Multiple Positive Solutions of Singular Nonlinear Sturm-Liouville Problems with Carathéodory Perturbed Term

Abstract

By employing a well-known fixed point theorem, we establish the existence of multiple positive solutions for the following fourth-order singular differential equation 𝐿𝑢=𝑝(𝑡)𝑓(𝑡,𝑢(𝑡),𝑢^′′(𝑡))−𝑔(𝑡,𝑢(𝑡),𝑢^′′(𝑡)),0<𝑡<1,𝛼₁𝑢(0)−𝛽₁𝑢′(0)=0,𝛾₁𝑢(1)+𝛿₁𝑢′(1)=0,𝛼₂𝑢^′′𝛽(0)−₂𝑢^′′′(0)=0,𝛾₂𝑢^′′(1)+𝛿₂𝑢^′′′(1)=0, with 𝛼_𝑖,𝛽_𝑖,𝛾_𝑖,𝛿_𝑖≥0 and 𝛽_𝑖𝛾_𝑖+𝛼_𝑖𝛾_𝑖+𝛼_𝑖𝛿_𝑖>0,𝑖=1,2, where 𝐿 denotes the linear operator 𝐿𝑢∶=(𝑟𝑢^′′′)′−𝑞𝑢^′′,𝑟∈𝐶¹([0,1],(0,+∞)), and 𝑞∈𝐶([0,1],[0,+∞)). This equation is viewed as a perturbation of the fourth-order Sturm-Liouville problem, where the perturbed term 𝑔∶(0,1)×[0,+∞)×(−∞,+∞)→(−∞,+∞) only satisfies the global Carathéodory conditions, which implies that the perturbed effect of 𝑔 on 𝑓 is quite large so that the nonlinearity can tend to negative infinity at some singular points.

1. Introduction

In this paper, we consider the existence of multiple positive solutions for the following fourth-order singular Sturm-Liouville boundary value problem involving a perturbed term𝐿𝑢=𝑝(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)−𝑔𝑡,𝑢(𝑡),𝑢^′′(𝑡)𝛼,0<𝑡<1,₁𝑢(0)−𝛽₁𝑢^′(0)=0,𝛾₁𝑢(1)+𝛿₁𝑢^′(1)𝛼=0,₂𝑢^′′(0)−𝛽₂𝑢^′′′(0)=0,𝛾₂𝑢^′′(1)+𝛿₂𝑢^′′′(1)=0, (1.1) where 𝛼_𝑖,𝛽_𝑖,𝛾_𝑖,𝛿_𝑖≥0 and 𝛽_𝑖𝛾_𝑖+𝛼_𝑖𝛾_𝑖+𝛼_𝑖𝛿_𝑖>0,𝑖=1,2, and 𝐿 denotes the linear operator 𝐿𝑢∶=(𝑟𝑢^′′′)^′−𝑞𝑢^′′,𝑟∈𝐶¹([0,1],(0,+∞)) and 𝑞∈𝐶([0,1],[0,+∞)) and 𝑞∈𝐶([0,1],[0,+∞)). The perturbed term, 𝑔∶(0,1)×[0,+∞)×(−∞,+∞)→[0,+∞), satisfies global Carathéodory’s conditions.

Equation (1.1) arises from many branches of applied mathematics and physics; for details, see [1–16]. It mainly describes the deformation of an elastic beam for 𝑔(𝑡,𝑢,𝑢^′′)≡0; for example, under the Lidstone boundary condition,𝑢(0)=𝑢(1)=𝑢^′′(0)=𝑢^′′(1)=0, (1.2) problem (1.1) is used to model such phenomena as the deflection of an elastic beam simply supported at the endpoints; see [1, 3, 5, 7–11]. Also, if the boundary condition of (1.1) is a Focal boundary condition, then it describes the deflection of an elastic beam having both end-points fixed, or having one end simply supported and the other end clamped with sliding clamps. In addition, the derivative 𝑢^′′ in 𝑓 is the bending moment term which represents the bending effect, see [1, 3, 5, 7–11, 13, 14, 16]. A brief discussion of the physical interpretation under some boundary conditions associated with the linear beam equation can be found in Zill and Cullen [17].

Recently, for the case where the nonlinearity 𝑓 does not contain the bending moment term 𝑢^′′, Ma and Wang [1] studied the existence of positive solutions for (1.1) subject to boundary conditions 𝑢(0)=𝑢(1)=𝑢^′′(0)=𝑢^′′(1)=0 and 𝑢(0)=𝑢^′(1)=𝑢^′′(0)=𝑢^′′′(1)=0 if 𝑓 is superlinear or sublinear. In the case where 𝑓 contains the bending moment term 𝑢^′′ and under the particular boundary conditions, the authors of papers [9, 12] studied the existence of positive solutions for (1.1) when 𝑓 satisfies the following growth condition:𝑓𝑡,𝑥,𝑦−𝛼𝑥−𝛽𝑦≤𝑎|𝑥|+𝑏𝑦+𝑐, (1.3) where 𝛼,𝛽∈ℝ𝑎,𝑏,𝑐>0,𝑎,𝑏 is enough small. But most of the above works were done on base of the assumptions that the nonlinearity is nonnegative and has no any singularity. In recent years, one found that the fourth-order changing-sign nonlinear problems also occur to the classical model for the elastic beam fixed at both ends, especially in the medium span or large span bridge constructions, this implies that it is necessary and quite natural to study fourth-order changing-sign boundary value problems.

In this paper, we focus on the particularly difficult and interesting situation, when (1.1) is singularly perturbed, so that the nonlinearity is allowed to change sign, even may tend to negative infinity. This problem has essential difference from those unperturbed problems of [1–16]. We quote in the sequel some papers from the relevant bibliography devoted to this subject. In [18], Loud considered the existence of 𝑇-periodic solutions for a first-order perturbed system of ordinary differential equations by employing the so-called bifurcation function𝑓₀(𝜃)=^𝑇₀𝑧₀(𝜏),𝜙𝜏−𝜃,𝑥₀(𝜏),0𝑑𝜏. (1.4) Moreover, the author of [18] also considered the case when 𝜃₀ is not a simple zero of 𝑓₀, and the existence of 𝑇-periodic solutions of the above problem is associated with the existence of the roots of a certain quadratic equation. Recently, by using the exponential dichotomies and contraction mapping principle, Xia et al. [19] established some sufficient conditions of the existence and uniqueness of almost periodic solution for a forced perturbed system with piecewise constant argument. The other works, such as Khanmamedov [20], Wu and Gan [21], Makarenkov and Nistri [22], Liu and Yang [23], Clavero et al. [24], and Cui and Geng [25], are rich sources for application of perturbed problems.

Our main tool used for the analysis here is known as Guo-Krasnoselskii’s fixed point theorem, for the convenience of the reader, we now state it as follows.

Lemma 1.1 (see, [26]). Let E be a real Banach space, 𝑃⊂𝐸 a cone. Assume Ω₁,Ω₂ are two bounded open subsets of 𝐸 with 𝜃∈Ω₁,Ω₁⊂Ω₂, and let 𝑇∶𝑃∩(Ω₂⧵Ω₁)→𝑃 be a completely continuous operator such that either(1)‖𝑇𝑥‖≤‖𝑥‖,𝑥∈𝑃∩𝜕Ω₁, and ‖𝑇𝑥‖≥‖𝑥‖,𝑥∈𝑃∩𝜕Ω₂, or(2)‖𝑇𝑥‖≥‖𝑥‖,𝑥∈𝑃∩𝜕Ω₁, and ‖𝑇𝑥‖≤‖𝑥‖,𝑥∈𝑃∩𝜕Ω₂.Then, 𝑇 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

2. Preliminaries and Lemmas

The following definition introduces global Carathéodory’s conditions imposed on a map.

Definition 2.1. A map 𝑔∶(𝑡,𝑥,𝑦)↦𝑔(𝑡,𝑥,𝑦) is said to satisfy global Crathéodory’s conditions if the following conditions hold: (i)for each (𝑥,𝑦)∈ℝ×ℝ, the mapping 𝑡↦𝑔(𝑡,𝑥,𝑦) is Lebesgue measurable; (ii)for a.e. 𝑡∈[0,1], the mapping (𝑥,𝑦)↦𝑔(𝑡,𝑥,𝑦) is continuous on ℝ×ℝ;(iii)there exists a 𝜌∈𝐿¹[0,1] such that, for a.e. 𝑡∈[0,1] and (𝑥,𝑦)∈ℝ×ℝ, we have 𝑔𝑡,𝑥,𝑦≤𝜌(𝑡). (2.1)

The following lemmas play an important role in proving our main results.

Lemma 2.2 (see, [27]). Let 𝜓₂ and 𝜙₂ be the solutions of the linear problems −𝑟(𝑡)𝜙^′₂(𝑡)′+𝑞(𝑡)𝜙₂(𝑡)𝜙=0,0<𝑡<1,₂(0)=𝛽₂,𝜙^′₂(0)=𝛼₂,−𝑟(𝑡)𝜓^′₂(𝑡)′+𝑞(𝑡)𝜓₂(𝑡)𝜓=0,0<𝑡<1,₂(0)=𝛿₂,𝜓^′₂(0)=−𝛾₂, (2.2) respectively. Then, (i)𝜙₂ is strictly increasing on [0,1] and 𝜙₂(𝑡)>0 on (0,1]; (ii)𝜓₂ is strictly decreasing on [0,1] and 𝜓₂(𝑡)>0 on [0,1).

Set𝑤₂=−𝑟(𝑡)𝜙₂(𝑡)𝜓^′₂(𝑡)−𝜓₂(𝑡)𝜙^′₂(𝑡), (2.3) by Liouville’s formula, one can easily show 𝑤₂= constant >0.

As [27], we define Green’s function for the BVP:−𝑟(𝑡)𝑢^′(𝑡)^′(𝑡)𝑢(𝑡)𝛼+𝑞=0,0<𝑡<1,₂𝑢(0)−𝛽₂𝑢^′(0)=0,𝛾₂𝑢(1)+𝛿₂𝑢^′(1)=0, (2.4) by𝐺₂()=1𝑡,𝑠𝑤₂𝜙₂(𝑡)𝜓₂(𝑠)𝜙,0≤𝑡≤𝑠≤1,₂(𝑠)𝜓₂(𝑡),0≤𝑠≤𝑡≤1, (2.5) then we have the following lemma.

Lemma 2.3. For any (𝑡,𝑠)∈[0,1]×[0,1],𝑖=1,2, we have 𝜃₂𝐺₂(𝑠,𝑠)𝐺₂(𝑡,𝑡)≤𝐺₂(𝑡,𝑠)≤𝐺₂(𝑠,𝑠),or𝐺₂(𝑡,𝑡), (2.6) where 𝜃₂=𝑤₂𝜙₂(1)𝜓₂(0). (2.7)

Proof. It follows from the monotonicity of 𝜙₂(𝑡) and 𝜓₂(𝑡) that the right-hand side of (2.6) holds. For the left hand side, by the monotonicity of 𝜙₂(𝑡) and 𝜓₂(𝑡), we have 𝐺₂()=1𝑡,𝑠𝑤₂𝜙₂(𝑡)𝜓₂(𝑠)𝜙,0≤𝑡≤𝑠≤1,₂(𝑠)𝜓₂(𝑡)≥1,0≤𝑠≤𝑡≤1,𝑤₂𝜙₂(𝑡)𝜓₂(𝑠)𝜙₂(𝑠)𝜓₂(𝑡)𝜙₂(1)𝜓₂(0)𝜙,0≤𝑡≤𝑠≤1,₂(𝑠)𝜓₂(𝑡)𝜙₂(𝑡)𝜓₂(𝑠)𝜙₂(1)𝜓₂(0)=𝑤,0≤𝑠≤𝑡≤1,₂𝜙₂(1)𝜓₂(0)𝐺₂(𝑡,𝑡)𝐺₂(𝑠,𝑠),=𝜃₂𝐺₂(𝑡,𝑡)𝐺₂(𝑠,𝑠). (2.8) The proof is completed.

Also, it is well known the Green function for the boundary value problem−𝑢^′′𝛼=0,0<𝑡<1,₁𝑢(0)−𝛽₁𝑢^′(0)=0,𝛾₁𝑢(1)+𝛿₁𝑢^′(1)=0, (2.9) is𝐺₁()=1𝑡,𝑠𝑤₁𝛽₁+𝛼₁𝑠𝛾₁+𝛿₁()𝛽1−𝑡,0≤𝑠≤𝑡≤1,₁+𝛼₁𝑡𝛾₁+𝛿₁(1−𝑠),0≤𝑡≤𝑠≤1, (2.10) where 𝑤₁=𝛽₁𝛾₁+𝛼₁𝛾₁+𝛼₁𝛿₁. Let𝑒(𝑡)=1𝑤₁𝛽₁+𝛼₁𝑡𝛾₁+𝛿₁(1−𝑡), (2.11) clearly,𝑒(𝑡)𝑒(𝑠)≤𝐺₁(𝑡,𝑠)≤𝑒(𝑠). (2.12) Now, we define an integral operator 𝑆∶𝐶[0,1]→[0,1] by𝑆𝑣(𝑡)=¹₀𝐺₁(𝑡,𝜏)𝑣(𝜏)𝑑𝜏, (2.13) and, then, by (2.9), we have(𝑆𝑣)^′′(𝑡)(𝑡)𝛼=−𝑣,0<𝑡<1,₁(𝑆𝑣)(0)−𝛽₁(𝑆𝑣)^′(0)=0,𝛾₁(𝑆𝑣)(1)+𝛿₁(𝑆𝑣)^′(1)=0. (2.14)

In order to obtain existence of positive solutions to problem (1.1), we will consider the existence of positive solutions to the following modified problem−𝑟𝑣^′′(𝑡)(𝑡)(𝑡)𝑓((𝑡)(𝑡((𝑡)(𝑡𝛼+𝑞𝑣=𝑝𝑡,𝑆𝑣,−𝑣))−𝑔𝑡,𝑆𝑣,−𝑣)),0<𝑡<1,₂𝑣(0)−𝛽₂𝑣^′(0)=0,𝛾₂𝑣(1)+𝛿₂𝑣^′(1)=0. (2.15)

Lemma 2.4. Let 𝑢(𝑡)=𝑆𝑣(𝑡),𝑣(𝑡)∈𝐶[0,1]. Then, we can transform (1.1) into (2.15). Moreover, if 𝑣∈𝐶([0,1],[0,+∞) is a solution of problem (2.15), then the function 𝑢(𝑡)=𝑆𝑣(𝑡) is a positive solution of problem (1.1).

Proof. It follows from (2.9) that 𝑢^′′(𝑡)=−𝑣(𝑡), put 𝑢^′′(𝑡)=−𝑣(𝑡) and 𝑢(𝑡)=𝑆𝑣(𝑡) into (1.1), we can transform (1.1) into (2.15).
Conversely, if 𝑣∈𝐶([0,1],[0,+∞)) is a solution of (2.15), let 𝑢(𝑡)=𝑆𝑣(𝑡), we have 𝑢^′′(𝑡)=−𝑣(𝑡), thus 𝑢=𝑆𝑣 is a solution of (1.1). The proof of Lemma 2.4 is completed.

In the rest of the paper, we always suppose that the following assumptions hold. (B1)𝑝∶(0,1)→[0,+∞) is continuous and satisfies 0<¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠<+∞. (2.16)(B2)𝑓∶[0,1]×[0,+∞)×(−∞,+∞)→[0,+∞) is continuous. (B3)𝑔∶[0,1]×[0,+∞)×(−∞,+∞)→[0,+∞) satisfies global Crathéodory’s condition and ¹₀𝜌(𝑠)𝑑𝑠>0. (2.17)

Remark 2.5. It follows from (B1), (B3) and from the monotonicity of 𝜙₂(𝑡),𝜓₂(𝑡) that there exists (𝑎,𝑏)⊂(0,1) such that 0<^𝑏_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤^𝑏_𝑎𝐺₂(𝑝(𝑠)(𝑠≤𝑠,𝑠)+𝜌)𝑑𝑠₁₀𝐺₂(𝑝(𝑠)(𝑠≤𝑠,𝑠)+𝜌)𝑑𝑠₁₀𝐺₂()𝑝(𝑠)𝜙𝑠,𝑠𝑑𝑠+²(1)𝜓₂(0)𝑤₂¹₀𝜌(𝑠)𝑑𝑠<+∞. (2.18) So for convenience, in the rest of this paper, we define serval constants as follows: 𝐾=¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)+𝜌(𝑠)𝑑𝑠,𝜇₂=𝜙₂(𝑎)𝜓₂(𝑏)𝜙₂(1)𝜓₂(0),𝑙=𝜇₂^𝑏_𝑎𝐺₂()𝑝(𝑠)𝜇𝑠,𝑠𝑑𝑠,₁=𝛽₁+𝛼₁𝑎𝛾₁+𝛿₁(1−𝑏)𝜃₂𝑤₁¹₀𝐺₁(𝑠,𝑠)𝐺₂(𝑠,𝑠)𝑑𝑠. (2.19)

Lemma 2.6. Assume (B3) is satisfied. Then, the boundary value problem −𝑟(𝑡)𝑦^′′(𝑡)(𝑡)𝑦(𝑡)(𝑡)𝛼+𝑞=𝜌,0<𝑡<1,₂𝑦(0)−𝛽₂𝑦^′(0)=0,𝛾₂𝑦(1)+𝛿₂𝑦^′(1)=0, (2.20) has a unique solution 𝑦(𝑡)=¹₀𝐺₂(𝑡,𝑠)𝜌(𝑠)𝑑𝑠, (2.21) which satisfies 𝑦(𝑡)≤𝐺₂(𝑡,𝑡)¹₀𝜌(𝑠)𝑑𝑠. (2.22)

Proof. First, 𝑦(𝑡)=∫¹₀𝐺₂(𝑡,𝑠)𝜌(𝑠)𝑑𝑠 solves the BVP (2.20), and it is the unique solution of the BVP (2.20), since −(𝑟(𝑡)𝑦^′)^′(𝑡)+𝑞(𝑡)𝑦(𝑡)=0 with boundary conditions 𝛼₂𝑦(0)−𝛽₂𝑦^′(0)=0,𝛾₂𝑦(1)+𝛿₂𝑦^′(1)=0 (2.23) only has a trivial solution. Finally, it follows from (2.6) and (B3) that (2.22) holds.

Define a modified function [⋅]^∗ for any 𝑧∈𝐶[0,1] by[𝑧(𝑡)]^∗=𝑧(𝑡)(𝑡)(𝑡),𝑧≥0,0,𝑧<0. (2.24) We consider the following approximating problem−𝑟(𝑡)𝑥^′′(𝑡)+𝑞(𝑡)𝑥(𝑡)=𝑝(𝑡)𝑓𝑡,𝑆𝑥(𝑡)−𝑦(𝑡)^∗,−𝑥(𝑡)−𝑦(𝑡)^∗−𝑔𝑡,𝑆𝑥(𝑡)−𝑦(𝑡)^∗,−𝑥(𝑡)−𝑦(𝑡)^∗(𝑡)𝛼+𝜌,0<𝑡<1,₂𝑥(0)−𝛽₂𝑥^′(0)=0,𝛾₂𝑥(1)+𝛿₂𝑥^′(1)=0. (2.25)

Lemma 2.7. If 𝑥(𝑡)≥𝑦(𝑡) for any 𝑡∈[0,1] is a positive solution of the BVP (2.25), then 𝑆(𝑥−𝑦) is a positive solution of the singular perturbed differential equation (1.1).

Proof. In fact, if 𝑥 is a positive solution of the BVP (2.25) such that 𝑥(𝑡)≥𝑦(𝑡) for any 𝑡∈[0,1], then, from (2.25) and the definition of [𝑧(𝑡)]^∗, we have −𝑟(𝑡)𝑥^′′(𝑡)(𝑡)𝑥(𝑡)(𝑡)𝑓𝑥(𝑡)(𝑡)𝑥(𝑡)(𝑡)𝑥(𝑡)(𝑡)𝑥(𝑡)(𝑡)(𝑡)𝛼+𝑞=𝑝𝑡,𝑆−𝑦,−−𝑦−𝑔𝑡,𝑆−𝑦,−−𝑦+𝜌,0<𝑡<1,₂𝑥(0)−𝛽₂𝑥^′(0)=0,𝛾₂𝑥(1)+𝛿₂𝑥^′(1)=0. (2.26) Let 𝑣=𝑥−𝑦, then −(𝑟(𝑡)𝑣^′)^′(𝑡)+𝑞(𝑡)𝑣(𝑡)=−(𝑟𝑥^′)^′(𝑡)+𝑞𝑥(𝑡)+(𝑟𝑦^′)^′(𝑡)−𝑞𝑦(𝑡), which implies that −𝑟(𝑡)𝑥^′′(𝑡)+𝑞(𝑡)𝑥(𝑡)=−𝑟(𝑡)𝑣^′′(𝑡)+𝑞(𝑡)𝑣(𝑡)−𝑟(𝑡)𝑦^′′(𝑡)(𝑡)𝑦(𝑡)𝑟(𝑡)𝑣+𝑞=−^′′(𝑡)(𝑡)𝑣(𝑡)(𝑡),𝛼+𝑞+𝜌₂𝑥(0)−𝛽₂𝑥^′(0)=0,𝛾₂𝑥(1)+𝛿₂𝑥^′(1)=0. (2.27) Thus, (2.26) becomes (2.15), that is, 𝑥−𝑦 is a positive solution of the differential equation (2.15). By Lemma 2.4, 𝑢=𝑆(𝑥−𝑦) is a positive solution of the singular perturbed differential equation (1.1). This completes the proof of Lemma 2.7.

Thus, the BVP (2.25) is equivalent to the integral equation𝑥(𝑡)=¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠. (2.28) Hence, we will look for fixed points 𝑥(𝑡)≥𝑦(𝑡),𝑡∈[0,1], for the mapping 𝑇 defined on 𝐸∶=𝐶([0,1],[0,+∞)) by(𝑇𝑥)(𝑡)=¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠. (2.29)

The basic space used in this paper is 𝐸=𝐶([0,1];ℝ), where ℝ is a real number set. Obviously, the space 𝐸 is a Banach space if it is endowed with the norm as follows:‖𝑢‖=max_𝑡∈[0,1]|𝑢(𝑡)|, (2.30) for any 𝑢∈𝐸. Let𝑃=𝑥∈𝐸∶𝑥(𝑡)≥𝜃₂𝐺₂(𝑡,𝑡)‖𝑥‖, (2.31) where 𝜃₂ is defined by (2.7), then 𝑃 is a cone of 𝐸.

Lemma 2.8. Assume that (B1)–(B3) hold. Then, 𝑇∶𝑃→𝑃 is well defined. Furthermore, 𝑇∶𝑃→𝑃 is a completely continuous operator.

Proof. For any fixed 𝑥∈𝑃, there exists a constant 𝐿>0 such that ‖𝑥‖≤𝐿. And then, 𝑥(𝑠)−𝑦(𝑠)^∗(𝑠)≤‖𝑥‖𝑆𝑥(𝑠)(𝑠≤𝑥≤𝐿,−y)^∗≤𝐿¹₀𝐺₁()𝛽𝑡,𝑠𝑑𝑠≤¹+𝛼₁𝛾₁+𝛿₁𝑤₁𝐿. (2.32)
On the other hand, since 𝑔 satisfies global Carathéodory’s condition, we have 𝑔(⋅,𝑢(⋅),𝑣(⋅))∈𝐿¹(0,1). Accordingly, 𝑇𝑥 in (2.29) is continuous on [0,1], and, by (2.32), (𝑇𝑥)(𝑡)=¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗(𝑠≤+𝜌)𝑑𝑠₁₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+2𝜌(𝑠)𝑑𝑠≤𝑁¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)+𝜌(𝑠)𝑑𝑠<+∞, (2.33) where 𝑁=max_{(𝑡,𝑢,𝑣)∈[0,1]×[0,((𝛽1+𝛼1)(𝛾1+𝛿1)/𝑤1)𝐿]×[−𝐿,0]}𝑓(𝑡,𝑢,𝑣)+2. (2.34) This implies that the operator 𝑇∶𝑃→𝐸 is well defined.
Next, for any 𝑥∈𝑃, by (2.6), we have ‖𝑇𝑥‖=max_0≤𝑡≤1¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗(𝑠≤+𝜌)𝑑𝑠₁₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠. (2.35)
On the other hand, from (2.6), we also have (𝑇𝑥)(𝑡)=¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠≥𝜃₂𝐺₂(𝑡,𝑡)¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠. (2.36) So (𝑇𝑥)(𝑡)≥𝜃₂𝐺(𝑡,𝑡)‖𝑇𝑥‖,𝑡∈[0,1], (2.37) which yields that 𝑇(𝑃)⊂𝑃.
At the end, according to the Ascoli-Arzela Theorem, using standard arguments, one can show 𝑇∶𝑃→𝑃 is a completely continuous operator.

3. Main Results

Theorem 3.1. Suppose (B1)–(B3) hold. In addition, assume that the following conditions are satisfied.(S1)There exists a constant ∫𝑟>max2𝐾,¹⁰𝜌(𝑠)𝑑𝑠𝜃₂ (3.1) such that for any (𝑡,𝑢,𝑣)∈[0,1]×[0,((𝛽₁+𝛼₁)(𝛾₁+𝛿₁)/𝑤₁)𝑟]×[−𝑟,0],𝑓(𝑡,𝑢,𝑣)≤(𝑟/𝐾)−2, where 𝜃₂ and 𝐾 are defined by (2.7) and (2.19), respectively.(S2) There exists a constant 𝑅>2𝑟 such that, for any (𝑡,𝑢,𝑣)∈[0,1]×[(1/2)𝜇₁𝑅,((𝛽₁+𝛼₁)(𝛾₁+𝛿₁)/𝑤₁)𝑅]×[−𝑅,−(1/2)𝜇₂𝑅], 𝑓()≥𝑅𝑡,𝑢,𝑣𝑙, (3.2) where 𝜇₁,𝜇₂,𝑙 are defined by (2.19).(S3)lim_{|𝑢|+|𝑣|→+∞}max_𝑡∈[0,1]𝑓(𝑡,𝑢,𝑣)|𝑢|+|𝑣|=0. (3.3)Then, the singular perturbed differential equation (1.1) has at least two positive solutions 𝑢₁,𝑢₂, and there exist two positive constants 𝑛₁,𝑛₂ such that 𝑢₁(𝑡)≥𝑛₁𝑒(𝑡),𝑢₂(𝑡)≥𝑛₂𝑒(𝑡), for any 𝑡∈[0,1].

Proof. Let Ω₁={𝑥∈𝑃∶‖𝑥‖<𝑟}. Then, for any 𝑥∈𝜕Ω₁,𝑠∈[0,1], we have 𝑥(𝑠)−𝑦(𝑠)^∗(𝑠)≤‖𝑥‖𝑆𝑥(𝑠)(𝑠≤𝑥≤𝑟,−𝑦)^∗≤𝑟¹₀𝐺₁()𝛽𝑡,𝑠𝑑𝑠≤¹+𝛼₁𝛾₁+𝛿₁𝑤₁𝑟. (3.4)
It follows from (S1) that ‖𝑇𝑥‖=max_[]𝑡∈0,1(𝑇𝑥)(𝑡)=max_0≤𝑡≤1¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗(𝑠≤+𝜌)𝑑𝑠₁₀𝐺₂(𝑟𝑠,𝑠)𝐾𝑝(𝑠)(𝑠≤𝑟−2+2𝜌)𝑑𝑠𝐾¹₀𝐺₂(𝑝(𝑠)(𝑠‖𝑥‖.𝑠,𝑠)+𝜌)𝑑𝑠=𝑟= (3.5) Therefore, ‖𝑇𝑥‖≤‖𝑥‖,𝑥∈𝑃∩𝜕Ω₁. (3.6)
On the other hand, let Ω₂={𝑥∈𝑃∶‖𝑥‖<𝑅} and 𝜕Ω₂={𝑥∈𝑃∶‖𝑥‖=𝑅}. Then, for any 𝑥∈𝜕Ω₂,𝑡∈[0,1], noticing 𝑅>2𝑟 and (2.22), we have 𝑥(𝑡)−𝑦(𝑡)≥𝑥(𝑡)−𝐺₂(𝑡,𝑡)¹₀𝜌(𝑠)(𝑡)−𝑥(𝑡)𝑑𝑠≥𝑥𝜃₂𝑅¹₀𝜌(𝑠)(𝑡)−𝑟𝑑𝑠≥𝑥𝑅𝑥(𝑡)≥12𝑥(𝑡)≥12𝜃₂𝐺₂(𝑡,𝑡)𝑅. (3.7) So by (3.7), for any 𝑥∈𝜕Ω₂,𝑡∈[𝑎,𝑏], we have 12𝜇₂𝜙𝑅=²(𝑎)𝜓₂(𝑏)2𝑤₂𝜃₂(𝑡)(𝑡)1𝑅≤𝑥−𝑦≤𝑅,2𝜇₁𝛽𝑅=¹+𝛼₁𝑎𝛾₁+𝛿₁(1−𝑏)𝜃₂𝑅2𝑤₁¹₀𝑒(𝑠)𝐺₂()≤1𝑠,𝑠𝑑𝑠2𝜃₂𝑅¹₀𝐺₁(𝑡,𝑠)𝐺₂()𝑥(𝑡)(𝑡)≤𝛽𝑠,𝑠𝑑𝑠≤𝑆−𝑦₁+𝛼₁𝛾₁+𝛿₁𝑤₁𝑅. (3.8)
It follows from (S2), (3.8), and (2.6) that, for any 𝑥∈𝜕Ω₂,𝑡∈[𝑎,𝑏], ‖𝑇𝑥‖≥¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠≥𝜃₂𝐺₂(𝑡,𝑡)¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗𝑑𝑠≥𝜃₂𝐺₂(𝑡,𝑡)^𝑏_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠),𝑥(𝑠)−𝑦(𝑠)𝑑𝑠≥𝜃₂𝜙₂(𝑎)𝜓₂(𝑏)𝑤₂^𝑏_𝑎𝐺₂()𝑝(𝑠)𝑅𝑠,𝑠𝑑𝑠𝑙=𝜇₂^𝑏_𝑎𝐺₂()𝑝(𝑠)𝑅𝑠,𝑠𝑑𝑠𝑙=𝑅=‖𝑥‖. (3.9) So we have ‖𝑇𝑥‖≥‖𝑥‖,𝑥∈𝑃∩𝜕Ω₂. (3.10)
Next, let us choose 𝜀>0 such that 𝜀¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠<1. (3.11) Then, for the above 𝜀, by (S3), there exists 𝑁>𝑅>0 such that, for any 𝑡∈[0,1] and for any |𝑢|+|𝑣|≥𝑁, 𝑓(𝑡,𝑢,𝑣)≤𝜀(|𝑢|+|𝑣|). (3.12) Let 𝜎=max_{(𝑡,𝑢,𝑣)∈[0,1]×[0,((𝛽1+𝛼1)(𝛾1+𝛿1)/𝑤1)𝑁]×[−𝑁,0]}𝑓(𝑡,𝑢,𝑣)+2, (3.13) take 𝑅^∗=𝜎∫¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)+𝜌(𝑠)𝑑𝑠+2∫¹₀𝐺₂(𝑠,𝑠)𝜌(𝑠)𝑑𝑠1−𝜀1+𝛽₁+𝛼₁𝛾₁+𝛿₁/𝑤₁∫¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠+𝑁, (3.14) then 𝑅^∗>𝑁>𝑅.
Now let Ω₃={𝑥∈𝑃∶‖𝑥‖<𝑅^∗} and 𝜕Ω₃={𝑥∈𝑃∶‖𝑥‖=𝑅^∗}. Then, for any 𝑥∈𝑃∩𝜕Ω₃, we have ‖𝑇𝑥‖=max_0≤𝑡≤1¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗(𝑠≤+𝜌)𝑑𝑠₁₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,𝑥(𝑠)−𝑦(𝑠)^∗(𝑠≤+2𝜌)𝑑𝑠max_{(𝑡,𝑢,𝑣)∈[0,1]×[0,((𝛽1+𝛼1)(𝛾1+𝛿1)/𝑤1)𝑁]×[−𝑁,0]}𝑓(𝑡,𝑢,𝑣)+2¹₀𝐺₂(𝑝(𝑠)(𝑠+𝑠,𝑠)+𝜌)𝑑𝑠₁₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝜀𝑆𝑥(𝑠)−𝑦(𝑠)^∗+𝑥(𝑠)−𝑦(𝑠)^∗+2𝜌(𝑠)𝑑𝑠≤𝜎¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)+𝜌(𝑠)𝑑𝑠+¹₀𝐺₂(𝑝(𝑠)𝜀𝛽𝑠,𝑠)1+¹+𝛼₁𝛾₁+𝛿₁𝑤₁‖𝑥‖+2𝜌(𝑠)𝑑𝑠=𝜎¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)+𝜌(𝑠)𝑑𝑠+2¹₀𝐺₂()𝜌(𝑠)𝛽𝑠,𝑠𝑑𝑠+𝜀1+¹+𝛼₁𝛾₁+𝛿₁𝑤₁×¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠‖𝑥‖<𝑅^∗=‖𝑥‖, (3.15) which implies that ‖𝑇𝑥‖≤‖𝑥‖,𝑥∈𝑃∩𝜕Ω₃. (3.16) By Lemma 1.1, 𝑇 has two fixed points 𝑥₁,𝑥₂ such that 𝑟≤‖𝑥₁‖≤𝑅≤‖𝑥₂‖.
It follows from ∫𝑟>¹⁰𝜌(𝑠)𝑑𝑠𝜃₂ (3.17) that 𝑥₁(𝑡)−𝑦(𝑡)≥𝑥₁(𝑡)−𝐺₂(𝑡,𝑡)¹₀𝜌(𝑠)𝑑𝑠≥𝑥₁(𝑡)−𝑥₁(𝑡)𝜃₂𝑟¹₀𝜌(𝑠)=∫𝑑𝑠1−¹⁰𝜌(𝑠)𝑑𝑠𝜃₂𝑟𝑥₁(𝑡)≥𝜃₂𝑟∫1−¹⁰𝜌(𝑠)𝑑𝑠𝜃₂𝑟𝐺₂(𝑡,𝑡)=𝑚₁𝐺₂(𝑡,𝑡)>0,𝑡∈(0,1). (3.18)
As for (3.18), we also find a positive constant 𝑚₂ such that 𝑥₂(𝑡)−𝑦(𝑡)≥𝑚₂𝐺₂(𝑡,𝑡)>0,𝑡∈(0,1). (3.19)
Let 𝑢_𝑖(𝑡)=𝑆(𝑥_𝑖−𝑦)(𝑡),(𝑖=1,2), then 𝑢_𝑖(𝑡)(),𝑢>0,𝑡∈0,1)(𝑖=1,2_𝑖(𝑡)=𝑆𝑥_𝑖−𝑦(𝑡)≥𝑚_𝑖¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝑠)𝑑𝑠≥𝑚_𝑖¹₀𝑒(𝑠)𝐺₂(𝑠,𝑠)𝑑𝑠𝑒(𝑡)=𝑛_𝑖𝑒(𝑡). (3.20) By Lemma 2.7, we know that the singular perturbed differential equation (1.1) has at least two positive solutions 𝑢₁,𝑢₂ satisfying 𝑢₁(𝑡)≥𝑛₁𝑒(𝑡),𝑢₁(𝑡)≥𝑛₂𝑒(𝑡),𝑡∈[0,1], (3.21) for some positive constants 𝑛₁,𝑛₂. The proof of Theorem 3.1 is completed.

Theorem 3.2. Suppose (B1)–(B3) hold. In addition, assume that the following conditions are satisfied.(S4) There exists a constant 𝑟>2∫¹₀𝜌(𝑠)𝑑𝑠𝜃₂ (3.22) such that, for any (𝑡,𝑢,𝑣)∈[0,1]×[(1/2)𝜇₁𝑟,((𝛽₁+𝛼₁)(𝛾₁+𝛿₁)/𝑤₁)𝑟]×[−𝑟,−(1/2)𝜇₂𝑟], 𝑓()≥𝑟𝑡,𝑢,𝑣𝑙, (3.23) where 𝜃₂ and 𝜇₁,𝜇₂,𝑙 are defined by (2.7) and (2.19), respectively.(S5) There exists a constant 𝑅>max{𝑟,((𝑟/𝑙)+2)𝐾} such that, for any (𝑡,𝑢,𝑣)∈[0,1]×[0,((𝛽₁+𝛼₁)(𝛾₁+𝛿₁)/𝑤₁)𝑅]×[−𝑅,0], 𝑓()≤𝑅𝑡,𝑢,𝑣𝐾−2, (3.24) where 𝐾,𝑙 are defined by (2.19) and 𝑟 is defined by (S4).
(S6)lim_{|𝑢|+|𝑣|→∞}min_{𝑡∈[𝑎,𝑏]}𝑓(𝑡,𝑢,𝑣)|𝑢|+|𝑣|=+∞. (3.25) Then, the singular perturbed differential equation (1.1) has at least two positive solutions 𝑢₁,𝑢₂, and there exist two positive constants 𝑛₁,𝑛₂ such that 𝑢₁(𝑡)≥𝑛₁𝑒(𝑡),𝑢₂(𝑡)≥𝑛₂𝑒(𝑡), for any 𝑡∈[0,1].

Proof. Firstly, let Ω₁={𝑥∈𝑃∶‖𝑥‖<𝑟}. Then, for any 𝑥∈𝜕Ω₁,𝑡∈[0,1], by (2.22), we have 𝑥(𝑡)−y(𝑡)≥𝑥(𝑡)−𝐺₂(𝑡,𝑡)¹₀𝜌(𝑠)(𝑡)−𝑥(𝑡)𝑑𝑠≥𝑥𝜃₂𝑟¹₀𝜌(𝑠)≥1𝑑𝑠2𝑥(𝑡)≥12𝜃₂𝐺₂(𝑡,𝑡)𝑟. (3.26) So, by (3.26), for any 𝑥∈𝜕Ω₁,𝑡∈[𝑎,𝑏], we have 12𝜇₂𝜙𝑟=²(𝑎)𝜓₂(𝑏)2𝑤₂𝜃₂(𝑡)(𝑡)1𝑟≤𝑥−𝑦≤𝑟,2𝜇₁𝛽𝑟=¹+𝛼₁𝑎𝛾₁+𝛿₁(1−𝑏)𝜃₂𝑟2𝑤₁¹₀𝑒(𝑠)𝐺₂()≤1𝑠,𝑠𝑑𝑠2𝜃₂𝑟¹₀𝐺₁(𝑡,𝑠)𝐺₂()𝑥(𝑡)(𝑡)≤𝛽𝑠,𝑠𝑑𝑠≤𝑆−𝑦₁+𝛼₁𝛾₁+𝛿₁𝑤₁𝑟. (3.27)
It follows from (S4), (3.27), and (2.6) that, for any 𝑥∈𝜕Ω₂,𝑡∈[𝑎,𝑏], ‖𝑇𝑥‖≥¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠≥𝜃₂𝐺₂(𝑡,𝑡)¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗𝑑𝑠≥𝜃₂𝐺₂(𝑡,𝑡)^𝑏_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠),𝑥(𝑠)−𝑦(𝑠)𝑑𝑠≥𝜃₂𝜙₂(𝑎)𝜓₂(𝑏)𝑤₂^𝑏_𝑎𝐺₂()𝑝(𝑠)𝑟𝑠,𝑠𝑑𝑠𝑙=𝜇₂^𝑏_𝑎𝐺₂()𝑝(𝑠)𝑟𝑠,𝑠𝑑𝑠𝑙=𝑅=‖𝑥‖. (3.28) Therefore, we have ‖𝑇𝑥‖≥‖𝑥‖,𝑥∈𝑃∩𝜕Ω₁. (3.29)
Next, by (S5), we have 𝑅>𝑟 and 𝑅𝐾𝑟−2>𝑙>0. (3.30)
Let Ω₂={𝑥∈𝑃∶‖𝑥‖<𝑅}. Then, for any 𝑥∈𝜕Ω₂,𝑠∈[0,1], we have 𝑥(𝑠)−𝑦(𝑠)^∗(𝑠)≤‖𝑥‖𝑆𝑥(𝑠)(𝑠≤𝑥≤𝑅,−𝑦)^∗≤𝑅¹₀𝐺₁()𝛽𝑡,𝑠𝑑𝑠≤¹+𝛼₁𝛾₁+𝛿₁𝑤₁𝑅. (3.31)
It follows from (S5) that ‖𝑇𝑥‖=max_𝑡∈[0,1](𝑇𝑥)(𝑡)=max_0≤𝑡≤1¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗(𝑠≤+𝜌)𝑑𝑠₁₀𝐺₂(𝑅𝑠,𝑠)𝐾𝑝(𝑠)(𝑠≤𝑅−2+2𝜌)𝑑𝑠𝐾¹₀𝐺₂(𝑝(𝑠)(𝑠‖𝑥‖.𝑠,𝑠)+𝜌)𝑑𝑠=𝑅= (3.32) Therefore, ‖𝑇𝑥‖≤‖𝑥‖,𝑥∈𝑃∩𝜕Ω₂. (3.33)
On the other hand, choose a large enough real number 𝑀>0 such that 12𝜇²₂𝑀^𝑏_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≥1. (3.34)
From (S6), there exists 𝑁>𝑅 such that, for any 𝑡∈[𝑎,𝑏], 𝑓(𝑡,𝑢,𝑣)≥𝑀(|𝑢|+|𝑣|),|𝑢|+|𝑣|≥𝑁. (3.35) Take 𝑅^∗>max2𝑁𝜇₂,𝑅, (3.36) then 𝑅^∗>𝑅>𝑟. Let Ω₃={𝑥∈𝑃∶‖𝑥‖<𝑅^∗}, for any 𝑥∈𝑃∩𝜕Ω₃ and for any 𝑡∈[𝑎,𝑏], we have 𝑥(𝑡)−𝑦(𝑡)≥𝑥(𝑡)−𝐺₂(𝑡,𝑡)¹₀𝜌(𝑠)(𝑡)−∫𝑑𝑠≥𝑥¹⁰𝜌(𝑠)𝑑𝑠𝜃₂𝑅^∗𝑥(𝑡)≥12𝑥(𝑡)≥𝜃₂𝐺₂(𝑡,𝑡)2𝑅^∗≥12𝜇₂𝑅^∗≥𝑁>0. (3.37) So, for any 𝑥∈𝑃∩𝜕Ω₃,𝑡∈[𝑎,𝑏], by (3.35), (3.37), we have ‖𝑇𝑥‖≥¹₀𝐺₂(𝑡,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗−𝑔𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗+𝜌(𝑠)𝑑𝑠≥𝜃₂𝐺₂(𝑡,𝑡)¹₀𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠)^∗,−𝑥(𝑠)−𝑦(𝑠)^∗𝑑𝑠≥𝜃₂𝐺₂(𝑡,𝑡)^𝑏_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑓𝑠,𝑆𝑥(𝑠)−𝑦(𝑠),𝑥(𝑠)−𝑦(𝑠)𝑑𝑠≥𝜃₂𝜙₂(𝑎)𝜓₂(𝑏)𝑤₂^𝑏_𝑎𝐺₂()𝑝(𝑠)𝑀𝑆𝑥(𝑠)(𝑠)+𝑥(𝑠)(𝑠)≥𝜃𝑠,𝑠−𝑦−𝑦𝑑𝑠₂𝜙₂(𝑎)𝜓₂(𝑏)𝑤₂^𝑏_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑀𝑥(𝑠)−𝑦(𝑠)𝑑𝑠=𝜇₂𝑀^𝑏_𝑎𝐺₂()𝑝(𝑠)1𝑠,𝑠𝑑𝑠×2𝜇₂𝑅^∗=12𝜇²₂𝑀^𝑏_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠×𝑅^∗≥𝑅^∗=‖𝑥‖. (3.38) Thus, ‖𝑇𝑥‖≥‖𝑥‖,𝑥∈𝑃∩𝜕Ω₃. (3.39)
By Lemma 1.1, 𝑇 has two fixed points 𝑥₁,𝑥₂ such that 𝑟≤‖𝑥₁‖≤𝑅≤‖𝑥₂‖.
Noticing that 𝑟>2∫¹₀𝜌(𝑠)𝑑𝑠𝜃₂, (3.40) we have 𝑥₁(𝑡)−𝑦(𝑡)≥𝑥₁(𝑡)−𝐺₂(𝑡,𝑡)¹₀𝜌(𝑠)𝑑𝑠≥𝑥₁(𝑡)−𝑥₁(𝑡)𝜃₂𝑟¹₀𝜌(𝑠)=∫𝑑𝑠1−¹⁰𝜌(𝑠)𝑑𝑠𝜃₂𝑟𝑥₁(𝑡)≥12𝑥₁(𝑡)≥12𝜃₂𝑟𝐺₂(𝑡,𝑡)=𝑚₁𝐺₂(𝑡,𝑡)>0,𝑡∈(0,1). (3.41) As for (3.41), we also can find a positive constant 𝑚₂ such that 𝑥₂(𝑡)−𝑦(𝑡)≥𝑚₂𝐺₂(𝑡,𝑡)>0,𝑡∈(0,1). (3.42) Let 𝑢_𝑖(𝑡)=𝑆(𝑥_𝑖−𝑦)(𝑡),(𝑖=1,2), then 𝑢_𝑖(𝑡)(),𝑢>0,𝑡∈0,1)(𝑖=1,2_𝑖(𝑡)=𝑆𝑥_𝑖−𝑦(𝑡)≥𝑚_𝑖¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝑠)𝑑𝑠≥𝑚_𝑖¹₀𝑒(𝑠)𝐺₂(𝑠,𝑠)𝑑𝑠𝑒(𝑡)=𝑛_𝑖𝑒(𝑡). (3.43) By Lemma 2.7, we know that the singular perturbed differential equation (1.1) has at least two positive solutions 𝑢₁,𝑢₂ satisfying 𝑢₁(𝑡)≥n₁𝑒(𝑡),𝑢₁(𝑡)≥𝑛₂𝑒(𝑡),𝑡∈[0,1], (3.44) for some positive constants 𝑛₁,𝑛₂. The proof of Theorem 3.2 is completed.

An example Consider the following singular perturbed boundary value problem−𝑒^𝑡𝑢^′′′′(𝑡)+2𝑒^𝑡𝑢^′′(𝑡)=1𝑡()𝑓(𝑡)1−𝑡𝑡,𝑢,𝑢^′′(𝑡)−sin(𝑢(𝑡))+arctan𝑢^′′(𝑡)(1+(𝜋/2))√𝑡,𝑢(0)=0,𝑢(1)=0,𝑢^′′(0)−2𝑢^′′′(0)=0,𝑢^′′(1)+𝑢^′′′(1)=0, (3.45) where𝑓=𝑦𝑡,𝑥,𝑦𝑡𝑥+²,∈[]×[]×[],∈[]×[]×[],100𝑡,𝑥,𝑦0,10,10−10,041.88𝑥+4.075𝑦,𝑡,𝑥,𝑦0,110,13.93−100,−102001+𝑡²𝑒^𝑥/13.93∈[]×[]×[],𝑦+sin𝑦,𝑡,𝑥,𝑦0,113.93,100−100,−422.8923.2469𝑥+^1/2,𝑡,𝑥,𝑦∈[0,1]×[100,+∞)×(−∞,−422.89]. (3.46) Then, the BVP (3.45) has at least two positive solutions 𝑢₁(𝑡) and 𝑢₂(𝑡) such that𝑢₁(𝑡)≥2.3924𝑡(1−𝑡),𝑡∈[0,1],𝑢₂(𝑡)≥24.833𝑡(1−𝑡),𝑡∈[0,1]. (3.47)

Proof. In fact, let 𝑟(𝑡)≡𝑒^𝑡,𝑞(𝑡)≡2𝑒^𝑡(𝑡)=1,𝑝𝑡()(𝑡)=11−𝑡,𝜌√𝑡,𝑔t,𝑥,𝑦=sin𝑥+arctan𝑦(1+(𝜋/2))√𝑡, (3.48) then 𝑔≤1𝑡,𝑥,𝑦√𝑡. (3.49)
The corresponding Green’s functions can be written by 𝐺₁()=𝑠()𝑡()𝐺𝑡,𝑠1−𝑡,0≤𝑠≤𝑡≤1,1−𝑠,0≤𝑡≤𝑠≤1,₂()=1𝑡,𝑠53𝑒3𝑒^𝑠+13𝑒^−2𝑠13𝑒^𝑡+23𝑒^−2𝑡5,0≤𝑠≤𝑡≤1,3𝑒^𝑡+13𝑒^−2𝑡13𝑒^𝑠+23𝑒^−2𝑠,0≤𝑡≤𝑠≤1, (3.50) and, for (𝑡,𝑠)∈[0,1]×[0,1], we have 𝑡(1−𝑡)𝑠(1−𝑠)≤𝐺₁(𝑡,𝑠)≤𝑠(1−𝑠) and 15𝑒²+𝑒⁻¹𝐺₂(𝑠,𝑠)𝐺₂(𝑡,𝑡)≤𝐺₂(𝑡,𝑠)≤𝐺₂(𝑠,𝑠)or𝐺₂(𝑡,𝑡). (3.51)

Now, take [1/4,3/4]⊂[0,1], then we have𝐾=¹₀𝐺₂(𝑝(𝑠)(𝑠=𝑠,𝑠)+𝜌)𝑑𝑠₁₀(5/3)𝑒^𝑠+(1/3)𝑒^−2𝑠(1/3)𝑒^𝑠+(2/3)𝑒^−2𝑠13𝑒𝑠()+11−𝑠√𝑠𝜇𝑑𝑠≈0.6206,₂=𝜙₂(𝑎)𝜓₂(𝑏)𝜙₂(1)𝜓₂(0)=𝜙₂(1/4)𝜓₂(3/4)𝜙₂(1)𝜓₂(0)=(5/3)𝑒^1/4+(1/3)𝑒^−1/2(1/3)𝑒^3/4+(2/3)𝑒^−3/2(5/3)𝑒+(1/3)𝑒⁻²≈0.4374,𝑙=𝜇₂^b_𝑎𝐺₂(𝑠,𝑠)𝑝(𝑠)𝑑𝑠=0.4374^3/4_1/4(5/3)𝑒^𝑠+(1/3)𝑒^−2𝑠(1/3)𝑒^𝑠+(2/3)𝑒^−2𝑠()𝜇3𝑒𝑠1−𝑠𝑑𝑠≈0.2786,₁=𝛽₁+𝛼₁𝑎𝛾₁+𝛿₁(1−𝑏)𝜃₂𝑤₁¹₀𝐺₁(𝑠,𝑠)𝐺₂()=1𝑠,𝑠𝑑𝑠165𝑒²+𝑒⁻¹¹₀𝑠())𝑒1−𝑠(5/3^𝑠+(1/3)𝑒^−2𝑠(1/3)𝑒^𝑠+(2/3)𝑒^−2𝑠𝜃3𝑒𝑑𝑠≈8.4578,₂=𝑤₂𝜙₂(1)𝜓₂(0)=𝑝(0)𝜓₂(0)𝜙^′₂(0)−𝜓^′₂(0)𝜙₂(0)𝜙₂(1)𝜓₂(0)≈4.938. (3.52) On the other hand, since¹₀𝜌(𝑠)𝑑𝑠=¹₀1√𝑠𝑑𝑠=2, (3.53) then (B1)–(B3) are satisfied.

Choose 𝑟=10, then∫10=𝑟>max2𝐾,¹⁰𝜌(𝑠)𝑑𝑠𝜃₂={1.2412,0.4050}, (3.54) and, for any (𝑡,𝑥,𝑦)∈[0,1]×[0,10]×[−10,0],𝑓𝑟𝑡,𝑥,𝑦≤11≤𝐾−2≈14.1134. (3.55) So the condition (S1) is satisfied.

On the other hand, we take 𝑅=100, then 𝑅>2𝑟=20, and, for any (𝑡,𝑥,𝑦)∈[0,1]×[(1/2)𝜇₁𝑅,((𝛽₁+𝛼₁)(𝛾₁+𝛿₁)/𝑤₁)𝑅]×[−𝑅,−(1/2)𝜇₂𝑅]=[0,1]×[13.93,100]×[−100,−422.89], we have 𝑓𝑅𝑡,𝑥,𝑦≥542.65≥𝑙≈358.93, (3.56) this implies that the condition (S2) holds. Next, we have lim_{|𝑥|+|𝑦|→+∞}max_𝑡∈[0,1]𝑓𝑡,𝑥,𝑦|𝑥|+𝑦=lim_{|𝑥|+𝑦→+∞}23.2469|𝑥|+𝑦^−1/2=0. (3.57) Thus, (S3) also holds. By Theorem 3.1, the BVP (3.45) has at least two positive solutions 𝑢₁(𝑡) and 𝑢₂(𝑡). Since𝑛₁=1𝜃3𝑒²𝑟−¹₀𝜌(𝑠)𝑑𝑠¹₀𝑠()51−𝑠3𝑒^𝑠+13𝑒^−2𝑠13𝑒^𝑠+23𝑒^−2𝑠𝑛𝑑𝑠≈2.3924,₂=1𝜃3𝑒²𝑅−¹₀𝜌(𝑠)𝑑𝑠¹₀𝑠()51−𝑠3𝑒^𝑠+13𝑒^−2𝑠13𝑒^𝑠+23𝑒^−2𝑠𝑑𝑠≈24.833, (3.58) thus 𝑢₁(𝑡)≥2.3924𝑡(1−𝑡),𝑡∈[0,1],𝑢₂(𝑡)≥24.833𝑡(1−𝑡),𝑡∈[0,1]. (3.59)

Acknowledgment

The authors were supported financially by the National Natural Science Foundation of China (11071141) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017).