1 Introduction

In this paper, we consider the following second-order p-Laplacian neutral functional differential equation:

${\left({\varphi }_{p}\left({x}^{\prime }\left(t\right)-c{x}^{\prime }\left(t-\sigma \right)\right)\right)}^{\prime }+f\left({x}^{\prime }\left(t\right)\right)+\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)=e\left(t\right),$
(1.1)

where ${\varphi }_{p}\left(x\right)={|x|}^{p-2}x$ for $x\ne 0$ and $p>1$; σ and c are given constants with $|c|\ne 1$; ${\varphi }_{p}\left(0\right)=0$, $f\left(0\right)=0$. The conjugate exponent of p is denoted by q, i.e. $\frac{1}{p}+\frac{1}{q}=1$. f, g, β, e, and τ are real continuous functions on ℝ; τ, β, and e are periodic with periodic T, $T>0$ is a constant; ${\int }_{0}^{T}e\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=0$, ${\int }_{0}^{T}\beta \left(t\right)\ne 0$.

As we know, the p-Laplace Rayleigh equation with a deviating argument $\tau \left(t\right)$ is applied in many fields such as physics, mechanics, engineering technique fields, and so on. The existence of a periodic solution for the second-order p-Laplacian Rayleigh equations with a deviating argument as follows:

${\left({\varphi }_{p}\left({x}^{\prime }\left(t\right)\right)\right)}^{\prime }+f\left(x\left(t\right)\right){x}^{\prime }\left(t\right)+g\left(x\left(t-\tau \left(t\right)\right)\right)=e\left(t\right)$
(1.2)

and

${\left({\varphi }_{p}\left({x}^{\prime }\left(t\right)\right)\right)}^{\prime }+f\left({x}^{\prime }\left(t\right)\right)+g\left(x\left(t-\tau \left(t\right)\right)\right)=e\left(t\right)$
(1.3)

has been extensively studied in [14]. In recent years, Lu et al. [57] used Mawhin’s continuation theory to do research to the existence of a periodic solution for p-Laplacian neutral Rayleigh equation. They obtained some existence results of periodic solutions to p-Laplacian neutral Rayleigh equations.

In the research mentioned above, the corresponding nonlinear terms of the second-order p-Laplacian Rayleigh equation did not include a variable coefficient. Only little literature discussed this kind of p-Laplacian Rayleigh equation. For more details refer to [811]. Here, we focus on [11] by Liang Feng et al. They discussed the existence of the solution to the following equation:

${\left({\varphi }_{p}\left({x}^{\prime }\left(t\right)-c{x}^{\prime }\left(t-r\right)\right)\right)}^{\prime }=f\left(x\left(t\right)\right){x}^{\prime }\left(t\right)+\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)+e\left(t\right).$
(1.4)

They established sufficient conditions for the existence of a T-periodic solution of (1.4). But their conclusions are founded on the prerequisite ${\int }_{0}^{T}\left(g\left(x\left(t-\tau \left(t\right)\right)\right)+e\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt=0$, which does not satisfy (1.1). Another significance of the paper is that the result is related to the deviating argument $\tau \left(t\right)$, while the conclusions in those existing papers mentioned above have no relation with $\tau \left(t\right)$.

2 Preliminary results

For convenience, throughout this paper, we will adopt the following assumptions:

(H1) ${\parallel x\parallel }_{p}={\left({\int }_{0}^{T}|x\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}$, ${\parallel x\parallel }_{\mathrm{\infty }}={max}_{t\in \left[0,T\right]}|x\left(t\right)|$, $\parallel x\parallel =max\left\{{\parallel x\parallel }_{\mathrm{\infty }},{\parallel {x}^{\prime }\left(t\right)\parallel }_{\mathrm{\infty }}\right\}$;

(H2) ${m}_{0}={min}_{t\in \left[0,T\right]}|\beta \left(t\right)|$, ${m}_{1}={max}_{t\in \left[0,T\right]}|\beta \left(t\right)|$;

(H3) ${C}_{T}=\left\{x|x\in C\left(\mathbb{R},\mathbb{R}\right),x\left(t+T\right)=x\left(t\right),\mathrm{\forall }t\in \mathbb{R}\right\}$;

(H4) ${C}_{T}^{1}=\left\{x|x\in {C}^{1}\left(R,R\right),x\left(t+T\right)=x\left(t\right),{x}^{\prime }\left(t+T\right)={x}^{\prime }\left(t\right),\mathrm{\forall }t\in \mathbb{R}\right\}$.

It is obvious that ${C}_{T}$ with norm ${\parallel x\parallel }_{\mathrm{\infty }}$ and ${C}_{T}^{1}$ with norm $\parallel x\parallel$ are two Banach spaces.

Now we define a linear operator $A:{C}_{T}⟶{C}_{T}$, $\left(Ax\right)\left(t\right)=x\left(t\right)-cx\left(t-\sigma \right)$.

According to [12, 13], we know that the operator A has the following properties.

Lemma 2.1 [12, 13]

If $|c|\ne 1$, then A has continuous bounded inverse on ${C}_{T}$ and

1. (1)

${\parallel {A}^{-1}x\parallel }_{\mathrm{\infty }}=\frac{{\parallel x\parallel }_{\mathrm{\infty }}}{|1-|C||}$, $\mathrm{\forall }x\in {C}_{T}$,

2. (2)

$\left({A}^{-1}x\right)\left(t\right)=\left\{\begin{array}{ll}{\sum }_{j\ge 0}{c}^{j}x\left(t-j\sigma \right),& |c|<1,\\ -{\sum }_{j\ge 0}{c}^{-j}x\left(t+j\sigma \right),& |c|>1,\end{array}$

3. (3)

${\int }_{0}^{T}|\left({A}^{-1}x\right)\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt\le \frac{1}{|1-|c||}{\int }_{0}^{T}|x\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt$, $\mathrm{\forall }x\in {C}_{T}$.

Lemma 2.2 If $|c|\ne 1$ and $p>1$, then

${\parallel {A}^{-1}x\left(t\right)\parallel }_{p}\le \frac{1}{1-|c|}{\parallel x\left(t\right)\parallel }_{p},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {C}_{T}.$
(2.1)

Proof We know that $x\left(t\right)$ is a periodic function. So ${\int }_{0}^{T}|x\left(t-j\sigma \right)|\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{T}|x\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt$ for $j\ge 0$. When $|c|<1$, from Lemma 2.1, we have

$\begin{array}{rcl}{\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt& =& {\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{p-1}|{A}^{-1}x\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{p-1}|\sum _{j\ge 0}{c}^{j}x\left(t-j\sigma \right)|\phantom{\rule{0.2em}{0ex}}dt\\ \le & \sum _{j\ge 0}|{c}^{j}|{\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{p-1}|x\left(t-j\sigma \right)|\phantom{\rule{0.2em}{0ex}}dt\\ \le & \sum _{j\ge 0}|{c}^{j}|{\left({\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{\left(p-1\right)q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}{\left({\int }_{0}^{T}|x\left(t-j\sigma \right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}\\ =& \sum _{j\ge 0}|{c}^{j}|{\left({\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}{\left({\int }_{0}^{T}|x\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}\\ =& \frac{1}{1-|c|}{\left({\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}{\left({\int }_{0}^{T}|x\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}\end{array}$

which implies ${\left({\int }_{0}^{T}|{A}^{-1}x\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}\le \frac{1}{1-|c|}{\left({\int }_{0}^{T}|x\left(t-j\sigma \right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}$. That is to say (2.1) holds. If $|c|>1$, we can also prove that (2.1) is true in the same way. Thus Lemma 2.2 is proved.

Now we consider the following equation in ${C}_{T}^{1}$:

${\left({\varphi }_{p}\left({u}^{\prime }\left(t\right)\right)\right)}^{\prime }=F\left(u\right).$
(2.2)

$F:{C}_{T}^{1}⟶{C}_{T}$ is continuous and takes a bounded set into bounded set.

Let us define $P:{C}_{T}^{1}⟶{C}_{T}$, $u|⟶u\left(0\right)$, $Q:{C}_{T}⟶{C}_{T}$, $h|⟶\frac{1}{T}{\int }_{0}^{T}h\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$ and

$H\left(h\left(t\right)\right)={\int }_{0}^{t}h\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}h\in {C}_{T}.$

It is clear that if $u\in {C}_{T}^{1}$ is the solution to (2.2), then u satisfies the abstract equation

$u=Pu+QF\left(u\right)+K\left(F\left(u\right)\right),$

where the operator $K:{C}_{T}⟶{C}_{T}^{1}$ is given by

$K\left(h\left(t\right)\right)=H\left\{{\varphi }_{q}\left[\alpha \left(\left(I-Q\right)h\right)+H\left(\left(I-Q\right)\right)\right]\right\}\left(t\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \mathbb{R},$

α is a continuous function which sends bounded sets of ${C}_{T}$ into bounded sets of ℝ, and it is a completely continuous mapping. For more details as regards the meaning of α, please refer to [14]. □

Lemma 2.3 [14]

Let Ω be an open bounded set in ${C}_{T}^{1}$. Suppose that the following conditions hold:

1. (i)

For each $\lambda \in \left(0,1\right)$, the equation ${\left({\varphi }_{p}\left({u}^{\prime }\right)\right)}^{\prime }=\lambda F\left(u\right)$ has no solution on Ω.

2. (ii)

The equation $Ϝ\left(u\right)=\frac{1}{T}{\int }_{0}^{T}F\left(u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt=0$ has no solution on $\partial \mathrm{\Omega }\cap \mathbb{R}$.

3. (iii)

The Brouwer degree of Ϝ, $deg\left\{Ϝ,\mathrm{\Omega }\cap \mathbb{R},0\right\}\ne 0$.

Then (2.2) has at least one T-periodic solution in $\overline{\mathrm{\Omega }}$.

Lemma 2.4 [15]

$\mathrm{\Omega }\subset {\mathbb{R}}^{n}$ is open bounded and symmetric with respect to $0\in \mathrm{\Omega }$. If $f\in C\left(\overline{\mathrm{\Omega }},{\mathbb{R}}^{n}\right)$ and $f\left(x\right)\ne \mu f\left(-x\right)$, $\mathrm{\forall }x\in \partial \mathrm{\Omega }$, $\mu \in \left[0,1\right]$, then $deg\left\{f,\mathrm{\Omega },0\right\}$ is an odd number.

3 Main results

Theorem 3.1 Suppose that the following conditions hold:

(A1) $\tau \left(t\right)\in {C}^{1}\left(\mathbb{R},\mathbb{R}\right)$, ${\tau }^{\prime }\left(t\right)>1$ or ${\tau }^{\prime }\left(t\right)<1$, ${m}_{2}={min}_{t\in \left[0,T\right]}\frac{1}{1-{\tau }^{\prime }\left(t\right)}$.

(A2) The sign of $\frac{\beta \left(t\right)}{1-{\tau }^{\prime }\left(t\right)}$ is unchanged in the interval $\left[0,T\right]$.

(A3) There exist constants ${r}_{1}\ge 0$, ${r}_{2}>0$ and $k>0$ such that

1. (1)

$|f\left(x\right)|\le k+{r}_{1}{|x|}^{p-1}$, $\mathrm{\forall }x\in \mathbb{R}$,

2. (2)

${lim}_{|x|⟶\mathrm{\infty }}\frac{|g\left(x\right)|}{{|x|}^{p-1}}\le {r}_{2}$.

(A4) There exists a constant $d>0$ such that $xg\left(x\right)>0$, $\mathrm{\forall }|x|>d$.

Then (1.1) has at least one solution with periodic T if there exists a constant $\epsilon >0$ such that the following condition holds:

$|1+|c||\left(a+{T}^{\frac{1}{q}}\right)\left[{m}_{1}T\left({r}_{2}+\epsilon \right)+{r}_{1}{T}^{\frac{1}{p}}\right]<|1-|c|{|}^{p},$
(3.1)

where a is defined in (3.8).

Proof Consider the homotopic equation of (1.1) as follows:

${\left({\varphi }_{p}\left({x}^{\prime }\left(t\right)-c{x}^{\prime }\left(t-\sigma \right)\right)\right)}^{\prime }+\lambda f\left({x}^{\prime }\left(t\right)\right)+\lambda \beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)=\lambda e\left(t\right),\phantom{\rule{1em}{0ex}}\lambda \in \left(0,1\right).$
(3.2)

Let $x\left(t\right)$ be a possible T-periodic solution to (3.2). By integrating both sides of (3.2) over [0, T], we have

${\int }_{0}^{T}\left[f\left({x}^{\prime }\left(t\right)\right)+\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt=0.$
(3.3)

Let $u\left(t\right)=t-\tau \left(t\right)$, by the condition (A1), we know that $u\left(t\right)$ has a unique inverse denoted by $t=\gamma \left(u\right)$; noting that $\tau \left(0\right)=\tau \left(T\right)$, we get

${\int }_{0}^{T}\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)\phantom{\rule{0.2em}{0ex}}dt={\int }_{-\tau \left(0\right)}^{T-\tau \left(T\right)}\frac{\beta \left(\gamma \left(u\right)\right)g\left(u\right)}{1-{\tau }^{\prime }\left(\gamma \left(u\right)\right)}\phantom{\rule{0.2em}{0ex}}du={\int }_{0}^{T}\frac{\beta \left(\gamma \left(u\right)\right)g\left(u\right)}{1-{\tau }^{\prime }\left(\gamma \left(u\right)\right)}\phantom{\rule{0.2em}{0ex}}du.$
(3.4)

Based on the condition of (A2) and the integral mean value theorem, there exists $\xi \in \left[0,T\right]$ such that

$g\left(x\left(\xi \right)\right){\int }_{0}^{T}\frac{\beta \left(\gamma \left(u\right)\right)}{1-{\tau }^{\prime }\left(\gamma \left(u\right)\right)}\phantom{\rule{0.2em}{0ex}}du=-{\int }_{0}^{T}f\left({x}^{\prime }\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt.$
(3.5)

By the condition (A3)(2) for a given $\epsilon >0$, $\mathrm{\exists }\rho >d>0$ when $|x\left(t\right)|>\rho$ such that

$\frac{|g\left(x\right)|}{{|x|}^{p-1}}\ge {r}_{2}-\epsilon >0,\phantom{\rule{2em}{0ex}}\frac{|g\left(x\right)|}{{|x|}^{p-1}}\le {r}_{2}+\epsilon .$
(3.6)

Now we can claim that there are two constants a and b such that

$|x\left(\xi \right)|\le a{\parallel x\parallel }_{p}+b,$
(3.7)

where

$a=\left\{\begin{array}{cc}{\left[\frac{{r}_{1}}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}{T}^{\frac{1}{q}}}\right]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}},\hfill & p-1<1,\hfill \\ {\left[\frac{{r}_{1}}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}{T}^{\frac{1}{q}}}\right]}^{\frac{1}{p-1}},\hfill & p-1>1,\hfill \end{array}$
(3.8)
$b=\left\{\begin{array}{cc}{\left[\frac{k}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}}\right]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}},\hfill & p-1<1,\hfill \\ {\left[\frac{k}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}}\right]}^{\frac{1}{p-1}},\hfill & p-1>1.\hfill \end{array}$
(3.9)

In the following, we prove the above claim in two cases.

Case 1. If $|x\left(\xi \right)|\le \rho$, $\xi \in \left[0,T\right]$, then (3.7) holds.

Case 2. If $|x\left(\xi \right)|>\rho$, $\xi \in \left[0,T\right]$, by (3.5), (3.6), and the condition (A3)(1), we have

$\begin{array}{r}|{r}_{2}-\epsilon ||x\left(\xi \right){|}^{p-1}{m}_{0}{m}_{2}T\le |g\left(x\left(\xi \right)\right)|{\int }_{0}^{T}\frac{\beta \left(\gamma \left(u\right)\right)}{1-{\tau }^{\prime }\left(\gamma \left(u\right)\right)}\phantom{\rule{0.2em}{0ex}}du\\ \phantom{|{r}_{2}-\epsilon ||x\left(\xi \right){|}^{p-1}{m}_{0}{m}_{2}T}\le {\int }_{0}^{T}|f\left({x}^{\prime }\left(t\right)\right)|\phantom{\rule{0.2em}{0ex}}dt\le kT+{r}_{1}{\int }_{0}^{T}|{x}^{\prime }\left(t\right){|}^{p-1}\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{|{r}_{2}-\epsilon ||x\left(\xi \right){|}^{p-1}{m}_{0}{m}_{2}T}\le kT+{r}_{1}{T}^{\frac{1}{p}}{\left({\int }_{0}^{T}|{x}^{\prime }\left(t\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{p-1}{p}}=kT+{r}_{1}{T}^{\frac{1}{p}}{\parallel {x}^{\prime }\parallel }_{p}^{p-1},\\ |x\left(\xi \right){|}^{p-1}\le \frac{k}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}}+\frac{{r}_{1}}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}{T}^{\frac{1}{q}}}{\parallel {x}^{\prime }\parallel }_{p}^{p-1}.\end{array}$

When $0, according to the Minkowski inequality, we have

$|x\left(\xi \right)|\le {\left[\frac{k}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}}\right]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}}+{\left[\frac{{r}_{1}}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}{T}^{\frac{1}{q}}}\right]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}}{\parallel {x}^{\prime }\parallel }_{p}.$

When $p-1>1$, from ${\left(a+b\right)}^{\frac{1}{m}}\le {\left(a\right)}^{\frac{1}{m}}+{\left(b\right)}^{\frac{1}{m}}$, $a,b\in \left[0,+\mathrm{\infty }\right)$, $m>1$, we have

$|x\left(\xi \right)|\le {\left[\frac{k}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}}\right]}^{\frac{1}{p-1}}+{\left[\frac{{r}_{1}}{\left({r}_{2}-\epsilon \right){m}_{0}{m}_{2}{T}^{\frac{1}{q}}}\right]}^{\frac{1}{p-1}}{\parallel {x}^{\prime }\parallel }_{p}.$

Therefore, (3.7) is also satisfied for case 2.

For any $t\in \mathbb{R}$, there exists ${t}_{0}\in \left[0,T\right]$, such that $t=kT+{t}_{0}$, where k is an integer. Then

$|x\left(t\right)|=|x\left({t}_{0}\right)|\le |x\left(\xi \right)|+{\int }_{0}^{T}|{x}^{\prime }\left(s\right)|\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\xi \in \left[0,T\right].$

By (3.7), we have

${\parallel x\parallel }_{\mathrm{\infty }}\le a{\parallel {x}^{\prime }\parallel }_{p}+b+{T}^{\frac{1}{q}}{\parallel {x}^{\prime }\parallel }_{p}=\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}+b.$
(3.10)

At first, we prove that there is a constant ${R}_{1}$ such that

${\parallel x\parallel }_{\mathrm{\infty }}\le {R}_{1}.$
(3.11)

By (3.10), we only need to prove that ${\parallel {x}^{\prime }\parallel }_{p}$ is bounded in order to prove (3.11).

If ${\parallel {x}^{\prime }\parallel }_{p}=0$, then ${\parallel {x}^{\prime }\parallel }_{p}$ is obviously bounded.

If $\frac{b}{a+{T}^{\frac{1}{q}}{\parallel {x}^{\prime }\parallel }_{p}}\ge h$, then ${\parallel {x}^{\prime }\parallel }_{p}\le \frac{b-ah}{h{T}^{\frac{1}{q}}}$, that is, ${\parallel {x}^{\prime }\parallel }_{p}$ is bounded as well.

If $\frac{b}{a+{T}^{\frac{1}{q}}{\parallel {x}^{\prime }\parallel }_{p}}, we prove that ${\parallel {x}^{\prime }\parallel }_{p}$ is bounded in the following.

By multiplying both sides of (3.2) by $A\left(x\left(t\right)\right)=x\left(t\right)-cx\left(t-\sigma \right)$ and integrating them over $\left[0,T\right]$, we have

$\begin{array}{r}|{\int }_{0}^{T}{\left[{\phi }_{p}\left(A{x}^{\prime }\right)\right]}^{\prime }A\left(x\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt|\\ \phantom{\rule{1em}{0ex}}=|{\phi }_{p}\left(A{x}^{\prime }\right)A\left(x\left(t\right)\right){|}_{0}^{T}-{\int }_{0}^{T}{\phi }_{p}\left(A{x}^{\prime }\right)A{x}^{\prime }\phantom{\rule{0.2em}{0ex}}dt|\\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{T}|A{x}^{\prime }{|}^{p}\phantom{\rule{0.2em}{0ex}}dt={\parallel A{x}^{\prime }\parallel }_{p}^{p}\\ \phantom{\rule{1em}{0ex}}=|\lambda {\int }_{0}^{T}A\left(x\left(t\right)\right)\left[f\left({x}^{\prime }\left(t\right)\right)+\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)-e\left(t\right)\right]\phantom{\rule{0.2em}{0ex}}dt|\\ \phantom{\rule{1em}{0ex}}\le \left(1+|c|\right){\parallel x\parallel }_{\mathrm{\infty }}{\int }_{0}^{T}\left[|f\left({x}^{\prime }\left(t\right)|+|\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)|+|e\left(t\right)|\right]\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(3.12)

Let ${E}_{1}=\left\{t\in \left[0,T\right]:|x\left(t-\tau \left(t\right)\right)|\le \rho \right\}$, ${E}_{2}=\left\{t\in \left[0,T\right]:|x\left(t-\tau \left(t\right)\right)|>\rho \right\}$, then

$\begin{array}{rl}{\int }_{0}^{T}|\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)|\phantom{\rule{0.2em}{0ex}}dt& ={\int }_{{E}_{1}}|\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)|\phantom{\rule{0.2em}{0ex}}dt+{\int }_{{E}_{2}}|\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)|\phantom{\rule{0.2em}{0ex}}dt\\ \le {m}_{1}{m}_{3}T+{m}_{1}T\left({r}_{2}+\epsilon \right){\parallel x\parallel }_{\mathrm{\infty }}^{p-1},\end{array}$
(3.13)

where

${m}_{3}=\underset{|x|\le \rho }{max}|g\left(x\right)|.$
(3.14)

By (3.12) and (3.13), we get

$\begin{array}{rl}{\parallel A{x}^{\prime }\parallel }_{p}^{p}\le & \left(1+|c|\right){\parallel x\parallel }_{\mathrm{\infty }}\\ ×\left[{m}_{1}{m}_{3}T+{m}_{1}T\left({r}_{2}+\epsilon \right){\parallel x\parallel }_{\mathrm{\infty }}^{p-1}+kT+{r}_{1}{T}^{\frac{1}{p}}{\parallel {x}^{\prime }\parallel }_{p}^{p-1}+{\int }_{0}^{T}|e\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt\right]\\ =& \left(1+|c|\right)\left({m}_{1}{m}_{3}T+kT+{\int }_{0}^{T}|e\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt\right){\parallel x\parallel }_{\mathrm{\infty }}+\left(1+|c|\right){m}_{1}T\left({r}_{2}+\epsilon \right){\parallel x\parallel }_{\mathrm{\infty }}^{p}\\ +\left(1+|c|\right){r}_{1}{T}^{\frac{1}{p}}{\parallel x\parallel }_{\mathrm{\infty }}{\parallel {x}^{\prime }\parallel }_{p}^{p-1}\\ \le & {a}_{1}\left[b+\left(a+{T}^{\frac{1}{q}}\right)\right]{\parallel {x}^{\prime }\parallel }_{p}+{a}_{2}{\left[b+\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}\right]}^{p}\\ +{a}_{3}\left[b+\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}\right]{\parallel {x}^{\prime }\parallel }_{p}^{p-1},\end{array}$
(3.15)

where ${a}_{1}=\left(1+|c|\right)\left({m}_{1}{m}_{3}T+kT+{\int }_{0}^{T}|e\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt\right)$, ${a}_{2}=\left(1+|c|\right){m}_{1}T\left({r}_{2}+\epsilon \right)$, ${a}_{3}=\left(1+|c|\right){r}_{1}{T}^{\frac{1}{p}}$.

By elementary analysis, we know that there is a constant $h>0$ which satisfies $b-ah>0$ such that

${\left(1+u\right)}^{p}\le 1+\left(1+p\right)u,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in \left(0,h\right].$
(3.16)

By (3.16), one has

$\begin{array}{rl}{\left[b+\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}\right]}^{p}& ={\left(a+{T}^{\frac{1}{q}}\right)}^{p}{\parallel {x}^{\prime }\parallel }_{p}^{p}{\left(1+\frac{b}{\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}}\right)}^{p}\\ \le {\left(a+{T}^{\frac{1}{q}}\right)}^{p}{\parallel {x}^{\prime }\parallel }_{p}^{p}+b\left(1+p\right){\left(a+{T}^{\frac{1}{q}}\right)}^{p-1}{\parallel {x}^{\prime }\parallel }_{p}^{p-1}.\end{array}$
(3.17)

By Lemma 2.2, together with (3.15) and (3.17), we can derive

$\begin{array}{r}|1-|c|{|}^{p}{\parallel {x}^{\prime }\parallel }_{p}^{p}\\ \phantom{\rule{1em}{0ex}}=|1-|c|{|}^{p}{\parallel {A}^{-1}A{x}^{\prime }\parallel }_{p}^{p}\le {\parallel A{x}^{\prime }\parallel }_{p}^{p}\\ \phantom{\rule{1em}{0ex}}\le {a}_{1}\left[b+\left(a+{T}^{\frac{1}{q}}\right)\right]{\parallel {x}^{\prime }\parallel }_{p}+{a}_{2}\left[{\left(a+{T}^{\frac{1}{q}}\right)}^{p}{\parallel {x}^{\prime }\parallel }_{p}^{p}+b\left(1+p\right){\left(a+{T}^{\frac{1}{q}}\right)}^{p-1}{\parallel {x}^{\prime }\parallel }_{p}^{p-1}\right]\\ \phantom{\rule{2em}{0ex}}+{a}_{3}\left[b+\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}\right]{\parallel {x}^{\prime }\parallel }_{p}^{p-1}\\ \phantom{\rule{1em}{0ex}}=\left({a}_{2}+{a}_{3}\right)\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}^{p}+\left[{a}_{2}b\left(1+p\right){\left(a+{T}^{\frac{1}{q}}\right)}^{p-1}+{a}_{3}b\right]{\parallel {x}^{\prime }\parallel }_{p}^{p-1}\\ \phantom{\rule{2em}{0ex}}+{a}_{1}\left(a+{T}^{\frac{1}{q}}\right){\parallel {x}^{\prime }\parallel }_{p}+{a}_{1}b.\end{array}$
(3.18)

From (3.1) and (3.18), we know that ${\parallel {x}^{\prime }\parallel }_{p}$ also is bounded in this case. Based on the above, we can derive the result that ${\parallel {x}^{\prime }\parallel }_{p}$ has a bound; therefore, (3.11) holds.

Secondly, we prove that there is a constant ${R}_{2}$ such that

${\parallel {x}^{\prime }\parallel }_{\mathrm{\infty }}\le {R}_{2}.$
(3.19)

Based on (1.1), together with (3.13) and the condition (A2)(1), we get

$\begin{array}{r}{\int }_{0}^{T}|{\left({\varphi }_{p}\left(\left(A{x}^{\prime }\left(t\right)\right)\right)\right)}^{\prime }|\\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{T}\left[|f\left({x}^{\prime }\left(t\right)\right)|+|\beta \left(t\right)g\left(x\left(t-\tau \left(t\right)\right)\right)|+|e\left(t\right)|\right)\right]\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}\le kT+{r}_{1}{T}^{\frac{1}{p}}{\parallel {x}^{\prime }\parallel }_{p}^{p-1}+{m}_{1}{m}_{3}T+{m}_{1}T\left({r}_{2}+\epsilon \right){\parallel x\parallel }_{\mathrm{\infty }}^{p-1}+{\int }_{0}^{T}|e\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt:={R}_{3}.\end{array}$

Because $A\left(x\left(0\right)\right)=A\left(x\left(T\right)\right)$, there exists ${t}_{0}\in \left[0,T\right]$ such that ${\left(Ax\left({t}_{0}\right)\right)}^{\prime }=A\left({x}^{\prime }\left({t}_{0}\right)\right)=0$. Noting that ${\varphi }_{p}\left(0\right)=0$, we have

$|{\varphi }_{p}\left(A{x}^{\prime }\left(t\right)\right)|=|{\int }_{{t}_{0}}^{t}{\left({\varphi }_{p}\left(A\left({x}^{\prime }\left(t\right)\right)\right)\right)}^{\prime }dt|\le {\int }_{0}^{T}|{\varphi }_{p}{\left(A\left({x}^{\prime }\left(t\right)\right)\right)}^{\prime }|\phantom{\rule{0.2em}{0ex}}dt\le {R}_{3},$

then

${\parallel A{x}^{\prime }\parallel }_{\mathrm{\infty }}\le {\varphi }_{q}\left({R}_{3}\right).$

From Lemma 2.1, we derive

${\parallel {x}^{\prime }\parallel }_{\mathrm{\infty }}={\parallel {A}^{-1}A{x}^{\prime }\parallel }_{\mathrm{\infty }}\le \frac{{\parallel A{x}^{\prime }\parallel }_{\mathrm{\infty }}}{|1-|c||}\le \frac{{\varphi }_{q}\left({R}_{3}\right)}{|1-|c||}:={R}_{2},$

therefore, (3.19) is satisfied.

Let $y\left(t\right)=\left(Ax\left(t\right)\right)$, then (3.2) is equivalent to the following equation:

${\left({\varphi }_{p}\left({y}^{\prime }\left(t\right)\right)\right)}^{\prime }+\lambda f\left({A}^{-1}{y}^{\prime }\left(t\right)\right)+\lambda g\left({A}^{-1}y\left(t-\tau \left(t\right)\right)\right)=\lambda e\left(t\right).$
(3.20)

Then $x={A}^{-1}y$ is a T-periodic solution of (3.2) if y is a T-periodic solution of (3.20). Let

$F\left(y\left(t\right)\right)=e\left(t\right)-f\left({\left({A}^{-1}y\left(t\right)\right)}^{\prime }\right)-g\left({A}^{-1}\left(y\left(t-\tau \left(t\right)\right)\right)\right),$
(3.21)

since f, g are continuous and A has a continuous inverse, the mapping $F:{C}_{T}^{1}⟶{C}_{T}$ in (3.21) is continuous and takes bounded sets into bounded sets.

In addition, (3.20) can be represented as

${\left({\varphi }_{p}\left({y}^{\prime }\left(t\right)\right)\right)}^{\prime }=\lambda F\left(y\left(t\right)\right).$
(3.22)

Let $R=Mmax\left\{{R}_{1},{R}_{2},\rho \right\}$; $M>1+|c|$ is a constant, $\mathrm{\Omega }=\left\{y\left(t\right)\in {C}_{T}^{1},{\parallel y\parallel }_{\mathrm{\infty }}, then (3.22) has no solution on Ω for $\lambda \in \left(0,1\right)$. In fact, if $y=Ax$ is a solution to (3.22) on Ω, then ${\parallel y\parallel }_{\mathrm{\infty }}=R$ or ${\parallel {y}^{\prime }\parallel }_{\mathrm{\infty }}=R$. If ${\parallel y\parallel }_{\mathrm{\infty }}=R$, then ${\parallel y\parallel }_{\mathrm{\infty }}={\parallel Ax\parallel }_{\mathrm{\infty }}={\parallel x\left(t\right)-cx\left(t-\sigma \right)\parallel }_{\mathrm{\infty }}\le \left(1+|c|\right){\parallel x\parallel }_{\mathrm{\infty }}$. That is to say, ${\parallel x\parallel }_{\mathrm{\infty }}\ge \frac{{\parallel y\parallel }_{\mathrm{\infty }}}{1+|c|}>{R}_{1}$. This is a contradiction with (3.11). Similarly, ${\parallel {y}^{\prime }\parallel }_{\mathrm{\infty }}\ne R$. Then (3.22) satisfies the condition (i) of Lemma 2.3.

If $y\in \partial \mathrm{\Omega }\cap \mathbb{R}$, y is a constant and $|y|={\parallel y\parallel }_{\mathrm{\infty }}=R$, then $f\left({\left({A}^{-1}y\right)}^{\prime }\right)=0$ and $|y|\le \left(1+|c|\right)|x|$, $|x|\ge \frac{|y|}{1+|c|}>\rho >d$. By (A4), we obtain $g\left({A}^{-1}\left(y\left(t-\tau \left(t\right)\right)\right)\right)=g\left({A}^{-1}y\right)=g\left(x\right)\ne 0$. Therefore

$Ϝ\left(y\right)=\frac{1}{T}{\int }_{0}^{T}F\left(y\right)\phantom{\rule{0.2em}{0ex}}dt=-g\left({A}^{-1}\left(y\left(t-\tau \left(t\right)\right)\right)\right)=-g\left(x\right)\ne 0$
(3.23)

on $\partial \mathrm{\Omega }\cap \mathbb{R}$. This indicates that (3.22) satisfies the condition (ii) of Lemma 2.3.

We know $\partial \left(\mathrm{\Omega }\cap \mathbb{R}\right)=\left\{-R,R\right\}$, then for $\mathrm{\forall }y\in \partial \left(\mathrm{\Omega }\cap \mathbb{R}\right)$, we have $y=R>d$ or $y=-R<-d$. By (3.23) and the condition (A4), we conclude that $Ϝ\left(y\right)\ne \mu Ϝ\left(-y\right)$, $\mu \in \left[0,1\right]$, $y\in \partial \left(\mathrm{\Omega }\cap \mathbb{R}\right)$. Based on Lemma 2.4, we get $deg\left\{Ϝ,\mathrm{\Omega }\cap \mathbb{R},0\right\}\ne 0$.

Based on the above, (3.22) satisfies all the conditions of Lemma 2.3. So does (3.20). By Lemma 2.3, (3.20) has at least one T-periodic solution, then (1.1) has also at least a periodic solution. □

4 Example

Consider the following equation:

${\left({\varphi }_{4}\left({x}^{\prime }\left(t\right)-\frac{1}{10}{x}^{\prime }\left(t-\frac{1}{2}\right)\right)\right)}^{\prime }+f\left({x}^{\prime }\left(t\right)\right)+\beta \left(t\right)g\left(x\left(t-\frac{1}{2}sint\right)\right)=\frac{1}{400}cost,$
(4.1)

where $p=4$, $c=\frac{1}{10}$, $\sigma =\frac{1}{2}$, $\tau \left(t\right)=\frac{1}{2}sint$, $e\left(t\right)=\frac{1}{400}cost$, $T=2\pi$. Obviously we get ${\tau }^{\prime }\left(t\right)<1$, ${m}_{2}=\frac{2}{3}$.

If we take $f\left(x\right)=\left\{\begin{array}{ll}\frac{1}{100}x,& |x|\le 1,\\ \frac{1}{100}{x}^{3},& |x|>1,\end{array}$ $g\left(x\right)=\frac{x}{100}+\frac{{x}^{3}}{10}$, $\beta \left(t\right)=\frac{1}{10\left(1+{sin}^{2}t\right)}$, then the condition (A4) of Theorem 3.1 is satisfied and

$|f\left(x\right)|\le \frac{1}{100}+\frac{1}{100}|{x}^{3}|,\phantom{\rule{2em}{0ex}}\underset{|x|⟶\mathrm{\infty }}{lim}\frac{|g\left(x\right)|}{|x{|}^{3}}\le \frac{1}{10},\phantom{\rule{2em}{0ex}}{m}_{0}=\frac{1}{20},\phantom{\rule{2em}{0ex}}{m}_{1}=\frac{1}{10}.$
(4.2)

By (4.2), we obtain $k=\frac{1}{100}$, ${r}_{1}=\frac{1}{100}$, ${r}_{2}=\frac{1}{10}$.

If we choose $\epsilon =0.01$, $\rho >1$, then when $|x|>\rho$, we have

$\frac{|g\left(x\right)|}{|x{|}^{3}}\ge {r}_{2}-\epsilon ,\phantom{\rule{2em}{0ex}}\frac{|g\left(x\right)|}{|x{|}^{3}}\le {r}_{2}+\epsilon .$

We calculate easily that

$a=0.843,\phantom{\rule{1em}{0ex}}|1+|c||\left(a+{T}^{\frac{1}{q}}\right)\left[{m}_{1}T\left({r}_{2}+\epsilon \right)+{r}_{1}{T}^{\frac{1}{p}}\right]=0.4496<|1-|c|{|}^{4}=0.6561.$

Based on the above, we know that (4.1) satisfies all conditions included in Theorem 3.1; therefore, (4.1) has at least one T-periodic solution.