Neutral operator with variable parameter and third-order neutral differential equation

Abstract

In this article, we discuss the properties of the neutral operator with variable parameter $(Ax)(t)=x(t)-c(t)x(t-\delta (t))$ and by applying Green’s function of a third-order differential equation and a fixed point theorem in cones, we obtain some sufficient conditions for existence, nonexistence, multiplicity of positive periodic solutions for a generalized third-order neutral differential equation.

1 Introduction

In [1], Zhang discussed the properties of the neutral operator $(A_{1}x)(t)=x(t)-cx(t-\delta )$, which became an effective tool for the research on differential equations with this prescribed neutral operator (see, e.g., [2–4]). Lu and Ge [5] investigated an extension of $A_1$, namely the neutral operator $(A_{2}x)(t)=x(t)-{\sum }_{i=1}^n{c}_{i}x(t-{\delta }_i)$, and obtained the existence of periodic solutions for the corresponding neutral differential equation. Afterwards, Du et al. [6] studied the neutral operator $(A_{3}x)(t)=x(t)-c(t)x(t-\delta )$, here $c(t)$ is ω-periodic functions. By means of Mawhin’s continuation theorem and the properties of $A_3$, they obtained sufficient conditions for the existence of periodic solutions to a Liénard neutral differential equation. Recently, in [7], Ren et al. investigated the neutral operator with variable delay $(A_4)x(t)-cx(t-\delta (t))$. By applying coincidence degree theory, they obtained sufficient conditions for the existence of periodic solutions to a Rayleigh neutral differential equation.

Motivated by [1, 5–7], in this paper, we consider the neutral operator $(Ax)(t)=x(t)-c(t)x(t-\delta (t))$, here $|c(t)|\ne 1$, $c,\delta \in C^1(\mathbb{R},\mathbb{R})$ and δ is an ω-periodic function for some $\omega >0$. Notice that here the neutral operator A is a natural generalization of the familiar operator $A_i$, $i=1,2,3,4$. But A possesses a more complicated nonlinearity than $A_i$, $i=1,2,3,4$. For example, the neutral operator $A_i$, $i=1,2$, is homogeneous in the following sense ${(A_{i}x)}^{\prime }(t)=(A_i{x}^{\prime })(t)$, $i=1,2$, 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 generalized neutral operator A which will be helpful for further studies of differential equations with this neutral operator; in Section 3, we consider a third-order neutral differential equation as follows:

$${(x(t)-c(t)x(t-\delta (t)))}^{‴}=a(t)x(t)-\lambda b(t)f\left(x(t-\tau (t))\right),$$

(1.1)

here λ is a positive parameter; $\delta (t)$ is said to be variable delay, $c,\delta \in C^1(\mathbb{R},\mathbb{R})$ and δ is an ω-periodic function for some $\omega >0$, $f\in C(\mathbb{R},[0,\mathrm{\infty }))$, and $f(x)>0$ for $x>0$; $a\in C(\mathbb{R},(0,\mathrm{\infty }))$ with $max\{a(t):t\in [0,\omega ]\}<\frac{64}{81\sqrt{3}}{(\frac{\pi }{\omega })}^3$, $b\in C(\mathbb{R},(0,\mathrm{\infty }))$, $\tau \in C(\mathbb{R},\mathbb{R})$, $a(t)$, $b(t)$ and $\tau (t)$ are ω-periodic functions. By applying Green’s function of a third-order differential equation and a fixed point theorem in cones, we obtain sufficient conditions for the existence, multiplicity and nonexistence of positive periodic solutions to the third-order neutral differential equation. We will give an example to illustrate our results, and an example is also given in this section. Our results improve and extend the results in [6–10].

2 Analysis of the generalized neutral operator with variable parameter

Let

$$c_{\mathrm{\infty }}=\underset{t\in [0,\omega ]}{max}|c(t)|,c_0=\underset{t\in [0,\omega ]}{min}|c(t)|.$$

Let $X=\{x\in C(\mathbb{R},\mathbb{R}):x(t+\omega )=x(t),t\in \mathbb{R}\}$ with the norm $\parallel x\parallel ={max}_{t\in [0,\omega ]}|x(t)|$, and let $C_{\omega }^{+}=\{x\in C(\mathbb{R},(0,\mathrm{\infty })):x(t+\omega )=x(t)\}$, $C_{\omega }^{-}=\{x\in C(\mathbb{R},(-\mathrm{\infty },0)):x(t+\omega )=x(t)\}$. Then $(X,\parallel \cdot \parallel )$ is a Banach space. A cone K in X is defined by $K=\{x\in X:x(t)\ge \alpha \parallel x\parallel ,\mathrm{\forall }t\in \mathbb{R}\}$, where α is a fixed positive number with $\alpha <1$. Moreover, define operators $A,B:C_{\omega }\to C_{\omega }$ by

$$(Ax)(t)=x(t)-c(t)x(t-\delta (t)),(Bx)(t)=c(t)x(t-\delta (t)).$$

Lemma 2.1 If $|c(t)|\ne 1$, then the operator A has a continuous inverse $A^{-1}$ on $C_{\omega }$, satisfying

• (1)
  $$\left(A^{-1}f\right)(t)=\{\begin{array}{cc}f(t)+{\sum }_{j=1}^{\mathrm{\infty }}{\prod }_{i=1}^{j}c(D_i)x(t-{\sum }_{i=1}^j\delta (D_i))\hfill & \mathit{\text{for }}|c(t)|<1,\mathrm{\forall }f\in C_{\omega },\hfill \\ -\frac{f(t+\delta (t))}{c(t+\delta (t))}-{\sum }_{j=1}^{\mathrm{\infty }}\frac{f(t+\delta (t)+{\sum }_{i=1}^j\delta (D_i^{\prime }))}{c(t+\delta (t)){\prod }_{i=1}^{j}c(D_i^{\prime })}\hfill & \mathit{\text{for }}|c(t)|>1,\mathrm{\forall }f\in C_{\omega }.\hfill \end{array}$$
• (2)
  $$|\left(A^{-1}f\right)(t)|\le \{\begin{array}{cc}\frac{\parallel f\parallel }{1-c_{\mathrm{\infty }}}\hfill & \mathit{\text{for }}{c}_{\mathrm{\infty }}<1\mathrm{\forall }f\in C_{\omega },\hfill \\ \frac{\parallel f\parallel }{c_0-1}\hfill & \mathit{\text{for }}{c}_0>1\mathrm{\forall }f\in C_{\omega }.\hfill \end{array}$$
• (3)
  $${\int }_0^{\omega }|\left(A^{-1}f\right)(t)|dt\le \{\begin{array}{cc}\frac{1}{1-c_{\mathrm{\infty }}}{\int }_0^{\omega }|f(t)|dt\hfill & \mathit{\text{for }}{c}_{\mathrm{\infty }}<1\mathrm{\forall }f\in C_{\omega },\hfill \\ \frac{1}{c_0-1}{\int }_0^{\omega }|f(t)|dt\hfill & \mathit{\text{for }}{c}_0>1\mathrm{\forall }f\in C_{\omega }.\hfill \end{array}$$

Proof Case 1: $|c(t)|\le c_{\mathrm{\infty }}<1$.

Let $t=D_1$ and $D_j=t-{\sum }_{i=1}^j\delta (D_i)$, $j=1,2,\dots$ .

$$\begin{array}{c}(Bx)(t)=c(t)x(t-\delta (t))=c(D_1)x(t-\delta (D_1));\hfill \\ \begin{array}{rl}\left(B^{2}x\right)(t)& =c(t)c(t-\delta (t))x(t-\delta (t)-\delta (t-\delta (t)))\\ =c(D_1)c(D_2)x(t-\delta (D_1)-\delta (D_2));\end{array}\hfill \\ \begin{array}{rl}\left(B^{3}x\right)(t)& =c(t)c(t-\delta (t))c(t-\delta (t)-\delta (t-\delta (t)))x(t-\delta (D_1)-\delta (D_2)-\delta (D_3))\\ =c(D_1)c(D_2)c(D_3)x(t-\sum _{i=1}^3\delta (D_i)).\end{array}\hfill \end{array}$$

Therefore

$$B^{j}x(t)=\prod _{i=1}^{j}c(D_i)x(t-\sum _{i=1}^j\delta (D_i)),$$

and

$$\sum _{j=0}^{\mathrm{\infty }}\left(B^{j}f\right)(t)=f(t)+\sum _{j=1}^{\mathrm{\infty }}\prod _{i=1}^{j}c(D_i)x(t-\sum _{i=1}^j\delta (D_i)).$$

Since $A=I-B$, we get from $\parallel B\parallel \le c_{\mathrm{\infty }}<1$ that A has a continuous inverse $A^{-1}:C_{\omega }\to C_{\omega }$ with

$$A^{-1}={(I-B)}^{-1}=I+\sum _{j=1}^{\mathrm{\infty }}{B}^j=\sum _{j=0}^{\mathrm{\infty }}{B}^j,$$

here $B^0=I$. Then

$$(A^{-1}f(t))=\sum _{j=0}^{\mathrm{\infty }}\left[B^{j}f\right](t)=f(t)+\sum _{j=1}^{\mathrm{\infty }}\prod _{i=1}^{j}c(D_i)x(t-\sum _{i=1}^j\delta (D_i)),$$

and consequently

$$\begin{array}{rl}|\left(A^{-1}f\right)(t)|& =|\sum _{j=0}^{\mathrm{\infty }}\left[B^{j}f\right](t)|\\ =|f(t)+\sum _{j=1}^{\mathrm{\infty }}\prod _{i=1}^{j}c(D_i)x(t-\sum _{i=1}^j\delta (D_i))|\\ \le (1+\sum _{j=1}^{\mathrm{\infty }}{c}_{\mathrm{\infty }}^j){|f|}_{\mathrm{\infty }}\\ \le \frac{{|f|}_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}}.\end{array}$$

Moreover,

$$\begin{array}{rl}{\int }_0^{\omega }|\left(A^{-1}f\right)(t)|dt& ={\int }_0^{\omega }|\sum _{j=0}^{\mathrm{\infty }}\left(B^{j}f\right)(t)|dt\le \sum _{j=0}^{\mathrm{\infty }}{\int }_0^{\omega }|\left(B^{j}f\right)(t)|dt\\ =\sum _{j=0}^{\mathrm{\infty }}{\int }_0^{\omega }|\prod _{i=1}^{j}c(D_i)x(t-\sum _{i=1}^j\delta (D_i))|dt\le \frac{1}{1-c_{\mathrm{\infty }}}{\int }_0^{\omega }|f(t)|dt.\end{array}$$

Case 2: $|c(t)|>c_0>1$.

Let $D_1^{\prime }=t$, $D_j^{\prime }=t+{\sum }_{i=1}^j\delta (D_i^{\prime })$, $j=1,2,\dots$ . And set

$$\begin{array}{c}E:C_{\omega }\to C_{\omega },(Ex)(t)=x(t)-\frac{1}{c(t)}x(t+\delta (t)),\hfill \\ B_1:C_{\omega }\to C_{\omega },(B_{1}x)(t)=\frac{1}{c(t)}x(t+\delta (t)).\hfill \end{array}$$

By the definition of the linear operator $B_1$, we have

$$\left(B_1^{j}f\right)(t)=\frac{1}{{\prod }_{i=1}^{j}c(D_i^{\prime })}f(t+\sum _{i=1}^j\delta \left(D_i^{\prime }\right)),$$

here $D_i$ is defined as in Case 1. Summing over j yields

$$\sum _{j=0}^{\mathrm{\infty }}\left(B_1^{j}f\right)(t)=f(t)+\sum _{j=1}^{\mathrm{\infty }}\frac{1}{{\prod }_{i=1}^{j}c(D_i^{\prime })}f(t+\sum _{i=1}^j\delta \left(D_i^{\prime }\right)).$$

Since $\parallel B_1\parallel <1$, we obtain that the operator E has a bounded inverse $E^{-1}$,

$$E^{-1}:C_{\omega }\to C_{\omega },E^{-1}={(I-B_1)}^{-1}=I+\sum _{j=1}^{\mathrm{\infty }}{B}_1^j,$$

and $\mathrm{\forall }f\in C_{\omega }$ we get

$$\left(E^{-1}f\right)(t)=f(t)+\sum _{j=1}^{\mathrm{\infty }}\left(B_1^{j}f\right)(t).$$

On the other hand, from $(Ax)(t)=x(t)-c(t)x(t-\delta (t))$, we have

$$(Ax)(t)=x(t)-c(t)x(t-\delta (t))=-c(t)[x(t-\delta (t))-\frac{1}{c(t)}x(t)],$$

i.e.,

$$(Ax)(t)=-c(t)(Ex)(t-\delta (t)).$$

Let $f\in C_{\omega }$ be arbitrary. We are looking for x such that

$$(Ax)(t)=f(t),$$

i.e.,

$$-c(t)(Ex)(t-\delta (t))=f(t).$$

Therefore

$$(Ex)(t)=-\frac{f(t+\delta (t))}{c(t+\delta (t))}=:f_1(t),$$

and hence

$$x(t)=\left(E^{-1}{f}_1\right)(t)=f_1(t)+\sum _{j=1}^{\mathrm{\infty }}\left(B_1^j{f}_1\right)(t)=-\frac{f(t+\delta (t))}{c(t+\delta (t))}-\sum _{j=1}^{\mathrm{\infty }}{B}_1^j\frac{f(t+\delta (t))}{c(t+\delta (t))},$$

proving that $A^{-1}$ exists and satisfies

$$\begin{array}{rl}\left[A^{-1}f\right](t)& =-\frac{f(t+\delta (t))}{c(t+\delta (t))}-\sum _{j=1}^{\mathrm{\infty }}{B}_1^j\frac{f(t+\delta (t))}{c(t+\delta (t))}\\ =-\frac{f(t+\delta (t))}{c(t+\delta (t))}-\sum _{j=1}^{\mathrm{\infty }}\frac{f(t+\delta (t)+{\sum }_{i=1}^j\delta (D_i^{\prime }))}{c(t+\delta (t)){\prod }_{i=1}^{j}c(D_i^{\prime })}\end{array}$$

and

$$|\left[A^{-1}f\right](t)|=|-\frac{f(t+\delta (t))}{c(t+\delta (t))}-\sum _{j=1}^{\mathrm{\infty }}\frac{f(t+\delta (t)+{\sum }_{i=1}^j\delta (D_i^{\prime }))}{c(t+\delta (t)){\prod }_{i=1}^{j}c(D_i^{\prime })}|\le \frac{\parallel f\parallel }{c_0-1}.$$

Statements (1) and (2) are proved. From the above proof, (3) can easily be deduced. □

Lemma 2.2 If $c(t)<0$ and $\sigma c_{\mathrm{\infty }}<\alpha$ here $\sigma =\frac{1-c_0^2}{1-c_{\mathrm{\infty }}^2}>1$, we have for $y\in K$ that

$$(\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})\parallel y\parallel \le \left(A^{-1}y\right)(t)\le \frac{1}{1-c_{\mathrm{\infty }}}\parallel y\parallel .$$

Proof Since $c(t)<0$ and $|c(t)|\le c_{\mathrm{\infty }}<\sigma c_{\mathrm{\infty }}<\alpha <1$, by Lemma 2.1, we have for $y\in K$ that

$$\begin{array}{rl}\left(A^{-1}y\right)(t)& =y(t)+\sum _{j=1}^{\mathrm{\infty }}\prod _{i=1}^{j}c(D_i)y(s-\sum _{i=1}^j\delta (D_i))\\ =y(t)+\sum _{j\ge 1\mathrm{even}}\prod _{i=1}^{j}c(D_i)y(t-\sum _{i=1}^j\delta (D_i))-\sum _{j\ge 1\mathrm{odd}}\prod _{i=1}^j|c(D_i)|y(t-\sum _{i=1}^j\delta (D_i))\\ \ge \alpha \parallel y\parallel +\alpha \sum _{j\ge 1\mathrm{even}}{c}_0^j\parallel y\parallel -\parallel y\parallel \sum _{j\ge 1\mathrm{odd}}{c}_{\mathrm{\infty }}^j\\ =\frac{\alpha }{1-c_0^2}\parallel y\parallel -\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}\parallel y\parallel \\ =(\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})\parallel y\parallel .\end{array}$$

 □

Lemma 2.3 If $c(t)>0$ and $c(t)<1$, then for $y\in K$ we have

$$\frac{\alpha }{1-c_0}\parallel y\parallel \le \left(A^{-1}y\right)(t)\le \frac{1}{1-c_{\mathrm{\infty }}}\parallel y\parallel .$$

Proof Since $c(t)>0$ and $c(t)<1$, $\alpha <1$, by Lemma 2.1, we have for $y\in K$ that

$$\left(A^{-1}y\right)(t)=y(t)+\sum _{j\ge 1}\prod _{i=1}^{j}c(D_i)y(t-\sum _{i=1}^j\delta (D_i))\ge \alpha \parallel y\parallel +\alpha \parallel y\parallel \sum _{j\ge 1}{c}_0^j=\frac{\alpha }{1-c_0}\parallel y\parallel .$$

 □

3 Positive periodic solutions for third-order neutral equations

At first, we introduce the following Green’s functions and properties of Green’s functions, which can be found in [11].

Theorem 3.1 For $\rho >0$ and $h\in X$, the equation

$$\{\begin{array}{c}{u}^{‴}-{\rho }^{3}u=h(t),\hfill \\ u(0)=u(\omega ),u^{\prime }(0)=u^{\prime }(\omega ),u^{″}(0)=u^{″}(\omega )\hfill \end{array}$$

(3.1)

has a unique solution which is of the form

$$u(t)={\int }_0^{\omega }{G}_1(t,s)(-h(s))ds,$$

(3.2)

where

$$G_1(t,s)=\{\begin{array}{c}\frac{2exp(\frac{1}{2}\rho (s-t))[sin(\frac{\sqrt{3}}{2}\rho (t-s)+\frac{\pi }{6})-exp(-\frac{1}{2}\rho \omega )sin(\frac{\sqrt{3}}{2}\rho (t-s-\omega )+\frac{\pi }{6})]}{3{\rho }^2(1+exp(-\rho \omega )-2exp(-\frac{\rho \omega }{2})cos(\frac{\sqrt{3}}{2}\rho \omega ))}+\frac{exp(\rho (t-s))}{3{\rho }^2(exp(\rho \omega )-1)},\hfill \\ 0\le s\le t\le \omega ,\hfill \\ \frac{2exp(\frac{1}{2}\rho (s-t-\omega ))[sin(\frac{\sqrt{3}}{2}\rho (t-s+\omega )+\frac{\pi }{6})-exp(-\frac{1}{2}\rho \omega )sin(\frac{\sqrt{3}}{2}\rho (t-s)+\frac{\pi }{6})]}{3{\rho }^2(1+exp(-\rho \omega )-2exp(-\frac{\rho \omega }{2})cos(\frac{\sqrt{3}}{2}\rho \omega ))}+\frac{exp(\rho (t+\omega -s))}{3{\rho }^2(exp(\rho \omega )-1)},\hfill \\ 0\le t\le s\le \omega .\hfill \end{array}$$

(3.3)

Theorem 3.2 For $\rho >0$ and $h\in X$, the equation

$$\{\begin{array}{c}{u}^{‴}+{\rho }^{3}u=h(t),\hfill \\ u(0)=u(\omega ),u^{\prime }(0)=u^{\prime }(\omega ),u^{″}(0)=u^{″}(\omega )\hfill \end{array}$$

(3.4)

has a unique ω-periodic solution

$$u(t)={\int }_0^{\omega }{G}_2(t,s)h(s)ds,$$

(3.5)

where

$$G_2(t,s)=\{\begin{array}{c}\frac{2exp(\frac{1}{2}\rho (t-s))[sin(\frac{\sqrt{3}}{2}\rho (t-s)-\frac{\pi }{6})-exp(\frac{1}{2}\rho \omega )sin(\frac{\sqrt{3}}{2}\rho (t-s-\omega )-\frac{\pi }{6})]}{3{\rho }^2(1+exp(\rho \omega )-2exp(\frac{1}{2}\rho \omega )cos(\frac{\sqrt{3}}{2}\rho \omega ))}+\frac{exp(\rho (s-t))}{3{\rho }^2(1-exp(-\rho \omega ))},\hfill \\ 0\le s\le t\le \omega ,\hfill \\ \frac{2exp(\frac{1}{2}\rho (t+\omega -s))[sin(\frac{\sqrt{3}}{2}\rho (t+\omega -s)-\frac{\pi }{6})-exp(\frac{1}{2}\rho \omega )sin(\frac{\sqrt{3}}{2}\rho (t-s)-\frac{\pi }{6})]}{3{\rho }^2(1+exp(\rho \omega )-2exp(\frac{1}{2}\rho \omega )cos(\frac{\sqrt{3}}{2}\rho \omega ))}+\frac{exp(\rho (s-t-\omega ))}{3{\rho }^2(1-exp(-\rho \omega ))},\hfill \\ 0\le t\le s\le \omega .\hfill \end{array}$$

(3.6)

Now we present the properties of the Green’s functions for (3.1), (3.4).

$$l=\frac{1}{3{\rho }^2(exp(\rho \omega )-1)},L=\frac{3+2exp(-\frac{\rho \omega }{2})}{3{\rho }^2{(1-exp(-\frac{\rho \omega }{2}))}^2}.$$

Theorem 3.3 ${\int }_0^{\omega }{G}_1(t,s)ds=\frac{1}{{\rho }^3}$ and if $\sqrt{3}\rho \omega <\frac{4}{3}\pi$ holds, then $0<l<G_1(t,s)\le L$ for all $t\in [0,\omega ]$ and $s\in [0,\omega ]$.

Theorem 3.4 ${\int }_0^{\omega }{G}_2(t,s)ds=\frac{1}{{\rho }^3}$ and if $\sqrt{3}\rho \omega <\frac{4}{3}\pi$ holds, then $0<l<G_2(t,s)\le L$ for all $[0,\omega ]$ and $s\in [0,\omega ]$.

Define the Banach space X as in Section 2. Denote

$$\begin{array}{r}M=max\{a(t):t\in [0,\omega ]\},m=min\{a(t):t\in [0,\omega ]\},{\rho }^3=M,\\ k=l(M+m)+\sigma LM,k_1=\frac{k-\sqrt{k^2-4\sigma LlMm}}{2\sigma LM},\alpha =\frac{l[m-(M+m)c_{\mathrm{\infty }}]}{LM(1-c_{\mathrm{\infty }})}.\end{array}$$

It is easy to see that $M,m,\beta ,k,k_1>0$.

Now we consider (1.1). First let

$${\overline{f}}_0=\underset{x\to 0}{\overline{lim}}\frac{f(x)}{x},{\overline{f}}_{\mathrm{\infty }}=\underset{x\to \mathrm{\infty }}{\overline{lim}}\frac{f(x)}{x},{\underline{f}}_0=\underset{x\to 0}{\underline{lim}}\frac{f(x)}{x},{\underline{f}}_{\mathrm{\infty }}=\underset{x\to \mathrm{\infty }}{\underline{lim}}\frac{f(x)}{x},$$

and denote

$$\begin{array}{c}{\overline{i}}_0=\text{number of 0’s in }({\overline{f}}_0,{\overline{f}}_{\mathrm{\infty }}),{\underline{i}}_0=\text{number of 0’s in }({\underline{f}}_0,{\underline{f}}_{\mathrm{\infty }});\hfill \\ {\overline{i}}_{\mathrm{\infty }}=\text{number of }\mathrm{\infty }\text{’s in }({\overline{f}}_0,{\overline{f}}_{\mathrm{\infty }}),{\underline{i}}_{\mathrm{\infty }}=\text{number of }\mathrm{\infty }\text{’s in }({\underline{f}}_0,{\underline{f}}_{\mathrm{\infty }}).\hfill \end{array}$$

It is clear that ${\overline{i}}_0,{\underline{i}}_0,{\overline{i}}_{\mathrm{\infty }},{\underline{i}}_{\mathrm{\infty }}\in \{0,1,2\}$. We will show that (1.1) has ${\overline{i}}_0$ or ${\underline{i}}_{\mathrm{\infty }}$ positive w-periodic solutions for sufficiently large or small λ, respectively.

In what follows, we discuss (1.1) in two cases, namely the case where $c(t)<0$ and $-c_{\mathrm{\infty }}>-min\{k_1,\frac{m}{M+m}\}$.

From $-c_{\mathrm{\infty }}>-\frac{m}{M+m}$, we have $\alpha =\frac{l[m-(M+m)c_{\mathrm{\infty }}]}{LM(1-c_{\mathrm{\infty }})}>\frac{l(m-(M+m)\cdot \frac{m}{M+m})}{LM(1-c_{\mathrm{\infty }})}=0$. So, we get $\alpha >0$. Moreover, we consider the equation

$$\sigma LM{x}^2-kx+lm=0.$$

Then the equation has a solution $x=k_1=\frac{k-\sqrt{k^2-4\sigma LlMm}}{2\sigma LM}$. From $c_{\mathrm{\infty }}<k_1$, we can get

$$\sigma LM{c}_{\mathrm{\infty }}^2-k{c}_{\mathrm{\infty }}+lm<0.$$

So, we have

$$\sigma LM{c}_{\mathrm{\infty }}^2-(l(M+m)+\sigma LM)c_{\mathrm{\infty }}+lm<0,$$

we get

$$\sigma c_{\mathrm{\infty }}>\frac{l[m-(M+m)c_{\mathrm{\infty }}]}{LM(1-c_{\mathrm{\infty }})}=\alpha .$$

On the other hand, the case where $c>0$ and $c_{\mathrm{\infty }}<min\{\frac{m}{M+m},\frac{LM-lm}{(L-l)M-lm}\}$ (note that $c_{\mathrm{\infty }}<\frac{m}{M+m}$ implies $\alpha >0$; $c_{\mathrm{\infty }}<\frac{LM-lm}{(L-l)M-lm}$ implies $\alpha <1$). Obviously, we have $c_{\mathrm{\infty }}<1$, which makes Lemma 2.1 applicable for both cases, and also Lemma 2.2 or 2.3, respectively.

Let $K=\{x\in X:x(t)\ge \alpha \parallel x\parallel \}$ denote the cone in X as defined in Section 2, where α is just as defined above. We also use $K_r=\{x\in K:\parallel x\parallel <r\}$ and $\partial K_r=\{x\in K:\parallel x\parallel =r\}$.

Let $y(t)=(Ax)(t)$, then from Lemma 2.1 we have $x(t)=(A^{-1}y)(t)$. Hence (1.1) can be transformed into

$$y^{‴}(t)-a(t)\left(A^{-1}y\right)(t)=-\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right),$$

(3.7)

which can be further rewritten as

$$y^{‴}(t)-a(t)y(t)+a(t)H(y(t))=-\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right),$$

(3.8)

where $H(y(t))=y(t)-(A^{-1}y)(t)=-c(t)(A^{-1}y)(t-\delta (t))$.

Now we discuss the two cases separately.

3.1 Case I: $c(t)<0$ and $-c_{\mathrm{\infty }}>-min\{k_1,\frac{m}{M+m}\}$

Now we consider

$$y^{‴}(t)-a(t)y(t)+a(t)H(y(t))=h(t),h\in C_{\omega }^{-},$$

(3.9)

and define the operators $T,\hat{H}:X\to X$ by

$$(Th)(t)={\int }_t^{t+\omega }{G}_1(t,s)(-h(s))ds,(\hat{H}y)(t)=-M+a(t)y(t)-a(t)H(y(t)).$$

Clearly T, $\hat{H}$ are completely continuous, $(Th)(t)>0$ for $h(t)<0$ and $\parallel \hat{H}\parallel \le (M-m+M\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}})$. By Theorem 3.1, the solution of (3.9) can be written in the form

$$y(t)=(Th))(t)+(T\hat{H}y)(t).$$

(3.10)

In view of $c(t)<0$ and $-c_{\mathrm{\infty }}>-min\{k_1,\frac{m}{M+m}\}$, we have

$$\parallel T\hat{H}\parallel \le \parallel T\parallel \parallel \hat{H}\parallel \le \frac{M-m+m{c}_{\mathrm{\infty }}}{M(1-c_{\mathrm{\infty }})}<1,$$

(3.11)

where we used the fact ${\int }_t^{t+\omega }{G}_1(t,s)ds=\frac{1}{M}$. Hence

$$y(t)={(I-T\hat{H})}^{-1}(Th)(t).$$

Define an operator $P:X\to X$ by

$$(Ph)(t)={(I-T\hat{H})}^{-1}(Th)(t).$$

Obviously, for any $h\in C_{\omega }^{-}$, if $max\{a(t):t\in [0,\omega ]\}<\frac{64}{81\sqrt{3}}{(\frac{\pi }{\omega })}^3$, $y(t)=(Ph)(t)$ is the unique positive ω-periodic solution of (3.9).

Lemma 3.1 P is completely continuous and

$$(Th)(t)\le (Ph)(t)\le \frac{M(1-c_{\mathrm{\infty }})}{m-(M+m)c_{\mathrm{\infty }}}\parallel Th\parallel \mathit{\text{for all }}h\in C_{\omega }^{-}.$$

(3.12)

Proof By the Neumann expansion of P, we have

$$\begin{array}{rl}P& ={(I-T\hat{H})}^{-1}T\\ =(I+T\hat{H}+{(T\hat{H})}^2+\cdots +{(T\hat{H})}^n+\cdots )T\\ =T+T\hat{H}T+{(T\hat{H})}^{2}T+\cdots +{(T\hat{H})}^{n}T+\cdots .\end{array}$$

(3.13)

Since T and $\hat{H}$ are completely continuous, so is P. Moreover, by (3.13), and recalling that $\parallel T\hat{H}\parallel \le \frac{M-m+m{c}_{\mathrm{\infty }}}{M(1-c_{\mathrm{\infty }})}<1$, we get

$$(Th)(t)\le (Ph)(t)\le \frac{M(1-c_{\mathrm{\infty }})}{m-(M+m)c_{\mathrm{\infty }}}\parallel Th\parallel .$$

 □

Define an operator $Q:X\to X$ by

$$Qy(t)=P(\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right)).$$

(3.14)

Lemma 3.2 $Q(K)\subset K$.

Proof From the definition of Q, it is easy to verify that $Qy(t+\omega )=Qy(t)$. For $y\in K$, we have from Lemma 3.1 that

$$\begin{array}{rl}Qy(t)& =P(\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right))\\ \ge T(\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right))\\ =\lambda {\int }_t^{t+\omega }{G}_1(t,s)b(s)f\left[\left(A^{-1}y\right)(s-\tau (s))\right]ds\\ \ge \lambda l{\int }_0^{\omega }b(s)f\left[\left(A^{-1}y\right)(s-\tau (s))\right]ds.\end{array}$$

On the other hand,

$$\begin{array}{rl}Qy(t)& =P(\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right))\\ \le \frac{M(1-c_{\mathrm{\infty }})}{m-(M+m)c_{\mathrm{\infty }}}\parallel T(\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right))\parallel \\ =\lambda \frac{M(1-c_{\mathrm{\infty }})}{m-(M+m)c_{\mathrm{\infty }}}\underset{t\in [0,\omega ]}{max}{\int }_t^{t+\omega }{G}_1(t,s)b(s)f\left(\left(A^{-1}y\right)(s-\tau (s))\right)ds\\ \le \lambda \frac{M(1-c_{\mathrm{\infty }})}{m-(M+m)c_{\mathrm{\infty }}}L{\int }_0^{\omega }b(s)f\left(\left(A^{-1}y\right)(s-\tau (s))\right)ds.\end{array}$$

Therefore

$$Qy(t)\ge \frac{l[m-(M+m)c_{\mathrm{\infty }}]}{LM(1-c_{\mathrm{\infty }})}\parallel Qy\parallel =\alpha \parallel Qy\parallel ,$$

i.e., $Q(K)\subset K$. □

From the continuity of P, it is easy to verify that Q is completely continuous in X. Comparing (3.8) to (3.9), it is obvious that the existence of periodic solutions for equation (3.8) is equivalent to the existence of fixed-points for the operator Q in X. Recalling Lemma 3.2, the existence of positive periodic solutions for (3.8) 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 $(A^{-1}y)(t)$ is a positive ω-periodic solution of (1.1).

Lemma 3.3 If there exists $\eta >0$ such that

$$f\left(\left(A^{-1}y\right)(t-\tau (t))\right)\ge \left(A^{-1}y\right)(t-\tau (t))\eta \mathit{\text{for }}t\in [0,\omega ]\mathit{\text{ and }}y\in K,$$

then

$$\parallel Qy\parallel \ge \lambda l\eta (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel ,y\in K.$$

Proof By Lemma 2.2 and Lemma 3.1, we have for $y\in K$ that

$$\begin{array}{rl}Qy(t)& =P(\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right))\\ \ge T(\lambda b(t)f\left(\left(A^{-1}y\right)(t-\tau (t))\right))\\ =\lambda {\int }_t^{t+\omega }{G}_1(t,s)b(s)f\left(\left(A^{-1}y\right)(s-\tau (s))\right)ds\\ \ge \lambda l\eta {\int }_0^{\omega }b(s)\left(A^{-1}y\right)(s-\tau (s))ds\\ \ge \lambda l\eta (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel .\end{array}$$

Hence

$$\parallel Qy\parallel \ge \lambda l\eta (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel ,y\in K.$$

 □

Lemma 3.4 If there exists $\epsilon >0$ such that

$$f\left(\left(A^{-1}y\right)(t-\tau (t))\right)\le \left(A^{-1}y\right)(t-\tau (t))\epsilon \mathit{\text{for }}t\in [0,\omega ]\mathit{\text{ and }}y\in K,$$

then

$$\parallel Qy\parallel \le \lambda \epsilon \frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}\parallel y\parallel ,y\in K.$$

Proof By Lemma 2.2 and Lemma 3.1, we have

$$\begin{array}{rl}\parallel Qy(t)\parallel & \le \lambda \frac{M(1-c_{\mathrm{\infty }})}{m-(M+m)c_{\mathrm{\infty }}}L{\int }_0^{\omega }b(s)f\left(\left(A^{-1}y\right)(s-\tau (s))\right)ds\\ \le \lambda \frac{M(1-c_{\mathrm{\infty }})}{m-(M+m)c_{\mathrm{\infty }}}L\epsilon {\int }_0^{\omega }b(s)\left(A^{-1}y\right)(s-\tau (s))ds\\ \le \lambda \epsilon \frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}\parallel y\parallel .\end{array}$$

 □

Define

$$\begin{array}{r}F(r)=max\{f(t):0\le t\le \frac{r}{1-c_{\mathrm{\infty }}}\},\\ f_1(r)=min\{f(t):(\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})r\le t\le \frac{r}{1-c_{\mathrm{\infty }}}\}.\end{array}$$

Lemma 3.5 If $y\in \partial K_r$, then

$$\parallel Qy\parallel \ge \lambda l{f}_1(r){\int }_0^{\omega }b(s)ds.$$

Proof By Lemma 2.2, we obtain $(\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})r\le (A^{-1}y)(t-\tau (t))\le \frac{r}{1-c_{\mathrm{\infty }}}$ for $y\in \partial K_r$, which yields $f((A^{-1}y)(t-\tau (t)))\ge f_1(r)$. The lemma now follows analogous to the proof of Lemma 3.3. □

Lemma 3.6 If $y\in \partial K_r$, then

$$\parallel Qy\parallel \le \lambda \frac{LM(1-c_{\mathrm{\infty }})F(r)}{m-(M+m)c_{\mathrm{\infty }}}{\int }_0^{\omega }b(s)ds.$$

Proof By Lemma 2.2, we can have $0\le (A^{-1}y)(t-\tau (t))\le \frac{r}{1-c_{\mathrm{\infty }}}$ for $y\in \partial K_r$, which yields $f((A^{-1}y)(t-\tau (t)))\le F(r)$. Similar to the proof of Lemma 3.4, we get the conclusion. □

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

Lemma 3.7 [12]

Let X be a Banach space and K be a cone in X. For $r>0$, define $K_r=\{u\in K:\parallel u\parallel <r\}$. Assume that $T:{\overline{K}}_r\to K$ is completely continuous such that $Tx\ne x$ for $x\in \partial K_r=\{u\in K:\parallel u\parallel =r\}$.

• (i) If $\parallel Tx\parallel \ge \parallel x\parallel$ for $x\in \partial K_r$, then $i(T,K_r,K)=0$;

• (ii) If $\parallel Tx\parallel \le \parallel x\parallel$ for $x\in \partial K_r$, then $i(T,K_r,K)=1$.

Now we give our main results on positive periodic solutions for (1.1).

Theorem 3.5

• (a) If ${\overline{i}}_0=1$ or 2, then (1.1) has ${\overline{i}}_0$ positive ω-periodic solutions for $\lambda >\frac{1}{f_1(1)l{\int }_0^{\omega }b(s)ds}>0$;

• (b) If ${\underline{i}}_{\mathrm{\infty }}=1$ or 2, then (1.1) has ${\underline{i}}_{\mathrm{\infty }}$ positive ω-periodic solutions for $0<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{LM(1-c_{\mathrm{\infty }})F(1){\int }_0^{\omega }b(s)ds}$;

• (c) If ${\overline{i}}_{\mathrm{\infty }}=0$ or ${\underline{i}}_0=0$, then (1.1) has no positive ω-periodic solutions for sufficiently small or sufficiently large $\lambda >0$, respectively.

Proof (a) Choose $r_1=1$. Take ${\lambda }_0=\frac{1}{f_1(r_1)l{\int }_0^{\omega }b(s)ds}>0$, then for all $\lambda >{\lambda }_0$, we have from Lemma 3.5 that

$$\parallel Qy\parallel >\parallel y\parallel \text{for }y\in \partial K_{r_1}.$$

(3.15)

Case 1. If ${\overline{f}}_0=0$, we can choose $0<{\overline{r}}_2<r_1$, so that $f(u)\le \epsilon u$ for $0\le u\le {\overline{r}}_2$, where the constant $\epsilon >0$ satisfies

$$\lambda \epsilon \frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}<1.$$

(3.16)

Let $r_2=(1-c_{\mathrm{\infty }}){\overline{r}}_2$, we have $f((A^{-1}y)(t-\tau (t)))\le \epsilon (A^{-1}y)(t-\tau (t))$ for $y\in K_{r_2}$. By Lemma 2.2, we have $0\le (A^{-1}y)(t-\tau (t))\le \frac{\parallel y\parallel }{1-c_{\mathrm{\infty }}}\le {\overline{r}}_2$ for $y\in \partial K_{r_2}$. In view of Lemma 3.4 and (3.16), we have for $y\in \partial K_{r_2}$ that

$$\parallel Qy\parallel \le \lambda \epsilon \frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}\parallel y\parallel <\parallel y\parallel .$$

It follows from Lemma 3.7 and (3.15) that

$$i(Q,K_{r_2},K)=1,i(Q,K_{r_1},K)=0,$$

thus $i(Q,K_{r_1}\mathrm{\setminus }{\overline{K}}_{r_2},K)=-1$ and Q has a fixed point y in $K_{r_1}\mathrm{\setminus }{\overline{K}}_{r_2}$, which means that $(A^{-1}y)(t)$ is a positive ω-positive solution of (1.1) for $\lambda >{\lambda }_0$.

Case 2. If ${\overline{f}}_{\mathrm{\infty }}=0$, there exists a constant $\tilde{H}>0$ such that $f(u)\le \epsilon u$ for $u\ge \tilde{H}$, where the constant $\epsilon >0$ satisfies

$$\lambda \epsilon \frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}<1.$$

(3.17)

Let $r_3=max\{2r_1,\frac{\tilde{H}(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)}\}$, we have $f((A^{-1}y)(t-\tau (t)))\le \epsilon (A^{-1}y)(t-\tau (t))$ for $y\in K_{r_3}$. By Lemma 2.2, we have $(A^{-1}y)(t-\tau (t))\ge (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})\parallel y\parallel \ge \tilde{H}$ for $y\in \partial K_{r_3}$. Thus by Lemma 3.4 and (3.17), we have for $y\in \partial K_{r_3}$ that

$$\parallel Qy\parallel \le \lambda \epsilon \frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}\parallel y\parallel <\parallel y\parallel .$$

Recalling Lemma 3.7 and (3.15) that

$$i(Q,K_{r_3},K)=1,i(Q,K_{r_1},K)=0,$$

then $i(Q,K_{r_3}\mathrm{\setminus }{\overline{K}}_{r_1},K)=1$ and Q has a fixed point y in $K_{r_3}\mathrm{\setminus }{\overline{K}}_{r_1}$, which means that $(A^{-1}y)(t)$ is a positive ω-positive solution of (1.1) for $\lambda >{\lambda }_0$.

Case 3. If ${\overline{f}}_0={\overline{f}}_{\mathrm{\infty }}=0$, from the above arguments, there exist $0<r_2<r_1<r_3$ such that Q has a fixed point $y_1(t)$ in $K_{r_1}\mathrm{\setminus }{\overline{K}}_{r_2}$ and a fixed point $y_2(t)$ in $K_{r_3}\mathrm{\setminus }{\overline{K}}_{r_1}$. Consequently, $(A^{-1}{y}_1)(t)$ and $(A^{-1}{y}_2)(t)$ are two positive ω-periodic solutions of (1.1) for $\lambda >{\lambda }_0$.

• (b) Let $r_1=1$. Take ${\lambda }_0=\frac{m-(M+m)c_{\mathrm{\infty }}}{LM(1-c_{\mathrm{\infty }})F(r_1){\int }_0^{\omega }b(s)ds}>0$, then by Lemma 3.6 we know if $\lambda <{\lambda }_0$ then

  $$\parallel Qy\parallel <\parallel y\parallel ,y\in \partial K_{r_1}.$$

  (3.18)

Case 1. If ${\underline{f}}_0=\mathrm{\infty }$, we can choose $0<{\overline{r}}_2<r_1$ so that $f(u)\ge \eta u$ for $0\le u\le {\overline{r}}_2$, where the constant $\eta >0$ satisfies

$$\lambda l\eta (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds>1.$$

(3.19)

Let $r_2=(1-c_{\mathrm{\infty }}){\overline{r}}_2$, we have $f((A^{-1}y)(t-\tau (t)))\ge \eta (A^{-1}y)(t-\tau (t))$ for $y\in K_{r_2}$. By Lemma 2.2, we have $0\le (A^{-1}y)(t-\tau (t))\le \frac{\parallel y\parallel }{1-c_{\mathrm{\infty }}}\le {\overline{r}}_2$ for $y\in \partial K_{r_2}$. Thus by Lemma 3.3 and (3.19),

$$\parallel Qy\parallel \ge \lambda l\eta (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel >\parallel y\parallel .$$

It follows from Lemma 3.7 and (3.18) that

$$i(Q,K_{r_2},K)=0,i(Q,K_{r_1},K)=1,$$

which implies $i(Q,K_{r_1}\mathrm{\setminus }{\overline{K}}_{r_2},K)=1$ and Q has a fixed point y in $K_{r_1}\mathrm{\setminus }{\overline{K}}_{r_2}$. Therefore $(A^{-1}y)(t)$ is a positive ω-periodic solution of (1.1) for $0<\lambda <{\lambda }_0$.

Case 2. If ${\underline{f}}_{\mathrm{\infty }}=\mathrm{\infty }$, there exists a constant $\tilde{H}>0$ such that $f(u)\ge \eta u$ for $u\ge \tilde{H}$, where the constant $\eta >0$ satisfies

$$\lambda l\eta (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds>1.$$

(3.20)

Let $r_3=max\{2r_1,\frac{\tilde{H}(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)}\}$, we have $f((A^{-1}y)(t-\tau (t)))\ge \eta (A^{-1}y)(t-\tau (t))$ for $y\in K_{r_3}$. By Lemma 2.2, we have $(A^{-1}y)(t-\tau (t))\ge (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})\parallel y\parallel \ge \tilde{H}$ for $y\in \partial K_{r_3}$. Thus by Lemma 3.3 and (3.20), we have for $y\in \partial K_{r_3}$ that

$$\parallel Qy\parallel \ge \lambda l\eta (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel >\parallel y\parallel .$$

It follows from Lemma 3.7 and (3.18) that

$$i(Q,K_{r_3},K)=0,i(Q,K_{r_1},K)=1,$$

i.e., $i(Q,K_{r_3}\mathrm{\setminus }{\overline{K}}_{r_1},K)=-1$ and Q has a fixed point y in $K_{r_3}\mathrm{\setminus }{\overline{K}}_{r_1}$. That means $(A^{-1}y)(t)$ is a positive ω-periodic solution of (1.1) for $0<\lambda <{\lambda }_0$.

Case 3. If ${\underline{f}}_0={\underline{f}}_{\mathrm{\infty }}=\mathrm{\infty }$, from the above arguments, Q has a fixed point $y_1$ in $K_{r_1}\mathrm{\setminus }{\overline{K}}_{r_2}$ and a fixed point $y_2$ in $K_{r_3}\mathrm{\setminus }{\overline{K}}_{r_1}$. Consequently, $(A^{-1}{y}_1)(t)$ and $(A^{-1}{y}_2)(t)$ are two positive ω-periodic solutions of (1.1) for $0<\lambda <{\lambda }_0$.

• (c) By Lemma 2.2, if $y\in K$, then $(A^{-1}y)(t-\tau (t))\ge (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})\parallel y\parallel >0$ for $t\in [0,\omega ]$.

Case 1. If ${\underline{i}}_0=0$, we have ${\underline{f}}_0>0$ and ${\underline{f}}_{\mathrm{\infty }}>0$. Let $b_1=min\{\frac{f(u)}{u};u>0\}>0$, then we obtain

$$f(u)\ge b_{1}u,u\in [0,+\mathrm{\infty }).$$

Assume that $y(t)$ is a positive ω-periodic solution of (1.1) for $\lambda >{\lambda }_0$, where ${\lambda }_0=\frac{(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{l{b}_1[\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}{(1-c_0)}^2]{\int }_0^{\omega }b(s)ds}>0$. Since $Qy(t)=y(t)$ for $t\in [0,\omega ]$, then by Lemma 3.3 if $\lambda >{\lambda }_0$, we have

$$\parallel y\parallel =\parallel Qy\parallel \ge \lambda l{b}_1(\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel >\parallel y\parallel ,$$

which is a contradiction.

Case 2. If ${\overline{i}}_{\mathrm{\infty }}=0$, we have ${\overline{f}}_0<\mathrm{\infty }$ and ${\overline{f}}_{\mathrm{\infty }}<\mathrm{\infty }$. Let $b_2=max\{\frac{f(u)}{u}:u>0\}>0$, then we obtain

$$f(u)\le b_{2}u,u\in [0,\mathrm{\infty }).$$

Assume that $y(t)$ is a positive ω-periodic solution of (1.1) for $0<\lambda <{\lambda }_0$, where ${\lambda }_0=\frac{m-(M+m)c_{\mathrm{\infty }}}{b_{2}LM{\int }_0^{\omega }b(s)ds}$. Since $Qy(t)=y(t)$ for $t\in [0,\omega ]$, it follows from Lemma 3.4 that

$$\parallel y\parallel =\parallel Qy\parallel \le \lambda b_2\frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}\parallel y\parallel <\parallel y\parallel ,$$

which is a contradiction. □

Theorem 3.6

• (a) If there exists a constant $b_1>0$ such that $f(u)\ge b_{1}u$ for $u\in [0,+\mathrm{\infty })$, then (1.1) has no positive ω-periodic solution for $\lambda >\frac{(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{l{b}_1[\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)]{\int }_0^{\omega }b(s)ds}$.

• (b) If there exists a constant $b_2>0$ such that $f(u)\le b_{2}u$ for $u\in [0,+\mathrm{\infty })$, then (1.1) has no positive ω-periodic solution for $0<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{b_{2}LM{\int }_0^{\omega }b(s)ds}$.

Proof From the proof of (c) in Theorem 3.5, we obtain this theorem immediately. □

Theorem 3.7 Assume ${\underline{i}}_0={\overline{i}}_0={\underline{i}}_{\mathrm{\infty }}={\overline{i}}_{\mathrm{\infty }}=0$, and that one of the following conditions holds:

• (1) ${\overline{f}}_0\le {\underline{f}}_{\mathrm{\infty }}$;

• (2) ${\underline{f}}_0>{\overline{f}}_{\mathrm{\infty }}$;

• (3) ${\underline{f}}_0\le {\underline{f}}_{\mathrm{\infty }}\le {\overline{f}}_0\le {\overline{f}}_{\mathrm{\infty }}$;

• (4) ${\underline{f}}_{\mathrm{\infty }}\le {\underline{f}}_0\le {\overline{f}}_{\mathrm{\infty }}\le {\overline{f}}_0$.

If

$$\begin{array}{r}\frac{(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{l[\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)]{\int }_0^{\omega }b(s)dsmax\{{\underline{f}}_0,{\overline{f}}_0,{\underline{f}}_{\mathrm{\infty }},{\overline{f}}_{\mathrm{\infty }}\}}\\ <\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{LM{\int }_0^{\omega }b(s)dsmin\{{\underline{f}}_0,{\overline{f}}_0,{\underline{f}}_{\mathrm{\infty }},{\overline{f}}_{\mathrm{\infty }}\}},\end{array}$$

then (1.1) has one positive ω-periodic solution.

Proof Case 1. If ${\overline{f}}_0\le {\underline{f}}_{\mathrm{\infty }}$, then

$$\frac{(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{l[\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)]{\int }_0^{\omega }b(s)ds}<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{LM{\int }_0^{\omega }b(s)ds}.$$

It is easy to see that there exists $0<\epsilon <f_{\mathrm{\infty }}$ such that

$$\frac{(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{({\overline{f}}_{\mathrm{\infty }}-\epsilon )l[\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)]{\int }_0^{\omega }b(s)ds}<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{({\underline{f}}_0+\epsilon )LM{\int }_0^{\omega }b(s)ds}.$$

For the above ε, we choose ${\overline{r}}_1>0$ such that $f(u)\le ({\underline{f}}_0+\epsilon )u$ for $0\le u\le {\overline{r}}_1$. Let $r_1=(1-c_{\mathrm{\infty }}){\overline{r}}_1$, we have $f((A^{-1}y)(t-\tau (t)))\le ({\underline{f}}_0+\epsilon )(A^{-1}y)(t-\tau (t))$ for $y\in K_{r_1}$. By Lemma 2.2, we have $0\le (A^{-1}y)(t-\tau (t))\le \frac{\parallel y\parallel }{1-c_{\mathrm{\infty }}}\le {\overline{r}}_1$ for $K\in \partial K_{r_1}$. Thus by Lemma 3.4 we have for $y\in \partial K_{r_1}$ that

$$\parallel Qy\parallel \le \lambda ({\underline{f}}_0+\epsilon )\frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}\parallel y\parallel <\parallel y\parallel .$$

On the other hand, there exists a constant $\tilde{H}>0$ such that $f(u)\ge ({\overline{f}}_{\mathrm{\infty }}-\epsilon )u$ for $u\ge \tilde{H}$. Let $r_2=max\{2r_1,\frac{\tilde{H}(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)}\}$, we have $f((A^{-1}y)(t-\tau (t)))\ge ({\overline{f}}_{\mathrm{\infty }}-\epsilon )(A^{-1}y)(t-\tau (t))$ for $y\in K_{r_2}$. By Lemma 2.2, we have $(A^{-1}y)(t-\tau (t))\ge (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})\parallel y\parallel \ge \tilde{H}$ for $y\in \partial K_{r_2}$. Thus by Lemma 3.3, for $y\in \partial K_{r_2}$,

$$\parallel Qy\parallel \ge \lambda l({\overline{f}}_{\mathrm{\infty }}-\epsilon )(\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel >\parallel y\parallel .$$

It follows from Lemma 3.7 that

$$i(Q,K_{r_1},K)=1,i(Q,K_{r_2},K)=0,$$

thus $i(Q,K_{r_2}\mathrm{\setminus }{\overline{K}}_{r_1},K)=-1$ and Q has a fixed point y in $K_{r_2}\mathrm{\setminus }{\overline{K}}_{r_1}$. So $(A^{-1}y)(t)$ is a positive ω-periodic solution of (1.1).

Case 2. If ${\underline{f}}_0>{\overline{f}}_{\mathrm{\infty }}$, in this case, we have

$$\frac{(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{{\overline{f}}_{0}l[\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)]{\int }_0^{\omega }b(s)ds}<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{{\underline{f}}_{\mathrm{\infty }}LM{\int }_0^{\omega }b(s)ds}.$$

It is easy to see that there exists $0<\epsilon <f_0$ such that

$$\frac{(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{({\overline{f}}_0-\epsilon )l[\alpha (1-c_{\mathrm{\infty }}^2)-c_{\mathrm{\infty }}(1-c_0^2)]{\int }_0^{\omega }b(s)ds}<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{({\underline{f}}_{\mathrm{\infty }}+\epsilon )LM{\int }_0^{\omega }b(s)ds}.$$

For the above ε, we choose ${\overline{r}}_1>0$ such that $f(u)\ge ({\overline{f}}_0-\epsilon )u$ for $0\le u\le {\overline{r}}_1$. Let $r_1=(1-c_{\mathrm{\infty }}){\overline{r}}_1$, we have $f((A^{-1}y)(t-\tau (t)))\ge ({\overline{f}}_0-\epsilon )(A^{-1}y)(t-\tau (t))$ for $y\in K_{r_1}$. By Lemma 2.2, we have $0\le (A^{-1}y)(t-\tau (t))\le \frac{\parallel y\parallel }{1-c_{\mathrm{\infty }}}\le {\overline{r}}_1$ for $y\in \partial K_{r_1}$. Thus we have by Lemma 3.3 that for $y\in \partial K_{r_1}$,

$$\parallel Qy\parallel \ge \lambda l({\overline{f}}_0-\epsilon )(\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}){\int }_0^{\omega }b(s)ds\parallel y\parallel >\parallel y\parallel .$$

On the other hand, there exists a constant $\tilde{H}>0$ such that $f(u)\le ({\underline{f}}_{\mathrm{\infty }}+\epsilon )u$ for $u\ge \tilde{H}$. Let $r_2=max\{2r_1,\frac{\tilde{H}(1-c_0^2)(1-c_{\mathrm{\infty }}^2)}{\alpha (1-c_{\mathrm{\infty }})-c_{\mathrm{\infty }}(1-c_0^2)}\}$, we have $f((A^{-1}y)(t-\tau (t)))\le ({\underline{f}}_{\mathrm{\infty }}+\epsilon )(A^{-1}y)(t-\tau (t))$ for $y\in K_{r_2}$. By Lemma 2.2 we have $(A^{-1}y)(t-\tau (t))\ge (\frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2})\parallel y\parallel \ge \tilde{H}$ for $y\in \partial K_{r_2}$. Thus by Lemma 3.4, for $y\in \partial K_{r_2}$,

$$\parallel Qy\parallel \le \lambda ({\underline{f}}_{\mathrm{\infty }}+\epsilon )\frac{LM{\int }_0^{\omega }b(s)ds}{m-(M+m)c_{\mathrm{\infty }}}\parallel y\parallel .$$

It follows from Lemma 3.7 that

$$i(Q,K_{r_1},K)=0,i(Q,K_{r_2},K)=1.$$

Thus $i(Q,K_{r_2}\mathrm{\setminus }{\overline{K}}_{r_1},K)=-1$ and Q has a fixed point y in $K_{r_2}\mathrm{\setminus }{\overline{K}}_{r_1}$, proving that $(A^{-1}y)(t)$ is a positive ω-periodic solution of (1.1).

Case 3. ${\underline{f}}_0\le {\underline{f}}_{\mathrm{\infty }}\le {\overline{f}}_0\le {\overline{f}}_{\mathrm{\infty }}$. The proof is the same as in Case 1.

Case 4. ${\underline{f}}_{\mathrm{\infty }}\le {\underline{f}}_0\le {\overline{f}}_{\mathrm{\infty }}\le {\overline{f}}_0$. The proof is the same as in Case 2. □

3.2 Case II: $c(t)>0$ and $c_{\mathrm{\infty }}<min\{\frac{m}{M+m},\frac{LM-lm}{(L-l)M-lm}\}$

Define

$$f_2(r)=min\{f(t):\frac{\alpha }{1-c_0}r\le t\le \frac{r}{1-c_{\mathrm{\infty }}}\}.$$

Similarly as in Section 3.1, we get the following results.

Theorem 3.8

• (a) If ${\overline{i}}_0=1$ or 2, then (1.1) has $i_0$ positive ω-periodic solutions for $\lambda >\frac{1}{f_2(1)l{\int }_0^{\omega }b(s)ds}>0$.

• (b) If ${\underline{i}}_{\mathrm{\infty }}=1$ or 2, then (1.1) has $i_{\mathrm{\infty }}$ positive ω-periodic solutions for $0<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{LM(1-c_{\mathrm{\infty }})F(1){\int }_0^{\omega }b(s)ds}$.

• (c) If ${\overline{i}}_{\mathrm{\infty }}=0$ or ${\underline{i}}_0=0$, then (1.1) has no positive ω-periodic solution for sufficiently small or large $\lambda >0$, respectively.

Theorem 3.9

• (a) If there exists a constant $b_1>0$ such that $f(u)\ge b_{1}u$ for $u\in [0,+\mathrm{\infty })$, then (1.1) has no positive ω-periodic solution for $\lambda >\frac{1-c_0}{l\alpha b_1{\int }_0^{\omega }b(s)ds}$.

• (b) If there exists a constant $b_2>0$ such that $f(u)\le b_{2}u$ for $u\in [0,+\mathrm{\infty })$, then (1.1) has no positive ω-periodic solution for $0<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{b_{2}LM{\int }_0^{\omega }b(s)ds}$.

Theorem 3.10 Assume that ${\underline{i}}_0={\overline{i}}_0={\underline{i}}_{\mathrm{\infty }}={\overline{i}}_{\mathrm{\infty }}=0$ hold, and that one of the following conditions holds:

• (1) ${\overline{f}}_0\le {\underline{f}}_{\mathrm{\infty }}$;

• (2) ${\underline{f}}_0>{\overline{f}}_{\mathrm{\infty }}$;

• (3) ${\underline{f}}_0\le {\underline{f}}_{\mathrm{\infty }}\le {\overline{f}}_0\le {\overline{f}}_{\mathrm{\infty }}$;

• (4) ${\underline{f}}_{\mathrm{\infty }}\le {\underline{f}}_0\le {\overline{f}}_{\mathrm{\infty }}\le {\overline{f}}_0$.

If

$$\frac{1-c_0}{l\alpha {\int }_0^{\omega }b(s)dsmax\{{\underline{f}}_0,{\overline{f}}_0,{\underline{f}}_{\mathrm{\infty }},{\overline{f}}_{\mathrm{\infty }}\}}<\lambda <\frac{m-(M+m)c_{\mathrm{\infty }}}{LM{\int }_0^{\omega }b(s)dsmin\{{\underline{f}}_0,{\overline{f}}_0,{\underline{f}}_{\mathrm{\infty }},{\overline{f}}_{\mathrm{\infty }}\}},$$

then (1.1) has one positive ω-periodic solution.

Remark 1 In a similar way, one can consider the third-order neutral functional differential equation ${(x(t)-c(t)x(t-\delta (t)))}^{‴}+a(t)x(t)=\lambda b(t)f(x(t-\tau (t)))$.

We illustrate our results with an example.

Example 3.1 Consider the following third-order neutral differential equation:

$$\begin{array}{r}{(u(t)+\frac{1}{300}(1-\frac{1}{2}sin2t)u(t-{cos}^{2}t))}^{‴}-\frac{1}{8}(1-\frac{1}{2}{sin}^{2}t)u(t)\\ =-\lambda (1-cos2t)u^2(t-\tau (t))a^{u(t-\tau (t))},\end{array}$$

(3.21)

where λ and $0<a<1$ are two positive parameters, $\tau (t+\pi )=\tau (t)$.

Comparing (3.21) to (1.1), we see that $\delta (t)={cos}^{2}t$, $c(t)=-\frac{1}{300}(1-\frac{1}{2}sin2t)$, $a(t)=\frac{1}{8}(1-\frac{1}{2}{sin}^{2}t)$, $b(t)=1-cos2t$, $\omega =\pi$, $f(u)=u^2{a}^u$. Clearly, $c_{\mathrm{\infty }}=\frac{1}{300}$, $c_0=\frac{1}{600}$, $M=\frac{1}{8}$, $m=\frac{1}{16}$, and we get $\rho =\frac{1}{2}$, noticing that $\frac{\sqrt{3}\pi }{2}<\frac{4\pi }{3}$ holds. ${\overline{f}}_0=0$, ${\overline{f}}_{\mathrm{\infty }}=0$, ${\overline{i}}_0=2$. By Theorem 3.5, we easily get the following conclusion: equation (3.21) has two positive π-periodic solutions for $\lambda >\frac{7}{40\pi r_1}$, where $r_1=min\{f(0.0016),f(\frac{300}{299})\}$.

In fact, by simple computations, we have

$$\begin{array}{r}l=\frac{1}{3\rho (exp(\rho \omega )-1)}=0.175,L=\frac{3+2exp(-\frac{\rho \omega }{2})}{3{\rho }^2{(1-exp(-\frac{\rho \omega }{2}))}^2}=17.62\\ k=2.235,k_1=0.0050,\alpha =0.0049,\\ c_{\mathrm{\infty }}=\frac{1}{300}<min\{k_1,\frac{m}{M+m}\}=0.0050,c_{\mathrm{\infty }}=\frac{1}{300}<0.0049=\alpha ,\end{array}$$

and

$$\begin{array}{r}{f}_1(1)=min\{f(t):0.0016\approx \frac{\alpha }{1-c_0^2}-\frac{c_{\mathrm{\infty }}}{1-c_{\mathrm{\infty }}^2}\le t\le \frac{300}{299}\}\\ =min\{f(0.0016),f\left(\frac{300}{299}\right)\}=r_1,\\ \frac{1}{f_1(1)l{\int }_0^{\pi }b(s)ds}=\frac{7}{40\pi r_1}.\end{array}$$