微分几何彭家贵课后题答案

1、习题一(P13)2.设a(t)是向量值函数,证实:(1) a 常数当且仅当a(t),a(t) 0;(2) a(t)的方向不变当且仅当 a(t) a (t)0.2(1)证实：a 常数 a 常数 a(t),a(t) 常数a (t),a(t) a(t),a (t)02a(t),a(t)0a(t),a(t)0.(2 )注意到：a(t) 0,所以a(t)的方向不变单位向量e(t) a(t) 常向量.a(t)假设单位向量e(t)a-常向量,贝U e(t) 0 e(t) e(t) 0.|a(t)|a(t)2a(t)a(t)dt (盘)孟a(t)a(t)反之,设e(t)为单位向量,假设e(t) e(t) 0,

2、那么e(t)/e(t).由e(t)为单位向量e(t),e(t)1e(t),e(t)0 e(t) e (t).e(t)/e (t)从而,由e(t) e (t)e (t)0 e(t)常向量.所以,a(t)的方向不变单位向量e(t) a(t)常向量 a(t)|e(t) e(t)a(t)a (t)d 10| (,)a(t)0a(t)a(t)dt a(t)a(t) a (t)0.即a(t)的方向不变当且仅当a(t) a (t)0.补充:定理r(t)平行于固定平面的充要条件是r(t),r (t),r (t)0.证实："":假设r(t)平行于固定平面,设n是平面的法向量,为一常向量.于是

3、, r(t),n 0 r (t),n0, r (t), n 0r(t),r (t),r (t)共面r(t),r (t),r (t)0."":假设 r(t),r (t),r (t)0,那么 r(t),r (t), r (t)共面.假设 r(t) r (t)0那么r(t)方向固定,从而平行于固定平面 .假设 r(t) r (t)0,那么 r (t) r(t) r (t).令 n(t) r(t) r (t),那么n(t) r (t) r (t) r(t) r (t)r(t) r (t) r(t) (t)r(t) (t)r (t)(t) r(t) r (t)(t)n(t)n(t)

4、n (t)0,又 n(t) 0n(t)有固定的方向,又n(t) r(t)r(t)平行于固定平面.3.证实性质与性质.性质1证实：设V,(X!,X2,X3),V2(yi, y2, y3), V3(Z, Z2,Z3),V2V3(Wi, W2, W3),ijkyiy2y3ZiZ2Z3V2V3y2y3y3yiyiy2Z2Z3Z3ZiZiZ2w, w2, w3wiy2Z3y3Z2, wysZiyiZ3,W3yiZ2y2Zi,左=场 VV3)i jkX2X3,X3Xi,XiX2XiX2X3W|w2W3W2W3W3WiWiW2X2W3X3W2, X3Wix,w3,x,w2X2W,Y3Z2X2【y,Z2 y2

5、Zi X3y3Zi y,Z3, X3【y2ZaX2Z2 X3Zayi X22 X3y3Zi,X3Z3 XiZiy2X2Z2 X3Zayi,X3Z3 XiZjy2,XiZ X2Z2】y3XiZiyi,X3Z3 XiN X2Z2】y2,XiNXiyiZ22乙,为卜3乙乙約3閥2Xiyiz2,XiZi約3匕,两3X3Z3y3X22 X3y3】Z3X2【y2Z3 Y3Z2x,yi X2y2Z3X2Z2y3XiyiZ2,Xiyi X2y2Z3XiZiX2Z2X3Z3 yi,y2,y3 伙2丫2 X33 Xiyi z,Z2,Z3Vi ,V3 V2Vi,V2 V3 右X2Z2X2Z2 X3Z3冋2沁* 为力

6、习,以3丫3 Xiyi X2y2】Z2,Xiyi(2)证实：设 Vi (X!,X2,X3),V2 (yi,y2, y3),V3 (Zi,Z2, Z3),V4 (23),那么Xi,X2,X3ijkViv XiX2X3yiy2y3Xi xyX3y2,X2ij kVV4ZiZ2Z3w-iw2w3X3XiXiX2y3yiyiy2X2X3讨2y3X3yiXiy3,X3Xiy2X2yiZ2Z3Z3Z2w2W ' W3w| ' wiw2Yi,Yz,丫3Y z2w3 z3w2,Y2 z3wi ziw3,丫5Z|W2 z2wi.左=vi V2,v3 v4XiYX2Y2X3Y3(ZiX2y3 yi

7、Z2X3 Xiy2Z3)(乙2X3 人乙2丫3 yiX2Z3)V3,Vi,V2(X2y3 X3y2)(Z2W3 Z3W2) (X3% Xiy3)(Z3W ziW3) (Xiy2 X2%)(乙W2 Z2WJx2Z2y3W3 y2W2X3Z3 yiWiX3Z3 Xiy3Ws %乙丫2可2 yiWiX2Z2X?W2y3Z3 y2Z2X3W3 yiZiX3W3 XiZiy3W3 XiWiy3Z3 XiWiy2Z2 yiZX2W2yiWiX2Z2yiZX2W2(XiyiZiWi X2y2Z2W2 XsysZsWs)y2W2X3Z3%WiX3Z3 %乙丫3可3 淬沖2(Xiy忆Wi X2y2Z2W? X3

8、y3Z3W3) X2W2y3Z3 y2Z2X3W3 yiZiX3W3 %那么丫3乙3 XiWiyzZ?(X忆X2Z2X3Z3)(yiWiy2W2y3W3) (MWX2W2><3可3)(%乙沖2丫3乙3)=Vi, V3V2 V4WV4V2V3 右(3)证实：设 Vi (xixx)" (yi,y2, y3), V3 (zi,Z2, Z3),那么ViV2XiX2X3yiy2y3X2y2X3y3X3y3XiyiyiX2y2Xi,X2,XXiX2y3X3y2,X2X3%Xiy3,X3x2X2yiV3,Vi,V2V3,Vi V2ZiXi Z2X2 Z3X3XiZ2y3%X2Z3)乙(

9、X2y3 X3y2) Z2(X3% Xiy3) Z3 (Xi y2 x?%) (ZiX2y3 yiZ2X3%y2Z3)(2X3同理,ZiXijZ2X2Z3X3丫Z2X3Z3X2,Y2Z2X2Z3X3Z3X3ZiXiZiZ2xix2论冷丫3ZsXiX3,YsZiX2Z2XiyiYy2丫2y3丫3V2,V3,ViV2,V3yi(z2X3 Z3X2) y2(Z3Xi 乙X3) y3(ZiX2 Z2XJViiffffffy zZyZXXZx y yX2f2f2f2f2f2fJJ(0,0,0) 0.y zz yZ XX Zx yy xijkRQPRQP证实：2FJJJJXyZyZZXXyPQRRQPRQ

10、px yZyZXZXy2r2q2p2r2q2P0x yX Zy zy xZ Xz y4.设 0；©6,3是正交标架,是1,2,3的一个置换,证实:V2V3yiZijy2Z2y3Z3y2Z2y3Z3y3Z3yiZiyiZiy2Z2Z1, Z2, Z3Vi,V2,V3Xi(y2Z3y3Z2)(ZiX2y3yiZ2X3Vi,V2 V3X2(y3ZiX2Z3)yiZ2XiZi X2Z2 X3Z3y2Zi)yiX2Z3)V3,Vi,V2%Z3)X3(%Z2(2X3XiZ2y3所以, Vi,V2, V3V3,Vi,V2V2,V3,M .性质证实：i fijkf,f)y zXyZ丄f丄XyZi O

11、;ei,e2,e 是正交标架;2O; ei , e2 ,e3与 O;e (i),e,e定向相同当且仅当是一个偶置换.i证实：当ij时,(i)(j)e (i) , e (j)0 ；当ij时,(i)(j)e (i),e (j) y ax1 2xx ,所以,0;e,e(2),e是正交标架.(2)证实：A)当 (12)(1) 2, (2)1, (3)301 0010e (1), e(2) , e e2 , e1, e3ei,e2, e310 0,det100 1;00 1001B)当(13)(1) 3, (2)2, (3)100 1001e (1),e,ee3, e2, eei,e2, e301 0,

12、det010 1;10 0101C)当(23)(2)3,2, (1)110 0100e (1), e(2) , e e,e3, eei,e2, e300 1,det001 1;01 0010D)当(1)(：12)(12),此时,O;e (1),e (2) ,e (3“O;e1,e2,e3；E)当(123)(12)(13)(1) 2,3,(：3) 1,001001e (1),e,ee2, e3, e1e1,e2,e310 (0 ,det100 1;01 (0010F)当(132)(13)(12)(1) 3,2,(：2) 1,01 (0001e (1),e,ee3, ei, e2e1, e2 ,e

13、3001 ,det100 1.10 (0010所以,O;e,e2,e3 与 O； e (1)：> e (2) ,e )疋向相同当且仅当是一个偶置换.习题二(P28)1.求以下曲线的弧长与曲率：令2|a|t tan , .1 4a tsec,那么2.1 4a2t2dt= asec3 dI sec d(sec tan2sec )dtan d secsec dtansec3secd sec dtan sec Isec»ansecIn |sectan | CA2 2|a|t .1所以,2 24a tIn 2|a|t: 1 4a2t2.14a2t2dt12asec3I2|a|2|a|tj

14、1 4a2t2 ln 2|a|t - 1 4a2t2(%l(x)r (t) dt 1 4a2t2dt-2|a|x.1 4a2x2 4|a|ln 2 |a | x 寸 1 4a2x22.设曲线r(t) (x(t), y(t),证实它的曲率为 x(t)y (t) x (t)y (t)3(x)2 (y)22证实：r(t) (x(t),y(t) r (t) (x (t), y (t) r (t)(x(t), y (t)t(s)n(s)drdtdtr (t)(x(t), y(t) dsdsds-JX(y(t),x(t)-ds&s)d 2r d ds2 dsds2dtr (t) rds&s

15、)r (t)(s)n(s)2鱼dsd2t(t)ds2-JX(s)( y ,x(t临x (t)乎ds(s)心y (t)乎dsy (畔ds-JX(s)x(吃dtd ty (t) y -dsdsdtx(t) ds(y)2 (x)2 罟dtdtd tx (t)x (t)/、dsds(s)亍dt y (t) dsdt 2dtx(t)y (t) 丁 x(t)y(t) 丁 dsdsdt(y)(x)ds由dS(t)l Ux)2 (y)2(s)x(t)y (t) x (t)y(t)即3(y)2 (x)2 2 (t) x(t)y (t) x (t)y (t)3(y)2 (x)2 23.设曲线C在极坐标下的表示为r

16、f(,证实曲线c的曲率表达式为2f2()2 fdf(2)d 232 2f2()乎d证实： x r cos f ( )cos , y rsin f( )sinr( )(f( )cos , f( )sin )r ( ) (f ( )cos f ( )sin , f ( )sinf ( )cos )r ( )(f ( )cos f ( )sin , f ( )sinf( )cos )(f ( )cos2 f ( )sinf ( )cos , f ( )sin2 f ( )cosf ( )sin )所以,xf ( )cosf ( )sin ；y f ( )sinf ( )cos ；xf ( )cos2

17、 f ( )sinf ( )cos ；yf ( )sin2 f ( )cosf ( )sin .因此,xyf ( )cosf ( )sinf ( )sin 2 f ( )cosf ( )sin(y)2(x)2f ( )sinf2( ) 2f ( )cos2f()f ()cos 2 f ( )sinf ( )cosf ( )cos2f ()f(2f()sinf ( )sinf ( )cosx ( )y(3_ (x)2 (y)22)y ()f2()f2()df df()等32 24.求以下曲线的曲率与挠率:(4) r(t) (at,、2aInt,¥)(a0)解：r (t)(a,-|,右)

18、,r(t)(0,2a 2at3 ),r(t)(0'学'昴);r (t)r (t)j2aa2at32a2V2a2r (t) r(t)2a4 4a4t8t62a4t4.2a2t41 2t2 t4x 2a2tt2 1r (t)2a2 a2t2t4t2 1r ,r ,r八2a22 a2壬(0,2 2a 6aF,F2 2a3 丁.所以,(t)r (t)r (t)r(t)3x2a23旦t2 1t211t2全t2 1f ' 1a323石t 1、2t2；at2 1 2(t)2r ,r ,r 2、2a3 2a2 t2 12T6/74 t 1r (t) r (t)|t / tt22.215

19、.证实:E3的正那么曲线r(t)的曲率与挠率分别为(t)r (t) r (t)3?r(t)3r r r(t)°证实：dsdr dt dt dst(s) &s)dt 2 ds 鹫s) r (t)芈 ds 根据弗雷内特标架运动方程t&s) r (t)dtr (t) ds r(t)勒/、dt d2t(t)2ds ds3rr (畔&s)1(s)t d一 n dsb(s)n(s)ds,得:n(s)b(s) t(s)n(s)1t(s)細(t)2dsdtds3r(t)r (t)1(s)3dtds3(t)r (t)/、 dt(s)dsr (t)r (t)&S) (s)

20、n (s)由 i&(s) =(s)t(s)(s)b(s)t&s)&(s) n(s)(s)(s)t(s)&(s) n(s) 2t&s),b(s)&&s)&(s)n (s)r (t) r (t)3dsdt(s)n(s)(s)b(s)(s)t(s)(s) (s)b(s)(s) (s)r (t)r (t) r (t)3由于t&(s),b(s)r (t)dt 3ds3r (碍兽 r(t)£, 1(s)d r(t)dsr (t)61 dt(s) dsr ,r,r所以, (s) (s)=(s)dt dsr ,rr ,r , r

21、命5s)dtdsr ,r , r2.6证实：曲线r(s)(133s)2(1 s 1)以s为弧长参数,并求出它的曲率,挠率与Frenet 标架.证实:1(1 s)21) r (s)21(1 s)212 .,21 s 1)所以,&S)r (s)(1 s)(1 s)4s 1)该曲线以s为弧长参数.1(1 sl!4,01)(s)(1 s) 116(1 s)11618(1 s2)n(s)(s)2(1 s)(11sf ,2(1s)(11sf,0b(s)t(s) n(s)2(1i1(1 s)2s)(11s)22(1.2(1 s)(1 s)2,J2(1s)(11s)?,4(1由 n(s)(121s)2

22、s)(112?s )21 1b(s) ,2(1 s)(1 s)2,2(1 s)(1s)4(1(s)I&(s),b(s)13s,13s,0,1s12(1s)(1s)3 .2(1s)(11s)P,4(1TP1s)(11s)21 3sJ(1s)(1 2(1所以,3s)(11s2)2.2(13s)(12)(s)18(1sv(1)；(s)3)所求Frenet标架是t(s)1(1 s)22(1n(s)2(1 s)(1b(s)2、2(12、. 2(11).r(s);t(s), n(s),b(s),其中1s)212 ',21),1s)2,2(1s)(11s)2,01),1s)(1s)°

23、;,4(11s2)2 (s 1).10.设T (X) XT P是E3中的一个合同变换,detT1.r(t)是E3中的正那么曲线.求曲线% Tor与曲线r的弧长参数、曲率、挠率之间的关系.t解：(1)S%t) %( )dd(Tor)dd(rT P)dtr ( )Td0tr ( )d S(t)0可见,% Tor与曲线r除相差一个常数外,有相同的弧长参数.可见,(3)%t)%(t)%(t)r (t)T r (t)T%(t)3r (t)Tsgn(detT) r (t) r (t) Tr (t)3% T or与曲线r有相同的曲率.%t)%,%2r T, r T,r%t)%(t)r (t) r (t)r

24、(t)T r (t)T(t)3(t)rT,r T r Tr (t) r (t)|2rT ,sg n(detT)(rr )Tsgn( detT)r (t) r (t)rT,(r r )Tsgn(detT) rT'sgn(detT)(rr )T2r (t) r (t)r ,(r r )2sgn(detT) r (t) r (t)2r (t) r (t)|r ,r ,rr (t) r (t)2r ,r ,r2(t)r (t) r (t)可见,% Tor与曲线r的曲率相差一个符号.a2-asgn( detT)13. (1)求曲率 (s)(s是弧长参数)的平面曲线 r (s). s解：设所求平面

25、曲线 r(s)x(s), y(s)由于s是弧长参数,所以可设x (s)|r (s)| 1cos , x(s)x(s)sina22a ssx (s)cos(arctan-), x (s)a2y(s) 1,由曲率的定义,知ds s2 a 2 ds a ss arcta nasin (arcta ns) as x(s)cos(arcta n-)dsads1 tans 21 tan (arctan )a (arcta n?)a_1a2ds aln(s . a2 s2)2sy(s)sin( arcta n§)dsas cos (arctan) ds a ds ds2 s 2sec (arcta

26、n )a-4rds .a2 s2a s所以,所求平面曲线r(s)aln(sa2 s2), -a2 s2).20. 证实：曲线 r(t) (t、3si nt,2cost,、3t si nt)与曲线 t) (2cos- ,2si n±, t)2 2是合同的.证实：1)对曲线C% % %t)作参数变换t 2u ,那么% (2cos u,2sin u, 2u).可知C是圆柱螺线(a 2,b2),它的曲率和挠率分别为 % 4 , % 0因此,只要证实曲线C: r r(t)的曲率 ；,挠率 4 ,从而根据曲 线论根本定理,它们可以通过刚体运动彼此重合.2)下面计算曲线C的曲率 与挠率.由 r (

27、t)(1、3 cost,进而 r (t) ( 3sint,2sin t,、3 cost)2cos t, si nt)|r (t)|2、2,r (t)r (t)(2、3 cost2, 4sint, 2、32cos t)2(1、3cost,2si nt,、为 cost)|r (t)r (t)|4.2104V (t)vvv(.3 cost, 2sin t, cost)r (t), r (t), r (t)18421. 证实：定理定理设(s)0是连续可微函数,那么(1) 存在平面E2的曲线r(s),它以s为弧长参数,(s)为曲率;(2) 上述曲线在相差一个刚体运动的意义下是唯一的.证实：先证实(1),

28、为此考虑下面的一阶微分方程组G(s)(s)e>(s)(s)e(s)drds(1.1)孚dsde2ds给定初值r0,e0,e°,其中e°,e°是E2中的一个与自然标架定向相同的正交标架,以及§(a,b),那么由微分方程组理论得,(1.1)有唯一一组解 r(s);e1 (s),e>(s)满足初 始条件：r(s)；e(s),q(s) Issr0；e0,e .假设r(s)为所求曲线,贝Ue(s),q(s)必是它的Frenet标架.因此,我们首先证实U(s),e2(s)s (a,b)均是与自然定向相同的正交标架.将微分方程组(1.1)改写成(1.2)d

29、eds2/(s), i 1,2-J-J-J(1.4)- gj (s)-e(s),ej(s)丁 e(s),ej(s)dsdsds2e(s),ajkQ(s)k 12丨ajkk 121ajkk 1Q(s),ej(s)24kek(s),ej(s)k 12akek(s),ej(s)k 12e(s),ek(s)ek(s),e(s)aik gkj (s) ajk gki (s)i, j(1.4)说明 gij(s)i,j 1,2是微分方程组(5)(1.5)ds fj(s)2aik fkj (s) ajk fki (s)k 1i,j 1,2e(s),j(s)其中0aij 22(s)(s)o0是一个反对称矩阵,即

30、aijaji0 i, j 1,2.令(1.3)gij(s)e(s),ej(s)( gj)i, j 1,2对(1.3)求导,并利用(1.2)有：k 1的解.定义ij1, i0, ij; i 1,2.那么 j.gdsij0,i, j1,2.且2(a1k k1a1kk1 ) 1a110, ij1k 122(a2k k2k 1 aik kja jk ki2k 1z1k k2a2kk2)a22a220,i j242 kk1)a12a210, i1,j2k 12(a2k k1a1kk2)a2120, i2,j1k 1旳d2即"ijaik kj ajk ki ,i, j1,2.所以,j,i 1,2

31、.是微分方程组(1.5)的解.注意到：gj(so) i,ji,2j ,2,所以gj(s) i,ji,2是微分方程组(佝满足初始条件gj(So)j的唯一解.从而i,j 1,2i,j 1,2gj(s) j, i, j 1,2.所以,e1 (s),e2(s)s (a,b)均是正交标架.由于F(s) u(s),e2(s),e1(s) e>(s)s (a,b)是关于s的连续 函数,且F(s) 1或-1.故由F(s.)e (so),e2(so),G(So) e2(so) =1 知,F(s) e (s),e2(s),e (s) e2(s) =1, s (a,b).可见,e(s),e2(s)s (a,b

32、)均是与自然定向相同的正交标架.于是由微分方程组(1.1)有：drds (s) =1,这说明s为弧长参数.从而由祟e(s)推出 t(s)e (s)是单位切向量.由亜ds(s)a(s)推出(s)兰1t&s)是曲线r(s)的曲率,从而 由ds(s)e?(s)推出由 n(s) - &s) -e2(s),即 e2(s)是单位正法向量.ds(s)(s) ds可见,微分方程组(1.1)的满足初始条件：r(s);e (s),q(s) |s s,r0;e°,e2唯一一组r(s);e)(s),e2(s)确实说明：存在平面E2的曲线r(s),它以s为弧长参数,(s)为曲率,当(s)是连续

33、可微函数时.再证实(2):设r)(s)与r2(s)是平面E2中两条以s为弧长参数的曲线,且定义在同一个参数区间(a,b)上, 1(s)2(s) 0 s (a,b).那么存在刚体运动T(X) XT P 把曲线 d(s)变为 ri(s),即? Tor?.证实开始：设0(a,b),考虑两条曲线在s 0处的Frenet标架A(O);ti(O), ni(0)与 a(0);t2(0),n2(0).那么存在 平面E2中一个刚体运动T把第二个标架变为第一个标架, 即ri与Tor?在s 0处 的Frenet标架重合.因此我们只须证实当曲线r2(s)与r1(s)在s 0处的Frenet标架重合时,r2曲线Fren

34、et标架的标架运动方程为(1.6)drdst(s)dtdsdnds(s)n (s)(s)t(s)这是一个关于向量值函数r,t, n的常微分方程.曲线r2 (s)的Frenet标架与?(s)的Frenet标架都是微分方程组(1.6)的解.它们在s 0处重合就意味着这两组解在 s 0的初值相等,由解对初值的唯一性定理立即得到2 r1.定理证实完成.习题三(P68)2( 1)r(u,v)a(u v),b(u v),4uv 是什么曲面?2 2笃笃z 马鞍面a bx a(u v) 解：y b(u v)z 4uv4.证实：曲面 F&U)0的切平面过原点. x x证实：无妨假定方程 F0Z)0确定一

35、个z f (x, y)的隐函数,于是x x0fZz)x xf12)fx 0fx-xx1(fy)0fyxxF2F1F2F1 ( 2)F2 (xF1 (-)F2xyFi ZF2设 r(x,y)x,y, f(x,y),那么rx1,0, fx1,0,yR ZF2xF2rxry0,1, fy0,1,FF2ryyFi ZF2yFi ZF2 FixF2F1F2,1xF2F2所以,P(x, y,z)处的切平面为yR ZF2xF2(Xx) F1(Y y)(Zz)易见,当(X,Y,Z)(0,0,0)时,有:yF1 ZF2xF2(0x) Fl(0F 2y)(0Z)yF1F2ZF2 yF1ZF2=0=右所以结论为真.

36、6.证实：曲面S在P点的切平面TpS等于曲面上过P点的曲线在P点的切向量的全体.证实：设曲面S的参数方程为r r(u,v), (u,v)D, P r(U0,V0),(u°,V0)D.令(u(t),v(t)为参数区域 D中过(u0,v0)那么的参数曲线,r(t) r (u(t), v(t)为曲面上过 P点的曲线.于是drdtpru(u°,v°)罟dtp (比当dt这说明曲线r(t) r(u(t),v(t)过P点的切向量drdt都可由与rv(U0,v°)线性表出.可见过drP点的切向量drdtP都在过P点的切平面上.另一方面,对于任意切向量wru(U0,v&

37、#176;)rv(U0,v°) TpS,在参数区域D中取过(u°,v°)且方向为I (,)的参数曲线(u(t), v(t)(U0t,v°t)那么此时,r(t) r(u(t),v(t) r(uo t,v. t)从而p ru(Uo,Vo)rv(Uo,Vo)w.dt这说明：在P点的切平面TpS中每一个向量都是过 P点的某一曲线的位于 P点的切向量.于是：曲面S在P点的切平面TpS等于曲面上过P点的曲线在P点的切向量的全体.25.求双曲抛物面r(u,v) a(u v), b(u v), 4uv的Gauss曲率K,平均曲率H ,主曲率i, 2和它们所对应的主方向E

38、 a2b22 216v , F ab216uv ,G2,2a brurv2 2b(uv), 2a(uv),ab , n2EG F2其中EGF282b2(uv)22a2(uv)2a2b2 o由ruu0 ,ruv(0,0, 4) , 1vv0 LN0, M解： 由 ru (a,b,4v), rv (a, b,4u)于是Gauss曲率K :216u o2b(u v), 2a(u v), ab8abJ' o.EGF 2LN MEG F264a2b2a2bEGF222b2(u v)22a2(u v)2a2b2 2平均曲率H :2 2MF 8ab(a b 16uv)EG F2 (EG F2)3/2

39、2 2ab(a b 16uv)4b2(u v)2 4a2(u v)22a2b2 3/2由于M2H2 K0,所以M2F2 (LNEGM2)( egF22所以主曲率1 :1 H . H2 KF2)M 2EGm、"EGEG F2 2EG F2M(F 、1G)EG F2ab (a2 b2 16uv) , (a2 b2 16u2)(a2 b2 16v2)4b2 (u v)2 4a2 (u v)22a2b2 3/2对应的主方向为du: dv ( 1F M):( 1E L)(1FM): 1E ,其中1F MMF(F .EG) M(EG F2)MF、EG MEG2 2EG FEG Fm/EG(f T

40、EG) egEG F21.所以du: dv同理,另一个主曲率G E.a2 b2 4u2 : . a2 b2 4v2.M(F 、EG)EG F22'ab (a2 b2 16uv)(ab16u2)(ab16v2),4b2(u v)2 4a2(u v)2 2a2b2 "对应的主方向为u: v.E .a2 b2 4u2: , a2 b2 4v2注：设W：TpS TpS为外恩格尔登变换,那么几 aru brvnv cru drvW ru ,W rvrudurvdvduW rudvW rvW ru ,Wdudvdudvrudududva c b ddudvru,rvdudva c b d

41、dudvdu dva cduduacdu0b ddvdvbddv0LGMFMGNFEGF2EGF2du0MELFNEMFdv0EGi F2EGF2(LGMF)(EGF2)MGNFdu0MELFNEMF(EGF2)dv0(LE)GF(MF)MGNFdu0MELF(NG)EF(MF)dv0rvdvrudurvdvGF LEMFdu0F EMF NGdv01GFLEMF du0EG F2FEMFNG dv0E F1LEMFdu0F GMFNGdv0E LFMdu0F MGNdv0du: dvFME LG N:F M.补充：定理1函数是主曲率的充要条件-日是ELF M0.FMG N2方向d - d u

42、:d v是主方向的充要条件是EduFdvLduMdv0 (WW).FduGdvMduNdv证实：1设du :dv是对应的主方向,那么有W drdr ,即nudu nvdvrudu rudv .分别用ru,rv与上式两边作内积,得Ldu MdvEdu Fdv , Mdu NdvFdu Gdv .所以主方向du : dv满足(E L)du ( F M)dv 0,(F M)du ( G N)dv 0.由于du, dv不全为零,可得E L F M0 F M G N2 在脐点,K H20, i 2 H. 从而由II HI可知L HE,M HF,N HG,WW中的两个方程成为恒等式.此时,任何方向都是主 方向.(E L)du ( F M)dv 0,(F M)du ( G N)dv 0.得到相应的主方向du : dv ( 1FM):( 1E L)(1GN):( 1FM )和u: v ( 2FM):( 2E L)(2GN):( 2