Open Access

Study on a kind of neutral Rayleigh equation with singularity

Boundary Value Problems20172017:92

https://doi.org/10.1186/s13661-017-0824-7

Received: 29 December 2016

Accepted: 8 June 2017

Published: 19 June 2017

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.

Keywords

periodic solutionneutral operatorRayleigh equationsingularity

MSC

34C2534K1334K40

1 Introduction

More recently, some classical tools have been used to study periodic solution for Rayleigh equation in the literature, including coincidence degree theory [14], the method of upper and lower solutions [5], the Manásevich-Mawhin continuation theorem [68], and the time map continuation theorem [911].

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.

Declarations

Acknowledgements

YX and ZBC would like to thank the referee for invaluable comments and insightful suggestions. This work was supported by NSFC Project (No. 11501170), China Postdoctoral Science Foundation funded project (2016M590886), Education Department of Henan Province project (No. 16B110006), Fundamental Research Funds for the Universities of Henan Provience (NSFRF140142), Henan Polytechnic University Outstanding Youth Fund (J2015-02) and Henan Polytechnic University Doctor Fund (B2013-055).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
College of Computer Science and Technology, Henan Polytechnic University
(2)
School of Mathematics and Information Science, Henan Polytechnic University
(3)
Department of Mathematics, Sichuan University

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