Positive Solutions for Fourth-Order Nonlinear Differential Equation with Integral Boundary Conditions

Abstract

This paper investigates the existence and nonexistence of positive solutions for a class of fourth-order nonlinear differential equation with integral boundary conditions. The associated Green's function for the fourth-order boundary value problems is first given, and the arguments are based on Krasnoselskii's fixed point theorem for operators on a cone.

1. Introduction

In this paper, we consider the following fourth-order boundary value problems (BVPs) with integral boundary conditions 𝑥⁽⁴⁾(𝑡)=𝑤(𝑡)𝑓𝑡,𝑥(𝑡),𝑥^′′(𝑡)(),𝑥(0),𝑡∈0,1=¹₀ℎ₁(𝑠)𝑥(𝑠)𝑥(1)=d𝑠,₁₀𝑘₁(𝑠)𝑥(𝑠)𝑥d𝑠,_′′(0)=¹₀ℎ₂(𝑠)𝑥^′′(𝑠)𝑥d𝑠,_′′(1)=¹₀𝑘₂(𝑠)𝑥^′′(𝑠)d𝑠, (1) where 𝑤 may be singular at 𝑡=0 and(or) 𝑡=1, 𝑓∶[0,1]×[0,+∞)×(−∞,0]→[0,+∞) is continuous, and ℎ₁,ℎ₂,𝑘₁,𝑘₂∈𝐿¹[0,1] are nonnegative.

The existence of solutions for nonlinear higher-order nonlocal BVPs has been studied by several authors, for example, see [1–9] and the references therein. However, there are a few papers dealing with the existence of positive solutions for the fourth-order boundary value problems with integral boundary conditions, see [7–15]. For the case of ℎ₁,ℎ₂,𝑘₁,𝑘₂=0, BVP (1) reduces to the following two-point BVP 𝑥⁽⁴⁾(𝑡)=𝑤(𝑡)𝑓𝑡,𝑥(𝑡),𝑥^′′(𝑡)(),𝑥(0)(1),𝑡∈0,1=𝑥=𝑥^′′(0)=𝑥^′′(1)=0. (2) A great deal of research has been devoted to the existence of solutions for problem (2) by using the Leray-Schauder continuation method, topological degree, and the method of lower and upper solutions.

For the case of ℎ₁=𝑔,ℎ₂=ℎ,𝑘₁=0,𝑘₂=0 or ℎ₁=0,ℎ₂=0,𝑘₁=𝑔, and 𝑘₂=ℎ, BVP (1) reduces to the following BVP with integral boundary conditions 𝑥⁽⁴⁾(𝑡)=𝑤(𝑡)𝑓𝑡,𝑥(𝑡),𝑥^′′(𝑡)(),𝑥(0),𝑡∈0,1=¹₀𝑔(𝑠)𝑥(𝑠)(1)𝑥d𝑠,𝑥=0,_′′(0)=¹₀ℎ(𝑠)𝑥^′′(𝑠)d𝑠,𝑥^′′(1)=0, (3) or 𝑥⁽⁴⁾(𝑡)=𝑤(𝑡)𝑓𝑡,𝑥(𝑡),𝑥^′′(𝑡)(),𝑥(0)(1),𝑡∈0,1=0,𝑥=¹₀𝑔(𝑠)𝑥(𝑠)𝑥d𝑠,_′′(0)=0,𝑥^′′(1)=¹₀ℎ(𝑠)𝑥^′′(𝑠)d𝑠. (4) By using the fixed point theorem of cone expansion and compression of norm type, Zhang and Ge [11] have obtained the existence and multiplicity of positive solutions for BVP (3) and (4).

The following assumptions will stand throughout this paper: (𝐴1)𝜔 is nonnegative, and 𝜔∈𝐿¹[0,1] may have singularities at 𝑡=0 and(or) 𝑡=1;  (𝐴2)𝑓∈𝐶([0,1]×[0,+∞)×(−∞,0],[0,+∞));  (𝐴3)ℎ₁,ℎ₂,𝑘₁,𝑘₂∈𝐿¹[0,1] are nonnegative and 𝜇₁=1−¹₀ℎ₁(𝑠)𝜈d𝑠>0,₁=1−¹₀𝑘₁(𝑠)𝜇d𝑠>0,₂=1−¹₀ℎ₂(𝑠)𝜈d𝑠>0,₂=1−¹₀𝑘₂(𝑠)d𝑠>0. (5)

The paper is organized as follows. In Section 2, we provide some necessary background material such as the Krasnoselskii’s fixed point theorem in cones. The associated Green’s function for the fourth-order boundary value problem is first given, and we also look at some properties of the Green’s function associated with the problem (1). In Section 3, the main results of problem (1) will be stated and proved. In Section 4, we give an example to illustrate how the main results can be used in practice.

2. Preliminary Results

In our main results, we will make use of the following lemmas.

Lemma 1 (in [16]). Let Ω₁ and Ω₂ be two bounded open sets in a real Banach space 𝐸, such that 𝜃∈Ω₁ and Ω₁⊂Ω₂. Let operator 𝐴∶𝑃∩(Ω₂⧵Ω₁)→𝑃 be continuous, where 𝜃 denotes the zero element of 𝐸 and 𝑃 is a cone in 𝐸. Suppose that one of the two conditions  (1)‖𝐴𝑥‖≤‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₁ and ‖𝐴𝑥‖≥‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₂ or  (2)‖𝐴𝑥‖≥‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₁ and ‖𝐴𝑥‖≤‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₂ is satisfied. Then, 𝐴 has at least one fixed point in 𝑃∩(Ω₂⧵Ω₁).

Lemma 2. If ℎ,𝑘∈𝐿¹[0,1] are nonnegative, and 𝜇=1−∫¹₀ℎ(𝑠)𝑑𝑠>0,𝜈=1−∫¹₀𝑘(𝑠)d𝑠>0, then for any 𝑦∈𝐶(0,1), the BVP −𝑥^′′(𝑡)(𝑡)(),𝑥(0)==𝑦,𝑡∈0,1₁₀ℎ(𝑠)𝑥(𝑠)𝑥(1)=d𝑠,₁₀𝑘(𝑠)𝑥(𝑠)d𝑠, (6) has a unique solution 𝑥 which is given by 𝑥(𝑡)=¹₀𝐻(𝑡,𝑠)𝑦(𝑠)d𝑠, (7) where 𝐻(𝑡,𝑠)=𝐺(𝑡,𝑠)+𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)𝐺()+𝑠,𝜏d𝜏𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)𝐺()𝐺()=𝑠()𝑡()𝑠,𝜏d𝜏,𝑡,𝑠1−𝑡,0≤𝑠≤𝑡≤1,1−𝑠,0≤𝑡≤𝑠≤1,𝑚=¹₀𝑠ℎ(𝑠)𝑑𝑠,𝑛=1−¹₀𝑠𝑘(𝑠)d𝑠. (8)(9)(10)

Proof. The general solution of −𝑥^′′(𝑡)=𝑦(𝑡) can be written as 𝑥(𝑡)=−^𝑡₀(𝑡−𝑠)𝑦(𝑠)d𝑠+𝐴𝑡+𝐵. (11) Now, we solve for 𝐴,𝐵 by 𝑥(0)=∫¹₀ℎ(𝑠)𝑥(𝑠)d𝑠 and 𝑥(1)=∫¹₀𝑘(𝑠)𝑥(𝑠)d𝑠, it follows that 𝐵=−¹₀ℎ(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏+𝐴¹₀𝜏ℎ(𝜏)d𝜏+𝐵¹₀ℎ(𝜏)−d𝜏,₁₀(1−𝑠)𝑦(𝑠)d𝑠+𝐴+𝐵=−¹₀𝑘(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏+𝐴¹₀𝜏𝑘(𝜏)d𝜏+𝐵¹₀𝑘(𝜏)d𝜏, (12) that is, 𝐴¹₀𝜏ℎ(𝜏)d𝜏−𝐵1−¹₀ℎ(𝜏)=d𝜏₁₀ℎ(𝜏)^𝜏₀()𝑦(𝑠)𝐴𝜏−𝑠d𝑠d𝜏,1−¹₀𝜏𝑘(𝜏)d𝜏+𝐵1−¹₀𝑘(𝜏)=d𝜏₁₀(1−𝑠)𝑦(𝑠)d𝑠−¹₀𝑘(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏. (13) Solving the above equations, we get 1𝐴=𝑚𝜈+𝑛𝜇𝜈¹₀ℎ(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏+𝜇¹₀()𝑦(𝑠)−1−𝑠d𝑠₁₀𝑘(𝜏)^𝜏₀()𝑦(𝑠),1𝜏−𝑠d𝑠d𝜏𝐵=𝑚𝑚𝜈+𝑛𝜇¹₀()𝑦(𝑠)−1−𝑠d𝑠₁₀𝑘(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏−𝑛¹₀ℎ(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏. (14) Therefore, (6) has a unique solution 𝑥(𝑡)=−^𝑡₀()𝑦(𝑠)1𝑡−𝑠d𝑠+×𝑚𝜈+𝑛𝜇𝜈𝑡¹₀ℎ(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏+𝜇𝑡¹₀(1−𝑠)𝑦(𝑠)d𝑠−¹₀𝑘(𝜏)×^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏+𝑚¹₀(1−𝑠)𝑦(𝑠)d𝑠−¹₀𝑘(𝜏)×^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏−𝑛¹₀ℎ(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏. (15) The unique solution of (6) be expressed as 𝑥(𝑡)=^𝑡₀𝑠(1−𝑡)𝑦(𝑠)d𝑠+¹_𝑡𝑡()𝑦(𝑠)11−𝑠d𝑠+×𝑚𝜈+𝑛𝜇𝜇𝑡¹₀𝜏𝑘(𝜏)d𝜏¹₀()𝑦(𝑠)×1−𝑠d𝑠−𝜇𝑡₁₀()𝑦(𝑠)×1−𝑠d𝑠−𝜈𝑡₁₀𝜏ℎ(𝜏)d𝜏¹₀(1−𝑠)𝑦(𝑠)d𝑠+𝜈𝑡¹₀ℎ(𝜏)^𝜏₀()𝑦(𝑠)×𝜏−𝑠d𝑠d𝜏+𝜇𝑡₁₀()𝑦(𝑠)×1−𝑠d𝑠−𝜇𝑡₁₀𝑘(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏+¹₀𝜏ℎ(𝜏)d𝜏¹₀(1−𝑠)𝑦(𝑠)d𝑠−¹₀(𝜏)×𝜏ℎd𝜏₁₀𝑘(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏−¹₀ℎ(𝜏)×^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏+¹₀(𝜏)×𝜏𝑘d𝜏₁₀ℎ(𝜏)^𝜏₀(𝜏−𝑠)𝑦(𝑠)d𝑠d𝜏−¹₀(𝜏)×𝜏𝑘d𝜏₁₀𝜏ℎ(𝜏)d𝜏¹₀()𝑦(𝑠)+1−𝑠d𝑠₁₀𝜏ℎ(𝜏)d𝜏¹₀𝜏𝑘(𝜏)d𝜏¹₀()𝑦(𝑠)=1−𝑠d𝑠_𝑡0𝑠(1−𝑡)𝑦(𝑠)d𝑠+¹_𝑡𝑡()𝑦(𝑠)11−𝑠d𝑠+×𝑚𝜈+𝑛𝜇𝜇𝑡¹₀𝑘(𝜏)^𝜏₀𝑠()𝑦(𝑠)+1−𝜏d𝑠d𝜏₁₀𝑘(𝜏)¹_𝜏𝜏(1−𝑠)𝑦(𝑠)d𝑠d𝜏−𝜈𝑡¹₀ℎ(𝜏)^𝜏₀𝑠()𝑦(𝑠)+1−𝜏d𝑠d𝜏₁₀ℎ(𝜏)¹_𝜏𝜏()𝑦(𝑠)+1−𝑠d𝑠d𝜏₁₀ℎ(𝜏)^𝜏₀𝑠()𝑦(𝑠)+1−𝜏d𝑠d𝜏₁₀ℎ(𝜏)¹_𝜏𝜏()𝑦(𝑠)−1−𝑠d𝑠d𝜏₁₀(𝜏)×𝜏𝑘d𝜏₁₀ℎ(𝜏)^𝜏₀𝑠()𝑦(𝑠)+1−𝜏d𝑠d𝜏₁₀ℎ(𝜏)¹_𝜏𝜏()𝑦(𝑠)+1−𝑠d𝑠d𝜏₁₀(𝜏)×𝜏ℎd𝜏₁₀𝑘(𝜏)^𝜏₀𝑠()𝑦(𝑠)+1−𝜏d𝑠d𝜏₁₀𝑘(𝜏)¹_𝜏𝜏()𝑦(𝑠)=1−𝑠d𝑠d𝜏₁₀𝐺()𝑦(𝑠)+1𝑡,𝑠d𝑠×𝑚𝜈+𝑛𝜇𝜇𝑡¹₀𝑘(𝜏)¹₀𝐺(𝑠,𝜏)𝑦(𝑠)d𝑠d𝜏−𝜈𝑡¹₀ℎ(𝜏)¹₀𝐺()𝑦(𝑠)+𝑠,𝜏d𝑠d𝜏₁₀ℎ(𝜏)¹₀𝐺(𝑠,𝜏)𝑦(𝑠)d𝑠d𝜏−¹₀𝜏𝑘(𝜏)d𝜏¹₀ℎ(𝜏)×¹₀𝐺()𝑦(𝑠)+𝑠,𝜏d𝑠d𝜏₁₀𝜏ℎ(𝜏)d𝜏¹₀𝑘(𝜏)×¹₀𝐺()𝑦(𝑠)=𝑠,𝜏d𝑠d𝜏₁₀𝐺()𝑦(𝑠)+𝑡,𝑠d𝑠𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑦(𝑠)¹₀𝑘(𝜏)𝐺()+𝑠,𝜏d𝜏d𝑠𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀𝑦(𝑠)¹₀ℎ(𝜏)𝐺(𝑠,𝜏)d𝜏d𝑠. (16) Therefore, the unique solution of (6) is 𝑥(𝑡)=∫¹₀𝐻(𝑡,𝑠)𝑦(𝑠)d𝑠.

Lemma 3 (in [11]). If ℎ,𝑘∈𝐿¹[0,1] are nonnegative, then,(1)𝑛>𝜈, (2)𝐺(𝑡,𝑠)>0, for all 𝑡,𝑠∈(0,1),𝐺(𝑡,𝑠)≥0, for all 𝑡,𝑠∈[0,1],(3) Letting 𝑒(𝑡)=𝑡(1−𝑡), for all 𝑡,𝑠∈[0,1], one has 𝑒(𝑡)𝑒(𝑠)≤𝐺(𝑡,𝑠)≤𝐺(𝑡,𝑡)=𝑒(𝑡)≤1/4,(4)  Letting 𝛿=(0,1/2),𝐽_𝛿=[𝛿,1−𝛿],forall𝑡∈𝐽_𝛿,𝑠∈[0,1], we have 𝐺(𝑡,𝑠)≥𝛿𝐺(𝑠,𝑠), where 𝐺(𝑡,𝑠) is defined by (9).

Lemma 4. If ℎ,𝑘∈𝐿¹[0,1] are nonnegative, and 𝜇=1−∫¹₀ℎ(𝑠)d𝑠>0,𝜈=1−∫¹₀𝑘(𝑠)d𝑠>0, for function 𝐻(𝑡,𝑠), we have the following properties:(1)𝐻(𝑡,𝑠)>0,forall𝑡,𝑠∈(0,1),𝐻(𝑡,𝑠)≥0,forall𝑡,𝑠∈[0,1],(2)𝜌𝑒(𝑡)𝑒(𝑠)≤𝐻(𝑡,𝑠)≤𝜂𝑒(𝑠),forall𝑡,𝑠∈[0,1],(3)𝐻(𝑡,𝑠)≥𝛿𝐻(𝑠,𝑠),𝑡∈𝐽_𝛿,𝑠∈[0,1],where 𝜌=(1/(𝑚𝜈+𝑛𝜇))(𝜇∫¹₀𝑒(𝜏)𝑘(𝜏)d𝜏+𝜈∫¹₀𝑒(𝜏)ℎ(𝜏)d𝜏), 𝜂=(𝑚+𝑛+𝜇(1−𝜈))/(𝑚𝜈+𝑛𝜇).

Proof.  (1) From (8) and Lemma 3, we get (1). (2) For all 𝑡,𝑠∈[0,1], by (8) and Lemma 3, we have 𝐻(𝑡,𝑠)=𝐺(𝑡,𝑠)+𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)𝐺()+𝑠,𝜏d𝜏𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)𝐺()≥𝑠,𝜏d𝜏𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)𝐺()+𝑠,𝜏d𝜏𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)𝐺()≥𝑒(𝑠)𝑠,𝜏d𝜏𝑚𝜈+𝑛𝜇𝜇𝑡¹₀𝑒(𝜏)𝑘(𝜏)+()𝜈d𝜏1−𝑡¹₀𝑒(𝜏)ℎ(𝜏)≥𝑒(𝑠)𝑒(𝑡)d𝜏𝜇𝑚𝜈+𝑛𝜇¹₀𝑒(𝜏)𝑘(𝜏)d𝜏+𝜈¹₀𝑒(𝜏)ℎ(𝜏)(𝑡)𝑒(𝑠),𝐻()()+d𝜏=𝜌𝑒𝑡,𝑠=𝐺𝑡,𝑠𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)𝐺()+𝑠,𝜏d𝜏𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)𝐺()()+𝑠,𝜏d𝜏≤𝐺𝑠,𝑠𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)𝐺()+𝑠,𝑠d𝜏𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)𝐺()(𝑠)𝑠,𝑠d𝜏=𝑒1+𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)+d𝜏𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)≤𝑒(𝑠)d𝜏×𝑚𝜈+𝑛𝜇𝑚𝜈+𝑛𝜇+𝑚+𝜇₁₀𝑘(𝜏)d𝜏+𝑛¹₀ℎ(𝜏)(𝑠).d𝜏=𝜂𝑒 (17)(3) For all 𝑡∈𝐽_𝛿,𝑠∈[0,1], by (8) and Lemma 3, we have 𝐻(𝑡,𝑠)=𝐺(𝑡,𝑠)+𝑚+𝜇𝑡𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)𝐺()+𝑠,𝜏d𝜏𝑛−𝜈𝑡𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)𝐺()()+𝛿𝑠,𝜏d𝜏≥𝛿𝐺𝑠,𝑠𝑚+𝜇𝑠𝑚𝜈+𝑛𝜇¹₀𝑘(𝜏)𝐺()+𝛿()𝑠,𝜏d𝜏𝑛−𝜈𝑠𝑚𝜈+𝑛𝜇¹₀ℎ(𝜏)𝐺()𝐻().𝑠,𝜏d𝜏=𝛿𝑠,𝑠 (18)

Lemma 5. Assume that (A1)–(A3) holds. If 𝑥(𝑡)∈𝐶²[0,1] is a solution of the following integral equation 𝑥(𝑡)=(𝑇𝑥)(𝑡)≜¹₀𝐻(𝑡,𝑠)𝑤(𝑠)𝑓𝑠,𝑥(𝑠),𝑥^′′(𝑠)d𝑠, (19) then 𝑥(𝑡)∈𝐶²[0,1]∩𝐶⁴(0,1) is a solution of BVP (1), where 𝐻(𝑡,𝑠)=¹₀𝐻₁(𝑡,𝜏)𝐻₂()𝐻𝜏,𝑠d𝜏,₁()()+𝑚𝑡,𝜏=𝐺𝑡,𝜏¹+𝜇₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁¹₀𝑘₁(𝜈)𝐺()+𝑛𝜏,𝜈d𝜈₁−𝜈₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁¹₀ℎ₁(𝜈)𝐺()𝐻𝜏,𝜈d𝜈,₂()()+𝑚𝜏,𝑠=𝐺𝜏,𝑠²+𝜇₂𝜏𝑚₂𝜈₂+𝑛₂𝜇₂¹₀𝑘₂(𝜈)𝐺()+𝑛𝑠,𝜈d𝜈₂−𝜈₂𝜏𝑚₂𝜈₂+𝑛₂𝜇₂¹₀ℎ₂(𝜈)𝐺()𝑚𝑠,𝜈d𝜈,₁=¹₀𝑠ℎ₁(𝑠)d𝑠,𝑛₁=1−¹₀𝑠𝑘₁(𝑠)𝑚d𝑠,₂=¹₀𝑠ℎ₂(𝑠)d𝑠,𝑛₂=1−¹₀𝑠𝑘₂(𝑠)d𝑠. (20)(21)(22)(23)

Proof. By using Lemma 2, the conclusion is obvious.

Lemma 6. If (A3) holds, then one has the following three properties:(1)𝐻(𝑡,𝑠)>0, for all 𝑡,𝑠∈(0,1),(2)0≤𝐻(𝑡,𝑠)≤𝜂₁𝜂₂/16,𝑡,𝑠∈[0,1],(3)𝐻(1/2,𝑠)≥(𝜌₁𝜌₂/120)𝑒(𝑠), for all 𝑠∈[0,1],where 𝜂₁=𝑚₁+𝑛₁+𝜇₁1−𝜈₁𝑚₁𝜈₁+𝑛₁𝜇₁,𝜂₂=𝑚₂+𝑛₂+𝜇₂1−𝜈₂𝑚₂𝜈₂+𝑛₂𝜇₂,𝜌₁=1𝑚₁𝜈₁+𝑛₁𝜇₁×𝜇₁¹₀𝑒(𝜏)𝑘₁(𝜏)d𝜏+𝜈₁¹₀𝑒(𝜏)ℎ₁(𝜏),𝜌d𝜏₂=1𝑚₂𝜈₂+𝑛₂𝜇₂×𝜇₂¹₀𝑒(𝜏)𝑘₂(𝜏)d𝜏+𝜈₂¹₀𝑒(𝜏)ℎ₂(𝜏)d𝜏. (24)

Proof. By using Lemma 3, the conclusions for (1) and the first part of (2) are obvious.
By using by (21) and Lemma 3, we have 𝐻₁()()+𝑚𝑡,𝜏=𝐺𝑡,𝜏¹+𝜇₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁¹₀𝑘₁(𝜈)𝐺()+𝑛𝜏,𝜈d𝜈₁−𝜈₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁¹₀ℎ₁(𝜈)𝐺()≤1𝜏,𝜈d𝜈4+𝑚₁+𝜇₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁¹₀14𝑘₁(𝑠)+𝑛d𝑠₁−𝜈₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁¹₀14ℎ₁(𝑠)≤𝜂d𝑠₁4. (25) Similarly, by (22), we get 𝐻₂()≤𝜂𝜏,𝑠²4. (26) So, by (20), (25), and (26), we concluded 𝐻(𝑡,𝑠)=¹₀𝐻₁(𝑡,𝜏)𝐻₂()𝜂𝜏,𝑠d𝜏≤¹𝜂₂.16 (27) By using Lemmas 3 and 4, we have 𝐻12,𝑠=¹₀𝐻₁12,𝜏𝐻₂()≥𝜌𝜏,𝑠d𝜏₁4¹₀𝑒(𝜏)𝐻₂()≥𝜌𝜏,𝑠d𝜏₁𝜌₂4𝑒(𝑠)¹₀𝑒²(𝜏)𝜌d𝜏=¹𝜌₂𝑒(𝑠).120 (28)

Let 𝐸=𝐶²[0,1]. Then 𝐸 is a Banach space with the norm ‖𝑥‖₂=‖𝑥‖+‖𝑥^′′‖, where ‖𝑥‖=max_𝑡∈[0,1]|𝑥(𝑡)|,‖𝑥^′′‖=max_𝑡∈[0,1]|𝑥^′′(𝑡)|.

For a fixed 𝛿=(0,1/2),𝐽_𝛿=[𝛿,1−𝛿], define the cone 𝐾⊂𝐸 by 𝐾≜𝑥∈𝐸∶𝑥≥0,𝑥^′′≤0,min_{𝑡∈𝐽𝛿}𝑥(𝑡)≥𝛿²‖𝑥‖,max_{𝑡∈𝐽𝛿}𝑥^′′(𝑡)≤−𝛿²𝑥^′′. (29) It's easy to prove that 𝐾 is a closed convex cone, and |𝑥(𝑡)|+𝑥^′′(𝑡)≥𝛿²‖𝑥‖₂,𝑡∈𝐽_𝛿,𝑥∈𝐾. (30)

Lemma 7. If (A1)–(A3) hold, then 𝑇(𝐾)⊂𝐾, and 𝑇∶𝐾→𝐾 is completely continuous.

Proof. By the properties of Green's function, if 𝑥∈𝐶²[0,1] then 𝑇𝑥∈𝐶²[0,1] and (𝑇𝑥)^′′(𝑡)=−¹₀𝐻₂(𝑡,𝑠)𝑤(𝑠)𝑓𝑠,𝑥(𝑠),𝑥^′′(𝑠)d𝑠. (31) For all 𝑥∈𝐾, we have by (19) and (31) (𝑇𝑥)(𝑡)≥0,(𝑇𝑥)^′′(𝑡)≤≤0,‖(𝑇𝑥)‖¹₀¹₀𝐺()+𝑚𝜏,𝜏¹+𝜇₁𝑚₁𝜈₁+𝑛₁𝜇₁×¹₀𝑘₁(𝜈)𝐺()+𝑛𝜈,𝜈d𝜈₁𝑚₁𝜈₁+𝑛₁𝜇₁¹₀ℎ₁(𝜈)𝐺(𝜈,𝜈)d𝜈×𝐻₂(𝜏,𝑠)𝜔(𝑠)𝑓𝑠,𝑥(𝑠),𝑥^′′(𝑠))d𝜏d𝑠,(𝑇𝑥^′′≤¹₀𝐺()+𝑚𝑠,𝑠²+𝜇₂𝑚₂𝜈₂+𝑛₂𝜇₂×¹₀𝑘₂(𝜏)𝐺()+𝑛𝜏,𝜏d𝜏₂𝑚₂𝜈₂+𝑛₂𝜇₂¹₀ℎ₂(𝜏)𝐺()(𝑠)𝑓(𝑠)𝜏,𝜏d𝜏×𝜔𝑠,𝑥,𝑥^′′(𝑠)d𝑠. (32)(33)(34) On the other hand, by (19), (33), and Lemma 3, we obtain min_{𝑡∈𝐽𝛿}(𝑇𝑥)(𝑡)=min_{𝑡∈𝐽𝛿}¹₀¹₀𝐺()+𝑚𝑡,𝜏¹+𝜇₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁×¹₀𝑘₁(𝜈)𝐺()𝑛𝜏,𝜈d𝜈+¹−𝜈₁𝑡𝑚₁𝜈₁+𝑛₁𝜇₁×¹₀ℎ₁(𝜈)𝐺(𝜏,𝜈)d𝜈×𝐻₂(𝜏,𝑠)𝜔(𝑠)𝑓𝑠,𝑥(𝑠),𝑥^′′(𝑠)d𝜏d𝑠≥𝛿¹₀¹₀𝐺()+𝑚𝜏,𝜏¹+𝜇₁𝛿𝑚₁𝜈₁+𝑛₁𝜇₁×¹₀𝑘₁(𝜈)𝐺()𝑛𝜈,𝜈d𝜈+¹−𝜈₁(1−𝛿)𝑚₁𝜈₁+𝑛₁𝜇₁×¹₀ℎ₁(𝜈)𝐺(𝜈,𝜈)d𝜈×𝐻₂(𝜏,𝑠)𝜔(𝑠)𝑓𝑠,𝑥(𝑠),𝑥^′′(𝑠)d𝜏d𝑠≥𝛿²¹₀¹₀𝐺()+𝑚𝜏,𝜏¹+𝜇₁𝑚₁𝜈₁+𝑛₁𝜇₁×¹₀𝑘₁(𝜈)𝐺()+𝑛𝜈,𝜈d𝜈₁𝑚₁𝜈₁+𝑛₁𝜇₁¹₀ℎ₁(𝜈)𝐺(𝜈,𝜈)d𝜈×𝐻₂(𝜏,𝑠)𝜔(𝑠)𝑓𝑠,𝑥(𝑠),𝑥^′′(𝑠)d𝜏d𝑠≥𝛿²‖𝑇𝑥‖. (35) Similarly, by (31), (34), and Lemma 3, we obtain max_{𝑡∈𝐽𝛿}(𝑇𝑥)^′′(𝑡)=−min_{𝑡∈𝐽𝛿}¹₀𝐻₂(𝑡,𝑠)𝑤(𝑠)𝑓𝑠,𝑥(𝑠),𝑥^′′(𝑠)d𝑠=−min_{𝑡∈𝐽𝛿}¹₀𝐺()+𝑚𝑡,𝑠²+𝜇₂𝑡𝑚₂𝜈₂+𝑛₂𝜇₂×¹₀𝑘₂(𝜏)𝐺()+𝑛𝑠,𝜏d𝜏₂−𝜈₂𝑡𝑚₂𝜈₂+𝑛₂𝜇₂¹₀ℎ₂(𝜏)𝐺()(𝑠)𝑓(𝑠)𝑠,𝜏d𝜏×𝜔𝑠,𝑥,𝑥^′′(𝑠)d𝑠≤−𝛿²¹₀𝐺()+𝑚𝑠,𝑠²+𝜇₂𝑚₂𝜈₂+𝑛₂𝜇₂×¹₀𝑘₂(𝜏)𝐺()+𝑛𝜏,𝜏d𝜏₂𝑚₂𝜈₂+𝑛₂𝜇₂¹₀ℎ₂(𝜏)𝐺()(𝑠)𝑓(𝑠)𝜏,𝜏d𝜏×𝜔𝑠,𝑥,𝑥^′′(𝑠)d𝑠≤−𝛿²(𝑇𝑥)^′′. (36) So 𝑇𝑥⊂𝐾, and then 𝑇(𝐾)⊂𝐾.
Clearly, the operator 𝑇 is continuous; thus by the Arzela-Ascoli theorem, it follows that 𝑇 is completely continuous.

3. Main Results

For convenience, we introduce the following notations: 𝐾_𝑟=𝑥∈𝐾∶‖𝑥‖₂≤𝑟,𝜕𝐾_𝑟=𝑥∈𝐾∶‖𝑥‖₂=𝑟,𝐾_𝑟,𝑅=𝑥∈𝐾∶𝑟≤‖𝑥‖₂,𝑓≤𝑅_𝛽=limsup_{|𝑥|+|𝑦|→𝛽}max_𝑡∈[0,1]𝑓𝑡,𝑥,𝑦|𝑥|+𝑦,𝑓_𝛽=liminf_{|𝑥|+|𝑦|→𝛽}min_𝑡∈[0,1]𝑓𝑡,𝑥,𝑦|𝑥|+𝑦,𝑀_𝛽=max_{[]|𝑥|+𝑦𝑡∈0,1,0<≤𝛽}𝑓,𝜂𝑡,𝑥,𝑦𝐿=¹𝜂₂+𝜂164²¹₀𝑤(𝑠)𝜌d𝑠,𝑁=¹𝜌₂+𝜌1204²𝛿²_𝛿^1−𝛿𝑒(𝑠)𝑤(𝑠)d𝑠. (37)

Theorem 8. Assume that (A1)–(A3) hold. In addition, one supposes that one of the following conditions is satisfied (𝐻₁)𝐿𝑓(𝑡,𝑥,𝑦)<|𝑥|+|𝑦|, for all |𝑥|+|𝑦|≠0,𝑡∈[0,1],(𝐻₂)𝑁𝑓(𝑡,𝑥,𝑦)>|𝑥|+|𝑦|, for all |𝑥|+|𝑦|≠0,𝑡∈[0,1].
Then, the BVP (1) has no positive solution.

Proof. Case 1. The condition (𝐻₁) holds. Assume towards a contraction that 𝑥^∗(𝑡) is a positive solution of BVP (1), from the proof of Lemma 7 and (29), then 𝑥^∗∈𝐾,𝑥^∗(𝑡)>0 for 0<𝑡<1. By Lemma 6 and (29), we have 𝑥^∗=max_[]𝑡∈0,1𝑥^∗(𝑡)=max_[]𝑡∈0,1𝑇𝑥^∗(𝑡)=max_[]𝑡∈0,1¹₀𝐻(𝑡,𝑠)𝑤(𝑠)𝑓𝑠,𝑥^∗(𝑠),𝑥^∗′′(𝑠)<𝜂d𝑠₁𝜂₂16𝐿¹₀𝑤(𝑠)𝑥^∗(𝑠)+𝑥^∗′′(𝑠≤𝜂)d𝑠₁𝜂₂𝑥16𝐿^∗21₀𝑤(𝑠)𝑥d𝑠,_∗′′=max_[]𝑡∈0,1𝑥^∗′′(𝑡)=max_[]𝑡∈0,1𝑇𝑥^∗′′(𝑡)=max_[]𝑡∈0,1¹₀𝐻₂(𝑡,𝑠)𝑤(𝑠)𝑓𝑠,𝑥^∗(𝑠),𝑥^∗′′(𝑠)<𝜂d𝑠₂4𝐿¹₀𝑤(𝑠)𝑥^∗(𝑠)+𝑥^∗′′(𝑠≤𝜂)d𝑠₂𝑥4𝐿^∗21₀𝑤(𝑠)d𝑠. (38) So, ‖𝑥^∗‖₂=‖𝑥^∗‖+‖𝑥^∗′′‖<(1/𝐿)(𝜂₁𝜂₂/16+𝜂₂/4)‖𝑥^∗‖₂∫¹₀𝑤(𝑠)d𝑠=‖𝑥^∗‖₂, which is a contradiction.
Case 2. The condition (𝐻₂) holds. Assume towards a contraction that 𝑥^∗∗(𝑡) is a positive solution of BVP (1), from the proof of Lemma 7 and (29), then 𝑥^∗∗∈𝐾, 𝑥^∗∗(𝑡)>0, for 0<𝑡<1. By Lemmas 4 and 6 and (30), we have 𝑥^∗∗=max_[]𝑡∈0,1𝑥^∗∗(𝑡)=max_[]𝑡∈0,1𝑇𝑥^∗∗(𝑡≥₁₀)𝐻12,𝑠𝑤(𝑠)𝑓𝑠,𝑥^∗∗(𝑠),𝑥^∗∗′′(𝑠)>𝜌d𝑠₁𝜌₂120𝑁¹₀𝑒(𝑠)𝑤(𝑠)𝑥^∗∗(𝑠)+𝑥^∗∗′′(𝑠≥𝜌)d𝑠₁𝜌₂120𝑁_𝛿^1−𝛿𝑒(𝑠)𝑤(𝑠)𝑥^∗∗(𝑠)+𝑥^∗∗′′(𝑠≥𝜌)d𝑠₁𝜌₂𝛿120𝑁²𝑥^∗∗_2𝛿^1−𝛿𝑒(𝑠)𝑤(𝑠)𝑥d𝑠,_∗∗′′=max_[]𝑡∈0,1𝑥^∗∗′′(𝑡)=max_[]𝑡∈0,1𝑇𝑥^∗∗′′(𝑡)=max_[]𝑡∈0,1¹₀𝐻₂(𝑡,𝑠)𝑤(𝑠)𝑓𝑠,𝑥^∗∗(𝑠),𝑥^∗∗′′(𝑠)d𝑠≥¹₀𝐻₂12,𝑠𝑤(𝑠)𝑓𝑠,𝑥^∗∗(𝑠),𝑥^∗∗′′(𝑠)>𝜌d𝑠₂4𝑁¹₀𝑒(𝑠)𝑤(𝑠)𝑥^∗∗(𝑠)+𝑥^∗∗′′(𝑠≥𝜌)d𝑠₂𝛿4𝑁²𝑥^∗∗_2𝛿^1−𝛿𝑒(𝑠)𝑤(𝑠)d𝑠. (39) So, ‖𝑥^∗∗‖₂=‖𝑥^∗∗‖+‖𝑥^∗∗′′‖>(1/𝑁)(𝜌₁𝜌₂/120+𝜌₂/4)𝛿²‖𝑥^∗∗‖₂∫_𝛿^1−𝛿𝑒(𝑠)𝑤(𝑠)d𝑠=‖𝑥^∗∗‖₂, which is a contradiction, and this completes the proof.

Theorem 9. Assume that (A1)–(A3) hold. In addition, one supposes that one of the following conditions is satisfied(𝐻₃)𝐿𝑓⁰<1<𝑁𝑓_∞,(𝐻₄)𝐿𝑓^∞<1<𝑁𝑓₀.
Then, the BVP (1) has at least one positive solution.

Proof. Case 1. The condition (𝐻₃) holds.
Considering 𝐿𝑓⁰<1, then there exists 𝑟₁>0, such that 𝑓(𝑡,𝑥,𝑦)≤(𝑓⁰+𝜀₁)(|𝑥|+|𝑦|), for 𝑡∈[0,1],(𝑥,𝑦)∈{(𝑥,𝑦)∶0<|𝑥|+|𝑦|≤𝑟₁}, where 𝜀₁>0, satisfies 𝐿(𝑓⁰+𝜀₁)≤1.
Then, for 𝑡∈[0,1],𝑥∈𝜕𝐾_𝑟1, by (19), (26), and Lemma 6, we have ‖𝑇𝑥‖₂=‖𝑇𝑥‖+(𝑇𝑥)^′′≤𝜂₁𝜂₂+𝜂164²¹₀𝑤(𝑠)𝑓𝑠,𝑥(𝑠),𝑠^′′(𝑠)≤𝜂d𝑠₁𝜂₂+𝜂164²𝑓⁰+𝜀₁×¹₀𝑤(𝑠)|𝑥(𝑠)|+𝑥^′′(𝑠≤𝜂)d𝑠₁𝜂₂+𝜂164²𝑓⁰+𝜀₁‖𝑥‖₂¹₀𝑤(𝑠)d𝑠≤‖𝑥‖₂, (40) which means that ‖𝑇𝑥‖₂≤‖𝑥‖₂,for𝑥∈𝜕𝐾_𝑟1. (41)
Next, turning to 𝑁𝑓_∞>1, then there exists 𝑅₁>0, such that 𝑓(𝑡,𝑥,𝑦)≥(𝑓_∞−𝜀₂)(|𝑥|+|𝑦|), for 𝑡∈[0,1],(𝑥,𝑦)∈{(𝑥,𝑦)∶0<|𝑥|+|𝑦|>𝑅₁}, where 𝜀₂>0, satisfies 𝑁(𝑓_∞−𝜀₂)≥1.
Let 𝑅₁=max{2𝑟₁,𝑅₁/𝛿²}, for all 𝑡∈𝐽_𝛿,𝑥∈𝜕𝐾_𝑅1. By (30) and Lemma 6, we have ‖𝑇𝑥‖₂=‖𝑇𝑥‖+(𝑇𝑥)^′′≥1𝑇𝑥2+(𝑇𝑥)^′′12≥𝜌₁𝜌₂+𝜌1204²¹₀𝑒(𝑠)𝑤(𝑠)𝑓𝑠,𝑥(𝑠),𝑠^′′(𝑠)≥𝜌d𝑠₁𝜌₂+𝜌1204²𝑓_∞−𝜀₂×¹₀𝑒(𝑠)𝑤(𝑠)|𝑥(𝑠)|+𝑥^′′(𝑠≥𝜌)d𝑠₁𝜌₂+𝜌1204²𝑓_∞−𝜀₂𝛿²‖𝑥‖₂×_𝛿^1−𝛿𝑒(𝑠)𝑤(𝑠)d𝑠≥‖𝑥‖₂ (42) which means that ‖𝑇𝑥‖₂≥‖𝑥‖₂,for𝑥∈𝜕𝐾_𝑅1. (43)
Case 2. The condition (𝐻₄) holds.
Considering 𝑁𝑓⁰>1, then there exists 𝑟₂>0, such that 𝑓(𝑡,𝑥,𝑦)≥(𝑓⁰−𝜀₃)(|𝑥|+|𝑦|), for 𝑡∈[0,1],(𝑥,𝑦)∈{(𝑥,𝑦)∶0<|𝑥|+|𝑦|≤𝑟₂}, where 𝜀₃>0, satisfies 𝑁(𝑓⁰−𝜀₃)≥1.
Then, for 𝑡∈𝐽_𝛿,𝑥∈𝜕𝐾_𝑟2, by (30) and Lemma 6, we have ‖𝑇𝑥‖₂≥1𝑇𝑥2+(𝑇𝑥)^′′12≥𝜌₁𝜌₂+𝜌1204²𝑓₀−𝜀₃×¹₀𝑒(𝑠)𝑤(𝑠)|𝑥(𝑠)|+𝑥^′′(𝑠≥𝜌)d𝑠₁𝜌₂+𝜌1204²𝑓₀−𝜀₃𝛿²‖𝑥‖₂×_𝛿^1−𝛿𝑒(𝑠)𝑤(𝑠)d𝑠≥‖𝑥‖₂, (44) which means that ‖𝑇𝑥‖₂≥‖𝑥‖₂,for𝑥∈𝜕𝐾_𝑟2. (45)
Next, turning to 𝐿𝑓^∞<1, then there exists 𝑅₂>0, such that 𝑓(𝑡,𝑥,𝑦)≤(𝑓^∞+𝜀₄)(|𝑥|+|𝑦|), for 𝑡∈[0,1],(𝑥,𝑦)∈{(𝑥,𝑦)∶|𝑥|+|𝑦|>𝑅₂}, where 𝜀₄>0, satisfies 𝐿(𝑓^∞+𝜀₄)≤1.
Let 𝑀_𝑅2=max_{[]|𝑥|+𝑦≤𝑡∈0,1,0<𝑅2}𝑓𝑡,𝑥,𝑦, (46) for 𝑡∈[0,1],𝑓(𝑡,𝑥,𝑦)≤𝑀_𝑅2+(𝑓^∞+𝜀₄)(|𝑥|+|𝑦|).
Let 𝑅₂=max{2𝑟₂,𝑅₂,𝐿𝑀_𝑅2/(1−𝐿(𝑓^∞+𝜀₄))}, for 𝑡∈[0,1],𝑥∈𝜕𝐾_𝑅2. By (26) and Lemma 6, we have ‖𝑇𝑥‖₂≤𝜂₁𝜂₂+𝜂164²¹₀𝑤(𝑠)𝑓𝑠,𝑥(𝑠),𝑠^′′(𝑠)≤𝜂d𝑠₁𝜂₂+𝜂164²×𝑀_𝑅2¹₀𝑤(𝑠)d𝑠+𝑓^∞+𝜀₄×¹₀𝑤(𝑠)|𝑥(𝑠)|+𝑥^′′(𝑠)d𝑠≤𝐿𝑀_𝑅2+‖𝑥‖₂Ł𝑓^∞+𝜀₄≤𝑅₂−𝐿𝑓^∞+𝜀₄𝑅₂+‖𝑥‖₂𝐿𝑓^∞+𝜀₄=‖𝑥‖₂, (47) which means that ‖𝑇𝑥‖₂≤‖𝑥‖₂,for𝑥∈𝜕𝐾_𝑅2. (48)
Applying Lemma 1 to (41) and (43), or (45), and (48) yields that 𝑇 has a fixed point 𝑥_∗∈𝐾_{𝑟𝑖,𝑅𝑖}(𝑖=1,2) with 𝑥_∗(𝑡)≥𝛿‖𝑥_∗‖₂, 𝑡∈(0,1). Thus, it follows that BVP (1) has at least one positive solution, and the theorem is proved.

Theorem 10. Assume that (A1)–(A3) hold, as do the following two conditions:(𝐻₅)  𝑁𝑓₀>1 and 𝑁𝑓_∞>1,(𝐻₆) there exists 𝑏>0, such that max_{𝑡∈[0,1],0<|𝑥|+|𝑦|≤𝑏}𝑓(𝑡,𝑥,𝑦)<𝑏/𝐿.
Then, the BVP (1) has at least two positive solutions 𝑥_∗(𝑡),𝑥_∗∗(𝑡), which satisfy 0<𝑥_∗2<𝑏<𝑥_∗∗2. (49)

Proof. We choose 𝑟,𝑅 with 0<𝑟<𝑏<𝑅.
If 𝑁𝑓₀>1, then by the proof of (45), we have ‖𝑇𝑥‖₂≥‖𝑥‖₂,for‖𝑥‖₂=𝑟. (50)
If 𝑁𝑓_∞>1, then by the proof of (43), we have ‖𝑇𝑥‖₂≥‖𝑥‖₂,for‖𝑥‖₂=𝑅. (51) On the other hand, by (𝐻₆), for 𝑥∈𝜕𝐾_𝑏, we have ‖𝑇𝑥‖₂≤𝜂₁𝜂₂+𝜂164²¹₀𝑤(𝑠)𝑓𝑠,𝑥(𝑠),𝑠^′′(𝑠)≤𝜂d𝑠₁𝜂₂+𝜂164²𝑀_𝑏¹₀𝑤(𝑠)d𝑠=𝐿𝑀_𝑏, (52) where 𝑀_𝑏=max_{𝑡∈[0,1],0<|𝑥|+|𝑦|≤𝑏}𝑓(𝑡,𝑥,𝑦)<𝑏/𝐿.
By (52), we have ‖𝑇𝑥‖₂<𝑏=‖𝑥‖₂. (53)
Applying Lemma 1 to (50), (51), and (53) yields that 𝑇 has a fixed point 𝑥_∗∈𝐾_𝑟,𝑏, and a fixed point 𝑥_∗∗∈𝐾_𝑏,𝑅. Thus it follows that BVP (1) has at least two positive solutions 𝑥_∗ and 𝑥_∗∗. Noticing (53), we have ‖𝑥_∗‖₂≠𝑏 and ‖𝑥_∗∗‖₂≠𝑏. Therefore, (49) holds, and the proof is complete.

Theorem 11. Assume that (A1)–(A3) hold, as do the following two conditions:(𝐻₇)  𝐿𝑓⁰<1 and 𝐿𝑓^∞<1,(𝐻₈) there exist 𝛿∈(0,1/2) and 𝐵>0 such that 𝑓(𝑡,𝑥,𝑦)>𝛿²𝐵/𝑁 for all 𝑡∈𝐽_𝛿,𝑥∈[𝛿²𝐵,𝐵] and 𝑦∈[−𝐵,−𝛿²𝐵].
Then, the BVP (1) has at least two positive solutions 𝑥^∗(𝑡),𝑥^∗∗(𝑡), which satisfy 0<𝑥^∗₂<𝐵<𝑥^∗∗₂. (54)

Proof. We choose 𝑟,𝑅 with 0<𝑟<𝐵<𝑅.
If 𝐿𝑓₀<1, then by the proof of (41), we have ‖𝑇𝑥‖₂≤‖𝑥‖₂,for‖𝑥‖₂=𝑟. (55)
If 𝐿𝑓_∞<1, then by the proof of (48), we have ‖𝑇𝑥‖₂≤‖𝑥‖₂,for‖𝑥‖₂=𝑅. (56) On the other hand, by (𝐻₈), for 𝑥∈𝜕𝐾_𝐵, we have ‖𝑇𝑥‖₂≥𝜌₁𝜌₂+𝜌1204²¹₀𝑒(𝑠)𝑤(𝑠)×𝑓𝑠,𝑥(𝑠),𝑠^′′(𝑠)d𝑠>𝐵=‖𝑥‖₂. (57)
Applying Lemma 1 to (55), (56), and (57) yields that 𝑇 has a fixed point 𝑥_∗∈𝐾_𝑟,𝐵, and a fixed point 𝑥_∗∗∈𝐾_𝐵,𝑅. Thus it follows that BVP (1) has at least two positive solutions 𝑥_∗ and 𝑥_∗∗. Noticing (57), we have ‖𝑥_∗‖₂≠𝐵 and ‖𝑥_∗∗‖₂≠𝐵. Therefore, (54) holds, and the proof is complete.

4. Example

Example 12. Consider the following fourth-order BVP 𝑥⁽⁴⁾(𝑡)=1√𝑡𝑓𝑡,𝑥(𝑡),𝑥^′′(𝑡(),𝑥(0)),𝑡∈0,1=¹₀𝑠𝑥(𝑠)d𝑠,𝑥(1)=¹₀(𝑠)𝑥𝑠𝑥d𝑠,_′′(0)=¹₀𝑠²𝑥^′′(𝑠)d𝑠,𝑥^′′(1)=¹₀𝑠²𝑥^′′(𝑠)d𝑠. (58) By calculation, we obtain 𝜇₁=1−¹₀1𝑠d𝑠=2,𝜈₁=1−¹₀1𝑠d𝑠=2,𝑚₁=¹₀𝑠²1d𝑠=3,𝑛₁=1−¹₀𝑠²2d𝑠=3,𝜇₂=1−¹₀𝑠²2d𝑠=3,𝜈₂=1−¹₀𝑠²2d𝑠=3,𝑚₂=¹₀𝑠³1d𝑠=4,𝑛₂=1−¹₀𝑠³3d𝑠=4,𝜂₁=52,𝜂₂=116,𝜌₁=16,𝜌₂=1,10𝐿=143196,𝛿=3,√𝑁=18133√21870002−2≈0.00032. (59) (1)  About the nonexistence of positive solution, we consider BVP (58) with 𝑓𝑡,𝑥,𝑦=𝑘₁𝑒^𝑡|𝑥|+𝑦+𝑘₂sin|𝑥|+𝑦, (60)where 𝑘₁ and 𝑘₂ are two positive real numbers, then 𝑓𝑡,𝑥,𝑦|𝑥|+𝑦<𝑒𝑘₁+𝑘₂,𝑓𝑡,𝑥,𝑦|𝑥|+𝑦>𝑘₁. (61)  If 𝑒𝑘₁+𝑘₂<1/𝐿 or 𝑘₁>1/𝑁, by Theorem 8, BVP (58) has no positive solution. (2) About the existence of positive solution, we consider BVP (58) with 𝑓𝑡,𝑥,𝑦=𝑘₁𝑒^𝑡−1|𝑥|+𝑦+𝑘₂sin|𝑥|+𝑦, (62)where 𝑘₁ and 𝑘₂ are two positive real numbers, then 𝑓₀=liminf_{|𝑥|+|𝑦|→0}min_𝑡∈[0,1]𝑓𝑡,𝑥,𝑦|𝑥|+𝑦=𝑘₁𝑒+𝑘₂,𝑓^∞=limsup_{|𝑥|+𝑦→∞}max_𝑡∈[0,1]𝑓𝑡,𝑥,𝑦|𝑥|+𝑦=𝑘₁. (63) If 𝑘₁<1/𝐿 and 𝑘₂>1/𝑁, then we have 𝐿𝑓^∞<1<𝑁𝑓₀. By Theorem 9, BVP (58) has at least one positive solution. (3) About the multiplicity of positive solution, we consider BVP (58) with 𝑓=𝑒𝑡,𝑥,𝑦_𝑡sin𝑘₁|𝑥|+𝑦,|𝑥|+𝑦𝑒≤𝑏,_𝑡sin𝑘₁|𝑥|+𝑦+𝑘₂(1+𝑡)|𝑥|+𝑦−𝑏²|𝑥|+𝑦,|𝑥|+𝑦+1>𝑏, (64)where 𝑘₁ and 𝑘₂ are two positive real numbers, then 𝑓₀=liminf_{|𝑥|+𝑦→0}min_𝑡∈[0,1]𝑓𝑡,𝑥,𝑦|𝑥|+𝑦=𝑘₁,𝑓_∞=limsup_{|𝑥|+𝑦→∞}max_𝑡∈[0,1]𝑓𝑡,𝑥,𝑦|𝑥|+𝑦=2𝑘₂. (65) If 𝑘₁>1/𝑁 and 𝑘₂>1/2𝑁, then we have 𝑁𝑓₀>1 and 𝑁𝑓_∞>1.
Furthermore, letting 𝑏>𝐿𝑀_𝑏>143𝑒/96, then we have 𝑀_𝑏<𝑏/𝐿. By Theorem 10, BVP (58) has at least two positive solutions.

Remark 13. In [4, 7–10], the existence of solutions for fourth-order ordinary differential equations BVP has been treated but did not discuss problems with singularities. Although, [11] seems to have considered the existence of positive solutions for the fourth-order BVP with integral boundary conditions for the singularity allowed to appear at 𝑡=0 and/or 𝑡=1. However, in [11], only the boundary conditions 𝑥(0)=0,𝑥^′′(0)=0, or 𝑥(1)=0,𝑥^′′(1)=0 have been considered. It is clear that the boundary conditions of Example 12 are 𝑥(0)≠0,𝑥^′′(0)≠0, and 𝑥(1)≠0,𝑥^′′(1)≠0. Hence, we generalize the fourth-order ordinary differential equations BVP.

Acknowledgments

This work is partially supported by the National Natural Science Foundation of PR China (10971045), the Natural Science Foundation of Hebei Province (A2013208147) and (A2011208012), and the Foundation of Hebei University of Science and Technology (XL201246). The authors thank the referee for his/her careful reading of the paper and useful suggestions that have greatly improved this paper.