# Neutral Operator and Neutral   Differential Equation

Authors: Jingli Ren; Zhibo Cheng; Stefan Siegmund
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

In this paper, we discuss the properties of the neutral operator(Ax)(t)=x(t)−cx(t−δ(t)), and by applying coincidence degree theory and fixed point index theory, we obtain sufficient conditions for the existence, multiplicity, and nonexistence of (positive) periodic solutions to two kinds of second-order differential equations with the prescribed neutral operator.


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## Body

## 1. Introduction

In [1], Zhang discussed the properties of the neutral operator (A1x)(t)=x(t)-cx(t-δ), which became an effective tool for the research on differential equations with this prescribed neutral operator, see, for example, [2–5]. Lu and Ge [6] investigated an extension of A1, namely, the neutral operator A2x(t)=x(t)-∑i=1ncix(t-δi) and obtained the existence of periodic solutions for a corresponding neutral differential equation.In this paper, we consider the neutral operator(Ax)(t)=x(t)-cx(t-δ(t)), where c is constant and |c|≠1, δ∈C1(ℝ,ℝ), and δ is an ω-periodic function for some ω>0. Although A is a natural generalization of the operator A1, the class of neutral differential equation with A typically possesses a more complicated nonlinearity than neutral differential equation with A1 or A2. For example, the neutral operators A1 and A2 are homogeneous in the following sense (Aix)′(t)=(Aix′)(t) for i=1,2, 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 neutral operator A which will be helpful for further studies of differential equations with this neutral operator; in Section 3, by Mawhin's continuation theorem, we obtain the existence of periodic solutions for a second-order Rayleigh-type neutral differential 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 second-order neutral differential equation. Several examples are also given to illustrate our results. Our results improve and extend the results in [1, 2, 4, 7].


## 2. Analysis of the Generalized Neutral Operator

LetCω={x∈C(ℝ,ℝ):x(t+ω)=x(t),t∈ℝ} with norm ∥x∥=maxt∈[0,ω]|x(t)|. Then (Cω,∥·∥) is a Banach space. A cone K in Cω is defined by K={x∈Cω:x(t)≥α∥x∥,forallt∈ℝ}, where α is a fixed positive number with α<1. Moreover, define operators A,B:Cω→Cω 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 Cω, satisfying
(1)
(2.2)(A-1f)(t)={f(t)+∑j=1∞cjf(s-∑i=1j-1δ(Di)),for|c|<1,∀f∈Cω,-f(t+δ(t))c-∑j=1∞(1/cj+1)f(s+δ(t)+∑i=1j-1δ(Di)),for|c|>1,∀f∈Cω.(2)
|(A-1f)(t)|≤∥f∥/|1-|c∥,forallf∈Cω.(3)
∫0ω|(A-1f)(t)|dt≤1/|1-|c||∫0ω|f(t)|dt,forallf∈Cω.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,
(2.3)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:Cω→Cω with
(2.4)A-1=(I-B)-1=I+∑j=1∞Bj=∑j=0∞Bj,
where B0=I. Then
(2.5)(A-1f(t))=∑j=0∞[Bjf](t)=∑j=0∞cjf(s-∑i=1j-1δ(Di)),
and consequently
(2.6)|(A-1f)(t)|=|∑j=0∞[Bjf](t)|=|∑j=0∞cjf(s-∑i=1j-1δ(Di))|≤‖f‖1-|c|.
Moreover,
(2.7)∫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(2.8)E:Cω⟶Cω,(Ex)(t)=x(t)-1cx(t+δ(t)),B1:Cω⟶Cω,(B1x)(t)=1cx(t+δ(t)).
By definition of the linear operator B1, we have
(2.9)(B1jf)(t)=1cjf(s+∑i=1j-1δ(Di)),
where Di is defined as in Case 1. Summing over j yields
(2.10)∑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,
(2.11)E-1:Cω⟶Cω,E-1=(I-B1)-1=I+∑j=1B1j,
and forallf∈Cω we get
(2.12)(E-1f)(t)=f(t)+∑j=1∞(B1jf)(t).
On the other hand, from (Ax)(t)=x(t)-cx(t-δ(t)), we have
(2.13)(Ax)(t)=x(t)-cx(t-δ(t))=-c[x(t-δ(t))-1cx(t)],
that is,
(2.14)(Ax)(t)=-c(Ex)(t-δ(t)).
Let f∈Cω be arbitrary. We are looking for x such that
(2.15)(Ax)(t)=f(t).
that is,
(2.16)-c(Ex)(t-δ(t))=f(t).
Therefore,
(2.17)(Ex)(t)=-f(t+δ(t))c=∶f1(t),
and hence
(2.18)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
(2.19)[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, one has for y∈K that
(2.20)α-|c|1-c2‖y‖≤(A-1y)(t)≤11-|c|‖y‖.Proof.
Sincec<0 and |c|<α<1, by Lemma 2.1, we have for y∈K that
(2.21)(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
(2.22)α1-c‖y‖≤(A-1y)(t)≤11-c‖y‖.Proof.
Sincec>0 and c<1, α<1, by Lemma 2.1, we have for y∈K that
(2.23)(A-1y)(t)=y(t)+∑j≥1cjy(s-∑i=1j-1δ(Di))≥α‖y‖+α‖y‖∑j≥1cj=α1-c‖y‖.


## 3. Periodic Solutions for Neutral Differential Equation

In this section, we consider the second-order neutral differential equation(3.1)(x(t)-cx(t-δ(t)))′′=f(t,x′(t))+g(t,x(t-τ(t)))+e(t),
where τ,e∈Cω and ∫0ωe(t)dt=0; f and g are continuous functions defined on ℝ2 and periodic in t with f(t,·)=f(t+ω,·),g(t,·)=g(t+ω,·), f(t,0)=0, f(t,u)≥0, or f(t,u)≤0 for all (t,u)∈ℝ2.We first recall Mawhin's continuation theorem which our study is based upon. LetX and Y be real Banach spaces and L:D(L)⊂X→Y a Fredholm operator with index zero, where D(L) denotes the domain of L. This means that ImL is closed in Y and dimKerL=dim(Y/ImL)<+∞. Consider supplementary subspaces X1, Y1, of X, Y respectively, such that X=KerL⊕X1, Y=ImL⊕Y1, and let P1:X→KerL and Q1:Y→Y1 denote the natural projections. Clearly, KerL∩(D(L)∩X1)={0}, thus the restriction LP1:=L|D(L)∩X1 is invertible. Let LP1-1 denote the inverse of LP1.LetΩ be an open bounded subset of X with D(L)∩Ω≠∅. A map N:Ω¯→Y is said to be L-compact in Ω¯ if Q1N(Ω¯) is bounded and the operator LP1-1(I-Q1)N:Ω¯→X is compact.Lemma 3.1 (Gaines and Mawhin [8]).
Suppose thatX and Y are two Banach spaces and L:D(L)⊂X→Y is a Fredholm operator with index zero. Furthermore, Ω⊂X is an open bounded set, and N:Ω¯→Y is L-compact on Ω¯. Assume that the following conditions hold:
(1)
Lx≠λNx,forallx∈∂Ω∩D(L),λ∈(0,1);(2)
Nx∉ImL,forallx∈∂Ω∩KerL;(3)
deg{JQ1N,Ω∩KerL,0}≠0, where J:ImQ1→KerL is an isomorphism.
Then the equationLx=Nx has a solution in Ω¯∩D(L).In order to use Mawhin's continuation theorem to study the existence ofω-periodic solutions for (3.1), we rewrite (3.1) in the following form:(3.2)(Ax1)′(t)=x2(t),x2′(t)=f(t,x1′(t))+g(t,x1(t-τ(t)))+e(t).
Clearly, if x(t)=(x1(t),x2(t))⊤ is an ω-periodic solution to (3.2), then x1(t) must be an ω-periodic solution to (3.1). Thus, the problem of finding an ω-periodic solution for (3.1) reduces to finding one for (3.2).Recall thatCω={ϕ∈C(ℝ,ℝ):ϕ(t+ω)≡ϕ(t)} with norm ∥ϕ∥=maxt∈[0,ω]|ϕ(t)|. Define X=Y=Cω×Cω={x=(x1(·),x2(·))∈C(ℝ,ℝ2):x(t)=x(t+ω),t∈ℝ} with norm ∥x∥=max{∥x1∥,∥x2∥}. Clearly, X and Y are Banach spaces. Moreover, define (3.3)L:D(L)={x∈C1(R,R2):x(t+ω)=x(t),t∈R}⊂X⟶Y
by (3.4)(Lx)(t)=((Ax1)′(t)x2′(t))
and N:X→Y by(3.5)(Nx)(t)=(x2(t)f(t,x1′(t))+g(t,x1(t-τ(t)))+e(t)).
Then (3.2) can be converted to the abstract equation Lx=Nx. From the definition of L, one can easily see that (3.6)Ker⁡L≅R2,Im⁡L={y∈Y:∫0ω(y1(s)y2(s))ds=(00)}.
So L is a Fredholm operator with index zero. Let P1:X→KerL and Q1:Y→ImQ1⊂ℝ2 be defined by (3.7)P1x=((Ax1)(0)x2(0));Q1y=1ω∫0ω(y1(s)y2(s))ds,
then ImP1=KerL, KerQ1=ImL. Setting LP1=L|D(L)∩KerP1 and LP1-1: ImL→D(L) denotes the inverse of LP1, then (3.8)[LP1-1y](t)=((A-1Fy1)(t)(Fy2)(t)),[Fy1](t)=∫0ty1(s)ds,[Fy2](t)=∫0ty2(s)ds.
From (3.5) and (3.8), it is clear that Q1N and LP1-1(I-Q1)N are continuous and Q1N(Ω¯) is bounded, and then LP1-1(I-Q1)N(Ω¯) is compact for any open bounded Ω⊂X which means N is L-compact on Ω¯.Now we give our main results on periodic solutions for (3.1).Theorem 3.2.
Suppose there exist positive constantsK1,D,M,b with M>∥e∥ such that:
(H1)|f(t,u)|≤K1|u|+b, for (t,u)∈ℝ×ℝ;
(H2) sgnx·g(t,x)>∥e∥, for |x|>D;
(H3)g(t,x)≥-M, for x≤-D and t∈ℝ.
Then (3.1) has at least one solution with period ω if 0<ω1/2(1+|c|)1/22K1/(|1-|c||-|c|δ1)<1, where δ1=maxt∈[0,ω]|δ′(t)|.Proof.
By construction (3.2) has an ω-periodic solution if and only if the following operator equation
(3.9)Lx=Nx
has an ω-periodic solution. From (3.8), we see that N is L-compact on Ω¯, where Ω is any open, bounded subset of Cω. For λ∈(0,1] define
(3.10)Ω1={x∈Cω:Lx=λNx}.
Then x=(x1,x2)⊤∈Ω1 satisfies
(3.11)(Ax1)′(t)=λx2(t),x2′(t)=λf(t,x1′(t))+λg(t,x1(t-τ(t)))+λe(t).
We first claim that there is a constant ξ∈ℝ such that
(3.12)|x1(ξ)|≤D.
In view of ∫0ω(Ax1)′(t)dt=0, we know that there exist two constants t1,t2∈[0,ω] such that (Ax1)′(t1)≥0,(Ax1)′(t2)≤0. From the first equation of (3.11), we have x2(t)=(1/λ)(Ax1)′(t), so
(3.13)x2(t1)=1λ(Ax1)′(t1)≥0,x2(t2)=1λ(Ax1)′(t2)≤0.
Let t3,t4∈[0,ω] be, respectively, a global maximum and minimum point of x2(t). Clearly, we have
(3.14)x2(t3)≥0,x2′(t3)=0,x2(t4)≤0,x2′(t4)=0.
Since f(t,x1′)≥0 or f(t,x1′)≤0, w.l.o.g., suppose f(t,x1′)≥0, for (t,x1′)∈[0,ω]×ℝ. Then
(3.15)-g(t3,x1(t3-τ(t3)))-e(t3)=f(t,x1′(t3))≥0,g(t3,x1(t3-τ(t3)))≤-e(t3)≤‖e‖.
From (H2) we see that
(3.16)x1(t3-τ(t3))<D.
Similarly, we have
(3.17)g(t4,x1(t4-τ(t4)))≥-e(t4)≥-‖e‖,
and again by (H2),
(3.18)x1(t4-τ(t4))<-D.
Case 1.
Ifx1(t3-τ(t3))∈(-D,D), define ξ=t3-τ(t3), obviously |x1(ξ)|≤D.
Case 2.
Ifx1(t3-τ(t3))<-D, from (3.18) and the fact that x is a continuous function in ℝ, there exists a constant ξ between x1(t3-τ(t3)) and x1(t4-τ(t4)) such that |x1(ξ)|=D. This proves (3.12).
Choose an integerk and a constant t5∈[0,ω] such that ξ=ωk+t5, then |x1(ξ)|=|x1(t5)|≤D. Hence
(3.19)|x1(t)|≤D+∫0ω|x1′(s)|ds.
Substituting x2(t)=(1/λ)(Ax1)′(t) into the second equation of (3.11) yields
(3.20)(1λ(Ax1)(t))′′=λf(t,x1′(t))+λg(t,x1(t-τ(t)))+λe(t),
that is,
(3.21)((Ax1)(t))′′=λ2f(t,x1′(t))+λ2g(t,x1(t-τ(t)))+λ2e(t).
Integrating both sides of (3.21) over [0,ω], we have
(3.22)∫0ω[f(t,x1′(t))+g(t,x1(t-τ(t)))]dt=0.
On the other hand, multiplying both sides of (3.21) by (Ax1)(t) and integrating over [0,ω], we get
(3.23)∫0ω((Ax1)(t))′′(Ax1(t))dt=-∫0ω|(Ax1)′(t)|2dt=-λ2∫0ωf(t,x1′(t))(Ax1)(t)dt-λ2∫0ωg(t,x1(t-τ(t)))(Ax1)(t)dt-λ2∫0ωe(t)(Ax1)(t)dt.
Using (H1), we have
(3.24)∫0ω|(Ax1)′(t)|2dt≤∫0ω|f(t,x1′(t))||[x1(t)-cx1(t-δ(t))]|dt+∫0ω|g(t,x1(t-τ(t)))||[x1(t)-cx1(t-δ(t))]|dt+∫0ω|e(t)||[x1(t)-cx1(t-δ(t))]|dt≤(1+|c|)‖x1‖[K1∫0ω|x1′(t)|dt+bω+∫0ω|g(t,x1(t-τ(t)))|dt+ω‖e‖].
Besides, we can assert that there exists some positive constant N1 such that
(3.25)∫0ω|g(t,x1(t-τ(t)))|dt≤2ωN1+ωb+K1∫0ω|x1′(t)|dt.
In fact, in view of condition (H1) and (3.22) we have
(3.26)∫0ω{g(t,x1(t-τ(t)))-K1|x1′(t)|-b}dt≤∫0ω{g(t,x1(t-τ(t)))-|f(t,x1′(t))|}dt≤∫0ω{g(t,x1(t-τ(t)))+f(t,x1′(t))}dt=0.
Define
(3.27)E1={t∈[0,ω]:x1(t-τ(t))>D};E2={t∈[0,ω]:|x1(t-τ(t))|≤D}∪{t∈[0,ω]:x1(t-τ(t))<-D}.
With these sets we get
(3.28)∫E2|g(t,x1(t-τ(t)))|dt≤ωmax⁡{M,sup⁡t∈[0,ω],|x1(t-τ(t))|≤D|g(t,x1)|}.∫E1{|g(t,x1(t-τ(t)))|-K1|x1′(t)|-b}dt=∫E1{g(t,x1(t-τ(t)))-K1|x1′(t)|-b}dt≤-∫E2{g(t,x1(t-τ(t)))-K1|x1′(t)|-b}dt≤∫E2{|g(t,x1(t-τ(t)))|+K1|x1′(t)|+b}dt,
which yields
(3.29)∫E1|g(t,x1(t-τ(t)))|dt≤∫E2|g(t,x1(t-τ(t)))|dt+∫E1∪E2(K1|x1′(t)|+b)dt=∫E2|g(t,x1(t-τ(t)))|dt+ωb+K1∫0ω|x1′(t)|dt.
That is,
(3.30)∫0ω|g(t,x1(t-τ(t)))|dt=∫E1|g(t,x1(t-τ(t)))|dt+∫E2|g(x1(t-τ(t)))|dt≤2∫E2|g(t,x1(t-τ(t)))|dt+ωb+K1∫0ω|x1′(t)|dt≤2ωmax⁡{M,sup⁡t∈[0,ω],|x1(t-τ(t))|<D|g(t,x1)|}+ωb+K1∫0ω|x1′(t)|dt=2ωD1+ωb+K1∫0ω|x1′(t)|dt,
where N1=max{M,supt∈[0,ω],|x1(t-τ(t))|<D|g(t,x1)|}, proving (3.25).
Substituting (3.25) into (3.24) and recalling (3.19), we get
(3.31)∫0ω|(Ax1)′(t)|2dt≤(1+|c|)|x1|0(2K1∫0ω|x1′(t)|dt+2ωb+2ωN1+ωmax⁡t∈[0,ω]|e(t)|)=(1+|c|)(2K1|x1|0∫0ω|x1′(t)|dt+2ωb|x1|0+2ωN1|x1|0+ω|x1|0max⁡t∈[0,ω]|e(t)|)≤(1+|c|)[2K1(D+∫0ω|x1′(t)|dt)∫0ω|x1′(t)|dt+(2ωb+2ωN1+ωmax⁡t∈[0,ω]|e(t)|)(D+∫0ω|x1′(t)|dt)]=(1+|c|)[2K1D∫0ω|x1′(t)|dt+2K1(∫0ω|x1′(t)|dt)2+N2∫0ω|x1′(t)|dt+N2D]=2K1(1+|c|)(∫0ω|x1′(t)|dt)2+(1+|c|)(N2+2K1D)∫0ω|x1′(t)|dt+(1+|c|)N2D,
where N2=2ωb+2ωN1+ω∥e∥. Since (Ax)(t)=x(t)-cx(t-δ(t)), we have
(3.32)(Ax1)′(t)=(x1(t)-cx1(t-δ(t)))′=x1′(t)-cx1′(t-δ(t))+cx1′(t-δ(t))δ′(t)=(Ax1′)(t)+cx1′(t-δ(t))δ′(t),(Ax1′)(t)=(Ax1)′(t)-cx1′(t-δ(t))δ′(t).
By applying Lemma 2.1, we have
(3.33)∫0ω|x1′(t)|dt=∫0ω|(A-1Ax1′)(t)|dt≤∫0ω|(Ax1′)(t)|dt|1-|c||=∫0ω|(Ax1)′(t)-cx1′(t-δ(t))δ′(t)|dt|1-|c||≤∫0ω|(Ax1′)(t)|dt+|c|δ1∫0ω|x1′(t)|dt|1-|c||,
where δ1=maxt∈[0,ω]|δ′(t)|. Since 0<ω1/2(1+|c|)1/22K1/(|1-|c||-|c|δ1), then |1-|c||-|c|δ1>0, so we get
(3.34)∫0ω|x1′(t)|dt≤∫0ω|(Ax1)′(t)|dt|1-|c||-|c|δ1≤ω1/2(∫0ω|(Ax1)′(t)|2dt)1/2|1-|c||-|c|δ1.
Applying the inequality (a+b)k≤ak+bk for a,b>0, 0<k<1, it follows from (3.31) and (3.34) that
(3.35)∫0ω|x1′(t)|dt≤ω1/2|1-|c||-|c|δ1[(1+|c|)1/22K1∫0ω|x1′(t)|dt+(1+|c|)1/2(∫0ω|x1′(t)|dt)1/2×(N2+2K1D)1/2+(1+|c|)1/2N2D1/2(1+|c|)1/22K1∫0ω|x1′(t)|dt+(1+|c|)1/2(∫0ω|x1′(t)|dt)1/2].
Since ω1/2(1+|c|)1/22K1/(|1-|c||-|c|δ1)<1, it is easy to see that there exists a constant M1>0 (independent of λ) such that
(3.36)∫0ω|x1′(t)|dt≤M1.
It follows from (3.19) that
(3.37)∥x1∥≤D+∫0ω|x1′(t)|dt≤D+M1∶=M2.
By the first equation of (3.11) we have ∫0ωx2(t)dt=∫0ω(Ax1)′(t)dt=0, which implies that there is a constant t1∈[0,ω] such that x2(t1)=0, hence ∥x2∥≤∫0ω|x2′(t)|dt. By the second equation of (3.11) we obtain
(3.38)x2′(t)=λf(t,x1′(t))+λg(x1(t-τ(t)))+λe(t).
So, from (H1) and (3.25), we have
(3.39)|x2|0≤∫0ω|f(t,x1′(t))|dt+∫0ω|g(t,x1(t-τ(t)))|dt+∫0ω|e(t)|dt≤2K1M1+2ωb+2ωN1+ω‖e‖∶=M3.
Let M4=M22+M32+1,Ω={x=(x1,x2)⊤:∥x1∥<M4,∥x2∥<M4}, then forallx∈∂Ω∩KerL(3.40)Q1Nx=1ω∫0ω(x2(t)f(t,x1′(t))+g(t,x1(t-τ(t)))+e(t))dt.
If Q1Nx=0, then x2(t)=0,x1=M4 or -M4. But if x1(t)=M4, we know
(3.41)0=∫0ωg(M4)dt,
that is, g(M4)=0. From assumption (H2), we know M4≤D, which yields a contradiction, one can argue similarly if  x1=-M4. We also have Q1Nx≠0, that is, forallx∈∂Ω∩KerL,x∉ImL, so conditions (1) and (2) of Lemma 3.1 are both satisfied. Define the isomorphism J:ImQ1→KerL as follows:
(3.42)J(x1,x2)⊤=(x2,x1)⊤.
Let H(μ,x)=μx+(1-μ)JQ1Nx,(μ,x)∈[0,1]×Ω, then, forall(μ,x)∈(0,1)×(∂Ω∩KerL),
(3.43)H(μ,x)=(μx1(t)+1-μω∫0ω[f(t,x1′(t))+g(t,x1(t-τ(t)))+e(t)]dt(μ+(1-μ))x2(t)).
We have ∫0ωe(t)dt=0. So, we can get
(3.44)H(μ,x)=(μx1(t)+1-μω∫0ω[f(t,x1′(t))+g(t,x1(t-τ(t)))]dt(μ+(1-μ))x2(t)),∀(μ,x)∈(0,1)×(∂Ω∩Ker⁡L).
From (H2), it is obvious that x⊤H(μ,x)>0, forall(μ,x)∈(0,1)×(∂Ω∩KerL). Hence
(3.45)deg⁡{JQ1N,Ω∩Ker⁡L,0}=deg⁡{H(0,x),Ω∩Ker⁡L,0}=deg⁡{H(1,x),Ω∩Ker⁡L,0}=deg⁡{I,Ω∩Ker⁡L,0}≠0.
So condition (3) of Lemma 3.1 is satisfied. By applying Lemma 3.1, we conclude that equation Lx=Nx has a solution x=(x1,x2)⊤ on Ω¯∩D(L), that is, (3.1) has an ω-periodic solution x1(t).By using a similar argument, we can obtain the following theorem.Theorem 3.3.
Suppose there exist positive constantsK1,D,M,b with M>∥e∥ such that:
(H1)
|f(t,u)|≤K1|u|+b, for (t,u)∈ℝ×ℝ;(H2)
sgnx·g(t,x)>∥e∥, for |x|>D,(H3)
g(t,x)≤M, for x≥D and t∈ℝ,
then (3.1) has at least one solution with period ω if 0<ω(1+|c|)1/22K1/(|1-|c||-|c|δ1)<1.Remark 3.4.
If∫0ωe(t)dt≠0 and f(t,0)≠0, the problem of existence of ω-periodic solutions to (3.1) can be converted to the existence of ω-periodic solutions to
(3.46)(x(t)-cx(t-δ(t)))′′=f1(t,x′(t))+g1(t,x(t-τ(t)))+e1(t),
where f1(t,x)=f(t,x)-f(t,0),g1(t,x)=g(t,x)+(1/ω)∫0ωe(t)dt+f(t,0), and e1(t)=e(t)-(1/ω)∫0ωe(t)dt. Clearly, ∫0ωe1(t)dt=0 and f1(t,0)=0, and (3.46) can be discussed by using Theorem 3.2 (or Theorem 3.3).


## 4. Positive Periodic Solutions for Neutral Equations

Consider the following second-order neutral functional differential equation:(4.1)(x(t)-cx(t-δ(t)))′′=-a(t)x(t)+λb(t)f(x(t-τ(t))),
where λ is a positive parameter; f∈C(ℝ,[0,∞)), and f(x)>0 for x>0; a∈C(ℝ,(0,∞)) with max{a(t):t∈[0,ω]}<(π/ω)2, b∈C(ℝ,(0,∞)), τ∈C(ℝ,ℝ), a(t), b(t), and τ(t) are ω-periodic functions.Define the Banach spaceX as in Section 2, and let Cω+={x∈C(ℝ,(0,∞)):x(t+ω)=x(t)}. Denote (4.2)M=max⁡{a(t):t∈[0,ω]},m=min⁡{a(t):t∈[0,ω]},β=M,L=12βsin⁡(βω/2),l=cos⁡⁡(βω/2)2βsin⁡(βω/2),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 (4.1). First let (4.3)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 (4.4)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 (4.1) has i¯0 or i̲∞ positive w-periodic solutions for sufficiently large or small λ, respectively.In the following we discuss (4.1) 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 as defined in Section 2, 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 (4.1) can be transformed into(4.5)y′′(t)+a(t)(A-1y)(t)=λb(t)f((A-1y)(t-τ(t))),
which can be further rewritten as(4.6)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

Assumec<0 and c>-min{k1,m/(M+m)}.Lemma 4.1 (see [7]).
The equation(4.7)y′′(t)+My(t)=h(t),h∈Cω+,
has a unique ω-periodic solution
(4.8)y(t)=∫tt+ωG(t,s)h(s)ds,
where
(4.9)G(t,s)=cos⁡β((ω/2)+t-s)2βsin⁡(βω/2),s∈[t,t+ω].Lemma 4.2 (see [7]).
One has∫tt+ωG(t,s)ds=1/M. Furthermore, if max{a(t):t∈[0,ω]}<(π/ω)2, then 0<l≤G(t,s)≤L for all t∈[0,ω] and s∈[t,t+ω].Now we consider(4.10)y′′(t)+a(t)y(t)-a(t)H(y(t))=h(t),h∈Cω+,
and define operators T,Ĥ:X→X by (4.11)(Th)(t)=∫tt+ωG(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 Lemma4.1, the solution of (4.10) can be written in the form(4.12)y(t)=(Th)(t)+(TĤy)(t).
In view of c<0 and c>-min{k1,m/(M+m)}, we have(4.13)‖TĤ‖≤‖T‖‖Ĥ‖≤M-m+m|c|M(1-|c|)<1,
and hence (4.14)y(t)=(I-TĤ)-1(Th)(t).
Define an operator P:X→X by (4.15)(Ph)(t)=(I-TĤ)-1(Th)(t).
Obviously, for any h∈Cω+, if max{a(t):t∈[0,ω]}<(π/ω)2, y(t)=(Ph)(t) is the unique positive ω-periodic solution of (4.10).Lemma 4.3.
P is completely continuous and
(4.16)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖,∀h∈Cω+.Proof.
By the Neumann expansion ofP, we have
(4.17)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 (4.17), and recalling that ∥TĤ∥≤(M-m+m|c|)/M(1-|c|)<1, we get
(4.18)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖.Define an operatorQ:X→X by(4.19)Qy(t)=P(λb(t)f((A-1y)(t-τ(t)))).Lemma 4.4.
One hasQ(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 4.3 that
(4.20)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG(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,
(4.21)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+ωG(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,
(4.22)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 (4.6) to (4.10), it is obvious that the existence of periodic solutions for (4.6) is equivalent to the existence of fixed points for the operator Q in X. Recalling Lemma 4.4, the existence of positive periodic solutions for (4.6) 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 (4.1).Lemma 4.5.
If there existsη>0 such that
(4.23)f((A-1y)(t-τ(t)))≥(A-1y)(t-τ(t))η,fort∈[0,ω],y∈K,
then
(4.24)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Proof.
By Lemmas2.2, 4.2, and 4.3, we have for y∈K that
(4.25)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG(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
(4.26)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Lemma 4.6.
If there existsɛ>0 such that
(4.27)f((A-1y)(t-τ(t)))≤(A-1y)(t-τ(t))ɛ,fort∈[0,ω],y∈K,
then
(4.28)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖,y∈K.Proof.
By Lemmas2.2, 4.2, and 4.3, we have
(4.29)‖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(4.30)F(r)=max⁡{f(t):0≤t≤r1-|c|},f1(r)=min⁡{f(t):α-|c|1-c2r≤t≤r1-|c|}.Lemma 4.7.
Ify∈∂Kr, then
(4.31)‖Qy‖≥λlf1(r)∫0ωb(s)ds.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 4.5.Lemma 4.8.
Ify∈∂Kr, then
(4.32)‖Qy‖≤λLM(1-|c|)F(r)m-(M+m)|c|∫0ωb(s)ds.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 4.6, we get the conclusion.We quote the fixed point theorem which our results will be based on.Lemma 4.9 (see [9]).
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 (4.1).Theorem 4.10.
(a) Ifi¯0=1 or 2, then (4.1) has i¯0 positive ω-periodic solutions for λ>1/f1(1)l∫0ωb(s)ds>0;
(b) Ifi̲∞=1 or 2, then (4.1) 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 (4.1) has no positive ω-periodic solutions for sufficiently small or sufficiently large λ>0, respectively.Proof.
(a) Chooser1=1. Take λ0=1/f1(r1)l∫0ωb(s)ds>0, then for all λ>λ0, we have from Lemma 4.7 that
(4.33)‖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
(4.34)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 4.6 and (4.34), we have for y∈∂Kr2 that
(4.35)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
It follows from Lemma 4.9 and (4.33) that
(4.36)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 (4.1) 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
(4.37)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 4.6 and (4.37), we have for y∈∂Kr3 that
(4.38)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
Recalling from Lemma 4.9 and (4.33) that
(4.39)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 (4.1) 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 (4.1) for λ>λ0.
(b) Letr1=1. Take λ0=(m-(M+m)|c|)/LM(1-|c|)F(r1)∫0ωb(s)ds>0; then by Lemma 4.8 we know if λ<λ0 then
(4.40)‖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
(4.41)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 4.5 and (4.41),
(4.42)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 and (4.40) that
(4.43)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 (4.1) 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
(4.44)λlηα-|c|1-c2∫0ωb(s)ds>1.
Letting 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 4.5 and (4.44), we have for y∈∂Kr3 that
(4.45)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 and (4.40) that
(4.46)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 (4.1) 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 (4.1) 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. Let b1=min{f(u)/u;u>0}>0, then we obtain
(4.47)f(u)≥b1u,u∈[0,+∞).
Assume y(t) is a positive ω-periodic solution of (4.1) 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 4.5, if λ>λ0 we have
(4.48)‖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¯∞<∞. Let b2=max{f(u)/u:u>0}>0, then we obtain
(4.49)f(u)≤b2u,u∈[0,∞).
Assume y(t) is a positive ω-periodic solution of (4.1) 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 4.6 that
(4.50)‖y‖=‖Qy‖≤λb2LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖,
which is a contradiction.Theorem 4.11.
(a) If there exists a constantb1>0 such that f(u)≥b1u for u∈[0,+∞), then (4.1) 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 (4.1) has no positive ω-periodic solution for 0<λ<(m-(M+m)|c|)/b2LM∫0ωb(s)ds.Proof.
From the proof of (c) in Theorem4.10, we obtain this theorem immediately.Theorem 4.12.
Assumei̲0=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(4.51)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 (4.1) has one positive ω-periodic solution.Proof.
We have the following cases.
Case 1.
Iff¯0≤f̲∞, then
(4.52)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 an 0<ɛ<f∞ such that
(4.53)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 4.6 we have for y∈∂Kr1 that
(4.54)‖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 4.5, for y∈∂Kr2(4.55)‖Qy‖≥λl(f¯∞-ɛ)α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 that
(4.56)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 (4.1).
Case 2.
Iff̲0>f¯∞, in this case, we have
(4.57)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 an 0<ɛ<f0 such that
(4.58)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 4.5 that for y∈∂Kr1(4.59)‖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 4.6, for y∈∂Kr2(4.60)‖Qy‖≤λ(f̲∞+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖.
It follows from Lemma 4.9 that
(4.61)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 (4.1).
Case 3.
One hasf̲0≤f̲∞≤f¯0≤f¯∞. The proof is the same as in Case 1.
Case 4.
One hasf̲∞≤f̲0≤f¯∞≤f¯0. The proof is the same as in Case 2.


### 4.2. Case  II

Assumec>0 and c<min{m/(M+m),(LM-lm)/(L-l)M-lm}.Define(4.62)f2(r)=min⁡{f(t):α1-cr≤t≤r1-c}.
Similarly as in Section 4.1, we get the following results.Theorem 4.13.
(a) Ifi¯0=1 or 2, then (4.1) has i0 positive ω-periodic solutions for λ>1/f2(1)l∫0ωb(s)ds>0.
(b) Ifi̲∞=1 or 2, then (4.1) 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 (4.1) has no positive ω-periodic solution for sufficiently small or large λ>0, respectively.Theorem 4.14.
(a) If there exists a constantb1>0 such that f(u)≥b1u for u∈[0,+∞), then (4.1) 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 (4.1) has no positive ω-periodic solution for 0<λ<(m-(M+m)c)/b2LM∫0ωb(s)ds.Theorem 4.15.
Assumei̲0=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(4.63)1-clα∫0ωb(s)dsmax⁡{f̲0,f¯0,f̲∞,f¯∞}<λ<m-(M+m)cLM∫0ωb(s)dsmin⁡{f̲0,f¯0,f̲∞,f¯∞},
then (4.1) has one positive ω-periodic solution.Remark 4.16.
In a similar way, one can consider the second-order neutral functional differential equation(x(t)-cx(t-δ(t)))′′-a(t)x(t)=-λb(t)f(x(t-τ(t))).


## 4.1. Case  I

Assumec<0 and c>-min{k1,m/(M+m)}.Lemma 4.1 (see [7]).
The equation(4.7)y′′(t)+My(t)=h(t),h∈Cω+,
has a unique ω-periodic solution
(4.8)y(t)=∫tt+ωG(t,s)h(s)ds,
where
(4.9)G(t,s)=cos⁡β((ω/2)+t-s)2βsin⁡(βω/2),s∈[t,t+ω].Lemma 4.2 (see [7]).
One has∫tt+ωG(t,s)ds=1/M. Furthermore, if max{a(t):t∈[0,ω]}<(π/ω)2, then 0<l≤G(t,s)≤L for all t∈[0,ω] and s∈[t,t+ω].Now we consider(4.10)y′′(t)+a(t)y(t)-a(t)H(y(t))=h(t),h∈Cω+,
and define operators T,Ĥ:X→X by (4.11)(Th)(t)=∫tt+ωG(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 Lemma4.1, the solution of (4.10) can be written in the form(4.12)y(t)=(Th)(t)+(TĤy)(t).
In view of c<0 and c>-min{k1,m/(M+m)}, we have(4.13)‖TĤ‖≤‖T‖‖Ĥ‖≤M-m+m|c|M(1-|c|)<1,
and hence (4.14)y(t)=(I-TĤ)-1(Th)(t).
Define an operator P:X→X by (4.15)(Ph)(t)=(I-TĤ)-1(Th)(t).
Obviously, for any h∈Cω+, if max{a(t):t∈[0,ω]}<(π/ω)2, y(t)=(Ph)(t) is the unique positive ω-periodic solution of (4.10).Lemma 4.3.
P is completely continuous and
(4.16)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖,∀h∈Cω+.Proof.
By the Neumann expansion ofP, we have
(4.17)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 (4.17), and recalling that ∥TĤ∥≤(M-m+m|c|)/M(1-|c|)<1, we get
(4.18)(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖.Define an operatorQ:X→X by(4.19)Qy(t)=P(λb(t)f((A-1y)(t-τ(t)))).Lemma 4.4.
One hasQ(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 4.3 that
(4.20)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG(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,
(4.21)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+ωG(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,
(4.22)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 (4.6) to (4.10), it is obvious that the existence of periodic solutions for (4.6) is equivalent to the existence of fixed points for the operator Q in X. Recalling Lemma 4.4, the existence of positive periodic solutions for (4.6) 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 (4.1).Lemma 4.5.
If there existsη>0 such that
(4.23)f((A-1y)(t-τ(t)))≥(A-1y)(t-τ(t))η,fort∈[0,ω],y∈K,
then
(4.24)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Proof.
By Lemmas2.2, 4.2, and 4.3, we have for y∈K that
(4.25)Qy(t)=P(λb(t)f((A-1y)(t-τ(t))))≥T(λb(t)f((A-1y)(t-τ(t))))=λ∫tt+ωG(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
(4.26)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.Lemma 4.6.
If there existsɛ>0 such that
(4.27)f((A-1y)(t-τ(t)))≤(A-1y)(t-τ(t))ɛ,fort∈[0,ω],y∈K,
then
(4.28)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖,y∈K.Proof.
By Lemmas2.2, 4.2, and 4.3, we have
(4.29)‖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(4.30)F(r)=max⁡{f(t):0≤t≤r1-|c|},f1(r)=min⁡{f(t):α-|c|1-c2r≤t≤r1-|c|}.Lemma 4.7.
Ify∈∂Kr, then
(4.31)‖Qy‖≥λlf1(r)∫0ωb(s)ds.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 4.5.Lemma 4.8.
Ify∈∂Kr, then
(4.32)‖Qy‖≤λLM(1-|c|)F(r)m-(M+m)|c|∫0ωb(s)ds.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 4.6, we get the conclusion.We quote the fixed point theorem which our results will be based on.Lemma 4.9 (see [9]).
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 (4.1).Theorem 4.10.
(a) Ifi¯0=1 or 2, then (4.1) has i¯0 positive ω-periodic solutions for λ>1/f1(1)l∫0ωb(s)ds>0;
(b) Ifi̲∞=1 or 2, then (4.1) 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 (4.1) has no positive ω-periodic solutions for sufficiently small or sufficiently large λ>0, respectively.Proof.
(a) Chooser1=1. Take λ0=1/f1(r1)l∫0ωb(s)ds>0, then for all λ>λ0, we have from Lemma 4.7 that
(4.33)‖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
(4.34)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 4.6 and (4.34), we have for y∈∂Kr2 that
(4.35)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
It follows from Lemma 4.9 and (4.33) that
(4.36)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 (4.1) 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
(4.37)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 4.6 and (4.37), we have for y∈∂Kr3 that
(4.38)‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
Recalling from Lemma 4.9 and (4.33) that
(4.39)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 (4.1) 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 (4.1) for λ>λ0.
(b) Letr1=1. Take λ0=(m-(M+m)|c|)/LM(1-|c|)F(r1)∫0ωb(s)ds>0; then by Lemma 4.8 we know if λ<λ0 then
(4.40)‖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
(4.41)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 4.5 and (4.41),
(4.42)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 and (4.40) that
(4.43)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 (4.1) 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
(4.44)λlηα-|c|1-c2∫0ωb(s)ds>1.
Letting 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 4.5 and (4.44), we have for y∈∂Kr3 that
(4.45)‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 and (4.40) that
(4.46)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 (4.1) 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 (4.1) 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. Let b1=min{f(u)/u;u>0}>0, then we obtain
(4.47)f(u)≥b1u,u∈[0,+∞).
Assume y(t) is a positive ω-periodic solution of (4.1) 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 4.5, if λ>λ0 we have
(4.48)‖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¯∞<∞. Let b2=max{f(u)/u:u>0}>0, then we obtain
(4.49)f(u)≤b2u,u∈[0,∞).
Assume y(t) is a positive ω-periodic solution of (4.1) 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 4.6 that
(4.50)‖y‖=‖Qy‖≤λb2LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖,
which is a contradiction.Theorem 4.11.
(a) If there exists a constantb1>0 such that f(u)≥b1u for u∈[0,+∞), then (4.1) 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 (4.1) has no positive ω-periodic solution for 0<λ<(m-(M+m)|c|)/b2LM∫0ωb(s)ds.Proof.
From the proof of (c) in Theorem4.10, we obtain this theorem immediately.Theorem 4.12.
Assumei̲0=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(4.51)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 (4.1) has one positive ω-periodic solution.Proof.
We have the following cases.
Case 1.
Iff¯0≤f̲∞, then
(4.52)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 an 0<ɛ<f∞ such that
(4.53)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 4.6 we have for y∈∂Kr1 that
(4.54)‖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 4.5, for y∈∂Kr2(4.55)‖Qy‖≥λl(f¯∞-ɛ)α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 that
(4.56)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 (4.1).
Case 2.
Iff̲0>f¯∞, in this case, we have
(4.57)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 an 0<ɛ<f0 such that
(4.58)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 4.5 that for y∈∂Kr1(4.59)‖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 4.6, for y∈∂Kr2(4.60)‖Qy‖≤λ(f̲∞+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖.
It follows from Lemma 4.9 that
(4.61)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 (4.1).
Case 3.
One hasf̲0≤f̲∞≤f¯0≤f¯∞. The proof is the same as in Case 1.
Case 4.
One hasf̲∞≤f̲0≤f¯∞≤f¯0. The proof is the same as in Case 2.


## 4.2. Case  II

Assumec>0 and c<min{m/(M+m),(LM-lm)/(L-l)M-lm}.Define(4.62)f2(r)=min⁡{f(t):α1-cr≤t≤r1-c}.
Similarly as in Section 4.1, we get the following results.Theorem 4.13.
(a) Ifi¯0=1 or 2, then (4.1) has i0 positive ω-periodic solutions for λ>1/f2(1)l∫0ωb(s)ds>0.
(b) Ifi̲∞=1 or 2, then (4.1) 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 (4.1) has no positive ω-periodic solution for sufficiently small or large λ>0, respectively.Theorem 4.14.
(a) If there exists a constantb1>0 such that f(u)≥b1u for u∈[0,+∞), then (4.1) 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 (4.1) has no positive ω-periodic solution for 0<λ<(m-(M+m)c)/b2LM∫0ωb(s)ds.Theorem 4.15.
Assumei̲0=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(4.63)1-clα∫0ωb(s)dsmax⁡{f̲0,f¯0,f̲∞,f¯∞}<λ<m-(M+m)cLM∫0ωb(s)dsmin⁡{f̲0,f¯0,f̲∞,f¯∞},
then (4.1) has one positive ω-periodic solution.Remark 4.16.
In a similar way, one can consider the second-order neutral functional differential equation(x(t)-cx(t-δ(t)))′′-a(t)x(t)=-λb(t)f(x(t-τ(t))).


## 5. Examples

Example 5.1.
Consider the following equation:(5.1)(x(t)-15x(t-160sin4t))′′=x′(t)sin4t+arctan(x(t-sin4t)1+cos⁡3(4t))+cos⁡4t.
Comparing (5.1) to (3.1), we have ω=π/2, f(t,x)=x(t)sin4t, g(t,x)=arctan(x/(1+cos3(4t))), c=15, δ(t)=(1/60)sin4t, τ(t)=sin4t, e(t)=cos4t and δ1=maxt∈[0,ω]|(1/15)cos4t|=1/15, and we can easily choose D>π/2 and M=π/2 such that (H2)and(H3) holds. Regarding assumption (H1) note that
(5.2)|f(t,x′(t))|≤|x′(t)|,
that is, (H1) holds with K1=1,b=0, and
(5.3)ω1/2(1+|c|)1/22K1|1-|c||-|c|δ1=π/2(1+15)1/22|1-15|-(1/15)⋅15=4π13<1.
Hence by Theorem 3.2, (5.1) has at least one π/2-periodic solution.Example 5.2.
Consider the following neutral functional differential equation:(5.4)(u(t)+730u(t-sint))′′+116u(t)=λ(1-sint)u2(t-τ(t))au(t-τ(t)),
where λ and 0<a<1 are two positive parameters, τ(t+2π)=τ(t).
Comparing (5.4) to (4.1), we see that δ(t)=sint, c=-7/30, a(t)≡1/16, b(t)=1-sint, ω=2π, f(u)=u2au. Clearly, M=1/16<(π/2π)2=1/4, f¯0=0, f¯∞=0, i¯0=2. By Theorem 4.10, we easily get the following conclusion: (5.4) has two positive ω-periodic solutions for λ>1/4πr1, where r1=min{f(0.27),f(30/23)}.
In fact, by simple computations, we have(5.5)M=m=116,β=14,L=12βsin⁡(β2π/2)=22,l=cos⁡(β2π/2)(2βsin(β2π/2))=2,k=2+28,k1=2+1-32,α=8232,|c|=730<min{k1,mM+m}=2+1-32,|c|=730<8232=α,f1(1)=min⁡{f(t):0.27≈(8/23)2-(7/30)1-(7/30)2≤t≤3023}=min⁡{f(0.27),f(3023)}=r1,1f1(1)l∫0ωb(s)ds=14πr1.



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*Source: 969276-2011-08-07.xml*