研究如下半线性波动方程的初值问题

(1.1) $ \begin{equation} \left \{ \begin{array}{ll} &u_{tt}-\Delta u = (1+|x|^2)^\alpha|u|^p, \; \; \; {\mathbb R}^n\times[0, \infty), \\ &u(x, 0) = \varepsilon f(x), \; \; u_t(x, 0) = \varepsilon g(x), \end{array} \right. \end{equation} $

其中$ \Delta = \sum\limits_{i = 1}^n\frac{\partial^2}{\partial x_i^2} $,且初值满足

(1.2) $ \begin{equation} \left \{ \begin{array}{ll} &(f, g)\in H^1({\mathbb R}^n)\times L^2({\mathbb R}^n), \\ &f(x)\geq0, g(x)\geq0, \\ &f(x) = g(x) = 0, \forall|x|>R>0, \\ &\alpha>-1, 0<\varepsilon \ll 1.\\ \end{array}\right. \end{equation} $

该研究问题是如下半线性波动方程小初值Cauchy问题(Strauss猜测)的拓展:

(1.3) $ \begin{equation} \left \{ \begin{array}{ll} &\square u(x, t) = |u|^p, \; \; \; {\mathbb R}^n\times[0, \infty), \\ &u(x, 0) = \varepsilon f(x), \; \; u_t(x, 0) = \varepsilon g(x). \end{array} \right. \end{equation} $

其中$ n\geq2 $, $ \varepsilon>0 $是小参数.由此定义(1.1)解的一个生命跨度为$ T(\varepsilon) $,

$T(\varepsilon) = \sup \{t>0:问题(1.1)的解在[0, t]\times {\mathbb R}^n中存在\}. $

当$ 1<p<2 $时,方程的解常用相应的积分方程来定义并用标准的Strichartz估计来处理,见Sideris^[1], Georgiev等^[2]. 1980年, Kato^[3]证明了当$ n = 1 $,对于任何$ p>1 $,都有$ T(\varepsilon)<\infty $. 1981年, Strauss^[4]对问题(1.3)当$ n\geq2 $有以下的Strauss猜想:当$ p>p_0(n) $时, $ T(\varepsilon) = \infty $,即存在整体解;当$ 1<p\leq p_0(n) $时, $ T(\varepsilon)<\infty $,即解在有限时间内破裂.其中, $ p_0(n) $为以下二次方程的正根:

(1.4) $ \begin{equation} \gamma(p.n): = 2+(n+1)p-(n-1)p^2, \end{equation} $

即

$ p_0(n) = \frac{n+1+\sqrt{n^2+10n-7}}{2(n-1)}. $

$ p_0(n) $为Strauss指数, $ p_0(n) $是单调递减的.这个猜想的部分结果已经被许多作者证实.

1992年, Zhou^[5]证明了当$ n = 1 $时,对于任何$ p>1 $有以下的生命跨度估计

(1.5) $ \begin{equation} \left \{ \begin{array}{ll} c\varepsilon^{-(p-1)/2}\leq T(\varepsilon)\leq C\varepsilon^{-(p-1)/2}, & \int_{\Bbb R} g(x){\rm d}x\neq0, \\ c\varepsilon^{-p(p-1)/(p+1)}\leq T(\varepsilon)\leq C\varepsilon^{-p(p-1)/(p+1)}, & \int_{\Bbb R} g(x){\rm d}x = 0, \end{array} \right. \end{equation} $

其中$ c $和$ C $代表正的常数,但独立于$ \varepsilon $.此外, 1990年, Lindblad^[6]给出了当$ p = 2 $时更精确的结果

(1.6) $ \begin{equation} \left \{ \begin{array}{ll} \exists \lim\limits_{\varepsilon\rightarrow +0}\varepsilon^{1/2}T(\varepsilon)>0, & \int_{\Bbb R} g(x){\rm d}x\neq0, \\ \exists \lim\limits_{\varepsilon\rightarrow +0}\varepsilon^{2/3}T(\varepsilon)>0, & \int_{\Bbb R} g(x){\rm d}x = 0. \end{array} \right. \end{equation} $

同样, Lindblad^[6]还得到当$ (n, p) = (2.2) $时的下列结果

(1.7) $ \begin{equation} \left \{ \begin{array}{ll} \exists \lim\limits_{\varepsilon\rightarrow +0}a(\varepsilon)^{-1}T(\varepsilon)>0, & \int_{\Bbb R} g(x){\rm d}x\neq0, \\ \exists \lim\limits_{\varepsilon\rightarrow +0}\varepsilon T(\varepsilon)>0, & \int_{\Bbb R} g(x){\rm d}x = 0, \end{array} \right. \end{equation} $

其中$ a = a(\varepsilon) $是满足下面等式的数字

(1.8) $ \begin{equation} \begin{array}{ll} a^2\varepsilon^2\log (1+a) = 1. \end{array} \end{equation} $

(1)对于次临界指标$ 1<p<p_0(n)(n\geq3) $或$ 2<p<p_0(2)(n = 2) $,有以下猜想

(1.9) $ \begin{equation} \begin{array}{ll} c\varepsilon^{-2p(p-1)/\gamma(p, n)}\leq T(\varepsilon)\leq C\varepsilon^{-2p(p-1)/\gamma(p, n)}, \end{array} \end{equation} $

其中$ \gamma(p, n) $由(1.4)式定义.注意到,将$ n = 1 $的$ \gamma(p, n) $代入(1.9)式中那么和(1.5)式的第二行相同.下表总结了验证这一猜想的结果.

注意到,对于$ n = 2, 3 $,有

$ \exists\lim\limits_{\varepsilon\rightarrow+0}\varepsilon^{2p(p-1)/\gamma(p, n)}T(\varepsilon)>0. $

已经在本表中建立.此外, $ n\geq4 $时的次临界情形生命跨度上界估计蕴含在Sideris^[1]的结果中,该文中的方法也被Georgiev等^[2]和Zhou, Han^[7]分别用来建立波动方程组和外区域上初边值问题次临界情形的生命跨度上界估计.最早的Kato型引理(Kato^[3])的思想是证明未知函数在全空间上的积分会在有限时间内破裂.

(2)对于临界指标$ p = p_0(n) $,有如下猜想

(1.10) $ \begin{equation} \exp(c\varepsilon^{-p(p-1)})\leq T(\varepsilon)\leq \exp(C\varepsilon^{-p(p-1)}) . \end{equation} $

下表总结了验证这一猜想的结果.

本文参照Takamura^[8]中的思想,利用波动方程解的表达式,改进了文献[9]中$ n = 2, $$ 1<p\leq2 $及$ n = 1, $$ p>1 $情形时解的生命跨度上界估计.主要结论为如下两个定理:

定理1.1  令$ n = 2, 1<p\leq2 $且$ f\equiv0 $.假设$ g\in C^1({\Bbb R}^2) $为非负且有紧支集$ g\subset\{x\in {\Bbb R}^2:|x|\leq R\} $.若问题(1.1)有解$ u\in C^1({\Bbb R}^2\times[0, T(\varepsilon))) $,且满足$ suppu(x, t)\subset \{(x, t)\in {\Bbb R}^2\times [0, \infty):|x|\leq t+R\} $.那么必定存在一个正常数$ \varepsilon_2 = \varepsilon_2(g, p, R) $,使得$ \bar{T}(\varepsilon) $满足

(1.11) $ \begin{equation} \bar{T}(\varepsilon)\leq \left\{ \begin{array}{ll} Ca(\varepsilon), &p = 2, \\ C\varepsilon^{-(p-1)/(3-p+2\alpha)}, \quad &1<p<2, \end{array}\right. \end{equation} $

其中$ a(\varepsilon) $满足$ a(\varepsilon)^{4\alpha+2}\varepsilon^2 \log (1+a) = 1, 0<\varepsilon\leq \varepsilon_2 $,且$ C $是一个独立于$ \varepsilon $的正常数.

定理1.2  当$ n = 1, p>1 $,假设$ f\in C^2({\Bbb R}), g\in C^1({\Bbb R}) $都有$ (f, g)\subset \{x\in {\Bbb R}^2:|x|\leq R\} $.若问题(1.1)有解$ u\in C^2({\Bbb R}\times[0, T(\varepsilon))) $,那么必定存在一个正常数$ \varepsilon_3 = \varepsilon_3(f, g, p, R) $,使得$ \bar{T}(\varepsilon) $满足

(1.12) $ \begin{equation} \bar{T}(\varepsilon)\leq \left\{ \begin{array}{ll} C\varepsilon^{-(p-1)/(2+2\alpha)}, & \int_{\Bbb R} g(x){\rm d}x>0, \\ C\varepsilon^{-(p-1)/(p+2p\alpha+1)}, &f\geq0(\not\equiv0), g\equiv0, \end{array}\right. \end{equation} $

其中$ 0<\varepsilon\leq \varepsilon_3 $,且$ C $是一个独立于$ \varepsilon $的正常数.

正文安排如下:第2节,利用构造辅助函数和球坐标变换,对半线性波动方程的加权非线性项进行估计;第3节,证明定理1.1,讨论$ \alpha $的取值范围,并与定理1.3的生命跨度估计作对比;第4节,证明定理1.2,并与定理1.3的生命跨度估计作对比.

定理1.3^[9]  设$ f $与$ g $满足(1.2)式,假设问题(1.1)有解

$ u\in C^2([0, T)\times {\mathbb R}^n), $

使得

$ supp(u, u_t)\subset\{(x, t):|x|\leq t+R\}. $

我们有$ T(\varepsilon)\leq C\varepsilon^{-\frac{p(p-1)}{(p-1)a-q+2}} $,其中$ a = n+1- \frac{(n-1)p}{2}+2\alpha $, $ q = n(p-1)-2\alpha $.

引理1.1^[8]  设$ p>1, a>0, q>0 $,满足

(1.13) $ \begin{equation} M: = \frac{p-1}{2}a- \frac{q}{2}+1>0. \end{equation} $

设$ F\in C^2([0, T)) $满足

(1.14) $ \begin{equation} F(t)\geq At^a, \; \; t\geq T_0, \end{equation} $

(1.15) $ \begin{equation} F''(t)\geq B(t+R)^{-q}|F(t)|^p, \; \; t\geq 0, \end{equation} $

(1.16) $ \begin{equation} F(0)\geq 0, \qquad F'(0)> 0, \end{equation} $

其中$ A, B, R, T_0 $为正的常数.当$ C_0 = C_0(p, a, q, B) $,使得

(1.17) $ \begin{equation} T<2^{2/M}T_1, \end{equation} $

(1.18) $ \begin{equation} T_1: = \max\bigg \{T_0, \frac{F(0)}{F'(0)}, R\bigg\}\geq C_0A^{-(p-1)/2M}. \end{equation} $

若假设(1.16)式用下式替换

(1.19) $ \begin{equation} F(0)>0, \qquad F'(0) = 0. \end{equation} $

同时$ t_0>0 $, $ F(t_0)\geq2F(0). $然后,将引理1.1的结论转化为存在一个正常数$ C_0 = C_0(p, a, q, B) $,

(1.20) $ \begin{equation} T<2^{2/M}T_2. \end{equation} $

从而

(1.21) $ \begin{equation} T_2: = \max\{T_0, t_0, R\}\geq C_0A^{-(p-1)/(2M)}. \end{equation} $

引理1.2^[8]  当$ n\geq2 $,且$ 1<p<p_0(n) $,假设假设$ f $与$ g $满足(1.2)式的函数,且问题(1.3)有解

$ (u, u_t)\in C([0, T(\varepsilon)), H^1({\mathbb R}^n)\times L^2({\mathbb R}^n)), $

使得

$ supp(u, u_t)\subset\{(x, t)\in {\mathbb R}^n\times[0, +\infty):|x|\leq t+R\}. $

那么存在$ \varepsilon_0 = \varepsilon_0(f, g, n, p, R) $,使得

(1.22) $ \begin{equation} T(\varepsilon)\leq C\varepsilon^{-2p(p-1)/\gamma(p, n)}, \end{equation} $

其中$ 0<\varepsilon<\varepsilon_0 $,且$ C $为正的常数独立于$ \varepsilon $.

在本文中,记号$ C $表示常数,在不同的地方可能表示不同的值.

构造辅助函数

(2.1) $ \begin{equation} F(t) = \int_{{\mathbb R}^n} u(x, t){\rm d}x, \end{equation} $

对于适当的$ a $和$ q $,函数$ F(t) $满足引理1.1的常微分不等式.首先,

(2.2) $ \begin{equation} F''(t) = \int_{{\mathbb R}^n} u_{tt}{\rm d}x = \int_{{\mathbb R}^n}[\Delta u+(1+|x|^2)^\alpha |u|^p]{\rm d}x = \int_{{\mathbb R}^n}(1+|x|^2)^\alpha |u|^p {\rm d}x, \end{equation} $

这里用到波动方程的有限传播速度以及如下事实:

$ \int_{{\mathbb R}^n}\Delta u {\rm d}x = \int_{|x| = t+R+1}Du\cdot n{\rm d}S $

为散度定理的推论.为了估计(2.2)式的右端项,直接应用Hölder不等式,得

$ \begin{eqnarray*} \bigg|\int_{{\mathbb R}^n} u(x, t){\rm d}x\bigg|& = &\bigg|\int_{|x|\leq t+R}(1+|x|^2)^{\frac{\alpha}{p}}u(x, t)(1+|x|^2)^{-\frac{\alpha}{p}}{\rm d}x\bigg| \\ &\leq &\bigg(\int_{|x|\leq t+R}(1+|x|^2)^{\alpha} |u(x, t)|^p {\rm d}x\bigg)^{\frac{1}{p}}\bigg (\int_{|x|\leq t+R}(1+|x|^2)^{-\frac{\alpha}{p-1}}{\rm d}x\bigg)^{ \frac{1}{p'}}, \end{eqnarray*} $

即

$ \begin{eqnarray*} \bigg|\int_{{\mathbb R}^n} u(x, t){\rm d}x\bigg|^p &\leq&\bigg(\int_{|x|\leq t+R}(1+|x|^2)^{\alpha}|u(x, t)|^p{\rm d}x\bigg)\bigg(\int_{|x|\leq t+R}(1+|x|^2)^{-\frac{\alpha}{p-1}}{\rm d}x\bigg)^{p-1} \\ &\leq & \bigg(\int_{{\mathbb R}^n}(1+|x|^2)^{\alpha} |u(x, t)|^p {\rm d}x \bigg)\bigg(\int_{|x|\leq t+R}(1+|x|^2)^{-\frac{\alpha}{p-1}}{\rm d}x\bigg)^{p-1}, \end{eqnarray*} $

从而有

(2.3) $ \begin{equation} \bigg (\int_{{\mathbb R}^n}(1+|x|^2)^{\alpha} |u(x, t)|^p {\rm d}x\bigg)\geq\bigg |\int_{{\mathbb R}^n}u(x, t){\rm d}x\bigg|^p \bigg(\int_{|x|\leq t+R}(1+|x|^2)^{-\frac{\alpha}{p-1}}{\rm d}x\bigg)^{1-p}. \end{equation} $

引理2.1^[9]  若$ \alpha<\frac{n(p-1)}{2} $,则

(2.4) $ \begin{equation} \int_{|x|\leq t+R}(1+|x|^2)^{-\frac{\alpha}{p-1}}{\rm d}x\leq C(t+R)^{n-\frac{2\alpha}{p-1}}, \end{equation} $

证  由球坐标变换,可得

(2.5) $ \begin{eqnarray} \int_{|x|\leq t+R}(1+|x|^2)^{-\frac{\alpha}{p-1}}{\rm d}x &\leq & C\int_0^{t+R}(1+r^2)^{n-\frac{2\alpha}{p-1}}r^{n-1}{\rm d}r \\ &\leq & C\int_0^{t+R}r^{n-\frac{2\alpha}{p-1}-1}{\rm d}r = C(t+R)^{n-\frac{2\alpha}{p-1}}, \end{eqnarray} $

引理证毕.

因此

$ \int_{{\mathbb R}^n}(1+|x|^2)^{\alpha} |u(x, t)|^p {\rm d}x\geq |F(t)|^p\cdot C^{p-1}(t+R)^{2\alpha-n(p-1)}, $

从而得到

(2.6) $ \begin{equation} F''(t)\geq C(t+R)^{2\alpha-n(p-1)}|F(t)|^p. \end{equation} $

在定理1.1的假设下,应用非齐次二维波动方程的Poisson公式可得

(3.1) $ \begin{equation} u(x, t) = \frac{\varepsilon}{2\pi}\int_{|x-y|\leq t}\frac{g(y)}{\sqrt{t-|x-y|^2}}{\rm d}y+\frac{1}{2\pi}\int_0^\tau {\rm d}\tau \int_{|x-y|\leq{t-\tau}}\frac{(1+|x|^2)^\alpha \cdot |u(y, \tau)|^p}{\sqrt{(t-\tau)^2-|x-y|^2}}{\rm d}y. \end{equation} $

注意到$ |y|\leq R, |x|\leq t+R $由于第一项的支集的特性,且第二项为非负,应用不等式

$ t-|x-y|\leq t-||x|-|y||\leq t-|x|+R, |x|\geq R, $

$ t+|x-y|\leq t+|x|+R\leq 2(t+R), $

可得

$ u(x, t)\geq \frac{\varepsilon}{2\sqrt2\pi\sqrt{t+R}\sqrt{t-|x|+R}}\int_{|x-y|\leq t}g(y){\rm d}y, \quad |x|\geq R. $

如果假设$ |x|+R\leq t $,当$ |y|\leq R $有$ |x-y|\leq t $,得到

$ \int_{|x-y|\leq t}g(y){\rm d}y = \| g\| _{L^1(R^2)}. $

从而

(3.2) $ \begin{equation} u(x, t)\geq\frac{\| g\| _{L^1(R^2)}\varepsilon}{2\sqrt2\pi\sqrt{t+R}\sqrt{t-|x|+R}}, \quad R\leq |x|\leq t-R. \end{equation} $

这与高维情况相同.结合(2.2)和(3.2)式,可得

$ \begin{eqnarray*} F''(t)& = &\int_{R^2}(1+|x|^2)^\alpha |u(x, t)|^p{\rm d}x\\ &\geq&\bigg(\frac{\| g\| _{L^1(R^2)}\varepsilon}{2\sqrt2\pi\sqrt{t+R}}\bigg)^p\int_{R\leq|x|\leq t-R}(1+|x|^2)^\alpha\cdot\frac1{(t-|x|+R)^{p/2}}{\rm d}x, \quad t\geq2R. \end{eqnarray*} $

由球坐标变换可知

(3.3) $ \begin{equation} F''(t)\geq \frac{2\pi\| g\|^p _{L^1(R^2)}\cdot\varepsilon^p}{(2\sqrt2\pi)^p(t+R)^{p/2}}\int_{\Bbb R} ^{t-R}\frac{r(1+r^2)^\alpha}{(t-r+R)^{p/2}}{\rm d}r, \quad t\geq2R. \end{equation} $

当$ 1<p<2 $时,且$ \alpha<0 $,由(3.3)式的右端项

(3.4) $ \begin{eqnarray} \int_{\Bbb R} ^{t-R}\frac{r(1+r^2)^\alpha}{(t-r+R)^{p/2}}{\rm d}r&\geq & \frac{[1+(t-R)^2]^\alpha}{2t^{p/2}}[(t-R)^2-R^2]\\ &\geq&\frac{(t-t/3)^{2\alpha}}{2t^{p/2}}[(t-t/3)^2-(t/3)^2]\\ &\geq&\frac16(\frac23)^{2\alpha}\cdot t^{2\alpha+2-p/2}, \; \; t\geq3R, \end{eqnarray} $

有

$ F''(t)\geq C\frac{\| g\|^p _{L^1(R^2)}\varepsilon^p}{\pi^{p-1}(t+R)^{p/2}}t^{2\alpha+2-p/2}\geq C\frac{\| g\| _{L^1(R^2)}}{\pi^{p-1}}\varepsilon^p\cdot t^{2\alpha+2-p}, \quad t\geq3R. $

对上面的不等式在$ [3R, t] $积分,利用$ F'(0) = \| g\| _{L^1(R^2)}\varepsilon>0 $,得到

$ F'(t)\geq C\frac{\| g\|^p _{L^1(R^2)}}{\pi^{p-1}}\varepsilon^p\cdot t^{2\alpha+3-p}, \quad t\geq4R. $

从而可得

(3.5) $ \begin{equation} F(t)\geq C\frac{\| g\|^p _{L^1(R^2)}}{\pi^{p-1}}\varepsilon^p\cdot t^{2\alpha+4-p}, \quad t\geq5R. \end{equation} $

结合(2.6)式可知

(3.6) $ \begin{equation} F''(t)\geq C(t+R)^{2\alpha-2(p-1)}|F(t)|^p. \end{equation} $

将当$ n = 2, 1<p<2 $时的情况应用到引理1.1,有

$ A = C\varepsilon^p, a = 2\alpha+4-p, q = 2(p-1)-2\alpha. $

那么

$ M: = \frac{p-1}2a-\frac{q}2+1 = -\frac12p^2+\frac32p+p\alpha>0. $

即$ \alpha>\frac{p-3}2 $.同时由(3.4)式中的条件可知$ \alpha<0 $,也使得引理2.1中的$ \alpha<\frac{n(p-1)}2 $成立.

综合上述, $ \frac{p-3}2<\alpha<0 $.

由改进的Kato引理1.1的生命跨度,可得$ \bar{T}(\varepsilon)\leq C_0A^{-(p-1)/(2M)}, $其中$ A = \varepsilon ^p $, $ M = -\frac12 p^2+\frac32p+p\alpha $.可得

(3.7) $ \begin{equation} \bar{T}(\varepsilon)\leq C_0\varepsilon ^\frac{-p(p-1)}{-p^2+3p+2p\alpha}, \end{equation} $

同时,将$ n = 2, 1<p<2 $的情况应用到定理1.3来估计生命跨度,可得

$ T(\varepsilon)\leq C_0\varepsilon ^\frac{-p(p-1)}{(p-1)a-q+2}, $

其中$ a = 3-\frac p2+2\alpha $, $ q = 2p-2-2\alpha $.从而

(3.8) $ \begin{equation} T(\varepsilon)\leq C\varepsilon ^\frac{-p(p-1)}{(p-1)(3-\frac p2+2\alpha)-2(p-1)+2\alpha+2} \leq C\varepsilon ^\frac{-p(p-1)}{-p^2/2+3p/2+2p\alpha+1}, \end{equation} $

比较(3.7)与(3.8)式,由于$ 0<\varepsilon \ll 1 $,且

$ -p^2+3p+2p\alpha-(-p^2/2+3p/2+2p\alpha+1) = -\frac12(p-1)(p-2)>0. $

从而

$ C_0\varepsilon ^\frac{-p(p-1)}{-p^2+3p+2p\alpha} <C\varepsilon ^\frac{-p(p-1)}{-p^2/2+3p/2+2p\alpha+1}. $

即当$ n = 2, 1<p<2 $的生命跨度$ \bar{T}(\varepsilon) $估计优于$ T(\varepsilon) $.

当$ p = 2 $时,由(3.3)式的右端项利用分部积分,可得

$ \int_{\Bbb R} ^{t-R}\frac{r(1+r^2)^\alpha}{t-r+R}{\rm d}r\geq [1+(t-R)^2]^\alpha\cdot\bigg [R\log t-(t-R)\log 2R+\int_r^{t-R}\log (t-r+R){\rm d}r\bigg]. $

为了不失一般性,假设

(3.9) $ \begin{equation} R\geq1. \end{equation} $

注意到$ (t-R)/2\geq R $等价于$ t\geq 3R $.因此,可以缩小积分区间为$ [R, (t-R)/2] $,可得

$ \int_{\Bbb R} ^{t-R}\log (t-r+R){\rm d}r\geq(\frac{t-R}2-R)\log \frac{t+3R}2, \quad t\geq3R. $

从而

$ \begin{eqnarray*} &&\int_{\Bbb R} ^{t-R}\frac{r(1+r^2)^\alpha}{t-r+R}{\rm d}r\\ &\geq&(\frac23)^{2\alpha}t^{2\alpha}\bigg[R(\log t-\log \frac{t+3R}2)+(t-R)(\frac12\log \frac{t+3R}2-\log 2R)\bigg], \quad t\geq3R. \end{eqnarray*} $

其中

$ \log \frac{t+3R}2\leq \log t, \; \; t\geq3R, $

$ \frac14\log \frac{t+3R}2\geq \log 2R, \quad t\geq32R^4-3R, $

可得

$ \int_{\Bbb R} ^{t-R}\frac{r(1+r^2)^\alpha}{t-r+R}{\rm d}r\geq \frac16(\frac23)^{2\alpha}\cdot t^{2\alpha+1}\log \frac{t+3R}2, \quad t\geq32R^4-3R. $

由(3.9)式可知, $ 32R^4-3R\geq3R $.结合(2.2)和(3.3)式,可得

$ \begin{eqnarray*} F''(t)&\geq & \frac{2\pi\| g\| _{L^1(R^2)}\varepsilon^2}{8\pi^2\cdot t}\cdot \frac16(\frac23)^{2\alpha}\cdot t^{2\alpha+1}\log \frac{t+3R}2 \\ &\geq& (\frac23)^{2\alpha}\frac{2\| g\| _{L^1(R^2)}\varepsilon^2}{48\pi(t+t/3)}\cdot t^{2\alpha+1}\log \frac{t}2\\ &\geq& (\frac23)^{2\alpha}\frac{\| g\| _{L^1(R^2)}\varepsilon^2}{32\pi}\cdot t^{2\alpha}\log \frac{t}2, \quad t\geq32R^4. \end{eqnarray*} $

这个式子出现了一个额外的项$ \log \frac {t}2 $.

对上面的不等式在$ [32R^4, t] $积分,利用$ F'(0) = \| g\| _{L^1(R^2)}\varepsilon>0 $,得到

$ \begin{eqnarray*} F'(t)&>&(\frac23)^{2\alpha}\frac{\| g\| _{L^1(R^2)}\varepsilon^2}{32\pi}\int_{t/2}^ts^{2\alpha}\log \frac s2{\rm d}s \\ &\geq& (\frac23)^{2\alpha}(1-\frac1{2^{2\alpha+1}})\cdot \frac1{2\alpha+1}\frac{\| g\| _{L^1(R^2)}\varepsilon^2}{32\pi}t^{2\alpha+1}\log \frac t4, \quad t\geq64R^4. \end{eqnarray*} $

同样步骤在$ [64R^4, t] $积分,由于$ F(0) = 0 $,可得

(3.10) $ \begin{eqnarray} F(t)&>&C\frac{\| g\| _{L^1(R^2)}\varepsilon^2}{32\pi}\int_{t/2}^ts^{2\alpha+1}\log \frac s4{\rm d}s\\ &\geq & (\frac23)^{2\alpha}(1-\frac1{2^{2\alpha+1}})(1-\frac1{2^{2\alpha+1}}) \\ &&\cdot \frac1{(2\alpha+1)(2\alpha+2)}\frac{\| g\| _{L^1(R^2)}\varepsilon^2}{32\pi}t^{2\alpha+2}\log \frac t8, \quad t\geq128R^4. \end{eqnarray} $

由(2.6)式当$ n = 2, p = 2 $时,可得

(3.11) $ \begin{equation} F''(t)\geq C(t+R)^{2\alpha-2}|F(t)|^p. \end{equation} $

将当$ n = 2, p = 2 $时的情况应用到引理1.1,有

$ a = 2\alpha+2, p = 2, q = 2-2\alpha, B = \pi^{-1}. $

由(1.13)式可知, $ M = 2\alpha+1 $.如果$ T_0 $满足

(3.12) $ \begin{equation} T_0\geq\max \bigg\{\frac{F(0)}{F'(0)}, 128R^4\bigg\}, \end{equation} $

将$ T_1 = T_0 $代入(1.18)式中,可得

(3.13) $ \begin{equation} T_0\geq C_0A^{(p-1)/(2M)}, \qquad A\geq C_0^{4\alpha +2}T_0^{-4\alpha-2}. \end{equation} $

同时,由(3.10)与(1.14)式

(3.14) $ \begin{equation} C\frac{\| g\| _{L^1(R^2)}\varepsilon^2}{32\pi}T_0^{2\alpha+2}\log \frac {T_0}8\geq AT_0^{2\alpha+2}, \end{equation} $

结合(3.13)与(3.14)式,可得

(3.15) $ \begin{equation} C\frac{\| g\| _{L^1(R^2)}\varepsilon^2}{32\pi}\log \frac {T_0}8\geq A, \qquad C\frac{8\| g\| _{L^1(R^2)}\varepsilon^2}{256\pi C_0^{4\alpha+2}}T_0^{4\alpha+2}\log \frac {T_0}8\geq 1. \end{equation} $

取适当的$ T_0 $,若$ \frac{\| g\| _{L^1(R^2)}}{\pi C_0^{4\alpha+2}}\geq 1 $,假设$ \frac{T_0}{16} = a(\varepsilon), $其中$ a(\varepsilon) $满足$ a(\varepsilon)^{4\alpha+2}\varepsilon^2 \log (1+a) = 1 $.因此存在一个正常数$ \varepsilon_{21} = \varepsilon_{21}(g, R) $,满足

$ a(\varepsilon)\geq \max\bigg\{\frac{F(0)}{16F'(0)}, 8R^4\bigg\}, \quad 0<\varepsilon\leq \varepsilon_{21}, $

由(3.12)式中的$ a(\varepsilon) $是单调递减的函数,且$ \lim\limits_{\varepsilon\rightarrow+0}a(\varepsilon) = \infty $.且$ R\geq 1 $可知

$ \frac{T_0}8 = 2a(\varepsilon)\geq8R^4+a(\varepsilon)\geq1+a(\varepsilon), \quad 0<\varepsilon\leq\varepsilon_{21}. $

即

(3.16) $ \begin{equation} T_0 = 16a(\varepsilon), \quad 0<\varepsilon\leq\varepsilon_{21}. \end{equation} $

结合(3.12)与(3.15)式,可得

$ \begin{eqnarray*} C\frac{8\| g\| _{L^1(R^2)}\varepsilon^2}{256\pi C_0^{4\alpha+2}}T_0^{4\alpha+2}\log \frac {T_0}8 &\geq& C\frac{8\| g\| _{L^1(R^2)}\varepsilon^2}{\pi C_0^{4\alpha+2}}(\frac{T_0}{16})^{4\alpha+2}\log \frac {T_0}8\\ &\geq & a(\varepsilon)^{4\alpha+2}\varepsilon^2\log (1+a(\varepsilon)) = 1, \quad 0<\varepsilon\leq\varepsilon_{21}. \end{eqnarray*} $

另一部分$ \frac{\| g\| _{L^1(R^2)}}{\pi C_0^{4\alpha+2}}\leq 1 $,假设

$ \frac{\| g\| _{L^1(R^2)}}{\sqrt{\pi}C_0^{2\alpha+1}}\cdot\frac {T_0}{16} = a(\varepsilon), $

由(3.12)式,且存在一个正常数$ \varepsilon_{22} = \varepsilon_{22}(g, R) $,满足

$ a(\varepsilon)\geq \frac{\| g\| _{L^1(R^2)}}{\sqrt{\pi}C_0^{2\alpha+1}}\max\bigg\{\frac{F(0)}{16F'(0)}, 8R^4\bigg\}, \quad 0<\varepsilon\leq \varepsilon_{22}. $

同时$ R\geq1 $,即

$ \frac{T_0}8 = \frac{2\sqrt{\pi}C_0^{2\alpha+1}}{\| g\| _{L^1(R^2)}}a(\varepsilon) \geq 8R^4+\frac{\sqrt{\pi}C_0^{2\alpha+1}}{\| g\| _{L^1(R^2)}}a(\varepsilon) \geq1+a(\varepsilon), \quad 0<\varepsilon\leq \varepsilon_{22}. $

结合(3.12)与(3.15)式,可得

(3.17) $ \begin{equation} T_0 = \frac{16\sqrt{\pi}C_0^{2\alpha+1}}{\| g\| _{L^1(R^2)}}a(\varepsilon), \quad 0<\varepsilon\leq\varepsilon_{22}. \end{equation} $

从而

$ \begin{eqnarray*} \frac{8\| g\| _{L^1(R^2)}\varepsilon^2}{256\pi C_0^{4\alpha+2}}T_0^{4\alpha+2}\log \frac {T_0}8 &\geq & \frac{16^{4\alpha}\pi^{2\alpha}\cdot8C}{\| g\|^{4\alpha} _{L^1(R^2)}}\bigg(\frac{\| g\| _{L^1(R^2)}}{\sqrt{\pi}C_0^{2\alpha+1}}\cdot \frac{T_0}{16}\bigg)^{4\alpha+2}\varepsilon^2\log \frac {T_0}8\\ &\geq & a(\varepsilon)^{4\alpha+2}\varepsilon^2\log (1+a(\varepsilon)) = 1, \quad 0<\varepsilon\leq\varepsilon_{22}. \end{eqnarray*} $

结合(3.16)和(3.17)式,由引理1.1可得

$ T_0 = 16\max\bigg\{1, \frac{\sqrt{\pi}C_0}{\| g\| _{L^1(R^2)}}\bigg\}a(\varepsilon), \quad 0<\varepsilon\leq\varepsilon_{2}, \ A = C^{4\alpha+2}_0T_0^{-4\alpha-2}, $

其中$ \varepsilon_2 = \min\{\varepsilon_{21}, \varepsilon_{22}\} $.同时$ 2^{2/M}T_1 = 2^{\frac2{2\alpha +1}}T_0 $,所以

$ T(\varepsilon)\leq16 \cdot 4^{\frac1{2\alpha +1}}\max\bigg\{1, \frac{\sqrt{\pi}C_0}{\| g\| _{L^1(R^2)}}\bigg\}a(\varepsilon)^{-2\alpha-1}, \quad 0<\varepsilon\leq\varepsilon_{2}. $

由改进的Kato引理1.1的生命跨度,可得

(3.18) $ \begin{equation} \bar{T}(\varepsilon)\leq Ca(\varepsilon) , \end{equation} $

其中$ a(\varepsilon) $满足$ a(\varepsilon)^{4\alpha+2}\varepsilon^2 \log (1+a) = 1 $,由(3.7)与(3.18)式完成了定理1.1的证明.

同时,将当$ n = 2, p = 2 $时的情况应用到定理1.3来估计生命跨度,可得$ T(\varepsilon)\leq C\varepsilon^\frac{-p(p-1)}{(p-1)a-q+2}, $其中$ a = n+1-(n-1)p/2+2\alpha, q = n(p-1)-2\alpha $.可得

(3.19) $ \begin{equation} T(\varepsilon)\leq C\varepsilon^\frac{-p(p-1)}{(p-1)(3-p/2+2\alpha)-2(p-1)+2\alpha+2}\leq C\varepsilon^\frac{-2}{4\alpha+2}. \end{equation} $

比较(3.18)与(3.19)式,有

$ \bar{T}(\varepsilon)\leq T(\varepsilon). $

即当$ n = 2, p = 2 $,的生命跨度$ \bar{T}(\varepsilon) $估计优于$ T(\varepsilon) $.

由于$ u\in C^2({\Bbb R}\times[0, T(\varepsilon))) $,且$ suppu(x, t)\subset \{x\in {\Bbb R}^2:|x|\leq R\} $,对(2.1)式在$ {\Bbb R} $积分

(4.1) $ \begin{equation} F''(t) = \int_{\Bbb R} (1+|x|^2)^\alpha|u|^p{\rm d}x = \int^{t+R}_{-(t+R)}(1+|x|^2)^\alpha|u|^p{\rm d}x. \end{equation} $

由(2.6)式当$ n = 1 $时,可得

(4.2) $ \begin{equation} F''(t)\geq C(t+R)^{2\alpha-(p-1)}|F(t)|^p, t\geq0. \end{equation} $

由一维的达朗贝尔公式可知

(4.3) $ \begin{equation} u(x, t) = \frac{f(x+t)+f(x-t)}2\varepsilon+\frac{\varepsilon}2\int^{x+t}_{x-t}g(\xi) {\rm d}\xi +\frac12\int^t_0{\rm d}\tau \int^{x+t-\tau}_{x-t+\tau}(1+|\xi|^2)^\alpha|u(\xi, \tau)|^p{\rm d}\xi. \end{equation} $

(1)当$ \int_{\Bbb R} g(x){\rm d}x>0 $,由(4.3)式可得

$ u(x, t)\geq G\varepsilon, \quad x+t\geq R, \quad x-t\leq -R, $

其中$ G = \frac12\int_{\Bbb R} g(x){\rm d}x>0 $.

结合(4.1)式,有

$ F''(t)\geq (G\varepsilon)^p\int^{t-R}_0(1+|x|^2)^\alpha {\rm d}x\geq \frac{G^p\varepsilon^p}{2\alpha+1}(t-R)^{2\alpha+1}, t\geq R. $

由$ F'(R)\geq F'(0)>0, F(R)>F(0)\geq0 $,并对上式在区间$ [R, t] $积分2次,有

(4.4) $ \begin{equation} \begin{array}{ll} F(t)>\frac{G^p\varepsilon^p(t-R)^{2\alpha+3}}{(2\alpha+1)(2\alpha+2)(2\alpha+3)}, t\geq R, \\ F(t)>\frac{(1/2)^{2\alpha+3}G^p\varepsilon^pt^{2\alpha+3}}{(2\alpha+1)(2\alpha+2)(2\alpha+3)}, t\geq 2R . \end{array} \end{equation} $

结合(4.2)与(4.4)式,由引理1.1可得

$ A = \frac{(1/2)^{2\alpha+3}G^p\varepsilon^p}{(2\alpha+1)(2\alpha+2)(2\alpha+3)}, a = 2\alpha+3, q = p-1-2\alpha. $

可得

$ M = \frac{p-1}2(2\alpha+3)-\frac{p-1-2\alpha}2+1 = p(1+\alpha), $

从而

(4.5) $ \begin{equation} T_0 = C_0A^{(p-1)/(2M)} = \frac{C_0(1/2)^{2\alpha+3}G^p}{(2\alpha+1)(2\alpha+2)(2\alpha+3)}\varepsilon^{-p(p-1)/(2p+2p\alpha)}. \end{equation} $

同时,将$ n = 1, p>1 $的情况应用到定理1.3中来估计生命跨度,可得

$ T(\varepsilon)\leq C\varepsilon^{\frac{-p(p-1)}{(p-1)a-q+2}}, $

其中$ a = 2+2\alpha, q = p-1-2\alpha $.

从而

(4.6) $ \begin{equation} T(\varepsilon)\leq C\varepsilon^{\frac{-p(p-1)}{(p-1)(2+2\alpha)-p+1+2\alpha+2}}\leq C\varepsilon^{\frac{-p(p-1)}{2p\alpha+p+1}}. \end{equation} $

比较(4.5)与(4.6)式,由于$ p>1 $且$ 0<\varepsilon\ll 1 $,可知

$ 2p+2p\alpha-(2p\alpha+p+1) = p-1>0, $

从而

$ \frac{C_0(1/2)^{2\alpha+3}G^p}{(2\alpha+1)(2\alpha+2)(2\alpha+3)}\varepsilon^{-p(p-1)/(2p+2p\alpha)}<C\varepsilon^{\frac{-p(p-1)}{2p\alpha+p+1}}. $

即当$ n = 1, p>1 $的生命跨度$ \bar{T}(\varepsilon) $估计优于$ T(\varepsilon) $.

(2)当$ f\geq0(\not\equiv0), g\equiv0 $由(4.3)式可得

$ u(x, t)\geq \frac{f(x-t)}2\varepsilon, x+t\geq R, -R\leq x-t\leq R. $

结合(4.1)式,可得

$ \begin{eqnarray*} F''(t)&\geq &\frac{\varepsilon^p}{2^p}\int^{t+R}_{t-R}(1+|x|^2)^\alpha f(x-t)^p{\rm d}x \geq \frac{\|f\|^p _{L^p(R)}}{2^p}\varepsilon^p\frac1{2\alpha+1}(t+R)^{2\alpha+1}, \quad t\geq R. \end{eqnarray*} $

由$ F'(R)\geq F'(0)>0, F(R)>F(0)\geq0 $,并对上式在区间$ [R, t] $积分2次,有

$ F(t)>\frac{\|f\|^p _{L^p(R)}\varepsilon^p\cdot t^{2\alpha+3}}{2^p(2\alpha+1)(2\alpha+2)(2\alpha+3)}, \quad t\geq R. $

与(1)同理,于是完成了定理1.2的证明.