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. (1)

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

  2. (2)

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

  3. (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.