Singular Positone and Semipositone Boundary Value Problems of Nonlinear Fractional Differential Equations

Abstract

We present some new existence results for singular positone and semipositone nonlinear fractional boundary value problem D^𝛼0+𝑢(𝑡)=𝜇𝑎(𝑡)𝑓(𝑡,𝑢(𝑡)),  0<𝑡<1, 𝑢(0)=𝑢(1)=𝑢^(0)=𝑢^(1)=0, where 𝜇>0, 𝑎, and 𝑓 are continuous, 𝛼∈(3,4] is a real number, and D^𝛼0+ is Riemann-Liouville fractional derivative. Throughout our nonlinearity may be singular in its dependent variable. Two examples are also given to illustrate the main results.

1. Introduction

Fractional calculus has played a significant role in engineering, science, economy, and other fields. Many papers and books on fractional calculus, and fractional differential equations have appeared recently, (see [1–9]). It should be noted that most of papers and books on fractional calculus, are devoted to the solvability of initial fractional differential equations (see [3, 4]). Here, we consider positive solutions of nonlinear fractional differential equation conjugate boundary value problem involving Riemann-Liouville derivative: 𝐃^𝛼0+𝑢𝑢((0𝑡))==𝑢𝜇𝑎(1()𝑡)=𝑓𝑢(_𝑡,𝑢(𝑡)),0<𝑡<1,(0)=𝑢^(1)=0,(1.1) where 𝜇>0, 𝑎, and 𝑓 are continuous. 𝛼∈(3,4] is a real number, and 𝐃^𝛼0+ is the Riemann-Liouville fractional derivative.

It is well known that in mechanics the boundary value problem (1.1) where 𝛼=4 describes the deflection of an elastic beam rigidly fixed at both ends. The integer order boundary value problem (1.2) has been studied extensively. For details, see for instance, the papers [10–13] and the references therein. In [10, 12], Yao considered 𝑢𝑢(^0()𝑡)==𝑢𝜆(𝑓1)(𝑡=,𝑢𝑢(_𝑡)),0<𝑡<1,(0)=𝑢^(1)=0,(1.2) and using a Krasnosel'skii fixed-point theorem, derived a 𝜆-interval such that, for any 𝜆 lying in this interval, the beam equation has existence and multiplicity on positive solution. In this paper, we will consider a more general situation, namely, the boundary value problem (1.1). To the best of our knowledge, there have been few papers which deal with the boundary value problem (1.1) for nonlinear fractional differential equation.

In this paper, in analogy with boundary value problems for differential equations of integer order, we firstly derive the corresponding Green's function named the fractional Green' function. Consequently problem (1.1) is reduced to an equivalent Fredholm integral equation of the second kind. Finally, using Krasnosel'skii's fixed-point theorems, the existence of positive solutions are obtained.

2. Preliminaries

For completeness, in this section, we will demonstrate and study the definitions and some fundamental facts of Riemann-Liouville derivatives of fractional order which can be found in [5].

Definition 2.1 (see [5, Definition 2.1]). The integral 𝐼^𝛼0+𝑓(𝑥)=1Γ(𝛼)^𝑥0𝑓(𝑡)(𝑥−𝑡)^1−𝛼𝑑𝑡,𝑥>0,(2.1) where 𝛼>0, is called the Riemann-Liouville fractional integral of order 𝛼.

Definition 2.2 (see [5, page 36-37]). For a function 𝑓(𝑥) given in the interval [0,∞), the expression 𝐃^𝛼0+𝑓(𝑥)=1Γ(𝑛−𝛼)𝑑𝑑𝑥^𝑛^x₀𝑓(𝑡)(𝑥−𝑡)^{𝛼−𝑛+1}𝑑𝑡,(2.2) where 𝑛=[𝛼]+1,[𝛼] denotes the integer part of number 𝛼, is called the Riemann-Liouville fractional derivative of order 𝛼.
From the definition of Riemann-Liouville derivative, for 𝜇>−1, we have 𝐃^𝛼0+𝑥^𝜇=Γ(1+𝜇)Γ(1+𝜇−𝛼)𝑥^𝜇−𝛼,(2.3) giving in particular 𝐃^𝛼0+𝑥^𝛼−𝑚=0,𝑚=1,2,3,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

Lemma 2.3. Let 𝛼>0, then the differential equation 𝐃^𝛼0+𝑢(𝑡)=0(2.4) has solutions 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛, 𝑐_𝑖∈ℝ, 𝑖=1,,2…,𝑛, as unique solutions, where 𝑛 is the smallest integer greater than or equal to 𝛼.
As 𝐃^𝛼0+𝐼^𝛼0+𝑢=𝑢, from Lemma 2.3, we deduce the following statement.

Lemma 2.4. Let 𝛼>0, then 𝐼^𝛼0+𝐃^𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛,(2.5) for some 𝑐_𝑖∈ℝ, 𝑖=1,2,…,𝑛, 𝑛 is the smallest integer greater than or equal to 𝛼.
The following Krasnosel'skii's fixed-point theorem will play a major role in our next analysis.

Theorem 2.5 (see [6]). Let 𝑋 be a Banach space, and let 𝑃⊂𝑋 be a cone in 𝑋. Assume that Ω₁,Ω₂ are open subsets of 𝑋 with 0∈Ω₁⊂Ω₁⊂Ω₂, and let 𝑆∶𝑃→𝑃 be a completely continuous operator such that, either (1)‖𝑆𝑤‖≤‖𝑤‖, 𝑤∈𝑃∩𝜕Ω₁, ‖𝑆𝑤‖≥‖𝑤‖, 𝑤∈𝑃∩𝜕Ω₂, or(2)‖𝑆𝑤‖≥‖𝑤‖, 𝑤∈𝑃∩𝜕Ω₁, ‖𝑆𝑤‖≤‖𝑤‖,𝑤∈𝑃∩𝜕Ω₂.Then 𝑆 has a fixed-point in 𝑃∩(Ω₂∖ Ω₁).

3. Green's Function and Its Properties

In this section, we derive the corresponding Green's function for boundary-value problem (1.1), and obtain some properties of Green's function.

Lemma 3.1. Let ℎ(𝑡)∈𝐶[0,1] be a given function, then the boundary-value problem, 𝐃^𝛼0+𝑢𝑢((0𝑡))==𝑢ℎ((1𝑡)),=𝑢0_<𝑡<1,3<𝛼≤4,(0)=𝑢^(1)=0,(3.1) has a unique solution 𝑢(𝑡)=¹⁰𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠,(3.2) where 𝐺(𝑡,𝑠)=1Γ(𝛼)𝑡^𝛼−2𝑡(1−𝑠)^𝛼−2[(𝑠−𝑡)+(𝛼−2)(1−𝑡)𝑠]+(𝑡−𝑠)^𝛼−1,0≤𝑠≤𝑡≤1,_𝛼−2(1−𝑠)^𝛼−2[(𝑠−𝑡)+(𝛼−2)(1−𝑡)𝑠],0≤𝑡≤𝑠≤1.(3.3) Here 𝐺(𝑡,𝑠) is called Green's function of boundary-value problem (3.1).

Proof. By means of the Lemma 2.4, we can reduce (3.1) to an equivalent integral equation 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+𝑐₃𝑡^𝛼−3+𝑐₄𝑡^𝛼−4+^𝑡0(𝑡−𝑠)^𝛼−1Γ(𝛼)ℎ(𝑠)𝑑𝑠.(3.4) From 𝑢(0)=𝑢(1)=𝑢^(0)=𝑢^(1)=0, we have 𝑐₃=𝑐₄=0 and 𝑐₁=¹⁰(1−𝑠)^𝛼−2(2𝑠−𝛼𝑠−1)𝑐Γ(𝛼)ℎ(𝑠)𝑑𝑠,₂=¹⁰(𝛼−1)(1−𝑠)^𝛼−2𝑠Γ(𝛼)ℎ(𝑠)𝑑𝑠.(3.5) Therefore, the unique solution of (3.1) is 𝑢(𝑡)=¹⁰𝑡^𝛼−1(1−𝑠)^𝛼−2(2𝑠−𝛼𝑠−1)Γ(𝛼)ℎ(𝑠)𝑑𝑠+¹⁰𝑡^𝛼−2(𝛼−1)(1−𝑠)^𝛼−2𝑠Γ(𝛼)ℎ(𝑠)𝑑𝑠+^𝑡0(𝑡−𝑠)^𝛼−1=1Γ(𝛼)ℎ(𝑠)𝑑𝑠Γ(𝛼)^𝑡0𝑡^𝛼−1+(1−𝑠)^𝛼−2(2𝑠−𝛼𝑠−1)+𝑡^𝛼−2(𝛼−1)(1−𝑠)^𝛼−2𝑠+(𝑡−𝑠)^𝛼−1ℎ(𝑠)𝑑𝑠_1𝑡=𝑡^𝛼−1(1−𝑠)^𝛼−2(2𝑠−𝛼𝑠−1)+𝑡^𝛼−2(𝛼−1)(1−𝑠)^𝛼−2𝑠ℎ(𝑠)𝑑𝑠₁₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠.(3.6) The proof is finished.

Lemma 3.2. The function 𝐺(𝑡,𝑠) defined by (3.3) has the following properties: (1)𝐺(𝑡,𝑠)=𝐺(1−𝑠,1−𝑡),𝑓𝑜𝑟𝑡,𝑠∈[0,1];(2)𝑡^𝛼−2(1−𝑡)²𝑞(𝑠)≤𝐺(𝑡,𝑠)≤(𝛼−1)𝑞(𝑠) and 𝐺(𝑡,𝑠)≤((𝛼−1)(𝛼−2)/Γ(𝛼))𝑡^𝛼−2(1−𝑡)² for 𝑡,𝑠∈[0,1],where 𝑞(𝑠)=((𝛼−2)/Γ(𝛼))𝑠²(1−𝑠)^𝛼−2.

Proof. Observing the expression of 𝐺(𝑡,𝑠), it is clear that 𝐺(𝑡,𝑠)=𝐺(1−𝑠,1−𝑡) for 𝑡,𝑠∈[0,1]. In the following, we consider Γ(𝛼)𝐺(𝑡,𝑠).
For 0≤𝑠≤𝑡≤1, we have
Γ(𝛼)𝐺(𝑡,𝑠)==((𝑡𝑡−−𝑠𝑠))(^𝛼−1𝑡−−𝑠(𝑡)−_𝛼−2𝑡𝑠)^𝛼−2(𝑡−𝑠)+(𝛼−2)(1−𝑡)(𝑡−𝑡𝑠)^𝛼−2𝑠=−(𝑡−𝑠)(𝛼−2)−_{𝑡−𝑡𝑠𝑡−𝑠}(𝑡−𝑡𝑠)^𝛼−2+(𝛼−2)(1−𝑡)(𝑡−𝑡𝑠)^𝛼−2𝑠≥−(𝑡−𝑠)(𝛼−2)(𝑡−𝑥𝑡𝑠)^𝛼−3𝑑_𝛼−3𝑥+(𝛼−2)(𝑡−𝑡𝑠)^𝛼−2(1−𝑡)𝑠=−(𝑡−𝑠)(𝛼−2)(𝑡−𝑡𝑠)_𝛼−3[(𝑡−𝑡𝑠)−(𝑡−𝑠)]+(𝛼−2)(𝑡−𝑡𝑠)^𝛼−2(1−𝑡)𝑠=(𝛼−2)(𝑡−𝑡𝑠)_𝛼−3(1−𝑡)𝑠+(𝛼−2)(𝑡−𝑡𝑠)^𝛼−2(1−𝑡)𝑠≥(𝛼−2)(𝑡−𝑡𝑠)_𝛼−2(1−𝑡)𝑠[−(𝑡−𝑠)+(𝑡−𝑡𝑠)](1−𝑡)²𝑠²=(𝛼−2)𝑡^𝛼−2(1−𝑡)²𝑠²(1−𝑠)^𝛼−2,(3.7) and Γ(𝛼)𝐺(𝑡,𝑠)=≤−−((𝑡𝑡−−𝑠𝑠))((𝛼𝛼−−22))(𝑡−^{𝑡−𝑡𝑠𝑡−𝑠}𝑥𝑠)^𝛼−3_𝛼−3𝑑𝑥+(𝛼−2)(𝑡−𝑡𝑠)^𝛼−2(1−𝑡)𝑠=−(𝑡−𝑠)(𝛼−2)(𝑡−𝑠)_𝛼−3[(𝑡−𝑡𝑠)−(𝑡−𝑠)]+(𝛼−2)(𝑡−𝑡𝑠)^𝛼−2(1−𝑡)𝑠=(𝛼−2)(1−𝑡)𝑠(𝑡−𝑡𝑠)_𝛼−2(1−𝑡)𝑠+(𝛼−2)(𝑡−𝑡𝑠)^𝛼−2(1−𝑡)𝑠≤(𝛼−2)₂−(𝑡−𝑠)^𝛼−2(1−𝑡)²𝑠²(𝑡−𝑡𝑠)^𝛼−3≤(𝛼−1)(𝛼−2)𝑡^𝛼−3(1−𝑡)²𝑠²(1−𝑠)^𝛼−3≤(𝛼−1)(𝛼−2)𝑠²(1−𝑠)^𝛼−2.(3.8)
For 0≤𝑡≤𝑠≤1, since 𝛼>3, we have
Γ(𝛼)𝐺(𝑡,𝑠)=≥((𝑡𝛼−−𝑡𝑠2))(𝑡^𝛼−2−[(𝑠𝑡𝑠−)𝑡)_𝛼−2+(𝛼−2)(1−𝑡)𝑠]=(𝛼−2)𝑡_𝛼−2(1−𝑡)𝑠≥(𝛼−2)𝑡_𝛼−2(1−𝑠)^𝛼−2(1−𝑡)𝑠Γ(𝛼)𝐺(𝑡,𝑠)=𝑡_𝛼−2(1−𝑡)²𝑠²(1−𝑠)^𝛼−2,≤𝑡_𝛼−2(1−𝑠)^𝛼−2[(𝑠−𝑡)+(𝛼−2)(1−𝑡)𝑠]≤(𝛼−(11−)𝑡(𝑠1)−^𝛼−2𝑠[)𝑠+_𝛼−2(𝛼−2)𝑠]≤(𝛼−1)𝑠₂𝑠(1−𝑠)^𝛼−2≤(𝛼−2)(𝛼−1)𝑠²(1−𝑠)^𝛼−2.(3.9) Thus 𝑡^𝛼−2(1−𝑡)²𝑞(𝑠)≤𝐺(𝑡,𝑠)≤(𝛼−1)𝑞(𝑠), for 𝑡,𝑠∈[0,1]. Combining 𝐺(𝑡,𝑠)=𝐺(1−𝑠,1−𝑡), we have 𝐺(𝑡,𝑠)≤(𝛼−1)𝑞(1−𝑡)=(𝛼−1)(𝛼−2)Γ(𝛼)𝑡^𝛼−2(1−𝑡)²,for𝑡,𝑠∈[0,1].(3.10) This completes the proof.

We note that 𝑢(𝑡) is a solution of (1.1) if and only if

𝑢(𝑡)=𝜇¹⁰𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠,0≤𝑡≤1.(3.11)

For our constructions, we will consider the Banach space 𝐸=𝐶[0,1] equipped with standard norm ‖𝑢‖=max_0≤𝑡≤1‖𝑢(𝑡)‖,𝑢∈𝑋. We define a cone 𝐾 by 𝐾=𝑢∈𝑋∣𝑢(𝑡)≥𝑡^𝛼−2(1−𝑡)²𝛼−1‖𝑢‖,𝑡∈[0,1],𝛼∈(3,4].(3.12) Define an integral operator 𝐴∶𝐾→𝑋 by 𝐴𝑢(𝑡)=𝜇¹⁰𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠,0≤𝑡≤1,𝑢∈𝐾.(3.13) Notice from (3.13) and Lemma 3.2 that, for 𝑢∈𝐾, 𝐴𝑢(𝑡)≥0 on [0,1] and 𝐴𝑢(𝑡)=𝜇¹⁰𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≤𝜇¹⁰(𝛼−1)𝑞(𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,(3.14) then ‖𝐴𝑢‖≤∫¹⁰(𝛼−1)𝑞(𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠 .

On the other hand, we have 𝐴𝑢(𝑡)=≥𝜇𝜇¹⁰₁₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≥𝑡𝑡^𝛼−2(1−𝑡)²𝑞(𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠_𝛼−2(1−𝑡)²𝛼−1𝜇¹⁰≥𝑡(𝛼−1)𝑞(𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠_𝛼−2(1−𝑡)²𝛼−1‖𝐴𝑢‖.(3.15) Thus, 𝐴(𝐾)⊂𝐾. In addition, standard arguments show that 𝐴 is completely continuous.

4. Singular Positone Problems

In this section we present some new result for the singular problem 𝐃^𝛼0+𝑢𝑢((0𝑡))==𝜇𝑢𝑎(1()𝑡)=𝑓𝑢(_𝑡,𝑢(𝑡)),0<𝑡<1,3<𝛼≤4,(0)=𝑢^(1)=0,(4.1) where 𝜇>0 and nonlinearity 𝑓 may be singular at 𝑢=0.

Using Theorem 2.5 we establish the following main result.

Theorem 4.1. Suppose that the following conditions are satisfied. ⎧⎪⎪⎨⎪⎪⎩𝑓𝑐𝑐𝑜𝑜(𝑛𝑛𝑡𝑡𝑡,𝑖𝑖𝑢𝑛𝑛)𝑢𝑢𝑓𝑎𝑜𝑜≤𝑢𝑢∶∈𝑠𝑠𝑔[0𝐶(𝑎𝑜𝑢,𝑛𝑛)1(𝑑0]+[,0×𝑛1ℎ,𝑜)∞𝑛((∩0𝑢𝑖)𝑛,)𝐿∞𝑐𝑎𝑟𝑜¹𝑛𝑒)𝑛[𝑎𝑑0⟶𝑠[,𝑖0ℎ1𝑛,]𝑔1(]𝑤0𝑜×,𝑛𝑖𝑡∞ℎ((00)𝑎,,∞𝑖∞>𝑠))𝑐0,𝑜𝑤𝑛𝑜ℎ𝑡𝑛𝑖𝑖𝑡𝑛≥ℎ(𝑢0𝑔𝑜0,𝑢1>𝑠),0((44..23))∃𝐾₀𝑔𝑛𝑜𝑛𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔𝑜𝑛(0,∞),(4.4)𝑎𝑤𝑖𝑡ℎ𝑔(𝑥𝑦)≤𝐾₀𝑔(𝑥)𝑔(𝑦)∀𝑥>0,𝑦>0,(4.5)₀=∃𝑟𝜇>(𝛼0−𝑤1𝑖)𝑡ℎ¹⁰𝑞(𝑠)𝑟𝑎(𝑠)𝑔𝑠^𝛼−2(1−𝑠)²𝑑𝑠<∞,(4.6)𝑔(𝑟)+ℎ(𝑟)>𝐾²⁰𝑎₀𝑔1⎧⎪⎪⎪⎨⎪⎪⎪⎩𝑇ℎ𝑒𝑟𝑒𝑒𝑥𝑖𝑠𝑡𝑠0<𝜃<1𝛼−1,(4.7)𝑛𝑜𝑛𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛2𝑔(₁𝑐ℎ𝑜𝑜𝑠𝑒𝑎𝑛𝑑𝑓𝑖𝑥𝑖𝑡)𝑎𝑛𝑑𝑎𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠,𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛ℎ₁∶(0,∞)⟶(0,∞),𝑎𝑛𝑑𝑎𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠∶[0,∞)⟶(0,∞)𝑤𝑖𝑡ℎℎ¹𝑔₁𝑎𝑛𝑑𝑤𝑖𝑡ℎ𝑓(𝑡,𝑢)≥𝑔₁𝑛𝑜𝑛𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔𝑜𝑛(0,∞)⎧⎪⎨⎪⎩∃0<𝑅₁(𝑢)+ℎ₁(𝑢)𝑓𝑜𝑟(𝑡,𝑢)∈[𝜃,1−𝜃]×(0,∞),(4.8)<𝑟<𝑅₂𝑅𝑤𝑖𝑡ℎ(𝑖=1,2),_𝑖𝑔₁(𝜃^𝛼/𝛼−1)𝑅_𝑖𝑔₁𝑅_𝑖𝑔₁(𝜃^𝛼/𝛼−1)𝑅_𝑖+𝑔₁𝑅_𝑖ℎ₁(𝜃^𝛼/𝛼−1)𝑅_𝑖<𝜇^𝜃1−𝜃𝐺(𝜎,𝑠)𝑎(𝑠)𝑑𝑠,(4.9) here 𝐺(𝑡,𝑠) is Green's function and ^𝜃1−𝜃𝐺(𝜎,𝑠)𝑎(𝑠)𝑑𝑠=sup_𝑡∈[0,1]^𝜃1−𝜃𝐺(𝑡,𝑠)𝑎(𝑠)𝑑𝑠.(4.10) Then (4.1) has two nonnegative solutions 𝑢_𝑖 with 𝑅₁<‖𝑢₁‖<𝑟<‖𝑢₂‖<𝑅₂ and 𝑢_𝑖(𝑡)>0 for 𝑡∈(0,1),𝑖=1,2.

Proof. First we will show that there exists a solution 𝑢₂ to (4.1) with 𝑢₂(𝑡)>0 for 𝑡∈(0,1) and 𝑟<‖𝑢₂‖<𝑅₂. Let Ω₁={𝑢∈𝐸∶‖𝑢‖<𝑟},Ω₂=𝑢∈𝐸∶‖𝑢‖<𝑅₂.(4.11) We now show ‖𝐴𝑢‖<‖𝑢‖for𝐾∩𝜕Ω₁.(4.12) To see this, let 𝑢∈𝐾∩𝜕Ω₁. Then ‖𝑢‖=||𝑢||_[0,1]=𝑟 and 𝑢(𝑡)≥(𝑡^𝛼−2(1−𝑡)²/(𝛼−1))𝑟 for 𝑡∈[0,1]. So for 𝑡∈[0,1], we have (𝐴𝑢)(𝑡)=≤𝜇𝜇¹⁰₁₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠=𝜇₁₀(𝛼−1)𝑞(𝑠)𝑎(𝑠)[𝑔(𝑢(s))+ℎ(𝑢(s))]𝑑𝑠(𝛼−1)𝑞(𝑠)𝑎(𝑠)𝑔(𝑢(𝑠))1+ℎ(𝑢(𝑠))≤𝜇₁₀𝑔(𝑢(𝑠))𝑑𝑠(𝛼−1)𝑞(𝑠)𝑎(𝑠)𝑔𝑠^𝛼−2(1−𝑠)²≤𝐾₀𝛼−1𝑟1+𝑔ℎ((𝑟r))𝑑𝑠𝑔𝑟𝛼−11+ℎ𝑔((𝑟𝑟))𝜇¹⁰=𝑎₀(𝛼−1)𝑞(𝑠)𝑎(𝑠)𝑔𝑠^𝛼−2(1−𝑠)²𝑑𝑠𝐾²⁰𝑔1𝛼−1[𝑔(𝑟)+ℎ(𝑟)].(4.13) This together with (4.7) yields ‖𝐴𝑢‖=‖𝐴𝑢‖_[0,1]<𝑟=‖𝑢‖,(4.14) so (4.12) is satisfied.
Next we show
‖𝐴𝑢‖>‖𝑢‖for𝐾∩𝜕Ω₂.(4.15) To see this, let 𝑢∈𝐾∩𝜕Ω₂ so ‖𝑢‖=‖𝑢‖_[0,1]=𝑅₂, and let 𝑢(𝑡)≥(𝑡^𝛼−2(1−𝑡)²/(𝛼−1))𝑅₂ for 𝑡∈[0,1].
We have
(𝐴𝑢)(𝜎)=≥𝜇𝜇¹⁰𝐺_𝜃1−𝜃(𝜎,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠=𝜇_𝜃1−𝜃𝐺(𝜎,𝑠)𝑎(𝑠)𝑔₁(𝑢(s))+ℎ₁(𝑢(s))𝑑𝑠𝐺(𝜎,𝑠)𝑎(𝑠)𝑔₁(𝑢(𝑠))1+ℎ¹(𝑢(𝑠))≥𝑔₁𝑔₁(𝑢(𝑠))𝑑𝑠𝑅₂𝜇^𝜃1−𝜃𝐺(𝜎,𝑠)𝑎(𝑠)1+ℎ¹𝑠^𝛼−2(1−𝑠)²/(𝛼−1)𝑅₂≥𝑔₁𝑔₁𝑠^𝛼−2(1−𝑠)²/(𝛼−1)𝑅₂𝑑𝑠𝑅₂𝜇^𝜃1−𝜃𝐺(𝜎,𝑠)𝑎(𝑠)1+ℎ¹(𝜃^𝛼/(𝛼−1))𝑅₂𝑔₁(𝜃^𝛼/(𝛼−1))𝑅₂𝑑𝑠.(4.16) This together with (4.9) yields (𝐴𝑢)(𝜎)>𝑅₂=‖𝑢‖.(4.17) Thus ||𝐴𝑢||>||𝑢||, so (4.15) is held.
Now Theorem 2.5 implies that 𝐴 has a fixed-point 𝑢₂∈𝐾∩(Ω₂⧵Ω₁), that is, 𝑟≤‖𝑢₂‖=‖𝑢₂‖_[0,1]≤𝑅 and 𝑢₂(𝑡)≥𝑞(𝑡)𝑟 for 𝑡∈[0,1]. It follows from (4.12) and (4.15) that ‖𝑢₂‖≠𝑟,‖𝑢₂‖≠𝑅₂, so we have 𝑟<||𝑢₂||<𝑅₂.
Similarly, if we put
Ω₁=𝑢∈𝐸∶‖𝑢‖<𝑅₁,Ω₂={𝑢∈𝐸∶‖𝑢‖<𝑟},(4.18) we can show that there exists a solution 𝑢₁ to (4.1) with 𝑢₁(𝑡)>0 for 𝑡∈(0,1) and 𝑅₁<||𝑢₁||<𝑟.
This completes the proof of Theorem 4.1.

Similar to the proof of Theorem 4.1, we have the following result.

Theorem 4.2. Suppose that (4.2)–(4.8) hold. In addition suppose ⎧⎪⎨⎪⎩∃0<𝑅¹<𝑟𝑤𝑖𝑡ℎ𝑅₁𝑔₁(𝜃^𝛼/(𝛼−1))𝑅₁𝑔₁𝑅₁𝑔₁(𝜃^𝛼/(𝛼−1))𝑅₁+𝑔₁𝑅₁ℎ₁(𝜃^𝛼/(𝛼−1))𝑅₁<𝜇^𝜃1−𝜃𝐺(𝜎,𝑠)𝑞(𝑠)𝑑𝑠.(4.19) Then (4.1) has a nonnegative solution 𝑢₁ with 𝑅₁<‖𝑢₁‖<𝑟 and 𝑢₁(𝑡)>0 for 𝑡∈(0,1).

Remark 4.3. If in (4.19) we have 𝑅₁>𝑟, then (4.1) a nonnegative solution 𝑢₂ with 𝑟<‖𝑢₂‖<𝑅₂ and 𝑢₂(𝑡)>0 for 𝑡∈(0,1).
It is easy to use Theorem 4.2 and Remark 4.3 to write theorems which guarantee the existence of more than two solutions to (4.1). We state one such result.

Theorem 4.4. Suppose that (4.2)–(4.6) and (4.8) hold. Assume that ∃𝑚∈{1,2,…} and constants 𝑅_𝑖,𝑟_𝑖(𝑖=1,…,𝑚), with 𝑟₁>𝑏₀, and 0<𝑅₁<𝑟₁<𝑅₂<𝑟₂<⋯<𝑅_𝑚<𝑟_𝑚.(4.20) In addition suppose for each 𝑖=1,…,𝑚 that 𝑟_𝑖𝑔𝑟_𝑖+ℎ𝑟_𝑖>𝐾²⁰𝑎₀𝑔1𝛼−1,(4.21) and 𝑅_𝑖𝑔₁(𝜃^𝛼/(𝛼−1))𝑅_𝑖𝑔₁𝑅_𝑖𝑔₁(𝜃^𝛼/(𝛼−1))𝑅_𝑖+𝑔₁𝑅_𝑖ℎ₁(𝜃^𝛼/(𝛼−1))𝑅_𝑖<𝜇^𝜃1−𝜃𝐺(𝜎,𝑠)𝑎(𝑠)𝑑𝑠(4.22) hold. Then (4.1) has nonnegative solutions 𝑦₁,…,𝑦_𝑚 with 𝑦_𝑖(𝑡)>0 for 𝑡∈(0,1).

Example 4.5. Consider the boundary value problem 𝐃^𝛼0+𝑢𝑢((0𝑡))==𝜎𝑢(1𝑢)=^−𝑎(𝑢𝑡)_+𝑢^𝑏(𝑡),𝑡∈(0,1),3<𝛼≤4,(0)=𝑢^(1)=0,0<𝑎<1<𝑏,(4.23) where 𝜎∈(0,𝜎₀) is such that 𝜎₀≤12𝑎₁,(4.24) here 𝑎₁=(𝛼−1)(𝛼−2)Γ(𝛼)¹⁰𝑠^𝛼−2(1−𝑠)²𝑠^𝛼−2(1−𝑠)²^−𝛼𝑑𝑠=¹⁰𝑠^{(𝛼−2)(1−𝑎)}(1−𝑠)^2(1−𝑎)𝑑𝑠<∞.(4.25) Then (4.23) has two solutions 𝑢₁,𝑢₂ with 𝑢₁(𝑡)>0,𝑢₂(𝑡)>0 for 𝑡∈(0,1),𝑖=1,2.
To see this we will apply Theorem 4.1 with (here 0<𝑅₁<1<𝑅₂ will be chosen below)
𝑔(𝑢)=𝑔₁(𝑢)=𝑢^−𝑎,ℎ(𝑢)=ℎ₁(𝑢)=𝑢^𝑏,𝑎(𝑡)=𝜎,𝐾₀=1,𝜃=14.(4.26) Clearly (4.2)–(4.6) and (4.8) hold, and 𝑎₀=(𝜎𝑎₁/(𝛼−1)^𝑎). Now (4.7) holds with 𝑟=1 since 𝑟𝑔(𝑟)+ℎ(𝑟)=21≥𝑎¹𝜎₀>𝐾²⁰𝑎₀(𝛼−1)^𝑎=𝐾²⁰𝑎₀𝑔1𝛼−1.(4.27) Finally notice (4.9) is satisfied for 𝑅₁ small and 𝑅₂ large since 𝑅_𝑖𝑔₁𝑅_𝑖1+ℎ₁(𝜃^𝛼/𝛼−1)𝑅_𝑖/g₁(𝜃^𝛼/𝛼−1)𝑅_𝑖=1+(𝛼−1)𝑅^{𝑖1+𝑎−(𝑎+𝑏)}𝜃^{𝛼(𝑎+𝑏)}𝑅^{𝑖𝑎+𝑏}⟶0,(4.28) as 𝑅₁→0,𝑅₂→∞, since 𝑏>1. Thus all the conditions of Theorem 4.1 are satisfied so existence is guaranteed.

5. Singular Semipositone Problems

In this section we present a new result for the singular semipositone problem: 𝐃^𝛼0+𝑢𝑢((0𝑡))==𝜇𝑢𝑎(1()𝑡)=𝑓𝑢(_𝑡,𝑢(𝑡)),0<𝑡<1,3<𝛼≤4,(0)=𝑢^(1)=0,(5.1) where 𝜇>0 and nonlinearity 𝑓 may be singular at 𝑢=0.

Before we prove our main result, we first state a result.

Lemma 5.1. Suppose 𝑎∈𝐿¹[0,1] with 𝑎>0 on (0,1). Then the boundary value problem, 𝐃^𝛼0+𝑢𝑢((0𝑡))==𝑢𝑎((1𝑡))𝑒=(𝑡𝑢)_,0<𝑡<1,3<𝛼≤4,(0)=𝑢^(1)=0,(5.2) has a solution 𝑤 with 𝑤(𝑡)≤𝑡^𝛼−2(1−𝑡)²𝛼−1𝐶⁰for𝑡∈[0,1],(5.3) here 𝐶₀=(𝛼−1)²(𝛼−2)Γ(𝛼)¹⁰𝑎(𝑠)𝑒(𝑠)𝑑𝑠.(5.4) In fact, from Lemma 3.1, (5.2) has solution 𝑤(𝑡)=¹⁰𝐺(𝑡,𝑠)𝑎(𝑠)𝑒(𝑠)𝑑𝑠.(5.5) According to Lemma 3.2, we have 𝑤(𝑡)≤¹⁰(𝛼−1)(𝛼−2)𝑡^𝛼−2(1−𝑡)²Γ(𝛼)𝑎(𝑠)𝑒(𝑠)𝑑𝑠=𝑡^𝛼−2(1−𝑡)²𝛼−1𝐶⁰.(5.6) The above Lemma together with Theorem 2.5 establish our main result.

Theorem 5.2. Suppose that the following conditions are satisfied. 𝑞∈𝐶(0,1)∩𝐿¹[0,1]𝑤𝑖𝑡ℎ𝑞>𝑜𝑛(0,1).(5.7)𝑓𝑎∶𝑓𝑢[𝑛0𝑐,𝑡1𝑖]𝑜𝑛×𝑒(0∈,∞𝐶)((⟶0,1)𝐑,(0𝑖𝑠,+𝑐𝑜∞𝑛𝑡)𝑖)𝑛𝑢𝑜𝑤𝑢𝑖𝑠𝑡ℎ𝑎𝑛𝑓𝑑(𝑡𝑡,ℎ𝑢𝑒)𝑟+𝑒𝑒𝑒𝑥(𝑖𝑡𝑠)𝑡≥𝑠𝑓𝑜𝑟(𝑡,𝑢)∈(0,1)×(0,∞),(5.8)⎧⎪⎪⎨⎪⎪⎩𝑐𝑐𝑓𝑜𝑜^∗𝑛𝑛(𝑡𝑡𝑡𝑖𝑖,𝑛𝑛𝑢𝑢𝑢)𝑜𝑜=𝑢𝑢𝑠𝑠𝑓𝑎𝑜𝑛𝑛(𝑡𝑑,[𝑢0𝑛),𝑜+∞𝑛𝑖𝑒)𝑛(𝑐𝑎𝑡𝑟)𝑛𝑒≤𝑎𝑑𝑠𝑔𝑖ℎ𝑛(𝑔𝑢)𝑜+𝑛ℎ(0(,𝑢∞))𝑜𝑛,ℎ[0≥,10]×(0,∞)𝑤𝑖𝑡ℎ𝑔>0𝑔𝑛𝑜𝑛𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔𝑜𝑛(0,∞),(5.9)∃𝐾₀𝑤𝑖𝑡ℎ𝑔(𝑥𝑦)≤𝐾₀𝑔(𝑥)𝑔(𝑦)∀𝑥>0,𝑦>0,(5.10)𝑎₀=∫¹⁰𝐺(𝜎,𝑠)𝑞(𝑠)𝑔𝑠^𝛼−2(1−𝑠)²𝑑𝑠<∞,(5.11)∃𝑟>𝜇𝐶₀𝑤𝑖𝑡ℎ𝑟𝑔𝑟−𝜇𝐶₀/(𝛼−1){1+(ℎ(𝑟)/𝑔(𝑟))}≥𝜇𝐾⁰𝑎₀,(5.12)⎧⎪⎪⎪⎨⎪⎪⎪⎩𝑇ℎ𝑒𝑟𝑒𝑒𝑥𝑖𝑠𝑡𝑠0<𝜃<1𝑛𝑜𝑛𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛2𝑔(₁𝑐ℎ𝑜𝑜𝑠𝑒𝑎𝑛𝑑𝑓𝑖𝑥𝑖𝑡)𝑎𝑛𝑑𝑎𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠,𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛ℎ₁∶(0,∞)⟶(0,∞),𝑎𝑛𝑑𝑎𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠∶[0,∞)⟶(0,∞)𝑤𝑖𝑡ℎℎ¹𝑔₁𝑎𝑛𝑑𝑤𝑖𝑡ℎ𝑓(𝑡,𝑢)+𝑒(𝑡)≥𝑔₁𝑛𝑜𝑛𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔𝑜𝑛(0,∞)(𝑢)+ℎ₁(𝑢)𝑓𝑜𝑟(𝑡,𝑢)∈[𝜃,1−𝜃]×(0,∞),(5.13)𝑎𝑛𝑑∃𝑅>𝑟𝑤𝑖𝑡ℎ𝑅𝑔₁((𝜖𝜃^𝛼/(𝛼−1))𝑅)𝑔₁(𝑅)𝑔₁((𝜖𝜃^𝛼/(𝛼−1))𝑅)+𝑔₁(𝑅)ℎ₁((𝜖𝜃^𝛼/(𝛼−1))𝑅)≤𝜇^𝜃1−𝜃𝐺(𝜎,𝑠)𝑞(𝑠)𝑑𝑠,(5.14) here 𝜖>0 is any constant (choose and fix it) so that 1−(𝜇𝐶₀/𝑅)≥𝜖 (note 𝜖 exists since 𝑅>𝑟>𝜇𝐶₀ in fact we can have 𝜖=1−(𝜇𝐶₀/𝑟) ) and 𝐺(𝑡,𝑠) is Green's function and ^𝜃1−𝜃𝐺(𝜎,𝑠)𝑎(𝑠)𝑑𝑠=sup_𝑡∈[0,1]^𝜃1−𝜃𝐺(𝑡,𝑠)𝑎(𝑠)𝑑𝑠.(5.15) Then (5.1) has a solution 𝑦∈𝐶[0,1]∩𝐶²(0,1) with 𝑦(𝑡)>0 for 𝑡∈(0,1).

Proof. To show that (5.1) has a nonnegative solution we will look at the boundary value problem 𝐃^𝛼0+𝑦𝑦((0𝑡))==𝑦𝜇𝑞(1()𝑡)=𝑓𝑦^∗_(𝑡,𝑦(𝑡)−𝜙(𝑡)),0<𝑡<1,3<𝛼≤4,(0)=𝑦^(1)=0,(5.16) where 𝜙(𝑡)=𝜇𝑀𝑤(𝑡),0≤𝑡≤1,(5.17) (𝑤 is as in Lemma 5.1).
We will show, using Theorem 2.5, that there exists a solution 𝑦₁ to (5.16) with 𝑦₁(𝑡)>𝜙(𝑡) for 𝑡∈(0,1). If this is true then 𝑢(𝑡)=𝑦₁(𝑡)−𝜙(𝑡),0≤𝑡≤1 is a nonnegative solution (positive on (0,1)) of (5.1), since
𝐃^𝛼0+𝑢(𝑡)=𝐃^𝛼0+𝑦₁(𝑡)−𝐃^𝛼0+𝜙(𝑡)==−−𝜇𝜇𝑞𝑞((𝑡𝑡))𝑓𝑓^∗𝑡𝑡,,𝑦𝑦₁₁(𝑡)−𝜙(𝑡)+𝜇𝑞(𝑡)𝑒(𝑡)=−𝜇𝑞(𝑡)𝑓𝑡,𝑦₁(𝑡)−𝜙(𝑡)+𝑒(𝑡)+𝜇𝑞(𝑡)𝑒(𝑡)=−𝜇𝑞(𝑡)𝑓(𝑡,𝑢(𝑡()𝑡)),−𝜙0(<𝑡)𝑡<1.(5.18) As a result, we will concentrate our study on (5.16). Let 𝐸,𝐾 as in Section 2, and let Ω₁={𝑢∈𝐸∶‖𝑢‖<𝑟},Ω₂={𝑢∈𝐸∶‖𝑢‖<𝑅}.(5.19) Next let 𝐴∶𝐾∩(Ω₂⧵Ω₁)→𝐸  be defined by (𝐴𝑦)(𝑡)=𝜇¹⁰𝐺(𝑡,𝑠)𝑞(𝑠)𝑓^∗(𝑠,𝑦(𝑠)−𝜙(𝑠))𝑑𝑠,0≤𝑡≤1.(5.20) In addition, standard argument shows that 𝐴(𝑃)⊂𝑃 and 𝐴 is completely continuous.
We now show
‖𝐴𝑦‖≤‖𝑦‖for𝐾∩𝜕Ω₁.(5.21) To see this, let 𝑦∈𝐾∩𝜕Ω₁. Then ‖𝑦‖=‖𝑦‖_[0,1]=𝑟 and 𝑦(𝑡)≥(𝑡^𝛼−2(1−𝑡)²/(𝛼−1))𝑟 for 𝑡∈[0,1]. Now for 𝑡∈(0,1), the Lemma 5.1 implies 𝑦(𝑡)−𝜙(𝑡)≥𝑡^𝛼−2(1−𝑡)²𝛼−1𝑟−𝜇𝑡^𝛼−2(1−𝑡)²𝛼−1𝐶⁰≥𝑡^𝛼−2(1−𝑡)²𝛼−1𝑟−𝜇𝐶⁰>0,(5.22) so for 𝑡∈[0,1] we have (𝐴𝑦)(𝑡)=≤𝜇𝜇¹⁰₁₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓^∗(𝑠,𝑦(𝑠)−𝜙(𝑠))𝑑𝑠=𝜇₁₀𝑞(𝑠)𝑎(𝑠)[𝑔(𝑦(𝑠)−𝜙(𝑠))+ℎ(𝑦(𝑠)−𝜙(𝑠))]𝑑𝑠𝑞(𝑠)𝑎(𝑠)𝑔(𝑦(𝑠)−𝜙(𝑠))1+ℎ(𝑦(𝑠)−𝜙(𝑠))≤𝜇₁₀𝑔(𝑦(𝑠)−𝜙(𝑠))𝑑𝑠𝑞(𝑠)𝑎(𝑠)𝑔𝑠^𝛼−2(1−𝑠)²≤𝜇𝐾₀𝛼−1𝑟−𝜇𝐶⁰1+ℎ𝑔((𝑟𝑟))𝑑𝑠𝑔𝑟−𝜇𝐶⁰𝛼−11+ℎ𝑔((𝑟𝑟))¹⁰=𝜇𝐾₀𝑞(𝑠)𝑎(𝑠)𝑔𝑠^𝛼−2(1−𝑠)²𝑑𝑠𝑎₀𝑔𝑟−𝜇𝐶⁰𝛼−11+ℎ𝑔{{𝑟𝑟}}.(5.23) This together with (5.12) yields ‖𝐴𝑦‖=‖𝐴𝑦‖_[0,1]≤𝑟=‖𝑦‖,(5.24) so (5.21) is satisfied.
Next we show
‖𝐴𝑦‖≥‖𝑦‖for𝐾∩𝜕Ω₂.(5.25) To see this let 𝑦∈𝐾∩𝜕Ω₂ so ‖𝑦‖=‖𝑦‖_[0,1]=𝑅 and 𝑦(𝑡)≥(𝑡^𝛼−2(1−𝑡)²/(𝛼−1))𝑅 for 𝑡∈[0,1]. Also for 𝑡∈[0,1] we have 𝑦(𝑡)−𝜙(𝑡)=𝑦(𝑡)−𝜇𝑤(𝑡)≥𝑡^𝛼−2(1−𝑡)²𝛼−1𝑅−𝜇𝐶⁰𝑡^𝛼−2(1−𝑡)²≥𝑡𝛼−1_𝛼−2(1−𝑡)²𝛼−1𝑅1−𝜇𝑅𝐶⁰≥𝜖𝑡^𝛼−2(1−𝑡)²𝛼−1𝑅.(5.26) As a result 𝑦(𝑡)−𝜙(𝑡)≥𝜖𝜃^𝛼𝛼−1𝑅for𝑡∈[𝜃,1−𝜃].(5.27) We have (𝐴𝑦)(𝜎)=≥𝜇𝜇¹⁰𝐺_01−𝜃(𝜎,𝑠)𝑞(𝑠)𝑓^∗(𝑠,𝑦(𝑠)−𝜙(𝑠))𝑑𝑠=𝜇_01−𝜃𝐺(𝜎,𝑠)𝑞(𝑠)𝑔₁(𝑦(𝑠)−𝜙(𝑠))+ℎ₁(𝑦(𝑠)−𝜙(𝑠))𝑑𝑠𝐺(𝜎,𝑠)𝑞(𝑠)𝑔₁(𝑦(𝑠)−𝜙(𝑠))1+ℎ¹(𝑦(𝑠)−𝜙(𝑠))≥𝜇𝑔₁𝑔₁(𝑦(𝑠)−𝜙(𝑠))𝑑𝑠(𝑅)^01−𝜃𝐺(𝜎,𝑠)𝑞(𝑠)1+ℎ¹((𝜖𝜃^𝛼/(𝛼−1))𝑅)𝑔₁((𝜖𝜃^𝛼/(𝛼−1))𝑅)𝑑𝑠.(5.28) This together with (5.14) yields (𝐴𝑦)(𝜎)≥𝑅=‖𝑦‖.(5.29) Thus ‖𝐴𝑦‖≥‖𝑦‖, so (5.25) is held.
Now Theorem 2.5 implies that 𝐴 has a fixed-point 𝑦∈𝐾∩(Ω₂⧵Ω₁), that is, 𝑟≤‖𝑦‖=‖𝑦‖_[0.1]≤𝑅 and 𝑦(𝑡)≥𝑡^𝛼−2(1−𝑡)²𝑟 for 𝑡∈[0,1]. Thus 𝑦(𝑡) is a solution of (5.16) with 𝑦(𝑡)>𝜙(𝑡) for 𝑡∈(0,1). Thus (5.1) has a positive solution 𝑢(𝑡)=𝑦(𝑡)−𝜙(𝑡)> for 𝑡∈(0,1).

Example 5.3. Consider the boundary value problem 𝐃^𝛼0+𝑦𝑦((0𝑡))==𝜇𝑦(1𝑦)=^−𝑎(𝑦𝑡)_+𝑦^𝑏(𝑡)−1,0<𝑡<1,3<𝛼≤4,(0)=𝑦^(1)=0,0<𝑎<1<𝑏,(5.30) where 𝜇∈(0,𝜇₀) is such that (𝛼−1)²(𝛼−2)𝜇₀2Γ(𝛼)+(𝛼−1)2𝜇⁰𝑎₀^1/𝛼≤1,(5.31) here 𝑎₀=¹⁰𝑞(𝑠)𝑎(𝑠)𝑔𝑠^𝛼−2(1−𝑠)²𝑑𝑠=¹⁰𝑠^{1−𝑎(𝛼−1)}(1−𝑠)^(𝛼−1)𝑑𝑠<∞.(5.32)
Then (5.30) has a solution 𝑦 with 𝑦(𝑡)>0 for 𝑡∈(0,1).
To see this we will apply Theorem 5.2 with (here 𝑅>1 will be chosen later, in fact here we choose 𝑅>1 so that 𝜖=1/2 works, i.e., we choose 𝑅 so that 1−(𝜇/(2Γ(𝛼)𝑅))≥1/2),
a𝐾₀𝑔(𝑦)=𝑔₁(𝑦)=𝑦^−𝑎,ℎ(𝑦)=ℎ₁(𝑦)=𝑦^𝑏,𝑎(𝑡)=1,𝑒(𝑡)=𝑡^−1/2,=1,𝜖=12,𝜃=14,𝐶⁰=(𝛼−1)²(𝛼−2)Γ(𝛼)¹⁰𝑠^−1/2𝑑𝑠=(𝛼−2)²(𝛼−2)2Γ(𝛼).(5.33) Clearly (5.7)–(5.11) and (5.13) hold. Now (5.12) holds with 𝑟=1 since 𝜇𝐾₀𝑎₀=𝜇𝑎₀<𝜇₀𝑎₀≤121−𝜇⁰𝐶₀𝛼−1^𝑎≤121𝛼−−𝜇1𝑐⁰^𝑎=𝑟{1+(ℎ(𝑟)/𝑔(𝑟))}𝑔𝑟−𝜇𝑀𝐶₀/(𝛼−1),(5.34) from (5.31). Finally notice (5.14) is satisfied for 𝑅 large since 𝑅𝑔₁(𝜖𝜃^𝛼/(𝛼−1))𝑅𝑔₁(𝑅)𝑔₁((𝜖𝜃^𝛼/(𝛼−1))𝑅)+𝑔₁(𝑅)ℎ₁((𝜖𝜃^𝛼/(𝛼−1))𝑅)=1+(𝛼−1)𝑅^{1+𝑎−(𝑎+𝑏)}𝜖𝜃^𝛼^𝑎+𝑏𝑅^𝑎+𝑏⟶0,(5.35) as 𝑅→∞, since 𝑏>1. Thus all the conditions of Theorem 5.2 are satisfied so existence is guaranteed.

Acknowledgments

This paper is supported by Key Subject of Chinese Ministry of Education (no. 109051) and Scientific Research Fund of Heilongjiang Provincial Education Department (no. 11544032).