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

(1.1) $ \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

$ E({\bf Z}) = \frac12\int_{{\Bbb R} ^n}|\nabla{\bf Z}|^2. $

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

(1.2) $ \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 $中$ \lambda_1 = 0, $那么形式地得到调和映照热流

(1.3) $ \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} $

当$ \lambda_2 = 0 $时,对应Schrödinger流

(1.4) $ \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.

当$ 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

(1.5) $ \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

(1.6) $ \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, $有

(1.7) $ \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, $有

(1.8) $ \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} $

(1.9) $ \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.

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 $和$ 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

(2.1) $ \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, $和$ j $满足

$ \frac{1}{p}-\frac{j}{3} = (1-\alpha)(\frac{1}{2}-\frac{m}{3})+\alpha(\frac{1}{2}-\frac{l}{3}). $

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

(2.2) $ \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

$ \begin{eqnarray} \lim\limits_{|x|\rightarrow\infty}{\bf u}_0(x) = {\bf 0}; \end{eqnarray} $

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

$ |{\bf u}(x, t)+{\bf Z}_\infty| = |{\bf Z}(x, t)| = 1. $

(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

(2.3) $ \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} $

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

(2.4) $ \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

(2.5) $ \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

$ \|{\bf u}\|_{L^\infty({\Bbb R} ^3)}\lesssim\|{\bf u}\|_{H^2({\Bbb R} ^3)}. $

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

(2.6) $ \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} $

证  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

(2.7) $ \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

$ \begin{eqnarray*} &&I_1 = \int_{{\Bbb R} ^3}\nabla {\bf u}\cdot\nabla[|\nabla {\bf u}|^2({\bf u}+{\bf Z}_\infty)]{\rm d}x, \\ &&I_2 = \int_{{\Bbb R} ^3}\nabla {\bf u}\cdot\nabla[({\bf u}+{\bf Z}_\infty)\times\Delta {\bf u}]{\rm d}x. \end{eqnarray*} $

对于$ I_1 $,由Sobolev不等式和Hölder不等式可知

$ \begin{eqnarray*} I_1&\leq&\int_{{\Bbb R} ^3}|\nabla {\bf u}|^2|\nabla^2{\bf u}|{\rm d}x+\int_{{\Bbb R} ^3}|\nabla {\bf u}|^4{\rm d}x\\ &\leq&\|\nabla {\bf u}\|_{L^3({\Bbb R} ^3)}\|\nabla {\bf u}\|_{L^6({\Bbb R} ^3)}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}+\|\nabla {\bf u}\|_{L^2({\Bbb R} ^3)}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^3\\ &\lesssim& \|{\bf u}\|_{H^2({\Bbb R} ^3)}\|\nabla^2{\bf u}\|_{L^2({\Bbb R} ^3)}^2. \end{eqnarray*} $

从而得到

(2.8) $ \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, $有

(2.9) $ \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

(2.10) $ \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} $

证  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

(2.11) $ \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

$ \begin{eqnarray*} &&I_3 = \int_{{\Bbb R} ^3}\nabla^2[({\bf u}+{\bf Z}_\infty)\times\Delta {\bf u}]\cdot\nabla^2{\bf u}{\rm d}x, \\ &&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. \end{eqnarray*} $

对于$ I_3, $有

(2.12) $ \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

(2.13) $ \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

(2.14) $ \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.

证  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.

把$ \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

(2.15) $ \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

$ \begin{eqnarray*} &&I_5 = \int_{{\Bbb R} ^3}\nabla^{m+1}[|\nabla {\bf u}|^2({\bf u}+{\bf Z}_\infty)]\cdot \nabla^{m+1}{\bf u}{\rm d}x;\\ &&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. \end{eqnarray*} $

对于$ I_5, $由分部积分,莱布尼兹公式以及Hölder不等式,可知

(2.16) $ \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

(2.17) $ \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

$ \|\nabla{\bf u}\|_{L^6}\lesssim\|\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}} $

和

$ \|\nabla^m{\bf u}\|_{L^6({\Bbb R} ^3)}\lesssim\|\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}}. $

对于$ I_{52}, $由引理$ 2.1 $和Hölder不等式,可得

(2.18) $ \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

$I_{53}\lesssim|| \nabla^{j+1}{\bf u}||_{L^6({\Bbb R} ^3)}||\nabla^{i+1-j}{\bf u}||_{L^6({\Bbb R} ^3)}||\nabla^{m-i}{\bf u}||_{L^6({\Bbb R} ^3)}||\nabla^{m+2}{\bf u}||_{L^2({\Bbb R} ^3)}.$

当$ 1\leq i\leq[\frac{m-1}{2}], $有

(2.19) $ \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)} $且$ 0<\theta_1<1. $

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

(2.20) $ \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} $且$ 0\leq \theta_2<1. $

因此,结合(2.19)和(2.20)式可得

(2.21) $ \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}, $当$ 0\leq j\leq[\frac{m}{2}] $时,由插值不等式以及Hölder不等式,可得

(2.22) $ \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} $且$ 1\leq\theta_3\leq\frac{3}{2}. $

当$ [\frac{m}{2}]+1\leq j\leq m $时,容易得到

(2.23) $ \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} $且$ 1\leq\theta_4<\frac{3}{2}. $

因此,由$ (2.22) $和$ (2.23) $式可以得到

(2.24) $ \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

(2.25) $ \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

(2.26) $ \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

(2.27) $ \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} $

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

(2.28) $ \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} $且$ 1\leq\theta_5\leq\frac{3}{2}. $

当$ [\frac{m}{2}]+1\leq l\leq m $时,可得

(2.29) $ \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} $且$ 1\leq\theta_6<\frac{3}{2}. $

因此,结合$ (2.28) $和$ (2.29) $式,我们得到

(2.30) $ \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).

证  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

(2.31) $ \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

$ {\bf u}\in C([0, T_0);H^{M+1}({\Bbb R} ^3)). $

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

(2.32) $ \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

$ 1-C(M)\|{\bf u}(\cdot, t)\|_{H^2({\Bbb R} ^3)}\geq0. $

另一方面,由$ (2.14) $式可知,对于任意的$ t>0, $有

(2.33) $ \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

(2.34) $ \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

$ \|{\bf u}(\cdot, t_0)\|_{H^{m+1}({\Bbb R} ^3)}^2\leq\|{\bf u}_0(\cdot)\|_{H^{m+1}({\Bbb R} ^3)}^2. $

特别地,有

$ \|{\bf u}(\cdot, t_0)\|_{H^{2}({\Bbb R} ^3)}^2\leq\|{\bf u}_0(\cdot)\|_{H^{2}({\Bbb R} ^3)}^2<\epsilon, $

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

(2.35) $ \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} $

记$ {\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.

本节将给出光滑解$ {\bf Z} $的衰减估计.由前一节可知

(3.1) $ \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

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

证  由引理$ 2.3 $可知

(3.3) $ \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

$ U(t) = \Big\{ \xi\in {\Bbb R} ^3| \ |\xi|\leq({\frac{a}{1+t}})^{\frac{1}{2}}\Big\}, $

其中$ a $是待定常数.利用Parseval恒等式,可知

$ \begin{eqnarray*} \int_{{\Bbb R} ^3}|\nabla^2{\bf u}|^2{\rm d}x& = &\int_{{\Bbb R} ^3}|\xi|^4|\widehat{{\bf u}}|^2{\rm d}\xi\geq\int_{{\Bbb R} ^3\setminus U(t)}|\xi|^4|\widehat{{\bf u}}|^2{\rm d}\xi\\ & \geq&\frac{a}{1+t}\int_{{\Bbb R} ^3\setminus U(t)}|\xi|^2|\widehat{{\bf u}}|^2{\rm d}\xi\\ & = &\frac{a}{1+t}\int_{{\Bbb R} ^3}|\xi|^2|\widehat{{\bf u}}|^2{\rm d}\xi-\frac{a}{1+t}\int_{U(t)}|\xi|^2|\widehat{{\bf u}}|^2{\rm d}\xi\\ & \geq&\frac{a}{1+t}\int_{{\Bbb R} ^3}|\xi|^2|\widehat{{\bf u}}|^2{\rm d}\xi-\frac{a^2}{{(1+t)}^2}\int_{{\Bbb R} ^3}|\widehat{{\bf u}}|^2{\rm d}\xi. \end{eqnarray*} $

再次利用Parseval恒等式,可得

(3.4) $ \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} $

把$ (3.4) $式代回$ (3.3) $式,可知

(3.5) $ \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

(3.6) $ \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} $

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

(3.7) $ \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} $

在$ [0, t] $上积分,可得

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

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, $有

(3.8) $ \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} $

证  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

(3.9) $ \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, $有

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

取$ m = k $可得

(3.10) $ \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

$ U(t) = \{\xi\in {\Bbb R} ^3| \ |\xi|\leq(\frac{a}{1+t})^{\frac{1}{2}}\}. $

我们可得

$ \begin{eqnarray*} \int_{{\Bbb R} ^3}|\nabla^{k+2}{\bf u}|^2{\rm d}x &\geq&\int_{{\Bbb R} ^3\setminus U(t)}|\xi|^4|\widehat{\nabla^{k}{\bf u}}|^2{\rm d}\xi \geq\frac{a}{1+t}\int_{{\Bbb R} ^3\setminus U(t)}|\xi|^2|\widehat{\nabla^{k}{\bf u}}|^2{\rm d}\xi\\ & = &\frac{a}{1+t}\int_{{\Bbb R} ^3}|\xi|^2|\widehat{\nabla^{k}{\bf u}}|^2{\rm d}\xi-\frac{a}{1+t}\int_{U(t)}|\xi|^2|\widehat{\nabla^{k}{\bf u}}|^2{\rm d}\xi\\ & \geq&\frac{a}{1+t}\int_{{\Bbb R} ^3}|\xi|^2|\widehat{\nabla^{k}{\bf u}}|^2{\rm d}\xi-\frac{a^2}{(1+t)^2}\int_{{\Bbb R} ^3}|\widehat{\nabla^{k}{\bf u}}|^2{\rm d}\xi. \end{eqnarray*} $

由Parseval恒等式,可知

(3.11) $ \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

(3.12) $ \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} $

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

(3.13) $ \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

(3.14) $ \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

(3.15) $ \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} $

证  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

(3.16) $ \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, $有

(3.17) $ \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.

记

$ \begin{eqnarray*} && II_1 = \int_{{\Bbb R} ^3}\nabla^m\Delta{\bf u}\cdot\nabla^m\partial_t{\bf u}{\rm d}x, \\ &&II_2 = \int_{{\Bbb R} ^3}\nabla^m[|\nabla {\bf u}|^2({\bf u}+{\bf Z}_{\infty})]\cdot\nabla^m\partial_t{\bf u}{\rm d}x, \\ &&II_3 = \int_{{\Bbb R} ^3}\nabla^m[({\bf u}+{\bf Z}_{\infty})\times\Delta {\bf u}]\cdot\nabla^m\partial_t{\bf u}{\rm d}x. \end{eqnarray*} $

对于$ m\geq1, $由引理3.2以及Young不等式可得

(3.18) $ \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

(3.19) $ \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, $有

(3.20) $ \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

(3.21) $ \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} $

取$ \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.

证  由引理3.2和引理3.3直接得到定理1.1(ⅱ) 的结论.