Table of Contents 
ISRN Applied Mathematics
Volume 2011, Article ID 698529, 28 pages
http://dx.doi.org/10.5402/2011/698529
Research Article

Existence and Lyapunov Stability of Positive Periodic Solutions for a Third-Order Neutral Differential Equation

Jingli Ren, Zhibo Cheng, and Yueli Chen

Department of Mathematics, Zhengzhou University, Zhengzhou 450001, China

Received 1 April 2011; Accepted 18 May 2011

Academic Editor: G. Scheuermann

Copyright © 2011 Jingli Ren 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 applying Green's function of third-order differential equation and a fixed point theorem in cones, we obtain some sufficient conditions for existence, nonexistence, multiplicity, and Lyapunov stability of positive periodic solutions for a third-order neutral differential equation.

1. Introduction

Neutral functional differential equations manifest themselves in many fields including biology, mechanics, and economics [1–4]. For example, in population dynamics, since a growing population consumes more (or less) food than a matured one, depending on individual species, this leads to neutral functional equations [1]. These equations also arise in classical “cobweb” models in economics where current demand depends on price but supply depends on the previous periodic solutions [2]. The study on neutral functional differential equations is more intricate than ordinary delay differential equations. In recent years, there has been a good amount of work on periodic solutions for neutral differential equations (see [5–12] and the references cited therein). For example, in [5], Wu and Wang discussed the second-order neutral delay differential equation(𝑥(𝑡)−𝑐𝑥(𝑡−𝛿))+𝑎(𝑡)𝑥(𝑡)=𝜆𝑏(𝑡)𝑓(𝑥(𝑡−𝜏(𝑡))).(1.1) By a fixed point theorem, they obtain some existence results of positive periodic solutions for (1.1). Recently, in [6], Cheung et al. considered second-order neutral functional differential equation(𝑥(𝑡)−𝑐𝑥(𝑡−𝜏(𝑡)))=𝑎(𝑡)𝑥(𝑡)−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡))).(1.2) By choosing available operators and applying Krasnoselskii's fixed point theorem, they obtained sufficient conditions for the existence of periodic solutions to (1.2).

In general, most of the existing results are concentrated on first-order and second-order neutral functional differential equations, while studies on third-order neutral functional differential equations are rather infrequent, especially on the positive periodic solutions for third-order neutral functional differential equations. In the study of high-order (in particular third-order) differential equations, the naive idea to translate the equation into a first-order differential system by defining 𝑥1=𝑥, 𝑥2=𝑥, 𝑥3=𝑥,…, works well for showing existence of periodic solutions, however, it does not obviously lead to existence proofs for positive periodic solutions, since the condition 𝑥=𝑥1≥0 of positivity for the higher order equation is different from the natural positivity condition (𝑥1,𝑥2,…)≥0 for the corresponding system. Another approach, which will be used in this paper, is to transform the third-order equation into a corresponding integral equation and to establish the existence of positive periodic solutions based on a fixed point theorem in cones. Following this path one needs an explicit representation of Green's function which is rather intricate to compute.

In this paper, we consider the following third-order neutral functional differential equation:(𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)))=−𝑎(𝑡)𝑥(𝑡)+𝜆𝑏(𝑡)𝑓(𝑥(𝑡−𝜏(𝑡))).(1.3) Here 𝜆 is a positive parameter; 𝑓∈𝐶(ℝ,[0,∞)), and 𝑓(𝑥)>0 for 𝑥>0; 𝑎∈𝐶(ℝ,(0,∞)), 𝑏∈𝐶(ℝ,(0,∞)), 𝜏,𝛿∈𝐶1(ℝ,ℝ), 𝑎(𝑡), 𝑏(𝑡), 𝛿(𝑡), and 𝜏(𝑡) are 𝜔-periodic functions.

Notice that here neutral operator (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)) is a natural generalization of the familiar operator (𝐴1𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿). But 𝐴 possesses a more complicated nonlinearity than 𝐴1. For example, the neutral operators 𝐴1 is homogeneous in the following senses (𝐴1𝑥)(𝑡)=(𝐴1𝑥)(𝑡), whereas the neutral operator 𝐴 in general is inhomogeneous. As a consequence many of the new results for differential equations with the neutral operator 𝐴 will not be a direct extension of known theorems for neutral differential equations.

The paper is organized as follows. In Section 2, we first analyze qualitative properties of the generalized neutral operator 𝐴 which will be helpful for further studies of differential equations with this neutral operator; in Section 3, we consider two types of third-order constant coefficient linear differential equations and present their Green's functions and properties for those equation; in Section 4, by an application of the fixed point index theorem we obtain sufficient conditions for the existence, multiplicity and nonexistence of positive periodic solutions to third-order neutral differential equation. We will give an example to illustrate our results; in Section 5, the Lyapunov stability of periodic solutions for the equation will then be established. And an example is also given in this section.

2. Analysis of the Generalized Neutral Operator

Let 𝑋={𝑥∈𝐶(ℝ,ℝ)∶𝑥(𝑡+𝜔)=𝑥(𝑡),𝑡∈ℝ} with norm ‖𝑥‖=max𝑡∈[0,𝜔]|𝑥(𝑡)|, and let 𝐶+𝜔={𝑥∈𝐶(ℝ,(0,∞))∶𝑥(𝑡+𝜔)=𝑥(𝑡)}, 𝐶−𝜔={𝑥∈𝐶(ℝ,(−∞,0))∶𝑥(𝑡+𝜔)=𝑥(𝑡)}. Then (𝑋,‖⋅‖) is a Banach space. A cone 𝐾 in 𝑋 is defined by 𝐾={𝑥∈𝑋∶𝑥(𝑡)≥𝛼‖𝑥‖, ∀𝑡∈ℝ}, where 𝛼 is a fixed positive number with 𝛼<1. Moreover, define operators 𝐴,𝐵∶𝑋→𝑋 by (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)),(𝐵𝑥)(𝑡)=𝑐𝑥(𝑡−𝛿(𝑡)).(2.1)

Lemma 2.1. If |𝑐|≠1, then the operator 𝐴 has a continuous inverse 𝐴−1 on 𝑋, satisfying (1)𝐴−1𝑓⎧⎪⎪⎨⎪⎪⎩(𝑡)=𝑓(𝑡)+∞𝑗=1𝑐𝑗𝑓𝑠−𝑗−1𝑖=1𝛿𝐷𝑖−,for|𝑐|<1,∀𝑓∈𝑋,𝑓(𝑡+𝛿(𝑡))𝑐−∞𝑗=11𝑐𝑗+1𝑓𝑠+𝛿(𝑡)+𝑗−1𝑖=1𝛿𝐷𝑖,for|𝑐|>1,∀𝑓∈𝑋,(2.2)(2)||𝐴−1𝑓||≤(𝑡)‖𝑓‖||||1−|𝑐|,∀𝑓∈𝑋,(2.3)(3)𝜔0||𝐴−1𝑓||1(𝑡)𝑑𝑡≤||||1−|𝑐|𝜔0||||𝑓(𝑡)𝑑𝑡,∀𝑓∈𝑋.(2.4)

Proof. We have the following cases. Case 1 (|𝑐|<1). Let 𝑡−𝛿(𝑡)=𝑠 and 𝐷𝑗∑=𝑠−𝑗−1𝑖=1𝛿(𝐷𝑖), 𝑗=1,2,…. Therefore 𝐵𝑗𝑥(𝑡)=𝑐𝑗𝑥𝑠−𝑗−1𝑖=1𝛿𝐷𝑖,∞𝑗=0𝐵𝑗𝑓(𝑡)=𝑓(𝑡)+∞𝑗=1𝑐𝑗𝑓𝑠−𝑗−1𝑖=1𝛿𝐷𝑖.(2.5) Since 𝐴=𝐼−𝐵, we get from ‖𝐵‖≤|𝑐|<1 that 𝐴 has a continuous inverse 𝐴−1∶𝑋→𝑋 with 𝐴−1=(𝐼−𝐵)−1=𝐼+∞𝑗=1𝐵𝑗=∞𝑗=0𝐵𝑗.(2.6) Here 𝐵0=𝐼. Then 𝐴−1=𝑓(𝑡)∞𝑗=0𝐵𝑗𝑓(𝑡)=∞𝑗=0𝑐𝑗𝑓𝑠−𝑗−1𝑖=1𝛿𝐷𝑖,(2.7) and consequently ||𝐴−1𝑓||=|||||(𝑡)∞𝑗=0𝐵𝑗𝑓|||||=|||||(𝑡)∞𝑗=0𝑐𝑗𝑓𝑠−𝑗−1𝑖=1𝛿𝐷𝑖|||||≤‖𝑓‖1−|𝑐|.(2.8) Moreover, 𝜔0||𝐴−1𝑓||(𝑡)𝑑𝑡=𝜔0|||||∞𝑗=0𝐵𝑗𝑓|||||≤(𝑡)𝑑𝑡∞𝑗=0𝜔0||𝐵𝑗𝑓(||=𝑡)𝑑𝑡∞𝑗=0𝜔0|||||𝑐𝑗𝑓𝑠−𝑗−1𝑖=1𝛿𝐷𝑖|||||≤1𝑑𝑡1−|𝑐|𝜔0||||𝑓(𝑡)𝑑𝑡.(2.9)Case 2 (|𝑐|>1). Let 1𝐸∶𝑋⟶𝑋,(𝐸𝑥)(𝑡)=𝑥(𝑡)−𝑐𝐵𝑥(𝑡+𝛿(𝑡)),1𝐵∶𝑋⟶𝑋,1𝑥1(𝑡)=𝑐𝑥(𝑡+𝛿(𝑡)).(2.10) By definition of the linear operator 𝐵1, we have 𝐵𝑗1𝑓1(𝑡)=𝑐𝑗𝑓𝑠+𝑗−1𝑖=1𝛿𝐷𝑖.(2.11) Here 𝐷𝑖 is defined as in Case 1. Summing over 𝑗 yields ∞𝑗=0𝐵𝑗1𝑓(𝑡)=𝑓(𝑡)+∞𝑗=11𝑐𝑗𝑓𝑠+𝑗−1𝑖=1𝛿𝐷𝑖.(2.12) Since ‖𝐵1‖<1, we obtain that the operator 𝐸 has a bounded inverse 𝐸−1, 𝐸−1∶𝑋⟶𝑋,𝐸−1=𝐼−𝐵1−1=𝐼+∞𝑗=1𝐵𝑗1,(2.13) and for all 𝑓∈𝑋 we get 𝐸−1𝑓(𝑡)=𝑓(𝑡)+∞𝑗=1𝐵𝑗1𝑓(𝑡).(2.14) On the other hand, from (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)), we have 1(𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡))=−𝑐𝑥(𝑡−𝛿(𝑡))−𝑐,𝑥(𝑡)(2.15) that is, (𝐴𝑥)(𝑡)=−𝑐(𝐸𝑥)(𝑡−𝛿(𝑡)).(2.16) Let 𝑓∈𝑋 be arbitrary. We are looking for 𝑥 such that (𝐴𝑥)(𝑡)=𝑓(𝑡),(2.17) that is −𝑐(𝐸𝑥)(𝑡−𝛿(𝑡))=𝑓(𝑡).(2.18) Therefore (𝐸𝑥)(𝑡)=−𝑓(𝑡+𝛿(𝑡))𝑐=∶𝑓1(𝑡),(2.19) and hence 𝐸𝑥(𝑡)=−1𝑓1(𝑡)=𝑓1(𝑡)+∞𝑗=1𝐵𝑗1𝑓1(𝑡)=−𝑓(𝑡+𝛿(𝑡))𝑐−∞𝑗=1𝐵𝑗1𝑓(𝑡+𝛿(𝑡))𝑐,(2.20) proving that 𝐴−1 exists and satisfies 𝐴−1𝑓(𝑡)=−𝑓(𝑡+𝛿(𝑡))𝑐−∞𝑗=1𝐵𝑗1𝑓(𝑡+𝛿(𝑡))𝑐=−𝑓(𝑡+𝛿(𝑡))𝑐−∞𝑗=11𝑐𝑗+1𝑓𝑠+𝛿(𝑡)+𝑗−1𝑖=1𝛿𝐷𝑖,||𝐴−1𝑓(||=|||||−𝑡)𝑓(𝑡+𝛿(𝑡))𝑐−∞𝑗=11𝑐𝑗+1𝑓𝑠+𝛿(𝑡)+𝑗−1𝑖=1𝛿𝐷𝑖|||||≤‖𝑓‖.|𝑐|−1(2.21) Statements (1) and (2) are proved. From the above proof, (3) can easily be deduced.

Lemma 2.2. If 𝑐<0 and |𝑐|<𝛼, we have for 𝑦∈𝐾 that 𝛼−|𝑐|1−𝑐2‖𝐴𝑦‖≤−1𝑦1(𝑡)≤‖1−|𝑐|𝑦‖.(2.22)

Proof. Since 𝑐<0 and |𝑐|<𝛼<1, by Lemma 2.1, one has for 𝑦∈𝐾 that 𝐴−1𝑦(𝑡)=𝑦(𝑡)+∞𝑗=1𝑐𝑗𝑦𝑠−𝑗−1𝑖=1𝛿𝐷𝑖=𝑦(𝑡)+𝑗≥1even𝑐𝑗𝑦𝑠−𝑗−1𝑖=1𝛿𝐷𝑖−𝑗≥1odd|𝑐|𝑗𝑦𝑠−𝑗−1𝑖=1𝛿𝐷𝑖≥𝛼‖𝑦‖+𝛼𝑗≥1even𝑐𝑗‖𝑦‖−‖𝑦‖𝑗≥1odd|𝑐|𝑗=𝛼1−𝑐2‖𝑦‖−|𝑐|1−𝑐2=‖𝑦‖𝛼−|𝑐|1−𝑐2‖𝑦‖.(2.23)

Lemma 2.3. If 𝑐>0 and 𝑐<1, then for 𝑦∈𝐾, one has 𝛼𝐴1−𝑐‖𝑦‖≤−1𝑦1(𝑡)≤1−𝑐‖𝑦‖.(2.24)

Proof. Since 𝑐>0, 𝑐<1, and 𝛼<1, by Lemma 2.1, we have for 𝑦∈𝐾 that 𝐴−1𝑦(𝑡)=𝑦(𝑡)+𝑗≥1𝑐𝑗𝑦𝑠−𝑗−1𝑖=1𝛿𝐷𝑖≥𝛼‖𝑦‖+𝛼‖𝑦‖𝑗≥1𝑐𝑗=𝛼1−𝑐‖𝑦‖.(2.25)

3. Green's Functions

Theorem 3.1. For 𝜌>0 and ℎ∈𝑋, the equation 𝑢−𝜌3𝑢=ℎ(𝑡),𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔)(3.1) has a unique solution which is of the form 𝑢(𝑡)=𝜔0𝐺1(𝑡,𝑠)(−ℎ(𝑠))d𝑠,(3.2) where 𝐺1⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩√(𝑡,𝑠)=2exp((1/2)𝜌(𝑠−𝑡))sin3/2𝜌(𝑡−𝑠)+𝜋/6−𝒲3𝜌2√1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡−𝑠))3𝜌2√(exp(𝜌𝜔)−1),0≤𝑠≤𝑡≤𝜔,2exp((1/2)𝜌(𝑠−𝑡−𝜔))sin3/2𝜌(𝑡−𝑠+𝜔)+𝜋/6−𝒴3𝜌2√1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡+𝜔−𝑠))3𝜌2(exp(𝜌𝜔)−1),0≤𝑡≤𝑠≤𝜔,(3.3) where 𝒲 denotes √exp(−(1/2)𝜌𝜔)sin((3/2)𝜌(𝑡−𝑠−𝜔)+𝜋/6)and𝒴 denotes √exp(−(1/2)𝜌𝜔)sin((3/2)𝜌(𝑡−𝑠)+𝜋/6).

Proof. It is easy to check that the associated homogeneous equation of (3.1) has the solution 𝑣(𝑡)=𝑐1exp(𝜌𝑡)+exp(−𝜌𝑡/2)(𝑐2√cos(3𝜌/2)𝑡+𝑐3√sin(3𝜌/2)𝑡). The only periodic solution of the associated homogeneous equation of (3.1) is the trivial solution, that is, 𝑐1,𝑐2,𝑐3=0. This follows by assuming that 𝑣(𝑡) is periodic; we immediately get that 𝑐1=0 and by assuming that 𝑐22+𝑐23>0 and choosing 𝜑 such that sin𝜑=𝑐2/𝑐22+𝑐23,  cos𝜑=𝑐3/𝑐22+𝑐23, we get 𝑣(𝑡)𝑐22+𝑐23−=exp𝜌𝑡2√sin𝜑cos3𝜌2√𝑡+cos𝜑sin3𝜌2𝑡−=exp𝜌𝑡2√sin𝜑+3𝜌2𝑡(3.4) which for 𝑡→∞ contradicts periodicity of 𝑣, proving that 𝑐2=𝑐3=0.
Applying the method of variation of parameters, we get 𝑐1(𝑡)=exp(−𝜌𝑡)3𝜌2ℎ𝑐(𝑡),2√(𝑡)=√3/3sin√3𝜌𝑡/2−(1/3)cos3𝜌𝑡/2𝜌2exp𝜌𝑡2𝑐ℎ(𝑡),3√(𝑡)=−(1/3)sin−√3𝜌𝑡/2√3/3cos3𝜌𝑡/2𝜌2exp𝜌𝑡2ℎ(𝑡),(3.5) and then 𝑐1(𝑡)=𝑐1(0)+𝑡0exp(−𝜌𝑠)3𝜌2𝑐ℎ(𝑠)d𝑠,2(𝑡)=𝑐2(0)+𝑡0√√3/3sin−√3𝜌𝑠/2(1/3)cos3𝜌𝑠/2𝜌2exp𝜌𝑠2𝑐ℎ(𝑠)d𝑠,3(𝑡)=𝑐3(0)+𝑡0√−(1/3)sin−√3𝜌𝑠/2√3/3cos3𝜌𝑠/2𝜌2exp𝜌𝑠2ℎ(𝑠)d𝑠,𝑢(𝑡)=𝑐1(−𝑡)exp(𝜌𝑡)+exp𝜌𝑡2𝑐2(√𝑡)cos3𝜌2𝑡+𝑐3(√𝑡)sin3𝜌2𝑡=𝑐1(0)exp(𝜌𝑡)+𝑐2−exp𝜌𝑡2√cos32𝜌𝑡+𝑐3−(0)exp𝜌𝑡2√sin32+𝜌𝑡𝑡0exp(𝜌(𝑡−𝑠))3𝜌2ℎ(𝑠)d𝑠+𝑡0√sin3/2𝜌(𝑠−𝑡)−𝜋/66𝜌2𝜌exp2(𝑠−𝑡)ℎ(𝑠)d𝑠.(3.6) Noting that 𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔), we obtain 𝑐1(0)=𝜔0exp(𝜌(𝜔−𝑠))3𝜌2𝑐(1−exp(𝜌𝜔))ℎ(𝑠)d𝑠,2(0)=𝜔0√2exp(𝜌(𝑠−𝜔)/2)exp(−𝜌𝜔/2)sin𝜋/6−3𝜌𝑠/2−sin𝒟3𝜌2√exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos𝑐3𝜌𝜔/2+1ℎ(𝑠)d𝑠,3(0)=𝜔0√2exp(𝜌(𝑠−𝜔)/2)exp(−𝜌𝜔/2)cos𝜋/6−3𝜌𝑠/2−cos𝒟3𝜌2√exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos3𝜌𝜔/2+1ℎ(𝑠)d𝑠,(3.7) where 𝒟 denotes √(𝜋/6−3𝜌(𝑠−𝜔)/2). Therefore 𝑢(𝑡)=𝑐1−(𝑡)exp(𝜌𝑡)+exp𝜌𝑡2𝑐2√(𝑡)cos3𝜌2𝑡+𝑐3√(𝑡)sin3𝜌2𝑡=𝑡0⎧⎪⎨⎪⎩√2exp((1/2)𝜌(𝑠−𝑡))sin3/2𝜌(𝑡−𝑠)+𝜋/6−𝒲3𝜌2√1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡−𝑠))3𝜌2⎫⎪⎬⎪⎭+(1−exp(𝜌𝜔))ℎ(𝑠)d𝑠𝜔𝑡⎧⎪⎨⎪⎩√2exp((1/2)𝜌(𝑠−𝑡−𝜔))sin3/2𝜌(𝑡−𝑠+𝜔)+𝜋/6−𝒴3𝜌2√1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡+𝜔−𝑠))3𝜌2(⎫⎪⎬⎪⎭=1−exp(𝜌𝜔))ℎ(𝑠)d𝑠𝜔0𝐺1(𝑡,𝑠)ℎ(𝑠)d𝑠,(3.8) where 𝐺1(𝑡,𝑠) is defined as in (3.3).
By direct calculation, we get the solution 𝑢 satisfies the periodic boundary value condition of the problem (3.1).

Theorem 3.2. For 𝜌>0 and ℎ∈𝑋, the equation 𝑢+𝜌3𝑢=ℎ(𝑡),𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔)(3.9) has a unique 𝜔-periodic solution 𝑢(𝑡)=𝜔0𝐺2(𝑡,𝑠)ℎ(𝑠)d𝑠,(3.10) where 𝐺2⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩√(𝑡,𝑠)=2exp((1/2)𝜌(𝑡−𝑠))sin3/2𝜌(𝑡−𝑠)−𝜋/6−𝒰3𝜌2√1+exp(𝜌𝜔)−2exp((1/2)𝜌𝜔)cos+3/2𝜌𝜔exp(𝜌(𝑠−𝑡))3𝜌2√(1−exp(−𝜌𝜔)),0≤𝑠≤𝑡≤𝜔,2exp((1/2)𝜌(𝑡+𝜔−𝑠))sin3/2𝜌(𝑡+𝜔−𝑠)−𝜋/6−𝒳3𝜌2√1+exp(𝜌𝜔)−2exp((1/2)𝜌𝜔)cos+3/2𝜌𝜔exp(𝜌(𝑠−𝑡−𝜔))3𝜌2(1−exp(−𝜌𝜔)),0≤𝑡≤𝑠≤𝜔.(3.11) where 𝒰 denotes √exp((1/2)𝜌𝜔)sin((3/2)𝜌(𝑡−𝑠−𝜔)−𝜋/6)and𝒳 denotes √−exp((1/2)𝜌𝜔)sin((3/2)𝜌(𝑡−𝑠)−𝜋/6).

Proof. It is similar to the proof of Theorem 3.1 and can therefore be omitted.

Now we present the properties of Green's functions for (3.1) and (3.9)1𝑙=3𝜌2(exp(𝜌𝜔)−1),𝐿=3+2exp(−𝜌𝜔/2)3𝜌2(1−exp(−𝜌𝜔/2))2.(3.12)

Theorem 3.3. ∫𝜔0𝐺1(𝑡,𝑠)d𝑠=1/𝜌3, and if √3𝜌𝜔<(4/3)𝜋 holds, then 0<𝑙<𝐺1(𝑡,𝑠)≤𝐿 for all 𝑡∈[0,𝜔] and 𝑠∈[0,𝜔].

Proof. One has the following: 𝐻1(𝑡,𝑠)=exp(𝜌(𝑡−𝑠))3𝜌2,𝐻exp(𝜌𝜔)−1∗1(𝑡,𝑠)=exp(𝜌(𝑡+𝜔−𝑠))3𝜌2,𝐻exp(𝜌𝜔)−12√(𝑡,𝑠)=2exp((1/2)𝜌(𝑠−𝑡))sin𝜌3/2(𝑡−𝑠)+𝜋/6−𝒲3𝜌2√1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos,𝐻3/2𝜌𝜔∗2√(𝑡,𝑠)=2exp((1/2)𝜌(𝑠−𝑡−𝜔))sin𝜌3/2(𝑡−𝑠+𝜔)+𝜋/6−𝒴3𝜌2√1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos.3/2𝜌𝜔(3.13) A direct computation shows that ∫𝜔0𝐺1(𝑡,𝑠)d𝑠=1/𝜌3. It is easy to see that 𝐻1(𝑡,𝑠)>0 for 𝑠∈[0,𝑡] and 𝐻∗1(𝑡,𝑠)>0 for 𝑠∈[𝑡,𝜔] and √exp(−𝜌𝜔)+1−2exp(−𝜌𝜔/2)cos(3𝜌𝜔/2)>[1−exp(−𝜌𝜔/2)]2>0.
For convenience, we denote √𝜃=(3/2)𝜌(𝑡−𝑠)+𝜋/6𝑔1√(𝑡,𝑠)=sin32𝜋𝜌(𝑡−𝑠)+6−−exp𝜌𝜔2√sin32𝜋𝜌(𝑡−𝑠−𝜔)+6−=sin(𝜃)−exp𝜌𝜔2√sin𝜃−32,𝑔𝜌𝜔∗1√(𝑡,𝑠)=sin32𝜌𝜋(𝑡−𝑠+𝜔)+6−−exp𝜌𝜔2√sin32𝜌𝜋(𝑡−𝑠)+6√=sin𝜃+32−𝜌𝜔−exp𝜌𝜔2sin𝜃.(3.14) If 𝑔1(𝑡,𝑠)>0 and 𝑔∗1(𝑡,𝑠)>0, then obviously 𝐻2(𝑡,𝑠)>0, 𝐻∗2(𝑡,𝑠)>0, and 𝐺1(𝑡,𝑠)>0.
For 0≤𝑠≤𝑡≤𝜔, Since √3𝜌𝜔<(4/3)𝜋, we have𝜋6√≤𝜃≤32𝜋𝜌𝜔+6<5𝜋6,−𝜋2<𝜋6−√32√𝜌𝜔≤𝜃−32𝜋𝜌𝜔≤6.(3.15)(i)For √−𝜋/2<𝜃−(3/2)𝜌𝜔≤0, then sin𝜃>0, √sin(𝜃−(3/2)𝜌𝜔)<0, we get 𝑔1(𝑡,𝑠)>0,(ii)For √0<𝜃−(3/2)𝜌𝜔≤𝜋/6, we have sin𝜃>0, √sin(𝜃−(3/2)𝜌𝜔)>0, and √0<34√𝜌𝜔≤𝜃−34𝜋𝜌𝜔≤6+√34𝜋𝜌𝜔<2,𝑔1−(𝑡,𝑠)=sin(𝜃)−exp𝜌𝜔2√sin𝜃−32√𝜌𝜔≥sin𝜃−sin𝜃−32√𝜌𝜔=2cos𝜃−34√𝜌𝜔sin34𝜌𝜔>0.(3.16)
For 0≤𝑡≤𝑠≤𝜔, −𝜋2√<−32𝜋𝜌𝜔+6𝜋≤𝜃≤6,𝜋6√≤𝜃+32𝜋𝜌𝜔≤6+√325𝜌𝜔<6𝜋.(3.17)(i)For −𝜋/2<𝜃≤0, we have sin𝜃<0, √sin(𝜃+(3/2)𝜌𝜔)>0, and then 𝑔∗1(𝑡,𝑠)>0.(ii)For 0<𝜃≤𝜋/6, we have sin𝜃>0, √sin(𝜃+(3/2)𝜌𝜔)>0, and √0<𝜃+34𝜋𝜌𝜔<2,𝑔∗1(√𝑡,𝑠)=sin𝜃+32−𝜌𝜔−exp𝜌𝜔2√sin𝜃≥sin𝜃+32√𝜌𝜔−sin𝜃=2cos𝜃+34√𝜌𝜔sin34𝜌𝜔>0.(3.18)
If √3𝜌𝜔<(4/3)𝜌𝜔, we get 𝑔1(𝑡,𝑠)>0 and 𝑔∗1(𝑡,𝑠)>0, proving that 𝐺(𝑡,𝑠)>0 for all 𝑡∈[0,𝜔] and 𝑠∈[0,𝜔].
Next we compute a lower and an upper bound for 𝐺1(𝑡,𝑠) for 𝑠∈[0,𝜔]. We have1𝑙=3𝜌2(≤exp(𝜌𝜔)−1)exp(𝜌(𝑡+𝜔−𝑠))3𝜌2(exp(𝜌𝜔)−1)<𝐺1≤(𝑡,𝑠)exp(𝜌(𝑡+𝜔−𝑠))3𝜌2+exp(𝜌𝜔)−1exp(𝜌(𝑠−𝑡−𝜔)/2)2+2exp(−𝜌𝜔/2)3𝜌2√exp(−𝜌𝜔)+1−2exp(−𝜌𝜔/2)cos≤3𝜌𝜔/2exp(𝜌𝜔)3𝜌2+exp(𝜌𝜔)−12+2exp(−𝜌𝜔/2)3𝜌2√exp(−𝜌𝜔)+1−2exp(−𝜌𝜔/2)cos≤13𝜌𝜔/23𝜌2+1−exp(−𝜌𝜔)2+2exp(−𝜌𝜔/2)3𝜌21−exp(−𝜌𝜔/2)2≤3+2exp(−𝜌𝜔/2)3𝜌21−exp(−𝜌𝜔/2)2=𝐿.(3.19) The proof is complete.

Similarly, the following dual theorem can be proved.

Theorem 3.4. ∫𝜔0𝐺2(𝑡,𝑠)d𝑠=1/𝜌3, and if √3𝜌𝜔<(4/3)𝜋 holds, then 0<𝑙<𝐺2(𝑡,𝑠)≤𝐿 for all [0,𝜔] and 𝑠∈[0,𝜔].

4. Positive Periodic Solutions for (1.3)

Define the Banach space 𝑋 as in Section 2. Denote [][]𝑀=max{𝑎(𝑡)∶𝑡∈0,𝜔},𝑚=min{𝑎(𝑡)∶𝑡∈0,𝜔},𝜌3=𝑀,𝑘=𝑙(𝑀+𝑚)+𝐿𝑀,𝑘1=√𝑘−𝑘2−4𝐿𝑙𝑀𝑚𝑙[]2𝐿𝑀,𝛼=𝑚−(𝑀+𝑚)|𝑐|.𝐿𝑀(1−|𝑐|)(4.1) It is easy to see that 𝑀,𝑚,𝛽,𝐿,𝑙,𝑘,𝑘1>0.

Now we consider (1.3). First let 𝑓0=lim𝑥→0𝑓(𝑥)𝑥,𝑓∞=lim𝑥→∞𝑓(𝑥)𝑥,𝑓0=lim𝑥→0𝑓(𝑥)𝑥,𝑓∞=lim𝑥→∞𝑓(𝑥)𝑥,(4.2) and denote 𝑖0:numberof0′sin𝑓0,𝑓∞,𝑖0𝑓:numberof0′sin0,𝑓∞,𝑖∞:numberof∞′sin𝑓0,𝑓∞,𝑖∞𝑓:numberof∞′sin0,𝑓∞.(4.3) It is clear that 𝑖0,𝑖0,𝑖∞,𝑖∞∈{0,1,2}. We will show that (1.3) has 𝑖0 or 𝑖∞ positive 𝑤-periodic solutions for sufficiently large or small 𝜆, respectively.

In the following we discuss (1.3) in two cases, namely, the case where 𝑐<0, and 𝑐>−min{𝑘1,𝑚/(𝑀+𝑚)} (note that 𝑐>−𝑚/(𝑀+𝑚) implies 𝛼>0, 𝑐>−𝑘1 implies |𝑐|<𝛼); and the case where 𝑐>0 and 𝑐<min{𝑚/(𝑀+𝑚),(𝐿𝑀−𝑙𝑚)/((𝐿−𝑙)𝑀−𝑙𝑚)}, (note that 𝑐<𝑚/(𝑀+𝑚) implies 𝛼>0, 𝑐<(𝐿𝑀−𝑙𝑚)/((𝐿−𝑙)𝑀−𝑙𝑚) implies 𝛼<1). Obviously, we have |𝑐|<1 which makes Lemma 2.1 applicable for both cases, and also Lemmas 2.2 or 2.3, respectively.

Let 𝐾={𝑥∈𝑋∶𝑥(𝑡)≥𝛼‖𝑥‖} denote the cone in 𝑋, where 𝛼 is just as defined above. We also use 𝐾𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟} and 𝜕𝐾𝑟={𝑥∈𝐾∶‖𝑥‖=𝑟}.

Let 𝑦(𝑡)=(𝐴𝑥)(𝑡), then from Lemma 2.1 we have 𝑥(𝑡)=(𝐴−1𝑦)(𝑡). Hence (1.3) can be transformed into𝑦𝐴(𝑡)+𝑎(𝑡)−1𝑦𝐴(𝑡)=𝜆𝑏(𝑡)𝑓−1𝑦,(𝑡−𝜏(𝑡))(4.4) which can be further rewritten as𝑦𝐴(𝑡)+𝑎(𝑡)𝑦(𝑡)−𝑎(𝑡)𝐻(𝑦(𝑡))=𝜆𝑏(𝑡)𝑓−1𝑦,(𝑡−𝜏(𝑡))(4.5) where 𝐻(𝑦(𝑡))=𝑦(𝑡)−(𝐴−1𝑦)(𝑡)=−𝑐(𝐴−1𝑦)(𝑡−𝛿(𝑡)).

Now we discuss the two cases separately.

4.1. Case I: 𝑐<0 and 𝑐>−min{𝑘1,𝑚/(𝑀+𝑚)}

Now we consider𝑦(𝑡)+𝑎(𝑡)𝑦(𝑡)−𝑎(𝑡)𝐻(𝑦(𝑡))=ℎ(𝑡),ℎ∈𝐶+𝜔,(4.6) and define operators 𝑇, 𝐻∶𝑋→𝑋 by (𝑇ℎ)(𝑡)=𝑡𝑡+𝜔𝐺2((𝑡,𝑠)ℎ(𝑠)d𝑠,𝐻𝑦𝑡)=𝑀−𝑎(𝑡)𝑦(𝑡)+𝑎(𝑡)𝐻(𝑦(𝑡)).(4.7) Clearly 𝑇, 𝐻 are completely continuous, (𝑇ℎ)(𝑡)>0 for ℎ(𝑡)>0 and ‖𝐻‖≤(𝑀−𝑚+𝑀(|𝑐|/(1−|𝑐|))). By Theorem 3.2, the solution of (4.6) can be written in the following form:𝑇𝑦(𝑡)=(𝑇ℎ)(𝑡)+𝐻𝑦(𝑡).(4.8) In view of 𝑐<0 and 𝑐>−min{𝑘1,𝑚/(𝑀+𝑚)}, we have‖‖𝑇𝐻‖‖‖‖𝐻‖‖≤≤‖𝑇‖𝑀−𝑚+𝑚|𝑐|𝑀(1−|𝑐|)<1,(4.9) and hence 𝐻𝑦(𝑡)=𝐼−𝑇−1(𝑇ℎ)(𝑡).(4.10) Define an operator 𝑃∶𝑋→𝑋 by 𝐻(𝑃ℎ)(𝑡)=𝐼−𝑇−1(𝑇ℎ)(𝑡).(4.11) Obviously, for any ℎ∈𝐶+𝜔, if (√3/2)𝜌𝜔<𝜋 hold, 𝑦(𝑡)=(𝑃ℎ)(𝑡) is the unique positive 𝜔-periodic solution of (4.6).

Lemma 4.1. 𝑃 is completely continuous, and (𝑇ℎ)(𝑡)≤(𝑃ℎ)(𝑡)≤𝑀(1−|𝑐|)‖𝑚−(𝑀+𝑚)|𝑐|𝑇ℎ‖,∀ℎ∈𝐶+𝜔.(4.12)

Proof. By the Neumann expansion of 𝑃, we have 𝐻𝑃=𝐼−𝑇−1𝑇=𝑇𝐻𝐼+𝑇𝐻+2𝑇𝐻+⋯+𝑛𝑇𝑇𝐻+⋯=𝑇+𝑇𝐻𝑇+2𝑇𝐻𝑇+⋯+𝑛𝑇+⋯.(4.13) Since 𝑇 and 𝐻 are completely continuous, so is 𝑃. Moreover, by (4.13) and recalling that ‖𝑇𝐻‖≤(𝑀−𝑚+𝑚|𝑐|)/𝑀(1−|𝑐|)<1, we get (𝑇ℎ)(𝑡)≤(𝑃ℎ)(𝑡)≤𝑀(1−|𝑐|)‖𝑚−(𝑀+𝑚)|𝑐|𝑇ℎ‖.(4.14)

Define an operator 𝑄∶𝑋→𝑋 by𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓−1𝑦.(𝑡−𝜏(𝑡))(4.15)

Lemma 4.2. One has that 𝑄(𝐾)⊂𝐾.

Proof. From the definition of 𝑄, it is easy to verify that 𝑄𝑦(𝑡+𝜔)=𝑄𝑦(𝑡). For 𝑦∈𝐾, we have from Lemma 4.1 that 𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓−1𝑦𝐴(𝑡−𝜏(𝑡))≥𝑇𝜆𝑏(𝑡)𝑓−1𝑦(𝑡−𝜏(𝑡))=𝜆𝑡𝑡+𝜔𝐺2𝐴(𝑡,𝑠)𝑏(𝑠)𝑓−1𝑦(𝑠−𝜏(𝑠))d𝑠≥𝜆𝑙𝜔0𝐴𝑏(𝑠)𝑓−1𝑦(𝑠−𝜏(𝑠))d𝑠.(4.16) On the other hand, 𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓−1𝑦≤(𝑡−𝜏(𝑡))𝑀(1−|𝑐|)‖‖𝑇𝐴𝑚−(𝑀+𝑚)|𝑐|𝜆𝑏(𝑡)𝑓−1𝑦‖‖(𝑡−𝜏(𝑡))=𝜆𝑀(1−|𝑐|)𝑚−(𝑀+𝑚)|𝑐|max[]𝑡∈0,𝜔𝑡𝑡+𝜔𝐺2𝐴(𝑡,𝑠)𝑏(𝑠)𝑓−1𝑦(𝑠−𝜏(𝑠))d𝑠≤𝜆𝑀(1−|𝑐|)𝐿𝑚−(𝑀+𝑚)|𝑐|𝜔0𝐴𝑏(𝑠)𝑓−1𝑦(𝑠−𝜏(𝑠))d𝑠.(4.17) Therefore 𝑙[]𝑄𝑦(𝑡)≥𝑚−(𝑀+𝑚)|𝑐|‖𝐿𝑀(1−|𝑐|)𝑄𝑦‖=𝛼‖𝑄𝑦‖,(4.18) that is, 𝑄(𝐾)⊂𝐾.

From the continuity of 𝑃, it is easy to verify that 𝑄 is completely continuous in 𝑋. Comparing (4.5) to (4.6), it is obvious that the existence of periodic solutions for (4.5) is equivalent to the existence of fixed points for the operator 𝑄 in 𝑋. Recalling Lemma 4.2, the existence of positive periodic solutions for (4.5) is equivalent to the existence of fixed points of 𝑄 in 𝐾. Furthermore, if 𝑄 has a fixed point 𝑦 in 𝐾, it means that (𝐴−1𝑦)(𝑡) is a positive 𝜔-periodic solutions of (1.3).

Lemma 4.3. If there exists 𝜂>0 such that 𝑓𝐴−1𝑦≥𝐴(𝑡−𝜏(𝑡))−1𝑦[](𝑡−𝜏(𝑡))𝜂,for𝑡∈0,𝜔,𝑦∈𝐾,(4.19) then ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠‖𝑦‖,𝑦∈𝐾.(4.20)

Proof. By Lemma 2.2, Theorem 3.4, and Lemma 4.1, we have for 𝑦∈𝐾 that 𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓−1𝑦𝐴(𝑡−𝜏(𝑡))≥𝑇𝜆𝑏(𝑡)𝑓−1𝑦(𝑡−𝜏(𝑡))=𝜆𝑡𝑡+𝜔𝐺2𝐴(𝑡,𝑠)𝑏(𝑠)𝑓−1𝑦(𝑠−𝜏(𝑠))d𝑠≥𝜆𝑙𝜂𝜔0𝐴𝑏(𝑠)−1𝑦(𝑠−𝜏(𝑠))d𝑠≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠‖𝑦‖.(4.21) Hence ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠‖𝑦‖,𝑦∈𝐾.(4.22)

Lemma 4.4. If there exists 𝜀>0 such that 𝑓𝐴−1𝑦≤𝐴(𝑡−𝜏(𝑡))−1𝑦[](𝑡−𝜏(𝑡))𝜀,for𝑡∈0,𝜔,𝑦∈𝐾,(4.23) then ∫‖𝑄𝑦‖≤𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖,𝑦∈𝐾.(4.24)

Proof. By Lemma 2.2, Theorem 3.4, and Lemma 4.1, we have ‖𝑄𝑦(𝑡)‖≤𝜆𝑀(1−|𝑐|)𝐿𝑚−(𝑀+𝑚)|𝑐|𝜔0𝐴𝑏(𝑠)𝑓−1𝑦(𝑠−𝜏(𝑠))d𝑠≤𝜆𝑀(1−|𝑐|)𝑚−(𝑀+𝑚)|𝑐|𝐿𝜀𝜔0𝐴𝑏(𝑠)−1𝑦∫(𝑠−𝜏(𝑠))d𝑠≤𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖.(4.25)

Define 𝑟𝐹(𝑟)=max𝑓(𝑡)∶0≤𝑡≤,𝑓1−|𝑐|1(𝑟)=min𝑓(𝑡)∶𝛼−|𝑐|1−𝑐2𝑟𝑟≤𝑡≤.1−|𝑐|(4.26)

Lemma 4.5. If 𝑦∈𝜕𝐾𝑟, then ‖𝑄𝑦‖≥𝜆𝑙𝑓1(𝑟)𝜔0𝑏(𝑠)d𝑠‖𝑦‖.(4.27)

Proof. By Lemma 2.2, we obtain ((𝛼−|𝑐|)/(1−𝑐2))𝑟≤(𝐴−1𝑦)(𝑡−𝜏(𝑡))≤𝑟/(1−|𝑐|) for 𝑦∈𝜕𝐾𝑟, which yields 𝑓((𝐴−1𝑦)(𝑡−𝜏(𝑡)))≥𝑓1(𝑟). The Lemma now follows analog to the proof of Lemma 4.3.

Lemma 4.6. If 𝑦∈𝜕𝐾𝑟, then ‖𝑄𝑦‖≤𝜆𝐿𝑀(1−|𝑐|)𝐹(𝑟)𝑚−(𝑀+𝑚)|𝑐|𝜔0𝑏(𝑠)d𝑠‖𝑦‖.(4.28)

Proof. By Lemma 2.2, we can have 0≤(𝐴−1𝑦)(𝑡−𝜏(𝑡))≤𝑟/(1−|𝑐|) for 𝑦∈𝜕𝐾𝑟, which yields 𝑓((𝐴−1𝑦)(𝑡−𝜏(𝑡)))≤𝐹(𝑟). Similar to the proof of Lemma 4.4, we get the conclusion.

We quote the fixed point theorem on which our results will be based.

Lemma 4.7 (see [13]). Let 𝑋 be a Banach space and 𝐾 a cone in 𝑋. For 𝑟>0, define 𝐾𝑟={𝑢∈𝐾∶‖𝑢‖<𝑟}. Assume that 𝑇∶𝐾𝑟→𝐾 is completely continuous such that 𝑇𝑥≠𝑥 for 𝑥∈𝜕𝐾𝑟={𝑢∈𝐾∶‖𝑢‖=𝑟}. (i)If ‖𝑇𝑥‖≥‖𝑥‖ for 𝑥∈𝜕𝐾𝑟, then 𝑖(T,𝐾𝑟,𝐾)=0.(ii)If ‖𝑇𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝐾𝑟, then 𝑖(𝑇,𝐾𝑟,𝐾)=1.

Now we give our main results on positive periodic solutions for (1.3).

Theorem 4.8. (a) If 𝑖0=1 or 2, then (1.3) has 𝑖0 positive 𝜔-periodic solutions for 𝜆>1/𝑓1∫(1)𝑙𝜔0𝑏(𝑠)d𝑠>0,
(b) if 𝑖∞=1 or 2, then (1.3) has 𝑖∞ positive 𝜔-periodic solutions for ∫0<𝜆<(𝑚−(𝑀+𝑚)|𝑐|)/𝐿𝑀(1−|𝑐|)𝐹(1)𝜔0𝑏(𝑠)d𝑠,
(c) if 𝑖∞=0 or 𝑖0=0, then (1.3) has no positive 𝜔-periodic solutions for sufficiently small or sufficiently large 𝜆>0, respectively.

Proof. (a) Choose 𝑟1=1. Taking 𝜆0=1/𝑓1(𝑟1∫)𝑙𝜔0𝑏(𝑠)d𝑠>0, then for all 𝜆>𝜆0, we have from Lemma 4.5 that ‖𝑄𝑦‖>‖𝑦‖,for𝑦∈𝜕𝐾𝑟1.(4.29)Case 1. If 𝑓0=0, we can choose 0<𝑟2<𝑟1, so that 𝑓(𝑢)≤𝜀𝑢 for 0≤𝑢≤𝑟2, where the constant 𝜀>0 satisfies ∫𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|<1.(4.30) Letting 𝑟2=(1−|𝑐|)𝑟2, we have 𝑓((𝐴−1𝑦)(𝑡−𝜏(𝑡)))≤𝜀(𝐴−1𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾𝑟2. By Lemma 2.2, we have 0≤(𝐴−1𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟2 for 𝑦∈𝜕𝐾𝑟2. In view of Lemma 4.4 and (4.30), we have for 𝑦∈𝜕𝐾𝑟2 that ∫‖𝑄𝑦‖≤𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖<‖𝑦‖.(4.31) It follows from Lemma 4.7 and (4.29) that 𝑖𝑄,𝐾𝑟2,𝐾=1,𝑖𝑄,𝐾𝑟1,𝐾=0.(4.32) thus 𝑖(𝑄,𝐾𝑟1⧵𝐾𝑟2,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟1⧵𝐾𝑟2, which means (𝐴−1𝑦)(𝑡) is a positive 𝜔-positive solution of (1.3) for 𝜆>𝜆0.Case 2. If 𝑓∞=0, there exists a constant 𝐻>0 such that 𝑓(𝑢)≤𝜀𝑢 for 𝐻𝑢≥, where the constant 𝜀>0 satisfies ∫𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|<1.(4.33) Letting 𝑟3=max{2𝑟1,𝐻(1−𝑐2)/(𝛼−|𝑐|)}, we have 𝑓((𝐴−1𝑦)(𝑡−𝜏(𝑡)))≤𝜀(𝐴−1𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾𝑟3. By Lemma 2.2, we have (𝐴−1𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐2𝐻))‖𝑦‖≥ for 𝑦∈𝜕𝐾𝑟3. Thus by Lemma 4.4 and (4.33), we have for 𝑦∈𝜕𝐾𝑟3 that ∫‖𝑄𝑦‖≤𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖<‖𝑦‖.(4.34) Recalling Lemma 4.7 and (4.29) and that 𝑖𝑄,𝐾𝑟3,𝐾=1,𝑖𝑄,𝐾𝑟1,𝐾=0,(4.35) then 𝑖(𝑄,𝐾𝑟3⧵𝐾𝑟1,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟3⧵𝐾𝑟1, which means (𝐴−1𝑦)(𝑡) is a positive 𝜔-positive solution of (1.3) for 𝜆>𝜆0.Case 3. If 𝑓0=𝑓∞=0, from the above arguments, there exist 0<𝑟2<𝑟1<𝑟3 such that 𝑄 has a fixed point 𝑦1(𝑡) in 𝐾𝑟1⧵𝐾𝑟2 and a fixed point 𝑦2(𝑡) in 𝐾𝑟3⧵𝐾𝑟1. Consequently, (𝐴−1𝑦1)(𝑡) and (𝐴−1𝑦2)(𝑡) are two positive 𝜔-periodic solutions of (1.3) for 𝜆>𝜆0.(b)Let 𝑟1=1. Taking 𝜆0=(𝑚−(𝑀+𝑚)|𝑐|)/𝐿𝑀(1−|𝑐|)𝐹(𝑟1)∫𝜔0𝑏(𝑠)d𝑠>0, then by Lemma 4.6 we know if 𝜆<𝜆0, then ‖𝑄𝑦‖<‖𝑦‖,𝑦∈𝜕𝐾𝑟1.(4.36)Case 1. If 𝑓0=∞, we can choose 0<𝑟2<𝑟1 so that 𝑓(𝑢)≥𝜂𝑢 for 0≤𝑢≤𝑟2, where the constant 𝜂>0 satisfies 𝜆𝑙𝜂𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠>1.(4.37) Letting 𝑟2=(1−|𝑐|)𝑟2, we have 𝑓((𝐴−1𝑦)(𝑡−𝜏(𝑡)))≥𝜂(𝐴−1𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾𝑟2. By Lemma 2.2, we have 0≤(𝐴−1𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟2 for 𝑦∈𝜕𝐾𝑟2. Thus by Lemma 4.3 and (4.37), ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠‖𝑦‖>‖𝑦‖.(4.38) It follows from Lemma 4.7 and (4.36) that 𝑖𝑄,𝐾𝑟2,𝐾=0,𝑖𝑄,𝐾𝑟1,𝐾=1,(4.39) which implies 𝑖(𝑄,𝐾𝑟1⧵𝐾𝑟2,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟1⧵𝐾𝑟2. Therefore (𝐴−1𝑦)(𝑡) is a positive 𝜔-periodic solution of (1.3) for 0<𝜆<𝜆0.Case 2. If 𝑓∞=∞, there exists a constant 𝐻>0 such that 𝑓(𝑢)≥𝜂𝑢 for 𝐻𝑢≥, where the constant 𝜂>0 satisfies 𝜆𝑙𝜂𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠>1.(4.40) Let 𝑟3=max{2𝑟1,𝐻(1−𝑐2)/(𝛼−|𝑐|)}, we have 𝑓((𝐴−1𝑦)(𝑡−𝜏(𝑡)))≥𝜂(𝐴−1𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾𝑟3. By Lemma 2.2, we have (𝐴−1𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐2𝐻))‖𝑦‖≥ for 𝑦∈𝜕𝐾𝑟3. Thus by Lemma 4.3 and (4.40), we have for 𝑦∈𝜕𝐾𝑟3 that ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠‖𝑦‖>‖𝑦‖.(4.41) It follows from Lemma 4.7 and (4.36) that 𝑖𝑄,𝐾𝑟3,𝐾=0,𝑖𝑄,𝐾𝑟1,𝐾=1,(4.42) that is, 𝑖(𝑄,𝐾𝑟3⧵𝐾𝑟1,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟3⧵𝐾𝑟1. That means (𝐴−1𝑦)(𝑡) is a positive 𝜔-periodic solution of (1.3) for 0<𝜆<𝜆0.Case 3. If 𝑓0=𝑓∞=∞, from the above arguments, 𝑄 has a fixed point 𝑦1 in 𝐾𝑟1⧵𝐾𝑟2 and a fixed point 𝑦2 in 𝐾𝑟3⧵𝐾𝑟1. Consequently, (𝐴−1𝑦1)(𝑡) and (𝐴−1𝑦2)(𝑡) are two positive 𝜔-periodic solutions of (1.3) for 0<𝜆<𝜆0.(c) By Lemma 2.2, if 𝑦∈𝐾, then (𝐴−1𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐2))‖𝑦‖>0 for 𝑡∈[0,𝜔].Case 1. If 𝑖0=0, we have 𝑓0>0 and 𝑓∞>0. Letting 𝑏1=min{𝑓(𝑢)/𝑢; 𝑢>0}>0, then we obtain 𝑓(𝑢)≥𝑏1[𝑢,𝑢∈0,+∞).(4.43) Assume 𝑦(𝑡) is a positive 𝜔-periodic solution of (1.3) for 𝜆>𝜆0, where 𝜆0=(1−𝑐2)/𝑙b1∫(𝛼−|𝑐|)𝜔0𝑏(𝑠)d𝑠>0. Since 𝑄𝑦(𝑡)=𝑦(𝑡) for 𝑡∈[0,𝜔], then by Lemma 4.3, if 𝜆>𝜆0 we have ‖𝑦‖=‖𝑄𝑦‖≥𝜆𝑙𝑏1𝛼−|𝑐|1−𝑐2𝜔0𝑏(𝑠)d𝑠‖𝑦‖>‖𝑦‖,(4.44) which is a contradiction.Case 2. If 𝑖∞=0, we have 𝑓0<∞ and 𝑓∞<∞. Letting 𝑏2=max{𝑓(𝑢)/𝑢∶𝑢>0}>0, then we obtain 𝑓(𝑢)≤𝑏2[𝑢,𝑢∈0,∞).(4.45) Assume 𝑦(𝑡) is a positive 𝜔-periodic solution of (1.3) for 0<𝜆<𝜆0, where 𝜆0=(𝑚−(𝑀+𝑚)|𝑐|)/𝑏2∫𝐿𝑀𝜔0𝑏(𝑠)d𝑠. Since 𝑄𝑦(𝑡)=𝑦(𝑡) for 𝑡∈[0,𝜔], it follows from Lemma 4.4 that ‖𝑦‖=‖𝑄𝑦‖≤𝜆𝑏2∫𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦