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

Authors: Jingli Ren; Zhibo Cheng; Yueli Chen
Journal: ISRN Applied Mathematics
 (2011)

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## 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.


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## Body

## 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(1.1)(x(t)-cx(t-δ))′′+a(t)x(t)=λb(t)f(x(t-τ(t))).
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(1.2)(x(t)-cx(t-τ(t)))′′=a(t)x(t)-f(t,x(t-τ(t))).
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 definingx1=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:(1.3)(x(t)-cx(t-δ(t)))′′′=-a(t)x(t)+λb(t)f(x(t-τ(t))).
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 Section2, 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

LetX={x∈C(ℝ,ℝ):x(t+ω)=x(t),t∈ℝ} with norm ∥x∥=maxt∈[0,ω]|x(t)|, and let Cω+={x∈C(ℝ,(0,∞)):x(t+ω)=x(t)}, Cω-={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 (2.1)(Ax)(t)=x(t)-cx(t-δ(t)),(Bx)(t)=cx(t-δ(t)).Lemma 2.1.
If|c|≠1, then the operator A has a continuous inverse A-1 on X, satisfying (1)
(2.2)(A-1f)(t)={f(t)+∑j=1∞cjf(s-∑i=1j-1δ(Di)),for|c|<1,∀f∈X,-f(t+δ(t))c-∑j=1∞1cj+1f(s+δ(t)+∑i=1j-1δ(Di)),for|c|>1,∀f∈X,(2)
(2.3)|(A-1f)(t)|≤‖f‖|1-|c||,∀f∈X,(3)
(2.4)∫0ω|(A-1f)(t)|dt≤1|1-|c||∫0ω|f(t)|dt,∀f∈X.Proof.
We have the following cases.Case 1 (|c|<1).
Lett-δ(t)=s  and Dj=s-∑i=1j-1δ(Di), j=1,2,…. Therefore
(2.5)Bjx(t)=cjx(s-∑i=1j-1δ(Di)),∑j=0∞(Bjf)(t)=f(t)+∑j=1∞cjf(s-∑i=1j-1δ(Di)).
Since A=I-B, we get from ∥B∥≤|c|<1 that A has a continuous inverse A-1:X→X with
(2.6)A-1=(I-B)-1=I+∑j=1∞Bj=∑j=0∞Bj.
Here B0=I. Then
(2.7)(A-1f(t))=∑j=0∞[Bjf](t)=∑j=0∞cjf(s-∑i=1j-1δ(Di)),
and consequently
(2.8)|(A-1f)(t)|=|∑j=0∞[Bjf](t)|=|∑j=0∞cjf(s-∑i=1j-1δ(Di))|≤‖f‖1-|c|.
Moreover,
(2.9)∫0ω|(A-1f)(t)|dt=∫0ω|∑j=0∞(Bjf)(t)|dt≤∑j=0∞∫0ω|(Bjf)(t)|dt=∑j=0∞∫0ω|cjf(s-∑i=1j-1δ(Di))|dt≤11-|c|∫0ω|f(t)|dt.Case 2 (|c|>1).
Let(2.10)E:X⟶X,(Ex)(t)=x(t)-1cx(t+δ(t)),B1:X⟶X,(B1x)(t)=1cx(t+δ(t)).
By definition of the linear operator B1, we have
(2.11)(B1jf)(t)=1cjf(s+∑i=1j-1δ(Di)).
Here Di is defined as in Case 1. Summing over j yields
(2.12)∑j=0∞(B1jf)(t)=f(t)+∑j=1∞1cjf(s+∑i=1j-1δ(Di)).
Since ∥B1∥<1, we obtain that the operator E has a bounded inverse E-1,
(2.13)E-1:X⟶X,E-1=(I-B1)-1=I+∑j=1∞B1j,
and for all f∈X we get
(2.14)(E-1f)(t)=f(t)+∑j=1∞(B1jf)(t).
On the other hand, from (Ax)(t)=x(t)-cx(t-δ(t)), we have
(2.15)(Ax)(t)=x(t)-cx(t-δ(t))=-c[x(t-δ(t))-1cx(t)],
that is,
(2.16)(Ax)(t)=-c(Ex)(t-δ(t)).
Let f∈X be arbitrary. We are looking for x such that
(2.17)(Ax)(t)=f(t),
that is
(2.18)-c(Ex)(t-δ(t))=f(t).
Therefore
(2.19)(Ex)(t)=-f(t+δ(t))c=:f1(t),
and hence
(2.20)x(t)=(E-1f1)(t)=f1(t)+∑j=1∞(B1jf1)(t)=-f(t+δ(t))c-∑j=1∞B1jf(t+δ(t))c,
proving that A-1 exists and satisfies
(2.21)[A-1f](t)=-f(t+δ(t))c-∑j=1∞B1jf(t+δ(t))c=-f(t+δ(t))c-∑j=1∞1cj+1f(s+δ(t)+∑i=1j-1δ(Di)),|[A-1f](t)|=|-f(t+δ(t))c-∑j=1∞1cj+1f(s+δ(t)+∑i=1j-1δ(Di))|≤‖f‖|c|-1.
Statements (1) and (2) are proved. From the above proof, (3) can easily be deduced.Lemma 2.2.
Ifc<0 and |c|<α, we have for y∈K that
(2.22)α-|c|1-c2‖y‖≤(A-1y)(t)≤11-|c|‖y‖.Proof.
Sincec<0 and |c|<α<1, by Lemma 2.1, one has for y∈K that
(2.23)(A-1y)(t)=y(t)+∑j=1∞cjy(s-∑i=1j-1δ(Di))=y(t)+∑j≥1evencjy(s-∑i=1j-1δ(Di))-∑j≥1odd|c|jy(s-∑i=1j-1δ(Di))≥α‖y‖+α∑j≥1evencj‖y‖-‖y‖∑j≥1odd|c|j=α1-c2‖y‖-|c|1-c2‖y‖=α-|c|1-c2‖y‖.Lemma 2.3.
Ifc>0 and c<1, then for y∈K, one has
(2.24)α1-c‖y‖≤(A-1y)(t)≤11-c‖y‖.Proof.
Sincec>0, c<1, and α<1, by Lemma 2.1, we have for y∈K that
(2.25)(A-1y)(t)=y(t)+∑j≥1cjy(s-∑i=1j-1δ(Di))≥α‖y‖+α‖y‖∑j≥1cj=α1-c‖y‖.


## 3. Green's Functions

Theorem 3.1.
Forρ>0 and h∈X, the equation
(3.1)u′′′-ρ3u=h(t),u(0)=u(ω),u′(0)=u′(ω),u′′(0)=u′′(ω)
has a unique solution which is of the form
(3.2)u(t)=∫0ωG1(t,s)(-h(s))ds,
where
(3.3)G1(t,s)={2exp⁡((1/2)ρ(s-t))[sin((3/2)ρ(t-s)+π/6)-W]3ρ2(1+exp⁡(-ρω)-2exp⁡(-ρω/2)cos⁡((3/2)ρω))+exp⁡(ρ(t-s))3ρ2(exp⁡(ρω)-1),0≤s≤t≤ω,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))+exp⁡(ρ(t+ω-s))3ρ2(exp⁡(ρω)-1),0≤t≤s≤ω,
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+c32>0 and choosing φ such that sinφ=c2/c22+c32,  cosφ=c3/c22+c32, we get
(3.4)v(t)c22+c32=exp⁡(-ρt2)(sinφcos⁡3ρ2t+cos⁡φsin3ρ2t)=exp⁡(-ρt2)sin(φ+3ρ2t)
which for t→∞ contradicts periodicity of v, proving that c2=c3=0.
Applying the method of variation of parameters, we get(3.5)c1′(t)=exp⁡(-ρt)3ρ2h(t),c2′(t)=(3/3)sin(3ρt/2)-(1/3)cos⁡(3ρt/2)ρ2exp⁡(ρt2)h(t),c3′(t)=-(1/3)sin(3ρt/2)-(3/3)cos⁡(3ρt/2)ρ2exp⁡(ρt2)h(t),
and then
(3.6)c1(t)=c1(0)+∫0texp⁡(-ρs)3ρ2h(s)ds,c2(t)=c2(0)+∫0t(3/3)sin(3ρs/2)-(1/3)cos⁡(3ρs/2)ρ2exp⁡(ρs2)h(s)ds,c3(t)=c3(0)+∫0t-(1/3)sin(3ρs/2)-(3/3)cos⁡(3ρs/2)ρ2exp⁡(ρs2)h(s)ds,u(t)=c1(t)exp⁡(ρt)+exp⁡(-ρt2)(c2(t)cos⁡3ρ2t+c3(t)sin3ρ2t)=c1(0)exp⁡(ρt)+c2exp⁡(-ρt2)cos⁡(32ρt)+c3(0)exp⁡(-ρt2)sin(32ρt)+∫0texp⁡(ρ(t-s))3ρ2h(s)ds+∫0tsin((3/2)ρ(s-t)-π/6)6ρ2exp⁡(ρ2(s-t))h(s)ds.
Noting that u(0)=u(ω),u′(0)=u′(ω),u′′(0)=u′′(ω), we obtain
(3.7)c1(0)=∫0ωexp⁡(ρ(ω-s))3ρ2(1-exp⁡(ρω))h(s)ds,c2(0)=∫0ω2exp⁡(ρ(s-ω)/2)[exp⁡(-ρω/2)sin(π/6-3ρs/2)-sinD]3ρ2(exp⁡(-ρω)-2exp⁡(-ρω/2)cos⁡(3ρω/2)+1)h(s)ds,c3(0)=∫0ω2exp⁡(ρ(s-ω)/2)[exp⁡(-ρω/2)cos⁡(π/6-3ρs/2)-cos⁡D]3ρ2(exp⁡(-ρω)-2exp⁡(-ρω/2)cos⁡(3ρω/2)+1)h(s)ds,
where 𝒟 denotes (π/6-3ρ(s-ω)/2). Therefore
(3.8)u(t)=c1(t)exp⁡(ρt)+exp⁡(-ρt2)(c2(t)cos⁡3ρ2t+c3(t)sin3ρ2t)=∫0t{2exp⁡((1/2)ρ(s-t))[sin((3/2)ρ(t-s)+π/6)-W]3ρ2(1+exp⁡(-ρω)-2exp⁡(-ρω/2)cos⁡((3/2)ρω))+exp⁡(ρ(t-s))3ρ2(1-exp⁡(ρω))[sin((3/2)ρ(t-s)+π/6)](((3/2)ρω))}h(s)ds+∫tω{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))+exp⁡(ρ(t+ω-s))3ρ2(1-exp⁡(ρω))[sin((3/2)ρ(t-s)+π/6)](((3/2)ρω))}h(s)ds=∫0ωG1(t,s)h(s)ds,
where G1(t,s) is defined as in (3.3).
By direct calculation, we get the solutionu satisfies the periodic boundary value condition of the problem (3.1).Theorem 3.2.
Forρ>0 and h∈X, the equation
(3.9)u′′′+ρ3u=h(t),u(0)=u(ω),u′(0)=u′(ω),u′′(0)=u′′(ω)
has a unique ω-periodic solution
(3.10)u(t)=∫0ωG2(t,s)h(s)ds,
where
(3.11)G2(t,s)={2exp⁡((1/2)ρ(t-s))[sin((3/2)ρ(t-s)-π/6)-U]3ρ2(1+exp⁡(ρω)-2exp⁡((1/2)ρω)cos⁡((3/2)ρω))+exp⁡(ρ(s-t))3ρ2(1-exp⁡(-ρω)),0≤s≤t≤ω,2exp⁡((1/2)ρ(t+ω-s))[sin((3/2)ρ(t+ω-s)-π/6)-X]3ρ2(1+exp⁡(ρω)-2exp⁡((1/2)ρω)cos⁡((3/2)ρω))+exp⁡(ρ(s-t-ω))3ρ2(1-exp⁡(-ρω)),0≤t≤s≤ω.
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 Theorem3.1 and can therefore be omitted.Now we present the properties of Green's functions for (3.1) and (3.9)(3.12)l=13ρ2(exp⁡(ρω)-1),L=3+2exp⁡(-ρω/2)3ρ2(1-exp⁡(-ρω/2))2.Theorem 3.3.
∫0ωG1(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:(3.13)H1(t,s)=exp⁡(ρ(t-s))3ρ2[exp⁡(ρω)-1],H1*(t,s)=exp⁡(ρ(t+ω-s))3ρ2[exp⁡(ρω)-1],H2(t,s)=2exp⁡((1/2)ρ(s-t))[sin((3/2)ρ(t-s)+π/6)-W]3ρ2(1+exp⁡(-ρω)-2exp⁡(-ρω/2)cos⁡((3/2)ρω)),H2*(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)).
A direct computation shows that ∫0ωG1(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)+1-2exp(-ρω/2)cos(3ρω/2)>[1-exp(-ρω/2)]2>0.
For convenience, we denoteθ=(3/2)ρ(t-s)+π/6(3.14)g1(t,s)=sin(32ρ(t-s)+π6)-exp⁡(-ρω2)sin(32ρ(t-s-ω)+π6)=sin(θ)-exp⁡(-ρω2)sin(θ-32ρω),g1*(t,s)=sin(32ρ(t-s+ω)+π6)-exp⁡(-ρω2)sin(32ρ(t-s)+π6)=sin(θ+32ρω)-exp⁡(-ρω2)sinθ.
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.
For0≤s≤t≤ω, Since 3ρω<(4/3)π, we have(3.15)π6≤θ≤32ρω+π6<5π6,-π2<π6-32ρω≤θ-32ρω≤π6.(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)
For0<θ-(3/2)ρω≤π/6, we have sinθ>0, sin(θ-(3/2)ρω)>0, and
(3.16)0<34ρω≤θ-34ρω≤π6+34ρω<π2,g1(t,s)=sin(θ)-exp⁡(-ρω2)sin(θ-32ρω)≥sinθ-sin(θ-32ρω)=2cos⁡(θ-34ρω)sin(34ρω)>0.
For0≤t≤s≤ω, (3.17)-π2<-32ρω+π6≤θ≤π6,π6≤θ+32ρω≤π6+32ρω<56π.(i)
For-π/2<θ≤0, we have sinθ<0, sin(θ+(3/2)ρω)>0, and then g1*(t,s)>0.(ii)
For0<θ≤π/6, we have sinθ>0, sin(θ+(3/2)ρω)>0, and
(3.18)0<θ+34ρω<π2,g1*(t,s)=sin(θ+32ρω)-exp⁡(-ρω2)sinθ≥sin(θ+32ρω)-sinθ=2cos⁡(θ+34ρω)sin(34ρω)>0.
If3ρω<(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 forG1(t,s) for s∈[0,ω]. We have(3.19)l=13ρ2(exp⁡(ρω)-1)≤exp⁡(ρ(t+ω-s))3ρ2(exp⁡(ρω)-1)<G1(t,s)≤exp⁡(ρ(t+ω-s))3ρ2[exp⁡(ρω)-1]+exp⁡(ρ(s-t-ω)/2)[2+2exp⁡(-ρω/2)]3ρ2[exp⁡(-ρω)+1-2exp⁡(-ρω/2)cos⁡(3ρω/2)]≤exp⁡(ρω)3ρ2[exp⁡(ρω)-1]+2+2exp⁡(-ρω/2)3ρ2[exp⁡(-ρω)+1-2exp⁡(-ρω/2)cos⁡(3ρω/2)]≤13ρ2[1-exp⁡(-ρω)]+2+2exp⁡(-ρω/2)3ρ2[1-exp⁡(-ρω/2)]2≤3+2exp⁡(-ρω/2)3ρ2[1-exp⁡(-ρω/2)]2=L.
The proof is complete.Similarly, the following dual theorem can be proved.Theorem 3.4.
∫0ωG2(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 spaceX as in Section 2. Denote (4.1)M=max⁡{a(t):t∈[0,ω]},m=min⁡{a(t):t∈[0,ω]},ρ3=M,k=l(M+m)+LM,k1=k-k2-4LlMm2LM,α=l[m-(M+m)|c|]LM(1-|c|).
It is easy to see that M,m,β,L,l,k,k1>0.Now we consider (1.3). First let (4.2)f¯0=lim⁡¯x→0f(x)x,f¯∞=lim⁡¯x→∞f(x)x,f̲0=lim⁡̲x→0f(x)x,f̲∞=lim⁡̲x→∞f(x)x,
and denote (4.3)i¯0:numberof0’sin(f¯0,f¯∞),i̲0:numberof0’sin(f̲0,f̲∞),i¯∞:numberof∞’sin(f¯0,f¯∞),i̲∞:numberof∞’sin(f̲0,f̲∞).
It is clear that i¯0,i̲0,i¯∞,i̲∞∈{0,1,2}. We will show that (1.3) has i¯0 or i̲∞ 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.LetK={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}.Lety(t)=(Ax)(t), then from Lemma 2.1 we have x(t)=(A-1y)(t). Hence (1.3) can be transformed into(4.4)y′′′(t)+a(t)(A-1y)(t)=λb(t)f((A-1y)(t-τ(t))),
which can be further rewritten as(4.5)y′′′(t)+a(t)y(t)-a(t)H(y(t))=λb(t)f((A-1y)(t-τ(t))),
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 consider(4.6)y′′′(t)+a(t)y(t)-a(t)H(y(t))=h(t),h∈Cω+,
and define operators T, Ĥ:X→X by (4.7)(Th)(t)=∫tt+ωG2(t,s)h(s)ds,(Ĥy)(t)=M-a(t)y(t)+a(t)H(y(t)).
Clearly T, Ĥ are completely continuous, (Th)(t)>0 for h(t)>0 and ∥Ĥ∥≤(M-m+M(|c|/(1-|c|))). By Theorem 3.2, the solution of (4.6) can be written in the following form:(4.8)y(t)=(Th)(t)+(TĤy)(t).
In view of c<0 and c>-min{k1,m/(M+m)}, we have(4.9)‖TĤ‖≤‖T‖‖Ĥ‖≤M-m+m|c|M(1-|c|)<1,
and hence (4.10)y(t)=(I-TĤ)-1(Th)(t).
Define an operator P:X→X by (4.11)(Ph)(t)=(I-TĤ)-1(Th)(t).
Obviously, for any h∈Cω+, 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
(4.12)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖,∀h∈Cω+.Proof.
By the Neumann expansion ofP, we have
(4.13)P=(I-TĤ)-1T=(I+TĤ+(TĤ)2+⋯+(TĤ)n+⋯)T=T+TĤT+(TĤ)2T+⋯+(TĤ)nT+⋯.
Since T and Ĥ are completely continuous, so is P. Moreover, by (4.13) and recalling that ∥TĤ∥≤(M-m+m|c|)/M(1-|c|)<1, we get
(4.14)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖.Define an operatorQ:X→X by(4.15)Qy(t)=P(λb(t)f((A-1y)(t-τ(t)))).Lemma 4.2.
One has thatQ(K)⊂K.Proof.
From the definition ofQ, it is easy to verify that Qy(t+ω)=Qy(t). For y∈K, we have from Lemma 4.1 that
(4.16)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG2(t,s)b(s)f[(A-1y)(s-τ(s))]ds≥λl∫0ωb(s)f[(A-1y)(s-τ(s))]ds.
On the other hand,
(4.17)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≤M(1-|c|)m-(M+m)|c|‖T(λb(t)f((A-1y)(t-τ(t))))‖=λM(1-|c|)m-(M+m)|c|max⁡t∈[0,ω]∫tt+ωG2(t,s)b(s)f((A-1y)(s-τ(s)))ds≤λM(1-|c|)m-(M+m)|c|L∫0ωb(s)f((A-1y)(s-τ(s)))ds.
Therefore
(4.18)Qy(t)≥l[m-(M+m)|c|]LM(1-|c|)‖Qy‖=α‖Qy‖,
that is, Q(K)⊂K.From the continuity ofP, 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 existsη>0 such that
(4.19)f((A-1y)(t-τ(t)))≥(A-1y)(t-τ(t))η,fort∈[0,ω],y∈K,
then
(4.20)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Proof.
By Lemma2.2, Theorem 3.4, and Lemma 4.1, we have for y∈K that
(4.21)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG2(t,s)b(s)f((A-1y)(s-τ(s)))ds≥λlη∫0ωb(s)(A-1y)(s-τ(s))ds≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖.
Hence
(4.22)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Lemma 4.4.
If there existsɛ>0 such that
(4.23)f((A-1y)(t-τ(t)))≤(A-1y)(t-τ(t))ɛ,fort∈[0,ω],y∈K,
then
(4.24)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖,y∈K.Proof.
By Lemma2.2, Theorem 3.4, and Lemma 4.1, we have
(4.25)‖Qy(t)‖≤λM(1-|c|)m-(M+m)|c|L∫0ωb(s)f((A-1y)(s-τ(s)))ds≤λM(1-|c|)m-(M+m)|c|Lɛ∫0ωb(s)(A-1y)(s-τ(s))ds≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖.Define(4.26)F(r)=max⁡{f(t):0≤t≤r1-|c|},f1(r)=min⁡{f(t):α-|c|1-c2r≤t≤r1-|c|}.Lemma 4.5.
Ify∈∂Kr, then
(4.27)‖Qy‖≥λlf1(r)∫0ωb(s)ds‖y‖.Proof.
By Lemma2.2, 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.
Ify∈∂Kr, then
(4.28)‖Qy‖≤λLM(1-|c|)F(r)m-(M+m)|c|∫0ωb(s)ds‖y‖.Proof.
By Lemma2.2, 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]).
LetX be a Banach space and K a cone in X. For r>0, define Kr={u∈K:∥u∥<r}. Assume that T:K¯r→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) Ifi¯0=1 or 2, then (1.3) has i¯0 positive ω-periodic solutions for λ>1/f1(1)l∫0ωb(s)ds>0,
(b) ifi̲∞=1 or 2, then (1.3) has i̲∞ positive ω-periodic solutions for 0<λ<(m-(M+m)|c|)/LM(1-|c|)F(1)∫0ωb(s)ds,
(c) ifi¯∞=0 or i̲0=0, then (1.3) has no positive ω-periodic solutions for sufficiently small or sufficiently large λ>0, respectively.Proof.
(a) Chooser1=1. Taking λ0=1/f1(r1)l∫0ωb(s)ds>0, then for all λ>λ0, we have from Lemma 4.5 that
(4.29)‖Qy‖>‖y‖,fory∈∂Kr1.Case 1.
Iff¯0=0, we can choose 0<r¯2<r1, so that f(u)≤ɛu for 0≤u≤r¯2, where the constant ɛ>0 satisfies
(4.30)λɛLM∫0ωb(s)dsm-(M+m)|c|<1.
Letting r2=(1-|c|)r¯2, we have f((A-1y)(t-τ(t)))≤ɛ(A-1y)(t-τ(t)) for y∈Kr2. By Lemma 2.2, we have 0≤(A-1y)(t-τ(t))≤∥y∥/(1-|c|)≤r¯2 for y∈∂Kr2. In view of Lemma 4.4 and (4.30), we have for y∈∂Kr2 that
(4.31)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
It follows from Lemma 4.7 and (4.29) that
(4.32)i(Q,Kr2,K)=1,i(Q,Kr1,K)=0.
thus i(Q,Kr1∖K¯r2,K)=-1 and Q has a fixed point y in Kr1∖K¯r2, which means (A-1y)(t) is a positive ω-positive solution of (1.3) for λ>λ0.Case 2.
Iff¯∞=0, there exists a constant H̃>0 such that f(u)≤ɛu for u≥H̃, where the constant ɛ>0 satisfies
(4.33)λɛLM∫0ωb(s)dsm-(M+m)|c|<1.
Letting  r3=max{2r1,H̃(1-c2)/(α-|c|)}, we have f((A-1y)(t-τ(t)))≤ɛ(A-1y)(t-τ(t)) for y∈Kr3. By Lemma 2.2, we have (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥≥H̃ for y∈∂Kr3. Thus by Lemma 4.4 and (4.33), we have for y∈∂Kr3 that
(4.34)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
Recalling Lemma 4.7 and (4.29) and that
(4.35)i(Q,Kr3,K)=1,i(Q,Kr1,K)=0,
then i(Q,Kr3∖K¯r1,K)=1 and Q has a fixed point y in Kr3∖K¯r1, which means (A-1y)(t) is a positive ω-positive solution of (1.3) for λ>λ0.Case 3.
Iff¯0=f¯∞=0, from the above arguments, there exist 0<r2<r1<r3 such that Q has a fixed point y1(t) in Kr1∖K¯r2 and a fixed point y2(t) in Kr3∖K¯r1. Consequently, (A-1y1)(t) and (A-1y2)(t) are two positive ω-periodic solutions of (1.3) for λ>λ0.(b)
Letr1=1. Taking λ0=(m-(M+m)|c|)/LM(1-|c|)F(r1)∫0ωb(s)ds>0, then by Lemma 4.6 we know if λ<λ0, then
(4.36)‖Qy‖<‖y‖,y∈∂Kr1.Case 1.
Iff̲0=∞, we can choose 0<r¯2<r1 so that f(u)≥ηu for 0≤u≤r¯2, where the constant η>0 satisfies
(4.37)λlηα-|c|1-c2∫0ωb(s)ds>1.
Letting r2=(1-|c|)r¯2, 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|)≤r¯2 for y∈∂Kr2. Thus by Lemma 4.3 and (4.37),
(4.38)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.7 and (4.36) that
(4.39)i(Q,Kr2,K)=0,i(Q,Kr1,K)=1,
which implies i(Q,Kr1∖K¯r2,K)=1 and Q has a fixed point y in Kr1∖K¯r2. Therefore (A-1y)(t) is a positive ω-periodic solution of (1.3) for 0<λ<λ0.Case 2.
Iff̲∞=∞, there exists a constant H̃>0 such that f(u)≥ηu for u≥H̃, where the constant η>0 satisfies
(4.40)λlηα-|c|1-c2∫0ωb(s)ds>1.
Let r3=max{2r1,H̃(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-c2))∥y∥≥H̃ for y∈∂Kr3. Thus by Lemma 4.3 and (4.40), we have for y∈∂Kr3 that
(4.41)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.7 and (4.36) that
(4.42)i(Q,Kr3,K)=0,i(Q,Kr1,K)=1,
that is, i(Q,Kr3∖K¯r1,K)=-1 and Q has a fixed point y in Kr3∖K¯r1. That means (A-1y)(t) is a positive ω-periodic solution of (1.3) for 0<λ<λ0.Case 3.
Iff̲0=f̲∞=∞, from the above arguments, Q has a fixed point y1 in Kr1∖K¯r2 and a fixed point y2 in Kr3∖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 Lemma2.2, if y∈K, then (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥>0 for t∈[0,ω].Case 1.
Ifi̲0=0, we have f̲0>0 and f̲∞>0. Letting b1=min{f(u)/u; u>0}>0, then we obtain
(4.43)f(u)≥b1u,u∈[0,+∞).
Assume y(t) is a positive ω-periodic solution of (1.3) for λ>λ0, where λ0=(1-c2)/lb1(α-|c|)∫0ωb(s)ds>0. Since Qy(t)=y(t) for t∈[0,ω], then by Lemma 4.3, if λ>λ0 we have
(4.44)‖y‖=‖Qy‖≥λlb1α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖,
which is a contradiction.Case 2.
Ifi¯∞=0, we have f¯0<∞ and f¯∞<∞. Letting b2=max{f(u)/u:u>0}>0, then we obtain
(4.45)f(u)≤b2u,u∈[0,∞).
Assume y(t) is a positive ω-periodic solution of (1.3) for 0<λ<λ0, where λ0=(m-(M+m)|c|)/b2LM∫0ωb(s)ds. Since Qy(t)=y(t) for t∈[0,ω], it follows from Lemma 4.4 that
(4.46)‖y‖=‖Qy‖≤λb2LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖,
which is a contradiction.Theorem 4.9.
(a) If there exists a constantb1>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|)∫0ωb(s)ds.
(b) If there exists a constantb2>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∫0ωb(s)ds.Proof.
From the proof of (c) in Theorem4.8, we obtain this theorem immediately.Theorem 4.10.
Assume thati̲0=i¯0=i̲∞=i¯∞=0 and that one of the following conditions holds:(1)
f¯0≤f̲∞;(2)
f̲0>f¯∞;(3)
f̲0≤f̲∞≤f¯0≤f¯∞;(4)
f̲∞≤f̲0≤f¯∞≤f¯0.If
(4.47)1-c2l(α-|c|)∫0ωb(s)dsmax⁡{f̲0,f¯0,f̲∞,f¯∞}<λ<m-(M+m)|c|LM∫0ωb(s)dsmin⁡{f̲0,f¯0,f̲∞,f¯∞},
then (1.3) has one positive ω-periodic solution.Proof.
We have the following cases.Case 1.
Iff¯0≤f̲∞, then
(4.48)1-c2f¯∞l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|f̲0LM∫0ωb(s)ds.
It is easy to see that there exists a  0<ɛ<f∞ such that
(4.49)1-c2(f¯∞-ɛ)l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|(f̲0+ɛ)LM∫0ωb(s)ds.
For the above ɛ,  we choose r¯1>0 such that f(u)≤(f̲0+ɛ)u for 0≤u≤r¯1. Letting r1=(1-|c|)r¯1, we have f((A-1y)(t-τ(t)))≤(f̲0+ɛ)(A-1y)(t-τ(t)) for y∈Kr1. By Lemma 2.2, we have 0≤(A-1y)(t-τ(t))≤∥y∥/(1-|c|)≤r¯1 for K∈∂Kr1. Thus by Lemma 4.4 we have for y∈∂Kr1 that
(4.50)‖Qy‖≤λ(f̲0+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
On the other hand, there exists a constantH̃>0 such that f(u)≥(f¯∞-ɛ)u for u≥H̃. Letting r2=max{2r1,H̃(1-c2)/(α-|c|)}, we have f((A-1y)(t-τ(t)))≥(f¯∞-ɛ)(A-1y)(t-τ(t)) for y∈Kr2. By Lemma 2.2, we have (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥≥H̃ for y∈∂Kr2. Thus by Lemma 4.3, for y∈∂Kr2, (4.51)‖Qy‖≥λl(f¯∞-ɛ)α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.7 that
(4.52)i(Q,Kr1,K)=1,i(Q,Kr2,K)=0.
Thus i(Q,Kr2∖K¯r1,K)=-1 and Q has a fixed point y in Kr2∖K¯r1. So (A-1y)(t) is a positive ω-periodic solution of (1.3).Case 2.
Iff̲0>f¯∞, in this case, we have
(4.53)1-c2f¯0l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|f̲∞LM∫0ωb(s)ds.
It is easy to see that there exists a  0<ɛ<f0 such that
(4.54)1-c2(f¯0-ɛ)l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|(f̲∞+ɛ)LM∫0ωb(s)ds.
For the above ɛ, we choose r¯1>0 such that f(u)≥(f¯0-ɛ)u for 0≤u≤r¯1. Letting r1=(1-|c|)r¯1, we have f((A-1y)(t-τ(t)))≥(f¯0-ɛ)(A-1y)(t-τ(t)) for y∈Kr1. By Lemma 2.2, we have 0≤(A-1y)(t-τ(t))≤∥y∥/(1-|c|)≤r¯1 for y∈∂Kr1. Thus we have by Lemma 4.3 that for y∈∂Kr1(4.55)‖Qy‖≥λl(f¯0-ɛ)α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
On the other hand, there exists a constant H̃>0 such that f(u)≤(f̲∞+ɛ)u for u≥H̃. Letting r2=max{2r1,H̃(1-c2)/(α-|c|)}, we have f((A-1y)(t-τ(t)))≤(f̲∞+ɛ)(A-1y)(t-τ(t)) for y∈Kr2. By Lemma 2.2 we have (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥≥H̃ for y∈∂Kr2. Thus by Lemma 4.4, for y∈∂Kr2,
(4.56)‖Qy‖≤λ(f̲∞+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖.
It follows from Lemma 4.7 that
(4.57)i(Q,Kr1,K)=0,i(Q,Kr2,K)=1.
Thus i(Q,Kr2∖K¯r1,K)=-1 and Q has a fixed point y in Kr2∖K¯r1, proving that (A-1y)(t) is a positive ω-periodic solution of (1.3).Case 3 (f̲0≤f̲∞≤f¯0≤f¯∞).
The proof is the same as in Case1.Case 4 (f̲∞≤f̲0≤f¯∞≤f¯0).
The proof is the same as in Case2.


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

Define(4.58)f2(r)=min⁡{f(t):α1-cr≤t≤r1-c}.
Similarly as in Section 4.1, we get the following results.Theorem 4.11.
(a) Ifi¯0=1 or 2, then (1.3) has i0 positive ω-periodic solutions for λ>1/f2(1)l∫0ωb(s)ds>0.
(b) Ifi̲∞=1 or 2, then (1.3) has i∞ positive ω-periodic solutions for 0<λ<(m-(M+m)c)/LM(1-c)F(1)∫0ωb(s)ds.
(c) Ifi¯∞=0 or i̲0=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 constantb1>0 such that f(u)≥b1u for u∈[0,+∞), then (1.3) has no positive ω-periodic solution for λ>(1-c)/lαb1∫0ωb(s)ds.
(b) If there exists a constantb2>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∫0ωb(s)ds.Theorem 4.13.
Assume thati̲0=i¯0=i̲∞=i¯∞=0 hold, and that one of the following conditions holds:(1)
f¯0≤f̲∞;(2)
f̲0>f¯∞;(3)
f̲0≤f̲∞≤f¯0≤f¯∞;(4)
f̲∞≤f̲0≤f¯∞≤f¯0.If
(4.59)1-clα∫0ωb(s)dsmax⁡{f̲0,f¯0,f̲∞,f¯∞}<λ<m-(M+m)cLM∫0ωb(s)dsmin⁡{f̲0,f¯0,f̲∞,f¯∞},
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:(4.60)(u(t)+1300u(t-cos⁡2t))′′′+11000(1-12sin2t)u(t)=λ(1-cos⁡2t)u2(t-τ(t))au(t-τ(t)),
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)=u2au. Clearly, M=1/1000, m=1/2000, and we get ρ=1/10, noticing that 3π/10<4π/3 holds. Moreover, we know that f¯0=0, f¯∞=0, i¯0=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(4.61)l=13ρ(exp⁡(ρω)-1)=90.3612,L=3+2exp⁡(-ρω/2)3ρ2(1-exp⁡(-ρω/2))2=7435.98,k=7.5715,k1=0.0060,α=0.0061,|c|=1300<min⁡{k1,mM+m}=0.0060,|c|=1300<0.0061=α,f1(1)=min⁡{f(t):0.0027≈0.0061-1/3001-(1/300)2≤t≤300299}=min⁡{f(0.0027),f(300299)}=r1,1f1(1)l∫0ωb(s)ds=251132πr1.


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

Now we consider(4.6)y′′′(t)+a(t)y(t)-a(t)H(y(t))=h(t),h∈Cω+,
and define operators T, Ĥ:X→X by (4.7)(Th)(t)=∫tt+ωG2(t,s)h(s)ds,(Ĥy)(t)=M-a(t)y(t)+a(t)H(y(t)).
Clearly T, Ĥ are completely continuous, (Th)(t)>0 for h(t)>0 and ∥Ĥ∥≤(M-m+M(|c|/(1-|c|))). By Theorem 3.2, the solution of (4.6) can be written in the following form:(4.8)y(t)=(Th)(t)+(TĤy)(t).
In view of c<0 and c>-min{k1,m/(M+m)}, we have(4.9)‖TĤ‖≤‖T‖‖Ĥ‖≤M-m+m|c|M(1-|c|)<1,
and hence (4.10)y(t)=(I-TĤ)-1(Th)(t).
Define an operator P:X→X by (4.11)(Ph)(t)=(I-TĤ)-1(Th)(t).
Obviously, for any h∈Cω+, 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
(4.12)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖,∀h∈Cω+.Proof.
By the Neumann expansion ofP, we have
(4.13)P=(I-TĤ)-1T=(I+TĤ+(TĤ)2+⋯+(TĤ)n+⋯)T=T+TĤT+(TĤ)2T+⋯+(TĤ)nT+⋯.
Since T and Ĥ are completely continuous, so is P. Moreover, by (4.13) and recalling that ∥TĤ∥≤(M-m+m|c|)/M(1-|c|)<1, we get
(4.14)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖.Define an operatorQ:X→X by(4.15)Qy(t)=P(λb(t)f((A-1y)(t-τ(t)))).Lemma 4.2.
One has thatQ(K)⊂K.Proof.
From the definition ofQ, it is easy to verify that Qy(t+ω)=Qy(t). For y∈K, we have from Lemma 4.1 that
(4.16)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG2(t,s)b(s)f[(A-1y)(s-τ(s))]ds≥λl∫0ωb(s)f[(A-1y)(s-τ(s))]ds.
On the other hand,
(4.17)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≤M(1-|c|)m-(M+m)|c|‖T(λb(t)f((A-1y)(t-τ(t))))‖=λM(1-|c|)m-(M+m)|c|max⁡t∈[0,ω]∫tt+ωG2(t,s)b(s)f((A-1y)(s-τ(s)))ds≤λM(1-|c|)m-(M+m)|c|L∫0ωb(s)f((A-1y)(s-τ(s)))ds.
Therefore
(4.18)Qy(t)≥l[m-(M+m)|c|]LM(1-|c|)‖Qy‖=α‖Qy‖,
that is, Q(K)⊂K.From the continuity ofP, 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 existsη>0 such that
(4.19)f((A-1y)(t-τ(t)))≥(A-1y)(t-τ(t))η,fort∈[0,ω],y∈K,
then
(4.20)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Proof.
By Lemma2.2, Theorem 3.4, and Lemma 4.1, we have for y∈K that
(4.21)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG2(t,s)b(s)f((A-1y)(s-τ(s)))ds≥λlη∫0ωb(s)(A-1y)(s-τ(s))ds≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖.
Hence
(4.22)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Lemma 4.4.
If there existsɛ>0 such that
(4.23)f((A-1y)(t-τ(t)))≤(A-1y)(t-τ(t))ɛ,fort∈[0,ω],y∈K,
then
(4.24)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖,y∈K.Proof.
By Lemma2.2, Theorem 3.4, and Lemma 4.1, we have
(4.25)‖Qy(t)‖≤λM(1-|c|)m-(M+m)|c|L∫0ωb(s)f((A-1y)(s-τ(s)))ds≤λM(1-|c|)m-(M+m)|c|Lɛ∫0ωb(s)(A-1y)(s-τ(s))ds≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖.Define(4.26)F(r)=max⁡{f(t):0≤t≤r1-|c|},f1(r)=min⁡{f(t):α-|c|1-c2r≤t≤r1-|c|}.Lemma 4.5.
Ify∈∂Kr, then
(4.27)‖Qy‖≥λlf1(r)∫0ωb(s)ds‖y‖.Proof.
By Lemma2.2, 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.
Ify∈∂Kr, then
(4.28)‖Qy‖≤λLM(1-|c|)F(r)m-(M+m)|c|∫0ωb(s)ds‖y‖.Proof.
By Lemma2.2, 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]).
LetX be a Banach space and K a cone in X. For r>0, define Kr={u∈K:∥u∥<r}. Assume that T:K¯r→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) Ifi¯0=1 or 2, then (1.3) has i¯0 positive ω-periodic solutions for λ>1/f1(1)l∫0ωb(s)ds>0,
(b) ifi̲∞=1 or 2, then (1.3) has i̲∞ positive ω-periodic solutions for 0<λ<(m-(M+m)|c|)/LM(1-|c|)F(1)∫0ωb(s)ds,
(c) ifi¯∞=0 or i̲0=0, then (1.3) has no positive ω-periodic solutions for sufficiently small or sufficiently large λ>0, respectively.Proof.
(a) Chooser1=1. Taking λ0=1/f1(r1)l∫0ωb(s)ds>0, then for all λ>λ0, we have from Lemma 4.5 that
(4.29)‖Qy‖>‖y‖,fory∈∂Kr1.Case 1.
Iff¯0=0, we can choose 0<r¯2<r1, so that f(u)≤ɛu for 0≤u≤r¯2, where the constant ɛ>0 satisfies
(4.30)λɛLM∫0ωb(s)dsm-(M+m)|c|<1.
Letting r2=(1-|c|)r¯2, we have f((A-1y)(t-τ(t)))≤ɛ(A-1y)(t-τ(t)) for y∈Kr2. By Lemma 2.2, we have 0≤(A-1y)(t-τ(t))≤∥y∥/(1-|c|)≤r¯2 for y∈∂Kr2. In view of Lemma 4.4 and (4.30), we have for y∈∂Kr2 that
(4.31)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
It follows from Lemma 4.7 and (4.29) that
(4.32)i(Q,Kr2,K)=1,i(Q,Kr1,K)=0.
thus i(Q,Kr1∖K¯r2,K)=-1 and Q has a fixed point y in Kr1∖K¯r2, which means (A-1y)(t) is a positive ω-positive solution of (1.3) for λ>λ0.Case 2.
Iff¯∞=0, there exists a constant H̃>0 such that f(u)≤ɛu for u≥H̃, where the constant ɛ>0 satisfies
(4.33)λɛLM∫0ωb(s)dsm-(M+m)|c|<1.
Letting  r3=max{2r1,H̃(1-c2)/(α-|c|)}, we have f((A-1y)(t-τ(t)))≤ɛ(A-1y)(t-τ(t)) for y∈Kr3. By Lemma 2.2, we have (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥≥H̃ for y∈∂Kr3. Thus by Lemma 4.4 and (4.33), we have for y∈∂Kr3 that
(4.34)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
Recalling Lemma 4.7 and (4.29) and that
(4.35)i(Q,Kr3,K)=1,i(Q,Kr1,K)=0,
then i(Q,Kr3∖K¯r1,K)=1 and Q has a fixed point y in Kr3∖K¯r1, which means (A-1y)(t) is a positive ω-positive solution of (1.3) for λ>λ0.Case 3.
Iff¯0=f¯∞=0, from the above arguments, there exist 0<r2<r1<r3 such that Q has a fixed point y1(t) in Kr1∖K¯r2 and a fixed point y2(t) in Kr3∖K¯r1. Consequently, (A-1y1)(t) and (A-1y2)(t) are two positive ω-periodic solutions of (1.3) for λ>λ0.(b)
Letr1=1. Taking λ0=(m-(M+m)|c|)/LM(1-|c|)F(r1)∫0ωb(s)ds>0, then by Lemma 4.6 we know if λ<λ0, then
(4.36)‖Qy‖<‖y‖,y∈∂Kr1.Case 1.
Iff̲0=∞, we can choose 0<r¯2<r1 so that f(u)≥ηu for 0≤u≤r¯2, where the constant η>0 satisfies
(4.37)λlηα-|c|1-c2∫0ωb(s)ds>1.
Letting r2=(1-|c|)r¯2, 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|)≤r¯2 for y∈∂Kr2. Thus by Lemma 4.3 and (4.37),
(4.38)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.7 and (4.36) that
(4.39)i(Q,Kr2,K)=0,i(Q,Kr1,K)=1,
which implies i(Q,Kr1∖K¯r2,K)=1 and Q has a fixed point y in Kr1∖K¯r2. Therefore (A-1y)(t) is a positive ω-periodic solution of (1.3) for 0<λ<λ0.Case 2.
Iff̲∞=∞, there exists a constant H̃>0 such that f(u)≥ηu for u≥H̃, where the constant η>0 satisfies
(4.40)λlηα-|c|1-c2∫0ωb(s)ds>1.
Let r3=max{2r1,H̃(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-c2))∥y∥≥H̃ for y∈∂Kr3. Thus by Lemma 4.3 and (4.40), we have for y∈∂Kr3 that
(4.41)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.7 and (4.36) that
(4.42)i(Q,Kr3,K)=0,i(Q,Kr1,K)=1,
that is, i(Q,Kr3∖K¯r1,K)=-1 and Q has a fixed point y in Kr3∖K¯r1. That means (A-1y)(t) is a positive ω-periodic solution of (1.3) for 0<λ<λ0.Case 3.
Iff̲0=f̲∞=∞, from the above arguments, Q has a fixed point y1 in Kr1∖K¯r2 and a fixed point y2 in Kr3∖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 Lemma2.2, if y∈K, then (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥>0 for t∈[0,ω].Case 1.
Ifi̲0=0, we have f̲0>0 and f̲∞>0. Letting b1=min{f(u)/u; u>0}>0, then we obtain
(4.43)f(u)≥b1u,u∈[0,+∞).
Assume y(t) is a positive ω-periodic solution of (1.3) for λ>λ0, where λ0=(1-c2)/lb1(α-|c|)∫0ωb(s)ds>0. Since Qy(t)=y(t) for t∈[0,ω], then by Lemma 4.3, if λ>λ0 we have
(4.44)‖y‖=‖Qy‖≥λlb1α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖,
which is a contradiction.Case 2.
Ifi¯∞=0, we have f¯0<∞ and f¯∞<∞. Letting b2=max{f(u)/u:u>0}>0, then we obtain
(4.45)f(u)≤b2u,u∈[0,∞).
Assume y(t) is a positive ω-periodic solution of (1.3) for 0<λ<λ0, where λ0=(m-(M+m)|c|)/b2LM∫0ωb(s)ds. Since Qy(t)=y(t) for t∈[0,ω], it follows from Lemma 4.4 that
(4.46)‖y‖=‖Qy‖≤λb2LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖,
which is a contradiction.Theorem 4.9.
(a) If there exists a constantb1>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|)∫0ωb(s)ds.
(b) If there exists a constantb2>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∫0ωb(s)ds.Proof.
From the proof of (c) in Theorem4.8, we obtain this theorem immediately.Theorem 4.10.
Assume thati̲0=i¯0=i̲∞=i¯∞=0 and that one of the following conditions holds:(1)
f¯0≤f̲∞;(2)
f̲0>f¯∞;(3)
f̲0≤f̲∞≤f¯0≤f¯∞;(4)
f̲∞≤f̲0≤f¯∞≤f¯0.If
(4.47)1-c2l(α-|c|)∫0ωb(s)dsmax⁡{f̲0,f¯0,f̲∞,f¯∞}<λ<m-(M+m)|c|LM∫0ωb(s)dsmin⁡{f̲0,f¯0,f̲∞,f¯∞},
then (1.3) has one positive ω-periodic solution.Proof.
We have the following cases.Case 1.
Iff¯0≤f̲∞, then
(4.48)1-c2f¯∞l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|f̲0LM∫0ωb(s)ds.
It is easy to see that there exists a  0<ɛ<f∞ such that
(4.49)1-c2(f¯∞-ɛ)l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|(f̲0+ɛ)LM∫0ωb(s)ds.
For the above ɛ,  we choose r¯1>0 such that f(u)≤(f̲0+ɛ)u for 0≤u≤r¯1. Letting r1=(1-|c|)r¯1, we have f((A-1y)(t-τ(t)))≤(f̲0+ɛ)(A-1y)(t-τ(t)) for y∈Kr1. By Lemma 2.2, we have 0≤(A-1y)(t-τ(t))≤∥y∥/(1-|c|)≤r¯1 for K∈∂Kr1. Thus by Lemma 4.4 we have for y∈∂Kr1 that
(4.50)‖Qy‖≤λ(f̲0+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
On the other hand, there exists a constantH̃>0 such that f(u)≥(f¯∞-ɛ)u for u≥H̃. Letting r2=max{2r1,H̃(1-c2)/(α-|c|)}, we have f((A-1y)(t-τ(t)))≥(f¯∞-ɛ)(A-1y)(t-τ(t)) for y∈Kr2. By Lemma 2.2, we have (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥≥H̃ for y∈∂Kr2. Thus by Lemma 4.3, for y∈∂Kr2, (4.51)‖Qy‖≥λl(f¯∞-ɛ)α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.7 that
(4.52)i(Q,Kr1,K)=1,i(Q,Kr2,K)=0.
Thus i(Q,Kr2∖K¯r1,K)=-1 and Q has a fixed point y in Kr2∖K¯r1. So (A-1y)(t) is a positive ω-periodic solution of (1.3).Case 2.
Iff̲0>f¯∞, in this case, we have
(4.53)1-c2f¯0l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|f̲∞LM∫0ωb(s)ds.
It is easy to see that there exists a  0<ɛ<f0 such that
(4.54)1-c2(f¯0-ɛ)l(α-|c|)∫0ωb(s)ds<λ<m-(M+m)|c|(f̲∞+ɛ)LM∫0ωb(s)ds.
For the above ɛ, we choose r¯1>0 such that f(u)≥(f¯0-ɛ)u for 0≤u≤r¯1. Letting r1=(1-|c|)r¯1, we have f((A-1y)(t-τ(t)))≥(f¯0-ɛ)(A-1y)(t-τ(t)) for y∈Kr1. By Lemma 2.2, we have 0≤(A-1y)(t-τ(t))≤∥y∥/(1-|c|)≤r¯1 for y∈∂Kr1. Thus we have by Lemma 4.3 that for y∈∂Kr1(4.55)‖Qy‖≥λl(f¯0-ɛ)α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
On the other hand, there exists a constant H̃>0 such that f(u)≤(f̲∞+ɛ)u for u≥H̃. Letting r2=max{2r1,H̃(1-c2)/(α-|c|)}, we have f((A-1y)(t-τ(t)))≤(f̲∞+ɛ)(A-1y)(t-τ(t)) for y∈Kr2. By Lemma 2.2 we have (A-1y)(t-τ(t))≥((α-|c|)/(1-c2))∥y∥≥H̃ for y∈∂Kr2. Thus by Lemma 4.4, for y∈∂Kr2,
(4.56)‖Qy‖≤λ(f̲∞+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖.
It follows from Lemma 4.7 that
(4.57)i(Q,Kr1,K)=0,i(Q,Kr2,K)=1.
Thus i(Q,Kr2∖K¯r1,K)=-1 and Q has a fixed point y in Kr2∖K¯r1, proving that (A-1y)(t) is a positive ω-periodic solution of (1.3).Case 3 (f̲0≤f̲∞≤f¯0≤f¯∞).
The proof is the same as in Case1.Case 4 (f̲∞≤f̲0≤f¯∞≤f¯0).
The proof is the same as in Case2.


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

Define(4.58)f2(r)=min⁡{f(t):α1-cr≤t≤r1-c}.
Similarly as in Section 4.1, we get the following results.Theorem 4.11.
(a) Ifi¯0=1 or 2, then (1.3) has i0 positive ω-periodic solutions for λ>1/f2(1)l∫0ωb(s)ds>0.
(b) Ifi̲∞=1 or 2, then (1.3) has i∞ positive ω-periodic solutions for 0<λ<(m-(M+m)c)/LM(1-c)F(1)∫0ωb(s)ds.
(c) Ifi¯∞=0 or i̲0=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 constantb1>0 such that f(u)≥b1u for u∈[0,+∞), then (1.3) has no positive ω-periodic solution for λ>(1-c)/lαb1∫0ωb(s)ds.
(b) If there exists a constantb2>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∫0ωb(s)ds.Theorem 4.13.
Assume thati̲0=i¯0=i̲∞=i¯∞=0 hold, and that one of the following conditions holds:(1)
f¯0≤f̲∞;(2)
f̲0>f¯∞;(3)
f̲0≤f̲∞≤f¯0≤f¯∞;(4)
f̲∞≤f̲0≤f¯∞≤f¯0.If
(4.59)1-clα∫0ωb(s)dsmax⁡{f̲0,f¯0,f̲∞,f¯∞}<λ<m-(M+m)cLM∫0ωb(s)dsmin⁡{f̲0,f¯0,f̲∞,f¯∞},
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:(4.60)(u(t)+1300u(t-cos⁡2t))′′′+11000(1-12sin2t)u(t)=λ(1-cos⁡2t)u2(t-τ(t))au(t-τ(t)),
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)=u2au. Clearly, M=1/1000, m=1/2000, and we get ρ=1/10, noticing that 3π/10<4π/3 holds. Moreover, we know that f¯0=0, f¯∞=0, i¯0=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(4.61)l=13ρ(exp⁡(ρω)-1)=90.3612,L=3+2exp⁡(-ρω/2)3ρ2(1-exp⁡(-ρω/2))2=7435.98,k=7.5715,k1=0.0060,α=0.0061,|c|=1300<min⁡{k1,mM+m}=0.0060,|c|=1300<0.0061=α,f1(1)=min⁡{f(t):0.0027≈0.0061-1/3001-(1/300)2≤t≤300299}=min⁡{f(0.0027),f(300299)}=r1,1f1(1)l∫0ωb(s)ds=251132πr1.


## 5. Lyapunov Stability

Whenδ(t)≡δ, then (1.3) is transformed into(5.1)(x(t)-cx(t-δ))′′′=-a(t)x(t)+λb(t)f(x(t-τ(t))).
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.
Assumeτ(t)>0 and τ′(t)<1 hold. And the following condition hold:(F1)
there exists a positive constantγ such that
(5.2)|fu(u)|≤γ,∀u∈R,herefu=dfdu.Then every ω-periodic solution of (5.1) is Lyapunov stable.Proof.
Letting(5.3)z1(t)=x(t),z2(t)=x′(t),z3(t)=x′′(t),
then system (5.1) can be transformed into
(5.4)z1′(t)=z2(t),z2′(t)=z3(t),z3′(t)=A1-1[-a(t)z1(t)+λb(t)f(z1(t-τ(t)))].
Suppose now thatz*(t)=(z1*(t),z2*(t),z3*(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 wk(t)=zk(t)-zk*(t). Then it follows from (5.4) that(5.5)w1′(t)=w2(t),w2′(t)=w3(t),w3′(t)=A1-1(-a(t)(z1(t)-z1*(t))+λb(t)(f(z1(t-τ(t)))-f(z1*(t-τ(t))))),
and so
(5.6)|w1′(t)|=|w2(t)|,|w2′(t)|=|w3(t)|,|w3′(t)|=|A1-1(-a(t)(z1(t)-z1*(t))+λb(t)(f(z1(t-τ(t)))-f(z1*(t-τ(t)))))|.
Letting uk(l)(t)=|wk(l)(t)|, l=0,1, k=1,2,3, then
(5.7)u1′(t)=u2(t),u2′(t)=u3(t),u3′(t)=|A1-1(-a(t)(z1(t)-z1*(t))+λb(t)(f(z1(t-τ(t)))-f(z1*(t-τ(t)))))|.
Take β=max{(|a|0+λ|b|0γ)/|1-|c∥,ln(1/(1-τ′*))/τ*,1}+1, here |a|0=maxt∈[0,ω]|a(t)|, and define a function V(·) by
(5.8)V(t,u1,u2,u3):=e-βt(u1+u2+u3)+λ|b|0γ|1-|c||∫t-τ(t)te-βsu1(s)ds.
There exists a sufficiently small positive constant ɛ such that e-βt≥ɛ. Let U(u1,u2,u3)=ɛ∑k=13uk(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
(5.9)V̇(t,u1,u2,u3)=-βe-βt(u1(t)+u2(t)+u3(t))+e-βt(u2(t)+u3(t))+e-βt|A1-1(-a(t)(z1(t)-z1*(t))+λb(t)(f(z1(t-τ(t)))-f(z1*(t-τ(t)))))|+λ|b|0γ|1-|c||(e-βtu1(t)-e-β(t-τ(t))(1-τ′(t))u1(t-τ(t)))≤-βe-βt(u1(t)+u2(t)+u3(t))+e-βt(u2(t)+u3(t))+e-βt|1-|c||[|a(t)||z1(t)-z1*(t)|+λ|b(t)|γ|(z1(t-τ(t)))-z1*(t-τ(t))|]+λ|b|0γ|1-|c||(e-βtu1(t)-e-β(t-τ(t))(1-τ′(t))u1(t-τ(t)))≤(-β+|a|0+λ|b|0γ|1-|c||)e-βtu1(t)+e-βt(-β+1)(u2(t)+u3(t))+λ|b|0γ|1-|c||(e-βt-e-β(t-τ(t))(1-τ′(t)))u1(t-τ(t))=(-β+|a|0+λ|b|0γ|1-|c||)e-βtu1(t)+e-βt(-β+1)(u2(t)+u3(t))+λ|b|0γ|1-|c||e-βt(1-eβτ(t)(1-τ′(t)))u1(t-τ(t))≤(-β+|a|0+λ|b|0γ|1-|c||)e-βtu1(t)+e-βt(-β+1)(u2(t)+u3(t))+λ|b|0γ|1-|c||e-βt(1-eβτ*(1-τ′*))u1(t-τ(t))<0.
HenceV 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 thatτ(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:(5.10)(u(t)+1500u(t-δ))′′′+18(1-12sin2t)u(t)=λ(1-cos⁡2t)(au(t-τ)+cos⁡2u(t-τ)),
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 f̲0=∞, f̲∞=a, i̲∞=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(5.11)l=13ρ(exp⁡(ρω)-1)=0.175,L=3+2exp⁡(-ρω/2)3ρ2(1-exp⁡(-ρω/2))2=17.62,k=2.235,k1=0.0050,α=0.0049,|c|=1500<min⁡{k1,mM+m}=0.0050,|c|=1500<0.0049=α,F(1)=min⁡{f(t):0≤t≤500499}=min⁡{f(0),f(500499)}=r2,m-(M-m)|c|LM(1-|c|)F(1)∫0ωb(s)ds≈125πr2.
Next, we consider that solution of (5.10) is Lyapunov stable. Since |fu(u)|≤a+2, then (F1) holds. By Corollary 5.2, we know solution of (5.10) is Lyapunov stable.



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*Source: 698529-2011-07-19.xml*