Linear difference operator with multiple variable parameters and applications to second-order differential equations

Abstract

In this article, we first investigate the linear difference operator $(Ax)(t):=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}(t))$ in a continuous periodic function space. The existence condition and some properties of the inverse of the operator A are explicitly pointed out. Afterwards, as applications of properties of the operator A, we study the existence of periodic solutions for two kinds of second-order functional differential equations with this operator. One is a kind of second-order functional differential equation, by applications of Krasnoselskii’s fixed point theorem, some sufficient conditions for the existence of positive periodic solutions are established. Another one is a kind of second-order quasi-linear differential equation, we establish the existence of periodic solutions of this equation by an extension of Mawhin’s continuous theorem.

1 Introduction

Difference operators play a very important role in solving functional differential equations, which derived from some practical problems, such as biology, economics and population models [11, 20, 25]. In the 1970s, Hale [10] gave a definition for a functional differential equations of an operator. Under the condition that the operator is stable, many researchers obtained the existence of periodic solutions for these functional differential equations by means of some fixed point theorems and topology degree theory. Zhang [26] in 1995 first introduced the properties of the linear autonomous difference operator $(A_{1}x)(t):=x(t)-cx(t- \delta )$, where c, δ are constants, which became an effective tool for the research on differential equation, since it relieved the above stability restriction. This work has attracted the attention of many scholars in differential equations, for example [2–4, 6–8, 13, 15, 17–19, 21–24, 27]. Lu and Ge [13] in 2004 investigated a linear autonomous difference operator with multiple parameters $(A_{2}x)(t):=x(t)- \sum_{i=1}^{n}c_{i}x(t-\delta _{i})$ which is an extension of $A_{1}$. And they obtained the existence of periodic solutions for the corresponding differential equation. Du et al. [8] in 2009 studied the difference operator $(A_{3}x)(t):=x(t)-c(t)x(t-\delta )$, where $c(t)$ is a periodic function. By applying Mawhin’s continuation theorem and the properties of $A_{3}$, they obtained sufficient conditions for the existence of periodic solutions to a kind of Liénard differential equation. Afterwards, Ren et al. [19] in 2011 considered a kind of second-order functional differential equation. By applications of the fixed point index theorem and the properties of the linear difference operator $(A_{4}x)(t):=x(t)-cx(t- \delta (t))$, where $\delta (t)$ is a periodic function, they obtained sufficient conditions for the existence, multiplicity and nonexistence of positive periodic solutions to the corresponding equation. Subsequently, Cheng and Li in [3] investigated the difference operator $(A_{5}x)(t):=x(t)-c(t)x(t-\delta (t))$, and applied it to a study of the corresponding functional differential equation.

Naturally, a new question arises: how does the linear difference operator work on multiple variable parameters? Besides practical interests, the topic has obvious intrinsic theoretical significance. To answer this question, in this paper, we discuss properties of the difference operator with multiple variable parameters $(Ax)(t):=x(t)- \sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i}(t))$, which is shown in Sect. 2, where $c_{i}(t)$, $\delta _{i}(t)\in C(\mathbb{R},\mathbb{R})$, and $c_{i}(t)$, $\delta _{i}(t)$ are ω-periodic functions on t, ω is a positive constant. As applications of properties of the difference operator A, we investigate the existence of periodic solutions for two kinds of second-order differential equations as follows.

In Sect. 3, we consider a kind of second-order differential equation with difference operator A:

$$ \bigl((Ax) (t) \bigr)''+a(t)x(t)=f \bigl(t,x\bigl(t-\tau (t)\bigr)\bigr), $$

(1.1)

where $\tau (t)\in C(\mathbb{R},\mathbb{R})$, $a(t)\in C(\mathbb{R},(0,+ \infty ))$, $f(t,x):=f(t,x(t-\tau (t)))\in C( \mathbb{R\times \mathbb{R}},\mathbb{R})$, and $\tau (t)$, $a(t)$, $f(t,x)$ are ω-periodic functions on t. By employing properties of A and Krasnoselskii’s fixed point theorem, some sufficient conditions for the existence of positive periodic solutions are established. Meanwhile, we obtain the $f(t,x)$ condition which is weaker than the condition $F(t,x):=f(t,x(t-\tau (t)))-ca(t)x(t-\tau (t))$ in [5, 14]. And we establish the existence of positive periodic solutions of Eq. (1.1) in the cases that $0<\sum_{i=1} ^{n}c_{i}(t)<1$ and $-1<\sum_{i=1}^{n}c_{i}(t)<0$, the authors in [19, 22] only discussed the existence of periodic solutions for equations in the case that $-1< c<0$.

In Sect. 4, by applications of the extension of Mawhin’s continuous theorem due to Ge and Ren [9], we study the following second-order quasi-linear differential equation:

$$ \bigl(\phi _{p}(Ax)'(t) \bigr)'=\tilde{f}\bigl(t,x(t),x'(t)\bigr), $$

(1.2)

where $\phi _{p}:\mathbb{R}\to \mathbb{R}$ is given by $\phi _{p}(s)=|s|^{p-2}s$, where $p>1$ is a constant, f̃: $[0,T]\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ is an $L^{2}$-Carathéodory function, i.e. it is measurable in the first variable and continuous in the second variable, and for every $0< r< s$ there exists $h_{r,s}\in L^{2}[0,T]$ such that $|\tilde{f}(t,x(t),x'(t))| \leq h_{r,s}$ for all $x\in [r,s]$ and a.e. $t\in [0,T]$. The obvious difficulty of Eq. (1.2) lies in the following two respects. First, although $(Ax)(t)=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i}(t))$ is a natural generalization of the operator $(A_{j}x)(t)$, $j=1,2,3,4,5$, this class of differential equation with A typically possesses a more complicated nonlinearity than differential equation with $(A_{j}x)(t)$. Second, we do not get $(Ax)'(t)=(Ax')(t)$, it means that the prior bounds of periodic solutions are not easy to estimate, we get over this problem here.

2 Properties of the difference operator A

In this section, we consider properties of the difference operator A. We first give the following notations which will be used in the proofs. Let

$$ C_{\omega }:=\bigl\{ x\in C(\mathbb{R},\mathbb{R}): x(t+\omega )=x(t), t \in \mathbb{R}\bigr\} $$

with norm $\|x\|:= \max_{t\in [0,\omega ]} |x(t)|$. Clearly, $(C_{\omega }, \|\cdot \|)$ is a Banach space. Define

$$\begin{aligned}& C_{\omega }^{+}:=\bigl\{ x \in C\bigl(\mathbb{R},(0,+\infty ) \bigr): x(t+\omega )=x(t), t\in \mathbb{R}\bigr\} , \\& c_{*}:=\min_{t\in [0,\omega ]}{ \Biggl\vert \sum _{i=1}^{n}c_{i}(t) \Biggr\vert },\qquad c^{*}:=\max_{t\in [0,\omega ]}{ \Biggl\vert \sum _{i=1}^{n}c_{i}(t) \Biggr\vert }, \\& \Vert c_{i} \Vert :=\max_{t\in [0,\omega ]} \bigl\vert c_{i}(t) \bigr\vert ,\qquad c_{\infty }:=\sum _{i=1}^{n} \Vert c_{i} \Vert ,\quad i=1,2,\ldots ,n, \\& k:=\bigl\{ \hat{k}\mid \Vert c_{\hat{k}} \Vert =\max \bigl\{ \Vert c_{1} \Vert , \Vert c _{2} \Vert ,\ldots, \Vert c_{n} \Vert \bigr\} \bigr\} . \end{aligned}$$

Lemma 2.1

([12])

If$c(t)\in C_{\omega }$, $\delta (t)\in C_{\omega }^{1}:=\{x\in C^{1}(\mathbb{R},\mathbb{R}): x(t+\omega )=x(t), t \in \mathbb{R}\}$and$\delta '(t)<1$, then$c(\mu (t))\in C_{\omega }$, where$\mu (t)$is the inverse function of$t-\delta (t)$.

Define operators $A,B:C_{\omega }\rightarrow C_{\omega }$ by

$$ (Ax) (t)=x(t)-\sum_{i=1}^{n}c_{i}(t)x \bigl(t-\delta _{i}(t)\bigr),\qquad (Bx) (t)=\sum _{i=1}^{n}c_{i}(t)x\bigl(t-\delta _{i}(t)\bigr), $$

then we have the following properties of the difference operator A.

Theorem 2.2

• (1) If$\sum_{i=1}^{n}\|c_{i}\|<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{1- \sum_{i=1}^{n} \Vert c_{i} \Vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned} $$

• (2) If$\sum_{i=1}^{n}\|e_{i}\|<1$and$\delta _{k}'(t)<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert e_{k} \Vert \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert e _{i} \Vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{ \Vert e_{k} \Vert }{1- \sum_{i=1}^{n} \Vert e_{i} \Vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt, \end{aligned} $$

  where$\sum_{i=1}^{n}\|e_{i}\|=\|\frac{1}{c_{k}}\| +\sum_{\substack{i=1\\ i\neq k}}^{n}\|\frac{c_{i}}{c_{k}}\|$, and for any$t_{0}\in \mathbb{R}$, $c_{k}(t_{0})\neq 0$.

Proof

Case 1: $\sum_{i=1}^{n}\|c_{i}\|<1$.

Let $t=D_{0}$ and $D_{j}=t-\sum_{i=1}^{j}\delta _{l_{i}}(D_{i-1})$, $j=1, 2, \ldots$ , then

$$\begin{aligned}& (Bx) (t)=\sum_{l_{1}=1}^{n}c_{l_{1}}(D_{0})x(D_{1}), \\& \bigl(B^{2}x\bigr) (t)=\sum_{l_{1}=1}^{n}c_{l_{1}}(D_{0}) \sum_{l_{2}=1}^{n}c_{l_{2}}(D_{1})x(D_{2}), \end{aligned}$$

therefore, we have

$$ \bigl(B^{j}x\bigr) (t)=\sum_{l_{1}=1}^{n}c_{l_{1}}(D_{0}) \sum_{l _{2}=1}^{n}c_{l_{2}}(D_{1}) \cdots \sum_{l_{j}=1}^{n}c_{l_{j}}(D _{j-1})x(D_{j}) $$

and

$$ \sum_{j=0}^{\infty }\bigl(B^{j}x \bigr) (t)=x(t)+\sum_{j=1}^{\infty } \sum _{l_{1}=1}^{n}c_{l_{1}}(D_{0}) \sum_{l_{2}=1}^{n}c _{l_{2}}(D_{1}) \cdots \sum_{l_{j}=1}^{n}c_{l_{j}}(D_{j-1})x(D _{j}), $$

where $B^{0}=I$. Since $A=I-B$ and $\|B\|\leq \sum_{i=1}^{n}\|c _{i}\|<1$, we see that A has a continuous inverse $A^{-1}$: $C_{\omega }\rightarrow C_{\omega }$ with

$$ \begin{aligned}[b] \bigl(A^{-1}x\bigr) (t)&= \bigl((I-B)^{-1}x\bigr) (t) =\sum_{j=0}^{\infty } \bigl(B^{j}x\bigr) (t) \\ &=x(t)+\sum_{j=1}^{\infty } \sum _{l_{1}=1}^{n}c_{l _{1}}(D_{0}) \sum_{l_{2}=1}^{n}c_{l_{2}}(D_{1}) \cdots \sum_{l_{j}=1}^{n}c_{l_{j}}(D_{j-1})x(D_{j}). \end{aligned} $$

(2.1)

Then

$$\begin{aligned} \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert &= \Biggl\vert x(t)+\sum_{j=1}^{\infty } \sum _{l_{1}=1}^{n}c_{l_{1}}(D_{0}) \sum_{l_{2}=1}^{n}c_{l _{2}}(D_{1}) \cdots \sum_{l_{j}=1}^{n}c_{l_{j}}(D_{j-1})x(D_{j}) \Biggr\vert \\ &\leq \Vert x \Vert +\sum_{j=1}^{\infty } \Biggl(\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr)^{j} \Vert x \Vert \\ &\leq \frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }. \end{aligned}$$

Moreover, we obtain

$$\begin{aligned} \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt &= \int ^{\omega }_{0} \Biggl\vert \sum _{j=0}^{\infty }\bigl(B^{j}x\bigr) (t) \Biggr\vert \,dt \\ &\leq \sum_{j=0}^{\infty } \int ^{\omega }_{0} \bigl\vert \bigl(B^{j}x \bigr) (t) \bigr\vert \,dt \\ &\leq \sum_{j=0}^{\infty } \int ^{\omega }_{0} \Biggl\vert \sum _{l_{1}=1}^{n}c_{l_{1}}(D_{0}) \sum_{l_{2}=1}^{n}c_{l _{2}}(D_{1}) \cdots \sum_{l_{j}=1}^{n}c_{l_{j}}(D_{j-1})x(D_{j}) \Biggr\vert \,dt \\ &\leq \sum_{j=0}^{\infty } \Biggl(\sum _{i=1}^{n} \Vert c _{i} \Vert \Biggr)^{j} \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt \\ &\leq \frac{1}{1-\sum_{i=1}^{n} \Vert c_{i} \Vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned}$$

Case 2: $\sum_{i=1}^{n}\|e_{i}\|<1$ and $\delta '_{k}(t)<1$.

The operator $(Ax)(t)=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i}(t))$ can be converted to

$$\begin{aligned} (Ax) (t)&=x(t)-c_{k}(t)x\bigl(t-\delta _{k}(t)\bigr)- \sum_{\substack{i=1\\i\neq k}}^{n}c_{i}(t)x \bigl(t-\delta _{i}(t)\bigr) \\ &=-c_{k}(t) \Biggl(x\bigl(t-\delta _{k}(t)\bigr)- \frac{x(t)}{c_{k}(t)} +\sum_{\substack{i=1\\i\neq k}}^{n} \frac{c_{i}(t)}{c_{k}(t)}x\bigl(t- \delta _{i}(t)\bigr) \Biggr). \end{aligned}$$

From Lemma 2.1, there exists an inverse function $\mu \in C(R,R)$, such that $\mu (t-\delta _{k}(t))=t$. Define

$$\begin{aligned}& (Ex) (t)=x(t)-\frac{1}{c_{k}(\mu (t))}x\bigl(\mu (t)\bigr) +\sum _{\substack{i=1 \\ i\neq k}}^{n}\frac{c_{i}(\mu (t))}{c_{k}(\mu (t))} x\bigl(\mu (t)- \delta _{i}\bigl(\mu (t)\bigr)\bigr), \\& e_{i}(t)= \textstyle\begin{cases} \frac{1}{c_{k}(\mu (t))},& \mbox{for } i=k, \\ -\frac{c_{i}(\mu (t))}{c_{k}(\mu (t))},& \mbox{for } i\neq k; \end{cases}\displaystyle \qquad \varepsilon _{i}(t)= \textstyle\begin{cases} \mu (t),& \mbox{for } i=k, \\ \mu (t)-\delta _{i}(\mu (t)),& \mbox{for } i\neq k. \end{cases}\displaystyle \end{aligned}$$

Then $(Ex)(t) =x(t)-\sum_{i=1}^{n}e_{i}(t)x(\varepsilon _{i}(t))$. Define $(\hat{B}x)(t)=\sum_{i=1}^{n}e_{i}(t)x(\varepsilon _{i}(t))$, let $\hat{D}_{0}=t$ and $\hat{D}_{j}=\varepsilon _{l_{j}}\cdots \varepsilon _{l_{2}}\varepsilon _{l_{1}}(t)$, $j=0, 1,2,\ldots$ , $l_{j}=1,2, \ldots ,n$, we have

$$\begin{aligned}& (\hat{B}x) (t) =\sum_{{l_{1}}=1}^{n}e_{l_{1}}( \hat{D}_{0})x( \hat{D}_{1}), \\& \bigl(\hat{B}^{2}x\bigr) (t) =\sum_{{l_{1}}=1}^{n}e_{l_{1}}( \hat{D}_{0}) \sum_{{l_{2}}=1}^{n}e_{l_{2}}( \hat{D}_{1}) x(\hat{D}_{2}), \\& \ldots, \\& \bigl(\hat{B}^{j}x\bigr) (t) =\sum_{{l_{1}}=1}^{n}e_{l_{1}}( \hat{D}_{0}) \sum_{{l_{2}}=1}^{n}e_{l_{2}}( \hat{D}_{1})\cdots \sum_{{l_{j}}=1}^{n}e_{l_{j}}( \hat{D}_{j-1}) x(\hat{D}_{j}). \end{aligned}$$

Since $\|\hat{B}\|\leq \sum_{i=1}^{n}\|e_{i}\| = \Vert \frac{1}{c _{k}} \Vert +\sum_{i\neq k}^{n} \Vert \frac{c_{i}}{c _{k}} \Vert <1$, we arrive at

$$ \bigl(E^{-1}x\bigr) (t)=\sum_{j=0}^{\infty } \bigl(\hat{B}^{j}x\bigr) (t) =\sum_{j=0}^{\infty } \Biggl(\sum_{{l_{1}}=1}^{n}e_{l_{1}}( \hat{D}_{0}) \sum_{{l_{2}}=1}^{n}e_{l_{2}}( \hat{D}_{1})\cdots \sum_{{l_{j}}=1}^{n}e_{l_{j}}( \hat{D}_{j-1}) x(\hat{D}_{j}) \Biggr). $$

Since $(Ax)(t)=-c_{k}(t)(Ex)(t-\delta _{k}(t)):=X(t)\in C_{\omega }$, we have $(Ex)(t)=-\frac{X(\mu (t))}{c_{k}(\mu (t))} =-e_{k}(t) X(\mu (t)):=X _{0}(t)\in C_{\omega }$. Therefore,

$$ \begin{aligned} \bigl(A^{-1}X\bigr) (t)&=x(t)= \bigl(E^{-1}X_{0}\bigr) (t) \\ &=\sum_{j=0}^{\infty } \Biggl(\sum _{{l_{1}}=1}^{n}e_{l_{1}}(\hat{D}_{0}) \sum_{{l_{2}}=1}^{n}e_{l_{2}}( \hat{D}_{1})\cdots \sum_{{l_{j}}=1}^{n}e_{l_{j}}( \hat{D}_{j-1}) X_{0}(\hat{D}_{j}) \Biggr). \end{aligned} $$

Similar to Case 1, we can get

$$\begin{aligned} \bigl\vert \bigl(A^{-1}X\bigr) (t) \bigr\vert &= \Biggl\vert \sum_{j=0}^{\infty } \Biggl(\sum _{{l_{1}}=1}^{n}e_{l_{1}}(\hat{D}_{0}) \sum_{{l_{2}}=1} ^{n}e_{l_{2}}( \hat{D}_{1})\cdots \sum_{{l_{j}}=1}^{n}e_{l_{j}}( \hat{D}_{j-1}) X_{0}(\hat{D}_{j}) \Biggr) \Biggr\vert \\ &\leq \sum_{j=0}^{\infty } \Biggl(\sum _{i=1}^{n} \Vert e _{i} \Vert \Biggr)^{j} \Vert X_{0} \Vert \\ &\leq \frac{ \Vert X \Vert \Vert e_{k} \Vert }{1-\sum_{i=1}^{n} \Vert e_{i} \Vert } \end{aligned}$$

and

$$ \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}X \bigr) (t) \bigr\vert \,dt \leq \frac{1}{1-\sum_{i=1} ^{n} \Vert e_{i} \Vert } \int ^{\omega }_{0} \bigl\vert X_{0}(t) \bigr\vert \,dt \leq \frac{ \Vert e_{k} \Vert }{1- \sum_{i=1}^{n} \Vert e_{i} \Vert } \int ^{\omega }_{0} \bigl\vert X(t) \bigr\vert \,dt. $$

 □

Remark 2.3

Theorem 2.2 extends and improves the corresponding lemmas in [3, 8, 13, 19, 26].

If $\delta _{i}(t)=\delta _{i}$, $i=1,2,\ldots ,n$, here $\delta _{i}$ are constants, then the operator A can be written as

$$ (Ax) (t)=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}), $$

then we have the following corollary.

Corollary 2.4

• (1) If$\sum_{i=1}^{n}\|c_{i}\|<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{1- \sum_{i=1}^{n} \Vert c_{i} \Vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned} $$

• (2) If$\sum_{i=1}^{n}\|e_{i}\|<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert e_{k} \Vert \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert e _{i} \Vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{ \Vert e_{k} \Vert }{1- \sum_{i=1}^{n} \Vert e_{i} \Vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt, \end{aligned} $$

  where$\sum_{i=1}^{n}\|e_{i}\|=\|\frac{1}{c_{k}}\| +\sum_{\substack{i=1 \\ i\neq k}}^{n}\|\frac{c_{i}}{c_{k}}\|$, and for any$t_{0}\in \mathbb{R}$, $c_{k}(t_{0})\neq 0$.

If $c_{i}(t)=c_{i}$, $i=1,2,\ldots ,n$, i.e., the $c_{i}$ are constants, then the operator A can be written as

$$ (Ax) (t)=x(t)-\sum_{i=1}^{n}c_{i}x \bigl(t-\delta _{i}(t)\bigr), $$

therefore, we have the following corollary.

Corollary 2.5

• (1) If$\sum_{i=1}^{n}|c_{i}|<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \vert c_{i} \vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{1- \sum_{i=1}^{n} \vert c_{i} \vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned} $$

• (2) If$\vert \frac{1}{c_{k}} \vert +\sum_{\substack{i=1 \\ i\neq k}}^{n} \vert \frac{c_{i}}{c_{k}} \vert <1$, and$\delta _{k}'(t)<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{ \vert c_{k} \vert -1-\sum_{\substack{i=1\\i\neq k}}^{n} \vert c_{i} \vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{ \vert c_{k} \vert -1- \sum_{\substack{i=1\\i\neq k}}^{n} \vert c_{i} \vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt, \end{aligned} $$

  where$|c_{k}|=\max \{|c_{1}|,|c_{2}|,\ldots, |c_{n}|\}$and$c_{k}\neq 0$.

If $c_{i}(t)=c_{i}$, $\delta _{i}(t)=\delta _{i}$, $i=1,2,\ldots ,n$, i.e. the $c_{i}$, $\delta _{i}$ are constants, then the operator A can be written as

$$ (Ax) (t)=x(t)-\sum_{i=1}^{n}c_{i}x(t- \delta _{i}), $$

then we obtain the following.

Corollary 2.6

• (1) If$\sum_{i=1}^{n}|c_{i}|<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \vert c_{i} \vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{1- \sum_{i=1}^{n} \vert c_{i} \vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned} $$

• (2) If$\vert \frac{1}{c_{k}} \vert +\sum_{\substack{i=1 \\ i\neq k}}^{n} \vert \frac{c_{i}}{c_{k}} \vert <1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$\begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{ \vert c_{k} \vert -1-\sum_{\substack{i=1\\i\neq k}}^{n} \vert c_{i} \vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{ \vert c_{k} \vert -1- \sum_{\substack{i=1\\i\neq k}}^{n} \vert c_{i} \vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt, \end{aligned}$$

  where$|c_{k}|=\max \{|c_{1}|,|c_{2}|,\ldots, |c_{n}|\}$and$c_{k}\neq 0$.

Remark 2.7

Corollary 2.6 can be found in [13].

If $n=1$, then the operator A can be written as

$$ (Ax) (t)=x(t)-c_{1}(t)x\bigl(t-\delta _{1}(t)\bigr), $$

therefore, we can get the following corollary.

Corollary 2.8

• (1) If$\|c_{1}\|<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{1- \Vert c_{1} \Vert }, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{1- \Vert c _{1} \Vert } \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned} $$

• (2) If$c_{1*}>1$and$\delta _{1}'(t)<1$, then the operatorAhas a continuous inverse$A^{-1}$on$C_{\omega }$, satisfying

  $$ \begin{aligned} (\mathrm{i}) &\quad \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{c_{1*}-1}, \\ (\mathrm{ii})&\quad \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \frac{1}{c_{1*}-1} \int ^{\omega }_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned} $$

Remark 2.9

Corollary 2.8 can be found in [3].

Remark 2.10

When $n=1$, (1) if $c_{1}(t)=c$, where c is constant, we can get the corresponding properties of $A_{4}$ in [19]; (2) if $\delta _{1}(t)=\delta$, where δ is constant, we can get the corresponding properties of $A_{3}$ in [8]; (3) if $c_{1}(t)=c$, $\delta _{1}(t)=\delta$, we can get the corresponding properties of $A_{1}$ in [26].

3 Periodic solutions for Eq. (1.1)

In this section, we discuss the existence of positive periodic solutions for Eq. (1.1) in the cases that $0<\sum_{i=1}^{n}c _{i}(t)<1$ and $-1<\sum_{i=1}^{n}c_{i}(t)<0$. Firstly, we recall Krasnoselskii’s fixed point theorem and some lemmas which our proofs are based on.

Theorem 3.1

(Krasnoselskii’s fixed point theorem [1])

Let$C_{\omega }$be a Banach space. Assume thatΩis a bounded closed convex subset of$C_{\omega }$. IfQ, $S:\varOmega \rightarrow C _{\omega }$satisfy

• (i) $Qx_{1}+Sx_{2}\in \varOmega$, $\forall x_{1}$, $x_{2} \in \varOmega$,

• (ii) Sis a contractive operator andQis a completely continuous operator.

Then$Q+S$has a fixed point inΩ.

Lemma 3.2

([5])

The equation

$$ \textstyle\begin{cases} y''(t)+My(t)=h(t), \\ y(0)=y(\omega ),\qquad y'(0)=y'(\omega ), \end{cases} $$

(3.1)

has a uniqueω-periodic solution

$$ y(t)= \int ^{\omega }_{0} G(t,s)h(s)\,ds, $$

where

$$ G(t,s)= \textstyle\begin{cases} \frac{\cos \sqrt{M}(t-s-\frac{\omega }{2})}{2\sqrt{M}\sin \frac{ \sqrt{M}\omega }{2}},& 0\leq s\leq t\leq \omega , \\ \frac{\cos \sqrt{M}(t-s+\frac{\omega }{2})}{2\sqrt{M}\sin \frac{\sqrt{M}\omega }{2}},& 0\leq t< s\leq \omega . \end{cases} $$

Lemma 3.3

([5])

$\int ^{\omega }_{0} G(t,s)\,ds=\frac{1}{M}$. And$G(t,s)$is a differentiable function witht.

Lemma 3.4

([22])

If$M<(\frac{\pi }{\omega })^{2}$, then$0< l\leq G(t,s)\leq L$for all$t\in [0,\omega ]$and$s\in [0,\omega]$.

Next, we consider the existence of positive periodic solutions for Eq. (1.1) in the case that $c_{\infty }\in (0,\frac{m}{M+m})$. Let $y(t)=(Ax)(t)$, from Theorem 2.2, we have $x(t)=(A^{-1}y)(t)$. Hence, Eq. (1.1) can be transformed into

$$ y''(t)+a(t)y(t)-a(t)H \bigl(y(t)\bigr)=f\bigl(t,x\bigl(t-\tau (t)\bigr)\bigr), $$

(3.2)

where $H(y(t))=- (\sum_{i=1}^{n}c_{i}(t)(A^{-1}y)(t-\delta _{i}(t)) ) =- (\sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i}(t)) )$.

We consider

$$ y''(t)+a(t)y(t)-a(t)H \bigl(y(t)\bigr)=h(t),\quad h\in C_{\omega }^{+}. $$

(3.3)

Define the operators $T,N:C_{\omega }\rightarrow C_{\omega }$ by

$$ (Th) (t)= \int ^{\omega }_{0} G(t,s)h(s)\,ds,\qquad (Ny) (t)= \bigl(M-a(t)\bigr)y(t)+a(t)H\bigl(y(t)\bigr). $$

(3.4)

Clearly, T is completely continuous and N is bounded in $C_{\omega }$. From Eq. (3.4) and Lemma 3.2, the solution for Eq. (3.3) can be written as

$$ y(t)=(Th) (t)+(TNy) (t). $$

(3.5)

On the other hand, since $H(y(t))=-\sum_{i=1}^{n}c_{i}(t)(A ^{-1}y)(t-\delta _{i}(t))$, from Lemma 2.2, it is clear that

$$\begin{aligned} \begin{aligned}[b] \bigl\vert Ny(t) \bigr\vert &\leq \bigl(M-a(t)\bigr) \bigl\vert y(t) \bigr\vert + \bigl\vert a(t) \bigr\vert \Biggl\vert -\sum _{i=1}^{n}c_{i}(t) \bigl(A ^{-1}y\bigr) \bigl(t-\delta _{i}(t)\bigr) \Biggr\vert \\ &\leq (M-m) \bigl\vert y(t) \bigr\vert +M\sum_{i=1}^{n} \Vert c_{i} \Vert \cdot \frac{1}{1- \sum_{i=1}^{n} \Vert c_{i} \Vert }\cdot \Vert y \Vert \\ &\leq \biggl(M-m+\frac{Mc_{\infty }}{1-c_{\infty }} \biggr) \Vert y \Vert . \end{aligned} \end{aligned}$$

(3.6)

And it follows that

$$ \Vert N \Vert \leq M-m+\frac{M c_{\infty }}{1-c_{\infty }}. $$

(3.7)

In view of $c_{\infty }\in (0,\frac{m}{M+m})$ and $\|T\|\leq \frac{1}{M}$ (see Lemma 3.3), we have from Eq. (3.7)

$$ \Vert TN \Vert \leq \Vert T \Vert \Vert N \Vert \leq \frac{1}{M} \biggl(M-m+\frac{Mc_{\infty }}{1-c _{\infty }} \biggr)\leq \frac{M-m(1-c_{\infty })}{M(1-c_{\infty })}< 1. $$

(3.8)

Therefore,

$$ y(t)=(I-TN)^{-1}(Th) (t). $$

(3.9)

Define an operator $P:C_{\omega }\rightarrow C_{\omega }$ by

$$ (Ph) (t)=(I-TN)^{-1}(Th) (t). $$

(3.10)

Obviously, if $M<(\frac{\pi }{\omega })^{2}$, for any $h\in C_{\omega }^{+}$, $y(t)=(Ph)(t)$ is the unique positive ω-periodic solution of Eq. (1.1). Let

$$ k_{0}:=\frac{\sqrt{(1-c_{*}^{2})^{2}+4\sigma ^{2}}-(1-c_{*}^{2})}{2 \sigma },\quad \sigma :=\frac{l}{L}. $$

Consider the equation

$$ \sigma c_{\infty }^{2}+ \bigl(1-c_{*}^{2}\bigr)c_{\infty }-\sigma =0, $$

(3.11)

it is easy to verify that $\sigma c_{\infty }^{2}+(1-c_{*}^{2})c_{ \infty }-\sigma \leq 0$ when $0< c_{\infty }\leq k_{0}$, and we have the following lemmas.

Lemma 3.5

Assume that$M<(\frac{\pi }{\omega })^{2}$, $c_{i}(t)\leq 0$, $c_{ \infty }\in (0,\frac{m}{M+m})$and$c_{\infty }\leq k_{0}$hold, where$i=1,2,\ldots , n$. Then

$$ (Th) (t)\leq (Ph) (t)\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }} \Vert Th \Vert ,\quad \textit{for all }h\in C_{\omega }^{+}. $$

Proof

From Eq. (3.8), for all $h\in C_{\omega }^{+}$, we obtain

$$\begin{aligned} (Ph) (t) &=(I-TN)^{-1}(Th) (t)\leq \bigl\Vert (I-TN)^{-1} \bigr\Vert \Vert Th \Vert \\ &\leq \frac{ \Vert Th \Vert }{1- \Vert TN \Vert } \leq \frac{M(1-c_{\infty })}{m-(M+m)c _{\infty }} \Vert Th \Vert . \end{aligned}$$

(3.12)

Since $\|TN\|<1$, by Neumann expansions of P, we have

$$ \begin{aligned}[b] P&=(I-TN)^{-1}T \\ &=\bigl(I+TN+(TN)^{2}+\cdots \bigr)T \\ &=T+TNT+(TN)^{2}T+ \cdots . \end{aligned} $$

(3.13)

From Lemma 3.4, for all $h(t)\in C_{\omega }^{+}$, we arrive at

$$\begin{aligned} (Th) (t)&= \int _{0}^{\omega }G(t,s)h(s)\,ds \\ &\geq l \int _{0}^{\omega }h(s)\,ds =\frac{l}{L}L \int _{0}^{\omega }h(s)\,ds \\ &\geq \sigma \max_{t\in [0,\omega ]} \int _{0}^{\omega }G(t,s)h(s)\,ds = \sigma \Vert Th \Vert >0. \end{aligned}$$

In view of $c_{i}(t)\leq 0$ ($i=1,2,\ldots , n$) and $c_{\infty } \leq k_{0}$, we get by Eq. (2.1)

$$ \begin{aligned} \bigl(A^{-1}Th\bigr) (t)&=\sum _{j=0}^{\infty } \Biggl(\sum _{{l_{1}}=1} ^{n}c_{l_{1}}(D_{0}) \sum_{{l_{2}}=1}^{n}c_{l_{2}}(D_{1}) \cdots \sum_{{l_{j}}=1}^{n}c_{l_{j}}(D_{j-1}) (Th) (D_{j}) \Biggr) \\ &=(Th) (t)+\sum_{j=\text{even}}^{\infty } \Biggl(\sum _{{l_{1}}=1}^{n}c_{l_{1}}(D_{0}) \sum_{{l_{2}}=1}^{n}c _{l_{2}}(D_{1}) \cdots \sum_{{l_{j}}=1}^{n}c_{l_{j}}(D_{j-1}) (Th) (D _{j}) \Biggr) \\ &\quad {}+ \sum_{j=\text{odd}}^{\infty } \Biggl(\sum _{{l_{1}}=1}^{n}c_{l_{1}}(D_{0}) \sum_{{l_{2}}=1}^{n}c _{l_{2}}(D_{1}) \cdots \sum_{{l_{j}}=1}^{n}c_{l_{j}}(D_{j-1}) (Th) (D _{j}) \Biggr) \\ &\geq \sigma \Vert Th \Vert +\sigma \Vert Th \Vert \sum _{j=\text{even}}^{\infty }c_{*}^{j} - \Vert Th \Vert \sum_{j=\text{odd}}^{\infty }c^{*j} \\ &\geq \sigma \Vert Th \Vert +\sigma \Vert Th \Vert \sum _{j=\text{even}}^{\infty }c_{*}^{j} - \Vert Th \Vert \sum_{j=\text{odd}}^{\infty }c_{\infty }^{j} \\ &\geq \frac{\sigma \Vert Th \Vert }{1-c_{*}^{2}}-\frac{c_{\infty } \Vert Th \Vert }{1-c_{\infty }^{2}} \geq 0, \end{aligned} $$

therefore, from equality (3.4), we can observe that

$$ (NTh) (t)= \bigl(M-a(t)\bigr) (Th) (t)- a(t) \Biggl(\sum _{i=1}^{n}c_{i}(t) \bigl(A ^{-1}Th\bigr) \bigl(t-\delta _{i}(t)\bigr) \Biggr) \geq 0, $$

clearly, $(TNTh)(t)\geq 0$. Then from the above analysis, we can get

$$ \begin{aligned} (Ph) (t)&=(Th) (t)+(TNTh) (t)+\bigl((TN)^{2}Th\bigr) (t)+ \bigl((TN)^{3}Th\bigr) (t)+\cdots \\ &\geq (Th) (t), \quad \mbox{for all } h\in C_{\omega }^{+}. \end{aligned} $$

 □

Lemma 3.6

Assume that$M<(\frac{\pi }{\omega })^{2}$, $c_{i}(t)\geq 0$and$c_{\infty }\in (0,\frac{m}{M+m})$hold, where$i=1,2,\ldots ,n$. Then

$$ \frac{m-(M+m)c_{\infty }}{M(1-c_{\infty })}(Th) (t) \leq (Ph) (t)\leq \frac{M(1-c _{\infty })}{m-(M+m)c_{\infty }} \Vert Th \Vert , \quad \textit{for all }h\in C_{\omega }^{+}. $$

Proof

Similarly as the proof of Lemma 3.5, it is easy to verify that

$$ (Ph) (t)\leq \frac{ \Vert Th \Vert }{1- \Vert TN \Vert } \leq \frac{M(1-c_{\infty })}{m-(M+m)c _{\infty }} \Vert Th \Vert . $$

From Eq. (3.13), we have

$$ \begin{aligned}[b] P &= \bigl(I+TN+(TN)^{2}+(TN)^{3}+\cdots \bigr)T \\ &=\bigl(I+(TN)^{2}+(TN)^{4}+\cdots \bigr)T+ \bigl(TN+(TN)^{3}+(TN)^{5}+\cdots \bigr)T \\ &=\bigl(I+(TN)^{2}+(TN)^{4}+\cdots \bigr)T+ \bigl(I+(TN)^{2}+(TN)^{4}+ \cdots \bigr)TNT \\ &=\bigl(I+(TN)^{2}+(TN)^{4}+\cdots \bigr) (I+TN)T. \end{aligned} $$

(3.14)

Then we get by Eq. (3.8)

$$ \begin{aligned} (Ph) (t) &\geq (I+TN) (Th) (t)\geq \bigl(I- \Vert TN \Vert \bigr) (Th) (t) \\ &\geq \frac{m-(m+M)c _{\infty }}{M(1-c_{\infty })}(Th) (t)>0,\quad \mbox{for all } h\in C_{\omega }^{+}. \end{aligned} $$

 □

When $n=1$, then $(Ax)(t)=x(t)-c_{1}(t)x(t-\delta _{1}(t))$, if $c_{1}(t)=c$, here c is a constant, then we have the following corollary.

Corollary 3.7

Assume that$M<(\frac{\pi }{\omega })^{2}$and$|c|\in (0, \frac{m}{M+m})$hold.

• (i) If$c<0$and$|c|\leq \sigma$, then

  $$ (Th) (t)\leq (Ph) (t)\leq \frac{M(1- \vert c \vert )}{m-(M+m) \vert c \vert } \Vert Th \Vert , \quad \textit{for all }h\in C_{\omega }^{+}. $$

• (ii) If$c>0$, then

  $$ \frac{m-(M+m)c}{M(1-c)}(Th) (t) \leq (Ph) (t)\leq \frac{M(1-c)}{m-(M+m)c} \Vert Th \Vert , \quad \textit{for all }h\in C_{\omega }^{+}. $$

Remark 3.8

If $\sum_{i=1}^{n}\|e_{i}\|<1$ and $\delta '_{k}(t)<1$, since

$$ \Vert TN \Vert \leq \Vert T \Vert \Vert N \Vert \leq 1- \frac{m}{M}+\frac{c_{\infty } \Vert \frac{1}{c _{k}} \Vert }{1-\sum_{i=1}^{n} \Vert e_{i} \Vert }, $$

we cannot get $\|TN\|<1$, therefore, we cannot get Lemma 3.5 and Lemma 3.6.

Next, we define operators $Q, S:C_{\omega }\rightarrow C_{\omega }$ by

$$ (Qx) (t)=P\bigl(f\bigl(t,x\bigl(t-\tau (t)\bigr) \bigr)\bigr),\qquad (Sx) (t)=\sum_{i=1}^{n}c_{i}(t)x \bigl(t-\delta _{i}(t)\bigr). $$

(3.15)

From the above analysis, the existence of periodic solutions for Eq. (1.1) is equivalent to the existence of solutions for the operator equation

$$ Qx+Sx=x $$

(3.16)

in $C_{\omega }$. Moreover, we have the following lemma.

Lemma 3.9

Qis completely continuous in$C_{\omega }$.

Proof

Since T is completely continuous and N is bounded in $C_{\omega }$, from Eq. (3.13), we see that P is completely continuous in $C_{\omega }$. By Eq. (3.15), it is easy to verify that Q is completely continuous in $C_{\omega }$. □

Now, we present our results of Eq. (1.1) in the case that $c_{\infty }\in (0,\frac{m}{M+m})$.

Case 1: $c_{i}(t)>0$, $i=1,2,\ldots ,n$.

Theorem 3.10

Assume that$M<(\frac{\pi }{\omega })^{2}$, $c_{i}(t)>0$and$0< c_{*}\leq \sum_{i=1}^{n}c_{i}(t)\leq c_{\infty }< \frac{m}{M+m}$hold. Furthermore, suppose the following condition is satisfied:

($F_{1}$):

There exist two positive constantsrandRsuch that

$$ \frac{M^{2}(1-c_{*})(1-c_{\infty })r}{(m-(M+m)c_{\infty })^{2}}< R $$

and

$$ \frac{M^{2}(1-c_{*})(1-c_{\infty })r}{m-(M+m)c_{\infty }}\leq f(t,x) \leq \bigl(m-(M+m)c_{\infty }\bigr)R, $$

for all$t\in [0,\omega ]$and$x\in [r,R]$.

Then Eq. (1.1) has at least one positiveω-periodic solution$x(t)$with$r\leq x(t)\leq R$.

Proof

Let

$$ \varOmega =\{x\in C_{\omega }: r\leq x\leq R, \mbox{for all } t\in \mathbb{R} \}. $$

Obviously, Ω is a bounded closed convex set in $C_{\omega }$.

For any $x\in \varOmega$, $t\in \mathbb{R}$, we get by Eq. (3.15)

$$ (Qx) (t+\omega ) =P\bigl(f\bigl(t+\omega ,x\bigl(t+\omega -\tau (t+\omega ) \bigr)\bigr)\bigr) =P\bigl(f\bigl(t,x\bigl(t- \tau (t)\bigr)\bigr)\bigr)=(Qx) (t) $$

and

$$ (Sx) (t+\omega ) =\sum_{i=1}^{n}c_{i}(t+ \omega )x\bigl(t+\omega - \delta _{i}(t+\omega )\bigr) =\sum _{i=1}^{n}c_{i}(t)x\bigl(t-\delta _{i}(t)\bigr)=(Sx) (t), $$

which show that $(Qx)(t)$ and $(Sx)(t)$ are ω-periodic. Thus, $Q(\varOmega )\subset C_{\omega }$, $S(\varOmega )\subset C_{\omega }$.

For all $x_{1}$, $x_{2}\in \varOmega$ and $t\in \mathbb{R}$, from Lemma 3.3, Lemma 3.6 and condition ($F_{1}$), we have

$$\begin{aligned} (Qx_{1}) (t)+(Sx_{2}) (t)&=P\bigl(f \bigl(t,x_{1}\bigl(t-\tau (t)\bigr)\bigr)\bigr)+\sum _{i=1} ^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }} \bigl\Vert T\bigl(f\bigl(t,x_{1} \bigl(t- \tau (t)\bigr)\bigr)\bigr) \bigr\Vert +\sum _{i=1}^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }}\max_{t\in [0, \omega ]} \int _{0}^{\omega } G(t,s)f\bigl(s,x_{1} \bigl(s-\tau (s)\bigr)\bigr)\,ds \\ &\quad {}+\sum_{i=1}^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }} \bigl(m-(M+m)c_{\infty }\bigr)R\cdot \frac{1}{M}+R\sum_{i=1}^{n}c_{i}(t) \\ &\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }} \bigl(m-(M+m)c_{\infty }\bigr)R\cdot \frac{1}{M}+c_{\infty }R \\ &=R \end{aligned}$$

and

$$\begin{aligned} (Qx_{1}) (t)+(Sx_{2}) (t) &=P\bigl(f \bigl(t,x_{1}\bigl(t-\tau (t)\bigr)\bigr)\bigr)+\sum _{i=1} ^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\geq \frac{m-(M+m)c_{\infty }}{M(1-c_{\infty })} \int _{0}^{\omega } G(t,s)f\bigl(s,x_{1} \bigl(s-\tau (s)\bigr)\bigr)\,ds \\ &\quad {}+\sum_{i=1}^{n}c_{i}(t)x_{2} \bigl(t- \delta _{i}(t)\bigr) \\ &\geq \frac{m-(M+m)c_{\infty }}{M(1-c_{\infty })}\cdot \frac{M^{2}(1-c _{\infty })}{m-(M+m)c_{\infty }}(1-c_{*})r \cdot \frac{1}{M} +r\sum_{i=1}^{n}c_{i}(t) \\ &\geq (1-c_{*})r+c_{*}r \\ &=r, \end{aligned}$$

which imply that $r\leq Qx_{1}+Sx_{2}\leq R$, for all $x_{1}$, $x_{2} \in \varOmega$. Therefore, $Qx_{1}+Sx_{2}\in \varOmega$.

For all $x_{1}$, $x_{2}\in \varOmega$, we obtain

$$\begin{aligned} \bigl\vert Sx_{1}(t)-Sx_{2}(t) \bigr\vert &= \Biggl\vert \sum_{i=1}^{n}c_{i}(t)x_{1} \bigl(t- \delta _{i}(t)\bigr) -\sum_{i=1}^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \Biggr\vert \\ &\leq \sum_{i=1}^{n} \bigl\vert c_{i}(t) \bigl(x_{1}\bigl(t-\delta _{i}(t) \bigr)-x _{2}\bigl(t-\delta _{i}(t)\bigr)\bigr) \bigr\vert \\ &\leq \sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x_{1}-x_{2} \Vert =c_{\infty } \Vert x _{1}-x_{2} \Vert , \end{aligned}$$

then from $c_{\infty }\in (0,\frac{m}{M+m})$, we conclude that S is contractive.

Since Q is completely continuous, by Theorem 3.1, there is an $x\in \varOmega$ such that $Qx+Sx=x$. Therefore, Eq. (1.1) has at least one positive ω-periodic solution $x(t)$ with $r\leq x(t)\leq R$. □

Case 2: $c_{i}(t)<0$, $i=1,2,\ldots ,n$.

We consider the existence of periodic solutions for Eq. (1.1) in the case that $-\frac{m}{M+m}<\sum_{i=1}^{n}c_{i}(t)<0$. To conclude the main result, firstly, we consider the equation

$$ Mc^{2}-(2M+m)c+m=0. $$

(3.17)

It is obvious that Eq. (3.17) has a solution $\zeta =\frac{2M+m- \sqrt{(2M+m)^{2}-4Mm}}{2M}$ and $0<\zeta <\frac{m}{m+M}$. If $c_{\infty }<\zeta$, we have $Mc_{\infty }^{2}-(2M+m)c_{\infty }+m>0$.

On the other hand, for any $0< c_{1}$, $c_{2}<\frac{m}{m+M}$, we obtain

$$ \begin{aligned} &(M+m)c_{1}c_{2}-mc_{1}-Mc_{2}+M \\ &\quad >(M+m)c_{1}c_{2}-m\frac{m}{m+M}-M \frac{m}{m+M}+M \\ &\quad =(M+m)c_{1}c_{2}-m+M>0. \end{aligned} $$

Then if $r>0$, we can get $\frac{(M+m)c_{\infty }c_{*}-Mc_{\infty }-mc _{*}+M}{Mc_{\infty }^{2}-(2M+m)c_{\infty }+m}r>0$, since $c_{*}\leq c _{\infty }<\frac{m}{M+m}$.

Therefore, we have the following theorem.

Theorem 3.11

Assume that$M<(\frac{\pi }{\omega })^{2}$, $c_{i}(t)<0$and$c_{\infty }<\min \{k_{0},\zeta \}$hold. Furthermore, suppose the following condition is satisfied:

($F_{2}$):

There exist two positive constantsr, Rsuch that

$$ \frac{(M+m)c_{\infty }c_{*}-Mc_{\infty }-mc_{*}+M}{Mc_{\infty }^{2}-(2M+m)c _{\infty }+m}r< R $$

and

$$ M(r+c_{\infty }R)\leq f(t,x)\leq \frac{m-(M+m)c_{\infty }}{1-c_{ \infty }}(R+c_{*}r), $$

for all$t\in [0,\omega ]$and$x\in [r,R]$.

Then Eq. (1.1) has at least one positiveω-periodic solution$x(t)$with$r\leq x(t)\leq R$.

Proof

We follow the same notations as in the proof of Theorem 3.10. For all $x_{1},x_{2}\in \varOmega$, from Lemma 3.3, Lemma 3.5 and condition ($F_{2}$), we see that

$$\begin{aligned} (Qx_{1}) (t)+(Sx_{2}) (t) &=P\bigl(f \bigl(t,x_{1}\bigl(t-\tau (t)\bigr)\bigr)\bigr)+\sum _{i=1} ^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }} \bigl\Vert T\bigl(f\bigl(t,x_{1} \bigl(t- \delta (t)\bigr)\bigr)\bigr) \bigr\Vert +\sum _{i=1}^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }}\max_{t\in [0, \omega ]} \int _{0}^{\omega } G(t,s)f\bigl(s,x_{1} \bigl(s-\delta (s)\bigr)\bigr)\,ds+r\sum_{i=1}^{n}c_{i}(t) \\ &\leq \frac{M(1-c_{\infty })}{m-(M+m)c_{\infty }}\cdot \frac{m-(M+m)c _{\infty }}{1-c_{\infty }}(R+c_{*}r) \cdot \frac{1}{M}-c_{*}r \\ &=R \end{aligned}$$

and

$$\begin{aligned} (Qx_{1}) (t)+(Sx_{2}) (t)&=P\bigl(f \bigl(t,x_{1}\bigl(t-\tau (t)\bigr)\bigr)\bigr)+\sum _{i=1} ^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\geq \int _{0}^{\omega } G(t,s)f\bigl(s,x_{1} \bigl(s-\tau (s)\bigr)\bigr)\,ds+\sum_{i=1}^{n}c_{i}(t)x_{2} \bigl(t-\delta _{i}(t)\bigr) \\ &\geq M(r+c_{\infty }R)\cdot \frac{1}{M}-c_{\infty }R \\ &=r. \end{aligned}$$

From the above two inequalities, it is clear that $Qx_{1}+Sx_{2} \in \varOmega$, for all $x_{1}$, $x_{2}\in \varOmega$.

We use a similar argument as in the proof of Theorem 3.10, we can observe that $Q(\varOmega )\subset C_{\omega }$, $S(\varOmega )\subset C _{\omega }$, S is contractive. Since Q is completely continuous, we get by a direct application of Theorem 3.1 that Eq. (1.1) has at least one positive ω-periodic solution $x(t)$ with $r\leq x(t)\leq R$. □

Remark 3.12

If $n=1$, then $(Ax)(t)=x(t)-c_{1}(t)x(t-\delta _{1}(t))$, we can also get Theorem 3.10 and Theorem 3.11 in a similar way.

If $n=1$ and $c_{1}(t)=c$, where c is a constant, from Corollary 3.7, we can get the following corollaries, which improve and extend the corresponding results from [5].

Corollary 3.13

Assume that$M<(\frac{\pi }{\omega })^{2}$and$0< c<\frac{m}{M+m}$hold. Furthermore, suppose the following condition is satisfied:

($F_{1}^{*}$):

There exist two positive constantsrandRsuch that

$$ \frac{M^{2}(1-c)^{2}r}{(m-(M+m)c)^{2}}< R $$

and

$$ M^{2}(1-c)^{2}r\leq f(t,x)\leq \bigl(m-(M+m)c \bigr)^{2}R, $$

for all$t\in [0,\omega ]$and$x\in [r,R]$.

Then Eq. (1.1) has at least one positiveω-periodic solution$x(t)$with$r\leq x(t)\leq R$.

Remark 3.14

Corollary 3.13 extends and improves Theorem 2.1 in [5].

Corollary 3.15

Suppose that$M<(\frac{\pi }{\omega })^{2}$, $c<0$and$|c|<\min \{ \sigma ,\zeta \}$hold. Furthermore, assume that the following condition is satisfied:

($F_{2}^{*}$):

There exist two non-negative constantsr, Rsuch that

$$ \frac{(M+m) \vert c \vert ^{2}-(M+m) \vert c \vert +M}{M \vert c \vert ^{2}-(2M+m) \vert c \vert +m}r< R $$

and

$$ M\bigl(r+ \vert c \vert R\bigr)\leq f(t,x)\leq \frac{m-(M+m) \vert c \vert }{1- \vert c \vert }\bigl(R+ \vert c \vert r\bigr), $$

for all$t\in [0,\omega ]$and$x\in [r,R]$.

Then Eq. (1.1) has at least oneω-periodic solution$x(t)$with$r\leq x(t)\leq R$.

Remark 3.16

Corollary 3.15 extends and improves Theorem 2.3 in [5].

4 Periodic solution for Eq. (1.2)

In this section, we investigative the existence of periodic solutions for Eq. (1.2) by applications of the extension of Mawhin’s continuous theorem [9], in order to use this theorem, we recall it first.

Let X̃ and Z̃ be Banach spaces with norms $\|\cdot \|_{\widetilde{X}}$ and $\|\cdot \|_{\widetilde{Z}}$, respectively. A continuous operator $\widetilde{M}:\widetilde{X} \cap \operatorname{dom} \widetilde{M}\to \widetilde{Z}$ is said to be quasi-linear if

• (1) $\operatorname{Im} \widetilde{M}:=\widetilde{M}(\widetilde{X}\cap \operatorname{dom} \widetilde{M})$ is a closed subset of Z̃;

• (2) $\operatorname{ker} \widetilde{M}:=\{x\in \widetilde{X}\cap \operatorname{dom} \widetilde{M}:\widetilde{M}x=0\}$ is a subspace of X̃ with $\operatorname{dim} \operatorname{ker}\widetilde{M}<+\infty$.

Let $\widetilde{X}_{1}=\operatorname{ker} \widetilde{M}$ and $\widetilde{X}_{2}$ be the complement space of $\widetilde{X}_{1}$ in X̃, then $\widetilde{X}=\widetilde{X}_{1}\oplus \widetilde{X}_{2}$. Meanwhile, $\widetilde{Z}_{1}$ is a subspace of Z̃ and $\widetilde{Z}_{2}$ is the complement space of $\widetilde{Z}_{1}$ in Z̃, so $\widetilde{Z}=\widetilde{Z}_{1}\oplus \widetilde{Z}_{2}$. Suppose that $\widetilde{P}:\widetilde{X}\to \widetilde{X}_{1}$ and $\widetilde{Q}: \widetilde{Z}\to \widetilde{Z} _{1}$ are two projects and $\widetilde{\varOmega }\subset \widetilde{X}$ is an open bounded set with the origin $\tilde{\theta }\in \widetilde{\varOmega }$.

Let $\widetilde{N}_{\tilde{\lambda }}:\overline{\widetilde{\varOmega }} \to \widetilde{Z}$, $\tilde{\lambda }\in [0,1]$ is a continuous operator. Denote $\widetilde{N}_{1}$ by Ñ, and let $\sum_{\lambda }=\{x\in \overline{\widetilde{\varOmega }}:\widetilde{M}x=\widetilde{N} _{\lambda }x\}$. $\widetilde{N}_{\tilde{\lambda }}$ is said to be M̃-compact in $\overline{\widetilde{\varOmega }}$ if

• (3) there is a vector subspace $\widetilde{Z}_{1}$ of Z̃ with $\operatorname{dim} \widetilde{Z}_{1}=$$\operatorname{dim} \widetilde{X}_{1}$ and an operator $\widetilde{R}:\overline{ \widetilde{\varOmega }}\times \widetilde{X}_{2}$ being continuous and compact such that, for $\tilde{\lambda }\in [0,1]$,

  $$\begin{aligned}& (\widetilde{I}-\widetilde{Q}) \widetilde{N}_{\tilde{\lambda }}(\overline{ \widetilde{\varOmega }})\subset \operatorname{Im} \widetilde{M}\subset ( \widetilde{I}-\widetilde{Q})\widetilde{Z}, \end{aligned}$$

  (4.1)
  $$\begin{aligned}& \widetilde{Q}\widetilde{N}_{\tilde{\lambda }} x=0,\qquad \tilde{\lambda }\in (0,1)\quad \Leftrightarrow \quad \widetilde{Q}\widetilde{N} \tilde{x}=0, \end{aligned}$$

  (4.2)
  $$\begin{aligned}& \widetilde{R}(\cdot ,0) \mbox{ is the zero operator}\quad \mbox{and}\quad \widetilde{R}(\cdot ,\tilde{\lambda })|_{\sum _{\tilde{\lambda }}} =( \widetilde{I}-\widetilde{P})|_{\sum _{\tilde{\lambda }}}, \end{aligned}$$

  (4.3)

  and

  $$ \widetilde{M}\bigl[\widetilde{P}+\widetilde{R}( \cdot ,\lambda )\bigr]=( \widetilde{I}-\widetilde{Q})\tilde{N}_{\tilde{\lambda }}. $$

  (4.4)

Lemma 4.1

([9])

LetX̃andZ̃be Banach space with norm$\|\cdot \|_{\widetilde{X}}$and$\|\cdot \|_{ \widetilde{Z}}$, respectively, and$\widetilde{\varOmega }\subset \widetilde{X}$be an open and bounded set with$\tilde{\theta }\in \widetilde{\varOmega }$. Suppose that$\widetilde{M}:\widetilde{X}\cap \operatorname{dom} \widetilde{M}\to \widetilde{Z}$is a quasi-linear operator and

$$ \widetilde{N}_{\tilde{\lambda }}:\overline{\widetilde{\varOmega }}\to \widetilde{Z},\quad \tilde{\lambda }\in (0,1) $$

is anM̃-compact mapping. In addition, if

• (a) $\widetilde{M}x\neq \widetilde{N}_{\tilde{\lambda }} x$, $\tilde{\lambda }\in (0,1)$, $x\in \partial \widetilde{\varOmega }$,

• (b) $\deg \{\widetilde{J}\widetilde{Q}\widetilde{N}, \widetilde{\varOmega }\cap \operatorname{ker} \widetilde{M},0\}\neq 0$,

where$\widetilde{N}=\widetilde{N}_{1}$, then the abstract equation$\widetilde{M}x=\widetilde{N}x$has at least one solution in$\overline{\widetilde{\varOmega }}$.

Let $\widetilde{J}:\widetilde{Z}_{1}\to \widetilde{X}_{1}$ be a homeomorphism with $\widetilde{J}(\tilde{\theta })=\tilde{\theta }$.

Theorem 4.2

Assume$\sum_{i=1}^{n}\|c_{i}\|<1$, or$\|\frac{1}{c_{k}}\| + \sum_{\substack{i=1 \\ i\neq k}}^{n}\|\frac{c_{i}}{c_{k}}\|<1$, Ω̃be open bounded set in$C^{1}_{\omega }$. Suppose the following conditions hold:

• (i) For each$\tilde{\lambda }\in (0,1)$, the equation

  $$ \bigl(\phi _{p}(Ax)'(t) \bigr)'=\tilde{\lambda }\tilde{f}\bigl(t,x(t),x'(t) \bigr) $$

  (4.5)

  has no solution on∂Ω̃.

• (ii) The equation

  $$ \widetilde{F}(a):=\frac{1}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(a,x(a),0\bigr)\,dt=0 $$

  has no solution on$\partial \widetilde{\varOmega }\cap \mathbb{R}$.

• (iii) The Brouwer degree

  $$ \deg \{\widetilde{F},\widetilde{\varOmega }\cap \mathbb{R},0\}\neq 0. $$

Then Eq. (1.2) has at least one periodic solution on$\overline{\widetilde{\varOmega }}$.

Proof

In order to use Lemma 4.1 to study the existence of periodic solution to Eq. (1.2). We can set $\widetilde{X}:=\{x\in C[0, \omega ]: x(0)=x(\omega )\}$ and $\widetilde{Z}:=C[0,\omega ]$,

$$ \widetilde{M}: \widetilde{X}\cap \operatorname{dom} \widetilde{M} \to \widetilde{Z}, \qquad (\widetilde{M}x) (t)=\bigl(\phi _{p}(Ax)'(t)\bigr)', $$

(4.6)

where $\operatorname{dom} \widetilde{M}:=\{u\in \widetilde{X}:\phi _{p}(Au)'\in C^{1}( \mathbb{R},\mathbb{R}) \}$. Then $\ker \widetilde{M}=\mathbb{R}$. In fact

$$\begin{aligned} \ker \widetilde{M}&=\bigl\{ x\in \widetilde{X}:\bigl(\phi _{p}(Ax)'(t) \bigr)'=0 \bigr\} \\ &=\bigl\{ x\in \widetilde{X}: \phi _{p}(Ax)'\equiv \tilde{c} \bigr\} \\ &=\bigl\{ x\in \widetilde{X}: (Ax)'\equiv \phi _{q}(c):=\tilde{c}_{1} \bigr\} \\ &=\bigl\{ x\in \widetilde{X}: (Ax) (t)\equiv \tilde{c}_{1}t+ \tilde{c}_{2} \bigr\} , \end{aligned}$$

where $q>1$ is a constant with $\frac{1}{p}+\frac{1}{q}=1$ and c̃, $\tilde{c}_{1}$, $\tilde{c}_{2}$ are constants in $\mathbb{R}$. Since $(Ax)(0)=(Ax)(\omega )$, we get $\ker \widetilde{M}=\{x\in \widetilde{X}: (Ax)(t)\equiv \tilde{c}_{2}\}$. In addition,

$$\begin{aligned} \operatorname{Im}\widetilde{M} =& \biggl\{ \tilde{y}\in\widetilde{Z}, \text{for }x(t)\in\widetilde{X}\cap \operatorname{dom}\widetilde{M}, \bigl(\phi_{p}(Ax)'(t)\bigr)'=\tilde{y}(t), \\ &\int^{\omega}_{0}\tilde{y}(t)\,dt=\int^{\omega}_{0}\bigl(\phi_{p}(Ax)'(t)\bigr)'\,dt=0 \biggr\} . \end{aligned}$$

So M̃ is quasi-linear. Let

$$\begin{aligned} &\widetilde{X}_{1}=\ker \widetilde{M},\qquad \widetilde{X}_{2}= \bigl\{ x\in \widetilde{X}: x(0)=x(\omega )=0\bigr\} , \\ &\widetilde{Z}_{1}=\mathbb{R},\qquad \widetilde{Z}_{2}= \operatorname{Im} \widetilde{M}. \end{aligned}$$

Clearly, $\operatorname{dim} \widetilde{X}_{1}= \operatorname{dim} \widetilde{Z} _{1}=1$, and $\widetilde{X}=\widetilde{X}_{1}\oplus \widetilde{X}_{2}$, $\widetilde{P}:\widetilde{X}\to \widetilde{X}_{1}$, $\widetilde{Q}: \widetilde{Z}\to \widetilde{Z}_{1}$, be defined by

$$ \widetilde{P}x=x(0),\qquad \widetilde{Q}y=\frac{1}{\omega } \int ^{\omega }_{0}y(s)\,ds. $$

For $\forall \overline{\widetilde{\varOmega }}\subset \widetilde{X}$, define $\widetilde{N}_{\tilde{\lambda }}:\overline{ \widetilde{\varOmega }}\to \widetilde{Z}$ by

$$ (\widetilde{N}_{\tilde{\lambda }} x) (t)=\tilde{\lambda } \tilde{f} \bigl(t,x(t),x'(t)\bigr). $$

We claim $(\widetilde{I}-\widetilde{Q})\widetilde{N}_{ \tilde{\lambda }} (\overline{\widetilde{\varOmega }})\subset \operatorname{Im} \widetilde{M}=(\widetilde{I}-\widetilde{Q})\widetilde{Z}$ holds. In fact, for $x\in \overline{\widetilde{\varOmega }}$, we see that

$$\begin{aligned} & \int ^{\omega }_{0}(\widetilde{I}-\widetilde{Q}) \widetilde{N}_{ \tilde{\lambda }} x(t)\,dt \\ &\quad = \int ^{\omega }_{0}(\widetilde{I}-\widetilde{Q}) \tilde{\lambda } \tilde{f}\bigl(t,x(t),x'(t)\bigr)\,dt \\ &\quad = \int ^{\omega }_{0}\tilde{\lambda } \tilde{f} \bigl(t,x(t),x'(t)\bigr)\,dt- \int ^{\omega }_{0}\frac{\tilde{\lambda }}{\omega } \int ^{\omega }_{0} \tilde{f}\bigl(s,x(s),x'(s) \bigr)\,ds\,dt \\ &\quad =0. \end{aligned}$$

Therefore, we have $(\widetilde{I}-\widetilde{Q})\widetilde{N}_{ \tilde{\lambda }} (\overline{\widetilde{\varOmega }})\subset \operatorname{Im} \widetilde{M}$. Moreover, for any $x\in \widetilde{Z}$, we get

$$\begin{aligned} & \int ^{\omega }_{0}(\widetilde{I}-\widetilde{Q})x(t)\,dt \\ &\quad = \int ^{\omega }_{0} \biggl(x(t)-\frac{1}{\omega } \int ^{\omega }_{0}x(s)\,ds \biggr)\,dt \\ &\quad = \int ^{\omega }_{0}x(t)\,dt- \int ^{\omega }_{0}\frac{1}{\omega } \int ^{\omega }_{0}x(s)\,ds\,dt \\ &\quad =0. \end{aligned}$$

So, $(\widetilde{I}-\widetilde{Q})\widetilde{Z}\subset \operatorname{Im} \widetilde{M}$. On the other hand, $x\in \operatorname{Im} \widetilde{M}$ and $\int ^{\omega }_{0}x(t)\,dt=0$, then we have $x(t)=x(t)-\int ^{\omega } _{0}x(t)\,dt$. Hence, we can get $x(t)\in (\widetilde{I}-\widetilde{Q}) \widetilde{Z}$. Therefore, $\operatorname{Im} \widetilde{M}=(\widetilde{I}- \widetilde{Q})\widetilde{Z}$.

From $\widetilde{Q}\widetilde{N}_{\tilde{\lambda }} x=0$, we obtain

$$ \frac{\tilde{\lambda }}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(t,x(t),x'(t) \bigr)\,dt=0. $$

Since $\tilde{\lambda }\in (0,1)$, then we have $\frac{1}{\omega } \int ^{\omega }_{0}\tilde{f}(t,x(t),x'(t))\,dt=0$. Therefore, $\widetilde{Q}\widetilde{N}x=0$, then Eq. (4.4) also holds.

Let $\widetilde{J}:\widetilde{Z}_{1}\to \widetilde{X}_{1}$, $\widetilde{J}(x)=x$, then $\widetilde{J}(0)=0$. Define $\widetilde{R}:\overline{ \widetilde{\varOmega }}\times [0,1]\to \widetilde{X}_{2}$,

$$\begin{aligned}& \widetilde{R}(x,\tilde{\lambda }) (t) \\& \quad =A^{-1} \int ^{t}_{0}\phi _{p}^{-1} \biggl(\tilde{a}+ \int ^{s}_{0}\lambda \tilde{f} \bigl(u,x(u),x'(u)\bigr)\,du -\frac{ \tilde{\lambda } s}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr)\,ds, \end{aligned}$$

(4.7)

where $\tilde{a}\in \widetilde{R}$ is a constant such that

$$ \begin{aligned}[b] \widetilde{R}(x,\tilde{ \lambda }) (\omega ) &=A^{-1} \int ^{\omega }_{0} \phi _{p}^{-1} \biggl(\tilde{a}+ \int ^{s}_{0}\tilde{\lambda } \tilde{f} \bigl(u,x(u),x'(u)\bigr)\,du-\frac{ \tilde{\lambda } s}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(t),x'(u) \bigr)\,du \biggr)\,ds \\ &=0. \end{aligned} $$

(4.8)

From Lemma 3.1 of [16], we know that ã is uniquely defined by

$$ \tilde{a}=\bar{\tilde{a}}(x,\lambda ), $$

where $\bar{\tilde{a}}(x,\lambda )$ is continuous on $\overline{ \widetilde{\varOmega }}\times [0,1]$ and bounded sets of $\overline{ \widetilde{\varOmega }}\times [0,1]$ into bounded sets of $\mathbb{R}$.

From Eq. (4.4), one can find that

$$ \widetilde{\mathrm{R}}:\overline{\widetilde{\varOmega }}\times [0,1] \to \widetilde{X}_{2}. $$

Now, for any $x\in \sum_{\tilde{\lambda }}=\{x\in \overline{ \widetilde{\varOmega }}: \widetilde{M}x=\widetilde{N}_{\tilde{\lambda }} x\}=\{x\in \overline{\widetilde{\varOmega }}: (\phi _{p}(Ax)'(t))'= \tilde{\lambda } \tilde{f}(t,x(t),x'(t))\}$, we have $\int ^{\omega } _{0}\tilde{f}(t,x(t),x'(t))\,dt=0$, together with Eq. (4.2) gives

$$\begin{aligned} \widetilde{R}(x,\tilde{\lambda }) (t)&=A^{-1} \int ^{t}_{0}\phi _{p}^{-1} (\tilde{a}+ \int ^{s}_{0}\tilde{\lambda } \tilde{f} \bigl(u,x(u),x'(u)\,du \bigr)\,ds \\ &=A^{-1} \int ^{t}_{0}\phi _{p}^{-1} \biggl(\tilde{a}+ \int ^{s}_{0} \bigl(\phi _{p}(Ax)'(u) \bigr)'\,du \biggr)\,ds \\ &=A^{-1} \int ^{t}_{0}\phi _{p}^{-1} \bigl(\tilde{a}+\phi _{p}(Ax)'(s)- \phi _{p}(Ax)'(0) \bigr)\,ds. \end{aligned}$$

Take $\tilde{a}=\phi _{p}(Ax)'(0)$, then we can get

$$\begin{aligned} \widetilde{R}(x,\tilde{\lambda }) (\omega )&=A^{-1} \int ^{\omega }_{0} \bigl(\phi _{p}^{-1} \bigl(\phi _{p}(Ax)'(s)\bigr) \bigr)\,ds \\ &=A^{-1} \int ^{\omega }_{0}(Ax)'(t)\,ds \\ &=A^{-1} \bigl((Ax) (\omega )-(Ax) (0) \bigr) \\ &=x(\omega )-x(0) \\ &=0, \end{aligned}$$

where ã is unique, we see that

$$ \tilde{a}=\bar{\tilde{a}}(x,\tilde{\lambda })=\phi _{p}(Ax)'(0),\quad \forall \tilde{\lambda }\in [0,1]. $$

So, we have

$$\begin{aligned} \tilde{R}(x,\tilde{\lambda }) (t)|_{x\in \sum _{\tilde{\lambda }}} &=A ^{-1} \int ^{t}_{0} \biggl(\phi _{p}^{-1} \biggl(\phi _{p}(Ax)'(0)+ \int ^{s} _{0} \tilde{\lambda } \tilde{f} \bigl(u,x(u),x'(u)\bigr)\,du \biggr) \biggr)\,ds \\ &=A^{-1} \int ^{t}_{0} \bigl(\phi _{p}^{-1} \bigl(\phi _{p}(Ax)'(s)\bigr) \bigr)\,ds \\ &=A^{-1} \int ^{t}_{0}(Ax)'(s)\,ds \\ &=x(t)-x(0) \\ &=(\widetilde{I}-\widetilde{P})x(t), \end{aligned}$$

which yields the second part of (4.3). Meanwhile, if $\tilde{\lambda }=0$, the

$$ \sum_{\tilde{\lambda }}=\{x\in \overline{\widetilde{\varOmega }}:{ \widetilde{M}}x =\widetilde{N}_{\tilde{\lambda }} x\}=\bigl\{ x\in \overline{ \widetilde{\varOmega }}: \bigl(\phi _{p}(Ax)'(t) \bigr)'=\tilde{\lambda } \tilde{f}\bigl(t,x(t),x'(t) \bigr) \bigr\} =\tilde{c}_{3}, $$

where $\tilde{c}_{3}\in \mathbb{R}$ is a constant. Thus, by the continuity of $\bar{\tilde{a}}(x,\tilde{\lambda })$ with respect to $(x,\tilde{\lambda })$, $\tilde{a}=\bar{\tilde{a}}(x,0)=\phi _{p}(A \tilde{c})'(0)=0$, we have

$$ \widetilde{R}(x,0) (t)=A^{-1} \int ^{t}_{0}\phi _{p}^{-1}(0)\,ds=0,\quad \forall x\in \overline{\widetilde{\varOmega }}, $$

which yields the first part of Eq. (4.3). Furthermore, we consider

$$\begin{aligned} & \widetilde{M}(\widetilde{P}+\widetilde{R}) =(\widetilde{I}- \widetilde{Q}) \widetilde{N}_{\tilde{\lambda }}, \end{aligned}$$

in fact,

$$ \frac{d}{dt}\phi _{p} \bigl(A(\widetilde{P}+\widetilde{R})\bigr)' =({\widetilde{I}}- { \widetilde{Q}})\widetilde{N}_{\tilde{\lambda }}. $$

(4.9)

Integrating both sides of Eq. (4.9) over $[0,s]$, we have

$$ \int ^{s}_{0}\frac{d}{dt}\phi _{p}\bigl(A(\widetilde{P}+\widetilde{R})\bigr)'\,ds = \int ^{s}_{0}(\widetilde{I}-\widetilde{Q}) \widetilde{N}_{ \tilde{\lambda }} \,ds. $$

Therefore,

$$\begin{aligned} \phi _{p}\bigl(A(\widetilde{P}+\widetilde{R})\bigr)'(s)- \tilde{a} &= \tilde{\lambda } \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du - \int ^{s}_{0}\frac{ \tilde{\lambda }}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)dudt \\ &=\tilde{\lambda } \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du-\frac{ \tilde{\lambda } s}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du, \end{aligned}$$

where $\tilde{a}:=\phi _{p}(A(\widetilde{P}+\widetilde{R}))'(0)$. Then we can get

$$ \bigl(A(\widetilde{P}+\widetilde{R}) \bigr)'(s) =\phi _{p}^{-1} \biggl( \tilde{a}+ \tilde{\lambda } \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du-\frac{ \tilde{\lambda } s}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr). $$

(4.10)

Integrating both sides of Eq. (4.10) over $[0,t]$, we arrive at

$$\begin{aligned}& \int ^{t}_{0}\bigl(A(\widetilde{P}+ \widetilde{R})\bigr)'(s)\,ds \\& \quad = \int ^{t}_{0}\phi _{p}^{-1} \biggl(\tilde{a}+\tilde{\lambda } \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du -\frac{\tilde{\lambda } s}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr)\,ds, \end{aligned}$$

then

$$\begin{aligned}& (\widetilde{P}+\widetilde{R}) (t)-(\widetilde{P}+\widetilde{R}) (0) \\& \quad =A ^{-1} \biggl( \int ^{t}_{0}\phi _{p}^{-1} \biggl(\tilde{a}+ \tilde{\lambda } \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du -\frac{ \tilde{\lambda } s}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr)\,ds \biggr). \end{aligned}$$

Since $\widetilde{R}(x,\tilde{\lambda })(0)=0$, $\widetilde{P}(t)= \widetilde{P}(0)=0$, we can get

$$\begin{aligned}& \widetilde{R}(x,\tilde{\lambda }) (t) \\& \quad =A^{-1} \biggl( \int ^{t}_{0}\phi _{p}^{-1} \biggl(\tilde{a}+\tilde{\lambda } \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du -\frac{ \tilde{\lambda } s}{\omega } \int ^{\omega }_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr)\,ds \biggr). \end{aligned}$$

Hence, $\widetilde{N}_{\tilde{\lambda }}$ is M-compact on $\overline{\widetilde{\varOmega }}$. Obviously, the equation

$$ \bigl(\phi _{p}(Ax)'(t)\bigr)'= \tilde{\lambda } \tilde{f}\bigl(t,x(t),x'(t)\bigr) $$

can be converted to

$$ \widetilde{M}x=\widetilde{N}_{\tilde{\lambda }} x, \quad \tilde{\lambda }\in (0,1), $$

where M̃ and $\widetilde{N}_{\tilde{\lambda }}$ are defined by Eqs. (4.2) and (4.6), respectively. As proved above,

$$ \widetilde{N}_{\tilde{\lambda }}:\overline{\widetilde{\varOmega }}\to \widetilde{Z},\quad \tilde{\lambda }\in (0,1) $$

is an M̃-compact mapping. From assumption (i), one finds

$$ \widetilde{M}x\neq \widetilde{N}_{\tilde{\lambda }} x,\quad \tilde{\lambda }\in (0,1), x\in \partial \widetilde{\varOmega }, $$

and assumptions (ii) and (iii) imply that $\deg \{\widetilde{J} \widetilde{Q}\widetilde{N},\widetilde{\varOmega }\cap \ker \widetilde{M}, \tilde{\theta }\}$ is valid and

$$ \deg \{\widetilde{J}\widetilde{Q}\widetilde{N},\widetilde{\varOmega } \cap \ker \widetilde{M},\tilde{\theta }\}\neq 0. $$

So by applications of Lemma 4.2, we see that Eq. (4.5) has one ω-periodic solution. □

4.1 Application of Theorem 4.2: quasi-linear equation

As an application, we consider the following p-Laplacian neutral equation:

$$ \Biggl(\phi _{p} \Biggl(x(t)-\sum _{i=1}^{n}c_{i}(t)x\bigl(t- \delta _{i}(t)\bigr) \Biggr)' \Biggr)' +g\bigl(t,x(t)\bigr)=p(t), $$

(4.11)

where $g(t,x(t))\in C(\mathbb{R}\times \mathbb{R},\mathbb{R})$ is an ω-periodic function about t, $p\in C(\mathbb{R},\mathbb{R})$ is an ω-periodic function and $\int ^{\omega }_{0}p(t)\,dt=0$. By application of Theorem 4.2, we will investigate the existence of periodic solution for Eq. (4.11) satisfying $\sum_{i=1}^{n}\|c_{i}\|<1$, or $\|\frac{1}{c_{k}}\| + \sum_{\substack{i=1 \\ i\neq k}}^{n}\|\frac{c_{i}}{c_{k}}\|<1$.

Theorem 4.3

Assume the following conditions are satisfied:

($H_{1}$):

There exist two positive constants$\widetilde{D}_{1}$and$\widetilde{D}_{2}$with$\widetilde{D}_{1}<\widetilde{D}_{2}$, such that$g(t,x(t))>0$for$x(t)>\widetilde{D}_{2}$and$g(t,x(t))<0$for$x(t)<-\widetilde{D}_{1}$.

($H_{2}$):

There exist positive constantsm, nandB̃such that

$$ \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \leq m \vert x \vert ^{p-1}+n, \quad \textit{for } \vert x \vert >\widetilde{B} \textit{ and } t\in \mathbb{R}. $$

Then Eq. (4.11) has at least one solution with periodωif

$$ \frac{\tilde{\sigma }\omega (m (1+\sum_{i=1}^{n} \Vert c _{i} \Vert ) )^{\frac{1}{p}}}{2} +\frac{\tilde{\sigma }\omega \sum_{i=1}^{n} \Vert c'_{i} \Vert }{2} +\tilde{\sigma }\sum _{i=1} ^{n}\frac{ \Vert c_{i} \Vert \Vert \delta '_{i} \Vert }{1-\delta '_{i}(t)}< 1, $$

where$\delta '_{i}(t)<1$for$i=1,2,\ldots ,n$and

$$ \tilde{\sigma }= \textstyle\begin{cases} \frac{1}{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }, & \textit{for } \sum_{i=1}^{n} \Vert c_{i} \Vert < 1, \\ \frac{1}{1- \Vert \frac{1}{c_{k}} \Vert -\sum_{\substack{i=1 \\[5pt] i\neq k}}^{n} \Vert \frac{c_{i}}{c_{k}} \Vert } , & \textit{for } \Vert \frac{1}{c_{k}} \Vert +\sum_{\substack{i=1 \\ i\neq k}}^{n} \Vert \frac{c_{i}}{c_{k}} \Vert < 1. \end{cases} $$

Proof

Consider the homotopic equation

$$ \Biggl(\phi _{p} \Biggl(x(t)-\sum _{i=1}^{n}c_{i}(t)x\bigl(t- \delta _{i}(t)\bigr) \Biggr)' \Biggr)'+ \tilde{\lambda } g\bigl(t,x(t)\bigr)= \tilde{\lambda } p(t). $$

(4.12)

Firstly, we claim that the set of all ω-periodic solutions of Eq. (4.12) is bounded. Let $x(t)\in C_{\omega }$ be an arbitrary ω-periodic solution of Eq. (4.12). Integrating both sides of Eq. (4.12) over $[0,\omega ]$, we have

$$ \int ^{\omega }_{0}g\bigl(t,x(t)\bigr)\,dt=0. $$

(4.13)

From the mean value theorem, there is a constant $\xi \in (0,\omega )$ such that

$$ g\bigl(\xi ,x(\xi )\bigr)=0, $$

then we get by condition ($H_{1}$)

$$ -\widetilde{D}_{1}\leq x(\xi )\leq \widetilde{D}_{2}. $$

Therefore,

$$\begin{aligned} \Vert x \Vert &=\max _{t\in [0,\omega ]} \bigl\vert x(t) \bigr\vert =\max _{t\in [\xi ,\xi +\omega ]} \bigl\vert x(t) \bigr\vert \\ &=\frac{1}{2}\max_{t\in [\xi ,\xi +\omega ]} \bigl( \bigl\vert x(t) \bigr\vert + \bigl\vert x(t- \omega ) \bigr\vert \bigr) \\ &=\frac{1}{2}\max_{t\in [\xi ,\xi +\omega ]} \biggl( \biggl\vert x( \xi )+ \int ^{\omega }_{\xi }x'(s)\,ds \biggr\vert + \biggl\vert x(\xi )- \int ^{\xi }_{t- \omega }x'(s)\,ds \biggr\vert \biggr) \\ & \leq \widetilde{D}_{2}+\frac{1}{2} \biggl( \int ^{t}_{\xi } \bigl\vert x'(s) \bigr\vert \,ds+ \int ^{\xi }_{t-\omega } \bigl\vert x'(s) \bigr\vert \,ds \biggr) \\ &\leq \widetilde{D}_{2}+\frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(s) \bigr\vert \,ds. \end{aligned}$$

(4.14)

Multiplying both sides of Eq. (4.12) by $(Ax)(t)$ and integrating over the interval $[0,\omega ]$, we get

$$ \begin{aligned} \int ^{\omega }_{0}\bigl(\phi _{p}(Ax)'(t) \bigr)'(Ax) (t)\,dt +\tilde{\lambda } \int ^{\omega }_{0}g\bigl(t,x(t)\bigr) (Ax) (t)\,dt = \tilde{\lambda } \int ^{\omega } _{0}p(t) (Ax) (t)\,dt. \end{aligned} $$

(4.15)

Substituting $\int ^{\omega }_{0}(\phi _{p}(Ax)'(t))'(Ax)(t)\,dt=- \int ^{\omega }_{0}|(Ax)'(t)|^{p}\,dt$ into Eq. (4.15), it is clear that

$$\begin{aligned} - \int ^{\omega }_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt= -\tilde{\lambda } \int ^{\omega } _{0}g\bigl(t,x(t)\bigr) (Ax) (t)\,dt+ \tilde{\lambda } \int ^{\omega }_{0}p(t) (Ax) (t)\,dt. \end{aligned}$$

So, we have

$$ \begin{aligned}[b] \int ^{\omega }_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt&\leq \int ^{\omega }_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \Biggl\vert x(t)-\sum_{i=1}^{n}c_{i}(t)x \bigl(t-\delta _{i}(t)\bigr) \Biggr\vert \,dt \\ &\quad {}+ \int ^{\omega }_{0} \bigl\vert p(t) \bigr\vert \Biggl\vert x(t)-\sum_{i=1}^{n}c_{i}(t)x \bigl(t- \delta _{i}(t)\bigr) \Biggr\vert \,dt \\ &\leq \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{ \omega }_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt \\ &\quad {}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{\omega } _{0} \bigl\vert p(t) \bigr\vert \,dt. \end{aligned} $$

(4.16)

Define

$$ E_{1}:=\bigl\{ t\in [0,\omega ]\mid \bigl\vert x(t) \bigr\vert \leq \widetilde{B}\bigr\} , \qquad E_{2}:=\bigl\{ t\in [0,\omega ]\mid \bigl\vert x(t) \bigr\vert >\widetilde{B}\bigr\} . $$

From condition ($H_{2}$), we obtain

$$ \begin{aligned}[b] \int ^{\omega }_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt&\leq \Biggl(1+\sum _{i=1} ^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int _{E_{1}+E_{2}} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt \\ &\quad {}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{\omega } _{0} \bigl\vert p(t) \bigr\vert \,dt \\ &\leq \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \bigl(m \Vert x \Vert ^{p-1}\omega +n\omega + \Vert g_{\tilde{B}} \Vert \omega \bigr) \\ &\quad {}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert p \Vert \omega \Vert x \Vert \\ &=m\omega \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert ^{p} + \widetilde{N}_{1} \Vert x \Vert , \end{aligned} $$

(4.17)

where $\|g_{\widetilde{B}}\|:=\max_{|x|\leq \widetilde{B}}|g(t,x(t))|$, $\|p\|:=\max_{t\in [0,\omega ]}|p(t)|$ and $\widetilde{N}_{1}:= (1+\sum_{i=1}^{n}\|c_{i}\| )(\|g_{\widetilde{B}}\| \omega +n\omega +\|p\|\omega )$. Substituting Eq. (4.14) into Eq. (4.17), we get

$$\begin{aligned} \int ^{\omega }_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt \leq& m\omega \Biggl(1+\sum _{i=1}^{n} \Vert c_{i} \Vert \Biggr) \biggl(\widetilde{D}_{2}+ \frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{p} \\ &{}+\widetilde{N}_{1} \biggl( \widetilde{D}_{2}+\frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr). \end{aligned}$$

(4.18)

Since $(Ax)(t)=x(t)-\sum_{i=1}^{n}x(t-\delta _{i}(t))$, we arrive at

$$ \begin{aligned} (Ax)'(t)&= \Biggl(x(t)-\sum _{i=1}^{n}c_{i}(t)x\bigl(t-\delta _{i}(t)\bigr) \Biggr)' \\ &=x'(t)-\sum_{i=1}^{n}c'_{i}(t)x \bigl(t-\delta _{i}(t)\bigr) -\sum_{i=1}^{n}c_{i}(t)x' \bigl(t-\delta _{i}(t)\bigr) \bigl(1-\delta '_{i}(t) \bigr) \\ &=x'(t)-\sum_{i=1}^{n}c'_{i}(t)x \bigl(t-\delta _{i}(t)\bigr) -\sum_{i=1}^{n}c_{i}(t)x' \bigl(t-\delta _{i}(t)\bigr) \\ &\quad {}+\sum_{i=1}^{n}c _{i}(t)x'\bigl(t-\delta _{i}(t)\bigr) \delta '_{i}(t) \end{aligned} $$

and

$$ \bigl(Ax'\bigr) (t)=x'(t)-\sum _{i=1}^{n}c_{i}(t)x' \bigl(t-\delta _{i}(t)\bigr). $$

Thus,

$$ \bigl(Ax'\bigr) (t)=(Ax)'(t)+\sum _{i=1}^{n}c'_{i}(t)x \bigl(t-\delta _{i}(t)\bigr) - \sum_{i=1}^{n}c_{i}(t)x' \bigl(t-\delta _{i}(t)\bigr)\delta '_{i}(t). $$

By applying Lemma 2.2 and the Hölder inequality, we have

$$\begin{aligned} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt =& \int ^{\omega }_{0} \bigl\vert \bigl(A^{-1}Ax' \bigr) (t) \bigr\vert \,dt \leq \tilde{\sigma } \int ^{\omega }_{0} \bigl\vert \bigl(Ax' \bigr) (t) \bigr\vert \,dt \\ =&\tilde{\sigma } \int ^{\omega }_{0} \Biggl\vert (Ax)'(t) +\sum_{i=1} ^{n}c'_{i}(t)x \bigl(t-\delta _{i}(t)\bigr) -\sum_{i=1}^{n}c_{i}(t)x' \bigl(t- \delta _{i}(t)\bigr)\delta '_{i}(t) \Biggr\vert \,dt \\ \leq& \tilde{\sigma } \int ^{\omega }_{0} \bigl\vert (Ax)'(t) \bigr\vert \,dt +\tilde{\sigma } \int ^{\omega }_{0} \Biggl\vert \sum _{i=1}^{n}c'_{i}(t)x \bigl(t-\delta _{i}(t)\bigr) \Biggr\vert \,dt \\ &{} +\tilde{\sigma } \int ^{\omega }_{0} \Biggl\vert \sum _{i=1}^{n}c_{i}(t)x' \bigl(t- \delta _{i}(t)\bigr)\delta '_{i}(t) \Biggr\vert \,dt \\ \leq& \tilde{\sigma } \omega ^{\frac{1}{q}} \biggl( \int ^{\omega }_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt \biggr) ^{\frac{1}{p}} +\tilde{\sigma }\omega \sum_{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert \Vert x \Vert \\ &{}+\tilde{\sigma }\sum_{i=1}^{n} \Vert c_{i} \Vert \bigl\Vert \delta '_{i} \bigr\Vert \int ^{\omega }_{0} \bigl\vert x' \bigl(t-\delta _{i}(t)\bigr) \bigr\vert \,dt \\ \leq& \tilde{\sigma } \omega ^{\frac{1}{q}} \biggl( \int ^{\omega }_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt \biggr) ^{\frac{1}{p}} +\tilde{\sigma }\omega \sum_{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert \Vert x \Vert \\ &{}+\tilde{\sigma }\sum_{i=1}^{n}\frac{ \Vert c_{i} \Vert \Vert \delta '_{i} \Vert }{1-\delta '_{i}} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt, \end{aligned}$$

(4.19)

where $\|c'_{i}\|=\max_{t\in [0,\omega ]}|c'_{i}(t)|$, $\| \delta '_{i}\|=\max_{t\in [0,\omega ]}|\delta '_{i}(t)|$, for $i=1,2,\ldots, n$. Substituting Eq. (4.14) and Eq. (4.18) into Eq. (4.19), since $(a+b)^{k}\leq a^{k}+b ^{k}$, $0< k<1$, we get

$$\begin{aligned} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \leq& \tilde{\sigma } \omega ^{\frac{1}{q}} \Biggl(m\omega \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Biggr) ^{\frac{1}{p}} \biggl( \widetilde{D}_{2}+\frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) \\ &{}+ \tilde{\sigma }\omega ^{\frac{1}{q}}( \widetilde{N}_{1})^{\frac{1}{p}} \biggl(\widetilde{D}_{2}+ \frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{\frac{1}{p}} \\ &{}+\tilde{\sigma }\sum_{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert \omega \biggl( \widetilde{D} _{2}+\frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) +\tilde{\sigma } \sum_{i=1}^{n} \frac{ \Vert c_{i} \Vert \Vert \delta '_{i} \Vert }{1-\delta '_{i}} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ \leq& \Biggl( \frac{\tilde{\sigma }\omega (m (1+\sum_{i=1}^{n} \Vert c_{i} \Vert ) )^{\frac{1}{p}}}{2} +\frac{ \tilde{\sigma }\sum_{i=1}^{n} \Vert c'_{i} \Vert \omega }{2} + \tilde{\sigma } \sum_{i=1}^{n}\frac{ \Vert c_{i} \Vert \Vert \delta '_{i} \Vert }{1- \delta '_{i}} \Biggr) \\ &{}\times\int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt+\tilde{\sigma } \omega ^{\frac{1}{q}} \Biggl(m\omega \Biggl(1+ \sum _{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Biggr)^{\frac{1}{p}} \widetilde{D}_{2} +\tilde{ \sigma } \omega ^{\frac{1}{q}}(\widetilde{N} _{1} \widetilde{D}_{2})^{\frac{1}{p}} \\ &{}+\tilde{\sigma } \omega ^{\frac{1}{q}}(\widetilde{N}_{1})^{ \frac{1}{p}} \biggl(\frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{ \frac{1}{p}} +\tilde{\sigma }\sum _{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert \omega \widetilde{D}_{2}. \end{aligned}$$

(4.20)

Since $\frac{\tilde{\sigma }\omega (m (1+\sum_{i=1} ^{n}\|c_{i}\| ) )^{\frac{1}{p}}}{2} +\frac{ \tilde{\sigma }\omega \sum_{i=1}^{n}\|c'_{i}\|}{2} + \tilde{\sigma }\sum_{i=1}^{n}\frac{\|c_{i}\|\|\delta '_{i}\|}{1- \delta '_{i}}<1$, it is easily seen that there exists a constant $M_{1}'>0$ (independent of λ̃) such that

$$ \int ^{\omega }_{0} \bigl\vert x'(t) \bigr\vert \,dt\leq M_{1}'. $$

(4.21)

From Eq. (4.14), we have

$$ \Vert x \Vert \leq \widetilde{D}_{2}+ \frac{1}{2} \int ^{\omega }_{0} \bigl\vert x'(s) \bigr\vert \,ds \leq \widetilde{D}_{2}+\frac{1}{2}M_{1}':=M_{1}. $$

(4.22)

As $(Ax)(0)=(Ax)(\omega )$, there exists a point $t_{0}\in (0,\omega )$ such that $(Ax)'(t_{0})=0$, while $\phi _{p}(0)=0$, from Eq. (4.12), we see that

$$\begin{aligned} \bigl\vert \phi _{p}(Ax)'(t) \bigr\vert &= \biggl\vert \int ^{t}_{t_{0}}\bigl(\phi _{p}(Ax)'(s) \bigr)'\,ds \biggr\vert \\ &\leq \tilde{\lambda } \int ^{\omega }_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+ \tilde{\lambda } \int ^{\omega }_{0} \bigl\vert p(t) \bigr\vert \,dt \\ &\leq \omega \Vert g_{M_{1}} \Vert +\omega \Vert p \Vert :=M_{2}', \end{aligned}$$

where $\|g_{M_{1}}\|:=\max_{|x(t)|\leq M_{1}}|g(t,x(t))|$. Next we claim that there exists a positive constant $M_{2}^{*}>M_{2}'+1$, such that, for all $t\in \mathbb{R}$, we obtain

$$ \bigl\Vert (Ax)' \bigr\Vert \leq M_{2}^{*}. $$

(4.23)

In fact, if $(Ax)'$ is not bounded, there exists a positive constant $M''_{2}$ such that $\|(Ax)'\|>M''_{2}$ for some $(Ax)'\in \mathbb{R}$, therefore, we have $\|\phi _{p}(Ax)'\|=\|(Ax)'\|^{p-1}\geq (M''_{2})^{p-1}$. Then it is a contradiction, so Eq. (4.23) holds. From Lemma 2.2 and Eq. (4.23), we arrive at

$$ \begin{aligned}[b] \bigl\Vert x' \bigr\Vert &= \bigl\Vert A^{-1}Ax' \bigr\Vert \\ &= \bigl\Vert A^{-1}\bigl(Ax'\bigr) (t) \bigr\Vert \\ &\leq \tilde{\sigma } \Biggl\Vert (Ax)'(t)+\sum _{i=1}^{n}c'_{i}(t)x \bigl(t- \delta _{i}(t)\bigr) -\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr\Vert \delta '_{i} \biggl\Vert \int ^{\omega }_{0} \bigl\vert x' \bigl(t-\delta _{i}(t)\bigr) \bigr\vert \,dt \biggr\Vert \\ &\leq \tilde{\sigma } \bigl\Vert (Ax)' \bigr\Vert +\tilde{ \sigma } \Biggl(\sum_{i=1} ^{n} \bigl\Vert c'_{i} \bigr\Vert \Vert x \Vert \Biggr) +\tilde{\sigma }\sum_{i=1}^{n} \Vert c _{i} \Vert \bigl\Vert \delta '_{i} \bigr\Vert \int ^{\omega }_{0} \bigl\vert x' \bigl(t-\delta _{i}(t)\bigr) \bigr\vert \,dt \\ &\leq \tilde{\sigma } M_{2}^{*}+\tilde{\sigma }\sum _{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert M_{1} +\tilde{\sigma }\sum_{i=1}^{n}\frac{ \Vert c_{i} \Vert \Vert \delta '_{i} \Vert }{1-\delta '_{i}}M'_{1}:=M_{2}. \end{aligned} $$

(4.24)

Set $M^{*}=\sqrt{M_{1}^{2}+M_{2}^{2}}+1$, we have

$$ \widetilde{\varOmega }=\bigl\{ x\in C^{1}_{\omega }(\mathbb{R}, \mathbb{R})\mid \Vert x \Vert \leq M^{*}+1, \bigl\Vert x' \bigr\Vert \leq M^{*}+1 \bigr\} , $$

and we know that Eq. (4.11) has no solution on ∂Ω̃ as $\tilde{\lambda }\in (0,1)$ and when $x(t)\in \partial \widetilde{\varOmega }\cap \mathbb{R}$, $x(t)=M^{*}+1$ or $x(t)=-M^{*}-1$. So, from condition ($H_{1}$), we see that

$$\begin{aligned}& \frac{1}{\omega } \int ^{\omega }_{0}g\bigl(M^{*}+1 \bigr)\,dt>0, \\& \frac{1}{\omega } \int ^{\omega }_{0}g\bigl(-M^{*}-1 \bigr)\,dt< 0, \end{aligned}$$

since $\int ^{\omega }_{0}e(t)\,dt=0$. So condition (ii) of Theorem 4.2 is also satisfied. Obviously, we can get

$$\begin{aligned} \deg \{\widetilde{F},\widetilde{\varOmega }\cap \mathbb{R},0\} &=\deg \biggl\{ \frac{1}{\omega } \int ^{\omega }_{0}g\bigl(t,x(t)\bigr)\,dt,\partial \widetilde{\varOmega }\cap \mathbb{R},0 \biggr\} \\ &=\deg \{x,\partial \widetilde{\varOmega }\cap \mathbb{R},0\}\neq 0. \end{aligned}$$

So condition (iii) of Theorem 4.2 is satisfied. In view of Theorem 4.2, there exists at least one ω-periodic solution. □

5 Conclusions

In this paper, we first investigate some properties of the neutral operator with multiple variable parameters $(Ax)(t)$. Afterwards, applying Krasnoselskii’s fixed point theorem and properties of the operator A, we prove the existence of a positive periodic solution for a second-order neutral differential equation with multiple variable parameters. On the other hand, we find that the second-order quasi-linear neutral differential equation has a periodic solution by using the extension of Mawhin’s continuous theorem.