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数学物理学报, 2019, 39(5): 1146-1157 doi:

论文

半线性波动方程Cauchy问题解的生命跨度估计

蒋红标,1, 汪海航,2

Lifespan Estimation of Solutions to Cauchy Problem of Semilinear Wave Equation

Jiang Hongbiao,1, Wang Haihang,2

通讯作者: 汪海航, E-mail: 1196443777@qq.com

收稿日期: 2018-09-27  

基金资助: 浙江省自然科学基金.  LY18A010008

Received: 2018-09-27  

Fund supported: the Natural Science Foundation of Zhejiang Province.  LY18A010008

作者简介 About authors

蒋红标,E-mail:hbj@126.com , E-mail:hbj@126.com

摘要

该文研究了Rn中半线性波动方程utt-△u=(1+|x|2α|u|p的小初值Cauchy问题解的生命跨度估计.主要利用了改进的Kato型引理,得到了当n=2,1 < p ≤ 2时及n=1,p > 1时改进的生命跨度上界估计.

关键词: 半线性波动方程 ; 初值问题 ; 常微分不等式 ; 生命跨度

Abstract

In this paper, the lifespan estimate to the Cauchy problem of the semi-linear wave equation utt utt-△u=(1+|x|2)α|u|p in Rn is studied. The upper bound of lifespan is improved for the cases n=2, 1 < p ≤ 2 and n=1, p > 1, by using the improved Kato's type lemma.

Keywords: Semilinear wave equations ; Initial value problems ; Ordinary differential inequality ; Lifespan

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本文引用格式

蒋红标, 汪海航. 半线性波动方程Cauchy问题解的生命跨度估计. 数学物理学报[J], 2019, 39(5): 1146-1157 doi:

Jiang Hongbiao, Wang Haihang. Lifespan Estimation of Solutions to Cauchy Problem of Semilinear Wave Equation. Acta Mathematica Scientia[J], 2019, 39(5): 1146-1157 doi:

1 引言

研究如下半线性波动方程的初值问题

{uttΔu=(1+|x|2)α|u|p,Rn×[0,),u(x,0)=εf(x),ut(x,0)=εg(x),
(1.1)

其中Δ=ni=12x2i,且初值满足

{(f,g)H1(Rn)×L2(Rn),f(x)0,g(x)0,f(x)=g(x)=0,|x|>R>0,α>1,0<ε1.
(1.2)

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:

{u(x,t)=|u|p,Rn×[0,),u(x,0)=εf(x),ut(x,0)=εg(x).
(1.3)

其中n2, ε>0是小参数.由此定义(1.1)解的一个生命跨度为T(ε),

T(ε)=sup

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:

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

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

\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}
(1.5)

其中 c 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

\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}
(1.6)

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

\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}
(1.7)

其中 a = a(\varepsilon) 是满足下面等式的数字

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

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

\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}
(1.9)

其中 \gamma(p, n) 由(1.4)式定义.注意到,将 n = 1 \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.

  

lower bound of T(\varepsilon)upper bound of T(\varepsilon)
n=2Zhou[10]Zhou[10]
n=3Lindblad[6]Lindblad[6]
n\geq4Lai, Zhou[11]Sideris[1]

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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

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

下表总结了验证这一猜想的结果.

  

lower bound of T(\varepsilon)upper bound of T(\varepsilon)
n=2Zhou[10]Zhou[10]
n=3Zhou[5]Zhou[5]
n\geq4Lindblad, Sogge [12]Takamura, Wakasa[13]

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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 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 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

\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}
(1.11)

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

\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}
(1.12)

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 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

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

F\in C^2([0, T)) 满足

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

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

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

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

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

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

若假设(1.16)式用下式替换

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

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) ,

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

从而

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

引理1.2[8]  当 n\geq2 ,且 1<p<p_0(n) ,假设假设 f 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

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

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.

2 二阶常微分不等式

构造辅助函数

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

对于适当的 a q ,函数 F(t) 满足引理1.1的常微分不等式.首先,

\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}
(2.2)

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*}

\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

\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}
(2.3)

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

\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}
(2.4)

  由球坐标变换,可得

\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}
(2.5)

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

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

3 定理1.1的证明

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

\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}
(3.1)

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 |x-y|\leq t ,得到

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

从而

\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}
(3.2)

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

\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}
(3.3)

1<p<2 时,且 \alpha<0 ,由(3.3)式的右端项

\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}
(3.4)

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

\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}
(3.5)

结合(2.6)式可知

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

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.

\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

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

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

\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}
(3.8)

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) .

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

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

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

\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}
(3.10)

由(2.6)式当 n = 2, p = 2 时,可得

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

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

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

T_1 = T_0 代入(1.18)式中,可得

\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}
(3.13)

同时,由(3.10)与(1.14)式

\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}
(3.14)

结合(3.13)与(3.14)式,可得

\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}
(3.15)

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 . R\geq 1 可知

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

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

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

\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}
(3.17)

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

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

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

\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}
(3.19)

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) .

4 定理1.2的证明

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

\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}
(4.1)

由(2.6)式当 n = 1 时,可得

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

由一维的达朗贝尔公式可知

\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}
(4.3)

(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.

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

\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}
(4.4)

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

\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}
(4.5)

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

\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}
(4.6)

比较(4.5)与(4.6)式,由于 p>1 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*}

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.

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