曲率流的拼挤估计

Squeezing Estimation of Curvature Flow

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∂/∂_tX(x,t)=σ_k^1/kn，当k=1时，∂/∂_tX(x,t)=Hn，此时H是平均曲率流；当k=2时，∂/∂_tX(x,t)=R^1/2n，R^1/2是常数量曲率流，本文得到k=2时的拼挤估计。

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 ∂/∂_tX(x,t)=σ_k^1/kn, when k=1, ∂/∂_tX(x,t)=Hn, where H is the average curvature flow; When k=2, ∂/∂_tX(x,t)=R^1/2n, R^1/2 is a constant number of curvature flows. In this paper, the squeezing estimate is obtained when k=2.

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)。