广义BBM-Burgers方程初边值问题的L²衰减估计

L² Attenuation Estimation for the Initial Boundary Value Problem of the Generalized BBM-Burgers Equation

摘要: 本文主要研究如下本文研究的是广义BBM-Burgers方程扩散波的衰减估计问题，利用已经证明过的广义BBM-Burgers方程的解关于扩散波的渐近稳定，对该解进行衰减估计，并且在本文将证明广义BBM-Burgers方程解在L²范数下的衰减速度为(1t)−12。即对方程：{utf(u)x=uxxuxxtu(x,t)|t=0=u0(x),u(0,t)=u−在本文中我们将证明在波的强度δ:=|u−u−|及初值u0(x)适当小的情况下，广义BBM-Burgers方程的扰动方程形式的解的衰减速度为(1t)−12。

Abstract: This paper mainly studies as follows: This paper studies the attenuation estimation of the diffusion wave of the generalized BBM-Burgers equation. The solution of the generalized BBM-Burgers equation, which has been proved to be about the asymptotic stability of the diffusion wave, is used to estimate the attenuation of the solution. In this paper, it is proved that the decay rate of the solution of the generalized BBM-Burgers equation is(1t)−12under the L² norm. Immediate pair equation:{utf(u)x=uxxuxxtu(x,t)|t=0=u0(x),u(0,t)=u−in this paper, we will prove that if the wave intensityδ:=|u−u−|and the initial valueu0(x)are appropriately small, the decay velocity of the perturbation equation of the generalized BBM-Burgers equation is(1t)−12.

1. 引言

本文考虑了以下广义BBM-Burgers方程

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

BBM方程是Benjamin等人 [1] 在对流体动力学的物理研究中，由Korteweg de Vries方程精炼而成。自1940年Burgers方程提出以来，对经典Burgers方程有了非常全面的研究结果，徐红梅 [2] 等人证明了多维空间中BBM-Burgers方程的全局解存在初始数据，并且利用能量估计和时频分解的方法，得到该解的衰减估计，证明了衰减速率与热传导方程的相同；余沛 [3] 等人研究了带有分数阶扩散的Burgers方程在初值属于 $L^p$ 情况下的弱解的衰减估计；易菊燕 [4] 等人讨论在半空间中广义KDV-Burgers方程解的收敛到稀疏波的收敛率，并且使用 $L^1$ -估计导出了解渐近衰减到稀疏波 $L^p$ -估计；关于Burgers方程的衰减估计以及其他模型衰减估计的相关研究成果读者可参考文献 [5] - [20] 。

在BBM-Burgers方程的解整体存在的基础上，本文主要证明广义BBM-Burgers方程非线性扩散波的解的衰减速率为 ${\left(1+t\right)}^{-\frac{1}{2}}$ ，观察广义BBM-Burgers方程中含有热传导方程

${\bar{u}}_t-{\bar{u}}_{xx}=0$ (1.3)

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，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)的解已经收敛到(1.5)的自相似解，本文将利用能量估计来证明扰动方程的衰减估计，因此定义如下扰动方程

$\{\begin{array}{l}\varphi \left(x,t\right)=u\left(x,t\right)-\bar{u}\left(x,t\right)\\ {\varphi }_0\left(x\right)=u_0\left(x\right)-\bar{u}\left(x,0\right),\varphi \left(0,t\right)=0.\end{array}$ (1.4)

则由(1.1) (1.3) (1.4)得到扰动方程为

$\{\begin{array}{l}{\varphi }_t-{\varphi }_{xx}+f{\left(\bar{u}+\varphi \right)}_x-{\left(\varphi +\bar{u}\right)}_{xxt}=0\\ {\varphi }_0\left(x\right)=0.\end{array}$ (1.5)

下面则是本文的定理。

定理1：对于问题(1.7)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 ${\Vert u_0\left(x\right)-\bar{u}\left(x,0\right)\Vert }_{H^2(R_{+})}+\left|u_{+}-u_{-}\right|\le \delta <\epsilon$ ，初边值问题(1.7)存在唯一解 $u\left(x,t\right)$ ，且满足

$u\left(x,t\right)-\bar{u}\left(x,t\right)\in C\left(\left[0,\infty \right),H^2\left(R_{+}\right)\right)\cap L^2\left(\left[0,\infty \right),H^3\left(R_{+}\right)\right).$ (1.6)

进一步解 $u\left(x,t\right)$ 的衰减速度为

${\displaystyle \underset{k=0}{\overset{1}{\sum }}{(1+t)}^{k+1}{\Vert {\partial }_x^k\left(\left(u-\bar{u}\right),{\left(u-\bar{u}\right)}_x\right)\left(t\right)\Vert }^2}+{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\displaystyle {\int }_0^t{\left(1+t\right)}^{k+1}{\Vert {\partial }_x^k\left(\left(u-\bar{u}\right),{\left(u-\bar{u}\right)}_x\right)\left(\tau \right)\Vert }^2}\text{d}\tau }\le C\left({\Phi }_0+\delta \right).$ (1.7)

2. 预备知识

引理2.1 (Sobolev不等式)对于任意函数 $f\left(x\right)\in H^1\left(R\right)$ ，有

${\Vert f\Vert }_{L^{\infty }}\le C{\Vert f\Vert }^{\frac{1}{2}}{\Vert f_x\Vert }^{\frac{1}{2}}.$ (2.1.1)

引理2.2(Young不等式)设 $1<p$ ， $q<\text{+}\infty$ ，满足 $\frac{1}{p}+\frac{1}{q}=1$ ，则有

$ab<\frac{a^p}{p}+\frac{b^q}{q}\left(a,b>0\right).$ (2.2.2)

特别地，当 $p=q=2$ 时，我们称之为Cauchy不等式。

以下均设 $\Omega \subset R^n$ 。

在本文中，对于函数空间而言， $L^p=L^p\left(R_{+}\right)\left(1\le p<\infty \right)$ 表示通常的Lebesgue空间，其范数可定义为

${\Vert f\Vert }_{L^p}={\left({\displaystyle {\int }_{R_{+}}{\left|f\right|}^p\text{d}x}\right)}^{\frac{1}{p}},1\le p<\infty ,{\Vert f\Vert }_{L^{\infty }}=\underset{x\in R^n}{ess\sup }\left|f\left(x\right)\right|.$ (2.2.3)

$H^l$ 表示通常的Sobolev空间，其范数为

${\Vert f\Vert }_{H^l}={\displaystyle \underset{k=0}{\overset{l}{\sum }}{\Vert {\partial }_x^{k}f\left(x\right)\Vert }_{L^2}^2}\text{d}x$ (2.2.4)

特别地，当 $l=0$ 时，记 ${\Vert \cdot \Vert }_0={\Vert \cdot \Vert }_{L^2}=\Vert \cdot \Vert$ ，在不会导致混淆的情况下，我们将区域 $R_{+}$ 省略不写，并且大写字母C和 $O\left(1\right)$ 表示一般常数。

3. 定理1的证明

为了证明定理1，先将给出先验假设，再将证明先验估计从而证明定理1。

命题3.1 (先验假设)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(1.7)的解，则我们作出如下的先验假设：

$\underset{0\le t\le T}{\sup }{\displaystyle \underset{k=0}{\overset{2}{\sum }}{\left(1+t\right)}^k{\Vert {\partial }_x^k\varphi \left(t\right)\Vert }_{L^2}^2}\le C\epsilon .$ (3.1)

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：

${\displaystyle \underset{k=0}{\overset{1}{\sum }}{\left(1+t\right)}^{k+1}{\Vert {\partial }^k\left(\varphi ,{\varphi }_x\right)\left(t\right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{1}{\sum }}{{\displaystyle {\int }_0^t{\left(1+t\right)}^{k+1}\Vert {\partial }_x^k\left(\varphi ,{\varphi }_x\right)\left(t\right)\Vert }}_{L^2}^2}\text{d}\tau \le C\left({\Vert {\varphi }_0\Vert }_2^2+\delta \right).$ (3.2)

引理3.2：在先验假设(3.1)的情况下，对于充分小的正数 $\delta$ ，有

${\displaystyle {\int }_0^{+\infty }\left(1+t\right)\left({\varphi }^2+{\varphi }_x^2\right)\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2}}\text{d}x\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2}}\text{d}x\text{d}\tau \le C\left({\Phi }_0+\delta \right).$ (3.2.1)

证明：在扰动方程(1.5)两边同时乘以 $\left(1+t\right)\varphi$ 可得

$\left(1+t\right)\varphi {\varphi }_t-\left(1+t\right)\varphi {\varphi }_{xx}+\left(1+t\right)\varphi f{\left(\bar{u}+\varphi \right)}_x-\left(1+t\right)\varphi {\left(\varphi +\bar{u}\right)}_{xxt}=0.$ (3.2.2)

在 $\left[0,t\right]\times \left[0,+\infty \right)$ 上对(3.2.2)进行积分可得

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_t}}\text{d}x\text{d}\tau -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_{xx}}}\text{d}x\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f{\left(\bar{u}+\varphi \right)}_x}}\text{d}x\text{d}\tau \\ -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\left(\varphi +\bar{u}\right)}_{xxt}}}\text{d}x\text{d}\tau =0.\end{array}$ (3.2.3)

对(3.2.3)的各项运用分部积分进行计算

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_t}}\text{d}x\text{d}\tau =\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\left(1+t\right){\varphi }^2|}_0^t}\text{d}x-\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}\tau ,$ (3.2.4)

$\begin{array}{l}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_{xx}}}\text{d}x\text{d}\tau =-{\displaystyle {\int }_0^t{\left(1+t\right)\varphi {\varphi }_x|}_0^{+\infty }}\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x^2}}\text{d}x\text{d}\tau \\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x^2}}\text{d}x\text{d}\tau ,\end{array}$ (3.2.5)

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f{\left(\bar{u}+\varphi \right)}_x}}\text{d}x\text{d}\tau \\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\left[f\left(\bar{u}+\varphi \right)-f\left(\bar{u}\right)\right]}_x}}\text{d}x\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f{\left(\bar{u}\right)}_x}}\text{d}x\text{d}\tau \\ =A+B.\end{array}$ (3.2.6)

对于式(3.2.6)中的A通过泰勒展开式(其中 $\bar{u}<\xi <\bar{u}+\varphi$ )则有

$\begin{array}{l}A={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\left[\varphi f^{\prime }\left(\xi \right)\right]}_x}}\text{d}x\text{d}\tau \\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_x{f}^{\prime }\left(\xi \right)\text{d}x\text{d}\tau }}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2{f}^{″}\left(\xi \right)\text{d}x\text{d}\tau }}，\end{array}$ (3.2.7)

$B={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}\tau .$ (3.2.8)

将(3.2.7)与(3.2.8)代入到(3.2.6)中可得

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f{\left(\bar{u}+\varphi \right)}_x}}\text{d}x\text{d}\tau ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_x{f}^{\prime }\left(\xi \right)\text{d}x\text{d}\tau }}\\ +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2{f}^{″}\left(\xi \right)\text{d}x\text{d}\tau }}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}\tau ,\end{array}$ (3.2.9)

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }(1+t)\varphi {\left(\varphi +\bar{u}\right)}_{xxt}}}\text{d}x\text{d}\tau ={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_{xxt}}}\text{d}x\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\bar{u}}_{xxt}}}\text{d}x\text{d}\tau .$ (3.2.10)

其中式(3.2.10)右侧的第一项分部积分可得

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_{xxt}}}\text{d}x\text{d}\tau ={\displaystyle {\int }_0^t{\left(1+t\right)\varphi {\varphi }_{xt}|}_0^{+\infty }}\text{d}\tau -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}\tau \\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x{\varphi }_{xt}}}\text{d}x\text{d}\tau \\ =-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }\left(1+t\right){{\varphi }_x^2|}_0^t}\text{d}x+\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}\tau .\end{array}$ (3.2.11)

式(3.2.10)右侧的第二项通过分部积分可得

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\bar{u}}_{xxt}}}\text{d}x\text{d}\tau ={\displaystyle {\int }_0^t{\left(1+t\right)\varphi {\bar{u}}_{xt}|}_0^{+\infty }}\text{d}\tau -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x{\bar{u}}_{xt}}}\text{d}x\text{d}\tau \\ =-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x{\bar{u}}_{xt}}}\text{d}x\text{d}\tau .\end{array}$ (3.2.12)

将(3.2.11)与(3.2.12)代入到(3.2.10)中可得

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\left(\varphi +\bar{u}\right)}_{xxt}}}\text{d}x\text{d}\tau =-\frac{1}{2}{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x^2{|}_0^t}\text{d}x+\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}\tau -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x{\bar{u}}_{xt}}}\text{d}x\text{d}\tau .$ (3.2.13)

将(3.2.4)~(3.2.5)(3.2.9)以及(3.2.13)代入到(3.2.3)中可得到

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\left(1+t\right){\varphi }^2|}_0^t}\text{d}x+\frac{1}{2}{\displaystyle {\int }_0^{+\infty }{\left(1+t\right){\varphi }_x^2|}_0^t}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x^2}}\text{d}x\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2{f}^{″}\left(\xi \right)\text{d}x\text{d}\tau }}\\ =\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}\tau -{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_x{f}^{\prime }\left(\xi \right)\text{d}x\text{d}\tau }}-{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}\tau \\ -\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x{\bar{u}}_{xt}}}\text{d}x\text{d}\tau .\end{array}$ (3.2.14)

对式(3.2.14)右侧的各项运用能量法进行估计

$\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }^2}}\text{d}x\text{d}\tau \le C\left({\Phi }_0+\delta \right),$ (3.2.15)

$\begin{array}{l}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi {\varphi }_x{f}^{\prime }\left(\xi \right)\text{d}x\text{d}\tau }}\le C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2}}\text{d}x\text{d}\tau +C\epsilon {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}\tau \\ \le C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2}}\text{d}x\text{d}\tau +C\left({\Phi }_0+\delta \right),\end{array}$ (3.2.16)

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\varphi f^{\prime }\left(\bar{u}\right){\bar{u}}_x}}\text{d}x\text{d}\tau \le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2}}\text{d}x\text{d}\tau ,$ (3.2.17)

$\frac{1}{2}{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }{\varphi }_x^2}}\text{d}x\text{d}\tau \le C\left({\Phi }_0+\delta \right),$ (3.2.18)

${\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x{\bar{u}}_{xt}}}\text{d}x\text{d}\tau \le C\delta +C\delta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x^2}}\text{d}x\text{d}\tau .$ (3.2.19)

将(3.2.15)~(3.2.19)代入到(3.2.14)中整理可得

${\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2}\text{d}x+{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }_x^2}}\text{d}x\text{d}\tau +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right){\varphi }^2}}\text{d}x\text{d}\tau \le C\left({\Phi }_0+\delta \right).$ (3.2.20)

因此引理3.2得证。

类似可得到更高阶的估计

${\displaystyle {\int }_0^{+\infty }{\left(1+t\right)}^2\left({\varphi }_x^2+{\varphi }_{xx}^2\right)\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^{+\infty }\left(1+t\right)\left({\varphi }_x^2+{\varphi }_{xx}^2\right)\text{d}x\text{d}\tau }}\le C\left({\Phi }_0+\delta \right).$ (3.2.21)

又因为 $\varphi \left(x,t\right)=u\left(x,t\right)-\bar{u}\left(x,t\right)$ ，所以我们有

${\displaystyle \underset{k=0}{\overset{1}{\sum }}{\left(1+t\right)}^{k+1}{\Vert {\partial }^k\left(\left(u-\bar{u}\right),{\left(u-\bar{u}\right)}_x\right)\left(t\right)\Vert }_{L^2}^2}+{\displaystyle \underset{k=0}{\overset{1}{\sum }}{{\displaystyle {\int }_0^t{\left(1+t\right)}^{k+1}\Vert {\partial }_x^k\left(\left(u-\bar{u}\right),{\left(u-\bar{u}\right)}_x\right)\left(\tau \right)\Vert }}_{L^2}^2}\text{d}\tau \le C\left({\Phi }_0+\delta \right).$ (3.2.22)

由此定理1的证明完成，即证明了广义BBM-Burgers方程初边值问题的L²衰减估计为 ${\left(1+t\right)}^{-\frac{1}{2}}$ 。