\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 154, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{7mm}}

\begin{document}
\title[\hfilneg EJDE-2018/154\hfil 
Monotone and oscillatory solutions]
{Monotone  and oscillation solutions to second-order differential equations
with asymptotic conditions modeling ocean flows}

\author[Y. Yang, Z. Liang \hfil EJDE-2018/154\hfilneg]
{Yanjuan Yang, Zaitao Liang}

\address{Yanjuan Yang \newline
College of Science,
Hohai University,
Nanjing 210098, China}
\email{yjyang90@163.com, jchuphd@126.com}

\address{Zaitao Liang \newline
 School of Mathematics and Big Data,
Anhui University of Science and Technology,
Huainan 232001,
Anhui, China}
\email{liangzaitao@sina.cn}


\dedicatory{Communicated by Adrian Constantin}

\thanks{Submitted March 20, 2018. Published August 21, 2018.}
\subjclass[2010]{34C10}
\keywords{Monotone solutions; oscillatory solutions; asymptotic condition;
\hfill\break\indent fixed point theorem}

\begin{abstract}
 In this article, we study the existence of monotone bounded solutions and
 of oscillatory solutions to a second-order differential equation with
 asymptotic conditions. Such asymptotic conditions arise in the study of
 the ocean flow in arctic gyres. Our approach relies on functional-analytic
 techniques.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

In this article, we study the existence of monotone bounded solutions
and of oscillatory solutions for the second-order differential equation
\begin{equation} \label{eq1.1}
x''+a(t)f(x)=h(t),\quad t\geq t_0,
\end{equation}
where the real-valued function $f\colon\mathbb R\to\mathbb R$ is continuous,
$a\colon[t_{0},+\infty)\to [0,\infty)$ and $h\colon[t_{0},+\infty)\to \mathbb R$
are continuous. From the view of physics, it is
interesting to consider the asymptotic conditions
\begin{equation}\label{xac}
\lim_{t \to \infty}x(t)=\psi_0\quad\text{and}\quad
\lim_{t \to \infty} \{ x'(t)\exp(t)\}=0,
\end{equation}
where $\psi_0\in\mathbb R$ is a constant.

As a special form of equation \eqref{eq1.1}, the  equation
\begin{equation}  \label{cms}
x''=\frac{F(x)}{\cosh^2(t)} - \frac{2\omega \sinh(t)}{\cosh^3(t)}\,,\quad t\geq t_0,
\end{equation}
with the asymptotic conditions
\begin{equation}\label{uac}
\lim_{t \to \infty} x(t)=\psi_0\quad\text{and}\quad
\lim_{t \to \infty} \{ x'(t) \cosh(t)\}=0\,,
\end{equation}
is a recently derived model for arctic gyres with a vanishing azimuthal
velocity (see the discussions in \cite{cj-prs}
and the discussions in \cite{cmh}). Recently, Chu has studied \eqref{cms}-\eqref{uac}
in a systematic way in the recent papers \cite{cmh,clip,cnonl,chu-na}.
 Note that the second condition in \eqref{uac} is equivalent to the second one in \eqref{xac}.
We point out that the specific form of \eqref{uac} and
of the associated differential equation is due to
physically relevant considerations (see the discussion \cite{cl-na}).

To prove the existence of monotone solutions and oscillatory solutions
of \eqref{eq1.1}-\eqref{xac}, we will apply Schauder
fixed point theorem. To do this, we transform the problem \eqref{eq1.1}-\eqref{xac}
into an integral equation. In fact, if $x(t)$ is a solution of the
 problem \eqref{eq1.1}-\eqref{xac}, integrating the equation \eqref{eq1.1} on $[t,\infty)$,
we have
\begin{equation} \label{eq1.4}
x'(t)=-\int^{\infty}_{t}h(s){\rm d}s +\int^{\infty}_{t}a(s)f(x(s)){\rm d}s,\quad
t\geq t_{0},
\end{equation}
then integrating \eqref{eq1.4} on $[t,\infty)$, we obtain
\begin{equation}\label{eq1.5}
x(t)=\psi_{0}+\int^{\infty}_{t}(s-t)h(s){\rm d}s
-\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s,\quad t\geq t_{0}.
\end{equation}
To make the integral equation \eqref{eq1.5} equivalent to problem \eqref{eq1.1}-\eqref{xac},
we assume that
\begin{equation}\label{eqf}
\lim_{t\to\infty}\{\exp(t)a(t)\}=0,\quad
\lim_{t\to\infty}\{\exp(t)h(t)\}=0.
\end{equation}
Indeed, suppose that $x\colon [t_{0},\infty)\to \mathbb R$ is a continuous function
satisfying \eqref{eq1.5}, and $\lim_{t \to \infty} x(t)=\psi_0$.
It is easy to show that $x$ satisfies \eqref{eq1.4} and the second
condition in \eqref{xac}, since
\begin{gather*}
\lim_{t \to \infty}\big\{\exp(t)\int^{\infty}_{t}h(s){\rm d}s\big\}
=\lim_{t \to \infty}\{\exp(t)h(t)\}=0,\\
\lim_{t \to \infty}\big\{\exp(t)\int^{\infty}_{t}a(s)f(x(s)){\rm d}s\big\}
=\lim_{t \to \infty}\{\exp(t)a(t)f(x(t))\}=0.
\end{gather*}
Therefore, in this paper, we shall study the equivalent integral equation
\eqref{eq1.5} of the problem \eqref{eq1.1}-\eqref{xac} under
condition \eqref{eqf}.

\section{Monotone solutions}

In this section, we study the existence  of monotone bounded solutions
for the integral equation \eqref{eq1.5} under suitable conditions.

\begin{theorem}\label{thm1}
Assume that $a,h\colon [t_{0},+\infty)\to [0,\infty)$
are continuous with
\begin{equation} \label{hc}\int^{\infty}_{t_0}h(s){\rm ds}>0.
\end{equation}
Suppose further that the limit
 \begin{equation} \label{nac}
J=\lim_{t\to\infty}\frac{a(t)}{h(t)}
\end{equation}
 exists and $J\neq 0$, and there exists a constant $\gamma > 0$ such that
\begin{equation} \label{sac}
\max_{x\in [\psi_{0}-\gamma,\psi_{0}+\gamma]}f(x)<\frac{1}{J}.
\end{equation}
Then there exists some $T_{\gamma}\geq t_{0}$ such that \eqref{eq1.5}
 has at least one decreasing bounded continuous solution
$x\colon [T_{\gamma},\infty)\to \mathbb R$ satisfying $\lim_{t \to \infty}  x(t)=\psi_0$.
More precisely, we have that
\begin{equation} \label{dac}
x(t)>\psi_{0},\quad x'(t)<0, \quad\text{for all } t>T_{\gamma}.
\end{equation}
\end{theorem}

\begin{proof}
Set
$$
M_{\gamma}=\max_{x\in [\psi_{0}-\gamma,\psi_{0}+\gamma]}|f(x)|.
$$
Obviously, $0\leq M_{\gamma} < \infty$ since $f$ is continuous.
From \eqref{eqf}, we have
\begin{equation} \label{tac}
\int^{\infty}_{t_{0}}sa(s){\rm d}s<\infty\quad \text{and}\quad
\int^{\infty}_{t_{0}}sh(s){\rm d}s<\infty.
\end{equation}
 By \eqref{tac}, we can choose $T_0 \ge \max\{t_0,0\}$ large enough such that
$$
M_{\gamma}\int^{\infty}_{T_{0}}sa(s){\rm d}s
<\frac{\gamma}{2}\quad \text{and}\quad
\int^{\infty}_{T_{0}}sh(s){\rm d}s<\frac{\gamma}{2}.
$$
Define the closed and convex subset
\[
X_0=\big\{x\in  C([T_0,\infty),\mathbb R):\lim_{t\to\infty}x(t)=\psi_0\big\}
\]
of the Banach space $X$ of all bounded functions $x\in  C([T_0,\infty),\mathbb R)$,
endowed with the supremum norm
$\| x \|=\sup_{t \ge T_0} \{ |x(t)|\}$.
Set
\[
\Omega=\big\{x\in X_{0}: \psi_0-\gamma\leq x(t)\leq \psi_0+\gamma,\quad
t\geq T_0\big\}.
\]
Let $\mathcal{T}\colon \Omega\to X_{0}$ be the operator defined as
\begin{equation} \label{oac}
[\mathcal{T}(x)](t)=\psi_{0}+\int^{\infty}_{t}(s-t)h(s){\rm d}s
-\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s,\quad t\geq T_{0}.
\end{equation}
Note that
\begin{gather*}
\big| \int_t^\infty (s-t)h(s)\,ds \big| \le \int_t^\infty s h(s)\,ds,\quad
t  \ge T_0\,, \\
\big| \int_t^\infty (s-t)a(s)f(x(s))\,ds \big|
\le M_{\gamma}\int_t^\infty s a(s)\,ds,\quad t  \ge T_0\,,
\end{gather*}
which confirms that ${\mathcal T}\colon \Omega \to X_{0}$.
Also, for any $x\in \Omega$, we have
$\lim_{t\to \infty}[{\mathcal T}(x)](t)=\psi_0$ since
\[
\lim_{t\to\infty}\int_t^\infty s h(s)\,ds=0,\quad \lim_{t\to\infty}\int_t^\infty s a(s)\,ds
=0.
\]
We shall apply the Schauder fixed point theorem \cite{z} to prove that there
exists a fixed point for the operator $\mathcal{T}$ in the nonempty closed
bounded convex set $\Omega$, and then we prove that \eqref{dac} holds.
It is divided into four steps.
\smallskip

\noindent\textbf{Step 1.} We prove that $\mathcal{T}(\Omega)\subset\Omega$.
For any $x\in \Omega$ and $t\geq T_0,$ we have
\begin{align*}
|[\mathcal{T}(x)](t)-\psi_0|
&=  \Big|\int^{\infty}_{t}(s-t)h(s){\rm d}s
 -\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s\Big|\\
&\leq  \int_t^\infty (s-t) h(s){\rm d}s+\int_t^\infty (s-t) |a(s)f(x(s))|{\rm d}s\\
&\leq  \int_t^\infty s h(s){\rm d}s+\int_t^\infty M_{\gamma}s a(s){\rm d}s\\
&\leq  \int_{T_{0}}^\infty s h(s){\rm d}s+M_{\gamma}\int_{T_{0}}^\infty s a(s)
 {\rm d}s
\leq  \gamma,
\end{align*}
which shows that $\mathcal{T}\colon \Omega\to\Omega$
is well-defined.
\smallskip

\noindent\textbf{Step 2.} We prove that $\mathcal{T}\colon \Omega\to\Omega$ is continuous.
For a given $\varepsilon>0,$ there exists a $T_*\geq T_0$ such that
\[
M_{\gamma}\int_{T_*}^\infty sa(s)ds<\frac{\varepsilon}{3}.
\]
By the fact that $f\colon [\psi_0-\gamma,\psi_0+\gamma]\to\mathbb R$ is continuous,
there exists a constant
$\delta>0$ such that for all $x,y\in[\psi_0-\gamma,\psi_0+\gamma]$ with
$|x-y|<\delta$, we have
\[
|f(x)-f(y)|<\frac{2\varepsilon}{3T^2_*a_{\ast}},\quad \text{for all }
 t\in[t_0,T_*],
\]
where $a_{\ast}=\max_{t\in[T_0,T_*]}a(t)$.
Therefore, for all $x_1,x_2\in\Omega$ with $\|x_1-x_2\|<\delta$, we obtain
\begin{align*}
|[\mathcal{T}(x_1)](t)-[\mathcal{T}(x_2)](t)|
& = \big|\int_t^{\infty}(s-t)a(s)[f(x_2(s))-f(x_1(s))]\big|\\
&\leq  \int_t^{\infty}(s-t)a(s)|f(x_2(s))-f(x_1(s))|ds\\
&\leq  \int_{T_0}^{T_*}(s-T_0)a(s)|f(x_2(s))-f(x_1(s))|ds\\
&\quad +\int_{T_*}^{\infty}(s-T_*)a(s)|f(x_2(s))-f(x_1(s))|ds\\
&= I_1+I_2.
\end{align*}
Since
\begin{gather*}
I_1\leq\frac{2\varepsilon}{3T^2_*a_{\ast}}a_{\ast}\int_{T_0}^{T_*}(s-T_0)ds
= \frac{2\varepsilon}{3T^2_*}\frac{(T_*-T_0)^2}{2}<\frac{\varepsilon}{3},\\
\begin{aligned}
I_2&\leq \int_{T_*}^{\infty}s a(s)\Big\{|f(x_1(s))|+|f(x_2(s))|\Big\}ds\\
& \leq  2M_{\gamma}\int_{T_*}^{\infty}sa(s) ds<\frac{2\varepsilon}{3},
\end{aligned}
\end{gather*}
we have
\[
\|[\mathcal{T}(x_1)]-[\mathcal{T}(x_2)]\|\leq \varepsilon.
\]
Therefore, $\mathcal{T}\colon \Omega\to\Omega$ is a continuous.
\smallskip

\noindent\textbf{Step 3.}
We prove that $\mathcal{T}(\Omega)$ is relatively compact in $X$.
Since $\mathcal{T}(\Omega)\subset\Omega,$ we know that
$\mathcal{T}(\Omega)$ is uniform bounded.
Differentiating two sides of \eqref{oac} with respect to $t$, we obtain
\[
[\mathcal{T}(x)]'(t)=-\int^{\infty}_{t}h(s){\rm d}s
+\int^{\infty}_{t}a(s)f(x(s)){\rm d}s,\quad t\geq T_{0}.
\]
For all $t\geq T_0,$ we have
\begin{align*}
|[\mathcal{T}(x)]'(t)|
&\leq \big|\int_t^\infty h(s)\,ds\big|
 + \big|\int_t^\infty a(s)f(x(s))\,ds\big|\\
&\leq \int_t^\infty h(s)\,ds+M_{\gamma}\int_t^\infty a(s)\,ds\\
&\leq \int_{T_{0}}^\infty h(s)\,ds
 +M_{\gamma}\int_{T_{0}}^\infty a(s)\,ds,
\end{align*}
which means that for all $x\in\Omega$, we have
\[
\Big|[\mathcal{T}(x)]'(t)\Big|\leq K,\quad t\geq T_0,
\]
where
$$
K=\int_{T_{0}}^\infty h(s)\,ds+M_{\gamma}\int_{T_{0}}^\infty a(s)\,ds.
$$
Let $\{x_n\}$ be an arbitrary sequence in $\Omega$. Then we have
\[
|[\mathcal{T}(x_n)]'(t)|\leq K,\quad t\geq T_0,\quad n\geq 1.
\]
Applying the mean value theorem, we obtain
\[
|[\mathcal{T}(x_n)](t_1)-[\mathcal{T}(x_n)](t_2)|
\leq K|t_1-t_2|,\quad t_1,t_2\geq T_0,\quad n\geq 1,
\]
which implies that $\{[\mathcal{T}(x_n)]\}$ is equicontinuous in $X$.

Furthermore, since
\[
\lim_{t\to\infty}\Big[\psi_{0}+\int^{\infty}_{t}(s-t)h(s){\rm d}s
-\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s\Big]=\psi_0,
\]
so for every $\epsilon>0$, there exists $t_{\epsilon}>T_0$ such that
\[
|[\mathcal{T}(x_n)](t)-\psi_0|\leq \epsilon,\quad t\geq t_\epsilon,\quad n\geq 1.
\]
Therefore, $\{[\mathcal{T}(x_n)]\}$ is equiconvergent in $X$.

By using the Arzela-Ascoli theorem \cite{z}, we obtain that
$\{[\mathcal{T}(x_n)]\}$ is relatively compact in $X$.

We have proved that all assumptions of the Schauder fixed point theorem
are satisfied. Therefore, the operator $\mathcal{T}$ has a fixed point
$x$ in $\Omega$, and this fixed point corresponds to a bounded solution
of \eqref{eq1.5} on $[T_0,\infty)$.
\smallskip

\noindent\textbf{Step 4.}
We show that the fixed point is decreasing.
Let $x$ be the fixed point of $\mathcal{T}$.
Define
\[
H(t)=\frac{\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s}
{\int^{\infty}_{t}(s-t)h(s){\rm d}s},\quad t>T_{0}.
\]
Then
\[
H(t)\leq \max_{x\in [\psi_{0}-\gamma,\psi_{0}+\gamma]}f(x)\cdot
\frac{\int^{\infty}_{t}(s-t)a(s){\rm d}s}{\int^{\infty}_{t}(s-t)h(s){\rm d}s}.
\]
Since
\[
\lim_{t\to \infty}\frac{\int^{\infty}_{t}(s-t)a(s){\rm d}s}
{\int^{\infty}_{t}(s-t)h(s){\rm d}s}=\lim_{t\to \infty}
\frac{\int^{\infty}_{t}a(s){\rm d}s}{\int^{\infty}_{t}h(s){\rm d}s}
=\lim_{t\to \infty}\frac{a(t)}{h(t)}=J,
\]
using the condition \eqref{sac}, we know that there exists $T_{1}\geq T_{0}$
such that
$H(t)<1$ for $t>T_{1}$, which yields
\[
\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s
< \int^{\infty}_{t}(s-t)h(s){\rm d}s,\quad t>T_{1},
\]
and hence for all $t>T_{1},$ we have
\[
x(t)=\psi_{0}+\int^{\infty}_{t}(s-t)h(s){\rm d}s
-\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s>\psi_{0}.
\]
Define
\[
L(t)=\frac{\int^{\infty}_{t}a(s)f(x(s)){\rm d}s}{\int^{\infty}_{t}h(s){\rm d}s},\quad
t>T_{0}.
\]
Then
\[
L(t) \leq \max_{x\in [\psi_{0}-\gamma,\psi_{0}
+\gamma]}f(x)\cdot\frac{\int^{\infty}_{t}a(s){\rm d}s}{\int^{\infty}_{t}h(s){\rm d}s}.
\]
Since \eqref{sac} holds, there exists $T_{2}\geq T_{0}$ such that
$L(t)<1$ for $t>T_{2}$,
which implies
\[
x'(t)= -\int^{\infty}_{t}h(s){\rm d}s +\int^{\infty}_{t}a(s)f(x(s)){\rm d}s<0,
\quad t>T_{2}.
\]
Let $T_{\gamma}=\max\{T_{1},T_{2}\}$, then \eqref{dac} holds.
\end{proof}

\begin{example} \label{xamp1} \rm
Consider the equation
\begin{equation}\label{eq1.6}
x''+\frac{1}{{\rm cosh}^{2}(t)}\frac{x}{8\psi_{0}}=e^{-2t},\quad t\geq t_{0}.
\end{equation}
It is easy to see that
 \begin{equation} \label{oc}
J=\lim_{t\to\infty}\frac{\frac{1}{{\rm cosh}^{2}(t)}}{e^{-t}}=4.
\end{equation}
We suppose that $\psi_{0}>0$, choose any $\gamma \in [0,\psi_{0})$, then
it is easy to check that
\[
\max_{x\in [\psi_{0}-\gamma,\psi_{0}+\gamma]}\frac{x}{8\psi_{0}}<\frac{1}{4}.
\]
We know that the solution of \eqref{eq1.6} is
\begin{equation}\label{jc}
x(t)=\psi_{0}+\int^{\infty}_{t}(s-t)e^{-s}{\rm d}s
-\int^{\infty}_{t}(s-t)\frac{x(s)}{8\psi_{0}}
\frac{1}{{\rm cosh}^{2}(s)}{\rm d}s,\quad t\geq t_{0}.
\end{equation}
Obviously, $x(t)>\psi_{0}$ for $t\geq t_{0}$. Indeed,
$\lim_{t \to \infty} \{ x(t)\}=\psi_0$ and
\[
x'(t)=-\int^{\infty}_{t}e^{-s}{\rm d}s+\int^{\infty}_{t}
\frac{x(s)}{8\psi_{0}}\frac{1}{{\rm cosh}^{2}(s)}{\rm d}s
<0.
\]
Therefore, $x(t)$ decreases towards $\psi_{0}$ as $t$ decreases towards infinity.
\end{example}

In fact, we can prove another result in a similar way.

\begin{theorem}\label{thm2}
Assume that $a,h\colon [t_{0},+\infty)\to [0,\infty)$
are continuous and \eqref{hc}, \eqref{nac} hold.
Suppose further that there exists a constant $\eta > 0$ such that
\begin{equation} \label{mac}
\min_{x\in [\psi_{0}-\eta,\psi_{0}+\eta]}f(x)>\frac{1}{J}.
\end{equation}
Then there exists some $T_{\eta}\geq t_{0}$ such that there exists a
increasing bounded continuous solution $x\colon [T_{\eta},\infty)\to \mathbb R$
to the equation {\rm \eqref{eq1.5}}, and $\lim_{t \to \infty} \{ x(t)\}=\psi_0$.
 More precisely, we have
\begin{equation} \label{vac}
x(t)<\psi_{0},\quad x'(t)>0, \quad\text{for all } t>T_{\eta}.
\end{equation}
\end{theorem}

\begin{proof}
Proceeding as in Steps 1--3 in the proof of Theorem \ref{thm1},
we know that the equation  \eqref{eq1.5} has at least one bounded continuous
solution $x\colon [T_{\eta},\infty)\to \mathbb R$ satisfying
$\lim_{t \to \infty} \{ x(t)\}=\psi_0$.

We only need to prove the solution above is increasing.
Define
\[
H(t)=\frac{\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s}
{\int^{\infty}_{t}(s-t)h(s){\rm d}s},\quad t>T_{0}.
\]
Then
\[
 H(t) \geq \min_{x\in [\psi_{0}-\eta,\psi_{0}+\eta]}f(x)
\frac{\int^{\infty}_{t}(s-t)a(s){\rm d}s}{\int^{\infty}_{t}(s-t)h(s){\rm d}s}.
\]
Since
\[
\lim_{t\to \infty}\frac{\int^{\infty}_{t}(s-t)a(s){\rm d}s}
{\int^{\infty}_{t}(s-t)h(s){\rm d}s}
=\lim_{t\to \infty}\frac{\int^{\infty}_{t}a(s){\rm d}s}
{\int^{\infty}_{t}h(s){\rm d}s}
=\lim_{t\to \infty}\frac{a(t)}{h(t)}=J,
\]
by \eqref{mac}, we know that there exists $T_{1}\geq T_{0}$ such that
$H(t)>1$ for $t>T_{1}$, which yields that
\[
\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s
> \int^{\infty}_{t}(s-t)h(s){\rm d}s,\quad t>T_{1},
\]
and hence for all $t>T_{1},$ we have
\[
x(t)=\psi_{0}+\int^{\infty}_{t}(s-t)h(s){\rm d}s
 -\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s<\psi_{0}.
\]
Define
\[
L(t)=\frac{\int^{\infty}_{t}a(s)f(x(s)){\rm d}s}{\int^{\infty}_{t}h(s){\rm d}s},
\quad t>T_{0}.
\]
Then
\[
 L(t)\geq \min_{x\in [\psi_{0}-\eta,\psi_{0}+\eta]}f(x)
\frac{\int^{\infty}_{t}a(s){\rm d}s}{\int^{\infty}_{t}h(s){\rm d}s}.
\]
Since \eqref{mac} holds, there exists $T_{2}\geq T_{0}$ such that
$L(t)>1$ for $t>T_{2}$, which implies
\[
x'(t)= -\int^{\infty}_{t}h(s){\rm d}s +\int^{\infty}_{t}a(s)f(x(s)){\rm d}s>0,\quad
t>T_{2}.
\]
Let $T_{\eta}=\max\{T_{1},T_{2}\}$, then \eqref{vac} holds.
\end{proof}


\begin{example} \label{examp2} \rm
Consider the equation
\begin{equation}\label{eq1.7}
x''+\frac{1}{{\rm sinh}^{2}(t)}\frac{x}{4\psi_{0}}=e^{-2t},\quad t\geq t_{0}.
\end{equation}
Then we know that
 \begin{equation} \label{lc}
J=\lim_{t\to\infty}\frac{\frac{1}{{\rm sinh}^{2}(t)}}{e^{-2t}}=4.
\end{equation}
Assume that $\psi_{0}>0$, take any $\gamma >0$, then we have
\[
\max_{x\in [\psi_{0}-\gamma,\psi_{0}+\gamma]}\frac{x}{4\psi_{0}}>\frac{1}{4}.
\]
We know that the solution of \eqref{eq1.7} is
\begin{equation}\label{jlc}
x(t)=\psi_{0}+\int^{\infty}_{t}(s-t)e^{-2s}{\rm d}s
-\int^{\infty}_{t}(s-t)\frac{x(s)}{4\psi_{0}}\frac{1}{{\rm sinh}^{2}(s)}{\rm d}s,
\quad t\geq t_{0}.
\end{equation}
Obviously, $x(t)<\psi_{0}$ for $t\geq t_{0}$. Indeed,
$\lim_{t \to \infty} \{ x(t)\}=\psi_0,$ and
\[
x'(t)= -\int^{\infty}_{t}e^{-2s}{\rm d}s+\int^{\infty}_{t}
\frac{x(s)}{4\psi_{0}}\frac{1}{{\rm sinh}^{2}(s)}{\rm d}s>0.
\]
Therefore, $x(t)$ increases towards $\psi_{0}$ as $t$ increases towards infinity.
\end{example}

\section{Oscillatory solutions}\label{s3}

In this section, we study the existence of oscillatory solutions for
the integral equation \eqref{eq1.5} under suitable conditions.
Define a function $H\colon [t_{0},\infty)\to \mathbb{R}$ as
\[
H(t)=\int^{\infty}_{t}(s-t)h(s){\rm d}s.
\]
For a fixed $\lambda> t_{0}$, we denote the upper bound of $H$ by
\[
\|H\|=\sup _{t\geq \lambda > t_{0}}|H(t)|.
\]
Fix a positive real number $R>\|H\|$ and define
\begin{gather*}
M_{R}=\sup _{x\in [-R,R]}|f(x)|,\quad
g(t)=M_{R}\int^{\infty}_{t}a(s){\rm d}s,\quad t\geq t_{0},\\
G(t)=\int^{\infty}_{t}g(s){\rm d}s.
\end{gather*}
Now we state and prove the main result of this section.

\begin{theorem}\label{thm3}
Assume that $G(t_0)<+\infty$ and
\begin{equation}\label{eq3.2}
\limsup _{t\to +\infty}\frac{H(t)}{G(t)}>1, \quad
\liminf _{t\to +\infty}\frac{H(t)}{G(t)}<-1.
\end{equation}
Then for every $\varepsilon$ with $0<\varepsilon< R-\|H\|$, there exist a real
 number $T(\varepsilon)>0$, a positive integer $N(\varepsilon)$, and two
increasing divergent sequences of positive numbers
$\{t_{n}\}_{n\geq1}$, $\{s_{n}\}_{n\geq1}$, such that \eqref{eq1.5}
 has a solution $x(t)$  defined on $[T(\varepsilon), +\infty)$ satisfying
$\lim_{t \to \infty}  x(t)=\psi_0$ and
\[
x(t_{n})> \psi _{0}\quad \text{and}\quad x(s_{n})< \psi _{0},\quad
\text{for all }  n\geq N(\varepsilon).
\]
\end{theorem}

\begin{proof}
To prove the above result, by  \eqref{eq1.5}, we just need to prove that the
equation
\begin{equation}\label{eq3.7}
x(t)=\int^{\infty}_{t}(s-t)h(s){\rm d}s-\int^{\infty}_{t}(s-t)a(s)f(x(s)){\rm d}s,
\quad t\geq t_{0},
\end{equation}
has a solution $x(t)$ such that $\lim_{t \to \infty} x(t)=0$ and
\[
x(t_{n})> 0\quad \text{and}\quad x(s_{n})< 0.
\]

Given a real number $\lambda >t_{0}$, choose an $\varepsilon$ with
$0<\varepsilon <R-\|H\|$. Since $G(t_0)<+\infty,$ there exists a number
$T(\varepsilon) > \lambda$ such that $G(t) <\varepsilon$ for all
$t\geq T(\varepsilon)$.
Define the closed and convex subset
\[
X_\varepsilon=\{x\in C([T(\varepsilon),+\infty),\mathbb{R}):
 \lim_{t \to \infty} x(t)=0\}
\]
of the Banach space $X$ of all functions
$x\in C([T(\varepsilon),+\infty),\mathbb{R})$, endowed with the supremum $\|\cdot\|$.
Set
\[
\Omega=\{x\in X_\varepsilon: \|x-H\|\leq\varepsilon\}.
\]
Define an operator $\mathcal{F}\colon \Omega\to \Omega$ as
\begin{equation}
[\mathcal{F}(x)](t)=H(t)-\int^{\infty}_{t}\int^{\infty}_{s}a(\tau)f(x(\tau)){\rm d}\tau{\rm d}s,\quad t\geq T(\varepsilon).
\end{equation}
Note that
\begin{equation}\label{eq3.8}
|[\mathcal{F}(x)](t)-H(t)|\leq \int^{\infty}_{t}M_{\|H\|+\varepsilon}
\int^{\infty}_{s}a(\tau){\rm d}\tau{\rm d}s\leq G(t)<\varepsilon,\quad
t\geq T(\varepsilon).
\end{equation}
Therefore, the operator $\mathcal{F}\colon \Omega\to \Omega$ is well-defined.

We shall apply the Schauder fixed point theorem to prove that there
exists a fixed point for the operator $\mathcal{F}$ in the nonempty closed
bounded convex set $\Omega$.

First, we prove that the operator $\mathcal{F}$ is uniformly continuous.
For a given constant $\xi>0$, there exists a $T(\xi)>T(\varepsilon)$ such that
\[
G(t)<\frac{\xi}{3},\quad t\geq T(\xi).
\]
Furthermore, there exists a $\delta(\xi)>0$ such that
\[
|a(t)f(x_{1})-a(t)f(x_{2})|<\frac{\xi}{3(T(\xi))^{2}},
\]
holds for all $ t\in[T(\varepsilon), T(\xi)]$ and
$x_{1}, x_{2}\in [-\|H\|-\varepsilon,\|H\|+\varepsilon]$ with
$\|x_{1}-x_{2}\|<\delta(\xi)$.
Now for all $x_{1},x_{2}\in \Omega$ satisfying $\|x_{1}-x_{2}\|<\delta(\xi)$,
we have
\begin{align*}
|[\mathcal{F}(x_{1})](t)-[\mathcal{F}(x_{2})](t)|
&\leq \int^{\infty}_{T(\varepsilon)}\int^{\infty}_{s} |
 a(\tau)f(x_{2}(\tau))-a(\tau)f(x_{1}(\tau))|{\rm d}\tau{\rm d}s\\
&= \int^{\infty}_{T(\varepsilon)}(s-T(\varepsilon))|a(s)f(x_{2}(s))-a(s)f(x_{1}(s))|
 {\rm d}s\\
&\leq |T(\xi)-T(\varepsilon)|\int^{T(\xi)}_{T(\varepsilon)}|a(s)f(x_{2}(s))
 -a(s)f(x_{1}(s))|{\rm d}s\\
&\quad +\int^{\infty}_{T(\xi)}\int^{\infty}_{s}|a(\tau)f(x_{2}(\tau))|{\rm d}
 \tau{\rm d}s\\
&\quad +\int^{\infty}_{T(\xi)}\int^{\infty}_{s}|a(\tau)f(x_{1}(\tau))|{\rm d}
 \tau{\rm d}s\\
&= I_{1}+I_{2}+I_{3}.
\end{align*}
Note that
\[
I_{1}<[T(\xi)-T(\varepsilon)]^{2}\frac{\xi}{3(T(\xi))^{2}}<\frac{\xi}{3},\quad
I_{2}+I_{3}<\frac{2}{3}\xi.
\]
Then we conclude that
\[
|\mathcal{F}(x_{1})](t)-\mathcal{F}(x_{2})](t)|<\xi.
\]
Therefore $\mathcal{F}$ is uniformly continuous.

Next, we apply the Arzela-Ascoli theorem to prove that the set
$\mathcal{F}(\Omega)$ is relatively compact.
Since $\mathcal{F}(\Omega)\subset \Omega$, we know that $\mathcal{F}(\Omega)$
is uniformly bounded. For any two real numbers $t_{1}, t_{2}$ with
$t_{2}\geq t_{1}\geq T(\varepsilon)$, we have
\begin{align*}
|[\mathcal{F}(x)](t_{2})-[\mathcal{F}(x)](t_{1})|
&\leq |H(t_{2})-H(t_{1})|+
\int^{t_{2}}_{t_{1}}\int^{\infty}_{s}|a(\tau)f(x(\tau))|{\rm d}\tau{\rm d}s\\
&\leq \int^{t_{2}}_{t_{1}}\int^{\infty}_{s}|h(\tau)|{\rm d}\tau{\rm d}s+
\int^{t_{2}}_{t_{1}}g(s){\rm d}s,\quad x\in \Omega,
\end{align*}
which shows that $\mathcal{F}(\Omega)$ is equicontinuous.

From the definition of $\mathcal{F}$, we have
\begin{equation}\label{eq3.3}
|\mathcal{F}(x)](t)|\leq |H(t)|+G(t),\quad t\geq T(\varepsilon),\quad\text{for all }
 x\in \Omega.
\end{equation}
By \eqref{eq3.3} and $\lim _{t\to\infty }H(t)=0$, we know that the set
$\mathcal{F}(\Omega)$ is equiconvergent. Therefore $\mathcal{F}(\Omega)$ is
relatively compact.

Up to now, all conditions of the Schauder fixed point theorem are established.
Therefore, the operator $\mathcal{F}$ has a fixed point in $\Omega$, that is,
the equation \eqref{eq3.7} has a solution $x(t)$, which satisfies
$\lim _{t\to\infty }x(t)=0$.

Finally, we prove that the solution $x(t)$ is oscillatory.
From \eqref{eq3.8}, we have
\[
|x(t)-H(t)|=|[\mathcal{F}(x)](t)-H(t)|\leq G(t),\quad t\geq T(\varepsilon),
\]
which yields
\begin{equation} \label{hg3.1}
H(t)-G(t)\leq x(t)\leq H(t)+G(t),\quad \text{for all } t\geq T(\varepsilon).
\end{equation}
By \eqref{eq3.2}, we know that there exist a positive integer $N(\varepsilon)$
and two sequences of positive numbers $\{t_{n}\}_{n\geq1}$, $\{s_{n}\}_{n\geq1}$,
$t_{n}$, $s_{n}\to \infty$ as $n\to \infty$, such that
\[
H(t_{n})-G(t_{n})> 0\quad\text{and} \quad  H(s_{n})+G(s_{n})<0,\quad \text{for all }
n\geq N(\varepsilon),
\]
it follows from \eqref{hg3.1} that
\[
x(t_{n})> 0\quad \text{and}\quad x(s_{n})< 0,\quad \text{for all }
n\geq N(\varepsilon).
\]
The proof is complete.
\end{proof}

\subsection*{Acknowledgements}
Y. Yang was supported by the Fundamental Research Funds for the Central
Universities (Grant No. 2017B715X14) and the Postgraduate Research
and Practice Innovation Program of Jiangsu Province (Grant No.
 KYCX17\_ 0508).
 Z. Liang was supported by the National Natural Science Foundation of China
(Grant No. 61773152).

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