\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 $11$. The existence and uniqueness of local solutions of the initial-value problem \label{e-star} \begin{gathered} t^2u''=u^n,\quad 10$:$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$v\in C^2[0,T]is the solution of \eqref{e1.1}, then \begin{gather} E(s) =v_s(s)^2-2\int_0^s v_s(r)^2dr-\frac{2}{n+1}v(s)^{n+1} =E(0),\label{e2.1}\\ (n+3)v_s(s)^2 =(n+1) E(0)+a''(s)-a'(s)+2(n+1) \int_0^s v_s(r)^2dr,\label{e2.2}\\ J''(s) =\frac{n^2-1}{4}J(s)^{\frac{n+3}{n-1}}(E(0)- \frac{a'(s)}{n+1}+2\int_0^s v_s(r)^2dr), \label{e2.3} \\ \begin{aligned} J'(s)^2 & =J'(0)^2 +\frac{(n-1)^2}{4}E(0)(J( s)^{\frac{2(n+1)}{n-1}}-J(0) ^{\frac{2(n+1)}{n-1}}) \\ & \quad +\frac{(n-1)^2}{2}J(s)^{\frac{2(n+1)}{n-1}} \int_0^s v_s(r)^2dr. \end{aligned} \label{e2.4} \end{gather} \end{lemma} \begin{lemma} \label{lem2} Foru_0>0$, 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,$v_s(s)0$, 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$a'(0)=0, hence we have \begin{align*} a'(s)e^{-s} & \geq\frac{8}{n+3}(v(r)^{(n+3)/2}e^{-r}\mid_{r=0}^s +\int_0^s v(r)^{(n+3)/2}e^{-r}dr)\\ & =\frac{8}{n+3}(v(s)^{(n+3)/2}e^{-s}-v( 0)^{(n+3)/2})+\frac{8}{n+3} \int_0^s v(r)^{(n+3)/2}e^{-r}dr. \end{align*} Sincea'(s)>0$for all$s>0$,$v$is increasing on$(0,\infty)and \label{e3.1} \begin{gathered} \begin{aligned} a'(s)e^{-s} & \geq\frac{8}{n+3}(v( s)^{(n+3)/2}e^{-s}-v(0)^{(n+3)/2}) +\frac{8}{n+3} \int_0^s v(0)^{(n+3)/2}e^{-r}dr \\ & =\frac{8}{n+3}(v(s)^{(n+3)/2}e^{-s}-v( 0)^{(n+3)/2})+\frac{8}{n+3}v(0) ^{(n+3)/2}(1-e^{-s}), \end{aligned}\\ a'(s) \geq\frac{8}{n+3}(v(s) ^{(n+3)/2}-v(0)^{(n+3)/2}) =\frac{8}{n+3}(v(s)^{(n+3)/2}-u_0^{(n+3)/2}). \end{gathered} Usingu_1=0$and integrating \eqref{e1.1}, we obtain $$\label{e3.2} \begin{gathered} v_s(s) =v(s)-u_0+\int_0^s v(r)^ndr, \\ v_s(s) \geq v(s)-u_0+\int_0^sv(0)^ndr=v(s)-u_0+u_0^n s, \\ (e^{-s}v(s))_s =e^{-s}(v_s(s)-v(s))\geq e^{-s}(u_0^n s-u_0), \\ a'(s) \geq\frac{8}{n+3}(v(s)^{(n+3)/2}-v(0)^{(n+3)/2}) =\frac{8}{n+3}(v(s)^{(n+3)/2}-u_0^{(n+3)/2}). \end{gathered}$$ 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$\epsilon\in(0,1), we obtain \begin{gather*} \epsilon v(s)^{(n+3)/2} \geq\epsilon(u_0+u_0^n(e^s-1-s))^{(n+3)/2},\\ \begin{aligned} \epsilon v(s)^{(n+3)/2}-8u_0^{(n+3)/2} & \geq\epsilon(u_0+u_0^n(e^s-1-s)) ^{(n+3)/2}-8u_0^{(n+3)/2}\\ & \geq\epsilon(u_0^{(n+3)/2}+u_0^{\frac{n(n+3)}{2}}(e^s-1-s)^{(n+3)/2}) -8u_0^{\frac{n+3} {2}}\\ & =(\epsilon-8)u_0^{(n+3)/2}+\epsilon u_0 ^{\frac{n(n+3)}{2}}(e^s-1-s)^{(n+3)/2}. \end{aligned} \end{gather*} Now, we want to find a numbers_0>0$such that $$e^{s_0}-s_0=1+\big(\frac{8-\epsilon}{\epsilon}u_0^{\frac{n+3} {2}(1-n)}\big)^{2/(n+3)}. \label{e3.3}$$ 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 alls\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$s>0and \begin{gather*} \begin{aligned} J'(s)& =-\frac{n-1}{2}\sqrt{\frac{2}{n+1}+E( 0)a(s)^{-\frac{n+1}{2}}+2a(s)^{-\frac{n+1}{2}} \int_0^s v_s(r)^2dr}\\ & \leq-\frac{n-1}{2}\sqrt{\frac{2}{n+1}+E(0)a( s)^{-\frac{n+1}{2}}}, \end{aligned}\\ J(s) \leq J(0)-\frac{n-1}{2}\int_0^s \sqrt{\frac{2}{n+1}+E(0)a(r)^{-\frac{n+1}{2}}}dr. \end{gather*} SinceE(0)<0$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 thatJ(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$vis 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) IfE(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}, $$\begin{gathered} v_{ss}(s)-v_s(s)=f(v),\\ v(0)=v_0,\quad v_s(0)=v_1, \end{gathered} \label{e6.1}$$ 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 $$\bar{E}(s):=v_s(s)^2-2 \int_0^s v_s(r)^2dr-2F(v(s))\label{e6.2}$$ 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 $$(v^2)_s-2v^2(s)\geq2v_0(v_1-v_0)+2\int_0^sv_s^2(r)dr. \label{e6.3}$$ 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{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} \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*} $$B(s)'' \geq2v_0(v_1-v_0) +2\bar{E}s+2^{2-\frac{p+1}{2}}kB(s)^{\frac{p+1}{2}}+B(s). \label{e6.6}$$ 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 somes_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 somes_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. \begin{thebibliography}{00} \bibitem{l1} M. R. Li; \emph{Nichtlineare Wellengleichungen 2. Ordnung auf beschr\"{a}nkten Gebieten}. PhD-Dissertation, T\"{u}bingen 1994. \bibitem{l2} M. R. Li; \emph{Estimates for the life-span of solutions of semilinear wave equations}. Comm. Pure Appl. Anal., 2008, \textbf{7}(2): 417-432. \bibitem{l3} M. R. 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