新疆师范大学，新疆 乌鲁木齐

收稿日期：2020年8月30日；录用日期：2020年9月20日；发布日期：2020年9月27日

摘要

我们通过对平均曲率流的拼挤估计，得到对一般数量曲率流的拼挤估计。对于 $\frac{\partial }{\partial t}X\left(x,t\right)={\sigma }_k^{1/k}n$，当 $k=1$ 时， $\frac{\partial }{\partial t}X\left(x,t\right)=Hn$，此时H是平均曲率流；当 $k=2$ 时， $\frac{\partial }{\partial t}X\left(x,t\right)=R^{\frac{1}{2}}n$，$R^{\frac{1}{2}}$ 是常数量曲率流，本文得到 $k=2$ 时的拼挤估计。

关键词

平均曲率流，拼挤估计，常数量

Squeezing Estimation of Curvature Flow

Jinpeng Liu, Adila·Abudureyimu, Yawei Mi

Xinjiang Normal University, Urumqi Xinjiang

Received: Aug. 30^th, 2020; accepted: Sep. 20^th, 2020; published: Sep. 27^th, 2020

ABSTRACT

In this paper, we focus on the estimation of the numerical curvature flow and some related problems. By means of the mean curvature flow squeezing estimation, we get the squeezing estimation of the general number of curvature flow squeezing. For $\frac{\partial }{\partial t}X\left(x,t\right)={\sigma }_k^{1/k}n$, when $k=1$,$\frac{\partial }{\partial t}X\left(x,t\right)=Hn$, where H is the average curvature flow; When $k=2$,$\frac{\partial }{\partial t}X\left(x,t\right)=R^{\frac{1}{2}}n$,$R^{\frac{1}{2}}$ is a constant number of curvature flows. In this paper, the squeezing estimate is obtained when $k=2$.

Keywords:Mean Curvature Flows, Crowding Estimation, Often the Number

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

在现代微分几何中偏微分方程是一种非常有力的工具。特别的，通过抛物极大值原理作为主要工具，抛物发展方程(几何热流)已经成功的应用于研究流行的几何量。一些学者已经对平均曲率流的拼挤估计 [1] [2] 进行了研究 [3]，得出的先验条件 [4] [5]，本文对满足先验条件的数量曲率流进行了拼挤估计。

2. 预备知识

设 $M^n$ 是一个n维光滑流形，是一个 $R^{n+1}$ 中的光滑浸入超曲面 [6]。对于 $\frac{\partial }{\partial t}X\left(x,t\right)={\sigma }_k^{1/k}n$，如果 $k=2$，则 $\frac{\partial }{\partial t}X\left(x,t\right)=R^{\frac{1}{2}}n$ ( $x\in M^n$，$t>0$ )。其中 $R^{\frac{1}{2}}$ 和 $n$ 分别是常数量曲率流和 $X\left(\text{·},t\right)$ 的单位法向量。在局部坐标系 $\left\{x_i\right\}$ 下， $X\left(\text{·},t\right)$ 的度量和第二基本形式为： $g_{ij}\left(x,t\right)=\left(\frac{\partial X\left(x,t\right)}{\partial x^i},\frac{\partial X\left(x,t\right)}{\partial x^j}\right)$ ； $h_{ij}\left(x,t\right)=\left(n\left(x,t\right),\frac{{\partial }^{2}X\left(x,t\right)}{\partial x^i\partial x^j}\right)$。在 $X\left(\text{·},t\right)$ 上的联络系数为 ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{kl}\left(\frac{\partial }{\partial x^i}{g}_{jl}+\frac{\partial }{\partial x^j}{g}_{jl}-\frac{\partial }{\partial x^l}{g}_{ij}\right)$ ；向量v在上的协变导数为 ${\nabla }_j{v}^i=\frac{\partial }{\partial x^j}{v}^i+{\Gamma }_{jk}^i{v}^k$。黎曼集合张量，里奇张量和标量曲率通过高斯方程给定： $R_{ijkl}=h_{ik}{h}_{jl}-h_{il}{h}_{jk}$ ； $R_{ik}=H{h}_{ik}-h_{il}{g}^{lj}{h}_{jk}$ ； $R=H^2-{\left|A\right|}^2$，其中。

回忆Guass-Weingarten关系式为 $\frac{{\partial }^{2}X}{\partial x^i\partial x^j}={\Gamma }_{ij}^k\frac{\partial X}{\partial x^k}+h_{ij}n$ ； $\frac{\partial n}{\partial x^j}=-h_{jl}{g}^{lm}\frac{\partial X}{\partial x^m}$。

引理1 (Hamiltion [4] [7] )：极大值原理是研究抛物方程的有用的工具。现在我们提出一个关于张量的极大值原理。设对称张量 $M_{ij}$，如果对所有向量 $v^i$ 有 $M_{ij}{v}^i{v}^j\ge 0$，则 $M_{ij}\ge 0$。让 $N_{ij}=P\left(M_{ij},g_{ij}\right)$ 是 $M_{ij}$ 中的多项式，是 $M_{ij}$ 和它自己的度量的乘积形成的。

假设 $0\le t\le T$，

$\frac{\partial }{\partial t}{M}_{ij}=\Delta M_{ij}+u^k{\nabla }_k{M}_{ij}+N_{ij}$

其中 $N_{ij}=P\left(M_{ij},g_{ij}\right)$ 满足

$N_{ij}{v}^i{v}^j\ge 0$，无论何时 $M_{ij}{v}^j=0$

如果在 $t=0$ 时 $M_{ij}\ge 0$，则在 $0\le t\le T$ 时 $M_{ij}\ge 0$ 仍然成立。

3. 主要结论及其证明

根据 $\frac{\partial }{\partial t}X\left(x,t\right)=R^{\frac{1}{2}}n$，为了得到关于超曲面 $X\left(\text{·},t\right)$ 的几何量发展方程，需要下面几个引理：

引理2： $\frac{1}{2}{\nabla }_i{\nabla }_{j}R=\square h_{ij}+{\nabla }_{i}H{\nabla }_{j}H-{\nabla }_i{h}_{kl}{\nabla }_j{h}_{mn}+\left(H{\left|A\right|}^2-C\right)h_{ij}-R{h}_{im}{h}_j^m$

证明：

$\begin{array}{c}\frac{1}{2}{\nabla }_i{\nabla }_{j}R=H{\nabla }_i{\nabla }_{j}H+{\nabla }_{i}H{\nabla }_{j}H-\frac{1}{2}{\nabla }_i{\nabla }_j\left(g^{km}{g}^{ln}{h}_{kl}{h}_{mn}\right)\\ =H{g}^{kl}{\nabla }_i\nabla {}_{j}h{}_{kl}-g^{km}{g}^{ln}{h}_{mn}{\nabla }_i{\nabla }_j{h}_{kl}+{\nabla }_{i}H{\nabla }_{j}H-g^{km}{g}^{ln}{\nabla }_i{h}_{kl}{\nabla }_j{h}_{mn}\\ =\left(H{g}^{kl}-h^{kl}\right){\nabla }_i{\nabla }_j{h}_{kl}+{\nabla }_{i}H{\nabla }_{j}H-g^{km}{g}^{ln}{\nabla }_i{h}_{kl}{\nabla }_j{h}_{mn}\\ =\left(H{g}^{kl}-h^{kl}\right){\nabla }_k{\nabla }_l{h}_{ij}+\left(H{g}^{kl}-h^{kl}\right)\left(R_{ikjm}{h}_l^m+R_{iklm}{h}_j^m\right)+{\nabla }_{i}H{\nabla }_{j}H-g^{km}{g}^{ln}{\nabla }_i{h}_{kl}{\nabla }_j{h}_{mn}\\ =\square h_{ij}+{\nabla }_i\nabla {}_{j}H-g^{km}{g}^{ln}{\nabla }_i{h}_{kl}{\nabla }_j{h}_{mn}+\ast \end{array}$

其中：

$\begin{array}{c}\ast =\left(H{g}^{kl}-h^{kl}\right)\left[\left(h_{ij}{h}_{km}-h_{im}{h}_{kj}\right)h_l^m+\left(h_{il}{h}_{km}-h_{im}{h}_{kl}\right)h_j^m\right]\\ =H{g}^{kl}{h}_{km}{h}_l^m{h}_{ij}-h^{kl}{h}_{ij}{h}_{km}{h}_l^m-H{g}^{kl}{h}_{im}{h}_{kj}{h}_l^m+h^{kl}{h}_{im}{h}_{kj}{h}_l^m+H{g}^{kl}{h}_{il}{h}_{km}{h}_j^m-h^{kl}{h}_{il}{h}_{km}{h}_j^m-H{g}^{kl}{h}_{im}{h}_{kl}{h}_j^m\\ =H{\left|A\right|}^2{h}_{ij}-h^{kl}{h}_l^m{h}_{mk}{h}_{ij}-H^2{h}_{im}{h}_j^m+{\left|A\right|}^2{h}_{im}{h}_j^m\\ =\left(H{\left|A\right|}^2-h^{kl}{h}_l^m{h}_{mk}\right)h_{ij}-R{h}_{im}{h}_j^m\\ =\left(H{\left|A\right|}^2-C\right)h_{ij}-R{h}_{im}{h}_j^m\end{array}$

引理3： $-H^2\nabla \left(\frac{{\left|A\right|}^2}{H^2}\right)+\frac{2R}{H}\nabla H=\nabla R$

证明： $\begin{array}{l}-H^2\nabla \left(\frac{{\left|A\right|}^2}{H^2}\right)+\frac{2R}{H}\nabla H\\ =-H^2\left(\frac{\nabla {\left|A\right|}^2}{H^2}-\frac{2{\left|A\right|}^2\nabla H}{H^3}\right)+\frac{2H^2\nabla H-2{\left|A\right|}^2\nabla H}{H}\\ =-\nabla {\left|A\right|}^2+2H\nabla H\\ =\nabla \left(H^2-{\left|A\right|}^2\right)\\ =\nabla R\end{array}$

引理4：

$-\frac{1}{H^2}\left(H{\nabla }_i{h}_{kl}-{\nabla }_{i}H{h}_{kl}\right)\left(H{\nabla }_j{h}_{kl}-{\nabla }_{j}H{h}_{kl}\right)=-{\nabla }_i{h}_{kl}{\nabla }_j{h}_{kl}-\frac{{\left|A\right|}^2}{H^2}{\nabla }_{i}H{\nabla }_{j}H+2\frac{{\nabla }_{j}H}{H}{h}_{kl}{\nabla }_i{h}_{kl}$

证明： $\begin{array}{l}-\frac{1}{H^2}\left(H{\nabla }_i{h}_{kl}-{\nabla }_{i}H{h}_{kl}\right)\left(H{\nabla }_j{h}_{kl}-{\nabla }_{j}H{h}_{kl}\right)\\ =-\frac{1}{H^2}\left(H^2{\nabla }_i{h}_{kl}{\nabla }_j{h}_{kl}-2H{\nabla }_{j}H{h}_{kl}{\nabla }_i{h}_{kl}+{\nabla }_{i}H{\nabla }_{j}H{\left|A\right|}^2\right)\\ =-{\nabla }_i{h}_{kl}{\nabla }_j{h}_{kl}-\frac{{\left|A\right|}^2}{H^2}{\nabla }_{i}H{\nabla }_{j}H+2\frac{{\nabla }_{j}H}{H}{h}_{kl}{\nabla }_i{h}_{kl}\end{array}$

引理5： $\frac{\partial }{\partial t}{H}^2=R^{-\frac{1}{2}}\left(\square H^2-2{\left|\nabla H\right|}_{Hg-h}^2-\frac{2}{H}{\left|H\nabla h_{kl}-\nabla H{h}_{kl}\right|}^2-\frac{H^5}{2R}{\left|\nabla \frac{{\left|A\right|}^2}{H^2}\right|}^2+2\left(H{\left|A\right|}^2-C\right)H^2\right)$

证明： $\begin{array}{c}\frac{\partial }{\partial t}H=\frac{\partial }{\partial t}{g}^{ij}{h}_{ij}+g^{ij}\frac{\partial }{\partial t}{h}_{ij}\\ =2R^{\frac{1}{2}}{\left|A\right|}^2+R^{-\frac{1}{2}}\left(\square H-\frac{1}{H^2}{\left|H\nabla h_{kl}-\nabla H{h}_{kl}\right|}^2-\frac{H^4}{4R}{\left|\nabla \frac{{\left|A\right|}^2}{H^2}\right|}^2+\left(H{\left|A\right|}^2-C\right)H-2R{\left|A\right|}^2\right)\\ =R^{-\frac{1}{2}}\left(\square H-\frac{1}{H^2}{\left|H\nabla h_{kl}-\nabla H{h}_{kl}\right|}^2-\frac{H^4}{4R}{\left|\nabla \frac{{\left|A\right|}^2}{H^2}\right|}^2+\left(H{\left|A\right|}^2-C\right)H\right)\end{array}$

因为，

$\begin{array}{c}\square H^2=\left(H{g}_{kl}-h_{kl}\right){\nabla }_k{\nabla }_l{H}^2\\ =\left(H{g}_{kl}-h_{kl}\right){\nabla }_k\left(2H{\nabla }_{l}H\right)\\ =2\left(H{g}_{kl}-h_{kl}\right)H{\nabla }_k{\nabla }_{l}H+2\left(H{g}_{kl}-h_{kl}\right){\nabla }_{k}H{\nabla }_{l}H\\ =2H\square H+2{\left|\nabla H\right|}_{Hg-h}\end{array}$

所以，

$\frac{\partial }{\partial t}{H}^2=2H\frac{\partial }{\partial t}H=R^{-\frac{1}{2}}\left(\square H^2-2{\left|\nabla H\right|}_{Hg-h}^2-\frac{2}{H}{\left|H\nabla h_{kl}-\nabla H{h}_{kl}\right|}^2-\frac{H^5}{2R}{\left|\nabla \frac{{\left|A\right|}^2}{H^2}\right|}^2+2\left(H{\left|A\right|}^2-C\right)H^2\right)$

定理1：根据数量曲率流，我们得到下面的发展方程：

1) $\frac{\partial }{\partial t}{g}_{ij}=-2R^{\frac{1}{2}}{h}_{ij}$

2) $\frac{\partial n}{\partial t}=-{\nabla }^j{R}^{\frac{1}{2}}\frac{\partial X}{\partial x^j}$

3) $\begin{array}{l}\frac{\partial h_{ij}}{\partial t}=R^{-\frac{1}{2}}(\square h_{ij}-\frac{1}{H^2}\left(H{\nabla }_i{h}_{kl}-{\nabla }_{i}H{h}_{kl}\right)\left(H{\nabla }_j{h}_{kl}-{\nabla }_{j}H{h}_{kl}\right)-\frac{H^4}{4R}{\nabla }_i\left(\frac{{\left|A\right|}^2}{H^2}\right){\nabla }_j\left(\frac{{\left|A\right|}^2}{H^2}\right)\\ \begin{array}{c}\stackrel{}{}\\ \end{array}+\left(H{\left|A\right|}^2-C\right)h_{ij}-2R{h}_{ik}{h}_{kj})\end{array}$

4) $\frac{\partial }{\partial t}{R}^{\frac{1}{2}}=R^{\frac{1}{2}}\square R^{\frac{1}{2}}+R\left(H{\left|A\right|}^2-C\right)$

证明：

根据度量的定义、超曲面 $X\left(\text{·},t\right)$ 第二基本形式和Guass-Weingarten关系式，我们得到：

1) $\begin{array}{c}\frac{\partial }{\partial t}{g}_{ij}=\frac{\partial }{\partial t}\left(\frac{\partial X}{\partial x^i},\frac{\partial X}{\partial x^j}\right)\\ =2R^{\frac{1}{2}}\left(\frac{\partial n}{\partial x^i},\frac{\partial X}{\partial x^j}\right)\\ =-2R^{\frac{1}{2}}\left(n,\frac{\partial X}{\partial x^i\partial x^j}\right)\\ =-2R^{\frac{1}{2}}{h}_{ij}\end{array}$

2) $\begin{array}{c}\frac{\partial n}{\partial t}=\left(\frac{\partial n}{\partial t},\frac{\partial X}{\partial x^i}\right)\frac{\partial X}{\partial x^j}{g}^{ij}\\ =-\left(n,\frac{\partial }{\partial t}\frac{\partial X}{\partial x^i}\right)\frac{\partial X}{\partial x^j}{g}^{ij}\\ =-\frac{\partial }{\partial x^i}{R}^{\frac{1}{2}}\frac{\partial X}{\partial x^j}{g}^{ij}\\ =-{\nabla }^j{R}^{\frac{1}{2}}\frac{\partial X}{\partial x^j}\end{array}$

3) $\begin{array}{c}\frac{\partial h_{ij}}{\partial t}=\frac{\partial }{\partial t}\left(\frac{{\partial }^{2}X}{\partial x_i\partial x_j},n\right)\\ =\langle \frac{{\partial }^2}{\partial x_i\partial x_j}\left(R^{\frac{1}{2}}n\right),n\rangle +\langle \frac{{\partial }^{2}X}{\partial x_i\partial x_j},-\frac{\partial R^{\frac{1}{2}}}{\partial x^i}\frac{\partial X}{\partial x^j}{g}^{ij}\rangle \\ =\langle \frac{\partial }{\partial x_i}\left(\frac{\partial }{\partial x_j}{R}^{\frac{1}{2}}n-R^{\frac{1}{2}}{h}_{jk}{g}^{kl}\frac{\partial X}{\partial x_l}\right),n\rangle +\langle {\Gamma }_{ij}^k\frac{\partial X}{\partial x_k}+h_{ij}n,-\frac{\partial R^{\frac{1}{2}}}{\partial x^i}\frac{\partial X}{\partial x^j}{g}^{ij}\rangle \\ =\frac{{\partial }^2{R}^{\frac{1}{2}}}{\partial x^i\partial x^j}-\left(h_{jk}{g}^{kl}\frac{\partial X}{\partial x^l}\frac{\partial }{\partial x_i},n\right)-\left({\Gamma }_{ij}^k\frac{\partial X}{\partial x_k},\frac{\partial R^{\frac{1}{2}}}{\partial x^i}\frac{\partial X}{\partial x^j}{g}^{ij}\right)-\left(h_{ij}n,\frac{\partial R^{\frac{1}{2}}}{\partial x^i}\frac{\partial X}{\partial x^j}{g}^{ij}\right)\end{array}$

$\begin{array}{l}=\frac{{\partial }^2{R}^{\frac{1}{2}}}{\partial x^i\partial x^j}-R^{\frac{1}{2}}{h}_{jk}{g}^{kl}{h}_{il}-{\Gamma }_{ij}^k\frac{\partial R^{\frac{1}{2}}}{\partial x^k}\\ ={\nabla }_i{\nabla }_j{R}^{\frac{1}{2}}-R^{\frac{1}{2}}{h}_{jk}{g}^{kl}{h}_{il}\\ =-\frac{1}{4}{R}^{-\frac{3}{2}}{\nabla }_{i}R{\nabla }_{j}R+\frac{1}{2}{R}^{-\frac{1}{2}}{\nabla }_i{\nabla }_{j}R-R^{\frac{1}{2}}{h}_{jk}{h}_{il}\\ =R^{-\frac{1}{2}}\left(\frac{1}{2}{\nabla }_{i}R{\nabla }_{j}R-\frac{1}{4R}{\nabla }_{i}R{\nabla }_{j}R-R{h}_{jk}{h}_{il}\right)\end{array}$ (3.1)

根据引理2 (3.1)括号里的式子可以表示为：

$\square h_{ij}+{\nabla }_{i}H{\nabla }_{j}H-{\nabla }_i{h}_{kl}{\nabla }_j{h}_{kl}+\left(H{\left|A\right|}^2-C\right)h_{ij}-\frac{1}{4R}{\nabla }_{i}R{\nabla }_{j}R-2R{h}_{ik}{h}_{kj}$ (3.2)

(3.2)式括号中的 $-\frac{1}{4R}{\nabla }_{i}R{\nabla }_{j}R$ 根据引理3可以计算为：

$\begin{array}{l}-\frac{1}{4R}{\nabla }_{i}R{\nabla }_{j}R\\ =-\frac{1}{4R}\left(-H^2{\nabla }_i\left(\frac{{\left|A\right|}^2}{H^2}\right)+\frac{2R}{H}{\nabla }_{i}H\right)\left(-H^2{\nabla }_j\left(\frac{{\left|A\right|}^2}{H^2}\right)+\frac{2R}{H}{\nabla }_{j}H\right)\\ =-\frac{1}{4R}\left(H^4{\nabla }_i\left(\frac{{\left|A\right|}^2}{H^2}\right){\nabla }_j\left(\frac{{\left|A\right|}^2}{H^2}\right)+\frac{4R^2}{H^2}{\nabla }_{i}H{\nabla }_{j}H-4HR{\nabla }_i\left(\frac{{\left|A\right|}^2}{H^2}\right){\nabla }_{j}H\right)\end{array}$

再根据引理4，我们可以计算出：

所以我们得到：

$\begin{array}{c}\frac{\partial }{\partial t}{h}_{ij}=R^{-\frac{1}{2}}\left(h_{ij}+{\nabla }_{i}H{\nabla }_{j}H-{\nabla }_i{h}_{kl}{\nabla }_j{h}_{kl}-\frac{1}{4R}{\nabla }_{i}R{\nabla }_{j}R+\left(H{\left|A\right|}^2-C\right)h_{ij}-2R{h}_{ik}{h}_{kj}\right)\\ =R^{-\frac{1}{2}}(\square h_{ij}-\frac{1}{H^2}\left(H{\nabla }_i{h}_{kl}-{\nabla }_{i}H{h}_{kl}\right)\left(H{\nabla }_j{h}_{kl}-{\nabla }_{j}H{h}_{kl}\right)-\frac{H^4}{4R}{\nabla }_i\left(\frac{{\left|A\right|}^2}{H^2}\right){\nabla }_j\left(\frac{{\left|A\right|}^2}{H^2}\right)\\ \begin{array}{c}\\ \end{array}+\left(H{\left|A\right|}^2-C\right)h_{ij}-2R{h}_{ik}{h}_{kj})\end{array}$

4) 根据引理4可得

$\begin{array}{c}\frac{\partial }{\partial t}{R}^{\frac{1}{2}}=\frac{1}{2}{R}^{-\frac{1}{2}}\frac{\partial }{\partial t}R\\ =\frac{1}{2}{R}^{-\frac{1}{2}}\frac{\partial }{\partial t}\left(H^2-{\left|A\right|}^2\right)\\ =\frac{1}{2}\left[\square R-2{\left|\nabla H\right|}_{Hg-h}^2+2{\left|\nabla A\right|}_{Hg-h}^2-\frac{2}{H^2}{\left|H\nabla h_{kl}-\nabla H{h}_{kl}\right|}_{Hg-h}^2-\frac{H^4}{2R}{\left|\nabla \frac{{\left|A\right|}^2}{H^2}\right|}_{Hg-h}^2+2R\left(H{\left|A\right|}^2-C\right)\right]\\ =\frac{1}{2}\square R-{\left|\nabla H\right|}_{Hg-h}^2+{\left|\nabla A\right|}_{Hg-h}^2-\frac{1}{H^2}{\left|H\nabla h_{kl}-\nabla H{h}_{kl}\right|}_{Hg-h}^2-\frac{H^4}{4R}{\left|\nabla \frac{{\left|A\right|}^2}{H^2}\right|}_{Hg-h}^2+R\left(H{\left|A\right|}^2-C\right)\end{array}$ (4.1)

$\therefore H^2{\nabla }_i\left(\frac{{\left|A\right|}^2}{H^2}\right)={\nabla }_{i}R-\frac{2R}{H}{\nabla }_{i}H$

则

$\begin{array}{c}-\frac{H^4}{4R}{\left|\nabla \frac{{\left|A\right|}^2}{H^2}\right|}_{Hg-h}^2=-\frac{1}{4R}{\langle \nabla R-\frac{2R}{H}\nabla H,\nabla R-\frac{2R}{H}\nabla H\rangle }_{Hg-h}\\ =-\frac{1}{4R}{\left|\nabla R\right|}_{Hg-h}^2+\frac{2}{4R}{\langle \nabla R,\frac{2R}{H}\nabla H\rangle }_{Hg-h}-\frac{1}{4R}\frac{4R^2}{H^2}{\left|\nabla H\right|}_{Hg-h}^2\\ =-\frac{1}{4R}{\left|\nabla R\right|}_{Hg-h}^2+\frac{1}{H}{\langle \nabla R,\nabla H\rangle }_{Hg-h}-\frac{R}{H^2}{\left|\nabla H\right|}_{Hg-h}^2\end{array}$ (4.2)

$\begin{array}{c}\square R^{\frac{1}{2}}=\left(H{g}_{ij}-h_{ij}\right){\nabla }_i{\nabla }_j{R}^{\frac{1}{2}}\\ =\left(H{g}_{ij}-h_{ij}\right){\nabla }_i\left(\frac{1}{2}{R}^{-\frac{1}{2}}{\nabla }_{j}R\right)\\ =\frac{1}{2}\left(H{g}_{ij}-h_{ij}\right)\left(-\frac{1}{2}{R}^{-\frac{3}{2}}{\nabla }_{i}R{\nabla }_{j}R+R^{-\frac{1}{2}}{\nabla }_i{\nabla }_{j}R\right)\\ =-\frac{1}{4}{R}^{-\frac{3}{2}}{\left|\nabla R\right|}_{Hg-h}^2+\frac{1}{2}{R}^{-\frac{1}{2}}\square R\end{array}$ (4.3)

$\begin{array}{c}{\left|\nabla A\right|}_{Hg-h}^2=\frac{1}{H^2}\left(H^2{\left|\nabla A\right|}_{Hg-h}^2-H{\langle \nabla {\left|A\right|}^2,\nabla H\rangle }_{Hg-h}+A^2{\left|\nabla H\right|}_{Hg-h}^2\right)+\frac{1}{H}{\langle \nabla {\left|A\right|}^2,\nabla H\rangle }_{Hg-h}-\frac{A^2}{H^2}{\left|\nabla H\right|}_{Hg-h}^2\\ =\frac{1}{H^2}{\left|H\nabla h_{kl}-\nabla H{h}_{kl}\right|}_{Hg-h}^2+\frac{1}{H}{\langle \nabla {\left|A\right|}^2,\nabla H\rangle }_{Hg-h}-\frac{A^2}{H^2}{\left|\nabla H\right|}_{Hg-h}^2\end{array}$ (4.4)

把(4.2)，(4.3)，(4.4)代入(4.1)得到

$\begin{array}{c}\frac{\partial }{\partial t}{R}^{\frac{1}{2}}=R^{\frac{1}{2}}\square R^{\frac{1}{2}}+\frac{1}{4R}{\left|\nabla R\right|}_{Hg-h}^2+\frac{1}{H}{\langle \nabla {\left|A\right|}^2,\nabla H\rangle }_{Hg-h}-\frac{{\left|A\right|}^2}{H^2}{\left|\nabla H\right|}_{Hg-h}^2-{\left|\nabla H\right|}_{Hg-h}^2-\frac{1}{4R}{\left|\nabla R\right|}_{Hg-h}^2\\ +\frac{1}{H}{\langle \nabla R,\nabla H\rangle }_{Hg-h}-\frac{R}{H^2}{\left|\nabla H\right|}_{Hg-h}^2+R\left(H{\left|A\right|}^2-C\right)\\ =R^{\frac{1}{2}}\square R^{\frac{1}{2}}+R\left(H{\left|A\right|}^2-C\right)\end{array}$

定理2 (拼挤估计)对于函数 $\frac{\partial }{\partial t}X\left(x,t\right)=R^{\frac{1}{2}}n$ ( $x\in M^n$，$t>0$ )，如果 $t=0$ 时 $R^{\frac{1}{2}}\ge 0$，则 $t>0$ 时 $R^{\frac{1}{2}}\ge 0$ 仍然成立。

证明：

根据命题1， $R^{\frac{1}{2}}$ 的发展方程是

$\begin{array}{c}\frac{\partial }{\partial t}{R}^{\frac{1}{2}}=R^{\frac{1}{2}}\square R^{\frac{1}{2}}+\frac{1}{4R}{\left|\nabla R\right|}_{Hg-h}^2+\frac{1}{H}{\langle \nabla {\left|A\right|}^2,\nabla H\rangle }_{Hg-h}-\frac{{\left|A\right|}^2}{H^2}{\left|\nabla H\right|}_{Hg-h}^2-{\left|\nabla H\right|}_{Hg-h}^2-\frac{1}{4R}{\left|\nabla R\right|}_{Hg-h}^2\\ +\frac{1}{H}{\langle \nabla R,\nabla H\rangle }_{Hg-h}-\frac{R}{H^2}{\left|\nabla H\right|}_{Hg-h}^2+R\left(H{\left|A\right|}^2-C\right)\end{array}$

$\because \frac{1}{H}{\langle \nabla {\left|A\right|}^2,\nabla H\rangle }_{Hg-h}-\frac{{\left|A\right|}^2}{H^2}{\left|\nabla H\right|}_{Hg-h}^2-{\left|\nabla H\right|}_{Hg-h}^2+\frac{1}{H}{\langle 2H\nabla H-\nabla A^2,\nabla H\rangle }_{Hg-h}-\frac{H^2-{\left|A\right|}^2}{H^2}{\left|\nabla H\right|}_{Hg-h}^2=0$

$\therefore$ 存在向量 $v^i$ 满足 $R^{\frac{1}{2}}{v}^j=0$

则 $\left(H{\left|A\right|}^2-C\right)R{v}^j=0$

由引理3我们可以得出结论。

基金项目

新疆师范大学重点实验室(XJNUSYS082018A02)。

文章引用

刘晋鹏,阿迪拉·阿布都热依木,米雅薇. 曲率流的拼挤估计
Squeezing Estimation of Curvature Flow[J]. 理论数学, 2020, 10(09): 906-913. https://doi.org/10.12677/PM.2020.109105

参考文献