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.

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 Section 2, 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

Let Cω={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 (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

Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M41"><mml:mo stretchy="false">|</mml:mo><mml:mi>c</mml:mi><mml:mo stretchy="false">|</mml:mo><mml:mo><</mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>).

Let t-δ(t)=s and Dj=s-∑i=1j-1δ(Di), j=1,2,…. Therefore,
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
A-1=(I-B)-1=I+∑j=1∞Bj=∑j=0∞Bj,
where B0=I. Then
(A-1f(t))=∑j=0∞[Bjf](t)=∑j=0∞cjf(s-∑i=1j-1δ(Di)),
and consequently
|(A-1f)(t)|=|∑j=0∞[Bjf](t)|=|∑j=0∞cjf(s-∑i=1j-1δ(Di))|≤‖f‖1-|c|.
Moreover,
∫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 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M55"><mml:mo stretchy="false">|</mml:mo><mml:mi>c</mml:mi><mml:mo stretchy="false">|</mml:mo><mml:mo>></mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>).

Let
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
(B1jf)(t)=1cjf(s+∑i=1j-1δ(Di)),
where Di is defined as in Case 1. Summing over j yields
∑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,
E-1:Cω⟶Cω,E-1=(I-B1)-1=I+∑j=1B1j,
and forallf∈Cω we get
(E-1f)(t)=f(t)+∑j=1∞(B1jf)(t).
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)],
that is,
(Ax)(t)=-c(Ex)(t-δ(t)).
Let f∈Cω be arbitrary. We are looking for x such that
(Ax)(t)=f(t).
that is,
-c(Ex)(t-δ(t))=f(t).
Therefore,
(Ex)(t)=-f(t+δ(t))c=∶f1(t),
and hence
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
[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.

If c<0 and |c|<α, one has for y∈K that
α-|c|1-c2‖y‖≤(A-1y)(t)≤11-|c|‖y‖.

Proof.

Since c<0 and |c|<α<1, by Lemma 2.1, we have for y∈K that
(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.

If c>0 and c<1 then for y∈K one has
α1-c‖y‖≤(A-1y)(t)≤11-c‖y‖.

Proof.

Since c>0 and c<1, α<1, by Lemma 2.1, we have for y∈K that
(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(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. Let X 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.

Let Ω 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 [<xref ref-type="bibr" rid="B1">8</xref>]).

Suppose that X 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:

Lx≠λNx,forallx∈∂Ω∩D(L),λ∈(0,1);

Nx∉ImL,forallx∈∂Ω∩KerL;

deg{JQ1N,Ω∩KerL,0}≠0, where J:ImQ1→KerL is an isomorphism.

Then the equation Lx=Nx has a solution in Ω¯∩D(L).

In order to use Mawhin's continuation theorem to study the existence of ω-periodic solutions for (3.1), we rewrite (3.1) in the following form:(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 that Cω={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 L:D(L)={x∈C1(R,R2):x(t+ω)=x(t),t∈R}⊂X⟶Y
by (Lx)(t)=((Ax1)′(t)x2′(t))
and N:X→Y by(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 KerL≅R2,ImL={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 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 [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 constants K1,D,M,b with M>∥e∥ such that:

(H_{1})|f(t,u)|≤K1|u|+b, for (t,u)∈ℝ×ℝ;

(H_{2}) sgnx·g(t,x)>∥e∥, for |x|>D;

(H_{3})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
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
Ω1={x∈Cω:Lx=λNx}.
Then x=(x1,x2)⊤∈Ω1 satisfies
(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
|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
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
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
-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
x1(t3-τ(t3))<D.
Similarly, we have
g(t4,x1(t4-τ(t4)))≥-e(t4)≥-‖e‖,
and again by (H2),
x1(t4-τ(t4))<-D.

Case 1.

If x1(t3-τ(t3))∈(-D,D), define ξ=t3-τ(t3), obviously |x1(ξ)|≤D.

Case 2.

If x1(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 integer k and a constant t5∈[0,ω] such that ξ=ωk+t5, then |x1(ξ)|=|x1(t5)|≤D. Hence
|x1(t)|≤D+∫0ω|x1′(s)|ds.
Substituting x2(t)=(1/λ)(Ax1)′(t) into the second equation of (3.11) yields
(1λ(Ax1)(t))′′=λf(t,x1′(t))+λg(t,x1(t-τ(t)))+λe(t),
that is,
((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
∫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
∫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
∫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
∫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
∫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
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
∫E2|g(t,x1(t-τ(t)))|dt≤ωmax{M,supt∈[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
∫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,
∫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,supt∈[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
∫0ω|(Ax1)′(t)|2dt≤(1+|c|)|x1|0(2K1∫0ω|x1′(t)|dt+2ωb+2ωN1+ωmaxt∈[0,ω]|e(t)|)=(1+|c|)(2K1|x1|0∫0ω|x1′(t)|dt+2ωb|x1|0+2ωN1|x1|0+ω|x1|0maxt∈[0,ω]|e(t)|)≤(1+|c|)[2K1(D+∫0ω|x1′(t)|dt)∫0ω|x1′(t)|dt+(2ωb+2ωN1+ωmaxt∈[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
(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
∫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
∫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
∫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
∫0ω|x1′(t)|dt≤M1.
It follows from (3.19) that
∥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
x2′(t)=λf(t,x1′(t))+λg(x1(t-τ(t)))+λe(t).
So, from (H1) and (3.25), we have
|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∩KerLQ1Nx=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
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:
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),
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
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∩KerL).
From (H2), it is obvious that x⊤H(μ,x)>0, forall(μ,x)∈(0,1)×(∂Ω∩KerL). Hence
deg{JQ1N,Ω∩KerL,0}=deg{H(0,x),Ω∩KerL,0}=deg{H(1,x),Ω∩KerL,0}=deg{I,Ω∩KerL,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 constants K1,D,M,b with M>∥e∥ such that:

|f(t,u)|≤K1|u|+b, for (t,u)∈ℝ×ℝ;

sgnx·g(t,x)>∥e∥, for |x|>D,

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
(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:(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 space X as in Section 2, and let Cω+={x∈C(ℝ,(0,∞)):x(t+ω)=x(t)}. Denote 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 f¯0=lim¯x→0f(x)x,f¯∞=lim¯x→∞f(x)x,f̲0=lim̲x→0f(x)x,f̲∞=lim̲x→∞f(x)x,
and denote 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.

Let K={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}.

Let y(t)=(Ax)(t), then from Lemma 2.1 we have x(t)=(A-1y)(t). Hence (4.1) can be transformed intoy′′(t)+a(t)(A-1y)(t)=λb(t)f((A-1y)(t-τ(t))),
which can be further rewritten asy′′(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

Assume c<0 and c>-min{k1,m/(M+m)}.

Lemma 4.1 (see [<xref ref-type="bibr" rid="B2">7</xref>]).

The equation
y′′(t)+My(t)=h(t),h∈Cω+,
has a unique ω-periodic solution
y(t)=∫tt+ωG(t,s)h(s)ds,
where
G(t,s)=cosβ((ω/2)+t-s)2βsin(βω/2),s∈[t,t+ω].

Lemma 4.2 (see [<xref ref-type="bibr" rid="B2">7</xref>]).

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 considery′′(t)+a(t)y(t)-a(t)H(y(t))=h(t),h∈Cω+,
and define operators T,Ĥ:X→X by (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 Lemma 4.1, the solution of (4.10) can be written in the formy(t)=(Th)(t)+(TĤy)(t).
In view of c<0 and c>-min{k1,m/(M+m)}, we have‖TĤ‖≤‖T‖‖Ĥ‖≤M-m+m|c|M(1-|c|)<1,
and hence y(t)=(I-TĤ)-1(Th)(t).
Define an operator P:X→X by (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
(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖,∀h∈Cω+.

Proof.

By the Neumann expansion of P, we have
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
(Th)(t)≤(Ph)(t)≤M(1-|c|)m-(M+m)|c|‖Th‖.

Define an operator Q:X→X byQy(t)=P(λb(t)f((A-1y)(t-τ(t)))).

Lemma 4.4.

One has 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 4.3 that
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,
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|maxt∈[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,
Qy(t)≥l[m-(M+m)|c|]LM(1-|c|)‖Qy‖=α‖Qy‖,
that is, Q(K)⊂K.

From the continuity of P, 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 exists η>0 such that
f((A-1y)(t-τ(t)))≥(A-1y)(t-τ(t))η,fort∈[0,ω],y∈K,
then
‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.

Proof.

By Lemmas 2.2, 4.2, and 4.3, we have for y∈K that
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
‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖,y∈K.

Lemma 4.6.

If there exists ɛ>0 such that
f((A-1y)(t-τ(t)))≤(A-1y)(t-τ(t))ɛ,fort∈[0,ω],y∈K,
then
‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖,y∈K.

Proof.

By Lemmas 2.2, 4.2, and 4.3, we have
‖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‖.

By Lemma 2.2, we obtain ((α-|c|)/(1-c2))r≤(A-1y)(t-τ(t))≤r/(1-|c|) for y∈∂Kr, which yields f((A-1y)(t-τ(t)))≥f1(r). The lemma now follows analog to the proof of Lemma 4.5.

Lemma 4.8.

If y∈∂Kr, then
‖Qy‖≤λLM(1-|c|)F(r)m-(M+m)|c|∫0ωb(s)ds.

Proof.

By Lemma 2.2, we can have 0≤(A-1y)(t-τ(t))≤r/(1-|c|) for y∈∂Kr, which yields f((A-1y)(t-τ(t)))≤F(r). Similar to the proof of Lemma 4.6, we get the conclusion.

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

Lemma 4.9 (see [<xref ref-type="bibr" rid="B3">9</xref>]).

Let X 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}.

If ∥Tx∥≥∥x∥ for x∈∂Kr, then i(T,Kr,K)=0.

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) If i¯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) If i̲∞=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) If i¯∞=0 or i̲0=0, then (4.1) has no positive ω-periodic solutions for sufficiently small or sufficiently large λ>0, respectively.

Proof.

(a) Choose r1=1. Take λ0=1/f1(r1)l∫0ωb(s)ds>0, then for all λ>λ0, we have from Lemma 4.7 that
‖Qy‖>‖y‖,fory∈∂Kr1.

Case 1.

If f¯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
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
‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
It follows from Lemma 4.9 and (4.33) that
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.

If f¯∞=0, there exists a constant H̃>0 such that f(u)≤ɛu for u≥H̃, where the constant ɛ>0 satisfies
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
‖Qy‖≤λɛLM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.
Recalling from Lemma 4.9 and (4.33) that
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.

If f¯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) Let r1=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
‖Qy‖<‖y‖,y∈∂Kr1.

Case 1.

If f̲0=∞, we can choose 0<r¯2<r1 so that f(u)≥ηu for 0≤u≤r¯2, where the constant η>0 satisfies
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),
‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 and (4.40) that
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.

If f̲∞=∞, there exists a constant H̃>0 such that f(u)≥ηu for u≥H̃, where the constant η>0 satisfies
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
‖Qy‖≥λlηα-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 and (4.40) that
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.

If f̲0=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 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 i̲0=0, we have f̲0>0 and f̲∞>0. Let b1=min{f(u)/u;u>0}>0, then we obtain
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
‖y‖=‖Qy‖≥λlb1α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖,
which is a contradiction.

Case 2.

If i¯∞=0, we have f¯0<∞ and f¯∞<∞. Let b2=max{f(u)/u:u>0}>0, then we obtain
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
‖y‖=‖Qy‖≤λb2LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖,
which is a contradiction.

Theorem 4.11.

(a) If there exists a constant b1>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 constant b2>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 Theorem 4.10, we obtain this theorem immediately.

Theorem 4.12.

Assume i̲0=i¯0=i̲∞=i¯∞=0 and that one of the following conditions holds:

f¯0≤f̲∞;

f̲0>f¯∞;

f̲0≤f̲∞≤f¯0≤f¯∞;

f̲∞≤f̲0≤f¯∞≤f¯0.

If
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.

If f¯0≤f̲∞, then
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
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
‖Qy‖≤λ(f̲0+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖<‖y‖.

On the other hand, there exists a constant H̃>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‖Qy‖≥λl(f¯∞-ɛ)α-|c|1-c2∫0ωb(s)ds‖y‖>‖y‖.
It follows from Lemma 4.9 that
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.

If f̲0>f¯∞, in this case, we have
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
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‖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‖Qy‖≤λ(f̲∞+ɛ)LM∫0ωb(s)dsm-(M+m)|c|‖y‖.
It follows from Lemma 4.9 that
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 has f̲0≤f̲∞≤f¯0≤f¯∞. The proof is the same as in Case 1.

Case 4.

One has f̲∞≤f̲0≤f¯∞≤f¯0. The proof is the same as in Case 2.

4.2. Case II

Assume c>0 and c<min{m/(M+m),(LM-lm)/(L-l)M-lm}.

Define 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) If i¯0=1 or 2, then (4.1) has i0 positive ω-periodic solutions for λ>1/f2(1)l∫0ωb(s)ds>0.

(b) If i̲∞=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) If i¯∞=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 constant b1>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 constant b2>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.

Assume i̲0=i¯0=i̲∞=i¯∞=0 hold and that one of the following conditions holds:

f¯0≤f̲∞;

f̲0>f¯∞;

f̲0≤f̲∞≤f¯0≤f¯∞;

f̲∞≤f̲0≤f¯∞≤f¯0.

If
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. ExamplesExample 5.1.

Consider the following equation:(x(t)-15x(t-160sin4t))′′=x′(t)sin4t+arctan(x(t-sin4t)1+cos3(4t))+cos4t.
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
|f(t,x′(t))|≤|x′(t)|,
that is, (H1) holds with K1=1,b=0, and
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:
(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
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.

Acknowledgment

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

ZhangM. R.Periodic solutions of linear and quasilinear neutral functional differential equationsLuS.GeW.Periodic solutions of neutral differential equation with multiple deviating argumentsWangQ.DaiB.Three periodic solutions of nonlinear neutral functional differential equationsWuJ.WangZ.Two periodic solutions of second-order neutral functional differential equationsWuJ.LiuY.Two periodic solutions of neutral difference systems depending on two parametersLuS.GeW.Existence of periodic solutions for a kind of second-order neutral functional differential equationCheungW. S.RenJ.HanW.Positive periodic solution of second-order neutral functional differential equationsGainesR. E.MawhinJ. L.KrasnoselskiiM.