上海理工大学，上海

收稿日期：2022年9月14日；录用日期：2022年10月5日；发布日期：2022年10月14日

摘要

本文研究了广义Kdv-burgers方程的初边值问题，说明了广义Kdv-burgers方程的解关于扩散波的渐近稳定性。即对方程： $u_t+f{\left(u\right)}_x+\delta u_{xxx}-\mu u_{xx}=0$，${u\left(x,y\right)|}_{t=0}=u_0\left(x\right),u\left(0,t\right)=u_{-}$，我们证明了在一些小性条件下，广义Kdv-burgers方程的解整体存在且当时间t趋于无穷时收敛到扩散波。

关键词

广义Kdv-Burgers方程，初边值问题，能量方法，扩散波，衰减速度

The Initial and Boundary Value Problem to the Generalized Kdv-Burgers Equation

Jie Zhang

University of Shanghai for Science and Technology, Shanghai

Received: Sep. 14^th, 2022; accepted: Oct. 5^th, 2022; published: Oct. 14^th, 2022

ABSTRACT

In this paper, we studied the initial and boundary value problem to the generalized KdV-Burgers equation; we will show the large time asymptotic stability of diffusion waves to the problem i.e. for the equation: $u_t+f{\left(u\right)}_x+\delta u_{xxx}-\mu u_{xx}=0$,${u\left(x,y\right)|}_{t=0}=u_0\left(x\right),u\left(0,t\right)=u_{-}$. It’s proved that under some smallness conditions the solution of the generalized Kdv-burgers equation exists globally and converges to the diffusion wave when the time t tends to infinity.

Keywords:Generalized KdV-Burgers Equation, Initial and Boundary Value Problem, Energy Methods, Diffusion Waves, Decay Rate

Copyright © 2022 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

本文考虑了以下广义Kdv-burgers方程

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

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

Kdv方程由荷兰科学家D. J. Korteweg和G. de Vries [1] 首次推导得出，得到以下Kdv方程：

$u_t+6u{u}_x+u_{xxx}=0$

Burgers方程是一类非线性偏微分方程，在流体力学的研究中有着非常重要的意义。Bateman在1915年首次提出如下方程 [2]：

$u_t+u{u}_x=\mu u_{xx}$

随着流体力学的不断研究发展，学者们相继提出了Kdv-Burgers方程：

$u_t+u{u}_x-\beta u_{xxx}-\alpha u_{xx}=0$

本文考虑的方程为广义Kdv-Burgers方程，关于该方程的主要研究成果有如下：

Bona. J. L.等人在 [3] [4] 中证明了该方程存在一个单调行波解，随后在 [5] 中进一步论述了该方程行波解的渐近行为，并得到了在初始扰动合适小的情况下，方程行波解是渐近稳定的结论。

Rajopadhye S. V.在 [6] 中证明了方程在小扰动下，方程解及其一阶导在 $L^2$ 范数空间内的渐近稳定性，并得到了 ${\left(1+t\right)}^{-\frac{1}{4}}$ 的衰减速度。

进一步的，Wang Z. A.及Zhu C. J. [7] 证明了该方程关于稀疏波的稳定性，得到了以下结论： $t\to \infty$ 时，有 ${\sup }_{x\in R}\left|u\left(x,t\right)-u^R\left(x,t\right)\right|\to 0$，其中 $u^R\left(x,t\right)$ 为黎曼问题的稀疏波解。

在 [8] 中，作者首先运用了相平面方法得到了单调行波解存在的限制条件。并由压缩映像原理证明了方程解的局部存在性。进一步通过加权能量方法得到柯西问题的解收敛到其相应行波解的代数衰减估计。

Duan R.，Zhao H. J.在 [9] 中研究了柯西问题(1.1)在增加了一个不等式条件的前提下，去除了初始扰动和波的强度充分小的条件，得到柯西问题解关于稀疏波的整体稳定性。

本文论述的是广义Kdv-Burgers方程关于扩散波的渐近稳定性，观察该方程可知，方程里包含了热传导方程：

$u_t-\mu u_{xx}=0$ (1.2)

我们知道热传导方程具有唯一的自相似解(即扩散波)，本文将证明广义Kdv-Burgers方程的初边值问题是非线性稳定的，也就是说在扩散波附件定义一个小扰动 ${\varphi }_0\left(x\right)$，我们证明了在波的强度充分小的情况下，问题(1.1)的解全局存在且随时间收敛到方程(1.2)的自相似解。

定义如下扰动：

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

则由问题(1.1)，(1.2)我们可以得到扰动方程：

$\{\begin{array}{l}{\varphi }_t+\delta {\varphi }_{xxx}-\mu {\varphi }_{xx}+f{\left(\bar{u}+\varphi \right)}_x+\delta {\bar{u}}_{xxx}=0\\ \varphi \left(x,0\right)={\varphi }_0\left(x\right),\varphi \left(0,t\right)=0\end{array}$ (1.3)

本文的主要定理如下：

定理1：假设问题(1.3)的初值 ${\Vert \varphi \left(x,0\right)\Vert }_2^2$ 和波的强度充分小，即 ${\Vert u_0\left(x\right)-\bar{u}\left(x,0\right)\Vert }_{H^2\left(R_{+}\right)}+\left|u_{+}-u_{-}\right|\le \delta <\epsilon$，则初边值问题(1.3)存在唯一全局解 $\varphi \left(x,t\right)\in L^{\infty }\left(H^2\right)\cap L^{\infty }\left(H^3\right)$，且满足：

${\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\Vert {\partial }^k\varphi \left(\tau \right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\displaystyle \underset{0}{\overset{t}{\int }}{\Vert {\partial }^{k+1}\varphi \left(x,\tau \right)\Vert }_{L^2}^2}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

记号：

1) 本文中的C，o (1)表示正常数。

2) 文中 $L^P\left(R_{+}\right)$ 为一般的Lebesgue空间，定义其范数为：

${\Vert f\Vert }_{L^p}={\left({\displaystyle \underset{R_{+}}{\int }{\left|f\left(x\right)\right|}^p\text{d}x}\right)}^{\frac{1}{p}},1\le p<\infty ,{\Vert f\Vert }_{L^{\infty }}=\underset{x\in R^{+}}{\sup }\left|f\left(x\right)\right|$

$H^l\left(R_{+}\right)$ 表示通常的Sobolev空间。其范数：

${\Vert f\Vert }_{H^l}={\Vert f\Vert }_l={\left({\displaystyle \underset{j=0}{\overset{l}{\sum }}{\Vert {\partial }_x^{l}f\left(x\right)\Vert }_{L^2}^2\text{d}x}\right)}^{\frac{1}{2}}$

2. 准备工作

引理2.1：假设 $\bar{u}$ 是问题(1.2)的自相似解，则有：

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

其中：

$\omega \left(x,t\right)={\text{e}}^{-\frac{\sigma x^2}{1+t}}$

$\sigma$ 是依赖于 $u_{+},u_{-}$ 的正常数。

为了得到定理1的结论，我们将在下文中给出扰动方程的局部存在性以及先验估计。然后由先验估计将方程解的局部存在性延拓到全局。

3. 定理1的证明

3.1. 命题3.1 (局部存在性)

考虑下列初始时间为 $\tau$ 的方程：

$\{\begin{array}{l}{\varphi }_t+\delta {\varphi }_{xxx}-\mu {\varphi }_{xx}+f{\left(\bar{u}+\varphi \right)}_x+\delta {\bar{u}}_{xxx}=0\\ {\varphi |}_{t=\tau }=\varphi \left(x,\tau \right)\end{array}$

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

3.2. 命题3.2 (先验估计)

在定理1的条件下，若使得 $\varphi \left(x,t\right)\in X\left(0,T\right)$ 为问题(1.2)的解，且我们做出如下先验假设：

$\underset{0\le t\le T}{\sup }{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\Vert {\partial }^k\varphi \left(t\right)\Vert }_{L^2}^2}\le c\epsilon$

C是正常数， $\epsilon$ 是依赖于初值及波的强度的一个充分小的正常数。根据以上先验假设，利用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。则我们可以得到先验估计：

${\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\Vert {\partial }^k\varphi \left(t\right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k}{\Vert {\partial }^{k+1}\varphi \left(x,t\right)\Vert }_{L^2}^2\le c{\Vert {\varphi }_0\Vert }_2^2$

3.3. 定理1的证明

由先验估计，我们可以将问题的局部存在性延拓到全局，我们接下来将进行具体的延拓证明。

首先，考虑初值 $\tau =0$ 时，则由定理条件及命题3.1可知存在合适小的正常数 $T_0$，使得：

$\underset{0\le t\le T_0}{\sup }{\Vert \varphi \left(t\right)\Vert }_2\le 2\delta$

我们令 $\delta$ 充分小。使得 $\delta <\frac{\epsilon }{2}$，则可得到：

$\underset{0\le t\le T_0}{\sup }{\Vert \varphi \left(t\right)\Vert }_2\le \epsilon ,0<t<T_0$

再由命题3.2，可知，在 $0<t<t_0$ 时，有：

${\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\Vert {\partial }^k\varphi \left(\tau \right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\displaystyle \underset{0}{\overset{t}{\int }}{\Vert {\partial }^{k+1}\varphi \left(x,\tau \right)\Vert }_{L^2}^2}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

故可得：

${\Vert \varphi \left(t_0\right)\Vert }_2\le \sqrt{C}{\Vert \varphi \left(x,0\right)\Vert }_2$

接下来，我们再将 $t=t_0$ 作为问题的初值，则由上式及命题3.1，存在一个充分小的正常数 $t_1$，使得方程在 $\left[t_0,t_0+t_1\right]$ 上有唯一解，且满足：

$\underset{t_0\le t\le t_0+t_1}{\sup }{\Vert \varphi \left(t\right)\Vert }_2\le 2\sqrt{c}\delta$

令 $\delta$ 合适小使其满足：

$\delta \le \frac{\epsilon }{2\sqrt{c}}$

则有：

$\underset{t_0\le t\le t_0+t_1}{\sup }{\Vert \varphi \left(t\right)\Vert }_2\le \epsilon$

再由先验估计可得在 $t\in \left[t_0,t_0+t_1\right]$ 有：

${\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\Vert {\partial }^k\varphi \left(\tau \right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\displaystyle \underset{0}{\overset{t}{\int }}{\Vert {\partial }^{k+1}\varphi \left(x,\tau \right)\Vert }_{L^2}^2}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

由此可得：

${\Vert \varphi \left(t_0+t_1\right)\Vert }_2\le \sqrt{C}{\Vert \varphi \left(x,0\right)\Vert }_2$

最后我们重复这个步骤，则只需要选取充分小的 $\delta$，我们就可以将方程解的局部存在性延拓到全局存在。且有：

${\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\Vert {\partial }^k\varphi \left(\tau \right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\displaystyle \underset{0}{\overset{t}{\int }}{\Vert {\partial }^{k+1}\varphi \left(x,\tau \right)\Vert }_{L^2}^2}}\le c{\Vert {\varphi }_0\Vert }_2^2$

则定理1得证。故此时我们只需得到命题3.2的证明。下个部分我们将用几个引理来证明先验估计。

4. 先验估计的证明

4.1. 引理4.1

在先验假设下，当问题(1.3)的初值及波的强度充分小的情况下，我们有：

${\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}+{\displaystyle {\int }_0^t{\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x\text{d}t}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

证明：在扰动方程(1.3)左右同时乘以 $2\varphi$，得到：

$2{\varphi }_t\varphi +2\delta {\varphi }_{xxx}\varphi -2\mu {\varphi }_{xx}\varphi +2\delta {\bar{u}}_{xxx}\varphi +2f{\left(\varphi +\bar{u}\right)}_x\varphi =0$

整理可得：

${\left({\varphi }^2\right)}_t+{\left[2\delta \varphi {\varphi }_x-2\mu \varphi {\varphi }_x-\delta {\varphi }_x^2\right]}_x+2\mu {\varphi }_x^2+2\delta {\bar{u}}_{xxx}\varphi +2\varphi f{\left(\varphi +\bar{u}\right)}_x=0$

${\left({\varphi }^2\right)}_t+{\left[2\delta \varphi {\varphi }_x-2\mu \varphi {\varphi }_x-\delta {\varphi }_x^2\right]}_x+2\mu {\varphi }_x^2+2\delta {\bar{u}}_{xxx}\varphi +2\varphi {\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_x+2\varphi f{\left(\bar{u}\right)}_x=0$

其中：

$\begin{array}{l}2\varphi {\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_x\\ =2\varphi \left[f^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right)-f^{\prime }\left(\bar{u}\right){\bar{u}}_x\right]\\ =2\varphi \left[f^{\prime }\left(\varphi +\bar{u}\right){\bar{u}}_x-f^{\prime }\left(\bar{u}\right){\bar{u}}_x+f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_x\right]\\ =2\varphi f^{″}\left(\xi \right)\varphi {\bar{u}}_x+2\varphi f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_x\\ =2\varphi f^{″}\left(\xi \right)\varphi {\bar{u}}_x+{\left(f^{\prime }\left(\varphi +\bar{u}\right){\varphi }^2\right)}_x-f^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }^2\end{array}$

故原式：

$\begin{array}{l}{\left({\varphi }^2\right)}_t+{\left[2\delta \varphi {\varphi }_x-2\mu \varphi {\varphi }_x-\delta {\varphi }_x^2\right]}_x+2\mu {\varphi }_x^2+2f^{″}\left(\xi \right){\bar{u}}_x{\varphi }^2\\ =f^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }^2-{\left(f^{\prime }\left(\varphi +\bar{u}\right){\varphi }^2\right)}_x-2\delta {\bar{u}}_{xxx}\varphi -2\varphi f{\left(\bar{u}\right)}_x\end{array}$

对上式关于 $x,t$ 在 $R_{+}\times \left[0,t\right]$ 上积分，可得：

$\begin{array}{l}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2}\text{d}x+2\mu {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}+2f^{″}\left(\xi \right){\bar{u}}_x{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2}}\text{d}x\text{d}t-{\displaystyle \underset{R^{+}}{\int }{\varphi }_0^2}\text{d}x\\ ={\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }^2\text{d}x\text{d}t}}-2\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\bar{u}}_{xxx}\varphi \text{d}x\text{d}t}}-{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }2\varphi f{\left(\bar{u}\right)}_x\text{d}x\text{d}t}}\\ ={\displaystyle \underset{i=1}{\overset{3}{\sum }}{I}_i}\end{array}$

我们首先来看 $2f^{″}\left(\xi \right){\bar{u}}_x{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2}}\text{d}x\text{d}t$ 项：

因为 $f\left(x\right)$ 是光滑函数，且由 $\bar{u}$ 的性质可知 $\bar{u}$ 有界，故：

$2f^{″}\left(\xi \right){\bar{u}}_x{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x\text{d}t}}\ge c{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2}}\text{d}x\text{d}t$

接下来我们将用先验估计及扩散波的性质对方程右侧的式子进行估计：

$\begin{array}{c}{I}_1={\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }^2\text{d}x\text{d}t}}\\ ={\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x{\varphi }^2\text{d}x\text{d}t}}+{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }^2\text{d}x\text{d}t}}\end{array}$

${\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x{\varphi }^2\text{d}x\text{d}t}}\le c{\Vert {\varphi }_x\Vert }_{L^{\infty }}{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{\varphi }^2\text{d}x\text{d}t}}\le c\epsilon {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{\varphi }^2\text{d}x\text{d}t}}$

${\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }^2\text{d}x\text{d}t}}\le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{\varphi }^2\text{d}x\text{d}t}}$

所以： $I_1\le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle {\int }_{R^{+}}{\varphi }^2\text{d}x\text{d}t}}$

$\begin{array}{c}{I}_2=2\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\bar{u}}_{xxx}\varphi \text{d}x\text{d}t}}\\ \le 2{\delta }^2{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\left(1+t\right)}^{-\frac{3}{2}}\omega \varphi \text{d}x\text{d}t}}\\ \le 2{\delta }^2{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\text{e}}^{-2\frac{\sigma x^2}{1+t}}}}\text{d}x\text{d}t+2{\delta }^2{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2}}\text{d}x\text{d}t\end{array}$

$\begin{array}{c}{I}_3={\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }2\varphi f{\left(\bar{u}\right)}_x\text{d}x\text{d}t}}\\ ={\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }2\varphi f^{\prime }\left(\bar{u}\right){\bar{u}}_x\text{d}x\text{d}t}}\\ \le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\left(1+t\right)}^{-\frac{1}{2}}{\text{e}}^{-\frac{\sigma x^2}{1+t}}\varphi \text{d}x\text{d}t}}\\ \le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\left(1+t\right)}^{-1}{\text{e}}^{-2\frac{\sigma x^2}{1+t}}\text{d}x\text{d}t}}+c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x\text{d}t}}\\ \le c\delta +c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x\text{d}t}}\end{array}$

将上述估计代入方程中：

${\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}+{\displaystyle {\int }_0^t{\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x\text{d}t}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

由此引理4.1得证。

4.2. 引理4.2

在先验假设下，当问题(1.3)的初值及波的强度充分小的情况下，我们有：

$\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x}+{\displaystyle {\int }_0^t\left(1+t\right)}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}+{\displaystyle {\int }_0^t\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_{xx}^2\text{d}x\text{d}t}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

证明：首先，我们对扰动方程关于x求导：

${\varphi }_{xt}+\delta {\varphi }_{xxxx}-\mu {\varphi }_{xxx}+\delta {\bar{u}}_{xxxx}+f{\left(\varphi +\bar{u}\right)}_{xx}=0$

在上式左右两边同乘 $2{\varphi }_x$ ：

$2{\varphi }_x{\varphi }_{xt}+2\delta {\varphi }_{xxxx}{\varphi }_x-2\mu {\varphi }_{xxx}{\varphi }_x+2\delta {\bar{u}}_{xxxx}{\varphi }_x+2f{\left(\varphi +\bar{u}\right)}_{xx}{\varphi }_x=0$

整理可得：

${\left({\varphi }_x^2\right)}_t+{\left[2\delta {\varphi }_{xxx}{\varphi }_x-2\mu {\varphi }_{xx}{\varphi }_x-\delta {\varphi }_{xx}^2\right]}_x+2\mu {\varphi }_{xx}^2+2\delta {\bar{u}}_{xxxx}{\varphi }_x+2f{\left(\varphi +\bar{u}\right)}_{xx}{\varphi }_x=0$

${\left({\varphi }_x^2\right)}_t+{\left[2\delta {\varphi }_{xxx}{\varphi }_x-2\mu {\varphi }_{xx}{\varphi }_x-\delta {\varphi }_{xx}^2\right]}_x+2\mu {\varphi }_{xx}^2+2\delta {\bar{u}}_{xxxx}{\varphi }_x+2{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_{xx}{\varphi }_x+2{\varphi }_{x}f{\left(\bar{u}\right)}_{xx}=0$

其中：

$\begin{array}{l}2{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_{xx}\\ =2f^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x^2-2f^{″}\left(\bar{u}\right){\bar{u}}_x^2+2f^{\prime }\left(\varphi +\bar{u}\right){\bar{u}}_{xx}-2f^{\prime }\left(\bar{u}\right){\bar{u}}_{xx}\\ +4f^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }_x+2f^{″}\left(\varphi +\bar{u}\right){\varphi }_x^2+2f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_{xx}\\ =2f^{‴}\left(\xi \right)\varphi {\bar{u}}_x^2+2f^{″}\left(\xi \right)\varphi {\bar{u}}_{xx}+4f^{″}(\varphi +\bar{u}){\bar{u}}_x{\varphi }_x+2f^{″}\left(\varphi +\bar{u}\right){\varphi }_x^2+2f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_{xx}\end{array}$

则：

$\begin{array}{l}2{\varphi }_x{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_{xx}\\ =2f^{‴}\left(\xi \right)\varphi {\bar{u}}_x^2{\varphi }_x+2f^{″}\left(\xi \right)\varphi {\bar{u}}_{xx}{\varphi }_x+4f^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }_x^2+2f^{″}\left(\varphi +\bar{u}\right){\varphi }_x^3+2f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_{xx}{\varphi }_x\end{array}$

对 $2f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_{xx}{\varphi }_x$ 有：

$2f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_{xx}{\varphi }_x={\left(f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_x^2\right)}_x-f^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }_x^2$

则原式：

$\begin{array}{l}{\left({\varphi }_x^2\right)}_t+{\left[2\delta {\varphi }_{xxx}{\varphi }_x-2\mu {\varphi }_{xx}{\varphi }_x-\delta {\varphi }_{xx}^2\right]}_x+2\mu {\varphi }_{xx}^2+4f^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }_x^2\\ =f^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }_x^2-{\left(f^{\prime }\left(\varphi +\bar{u}\right){\varphi }_x^2\right)}_x-2f^{‴}\left(\xi \right)\varphi {\bar{u}}_x^2{\varphi }_x\\ -2f^{″}\left(\xi \right)\varphi {\bar{u}}_{xx}{\varphi }_x-2f^{″}\left(\varphi +\bar{u}\right){\varphi }_x^3-2\delta {\bar{u}}_{xxxx}{\varphi }_x-{\varphi }_{x}f{\left(\bar{u}\right)}_{xx}\end{array}$

同样的，上式乘 $\left(1+t\right)$ 后关于 $x,t$ 在 $R_{+}\times \left[0,t\right]$ 上积分：

$\begin{array}{l}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2}\text{d}x+2\mu {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_{xx}^2\text{d}x\text{d}t}}+4{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }_x^2\text{d}x\text{d}t}}\\ ={\displaystyle \underset{R^{+}}{\int }{\varphi }_{0x}^2}\text{d}x+{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}+{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }_x^2\text{d}x\text{d}t}}-2{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{‴}\left(\xi \right)\varphi {\bar{u}}_x^2{\varphi }_x\text{d}x\text{d}t}}\\ -2{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\xi \right)\varphi {\bar{u}}_{xx}{\varphi }_x\text{d}x\text{d}t}}-2{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x^3\text{d}x\text{d}t}}-2\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\bar{u}}_{xxxx}{\varphi }_x\text{d}x\text{d}t}}\\ -{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_{x}f{\left(\bar{u}\right)}_{xx}\text{d}x\text{d}t}}\\ ={\displaystyle \underset{i=4}{\overset{11}{\sum }}{I}_i}\end{array}$

相同的，我们有：

$4{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }_x^2\text{d}x\text{d}t}}\ge c{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}$

下面我们估计方程右侧：

由引理4.1的结论可知： $I_5={\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$ ；

$\begin{array}{c}{I}_6={\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right)\left({\varphi }_x+{\bar{u}}_x\right){\varphi }_x^2\text{d}x\text{d}t}}\\ ={\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x{\varphi }_x^2\text{d}x\text{d}t}}+{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }_x^2\text{d}x\text{d}t}}\end{array}$

其中：

$\begin{array}{l}{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x{\varphi }_x^2\text{d}x\text{d}t}}\\ \le {\Vert {\varphi }_x\Vert }_{L^{\infty }}{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x^2\text{d}x\text{d}t}}\le c\epsilon {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\end{array}$

${\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\bar{u}}_x{\varphi }_x^2\text{d}x\text{d}t}}\le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}$

所以有：

$I_6\le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}$

$\begin{array}{c}{I}_7=2{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{‴}\left(\xi \right)\varphi {\bar{u}}_x^2{\varphi }_x\text{d}x\text{d}t}}\\ \le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\left(1+t\right)}^{-1}{\text{e}}^{-\frac{\sigma x^2}{1+t}}\varphi {\varphi }_x\text{d}x\text{d}t}}\\ \le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x\text{d}t}}+\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\end{array}$

$\begin{array}{c}{I}_8=2{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\xi \right)\varphi {\bar{u}}_{xx}{\varphi }_x\text{d}x\text{d}t}}\\ \le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\left(1+t\right)}^{-1}{\text{e}}^{-\frac{\sigma x^2}{1+t}}\varphi {\varphi }_x\text{d}x\text{d}t}}\\ \le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }^2\text{d}x\text{d}t}}+\delta {\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\end{array}$

$\begin{array}{c}{I}_9=2{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{f}^{″}\left(\varphi +\bar{u}\right){\varphi }_x^3\text{d}x\text{d}t}}\\ \le c{\Vert {\varphi }_x\Vert }_{L^{\infty }}{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\\ \le c\epsilon {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\end{array}$

$\begin{array}{c}{I}_{10}=2\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\bar{u}}_{xxxx}{\varphi }_x\text{d}x\text{d}t}}\\ \le c{\delta }^2{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\left(1+t\right)}^{-\frac{3}{2}}{\text{e}}^{-\frac{\sigma x^2}{1+t}}{\varphi }_x\text{d}x\text{d}t}}\\ \le c{\delta }^2{\displaystyle \underset{0}{\overset{t}{\int }}{\left(1+t\right)}^{-1}{\displaystyle \underset{R^{+}}{\int }{\text{e}}^{-2\frac{\sigma x^2}{1+t}}\text{d}x\text{d}t}}+c{\delta }^2{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\\ \le c\delta +c{\delta }^2{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\end{array}$

$\begin{array}{c}{I}_{11}={\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_{x}f{\left(\bar{u}\right)}_{xx}\text{d}x\text{d}t}}\\ ={\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x{f}^{\prime }\left(\bar{u}\right){\bar{u}}_{xx}\text{d}x\text{d}t}}+{\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x{f}^{″}\left(\bar{u}\right){\bar{u}}_x^2\text{d}x\text{d}t}}\\ \le c\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x{\left(1+t\right)}^{-1}\text{d}x\text{d}t}}\\ \le c\delta +c\delta {\displaystyle \underset{0}{\overset{t}{\int }}\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}\end{array}$

综上有：

$\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x}+{\displaystyle {\int }_0^t\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_x^2\text{d}x\text{d}t}}+{\displaystyle {\int }_0^t\left(1+t\right){\displaystyle \underset{R^{+}}{\int }{\varphi }_{xx}^2\text{d}x\text{d}t}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

则引理4.2成立。

4.3. 引理4.3

在先验假设下，当问题(1.3)的初值及波的强度充分小的情况下，我们有：

${\left(1+t\right)}^2{\displaystyle \underset{R^{+}}{\int }{\varphi }_{xx}^2\text{d}x}+{\displaystyle {\int }_0^t{\left(1+t\right)}^2{\displaystyle \underset{R^{+}}{\int }{\varphi }_{xxx}^2\text{d}x\text{d}t}}+{\displaystyle {\int }_0^t{\left(1+t\right)}^2{\displaystyle \underset{R^{+}}{\int }{\varphi }_{xx}^2\text{d}x\text{d}t}}\le c\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right)$

证明：扰动方程关于x求导：

${\varphi }_{xt}+\delta {\varphi }_{xxxx}-\mu {\varphi }_{xxx}+\delta {\bar{u}}_{xxxx}+f{\left(\varphi +\bar{u}\right)}_{xx}=0$

上式乘以 $-2{\varphi }_{xxx}$ ：

$-2{\varphi }_{xxx}{\varphi }_{xt}-2\delta {\varphi }_{xxxx}{\varphi }_{xxx}+2\mu {\varphi }_{xxx}^2-2\delta {\bar{u}}_{xxxx}{\varphi }_{xxx}-2f{\left(\varphi +\bar{u}\right)}_{xx}{\varphi }_{xxx}=0$

整理得：

${\left({\varphi }_{xx}^2\right)}_t-\delta {\left({\varphi }_{xxx}^2\right)}_x+2\mu {\varphi }_{xxx}^2-2\delta {\bar{u}}_{xxxx}{\varphi }_{xxx}-2{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_{xx}{\varphi }_{xxx}+2f{\left(\bar{u}\right)}_{xx}{\varphi }_{xxx}=0$

${\left({\varphi }_{xx}^2\right)}_t-\delta {\left({\varphi }_{xxx}^2\right)}_x+2\mu {\varphi }_{xxx}^2=2{\left[f\left(\varphi +\bar{u}\right)-f\left(\bar{u}\right)\right]}_{xx}{\varphi }_{xxx}-2f{\left(\bar{u}\right)}_{xx}{\varphi }_{xxx}+2\delta {\bar{u}}_{xxxx}{\varphi }_{xxx}$