Study on a kind of neutral Rayleigh equation with singularity

Abstract

In this paper, we consider a kind of neutral Rayleigh equation with singularity,

$$\bigl(x(t)-cx(t-\delta)\bigr)''+f \bigl(t,x'(t)\bigr)+g\bigl(t,x(t)\bigr)=e(t), $$

where g has a singularity at $x=0$. By applications of coincidence degree theory, we find that the existence of positive periodic solution for this equation.

1 Introduction

More recently, some classical tools have been used to study periodic solution for Rayleigh equation in the literature, including coincidence degree theory [1–4], the method of upper and lower solutions [5], the Manásevich-Mawhin continuation theorem [6–8], and the time map continuation theorem [9–11].

From then on, the study of the existence of positive periodic solutions for Rayleigh equations with singularity has attracted many researchers’ attention [12, 13]. In 2015, Wang and Ma [12] investigated the following singular Rayleigh equation:

$$x''+f\bigl(t,x'\bigr)+g(x)=p(t), $$

where g had a singularity at the origin, i.e., $\lim_{x\to +\infty} g(x)=+\infty$. By applications of the limit properties of time map, the authors found that the existence of periodic solution for this equation. Afterwards, by using Manásevich-Mawhin continuation theorem, Lu, Zhong and Chen [13] discussed the existence of periodic solution for the following two kinds of p-Laplacian singular Rayleigh equations:

$$\bigl(\big|x'\big|^{p-2}x'\bigr)'+f \bigl(x'\bigr)-g_{1}^{*}(x)+g_{2}^{*}(x)=h(t) $$

and

$$\bigl(\big|x'\big|^{p-2}x'\bigr)'+f \bigl(x'\bigr)+g_{1}^{*}(x)-g_{2}^{*}(x)=h(t), $$

where $g_{1}, g_{2}:(0,+\infty)\to\mathbb{R}$ were continuous and $g_{1}(x)$ was unbounded as $x\to0^{+}$.

In the above papers, the authors investigated several kinds of Rayleigh equations or singular Rayleigh equations. However, the study of the neutral Rayleigh equation with singularity is relatively rare. Motivated by [12, 13], we consider the neutral Rayleigh equation with singularity

$$ { } \bigl(x(t)-cx(t-\delta)\bigr)''+f \bigl(t,x'(t)\bigr)+g\bigl(t,x(t)\bigr)=e(t), $$

(1.1)

where $|c|\neq1$, δ is a constant, $e \in C[0,T]$ and $\int^{T}_{0}e(t)\,dt=0$; f is continuous functions defined on $\mathbb{R}^{2}$ and periodic in t with $f(t,\cdot)=f(t+T,\cdot)$, and $f(t,0)=0$; $g:\mathbb{R}\times(0,+\infty)\to\mathbb{R}$ is an $L^{2}$-Carathéodory function and $g(t,\cdot)=g(t+T,\cdot)$, $g(t,x)=g_{0}(x)+g_{1}(t,x)$, here $g_{1}:\mathbb{R}\times(0,+\infty)\to \mathbb{R}$ 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 strong singularity at the origin such that

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

(1.2)

By application of coincidence degree theory, we find the existence of positive periodic solutions of (1.1). Our results improve and extend the results in [12, 13].

2 Preparation

In this section, we give some lemmas, which will be used in this paper.

Lemma 2.1

see [14]

If $|c|\neq1$, then the operator $(Ax)(t):=x(t)-cx(t-\delta)$ has a continuous inverse $A^{-1}$ on the space

$$C_{T}:=\bigl\{ x|x\in(\mathbb{R},\mathbb{R}), x(t+T)- x(t)\equiv0, \forall t\in\mathbb{R}\bigr\} , $$

and satisfying

$$\bigl\vert \bigl(A^{-1}x \bigr) (t) \bigr\vert \leq \frac{\|x\|}{|1-|c||}, $$

where $\|x\| =\max_{t\in[0,T]}|x(t)|, \forall x\in C_{T}$.

Lemma 2.2

Gaines and Mawhin [1]

Suppose that X and Y are two Banach spaces, and $L:D(L)\subset X\rightarrow Y$ is a Fredholm operator with index zero. Let $\Omega\subset X$ be an open bounded set and $N:\overline{\Omega}\rightarrow Y$ 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 $\overline{\Omega}\cap D(L)$.

Set

$$\begin{gathered} X:=\bigl\{ x\in C^{1}(\mathbb{R},\mathbb{R}): x(t+T)- x(t)\equiv0, \forall t\in\mathbb{R}\bigr\} , \\Y:=\bigl\{ y\in C(\mathbb{R},\mathbb{R}): y(t+T)- y(t)\equiv0, \forall t\in \mathbb{R}\bigr\} , \end{gathered}$$

with the norm

$$\|x\|_{X}=\max\bigl\{ \|x\|,\big\| x'\big\| \bigr\} ,\qquad \|y \|_{Y}=\|y\|. $$

Clearly, X and Y are both Banach spaces. Meanwhile, define

$$L:D(L)=\bigl\{ x\in X: x''\in C(\mathbb{R},\mathbb{R}) \bigr\} \subset X\rightarrow Y $$

by

$$ (Lx) (t)=(Ax)''(t) $$

and $N: X\rightarrow Y$ by

$$ { } (Nx) (t)=-f\bigl(t,x'(t)\bigr)-g\bigl(t,x(t) \bigr)+e(t). $$

(2.1)

Then (1.1) can be converted to the abstract equation $Lx=Nx$. From the definition of L, one can easily see that

$$\operatorname{Ker} L \cong\mathbb {R},\qquad \operatorname{Im} L= \biggl\{ y\in Y: \large \int_{0}^{T}y(s)\,ds=0 \biggr\} . $$

So L is a Fredholm operator with index zero. Let $P:X\rightarrow\operatorname{Ker} L$ and $Q:Y\rightarrow\operatorname{Im} Q\subset\mathbb {R}$ be defined by

$$Px=(Ax) (0);\qquad Qy=\frac{1}{T} \int_{0}^{T} y(s)\,ds, $$

then $\operatorname{Im} P=\operatorname{Ker} L$, $\operatorname{Ker} Q=\operatorname{Im} L$. Let K denote the inverse of $L|_{\operatorname{Ker} p\cap D(L)}$. It is easy to see that $\operatorname{Ker} L= \operatorname{Im} Q=\mathbb{R}$ and

$$[Ky](t)= \int^{T}_{0}G(t,s)y(s)\,ds, $$

where

$$ { } G(t,s)= \textstyle\begin{cases} \frac{-s(T-t)}{T}, &0\leq s < t\leq T;\\ \frac{-t(T-s)}{T},& 0\leq t\leq s \leq T. \end{cases} $$

(2.2)

From (2.1) and (2.2), it is clear that QN and $K(I-Q)N$ are continuous, $QN(\overline{\Omega})$ is bounded and then $K(I-Q)N(\overline{\Omega})$ is compact for any open bounded $\Omega\subset X$, which means N is L-compact on Ω̄.

3 Positive periodic solution for (1.1)

For the sake of convenience, we list the following assumptions, which will be used repeatedly in the sequel:

$(H_{1})$ :

there exists a positive constant K such that $|f(t,u)|\leq K$, for $(t,u)\in \mathbb{R}\times\mathbb{R}$;

$(H_{2})$ :

there exist positive constants α and β such that $|f(t,u)|\leq\alpha|u|+\beta$, for $(t,u)\in\mathbb{R}\times\mathbb{R}$;

$(H_{3})$ :

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

$(H_{4})$ :

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

$(H_{5})$ :

there exists a positive constant $D_{1}$ such that $g(t,x)>\|e\|$ for $x>D_{1}$;

$(H_{6})$ :

there exist positive constants m, n such that

$$ g(t,x)\leq m x+n,\quad \mbox{for all } x>0. $$

Now we give our main results on periodic solutions for (1.1).

Theorem 3.1

Assume that conditions $(H_{1})$, $(H_{4})$, $(H_{6})$ hold. Then (1.1) has at least one solution with period T if $mT^{2}<\pi|1-|c||$.

Proof

By construction (1.1) has an T-periodic solution if and only if the operator equation

$$Lx=Nx $$

has an T-periodic solution. To use Lemma 2.1, we embed this operator equation into an equation family with a parameter $\lambda\in(0,1)$,

$$Lx=\lambda Nx, $$

which is equivalent to the following equation:

$$ { } \bigl((Ax) (t)\bigr)''+\lambda f \bigl(t,x'(t)\bigr)+\lambda g\bigl(t,x(t)\bigr)=\lambda e(t), $$

(3.1)

where $(Ax)(t)=x(t)-cx(t-\delta)$ in Section 2.

We first claim that there is a point $\xi\in(0,T)$ such that

$$ { } 0< x(\xi)\leq D. $$

(3.2)

Integrating both sides of (3.1) over $[0,T]$, we have

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

(3.3)

This shows that there at least exists a point $\xi\in(0,T)$ such that

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

then by $(H_{1})$, we have

$$g\bigl(\xi,x(\xi)\bigr)=\bigl|-f\bigl(\xi,x'(\xi)\bigr)\bigr|\leq K , $$

and in view to $(H_{4})$ we get $x(\xi)\leq D$. Since $x(t)$ is periodic with periodic T and $x(t)>0$, for $t\in[0,T]$. Then $0< x(\xi)\leq D$. (3.2) is proved. Therefore, we have

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

(3.4)

For $|c|\neq1$, by applying Lemma 2.1, we have

$$ { } \begin{aligned}[b] \big\| x'\big\| &=\max _{t\in[0,T]} \bigl\vert A^{-1}Ax'(t) \bigr\vert \\ &\leq\frac{\max_{t\in[0,T]} \vert Ax'(t) \vert }{|1-|c||} \\ &=\frac{|(Ax)'(t)|}{|1-|c||}, \end{aligned} $$

(3.5)

since $(Ax)'(t)=(x(t)-cx(t-\delta))'=x'(t)-cx'(t-\delta)=(Ax')(t)$ (see [15, 16]).

On the other hand, from $\int^{T}_{0}(Ax)'(t)\,dt=0$, there exists a point $t_{2}\in(0,T)$ such that $(Ax)'(t_{2})=0$, which together with the integration of (3.1) on interval $[0,T]$ yields

$$ { } \begin{aligned}[b] 2\big|(Ax)'(t)\big|\leq{}&2 \biggl((Ax)'(t_{2})+\frac{1}{2} \int^{T}_{0}\big|(Ax)''(t)\big|\,dt \biggr) \\ \leq{}&\lambda \int^{T}_{0}\bigl|-f\bigl(t,x'(t)\bigr)-g \bigl(t,x(t)\bigr)+e(t)\bigr|\,dt \\ \leq{}& \int^{T}_{0}\big|f\bigl(t,x'(t)\bigr)\big|\,dt+ \int^{T}_{0}\big|g\bigl(t,x(t)\bigr)\big|\,dt+ \int^{T}_{0}\big|e(t)\big|dt. \end{aligned} $$

(3.6)

Write

$$I_{+}=\bigl\{ t\in[0,T]:g\bigl(t,x(t)\bigr)\geq0\bigr\} ;\qquad I_{-}=\bigl\{ t\in[0,T]:g \bigl(t,x(t)\bigr)\leq0\bigr\} . $$

Then we get from $(H_{1})$, $(H_{6})$ and (3.3)

$$ { } \begin{aligned}[b] \int^{T}_{0}\big|g\bigl(t,x(t)\bigr)\big|\,dt&= \int_{I_{+}}g\bigl(t,x(t)\bigr)\,dt- \int_{I_{-}}g\bigl(t,x(t)\bigr)\,dt \\ &=2 \int_{I_{+}}g\bigl(t,x(t)\bigr)\,dt+ \int^{T}_{0}f\bigl(t,x'(t)\bigr)\,dt \\ &\leq2 \int^{T}_{0}\bigl(mx(t)+n\bigr)\,dt+ \int^{T}_{0}\big|f\bigl(t,x'(t)\bigr)\big|\,dt \\ &\leq2m \int^{T}_{0}\big|x(t)\big|\,dt+2nT+KT. \end{aligned} $$

(3.7)

Substituting (3.7) into (3.6), and from $(H_{1})$, we have

$$ { } \begin{aligned}[b] 2\big|(Ax)'(t)\big|\leq{}&2m \int^{T}_{0}\big|x(t)\big|\,dt+2nT+2KT+\|e\|T \\ \leq{}&2mT^{\frac{1}{2}} \biggl( \int^{T}_{0}\big|x(t)\big|\,dt \biggr)^{\frac{1}{2}}+N_{1}, \end{aligned} $$

(3.8)

where $N_{1}=2T(n+K)+\|e\| T$. In view of an inequality (found in [17], Lemma 2.3) and (3.1), we have

$$ { } \biggl( \int^{T}_{0}\big|x(t)\big|^{2}\,dt \biggr)^{\frac{1}{2}}\leq \biggl(\frac{T}{\pi } \biggr) \biggl( \int^{T}_{0}\big|x'(t)\big|^{p}dt \biggr)^{\frac{1}{2}}+DT^{\frac{1}{2}}. $$

(3.9)

Substituting (3.9) into (3.8), we have

$$ { } 2\big|(Ax)'(t)\big|\leq2mT^{\frac{1}{2}} \biggl( \biggl( \frac{T}{\pi} \biggr) \biggl( \int^{T}_{0}\big|x'(t)\big|^{p}dt \biggr)^{\frac {1}{2}}+DT^{\frac{1}{2}} \biggr)+N_{1}. $$

(3.10)

Substituting (3.10) into (3.5), we have

$$ \begin{aligned} \big\| x'\big\| \leq{}& \frac{mT^{\frac{1}{2}} ( (\frac{T}{\pi} ) (\int ^{T}_{0}|x'(t)|^{2}\,dt )^{\frac{1}{2}}+DT^{\frac{1}{2}} )+\frac {N_{1}}{2}}{|1-|c||} \\ \leq{}&\frac{mT (\frac{T}{\pi} )\|x'\|+mTD+\frac{N_{1}}{2}}{|1-|c||}. \end{aligned} $$

Since $\frac{mT^{2}}{\pi|1-|c||}<1$, it is easy to see that there exists a positive constant $M_{2}$ such that

$$ { } \big\| x'\big\| \leq M_{2}. $$

(3.11)

Substituting (3.11) into (3.4), we have

$$ { } \|x\|\leq D+\frac{1}{2} \int^{T}_{0}\big|x'(t)\big|\,dt\leq D+ \frac{1}{2}TM_{2}:=M_{1}. $$

(3.12)

Next, it follows from (3.1) that

$$ { } (Ax)''(t)+\lambda f \bigl(t,x'(t)\bigr)+\lambda\bigl(g_{0}\bigl(x(t) \bigr)+g_{1}\bigl(t,x(t)\bigr)\bigr)=\lambda e(t). $$

(3.13)

Multiplying both sides of (3.13) by $x'(t)$, we get

$$ { } (Ax)''(t)x'(t)+\lambda f \bigl(t,x'(t)\bigr)x'(t)+\lambda g_{0} \bigl(x(t)\bigr)x'(t)+\lambda g_{1}\bigl(t,x(t) \bigr)x'(t)=\lambda e(t)x'(t). $$

(3.14)

Let $\tau\in[0,T]$, for any $\tau\leq t\leq T$, we integrate (3.14) on $[\tau, t]$ and get

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

(3.15)

By (3.1), (3.7), (3.12) and $(H_{1})$, we have

$$\begin{aligned} \biggl\vert \int^{t}_{\tau}(Ax)''(s)x'(s)\,ds \biggr\vert \leq{}& \int^{t}_{\tau }\big|(Ax)''(s)\big|\big|x'(s)\big|\,ds \\ \leq{}&\big\| x'\big\| \int^{T}_{0}\big|(Ax)''(s)\big|\,ds \\ \leq{}&\lambda M_{2} \biggl( \int^{T}_{0}\big|f\bigl(t,x'(s)\bigr)\big|\,ds+ \int^{T}_{0}\big|g\bigl(s,x(s)\bigr)\big|\,ds+ \int ^{T}_{0}\big|e(s)\big|\,ds \biggr) \\ \leq{}&\lambda M_{2} \bigl( 2mTM_{1}+2nT +2KT+T\|e\| \bigr). \end{aligned}$$

We have

$$\begin{gathered} \biggl\vert \int^{t}_{\tau}f\bigl(t,x(s)\bigr)x'(s)\,ds \biggr\vert \leq\big\| x'\big\| \int ^{T}_{0}\big|f\bigl(t,x(s)\bigr)\big|\,ds\leq M_{2} KT. \\ \biggl\vert \int^{t}_{\tau}g_{1}\bigl(s,x(s) \bigr)x'(s)\,ds \biggr\vert \leq\big\| x'\big\| \int ^{T}_{0}\big|g_{1}\bigl(t,x(t)\bigr)\big|\,dt\leq M_{2}\sqrt{T}|g_{M_{1}}|_{2}, \end{gathered} $$

where $g_{M_{1}}=\max_{0\leq x\leq M_{1}}|g_{1}(t,x)|\in L^{2}(0,T)$.

$$\biggl\vert \int^{t}_{\tau}e(s)x'(s)\,dt \biggr\vert \leq M_{2}T\|e\|. $$

From these inequalities we can derive from (3.15) that

$$ \biggl\vert \int^{x(t)}_{x(\tau)}g_{0}(u)\,du \biggr\vert \leq M_{2} \bigl(2mTM_{1}+2nT+\sqrt{T}|g_{M_{1}}|_{2}+3KT+2T \|e\| \bigr). $$

In view of the strong force condition (1.2), we know that there exists a constant $M_{3}>0$ such that

$$ { } x(t)\geq M_{3}, \quad\forall t\in[\tau,T]. $$

(3.16)

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

From (3.11), (3.12) and (3.16), we let

$$\Omega=\bigl\{ x: E_{1}\leq x(t)\leq E_{2}, \big\| x' \big\| \leq E_{3}, \forall t\in[0,T]\bigr\} , $$

where $0< E_{1}< M_{3}$, $E_{2}>\max\{M_{1}, D\}$, $E_{3}>M_{2}$. Then condition (1) of Lemma 2.1 is satisfied. If $x\in\partial\Omega\cap\operatorname{Ker} L$, then $x(t)=E_{1}$ or $(E_{2})$. In this case

$$QNx= \frac{1}{T} \int^{T}_{0}g(t,E_{1})\,dt:=- \bar{g}(E_{1}), $$

or

$$QNx= \frac{1}{T} \int^{T}_{0}g(t,E_{2})\,dt:=- \bar{g}(E_{2}), $$

since $f(t,0)=0$. According to the condition $(H_{4})$, we get $QNx\neq0$, which implies $Nx\neq\operatorname{Im} L$ for $x\in\partial \Omega\cap\operatorname{Ker} L$. Hence, condition (2) of Lemma 2.1 holds. To check condition (3) of Lemma 2.1, we define an isomorphism $J: \operatorname{Im} Q\to\operatorname{Ker} L=R$, $J(u)=u$. It is noted that if $x\in \Omega\cap\operatorname{Ker} L$, then $x(t)=c$ with $E_{1}< c< E_{2}$,

$$JQNx=- \int^{T}_{0}g(t,c)\,dt. $$

From $(H_{4})$, we can derive

$$\deg(JQN,\Omega\cap\operatorname{Ker} L,0)=-1. $$

So condition (3) of Lemma 2.1 is satisfied. By applying Lemma 2.1, we conclude that the equation $Lx=Nx$ has a solution x on $\bar{\Omega}\cap D(L)$, i.e., (1.1) has at least one positive T-periodic solution $x(t)$. □

Theorem 3.2

Assume that conditions $(H_{2})$, $(H_{3})$, $(H_{5})$-$(H_{6})$ hold. Then (1.1) has at least one positive solution with period T if $\frac{mT (\frac{T}{\pi} )+\alpha T}{|1-|c||}<1$.

Proof

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

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

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

(3.17)

In view of $\int^{T}_{0}(Ax)'(t)\,dt=0$, we know that there exist two constants $t_{3},t_{4}\in[0,\omega]$ such that $(Ax)'(t_{3})\geq 0$, $(Ax)'(t_{4})\leq0$. Let $\xi^{*}\in(0,T)$ be a global maximum point of $(Ax)'(t)$. Clearly, we have

$$(Ax)'\bigl(\xi^{*}\bigr)\geq0, \qquad(Ax)''\bigl( \xi^{*}\bigr)=0. $$

From $(H_{3})$, we know $f(\xi^{*},x'(\xi^{*}))\geq0$. Therefore, we see that

$$g\bigl(\xi^{*},x\bigl(\xi^{*}\bigr)\bigr)-e\bigl(\xi^{*}\bigr)=-f\bigl( \xi^{*},x'\bigl(\xi^{*}\bigr)\bigr)\leq0, $$

i.e.

$$g\bigl(\xi^{*},x'\bigl(\xi^{*}\bigr)\bigr)\leq e\bigl(\xi^{*}\bigr)\leq \|e\|. $$

From $(H_{5})$, we have

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

Since $x(t)>0$, hence, we can get $0< x(\xi^{*})\leq D_{1}$. This proves (3.17).

Similarly, from (3.4), we have

$$ { } \big|x(t)\big|\leq D_{1}+\frac{1}{2} \int^{T}_{0}\big|x'(t)\big|\,dt. $$

(3.18)

From (3.7), $(H_{2})$ and $(H_{6})$, we have

$$ { } \begin{aligned}[b] \int^{T}_{0}\big|g\bigl(t,x(t)\bigr)\big|\,dt&= \int_{I_{+}}g\bigl(t,x(t)\bigr)\,dt- \int_{I_{-}}g\bigl(t,x(t)\bigr)\,dt \\ &=2 \int_{I_{+}}g\bigl(t,x(t)\bigr)\,dt+ \int^{T}_{0}f\bigl(t,x'(t)\bigr)\,dt \\ &\leq2 \int_{I_{+}}\bigl(mx(t)+n\bigr)\,dt+ \int^{T}_{0}\big|f\bigl(t,x'(t)\bigr)\big|\,dt \\ &\leq2m \int^{T}_{0}\big|x(t)\big|\,dt+2nT+\alpha \int^{T}_{0}\big|x'(t)\big|\,dt+\beta T. \end{aligned} $$

(3.19)

Substituting (3.19) into (3.6), and from $(H_{2})$, we have

$$ { } \begin{aligned}[b] 2\big|(Ax)'(t)\big|\leq{}&2m \int^{T}_{0}\big|x(t)\big|\,dt+2nT+2\alpha \int^{T}_{0}\big|x'(t)\big|\,dt+ 2\beta T+\|e\| T \\ \leq{}&2mT^{\frac{1}{2}} \biggl( \int^{T}_{0}\big|x(t)\big|^{2}\,dt \biggr)^{\frac {1}{2}}+2\alpha \big\| x'\big\| T+N_{2}, \end{aligned} $$

(3.20)

where $N_{2}=2T(n+\beta)+\|e\| T$. Substituting (3.9) into (3.20), we have

$$ { } \begin{aligned}[b] 2\big|(Ax)'(t)\big| \leq{}&2mT^{\frac{1}{2}} \biggl( \biggl(\frac{T}{\pi} \biggr) \biggl( \int^{T}_{0}\big|x'(t)\big|^{2}\,dt \biggr)^{\frac{1}{2}}+D_{1} T^{\frac{1}{2}} \biggr)+2\alpha \big\| x'\big\| T+N_{2} \\ \leq{}&2mT \biggl(\frac{T}{\pi} \biggr)\big\| x'\big\| +2\alpha \big\| x'\big\| T+2m D_{1} T+N_{2} \\ ={}& \biggl(2mT \biggl(\frac{T}{\pi} \biggr)+2\alpha T \biggr) \big\| x'\big\| +2m D_{1}T+N_{2}. \end{aligned} $$

(3.21)

Similarly, for $|c|\neq1$, we can get

$$\big\| x'\big\| \leq\frac{ (mT (\frac{T}{\pi} )+\alpha T )\|x'\|}{|1-|c||}+\frac{mD_{1} T+\frac{N_{2}}{2}}{|1-|c||}. $$

Since $\frac{mT (\frac{T}{\pi} )+\alpha T}{|1-|c||}<1$, it is easy to see that there exists a positive constant $M_{2}$ such that

$$ \big\| x'\big\| \leq M_{2}. $$

The proof left is as same as Theorem 3.1. □

We illustrate our results with some examples.

Example 3.1

Consider the following neutral Rayleigh equation with singularity:

$$ { } \biggl(x(t)-\frac{1}{10}x(t- \delta) \biggr)''+\cos^{2}(2t)\sin x'(t)+\frac{1}{6\pi}\bigl(\sin(4t)+5\bigr)x(t)- \frac{1}{u^{\mu}}=\cos^{2}(2t), $$

(3.22)

where $\mu\geq1$ and δ is a constant.

It is clear that $T=\frac{\pi}{2}$, $c=\frac{1}{10}$, $e(t)=\cos^{2}(2t)$, $f(t,u)=\cos^{2}(2t)\sin u$, $g(t,x)=\frac{1}{6\pi} {(\sin (4t)+5)x(t)-\frac{1}{x^{\mu}(t)}}$. Choose $K=1$, $D=2$, $m=\frac{1}{\pi}$, it is obvious that $(H_{1})$, $(H_{4})$ and $(H_{6})$ hold. Next, we consider

$$\begin{aligned} \frac{mT^{2}}{\pi|1-|c||} &=\frac{\frac{1}{\pi}\times (\frac{\pi}{2} )^{2}}{\pi \vert 1-\frac{1}{10} \vert } \\ &=\frac{5}{18}< 1. \end{aligned}$$

Therefore, by Theorem 3.1, (3.22) has at least one $\frac{\pi}{2}$-periodic solution.

Example 3.2

Consider the following a kind of neutral Rayleigh equation:

$$ { } \bigl(x(t)-100x (t-\delta ) \bigr)'' +\frac{1}{5\pi} \bigl(\sin^{2} t+4 \bigr)x'(t)+\bigl(\cos^{2} t+4\bigr) x(t)- \frac{1}{x^{\mu}}=\sin(2t), $$

(3.23)

where $\mu\geq1$ and δ is a constant.

It is clear that $T=\pi$, $c=100$, $e(t)=\sin(2t)$, $f(t,u)=\frac{1}{5\pi} (\sin^{2} t+4 )u(t)$, $g(t,x)= (\cos^{2} t+4) x(t)-\frac{1}{x^{\mu}(t)}$. Choose $m=5$, $D_{1}=3$, $a=\frac{1}{\pi}$, it is obvious that $(H_{1})$, $(H_{2})$, $(H_{5})$ and $(H_{6})$ hold. Next, we consider

$$\begin{aligned} \frac{m T (\frac{T}{\pi} )+\alpha T}{|1-|c||} &=\frac{5\times \pi (\frac{\pi}{\pi} )+\frac{1}{\pi}\times\pi}{ 1-100} \\ &= \frac{5\pi+1}{99}< 1. \end{aligned}$$

So, (3.23) has at least one nonconstant π-periodic solution by application of Theorem 3.2.