The Existence of Positive Solutions for Fractional Differential Equations with Sign Changing Nonlinearities

Abstract

We investigate the existence of at least two positive solutions to eigenvalue problems of fractional differential equations with sign changing nonlinearities in more generalized boundary conditions. Our analysis relies on the Avery-Peterson fixed point theorem in a cone. Some examples are given for the illustration of main results.

1. Introduction

The theory of fractional differential equations has become an important aspect of differential equations (see [1–8]). Boundary value problems of fractional differential equations have been investigated in many papers (see [9–46]). The existence of positive solutions to boundary value problems of fractional differential equations has been studied by many authors when nonlinearities are positive (see [9–24]). There are a few papers to study the existence of positive solutions of semipositone fractional differential equations. For example, using the Krasnoselskii fixed point theorem, Yuan et al. [9] discussed the existence of positive solutions for the singular positone and semipositone two-point boundary value problems 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑓((𝑡,𝑢(0)=𝜇𝑎𝑡,𝑢))=𝑢^′(0)=𝑢(1)=𝑢^′(1)=0,3<𝛼≤4, (1.1) where 𝜇>0,𝑎 and 𝑓 are continuous. In [10], Wang et al. studied the existence of positive solutions for the singular semipositone two-point boundary value problems 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)+𝜆𝑓𝑡,𝑢))=0,=𝑢^′(0)=𝑢(1)=0,2<𝛼≤3, (1.2)(1.3) where 𝜆>0, 𝑓 is continuous.

In [11], using Krasnoselskii fixed point theorem, Goodrich discussed the existence of at least one positive solutions for the system of fractional boundary value problems −𝐷^𝜈1₀₊𝑦₁(𝑡)=𝜆₁𝑎₁(𝑡)𝑓𝑦₁(𝑡),𝑦₂(𝑡),−𝐷^𝜈2₀₊𝑦₂(𝑡)=𝜆₂𝑎₂(𝑡)𝑔𝑦₁(𝑡),𝑦₂(𝑡),𝑦₁^(𝑖)(0)=𝑦₂^(𝑖)(0)=0,0≤𝑖≤𝑛−2,𝐷^𝛼₀₊𝑦₁(𝑡)∣_𝑡=1=𝜙₁𝑦,𝐷^𝛼₀₊𝑦₂(𝑡)∣_𝑡=1=𝜙₂𝑦, (1.4) where 𝑎₁, 𝑎₂, 𝑓, and 𝑔 are nonnegative for 𝑡∈[0,1].

Motivated by the excellent results mentioned above, in this paper, we investigate the existence of at least two positive solutions for the problem 𝐷^𝛼₀₊𝑦(𝑡)(𝑡[],𝑦+𝜆𝑓𝑡,𝑦)=0,𝑡∈0,1_(𝑖)(0)𝐷=0,0≤𝑖≤𝑛−2,_𝛽0+𝑦(𝑡)∣_𝑡=1=ℎ𝑦, (1.5) where 𝜆>0,𝛼∈(𝑛−1,𝑛],𝑛≥3,1≤𝛽≤𝑛−2<𝛼−1, 𝑓∈𝐶([0,1]×ℝ⁺,[−𝑀,∞)),𝑀>0, ℎ∈𝐶(𝐶(ℝ⁺),ℝ⁺),ℝ⁺=[0,∞). The main tool is the Avery-Peterson theorem. To the best of our knowledge, this is the first paper dealing with eigenvalue problems of fractional differential equations with sign changing nonlinearities involving more general boundary conditions. Our results improve some of the earlier work presented in [10, 17, 46].

2. Preliminaries

For the convenience of the readers, we present here some necessary definitions and lemmas from fractional calculus theory. For more details see [1, 2].

Definition 2.1. The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑦∶(0,∞)→ℝ is given by 𝐼^𝛼₀₊𝑦(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠, (2.1) provided that the right-hand side is pointwise defined on (0,∞).

Definition 2.2. The Riemann-Liouville fractiontal derivative of order 𝛼>0 of a function 𝑦∶(0,∞)→ℝ is given by 𝐷^𝛼₀₊𝑦(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑦(𝑠)𝑑𝑠, (2.2) provided that the right-hand side is pointwise defined on (0,∞), where 𝑛=[𝛼]+1.

Lemma 2.3. Assume 𝑓∈𝐶[0,1], 𝑞≥𝑝≥0, then 𝐷^𝑝₀₊𝐼^𝑞₀₊𝑓(𝑡)=𝐼₀^𝑞−𝑝₊𝑓(𝑡). (2.3)

Lemma 2.4. Assume 𝛼>0, then(1)If 𝜆>−1, 𝜆≠𝛼−𝑖, 𝑖=1,2,…,[𝛼]+1, 𝑡>0, then 𝐷^𝛼₀₊𝑡^𝜆=Γ(𝜆+1)Γ()𝑡𝜆−𝛼+1^𝜆−𝛼. (2.4)(2)𝐷^𝛼₀₊𝑡^𝛼−𝑖=0, 𝑖=1,2,…,𝑛.(3)𝐷^𝛼₀₊𝐼^𝛼₀₊𝑢(𝑡)=𝑢(𝑡),  for a.e. 𝑡∈[0,1], where 𝑢∈𝐿¹[0,1].(4)𝐷^𝛼₀₊𝑢(𝑡)=0 if and only if 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛, (2.5)for some 𝑐_𝑖∈ℝ, 𝑖=1,2,…,𝑛, where 𝑛 is the least integer greater than or equal to 𝛼.

Lemma 2.5 (see[11]). Given 𝑔∈𝐶[0,1], 𝑦 is a solution of the problem 𝐷^𝛼₀₊𝑦(𝑡)(𝑡)[],𝑦+𝑔=0,𝑡∈0,1_(𝑖)(0)𝐷=0,0≤𝑖≤𝑛−2,_𝛽0+𝑦(𝑡)∣_𝑡=1=ℎ𝑦, (2.6) if and only if it satisfies 𝑦(𝑡)=Γ𝛼−𝛽Γ(𝛼)𝑡^𝛼−1ℎ𝑦+¹₀𝐺(𝑡,𝑠)𝑔(𝑠)𝑑𝑠, (2.7) where 𝐺()=𝑡𝑡,𝑠^𝛼−1(1−𝑠)^{𝛼−𝛽−1}−(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑡,0≤𝑠≤𝑡≤1,_𝛼−1(1−𝑠)^{𝛼−𝛽−1}Γ(𝛼),0≤𝑡≤𝑠≤1. (2.8)

Lemma 2.6 (see[11]). 𝐺(𝑡,𝑠) is continuous on [0,1]×[0,1] and 0≤𝐺(𝑡,𝑠)≤𝐺(1,𝑠),𝑡,𝑠∈[0,1]. (2.9)

Lemma 2.7. 𝐺(𝑡,𝑠)≥𝑡^𝛼−1𝐺(1,𝑠),𝑡,𝑠∈[0,1].

Proof. For 𝑠≤𝑡, 𝐺()=𝑡𝑡,𝑠^𝛼−1(1−𝑠)^{𝛼−𝛽−1}−(𝑡−𝑠)^𝛼−1Γ(𝛼)=𝑡^𝛼−1(1−𝑠)^{𝛼−𝛽−1}−(1−𝑠/𝑡)^𝛼−1Γ(𝛼)≥𝑡^𝛼−1(1−𝑠)^{𝛼−𝛽−1}−(1−𝑠)^𝛼−1Γ(𝛼)=𝑡^𝛼−1𝐺(1,𝑠). (2.10) For 𝑠>𝑡, 𝐺()=𝑡𝑡,𝑠^𝛼−1(1−𝑠)^{𝛼−𝛽−1}Γ(𝛼)≥𝑡^𝛼−1(1−𝑠)^{𝛼−𝛽−1}−(𝑡−𝑠)^𝛼−1Γ(𝛼)=𝑡^𝛼−1𝐺(1,𝑠). (2.11)

By simple calculation, we can get ¹₀𝐺()𝛽1,𝑠𝑑𝑠=Γ(),𝛼−𝛽𝛼+1¹_1/2𝐺()21,𝑠𝑑𝑠=^𝛽𝛼−𝛼+𝛽2^𝛼Γ().𝛼−𝛽𝛼+1 (2.12)

By Lemma 2.5, we can easily get the following lemma.

Lemma 2.8. The boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)𝑢+1=0,0<𝑡<1,_(𝑖)(0)=𝐷^𝛽₀₊𝑢(𝑡)∣_𝑡=1=0,0≤𝑖≤𝑛−2, (2.13) has a unique solution 𝑢(𝑡)=𝑡^𝛼−1Γ(𝛼)1−𝑡𝛼−𝛽𝛼. (2.14)
Obviously, 𝑢 satisfies 𝛽𝑡^𝛼−1Γ()(𝑡)≤𝑡𝛼−𝛽𝛼+1≤𝑢^𝛼−1Γ(𝛼)[].𝛼−𝛽,𝑡∈0,1 (2.15)

Lemma 2.9. 𝑦≥𝜆𝑀𝑢 is a solution of the following problem: 𝐷^𝛼₀₊𝑦+𝜆𝑓𝑡,𝑦−𝜆𝑀𝑢+𝑀=0,𝑡∈[0,1],𝑦^(𝑖)(0)𝐷=0,0≤𝑖≤𝑛−2,_𝛽0+𝑦(𝑡)∣_𝑡=1=ℎ𝑦−𝜆𝑀𝑢, (2.16) if and only if 𝑦=𝑦−𝜆𝑀𝑢 is a positive solution of (1.5).

Proof. In fact, if 𝑦 is a positive solution of the problem (1.5), by Lemma 2.8, we get that 𝑦 satisfies 𝐷^𝛼₀₊𝑦+𝜆𝑀𝑢+𝜆𝑓𝑡,𝑦+𝑀=0,𝑡∈[0,1],𝑦+𝜆𝑀𝑢^(𝑖)(0)𝐷=0,0≤𝑖≤𝑛−2,_𝛽0+𝑦+𝜆𝑀𝑢∣_𝑡=1=ℎ𝑦. (2.17) Take 𝑦=𝑦+𝜆𝑀𝑢. Then 𝑦 satisfies (2.16) and 𝑦≥𝜆𝑀𝑢.
On the other hand, if 𝑦 is a solution of (2.16) and 𝑦≥𝜆𝑀𝑢. Take 𝑦=𝑦−𝜆𝑀𝑢. By Lemma 2.8, we can easily get that 𝑦 satisfies (1.5). Clearly, 𝑦≥0.

Define functions ℎ, 𝑓 and an operator 𝑇∶𝐶[0,1]→𝐶[0,1] by ℎ𝑦,𝑓(𝑡)=Γ=ℎmax𝑦−𝜆𝑀𝑢,0𝑡,𝑦=𝑓𝑡,max𝑦−𝜆𝑀𝑢,0+𝑀.𝑇𝑦𝛼−𝛽Γ(𝛼)𝑡^𝛼−1ℎ𝑦+𝜆¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠. (2.18)

Obviously, 𝑦≥𝜆𝑀𝑢 is a fixed point of the operator 𝑇 if and only if 𝑦−𝜆𝑀𝑢 is a positive solution of the problem (1.5).

Take 𝑋=𝐶[0,1] with norm ‖𝑥‖=max_𝑡∈[0,1]|𝑥(𝑡)|. Define a cone 𝑃 by 𝑃=𝑦∈𝐶[0,1]∣𝑦(𝑡)≥𝑡^𝛼−1𝑦,𝑡∈[0,1]. (2.19)

Lemma 2.10. 𝑇∶𝑃→𝑃 is a completely continuous operator.

Proof.. Take 𝑦∈𝑃. By Lemmas 2.6 and 2.7, we get 𝑇𝑦(𝑡)≥𝑡^𝛼−1Γ𝛼−𝛽Γ(𝛼)ℎ𝑦+𝜆¹₀𝐺(1,𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠≥𝑡^𝛼−1𝑇𝑦. (2.20) So, 𝑇∶𝑃→𝑃. Let Ω⊂𝑃 be bounded. It follows from the continuity of ℎ, 𝑓 that there exist constants 𝑀₁ and 𝑀₂ such that ℎ(𝑦)≤𝑀₁ and 𝑓(𝑡,𝑦)≤𝑀₂ for 𝑡∈[0,1],𝑦∈Ω. Thus, ≤Γ𝑇𝑦𝛼−𝛽Γ(𝛼)𝑀₁+𝜆𝑀₂¹₀𝐺(1,𝑠)𝑑𝑠. (2.21) That is 𝑇(Ω) is bounded. For 𝑦∈Ω,𝑡₁,𝑡₂∈[0,1], 𝑇𝑦𝑡₁−𝑇𝑦𝑡₂≤𝑀₁Γ𝛼−𝛽Γ(𝛼)𝑡₁^𝛼−1−𝑡₂^𝛼−1+𝜆𝑀₂¹₀𝐺𝑡₁,𝑠−𝐺𝑡₂,𝑠𝑑𝑠. (2.22) By the uniform continuity of 𝑡^𝛼−1 and 𝐺(𝑡,𝑠), we get that 𝑇(Ω) is equicontinuous. Obviously, 𝑇∶𝑃→𝑃 is continuous. By the Arzela-Ascoli theorem, we get that 𝑇∶𝑃→𝑃 is completely continuous.

Definition 2.11. A map 𝜙 is said to be a nonnegative, continuous, and concave functional on a cone 𝑃 of a real Banach space 𝐸 if and only if 𝜙∶𝑃→ℝ⁺ is continuous and 𝜙𝑡𝑥+(1−𝑡)𝑦≥𝑡𝜙(𝑥)+(1−𝑡)𝜙𝑦, (2.23) for all 𝑥,𝑦∈𝑃 and 𝑡∈[0,1].

Definition 2.12. A map Φ is said to be a nonnegative, continuous, and convex functional on a cone 𝑃 of a real Banach space 𝐸 iff Φ∶𝑃→ℝ⁺ is continuous and Φ𝑡𝑥+(1−𝑡)𝑦≤𝑡Φ(𝑥)+(1−𝑡)Φ𝑦, (2.24) for all 𝑥,𝑦∈𝑃 and 𝑡∈[0,1].

Let 𝜑 and Θ be nonnegative, continuous, and convex functional on 𝑃, Φ a nonnegative, continuous, and concave functional on 𝑃, and Ψ a nonnegative continuous functional on 𝑃. Then, for positive numbers 𝑎, 𝑏, 𝑐, and 𝑑, we define the following sets: 𝑃=(𝑥),𝑃=(𝑥)(𝑥),𝑃=(𝑥)(𝑥)(𝑥),𝑅=(𝑥)(𝑥).𝜑,𝑑𝑥∈𝑃∶𝜑<𝑑𝜑,Φ,𝑏,𝑑𝑥∈𝑃∶𝑏≤Φ,𝜑≤𝑑𝜑,Θ,Φ,𝑏,𝑐,𝑑𝑥∈𝑃∶𝑏≤Φ,Θ≤𝑐,𝜑≤𝑑𝜑,Ψ,𝑎,𝑑𝑥∈𝑃∶𝑎≤Ψ,𝜑≤𝑑 (2.25)

We will use the following fixed point theorem of Avery and Peterson to study the problem (1.5).

Theorem 2.13 (see [47]). Let 𝑃 be a cone in a real Banach space 𝐸. Let 𝜑 and Θ be nonnegative, continuous, and convex functionals on 𝑃, Φ a nonnegative, continuous, and concave functional on 𝑃, and Ψ a nonnegative continuous functional on 𝑃 satisfying Ψ(𝑘𝑥)≤𝑘Ψ(𝑥) for 0≤𝑘≤1, such that for some positive numbers 𝑀 and 𝑑, Φ(𝑥)≤Ψ(𝑥),‖𝑥‖≤𝑀𝜑(𝑥) (2.26) for all 𝑥∈𝑃(𝜑,𝑑). Suppose that 𝑇∶𝑃𝜑,𝑑⟶𝑃𝜑,𝑑 (2.27) is completely continuous and there exist positive numbers 𝑎, 𝑏, 𝑐 with 𝑎<𝑏, such that the following conditions are satisfied:(S1){𝑥∈𝑃(𝜑,Θ,Φ,𝑏,𝑐,𝑑)∶Φ(𝑥)>𝑏}≠∅ and Φ(𝑇𝑥)>𝑏 for 𝑥∈𝑃(𝜑,Θ,Φ,𝑏,𝑐,𝑑);(S2)Φ(𝑇𝑥)>𝑏 for 𝑥∈𝑃(𝜑,Φ,𝑏,𝑑) with Θ(𝑇𝑥)>𝑐;(S3)0∉𝑅(𝜑,Ψ,𝑎,𝑑) and Ψ(𝑇𝑥)<𝑎 for 𝑥∈𝑅(𝜑,Ψ,𝑎,𝑑) with Ψ(𝑥)=𝑎.
Then 𝑇 has at least three fixed points 𝑥₁,𝑥₂,𝑥₃∈𝑃(𝜑,𝑑), such that 𝜑𝑥_𝑖𝑥≤𝑑,for𝑖=1,2,3,𝑏<Φ₁,𝑎<Ψ𝑥₂,Φ𝑥₂<𝑏,Ψ𝑥₃<𝑎. (2.28)

3. Main Results

We define a concave function Φ(𝑥)=min_{𝑡∈[1/2,1]}|𝑥(𝑡)| and convex functions Ψ(𝑥)=Θ(𝑥)=𝜑(𝑥)=‖𝑥‖.

Theorem 3.1. Assume that there exists a constant 0<𝑙<Γ(𝛼)/Γ(𝛼−𝛽), such that ℎ(𝑦)≤𝑙‖𝑦‖ for 𝑦∈𝑃. In addition, suppose that there exist constants 𝑘, 𝑎, 𝑏, 𝑐, 𝑑 with 𝑘>2^2𝛼−1𝛽Γ(𝛼)/(Γ(𝛼)−Γ(𝛼−𝛽)𝑙), [Γ(𝛼)−Γ(𝛼−𝛽)𝑙]𝛼𝑀/𝛽Γ(𝛼)<𝑎<𝑏−[Γ(𝛼)−Γ(𝛼−𝛽)𝑙]𝛼𝑀/𝛽Γ(𝛼)<2^𝛼−1𝑏<𝑐<𝑑, such that the following conditions hold:(C1)𝑓(𝑡,𝑦)≤𝑑−𝑀,  for (𝑡,𝑦)∈[0,1]×[0,𝑑];(C2)𝑓(𝑡,𝑦)≥𝑘𝑏−𝑀,  for (𝑡,𝑦)∈[1/2,1]×[𝑏−[Γ(𝛼)−Γ(𝛼−𝛽)𝑙]𝛼𝑀/𝛽Γ(𝛼),𝑐];(C3)𝑓(𝑡,𝑦)≤𝑎−𝑀,  for (𝑡,𝑦)∈[0,1]×[0,𝑎].
Then the problem (1.5) has at least two positive solutions for 2^2𝛼−1𝛼−𝛽Γ(𝛼+1)𝑘𝛼2^𝛽𝛼Γ(𝛼)𝑙−1+𝛽<𝜆<𝛼−𝛽−Γ𝛼−𝛽𝛽. (3.1)

Proof. Take 𝑦∈𝑃(𝜑,𝑑). By ‖max{𝑦−𝜆𝑀𝑢,0}‖≤𝑑, (C1), Lemma 2.6, (2.12), and (3.1), we have ≤Γ𝑇𝑦𝛼−𝛽Γ(𝛼)𝑙𝑑+𝜆𝑑¹₀𝐺()=Γ1,𝑠𝑑𝑠𝛼−𝛽Γ(𝛼)𝑙+𝜆¹₀𝐺(1,𝑠)𝑑𝑠𝑑≤𝑑. (3.2) This means that 𝑇∶𝑃(𝜑,𝑑)→𝑃(𝜑,𝑑).
It is easy to see that {𝑦∈𝑃(𝜑,Θ,Φ,𝑏,𝑐,𝑑)∶Φ(𝑦)>𝑏}≠∅. 𝑦∈𝑃(𝜑,Θ,Φ,𝑏,𝑐,𝑑) implies min_{𝑡∈[1/2,1]}𝑦(𝑡)≥𝑏, ‖𝑦‖≤𝑐. It follows from (2.15) and (3.1) that min_{𝑡∈[1/2,1]}(𝑦−𝜆𝑀𝑢)≥𝑏−𝛼[Γ(𝛼)−Γ(𝛼−𝛽)𝑙]𝑀/𝛽Γ(𝛼). By (C2), (2.12), (3.1), and Lemma 2.7, we get Φ𝑇𝑦=min_{[]𝑡∈1/2,1}𝑇𝑦(𝑡)≥𝜆min_{[]𝑡∈1/2,1}¹₀𝐺()𝑓(𝑠)1𝑡,𝑠𝑠,𝑦𝑑𝑠≥2^𝛼−1𝜆𝑘𝑏¹_1/2𝐺(1,𝑠)𝑑𝑠>𝑏. (3.3) So, the condition (S1) of Theorem 2.13 holds.
Take 𝑦∈𝑃(𝜑,Φ,𝑏,𝑑) with Θ(𝑇𝑦)>𝑐. By 𝑇𝑦∈𝑃, we get min_{[]𝑡∈1/2,1}𝑇𝑦(𝑡)≥min_{[]𝑡∈1/2,1}𝑡^𝛼−1≥1𝑇𝑦2^𝛼−1>1𝑇𝑦2^𝛼−1𝑐>𝑏. (3.4) Thus, (S2) holds.
By 𝑎>0, we have 0∉𝑅(𝜑,Ψ,𝑎,𝑑). Take 𝑦∈𝑅(𝜑,Ψ,𝑎,𝑑) with Ψ(𝑦)=𝑎. By (C3), we get Ψ=≤Γ𝑇𝑦𝑇𝑦𝛼−𝛽Γ(𝛼)𝑙𝑎+𝜆𝑎¹₀𝐺()=Γ1,𝑠𝑑𝑠𝛼−𝛽Γ(𝛼)𝑙+𝜆¹₀𝐺(1,𝑠)𝑑𝑠𝑎≤𝑎. (3.5) By Theorem 2.13, we get that 𝑇 has at least three fixed points 𝑦₁,𝑦₂,𝑦₃∈𝑃(𝜑,𝑑) such that ‖𝑦_𝑖‖≤𝑑,𝑖=1,2,3, and 𝑏<Φ𝑦₁,𝑎<Ψ𝑦₂,Φ𝑦₂<𝑏,Ψ𝑦₃<𝑎. (3.6)
If 𝑦∈𝑃 and ‖𝑦‖>𝑎, by (2.15) and (3.1), we have 𝑦(𝑡)≥𝑡^𝛼−1𝑦>𝑎𝑡^𝛼−1>𝜆𝑀𝑢(𝑡). (3.7)
Obviously, ‖𝑦₁‖>𝑏>𝑎 and ‖𝑦₂‖>𝑎. So, 𝑦₁−𝜆𝑀𝑢,𝑦₂−𝜆𝑀𝑢 are two positive solutions of (1.5). The proof is completed.

4. Example

For convenience, we define the following notations: [𝑎,𝑏]∶={𝑥∶𝑥∈ℝ,𝑎≤𝑥≤𝑏},(𝑎,𝑏]∶={𝑥∶𝑥∈ℝ,𝑎<𝑥≤𝑏},for𝑎,𝑏∈ℝ,𝑎<𝑏. (4.1)

Example 4.1. Consider the following boundary value problem: 𝐷₀^5/2₊𝑦(𝑡)(𝑡)[],𝑦(0)+𝜆𝑓𝑡,𝑦=0,𝑡∈0,1=𝑦^′(0)𝑦=0,_′(1)=ℎ𝑦, (4.2) where ℎ(𝑦)=∫¹₀𝑦(𝑠)𝑑𝑔(𝑠),𝑔(𝑠) is a bounded variation function on [0,1] with 0<⋁¹₀(𝑔)≤1<3/2, 𝑓=1𝑡,𝑦sin𝑡−2𝜋−1−√∈[]×[],1𝑦,𝑡,𝑦0,10,6sin𝑡−2𝜋−1+601𝑦−6−√∈[]×(],1𝑦,𝑡,𝑦0,16,7sin𝑡−2𝜋+600−√∈[]×(],1𝑦,𝑡,𝑦0,17,900sin𝑡−2𝜋+570,𝑡,𝑦∈[0,1]×(900,+∞). (4.3) Corresponding to the problem (1.5), we get that 𝛼=5/2,𝛽=1,𝑛=3,ℎ(𝑦)≤‖𝑦‖⋁¹₀(𝑔). Take 𝑙=1,𝑘=50,𝑀=𝑎=6,𝑏=12,𝑐=36,𝑑=620.
By simple calculation, we can get that the conditions of Theorem 3.1 are satisfied. So, when (9/35)√𝜋<𝜆<(15/16)√𝜋, the problem (4.2) has at least two positive solutions.

Example 4.2. Consider the following boundary value problem: 𝐷₀^7/2₊𝑦(𝑡)(𝑡)[],𝑦(0)+𝜆𝑓𝑡,𝑦=0,𝑡∈0,1=𝑦^′(0)=𝑦^′′(0)𝐷=0,_03/2+𝑦(𝑡)∣_𝑡=1=ℎ𝑦, (4.4) where ℎ(𝑦)=∫¹₀𝑦(𝑠)𝑑𝑔(𝑠),𝑔(𝑠) is a bounded variation function on [0,1] with 0<⋁¹₀(𝑔)≤(15/8)√𝜋−1, 𝑓=−3𝑡,𝑦4𝜋cos21𝑡−4√∈[]×[],−3𝑦,𝑡,𝑦0,10,14𝜋cos2−1𝑡+3205𝑦−14√∈[]×(],−3𝑦,𝑡,𝑦0,11,1.24𝜋cos21𝑡+641−4√∈[]×(],−3𝑦,𝑡,𝑦0,11.2,124𝜋cos21𝑡+641−2√3,𝑡,𝑦∈[0,1]×(12,+∞). (4.5) Corresponding to the problem (1.5), we get that 𝛼=7/2,𝛽=3/2,𝑛=4,ℎ(𝑦)≤‖𝑦‖⋁¹₀(𝑔). Take 𝑙=(15/8)√𝜋−1,𝑘=320,𝑀=𝑎=1,𝑏=2,𝑐=12,𝑑=642.
Obviously, 𝑓∈𝐶([0,1]×ℝ⁺,[−1,∞)),ℎ(𝑦)≤𝑙‖𝑦‖. By simple calculation, we can get that 𝑘,𝑙,𝑎,𝑏,𝑐,𝑑,𝛼,𝛽,𝑀 satisfy 0<𝑙<Γ(𝛼)/Γ(𝛼−𝛽),𝑘>2^2𝛼−1𝛽Γ(𝛼)/(Γ(𝛼)−Γ(𝛼−𝛽)𝑙), and Γ(𝛼)−Γ𝛼−𝛽𝑙𝛼𝑀(𝛼)Γ(𝛼)𝑙𝛽Γ<𝑎<𝑏−−Γ𝛼−𝛽𝛼𝑀𝛽Γ(𝛼)<2^𝛼−1𝑏<𝑐<𝑑. (4.6)
It is easy to see that 𝑓(𝑡,𝑦)+1≤642, for (𝑡,𝑦)∈[0,1]×[0,642], and 𝑓(𝑡,𝑦)+1≤1, for (𝑡,𝑦)∈[0,1]×[0,1]. So, conditions (C1) and (C3) of Theorem 3.1 hold.
For (𝑡,𝑦)∈[1/2,1]×[2−56/45√𝜋,12],𝑓(𝑡,𝑦)+1≥𝑘𝑏=640. Therefore, condition (C2) of Theorem 3.1 holds. So, for (21/8(7√2−2))√𝜋<𝜆<14/3, the problem (4.4) has at least two positive solutions.
Specially, in Example 4.2, we take 𝑔(𝑡)=0,𝑡<𝜉,1,𝑡≥𝜉, (4.7) where 0<𝜉<1 and all other conditions remain unchanged. Then ℎ(𝑦)=𝑦(𝜉). Clearly, ∨¹₀(𝑔)=1<(15/8)√𝜋−1. The problem (4.4) has at least two positive solutions for (21/8(7√2−2))√𝜋<𝜆<14/3.

Acknowledgments

The authors are grateful to Editor Bashir Ahmad and anonymous referees for their constructive comments and suggestions which led to improvement of the paper. This work is supported by the Natural Science Foundation of China (11171088), the Doctoral Program Foundation of Hebei University of Science and Technology (QD201020), and the Foundation of Hebei University of Science and Technology (XL201236).