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Discrete Dynamics in Nature and Society
VolumeΒ 2008, Article IDΒ 257680, 19 pages
http://dx.doi.org/10.1155/2008/257680
Research Article

Twin Positive Solutions of a Nonlinear π‘š-Point Boundary Value Problem for Third-Order 𝑝-Laplacian Dynamic Equations on Time Scales

1Department of Mathematics, North University of China, Taiyuan 030051, China
2Department of Mathematics, School of Science, Tianjin University of Commerce, Tianjin 300134, China

Received 20 April 2008; Accepted 20 June 2008

Academic Editor: BinggenΒ Zhang

Copyright Β© 2008 Wei Han and Guang Zhang. 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

Several existence theorems of twin positive solutions are established for a nonlinear π‘š-point boundary value problem of third-order 𝑝-Laplacian dynamic equations on time scales by using a fixed point theorem. We present two theorems and four corollaries which generalize the results of related literature. As an application, an example to demonstrate our results is given. The obtained conditions are different from some known results.

1. Introduction

A time scale 𝐓 is a nonempty closed subset of ℝ. We make the blanket assumption that 0 and 𝑇 are points in 𝐓. By an interval (0,𝑇), we always mean the intersection of the real interval (0,𝑇) with the given time scale, that is, (0,𝑇)βˆ©π“.

In this paper, we will be concerned with the existence of positive solutions of the 𝑝-Laplacian dynamic equations on time scales:(πœ™π‘(π‘’Ξ”βˆ‡))βˆ‡+π‘Ž(𝑑)𝑓(𝑑,𝑒(𝑑))=0,π‘‘βˆˆ(0,𝑇),(1.1)πœ™π‘(π‘’Ξ”βˆ‡(0))=π‘šβˆ’2𝑖=1π‘Žπ‘–πœ™π‘(π‘’Ξ”βˆ‡(πœ‰π‘–)),𝑒Δ(0)=0,𝑒(𝑇)=π‘šβˆ’2𝑖=1𝑏𝑖𝑒(πœ‰π‘–),(1.2)πœ™π‘(𝑠)where 𝑝 is πœ™π‘(𝑠)=-Laplacian operator; that is, |𝑠, and

(H1)|π‘βˆ’2 satisfy 𝑠,𝑝>1, and πœ™π‘βˆ’1(H2)=πœ™π‘ž,1/𝑝+1/π‘ž= and there exists 1,0<πœ‰1 such that <β‹―<πœ‰π‘šβˆ’2(H3)<𝜌(𝑇).We point out that the π‘Žπ‘–,π‘π‘–βˆˆ[0,+∞),𝑖=1,2,…,-derivative and the βˆ‘0<π‘šβˆ’2𝑖=1π‘Žπ‘–<1-derivative in (1.2) and the βˆ‘π‘šβˆ’2𝑖=1𝑏𝑖<1; space in π‘Ž(𝑑)∈𝐢ld([0,𝑇],[0,+∞)) are defined in Section 2.

Recently, there has been much attention paid to the existence of positive solutions for third-order nonlinear boundary value problems of differential equations. For example, see [1–10] and the listed references. Anderson [2] considered the following third-order nonlinear problem:𝑑0∈(πœ‰π‘šβˆ’2,𝑇)He used the Krasnoselskii and the Leggett and Williams fixed-point theorems to prove the existence of solutions to the nonlinear problem (1.3). Li [6] considered the existence of single and multiple positive solutions to the nonlinear singular third-order two-point boundary value problem:π‘Ž(𝑑0)>0;Under various assumptions on π‘“βˆˆπΆ([0,𝑇]Γ—[0,+∞),[0,+∞)) and Ξ”, they established intervals of the parameter βˆ‡ which yield the existence of at least two and infinitely many positive solutions of the boundary value problem by using Krasnoselski's fixed-point theorem of cone expansion-compression type. Liu et al. [7] discussed the existence of at least one or two nondecreasing positive solutions for the following singular nonlinear third-order differential equations:𝐢𝑙𝑑Green's function and the fixed-point theorem of cone expansion-compression type are utilized in their paper. In [8], Sun considered the following nonlinear singular third-order three-point boundary value problem:(H2)He obtained various results on the existence of single and multiple positive solutions to the boundary value problem (1.6) by using a fixed-point theorem of cone expansion-compression type due to Krasnosel'skii. In [10], Zhou and Ma studied the existence and iteration of positive solutions for the following third-order generalized right-focal boundary value problem with π‘₯ξ…žξ…žξ…ž(𝑑)=𝑓(𝑑,π‘₯(𝑑)),𝑑1≀𝑑≀𝑑3,π‘₯(𝑑1)=π‘₯ξ…ž(𝑑2)=0,𝛾π‘₯(𝑑3)+𝛿π‘₯ξ…žξ…ž(𝑑3)=0.(1.3)-Laplacian operator:π‘’ξ…žξ…žξ…ž(𝑑)+πœ†π‘Ž(𝑑)𝑓(𝑒(𝑑))=0,0<𝑑<1,𝑒(0)=π‘’ξ…ž(0)=π‘’ξ…žξ…ž(1)=0.(1.4)They established a corresponding iterative scheme for (1.7) by using the monotone iterative technique.

On the other hand, the existence of positive solutions for third-order nonlinear boundary value problems of difference equations is also extensively studied by a number of authors (see [1, 3, 5, 9] and the listed references). The present work is motivated by a recent paper [4]. In [4], Henderson and Yin considered the existence of solutions for a third-order boundary value problem on a time-scale equation of the formπ‘Žwhich is uniform for the third-order difference equation and the third-order differential equation.

2. Preliminaries and Lemmas

For convenience, we list the following definitions which can be found in [4, 11–15].

Definition 2.1. Let 𝑓 be a time scale. For πœ† and π‘₯ξ…žξ…žξ…ž(𝑑)+πœ†π›Ό(𝑑)𝑓(𝑑,π‘₯(𝑑))=0,π‘Ž<𝑑<𝑏,π‘₯(π‘Ž)=π‘₯ξ…žξ…ž(π‘Ž)=π‘₯ξ…ž(𝑏)=0.(1.5), define the forward jump operator π‘’ξ…žξ…žξ…ž(𝑑)βˆ’πœ†π‘Ž(𝑑)𝐹(𝑑,𝑒(𝑑))=0,0<𝑑<1,𝑒(0)=π‘’ξ…ž(πœ‚)=π‘’ξ…žξ…ž(1)=0.(1.6) and the backward jump operator 𝑝, respectively, by(πœ™π‘(π‘’ξ…žξ…ž))ξ…ž(𝑑)=π‘ž(𝑑)𝑓(𝑑,𝑒(𝑑)),0≀𝑑≀1,𝑒(0)=π‘šξ“π‘–=1𝛼𝑖𝑒(πœ‰π‘–),π‘’ξ…ž(πœ‚)=0,π‘’ξ…žξ…ž(1)=𝑛𝑖=1π›½π‘–π‘’ξ…žξ…ž(πœƒπ‘–).(1.7)for all 𝑒Δ3ξ€·=𝑓𝑑,𝑒,𝑒Δ,𝑒ΔΔ,π‘‘βˆˆπ“,(1.8). If 𝐓, 𝑑<sup𝐓 is said to be right-scattered, and if π‘Ÿ>inf𝐓 is said to be left-scattered; if 𝜎, 𝜌 is said to be right-dense, and if ξ€½ξ€Ύξ€½ξ€ΎπœŽ(𝑑)=infπœβˆˆπ“βˆ£πœ>π‘‘βˆˆπ“,𝜌(π‘Ÿ)=supπœβˆˆπ“βˆ£πœ<π‘Ÿβˆˆπ“(2.1) is said to be left-dense. If 𝑑,π‘Ÿβˆˆπ“ has a right-scattered minimum 𝜎(𝑑)>𝑑, define 𝑑; otherwise set 𝜌(π‘Ÿ)<π‘Ÿ,π‘Ÿ. If 𝜎(𝑑)=𝑑 has a left-scattered maximum 𝑑, define 𝜌(π‘Ÿ)=π‘Ÿ,π‘Ÿ; otherwise set 𝐓.

Definition 2.2. For π‘š and π“π‘˜=π“βˆ’{π‘š}, the delta derivative of π“π‘˜=𝐓 at the point 𝐓 is defined to be the number 𝑀 (provided that it exists), with the property that for each π“π‘˜=π“βˆ’{𝑀} there is a neighborhood π“π‘˜=𝐓 of π‘“βˆΆπ“β†’π‘… such thatπ‘‘βˆˆπ“π‘˜for all 𝑓.

For 𝑑 and 𝑓Δ(𝑑), the nabla derivative of πœ–>0 at π‘ˆ is denoted by 𝑑 (provided that it exists), with the property that for each ||𝑓(𝜎(𝑑))βˆ’π‘“(𝑠)βˆ’π‘“Ξ”||||||(𝑑)(𝜎(𝑑)βˆ’π‘ )β‰€πœ–πœŽ(𝑑)βˆ’π‘ (2.2) there is a neighborhood π‘ βˆˆπ‘ˆ of π‘“βˆΆπ“β†’π‘… such thatπ‘‘βˆˆπ“π‘˜for all 𝑓.

Definition 2.3. A function 𝑑 is left-dense continuous (i.e., π‘“βˆ‡(𝑑)-continuous) if πœ–>0 is continuous at each left-dense point in π‘ˆ, and its right-sided limit exists at each right-dense point in 𝑑.

Definition 2.4. If ||𝑓(𝜌(𝑑))βˆ’π‘“(𝑠)βˆ’π‘“βˆ‡||||||(𝑑)(𝜌(𝑑)βˆ’π‘ )β‰€πœ–πœŒ(𝑑)βˆ’π‘ (2.3), then one defines the delta integral byπ‘ βˆˆπ‘ˆ If 𝑓, then one defines the nabla integral by𝑙𝑑

To prove the main results in this paper, we will employ several lemmas. These lemmas are based on the linear BVP𝑓𝐓

Lemma 2.5. If 𝐓 and πœ™Ξ”(𝑑)=𝑓(𝑑), then for ξ€œπ‘π‘Žπ‘“(𝑑)Δ𝑑=πœ™(𝑏)βˆ’πœ™(π‘Ž).(2.4) the BVP (2.6)-(2.7) has the unique solutionπΉβˆ‡(𝑑)=𝑓(𝑑)where ξ€œπ‘π‘Žπ‘“(𝑑)βˆ‡π‘‘=𝐹(𝑏)βˆ’πΉ(π‘Ž).(2.5)

Proof. (i) Let (πœ™π‘(π‘’Ξ”βˆ‡))βˆ‡+β„Ž(𝑑)=0,π‘‘βˆˆ(0,𝑇),(2.6)πœ™π‘(π‘’Ξ”βˆ‡(0))=π‘šβˆ’2𝑖=1π‘Žπ‘–πœ™π‘(π‘’Ξ”βˆ‡(πœ‰π‘–)),𝑒Δ(0)=0,𝑒(𝑇)=π‘šβˆ’2𝑖=1𝑏𝑖𝑒(πœ‰π‘–).(2.7) be a solution, then we will show that (2.8) holds. By taking the nabla integral of problem (2.6) on βˆ‘π‘šβˆ’2𝑖=1π‘Žπ‘–β‰ 1, we haveβˆ‘π‘šβˆ’2𝑖=1𝑏𝑖≠1thenβ„ŽβˆˆπΆπ‘™π‘‘[0,𝑇]By taking the nabla integral of (2.11) on ξ€œπ‘’(𝑑)=βˆ’π‘‘0(π‘‘βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ +𝐢,(2.8), we can getβˆ‘π΄=βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0β„Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–,∫𝐢=𝑇0(π‘‡βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0ξ€Έβˆ‘β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0ξ€Έβ„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖.(2.9)By taking the delta integral of (2.12) on 𝑒, we can get(0,𝑑)Similarly, let πœ™π‘(π‘’Ξ”βˆ‡ξ€œ(𝑑))=βˆ’π‘‘0β„Ž(𝜏)βˆ‡πœ+𝐴(2.10) on (2.10), then we have π‘’Ξ”βˆ‡(𝑑)=πœ™π‘žξ‚€βˆ’ξ€œπ‘‘0ξ‚β„Ž(𝜏)βˆ‡πœ+𝐴=βˆ’πœ™π‘žξ‚€ξ€œπ‘‘0.β„Ž(𝜏)βˆ‡πœβˆ’π΄(2.11); let (0,𝑑) on (2.10), then we haveπ‘’Ξ”ξ€œ(𝑑)=βˆ’π‘‘0πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ +𝐡.(2.12) Let (0,𝑑) on (2.12), then we haveξ€œπ‘’(𝑑)=βˆ’π‘‘0(π‘‘βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ +𝐡𝑑+𝐢.(2.13)Let 𝑑=0 on (2.13), then we haveπœ™π‘(π‘’Ξ”βˆ‡(0))=𝐴Similarly, let 𝑑=πœ‰π‘– on (2.13), then we haveπœ™π‘(π‘’Ξ”βˆ‡(πœ‰π‘–ξ€œ))=βˆ’πœ‰π‘–0β„Ž(𝜏)βˆ‡πœ+𝐴.(2.14)By the boundary condition (2.7), we can get𝑑=0𝑒Δ(0)=𝐡.(2.15)Solving (2.19), we get𝑑=𝑇By the boundary condition (2.7), we can obtainξ€œπ‘’(𝑇)=βˆ’π‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ +𝐡𝑇+𝐢.(2.16)Substituting (2.20) in the above expression, one has𝑑=πœ‰π‘–
(ii) We show that the function 𝑒(πœ‰π‘–ξ€œ)=βˆ’πœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ +π΅πœ‰π‘–+𝐢.(2.17) given in (2.8) is a solution.
Let 𝐡=0,(2.18)𝐴=π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚€βˆ’ξ€œπœ‰π‘–0.β„Ž(𝜏)βˆ‡πœ+𝐴(2.19) be as in (2.8). By [12, Theorem 2.10(iii)] and taking the delta derivative of (2.8), we haveβˆ‘π΄=βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0β„Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–.(2.20)moreover, we get
βˆ’ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ +𝐢=π‘šβˆ’2𝑖=1π‘π‘–ξ‚ƒβˆ’ξ€œπœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0.β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ +𝐢(2.21)Taking the nabla derivative of this expression yields ∫𝐢=𝑇0(π‘‡βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0ξ€Έβˆ‘β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0ξ€Έβ„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖.(2.22). Also, routine calculation verifies that 𝑒 satisfies the boundary value conditions in (2.7) so that 𝑒 given in (2.8) is a solution of (2.6) and (2.7). The proof is complete.

Lemma 2.6. Assume  π‘’Ξ”ξ€œ(𝑑)=βˆ’π‘‘0πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚β„Ž(𝜏)βˆ‡πœβˆ’π΄βˆ‡π‘ ;(2.23)   holds. For π‘’Ξ”βˆ‡(𝑑)=βˆ’πœ™π‘žξ‚€ξ€œπ‘‘0,πœ™β„Ž(𝜏)βˆ‡πœβˆ’π΄π‘(π‘’Ξ”βˆ‡ξ‚€ξ€œ)=βˆ’π‘‘0.β„Ž(𝜏)βˆ‡πœβˆ’π΄(2.24) and (πœ™π‘(π‘’Ξ”βˆ‡))βˆ‡=βˆ’β„Ž(𝑑), the unique solution 𝑒 of (2.6) and (2.7) satisfies𝑒

Proof. Let(H1)Sinceβ„ŽβˆˆπΆπ‘™π‘‘[0,𝑇]then β„Žβ‰₯0.
According to Lemma 2.5, we get𝑒If 𝑒(𝑑)β‰₯0forπ‘‘βˆˆ[0,𝑇].(2.25), we haveπœ‘0(𝑠)=πœ™π‘žξ‚€ξ€œπ‘ 0.β„Ž(𝜏)βˆ‡πœβˆ’π΄(2.26) So βˆ«π‘ 0βˆ«β„Ž(𝜏)βˆ‡πœβˆ’π΄=𝑠0βˆ‘β„Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0β„Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–β‰₯0,(2.27).

Lemma 2.7. Assume  πœ‘0(𝑠)β‰₯0  holds. If βˆ«π‘’(0)=𝐢=𝑇0(π‘‡βˆ’π‘ )πœ‘0βˆ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖β‰₯βˆ«π‘‡0(π‘‡βˆ’π‘ )πœ‘0βˆ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖=βˆ«π‘‡0(π‘‡βˆ’π‘ )πœ‘0βˆ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€·βˆ«π‘‡0(π‘‡βˆ’π‘ )πœ‘0∫(𝑠)βˆ‡π‘ βˆ’π‘‡πœ‰π‘–(π‘‡βˆ’π‘ )πœ‘0ξ€Έ(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖=ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘0βˆ‘(𝑠)βˆ‡π‘ +π‘šβˆ’2𝑖=1π‘π‘–βˆ«π‘‡πœ‰π‘–(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œβ‰₯0,𝑒(𝑇)=βˆ’π‘‡0(π‘‡βˆ’π‘ )πœ‘0ξ€œ(𝑠)βˆ‡π‘ +𝐢=βˆ’π‘‡0(π‘‡βˆ’π‘ )πœ‘0∫(𝑠)βˆ‡π‘ +𝑇0(π‘‡βˆ’π‘ )πœ‘0βˆ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œβ‰₯βˆ’π‘‡0(π‘‡βˆ’π‘ )πœ‘0∫(𝑠)βˆ‡π‘ +𝑇0(π‘‡βˆ’π‘ )πœ‘0βˆ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖=βˆ‘π‘šβˆ’2𝑖=1π‘π‘–βˆ«π‘‡πœ‰π‘–(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖β‰₯0.(2.28) and π‘‘βˆˆ(0,𝑇), then the unique solution ξ€œπ‘’(𝑑)=βˆ’π‘‘0(π‘‘βˆ’π‘ )πœ‘01(𝑠)βˆ‡π‘ +βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖[ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ‘0ξ€œ(𝑠)βˆ‡π‘ ]β‰₯βˆ’π‘‡0(π‘‡βˆ’π‘ )πœ‘01(𝑠)βˆ‡π‘ +βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖[ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‰π‘–0(π‘‡βˆ’π‘ )πœ‘0=1(𝑠)βˆ‡π‘ ]βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖[βˆ’1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘0ξ€œ(𝑠)βˆ‡π‘ +𝑇0(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‰π‘–0(π‘‡βˆ’π‘ )πœ‘0=1(𝑠)βˆ‡π‘ ]βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡πœ‰π‘–(π‘‡βˆ’π‘ )πœ‘0(𝑠)βˆ‡π‘ β‰₯0.(2.29) of (2.6) and (2.7) satisfies𝑒(𝑑)β‰₯0,π‘‘βˆˆ[0,𝑇]where (𝐻1)

Proof. It is easy to check that β„ŽβˆˆπΆπ‘™π‘‘[0,𝑇]; this implies thatβ„Žβ‰₯0It is easy to see that 𝑒 for any infπ‘‘βˆˆ[0,𝑇]‖‖𝑒‖‖,𝑒(𝑑)β‰₯𝛾(2.30) with βˆ‘π›Ύ=π‘šβˆ’2𝑖=1𝑏𝑖(π‘‡βˆ’πœ‰π‘–)βˆ‘π‘‡βˆ’π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–,‖‖𝑒‖‖=maxπ‘‘βˆˆ[0,𝑇]||||.𝑒(𝑑)(2.31). Hence, π‘’Ξ”βˆ«(𝑑)=βˆ’π‘‘0πœ‘0(𝑠)βˆ‡π‘ β‰€0 is a decreasing function on ‖‖𝑒‖‖=𝑒(0),minπ‘‘βˆˆ[0,𝑇]𝑒(𝑑)=𝑒(𝑇).(2.32). This means that the graph of 𝑒Δ(𝑑2)≀𝑒Δ(𝑑1) is concave down on 𝑑1,𝑑2∈[0,𝑇].
For each 𝑑1≀𝑑2, we have𝑒Δ(𝑑)that is,[0,𝑇]so that𝑒(𝑑)With the boundary condition (0,𝑇), we haveπ‘–βˆˆ{1,2,…,π‘šβˆ’2}This completes the proof.

Let the norm on 𝑒(𝑇)βˆ’π‘’(0)β‰₯π‘‡βˆ’0𝑒(𝑇)βˆ’π‘’(πœ‰π‘–)π‘‡βˆ’πœ‰π‘–,(2.33) be the maximum norm. Then, the 𝑇𝑒(πœ‰π‘–)βˆ’πœ‰π‘–π‘’(𝑇)β‰₯(π‘‡βˆ’πœ‰π‘–)𝑒(0),(2.34) is a Banach space. It is easy to see that BVP (1.1)-(1.2) has a solution π‘‡π‘šβˆ’2𝑖=1𝑏𝑖𝑒(πœ‰π‘–)βˆ’π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–π‘’(𝑇)β‰₯π‘šβˆ’2𝑖=1𝑏𝑖(π‘‡βˆ’πœ‰π‘–)𝑒(0).(2.35) if and only if βˆ‘π‘’(𝑇)=π‘šβˆ’2𝑖=1𝑏𝑖𝑒(πœ‰π‘–) is a fixed point of the operatorβˆ‘π‘’(𝑇)β‰₯π‘šβˆ’2𝑖=1𝑏𝑖(π‘‡βˆ’πœ‰π‘–)βˆ‘π‘‡βˆ’π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–π‘’(0).(2.36)where𝐢𝑙𝑑[0,𝑇]Denote𝐢𝑙𝑑[0,𝑇]where 𝑒=𝑒(𝑑) is the same as in Lemma 2.7. It is obvious that 𝑒 is a cone in ξ€œ(𝐴𝑒)(𝑑)=βˆ’π‘‘0(π‘‘βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚π΄ξ‚ξ‚π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’βˆ‡π‘ +𝐢,(2.37). By Lemma 2.7, ξ‚βˆ‘π΄=βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–,ξ‚βˆ«πΆ=𝑇0(π‘‡βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0ξ‚π΄ξ€Έβˆ‘π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0ξ‚π΄ξ€Έπ‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖.(2.38). So by applying Arzela-Ascoli theorem on time scales [16], we can obtain that 𝐾=π‘’βˆ£π‘’βˆˆπΆπ‘™π‘‘[0,𝑇],𝑒(𝑑)β‰₯0,infπ‘‘βˆˆ[0,𝑇]‖‖𝑒‖‖,𝑒(𝑑)β‰₯𝛾(2.39) is relatively compact. In view of Lebesgue's dominated convergence theorem on time scales [13], it is easy to prove that 𝛾 is continuous. Hence, 𝐾 is completely continuous.

Lemma 2.8. 𝐢𝑙𝑑[0,𝑇] is completely continuous.

Proof. First, we show that 𝐴(𝐾)βŠ‚πΎ maps bounded set into bounded set.
Assume 𝐴(𝐾) is a constant and 𝐴. Note that the continuity of π΄βˆΆπΎβ†’πΎ guarantees that there is π΄βˆΆπΎβ†’πΎ such that 𝐴 for 𝑐>0. Soπ‘’βˆˆπΎπ‘={π‘’βˆˆπΎβˆΆβ€–π‘’β€–β‰€π‘}That is, 𝑓 is uniformly bounded.
In addition, notice that for any π‘ξ…ž>0, we have𝑓(𝑑,𝑒(𝑑))β‰€πœ™π‘(π‘ξ…ž)So, by applying Arzela-Ascoli theorem on time scales [16], we obtain that π‘‘βˆˆ[0,𝑇] is relatively compact.
Finally, we prove that ‖‖‖‖𝐴𝑒=maxπ‘‘βˆˆ[0,𝑇]||||β‰€ξ‚πΆβ‰€βˆ«π΄π‘’(𝑑)𝑇0(π‘‡βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0ξ‚π΄ξ€Έπ‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–β‰€π‘ξ…žβˆ«π‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0ξ€·βˆ‘π‘Ž(𝜏)βˆ‡πœ/1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ€Έξ€Έβˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖.(2.40) is continuous. Suppose that 𝐴𝐾𝑐 and 𝑑1,𝑑2∈[0,𝑇] converges to ||𝐴𝑒(𝑑1)βˆ’π΄π‘’(𝑑2)||=|||ξ€œπ‘‘10(𝑑2βˆ’π‘‘1)πœ™π‘žξ‚€ξ€œπ‘ 0ξ‚π΄ξ‚ξ€œπ‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’βˆ‡π‘ +𝑑2𝑑1(𝑑2βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0𝐴|||π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’βˆ‡π‘ β‰€π‘ξ…ž||𝑑1βˆ’π‘‘2||ξ‚ƒξ€œπ‘‡0πœ™π‘žξ‚€ξ€œπ‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚βˆ‡π‘ +𝑇maxπ‘ βˆˆ[0,𝑇]πœ™π‘žξ‚€ξ€œπ‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–.(2.41) uniformly on 𝐴𝐾𝑐 Hence, π΄βˆΆπΎπ‘β†’πΎ is uniformly bounded and equicontinuous on {𝑒𝑛}βˆžπ‘›=1βŠ‚πΎπ‘ The Arzela-Ascoli theorem on time scales [16] tells us that there exists uniformly convergent subsequence in 𝑒𝑛(𝑑) Let π‘’βˆ—(𝑑) be a subsequence which converges to [0,𝑇]. uniformly on {𝐴𝑒𝑛(𝑑)}βˆžπ‘›=1 In addition,[0,𝑇].Observe the expression of {𝐴𝑒𝑛(𝑑)}βˆžπ‘›=1. and then letting {𝐴𝑒𝑛(π‘š)(𝑑)}βˆžπ‘š=1 we obtain𝑣(𝑑)where[0,𝑇]. Here, we have used the Lebesgue dominated convergence theorem on time scales [13]. From the definition of 0≀𝐴𝑒𝑛𝑐(𝑑)β‰€ξ…žβˆ«π‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ€·βˆ«π‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0ξ€·βˆ‘π‘Ž(𝜏)βˆ‡πœ/1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ€Έξ€Έβˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖.(2.42), we know that {𝐴𝑒𝑛(π‘š)(𝑑)}, on π‘šβ†’βˆž, This shows that each subsequence of ξ€œπ‘£(𝑑)=βˆ’π‘‘0(π‘‘βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0π‘Ž(𝜏)𝑓(𝜏,π‘’βˆ—ξ‚‹π΄(𝜏))βˆ‡πœβˆ’βˆ—ξ‚ξ‚‹πΆβˆ‡π‘ +βˆ—,(2.43) uniformly converges to ξ‚‹π΄βˆ—βˆ‘=βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)𝑓(𝜏,π‘’βˆ—(𝜏))βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–,ξ‚‹πΆβˆ—=1βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ‚ƒξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0π‘Ž(𝜏)𝑓(𝜏,π‘’βˆ—ξ‚‹π΄(𝜏))βˆ‡πœβˆ’βˆ—ξ‚βˆ’βˆ‡π‘ π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ™π‘žξ‚€ξ€œπ‘ 0π‘Ž(𝜏)𝑓(𝜏,π‘’βˆ—ξ‚‹π΄(𝜏))βˆ‡πœβˆ’βˆ—ξ‚ξ‚„.βˆ‡π‘ (2.44) Therefore, the sequence 𝐴 uniformly converges to 𝑣(𝑑)=π΄π‘’βˆ—(𝑑) This means that [0,𝑇]. is continuous at {𝐴𝑒𝑛(𝑑)}βˆžπ‘›=1. So, π΄π‘’βˆ—(𝑑). is continuous on {𝐴𝑒𝑛(𝑑)}βˆžπ‘›=1 since π΄π‘’βˆ—(𝑑). is arbitrary. Thus, 𝐴 is completely continuous. This proof is complete.

Lemma 2.9. Letπ‘’βˆ—βˆˆπΎπ‘For 𝐴, 𝐾𝑐

Proof. Sinceπ‘’βˆ—then 𝐴. For all πœ‘(𝑠)=πœ™π‘žξ‚€ξ€œπ‘ 0𝐴.π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’(2.45), we haveπœ‰π‘–(𝑖=1,…,π‘šβˆ’2)In fact, let ξ€œπœ‰π‘–0(πœ‰π‘–πœ‰βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ β‰€π‘–π‘‡ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ .(2.46); taking the nabla derivative of this expression, we haveξ€œπ‘ 0ξ‚ξ€œπ‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’π΄=𝑠0βˆ‘π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–β‰₯0,(2.47)Hence, πœ‘(𝑠)β‰₯0 is a nondecreasing function on π‘‘βˆˆ(0,𝑇]. That is,ξ‚€βˆ«π‘‘0(π‘‘βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ π‘‘ξ‚βˆ‡=π‘‘βˆ«π‘‘0βˆ«πœ‘(𝑠)βˆ‡π‘ βˆ’π‘‘0(π‘‘βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ π‘‘πœŒ(𝑑)β‰₯0.(2.48)For all βˆ«πœ“(𝑑)=𝑑𝑑0βˆ«πœ‘(𝑠)βˆ‡π‘ βˆ’π‘‘0(π‘‘βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ ,πœ“βˆ‡ξ€œ(𝑑)=𝑑0ξ€œπœ‘(𝑠)βˆ‡π‘ +π‘‘πœ‘(𝑑)βˆ’π‘‘0πœ‘(𝑠)βˆ‡π‘ =π‘‘πœ‘(𝑑)β‰₯0.(2.49)By (2.51), for πœ“(𝑑), we have[0,𝑇]

Lemma 2.10 (See [17]). Let πœ“(𝑑)β‰₯0.(2.50) be a Banach space, and let π‘‘βˆˆ(0,𝑇] be a cone. Assume βˆ«π‘‘0(π‘‘βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ π‘‘β‰€βˆ«π‘‡0(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ π‘‡.(2.51) are open bounded subsets of πœ‰π‘–(𝑖=1,…,π‘šβˆ’2) with ξ€œπœ‰π‘–0(πœ‰π‘–πœ‰βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ β‰€π‘–π‘‡ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ .(2.52) and let
𝐸 be a completely continuous operator such that
(i)πΎβŠ‚πΈ and Ξ©1,Ξ©2, or(ii)𝐸, and 0∈Ω1,Ξ©1βŠ‚Ξ©2,Then, 𝐹∢𝐾∩(Ξ©2⧡Ω1)→𝐾 has a fixed point in ‖𝐹𝑒‖≀‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©1,

Now, we introduce the following notations. Let‖𝐹𝑒‖β‰₯‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©2For ‖𝐹𝑒‖β‰₯‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©1, and ‖𝐹𝑒‖≀‖𝑒‖,π‘’βˆˆπΎβˆ©πœ•Ξ©2.,𝐹by Lemma 2.6, where 𝐾∩(Ξ©2⧡Ω1). and 𝐴0=1βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„ξ‚‡βˆ‡π‘ βˆ’1,𝐡0=ξ‚†π‘‡βˆ‘π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„ξ‚‡βˆ‡π‘ βˆ’1.(2.53) are well defined.

3. Main Results

Theorem 3.1. Assume  𝑙>0,Ω𝑙={π‘’βˆˆπΎβˆΆβ€–π‘’β€–<𝑙}, πœ•Ξ©π‘™={π‘’βˆˆπΎβˆΆβ€–π‘’β€–=𝑙}, and ‖‖‖‖𝛼(𝑙)=supπ΄π‘’βˆΆπ‘’βˆˆπœ•Ξ©π‘™ξ€Ύξ€½β€–β€–β€–β€–,𝛽(𝑙)=infπ΄π‘’βˆΆπ‘’βˆˆπœ•Ξ©π‘™ξ€Ύ,(2.54)  hold, and assume that the following conditions hold:
(A1)𝛼, and 𝛽(A2)(𝐻1);(A3)there exist (𝐻2) and (𝐻3) such that π‘π‘–βˆˆπΆ([0,+∞),[0,+∞)),𝑖=1,2(A4)there exists lim𝑙→0+𝑝1(𝑙)π‘™π‘βˆ’1<𝐴0π‘βˆ’1,limπ‘™β†’βˆžπ‘2(𝑙)π‘™π‘βˆ’1<𝐴0π‘βˆ’1;(3.1) such that π‘˜π‘–βˆˆπΏ1([0,𝑇],[0,+∞)),𝑖=1,2Then, problem (1.1)-(1.2) has at least two positive solutions 0<𝑐1≀𝑐2 satisfying 0β‰€πœ†2<π‘βˆ’1<πœ†1.

Theorem 3.2. Assume  𝑓(𝑑,𝑙)≀𝑝1(𝑙)+π‘˜1(𝑑)π‘™πœ†1,(𝑑,𝑙)∈[0,𝑇]Γ—[0,𝑐1],𝑓(𝑑,𝑙)≀𝑝2(𝑙)+π‘˜2(𝑑)π‘™πœ†2,(𝑑,𝑙)∈[0,𝑇]Γ—[𝑐2,+∞);(3.2), 𝑏>0, and ξ€½ξ€Ύmin𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[𝛾𝑏,𝑏]β‰₯(𝑏𝐡0)π‘βˆ’1.(3.3)  hold, and assume that the following conditions hold:
(B1)π‘’βˆ—1,π‘’βˆ—2, and 0<β€–π‘’βˆ—1β€–<𝑏<β€–π‘’βˆ—2β€–(B2)(𝐻1); (B3)there exist (𝐻2) and (𝐻3) such that π‘π‘–βˆˆπΆ([0,+∞),[0,+∞)),𝑖=3,4(B4)there exists lim𝑙→0+𝑝3(𝑙)π‘™π‘βˆ’1>𝐡0π›Ύξ‚π‘βˆ’1,limπ‘™β†’βˆžπ‘4(𝑙)π‘™π‘βˆ’1>𝐡0π›Ύξ‚π‘βˆ’1;(3.4) such that π‘˜π‘–βˆˆπΏ1([0,𝑇],[0,+∞)),𝑖=3,4Then, problem (1.1)-(1.2) has at least two positive solutions 0<𝑐3≀𝑐4 satisfying 0β‰€πœ†4<π‘βˆ’1<πœ†3.

Proof of Theorem 3.1. Let𝑓(𝑑,𝑙)β‰₯𝑝3(𝑙)βˆ’π‘˜3(𝑑)π‘™πœ†3,(𝑑,𝑙)∈[0,𝑇]Γ—[0,𝑐3],𝑓(𝑑,𝑙)β‰₯𝑝4(𝑙)βˆ’π‘˜4(𝑑)π‘™πœ†4,(𝑑,𝑙)∈[0,𝑇]Γ—[𝑐4,+∞);(3.5)then there exist π‘Ž>0 and ξ€½ξ€Ύmax𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[0,π‘Ž]≀(π‘Žπ΄0)π‘βˆ’1.(3.6) such thatπ‘’βˆ—3,π‘’βˆ—4If 0<β€–π‘’βˆ—3β€–<π‘Ž<β€–π‘’βˆ—4β€–, then 1πœ–=2𝐴min0π‘βˆ’1βˆ’lim𝑙→0+𝑝1(𝑙)π‘™π‘βˆ’1,𝐴0π‘βˆ’1βˆ’limπ‘™β†’βˆžπ‘2(𝑙)π‘™π‘βˆ’1ξ‚„,(3.7). By condition 0<π‘Ž1≀𝑐1, we have𝑐2β‰€π‘Ž2<+∞so that𝑝1(𝑙)≀(𝐴0π‘βˆ’1βˆ’πœ–)π‘™π‘βˆ’1,0β‰€π‘™β‰€π‘Ž1,𝑝2(𝑙)≀(𝐴0π‘βˆ’1βˆ’πœ–)π‘™π‘βˆ’1,π‘Ž2≀𝑙≀+∞.(3.8)Therefore,0β‰€π‘™β‰€π‘Ž1,π‘’βˆˆπœ•Ξ©π‘™It follows that0≀𝑒(𝑑)≀𝑙,0≀𝑑≀𝑇Noticing (A3), we have𝑓(𝑑,𝑒(𝑑))≀𝑝1(𝑒(𝑑))+π‘˜1(𝑑)π‘’πœ†1(𝑑)≀(𝐴0π‘βˆ’1βˆ’πœ–)π‘’π‘βˆ’1(𝑑)+π‘˜1(𝑑)π‘’πœ†1(𝑑)≀(𝐴0π‘βˆ’1β€–β€–π‘’β€–β€–βˆ’πœ–)π‘βˆ’1+π‘˜1‖‖𝑒‖‖(𝑑)πœ†1=(𝐴0π‘βˆ’1βˆ’πœ–)π‘™π‘βˆ’1+π‘˜1(𝑑)π‘™πœ†1(3.9)Therefore, there exist ξ€œπ‘ 0𝐴=ξ€œπ‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ’π‘ 0βˆ‘π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)𝑓(𝜏,𝑒(𝜏))βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–β‰€ξ€œπ‘ 0ξ€Ίπ‘Ž(𝜏)(𝐴0π‘βˆ’1βˆ’πœ–)π‘™π‘βˆ’1+π‘˜1(𝜏)π‘™πœ†1ξ€»βˆ‘βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0ξ€Ίπ‘Ž(𝜏)(𝐴0π‘βˆ’1βˆ’πœ–)π‘™π‘βˆ’1+π‘˜1(𝜏)π‘™πœ†1ξ€»βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–.(3.10) such that ‖‖‖‖≀1𝐴𝑒𝐢=βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖(ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‰π‘–0(πœ‰π‘–β‰€1βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ )βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡0≀1(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–Γ—ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚†ξ€œπ‘ 0π‘Ž(𝜏)[(𝐴0π‘βˆ’1βˆ’πœ–)π‘™π‘βˆ’1+π‘˜1(𝜏)π‘™πœ†1+βˆ‘]βˆ‡πœπ‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)[(𝐴0π‘βˆ’1βˆ’πœ–)π‘™π‘βˆ’1+π‘˜1(𝜏)π‘™πœ†1]βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚‡βˆ‡π‘ .(3.11). It implies that 𝛼(𝑙)𝑙≀1βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–Γ—ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚†ξ€œπ‘ 0π‘Ž(𝜏)[𝐴0π‘βˆ’1βˆ’πœ–+π‘˜1(𝜏)π‘™πœ†1βˆ’π‘+1+βˆ‘]βˆ‡πœπ‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)[𝐴0π‘βˆ’1βˆ’πœ–+π‘˜1(𝜏)π‘™πœ†1βˆ’π‘+1]βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚‡βˆ‡π‘ .(3.12).
If πœ†1βˆ’π‘+1>0 and lim𝑙→0+𝛼(𝑙)𝑙≀1βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0π‘Ž(𝜏)(𝐴0π‘βˆ’1βˆ‘βˆ’πœ–)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)(𝐴0π‘βˆ’1βˆ’πœ–)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„=βˆ‡π‘ (𝐴0π‘βˆ’1βˆ’πœ–)1/ξ€·π‘βˆ’1ξ€Έβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„βˆ‡π‘ =(𝐴0π‘βˆ’1βˆ’πœ–)1/ξ€·π‘βˆ’1𝐴0βˆ’1=(1βˆ’π΄0βˆ’(π‘βˆ’1)πœ–)1/ξ€·π‘βˆ’1ξ€Έ<1.(3.13), then 0<π‘Ž1<π‘Ž1. Similar to the above argument, noticing that 𝛼(π‘Ž1)<π‘Ž1, we can get ‖𝐴𝑒‖<‖𝑒‖,π‘’βˆˆπœ•Ξ©π‘Ž1. Therefore, there exist π‘Ž2≀𝑙<+∞ such that π‘’βˆˆπœ•Ξ©π‘™. It implies that 0≀𝑒(𝑑)≀𝑙.
On the other hand, since πœ†2βˆ’π‘+1<0 is continuous, by condition limπ‘™β†’βˆž(𝛼(𝑙)/𝑙)<1, there exist 0<π‘Ž2<π‘Ž2 such that𝛼(π‘Ž2)<π‘Ž2If ‖𝐴𝑒‖<‖𝑒‖,π‘’βˆˆπœ•Ξ©π‘Ž2, then π‘“βˆΆ[0,𝑇]Γ—[0,+∞)β†’[0,+∞). Applying Lemma 2.9, it follows that(A4) In the same way, we can prove that if π‘Ž1<𝑏1<𝑏<𝑏2<π‘Ž2, then ξ€½min𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[𝛾𝑏𝑖,𝑏𝑖]ξ€Ύβ‰₯(𝑏𝑖𝐡0)π‘βˆ’1,𝑖=1,2.(3.14).

Now, we consider the operator π‘’βˆˆπœ•Ξ©π‘1 on 𝛾𝑏1≀𝑒(𝑑)≀𝑏1,0≀𝑑≀𝑇 and ‖‖‖‖𝐴𝑒=max0≀𝑑≀𝑇||||ξ€œ(𝐴𝑒)(𝑑)β‰₯βˆ’π‘‡01(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ +βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖(ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‰π‘–0(πœ‰π‘–=βˆ‘βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ )π‘šβˆ’2𝑖=1π‘π‘–βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡0βˆ‘(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–βˆ«πœ‰π‘–0(πœ‰π‘–βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖β‰₯βˆ‘π‘šβˆ’2𝑖=1π‘π‘–βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡0βˆ‘(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)ξ€œπ‘‡0=π‘‡βˆ‘(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)ξ€œπ‘‡0β‰₯π‘‡βˆ‘(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0π‘Ž(𝜏)(𝑏1𝐡0)π‘βˆ’1βˆ‘βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)(𝑏1𝐡0)π‘βˆ’1βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„=π‘‡βˆ‘βˆ‡π‘ π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)𝑏1𝐡0ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„βˆ‡π‘ =𝑏1𝐡0𝐡0βˆ’1=𝑏1=‖‖𝑒‖‖.(3.15), respectively. By Lemma 2.10, we assert that the operator π‘’βˆˆπœ•Ξ©π‘2 has two fixed points ‖𝐴𝑒‖β‰₯‖𝑒‖ such that 𝐴 and Ω𝑏1β§΅Ξ©π‘Ž1. Therefore, Ξ©π‘Ž2⧡Ω𝑏2, are positive solutions of problem (1.1)-(1.2).

Proof of Theorem 3.2. Let𝐴then there exist π‘’βˆ—1,π‘’βˆ—2∈𝐾 and π‘Ž1β‰€β€–π‘’βˆ—1‖≀𝑏1 such that𝑏2β‰€β€–π‘’βˆ—2β€–β‰€π‘Ž2 If π‘’βˆ—π‘–,𝑖=1,2, then 1πœ–=2minlim𝑙→0+𝑝3(𝑙)π‘™π‘βˆ’1βˆ’ξ‚€π΅0π›Ύξ‚π‘βˆ’1,limπ‘™β†’βˆžπ‘4(𝑙)π‘™π‘βˆ’1βˆ’ξ‚€π΅0π›Ύξ‚π‘βˆ’1ξ‚„,(3.16). By Lemma 2.9 and condition 0<𝑏3≀𝑐3, we have𝑐4≀𝑏4<+∞It follows that𝑝3𝐡(𝑙)β‰₯0π›Ύξ‚π‘βˆ’1𝑙+πœ–π‘βˆ’1,0≀𝑙≀𝑏3,𝑝4𝐡(𝑙)β‰₯0π›Ύξ‚π‘βˆ’1𝑙+πœ–π‘βˆ’1,𝑏4≀𝑙≀+∞.(3.17)Noticing 0≀𝑙≀𝑏3,π‘’βˆˆπœ•Ξ©π‘™, we get𝛾𝑙≀𝑒(𝑑)≀𝑙,0≀𝑑≀𝑇Therefore, there exists (B3) with ‖‖‖‖𝐴𝑒=max0≀𝑑≀𝑇||||β‰₯π‘‡βˆ‘(𝐴𝑒)(𝑑)π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)ξ€œπ‘‡0β‰₯π‘‡βˆ‘(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)Γ—ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚†ξ€œπ‘ 0ξ€Ίπ‘π‘Ž(𝜏)3(𝑒(𝜏))βˆ’π‘˜3(𝜏)π‘’πœ†3ξ€»+βˆ‘βˆ‡πœπ‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)[𝑝3(𝑒(𝜏))βˆ’π‘˜3(𝜏)π‘’πœ†3]βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚‡β‰₯π‘‡βˆ‘βˆ‡π‘ π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)Γ—ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚†ξ€œπ‘ 0π΅π‘Ž(𝜏)0π›Ύξ‚π‘βˆ’1+πœ–(𝛾𝑙)π‘βˆ’1βˆ’π‘˜3(𝜏)π‘™πœ†3ξ‚„+βˆ‘βˆ‡πœπ‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)ξ€Ίξ€·(𝐡0/𝛾)π‘βˆ’1ξ€Έ+πœ–(𝛾𝑙)π‘βˆ’1βˆ’π‘˜3(𝜏)π‘™πœ†3ξ€»βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚‡βˆ‡π‘ .(3.18) such that 𝛽(𝑙)𝑙β‰₯π‘‡βˆ‘π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)Γ—ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚†ξ€œπ‘ 0π‘Ž(𝜏)[𝐡0π‘βˆ’1+π›Ύπ‘βˆ’1πœ–βˆ’π‘˜3(𝜏)π‘™πœ†3βˆ’π‘+1+βˆ‘]βˆ‡πœπ‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)[𝐡0π‘βˆ’1+π›Ύπ‘βˆ’1πœ–βˆ’π‘˜3(𝜏)π‘™πœ†3βˆ’π‘+1]βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚‡βˆ‡π‘ .(3.19). It implies that πœ†3βˆ’π‘+1>0 for lim𝑙→0+𝛽(𝑙)𝑙β‰₯π‘‡βˆ‘π‘šβˆ’2𝑖=1π‘π‘–βˆ’βˆ‘π‘šβˆ’2𝑖=1π‘π‘–πœ‰π‘–βˆ‘π‘‡(1βˆ’π‘šβˆ’2𝑖=1𝑏𝑖)Γ—ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0π‘Ž(𝜏)(𝐡0π‘βˆ’1+π›Ύπ‘βˆ’1βˆ‘πœ–)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)(𝐡0π‘βˆ’1+π›Ύπ‘βˆ’1πœ–)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„βˆ‡π‘ =(𝐡0π‘βˆ’1+π›Ύπ‘βˆ’1πœ–)1/ξ€·π‘βˆ’1𝐡0βˆ’1=(1+π›Ύπ‘βˆ’1𝐡0βˆ’(π‘βˆ’1)πœ–)1/ξ€·π‘βˆ’1ξ€Έ>1.(3.20).
If 𝑏3 and 0<𝑏3<π‘Ž, then 𝛽(𝑏3)>𝑏3. Similar to the above argument, noticing that ‖𝐴𝑒‖>‖𝑒‖, we can get π‘’βˆˆπœ•Ξ©π‘3.
Therefore, there exist 𝑏4≀𝛾𝑙<+∞ with π‘’βˆˆπœ•Ξ©π‘™ such that 𝑏4≀𝛾𝑙≀𝑒(𝑑)≀𝑙,0≀𝑑≀𝑇. It implies that πœ†4βˆ’π‘+1<0 for lim𝑙→+∞(𝛽(𝑙)/𝑙)>1.
By condition 𝑏4, we can see that there exist 0<𝑏4<+∞ such that𝛽(𝑏4)>𝑏4If ‖𝐴𝑒‖>‖𝑒‖, then π‘’βˆˆπœ•Ξ©π‘4, (B4), and 𝑏3<π‘Ž3<π‘Ž<π‘Ž4<𝑏4. It follows thatξ€½max𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[0,π‘Žπ‘–]≀(π‘Žπ‘–π΄0)π‘βˆ’1,𝑖=3,4.(3.21)Similarly, if π‘’βˆˆπœ•Ξ©π‘Ž3, then 0≀𝑒(𝑑)β‰€π‘Ž3.

Now, we study the operator 0≀𝑑≀𝑇 on 𝑓(𝑑,𝑒(𝑑))≀(π‘Ž3𝐴0)π‘βˆ’1 and ‖‖‖‖≀1π΄π‘’βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπ‘‡0≀1(π‘‡βˆ’π‘ )πœ‘(𝑠)βˆ‡π‘ βˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘π‘–π‘Ž3𝐴0ξ€œπ‘‡0(π‘‡βˆ’π‘ )πœ™π‘žξ‚ƒξ€œπ‘ 0βˆ‘π‘Ž(𝜏)βˆ‡πœ+π‘šβˆ’2𝑖=1π‘Žπ‘–βˆ«πœ‰π‘–0π‘Ž(𝜏)βˆ‡πœβˆ‘1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ‚„βˆ‡π‘ =π‘Ž3=‖‖𝑒‖‖.(3.22), respectively. By Lemma 2.10, we assert that the operator π‘’βˆˆπœ•Ξ©π‘Ž4 has two fixed points ‖𝐴𝑒‖≀‖𝑒‖ such that 𝐴 and Ξ©π‘Ž3⧡Ω𝑏3. Therefore, Ω𝑏4β§΅Ξ©π‘Ž4, are positive solutions of problem (1.1)-(1.2).

4. Further Discussion

If the conditions of Theorems 3.1 and 3.2 are weakened, we will get the existence of single positive solution of problem (1.1)-(1.2).

Corollary 4.1. Assume  𝐴, π‘’βˆ—3,π‘’βˆ—4∈𝐾, and 𝑏3β‰€β€–π‘’βˆ—3β€–β‰€π‘Ž3  hold, and assume that the following conditions hold:
(C1)π‘Ž4β‰€β€–π‘’βˆ—4‖≀𝑏4, and π‘’βˆ—π‘–,𝑖=3,4(C2)(𝐻1);(C3)there exist (𝐻2) and (𝐻3) such that 𝑝1∈𝐢([0,+∞),[0,+∞))(C4)there exists lim𝑙→0+(𝑝1(𝑙)/π‘™π‘βˆ’1)<𝐴0π‘βˆ’1; such that π‘˜1∈𝐿1([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

Corollary 4.2. Assume  𝑐1>0, πœ†1>π‘βˆ’1, and 𝑓(𝑑,𝑙)≀𝑝1(𝑙)+π‘˜1(𝑑)π‘™πœ†1,(𝑑,𝑙)∈[0,𝑇]Γ—[0,𝑐1];(4.1)  hold, and assume that the following conditions hold:
(D1)𝑏>0, and ξ€½ξ€Ύmin𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[𝛾𝑏,𝑏]β‰₯(𝑏𝐡0)π‘βˆ’1.(4.2)(D2)(𝐻1);(C3)there exist (𝐻2) and (𝐻3) such that 𝑝2∈𝐢([0,+∞),[0,+∞))(D4)there exists limπ‘™β†’βˆž(𝑝2(𝑙)/π‘™π‘βˆ’1)<𝐴0π‘βˆ’1; such that π‘˜2∈𝐿1([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

Corollary 4.3. Assume  𝑐2>0, 0β‰€πœ†2<π‘βˆ’1, and 𝑓(𝑑,𝑙)≀𝑝2(𝑙)+π‘˜2(𝑑)π‘™πœ†2,(𝑑,𝑙)∈[0,𝑇]Γ—[𝑐2,+∞);(4.3)   hold, and assume that the following conditions hold:
(E1)𝑏>0, and ξ€½ξ€Ύmin𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[𝛾𝑏,𝑏]β‰₯(𝑏𝐡0)π‘βˆ’1.(4.4)(E2)(𝐻1);(E3)there exist (𝐻2) and (𝐻3) such that 𝑝3∈𝐢([0,+∞),[0,+∞))(E4)there exists lim𝑙→0+(𝑝3(𝑙)/π‘™π‘βˆ’1)>(𝐡0/𝛾)π‘βˆ’1; such that π‘˜3∈𝐿1([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

Corollary 4.4. Assume  𝑐3>0, πœ†3>π‘βˆ’1, and 𝑓(𝑑,𝑙)β‰₯𝑝3(𝑙)βˆ’π‘˜3(𝑑)π‘™πœ†3,(𝑑,𝑙)∈[0,𝑇]Γ—[0,𝑐3];(4.5)  hold, and assume that the following conditions hold:
(F1)π‘Ž>0, and ξ€½ξ€Ύmax𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[0,π‘Ž]≀(π‘Žπ΄0)π‘βˆ’1.(4.6)(F2)(𝐻1);(F3)there exist (𝐻2) and (𝐻3) such that 𝑝4∈𝐢([0,+∞),[0,+∞))(F4)there exists limπ‘™β†’βˆž(𝑝4(𝑙)/π‘™π‘βˆ’1)>(𝐡0/𝛾)π‘βˆ’1; such that π‘˜4∈𝐿1([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

The proof of the above results is similar to those of Theorems 3.1 and 3.2; thus we omit it.

5. Some Examples

In this section, we present a simple example to explain our results. We only study the case 𝑐4>0.

Let 0β‰€πœ†4<π‘βˆ’1. Consider the following BVP:𝑓(𝑑,𝑙)β‰₯𝑝4(𝑙)βˆ’π‘˜4(𝑑)π‘™πœ†4,(𝑑,𝑙)∈[0,𝑇]Γ—[𝑐4,+∞);(4.7) whereπ‘Ž>0It is easy to check that ξ€½ξ€Ύmax𝑓(𝑑,𝑙)∢(𝑑,𝑙)∈[0,𝑇]Γ—[0,π‘Ž]≀(π‘Žπ΄0)π‘βˆ’1.(4.8) is continuous. In this case, 𝐓=𝑅,(0,𝑇)=(0,1), and 𝑓(𝑑,0)≑0, and it follows from a direct calculation that(πœ™3(π‘’ξ…žξ…ž))ξ…žπœ™+𝑓(𝑑,𝑒(𝑑))=0,π‘‘βˆˆ(0,1),3(π‘’ξ…žξ…žξ€·0ξ€Έ1)=2πœ™3ξ‚€π‘’ξ…žξ…žξ‚€12,π‘’ξ…ž1(0)=0,𝑒(1)=2𝑒12,(5.1)We have⎧βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽ©π‘“(𝑑,𝑒)=223𝑒31+min√,2𝑑(1βˆ’π‘‘)π‘’ξ‚‡βˆšπ‘’5√,(𝑑,𝑒)∈[0,1]Γ—[0,1],225,(𝑑,𝑒)∈[0,1]Γ—[1,3],74𝑒+361min√,𝑑(1βˆ’π‘‘)2𝑒3ξ‚‡βˆšπ‘’3,(𝑑,𝑒)∈[0,1]Γ—[3,+∞).(5.2)Choosing π‘“βˆΆ[0,1]Γ—[0,+∞)β†’[0,+∞), and 𝑝=3,π‘Ž(𝑑)≑1,π‘š=3,π‘Ž1=𝑏1=1/2, it is easy to check thatπœ‰1=1/2It follows that 𝐴0=1∫1βˆ’1/210(1βˆ’π‘ )πœ™π‘žξ‚€ξ€·ξ€Έβ‹…ξ€·ξ€Έπ‘ +1/21/21βˆ’1/2π‘‘π‘ βˆ’1𝑏=1.1062,𝛾=1(1βˆ’πœ‰1)1βˆ’π‘1πœ‰1=ξ€·ξ€Έ1/2(1βˆ’1/2)ξ€·ξ€Έβ‹…ξ€·ξ€Έ=11βˆ’1/21/23.(5.3) satisfies the conditions 𝐡0=𝑏1βˆ’π‘1πœ‰11βˆ’π‘1ξ€œπ‘‡0(1βˆ’π‘ )πœ™π‘žξ‚€π‘Žπ‘ +1πœ‰11βˆ’π‘Ž1ξ‚ξ‚„π‘‘π‘ βˆ’1=⋅1/2βˆ’1/21/2ξ€œ1βˆ’1/210ξ‚€ξ€·ξ€Έβ‹…ξ€·ξ€Έ(1βˆ’π‘ )𝑠+1/21/21βˆ’1/21/2ξ‚„π‘‘π‘ βˆ’1=4.4248<5.(5.4) of Theorem 3.1; then problem (5.1) has at least two positive solutions.

Acknowledgment

Project supported by the National Natural Science Foundation of China (Grant NO. 10471040).

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