Positive periodic solution for higher-order p-Laplacian neutral singular Rayleigh equation with variable coefficient

Abstract

In this paper, we consider the existence of a positive periodic solution for the following kind of high-order p-Laplacian neutral singular Rayleigh equation with variable coefficient:

$$ \bigl(\varphi_{p}\bigl(x(t)-c(t)x(t-\sigma)\bigr)^{(n)} \bigr)^{(m)}+f\bigl(t,x'(t)\bigr)+g\bigl(t,x(t)\bigr)=e(t). $$

Our proof is based on Mawhin’s continuation theory.

1 Introduction

In this paper, we consider the following high-order p-Laplacian neutral singular Rayleigh equation with variable coefficient:

$$ { } \bigl(\varphi_{p}\bigl(x(t)-c(t)x(t-\sigma) \bigr)^{(n)}\bigr)^{(m)}+f\bigl(t,x'(t)\bigr)+g \bigl(t,x(t)\bigr)=e(t), $$

(1.1)

where $p>1$, $\varphi_{p}(x)= \vert x \vert ^{p-2}x$ for $x\neq 0$ and $\varphi_{p}(0)=0$, $c\in C^{n} (\mathbb{R},\mathbb{R})$ and $c(t+T)\equiv c(t)$, f is a continuous function defined in $\mathbb{R}^{2}$ and periodic in t with $f(t,\cdot)=f(t+T,\cdot)$ and $f(t,0)=0$, $g(t,x)=g_{0}(x)+g_{1}(t,x)$, where $g_{1}:\mathbb{R}\times(0,+\infty)\to \mathbb{R}$ is an $L^{2}$-Carathéodory function, $g_{1}(t,\cdot)=g_{1}(t+T,\cdot)$, $g_{0}\in C((0,\infty);\mathbb{R})$ has a singularity at $x=0$, $e:\mathbb{R}\rightarrow\mathbb{R}$ is a continuous periodic function with $e(t+T)\equiv e(t)$ and $\int^{T}_{0}e(t)\,dt=0$, T is a positive constant, and n and m are positive integers.

In recent years, there are many works on periodic solutions for high-order neutral differential equations (see [1–11] and the references therein). Wang and Lu [5] in 2007 investigated the existence of periodic solution for the following high-order neutral functional differential equation with distributed delay:

$$ { } \bigl(x(t)-cx(t-\sigma)\bigr)^{(n)}+f\bigl(x(t) \bigr)x'(t)+g \biggl( \int^{0}_{-r}x(t+s)\,d\alpha (s) \biggr)=p(t). $$

(1.2)

Using the continuation theorem of coincidence degree theory, they obtained the existence of periodic solutions for (1.2). Afterwards, Ren et al. considered the following high-order p-Laplacian neutral differential equation

$$ { } \bigl(\varphi_{p}\bigl(x(t)-cx(t-\sigma) \bigr)^{(l)}\bigr)^{(n-l)}=F\bigl(t,x(t),x'(t), \ldots ,x^{(l-1)}(t)\bigr). $$

(1.3)

They obtained the existence of periodic solutions for (1.3) in the general case ($\vert c \vert \neq1$) in [10] and in the critical case ($\vert c \vert =1$) in [9], respectively.

At the same time, some authors began to consider high-order neutral differential equation with singularity. Recently, applying the coincidence degree theory and some analysis skills, Xin et al. [11] discussed the existence of a positive periodic solution for the following neutral Liénard equation with singularity:

$$ { } \bigl(\varphi_{p}\bigl(x(t)-cx(t-\tau) \bigr)^{(n)}\bigr)^{(m)}+f\bigl(x(t)\bigr)x'(t)+g \bigl(t,x(t-\sigma)\bigr)=e(t). $$

(1.4)

Inspired by these results in [5, 9–11], in this paper, we consider the existence of a positive periodic solution for (1.1) with singularity by applications of Mawhin’s continuation theory. The obvious difficulty lies in the following two respects. Firstly, $(x(t)-c(t)x(t-\sigma))^{(n)}\neq x^{(n)}(t)-c(t)x^{(n)}(t-\sigma)$, and the calculation of $(x(t)-c(t)x(t-\sigma))^{(n)}$ is very complicated. Secondly, a priori bounds of periodic solutions are not easy to estimate.

2 Preparation

Firstly, we give qualitative properties of the neutral operator $(Ax)(t):=x(t)-c(t)x(t-\sigma)$.

Lemma 2.1

(see [12])

If $\vert c(t) \vert \neq1$, then the operator A has a continuous inverse $A^{-1}$ on $C_{T}:=\{\phi\in C(\mathbb{R},\mathbb{R}):\phi(t+T)\equiv\phi(t)\}$, satisfying

$$\bigl\vert \bigl(A^{-1}f \bigr) (t) \bigr\vert \leq \frac{ \vert f \vert _{\infty}}{\Gamma},\quad\forall f\in C_{T}, $$

where $\Gamma:= \bigl\{ \scriptsize{ \begin{array}{l@{\quad}l} 1- \vert c \vert _{\infty}&\textit{for } \vert c \vert _{\infty}:=\max_{t\in[0,T]} \vert c(t) \vert <1,\\ \vert c \vert _{0}-1&\textit{for } \vert c \vert _{0}:=\min_{t\in[0,T]} \vert c(t) \vert >1. \end{array} }$

Lemma 2.2

(Gaines and Mawhin [13])

Let X and Y be two Banach spaces, and let $L:D(L)\subset X\rightarrow Y$ be a Fredholm operator with index zero. Let $\Omega\subset X$ be an open bounded set, and let $N:\overline{\Omega}\rightarrow Y$ be L-compact on Ω̅. Assume that the following conditions hold:

• (1) $Lx\neq\lambda Nx$, $\forall x\in\partial\Omega\cap D(L)$, $\lambda\in(0,1)$;

• (2) $Nx\notin\operatorname{Im} L$, $\forall x\in\partial \Omega\cap\operatorname{Ker} L$;

• (3) $\deg\{JQN,\Omega\cap\operatorname{Ker} L,0\}\neq0$, where $J:\operatorname{Im} Q\rightarrow\operatorname{Ker} L$ is an isomorphism.

Then the equation $Lx=Nx$ has a solution in $\overline{\Omega}\cap D(L)$.

Lemma 2.3

(see [11])

If $x\in C^{1}_{T}:=\{x\in C^{1}(\mathbb{R},\mathbb{R}):x(t+T)\equiv x(t)\}$ and there exists a point $t_{0}\in(0,T)$ such that $\vert x(t_{0}) \vert < d$, then

$$\vert x \vert _{\infty}\leq d+\frac{1}{2} \int^{T}_{0} \bigl\vert x'(t) \bigr\vert \,dt, $$

where $\vert x \vert _{\infty}:=\max_{t\in\mathbb{R}} \vert x(t) \vert$.

To use the continuation degree theorem, we rewrite (1.1) in the form

$$ { } \textstyle\begin{cases} (Ax_{1})^{(n)}(t)=\varphi_{q}(x_{2}(t)),\\ x_{2}^{(m)}(t)=-f(t,x_{1}'(t))+g(t,x_{1}(t))+e(t), \end{cases} $$

(2.1)

where $\frac{1}{p}+\frac{1}{q}=1$. Clearly, if a periodic solution of (2.1) is $x(t):= \bigl ({\scriptsize\begin{matrix}{} x_{1}\cr x_{2} \end{matrix}} \bigr )$, then $x_{1}(t)$ must be a periodic solution of (1.1). Thus, the problem of finding a periodic solution for (1.1) reduces to finding a periodic solution for (2.1).

Now, set

$$X:=\bigl\{ x\in C\bigl(\mathbb{R},\mathbb{R}^{2}\bigr): x(t+T)\equiv x(t)\bigr\} $$

with the norm $\vert x \vert _{\infty}=\max\{ \vert x_{1} \vert _{\infty}, \vert x_{2} \vert _{\infty}\}$ and

$$Y:=\bigl\{ x\in C^{1}\bigl(\mathbb{R},\mathbb{R}^{2}\bigr): x(t+T)\equiv x(t)\bigr\} $$

with the norm $\Vert x \Vert =\max\{ \vert x \vert _{\infty}, \vert x' \vert _{\infty}\}$. Clearly, both X and Y are Banach spaces. Meanwhile, define

$$L:D(L)=\bigl\{ x\in C^{n+m}\bigl(\mathbb{R},\mathbb{R}^{2} \bigr): x(t+T) = x(t), t \in \mathbb{R}\bigr\} \subset X\rightarrow Y $$

by

$$ (Lx) (t)=\left ( \textstyle\begin{array}{c} (A x_{1})^{(n)}(t)\\ x_{2}^{(m)}(t) \end{array}\displaystyle \right ) $$

and $N: X\rightarrow Y$ by

$$ { } (Nx) (t)=\left ( \textstyle\begin{array}{c} \varphi_{q}(x_{2}(t))\\ -f(t,x_{1}'(t))-g(t,x_{1}(t))+e(t) \end{array}\displaystyle \right ). $$

(2.2)

Then (2.1) can be converted into the abstract equation $Lx=Nx$.

If $x= \bigl ({\scriptsize\begin{matrix}{} x_{1}\cr x_{2} \end{matrix}} \bigr ) \in \operatorname{Ker} L$, that is, $\bigl\{ \scriptsize{ \begin{array}{l@{\quad}l} (x_{1}(t)-c(t)x_{1}(t-\sigma))^{(n)}=0,\\ x_{2}^{(m)}(t)=0, \end{array} }$ then we have

$$ \textstyle\begin{cases} x_{1}(t)-c(t)x_{1}(t-\sigma)=a_{n-1}t^{n-1}+a_{n-2}t^{n-2}+\cdots +a_{1}t+a_{0},\\ x_{2}(t)=b_{m-1}t^{m-1}+b_{m-2}t^{m-2}+\cdots+b_{1}t+b_{0}, \end{cases} $$

where $a_{0},\ldots,a_{n-1}, b_{0},\ldots,b_{m-1}\in\mathbb{R}$ are constant. From $x_{1}(t)-c(t)x_{1}(t-\sigma)\in C_{T}$ and $x_{2}(t)\in C_{T}$ we have $a_{1}=\cdots=a_{n-1}=0$ and $b_{1}=b_{2}=\cdots=b_{m-1}=0$. Let $\phi(t)\neq0$ be a solution of $x(t)-c(t)x(t-\sigma)=1$. Then $\operatorname{Ker} L=x= \bigl ({\scriptsize\begin{matrix}{} a\phi(t),a\in\mathbb{R}\cr b,b\in\mathbb{R} \end{matrix}} \bigr )$. From the definition of L we can easily see that

$$\operatorname{Ker} L \cong\mathbb {R}^{2},\qquad \operatorname{Im} L= \left\{y\in Y: \large \int_{0}^{T} \begin{pmatrix} y_{1}(s)\\ y_{2}(s) \end{pmatrix} \,ds= \begin{pmatrix} 0\\ 0 \end{pmatrix} \right\}. $$

So L is a Fredholm operator with index zero.

Next, we will consider L-compact N. Let $P:X\rightarrow\operatorname{Ker} L$ and $Q:Y\rightarrow \operatorname{Im} Q\subset\mathbb {R}^{2}$ be defined by

$$Px= \begin{pmatrix} (Ax_{1})(0)\\x_{2}(0) \end{pmatrix} \quad\mbox{and}\quad Qy= \frac{1}{T} \int_{0}^{T} \begin{pmatrix} y_{1}(s)\\ y_{2}(s) \end{pmatrix} \,ds. $$

Then $\operatorname{Im} P=\operatorname{Ker} L$ and $\operatorname{Ker} Q=\operatorname{Im} L$. Denote $L_{P}=L\vert_{D(L)\cap\operatorname {Ker} P}$ and let $L_{P}^{-1}: \operatorname{Im} L\rightarrow D(L)$ be the inverse of $L_{P}$. Then

$$ { } \begin{aligned} &\bigl[L_{P}^{-1}y \bigr](t)= \begin{pmatrix} (A^{-1}Gy_{1})(t)\\ (Gy_{2})(t) \end{pmatrix} , \\ &[Gy_{1}](t)=\sum _{i=1}^{n-1}\frac{1}{i!}a_{i}t^{i}+ \frac{1}{(n-1)!} \int ^{t}_{0}(t-s)^{n-1}y_{1}(s)\,ds, \\ &[Gy_{2}](t)=\sum_{i=1}^{m-1} \frac{1}{i!}b_{i}t^{i}+\frac{1}{(m-1)!} \int _{0}^{t}(t-s)^{m-1}y_{2}(s)\,ds, \end{aligned} $$

(2.3)

where $a_{i}:=(Ax_{1})^{(i)}(0)$ are defined as follows:

$$E_{1}Z=C,\quad \mbox{where }E_{1}= \begin{pmatrix}1 & 0 & 0 & \cdots& 0 & 0 \\ e_{1} & 1 & 0 & \cdots& 0 & 0 \\ e_{2} & e_{1} & 1 & \cdots& 0 & 0 \\ \cdots\\ e_{n-3} & e_{n-4} & e_{n-5} & \cdots& 1 & 0 \\ e_{n-2} & e_{n-3} & e_{n-4} & \cdots& e_{1} &0 \end{pmatrix} _{(n-1)\times(n-1)}, $$

$Z=(a_{n-1},a_{n-2}\cdots,a_{1})^{\top}$, $C=(c_{1},c_{2},\ldots,c_{n-1})^{\top }$, $c_{i}=-\frac{1}{i!T}\int^{T}_{0}(T-s)^{i}y_{1}(s)\,ds$, and $e_{j}=\frac{T^{j}}{(j+1)!}$, $j=1,2,\ldots,n-2$. Similarly, we can get $b_{1}:=x_{2}^{(i)}(0)$, $i=1,2,\ldots,m-1$. Therefore, from (2.2) and (2.3) we get that N is L-compact on Ω̄.

3 Periodic solutions for (1.1) with repulsive singularity

For convenience, we list the following assumptions, which will further used repeatedly:

(H₁):

There exists a positive constant K such that $\vert f(t,u) \vert \leq K$ for $(t,u)\in \mathbb{R}\times\mathbb{R}$.

(H₂):

There exist positive constants α and β such that $\vert f(t,u) \vert \leq\alpha \vert u \vert ^{p-1}+\beta$ for $(t,u)\in\mathbb{R}\times\mathbb{R}$.

(H₃):

$f(t,u)\geq0$ for $(t,u)\in\mathbb{R}\times\mathbb{R}$;

(H₄):

There exists a positive constant D such that $g(t,x)>K$ for $x>D$.

(H₅):

There exists a positive constant $D_{1}$ such that $g(t,x)> \vert e \vert _{\infty}$ for $x>D_{1}$.

(H₆):

There exist positive constants γ, ζ such that

$$ g(t,x)\leq\gamma x^{p-1}+\zeta\quad \mbox{for all } x>0. $$

(H₇):

(Repulsive singularity) $\int^{1}_{0}g_{0}(s)\,ds=-\infty$.

Theorem 3.1

Assume that (H₁), (H₄), and (H₆)-(H₇) hold. Then (1.1) has at least one T-periodic solution if

$$0< \frac{T^{2p}}{2^{2p-1}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)+(m-2)} \frac{\gamma}{ (\Gamma-\frac{T}{2} \sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} )^{p-1}}< 1, $$

where $c_{n-k}:=\max_{t\in[0,\omega]} \vert c^{(n-k)}(t) \vert$.

Proof

Consider the abstract equation

$$Lx=\lambda Nx,\quad \lambda\in(0,1). $$

Set $\Omega_{1}=\{x:Lx=\lambda Nx,\lambda\in (0,1)\}$. If $x(t)=(x_{1}(t),x_{2}(t))^{\top}\in\Omega_{1}$, then

$$ { } \textstyle\begin{cases} (Ax_{1})^{(n)}(t)=\lambda\varphi_{q}(x_{2}(t)),\\ x_{2}^{(m)}(t)=-\lambda f(t,x_{1}'(t))-\lambda g(t,x_{1}(t))+\lambda e(t). \end{cases} $$

(3.1)

Substituting $x_{2}(t)=\lambda^{1-p}\varphi_{p}[(Ax_{1})^{(n)}(t)]$ into the second equation of (3.1), we have

$$ { } \bigl(\varphi_{p}(Ax_{1})^{(n)}(t) \bigr)^{(m)}+\lambda^{p} f\bigl(t,x_{1}'(t) \bigr)+\lambda^{p}g\bigl(t,x_{1}(t)\bigr)= \lambda^{p}e(t). $$

(3.2)

Integrating both sides of (3.2) from 0 to T, we have

$$ { } \int^{T}_{0}\bigl(f\bigl(t,x_{1}'(t) \bigr)+g\bigl(t,x_{1}(t)\bigr)\bigr)\,dt=0. $$

(3.3)

From the mean value theorem, there exists a point $\xi\in(0,T)$ such that

$$f\bigl(\xi,x_{1}'(\xi)\bigr)+g\bigl(\xi,x_{1}( \xi)\bigr)=0. $$

Then by (H₁) we have

$$g\bigl(\xi,x_{1}(\xi)\bigr)= \bigl\vert -f\bigl( \xi,x_{1}'(\xi)\bigr) \bigr\vert \leq K , $$

and in view of (H₄), we get that $x_{1}(\xi)\leq D$. Since $x_{1}(t)$ is periodic with period T and $x_{1}(t)>0$ for $t\in[0,T]$. Then $0< x_{1}(\xi)\leq D$. Therefore, from Lemma 2.3 we can get

$$ { } \begin{aligned} \vert x_{1} \vert _{\infty}&\leq D+\frac{1}{2} \int^{T}_{0} \bigl\vert x_{1}'(s) \bigr\vert \,ds. \end{aligned} $$

(3.4)

From (3.4) and the Wirtinger inequality (see [14], Lemma 2.4) we get

$$ { } \begin{aligned}[b] \vert x_{1} \vert _{\infty}&\leq D+\frac{1}{2}T^{\frac{1}{2}} \biggl( \int^{T}_{0} \bigl\vert x_{1}'(s) \bigr\vert ^{2}\,ds \biggr)^{\frac{1}{2}} \\ &\leq D+\frac{1}{2}T^{\frac{1}{2}} \biggl(\frac{T}{2\pi} \biggr)^{n-1} \biggl( \int ^{T}_{0} \bigl\vert x_{1}^{(n)}(s) \bigr\vert ^{2}\,ds \biggr)^{\frac{1}{2}} \\ &\leq D+\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{n-1} \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}. \end{aligned} $$

(3.5)

Since $x_{1}^{(i-1)}(0)=x_{1}^{(i-1)}(T)$, $i=1,2\ldots,n-1$, there exists a point $t_{i}^{*}\in[0,T]$ such that $x_{1}^{(i)}(t_{i}^{*})=0$. From the Hölder and Wirtinger inequalities, we can easily get

$$ { } \begin{aligned}[b] \bigl\vert x_{1}^{(i)} \bigr\vert _{\infty}&\leq \frac{1}{2} \int^{T}_{0} \bigl\vert x_{1}^{(i+1)}(t) \bigr\vert \,dt \\ &\leq \frac{T^{\frac{1}{2}}}{2} \biggl( \int^{T}_{0} \bigl\vert x_{1}^{(i+1)}(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\ &\leq \frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{(n-i-1)} \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}. \end{aligned} $$

(3.6)

On the other hand, since $(Ax_{1})(t)=x_{1}(t)-c(t)x_{1}(t-\sigma)$, we have

$$\begin{aligned} \bigl(Ax_{1}(t)\bigr)^{(n)}&= \bigl(x_{1}(t)-c(t)x_{1}(t- \sigma)\bigr)^{(n)} \\ &= x_{1}^{(n)}(t)- \biggl(c^{(n)}(t)x_{1}(t- \sigma)+nc^{(n-1)}(t)x_{1}'(t-\sigma )\\ &\quad {}+ \frac{n(n-1)}{2!}c^{(n-2)}x_{1}''(t- \sigma)+\cdots +c(t)x_{1}^{(n)}(t-\sigma) \biggr) \\ &= x_{1}^{(n)}(t)-c(t)x_{1}^{(n)}(t- \sigma)- \biggl(c^{(n)}(t)x_{1}(t-\sigma )+nc^{(n-1)}(t)x_{1}'(t- \sigma) \\ &\quad{}+\frac{n(n-1)}{2!}c^{(n-2)}x_{1}''(t- \sigma)+\cdots +nc'(t)x_{1}^{(n-1)}(t-\sigma) \biggr). \end{aligned}$$

So, we can get

$$\begin{aligned} Ax_{1}^{(n)}(t)&= \bigl(Ax_{1}(t) \bigr)^{(n)}+ \biggl(c^{(n)}(t)x_{1}(t-\sigma )+nc^{(n-1)}(t)x_{1}'(t-\sigma) \\ &\quad{}+\frac{n(n-1)}{2!}c^{(n-2)}x_{1}''(t- \sigma)+\cdots +nc'(t)x_{1}^{(n-1)}(t-\sigma) \biggr). \end{aligned}$$

Applying Lemma 2.2, (3.5), and (3.6), we have

$$\begin{aligned} \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}&= \max _{t\in[0,T]} \bigl\vert A^{-1}Ax_{1}^{(n)}(t) \bigr\vert \\ &\leq \Bigl(\max_{t\in[0,T]}\bigl\vert (Ax_{1})^{(n)}(t)+c^{(n)}(t)x_{1}(t-\sigma )\\ &\quad {}+nc^{(n-1)}(t)x_{1}'(t-\sigma) +\cdots+nc'(t)x_{1}^{(n-1)}(t-\sigma)\bigr\vert \Bigr)/{\Gamma} \\ &\leq \frac{\varphi_{q}( \vert x_{2} \vert _{\infty})+c_{n} \vert x_{1} \vert _{\infty}+nc_{n-1} \vert x_{1}' \vert _{\infty}+\cdots +nc_{1} \vert x_{1}^{(n-1)} \vert _{\infty}}{\Gamma} \\ &\leq \biggl(\varphi_{q}\bigl( \vert x_{2} \vert _{\infty}\bigr)+c_{n} \biggl(D+\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{n-1} \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}\biggr)\\ &\quad {} +nc_{n-1} \biggl(\frac{1}{2}T \biggl(\frac{T}{2\pi} \biggr)^{n-2} \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}\biggr)+\cdots+nc_{1}\frac{T}{2} \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}\biggr)\big/{\Gamma} \\ &\begin{aligned} &\leq \biggl(\varphi_{q}\bigl( \vert x_{2} \vert _{\infty}\bigr)+\frac{T}{2} \biggl( \biggl(\frac{T}{2\pi} \biggr)^{n-1}c_{n}+nc_{n-1} \biggl(\frac{T}{2\pi} \biggr)^{n-2}\\ &\quad {}+ \frac{n(n-1)}{2!}c_{n-2} \biggl(\frac{T}{2\pi} \biggr)^{n-3}+\cdots +nc_{1} \biggr) \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}+c_{n}D\biggr)\big/{\Gamma} \\ &\leq \frac{\varphi_{q}( \vert x_{2} \vert _{\infty})+\frac{T}{2} (\sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} ) \vert x_{1}^{(n)} \vert _{\infty}+c_{n}D}{\Gamma}. \end{aligned} \end{aligned}$$

Since $\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi } )^{n-1-k} )>0$, we have

$$ { } \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}\leq\frac{\varphi_{q}( \vert x_{2} \vert _{\infty})+c_{n}D}{\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} )}. $$

(3.7)

In view of $\int^{T}_{0}(\varphi_{q}(x_{2}(t)))\,dt=\int^{T}_{0}(Ax_{1}(t))^{(n)}(t)\,dt=0$, there exists a point $t_{2}\in(0,T)$ such that $x_{2}(t_{2})=0$. From the Wirtinger inequality and from (3.4) we easily get

$$ { } \begin{aligned}[b] \vert x_{2} \vert _{\infty}&\leq \frac{1}{2} \int^{T}_{0} \bigl\vert x_{2}'(t) \bigr\vert \,dt\leq \frac{\sqrt{T}}{2} \biggl( \int^{T}_{0} \bigl\vert x_{2}'(t) \bigr\vert ^{2} \biggr)^{\frac{1}{2}} \\ &\leq \frac{\sqrt{T}}{2} \biggl(\frac{T}{2\pi} \biggr)^{m-2} \biggl( \int ^{T}_{0} \bigl\vert x_{2}^{(m-1)}(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\ &\leq \frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{m-2} \bigl\vert x_{2}^{(m-1)} \bigr\vert _{\infty}. \end{aligned} $$

(3.8)

Besides, from $x_{2}^{(m-2)}(0)=x_{2}^{(m-2)}(T)$, there exists a point $t_{3}\in(0,T)$ such that $x_{2}^{(m-1)}(t_{3})=0$, which, together with the integration of the second equation of (3.1) on interval $[0,T]$, yield

$$ { } \begin{aligned}[b] 2 \bigl\vert x_{2}^{(m-1)}(t) \bigr\vert &\leq 2 \biggl(x_{2}^{(m-1)}(t_{3})+ \frac{1}{2} \int^{T}_{0} \bigl\vert x_{2}^{(m)}(t) \bigr\vert \,dt \biggr) \\ &\leq \lambda \int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr)-g\bigl(t,x_{1}(t)\bigr)+e(t) \bigr\vert \,dt \\ &\leq \int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr) \bigr\vert \,dt+ \int^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+ \int^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\leq KT+ \int^{T}_{0} \bigl\vert g\bigl(t,x_{1}(t) \bigr) \bigr\vert \,dt+T \vert e \vert _{\infty}, \end{aligned} $$

(3.9)

since $\vert f(t,u) \vert \leq K$ form (H₁). From (H₁) and (H₆) we have

$$ { } \begin{aligned}[b] \int^{T}_{0} \bigl\vert g\bigl(t,x_{1}(x) \bigr) \bigr\vert \,dt&= \int_{g(t,x_{1}(t))\geq 0}g\bigl(t,x_{1}(t)\bigr)\,dt- \int_{g(t,x_{1}(t))\leq0}g\bigl(t,x_{1}(t)\bigr)\,dt \\ &= 2 \int_{g(t,x_{1}(t))\geq0}g\bigl(t,x_{1}(t)\bigr)\,dt+ \int^{T}_{0}f(t,x_{1}'(t)\,dt \\ &\leq 2 \int^{T}_{0}\bigl(\gamma x_{1}^{p-1}+ \zeta\bigr)\,dt+ \int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr) \bigr\vert \,dt \\ &\leq 2\gamma \vert x_{1} \vert _{\infty}^{p-1}T+2 \zeta T+KT. \end{aligned} $$

(3.10)

Since $(1+x)^{k}\leq1+(1+k)x$ for $x\in[0,\delta]$, where δ is a constant, which depends only on $k>0$, substituting (3.10) into (3.9), we have

$$ { } \begin{aligned}[b] 2 \bigl\vert x_{2}^{(m-1)}(t) \bigr\vert &\leq 2T\gamma \vert x_{1} \vert _{\infty}^{p-1}+2 \zeta T+2KT+T \vert e \vert _{\infty}\\ &\leq 2T\gamma \biggl(D+\frac{1}{2} \int^{T}_{0} \bigl\vert x_{1}'(t) \bigr\vert \,dt \biggr)^{p-1}+N_{1} \\ &= 2T\gamma \biggl(1+\frac{D}{\frac{1}{2}\int^{T}_{0} \vert x_{1}'(t) \vert \,dt} \biggr)^{p-1} \biggl( \frac{1}{2} \biggr)^{p-1} \biggl( \int^{T}_{0} \bigl\vert x_{1}'(t) \bigr\vert \,dt \biggr)^{p-1}+N_{1} \\ &\leq \frac{1}{2^{p-2}}T\gamma \biggl(1+\frac{2Dp}{\int^{T}_{0} \vert x_{1}'(t) \vert \,dt} \biggr) \biggl( \int^{T}_{0} \bigl\vert x_{1}'(t) \bigr\vert \,dt \biggr)^{p-1}+N_{1}, \end{aligned} $$

(3.11)

where $N_{1}:=2\zeta T+2KT+T \vert e \vert _{\infty}$. Substituting (3.6) and (3.7) into (3.11), we have

$$ { } \begin{aligned}[b] &2 \bigl\vert x_{2}^{(m-1)}(t) \bigr\vert \\ &\quad \leq \frac{T^{p} \gamma}{2^{p-2}} \bigl\vert x_{1}' \bigr\vert ^{p-1}_{\infty}+\frac {DpT^{p-1}\gamma}{2^{p-3}} \bigl\vert x_{1}' \bigr\vert ^{p-2}_{\infty}+N_{1} \\ &\quad \leq \frac{T^{p} \gamma}{2^{p-2}}\cdot \biggl(\frac{T}{2} \biggr)^{p-1} \biggl(\frac{T}{2\pi} \biggr)^{(n-2))p-1)} \bigl\vert x_{1}^{(n)} \bigr\vert ^{p-1}_{\infty}\\ &\qquad {}+\frac{DpT^{p-1}\gamma}{2^{p-3}} \biggl( \frac{T}{2} \biggr)^{p-2} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-2)} \bigl\vert x_{1}^{(n)} \bigr\vert ^{p-2}_{\infty}+N_{1} \\ &\quad \leq \frac{T^{2p-1}\gamma}{2^{2p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)} \frac{(\varphi_{q}( \vert x_{2} \vert _{\infty})+C_{n} D)^{p-1}}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} ) )^{p-1}} \\ &\qquad{}+\frac{DpT^{2p-3}\gamma}{2^{2p-5}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-2)} \frac{(\varphi_{q}( \vert x_{2} \vert _{\infty})+C_{n} D)^{p-2}}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} ) )^{p-2}}+N_{1}. \end{aligned} $$

(3.12)

Combining of (3.8) and (3.12) implies

$$ { } \begin{aligned}[b] \vert x_{2} \vert _{\infty}&\leq \frac{T}{2} \biggl(\frac{T}{2\pi } \biggr)^{m-2} \bigl\vert x_{2}^{(m-1)} \bigr\vert _{\infty}\\ &\leq \frac{T}{4} \biggl(\frac{T}{2\pi} \biggr)^{m-2} \biggl[ \frac{T^{2p-1}\gamma}{2^{2p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)} \frac{(\varphi_{q}( \vert x_{2} \vert _{\infty})+C_{n} D)^{p-1}}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} ) )^{p-1}}\hspace{-20pt} \\ &\quad{}+\frac{DpT^{2p-3}\gamma}{2^{2p-5}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-2)} \frac{(\varphi_{q}( \vert x_{2} \vert _{\infty})+C_{n} D)^{p-2}}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} ) )^{p-2}}+N_{1} \biggr].\hspace{-20pt} \end{aligned} $$

(3.13)

So, we have

$$\begin{aligned} \vert x_{2} \vert _{\infty}&\leq \frac{T^{2p}\gamma}{2^{2p-1}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)+(m-2)}\frac{ \vert x_{2} \vert _{\infty}}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac {T}{2\pi} )^{n-1-k} ) )^{p-1}} \\ &\quad{}+\frac{T^{2p}\gamma}{2^{2p-1}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)+(m-2)} \frac{ (\sum_{i=0}^{p-1}C^{i}_{p-1}( \vert x_{2} \vert _{\infty}^{q-1})^{p-1-i}(c_{n}D)^{i} )}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac {T}{2\pi} )^{n-1-k} ) )^{p-1}} \\ &\begin{aligned}[b] &\quad{}+\frac{DpT^{2p-2}\gamma}{2^{2p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-2)+(m-2)}\frac{ (\sum_{i=0}^{p-1}C^{i}_{p-2}( \vert x_{2} \vert _{\infty}^{q-1})^{p-2-i}(c_{n}D)^{i} )}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac {T}{2\pi} )^{n-1-k} ) )^{p-2}}\\ &\quad {} +\frac{T}{4} \biggl(\frac{T}{2\pi} \biggr)^{m-2}N_{1}. \end{aligned} { } \end{aligned}$$

(3.14)

Since

$$\frac{T^{2p}}{2^{2p-1}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)+(m-2)} \frac{\gamma}{ (\Gamma-\frac{T}{2} \sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} )^{p-1}}< 1, $$

there exists a positive constant $M_{1}$ such that

$$ \vert x_{2} \vert _{\infty}\leq M_{1}. $$

(3.15)

Therefore, from (3.7) we have

$$ { } \begin{aligned}[b] \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}&\leq \frac{\varphi_{q}( \vert x_{2} \vert _{\infty})+c_{n}D}{\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} )} \\ &\leq \frac{M_{1}^{q-1}+c_{n}D}{\Gamma-\frac{T}{2}\sum^{n-1}_{k=0}C^{k}_{n}c_{n-k} (\frac{T}{2\pi} )^{n-1-k}}:=M_{n}'. \end{aligned} $$

(3.16)

From (3.6) we have

$$ \bigl\vert x_{1}' \bigr\vert _{\infty}\leq\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{n-2} \bigl\vert x_{1}^{(n)} \bigr\vert _{\infty}\leq\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{n-2} M_{n}':=M_{2}. $$

(3.17)

Hence, from (3.4) we have

$$ { } \vert x_{1} \vert _{\infty}\leq D+ \frac{1}{2} \int^{T}_{0} \bigl\vert x_{1}'(t) \bigr\vert \,dt\leq D+\frac{TM_{2}}{2}:=M_{3}. $$

(3.18)

From (3.6), (3.9), and (3.10) we have

$$\begin{aligned} \bigl\vert x_{2}^{(m-1)} \bigr\vert _{\infty}&\leq \frac{1}{2}\max \biggl\vert \int^{T}_{0}x_{2}^{(m)}(t)\,dt \biggr\vert \\ &\leq \frac{1}{2} \int^{T}_{0} \bigl\vert -f\bigl(t,x_{1}'(t) \bigr)-g\bigl(t,x_{1}(t)\bigr)+e(t) \bigr\vert \,dt \\ &\leq \frac{1}{2} \int^{T}_{0} \bigl\vert f(t,x_{1}'(t) \bigr\vert \,dt+\frac {1}{2} \int^{T}_{0} \bigl\vert g(t,x_{1}(t) \bigr\vert \,dt+\frac{1}{2} \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\leq KT+m M_{3}^{p-1}T+nT+\frac{1}{2} \vert e \vert _{\infty}T:=M_{m-1}. \end{aligned}$$

From (3.8) we get

$$\bigl\vert x_{2}' \bigr\vert _{\infty}\leq \frac{T}{2} \biggl(\frac{T}{2\pi } \biggr)^{m-3} \bigl\vert x_{2}^{(m-1)} \bigr\vert _{\infty}\leq \frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{m-3}M_{m-1}:=M_{4}. $$

On the other hand, since $g(t,x_{1})=g_{1}(t,x_{1}(t))+g_{0}(x_{1}(t))$, (3.2) can be rewritten as

$$ { } \bigl(\varphi_{p}(Ax_{1})^{(n)} \bigr)^{(m)}+\lambda^{p} f\bigl(t,x_{1}'(t) \bigr)+\lambda^{p}\bigl(g_{1}\bigl(t,x_{1}(t) \bigr)+g_{0}\bigl(x(t)\bigr)\bigr)=\lambda^{p} e(t). $$

(3.19)

Let $\tau\in[0,T]$ for any $\tau\leq t\leq T$. Multiplying both sides of (3.19) by $x_{1}'(t)$ and integrating on $[\tau,t]$, we have

$$ { } \begin{aligned}[b] \lambda^{p} \int^{x_{1}(t)}_{x_{1}(\tau)}g_{0}(u)\,du&= \lambda^{p} \int^{t}_{\tau }g_{0}\bigl(x_{1}(s) \bigr)x_{1}'(s)\,ds \\ &= - \int^{t}_{\tau}\bigl(\varphi_{p}(Ax_{1})^{(n)}(s) \bigr)^{(m)}x_{1}'(s)\,ds-\lambda^{p} \int ^{t}_{\tau}f\bigl(s,x_{1}'(s) \bigr)x_{1}'(s)\,ds \\ &\quad{}-\lambda^{p} \int^{t}_{\tau}g_{1}\bigl(s,x_{1}(s) \bigr)x_{1}'(s)\,ds+\lambda^{p} \int^{t}_{\tau }e(s)x_{1}'(s)\,ds. \end{aligned} $$

(3.20)

By (3.2), (3.12), (3.17), and (3.18) we have

$$ { } \begin{aligned}[b] &\biggl\vert \int^{t}_{\tau}\bigl(\varphi_{p}(Ax_{1})^{(n)}(s) \bigr)^{(m)}x_{1}(s)\,ds \biggr\vert \\ &\quad \leq \int^{t}_{\tau}\bigl\vert \bigl( \varphi_{p}(Ax_{1})^{(n)}(s)\bigr)^{(m)} \bigr\vert \bigl\vert x_{1}'(s) \bigr\vert \,ds \\ &\quad \leq \bigl\vert x_{1}' \bigr\vert _{\infty}\int^{t}_{\tau}\bigl\vert \bigl(\varphi _{p}(Ax_{1})^{(n)}(s)\bigr)^{(m)} \bigr\vert \,ds \\ &\quad \leq \lambda^{p} M_{2} \biggl( \int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr) \bigr\vert \,dt+ \int^{T}_{0} \bigl\vert g\bigl(t,x_{1}(t) \bigr) \bigr\vert \,dt+ \int^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \biggr) \\ &\quad \leq \lambda^{p} M_{2}\bigl(2KT+2mM_{3}^{p-1}T+2nT+ \vert e \vert _{\infty}T\bigr). \end{aligned} $$

(3.21)

Moreover, we have

$$ { } \begin{aligned} & \biggl\vert \int^{t}_{\tau}f\bigl(s,x_{1}'(s) \bigr)x_{1}'(s)\,ds \biggr\vert \leq \bigl\vert x_{1}' \bigr\vert _{\infty}\int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr) \bigr\vert \,dt\leq M_{2} KT, \\ & \biggl\vert \int^{t}_{\tau}g_{1}\bigl(s,x_{1}(s) \bigr)x_{1}'(s)\,ds \biggr\vert \leq \bigl\vert x_{1}' \bigr\vert _{\infty}\int^{T}_{0} \bigl\vert g_{1} \bigl(t,x_{1}(t)\bigr) \bigr\vert \,dt\leq M_{2}\sqrt{T} \Vert g_{M_{3}} \Vert _{2}, \\ & \biggl\vert \int^{t}_{\tau}e(s)x_{1}'(s)\,ds \biggr\vert \leq \bigl\vert x_{1}' \bigr\vert _{\infty}\int^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt\leq M_{2} \vert e \vert _{\infty}T, \end{aligned} $$

(3.22)

where $g_{M_{3}}:=\max_{0< x\leq M_{3}} \vert g_{1}(t,u) \vert \in L^{2}(0,T)$ and $\Vert g_{M_{3}} \Vert _{2}:= (\int^{T}_{0} \vert g_{1}(t,x_{1}'(t)) \vert ^{2}\,dt )^{\frac{1}{2}}$. Substituting (3.21) and (3.22) into (3.20), we have

$$ \biggl\vert \int^{x_{1}(t)}_{x_{1}(\tau)}g_{0}(x)\,dx \biggr\vert \leq M_{2} \bigl(3KT +2m M_{3}^{p-1}T+2nT+\sqrt{T} \Vert g_{M_{3}} \Vert _{2}+2 \vert e \vert _{\infty}T \bigr):=M_{5}^{*}. $$

From repulsive singular condition (H₇) we know that there exists a constant $M_{5}>0$ such that

$$ { } x_{1}(t)\geq M_{5},\quad\forall t\in[ \tau,T]. $$

(3.23)

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

Let

$$\Omega_{2}=\bigl\{ x=(x_{1},x_{2})^{\top}: E_{5}< x_{1}(t)< E_{1}, \bigl\vert x_{1}' \bigr\vert _{\infty}< E_{2}, \vert x_{2} \vert _{\infty}< E_{3}, \bigl\vert x_{2}' \bigr\vert _{\infty}< E_{4}\bigr\} , $$

where $0< E_{5}< M_{5}$, $E_{1}>\max\{D,M_{3}\}$, $E_{2}>M_{2}$, $E_{3}>M_{1}$, and $E_{4}>M_{4}$. Next, we shall prove that $\Omega_{2}$ is a bounded set. In fact, for all $x\in\Omega_{2}$, $x_{2}=0$, $x_{1}=a_{0}\phi(t)$, and $a_{0}\in\mathbb{R}^{+}$, we have

$$0= \int^{T}_{0}g\bigl(t,a_{0}\phi(t) \bigr)\,dt. $$

From assumption (H₁) we have $0< a_{0}\phi(t)\leq D$. So $\Omega_{2}$ is a bounded set.

Let $\Omega=\{x\in(x_{1},x_{2})^{\top}: \Vert x \Vert \leq M\}$, where $M=\max\{E_{1},E_{2},E_{3},E_{4}\}$. Then $\Omega_{1}\cup\Omega_{2}\subset\Omega$, and, as it follows from the above proof, $Lx\neq\lambda Nx$ for all $(x,\lambda)\in\partial \Omega\times(0,1)$, so that conditions (1) and (2) of Lemma 2.2 are both satisfied. Define the isomorphism $J:\operatorname{Im} Q\rightarrow\operatorname{Ker} L$ as follows:

$$J(x_{1},x_{2})^{\top}=(x_{2},-x_{1})^{\top}. $$

Let $H(\mu,x)=-\mu x+(1-\mu)JQNx$, $(\mu,x)\in[0,1]\times\Omega$. Then, for all $(\mu,x)\in(0,1)\times(\partial\Omega\cap\operatorname {Ker} L)$,

$$ H(\mu,x)= \begin{pmatrix}-\mu x_{1}(t)-\frac{1-\mu}{T}\int^{T}_{0}g(t,x_{1}(t))\,dt\\ -\mu x_{2}(t)-(1-\mu)\varphi_{q}(x_{2}(t)) \end{pmatrix} , $$

since $\int^{T}_{0}e(t)\,dt=0$ and $f(t,0)=0$. From (H₄) it is obvious that $x^{\top}H(\mu,x)<0$ for all $(\mu,x)\in(0,1)\times (\partial\Omega\cap\operatorname{Ker} L)$. Hence

$$\begin{aligned} \deg\{JQN,\Omega\cap\operatorname{Ker} L,0\}&=\deg\bigl\{ H(0,x),\Omega\cap \operatorname{Ker} L,0\bigr\} \\ &=\deg\bigl\{ H(1,x),\Omega\cap\operatorname{Ker} L,0\bigr\} \\ &=\deg\{I,\Omega\cap\operatorname{Ker} L,0\}\neq0. \end{aligned}$$

So condition (3) of Lemma 2.2 is satisfied. Applying Lemma 2.2, we conclude that equation $Lx=Nx$ has a solution $x=(x_{1},x_{2})^{\top}$ on $\bar{\Omega}\cap D(L)$, that is, (1.1) has a T-periodic solution $x_{1}(t)$. □

Theorem 3.2

Assume that (H₂)-(H₃) and (H₅)-(H₇) hold. Then (1.1) has at least a nonconstant T-periodic solution if

$$0< \frac{ (\frac{T^{2p}\gamma}{2^{2p-1}}+\frac{T^{p+1} \alpha}{2^{p}} ) (\frac{T}{2\pi} )^{m+n-4}}{ (\Gamma -\frac{T}{2} \sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} )^{p-1}}< 1. $$

Proof

We follow the same strategy and notation as the proof of Theorem 3.1. Now, we consider $\Vert x' \Vert \leq M_{2}$.

We first claim that there is a constant $\xi^{*}\in[0,T]$ such that

$$ { } 0< x_{1}\bigl(\xi^{*}\bigr)\leq D_{1}. $$

(3.24)

Since $\int^{T}_{0}(\varphi_{p}(Ax_{1})'(t))'dt=0$, there exist two points $\xi^{*}, \xi_{*}\in[0,T]$ such that

$$\bigl(\varphi_{p}(Ax_{1})'\bigl(\xi^{*}\bigr) \bigr)'\geq0\quad\mbox{and}\quad\bigl(\varphi _{p}(Ax_{1})'( \xi_{*})\bigr)'\leq0. $$

From (H₃) and (3.2) we have

$$g\bigl(\xi^{*},x_{1}\bigl(\xi^{*}\bigr)\bigr)-e\bigl(\xi^{*}\bigr)\leq-f \bigl(\xi^{*},x_{1}'\bigl(\xi^{*}\bigr)\bigr)\leq0, $$

since $f(\xi^{*},x_{1}'(\xi^{*}))>0$. Therefore, we get

$$g\bigl(\xi^{*},x_{1}'\bigl(\xi^{*}\bigr)\bigr)\leq e\bigl( \xi^{*}\bigr)\leq \vert e \vert _{\infty}. $$

From (H₅) we have

$$x_{1}(\xi)\leq D_{1}. $$

Since $x(t)>0$, we get $0< x_{1}(\xi^{*})\leq D_{1}$. This proves (3.24).

Similarly, from (3.4) we have

$$ { } \bigl\vert x_{1}(t) \bigr\vert \leq D_{1}+\frac{1}{2} \int^{T}_{0} \bigl\vert x_{1}'(t) \bigr\vert \,dt. $$

(3.25)

From (3.9) and (H₂) we get

$$ { } \begin{aligned}[b] 2 \bigl\vert x_{2}^{(m-1)}(t) \bigr\vert &\leq 2 \biggl(x_{2}^{(m-1)}(t_{3})+ \frac{1}{2} \int^{T}_{0} \bigl\vert x_{2}^{(m)}(t) \bigr\vert \,dt \biggr) \\ &\leq \lambda \int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr)-g\bigl(t,x_{1}(t)\bigr)+e(t) \bigr\vert \,dt \\ &\leq \int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr) \bigr\vert \,dt+ \int^{T}_{0} \bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+ \int^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\leq \alpha \int^{T}_{0} \bigl\vert x_{1}'(t) \bigr\vert ^{p-1}\,dt+\beta T+ \int^{T}_{0} \bigl\vert g\bigl(t,x_{1}(t) \bigr) \bigr\vert \,dt+T \vert e \vert _{\infty}. \end{aligned} $$

(3.26)

From (3.10), (H₂), and (H₇) we have

$$ { } \begin{aligned}[b] \int^{T}_{0} \bigl\vert g\bigl(t,x_{1}(t) \bigr) \bigr\vert \,dt&= \int_{g(t,x_{1}(t))\geq 0}g\bigl(t,x_{1}(t)\bigr)\,dt- \int_{g(t,x_{1}(t))< 0}g\bigl(t,x_{1}(t)\bigr)\,dt \\ &= 2 \int_{g(t,x_{1}(t))\geq0}g\bigl(t,x_{1}(t)\bigr)\,dt+ \int^{T}_{0}f\bigl(t,x_{1}'(t) \bigr)\,dt \\ &\leq 2 \int_{g(t,x_{1}(t))\geq0}\bigl(\gamma x_{1}^{p-1}(t)+\zeta \bigr)\,dt+ \int^{T}_{0} \bigl\vert f\bigl(t,x_{1}'(t) \bigr) \bigr\vert \,dt \\ &\leq 2\gamma \vert x_{1} \vert ^{p-1}T+2\zeta T+\alpha \int^{T}_{0} \bigl\vert x_{1}'(t) \bigr\vert ^{p-1}\,dt+\beta T. \end{aligned} $$

(3.27)

Substituting (3.27) into (3.26), from (3.11) we have

$$ { } \begin{aligned}[b] 2 \bigl\vert x_{2}^{(m-1)}(t) \bigr\vert &\leq 2\gamma \int^{T}_{0} \bigl\vert x(t) \bigr\vert ^{p-1}\,dt+2\zeta T+2\alpha \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{p-1}\,dt+2\beta T+ \vert e \vert _{\infty}T \\ &\leq \biggl(\frac{T^{p} \gamma}{2^{p-2}}+2\alpha T \biggr) \bigl\vert x_{1}' \bigr\vert _{\infty}^{p-1}+ \frac{DpT^{p-1}\gamma }{2^{p-3}} \bigl\vert x_{1}' \bigr\vert _{\infty}^{p-2}+N_{2}, \end{aligned} $$

(3.28)

where $N_{2}=2T(\zeta+\beta)+ \Vert e \Vert T$. From (3.12), (3.13), and (3.14) we get

$$ \begin{aligned} \vert x_{2} \vert _{\infty}&\leq \biggl(\frac{T^{2p}\gamma }{2^{2p-1}}+\frac{T^{p+1} \alpha}{2^{p}} \biggr) \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)+(m-2)}\frac { \vert x_{2} \vert _{\infty}}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac {T}{2\pi} )^{n-1-k} ) )^{p-1}} \\ &\quad{}+\frac{T^{2p}\gamma}{2^{2p-1}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)+(m-2)} \frac{ (\sum_{i=0}^{p-1}C^{i}_{p-1}( \vert x_{2} \vert _{\infty}^{q-1})^{p-1-i}(c_{n}D)^{i} )}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac {T}{2\pi} )^{n-1-k} ) )^{p-1}} \\ &\quad{}+\frac{DpT^{2p-2}\gamma}{2^{2p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-2)+(m-2)} \frac{ (\sum_{i=0}^{p-1}C^{i}_{p-2}( \vert x_{2} \vert _{\infty}^{q-1})^{p-2-i}(c_{n}D)^{i} )}{ (\Gamma-\frac{T}{2} (\sum_{k=0}^{n-1}C^{k}_{n}c_{n-k} (\frac {T}{2\pi} )^{n-1-k} ) )^{p-2}}\\ &\quad {}+\frac{T}{4} \biggl(\frac{T}{2\pi} \biggr)^{m-2}N_{2}. \end{aligned} $$

Since $\frac{ (\frac{T^{2p}\gamma}{2^{2p-1}}+\frac{T^{p+1} \alpha}{2^{p}} ) (\frac{T}{2\pi} )^{(n-2)(p-1)+(m-2)}}{ (\Gamma-\frac{T}{2} \sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} )^{p-1}}<1$, it is easy to see that there exists a positive constant $M_{2}$ such that

$$ \bigl\Vert x' \bigr\Vert \leq M_{2}. $$

The rest of the proof is the same as in Theorem 3.1. □

We illustrate our results with an example.

Example 3.1

Consider the neutral functional differential

$$ { } \begin{aligned}[b] &\biggl(\varphi_{p} \biggl(x(t)-\frac{1}{64}\sin(4t) x (t-\sigma ) \biggr)''' \biggr)'''+\cos^{2}(2t)\sin x'(t)+\frac{1}{4\pi}\bigl(\sin(4t)+3\bigr)x^{3}(t)- \frac{1}{x^{\mu}}\hspace{-30pt}\\ &\quad =5\cos(4t), \end{aligned} $$

(3.29)

where $p=4$, σ and μ are constants, and $0<\sigma<T$. It is clear that $T=\frac{\pi}{2}$, $n=3$, $m=3$, $c(t)=\frac{1}{64}\sin4t$, $e(t)=5\cos4t$, $c_{1}=\max_{t\in[0,T]}\vert\frac{1}{16}\cos 4t\vert=\frac{1}{16}$, $c_{2}=\max_{t\in[0,T]}\vert-\frac{1}{4}\sin 4t\vert=\frac{1}{4}$, and $c_{3}=\max_{t\in[0,T]}\vert-\cos4t\vert=1$. In this case, $f(t,u)=\cos^{2}(2t)\sin u$, $f(t,0)=0$, $\vert f(t,u) \vert =\vert\cos^{2}(2t)\sin u\vert\leq1$, $K=1$; and $g(t,x)=\frac{1}{4\pi}(\sin4t+3)x^{3}(t)-\frac{1}{x^{\mu}}\leq\frac {1}{\pi}x^{3}+1$, $\gamma=\frac{1}{\pi}$, $\zeta=1$; Obviously, conditions (H₁) and (H₆)-(H₇) hold. Choose $D=4\pi$ such that (H₄) holds. Now we consider the following condition:

$$\begin{aligned} &\frac{T^{2p}}{2^{2p-1}} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)(p-1)+(m-2)} \frac{\gamma}{ (\Gamma-\frac{T}{2} \sum_{k=0}^{n-1}C_{n}^{k}c_{n-k} (\frac{T}{2\pi} )^{n-1-k} )^{p-1}} \\ &\quad = \frac{ (\frac{\pi}{2} )^{8}}{2^{7}} \biggl(\frac{\frac{\pi }{2}}{2\pi} \biggr)^{4} \frac{\frac{1}{\pi}}{ (\frac{63}{64} -\frac{\pi}{4}\times (1\times\frac{1}{16}+3\times\frac{1}{4}\times \frac{1}{4}+3\times\frac{1}{16} ) )^{3}} \\ &\quad \approx \frac{\pi^{8}}{2^{26}}< 1. \end{aligned}$$

So, by Theorem 3.1, (3.29) has at least one nonconstant $\frac{\pi}{2}$-periodic solution.

4 Conclusions

In summary, a periodic solution of (1.1) with singularity is illustrated by Theorems 3.1 and 3.2. In Theorem 3.1, we consider the existence of a periodic solution for (1.1) in the case $\vert f(t,u) \vert \leq K$. Furthermore, in Theorem 3.2, we give a condition on $f(t,u)$ that is weaker than $\vert f(,u) \vert \leq K$ in Theorem 3.1, that is, we obtain the existence of periodic solution for (1.1) in the case where $\vert f(t,u) \vert \leq\alpha \vert u \vert ^{p-1}+\beta$. From the mathematical point of view, the results are valuable to understand the periodic solutions for high-order neutral differential equations.