Discrete Dynamics in Nature and Society

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Discrete Dynamics in Nature and Society / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 293734 | 19 pages | https://doi.org/10.1155/2012/293734

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

Jiqiang Jiang ,1 Lishan Liu ,1,2 and Yonghong Wu2

1School of Mathematical Sciences, Qufu Normal University, Qufu 273165, China

2Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia

Academic Editor: Cengiz Çinar
Received25 Mar 2012
Accepted15 May 2012
Published08 Aug 2012

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 𝜙𝑝𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),0<𝑡<1,𝑢(0)=𝑢(1)−𝑎𝑢(𝜉)=𝑢(0)=𝑢(1)−𝑏𝑢(𝜂)=0,(1.1) where 𝜙𝑝(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: 𝜙𝑝𝑢(𝑡)=𝑤(𝑡)𝑓(𝑡,𝑢(𝑡)),0<𝑡<1,𝑢(0)=𝑢(1)=10𝜙𝑔(𝑠)𝑢(𝑠)𝑑𝑠,𝑝𝑢(0)=𝜙𝑝𝑢=(1)10ℎ(𝑠)𝜙𝑝𝑢(𝑠)𝑑𝑠,(1.2) where 𝜙𝑝(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: 𝜙𝑝1𝑢(𝑡)=𝜆𝑝1−1𝑎1(𝑡)𝑓1𝜙(𝑡,𝑢(𝑡),𝑣(𝑡)),0<𝑡<1,𝑝2𝑣(𝑡)=𝜇𝑝2−1𝑎2(𝑡)𝑓2(𝑡,𝑢(𝑡),𝑣(𝑡)),𝑢(0)=𝑢(1)=10𝑢(𝑠)𝑑𝜉1𝜙(𝑠),𝑝1𝑢(0)=𝜙𝑝1𝑢=(1)10𝜙𝑝1𝑢(𝑠)𝑑𝜂1(𝑠),𝑣(0)=𝑣(1)=10𝑣(𝑠)𝑑𝜉2𝜙(𝑠),𝑝2𝑣(0)=𝜙𝑝2𝑣=(1)10𝜙𝑝2𝑣(𝑠)𝑑𝜂2(𝑠),(1.3) where 𝜆 and 𝜇 are positive parameters, 𝜙𝑝𝑖(𝑥)=|𝑥|𝑝𝑖−2𝑥, 𝑝𝑖>1, 𝜙𝑞𝑖=𝜙𝑝−1𝑖, 1/𝑝𝑖+1/𝑞𝑖=1, 𝜉𝑖,𝜂𝑖∶[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 𝐾𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟}, 𝜕𝐾𝑟={𝑥∈𝐾∶‖𝑥‖=𝑟}, and 𝐾𝑟,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∶𝐾𝑟,𝑅→𝐾 a completely continuous operator. If the following conditions hold(i)‖𝑇𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝐾𝑅;(ii)there exists 𝑒∈𝜕𝐾1 such that 𝑥≠𝑇𝑥+𝑚𝑒 for any 𝑥∈𝜕𝐾𝑟 and 𝑚>0. Then 𝑇 has a fixed point in 𝐾𝑟,𝑅.

Remark 1.3. If (i) and (ii) are satisfied for 𝑥∈𝜕𝐾𝑟 and 𝑥∈𝜕𝐾𝑅, 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‖=max0≤𝑡≤1||||𝑢(𝑡),‖𝑣‖=max0≤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<𝐿𝑖∶=10𝑒(𝑠)𝑎𝑖(𝑠)𝑑𝑠𝑞𝑖−1<+∞,𝑖=1,2,(2.2) where 𝑒(𝑠)=𝑠(1−𝑠), 𝑠∈[0,1].(𝐻2)𝜉𝑖,𝜂𝑖∶[0,1]→ℝ+(𝑖=1,2) are nondecreasing functions of bounded variation, and 𝛼𝑖∈[0,1), 𝛽𝑖∈[0,1), where 𝛼𝑖=10𝑑𝜉𝑖(𝑠),𝛽𝑖=10𝑑𝜂𝑖(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 𝑓1(𝑡,𝑥,𝑦)≤𝑔1(𝑡,𝑥)+ℎ1[](𝑡,𝑦),∀(𝑡,𝑥,𝑦)∈0,1×ℝ+0×ℝ+,𝑓2(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 −𝑥(𝑡)=𝜙𝑞𝑖(𝑦(𝑡)),0<𝑡<1,𝑥(0)=𝑥(1)=10𝑥(𝑠)𝑑𝜉𝑖(𝑠)(2.6) has a unique solution 𝑥(𝑡)=10𝐻𝑖(𝑡,𝑠)𝜙𝑞𝑖(𝑦(𝑠))𝑑𝑠,(2.7) where 𝐻𝑖1(𝑡,𝑠)=𝐺(𝑡,𝑠)+1−𝛼𝑖10𝐺(𝜏,𝑠)𝑑𝜉𝑖(𝜏),𝑖=1,2,𝐺(𝑡,𝑠)=𝑠(1−𝑡),0≤𝑠≤𝑡≤1,𝑡(1−𝑠),0≤𝑡≤𝑠≤1.(2.8)

Lemma 2.2. If (𝐻2) holds, then for any 𝑧∈𝐿(0,1), the boundary value problem −𝑦(𝑡)=𝑧(𝑡),0<𝑡<1,𝑦(0)=𝑦(1)=10𝑦(𝑠)𝑑𝜂𝑖(𝑠)(2.9) has a unique solution 𝑦(𝑡)=10𝐾𝑖(𝑡,𝑠)𝑧(𝑠)𝑑𝑠,(2.10) where 𝐾𝑖1(𝑡,𝑠)=𝐺(𝑡,𝑠)+1−𝛽𝑖10𝐺(𝜏,𝑠)𝑑𝜂𝑖(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∈0,11𝑒(𝑡)=4.(2.12)

Remark 2.4. If (𝐻2) holds, it is easy to testify 𝐻𝑖(𝑡,𝑠) defined by (2.8) that: 𝜌𝑖𝑒(𝑠)≤𝐻𝑖(𝑡,𝑠)≤𝛾𝑖1𝑒(𝑠)≤4𝛾𝑖<𝛾𝑖[],𝑡,𝑠∈0,1,𝑖=1,2,(2.13) where 𝛾𝑖=11−𝛼𝑖,𝜌𝑖=∫10𝑒(𝜏)𝑑𝜉𝑖(𝜏)1−𝛼𝑖,𝑖=1,2.(2.14)

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

Lemma 2.6. For 𝑥>0, 𝑦>0, we have 𝜙𝑞𝑖2(𝑥+𝑦)≤𝑞𝑖−1𝜙𝑞𝑖(𝑥)+𝜙𝑞𝑖(𝑦),𝑞𝑖𝜙≥2𝑞𝑖(𝑥)+𝜙𝑞𝑖(𝑦),1<𝑞𝑖,<2≤2𝑞𝑖−1𝜙𝑞𝑖(𝑥)+𝜙𝑞𝑖(𝑦),𝑞𝑖𝜙>1,𝑖=1,2,(2.15)𝑞𝑖(𝑥)>𝜙𝑞𝑖(𝑦)>𝜙𝑞𝑖(0)=0,𝑥>𝑦>0,𝑞𝑖>1,𝑖=1,2.(2.16)

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

Let 𝐾=(𝑢,𝑣)∈𝐶+[]0,1×𝐶+[][],0,1∶𝑢,𝑣areconcaveon0,1min[]𝑡∈0,1𝑢(𝑡)≥Λ‖𝑢‖,min[]𝑡∈0,1,𝑣(𝑡)≥Λ‖𝑣‖(2.17) where 𝜌Λ=min1𝜎𝑞11−1𝛾1𝜈𝑞11−1,𝜌2𝜎𝑞22−1𝛾2𝜈𝑞22−1,𝜎𝑖=∫10𝑒(𝑠)𝑑𝜂𝑖(𝑠)1−𝛽𝑖,𝜈𝑖=11−𝛽𝑖,𝑖=1,2.(2.18)

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

Remark 2.7. By the definition of 𝜌𝑖, 𝜎𝑖, 𝛾𝑖, 𝜈𝑖(𝑖=1,2), we have 0<Λ<1.
To overcome singularity, we consider the following approximate problem of (1.3): 𝜙𝑝1𝑢(𝑡)=𝜆𝑝1−1𝑎1(𝑡)𝑓1𝑛𝜙(𝑡,𝑢(𝑡),𝑣(𝑡)),0<𝑡<1,𝑝2𝑣(𝑡)=𝜇𝑝2−1𝑎2(𝑡)𝑓2𝑛(𝑡,𝑢(𝑡),𝑣(𝑡)),𝑢(0)=𝑢(1)=10𝑢(𝑠)𝑑𝜉1𝜙(𝑠),𝑝1𝑢(0)=𝜙𝑝1𝑢=(1)10𝜙𝑝1𝑢(𝑠)𝑑𝜂1(𝑠),𝑣(0)=𝑣(1)=10𝑣(𝑠)𝑑𝜉2𝜙(𝑠),𝑝2𝑣(0)=𝜙𝑝2𝑣=(1)10𝜙𝑝2𝑣(𝑠)𝑑𝜂2(𝑠),(2.19) where 𝑛 is a positive integer and 𝑓1𝑛(𝑡,𝑢,𝑣)=𝑓1𝑡,max𝑢,𝑛−1,𝑣,𝑓2𝑛(𝑡,𝑢,𝑣)=𝑓2𝑡,𝑢,max𝑣,𝑛−1.(2.20)
Clearly, 𝑓𝑖𝑛∈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 𝐴𝜆𝑛∶𝐾→𝐶[0,1], 𝐵𝜇𝑛∶𝐾→𝐶[0,1], and 𝑇𝑛∶𝐾→𝐸 by 𝐴𝜆𝑛(𝑢,𝑣)(𝑡)=𝜆10𝐻1(𝑡,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝐵(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠,(2.21)𝜇𝑛(𝑢,𝑣)(𝑡)=𝜇10𝐻2(𝑡,𝑠)𝜙𝑞210𝐾2(𝑠,𝜏)𝑎2(𝜏)𝑓2𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠,(2.22) and 𝑇𝑛(𝑢,𝑣)=(𝐴𝜆𝑛(𝑢,𝑣),𝐵𝜇𝑛(𝑢,𝑣)), respectively.

Lemma 2.8. Assume that (𝐻1)–(𝐻3) hold, then for each 𝜆>0, 𝜇>0, 𝑛∈ℕ, 𝑇𝑛∶𝐾𝑟,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 𝐴𝜆𝑛(𝑢,𝑣)(𝑡)=−𝜆𝜙𝑞110𝐾1(𝑡,𝜏)𝑎1(𝜏)𝑓1𝑛𝐴(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏≤0,𝜆𝑛(𝑢,𝑣)(0)=𝐴𝜆𝑛(𝑢,𝑣)(1)=𝜆10𝐻1(0,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠≥0,(2.23) which implies that 𝐴𝜆𝑛 is nonnegative and concave on [0,1]. Similarly, by (2.22) we can obtain that 𝐵𝜇𝑛 is nonnegative and concave on [0,1]. For any (𝑢,𝑣)∈𝐾 and 𝑡∈[0,1], it follows from (2.13) that 𝐴𝜆𝑛(𝑢,𝑣)(𝑡)=𝜆10𝐻1(𝑡,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠≤𝜆𝛾1𝜈𝑞11−110𝑒(𝑠)𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠.(2.24)
Thus ‖‖𝐴𝜆𝑛‖‖(𝑢,𝑣)≤𝜆𝛾1𝜈𝑞11−110𝑒(𝑠)𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠.(2.25)
On the other hand, by (2.13) and (2.18), we have 𝐴𝜆𝑛(𝑢,𝑣)(𝑡)=𝜆10𝐻1(𝑡,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠≥𝜆𝜌1𝜎𝑞11−110𝑒(𝑠)𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)𝑓1𝑛≥𝜌(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠1𝜎𝑞11−1𝛾1𝜈𝑞11−1‖‖𝐴𝜆𝑛‖‖‖‖𝐴(𝑢,𝑣)≥Λ𝜆𝑛‖‖.(𝑢,𝑣)(2.26)
This implies that min[]𝑡∈0,1𝐴𝜆𝑛‖‖𝐴(𝑢,𝑣)(𝑡)≥Λ𝜆𝑛‖‖.(𝑢,𝑣)(2.27)
Similar to (2.27), we also have min[]𝑡∈0,1𝐵𝜇𝑛‖‖𝐵(𝑢,𝑣)(𝑡)≥Λ𝜇𝑛‖‖.(𝑢,𝑣)(2.28)
Therefore, 𝑇𝑛(𝐾)⊂𝐾.
Next, we prove that 𝑇𝑛∶𝐾𝑟,𝑅→𝐾 is completely continuous. Suppose (𝑢𝑚,𝑣𝑚)∈𝐾𝑟,𝑅 and (𝑢0,𝑣0)∈𝐾𝑟,𝑅 with ‖(𝑢𝑚,𝑣𝑚)−(𝑢0,𝑣0)‖→0(𝑚→∞). We notice that 𝑡∈[0,1]𝑓𝑖𝑛(𝑡,𝑢𝑚(𝑡),𝑣𝑚(𝑡))−𝑓𝑖𝑛(𝑡,𝑢0(𝑡),𝑣0(𝑡))→0(𝑚→∞). Using the Lebesgue dominated convergence theorem, we have ||||𝜙𝑝1𝑞−1110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢𝑚(𝜏),𝑣𝑚(𝜏)𝑑𝜏−𝜙𝑝1𝑞−1110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢0(𝜏),𝑣0(||||𝜏)𝑑𝜏≤𝜈110𝑒(𝜏)𝑎1||𝑓(𝜏)1𝑛𝜏,𝑢𝑚(𝜏),𝑣𝑚(𝜏)−𝑓1𝑛𝜏,𝑢0(𝜏),𝑣0||(𝜏)𝑑𝜏⟶0,𝑚⟶∞.(2.29)
Therefore, ‖‖𝐴𝜆𝑛𝑢𝑚,𝑣𝑚−𝐴𝜆𝑛𝑢0,𝑣0‖‖≤𝜆𝛾110||||𝜙𝑒(𝑠)𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢𝑚(𝜏),𝑣𝑚(𝜏)𝑑𝜏−𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢0(𝜏),𝑣0||||(𝜏)𝑑𝜏𝑑𝑠⟶0,𝑚⟶∞.(2.30)
Similarly, we also have ‖‖𝐵𝜇𝑛𝑢𝑚,𝑣𝑚−𝐵𝜇𝑛𝑢0,𝑣0‖‖⟶0,𝑚⟶∞.(2.31)
So 𝐴𝜆𝑛∶𝐾𝑟,𝑅→𝐶[0,1] and 𝐵𝛽𝑛∶𝐾𝑟,𝑅→𝐶[0,1] are continuous. Therefore, 𝑇𝑛∶𝐾𝑟,𝑅→𝐾 is also continuous.
Let 𝐷⊂𝐾𝑟,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 𝐿𝑟∶=10𝑒(𝜏)𝑎1(𝜏)𝑔1(𝑠,𝑟Λ)𝑑𝜏𝑞1−1<+∞.(2.32)
It is easy to show that 𝐴𝜆𝑛(𝐷) is uniformly bounded. In order to show that 𝑇𝑛 is a compact operator, we only need to show that 𝐴𝜆𝑛(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 ||𝐻1𝑡1,𝑠−𝐻1𝑡2||,𝑠<𝜀.(2.33)
This together with (2.15) and (2.32) implies ||𝐴𝜆𝑛𝑡(𝑢,𝑣)1−𝐴𝜆𝑛𝑡(𝑢,𝑣)2||≤𝜆10||𝐻1𝑡1,𝑠−𝐻1𝑡2||𝜙,𝑠𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠<𝜀𝜆𝜈𝑞11−1𝜙𝑞110𝑒(𝜏)𝑎1𝑔(𝜏)1𝜏,max𝑢(𝜏),𝑛−1+ℎ1(𝜏,𝑣(𝜏))𝑑𝜏≤𝜀𝜆𝜈𝑞11−1𝜙𝑞110𝑒(𝜏)𝑎1𝑔(𝜏)1(𝜏,𝑟Λ)+ℎ1(𝜏,𝑣(𝜏))𝑑𝜏≤𝜀𝜆𝜈𝑞11−12𝑞1−1𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)𝑔1(𝜏,𝑟Λ)𝑑𝜏+𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)ℎ1(𝜏,𝑣(𝜏))𝑑𝜏≤𝜀𝜆𝜈𝑞11−12𝑞1−1⎡⎢⎢⎣𝐿𝑟+𝐿1max[]𝜏∈[0,1]𝑦∈𝑟Λ,𝑅ℎ1(𝜏,𝑦)𝑞1−1⎤⎥⎥⎦,||𝑡1−𝑡2||<𝛿,(𝑢,𝑣)∈𝐷.(2.34)
This means that 𝐴𝜆𝑛(𝐷) is equicontinuous. By the Arzela-Ascoli theorem, 𝐴𝜆𝑛(𝐷) is a relatively compact set and that 𝐴𝜆𝑛∶𝐾𝑟,𝑅→𝐶[0,1] is a completely continuous operator.
In the same way, we can show that 𝐵𝜇𝑛∶𝐾𝑟,𝑅→𝐶[0,1] is also completely continuous, and so 𝑇𝑛∶𝐾𝑟,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 𝑀𝑖𝜌=6𝑖𝜎𝑞𝑖−1Λ𝐿𝑖−1,𝑁𝑖=𝛾𝑖𝜈𝑞𝑖𝑖−1𝐿𝑖−1𝑓,𝑖=1,2,𝛼1=⎛⎜⎜⎝limsup𝑥→𝛼sup+𝑡∈[0,1]𝑦∈ℝ𝑓1(𝑡,𝑥,𝑦)𝜙𝑝1⎞⎟⎟⎠(𝑥)𝑞1−1,𝑓𝛼2=⎛⎜⎜⎝limsup𝑦→𝛼sup+𝑡∈[0,1]𝑥∈ℝ𝑓2(𝑡,𝑥,𝑦)𝜙𝑝2⎞⎟⎟⎠(𝑦)𝑞2−1,𝑓1𝛼=⎛⎜⎜⎝liminf𝑥→𝛼inf+𝑡∈[0,1]𝑦∈ℝ𝑓1(𝑡,𝑥,𝑦)𝜙𝑝1⎞⎟⎟⎠(𝑥)𝑞1−1,𝑓2𝛼=liminf𝑦→𝛼inf+𝑡∈[0,1]𝑥∈ℝ𝑓2(𝑡,𝑥,𝑦)𝜙𝑝2(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, 𝑓1∞, 𝑓02∈(0,∞) and 𝑀1/𝑓1∞<𝑁1/𝑓01, then for each 𝜆∈(𝑀1/𝑓1∞,𝑁1/𝑓01), 𝜇∈(0,𝑁2/𝑓02), the system (1.3) has at least one positive solution.(𝐶2) If 𝑓01, 𝑓02, 𝑓2∞∈(0,∞) and 𝑀2/𝑓2∞<𝑁2/𝑓02, then for each 𝜆∈(0,𝑁1/𝑓01), 𝜇∈(𝑀2/𝑓2∞,𝑁2/𝑓02), the system (1.3) has at least one positive solution.(𝐶3) If 𝑓01=0, 𝑓1∞=∞, 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, 𝑓2∞=∞, then for each 𝜆∈(0,𝑁1/𝑓01), 𝜇∈(0,∞), the system (1.3) has at least one positive solution.(𝐶5) If 𝑓0𝑖=0, 𝑓𝑖∞=∞(𝑖=1,2), then for each 𝜆∈(0,∞), 𝜇∈(0,∞), the system (1.3) has at least one positive solution.(𝐶6) If 0<𝑓01<∞, 𝑓1∞=∞ or 𝑓2∞=∞, 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<𝑓1∞<∞, and 𝑓01=0, 0<𝑓2∞<∞, then for each 𝜆∈(𝑀1/𝑓1∞,∞), 𝜇∈(0,∞) or 𝜆∈(0,∞), 𝜇∈(𝑀2/𝑓2∞,∞), 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/𝑓1∞,𝑁1/𝑓01), 𝜇∈(0,𝑁2/(𝑓02)), choose 𝜀1>0 such that 𝑓1∞−𝜀1>0 and 𝑀1𝑓1∞−𝜀1𝑁≤𝜆≤1𝑓01+𝜀1𝑁,0<𝜇≤2𝑓02+𝜀1.(3.2)
It follows from 𝑓0𝑖∈(0,∞) of (𝐶1) that there exists 𝑟1>0 such that for any 𝑡∈[0,1], 𝑓1(𝑓𝑡,𝑥,𝑦)≤01+𝜀1𝑝1−1𝜙𝑝1(𝑟𝑥)≤1𝑓01+𝜀1𝑝1−1,0<𝑥≤𝑟1𝑓,𝑦≥0,(3.3)2𝑓(𝑡,𝑥,𝑦)≤02+𝜀1𝑝2−1𝜙𝑝2𝑟(𝑦)≤1𝑓02+𝜀1𝑝2−1,𝑥≥0,0<𝑦≤𝑟1.(3.4)
Let 𝐾𝑟1={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑟1,‖𝑣‖<𝑟1}. For any (𝑢,𝑣)∈𝜕𝐾𝑟1, 𝑛>1/𝑟1, by (2.13), (3.3), we have ‖‖𝐴𝜆𝑛‖‖(𝑢,𝑣)=max[]𝑡∈0,1𝜆10𝐻1(𝑡,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠≤𝜆𝛾1𝜈𝑞11−1𝜙𝑞110𝑒(𝜏)𝑎1𝑓(𝜏)01+𝜀1𝑝1−1𝜙𝑝1(𝑢(𝜏))𝑑𝜏≤𝜆𝛾1𝜈𝑞11−1𝑓01+𝜀1𝐿1𝑟1=𝜆𝑁1−1𝑓01+𝜀1𝑟1.(3.5)
Similarly, we also have ‖‖𝐵𝜇𝑛‖‖(𝑢,𝑣)≤𝜇𝑁2−1𝑓02+𝜀1𝑟1.(3.6)
Therefore, we have ‖‖𝑇𝑛‖‖=‖‖𝐴(𝑢,𝑣)𝜆𝑛‖‖+‖‖𝐵(𝑢,𝑣)𝜇𝑛‖‖≤(𝑢,𝑣)𝜆𝑁1−1𝑓01+𝜀1+𝜇𝑁2−1𝑓02+𝜀1𝑟1≤2𝑟1=‖(𝑢,𝑣)‖.(3.7)
On the other hand, by 𝑓1∞>𝑓1∞−𝜀1>0, there exists 𝑅0>0 such that 𝑓1𝑓(𝑡,𝑥,𝑦)≥1∞−𝜀1𝑝1−1𝜙𝑝1[](𝑥),𝑡∈0,1,𝑥≥𝑅0,𝑦≥0.(3.8)
Let 𝑅1>max{2𝑟1,Λ−1𝑅0}, 𝐾𝑅1={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑅1,‖𝑣‖<𝑅1}. Next, we take (𝜑1,𝜑2)=(1,1)∈𝜕𝐾1, and for any (𝑢,𝑣)∈𝜕𝐾𝑅1, 𝑚>0, 𝑛∈ℕ, we will show (𝑢,𝑣)≠𝐴𝜆𝑛𝜑(𝑢,𝑣)+𝑚1,𝜑2.(3.9)
Otherwise, there exist (𝑢0,𝑣0)∈𝜕𝐾𝑅1 and 𝑚0>0 such that 𝑢0,𝑣0=𝐴𝜆𝑛𝑢0,𝑣0+𝑚0𝜑1,𝜑2.(3.10)
From (𝑢0,𝑣0)∈𝜕𝐾𝑅1, 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)=𝜆10𝐻1(𝑡,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢0(𝜏),𝑣0(𝜏)𝑑𝜏𝑑𝑠+𝑚0≥𝜆𝜌1𝜎𝑞11−110𝑒(𝑠)𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢0(𝜏),𝑣0(𝜏)𝑑𝜏𝑑𝑠+𝑚0≥𝜆𝜌1𝜎𝑞11−110𝑒(𝑠)𝜙𝑞110𝑒(𝜏)𝑎1𝑓(𝜏)1∞−𝜀1𝑝1−1𝜙𝑝1𝑢0(𝜏)𝑑𝜏𝑑𝑠+𝑚0≥𝜆𝜌1𝜎𝑞11−110𝑒(𝑠)𝜙𝑞110𝑒(𝜏)𝑎1𝑓(𝜏)1∞−𝜀1𝑝1−1Λ𝑅1𝑝1−1𝑑𝜏𝑑𝑠+𝑚0=16𝜆𝜌1𝜎𝑞11−1𝑓1∞−𝜀1Λ𝑅1𝐿1+𝑚0=𝜆𝑀1−1𝑓1∞−𝜀1𝑅1+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/𝑓1∞,𝑁1/𝑓01), 𝜇∈(0,𝑁2/𝑓02), we obtain that 𝑇𝑛 has a fixed point (𝑢𝑛,𝑣𝑛) in 𝐾𝑟1,𝑅1 satisfying 𝑟1<‖𝑢𝑛‖<𝑅1,𝑟1<‖𝑣𝑛‖<𝑅1.
Let {(𝑢𝑛,𝑣𝑛)}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 {𝑢𝑛}𝑛≥𝑛1 are equicontinuous on [0,1]. From (𝑢𝑛,𝑣𝑛)∈𝐾𝑟1,𝑅1, we know that 𝑅1≥𝑢𝑛(𝜏)≥Λ‖𝑢𝑛‖≥Λ𝑟1, 𝑅1≥𝑣𝑛(𝜏)≥Λ‖𝑣𝑛‖≥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 ||𝐻1𝑡1,𝑠−𝐻1𝑡2||,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 ||𝑢𝑛𝑡1−𝑢𝑛𝑡2||≤𝜆10||𝐻1𝑡1,𝑠−𝐻1𝑡2||𝜙,𝑠𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢𝑛(𝜏),𝑣𝑛(𝜏)𝑑𝜏𝑑𝑠<𝜀𝜆𝜈𝑞11−1𝜙𝑞110𝑒(𝜏)𝑎1𝑔(𝜏)1𝑢𝜏,max𝑛(𝜏),𝑛−1+ℎ1𝜏,𝑣𝑛(𝜏)𝑑𝜏≤𝜀𝜆𝜈𝑞11−12𝑞1−1𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)𝑔1(𝜏,𝑟Λ)𝑑𝜏+𝜙𝑞110𝑒(𝜏)𝑎1(𝜏)ℎ1(𝜏,𝑣(𝜏))𝑑𝜏≤𝜀𝜆𝜈𝑞11−12𝑞1−1⎡⎢⎢⎣𝐿𝑟+𝐿1⎛⎜⎜⎝max𝑟11𝜏∈[0,1]𝑦∈Λ,𝑅ℎ1⎞⎟⎟⎠(𝜏,𝑦)𝑞1−1⎤⎥⎥⎦.(3.13)
Similarly, {𝑣𝑛}𝑛≥𝑛1 are also equicontinuous on [0,1]. By the Ascoli-Arzela theorem, the sequence {(𝑢𝑛,𝑣𝑛)}𝑛≥𝑛1 has a subsequence being uniformly convergent on [0,1]. From Lemma 2.2, we know that 𝑢𝑛(𝑠)=𝜆𝑝1−110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢𝑛(𝜏),𝑣𝑛𝑣(𝜏)𝑑𝜏,𝑛(𝑠)=𝜇𝑝2−110𝐾2(𝑠,𝜏)𝑎2(𝜏)𝑓2𝑛𝜏,𝑢𝑛(𝜏),𝑣𝑛(𝜏)𝑑𝜏.(3.14)
Since the properties of 𝐾𝑖(𝑡,𝑠)(𝑖=1,2) are similar to those of 𝐻𝑖(𝑡,𝑠)(𝑖=1,2), so (𝑢𝑛,𝑣𝑛) have the similar properties of (𝑢𝑛,𝑣𝑛), that is, (𝑢𝑛,𝑣𝑛) also has a subsequence being uniformly convergent on [0,1]. Without loss of generality, we still assume that {(𝑢𝑛,𝑣𝑛)}𝑛≥𝑛1 itself uniformly converges to (𝑢,𝑣) on [0,1] and {(𝑢𝑛,𝑣𝑛)}𝑛≥𝑛1 itself uniformly converges to (𝑢,𝑣) on [0,1], respectively. Since {(𝑢𝑛,𝑣𝑛)}𝑛≥𝑛1∈𝐾𝑟1,𝑅1⊂𝐾, so we have 𝑢𝑛≥0, 𝑣𝑛≥0. By (2.19), we have 𝑢𝑛(𝑡)=𝑢𝑛12+1𝑡−2𝑢𝑛12−𝑡1/2𝑑𝑠𝑠1/2𝜙𝑞1𝑢𝑝1𝑛−112+𝑠2−12𝑢𝑝1𝑛−112−𝑠21/2𝑑𝑠1𝑠11/2𝜆𝑝1−1𝑎1(𝜏)𝑓1𝑛𝜏,𝑢𝑛(𝜏),𝑣𝑛(𝜏)𝑑𝜏𝑑𝑠2𝑣,𝑡∈(0,1),(3.15)𝑛(𝑡)=𝑣𝑛12+1𝑡−2𝑣𝑛12−𝑡1/2𝑑𝑠𝑠1/2𝜙𝑞2𝑣𝑝2𝑛−112+𝑠2−12𝑣𝑝2𝑛−112−𝑠21/2𝑑𝑠1𝑠11/2𝜇𝑝2−1𝑎2(𝜏)𝑓2𝑛𝜏,𝑢𝑛(𝜏),𝑣𝑛(𝜏)𝑑𝜏𝑑𝑠2,𝑡∈(0,1).(3.16)
From (3.15) and (3.16), we know that {𝑢𝑛(1/2)}𝑛≥𝑛1, {𝑣𝑛(1/2)}𝑛≥𝑛1, {𝑢𝑛(1/2)}𝑛≥𝑛1, {𝑣𝑛(1/2)}𝑛≥𝑛1, {𝑢𝑛(1/2)}𝑛≥𝑛1, {𝑣𝑛(1/2)}𝑛≥𝑛1 are bounded sets. Without loss of generality, we may assume (𝑢𝑛(1/2),𝑣𝑛(1/2))→(𝑐1,𝑑1),(𝑢𝑛(1/2),𝑣𝑛(1/2))→(𝑐2,𝑑2),(𝑢𝑛(1/2),𝑣𝑛(1/2))→(𝑐3,𝑑3) as 𝑛→∞. Then by (3.15), (3.16), and the Lebesgue dominated convergence theorem, we have 1𝑢(𝑡)=𝑢2+𝑐11𝑡−2−𝑡1/2𝑑𝑠𝑠1/2𝜙𝑞1𝑐𝑝12−1+𝑐𝑝13−1𝑠2−12−𝑠21/2𝑑𝑠1𝑠11/2𝜆𝑝1−1𝑎1(𝜏)𝑓1(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠21,𝑡∈(0,1),(3.17)𝑣(𝑡)=𝑣2+𝑑11𝑡−2−𝑡1/2𝑑𝑠𝑠1/2𝜙𝑞2(𝑑𝑝22−1+𝑑𝑝23−1𝑠2−12−𝑠21/2𝑑𝑠1𝑠11/2𝜇𝑝2−1𝑎2(𝜏)𝑓2(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏)𝑑𝑠2𝑡∈(0,1).(3.18)
By (3.17) and (3.18), direct computation shows that 𝜙𝑝1𝑢(𝑡)=𝜆𝑝1−1𝑎1(𝑡)𝑓1𝜙(𝑡,𝑢(𝑡),𝑣(𝑡)),𝑝2𝑣(𝑡)=𝜇𝑝2−1𝑎2(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, 𝑢𝑛(0)=𝑢𝑛∫(1)=10𝑢𝑛(𝑠)𝑑𝜉1(𝑠), 𝑣𝑛(0)=𝑣𝑛∫(1)=10𝑣𝑛(𝑠)𝑑𝜉2(𝑠),  𝜙𝑝1(𝑢𝑛(0))=𝜙𝑝1(𝑢𝑛∫(1))=10𝜙𝑝1(𝑢𝑛(𝑠))𝑑𝜂1(𝑠),  𝜙𝑝2(𝑣𝑛(0))=𝜙𝑝2(𝑣𝑛∫(1))=10𝜙𝑝2(𝑣𝑛(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 𝑓𝑖0=∞, 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]𝑓11(𝑡,𝑥,𝑦)≤𝜆𝑁1−𝜀2𝑝1−1𝜙𝑝1(𝑥),𝑥≥𝑅∗2𝑓,𝑦≥0,(3.21)21(𝑡,𝑥,𝑦)≤𝜆𝑁2−𝜀2𝑝2−1𝜙𝑝2(𝑦),𝑥≥0,𝑦≥𝑅∗2.(3.22)
Let 𝑅2=Λ−1𝑅∗2, 𝐾𝑅2={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑅2,‖𝑣‖<𝑅2}. For any (𝑢,𝑣)∈𝜕𝐾𝑅2, 𝑛∈ℕ, by (2.13), (3.21), we have ‖‖𝐴𝜆𝑛‖‖(𝑢,𝑣)=max[]𝑡∈0,1𝜆10𝐻1(𝑡,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛(𝜏,𝑢(𝜏),𝑣(𝜏))𝑑𝜏𝑑𝑠≤𝜆𝛾1𝜈𝑞11−1𝜙𝑞110𝑒(𝜏)𝑎11(𝜏)𝜆𝑁1−𝜀2𝑝1−1𝜙𝑝1(𝑢(𝜏))𝑑𝜏≤𝜆𝛾1𝜈𝑞11−11𝜆𝑁1−𝜀2𝐿1𝑅2<𝑅2.(3.23)
Similarly, by (3.22) we have ‖𝐵𝜇𝑛(𝑢,𝑣)‖<𝑅2. Therefore, ‖‖𝑇𝑛‖‖=‖‖𝐴(𝑢,𝑣)𝜆𝑛‖‖+‖‖𝐵(𝑢,𝑣)𝜇𝑛‖‖(𝑢,𝑣)≤2𝑅2=‖(𝑢,𝑣)‖,(𝑢,𝑣)∈𝜕𝐾𝑅2,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 𝑓11(𝑡,𝑥,𝑦)≥𝜆𝑀1+𝜀3𝑝1−1𝜙𝑝1[](𝑥),𝑡∈0,1,0<𝑥≤𝑟∗2,𝑦≥0.(3.25)
Let 0<𝑟2<min{𝑅2,𝑟∗2},  𝐾𝑟2={(𝑢,𝑣)∈𝐾∶‖𝑢‖<𝑟2,‖𝑣‖<𝑟2}. Next, we take (𝜑1,𝜑2)=(1,1)∈𝜕𝐾1, 𝑛>1/𝑟2, and for any (𝑢,𝑣)∈𝜕𝐾𝑟2, 𝑚>0, we will show (𝑢,𝑣)≠𝐴𝜆𝑛𝜑(𝑢,𝑣)+𝑚1,𝜑2.(3.26)
Otherwise, there exist (𝑢0,𝑣0)∈𝜕𝐾𝑟2 and 𝑚0>0 such that 𝑢0,𝑣0=𝐴𝜆𝑛𝑢0,𝑣0+𝑚0𝜑1,𝜑2.(3.27)
From (𝑢0,𝑣0)∈𝜕𝐾𝑟2, 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)=𝜆10𝐻1(𝑡,𝑠)𝜙𝑞110𝐾1(𝑠,𝜏)𝑎1(𝜏)𝑓1𝑛𝜏,𝑢0(𝜏),𝑣0(𝜏)𝑑𝜏𝑑𝑠+𝑚0≥𝜆𝜌1𝜎𝑞11−110𝑒(𝑠)𝜙𝑞110𝑒(𝜏)𝑎1(