数学杂志  2018, Vol. 38 Issue (6): 1049-1053   PDF    
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徐红梅
朱丽丽
高维BBM-Burgers方程解的衰减估计
徐红梅, 朱丽丽    
河海大学理学院, 江苏 南京 211100
摘要:本文研究了多维空间的Benjamin-Bona-Mahony-Burgers方程.在解整体存在的前提下,利用能量积分、时频分解等方法,得到此方程柯西问题解的衰减估计,并且此方程的衰减速率与热传导方程的相同.
关键词BBM-Burgers方程    时频分解    衰减估计    
THE DECAY ESTIMATE OF SOLUTIONS FOR BBM-BURGERS EQUATION IN MULTI DIMENSION
XU Hong-mei, ZHU Li-li    
College of Science, Hohai University, Nanjing 211100, China
Abstract: In this paper, Benjamin-Bona-Mahony-Burgers equation in multi-dimensional space is investigated. When the global solution of this equation with initial data exists, by using methods of energy estimate and time-frequency decomposition, we obtain the decay estimate of the solution, and the decay rate is the same as that of the heat conduction equation.
Keywords: BBM-Burgers equation     time-frequency decomposition     decay estimate    
1 引言

本文研究了多维空间Benjamin-Bona-Mahony-Burgers方程

$ \begin{equation} \left\{ \begin{aligned} \ u_t + \sum\limits_{j=1}^n (u^2)_{x_j} &= \delta \sum\limits_{j=1}^n u_{x_j x_j t}+\sum\limits_{j=1}^n (\sum\limits_{l=1}^N (-1)^{l+1} \gamma_l \partial_{x_j}^{2l} u) , \ x \in \mathbb{R}^n , \ t > 0 \ , \\ \ u(x , 0) = u_0(x) \end{aligned} \right. \end{equation} $ (1.1)

解的$\ L^2$衰减估计.此处$ n$表示空间维数, $n \geq 2$, $ N \geq 2$正整数, $\delta, \gamma_l > 0$常数, $ u_0 \in H^{1+[\frac{n}{2}]}$, $ H^l (\mathbb{R}^n)$是一般的Sobolev空间.

BBM方程是Benjamin, Bona和$Mahony [1]在对流体动力学的物理研究中, 由Kortewey-deVries (Kdv)方程精炼而成.在描述非线性色散系统中小振幅长波的传播时, 需要考虑耗散机制, 由此产生了BBM-Burgers方程(1.1).如文献[1, 2]指出, 此方程中$\ \gamma_1 \sum\limits_{j=1}^n \partial_{x_j x_j} u$$\ \gamma_2 \sum\limits_{j=1}^n \partial_{x_j}^4 u$有不同的物理背景, 称为粘性项和耗散项.关于此方程解的存在性和大时间状态得到了广泛的关注, 很多数学工作者对其作出了详细研究.

$N=1$时, Schonbek [3]讨论了方程(1.1)解的存在性和$ u(x, t; \delta_1, \gamma_1)$$\delta \rightarrow 0, \gamma_1 \rightarrow 0$时的收敛性.在文献[4]中, 赵等人得到了$N=2$时方程(1.1)解的存在性和收敛性.在文献[5]中, 王和张得到了$ N=2$, 空间维数$ 2 \leq n \leq 6$时解的整体存在性和衰减估计.在文献[6, 7]中, Kondo和Webler分别给出了一维和多维空间情况下$(1.1)$解的存在性和收敛性.本文中, 延续文[7]的结论, 在解整体存在的前提下, 研究解的衰减估计.

本文中, $\ C$表示一般常数, $\ L_p$为Lebesgue可测空间. $F(f)$$\widehat{f}$代表函数$f(x)$的傅里叶变换, 且$F(f)=\displaystyle\int e^{-i x \xi} f(x)\, dx$. $F^{-1}(\widehat{f})$$f$表示函数$\widehat{f}$的逆傅里叶变换.

本文安排如下, 在第二节中给出一些准备工作, 如方程(1.1)的解的存在性结论, 解的表达式等.第三节, 用能量估计、时频分解等工具给出解的衰减估计.

2 准备工作

本文是在解的整体存在前提下做出的, 为了读者方便, 先列出文献[7]的结论.

定理2.1  若$\ u_0 \in H^{1+[\frac{n}{2}]}$, 则(1.1)式存在整体解$\ u \in C((0, \infty); H^s (\mathbb{R}^n)), s\geq 1+[\frac{n}{2}]$.对方程(1.1)的变量$\ x$作傅里叶变换, 得到

$ \begin{equation*} \widehat{u_t}+\delta |\xi|^2 \widehat{u_t}-\sum\limits_{j=1}^n (\sum\limits_{l=1}^N (-1)^{l+1} \gamma_l \cdot (-1)^l \xi_j^{2l}) \widehat u = -\sum\limits_{j=1}^n i \xi_j F(u^2).\\ \end{equation*} $

若令

$ \begin{equation} \widehat{G} = e^{- \frac{\sum\limits_{l=1}^N \gamma_l \sum\limits_{j=1}^n \xi_j^{2l}}{1+\delta |\xi|^2} t} \ , \ \widehat{H} = \frac{1}{1+\delta |\xi|^2} \widehat{G}\ , \\ \end{equation} $ (2.1)

$ \begin{equation} \widehat{u}(\xi , t) = \widehat{G} \widehat{u_0}-\sum\limits_{j=1}^n \displaystyle\int_0^t \widehat{H}(t-s) \cdot i \xi_j F(u^2)(s)\, ds.\\ \end{equation} $ (2.2)

下面分析$G$的衰减.

定理2.2  存在常数$C$, 有$\ \Vert{\partial_x^\alpha G}\Vert_{L_2} \leq C(1+t)^{-\frac{n}{4}- \frac{|\alpha|}{2}}$, $\Vert{\partial_x^\alpha H}\Vert_{L_2} \leq C(1+t)^{-\frac{n}{4}- \frac{|\alpha|}{2}}$.

  由(2.1)式和Parseval等式, 得

$ \begin{eqnarray} \Vert{\partial_x^\alpha G}\Vert_{L_2} &=& \Vert{\xi^\alpha \widehat{G} }\Vert_{L_2} \notag\\ &\leq& (\displaystyle\int_{|\xi|^2 \leq \frac{1}{\delta}} |\xi|^{2|\alpha|} e^{- \frac{2\gamma_1 \sum\limits_{j=1}^n \xi_j^2}{2} t}\, d\xi)^{\frac{1}{2}} + (\displaystyle\int_{|\xi|^2 \geq \frac{1}{\delta}} |\xi|^{2|\alpha|} e^{- \frac{2 \sum\limits_{l=1}^2 \gamma_l \sum\limits_{j=1}^n \xi_j^{2l}}{2 \delta |\xi|^2} t}\, d\xi)^{\frac{1}{2}} \notag\\ &\leq& (\displaystyle\int_{|\xi|^2 \leq \frac{1}{\delta}} (|\xi| \sqrt{t})^{2|\alpha|} e^{- \gamma_1 |\xi|^2 t} t^{-|\alpha| - \frac{n}{2}}\, d{\xi \sqrt{t}})^{\frac{1}{2}} + (\displaystyle\int_{|\xi|^2 \geq \frac{1}{\delta}} |\xi|^{2|\alpha|} e^{- \frac{\gamma_1}{\delta} t} e^{-\frac{\gamma_2 |\xi|^4}{\delta |\xi|^2} t} \, d\xi)^{\frac{1}{2}} \notag\\ &\leq& C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}. \end{eqnarray} $ (2.3)

由(2.1), (2.3)式和Parseval等式, 得

$ \begin{equation*} \Vert{\partial_x^\alpha} H \Vert_{L_2}=\Vert{\xi^\alpha \frac{1}{1+\alpha |\xi|^2} \widehat G}\Vert_{L_2} \leq \Vert{\xi^\alpha \widehat G}\Vert_{L_2} \leq C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}.\notag\\ \end{equation*} $

定理得证.

3 衰减估计

由能量积分, 可得下述定理.

定理3.1  若$\ u_0 \in H^{1+[\frac{n}{2}]}(\mathbb{R}^n)$, 可得$\ u \in H^1 (\mathbb{R}^n)$.

  在(1.1)式两边同乘以$\ u$, 再关于变量$\ x \ $积分得

$ \begin{equation} \frac{1}{2} \frac{d}{dt} \Vert{u}\Vert_{L_2}^2 + \frac{\delta}{2} \frac{d}{dt} \sum\limits_{j=1}^n \Vert{u_{x_j}}\Vert_{L_2}^2+\sum\limits_{j=1}^n {\sum\limits_{l=1}^N {\gamma_l \Vert{\partial_{x_j}^l u}\Vert_{L_2}^2}}=-\displaystyle\int_{R^n} {u \sum\limits_{j=1}^n (u^2)_{x_j} }\, dx=0. \end{equation} $ (3.1)

于是$\Vert{u}\Vert_{L_2}^2+ \delta \sum\limits _{j=1}^n \Vert{u_{x_j}}\Vert_{L_2}^2 \leq \Vert{u_0}\Vert_{L_2}^2+ \delta \sum\limits _{j=1}^n \Vert{\partial_{x_j} u_0}\Vert_{L_2}^2.$定理得证.

作光滑截断函数$\chi_0(\eta)= \begin{cases} 1, & \ {|\eta| \leq 1}\\ 0, & \ {|\eta| > 2}. \end{cases}$并令$\ \chi (t, \xi)=\chi_0 ((1+t)|\xi|^2).\ $定义时频算子$\ \chi (t, D)$, 它的特征$\ \chi (t, \xi).\ $$\ u_L (x, t)=\chi (t, D) u(x, t)$, 则对$\ u_L (x, t)\ $有下述估计.

定理3.2  $\ u_0 \in L_1$, 有$\ \Vert{\partial_x^\alpha u_L (x, t)}\Vert_{L_2} \leq C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}.$

  由(2.2)式和Minkowski不等式, 得

$ \begin{equation} \Vert{\partial_x^\alpha u_L (x, t)}\Vert_{L_2} \leq \Vert{\partial_x^\alpha G}\Vert_{L_2} \Vert {u_0}\Vert_{L_1}+(\sum\limits_{j=1}^n {\displaystyle\int_0^t \Vert{\chi (t, D) \partial_x^\alpha \partial_{x_j} H(t-s)}\Vert_{L_2}^2 \Vert{u^2}\Vert_{L_1}^2 (s)ds})^{\frac{1}{2}}. \end{equation} $ (3.2)

$\ u_0 \in L_1 \ $和定理2.2,

$ \begin{equation} \Vert{\partial_x^\alpha G} \Vert_{L_2} \Vert u_0 \Vert_{L_1} \leq C(1+t)^{-\frac{n}{4} -\frac{|\alpha|}{2}}. \end{equation} $ (3.3)

由定理3.1和(2.1)式, 得

$ \begin{array}{l} \;\;\;\;(\sum\limits_{j=1}^n {\displaystyle\int_0^t \Vert{\chi (t, D) \partial_x^\alpha \partial_{x_j} H(t-s)}\Vert_{L_2}^2 \Vert{u^2}\Vert_{L_1}^2 (s)\, ds})^{\frac{1}{2}} \notag\\ \leq (\sum\limits_{j=1}^n {\displaystyle\int_0^t {(\displaystyle\int_{{|\xi|^2 t} \leq 2} {|\xi|^{2 |\alpha|+2}} \, d\xi) \Vert u \Vert_{L_2}^4} \, ds})^{\frac{1}{2}}\notag\\ \leq (\sum\limits_{j=1}^n {\displaystyle\int_0^t {(\displaystyle\int_{{{|\xi|}^2 t} \leq 2} {(|\xi| \sqrt t)^{2 |\alpha|+2} t^{-|\alpha|-1-{\frac{n}{2}}}} \, d{\xi \sqrt t}) \Vert u \Vert_{L_2}^4} \, ds})^{\frac{1}{2}} \leq C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}. \end{array} $ (3.4)

由(3.2)-(3.4)式, 定理得证.

定理3.3  $\Vert u \Vert_{H^1 (\mathbb{R}^n)} \leq C(1+t)^{-\frac{n}{4}}.$

  令$\ \eta (t) = \sqrt{\frac{\mu}{1+t}}$, $\mu > 0$待定.由Parseval等式,

$ \begin{eqnarray} \sum\limits_{j=1}^n {\Vert{ \partial_{x_j} u }\Vert_{L_2}^2 } &=& \displaystyle\int {|\xi|^2 |\widehat{u}|^2}\, d\xi \notag \geq \displaystyle\int_{|\xi| \geq \eta (t)} {|\xi|^2 |\widehat{u}|^2}\, d\xi \notag\\ & \geq&\eta^2 (t) ({\Vert u \Vert_{L_2}^2} - \displaystyle\int_{|\xi| \leq \eta (t)} {|\widehat{u}|^2}\, d\xi) = \eta^2 (t) ({\Vert u \Vert_{L_2}^2} - \Vert{ \chi (t, D) u }\Vert_{L_2}^2 ). \end{eqnarray} $ (3.5)

同理

$ \begin{eqnarray} \sum\limits_{j=1}^n {\Vert{\partial_{x_j x_j} u}\Vert_{L_2}^2} & \geq&C_1 \displaystyle\int {|\widehat{u}|^2 |\xi|^4}\, d\xi \notag\\ & \geq&C_1 \eta^2 (t) (\sum\limits_{j=1}^n {\Vert {\partial _{x_j} u} \Vert_{L_2}^2} - \sum\limits_{j=1}^n \Vert{ \chi (t, D) \partial_{x_j} u }\Vert_{L_2}^2 ). \end{eqnarray} $ (3.6)

由(3.5), (3.6)和(3.1)式, 得

$ \begin{array}{l} \;\;\;\;\;\frac{d}{dt} \Vert{u}\Vert_{L_2}^2 + \delta \frac{d}{dt} (\sum\limits_{j=1}^n {\Vert{\partial_{x_j} u} \Vert_{L_2}^2}) + \gamma_1 \eta^2 (t) {\Vert u\Vert_{L_2}^2} + \gamma_2 \eta^2 (t) (\sum\limits_{j=1}^n {\Vert{\partial_{x_j} u}\Vert_{L_2}^2}) \notag\\ \leq\gamma_1 \eta^2 (t) {\Vert {\chi (x, D) u}\Vert_{L_2}^2} + \gamma_2 \eta^2 (t) \sum\limits_{j=1}^n {\Vert{\chi (x, D) \partial_{x_j} u}\Vert_{L_2}^2}. \end{array} $ (3.7)

$\ P=\Vert{u}\Vert_{L_2}^2+\delta \sum\limits_{j=1}^n \Vert{\partial_{x_j} u}\Vert_{L_2}^2$, 由定理3.2和(3.7)式, 得

$ \begin{eqnarray*} &&\frac{dP}{dt}+\gamma_1 \frac{\mu}{1+t} \Vert{u}\Vert_{L_2}^2 +\gamma_2 \frac{\mu}{1+t} (\sum\limits_{j=1}^n \Vert{\partial_{x_j} u}\Vert_{L_2}^2) \leq C(1+t)^{-1-\frac{n}{2}}, \\&& \frac{dP}{dt}+ \min(\gamma_1, \frac{\gamma_2}{\delta}) \frac{\mu}{1+t} P \leq C(1+t)^{-1-\frac{n}{2}}. \end{eqnarray*} $

$\ a = \min(\gamma_1, \frac{\gamma_2}{\delta}) \mu > 0$, 则

$ \begin{eqnarray*} &&\frac{d(e^{\displaystyle\displaystyle\int_0^t \frac{a}{1+s} \, ds} P(t))}{dt} = e^{\displaystyle\displaystyle\int_0^t \frac{a}{1+s} \, ds} (\frac{a}{1+t} P(t)+\frac{dP}{dt})\\ & \leq&C e^{\displaystyle\displaystyle\int_0^t \frac{a}{1+s} \, ds} (1+t)^{-1-\frac{n}{2}} = C (1+t)^{a-1-\frac{n}{2}}. \end{eqnarray*} $

$\ a-\frac{n}{2} > 1$, 则$P(t) \leq C(1+t)^{-\frac{n}{2}}, $定理得证.

再由数学归纳法可得本文结论.

定理3.4  当$\ u_0 \in {L_1 \cap H^{1+[\frac{n}{2}]}} (\mathbb{R}^n)$, 有$\ \Vert{\partial_x^\alpha u}\Vert_{L_2} \leq C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}.$

  当$\ |\alpha| = 0$时, 由定理3.3可得.由(2.2)式和Minkowski不等式, 有

$ \begin{equation} \Vert{\partial_x^\alpha u}\Vert_{L_2}^2 \leq \Vert{\partial_x^\alpha G}\Vert_{L_2}^2 \Vert u_0 \Vert_{L_1}^2 + \sum\limits_{j=1}^n {\displaystyle\displaystyle\int_0^t {\Vert{\partial_x^\alpha \partial_{x_j} H(t-s)}\Vert_{L_2}^2 \Vert{u^2 (s)}\Vert_{L_1}^2}\, ds.} \end{equation} $ (3.8)

$\ |\alpha|=1$时, 由定理2.2和定理3.3得

$ \begin{eqnarray} &&\displaystyle\displaystyle\int_0^t {\Vert{\partial_x^\alpha \partial_{x_j} H(t-s)}\Vert_{L_2}^2 \Vert{u^2(s)}\Vert_{L_1}^2} \, ds \leq \displaystyle\displaystyle\int_0^t {(1+t-s)^{-\frac{n}{2}-1-1} \Vert u \Vert_{L_2}^4} \, ds \notag\\ &\leq&\displaystyle\displaystyle\int_0^t (1+t-s)^{-\frac{n}{2}-2} (1+s)^{-n} \, ds \leq C(1+t)^{-1-\frac{n}{2}}. \end{eqnarray} $ (3.9)

由(3.3), (3.8)和(3.9)式得, 当$\ |\alpha|=1$时, 定理成立.

假设当$\ |\alpha|=k$时, 定理成立.则当$\ |\alpha|=k+1$时, 不妨设$\ \alpha=(\alpha_1, \alpha_2, \cdots, \alpha_n)$$\ \alpha_1 > 1$.令$\ \beta=(\alpha_1 -1, \alpha_2, \cdots, \alpha_n)$, 则$\ |\beta|=k$, 且

$ \begin{eqnarray*} &\ &\displaystyle\displaystyle\int_0^t {\Vert{\partial_x^\alpha \partial_{x_j} H(t-s)}\Vert_{L_2}^2 \Vert{u^2(s)}\Vert_{L_1}^2} \, ds\\ & =&\displaystyle\displaystyle\int_0^{\frac{t}{2}} {\Vert{\partial_x^\alpha \partial_{x_j} H(t-s)}\Vert_{L_2}^2 \Vert{u^2(s)}\Vert_{L_1}^2} \, ds + \displaystyle\displaystyle\int_{\frac{t}{2}}^t {\Vert{\partial_{x_j} \partial_{x_1} H(t-s)}\Vert_{L_2}^2 \Vert{\partial_x^{\beta} u^2 (s)}\Vert_{L_1}^2}\, ds\\ & =&\displaystyle\displaystyle\int_0^{\frac{t}{2}} {(1+t-s)^{-\frac{n}{2}-|\alpha|-1} (1+s)^{-n}}\, ds \\&&+ \sum\limits_{{\beta}_1 + {\beta}_2 = \beta} {\displaystyle\displaystyle\int_{\frac{t}{2}}^t {(1+t-s)^{-\frac{n}{2}-2} {\Vert{\partial_x^{{\beta}_1} u(s)}\Vert_{L_2}^2} {\Vert{\partial_x^{{\beta}_2} u(s)}\Vert_{L_2}^2}}\, ds}\\ & \leq&C(1+t)^{-\frac{n}{2}-|\alpha|}+C \sum\limits_{{\beta}_1 + {\beta}_2 = {\beta}} {\displaystyle\displaystyle\int_{\frac{t}{2}}^t {(1+t-s)^{-\frac{n}{2}-2} (1+s)^{-\frac{n}{2}-|{\beta}_1|} (1+s)^{-\frac{n}{2}-|{\beta}_2|} }\, ds}\\ & \leq&C(1+t)^{-\frac{n}{2}-|\alpha|} + C(1+t)^{-n-|\alpha|+1} \leq C(1+t)^{-\frac{n}{2}-|\alpha|}. \end{eqnarray*} $

由数学归纳法

$ \begin{equation*} (\displaystyle\displaystyle\int_0^t {\Vert{\partial_x^\alpha \partial_{x_j} H(t-s)}\Vert_{L_2}^2 \Vert{u^2 (s)}\Vert_{L_1}^2} \, ds)^{\frac{1}{2}} \leq C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}. \end{equation*} $ (3.10)

由(3.3), (3.8)和(3.10)式, 定理得证.

参考文献
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