曲率流的拼挤估计
Squeezing Estimation of Curvature Flow
DOI: 10.12677/PM.2020.109105, PDF, HTML, XML,    科研立项经费支持
作者: 刘晋鹏, 阿迪拉·阿布都热依木, 米雅薇:新疆师范大学,新疆 乌鲁木齐
关键词: 平均曲率流拼挤估计常数量Mean Curvature Flows Crowding Estimation Often the Number
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)=σk1/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)=R1/2n,R1/2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)=σk1/kn, when k=1, ∂/∂tX(x,t)=Hn, where H is the average curvature flow; When k=2, ∂/∂tX(x,t)=R1/2n, R1/2 is a constant number of curvature flows. In this paper, the squeezing estimate is obtained when k=2.
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[J]. 理论数学, 2020, 10(9): 906-913. https://doi.org/10.12677/PM.2020.109105

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

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 t X ( x , t ) = σ k 1 / k n ,如果 k = 2 ,则 t X ( x , t ) = R 1 2 n ( x M n t > 0 )。其中 R 1 2 n 分别是常数量曲率流和 X ( · , t ) 的单位法向量。在局部坐标系 { x i } 下, X ( · , t ) 的度量和第二基本形式为: g i j ( x , t ) = ( X ( x , t ) x i , X ( x , t ) x j ) h i j ( x , t ) = ( n ( x , t ) , 2 X ( x , t ) x i x j ) 。在 X ( · , t ) 上的联络系数为 Γ i j k = 1 2 g k l ( x i g j l + x j g j l x l g i j ) ;向量v在上的协变导数为 j v i = x j v i + Γ j k 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 i j k l = h i k h j l h i l h j k R i k = H h i k h i l g l j h j k R = H 2 | A | 2 ,其中

回忆Guass-Weingarten关系式为 2 X x i x j = Γ i j k X x k + h i j n n x j = h j l g l m X 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 i j ,如果对所有向量 v i M i j v i v j 0 ,则 M i j 0 。让 N i j = P ( M i j , g i j ) M i j 中的多项式,是 M i j 和它自己的度量的乘积形成的。

假设 0 t T

t M i j = Δ M i j + u k k M i j + N i j

其中 N i j = P ( M i j , g i j ) 满足

N i j v i v j 0 ,无论何时 M i j v j = 0

如果在 t = 0 M i j 0 ,则在 0 t T M i j 0 仍然成立。

3. 主要结论及其证明

根据 t X ( x , t ) = R 1 2 n ,为了得到关于超曲面 X ( · , t ) 的几何量发展方程,需要下面几个引理:

引理2: 1 2 i j R = h i j + i H j H i h k l j h m n + ( H | A | 2 C ) h i j R h i m h j m

证明:

1 2 i j R = H i j H + i H j H 1 2 i j ( g k m g l n h k l h m n ) = H g k l i h j k l g k m g l n h m n i j h k l + i H j H g k m g l n i h k l j h m n = ( H g k l h k l ) i j h k l + i H j H g k m g l n i h k l j h m n = ( H g k l h k l ) k l h i j + ( H g k l h k l ) ( R i k j m h l m + R i k l m h j m ) + i H j H g k m g l n i h k l j h m n = h i j + i H j g k m g l n i h k l j h m n +

其中:

= ( H g k l h k l ) [ ( h i j h k m h i m h k j ) h l m + ( h i l h k m h i m h k l ) h j m ] = H g k l h k m h l m h i j h k l h i j h k m h l m H g k l h i m h k j h l m + h k l h i m h k j h l m + H g k l h i l h k m h j m h k l h i l h k m h j m H g k l h i m h k l h j m = H | A | 2 h i j h k l h l m h m k h i j H 2 h i m h j m + | A | 2 h i m h j m = ( H | A | 2 h k l h l m h m k ) h i j R h i m h j m = ( H | A | 2 C ) h i j R h i m h j m

引理3: H 2 ( | A | 2 H 2 ) + 2 R H H = R

证明: H 2 ( | A | 2 H 2 ) + 2 R H H = H 2 ( | A | 2 H 2 2 | A | 2 H H 3 ) + 2 H 2 H 2 | A | 2 H H = | A | 2 + 2 H H = ( H 2 | A | 2 ) = R

引理4:

1 H 2 ( H i h k l i H h k l ) ( H j h k l j H h k l ) = i h k l j h k l | A | 2 H 2 i H j H + 2 j H H h k l i h k l

证明: 1 H 2 ( H i h k l i H h k l ) ( H j h k l j H h k l ) = 1 H 2 ( H 2 i h k l j h k l 2 H j H h k l i h k l + i H j H | A | 2 ) = i h k l j h k l | A | 2 H 2 i H j H + 2 j H H h k l i h k l

引理5: t H 2 = R 1 2 ( H 2 2 | H | H g h 2 2 H | H h k l H h k l | 2 H 5 2 R | | A | 2 H 2 | 2 + 2 ( H | A | 2 C ) H 2 )

证明: t H = t g i j h i j + g i j t h i j = 2 R 1 2 | A | 2 + R 1 2 ( H 1 H 2 | H h k l H h k l | 2 H 4 4 R | | A | 2 H 2 | 2 + ( H | A | 2 C ) H 2 R | A | 2 ) = R 1 2 ( H 1 H 2 | H h k l H h k l | 2 H 4 4 R | | A | 2 H 2 | 2 + ( H | A | 2 C ) H )

因为,

H 2 = ( H g k l h k l ) k l H 2 = ( H g k l h k l ) k ( 2 H l H ) = 2 ( H g k l h k l ) H k l H + 2 ( H g k l h k l ) k H l H = 2 H H + 2 | H | H g h

所以,

t H 2 = 2 H t H = R 1 2 ( H 2 2 | H | H g h 2 2 H | H h k l H h k l | 2 H 5 2 R | | A | 2 H 2 | 2 + 2 ( H | A | 2 C ) H 2 )

定理1:根据数量曲率流,我们得到下面的发展方程:

1) t g i j = 2 R 1 2 h i j

2) n t = j R 1 2 X x j

3) h i j t = R 1 2 ( h i j 1 H 2 ( H i h k l i H h k l ) ( H j h k l j H h k l ) H 4 4 R i ( | A | 2 H 2 ) j ( | A | 2 H 2 ) + ( H | A | 2 C ) h i j 2 R h i k h k j )

4) t R 1 2 = R 1 2 R 1 2 + R ( H | A | 2 C )

证明:

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

1) t g i j = t ( X x i , X x j ) = 2 R 1 2 ( n x i , X x j ) = 2 R 1 2 ( n , X x i x j ) = 2 R 1 2 h i j

2) n t = ( n t , X x i ) X x j g i j = ( n , t X x i ) X x j g i j = x i R 1 2 X x j g i j = j R 1 2 X x j

3) h i j t = t ( 2 X x i x j , n ) = 2 x i x j ( R 1 2 n ) , n + 2 X x i x j , R 1 2 x i X x j g i j = x i ( x j R 1 2 n R 1 2 h j k g k l X x l ) , n + Γ i j k X x k + h i j n , R 1 2 x i X x j g i j = 2 R 1 2 x i x j ( h j k g k l X x l x i , n ) ( Γ i j k X x k , R 1 2 x i X x j g i j ) ( h i j n , R 1 2 x i X x j g i j )

= 2 R 1 2 x i x j R 1 2 h j k g k l h i l Γ i j k R 1 2 x k = i j R 1 2 R 1 2 h j k g k l h i l = 1 4 R 3 2 i R j R + 1 2 R 1 2 i j R R 1 2 h j k h i l = R 1 2 ( 1 2 i R j R 1 4 R i R j R R h j k h i l ) (3.1)

根据引理2 (3.1)括号里的式子可以表示为:

h i j + i H j H i h k l j h k l + ( H | A | 2 C ) h i j 1 4 R i R j R 2 R h i k h k j (3.2)

(3.2)式括号中的 1 4 R i R j R 根据引理3可以计算为:

1 4 R i R j R = 1 4 R ( H 2 i ( | A | 2 H 2 ) + 2 R H i H ) ( H 2 j ( | A | 2 H 2 ) + 2 R H j H ) = 1 4 R ( H 4 i ( | A | 2 H 2 ) j ( | A | 2 H 2 ) + 4 R 2 H 2 i H j H 4 H R i ( | A | 2 H 2 ) j H )

再根据引理4,我们可以计算出:

所以我们得到:

t h i j = R 1 2 ( h i j + i H j H i h k l j h k l 1 4 R i R j R + ( H | A | 2 C ) h i j 2 R h i k h k j ) = R 1 2 ( h i j 1 H 2 ( H i h k l i H h k l ) ( H j h k l j H h k l ) H 4 4 R i ( | A | 2 H 2 ) j ( | A | 2 H 2 ) + ( H | A | 2 C ) h i j 2 R h i k h k j )

4) 根据引理4可得

t R 1 2 = 1 2 R 1 2 t R = 1 2 R 1 2 t ( H 2 | A | 2 ) = 1 2 [ R 2 | H | H g h 2 + 2 | A | H g h 2 2 H 2 | H h k l H h k l | H g h 2 H 4 2 R | | A | 2 H 2 | H g h 2 + 2 R ( H | A | 2 C ) ] = 1 2 R | H | H g h 2 + | A | H g h 2 1 H 2 | H h k l H h k l | H g h 2 H 4 4 R | | A | 2 H 2 | H g h 2 + R ( H | A | 2 C ) (4.1)

H 2 i ( | A | 2 H 2 ) = i R 2 R H i H

H 4 4 R | | A | 2 H 2 | H g h 2 = 1 4 R R 2 R H H , R 2 R H H H g h = 1 4 R | R | H g h 2 + 2 4 R R , 2 R H H H g h 1 4 R 4 R 2 H 2 | H | H g h 2 = 1 4 R | R | H g h 2 + 1 H R , H H g h R H 2 | H | H g h 2 (4.2)

R 1 2 = ( H g i j h i j ) i j R 1 2 = ( H g i j h i j ) i ( 1 2 R 1 2 j R ) = 1 2 ( H g i j h i j ) ( 1 2 R 3 2 i R j R + R 1 2 i j R ) = 1 4 R 3 2 | R | H g h 2 + 1 2 R 1 2 R (4.3)

| A | H g h 2 = 1 H 2 ( H 2 | A | H g h 2 H | A | 2 , H H g h + A 2 | H | H g h 2 ) + 1 H | A | 2 , H H g h A 2 H 2 | H | H g h 2 = 1 H 2 | H h k l H h k l | H g h 2 + 1 H | A | 2 , H H g h A 2 H 2 | H | H g h 2 (4.4)

把(4.2),(4.3),(4.4)代入(4.1)得到

t R 1 2 = R 1 2 R 1 2 + 1 4 R | R | H g h 2 + 1 H | A | 2 , H H g h | A | 2 H 2 | H | H g h 2 | H | H g h 2 1 4 R | R | H g h 2 + 1 H R , H H g h R H 2 | H | H g h 2 + R ( H | A | 2 C ) = R 1 2 R 1 2 + R ( H | A | 2 C )

定理2 (拼挤估计)对于函数 t X ( x , t ) = R 1 2 n ( x M n t > 0 ),如果 t = 0 R 1 2 0 ,则 t > 0 R 1 2 0 仍然成立。

证明:

根据命题1, R 1 2 的发展方程是

t R 1 2 = R 1 2 R 1 2 + 1 4 R | R | H g h 2 + 1 H | A | 2 , H H g h | A | 2 H 2 | H | H g h 2 | H | H g h 2 1 4 R | R | H g h 2 + 1 H R , H H g h R H 2 | H | H g h 2 + R ( H | A | 2 C )

1 H | A | 2 , H H g h | A | 2 H 2 | H | H g h 2 | H | H g h 2 + 1 H 2 H H A 2 , H H g h H 2 | A | 2 H 2 | H | H g h 2 = 0

存在向量 v i 满足 R 1 2 v j = 0

( H | A | 2 C ) 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)。

参考文献

[1] Hamilton, R.S. (1995) Isoperimtric Estimates for the Curve Shrinking Flow in the Plan. Annals of Mathematics Studies, 137, 201-222.
https://doi.org/10.1515/9781400882571-014
[2] Hamilton, R.S. (1995) Harnack Estimates for the Mean Curvature Flow. Journal of Differential Geometry, 41, 215-226.
https://doi.org/10.4310/jdg/1214456010
[3] 陈旭忠. 关于曲率流的某些问题[D]: [博士学位论文]. 杭州: 浙江大学, 2006.
[4] Hamilton, R.S. (1982) Three-Manifolds with Positive Ricci Curvature. Journal of Differential Geometry, 17, 255-306.
https://doi.org/10.4310/jdg/1214436922
[5] Hamilton, R.S. (1986) Four-Manifolds with Positive Curvature Oper-ator. Journal of Differential Geometry, 24, 153-179.
https://doi.org/10.4310/jdg/1214440433
[6] Hamilton, R.S. (1994) Convex Hypersurfaces with Pinched Second Fundamental Form. Communications in Analysis and Geometry, 2, 167-172.
https://doi.org/10.4310/CAG.1994.v2.n1.a10
[7] Hamilton, R.S. (1995) A Compactness Property for Solution of the Ricci Flow. American Journal of Mathematics, 117, 545-572.
https://doi.org/10.2307/2375080