考虑如下一维半线性薛定谔方程初边值问题

$ \begin{equation}\label{eq:a1} \left\{\begin{array}{ll} {\rm i}u _{t}+u_{xx} =\lambda |u|^{p}, \quad &x>0, t>0, \\ u(0, x)= \varepsilon f(x), \quad\quad & x>0, \\ u(t, 0)=0, & t>0, \end{array}\right. \end{equation} $

(1.1)

其中$\lambda=\lambda_{1}+{\rm i}\lambda_{2}\in{\Bbb C}\setminus\{0\}$, 并且$\lambda_{1}>0.$假设初值$f(x)=f_{1}(x)+{\rm i}f_{2}(x)$满足

$ \begin{equation}\label{eq:a2} f_{2}(x)<0, ~~~~-\int_{0<x\leq1}xf_{2}(x){\rm d}x\geq C, \end{equation} $

(1.2)

这里和后面的介绍中, $C$指的是一个正常数, 但不是固定不变的.

方程(1.1)在Sobolev空间$H^{s}(s\geq0)$中局部解的存在性, 可参见文献[1, 16]及其中相关的参考文献.如果想要研究解的大时间行为:解是整体存在还是在有限时间内破裂, 那么非线性项的指数$p$起着重要的作用.首先介绍两个重要的指数:一个是$p_{c}(n)=1+\frac{2}{n}$, 它是由Germain等^[4]引入的; 另一个是$p_{0}(n)$, 它是方程$(n-1)p^{2}-(n+1)p-2=0$的正根, 是由Strauss^[13]引入的, 文中作者证明了当$p>p_{0}(n+1)$时, ${\Bbb R}^n$中的小初值Cauchy问题(1.1)存在整体解.指数$p_c(n)=1+\frac{2}{n}$被称作Fujita指数, 是小初值半线性热传导方程和耗散半线性波动方程Cauchy问题的临界指标, 相关结果见文献[2, 10, 15].更多关于小初值半线性薛定谔方程解在有限时间内破裂的结果, 见文献[11-12].

Ikeda和Wakasugi^[6]研究了方程(1.1)在${\Bbb R}^n$中的Cauchy问题, 当$p$满足$1<p\leq p_{c}(n)$时, 选择合适的初值, 得到了解在有限时间内破裂的结果, 证明过程运用了反证法, 该方法由Zhang^[17-18]引入.此外, 运用Kuiper^[9]中的试探函数方法, Ikeda^[5]证明了$1<p<1+\frac 2n$时的生命跨度上界满足如下形式

$ \begin{equation}\label{2aa} T_{\varepsilon}\leq C\varepsilon^{\frac{1}{\alpha}}, \end{equation} $

(1.3)

其中$\alpha=\frac k2-\frac{1}{p-1}$且$n<k<\frac {2}{p-1}$.最近, Ikeda和Inui^[7]证明了当$1<p<1+\frac4n$时小初值$L^2$和$H^1$解将在有限时间内破裂，并进一步给出了解的生命跨度上界估计.

本论文研究一维空间中半无界区域上的初边值问题(1.1), 当$1<p\leq2$时将证明解在有限时间内破裂; 进一步, 当$1<p<2$时, 将建立解的生命跨度上界估计.证明的主要方法是反正法和试探函数法.与文献[5-6, 9, 14]相比, 我们借鉴了文献[3]中的想法, 引入黎曼-刘维尔分数阶导数来构造新的试探函数.

注1.1 我们猜测初边值问题(1.1)存在临界指标$p=2$, 该指标相应于二维Cauchy问题的临界指标.初边值问题(1.1)在$p>2$时的整体存在性是一个非常有趣的问题.

注1.2 对于临界情性$p=2$, 本文中的方法无法建立其生命跨度上界估计, 这也是我们接下来要研究的问题.

本文的主要结论为如下两个定理：

定理1.1 当$1<p\leq 2$时, 假设初始值满足条件(1.2), 则初边值问题(1.1)的解将在有限时间内破裂.

定理1.2 当$1<p\leq 2$时, 假设初始值满足条件(1.2), 则存在$\varepsilon_{0}>0$和$C=C(p, \lambda)$, 使得方初边值问题(1.1)解的上生命跨度上界估计满足

$ \begin{equation} T(\varepsilon)\leq C\varepsilon^{\frac{p-1}{p-2}}, \end{equation} $

(1.4)

其中$\varepsilon\in(0, \varepsilon_{0})$.

正文内容安排如下:在第二部分介绍黎曼-刘维尔分数阶导数并构造试探函数.在第三部分证明定理1.1.在第四部分证明定理1.2.

在这部分, 我们将引入黎曼-刘维尔分数阶导数来构造一个试探函数.下文中$AC[0, T]$指的是定义在$[0, T]$上绝对连续的函数, 其中$0<T<\infty$, 则对任意的$g\in AC[0, T]$, 其$\alpha$阶的右手黎曼-刘维尔分数阶导数定义如下

$ \begin{equation}\label{eq:a3} D_{t|T}^{\alpha}g(t)=-\frac{1}{\Gamma(1-\alpha)}\frac{\rm d}{{\rm d}t}\int_t^T(s-t)^{-\alpha}g(s){\rm d}s, ~~~t\in [0, T], \end{equation} $

(2.1)

其中$\alpha\in (0, 1)$且$\Gamma$是欧拉函数.关于黎曼-刘维尔分数阶导数的详细定义, 见文献[8].

引理2.1^[8] 令$g\in AC^{N+1}[0, T]:=\{g: [0, T]\rightarrow {\Bbb R}$, $ \partial_t^Ng\in AC[0, T]\}$, 则对任意指数$N\geq 0$, 有

$ \begin{equation}\label{eq:a4} (-1)^N\partial_t^N\big(D_{t|T}^{\alpha}g(t)\big)=D_{t|T}^{N+\alpha}g(t), \end{equation} $

(2.2)

其中$\partial_t^N$指的是对$t$的$N$阶偏导.

引理2.2 令$h(t)=\big(1-\frac tT\big)_{+}^{\sigma}$其中$t\geq 0$, $T>0$且$\sigma\gg 1$, 则有

$ \begin{equation}\label{eq:a5} D_{t|T}^{\alpha}h(t)=\frac{(1-\alpha+\sigma)\Gamma(\sigma+1)}{\Gamma(2-\alpha+\sigma)}T^{-\sigma}(T-t)_{+}^{\sigma-\alpha} =CT^{-\sigma}(T-t)_{+}^{\sigma-\alpha}, \end{equation} $

(2.3)

$ \begin{eqnarray}\label{eq:a6} D_{t|T}^{\alpha+1}h(t)&=&\frac{(1-\alpha+\sigma)(\sigma-\alpha)\Gamma(\sigma+1)}{\Gamma(2-\alpha+\sigma)} T^{-\sigma}(T-t)_{+}^{\sigma-\alpha-1}\\ &=&CT^{-\sigma}(T-t)_{+}^{\sigma-\alpha-1}, \end{eqnarray} $

(2.4)

因此进一步可以得到

$ \begin{equation}\label{eq:a7} \big(D_{t|T}^{\alpha}h\big)(T)=0, ~~\big(D_{t|T}^{\alpha+1}h\big)(T)=0, \end{equation} $

(2.5)

$ \begin{equation}\label{eq:a8} \big(D_{t|T}^{\alpha}h\big)(0)=CT^{-\alpha}, ~~\big(D_{t|T}^{\alpha+1}h\big)(0)=CT^{-\alpha-1}. \end{equation} $

(2.6)

引理$2.2$的证明可以通过变量替换得到, 具体可见文献[3].

设$\Phi(x)$是一个光滑、单调递减的函数, 且满足以下条件

$ \Phi \left( x \right) = \left\{ {\begin{array}{*{20}{l}} {1,}&{{\text{0}} \leqslant {\text{x}} \leqslant {\text{1}},} \\ {0,}&{{\text{x}} \geqslant {\text{2}}.} \end{array}} \right. $

记$\phi_{1}(x)=\Phi(\frac{x}{B})$, $\phi_{2}(t)=(1-\frac{t}{T})_{+}$, 我们构造如下形式的试探函数

$ \begin{eqnarray}\label{8} \phi(t, x)=x\phi_{1}^{l}(x)D^{\alpha}_{t\mid T}(\phi_{2}^{k}(t)) =Cx\phi_{1}^{l}(x)T^{-\alpha} (1-\frac{t}{T})_{+}^{k-\alpha}, ~~~~~l, k\gg1. \end{eqnarray} $

(2.7)

则由引理$2.2$可以得到

$ \begin{eqnarray}\label{9} \phi_{t}(t, x)=-x\phi_{1}^{l}(x)D^{\alpha+1}_{t\mid T}(\phi_{2}^{k}(t)) =-Cx\phi_{1}^{l}(x)T^{-\alpha-1} (1-\frac{t}{T})_{+}^{k-\alpha-1}, \end{eqnarray} $

(2.8)

$ \phi(T, x)=0, \phi(t, 0)=0, \phi_{t}(T, x)=0, $

$ \phi(0, x)=Cx\phi_{1}^{l}(x)T^{-\alpha}, \phi_{t}(0, x)=-Cx\phi_{1}^{l}(x)T^{-\alpha-1}. $

方程(1.1)两边同时乘以$\phi(t, x)$, 并在$[0, T]\times[0, \infty)$上积分, 得

$ \begin{eqnarray}\label{10} &&\int_{0}^{T}\int_{0}^{+\infty}\lambda |u|^{p}\phi(t, x){\rm d}x{\rm d}t\\ &=&\int_{0}^{T}\int_{0}^{+\infty}({\rm i}u_{t}+u_{xx})\phi(t, x){\rm d}x{\rm d}t\\ &=&\int_{0}^{T}\int_{0}^{+\infty}\Big(({\rm i}u\phi )_{t}-u({\rm i}\phi_{t})\Big){\rm d}x{\rm d}t+\int_{0}^{T}\int_{0}^{+\infty}\Big((u_{x}\phi)_{x}-u_{x}\phi_{x}\Big){\rm d}x{\rm d}t\\ & =&-\int_{0}^{+\infty}{\rm i}u(0, x)\phi(0, x){\rm d}x-\int_{0}^{T}\int_{0}^{+\infty}u({\rm i}\phi_{t}){\rm d}x{\rm d}t\\ &&+\int_{0}^{T}\int_{0}^{+\infty} u\phi_{xx}{\rm d}x{\rm d}t. \end{eqnarray} $

(3.1)

式(3.1)两边取实部, 得到

$ \begin{eqnarray}\label{11} &&\lambda_{1}\int_{0}^{T}\int_{0}^{+\infty} |u|^{p}\phi(t, x){\rm d}x{\rm d}t- \varepsilon\int_{0}^{+\infty}f_{2}(x)\phi(0, x){\rm d}x\\ &=&\int_{0}^{T}\int_{0}^{+\infty} {\rm Re} u(-{\rm i}\phi_{t}+\phi_{xx}){\rm d}x{\rm d}t\\ &\leq& \int_{0}^{T}\int_{0}^{+\infty}|u|(|\phi_{t}|+|\phi_{xx}|){\rm d}x{\rm d}t. \end{eqnarray} $

(3.2)

令

$ \begin{eqnarray}\label{12} I&=&\lambda_{1}\int_{0}^{T}\int_{0}^{+\infty} |u|^{p}\phi(t, x){\rm d}x{\rm d}t\\ &=&C\lambda_{1}\int_{0}^{T}\int_{0}^{+\infty} |u|^{p}x\phi_{1}^{l}(x)T^{-\alpha} (1-\frac{t}{T})_{+}^{k-\alpha}{\rm d}x{\rm d}t, \\ J&=&-\varepsilon\int_{0}^{+\infty}f_{2}(x)\phi(0, x){\rm d}x, \\ K_{1}&=&\int_{0}^{T}\int_{0}^{+\infty}|u||\phi_{t}|{\rm d}x{\rm d}t, \\ K_{2}&=&\int_{0}^{T}\int_{0}^{+\infty}|u||\phi_{xx}|{\rm d}x{\rm d}t, \end{eqnarray} $

(3.3)

则(3.2)式可以简写为

$ \begin{equation}\label{13} I+J\leq K_{1}+K_{2}. \end{equation} $

(3.4)

由(1.2)和(2.8)式易得

$ \begin{eqnarray}\label{14} J&=&-\varepsilon\int_{0}^{+\infty}f_{2}(x)\phi(0, x){\rm d}x\\ &=&-C\varepsilon T^{-\alpha}\int_{0}^{+\infty}xf_{2}(x)\phi^{l}_{1}(x){\rm d}x\\ &\geq&-C\varepsilon T^{-\alpha}\int_{x\leq1}xf_{2}(x){\rm d}x\\ &\geq& C\varepsilon T^{-\alpha}. \end{eqnarray} $

(3.5)

结合(3.4)和(3.5)式可得

$ \begin{equation}\label{15} I\leq K_{1}+K_{2}. \end{equation} $

(3.6)

接下来我们将对$K_{1}$和$K_{2}$分别进行估计.对于$K_{1}$, 注意到$l, k\gg1$, 再结合(2.8)式、Hölder不等式和Young不等式, 可得

$ \begin{eqnarray}\label{16} K_{1}&=&\int_{0}^{T}\int_{0}^{+\infty}|u||\phi_{t}|{\rm d}x{\rm d}t\\ &=&C\int_{0}^{T}\int_{0}^{+\infty}x|u|\phi_{1}^{l}T^{-\alpha-1} (1-\frac{t}{T})_{+}^{k-\alpha-1}{\rm d}x{\rm d}t\\ &=&C\int_{0}^{T}\int_{0}^{+\infty}|u|x^{\frac{1}{p}}\phi_{1}^{\frac{l}{p}}T^{-\frac{\alpha}{p}} (1-\frac{t}{T})_{+}^{\frac{k-\alpha}{p}}x^{1-\frac{1}{p}}\phi_{1}^{l-\frac{l}{p}}\\ &&\times T^{-\alpha-1+\frac{\alpha}{p}} (1-\frac{t}{T})_{+}^{k-\alpha-1-\frac{k-\alpha}{p}}{\rm d}x{\rm d}t\\ & \leq & CI^{\frac{1}{p}}\Big(\int_{0}^{T}\int_{0}^{+\infty}x\phi_{1}^{l}T^{-\frac{(\alpha+1)p}{p-1}+\frac{\alpha}{p-1}} (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t\Big)^{\frac{p-1}{p}}\\ & \leq &\frac{1}{3}I+C\int_{0}^{T}\int_{0}^{+\infty}x\phi_{1}^{l}T^{-\frac{(\alpha+1)p}{p-1}+\frac{\alpha}{p-1}} (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t \\ &\triangleq &\frac{1}{3}I+K_{11}. \end{eqnarray} $

(3.7)

当$1<p<2$时, 引入如下变量替换: $s=\frac{t}{T}, y=\frac{x}{T^{\frac{1}{2}}}, B=T^{\frac{1}{2}}$, 则对$K_{11}$有如下估计

$ \begin{eqnarray}\label{17} K_{11}&=&C\int_{0}^{T}\int_{0}^{+\infty}x\phi_{1}^{l}T^{-\frac{(\alpha+1)p}{p-1}+\frac{\alpha}{p-1}} (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t\\ &=&CT^{-\alpha-\frac{p}{p-1}}\int_{0}^{T}\int_{0}^{+\infty}x\phi_{1}^{l} (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t\\ &\leq& CT^{-\alpha-\frac{p}{p-1}}\int_{0}^{1}\int_{0}^{2}T^{\frac{1}{2}}y (1-s)_{+}^{k-\alpha-\frac{p}{p-1}}T^{\frac{1}{2}+1}{\rm d}y{\rm d}s\\ &\leq &CT^{-\alpha-\frac{p}{p-1}+2}, \end{eqnarray} $

(3.8)

结合(3.7)式, 得

$ \begin{equation}\label{18} K_{1}\leq \frac{1}{3}I+CT^{-\alpha-\frac{p}{p-1}+2}. \end{equation} $

(3.9)

根据试探函数$\phi(t, x)$的构造易得

$ \begin{eqnarray}\label{19} \phi_{x}(t, x)&=&CT^{-\alpha}(1-\frac{t}{T})_{+}^{k-\alpha}\Big( \phi_{1}^{l}+lx\phi_{1}^{l-1}\phi_{1x}\Big), \\ \phi_{xx}(t, x)&=&CT^{-\alpha}(1-\frac{t}{T})_{+}^{k-\alpha}\Big( 2l\phi_{1}^{l-1}\phi_{1x} +l(l-1)x\phi_{1}^{l-2}(\phi_{1x})^{2} +lx\phi_{1}^{l-1}\phi_{1xx}\Big), \end{eqnarray} $

(3.10)

利用和估计$K_1$相同的办法, 我们可以得到

$ \begin{eqnarray}\label{20} K_{2}&=&\int_{0}^{T}\int_{0}^{+\infty}|u||\phi_{xx}|{\rm d}x{\rm d}t\\ &\leq &CI^{\frac{1}{p}}\Big(\int_{0}^{T}\int_{0}^{+\infty}T^{-\alpha}(1-\frac{t}{T})_{+}^{k-\alpha} x^{-\frac{1}{p-1}}\phi_{1}^{l-\frac{2p}{p-1}}\\ &&\times(\phi_{1x}+x(\phi_{1x})^{2}+x\phi_{1xx})^{\frac{p}{p-1}}{\rm d}x{\rm d}t\Big)^{\frac{p-1}{p}}\\ &\leq &\frac{1}{3}I+C\int_{0}^{T}\int_{0}^{+\infty}T^{-\alpha}(1-\frac{t}{T})_{+}^{k-\alpha} x^{-\frac{1}{p-1}}\phi_{1}^{l-\frac{2p}{p-1}}\\ &&\times(\phi_{1x}+x(\phi_{1x})^{2}+x\phi_{1xx})^{\frac{p}{p-1}}{\rm d}x{\rm d}t\\ &\triangleq&\frac{1}{3}I+K_{21}. \end{eqnarray} $

(3.11)

再次利用(3.8)式中相同的变量替换, 得

$ \begin{eqnarray}\label{21} K_{21}&=&C\int_{0}^{T}\int_{0}^{+\infty}T^{-\alpha}(1-\frac{t}{T})_{+}^{k-\alpha} x^{-\frac{1}{p-1}}\phi_{1}^{l-\frac{2p}{p-1}}\\ &&\times(\phi_{1x}+x(\phi_{1x})^{2}+x\phi_{1xx})^{\frac{p}{p-1}}{\rm d}x{\rm d}t\\ &\leq &CT^{-\alpha}\int_{0}^{1}\int_{1}^{2}(1-s)_{+}^{k-\alpha} (T^{\frac{1}{2}}y)^{-\frac{1}{p-1}}\\ &&\times(T^{-\frac{p}{2(p-1)}}+T^{-\frac{p}{2(p-1)}}y^{\frac{p}{p-1}}) T^{\frac{1}{2}+1}{\rm d}y{\rm d}s\\ &\leq &CT^{-\alpha-\frac{p+1}{2(p-1)}+\frac{3}{2}}, \end{eqnarray} $

(3.12)

联立(3.11)式, 得到

$ \begin{equation}\label{22} K_{2}\leq \frac{1}{3}I+CT^{-\alpha-\frac{p+1}{2(p-1)}+\frac{3}{2}}. \end{equation} $

(3.13)

因此, 结合(3.6), (3.9)和(3.13)式, 得到

$ \begin{eqnarray}\label{23} CT^{-\alpha}\int_{0}^{T}\int_{0}^{+\infty} |u|^{p}x\phi_{1}^{l}(1-\frac{t}{T})_{+}^{k-\alpha}{\rm d}x{\rm d}t \leq C(T^{-\alpha-\frac{p}{p-1}+2}+T^{-\alpha-\frac{p+1}{2(p-1)}+\frac{3}{2}}). \end{eqnarray} $

(3.14)

令$T\rightarrow +\infty$, 则当$1<p<2$时有

$ \begin{equation}\label{24} \int_{0}^{+\infty}\int_{0}^{+\infty}|u|^{p}x{\rm d}x{\rm d}t=0, \end{equation} $

(3.15)

由此可得对于任意的$t, x>0$, 有$u(t, x)=0$成立, 这样我们就得到了一个矛盾.于是, 解将在有限时间内破裂.

当$p=2$时, 引入如下变量替换

$ B=M^{-\frac{1}{2}}T^{\frac{1}{2}}, \quad y=M^{\frac{1}{2}}T^{-\frac{1}{2}}x=B^{-1}x, \quad s=\frac{t}{T}, $

其中$1\ll M<T$且当$T\rightarrow \infty$时就不会同时有$M\rightarrow \infty$.则可以估计$K_{11}$如下

$ \begin{eqnarray}\label{25} K_{11}&=&CT^{-\alpha-\frac{p}{p-1}}\int_{0}^{T}\int_{0}^{+\infty}x\phi_{1}^{l} (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t\\ &\leq &CT^{-\alpha-\frac{p}{p-1}}\int_{0}^{1}\int_{0}^{2}M^{-\frac{1}{2}}T^{\frac{1}{2}}y (1-s)_{+}^{k-\alpha-\frac{p}{p-1}}M^{-\frac{1}{2}}T^{\frac{1}{2}+1}{\rm d}y{\rm d}s\\ &\leq &CT^{-\alpha-\frac{p}{p-1}+2}M^{-1}\int_{0}^{1}\int_{0}^{2}y(1-s)_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}y{\rm d}s\\ &\leq& CT^{-\alpha}M^{-1}, \end{eqnarray} $

(3.16)

估计$K_{21}$如下

$ \begin{eqnarray}\label{26} K_{21}&=&CT^{-\alpha}\int_{0}^{T}\int_{0}^{+\infty}(1-\frac{t}{T})_{+}^{k-\alpha} x^{-\frac{1}{p-1}}\phi_{1}^{l-\frac{2p}{p-1}}\\ &&\times(\phi_{1x}+x(\phi_{1x})^{2}+x\phi_{1xx})^{\frac{p}{p-1}}{\rm d}x{\rm d}t\\ &\leq &CT^{-\alpha}\int_{0}^{1}\int_{1}^{2}(1-s)_{+}^{k-\alpha} M^{\frac{1}{2(p-1)}}T^{-\frac{1}{2(p-1)}}y^{-\frac{1}{p-1}}\\ &&\times M^{\frac{p}{2(p-1)}}T^{-\frac{p}{2(p-1)}}(1+y^{\frac{p}{p-1}})M^{-\frac{1}{2}} T^{\frac{1}{2}+1}{\rm d}y{\rm d}s\\ &\leq &CT^{-\alpha-\frac{p+1}{2(p-1)}+\frac{3}{2}}M^{\frac{p+1}{2(p-1)}-\frac{1}{2}}\\ &&\times\int_{0}^{1}\int_{0}^{2}(1-s)_{+}^{k-\alpha}y^{-\frac{1}{p-1}}(1+y^{\frac{p}{p-1}}){\rm d}y{\rm d}s\\ &\leq &CT^{-\alpha}M, \end{eqnarray} $

(3.17)

上述两式结合(3.6)式可得

$ CT^{-\alpha}\int_{0}^{T}\int_{0}^{+\infty} |u|^{p}x\phi_{1}^{l}(1-\frac{t}{T})_{+}^{k-\alpha}{\rm d}x{\rm d}t\leq CT^{-\alpha}M^{-1}+CT^{-\alpha}M, $

这等价于

$ \begin{equation}\label{27} \int_{0}^{T}\int_{0}^{+\infty} |u|^{p}x\phi_{1}^{l}(1-\frac{t}{T})_{+}^{k-\alpha}{\rm d}x{\rm d}t\leq CM^{-1}+CM. \end{equation} $

(3.18)

上式中令$T\rightarrow \infty$得

$ \begin{equation}\label{28} \int_{0}^{+\infty}\int_{0}^{+\infty}|u|^{p}x{\rm d}x{\rm d}t\leq C, \end{equation} $

(3.19)

这意味着

$ \begin{equation}\label{29} \int_{0}^{T}\int_{M^{-\frac{1}{2}}T^{\frac{1}{2}}}^{2M^{-\frac{1}{2}}T^{\frac{1}{2}}} |u|^{p}x\phi_{1}^{l}(1-\frac{t}{T})_{+}^{k}{\rm d}x{\rm d}t \rightarrow 0, ~~~~T \rightarrow +\infty. \end{equation} $

(3.20)

另外通过计算我们有

$ \begin{equation}\label{30} (x\phi_{1}^{l})_{xx}=l\phi_{1}^{l-1}\phi_{1x}+xl(l-1)\phi_{1}^{l-2}(\phi_{1x})^{2} +xl\phi_{1}^{l-1}\phi_{1xx}, \end{equation} $

(3.21)

则可对$K_{2}$进行如下估计

$ \begin{eqnarray}\label{31} K_{2}&=&\int_{0}^{T}\int_{0}^{+\infty}|u||\phi_{xx}|{\rm d}x{\rm d}t\\ &=&C\int_{0}^{T}\int_{0}^{+\infty}|u|x^{\frac{1}{p}}\phi_{1}^{\frac{l}{p}} (1-\frac{t}{T})_{+}^{\frac{k}{p}}x^{-\frac{1}{p}}\phi_{1}^{-\frac{l}{p}}\\ &&\times(1-\frac{t}{T})_{+}^{-\frac{k}{p}}T^{-\alpha}(1-\frac{t}{T})_{+}^{k-\alpha} (x\phi_{1}^{l})_{xx}{\rm d}x{\rm d}t\\ &\leq &CT^{-\alpha}\Big(\int_{0}^{T}\int_{B}^{2B}|u|^{p}x\phi_{1}^{l} (1-\frac{t}{T})_{+}^{k}{\rm d}x{\rm d}t\Big)^{\frac{1}{p}}\\ &&\times\Big(\int_{0}^{1}\int_{1}^{2}T^{-\frac{1}{2(p-1)}}M^{\frac{1}{2(p-1)}}\\ &&\times y^{-\frac{1}{p-1}} (1-s)_{+}^{k-\frac{\alpha p}{p-1}}T^{-\frac{p}{2(p-1)}}M^{\frac{p}{2(p-1)}}\\ &&\times(1+y^{\frac{p}{p-1}}) (T^{\frac{3}{2}}M^{-\frac{1}{2}})^{\frac{p}{p-1}}{\rm d}y{\rm d}s\Big)^{\frac{p-1}{p}}\\ &\leq &CT^{-\alpha}M^{\frac{1}{2}} \Big(\int_{0}^{T}\int_{T^{\frac{1}{2}}M^{-\frac{1}{2}}}^{2T^{\frac{1}{2}}M^{-\frac{1}{2}}} |u|^{p}x\phi_{1}^{l} (1-\frac{t}{T})_{+}^{k}{\rm d}x{\rm d}t\Big)^{\frac{1}{p}}. \end{eqnarray} $

(3.22)

联立(3.6), (3.7), (3.16)和(3.22)式可得

$ \begin{eqnarray*} &&T^{-\alpha}\int_{0}^{T}\int_{0}^{+\infty} |u|^{p}x\phi_{1}^{l}(1-\frac{t}{T})_{+}^{k-\alpha}{\rm d}x{\rm d}t\\ &\leq &CT^{-\alpha}M^{-1}+CT^{-\alpha}M^{\frac{1}{2}} \Big(\int_{0}^{T}\int_{T^{\frac{1}{2}}M^{-\frac{1}{2}}}^{2T^{\frac{1}{2}}M^{-\frac{1}{2}}} |u|^{p}x\phi_{1}^{l} (1-\frac{t}{T})_{+}^{k}{\rm d}x{\rm d}t\Big)^{\frac{1}{p}}, \end{eqnarray*} $

上式等价于

$ \begin{eqnarray}\label{32} &&\int_{0}^{T}\int_{0}^{+\infty} |u|^{p}x\phi_{1}^{l}(1-\frac{t}{T})_{+}^{k-\alpha}{\rm d}x{\rm d}t\\ &\leq& CM^{-1}+CM^{\frac{1}{2}} \Big(\int_{0}^{T}\int_{T^{\frac{1}{2}}M^{-\frac{1}{2}}}^{2T^{\frac{1}{2}}M^{-\frac{1}{2}}} |u|^{p}x\phi_{1}^{l} (1-\frac{t}{T})_{+}^{k}{\rm d}x{\rm d}t\Big)^{\frac{1}{p}}. \end{eqnarray} $

(3.23)

结合(3.20)和(3.23)式并令$T\rightarrow +\infty$得

$ \begin{equation}\label{33} \int_{0}^{+\infty}\int_{0}^{+\infty} |u|^{p}x{\rm d}x{\rm d}t\leq CM^{-1}. \end{equation} $

(3.24)

再令$M\rightarrow +\infty$, 那么可以得到

$ \begin{equation}\label{34}\int_{0}^{+\infty}\int_{0}^{+\infty} |u|^{p}x{\rm d}x{\rm d}t=0. \end{equation} $

(3.25)

同样导致矛盾, 因此完成了定理1.1的证明.

首先, 由Hölder可对$K_{1}$和$K_{2}$进行如下估计

$ \begin{eqnarray}\label{35} K_{1}&=&\int_{0}^{T}\int_{0}^{+\infty}|u||\phi_{t}|{\rm d}x{\rm d}t\\ &=&C\int_{0}^{T}\int_{0}^{2B}x|u|\phi_{1}^{l}T^{-\alpha-1} (1-\frac{t}{T})_{+}^{k-\alpha-1}{\rm d}x{\rm d}t\\ &\leq& CI^{\frac{1}{p}}\Big(\int_{0}^{T}\int_{0}^{2B}T^{-\alpha-\frac{p}{p-1}}x\phi_{1}^{l} (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t\Big)^{\frac{p-1}{p}}\\ &\leq& CT^{\frac{\alpha}{p}-(\alpha+1)}I^{\frac{1}{p}}\Big(\int_{0}^{T}\int_{0}^{2B}x (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t\Big)^{\frac{p-1}{p}}\\ &\triangleq& CT^{\frac{\alpha}{p}-(\alpha+1)}I^{\frac{1}{p}}K_{11}, \end{eqnarray} $

(4.1)

$ \begin{eqnarray}\label{36} K_{2}&=&\int_{0}^{T}\int_{0}^{+\infty}|u||\phi_{xx}|{\rm d}x{\rm d}t\\ &\leq &CT^{\frac{\alpha}{p}-\alpha}I^{\frac{1}{p}}\Big(\int_{0}^{T}\int_{B}^{2B} (1-\frac{t}{T})_{+}^{k-\alpha}x^{-\frac{1}{p-1}} (\phi_{1x}+x(\phi_{1x})^{2}+x\phi_{1xx})^{\frac{p}{p-1}}{\rm d}x{\rm d}t\Big)^{\frac{p-1}{p}}\\ &\triangleq &CT^{\frac{\alpha}{p}-\alpha}I^{\frac{1}{p}}K_{21}. \end{eqnarray} $

(4.2)

另一方面, 引入如下的变量替换

$ s=\frac{t}{T}, \quad y=\frac{x}{T^{\frac{1}{2}}}, \quad B=T^{\frac{1}{2}}, $

则通过直接计算可估计$K_{11}$和$K_{21}$如下

$ \begin{eqnarray}\label{37} K_{11}&=&\Big(\int_{0}^{T}\int_{0}^{2B}x (1-\frac{t}{T})_{+}^{k-\alpha-\frac{p}{p-1}}{\rm d}x{\rm d}t\Big)^{\frac{p-1}{p}}\\ &=&\Big(\int_{0}^{1}\int_{0}^{2}T^{\frac{1}{2}}y(1-s)_{+}^{k-\alpha-\frac{p}{p-1}} T^{\frac{1}{2}+1}{\rm d}y{\rm d}s\Big)^{\frac{p-1}{p}}\\ &\leq& CT^{\frac{2(p-1)}{p}}, \end{eqnarray} $

(4.3)

$ \begin{eqnarray}\label{38} K_{21}&=& \Big(\int_{0}^{T}\int_{B}^{2B} (1-\frac{t}{T})_{+}^{k-\alpha}x^{-\frac{1}{p-1}}\\ &&\times (\phi_{1x}+x(\phi_{1x})^{2}+x\phi_{1xx})^{\frac{p}{p-1}}{\rm d}x{\rm d}t\Big)^{\frac{p-1}{p}}\\ &=&\Big(\int_{0}^{1}\int_{1}^{2}(T^{\frac{1}{2}}y)^{-\frac{1}{p-1}} (1-s)_{+}^{k-\alpha}\\ &&\times(T^{-\frac{p}{2(p-1)}} +T^{-\frac{p}{2(p-1)}} y^{\frac{p}{p-1}})T^{\frac{1}{2}+1}{\rm d}y{\rm d}s\Big)^{\frac{p-1}{p}}\\ &\leq &CT^{-\frac{p+1}{2p}+\frac{3(p+1)}{2p}}. \end{eqnarray} $

(4.4)

联立(4.1), (4.2), (4.3)和(4.4)式可得

$ \begin{eqnarray}\label{39} K_{1}&\leq &CT^{\frac{\alpha}{p}-(\alpha+1)+\frac{2(p-1)}{p}}I^{\frac{1}{p}}, \\ K_{2}&\leq & CT^{\frac{\alpha}{p}-\alpha-\frac{p+1}{2p}+\frac{3(p-1)}{2p}}I^{\frac{1}{p}} =CT^{\frac{\alpha}{p}-(\alpha+1)+\frac{2(p-1)}{p}}I^{\frac{1}{p}}. \end{eqnarray} $

(4.5)

由(3.4), (3.5)和(4.5)式可推出

$ \begin{equation}\label{40} \varepsilon T^{-\alpha}\leq CT^{\frac{\alpha}{p}-(\alpha+1)+\frac{2(p-1)}{p}}I^{\frac{1}{p}}-I. \end{equation} $

(4.6)

由于对任意的$a>0$, $0<b<1$及$c\geq0$有

$ \begin{equation}\label{41} ac^{b}-c\leq(1-b)b^{\frac{b}{1-b}}a^{\frac{1}{1-b}}, \end{equation} $

(4.7)

因此由(4.6)和(4.7)式可得

$ \begin{eqnarray}\label{42} C\varepsilon T^{-\alpha} &\leq &CT^{\frac{\alpha}{p}-(\alpha+1)+\frac{2(p-1)}{p}}I^{\frac{1}{p}}-I\\ &\leq &(1-\frac{1}{p})(\frac{1}{p})^{\frac{1}{p-1}}(CT^{\frac{\alpha}{p}-(\alpha+1)+\frac{2(p-1)}{p}})^{\frac{p}{p-1}}\\ &\leq& CT^{-\alpha-\frac{p}{p-1}+2}. \end{eqnarray} $

(4.8)

从(4.8)式可以得出我们预期的解的生命跨度上界估计

$ \begin{equation}\label{43} T\leq C\varepsilon^{\frac{p-1}{p-2}}, \end{equation} $

(4.9)

于是完成了定理1.2的证明.