Table of Contents
ISRN Applied Mathematics
VolumeΒ 2011, Article IDΒ 612591, 13 pages
http://dx.doi.org/10.5402/2011/612591
Research Article

Positive Solutions for Second-Order Nonlinear Ordinary Differential Systems with Two Parameters

Lan Sun,1Β Yukun An,1Β and Min Jiang2

1Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China
2College of Computer Science and Technology, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China

Received 7 September 2011; Accepted 23 October 2011

Academic Editor: G. C.Β Georgiou

Copyright Β© 2011 Lan Sun 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

By using fixed-point theorem and under suitable conditions, we investigate the existence and multiplicity positive solutions to the following systems: π‘’ξ…žξ…ž(𝑑)+π‘Žπ‘’(𝑑)+𝑏𝑣(𝑑)+πœ†β„Ž1(𝑑)𝑓(𝑒(𝑑),𝑣(𝑑))=0,π‘‘βˆˆ[0,1],π‘£ξ…žξ…ž(𝑑)+𝑐𝑒(𝑑)+𝑑𝑣(𝑑)+πœ‡β„Ž2(𝑑)𝑔(𝑒(𝑑),𝑣(𝑑))=0,π‘‘βˆˆ[0,1],𝑒(0)=𝑒(1)=0,𝑣(0)=𝑣(1)=0, where π‘Ž,𝑏,𝑐,𝑑 are four positive constants and πœ†>0, πœ‡>0, 𝑓(𝑒,𝑣),𝑔(𝑒,𝑣)∈𝐢(𝑅+×𝑅+,𝑅+) and β„Ž1,β„Ž2∈𝐢([0,1],𝑅+). We derive two explicit intervals of πœ† and πœ‡, such that the existence and multiplicity of positive solutions for the systems is guaranteed.

1. Introduction

In this paper, we are concerned with the existence and multiplicity of positive solutions to the following boundary value problem of second-order nonlinear differential systems with two parameters: π‘’ξ…žξ…ž(𝑑)+π‘Žπ‘’(𝑑)+𝑏𝑣(𝑑)+πœ†β„Ž1[],𝑣(𝑑)𝑓(𝑒(𝑑),𝑣(𝑑))=0,π‘‘βˆˆ0,1ξ…žξ…ž(𝑑)+𝑐𝑒(𝑑)+𝑑𝑣(𝑑)+πœ‡β„Ž2[],𝑣(𝑑)𝑔(𝑒(𝑑),𝑣(𝑑))=0,π‘‘βˆˆ0,1𝑒(0)=𝑒(1)=0,(0)=𝑣(1)=0,(1.1) where π‘Ž,𝑏,𝑐,𝑑 are four positive constants.

In addition, we assume the following conditions throughout this paper:(𝐻1)β„Ž1(𝑑),β„Ž2(𝑑)∈𝐢([0,1],𝑅+) does not vanish identically on any subinterval of [0,1];(𝐻2)0<π‘Ž<πœ‹2,0<𝑑<πœ‹2,𝑐>0,𝑏>0; (𝐻3)𝑓,π‘”βˆˆπΆ(𝑅+×𝑅+,𝑅+).

The boundary value problem of ordinary differential systems has attracted much attention, see [1–6] and the references therein. Recently, Fink and Gatica [4] and Ma [7] have studied the existence of positive solutions of the following systems: π‘₯ξ…žξ…ž[],𝑦(𝑑)+πœ†π‘Ž(𝑑)𝑓(π‘₯(𝑑),𝑦(𝑑))=0,π‘‘βˆˆ0,1ξ…žξ…ž[],(𝑑)+πœ‡π‘(𝑑)𝑔(π‘₯(𝑑),𝑦(𝑑))=0,π‘‘βˆˆ0,1π‘₯(0)=π‘₯(1)=𝑦(0)=𝑦(1)=0.(1.2) In [4], a multiplicity result has been established when 𝑓(0,0)>0,𝑔(0,0)>0. In [7], a multiplicity result has been given for the more general case 𝑓(0,0)β‰₯0,𝑔(0,0)β‰₯0.

Also, by using Krasnoselβ€²skill fixed-point theorem, Ru and An [8] considered the existence of positive solutions of the following systems: (βˆ’)𝑝𝑒(2𝑝)[],=πœ†π‘Ž(𝑑)𝑓(𝑒(𝑑),𝑣(𝑑)),π‘‘βˆˆ0,1(βˆ’)π‘žπ‘£(2π‘ž)[],𝑒=πœ‡π‘(𝑑)𝑔(𝑒(𝑑),𝑣(𝑑)),π‘‘βˆˆ0,1(2𝑖)(0)=𝑒(2𝑖)𝑣(1)=0,0β‰€π‘–β‰€π‘βˆ’1,(2𝑗)(0)=𝑣(2𝑗)(1)=0,0β‰€π‘—β‰€π‘žβˆ’1,(1.3) where πœ†>,πœ‡>0, 𝑝,π‘žβˆˆπ‘.𝑓,π‘”βˆΆ[0,∞)Γ—[0,∞)β†’[0,∞).

More recently, Dalbono and Mckenna [5] proved the existence and multiplicity of solutions to a class of asymmetric weakly coupled systems as follows: βŽ‘βŽ’βŽ’βŽ£π‘’1ξ…žξ…žπ‘’(𝑑)2ξ…žξ…ž(⎀βŽ₯βŽ₯⎦+βŽ‘βŽ’βŽ’βŽ£π‘π‘‘)1πœ€πœ€π‘2⎀βŽ₯βŽ₯βŽ¦βŽ‘βŽ’βŽ’βŽ£π‘’+1𝑒(𝑑)+2(⎀βŽ₯βŽ₯⎦=⎑⎒⎒⎣⎀βŽ₯βŽ₯⎦,𝑒𝑑)sin(𝑑)sin(𝑑)1(0)=𝑒2(0)=0=𝑒1(πœ‹)=𝑒2(πœ‹),(1.4) where πœ€ is suitably small and the positive numbers 𝑏1, 𝑏2 satisfy β„Ž2<𝑏1<(β„Ž+1)2,π‘˜2<𝑏2<(π‘˜+1)2forsomeβ„Ž,π‘˜βˆˆπ‘.(1.5) Applying a classical change of variables, the authors transformed the initial problem into an equivalent problem whose solutions can be characterized by their nodal properties. Meanwhile, in [5], there are two open questions. (1)Can one replace the near-diagonal matrix with something more general and use information on the eigenvalues of matrix?(2)Can one replace the near-diagonal matrix with something more general and use information on the eigenvalues of matrix?

Inspired by the above works and the two open questions, we consider the existence and multiplicity of positive solutions to (1.1). The paper is organized as follows. In Section 2, we state some preliminaries. In Sections 3 and 4, we prove the existence and multiplicity results of (1.1).

2. Preliminaries

In order to prove our results, we state the well-known fixed-point theorem [9]:

Lemma 2.1. Let 𝐸 be a Banach space and let πΎβŠ‚πΈ be a cone in 𝐸. Assume Ξ©1, Ξ©2 are two open subsets of 𝐸 with 0∈Ω1, Ξ©1βŠ‚Ξ©2, and let π‘‡βˆΆπΎβˆ©(Ξ©2⧡Ω1)→𝐾 be a completely continuous operator such that either(i)‖𝑇𝑒‖≀‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©1 and ‖𝑇𝑒‖β‰₯‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©2; or(ii)‖𝑇𝑒‖β‰₯‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©1 and ‖𝑇𝑒‖≀‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©2.Then, 𝑇 has a fixed point in 𝐾∩(Ξ©2⧡Ω1).
To be convenient, we introduce the following notations: 𝑓0=lim𝑒+𝑣→0𝑓(𝑒,𝑣)𝑒+𝑣,π‘“βˆž=lim𝑒+π‘£β†’βˆžπ‘“(𝑒,𝑣),𝑔𝑒+𝑣0=lim𝑒+𝑣→0𝑔(𝑒,𝑣)𝑒+𝑣,π‘”βˆž=lim𝑒+π‘£β†’βˆžπ‘”(𝑒,𝑣),𝑒+𝑣(2.1) and suppose that 𝑓0,𝑔0,π‘“βˆž,π‘”βˆžβˆˆ[0,+∞].
Let πœ”1=βˆšπ‘Ž,πœ”2=βˆšπ‘‘,𝐻(𝑑)∈𝐢[0,1], 𝐺𝑖(𝑑,𝑠),𝑖=1,2 be πΊπ‘Ÿπ‘’π‘’π‘›ξ…žπ‘  function of the corresponding to linear boundary value problem: βˆ’π‘’ξ…žξ…ž(𝑑)βˆ’πœ”2𝑖𝑒(𝑑)=𝐻(𝑑),𝑒(0)=𝑒(1)=0.(2.2) Then, the solution of (2.2) is given by ξ€œπ‘’(𝑑)=10𝐺𝑖(𝑑,𝑠)𝐻(𝑠)𝑑𝑠.(2.3) It is well known that 𝐺𝑖(𝑑,𝑠) can be expressed by πΊπ‘–βŽ§βŽͺβŽͺ⎨βŽͺβŽͺ⎩(𝑑,𝑠)=sinπœ”π‘–π‘‘sinπœ”π‘–(1βˆ’π‘ )πœ”π‘–sinπœ”π‘–0≀𝑑≀𝑠≀1,sinπœ”π‘–π‘ sinπœ”π‘–(1βˆ’π‘‘)πœ”π‘–sinπœ”π‘–0≀𝑠≀𝑑≀1.(2.4) In addition, it can be easily to be checked that 𝑀𝑖=max[]π‘‘βˆˆ0,1𝐺𝑖(𝑑,𝑑)>0,π‘šπ‘–=min[]π‘‘βˆˆ1/4,3/4𝐺𝑖(𝑑,𝑑)>0,𝑖=1,2.(2.5)

See [6], we have the following lemma.

Lemma 2.2. 𝐺𝑖(𝑑,𝑠) has the following properties:(i)𝐺𝑖(𝑑,𝑠)>0,forall𝑑,π‘ βˆˆ(0,1), (ii)𝐺𝑖(𝑑,𝑠)≀𝐢𝑖𝐺𝑖(𝑠,𝑠),forall𝑑,π‘ βˆˆ[0,1],𝐢𝑖=1/sinπœ”π‘–, (iii)𝐺𝑖(𝑑,𝑠)β‰₯𝛿𝑖𝐺𝑖(𝑑,𝑑)𝐺𝑖(𝑠,𝑠),forall𝑑,π‘ βˆˆ[0,1],𝛿𝑖=πœ”π‘–sinπœ”π‘–,𝑖=1,2. It is obvious that problem (1.1) is equivalent to the equation: ξ‚΅ξ€œ(𝑒(𝑑),𝑣(𝑑))=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑s,10𝐺2ξ€·(𝑑,𝑠)πœ‡β„Ž2ξ€Έξ‚Ά,(𝑠)𝑔(𝑒(𝑠),𝑣(𝑠))+𝑐𝑒(𝑠)𝑑𝑠(2.6) and consequently it is equivalent to the fixed-point problem: (𝑒,𝑣)=𝐀(𝑒,𝑣)(2.7) with π€βˆΆπΆ[0,1]×𝐢[0,1]→𝐢[0,1]×𝐢[0,1] given by =ξ‚΅ξ€œπ€(𝑒,𝑣)10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠,10𝐺2ξ€·(𝑑,𝑠)πœ‡β„Ž2ξ€Έξ‚Ά.(𝑠)𝑔(𝑒(𝑠),𝑣(𝑠))+𝑐𝑒(𝑠)𝑑𝑠(2.8) For convenience, denote π€πœ†ξ€œ(𝑒,𝑣)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1𝐀(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠,πœ‡ξ€œ(𝑒,𝑣)=10𝐺2ξ€·(𝑑,𝑠)πœ‡β„Ž2ξ€Έ(𝑠)𝑔(𝑒(𝑠),𝑣(𝑠))+𝑐𝑒(𝑠)𝑑𝑠.(2.9) It is obvious that 𝐀 is completely continuous. Let πœŽπ‘–=π›Ώπ‘–π‘šπ‘–/𝐢𝑖=(sinπœ”π‘–/4)(sin3πœ”π‘–/4)(sinπœ”π‘–), 𝑖=1,2,𝜎=min{𝜎1,𝜎2} and define a cone in 𝐢[0,1]×𝐢[0,1] by ξ‚»[][]𝐊=(𝑒,𝑣)∈𝐢0,1×𝐢0,1βˆΆπ‘’(𝑑)β‰₯0,𝑣(𝑑)β‰₯0,min1/4≀𝑑≀3/4β€–β€–ξ‚Ό(𝑒(𝑑)+𝑣(𝑑))β‰₯𝜎(𝑒,𝑣),(2.10) where β€–(𝑒,𝑣)β€–=‖𝑒‖+‖𝑣‖=supπ‘‘βˆˆ[0,1]𝑒(𝑑)+supπ‘‘βˆˆ[0,1]𝑣(𝑑).

Lemma 2.3. 𝐴(𝐾)βŠ‚πΎ. Proof. For any (𝑑,𝑠)∈[1/4,3/4]Γ—[0,1], by Lemma 2.2, we have π΄πœ†ξ€œ(𝑒,𝑣)(𝑑)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έβ‰€ξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠10𝐢1𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1‖‖𝐴(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠,πœ†β€–β€–β‰€ξ€œ(𝑒,𝑣)10𝐢1𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1𝐴(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠,πœ†ξ€œ(𝑒,𝑣)(𝑑)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝛿1𝐺1ξ€œ(𝑑,𝑑)10𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έβ‰₯𝛿(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠1𝐺1(𝑑,𝑑)𝐢1β€–β€–π΄πœ†(β€–β€–.𝑒,𝑣)(2.11) Similarly, for any (𝑑,𝑠)∈[1/4,3/4]Γ—[0,1], we have π΄πœ‡π›Ώ(𝑒,𝑣)(𝑑)β‰₯2𝐺2(𝑑,𝑑)𝐢2β€–β€–π΄πœ‡β€–β€–.(𝑒,𝑣)(2.12) Then, for any (𝑑,𝑠)∈[1/4,3/4]Γ—[0,1], we have π΄πœ†(𝑒,𝑣)(𝑑)+π΄πœ‡ξ€œ(𝑒,𝑣)(𝑑)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έξ€œπ‘“(𝑒,𝑣)+𝑏𝑣𝑑𝑠+10𝐺2ξ€·(𝑑,𝑠)πœ‡β„Ž2ξ€Έβ‰₯𝛿𝑔(𝑒,𝑣)+𝑐𝑒𝑑𝑠1𝐺1(𝑑,𝑑)𝐢1β€–β€–π΄πœ†β€–β€–+𝛿(𝑒,𝑣)2𝐺2(𝑑,𝑑)𝐢2β€–β€–π΄πœ‡β€–β€–(𝑒,𝑣)β‰₯πœŽβ€–π΄(𝑒,𝑣)β€–.(2.13) Thus, min1/4≀𝑑≀3/4π΄πœ†(𝑒,𝑣)+π΄πœ‡(𝑒,𝑣)β‰₯πœŽβ€–π΄(𝑒,𝑣)β€–. Therefore, 𝐴(𝐾)βŠ‚πΎ.

3. Existence Results

We assume the following:(𝐻4)0<𝑏<2πœ”1cos2(πœ”1/2),0<𝑐<2πœ”2cos2(πœ”2/2); (𝐻5)π΄π‘–βˆ«=β€–10β„Žπ‘–(𝑠)𝑑𝑠‖,𝐡𝑖=∫3/41/4β„Žπ‘–(𝑠)𝑑𝑠>0,𝑖=1,2; (𝐻6)𝑃𝑖=𝐢𝑖𝑀𝑖=(sec2πœ”π‘–/2)/4πœ”π‘–,𝑄𝑖=π›Ώπ‘–π‘š2𝑖=sin(πœ”π‘–/4)sin(3πœ”π‘–/4),𝑖=1,2.

Theorem 3.1. Assume (𝐻1)–(𝐻6) hold. Then one has the following:
(1)If 0<𝑓0,π‘“βˆž,𝑔0,π‘”βˆž<∞,2𝑃1𝐴1𝑓0<𝑄1𝜎𝐡1π‘“βˆž(1βˆ’2𝑃1𝑏), then for each πœ†βˆˆ(1/𝑄1𝜎𝐡1π‘“βˆž,1βˆ’2𝑃1𝑏/2𝑃1𝐴1𝑓0) and πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑔0), (1.1) has at least one positive solution. (2)If 0<𝑓0,π‘“βˆž,𝑔0,π‘”βˆž<∞,2𝑃2𝐴2𝑔0<𝑄2𝜎𝐡2π‘”βˆž(1βˆ’2𝑃2𝑐), then for each πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑓0) and πœ‡βˆˆ(1/𝑄2𝜎𝐡2π‘”βˆž,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑔0), (1.1) has at least one positive solution. (3)If 𝑓0=0,π‘“βˆž=∞, then for each πœ†βˆˆ(0,∞) and πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑔0), (1.1) has at least one positive solution.(4)If 𝑔0=0,π‘”βˆž=∞, then for each πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑓0) and πœ‡βˆˆ(0,∞), (1.1) has at least one positive solution.(5)If 𝑓0=0,π‘“βˆž=∞ and 𝑔0=0,π‘”βˆž=∞, then for each πœ†βˆˆ(0,∞) and πœ‡βˆˆ(0,∞), (1.1) has at least one positive solution.(6)If π‘“βˆž=∞,0<𝑓0<∞ or π‘”βˆž=∞,0<𝑔0<∞, then for each πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑓0) and πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑔0), (1.1) has at least one positive solution. (7)If 𝑓0=0,0<π‘“βˆž<∞ and 𝑔0=0,0<π‘”βˆž<∞, then for each πœ†βˆˆ(1/𝑄1𝜎𝐡1π‘“βˆž,∞),πœ‡βˆˆ(0,∞) or πœ†βˆˆ(0,∞),πœ‡βˆˆ((1/𝑄2𝜎𝐡2π‘”βˆž)∞),(1.1) has at least one positive solution.

Proof. We only prove case (1.1). The other cases can be proved similarly. In order to apply the Lemma 2.1, we construct the sets Ξ©1,Ξ©2.
Let πœ†βˆˆ(1/𝑄1𝜎𝐡1π‘“βˆž,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑓0),πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑔0), and we choose πœ€>0 such that 1πœŽπ‘„1𝐡1ξ€·π‘“βˆžξ€Έβˆ’πœ€β‰€πœ†β‰€1βˆ’2𝑃1𝑏2𝑃1𝐴1𝑓0ξ€Έ+πœ€,0<πœ‡β‰€1βˆ’2𝑃2𝑐2𝑃2𝐴2𝑔0ξ€Έ+πœ€.(3.1)
By the definition of 𝑓0 and 𝑔0, there exists 𝑅1>0, such that 𝑓𝑓(𝑒,𝑣)≀0𝑔+πœ–(𝑒+𝑣),𝑔(𝑒,𝑣)≀0ξ€Έξ€Ί+πœ–(𝑒+𝑣),for𝑒+π‘£βˆˆ0,𝑅1ξ€».(3.2) Choosing (𝑒,𝑣)∈𝐾 with β€–(𝑒,𝑣)β€–=𝑅1, we have π΄πœ†ξ€œ(𝑒,𝑣)(𝑑)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠≀𝐢1𝑀1ξ€œ10ξ€·πœ†β„Ž1𝑓(𝑠)0ξ€Έξ€Έ+πœ€(𝑒(𝑠)+𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠≀𝐢1𝑀1β€–β€–ξ€œ(𝑒,𝑣)10ξ€·πœ†β„Ž1𝑓(𝑠)0ξ€Έξ€Έ+πœ€+𝑏𝑑𝑠=𝑃1ξ€·β€–(𝑒,𝑣)β€–πœ†π΄1𝑓0≀1+πœ€+𝑏2β€–(𝑒,𝑣)β€–,(3.3) namely, β€–π΄πœ†(𝑒,𝑣)(𝑑)‖≀(1/2)β€–(𝑒,𝑣)β€–. In the same way, we also have β€–β€–π΄πœ‡β€–β€–β‰€1(𝑒,𝑣)(𝑑)2β€–(𝑒,𝑣)β€–.(3.4) then β€–π΄πœ†(𝑒,𝑣)(𝑑)‖≀‖(𝑒,𝑣)(𝑑)β€–. Thus, if we set Ξ©1={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<𝑅1}, then ‖𝐴(𝑒,𝑣)(𝑑)‖≀‖(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©1.(3.5) On the other hand, by the definition of π‘“βˆž, there exists 𝑅2β€²>0, such that 𝑓(𝑒,𝑣)β‰₯(π‘“βˆžβˆ’πœ–)(𝑒+𝑣), for 𝑒+π‘£βˆˆ[π‘…ξ…ž2,+∞). Let 𝑅2=max{2𝑅1,πœŽβˆ’1π‘…ξ…ž2} and Ξ©2={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<𝑅2}. If we choose (𝑒,𝑣)∈𝐾 with β€–(𝑒,𝑣)β€–=𝑅2, such that min1/4≀𝑑≀3/4(𝑒(𝑑)+𝑣(𝑑))β‰₯πœŽβ€–(𝑒,𝑣)β€–β‰₯𝑅2, then we have π΄πœ†ξ‚€1(𝑒,𝑣)4=ξ€œ10𝐺1ξ‚€14,π‘ πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝛿1𝐺1ξ‚€14,14ξ‚ξ€œ3/41/4𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝛿1π‘š21ξ€œ3/41/4ξ€·πœ†β„Ž1𝑓(𝑠)βˆžβˆ’πœ€ξ€Έξ€Έ(𝑒(𝑠)+𝑣(𝑠))𝑑𝑠β‰₯𝑄1𝜎𝐡1πœ†ξ€·π‘“βˆžξ€Έβˆ’πœ€β€–(𝑒,𝑣)β€–β‰₯β€–(𝑒,𝑣)β€–.(3.6) Hence ‖‖𝐴‖𝐴(𝑒,𝑣)(𝑑)β€–β‰₯πœ†β€–β€–β‰₯β€–(𝑒,𝑣)(𝑑)(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©2.(3.7) Therefore, it follows from (3.5) to (3.7) and Lemma 2.2, 𝐴 has a fixed point (𝑒,𝑣)∈𝐾∩(Ξ©2⧡Ω1), which is a positive solution of (1.1).

Similarly, we also have the following results.

Theorem 3.2. Assume (𝐻1)–(𝐻6) hold. Then one has the following:(1)If 0<𝑓0,π‘“βˆž,𝑔0,π‘”βˆž<∞,2𝑃1𝐴1π‘“βˆž<𝑄1𝜎𝐡1𝑓0(1βˆ’2𝑃1𝑏), then for each πœ†βˆˆ(1/𝑄1𝜎𝐡1𝑓0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1π‘“βˆž) and πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2π‘”βˆž), (1.1) has at least one positive solution.(2)If 0<𝑓0,π‘“βˆž,𝑔0,π‘”βˆž<∞,2𝑃2𝐴2π‘”βˆž<𝑄2𝜎𝐡2𝑔0(1βˆ’2𝑃2𝑐), then for each πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1π‘“βˆž) and πœ‡βˆˆ(1/𝑄2𝜎𝐡2𝑔0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2π‘”βˆž), (1.1) has at least one positive solution.(3)If 𝑓0=∞,π‘“βˆž=0, then for each πœ†βˆˆ(0,∞) and πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑔0), (1.1) has at least one positive solution.(4)If 𝑔0=∞,π‘”βˆž=0, then for each πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑓0) and πœ‡βˆˆ(0,∞), (1.1) has at least one positive solution.(5)If 𝑓0=∞,π‘“βˆž=0 and 𝑔0=∞,π‘”βˆž=0, then for each πœ†βˆˆ(0,∞) and πœ‡βˆˆ(0,∞), (1.1) has at least one positive solution.(6)If 𝑓0=∞,0<π‘“βˆž<∞ or 𝑔0=∞,0<π‘”βˆž<∞ then for each πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1π‘“βˆž) and πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2π‘”βˆž), (1.1) has at least one positive solution.(7)If π‘“βˆž=0,0<𝑓0<∞ and π‘”βˆž=0,0<𝑔0<∞ then for each πœ†βˆˆ(1/𝑄1𝜎𝐡1𝑓0,∞),πœ‡βˆˆ(0,∞) or πœ†βˆˆ(0,∞),πœ‡βˆˆ(1/𝑄2𝜎𝐡2𝑔0), (1.1) has at least one positive solution.

Proof. We only prove case (1). The other cases can be proved similarly.
Let πœ†βˆˆ(1/𝑄1𝜎𝐡1𝑓0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1π‘“βˆž),πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2π‘”βˆž), and we choose πœ€>0 such that 1πœŽπ‘„1𝐡1𝑓0ξ€Έβˆ’πœ€β‰€πœ†β‰€1βˆ’2𝑃1𝑏2𝑃1𝐴1ξ€·π‘“βˆžξ€Έ+πœ€,0<πœ‡β‰€1βˆ’2𝑃2𝑐2𝑃2𝐴2ξ€·π‘”βˆžξ€Έ+πœ€,(3.8)
by the definition of 𝑓0, there exists 𝑅1>0, such that 𝑓𝑓(u,𝑣)β‰₯0ξ€Έξ€Ίβˆ’πœ–(𝑒+𝑣),for𝑒+π‘£βˆˆ0,𝑅1ξ€».(3.9) Choosing (𝑒,𝑣)∈𝐾 with β€–(𝑒,𝑣)β€–=𝑅1, such that min1/4≀𝑑≀3/4(𝑒(𝑑)+𝑣(𝑑))β‰₯πœŽβ€–(𝑒,𝑣)β€–, then we have π΄πœ†ξ‚€1(𝑒,𝑣)4=ξ€œ10𝐺1ξ‚€14,π‘ πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝛿1𝐺1ξ‚€14,14ξ‚ξ€œ3/41/4𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝛿1π‘š21ξ€œ3/41/4ξ€·πœ†β„Ž1𝑓(𝑠)0βˆ’πœ€ξ€Έξ€Έ(𝑒(𝑠)+𝑣(𝑠))𝑑𝑠β‰₯𝑄1𝜎𝐡1πœ†ξ€·π‘“0ξ€Έβˆ’πœ€β€–(𝑒,𝑣)β€–β‰₯β€–(𝑒,𝑣)β€–.(3.10) So, if we set Ξ©1={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<𝑅1}, then ‖𝐴(𝑒,𝑣)(𝑑)β€–β‰₯β€–(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©1.(3.11) On the other hand, by the definition of π‘“βˆž and π‘”βˆž, there exists 𝑅2>2𝑅1, such that 𝑓𝑓(𝑒,𝑣)β‰€βˆžξ€Έξ€·π‘”+πœ–(𝑒+𝑣),𝑔(𝑒,𝑣)β‰€βˆžξ€Έξ€Ίπ‘…+πœ–(𝑒+𝑣)for𝑒+π‘£βˆˆξ…ž2,ξ€Έ+∞.(3.12) Choosing (𝑒,𝑣)∈𝐾 with β€–(𝑒,𝑣)β€–=𝑅2, we have π΄πœ†ξ€œ(𝑒,𝑣)(𝑑)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έβ‰€ξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠10𝐢1𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠≀𝐢1𝑀1ξ€œ10ξ€·πœ†β„Ž1𝑓(𝑠)βˆžξ€Έξ€Έ+πœ€(𝑒(𝑠)+𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠≀𝑃1ξ€·πœ†π΄1ξ€·π‘“βˆžξ€Έξ€Έβ‰€1+πœ€+𝑏‖(𝑒,𝑣)β€–2β€–(𝑒,𝑣)β€–.(3.13) In the same way, we also have β€–β€–π΄πœ‡β€–β€–β‰€1(𝑒,𝑣)(𝑑)2β€–(𝑒,𝑣)β€–.(3.14) Hence, if we set Ξ©2={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<𝑅2}, then ‖𝐴(𝑒,𝑣)(𝑑)‖≀‖(𝑒,𝑣)(𝑑)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©2.(3.15) Therefore, it follows from (3.11) to (3.15) and Lemma 2.2 that 𝐴 has a fixed point (𝑒,𝑣)∈𝐾∩(Ξ©2⧡Ω1), which is a positive solution of (1.1).

4. Multiplicity Results

Theorem 4.1. Assume (𝐻1)–(𝐻6) hold. In addition, assume that there exist three constants π‘Ÿ,𝑀,𝑁, where 𝑁 is sufficient small with 2𝑃1𝑁𝐴1≀𝑄1𝐡1𝑀(1βˆ’2𝑃1𝑏),2𝑃2𝑁𝐴2≀𝑄2𝐡2𝑀(1βˆ’2𝑃2𝑐) such that(i)𝑓0=π‘“βˆž=0,𝑔0=π‘”βˆž=0, (ii)𝑓(𝑒,𝑣)β‰₯π‘€π‘Ÿπ‘œπ‘Ÿπ‘”(𝑒,𝑣)β‰₯π‘€π‘Ÿ,  for πœŽπ‘Ÿβ‰€β€–(𝑒,𝑣)β€–β‰€π‘Ÿ.Then, for any πœ†βˆˆ(1/𝑄1𝐡1𝑀,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑁), πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑁), or πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑁), πœ‡βˆˆ(1/𝑄2𝐡2𝑀,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑁), the problem (1.1) has at least two positive solutions.

Proof. We only prove the case of πœ†βˆˆ(1/𝑄1𝜎𝐡1𝑀,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑁), πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐/2𝑃2𝐴2𝑁)), The other case is similar.
Step 1. Since 𝑓0=𝑔0=0, there exists π‘Ÿ1∈(0,π‘Ÿ) such that 𝑓(𝑒,𝑣)≀𝑁(𝑒+𝑣),𝑔(𝑒,𝑣)≀𝑁(𝑒+𝑣),for0<𝑒+𝑣<π‘Ÿ1.(4.1) Set Ξ©1={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<π‘Ÿ1},forall(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©1, then we have π΄πœ†ξ€œ(𝑒,𝑣)(𝑑)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έβ‰€ξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠10𝐢1𝑀1ξ€·πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠≀𝐢1𝑀1ξ€œ10ξ€·πœ†β„Ž1ξ€Έ(𝑠)𝑁(𝑒(𝑠)+𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠≀𝑃1ξ€·πœ†π΄1≀1𝑁+𝑏‖(𝑒,𝑣)β€–2β€–(𝑒,𝑣)β€–.(4.2) In the same way, we also have β€–β€–π΄πœ‡β€–β€–β‰€1(𝑒,𝑣)(𝑑)2β€–(𝑒,𝑣)β€–.(4.3) Hence, ‖𝐴(𝑒,𝑣)‖≀‖(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©1.(4.4)
Step 2. Since π‘“βˆž=π‘”βˆž=0, there exists 𝑅>π‘Ÿ such that 𝑓(𝑒,𝑣)≀𝑁(𝑒+𝑣),𝑔(𝑒,𝑣)≀𝑁(𝑒+𝑣),for𝑒+𝑣β‰₯𝑅.(4.5) Similarly, set Ξ©2={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<π‘Ÿ2}, then ‖𝐴(𝑒,𝑣)‖≀‖(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©2.(4.6)
Step 3. Let Ξ©3={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<π‘Ÿ}, we can see that π΄πœ†ξ‚€1(𝑒,𝑣)4=ξ€œ10𝐺1ξ‚€14,π‘ πœ†β„Ž1ξ€Έβ‰₯ξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠10𝛿1𝐺1ξ‚€14,14𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝛿1𝐺1ξ‚€14,14ξ‚ξ€œ3/41/4𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝑄1ξ€œ3/41/4ξ€·πœ†β„Ž1ξ€Έ(𝑠)(𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠))𝑑𝑠β‰₯𝑄1πœ†π΅1π‘€π‘Ÿβ‰₯π‘Ÿ.(4.7) Then, ‖𝐴(𝑒,𝑣)β€–β‰₯β€–(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©3.(4.8)
Consequently, by Lemma 2.1 and from (4.4)–(4.8), 𝐴 has two fixed points: (𝑒1,𝑣1)∈𝐾∩(Ξ©3⧡Ω1) and (𝑒2,𝑣2)∈𝐾∩(Ξ©2⧡Ω3), so (1.1) has at least two positive solutions satisfying 0<(𝑒1,𝑣1)<π‘Ÿ<(𝑒2,𝑣2).

Theorem 4.2. Assume (𝐻1)–(𝐻6) hold. In addition, assume that there exist constants π‘Ÿ,𝑀,𝑁, where 𝑁 is sufficient large with 2𝑃1𝑀𝐴1≀𝑄1𝜎𝐡1𝑁(1βˆ’2𝑃1𝑏),2𝑃2𝑀𝐴2≀𝑄2𝜎𝐡2𝑁(1βˆ’2𝑃2𝑐) such that(i)𝑓0=π‘“βˆž=∞ or 𝑔0=π‘”βˆž=∞,(ii)𝑓(𝑒,𝑣)β‰€π‘€π‘Ÿ or 𝑔(𝑒,𝑣)β‰€π‘€π‘Ÿ,  for πœŽπ‘Ÿβ‰€β€–(𝑒,𝑣)β€–β‰€π‘Ÿ,then, for any πœ†βˆˆ(1/𝑄1𝜎𝐡1𝑁,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑀), πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑀), or πœ†βˆˆ(0,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑀), πœ‡βˆˆ(1/𝑄2𝜎𝐡2𝑁,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑀), the problem (1.1) has at least two positive solutions.

Proof. We only prove the case of πœ†βˆˆ(1/𝑄1𝜎𝐡1𝑁,(1βˆ’2𝑃1𝑏)/2𝑃1𝐴1𝑀), πœ‡βˆˆ(0,(1βˆ’2𝑃2𝑐)/2𝑃2𝐴2𝑀), The other case is similar.
Step 1. Since 𝑓0=∞, there exists π‘Ÿ1∈(0,π‘Ÿ) such that 𝑓(𝑒,𝑣)β‰₯𝑁(𝑒+𝑣),for0<𝑒+π‘£β‰€π‘Ÿ1.(4.9) Set Ξ©1={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<π‘Ÿ1}, then we have π΄πœ†ξ‚€1(𝑒,𝑣)4=ξ€œ10𝐺1ξ‚€14,π‘ πœ†β„Ž1ξ€Έβ‰₯ξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠10𝛿1𝐺1ξ‚€14,14𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝛿1𝐺1ξ‚€14,14ξ‚ξ€œ3/41/4𝐺1ξ€·(𝑠,𝑠)πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠β‰₯𝑄1ξ€œ3/41/4ξ€·πœ†β„Ž1ξ€Έ(𝑠)(𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠))𝑑𝑠β‰₯𝑄1πœ†π΅1π‘πœŽβ€–(𝑒,𝑣)β€–β‰₯π‘Ÿ.(4.10) Hence, ‖𝐴(𝑒,𝑣)β€–β‰₯β€–(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©1.(4.11)
Step 2. Since π‘“βˆž=∞, there exists 𝑅>π‘Ÿ such that 𝑓(𝑒,𝑣)β‰₯𝑁(𝑒+𝑣),for𝑒+𝑣β‰₯𝑅.(4.12) Similarly, set Ξ©2={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<π‘Ÿ2}, then ‖𝐴(𝑒,𝑣)β€–β‰₯β€–(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©2.(4.13)
Step 3. Let Ξ©3={(𝑒,𝑣)βˆˆπΎβˆΆβ€–(𝑒,𝑣)β€–<π‘Ÿ},forall(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©3, we can see that π΄πœ†ξ€œ(𝑒,𝑣)(𝑑)=10𝐺1ξ€·(𝑑,𝑠)πœ†β„Ž1ξ€Έβ‰€ξ€œ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠10𝐢1𝑀1ξ€·πœ†β„Ž1ξ€Έ(𝑠)𝑓(𝑒(𝑠),𝑣(𝑠))+𝑏𝑣(𝑠)𝑑𝑠≀𝑃1ξ€·πœ†π΄1‖‖≀1𝑀+𝑏(𝑒,𝑣)2β€–β€–.(𝑒,𝑣)(4.14) In the same way, we also have β€–β€–π΄πœ‡β€–β€–β‰€1(𝑒,𝑣)(𝑑)2β€–(𝑒,𝑣)β€–.(4.15) Hence, ‖𝐴(𝑒,𝑣)‖≀‖(𝑒,𝑣)β€–,βˆ€(𝑒,𝑣)βˆˆπΎβˆ©πœ•Ξ©3.(4.16)

Consequently, by Lemma 2.1 and from (4.11)–(4.16), 𝐴 has two fixed points: (𝑒1,𝑣1)∈𝐾∩(Ξ©3⧡Ω1) and (𝑒2,𝑣2)∈𝐾∩(Ξ©2⧡Ω3), so (1.1) has at least two positive solutions satisfying 0<(𝑒1,𝑣1)<π‘Ÿ<(𝑒2,𝑣2).

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