Positive Solutions for p-Laplacian Fourth-Order Differential System with Integral Boundary Conditions

Abstract

This paper investigates the existence of positive solutions for a class of singular 𝑝-Laplacian fourth-order differential equations with integral boundary conditions. By using the fixed point theory in cones, explicit range for 𝜆 and 𝜇 is derived such that for any 𝜆 and 𝜇 lie in their respective interval, the existence of at least one positive solution to the boundary value system is guaranteed.

1. Introduction

Boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics and so on. Fourth-order differential equations boundary value problems, including those with the 𝑝-Laplacian operator, have their origin in beam theory [1, 2], ice formation [3, 4], fluids on lungs [5], brain warping [6, 7], designing special curves on surfaces [8], and so forth. In beam theory, more specifically, a beam with a small deformation, a beam of a material that satisfies a nonlinear power-like stress and strain law, and a beam with two-sided links that satisfies a nonlinear powerlike elasticity law can be described by fourth order differential equations along with their boundary value conditions. For more background and applications, we refer the reader to the work by Timoshenko [9] on elasticity, the monograph by Soedel [10] on deformation of structure, and the work by Dulcska [11] on the effects of soil settlement. Due to their wide applications, the existence and multiplicity of positive solutions for fourth-order (including 𝑝-Laplacian operator) boundary value problems has also attracted increasing attention over the last decades; see [12–33] and references therein. In [28], Zhang and Liu studied the following singular fourth-order four-point boundary value problem 𝑢(0)=𝜙𝑢_𝑝(1𝑢)−^(𝑎𝑡)𝑢(𝜉^)==𝑓𝑢(_𝑡,𝑢(𝑡)),0<𝑡<1,(0)=𝑢^(1)−𝑏𝑢^(𝜂)=0,(1.1) where 𝜙_𝑝(𝑥)=|𝑥|^𝑝−2𝑥, 𝑝>1,  0<𝜉, 𝜂<1,  0≤𝑎, 𝑏<1, 𝑓∈𝐶((0,1)×(0,∞),(0,∞)), 𝑓(𝑡,𝑥) may be singular at 𝑡=0 and/or 𝑡=1 and 𝑥=0. The authors gave sufficient conditions for the existence of one positive solution by using the upper and lower solution method, fixed point theorems, and the properties of the Green function.

In [32], Zhang et al. discussed the existence and nonexistence of symmetric positive solutions of the following fourth-order boundary value problem with integral boundary conditions: 𝜙_𝑝𝑢^𝑢(𝑡()0)=^=𝑢(𝑤1)(=𝑡)𝑓₁₀(𝑡,𝑢(𝑡)),0<𝑡<1,𝜙𝑔(𝑠)𝑢(𝑠)𝑑𝑠,_𝑝𝑢^(0)=𝜙_𝑝𝑢^(1)=¹⁰ℎ(𝑠)𝜙_𝑝𝑢^(𝑠)𝑑𝑠,(1.2) where 𝜙_𝑝(𝑥)=|𝑥|^𝑝−2𝑥, 𝑝>1, 𝑤∈𝐿¹[0,1] is nonnegative, symmetric on the interval [0,1],𝑓∶[0,1]×[0,+∞)→[0,+∞) is continuous, 𝑓(1−𝑡,𝑥)=𝑓(𝑡,𝑥) for all (𝑡,𝑥)∈[0,1]×[0,+∞), and 𝑔,ℎ∈𝐿¹[0,1] are nonnegative, symmetric on [0,1].

Motivated by the work of the above papers, in this paper, we study the existence of positive solutions of the following singular fourth-order boundary value system with integral boundary conditions: 𝜙_𝑝1𝑢𝜙^(𝑡)^=𝜆^𝑝1−1𝑎₁(𝑡)𝑓₁(𝑡,𝑢(𝑡),𝑣(𝑡)),0<𝑡<1,_𝑝2𝑣𝑢^((𝑡0))=^𝑢=(1𝜇)=^𝑝2−1𝑎₂₁₀(𝑡)𝑓₂(𝑡,𝑢(𝑡),𝑣(𝑡)),𝜙𝑢(𝑠)𝑑𝜉₁(𝑠),_𝑝1𝑢^(0)𝑣=(0)𝜙=_𝑝1𝑣𝑢(1^)(1=)₁₀=¹⁰𝜙_𝑝1𝑢^(𝑠)𝑑𝜂₁(𝑠),𝜙𝑣(𝑠)𝑑𝜉₂(𝑠),_𝑝2𝑣^(0)=𝜙_𝑝2𝑣^(1)=¹⁰𝜙_𝑝2𝑣^(𝑠)𝑑𝜂₂(𝑠),(1.3) where 𝜆 and 𝜇 are positive parameters, 𝜙_𝑝𝑖(𝑥)=|𝑥|^𝑝𝑖−2𝑥, 𝑝_𝑖>1, 𝜙_𝑞𝑖=𝜙^𝑝−1_𝑖, 1/𝑝_𝑖+1/𝑞_𝑖=1, 𝜉_𝑖,𝜂_𝑖∶[0,1]→ℝ⁺(𝑖=1,2) are nondecreasing functions of bounded variation, and the integrals in (1.3) are Riemann-Stieltjes integrals, 𝑓₁∶[0,1]×ℝ⁺⁰×ℝ⁺→ℝ⁺ and 𝑓₂∶[0,1]×ℝ⁺×ℝ⁺⁰→ℝ⁺ are two continuous functions, and 𝑓₁(𝑡,𝑥,𝑦) may be singular at 𝑥=0 while 𝑓₂(𝑡,𝑥,𝑦) may be singular at 𝑦=0;    𝑎₁, 𝑎₂∶(0,1)→ℝ⁺ are continuous and may be singular at 𝑡=0 and/or 𝑡=1, in which ℝ⁺=[0,+∞), ℝ⁺⁰=(0,+∞).

Compared to previous results, our work presented in this paper has the following new features. Firstly, our study is on singular nonlinear differential systems, that is, 𝑎₁ and 𝑎₂ in (1.3) are allowed to be singular at 𝑡=0 and/or 𝑡=1, meanwhile 𝑓₁(𝑡,𝑥,𝑦) is allowed to be singular at 𝑥=0 while 𝑓₂(𝑡,𝑥,𝑦) is allowed to be singular at 𝑦=0, which bring about many difficulties. Secondly, the main tools used in this paper is a fixed-point theorem in cones, and the results obtained are the conditions for the existence of solutions to the more general system (1.3). Thirdly, the techniques used in this paper are approximation methods, and a special cone has been developed to overcome the difficulties due to the singularity and to apply the fixed-point theorem. Finally, we discuss the boundary value problem with integral boundary conditions, that is, system (1.3) including fourth-order three-point, multipoint and nonlocal boundary value problems as special cases. To our knowledge, very few authors studied the existence of positive solutions for 𝑝-Laplacian fourth-order differential equation with boundary conditions involving Riemann-Stieltjes integrals. Hence we improve and generalize the results of previous papers to some degree, and so it is interesting and important to study the existence of positive solutions for system (1.3).

The rest of this paper is organized as follows. In Section 2, we present some lemmas that are used to prove our main results. In Section 3, the existence of positive solution for system (1.3) is established by using the fixed point theory in cone. Finally, in Section 4, one example is also included to illustrate the main results.

Definition 1.1. A vector (𝑢,𝑣)∈(𝐶²[0,1]∩𝐶⁴(0,1))×(𝐶²[0,1]∩𝐶⁴(0,1)) is said to be a positive solution of system (1.3) if and only if (𝑢,𝑣) satisfies (1.3) and 𝑢(𝑡)>0, 𝑣(𝑡)≥0 or 𝑢(𝑡)≥0, 𝑣(𝑡)>0 for any 𝑡∈(0,1).

Let 𝐾 be a cone in a Banach space 𝐸. For 0<𝑟<𝑅<+∞, let 𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟}, 𝜕𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖=𝑟}, and 𝐾_𝑟,𝑅={𝑥∈𝐾∶𝑟≤‖𝑥‖≤𝑅}. The proof of the main theorem of this paper is based on the fixed point theory in cone. We list one lemma [34, 35] which is needed in our following argument.

Lemma 1.2. Let 𝐾 be a positive cone in real Banach space 𝐸 and 𝑇∶𝐾_𝑟,𝑅→𝐾 a completely continuous operator. If the following conditions hold(i)‖𝑇𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝐾_𝑅;(ii)there exists 𝑒∈𝜕𝐾₁ such that 𝑥≠𝑇𝑥+𝑚𝑒 for any 𝑥∈𝜕𝐾_𝑟 and 𝑚>0. Then 𝑇 has a fixed point in 𝐾_𝑟,𝑅.

Remark 1.3. If (i) and (ii) are satisfied for 𝑥∈𝜕𝐾_𝑟 and 𝑥∈𝜕𝐾_𝑅, respectively. Then Lemma 1.2 is still true.

2. Preliminaries and Lemmas

The basic space used in this paper is 𝐸=𝐶[0,1]×𝐶[0,1]. Obviously, the space 𝐸 is a Banach space if it is endowed with the norm as follows: ‖(𝑢,𝑣)‖∶=‖𝑢‖+‖𝑣‖,‖𝑢‖=max_0≤𝑡≤1||𝑢(𝑡)||,‖𝑣‖=max_0≤𝑡≤1||𝑣(𝑡)||(2.1)

for any (𝑢,𝑣)∈𝐸. Denote 𝐶⁺[0,1]={𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0,0≤𝑡≤1}. For convenience, we list the following assumptions:(𝐻₁)𝑎₁,𝑎₂∶(0,1)→ℝ⁺ are continuous and 0<𝐿_𝑖∶=¹⁰𝑒(𝑠)𝑎_𝑖(𝑠)𝑑𝑠^𝑞𝑖−1<+∞,𝑖=1,2,(2.2) where 𝑒(𝑠)=𝑠(1−𝑠), 𝑠∈[0,1].(𝐻₂)𝜉_𝑖,𝜂_𝑖∶[0,1]→ℝ⁺(𝑖=1,2) are nondecreasing functions of bounded variation, and 𝛼_𝑖∈[0,1), 𝛽_𝑖∈[0,1), where 𝛼_𝑖=¹⁰𝑑𝜉_𝑖(𝑠),𝛽_𝑖=¹⁰𝑑𝜂_𝑖(𝑠),𝑖=1,2.(2.3)(𝐻₃)𝑓₁∶[0,1]×ℝ⁺⁰×ℝ⁺→ℝ⁺,𝑓₂∶[0,1]×ℝ⁺×ℝ⁺⁰→ℝ⁺ are continuous and satisfy 𝑓𝑓₁₂(𝑡,𝑥,𝑦)≤𝑔₁(𝑡,𝑥)+ℎ₁(𝑡,𝑦),∀(𝑡,𝑥,𝑦)∈[0,1]×ℝ⁺⁰×ℝ⁺,(𝑡,𝑥,𝑦)≤𝑔₂(𝑡,𝑥)+ℎ₂(𝑡,𝑦),∀(𝑡,𝑥,𝑦)∈[0,1]×ℝ⁺×ℝ⁺⁰,(2.4)where 𝑔₁,ℎ₂∶[0,1]×𝑅⁺⁰→𝑅⁺ are continuous and nonincreasing in the second variable, and 𝑔₂,ℎ₁∶[0,1]×𝑅⁺→𝑅⁺ are continuous and for any constant 𝑟>0, 0<¹⁰𝑒(𝑠)𝑎₁(𝑠)𝑔₁(𝑠,𝑟)𝑑𝑠<+∞,0<¹⁰𝑒(𝑠)𝑎₂(𝑠)ℎ₂(𝑠,𝑟)𝑑𝑠<+∞.(2.5)

Similar to the proof of Lemmas 2.1 and  2.2 in [32], the following two lemmas are valid.

Lemma 2.1. If (𝐻₂) holds, then for any 𝑦∈𝐿(0,1), the boundary value problem −𝑥𝑥(^0()𝑡)==𝑥𝜙(1_𝑞𝑖)(𝑦=(𝑡)₁₀),0<𝑡<1,𝑥(𝑠)𝑑𝜉_𝑖(𝑠)(2.6) has a unique solution 𝑥(𝑡)=¹⁰𝐻_𝑖(𝑡,𝑠)𝜙_𝑞𝑖(𝑦(𝑠))𝑑𝑠,(2.7) where 𝐻_𝑖(𝑡,𝑠)=𝐺(𝑡,𝑠)+11−𝛼_𝑖¹⁰𝐺(𝑡,𝑠)=𝑠𝑡((11−−𝑠𝑡)),,𝐺(00𝜏≤≤,𝑠)𝑠𝑡𝑑≤≤𝜉_𝑖𝑠𝑡(𝜏≤≤),11,.𝑖=1,2,(2.8)

Lemma 2.2. If (𝐻₂) holds, then for any 𝑧∈𝐿(0,1), the boundary value problem 𝑦−(𝑦0)^=(𝑡)𝑦=(1𝑧)(=𝑡),₁₀0<𝑡<1,𝑦(𝑠)𝑑𝜂_𝑖(𝑠)(2.9) has a unique solution 𝑦(𝑡)=¹⁰𝐾_𝑖(𝑡,𝑠)𝑧(𝑠)𝑑𝑠,(2.10) where 𝐾_𝑖(𝑡,𝑠)=𝐺(𝑡,𝑠)+11−𝛽_𝑖¹⁰𝐺(𝜏,𝑠)𝑑𝜂_𝑖(𝜏),𝑖=1,2.(2.11)

Remark 2.3. For 𝑡, 𝑠∈[0,1], we have 𝑒(𝑡)𝑒(𝑠)≤𝐺(𝑡,𝑠)≤𝑒(𝑠)or𝑒(𝑡)≤max_[]𝑡∈0,1𝑒(𝑡)=14.(2.12)

Remark 2.4. If (𝐻₂) holds, it is easy to testify 𝐻_𝑖(𝑡,𝑠) defined by (2.8) that: 𝜌_𝑖𝑒(𝑠)≤𝐻_𝑖(𝑡,𝑠)≤𝛾_𝑖𝑒(𝑠)≤14𝛾^𝑖<𝛾_𝑖,𝑡,𝑠∈[0,1],𝑖=1,2,(2.13) where 𝛾_𝑖=11−𝛼_𝑖,𝜌_𝑖=∫¹⁰𝑒(𝜏)𝑑𝜉_𝑖(𝜏)1−𝛼_𝑖,𝑖=1,2.(2.14)

Remark 2.5. From (2.11), we can prove that the properties of 𝐾_𝑖(𝑡,𝑠)(𝑖=1,2) are similar to those of 𝐻_𝑖(𝑡,𝑠)(𝑖=1,2).

Lemma 2.6. For 𝑥>0, 𝑦>0, we have 𝜙_𝑞𝑖(𝑥+𝑦)≤2^𝑞𝑖−1𝜙_𝑞𝑖𝜙(𝑥)+𝜙_𝑞𝑖(𝑦),𝑞_𝑖≥2_𝑞𝑖≤2_𝑞𝑖−1(𝑥)+𝜙_𝑞𝑖(𝑦),1<𝑞_𝑖<2,𝜙𝜙_𝑞𝑖(𝑥)+𝜙_𝑞𝑖(𝑦),𝑞_𝑖>1,𝑖=1,2,(2.15)_𝑞𝑖(𝑥)>𝜙_𝑞𝑖(𝑦)>𝜙_𝑞𝑖(0)=0,𝑥>𝑦>0,𝑞_𝑖>1,𝑖=1,2.(2.16)

Proof. The proof of this lemma is easy, and we omit it.

Let 𝐾=m(𝑢i,n𝑣_[]𝑡∈0,1)∈𝐶⁺[0,1]×𝐶⁺[0,1]∶𝑢,𝑣areconcaveon[0,1],𝑢(𝑡)≥Λ‖𝑢‖,min_[]𝑡∈0,1𝑣(𝑡)≥Λ‖𝑣‖,(2.17) where Λ=min𝜌¹𝜎^𝑞11−1𝛾₁𝜈^𝑞11−1,𝜌²𝜎^𝑞22−1𝛾₂𝜈^𝑞22−1,𝜎^𝑖=∫¹⁰𝑒(𝑠)𝑑𝜂_𝑖(𝑠)1−𝛽_𝑖,𝜈_𝑖=11−𝛽_𝑖,𝑖=1,2.(2.18)

It is easy to see that 𝐾 is a cone of 𝐸. For any 0<𝑟<𝑅, let 𝐾_𝑟,𝑅={(𝑢,𝑣)∈𝐾∶𝑟<‖𝑢‖<𝑅,𝑟<‖𝑣‖<𝑅}.

Remark 2.7. By the definition of 𝜌_𝑖, 𝜎_𝑖, 𝛾_𝑖, 𝜈_𝑖(𝑖=1,2), we have 0<Λ<1.
To overcome singularity, we consider the following approximate problem of (1.3): 𝜙_𝑝1𝑢𝜙^(𝑡)^=𝜆^𝑝1−1𝑎₁(𝑡)𝑓_1𝑛(𝑡,𝑢(𝑡),𝑣(𝑡)),0<𝑡<1,_𝑝2𝑣^𝑢((𝑡0))=^=𝑢(𝜇1)=^𝑝2−1𝑎₂₁₀(𝑡)𝑓_2𝑛(𝑡,𝑢(𝑡),𝑣(𝑡)),𝜙𝑢(𝑠)𝑑𝜉₁(𝑠),_𝑝1𝑢^(0)𝑣=(0)𝜙=_𝑝1𝑣𝑢(1^)(1=)₁₀=¹⁰𝜙_𝑝1𝑢^(𝑠)𝑑𝜂₁(𝑠),𝜙𝑣(𝑠)𝑑𝜉₂(𝑠),_𝑝2𝑣^(0)=𝜙_𝑝2𝑣^(1)=¹⁰𝜙_𝑝2𝑣^(𝑠)𝑑𝜂₂(𝑠),(2.19) where 𝑛 is a positive integer and 𝑓_1𝑛(𝑡,𝑢,𝑣)=𝑓₁𝑡,max𝑢,𝑛⁻¹,𝑣,𝑓_2𝑛(𝑡,𝑢,𝑣)=𝑓₂𝑡,𝑢,max𝑣,𝑛⁻¹.(2.20)
Clearly, 𝑓_𝑖𝑛∈𝐶([0,1]×ℝ⁺×ℝ⁺,ℝ⁺)(𝑖=1,2).
By Lemmas 2.1 and 2.2, for each 𝑛∈ℕ, 𝜆>0, 𝜇>0, let us define operators 𝐴^𝜆𝑛∶𝐾→𝐶[0,1], 𝐵^𝜇𝑛∶𝐾→𝐶[0,1], and 𝑇_𝑛∶𝐾→𝐸 by 𝐴^𝜆𝑛(𝑢,𝑣)(𝑡)=𝜆¹⁰𝐻₁(𝑡,𝑠)𝜙_𝑞1¹⁰𝐵𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠,(2.21)_𝜇𝑛(𝑢,𝑣)(𝑡)=𝜇¹⁰𝐻₂(𝑡,𝑠)𝜙_𝑞2¹⁰𝐾₂(𝑠,𝜏)𝑎₂(𝜏)𝑓_2𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠,(2.22) and 𝑇_𝑛(𝑢,𝑣)=(𝐴^𝜆𝑛(𝑢,𝑣),𝐵^𝜇𝑛(𝑢,𝑣)), respectively.

Lemma 2.8. Assume that (𝐻₁)–(𝐻₃) hold, then for each 𝜆>0, 𝜇>0, 𝑛∈ℕ, 𝑇_𝑛∶𝐾_𝑟,𝑅→𝐾 is a completely continuous operator.

Proof. Let 𝜆>0, 𝜇>0, and 𝑛∈ℕ be fixed. For any (𝑢,𝑣)∈𝐾, by (2.21), we have 𝐴𝐴^𝜆𝑛(𝑢,𝑣)^(𝑡)=−𝜆𝜙_𝑞1¹⁰𝐾₁(𝑡,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏≤0,_𝜆𝑛(𝑢,𝑣)(0)==𝜆𝐴^𝜆𝑛₁₀(𝑢,𝑣)(1)𝐻₁(0,𝑠)𝜙_𝑞1¹⁰𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠≥0,(2.23) which implies that 𝐴^𝜆𝑛 is nonnegative and concave on [0,1]. Similarly, by (2.22) we can obtain that 𝐵^𝜇𝑛 is nonnegative and concave on [0,1]. For any (𝑢,𝑣)∈𝐾 and 𝑡∈[0,1], it follows from (2.13) that 𝐴^𝜆𝑛(𝑢,𝑣)(𝑡)=𝜆¹⁰𝐻₁(𝑡,𝑠)𝜙_𝑞1¹⁰≤𝜆𝛾₁𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠𝜈^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠.(2.24)
Thus ‖‖𝐴^𝜆𝑛(𝑢,𝑣)‖‖≤𝜆𝛾₁𝜈^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠.(2.25)
On the other hand, by (2.13) and (2.18), we have 𝐴^𝜆𝑛(𝑢,𝑣)(𝑡)=𝜆¹⁰𝐻₁(𝑡,𝑠)𝜙_𝑞1¹⁰≥𝜆𝜌₁𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠𝜎^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰≥𝜌𝑒(𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠₁𝜎^𝑞11−1𝛾₁𝜈^𝑞11−1‖‖𝐴^𝜆𝑛(𝑢,𝑣)‖‖≥Λ‖‖𝐴^𝜆𝑛(𝑢,𝑣)‖‖.(2.26)
This implies that min_[]𝑡∈0,1𝐴^𝜆𝑛(𝑢,𝑣)(𝑡)≥Λ‖‖𝐴^𝜆𝑛(𝑢,𝑣)‖‖.(2.27)
Similar to (2.27), we also have min_[]𝑡∈0,1𝐵^𝜇𝑛(𝑢,𝑣)(𝑡)≥Λ‖‖𝐵^𝜇𝑛(𝑢,𝑣)‖‖.(2.28)
Therefore, 𝑇_𝑛(𝐾)⊂𝐾.
Next, we prove that 𝑇_𝑛∶𝐾_𝑟,𝑅→𝐾 is completely continuous. Suppose (𝑢_𝑚,𝑣_𝑚)∈𝐾_𝑟,𝑅 and (𝑢₀,𝑣₀)∈𝐾_𝑟,𝑅 with ‖(𝑢_𝑚,𝑣_𝑚)−(𝑢₀,𝑣₀)‖→0(𝑚→∞). We notice that 𝑡∈[0,1]𝑓_𝑖𝑛(𝑡,𝑢_𝑚(𝑡),𝑣_𝑚(𝑡))−𝑓_𝑖𝑛(𝑡,𝑢₀(𝑡),𝑣₀(𝑡))→0(𝑚→∞). Using the Lebesgue dominated convergence theorem, we have ||||𝜙_{𝑝1𝑞−11}¹⁰−𝜙_{𝑝1𝑞−11}𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢_𝑚(𝜏),𝑣_𝑚(𝜏)𝑑𝜏¹⁰≤𝜈₁𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢₀(𝜏),𝑣₀(𝜏)𝑑𝜏||||¹⁰𝑒(𝜏)𝑎₁(𝜏)||𝑓_1𝑛𝜏,𝑢_𝑚(𝜏),𝑣_𝑚(𝜏)−𝑓_1𝑛𝜏,𝑢₀(𝜏),𝑣₀(𝜏)||𝑑𝜏⟶0,𝑚⟶∞.(2.29)
Therefore, ‖‖𝐴≤^𝜆𝑛𝜆𝑢𝛾_𝑚1,𝑣_𝑚−𝐴^𝜆𝑛𝑢₀,𝑣₀‖‖¹⁰𝑒(𝑠)||||𝜙_𝑞1¹⁰−𝜙_𝑞1𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢_𝑚(𝜏),𝑣_𝑚(𝜏)𝑑𝜏¹⁰𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢₀(𝜏),𝑣₀(𝜏)𝑑𝜏||||𝑑𝑠⟶0,𝑚⟶∞.(2.30)
Similarly, we also have ‖‖𝐵^𝜇𝑛𝑢_𝑚,𝑣_𝑚−𝐵^𝜇𝑛𝑢₀,𝑣₀‖‖⟶0,𝑚⟶∞.(2.31)
So 𝐴^𝜆𝑛∶𝐾_𝑟,𝑅→𝐶[0,1] and 𝐵^𝛽𝑛∶𝐾_𝑟,𝑅→𝐶[0,1] are continuous. Therefore, 𝑇_𝑛∶𝐾_𝑟,𝑅→𝐾 is also continuous.
Let 𝐷⊂𝐾_𝑟,𝑅 be any bounded set, then for any (𝑢,𝑣)∈𝐷, we have (𝑢,𝑣)∈𝐾, 𝑟≤‖𝑢‖≤𝑅, 𝑟≤‖𝑣‖≤𝑅, and then 0<Λ𝑟≤𝑢(𝜏)≤𝑅, 0<Λ𝑟≤𝑣(𝜏)≤𝑅 for any 𝜏∈[0,1]. By (𝐻₃), we have 𝐿_𝑟∶=¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑔₁(𝑠,𝑟Λ)𝑑𝜏^𝑞1−1<+∞.(2.32)
It is easy to show that 𝐴^𝜆𝑛(𝐷) is uniformly bounded. In order to show that 𝑇_𝑛 is a compact operator, we only need to show that 𝐴^𝜆𝑛(𝐷) is equicontinuous. By the uniformly continuity of 𝐻₁(𝑡,𝑠) on [0,1]×[0,1], for all 𝜀>0, there is 𝛿>0 such that for any 𝑡₁, 𝑡₂, 𝑠∈[0,1] and |𝑡₁−𝑡₂|<𝛿, we have ||𝐻₁𝑡₁,𝑠−𝐻₁𝑡₂,𝑠||<𝜀.(2.33)
This together with (2.15) and (2.32) implies ||𝐴^𝜆𝑛(≤𝑢,𝜆𝑣)₁₀𝑡₁−𝐴^𝜆𝑛(𝑢,𝑣)𝑡₂||||𝐻₁𝑡₁,𝑠−𝐻₁𝑡₂,𝑠||𝜙_𝑞1¹⁰<𝜀𝜆𝜈_𝑞11−1𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠𝜙_𝑞1¹⁰≤𝜀𝜆𝜈_𝑞11−1𝑒(𝜏)𝑎₁(𝜏)𝑔₁𝜏,max𝑢(𝜏),𝑛⁻¹+ℎ₁(𝜏,𝑣(𝜏))𝑑𝜏𝜙_𝑞1¹⁰≤𝜀𝜆𝜈_𝑞11−1𝑒(𝜏)𝑎₁(𝜏)𝑔₁(𝜏,𝑟Λ)+ℎ₁(𝜏,𝑣(𝜏))𝑑𝜏2^𝑞1−1𝜙^𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑔₁(𝜏,𝑟Λ)𝑑𝜏+𝜙_𝑞1¹⁰≤𝜀𝜆𝜈_𝑞11−1𝑒(𝜏)𝑎₁(𝜏)ℎ₁(𝜏,𝑣(𝜏))𝑑𝜏2^𝑞1−1⎡⎢⎢⎣𝐿_𝑟+𝐿₁max^{[]𝜏∈[0,1]𝑦∈𝑟Λ,𝑅}ℎ₁(𝜏,𝑦)^𝑞1−1⎤⎥⎥⎦,||𝑡₁−𝑡₂||<𝛿,(𝑢,𝑣)∈𝐷.(2.34)
This means that 𝐴^𝜆𝑛(𝐷) is equicontinuous. By the Arzela-Ascoli theorem, 𝐴^𝜆𝑛(𝐷) is a relatively compact set and that 𝐴^𝜆𝑛∶𝐾_𝑟,𝑅→𝐶[0,1] is a completely continuous operator.
In the same way, we can show that 𝐵^𝜇𝑛∶𝐾_𝑟,𝑅→𝐶[0,1] is also completely continuous, and so 𝑇_𝑛∶𝐾_𝑟,𝑅→𝐾 is completely continuous. Now since 𝜆, 𝜇, and 𝑛 are given arbitrarily, the conclusion of this lemma is valid.

3. Main Results

For notational convenience, we denote by 𝑓𝑀_𝑖=6𝜌_𝑖𝜎^𝑞𝑖−1Λ𝐿_𝑖⁻¹,𝑁_𝑖=𝛾_𝑖𝜈^{𝑞𝑖𝑖−1}𝐿_𝑖⁻¹,𝑖=1,2,_𝛼1=⎛⎜⎜⎝limsup_𝑥→𝛼sup_{+𝑡∈[0,1]𝑦∈ℝ}𝑓₁(𝑡,𝑥,𝑦)𝜙_𝑝1(𝑥)⎞⎟⎟⎠^𝑞1−1,𝑓^𝛼2=⎛⎜⎜⎝limsup_𝑦→𝛼sup_{+𝑡∈[0,1]𝑥∈ℝ}𝑓₂(𝑡,𝑥,𝑦)𝜙_𝑝2(𝑦)⎞⎟⎟⎠^𝑞2−1𝑓_1𝛼,=⎛⎜⎜⎝liminf_𝑥→𝛼inf_{+𝑡∈[0,1]𝑦∈ℝ}𝑓₁(𝑡,𝑥,𝑦)𝜙_𝑝1(𝑥)⎞⎟⎟⎠^𝑞1−1,𝑓_2𝛼=liminf_𝑦→𝛼inf_{+𝑡∈[0,1]𝑥∈ℝ}𝑓₂(𝑡,𝑥,𝑦)𝜙_𝑝2(𝑦)^𝑞2−1,(3.1) where 𝛼 denotes 0 or ∞. The main results of this paper are the following.

Theorem 3.1. Assume that (𝐻₁)–(𝐻₃) hold. Then we have:(𝐶₁) If 𝑓⁰¹, 𝑓_1∞, 𝑓⁰²∈(0,∞) and 𝑀₁/𝑓_1∞<𝑁₁/𝑓⁰¹, then for each 𝜆∈(𝑀₁/𝑓_1∞,𝑁₁/𝑓⁰¹), 𝜇∈(0,𝑁₂/𝑓⁰²), the system (1.3) has at least one positive solution.(𝐶₂) If 𝑓⁰¹, 𝑓⁰², 𝑓_2∞∈(0,∞) and 𝑀₂/𝑓_2∞<𝑁₂/𝑓⁰², then for each 𝜆∈(0,𝑁₁/𝑓⁰¹), 𝜇∈(𝑀₂/𝑓_2∞,𝑁₂/𝑓⁰²), the system (1.3) has at least one positive solution.(𝐶₃) If 𝑓⁰¹=0, 𝑓_1∞=∞, 0<𝑓⁰²<∞, then for each 𝜆∈(0,∞), 𝜇∈(0,𝑁₂/𝑓⁰²), the system (1.3) has at least one positive solution.(𝐶₄) If 0<𝑓⁰¹<∞, 𝑓⁰²=0, 𝑓_2∞=∞, then for each 𝜆∈(0,𝑁₁/𝑓⁰¹), 𝜇∈(0,∞), the system (1.3) has at least one positive solution.(𝐶₅) If 𝑓^0𝑖=0, 𝑓_𝑖∞=∞(𝑖=1,2), then for each 𝜆∈(0,∞), 𝜇∈(0,∞), the system (1.3) has at least one positive solution.(𝐶₆) If 0<𝑓⁰¹<∞, 𝑓_1∞=∞ or 𝑓_2∞=∞, 0<𝑓⁰²<∞, then for each 𝜆∈(0,𝑁₁/𝑓⁰¹), 𝜇∈(0,𝑁₂/𝑓⁰²), the system (1.3) has at least one positive solution.(𝐶₇) If 𝑓⁰¹=0, 0<𝑓_1∞<∞, and 𝑓⁰¹=0, 0<𝑓_2∞<∞, then for each 𝜆∈(𝑀₁/𝑓_1∞,∞), 𝜇∈(0,∞) or 𝜆∈(0,∞), 𝜇∈(𝑀₂/𝑓_2∞,∞), the system (1.3) has at least one positive solution.

Proof. We only prove the condition in which (𝐶₁) holds. The other cases can be proved similarly.
Let 𝜆∈(𝑀₁/𝑓_1∞,𝑁₁/𝑓⁰¹), 𝜇∈(0,𝑁₂/(𝑓⁰²)), choose 𝜀₁>0 such that 𝑓_1∞−𝜀₁>0 and 𝑀₁𝑓_1∞−𝜀₁≤𝜆≤𝑁¹𝑓⁰¹+𝜀₁,0<𝜇≤𝑁²𝑓⁰²+𝜀₁.(3.2)
It follows from 𝑓^0𝑖∈(0,∞) of (𝐶₁) that there exists 𝑟₁>0 such that for any 𝑡∈[0,1], 𝑓𝑓₁(𝑡,𝑥,𝑦)≤𝑓⁰¹+𝜀₁^𝑝1−1𝜙_𝑝1(𝑥)≤𝑟₁𝑓⁰¹+𝜀₁^𝑝1−1,0<𝑥≤𝑟₁,𝑦≥0,(3.3)₂(𝑡,𝑥,𝑦)≤𝑓⁰²+𝜀₁^𝑝2−1𝜙_𝑝2(𝑦)≤𝑟₁𝑓⁰²+𝜀₁^𝑝2−1,𝑥≥0,0<𝑦≤𝑟₁.(3.4)
Let 𝐾_𝑟1={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑟₁,‖𝑣‖<𝑟₁}. For any (𝑢,𝑣)∈𝜕𝐾_𝑟1, 𝑛>1/𝑟₁, by (2.13), (3.3), we have ‖‖𝐴^𝜆𝑛(𝑢,𝑣)‖‖=max_[]𝑡∈0,1𝜆¹⁰𝐻₁(𝑡,𝑠)𝜙_𝑞1¹⁰≤𝜆𝛾₁𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠𝜈^𝑞11−1𝜙_𝑞1≤𝜆𝛾₁¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑓⁰¹+𝜀₁^𝑝1−1𝜙_𝑝1(𝑢(𝜏))𝑑𝜏𝜈^𝑞11−1𝑓⁰¹+𝜀₁𝐿₁𝑟₁=𝜆𝑁¹⁻¹𝑓⁰¹+𝜀₁𝑟₁.(3.5)
Similarly, we also have ‖‖𝐵^𝜇𝑛(𝑢,𝑣)‖‖≤𝜇𝑁²⁻¹𝑓⁰²+𝜀₁𝑟₁.(3.6)
Therefore, we have ‖‖𝑇_𝑛(𝑢,𝑣)‖‖≤=‖‖𝜆𝐴𝑁^𝜆𝑛(𝑢₁₋₁,𝑣)‖‖+‖‖𝐵^𝜇𝑛(𝑢,𝑣)‖‖𝑓⁰¹+𝜀₁+𝜇𝑁²⁻¹𝑓⁰²+𝜀₁𝑟₁≤2𝑟₁=‖(𝑢,𝑣)‖.(3.7)
On the other hand, by 𝑓_1∞>𝑓_1∞−𝜀₁>0, there exists 𝑅₀>0 such that 𝑓₁(𝑡,𝑥,𝑦)≥𝑓_1∞−𝜀₁^𝑝1−1𝜙_𝑝1(𝑥),𝑡∈[0,1],𝑥≥𝑅₀,𝑦≥0.(3.8)
Let 𝑅₁>max{2𝑟₁,Λ⁻¹𝑅₀}, 𝐾_𝑅1={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑅₁,‖𝑣‖<𝑅₁}. Next, we take (𝜑₁,𝜑₂)=(1,1)∈𝜕𝐾₁, and for any (𝑢,𝑣)∈𝜕𝐾_𝑅1, 𝑚>0, 𝑛∈ℕ, we will show (𝑢,𝑣)≠𝐴^𝜆𝑛(𝑢,𝑣)+𝑚𝜑₁,𝜑₂.(3.9)
Otherwise, there exist (𝑢₀,𝑣₀)∈𝜕𝐾_𝑅1 and 𝑚₀>0 such that 𝑢₀,𝑣₀=𝐴^𝜆𝑛𝑢₀,𝑣₀+𝑚₀𝜑₁,𝜑₂.(3.10)
From (𝑢₀,𝑣₀)∈𝜕𝐾_𝑅1, we know that ‖𝑢₀‖=𝑅₁ or ‖𝑣₀‖=𝑅₁. Without loss of generality, we may suppose that ‖𝑢₀‖=𝑅₁, then 𝑢₀(𝜏)≥Λ‖𝑢₀‖=Λ𝑅₁>𝑅₀ for any 𝜏∈[0,1]. So, by (2.13), (3.8), we have 𝑢₀(𝑡)=𝜆¹⁰𝐻₁(𝑡,𝑠)𝜙_𝑞1¹⁰𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢₀(𝜏),𝑣₀(𝜏)𝑑𝜏𝑑𝑠+𝑚₀≥𝜆𝜌₁𝜎^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢₀(𝜏),𝑣₀(𝜏)𝑑𝜏𝑑𝑠+𝑚₀≥𝜆𝜌₁𝜎^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑓_1∞−𝜀₁^𝑝1−1𝜙_𝑝1𝑢₀(𝜏)𝑑𝜏𝑑𝑠+𝑚₀≥𝜆𝜌₁𝜎^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑓_1∞−𝜀₁^𝑝1−1Λ𝑅₁^𝑝1−1𝑑𝜏𝑑𝑠+𝑚₀=16𝜆𝜌¹𝜎^𝑞11−1𝑓_1∞−𝜀₁Λ𝑅₁𝐿₁+𝑚₀=𝜆𝑀¹⁻¹𝑓_1∞−𝜀₁𝑅₁+𝑚₀>𝑅₁.(3.11)
This implies that 𝑅₁>𝑅₁, which is a contradiction. This yields that (3.9) holds. By (3.7), (3.9), and Lemma 1.2, for any 𝑛>1/𝑟₁and 𝜆∈(𝑀₁/𝑓_1∞,𝑁₁/𝑓⁰¹), 𝜇∈(0,𝑁₂/𝑓⁰²), we obtain that 𝑇_𝑛 has a fixed point (𝑢_𝑛,𝑣_𝑛) in 𝐾_𝑟1,𝑅1 satisfying 𝑟₁<‖𝑢_𝑛‖<𝑅₁,𝑟₁<‖𝑣_𝑛‖<𝑅₁.
Let {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥𝑛1 be the sequence of solutions of boundary value problems (2.19), where 𝑛₁>1/𝑟₁is a fixed integer. It is easy to see that they are uniformly bounded. Next we show that {𝑢_𝑛}_𝑛≥𝑛1 are equicontinuous on [0,1]. From (𝑢_𝑛,𝑣_𝑛)∈𝐾_𝑟1,𝑅1, we know that 𝑅₁≥𝑢_𝑛(𝜏)≥Λ‖𝑢_𝑛‖≥Λ𝑟₁, 𝑅₁≥𝑣_𝑛(𝜏)≥Λ‖𝑣_𝑛‖≥Λ𝑟₁, 𝜏∈[0,1]. For any 𝜀>0, by the continuous of 𝐻₁(𝑡,𝑠) in [0,1]×[0,1], there exists 𝛿₁>0 such that for any 𝑡₁, 𝑡₂, 𝑠∈[0,1] and |𝑡₁−𝑡₂|<𝛿₁, we have ||𝐻₁𝑡₁,𝑠−𝐻₁𝑡₂,𝑠||<𝜀.(3.12)
This combining with (2.15), (2.32) implies that for any 𝑡₁, 𝑡₂∈[0,1] and |𝑡₁−𝑡₂|<𝛿₁, we have ||𝑢_𝑛𝑡≤₁𝜆−₁₀𝑢_𝑛𝑡₂||||𝐻₁𝑡₁,𝑠−𝐻₁𝑡₂,𝑠||𝜙_𝑞1¹⁰<𝜀𝜆𝜈_𝑞11−1𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢_𝑛(𝜏),𝑣_𝑛(𝜏)𝑑𝜏𝑑𝑠𝜙_𝑞1¹⁰≤𝜀𝜆𝜈_𝑞11−1𝑒(𝜏)𝑎₁(𝜏)𝑔₁𝜏,max𝑢_𝑛(𝜏),𝑛⁻¹+ℎ₁𝜏,𝑣_𝑛(𝜏)𝑑𝜏2^𝑞1−1𝜙^𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑔₁(𝜏,𝑟Λ)𝑑𝜏+𝜙_𝑞1¹⁰≤𝜀𝜆𝜈_𝑞11−1𝑒(𝜏)𝑎₁(𝜏)ℎ₁(𝜏,𝑣(𝜏))𝑑𝜏2^𝑞1−1⎡⎢⎢⎣𝐿_𝑟+𝐿₁⎛⎜⎜⎝max_{𝑟11𝜏∈[0,1]𝑦∈Λ,𝑅}ℎ₁(𝜏,𝑦)⎞⎟⎟⎠^𝑞1−1⎤⎥⎥⎦.(3.13)
Similarly, {𝑣_𝑛}_𝑛≥𝑛1 are also equicontinuous on [0,1]. By the Ascoli-Arzela theorem, the sequence {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥𝑛1 has a subsequence being uniformly convergent on [0,1]. From Lemma 2.2, we know that 𝑢^𝑛(𝑠)=𝜆^𝑝1−1¹⁰𝑣𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢_𝑛(𝜏),𝑣_𝑛(𝜏)𝑑𝜏,_𝑛(𝑠)=𝜇^𝑝2−1¹⁰𝐾₂(𝑠,𝜏)𝑎₂(𝜏)𝑓_2𝑛𝜏,𝑢_𝑛(𝜏),𝑣_𝑛(𝜏)𝑑𝜏.(3.14)
Since the properties of 𝐾_𝑖(𝑡,𝑠)(𝑖=1,2) are similar to those of 𝐻_𝑖(𝑡,𝑠)(𝑖=1,2), so (𝑢^𝑛,𝑣^𝑛) have the similar properties of (𝑢_𝑛,𝑣_𝑛), that is, (𝑢^𝑛,𝑣^𝑛) also has a subsequence being uniformly convergent on [0,1]. Without loss of generality, we still assume that {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥𝑛1 itself uniformly converges to (𝑢,𝑣) on [0,1] and {(𝑢^𝑛,𝑣^𝑛)}_𝑛≥𝑛1 itself uniformly converges to (𝑢^,𝑣^) on [0,1], respectively. Since {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥𝑛1∈𝐾_𝑟1,𝑅1⊂𝐾, so we have 𝑢_𝑛≥0, 𝑣_𝑛≥0. By (2.19), we have 𝑢_𝑛(𝑡)=𝑢_𝑛1−_𝑡1/22+𝑡−12𝑢^𝑛21𝑑𝑠^𝑠1/2𝜙_𝑞1𝑢^{𝑝1𝑛−1}12+𝑠²−1−_𝑠21/22𝑢^{𝑝1𝑛−1}21𝑑𝑠₁^𝑠1_1/2𝑣𝜆^𝑝1−1𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢_𝑛(𝜏),𝑣_𝑛(𝜏)𝑑𝜏𝑑𝑠₂,𝑡∈(0,1),(3.15)_𝑛(𝑡)=𝑣_𝑛1−2_𝑡1/2+𝑡−12𝑣^𝑛21𝑑𝑠^𝑠1/2𝜙_𝑞2𝑣^{𝑝2𝑛−1}12+𝑠²−1−_𝑠21/22𝑣^{𝑝2𝑛−1}21𝑑𝑠₁^𝑠1_1/2𝜇^𝑝2−1𝑎₂(𝜏)𝑓_2𝑛𝜏,𝑢_𝑛(𝜏),𝑣_𝑛(𝜏)𝑑𝜏𝑑𝑠₂,𝑡∈(0,1).(3.16)
From (3.15) and (3.16), we know that {𝑢^𝑛(1/2)}_𝑛≥𝑛1, {𝑣^𝑛(1/2)}_𝑛≥𝑛1, {𝑢^𝑛(1/2)}_𝑛≥𝑛1, {𝑣^𝑛(1/2)}_𝑛≥𝑛1, {𝑢^{𝑛}(1/2)}_𝑛≥𝑛1, {𝑣^{𝑛}(1/2)}_𝑛≥𝑛1 are bounded sets. Without loss of generality, we may assume (𝑢^𝑛(1/2),𝑣^𝑛(1/2))→(𝑐₁,𝑑₁),(𝑢^𝑛(1/2),𝑣^𝑛(1/2))→(𝑐₂,𝑑₂),(𝑢^{𝑛}(1/2),𝑣^{𝑛}(1/2))→(𝑐₃,𝑑₃) as 𝑛→∞. Then by (3.15), (3.16), and the Lebesgue dominated convergence theorem, we have 𝑢(𝑡)=𝑢12+𝑐¹𝑡−12−^𝑡1/2𝑑𝑠^𝑠1/2𝜙_𝑞1𝑐^𝑝12−1+𝑐^𝑝13−1𝑠²−1−_𝑠21/22𝑑𝑠₁^𝑠1_1/2𝑣(𝑡)=𝑣1𝜆^𝑝1−1𝑎₁(𝜏)𝑓₁(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠₂,𝑡∈(0,1),(3.17)−2_𝑡1/2+𝑑¹𝑡−12𝑑𝑠^𝑠1/2𝜙_𝑞2(𝑑^𝑝22−1+𝑑^𝑝23−1𝑠²−1−_𝑠21/22𝑑𝑠₁^𝑠1_1/2𝜇^𝑝2−1𝑎₂(𝜏)𝑓₂(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏)𝑑𝑠₂𝑡∈(0,1).(3.18)
By (3.17) and (3.18), direct computation shows that 𝜙𝜙_𝑝1𝑢^(𝑡)^=𝜆^𝑝1−1𝑎₁(𝑡)𝑓₁(𝑡,𝑢(𝑡),𝑣(𝑡)),_𝑝2𝑣^(𝑡)^=𝜇^𝑝2−1𝑎₂(𝑡)𝑓₂(𝑡,𝑢(𝑡),𝑣(𝑡)),0<𝑡<1.(3.19)
On the other hand, (𝑢,𝑣) satisfies the boundary condition of (1.3). In fact, 𝑢_𝑛(0)=𝑢_𝑛(1)=∫¹⁰𝑢_𝑛(𝑠)𝑑𝜉₁(𝑠), 𝑣_𝑛(0)=𝑣_𝑛(1)=∫¹⁰𝑣_𝑛(𝑠)𝑑𝜉₂(𝑠),  𝜙_𝑝1(𝑢^𝑛(0))=𝜙_𝑝1(𝑢^𝑛(1))=∫¹⁰𝜙_𝑝1(𝑢^𝑛(𝑠))𝑑𝜂₁(𝑠),  𝜙_𝑝2(𝑣^𝑛(0))=𝜙_𝑝2(𝑣^𝑛(1))=∫¹⁰𝜙_𝑝2(𝑣^𝑛(𝑠))𝑑𝜂₂(𝑠), and so the conclusion holds by letting 𝑛→∞.

Theorem 3.2. Assume that (𝐻₁)–(𝐻₃) hold. Then we have:(𝐷₁) If 𝑓₁₀, 𝑓^∞1, 𝑓^∞2∈(0,∞) and 𝑀₁/𝑓₁₀<𝑁₁/𝑓^∞1, then for each 𝜆∈(𝑀₁/𝑓₁₀,𝑁₁/𝑓^∞1), 𝜇∈(0,𝑁₂/𝑓^∞2), the system (1.3) has at least one positive solution.(𝐷₂) If 𝑓^∞1, 𝑓₂₀, 𝑓^∞2∈(0,∞) and 𝑀₂/𝑓₂₀<𝑁₂/𝑓^∞2, then for each 𝜆∈(0,𝑁₁/𝑓^∞1), 𝜇∈(𝑀₂/𝑓₂₀,𝑁₂/𝑓^∞2), the system (1.3) has at least one positive solution.(𝐷₃) If 𝑓₁₀=∞, 𝑓^∞1=0, 0<𝑓^∞2<∞, then for each 𝜆∈(0,∞), 𝜇∈(0,𝑁₂/(𝑓^∞2)), the system (1.3) has at least one positive solution.(𝐷₄) If 0<𝑓^∞1<∞, 𝑓₂₀=∞, 𝑓^∞2=0, then for each 𝜆∈(0,𝑁₁/𝑓^∞1), 𝜇∈(0,∞), the system (1.3) has at least one positive solution.(𝐷₅) If 𝑓_𝑖0=∞, 𝑓^∞𝑖=0(𝑖=1,2), then for each 𝜆∈(0,∞), 𝜇∈(0,∞), the system (1.3) has at least one positive solution.(𝐷₆) If 0<𝑓^∞1<∞, 𝑓₁₀=∞ or 𝑓₂₀=∞, 0<𝑓^∞2<∞, then for each 𝜆∈(0,𝑁₁/𝑓^∞1), 𝜇∈(0,𝑁₂/𝑓^∞2), the system (1.3) has at least one positive solution.(𝐷₇) If 𝑓^∞1=0, 0<𝑓₁₀<∞, and 𝑓^∞2=0, 0<𝑓₂₀<∞, then for each 𝜆∈(𝑀₁/𝑓₁₀,∞), 𝜇∈(0,∞) or 𝜆∈(0,∞), 𝜇∈(𝑀₂/𝑓₂₀,∞), the system (1.3) has at least one positive solution.

Proof. We may suppose that condition (𝐷₁) holds. Similarly, we can prove the other cases.
Let 𝜆∈(𝑀₁/𝑓₁₀,𝑁₁/𝑓^∞1), 𝜇∈(0,𝑁₂/𝑓^∞2). We can choose 𝜀₂>0 such that 𝑁₁−𝜀₂>0, 𝑁₂−𝜀₂>0 and 𝜆𝑓^∞1<𝑁₁−𝜀₂,𝜇𝑓^∞2<𝑁₂−𝜀₂.(3.20)
It follows from (𝐷₁) and (2.16) that there exists 𝑅^∗2>0 such that for any 𝑡∈[0,1]𝑓₁(𝑡,𝑥,𝑦)≤1𝜆𝑁¹𝑓−𝜀₂^𝑝1−1𝜙_𝑝1(𝑥),𝑥≥𝑅^∗2,𝑦≥0,(3.21)₂(𝑡,𝑥,𝑦)≤1𝜆𝑁²−𝜀₂^𝑝2−1𝜙_𝑝2(𝑦),𝑥≥0,𝑦≥𝑅^∗2.(3.22)
Let 𝑅₂=Λ⁻¹𝑅^∗2, 𝐾_𝑅2={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑅₂,‖𝑣‖<𝑅₂}. For any (𝑢,𝑣)∈𝜕𝐾_𝑅2, 𝑛∈ℕ, by (2.13), (3.21), we have ‖‖𝐴^𝜆𝑛(𝑢,𝑣)‖‖=max_[]𝑡∈0,1𝜆¹⁰𝐻₁(𝑡,𝑠)𝜙_𝑞1¹⁰≤𝜆𝛾₁𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠𝜈^𝑞11−1𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)1𝜆𝑁¹≤𝜆𝛾₁−𝜀₂^𝑝1−1𝜙_𝑝1(𝑢(𝜏))𝑑𝜏𝜈^𝑞11−11𝜆𝑁¹−𝜀₂𝐿₁𝑅₂<𝑅₂.(3.23)
Similarly, by (3.22) we have ‖𝐵^𝜇𝑛(𝑢,𝑣)‖<𝑅₂. Therefore, ‖‖𝑇_𝑛(𝑢,𝑣)‖‖=‖‖𝐴^𝜆𝑛(𝑢,𝑣)‖‖+‖‖𝐵^𝜇𝑛(𝑢,𝑣)‖‖≤2𝑅₂=‖(𝑢,𝑣)‖,(𝑢,𝑣)∈𝜕𝐾_𝑅2,𝑛∈ℕ.(3.24)
On the other hand, choose 𝜀₃>0 such that 𝑀₁+𝜀₃<𝜆𝑓₁₀. By the condition 𝑓₁₀∈(0,∞) of (𝐷₁) and (2.16), there exists 𝑟^∗2>0 such that 𝑓₁(𝑡,𝑥,𝑦)≥1𝜆𝑀¹+𝜀₃^𝑝1−1𝜙_𝑝1(𝑥),𝑡∈[0,1],0<𝑥≤𝑟^∗2,𝑦≥0.(3.25)
Let 0<𝑟₂<min{𝑅₂,𝑟^∗2},  𝐾_𝑟2={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑟₂,‖𝑣‖<𝑟₂}. Next, we take (𝜑₁,𝜑₂)=(1,1)∈𝜕𝐾₁, 𝑛>1/𝑟₂, and for any (𝑢,𝑣)∈𝜕𝐾_𝑟2, 𝑚>0, we will show (𝑢,𝑣)≠𝐴^𝜆𝑛(𝑢,𝑣)+𝑚𝜑₁,𝜑₂.(3.26)
Otherwise, there exist (𝑢₀,𝑣₀)∈𝜕𝐾_𝑟2 and 𝑚₀>0 such that 𝑢₀,𝑣₀=𝐴^𝜆𝑛𝑢₀,𝑣₀+𝑚₀𝜑₁,𝜑₂.(3.27)
From (𝑢₀,𝑣₀)∈𝜕𝐾_𝑟2, we know that ‖𝑢₀‖=𝑟₂ or ‖𝑣₀‖=𝑟₂. Without loss of generality, we may suppose that ‖𝑢₀‖=𝑟₂, then 𝑢₀(𝜏)≥Λ‖𝑢₀‖≥Λ𝑟₂ for any 𝜏∈[0,1]. So, we have 𝑢₀(𝑡)=𝜆¹⁰𝐻₁(𝑡,𝑠)𝜙_𝑞1¹⁰𝐾₁(𝑠,𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢₀(𝜏),𝑣₀(𝜏)𝑑𝜏𝑑𝑠+𝑚₀≥𝜆𝜌₁𝜎^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑓_1𝑛𝜏,𝑢₀(𝜏),𝑣₀(𝜏)𝑑𝜏𝑑𝑠+𝑚₀≥𝜆𝜌₁𝜎^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)1𝜆𝑀¹+𝜀₃^𝑝1−1𝜙_𝑝1𝑢₀(𝜏)𝑑𝜏𝑑𝑠+𝑚₀≥𝜆𝜌₁𝜎^𝑞11−1¹⁰𝑒(𝑠)𝜙_𝑞1¹⁰𝑒(𝜏)𝑎₁(𝜏)𝑀₁+𝜀₃^𝑝1−1Λ𝑟₂^𝑝1−1𝑑𝜏𝑑𝑠+𝑚₀=16𝜆𝜌¹𝜎^𝑞11−11𝜆𝑀¹+𝜀₃Λ𝑟₂𝐿₁+𝑚₀>𝑟₂.(3.28)
This implies that 𝑟₂>𝑟₂, which is a contradiction. This yields that (3.26) holds. By (3.24), (3.26), and Lemma 1.2, for any 𝑛>1/𝑟₂and 𝜆∈(𝑀₁/𝑓₁₀,𝑁₁/𝑓^∞1), 𝜇∈(0,𝑁₂/𝑓^∞2), we obtain that 𝑇_𝑛 has a fixed point (𝑢_𝑛,𝑣_𝑛) in 𝐾_𝑟2,𝑅2 and 𝑟₂<‖𝑢_𝑛‖<𝑅₂,  𝑟₂<‖𝑣_𝑛‖<𝑅₂. The rest of proof is similar to Theorem 3.1.

4. An Example

Example 4.1. We consider system (1.3) with 𝑝₁=3/2,   𝑝₂=7/3,  𝑎₁(𝑡)=1/(𝑡√(1−𝑡),  𝑎₂(𝑡)=1/((1−𝑡)√𝑡), 𝑓₁(𝑡,𝑢,𝑣)=𝑡²+1√𝑢+1+sin𝑣²𝑓₂+𝑣+𝑡,(𝑡,𝑢,𝑣)∈[0,1]×ℝ⁺⁰×ℝ⁺,(𝑡,𝑢,𝑣)=2+sin(𝑢+ln(𝑡+1))+𝑡⁴+𝑡+3√𝑣,(𝑡,𝑢,𝑣)∈[0,1]×ℝ⁺×ℝ⁺⁰.(4.1)

Obviously, 𝑎₁,  𝑎₂ are singular at 𝑡=0 and 𝑡=1, 𝑓₁(𝑡,𝑢,𝑣) is singular at 𝑢=0 and 𝑓₂(𝑡,𝑢,𝑣) is singular at 𝑣=0. Choose 𝑔₁(𝑡,𝑢)=(𝑡²+1)/√𝑢, ℎ₁(𝑡,𝑣)=1+sin(𝑣²+𝑣+𝑡), 𝑔₂(𝑡,𝑢)=2+sin(𝑢+ln(𝑡+1)), and ℎ₂(𝑡,𝑣)=(𝑡⁴+𝑡+3)/√𝑣. Let 𝜉₁(𝑠)=⎧⎪⎪⎨⎪⎪⎩0,𝑠∈0,1