本文研究了以下非线性Klein-Gordon方程的初边值问题

$ \begin{equation} \begin{cases} u_{tt}-\Delta u+u=\left | u \right |^{\alpha -1}u-\left | u \right |^{^{\beta -1}}u,&\text{ } x\in\Omega \subset R^{^{n}}, \\ u|_{\Gamma }=0, \ \ \ \Gamma=\partial \Omega \times \left [ 0, T \right ], \\ u\left ( x, 0 \right )=u_{0}\left ( x \right ), u_{t}\left ( x, 0 \right )=u_{1}\left ( x \right ),&\text{ } \end{cases} \end{equation}$

(1.1)

其中$u=u\left (t, x \right)$是复值函数, $\Delta$是$\Omega$上的Laplace算子, $\Omega$是$R^{n}$中带有光滑边界$\Gamma$的有界域, $1\leq\beta <\alpha$, $\alpha =\beta +2$, 且$\alpha $, $\beta $是常数.当$n>2$时, $1< \alpha \leq \frac{2n}{n-2}$; 当$n\leq 2 $时, $1< \alpha < \infty$.

Klein-Gordon方程是相对论量子力学和量子场论中用于描述自旋为零的粒子的基本方程.对于该类方程的研究已有一些文献[1-9].值得特别指出的是: Shatah^[1]证明了对非线性项$f\left (u \right)$ (其中$x\in R^{n}$, $n\geq 3$, $\displaystyle\int_{0}^{s}f\left (s \right)ds> +\infty)$基态的存在性和不稳定性, 得到了在不稳定基态下解不爆破的结果.黄文毅^[2]对一类带有阻尼项和非负势能的非线性Klein-Gordon方程

$ \begin{equation} \begin{cases} u_{tt}-\Delta u+T\left ( x \right )u+u+u_{t}=h\left ( x \right ), x\in R^{n}, \\ u\left ( 0, x \right )=u_{0},&\text{ } \\ u_{t}\left ( 0, x \right )=u_{1},&\text{ } \end{cases} \end{equation}$

(1.2)

其中$\displaystyle\int_{0}^{u}h\left (u \right)du> +\infty$.通过引进势井的方法研究了其柯西问题, 得到了解爆破和整体存在的最佳条件.文献[9]采用Galerkin方法和变分法研究了以下方程的初值问题

$ \begin{equation} \begin{cases} u_{tt}-\Delta u+u=u^{2}+u^{3}, x\in R^{n}, \\ u\left ( 0, x \right )=u_{0},&\text{ } \\ u_{t}\left ( 0, x \right )=u_{1},&\text{ } \end{cases} \end{equation}$

(1.3)

给出了当空间维数限制为$n\leq 3$时解整体存在的充分必要条件, 解爆破时的生命跨度的估计等等.李考虑了不同于文献[1]的变分问题, 对$\displaystyle\int \left (u ^{2}+u^{3} \right)du=\frac{1}{3}u^{3}+\frac{1}{4}u^{4}$在$\left (-\infty, +\infty \right)$上不定号的情形加以了研究, Shatah的方法对此情形不适用, 这也是该文的创新之处.

本文也有类似的困难, 但李^[9]中方程的非线性项确定了, 势井深度的正性容易通过显性的方程求解来确定, 而本文由于非线性项中的指数不是确定的, 在确定势井深度的正性时, 需要精细巧妙的讨论和估计才能确定, 且文献[9]没有考虑初始能量等于势井深度临界情形下的生命跨度, 这也是本文的意义所在.

对$u\in H^{1}\left (\Omega \right)$, $n\leq 6$, 定义如下能量泛函

$ \begin{eqnarray}J\left ( u \right )&=&\frac{1}{2}\left \| u \right \|_{H^{1}\left ( \Omega \right )}^{2}+\frac{1}{\beta +1}\displaystyle\int_{\Omega }\left | u \right |^{\beta +1}dx-\frac{1}{\alpha +1}\displaystyle\int_{\Omega }\left | u \right |^{\alpha +1}dx\\ &=&\frac{1}{2}a\left ( u \right )+\frac{1}{\beta +1}b_{1}\left ( u \right )-\frac{1}{\alpha +1}b_{2}\left ( u \right ).\end{eqnarray}$

定义势井

$ W_{1}=\left \{ u\in H^{1}\left ( \Omega \right )|a\left ( u \right )> b_{2}\left ( u \right )-b_{1}\left ( u \right ) , J\left ( u \right )< d\right \}\cup \left \{ 0 \right \}, $

及其对应的势井外集

$ W_{2}=\left \{ u\in H^{1}\left ( \Omega \right )|a\left ( u \right )< b_{2}\left ( u \right )-b_{1}\left ( u \right ), J\left ( u \right )< d \right \}, $

其中

$ \begin{equation} d=\inf\limits_{u\in H^{1}\left ( \Omega \right ), u\neq 0}\sup\limits_{\lambda \geq 0}J\left ( \lambda u \right ) \end{equation}$

(2.1)

为势井深度.

下面将证明$d$始终大于0, 即有如下引理.

引理2.1   若$d$由(2.1)式给出, 则$d> 0$.

证  当$\lambda \geq 0$时,

$ J\left (\lambda u \right )=\frac{1}{2}\lambda ^{2}a\left ( u \right )+\frac{1}{\beta +1}\lambda ^{\beta +1}b_{1}\left ( u \right )-\frac{1}{\alpha +1}\lambda ^{\alpha +1}b_{2}\left ( u \right ), $

注意到$\begin{aligned}{J}'\left (\lambda u \right)=\lambda a\left (u \right)+\lambda ^{\beta }b_{1}\left (u \right)-\lambda ^{\alpha }b_{2}\left (u \right)=0\end{aligned}$存在零根.

下面将证明${J}'\left (\lambda u \right)=0$存在正根, 令

$ f\left ( \lambda \right )=a\left ( u \right ) +\lambda ^{\beta -1}b_{1}\left ( u \right )-\lambda ^{\alpha -1}b_{2}\left ( u \right ), $

注意到$f\left (0 \right)=a\left (u \right)> 0, $当$\lambda \rightarrow +\infty $时, $f\left (\lambda \right)\rightarrow -\infty $, 因此由介值定理必然存在正根$\lambda _{0}$, 使得$f\left (\lambda _{0} \right)=0$, 即

$ \begin{eqnarray}a\left ( u \right )+\lambda _{0}^{\beta -1}b_{1}\left ( u \right )&=&\lambda _{0}^{\alpha -1}b_{2}\left ( u \right ), \\ J\left ( \lambda _{0}u \right )&=&\lambda _{0}^{2}\left [ \left ( \frac{1}{2}-\frac{1}{\beta +1} \right )a\left ( u \right )+\left ( \frac{1}{\beta +1}-\frac{1}{\alpha +1} \right )\lambda _{0}^{\alpha -1}b_{2}\left ( u \right ) \right ]\\ &\geq& \left ( \frac{1}{2}-\frac{1}{\beta +1} \right )\lambda _{0}^{2}a\left ( u \right ).\end{eqnarray}$

又因为

$ \frac{a\left ( u \right )}{\lambda _{0}^{\beta -1}}+b_{1}\left ( u \right )=\lambda _{0}^{2}b_{2}\left ( u \right ), ~~\frac{a\left ( u \right )}{\lambda _{0}^{\beta -1}b_{2}\left ( u \right )}+\frac{b_{1}\left ( u \right )}{b_{2}\left ( u \right )}=\lambda _{0}^{2}, $

所以

$ \begin{aligned}J\left (\lambda _{0}u \right )&\geq \left ( \frac{1}{2}-\frac{1}{\beta +1} \right )\left ( \frac{a\left ( u \right )}{\lambda _{0}^{\beta -1}b_{2}\left ( u \right )}+\frac{b_{1}\left ( u \right )}{b_{2}\left ( u \right )} \right )a\left ( u \right )\\ &\geq \left ( \frac{1}{2}-\frac{1}{\beta +1} \right )\frac{b_{1}\left ( u \right )}{b_{2}\left ( u \right )}a\left ( u \right ).\end{aligned} $

因为$\mu \left (\Omega \right)< \infty$ ($\mu$为$\Omega$的Lebesgue测度), 由$\beta <\alpha$有

$ \begin{eqnarray}&&\left \| u \right \|_{L^{\beta +1}\left ( \Omega \right )}\leq C_{1}\left \| u \right \|_{L^{\alpha +1}\left ( \Omega \right )}, ~~b_{1}^{\frac{\alpha +1}{\beta +1}}\left ( u \right )\leq C_{1}^{\alpha +1}b_{2}\left ( u \right ), ~~b_{1}^{-\frac{\alpha +1}{\beta +1}}\left ( u \right )\geq C_{1}^{-\left ( \alpha +1 \right )}b_{2}^{-1}\left ( u \right ), \nonumber\\ &&b_{1}\left ( u \right )\geq \left (C _{1}^{-\left ( \alpha +1 \right )}\right )^{-\frac{\beta +1}{\alpha +1}}\left ( b_{2}^{-1} \right )^{-\frac{\beta +1}{\alpha +1}}\left ( u \right ) =C_{1}^{\beta +1}b_{2}^{\frac{\beta +1}{\alpha +1}}\left ( u \right ), \nonumber\\ &&J\left ( \lambda _{0}u \right )\geq \left ( \frac{1}{2}-\frac{1}{\beta +1} \right )C_{1}^{\beta +1}\frac{b_{2}^{\frac{\beta +1}{\alpha +1}}\left ( u\right )}{b_{2}\left ( u \right )}a\left ( u \right )=\left ( \frac{1}{2}-\frac{1}{\beta +1} \right )C_{1}^{\beta +1}\frac{a\left ( u \right )}{b_{2}^{\frac{\alpha -\beta }{\alpha +1}}\left ( u \right )}. \end{eqnarray}$

(2.2)

由于$H^{1}\left (\Omega \right)$嵌入到$L^{\alpha +1}\left (\Omega \right)$, 有$\left \| u \right \|_{L^{\alpha +1}\left (\Omega \right)}\leq C_{2}\left \| u \right \|_{H^{1}\left (\Omega \right)}$, 结合(2.2)式有

$ J\left ( \lambda _{0}u \right )\geq \left ( \frac{1}{2}-\frac{1}{\beta +1} \right )\frac{C_{1}^{\beta +1}}{C_{2}^{2}}> 0, $

即$d>0, $其中$c_{1}$, $c_{2}$为常数.

为了得到解整体存在和爆破的条件, 这一节将介绍不变集$W_{1}$, $W_{2}$.接下来, 将利用如下事实$a\left (u \right)> b_{2}\left (u \right)-b_{1}\left (u \right)$有效等价于$ \lambda _{0}\left (u \right)> 1, $ $\ a\left (u \right)< b_{2}\left (u \right)-b_{1}\left (u \right)$有效等价于$\ \lambda _{0}\left (u \right)< 1.$

记$W=\left \{ u|u\in H^{1}\left (\Omega), J\left (u \right)< d \right) \right \}, $将有如下引理.

引理3.1    $W=W_{1}\cup W_{2}$, $W_{1}\cap W_{2}=\varnothing$.

证   $W_{1}\cap W_{2}=\varnothing$是显然的.接下来将证明$W=W_{1}\cup W_{2}$.实际上, 只需证明$W\subseteq W_{1}\cup W_{2}$.容易看到$J\left (\lambda _{0}u \right)=\sup\limits_{\lambda \geq 0}J\left (\lambda u \right), \lambda _{0}\left (u \right)=1$等价于$a\left (u \right)=b_{2}\left (u \right)-b_{1}\left (u \right)> 0.$因此, 若$\lambda _{0}\left (u \right)=1$, 则有

$ J\left ( u \right )=J\left (\lambda _{0}\left ( u \right )u \right )=\sup\limits_{\lambda \geq 0}J\left ( \lambda u \right )\geq d.$

于是, 若$u\in W$, $u\neq 0$, 则$\lambda _{0}\left (u \right)\neq 1$等价于$a\left (u \right)\neq b_{2}\left (u \right)-b_{1}\left (u \right)$.这意味着$u\in W_{1}\setminus \left \{ 0 \right \}$或$u\in W_{2}$, 即$W\subseteq W_{1}\cup W_{2}$.

(不变集)  若$u_{0}, u_{1}\in\Sigma $ (其中$\Sigma\subseteq H^{1}\left (\Omega \right)$为集合), 则(1.1)式的解$u\left (t, x \right)\in \Sigma$, 把$\Sigma $叫做问题(1.1)的解的不变集.

引理3.2   若

$ \begin{equation} E\left ( 0 \right )=\frac{1}{2}\left \| u_{1} \right \|_{L^{2}\left ( \Omega \right )}^{2}+J\left ( u_{0} \right )< d, \end{equation}$

(3.1)

则$W_{1}$和$W_{2}$是问题(1.1)解的不变集.

证  由方程(1.1)有

$ \begin{equation} \frac{1}{2}\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+J\left ( u\right )=E\left ( 0 \right )< d, \end{equation}$

(3.2)

这意味着$u\in W$.通常把$u\left (x, 0, u_{0}, u_{1} \right)$简记为$u\left (t \right)$.

(1) 令$u_{0}\in W_{1}$, 可以断言$u\left (t \right)\in W_{1}$.若不成立, 则存在$t_{0}> 0$, 使得$u\left (t_{0} \right)\notin W_{1}$.一方面, 有$u\left (t_{0} \right)\in W_{2}$; 另一方面, 因为$a\left (u\left (t \right) \right)$, $b_{1}\left (u\left (t \right) \right)$和$b_{2}\left (u\left (t \right) \right)$关于$t$连续, 由$W_{1}$和$W_{2}$的定义, 知存在时间$t^{*}$($0<t ^{*}< t_{0}$), 使得

$ a\left ( u\left (t ^{*} \right ) \right )=b_{2}\left ( u\left (t ^{*} \right ) \right )-b_{1}\left ( u\left (t ^{*} \right ) \right ).$

若$a\left (u\left (t^{*} \right) \right)\neq 0$, 则$\lambda _{0}\left (u\left (t^{*} \right) \right)=1$.因此$J\left (u\left (t ^{*} \right) \right)\geq d$, 即$u\left (t^{*} \right)\notin W$, 这与(3.2)式矛盾.

若$a\left (u\left (t^{*} \right) \right)=0$, 假设$0< a\left (u\left (t \right) \right)< b_{2}\left (u\left (t \right) \right)-b_{1}\left (u\left (t \right)\right)$, 对$t^{*}< t\leq t_{0}$, 有$\lim\limits_{t\rightarrow t^{*+}} a\left (u\left (t \right) \right)=0.$

因为当$2< n\leq 6$时, $1< \alpha \leq \frac{2n}{n-2}$; 当$n\leq 2$时, $1< \alpha < \infty$, 因此由Sobolev嵌入不等式知

$ 1< \frac{b_{2}\left ( u\left ( t \right )-b_{1}\left ( u\left ( t \right ) \right ) \right )}{a\left ( u\left ( t \right ) \right )}< C_{1}^{\alpha +1}\left ( a\left ( u \right ) \right )^{\frac{\alpha +1}{2}}+C_{2}^{\beta +1}\left ( a\left ( u \right ) \right )^{\frac{\beta +1}{2}}, $

结果

$ 1\leq \lim\limits_{t\rightarrow t^{*+}}\frac{b_{2}\left ( u\left ( t \right ) \right )-b_{1}\left ( u\left ( t \right ) \right )}{a\left ( u\left ( t \right ) \right )}=0.$

这个矛盾说明$u\left (t \right)\in W_{1}$.

(2) 令$u_{0}\in W_{2}$, 可以断言$u\left (t \right)\in W_{2}$.假设不成立, 即存在$t_{0}$, 使得$u\left (t_{0} \right)\in W_{1}$, 则存在$t^{*}$, 使得$a\left (u\left (t^{*} \right) \right)=b_{2}\left (u\left (t^{*} \right) \right)-b_{1}\left (u\left (t^{*} \right) \right)$ ($0< t^{*}\leq t_{0}$), 且对$0\leq t< t^{*}$, $u\left (t \right)\in W_{2}$.

首先考虑$a\left (u\left (t ^{*} \right) \right)=0$的情况.一方面, 因为$a\left (u\left (t \right) \right)$, $b_{1}\left (u\left (t \right) \right)$和$b_{2}\left (u\left (t \right) \right)$关于$t$连续, 可以断言

$ \lim\limits_{t\rightarrow t^{*-}}a\left ( u\left ( t \right ) \right )=0.$

另一方面, 因为对$0\leq t< t^{*}$, $u\left (t \right)\in W_{2}$, 则$a\left (u\left (t \right) \right)\neq 0$, $a\left (u\left (t \right) \right)< b_{2}\left (u\left (t \right) \right)-b_{1}\left (u\left (t \right) \right)$, $0\leq t< t^{*}$.由(1)知

$ 1< \frac{b_{2}\left ( u\left ( t \right )-b_{1}\left ( u\left ( t \right ) \right ) \right )}{a\left ( u\left (t \right ) \right )}\leq C_{1}^{\alpha +1}\left ( a\left ( u \right ) \right )^{\frac{\alpha +1}{2}}+ C_{2}^{\beta +1}\left ( a\left ( u \right ) \right )^{\frac{\beta +1}{2}}, $

结果

$ 1\leq \lim\limits_{t\rightarrow t^{*-}}\frac{b_{2}\left ( u\left ( t \right ) \right )-b_{1}\left ( u\left ( t \right ) \right )}{a\left ( u\left ( t \right ) \right )}=0, $

然而这是不可能的.

若$a\left (u\left (t ^{*} \right) \right)\neq 0$, 则与(1)相同的讨论有$J\left (u\left (t^{*} \right) \right)\geq d$, 这与(3.2)式矛盾.所以$u\left (t \right)\in W_{2}$.

首先研究当初始能量$E\left (0 \right)= d$时解整体存在和爆破的条件.

定理4.1   假设$E\left (0 \right)= d$, $u_{0}\in H^{1}\left (\Omega \right)$, $u_{1}\in L^{2}\left (\Omega \right)$.

1) 若$a\left (u_{0} \right)=b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, 且$a\left (u_{0} \right)\neq 0$, 则问题(1.1)存在整体弱解$u\left (t, x \right)\in C\left ([0, \infty); H_{0}^{1}\left (\Omega \right) \right), $ $u_{t}\left (t, x \right)\in C\left ([0, \infty); L^{2}\left (\Omega \right) \right).$

2) 若$a\left (u_{0} \right)> b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$或$a\left (u_{0} \right)= 0$, 则问题(1.1)存在整体弱解$u\left (t, x \right)\in C\left ([0, \infty); H_{0}^{1}\left (\Omega \right) \right), $ $u_{t}\left (t, x \right)\in C\left ([0, \infty); L^{2}\left (\Omega \right) \right).$

3) 若$\left (u_{0}, u_{1} \right)\geq 0$, $a\left (u_{0} \right)< b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, 则问题(1.1)的解在有限时间内爆破, 即存在常数$T>0$, 使得$\lim\limits_{t\rightarrow T^{-}}\left \| u \right \|_{L^{2}\left (\Omega \right)}^{2}=+\infty.$

注  当$\left (u_{0}, u_{1} \right)<0$时, 添上$\left \| u_{0} \right \|_{L^{2}\left (\Omega \right)}^{2}<\left \| \bar{u} \right \|_{L^{2}\left (\Omega \right)}^{2}$ ($\bar{u}$为(1.1)式的平衡态)这个条件, 同理可证, 证明过程略.

证   1)若$a\left (u_{0} \right)=b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$且$a\left (u_{0} \right)\neq 0$.令$\left \{ w_{j}\left (x \right) \right \}$是$H^{1}\left (\Omega \right)$的一个基, 构造问题(1.1)的近似解$u_{m}\left (t, x \right)$, 使得

$ u_{m}\left ( t, x \right )=\sum\limits_{j=1}^{m}g_{im}\left ( t \right )w_{j}\left ( x \right ), m=1, 2, \cdots$

满足

$ \begin{eqnarray} \left (u _{mtt}, w_{s} \right )-\left ( \Delta u_{m, }, w_{s} \right )+\left ( u_{m}, w_{s} \right )=\left ( \left | u_{m} \right |^{\alpha -1}u_{m}, w_{s} \right )-\left ( \left | u_{m} \right |^{\beta -1}u_{m}, w_{s} \right ), \end{eqnarray} $

(4.1)

$ \begin{eqnarray} u_{m}\left ( x, 0 \right )=\sum\limits_{j=1}^{m}a_{jm}w_{j}\left ( x \right )\rightarrow u_{0}\left ( x \right ), ~~\mbox{在}~~H^{1}\left ( \Omega \right )~\mbox{中}, \end{eqnarray} $

(4.2)

$\begin{eqnarray} u_{mt}\left ( x, 0 \right )=\sum\limits_{j=1}^{m}b_{jm}w_{j}\rightarrow u_{1}\left ( x \right ), ~~\mbox{在}~L^{2}\left ( \Omega \right )~\mbox{中}. \end{eqnarray} $

(4.3)

将(4.1)式与$g{}'_{im}$相乘, 并关于$s$求和得

$ \begin{eqnarray}&&\frac{1}{2}\frac{d }{d t}\left \| u_{mt} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{1}{2}\frac{d }{d t}\left \| \triangledown u_{m} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{1}{2}\frac{d }{d t}\left \| u_{m} \right \|_{L^{2}\left ( \Omega \right )}^{2}\\ &=&\frac{1}{\alpha +1}\frac{d }{d t}\left \| u_{m} \right \|_{L^{\alpha +1}\left ( \Omega \right )}^{\alpha +1}-\frac{1}{\beta +1}\frac{d }{d t}\left \| u_{m} \right \|_{L^{\beta +1}\left ( \Omega \right )}^{\beta +1}.\end{eqnarray}$

关于$t$积分得

$ \begin{aligned}&\frac{1}{2}\left \| u_{mt} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{1}{2}\left \| \bigtriangledown u_{m} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{1}{2}\left \| u_{m} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{1}{\beta +1}\left \| u_{m} \right \|_{L^{\beta +1}\left ( \Omega \right )}^{\beta +1} -\frac{1}{\alpha +1}\left \| u_{m} \right \|_{L^{\alpha +1}\left ( \Omega \right )}^{\alpha +1} \\ =&\frac{1}{2}\left \| u_{mt}\left ( 0 \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{1}{2}\left \| \bigtriangledown u_{m}\left ( 0 \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{1}{2}\left \| u_{m}\left ( 0 \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} -\frac{1}{\alpha +1}\left \| u_{m}\left ( 0 \right ) \right \|_{L^{\alpha +1}\left ( \Omega \right )}^{\alpha +1}\\&+\frac{1}{\beta +1}\left \| u_{m}\left ( 0 \right ) \right \|_{L^{\beta +1}\left ( \Omega \right )}^{\beta +1}.\end{aligned} $

对于充分大的$m$, 有

$ \frac{1}{2}\left \| u_{mt} \right \|_{L^{2}}^{2}\left ( \Omega \right )+J\left ( u_{m} \right )=E_{m}\left ( 0 \right )=d, \quad 0\leq t< \infty $

且$u_{m}\left (0 \right)\in W_{1}$,

$ \begin{aligned}J\left ( u_{m} \right )=&\frac{1}{2}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}+\frac{1}{\beta +1}\left \| u_{m} \right \|_{^{\beta +1}\left ( \Omega \right )}^{\beta +1}-\frac{1}{\alpha +1}\left \| u_{m} \right \|_{L^{\alpha +1}\left ( \Omega \right )}^{\alpha +1}\\ =&\left ( \frac{1}{2}-\frac{1}{\beta +1} \right )\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}+\left ( \frac{1}{\beta +1}-\frac{1}{\alpha +1} \right )\left \| u_{m} \right \|_{L^{\alpha +1}\left ( \Omega \right )}^{\alpha +1}\\ &+\frac{1}{\beta +1}\left ( a\left ( u_{m} \right )-b_{2}\left ( u_{m} \right )+b_{1}\left ( u_{m} \right ) \right )\\ \geq& \frac{\beta -1}{2\left ( \beta +1 \right )}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}.\end{aligned} $

所以

$ \frac{1}{2}\left \| u_{mt} \right \|_{L^{2}\left ( \Omega \right )}^{2}<d, ~~J\left ( u_{m} \right )\leq \frac{\beta -1}{2\left ( \beta +1 \right )}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}< d, $

即$u\left (t, x \right)\in C\left ([0, \infty); H_{0}^{1}\left (\Omega \right) \right)$, $u_{t}\left (t, x \right)\in C\left ([0, \infty); L^{2}\left (\Omega \right) \right).$

2) 若$a\left (u_{0} \right)> b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$或$a\left (u_{0} \right)=0$, 可以断言

$ \begin{equation} a\left ( u\left ( t \right ) \right )> b_{2}\left ( u\left ( t \right ) \right )-b_{1}\left ( u\left ( t \right ) \right ) \end{equation}$

(4.4)

或$\ a\left (u \right)=0, $对所有的$t\geq 0$成立.

实际上, 如果存在$t_{1}$, 使得

$ \begin{equation} a\left ( u\left (t_{1} \right ) \right )> b_{2}\left ( u\left (t_{1} \right ) \right )-b_{1}\left ( u\left ( t_{1} \right ) \right ), \end{equation}$

(4.5)

那么由于$a\left (u\left (t \right) \right)$, $b_{1}\left (u\left (t \right) \right)$和$b_{2}\left (u\left (t \right) \right)$关于$t$连续, 则存在$t_{0}$, 使得

$ \begin{equation} a\left ( u\left (t_{0} \right ) \right )= b_{2}\left ( u\left (t_{0} \right ) \right )-b_{1}\left ( u\left ( t_{0} \right ) \right ) \end{equation}$

(4.6)

且有

$ \begin{equation} a\left ( u\left (t \right ) \right )< b_{2}\left ( u\left (t \right ) \right )-b_{1}\left ( u\left ( t \right ) \right ), \end{equation}$

(4.7)

对$t_{0}< t\leq t_{1}$成立.

由于$a\left (u\left (t_{0} \right) \right)\neq 0$且$a\left (u_{0} \right)\neq b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, 所以有$u\left (t \right)\not\equiv u\left (t_{0} \right)$, 对$0\leq t\leq t_{0}$成立.令$v\left (t \right)=u\left (t_{0}-t \right)\not\equiv u\left (t_{0} \right)$, 则$v\left (0 \right)=u\left (t_{0} \right)$, $v\left (t_{0} \right)=u_{0}$, $v_{t}\left (0 \right)=\lim\limits_{t\rightarrow 0}-u_{t}\left (t-t_{0} \right)=0$, $v\left (t \right)$满足

$ v_{tt}-\Delta v+v=\left | v \right |^{\alpha -1}v-\left | v \right |^{\beta -1}v, ~~v\left ( 0 \right )=u\left ( t_{0} \right ), ~~v_{t}\left ( 0 \right )=0.$

然而$u\left (t_{0} \right)$也是以上问题的一个解, 这和解的唯一性矛盾.因此(4.6)式是有效的, 类似1)的证明得$u\left (t, x \right)\in C\left ([0, \infty); H_{0}^{1}\left (\Omega \right) \right)$, $u_{t}\left (t, x \right)\in C\left ([0, \infty); L^{2}\left (\Omega \right) \right).$

3) 与2)的证明类似, 若$a\left (u_{0} \right)< b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, 可以断言

$ \begin{eqnarray} &&a\left ( u\left ( t \right ) \right )< b_{2}\left ( u\left ( t \right ) \right )-b_{1}\left ( u\left ( t \right ) \right ), \end{eqnarray}$

(4.8)

$ \begin{eqnarray} \frac{1}{2}{\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''=\left \| u_{t} \right \|_{L^{2}}^{2}+b_{2}\left ( u \right )-b_{1}\left ( u \right )-a\left ( u \right ). \end{eqnarray} $

(4.9)

若$\left (u_{0}, u_{1} \right)\geq 0$, 则存在使得$t_{0}> 0$, 使得$\left (u\left (t_{0} \right), u_{t}\left (t_{0} \right) \right)=\frac{1}{2}\left (\left \| u\left (t_{0} \right) \right \|_{L^{2}\left (\Omega \right)}^{2} \right)_{t}^{{}'}\geq 0$.实际上, 若${\left (\left \| u\left (t \right) \right \|_{L^{2}\Omega }^{2} \right)}'=2\left (u\left (t \right), u_{t}\left (t \right) \right) < 0$, 由(4.8)和(4.9)式有

$ \displaystyle\int_{0}^{t}\left [ \left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+b_{2}\left ( u \right )-b_{1}\left ( u \right )-a\left ( u \right ) \right ]d\tau < \left ( u_{0}, u_{1} \right ), $

对所有$t\geq 0$成立, 即

$ \displaystyle\int_{0}^{+\infty }\left [ \left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+b_{2}\left ( u \right )-b_{1}\left ( u \right )-a\left ( u \right ) \right ]d\tau$

是有限的.因此

$ \begin{eqnarray}\lim\limits_{t\rightarrow +\infty }\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}&=&0, \\ \lim\limits_{t\rightarrow +\infty }\left [ b_{2}\left ( u \right )-b_{1}\left ( u \right )-a\left ( u \right ) \right ]&=&0, \\ \lim\limits_{t\rightarrow +\infty }J\left ( u \right )&=&d, \\ \lim\limits_{t\rightarrow +\infty }\left [ b_{2}\left ( u \right )-b_{1}\left ( u \right )-a\left ( u \right )+2J\left ( u \right ) \right ]&=&\lim\limits_{t\rightarrow +\infty }\left [ \left ( 1-\frac{2}{\alpha +1}\right )b_{2}+\left ( \frac{2}{\beta +1}-1 \right )b_{1} \right ]\\ &=&2d.\end{eqnarray}$

因为

$ \begin{aligned}J\left ( u \right )=&\frac{1}{2}a\left ( u \right )+\frac{1}{\beta +1}b_{1}\left ( u \right )-\frac{1}{\alpha +1}b_{2}\left ( u\right )\\ =&\frac{1}{2}a\left ( u \right )+\frac{1}{2\left ( \beta +1 \right )}b_{1}\left ( u \right )+\left [ \frac{\alpha -2\beta +1}{2\left ( \beta -1 \right )\left ( \alpha +1 \right )}\right ]b_{2}\left ( u \right )\\ &-\frac{1}{2\left ( \beta -1 \right )}\left [ \left ( 1-\frac{2}{\alpha +1} \right )b_{2}\left ( u \right ) +\left ( \frac{2}{\beta +1}-1 \right )b_{1}\left ( u \right ) \right ], \end{aligned} $

所以

$ \lim\limits_{t\rightarrow \infty }\left \{ \frac{1}{2}a\left ( u \right )+\frac{1}{2\left ( \beta +1 \right )}b_{1}\left ( u \right )+\left [ \frac{\alpha -2\beta +1}{2\left ( \beta -1 \right )\left ( \alpha +1 \right )}+\frac{1}{\alpha +1} \right ]b_{2}\left ( u \right ) \right \}=\frac{\beta }{\beta -1}d. $

由$a\left (u \right)$, $b_{1}\left (u \right)$, $b_{2}\left (u \right)\geq 0$知对所有的$t\geq 0 $, $a\left (u \right)$, $b_{1}\left (u \right)$, $b_{2}\left (u \right)$有限.由引理3.2的证明知存在序列$\left \{ t_{i} \right \}_{i=1, 2, \cdots}$使得$\lim\limits_{t_{i}\rightarrow t^{*-}}a\left (u\left (t_{i} \right) \right)=0$, 又由(4.8)式知

$ 1>\frac{a\left ( u\left ( t_{i} \right ) \right )}{b_{2}\left ( u\left ( t_{i} \right ) \right )-b_{1}\left ( u\left ( t_{i} \right ) \right )}\geq \frac{1}{C_{1}^{\alpha +1}a\left ( u\left ( t_{i} \right ) \right )^{\frac{\alpha -1}{2}}+C_{2}^{\beta +1}a\left ( u\left ( t_{i} \right ) \right )^{\frac{\beta -1}{2}}}\rightarrow \infty \ (t_{i}\rightarrow t^{*-}), $

上式矛盾.这个矛盾说明存在$t_{1}> 0$, 使得${\left (\left \| u\left (t_{1} \right) \right \|_{L^{2}\left (\Omega \right)}^{2} \right)}'\geq 0$.因为${\left (\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2} \right)}''> 0$, 对所有的$t> 0$成立, 因此有${\left (\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2} \right)}'> 0$, 对$t> t_{1}$.当然, 存在$t_{2}> t_{1}$, 使得$\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2}> \frac{2\left (\beta +1 \right)}{\beta -1}E\left (0 \right)$.因此有

$ \begin{equation} {\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''> \left ( \beta +3 \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}, \end{equation}$

(4.10)

对$t\geq t_{_{2}}$.结果

$ \begin{eqnarray} &&\left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2}{\left (\left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''-\frac{\beta +3}{4}\left [ {\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}' \right ]^{2}\nonumber\\ &>& \left ( \beta +3 \right )\left [ \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2}\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}-\left (u _{t}, u \right )^{2} \right ] \end{eqnarray}$

(4.11)

且

$ {\left (\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{-\frac{1}{2}} \right)}''=-\frac{1}{4}\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{-\frac{9}{2}}\left [{\left (\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2} \right)}'' -\frac{5}{4}{\left (\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2} \right)}'\right]< 0, $

对$t> t_{2}$.因为对$t> t_{1}$, ${\left (\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{-\frac{1}{2}} \right)}''=-\frac{1}{4}\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{-\frac{5}{2}}{\left (\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2} \right)}'< 0$.对$t\geq t_{2}$, $\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{-\frac{1}{2}}$是递减的凹函数, 因此存在$T< +\infty $, 使得$\lim\limits_{t\rightarrow T^{-}}\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{-\frac{1}{2}}=0$, 即$\lim\limits_{t\rightarrow T^{-}}\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2}=+\infty $.

其次考虑当初始能量$E\left (0 \right)< d$时解整体存在和爆破的条件.

定理4.2   假设$n\leq 6$, $0< E\left (0 \right)< d$, $u_{0}\in H^{1}\left (\Omega \right)$, $u_{1}\in L^{2}\left (\Omega \right)$,

1) 若$a\left (u_{0} \right)- b_{2}\left (u_{0}\right)+b_{1}\left (u_{0} \right)> 0$ (或$a\left (u_{0} \right)- b_{2}\left (u_{0} \right)+b_{1}\left (u_{0} \right)= 0$), 则问题(1.1)有整体弱解$u\left (t, x \right)\in C\left ([0, \infty); H_{0}^{1}\left (\Omega \right) \right)$, $u_{t}\left (t, x \right)\in C\left ([0, \infty); L^{2}\left (\Omega \right) \right).$

2) 若$a\left (u_{0}\right)- b_{2}\left (u_{0}\right)+b_{1}\left (u_{0}\right)< 0$, 则问题(1.1)的解在有限时间内爆破.

证   1)同定理4.1中1)的证明, 由$E\left (0 \right)< d, ~ a\left (u_{0} \right)-b_{2}\left (u_{0} \right)+b_{1}\left (u_{0} \right)> 0$知

$ \frac{1}{2}\left \| u_{mt} \right \|_{L^{2}\left ( \Omega \right )}^{2}+J\left ( u_{m} \right )=E_{m}\left ( 0 \right )< d, 0\leq t< \infty .$

因为

$ \frac{1}{2}\left \| u_{1} \right \|_{L^{2}\left ( \Omega \right )}^{2}+J\left ( u_{0} \right )=E\left ( 0 \right )< d, 0\leq t< \infty , $

所以$J\left (u_{0} \right)< d$, 又由$a\left (u_{0} \right)-b_{2}\left (u_{0} \right)+b_{1}\left (u_{0} \right)> 0$, 得$u_{0}\in W_{1}$.对于充分大的$m$, 有

$ \begin{equation} \frac{1}{2}\left \| u_{mt} \right \|_{L^{2}\left ( \Omega \right )}^{2}+J\left ( u_{m} \right )=E_{m}\left ( 0 \right )< d, 0\leq t< \infty, \end{equation}$

(4.12)

且$u_{m}\left (0 \right)\in W_{1}$.同引理3.2, 由(4.12)式对于充分大的$m$和$0\leq t< \infty $, 可以证明$u_{m}\left (t \right)\in W_{1}$且

$ \begin{eqnarray}J\left ( u_{m} \right )&=& \frac{1}{2}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}+\frac{1}{\beta +1}\left \| u_{m} \right \|_{L^{\beta +1}\left ( \Omega \right )}^{\beta +1}-\frac{1}{\alpha +1}\left \| u_{m} \right \|_{L^{\alpha +1}\left ( \Omega \right )}^{\alpha +1} \\ &=&\left ( \frac{1}{2}-\frac{1}{\alpha +1} \right )\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}+\left ( \frac{1}{\beta +1}-\frac{1}{\alpha +1} \right )\left \| u_{m} \right \|_{L^{\beta +1}\left ( \Omega \right )}^{\beta +1}\\ &&+\frac{1}{\alpha +1}\left ( a\left ( u_{m} \right )-b_{2}\left ( u_{m} \right )+b_{1}\left ( u_{m} \right ) \right )\\ &\geq& \frac{\alpha -1}{2\left ( \alpha +1 \right )}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}.\end{eqnarray}$

因此

$ \begin{eqnarray}&&\frac{1}{2}\left \| u_{mt} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\frac{\alpha -1}{2\left ( \alpha +1 \right )}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}< d, 0\leq t< \infty, \\ &&\left \| u_{mt} \right \|_{L^{2}\left ( \Omega \right )}^{2}\leq 2d, 0\leq t< \infty, \\ &&\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{2}\leq \frac{2\left ( \alpha +1 \right )}{\alpha-1 }d, 0\leq t< \infty, \\ &&\left \| u_{m} \right \|_{L^{\alpha +1}\left ( \Omega \right )}^{\alpha +1}\leq C_{*}^{\alpha +1}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{\alpha +1}\leq C_{*}^{\alpha +1}\left ( \frac{2\left ( \alpha +1 \right )}{\alpha -1}d \right )^{\frac{\alpha +1}{2}}, \\ &&\left \| u_{m} \right \|_{L^{\beta +1}\left ( \Omega \right )}^{\beta +1}\leq C_{*}^{\beta +1}\left \| u_{m} \right \|_{H^{1}\left ( \Omega \right )}^{\beta +1}\leq C_{*}^{\beta +1}\left ( \frac{2\left ( \beta +1 \right )}{\beta -1}d \right )^{\frac{\beta +1}{2}}, \end{eqnarray}$

其中$c_{*}$为常数, 所以$u\left (t, x \right)\in C\left ([0, \infty); H_{0}^{1}\left (\Omega \right) \right)$, $u_{t}\left (t, x \right)\in C\left ([0, \infty); L^{2}\left (\Omega \right) \right).$

2) 当$a\left (u_{0} \right)-b_{2}\left (u_{0} \right)+b_{1}\left (u_{0}\right)< 0$时, 类似定理4.1中3)的证明, 易证$\left \| u \right \|_{L^{2}\left (\Omega \right)}^{2}$在有限时间内爆破.

本文最后一部分将给出定理4.1和定理4.2中爆破解的生命跨度的上界估计.

定理5.1   若$E\left (0 \right)= d$, $\left (u_{0}, u_{1} \right)\geq 0$, $a\left (u_{0} \right)< b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, $T_{0}$是(1.1)的生命跨度, 则

$ T_{0}\leq\frac{2\left \| u_{0} \right \|_{L^{2}}^{2}}{\left ( \beta +3 \right )\left ( u_{0}, u_{1}\right )}.$

证  由定理4.2, 对所有$t\geq 0$,

$ \begin{eqnarray}&&a\left ( u\left ( t \right ) \right )< b_{2}\left ( u\left ( t \right ) \right )-b_{1}\left ( u\left ( t \right ) \right ), \\ &&\frac{1}{2}{\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''=\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+b_{2}\left ( u \right )-b_{1}\left ( u \right )-a\left ( u \right ).\end{eqnarray}$

因为

$ \frac{1}{2}\left ( \left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+a\left ( u \right ) \right )=-\frac{1}{\beta +1}b_{1}\left ( u \right )+\frac{1}{\alpha +1}b_{2}\left ( u \right )+E\left ( 0 \right ), $

所以

$ \begin{equation} {\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''=\left ( \beta +3 \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\left [ 2-\frac{2\left ( \beta +1 \right )}{\alpha +1} \right ]b_{2}\left ( u \right )+\left ( \beta -1 \right )a\left ( u \right )-2\left ( \beta +1 \right )E\left ( 0 \right ). \end{equation}$

(5.1)

又由引理2.1的证明有

$ J\left ( \lambda _{0}\left ( u \right ) \right )=\lambda _{0}^{2}\left [ \left ( \frac{1}{2}-\frac{1}{\beta +1} \right )a\left ( u \right )+\left ( \frac{1}{\beta +1}-\frac{1}{\alpha +1} \right )\lambda _{0}^{\alpha -1}b_{2}\left ( u \right ) \right ]\geq d.$

因此

$ a\left ( u \right )\geq \frac{\frac{2\left ( \beta +1 \right )}{\beta -1}}{\lambda _{0}^{2}}d-\frac{2\left ( \alpha -\beta \right )}{\left ( \alpha +1 \right )\left ( \beta -1 \right )}\lambda _{0}^{\alpha -1}b_{2}\left ( u \right ).$

当$\lambda _{0}^{2}< 1$时,

$ \begin{eqnarray}&&a\left ( u \right )\geq \frac{2\left ( \beta +1 \right )}{\beta -1}d-\frac{2\left ( \alpha -\beta \right )}{\left ( \alpha +1 \right )\left ( \beta -1 \right )}\lambda _{0}^{\alpha -1}b_{2}\left ( u \right ), \\ &&{\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^2{} \right )}''> \left ( \beta +3 \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+2\left ( \beta +1 \right )\left ( d-E\left ( 0 \right ) \right ).\end{eqnarray}$

由$E\left (0 \right)=d$, 得

$ \begin{equation} {\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''\geq \left ( \beta +3 \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}. \end{equation}$

(5.2)

令$y\left (t \right)=\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2}$, 由(5.2)式得

$ \begin{equation} {y}''\geq \frac{\beta +3}{4}\frac{\left ( {y}'\left ( t \right ) \right )^{2}}{y\left ( t \right )}. \end{equation}$

(5.3)

因为$a\left (u_{0} \right)< b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, 有$a\left (u_{0} \right)\neq 0$.当$\left (u_{0}, u_{1} \right)\geq 0$时, 令$T_{1}> 0$, 使得对$0\leq t\leq T_{1}$, ${y}'\left (t \right)> 0$,

$ {y}''\geq \frac{\beta +3}{4}\frac{\left ( {y}'\left ( t \right ) \right )^{2}}{y\left ( 0 \right )}.$

解上式并在$0$到$T_{1}$上积分得

$ \begin{equation} T_{0}=T_{1}\leq\frac{2\left \| u_{0} \right \|_{L^{2}}^{2}}{\left ( \beta +3 \right )\left ( u_{0}, u_{1}\right )}. \end{equation}$

(5.4)

定理5.2   若$E\left (0 \right)< d$, $a\left (u_{0} \right)< b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, $T_{2}$是(1.1)的生命跨度, 则有如下估计:

1) 若$\left (u_{0}, u_{1} \right)> 0$, 则

$ T_{2}\leq \sqrt{\frac{2\left ( \beta +1 \right )}{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}}\arcsin\left \{ \sqrt{\frac{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}{\frac{\left ( \beta +1 \right )\left ( 2-q \right )^{2}}{2}\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}^{\frac{q-6}{2}}\left ( u_{0}, u_{1} \right )^{2}+\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}} \right \};$

2) 若$\left (u_{0}, u_{1} \right)\leq 0$, 则

$ \begin{eqnarray}T_{2}&\leq& -\frac{1}{c\left ( \beta +1 \right )}\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}\sqrt{\frac{2\left ( \beta +1 \right )c}{\beta +3}}\arctan\frac{1}{\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}}\sqrt{\frac{\beta +3}{8c\left ( \beta +1 \right )}}\left ( u_{0}, u_{1} \right )\\ &&+\sqrt{\frac{2\left ( \beta +1 \right )}{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}}\times \frac{\pi }{2}.\end{eqnarray}$

这里$q=\max\left \{ \frac{2\left (\beta +1 \right)d}{\left (\beta +1 \right)d-\left (\beta -1 \right)E\left (0 \right)}, \frac{5}{2} \right \}, $ $c=d-E\left (0 \right)$.

证  由定理4.2, 对所有$t\geq 0$, 同定理5.1的证明, 由$E\left (0 \right)< d$, 知$\left (d-E\left (0 \right) \right)\doteq c> 0$, 其中$\doteq$表示“等价于”.所以

$ \begin{equation} {\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''\geq \left ( \beta +3 \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+2\left ( \beta +1 \right )c. \end{equation}$

(5.5)

令$y\left (t \right)=\left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2}$, 由(5.5)式得

$ \begin{equation} {y}''\geq \frac{\beta +3}{4}\frac{\left ( {y}'\left ( t \right ) \right )^{2}}{y\left ( t \right )}+2\left ( \beta +1 \right )c. \end{equation}$

(5.6)

因为$a\left (u_{0} \right)< b_{2}\left (u_{0} \right)-b_{1}\left (u_{0} \right)$, 有$a\left (u_{0} \right)\neq 0$.当$\left (u_{0}, u_{1} \right)< 0$时, 令$T_{3}> 0$, 使得对$0\leq t\leq T_{3}$, ${y}'\left (t \right)\leq 0$,

$ {y}''\geq \frac{\beta +3}{4}\frac{\left ( {y}'\left ( t \right ) \right )^{2}}{y\left ( 0 \right )}+2\left ( \beta +1 \right )c.$

将上式在$0$到$T_{3}$上积分得

$ \begin{equation} \begin{aligned} T_{3}&\leq -\frac{1}{c\left ( \beta +1 \right )}\sqrt{\frac{2\left ( \beta +1 \right )cy\left ( 0 \right )}{\beta +3}}\arctan\sqrt{\frac{\beta +3}{8c\left ( \beta +1 \right )y\left ( 0 \right )}}{y}'\left ( 0 \right )\\ &\doteq -\frac{1}{c\left ( \beta +1 \right )}\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}\sqrt{\frac{2\left ( \beta +1 \right )c}{\beta +3}}\arctan\left [ \frac{1}{\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}}\sqrt{\frac{\beta +3}{8c\left ( \beta +1 \right )}}\left ( u_{0}, u_{1} \right ) \right ]. \end{aligned} \end{equation}$

(5.7)

现在考虑当$\left (u_{0}, u_{1} \right)> 0$时, 解的生命跨度.

注意到

$ \frac{q}{2}\left ( \left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+a\left ( u \right ) \right )+\frac{q}{\beta +1}b_{1}\left ( u \right )-\frac{q}{\alpha +1}b_{2}\left ( u \right )-qE\left ( 0 \right )=0, $

与(5.1)式类似可以得

$ \begin{eqnarray}\frac{1}{2}{\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''&=&\left ( 1+\frac{q}{2} \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\left ( -1+\frac{q}{\beta +1} \right )b_{1}\left ( u \right )+\left ( 1-\frac{q}{\beta +1} \right )b_{2}\left ( u \right )\\ &&+\left ( \frac{q}{2}-1 \right )a\left ( u \right )-qE\left ( 0 \right ).\end{eqnarray}$

因为

$ a\left ( u \right )\geq \frac{2\left ( \beta +1 \right )}{\beta -1}d-\frac{2\left ( \alpha -\beta \right )}{\left ( \alpha +1 \right )\left ( \beta -1 \right )}b_{2}\left ( u \right ), $

有

$ \begin{eqnarray}\frac{1}{2}{\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''&=&\left ( 1+\frac{q}{2} \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+\left ( -1+ \frac{q}{\beta +1}\right )b_{1}\left ( u \right )\\ &&+\left [ 1-\frac{q}{\alpha +1}-\left ( \frac{q}{2}-1 \right )\frac{2\left ( \alpha -\beta \right )}{\left ( \alpha +1 \right )\left ( \beta -1 \right )} \right ]b_{2}\left ( u \right )\\ &&+\left ( \frac{q}{2}-1 \right )\frac{2\left ( \beta +1 \right )}{\beta -1}d- qE\left ( 0 \right ).\end{eqnarray}$

若取$q=\max\left \{ \frac{2\left (\beta +1 \right)d}{\left (\beta +1 \right)d-\left (\beta -1 \right)E\left (0 \right)}, \frac{5}{2} \right \}$, 由$E\left (0 \right)< d$, 则有$2< q< \beta +1$,

$ \begin{eqnarray}&&1-\frac{q}{\beta +1}< 1-\frac{q}{\alpha +1}-\left ( \frac{q}{2}-1 \right )\frac{2\left ( \alpha -\beta \right )}{\left ( \alpha +1 \right )\left ( \beta -1 \right )}, \\ &&\left ( \frac{q}{2}-1 \right )\cdot \frac{2\left ( \beta +1 \right )}{\beta -1}d\geq qE\left ( 0 \right ).\end{eqnarray}$

由$b_{1}\left (u \right), b_{2}\left (u \right)\geq 0$知

$ \begin{eqnarray}{\left ( \left \| u\left ( t \right ) \right \|_{L^{2}\left ( \Omega \right )}^{2} \right )}''&\geq& \left ( 2+q \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+2\left ( 1-\frac{q}{\beta +1} \right )\left ( b_{2} \left ( u \right )-b_{1}\left ( u \right )\right )\\ &\geq& \left ( 2+q \right )\left \| u_{t} \right \|_{L^{2}\left ( \Omega \right )}^{2}+2\left ( 1-\frac{q}{\beta +1} \right )a\left ( u \right ).\end{eqnarray}$

因为$a\left (u \right)\geq \left \| u\left (t \right) \right \|_{L^{2}\left (\Omega \right)}^{2}=y\left (t \right)$,

$ \begin{equation} {y}''y\left ( t \right )\geq \frac{2+q}{4}\left ( {y}'\left ( t \right ) \right )^{2}+2\left ( 1-\frac{q}{\beta +1} \right )y^{2}. \end{equation}$

(5.8)

令$Z\left (t \right)=y^{-\frac{q-2}{4}}\left (t \right)$, 则

$ \begin{equation} {Z}''\leq -\frac{1}{2\left ( \beta +1 \right )}\left \{ \left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ] \right \}Z\left ( \left ( t \right ) \right ). \end{equation}$

(5.9)

因为${Z}'\left (t \right) < 0$, 把(5.8)式乘以${Z}'$, 积分得

$ \begin{equation} {Z}'\leq -\sqrt{\frac{1}{2\left ( \beta +1 \right )}\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]\left ( Z^{2}\left ( 0 \right )-Z^{2}\left ( t \right ) \right )+\left ( {Z}'\left ( 0 \right ) \right )^{2}}. \end{equation}$

(5.10)

存在$t_{0}$, 使得$Z\left (t_{0} \right)=0$, 因此上式积分得

$ \begin{aligned}t_{0}&\leq -\displaystyle\int_{Z\left ( 0 \right )}^{0}\frac{dZ}{\sqrt{\left ( {Z}'\left ( 0 \right ) \right )^{2}}+\frac{1}{2\left ( \beta +1 \right )}\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]\left ( Z^{2}\left ( 0 \right )-Z^{2}\left ( t \right ) \right )}\\ &=\sqrt{\frac{2\left ( \beta +1 \right )}{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}}\arcsin\left \{ \sqrt{\frac{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}{2\left ( \beta +1 \right )\left ( {Z}'\left ( 0 \right ) \right )^{2}+\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]Z^{2}\left ( 0 \right )}} Z\left ( 0 \right )\right \}.\end{aligned} $

特别地, 若$\left (u_{0}, u_{1} \right)=0$, 即${Z}'\left (0 \right)$, 则

$ t_{0}\leq \sqrt{\frac{2\left ( \beta +1 \right )}{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}}\times \frac{\pi }{2}.$

将这个结果结合(5.7)式有

1) 若$\left (u_{0}, u_{1} \right)> 0$, 则

$ T_{2}\leq \sqrt{\frac{2\left ( \beta +1 \right )}{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}}\arcsin\left \{ \sqrt{\frac{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}{\frac{\left ( \beta +1 \right )\left ( 2-q \right )^{2}}{2}\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}^{\frac{q-6}{2}}\left ( u_{0}, u_{1} \right )^{2}+\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}} \right \};$

2) 若$\left (u_{0}, u_{1} \right)\leq 0$, 则

$ \begin{eqnarray}T_{2}&\leq& -\frac{1}{c\left ( \beta +1 \right )}\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}\sqrt{\frac{2\left ( \beta +1 \right )c}{\beta +3}}\arctan\frac{1}{\left \| u_{0} \right \|_{L^{2}\left ( \Omega \right )}}\sqrt{\frac{\beta +3}{8c\left ( \beta +1 \right )}}\left ( u_{0}, u_{1} \right )\\ &&+\sqrt{\frac{2\left ( \beta +1 \right )}{\left ( q-2 \right )\left [ \left ( \beta +1 \right )-q \right ]}}\times \frac{\pi }{2}.\end{eqnarray}$