本文主要研究如下低维空间中半线性波动方程的柯西问题

$ \begin{equation}\label{1}\left\{\begin{array}{ll}\square u=F(u), ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x, t)\in{{R}}^{n}\times(0, +\infty), \\ t=0:~u=\varepsilon f(x), ~~u_t=\varepsilon g(x), ~~~~x\in{{R}}^{n}, \end{array}\right. \end{equation} $

(1.1)

其中$\square =\partial_{t} ^{2}-\sum\limits_{i=1}^n \partial_{x_i} ^{2}$是波算子,

$ \begin{equation} \begin{aligned}F(u)\geq Au^{p}, ~~~~u\geq0, \end{aligned}\end{equation} $

(1.2)

$A>0, p>1, \varepsilon>0$是小参数.考虑$n=2$和$n=3$情形, 且$p>p_{0}(n)$, 其中$p_{0}(n)$是下列二次方程的正根

$ \begin{equation*}\begin{aligned}(n-1)p^{2}-(n+1)p-2=0. \end{aligned}\end{equation*} $

对于半线性波动方程柯西问题$(1.1)$, 已经有了很多关于解整体存在和爆破的相关结果.当初值$f(x)$, $g(x)$具有紧支集且$F(u)=|u|^{p}$时, 柯西问题$(1.1)$有Strauss猜想.即:初值“充分小”, $p>p_{0}(n)$, 则式$(1.1)$有整体解; 初值在某种意义下为正, 且$1 <p\leq p_{0}(n)$, 则式$(1.1)$没有整体解.

对于$n=3$情形, John^{[1, 2]}得到如下结果:设$F(u)$满足

$ \begin{equation}\begin{aligned}F(u)\geq A|u|^{p}, ~~~A>0, \end{aligned}\end{equation} $

(1.3)

初值是具有紧支集的光滑函数且充分小, 若$p>p_{0}(3)=1+\sqrt{2}$, 则式$(1.1)$存在整体光滑解.若初值不恒为0, 且$1<p<p_{0}(3)$, 则式(1.1)不存在整体解.对于$n=2$情形, Glassey^{[3, 4]}验证了Strauss猜想, 但是临界情形$p=p_{0}(2)$没有解决.

对于临界情形$p=p_{0}(3)$和$p=p_{0}(2)$, Schaeffer^[5]已经证明了即使对于小初值, 式$(1.1)$不存在整体解.

至此, 对于$n=2$和$n=3$的情形, Strauss猜想已经得到完整验证.

对于高维情形$n\geq4$, 次临界情形$1<p<p_{0}(n)$由Sideris^[6]给出证明; 超临界情形$p>p_{0}(n)$被Georgive, Lindblad和Sogge^[7]证明.而临界情形$p=p_{0}(n)$ ($n\geq4$)被Yordanov和Zhang^[8]以及Zhou^[9]分别独立验证, 并且对于情形$n\geq2$, Zhou和Han^[10]给出了临界情形$p=p_{0}(n)$解的生命跨度的上界估计, 有关最新结果也可参见Rammaha, Takamura, Uesaka和Wakasa^[11].

若初值不具有紧支集, 则式$(1.1)$解的情况会发生变化.事实上, 对于$n=3$情形, Asakura^[12]设$F(u)$满足式$(1.3)$, 且$p>p_{0}(3)$, 初值$f(x)\in{C}^{3}({R}^{3}), g(x)\in{C}^{2}({R}^{3})$.如果初值满足

$ \begin{equation} \begin{aligned} D_{x}^{\alpha}f(x), D_{x}^{\beta}g(x)={O}((1+|x|)^{-1-k}), ~~~|x|\rightarrow\infty, ~~~|\alpha|\leq3, ~~~|\beta|\leq2, \end{aligned} \end{equation} $

(1.4)

其中$k>\frac{2}{p-1}$, 那么柯西问题$(1.1)$的整体解存在; 如果初值满足

$ \begin{equation}\begin{aligned}f(x)=0, g(x)\geq\frac{\epsilon}{(1+|x|)^{1+k}}, ~~~\epsilon>0, \end{aligned}\end{equation} $

(1.5)

其中$0<k<\frac{2}{p-1}$, 则方程组$(1.1)$的解在有限时间内一定会产生爆破.对于$n=2$情形, Tsutaya^[14]设$F(u)$满足式$(1.3)$, 且$p>p_{0}(2)$, 初值$f(x)\in{C}^{3}({R}^{2})$, $g(x)\in{C}^{2}({R}^{2})$满足式$(1.4)$, 且$k>\frac{2}{p-1}$, 那么柯西问题$(1.1)$存在整体解; 如果$F(u)$满足式$(1.2)$, 且$0<p<p_{0}(2)$初值满足式$(1.5)$, 且$0<k<\frac{2}{p-1}$, 则方程组$(1.1)$不存在整体解.

最近, Kong和Liu^[15]提出双曲Yamabe问题, 考虑$(1+n)$维Minkowski空间中的Yamabe方程解的整体存在与爆破. Minkowski时空中的双曲Yamabe问题为如下半线性波动方程柯西问题所刻画

$ \left\{ {\begin{array}{*{20}{l}} {\square u = {u^{\frac{{n + 3}}{{n - 1}}}},}\\ {t = 0:u = \varepsilon f(x) > 0,{u_t} = \varepsilon g(x).} \end{array}} \right. $

(1.6)

在文^[15]中, Kong和Liu证明:若初值$f(x)\in{C}^{3}({R}^{n})$, $g(x)\in{C}^{2}({R}^{n})$($n$=2或3), 且满足

$ \left\{ {\begin{array}{*{20}{l}} \Sigma_{|\alpha|\leq3}|D_{\alpha}^{x}f(x)|+\Sigma_{|\alpha|\leq2}|D_{\alpha}^{x}g(x)| \leq\frac{\epsilon}{(1+|x|)^{1+k}} ~~~~(k\geq\frac{n-1}{2}), \\ g(x)-|\nabla f(x)|>0, \end{array}} \right. $

则柯西问题$(1.6)$的光滑解整体存在; 若初值$f(x)\in{C}^{3}({R}^{n}), g(x)\in{C}^{2}({R}^{n})$ ($n$ =2或3), 且满足

$ \begin{equation}\begin{aligned}g(x)-|\nabla f(x)|\geq\frac{1}{(1+|x|)^{1+k}}, \end{aligned}\nonumber\end{equation} $

其中$0\leq k<\frac{n-1}{2}$, 则式$(1.6)$的解不会整体存在.

本文研究一类具有衰减初值的半线性波动方程的柯西问题$(1.1)$.不同于Asakura^[12]和Kong和Liu^[15], 假设初值是具有形如式$(2.1)$的对数型的衰减, 比代数衰减更缓慢一些.得到了半线性方程的柯西问题$(1.1)$的解在有限的时间内爆破, 并得到了生命区间的下界.

下面给出本文主要结果.

定理2.1  设$f(x)\in{C}^{3}({R}^{n})$, $g(x)\in{C}^{2}({{R}}^{n})$($n$ =2或3).若初值满足如下条件

$ \begin{equation}\label{7}f(x)>0, ~~~ g(x)-|\nabla f(x)|\geq\frac{C}{(1+\log(1+|x|))^{1+k}}, ~~~~~~C>0, \end{equation} $

(2.1)

且$0<k<\frac{2}{p-1}$, 则柯西问题$(1.1)$的正解必在有限时间内爆破, 即解存在区间为$[0, T^{*})$, 且生命跨度$T^{*}$满足

$ \begin{equation}\begin{aligned}T^{*}\geq C_{0}\varepsilon^{\frac{p-1}{k(p-1)-2}}, \end{aligned}\end{equation} $

(2.2)

这里$ C_{0}$是与$\varepsilon$无关的正常数.

定理$2.1$中解的正性可由引理$3.3$保证.由定理$2.1$, 可得Yamabe问题$(1.6)$的结论.

推论2.1  对于$n$=2和$n$=3情形的Yamabe问题$(1.6)$.设$f(x)\in{C}^{3}({R}^{n}), g(x)\in{C}^{2}({{R}}^{n})$且满足式$(2.1)$, $0<k<\frac{2}{p-1}$ ($n=2$, $0<k<\frac{1}{2}$; $n=3$, $0<k<1$), 则柯西问题$(1.6)$的解存在区间为$ [0, T^{*})$, 生命跨度$T^{*}$的下界由式$(2.2)$给出.

注2.1  Kong和Liu^[15]未明确给出Yamabe问题$(1.6)$解的生命跨度的估计.

注2.2 对于$F(u)$满足式$(1.3)$, 初值满足式$(1.5)$, 且$0<k<\frac{2}{p-1}$, Tsutaya^[13]指出柯西问题$(1.1)$的解必发生爆破, 并给出同样的下界估计式$(2.2)$.由此, 初值的对数型衰减虽比代数型衰减慢, 但不影响生命跨度的下界估计.

注2.3  在文献[13]中, Tsutaya设$F(u)=A|u|^{p}, A>0$.若初值满足$f(x)=0$, $g(x)\geq\frac{C}{(1+|x|)^{1+k}}$, 且$0<k<\frac{2}{p-1}$和$p>1+\sqrt{2}$, 则生命跨度$T^{*}$满足$T^{*}\leq c_{0}C^{\frac{p-1}{k(p-1)-2}}$, 其中$c_{0}$为与$C$无关的正常数.因此, 猜测柯西问题$(1.1)$的生命跨度$T^{*}$也满足$T^{*}\leq c_{1}\varepsilon^{\frac{p-1}{k(p-1)-2}}$, 其中$c_{1}$为与$\varepsilon$无关的正常数.

在这一节中, 将给出一些引理以及定理$2.1$的证明.尽管所需引理在Kong和Liu^[15]和Kubota^[16]中已经给出, 但是为了完整起见, 在此依然将其列出, 证明可参见Kong和Liu^[15]和Kubota^[16].

引理3.1  考虑下列柯西问题

$ \left\{ {\begin{array}{*{20}{l}}\square u=w(t, x), ~~~~~~~~~~~~~~~\quad\quad~~~~~~~(x, t)\in{{R}}^{n}\times(0, +\infty)\\ t=0:u=\varepsilon f(x), ~~u_t=\varepsilon g(x), ~\quad~~x\in{{R}}^{n} \end{array}} \right. $

其中$n$ =2或3.当$f(x)\in{C}^{3}({{R}}^{n})$, $g(x)\in{C}^{2}({{R}}^{n})$, $w(t, x)\in{C}^{2}({{R}}^{n})$, 有

$ \begin{equation}\begin{aligned}u(t, x)=u^{0}(t, x)+Lw(t, x), \end{aligned}\end{equation} $

(3.1)

其中

$ \begin{equation} u^{0}(t, x)=\left\{\begin{array}{ll}\frac{t}{4\pi}\int_{|\xi|=1}\varepsilon g(x+t\xi)dw_{\xi}+\frac{1}{4\pi}\frac{\partial}{\partial t}(t\int_{|\xi|=1}\varepsilon f(x+t\xi)dw_{\xi})~~~~~~~(n=3), \\ \frac{1}{2\pi}\int_{0}^{t}\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\int_{|\xi=1|}\varepsilon g(x+\rho\xi)dw_{\xi}d\rho\\ + \frac{1}{2\pi}\frac{\partial}{\partial_{t}}(\int_{0}^{t}\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\int_{|\xi|=1 }\varepsilon f(x+\rho\xi)dw_{\xi}d\rho)~~~~(n=2) \end{array}\right. \end{equation} $

(3.2)

是柯西问题

$ \left\{ {\begin{array}{*{20}{l}}\square u=0, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x, t)\in{{R}}^{n} \times (0, +\infty)\\ t=0:u=\varepsilon f(x), ~~u_t=\varepsilon g(x), ~~x\in{{R}}^{n} \end{array}} \right. $

的解;

$ \begin{equation}Lw(t, x)=\left\{\begin{array}{ll}\frac{1}{4\pi}\int_{0}^{t}(t-\tau)\int_{|\xi|=1}w(x+(t-\tau)\xi, \tau)dw_{\xi}d\tau~~~~~~~(n=3), \\ \frac{1}{2\pi}\int_{0}^{t}\int_{|x-y|\leq t-\tau}\frac{w(\tau, y)}{\sqrt{(t-\tau)^{2}-(x-y)^{2}}}dyd\tau~~\quad\quad\quad~~~~~(n=2), \end{array}\right. \end{equation} $

(3.3)

是柯西问题

$ \left\{ {\begin{array}{*{20}{l}}\square u=w(t, x), ~(x, t)\in {{R}}^{n} \times (0, +\infty), \\ t=0:u=0, ~~u_t=0 \end{array}} \right. $

的解.

引理3.2  若$b(\lambda)$是$(-\infty, \infty)$上连续函数, 则

$ \int_{|w|=1}b(y\cdot w)dS_{w}=w_{n-1}\int_{-1}^{1} b(|y|\eta)(1-\eta^{2})^{\frac{n-3}{2}}d\eta, $

其中$w_{k}=\frac{2\pi^{\frac{k}{2}}}{\Gamma(\frac{k}{2})}~~(k\geq1). $

引理3.3  若$f(x)\in{C}^{3}({R}^{n}), g(x)\in{C}^{2}({R}^{n})$满足式$(2.1)$, 则当$\varepsilon$充分小时, 柯西问题$(1.1)$在$[0, +\infty)$上有正的${C}^{2}$解.

证明参见Kong和Liu^[15], 此处从略.

定理2.1的证明  利用John^[1]和Kong和Liu^[15]中的迭代法来证明.由引理$3.3$知, 只要$u$存在, 则$u>0$.当$n=3$时, 由式$(3.2)$和$(2.1)$有

$ \begin{equation*}\begin{aligned}u^{0}(t, x)&={\frac{t}{4\pi}}\int_{|\xi|=1}\varepsilon g(x+t\xi)dw_{\xi}+\frac{1}{4\pi}\frac{\partial}{\partial t}(t\int_{|\xi|=1}\varepsilon f(x+t\xi)dw_{\xi})\\ &=\frac{\varepsilon t}{4\pi}\int_{|\xi|=1}[g(x+t\xi)+\nabla f(x+t\xi)\xi+\frac{f(x+t\xi)}{t}]dw_{\xi}\\ &\geq\frac{\varepsilon t}{4\pi}\int_{|\xi|=1}[g(x+t\xi)-|\nabla f(x+t\xi)|+\frac{f(x+t\xi)}{t}]dw_{\xi}\\ &\geq\frac{\varepsilon t}{4\pi}\int_{|\xi|=1}[g(x+t\xi)-|\nabla f(x+t\xi)|]dw_{\xi}\\ &\geq\frac{\varepsilon t}{4\pi}\int_{|\xi|=1}\frac{C}{(1+\log(1+|x+t\xi|))^{1+k}}dw_{\xi}, \end{aligned}\end{equation*} $

上式右端项是方程$\square u=0$带有初值$f=0$, $g=(1+\log(1+r))^{-1-k}$的解.令$r=|x|$, 则$|x+t\xi|=\sqrt{(x+t\xi)^{2}}=\sqrt{r^{2}+t^{2}+2tx\xi}$, 由引理$3.2$, 取$y=2tx$, 则

$ \begin{equation*}\begin{aligned}\varphi(t, x)&\triangleq \int_{|\xi|=1}\frac{C}{(1+\log(1+|x+t\xi|))^{1+k}}dw_{\xi}\\ &=w_{2}\int_{-1}^{1}\frac{1}{[1+\log(1+\sqrt{r^{2}+t^{2}+2tr\eta})]^{1+k}}(1-\eta^{2})^{\frac{n-3}{2}}d\eta\\ &=\frac{2\pi}{tr}\int_{|r-t|}^{r+t}\frac{\rho}{[1+\log(1+\rho)]^{1+k}}d\rho. \end{aligned}\end{equation*} $

故

$ \begin{equation*}\begin{aligned}u^{0}(t, x)\geq\frac{C\varepsilon}{2r}\int_{|r-t|}^{r+t}\frac{\rho}{[1+\log(1+\rho)]^{1+k}}d\rho. \end{aligned}\end{equation*} $

对于$(t, x)\in\Sigma\triangleq\{(t, x):|x|-t\geq R>0\}$, 有

$ \begin{equation*}\begin{aligned}u^{0}(t, x)\geq\frac{C\varepsilon}{2r}\int_{r}^{r+t}\frac{\rho}{[1+\log(1+\rho)]^{1+k}}d\rho &\geq\frac{C\varepsilon}{2r}\frac{1}{[1+\log(1+r+t)]^{1+k}}\int_{r}^{r+t}\rho d\rho\\ &\geq\frac{C\varepsilon t}{2}\frac{1}{[1+\log(1+r+t)]^{1+k}}. \end{aligned}\end{equation*} $

若函数$p(t, x)$是$\bf{C}^{m}(m\geq2)$光滑的, 记$\overline{p}(t, r)$为$p(t, x)$在半径为$r$的球面上的平均值, 即$\overline{p}(t, r)=\frac{1}{4\pi}\int_{|\xi|=1}p(t, r\xi)dw_{\xi}. $由式$(3.3)$得

$ \begin{equation*}\begin{aligned}\overline{Lw}(t, r)&=\frac{1}{(4\pi)^{2}}\int_{0}^{t}(t-\tau)d\tau \int_{|\xi|=1}dw_{\xi}\int_{|\eta|=1}w(r\xi+(t-\tau)\eta, \tau)dw_{\eta}\\ &=\int_{0}^{t}d\tau\int_{|r-t+\tau|}^{r+t-\tau}\frac{\lambda}{8\pi r}d\lambda\int_{|\zeta|=1}w(\lambda\zeta, \tau)dw_{\zeta}\\ &=\int_{0}^{t}d\tau\int_{|r-t+\tau|}^{r+t-\tau}\frac{\lambda}{2r}\overline{w}(\tau, \lambda)d\lambda. \end{aligned}\end{equation*} $

因此有$\overline{Lw}=P\overline{w}, $其中$P$算子定义为$P\overline{w}=\int_{0}^{t}d\tau\int_{|r-t+\tau|}^{r+t-\tau}\frac{\lambda}{2r}\overline{w}(\lambda, \tau)d\tau. $假设式$(1.1)$存在整体的$\bf{C}^{2}$解, 接下来通过推出矛盾来证明定理$2.1$.由式$(3.1)$有

$ \begin{equation}\begin{aligned}u(t, x)=u^{0}(t, x)+LF(u)(t, x). \end{aligned}\end{equation} $

(3.4)

令$\overline{u}$是$u(t, x)$在半径为$r$的球面上的平均值, 则有$\overline{u}(t, r)=\overline{u^{0}}(t, r)+\overline{LF}(t, r)\geq\overline{u^{0}}(t, r)+A\overline{Lu^{p}}(t, r), $又$p> p_{0}=1+\sqrt{2}$.故$u^{p}$是凸的, 所以$\overline{u^{p}}(t, r)\geq\overline{u}^{p}(t, r), $故

$ \begin{equation}\begin{aligned}\overline{u}(t, r)\geq\overline{u^{0}}(t, r)+AL\overline{u}^{p}(t, r). \end{aligned}\end{equation} $

(3.5)

又由$u>0$, 成立

$ \begin{equation*}\begin{aligned}\overline{u}(t, r)\geq\overline{u^{0}}(t, r) &\geq\frac{1}{4\pi}\int_{|\xi|=1}\frac{C\varepsilon t}{2[1+\log(1+r+t)]^{1+k}}dw_{\xi}\\ &=\frac{C\varepsilon t}{2[1+\log(1+r+t)]^{1+k}}, ~~~\forall(t, x)\in\Sigma. \end{aligned}\end{equation*} $

进一步设$\overline{u}$有下列估计

$ \begin{equation}\begin{aligned}\overline{u}(t, r)\geq\frac{ct^{a}}{[1+\log(1+r+t)]^{b}}, ~~~\forall(t, x)\in\Sigma, \end{aligned}\end{equation} $

(3.6)

其中$a, b, c$均为正常数.显然当$a=1, b=1+k, c=\frac{C\varepsilon}{2}$, 式$(3.6)$是满足的.由式$(3.5)$-$(3.6)$,

$ \begin{equation}\begin{aligned}\overline{u}(t, r)\geq AL\overline{u}^{p}(t, r) &\geq A\int_{0}^{t}d\tau\int_{|r-t+\tau|}^{r+t-\tau}\frac{\lambda}{2r}[\frac{c\tau^{a}}{(1+\log(1+\lambda+\tau))^{b}}]^{p}d\lambda \\ &\geq\frac{Ac^{p}}{2r}\frac{1}{(1+\log(1+r+t))^{bp}}\int_{0}^{t}\tau^{ap}d\tau\int_{r}^{r+t-\tau}\lambda d\lambda\\ &\geq\frac{Ac^{p}}{2(ap+2)^{2}}\frac{t^{ap+2}}{(1+\log(1+r+t))^{bp}}, ~~~\forall(t, x)\in\Sigma. \end{aligned}\end{equation} $

(3.7)

一直迭代下去, 并比较式$(3.6)$和$(3.7)$, 可定义如下序列$\{a_{j}\}, \{b_{j}\}, \{c_{j}\}$,

$ \left\{ {\begin{array}{*{20}{l}}a_{j}=pa_{j-1}+2, ~~~~~a_{0}=1, \\ b_{j}=pb_{j-1}, ~~~~~b_{0}=1+k, \\ c_{j}=\frac{Ac_{j-1}^{p}}{2(pa_{j-1}+2)^{2}}, ~~~~~c_{0}=\frac{C\varepsilon}{2}. \end{array}} \right. $

从而有

$ \begin{equation}\begin{aligned}a_{j}=\frac{p+1}{p-1}p^{j}-\frac{2}{p-1}, ~~b_{j}=(1+k)p^{j} \end{aligned}\end{equation} $

(3.8)

和

$ \begin{equation*}\begin{aligned}c_{j}>\frac{Ac_{j-1}^{p}}{2(\frac{p+1}{p-1})^{2}p^{2j}}. \end{aligned}\end{equation*} $

因此可以得到

$ \begin{equation}\begin{aligned}c_{j}>(\frac{A}{2}(\frac{P-1}{P+1})^{2})^{\frac{p^{j}-1}{p-1}}\frac{c_{0}^{p^{j}}}{p^{2s_{j}}}, ~~~s_{j}=p^{j}\sum\limits_{k=1}^{j}\frac{k}{p^{k}}. \end{aligned}\end{equation} $

(3.9)

联立式$(3.6)$, $(3.8)$和$(3.9)$得到$\overline{u}(t, r)\geq(\frac{A}{2}(\frac{p-1}{p+1})^{2})^{\frac{1}{1-p}}t^{\frac{2}{1-p}}e^{p^{j}L(t, r)}, ~\forall(t, x)\in\Sigma, $其中

$ \begin{equation*}\begin{aligned} L(t, r)=&\frac{1}{p-1}\log\frac{A}{2}(\frac{p-1}{p+1})^{2}+\log c_{0}-2\sum\limits_{k=1}^{\infty}\frac{k}{p^{k}}\log p+\frac{p+1}{p-1}\log t\\ &-(1+k)\log(1+\log(1+r+t)). \end{aligned}\end{equation*} $

由于$0 <k\leq\frac{2}{p-1}$, 故存在$(t_{0}, r_{0})\in\Sigma$, 使得$L(t_{0}, r_{0})>0$.当$j\rightarrow\infty$时, 有$\overline{u}(t_{0}, r_{0})\rightarrow\infty $与假设矛盾.由$L(t, r)>0$知

$ \begin{equation*}\begin{aligned}\{\frac{A}{2}(\frac{p-1}{p+1})^{2}\}^{\frac{1}{p-1}}c_{0}\frac{1}{p^{2\sum\limits_{k=1}^{\infty}\frac{k}{p^{k}}}}\frac{t^{\frac{p+1}{p-1}}}{(1+\log(1+r+t))^{1+k}}>1. \end{aligned}\end{equation*} $

固定$p$, 从而有

$ \begin{equation}\begin{aligned}C_{0}\varepsilon\frac{t^{\frac{p+1}{p-1}}}{(1+\log(1+r+t))^{1+k}}>1, \end{aligned}\end{equation} $

(3.10)

其中$C_{0}=\frac{1}{2}\{\frac{A}{2}(\frac{P-1}{P+1})^{2}\}^{\frac{1}{P-1}}\frac{1}{p^{2\sum\limits_{k=1}^{\infty}\frac{k}{p^{k}}}}. $从而由式$(3.10)$可得生命区间的下界估计

$ \begin{equation}\begin{aligned}T^{*}\geq C_{0}\varepsilon^{\frac{p-1}{k(p-1)-2}}. \end{aligned}\end{equation} $

(3.11)

当$n=2$时, 由式$(2.1)$和$(3.2)$有

$ \begin{equation*} \begin{aligned} u^{0}(t, x)=&\frac{\varepsilon}{2\pi}\int_{0}^{t}\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\int_{|\xi|=1}g(x+\rho\xi)dw_{\xi} d\rho\\ &+\frac{\varepsilon}{2\pi t}\int_{0}^{t}\{\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\int_{|\xi|=1}f(x+\rho\xi)dw_{\xi}+\frac{\rho^{2}}{\sqrt{t^{2}-\rho^{2}}}\int_{|\xi|=1}\xi\nabla f(x+\rho\xi)dw_{\xi}\}d\rho\\ =&\frac{\varepsilon}{2\pi}\int_{0}^{t}\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\int_{|\xi|=1}(g(x+\rho\xi)+\frac{\rho}{t}\xi\nabla f(x+\rho\xi))dw_{\xi}d\rho\\ \geq&\frac{\varepsilon}{2\pi}\int_{0}^{t}\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\int_{|\xi|=1}\frac{C}{(1+\log(1+|x+\rho\xi|))^{1+k}}dw_{_{\xi}}d\rho\\ \end{aligned} \end{equation*} $

$ \begin{equation*} \begin{aligned} \geq&\frac{4C\varepsilon}{2\pi}\int_{0}^{t}\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\int_{|\rho-r|}^{\rho+r}\frac{1}{(1+\log(1+\lambda))^{1+k}}\frac{\lambda}{\sqrt{\lambda^{2}-(\rho-r)^{2}}\sqrt{(\rho+r)^{2}-\lambda^{2}}}d\lambda d\rho\\ \geq&\frac{4C\varepsilon}{2\pi}\int_{0}^{t}\frac{\rho}{\sqrt{t^{2}-\rho^{2}}}\frac{1}{(1+\log(1+\rho+r))^{1+k}}\frac{1}{\sqrt{(\rho+r)^{2}-(\rho-r)^{2}}\sqrt{(\rho+r)^{2}-(\rho-r)^{2}}}\\ &\int_{|\rho-r|}^{\rho+r}\lambda d\lambda d\rho\\ \geq&\frac{C\varepsilon}{2\pi}\frac{t}{(1+\log(1+r+t))^{1+k}}, ~~~\forall t>0. \end{aligned} \end{equation*} $

类似地, 设$u$有下列形式的估计

$ \begin{equation}\begin{aligned}u(t, x)\geq\frac{ct^{a}}{(1+\log(1+r+t))^{b}}, \end{aligned}\end{equation} $

(3.12)

其中$a, b, c$均为正常数.显然当$a=1, b=1+k, c=\frac{C\varepsilon}{2\pi}$时, 式$(3.12)$是满足的.由式$(3.4)$和$(3.12)$, 可知

$ \begin{equation}\begin{aligned}u(t, x)\geq&\frac{1}{2\pi}\int_{0}^{t}\int_{|x-y|\leq t-\tau}\frac{Au^{p}}{\sqrt{(t-\tau)^{2}-|x-y|^{2}}}dyd\tau\\ =&\frac{1}{2\pi}\int_{0}^{t}\int_{0}^{t-\tau}\frac{\rho}{\sqrt{(t-\tau)^{2}-\rho^{2}}}\int_{|\xi|=1}Au^{p}(t, x+\rho\xi)dw_{\xi}d\rho d\tau\\ \geq&\frac{Ac^{p}}{2\pi}\int_{0}^{t}\int_{0}^{t-\tau}\frac{\rho}{\sqrt{(t-\tau)^{2}-\rho^{2}}}\int_{|\xi|=1}\frac{\tau^{ap}}{(1+\log(1+\tau+|x+\rho\xi|))^{bp}}dw_{\xi}d\rho d\tau\\ \geq&\frac{Ac^{p}}{(1+\log(1+r+t))^{bp}}\int_{0}^{t}\tau^{ap}\int_{0}^{t-\tau}\frac{\rho}{\sqrt{(t-\tau)^{2}-\rho^{2}}}d\rho d\tau\\ =&\frac{Ac^{p}}{(1+\log(1+r+t))^{bp}}\int_{0}^{t}\tau^{ap}(t-\tau)d\tau\\ \geq&\frac{Ac^{p}}{(ap+2)^{2}}\frac{t^{ap+2}}{(1+\log(1+r+t))^{bp}}. \end{aligned}\end{equation} $

(3.13)

一直迭代下去, 并比较式$(3.12)$和$(3.13)$, 可定义如下序列$\{a_{j}\}, \{b_{j}\}, \{c_{j}\}$,

$ \left\{ {\begin{array}{*{20}{l}}a_{j}=pa_{j-1}+2, ~~~~~a_{0}=1, \\ b_{j}=pb_{j-1}, ~~~~~b_{0}=1+k, \\ c_{j}=\frac{Ac_{j-1}^{p}}{(pa_{j-1}+2)^{2}}, ~~~~~c_{0}=\frac{C\varepsilon}{2\pi}. \end{array}} \right. $

从而可求得

$ \begin{equation}\begin{aligned}a_{j}=\frac{p+1}{p-1}p^{j}-\frac{2}{p-1}, ~~b_{j}=(1+k)p^{j} \end{aligned}\end{equation} $

(3.14)

和

$ \begin{equation*}\begin{aligned}c_{j}>\frac{Ac_{j-1}^{p}}{(\frac{p+1}{p-1})^{2}p^{2j}}. \end{aligned}\end{equation*} $

因此可以得到

$ \begin{equation}\begin{aligned}c_{j}>(A(\frac{P-1}{P+1})^{2})^{\frac{p^{j}-1}{p-1}}\frac{c_{0}^{p^{j}}}{p^{2s_{j}}}, ~~~s_{j}=p^{j}\sum\limits_{k=1}^{j}\frac{k}{p^{k}}. \end{aligned}\end{equation} $

(3.15)

联立式$(3.12)$, $(3.14)$和$(3.15)$, 可得

$ \begin{equation*}\begin{aligned}u(t, r)\geq(A(\frac{p-1}{p+1})^{2})^{\frac{1}{1-p}}t^{\frac{2}{1-p}}e^{p^{j}L(t, r)}, ~~~\forall(t, x)\in\Sigma, \end{aligned}\end{equation*} $

其中

$ \begin{equation*}\begin{aligned}L(t, r)&=\frac{1}{p-1}\log A(\frac{p-1}{p+1})^{2}+\log c_{0}-2\sum\limits_{k=1}^{\infty}\frac{k}{p^{k}}\log p+\frac{p+1}{p-1}\log t\\&-(1+k)\log(1+\log(1+r+t)). \end{aligned}\end{equation*} $

由于$0<k\leq\frac{2}{p-1}$, 故存在$(t_{0}, r_{0})\in\Sigma$, 使得$L(t_{0}, r_{0})>0$.于是当$j\rightarrow\infty$时, 有$u(t_{0}, x_{0})\rightarrow\infty. $与假设矛盾.由$L(t, r)>0$有

$ \begin{equation*}\begin{aligned}\{A(\frac{p-1}{p+1})^{2}\}^{\frac{1}{p-1}}c_{0}\frac{1}{p^{2\sum\limits_{k=1}^{\infty}\frac{k}{p^{k}}}}\frac{t^{\frac{p+1}{p-1}}}{(1+\log(1+r+t))^{1+k}}>1. \end{aligned}\end{equation*} $

固定$p$, 从而有

$ \begin{equation}\begin{aligned}C_{0}\varepsilon\frac{t^{\frac{p+1}{p-1}}}{(1+\log(1+r+t))^{1+k}}>1, \end{aligned}\end{equation} $

(3.16)

其中

$ \begin{equation*}\begin{aligned}C_{0}=\frac{1}{2}\{A(\frac{P-1}{P+1})^{2}\}^{\frac{1}{P-1}}\frac{1}{p^{2\sum\limits_{k=1}^{\infty}\frac{k}{p^{k}}}}. \end{aligned}\end{equation*} $

从而由式$(3.16)$可得$n=2$情形下生命区间的下界估计

$ \begin{equation}\begin{aligned}T^{*}\geq C_{0}\varepsilon^{\frac{p-1}{k(p-1)-2}}. \end{aligned}\end{equation} $

(3.17)

因此柯西问题$(1.1)$的解不会整体存在, 解的生命跨度由式$(3.11)$和$(3.17)$给出.证毕.