非线性扩散波关于可压缩微极流体方程解的渐近稳定性

Asymptotic Stability of Nonlinear Diffusion Wave Solutions for Compressible Micropolar Fluid Equations

摘要: 本文研究一维可压缩等熵微极流体方程解的大时间行为，我们证明在初始扰动和波的强度适当小的条件下，当时间时，该方程的解收敛于平面扩散波，并得到了相应的衰减速度，本文主要的研究方法为能量方法，结合了反导数法、Cauchy不等式和Young-不等式证明了非线性扩散波关于可压缩微极流体方程解的渐近稳定性。

Abstract: In this paper, we study the large time behavior of the solution of one dimensional compressible isentropic micropolar fluid equation. Under the condition of the initial disturbance and wave intensity being appropriately small, at time , we prove that the solution of the equation converges to plane diffusion wave , and the corresponding decay rate is obtained. The main research method in this paper is the energy method, which combines the inverse derivative method, Cauchy inequality and Young inequality. By using these methods, we prove the asymptotic stability of the solution of the nonlinear diffusion wave to the compressible micropolar fluid equation.

1. 引言

在本文中我们研究如下具有粘度相关系数的一维可压缩等熵微极流体方程的大时间行为

$\{\begin{array}{l}{v}_t-u_x=0\hfill \\ u_t+p{\left(v\right)}_x=-\alpha ux\in ℝ,t>0\hfill \\ {\omega }_t={\left(\frac{A\left(v\right){\omega }_x}{v}\right)}_x-4{\mu }_r\left(v\right)v\omega ,\hfill \end{array}$ (1.1)

给定初值

$\left(v,u,\omega \right)\left(x,0\right)=\left(v_0,u_0,{\omega }_0\right)\left(x\right),$ (1.2)

且满足条件

$\underset{x\to \pm \infty }{\lim }\left(v_0,u_0,{\omega }_0\right)\left(x\right)=\left(v_{\pm },u_{\pm },{\omega }_{\pm }\right),$ (1.3)

此处t和x分别代表时间和空间变量。这里v表示比容，u表示速度， $\omega$ 表示微旋转速度， $p\left(v\right)$ 表示压力， $\alpha >0$ 表示阻尼系数。

在这篇文章中

$p\left(v\right)=v^{-\gamma },\gamma \ge 1,\alpha \ge 0,$

并且 $A\left(v\right),{\mu }_r\left(v\right)$ 是关于v的光滑的正函数 $\left(v>0\right)$，表示微粘度系数。

如果忽略流体的微观结构，即 $\omega =0$，系统(1.1)可视为拉格朗日坐标下带阻尼的等熵欧拉方程，动量方程中带有阻尼项 $\alpha u$，可用于模拟多孔介质中的可压缩流。

此时，系统(1.1)等同系统

$\{\begin{array}{l}{v}_t-u_x=0\hfill \\ u_t+p{\left(v\right)}_x=-\alpha u,\hfill \end{array}$ (1.4)

初值满足

$\left(u\left(x,0\right),v\left(x,0\right)\right)=\left(u_0\left(x\right),v_0\left(x\right)\right)\to \left(u_{\pm },v_{\pm }\right),\text{as}x\to \pm \infty$ (1.5)

当 $v_{-}\ne v_{+}$，此时方程有波的现象。刘太平 [1] [2] 证明了在初始扰动和波的强度 $\left|u_{+}-u_{-}\right|$ 较小的情况下，当 $t\to \infty$ 时，

$\left|v_{+}-v_{-}\right|<\delta ,\left(v_{+}\ne v_{-}\right),$

方程(1.4)的解趋向于非线性扩散波 $\left(\bar{v},\bar{u}\right)$。其中 $\left(\bar{v},\bar{u}\right)$ 满足方程

$\{\begin{array}{l}{\bar{v}}_t-{\bar{u}}_x=0,\hfill \\ p{\left(\bar{v}\right)}_x=-\alpha \bar{u},x\in ℝ,t>0\hfill \end{array}$ (1.6)

而对于带阻尼的非线性发展方程的柯西问题，已经有许多关于基本波的渐近稳定性的研究，参考文献 [3] [4] [5] [6]。如在一维情况下，刘太平 [5] [6] 先证明了在初始扰动和波的强度满足小性条件的情况下，非线性扩散波关于带阻尼的双曲守恒律方程的渐近稳定性。对于方程(1.4)，Nishihara在 [7] [8] 进一步利用能量方法得到了和刘太平相同的结果，并获得在 $L^2$ 和 $L^{\infty }$ 范数下的收敛速度。对于小性条件下收敛速度其他相关结论，我们参考 [9] [10]，与此同时，也有很多作者针对初始扰动较大的情况进行了研究，如Zhao [11] 证明了对于给定的大初始值情况下，带摩擦阻尼的p系统存在唯一的整体光滑解，且当时间 $t\to \pm \infty$ 时，该方程的解渐近趋向于对应的非线性扩散波 $\left(\bar{v},\bar{u}\right)\left(x,t\right)$，并且他还得到了 $L^p\left(2\le p\le \infty \right)$ 范数下的衰减速率。对于具有非线性阻尼的p系统的柯西问题的其他结果，我们参考 [12] - [23]。

我们的方程与(1.4)相比，增加一个方程，考虑了微旋转速度 $\omega$，即思考了微观结构该模型的渐近行为，增加了耦合性和研究该模型的难度，这也是本文的创新之处。

本文研究了一维可压缩微极流体模型(1.1)问题解的大时间行为，该模型具有一般的微粘度相关系数。我们将证明，在粘度系数 $A\left(v\right),{\mu }_r\left(v\right),\left(v>0\right)$ 是关于v的光滑的正函数，并且初始扰动和波的强度足够小的情况下，那么方程(1.1)~(1.3)的解整体存在，并且当 $t\to \infty$ 时，该解收敛到非线性扩散波 $\left(\bar{v},\bar{u},0\right)\left(x,t\right)$。文章主要结果的精确陈述在下面的定理1.1。

定理1.1设 $A\left(v\right),{\mu }_r\left(v\right),\left(v>0\right)$ 是关于v的光滑正函数，设 $\left(v_0,u_0,{\omega }_0\right)\left(x\right)\in H^2\times H^2\times H^2$，则存在适当小的 $\delta >0$，当 $\left|u_{+}-u_{-}\right|+\left|v_{+}-v_{-}\right|+{\Vert \left(v-\bar{v}\right)\left(x\right)\Vert }_2+{\Vert \left(u-\bar{u}\right)\left(x\right)\Vert }_2+{\Vert \omega \left(x\right)\Vert }_2\le \delta$ 时，方程(1.1)~(1.3)的解 $\left(v,u,\omega \right)$ 全局存在，并且满足

$v-\bar{v}-m\in C\left(0,\infty ,H^2\right),u-\bar{u}-\hat{u}\in C\left(0,\infty ,H^2\right),\omega \in C\left(0,\infty ,H^2\right),$

$v_x-{\bar{v}}_x-m_x\in L^2\left(0,\infty ,H^1\right),u_x-{\bar{u}}_x-\hat{u}\in L^2\left(0,\infty ,H^1\right),{\omega }_x\in L^2\left(0,\infty ,H^1\right),$

当t趋于无穷时，进一步有以下结论：

$\begin{array}{l}1){\Vert {\partial }_x^k\left(v\left(x,t\right)-\bar{v}\left(x+x_0,t\right)-m\right)\Vert }_{L^2}=O\left(1\right)\delta {\left(1+t\right)}^{-\frac{k}{2}},\\ 2){\Vert {\partial }_x^k\left(u-\bar{u}-\hat{u}\right)\left(x,t\right)\Vert }_{L^2}=O\left(1\right)\delta {\left(1+t\right)}^{-\frac{k}{2}},\\ 3){\Vert {\partial }_x^k\omega \left(x,t\right)\Vert }_{L^2}=O\left(1\right)\delta {\left(1+t\right)}^{-\frac{k}{2}},k=1,2\end{array}$ (1.9)

$\begin{array}{l}4){\Vert \left(v-\bar{v}-m\right)\left(t\right)\Vert }_{L^{\infty }}=O\left(1\right)\delta {\left(1+t\right)}^{-\frac{1}{4}},\\ 5){\Vert \left(u-\bar{u}-\hat{u}\right)\left(t\right)\Vert }_{L^{\infty }}=O\left(1\right)\delta {\left(1+t\right)}^{-\frac{1}{4}},\\ 6){\Vert \omega \left(t\right)\Vert }_{L^{\infty }}=O\left(1\right)\delta {\left(1+t\right)}^{-\frac{1}{4}}.\end{array}$ (1.10)

其中 $x_0$ 满足

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(v\left(x,0\right)-\bar{v}\left(x+x_0,t\right)\right)\text{d}x}=\frac{u_{+}-u_{-}}{-\alpha },$ (1.11)

并且

$\begin{array}{l}\hat{u}\equiv u_{-}{\text{e}}^{-\alpha t}+{\displaystyle {\int }_{-\infty }^x{m}_t\left(\eta ,t\right)\text{d}\eta },\\ {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{m}_0\left(x\right)\text{d}x}=1,\\ m\left(x,t\right)\equiv -\frac{u_{+}-u_{-}}{-\alpha }{m}_0\left(x,t\right){\text{e}}^{-\alpha t}.\end{array}$ (1.12)

本文的安排如下：在第二部分，我们给出一些在证明过程中所需相关引理；在第三部分，我们对问题(1.1)~(1.3)的扰动方程进行基本能量估计和时间加权估计，从而证明定理1.1。

标记在文章中，我们用 $C,O\left(1\right)$ 表示正常数。对于函数空间来说， $L^P=L^P\left({ℝ}^n\right)\left(1\le p\le \infty \right)$ 表示一般的Lebesgue空间，其范数为

${\Vert f\Vert }_{L^p}={\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^P}\text{d}x\right)}^{\frac{1}{p}},$ (1.13)

${\left|f\right|}_{L^{\infty }}=\underset{,-\infty <x<\infty }{\sup }\left|f\left(x\right)\right|,1\le p<\infty ,$ (1.14)

并且积分区域R将会被省略，而不产生混淆概念的后果，用 $H^m\left(m\ge 0\right)$ 用来表示m阶Sobolev空间，其范数为

$\Vert f\Vert ={\left({\displaystyle \underset{k=0}{\overset{m}{\sum }}{\Vert {\partial }_x^{k}f\Vert }^2}\right)}^{\frac{1}{2}},\Vert \cdot \Vert ={\Vert \cdot \Vert }_0={\Vert \cdot \Vert }_{L^2}.$ (1.15)

2. 准备工作

非线性扩散方程，其中一自相似解 $\left(\bar{v},\bar{u}\right)\left(\frac{x}{\sqrt{1+t}}\right)$ 满足

$\{\begin{array}{l}{\bar{v}}_t-{\bar{u}}_x=0,\hfill \\ p{\left(\bar{v}\right)}_x=-\alpha \bar{u},x\in ℝ,t>0\hfill \end{array}$ (2.1)

命题2.1 (参考 [24] )对于上述该自相似解，当 $x\to \pm \infty$ 时，有 $v\left(\pm \infty ,t\right)=v_{\pm }$，让 $\zeta =\frac{x}{\sqrt{1+t}}$，则有以下性质：

$1){\left|\bar{v}\left(\zeta \right)-v_{-}\right|}_{\zeta >0}+{\left|\bar{v}\left(\zeta \right)-v_{+}\right|}_{\zeta <0}\le C\left|v_{+}-v_{-}\right|{\text{e}}^{-\mu {\zeta }^2},$ (2.2)

$2)\left|{\partial }_x^r{\partial }_t^l\bar{v}\right|\le C\left|v_{+}-v_{-}\right|{\left(1+t\right)}^{-\frac{r+2l}{2}}{\text{e}}^{-\frac{\mu {\zeta }^2}{1+t}},r+l\ge 1,r,l\ge 0,$ (2.3)

$3)\left|{\partial }_x^r{\partial }_t^l\bar{u}\right|\le C\left|v_{+}-v_{-}\right|{\left(1+t\right)}^{-\frac{r+2l+1}{2}}{\text{e}}^{-\frac{\mu {\zeta }^2}{1+t}},r,l\ge 0.$ (2.4)

其中C是大于0的常数， $r,l$ 是正整数，该引理在下面第三部分的证明中起了一定的作用。

3. 定理1.1的证明

为了证明定理1.1，我们首先构造扰动方程

令

$\varphi \left(x,t\right)=v\left(x,t\right)-\bar{v}\left(x+x_0,t\right)-m\left(x,t\right),$

$\psi \left(x,t\right)=u\left(x,t\right)-\bar{u}\left(x+x_0,t\right)-\hat{u}\left(x,t\right),$

$\omega \left(x,t\right)=\omega \left(x,t\right),$

其中

$\begin{array}{l}\hat{u}\equiv u_{-}{\text{e}}^{-\alpha t}+{\displaystyle \underset{-\infty }{\overset{x}{\int }}{m}_t\left(\eta ,t\right)\text{d}\eta },\\ {\displaystyle {\int }_{-\infty }^{\infty }{m}_0\left(x\right)\text{d}x}=1,\\ m\left(x,t\right)\equiv -\frac{u_{+}-u_{-}}{-\alpha }{m}_0\left(x,t\right){\text{e}}^{-\alpha t},\end{array}$

我们得到以下扰动方程

$\{\begin{array}{l}{\varphi }_t-{\psi }_x=0\\ {\psi }_t+{\left[p\left(\varphi +\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_x+{\bar{u}}_t+\alpha \psi =0\\ {\omega }_t={\left(\frac{A\left(v\right){\omega }_x}{v}\right)}_x-4{\mu }_r\left(v\right)v\omega ,\end{array}$ (3.1)

初值满足

$\begin{array}{c}{\left(\varphi \left(t,x\right),\psi \left(t,x\right),\omega \left(t,x\right)\right)|}_{t=0}=\left({\varphi }_0\left(x\right),{\psi }_0\left(x\right),{\omega }_0\left(x\right)\right)\\ =\left(v_0\left(t\right)-{\bar{v}}_0\left(t\right)-m_0\left(t\right),u_0\left(t\right)-{\bar{u}}_0\left(t\right)-{\hat{u}}_0\left(t\right),{\omega }_0\left(x\right)\right)\end{array}$ (3.2)

我们令

$y\left(x,t\right)={\displaystyle \underset{-\infty }{\overset{x}{\int }}\varphi \left(\eta ,t\right)\text{d}\eta },$

注意到

$y\left(\infty ,t\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\varphi \left(\eta ,t\right)\text{d}\eta }=0=y\left(-\infty ,t\right),$

那么根据(3.1)₁，可以得到

$y_x=\varphi ,$ (3.3)

$y_t={\displaystyle \underset{-\infty }{\overset{x}{\int }}{\varphi }_t\left(\eta ,t\right)\text{d}\eta }={\displaystyle \underset{-\infty }{\overset{x}{\int }}{\psi }_x\left(\eta ,t\right)\text{d}\eta }=\psi ,$ (3.4)

则扰动方程(3.1)转换为以下形式

$\{\begin{array}{l}{y}_{tt}+{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_x+\alpha y_t-\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xt}=0,\hfill \\ {\omega }_t={\left(\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right)}_x-4{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega ,\hfill \end{array}$ (3.5)

初值为

$\left(y,\omega \right)\left(x,0\right)=\left(y_0\left(x\right),{\omega }_0\left(x\right)\right),y_t\left(x,0\right)=y_1,$ (3.6)

由于定理1.1的条件以及(3.2)知

$y\left(x,0\right)\in H^2\left(ℝ\right),y_t\left(x,0\right)\in H^1\left(ℝ\right),\omega \left(x,0\right)\in H^2\left(ℝ\right),$ (3.7)

通过以上分析，我们知道，只要证明问题(3.5)~(3.6)的解的整体存在性，即可证明问题(1.1)~(1.2)的解的整体存在性，因此问题转化为如何证明(3.5)~(3.6)解的整体存在性并得到相应的衰减估计。

定义初值问题(3.5)~(3.6)的解空间如下：

$X\left(0,T\right)=\{y\in L^{\infty }\left(0,T;H^3\right),\omega \in L^{\infty }\left(0,T;H^2\right)\cap L^2\left(0,T;H^1\right),y_x\in L^2\left(0,T;H^2\right)\}$

定理3.1 (整体存在性)设初值满足(3.7)，则存在与 $u_{\pm },v_{\pm }$ 无关的 $\tilde{\delta }$，使得当

$\left|u_{+}-u_{-}\right|+\left|v_{+}-v_{-}\right|+N_0\le \tilde{\delta }$ (3.8)

时，初值问题(3.5)~(3.6)的解 $\left(y,\omega \right)$ 在 $X\left[0,\infty \right]$ 中全局存在且唯一。

根据偏微分方程的经典理论，初值问题(3.5)~(3.6)的解局部存在且唯一。

定理3.2 (局部存在性)假设 $\left(y_0,{\omega }_0\right)\in H^3\times H^2$，且 $N_0\le \frac{{\epsilon }_0}{2}$，则存在正常数 $T_0$，其中

$N_0={\Vert y_0\Vert }_{H^2}+{\Vert {\omega }_0\Vert }_{H^2}+{\Vert y_1\Vert }_{H^1},$ (3.9)

使得问题(3.5)~(3.6)在 $\left[0,T_0\right]$ 上有唯一解 $\left(y,\omega \right)\in X\left(0,T_0\right)$，且满足

${\Vert \left(y,\omega \right)\Vert }_{H^2}^2\le C_1\left({\Vert \left(y_0,{\omega }_0\right)\Vert }_{H^2}^2+{\Vert y_t\left(0\right)\Vert }_{H^1}^2\right),$ (3.10)

为了研究解的全局存在性和解的大时间渐近行为，我们只需要证明解的一致先验估计。为此，我们做如下先验假设：

先验假设3.1对于时间 $T>0$，设 $\left(y,\omega \right)$ 是问题(3.5)~(3.6)问题的解，并且满足

$N\left(T\right)=:{\displaystyle \underset{k=0}{\overset{1}{\sum }}{\left(1+k\right)}^k{\Vert {\partial }_t^{k}y\left(\cdot ,t\right)\Vert }^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+k\right)}^k{\Vert {\partial }_x^{k}y\left(\cdot ,t\right)\Vert }^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+k\right)}^k{\Vert {\partial }^k\omega \left(\cdot ,t\right)\Vert }^2}\le C{\epsilon }^2,$ (3.11)

其中C是大于0的常数， $\epsilon$ 是依赖于初值和波的强度的某个充分小的正常数。

根据先验假设3.1以及Sobolev引理，有

$\begin{array}{l}{\Vert y\Vert }_{L^{\infty }}\le C{\Vert y\Vert }_{L^2}^{\frac{1}{2}}{\Vert y_x\Vert }_{L^2}^{\frac{1}{2}}\le C\delta {\left(1+t\right)}^{-\frac{1}{4}},\\ {\Vert y_x\Vert }_{L^{\infty }}\le C{\Vert y_x\Vert }_{L^2}^{\frac{1}{2}}{\Vert y_{xx}\Vert }_{L^2}^{\frac{1}{2}}\le C\delta {\left(1+t\right)}^{-\frac{3}{4}},\end{array}$ (3.12)

同样地，

${\Vert \omega \Vert }_{L^{\infty }}\le C\delta {\left(1+t\right)}^{-\frac{1}{4}},{\Vert {\omega }_x\Vert }_{L^{\infty }}\le C\delta {\left(1+t\right)}^{-\frac{3}{4}}.$

在先验假设3.1下，可证明如下先验估计：

定理3.3 (先验估计)设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，则存在与 $u_{\pm },v_{\pm }$ 无关的 ${\delta }_2$，使得 $\left|u_{+}-u_{-}\right|+\left|v_{+}-v_{-}\right|+N_0\le {\delta }_2$ 时，下面估计对 $t\in \left[0,T\right]$ 成立

$N{\left(t\right)}^2+{\displaystyle {\int }_0^t\left({\Vert y_x\Vert }_{H^1}^2+{\Vert y_t\Vert }_{H^1}^2+{\Vert {\omega }_x\Vert }_{H^1}^2\right)\text{d}t}\le C_2\left(N_0^2+{\delta }^2\right).$ (3.13)

其中 $N\left(t\right)={\Vert \left(y,\omega \right)\Vert }_{H^2}+{\Vert y_t\Vert }_{H^1}$。

下面我们将通过对 $\left(y,\omega \right)$ 进行低阶、高阶以及时间衰减估计来证明(3.13)式。

证明过程由下面几个引理构成。

引理3.1设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y^2+y_x^2+y_t^2\right)}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y_t^2+y_x^2\right)\text{d}x\text{d}t}}\le C{N}_0^2,$ (3.14)

证明：

在(3.5)₁两端同时乘以y，并在 $\left[0,T\right]\times R$ 上进行积分，得到

$\begin{array}{l}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\alpha \frac{y^2}{2}\text{d}x|{}_0^T}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}y{y}_t\text{d}x|{}_0^T}-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_t^2\text{d}x}}\text{d}t\\ -{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]\text{d}x}}\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\alpha }{y}_{x}p{\left(\bar{v}\right)}_t\text{d}x}}\text{d}t=0\end{array}$ (3.15)

由于 $p^{\prime }<0$，当 $\delta$ 足够小时，利用Cauchy-Schwarz不等式，由上式可得

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^2\left(x,T\right)\text{d}x}+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2\text{d}x}}\text{d}t=O\left(1\right)\left[{\delta }^2+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_t^2\left(x,T\right)\text{d}x}+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_t^2\text{d}x}}\text{d}t\right]$ (3.16)

在(3.5)₁两端同时乘以 $y_t$，并在 $\left[0,T\right]\times R$ 进行积分，得到

$\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_t^2\text{d}x|{}_0^T}+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\alpha y_t^2\text{d}x}}\text{d}t-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]\text{d}x}}\text{d}t-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xt}{y}_t\text{d}x}}\text{d}t=0.$ (3.17)

令

$q\left(x,t\right)\equiv -{\displaystyle \underset{0}{\overset{y_x}{\int }}\left[p\left(\eta +\bar{v}\left(x,t\right)\right)-p\left(\bar{v}\left(x,t\right)\right)\right]\text{d}\eta },$

$-q_t\equiv y_{xt}\left[p\left(y_x+\bar{v}\right)-p\left(\bar{v}\right)\right]+{\displaystyle \underset{0}{\overset{y_x}{\int }}\left[p^{\prime }\left(\eta +\bar{v}\right)-p^{\prime }\left(\bar{v}\right)\right]{\bar{v}}_t\text{d}\eta },$

由于 $p^{\prime }<0$，则存在 ${\tilde{C}}_1,{\tilde{C}}_2>0$，使得

${\tilde{C}}_2{\left(y_x\right)}^2\le q\le {\tilde{C}}_1{\left(y_x\right)}^2$

从而

$\begin{array}{l}-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}\left[p\left(y_x+\bar{v}\right)-p\left(\bar{v}\right)\right]\text{d}x}}\text{d}t\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}q\text{d}x}|{}_0^T+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{y_x}{\int }}\left[p^{\prime }\left(\eta +\bar{v}\right)-p^{\prime }\left(\bar{v}\right)\right]\left|{\bar{v}}_t\right|\text{d}\eta \text{d}x\text{d}t}}}\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}q\text{d}x}|{}_0^T+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}O\left(1\right)y_x^2\left|{\bar{v}}_t\right|\text{d}x\text{d}t}}\\ \le {\tilde{C}}_1{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2\left(x,T\right)\text{d}x}-{\tilde{C}}_2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2\left(x,0\right)\text{d}x}+O\left(1\right)\delta {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2}\text{d}x\text{d}t},\end{array}$ (3.18)

根据(3.13)，(3.17)满足

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y_x^2+y_t^2\right)}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_t^2}\text{d}x\text{d}t}=O\left(1\right){\delta }^2+O\left(1\right)\delta {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2}\text{d}x\text{d}t},$ (3.19)

则结合等式(3.16)得到基本估计

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y^2+y_x^2+y_t^2\right)}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y_t^2+y_x^2\right)\text{d}x\text{d}t}}\le C{N}_0^2,$ (3.20)

引理3.2设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_0^2}\text{d}x,$ (3.21)

证明：

将方程(3.5)₂乘上 $\omega \left(x,t\right)$，得到

$\omega {\omega }_t=\omega {\left(\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right)}_x-4{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right){\omega }^2,$ (3.22)

将(3.22)关于 $\left(t,x\right)$ 在 $\left[0,T\right]\times R$ 上积分，我们得到

$\begin{array}{l}\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_x^2}}\text{d}x\text{d}t\\ +4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right){\omega }^2}}\text{d}x\text{d}t=\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_0^2}\text{d}x,\end{array}$ (3.23)

根据先验假设(3.10)以及引理2.1，存在一个大于0的常数 $M^{\prime }$，使得

$\frac{1}{M^{\prime }}\le v=y_x+\bar{v}+m\le M^{\prime },$

那么在方程(3.23)中，我们有

${\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_x^2}}\text{d}x\text{d}t\ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t,$ (3.24)

$4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right){\omega }^2}}\text{d}x\text{d}t\ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}}\text{d}x\text{d}t,$ (3.25)

根据(3.24)、(3.25)，并结合(3.23)，整理得到

${\displaystyle \underset{R}{\int }{\omega }^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_0^2}\text{d}x,$ (3.26)

则有

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t\le C{N}_0^2,$ (3.27)

现在我们来证明 ${\omega }_x$ 的高阶估计。

引理3.3设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x,$ (3.28)

证明：

将方程(3.5)₂对x求导，再乘上 ${\omega }_x$

${\omega }_x{\omega }_{xt}={\omega }_x{\left[\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right]}_{xx}-4{\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_x,$ (3.29)

将(3.29)关于 $\left(t,x\right)$ 在 $\left[0,T\right]\times R$ 上积分得到，有

$\begin{array}{l}\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}-\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}}}{\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_x\right]}_x\text{d}x\text{d}t\\ +4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_x}}\text{d}x\text{d}t=\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x,\end{array}$ (3.30)

其中

$\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}}}{\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_x\right]}_x\text{d}x\text{d}t\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right]}_x{\omega }_x{\omega }_{xx}}}+\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_{xx}^2\text{d}x\text{d}t\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[\frac{\left(y_{xx}+{\bar{v}}_x+m_x\right)A^{\prime }\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}-\frac{\left(y_{xx}+{\bar{v}}_x+m_x\right)A\left(y_x+\bar{v}+m\right)}{{\left(y_x+\bar{v}+m\right)}^2}\right]{\omega }_x{\omega }_{xx}\text{d}x\text{d}t}}\\ +{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_{xx}^2\text{d}x\text{d}t}}\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}}\text{d}x\text{d}t-C\left(\epsilon +\delta \right){\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}}\text{d}x\text{d}t-C\left(\epsilon +\delta \right){\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t,\end{array}$ (3.31)

并且

$\begin{array}{l}4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_x}}\text{d}x\text{d}t\\ \ge 4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\omega \right]}_x}}\text{d}x\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\mu }_r\left(y_x+\bar{v}+m\right){\omega }_x^2+{{\mu }^{\prime }}_r\left(y_x+\bar{v}+m\right)}}\left(y_{xx}+{\bar{v}}_x+m_x\right)\omega {\omega }_x\text{d}x\text{d}t\\ \ge C\left(\epsilon +\delta \right){\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2\text{d}x\text{d}t}}+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2\text{d}x\text{d}t}},\end{array}$ (3.32)

把(3.31)、(3.32)，代入(3.30)，取 $\epsilon ,\delta$ 适当小，可得

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x\text{d}t-C\left(\epsilon +\delta \right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\text{d}x\text{d}t,$ (3.33)

应用柯西不等式，并利用(3.21)有

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_0^2}\text{d}x,$ (3.34)

由初值条件以及引理3.2则有

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}}\text{d}x\text{d}t\le C{N}_0^2,$ (3.35)

引理3.4设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\left(x,T\right)\text{d}x}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\text{d}x}}\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2\text{d}x}}\text{d}t\le C{\delta }^2,$ (3.36)

证明：

对(3.5)₁有关x求导，同时乘以 $y_x$ 得到

$y_{xtt}{y}_x+{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_{xx}{y}_x+\alpha y_x{y}_{xt}-\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xxt}{y}_x=0$

并在 $\left[0,T\right]\times R$ 上进行积分，得到

$\begin{array}{l}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}{y}_x\text{d}x}|{}_0^T-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\text{d}x\text{d}t}}+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_{xx}{y}_x\text{d}x}}\text{d}t\\ +\frac{\alpha }{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2}\left(x,T\right)\text{d}x-\frac{\alpha }{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2}\left(x,0\right)\text{d}x-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xxt}{y}_x\text{d}x\text{d}t}}=0,\end{array}$ (3.37)

其中

$\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_{xx}{y}_x\text{d}x}}\text{d}t\\ =-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(y_x+\bar{v}+m\right)}_x{y}_{xx}\text{d}x}}\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_x{y}_{xx}\text{d}x}}\text{d}t\\ =-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{p}^{\prime }\left(y_x+\bar{v}+m\right)\left(y_{xx}+{\bar{v}}_x+m_x\right)y_{xx}\text{d}x}}\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_x{y}_{xx}\text{d}x}}\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y_{xx}+{\bar{v}}_x+m_x\right)y_{xx}\text{d}x}}\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_x{y}_{xx}\text{d}x}}\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2\text{d}x}}\text{d}t,\end{array}$ (3.38)

$\begin{array}{l}-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xxt}{y}_x\text{d}x\text{d}t}}\\ =-\frac{1}{\alpha }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_{xx}{y}_{xt}\text{d}x|{}_0^T}+\frac{1}{\alpha }{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_{xx}{y}_{xt}\text{d}x\text{d}t}}\\ \ge C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_{xx}{y}_{xt}\text{d}x|{}_0^T}+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[p^{″}\left(\bar{v}\right){\bar{v}}_x^2+p^{\prime }\left(\bar{v}\right){\bar{v}}_{xx}\right]y_{xt}\text{d}x\text{d}t}}\\ \ge C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_{xx}{y}_{xt}\text{d}x|{}_0^T}+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\text{d}x\text{d}t}},\end{array}$ (3.39)

同样地对(3.5)₁有关x求导并乘以 $y_{xt}$，在 $\left[0,T\right]\times R$ 上进行积分，得到

$\begin{array}{l}\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\left(x,t\right)\text{d}x}|{}_0^T+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_{xx}{y}_{xt}\text{d}x}}\text{d}t\\ +{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\alpha y_{xt}^2\text{d}x\text{d}t}}-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xxt}{y}_{xt}\text{d}x\text{d}t}}=0,\end{array}$ (3.40)

其中

$\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_{xx}{y}_{xt}\text{d}x}}\text{d}t\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(y_x+\bar{v}+m\right)}_{xx}{y}_{xt}\text{d}x}}\text{d}t-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_{xx}{y}_{xt}\text{d}x}}\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y_{xx}+{\bar{v}}_x+m_x\right)y_{xxt}\text{d}x}}\text{d}t-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_{xx}{y}_{xt}\text{d}x}}\text{d}t\\ \ge C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2}\text{d}x|{}_0^T-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left({\bar{v}}_{xx}+m_{xx}\right)y_{xt}\text{d}x}}\text{d}t-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}p{\left(\bar{v}\right)}_{xx}{y}_{xt}\text{d}x}}\text{d}t\\ \ge C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2}\text{d}x|{}_0^T+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\text{d}x}}\text{d}t,\end{array}$ (3.41)

$-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xxt}{y}_{xt}\text{d}x\text{d}t}}\ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\text{d}x\text{d}t}},$ (3.42)

将(3.38)~(3.39)代入(3.37)，同样地将(3.41)~(3.42)代入(3.40)，结合得到

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\left(x,T\right)\text{d}x}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\text{d}x}}\text{d}t\le {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\left(x,0\right)\text{d}x}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2\left(x,0\right)\text{d}x}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2\left(x,0\right)\text{d}x},$ (3.43)

则得到以下了基本衰减估计：

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\left(x,T\right)\text{d}x}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2\text{d}x}}\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xx}^2\text{d}x}}\text{d}t\le C{N}_0^2,$ (3.44)

接下来，我们进行有关时间衰减估计的证明。

引理3.5设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

$\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y_x^2+y_t^2\right)\left(x,T\right)\text{d}x}\le C{\delta }^2,$ (3.45)

证明：

在(3.5)₁两端同时乘以 $\left(1+t\right)y_t$，并在 $\left[0,T\right]\times R$ 上进行积分，得到

${\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{y}_{tt}\text{d}x\text{d}t}}+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{\left[p^{\prime }\left(\bar{v}\right)y_x\right]}_x\text{d}x}}\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\alpha \left(1+t\right)y_t^2\text{d}x\text{d}t}}={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}F\left(1+t\right)y_t\text{d}x\text{d}t}}$ (3.46)

这里

$F=\frac{1}{\alpha }p{\left(\bar{v}\right)}_{xt}+{\left[p^{\prime }\left(\bar{v}\right)y_x\right]}_x-{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)\right]}_x$

将(3.46)整合得到

$\begin{array}{l}\frac{1}{2}\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_t^2}\left(x,T\right)\text{d}x+\frac{1}{2}\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{p}^{\prime }\left(\bar{v}\right)y_x^2\left(x,T\right)\text{d}x}+\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t^2}\text{d}x\text{d}t}\\ =-\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{p}^{\prime }\left(\bar{v}\right)y_x^2}\text{d}x\text{d}t}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1+t}{2}{p}^{″}\left(\bar{v}\right)y_x^2{\bar{v}}_t}\text{d}x+\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_{t}F}\text{d}x\text{d}t}\\ \le C{\delta }^2+\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_{t}F}\text{d}x\text{d}t},\end{array}$ (3.47)

其中

$\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_{t}F}\text{d}x\text{d}t}={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_{t}p{\left(\bar{v}\right)}_{xt}}\text{d}x\text{d}t}-\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)-p^{\prime }\left(\bar{v}\right)y_x\right]}_x}\text{d}x\text{d}t},$ (3.48)

在上式中

$\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_{t}p{\left(\bar{v}\right)}_{xt}}\text{d}x\text{d}t}\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t\left[p^{\prime }\left(\bar{v}\right){\bar{v}}_{xt}+p^{″}\left(\bar{v}\right){\bar{v}}_t{\bar{v}}_x\right]\text{d}x\text{d}t}}\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{p}^{\prime }\left(\bar{v}\right){\bar{v}}_{xt}}\text{d}x\text{d}t}+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{p}^{″}\left(\bar{v}\right){\bar{v}}_t{\bar{v}}_x}\text{d}x\text{d}t}\\ \le C\delta {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^{-\frac{1}{2}}{y}_t{\text{e}}^{-\frac{\mu {\zeta }^2}{1+t}}}\text{d}x\text{d}t}+C{\delta }^2{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^{-\frac{1}{2}}{y}_t{\text{e}}^{-\frac{2\mu {\zeta }^2}{1+t}}}\text{d}x\text{d}t}\\ \le C\delta {\displaystyle \underset{0}{\overset{T}{\int }}\left(1+t\right)y_t^2\text{d}t}+C{\delta }^2,\end{array}$ (3.49)

$\begin{array}{l}\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{\left[p\left(y_x+\bar{v}+m\right)-p\left(\bar{v}\right)-p^{\prime }\left(\bar{v}\right)y_x\right]}_x}\text{d}x\text{d}t}\\ =\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{\left[p^{\prime }\left(\bar{v}\right)m+o\left(1\right){\left(y_x+m\right)}^2\right]}_x}\text{d}x\text{d}t}\\ =\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{p}^{″}\left(\bar{v}\right){\bar{v}}_{x}m\text{d}x\text{d}t}}+\alpha {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{p}^{\prime }\left(\bar{v}\right)m_x}\text{d}x\text{d}t}+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t{\left[{\left(y_x+m\right)}^2\right]}_x}\text{d}x\text{d}t}\\ \le C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_t^{2}m\text{d}x\text{d}t}}+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)y_{xt}{\left(y_x+m\right)}^2}\text{d}x\text{d}t}\\ \le C{\displaystyle \underset{0}{\overset{T}{\int }}\left(1+t\right)y_t^2\text{d}t}+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}_{xt}^2}\text{d}x\text{d}t},\end{array}$ (3.50)

同样地将(3.48)~(3.50)代入(3.47)，则得到以下了基本衰减估计：

$\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(y_x^2+y_t^2\right)\left(x,T\right)\text{d}x}\le C{N}_0^2,$ (3.51)

引理3.6设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

$\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\left(x,t\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x,$ (3.52)

证明：

对方程(3.5)₂两边同时乘以 $\left(1+t\right)\omega$，并在 $\left[0,T\right]\times R$ 上积分，得到

$\begin{array}{l}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+T\right){\omega }^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)\frac{A\left(y_x+\bar{v}+m\right){\omega }_x^2}{y_x+\bar{v}+m}\text{d}x\text{d}t}}\\ +4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right){\omega }^2}}\text{d}x\text{d}t=\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_0^2}\text{d}x+\frac{1}{2}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}}\text{d}x\text{d}t,\end{array}$ (3.53)

由于 $y_x+\bar{v}+m$ 有正下界，且 $A\left(v\right),{\mu }_r\left(v\right)$ 为大于0的光滑函数，所以有

${\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)\frac{A\left(y_x+\bar{v}+m\right){\omega }_x^2}{y_x+\bar{v}+m}\text{d}x\text{d}t}}\ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2\text{d}x\text{d}t}},$ (3.54)

$4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right){\omega }^2}}\text{d}x\text{d}t\ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }^2}}\text{d}x\text{d}t,$ (3.55)

把(3.54)、(3.55)，代入至(3.53)，整理得到

$\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_0^2}\text{d}x+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}}\text{d}x\text{d}t,$ (3.56)

由初值条件及引理3.2得到

$\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }^2}}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_0^2}\text{d}x,$ (3.57)

引理3.7设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

$\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x.$ (3.58)

证明：

将方程(3.5)₂对x求导，两边同时再乘以 $\left(1+t\right){\omega }_x$，得到

$\left(1+t\right){\omega }_x{\omega }_{xt}=\left(1+t\right){\omega }_x{\left[\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right]}_{xx}-4\left(1+t\right){\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_x,$ (3.59)

将(3.59)关于 $\left(t,x\right)$ 在 $\left[0,T\right]\times R$ 上分部积分，得到

$\begin{array}{l}\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+T\right){\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}{\left[\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right]}_x}}\text{d}x\text{d}t\\ +4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_x}}\text{d}x\text{d}t\\ =\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x+\frac{1}{2}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t,\end{array}$ (3.60)

其中

$\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}{\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_x\right]}_x}}\text{d}x\text{d}t\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)}}\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right]{\omega }_{xx}^2+\left(1+t\right){\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right]}_x{\omega }_x{\omega }_{xx}\text{d}x\text{d}t\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)}}\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right]{\omega }_{xx}^2\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{R}{\int }\left(1+t\right)\frac{\left(y_{xx}+{\bar{v}}_x+m_x\right)A^{\prime }\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_x{\omega }_{xx}}}}\text{d}x\text{d}t\\ -{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)\frac{\left(y_{xx}+{\bar{v}}_x+m_x\right)A\left(y_x+\bar{v}+m\right)}{{\left(y_x+\bar{v}+m\right)}^2}{\omega }_x{\omega }_{xx}}}\text{d}x\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t-C\left(\epsilon +\delta \right){\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}}\text{d}x\text{d}t-C\left(\epsilon +\delta \right){\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t,\\ \ge C\delta {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t-C\left(\epsilon +\delta \right){\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}}\text{d}x\text{d}t.\end{array}$ (3.61)

并且

$\begin{array}{l}4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_x}}\text{d}x\text{d}t\\ \ge 4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\omega \right]}_x}}\text{d}x\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\mu }_r\left(y_x+\bar{v}+m\right){\omega }_x^2\text{d}x\text{d}t}}+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right)\left(y_{xx}+{\bar{v}}_x+m_x\right){{\mu }^{\prime }}_r\left(y_x+\bar{v}+m\right)\omega {\omega }_x\text{d}x\text{d}t}}\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2\text{d}x\text{d}t}}-C\left(\epsilon +\delta \right){\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}}\text{d}x\text{d}t-C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\text{d}x\text{d}t}\\ \ge C\delta {\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}\text{d}x\text{d}t}-C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}\text{d}x\text{d}t},\end{array}$ (3.62)

把(3.61)、(3.62)，代入至(3.60)，整理得到

$\begin{array}{l}\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t\\ \le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0x}^2}\text{d}x+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_x^2}}\text{d}x\text{d}t,\end{array}$ (3.63)

结合公式(3.28)和(3.41)，我们有

$\left(1+T\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t\le C{N}_0^2,$ (3.64)

引理3.8 设 $\left(y,\omega \right)\in X\left(0,T\right)$ 是初值问题(3.5)~(3.6)的解，那么有以下估计

${\left(1+T\right)}^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2}}{\omega }_{xxx}^2\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0xx}^2}\text{d}x.$ (3.65)

证明：

将方程(3.5)₂对x求二次导，再乘上 ${\left(1+t\right)}^2{\omega }_{xx}$，则有

$\begin{array}{c}{\left(1+t\right)}^2{\omega }_{xx}{\omega }_{xxt}={\left(1+t\right)}^2{\omega }_{xx}{\left(\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right)}_{xxx}\\ -4{\left(1+t\right)}^2{\omega }_{xx}{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_{xx},\end{array}$ (3.66)

将(3.66)对 $\left(t,x\right)$ 在 $\left[0,T\right]\times R$ 上积分得到，有

$\begin{array}{l}\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+T\right)}^2{\omega }_{xx}^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xxx}{\left[\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right]}_{xx}}}\text{d}x\text{d}t\\ +4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xx}{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_{xx}}}\text{d}x\text{d}t\\ =\frac{1}{2}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0xx}^2}\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t,\end{array}$ (3.67)

其中

$\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xxx}{\left[\frac{A\left(y_x+\bar{v}+m\right){\omega }_x}{y_x+\bar{v}+m}\right]}_{xx}}}\text{d}x\text{d}t\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xxx}\left[\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}{\omega }_{xxx}+2{\left(\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right)}_x{\omega }_{xx}+{\left(\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right)}_{xx}{\omega }_x\right]\text{d}x\text{d}t}}\\ ={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}}}{\left(1+t\right)}^2{\omega }_{xxx}^2\text{d}x\text{d}t+2{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right)}_x}}{\left(1+t\right)}^2{\omega }_{xx}{\omega }_{xxx}\text{d}x\text{d}t\\ +{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(\frac{A\left(y_x+\bar{v}+m\right)}{y_x+\bar{v}+m}\right)}_{xx}}}{\left(1+t\right)}^2{\omega }_x{\omega }_{xxx}\text{d}x\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2}}{\omega }_{xxx}^2\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2}}{\omega }_{xx}{\omega }_{xxx}\text{d}x\text{d}t+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2}}{\omega }_x{\omega }_{xxx}\text{d}x\text{d}t,\end{array}$ (3.68)

并且

$\begin{array}{l}4{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xx}{\left[{\mu }_r\left(y_x+\bar{v}+m\right)\left(y_x+\bar{v}+m\right)\omega \right]}_{xx}}}\text{d}x\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xx}{\left[{\mu }_r\left(\bar{v}\right)\left(y_x+\bar{v}+m\right)\omega \right]}_{xx}}}\text{d}x\text{d}t\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xx}\left[{\left(y_x+\bar{v}+m\right)}_{xx}\omega +2{\left(y_x+\bar{v}+m\right)}_x{\omega }_x+\left(y_x+\bar{v}+m\right){\omega }_{xx}\right]\text{d}x\text{d}t}}\\ \ge C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xx}^2}}\text{d}x\text{d}t+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_x{\omega }_{xx}}}\text{d}x\text{d}t+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2\omega {\omega }_{xx}\text{d}x\text{d}t}},\end{array}$ (3.69)

则将上式整理，得到

$\frac{1}{2}{\left(1+T\right)}^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}\left(x,T\right)\text{d}x+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xxx}^2}}\text{d}x\text{d}t\le C{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{0xx}^2}\text{d}x+H,$ (3.70)

其中

$\begin{array}{c}H=-C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_x^2\text{d}x\text{d}t}-C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2\omega {\omega }_{xx}}\text{d}x\text{d}t}}\\ -{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2}}{\omega }_{xx}{\omega }_{xxx}\text{d}x\text{d}t-{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2}}{\omega }_x{\omega }_{xxx}\text{d}x\text{d}t\\ \le C^{\prime }{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xxx}^2}}\text{d}x\text{d}t+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(1+t\right){\omega }_{xx}^2}}\text{d}x\text{d}t+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_x^2}}\text{d}x\text{d}t+C{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }^2}}\text{d}x\text{d}t\\ \le C^{\prime }{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xxx}^2}}\text{d}x\text{d}t+C{\delta }^2,\end{array}$ (3.71)

根据先验估计以及上述引理，有

${\left(1+T\right)}^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\omega }_{xx}^2}\left(x,T\right)\text{d}x+{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left(1+t\right)}^2{\omega }_{xxx}^2}}\text{d}x\text{d}t\le C{\epsilon }^2.$ (3.72)

下面我们开始定理3.1的证明。

令 $\delta =\min \{\frac{\epsilon }{2\sqrt{C_1+1}},\frac{\epsilon }{\left(2\sqrt{C_1+1}\right)\left(2\sqrt{C_2+1}\right)}\}$，

则在定理1.1条件下， $N\left(t\right)\le \delta \le \frac{\epsilon }{2}\le \frac{{\epsilon }_0}{2}$。

由局部存在性定理3.2，在 $\left[0,T_0\right]$ 上，令 $N\left(t\right)={\Vert \left(y,\omega \right)\Vert }_{H^2}+{\Vert y_t\Vert }_{H^1}$，问题(3.5)~(3.6)的解存在，且满足

$N\left(t\right)\le \sqrt{C_1},\Vert N_0\Vert \le \sqrt{C_1}\delta \le \frac{\epsilon }{2\sqrt{C_1}+1}<\epsilon ,$ (3.73)

从而满足先验假设3.1，由先验估计定理2.2知，在 $\left[0,T_0\right]$ 上，

$N\left(t\right)\le \sqrt{C_2}{N}_0\le \sqrt{C_2}\delta \le \frac{\epsilon }{2\sqrt{C_1}+1}<\epsilon <{\epsilon }_0,$ (3.74)

特别地，上式对 $T=T_0$ 也成立。以 $T=T_0$ 为初始，局部存在性定理，可延拓到 $\left[T_0,2T_0\right]$，依次进行下去，可以将解延拓到 $T=+\infty$。

根据引理3.1至引理3.8我们已经证明先验估计(3.13)成立，由解的局部存在性定理和先验估计(3.13)，我们通过以上可以证明(3.5)~(3.6)的解整体存在，进一步证明了(3.1)~(3.2)的解整体存在，并满足如下估计：

${\Vert \left(u-\bar{u}-\hat{u}\right)\left(\cdot ,t\right)\Vert }_{L^2}+{\Vert \left(v-\bar{v}-m\right)\left(\cdot ,t\right)\Vert }_{L^2}+{\Vert \omega \left(\cdot ,t\right)\Vert }_{L^2}=O\left(1\right)\delta {\left(t+1\right)}^{-\frac{1}{2}},$ (3.75)

对于衰减估计(3.49)，我们也可以推广到更高阶导数，如此我们有

${\Vert {\partial }_x^k\left(u-\bar{u}-\hat{u}\right)\left(\cdot ,t\right)\Vert }_{L^2}+{\Vert {\partial }_x^k\left(v-\bar{v}-m\right)\left(\cdot ,t\right)\Vert }_{L^2}+{\Vert {\partial }_x^k\omega \left(\cdot ,t\right)\Vert }_{L^2}=O\left(1\right)\delta {\left(t+1\right)}^{-\frac{k}{2}},k=1,2$ (3.76)

基于上式，从Sobolev引理得到逐点估计

$\underset{t\in \left[0,T\right]}{\sup }\left({\Vert u-\bar{u}-\hat{u}\Vert }_{L^{\infty }}+{\Vert v-\bar{v}-m\Vert }_{L^{\infty }}+{\Vert \omega \Vert }_{L^{\infty }}\right)=O\left(1\right)\delta {\left(t+1\right)}^{-\frac{1}{4}}.$ (3.77)

从而定理1.1得到证明。