0 引言

设n≥5, □=u_tt-Δ为Rⁿ上的波动算子，其中$ \Delta = \sum\limits_{i = 1}^n {\frac{{{\partial ^2}}}{{\partial x_i^2}}} $.研究以下半线性波动方程Cauchy问题：

$ \left\{ \begin{gathered} \square u\left( {x,t} \right) = {\left( {1 + {{\left| x \right|}^2}} \right)^\alpha }{\left| u \right|^p}, \hfill \\ u\left( {0,x} \right) = \varepsilon f\left( x \right),{u_t} = \varepsilon g\left( x \right), \hfill \\ \end{gathered} \right. $

(1)

其中初值满足

$ \left\{ \begin{array}{l} \left( {f,g} \right) \in {H^1}\left( {{R^n}} \right) \times {L^2}\left( {{R^n}} \right),\\ f\left( x \right) \ge 0,g\left( x \right) \ge 0,\\ f\left( x \right) = g\left( x \right) = 0,\forall \left| x \right| > R > 0,\\ \alpha > 0,0 < \varepsilon \le 1. \end{array} \right. $

(2)

且p=p_c(n), p_c(n)为二次代数方程:

$ \left( {n - 1} \right){p^2} - \left( {n + 4\alpha + 1} \right)p - 2 = 0 $

的正根，即

$ {p_c}\left( n \right) = \frac{{n + 4\alpha + 1 + \sqrt {{n^2} + \left( {8\alpha + 10} \right)n + 16{\alpha ^2} + 8\alpha - 7} }}{{2\left( {n - 1} \right)}}, $

该问题为以下半线性波动方程小初值Cauchy问题(Strauss猜测)的拓展：

$ \left\{ \begin{array}{l} \square u\left( {x,t} \right) = {\left| u \right|^p}, \hfill \\ u\left( {0,x} \right) = \varepsilon f\left( x \right),{u_t} = \varepsilon g\left( x \right), \hfill \\ \end{array} \right. $

(3)

其中，n≥2, ε＞0为小参数.已知结果表明，问题(3)存在一个临界指数p₀(n), 为下述二次方程的正根：

$ \left( {n - 1} \right){p^2} - \left( {n + 1} \right)p - 2 = 0. $

临界指标是指正实数p₀(n)将(1, +∞)分成(1, p₀(n)]和(p₀(n), +∞)2个区间.当p∈(1, p₀(n)]时，问题(3)不存在整体解；当p∈(p₀(n), +∞)时，问题(3)对小初值有整体解.

下面简单介绍Strauss猜测的研究历史.1979年，JOHN^[1]证明了当1＜p＜p₀(3)=$ 1 + \sqrt 2 $时，问题(3)不存在整体解；当p＞p₀(3)=$ 1 + \sqrt 2 $时，存在小初值整体解.当n=2时，问题的整体解部分由GLASSEY^[2]证实.随后，一些学者对高维波动方程研究取得了若干进展.1995年，ZHOU^[3]证明了n=4时的整体解.1996年，LINDBLAD等^[4]解决了3≤n≤8时的整体解.1997年，GEORGIEV等^[5]证明了当n≥4且$ {p_0}\left( n \right) < p \le \frac{{n + 3}}{{n - 1}} $时，问题(3)存在整体解.对于次临界情形1＜p＜p₀(n)，GLASSEY^[6]和SIDERIS^[7]分别得到n=2和n≥4时的爆破结果.对于临界情形p=p₀(n), SCHAEFFER^[8]得到n=2, 3时(对小初值)解的破裂结果.YORDANOV等^[9]和ZHOU^[10]分别用不同方法独立解决了n≥4时临界情形解的破裂结果.更多关于Strauss猜测的相关问题与研究进展可参考文献[11-14].

此后，一些学者研究了问题(3)在外区域上的初边值问题，得到次临界情形、临界情形的若干破裂结果、生命跨度估计及超临界情形的整体存在性结果，相关进展可参考文献[15-20].

最近，李卓然^[21]研究了带次临界指标(1＜p＜p_c(n))的Cauchy问题(1)，得到了解的破裂结果和生命跨度上界估计.

本文主要研究带临界指标p=p_c(n)的Cauchy问题(1)，证明了解将会在有限时间内破裂，主要结果如下：

定理1  设n≥5, 且初值f与g满足式(2).假设问题(1)有解

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

使得

$ {\rm{supp}}\left( {u,{u_t}} \right) \subset \left\{ {\left( {x,t} \right):\left| x \right| \le t + R} \right\}. $

若$ 0 < \alpha < \min \left( {\frac{{n - 4}}{4}, \frac{{n\left( {p - 1} \right)}}{2}} \right) $, 且p=p_c(n)，则T＜∞.

文中，记号C表示常数，在不同的地方可表示不同的值.

1 预备知识

为证明本文的主要结论，需要应用以下引理：

引理1^[9]  设p＞1, a≥1, 且(p-1)a=q-2.t≥T₀＞0, 若F∈C²([0, T))满足

$ F\left( t \right) \ge {K_0}{\left( {t + R} \right)^a}, $

(4)

且

$ F''\left( t \right) \ge {K_1}{\left( {t + R} \right)^{ - q}}{\left[ {F\left( t \right)} \right]^p}, $

(5)

其中K₀, K₁, T₀和R是正常数.当固定K₁, 存在C₀＞0且C₀不依赖于T₀和R, 使得K₀≥C₀, 则T＜∞, 即F(t)在有限时间内破裂.

为了叙述引理2，参照文献[9]引入试验函数：

$ {\varphi _1}\left( x \right) = \int_{{S^{n - 1}}} {{{\rm{e}}^{x \cdot \omega }}{\rm{d}}\omega } , $

(6)

且定义函数

$ {\psi _1}\left( {x,t} \right) = {\varphi _1}\left( x \right){{\rm{e}}^{ - t}}. $

(7)

现引入2个泛函：

$ \left\{ \begin{array}{l} {F_0}\left( t \right) = \int_{{R^n}} {u\left( {x,t} \right){\rm{d}}x} ,\\ {F_1}\left( t \right) = \int_{{R^n}} {u\left( {x,t} \right){\psi _1}\left( {x,t} \right){\rm{d}}x} , \end{array} \right. $

(8)

其中ψ₁(x, t)为由式(7)定义的函数, u为问题(1)的解.u所满足的假设条件蕴含了F₀(t)与F₁(t)关于t连续可微.

引理2^[21]  设$ \alpha < \frac{{n\left( {p - 1} \right)}}{2} $，则

$ \frac{{{{\rm{d}}^2}{F_0}\left( t \right)}}{{{\rm{d}}{t^2}}} \ge {K_1}{\left( {t + R} \right)^{ - n\left( {p - 1} \right) + 2\alpha }}{\left| {{F_0}\left( t \right)} \right|^p}. $

(9)

引理3^[21]  设p＞1，则

$ \begin{array}{*{20}{c}} {\int_{{R^n}} {{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{{\left| {u\left( {x,t} \right)} \right|}^p}{\rm{d}}x} \ge }\\ {C{\varepsilon ^p}{{\left( {t + R} \right)}^{ - \left( {n - 1} \right)\left( {\frac{p}{2} - 1} \right) + 2\alpha }}.} \end{array} $

(10)

2 定理的证明

为了得到临界情形的破裂结果，采用文献[9]的方法建立一个关于非线性项的改进的下界估计.不失一般性, 假设u(·, t)为径向函数.否则可定义

$ \bar u\left( {t,r} \right) = \frac{1}{{{\mathit{\boldsymbol{\omega }}_n}}}\int_{\left| \mathit{\boldsymbol{\omega }} \right| = 1} {u\left( {t,r\mathit{\boldsymbol{\omega }}} \right){\rm{d}}\omega } ,\;\;\;\left( {t,r} \right) \in \left[ {0,T} \right) \times R. $

由达布公式可知：

$ \begin{array}{l} {\left( {\bar u} \right)_{tt}} - \Delta \bar u = \overline {\left( {{u_{tt}} - \Delta u} \right)} = {\left| {\bar u} \right|^p} = \\ \;\;\;\;\;\;\frac{1}{{{\mathit{\boldsymbol{\omega }}_n}}}\int_{\left| \mathit{\boldsymbol{\omega }} \right| = 1} {{{\left| {u\left( {t,r\mathit{\boldsymbol{\omega }}} \right)} \right|}^p}{\rm{d}}\omega } \ge {\left| {\bar u} \right|^p}. \end{array} $

其中, ω∈Rⁿ为单位向量^[22].引入u关于空间变量的Radon变换：

$ R\left( u \right)\left( {\rho ,t} \right) = \int_{x \cdot \omega = \rho } {u\left( {x,t} \right){\rm{d}}{S_x}} , $

(11)

其中, dS_x为在超平面{x|x·ω=ρ}上的Lebesgue测度，容易证明R(u)仅依赖于ρ和t，而与ω无关.由式(11)和径向对称条件可得

$ \begin{array}{*{20}{c}} {R\left( u \right)\left( {\rho ,t} \right) = \int_{\left\{ {x'\left| {x' \cdot \mathit{\boldsymbol{\omega }}} \right. = 0} \right\}} {u\left( {\rho \mathit{\boldsymbol{\omega }} + x',t} \right){\rm{d}}{S_{x'}}} = }\\ {C\int_0^\infty {u\left( {\sqrt {{\rho ^2} + {{\left| {x'} \right|}^2}} ,t} \right)} \;\;{{\left| {x'} \right|}^{n - 2}}{\rm{d}}\left| {x'} \right|,} \end{array} $

令

$ r = \sqrt {{\rho ^2} + {{\left| {x'} \right|}^2}} , $

$ R\left( u \right)\left( {\rho ,t} \right) = C\int_{\left| \rho \right|}^\infty {u\left( {r,t} \right){{\left( {{r^2} - {\rho ^2}} \right)}^{\frac{{n - 3}}{2}}}r{\rm{d}}r} , $

(12)

这意味着R(u)(ρ, t)不依赖于ω.

由于u是问题(1)的解，显然R(u)满足一维波动方程:

$ \begin{array}{l} \partial _t^2R\left( u \right)\left( {\rho ,t} \right) - \partial _\rho ^2R\left( u \right)\left( {\rho ,t} \right) = \\ \;\;\;\;\;\;\;\;\;\;R\left( {{{\left| u \right|}^p}\left( {^1 + {{\left| \cdot \right|}^2}} \right)\alpha } \right)\left( {\rho ,t} \right), \end{array} $

(13)

由D'Alembert公式和假设条件: u的初始值是非负的，可得

$ \begin{array}{l} R\left( u \right)\left( {\rho ,t} \right) \ge \\ \frac{1}{2}\int_0^t {\int_{\rho - \left( {t - s} \right)}^{\rho + \left( {t - s} \right)} {R\left( {{{\left| u \right|}^p}{{\left( {1 + {{\left| \cdot \right|}^2}} \right)}^\alpha }} \right)\left( {{\rho _1},s} \right){\rm{d}}{\rho _1}{\rm{d}}s} } . \end{array} $

(14)

注意到supp u(·, s)包含于半径为s+R、圆心为原点的球B(0, s+R)内，若|ρ₁|＞s+R，由于向量y垂直于单位向量ω，则可得

$ \left| {{\rho _1}\mathit{\boldsymbol{\omega }} + \mathit{\boldsymbol{y}}} \right| = \sqrt {{{\left| {{\rho _1}} \right|}^2} + {{\left| \mathit{\boldsymbol{y}} \right|}^2}} \ge \left| {{\rho _1}} \right| > s + R. $

因此

$ \begin{array}{l} R\left( {{{\left| u \right|}^p}{{\left( {1 + {{\left| \cdot \right|}^2}} \right)}^\alpha }} \right)\left( {{\rho _1},s} \right) = \\ \int_{\left\{ {\mathit{\boldsymbol{y}}\left| {\mathit{\boldsymbol{y}} \cdot \mathit{\boldsymbol{\omega }}} \right. = 0} \right\}} {{{\left| {u\left( {{\rho _1}\mathit{\boldsymbol{\omega }} + \mathit{\boldsymbol{y}},s} \right)} \right|}^p}{{\left( {1 + {{\left| {{\rho _1}\mathit{\boldsymbol{\omega }} + \mathit{\boldsymbol{y}}} \right|}^2}} \right)}^\alpha }{\rm{d}}{S_y}} = 0, \end{array} $

即

$ {\rm{supp}}\;R\left( {{{\left| u \right|}^p}} \right)\left( { \cdot ,s} \right) \subset B\left( {0,s + R} \right). $

(15)

除非另有说明，文中均假设ρ≥0.若$ s \le \frac{{t - \rho - R}}{2} $，则

$ \rho + \left( {t - s} \right) \ge s + R,\;\;\;\;\;\rho - \left( {t - s} \right) \le - \left( {s + R} \right), $

由式(14)和(15)，得到

$ \begin{array}{l} R\left( u \right)\left( {\rho ,t} \right) \ge \frac{1}{2}\int_0^{\frac{{t - \rho - R}}{2}} {\int_{\rho - \left( {t - s} \right)}^{\rho + \left( {t - s} \right)} {R\left( {{{\left| u \right|}^p}{{\left( {1 + {{\left| \cdot \right|}^2}} \right)}^\alpha }} \right)\left( {{\rho _1},s} \right){\rm{d}}{\rho _1}{\rm{d}}s} } = \\ \;\;\;\;\;\;\;\;\;\;\;\frac{1}{2}\int_0^{\frac{{t - \rho - R}}{2}} {\int_{ - \infty }^{ + \infty } {R\left( {{{\left| u \right|}^p}{{\left( {1 + {{\left| \cdot \right|}^2}} \right)}^\alpha }} \right)\left( {{\rho _1},s} \right){\rm{d}}{\rho _1}{\rm{d}}s} } = \\ \;\;\;\;\;\;\;\;\;\;\;\frac{1}{2}\int_0^{\frac{{t - \rho - R}}{2}} {\int_{{R^n}} {{{\left| {u\left( {\mathit{\boldsymbol{y}},s} \right)} \right|}^p}{{\left( {1 + {{\left| \mathit{\boldsymbol{y}} \right|}^2}} \right)}^\alpha }{\rm{d}}\mathit{\boldsymbol{y}}{\rm{d}}s} } , \end{array} $

(16)

由式(10)可得

$ \int_{{R^n}} {{{\left| {u\left( {\mathit{\boldsymbol{y}},s} \right)} \right|}^p}{{\left( {1 + {{\left| \mathit{\boldsymbol{y}} \right|}^2}} \right)}^\alpha }{\rm{d}}\mathit{\boldsymbol{y}}} \ge C{s^{\left( {n - 1} \right)\left( {1 - \frac{p}{2}} \right) + 2\alpha }}. $

当$ 2\alpha > \left( {n - 1} \right)\left( {\frac{p}{2} - 1} \right) $时, 有

$ \left( {n - 1} \right)\left( {1 - \frac{p}{2}} \right) + 2\alpha > 0, $

结合式(16)及ρ≥0，得到

$ R\left( u \right)\left( {\rho ,t} \right) \ge C{\left( {t - \rho - R} \right)^{n - \left( {n - 1} \right)\frac{p}{2} + 2\alpha }}. $

(17)

下面引入函数f∈L^p(R)的一个变换：

$ T\left( f \right)\left( \rho \right) = \frac{1}{{{{\left| {\left( {t - \rho + R} \right)} \right|}^{\left( {n - 1} \right)/2}}}}\int_\rho ^{t + R} {f\left( r \right){{\left| {r - \rho } \right|}^{\left( {n - 3} \right)/2}}{\rm{d}}r} , $

(18)

注意到

$ \begin{array}{l} T\left( f \right)\left( \rho \right) \le \frac{1}{{\left| {\left( {t - \rho + R} \right)} \right|}}\left| {\int_\rho ^{t + R} {\left| {f\left( r \right)} \right|{\rm{d}}r} } \right| \le \\ \frac{2}{{2\left| {\left( {t - \rho + R} \right)} \right|}}\left| {\int_{ - \left( {t + R} \right) + 2\rho }^{t + R} {\left| {f\left( r \right)} \right|{\rm{d}}r} } \right| \le 2M\left( {\left| f \right|} \right)\left( \rho \right), \end{array} $

其中M(|f|)为f的极大值函数.因此，存在C＞0,

$ {\left\| {T\left( f \right)} \right\|_p} \le C{\left\| f \right\|_p}, $

(19)

令

$ f\left( r \right) = \left\{ \begin{array}{l} \left| {u\left( {r,t} \right)} \right|{\left( {1 + {r^2}} \right)^{\frac{\alpha }{p}}}{r^{\frac{{n - 1}}{p}}},\;\;\;\;\;r \ge 0,\\ 0,\;\;\;\;r < 0, \end{array} \right. $

则有

$ \begin{array}{l} \int_0^{t + R} {\left[ {\frac{1}{{{{\left| {\left( {t - \rho + R} \right)} \right|}^{\left( {n - 1} \right)/2}}}}\int_\rho ^{t + R} {\left| {u\left( {r,t} \right)} \right|{{\left( {1 + {r^2}} \right)}^{\frac{\alpha }{p}}}} \times } \right.} \\ \;\;\;\;\;\;{\left. {{r^{\frac{{n - 1}}{p}}}{{\left| {r - \rho } \right|}^{\left( {n - 3} \right)/2}}{\rm{d}}r} \right]^p}{\rm{d}}\rho \le \\ \;\;\;\;\;\;C\int_0^\infty {{{\left| {u\left( {r,t} \right)} \right|}^p}{{\left( {1 + {r^2}} \right)}^\alpha }{r^{n - 1}}{\rm{d}}r} = \\ \;\;\;\;\;\;C\int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} , \end{array} $

(20)

当r≥ρ且1＜p≤2时，有

$ {r^{\frac{{n - 1}}{p}}} = {r^{\frac{{n - 1}}{2}}}{r^{\frac{{n - 1}}{p} - \frac{{n - 1}}{2}}} \ge {r^{\frac{{n - 1}}{2}}}{r^{\frac{{n - 1}}{p} - \frac{{n - 1}}{2}}}, $

(21)

由式(20)，若α＞0, 则$ {\rho ^{\frac{{2\alpha }}{p}}} \le {\left( {1 + {r^2}} \right)^{\frac{\alpha }{p}}} $, 所以

$ \begin{array}{l} \int_0^{t + R} {{{\left[ {\frac{1}{{{{\left| {\left( {t - \rho + R} \right)} \right|}^{\left( {n - 1} \right)/2}}}}\int_\rho ^{t + R} {\left| {u\left( {r,t} \right)} \right|{r^{\frac{{n - 1}}{2}}}{{\left| {r - \rho } \right|}^{\frac{{n - 3}}{2}}}{\rm{d}}r} } \right]}^p}} \times \\ \;\;\;\;\;{\rho ^{\frac{{2\alpha }}{p}}}{\rho ^{n - 1 - \frac{{{{\left( {n - 1} \right)}^p}}}{2}}}{\rm{d}}\rho \le \int_0^{t + R} {\left[ {\frac{1}{{{{\left| {\left( {t - \rho + R} \right)} \right|}^{\left( {n - 1} \right)/2}}}} \times } \right.} \\ \;\;\;\;\;{\left. {\int_\rho ^{t + R} {\left| {u\left( {r,t} \right)} \right|{{\left( {1 + {r^2}} \right)}^{\frac{\alpha }{p}}}{r^{\frac{{n - 1}}{2}}}{{\left| {r - \rho } \right|}^{\frac{{n - 3}}{2}}}{\rm{d}}r} } \right]^p} \times \\ \;\;\;\;\;{\rho ^{n - 1 - \frac{{\left( {n - 1} \right)p}}{2}}}{\rm{d}}\rho \le C\int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} , \end{array} $

(22)

由式(12)且注意到supp u(·, t)⊂B(0, t+R), 可得

$ \begin{array}{l} R\left( {\left| u \right|} \right)\left( {\rho ,t} \right) = C\int_\rho ^{t + R} {\left| {u\left( {r,t} \right)} \right|{{\left( {{r^2} - {\rho ^2}} \right)}^{\frac{{n - 3}}{2}}}r{\rm{d}}r} \le \\ \;\;\;\;\;C\int_\rho ^{t + R} {\left| {u\left( {r,t} \right)} \right|{{\left( {r + \rho } \right)}^{\frac{{n - 3}}{2}}}{{\left( {r - \rho } \right)}^{\frac{{n - 3}}{2}}}r{\rm{d}}r} \le \\ \;\;\;\;\;C\int_\rho ^{t + R} {{{\left| {u\left( {r,t} \right)} \right|}^{\frac{{n - 1}}{2}}}{{\left( {r - \rho } \right)}^{\frac{{n - 3}}{2}}}{\rm{d}}r} . \end{array} $

(23)

结合式(21)与(23)，有

$ \begin{array}{l} \int_0^{t + R} {\frac{{{{\left[ {R\left( {\left| u \right|} \right)\left( {\rho ,t} \right)} \right]}^p}}}{{{{\left| {\left( {t - \rho + R} \right)} \right|}^{\left( {n - 1} \right)/2}}}}{\rho ^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}{\rm{d}}\rho \le } \\ \;\;\;\;\;C\int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} . \end{array} $

(24)

由式(17) R(|u|)的下界，并结合式(24)及r≥ρ, 得到

$ \begin{array}{l} \int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} \ge \\ \;\;\;\;\;C\int_0^{t - R - 1} {\frac{{{{\left[ {R\left( {\left| u \right|} \right)\left( {\rho ,t} \right)} \right]}^p}}}{{{{\left| {\left( {t - \rho + R} \right)} \right|}^{\left( {n - 1} \right)p/2}}}}{\rho ^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}{\rm{d}}\rho } \ge \\ \;\;\;\;\;C\int_0^{t - R - 1} {\frac{{{{\left( {t - \rho - R} \right)}^{np - \left( {n - 1} \right){p^2}/2 + 2\alpha p}}}}{{{{\left( {t - \rho + R} \right)}^{\left( {n - 1} \right)p/2}}}}{\rho ^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}{\rm{d}}\rho } . \end{array} $

当ρ∈(0, t-R-1)时, 显然存在常数C_n＞0，对所有t＞2(R+1), 有

$ \left( {t - \rho + R} \right) \le {C_n}\left( {t - \rho - R} \right), $

由此可得

$ \begin{array}{l} \int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} \ge \\ \;\;\;\;C\int_0^{t - R - 1} {\frac{{{\rho ^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}}}{{{{\left( {t - \rho + R} \right)}^{\frac{{\left( {n - 1} \right)p}}{2} - np + \frac{{\left( {n - 1} \right){p^2}}}{2} - 2\alpha p}}}}{\rm{d}}\rho } , \end{array} $

(25)

由于p是满足式(1)的临界指数，即

$ \left( {n - 1} \right){p^2} - \left( {n + 4\alpha + 1} \right)p - 2 = 0, $

从而有

$ \begin{array}{*{20}{c}} {\left( {n - 1} \right)p/2 - np + \left( {n - 1} \right){p^2}/2 - 2\alpha p = }\\ {\frac{{\left( {n - 1} \right){p^2} - \left( {n + 4\alpha + 1} \right)p}}{2} = 1,} \end{array} $

由式(25)，

$ \int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} \ge {C_n}\int_0^{t - R - 1} {\frac{{{\rho ^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}}}{{\left( {t - \rho + R} \right)}}{\rm{d}}\rho } , $

(26)

由此，

$ \begin{array}{*{20}{c}} {\int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} \ge {C_n}\int_0^{t - R - 1} {\frac{{{\rho ^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}}}{{\left( {t - \rho + R} \right)}}{\rm{d}}\rho } \ge }\\ {{C_n}{{\left( {t - R - 1} \right)}^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}\int_{\left( {t - R - 1} \right)/2}^{t - R - 1} {\frac{1}{{\left( {t - \rho - R} \right)}}{\rm{d}}\rho } ,} \end{array} $

根据上式，重新定义下界

$ \begin{array}{l} \int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} \ge \\ \;\;\;\;C{\left( {t - R} \right)^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}\ln \left( {t - R - 1 - \left( {t - R - 1} \right)/2} \right) \ge \\ \;\;\;\;C{\left( {t - R} \right)^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}\ln \left( {t - R} \right)/2, \end{array} $

结合式(1)与式(8)，有

$ \begin{array}{*{20}{c}} {\frac{{{{\rm{d}}^2}{F_0}\left( t \right)}}{{{\rm{d}}{t^2}}} = \int_{{R^n}} {{{\left| {u\left( {x,t} \right)} \right|}^p}{{\left( {1 + {{\left| x \right|}^2}} \right)}^\alpha }{\rm{d}}x} \ge }\\ {C{{\left( {t - R} \right)}^{n - 1 - \left( {n - 1} \right)p/2 + 2\alpha }}\ln \left( {t - R} \right)/2,} \end{array} $

(27)

可知，式(27)的下界较式(10)的下界多了一项ln t.

由于n-1-(n-1)p/2+2α≥0，当n≥4时, 对式(27)积分2次，得到

$ {F_0}\left( t \right) \ge C{\left( {t - R} \right)^{n + 1 - \left( {n - 1} \right)p/2 + 2\alpha }}\ln t, $

有

$ \begin{array}{*{20}{c}} {{F_0}\left( t \right) \ge C{{\left( {t + R} \right)}^{n + 1 - \left( {n - 1} \right)p/2 + 2\alpha }} \times }\\ {{{\left( {\frac{{t - R}}{{t + R}}} \right)}^{n + 1 - \left( {n - 1} \right)p/2 + 2\alpha }}\ln t,} \end{array} $

由于t充分大，显然

$ \mathop {\lim }\limits_{t \to \infty } {\left( {\frac{{t - R}}{{t + R}}} \right)^{n + 1 - \left( {n - 1} \right)p/2 + 2\alpha }}\ln t = \infty , $

因此

$ {F_0}\left( t \right) \ge {K_0}{\left( {t + R} \right)^{n + 1 - \left( {n - 1} \right)p/2 + 2\alpha }}. $

(28)

其中当t充分大时, K₀＞0为任意大数.结合式(9)与(28)，在引理1中取参数

$ \begin{array}{*{20}{c}} {a = n + 1 - \left( {n - 1} \right)p/2 + 2\alpha ,}\\ {q = n\left( {p - 1} \right) - 2\alpha ,} \end{array} $

即得对满足条件(p-1)[n+1-(n-1)p/2+2α]= n(p-1)-2α-2且p＞1的p有p=p_c, 定理1成立.当p=p_c(n)且初始条件满足式(2)时，问题(1)的解在有限时间内破裂.

3 α的取值范围

由式(22)中的条件可知α＞0.下面验证

$ \alpha < \frac{{n\left( {{p_c}\left( n \right) - 1} \right)}}{2}. $

(29)

由p_c(n)的表达式可知，式(29)等价于

$ \begin{array}{*{20}{c}} {2\alpha < n\left( {\frac{{n + 4\alpha + 1 + \sqrt {{n^2} + \left( {8\alpha + 10} \right)n + 16{\alpha ^2} + 8\alpha - 7} }}{{2\left( {n - 1} \right)}} - 1} \right),}\\ {\frac{{4\alpha \left( {n - 1} \right)}}{n} < - n + 4\alpha + 3 + \sqrt {{n^2} + \left( {8\alpha + 10} \right)n + 16{\alpha ^2} + 8\alpha - 7} ,}\\ {\frac{{{{\left( {{n^2} - 3n - 4\alpha } \right)}^2}}}{{{n^2}}} < {n^2} + \left( {8\alpha + 10} \right)n + 16{\alpha ^2} + 8\alpha - 7,}\\ {2{\alpha ^2}\left( {1 - {n^2}} \right) + \left( {3n - 2{n^2} - {n^3}} \right)\alpha + 2{n^2}\left( {1 - n} \right) < 0,n \ge 5,}\\ {2{\alpha ^2}\left( {1 + n} \right) + n\left( {n + 3} \right)\alpha + 2{n^2} > 0.} \end{array} $

即验证了式(29).

另一方面，式(21)中要求1＜p_c(n)≤2，

即

$ 1 < \frac{{n + 4\alpha + 1 + \sqrt {{n^2} + \left( {8\alpha + 10} \right)n + 16{\alpha ^2} + 8\alpha - 7} }}{{2\left( {n - 1} \right)}} \le 2, $

$ \begin{array}{l} n - \left( {4\alpha + 3} \right) < \sqrt {{n^2} + \left( {8\alpha + 10} \right)n + 16{\alpha ^2} + 8\alpha - 7} \le \\ \;\;\;\;\;3n - \left( {4\alpha + 5} \right). \end{array} $

亦即

$ - 1 < \alpha \le \frac{{\left( {n - 4} \right)}}{4}, $

综合可得

$ 0 < \alpha < \min \left( {\frac{{n - 4}}{4},\frac{{n\left( {p - 1} \right)}}{2}} \right). $