢0,10,1𝑡,𝑢√𝑒𝑡²1+𝑡²,for(𝑡,𝑢)∈[0,1]×[283,550]. (3.39) Taking 𝜓₁(𝑡)=𝑡² and 𝜓₂(𝑡)=𝑡²/(1+𝑡²). For 𝜉=1/4, 𝜏=3/4, by computations, we find that 𝑐_𝛼=1/9, 𝑐_𝛽=2/9, 𝐼^𝛿_𝜉+𝜓₂(𝜏)=(1/60√𝜋)(49√7tan⁻¹(2/7)−tan⁻¹(√2)−1132√2/21)≈0.001291, also ℐ^𝛿₀₊𝜓₁(1)=128/10395√𝜋. Therefore, 𝒦¹_{𝛼,𝛽,𝛿}≈0.583566 and 𝒦²_{𝜉,𝛽,𝜏,𝛿}≈0.000993874. Hence, 𝑓(𝑡,𝑢)≤𝜌₁𝒦¹_{𝛼,𝛽,𝛿}𝜓₁(𝑡)≈0.583566𝑡²,for()∈[]×[],𝑓()𝑡,𝑢0,10,1𝑡,𝑢≥𝜌₂𝒦²_{𝜉,𝛽,𝜏,𝛿}𝜓₂(𝑡)𝑡≈0.546631²1+𝑡²,for(𝑡,𝑢)∈[0,1]×[283,550]. (3.40) Assumptions (i) and (ii) of the Theorem 3.5 are satisfied. Therefore, the boundary value problem (3.38) has a positive solution for 𝜆₁∈[0,1/9], 𝜆₂∈[0,2/9] and no positive solution for 𝜆₁>1/9, 𝜆₂>2/9.

Example 3.8. Consider the boundary value problem ^𝑐𝒟^𝛿₀₊𝑢(𝑡)=𝑒^−33𝑡11√𝜋𝑒^33𝑡cos(25𝑡)+500sin²𝑡𝑢²14√𝜋+425𝑒^(1/8)𝑡(),1+𝑢𝑢(1)=𝛽𝑢𝜂+𝜆₂,𝑢^′(0)=𝛼𝑢^′𝜂−𝜆₁,𝑢^′′(0)=0,𝑢^′′′(0)=0, (3.41)(3.42) where 𝛿=7/2, 𝛼=2/3, 𝛽=3/5, 𝜂=1/2, and 𝜆₁,𝜆₂∈ℝ. Let 𝑓(𝑡,𝑢)=𝑒^−33𝑡(11√𝜋𝑒^33𝑡cos(25𝑡)+500sin²𝑡)𝑢²/(14√𝜋+425𝑒^(1/8)𝑡)(1+𝑢), (𝑡,𝑢)∈[0,1]×[0,∞]. For 𝑢,𝑣∈𝒫, 𝑡∈[0,1], we have the following estimate: 𝑓()(≤𝑒𝑡,𝑢−𝑓𝑡,𝑣)^−33𝑡11√𝜋𝑒^33𝑡+50014√𝜋+425𝑒^(1/8)𝑡𝑢²+𝑣1+𝑢²≤𝑒1+𝑣_−33𝑡11√𝜋𝑒^33𝑡+50014√𝜋+425𝑒^(1/8)𝑡|𝑢−𝑣|(𝑢+𝑣+𝑢𝑣)()≤𝑒1+𝑢)(1+𝑣_−33𝑡11√𝜋𝑒^33𝑡+50014√𝜋+425𝑒^(1/8)𝑡(|𝑢−𝑣|). (3.43) Let 𝜙(𝑡)=𝑒^−33𝑡(11√𝜋𝑒^33𝑡+500)/(14√𝜋+425𝑒^(1/8)𝑡), then by computations, we have ℐ^𝛿−1₀₊𝜙(1)≈0.013469, (2𝛿−𝛼−1)ℐ^𝛿−1₀₊𝜙(1)/(𝛿−1)(1−𝛼)(1−𝛽)≈0.215504<1. All the conditions of Theorem 3.6 are satisfied. Therefore, by Theorem 3.6, the boundary value problem (3.41), (3.41) has a unique positive solution.