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

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 [14]. 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 [512] and the references cited therein). For example, in [5], Wu and Wang discussed the second-order neutral delay differential equation(𝑥(𝑡)𝑐𝑥(𝑡𝛿))+𝑎(𝑡)𝑥(𝑡)=𝜆𝑏(𝑡)𝑓(𝑥(𝑡𝜏(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(𝑡𝜏(𝑡)))=𝑎(𝑡)𝑥(𝑡)𝑓(𝑡,𝑥(𝑡𝜏(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=𝑥10 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(𝑡𝛿(𝑡)))=𝑎(𝑡)𝑥(𝑡)+𝜆𝑏(𝑡)𝑓(𝑥(𝑡𝜏(𝑡))).(1.3) Here 𝜆 is a positive parameter; 𝑓𝐶(,[0,)), and 𝑓(𝑥)>0 for 𝑥>0; 𝑎𝐶(,(0,)), 𝑏𝐶(,(0,)), 𝜏,𝛿𝐶1(,), 𝑎(𝑡), 𝑏(𝑡), 𝛿(𝑡), and 𝜏(𝑡) are 𝜔-periodic functions.

Notice that here neutral operator (𝐴𝑥)(𝑡)=𝑥(𝑡)𝑐𝑥(𝑡𝛿(𝑡)) is a natural generalization of the familiar operator (𝐴1𝑥)(𝑡)=𝑥(𝑡)𝑐𝑥(𝑡𝛿). 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={𝑥𝐶(,)𝑥(𝑡+𝜔)=𝑥(𝑡),𝑡} with norm 𝑥=max𝑡[0,𝜔]|𝑥(𝑡)|, and let 𝐶+𝜔={𝑥𝐶(,(0,))𝑥(𝑡+𝜔)=𝑥(𝑡)}, 𝐶𝜔={𝑥𝐶(,(,0))𝑥(𝑡+𝜔)=𝑥(𝑡)}. Then (𝑋,) is a Banach space. A cone 𝐾 in 𝑋 is defined by 𝐾={𝑥𝑋𝑥(𝑡)𝛼𝑥, 𝑡}, where 𝛼 is a fixed positive number with 𝛼<1. Moreover, define operators 𝐴,𝐵𝑋𝑋 by (𝐴𝑥)(𝑡)=𝑥(𝑡)𝑐𝑥(𝑡𝛿(𝑡)),(𝐵𝑥)(𝑡)=𝑐𝑥(𝑡𝛿(𝑡)).(2.1)

Lemma 2.1. If |𝑐|1, then the operator 𝐴 has a continuous inverse 𝐴1 on 𝑋, satisfying (1)𝐴1𝑓(𝑡)=𝑓(𝑡)+𝑗=1𝑐𝑗𝑓𝑠𝑗1𝑖=1𝛿𝐷𝑖,for|𝑐|<1,𝑓𝑋,𝑓(𝑡+𝛿(𝑡))𝑐𝑗=11𝑐𝑗+1𝑓𝑠+𝛿(𝑡)+𝑗1𝑖=1𝛿𝐷𝑖,for|𝑐|>1,𝑓𝑋,(2.2)(2)||𝐴1𝑓||(𝑡)𝑓||||1|𝑐|,𝑓𝑋,(2.3)(3)𝜔0||𝐴1𝑓||1(𝑡)𝑑𝑡||||1|𝑐|𝜔0||||𝑓(𝑡)𝑑𝑡,𝑓𝑋.(2.4)

Proof. We have the following cases. Case 1 (|𝑐|<1). Let 𝑡𝛿(𝑡)=𝑠 and 𝐷𝑗=𝑠𝑗1𝑖=1𝛿(𝐷𝑖), 𝑗=1,2,. Therefore 𝐵𝑗𝑥(𝑡)=𝑐𝑗𝑥𝑠𝑗1𝑖=1𝛿𝐷𝑖,𝑗=0𝐵𝑗𝑓(𝑡)=𝑓(𝑡)+𝑗=1𝑐𝑗𝑓𝑠𝑗1𝑖=1𝛿𝐷𝑖.(2.5) Since 𝐴=𝐼𝐵, we get from 𝐵|𝑐|<1 that 𝐴 has a continuous inverse 𝐴1𝑋𝑋 with 𝐴1=(𝐼𝐵)1=𝐼+𝑗=1𝐵𝑗=𝑗=0𝐵𝑗.(2.6) Here 𝐵0=𝐼. Then 𝐴1=𝑓(𝑡)𝑗=0𝐵𝑗𝑓(𝑡)=𝑗=0𝑐𝑗𝑓𝑠𝑗1𝑖=1𝛿𝐷𝑖,(2.7) and consequently ||𝐴1𝑓||=|||||(𝑡)𝑗=0𝐵𝑗𝑓|||||=|||||(𝑡)𝑗=0𝑐𝑗𝑓𝑠𝑗1𝑖=1𝛿𝐷𝑖|||||𝑓1|𝑐|.(2.8) Moreover, 𝜔0||𝐴1𝑓||(𝑡)𝑑𝑡=𝜔0|||||𝑗=0𝐵𝑗𝑓|||||(𝑡)𝑑𝑡𝑗=0𝜔0||𝐵𝑗𝑓(||=𝑡)𝑑𝑡𝑗=0𝜔0|||||𝑐𝑗𝑓𝑠𝑗1𝑖=1𝛿𝐷𝑖|||||1𝑑𝑡1|𝑐|𝜔0||||𝑓(𝑡)𝑑𝑡.(2.9)Case 2 (|𝑐|>1). Let 1𝐸𝑋𝑋,(𝐸𝑥)(𝑡)=𝑥(𝑡)𝑐𝐵𝑥(𝑡+𝛿(𝑡)),1𝐵𝑋𝑋,1𝑥1(𝑡)=𝑐𝑥(𝑡+𝛿(𝑡)).(2.10) By definition of the linear operator 𝐵1, we have 𝐵𝑗1𝑓1(𝑡)=𝑐𝑗𝑓𝑠+𝑗1𝑖=1𝛿𝐷𝑖.(2.11) Here 𝐷𝑖 is defined as in Case 1. Summing over 𝑗 yields 𝑗=0𝐵𝑗1𝑓(𝑡)=𝑓(𝑡)+𝑗=11𝑐𝑗𝑓𝑠+𝑗1𝑖=1𝛿𝐷𝑖.(2.12) Since 𝐵1<1, we obtain that the operator 𝐸 has a bounded inverse 𝐸1, 𝐸1𝑋𝑋,𝐸1=𝐼𝐵11=𝐼+𝑗=1𝐵𝑗1,(2.13) and for all 𝑓𝑋 we get 𝐸1𝑓(𝑡)=𝑓(𝑡)+𝑗=1𝐵𝑗1𝑓(𝑡).(2.14) On the other hand, from (𝐴𝑥)(𝑡)=𝑥(𝑡)𝑐𝑥(𝑡𝛿(𝑡)), we have 1(𝐴𝑥)(𝑡)=𝑥(𝑡)𝑐𝑥(𝑡𝛿(𝑡))=𝑐𝑥(𝑡𝛿(𝑡))𝑐,𝑥(𝑡)(2.15) that is, (𝐴𝑥)(𝑡)=𝑐(𝐸𝑥)(𝑡𝛿(𝑡)).(2.16) Let 𝑓𝑋 be arbitrary. We are looking for 𝑥 such that (𝐴𝑥)(𝑡)=𝑓(𝑡),(2.17) that is 𝑐(𝐸𝑥)(𝑡𝛿(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))𝑐𝑗=1𝐵𝑗1𝑓(𝑡+𝛿(𝑡))𝑐,(2.20) proving that 𝐴1 exists and satisfies 𝐴1𝑓(𝑡)=𝑓(𝑡+𝛿(𝑡))𝑐𝑗=1𝐵𝑗1𝑓(𝑡+𝛿(𝑡))𝑐=𝑓(𝑡+𝛿(𝑡))𝑐𝑗=11𝑐𝑗+1𝑓𝑠+𝛿(𝑡)+𝑗1𝑖=1𝛿𝐷𝑖,||𝐴1𝑓(||=|||||𝑡)𝑓(𝑡+𝛿(𝑡))𝑐𝑗=11𝑐𝑗+1𝑓𝑠+𝛿(𝑡)+𝑗1𝑖=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 𝑦𝐾 that 𝛼|𝑐|1𝑐2𝐴𝑦1𝑦1(𝑡)1|𝑐|𝑦.(2.22)

Proof. Since 𝑐<0 and |𝑐|<𝛼<1, by Lemma 2.1, one has for 𝑦𝐾 that 𝐴1𝑦(𝑡)=𝑦(𝑡)+𝑗=1𝑐𝑗𝑦𝑠𝑗1𝑖=1𝛿𝐷𝑖=𝑦(𝑡)+𝑗1even𝑐𝑗𝑦𝑠𝑗1𝑖=1𝛿𝐷𝑖𝑗1odd|𝑐|𝑗𝑦𝑠𝑗1𝑖=1𝛿𝐷𝑖𝛼𝑦+𝛼𝑗1even𝑐𝑗𝑦𝑦𝑗1odd|𝑐|𝑗=𝛼1𝑐2𝑦|𝑐|1𝑐2=𝑦𝛼|𝑐|1𝑐2𝑦.(2.23)

Lemma 2.3. If 𝑐>0 and 𝑐<1, then for 𝑦𝐾, one has 𝛼𝐴1𝑐𝑦1𝑦1(𝑡)1𝑐𝑦.(2.24)

Proof. Since 𝑐>0, 𝑐<1, and 𝛼<1, by Lemma 2.1, we have for 𝑦𝐾 that 𝐴1𝑦(𝑡)=𝑦(𝑡)+𝑗1𝑐𝑗𝑦𝑠𝑗1𝑖=1𝛿𝐷𝑖𝛼𝑦+𝛼𝑦𝑗1𝑐𝑗=𝛼1𝑐𝑦.(2.25)

3. Green's Functions

Theorem 3.1. For 𝜌>0 and 𝑋, the equation 𝑢𝜌3𝑢=(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)((𝑠))d𝑠,(3.2) where 𝐺1(𝑡,𝑠)=2exp((1/2)𝜌(𝑠𝑡))sin3/2𝜌(𝑡𝑠)+𝜋/6𝒲3𝜌21+exp(𝜌𝜔)2exp(𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡𝑠))3𝜌2(exp(𝜌𝜔)1),0𝑠𝑡𝜔,2exp((1/2)𝜌(𝑠𝑡𝜔))sin3/2𝜌(𝑡𝑠+𝜔)+𝜋/6𝒴3𝜌21+exp(𝜌𝜔)2exp(𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡+𝜔𝑠))3𝜌2(exp(𝜌𝜔)1),0𝑡𝑠𝜔,(3.3) where 𝒲 denotes exp((1/2)𝜌𝜔)sin((3/2)𝜌(𝑡𝑠𝜔)+𝜋/6)and𝒴 denotes exp((1/2)𝜌𝜔)sin((3/2)𝜌(𝑡𝑠)+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)(𝑐2cos(3𝜌/2)𝑡+𝑐3sin(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=exp𝜌𝑡2sin𝜑cos3𝜌2𝑡+cos𝜑sin3𝜌2𝑡=exp𝜌𝑡2sin𝜑+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)3𝜌2𝑐(𝑡),2(𝑡)=3/3sin3𝜌𝑡/2(1/3)cos3𝜌𝑡/2𝜌2exp𝜌𝑡2𝑐(𝑡),3(𝑡)=(1/3)sin3𝜌𝑡/23/3cos3𝜌𝑡/2𝜌2exp𝜌𝑡2(𝑡),(3.5) and then 𝑐1(𝑡)=𝑐1(0)+𝑡0exp(𝜌𝑠)3𝜌2𝑐(𝑠)d𝑠,2(𝑡)=𝑐2(0)+𝑡03/3sin3𝜌𝑠/2(1/3)cos3𝜌𝑠/2𝜌2exp𝜌𝑠2𝑐(𝑠)d𝑠,3(𝑡)=𝑐3(0)+𝑡0(1/3)sin3𝜌𝑠/23/3cos3𝜌𝑠/2𝜌2exp𝜌𝑠2(𝑠)d𝑠,𝑢(𝑡)=𝑐1(𝑡)exp(𝜌𝑡)+exp𝜌𝑡2𝑐2(𝑡)cos3𝜌2𝑡+𝑐3(𝑡)sin3𝜌2𝑡=𝑐1(0)exp(𝜌𝑡)+𝑐2exp𝜌𝑡2cos32𝜌𝑡+𝑐3(0)exp𝜌𝑡2sin32+𝜌𝑡𝑡0exp(𝜌(𝑡𝑠))3𝜌2(𝑠)d𝑠+𝑡0sin3/2𝜌(𝑠𝑡)𝜋/66𝜌2𝜌exp2(𝑠𝑡)(𝑠)d𝑠.(3.6) Noting that 𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔),𝑢(0)=𝑢(𝜔), we obtain 𝑐1(0)=𝜔0exp(𝜌(𝜔𝑠))3𝜌2𝑐(1exp(𝜌𝜔))(𝑠)d𝑠,2(0)=𝜔02exp(𝜌(𝑠𝜔)/2)exp(𝜌𝜔/2)sin𝜋/63𝜌𝑠/2sin𝒟3𝜌2exp(𝜌𝜔)2exp(𝜌𝜔/2)cos𝑐3𝜌𝜔/2+1(𝑠)d𝑠,3(0)=𝜔02exp(𝜌(𝑠𝜔)/2)exp(𝜌𝜔/2)cos𝜋/63𝜌𝑠/2cos𝒟3𝜌2exp(𝜌𝜔)2exp(𝜌𝜔/2)cos3𝜌𝜔/2+1(𝑠)d𝑠,(3.7) where 𝒟 denotes (𝜋/63𝜌(𝑠𝜔)/2). Therefore 𝑢(𝑡)=𝑐1(𝑡)exp(𝜌𝑡)+exp𝜌𝑡2𝑐2(𝑡)cos3𝜌2𝑡+𝑐3(𝑡)sin3𝜌2𝑡=𝑡02exp((1/2)𝜌(𝑠𝑡))sin3/2𝜌(𝑡𝑠)+𝜋/6𝒲3𝜌21+exp(𝜌𝜔)2exp(𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡𝑠))3𝜌2+(1exp(𝜌𝜔))(𝑠)d𝑠𝜔𝑡2exp((1/2)𝜌(𝑠𝑡𝜔))sin3/2𝜌(𝑡𝑠+𝜔)+𝜋/6𝒴3𝜌21+exp(𝜌𝜔)2exp(𝜌𝜔/2)cos+3/2𝜌𝜔exp(𝜌(𝑡+𝜔𝑠))3𝜌2(=1exp(𝜌𝜔))(𝑠)d𝑠𝜔0𝐺1(𝑡,𝑠)(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 𝑋, the equation 𝑢+𝜌3𝑢=(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)(𝑠)d𝑠,(3.10) where 𝐺2(𝑡,𝑠)=2exp((1/2)𝜌(𝑡𝑠))sin3/2𝜌(𝑡𝑠)𝜋/6𝒰3𝜌21+exp(𝜌𝜔)2exp((1/2)𝜌𝜔)cos+3/2𝜌𝜔exp(𝜌(𝑠𝑡))3𝜌2(1exp(𝜌𝜔)),0𝑠𝑡𝜔,2exp((1/2)𝜌(𝑡+𝜔𝑠))sin3/2𝜌(𝑡+𝜔𝑠)𝜋/6𝒳3𝜌21+exp(𝜌𝜔)2exp((1/2)𝜌𝜔)cos+3/2𝜌𝜔exp(𝜌(𝑠𝑡𝜔))3𝜌2(1exp(𝜌𝜔)),0𝑡𝑠𝜔.(3.11) where 𝒰 denotes exp((1/2)𝜌𝜔)sin((3/2)𝜌(𝑡𝑠𝜔)𝜋/6)and𝒳 denotes exp((1/2)𝜌𝜔)sin((3/2)𝜌(𝑡𝑠)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(1exp(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(𝑡𝑠))3𝜌2,𝐻exp(𝜌𝜔)11(𝑡,𝑠)=exp(𝜌(𝑡+𝜔𝑠))3𝜌2,𝐻exp(𝜌𝜔)12(𝑡,𝑠)=2exp((1/2)𝜌(𝑠𝑡))sin𝜌3/2(𝑡𝑠)+𝜋/6𝒲3𝜌21+exp(𝜌𝜔)2exp(𝜌𝜔/2)cos,𝐻3/2𝜌𝜔2(𝑡,𝑠)=2exp((1/2)𝜌(𝑠𝑡𝜔))sin𝜌3/2(𝑡𝑠+𝜔)+𝜋/6𝒴3𝜌21+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 𝐻1(𝑡,𝑠)>0 for 𝑠[𝑡,𝜔] and exp(𝜌𝜔)+12exp(𝜌𝜔/2)cos(3𝜌𝜔/2)>[1exp(𝜌𝜔/2)]2>0.
For convenience, we denote 𝜃=(3/2)𝜌(𝑡𝑠)+𝜋/6𝑔1(𝑡,𝑠)=sin32𝜋𝜌(𝑡𝑠)+6exp𝜌𝜔2sin32𝜋𝜌(𝑡𝑠𝜔)+6=sin(𝜃)exp𝜌𝜔2sin𝜃32,𝑔𝜌𝜔1(𝑡,𝑠)=sin32𝜌𝜋(𝑡𝑠+𝜔)+6exp𝜌𝜔2sin32𝜌𝜋(𝑡𝑠)+6=sin𝜃+32𝜌𝜔exp𝜌𝜔2sin𝜃.(3.14) If 𝑔1(𝑡,𝑠)>0 and 𝑔1(𝑡,𝑠)>0, then obviously 𝐻2(𝑡,𝑠)>0, 𝐻2(𝑡,𝑠)>0, and 𝐺1(𝑡,𝑠)>0.
For 0𝑠𝑡𝜔, Since 3𝜌𝜔<(4/3)𝜋, we have𝜋6𝜃32𝜋𝜌𝜔+6<5𝜋6,𝜋2<𝜋632𝜌𝜔𝜃32𝜋𝜌𝜔6.(3.15)(i)For 𝜋/2<𝜃(3/2)𝜌𝜔0, then sin𝜃>0, sin(𝜃(3/2)𝜌𝜔)<0, we get 𝑔1(𝑡,𝑠)>0,(ii)For 0<𝜃(3/2)𝜌𝜔𝜋/6, we have sin𝜃>0, sin(𝜃(3/2)𝜌𝜔)>0, and 0<34𝜌𝜔𝜃34𝜋𝜌𝜔6+34𝜋𝜌𝜔<2,𝑔1(𝑡,𝑠)=sin(𝜃)exp𝜌𝜔2sin𝜃32𝜌𝜔sin𝜃sin𝜃32𝜌𝜔=2cos𝜃34𝜌𝜔sin34𝜌𝜔>0.(3.16)
For 0𝑡𝑠𝜔, 𝜋2<32𝜋𝜌𝜔+6𝜋𝜃6,𝜋6𝜃+32𝜋𝜌𝜔6+325𝜌𝜔<6𝜋.(3.17)(i)For 𝜋/2<𝜃0, we have sin𝜃<0, sin(𝜃+(3/2)𝜌𝜔)>0, and then 𝑔1(𝑡,𝑠)>0.(ii)For 0<𝜃𝜋/6, we have sin𝜃>0, sin(𝜃+(3/2)𝜌𝜔)>0, and 0<𝜃+34𝜋𝜌𝜔<2,𝑔1(𝑡,𝑠)=sin𝜃+32𝜌𝜔exp𝜌𝜔2sin𝜃sin𝜃+32𝜌𝜔sin𝜃=2cos𝜃+34𝜌𝜔sin34𝜌𝜔>0.(3.18)
If 3𝜌𝜔<(4/3)𝜌𝜔, we get 𝑔1(𝑡,𝑠)>0 and 𝑔1(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+𝜔𝑠))3𝜌2(exp(𝜌𝜔)1)<𝐺1(𝑡,𝑠)exp(𝜌(𝑡+𝜔𝑠))3𝜌2+exp(𝜌𝜔)1exp(𝜌(𝑠𝑡𝜔)/2)2+2exp(𝜌𝜔/2)3𝜌2exp(𝜌𝜔)+12exp(𝜌𝜔/2)cos3𝜌𝜔/2exp(𝜌𝜔)3𝜌2+exp(𝜌𝜔)12+2exp(𝜌𝜔/2)3𝜌2exp(𝜌𝜔)+12exp(𝜌𝜔/2)cos13𝜌𝜔/23𝜌2+1exp(𝜌𝜔)2+2exp(𝜌𝜔/2)3𝜌21exp(𝜌𝜔/2)23+2exp(𝜌𝜔/2)3𝜌21exp(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)𝑡0,𝜔},𝑚=min{𝑎(𝑡)𝑡0,𝜔},𝜌3=𝑀,𝑘=𝑙(𝑀+𝑚)+𝐿𝑀,𝑘1=𝑘𝑘24𝐿𝑙𝑀𝑚𝑙[]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=lim𝑥0𝑓(𝑥)𝑥,𝑓=lim𝑥𝑓(𝑥)𝑥,𝑓0=lim𝑥0𝑓(𝑥)𝑥,𝑓=lim𝑥𝑓(𝑥)𝑥,(4.2) and denote 𝑖0:numberof0sin𝑓0,𝑓,𝑖0𝑓:numberof0sin0,𝑓,𝑖:numberofsin𝑓0,𝑓,𝑖𝑓:numberofsin0,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),(𝐿𝑀𝑙𝑚)/((𝐿𝑙)𝑀𝑙𝑚)}, (note that 𝑐<𝑚/(𝑀+𝑚) implies 𝛼>0, 𝑐<(𝐿𝑀𝑙𝑚)/((𝐿𝑙)𝑀𝑙𝑚) 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={𝑥𝑋𝑥(𝑡)𝛼𝑥} denote the cone in 𝑋, where 𝛼 is just as defined above. We also use 𝐾𝑟={𝑥𝐾𝑥<𝑟} and 𝜕𝐾𝑟={𝑥𝐾𝑥=𝑟}.

Let 𝑦(𝑡)=(𝐴𝑥)(𝑡), then from Lemma 2.1 we have 𝑥(𝑡)=(𝐴1𝑦)(𝑡). Hence (1.3) can be transformed into𝑦𝐴(𝑡)+𝑎(𝑡)1𝑦𝐴(𝑡)=𝜆𝑏(𝑡)𝑓1𝑦,(𝑡𝜏(𝑡))(4.4) which can be further rewritten as𝑦𝐴(𝑡)+𝑎(𝑡)𝑦(𝑡)𝑎(𝑡)𝐻(𝑦(𝑡))=𝜆𝑏(𝑡)𝑓1𝑦,(𝑡𝜏(𝑡))(4.5) where 𝐻(𝑦(𝑡))=𝑦(𝑡)(𝐴1𝑦)(𝑡)=𝑐(𝐴1𝑦)(𝑡𝛿(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))=(𝑡),𝐶+𝜔,(4.6) and define operators 𝑇, 𝐻𝑋𝑋 by (𝑇)(𝑡)=𝑡𝑡+𝜔𝐺2((𝑡,𝑠)(𝑠)d𝑠,𝐻𝑦𝑡)=𝑀𝑎(𝑡)𝑦(𝑡)+𝑎(𝑡)𝐻(𝑦(𝑡)).(4.7) Clearly 𝑇, 𝐻 are completely continuous, (𝑇)(𝑡)>0 for (𝑡)>0 and 𝐻(𝑀𝑚+𝑀(|𝑐|/(1|𝑐|))). By Theorem 3.2, the solution of (4.6) can be written in the following form:𝑇𝑦(𝑡)=(𝑇)(𝑡)+𝐻𝑦(𝑡).(4.8) In view of 𝑐<0 and 𝑐>min{𝑘1,𝑚/(𝑀+𝑚)}, we have𝑇𝐻𝐻𝑇𝑀𝑚+𝑚|𝑐|𝑀(1|𝑐|)<1,(4.9) and hence 𝐻𝑦(𝑡)=𝐼𝑇1(𝑇)(𝑡).(4.10) Define an operator 𝑃𝑋𝑋 by 𝐻(𝑃)(𝑡)=𝐼𝑇1(𝑇)(𝑡).(4.11) Obviously, for any 𝐶+𝜔, if (3/2)𝜌𝜔<𝜋 hold, 𝑦(𝑡)=(𝑃)(𝑡) is the unique positive 𝜔-periodic solution of (4.6).

Lemma 4.1. 𝑃 is completely continuous, and (𝑇)(𝑡)(𝑃)(𝑡)𝑀(1|𝑐|)𝑚(𝑀+𝑚)|𝑐|𝑇,𝐶+𝜔.(4.12)

Proof. By the Neumann expansion of 𝑃, we have 𝐻𝑃=𝐼𝑇1𝑇=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(𝑀𝑚+𝑚|𝑐|)/𝑀(1|𝑐|)<1, we get (𝑇)(𝑡)(𝑃)(𝑡)𝑀(1|𝑐|)𝑚(𝑀+𝑚)|𝑐|𝑇.(4.14)

Define an operator 𝑄𝑋𝑋 by𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓1𝑦.(𝑡𝜏(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 𝑦𝐾, we have from Lemma 4.1 that 𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓1𝑦𝐴(𝑡𝜏(𝑡))𝑇𝜆𝑏(𝑡)𝑓1𝑦(𝑡𝜏(𝑡))=𝜆𝑡𝑡+𝜔𝐺2𝐴(𝑡,𝑠)𝑏(𝑠)𝑓1𝑦(𝑠𝜏(𝑠))d𝑠𝜆𝑙𝜔0𝐴𝑏(𝑠)𝑓1𝑦(𝑠𝜏(𝑠))d𝑠.(4.16) On the other hand, 𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓1𝑦(𝑡𝜏(𝑡))𝑀(1|𝑐|)𝑇𝐴𝑚(𝑀+𝑚)|𝑐|𝜆𝑏(𝑡)𝑓1𝑦(𝑡𝜏(𝑡))=𝜆𝑀(1|𝑐|)𝑚(𝑀+𝑚)|𝑐|max[]𝑡0,𝜔𝑡𝑡+𝜔𝐺2𝐴(𝑡,𝑠)𝑏(𝑠)𝑓1𝑦(𝑠𝜏(𝑠))d𝑠𝜆𝑀(1|𝑐|)𝐿𝑚(𝑀+𝑚)|𝑐|𝜔0𝐴𝑏(𝑠)𝑓1𝑦(𝑠𝜏(𝑠))d𝑠.(4.17) Therefore 𝑙[]𝑄𝑦(𝑡)𝑚(𝑀+𝑚)|𝑐|𝐿𝑀(1|𝑐|)𝑄𝑦=𝛼𝑄𝑦,(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 (𝐴1𝑦)(𝑡) is a positive 𝜔-periodic solutions of (1.3).

Lemma 4.3. If there exists 𝜂>0 such that 𝑓𝐴1𝑦𝐴(𝑡𝜏(𝑡))1𝑦[](𝑡𝜏(𝑡))𝜂,for𝑡0,𝜔,𝑦𝐾,(4.19) then 𝑄𝑦𝜆𝑙𝜂𝛼|𝑐|1𝑐2𝜔0𝑏(𝑠)d𝑠𝑦,𝑦𝐾.(4.20)

Proof. By Lemma 2.2, Theorem 3.4, and Lemma 4.1, we have for 𝑦𝐾 that 𝐴𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓1𝑦𝐴(𝑡𝜏(𝑡))𝑇𝜆𝑏(𝑡)𝑓1𝑦(𝑡𝜏(𝑡))=𝜆𝑡𝑡+𝜔𝐺2𝐴(𝑡,𝑠)𝑏(𝑠)𝑓1𝑦(𝑠𝜏(𝑠))d𝑠𝜆𝑙𝜂𝜔0𝐴𝑏(𝑠)1𝑦(𝑠𝜏(𝑠))d𝑠𝜆𝑙𝜂𝛼|𝑐|1𝑐2𝜔0𝑏(𝑠)d𝑠𝑦.(4.21) Hence 𝑄𝑦𝜆𝑙𝜂𝛼|𝑐|1𝑐2𝜔0𝑏(𝑠)d𝑠𝑦,𝑦𝐾.(4.22)

Lemma 4.4. If there exists 𝜀>0 such that 𝑓𝐴1𝑦𝐴(𝑡𝜏(𝑡))1𝑦[](𝑡𝜏(𝑡))𝜀,for𝑡0,𝜔,𝑦𝐾,(4.23) then 𝑄𝑦𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚(𝑀+𝑚)|𝑐|𝑦,𝑦𝐾.(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)𝑓1𝑦(𝑠𝜏(𝑠))d𝑠𝜆𝑀(1|𝑐|)𝑚(𝑀+𝑚)|𝑐|𝐿𝜀𝜔0𝐴𝑏(𝑠)1𝑦(𝑠𝜏(𝑠))d𝑠𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚(𝑀+𝑚)|𝑐|𝑦.(4.25)

Define 𝑟𝐹(𝑟)=max𝑓(𝑡)0𝑡,𝑓1|𝑐|1(𝑟)=min𝑓(𝑡)𝛼|𝑐|1𝑐2𝑟𝑟𝑡.1|𝑐|(4.26)

Lemma 4.5. If 𝑦𝜕𝐾𝑟, then 𝑄𝑦𝜆𝑙𝑓1(𝑟)𝜔0𝑏(𝑠)d𝑠𝑦.(4.27)

Proof. By Lemma 2.2, we obtain ((𝛼|𝑐|)/(1𝑐2))𝑟(𝐴1𝑦)(𝑡𝜏(𝑡))𝑟/(1|𝑐|) for 𝑦𝜕𝐾𝑟, which yields 𝑓((𝐴1𝑦)(𝑡𝜏(𝑡)))𝑓1(𝑟). The Lemma now follows analog to the proof of Lemma 4.3.

Lemma 4.6. If 𝑦𝜕𝐾𝑟, then 𝑄𝑦𝜆𝐿𝑀(1|𝑐|)𝐹(𝑟)𝑚(𝑀+𝑚)|𝑐|𝜔0𝑏(𝑠)d𝑠𝑦.(4.28)

Proof. By Lemma 2.2, we can have 0(𝐴1𝑦)(𝑡𝜏(𝑡))𝑟/(1|𝑐|) for 𝑦𝜕𝐾𝑟, which yields 𝑓((𝐴1𝑦)(𝑡𝜏(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={𝑢𝐾𝑢<𝑟}. Assume that 𝑇𝐾𝑟𝐾 is completely continuous such that 𝑇𝑥𝑥 for 𝑥𝜕𝐾𝑟={𝑢𝐾𝑢=𝑟}. (i)If 𝑇𝑥𝑥 for 𝑥𝜕𝐾𝑟, then 𝑖(T,𝐾𝑟,𝐾)=0.(ii)If 𝑇𝑥𝑥 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 0𝑢𝑟2, where the constant 𝜀>0 satisfies 𝜆𝜀𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚(𝑀+𝑚)|𝑐|<1.(4.30) Letting 𝑟2=(1|𝑐|)𝑟2, we have 𝑓((𝐴1𝑦)(𝑡𝜏(𝑡)))𝜀(𝐴1𝑦)(𝑡𝜏(𝑡)) for 𝑦𝐾𝑟2. By Lemma 2.2, we have 0(𝐴1𝑦)(𝑡𝜏(𝑡))𝑦/(1|𝑐|)𝑟2 for 𝑦𝜕𝐾𝑟2. In view of Lemma 4.4 and (4.30), we have for 𝑦𝜕𝐾𝑟2 that 𝑄𝑦𝜆𝜀𝐿𝑀𝜔0𝑏(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,𝐾𝑟1𝐾𝑟2,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟1𝐾𝑟2, which means (𝐴1𝑦)(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(1𝑐2)/(𝛼|𝑐|)}, we have 𝑓((𝐴1𝑦)(𝑡𝜏(𝑡)))𝜀(𝐴1𝑦)(𝑡𝜏(𝑡)) for 𝑦𝐾𝑟3. By Lemma 2.2, we have (𝐴1𝑦)(𝑡𝜏(𝑡))((𝛼|𝑐|)/(1𝑐2𝐻))𝑦 for 𝑦𝜕𝐾𝑟3. Thus by Lemma 4.4 and (4.33), we have for 𝑦𝜕𝐾𝑟3 that 𝑄𝑦𝜆𝜀𝐿𝑀𝜔0𝑏(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,𝐾𝑟3𝐾𝑟1,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟3𝐾𝑟1, which means (𝐴1𝑦)(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 𝐾𝑟1𝐾𝑟2 and a fixed point 𝑦2(𝑡) in 𝐾𝑟3𝐾𝑟1. Consequently, (𝐴1𝑦1)(𝑡) and (𝐴1𝑦2)(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 0𝑢𝑟2, where the constant 𝜂>0 satisfies 𝜆𝑙𝜂𝛼|𝑐|1𝑐2𝜔0𝑏(𝑠)d𝑠>1.(4.37) Letting 𝑟2=(1|𝑐|)𝑟2, we have 𝑓((𝐴1𝑦)(𝑡𝜏(𝑡)))𝜂(𝐴1𝑦)(𝑡𝜏(𝑡)) for 𝑦𝐾𝑟2. By Lemma 2.2, we have 0(𝐴1𝑦)(𝑡𝜏(𝑡))𝑦/(1|𝑐|)𝑟2 for 𝑦𝜕𝐾𝑟2. Thus by Lemma 4.3 and (4.37), 𝑄𝑦𝜆𝑙𝜂𝛼|𝑐|1𝑐2𝜔0𝑏(𝑠)d𝑠𝑦>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,𝐾𝑟1𝐾𝑟2,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟1𝐾𝑟2. Therefore (𝐴1𝑦)(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|1𝑐2𝜔0𝑏(𝑠)d𝑠>1.(4.40) Let 𝑟3=max{2𝑟1,𝐻(1𝑐2)/(𝛼|𝑐|)}, we have 𝑓((𝐴1𝑦)(𝑡𝜏(𝑡)))𝜂(𝐴1𝑦)(𝑡𝜏(𝑡)) for 𝑦𝐾𝑟3. By Lemma 2.2, we have (𝐴1𝑦)(𝑡𝜏(𝑡))((𝛼|𝑐|)/(1𝑐2𝐻))𝑦 for 𝑦𝜕𝐾𝑟3. Thus by Lemma 4.3 and (4.40), we have for 𝑦𝜕𝐾𝑟3 that 𝑄𝑦𝜆𝑙𝜂𝛼|𝑐|1𝑐2𝜔0𝑏(𝑠)d𝑠𝑦>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,𝐾𝑟3𝐾𝑟1,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾𝑟3𝐾𝑟1. That means (𝐴1𝑦)(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 𝐾𝑟1𝐾𝑟2 and a fixed point 𝑦2 in 𝐾𝑟3𝐾𝑟1. Consequently, (𝐴1𝑦1)(𝑡) and (𝐴1𝑦2)(𝑡) are two positive 𝜔-periodic solutions of (1.3) for 0<𝜆<𝜆0.(c) By Lemma 2.2, if 𝑦𝐾, then (𝐴1𝑦)(𝑡𝜏(𝑡))((𝛼|𝑐|)/(1𝑐2))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,𝑢0,+).(4.43) Assume 𝑦(𝑡) is a positive 𝜔-periodic solution of (1.3) for 𝜆>𝜆0, where 𝜆0=(1𝑐2)/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=𝑄𝑦𝜆𝑙𝑏1𝛼|𝑐|1𝑐2𝜔0𝑏(𝑠)d𝑠𝑦>𝑦,(4.44) which is a contradiction.Case 2. If 𝑖=0, we have 𝑓0< and 𝑓<. Letting 𝑏2=max{𝑓(𝑢)/𝑢𝑢>0}>0, then we obtain 𝑓(𝑢)𝑏2[𝑢,𝑢0,).(4.45) Assume 𝑦(𝑡) is a positive 𝜔-periodic solution of (1.3) for 0<𝜆<𝜆0, where 𝜆0=(𝑚(𝑀+𝑚)|𝑐|)/𝑏2𝐿𝑀𝜔0𝑏(𝑠)d𝑠. Since 𝑄𝑦(𝑡)=𝑦(𝑡) for 𝑡[0,𝜔], it follows from Lemma 4.4 that 𝑦=𝑄𝑦𝜆𝑏2𝐿𝑀𝜔0𝑏(𝑠)d𝑠𝑚(𝑀+𝑚)|𝑐|𝑦