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Discrete Dynamics in Nature and Society
VolumeΒ 2012Β (2012), Article IDΒ 293734, 19 pages
http://dx.doi.org/10.1155/2012/293734
Research Article

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

1School of Mathematical Sciences, Qufu Normal University, Qufu 273165, China
2Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia

Received 25 March 2012; Accepted 15 May 2012

Academic Editor: CengizΒ Γ‡inar

Copyright Β© 2012 Jiqiang Jiang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 πœ† and πœ‡ is derived such that for any πœ† and πœ‡ 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 ξ€·πœ™π‘ξ€·π‘’ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…ž=𝑓(𝑑,𝑒(𝑑)),0<𝑑<1,𝑒(0)=𝑒(1)βˆ’π‘Žπ‘’(πœ‰)=π‘’ξ…žξ…ž(0)=π‘’ξ…žξ…ž(1)βˆ’π‘π‘’ξ…žξ…ž(πœ‚)=0,(1.1) where πœ™π‘(π‘₯)=|π‘₯|π‘βˆ’2π‘₯, 𝑝>1,  0<πœ‰, πœ‚<1,  0β‰€π‘Ž, 𝑏<1, π‘“βˆˆπΆ((0,1)Γ—(0,∞),(0,∞)), 𝑓(𝑑,π‘₯) may be singular at 𝑑=0 and/or 𝑑=1 and π‘₯=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: ξ€·πœ™π‘ξ€·π‘’ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…žξ€œ=𝑀(𝑑)𝑓(𝑑,𝑒(𝑑)),0<𝑑<1,𝑒(0)=𝑒(1)=10πœ™π‘”(𝑠)𝑒(𝑠)𝑑𝑠,π‘ξ€·π‘’ξ…žξ…žξ€Έ(0)=πœ™π‘ξ€·π‘’ξ…žξ…žξ€Έ=ξ€œ(1)10β„Ž(𝑠)πœ™π‘ξ€·π‘’ξ…žξ…žξ€Έ(𝑠)𝑑𝑠,(1.2) where πœ™π‘(π‘₯)=|π‘₯|π‘βˆ’2π‘₯, 𝑝>1, π‘€βˆˆπΏ1[0,1] is nonnegative, symmetric on the interval [0,1],π‘“βˆΆ[0,1]Γ—[0,+∞)β†’[0,+∞) is continuous, 𝑓(1βˆ’π‘‘,π‘₯)=𝑓(𝑑,π‘₯) for all (𝑑,π‘₯)∈[0,1]Γ—[0,+∞), and 𝑔,β„ŽβˆˆπΏ1[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: ξ€·πœ™π‘1ξ€·π‘’ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…ž=πœ†π‘1βˆ’1π‘Ž1(𝑑)𝑓1ξ€·πœ™(𝑑,𝑒(𝑑),𝑣(𝑑)),0<𝑑<1,𝑝2ξ€·π‘£ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…ž=πœ‡π‘2βˆ’1π‘Ž2(𝑑)𝑓2ξ€œ(𝑑,𝑒(𝑑),𝑣(𝑑)),𝑒(0)=𝑒(1)=10𝑒(𝑠)π‘‘πœ‰1πœ™(𝑠),𝑝1ξ€·π‘’ξ…žξ…žξ€Έ(0)=πœ™π‘1ξ€·π‘’ξ…žξ…žξ€Έ=ξ€œ(1)10πœ™π‘1ξ€·π‘’ξ…žξ…žξ€Έ(𝑠)π‘‘πœ‚1ξ€œ(𝑠),𝑣(0)=𝑣(1)=10𝑣(𝑠)π‘‘πœ‰2πœ™(𝑠),𝑝2ξ€·π‘£ξ…žξ…žξ€Έ(0)=πœ™π‘2ξ€·π‘£ξ…žξ…žξ€Έ=ξ€œ(1)10πœ™π‘2ξ€·π‘£ξ…žξ…žξ€Έ(𝑠)π‘‘πœ‚2(𝑠),(1.3) where πœ† and πœ‡ are positive parameters, πœ™π‘π‘–(π‘₯)=|π‘₯|π‘π‘–βˆ’2π‘₯, 𝑝𝑖>1, πœ™π‘žπ‘–=πœ™π‘βˆ’1𝑖, 1/𝑝𝑖+1/π‘žπ‘–=1, πœ‰π‘–,πœ‚π‘–βˆΆ[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(𝑑,π‘₯,𝑦) may be singular at π‘₯=0 while 𝑓2(𝑑,π‘₯,𝑦) may be singular at 𝑦=0;    π‘Ž1, π‘Ž2∢(0,1)→ℝ+ are continuous and may be singular at 𝑑=0 and/or 𝑑=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, π‘Ž1 and π‘Ž2 in (1.3) are allowed to be singular at 𝑑=0 and/or 𝑑=1, meanwhile 𝑓1(𝑑,π‘₯,𝑦) is allowed to be singular at π‘₯=0 while 𝑓2(𝑑,π‘₯,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 (𝑒,𝑣)∈(𝐢2[0,1]∩𝐢4(0,1))Γ—(𝐢2[0,1]∩𝐢4(0,1)) is said to be a positive solution of system (1.3) if and only if (𝑒,𝑣) satisfies (1.3) and 𝑒(𝑑)>0, 𝑣(𝑑)β‰₯0 or 𝑒(𝑑)β‰₯0, 𝑣(𝑑)>0 for any π‘‘βˆˆ(0,1).

Let 𝐾 be a cone in a Banach space 𝐸. For 0<π‘Ÿ<𝑅<+∞, let πΎπ‘Ÿ={π‘₯βˆˆπΎβˆΆβ€–π‘₯β€–<π‘Ÿ}, πœ•πΎπ‘Ÿ={π‘₯βˆˆπΎβˆΆβ€–π‘₯β€–=π‘Ÿ}, and πΎπ‘Ÿ,𝑅={π‘₯βˆˆπΎβˆΆπ‘Ÿβ‰€β€–π‘₯‖≀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 π‘‡βˆΆπΎπ‘Ÿ,𝑅→𝐾 a completely continuous operator. If the following conditions hold(i)‖𝑇π‘₯‖≀‖π‘₯β€– for π‘₯βˆˆπœ•πΎπ‘…;(ii)there exists π‘’βˆˆπœ•πΎ1 such that π‘₯≠𝑇π‘₯+π‘šπ‘’ for any π‘₯βˆˆπœ•πΎπ‘Ÿ and π‘š>0. Then 𝑇 has a fixed point in πΎπ‘Ÿ,𝑅.

Remark 1.3. If (i) and (ii) are satisfied for π‘₯βˆˆπœ•πΎπ‘Ÿ and π‘₯βˆˆπœ•πΎπ‘…, respectively. Then Lemma 1.2 is still true.

2. Preliminaries and Lemmas

The basic space used in this paper is 𝐸=𝐢[0,1]×𝐢[0,1]. Obviously, the space 𝐸 is a Banach space if it is endowed with the norm as follows: β€–β€–(𝑒,𝑣)∢=‖𝑒‖+‖𝑣‖,‖𝑒‖=max0≀𝑑≀1||||𝑒(𝑑),‖𝑣‖=max0≀𝑑≀1||||𝑣(𝑑)(2.1)

for any (𝑒,𝑣)∈𝐸. Denote 𝐢+[0,1]={π‘’βˆˆπΆ[0,1]βˆΆπ‘’(𝑑)β‰₯0,0≀𝑑≀1}. For convenience, we list the following assumptions:(𝐻1)π‘Ž1,π‘Ž2∢(0,1)→ℝ+ are continuous and 0<πΏπ‘–ξ‚΅ξ€œβˆΆ=10𝑒(𝑠)π‘Žπ‘–ξ‚Ά(𝑠)π‘‘π‘ π‘žπ‘–βˆ’1<+∞,𝑖=1,2,(2.2) where 𝑒(𝑠)=𝑠(1βˆ’π‘ ), π‘ βˆˆ[0,1].(𝐻2)πœ‰π‘–,πœ‚π‘–βˆΆ[0,1]→ℝ+(𝑖=1,2) are nondecreasing functions of bounded variation, and π›Όπ‘–βˆˆ[0,1), π›½π‘–βˆˆ[0,1), where 𝛼𝑖=ξ€œ10π‘‘πœ‰π‘–(𝑠),𝛽𝑖=ξ€œ10π‘‘πœ‚π‘–(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(𝑑,π‘₯,𝑦)≀𝑔1(𝑑,π‘₯)+β„Ž1[](𝑑,𝑦),βˆ€(𝑑,π‘₯,𝑦)∈0,1×ℝ+0×ℝ+,𝑓2(𝑑,π‘₯,𝑦)≀𝑔2(𝑑,π‘₯)+β„Ž2([]𝑑,𝑦),βˆ€(𝑑,π‘₯,𝑦)∈0,1×ℝ+×ℝ+0,(2.4)where 𝑔1,β„Ž2∢[0,1]×𝑅+0→𝑅+ are continuous and nonincreasing in the second variable, and 𝑔2,β„Ž1∢[0,1]×𝑅+→𝑅+ are continuous and for any constant π‘Ÿ>0, ξ€œ0<10𝑒(𝑠)π‘Ž1(𝑠)𝑔1ξ€œ(𝑠,π‘Ÿ)𝑑𝑠<+∞,0<10𝑒(𝑠)π‘Ž2(𝑠)β„Ž2(𝑠,π‘Ÿ)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 π‘¦βˆˆπΏ(0,1), the boundary value problem βˆ’π‘₯ξ…žξ…ž(𝑑)=πœ™π‘žπ‘–ξ€œ(𝑦(𝑑)),0<𝑑<1,π‘₯(0)=π‘₯(1)=10π‘₯(𝑠)π‘‘πœ‰π‘–(𝑠)(2.6) has a unique solution ξ€œπ‘₯(𝑑)=10𝐻𝑖(𝑑,𝑠)πœ™π‘žπ‘–(𝑦(𝑠))𝑑𝑠,(2.7) where 𝐻𝑖1(𝑑,𝑠)=𝐺(𝑑,𝑠)+1βˆ’π›Όπ‘–ξ€œ10𝐺(𝜏,𝑠)π‘‘πœ‰π‘–ξ‚»(𝜏),𝑖=1,2,𝐺(𝑑,𝑠)=𝑠(1βˆ’π‘‘),0≀𝑠≀𝑑≀1,𝑑(1βˆ’π‘ ),0≀𝑑≀𝑠≀1.(2.8)

Lemma 2.2. If (𝐻2) holds, then for any π‘§βˆˆπΏ(0,1), the boundary value problem βˆ’π‘¦ξ…žξ…žξ€œ(𝑑)=𝑧(𝑑),0<𝑑<1,𝑦(0)=𝑦(1)=10𝑦(𝑠)π‘‘πœ‚π‘–(𝑠)(2.9) has a unique solution ξ€œπ‘¦(𝑑)=10𝐾𝑖(𝑑,𝑠)𝑧(𝑠)𝑑𝑠,(2.10) where 𝐾𝑖1(𝑑,𝑠)=𝐺(𝑑,𝑠)+1βˆ’π›½π‘–ξ€œ10𝐺(𝜏,𝑠)π‘‘πœ‚π‘–(𝜏),𝑖=1,2.(2.11)

Remark 2.3. For 𝑑, π‘ βˆˆ[0,1], we have 𝑒(𝑑)𝑒(𝑠)≀𝐺(𝑑,𝑠)≀𝑒(𝑠)or𝑒(𝑑)≀max[]π‘‘βˆˆ0,11𝑒(𝑑)=4.(2.12)

Remark 2.4. If (𝐻2) holds, it is easy to testify 𝐻𝑖(𝑑,𝑠) defined by (2.8) that: πœŒπ‘–π‘’(𝑠)≀𝐻𝑖(𝑑,𝑠)≀𝛾𝑖1𝑒(𝑠)≀4𝛾𝑖<𝛾𝑖[],𝑑,π‘ βˆˆ0,1,𝑖=1,2,(2.13) where 𝛾𝑖=11βˆ’π›Όπ‘–,πœŒπ‘–=∫10𝑒(𝜏)π‘‘πœ‰π‘–(𝜏)1βˆ’π›Όπ‘–,𝑖=1,2.(2.14)

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

Lemma 2.6. For π‘₯>0, 𝑦>0, we have πœ™π‘žπ‘–ξ‚»2(π‘₯+𝑦)β‰€π‘žπ‘–βˆ’1ξ€Ίπœ™π‘žπ‘–(π‘₯)+πœ™π‘žπ‘–ξ€»(𝑦),π‘žπ‘–πœ™β‰₯2π‘žπ‘–(π‘₯)+πœ™π‘žπ‘–(𝑦),1<π‘žπ‘–,<2≀2π‘žπ‘–βˆ’1ξ€Ίπœ™π‘žπ‘–(π‘₯)+πœ™π‘žπ‘–ξ€»(𝑦),π‘žπ‘–πœ™>1,𝑖=1,2,(2.15)π‘žπ‘–(π‘₯)>πœ™π‘žπ‘–(𝑦)>πœ™π‘žπ‘–(0)=0,π‘₯>𝑦>0,π‘žπ‘–>1,𝑖=1,2.(2.16)

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

Let 𝐾=(𝑒,𝑣)∈𝐢+[]0,1×𝐢+[][],0,1βˆΆπ‘’,𝑣areconcaveon0,1min[]π‘‘βˆˆ0,1𝑒(𝑑)β‰₯Λ‖𝑒‖,min[]π‘‘βˆˆ0,1ξ‚Ό,𝑣(𝑑)β‰₯Λ‖𝑣‖(2.17) where ξƒ―πœŒΞ›=min1πœŽπ‘ž11βˆ’1𝛾1πœˆπ‘ž11βˆ’1,𝜌2πœŽπ‘ž22βˆ’1𝛾2πœˆπ‘ž22βˆ’1ξƒ°,πœŽπ‘–=∫10𝑒(𝑠)π‘‘πœ‚π‘–(𝑠)1βˆ’π›½π‘–,πœˆπ‘–=11βˆ’π›½π‘–,𝑖=1,2.(2.18)

It is easy to see that 𝐾 is a cone of 𝐸. For any 0<π‘Ÿ<𝑅, let πΎπ‘Ÿ,𝑅={(𝑒,𝑣)βˆˆπΎβˆΆπ‘Ÿ<‖𝑒‖<𝑅,π‘Ÿ<‖𝑣‖<𝑅}.

Remark 2.7. By the definition of πœŒπ‘–, πœŽπ‘–, 𝛾𝑖, πœˆπ‘–(𝑖=1,2), we have 0<Ξ›<1.
To overcome singularity, we consider the following approximate problem of (1.3): ξ€·πœ™π‘1ξ€·π‘’ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…ž=πœ†π‘1βˆ’1π‘Ž1(𝑑)𝑓1π‘›ξ€·πœ™(𝑑,𝑒(𝑑),𝑣(𝑑)),0<𝑑<1,𝑝2ξ€·π‘£ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…ž=πœ‡π‘2βˆ’1π‘Ž2(𝑑)𝑓2π‘›ξ€œ(𝑑,𝑒(𝑑),𝑣(𝑑)),𝑒(0)=𝑒(1)=10𝑒(𝑠)π‘‘πœ‰1πœ™(𝑠),𝑝1ξ€·π‘’ξ…žξ…žξ€Έ(0)=πœ™π‘1ξ€·π‘’ξ…žξ…žξ€Έ=ξ€œ(1)10πœ™π‘1ξ€·π‘’ξ…žξ…žξ€Έ(𝑠)π‘‘πœ‚1ξ€œ(𝑠),𝑣(0)=𝑣(1)=10𝑣(𝑠)π‘‘πœ‰2πœ™(𝑠),𝑝2ξ€·π‘£ξ…žξ…žξ€Έ(0)=πœ™π‘2ξ€·π‘£ξ…žξ…žξ€Έ=ξ€œ(1)10πœ™π‘2ξ€·π‘£ξ…žξ…žξ€Έ(𝑠)π‘‘πœ‚2(𝑠),(2.19) where 𝑛 is a positive integer and 𝑓1𝑛(𝑑,𝑒,𝑣)=𝑓1𝑑,max𝑒,π‘›βˆ’1ξ€Ύξ€Έ,𝑣,𝑓2𝑛(𝑑,𝑒,𝑣)=𝑓2𝑑,𝑒,max𝑣,π‘›βˆ’1.ξ€Ύξ€Έ(2.20)
Clearly, π‘“π‘–π‘›βˆˆπΆ([0,1]×ℝ+×ℝ+,ℝ+)(𝑖=1,2).
By Lemmas 2.1 and 2.2, for each π‘›βˆˆβ„•, πœ†>0, πœ‡>0, let us define operators π΄πœ†π‘›βˆΆπΎβ†’πΆ[0,1], π΅πœ‡π‘›βˆΆπΎβ†’πΆ[0,1], and π‘‡π‘›βˆΆπΎβ†’πΈ by π΄πœ†π‘›ξ€œ(𝑒,𝑣)(𝑑)=πœ†10𝐻1(𝑑,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1𝑛𝐡(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ ,(2.21)πœ‡π‘›ξ€œ(𝑒,𝑣)(𝑑)=πœ‡10𝐻2(𝑑,𝑠)πœ™π‘ž2ξ‚΅ξ€œ10𝐾2(𝑠,𝜏)π‘Ž2(𝜏)𝑓2𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ ,(2.22) and 𝑇𝑛(𝑒,𝑣)=(π΄πœ†π‘›(𝑒,𝑣),π΅πœ‡π‘›(𝑒,𝑣)), respectively.

Lemma 2.8. Assume that (𝐻1)–(𝐻3) hold, then for each πœ†>0, πœ‡>0, π‘›βˆˆβ„•, π‘‡π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐾 is a completely continuous operator.

Proof. Let πœ†>0, πœ‡>0, and π‘›βˆˆβ„• be fixed. For any (𝑒,𝑣)∈𝐾, by (2.21), we have ξ€·π΄πœ†π‘›ξ€Έ(𝑒,𝑣)ξ…žξ…ž(𝑑)=βˆ’πœ†πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑑,𝜏)π‘Ž1(𝜏)𝑓1𝑛𝐴(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœβ‰€0,πœ†π‘›(𝑒,𝑣)(0)=π΄πœ†π‘›ξ€œ(𝑒,𝑣)(1)=πœ†10𝐻1(0,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ β‰₯0,(2.23) which implies that π΄πœ†π‘› is nonnegative and concave on [0,1]. Similarly, by (2.22) we can obtain that π΅πœ‡π‘› is nonnegative and concave on [0,1]. For any (𝑒,𝑣)∈𝐾 and π‘‘βˆˆ[0,1], it follows from (2.13) that π΄πœ†π‘›ξ€œ(𝑒,𝑣)(𝑑)=πœ†10𝐻1(𝑑,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ β‰€πœ†π›Ύ1πœˆπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ .(2.24)
Thus β€–β€–π΄πœ†π‘›β€–β€–(𝑒,𝑣)β‰€πœ†π›Ύ1πœˆπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ .(2.25)
On the other hand, by (2.13) and (2.18), we have π΄πœ†π‘›ξ€œ(𝑒,𝑣)(𝑑)=πœ†10𝐻1(𝑑,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ β‰₯πœ†πœŒ1πœŽπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)𝑓1𝑛β‰₯𝜌(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ 1πœŽπ‘ž11βˆ’1𝛾1πœˆπ‘ž11βˆ’1β€–β€–π΄πœ†π‘›β€–β€–β€–β€–π΄(𝑒,𝑣)β‰₯Ξ›πœ†π‘›β€–β€–.(𝑒,𝑣)(2.26)
This implies that min[]π‘‘βˆˆ0,1π΄πœ†π‘›β€–β€–π΄(𝑒,𝑣)(𝑑)β‰₯Ξ›πœ†π‘›β€–β€–.(𝑒,𝑣)(2.27)
Similar to (2.27), we also have min[]π‘‘βˆˆ0,1π΅πœ‡π‘›β€–β€–π΅(𝑒,𝑣)(𝑑)β‰₯Ξ›πœ‡π‘›β€–β€–.(𝑒,𝑣)(2.28)
Therefore, 𝑇𝑛(𝐾)βŠ‚πΎ.
Next, we prove that π‘‡π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐾 is completely continuous. Suppose (π‘’π‘š,π‘£π‘š)βˆˆπΎπ‘Ÿ,𝑅 and (𝑒0,𝑣0)βˆˆπΎπ‘Ÿ,𝑅 with β€–(π‘’π‘š,π‘£π‘š)βˆ’(𝑒0,𝑣0)β€–β†’0(π‘šβ†’βˆž). We notice that π‘‘βˆˆ[0,1]𝑓𝑖𝑛(𝑑,π‘’π‘š(𝑑),π‘£π‘š(𝑑))βˆ’π‘“π‘–π‘›(𝑑,𝑒0(𝑑),𝑣0(𝑑))β†’0(π‘šβ†’βˆž). Using the Lebesgue dominated convergence theorem, we have ||||πœ™π‘1π‘žβˆ’11ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,π‘’π‘š(𝜏),π‘£π‘šξ€Έξ‚Ά(𝜏)π‘‘πœβˆ’πœ™π‘1π‘žβˆ’11ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒0(𝜏),𝑣0(ξ€Έξ‚Ά||||𝜏)π‘‘πœβ‰€πœˆ1ξ€œ10𝑒(𝜏)π‘Ž1||𝑓(𝜏)1π‘›ξ€·πœ,π‘’π‘š(𝜏),π‘£π‘šξ€Έ(𝜏)βˆ’π‘“1π‘›ξ€·πœ,𝑒0(𝜏),𝑣0ξ€Έ||(𝜏)π‘‘πœβŸΆ0,π‘šβŸΆβˆž.(2.29)
Therefore, β€–β€–π΄πœ†π‘›ξ€·π‘’π‘š,π‘£π‘šξ€Έβˆ’π΄πœ†π‘›ξ€·π‘’0,𝑣0ξ€Έβ€–β€–β‰€πœ†π›Ύ1ξ€œ10||||πœ™π‘’(𝑠)π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,π‘’π‘š(𝜏),π‘£π‘šξ€Έξ‚Ά(𝜏)π‘‘πœβˆ’πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒0(𝜏),𝑣0ξ€Έξ‚Ά||||(𝜏)π‘‘πœπ‘‘π‘ βŸΆ0,π‘šβŸΆβˆž.(2.30)
Similarly, we also have β€–β€–π΅πœ‡π‘›ξ€·π‘’π‘š,π‘£π‘šξ€Έβˆ’π΅πœ‡π‘›ξ€·π‘’0,𝑣0ξ€Έβ€–β€–βŸΆ0,π‘šβŸΆβˆž.(2.31)
So π΄πœ†π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐢[0,1] and π΅π›½π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐢[0,1] are continuous. Therefore, π‘‡π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐾 is also continuous.
Let π·βŠ‚πΎπ‘Ÿ,𝑅 be any bounded set, then for any (𝑒,𝑣)∈𝐷, we have (𝑒,𝑣)∈𝐾, π‘Ÿβ‰€β€–π‘’β€–β‰€π‘…, π‘Ÿβ‰€β€–π‘£β€–β‰€π‘…, and then 0<Ξ›π‘Ÿβ‰€π‘’(𝜏)≀𝑅, 0<Ξ›π‘Ÿβ‰€π‘£(𝜏)≀𝑅 for any 𝜏∈[0,1]. By (𝐻3), we have πΏπ‘Ÿξ‚΅ξ€œβˆΆ=10𝑒(𝜏)π‘Ž1(𝜏)𝑔1ξ‚Ά(𝑠,π‘ŸΞ›)π‘‘πœπ‘ž1βˆ’1<+∞.(2.32)
It is easy to show that π΄πœ†π‘›(𝐷) is uniformly bounded. In order to show that 𝑇𝑛 is a compact operator, we only need to show that π΄πœ†π‘›(𝐷) is equicontinuous. By the uniformly continuity of 𝐻1(𝑑,𝑠) on [0,1]Γ—[0,1], for all πœ€>0, there is 𝛿>0 such that for any 𝑑1, 𝑑2, π‘ βˆˆ[0,1] and |𝑑1βˆ’π‘‘2|<𝛿, we have ||𝐻1𝑑1ξ€Έ,π‘ βˆ’π»1𝑑2ξ€Έ||,𝑠<πœ€.(2.33)
This together with (2.15) and (2.32) implies ||π΄πœ†π‘›ξ€·π‘‘(𝑒,𝑣)1ξ€Έβˆ’π΄πœ†π‘›ξ€·π‘‘(𝑒,𝑣)2ξ€Έ||ξ€œβ‰€πœ†10||𝐻1𝑑1ξ€Έ,π‘ βˆ’π»1𝑑2ξ€Έ||πœ™,π‘ π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ <πœ€πœ†πœˆπ‘ž11βˆ’1πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1𝑔(𝜏)1ξ€·ξ€½πœ,max𝑒(𝜏),π‘›βˆ’1ξ€Ύξ€Έ+β„Ž1ξ€»ξ‚Ά(𝜏,𝑣(𝜏))π‘‘πœβ‰€πœ€πœ†πœˆπ‘ž11βˆ’1πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1𝑔(𝜏)1(𝜏,π‘ŸΞ›)+β„Ž1ξ€»ξ‚Ά(𝜏,𝑣(𝜏))π‘‘πœβ‰€πœ€πœ†πœˆπ‘ž11βˆ’12π‘ž1βˆ’1ξ‚Έπœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)𝑔1ξ‚Ά(𝜏,π‘ŸΞ›)π‘‘πœ+πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)β„Ž1(𝜏,𝑣(𝜏))π‘‘πœξ‚Άξ‚Ήβ‰€πœ€πœ†πœˆπ‘ž11βˆ’12π‘ž1βˆ’1βŽ‘βŽ’βŽ’βŽ£πΏπ‘Ÿ+𝐿1max[]𝜏∈[0,1]π‘¦βˆˆπ‘ŸΞ›,π‘…β„Ž1ξƒͺ(𝜏,𝑦)π‘ž1βˆ’1⎀βŽ₯βŽ₯⎦,||𝑑1βˆ’π‘‘2||<𝛿,(𝑒,𝑣)∈𝐷.(2.34)
This means that π΄πœ†π‘›(𝐷) is equicontinuous. By the Arzela-Ascoli theorem, π΄πœ†π‘›(𝐷) is a relatively compact set and that π΄πœ†π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐢[0,1] is a completely continuous operator.
In the same way, we can show that π΅πœ‡π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐢[0,1] is also completely continuous, and so π‘‡π‘›βˆΆπΎπ‘Ÿ,𝑅→𝐾 is completely continuous. Now since πœ†, πœ‡, and 𝑛 are given arbitrarily, the conclusion of this lemma is valid.

3. Main Results

For notational convenience, we denote by π‘€π‘–ξ€·πœŒ=6π‘–πœŽπ‘žπ‘–βˆ’1Ξ›πΏπ‘–ξ€Έβˆ’1,𝑁𝑖=ξ‚€π›Ύπ‘–πœˆπ‘žπ‘–π‘–βˆ’1πΏπ‘–ξ‚βˆ’1𝑓,𝑖=1,2,𝛼1=βŽ›βŽœβŽœβŽlimsupπ‘₯→𝛼sup+π‘‘βˆˆ[0,1]π‘¦βˆˆβ„π‘“1(𝑑,π‘₯,𝑦)πœ™π‘1⎞⎟⎟⎠(π‘₯)π‘ž1βˆ’1,𝑓𝛼2=βŽ›βŽœβŽœβŽlimsup𝑦→𝛼sup+π‘‘βˆˆ[0,1]π‘₯βˆˆβ„π‘“2(𝑑,π‘₯,𝑦)πœ™π‘2⎞⎟⎟⎠(𝑦)π‘ž2βˆ’1,𝑓1𝛼=βŽ›βŽœβŽœβŽliminfπ‘₯→𝛼inf+π‘‘βˆˆ[0,1]π‘¦βˆˆβ„π‘“1(𝑑,π‘₯,𝑦)πœ™π‘1⎞⎟⎟⎠(π‘₯)π‘ž1βˆ’1,𝑓2𝛼=liminf𝑦→𝛼inf+π‘‘βˆˆ[0,1]π‘₯βˆˆβ„π‘“2(𝑑,π‘₯,𝑦)πœ™π‘2ξƒͺ(𝑦)π‘ž2βˆ’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:(𝐢1) If 𝑓01, 𝑓1∞, 𝑓02∈(0,∞) and 𝑀1/𝑓1∞<𝑁1/𝑓01, then for each πœ†βˆˆ(𝑀1/𝑓1∞,𝑁1/𝑓01), πœ‡βˆˆ(0,𝑁2/𝑓02), the system (1.3) has at least one positive solution.(𝐢2) If 𝑓01, 𝑓02, 𝑓2∞∈(0,∞) and 𝑀2/𝑓2∞<𝑁2/𝑓02, then for each πœ†βˆˆ(0,𝑁1/𝑓01), πœ‡βˆˆ(𝑀2/𝑓2∞,𝑁2/𝑓02), the system (1.3) has at least one positive solution.(𝐢3) If 𝑓01=0, 𝑓1∞=∞, 0<𝑓02<∞, then for each πœ†βˆˆ(0,∞), πœ‡βˆˆ(0,𝑁2/𝑓02), the system (1.3) has at least one positive solution.(𝐢4) If 0<𝑓01<∞, 𝑓02=0, 𝑓2∞=∞, then for each πœ†βˆˆ(0,𝑁1/𝑓01), πœ‡βˆˆ(0,∞), the system (1.3) has at least one positive solution.(𝐢5) If 𝑓0𝑖=0, π‘“π‘–βˆž=∞(𝑖=1,2), then for each πœ†βˆˆ(0,∞), πœ‡βˆˆ(0,∞), the system (1.3) has at least one positive solution.(𝐢6) If 0<𝑓01<∞, 𝑓1∞=∞ or 𝑓2∞=∞, 0<𝑓02<∞, then for each πœ†βˆˆ(0,𝑁1/𝑓01), πœ‡βˆˆ(0,𝑁2/𝑓02), the system (1.3) has at least one positive solution.(𝐢7) If 𝑓01=0, 0<𝑓1∞<∞, and 𝑓01=0, 0<𝑓2∞<∞, then for each πœ†βˆˆ(𝑀1/𝑓1∞,∞), πœ‡βˆˆ(0,∞) or πœ†βˆˆ(0,∞), πœ‡βˆˆ(𝑀2/𝑓2∞,∞), the system (1.3) has at least one positive solution.

Proof. We only prove the condition in which (𝐢1) holds. The other cases can be proved similarly.
Let πœ†βˆˆ(𝑀1/𝑓1∞,𝑁1/𝑓01), πœ‡βˆˆ(0,𝑁2/(𝑓02)), choose πœ€1>0 such that 𝑓1βˆžβˆ’πœ€1>0 and 𝑀1𝑓1βˆžβˆ’πœ€1π‘β‰€πœ†β‰€1𝑓01+πœ€1𝑁,0<πœ‡β‰€2𝑓02+πœ€1.(3.2)
It follows from 𝑓0π‘–βˆˆ(0,∞) of (𝐢1) that there exists π‘Ÿ1>0 such that for any π‘‘βˆˆ[0,1], 𝑓1(𝑓𝑑,π‘₯,𝑦)≀01+πœ€1𝑝1βˆ’1πœ™π‘1(ξ€Ίπ‘Ÿπ‘₯)≀1𝑓01+πœ€1𝑝1βˆ’1,0<π‘₯β‰€π‘Ÿ1𝑓,𝑦β‰₯0,(3.3)2𝑓(𝑑,π‘₯,𝑦)≀02+πœ€1𝑝2βˆ’1πœ™π‘2ξ€Ίπ‘Ÿ(𝑦)≀1𝑓02+πœ€1𝑝2βˆ’1,π‘₯β‰₯0,0<π‘¦β‰€π‘Ÿ1.(3.4)
Let πΎπ‘Ÿ1={(𝑒,𝑣)βˆˆπΎβˆΆβ€–π‘’β€–<π‘Ÿ1,‖𝑣‖<π‘Ÿ1}. For any (𝑒,𝑣)βˆˆπœ•πΎπ‘Ÿ1, 𝑛>1/π‘Ÿ1, by (2.13), (3.3), we have β€–β€–π΄πœ†π‘›β€–β€–(𝑒,𝑣)=max[]π‘‘βˆˆ0,1πœ†ξ€œ10𝐻1(𝑑,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ β‰€πœ†π›Ύ1πœˆπ‘ž11βˆ’1πœ™π‘ž1ξ€œ10𝑒(𝜏)π‘Ž1𝑓(𝜏)01+πœ€1𝑝1βˆ’1πœ™π‘1(𝑒(𝜏))π‘‘πœβ‰€πœ†π›Ύ1πœˆπ‘ž11βˆ’1𝑓01+πœ€1𝐿1π‘Ÿ1=πœ†π‘1βˆ’1𝑓01+πœ€1ξ€Έπ‘Ÿ1.(3.5)
Similarly, we also have β€–β€–π΅πœ‡π‘›β€–β€–(𝑒,𝑣)β‰€πœ‡π‘2βˆ’1𝑓02+πœ€1ξ€Έπ‘Ÿ1.(3.6)
Therefore, we have ‖‖𝑇𝑛‖‖=‖‖𝐴(𝑒,𝑣)πœ†π‘›β€–β€–+‖‖𝐡(𝑒,𝑣)πœ‡π‘›β€–β€–β‰€ξ€Ί(𝑒,𝑣)πœ†π‘1βˆ’1𝑓01+πœ€1ξ€Έ+πœ‡π‘2βˆ’1𝑓02+πœ€1π‘Ÿξ€Έξ€»1≀2π‘Ÿ1=β€–(𝑒,𝑣)β€–.(3.7)
On the other hand, by 𝑓1∞>𝑓1βˆžβˆ’πœ€1>0, there exists 𝑅0>0 such that 𝑓1𝑓(𝑑,π‘₯,𝑦)β‰₯1βˆžβˆ’πœ€1𝑝1βˆ’1πœ™π‘1[](π‘₯),π‘‘βˆˆ0,1,π‘₯β‰₯𝑅0,𝑦β‰₯0.(3.8)
Let 𝑅1>max{2π‘Ÿ1,Ξ›βˆ’1𝑅0}, 𝐾𝑅1={(𝑒,𝑣)βˆˆπΎβˆΆβ€–π‘’β€–<𝑅1,‖𝑣‖<𝑅1}. Next, we take (πœ‘1,πœ‘2)=(1,1)βˆˆπœ•πΎ1, and for any (𝑒,𝑣)βˆˆπœ•πΎπ‘…1, π‘š>0, π‘›βˆˆβ„•, we will show (𝑒,𝑣)β‰ π΄πœ†π‘›ξ€·πœ‘(𝑒,𝑣)+π‘š1,πœ‘2ξ€Έ.(3.9)
Otherwise, there exist (𝑒0,𝑣0)βˆˆπœ•πΎπ‘…1 and π‘š0>0 such that 𝑒0,𝑣0ξ€Έ=π΄πœ†π‘›ξ€·π‘’0,𝑣0ξ€Έ+π‘š0ξ€·πœ‘1,πœ‘2ξ€Έ.(3.10)
From (𝑒0,𝑣0)βˆˆπœ•πΎπ‘…1, we know that ‖𝑒0β€–=𝑅1 or ‖𝑣0β€–=𝑅1. Without loss of generality, we may suppose that ‖𝑒0β€–=𝑅1, then 𝑒0(𝜏)β‰₯Λ‖𝑒0β€–=Λ𝑅1>𝑅0 for any 𝜏∈[0,1]. So, by (2.13), (3.8), we have 𝑒0ξ€œ(𝑑)=πœ†10𝐻1(𝑑,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒0(𝜏),𝑣0ξ€Έξ‚Ά(𝜏)π‘‘πœπ‘‘π‘ +π‘š0β‰₯πœ†πœŒ1πœŽπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒0(𝜏),𝑣0ξ€Έξ‚Ά(𝜏)π‘‘πœπ‘‘π‘ +π‘š0β‰₯πœ†πœŒ1πœŽπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1𝑓(𝜏)1βˆžβˆ’πœ€1𝑝1βˆ’1πœ™π‘1𝑒0ξ€Έξ‚Ά(𝜏)π‘‘πœπ‘‘π‘ +π‘š0β‰₯πœ†πœŒ1πœŽπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1𝑓(𝜏)1βˆžβˆ’πœ€1𝑝1βˆ’1Λ𝑅1𝑝1βˆ’1ξ‚Άπ‘‘πœπ‘‘π‘ +π‘š0=16πœ†πœŒ1πœŽπ‘ž11βˆ’1𝑓1βˆžβˆ’πœ€1Λ𝑅1𝐿1+π‘š0=πœ†π‘€1βˆ’1𝑓1βˆžβˆ’πœ€1𝑅1+π‘š0>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/π‘Ÿ1and πœ†βˆˆ(𝑀1/𝑓1∞,𝑁1/𝑓01), πœ‡βˆˆ(0,𝑁2/𝑓02), we obtain that 𝑇𝑛 has a fixed point (𝑒𝑛,𝑣𝑛) in πΎπ‘Ÿ1,𝑅1 satisfying π‘Ÿ1<‖𝑒𝑛‖<𝑅1,π‘Ÿ1<‖𝑣𝑛‖<𝑅1.
Let {(𝑒𝑛,𝑣𝑛)}𝑛β‰₯𝑛1 be the sequence of solutions of boundary value problems (2.19), where 𝑛1>1/π‘Ÿ1is a fixed integer. It is easy to see that they are uniformly bounded. Next we show that {𝑒𝑛}𝑛β‰₯𝑛1 are equicontinuous on [0,1]. From (𝑒𝑛,𝑣𝑛)βˆˆπΎπ‘Ÿ1,𝑅1, we know that 𝑅1β‰₯𝑒𝑛(𝜏)β‰₯Λ‖𝑒𝑛‖β‰₯Ξ›π‘Ÿ1, 𝑅1β‰₯𝑣𝑛(𝜏)β‰₯Λ‖𝑣𝑛‖β‰₯Ξ›π‘Ÿ1, 𝜏∈[0,1]. For any πœ€>0, by the continuous of 𝐻1(𝑑,𝑠) in [0,1]Γ—[0,1], there exists 𝛿1>0 such that for any 𝑑1, 𝑑2, π‘ βˆˆ[0,1] and |𝑑1βˆ’π‘‘2|<𝛿1, we have ||𝐻1𝑑1ξ€Έ,π‘ βˆ’π»1𝑑2ξ€Έ||,𝑠<πœ€.(3.12)
This combining with (2.15), (2.32) implies that for any 𝑑1, 𝑑2∈[0,1] and |𝑑1βˆ’π‘‘2|<𝛿1, we have ||𝑒𝑛𝑑1ξ€Έβˆ’π‘’π‘›ξ€·π‘‘2ξ€Έ||ξ€œβ‰€πœ†10||𝐻1𝑑1ξ€Έ,π‘ βˆ’π»1𝑑2ξ€Έ||πœ™,π‘ π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒𝑛(𝜏),𝑣𝑛(ξ€Έξ‚Άπœ)π‘‘πœπ‘‘π‘ <πœ€πœ†πœˆπ‘ž11βˆ’1πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1𝑔(𝜏)1ξ€·ξ€½π‘’πœ,max𝑛(𝜏),π‘›βˆ’1ξ€Ύξ€Έ+β„Ž1ξ€·πœ,𝑣𝑛(𝜏)ξ€Έξ€»π‘‘πœβ‰€πœ€πœ†πœˆπ‘ž11βˆ’12π‘ž1βˆ’1ξ‚Έπœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)𝑔1ξ‚Ά(𝜏,π‘ŸΞ›)π‘‘πœ+πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)β„Ž1(𝜏,𝑣(𝜏))π‘‘πœξ‚Άξ‚Ήβ‰€πœ€πœ†πœˆπ‘ž11βˆ’12π‘ž1βˆ’1βŽ‘βŽ’βŽ’βŽ£πΏπ‘Ÿ+𝐿1βŽ›βŽœβŽœβŽmaxξ€Ίπ‘Ÿ11ξ€»πœβˆˆ[0,1]π‘¦βˆˆΞ›,π‘…β„Ž1⎞⎟⎟⎠(𝜏,𝑦)π‘ž1βˆ’1⎀βŽ₯βŽ₯⎦.(3.13)
Similarly, {𝑣𝑛}𝑛β‰₯𝑛1 are also equicontinuous on [0,1]. By the Ascoli-Arzela theorem, the sequence {(𝑒𝑛,𝑣𝑛)}𝑛β‰₯𝑛1 has a subsequence being uniformly convergent on [0,1]. From Lemma 2.2, we know that π‘’π‘›ξ…žξ…ž(𝑠)=πœ†π‘1βˆ’1ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒𝑛(𝜏),𝑣𝑛𝑣(𝜏)π‘‘πœ,π‘›ξ…žξ…ž(𝑠)=πœ‡π‘2βˆ’1ξ€œ10𝐾2(𝑠,𝜏)π‘Ž2(𝜏)𝑓2π‘›ξ€·πœ,𝑒𝑛(𝜏),𝑣𝑛(𝜏)π‘‘πœ.(3.14)
Since the properties of 𝐾𝑖(𝑑,𝑠)(𝑖=1,2) are similar to those of 𝐻𝑖(𝑑,𝑠)(𝑖=1,2), so (π‘’π‘›ξ…žξ…ž,π‘£π‘›ξ…žξ…ž) have the similar properties of (𝑒𝑛,𝑣𝑛), that is, (π‘’π‘›ξ…žξ…ž,π‘£π‘›ξ…žξ…ž) also has a subsequence being uniformly convergent on [0,1]. Without loss of generality, we still assume that {(𝑒𝑛,𝑣𝑛)}𝑛β‰₯𝑛1 itself uniformly converges to (𝑒,𝑣) on [0,1] and {(π‘’π‘›ξ…žξ…ž,π‘£π‘›ξ…žξ…ž)}𝑛β‰₯𝑛1 itself uniformly converges to (π‘’ξ…žξ…ž,π‘£ξ…žξ…ž) on [0,1], respectively. Since {(𝑒𝑛,𝑣𝑛)}𝑛β‰₯𝑛1βˆˆπΎπ‘Ÿ1,𝑅1βŠ‚πΎ, so we have 𝑒𝑛β‰₯0, 𝑣𝑛β‰₯0. By (2.19), we have 𝑒𝑛(𝑑)=𝑒𝑛12+ξ‚€1π‘‘βˆ’2ξ‚π‘’ξ…žπ‘›ξ‚€12ξ‚βˆ’ξ€œπ‘‘1/2ξ€œπ‘‘π‘ π‘ 1/2πœ™π‘ž1ξ‚€π‘’ξ…žξ…žπ‘1π‘›βˆ’1ξ‚€12+𝑠2βˆ’12ξ‚π‘’ξ…žξ…žξ…žπ‘1π‘›βˆ’1ξ‚€12ξ‚βˆ’ξ€œπ‘ 21/2𝑑𝑠1ξ€œπ‘ 11/2πœ†π‘1βˆ’1π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒𝑛(𝜏),𝑣𝑛(𝜏)π‘‘πœπ‘‘π‘ 2𝑣,π‘‘βˆˆ(0,1),(3.15)𝑛(𝑑)=𝑣𝑛12+ξ‚€1π‘‘βˆ’2ξ‚π‘£ξ…žπ‘›ξ‚€12ξ‚βˆ’ξ€œπ‘‘1/2ξ€œπ‘‘π‘ π‘ 1/2πœ™π‘ž2ξ‚€π‘£ξ…žξ…žπ‘2π‘›βˆ’1ξ‚€12+𝑠2βˆ’12ξ‚π‘£ξ…žξ…žξ…žπ‘2π‘›βˆ’1ξ‚€12ξ‚βˆ’ξ€œπ‘ 21/2𝑑𝑠1ξ€œπ‘ 11/2πœ‡π‘2βˆ’1π‘Ž2(𝜏)𝑓2π‘›ξ€·πœ,𝑒𝑛(𝜏),𝑣𝑛(𝜏)π‘‘πœπ‘‘π‘ 2,π‘‘βˆˆ(0,1).(3.16)
From (3.15) and (3.16), we know that {π‘’ξ…žπ‘›(1/2)}𝑛β‰₯𝑛1, {π‘£ξ…žπ‘›(1/2)}𝑛β‰₯𝑛1, {π‘’π‘›ξ…žξ…ž(1/2)}𝑛β‰₯𝑛1, {π‘£π‘›ξ…žξ…ž(1/2)}𝑛β‰₯𝑛1, {π‘’π‘›ξ…žξ…žξ…ž(1/2)}𝑛β‰₯𝑛1, {π‘£π‘›ξ…žξ…žξ…ž(1/2)}𝑛β‰₯𝑛1 are bounded sets. Without loss of generality, we may assume (π‘’ξ…žπ‘›(1/2),π‘£ξ…žπ‘›(1/2))β†’(𝑐1,𝑑1),(π‘’π‘›ξ…žξ…ž(1/2),π‘£π‘›ξ…žξ…ž(1/2))β†’(𝑐2,𝑑2),(π‘’π‘›ξ…žξ…žξ…ž(1/2),π‘£π‘›ξ…žξ…žξ…ž(1/2))β†’(𝑐3,𝑑3) as π‘›β†’βˆž. Then by (3.15), (3.16), and the Lebesgue dominated convergence theorem, we have ξ‚€1𝑒(𝑑)=𝑒2+𝑐1ξ‚€1π‘‘βˆ’2ξ‚βˆ’ξ€œπ‘‘1/2ξ€œπ‘‘π‘ π‘ 1/2πœ™π‘ž1𝑐𝑝12βˆ’1+𝑐𝑝13βˆ’1𝑠2βˆ’12ξ‚βˆ’ξ€œπ‘ 21/2𝑑𝑠1ξ€œπ‘ 11/2πœ†π‘1βˆ’1π‘Ž1(𝜏)𝑓1ξ€Έ(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ 2ξ‚€1,π‘‘βˆˆ(0,1),(3.17)𝑣(𝑑)=𝑣2+𝑑1ξ‚€1π‘‘βˆ’2ξ‚βˆ’ξ€œπ‘‘1/2ξ€œπ‘‘π‘ π‘ 1/2πœ™π‘ž2(𝑑𝑝22βˆ’1+𝑑𝑝23βˆ’1𝑠2βˆ’12ξ‚βˆ’ξ€œπ‘ 21/2𝑑𝑠1ξ€œπ‘ 11/2πœ‡π‘2βˆ’1π‘Ž2(𝜏)𝑓2(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœ)𝑑𝑠2π‘‘βˆˆ(0,1).(3.18)
By (3.17) and (3.18), direct computation shows that ξ€·πœ™π‘1ξ€·π‘’ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…ž=πœ†π‘1βˆ’1π‘Ž1(𝑑)𝑓1ξ€·πœ™(𝑑,𝑒(𝑑),𝑣(𝑑)),𝑝2ξ€·π‘£ξ…žξ…ž(𝑑)ξ€Έξ€Έξ…žξ…ž=πœ‡π‘2βˆ’1π‘Ž2(𝑑)𝑓2(𝑑,𝑒(𝑑),𝑣(𝑑)),0<𝑑<1.(3.19)
On the other hand, (𝑒,𝑣) satisfies the boundary condition of (1.3). In fact, 𝑒𝑛(0)=π‘’π‘›βˆ«(1)=10𝑒𝑛(𝑠)π‘‘πœ‰1(𝑠), 𝑣𝑛(0)=π‘£π‘›βˆ«(1)=10𝑣𝑛(𝑠)π‘‘πœ‰2(𝑠),β€‰β€‰πœ™π‘1(π‘’π‘›ξ…žξ…ž(0))=πœ™π‘1(π‘’π‘›ξ…žξ…žβˆ«(1))=10πœ™π‘1(π‘’π‘›ξ…žξ…ž(𝑠))π‘‘πœ‚1(𝑠),β€‰β€‰πœ™π‘2(π‘£π‘›ξ…žξ…ž(0))=πœ™π‘2(π‘£π‘›ξ…žξ…žβˆ«(1))=10πœ™π‘2(π‘£π‘›ξ…žξ…ž(𝑠))π‘‘πœ‚2(𝑠), and so the conclusion holds by letting π‘›β†’βˆž.

Theorem 3.2. Assume that (𝐻1)–(𝐻3) hold. Then we have:(𝐷1) If 𝑓10, π‘“βˆž1, π‘“βˆž2∈(0,∞) and 𝑀1/𝑓10<𝑁1/π‘“βˆž1, then for each πœ†βˆˆ(𝑀1/𝑓10,𝑁1/π‘“βˆž1), πœ‡βˆˆ(0,𝑁2/π‘“βˆž2), the system (1.3) has at least one positive solution.(𝐷2) If π‘“βˆž1, 𝑓20, π‘“βˆž2∈(0,∞) and 𝑀2/𝑓20<𝑁2/π‘“βˆž2, then for each πœ†βˆˆ(0,𝑁1/π‘“βˆž1), πœ‡βˆˆ(𝑀2/𝑓20,𝑁2/π‘“βˆž2), the system (1.3) has at least one positive solution.(𝐷3) If 𝑓10=∞, π‘“βˆž1=0, 0<π‘“βˆž2<∞, then for each πœ†βˆˆ(0,∞), πœ‡βˆˆ(0,𝑁2/(π‘“βˆž2)), the system (1.3) has at least one positive solution.(𝐷4) If 0<π‘“βˆž1<∞, 𝑓20=∞, π‘“βˆž2=0, then for each πœ†βˆˆ(0,𝑁1/π‘“βˆž1), πœ‡βˆˆ(0,∞), the system (1.3) has at least one positive solution.(𝐷5) If 𝑓𝑖0=∞, π‘“βˆžπ‘–=0(𝑖=1,2), then for each πœ†βˆˆ(0,∞), πœ‡βˆˆ(0,∞), the system (1.3) has at least one positive solution.(𝐷6) If 0<π‘“βˆž1<∞, 𝑓10=∞ or 𝑓20=∞, 0<π‘“βˆž2<∞, then for each πœ†βˆˆ(0,𝑁1/π‘“βˆž1), πœ‡βˆˆ(0,𝑁2/π‘“βˆž2), the system (1.3) has at least one positive solution.(𝐷7) If π‘“βˆž1=0, 0<𝑓10<∞, and π‘“βˆž2=0, 0<𝑓20<∞, then for each πœ†βˆˆ(𝑀1/𝑓10,∞), πœ‡βˆˆ(0,∞) or πœ†βˆˆ(0,∞), πœ‡βˆˆ(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 πœ†βˆˆ(𝑀1/𝑓10,𝑁1/π‘“βˆž1), πœ‡βˆˆ(0,𝑁2/π‘“βˆž2). We can choose πœ€2>0 such that 𝑁1βˆ’πœ€2>0, 𝑁2βˆ’πœ€2>0 and πœ†π‘“βˆž1<𝑁1βˆ’πœ€2,πœ‡π‘“βˆž2<𝑁2βˆ’πœ€2.(3.20)
It follows from (𝐷1) and (2.16) that there exists π‘…βˆ—2>0 such that for any π‘‘βˆˆ[0,1]𝑓1ξ‚€1(𝑑,π‘₯,𝑦)β‰€πœ†ξ€·π‘1βˆ’πœ€2𝑝1βˆ’1πœ™π‘1(π‘₯),π‘₯β‰₯π‘…βˆ—2𝑓,𝑦β‰₯0,(3.21)2ξ‚€1(𝑑,π‘₯,𝑦)β‰€πœ†ξ€·π‘2βˆ’πœ€2𝑝2βˆ’1πœ™π‘2(𝑦),π‘₯β‰₯0,𝑦β‰₯π‘…βˆ—2.(3.22)
Let 𝑅2=Ξ›βˆ’1π‘…βˆ—2, 𝐾𝑅2={(𝑒,𝑣)βˆˆπΎβˆΆβ€–π‘’β€–<𝑅2,‖𝑣‖<𝑅2}. For any (𝑒,𝑣)βˆˆπœ•πΎπ‘…2, π‘›βˆˆβ„•, by (2.13), (3.21), we have β€–β€–π΄πœ†π‘›β€–β€–(𝑒,𝑣)=max[]π‘‘βˆˆ0,1πœ†ξ€œ10𝐻1(𝑑,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1𝑛(𝜏,𝑒(𝜏),𝑣(𝜏))π‘‘πœπ‘‘π‘ β‰€πœ†π›Ύ1πœˆπ‘ž11βˆ’1πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1ξ‚€1(𝜏)πœ†ξ€·π‘1βˆ’πœ€2𝑝1βˆ’1πœ™π‘1ξ‚Ά(𝑒(𝜏))π‘‘πœβ‰€πœ†π›Ύ1πœˆπ‘ž11βˆ’11πœ†ξ€·π‘1βˆ’πœ€2𝐿1𝑅2<𝑅2.(3.23)
Similarly, by (3.22) we have β€–π΅πœ‡π‘›(𝑒,𝑣)β€–<𝑅2. Therefore, ‖‖𝑇𝑛‖‖=‖‖𝐴(𝑒,𝑣)πœ†π‘›β€–β€–+‖‖𝐡(𝑒,𝑣)πœ‡π‘›β€–β€–(𝑒,𝑣)≀2𝑅2=β€–(𝑒,𝑣)β€–,(𝑒,𝑣)βˆˆπœ•πΎπ‘…2,π‘›βˆˆβ„•.(3.24)
On the other hand, choose πœ€3>0 such that 𝑀1+πœ€3<πœ†π‘“10. By the condition 𝑓10∈(0,∞) of (𝐷1) and (2.16), there exists π‘Ÿβˆ—2>0 such that 𝑓1ξ‚€1(𝑑,π‘₯,𝑦)β‰₯πœ†ξ€·π‘€1+πœ€3𝑝1βˆ’1πœ™π‘1[](π‘₯),π‘‘βˆˆ0,1,0<π‘₯β‰€π‘Ÿβˆ—2,𝑦β‰₯0.(3.25)
Let 0<π‘Ÿ2<min{𝑅2,π‘Ÿβˆ—2},β€‰β€‰πΎπ‘Ÿ2={(𝑒,𝑣)βˆˆπΎβˆΆβ€–π‘’β€–<π‘Ÿ2,‖𝑣‖<π‘Ÿ2}. Next, we take (πœ‘1,πœ‘2)=(1,1)βˆˆπœ•πΎ1, 𝑛>1/π‘Ÿ2, and for any (𝑒,𝑣)βˆˆπœ•πΎπ‘Ÿ2, π‘š>0, we will show (𝑒,𝑣)β‰ π΄πœ†π‘›ξ€·πœ‘(𝑒,𝑣)+π‘š1,πœ‘2ξ€Έ.(3.26)
Otherwise, there exist (𝑒0,𝑣0)βˆˆπœ•πΎπ‘Ÿ2 and π‘š0>0 such that 𝑒0,𝑣0ξ€Έ=π΄πœ†π‘›ξ€·π‘’0,𝑣0ξ€Έ+π‘š0ξ€·πœ‘1,πœ‘2ξ€Έ.(3.27)
From (𝑒0,𝑣0)βˆˆπœ•πΎπ‘Ÿ2, we know that ‖𝑒0β€–=π‘Ÿ2 or ‖𝑣0β€–=π‘Ÿ2. Without loss of generality, we may suppose that ‖𝑒0β€–=π‘Ÿ2, then 𝑒0(𝜏)β‰₯Λ‖𝑒0β€–β‰₯Ξ›π‘Ÿ2 for any 𝜏∈[0,1]. So, we have 𝑒0ξ€œ(𝑑)=πœ†10𝐻1(𝑑,𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝐾1(𝑠,𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒0(𝜏),𝑣0ξ€Έξ‚Ά(𝜏)π‘‘πœπ‘‘π‘ +π‘š0β‰₯πœ†πœŒ1πœŽπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1(𝜏)𝑓1π‘›ξ€·πœ,𝑒0(𝜏),𝑣0ξ€Έξ‚Ά(𝜏)π‘‘πœπ‘‘π‘ +π‘š0β‰₯πœ†πœŒ1πœŽπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1ξ‚€1(𝜏)πœ†ξ€·π‘€1+πœ€3𝑝1βˆ’1πœ™π‘1𝑒0ξ€Έξ‚Ά(𝜏)π‘‘πœπ‘‘π‘ +π‘š0β‰₯πœ†πœŒ1πœŽπ‘ž11βˆ’1ξ€œ10𝑒(𝑠)πœ™π‘ž1ξ‚΅ξ€œ10𝑒(𝜏)π‘Ž1𝑀(𝜏)1+πœ€3𝑝1βˆ’1ξ€·Ξ›π‘Ÿ2𝑝1βˆ’1ξ‚Άπ‘‘πœπ‘‘π‘ +π‘š0=16πœ†πœŒ1πœŽπ‘ž11βˆ’11πœ†ξ€·π‘€1+πœ€3ξ€ΈΞ›π‘Ÿ2𝐿1+π‘š0>π‘Ÿ2.(3.28)
This implies that π‘Ÿ2>π‘Ÿ2, which is a contradiction. This yields that (3.26) holds. By (3.24), (3.26), and Lemma 1.2, for any 𝑛>1/π‘Ÿ2and πœ†βˆˆ(𝑀1/𝑓10,𝑁1/π‘“βˆž1), πœ‡βˆˆ(0,𝑁2/π‘“βˆž2), we obtain that 𝑇𝑛 has a fixed point (𝑒𝑛,𝑣𝑛) in πΎπ‘Ÿ2,𝑅2 and π‘Ÿ2<‖𝑒𝑛‖<𝑅2,β€‰β€‰π‘Ÿ2<‖𝑣𝑛‖<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,β€‰β€‰π‘Ž1√(𝑑)=1/(𝑑(1βˆ’π‘‘),β€‰β€‰π‘Ž2√(𝑑)=1/((1βˆ’π‘‘)𝑑), 𝑓1𝑑(𝑑,𝑒,𝑣)=2+1βˆšπ‘’ξ€·π‘£+1+sin2ξ€Έ[]+𝑣+𝑑,(𝑑,𝑒,𝑣)∈0,1×ℝ+0×ℝ+,𝑓2𝑑(𝑑,𝑒,𝑣)=2+sin(𝑒+ln(𝑑+1))+4+𝑑+3βˆšπ‘£[],(𝑑,𝑒,𝑣)∈0,1×ℝ+×ℝ+0.(4.1)

Obviously, π‘Ž1,β€‰β€‰π‘Ž2 are singular at 𝑑=0 and 𝑑=1, 𝑓1(𝑑,𝑒,𝑣) is singular at 𝑒=0 and 𝑓2(𝑑,𝑒,𝑣) is singular at 𝑣=0. Choose 𝑔1(𝑑,𝑒)=(𝑑2√+1)/𝑒, β„Ž1(𝑑,𝑣)=1+sin(𝑣2+𝑣+𝑑), 𝑔2(𝑑,𝑒)=2+sin(𝑒+ln(𝑑+1)), and β„Ž2(𝑑,𝑣)=(𝑑4βˆšπ‘£+𝑑+3)/. Let πœ‰1⎧βŽͺβŽͺ⎨βŽͺβŽͺβŽ©ξ‚ƒ1(𝑠)=0,π‘ βˆˆ0,3,151,π‘ βˆˆ3,23,142,π‘ βˆˆ3ξ‚„,πœ‰,12⎧βŽͺβŽͺ⎨βŽͺβŽͺβŽ©ξ‚ƒ1(𝑠)=0,π‘ βˆˆ0,2,171,π‘ βˆˆ2,34,133,π‘ βˆˆ4ξ‚„,πœ‚,11⎧βŽͺ⎨βŽͺβŽ©ξ‚ƒ1(𝑠)=0,π‘ βˆˆ0,2,351,π‘ βˆˆ2ξ‚„,πœ‚,12⎧βŽͺ⎨βŽͺβŽ©ξ‚ƒ1(𝑠)=0,π‘ βˆˆ0,2,471,π‘ βˆˆ2ξ‚„.,1(4.2)

By direct calculation, we have 𝛼1=1/4,  𝛼2=1/3,