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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 142, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/142\hfil Long term behavior of solutions]
{Long term behavior of solutions for Riccati initial-value
problems}

\author[S. Y. Bahk, N. E. Dyakevich, S. C. Johnson\hfil EJDE-2008/142\hfilneg]
{Sarah Y. Bahk, Nadejda E. Dyakevich, Stefan C. Johnson} % in alphabetical order

\address{
Department of Mathematics, California State University San
Bernardino, 5500 University Parkway, San Bernardino, CA
92407-2397, USA} 
\email[Sarah Y. Bahk]{sbahk@csusb.edu}
\email[Nadejda E. Dyakevich]{dyakevic@csusb.edu} 
\email[Stefan C. Johnson]{steve@sevej.us}

\thanks{Submitted May 23, 2008. Published October 24, 2008.}
\subjclass[2000]{34C11}
\keywords{Riccati equation; unbounded growth in finite time; comparison}

\begin{abstract}
 The Riccati equation has been known since the early 1700s.
 Numerous papers have been written on the solvability of its
 special cases. However, to the best of our knowledge, there are no
 papers that investigate the exact (equation specific) conditions
 for unbounded growth in finite time of solutions for Riccati
 initial-value problems. In this paper, we first derive conditions
 that are necessary and sufficient for the solutions of  Riccati
 problems with constant coefficients to grow unbounded in finite
 time. Then we use a comparison method to extend these results to
 Riccati problems with variable coefficients.
\end{abstract}

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

\section{Introduction}

Count Jacopolo Francesco Riccati (May 28, 1676 - April 15, 1754)
is famous for introducing and researching the solvability of the
equation that now bears his name:
\begin{equation}
y'(t)=a(t)y^{2}+b(t)y+c(t). \label{e1.1}
\end{equation}
The matrix form of this equation is very important in modern times
since it is used extensively in design problems in filtering and
control \cite{Abo,Bitt}. Even though the Riccati equation
\eqref{e1.1} is not solvable in general, numerous methods have been
developed for finding solutions for many of its special cases
\cite{All}, \cite{Pol}-\cite{Whi}.

In Section 2, we consider real solutions of the initial-value problem
\begin{equation}
\begin{gathered}
y'(t)=ay^{2}+by+c,\\
y(0)=d,
\end{gathered}\label{e1.2}
\end{equation}
where $a$, $b$, $c$, and $d$ are real numbers and $t\geq0$
represents time. We determine conditions on the constants $a$,
$b$, $c$, and $d$ that are necessary and sufficient for $y(t)$ to
approach either $+\infty$ or $-\infty $ as $t$ approaches some
finite value $t_{b}$. We provide exact values for the time $t_{b}$
for the cases when $4ac-b^{2}$ is positive, negative, or zero. In
particular, we are interested in the first occurrence of blow-up.
We do not consider behavior of $y(t)$ for $t>t_{b}$.

In Section 3, we use a comparison theorem to extend the results to
the more general initial-value problem
\begin{gather*}
y'(t)=a(t)y^{2}+b(t)y+c(t),\\
y(0)=d,
\end{gather*}
where $a(t)$, $b(t)$, $c(t)$ are continuous and differentiable functions for
$t\geq0$, and $d$ is a real number.

\section{Riccati Problems with Constant Coefficients}

\begin{theorem} \label{thm1.1}
The following is true for the solution
$y(t)$  of \eqref{e1.2}:
\begin{enumerate}
\item Let $4ac-b^{2}>0$. If $a>0$, then
$y(t)\to+\infty$, while if $a<0$, then
$y(t)\to-\infty$.

\item  Let $4ac-b^{2}=0$. If $a>0$ and $d>-\frac{b}{2a}$,
  then $y(t)\to+\infty$. If $a<0$
  and $d<-\frac{b}{2a}$, then $y(t)\to-\infty$. Otherwise,
  $y(t)$ is bounded for any finite $t>0$.
  If $d=-\frac{b}{2a}$, then $y(t)\equiv d$.

\item Let $4ac-b^{2}<0$. If $a>0$ and
 $d>\frac{-b+\sqrt{b^{2}-4ac}}{2a}$, then $y(t)\to+\infty$. If
 $a<0$ and $d<\frac{-b+\sqrt{b^{2}-4ac}}{2a}$, then
 $y(t)\to-\infty$. Otherwise, $y(t)$ is bounded for
 any finite $t>0$. If $d=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$,
 then $y(t)\equiv d$.

\item  If $a=0$, then $y(t)$ is bounded for all $t>0$.
\end{enumerate}
\end{theorem}

\begin{proof}
(1) Let $4ac-b^{2}>0$. The solution of the initial value problem
\eqref{e1.2} can be found using separation of variables:
\begin{equation}
y(t)   =\frac{\sqrt{4ac-b^{2}}}{2a}\tan\Big[  \frac{t\sqrt{4ac-b^{2}}}
{2}+\arctan\Big(  \frac{b+2ad}{\sqrt{4ac-b^{2}}}\Big)  \Big]
 -\frac{b}{2a}. \label{e2.1}
\end{equation}
We can find the blow-up time $t_{b}$ by solving for $t$ in equation
\begin{gather*}
\frac{t\sqrt{4ac-b^{2}}}{2}+\arctan\Big(  \frac{b+2ad}{\sqrt{4ac-b^{2}}
}\Big)  =\frac{\pi}{2},\\
t_{b}=\frac{\pi}{\sqrt{4ac-b^{2}}}-\frac{2}{\sqrt{4ac-b^{2}}}\arctan\big[
\frac{b+2ad}{\sqrt{4ac-b^{2}}}\big]  .
\end{gather*}
Also,
\[
-\frac{\pi}{2}<\arctan\big[  \frac{b+2ad}{\sqrt{4ac-b^{2}}}\big]
<\frac{\pi}{2}.
\]
This implies that $t_{b}$ is always positive and that the
solution $y(t)$ of \eqref{e1.2} is guaranteed to blow-up as $t$
approaches $t_{b}$. Also, from equation \eqref{e2.1} we notice that
if $a>0$, then $y(t)\to+\infty$, while if $a<0$, then
$y(t)\to-\infty$. Changing the initial value $d$ cannot
prevent blow-up from occurring. However, $d$ influences the
blow-up time $t_{b}$. For example, if $a<0$, then decreasing $d$
will accelerate the blow-up. If $a>0$, then increasing $d$ will
accelerate the blow-up.

(2) Let $4ac-b^{2}=0$. Using separation of variables, we obtain
\[
\int{\frac{dy}{a(  y+\frac{b}{2a})  ^{2}}}=\int{dt}.
\]
Integration leads to the solution:
\begin{equation}
y(t)=\frac{2ad+b}{a(  2-2adt-bt)  }-\frac{b}{2a}. \label{e2.2}
\end{equation}
To find the blow-up time, we set the denominator of the first term
in \eqref{e2.2} equal to $0$ and solve for $t$  in
$2-2adt-bt=0$,
\[
t_{b}=\frac{2}{2ad+b}.
\]
From the inequality $t_{b}>0$,
and from \eqref{e2.2} we obtain the following:
If $a>0$ and $d>-\frac{b}{2a}$, then $y(t)\to+\infty$, while if $a<0$ and
$d<-\frac{b}{2a}$, then $y(t)\to-\infty$. Initial value
$d$ is very important since certain values can prevent blow-up
from occurring. Also, $d$ influences the blow-up time $t_{b}$. If
blow-up occurs for some value $d$, then decreasing $d$ (if $a<0$)
or increasing $d$ (if $a>0$) will accelerate the blow-up. If
$d=-\frac{b}{2a}$, then $y(t)\equiv d$ satisfies the
initial-value problem \eqref{e1.2}. Therefore, in this special
case, $y(t)$ is bounded for all finite $t>0$.

(3) Let $4ac-b^{2}<0$. We notice that if
\[
d=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a},
\]
then $y\equiv d$ is the solution of the initial-value problem
\eqref{e1.2}. Therefore, in this case $y(t)$ is bounded for all
$t>0$. Now let us consider the case when
\[
d\neq\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.
\]
 Using separation of variables, we have
\begin{gather}
\int{\frac{dy}{ay^{2}+by+c}}=\int{dt},\nonumber\\
\frac{1}{\sqrt{b^{2}-4ac}}\ln\big| \frac{2ay+b-\sqrt{b^{2}-4ac}
}{2ay+b+\sqrt{b^{2}-4ac}}\big| =t+C_{1},  \label{e2.3}
\end{gather}
where $C_{1}$ is a constant of integration. We can find $C_{1}$
by substituting the initial condition $y(0)=d$ into equation \eqref{e2.3}.
Thus,
\begin{equation}
C_{1}=\frac{1}{\sqrt{b^{2}-4ac}}\ln\big| \frac{2ad+b-\sqrt{b^{2}-4ac}
}{2ad+b+\sqrt{b^{2}-4ac}}\big| . \label{e2.4}
\end{equation}
We will consider the two possible cases:
\begin{equation}
\frac{2ad+b-\sqrt{b^{2}-4ac}}{2ad+b+\sqrt{b^{2}-4ac}}>0 \label{e2.5}
\end{equation}
and
\begin{equation}
\frac{2ad+b-\sqrt{b^{2}-4ac}}{2ad+b+\sqrt{b^{2}-4ac}}<0. \label{e2.6}
\end{equation}
We now substitute \eqref{e2.4} into \eqref{e2.3} and solve for
$y(t)$. In case \eqref{e2.5}, we can omit the absolute value symbol:
\[
\frac{1}{\sqrt{b^{2}-4ac}}\ln\Big(  \frac{2ay+b-\sqrt{b^{2}-4ac}}
{2ay+b+\sqrt{b^{2}-4ac}}\Big)
=t+\frac{1}{\sqrt{b^{2}-4ac}}\ln\Big(  \frac{2ad+b-\sqrt{b^{2}-4ac}
}{2ad+b+\sqrt{b^{2}-4ac}}\Big).
\]
In the case \eqref{e2.6}, we have
\begin{align*}
&\frac{1}{\sqrt{b^{2}-4ac}}\ln\Big(  -\frac{2ay+b-\sqrt{b^{2}-4ac}
}{2ay+b+\sqrt{b^{2}-4ac}}\Big)\\
&= t+\frac{1}{\sqrt{b^{2}-4ac}}\ln\Big(  -\frac{2ad+b-\sqrt{b^{2}-4ac}
}{2ad+b+\sqrt{b^{2}-4ac}}\Big)  .
\end{align*}
In both cases \eqref{e2.5} and \eqref{e2.6}, the solution of
\eqref{e1.2} is given by the  formula
\begin{equation}
y(t)=\frac{-bd+d\sqrt{b^{2}-4ac}-2c+(bd+d\sqrt{b^{2}-4ac}+2c)e^{t\sqrt
{b^{2}-4ac}}}{2ad+b+\sqrt{b^{2}-4ac}-(2ad+b-\sqrt{b^{2}-4ac})e^{t\sqrt
{b^{2}-4ac}}}. \label{e2.7}
\end{equation}
To find the blow-up time we set the denominator equal to $0$,
\begin{equation}
2ad+b+\sqrt{b^{2}-4ac}-(2ad+b-\sqrt{b^{2}-4ac})e^{t\sqrt
{b^{2}-4ac}}=0
\label{e2.8}
\end{equation}
and solve for $t$ to obtain
\[
t_{b}=\frac{1}{\sqrt{b^{2}-4ac}}\ln\Big(  \frac{2ad+b+\sqrt{b^{2}-4ac}
}{2ad+b-\sqrt{b^{2}-4ac}}\Big).
\]
Since the blow-up time $t_{b}$ must be positive, we have
\begin{equation}
\frac{2ad+b+\sqrt{b^{2}-4ac}}{2ad+b-\sqrt{b^{2}-4ac}}>1. \label{e2.9}
\end{equation}
Let us observe that if  \eqref{e2.6} holds, then
\eqref{e2.8} can never be satisfied, thus, there is no blowup. On
the other hand, if  \eqref{e2.5} holds,
then there are two possibilities: either
\begin{equation}
2ad+b-\sqrt{b^{2}-4ac}>0\quad\text{and}\quad
2ad+b+\sqrt{b^{2}-4ac}>0 \label{e2.10}
\end{equation}
or
\begin{equation}
2ad+b-\sqrt{b^{2}-4ac}<0\quad\text{and}\quad
2ad+b+\sqrt{b^{2}-4ac}<0.\label{e2.11}
\end{equation}
Solving \eqref{e2.10} and \eqref{e2.9} simultaneously, we obtain the
conditions on $d$ that lead to blow-up in finite time:
\begin{gather*}
d>\frac{-b+\sqrt{b^{2}-4ac}}{2a},\quad \text{if }a>0,\\\\
d<\frac{-b+\sqrt{b^{2}-4ac}}{2a},\quad \text{if }a<0.
\end{gather*}
Solving \eqref{e2.11} and \eqref{e2.9} simultaneously, we obtain a
contradiction which implies that there is no blow-up in this case.
We notice from \eqref{e2.7} that if $a>0$ and
$d>\frac{-b+\sqrt{b^{2}-4ac}}{2a}$, then
$y(t)\to+\infty$. If $a<0$ and
$d<\frac{-b+\sqrt{b^{2}-4ac}}{2a}$, then
$y(t)\to-\infty$. Thus, initial value $d$ is very
important since certain values can prevent blow-up from occurring.
Also, $d$ influences the value of $t_{b}$ at which blow-up
occurs.

(4) If $a=0$, then the equation is linear. Using separation of variables, we
obtain
\begin{gather*}
\int{\frac{dy}{by+c}}    =\int{dt},\\
y(t)    =\frac{(bd+c)  e^{bt}-c}{b}.
\end{gather*}
Here we find that $y(t)$ is bounded for any finite time
$t>0$.
\end{proof}

\subsection*{Example}
Let us investigate the blow-up property of the
solution for the initial-value problem
\begin{equation}
\begin{gathered}
y'(t)=-4y^{2}+5y-1,\\
y(0)=d
\end{gathered} \label{e2.12}
\end{equation}
with three different values of $d$ as indicated below.

First, we note that $a=-4<0$, $b=5$, $c=-1$, and $4ac-b^{2}=-9<0$.
Also, $(-b+\sqrt{b^{2}-4ac})/(2a)=0.25$. According to Theorem
\ref{thm1.1}, we expect that the solution of problem \eqref{e2.12}
blows up for any $d<0.25$ and is bounded otherwise. The solution
of problem \eqref{e2.12} is given by formula \eqref{e2.7}:
\[
y(t)=\frac{-2d+2+( 8d-2) e^{3t}}{-8d+8-(-8d+2) e^{3t}}.
\]
Let us notice that if we differentiate function $y$ with respect
to $d$, then the corresponding derivative is as follows:
\[
\frac{9e^{3t}}{(-4d+4+4de^{3t}-e^{3t}) ^{2}}.
\]
Therefore, the function $y$ and its derivative with respect to $d$ are
both discontinuous when
\[
t=\frac{1}{3}\ln \big( \frac{4d-4}{4d-1}\big) ,
\]
which holds only for $d<0.25$.

If $d=2$,  we have
\[
y(t)=\frac{-2+14e^{3t}}{-8+14e^{3t}}.
\]
This function is bounded for any finite time $t>0$.

If $d=0.5$,  we have
\[
y(t)=\frac{1+2e^{3t}}{4+2e^{3t}}.
\]
This function is bounded for any finite time $t>0$.

If $d=0$,  we have
\[
y(t)=\frac{1-e^{3t}}{4-e^{3t}}.
\]
Here $y(t)\to-\infty$ when $t_{b}=\ln(4)/3$.

\section{Riccati Problems with Variable Coefficients}

Let $a(t)$, $b(t)$, and $c(t)$ be continuous and differentiable
functions for $t\geq0$. We consider the
 initial-value problem
\begin{equation}
\begin{gathered}
y'(t)=a(t)y^{2}+b(t)y+c(t),\\
y(0)=d.
\end{gathered} \label{e3.1}
\end{equation}
We notice that $\varphi(t)=\exp\big(\int_0^t b(\hat{t})d\hat{t}\big)$
is the unique solution of the initial-value problem
\begin{gather*}
\varphi'(t)=b(t)\varphi(t),\\
\varphi(0)=1.
\end{gather*}
Also, $\varphi(t)$ is bounded for any finite
$t\geq0$. Let $\psi(t)$ satisfy the initial-value problem
\begin{equation}
\begin{gathered}
\psi'(t)=a(t)\psi^{2}(t)\varphi(t)+\frac{c(t)}{\varphi(t)},\\
\psi(0)=d.
\end{gathered} \label{e3.2}
\end{equation}
Then $y(t)=\varphi(t)\psi(t)$ satisfies the Riccati initial-value
problem \eqref{e3.1}. We now investigate conditions on $a(t)$,
$c(t)$, and $d$ that lead to unbounded growth of $\psi(t)$, and
therefore, of $y(t)$.

\begin{theorem} \label{thm2.1}
Let $a(t)c(t)\geq0$. Then the
following is true for the solution $y(t)$ of \eqref{e3.1}:
\begin{itemize}
\item[(1)] If $a(t)\exp\big(\int_0^t b(\hat{t})d\hat{t}\big)
\geq k_{1}>0$  and $c(t)\exp\big(-\int_0^t b(\hat{t})d\hat{t}\big)\geq k_{2}>0$
 for all $t>0$, then $y(t)\to+\infty$ for any initial condition $d$.

\item[(2)] If $a(t)\exp\big( \int_0^t b(\hat{t})d\hat{t}\big)\leq k_{3}<0$
 and $c(t)\exp\big(-\int_0^t  b(\hat{t})d\hat{t}\big)\leq k_{4}<0$
 for all $t>0$, then $y(t)\to-\infty$ for any initial condition $d$.

\item[(3)] Let $a(t)c(t)$ have zeroes at $t=t_{1},t_{2},t_{3},\dots $ and let
$g_{1}(t)$ and $g_{2}(t)$ be nontrivial,
continuous, and differentiable functions such that for all $t\geq0$,
\begin{gather*}
g_{1}(t)   \leq\min\big\{a(t)e^{\int_{0}^{t}b(\hat{t})d\hat{t}
},c(t)e^{-\int_{0}^{t}b(\hat{t})d\hat{t}}\big\}  ,\\
g_{2}(t)   \geq\max\big\{a(t)e^{\int_{0}^{t}b(\hat{t})d\hat{t}},
c(t)e^{-\int_{0}^{t}b(\hat{t})d\hat{t}}\big\}  .
\end{gather*}

\end{itemize}
Then one of the following three statements is true:
\begin{itemize}

\item[(a)] $y(t)\to+\infty$ if for some $t_{b_{1}}>0$,
\begin{equation}
\int_{0}^{t_{b_{1}}}g_{1}(t)dt=\frac{\pi}{2}-\mathop{\rm arctan}(d). \label{e3.3}
\end{equation}

\item[(b)] $y(t)\to-\infty$  if for some $t_{b_{2}}>0$,
\begin{equation}
\int_{0}^{t_{b_{2}}}g_{2}(t)dt=-\frac{\pi}{2}-\mathop{\rm arctan}(d). \label{e3.4}
\end{equation}

\item[(c)] $y(t)$ is bounded for all $t>0$ if the following two
inequalities hold simultaneously:
\[
\int_{0}^{t}g_{2}(\hat{t})d\hat{t}
<\frac{\pi}{2}-\mathop{\rm arctan}(d), \quad
\int_{0}^{t}g_{1}(\hat{t})d\hat{t} >-\frac{\pi}{2}-\mathop{\rm arctan}(d).
\]
\end{itemize}
\end{theorem}

\begin{proof}
(1) By the comparison theorem \cite[pp. 221-223]{Cole}, the
solution $\bar{\psi }(t)$ of the initial-value problem
\begin{gather*}
\bar{\psi}'(t)=k_{1}\bar{\psi}^{2}(t)+k_{2},\\
\bar{\psi}(0)=\bar{d},
\end{gather*}
where $\bar{d}\leq d$, is a lower solution for \eqref{e3.2}. By
Theorem \ref{thm1.1}, $\bar{\psi}(t)$ approaches $+\infty$ as $t$
approaches $\bar{t}_{b}$. Therefore, $\psi(t)\geq\bar{\psi}(t)$
also approaches $+\infty$ as $t$ approaches some
$t_{b}\leq\bar{t}_{b}$.

(2) By the comparison theorem, solution $\tilde{\psi}(t)$ of the
initial-value problem
\begin{gather*}
\tilde{\psi}'(t)=k_{3}\tilde{\psi}^{2}(t)+k_{4},\\
\tilde{\psi}(0)=\tilde{d},
\end{gather*}
where $\tilde{d}\geq d$, is an upper solution for \eqref{e3.2}. By
Theorem \ref{thm1.1}, $\tilde{\psi}(t)$ approaches $-\infty$ as $t$
approaches $\tilde{t}_{b}$. Therefore,
$\psi(t)\leq\tilde{\psi}(t)$ also approaches $-\infty$ as $t$
approaches $t_{b}\leq\tilde{t}_{b}$.

(3)(a). By the comparison theorem, the solution
$\hat{\psi}(t)$ of the initial-value problem
\begin{gather*}
\hat{\psi}'(t)=g_{1}(t)\hat{\psi}^{2}(t)+g_{1}(t),\\
\hat{\psi}(0)=\hat{d},
\end{gather*}
where $\hat{d}\geq d$, is a lower solution for \eqref{e3.2}. Using the
separation of variables method, we obtain that $\hat{\psi}(t)$ approaches
$+\infty$ as $t$ approaches $\hat{t}_{b}$ provided that \eqref{e3.3} holds.
Therefore, $\psi(t)\geq\hat{\psi}(t)$ also approaches $+\infty$ as $t$
approaches $t_{b_{1}}\leq\hat{t}_{b}$. If condition \eqref{e3.3} does not hold,
then $\hat{\psi}(t)$ and $\psi(t)$ are bounded from above for all $t>0$.

(3)(b) By the comparison theorem, the solution $\breve {\psi}(t)$ of
the initial-value problem
\begin{gather*}
\breve{\psi}'(t)=g_{2}(t)\breve{\psi}^{2}(t)+g_{2}(t),\\
\breve{\psi}(0)=\breve{d},
\end{gather*}
where $\breve{d}\geq d$, is an upper solution for \eqref{e3.2}. Using the
separation of variables method, we obtain that $\breve{\psi}(t)$ approaches
$-\infty$ as $t$ approaches $\breve{t}_{b}$ provided that \eqref{e3.4} holds.
Therefore, $\psi(t)\leq\breve{\psi}(t)$ also approaches $-\infty$ as $t$
approaches $t_{b_{2}}\leq\breve{t}_{b}$. If condition \eqref{e3.4} does not
hold, then $\breve{\psi}(t)$ and $\psi(t)$ are bounded from below for all
$t>0$.

(3)(c) The proof of this part follows directly from the proofs described in parts (a)
and (b) above.
\end{proof}

\begin{theorem} \label{thm2.2}
Let $c(t)=0$. One of the
following three statements is true for the solution $y(t)$ of problem
\eqref{e3.1}:
\begin{itemize}

\item[(1)] $y(t)\to+\infty$ provided $d>0$  and
\begin{equation}
\int_{0}^{t_{b_{1}}}a(t)e^{\int_{0}^{t}b(\hat{t})d\hat{t}}dt
=\frac{1}{d} \label{e3.5}
\end{equation}
for some $t_{b_{1}}>0$.

\item[(2)] $y(t)\to-\infty$ provided $d<0$ and
\begin{equation}
\int_{0}^{t_{b_{2}}}a(t)e^{\int_{0}^{t}b(\hat{t})d\hat{t}}dt=\frac{1}{d}
\label{e3.6}
\end{equation}
for some $t_{b_{2}}>0$.

\item[(3)] If
\[
-| \frac{1}{d}| <\int_{0}^{t}a(\bar{t})e^{\int_{0}^{\bar
{t}}b(\hat{t})d\hat{t}}d\bar{t}<| \frac{1}{d}|
\]
for all $t\geq0$, then $y(t)$ is bounded for all $t\geq0$.
\end{itemize}
\end{theorem}

\begin{proof}
(1) Applying the separation of variables method to the problem
\begin{gather*}
\psi'(t)=a(t)\psi^{2}(t),\\
\psi(0)=d,
\end{gather*}
we obtain that $\psi(t)$ approaches $+\infty$ as $t$ approaches $t_{b_{1}}$
provided that $d>0$, and \eqref{e3.5} holds. Otherwise, $\psi(t)$ is bounded
from above for all $t>0$.

(2) Similarly, we obtain that $\psi(t)$ approaches $-\infty$ as $t$
approaches $t_{b_{2}}$ provided that $d<0$, and \eqref{e3.6} holds.
Otherwise, $\psi(t)$ is bounded from below for all $t>0$.

(3) The proof of this part follows directly from the proofs of parts 1 and 2 above.
\end{proof}

\begin{theorem} \label{thm2.3}
For the case $a(t)c(t)<0$, one of the following two statements is true
for the solution $y(t)$ of \eqref{e3.1}:
\begin{itemize}
\item[(1)] If $a(t)\exp\big(\int_0^t b(\hat{t})d\hat{t}\big)\geq k_{5}>0$
 and $0>c(t)\exp\big(-\int_0^t b(\hat{t})d\hat{t}\big)\geq k_{6}$
for all $t>0$, then $y(t)\to+\infty$ for any
$d>\sqrt{|k_{6}/k_{5}|}$.

\item[(2)] If $a(t)\exp\big(\int_0^t b(\hat{t})d\hat{t}\big)\leq
k_{7}<0$
 and $0<c(t)\exp\big(-\int_0^t b(\hat{t})d\hat{t}\big)\leq k_{8}$
 for all $t>0$, then $y(t)\to-\infty$ for any
$d<-\sqrt{| k_{8}/k_{7}|}$.
\end{itemize}
\end{theorem}

\begin{proof}
(1) By the comparison theorem \cite[pp. 221-223]{Cole}, the
solution $\bar{\psi}(t)$ of the initial-value problem
\begin{gather*}
\bar{\psi}'(t)=k_{5}\bar{\psi}^{2}(t)+k_{6},\\
\bar{\psi}(0)=\bar{d},
\end{gather*}
where $\sqrt{|k_{6}/k_{5}|}<\bar{d}\leq d$, is a lower solution
for \eqref{e3.2}. By Theorem \ref{thm1.1},
$\bar{\psi}(t)$ approaches $+\infty$ as $t$ approaches
$\bar{t}_{b}$. Therefore, $\psi(t)\geq\bar{\psi }(t)$ also
approaches $+\infty$ as $t$ approaches some
$t_{b}\leq\bar{t}_{b}$.

(2) By the comparison theorem, the solution $\tilde{\psi}(t)$ of the
initial-value problem
\begin{gather*}
\tilde{\psi}'(t)=k_{7}\tilde{\psi}^{2}(t)+k_{8},\\
\tilde{\psi}(0)=\tilde{d},
\end{gather*}
where $-\sqrt{|k_{8}/k_{7}|}>\tilde{d}\geq d$ is an upper solution
for \eqref{e3.2}. By Theorem \ref{thm1.1},
$\tilde{\psi}(t)$ approaches $-\infty$ as $t$
approaches $\tilde{t}_{b}$. Therefore,
$\psi(t)\leq\tilde{\psi}(t)$ also approaches $-\infty$ as $t$
approaches $t_{b}\leq\tilde{t}_{b}$.
\end{proof}

\subsection*{Acknowledgements}
The authors are grateful to the College of Natural Sciences
of California State University San Bernardino for its support of
this project.
The authors would like to thank the
anonymous referee for his or her helpful suggestions.


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\end{document}