数学物理学报, 2024, 44(3): 661-669

三维稳态向列型液晶方程各向异性的Liouville定理

陈浩,1, 邓雪梅1,2, 别群益,1,2,*

1.三峡大学理学院 湖北宜昌 443002

2.三峡大学三峡数学研究中心 湖北宜昌 443002

Anisotropic Liouville Type Theorem for the Stationary Nematic Liquid Crystal Equations in $\mathbb{R}^{3}$

Chen Hao,1, Deng Xuemei1,2, Bie Qunyi,1,2,*

1. College of Science, China Three Gorges University, Hubei Yichang 443002

2. Three Gorges Mathematical research Center, China Three Gorges University, Hubei Yichang 443002

通讯作者: *别群益，E-mail：qybie@126.com

收稿日期: 2023-05-17   修回日期: 2023-10-7  

基金资助: 国家自然科学基金(11871305)

Received: 2023-05-17   Revised: 2023-10-7  

Fund supported: NSFC(11871305)

作者简介 About authors

陈浩，E-mail：chenhao215@outlook.com

摘要

该文研究了三维稳态向列型液晶方程的Liouville定理, 证明了如果 $\nabla d\in{L^2}({\mathbb{R}^3}) \cap {L^q}({\mathbb{R}^3})$, $u\in{L^6}({\mathbb{R}^3}) \cap {L^q}({\mathbb{R}^3})$, 以及 $u_i$ 满足各向异性的可积条件 ${u_i} \in L_{{x_i}}^{\frac{q}{{q - 2}}}L_{{{\tilde x}_i}}^s(\mathbb{R} \times {\mathbb{R}^2}),\forall i = 1,2,3$, 其中 $2 < q < \infty, 1 \le s \le \infty $ 且 $\frac{2}{q} + \frac{1}{s} \ge \frac{1}{2}$, 则 $u=0, \nabla d=0$.

关键词： 向列型液晶方程; Liouville定理; 各向异性

Abstract

This paper investigates a Liouville type theorem for three-dimensional stationary liquid crystal equations. We show that if $u\in{L^6}({\mathbb{R}^3}) \cap {L^ q}({\mathbb{R}^3})$, $\nabla d\in{L^2}({\mathbb{R}^3}) \cap {L^q}({\mathbb{R}^3})$ and the anisotropic integrability conditions of ${u_i} \in L_{{x_i}}^{\frac{q}{{q - 2}}}L_{{{\tilde x}_i}}^s(\mathbb{R} \times {\mathbb{R}^2}), \forall i = 1,2,3$, $\frac{2}{q} + \frac{1}{s} \ge \frac{1}{2}$, $2 < q < \infty,1 \le s \le \infty $ are satisfied, then $u=0, \nabla d = 0$.

Keywords： Nematic liquid crystal equations; The Liouville problem; Anisotropy

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本文引用格式

陈浩, 邓雪梅, 别群益. 三维稳态向列型液晶方程各向异性的Liouville定理[J]. 数学物理学报, 2024, 44(3): 661-669

Chen Hao, Deng Xuemei, Bie Qunyi. Anisotropic Liouville Type Theorem for the Stationary Nematic Liquid Crystal Equations in $\mathbb{R}^{3}$[J]. Acta Mathematica Scientia, 2024, 44(3): 661-669

1 引言

液晶介于液体与晶体之间, 其性质涉及物理、化学、材料等众多学科. 向列型液晶的特点是分子具有长程取向有序, 局部区域的分子取沿同一方向排列, 其应用最为广泛, 理论发展也较为完善, 其中最为著名的就是Ericksen-Lesile系统. Lin[1]提出一般Ericksen-Lesile系统的简化, 该模型成功模拟向列型液晶的各种动力学行为, 宏观地描述了材料在流体影响下的演变速度场; 此外, 它还提供了棒状液晶的流体微观取向的宏观描述. 本文将从数学角度研究如下简化的Ericksen-Lesile系统

$\begin{equation} \left\{\begin{array}{lr} {u \cdot \nabla u + \nabla P = \Delta u - \nabla \cdot (\nabla d \odot \nabla d)},\label{YJ1}\\ {u \cdot \nabla d = \Delta d + |\nabla d{|^2}d,}\\ {\nabla \cdot u = 0}, \end{array} \right. \end{equation}$

其中 $u:{\mathbb{R}^3} \to {\mathbb{R}^3}$ 表示速度场, $d:{\mathbb{R}^3} \to {\mathbb{S}^2}$ 表示液晶内部分子朝向, $\mathbb{S}^2$ 是 $\mathbb{R}^3$ 中球心在原点的单位球面. 向量 ${\nabla \cdot (\nabla d \odot \nabla d)}$ 的第 $i$ 个分量为 $\sum\limits_{j = 1}^3 {{\partial _j}({\partial _i}d{\partial _j}d)} $.

该系统可以看作一个不可压缩Navier-Stokes方程与一个调和映照热流方程的耦合, Navier-Stokes方程决定了液晶的运动, 调和映照方程则控制内部分子的朝向. 当 $\nabla d = 0$ 时, 这个系统简化为Navier-Stokes方程

$\begin{matrix} \left\{ \begin{array}{l} u \cdot \nabla u + \nabla P = \Delta u,\\ \nabla \cdot u = 0. \end{array} \right.\label{ns} \end{matrix}$

目前关于稳态流体的Liouville定理引起了广泛的关注. 对于Navier-Stokes方程的一个公开问题是如果方程(1.2)的解满足条件

$\begin{matrix} \left\{ \begin{array}{l} \mathop {\lim }\limits_{|x| \to \infty } |u| = 0,\\[3mm] \int_{{\mathbb{\mathbb{R}}^3}} {|\nabla u{|^2}} {\rm d}x < \infty,\label{D-solution} \end{array} \right. \end{matrix}$

那么是否有$u = 0$?需要说明的是满足条件(1.3)的解称为D-解. 关于这个公开问题已经有了一些结果[2⇓⇓-5]. 其中, Galdi[2]证明了当 $u \in {L^{\frac{9}{2}}}({\mathbb{R}^3})$时, $u \equiv 0$.Chae-Wolf[3]将文献[2]条件推广到了如下情形

$\int_{{\mathbb{R}^3}} |u|{^{\frac{9}{2}}}{\left\{ \log \left(2 + \frac{1}{{|u|}}\right) \right\} ^{ - 1}}{\rm d}x < \infty.$

Seregin[4]证明了速度场满足 $u \in {L^6}({\mathbb{R}^3}) \cap BM{O^{ - 1}}({\mathbb{R}^3})$, 则Liouville定理也成立. 最近, Chae[5]证明了若

$u \in {L^6}({\mathbb{R}^3}) \cap {L^q}({\mathbb{R}^3}), {u_i} \in L_{{x_i}}^{\frac{q}{{q - 2}}}L_{{{\tilde x}_i}}^s(\mathbb{R} \times {\mathbb{R}^2}), \forall i = 1,2,3,$

其中

$\frac{2}{q} + \frac{1}{s} \ge \frac{1}{2}, 2 < q < \infty, 1 \le s \le \infty.$

则$u = 0$.

对于液晶系统(1.1), 在二维空间下, 文献[6⇓⇓-9]得到了关于初边值问题Leray-Hopf弱解的全局存在性. 对于三维空间, Lin-Wang[10]证明了在条件满足${d_0} \in S_ + ^2 $(${d_0}$表示在上半球面取值)时, 初边值问题Leray-Hopf弱解的全局存在性. 目前, 对于系统(1.1)Liouville定理的研究相对较少[11,12]. Jarríin[11]证明了此系统在局部Morrey空间中的Liouville定理. Hao[12]证明了 $u \in D_0^1, d \in {L^\infty } \cap \mathring{H} $, 且满足条件 $|u| + |\nabla d| \in {L^q}({\mathbb{R}^3})$ 时, 可以得到 $u = 0, \nabla d = 0$.

其中

$\begin{align*} D_0^1={\{u \in L^{6} ({\mathbb{R}^3})| \|\nabla u \| _{L^{2}({\mathbb{R}^3})} < \infty\}},\\ \mathring{H}={\{d \in L_{loc}^1({\mathbb{R}^3})} | \|\nabla d \| _{L^{2}({\mathbb{R}^3})} < \infty\}. \end{align*}$

文献[11,12]对于液晶系统的研究结果是各向同性的. 受文献[5,11,12]启发, 本文主要研究关于速度各分量各向异性的Liouville定理.

本文的主要结果如下.

定理 1.1 设 $2 < q < \infty,1 \le s \le \infty $ 且 $\frac{2}{q} + \frac{1}{s} \ge \frac{1}{2}$. 若 $u$ 和 $d$ 是系统(1.1)的解, 满足条件

$\begin{align*} & u \in{{L^6}({\mathbb{R}^3})}\cap {L^q}({\mathbb{R}^3}), \nabla d \in {{L^2}({\mathbb{R}^3})}\cap{L^q}({\mathbb{R}^3}),\\ &{u_i} \in L_{{x_i}}^{\frac{q}{{q - 2}}}L_{{{\tilde x}_i}}^s(\mathbb{R} \times {\mathbb{R}^2}), \forall i = 1,2,3, \end{align*}$

则

$u = 0, \nabla d = 0.$

注 1.1 上述情况的一个特例, 当 $s=q=6$ 时, 定理 1.1 中关于 $u, \nabla d$ 的条件转化为

$\begin{aligned} {u_i} \in (L_{{x_i}}^{\frac{{\rm{3}}}{{\rm{2}}}} \cap L_{{x_i}}^{\rm{6}})L_{{{\tilde x}_i}}^{\rm{6}}(\mathbb{R} \times {\mathbb{R}^2}), \nabla d\in {L^{\rm{6}}}({\mathbb{R}^3}) \cap {L^2}({\mathbb{R}^3}). \end{aligned}$

相比于Hao[12]的结果

$\begin{aligned}{u \in {L^6}({\mathbb{R}^3}) \cap {L^{\frac{{\rm{9}}}{{\rm{2}}}}}({\mathbb{R}^3}), \nabla u \in {L^2}({\mathbb{R}^3}), \nabla d\in{L^{\frac{9}{2}}}({\mathbb{R}^3}) \cap {L^2}({\mathbb{R}^3})\cap L_{loc}^1} ({\mathbb{R}^3}),\end{aligned}$

条件(1.4)与条件(1.5)无包含关系.

2 预备知识

下面对本文使用的符号进行说明. 记

${\delta _{jk}} = \left\{ \begin{array}{l} 1, {j} = {k},\\ 0, {j} \ne {k}. \end{array} \right.$

对于 $\Omega \subset {\mathbb{R}^2}$, 记

$\int_\Omega f {\rm d}{{\tilde x}_{\rm{1}}} = \int_\Omega f {\rm d}{x_2}{\rm d}{x_3}, \int_\Omega f {\rm d}{{\tilde x}_{\rm{2}}} = \int_\Omega f {\rm d}{x_3}{\rm d}{x_1}, \int_\Omega f {\rm d}{{\tilde x}_{\rm{3}}} = \int_\Omega f {\rm d}{x_1}{\rm d}{x_2},$

其中${{\tilde x}_{\rm{1}}}: = ({x_2},{x_3}), {{\tilde x}_{\rm{2}}}: = ({x_3},{x_1}), {{\tilde x}_{\rm{3}}}: = ({x_1},{x_2}).$

定义 2.1 设 $1 \leq r,s \le + \infty,i \in \{ 1,2,3\}$, 若

$\| f \| _{L_{{x_i}}^rL_{{{\tilde x}_i}}^s}: = {\left\{ {\int_\mathbb{R} {{{\left( {\int_{{\mathbb{R}^2}} | f{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{r}{s}}}} {\rm d}x_i} \right\}^{\frac{1}{r}}} < + \infty. $

则称$f \in L_{{x_i}}^rL_{{{\tilde x}_i}}^s(\mathbb{R} \times {\mathbb{R}^2}). $ 此处, $r$ 或 $s=\infty$ 时, 可以理解为本性上界.

定义 2.2 取一个光滑的实函数 $\psi :[\infty ) \to [0,1]$, 满足

$\psi (s) = \left\{ \begin{array}{l} 1, 0 \le s \le 1,\\ 0, s \ge 4. \end{array} \right.$

对于集合${{\cal J}_1} = \{ 2,3\}, \quad {{\cal J}_2} = \{ {\rm{1}},3\}, \quad {{\cal J}_3} = \{ {\rm{1,2}}\}, $当 $R > 0$, 定义截断函数

${\varphi _R}(x) = \prod\limits_{j = 1}^3 \psi \left( {\frac{{x_j^2}}{{{R^2}}}} \right), {\tilde \varphi _{i,R}}(x) = \prod\limits_{j \in {{\cal J}_i}} \psi \left( {\frac{{x_j^2}}{{{R^2}}}} \right),\quad i \in \{ 1,2,3\}. $

记

${D_i}: = \left\{ {{{\tilde x}_i} \in {\mathbb{R}^2}\big||{x_j}| < 2R,\quad \forall j \in {{\cal J}_i}} \right\},\quad i \in \left\{ {1,2,3} \right\}.$

3 定理1.1的证明

证 将 (1.1)1 式两边同乘以 $u{\varphi _R}$ 并在 $\mathbb{R}^3$ 上积分, 得

$\begin{equation} \underbrace{- \int_{{\mathbb{R}^3}} \Delta u \cdot {\varphi _R}u{\rm d}x}_{ J_1}+ \underbrace{\int_{{\mathbb{R}^3}} {{\mathop{\rm div}\nolimits} } (\nabla d \odot \nabla d) \cdot {\varphi _R}u{\rm d}x}_{J_2}+ \underbrace{\int_{{\mathbb{R}^3}} {u \cdot \nabla u \cdot {\varphi _R}u{\rm d}x}}_{J_3}+ \underbrace{\int_{{\mathbb{R}^3}} {\nabla P} \cdot {\varphi _R}u{\rm d}x}_{J_4}=0, \label{1} \end{equation}$

下面分别处理 $J_{1},J_{2},J_{3}$ 和 $J_{4}.$ 首先, 由分部积分公式, 得

$\begin{matrix} J_1 &= - \sum\limits_{i,j = 1}^3 {\int_{{\mathbb{R}^3}} {(\partial _j^2{u_i})({\varphi _R}{u_i})} } {\rm d}x \\ &= \frac{1}{2}\sum\limits_{i,j = 1}^3 {\int_{{\mathbb{R}^3}} {({\partial _j}{\varphi _R})} } {\partial _j}(u_i^2){\rm d}x + \int_{\mathbb{R}^3} | \nabla u{|^2}{\varphi _R}{\rm d}x \\ &= - \frac{1}{2}\int_{{\mathbb{R}^3}} | u{|^2}\Delta {\varphi _R}{\rm d}x + \int_{{\mathbb{R}^3}} | \nabla u{|^2}{\varphi _R}{\rm d}x. \label{x1} \end{matrix}$

对于 $J_{2}$, 由 $\nabla \cdot u = 0$, 得到

$\begin{matrix} J_2 &= \sum\limits_{i,j = 1}^3 \int_{{\mathbb{R}^3}} {{u_i}{\partial _j}({\partial _i}d \cdot {\partial _j}d){\varphi _R}{\rm d}x} \\ &=\sum\limits_{i,j = 1}^3 \int_{{\mathbb{R}^3}} {{u_i}({\partial _i}{\partial _j}d \cdot {\partial _j}d + {\partial _i}d \cdot \partial _j^2d){\varphi _R}{\rm d}x} \\ &=\sum\limits_{i,j = 1}^3 \int_{{\mathbb{R}^3}} {\frac{1}{2}{u_i}{\partial _i}|\nabla d{|^2}{\varphi _R} + (u \cdot \nabla d) \cdot \Delta d{\varphi _R}{\rm d}x}.\label{x2} \end{matrix}$

对于 $J_{3}$, 同样由分部积分公式及 $\nabla \cdot u = 0$, 可得

$\begin{equation} J_3 = \frac{1}{2}\int_{{\mathbb{R}^3}} u \cdot \nabla |u{|^2}{\varphi _R}{\rm d}x =- \frac{1}{2}\int_{{\mathbb{R}^3}}| u{|^2}u \cdot \nabla {\varphi _R}{\rm d}x.\label{x3} \end{equation}$

类似地, 对于 $J_{4}$, 有

$\begin{equation} J_4={ \sum\limits_{i = 1}^3 {\int_{{\mathbb{R}^3}} {({\partial _i}P)} } {\varphi _R}{u_i}{\rm d}x = - \int_{{\mathbb{R}^3}} P (u \cdot \nabla {\varphi _R}){\rm d}x}. \label{x4} \end{equation}$

将(3.2), (3.3), (3.4)和(3.5)式代入(3.1)式得

$\begin{matrix} \int_{{\mathbb{R}^3}} {(|\nabla u{|^2} + |\Delta d{|^2})} {\varphi _R}{\rm d}x = &\int_{{\mathbb{R}^3}} {\left(\frac{1}{2}|u{|^2} + \frac{1}{2}|\nabla d{|^2} + P\right)(u \cdot \nabla {\varphi _R})} {\rm d}x \\ &+ \int_{{\mathbb{R}^3}} {\frac{1}{2}|u{|^2}\Delta {\varphi _R}{\rm d}x} - \int_{{\mathbb{R}^3}} {(u \cdot \nabla d) \cdot \Delta {\rm d}{\varphi _R}} {\rm d}x.\label{4} \end{matrix}$

将(1.1)2式两边同乘以 $ - \Delta d{\varphi _R}$ 并在 $\mathbb{R}^3$ 上积分, 得

$\begin{equation} \int_{{\mathbb{R}^3}} {|\Delta d{|^2}{\varphi _R}{\rm d}x} = - \int_{{\mathbb{R}^3}} {|\nabla d{|^2}d \cdot \Delta {\rm d}{\varphi _R}{\rm d}x} + \int_{{\mathbb{R}^3}} {u \cdot \nabla d \cdot \Delta {\rm d}{\varphi _R}{\rm d}x}. \label{2} \end{equation}$

由(3.6)和(3.7)式, 得

$\begin{matrix} \int_{{\mathbb{R}^3}} {(|\nabla u{|^2} + |\Delta d{|^2})} {\varphi _R}{\rm d}x = & \int_{{\mathbb{R}^3}} {\left(\frac{1}{2}|u{|^2} +\frac{1}{2}|\nabla d{|^2} + P\right)(u \cdot \nabla {\varphi _R})} {\rm d}x \\ & + \int_{{\mathbb{R}^3}} {\frac{1}{2}|u{|^2}\Delta {\varphi _R}{\rm d}x} - \int_{{\mathbb{R}^3}} | \nabla |d{|^2}(d \cdot \Delta d){\varphi _R}{\rm d}x. \label{3} \end{matrix}$

由于 $|d|{\rm{ = 1}}$, 可得 $\Delta |d{|^{\rm{2}}}{\rm{ = 0}}$, 从而

$\begin{matrix} - |\nabla d{|^2} &= - \sum\limits_{i,j = 1}^3 {({\partial _i}{d_j})^2} + \frac{1}{2}\Delta |d{|^2} = - \sum\limits_{i,j = 1}^3 {({\partial _i}{d_j})^2} + \sum\limits_{i,j = 1}^3 \frac{1}{2}{\partial _j}{\partial _j}({d_i}^2) \\ &= - \sum\limits_{i,j = 1}^3 {({\partial _i}{d_j})^2} +\sum\limits_{i,j = 1}^3 \frac{1}{2}{\partial _j}(2({\partial _j}{d_i}){d_i}) \\ & = - \sum\limits_{i,j = 1}^3 {({\partial _i}{d_j})^2} +\sum\limits_{i,j = 1}^3 (({\partial _j}{\partial _j}{d_i}){d_i} + {({\partial _j}d)^2}) \\ &= \sum\limits_{i,j = 1}^3 ({\partial _j}{\partial _j}{d_i}){d_i}= d \cdot\Delta d.\label{d} \end{matrix}$

又由(3.9)式, 可得

$\begin{matrix} - \int_{{\mathbb{R}^3}} {|\nabla d{|^2}(d \cdot \Delta d)} {\varphi _R}{\rm d}x &= \int_{{\mathbb{R}^3}} {|d \cdot \Delta d{|^2}} {\varphi _R}{\rm d}x \\ & \le \int_{{\mathbb{R}^3}} {|d{|^2}|\Delta d{|^2}{\varphi _R}{\rm d}x \le } \int_{{\mathbb{R}^3}} {|\Delta d{|^2}{\varphi _R}{\rm d}x}.\label{e} \end{matrix}$

将(3.10)式代入(3.8)式, 有

$\begin{matrix} \int_{{\mathbb{R}^3}} {|\nabla u{|^2}} {\varphi _R}{\rm d}x \le &\int_{{\mathbb{R}^3}} {\left(\frac{{|u{|^2}}}{2} + P\right)(u \cdot \nabla {\varphi _R}){\rm d}x} \\ &+ \frac{1}{2}\int_{{\mathbb{R}^3}} {|u{|^2}\Delta {\varphi _R}} {\rm d}x + \frac{1}{2}\int_{{\mathbb{R}^3}} {|\nabla d{|^2}(u \cdot \nabla {\varphi _R}){\rm d}x}.\label{Z} \end{matrix}$

下面对压力项 $P$ 作估计. 将(1.1)1式同时作用散度算子, 由于 $\nabla \cdot u = 0$, 可得

${\rm{div}}\left( {\left( {u \cdot \nabla } \right)u} \right) + {\rm{div}}\left( {{\rm{div}}\left( {\nabla d \odot \nabla d} \right)} \right) + \Delta P = 0.$

将 $(u \cdot \nabla )u$ 和 ${\mathop{\rm div}\nolimits} (\nabla d \odot \nabla d)$ 分别用分量表示为

${[(u \cdot \nabla )u]_i} = \sum\limits_{j = 1}^3 {{\partial _j}} ({u_j}{u_i}), {[{\mathop{\rm div}\nolimits} (\nabla d \odot \nabla d)]_i} = \sum\limits_{k = 1}^3 {\sum\limits_{j = 1}^3 {{\partial _j}({\partial _i}{d_k}{\partial _j}{d_k}), i = 1,2,3} }.$

因此, 由Riesz算子 ${{\cal R}_i}=\frac{{{\partial _i}}}{{\sqrt { - \Delta } }}$, 有

$\begin{matrix} P &= \sum\limits_{i = 1}^3 {\sum\limits_{i = j}^3 {\frac{1}{{ - \Delta }}} } {{\partial _i}{\partial _j}({u_j}{u_i})} + \sum\limits_{i = 1}^3 {\sum\limits_{j = 1}^3 {\sum\limits_{k = 1}^3 {\frac{1}{{ - \Delta }} {{\partial _i}{\partial _j}({\partial _i}{d_k}{\partial _j}{d_k})} } } } \\ &= \sum\limits_{i = 1}^3 {\sum\limits_{i = j}^3 { {\frac{{{\partial _i}}}{{\sqrt { - \Delta } }}\frac{{{\partial _j}}}{{\sqrt { - \Delta } }}({u_j}{u_i})} } } + \sum\limits_{i = 1}^3 {\sum\limits_{j = 1}^3 {\sum\limits_{k = 1}^3 { {\frac{{{\partial _i}}}{{\sqrt { - \Delta } }}\frac{{{\partial _j}}}{{\sqrt { - \Delta } }}({\partial _i}{d_k}{\partial _j}{d_k})} } } } \\ &= \sum\limits_{i = 1}^3 {\sum\limits_{i = j}^3 { {{{\cal R}_i}{{\cal R}_j}({u_j}{u_i})} } } + \sum\limits_{i = 1}^3 {\sum\limits_{j = 1}^3 {\sum\limits_{k = 1}^3 { {{{\cal R}_i}{{\cal R}_j}({\partial _i}{d_k}{\partial _j}{d_k})} } } } \\ & =: P_{1} + P_{2}\label{P}. \end{matrix}$

由标准的Calderon-Zygmund不等式[13], 存在常数 $C$, 使得

$\begin{array}{l} \| {P_1}\| _{{L^q}({\mathbb{R}^3})} \le C\| u\| _{{L^{2q}}({\mathbb{R}^3})}^2\quad \forall q \in (1, + \infty ),\\ \| {P_2}\| _{{L^q}({\mathbb{R}^3})} \le C\| \nabla d\| _{{L^{2q}}({\mathbb{R}^3})}^2\quad \forall q \in (1, + \infty ). \end{array}$

又由Minkowski不等式, 有

$\| P\| _{{L^q}({\mathbb{R}^3})} \le \| P_1\| _{{L^q}({\mathbb{R}^3})}{\rm{ + }}\| {P_2}\| _{{L^q}({\mathbb{R}^3})} \le C\| u\| _{{L^{2q}}({\mathbb{R}^3})}^2{\rm{ + }}C\| \nabla d\| _{{L^{2q}}({\mathbb{R}^3})}^2, \forall q \in (1, + \infty ).$

记 $Q=\frac{{|u{|^2}}}{2} + {P_1}$, 则

$\begin{array}{l} \| {Q}{\| _{{L^q}({\mathbb{R}^3})}} \le C\| u\| _{{L^{2q}}({\mathbb{R}^3})}^2, \quad \forall q \in (1, + \infty ).\\ \end{array}$

将(3.12)式代入(3.11)式, 有

$\begin{matrix} \int_{{\mathbb{R}^3}} {|\nabla u{|^2}} {\varphi _R}{\rm d}x & \le \frac{1}{2}\int_{{\mathbb{R}^3}} {|u{|^2}\Delta {\varphi _R}} {\rm d}x + \int_{{\mathbb{R}^3}} {\left(\frac{{|u{|^2}}}{2} + {P_1}\right)(u \cdot \nabla {\varphi _R}){\rm d}x} \\ & \ + \int_{{\mathbb{R}^3}} {{P_2}(u \cdot \nabla {\varphi _R}){\rm d}x} + \frac{1}{2}\int_{{\mathbb{R}^3}} {|\nabla d{|^2}(u \cdot \nabla {\varphi _R}){\rm d}x} \\ & \le \frac{1}{2}\int_{{\mathbb{R}^3}} {|u{|^2}\Delta {\varphi _R}} {\rm d}x + \int_{{\mathbb{R}^3}} {Q(u \cdot \nabla {\varphi _R}){\rm d}x} + \int_{{\mathbb{R}^3}} {{P_2}(u \cdot \nabla {\varphi _R}){\rm d}x} \\ & \ + \frac{1}{2}\int_{{\mathbb{R}^3}} {|\nabla d{|^2}(u \cdot \nabla {\varphi _R}){\rm d}x} =: \sum\limits_{i = 1}^4 {{I_i}}. \end{matrix}$

下面分别对 ${I_i}\ (i = 1,2,3,4)$ 各项进行估计. 对于 ${I_1}$, 利用Hölder不等式, 有

$\begin{align*} {I_1} & = \frac{1}{2}\sum\limits_{i = 1}^3 {\int_{_{ R \le |{x_i}| \le 2R }} {\int_{{D_i}} {|u{|^2}{{\tilde \varphi }_{i,R}}\left\{ {\frac{2}{{{R^2}}}\psi '\left( {\frac{{x_i^2}}{{{R^2}}}} \right) + \frac{{4\tilde x_i^2}}{{{R^4}}}\psi ''\left( {\frac{{x_i^2}}{{{R^2}}}} \right)} \right\}{\rm d}{{\tilde x}_i}{\rm d}x_i} } } \\ & \le \frac{C}{{{R^2}}}\sum\limits_{i = 1}^3 {{{\left( {\int_{ R \le |{x_i}| \le 2R } {\int_{{D_i}} {|u{|^6}} } {\rm d}x} \right)}^{\frac{1}{3}}}} {\left( {\int_{ R \le |{x_i}| \le 2R } {\int_{{D_i}} 1{\rm d}{{\tilde x}_i}{\rm d}x_i } } \right)^{\frac{2}{3}}}\\ & \le \frac{C}{{{R^2}}}\sum\limits_{i = 1}^3 {{{\left( {\int_{ R \le |{x_i}| \le 2R } {\int_{{\mathbb{R}^2}} {|u{|^6}} } {\rm d}x} \right)}^{\frac{1}{3}}}}. \end{align*}$

当 $R\rightarrow+\infty$ 时, 则 $I_{1}\rightarrow0.$

对于 ${I_2}$ 考虑各向异性, 有

$\begin{align*} {I_2} &\le \sum\limits_{i = 1}^3 {\frac{2}{{{R^2}}}} \left| {\int_{{\mathbb{R}^3}} {{x_i}} {{\tilde \varphi }_{i,R}}Q{u_i} \cdot \psi '\left( {\frac{{x_i^2}}{{{R^2}}}} \right){\rm d}x} \right|\\ & \le \frac{C}{R}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {\int_{{D_i}} | } } Q||{u_i}|{\rm d}{{\tilde x}_i}{\rm d}x_i\\ & \le \frac{C}{R}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | Q{|^{\frac{q}{2}}}{\rm d}{{\tilde x}_i}} \right)}^{\frac{2}{q}}}} } {\left( {\int_{{D_i}} {|{u_i}{|^s}{\rm d}{{\tilde x}_i}} } \right)^{\frac{1}{s}}}{\left( {\int_{D_{i}} {1{\rm d}{{\tilde x}_i}} } \right)^{\frac{{qs - q - 2s}}{{qs}}}}{\rm d}x_i\\ &\le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | Q{|^{\frac{q}{2}}}{\rm d}{{\tilde x}_i}} \right)}^{\frac{2}{q}}}} } {\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)^{\frac{1}{s}}}{\rm d}x_i\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}}\\ & \ \times \sum\limits_{i = 1}^3 {{{\left( {\int_{ R < |{x_i}| < 2R} {\int_{{D_i}} | } Q{|^{\frac{q}{2}}}{\rm d}{{\tilde x}_i}{\rm d}x_i} \right)}^{\frac{2}{q}}}} {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}^{\frac{{q - 2}}{q}}}\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}} {{{\left( {\int_{{\mathbb{R}^3}} {|Q{|^{\frac{q}{2}}}{\rm d}x} } \right)}^{\frac{2}{q}}}} {\sum\limits_{i = 1}^3 {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{\mathbb{R}^2}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}} ^{\frac{{q - 2}}{q}}}\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}} {{{\left( {\int_{{\mathbb{R}^3}} {|u{|^q}{\rm d}x} } \right)}^{\frac{2}{q}}}} {\sum\limits_{i = 1}^3 {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{\mathbb{R}^2}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}} ^{\frac{{q - 2}}{q}}}. \end{align*}$

对于 ${I_3}$ 和 ${I_4}$, 类似于 ${I_2}$ 的估计, 有

$\begin{align*} {I_3} &\le \sum\limits_{i = 1}^3 {\frac{2}{{{R^2}}}} \left| {\int_{{\mathbb{R}^3}} {{x_i}} {{\tilde \varphi }_{i,R}}{P_2}{u_i} \cdot \psi '\left( {\frac{{x_i^2}}{{{R^2}}}} \right){\rm d}x} \right|\\ & \le \frac{C}{R}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {\int_{{D_i}} | } } {P_2}||{u_i}|{\rm d}{{\tilde x}_i}{\rm d}x_i\\ & \le \frac{C}{R}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | {P_2}{|^{\frac{q}{2}}}{\rm d}{{\tilde x}_i}} \right)}^{\frac{2}{q}}}} } {\left( {\int_{{D_i}} {|{u_i}{|^s}{\rm d}{{\tilde x}_i}} } \right)^{\frac{1}{s}}}{\left( {\int_{D_{i}} {1{\rm d}{{\tilde x}_i}} } \right)^{\frac{{qs - q - 2s}}{{qs}}}}{\rm d}x_i\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | {P_2}{|^{\frac{q}{2}}}{\rm d}{{\tilde x}_i}} \right)}^{\frac{2}{q}}}} } {\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)^{\frac{1}{s}}}{\rm d}x_i\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}}\\ & \ \times \sum\limits_{i = 1}^3 {{{\left( {\int_{ R < |{x_i}| < 2R} {\int_{{D_i}} | } {P_2}{|^{\frac{q}{2}}}|{\rm d}{{\tilde x}_i}{\rm d}x_i} \right)}^{\frac{2}{q}}}} {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}^{\frac{{q - 2}}{q}}}\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}} {{{\left( {\int_{{\mathbb{R}^3}} {|{P_2}{|^{\frac{q}{2}}}{\rm d}x} } \right)}^{\frac{2}{q}}}} {\sum\limits_{i = 1}^3 {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}} ^{\frac{{q - 2}}{q}}}\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}} {{{\left( {\int_{{\mathbb{R}^3}} {|\nabla d{|^q}{\rm d}x} } \right)}^{\frac{2}{q}}}} {\sum\limits_{i = 1}^3 {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}} ^{\frac{{q - 2}}{q}}}. \\ {I_4}&\le \sum\limits_{i = 1}^3 {\frac{2}{{{R^2}}}} \left| {\int_{{\mathbb{R}^3}} {{x_i}} {{\tilde \varphi }_{i,R}}|\nabla d{|^2}{u_i} \cdot \psi '\left( {\frac{{x_i^2}}{{{R^2}}}} \right){\rm d}x} \right|\\ & \le \frac{C}{R}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {\int_{{D_i}} | } } \nabla d{|^2}|{u_i}|{\rm d}{{\tilde x}_i}{\rm d}x_i\\ &\le \frac{C}{R}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | \nabla d{|^q}{\rm d}{{\tilde x}_i}} \right)}^{\frac{2}{q}}}} } {\left( {\int_{{D_i}} {|{u_i}{|^s}{\rm d}{{\tilde x}_i}} } \right)^{\frac{1}{s}}}{\left( {\int_{D_{i}} {1{\rm d}{{\tilde x}_i}} } \right)^{\frac{{qs - q - 2s}}{{qs}}}}{\rm d}x_i\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}}\sum\limits_{i = 1}^3 {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | \nabla d{|^q}{\rm d}{{\tilde x}_i}} \right)}^{\frac{2}{q}}}} } {\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)^{\frac{1}{s}}}{\rm d}x_i\\ &\le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}}\\ & \ \times\sum\limits_{i = 1}^3 {{{\left( {\int_{ R < |{x_i}| < 2R} {\int_{{D_i}} | } \nabla d{|^q}{\rm d}{{\tilde x}_i}{\rm d}x_i} \right)}^{\frac{2}{q}}}} {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{D_i}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}^{\frac{{q - 2}}{q}}}\\ & \le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}} {{{\left( {\int_{{\mathbb{R}^3}} {|\nabla d{|^q}{\rm d}x} } \right)}^{\frac{2}{q}}}} {\sum\limits_{i = 1}^3 {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{\mathbb{R}^2}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}} ^{\frac{{q - 2}}{q}}}\\ &\le C{R^{\frac{{2(qs - q - 2s)}}{{qs}} - 1}} {{{\left( {\int_{{\mathbb{R}^3}} {|\nabla d{|^q}{\rm d}x} } \right)}^{\frac{2}{q}}}} {\sum\limits_{i = 1}^3 {\left\{ {\int_{ R < |{x_i}| < 2R} {{{\left( {\int_{{\mathbb{R}^2}} | {u_i}{|^s}{\rm d}{{\tilde x}_i}} \right)}^{\frac{q}{{s(q - 2)}}}}} {\rm d}x_i} \right\}} ^{\frac{{q - 2}}{q}}}. \end{align*}$

由定理 1.1 中的条件 $\frac{2}{q} + \frac{1}{s} \ge \frac{1}{2}$ 可知

$\frac{{2(qs - q - 2s)}}{{qs}} - 1 \le 0.$

当 $R\rightarrow+\infty$ 时, 则 $I_{2},I_{3},I_{4}\rightarrow0$. 从而,

$\int_{{\mathbb{R}^3}} {|\nabla u{|^2}}{\rm d}x=0.$

又 $u\in{L^6}({\mathbb{R}^3})$, 则

$u=0.$

记 $d_{i,j}=\frac{{\partial {d_i}}}{{\partial {x_j}}}$, 将 $u = 0$ 代入(1.1)2式可得

$\Delta d = -|\nabla d{|^2}d.$

将(3.14)式两边同时乘以 $x \cdot \nabla d{\varphi _R}$ 并在 ${\mathbb{R}^3}$ 积分, 由于 ${\partial _{{x_j}}}|d{|^2} =\sum\limits_{i = 1}^3 2 d_{i}d_{i,j} = 0 (j = 1,2,3),$ 可知

$\sum\limits_{i,j = 1}^3\int_{{\mathbb{R}^3}} {\Delta {d_i}{x_j}{d_{i,j}}{\varphi _R}{\rm d}x} = -\sum\limits_{i,j = 1}^3\int_{{\mathbb{R}^3}} {|\nabla d{|^2}{x_j}{d_i}{d_{i,j}}{\varphi _R}{\rm d}x = 0,} $

由分部积分公式, 有

$\begin{align*} 0&=\sum\limits_{i,j = 1}^3\int_{{\mathbb{R}^3}} \Delta {d_i}{x_j}{d_{i,j}}{\varphi _R}{\rm d}x\\ &= - \sum\limits_{i,j,k = 1}^3\int_{{\mathbb{R}^3}} {{d_{i,k}}{\delta _{j,k}}{d_{i,j}}{\varphi _R} +{d_{i,k}}{x_j}{d_{i,j}}{\partial _{{x_k}}}{\varphi _R}} + {d_{i,k}}{x_j}{d_{i,jk}}{\varphi _R}{\rm d}x\\ &{\rm{ = }} - \sum\limits_{i,j,k = 1}^3\int_{{\mathbb{R}^3}} {{d_{i,k}}{\delta _{j,k}}{d_{i,j}}{\varphi _R} +{d_{i,k}}{x_j}{d_{i,j}}{\partial _{{x_k}}}{\varphi _R}} + {\frac{{\rm{1}}}{{\rm{2}}}\partial_{x_j}{{({d_{i,k}})}^2}{x_j}{\varphi _R}}{\rm d}x\\ &{\rm{ = }} -\sum\limits_{i,j,k = 1}^3 \int_{{\mathbb{R}^3}} {\left( {|\nabla d{|^2}{\varphi _R} + {d_{i,k}}{x_j}{d_{i,j}}{\partial _{{x_k}}}{\varphi _R}} \right)} {\rm d}x\\ & \ + \sum\limits_{i,j,k = 1}^3\int_{{\mathbb{R}^3}} {\left( {\frac{3}{2}|\nabla d{|^2}{\varphi _R} + \frac{1}{2}|\nabla d{|^2}{x_j}{\partial _{{x_j}}}{\varphi _R}} \right)} {\rm d}x\\ &{\rm{ = }} \frac{1}{2}\int_{{\mathbb{R}^3}} | \nabla d{|^2}{\varphi _R}{\rm d}x + \int_{{\mathbb{R}^3}} \sum\limits_{i,j,k = 1}^3{\left(\frac{1}{2}|\nabla d{|^2}{x_j}{\partial _{{x_j}}}{\varphi _R} - {d_{i,k}}{x_j}{d_{i,j}}{\partial _{{x_k}}}{\varphi _R}\right){\rm d}x}, \end{align*}$

所以

$\int_{{\mathbb{R}^3}} | \nabla d{|^2}{\varphi _R}{\rm d}x =- \sum\limits_{i,j,k = 1}^3\int_{{\mathbb{R}^3}} {\left(|\nabla d{|^2}{x_j}{\partial _{{x_j}}}{\varphi _R} - 2 {d_{i,k}}{x_j}{d_{i,j}}{\partial _{{x_k}}}{\varphi _R}\right){\rm d}x},$

又因为

$|{x_j}{\partial _{{x_i}}} {\varphi _k}| \le \frac{C}{R}|{x_j}| \le C, |{x_j}{\partial _{{x_k}}} {\varphi _k}| \le \frac{C}{R}|{x_j}| \le C.$

故当 $R \to \infty $ 时, 有

$\int_{{\mathbb{R}^3}} {|\nabla d{|^2}{\rm d}x} \le C\int_{R \le |x| \le 2R} {|\nabla d{|^2}{\rm d}x} \to 0.$

由 $\nabla d \in L^{2}(\mathbb{R}^3)$, 则 $\nabla d = 0.$

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