Existence of periodic solutions for p-Laplacian neutral Rayleigh equation

Abstract

In this paper, we use topological degree theory to establish new results on the existence of periodic solutions for a p-Laplacian neutral Rayleigh equation.

1 Introduction

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

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

(1.1)

where ${\varphi }_p(x)={|x|}^{p-2}x$ for $x\ne 0$ and $p>1$; σ and c are given constants with $|c|\ne 1$; ${\varphi }_p(0)=0$, $f(0)=0$. 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; ${\int }_0^{T}e(t)dt=0$, ${\int }_0^T\beta (t)\ne 0$.

As we know, the p-Laplace Rayleigh equation with a deviating argument $\tau (t)$ 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:

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

(1.2)

and

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

(1.3)

has been extensively studied in [1–4]. In recent years, Lu et al. [5–7] 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 [8–11]. Here, we focus on [11] by Liang Feng et al. They discussed the existence of the solution to the following equation:

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

(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 ${\int }_0^T(g(x(t-\tau (t)))+e(t))dt=0$, which does not satisfy (1.1). Another significance of the paper is that the result is related to the deviating argument $\tau (t)$, while the conclusions in those existing papers mentioned above have no relation with $\tau (t)$.

2 Preliminary results

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

(H₁) ${\parallel x\parallel }_p={({\int }_0^T|x(t){|}^{p}dt)}^{\frac{1}{p}}$, ${\parallel x\parallel }_{\mathrm{\infty }}={max}_{t\in [0,T]}|x(t)|$, $\parallel x\parallel =max\{{\parallel x\parallel }_{\mathrm{\infty }},{\parallel x^{\prime }(t)\parallel }_{\mathrm{\infty }}\}$;

(H₂) $m_0={min}_{t\in [0,T]}|\beta (t)|$, $m_1={max}_{t\in [0,T]}|\beta (t)|$;

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

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

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 $A:C_T⟶C_T$, $(Ax)(t)=x(t)-cx(t-\sigma )$.

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

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

• (3) ${\int }_0^T|(A^{-1}x)(t)|dt\le \frac{1}{|1-|c||}{\int }_0^T|x(t)|dt$, $\mathrm{\forall }x\in C_T$.

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

$${\parallel A^{-1}x(t)\parallel }_p\le \frac{1}{1-|c|}{\parallel x(t)\parallel }_p,\mathrm{\forall }x\in C_T.$$

(2.1)

Proof We know that $x(t)$ is a periodic function. So ${\int }_0^T|x(t-j\sigma )|dt={\int }_0^T|x(t)|dt$ for $j\ge 0$. When $|c|<1$, from Lemma 2.1, we have

$$\begin{array}{rcl}{\int }_0^T|A^{-1}x(t){|}^{p}dt& =& {\int }_0^T|A^{-1}x(t){|}^{p-1}|A^{-1}x(t)|dt\\ =& {\int }_0^T|A^{-1}x(t){|}^{p-1}|\sum _{j\ge 0}{c}^{j}x(t-j\sigma )|dt\\ \le & \sum _{j\ge 0}\left|c^j\right|{\int }_0^T|A^{-1}x(t){|}^{p-1}|x(t-j\sigma )|dt\\ \le & \sum _{j\ge 0}\left|c^j\right|{\left({\int }_0^T\right|A^{-1}x(t){|}^{(p-1)q}dt)}^{\frac{1}{q}}{\left({\int }_0^T\right|x(t-j\sigma ){|}^{p}dt)}^{\frac{1}{p}}\\ =& \sum _{j\ge 0}\left|c^j\right|{\left({\int }_0^T\right|A^{-1}x(t){|}^{p}dt)}^{\frac{1}{q}}{\left({\int }_0^T\right|x(t){|}^{p}dt)}^{\frac{1}{p}}\\ =& \frac{1}{1-|c|}{\left({\int }_0^T\right|A^{-1}x(t){|}^{p}dt)}^{\frac{1}{q}}{\left({\int }_0^T\right|x(t){|}^{p}dt)}^{\frac{1}{p}}\end{array}$$

which implies ${({\int }_0^T|A^{-1}x(t){|}^{p}dt)}^{\frac{1}{p}}\le \frac{1}{1-|c|}{({\int }_0^T|x(t-j\sigma ){|}^{p}dt)}^{\frac{1}{p}}$. 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 $C_T^1$:

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

(2.2)

$F:C_T^1⟶C_T$ is continuous and takes a bounded set into bounded set.

Let us define $P:C_T^1⟶C_T$, $u|⟶u(0)$, $Q:C_T⟶C_T$, $h|⟶\frac{1}{T}{\int }_0^{T}h(s)ds$ and

$$H(h(t))={\int }_0^{t}h(s)ds,h\in C_T.$$

It is clear that if $u\in C_T^1$ is the solution to (2.2), then u satisfies the abstract equation

$$u=Pu+QF(u)+K(F(u)),$$

where the operator $K:C_T⟶C_T^1$ is given by

$$K(h(t))=H\left\{{\varphi }_q[\alpha ((I-Q)h)+H((I-Q))]\right\}(t),\mathrm{\forall }t\in \mathbb{R},$$

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:

• (i) For each $\lambda \in (0,1)$, the equation ${({\varphi }_p(u^{\prime }))}^{\prime }=\lambda F(u)$ has no solution on ∂ Ω.

• (ii) The equation $Ϝ(u)=\frac{1}{T}{\int }_0^{T}F(u(t))dt=0$ has no solution on $\partial \mathrm{\Omega }\cap \mathbb{R}$.

• (iii) The Brouwer degree of Ϝ, $deg\{Ϝ,\mathrm{\Omega }\cap \mathbb{R},0\}\ne 0$.

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 $f\in C(\overline{\mathrm{\Omega }},{\mathbb{R}}^n)$ and $f(x)\ne \mu f(-x)$, $\mathrm{\forall }x\in \partial \mathrm{\Omega }$, $\mu \in [0,1]$, then $deg\{f,\mathrm{\Omega },0\}$ is an odd number.

3 Main results

Theorem 3.1 Suppose that the following conditions hold:

(A₁) $\tau (t)\in C^1(\mathbb{R},\mathbb{R})$, ${\tau }^{\prime }(t)>1$ or ${\tau }^{\prime }(t)<1$, $m_2={min}_{t\in [0,T]}\frac{1}{1-{\tau }^{\prime }(t)}$.

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

(A₃) There exist constants $r_1\ge 0$, $r_2>0$ and $k>0$ such that

• (1) $|f(x)|\le k+r_1{|x|}^{p-1}$, $\mathrm{\forall }x\in \mathbb{R}$,

• (2) ${lim}_{|x|⟶\mathrm{\infty }}\frac{|g(x)|}{{|x|}^{p-1}}\le r_2$.

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

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

$$|1+|c||(a+T^{\frac{1}{q}})[m_{1}T(r_2+\epsilon )+r_1{T}^{\frac{1}{p}}]<|1-|c|{|}^p,$$

(3.1)

where a is defined in (3.8).

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

$${\left({\varphi }_p(x^{\prime }(t)-c{x}^{\prime }(t-\sigma ))\right)}^{\prime }+\lambda f(x^{\prime }(t))+\lambda \beta (t)g\left(x(t-\tau (t))\right)=\lambda e(t),\lambda \in (0,1).$$

(3.2)

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

$${\int }_0^T[f(x^{\prime }(t))+\beta (t)g\left(x(t-\tau (t))\right)]dt=0.$$

(3.3)

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

$${\int }_0^T\beta (t)g\left(x(t-\tau (t))\right)dt={\int }_{-\tau (0)}^{T-\tau (T)}\frac{\beta (\gamma (u))g(u)}{1-{\tau }^{\prime }(\gamma (u))}du={\int }_0^T\frac{\beta (\gamma (u))g(u)}{1-{\tau }^{\prime }(\gamma (u))}du.$$

(3.4)

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

$$g(x(\xi )){\int }_0^T\frac{\beta (\gamma (u))}{1-{\tau }^{\prime }(\gamma (u))}du=-{\int }_0^{T}f(x^{\prime }(t))dt.$$

(3.5)

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

$$\frac{|g(x)|}{{|x|}^{p-1}}\ge r_2-\epsilon >0,\frac{|g(x)|}{{|x|}^{p-1}}\le r_2+\epsilon .$$

(3.6)

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

$$|x(\xi )|\le a{\parallel x\parallel }_p+b,$$

(3.7)

where

$$a=\{\begin{array}{cc}{[\frac{r_1}{(r_2-\epsilon )m_0{m}_2{T}^{\frac{1}{q}}}]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}},\hfill & p-1<1,\hfill \\ {[\frac{r_1}{(r_2-\epsilon )m_0{m}_2{T}^{\frac{1}{q}}}]}^{\frac{1}{p-1}},\hfill & p-1>1,\hfill \end{array}$$

(3.8)

$$b=\{\begin{array}{cc}{[\frac{k}{(r_2-\epsilon )m_0{m}_2}]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}},\hfill & p-1<1,\hfill \\ {[\frac{k}{(r_2-\epsilon )m_0{m}_2}]}^{\frac{1}{p-1}},\hfill & p-1>1.\hfill \end{array}$$

(3.9)

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

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

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

$$\begin{array}{r}|r_2-\epsilon ||x(\xi ){|}^{p-1}{m}_0{m}_{2}T\le |g(x(\xi ))|{\int }_0^T\frac{\beta (\gamma (u))}{1-{\tau }^{\prime }(\gamma (u))}du\\ \phantom{|r_2-\epsilon ||x(\xi ){|}^{p-1}{m}_0{m}_{2}T}\le {\int }_0^T\left|f(x^{\prime }(t))\right|dt\le kT+r_1{\int }_0^T|x^{\prime }(t){|}^{p-1}dt\\ \phantom{|r_2-\epsilon ||x(\xi ){|}^{p-1}{m}_0{m}_{2}T}\le kT+r_1{T}^{\frac{1}{p}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{p}dt)}^{\frac{p-1}{p}}=kT+r_1{T}^{\frac{1}{p}}{\parallel x^{\prime }\parallel }_p^{p-1},\\ |x(\xi ){|}^{p-1}\le \frac{k}{(r_2-\epsilon )m_0{m}_2}+\frac{r_1}{(r_2-\epsilon )m_0{m}_2{T}^{\frac{1}{q}}}{\parallel x^{\prime }\parallel }_p^{p-1}.\end{array}$$

When $0<p-1\le 1$, according to the Minkowski inequality, we have

$$|x(\xi )|\le {\left[\frac{k}{(r_2-\epsilon )m_0{m}_2}\right]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}}+{\left[\frac{r_1}{(r_2-\epsilon )m_0{m}_2{T}^{\frac{1}{q}}}\right]}^{\frac{1}{p-1}}{2}^{\frac{2-p}{p-1}}{\parallel x^{\prime }\parallel }_p.$$

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

$$|x(\xi )|\le {\left[\frac{k}{(r_2-\epsilon )m_0{m}_2}\right]}^{\frac{1}{p-1}}+{\left[\frac{r_1}{(r_2-\epsilon )m_0{m}_2{T}^{\frac{1}{q}}}\right]}^{\frac{1}{p-1}}{\parallel x^{\prime }\parallel }_p.$$

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

For any $t\in \mathbb{R}$, there exists $t_0\in [0,T]$, such that $t=kT+t_0$, where k is an integer. Then

$$|x(t)|=|x(t_0)|\le |x(\xi )|+{\int }_0^T|x^{\prime }(s)|ds,\xi \in [0,T].$$

By (3.7), we have

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

(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 $\frac{b}{a+T^{\frac{1}{q}}{\parallel x^{\prime }\parallel }_p}<h$, we prove that ${\parallel x^{\prime }\parallel }_p$ is bounded in the following.

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

$$\begin{array}{r}\left|{\int }_0^T{\left[{\phi }_p\left(A{x}^{\prime }\right)\right]}^{\prime }A(x(t))dt\right|\\ =|{\phi }_p\left(A{x}^{\prime }\right)A(x(t)){|}_0^T-{\int }_0^T{\phi }_p\left(A{x}^{\prime }\right)A{x}^{\prime }dt|\\ ={\int }_0^T|A{x}^{\prime }{|}^{p}dt={\parallel A{x}^{\prime }\parallel }_p^p\\ =\left|\lambda {\int }_0^{T}A(x(t))[f(x^{\prime }(t))+\beta (t)g\left(x(t-\tau (t))\right)-e(t)]dt\right|\\ \le (1+|c|){\parallel x\parallel }_{\mathrm{\infty }}{\int }_0^T\left[\right|f(x^{\prime }(t)|+|\beta (t)g\left(x(t-\tau (t))\right)|+|e(t)|]dt.\end{array}$$

(3.12)

Let $E_1=\{t\in [0,T]:|x(t-\tau (t))|\le \rho \}$, $E_2=\{t\in [0,T]:|x(t-\tau (t))|>\rho \}$, then

$$\begin{array}{rl}{\int }_0^T|\beta (t)g\left(x(t-\tau (t))\right)|dt& ={\int }_{E_1}|\beta (t)g\left(x(t-\tau (t))\right)|dt+{\int }_{E_2}|\beta (t)g\left(x(t-\tau (t))\right)|dt\\ \le m_1{m}_{3}T+m_{1}T(r_2+\epsilon ){\parallel x\parallel }_{\mathrm{\infty }}^{p-1},\end{array}$$

(3.13)

where

$$m_3=\underset{|x|\le \rho }{max}|g(x)|.$$

(3.14)

By (3.12) and (3.13), we get

$$\begin{array}{rl}{\parallel A{x}^{\prime }\parallel }_p^p\le & (1+|c|){\parallel x\parallel }_{\mathrm{\infty }}\\ \times [m_1{m}_{3}T+m_{1}T(r_2+\epsilon ){\parallel x\parallel }_{\mathrm{\infty }}^{p-1}+kT+r_1{T}^{\frac{1}{p}}{\parallel x^{\prime }\parallel }_p^{p-1}+{\int }_0^T|e(t)\left|dt\right]\\ =& (1+|c|)(m_1{m}_{3}T+kT+{\int }_0^T|e(t)\left|dt\right){\parallel x\parallel }_{\mathrm{\infty }}+(1+|c|)m_{1}T(r_2+\epsilon ){\parallel x\parallel }_{\mathrm{\infty }}^p\\ +(1+|c|)r_1{T}^{\frac{1}{p}}{\parallel x\parallel }_{\mathrm{\infty }}{\parallel x^{\prime }\parallel }_p^{p-1}\\ \le & a_1[b+(a+T^{\frac{1}{q}})]{\parallel x^{\prime }\parallel }_p+a_2{[b+(a+T^{\frac{1}{q}}){\parallel x^{\prime }\parallel }_p]}^p\\ +a_3[b+(a+T^{\frac{1}{q}}){\parallel x^{\prime }\parallel }_p]{\parallel x^{\prime }\parallel }_p^{p-1},\end{array}$$

(3.15)

where $a_1=(1+|c|)(m_1{m}_{3}T+kT+{\int }_0^T|e(t)|dt)$, $a_2=(1+|c|)m_{1}T(r_2+\epsilon )$, $a_3=(1+|c|)r_1{T}^{\frac{1}{p}}$.

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

$${(1+u)}^p\le 1+(1+p)u,\mathrm{\forall }u\in (0,h].$$

(3.16)

By (3.16), one has

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

(3.17)

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

$$\begin{array}{r}|1-|c|{|}^p{\parallel x^{\prime }\parallel }_p^p\\ =|1-|c|{|}^p{\parallel A^{-1}A{x}^{\prime }\parallel }_p^p\le {\parallel A{x}^{\prime }\parallel }_p^p\\ \le a_1[b+(a+T^{\frac{1}{q}})]{\parallel x^{\prime }\parallel }_p+a_2[{(a+T^{\frac{1}{q}})}^p{\parallel x^{\prime }\parallel }_p^p+b(1+p){(a+T^{\frac{1}{q}})}^{p-1}{\parallel x^{\prime }\parallel }_p^{p-1}]\\ +a_3[b+(a+T^{\frac{1}{q}}){\parallel x^{\prime }\parallel }_p]{\parallel x^{\prime }\parallel }_p^{p-1}\\ =(a_2+a_3)(a+T^{\frac{1}{q}}){\parallel x^{\prime }\parallel }_p^p+[a_{2}b(1+p){(a+T^{\frac{1}{q}})}^{p-1}+a_{3}b]{\parallel x^{\prime }\parallel }_p^{p-1}\\ +a_1(a+T^{\frac{1}{q}}){\parallel x^{\prime }\parallel }_p+a_{1}b.\end{array}$$

(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

$$\begin{array}{r}{\int }_0^T\left|{\left({\varphi }_p\left((A{x}^{\prime }(t))\right)\right)}^{\prime }\right|\\ \le {\int }_0^T\left[\right|f(x^{\prime }(t))|+|\beta (t)g\left(x(t-\tau (t))\right)|+|e(t)\left|\right)]dt\\ \le kT+r_1{T}^{\frac{1}{p}}{\parallel x^{\prime }\parallel }_p^{p-1}+m_1{m}_{3}T+m_{1}T(r_2+\epsilon ){\parallel x\parallel }_{\mathrm{\infty }}^{p-1}+{\int }_0^T|e(t)|dt:=R_3.\end{array}$$

Because $A(x(0))=A(x(T))$, there exists $t_0\in [0,T]$ such that ${(Ax(t_0))}^{\prime }=A(x^{\prime }(t_0))=0$. Noting that ${\varphi }_p(0)=0$, we have

$$\left|{\varphi }_p(A{x}^{\prime }(t))\right|=\left|{\int }_{t_0}^t{\left({\varphi }_p\left(A(x^{\prime }(t))\right)\right)}^{\prime }dt\right|\le {\int }_0^T\left|{\varphi }_p{\left(A(x^{\prime }(t))\right)}^{\prime }\right|dt\le R_3,$$

then

$${\parallel A{x}^{\prime }\parallel }_{\mathrm{\infty }}\le {\varphi }_q(R_3).$$

From Lemma 2.1, we derive

$${\parallel x^{\prime }\parallel }_{\mathrm{\infty }}={\parallel A^{-1}A{x}^{\prime }\parallel }_{\mathrm{\infty }}\le \frac{{\parallel A{x}^{\prime }\parallel }_{\mathrm{\infty }}}{|1-|c||}\le \frac{{\varphi }_q(R_3)}{|1-|c||}:=R_2,$$

therefore, (3.19) is satisfied.

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

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

(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

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

(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

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

(3.22)

Let $R=Mmax\{R_1,R_2,\rho \}$; $M>1+|c|$ is a constant, $\mathrm{\Omega }=\{y(t)\in C_T^1,{\parallel y\parallel }_{\mathrm{\infty }}<R,{\parallel y^{\prime }\parallel }_{\mathrm{\infty }}<R\}$, then (3.22) has no solution on ∂ Ω for $\lambda \in (0,1)$. 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 ${\parallel y\parallel }_{\mathrm{\infty }}={\parallel Ax\parallel }_{\mathrm{\infty }}={\parallel x(t)-cx(t-\sigma )\parallel }_{\mathrm{\infty }}\le (1+|c|){\parallel x\parallel }_{\mathrm{\infty }}$. 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 $f({(A^{-1}y)}^{\prime })=0$ and $|y|\le (1+|c|)|x|$, $|x|\ge \frac{|y|}{1+|c|}>\rho >d$. By (A₄), we obtain $g(A^{-1}(y(t-\tau (t))))=g(A^{-1}y)=g(x)\ne 0$. Therefore

$$Ϝ(y)=\frac{1}{T}{\int }_0^{T}F(y)dt=-g\left(A^{-1}\left(y(t-\tau (t))\right)\right)=-g(x)\ne 0$$

(3.23)

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

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

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:

$${\left({\varphi }_4(x^{\prime }(t)-\frac{1}{10}{x}^{\prime }(t-\frac{1}{2}))\right)}^{\prime }+f(x^{\prime }(t))+\beta (t)g\left(x(t-\frac{1}{2}sint)\right)=\frac{1}{400}cost,$$

(4.1)

where $p=4$, $c=\frac{1}{10}$, $\sigma =\frac{1}{2}$, $\tau (t)=\frac{1}{2}sint$, $e(t)=\frac{1}{400}cost$, $T=2\pi$. Obviously we get ${\tau }^{\prime }(t)<1$, $m_2=\frac{2}{3}$.

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

$$|f(x)|\le \frac{1}{100}+\frac{1}{100}\left|x^3\right|,\underset{|x|⟶\mathrm{\infty }}{lim}\frac{|g(x)|}{|x{|}^3}\le \frac{1}{10},m_0=\frac{1}{20},m_1=\frac{1}{10}.$$

(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

$$\frac{|g(x)|}{|x{|}^3}\ge r_2-\epsilon ,\frac{|g(x)|}{|x{|}^3}\le r_2+\epsilon .$$

We calculate easily that

$$a=0.843,|1+|c||(a+T^{\frac{1}{q}})[m_{1}T(r_2+\epsilon )+r_1{T}^{\frac{1}{p}}]=0.4496<|1-|c|{|}^4=0.6561.$$

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.