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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 250, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/250\hfil Asymptotic behavior]
{Asymptotic behavior of positive solutions of the
nonlinear differential equation $t^2u''=u^n$}

\author[M.-R. Li, H.-Y. Yao, Y.-T. Li \hfil EJDE-2013/250\hfilneg]
{Meng-Rong Li, Hsin-Yu Yao, Yu-Tso Li}  % in alphabetical order

\address{Meng-Rong Li \newline
Department of Mathematical Sciences,
National Chengchi University, Taipei, Taiwan}
\email{liwei@math.nccu.edu.tw}

\address{Hsin-Yu Yao \newline
Department of Mathematical Sciences,
National Chengchi University, Taipei, Taiwan}
\email{diadia0914@gmail.com}

\address{Yu-Tso Li \newline
Department of Aerospace and Systems Engineering,
Feng Chia University, Taichung, Taiwan}
\email{joycelion74@gmail.com}

\thanks{Submitted October 5, 2013. Published November 20, 2013.}
\subjclass[2000]{34A34, 34C11, 34C60}
\keywords{Nonlinear differential equation; Emden-Fowler equation; blow-up rate}

\begin{abstract}
 In this article we study properties of positive solutions
 of the ordinary differential equation $t^2u''=u^n$ for
 $1<n\in\mathbb{N}$, we obtain conditions for their blow-up
 in finite time, and some properties for global solutions.
 Equations containing more general nonlinear terms are also considered.
\end{abstract}

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



\section{Introduction}

Some interesting results on the blow-up, blow-up rates, and estimates 
for the life-span of solutions of the Emden-Fowler
equation and the semi-linear wave equation $\square u+f(u)=0$
have been obtained, as shown in the references. 

Here we wish to study the Emden-Fowler type wave equation, i.e.
solutions, independent of the space variable $x$, of the equation
$t^2 u_{tt}-\Delta u=u^n$ for $n>1$.

The  existence and uniqueness of local solutions of the initial-value problem
\begin{equation} \label{e-star}
\begin{gathered}
t^2u''=u^n,\quad 1<n\in\mathbb{N},\\
u(1)=u_0,\quad u'(1)=u_1,
\end{gathered}
\end{equation}
follow by standard arguments. Considering the transformation 
$t=e^s$, $u(t)=v(s)$, we have
$t^2u''(t)=-v_s(s)+v_{ss}(s)$,
$v( s)^n=-v_s(s)+v_{ss}(s)$ and
$v(0)=u(1)=u_0$; $v_s(0)=u'(1)=u_1$, the problem \eqref{e-star}
can be transformed into 
\begin{equation}
\begin{gathered}
v_{ss}(s)-v_s(s)=v^n(s),\\
v(0)=u_0,\quad v_s(0)=u_1.
\end{gathered}\label{e1.1}
\end{equation}

Thus, the  existence of local solutions $u$ for \eqref{e-star} in
$(1,T)$ is equivalent to the  existence of local solutions $v$
for \eqref{e1.1} in $(0,\ln T)$. In this article, we
give estimates for the life-span $T^{\ast}$ of positive solutions $u$ of
\eqref{e-star} in three different cases. The main results are as follows:

\begin{itemize}
\item[(a)] $u_1=0$, $u_0>0$:  $T^{\ast}\leq e^{k_1}$, 
$k_1:=s_0 +\frac{2(n+3)}{8-\epsilon}\frac{2}{n-1}v(s_0)
^{\frac{1-n}{2}}$, $\varepsilon\in(0,1)$;

\item[(b)] $u_1>0$, $u_0>0$:

\begin{itemize}
\item[(i)] $E(0)\geq0$, $T^{\ast}\leq e^{k_2}$, $k_2
:=\frac{2}{n-1}\sqrt{\frac{n+1}{2}}u_0^{\frac{1-n}{2}}$;

\item[(ii)] $E(0)<0$, $T^{\ast}\leq e^{k_3}$, 
$k_3 :=\frac{2}{n-1}\frac{u_0}{u_1}$;
\end{itemize}

\item[(c)] $u_1<0$, $u_0\in(0,(-u_1)^{\frac{1}{n}})$:
 $u(t)\leq(u_0-u_1-u_0^n) +(u_1+u_0^n)t-u_0^n\ln t$.
\end{itemize}
where $E(0)$ is defined in the next section and $s_0$ 
is given by \eqref{e3.3}.

In Section 6, we replace the nonlinear term $v^n$ by a more general 
increasing function $f(v)$.

\section{Notation and some lemmas}

For a given function $v$,  we use the following functions
\[
a(s): =v(s)^2,\quad 
E(0): =u_1^2-\frac{2}{n+1}u_0^{n+1},\quad
J(s):=a(s) ^{-\frac{n-1}{4}},
\]
where $u_0$ and $u_1$ are the given initial conditions.

By an easy calculation we can obtain the following two Lemmas; we
shall omit the proof of the first lemma.


\begin{lemma} \label{lem1}
Suppose that $v\in C^2[0,T]$ is the
solution of \eqref{e1.1}, then
\begin{gather}
E(s)  =v_s(s)^2-2\int_0^s v_s(r)^2dr-\frac{2}{n+1}v(s)^{n+1}
=E(0),\label{e2.1}\\
(n+3)v_s(s)^2   =(n+1) E(0)+a''(s)-a'(s)+2(n+1)
\int_0^s v_s(r)^2dr,\label{e2.2}\\
J''(s)  =\frac{n^2-1}{4}J(s)^{\frac{n+3}{n-1}}(E(0)-
\frac{a'(s)}{n+1}+2\int_0^s v_s(r)^2dr), \label{e2.3} \\
\begin{aligned}
J'(s)^2 & =J'(0)^2
+\frac{(n-1)^2}{4}E(0)(J(
s)^{\frac{2(n+1)}{n-1}}-J(0)
^{\frac{2(n+1)}{n-1}}) \\
& \quad  +\frac{(n-1)^2}{2}J(s)^{\frac{2(n+1)}{n-1}}
\int_0^s v_s(r)^2dr.
\end{aligned} \label{e2.4} 
\end{gather}
\end{lemma}


\begin{lemma} \label{lem2}
For $u_0>0$, the positive solution $v$ of \eqref{e1.1} satisfies:
\begin{gather}
\text{(i) If $u_1\geq0$, then $v_s(s)>0$ for  all 
 $s>0$}  \label{e2.5}\\
\text{(ii) If $u_1 <0$, $u_0\in(0,(-u_1)^{\frac{1}{n}})$,
 then $v_s(s)<0$  for  all $s>0$.} \label{e2.6}
\end{gather}
\end{lemma}

\begin{proof}
(i) Since $v_{ss}(0)=u_1+u_0^n>0$, we know
that $v_{ss}(s)>0$ in $[0,s_1)$ and
$v_s(s)$ is increasing in $[0,s_1)$ for some
$s_1>0$. Moreover, since $v$ and $v_s$ are increasing in 
$[0,s_1)$, $v_{ss}(s_1)=v_s(s_1) +v(s_1)^n>v_s(0)+v(0)
^n>0$ for all $s\in[0,s_1)$ and $v_s(s_1)>v_s(s)>0$ for all 
$s\in[0,s_1)$, we know that there exists a positive number $s_2>0$, 
such that $v_s(s)>0$ for all $s\in[0,s_1+s_2)$.
Continuing this process, we obtain $v_s(s)>0$ for all $s>0$,
for which the solution exists. 

(ii) Since $v_{ss}(0)=v_s(0)+v(0)^n=u_1+u_0^n<0$, there exists a 
positive number $s_1>0$
such that $v_{ss}(s)<0$ in $[0,s_1),v_s(s)$ is decreasing in $[0,s_1)$;
therefore, $v_s(s)<v_s(0)=u_1<0$ for all
$s\in[0,s_1)$ and $v(s)$ is decreasing in
$[0,s_1)$. Moreover, since $v$ and $v_s$ are decreasing in
$[0,s_1)$, $v_{ss}(s)=v_s(s)+v(s)^n<v_s(0)+v(0)^n<0$
 for  all $s\in[0,s_1)$ and $v_s(s_1)<v_s(s)<0$ for all $s\in[0,s_1)$, we know
that there exists a positive number $s_2>0$, such that 
$v_s(s)<0$ for all $s\in[0,s_1+s_2)$. Continuing this
process, we obtain $v_s(s)<0$ for all $s>0$ in the interval
of existence.
\end{proof}

\section{Life-span of positive solutions of \eqref{e-star} when $u_1=0$, $u_0>0$}

In this section we want to estimate the life-span of a positive solution
$u$ of \eqref{e-star} if $u_1=0$, $u_0>0$. Here the life-span
$T^{\ast}$ of $u$ means that $u$ is the solution of equation 
\eqref{e-star} and $u$ exists only in $[0,T^{\ast})$ so that
the problem \eqref{e-star} possesses a positive solution $u\in
C^2[0,T^{\ast})$.


\begin{theorem} \label{thm3}
 For $u_1=0$, $u_0>0$, the positive solution $u$ of
\eqref{e-star} blows up in finite time; that is, there exists
$T^{\ast}<\infty$ so that
\[
u(t)^{-1}\to 0 \quad\text{as }t\to  T^{\ast}.
\]
\end{theorem}

\begin{proof}
 By \eqref{e2.5}, we know that $v_s(s)>0$, $a'(s)>0$ for all $s>0$ provided that
$u_1=0$, $u_0>0$.
By Lemma \ref{lem1}, 
\begin{gather*}
a''(s)-a'(s) =2(v_s(s)^2+v(s)^{n+1}), \\
(a'(s)e^{-s})' =e^{-s}(a''(s)-a'(s) )=2e^{-s}(v_s(s)^2+v(s) ^{n+1}),\\
a'(s)e^{-s} =2\int_0^s e^{-r}(v_s(r)^2+v(r)^{n+1}) dr
\geq4 \int_0^se^{-r}v_s(r)v(r)^{\frac{n+1}{2}}dr, 
\end{gather*}
and $a'(0)=0$, hence we have
\begin{align*}
a'(s)e^{-s}  
&  \geq\frac{8}{n+3}(v(r)^{(n+3)/2}e^{-r}\mid_{r=0}^s
+\int_0^s v(r)^{(n+3)/2}e^{-r}dr)\\
&  =\frac{8}{n+3}(v(s)^{(n+3)/2}e^{-s}-v(
0)^{(n+3)/2})+\frac{8}{n+3}
\int_0^s v(r)^{(n+3)/2}e^{-r}dr.
\end{align*}
Since $a'(s)>0$ for all $s>0$, $v$ is increasing on $(0,\infty)$ and
\begin{equation} \label{e3.1}
\begin{gathered}
\begin{aligned}
a'(s)e^{-s}
&  \geq\frac{8}{n+3}(v( s)^{(n+3)/2}e^{-s}-v(0)^{(n+3)/2})
+\frac{8}{n+3} \int_0^s v(0)^{(n+3)/2}e^{-r}dr \\
& =\frac{8}{n+3}(v(s)^{(n+3)/2}e^{-s}-v(
0)^{(n+3)/2})+\frac{8}{n+3}v(0)
^{(n+3)/2}(1-e^{-s}),
\end{aligned}\\
a'(s) \geq\frac{8}{n+3}(v(s) ^{(n+3)/2}-v(0)^{(n+3)/2})
=\frac{8}{n+3}(v(s)^{(n+3)/2}-u_0^{(n+3)/2}).
\end{gathered}
\end{equation}
Using $u_1=0$ and integrating \eqref{e1.1}, we obtain
\begin{equation} \label{e3.2}
\begin{gathered}
v_s(s)  =v(s)-u_0+\int_0^s v(r)^ndr, \\
v_s(s) \geq v(s)-u_0+\int_0^sv(0)^ndr=v(s)-u_0+u_0^n s, \\
(e^{-s}v(s))_s   =e^{-s}(v_s(s)-v(s))\geq e^{-s}(u_0^n s-u_0), \\
a'(s)  \geq\frac{8}{n+3}(v(s)^{(n+3)/2}-v(0)^{(n+3)/2})
=\frac{8}{n+3}(v(s)^{(n+3)/2}-u_0^{(n+3)/2}).
\end{gathered}
\end{equation}
According to \eqref{e3.2}, and since $v'(s) >0$,
$v(s)^{(n+3)/2}\geq(u_0+u_0^n(e^s-1-s))^{(n+3)/2}$, for all
$\epsilon\in(0,1)$, we obtain
\begin{gather*}
\epsilon v(s)^{(n+3)/2}   \geq\epsilon(u_0+u_0^n(e^s-1-s))^{(n+3)/2},\\
\begin{aligned}
\epsilon v(s)^{(n+3)/2}-8u_0^{(n+3)/2}
& \geq\epsilon(u_0+u_0^n(e^s-1-s)) ^{(n+3)/2}-8u_0^{(n+3)/2}\\
&  \geq\epsilon(u_0^{(n+3)/2}+u_0^{\frac{n(n+3)}{2}}(e^s-1-s)^{(n+3)/2})
-8u_0^{\frac{n+3} {2}}\\
&  =(\epsilon-8)u_0^{(n+3)/2}+\epsilon u_0
^{\frac{n(n+3)}{2}}(e^s-1-s)^{(n+3)/2}.
\end{aligned}
\end{gather*}
Now, we want to find a number $s_0>0$ such that
\begin{equation}
e^{s_0}-s_0=1+\big(\frac{8-\epsilon}{\epsilon}u_0^{\frac{n+3}
{2}(1-n)}\big)^{2/(n+3)}. \label{e3.3}
\end{equation}


This means that there exists a number $s_0>0$  satisfying \eqref{e3.3} with
$\epsilon\in(0,1)$ such that
\[
\epsilon v(s)^{(n+3)/2}-8u_0^{(n+3)/2}\geq0\quad\text{for  all }s\geq s_0.
\]
From \eqref{e3.1}, it follows that
\begin{align*}
a'(s)
& \geq\frac{8}{n+3}v(s) ^{(n+3)/2}-\frac{8}{n+3}u_0^{(n+3)/2}\\
&=\frac{8-\epsilon}{n+3}v(s)^{(n+3)/2}+\frac{\epsilon v(s)
^{(n+3)/2}-8u_0^{(n+3)/2}}{n+3}\\
&  \geq\frac{8-\epsilon}{n+3}v(s)^{(n+3)/2}, \quad \text{for all }
s\geq s_0.
\end{align*}
For all $s\geq s_0$, $\epsilon\in(0,1)$, we obtain that
\begin{gather*}
2v(s)v_s(s)\geq\frac{8-\epsilon}{n+3}v(s)^{(n+3)/2}, \\
v(s)^{-\frac{n+1}{2}}v_s(s)\geq\frac {8-\epsilon}{2(n+3)}, \\
\frac{2}{1-n}(v(s)^{\frac{1-n}{2}})_s \geq\frac{8-\epsilon}{2(n+3)}
\end{gather*}
and hence
\[
(v(s)^{\frac{1-n}{2}})_s\leq\frac{8-\epsilon }{2(n+3)}\frac{1-n}{2}.
\]
Integrating the above inequality, we conclude that
\[
v(s)^{\frac{1-n}{2}}\leq v(s_0)^{\frac{1-n}
{2}}-\frac{8-\epsilon}{2(n+3)}\frac{n-1}{2}(s-s_0).
\]
Thus, there exists a number
\[
s_1^{\ast}\leq s_0+\frac{2(n+3)}{8-\epsilon}\frac{2}
{n-1}v(s_0)^{\frac{1-n}{2}}=:k_1
\]
such that $v(s)^{-1}\to 0$ for $s\to  s_1^{\ast}$, that is, 
$u(t)^{-1}\to 0$ as $t\to  e^{k_1}$, which implies that the
 life-span $T^{\ast}$ of a positive solution $u$ is finite and 
$T^{\ast}\leq e^{k_1}$.
\end{proof}

\section{Life-span of positive solutions of \eqref{e-star} when $u_1>0$, $u_0>0$}

In this section we estimate the life-span of a positive solution $u$ of
\eqref{e-star} whenever $u_1>0$, $u_0>0$.

\begin{theorem} \label{thm4}
 For $u_1>0$, $u_0>0$, the positive solution $u$ of
\eqref{e-star} blows up in finite time; that is, there exists a
number $T^{\ast}<\infty$ so that
\[
u(t)^{-1}\to 0\quad\text{as }t\to  T^{\ast}.
\]
\end{theorem}

\begin{proof}
We separate the proof into two parts depending on whether
$E(0)\geq0$ or $E(0)<0$.

(i) Assume that $E(0)\geq0$. By \eqref{e2.1} and
\eqref{e2.5} we have
\begin{gather*}
v_s(s)^2-\frac{2}{n+1}v(s)^{n+1}\geq E(0), \\
v_s(s)^2\geq\frac{2}{n+1}v(s)^{n+1}
+E(0),v_s(s)\geq\sqrt{\frac{2}{n+1}v(s)^{n+1}+E(0)}.
\end{gather*}
Since $E(0)\geq0$, we obtain
\begin{gather*}
v_s(s)\geq\sqrt{\frac{2}{n+1}}v(s)^{\frac{n+1}{2}}, \\
v(s)^{-\frac{n+1}{2}} \cdot v_s(s)\geq \sqrt{\frac{2}{n+1}},\\
(v(s)^{\frac{1-n}{2}})_s\leq\frac{1-n}{2}\sqrt{\frac{2}{n+1}}.
\end{gather*}
Integrating the above inequality, we obtain
\[
v(s)^{\frac{1-n}{2}}\leq u_0^{\frac{1-n}{2}}+\frac{1-n}
{2}\sqrt{\frac{2}{n+1}}s.
\]
Thus, there exists
\[
s_2^{\ast}\leq\frac{2}{n-1}\sqrt{\frac{n+1}{2}}u_0^{\frac{1-n}{2}}=:k_2
\]
such that $v(s)^{-1}\to 0$ for $s\to  s_2^{\ast}$;
 that is, $u(t)^{-1}\to 0$  as  $t\to  e^{k_2}$,
 which means that the life-span $T^{\ast}$ of a
positive solution $u$ is finite and 
$T^{\ast}\leq e^{k_2}$.
\medskip

(ii ) Assume that $E(0)<0$. From \eqref{e2.1} and
\eqref{e2.5} we obtain that $J'(s)
=-\frac{n-1}{4}a(s)^{-\frac{n+3}{4}}a'(
s)$, $a'(s)>0$, $v_s(s)>0$ for all $s>0$ and
\begin{gather*}
\begin{aligned}
J'(s)&  =-\frac{n-1}{2}\sqrt{\frac{2}{n+1}+E(
0)a(s)^{-\frac{n+1}{2}}+2a(s)^{-\frac{n+1}{2}}
\int_0^s v_s(r)^2dr}\\
&  \leq-\frac{n-1}{2}\sqrt{\frac{2}{n+1}+E(0)a(
s)^{-\frac{n+1}{2}}},
\end{aligned}\\
J(s)  \leq J(0)-\frac{n-1}{2}\int_0^s
\sqrt{\frac{2}{n+1}+E(0)a(r)^{-\frac{n+1}{2}}}dr.
\end{gather*}
Since $E(0)<0$ and $a'(s)>0$ for all $s>0$, 
\begin{align*}
J(s)&  \leq J(0)-\frac{n-1}{2}
\int_0^s \sqrt{\frac{2}{n+1}+E(0)a(0)^{-\frac{n+1}{2}} }dr\\
&  =a(0)^{-\frac{n-1}{4}}-\frac{n-1}{2}\sqrt{\frac{2}
{n+1}+E(0)a(0)^{-\frac{n+1}{2}}}s.
\end{align*}
Thus, there exists a number
\[
s_3^{\ast}\leq\frac{2}{n-1}a(0)^{-\frac{n-1}{4}}(
\frac{2}{n+1}+E(0)a(0)^{-\frac{n+1}{2}})^{-\frac{1}{2}}=:k_3
\]
such that $J(s_3^{\ast})=0$ and $a(s)^{-1}\to 0$ for $s\to  s_3^{\ast}$;
 that is, $u(t)^{-1}\to 0$  as $t\to  e^{k_3}$. This means
that the life-span $T^{\ast}$ of $u$ is finite and 
$T^{\ast}\leq e^{k_3}$.
\end{proof}


\section{Life-span of positive solutions of \eqref{e-star} when $u_1<0$}

Finally, we estimate the life-span of a positive solution $u$ of 
\eqref{e-star} when $u_1<0$. 

\begin{theorem} \label{thm5}
For $u_1<0$, $u_0\in(0,(-u_1)^{\frac{1}{n}})$ we have
\[
u(t)\leq(u_0-u_1-u_0^n)+(u_1+u_0^n)t-u_0^n\ln t,
\]
and in particular, if $E(0)\geq0$, we have
\[
u(t)\leq(u_0^{\frac{1-n}{2}}+\frac{n-1}{2}
\sqrt{\frac{2}{n+1}}\ln t)^{\frac{2}{1-n}}.
\]
\end{theorem}

\begin{proof}
(i) By \eqref{e1.1} and
integrating this equation with respect to $s$, we get 
$v_s(s) =(u_1-u_0)+v(s)+\int_0^sv(r)^ndr$.
 By \eqref{e2.6}, we have that $v$ is decreasing and
\[
v_s(s)\leq(u_1-u_0)+v(s)+\int_0^sv(0)^ndr=(u_1-u_0)+v(s)+u_0^ns,
\]
\begin{align*}
e^{-s}v(s)-u_0  &  \leq(u_1-u_0)
\int_0^se^{-r}dr+u_0^n\int_0^s re^{-r}dr\\
&  =(u_1-u_0)(1-e^{-s})+u_0^n(-se^{-s}-e^{-s}+1);
\end{align*}
that is,
\begin{align*}
u(t)&  \leq(u_0-u_1)+u_1t+u_0 ^n(t-1-\ln t)\\
&  =(u_0-u_1-u_0^n)+(u_1+u_0^n) t-u_0^n\ln t.
\end{align*}


(ii) If $E(0)\geq0, $ by \eqref{e2.1}, we have
\begin{gather*}
v_s(s)^2-\frac{2}{n+1}v(s)^{n+1} =E(0)+2\int_0^sv_s(r)^2dr\geq E(0),\\
v_s(s)^2  \geq E(0)+\frac{2} {n+1}v(s)^{n+1}
\geq\frac{2}{n+1}v(s)^{n+1}.
\end{gather*}
By \eqref{e2.6}, we obtain that 
$-v_s(s)\geq \sqrt{\frac{2}{n+1}}v(s)^{\frac{n+1}{2}},
\frac{2}{n-1}(v(s)^{\frac{1-n}{2}})_s\geq\sqrt{\frac{2}{n+1}}$ 
and
\begin{gather*}
\sqrt{\frac{2}{n+1}}s  \leq\frac{2}{n-1}(v(s)
^{\frac{1-n}{2}}-v(0)^{\frac{1-n}{2}}),\\
v(s)^{\frac{1-n}{2}} \geq(u_0^{\frac{1-n}{2}
}+\frac{n-1}{2}\sqrt{\frac{2}{n+1}}s).
\end{gather*}
Then, we know that 
\[
v(s)\leq(u_0^{\frac{1-n}{2}
}+\frac{n-1}{2}\sqrt{\frac{2}{n+1}}s)^{\frac{2}{1-n}},
\quad\text{for all  }s\geq0;
\]
 that is,
\[
u(t)\leq(u_0^{\frac{1-n}{2}}+\frac{n-1}{2}
\sqrt{\frac{2}{n+1}}\ln t)^{\frac{2}{1-n}}\quad \text{for all } t\geq1.
\]
\end{proof}


\section{A generalization of Theorem \ref{thm4}}

In this section we want to extent the blow-up result for the following 
generalization of \eqref{e1.1}, 
\begin{equation}
\begin{gathered}
v_{ss}(s)-v_s(s)=f(v),\\
v(0)=v_0,\quad v_s(0)=v_1,
\end{gathered}  \label{e6.1}
\end{equation}
where $f$ is an increasing continuous function with $f(0)=0$.
We have the following result.

\begin{theorem} \label{thm6}
Suppose that $f$ is an increasing function with $f(0)=0$ and suppose $v$ 
is a positive solution of \eqref{e6.1}.
 If $F(v):=\int_0^{v}f(r)dr$, then
\begin{equation}
\bar{E}(s):=v_s(s)^2-2
\int_0^s v_s(r)^2dr-2F(v(s))\label{e6.2}
\end{equation}
is  constant. 
Furthermore, if there exists a positive constant $k$  such that
 $F( s)\geq ks^{p+1}$, $p>1$ for all $s\geq0$, and $ v_1>0$, 
then the life span of $v$ is finite.
\end{theorem}

\begin{proof}
 By an argument similar to that used in proving \eqref{e2.1},
we easily obtain that $\bar{E}(s)$ is a constant. Since $f$
is increasing, we have
\[
vf(v)=(v-0)\cdot(f(v)-f(0))\geq0\quad \text{for }v\geq0,
\]
thus
\begin{equation}
(v^2)_s-2v^2(s)\geq2v_0(v_1-v_0)+2\int_0^sv_s^2(r)dr. \label{e6.3}
\end{equation}
By \eqref{e6.2} and \eqref{e6.3}, we have 
$\bar{E}(s)=v_1^2-2F(v_0):=\bar{E}$, and
\begin{gather}
v^2(s)\geq v_0v_1e^{2s}-v_0(v_1-v_0), \label{e6.4.1}\\
\begin{aligned}
(v^2)_s-2v^2(s)&  \geq2v_0(v_1-v_0)+2\int_0^sv_s^2(r) dr \\
&  =2v_0(v_1-v_0)+2\int_0^s(\bar{E}+2F(v(r))+2
\int_0^{r} v_s(\eta)^2d\eta)dr \\
&  \geq2v_0(v_1-v_0)+2\bar{E}s+4k\int_0^s v^{p+1}(r)dr \\
&  \geq2v_0(v_1-v_0)+2\bar{E}s+4ks^{1-p}(\int_0^sv^2(r)dr)^{\frac{p+1}{2}} .
\label{e6.4.2}
\end{aligned}
\end{gather}

Let $\int_0^sv^2(r)dr:=b(s)$, $e^{-s}b(s)=B(s)$. Then
\[
b(s)''-2b(s)'\geq 2v_0(v_1-v_0)+2\bar{E}s+4ks^{1-p}b(s)
^{\frac{p+1}{2}}\quad \text{for } s>0
\]
and by \eqref{e6.4.2}, we have
\begin{equation}
\begin{aligned}
&  (e^{-s}(b(s)'-b(s)))' \\
&  =e^{-s}(b(s)''-2b(s)'+b(s)) \\
&  \geq2v_0(v_1-v_0)+2\bar{E}s+4ks^{1-p}e^{-s}
b^{\frac{p+1}{2}}+e^{-s}b(s) \\
&  =2v_0(v_1-v_0)+2\bar{E}s+4ks^{1-p}(
e^{-s}b(s))^{\frac{p+1}{2}}e^{\frac{p-1}{2}s}
+e^{-s}b(s)\geq0,
\end{aligned}  \label{e6.5}
\end{equation}
\begin{align*}
(e^{-s}b(s))''
&  =(e^{-s}(b(s)'-b(s)))' \\
&  \geq2v_0(v_1-v_0)+2\bar{E}s+4k(\frac{e^s
}{s^2})^{\frac{p+1}{2}}(e^{-s}b(s))
^{\frac{p+1}{2}}+e^{-s}b(s) \\
&  \geq2v_0(v_1-v_0)+2\bar{E}s+2^{2-\frac{p+1}{2}
}k(e^{-s}b(s))^{\frac{p+1}{2}}+e^{-s}b(s),
\end{align*}
\begin{equation}
B(s)''   \geq2v_0(v_1-v_0)
+2\bar{E}s+2^{2-\frac{p+1}{2}}kB(s)^{\frac{p+1}{2}}+B(s). \label{e6.6}
\end{equation}

From \eqref{e6.4.1} it follows that
\begin{gather*}
b(s) \geq\frac{v_0v_1}{2}(e^{2s}-1) -v_0(v_1-v_0)s,\\
B(s) \geq\frac{v_0v_1}{2}(e^s-e^{-s}) -v_0(v_1-v_0)se^{-s}, \\
2v_0(v_1-v_0)+2\bar{E}s+\frac{B(s)}{2} \geq0,\quad s\geq s_0
\end{gather*}
for some $s_0>0$. Therefore,
\begin{gather*}
B(s)''\geq\frac{B(s)}{2}\geq \frac{v_0v_1}{5}e^s,\quad s\geq s_0, \\
B'(s)\geq\frac{v_0v_1}{5}(e^s-e^{s_0})+B'(s_0)>0,\ s\geq s_1
\end{gather*}
for some $s_1>s_0$.

By \eqref{e6.6}, for all $s\geq s_1$,
\begin{gather*}
\begin{aligned}
((B(s)')^2)'
&  =2B(s)'B(s)''\\
&  \geq2^{2-\frac{p-1}{2}}kB(s)^{\frac{p+1}{2}}B(
s)'\\
&  =2^{2-\frac{p-1}{2}}k\frac{2}{p+3}(B(s)^{\frac
{p+3}{2}})',
\end{aligned}\\
(B')^2-B'(s_1)^2  
\geq\frac{2^{3-\frac{p-1}{2}}k}{p+3}(B^{\frac{p+3}{2}}-B(
s_1)^{\frac{p+3}{2}}),\\
\begin{aligned}
(B')^2  
& \geq\frac{2^{3-\frac{p-1}{2}}k}
{p+3}(B^{\frac{p+3}{2}}-B(s_1)^{\frac{p+3}{2}})+B'(s_1)^2\\
&  =\frac{2^{3-\frac{p-1}{2}}k}{2(p+3)}B^{\frac{p+3}{2}
}+\Big(\frac{2^{3-\frac{p-1}{2}}k}{2(p+3)}B^{\frac{p+3}{2}
}-2B(s_1)^{\frac{p+3}{2}}\Big)+B'(s_1)^2,
\end{aligned}\\
B'\geq\frac{2^{\frac{7-p}{4}}}{\sqrt{p+3}}B^{\frac{p+3}{4}}\quad
\text{for }s\geq s_2,
\end{gather*}
for some $s_2>s_1$; hence, for $ s\geq s_2$,
\begin{gather*}
\frac{4}{1-p}(B^{\frac{1-p}{4}})'=B^{\frac{p+3}{-4}
}B'(s)\geq\frac{2^{\frac{7-p}{4}}}{\sqrt{p+3}},
\\
B(s)^{\frac{1-p}{4}}\leq B(s_2)^{\frac{1-p}
{4}}-\frac{p-1}{4}\frac{2^{\frac{7-p}{4}}}{\sqrt{p+3}}(s-s_2)\quad
\text{for all }s\geq s_2>0.
\end{gather*}
Thus $B(s)$ blows up at a finite $s^{\ast}$. Since 
$b(s)=e^sB(s)$, $b(s)$ also blows up at
$s^{\ast}$. Further, since $v^2(s)=b'(s)\geq2b(s)$, $v(s)$ blows up at
$s^{\ast}$, as well.
\end{proof}


\subsection*{Acknowledgements}
Thanks are due to Professors Tai-Ping Liu, Ton Yang and  Shih-Shien Yu 
for their continuous encouragement;
to the anonymous referee for his/her helpful comments;
and  to Professor K. Schmitt for his comments and suggestions on 
Theorem \ref{thm6}.
The authors want to thank Metta Education, Grand Hall and Auria Solar 
for their financial assistance.

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\end{document}