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

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‖=max_{𝑡∈[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=1𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1−𝑓(𝑡+𝛿(𝑡))𝛿𝐷_𝑖,for|𝑐|<1,∀𝑓∈𝑋,𝑐−^∞_𝑗=11𝑐^𝑗+1𝑓𝑠+𝛿(𝑡)+^𝑗−1_𝑖=1𝛿𝐷_𝑖,for|𝑐|>1,∀𝑓∈𝑋,(2.2)(2)||𝐴⁻¹𝑓(𝑡)||≤‖𝑓‖||1−|𝑐|||,∀𝑓∈𝑋,(2.3)(3)^𝜔0||𝐴⁻¹𝑓(𝑡)||𝑑𝑡≤1||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 𝐷_𝑗=𝑠−∑^{𝑗−1𝑖=1}𝛿(𝐷_𝑖), 𝑗=1,2,…. Therefore 𝐵^𝑗𝑥(𝑡)=𝑐^𝑗𝑥𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖,_∞_𝑗=0𝐵^𝑗𝑓(𝑡)=𝑓(𝑡)+^∞_𝑗=1𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖.(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=1𝐵^𝑗=^∞_𝑗=0𝐵^𝑗.(2.6) Here 𝐵⁰=𝐼. Then 𝐴⁻¹𝑓(𝑡)=^∞_𝑗=0𝐵^𝑗𝑓(𝑡)=^∞_𝑗=0𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖,(2.7) and consequently ||𝐴⁻¹𝑓(𝑡)||=|||||^∞_𝑗=0𝐵^𝑗𝑓(𝑡)|||||=|||||^∞_𝑗=0𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖|||||≤‖𝑓‖1−|𝑐|.(2.8) Moreover, ^𝜔0||𝐴⁻¹𝑓(𝑡)||𝑑𝑡=^𝜔0|||||_∞_𝑗=0𝐵^𝑗≤𝑓(𝑡)|||||𝑑𝑡_∞_𝑗=0^𝜔0=||𝐵^𝑗𝑓(𝑡)||𝑑𝑡_∞_𝑗=0^𝜔0|||||𝑐_𝑗𝑓𝑠−^𝑗−1_𝑖=1≤1𝛿𝐷_𝑖|||||𝑑𝑡1−|𝑐|^𝜔0||𝑓(𝑡)||𝑑𝑡.(2.9)Case 2 (|𝑐|>1). Let 𝐸∶𝑋⟶𝑋,(𝐸𝑥)(𝑡)=𝑥(𝑡)−1𝐵𝑐𝑥(𝑡+𝛿(𝑡)),₁∶𝑋⟶𝑋,𝐵₁𝑥(𝑡)=1𝑐𝑥(𝑡+𝛿(𝑡)).(2.10) By definition of the linear operator 𝐵₁, we have 𝐵^𝑗1𝑓(𝑡)=𝑐1_𝑗𝑓𝑠+^𝑗−1_𝑖=1𝛿𝐷_𝑖.(2.11) Here 𝐷_𝑖 is defined as in Case 1. Summing over 𝑗 yields ^∞_𝑗=0𝐵^𝑗1𝑓(𝑡)=𝑓(𝑡)+^∞_𝑗=11𝑐^𝑗𝑓𝑠+^𝑗−1_𝑖=1𝛿𝐷_𝑖.(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=1𝐵^𝑗1,(2.13) and for all 𝑓∈𝑋 we get 𝐸⁻¹𝑓(𝑡)=𝑓(𝑡)+^∞_𝑗=1𝐵^𝑗1𝑓(𝑡).(2.14) On the other hand, from (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)), we have (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡))=−𝑐𝑥(𝑡−𝛿(𝑡))−1𝑐𝑥(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 𝑥(𝑡)=𝐸⁻¹𝑓₁(𝑡)=𝑓₁(𝑡)+^∞_𝑗=1𝐵^𝑗1𝑓₁(𝑡)=−𝑓(𝑡+𝛿(𝑡))𝑐−^∞_𝑗=1𝐵^𝑗1𝑓(𝑡+𝛿(𝑡))𝑐,(2.20) proving that 𝐴⁻¹ exists and satisfies 𝐴⁻¹𝑓(𝑡)=−𝑓(𝑡+𝑐𝛿(𝑡))−^∞_𝑗=1𝐵^𝑗1𝑓(𝑡+𝛿(𝑡))=−𝑓(𝑡+𝛿(𝑡))𝑐𝑐−^∞_𝑗=11𝑐^𝑗+1𝑓𝑠+𝛿(𝑡)+^𝑗−1_𝑖=1||𝐴₋₁𝛿𝐷_𝑖,𝑓(𝑡)||=|||||−𝑓(𝑡+𝛿(𝑡))𝑐−^∞_𝑗=11𝑐^𝑗+1𝑓𝑠+𝛿(𝑡)+^𝑗−1_𝑖=1𝛿𝐷_𝑖|||||≤‖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‖𝑦‖≤𝐴⁻¹𝑦(𝑡)≤11−|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)==𝑦𝑦((𝑡𝑡))++^∞_𝑗=1𝑐^𝑗𝑦_𝑗≥1even𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖𝑐^𝑗𝑦𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖−_𝑗≥1odd|𝑐|^𝑗𝑦𝑠−^𝑗−1_𝑖=1≥𝛼‖𝑦‖+𝛼_𝑗≥1even𝛿𝐷_𝑖𝑐^𝑗‖𝑦‖−‖𝑦‖_𝑗≥1odd|𝑐|^𝑗=𝛼1−𝑐²‖𝑦‖−|𝑐|1−𝑐²=𝛼−|𝑐|‖𝑦‖1−𝑐²‖𝑦‖.(2.23)

Lemma 2.3. If 𝑐>0 and 𝑐<1, then for 𝑦∈𝐾, one has 𝛼1−𝑐‖𝑦‖≤𝐴⁻¹𝑦(𝑡)≤11−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)=≥𝑦𝛼(‖𝑡𝑦)‖++𝛼_𝑗≥1𝑐‖^𝑗𝑦𝑦‖𝑠_𝑗≥1−^𝑗−1_𝑖=1𝛿𝐷_𝑖𝑐^𝑗=𝛼1−𝑐‖𝑦‖.(2.25)

3. Green's Functions

Theorem 3.1. For 𝜌>0 and ℎ∈𝑋, the equation 𝑢(0)=𝑢(𝜔),𝑢𝑢^_−𝜌³𝑢=ℎ(𝑡),(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))sin√3/2𝜌(𝑡−𝑠)+𝜋/6−𝒲3𝜌+²1ex+pe(x𝜌p(𝑡(−−𝜌𝑠𝜔)))−2exp(−𝜌𝜔/2)cos√3/2𝜌𝜔2ex3p𝜌(²((1e/x2p)𝜌(𝜌(𝑠𝜔−)−𝑡−1)𝜔,))0s≤in𝑠≤√𝑡≤𝜔,3/2𝜌(𝑡−𝑠+𝜔)+𝜋/6−𝒴+e3x𝜌p²(𝜌1(𝑡++e𝜔xp−(−𝑠)𝜌)𝜔)−2exp(−𝜌𝜔/2)cos√3/2𝜌𝜔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)(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=exp−𝜌𝑡2sin𝜑cos√3𝜌2𝑡+cos𝜑sin√3𝜌2𝑡=exp−𝜌2𝑡sin𝜑+√3𝜌2𝑡(3.4) which for 𝑡→∞ contradicts periodicity of 𝑣, proving that 𝑐₂=𝑐₃=0.
Applying the method of variation of parameters, we get 𝑐^1(𝑡)=exp(−𝜌𝑡)3𝜌²𝑐ℎ(𝑡),_2(𝑡)=√3/3sin√3𝜌𝑡/2−(1/3)cos√3𝜌𝑡/2𝜌²exp𝜌𝑡𝑐2ℎ(𝑡),_3(𝑡)=−(1/3)sin√3𝜌𝑡/2−√3/3cos√3𝜌𝑡/2𝜌²exp𝜌𝑡2ℎ(𝑡),(3.5) and then 𝑐₁(𝑡)=𝑐₁(0)+^𝑡0exp(−𝜌𝑠)3𝜌²𝑐ℎ(𝑠)d𝑠,₂(𝑡)=𝑐₂(0)+^𝑡0√3/3sin√3𝜌𝑠/2−(1/3)cos√3𝜌𝑠/2𝜌²exp𝜌𝑠𝑐2ℎ(𝑠)d𝑠,₃(𝑡)=𝑐₃(0)+^𝑡0−(1/3)sin√3𝜌𝑠/2−√3/3cos√3𝜌𝑠/2𝜌²exp𝜌𝑠𝑢(𝑡)=𝑐₁2ℎ(𝑠)d𝑠,(𝑡)exp(𝜌𝑡)+exp−𝜌𝑡2𝑐²(𝑡)cos√3𝜌2𝑡+𝑐³(𝑡)sin√3𝜌=𝑐₁2𝑡(0)exp(𝜌𝑡)+𝑐₂exp−𝜌𝑡2cos√32𝜌𝑡+𝑐³(0)exp−𝜌𝑡2sin√3+2𝜌𝑡_𝑡0exp(𝜌(𝑡−𝑠))3𝜌²ℎ(𝑠)d𝑠+^𝑡0sin√3/2𝜌(𝑠−𝑡)−𝜋/66𝜌²exp𝜌2(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))𝑐3𝜌²(1−exp(𝜌𝜔))ℎ(𝑠)d𝑠,₂(0)=^𝜔02exp(𝜌(𝑠−𝜔)/2)exp(−𝜌𝜔/2)sin𝜋/6−√3𝜌𝑠/2−sin𝒟𝑐3𝜌²exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos√3𝜌𝜔/2+1ℎ(𝑠)d𝑠,₃(0)=^𝜔02exp(𝜌(𝑠−𝜔)/2)exp(−𝜌𝜔/2)cos𝜋/6−√3𝜌𝑠/2−cos𝒟3𝜌²exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos√3𝜌𝜔/2+1ℎ(𝑠)d𝑠,(3.7) where 𝒟 denotes (𝜋/6−√3𝜌(𝑠−𝜔)/2). Therefore 𝑢(𝑡)=𝑐₁(𝑡)exp(𝜌𝑡)+exp−𝜌𝑡2𝑐²(𝑡)cos√3𝜌2𝑡+𝑐³(𝑡)sin√3𝜌=_𝑡02𝑡⎧⎪⎨⎪⎩2exp((1/2)𝜌(𝑠−𝑡))sin√3/2𝜌(𝑡−𝑠)+𝜋/6−𝒲3+𝜌²e1x+p(e𝜌xp(𝑡(−−𝑠𝜌)𝜔))−2exp(−𝜌𝜔/2)cos√3/2𝜌𝜔+3𝜌²(1−exp(𝜌𝜔))⎫⎪⎬⎪⎭ℎ(𝑠)d𝑠_𝜔𝑡⎧⎪⎨⎪⎩2exp((1/2)𝜌(𝑠−𝑡−𝜔))sin√3/2𝜌(𝑡−𝑠+𝜔)+𝜋/6−𝒴+exp3(𝜌𝜌²(𝑡+1+𝜔e−xp𝑠))(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos√3/2𝜌𝜔=3𝜌²(1−exp(𝜌𝜔))⎫⎪⎬⎪⎭ℎ(𝑠)d𝑠_𝜔0𝐺₁(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(0)=𝑢(𝜔),𝑢𝑢^_+𝜌³𝑢=ℎ(𝑡),(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))sin√3/2𝜌(𝑡−𝑠)−𝜋/6−𝒰3𝜌+²1e+xpex(p𝜌((𝑠𝜌𝜔−)𝑡−))2exp((1/2)𝜌𝜔)cos√3/2𝜌𝜔2ex3p𝜌(²((11/2−)𝜌ex(𝑡p+(−𝜔𝜌−𝜔)𝑠)),)s0in≤𝑠≤√𝑡≤𝜔,3/2𝜌(𝑡+𝜔−𝑠)−𝜋/6−𝒳+e3x𝜌p²(𝜌1(+𝑠−ex𝑡p−(𝜌𝜔𝜔)))−2exp((1/2)𝜌𝜔)cos√3/2𝜌𝜔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)𝑙=13𝜌²(exp(𝜌𝜔)−1),𝐿=3𝜌3²(+1−2eexxpp((−−𝜌𝜌𝜔𝜔//22)))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))𝐻3𝜌²exp(𝜌𝜔)−1_∗1,(𝑡,𝑠)=exp(𝜌(𝑡+𝜔−𝑠))𝐻3𝜌²exp(𝜌𝜔)−1₂,(𝑡,𝑠)=2exp((1/2)𝜌(𝑠−𝑡))sin√3/2𝜌(𝑡−𝑠)+𝜋/6−𝒲𝐻3𝜌²1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos√3/2𝜌𝜔_∗2,(𝑡,𝑠)=2exp((1/2)𝜌(𝑠−𝑡−𝜔))sin√3/2𝜌(𝑡−𝑠+𝜔)+𝜋/6−𝒴3𝜌²1+exp(−𝜌𝜔)−2exp(−𝜌𝜔/2)cos√3/2𝜌𝜔.(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)=sin√32𝜌(𝑡−𝑠)+𝜋6−exp−𝜌2𝜔sin√3=sin(𝜃)−exp−𝜌𝜔2𝜌(𝑡−𝑠−𝜔)+𝜋62sin𝜃−√3𝑔2𝜌𝜔_∗1,(𝑡,𝑠)=sin√32𝜌(𝑡−𝑠+𝜔)+𝜋6−exp−𝜌2𝜔sin√3=sin𝜃+√2𝜌(𝑡−𝑠)+𝜋632𝜌𝜔−exp−𝜌2𝜔sin𝜃.(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≤√3−𝜋2𝜌𝜔+𝜋6<56𝜋,2<𝜋6−√32𝜌𝜔≤𝜃−√32𝜌𝜔≤𝜋6.(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<√34𝜌𝜔≤𝜃−√34𝜌𝜔≤𝜋6+√3𝑔₁4𝜌𝜔<𝜋2,(𝑡,𝑠)=sin(𝜃)−exp−𝜌𝜔2sin𝜃−√3≥sin𝜃−sin𝜃−√2𝜌𝜔3=2cos𝜃−√2𝜌𝜔34𝜌𝜔sin√3>0.4𝜌𝜔(3.16)
For 0≤𝑡≤𝑠≤𝜔, −𝜋2<−√3𝜋2𝜌𝜔+𝜋6≤𝜃≤𝜋6,6≤𝜃+√32𝜌𝜔≤𝜋6+√32𝜌𝜔<65𝜋.(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+√3𝑔_∗14𝜌𝜔<𝜋2,(𝑡,𝑠)=sin𝜃+√3≥sin𝜃+√2𝜌𝜔−exp−𝜌2𝜔sin𝜃3=2cos𝜃+2√𝜌𝜔−sin𝜃34𝜌𝜔sin√3>0.4𝜌𝜔(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 have𝑙=1≤3e𝜌x²p(e(x𝜌p(𝑡(𝜌+𝜔𝜔)−−1𝑠)))3𝜌²(exp(𝜌𝜔)−1)<𝐺¹≤exp(𝜌(𝑡+𝜔−𝑠))(𝑡,𝑠)3𝜌²exp(𝜌𝜔)−1+3𝜌²exp(𝜌(𝑠−𝑡−𝜔)/2)2+2exp(−𝜌𝜔/2)≤exp(𝜌𝜔)exp(−𝜌𝜔)+1−2exp(−𝜌𝜔/2)cos√3𝜌𝜔/23𝜌²exp(𝜌𝜔)−1+3𝜌²2+2exp(−𝜌𝜔/2)≤1exp(−𝜌𝜔)+1−2exp(−𝜌𝜔/2)cos√3𝜌𝜔/23𝜌²1−exp(−𝜌𝜔)+3𝜌2²+1−2eexxpp((−−𝜌𝜌𝜔𝜔//22))²≤3+2exp(−𝜌𝜔/2)3𝜌²1−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 𝑘𝑀=𝑙=(𝑀ma+x𝑚{)𝑎+(𝑡)𝐿∶𝑀𝑡,∈[0𝑘,₁𝜔]},𝑚=min{𝑎(𝑡)∶𝑡∈[0,𝜔]},𝜌³=𝑀,=𝑘−√𝑘²−4𝐿𝑙𝑀𝑚2𝐿𝑀,𝛼=𝑙[𝑚𝐿−𝑀(𝑀(1−+|𝑚𝑐|))|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=lim_𝑥→0𝑓(𝑥)𝑥,𝑓_∞=lim_𝑥→∞𝑓(𝑥)𝑥,𝑓⁰=lim_𝑥→0𝑓(𝑥)𝑥,𝑓^∞=lim_𝑥→∞𝑓(𝑥)𝑥,(4.2) and denote 𝑖₀:numberof0′sin𝑓₀,𝑓_∞,𝑖⁰:numberof0′sin𝑓₀,𝑓_∞,𝑖_∞:numberof∞′sin𝑓₀,𝑓_∞,𝑖^∞:numberof∞′sin𝑓₀,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 𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟} and 𝜕𝐾_𝑟={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 into𝑦^(𝑡)+𝑎(𝑡)𝐴⁻¹𝑦(𝑡)=𝜆𝑏(𝑡)𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡)),(4.4) which can be further rewritten as𝑦^(𝑡)+𝑎(𝑡)𝑦(𝑡)−𝑎(𝑡)𝐻(𝑦(𝑡))=𝜆𝑏(𝑡)𝑓𝐴⁻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+𝜔}𝐺₂(𝑡,𝑠)ℎ(𝑠)d𝑠,𝐻𝑦(𝑡)=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:𝑦(𝑡)=(𝑇ℎ)(𝑡)+𝑇𝐻𝑦(𝑡).(4.8) In view of 𝑐<0 and 𝑐>−min{𝑘₁,𝑚/(𝑀+𝑚)}, we have‖‖𝑇𝐻‖‖≤‖𝑇‖‖‖𝐻‖‖≤𝑀−𝑚+𝑚|𝑐|𝑀(1−|𝑐|)<1,(4.9) and hence 𝑦(𝑡)=𝐼−𝑇𝐻⁻¹(𝑇ℎ)(𝑡).(4.10) Define an operator 𝑃∶𝑋→𝑋 by (𝑃ℎ)(𝑡)=𝐼−𝑇𝐻⁻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 𝑃==𝐼𝐼−+𝑇𝑇𝐻𝐻+⁻¹𝑇𝑇𝐻₂=𝑇+𝑇𝐻𝑇+𝑇𝐻₂+⋯+𝑇𝐻^𝑛+⋯𝑇𝑇+⋯+𝑇𝐻^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 by𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓𝐴⁻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 𝑄𝑦(𝑡)=≥𝑇𝑃𝜆𝜆𝑏𝑏((𝑡𝑡))𝑓𝑓𝐴𝐴⁻¹₋₁𝑦(𝑡−𝜏(𝑡))=𝜆_{𝑡𝑡+𝜔}𝑦(𝑡−𝜏(𝑡))≥𝜆𝑙_𝜔0𝐺₂(𝑡,𝑠)𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠.(4.16) On the other hand, 𝑄𝑦(𝑡)=≤𝑃𝑀𝜆𝑏((1𝑡)−𝑓|𝑐|)𝐴⁻¹𝑦(𝑡−𝜏(𝑡))𝑚−(𝑀+𝑚)|𝑐|‖‖𝑇𝜆𝑏(𝑡)𝑓𝐴⁻¹=𝜆𝑀(1−|𝑐|)𝑦(𝑡−𝜏(𝑡))‖‖𝑚−(𝑀+𝑚)|𝑐|max^{[]𝑡∈0,𝜔}^{𝑡𝑡+𝜔}≤𝜆𝑀(1−|𝑐|)𝐺₂(𝑡,𝑠)𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠𝑚−(𝑀+𝑚)|𝑐|𝐿^𝜔0𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠.(4.17) Therefore 𝑄𝑦(𝑡)≥𝑙[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 𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡))≥𝐴⁻¹𝑦(𝑡−𝜏(𝑡))𝜂,for𝑡∈[0,𝜔],𝑦∈𝐾,(4.19) then ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐²^𝜔0𝑏(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 𝑄𝑦(𝑡)=≥𝑃𝑇𝜆𝜆𝑏𝑏((𝑡𝑡))𝑓𝑓𝐴𝐴⁻¹₋₁𝑦(𝑡−𝜏(𝑡))=𝜆_{𝑡𝑡+𝜔}𝑦(𝑡−𝜏(𝑡))≥𝜆𝑙𝜂_𝜔0𝐺₂(𝑡,𝑠)𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠≥𝜆𝑙𝜂𝛼−𝑏|𝑐(𝑠|)𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠1−𝑐²^𝜔0𝑏(𝑠)d𝑠‖𝑦‖.(4.21) Hence ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐²^𝜔0𝑏(𝑠)d𝑠‖𝑦‖,𝑦∈𝐾.(4.22)

Lemma 4.4. If there exists 𝜀>0 such that 𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡))≤𝐴⁻¹𝑦(𝑡−𝜏(𝑡))𝜀,for𝑡∈[0,𝜔],𝑦∈𝐾,(4.23) then ‖𝑄𝑦‖≤𝜆𝜀𝐿𝑀∫^𝜔0𝑏(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|)𝑚−(𝑀+𝑚)|𝑐|𝐿^𝜔0≤𝜆𝑀(1−|𝑐|)𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠𝑚−(𝑀+𝑚)|𝑐|𝐿𝜀^𝜔0≤𝜆𝜀𝐿𝑀∫_𝜔0𝑏(𝑠)𝐴⁻¹𝑦(𝑠−𝜏(𝑠))d𝑠𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖.(4.25)

Define 𝐹(𝑟)=max𝑓(𝑡)∶0≤𝑡≤𝑟𝑓1−|𝑐|₁,(𝑟)=min𝑓(𝑡)∶𝛼−|𝑐|1−𝑐²𝑟≤𝑡≤𝑟1−|𝑐|.(4.26)

Lemma 4.5. If 𝑦∈𝜕𝐾_𝑟, 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∈𝜕𝐾_𝑟, 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∈𝜕𝐾_𝑟, 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∈𝜕𝐾_𝑟, 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 𝐾_𝑟={𝑢∈𝐾∶‖𝑢‖<𝑟}. Assume that 𝑇∶𝐾_𝑟→𝐾 is completely continuous such that 𝑇𝑥≠𝑥 for 𝑥∈𝜕𝐾_𝑟={𝑢∈𝐾∶‖𝑢‖=𝑟}. (i)If ‖𝑇𝑥‖≥‖𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖(T,𝐾_𝑟,𝐾)=0.(ii)If ‖𝑇𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖(𝑇,𝐾_𝑟,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)𝑙∫^𝜔0𝑏(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)𝑙∫^𝜔0𝑏(𝑠)d𝑠>0, then for all 𝜆>𝜆₀, we have from Lemma 4.5 that ‖𝑄𝑦‖>‖𝑦‖,for𝑦∈𝜕𝐾_𝑟1.(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 𝜆𝜀𝐿𝑀∫^𝜔0𝑏(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∈𝐾_𝑟2. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₂ for 𝑦∈𝜕𝐾_𝑟2. In view of Lemma 4.4 and (4.30), we have for 𝑦∈𝜕𝐾_𝑟2 that ‖𝑄𝑦‖≤𝜆𝜀𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖<‖𝑦‖.(4.31) It follows from Lemma 4.7 and (4.29) that 𝑖𝑄,𝐾_𝑟2,𝐾=1,𝑖𝑄,𝐾_𝑟1,𝐾=0.(4.32) thus 𝑖(𝑄,𝐾_𝑟1⧵𝐾_𝑟2,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟1⧵𝐾_𝑟2, 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 𝑢≥𝐻, where the constant 𝜀>0 satisfies 𝜆𝜀𝐿𝑀∫^𝜔0𝑏(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∈𝐾_𝑟3. By Lemma 2.2, we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟3. Thus by Lemma 4.4 and (4.33), we have for 𝑦∈𝜕𝐾_𝑟3 that ‖𝑄𝑦‖≤𝜆𝜀𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖<‖𝑦‖.(4.34) Recalling Lemma 4.7 and (4.29) and that 𝑖𝑄,𝐾_𝑟3,𝐾=1,𝑖𝑄,𝐾_𝑟1,𝐾=0,(4.35) then 𝑖(𝑄,𝐾_𝑟3⧵𝐾_𝑟1,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟3⧵𝐾_𝑟1, 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 𝐾_𝑟1⧵𝐾_𝑟2 and a fixed point 𝑦₂(𝑡) in 𝐾_𝑟3⧵𝐾_𝑟1. 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∈𝜕𝐾_𝑟1.(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−𝑐²^𝜔0𝑏(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∈𝐾_𝑟2. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₂ for 𝑦∈𝜕𝐾_𝑟2. Thus by Lemma 4.3 and (4.37), ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐²^𝜔0𝑏(𝑠)d𝑠‖𝑦‖>‖𝑦‖.(4.38) It follows from Lemma 4.7 and (4.36) that 𝑖𝑄,𝐾_𝑟2,𝐾=0,𝑖𝑄,𝐾_𝑟1,𝐾=1,(4.39) which implies 𝑖(𝑄,𝐾_𝑟1⧵𝐾_𝑟2,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟1⧵𝐾_𝑟2. 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 𝑢≥𝐻, where the constant 𝜂>0 satisfies 𝜆𝑙𝜂𝛼−|𝑐|1−𝑐²^𝜔0𝑏(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∈𝐾_𝑟3. By Lemma 2.2, we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟3. Thus by Lemma 4.3 and (4.40), we have for 𝑦∈𝜕𝐾_𝑟3 that ‖𝑄𝑦‖≥𝜆𝑙𝜂𝛼−|𝑐|1−𝑐²^𝜔0𝑏(𝑠)d𝑠‖𝑦‖>‖𝑦‖.(4.41) It follows from Lemma 4.7 and (4.36) that 𝑖𝑄,𝐾_𝑟3,𝐾=0,𝑖𝑄,𝐾_𝑟1,𝐾=1,(4.42) that is, 𝑖(𝑄,𝐾_𝑟3⧵𝐾_𝑟1,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟3⧵𝐾_𝑟1. 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 𝐾_𝑟1⧵𝐾_𝑟2 and a fixed point 𝑦₂ in 𝐾_𝑟3⧵𝐾_𝑟1. 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)≥𝑏₁𝑢,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−𝑐²^𝜔0𝑏(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)≤𝑏₂𝑢,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|)/𝑏₂𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠. Since 𝑄𝑦(𝑡)=𝑦(𝑡) for 𝑡∈[0,𝜔], it follows from Lemma 4.4 that ‖𝑦‖=‖𝑄𝑦‖≤𝜆𝑏₂𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖<‖𝑦‖,(4.46) which is a contradiction.

Theorem 4.9. (a) If there exists a constant 𝑏₁>0 such that 𝑓(𝑢)≥𝑏₁𝑢 for 𝑢∈[0,+∞), then (1.3) has no positive 𝜔-periodic solution for 𝜆>(1−𝑐²)/𝑙𝑏₁(𝛼−|𝑐|)∫^𝜔0𝑏(𝑠)d𝑠.
(b) If there exists a constant 𝑏₂>0 such that 𝑓(𝑢)≤𝑏₂𝑢 for 𝑢∈[0,+∞), then (1.3) has no positive 𝜔-periodic solution for 0<𝜆<(𝑚−(𝑀+𝑚)|𝑐|)/𝑏₂𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠.

Proof. From the proof of (c) in Theorem 4.8, we obtain this theorem immediately.

Theorem 4.10. Assume that 𝑖₀=𝑖₀=𝑖_∞=𝑖_∞=0 and that one of the following conditions holds: (1)𝑓₀≤𝑓_∞;(2)𝑓₀>𝑓_∞;(3)𝑓₀≤𝑓_∞≤𝑓₀≤𝑓_∞;(4)𝑓_∞≤𝑓₀≤𝑓_∞≤𝑓₀.If 1−𝑐²𝑙(𝛼−|𝑐|)∫^𝜔0𝑏(𝑠)d𝑠max𝑓₀,𝑓₀,𝑓_∞,𝑓_∞<𝜆<𝐿𝑀∫^𝜔0𝑏(𝑠)𝑚d𝑠−m(𝑀in+𝑓𝑚₀),|𝑓𝑐|⁰,𝑓_∞,𝑓_∞,(4.47) then (1.3) has one positive 𝜔-periodic solution.

Proof. We have the following cases.Case 1. If 𝑓₀≤𝑓_∞, then 1−𝑐²𝑓_∞𝑙(𝛼−|𝑐|)∫^𝜔0𝑏(𝑠)d𝑠<𝜆<𝑓𝑚₀𝐿−𝑀(𝑀∫^𝜔0+𝑏𝑚(𝑠))|𝑐d|𝑠.(4.48) It is easy to see that there exists a 0<𝜀<𝑓_∞ such that 1−𝑐²𝑓_∞−𝜀𝑙(𝛼−|𝑐|)∫^𝜔0𝑏(𝑠)d𝑠<𝜆<𝑓₀𝑚+−𝜀(𝑀𝐿𝑀+∫𝑚^𝜔0)𝑏|𝑐(|𝑠)d𝑠.(4.49) For the above 𝜀, we choose 𝑟₁>0 such that 𝑓(𝑢)≤(𝑓₀+𝜀)𝑢 for 0≤𝑢≤𝑟₁. Letting 𝑟₁=(1−|𝑐|)𝑟₁, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≤(𝑓₀+𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟1. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₁ for 𝐾∈𝜕𝐾_𝑟1. Thus by Lemma 4.4 we have for 𝑦∈𝜕𝐾_𝑟1 that ‖𝑄𝑦‖≤𝜆𝑓₀+𝜀𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖<‖𝑦‖.(4.50)
On the other hand, there exists a constant 𝐻>0 such that 𝑓(𝑢)≥(𝑓_∞−𝜀)𝑢 for 𝑢≥𝐻. Letting 𝑟₂=max{2𝑟₁,𝐻(1−𝑐²)/(𝛼−|𝑐|)}, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≥(𝑓_∞−𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟2. By Lemma 2.2, we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟2. Thus by Lemma 4.3, for 𝑦∈𝜕𝐾_𝑟2, ‖𝑄𝑦‖≥𝜆𝑙𝑓_∞−𝜀𝛼−|𝑐|1−𝑐²^𝜔0𝑏(𝑠)d𝑠‖𝑦‖>‖𝑦‖.(4.51) It follows from Lemma 4.7 that 𝑖𝑄,𝐾_𝑟1,𝐾=1,𝑖𝑄,𝐾_𝑟2,𝐾=0.(4.52) Thus 𝑖(𝑄,𝐾_𝑟2⧵𝐾_𝑟1,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟2⧵𝐾_𝑟1. So (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-periodic solution of (1.3).Case 2. If 𝑓₀>𝑓_∞, in this case, we have 1−𝑐²𝑓₀𝑙(𝛼−|𝑐|)∫^𝜔0𝑏(𝑠)d𝑠<𝜆<𝑓𝑚_∞−𝐿𝑀(𝑀∫^𝜔0+𝑏𝑚(𝑠))|𝑐d|𝑠.(4.53) It is easy to see that there exists a 0<𝜀<𝑓₀ such that 1−𝑐²𝑓₀−𝜀𝑙(𝛼−|𝑐|)∫^𝜔0𝑏(𝑠)d𝑠<𝜆<𝑓_∞𝑚+−𝜀(𝑀𝐿𝑀+∫𝑚^𝜔0)𝑏|𝑐(|𝑠)d𝑠.(4.54) For the above 𝜀, we choose 𝑟₁>0 such that 𝑓(𝑢)≥(𝑓₀−𝜀)𝑢 for 0≤𝑢≤𝑟₁. Letting 𝑟₁=(1−|𝑐|)𝑟₁, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≥(𝑓₀−𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟1. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₁ for 𝑦∈𝜕𝐾_𝑟1. Thus we have by Lemma 4.3 that for 𝑦∈𝜕𝐾_𝑟1‖𝑄𝑦‖≥𝜆𝑙𝑓₀−𝜀𝛼−|𝑐|1−𝑐²^𝜔0𝑏(𝑠)d𝑠‖𝑦‖>‖𝑦‖.(4.55) On the other hand, there exists a constant 𝐻>0 such that 𝑓(𝑢)≤(𝑓_∞+𝜀)𝑢 for 𝑢≥𝐻. Letting 𝑟₂=max{2𝑟₁,𝐻(1−𝑐²)/(𝛼−|𝑐|)}, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≤(𝑓_∞+𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟2. By Lemma 2.2 we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟2. Thus by Lemma 4.4, for 𝑦∈𝜕𝐾_𝑟2, ‖𝑄𝑦‖≤𝜆𝑓_∞+𝜀𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠𝑚−(𝑀+𝑚)|𝑐|‖𝑦‖.(4.56) It follows from Lemma 4.7 that 𝑖𝑄,𝐾_𝑟1,𝐾=0,𝑖𝑄,𝐾_𝑟2,𝐾=1.(4.57) Thus 𝑖(𝑄,𝐾_𝑟2⧵𝐾_𝑟1,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟2⧵𝐾_𝑟1, proving that (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-periodic solution of (1.3).Case 3 (𝑓₀≤𝑓_∞≤𝑓₀≤𝑓_∞). The proof is the same as in Case 1.Case 4 (𝑓_∞≤𝑓₀≤𝑓_∞≤𝑓₀). The proof is the same as in Case 2.

4.2. Case II: 𝑐>0 and 𝑐<min{𝑚/(𝑀+𝑚),(𝐿𝑀−𝑙𝑚)/((𝐿−𝑙)𝑀−𝑙𝑚)}

Define 𝑓₂(𝑟)=min𝑓(𝑡)∶𝛼1−𝑐𝑟≤𝑡≤1−𝑟𝑐.(4.58) Similarly as in Section 4.1, we get the following results.

Theorem 4.11. (a) If 𝑖₀=1 or 2, then (1.3) has 𝑖₀ positive 𝜔-periodic solutions for 𝜆>1/𝑓₂(1)𝑙∫^𝜔0𝑏(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 solution for sufficiently small or large 𝜆>0, respectively.

Theorem 4.12. (a) If there exists a constant 𝑏₁>0 such that 𝑓(𝑢)≥𝑏₁𝑢 for 𝑢∈[0,+∞), then (1.3) has no positive 𝜔-periodic solution for 𝜆>(1−𝑐)/𝑙𝛼𝑏₁∫^𝜔0𝑏(𝑠)d𝑠.
(b) If there exists a constant 𝑏₂>0 such that 𝑓(𝑢)≤𝑏₂𝑢 for 𝑢∈[0,+∞), then (1.3) has no positive 𝜔-periodic solution for 0<𝜆<(𝑚−(𝑀+𝑚)𝑐)/𝑏₂𝐿𝑀∫^𝜔0𝑏(𝑠)d𝑠.

Theorem 4.13. Assume that 𝑖₀=𝑖₀=𝑖_∞=𝑖_∞=0 hold, and that one of the following conditions holds: (1)𝑓₀≤𝑓_∞;(2)𝑓₀>𝑓_∞;(3)𝑓₀≤𝑓_∞≤𝑓₀≤𝑓_∞;(4)𝑓_∞≤𝑓₀≤𝑓_∞≤𝑓₀.If 1−𝑐𝑙𝛼∫^𝜔0𝑏(𝑠)d𝑠max𝑓₀,𝑓₀,𝑓_∞,𝑓_∞<𝜆<𝐿𝑀∫^𝜔0𝑏(𝑠)𝑚d𝑠−m(i𝑀n+𝑓𝑚₀,)𝑓𝑐⁰,𝑓_∞,𝑓_∞,(4.59) then (1.3) has one positive 𝜔-periodic solution.

Remark 4.14. In a similar way, one can consider the third-order neutral functional differential equation (𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)))^−𝑎(𝑡)𝑥(𝑡)=−𝜆𝑏(𝑡)𝑓(𝑥(𝑡−𝜏(𝑡))).

We illustrate our results with an example.

Example 4.15. Consider the following third-order neutral functional differential equation: 𝑢(𝑡)+3100𝑢(𝑡−cos²𝑡)^+101001−21sin²𝑡𝑢(𝑡)=𝜆1−cos²𝑡𝑢²(𝑡−𝜏(𝑡))𝑎^{𝑢(𝑡−𝜏(𝑡))},(4.60) where 𝜆 and 0<𝑎<1 are two positive parameters, and 𝜏(𝑡+𝜋)=𝜏(𝑡).

Comparing (4.60) to (1.3), we see that 𝛿(𝑡)=cos²𝑡, 𝑐=−1/300, 𝑎(𝑡)=(1/1000)(1−(1/2)sin²𝑡), 𝑏(𝑡)=1−cos²𝑡, 𝜔=𝜋, 𝑓(𝑢)=𝑢²𝑎^𝑢. Clearly, 𝑀=1/1000, 𝑚=1/2000, and we get 𝜌=1/10, noticing that √3𝜋/10<4𝜋/3 holds. Moreover, we know that 𝑓₀=0, 𝑓_∞=0, 𝑖₀=2. By Theorem 4.8, we easily get the following conclusion: (4.60) has two positive 𝜔-periodic solutions for 𝜆>25/1132𝜋𝑟₁, where 𝑟₁=min{𝑓(0.0027),𝑓(300/299)}.

In fact, by simple computations, we have 𝑙=13𝜌(exp(𝜌𝜔)−1)=90.3612,𝐿=3𝜌3²(+12−eexxpp((−−𝜌𝜌𝜔𝜔//22)))²𝑘=7.5715,𝑘₁=7435.98,|𝑐|=1=0.0060,𝛼=0.0061,300<min𝑘¹,𝑚𝑓𝑀+𝑚=0.0060,|𝑐|=3100<0.0061=𝛼,₁(1)=min𝑓(𝑡)∶0.0027≈0.0061−1/3001−(1/300)²≤𝑡≤300299=min𝑓(0.0027),𝑓329009=𝑟¹1,𝑓₁(1)𝑙∫^𝜔0𝑏(𝑠)d𝑠=113225𝜋𝑟₁.(4.61)

5. Lyapunov Stability

When 𝛿(𝑡)≡𝛿, then (1.3) is transformed into(𝑥(𝑡)−𝑐𝑥(𝑡−𝛿))^=−𝑎(𝑡)𝑥(𝑡)+𝜆𝑏(𝑡)𝑓(𝑥(𝑡−𝜏(𝑡))).(5.1) Define operator 𝐴₁∶𝑋→𝑋 by (𝐴₁𝑥)(𝑡)=(𝑥(𝑡)−𝑐𝑥(𝑡−𝛿)). Obviously, 𝐴₁ satisfies Lemma 2.1. Denote by 𝜏_∗ and 𝜏^∗ the essential infimum and supremum of a given function 𝜏∈𝐶¹[0,𝜔], if they exist.

Now, we study the Lyapunov stability of the periodic solutions of (5.1).

Theorem 5.1. Assume 𝜏(𝑡)>0 and 𝜏^(𝑡)<1 hold. And the following condition hold: (𝐹₁) there exists a positive constant 𝛾 such that ||𝑓_𝑢(𝑢)||≤𝛾,∀𝑢∈ℝ,here𝑓_𝑢=𝑑𝑓𝑑𝑢.(5.2)Then every 𝜔-periodic solution of (5.1) is Lyapunov stable.

Proof. Letting 𝑧₁(𝑡)=𝑥(𝑡),𝑧₂(𝑡)=𝑥^(𝑡),𝑧₃(𝑡)=𝑥^(𝑡),(5.3) then system (5.1) can be transformed into 𝑧𝑧^1(𝑡)=𝑧₂(𝑡),_2𝑧(𝑡)=𝑧₃(𝑡),_3(𝑡)=𝐴¹⁻¹−𝑎(𝑡)𝑧₁(𝑡)+𝜆𝑏(𝑡)𝑓𝑧₁(𝑡−𝜏(𝑡)).(5.4)
Suppose now that 𝑧^∗(𝑡)=(𝑧^∗1(𝑡),𝑧^∗2(𝑡),𝑧^∗3(𝑡))^⊤ is a 𝜔-periodic solution of (5.4). Let 𝑧(𝑡)=(𝑧₁(𝑡),𝑧₂(𝑡),𝑧₃(𝑡))^⊤ be any arbitrary solution of (5.4). For any 𝑘=1,2,3, write 𝑤_𝑘(𝑡)=𝑧_𝑘(𝑡)−𝑧^∗𝑘(𝑡). Then it follows from (5.4) that𝑤𝑤^1(𝑡)=𝑤₂(𝑡),_2𝑤(𝑡)=𝑤₃(𝑡),_3(𝑡)=𝐴¹⁻¹−𝑎(𝑡)𝑧₁(𝑡)−𝑧^∗1(𝑡)+𝜆𝑏(𝑡)𝑓𝑧₁(𝑡−𝜏(𝑡))−𝑓𝑧^∗1(𝑡−𝜏(𝑡)),(5.5) and so ||||𝑤𝑤^1(𝑡)||=||𝑤₂(𝑡)_2||,||𝑤(𝑡)||=||𝑤₃(𝑡)_3||,(𝑡)||=||𝐴¹⁻¹−𝑎(𝑡)𝑧₁(𝑡)−𝑧^∗1(𝑡)+𝜆𝑏(𝑡)𝑓𝑧₁(𝑡−𝜏(𝑡))−𝑓𝑧^∗1(𝑡−𝜏(𝑡))||.(5.6) Letting 𝑢^𝑘(𝑙)(𝑡)=|𝑤^𝑘(𝑙)(𝑡)|, 𝑙=0,1, 𝑘=1,2,3, then 𝑢𝑢^1(𝑡)=𝑢₂(𝑡),_2𝑢(𝑡)=𝑢₃(𝑡),_3(𝑡)=||𝐴¹⁻¹−𝑎(𝑡)𝑧₁(𝑡)−𝑧^∗1(𝑡)+𝜆𝑏(𝑡)𝑓𝑧₁(𝑡−𝜏(𝑡))−𝑓𝑧^∗1(𝑡−𝜏(𝑡))||.(5.7) Take 𝛽=max{(|𝑎|₀+𝜆|𝑏|₀𝛾)/|1−|𝑐‖,ln(1/(1−𝜏^∗))/𝜏_∗,1}+1, here |𝑎|₀=max_{𝑡∈[0,𝜔]}|𝑎(𝑡)|, and define a function 𝑉(⋅) by 𝑉𝑡,𝑢₁,𝑢₂,𝑢₃∶=𝑒^−𝛽𝑡𝑢₁+𝑢₂+𝑢₃+𝜆||𝑏||⁰𝛾||1−|𝑐|||^{𝑡𝑡−𝜏(𝑡)}𝑒^−𝛽𝑠𝑢₁(𝑠)d𝑠.(5.8) There exists a sufficiently small positive constant 𝜀 such that 𝑒^−𝛽𝑡≥𝜀. Let 𝑈(𝑢₁,𝑢₂,𝑢₃)=𝜀∑^3𝑘=1𝑢_𝑘(𝑡). It is obvious that 𝑉(𝑡,𝑢₁,𝑢₂,𝑢₃)>0 and 𝑉(𝑡,𝑢₁,𝑢₂,𝑢₃)≥𝑈(𝑢₁,𝑢₂,𝑢₃)>0. From (𝐹₁) and Lemma 2.1, we get 𝑉̇𝑡,𝑢₁,𝑢₂,𝑢₃=+−𝛽𝑒𝑒^−𝛽𝑡_−𝛽𝑡𝑢₁(𝑡)+𝑢₂(𝑡)+𝑢₃(𝑡)+𝑒^−𝛽𝑡𝑢₂(𝑡)+𝑢₃(𝑡)+𝜆||𝑏||||𝐴¹⁻¹−𝑎(𝑡)𝑧₁(𝑡)−𝑧^∗1(𝑡)+𝜆𝑏(𝑡)𝑓𝑧₁(𝑡−𝜏(𝑡))−𝑓𝑧^∗1(𝑡−𝜏(𝑡))₀||𝛾||1−|𝑐|||𝑒^−𝛽𝑡≤−𝛽𝑒_−𝛽𝑡𝑢₁(𝑡)−𝑒^{−𝛽(𝑡−𝜏(𝑡))}1−𝜏^(𝑡)𝑢₁(𝑡−𝜏(𝑡))+𝑒𝑢₁(𝑡)+𝑢₂(𝑡)+𝑢₃(𝑡)+𝑒^−𝛽𝑡𝑢₂(𝑡)+𝑢₃(𝑡)_−𝛽𝑡||1−|𝑐|||||𝑎(𝑡)||||𝑧¹+𝜆||𝑏||(𝑡)−𝑧^∗1(𝑡)||+𝜆||𝑏(𝑡)||𝛾||𝑧₁(𝑡−𝜏(𝑡))−𝑧^∗1(𝑡−𝜏(𝑡))₀||𝛾||1−|𝑐|||𝑒^−𝛽𝑡≤−𝛽+|𝑎|₀𝑢₁(𝑡)−𝑒^{−𝛽(𝑡−𝜏(𝑡))}1−𝜏^(𝑡)𝑢₁(𝑡−𝜏(𝑡))+𝜆||𝑏||₀𝛾||1−|𝑐|||𝑒^−𝛽𝑡+𝜆||𝑏||𝑢₁(𝑡)+𝑒^−𝛽𝑡(−𝛽+1)𝑢₂(𝑡)+𝑢₃(𝑡)₀𝛾||1−|𝑐|||𝑒^−𝛽𝑡=−𝛽+|𝑎|₀−𝑒^{−𝛽(𝑡−𝜏(𝑡))}1−𝜏^(𝑡)𝑢₁(𝑡−𝜏(𝑡))+𝜆||𝑏||₀𝛾||1−|𝑐|||𝑒^−𝛽𝑡+𝜆||𝑏||𝑢₁(𝑡)+𝑒^−𝛽𝑡(−𝛽+1)𝑢₂(𝑡)+𝑢₃(𝑡)₀𝛾||1−|𝑐|||𝑒^−𝛽𝑡1−𝑒^{𝛽𝜏(𝑡)}1−𝜏^≤−𝛽+|𝑎|₀(𝑡)𝑢₁(𝑡−𝜏(𝑡))+𝜆||𝑏||₀𝛾||1−|𝑐|||𝑒^−𝛽𝑡+𝜆||𝑏||𝑢₁(𝑡)+𝑒^−𝛽𝑡(−𝛽+1)𝑢₂(𝑡)+𝑢₃(𝑡)₀𝛾||1−|𝑐|||𝑒^−𝛽𝑡1−𝑒^𝛽𝜏∗1−𝜏^∗𝑢₁<0.(𝑡−𝜏(𝑡))(5.9)
Hence 𝑉 is a Lyapunov function for nonautonomous (5.1) (see [14, page 50]), and so the 𝜔-periodic solution 𝑧^∗ of (5.1) is Lyapunov stable.

Corollary 5.2. Assume that 𝜏(𝑡)=𝜏>0 and (𝐹₁) holds. Then every 𝜔-periodic solution of (5.1) is Lyapunov stable.

We illustrate our results with an example.

Example 5.3. Consider the following third-order neutral functional differential equation: 𝑢(𝑡)+5010𝑢(𝑡−𝛿)^+811−12sin²𝑡𝑢(𝑡)=𝜆1−cos²𝑡𝑎𝑢(𝑡−𝜏)+cos²𝑢(𝑡−𝜏),(5.10) where 𝜆, 𝛿, 𝜏, and 𝑎 are positive parameters.

Comparing (5.10) to (5.1), we see that 𝑐=−1/500, 𝑎(𝑡)=(1/8)(1−(1/2)sin²𝑡), 𝑏(𝑡)=1−cos²𝑡, 𝜔=𝜋, 𝑓(𝑢)=𝑎𝑢+cos²𝑢. Clearly, 𝑀=1/8,𝑚=1/16, and we get 𝜌=1/2. Noticing that √3𝜋/2<4𝜋/3 holds. Moreover, we know that 𝑓₀=∞, 𝑓_∞=𝑎, 𝑖_∞=1. By Theorem 4.8, we easily get the following conclusion: (5.10) has at least one positive 𝜔-periodic solutions for 0<𝜆<1/25𝜋𝑟₂, where 𝑟₂=min{𝑓(0),𝑓(500/499)}.

In fact, by simple computations, we have 𝑙=13𝜌(exp(𝜌𝜔)−1)=0.175,𝐿=3𝜌3²(+1−2eexxpp((−−𝜌𝜌𝜔𝜔//22)))²𝑘=2.235,𝑘₁=17.62,|𝑐|=1=0.0050,𝛼=0.0049,500<min𝑘¹,𝑚𝐹(1)=min𝑓(𝑡)∶𝑀0≤+𝑡𝑚≤5=000.0050,|𝑐|=5100<0.0049=𝛼,499=min𝑓(0),𝑓450909=𝑟²𝑚−(𝑀−𝑚)|𝑐|,𝐿𝑀(1−|𝑐|)𝐹(1)∫^𝜔0𝑏(𝑠)d𝑠≈251𝜋𝑟₂.(5.11) Next, we consider that solution of (5.10) is Lyapunov stable. Since |𝑓_𝑢(𝑢)|≤𝑎+2, then (𝐹₁) holds. By Corollary 5.2, we know solution of (5.10) is Lyapunov stable.

Acknowledgment

This research is supported by the National Natural Science Foundation of China (no. 10971202).