0 引言

考虑变系数耗散波动方程Cauchy问题：

$ \begin{array}{*{20}{c}} {{u_u} - \sum\limits_{i,j = 1}^n {{\partial _{{x_i}}}\left( {{a^{ij}}\left( x \right){\partial _{{x_j}}}u} \right)} + {u_t} = 0,}\\ {\left( {t,x} \right) \in \left( {0,\infty } \right) \times {R^n},} \end{array} $

(1)

其初值条件为：

$ \left\{ \begin{array}{l} u\left( {0,x} \right) = {u_0}\left( x \right),\\ {u_t}\left( {0,x} \right) = {u_1}\left( x \right), \end{array} \right. $

(2)

其中，系数a^ij(x)属于C^∞(Rⁿ)(满足一致椭圆条件), 且对于任意的x∈Rⁿ，A (x)=(a^ij(x))是对称正定矩阵.

假设{u₀, u₁}属于能量空间且具有紧支集，即

$ {u_0} \in {H^1}\left( {{R^n}} \right),{u_1} \in {L^2}\left( {{R^n}} \right), $

(3)

$ {\rm{supp}}\;{u_0} \cup {\rm{supp}}\;{u_1} \subset B\left( L \right), $

(4)

其中, L>0，B(L)是以原点O为圆心，L为半径的一个球：B(L)≡{x∈Rⁿ||x| < L}.

当a^ij=δ^ij(Kroneckerδ函数)时，问题(1)是常系数耗散波动方程；此时具初值(2)的Cauchy问题在(0, ∞)内存在整体解u(t)^[1-2]：

$ u\left( t \right) \in C\left( {\left[ {0,\infty } \right);{H^1}\left( {{R^n}} \right)} \right) \cap {C^1}\left( {\left[ {0,\infty } \right);{L^2}\left( {{R^n}} \right)} \right), $

(5)

$ {\rm{supp}}\;u\left( t \right) \subset B\left( {t + L} \right),\;\;\;t \geqslant 0, $

(6)

而且，通过标准的能量方法，一方面可以得到以下能量估计^[3-5]：

$ E\left( t \right) \leqslant CE\left( 0 \right){\left( {1 + t} \right)^{ - 1}},t \geqslant 0, $

其中,

$ E\left( t \right) = \frac{1}{2}\int_{{R^n}} {\left( {{{\left| {u\left( {t,x} \right)} \right|}^2} + {{\left| {\nabla u\left( {t,x} \right)} \right|}^2}} \right){\rm{d}}x} , $

$ E\left( 0 \right) = \frac{1}{2}\left\| {{u_1}} \right\|_{{L^2}\left( {{R^n}} \right)}^2 + \frac{1}{2}\left\| {\nabla {u_0}} \right\|_{{L^2}\left( {{R^n}} \right)}^2. $

另一方面，TODOROVA等^[6]已得到以下衰减估计：

$ \begin{array}{*{20}{c}} {\int_{B{{\left( {{t^{\frac{1}{2} + \delta }}} \right)}^c}} {\left( {{{\left| {{u_t}} \right|}^2} + {{\left| {\nabla u} \right|}^2}} \right){\rm{d}}x} \leqslant }\\ {CE\left( 0 \right)\exp \left( { - \frac{1}{2}{t^{2\delta }}} \right),} \end{array} $

(7)

其中，δ>0为常数，证明过程用到了假设(3)和(4).B(L)^c是B(L)的补集，即

$ B{\left( L \right)^c} \equiv {R^n}\backslash B\left( L \right) = \left\{ {x \in {R^n}\left| {\left| x \right| \geqslant L} \right.} \right\}. $

ONO^[7]研究了当区域大于$B{({t^{\frac{1}{2} + \delta }})^c}$时，能量衰减估计为：

$ \begin{array}{*{20}{c}} {\int_{B{{\left( {{{\left( {t + L} \right)}^{\frac{1}{2}}}\log {{\left( {1 + t} \right)}^m}} \right)}^c}} {\left( {{{\left| {{u_t}} \right|}^2} + {{\left| {\nabla u} \right|}^2}} \right){\rm{d}}x} \leqslant }\\ {{{\rm{e}}^R}E\left( 0 \right){{\left( {1 + t} \right)}^{ - m}},t \geqslant 0,} \end{array} $

其中m>0为常数.当m>0和δ>0时，对于充分大的t>0，有

$ {\left( {t + L} \right)^{\frac{1}{2}}}\log {\left( {1 + t} \right)^m} < {t^{\frac{1}{2} + \delta }}, $

因此, $B{({t^{\frac{1}{2} + \delta }})^c} \subset B{({\left( {t + L} \right)^{\frac{1}{2}}}{\rm{log}}{\left( {1 + t} \right)^m})^c}$.

本文研究的是变系数问题，目的是在初值(2)满足条件(3)和(4)时，建立一个比ONO^[7]的研究结果更大的外区域上Cauchy问题(1)的能量衰减估计.

方程(1)两边同乘以2u_t，利用a^ij(x)=a^ji(x)得

$ \begin{array}{*{20}{c}} {{\partial _t}{{\left( {{u_t}} \right)}^2} - \sum\limits_{i,j = 1}^n {\partial {x_i}\left( {{a^{ij}}\left( x \right){\partial _{{x_j}}}u{\partial _t}u} \right)} + }\\ {{\partial _t}\left( {{a^{ij}}\left( x \right){\partial _{{x_i}}}u{\partial _{{x_j}}}u} \right) + 2u_t^2 = 0.} \end{array} $

在Rⁿ上积分并利用有限传播速度，可得

$ \frac{{{\rm{d}}E\left( t \right)}}{{{\rm{d}}t}} = - 2\int_{{R^n}} {u_t^2{\rm{d}}x} \leqslant 0, $

(8)

其中, $E(t) = \int_{{R^n}} {(|{u_t}{|^2} + {a^{ij}}{u_{{x_i}}}{u_{{x_j}}}){\rm{d}}x} $为问题(1)的能量.意味着时间变量的能量函数是递减的.

对于变系数波动方程，有着与常系数情形类似的的有限传播速度：固定一点(t₀, x₀)∈(0, ∞)×Rⁿ，可以找到以(t₀, x₀)为顶点的弯曲的“锥状”区域Q，若在Q₀=Q∩{t=0}上，有u≡u_t≡0，那么在Q内u≡0.此区域边界由水平集p=0给定，其中p满足Hamilton-Jacobi偏微分方程

$ {p_t} - {\left( {{a^{ij}}\left( x \right){p_{{x_i}}}{p_{{x_j}}}} \right)^{\frac{1}{2}}} = 0,\;\;\;\left( {t,x} \right) \in \left( {0,\infty } \right) \times {R^n}. $

(9)

通过分离变量，简化式(9)，得到

$ p\left( {t,x} \right) = q\left( x \right) + t - {t_0},\;\;\;\;0 \leqslant t \leqslant {t_0},x \in {R^n}, $

(10)

其中q满足

$ \left\{ \begin{array}{l} \sum\limits_{i,j = 1}^n {{a^{ij}}\left( x \right){q_{{x_i}}}{q_{{x_j}}}} = 1,\;\;\;\;\;q > 0,\\ q\left( {{x_0}} \right) = 0, \end{array} \right. $

(11)

事实上，q(x)是以a^ij(x)为度量的黎曼流形上x到x₀的距离.于是可定义倒向的“弯曲锥”：

$ \begin{array}{*{20}{c}} {Q = :\left\{ {\left( {t,x} \right)\left| {p\left( {t,x} \right) < 0} \right.} \right\} = }\\ {\left\{ {\left( {t,x} \right)\left| {q\left( x \right) < {t_0} - t} \right.,} \right.} \end{array} $

(12)

时间t方向的锥截面：

$ {Q_t} = :\left\{ {x\left| {q\left( x \right) < {t_0} - t} \right.} \right\}. $

(13)

相关内容的详细介绍可参阅文献[8].

下文中将顶点(t₀, x₀)移到原点，方便起见，用B(L)定义区域

$ \left\{ {x\left| {q\left( {x,0} \right) < L} \right.} \right\}. $

1 主要结果

定理1 令m>0为常数，假设初值条件{u₀, u₁}满足条件(3)和(4)，那么问题(1)的解满足:

$ \begin{array}{l} \int_{B{{\left( {{{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } \right)}^c}} {\left( {{{\left| {{u_t}} \right|}^2} + {a^{ij}}\left( x \right){u_{{x_i}}}{u_{{x_j}}}} \right){\rm{d}}x} \leqslant \\ \;\;\;\;\;\;{{\rm{e}}^R}E\left( 0 \right){\left( {1 + t} \right)^{ - m}}, \end{array} $

(14)

其中t≥0.进一步，如果u₀+u₁=0，则

$ \begin{array}{*{20}{c}} {\int_{B{{\left( {{{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } \right)}^c}} {{{\left| u \right|}^2}{\rm{d}}x} \leqslant }\\ {{{\rm{e}}^R}\left\| {{u_0}} \right\|_{{L^2}\left( {{R^n}} \right)}^2{{\left( {1 + t} \right)}^{ - m}}.} \end{array} $

(15)

2 衰减估计

根据文献[6]中的方法，由距离函数q(x)和a^ij(x)之间的关系，构造权函数：

$ \varphi \left( {t,x} \right) = \frac{1}{2}\left( {t + L - \sqrt {{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}} } \right), $

通过计算易得

$ \varphi = \frac{1}{2}\frac{{q{{\left( x \right)}^2}}}{{t + L + \sqrt {{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}} }}, $

$ \begin{array}{*{20}{c}} {{\varphi _t} = \frac{1}{2}\left( {1 - \frac{{t + L}}{{\sqrt {{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}} }}} \right) = }\\ { - \frac{\varphi }{{\sqrt {{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}} }},} \end{array} $

$ \begin{array}{*{20}{c}} {\varphi _t^2 = \frac{1}{4}\left( {1 + \frac{{{{\left( {t + L} \right)}^2}}}{{{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}}} - } \right.}\\ {\left. {\frac{{2\left( {t + L} \right)}}{{\sqrt {{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}} }}} \right),} \end{array} $

$ {\varphi _{{x_i}}} = \frac{1}{2}\frac{{q{q_{{x_i}}}}}{{\sqrt {{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}} }}, $

$ {a^{ij}}\left( x \right){\varphi _{{x_i}}}{\varphi _{{x_j}}} = \frac{1}{4}\frac{{q{{\left( x \right)}^2}}}{{{{\left( {t + L} \right)}^2} - q{{\left( x \right)}^2}}}. $

(16)

引理1 对于q(x) < t+L，函数φ(t, x)= $\frac{1}{2}(t + L - \sqrt {{{\left( {t + L} \right)}^2} - q{{(x)}^2})} $满足：

$ \left\{ \begin{array}{l} \varphi \left( {t,x} \right) \geqslant 0,\\ {\varphi _t}\left( {t,x} \right) = {\varphi _t}{\left( {t,x} \right)^2} - {a^{ij}}\left( x \right){\varphi _{{x_i}}}{\varphi _{{x_j}}},\\ \frac{{q{{\left( x \right)}^2}}}{{4\left( {t + L} \right)}} \leqslant \varphi \left( {t,x} \right) \leqslant \frac{{t + L}}{2}, \end{array} \right. $

(17)

此引理可通过计算直接得到，此证略.

引理2 (柯西-施瓦茨不等式)^[8]如果A 是一个n×n的非负对称矩阵，那么对于x, y∈Rⁿ，有

$ \left| {\sum\limits_{i,j = 1}^n {{a^{ij}}{x_i}{y_j}} } \right| \leqslant {\left( {\sum\limits_{i,j = 1}^n {{a^{ij}}{x_i}{x_j}} } \right)^{\frac{1}{2}}}{\left( {\sum\limits_{i,j = 1}^n {{a^{ij}}{y_i}{y_j}} } \right)^{\frac{1}{2}}}. $

(18)

下述衰减估计即为定理1中的式(14).

定理2 令m>0.假定初值条件u₀, u₁满足条件(3)和(4).那么问题(1)的解满足：

$ \begin{array}{l} \int_{B{{\left( {{{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } \right)}^c}} {\left( {{{\left| {{u_t}} \right|}^2} + {a^{ij}}\left( x \right){u_{{x_i}}}{u_{{x_j}}}} \right){\rm{d}}x} \leqslant \\ \;\;\;\;\;\;\;\;\;\;I_1^2{\left( {1 + t} \right)^{ - m}}, \end{array} $

(19)

其中，t≥0，

$ \begin{array}{l} I_1^2 \equiv \int_{{R^n}} {{{\rm{e}}^{2\varphi \left( {0,x} \right)}}\left( {{{\left| {{u_1}\left( x \right)} \right|}^2} + {a^{ij}}\left( x \right){u_{0{x_i}}}{u_{0{x_j}}}} \right){\rm{d}}x} \leqslant \\ \;\;\;\;\;\;\;{{\rm{e}}^L}E\left( 0 \right). \end{array} $

(20)

证明 在方程(1)两边同乘以2e^2φu_t，得到

$ \begin{array}{l} 0 = {{\rm{e}}^{2\varphi }}\left( {\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {u_t^2 + {a^{ij}}{\mathit{u}_{{x_i}}}{\mathit{u}_{{x_j}}}} \right) - } \right.\\ \;\;\;\;\;\left. {2{\partial _{{x_i}}}\left( {{a^{ij}}{\partial _{{x_j}}}u{u_t}} \right) + 2u_t^2} \right) = \\ \;\;\;\;\;\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {{{\rm{e}}^{2\varphi }}\left( {u_t^2 + {a^{ij}}{\mathit{u}_{{x_i}}}{\mathit{u}_{{x_j}}}} \right)} \right) - 2{\partial _{{x_i}}}\left( {{{\rm{e}}^{2\varphi }}{a^{ij}}{\partial _{{x_j}}}u{u_t}} \right) - \\ \;\;\;\;\;\frac{{2{{\rm{e}}^{2\varphi }}}}{{{\varphi _t}}}Y\left( {t,x} \right), \end{array} $

(21)

其中，

$ \begin{array}{l} Y\left( {t,x} \right) = {a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}u_t^2 - 2{\varphi _t}{u_t}{a^{ij}}{\varphi _{{x_i}}}{u_{{x_j}}} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\varphi _t^2{a^{ij}}{\mathit{u}_{{x_i}}}{\mathit{u}_{{x_j}}}. \end{array} $

(22)

然后根据引理2中的柯西-施瓦茨不等式，有

$ \begin{array}{l} Y\left( {t,x} \right) \geqslant {a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}u_t^2 - \\ \left| {2{\varphi _t}{u_t}\sqrt {{a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}} \sqrt {{a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} } \right| + \varphi _t^2{a^{ij}}{u_{{x_i}}}{u_{{x_j}}} = \\ {\left( {\sqrt {{a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}} \left| {{u_t}} \right| - \sqrt {{a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} \left| {{\varphi _t}} \right|} \right)^2} \geqslant 0. \end{array} $

(23)

当x≠0，0代表原点，因为式(16)中φ_t < 0，因此$\frac{{ - Y\left( {t, x} \right)}}{{{\varphi _t}}} \geqslant 0$.当x=0，即x在原点时，有φ_t=0和φ_xi=0，然后结合式(17)的第2个等式，得到

$ - \frac{Y}{{{\varphi _t}}} = u_t^2 \geqslant 0. $

对式(21)在Rⁿ上积分，可得

$ \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\left\| {{{\rm{e}}^\varphi }{u_t}} \right\|_{{L^2}\left( {{R^n}} \right)}^2 + \left\| {{{\rm{e}}^\varphi }\sqrt {{a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} } \right\|_{{L^2}\left( {{R^n}} \right)}^2} \right) \leqslant 0, $

(24)

其中，

$ \begin{array}{l} \left\| {{{\rm{e}}^\varphi }{u_t}} \right\|_{{L^2}\left( {{R^n}} \right)}^2 + \left\| {{{\rm{e}}^\varphi }\sqrt {{a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} } \right\|_{{L^2}\left( {{R^n}} \right)}^2 \leqslant \\ \left\| {{{\rm{e}}^{\varphi \left( {0, \cdot } \right)}}{u_1}} \right\|_{{L^2}\left( {{R^n}} \right)}^2 + \left\| {{{\rm{e}}^{\varphi \left( {0, \cdot } \right)}}\sqrt {{a^{ij}}{u_{0{x_i}}}{u_{0{x_j}}}} } \right\|_{{L^2}\left( {{R^n}} \right)}^2 \equiv I_1^2, \end{array} $

(25)

这里t≥0，由式(17)第3个不等式易知：

$ I_1^2 \leqslant {{\rm{e}}^L}E\left( 0 \right). $

另一方面，由有限传播速度和式(17)第3个不等式，对所有t>0，有

$ \begin{array}{l} \left\| {{{\rm{e}}^\varphi }{u_t}} \right\|_{{L^2}\left( {{R^n}} \right)}^2 + \left\| {{{\rm{e}}^\varphi }\sqrt {{a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} } \right\|_{{L^2}\left( {{R^n}} \right)}^2 = \\ \int_{q\left( x \right) < t + L} {{{\rm{e}}^{2\varphi }}\left( {{{\left| {{u_t}} \right|}^2} + {a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} \right){\rm{d}}x} \geqslant \\ \int_{q\left( x \right) < t + L} {{{\rm{e}}^{\frac{{q{{\left( x \right)}^2}}}{{2\left( {t + L} \right)}}}}\left( {{{\left| {{u_t}} \right|}^2} + {a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} \right){\rm{d}}x} \geqslant \\ \int_{q\left( x \right) \geqslant {{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } {{{\rm{e}}^{\frac{{q{{\left( x \right)}^2}}}{{2\left( {t + L} \right)}}}}\left( {{{\left| {{u_t}} \right|}^2} + {a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} \right){\rm{d}}x} \geqslant \\ {\left( {1 + t} \right)^m}\int_{q\left( x \right) \geqslant {{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } {\left( {{{\left| {{u_t}} \right|}^2} + {a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} \right){\rm{d}}x} . \end{array} $

(26)

因此，由式(25)和(26)可得

$ \begin{array}{l} \int_{q\left( x \right) \geqslant {{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } {\left( {{{\left| {{u_t}} \right|}^2} + {a^{ij}}{u_{{x_i}}}{u_{{x_j}}}} \right){\rm{d}}x} \leqslant \\ \;\;\;\;\;I_1^2{\left( {1 + t} \right)^{ - m}}, \end{array} $

(27)

其中t≥0，式(19)得证.

下述衰减估计即为定理1中的式(15).

定理3  令m>0，假定初值条件{u₀, u₁}满足条件(3)和(4).那么问题(1)的解满足:

$ \int_{B{{\left( {{{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } \right)}^c}} {{{\left| u \right|}^2}{\rm{d}}x} \leqslant I_0^2{\left( {1 + t} \right)^{ - m}}, $

(28)

其中t≥0，

$ I_0^2 \equiv \int_{{R^n}} {{{\rm{e}}^{2\varphi \left( {0,x} \right)}}{u_0}{{\left( x \right)}^2}{\rm{d}}x} \leqslant {{\rm{e}}^R}\left\| {{u_0}} \right\|_{{L^2}\left( {{R^n}} \right)}^2. $

(29)

证明 令

$ v\left( {t,x} \right) = \int_0^t {u\left( {s,x} \right){\rm{d}}s} , $

其中u=u(t, x)是带初值(2)的问题(1)的解.则有v_t=u, v(0, x)=0和

$ {u_t} + u - {\partial _{{x_i}}}\left( {{a^{ij}}{\partial _{{x_j}}}v} \right) = {u_0} + {u_1}, $

(30)

式(30)两边同乘2e^2φu，可得

$ \begin{array}{l} 2{{\rm{e}}^{2\varphi }}\left( {{u_0} + {u_1}} \right)u = {{\rm{e}}^{2\varphi }}\left( {\frac{{\rm{d}}}{{{\rm{d}}t}}{u^2} + 2{u^2} - } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\left. {2{\partial _{{x_i}}}\left( {u{a^{ij}}{\partial _{{x_j}}}v} \right) + \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {{a^{ij}}{\partial _{{x_i}}}{\partial _{{x_j}}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {{{\rm{e}}^{2\varphi }}\left( {{u^2} + {a^{ij}}{v_{{x_i}}}{v_{{x_j}}}} \right)} \right) - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;2{\partial _{{x_i}}}\left( {{{\rm{e}}^{2\varphi }}u{a^{ij}}{\partial _{{x_j}}}v} \right) - \frac{{2{{\rm{e}}^{2\varphi }}}}{{{\varphi _t}}}Z\left( {t,x} \right), \end{array} $

(31)

其中，

$ \begin{array}{l} Z\left( {t,x} \right) = \left( {\varphi _t^2 - {\varphi _t}} \right){u^2} - \\ \;\;\;\;\;\;\;2{\varphi _t}u{a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}} + \varphi _t^2{a^{ij}}{v_{{x_i}}}{v_{{x_j}}} = {a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}{u^2} - \\ \;\;\;\;\;\;\;2{\varphi _t}u{a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}} + \varphi _t^2{a^{ij}}{v_{{x_i}}}{v_{{x_j}}} \geqslant {a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}{u^2} - \\ \;\;\;\;\;\;\;\left| {2{\varphi _t}u\sqrt {{a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}} \sqrt {{a^{ij}}{v_{{x_i}}}{v_{{x_j}}}} } \right| + \varphi _t^2{a^{ij}}{v_{{x_i}}}{v_{{x_j}}} = \\ \;\;\;\;\;\;\;{\left( {\sqrt {{a^{ij}}{\varphi _{{x_i}}}{\varphi _{{x_j}}}} \left| u \right| - \sqrt {{a^{ij}}{v_{{x_i}}}{v_{{x_j}}}} \left| {{\varphi _t}} \right|} \right)^2} \geqslant 0, \end{array} $

(32)

这里用到了柯西-施瓦茨不等式.类似地，当x≠0, 即x不在原点时，因为式(16)中φ_t<0，因此$\frac{{ - Z\left( {t, x} \right)}}{{{\varphi _t}}} \geqslant 0$；当x在原点时，因为φ_t=0，φ_xi=0，所以$\frac{{ - Z}}{{{\varphi _t}}} = {u^2} \geqslant 0$.

因此式(31)两边在Rⁿ上积分并利用有限传播速度，可得

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\left\| {{{\rm{e}}^\varphi }u} \right\|_{{L^2}\left( {{R^n}} \right)}^2 + \left\| {{{\rm{e}}^\varphi }\sqrt {{a^{ij}}{v_{{x_i}}}{v_{{x_j}}}} } \right\|_{{L^2}\left( {{R^n}} \right)}^2} \right) \leqslant \\ \;\;\;\;\;\;\;\;2\int_{{R^n}} {{{\rm{e}}^{2\varphi }}\left( {{u_0} + {u_1}} \right)u{\rm{d}}x} , \end{array} $

(33)

若u₀+u₁=0，可得

$ \left\| {{{\rm{e}}^\varphi }u} \right\|_{{L^2}\left( {{R^n}} \right)}^2 \leqslant \left\| {{{\rm{e}}^{\varphi \left( {0, \cdot } \right)}}{u_0}} \right\|_{{L^2}\left( {{R^n}} \right)}^2 \equiv I_0^2, $

(34)

其中t≥0.由式(17)第3个不等式，易知

$ I_0^2 \leqslant {{\rm{e}}^R}\left\| {{u_0}} \right\|_{{L^2}\left( {{R^n}} \right)}^2. $

另一方面，与定理2的证明类似，可得

$ \begin{array}{l} \left\| {{{\rm{e}}^\varphi }u} \right\|_{{L^2}\left( {{R^n}} \right)}^2 = \int_{q\left( x \right) < t + L} {{{\rm{e}}^{2\varphi }}{{\left| u \right|}^2}{\rm{d}}x} \geqslant \\ \;\;\;\;\;\int_{q\left( x \right) < t + L} {{{\rm{e}}^{\frac{{q{{\left( x \right)}^2}}}{{2\left( {t + L} \right)}}}}{{\left| u \right|}^2}{\rm{d}}x} \geqslant \\ \;\;\;\;\;\int_{q\left( x \right) \geqslant {{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } {{{\rm{e}}^{\frac{{q{{\left( x \right)}^2}}}{{2\left( {t + L} \right)}}}}{{\left| u \right|}^2}{\rm{d}}x} \geqslant \\ \;\;\;\;\;{\left( {1 + t} \right)^m}\int_{q\left( x \right) \geqslant {{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } {{{\left| u \right|}^2}{\rm{d}}x} . \end{array} $

(35)

于是，由式(34)和(35)可得

$ \int_{q\left( x \right) \geqslant {{\left( {t + L} \right)}^{\frac{1}{2}}}\sqrt {2\log {{\left( {1 + t} \right)}^m}} } {{{\left| u \right|}^2}{\rm{d}}x} \leqslant I_0^2{\left( {1 + t} \right)^{ - m}}, $

(36)

其中t≥0.定理3得证.

定理1证明 由定理2和定理3可得定理1.