Existence of periodic solution for generalized neutral Rayleigh equation with variable parameter

Abstract

In this paper, we consider a generalized neutral Rayleigh equation with variable parameter

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

where $|c(t)|\neq1$, $c,~\delta\in C^{1}(\mathbb{R},\mathbb{R})$ and c, δ are ω-periodic functions for some $\omega> 0$. By applications of coincidence degree theory and some analysis skills, sufficient conditions for the existence of periodic solution is established.

1 Introduction

In the paper, we consider the generalized neutral Rayleigh differential equation with variable parameter

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

(1.1)

where $|c(t)|\neq1$, $c, \delta\in C^{1}(\mathbb{R},\mathbb{R})$ and c, δ are ω-periodic functions for some $\omega> 0$, $\tau, e \in C[0,\omega]$ and $\int^{\omega}_{0}e(t)\,dt=0$; f and g are continuous functions defined on $\mathbb{R}^{2}$ and periodic in t with $f(t,\cdot)=f(t+\omega,\cdot)$, $g(t,\cdot)=g(t+\omega,\cdot)$, and $f(t,0)=0$.

Neutral differential equations manifest themselves in many fields including biology, mechanics and economics [1–4]. For example, in population dynamics, since a growing population consumes more (or less) food than a matured one, depending on individual species, this leads to neutral equations [2]. These equations also arise in classical ‘cobweb’ models in economics where current demand depends on price, but supply depends on the previous periodic [4]. The study on neutral differential equations is more intricate than that on ordinary delay differential equations. In recent years, there is a good amount of work on periodic solutions for neutral differential equations (see [5–14] and the references cited therein). For example, in [5], Lu et al. considered the following neutral differential equation with deviating arguments:

$$\bigl(x(t)-cx(t-\delta)\bigr)''+f\bigl(x'(t) \bigr)+f\bigl(x\bigl(t-\tau(t)\bigr)\bigr)=p(t). $$

By using the continuation theorem and some analysis techniques, some new results on the existence of periodic solutions are obtained. Afterwards, Du et al. [10] investigated the second order neutral equation

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

(1.2)

by using Mawhin’s continuous theorem, the authors obtained the existence of periodic solution to (1.2). Recently, Ren et al. [12] studied the neutral equation with variable delay

$$ { } \bigl(x(t)-cx\bigl(t-\delta(t)\bigr)\bigr)''=-a(t)x(t)+ \lambda b(t)f\bigl(x\bigl(t-\tau(t)\bigr)\bigr), $$

(1.3)

by an application of the fixed-point index theorem, the authors obtained sufficient conditions for the existence, multiplicity and nonexistence of positive periodic solutions to (1.3).

Motivated by [5, 10, 12], in this paper, we consider the generalized neutral Rayleigh equation (1.1). Notice that here the neutral operator A is a natural generalization of the familiar operator $A_{1}=x(t)-cx(t-\delta)$, $A_{2}=x(t)-c(t)x(t-\delta)$, $A_{3}=x(t)-cx(t-\delta(t))$. But A possesses a more complicated nonlinearity than $A_{i}$, $i=1,2,3$. For example, the neutral operator $A_{1}$ is homogeneous in the following sense $(A_{1} x)'(t)=(A_{1} x')(t)$, whereas the neutral operator A in general is inhomogeneous. As a consequence, many of the new results for differential equations with the neutral operator A will not be a direct extension of known theorems for neutral differential equations.

The paper is organized as follows. In Section 2, we first analyze qualitative properties of the generalized neutral operator A which will be helpful for further studies of differential equations with this neutral operator; in Section 3, by Mawhin’s continuation theorem, we obtain the existence of periodic solutions for the generalized neutral Rayleigh equation with variable parameter. We will give an example to illustrate our results, and an example is also given in this section. Our results improve and extend the results in [5, 10, 12–14].

2 Analysis of the generalized neutral operator with variable parameter

Let

$$c_{\infty}=\max_{t\in[0,\omega]}\bigl|c(t)\bigr|,\qquad c_{0}=\min _{t\in [0,\omega]}\bigl|c(t)\bigr|. $$

Let $X=\{x \in C(\mathbb{R},\mathbb{R}): x(t+\omega)=x(t), t\in \mathbb{R}\}$ with norm $\|x\|= \max_{t\in[0,\omega]} |x(t)|$. Then $(X, \|\cdot\|)$ is a Banach space. Moreover, define operators $A, B: C_{\omega}\rightarrow C_{\omega}$ by

$$(Ax) (t)=x(t)-c(t)x\bigl(t-\delta(t)\bigr), \qquad (Bx) (t)=c(t)x\bigl(t- \delta(t)\bigr). $$

Lemma 2.1

If $|c(t)|\neq1$, then the operator A has a continuous inverse $A^{-1}$ on $C_{\omega}$, satisfying

• (1)
  $$\bigl(A^{-1}f \bigr) (t)= \textstyle\begin{cases} f(t)+\sum^{\infty}_{j=1}\prod^{j}_{i=1}c(D_{i}) x (t-\sum^{j}_{i=1}\delta(D_{i}) ) & \textit{for } |c(t)|< 1, \forall f\in C_{\omega},\\ -\frac{f(t+\delta(t))}{c(t+\delta(t))}-\sum^{\infty}_{j=1}\frac {f (t+\delta(t)+\sum^{j}_{i=1} \delta(D_{i}') )}{c(t+\delta(t))\prod^{j}_{i=1}c(D_{i}')} & \textit{for } |c(t)|>1, \forall f\in C_{\omega}. \end{cases} $$
• (2)
  $$\bigl\vert \bigl(A^{-1}f \bigr) (t)\bigr\vert \leq \textstyle\begin{cases} \frac{\|f\|}{1-c_{\infty}} &\textit{for } c_{\infty}< 1 \forall f\in C_{\omega}, \\ \frac{\|f\|}{c_{0}-1} & \textit{for } c_{0}>1 \forall f\in C_{\omega}. \end{cases} $$
• (3)
  $$\int^{\omega}_{0}\bigl\vert \bigl(A^{-1}f \bigr) (t)\bigr\vert \,dt\leq \textstyle\begin{cases}\frac{1}{1-c_{\infty}}\int^{\omega}_{0}|f(t)|\,dt &\textit{for } c_{\infty}< 1 \forall f\in C_{\omega},\\ \frac{1}{c_{0}-1}\int^{\omega}_{0}|f(t)|\,dt &\textit{for } c_{0}>1 \forall f\in C_{\omega}, \end{cases} $$

where $D_{1}=t$, $D_{i}=t-\sum_{k=1}^{i} \delta(D_{k})$, $k=1,2,\ldots$ , and $D_{1}'=t$, $D_{i}'=t+\sum_{k=1}^{i} \delta(D_{k}')$, $k=1,2,\ldots$ .

Proof

Case 1: $|c(t)|\leq c_{\infty}<1$.

Let $t=D_{1}$ and $D_{j}=t-\sum^{j}_{i=1}\delta(D_{i})$, $j=1,2,\ldots$ .

$$\begin{aligned}& (Bx) (t)=c(t)x\bigl(t-\delta(t)\bigr)=c(D_{1})x\bigl(t- \delta(D_{1})\bigr); \\& \bigl(B^{2}x\bigr) (t)=c(t)c\bigl(t-\delta(t) \bigr)x\bigl(t-\delta(t)-\delta\bigl(t-\delta(t)\bigr)\bigr) =c(D_{1})c(D_{2}))x\bigl(t-\delta(D_{1})- \delta(D_{2})\bigr); \\& \begin{aligned}[b] \bigl(B^{3}x\bigr) (t)&=c(t)c\bigl(t-\delta(t) \bigr)c\bigl(t-\delta(t)-\delta\bigl(t-\delta (t)\bigr)\bigr)x\bigl(t- \delta(D_{1})-\delta(D_{2})-\delta(D_{3})\bigr) \\ &=c(D_{1})c(D_{2})c(D_{3})x \Biggl(t-\sum ^{3}_{i=1}\delta(D_{i}) \Biggr). \end{aligned} \end{aligned}$$

Therefore

$$B^{j}x(t)=\prod^{j}_{i=1}c(D_{i})x \Biggl(t-\sum^{j}_{i=1} \delta(D_{i}) \Biggr), $$

and

$$\sum^{\infty}_{j=0} \bigl(B^{j}f \bigr) (t)=f(t)+\sum^{\infty}_{j=1}\prod ^{j}_{i=1}c(D_{i})x \Biggl(t-\sum ^{j}_{i=1}\delta(D_{i}) \Biggr). $$

Since $A=I-B$, we get from $\|B\|\leq c_{\infty}<1$ that A has a continuous inverse $A^{-1}: C_{\omega}\rightarrow C_{\omega}$ with

$$A^{-1}=(I-B)^{-1}=I+\sum^{\infty}_{j=1}B^{j}= \sum^{\infty}_{j=0}B^{j}, $$

here $B^{0}=I$. Then

$$\bigl(A^{-1}f(t) \bigr)=\sum^{\infty}_{j=0} \bigl[B^{j}f \bigr](t)=f(t)+\sum^{\infty}_{j=1} \prod^{j}_{i=1}c(D_{i})x \Biggl(t- \sum^{j}_{i=1}\delta(D_{i}) \Biggr), $$

and consequently

$$\begin{aligned} \bigl\vert \bigl(A^{-1}f \bigr) (t)\bigr\vert &=\Biggl\vert \sum ^{\infty}_{j=0} \bigl[B^{j}f \bigr](t) \Biggr\vert \\ &=\Biggl\vert f(t)+\sum^{\infty}_{j=1}\prod ^{j}_{i=1}c(D_{i})x \Biggl(t-\sum ^{j}_{i=1}\delta(D_{i}) \Biggr) \Biggr\vert \\ &\leq \Biggl(1+\sum^{\infty}_{j=1}c_{\infty}^{j} \Biggr)|f|_{\infty}\\ &\leq\frac{|f|_{\infty}}{1-c_{\infty}}. \end{aligned}$$

Moreover,

$$\begin{aligned} \int^{\omega}_{0}\bigl\vert \bigl(A^{-1}f \bigr) (t)\bigr\vert \,dt &=\int^{\omega}_{0}\Biggl\vert \sum_{j=0}^{\infty}\bigl(B^{j}f \bigr) (t)\Biggr\vert \,dt \\ &\leq\sum_{j=0}^{\infty}\int ^{\omega}_{0}\bigl\vert \bigl(B^{j}f \bigr) (t)\bigr\vert \,dt \\ &=\sum_{j=0}^{\infty}\int^{\omega}_{0} \Biggl\vert \prod^{j}_{i=1}c(D_{i})x \Biggl(t-\sum^{j}_{i=1} \delta(D_{i}) \Biggr)\Biggr\vert \,dt \\ &\leq\frac{1}{1-c_{\infty}}\int^{\omega}_{0}\bigl|f(t)\bigr|\,dt. \end{aligned}$$

Case 2: $|c(t)|>c_{0}>1$.

Let $D_{1}'=t$, $D_{j}'=t+\sum^{j}_{i=1}\delta(D_{i}')$, $j=1,2,\ldots$ . And set

$$\begin{aligned}& E: C_{\omega}\rightarrow C_{\omega}, \quad (Ex) (t)=x(t)- \frac{1}{c(t)}x\bigl(t+\delta(t)\bigr), \\& B_{1}: C_{\omega}\rightarrow C_{\omega}, \quad (B_{1}x) (t)=\frac{1}{c(t)}x\bigl(t+\delta(t)\bigr). \end{aligned}$$

By definition of the linear operator $B_{1}$, we have

$$\bigl(B_{1}^{j}f \bigr) (t)=\frac{1}{\prod^{j}_{i=1}c(D_{i}')}f \Biggl(t+ \sum^{j}_{i=1}\delta\bigl(D_{i}' \bigr) \Biggr), $$

here $D_{i}$ is defined as in Case 1. Summing over j yields

$$\sum^{\infty}_{j=0} \bigl(B_{1}^{j}f \bigr) (t)=f(t)+\sum^{\infty}_{j=1} \frac{1}{ \prod^{j}_{i=1}c(D_{i}')}f \Biggl(t+\sum^{j}_{i=1} \delta\bigl(D_{i}'\bigr) \Biggr). $$

Since $\|B_{1}\|<1$, we obtain that the operator E has a bounded inverse $E^{-1}$,

$$E^{-1}: C_{\omega}\rightarrow C_{\omega}, \quad E^{-1}=(I-B_{1})^{-1}=I+\sum ^{\infty}_{j=1}B_{1}^{j}, $$

and $\forall f\in C_{\omega}$ we get

$$\bigl(E^{-1}f \bigr) (t)=f(t)+\sum^{\infty}_{j=1} \bigl(B_{1}^{j}f \bigr) (t). $$

On the other hand, from $(Ax)(t)=x(t)-c(t)x(t-\delta(t))$, we have

$$(Ax) (t)=x(t)-c(t)x\bigl(t-\delta(t)\bigr)=-c(t) \biggl[x\bigl(t-\delta(t) \bigr)-\frac {1}{c(t)}x(t) \biggr], $$

i.e.,

$$(Ax) (t)=-c(t) (Ex) \bigl(t-\delta(t)\bigr). $$

Let $f\in C_{\omega}$ be arbitrary. We are looking for x such that

$$(Ax) (t)=f(t), $$

i.e.,

$$-c(t) (Ex) \bigl(t-\delta(t)\bigr)=f(t). $$

Therefore

$$(Ex) (t)=-\frac{f(t+\delta(t))}{c(t+\delta(t))}=:f_{1}(t), $$

and hence

$$x(t)=\bigl(E^{-1}f_{1}\bigr) (t)=f_{1}(t)+\sum ^{\infty}_{j=1} \bigl(B_{1}^{j}f_{1} \bigr) (t) =-\frac{f(t+\delta(t))}{c(t+\delta(t))}-\sum^{\infty}_{j=1}B_{1}^{j} \frac {f(t+\delta(t))}{c(t+\delta(t))}, $$

proving that $A^{-1}$ exists and satisfies

$$\begin{aligned} \bigl[A^{-1}f \bigr](t)&=-\frac{f(t+\delta(t))}{c(t+\delta(t))}-\sum ^{\infty}_{j=1}B_{1}^{j} \frac{f(t+\delta(t))}{c(t+\delta(t))} \\ &=-\frac{f(t+\delta(t))}{c(t+\delta(t))}-\sum^{\infty}_{j=1} \frac {f (t+\delta(t)+\sum^{j}_{i=1}\delta(D_{i}') )}{ c(t+\delta(t))\prod^{j}_{i=1}c(D_{i}')} \end{aligned}$$

and

$$\bigl\vert \bigl[A^{-1}f \bigr](t)\bigr\vert =\Biggl\vert - \frac{f(t+\delta (t))}{c(t+\delta(t))}-\sum^{\infty}_{j=1} \frac{f (t+\delta(t)+\sum^{j}_{i=1}\delta(D_{i}') )}{ c(t+\delta(t))\prod^{j}_{i=1}c(D_{i}')}\Biggr\vert \leq\frac{\|f\|}{c_{0}-1}. $$

Statements (1) and (2) are proved. From the above proof, (3) can easily be deduced. □

3 Periodic solution for (1.1)

We first recall Mawhin’s continuation theorem which our study is based upon. Let X and Y be real Banach spaces and $L:D(L)\subset X\rightarrow Y$ be a Fredholm operator with index zero, here $D(L)$ denotes the domain of L. This means that ImL is closed in Y and $\dim \operatorname{Ker} L=\dim(Y/\operatorname{Im} L)<+\infty$. Consider supplementary subspaces $X_{1}$, $Y_{1}$, of X, Y, respectively, such that $X=\operatorname{Ker} L \oplus X_{1}$, $Y=\operatorname{Im} L\oplus Y_{1}$, and let $P_{1}:X\rightarrow \operatorname{Ker} L$ and $Q_{1}:Y\rightarrow Y_{1}$ denote the natural projections. Clearly, $\operatorname{Ker} L\cap(D(L)\cap X_{1})=\{0\}$, thus the restriction $L_{P_{1}}:=L|_{D(L)\cap X_{1}}$ is invertible. Let $L_{P_{1}}^{-1}$ denote the inverse of $L_{P_{1}}$.

Let Ω be an open bounded subset of X with $D(L)\cap\Omega\neq\emptyset$. A map $N:\overline{\Omega}\rightarrow Y$ is said to be L-compact in $\overline{\Omega}$ if $Q_{1}N(\overline{\Omega})$ is bounded and the operator $L_{P_{1}}^{-1}(I-Q_{1})N:\overline{\Omega}\rightarrow X$ is compact.

Lemma 3.1

(Gaines and Mawhin [15])

Suppose that X and Y are two Banach spaces, and $L:D(L)\subset X\rightarrow Y$ is a Fredholm operator with index zero. Furthermore, $\Omega\subset X$ is an open bounded set and $N:\overline{\Omega}\rightarrow Y$ is L-compact on $\overline{\Omega}$. 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) $\deg\{JQ_{1}N,\Omega\cap \operatorname{Ker} L,0\}\neq0$, where $J:\operatorname{Im} Q_{1}\rightarrow \operatorname{Ker} L$ is an isomorphism.

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

In order to use Mawhin’s continuation theorem to study the existence of ω-periodic solutions for (1.1), we rewrite (1.1) in the following form:

$$ { } \left \{ \textstyle\begin{array}{@{}l} (Ax_{1})'(t)=x_{2}(t),\\ x_{2}'(t)=-f(t,x_{1}'(t))-g(t,x_{1}(t-\tau(t)))+e(t). \end{array}\displaystyle \right . $$

(3.1)

Clearly, if $x(t)=(x_{1}(t),x_{2}(t))^{\top}$ is an ω-periodic solution to (3.1), then $x_{1}(t)$ must be an ω-periodic solution to (1.1). Thus, the problem of finding an ω-periodic solution for (1.1) reduces to finding one for (3.1).

Recall that $C_{\omega}=\{\phi\in C(\mathbb{R},\mathbb{R}): \phi(t+\omega)\equiv\phi(t)\}$ with norm $\|\phi\|=\max_{t\in[0,\omega]}|\phi(t)|$. Define $X=Y=C_{\omega}\times C_{\omega}= \{x=(x_{1}(\cdot),x_{2}(\cdot))\in C(\mathbb{R},\mathbb{R}^{2}) : x(t) = x(t+\omega), t \in\mathbb{R}\}$ with norm $\|x\|=\max\{\|x_{1}\|,\|x_{2}\|\}$. Clearly, X and Y are Banach spaces. Moreover, define

$$L:D(L)=\bigl\{ x\in C^{1}\bigl(\mathbb{R},\mathbb{R}^{2} \bigr): x(t+\omega) = x(t), t \in \mathbb{R}\bigr\} \subset X\rightarrow Y $$

by

$$ (Lx) (t)= \begin{pmatrix} (A x_{1})'(t)\\ x_{2}'(t) \end{pmatrix} $$

and $N: X\rightarrow Y$ by

$$ { } (Nx) (t)= \begin{pmatrix} x_{2}(t)\\ -f(t,x_{1}'(t))-g(t,x_{1}(t-\tau(t)))+e(t) \end{pmatrix}. $$

(3.2)

Then (3.1) can be converted to the abstract equation $Lx=Nx$. From the definition of L, one can easily see that

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

So L is a Fredholm operator with index zero. Let $P_{1}:X\rightarrow \operatorname{Ker} L$ and $Q_{1}:Y\rightarrow \operatorname{Im} Q_{1}\subset\mathbb{R}^{2}$ be defined by

$$P_{1}x= \begin{pmatrix} (Ax_{1})(0)\\ x_{2}(0) \end{pmatrix} ; \qquad Q_{1}y= \frac{1}{\omega}\int_{0}^{\omega}\begin{pmatrix} y_{1}(s)\\ y_{2}(s) \end{pmatrix}\,ds, $$

then $\operatorname{Im} P_{1}=\operatorname{Ker} L$, $\operatorname{Ker} Q_{1}=\operatorname{Im} L$. Set $L_{P_{1}}=L|_{D(L)\cap \operatorname{Ker} P_{1}}$ and $L_{P_{1}}^{-1}: \operatorname{Im} L\rightarrow D(L)$ denotes the inverse of $L_{P_{1}}$, then

$$\begin{aligned}& \bigl[L_{P_{1}}^{-1}y\bigr](t)= \begin{pmatrix} (A^{-1}Fy_{1})(t)\\ (Fy_{2})(t) \end{pmatrix}, \\& {[}Fy_{1}](t)=\int^{t}_{0}y_{1}(s)\,ds, \qquad [Fy_{2}](t)=\int^{t}_{0}y_{2}(s)\,ds. \end{aligned}$$

(3.3)

From (3.2) and (3.3), it is clear that $Q_{1}N$ and $L_{P_{1}}^{-1}(I-Q_{1})N$ are continuous, and $Q_{1}N(\overline{\Omega})$ is bounded, and then $L_{P_{1}}^{-1}(I-Q_{1})N(\overline{\Omega})$ is compact for any open bounded $\Omega\subset X$, which means N is L-compact on $\bar{\Omega}$.

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

(H₁):

there exists a positive constant $K_{1}$ such that $|f(t,u)|\leq K_{1}$ for $(t,u)\in \mathbb{R}\times\mathbb{R}$;

(H₂):

there exists a positive constant D such that $x\cdot g(t,x)>0$ and $|g(x,x)|>K_{1}$ for $|x|>D$;

(H₃):

there exists a positive constant M and $M>\|e\|$ such that $g(t,x)\geq-M$ for $x\leq-D$ and $t\in\mathbb{R}$;

(H₄):

there exists a positive constant M and $M>\|e\|$ such that $g(t,x)\leq M$ for $x\geq D$ and $t\in\mathbb{R}$;

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

Theorem 3.1

Assume that conditions (H₁)-(H₃) hold. Suppose that one of the following conditions is satisfied:

• (i) If $c_{\infty}<1$ and $1-c_{\infty}-\delta_{1}c_{\infty}-\frac{1}{2}c_{1}\omega>0$;

• (ii) If $c_{0}>1$ and $c_{0}-1-\delta_{1}c_{\infty}-\frac{1}{2}c_{1}\omega>0$;

where $\delta_{1}=\max_{t\in[0,\omega]}|\delta'(t)|$, $c_{1}=\max_{t\in[0,\omega]}|c'(t)|$.

Then (1.1) has at least one solution with period ω.

Proof

By construction, (3.1) has an ω-periodic solution if and only if the following operator equation

$$Lx=Nx $$

has an ω-periodic solution. From (3.3), we see that N is L-compact on $\bar{\Omega}$, where Ω is any open, bounded subset of $C_{\omega}$. For $\lambda\in(0,1]$, define

$$\Omega_{1}=\{x \in C_{\omega} : Lx=\lambda Nx\}. $$

Then $x = (x_{1},x_{2})^{\top}\in\Omega_{1}$ satisfies

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

(3.4)

Substituting $x_{2}(t)=\frac{1}{\lambda}(Ax_{1})'(t)$ into the second equation of (3.4) yields

$$\biggl(\frac{1}{\lambda}(Ax_{1}) (t) \biggr)''=- \lambda f\bigl(t,x_{1}'(t)\bigr)-\lambda g \bigl(t,x_{1}\bigl(t-\tau(t)\bigr)\bigr)+\lambda e(t), $$

i.e.,

$$ \bigl((Ax_{1}) (t)\bigr)''=- \lambda^{2} f\bigl(t,x_{1}'(t)\bigr)- \lambda^{2} g\bigl(t,x_{1}\bigl(t-\tau(t)\bigr)\bigr)+ \lambda^{2} e(t). $$

(3.5)

We first claim that there is a constant $\xi\in\mathbb{R}$ such that

$$ \bigl|x_{1}(\xi)\bigr|\leq D. $$

(3.6)

Integrating both sides of (3.5) over $[0,\omega]$, we have

$$ \int^{\omega}_{0}\bigl[f \bigl(t,x_{1}'(t)\bigr)+g\bigl(t,x_{1}\bigl(t- \tau(t)\bigr)\bigr)\bigr]\,dt=0, $$

(3.7)

which yields that there at least exists a point $t_{1}$ such that

$$f\bigl(t_{1},x'_{1}(t_{1})\bigr)+g \bigl(t_{1},x_{1}\bigl(t_{1}-\tau(t_{1}) \bigr)\bigr)=0, $$

then by (H₁) we have

$$\bigl|g\bigl(t_{1},x_{1}\bigl(t_{1}- \tau(t_{1})\bigr)\bigr)\bigr|=\bigl|-f\bigl(t_{1},x'_{1}(t_{1}) \bigr)\bigr|\leq K , $$

and in view of (H₂) we get that $|x_{1}(t_{1}-\tau(t_{1}))|\leq D$. Since $x_{1}(t)$ is periodic with periodic ω. So $t_{1}-\tau(t_{1})=n\omega+\xi$, $\xi\in[0,\omega]$, where n is some integer, then $|x_{1}(\xi)|\leq D$. Equation (3.6) is proved.

Then we have

$$\bigl|x(t)\bigr|=\biggl\vert x(\xi)+\int^{t}_{\xi}x'(s)\,ds\biggr\vert \leq D+\int^{t}_{\xi}\bigl|x'(s)\bigr|\,ds, \quad t\in[\xi,\xi+\omega], $$

and

$$\bigl|x(t)\bigr|=\bigl|x(t-\omega)\bigr|=\biggl\vert x(\xi)-\int_{t-\omega}^{\xi}x'(s)\,ds\biggr\vert \leq D +\int_{t-\omega}^{\xi}\bigl|x'(s)\bigr|\,ds, \quad t\in[\xi,\xi+\omega]. $$

Combining the above two inequalities, we obtain

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

(3.8)

On the other hand, multiplying both sides of (3.5) by $(Ax_{1})(t)$ and integrating over $[0,\omega]$, we get

$$\begin{aligned} \int^{\omega}_{0}\bigl((Ax_{1}) (t) \bigr)''\bigl(Ax_{1}(t)\bigr)\,dt ={}&{-}\int ^{\omega }_{0}\bigl|(Ax_{1})'(t)\bigr|^{2}\,dt =-\lambda^{2}\int^{\omega}_{0}f \bigl(t,x_{1}'(t)\bigr) (Ax_{1}) (t)\,dt \\ &{}- \lambda^{2}\int^{\omega}_{0}g \bigl(t,x_{1}\bigl(t-\tau(t)\bigr)\bigr) (Ax_{1}) (t)\,dt\\ &{}+\lambda^{2}\int^{\omega}_{0}e(t) (Ax_{1}) (t)\,dt. \end{aligned}$$

Using (H₁), we have

$$\begin{aligned} \int^{\omega}_{0}\bigl|(Ax_{1})'(t)\bigr|^{2}\,dt \leq{}& \int^{\omega}_{0}\bigl|f\bigl(t,x_{1}'(t) \bigr)\bigr|\bigl\vert \bigl[x_{1}(t)-c(t)x_{1}\bigl(t-\delta(t) \bigr)\bigr]\bigr\vert \,dt \\ &{}+\int^{\omega}_{0}\bigl|g\bigl(t,x_{1}\bigl(t- \tau(t)\bigr)\bigr)\bigr|\bigl|\bigl[x_{1}(t)-c(t)x_{1}\bigl(t-\delta(t) \bigr)\bigr]\bigr|\,dt \\ &{}+\int^{\omega}_{0}\bigl|e(t)\bigr|\bigl| \bigl[x_{1}(t)-c(t)x_{1}\bigl(t-\delta(t)\bigr)\bigr]\bigr|\,dt \\ \leq{}&(1+c_{\infty})\|x_{1}\| \biggl[K_{1}\omega+\int ^{\omega}_{0}\bigl|g\bigl(t,x_{1}\bigl(t-\tau (t) \bigr)\bigr)\bigr|\,dt+\omega\|e\| \biggr]. \end{aligned}$$

(3.9)

Besides, we can assert that there exists some positive constant $N_{1}$ such that

$$ { } \int^{\omega}_{0}\bigl|g \bigl(t,x_{1}\bigl(t-\tau(t)\bigr)\bigr)\bigr|\,dt\leq2\omega N_{1}+K_{1}\omega. $$

(3.10)

In fact, in view of condition (H₁) and (3.7), we have

$$\begin{aligned} &\int_{0}^{\omega}\bigl\{ g\bigl(t,x_{1} \bigl(t-\tau(t)\bigr)\bigr)-K_{1}\bigr\} \,dt\\ &\quad\leq\int_{0}^{\omega } \bigl\{ g\bigl(t,x_{1}\bigl(t-\tau(t)\bigr)\bigr)-\bigl|f \bigl(t,x_{1}'(t)\bigr)\bigr|\bigr\} \,dt \\ &\quad\leq\int_{0}^{\omega}\bigl\{ g\bigl(t,x_{1} \bigl(t-\tau(t)\bigr)\bigr)+f\bigl(t,x_{1}'(t)\bigr)\bigr\} \,dt \\ &\quad=0. \end{aligned}$$

Define

$$\begin{aligned} &E_{1}=\bigl\{ t\in[0,\omega] : x_{1}\bigl(t-\tau(t)\bigr)>D \bigr\} ; \\ &E_{2}=\bigl\{ t\in[0,\omega] : \bigl|x_{1}\bigl(t-\tau(t)\bigr)\bigr| \leq D\bigr\} \cup\bigl\{ t\in[0,\omega] : x_{1}\bigl(t-\tau(t)\bigr)< -D \bigr\} . \end{aligned}$$

With these sets we get

$$\begin{aligned}& \int_{E_{2}}\bigl|g\bigl(t,x_{1}\bigl(t-\tau(t)\bigr) \bigr)\bigr|\,dt\leq\omega\max\Bigl\{ M,\sup_{t\in[0,\omega], |x_{1}(t-\tau(t))|\leq D}\bigl|g(t,x_{1})\bigr| \Bigr\} , \\& \begin{aligned}[b] \int_{E_{1}}\bigl\{ \bigl|g\bigl(t,x_{1} \bigl(t-\tau(t)\bigr)\bigr)\bigr|-K_{1}\bigr\} \,dt&=\int_{E_{1}} \bigl\{ g\bigl(t,x_{1}\bigl(t-\tau(t)\bigr)\bigr)-K_{1}\bigr\} \,dt \\ &\leq-\int_{E_{2}}\bigl\{ g\bigl(t,x_{1}\bigl(t- \tau(t)\bigr)\bigr)-K_{1}\bigr\} \,dt \\ &\leq\int_{E_{2}}\bigl\{ \bigl|g\bigl(t,x_{1}\bigl(t- \tau(t)\bigr)\bigr)\bigr|+K_{1}\bigr\} \,dt, \end{aligned} \end{aligned}$$

which yields

$$\begin{aligned}[b] \int_{E_{1}}\bigl|g\bigl(t,x_{1}\bigl(t-\tau(t)\bigr) \bigr)\bigr|\,dt&\leq\int_{E_{2}}\bigl|g\bigl(t,x_{1}\bigl(t-\tau (t)\bigr)\bigr)\bigr|\,dt+\int_{E_{1} \cup E_{2}}K_{1}\,dt \\ &=\int_{E_{2}}\bigl|g\bigl(t,x_{1}\bigl(t-\tau(t)\bigr) \bigr)\bigr|\,dt+K_{1}\omega. \end{aligned} $$

That is,

$$\begin{aligned} \int_{0}^{\omega}\bigl|g\bigl(t,x_{1}\bigl(t- \tau(t)\bigr)\bigr)\bigr|\,dt&=\int_{E_{1}}\bigl|g\bigl(t,x_{1} \bigl(t-\tau(t)\bigr)\bigr)\bigr|\,dt+\int_{E_{2}}\bigl|g \bigl(x_{1}\bigl(t-\tau(t)\bigr)\bigr)\bigr|\,dt \\ &\leq2\int_{E_{2}}\bigl|g\bigl(t,x_{1}\bigl(t-\tau(t) \bigr)\bigr)\bigr|\,dt+K_{1}\omega \\ &\leq2 \omega\max\Bigl\{ M,\sup_{t\in[0,\omega],|x_{1}(t-\tau (t))|< D}\bigl|g(t,x_{1})\bigr| \Bigr\} +K_{1}\omega \\ &=2\omega N_{1}+K_{1}\omega, \end{aligned}$$

where $N_{1}=\max\{M,\sup_{t\in[0,\omega],|x_{1}(t-\tau (t))|< D}|g(t,x_{1})|\}$, proving (3.10).

Substituting (3.10) into (3.9) and recalling (3.8), we get

$$\begin{aligned} \int^{\omega}_{0}\bigl|(Ax_{1})'(t)\bigr|^{2}\,dt& \leq(1+c_{\infty})\|x_{1}\| \bigl(2K_{1}\omega +2\omega N_{1}+\omega\|e\| \bigr) \\ &\leq(1+c_{\infty}) \bigl(2K_{1}\omega+2\omega N_{1}+\omega \|e\| \bigr) \biggl(D+\frac{1}{2}\int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr) \\ &=(1+c_{\infty})N_{2}D+(1+c_{\infty})\frac{N_{2}}{2} \int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt, \end{aligned}$$

(3.11)

where $N_{2}=2K_{1}\omega+2\omega N_{1}+\omega\|e\|$. Since $(Ax)(t)=x(t)-c(t)x(t-\delta(t))$, we have

$$\begin{aligned} (Ax_{1})'(t)&=\bigl(x_{1}(t)-c(t)x_{1} \bigl(t-\delta(t)\bigr)\bigr)' \\ &=x_{1}'(t)-c'(t)x_{1}\bigl(t- \delta(t)\bigr)-c(t)x_{1}'\bigl(t-\delta(t)\bigr) \bigl(1- \delta'(t)\bigr) \\ &=x_{1}'(t)-c'(t)x_{1}\bigl(t- \delta(t)\bigr)-c(t)x_{1}'\bigl(t-\delta(t) \bigr)+c(t)x_{1}'\bigl(t-\delta (t)\bigr) \delta'(t) \\ &=\bigl(Ax_{1}'\bigr) (t)-c'(t)x_{1} \bigl(t-\delta(t)\bigr)+c(t)x_{1}'\bigl(t-\delta(t)\bigr) \delta'(t), \end{aligned}$$

and

$$\bigl(Ax_{1}'\bigr) (t)=(Ax_{1})'(t)+c'(t)x_{1} \bigl(t-\delta(t)\bigr)-c(t)x_{1}'\bigl(t-\delta(t)\bigr) \delta'(t). $$

Case (i): If $c_{\infty}<1$, by applying Lemma 2.1, we have

$$\begin{aligned} \int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt&= \int^{\omega }_{0}\bigl|\bigl(A^{-1}Ax_{1}' \bigr) (t)\bigr|\,dt \\ &\leq\frac{\int^{\omega}_{0}|(Ax_{1}')(t)|\,dt}{1-c_{\infty}} \\ &=\frac{\int^{\omega}_{0}|(Ax_{1})'(t)+c'(t)x_{1}(t-\delta (t))-c(t)x_{1}'(t-\delta(t))\delta'(t)|\,dt}{1-c_{\infty}} \\ &\leq\frac{\int^{\omega}_{0}|(Ax_{1}')(t)|\,dt+\int^{\omega}_{0}|c'(t)x_{1}(t-\delta (t))|\,dt+\int^{\omega}_{0}|c(t)x_{1}'(t-\delta(t))\delta'(t)|\,dt}{1-c_{\infty}} \\ &\leq\frac{\int^{\omega}_{0}|(Ax_{1}')(t)|\,dt+c_{1}\omega\|x_{1}\|+c_{\infty}\delta _{1}\int^{\omega}_{0}|x_{1}'(t)|\,dt}{1-c_{\infty}} \\ &\leq\frac{\int^{\omega}_{0}|(Ax_{1}')(t)|\,dt+c_{1}\omega D+(\frac {1}{2}c_{1}\omega+c_{\infty}\delta_{1})\int^{\omega}_{0}|x_{1}'(t)|\,dt}{1-c_{\infty}}, \end{aligned}$$

where $c_{1}=\max_{t\in[0,\omega]}|c'(t)|$, $\delta_{1}=\max_{t\in [0,\omega]}|\delta'(t)|$. Since $1-c_{\infty}-\frac{1}{2}c_{1}\omega-c_{\infty}\delta_{1}>0$, so we get

$$\begin{aligned} \int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt \leq\frac{\int^{\omega}_{0}\bigl|(Ax_{1})'(t)\bigr|\,dt+c_{1}\omega D}{1-c_{\infty}-\frac{1}{2}c_{1}\omega-c_{\infty}\delta_{1}} \leq\frac{\omega^{\frac{1}{2}} (\int^{\omega}_{0}\bigl|(Ax_{1})'(t)\bigr|^{2}\,dt )^{\frac{1}{2}}+c_{1}\omega D}{1-c_{\infty}-\frac{1}{2}c_{1}\omega-c_{\infty}\delta_{1}}. \end{aligned}$$

(3.12)

Applying the inequality $(a+b)^{k}\leq a^{k}+b^{k}$ for $a,b>0$, $0< k<1$, it follows from (3.11) and (3.12) that

$$\begin{aligned} \int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt \leq{}&\frac{\omega^{\frac {1}{2}}}{1-c_{\infty}-\frac{1}{2}c_{1}\omega-c_{\infty}\delta_{1}} \biggl[(1+c_{\infty})^{\frac{1}{2}} \biggl( \frac{N_{2}}{2} \biggr)^{\frac{1}{2}} \biggl(\int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{\frac{1}{2}}\\ &{}+(1+c_{\infty})^{\frac {1}{2}}(N_{2}D)^{\frac{1}{2}}+c_{1} \omega^{\frac{1}{2}}D \biggr]. \end{aligned}$$

It is easy to see that there exists a constant $M_{1}>0$ (independent of λ) such that

$$\int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt \leq M_{1}. $$

It follows from (3.8) that

$$\|x_{1}\|\leq D+ \frac{1}{2}\int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt \leq D+\frac{1}{2}M_{1}:=M_{2}. $$

Case (ii): If $c_{0}>1$,we have

$$\begin{aligned} \int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt&= \int^{\omega }_{0}\bigl|\bigl(A^{-1}Ax_{1}' \bigr) (t)\bigr|\,dt \\ &\leq\frac{\int^{\omega}_{0}|(Ax_{1}')(t)|\,dt}{c_{0}-1} \\ &=\frac{\int^{\omega}_{0}|(Ax_{1})'(t)+c'(t)x_{1}(t-\delta (t))-c(t)x_{1}'(t-\delta(t))\delta'(t)|\,dt}{c_{0}-1} \\ &\leq\frac{\int^{\omega}_{0}|(Ax_{1}')(t)|\,dt+\int^{\omega}_{0}|c'(t)x_{1}(t-\delta (t))|\,dt+\int^{\omega}_{0}|c(t)x_{1}'(t-\delta(t))\delta'(t)|\,dt}{c_{0}-1} \\ &\leq\frac{\int^{\omega}_{0}|(Ax_{1}')(t)|\,dt+c_{1}\omega D+(\frac {1}{2}c_{1}\omega+c_{\infty}\delta_{1})\int^{\omega}_{0}|x_{1}'(t)|\,dt}{c_{0}-1}. \end{aligned}$$

Since $c_{0}-1-\frac{1}{2}c_{1}\omega-c_{\infty}\delta_{1}>0$, so we get

$$ \int^{\omega}_{0}\bigl|x_{1}'(t)\bigr|\,dt \leq\frac{\int^{\omega}_{0}|(Ax_{1})'(t)|\,dt+c_{1}\omega D}{c_{0}-1-\frac{1}{2}c_{1}\omega-c_{\infty}\delta_{1}} \leq\frac{\omega^{\frac{1}{2}} (\int^{\omega}_{0}|(Ax_{1})'(t)|^{2}\,dt )^{\frac{1}{2}}+c_{1}\omega D}{c_{0}-1-\frac{1}{2}c_{1}\omega-c_{\infty}\delta_{1}}. $$

(3.13)

Similarly, we can get $\|x_{1}\|\leq M_{2}$.

By the first equation of (3.4) we have $\int^{\omega}_{0}x_{2}(t)\,dt=\int^{\omega}_{0}(Ax_{1})'(t)\,dt=0$, which implies that there is a constant $t_{1}\in[0,\omega]$ such that $x_{2}(t_{1})=0$, hence $\|x_{2}\|\leq\int^{\omega}_{0}|x_{2}'(t)|\,dt$. By the second equation of (3.4) we obtain

$$x_{2}'(t)=-\lambda f\bigl(t,x_{1}'(t) \bigr)-\lambda g\bigl(x_{1}\bigl(t-\tau(t)\bigr)\bigr)+\lambda e(t). $$

So, from (H₁) and (3.10), we have

$$\begin{aligned} \|x_{2}\|&\leq\int^{\omega}_{0}\bigl|f \bigl(t,x_{1}'(t)\bigr)\bigr|\,dt+\int^{\omega }_{0}\bigl|g \bigl(t,x_{1}\bigl(t-\tau(t)\bigr)\bigr)\bigr|\,dt+\int^{\omega}_{0}\bigl|e(t)\bigr|\,dt \\ &\leq2K_{1}\omega+2\omega N_{1}+\omega\|e \|:=M_{3}. \end{aligned}$$

Let $M_{4}=\sqrt{M_{2}^{2}+M_{3}^{2}}+1$, $\Omega=\{x=(x_{1},x_{2})^{\top}: \|x_{1}\|< M_{4}, \|x_{2}\|< M_{4}\}$, then $\forall x\in \partial\Omega\cap \operatorname{Ker} L$

$$ Q_{1}Nx=\frac{1}{\omega}\int^{\omega}_{0} \begin{pmatrix} x_{2}(t)\\ -f(t,x_{1}'(t))-g(t,x_{1}(t-\tau(t)))+e(t) \end{pmatrix}\,dt. $$

If $Q_{1}Nx=0$, then $x_{2}(t)=0$, $x_{1}=M_{4}$ or $-M_{4}$. But if $x_{1}(t)=M_{4}$, we know

$$0=\int^{\omega}_{0}g(t,M_{4})\,dt, $$

there exists a point $t_{2}$ such that $g(t_{2},M_{4})=0$. From assumption (H₂), we know $M_{4}\leq D$, which yields a contradiction. Similarly if $x_{1}=-M_{4}$. We also have $Q_{1}Nx\neq0$, i.e., $\forall x\in\partial\Omega \cap \operatorname{Ker} L$, $x\notin \operatorname{Im} L$, so conditions (1) and (2) of Lemma 3.1 are both satisfied. Define the isomorphism $J:\operatorname{Im} Q_{1}\rightarrow \operatorname{Ker} L$ as follows:

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

Let $H(\mu,x)=\mu x+(1-\mu)JQ_{1}Nx$, $(\mu,x)\in[0,1]\times\Omega$, then $\forall (\mu,x)\in(0,1)\times(\partial\Omega\cap \operatorname{Ker} L)$,

$$ H(\mu,x)= \begin{pmatrix} \mu x_{1}(t)+\frac{1-\mu}{\omega}\int^{\omega }_{0}[f(t,x_{1}'(t))+g(t,x_{1}(t-\tau(t)))-e(t)]\,dt\\ (\mu+(1-\mu))x_{2}(t) \end{pmatrix}. $$

We have $\int^{\omega}_{0}e(t)\,dt=0$. So, we can get

$$\begin{aligned}& H(\mu,x)= \begin{pmatrix}\mu x_{1}(t)+\frac{1-\mu}{\omega}\int^{\omega }_{0}[f(t,x_{1}'(t))+g(t,x_{1}(t-\tau(t)))]\,dt\\ (\mu+(1-\mu))x_{2}(t) \end{pmatrix}, \\& \quad\forall (\mu,x)\in(0,1)\times(\partial\Omega\cap \operatorname{Ker} L). \end{aligned}$$

From (H₂), it is obvious that $x^{\top}H(\mu,x)>0$, $\forall (\mu,x)\in(0,1)\times(\partial\Omega\cap \operatorname{Ker} L)$. Hence

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

So condition (3) of Lemma 3.1 is satisfied. By applying Lemma 3.1, we conclude that equation $Lx=Nx$ has a solution $x=(x_{1},x_{2})^{\top}$ on $\bar{\Omega}\cap D(L)$, i.e., (1.1) has an ω-periodic solution $x_{1}(t)$. □

By using a similar argument, we can obtain the following theorem.

Theorem 3.2

Assume that conditions (H₁), (H₂), (H₄) hold. Suppose one of the following conditions is satisfied:

• (i) If $c_{\infty}<1$ and $1-c_{\infty}-\delta_{1}c_{\infty}-\frac{1}{2}c_{1}\omega>0$;

• (ii) If $c_{0}>1$ and $c_{0}-1-\delta_{1}c_{\infty}-\frac{1}{2}c_{1}\omega>0$.

Then (1.1) has at least one solution with period ω.

Remark 3.1

If $\int_{0}^{\omega}e(t)\,dt\neq0$ and $f(t,0)\neq0$, the problem of existence of ω-periodic solutions to (1.1) can be converted to the existence of ω-periodic solutions to the equation

$$ { } \bigl(x(t)-c(t)x\bigl(t-\delta(t)\bigr)\bigr)''+f_{1} \bigl(t,x'(t)\bigr)+g_{1}\bigl(t,x\bigl(t-\tau(t)\bigr) \bigr)=e_{1}(t), $$

(3.14)

where $f_{1}(t,x)=f(t,x)-f(t,0)$, $g_{1}(t,x)=g(t,x)+\frac{1}{\omega }\int_{0}^{\omega}e(t)\,dt+f(t,0)$ and $e_{1}(t)=e(t)-\frac{1}{\omega}\int_{0}^{\omega}e(t)\,dt$. Clearly, $\int_{0}^{\omega}e_{1}(t)\,dt=0$ and $f_{1}(t,0)=0$, (3.14) can be discussed by using Theorem 3.1 (or Theorem 3.2).

Example 3.1

Consider the following equation:

$$\begin{aligned} &\biggl(x(t)-\frac{1}{150}\sin(16t)x \biggl(t- \frac{1}{160}\sin16t \biggr) \biggr)''+\cos 16t\sin x'(t) \\ &\quad{}+\arctan \biggl(\frac{x(t-\sin16t)}{1+\cos^{2}(8t)} \biggr)=\cos16t. \end{aligned}$$

(3.15)

Comparing (3.15) to (1.1), we have $\omega=\frac{\pi}{8}$, $f(t,u)=\cos16t\sin u$, $g(t,x)=\arctan\frac{x}{1+\cos^{2}(8t)}$, $c(t)=\frac{1}{150}\sin16t$, $\delta(t)=\frac{1}{160}\sin16t$, $\tau(t)=\sin16t$, $e(t)=\cos16t$ and $\delta_{1}=\max_{t\in[0,\frac{\pi}{8}]}|\frac{1}{10} \cos16t|=\frac {1}{10}$, $c_{\infty}=\max_{t\in[0,\frac{\pi}{8}]}\vert \frac{1}{150}\sin 16t\vert =\frac{1}{150}<1$, $c_{1}=\max_{t\in[0,\frac{\pi}{8}]}\vert \frac{16}{150}\cos 16t\vert =\frac{8}{75}$. We can easily choose $K_{1}=1$, $D>\frac{\pi}{8}$ and $M=\frac{\pi}{8}$ such that (H₁)-(H₃) hold. And

$$1-c_{\infty}-\delta_{1}c_{\infty}-\frac{1}{2}c_{1} \omega=1-\frac{1}{150}-\frac {1}{10}\times\frac{1}{150}- \frac{1}{2}\times\frac{8}{75}\times\frac{\pi}{8}>0. $$

Hence, by Theorem 3.1, (3.15) has at least one $\frac{\pi}{8}$-periodic solution.