数学物理学报, 2019, 39(6): 1443-1455 doi:

论文

三维Landau-Lifshitz-Gilbert方程的解的衰减估计

林俊宇,, 万明练,, 曹木花,

Decay Estimates of the Global Solution for the Landau-Lifshitz-Gilbert Equation in Three Dimensions

Lin Junyu,, Wan Minglian,, Cao Muhua,

收稿日期: 2018-07-23  

基金资助: 国家自然科学基金.  11571117
广东省自然科学基金.  2016A030313451
广州市科技计划项目.  201707010040

Received: 2018-07-23  

Fund supported: NSFC.  11571117
the Natural Science Foundation of Guangdong Province.  2016A030313451
the Guangzhou Science and Technology Plan Projects.  201707010040

作者简介 About authors

林俊宇,E-mail:scjylin@scut.edu.cn , E-mail:scjylin@scut.edu.cn

万明练,E-mail:1841435355@qq.com , E-mail:1841435355@qq.com

曹木花,E-mail:244901795@qq.com , E-mail:244901795@qq.com

摘要

该文关注三维Landau-Lifshitz-Gilbert方程的柯西问题.首先,该文通过能量方法以及连续性办法,得到在适当小的条件下整体光滑解的存在唯一性,并得到解的单调不等式.最后利用该单调不等式以及傅里叶分片法,得到解的衰减估计.

关键词: Landau-Lifshitz-Gilbert方程 ; 能量估计 ; 傅里叶分片法

Abstract

In this paper, the authors consider the Cauchy problem for the Landau-Lifshitz-Gilbert equation in $\mathbb{R}$3. By energy method and standard continuity argument, the authors obtain the global existence and uniqueness of smooth solution under suitable small initial data firstly. After establishing a monotone inequality for this solution, the authors built the time decay rates for this solution by the Fourier splitting method.

Keywords: Landau-Lifshitz-Gilbert equation ; Energy estimate ; Fourier splitting method

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本文引用格式

林俊宇, 万明练, 曹木花. 三维Landau-Lifshitz-Gilbert方程的解的衰减估计. 数学物理学报[J], 2019, 39(6): 1443-1455 doi:

Lin Junyu, Wan Minglian, Cao Muhua. Decay Estimates of the Global Solution for the Landau-Lifshitz-Gilbert Equation in Three Dimensions. Acta Mathematica Scientia[J], 2019, 39(6): 1443-1455 doi:

1 引言

Landau-Lifshitz-Gilbert方程是铁磁材料连续理论中的重要方程.它首次由Landau和Lifshitz提出[1].该方程为以下形式

$ \begin{eqnarray} \left\{\begin{array}{ll} {\bf Z}_t = \lambda_1{\bf Z}\times {\bf H}_{eff}-\lambda_2{\bf Z}\times({\bf Z}\times {\bf H}_{eff}), \ \ &\mbox{in}\ {\Bbb R} ^n\times (0, +\infty), \\ {\bf Z}(x, t) = {\bf Z}_0(x), \ \ & \mbox{in}\ {\Bbb R} ^n\times \{t = 0\}, \end{array}\right. \end{eqnarray} $

其中$ {\bf Z}: {\Bbb R} ^n\times (0, +\infty)\rightarrow{\mathbb S}^2 $为单位方向向量场, $ {\mathbb S}^2 $为单位球面, $ \times $是中向量积$ \lambda_1\in {\Bbb R} , $$ \lambda_2>0 $表示Gilbert阻尼系数, $ {\bf Z}_0: {\Bbb R} ^n \rightarrow{\mathbb S}^2 $为初始方向向量, $ {\bf H}_{eff} $表示有效场,方便起见,仅考虑Dirichlet能为

那么对应的有$ {\bf H}_{eff} = \triangle{\bf Z}. $我们将得到简单的Landau-Lifshitz-Gilbert方程

$ \begin{eqnarray} \left\{\begin{array}{ll} {\bf Z}_t = \lambda_1{\bf Z}\times \triangle{\bf Z}-\lambda_2{\bf Z}\times({\bf Z}\times \triangle{\bf Z}), \ \ &\mbox{in}\ {\Bbb R} ^n\times (0, +\infty), \\ {\bf Z}(x, t) = {\bf Z}_0(x), & \mbox{in}\ {\Bbb R} ^n\times \{t = 0\}. \end{array}\right. \end{eqnarray} $

易知$ -{\bf Z}\times({\bf Z}\times \triangle{\bf Z}) = \triangle{\bf Z}+|\nabla{\bf Z}|^2{\bf Z}. $如果方程$ (1.2)_1 $$ \lambda_1 = 0, $那么形式地得到调和映照热流

$ \begin{eqnarray} \left\{\begin{array}{ll} {\bf Z}_t = \lambda_2(\triangle{\bf Z}+|\nabla{\bf Z}|^2{\bf Z}), \end{array}\right. \end{eqnarray} $

$ \lambda_2 = 0 $时,对应Schrödinger流

$ \begin{eqnarray} \left\{\begin{array}{ll} {\bf Z}_t = \lambda_1{\bf Z}\times\triangle {\bf Z}.\end{array}\right. \end{eqnarray} $

因此,方程(1.2)可以看作是调和映照热流以及Schrödinger流的结合体.

$ n = 2 $时,基于研究调和映照热流的方法,文献[2-5]研究了几乎处处光滑的弱解以及能量递减解的唯一性.然而, $ n = 2 $时有限时刻爆破解的存在性还没解决.

对于$ n\geq 3, $文献[6]得到弱解的存在性.由于方程(1.2)目前没能得到类似调和映照热流的Bochner公式以及抛物能量不等式,因此方程(1.2)弱解的正则性以及唯一性需要利用其它办法研究. Moser[7-8], Melcher[10]以及Wang[11]建立了$ n = 3, 4 $时方程(1.2)的弱解的部分正则性(也可参考文献[9]).文献[12]得到了$ n = 3, 4 $时方程(1.2)有限时刻爆破解存在性的部分结果.

对于小初值问题,文献[13-14]研究了$ \dot{W}^{1, n}({\Bbb R} ^n) $空间以及Morrey空间中的整体适定性.文献[15]研究了BMO空间中小初值的适定性,但是需要初始方向向量落在$ {\mathbb S}^2\setminus\{(0, 0, -1)\}. $

文献[16]利用半离散办法得到三维方程(1.2)小初值的整体光滑解的存在性.本文将利用能量办法以及连续性办法得到小初值整体光滑解的存在性,并建立单调不等式,该单调不等式时建立衰减估计的关键.本文的主要结果如下.

定理1.1  设$ {\bf Z}_0: {\Bbb R} ^3\rightarrow{\mathbb S}^2, $对于任意的$ M\geq 1, $满足$ ({\bf Z}_0-{\bf Z}_\infty)\in H^{M+1}({\Bbb R} ^3), $其中$ {\bf Z}_\infty\in{\mathbb S}^2 $是一个给定的常值单位向量,满足

$ \begin{equation} \lim\limits_{\mid x\mid\rightarrow\infty}[{\bf Z}_0(x)-{\bf Z}_\infty] = {\bf 0}. \end{equation} $

(ⅰ) 存在常数$ \epsilon>0 $使得若

$ \begin{equation} ||{\bf Z}_0-{\bf Z}_\infty||_{H^2({\Bbb R} ^3)}^2<\epsilon, \end{equation} $

则方程(1.2)存在唯一整体解$ {\bf Z} $满足下面的单调不等式:对于任意的$ t>0, $

$ \begin{equation} \|({\bf Z}-{\bf Z}_\infty)(\cdot, t)\|_{H^{M+1}({\Bbb R} ^3)}^2+\int_{0}^{t}\|\nabla({\bf Z}-{\bf Z}_\infty)(\cdot, \tau)\|_{H^{M+1}({\Bbb R} ^3)}^2{\rm d}\tau\leq\|{\bf Z}_0-{\bf Z}_\infty\|_{H^{M+1}({\Bbb R} ^3)}^2; \end{equation} $

(ⅱ) 进一步,如果$ ({\bf Z}_0-{\bf Z}_\infty)\in L^1({\Bbb R} ^3), $那么由(ⅰ) 得到的整体解$ {\bf Z} $有下面的衰减估计:对于任意的$ t>0, $

$ \begin{eqnarray} &&\|\nabla^m({\bf Z}-{\bf Z}_\infty)(\cdot, t)\|_{L^2({\Bbb R} ^3)}\lesssim(1+t)^{-\frac{m}{2}}, \ \forall\ m = 1, 2, \cdot\cdot\cdot, M+1; \end{eqnarray} $

$ \begin{eqnarray} && \|\nabla^m\partial_t({\bf Z}-{\bf Z}_\infty)(\cdot, t)\|_{L^2({\Bbb R} ^3)}\lesssim(1+t)^{-1-\frac{m}{2}}, \ \forall\ m = 0, 1, \cdot\cdot\cdot, M. \end{eqnarray} $

注1.1  (ⅰ) 对比文献[16],本文建立了更高阶的估计,得到单调不等式估计(1.7).该单调不等式是建立所需的衰减估计的关键.

(ⅱ) 小初值条件(1.5)是必需的,因为文献[12]得到了三维方程(1.2)的有限时刻爆破解.

(ⅲ) 类似Navier-Stokes方程,方程(1.2)解的长时间行为或者衰减估计是重要的研究内容之一.受文献[17-18]的启发,本文建立了方程(1.2)的一些衰减估计.主要困难在于处理含有向量积的项: $ {\bf Z}\times\triangle {\bf Z}. $

本文将用到以下的记号. $ A\lesssim B $表示存在绝对正常数$ C $使得$ A\leq CB. $$ L^p ({\Bbb R} ^3) $表示Lebesgue空间. $ H^m({\Bbb R} ^3) = W^{m, 2}({\Bbb R} ^3) $表示Sobolev空间.方便起见,令$ \lambda_1 = \lambda_2 = 1. $

在第二节,建立小初值条件下的单调不等式估计,并证明整体光滑解的存在性;在第三节,利用傅里叶分片办法,建立时间衰减估计.

2 能量估计以及整体存在性

在本节,我们先建立能量估计,然后利用能量估计以及连续性方法得到整体解的存在性.

首先给出以下的Gagliardo-Nirenberg不等式.

引理2.1  设$ p\in[2, +\infty] $为实数, $ j, m, l $为整数.当$ p = \infty, $满足$ m\leq j+1 $$ j+2\leq l. $那么,对于任意的$ u\in C_0^\infty({\Bbb R} ^3), $下面估计成立

$ \begin{equation} ||\nabla^ju||_{L^p({\Bbb R} ^3)}\lesssim ||\nabla^mu||_{L^2({\Bbb R} ^3)}^{1-\alpha}||\nabla^lu||_{L^2({\Bbb R} ^3)}^\alpha, \end{equation} $

其中$ 0\leq \alpha\leq 1, $$ j $满足

$ {\bf u} = {\bf Z}-{\bf Z}_\infty, $则方程(1.2)可以写成以下形式

$ \begin{eqnarray} \left\{\begin{array}{ll} {\bf u}_t = \Delta {\bf u}+|\nabla {\bf u}|^2({\bf u}+{\bf Z}_\infty)+({\bf u}+{\bf Z}_\infty)\times \Delta {\bf u}, \ \ & \mbox{in}\ {\Bbb R} ^3\times (0, +\infty), \\ {\bf u}(x, t) = {\bf u}_0(x)\equiv{\bf Z}_0(x)-{\bf Z}_\infty, & \mbox{in}\ {\Bbb R} ^3\times \{t = 0\}. \end{array}\right. \end{eqnarray} $

注2.1  (ⅰ) 由(1.5)式可知

(ⅱ) 由$ |{\bf Z}_0(x, t)| = 1 $容易得到$ |{\bf Z}(x, t)| = 1. $因此

(ⅲ) 为了得到定理1.1的结果,只需研究方程(2.2)的整体存在性以及衰减估计.

下面将建立方程(2.2)的一些估计.

引理2.2  设$ {\bf u} $是方程$ (2.2) $的光滑解,那么对于任意的$ t>0, $有估计

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|{\bf u}(x, t)|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla {\bf u}(x, t)|^2{\rm d}x\lesssim \|{\bf u}(\cdot, t)\|_{H^2({\Bbb R} ^3)}\int_{{\Bbb R} ^3}|\nabla {\bf u}(x, t)|^2{\rm d}x. \end{eqnarray} $

  首先,用$ {\bf u} $乘以$ (2.2)_1 $式并在$ {\Bbb R} ^3 $积分,可得

$ \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|{\bf u}|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x = \int_{{\Bbb R} ^3}{\bf u}\cdot[\nabla\cdot({\bf Z}_\infty\times\nabla {\bf u})]{\rm d}x+\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2({\bf u}+{\bf Z}_\infty)\cdot {\bf u} {\rm d}x. \end{equation} $

由Sobolev不等式可得

$ \begin{eqnarray} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|{\bf u}|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x&\leq&\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2|{\bf u}| {\rm d}x\\ &\leq&\|{\bf u}\|_{L^\infty}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x\lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla {\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

其中用到

从而可以得到(2.3)式.

引理2.3  设$ {\bf u} $是方程$ (2.2) $的光滑解,那么对任意的$ t>0, $有估计

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla {\bf u}(x, t)|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla^2{\bf u}(x, t)|^2{\rm d}x\lesssim \|{\bf u}(\cdot, t)\|_{H^2({\Bbb R} ^3)}\int_{{\Bbb R} ^3}|\nabla^2 {\bf u}(x, t)|^2{\rm d}x. \end{eqnarray} $

  把$ \nabla $作用在$ (2.2)_1 $式两边,然后乘以$ \nabla {\bf u} $并在$ {\Bbb R} ^3 $上积分,则有

$ \begin{eqnarray} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla^2 {\bf u}|^2{\rm d}x = I_1+I_2. \end{eqnarray} $

其中

对于$ I_1 $,由Sobolev不等式和Hölder不等式可知

从而得到

$ \begin{eqnarray} \int_{{\Bbb R} ^3}\nabla {\bf u}\cdot\nabla[|\nabla {\bf u}|^2({\bf u}+{\bf Z}_\infty)]{\rm d}x\lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

对于$ I_2, $

$ \begin{eqnarray} I_2 = \int_{{\Bbb R} ^3}\nabla {\bf u}\cdot\nabla[({\bf u}+{\bf Z}_\infty)\times\Delta {\bf u}]{\rm d}x = -\int_{{\Bbb R} ^3}\triangle{\bf u}\cdot [({\bf u}+{\bf Z}_\infty)\times\triangle {\bf u}]{\rm d}x = 0. \end{eqnarray} $

把(2.8)式和(2.9)式代回(2.7)式,则得到该引理的证明.

引理2.4  设$ {\bf u} $是方程$ (2.2) $的光滑解,那么对任意的$ t>0, $有估计

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla^2{\bf u}(x, t)|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla^3{\bf u}(x, t)|^2{\rm d}x\lesssim \|{\bf u}\|_{H^2({\Bbb R} ^3)}\int_{{\Bbb R} ^3}|\nabla^3{\bf u}(x, t)|^2{\rm d}x. \end{eqnarray} $

  把$ \triangle $作用在$ (2.2)_1 $式两边,然后两边乘以$ \triangle{\bf u} $并在$ {\Bbb R} ^3 $上积分,可得

$ \begin{eqnarray} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla^2{\bf u}|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla^3{\bf u}|^2{\rm d}x = I_3+I_4, \end{eqnarray} $

其中

对于$ I_3, $

$ \begin{eqnarray} I_3& = &-\int_{{\Bbb R} ^3}\nabla^2[({\bf u}+{\bf Z}_\infty)\times\nabla {\bf u}]\cdot\nabla^3 {\bf u}{\rm d}x\\ &\leq&\|\nabla {\bf u}\|_{L^3({\Bbb R} ^3)}\|\nabla^2{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^3{\bf u}\|_{L^2({\Bbb R} ^3)}\\ &\lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^3{\bf u}\|_{L^2{({\Bbb R} ^3})}^2. \end{eqnarray} $

对于$ I_4 $,由Sobolev和Hölder不等式可知

$ \begin{eqnarray} I_4& = &\int_{{\Bbb R} ^3}\nabla^2[|\nabla {\bf u}|^2({\bf u}+{\bf Z}_\infty)]\cdot\nabla^2{\bf u}{\rm d}x\\ & = &-\int_{{\Bbb R} ^3}\nabla[|\nabla {\bf u}|^2({\bf u}+{\bf Z}_\infty)]\cdot\nabla^3{\bf u}{\rm d}x\\ &\leq&\|\nabla {\bf u}\|_{L^3({\Bbb R} ^3)}\|\nabla^2{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^3{\bf u}\|_{L^2({\Bbb R} ^3)}+\|\nabla {\bf u}\|_{L^6({\Bbb R} ^3)}^3\|\nabla^3{\bf u}\|_{L^2({\Bbb R} ^3)}\\ &\lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^3{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

结合(2.11), (2.12)和(2.13)式,我们可以得到估计(2.10).

引理2.5  设$ {\bf u} $是方程$ (2.2) $的光滑解,那么对任意的$ m = 0, 1, 2, \cdot\cdot\cdot, M, $下面的估计对任意的$ t>0 $成立

$ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla^{m+1}{\bf u}(x, t)|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla^{m+2}{\bf u}(x, t)|^2{\rm d}x\\ &\leq & C(M)\|{\bf u}(\cdot, t)\|_{H^2({\Bbb R} ^3)}\int_{{\Bbb R} ^3}|\nabla^{m+2}{\bf u}(x, t)|^2{\rm d}x, \end{eqnarray} $

其中$ C(M)>0 $是依赖于$ M $的常数.

  首先,由引理2.3和引理2.4知,当$ m = 0, 1 $时, (2.14)式成立.

下面证明(2.14)式对于$ 2\leq m\leq M $也成立.

$ \nabla^{m-1}\triangle $作用在$ (2.2)_1 $式两边,两边乘以$ \nabla^{m-1}\triangle{\bf u} $以及在$ {\Bbb R} ^3 $积分,可得

$ \begin{eqnarray} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla^{m+1}{\bf u}|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla^{m+2}{\bf u}|^2{\rm d}x = I_5+I_6, \end{eqnarray} $

其中

对于$ I_5, $由分部积分,莱布尼兹公式以及Hölder不等式,可知

$ \begin{eqnarray} I_5& = &-\int_{{\Bbb R} ^3}\sum\limits_{i = 0}^{m}C_m^i\nabla^i(|\nabla {\bf u}|^2)\nabla^{m-i}({\bf u}+{\bf Z}_\infty)\nabla^{m+2}{\bf u}{\rm d}x\\ & = &-\int_{{\Bbb R} ^3}\sum\limits_{i = 0}^{m}\sum\limits_{j = 0}^{i}C_m^iC_i^j\nabla ^{j+1}{\bf u}\nabla^{i+1-j}{\bf u}\nabla^{m-i}({\bf u}+{\bf Z}_\infty)\nabla^{m+2}{\bf u}{\rm d}x\\ & = &-\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2\nabla^m {\bf u}\nabla^{m+2}{\bf u} {\rm d}x-\int_{R^3}\sum\limits_{i = 1}^{m-1}C_m^i\nabla^{i+1}{\bf u}\nabla {\bf u}\nabla^{m-i}{\bf u}\nabla^{m+2}{\bf u} {\rm d}x\\ &&-\int_{{\Bbb R} ^3}\sum\limits_{i = 1}^{m-1}\sum\limits_{j = 0}^{i-1}C_m^iC_i^j\nabla^{j+1}{\bf u}\nabla^{i+1-j}{\bf u}\nabla^{m-i}{\bf u}\nabla^{m+2}{\bf u}{\rm d}x\\ && -\int_{{\Bbb R} ^3}\sum\limits_{j = 0}^{m}C_m^j\nabla^{j+1}{\bf u}\nabla^{m+1-j}{\bf u} ({\bf u}+{\bf Z}_\infty)\nabla^{m+2}{\bf u}{\rm d}x\\ & = &I_{51}+I_{52}+I_{53}+I_{54}, \end{eqnarray} $

其中$ C_m^i = \frac{m!}{i!(m-i)!} $是组合数.

由Hölder不等式和Sobolev不等式,可得

$ \begin{eqnarray} I_{51}&\lesssim &\|\nabla {\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla {\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^m {\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&||\nabla^2{\bf u}||_{L^2({\Bbb R} ^3)}||\nabla {\bf u}||_{L^2({\Bbb R} ^3)}^{1-\frac{1}{m+1}}||\nabla^{m+2}{\bf u}||_{L^2({\Bbb R} ^3)}^{\frac{1}{m+1}} \|\nabla {\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{1}{m+1}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{1}{m+1}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}||\nabla {\bf u}\|_{L^2({\Bbb R} ^3)}||\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2\\ & \lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

其中用到(2.3)式以及

对于$ I_{52}, $由引理$ 2.1 $和Hölder不等式,可得

$ \begin{eqnarray} I_{52}&\lesssim&\sum\limits_{i = 1}^{m-1}\|\nabla^{i+1}{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla {\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{m-i}{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\sum\limits_{i = 1}^{m-1}\|\nabla {\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{i+1}{m+1}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{i+1}{m+1}}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}\|\nabla {\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{i+1}{m+1}} \|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{i+1}{m+1}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\nonumber\\ & \lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

对于$ I_{53}, $利用Hölder不等式,可得

$ 1\leq i\leq[\frac{m-1}{2}], $

$ \begin{eqnarray} && ||\nabla^{j+1}{\bf u}\|_{L^6({\Bbb R} ^3)}||\nabla^{i+1-j}{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{m-i}{\bf u}\|_{L^6({\Bbb R} ^3)}||\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|\nabla^{\theta_1} {\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{j+1}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{j+1}{m}}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{i-j}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2}^{\frac{i-j}{m}}\\ && \cdot \|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{i+1}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{i+1}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & = &\|\nabla^{\theta_1} {\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{j+1}{m}}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{1+\frac{j+1}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2\\ && \lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

这里$ \theta_1 = 2-\frac{m}{m-(j+1)} $$ 0<\theta_1<1. $

$ [\frac{m-1}{2}]+1\leq i\leq m-1 $时,有

$ \begin{eqnarray} I_{53}&\lesssim&\|\nabla^{j+1}{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{i+1-j}{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{m-i}{\bf u}\|_{L^6({\Bbb R} ^3)}||\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{j}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{j}{m}}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{i-j}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{i-j}{m}}\\ &&\cdot\|\nabla^{\theta_2} {\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{i}{m}}||\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{i}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2}\\ & = &\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{2-\frac{i}{m}}\|\nabla^{\theta_2} {\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{i}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2\\ & \lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

这里$ \theta_2 = 2-\frac{m}{i} $$ 0\leq \theta_2<1. $

因此,结合(2.19)和(2.20)式可得

$ \begin{eqnarray} I_{53}\lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

对于$ I_{54}, $$ 0\leq j\leq[\frac{m}{2}] $时,由插值不等式以及Hölder不等式,可得

$ \begin{eqnarray} I_{54}&\lesssim&\|\nabla ^{j+1}{\bf u}||_{L^3({\Bbb R} ^3)}||\nabla^{m+1-j}{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|\nabla^{\theta_3} {\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{j}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{j}{m}}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{j}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2}^{1-\frac{j}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim &\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

这里$ \theta_3 = 2-\frac{m}{2m-2j} $$ 1\leq\theta_3\leq\frac{3}{2}. $

$ [\frac{m}{2}]+1\leq j\leq m $时,容易得到

$ \begin{eqnarray} I_{54}&\lesssim&\|\nabla ^{j+1}{\bf u}||_{L^3({\Bbb R} ^3)}\|\nabla^{m+1-j}{\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{j-\frac{1}{2}}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{j-\frac{1}{2}}{m}}\|\nabla^{\theta_4} {\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{j-\frac{1}{2}}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{j-\frac{1}{2}}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

这里$ \theta_4 = 2-\frac{m}{2j-1} $$ 1\leq\theta_4<\frac{3}{2}. $

因此,由$ (2.22) $$ (2.23) $式可以得到

$ \begin{eqnarray} I_{54}\lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

所以,结合$ (2.17), (2.21), (2.18), (2.24) $以及$ (2.16) $式,可得

$ \begin{eqnarray} I_5\lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

下面将估计$ I_6. $

利用莱布尼兹公式,可知

$ \begin{eqnarray} I_6& = &\int_{{\Bbb R} ^3}\nabla^{m+1}[({\bf u}+{\bf Z}_\infty)\times\Delta {\bf u}]\cdot\nabla^{m+1}{\bf u}{\rm d}x\\ & = &-\int_{{\Bbb R} ^3}\nabla^{m+1}[({\bf u}+{\bf Z}_\infty)\times\nabla {\bf u}]\cdot\nabla^{m+2}{\bf u}{\rm d}x\\ & = &-\int_{{\Bbb R} ^3}[\sum\limits_{l = 0}^{m}C_{m+1}^l\nabla^{m-l+1}({\bf u}+{\bf Z}_\infty)\times\nabla^{l+1}{\bf u}]\cdot\nabla^{m+2}{\bf u}{\rm d}x\\ && -\int_{{\Bbb R} ^3}[({\bf u}+{\bf Z}_\infty)\times\nabla^{m+2}{\bf u}]\cdot\nabla^{m+2}{\bf u}{\rm d}x\\ & = &-\int_{{\Bbb R} ^3}[\sum\limits_{l = 0}^{m}C_{m+1}^l\nabla^{m-l+1}({\bf u}+{\bf Z}_\infty)\times\nabla^{l+1}{\bf u}]\cdot\nabla^{m+2}{\bf u}{\rm d}x. \end{eqnarray} $

由Hölder不等式,可知

$ \begin{eqnarray} I_6\lesssim\|\nabla^{l+1}{\bf u}||_{L^3({\Bbb R} ^3)}\|\nabla^{m+1-l}{\bf u}\|_{L^6({\Bbb R} ^3)}||\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}. \end{eqnarray} $

$ 0\leq l\leq[\frac{m}{2}] $时,利用插值不等式,可得

$ \begin{eqnarray} I_6&\lesssim&\|\nabla^{l+1}{\bf u}\|_{L^3({\Bbb R} ^3)}\|\nabla^{m+1-l}{\bf u}\|_{L^6({\Bbb R} ^3)}||\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|\nabla^{\theta_5} {\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{l}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{l}{m}}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{l}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{l}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ &\lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

这里$ \theta_5 = 2-\frac{m}{2m-2l} $$ 1\leq\theta_5\leq\frac{3}{2}. $

$ [\frac{m}{2}]+1\leq l\leq m $时,可得

$ \begin{eqnarray} I_6&\lesssim&\|\nabla^{l+1}{\bf u}\|_{L^3({\Bbb R} ^3)}\|\nabla^{m+1-l}{\bf u}\|_{L^6({\Bbb R} ^3)}||\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{l-\frac{1}{2}}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{l-\frac{1}{2}}{m}}\cdot||\nabla^{\theta_6} {\bf u}\|_{L^2({\Bbb R} ^3)}^{\frac{l-\frac{1}{2}}{m}}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^{1-\frac{l-\frac{1}{2}}{m}}\cdot\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}\\ & \lesssim&\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2, \end{eqnarray} $

这里$ \theta_6 = 2-\frac{m}{2l-1} $$ 1\leq\theta_6<\frac{3}{2}. $

因此,结合$ (2.28) $$ (2.29) $式,我们得到

$ \begin{eqnarray} I_6\lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^{m+2}{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

把(2.25)式和(2.30)式代回(2.15)式,我们得到本引理的证明.

下面将证明定理1.1中(ⅰ).

  令$ \epsilon\leq \frac{1}{C(M)+1}, $其中$ C(M)>0 $是引理2.5中的常数.在定理1.1条件下,可知$ {\bf u}_0 = {\bf Z}_0-{\bf Z}_\infty\in H^{M+1}({\Bbb R} ^3) $以及

$ \begin{equation} \|{\bf u}_0\|_{H^2({\Bbb R} ^3)}^2<\epsilon. \end{equation} $

由文献[16]可知,带有初始值$ {\bf u}_0\in H^{M+1}({\Bbb R} ^3) $的柯西问题(2.2)有局部光滑解

断言:在(2.31)式条件下,对于任意的$ 0<t<T_0, $$ {\bf u} $满足

$ \begin{eqnarray} \|{\bf u}(\cdot, t)\|_{H^2({\Bbb R} ^3)}^2<\epsilon. \end{eqnarray} $

否则,设$ t_0 $是使得$ \|{\bf u}(\cdot, t_0)\|_{H^2({\Bbb R} ^3)}^2\geq\epsilon $成立的第一个时间.那么对于任意的$ t<t_0, $都有

另一方面,由$ (2.14) $式可知,对于任意的$ t>0, $

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\|{\bf u}(\cdot, t)\|_{H^{m+1}({\Bbb R} ^3)}^2+[1-C(M)\|{\bf u}(\cdot, t)\|_{H^2({\Bbb R} ^3)}]\|{\bf u}(\cdot, t)\|_{H^{m+2}({\Bbb R} ^3)}^2\leq 0. \end{eqnarray} $

从而对任意的$ t<t_0, $有估计

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\|{\bf u}(\cdot, t)\|_{H^{m+1}({\Bbb R} ^3)}^2\leq0, \end{eqnarray} $

这表明

特别地,有

这与$ t_0 $的定义矛盾.因此(2.32)式成立.

结合连续性方法[20],我们得到方程(2.2)整体光滑解的存在性.同时,不难得到唯一性,我们省略其证明.

由(2.14)式并在$ [0, t] $上积分,结合Gronwall不等式可得

$ \begin{eqnarray} \|{\bf u}(\cdot, t)\|_{H^{M+1}({\Bbb R} ^3)}^2+\int_0^t\|\nabla{\bf u}(\cdot, s)\|_{H^{M+1}({\Bbb R} ^3)}^2{\rm d}s\leq \|{\bf u}_0(\cdot)\|_{H^{M+1}({\Bbb R} ^3)}^2. \end{eqnarray} $

$ {\bf Z} = {\bf u}+{\bf Z}_\infty. $则可得到条件(1.6)下整体光滑解的存在性.由(2.35)式可得单调不等式(1.7).

定理1.1(ⅰ) 证明完毕.

3 解的衰减估计

本节将给出光滑解$ {\bf Z} $的衰减估计.由前一节可知

$ \begin{eqnarray} \|{\bf u}(\cdot, t)\|_{H^2({\Bbb R} ^3)}^2 = \|{\bf Z}(\cdot, t)-{\bf Z}_\infty\|_{H^2({\Bbb R} ^3)}^2<\epsilon, \ \ \forall t>0. \end{eqnarray} $

下面将利用傅里叶分片法[19]建立在(3.1)式条件下的衰减估计.

引理3.1  设$ {\bf u}_0\in H^{M+1}({\Bbb R} ^3)\cap L^1({\Bbb R} ^3) $并满足(3.1)式,则方程(2.2)的整体光滑解有下面的衰减估计

$ \begin{eqnarray} \|\nabla {\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2\leq C(1+t)^{-1}, \ \forall\ t> 0. \end{eqnarray} $

  由引理$ 2.3 $可知

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x+\int_{{\Bbb R} ^3}|\nabla^2{\bf u}|^2{\rm d}x\leq 0. \end{eqnarray} $

类似文献[21],定义依赖时间的球体

其中$ a $是待定常数.利用Parseval恒等式,可知

再次利用Parseval恒等式,可得

$ \begin{eqnarray} \int_{{\Bbb R} ^3}|\nabla^2{\bf u}|^2{\rm d}x\geq\frac{a}{1+t}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x-\frac{a^2}{{(1+t)}^2}\int_{{\Bbb R} ^3}|{\bf u}|^2{\rm d}x. \end{eqnarray} $

$ (3.4) $式代回$ (3.3) $式,可知

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x+\frac{a}{1+t}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x\leq\frac{a^2}{{(1+t)}^2}\int_{{\Bbb R} ^3}|{\bf u}|^2{\rm d}x. \end{eqnarray} $

由(3.1)式可知

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x+\frac{a}{1+t}\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2{\rm d}x\lesssim(1+t)^{-2}. \end{eqnarray} $

$ a = 3 $并且在(3.6)式两边同时乘以$ (1+t)^3, $可得

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}[(1+t)^3\|\nabla {\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2]\lesssim(1+t). \end{eqnarray} $

$ [0, t] $上积分,可得

证毕.

引理3.2  设$ {\bf u}_0\in H^{M+1}({\Bbb R} ^3)\cap L^1({\Bbb R} ^3) $并满足(3.1)式,那么方程(2.2)整体光滑解$ {\bf u} $有下面的衰减估计:对于任意的$ t>0, $

$ \begin{eqnarray} \|\nabla^{m}{\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2\leq C(1+t)^{-m}, \ \ \ \forall\ m = 1, \cdot\cdot\cdot, M+1. \end{eqnarray} $

  首先,引理3.1表明(3.8)式在$ m = 1 $是成立的.

假设$ (3.8) $式对于$ m = k, (1\leq k\leq M) $成立,即

$ \begin{eqnarray} \|\nabla^{k}{\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2\leq C(1+t)^{-k}. \end{eqnarray} $

下面要证明$ (3.8) $式对于$ m = k+1 $成立.

由引理2.5可知,对于$ m = 0, 1, 2, \cdot\cdot\cdot, M, $

$ m = k $可得

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla^{k+1}{\bf u}(x, t)|^2{\rm d}x+\|\nabla^{k+2}{\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2\leq 0. \end{eqnarray} $

注意到

我们可得

由Parseval恒等式,可知

$ \begin{eqnarray} \int_{{\Bbb R} ^3}|\nabla^{k+2}{\bf u}|^2{\rm d}x\geq\frac{a}{1+t}\int_{{\Bbb R} ^3}|\nabla^{k+1}{\bf u}|^2{\rm d}x-\frac{a^2}{(1+t)^2}\int_{{\Bbb R} ^3}|\nabla^{k}{\bf u}|^2{\rm d}x. \end{eqnarray} $

把(3.11)式代回(3.10)式并结合(3.9)式,可得

$ \begin{equation} \frac{\rm d}{{\rm d}t}\int_{{\Bbb R} ^3}|\nabla^{k+1}{\bf u}|^2{\rm d}x+\frac{a}{1+t}\int_{{\Bbb R} ^3}|\nabla^{k+1}{\bf u}|^2{\rm d}x \leq\frac{a^2}{(1+t)^2}\int_{{\Bbb R} ^3}|\nabla^{k}{\bf u}|^2{\rm d}x\lesssim(1+t)^{-k-2}. \end{equation} $

$ a = k+3 $并在(3.12)式两边同时乘以$ (1+t)^{k+3}, $可得

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}[(1+t)^{k+3}\|\nabla^{k+1}{\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2]\lesssim(1+t). \end{eqnarray} $

最后在$ [0, t] $积分,可得

$ \begin{eqnarray} \|\nabla^{k+1}{\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2\lesssim(1+t)^{-k-1}. \end{eqnarray} $

从而$ (3.8) $式对于$ m = k+1 $成立.由归纳法可知,该引理成立.

引理3.3  设$ {\bf u}_0\in H^{M+1}({\Bbb R} ^3)\cap L^1({\Bbb R} ^3) $并且满足(3.1)式,那么方程(2.2)的整体光滑解$ {\bf u} $由下面的衰减估计

$ \begin{eqnarray} \|\nabla^{m}\partial_t{\bf u}(\cdot, t)\|_{L^2({\Bbb R} ^3)}^2\leq C(1+t)^{-2-m}, \ \ \ \mbox{对}\ m = 0, 1, \cdot\cdot\cdot, M-1. \end{eqnarray} $

  方程$ (2.2)_1 $两边同时求空间的$ m $偏导数,然后两边同时乘以$ \nabla^m\partial_t{\bf u} $并在$ {\Bbb R} ^3 $上积分,可得

$ \begin{eqnarray} \|\nabla^{m}\partial_t{\bf u}\|_{L^2({\Bbb R} ^3)}^2 = \int_{{\Bbb R} ^3}\nabla^{m}[\Delta{\bf u}+|\nabla{\bf u}|^2({\bf u}+{\bf Z}_{\infty})+({\bf u}+{\bf Z}_{\infty})\times\Delta{\bf u}]\cdot\nabla^m\partial_t{\bf u}{\rm d}x. \end{eqnarray} $

对于$ m = 0, $

$ \begin{eqnarray} \|\partial_t{\bf u}\|_{L^2({\Bbb R} ^3)}^2&\lesssim&(\|\Delta{\bf u}\|_{L^2({\Bbb R} ^3)}+\|\nabla{\bf u}\|_{L^3({\Bbb R} ^3)}\|\nabla{\bf u}\|_{L^6({\Bbb R} ^3)})\|\partial_t{\bf u}\|_{L^2({\Bbb R} ^3)}\\ &\lesssim&\|\Delta{\bf u}\|_{L^2({\Bbb R} ^3)}^2+\|{\bf u}\|_{H^2({\Bbb R} ^3)}^2\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^2+\eta\|\partial_t{\bf u}\|_{L^2({\Bbb R} ^3)}^2\\ & \lesssim&(1+t)^{-2}+\eta\|\partial_t{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray} $

因此,取$ \eta $充分小,则(3.15)式对于$ m = 0 $成立.

对于$ m\geq1, $由引理3.2以及Young不等式可得

$ \begin{eqnarray} II_1\lesssim\|\nabla^{m+2}{\bf u}\|_{L^2}^2+\eta\|\nabla^m\partial_t{\bf u}\|_{L^2}^2\lesssim(1+t)^{-m-2}+\eta\|\nabla^m\partial_t{\bf u}\|_{L^2}^2. \end{eqnarray} $

对于$ II_2, $由文献[18]可得

$ \begin{eqnarray} II_2&\lesssim&\sum\limits_{j = 0}^{m-1}\sum\limits_{i = 0}^{j}\|\nabla^{i+1}{\bf u}\|_{L^6}\|\nabla^{j+1-i}{\bf u}\|_{L^6}\|\nabla^{m-j}{\bf u}\|_{L^6}\|\nabla^{m}\partial_t{\bf u}\|_{L^2}\\ && +\sum\limits_{i = 0}^{m}\|\nabla^{i+1}{\bf u}\|_{L^3}\|\nabla^{m+1-i}{\bf u}\|_{L^6}\|\nabla^{m}\partial_t{\bf u}\|_{L^2}\\ & \lesssim&\sum\limits_{j = 0}^{m-1}\sum\limits_{i = 0}^{j}\|\nabla^{i+2}{\bf u}\|_{L^2}^2\|\nabla^{j+2-i}{\bf u}\|_{L^2}^2\|\nabla^{m+1-j}{\bf u}\|_{L^2}^2\\ && +\sum\limits_{i = 0}^{m}\|\nabla^{i+1}{\bf u}\|_{H^1}^2\|\nabla^{m+2-i}{\bf u}\|_{L^2}^2+\eta\|\nabla^{m}\partial_t{\bf u}\|_{L^2}^2 \\ & \lesssim&\sum\limits_{j = 0}^{m-1}\sum\limits_{i = 0}^{j}(1+t)^{-2-i}(1+t)^{-2-j+i}(1+t)^{-1-m+j}\\ && +\sum\limits_{i = 0}^{m}(1+t)^{-i-1}(1+t)^{-m-2+i}+\eta\|\nabla^{m}\partial_t{\bf u}\|_{L^2}^2 \\ & \lesssim&(1+t)^{-3-m}+\eta\|\nabla^{m}\partial_t{\bf u}\|_{L^2}^2. \end{eqnarray} $

对于$ II_3, $

$ \begin{eqnarray} II_3&\lesssim&\sum\limits_{i = 0}^{m-1}\|\nabla^{m-i}{\bf u}\|_{L^3}|\nabla^{i}\Delta{\bf u}\|_{L^6}\|\nabla^{m}\partial_t{\bf u}\|_{L^2}+\|\nabla^{m}\Delta{\bf u}\|_{L^2}\|\nabla^{m}\partial_t{\bf u}\|_{L^2}\\ &\lesssim&\sum\limits_{i = 0}^{m-1}\|\nabla^{m-i}{\bf u}\|_{H^1}^2\|\nabla^{i}\Delta{\bf u}\|_{L^2}^2+\|\nabla^{m}\Delta{\bf u}\|_{L^2}^2+\eta\|\nabla^{m}\partial_t{\bf u}\|_{L^2}^2\\ &\lesssim&\sum\limits_{i = 0}^{m-1}(1+t)^{-m+i}(1+t)^{-i-2}+(1+t)^{-m-2}+\eta\|\nabla^{m}\partial_t{\bf u}\|_{L^2}^2\\ & \lesssim&(1+t)^{-2-m}+\eta\|\nabla^{m}\partial_t{\bf u}\|_{L^2}^2. \end{eqnarray} $

结合(3.18), (3.19), (3.20)以及(3.16)式,可得

$ \begin{eqnarray} \|\nabla^{m}{\bf u}_t\|_{L^2({\Bbb R} ^3)}^2\leq (1+t)^{-2-m}+\eta\|\nabla^{m}\partial_t{\bf u}\|_{L^2}^2. \end{eqnarray} $

$ \eta>0 $充分小并结合(3.17)和(3.21)式,我们可以得到引理的结论.

下面给出定理1.1中(ⅱ) 的证明.

  由引理3.2和引理3.3直接得到定理1.1(ⅱ) 的结论.

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