Advances in Difference Equations

, 2019:106

# Periodic solutions for p-Laplacian neutral differential equation with multiple delay and variable coefficients

Open Access
Research

## Abstract

In this paper, we first discuss some properties of the neutral operator with multiple delays and variable coefficients $$(Ax)(t):=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i})$$. Afterwards, by using an extension of Mawhin’s continuation theorem, a second order p-Laplacian neutral differential equation
$$\Biggl(\phi _{p} \Biggl(x(t)-\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}) \Biggr)' \Biggr)'=\tilde{f} \bigl(t,x(t),x'(t)\bigr)$$
is studied. Some new results on the existence of a periodic solution are obtained. Meanwhile, the approaches to estimate a priori bounds of periodic solutions are different from those known in the literature.

## Keywords

Neutral operator p-Laplacian Periodic solution Extension of Mawhin’s continuation theorem Singularity

34C25 34K14

## 1 Introduction

In this paper, we consider a second order p-Laplacian neutral differential equation
$${ } \Biggl(\phi _{p} \Biggl(x(t)-\sum _{i=1}^{n}c_{i}(t)x(t-\delta _{i}) \Biggr)' \Biggr)'=\tilde{f} \bigl(t,x(t),x'(t)\bigr),$$
(1.1)
where $$\phi _{p}:\mathbb{R}\to \mathbb{R}$$ is given by $$\phi _{p}(s)=|s|^{p-2}s$$, here $$p>1$$ is a constant, $$c_{i}(t)\in C ^{1}(\mathbb{R},\mathbb{R})$$ and $$c_{i}(t+T)=c_{i}(t)$$ and $$\delta _{i}$$ are constants in $$[0,T)$$ for $$i=1,2,\dots ,n$$; : $$[0,T]\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$$ 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 study of the properties of the neutral operator $$(A_{1}x)(t):=x(t)-cx(t- \delta )$$ began with the paper of Zhang . In 2004, Lu and Ge  investigated an extension of $$A_{1}$$, namely the neutral operator $$(A_{2}x)(t):=x(t)-\sum_{i=1}^{n}c_{i}x(t- \delta _{i})$$. Afterwards, Du  discussed the neutral operator $$(A_{3}x)(t):=x(t)-c(t)x(t-\delta )$$, here $$c(t)$$ is a T-periodic function. And by using Mawhin’s continuation theorem and the properties of $$A_{3}$$, they obtained sufficient conditions for the existence of periodic solutions to the following Liénard neutral differential equation:
$$\bigl(x(t)-c(t)x(t-\tau )\bigr)''+f\bigl(x(t) \bigr)x'(t)+g\bigl(x\bigl(t-\gamma (t)\bigr)\bigr)=e(t).$$
In recent years, many works have been published on the existence of periodic solutions of second-order neutral differential equations (see [1, 3, 4, 5, 7, 9, 11, 12, 13, 16, 17, 18, 19]). In 2007, Zhu and Lu  discussed the existence of periodic solutions for a p-Laplacian neutral differential equation
$$\bigl(\phi _{p}\bigl(x(t)-cx(t-\tau )\bigr)' \bigr)'+g\bigl(t,x\bigl(t-\delta (t)\bigr)\bigr)=p(t).$$
Since $$(\phi _{p}(x'(t)))'$$ is nonlinear (i.e., quasilinear), Mawhin’s continuation theorem  cannot be applied directly. In order to get around this difficulty, Zhu and Lu translated the p-Laplacian neutral differential equation into a two-dimensional system
$$\textstyle\begin{cases} (x_{1}(t)-cx_{1}(t-\tau ))'(t)=\phi _{q}(x_{2}(t))= \vert x_{2}(t) \vert ^{q-2}x _{2}(t), \\ x_{2}'(t)=-g(t,x_{1}(t-\delta (t)))+p(t), \end{cases}$$
where $$\frac{1}{p}+\frac{1}{q}=1$$, for which Mawhin’s continuation theorem can be applied. Afterwards, Du  discussed the existence of a periodic solution for a p-Laplacian neutral differential equation
$$\bigl(\phi _{p}\bigl(x(t)-c(t)x(t-\tau )\bigr)' \bigr)'+f\bigl(x(t)\bigr)x'(t)+g\bigl(x\bigl(t-\gamma (t)\bigr)\bigr)=e(t),$$
by applying Mawhin’s continuation theorem.

However, the existence of a periodic solution for p-Laplacian neutral differential equation (1.1) has not been studied until now. The obvious difficulty lies in the following two respects. First, although $$(Ax)(t)=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i})$$ is a natural generalization of the operators $$A_{1}$$, $$A_{2}$$ and $$A_{3}$$, the class of neutral differential equations with A typically possesses a more complicated nonlinearity than neutral differential equations with $$A_{1}$$, $$A_{2}$$ and $$A_{3}$$. Second, we do not get $$(Ax)'(t)=(Ax')(t)$$, meanwhile a priori bounds of periodic solutions are not easy to estimate.

The remaining part of the paper is organized as follows. In Sect. 2, we analyze qualitative properties of the generalized neutral operator A. In Sect. 3, by employing an extension of Mawhin’s continuation theorem, we state and prove the existence of periodic solutions for Eq. (1.1). In Sect. 4, we investigate the existence of periodic solutions for a p-Laplacian neutral differential equation by applying Theorem 3.2. In comparison to  and , we avoid translating the equation into a two-dimensional system. In Sect. 5, we discuss the existence of periodic solutions for a p-Laplacian neutral differential equation with singularity by applying Theorem 3.2. In Sect. 6, we give four examples to demonstrate the validity of the methods.

## 2 Analysis of the generalized neutral operator

Let
$$\Vert c_{i} \Vert :=\max_{t\in [0,T]} \bigl\vert c_{i}(t) \bigr\vert ,\quad i=1,2,\dots n; \qquad \Vert c_{k} \Vert :=\max \bigl\{ \Vert c_{1} \Vert , \Vert c_{2} \Vert ,\dots , \Vert c_{n} \Vert \bigr\} .$$
Set $$C_{T}:=\{x\in C(\mathbb{R},\mathbb{R}):x(t+T)=x(t),t\in \mathbb{R}\}$$, then $$(C_{T},\|\cdot \|)$$ is a Banach space. Define operators $$A,B:C_{T}\rightarrow C_{T}$$, by
$$(Ax) (t)=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}), \qquad (Bx) (t)=\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}).$$

### Lemma 2.1

If$$\sum_{i=1}^{n}\|c_{i}\|\neq 1$$, then operatorAhas a continuous inverse$$A^{-1}$$on$$C_{T}$$, satisfying
1. (1)
$$\bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert \leq \textstyle\begin{cases} \frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }, \quad \textit{for }\sum_{i=1}^{n} \Vert c_{i} \Vert < 1; \\ \frac{\frac{1}{ \Vert c_{k} \Vert } \Vert x \Vert }{1-\frac{1}{ \Vert c_{k} \Vert }-\sum_{i=1,i\neq k}^{n} \Vert \frac{c_{i}}{c_{k}} \Vert }, \quad \textit{for }\sum_{i=1}^{n} \Vert c_{i} \Vert >1; \end{cases}$$

2. (2)
$$\int ^{T}_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt\leq \textstyle\begin{cases} \frac{1}{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }\int ^{T}_{0} \vert x(t) \vert \,dt, \quad \textit{for }\sum_{i=1}^{n} \Vert c_{i} \Vert < 1; \\ \frac{\frac{1}{ \Vert c_{k} \Vert }}{1-\frac{1}{ \Vert c_{k} \Vert }-\sum_{i=1,i \neq k}^{n} \Vert \frac{c_{i}}{c_{k}} \Vert }\int ^{T}_{0} \vert x(t) \vert \,dt, \quad \textit{for }\sum_{i=1}^{n} \Vert c_{i} \Vert >1. \end{cases}$$

### Proof

Case 1:
\begin{aligned} &\sum_{i=1}^{n} \Vert c_{i} \Vert < 1. \\ &(Bx) (t)=\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}); \\ &\bigl(B^{2}x\bigr) (t)=\sum_{l_{1}=1}^{n}c_{l_{1}}(t) \sum_{l_{2}=1}^{n}c_{l_{2}}(t-\delta _{l_{1}}) x(t-\delta _{l_{1}}- \delta _{l_{2}}); \\ &\bigl(B^{3}x\bigr) (t)=\sum_{l_{1}=1}^{n}c_{l_{1}}(t) \sum_{l_{2}=1}^{n}c_{l_{2}}(t-\delta _{l_{1}}) \sum_{l_{3}=1} ^{n}c_{l_{2}}(t- \delta _{l_{1}}-\delta _{l_{2}})x(t-\delta _{l_{1}}- \delta _{l_{2}}-\delta _{l_{3}}). \end{aligned}
Therefore, we have
$$\bigl(B^{j}x\bigr) (t)=\sum_{l_{1}=1}^{n}c_{l_{1}}(t) \sum_{l_{2}=1} ^{n}c_{l_{2}}(t-\delta _{l_{1}})\cdots \sum_{l_{j}=1}^{n}c_{l _{j}}(t- \delta _{l_{1}}-\delta _{l_{2}}\cdots -\delta _{l_{j-1}})x(t- \delta _{l_{1}}-\delta _{l_{2}}-\cdots -\delta _{l_{j}}),$$
and
\begin{aligned} \sum_{j=0}^{\infty }\bigl(B^{j}x \bigr) (t) =&x(t)+\sum_{j=1}^{\infty }\sum _{l_{1}=1}^{n}c_{l_{1}}(t)\sum _{l_{2}=1}^{n}c_{l _{2}}(t-\delta _{l_{1}}) \cdots \\ &{}\times\sum_{l_{j}=1}^{n}c_{l_{j}}(t- \delta _{l_{1}}-\delta _{l_{2}}-\cdots -\delta _{l_{j-1}})x(t- \delta _{l _{1}}-\delta _{l_{2}}-\cdots -\delta _{l_{j}}). \end{aligned}
Since $$A=I-B$$ and $$\|B\|<1$$, we get that A has a continuous inverse $$A^{-1}$$: $$C_{T}\rightarrow C_{T}$$ with
$$A^{-1}=(I-B)^{-1}=I+\sum_{j=1}^{\infty }B^{j}= \sum_{j=0} ^{\infty }B^{j},$$
where $$B^{0}=I$$. Then
\begin{aligned} \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert ={}& \Biggl\vert \sum_{j=0}^{\infty }\bigl(B^{j}x\bigr) (t) \Biggr\vert \\ ={}& \Biggl\vert x(t)+\sum_{j=1}^{\infty } \bigl(B^{j}x\bigr) (t) \Biggr\vert \\ ={}& \Biggl\vert x(t)\\ &{}+\sum_{l_{1}=1}^{n}c_{l_{1}}(t) \cdots \sum_{l_{j}=1}^{n}c_{l_{j}}(t- \delta _{l_{1}}-\delta _{l_{2}}-\cdots -\delta _{l_{j-1}})x(t- \delta _{l_{1}}-\delta _{l_{2}}-\cdots -\delta _{l_{j}}) \Biggr\vert \\ \leq{}&\frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }. \end{aligned}
Moreover,
\begin{aligned} &\int ^{T}_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt \\ &\quad = \int ^{T}_{0} \Biggl\vert \sum _{j=0} ^{\infty }\bigl(B^{j}x\bigr) (t) \Biggr\vert \,dt \\ &\quad \leq\sum_{j=0}^{\infty } \int ^{T}_{0} \bigl\vert \bigl(B^{j}x \bigr) (t) \bigr\vert \,dt \\ &\quad \leq\sum_{j=0}^{\infty } \int ^{T}_{0} \Biggl\vert \sum _{l _{1}=1}^{n}c_{l_{1}}(t)\sum _{l_{2}=1}^{n}c_{l_{2}}(t-\delta _{l_{1}}) \cdots \\ &\qquad {}\times\sum_{l_{j}=1}^{n}c_{l_{j}}(t- \delta _{l_{1}}- \delta _{l_{2}}-\cdots -\delta _{l_{j-1}})x(t- \delta _{l_{1}}-\delta _{l _{2}}-\cdots -\delta _{l_{j}}) \Biggr\vert \,dt \\ &\quad \leq\frac{1}{1-\sum_{i=1}^{n} \Vert c_{i} \Vert } \int ^{T}_{0} \bigl\vert x(t) \bigr\vert \,dt. \end{aligned}

Case 2: $$\sum_{i=1}^{n}\|c_{i}\|>1$$.

The operator $$(Ax)(t)=x(t)-\sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i})$$ can be converted to
\begin{aligned} (Ax) (t)&=x(t)-c_{k}(t)x(t-\delta _{k})-\sum _{i=1,i\neq k}^{n}c _{i}(t)x(t-\delta _{i}) \\ &=-c_{k}(t) \Biggl(-\frac{x(t)}{c_{k}(t)}+x(t-\delta _{k})+ \sum_{i=1,i\neq k}^{n}\frac{c_{i}(t)}{c_{k}(t)}x(t-\delta _{i}) \Biggr) \\ &=-c_{k}(t) \Biggl(x(t-\delta _{k})-\frac{x(t)}{c_{k}(t)}+ \sum_{i=1,i\neq k}^{n}\frac{c_{i}(t)}{c_{k}(t)}x(t-\delta _{i}) \Biggr). \end{aligned}
Let $$t_{1}=t-\delta _{k}$$, it is clear that
$$(Ax) (t_{1}+\delta _{k})=-c_{k}(t_{1}+ \delta _{k}) \Biggl(x(t_{1})-\frac{x(t _{1}+\delta _{k})}{c_{k}(t_{1}+\delta _{k})} +\sum _{i=1,i\neq k} ^{n}\frac{c_{i}(t_{1}+\delta _{k})}{c_{k}(t_{1}+\delta _{k})}x(t_{1}+ \delta _{k}-\delta _{i}) \Biggr).$$
Define
\begin{aligned}& (Ex) (t)=-c_{k}(t_{1}+\delta _{k}) \Biggl(x(t_{1})-\frac{x(t_{1}+\delta _{k})}{c_{k}(t_{1}+\delta _{k})} +\sum_{i=1,i\neq k}^{n} \frac{c _{i}(t_{1}+\delta _{k})}{c_{k}(t_{1}+\delta _{k})}x(t_{1}+\delta _{k}- \delta _{i}) \Biggr), \\& e_{i}= \textstyle\begin{cases} \frac{1}{c_{k}(t_{1}+\delta _{k})}, \quad \mbox{for } i=k; \\ -\frac{c_{i}(t_{1}+\delta _{k})}{c_{k}(t_{1}+\delta _{k})}, \quad \mbox{for } i\neq k. \end{cases}\displaystyle \qquad \varepsilon _{i}= \textstyle\begin{cases} -\delta _{k}, \quad \mbox{for } i=k; \\ \delta _{i}-\delta _{k}, \quad \mbox{for } i\neq k. \end{cases}\displaystyle \end{aligned}
Therefore, $$(Ex)(t_{1}+\delta _{k})=x(t_{1}+\delta _{k})-\sum_{i=1}^{n}e_{i}(t_{1}+\delta _{k})x(t_{1}-\varepsilon _{i})$$ and, from Case 1, we get
$$\bigl\vert \bigl(E^{-1}x\bigr) (t) \bigr\vert \leq \frac{ \Vert x \Vert }{1-\sum_{i=1}^{n} \Vert e_{i} \Vert }.$$
Moreover, since $$(A^{-1}x)(t)=-\frac{1}{c_{k}(t)}(E^{-1}x)(t)$$, we have
\begin{aligned} \bigl\vert \bigl(A^{-1}x\bigr) (t) \bigr\vert &\leq \biggl\vert - \frac{1}{c_{k}(t)}\bigl(E^{-1}x\bigr) (t) \biggr\vert \\ &\leq \frac{\frac{1}{ \Vert c_{k} \Vert } \Vert x \Vert }{1-\frac{1}{ \Vert c_{k} \Vert }-\sum_{i=1,i\neq k}^{n} \Vert \frac{c_{i}}{c_{k}} \Vert }. \end{aligned}
Meanwhile, we obtain
$$\int ^{T}_{0} \bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \,dt \leq \frac{\frac{1}{ \Vert c_{k} \Vert }}{1-\frac{1}{ \Vert c_{k} \Vert }-\sum_{i=1,i\neq k}^{n} \Vert \frac{c_{i}}{c_{k}} \Vert } \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt.$$
□

## 3 Periodic solutions for equation (1.1)

In order to use an extension of Mawhin’s continuation theorem , we recall it firstly.

Let X and Z be Banach spaces with norms $$\|\cdot \|_{X}$$ and $$\|\cdot \|_{Z}$$, respectively. A continuous operator $$M:X\cap \operatorname{dom} M\to Z$$ is said to be quasilinear if
1. (1)

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

2. (2)

$$\ker M:=\{x\in X\cap \operatorname{dom}M: Mx=0\}$$ is a subspace of X with $$\dim \ker M<+\infty$$.

Let $$X_{1}=\ker M$$ and $$X_{2}$$ be the complement space of $$X_{1}$$ in X, then $$X=X_{1}\oplus X_{2}$$. Furthermore, $$Z_{1}$$ is a subspace of Z and $$Z_{2}$$ is the complement space of $$Z_{1}$$ in Z, so $$Z=Z_{1}\oplus Z_{2}$$. Suppose that $$P:X\rightarrow X_{1}$$ and $$Q: Z\rightarrow Z_{1}$$ are two projections and $$\varOmega \subset X$$ is an open and bounded set with the origin $$\theta \in \varOmega$$.

Let $$N_{\lambda }:\bar{\varOmega }\to Z$$, $$\lambda \in [0,1]$$ be a continuous operator. Denote $$N_{1}$$ by N, and let $$\sum_{\lambda }=\{x\in \bar{ \varOmega }:Mx=N_{\lambda }x\}$$. Then $$N_{\lambda }$$ is said to be M-compact in Ω̄ if
1. (3)
there is a vector subspace $$Z_{1}$$ of Z with $$\dim Z_{1}=\dim X _{1}$$ and an operator $$R:\bar{\varOmega }\times X_{2}$$ being continuous and compact such that for $$\lambda \in [0,1]$$,
\begin{aligned}& (I-Q)N_{\lambda }(\bar{\varOmega })\subset \operatorname{Im}M\subset (I-Q)Z, \end{aligned}
(3.1)
\begin{aligned}& QN_{\lambda }x=0, \quad \lambda \in (0,1) \Leftrightarrow QNx=0, \end{aligned}
(3.2)
\begin{aligned}& R(\cdot ,0) \mbox{ is the zero operator and } R(\cdot ,\lambda )|_{ \sum _{\lambda }}=(I-P)|_{\sum _{\lambda }}, \end{aligned}
(3.3)
and
$$M\bigl[P+R(\cdot ,\lambda )\bigr]=(I-Q)N_{\lambda }.$$
(3.4)
Let $$J:Z_{1}\to X_{1}$$ be a homeomorphism with $$J(\theta )=\theta$$.

Next, we investigate existence of periodic solutions for Eq. (1.1) by applying the extension of Mawhin’s continuation theorem.

### Lemma 3.1

()

LetXandZbe Banach spaces with norm$$\|\cdot \|_{X}$$and$$\|\cdot \|_{Z}$$, respectively, and$$\varOmega \subset X$$be an open and bounded set with$$\theta \in \varOmega$$. Suppose that$$M:X\cap \operatorname{dom}M\to Z$$is a quasilinear operator and
$$N_{\lambda }:\bar{\varOmega }\to Z,\quad \lambda \in (0,1)$$
is anM-compact mapping. In addition, if
1. (a)

$$Mx\neq N_{\lambda }x$$, $$\lambda \in (0,1)$$, $$x\in \partial \varOmega$$,

2. (b)

$$\deg \{JQN,\varOmega \cap \ker M,0\}\neq 0$$,

where$$N=N_{1}$$, then the abstract equation$$Mx=Nx$$has at least one solution inΩ̄.

### Theorem 3.2

Assume$$\sum_{i=1}^{n}\|c_{i}\|\neq 1$$, Ωis an open bounded set in$$C^{1}_{T}$$. Suppose the following conditions hold:
1. (i)
For each$$\lambda \in (0,1)$$, the equation
$$\bigl(\phi _{p}(Ax)'(t) \bigr)'=\lambda \tilde{f}\bigl(t,x(t),x'(t)\bigr)$$
(3.5)
has no solution on∂Ω.

2. (ii)
The equation
$$F(a):=\frac{1}{T} \int ^{T}_{0}\tilde{f}(t,a,0)\,dt=0$$
has no solution on$$\partial \varOmega \cap \mathbb{R}$$.

3. (iii)
The Brouwer degree
$$\deg \{F,\varOmega \cap \mathbb{R},0\}\neq 0.$$
Then Eq. (1.1) has at least oneT-periodic solution onΩ̄.

### Proof

In order to use Lemma 3.1 we study the existence of periodic solutions to Eq. (1.1). We set $$X:=\{x\in C[0,T]: x(0)=x(T)\}$$ and $$Z:=C[0,T]$$,
$$M:X\cap \operatorname{dom}M \to Z, \qquad (Mx) (t)=\bigl(\phi _{p}(Ax)'(t)\bigr)',$$
(3.6)
where $$\operatorname{dom}M:=\{u\in X:\phi _{p}(Au)'\in C^{1}(\mathbb{R},\mathbb{R}) \}$$. Then $$\ker M=\mathbb{R}$$. In fact,
\begin{aligned} \ker M&=\bigl\{ x\in X:\bigl(\phi _{p}(Ax)'(t) \bigr)'=0 \bigr\} \\ &=\bigl\{ x\in X: \phi _{p}(Ax)'\equiv c \bigr\} \\ &=\bigl\{ x\in X: (Ax)'\equiv \phi _{q}(c):=c_{1} \bigr\} \\ &=\bigl\{ x\in X: (Ax) (t)\equiv c_{1}t+c_{2} \bigr\} , \end{aligned}
where $$q>1$$ is a constant with $$\frac{1}{p}+\frac{1}{q}=1$$ and c, $$c_{1}$$, $$c_{2}$$ are constants in $$\mathbb{R}$$. Since $$(Ax)(0)=(Ax)(T)$$, then we get $$\ker M=\{x\in X: (Ax)(t)\equiv c_{2} \}$$. In addition,
\begin{aligned} \operatorname{Im} M ={}&\biggl\{ y\in Z, \mbox{ for }x(t)\in X\cap \operatorname{dom} M, \bigl( \phi _{p}(Ax)'\bigr)'=y(t),\\ &{} \int ^{T}_{0}y(t)\,dt= \int ^{T}_{0}(\phi _{p} \bigl((Ax)'\bigr)'\,dt=0 \biggr\} . \end{aligned}
So M is quasilinear. Let
\begin{aligned} & X_{1}=\ker M, \qquad X_{2}=\bigl\{ x\in X: x(0)=x(T)=0\bigr\} , \\ & Z_{1}=\mathbb{R}, \qquad Z_{2}=\operatorname{Im} M. \end{aligned}
Clearly, $$\operatorname{dim} X_{1}= \operatorname{dim} Z_{1}=1$$, and $$X=X_{1}\oplus X_{2}$$, $$P:X\to X_{1}$$, $$Q:Z\to Z_{1}$$, are defined by
$$Px=x(0), \qquad Qy=\frac{1}{T}\int ^{T}_{0}y(s)\,ds.$$
For $$\forall \bar{\varOmega }\subset X$$, define $$N_{\lambda }:\bar{ \varOmega }\to Z$$ by
$$(N_{\lambda }x) (t)=\lambda \tilde{f}\bigl(t,x(t),x'(t)\bigr).$$
We claim that $$(I-Q)N_{\lambda }(\bar{\varOmega })\subset \operatorname{Im}M=(I-Q)Z$$ holds. In fact, for $$x\in \bar{\varOmega }$$, we observe that
\begin{aligned} & \int ^{T}_{0}(I-Q)N_{\lambda }x(t)\,dt \\ &\quad = \int ^{T}_{0}(I-Q)\lambda \tilde{f} \bigl(t,x(t),x'(t)\bigr)\,dt \\ &\quad = \int ^{T}_{0}\lambda \tilde{f}\bigl(t,x(t),x'(t) \bigr)\,dt- \int ^{T}_{0}\frac{ \lambda }{T} \int ^{T}_{0}\tilde{f}\bigl(s,x(s),x'(s) \bigr)\,ds\,dt \\ &\quad =0. \end{aligned}
Therefore, we have $$(I-Q)N_{\lambda }(\bar{\varOmega })\subset \operatorname{Im} M$$.
Moreover, for any $$x\in Z$$, it is obvious that
\begin{aligned} \int ^{T}_{0}(I-Q)x(t)\,dt & = \int ^{T}_{0} \biggl(x(t)- \int ^{T}_{0}\frac{1}{T} \int ^{T}_{0}x(t)\,dt \biggr)\,dt =0. \end{aligned}
So, we have $$(I-Q)Z\subset \operatorname{Im}M$$. On the other hand, $$x\in \operatorname{Im}M$$ and $$\int ^{T}_{0}x(t)\,dt=0$$, so we have $$x(t)=x(t)- \int ^{T}_{0}x(t)\,dt$$. Hence, we get $$x(t)\in (I-Q)Z$$. Therefore, $$\operatorname{Im}M=(I-Q)Z$$.

From $$QN_{\lambda }x=0$$, we get $$\frac{\lambda }{T}\int ^{T}_{0} \tilde{f}(t,x(t),x'(t))\,dt=0$$. Since $$\lambda \in (0,1)$$, we have $$\frac{1}{T}\int ^{T}_{0}\tilde{f}(t,x(t), x'(t))\,dt=0$$. Therefore, $$QNx=0$$, and so Eq. (3.4) also holds.

Let $$J:Z_{1}\to X_{1}$$, $$J(x)=x$$, then $$J(0)=0$$. Define $$R:\bar{\varOmega }\times [0,1]\to X_{2}$$,
\begin{aligned} R(x,\lambda ) (t) = & A^{-1} \int ^{t}_{0}\phi _{p}^{-1} \biggl(a+ \int ^{s} _{0}\lambda \tilde{f} \bigl(u,x(u),x'(u)\bigr)\,du \\ &{}-\frac{\lambda s}{T} \int ^{T}_{0} \tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr)\,ds, \end{aligned}
(3.7)
where $$a\in R$$ is a constant such that
\begin{aligned} R(x,\lambda ) (T) =&A^{-1} \int ^{T}_{0}\phi _{p}^{-1} \biggl(a+ \int ^{s} _{0}\lambda \tilde{f} \bigl(u,x(u),x'(u)\bigr)\,du-\frac{\lambda s}{T} \int ^{T}_{0} \tilde{f}\bigl(u,x(t),x'(u) \bigr)\,du \biggr)\,ds \\ =&0. \end{aligned}
(3.8)
From Lemma 2.3 of , we know that a is uniquely defined by
$$a=\tilde{a}(x,\lambda ),$$
where $$\tilde{a}(x,\lambda )$$ is continuous on $$\bar{\varOmega }\times [0,1]$$ and maps bounded sets of $$\bar{\varOmega } \times [0,1]$$ into bounded sets of $$\mathbb{R}$$.
From Eq. (3.4), one can find that
$$\mbox{R}:\bar{\varOmega }\times [0,1] \to X_{2}.$$
Now, for any $$x\in \sum_{\lambda }=\{x\in \bar{\varOmega }: Mx=N_{\lambda }x\}=\{x\in \bar{\varOmega }: (\phi _{p}(Ax)'(t))'=\lambda \tilde{f}(t,x(t),x'(t)) \}$$, we have $$\int ^{T}_{0}\tilde{f}(t,x(t),x'(t))\,dt=0$$, which, together with Eq. (3.7), gives
\begin{aligned} R(x,\lambda ) (t) &=A^{-1} \int ^{t}_{0}\phi _{p}^{-1} (a+ \int ^{s} _{0}\lambda \tilde{f} \bigl(u,x(u),x'(u)\,du \bigr)\,ds \\ &=A^{-1} \int ^{t}_{0}\phi _{p}^{-1} \biggl(a+ \int ^{s}_{0} \bigl(\phi _{p}(Ax)'(u) \bigr)'\,du \biggr)\,ds \\ &=A^{-1} \int ^{t}_{0}\phi _{p}^{-1} \bigl(a+\phi _{p}(Ax)'(s)-\phi _{p}(Ax)'(0) \bigr)\,ds. \end{aligned}
Taking $$a=\phi _{p}(Ax)'(0)$$, we then have
\begin{aligned} R(x,\lambda ) (T) &=A^{-1} \int ^{T}_{0} \bigl(\phi _{p}^{-1} \bigl(\phi _{p}(Ax)'(s)\bigr) \bigr)\,ds \\ &=A^{-1} \int ^{T}_{0}(Ax)'(t)\,ds \\ &=A^{-1} \bigl((Ax) (T)-(Ax) (0) \bigr) \\ &=x(T)-x(0) \\ &=0, \end{aligned}
where a is unique, and we see that
$$a=\tilde{a}(x,\lambda )=\phi _{p}(Ax)'(0), \quad \forall \lambda \in [0,1].$$
Thus, we derive
\begin{aligned} R(x,\lambda ) (t)|_{x\in \sum _{\lambda }} &=A^{-1} \int ^{t}_{0} \biggl(\phi _{p}^{-1} \biggl(\phi _{p}(Ax)'(0)+ \int ^{s}_{0} \lambda \tilde{f} \bigl(t,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) \\ &=(I-P)x(t), \end{aligned}
which yields the second part of Eq. (3.3). Meanwhile, if $$\lambda =0$$, then
$$\sum_{\lambda }=\{x\in \bar{\varOmega }:Mx=N_{\lambda }x\}=\{x\in \bar{ \varOmega }: (\phi _{p}(Ax)'(t))'=\lambda \tilde{f}(t,x(t),x'(t))\}=c_{3},$$
where $$c_{3}\in \mathbb{R}$$ is a constant, so by the continuity of $$\tilde{a}(x,\lambda )$$ with respect to $$(x,\lambda )$$, $$a=\tilde{a}(x,0)= \phi _{p}(Ac)'(0)=0$$. Hence,
$$R(x,0) (t)=A^{-1} \int ^{t}_{0}\phi _{p}^{-1}(0)\,ds=0, \quad \forall x\in \bar{\varOmega },$$
which yields the first part of Eq. (3.3). Furthermore, we consider
\begin{aligned} & M(P+R)=(I-Q)N_{\lambda }, \end{aligned}
and, in fact,
$${ } \frac{d}{\,dt}\phi _{p}\bigl(A(P+R) \bigr)'=(I-Q)N_{\lambda }.$$
(3.9)
Integrating both sides of (3.9) over $$[0,s]$$, we have
$$\int ^{s}_{0}\frac{d}{\,dt}\phi _{p} \bigl(A(P+R)\bigr)'\,ds= \int ^{s}_{0}(I-Q)N_{ \lambda }\,ds.$$
Therefore, we arrive at
\begin{aligned} \phi _{p}\bigl(A(P+R)\bigr)'(s)-a &=\lambda \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du- \int ^{s}_{0}\frac{\lambda }{T} \int ^{T}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du\,dt \\ &=\lambda \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du-\frac{\lambda s}{T} \int ^{T}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du, \end{aligned}
where $$a:=\phi _{p}(A(P+R))'(0)$$. Then, we get
\begin{aligned} &\bigl(A(P+R)\bigr)'(s) \\ &\quad =\phi _{p}^{-1} \biggl(a+\lambda \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du -\frac{ \lambda s}{T} \int ^{T}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr). \end{aligned}
(3.10)
Integrating both sides of (3.10) over $$[0,t]$$, we derive
\begin{aligned} &\int ^{t}_{0}\bigl(A(P+R)\bigr)'(s)\,ds\\ &\quad = \int ^{t}_{0}\phi _{p}^{-1} \biggl(a+\lambda \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du-\frac{\lambda s}{T} \int ^{T}_{0} \tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr)\,ds, \end{aligned}
i.e.,
\begin{aligned} &(P+R) (t)-(P+R) (0)\\ &\quad =A^{-1} \biggl( \int ^{t}_{0}\biggl(\phi _{p}^{-1} \biggl(\biggl(a+\lambda \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du\\ &\qquad {}-\frac{\lambda s}{T} \int ^{T}_{0} \tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du\biggr)\biggr)\biggr)\,ds \biggr). \end{aligned}
Since $$R(x,\lambda )(0)=0$$, $$P(t)=P(0)$$, we obtain
\begin{aligned} &R(x,\lambda ) (t)\\ &\quad =A^{-1} \biggl( \int ^{t}_{0}\phi _{p}^{-1} \biggl(a+ \lambda \int ^{s}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du-\frac{\lambda s}{T} \int ^{T}_{0}\tilde{f}\bigl(u,x(u),x'(u) \bigr)\,du \biggr)\,dt \biggr). \end{aligned}
Hence, we have that $$N_{\lambda }$$ is M-compact on Ω̄. Obviously, the equation
$$\bigl(\phi _{p}(Ax)'(t)\bigr)'=\lambda \tilde{f}\bigl(t,x(t),x'(t)\bigr)$$
can be converted to
$$Mx=N_{\lambda }x, \quad \lambda \in (0,1),$$
where M and $$N_{\lambda }$$ are defined by Eqs. (3.6) and (3.7), respectively. As proved above,
$$N_{\lambda }:\bar{\varOmega }\to Z, \quad \lambda \in (0,1),$$
is an M-compact mapping. From assumption (i), one finds
$$Mx\neq N_{\lambda }x, \quad \lambda \in (0,1), x\in \partial \varOmega ,$$
and assumptions (ii) and (iii) imply that $$\deg \{JQN,\varOmega \cap \ker M,\theta \}$$ is valid and
$$\deg \{JQN,\varOmega \cap \operatorname{ker}M,\theta \}\neq 0.$$
So by applications of Lemma 3.1, we see that Eq. (1.1) has a T-periodic solution. □

## 4 Application of Theorem 3.2: p-Laplacian equation

As an application, we consider the following p-Laplacian neutral Liénard equation:
$$\Biggl(\phi _{p} \Biggl(x(t)-\sum _{i=1}^{n}c_{i}(t)x(t-\delta _{i}) \Biggr)' \Biggr)'+f\bigl(x(t) \bigr)x'(t)+g\bigl(t,x(t)\bigr)=e(t),$$
(4.1)
where $$\phi _{p}:\mathbb{R}\to \mathbb{R}$$ is given by $$\phi _{p}(s)=|s|^{p-2}s$$, here $$p>1$$ is a constant, g is a continuous function defined on $$\mathbb{R}^{2}$$ and periodic in t with $$g(t,\cdot )=g(t+T,\cdot )$$, $$f\in C(\mathbb{R},\mathbb{R})$$, e is a continuous periodic function defined on $$\mathbb{R}$$ with period T and $$\int ^{T}_{0}e(t)\,\,dt=0$$. Next, by applications of Theorem 3.2, we will investigate the existence of periodic solution for Eq. (4.1) in the case that $$\sum_{i=1}^{n}\|c_{i}\|\neq 1$$.
Define
$$\sigma := \textstyle\begin{cases} \frac{1}{1-\sum_{i=1}^{n} \Vert c_{i} \Vert }, \quad \mbox{for }\sum_{i=1}^{n} \Vert c_{i} \Vert < 1; \\ \frac{\frac{1}{ \Vert c_{k} \Vert }}{1-\frac{1}{ \Vert c_{k} \Vert }- \Vert \frac{c_{i}}{c_{k}} \Vert }, \quad \mbox{for }\sum_{i=1}^{n} \Vert c_{i} \Vert >1. \end{cases}$$

### Theorem 4.1

Suppose$$\sum_{i=1}^{n}\|c_{i}\|\neq 1$$holds. Assume the following conditions hold:
$$(H_{1})$$
There exists a constant $$D>0$$ such that
$$xg(t,x)>0, \quad \forall (t,x)\in [0,T]\times \mathbb{R},\textit{ with } \vert x \vert >D.$$
$$(H_{2})$$
There exist positive constantsm, ñsuch that
$$\bigl\vert f(x) \bigr\vert \leq m \vert x \vert ^{p-2}+ \tilde{n}, \quad x\in \mathbb{R}.$$
$$(H_{3})$$
There exist positive constantsa, b, Bsuch that
$$\bigl\vert g(t,x) \bigr\vert \leq a \vert x \vert ^{p-1}+b, \quad \textit{for } \vert x \vert >B \textit{ and }t\in [0,T].$$
Then Eq. (4.1) has at least oneT-periodic solution, if
$$\sigma T^{\frac{1}{q}} \biggl(\frac{m\sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}}+\frac{aT (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p}} \biggr)^{ \frac{1}{p}} +\frac{\sigma T\sum_{i=1}^{n} \Vert c'_{i} \Vert }{2}< 1.$$

### Proof

Consider the homotopic equation
$$\Biggl(\phi _{p} \Biggl(x(t)-\sum _{i=1}^{n}c_{i}(t)x(t-\delta _{i}) \Biggr)' \Biggr)'+\lambda f\bigl(x(t) \bigr)x'(t)+\lambda g\bigl(t,x(t)\bigr)=\lambda e(t).$$
(4.2)
Firstly, we claim that the set of all T-periodic solutions of Eq. (4.2) is bounded. Let $$x(t)\in C_{T}$$ be an arbitrary T-periodic solution of Eq. (4.2). Integrating both sides of (4.2) over $$[0,T]$$, we have
$$\int ^{T}_{0}g\bigl(t,x(t)\bigr)\,dt=0.$$
(4.3)
From the mean-value theorem for integrals, there is a constant $$\xi \in [0,T]$$ such that
$$g\bigl(\xi ,x(\xi )\bigr)=0.$$
In view of condition $$(H_{1})$$, we obtain
$$\bigl\vert x(\xi ) \bigr\vert \leq D.$$
Then, we have
\begin{aligned}[b] \Vert x \Vert &=\max _{t\in [0,T]} \bigl\vert x(t) \bigr\vert =\max_{t\in [\xi ,\xi +T]} \bigl\vert x(t) \bigr\vert \\ &=\frac{1}{2}\max_{t\in [\xi ,\xi +T]} \bigl( \bigl\vert x(t) \bigr\vert + \bigl\vert x(t-T) \bigr\vert \bigr) \\ &=\frac{1}{2}\max_{t\in [\xi ,\xi +T]} \biggl( \biggl\vert x(\xi )+ \int ^{T} _{\xi }x'(s)\,ds \biggr\vert + \biggl\vert x(\xi )- \int ^{\xi }_{t-T}x'(s)\,ds \biggr\vert \biggr) \\ & \leq D+\frac{1}{2} \biggl( \int ^{t}_{\xi } \bigl\vert x'(s) \bigr\vert \,ds+ \int ^{\xi }_{t-T} \bigl\vert x'(s) \bigr\vert \,ds \biggr) \\ &\leq D+\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(s) \bigr\vert \,ds. \end{aligned}
(4.4)
Multiplying both sides of Eq. (4.2) by $$(Ax)(t)$$ and integrating over the interval $$[0,T]$$, we get
\begin{aligned} &\int ^{T}_{0}\bigl(\phi _{p}(Ax)'(t) \bigr)'(Ax) (t)\,dt +\lambda \int ^{T}_{0}f\bigl(x(t)\bigr)x'(t) (Ax) (t)\,dt+ \lambda \int ^{T}_{0}g\bigl(t,x(t)\bigr) (Ax) (t)\,dt \\ &\quad =\lambda \int ^{T}_{0}e(t) (Ax) (t)\,dt. \end{aligned}
(4.5)
Substituting $$\int ^{T}_{0}(\phi _{p}(Ax)'(t))'(Ax)(t)\,dt=-\int ^{T}_{0}|(Ax)'(t)|^{p}\,dt$$, $$\int ^{T}_{0}f(x(t))x'(t)x(t)\,dt=0$$ into Eq. (4.5), we see that
\begin{aligned} - \int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt ={}&\lambda \int ^{T}_{0}f\bigl(x(t)\bigr)x'(t) \Biggl(\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}) \Biggr)\,dt\\ &{}-\lambda \int ^{T} _{0}g\bigl(t,x(t)\bigr) (Ax) (t)\,dt \\ &{}+\lambda \int ^{T}_{0}e(t) (Ax) (t)\,dt. \end{aligned}
Thus, we have
\begin{aligned} &\int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt\\ &\quad \leq \int ^{T}_{0} \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \Biggl\vert \sum _{i=1}^{n}c_{i}(t)x(t-\delta _{i}) \Biggr\vert \,dt \\ &\qquad {}+ \int ^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \Biggl\vert x(t)-\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}) \Biggr\vert \,dt \\ &\qquad {}+ \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \Biggl\vert x(t)-\sum_{i=1}^{n}c_{i}(t)x(t- \delta _{i}) \Biggr\vert \,dt \\ &\quad \leq\sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert \int ^{T}_{0} \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt + \Biggl(1+\sum _{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt \\ &\qquad {}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt. \end{aligned}
Define
$$E_{1}:=\bigl\{ t\in [0,T]| \bigl\vert x(t) \bigr\vert \leq B\bigr\} , \qquad E_{2}:=\bigl\{ t\in [0,T]| \bigl\vert x(t) \bigr\vert >B\bigr\} .$$
Using conditions $$(H_{2})$$ and $$(H_{3})$$, we arrive at
\begin{aligned} &\int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt \\ &\quad \leq \sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert \int ^{T}_{0} \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt + \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 \\ &\qquad {}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\quad \leq \sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert \bigl(m \Vert x \Vert ^{p-2}+\tilde{n}\bigr) \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt + \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr)T \Vert g _{B} \Vert \Vert x \Vert \\ &\qquad {}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr)bT \Vert x \Vert +aT \Biggl(1+ \sum _{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert ^{p}+ \Vert e \Vert T \Biggl(1+ \sum _{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \\ &\quad \leq m\sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert ^{p-1} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt+ \tilde{n}\sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &\qquad {}+aT \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert ^{p}+N_{1} \Vert x \Vert , \end{aligned}
(4.6)
where $$\|e\|:=\max_{t\in [0,T]}|e(t)|$$, $$\|g_{B}\|:=\max_{|x(t)|\leq B}|g(t,x(t))|$$ and $$N_{1}:= (1+\sum_{i=1}^{n}\|c_{i}\| )T(\|g_{B}\|+b+\|e\|)$$. Substituting Eq. (4.4) into Eq. (4.6), we get
\begin{aligned} \int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt\leq{}& m\sum_{i=1}^{n} \Vert c_{i} \Vert \biggl(D+\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{p-1} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &{}+aT \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \biggl(D+ \frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{p} \\ &{}+\tilde{n}\sum_{i=1}^{n} \Vert c_{i} \Vert \biggl(D+\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &{}+N_{1} \biggl(D+ \frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr). \end{aligned}
(4.7)
Next, we introduce a classical inequality: there exists a $$\kappa (p)>0$$, which is depends on p only, such that
$$(1+x)^{p}\leq 1+(1+p)x, \quad \mbox{for } x\in \bigl[0,\kappa (p)\bigr].$$
(4.8)
Then, we consider the following two cases:
Case 1: If $$\frac{D}{\frac{1}{2}\int ^{T}_{0}|x'(t)|\,dt}> \kappa (p)$$, we deduce
$$\int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt< \frac{2D}{\kappa (p)}.$$
From Eq. (4.4), it is clear that
\begin{aligned}[b] \Vert x \Vert &\leq D+ \frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &\leq D+\frac{1}{2}\frac{2D}{\kappa (p)} \\ &=D+\frac{D}{\kappa (p)}:=M_{11}. \end{aligned}
(4.9)
Case 2: If $$\frac{D}{\frac{1}{2}\int ^{T}_{0}|x'(t)|\,dt}< \kappa (p)$$, then
\begin{aligned} &\int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt \\ &\quad \leq aT \Biggl(1+\sum _{i=1}^{n} \Vert c _{i} \Vert \Biggr) \biggl(\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{p} \\ &\qquad {} +m \sum_{i=1}^{n} \Vert c_{i} \Vert \biggl(\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p-1} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &\qquad {}+aT \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) (1+p)D \biggl(\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{p-1} \\ &\qquad {}+mpD\sum_{i=1}^{n} \Vert c_{i} \Vert \biggl(\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p-2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &\qquad {}+\frac{\tilde{n}\sum_{i=1}^{n} \Vert c_{i} \Vert }{2} \biggl( \int ^{T} _{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{2} + \Biggl(\tilde{n}\sum _{i=1}^{n} \Vert c _{i} \Vert + \frac{N_{1}}{2} \Biggr) \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt+N_{1}D \\ &\quad = \biggl(\frac{aT (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p}}+ \frac{m \sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}} \biggr) \biggl( \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p} \\ &\qquad {}+ \biggl(\frac{aT(1+p)D (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p-1}}+ \frac{mpD \sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-2}} \biggr) \biggl( \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p-1} \\ &\qquad {}+\frac{\tilde{n}\sum_{i=1}^{n} \Vert c_{i} \Vert }{2} \biggl( \int ^{T} _{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{2} + \Biggl(\tilde{n}\sum _{i=1}^{n} \Vert c _{i} \Vert + \frac{N_{1}}{2} \Biggr) \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt+N_{1}D. \end{aligned}
(4.10)
Since $$(Ax)(t)=x(t)-\sum_{i=1}^{n}x(t-\delta _{i})$$, we have
\begin{aligned} (Ax)'(t) &= \Biggl(x(t)-\sum _{i=1}^{n}c_{i}(t)x(t-\delta _{i}) \Biggr)' \\ &=x'(t)-\sum_{i=1}^{n}c'_{i}(t)x(t- \delta _{i})-\sum_{i=1}^{n}c_{i}(t)x'(t- \delta _{i}) \\ &=\bigl(Ax'\bigr) (t)-\sum_{i=1}^{n}c'_{i}(t)x(t- \delta _{i}) \end{aligned}
and
$$\bigl(Ax'\bigr) (t)=(Ax)'(t)+\sum _{i=1}^{n}c'_{i}(t)x(t-\delta _{i}).$$
By applying Lemma 2.1 and Hölder inequality, we get
\begin{aligned}[b] \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt &= \int ^{T}_{0} \bigl\vert \bigl(A^{-1}Ax' \bigr) (t) \bigr\vert \,dt \\ &\leq \sigma \int ^{T}_{0} \bigl\vert \bigl(Ax' \bigr) (t) \bigr\vert \,dt \\ &=\sigma \int ^{T}_{0} \Biggl\vert (Ax)'(t)+ \sum_{i=1}^{n}c'_{i}(t)x(t- \delta _{i}) \Biggr\vert \,dt \\ &\leq \sigma \int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert \,dt+\sigma \int ^{T}_{0} \Biggl\vert \sum _{i=1}^{n}c'_{i}(t)x(t-\delta _{i}) \Biggr\vert \,dt \\ &\leq \sigma T^{\frac{1}{q}} \biggl( \int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt \biggr) ^{\frac{1}{p}}+\sigma T\sum _{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert \Vert x \Vert , \end{aligned}
(4.11)
where $$\|c'_{i}\|:=\max_{t\in [0,T]}|c'_{i}(t)|$$, for $$i=1,2,\dots ,n$$. Substituting Eq. (4.10) into Eq. (4.11), since $$(\tilde{a}+\tilde{b})^{k}\leq \tilde{a}^{k}+\tilde{b}^{k}$$, $$0< k \leq 1$$, we have
\begin{aligned} &\int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &\quad \leq \sigma T^{\frac{1}{q}} \biggl(\frac{aT (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p}}+ \frac{m \sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}} \biggr)^{\frac{1}{p}} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &\qquad {}+\sigma T^{\frac{1}{q}} \biggl(\frac{aT (1+\sum_{i=1} ^{n} \Vert c_{i} \Vert )(1+p)D}{2^{p-1}}+\frac{m\sum_{i=1}^{n} \Vert c _{i} \Vert pD}{2^{p-2}} \biggr)^{\frac{1}{p}} \biggl( \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{\frac{p}{p-1}} \\ &\qquad {}+\sigma T^{\frac{1}{q}} \biggl(\frac{\tilde{n}\sum_{i=1}^{n} \Vert c_{i} \Vert }{2} \biggr)^{\frac{1}{p}} \biggl( \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{\frac{2}{p}} \\ &\qquad {} +\sigma T^{\frac{1}{q}} \Biggl(\tilde{n}\sum _{i=1} ^{n} \Vert c_{i} \Vert D+\frac{N_{1}}{2} \Biggr)^{\frac{1}{p}} \biggl( \int ^{T} _{0} \bigl\vert x'(t) \bigr\vert \,dt \biggr)^{\frac{1}{p}} \\ &\qquad {}+\sigma T^{\frac{1}{q}}(N_{1}D)^{\frac{1}{p}}+\frac{\sigma T\sum_{i=1}^{n} \Vert c'_{i} \Vert }{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt+\sigma T\sum_{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert D. \end{aligned}
(4.12)
Since $$\sigma T^{\frac{1}{q}} (\frac{m\sum_{i=1}^{n}\|c _{i}\|}{2^{p-1}}+ \frac{aT (1+\sum_{i=1}^{n}\|c_{i}\| )}{2^{p}} ) ^{\frac{1}{p}} +\frac{\sigma T\sum_{i=1}^{n}\|c'_{i}\|}{2}<1$$, it is easily see that there exists a constant $$M_{1}'>0$$ (independent of λ) such that
$${ } \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt\leq M_{1}'.$$
(4.13)
From Eq. (4.4), we obtain
$${ } \Vert x \Vert \leq D+\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(s) \bigr\vert \,ds\leq D+\frac{1}{2}M_{1}':=M _{12}.$$
(4.14)
Let $$M_{1}=\sqrt{M_{11}^{2}+M_{12}^{2}}+1$$. As $$(Ax)(0)=(Ax)(T)$$, there exists a point $$t_{0}\in (0,T)$$ such that $$(Ax)'(t_{0})=0$$. Moreover, since $$\phi _{p}(0)=0$$, due to Eq. (4.14), it is obvious 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 \lambda \int ^{T}_{0} \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+\lambda \int ^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+ \lambda \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\leq \Vert f_{M_{1}} \Vert \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt+T \Vert g_{M_{1}} \Vert +T \Vert e \Vert \\ \ &\leq \Vert f_{M_{1}} \Vert M'_{1}+T \Vert g_{M_{1}} \Vert +T \Vert e \Vert :=M_{2}', \end{aligned}
where $$\|f_{M_{1}}\|:=\max_{|x(t)|\leq M_{1}}|f(x(t))|$$ and $$\|g_{M_{1}}\|:=\max_{|x(t)|\leq M_{1}}|g(t,x(t))|$$. Next we claim that there exists a positive constant $$M_{2}^{*}>M_{2}'+1$$, such that, for all $$t\in \mathbb{R}$$,
$${ } \bigl\Vert (Ax)' \bigr\Vert \leq M_{2}^{*}.$$
(4.15)
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 $$\|\phi _{p}(Ax)'\|=\|(Ax)^{\prime \,p-1}\|\geq M''_{2}$$, which is a contradiction. Hence, Eq. (4.15) holds. From Lemma 2.1 and Eq. (4.15), we have
{ } \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 \sigma \Biggl\Vert (Ax)'(t)+\sum _{i=1}^{n}c'_{i}(t)x(t- \delta _{i}) \Biggr\Vert \\ &\leq \sigma \bigl\Vert (Ax)' \bigr\Vert +\sigma \Biggl(\sum _{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert \Vert x \Vert \Biggr) \\ &\leq \sigma M_{2}^{*}+\sigma \sum _{i=1}^{n} \bigl\Vert c'_{i} \bigr\Vert M_{1}:=M _{2}. \end{aligned}
(4.16)
Setting $$M=\sqrt{M_{1}^{2}+M_{2}^{2}}+1$$, we get
$$\varOmega =\bigl\{ x\in C^{1}_{T}(\mathbb{R},\mathbb{R})| \Vert x \Vert \leq M+1, \bigl\Vert x' \bigr\Vert \leq M+1 \bigr\} ,$$
and we know that Eq. (4.1) has no solution on ∂Ω as $$\lambda \in (0,1)$$. When $$x(t)\in \partial \varOmega \cap \mathbb{R}$$, $$x(t)=M+1$$ or $$x(t)=-M-1$$, and from Eq. (4.4) we know that $$M+1>D$$. Thus, from condition $$(H_{1})$$, we see that
\begin{aligned}& \frac{1}{T} \int ^{T}_{0}g(t,M+1)\,dt>0, \\& \frac{1}{T} \int ^{T}_{0}g(t,-M-1)\,dt< 0, \end{aligned}
since $$\int ^{T}_{0}e(t)\,dt=0$$. So condition (ii) of Theorem 3.2 is also satisfied. Set
$$H(x,\mu )=\mu x+(1-\mu )\frac{1}{T} \int ^{T}_{0}g(t,x)\,dt, \quad x\in \partial \varOmega \cap \mathbb{R}, \mu \in [0,1].$$
Obviously, from condition $$(H_{1})$$, we can get $$xH(x,\mu )>0$$ and thus $$H(x,\mu )$$ is a homotopic transformation, as well as
\begin{aligned} \deg \{F,\varOmega \cap \mathbb{R},0\} &=\deg \biggl\{ \frac{1}{T} \int ^{T} _{0}g(t,x)\,dt,\varOmega \cap \mathbb{R},0 \biggr\} \\ &=\deg \{x,\varOmega \cap \mathbb{R},0\}\neq 0. \end{aligned}
So condition (iii) of Theorem 3.2 is satisfied. In view of Theorem 3.2, there exists at least one T-periodic solution. □

## 5 Application of Theorem 3.2: p-Laplacian equation with singularity

In this section, we consider Eq. (4.1) with a singularity. Here $$g(t,x(t))=g_{0}(x)+g_{1}(t,x(t))$$, $$g_{0}\in C((0,\infty );R)$$ and $$g_{1}$$ is an $$L^{2}$$-Carathéodory function, and $$g_{0}$$ has a singularity at $$x=0$$, i.e.,
$${ } \int ^{1}_{0}g_{0}(x)\,dx=-\infty .$$
(5.1)
Next, we consider the existence of periodic solutions for Eq. (4.1) with singularity by applying Theorem 3.2.

### Theorem 5.1

Suppose$$\sum_{i=1}^{n}\|c_{i}\|\neq 1$$and condition$$(H_{2})$$hold. Assume that the following conditions hold:
$$(H_{4})$$

There exist positive constants$$0< D_{1}< D_{2}$$such thatxis a positive continuousT-periodic function satisfying$$\int ^{T}_{0}g(t,x(t))\,dt<0$$, for some$$x\in (0,D_{1})$$and$$\int ^{T} _{0}g(t,x(t))\,dt>0$$, for some$$x\in (D_{2},\infty )$$.

$$(H_{5})$$
There exist positive constants α and β such that
$$g(t,x)\leq \alpha x^{p-1}+\beta , \quad \textit{for }t\in [0,T], \textit{ and }x>0.$$
(5.2)
Then Eq. (4.1) has at least oneT-periodic solution if
$$\sigma T^{\frac{1}{q}} \biggl(\frac{m\sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}}+\frac{ \alpha T (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p-1}} \biggr) ^{\frac{1}{p}}+\frac{\sigma T\sum_{i=1}^{n} \Vert c'_{i} \Vert }{2}< 1.$$

### Proof

Consider the homotopic equation
$${ } \bigl(\phi _{p}(Ax)'(t) \bigr)'+\lambda f\bigl(x(t)\bigr)x'(t)+\lambda g \bigl(t,x(t)\bigr)=\lambda e(t).$$
(5.3)
We follow the same strategy and notation as in the proof of Theorem 4.1. From condition $$(H_{4})$$, we know that there exists a constant $$D_{2}>0$$ such that
$${ } \bigl\vert x(t) \bigr\vert \leq D_{2}+\frac{1}{2} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt.$$
(5.4)
From Eq. (4.5), we have
\begin{aligned} \int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt\leq{}& \sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert \int ^{T}_{0} \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt\\ &{}+ \Biggl(1+\sum _{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt \\ &{}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt. \end{aligned}
From Eq. (4.3) and condition $$(H_{5})$$, we get
\begin{aligned}[b] \int ^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt &= \int _{g(t,x(t))\geq 0}g^{+}\bigl(t,x(t)\bigr)\,dt- \int _{g(t,x(t))< 0}g^{-}\bigl(t,x(t)\bigr)\,dt \\ &=2 \int _{g(t,x(t))\geq 0}g^{+}\bigl(t,x(t)\bigr)\,dt \\ &\leq 2 \int ^{T}_{0}\bigl(\alpha x^{p-1}+\beta \bigr)\,dt \\ &\leq 2\alpha T \Vert x \Vert ^{p-1}+2\beta T, \end{aligned}
(5.5)
where $$g^{+}:=\max \{g(t,x), 0\}$$. Using condition $$(H_{2})$$ and Eq. (5.5), we derive
\begin{aligned}[b] \int ^{T}_{0} \bigl\vert (Ax)'(t) \bigr\vert ^{p}\,dt\leq{}& m\sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert ^{p-1} \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt+\tilde{n}\sum_{i=1}^{n} \Vert c_{i} \Vert \Vert x \Vert \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt \\ &{}+2\alpha T \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert ^{p}+2 \beta \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert T \\ &{}+ \Biggl(1+\sum_{i=1}^{n} \Vert c_{i} \Vert \Biggr) \Vert x \Vert \Vert e \Vert T. \end{aligned}
(5.6)
Following the same strategy and notation as in the proof of Theorem 4.1, we can obtain, since $$\sigma T^{\frac{1}{q}} (\frac{m \sum_{i=1}^{n}\|c_{i}\|}{2^{p-1}}+\frac{\alpha T (1+ \sum_{i=1}^{n}\|c_{i}\| )}{2^{p-1}} )^{\frac{1}{p}}+\frac{ \sigma T\sum_{i=1}^{n}\|c'_{i}\|}{2}<1$$, that there exists a constant $$M_{3}'>0$$ (independent of λ) such that
$$\int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt\leq M_{3}'.$$
(5.7)
From Eq. (5.7), we get
$$\Vert x \Vert \leq D_{2}+ \frac{1}{2} \int ^{T}_{0} \bigl\vert x'(s) \bigr\vert \,ds\leq D_{2}+ \frac{1}{2}M_{3}':=M_{3}.$$
(5.8)
From Eqs. (4.15), (4.16) and (5.8), we get that there exists a constant $$M_{3}^{*}$$, such that, for all $$t\in \mathbb{R}$$,
$$\bigl\Vert x' \bigr\Vert \leq M_{3}^{*}.$$
(5.9)
On the other hand, multiplying both sides of (5.3) by $$x'(t)$$, we get
$$\bigl(\phi _{p}(Ax)'(t) \bigr)x'(t)+\lambda f(x(t)x'(t)x'(t)+ \lambda \bigl(g_{1}\bigl(t,x(t)\bigr)+g _{0}\bigl(x(t)\bigr) \bigr)x'(t)=\lambda e(t)x'(t),$$
(5.10)
since $$g(t,x(t))=g_{0}(x)+g_{1}(t,x(t))$$. Letting $$\tau \in [0,T]$$, for any $$\tau \leq t\leq T$$, we integrate Eq. (5.10) on $$[\tau ,t]$$ and get
\begin{aligned} \lambda \int ^{t}_{\tau }g_{0}(x)\,dt ={}& \int ^{t}_{\tau }\bigl(\phi _{p}(Ax)'(t) \bigr)'x'(t)\,dt- \lambda \int ^{t}_{\tau }f\bigl(x(t)\bigr)x'(t)x'(t)\,dt \\ &{}-\lambda \int ^{t}_{\tau }g_{1}\bigl(t,x(t) \bigr)x'(t)\,dt+\lambda \int ^{t}_{ \tau }e(t)x'(t)\,dt. \end{aligned}
Furthermore,
\begin{aligned} \lambda \biggl\vert \int ^{x(t)}_{x(\tau )} g_{0}(u)\,du \biggr\vert \leq{}& \int ^{t}_{\tau } \bigl\vert \bigl(\phi _{p}(Ax)'(t)\bigr)' \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+\lambda \int ^{t}_{\tau } \bigl\vert f(x(t) \bigr\vert \bigl\vert x'(t) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt \\ &{}+\lambda \int ^{t}_{\tau } \bigl\vert g_{1} \bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+ \lambda \int ^{t} _{\tau } \bigl\vert e(t) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt. \end{aligned}
From Eq. (5.3), we have
\begin{aligned} &\int ^{t}_{\tau } \bigl\vert \bigl(\phi _{p}(Ax)'(t)\bigr)' \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt \\ &\quad = \bigl\Vert x' \bigr\Vert \int ^{t}_{ \tau } \bigl\vert \bigl(\phi _{p}(Ax)'(t)\bigr)' \bigr\vert \,dt \\ &\quad \leq \bigl\Vert x' \bigr\Vert \biggl(\lambda \int ^{T}_{0} \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+\lambda \int ^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+\lambda \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \biggr) \\ &\quad \leq \lambda M_{3}^{*} \biggl( \Vert f_{M_{3}} \Vert \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt+2 \alpha T \Vert x \Vert ^{p-1}+2\beta T+T \Vert e \Vert \biggr) \\ &\quad \leq \lambda M_{3}^{*}\bigl( \Vert f_{M_{3}} \Vert M'_{3}+2\alpha T(M_{3})^{p-1}+2 \beta T+T \Vert e \Vert \bigr), \end{aligned}
where $$\|f_{M_{3}}\|:=\max_{|x(t)|\leq M_{3}}|f(x(t))|$$. From Eqs. (5.7) and (5.8), we obtain
\begin{aligned}& \begin{aligned} \lambda \int ^{t}_{\tau } \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt &\leq \lambda \int ^{T}_{0} \bigl\vert f\bigl(x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt \\ &\leq \lambda \Vert f_{M_{3}} \Vert \biggl( \int ^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt \biggr) \\ &\leq \lambda \Vert f_{M_{3}} \Vert \bigl(M^{*}_{3} \bigr)^{2}T, \end{aligned} \\& \begin{aligned} \lambda \int ^{t}_{\tau } \bigl\vert g_{1}(t,x) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt &\leq \lambda \int ^{T} _{0} \bigl\vert g_{1}(t,x) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt \\ &\leq \lambda \Vert g_{1M_{3}} \Vert M'_{3}, \end{aligned} \end{aligned}
where $$\|g_{1M_{3}}\|:=\max_{|x(t)|\leq M_{3}}|g_{1}(t,x)|$$,
\begin{aligned} \lambda \int ^{t}_{\tau } \bigl\vert e(t) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt &\leq \lambda \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt \\ &\leq \lambda \Vert e \Vert M'_{3}. \end{aligned}
From these inequalities, we get
\begin{aligned} \biggl\vert \int ^{x(t)}_{x(\tau )} g_{0}(u)\,du \biggr\vert \leq{} &M_{3}^{*}\bigl( \Vert f _{M_{3}} \Vert M'_{3}+2\alpha T(M_{3})^{p-1}+2\beta T+T \Vert e \Vert \bigr) \\ &{}+ \Vert f_{M_{3}} \Vert \bigl(M^{*}_{3} \bigr)^{2}T+ \Vert g_{1M_{3}} \Vert M'_{3}+ \Vert e \Vert M'_{3}. \end{aligned}
In view of Eq. (5.1), we know that there exists a constant $$M_{4}>0$$ such that
$$x(t)\geq M_{4}, \quad \forall t\in [\tau ,T].$$
(5.11)

The case $$t\in [0,\tau ]$$ can be treated similarly.

From Eqs. (5.8), (5.9) and (5.11), we have
$$\varOmega =\bigl\{ x\in C^{1}_{T}(\mathbb{R},\mathbb{R})| E_{1}\leq x\leq E _{2}, \bigl\Vert x' \bigr\Vert \leq M_{3}^{*},\forall t\in [0,T]\bigr\} ,$$
where $$0< E_{1}<\min (M_{4},D_{1})$$, $$E_{2}>\max (M_{3},D_{2})$$. This proves the claim, and the rest of the proof is identical to that of Theorem 4.1. □

## 6 Examples

### Example 6.1

Consider the p-Laplacian Liénard equation in the case $$\sum_{i=1}^{n}\|c_{i}\|<1$$:
\begin{aligned} & \biggl(\phi _{p} \biggl(x(t)- \biggl(\frac{1}{40}\sin (4t)x(t-\delta _{1})+ \frac{1}{60}\cos \biggl(4t-\frac{\pi }{3} \biggr)x(t-\delta _{2}) \biggr) \biggr)' \biggr)' \\ &\quad {}+ \biggl(\frac{1}{20}x \biggr)x'(t)+\frac{1}{40} \bigl(2+\sin (4t) \bigr)x ^{2}=\sin (4t), \end{aligned}
(6.1)
where $$p=3$$, $$\delta _{1}$$, $$\delta _{2}$$ are constants and $$0<\delta _{1}$$, $$\delta _{2}< T$$.
Comparing Eq. (6.1) with Eq. (4.1), it is easy to see that $$c_{1}(t)=\frac{1}{40}\sin (4t)$$, $$c_{2}(t)=\frac{1}{60}\cos (4t-\frac{ \pi }{3})$$, $$f(x)=\frac{1}{20}x$$, $$g(t,x)=\frac{1}{40}(2+\sin 4t)x ^{2}$$, $$e(t)=\sin (4t)$$, $$T=\frac{\pi }{2}$$. It is easy to see that there exists a constant $$D=1$$ such that condition $$(H_{1})$$ holds. Obviously, we get $$|f(x)|=|\frac{1}{20}x|\leq \frac{1}{20}|x|+3$$, here $$m=\frac{1}{20}$$, $$\tilde{n}=3$$, and condition $$(H_{2})$$ holds. Consider $$|g(t,x)|=|\frac{1}{40}(2+\sin (4t))x^{2}|\leq \frac{3}{40}|x|^{2}+1$$, here $$a=\frac{3}{40}$$, $$b=1$$. So, condition $$(H_{3})$$ is satisfied. Moreover, $$\|c_{i}\|=\frac{1}{40}$$, $$\|c_{2}\|=\frac{1}{60}$$. So, we have $$\sum_{i=1}^{2}\|c_{i}\|=\|c_{1}\|+\|c_{2}\|=\frac{1}{24}<1$$. Also $$\sigma =\frac{1}{1-\|c_{1}\|-\|c_{2}\|}=\frac{24}{23}$$, $$\|c'_{1}\|= \frac{1}{10}$$ and $$\|c'_{2}\|=\frac{1}{15}$$. Next, we consider the condition
\begin{aligned} &\sigma T^{\frac{1}{q}} \biggl(\frac{m\sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}}+\frac{a T (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p}} \biggr)^{ \frac{1}{p}}+\frac{\sigma T\sum_{i=1}^{n} \Vert c'_{i} \Vert }{2} \\ &\quad =\frac{24}{23}\times \biggl(\frac{\pi }{2} \biggr)^{\frac{2}{3}} \times \biggl(\frac{\frac{1}{20}\times \frac{1}{24}}{2^{2}}+\frac{ \frac{3}{40}\times \frac{\pi }{2}\times \frac{25}{24}}{2^{3}} \biggr) ^{\frac{1}{3}} + \frac{\frac{24}{23}\times \frac{\pi }{2}\times \frac{1}{6}}{2} \\ &\quad \approx 0.4909< 1. \end{aligned}

Therefore, by Theorem 4.1, we know that Eq. (6.1) has at least one positive $$\frac{\pi }{2}$$-periodic solution.

### Example 6.2

Consider the p-Laplacian Liénard equation in the case $$\sum_{i=1}^{n}\|c_{i}\|>1$$:
\begin{aligned}[b] & \biggl(\phi _{p} \biggl(x(t)- \biggl( \biggl(\frac{1}{8}\cos (8t)+ \frac{15}{8} \biggr)x(t-\delta _{3})+\frac{1}{64}\sin \biggl(8t-\frac{ \pi }{6} \biggr)x(t-\delta _{4}) \biggr) \biggr)' \biggr)' \\ &\quad {}+ \biggl(\frac{1}{24}x^{2}+1 \biggr)x'(t)+ \frac{1}{16}(2+\sin 8t)x ^{3}=\cos \biggl(8t-\frac{\pi }{4} \biggr), \end{aligned}
(6.2)
where $$p=5$$, $$\delta _{3}$$, $$\delta _{4}$$ are constants and $$0<\delta _{3}$$, $$\delta _{4}< T$$.
Comparing Eq. (6.2) with Eq. (4.1), it is easy to see that $$c_{1}(t)=\frac{1}{8}\cos (8t)+\frac{15}{8}$$, $$c_{2}(t)= \frac{1}{64}\sin (8t-\frac{\pi }{6})$$, $$f(x)=\frac{1}{24}x^{2}+1$$, $$g(t,x)=\frac{1}{16}(2+\sin 8t)x^{3}$$, $$e(t)=\cos (8t-\frac{\pi }{4})$$. $$T=\frac{\pi }{4}$$. It is easy to see that there exists a constant $$D=1$$ such that $$(H_{1})$$ holds. Obviously, we get $$|f(x)|=| \frac{1}{24}x^{2}+1|\leq \frac{1}{24}|x|^{2}+2$$, here $$m= \frac{1}{24}$$, $$\tilde{n}=2$$, and condition $$(H_{2})$$ holds. Consider $$|g(t,x)|=|\frac{1}{16}(2+\sin 8t)x^{3}|\leq \frac{3}{16}|x|^{3}+1$$, here $$a=\frac{3}{16}$$, $$b=1$$. So, condition $$(H_{3})$$ is satisfied. Furthermore, $$\|c_{1}\|=\frac{1}{8}+\frac{15}{8}=2$$, $$\|c_{2}\|= \frac{1}{64}$$, so we have $$\sum_{i=1}^{2}\|c_{i}\|=\|c_{1}\|+ \|c_{2}\|=\frac{129}{64}>1$$, $$\sigma =\frac{\frac{1}{\|c_{k}\|}}{1-\frac{1}{ \|c_{k}\|}-\sum_{i=1,i\neq k}^{n}\|\frac{c_{i}}{c_{k}}\|}=\frac{ \frac{1}{2}}{1-\frac{1}{2}-\frac{\frac{1}{2}}{\frac{1}{64}}}= \frac{64}{63}$$, $$\|c'_{1}\|=1$$ and $$\|c'_{2}\|=\frac{1}{8}$$. Next, we consider the condition
\begin{aligned} &\sigma T^{\frac{1}{q}} \biggl(\frac{m\sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}}+\frac{a T(1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p}} \biggr)^{\frac{1}{p}}+\frac{ \sigma T\sum_{i=1}^{n} \Vert c'_{i} \Vert }{2} \\ &\quad =\frac{64}{63}\times \biggl(\frac{\pi }{4}\biggr)^{\frac{4}{5}}\times \biggl(\frac{ \frac{1}{24}\times \frac{129}{64}}{2^{4}}+\frac{\frac{3}{16}\times \frac{ \pi }{4}\times \frac{193}{64}}{2^{5}} \biggr)^{\frac{1}{5}}+ \frac{ \frac{64}{63}\times \frac{\pi }{4}\times (1+\frac{1}{8})}{2} \\ &\quad \approx 0.8283< 1. \end{aligned}

Therefore, by Theorem 4.1, we know that Eq. (6.2) has at least one positive $$\frac{\pi }{4}$$-periodic solution.

### Example 6.3

Consider the following p-Laplacian Liénard equation with singularity
\begin{aligned} & \biggl(\phi _{p} \biggl(x(t)- \biggl( \biggl(\frac{1}{16}\cos \biggl(8t- \frac{ \pi }{16} \biggr) \biggr)x(t-\delta _{1})+\frac{1}{24} \sin (8t)x(t- \delta _{2}) \\ &\qquad {}+\frac{1}{48}\cos \biggl(8t+ \frac{\pi }{6}\biggr)x(t-\delta _{3}) \biggr) \biggr)' \biggr)' + \biggl(\frac{1}{10}x^{3} \biggr)x'(t)+ \frac{1}{32} \biggl(\frac{1}{2}+2 \sin 8t \biggr)x^{4}- \frac{1}{x^{\mu }} \\ &\quad =\cos \biggl(8t+\frac{\pi }{4} \biggr), \end{aligned}
(6.3)
where $$p=5$$, $$\mu \geq 1$$, $$\delta _{1}$$, $$\delta _{2}$$ and $$\delta _{3}$$ are constants and $$0<\delta _{1}$$, $$\delta _{2},\delta _{3}< T$$.
Comparing Eq. (6.3) with Eq. (4.1), it is easy to see that $$c_{1}(t)=\frac{1}{16}\cos (8t-\frac{\pi }{16}))$$, $$c_{2}(t)= \frac{1}{24}\sin (8t)$$, $$c_{3}(t)=\frac{1}{48}\cos (8t+\frac{\pi }{6})$$, $$f(x)=\frac{1}{10}x^{3}$$, $$g(t,x)=\frac{1}{32} (\frac{1}{2}+2 \sin (8t) )x^{4}-\frac{1}{x^{\mu }}$$, $$e(t)=\cos (8t+\frac{ \pi }{4})$$. $$T=\frac{\pi }{4}$$. It is obvious that $$(H_{2})$$, $$(H_{4})$$ and $$(H_{5})$$ hold. $$\|c_{1}\|=\frac{1}{16}$$, $$\|c_{2}\|= \frac{1}{24}$$ and $$\|c_{3}\|=\frac{1}{48}$$, so we have $$\sum_{i=1}^{3}\|c_{i}\|=\frac{1}{8}<1$$, $$\sigma =\frac{1}{1-\sum_{i=1}^{3}\|c_{i}\|}=\frac{1}{1-\frac{1}{16}-\frac{1}{24}- \frac{1}{48}}=\frac{8}{7}$$. Furthermore, $$\|c'_{1}\|=\frac{1}{2}$$, $$\|c'_{2}\|=\frac{1}{4}$$ and $$\|c'_{3}\|=\frac{1}{6}$$. Next, we consider the condition
\begin{aligned} &\sigma T^{\frac{1}{q}} \biggl(\frac{m\sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}}+\frac{ \alpha T (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p-1}} \biggr) ^{\frac{1}{p}}+\frac{\sigma T\sum_{i=1}^{n} \Vert c'_{i} \Vert }{2} \\ &\quad =\frac{8}{7}\times \biggl(\frac{\pi }{4}\biggr)^{\frac{4}{5}}\times \biggl(\frac{ \frac{1}{10}\times \frac{1}{8}}{2^{4}}+\frac{\frac{5}{64}\times \frac{ \pi }{4}\times \frac{9}{8}}{2^{4}} \biggr)^{\frac{1}{4}} + \frac{ \frac{8}{7}\times \frac{\pi }{4}\times \frac{11}{12}}{2} \\ &\quad \approx 0.7391< 1. \end{aligned}

Therefore, by Theorem 5.1, we know that Eq. (6.3) has at least one positive $$\frac{\pi }{4}$$-periodic solution.

### Example 6.4

Consider the following p-Laplacian Liénard equation with singularity
\begin{aligned}[b] & \biggl(\phi _{p} \biggl(x(t)- \biggl( \biggl(\frac{1}{6}\cos \biggl(6t+\frac{ \pi }{5} \biggr)+\frac{11}{6} \biggr)x(t-\delta _{1}) +\frac{1}{36} \cos \biggl(6t+\frac{\pi }{3} \biggr)x(t-\delta _{2})\\ &\quad {}- \frac{1}{24} \cos \biggl(6t+\frac{\pi }{6} \biggr)x(t-\delta _{3}) \biggr) \biggr)' \biggr)' \\ &\quad {}+ \biggl(\frac{1}{149}x^{4}+2 \biggr)x'(t)+ \frac{1}{256}\bigl(1+\sin (6t)\bigr)x ^{3}-\frac{1}{x^{\mu }}=\cos \biggl(6t+\frac{\pi }{4} \biggr), \end{aligned}
(6.4)
where $$p=3$$, $$\mu \geq 1$$, $$\delta _{1}$$, $$\delta _{2}$$, and $$\delta _{3}$$ are constants and $$0<\delta _{1}$$, $$\delta _{2},\delta _{3}< T$$.
Comparing Eq. (6.4) with Eq. (4.1), it is easy to see that $$c_{1}(t)=\frac{1}{6}\cos (6t+\frac{\pi }{5})+\frac{11}{6}$$, $$c_{2}(t)=\frac{1}{36}\cos (6t+\frac{\pi }{3})$$, $$c_{3}(t)=- \frac{1}{24}\cos (6t+\frac{\pi }{6})$$. $$f(x)=\frac{1}{149}x^{4}+2$$, $$g(t,x)=\frac{1}{256}(1+\sin (6t))x^{5}-\frac{1}{x^{\mu }}$$, $$e(t)=\sin (6t-\frac{\pi }{4})$$, $$T=\frac{\pi }{3}$$. It is obvious that $$(H_{2})$$, $$(H_{4})$$ and $$(H_{5})$$ hold. Furthermore, $$\|c_{1}\|= \frac{1}{6}+\frac{11}{6}=2$$, $$\|c_{2}\|=\frac{1}{36}$$ and $$\|c_{3}\|= \frac{1}{24}$$, so we have $$\sum_{i=1}^{3}\|c_{i}\|= \frac{149}{72}>1$$, $$\sigma =\frac{\frac{1}{\|c_{k}\|}}{1-\frac{1}{\|c _{k}\|}-\sum_{i=1,i\neq k}^{n}\|\frac{c_{i}}{c_{k}}\|}=\frac{ \frac{1}{2}}{1-\frac{1}{2}-\frac{\frac{1}{2}}{\frac{1}{36}}-\frac{ \frac{1}{2}}{\frac{1}{24}}}=\frac{576}{571}$$, $$\|c'_{1}\|=1$$, $$\|c'_{2}\|=\frac{1}{6}$$ and $$\|c'_{3}\|=\frac{1}{4}$$. Next, we consider the condition
\begin{aligned} &\sigma T^{\frac{1}{q}} \biggl(\frac{m\sum_{i=1}^{n} \Vert c_{i} \Vert }{2^{p-1}}+\frac{ \alpha T (1+\sum_{i=1}^{n} \Vert c_{i} \Vert )}{2^{p-1}} \biggr) ^{\frac{1}{p}}+\frac{\sigma T\sum_{i=1}^{n} \Vert c'_{i} \Vert }{2} \\ &\quad =\frac{576}{571}\times \biggl(\frac{\pi }{3}\biggr)^{\frac{2}{3}} \times \biggl(\frac{ \frac{1}{149}\times \frac{149}{72}}{2^{2}}+\frac{\frac{1}{128}\times \frac{\pi }{3}\times \frac{221}{72}}{2^{2}} \biggr)^{\frac{1}{3}} + \frac{ \frac{576}{571}\times \frac{\pi }{3}\times (1+\frac{7}{6}+ \frac{1}{4} )}{2} \\ &\quad \approx 0.9705< 1. \end{aligned}

Therefore, by Theorem 5.1, we know that Eq. (6.4) has at least one positive $$\frac{\pi }{3}$$-periodic solution.

## 7 Conclusions

In this paper, we first investigated some properties of the neutral operator with multiple delays and variable coefficients $$(Ax)(t):=x(t)- \sum_{i=1}^{n}c_{i}(t)x(t-\delta _{i})$$. Afterwards, by using an extension of Mawhin’s continuation theorem due to Ge and Ren, properties of the neutral operator A, we studied the existence of a periodic solution for equation (1.1). At last, by applying Theorem 3.2, we discussed the existence of a periodic solution for two p-Laplacian neutral differential equations. In comparison to  and , we avoided translating the equation into a two-dimensional system.

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