Existence of Positive Solutions for Fractional Differential Equation with Nonlocal Boundary Condition

Abstract

By using the fixed point theorem, existence of positive solutions for fractional differential equation with nonlocal boundary condition 𝐷^𝛼₀₊𝑢(𝑡)+𝑎(𝑡)𝑓(𝑡,𝑢(𝑡))=0, 0<𝑡<1, 𝑢(0)=0, 𝑢(1)=∑^∞_𝑖=1𝛼_𝑖𝑢(𝜉_𝑖) is considered, where 1<𝛼≤2 is a real number, 𝐷^𝛼₀₊ is the standard Riemann-Liouville differentiation, and 𝜉_𝑖∈(0,1),𝛼_𝑖∈[0,∞) with ∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1<1, 𝑎(𝑡)∈𝐶([0,1],[0,∞)),𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,∞),[0,∞)).

1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various sciences such as physics, mechanics, chemistry, and engineering. For details, see [1–6] and references therein.

It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations in terms of special functions [6–8]. Recently, there are some papers that deal with the existence and multiplicity of solution (or positive solution) of nonlinear initial fractional differential equation by the use of techniques of nonlinear analysis (fixed-point theorems, Leray-Shauder theory, etc.); see [9–17].

Recently, Bai and Lü [15] studied the existence of positive solutions of nonlinear fractional differential equation 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)(1)+𝑓𝑡,𝑢))=0,0<𝑡<1,=𝑢=0, (1.1) where 1<𝛼≤2 is a real number, 𝐷^𝛼₀₊ is the standard Riemann-Liouville differentiation, and 𝑓∶[0,1]×[0,∞)→[0,∞) is continuous.

In this paper, we study the existence of positive solutions for fractional differential equation with nonlocal boundary condition 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑓((𝑡𝑢(0)(1)=+𝑎𝑡,𝑢))=0,0<𝑡<1,=0,𝑢^∞_𝑖=1𝛼_𝑖𝑢𝜉_𝑖, (1.2) where 1<𝛼≤2 is a real number, 𝐷^𝛼₀₊ is the standard Riemann-Liouville differentiation, and 𝜉_𝑖∈(0,1),𝛼_𝑖∈[0,∞) with ∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1<1, 𝑎(𝑡)∈𝐶([0,1],[0,∞)),𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,∞),[0,∞)).

We assume the following conditions hold throughout the paper:(H1)𝜉_𝑖∈(0,1),𝛼_𝑖∈[0,∞) is both constants with ∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1<1,(H2)𝑎(𝑡)∈𝐶([0,1],[0,∞)), 𝑎(𝑡)≢0 on [𝑎,𝑏]⊂(0,1),(H3)𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,∞),[0,∞)).

Remark 1.1. To our knowledge, there are no results about the existence of positive solutions for problem (1.2).

2. The Preliminary Lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature.

Definition 2.1. The 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 the right side is pointwise defined on (0,∞).

Definition 2.2. The fractional derivative of order 𝛼>0 of a function 𝑦∶(0,∞)→𝑅 is given by 𝐷^𝛼₀₊𝑦(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀𝑦(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠, (2.2) where 𝑛=[𝛼]+1, provided the right side is pointwise defined on (0,∞).

Definition 2.3. The map 𝜃 is said to be a nonnegative continuous concave functional on a cone 𝑃 of a real Banach space 𝐸, provided that 𝜃∶𝑃→[0,∞) is continuous and 𝜃𝑡𝑥+(1−𝑡)𝑦≥𝑡𝜃(𝑥)+(1−𝑡)𝜃𝑦, (2.3) for all 𝑥,𝑦∈𝑃 and 0≤𝑡≤1.

Remark 2.4. As a basic example, we quote for 𝜆>−1, 𝐷^𝛼₀₊𝑡^𝜆=Γ(𝜆+1)Γ()𝑡𝜆−𝛼+1^𝜆−𝛼, (2.4) giving in particular 𝐷^𝛼₀₊𝑡^𝛼−𝑚=0,𝑚=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.
From Definition 2.2 and Remark 2.4, we then obtain the following.

Lemma 2.5. Let 𝛼>0. If one assumes 𝑢∈𝐶(0,1)∩𝐿(0,1), then the fractional differential equation 𝐷^𝛼₀₊𝑢(𝑡)=0 (2.5) has 𝑢(𝑡)=𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯+𝐶_𝑁𝑡^𝛼−𝑁,𝐶_𝑖∈𝑅,𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼, as unique solutions.

Lemma 2.6. Assume that 𝑢∈𝐶(0,1)∩𝐿(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝑢∈𝐶(0,1)∩𝐿(0,1). Then, 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)=u(𝑡)+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯+𝐶_𝑁𝑡^𝛼−𝑁, (2.6) for some 𝐶_𝑖∈𝑅,𝑖=1,2,…,𝑁.

Lemma 2.7 (see [15]). Given 𝑦∈𝐶[0,1] and 1<𝛼≤2, the unique solution of 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)(1)+𝑦=0,0<𝑡<1,=𝑢=0 (2.7) is 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (2.8) where 𝐺()=[𝑡(𝑡,𝑠1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1Γ(𝛼)[𝑡(,0≤𝑠≤𝑡≤1,1−𝑠)]^𝛼−1Γ(𝛼),0≤𝑡≤𝑠≤1. (2.9)

Lemma 2.8. Suppose (𝐻1) holds. Given 𝑦∈𝐶[0,1] and 1<𝛼≤2, the unique solution of 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)(1)=+𝑦=0,0<𝑡<1,=0,𝑢^∞_𝑖=1𝛼_𝑖𝑢𝜉_𝑖 (2.10) is 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠+𝐵𝑦𝑡^𝛼−1, (2.11) where 𝐺()=[𝑡(𝑡,𝑠1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1Γ(𝛼)[𝑡(,0≤𝑠≤𝑡≤1,1−𝑠)]^𝛼−1Γ(𝛼)𝐵𝑦=∑,0≤𝑡≤𝑠≤1,_∞𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑦(𝑠)𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1. (2.12)

Proof. By applying Lemmas 2.6 and 2.7, we have 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2. (2.13) Because ^∞_𝑖=1𝛼_𝑖¹₀𝐺𝜉_𝑖∑,𝑠𝑑𝑠=^∞𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−11−𝜉_𝑖𝛼Γ(𝛼),𝛼_𝑖𝜉_𝑖^𝛼−11−𝜉_𝑖<𝛼_𝑖𝜉_𝑖^𝛼−1, (2.14) by (H1), ∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1(1−𝜉_𝑖) is converge. Therefore, ∑^∞_𝑖=1𝛼_𝑖∫¹₀𝐺(𝜉_𝑖,𝑠)𝑑𝑠 is converge. 𝑦(𝑡) is continuous function on [0,1], so ∑^∞_𝑖=1𝛼_𝑖∫¹₀𝐺(𝜉_𝑖,𝑠)𝑦(𝑠)𝑑𝑠 is converge.
By 𝑢(0)=0, 𝑢(1)=∑^∞_𝑖=1𝛼_𝑖𝑢(𝜉_𝑖), there are 𝐶₂=0,𝐶₁=(∑^∞_𝑖=1𝛼_𝑖∫¹₀𝐺(𝜉_𝑖,𝑠)𝑦(𝑠)𝑑𝑠)/(1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1). Therefore, 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠+𝐵𝑦𝑡^𝛼−1,𝐵𝑦=∑^∞_𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑦(𝑠)𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1. (2.15)

Lemma 2.9 (see [15]). The function 𝐺(𝑡,𝑠) defined by (2.9) satisfies the following conditions:(1)𝐺(𝑡,𝑠)>0, for 𝑡,𝑠∈(0,1),(2)there exists a positive function 𝛾∈𝐶(0,1) such thatmin_{()()1/4≤𝑡≤3/4}𝐺(𝑡,𝑠)≥𝛾(𝑠)max_0≤𝑡≤1𝐺(𝑡,𝑠)=𝛾(𝑠)𝐺(𝑠,𝑠),0<𝑠<1. (2.16)

Lemma 2.10 (see [18]). Let E be a Banach space, 𝑃⊆𝐸 a cone and Ω₁,Ω₂ two bounded open sets of 𝐸 with 0∈Ω₁⊂Ω₁⊂Ω₂. Suppose that 𝐴∶𝑃∩(Ω₂⧵Ω₁)→𝑃 is a completely continuous operator such that either(i)‖𝐴𝑥‖≤‖𝑥‖,𝑥∈𝑃∩𝜕Ω₁ and ‖𝐴𝑥‖≥‖𝑥‖,𝑥∈𝑃∩𝜕Ω₂, or(ii)‖𝐴𝑥‖≥‖𝑥‖,𝑥∈𝑃∩𝜕Ω₁ and ‖𝐴𝑥‖≤‖𝑥‖,𝑥∈𝑃∩𝜕Ω₂holds. Then, 𝐴 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

Lemma 2.11 (see [19]). Let 𝑃 be a cone in real Banach space 𝐸,𝑃_𝑐={𝑥∈𝑃∣‖𝑥‖≤𝑐},𝜃 a nonnegative continuous concave functional on 𝑃 such that 𝜃(𝑥)≤‖𝑥‖, for all 𝑥∈𝑃_𝑐, and 𝑃(𝜃,𝑏,𝑑)={𝑥∈𝑃∣𝑏≤𝜃(𝑥),‖𝑥‖≤𝑑}. Suppose that 𝐴∶𝑃_𝑐→𝑃_𝑐 is completely continuous, and there exist constants 0<𝑎<𝑏<𝑑≤𝑐 such that(C1){𝑥∈𝑃(𝜃,𝑏,𝑑)∣𝜃(𝑥)>𝑏}≠∅, and 𝜃(𝐴𝑥)>𝑏,𝑥∈𝑃(𝜃,𝑏,𝑑),(C2)‖𝐴𝑥‖≤𝑎, for 𝑥≤𝑎,(C3)𝜃(𝐴𝑥)>𝑏 for 𝑥∈𝑃(𝜃,𝑏,𝑐) with ‖𝐴𝑥‖>𝑑.Then, 𝐴 has at least three fixed points 𝑥₁,𝑥₂,𝑥₃ with 𝑥₁<𝑎,𝑏<𝜃𝑥₂,𝑎<𝑥₃,𝜃𝑥₃<𝑏. (2.17)

Remark 2.12. If there holds 𝑑=𝑐, then condition (C1) of Lemma 2.11 implies condition (C3) of Lemma 2.11.

3. The Main Results

Let 𝐸=𝐶[0,1] be endowed with the ordering 𝑢≤𝑣 if 𝑢(𝑡)≤𝑣(𝑡) for all 𝑡∈[0,1], and the maximum norm, ‖𝑢‖=max_0≤𝑡≤1|𝑢(𝑡)|. Define the cone 𝑃⊂𝐸 by 𝑃={𝑢∈𝐸∣𝑢(𝑡)≥0}.

Let the nonnegative continuous concave functional 𝜃 on the cone 𝑃 be defined by 𝜃(𝑢)=min_{(1/4)≤𝑡≤(3/4)}|𝑢(𝑡)|.

Lemma 3.1 (see [15]). Let 𝑇∶𝑃→𝐸 be the operator defined by 𝑇𝑢(𝑡)∶=∫¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠, then 𝑇∶𝑃→𝑃 is completely continuous.

Lemma 3.2. Let 𝐴∶𝑃→𝐸 be the operator defined by 𝐴𝑢(𝑡)∶=¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝐵𝑎(⋅)𝑓(⋅,𝑢(⋅))𝑡^𝛼−1, (3.1) then 𝐴∶𝑃→𝑃 is completely continuous.

Proof. The proof is similar to Lemma 3.1, so we omit.
Denote 𝑀=¹₀𝐺()𝑎(𝑠)∑𝑠,𝑠𝑑𝑠+^∞𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑎(𝑠)𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1−1,𝑁=^3/4_1/4𝛾(𝑠)𝐺(𝑠,𝑠)𝑎(𝑠)𝑑𝑠⁻¹. (∗)

Theorem 3.3. Assume (H1)–(H3) hold, and there exist two positive constants 𝑟₂>𝑟₁>0 such that(1)𝑓(𝑡,𝑢)≤𝑀𝑟₂,forall𝑡∈[0,1],𝑢∈[0,𝑟₂], (2)𝑓(𝑡,𝑢)≥𝑁𝑟₁,forall𝑡∈[0,1],𝑢∈[0,𝑟₁], where 𝑀,𝑁 is defined in (*),then problem (1.2) has at least one positive solution 𝑢 such that 𝑟₁≤‖𝑢‖≤𝑟₂.

Proof. By Lemmas 2.8 and 3.2, we know 𝐴∶𝑃→𝑃 is completely continuous, and problem (1.2) has a solution 𝑢=𝑢(𝑡) if and only if 𝑢 solves the operator equation 𝑢=𝐴𝑢. In order to apply Lemma 2.10, we separate the proof into the following two steps.
Step 1. Let Ω₂={𝑢∈𝑃∣‖𝑢‖≤𝑟₂}. For 𝑢∈𝜕Ω₂, we have 0≤𝑢(𝑡)≤𝑟₂ for all 𝑡∈[0,1]. It follows from (1) that for 𝑡∈[0,1], ‖𝐴𝑢‖≤¹₀𝐺()𝑎(𝑠)𝑓((𝑠∑𝑠,𝑠𝑠,𝑢))𝑑𝑠+^∞𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠1−∑^∞_i=1𝛼_𝑖𝜉_𝑖^𝛼−1≤𝑀𝑟₂¹₀𝐺()𝑎(𝑠)∑𝑠,𝑠𝑑𝑠+^∞𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑎(𝑠)𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1=𝑟₂=‖𝑢‖. (3.2) Therefore, ‖𝐴𝑢‖≤‖𝑢‖,𝑢∈𝑃∩𝜕Ω₂. (3.3)Step 2. Let Ω₁={𝑢∈𝑃∣‖𝑢‖≤𝑟₁}. For 𝑢∈𝜕Ω₁, we have 0≤𝑢(𝑡)≤𝑟₁ for all 𝑡∈[0,1]. By assumption (2), for 𝑡∈[1/4,3/4], there is (𝐴𝑢)(𝑡)=¹₀𝐺()𝑎(𝑠)𝑓((𝑠𝑡𝑡,𝑠𝑠,𝑢))𝑑𝑠+^𝛼−1∑^∞_𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1≥¹₀𝛾(𝑠)𝐺(𝑠,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≥𝑁𝑟^3/4_1/4𝛾(𝑠)𝐺(𝑠,𝑠)𝑎(𝑠)𝑑𝑠=𝑟₁=‖𝑢‖. (3.4) So, ‖𝐴𝑢‖≥‖𝑢‖,𝑢∈𝑃∩𝜕Ω₁. (3.5) Therefore, by (ii) of Lemma 2.10, we complete the proof.

Example 3.4. Consider the problem 𝐷^3/2₀₊𝑢(𝑡)+𝑢²+sin𝑡4+15𝑢(0)(1)==0,0<𝑡<1,=0,𝑢^∞_𝑖=1𝛼_𝑖𝑢𝜉_𝑖, (3.6) where ∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^1/2=1/5.
A simple computation showed 𝑀≥1.4,𝑁≈13.6649. Choosing 𝑟₁=(1/70),𝑟₂=(9/10), we have 𝑓(𝑡,𝑢)=𝑢²+sin𝑡4+15≤1.2207≤𝑀𝑟₂,()∈[]×9𝑡,𝑢0,10,,𝑓()10𝑡,𝑢=𝑢²+sin𝑡4+15≥15≥𝑁𝑟₁,()∈[]×1𝑡,𝑢0,10,.70 (3.7)
With the use of Theorem 3.3, problem (3.6) has at least one positive solutions 𝑢 such that (1/70)≤‖𝑢‖≤(9/10).

Theorem 3.5. Assume (H1)–(H3) hold, and there exist constants 0<𝑎<𝑏<𝑐 such that the following assumptions hold:(A1)𝑓(𝑡,𝑢)<𝑀𝑎 for (𝑡,𝑢)∈[0,1]×[0,𝑎],(A2)𝑓(𝑡,𝑢)≥𝑁𝑏 for (𝑡,𝑢)∈[1/4,3/4]×[𝑏,𝑐],(A3)𝑓(𝑡,𝑢)≤𝑀𝑐 for (𝑡,𝑢)∈[0,1]×[0,𝑐], where 𝑀,𝑁 is defined in (*).Then, the boundary value problem (1.2) has at least three positive solutions 𝑢₁,𝑢₂,𝑢₃ with 𝑢₁<𝑎,𝑏<min_{()()1/4≤𝑡≤3/4}𝑢₂<𝑢₂𝑢≤𝑐,𝑎<₃≤𝑐,min_{()()1/4≤𝑡≤3/4}𝑢₃<𝑏. (3.8)

Proof. We show that all the conditions of Lemma 2.9 are satisfied.
If 𝑢∈𝑃_𝑐, then ‖𝑢‖≤𝑐. Assumption (A3) implies 𝑓(𝑡,𝑢(𝑡))≤𝑀𝑐 for 0≤𝑡≤1. Consequently, ‖𝐴𝑢‖=max_0≤𝑡≤1¹₀𝐺()𝑎(𝑠)𝑓((𝑠𝑡𝑡,𝑠𝑠,𝑢))𝑑𝑠+^𝛼−1∑^∞_𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1≤¹₀𝐺()𝑎(𝑠)𝑓((𝑠𝑡𝑠,𝑠𝑠,𝑢))𝑑𝑠+^𝛼−1∑^∞_𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1≤¹₀𝐺()𝑎(𝑠)∑𝑠,𝑠𝑑𝑠+^∞𝑖=1𝛼_𝑖∫¹₀𝐺𝜉_𝑖,𝑠𝑎(𝑠)𝑑𝑠1−∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^𝛼−1‖𝑢‖≤‖𝑢‖. (3.9) Hence, 𝐴∶𝑃_𝑐→𝑃_𝑐. In the same way, if 𝑢∈𝑃_𝑎, then assumption (A1) yields 𝑓(𝑡,𝑢(𝑡))<𝑀𝑎,0≤𝑡≤1. Therefore, condition (C2) of Lemma 2.11 is satisfied.
To check condition (C1) of Lemma 2.11, we choose 𝑢(𝑡)=(𝑏+𝑐)/2,0≤𝑡≤1. It is easy to see that 𝑢(𝑡)=(𝑏+𝑐)/2∈𝑃(𝜃,𝑏,𝑐),𝜃(𝑢)=(𝜃(𝑏+𝑐))/2>𝑏, and consequently, {𝑢∈𝑃(𝜃,𝑏,𝑑)∣𝜃(𝑢)>𝑏}≠∅ Hence, if 𝑢∈𝑃(𝜃,𝑏,𝑐), then 𝑏≤𝑢(𝑡)≤𝑐 for (1/4)≤𝑡≤(3/4). From assumption (A2), we have 𝑓(𝑡,𝑢(𝑡))≥𝑁𝑏 for (1/4)≤𝑡≤(3/4). So, 𝜃(𝐴𝑢)=min_{()()1/4≤𝑡≤3/4}|(𝐴𝑢)(𝑡)|≥¹₀𝛾(𝑠)𝐺(𝑠,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠>𝑁𝑏^3/4_1/4𝛾(𝑠)𝐺()𝑎(𝑠)‖𝑢‖𝑠,𝑠𝑑𝑠=𝑏= (3.10)𝜃(𝐴𝑢)>𝑏, for all 𝑢∈𝑃(𝜃,𝑏,𝑐).
This shows that condition (C1) of Lemma 2.11 is also satisfied.
By Lemma 2.11 and Remark 2.12, the boundary value problem (1.2) has at least three positive solutions 𝑢₁,𝑢₂,and𝑢₃ with 𝑢₁<𝑎,𝑏<min_{()1/4≤𝑡≤(3/4)}𝑢₂,𝑎<𝑢₃,min_{()1/4≤𝑡≤(3/4)}𝑢₃<𝑏. (3.11) The proof is complete.

Example 3.6. Consider the problem 𝐷^3/2₀₊𝑢(𝑡)()𝑢(0)(1)=+𝑓𝑡,𝑢=0,0<𝑡<1,=0,𝑢^∞_𝑖=1𝛼_𝑖𝑢𝜉_𝑖, (3.12) where ∑^∞_𝑖=1𝛼_𝑖𝜉_𝑖^1/2=(1/5), 𝑓()=𝑡𝑡,𝑢40+14𝑢²𝑡,𝑢≤1,13+40+𝑢,𝑢>1. (3.13)
We have 𝑀≥1.4,𝑁≈13.6649. Choosing 𝑎=(1/14),𝑏=1,𝑐=36, there hold 𝑓()=𝑡𝑡,𝑢40+14𝑢²()∈[]×1≤0.097≤𝑀𝑎,𝑡,𝑢0,10,,𝑓()𝑡14𝑡,𝑢=13+()∈140+𝑢≥14.025≥𝑁𝑏≈13.7,𝑡,𝑢4,34×[],𝑓()𝑡1,36𝑡,𝑢≤13+()∈[]×[].40+𝑢≤48.025≤𝑀𝑐≈50.4,𝑡,𝑢0,10,36 (3.14) With the use of Theorem 3.5, problem (3.12) has at least three positive solutions 𝑢₁,𝑢₂, and 𝑢₃ with max_0≤𝑡≤1𝑢₁(𝑡<1)14,1<min_{()()1/4≤𝑡≤3/4}𝑢₂(𝑡)<max_0≤𝑡≤1𝑢₂(𝑡1)≤36,14<max_0≤𝑡≤1𝑢₃(𝑡)≤36,min_{()()1/4≤𝑡≤3/4}𝑢₃(𝑡)<1. (3.15)

Acknowledgment

This work is sponsored by the NSFC (11101335, 11026060, Gansu Provincial Department of Education Fund (1101-02), NWNU-KJCXGC (03-69).