广义BBM-Burgers方程扩散波的初边值问题

Initial Boundary Value Problem of Diffused Waves in Generalized BBM-Burgers Equation

摘要: 本文主要研究的是广义BBM-Burgers方程的初边值问题，利用能量估计的方法证明了广义BBM-Burgers方程的解关于扩散波的渐近稳定性。即对方程：在本文中我们将证明在波的强度及初值u₀(x)适当小的情况下，广义BBM-Burgers方程的解整体存在且当时间t趋于无穷时收敛到非线性扩散波。

Abstract: This paper studies the initial boundary value problem of the generalized BBM-Burgers equation, and proves the asymptotic stability of the solution of the generalized BBM-Burgers equation with respect to the diffusion wave by using the method of energy estimation. To the equation: , in this paper, we will prove that the solution of the generalized BBM-Burgers equation exists as a whole and converges to the nonlinear diffusion waveas time t approaches infinity when the wave intensityand initial value u₀(x) are appropriately small.

1. 引言

本文考虑了以下广义BBM-Burgers方程

$\{\begin{array}{l}{u}_t+f{\left(u\right)}_x=u_{xx}+u_{xxt},\\ {u\left(x,t\right)|}_{t=0}=u_0\left(x\right),u\left(0,t\right)=u_{-}.\end{array}$ (1.1)

其中， $\left(x,t\right)\in R_{+}\times R_{+}$ ， $f\left(u\right)$ 是一个充分光滑的凸函数， $\delta$ 表示色散系数且 $\delta \ne 0$ ，常数 $\mu >0$ 为耗散系数。此外，我们假设 $u_0\left(x\right)$ 在边界 $x=0$ ， $x=+\infty$ 上有：

$u\left(0,0\right)=u_{-},u\left(+\infty ,0\right)=u_{+},u_{+}\ne u_{-}.$ (1.2)

一些基本的波，比如稀疏波、扩张波、行波等均可描述上述初边值问题以及相应的大时间性态的渐近稳定性。本文主要研究的方向是非线性扩散波，广义BBM-Burgers方程最早是Benjamin，Bona和Mahony [1] 提出的，它是用来描述浅水波现象的KdV方程 $u_t+u{u}_x+u_{xxx}=0$ 的一个改变，能够更好地解决热传导问题、热学中的双温问题等，因此对于该类问题具有很好的理论以及实际意义，对于BBM-Burgers方程

$u_t+f\left(u\right)=u_{xx}+u_{xxt}.$ (1.3)

解的存在性和大时间行为已经有广泛的研究了，Tang Y [2] 研究从孤子解的演化角度去证明了Burgers方程在扰动作用下激波研究的一般途径；Yang Xiaojia [3] 等人研究了burgers方程的两层高阶紧致有限差分隐式，利用傅里叶分析法对方案的稳定性进行了分析，最后通过数值实验验证了所提方案的准确性与可靠性。

读者可阅读 [4] - [11] ，下面介绍一些学者对于该问题的研究成果。

刘太平等人已经研究了Burgers方程的初边值问题 [12] ；

$\{\begin{array}{l}{u}_t+f{\left(u\right)}_x=u_{xx},x\in R^{+},t>0,\\ {u\left(x,t\right)|}_{x=0}=u_{-},t>0,\\ {u\left(x,t\right)|}_{t=0}=u_0\left(x\right)=\{\begin{array}{l}{u}_{-},x=0,\\ u_{+},x\to \infty .\end{array}\end{array}$ (1.4)

其中 $f\left(u\right)$ 为R上的光滑函数。

蒋米娜，徐艳玲 [13] 证明广义BBM-Burgers方程在假设条件 $u_{-}<u_{+}$ 下，依据特征速度 $f^{\prime }\left(u_{\pm }\right)$ 的符号不同，将问题分成了五种情况，并且证明了解的整体存在性，分别找到了它们的渐近状态；陈琴，刘艳 [14] 研究了带有一般边界条件的广义BBM-Burgers方程的初边值问题，运用 $L^2$ -能量法证明其强边界层解具有非线性稳定性。对于BBM-Burgers方程的研究成果，读者可参考文献 [15] - [20] 。

在前人研究的基础上，本文主要研究广义BBM-Burgers方程非线性扩散波的解的渐近稳定性，观察广义BBM-Burgers方程中含有热传导方程

${\bar{u}}_t-{\bar{u}}_{xx}=0$ (1.5)

对于热传导方程，我们知道该方程的解是具有唯一的自相似解，即扩散波在本文中将证明广义BBM-Burgers方程的初边值问题在大时间里的解是非线性稳定的。我们将在扩散波附近定义一个扰动 $\varphi =u-\bar{u}$ ，证明在波 $\delta :=\left|u_{+}-u_{-}\right|$ 足够小的

情况下，(1.1)的解收敛到(1.5)的自相似解。

定义如下的扰动

$\{\begin{array}{l}\varphi \left(x,t\right)=u\left(x,t\right)-\bar{u}\left(x,t\right),\\ {\varphi }_0\left(x\right)=u_0\left(x\right)-\bar{u}\left(x,0\right),\varphi \left(0,t\right)=0.\end{array}$ (1.6)

则由(1.1) (1.5) (1.6)得到扰动方程为

$\{\begin{array}{l}{\varphi }_t-{\varphi }_{xx}+f{\left(\bar{u}+\varphi \right)}_x-{\left(\varphi +\bar{u}\right)}_{xxt}=0,\\ {\varphi }_0\left(x\right)=0.\end{array}$ (1.7)

下面则是本文的定理。

定理1：对于问题(1.6)假设存在足够小的波的强度 $\delta$ 和初值 ${\Phi }_0={\Vert {\varphi }_0\Vert }_2^2$ ，且 $\delta$ 满足 ${\Vert u_0\left(x\right)-\bar{u}\left(x,0\right)\Vert }_{H^2\left(R_{+}\right)}+\left|u_{+}-u_{-}\right|\le \delta <\epsilon$ ，初边值问题(1.6)存在唯一解 $u\left(x,t\right)$ ，且满足

$u\left(x,t\right)-\bar{u}\left(x,t\right)\in C\left(\left[0,\infty \right),H^2\left(R_{+}\right)\right)\cap L^2\left(\left[0,\infty \right),H^3\left(R_{+}\right)\right).$ (1.8)

进一步则有

${\displaystyle \underset{k=0}{\overset{2}{\sum }}{\Vert {\partial }_x^k\left(\varphi ,{\varphi }_x\right)\left(t\right)\Vert }^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\displaystyle {\int }_0^t{\Vert {\partial }_x^k\left(\varphi ,{\varphi }_x\right)\left(t\right)\Vert }^2\text{d}t}}\le C\left({\Phi }_0+\delta \right).$ (1.9)

记号：

在本文中，回忆一些关于勒贝格空间和分数阶索伯列夫空间的一些必要的基础知识，对于更多细节的地方，可以参考 [21] [22] [23] 。 $L^p=L^p\left(R_{+}\right)$ 表示为一般的Lebesgue空间，其定义的范数为

${\Vert v\Vert }_{L^P}={\left({\displaystyle {\int }_0^{+\infty }{\left|v\left(x\right)\right|}^p}\text{d}x\right)}^{\frac{1}{p}},1\le p<+\infty ,{\Vert v\Vert }_{L^{\infty }}=\text{esssuup}\left|v\left(x\right)\right|.$

$u\in W_{l,2}=H^l\left(\Omega \right)$ ， $H^l$ 表示一般的Sobolev空间，u的范数为

${\Vert u\Vert }_{H^l}={\left({\displaystyle \underset{\left|\alpha \right|\le l}{\sum }{{\displaystyle \int \left|D^{\alpha }u\right|}}^2\text{d}x}\right)}^{\frac{1}{2}}$

在不会导致混淆的情况下，我们将省略积分区域 $R_{+}$ ，并且 $C>0$ ， $O\left(1\right)$ ，来表示一般常数。

2. 基础知识

引理2.1：假设 $\bar{u}\left(x,t\right)$ 为问题(1.5)的自相似解，则 $\bar{u}\left(x,t\right)$ 满足：

$\left|{\partial }_x^k{\partial }_t^l\bar{u}\right|=E\left|u_{+}-u_{-}\right|{\left(1+t\right)}_{}^{-\frac{k}{2}-l}\omega \left(x,t\right).$ (2.1)

其中 $k,l=1,2,\cdots$ ， $\omega \left(x,t\right)=\exp \left\{-\frac{\sigma x^2}{1+t}\right\}$ 。 $\sigma$ 是正常数，这里的 $\sigma$ 仅与 $u_{+}$ 和 $u_{-}$ 有关。

为了得到定理1.1的证明，我们下面将先给出扰动方程(1.7)的局部存在性以及先验估计，根据先验估计的证明相关引理，再利用靴带原理，将解延拓到全局，最终证明广义BBM-Burgers方程的初边值问题在大时间里的解是非线性稳定的，即证明定理1.1。

引理2.2 (Sobolev不等式)对于任意函数 $f\left(x\right)\in H^1\left(R\right)$ ，有

${\Vert f\Vert }_{L^{\infty }}\le C{\Vert f\Vert }^{\frac{1}{2}}{\Vert f_x\Vert }^{\frac{1}{2}}.$ (2.2)

引理2.3 (Young不等式)设 $1<p,q<\infty$ ，满足 $\frac{1}{p}+\frac{1}{q}=1$ ，则有

$ab<\frac{a^p}{p}+\frac{b^q}{q}\left(a,b>0\right).$ (2.3)

特别地，当 $p=q=2$ 时，我们称之为Cauchy不等式。

3. 定理1.1的证明

1) 命题3.1 (局部存在性)考虑初始时间为t的扰动方程

$\{\begin{array}{l}{\varphi }_t-{\varphi }_{xx}+f{\left(\bar{u}+\varphi \right)}_x-{\left(\varphi +\bar{u}\right)}_{xxt}=0,\\ {\varphi }_0\left(x\right)=0.\end{array}$ (3.1)

若 $\varphi \left(x,t\right)\in H^2$ ，且有 ${\Vert \varphi \left(t\right)\Vert }_2\le M$ 存在适当小的 $t_0$ ，使得方程在 $\left[t,t+t_0\right]$ 上有唯一解。

2) 命题3.2 (先验估计)在定理1的条件下，若使得 $\varphi \left(x,t\right)\in X\left(x,T\right)$ 为问题(3.1)的解，则我们作出如下的先验假设：

$\underset{0\le t\le T}{\sup }{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\Vert {\partial }^k\varphi \left(t\right)\Vert }_{L^2}^2}\le C\epsilon .$ (3.2)

其中C为正常数， $\epsilon$ 是依赖于初值以及波的强度的一个充分小的正常数。根据以上先验假设，我们用Sobolev不等式 ${\Vert f\Vert }_{L^{\infty }}\le {\left({\Vert f\Vert }_{L^2}\right)}^{\frac{1}{2}}{\left({\Vert {\partial }_{x}f\Vert }_{L^2}\right)}^{\frac{1}{2}}$ 可以得到相应函数的 $L^{\infty }$ 范数，则我们可以得到如下的先验估计：

${\displaystyle \underset{k=0}{\overset{2}{\sum }}{\Vert {\partial }^k\varphi \left(t\right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\displaystyle {\int }_0^{+\infty }{\Vert {\partial }^k\varphi (x,t)\Vert }_{L^2}^2}\text{d}x}\le C{\Vert {\varphi }_0\Vert }_2^2.$ (3.3)

3) 定理1.1的证明

为了证明本文的主要结果，首先证明一些基本的引理。下面将证明 $\varphi \left(x,t\right)$ 的低阶、高阶估计。证明过程由下面几个引理组成。

4. 先验估计的证明

引理4.1：在先验假设的情况下，对于充分小的正数 $\delta$ ，有

${\Vert \varphi \Vert }^2+{\Vert {\varphi }_x\Vert }^2+{\displaystyle {\int }_0^t\left({\Vert \varphi \Vert }^2+{\Vert {\varphi }_x\Vert }^2\right)\text{d}t}\le C\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right).$ (4.1.1)

证明：在扰动方程(1.6)两边同时乘以 $\varphi$ 后，在 $\left[0,t\right]\times \left[0,+\infty \right)$ 上积分，

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi {\varphi }_t}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi {\varphi }_{xx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi f{\left(\bar{u}+\varphi \right)}_x}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi {\left(\varphi +\bar{u}\right)}_{xxt}}}\text{d}x\text{d}t=0.$ (4.1.2)

将(4.1.2)整理可得

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi {\varphi }_t}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi {\varphi }_{xx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }-{\varphi }_{xxt}\varphi }}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\bar{u}}_{xxt}\varphi }}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\varphi +\bar{u}\right)}_x\varphi }}\text{d}x\text{d}t.$ (4.1.3)

则(4.1.3)通过分部积分可得

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi {\varphi }_t}\text{d}x}\text{d}t=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }^2}\text{d}x-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_0^2}\text{d}x.$ (4.1.4)

$\begin{array}{c}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }-\varphi {\varphi }_{xx}}}\text{d}x\text{d}t=-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi }}\text{d}\left({\varphi }_x\right)x\text{d}t\\ =-{\displaystyle {\int }_0^t{\varphi {\varphi }_x|}_0^{+\infty }\text{d}t}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t.\end{array}$ (4.1.5)

$\begin{array}{c}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }-{\varphi }_{xxt}\varphi }}\text{d}x\text{d}t=-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi }}\text{d}\left({\varphi }_{xt}\right)\text{d}t\\ =-{\displaystyle {\int }_0^t{\varphi {\varphi }_{xt}|}_0^{+\infty }\text{d}t}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xt}{\varphi }_x}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xt}{\varphi }_x}}\text{d}x\text{d}t,\end{array}$ (4.1.6)

其中

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x\text{d}\left({\varphi }_x\right)\text{d}x}}={\displaystyle {\int }_0^{+\infty }{{\varphi }_x^2|}_0^t}\text{d}x-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}t.$ (4.1.7)

所以

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}t=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{{\varphi }_x^2|}_0^t}\text{d}x=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2|\text{d}x}-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{x0}^2|\text{d}x}.$ (4.1.8)

将(4.1.8)代入到(4.1.6)中可得

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }-\varphi {\varphi }_{xxt}}}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}t=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}\text{d}x-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{x0}^2}\text{d}x.$ (4.1.9)

再将(4.1.4) (4.1.5) (4.1.9)代入到(4.1.3)中并整理可得

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}\text{d}x\\ =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_0^2}\text{d}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{x0}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\bar{u}}_{xxt}\varphi }}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\varphi +\bar{u}\right)}_x\varphi }}\text{d}x\text{d}t\end{array}$ (4.1.10)

因为 ${\bar{u}}_x$ 是有界的，所以对方程(4.1.10)的右侧进行估计

$\begin{array}{c}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\bar{u}}_{xxt}\varphi }}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi \text{d}\left({\bar{u}}_{xt}\right)\text{d}t}}={\displaystyle {\int }_0^t{\varphi {\bar{u}}_{xt}|}_0^{+\infty }}\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\bar{u}}_{xt}}\text{d}x}\text{d}t\\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\bar{u}}_{xt}}\text{d}x}\text{d}t\le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}\text{d}x}\text{d}t.\end{array}$ (4.1.11)

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\varphi +\bar{u}\right)}_x\varphi }}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_x\varphi }}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\bar{u}\right)}_x\varphi }}\text{d}x\text{d}t.$ (4.1.12)

则有

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_x\varphi }}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\xi \right)\varphi {\bar{u}}_x\text{d}x\text{d}t}}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\varphi {\varphi }_x}{f}^{\prime }\left(\varphi +\bar{u}\right)\text{d}x\text{d}t}\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\xi \right)\varphi {\bar{u}}_x\text{d}x\text{d}t}}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\varphi +\bar{u}\right)}}\left({\varphi }_x+{\bar{u}}_x\right){\varphi }^2\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\xi \right)\varphi {\bar{u}}_x\text{d}x\text{d}t}}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x{\varphi }^2\text{d}x\text{d}t}}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }^2\text{d}x\text{d}t}}.\end{array}$ (4.1.13)

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_x\varphi }}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }^2\text{d}x\text{d}t}}\\ \le C\delta +C\left(\delta +\epsilon \right){\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}t+C\epsilon {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t.\end{array}$ (4.1.14)

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\bar{u}\right)}_x\varphi }}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{\prime }{\left(\bar{u}\right)}_x{\bar{u}}_x\varphi }}\text{d}x\text{d}t\le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}t.$ (4.1.15)

将(4.1.13)和(4.1.15)代入到(4.1.12)中得

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\varphi +\bar{u}\right)}_x\varphi }}\text{d}x\text{d}t\le C\delta +C\left(\delta +\epsilon \right){\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}t+C\epsilon {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t.$ (4.1.16)

再将(4.1.11)和(4.1.16)代入到(4.1.10)中得

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }^2\text{d}x\text{d}t}}\\ \le C\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)+C\left(\delta +\epsilon \right){\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left({\varphi }^2+{\varphi }_x^2\right)\text{d}x\text{d}t}}.\end{array}$ (4.1.17)

由于 ${\bar{u}}_x$ 是有界的，且f是光滑函数，所以

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }^2\text{d}x\text{d}t}}\ge C{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}t.$ (4.1.18)

将(4.1.18)代入到(4.1.17)中并且整理可得

${\displaystyle {\int }_0^{+\infty }{\varphi }^2}\text{d}x+{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left({\varphi }^2+{\varphi }_x^2\right)\text{d}x\text{d}t}}\le C\left({\Phi }_0+\delta \right).$ (4.1.19)

即引理4.1证毕。

引理4.2：在先验假设的情况下，对于充分小的正数 $\delta$ ，有

${\Vert {\varphi }_x\Vert }^2\text{+}{\Vert {\varphi }_{xx}\Vert }^2+{\displaystyle {\int }_0^t\left({\Vert {\varphi }_x\Vert }^2+{\Vert {\varphi }_{xx}\Vert }^2\right)\text{d}t}\le C\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right).$

证明：在扰动方程(1.6)两边关于x求导，再在两边同时乘以 ${\varphi }_x$ 后，在 $\left[0,t\right]\times \left[0,+\infty \right)$ 上积分，

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xxx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{x}f{\left(\bar{u}+\varphi \right)}_{xx}}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\left(\varphi +\bar{u}\right)}_{xxxt}}}\text{d}x\text{d}t=0.$ (4.2.1)

下面计算(4.2.1)的各项

$\begin{array}{c}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x\text{d}\left({\varphi }_x\right)\text{d}x}}\\ =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2\text{d}}x-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{x0}^2\text{d}}x.\end{array}$ (4.2.2)

$\begin{array}{c}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }-{\varphi }_x{\varphi }_{xxx}}}\text{d}x\text{d}t=-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x\text{d}\left({\varphi }_{xx}\right)\text{d}t}}\\ =-{\displaystyle {\int }_0^t{{\varphi }_x{\varphi }_{xx}|}_0^{+\infty }}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2\text{d}x}}\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2\text{d}x}}\text{d}t.\end{array}$ (4.2.3)

$\begin{array}{c}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\varphi }_{xxxt}}}\text{d}x\text{d}t=-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x\text{d}\left({\varphi }_{xxt}\right)\text{d}t}}\\ =-{\displaystyle {\int }_0^t{{\varphi }_x{\varphi }_{xxt}|}_0^{+\infty }}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\varphi }_{xxt}}\text{d}x}\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}\text{d}\left({\varphi }_{xx}\right)\text{d}x}}\\ =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{{\varphi }_{xx}^2|}_0^t}\text{d}x\\ =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}\text{d}x-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx0}^2}\text{d}x.\end{array}$ (4.2.4)

将(4.2.2)~(4.2.4)代入到(4.2.1)中整理可得

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2\text{d}}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2\text{d}x}}\text{d}t\\ =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{x0}^2\text{d}}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx0}^2}\text{d}x-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{x}f{\left(\bar{u}+\varphi \right)}_{xx}}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\bar{u}}_{xxxt}}}\text{d}x\text{d}t.\end{array}$ (4.2.5)

下面对(4.2.5)的右侧进行估计

$\begin{array}{c}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x{\bar{u}}_{xxxt}}}\text{d}x\text{d}t={\displaystyle {\int }_0^t{{\varphi }_x{\bar{u}}_{xxt}|}_0^{+\infty }}\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\bar{u}}_{xxt}}\text{d}x\text{d}t}\\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\bar{u}}_{xxt}}\text{d}x\text{d}t}\\ \le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}}\text{d}x.\end{array}$ (4.2.6)

$\begin{array}{l}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{x}f{\left(\bar{u}+\varphi \right)}_{xx}}}\text{d}x\text{d}t\\ =-{\displaystyle {\int }_0^t{{\varphi }_{x}f{\left(\bar{u}+\varphi \right)}_x|}_0^{+\infty }}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}f{\left(\bar{u}+\varphi \right)}_x}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}f{\left(\bar{u}+\varphi \right)}_x}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}f{\left(\bar{u}\right)}_x}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}f{\left(\bar{u}\right)}_x}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}+\varphi \right)\left({\varphi }_x+{\bar{u}}_x\right)\text{d}x\text{d}t}}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}t\\ =I_1-I_2+I_3.\end{array}$ (4.2.7)

下面对(4.2.7)的各项进行计算

$\begin{array}{c}{I}_1-I_2={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}+\varphi \right)\left({\varphi }_x+{\bar{u}}_x\right)\text{d}x\text{d}t}}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}+\varphi \right){\varphi }_x}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}+\varphi \right){\bar{u}}_x}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}t,\end{array}$ (4.2.8)

利用泰勒展开式以及分部积分可将(4.2.8)写成

$\begin{array}{c}{I}_1-I_2={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\xi \right)\varphi {\varphi }_{xx}{\bar{u}}_x}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{\prime }\left(\bar{u}+\varphi \right){\varphi }_x{\varphi }_{xx}}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{f^{″}\left(\xi \right)\varphi {\varphi }_x{\bar{u}}_x|}_0^{+\infty }}\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\left(f^{″}\left(\xi \right)\varphi {\bar{u}}_x\right)}_x{\varphi }_x}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{\prime }\left(\bar{u}+\varphi \right){\varphi }_x{\varphi }_{xx}}}\text{d}x\text{d}t\\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(f^{‴}\left(\xi \right)\varphi {\bar{u}}_x+f^{″}\left(\xi \right){\varphi }_x{\bar{u}}_x+f^{″}\left(\xi \right)\varphi {\bar{u}}_{xx}\right){\varphi }_x}}\text{d}x\text{d}t\\ -\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\varphi }_x^2}}\text{d}x\text{d}t-\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x{\varphi }_x}}\text{d}x\text{d}t.\end{array}$ (4.2.9)

将 $-\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x{\varphi }_x}}\text{d}x\text{d}t$ 移到等式(4.2.5)左边，则

$I_1-I_2\le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t+C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}t.$ (4.2.10)

$I_3=-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}t\le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}}\text{d}x\text{d}t.$ (4.2.11)

将(4.2.6) (4.2.10)~(4.2.11)代入到(4.2.5)中并且整理可得

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2\text{d}}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2\text{d}x}}\text{d}t+\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x{\varphi }_x}}\text{d}x\text{d}t\\ \le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}}\text{d}x+C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t+C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}t.\end{array}$ (4.2.12)

由于 ${\bar{u}}_x$ 是有界的，且f是光滑函数，所以

$\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x{\varphi }_x}}\text{d}x\text{d}t\ge C{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t.$ (4.2.13)

将(4.2.13)代入到(4.2.12)中，并且运用已得证的引理4.1可得

${\displaystyle {\int }_0^{+\infty }{\varphi }_x^2\text{d}}x+{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2\text{d}x}}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}t\le C\left({\Phi }_0+\delta \right).$ (4.2.14)

即引理4.2证毕。

引理4.3：在先验假设的情况下，对于充分小的正数 $\delta$ ，有

${\Vert {\varphi }_{xx}\Vert }^2+{\Vert {\varphi }_{xxx}\Vert }^2+{\displaystyle {\int }_0^t\left({\Vert {\varphi }_{xx}\Vert }^2+{\Vert {\varphi }_{xxx}\Vert }^2\right)\text{d}t}\le C\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right).$

证明：在扰动方程(1.6)两边关于x求导，再在两边同时乘以 ${\varphi }_{xx}$ 后，在 $\left[0,t\right]\times \left[0,+\infty \right)$ 上积分，

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\varphi }_{xxt}}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\varphi }_{xxxx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}f{\left(\bar{u}+\varphi \right)}_{xxx}}}\text{d}x\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\left(\varphi +\bar{u}\right)}_{xxxxt}}}\text{d}x\text{d}t=0.$ (4.3.1)

下面则对(4.3.1)的各项进行计算

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\varphi }_{xxt}}}\text{d}x\text{d}t=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{{\varphi }_{xx}^2|}_0^t}\text{d}x=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}\text{d}x-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx0}^2}\text{d}x.$ (4.3.2)

$-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\varphi }_{xxxx}}}\text{d}x\text{d}t=-{\displaystyle {\int }_0^t{{\varphi }_{xx}{\varphi }_{xxx}|}_0^{+\infty }}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}}\text{d}x\text{d}t={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}}\text{d}x\text{d}t.$ (4.3.3)

$\begin{array}{c}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\varphi }_{xxxxt}}}\text{d}x\text{d}t=-{\displaystyle {\int }_0^t{{\varphi }_{xx}{\varphi }_{xxxt}|}_0^{+\infty }}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{\varphi }_{xxxt}}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{\varphi }_{xxxt}}}\text{d}x\text{d}t=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{{\varphi }_{xxx}^2|}_0^t}\text{d}x\\ =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}\text{d}x-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx0}^2}\text{d}x.\end{array}$ (4.3.4)

将(4.3.2)~(4.3.4)代入到(4.3.1)中可得

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}\text{d}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}}\text{d}x\text{d}t\\ =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx0}^2}\text{d}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx0}^2}\text{d}x-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}f{\left(\bar{u}+\varphi \right)}_{xxx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\bar{u}}_{xxxxt}}}\text{d}x\text{d}t.\end{array}$ (4.3.5)

对等式(4.3.5)的右侧进行估计

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{\bar{u}}_{xxxxt}}}\text{d}x\text{d}t\le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}}\text{d}x\text{d}t.$ (4.3.6)

$\begin{array}{c}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}f{\left(\bar{u}+\varphi \right)}_{xxx}}}\text{d}x\text{d}t=-{\displaystyle {\int }_0^t{{\varphi }_{xx}f{\left(\bar{u}+\varphi \right)}_{xx}|}_0^{+\infty }}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}f{\left(\bar{u}+\varphi \right)}_{xx}}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}f{\left(\bar{u}+\varphi \right)}_{xx}}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{\left[f^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_x+{\varphi }_x\right)\right]}_x}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right){({\bar{u}}_x+{\varphi }_x)}^2\text{d}x\text{d}t}}\\ +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{f}^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xx}+{\varphi }_{xx}\right)\text{d}x\text{d}t}}\\ =I_1+I_2.\end{array}$ (4.3.7)

$\begin{array}{c}{I}_1={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x^2+{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right){\varphi }_x^2+{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right)2{\bar{u}}_x{\varphi }_x\text{d}x\text{d}t}}\\ -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\bar{u}\right)}_{xx}{\varphi }_{xxx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\bar{u}\right)}_{xx}{\varphi }_{xxx}}}\text{d}x\text{d}t\\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{\bar{u}}_x^2{f}^{‴}\left(\xi \right)\varphi }}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left[{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right){\varphi }_x^2+{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right)2{\bar{u}}_x{\varphi }_x\right]\text{d}x\text{d}t}}\\ -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{\prime }\left(\bar{u}\right){\bar{u}}_{xx}{\varphi }_{xxx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\bar{u}\right)}_{xx}{\varphi }_{xxx}}}\text{d}x\text{d}t.\end{array}$ (4.3.8)

所以对其估计为

$\begin{array}{c}{I}_1={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}{\bar{u}}_x^2{f}^{‴}\left(\xi \right)\varphi }}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left[{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right){\varphi }_x^2+{\varphi }_{xxx}{f}^{″}\left(\bar{u}+\varphi \right)2{\bar{u}}_x{\varphi }_x\right]\text{d}x\text{d}t}}\\ -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{\prime }\left(\bar{u}\right){\bar{u}}_{xx}{\varphi }_{xxx}}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }f{\left(\bar{u}\right)}_{xx}{\varphi }_{xxx}}}\text{d}x\text{d}t\\ \le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left({\varphi }_{xxx}^2+{\varphi }_x^2+{\varphi }^2\right)\text{d}x\text{d}t}}\end{array}$ (4.3.9)

$\begin{array}{c}{I}_2={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xx}+{\varphi }_{xx}\right)\text{d}\left({\varphi }_{xx}\right)\text{d}t}},\\ ={\displaystyle {\int }_0^t{f}^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xx}+{\varphi }_{xx}\right)}{{\varphi }_{xx}|}_0^{+\infty }\text{d}t-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\left[f^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xx}+{\varphi }_{xx}\right)\right]}_x}}{\varphi }_{xx}\text{d}x\text{d}t\\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\left[f^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xx}+{\varphi }_{xx}\right)\right]}_x}}{\varphi }_{xx}\text{d}x\text{d}t\\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left[f^{″}\left(\bar{u}+\varphi \right)\left({\bar{u}}_x+{\varphi }_x\right)\left({\bar{u}}_{xx}+{\varphi }_{xx}\right)+f^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xxx}+{\varphi }_{xxx}\right)\right]{\varphi }_{xx}\text{d}x\text{d}t}}\\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left[f^{″}\left(\bar{u}+\varphi \right)\left({\bar{u}}_x+{\varphi }_x\right){\bar{u}}_{xx}+f^{″}\left(\bar{u}+\varphi \right){\varphi }_x{\varphi }_{xx}+f^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xxx}+{\varphi }_{xxx}\right)\right]{\varphi }_{xx}\text{d}x\text{d}t}}\\ -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x}}{\varphi }_{xx}^2\text{d}x\text{d}t\end{array}$ (4. 3.10)

将 $-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x}}{\varphi }_{xx}^2\text{d}x\text{d}t$ 移至(4.3.5)的左侧，则对(4.3.10)的估计为

$\begin{array}{c}{I}_2=-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left[f^{″}\left(\bar{u}+\varphi \right)\left({\bar{u}}_x+{\varphi }_x\right){\bar{u}}_{xx}+f^{″}\left(\bar{u}+\varphi \right){\varphi }_x{\varphi }_{xx}\right]{\varphi }_{xx}\text{d}x\text{d}t}}\\ +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}{f}^{\prime }\left(\bar{u}+\varphi \right)\left({\bar{u}}_{xxx}+{\varphi }_{xxx}\right)\text{d}x\text{d}t}}\\ \le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left({\varphi }^2+{\varphi }_x^2+{\varphi }_{xx}^2+{\varphi }_{xxx}^2\right)\text{d}x\text{d}t}}.\end{array}$ (4.3.11)

将(4.3.6) (4.3.10)以及(4.3.11)代入到(4.3.5)中可得

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}\text{d}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}}\text{d}x\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x}}{\varphi }_{xx}^2\text{d}x\text{d}t\\ \le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left({\varphi }^2+{\varphi }_x^2+{\varphi }_{xx}^2+{\varphi }_{xxx}^2\right)\text{d}x\text{d}t}}.\end{array}$ (4.3.12)

由于 ${\bar{u}}_x$ 是有界的，且f是光滑函数，所以

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{f}^{″}\left(\bar{u}+\varphi \right){\bar{u}}_x}}{\varphi }_{xx}^2\text{d}x\text{d}t\ge C{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}}\text{d}x\text{d}t.$ (4.3.13)

将(4.3.13)代入到(4.3.12)中，并且运用已得证的引理4.1和引理4.2可得

${\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2\text{d}}x+{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xxx}^2\text{d}x}}\text{d}t+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_{xx}^2}}\text{d}x\text{d}t\le C\left({\Phi }_0+\delta \right).$ (4.3.14)

则引理4.3证毕。

由此引理全部证明完毕，故先验估计得证。