Existence of Positive Solutions for a Functional Fractional Boundary Value Problem

Abstract

We study the existence of positive solutions for a boundary value problem of fractional-order functional differential equations. Several new existence results are obtained.

1. Introduction

Fractional differential equations can describe many phenomena in various fields of science and engineering such as physics, mechanics, chemistry, control, and engineering. Due to their considerable importance and application, significant progress has been made in there are a great number of excellent works about ordinary and partial differential equations involving fractional derivatives; see, for instance, [1–15].

As pointed out in [16], boundary value problems associated with functional differential equations have arisen from problems of physics and variational problems of control theory appeared early in the literature; see [17, 18]. Since then many authors (see, e.g., [19–23]) investigated the existence of solutions for boundary value problems concerning functional differential equations. Recently an increasing interest in studying the existence of solutions for boundary value problems of fractional-order functional differential equations is observed; see for example, [24–26].

For 𝜏>0, we denote by 𝐶_𝜏 the Banach space of all continuous functions 𝜓∶[−𝜏,0]→𝑅 endowed with the sup-norm‖𝜓‖_[−𝜏,0]∶=sup{|𝜓(𝑠)|∶𝑠∈[−𝜏,0]}. (1.1)

If 𝑢∶[−𝜏,1]→𝑅, then for any 𝑡∈[0,1], we denote by 𝑢_𝑡 the element of 𝐶_𝜏 defined by𝑢_𝑡(𝜃)=𝑢(𝑡+𝜃),for𝜃∈[−𝜏,0]. (1.2)

In this paper we investigate a fractional-order functional differential equation of the form𝐷^𝜌𝑢(𝑡)=𝑓(𝑡,𝑢_𝑡),𝑡∈[0,1], (1.3) where 𝜌∈(𝑚−1,𝑚] (𝑚≥3 is a natural number), 𝑓(𝑡,𝑢_𝑡)∶[0,1]×𝐶_𝜏→𝑅 is a continuous function, associated with the boundary condition𝑢^′(0)=⋯=𝑢^(𝑚−2)(0)=0,𝑢^(𝑚−2)(1)=0, (1.4) and the initial condition𝑢₀=𝜙, (1.5) where 𝜙 is an element of the space𝐶⁺_𝜏(0)∶={𝜓∈𝐶_𝜏∶𝜓(𝑠)≥0,𝑠∈[−𝜏,0],𝜓(0)=0}. (1.6)

To the best of the authors knowledge, no one has studied the existence of positive solutions for problem (1.3)–(1.5). The aim of this paper is to fill the gap in the relevant literatures. The key tool in finding our main results is the following well-known fixed point theorem due to Krasnoselski [27].

Theorem 1.1. Let 𝐵 be a Banach space and let 𝐾 be a cone in 𝐵. Assume that Ω₁ and Ω₂ are open subsets of 𝐵, with 0∈Ω₁⊂Ω₁⊂Ω₂, and let 𝐴∶𝐾∩(Ω₂⧵Ω₁)→𝐾 be a completely continuous operator such that either ‖𝐴𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₁,‖𝐴𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₂, (1.7) or ‖𝐴𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₁,‖𝐴𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₂, (1.8) Then 𝐴 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

2. Preliminaries

Firstly, we recall some definitions of fractional calculus, which can be found in [11–14].

Definition 2.1. The Riemann-Liouville fractional derivative of order 𝛼>0 of a continuous function 𝑓∶(0,∞)→𝑅 is given by 𝐷^𝛼1𝑓(𝑡)=(𝑑Γ(𝑛−𝛼))𝑑𝑡^𝑛𝑡₀𝑓(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠, (2.1) where 𝑛=[𝛼]+1 and [𝛼] denotes the integer part of number 𝛼, provided that the right side is pointwise defined on (0,∞).

Definition 2.2. The Riemann-Liouville fractional integral of order 𝛼 of a function 𝑓∶(0,∞)→𝑅 is defined as 𝐼^𝛼1𝑓(𝑡)=Γ(𝛼)^𝑡₀𝑓(s)(𝑡−𝑠)^1−𝛼𝑑𝑠, (2.2) provided that the integral exists.

Thefollowing lemma is crucial in finding an integral representation of the boundary value problem (1.3)–(1.5).

Lemma 2.3 (see [4]). Suppose that 𝑢∈𝐶(0,1)∩𝐿(0,1) with a fractional derivative of order 𝛼>0. Then 𝐼^𝛼𝐷^𝛼𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛, (2.3) for some 𝑐_𝑖∈ℝ, 𝑖=1,2,…,𝑛, where 𝑛=[𝛼]+1.

From Lemma 2.3, we now give an integral representation of the solution of the linearized problem.

Lemma 2.4. If ℎ∈𝐶[0,1], then the boundary value problem 𝐷^𝜌𝑢(𝑡)+ℎ(𝑡)=0,0<𝑡<1,𝑚−1<𝜌≤𝑚,𝑢(0)=𝑢^′(0)=⋯=𝑢^(𝑚−2)(0)=0,𝑢^(𝑚−2)(1)=0 (2.4)(2.5) has a unique solution 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠, (2.6) where 1𝐺(𝑡,𝑠)=𝑡Γ(𝜌)^𝜌−1(1−𝑠)^{𝜌−𝑚+1}−(𝑡−𝑠)^𝜌−1𝑡,𝑠<𝑡,_𝜌−1(1−𝑠)^{𝜌−𝑚+1},𝑡≤𝑠. (2.7)

Proof. We may apply Lemma 2.3 to reduce BVP (2.4), (2.5) to an equivalent integral equation 𝑢(𝑡)=𝑐₁𝑡^𝜌−1+𝑐₂𝑡^𝜌−2+⋯+𝑐_𝑚𝑡^𝜌−𝑚−^𝑡₀(𝑡−𝑠)^𝜌−1Γ(𝜌)ℎ(𝑠)𝑑𝑠. (2.8) By the boundary condition (2.5), we easily obtain that 𝑐₂=𝑐₃=⋯=𝑐_𝑚𝑐=0,₁=1Γ(𝜌)¹₀(1−𝑠)^{𝜌−𝑚+1}ℎ(𝑠)𝑑𝑠. (2.9) Hence, the unique solution of BVP (2.4), (2.5) is 𝑢(𝑡)=¹₀1𝑡Γ(𝜌)^𝜌−1(1−𝑠)^{𝜌−𝑚+1}ℎ(𝑠)𝑑𝑠−^𝑡₀(𝑡−𝑠)^𝜌−1Γ(𝜌)ℎ(𝑠)𝑑𝑠=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠. (2.10) The proof is complete.

Lemma 2.5. 𝐺(𝑡,𝑠) has the following properties.(i)0≤𝐺(𝑡,𝑠)≤𝐵(𝑠), 𝑡,𝑠∈[0,1], where 𝐵(𝑠)=(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1Γ(𝜌); (2.11)(ii)𝐺(𝑡,𝑠)≥(𝑡^𝜌−1/(𝑚−2))𝐵(𝑠), for 0≤𝑡, 𝑠≤1.

Proof. It is easy to check that (i) holds. Next, we prove that (ii) holds. If 𝑡>𝑠, then 𝐺(𝑡,𝑠)=𝑡𝐵(𝑠)^𝜌−1(1−𝑠)^{𝜌−𝑚+1}−(𝑡−𝑠)^𝜌−1(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1=𝑡^𝑚−2(𝑡−𝑡𝑠)^{𝜌−𝑚+1}−(𝑡−𝑠)^𝜌−1(𝑡−𝑡𝑠)^{𝜌−𝑚+1}(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1≥𝑡^𝑚−2(𝑡−𝑡𝑠)^{𝜌−𝑚+1}−(𝑡−𝑠)^𝑚−2(𝑡−𝑡𝑠)^{𝜌−𝑚+1}(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1=𝑡^{𝜌−𝑚+1}(1−𝑠)^{𝜌−𝑚+1}[𝑡^𝑚−2−(𝑡−𝑠)^𝑚−2](1−𝑠)^{𝜌−𝑚+1}[1−(1−𝑠)^𝑚−2]=𝑡^{𝜌−𝑚+1}[𝑡^𝑚−3+𝑡^𝑚−4(𝑡−𝑠)+⋯+(𝑡−𝑠)^𝑚−3]1+(1−𝑠)+⋯+(1−𝑠)^𝑚−3≥𝑡^{𝜌−𝑚+1}𝑡^𝑚−31+(1−𝑠)+⋯+(1−𝑠)^𝑚−3≥𝑡^{𝜌−𝑚+1}𝑡^𝑚−3=𝑡𝑚−2^𝜌−2≥𝑡𝑚−2^𝜌−1.𝑚−2 (2.12) If 𝑡≤𝑠, then 𝐺(𝑡,𝑠)=𝑡𝐵(𝑠)^𝜌−1(1−𝑠)^{𝜌−𝑚+1}(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1≥𝑡^𝜌−1(1−𝑠)^{𝜌−𝑚+1}(1−𝑠)^{𝜌−𝑚+1}=𝑡^𝜌−1≥𝑡^𝜌−1.𝑚−2 (2.13) The proof is complete.

3. Main Result

In the sequel we will denote by 𝐶₀[0,1] the space of all continuous functions 𝑥∶[0,1]→𝑅 with 𝑥(0)=0. This is a Banach space when it is furnished with the usual sup-norm‖𝑢‖₀=max_𝑡∈[0,1]|𝑢(𝑡)|. (3.1) For each 𝜙∈𝐶⁺_𝜏(0) and 𝑥∈𝐶₀[0,1], we define𝑥_𝑡(𝑠;𝜙)∶=𝑥(𝑡+𝑠),𝑡+𝑠≥0,𝜙(𝑡+𝑠),𝑡+𝑠≤0,𝑠∈[−𝜏,0], (3.2) and observe that 𝑥_𝑡(;𝜙)∈𝐶_𝜏.

By a solution of the boundary value problem (1.3)–(1.5), we mean a function 𝑢∈𝐶₀[0,1] such that 𝐷^𝜌𝑢 exists on [0,1] and 𝑢 satisfies boundary condition (1.4), and for a certain 𝜙, the relation𝐷^𝜌𝑢(𝑡)=𝑓(𝑡,𝑢_𝑡(⋅;𝜙)) (3.3) holds for all 𝑡∈[0,1].

By Lemma 2.4 we know that a function 𝑢 is a solution of the boundary value problem (1.3)–(1.5) if and only if it satisfies𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢_𝑠(⋅;𝜙))𝑑𝑠∶=(𝑇_𝜙𝑢)(𝑡),𝑠∈[0,1]. (3.4)

We set𝐶⁺₀[0,1]={𝑢∈𝐶₀[0,1]∶𝑢(𝑡)≥0,𝑡∈[0,1]}. (3.5)

Define the cone 𝑃⊂𝐶₀[0,1] by𝑃={𝑦∈𝐶⁺₀[0,1]∶min_{𝜏≤𝑡≤1}1𝑦(𝑡)≥𝜏𝑚−2^𝜌−1‖𝑦‖₀}, (3.6) where 0<𝜏<1.

By Lemma 2.4, the boundary value problem (1.3)–(1.5) is equivalent to the integral equation𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢_𝑠)𝑑𝑠,𝑡∈[0,1]. (3.7) In this paper, we assume that 0<𝜏<1, 𝜙∈𝐶⁺_𝜏(0), and we make use of the following assumptions.(H₁)𝑓(𝑡,𝜓)≥0 for 𝑡∈[0,1] and 𝜓∈𝐶⁺_𝜏(0).(H₂)There exist constants 𝑀>‖𝜙‖_[−𝜏,0] and 𝜏_𝑖, 𝑖=1,2,…,𝑞, with 0<𝜏₁<𝜏₂<⋯<𝜏_𝑞<𝜏, as well as continuous functions 𝑎_𝑘∈𝐶[0,1] and nondecreasing continuous functions 𝑔_𝑘∶𝑅⁺→𝑅⁺ such that 𝑓(𝑡,𝜓)≤^𝑞_𝑘=1𝑎_𝑘(𝑡)𝑔_𝑘(‖𝜓‖_{[−𝜏𝑘,0]}),𝜓∈𝐶⁺_𝜏(0),‖𝜓‖_[−𝜏,0]≤𝑀. (3.8)(H₃)There exist functions 𝜔∶[0,1]→[0,𝜏], continuous 𝑐∶[0,1]→𝑅⁺, and nondecreasing 𝜂∶𝑅⁺→𝑅⁺ such that 𝑓(𝑡,𝜓)≥𝑐(𝑡)𝜂(𝜓(−𝜔(𝑡))),𝜓∈𝐶⁺_𝜏(0). (3.9)

Similar to the proof of Lemma 3.2 in [4], we have the following

Lemma 3.1. Let (𝐻₁) holds. Then 𝑇_𝜙∶𝑃→𝑃 is completely continuous.

Lemma 3.2. If 0<𝜏<1 and 𝑢∈𝑃, then we have ‖𝑢_𝑡(⋅;𝜙)‖_[−𝜏,0]≥𝛾‖𝑢‖₀,𝑡∈[𝜏,1], (3.10) where 𝜏𝛾=^𝜌−1𝑚−2. (3.11)

Proof. From the definition of 𝑢_𝑡(𝑠;𝜙), for 𝑡≥𝜏, we have 𝑢_𝑡(𝑠;𝜙)=𝑢(𝑡+𝑠),𝑠∈[−𝜏,0]. (3.12) Thus, we get for 𝑢∈𝑃 that ‖𝑢_𝑡(⋅;𝜙)‖_[−𝜏,0]=max_{𝑠∈[−𝜏,0]}𝑢(t+𝑠)≥𝑢(𝑡)≥𝛾‖𝑢‖₀,𝑡≥𝜏. (3.13) We are now in a position to present and prove our main result.

Theorem 3.3. Let (H₁), (H₂), and (H₃) hold. If 𝑀 (as in (H₂)), satisfying ^𝑞_𝑘=1𝑔_𝑘(𝑀)¹₀[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1]𝑎_𝑘(𝑠)𝑑𝑠≤Γ(𝜌)𝑀, (3.14) and there exists a constant 𝑚>0 (𝑚≠𝑀) satisfying 𝛾𝜂(𝛾𝑚)Γ(𝜌)¹_𝜏[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1]𝑐(𝑠)𝑑𝑠≥𝑚, (3.15) then (1.3)–(1.5) has a positive solution.

Proof. If 𝑢∈𝑃 with ‖𝑢‖₀=𝑀, then from (3.8), (3.14), and Lemma 2.4 (i), we get for any 𝑡∈[0,1] that (𝑇_𝜙𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢_𝑠(⋅;𝜙))𝑑𝑠≤¹₀𝐵(𝑠)𝑓(𝑠,𝑢_𝑠≤(⋅;𝜙))𝑑𝑠_𝑞𝑘=110𝐵(𝑠)𝑎_𝑘(𝑠)𝑔_𝑘(‖𝑢_𝑠(⋅;𝜙)‖_{[−𝜏𝑘,0]}≤)𝑑𝑠_𝑞𝑘=1𝑔_𝑘(𝑀)¹₀𝐵(𝑠)𝑎_𝑘=1(𝑠)𝑑𝑠Γ(𝜌)^𝑞_𝑘=1𝑔_𝑘(𝑀)¹₀[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1]𝑎_𝑘(𝑠)𝑑𝑠≤𝑀. (3.16) Now if we set Ω₁={𝑢∈𝐶[0,1]∶‖𝑢‖₀<𝑀}, (3.17) then (3.16) shows that ‖𝑇_𝜙𝑢‖₀≤‖𝑢‖₀ for 𝑢∈𝑃∩𝜕Ω₁.
Without loss of generality, we suppose that 𝑚<𝑀. For 𝑢∈𝑃 with ‖𝑢‖₀=𝑚, we have from Lemma 3.2 and (3.9) that(𝑇_𝜙𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢_𝑠(⋅;𝜙))𝑑𝑠≥¹_𝜏𝛾𝐵(𝑠)𝑓(𝑠,𝑢_𝑠≥(⋅;𝜙))𝑑𝑠_1𝜏𝛾𝐵(𝑠)𝑐(𝑠)𝜂(𝑢_𝑠(−𝜔(𝑠);𝜙))𝑑𝑠=¹_𝜏𝛾𝐵(𝑠)𝑐(𝑠)𝜂(𝑢(𝑠−𝜔(𝑠)))𝑑𝑠≥𝜂(𝛾‖𝑢‖₀)¹_𝜏𝛾𝛾𝐵(𝑠)𝑐(𝑠)𝑑𝑠=𝜔(𝛾𝑚)Γ(𝜌)¹_𝜏[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1]𝑐(𝑠)𝑑𝑠≥𝑚. (3.18) Now if we set Ω₂={𝑢∈𝐶[0,1]∶‖𝑢‖₀<𝑚}, (3.19) then (3.18) shows that ‖𝑇_𝜙𝑢‖₀≥‖𝑢‖₀ for 𝑢∈𝑃∩𝜕Ω₂.
Hence by the first part of Theorem 1.1, 𝑇_𝜙 has a fixed point 𝑢∈𝑃∩(Ω₁⧵Ω₂), and accordingly, 𝑢 is a solution of (1.3)–(1.5).

Having in mind the proof of Theorem 3.3, one can easily conclude the following results.

Theorem 3.4. Let (H₁), (H₂), (H₃), and (3.14) hold. If the function 𝜂 satisfies the condition limsup_𝑥→+∞𝜂(𝑥)𝑥>Γ(𝜌)𝛾²∫¹_𝜏[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1]𝑐(𝑠)𝑑𝑠, (3.20) where 𝛾 is as in (3.11). Then (1.3)–(1.5) has a positive solution.

Theorem 3.5. Let (H₁), (H₂), (H₃), and (3.14) hold. If the function 𝜂 satisfies the condition limsup_𝑥→0+𝜂(𝑥)𝑥>Γ(𝜌)𝛾²∫¹_𝜏[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1]𝑐(𝑠)𝑑𝑠. (3.21) Then (1.3)–(1.5) has a positive solution.

Theorem 3.6. Let (H₁), (H₃) and (3.15) hold. Assume that(H₄)𝑓_∞=0, uniformly for 𝑡∈[0,1], where 𝑓_∞∶=lim_{𝜓∈𝐶+𝜏,‖𝜓‖[−𝜏,0]→∞}𝑓(𝑡,𝜓)‖𝜓‖_[−𝜏,0]. (3.22) Then (1.3)–(1.5) has a positive solution.

Proof. Since (H₃) and (3.15) hold, we have from the proof of Theorem 3.3 that ‖𝑇_𝜙𝑢‖₀≥‖𝑢‖₀for𝑢∈𝑃∩𝜕Ω₂. (3.23) From (𝐻₄), we will consider two cases in the following.Case 1 (𝑓 is bounded). In this case, there exists a positive constant 𝐿>0 such that |𝑓(𝑡,𝜓)|≤𝐿,𝑡∈[0,1],𝜓∈𝐶⁺_𝜏. (3.24) We choose a positive constant 𝑀₁≥𝐿(𝑚−2)Γ(𝜌+1)(𝜌−𝑚+2). (3.25) For 𝑢∈𝑃 with ‖𝑢‖₀=𝑀₁, we have (𝑇_𝜙𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢_𝑠(⋅;𝜙))𝑑𝑠≤¹₀𝐵(𝑠)𝑓(𝑠,𝑢_𝑠≤(⋅;𝜙))𝑑𝑠₁₀𝐿𝐿𝐵(𝑠)𝑑𝑠=Γ(𝜌)¹₀[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1=𝐿]𝑑𝑠(1Γ(𝜌)−1𝜌−𝑚+2𝜌)≤𝑀₁. (3.26)Case 2 (𝑓 is unbounded). In this case, there exists a positive constant 𝑀₂≥‖𝜙‖_[−𝜏,0] such that 𝑓(𝑡,𝜓)≤𝜖‖𝜓‖_[−𝜏,0],𝑡∈[0,1],𝜓∈𝐶⁺_𝜏with‖𝜓‖_[−𝜏,0]≥𝑀₂, (3.27) where 𝜖>0 is a constant satisfying 𝜖≤Γ(𝜌+1)(𝜌−𝑚+2)𝑚−2. (3.28) By the definition of 𝑢_𝑠(⋅;𝜙), we easily obtain that ‖𝑢_𝑠(⋅;𝜙)‖_[−𝜏,0]≤max{‖𝑢‖₀,‖𝜙‖_[−𝜏,0]}. (3.29)
If 𝑢∈𝑃 with ‖𝑢‖₀=𝑀₂, then from (3.8), (3.14), and Lemma 2.4 (i), we get for any 𝑡∈[0,1] that(𝑇_𝜙𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢_𝑠(⋅;𝜙))𝑑𝑠≤¹₀𝐵(𝑠)𝑓(𝑠,𝑢_𝑠(⋅;𝜙))𝑑𝑠=𝜖𝑀₂¹₀𝐵(𝑠)𝑑𝑠=𝜖𝑀₂Γ(𝜌)¹₀[(1−𝑠)^{𝜌−𝑚+1}−(1−𝑠)^𝜌−1=]𝑑𝑠𝜖𝑀₂(𝑚−2)Γ(𝜌+1)(𝜌−𝑚+2)≤𝑀₂. (3.30) Set 𝑀₃=max{𝑀₁,𝑀₂,2𝑚}. Then in either case we may put Ω₃={𝑢∈𝐶[0,1]∶‖𝑢‖₀<𝑀₃}, (3.31) and for 𝑢∈𝑃∩𝜕Ω₃, ‖𝑇_𝜙𝑢‖₀≥‖𝑢‖₀. By the first part of Theorem 1.1, 𝑇_𝜙 has a fixed point 𝑢∈𝑃∩(Ω₃⧵Ω₂), and accordingly, 𝑢 is a solution of (1.3)–(1.5).

4. An Example

To illustrate our results, we present the following example.

Example 4.1. Consider the boundary value problem of fractional-order functional differential equations 𝐷^2.6𝑢(𝑡)=√11−𝑡|𝑢(𝑡−3)|²1+2exp(|𝑢(𝑡−2)|^1/2𝑢),𝑡∈[0,1],_′(0)=𝑢^′1(1)=0,𝑢(𝑡)=𝜙(𝑡),𝑡∈[−2,0], (4.1) where 𝜙∈𝐶⁺_1/2(0) with ‖𝜙‖_[−1/2,0]≤4.
Observe that𝑓(𝑡,𝜓)=√1−𝑡|𝜓(−𝛼)|²+2exp(|𝜓(−𝛽)|^1/2)≤𝑎₁(𝑡)𝑔₁(‖𝜓‖_[−𝜏1,0])+𝑎₂(𝑡)𝑔₂(‖𝜓‖_[−𝜏2,0]), (4.2) where we have set 𝑎₁(𝑡)∶=√1−𝑡,𝑎₂(𝑡)∶=2,𝜏₁=𝛼,𝜏₂=𝛽,𝑔₁(𝑥)∶=𝑥²,𝑔₂(𝑥)∶=exp(√𝑥). (4.3) Also we obtain 𝑓(𝑡,𝜓)≥𝑐(𝑡)𝜂(𝜓(−𝜔(𝑡))), (4.4) where 𝑐(𝑡)∶=2,𝜔(𝑡)=𝛽,𝜂(𝑥)=exp(√𝑥),𝑥≥0. (4.5) Thus, (H₁) and (H₂) hold. Note that here 𝜌=2.6, 𝑚=3, we have ²_𝑘=1𝑔_𝑘(4)¹₀[(1−𝑠)^0.6−(1−𝑠)^1.6]𝑎_𝑘(𝑠)𝑑𝑠=5.2615<5.7184=4Γ(2.6), (4.6) and ‖𝜙‖_[−1/2,0]≤4, which implies that condition (3.14) holds with 𝑀=4. Finally, we observe that lim_𝑥→+∞(𝜂(𝑥)/𝑥)=+∞, and therefore condition (3.20) is satisfied. Then all assumptions of Theorem 3.4 hold. Thus, with Theorem 3.4, problem (1.3)–(1.5) has at least one positive solution.

Acknowledgments

The author would like to thank the referees for their valuable suggestions and comments. The project is supported by the National Natural Science Foundation of China (10771212) and the Natural Science Foundation of Jiangsu Education Office (06KJB110010).