Đánh giá lớp phép biến hình á bảo giác lên hình vành khăn bị cắt theo các cung tròn đối xứng quay 4

32

CHUaNG 3

cAc nANH GIA CHOLOP HAM G
Trang chu'dng nay, chung Wi se tie'n hanh danh gia cac d(;liIu'Qngd~c tru'ng
cho mi~n chu§'n A va modun cua cac ham g EG. E>~thie't l~p cac danh gia cho
lOp ham G, chung ta c~n d1!avao cac danh gia cac d(;liIu'Qnghinh hQc cu~ lOp
ham

f

E

F,

voi

dii

f=g-l,gEG

M'(oo,f)=m*(oo,gfX=1,

g=f-l,

neu

a

ph~n

1.2

voi

chu

Y

fEF.

3.1 Danh ghi M* (0, g)

Dinh ly 3.1: Du'oi cac ky hi~u va gia thie't dii neu a ph~n 1.2, Vg EG ta co:
M*

(0, g)

M' (O,g);'

E>~ngthuc

2::

(3.1)

1,

2-';'

(3.2)

(~r

a (3.1) xay ra khi va chi khi B = Bo voi Bo la m~t ph~ng phuc ma rQng

w bi celt dQc p cling troll d6ng Him t(;li 0 sao cho Bo bie'n thanh chinh no bai

phep quay ;=/~w

va g(w)=awlwr-1 voi lal=1.

Chung minh:
Xet PBHKABG f

E

F mi~n A len mi~n B, theo (2.24)ta co:
P\ :rRK

D1!avao (1.11) va (2.20) ta du'Qc
2

~Sl

:rRK(g)

M* (0, g fK , g

E

G,

trang do d~ng thuc xay ra khi va chi khi w=f(z)=g-l(z)=bzlzr~-l
B = f (A) = Bo.

(3.3)

voi Ibl=1, tuc


33

Tif (3.3), ta co:
*

M

-1:

(O,g)

PSI

~1-~,gEG,
7rRK

hay
M*(O,g}~ ~

1
~1,gEG.
1- PSI2
TrRK

Nhu' v~y ta co (3.1) voi tru'ong hQp d~ng thue xciy ra khi va em khi (3.3) xciy ra
d~ng thue, tue Ia B = Bo va z = g( w) = I-I (w) = awlwr-I voi lal= 1.
M~t khae, dl!a vao eong thue (2.32), ta co:
'

m (0,I ) ~

c(R,/)
4

-1-

1-'

I

E

F

P RK

2

=>m'(0,/) ~ 4"c(R,/)
d(R,/) ,/EF,
ke"t hQp voi (2.20), ta suy ra
M* (O,grX ~4* ~,gEG.
d

V~y ta co (3.2).
3.2 Danh gia Ig(w)1
Dinh Iy 3.2: Du'oi cae ky hi~u va gicithie"t(j phffn 1.2, Vg E G, WEB ta co:
4-~lwIK~lg(w)I~4~ M*(O,g)lwIK.
Chung minh:
Theo(2.31),

V/EF,zEA,taeo:
4-~ m'(O,f)lzIX

Thay z=g(w) va f(z)=w

~

I/(z)1 ~ 4~ Izlx.

, ta du'Qe

1-

4-p m'(O,/)lg(

1

1-

I

W)IK ~ Iwl ~ 4p Ig( W)IK,

(3.4)


34

tlido
....

Ig(w)l~

va Ig(w)I~4-~lwIK,

4pIWlK K

m'(O,j)

ke't hK
K

K

<

(

I

4-plwl

4p

~ g w

K

Iw
I

)1
- M*(O,g

-1'

)

suy ra (3.4)8
3.3 Danh gia ban klnh R(g)

Djnh ly 3.3: Du'oi cae ky hi~u va gia thie't d phftn 1.2, Vg E G ta co cae danh gia:
K

K

4-PdK ~R(g)~4P

(3.5)

M*(O,g)CK,
K

R(g»

PSI
[ ~(1-M*(0,g)f)

-2

]

voi SI >0.

Chung minh:

Tli (2.32), chung minh tu'dng tv' (3.4) ta co
Rt ~

~(R,j)

va Rt ~4-;d(R,j),j

E

4-" m'(O,j)
hay
K

K
4pc

K

K,jEF.

4-PdK~R~

m'(O,j)
Ke't h
Tli cong thuc (3.3) ta d~ dang suy ra du'R-i (g)~~(l-M*
PSI

tuc ta co (3.6)8

SI > 0,

(O,gri), Vg E G,

F,

(3.6)

35

Danh gia (3.1) co th€ lam cho s~c han nho
H~ qua 3.1:
D~t

E=

PSI c!(;~~O),ta co:

7rC22 p

M*(O,g)~(l+E)t

,VgEG.

(3.7)

D~ng thuc xiiy ra khi va chI khi B=Bo va g(w)=awlwIK-1voi lal=1.
Chung minh:
Ke't h<;5p(3.3) va (3.5) ta suy ra
PSI
7r4; C2M* (O,g)t

~ 1- M* (O,g)-t

hay
M* (O,g)t ~ 1+

P~I

7r4P C2

.

Tli do ta co (3.7).
D~ng thuc (3.7) xiiy ra khi va chI khi d~ng thuc (3.3) xiiy ra hIe do SI= 0 keD
theo E=O, tuc 1a B=Bo va z=g(W)=f-I(W)=awlwIK-I voi lal=1..
H~ qua 3.2:
Trang tru'ong h<;5pK = 1, M*(O,g)=Ig'(O)1nen (3.7) trd thanh
Ig' (0)1 ~ ../1+ E, Vg E G .

(3.8)

D~ng thuc xiiy ra khi va chI khi B = Bo va g (w)= aw voi lal= 1, ba'"td~ng thuc
nay s~c han ba'"td~ng thuc c6 di€n Ig'(O)1~ 1,Vg E G voi K = 1 (xem [10], IT.350).
3.4 Dauh gia g6c md 2~(g)
Nhu'ta dff bie't 0 < ~(g) < 7r,Vg E G. Bay gio, ta tim cae danh gia co th€ s~c
P
han cho ~(g) trong mQt s6tru'ong h<;5pnaG do.


36

3.4.1 C:}n dum cua r3(g): (Dung phuong phap dQ dai qie tri)
Dinh Iy 3.4: Voi cae ky hi~u va gia thie't trong ml,le 1.2, gia sa c < d, khi do
VgEGtaeo:
n

f3(g)

1£

ITK21n4' dM* (O,g)t
c
'2 IT _
d
p
2p f- dr
c rO(r)

'2 IT -

P

nO.oK2In 4 P dM* (O,g)t

2pln~ c
c

~/

.

(3.9)

Chung minh:

~

A

..R......
"""'"
""""::::::::"':§~::"':::::~

'/

(T
2

O
.

:

\

'"

0

..(""""""""""""""""",~

\..

o

:

(

(Tj

"'~

>""~:::::::::4.::::"""

:

/t

"""'"

:

(.. "

\..::::':.:::::::=::::<::/

0

""""":~
..., '. L I

~:::::::::::::::::./

\

z

j

)
i

,)

Hinh 3.1

Ap dl,lngb6 d~ 2.10 vao bai loan dang xet voi

Bo

la tu giae eong co hai

e~nh n~m lIen hai duang tron Iwl= c va Iwl= d; hai eanh con l~i la cae eung cua

(TIva (T2va dQ do p(z) = 1~I'zE Ao, Ao=g(Bo), ta co:
Ip

(

Cr = fp(z) Idz 1=f~,
) - zI
1

co'

voi Cr={zllzl=r}nBo,c~r~d,Cr

c,

=g(CJ.

Iz I

Ir I

= dr + irdqJ I

r

'2

1

+ ire'
1

I dz I = I e'
1

B<)tz=re'irdqJ I =1 dqJ I.

r


37

(Ba't d~ng thuc tren co duQc VIc<;lnhhuy€n cua tam ghic vuong khong nho hon
c<;lnhgoc vuong).
VI v~y, ta co:

lp(c,);, c,fld~;'2a~t
M~t

khac,

do

tinh

d6i

xung

H={wlcm(c,g)~lzl~M(d,g),gEG,

ta

-P).

quay

(1.4)

A'

tha'y

va

n~m

trong

d~t

B' =BnH

hlnh

vanh

taco:

S p ( Ao ) = Jf p2
Au

( Z ) dxdy =! H p2 ( Z ) dxdy
P A'

=! Ifdxt =! If ~dY2 =! Ifrdr~qJ

P A' Izl
P A' X + y
P A' r
~!

P

2] dqJ1dr
0

m

r

= 21l:1rlM(d,g).
P

m(c,g)

Tli do theo b6 d€ 2.10 ta suy ra
21l:In M(d,g) ~.l(2a)2
P
m(c,g)
K

Jc rQ(r)
dr ~.l(2a)2 ~ Jdr ,
K
Qo c r

tuc
M(d,g)
a~

~Kln

1l:Kln m(c,g) ~
d

dr

M(d,g)
m(c,g)
2pIn d
c

2p cf rQ(r)

Ngoai fa, theo (3.4) k€t hQp voi (1.8) va (1.9), ta co:
-K

K

m(c,g)=4PcK va M(d,g)=4PM*(O,g)dK,gEG.

voi
khan


38

Suy fa
2K

2K

n-Kln 4" M*(O,g)dK
~Kln
I
cK
d
<

a:::::

2p

f rO(r)
dr

4" M*(O,g)dK
cK
2pln-

-

c

d
c

VI f3= TC-a ta c6 (3.9) .
p
Nhan xet:
Ne'u c=const, d=const va cho Do~O ma M*(O,g):::::M~=constthl a~O
TC

~

We f3 ~ - .
p

Vi dV 3*1:

--

B

~

/"""'

\

:;

I

-

=k (z )

/

;..

/.

\\

(:_~) red

K-l

~A

2J

CS
r

~

z=zz '

;=h(w)

\R

------

(

\

""""""""\

~

y/

t"");
:

-,)

/""""""""'~"""

((

Ii

~:::

gEG

~

'......

(-';

\

:.::::~

)\
R

j

Hinh 3.2
Gia sa Bo=Bn{wlr
e6d<;lngnhuhinh3.2(p=2),c6dinh

eho Do~ 0 ta chung rninh M* (0, g), g E G kh6ng dffn de'n
M* (O,g):::::Mo =const.

00

c, d va

khi Do ~ 0 tuc la


39

GQi

; = h (w)

la PBHBG don di<%pmi€n BIen mi€n A la m~t ph~ng md

rQng bi ca:t dQc p cung troll tam 0 thai h (0) = 0, h (00 ) = 00 va khai tri€n Laurent

cua h(w) trong Ian c~n w= 00 c6 dc,lllg
a1 az

(3.9a)

h( w) = w+ao +-+2+'"
w w

tuc la anh cua duong trOll Iwl = R voi R ra't IOnbdi h g~n trung voi duong troll
1;1

= R . N6i cach khac
m* (oo,h) = Ih' (00)1= Hm Ih( w)lw-+ooIwl - 1.

Theo Thao[ll,

va do d6 mi€n

tr. 109], ham h(w) Ia PBHBG don di<%pmi€n B !en mi€n

A

c6 tfnh d6i xung quay ca'p p. GQi z =k(;)

PBHKABG mi€n A len mi€n A trong d6 m6i duong troll
duong troll

Izi

1;1

=R

A

=;VIK-lla

duQc bie'n thanh

= RK . VI argz = arg; nen k(;) cling c6 tfnh d6i xung quay ca'p p.

Khi d6 z = g (w)= ko(h (w)) la PBHKABG mi€n BIen mi€n chufin A c6 tinh d6i
/

,.('

xl1ngquaycap

-

p. Honnua

*

. M(R,g)

I1m
m (oo,g) = R-+oo
K
R

. RK
= IR-+oo
1m~=.
R

"

1 V ~y gE G .

GQi C, la anh cua C, = {wllwl= r} voi r ra't be bdi h va C~ Ia anh cua C,
bdi k;

Zl EC~

sao cho Izll=M(r,g),

;1 EC, sao cho k(~)=z,

va w,EC, sao cho

h(w,)=~. Ta c6:
M. (0 , g) = lim
M (r , g) = lim

1:J = lim
k (~ )1= lim ~ K
,-+0 rK
,-+0rK
,-+0 rK
,-+0 rK
I

= lim Ih(WI)IK= lim h(Wl)

K
I

H-+o

Iw,IK

H-+o

W,

I

I

= h'(O) K -:I:-0 (VI h Ia PBHBG)
l

I

40

Mi;it khac, nSu r~O

thl ta colh(w)I~lh'(O)llwl=lh'(O)1r

tuc Cr g~n trung

; =Ih'(O)fr.

du'ong tron 1;1=;, voi

NSu n6i hai cung cua nhat cfit trong mi€n anh

A d€ du'Qc du'ong troll

1;1

= R)

thl trong mi€n B cling se co hai cung n6i tu'dng ling. Nhu' da: neu tren, anh cac
du'ong troll Iwl= R,
du'ong troll

Iwl

= R va

1;1

= r voi R d't IOnva r ra't be bdi h g~n trling voi cac~

1;1

=Ih'(O)lr.

Khi cho Qo ~ 0 do tinh ba't biSn cua modun hai mi€n nh! lien qua PBRBG

; = h (w) ta co

R

R

R)

R) ~

d'

Ih' (0)1 r

~ .:.
r

A,.,!'
,.,!' /
,
,.,!'

/
' V <;lyneu R rat 1on va r rat b e th 1 R) ~ d , Ih '0( )1~ -R)
c

R~

00

ta co

I

~

'
d
-.c Ch0 r ~ 0 va

h ' (0)1= dc .

T6m I~i, ta c6 M' (O,g)=(

~r

<00.

V~y danh gia (3.9) la co y nghia va tic$mc~n dung trong tru'ong hQpnay.
3.4.2 C~n tren cua ~(g):
D~u lien, ta chia mi€n B lam p ph~n b~ng nhau b~ng p du'ong cong
JordanYj(J=1,2,...,p)

n6i 0 va 00, du'ong nQ chuy€n thanh du'ong Ida bdi phep

quay mOt goc 2nj. Cac du'ong cong Yj nay chia mi€n B thanh p mi€n nh! lien
p

B~,(J = 1,2,...,p) voi bien trong la mOtthanh ph~n bien G"jcua B.
Ki hic$uC(a,r) chi du'ong trOlltam t<;lidi€m a va ban kinh la r.
Tren

B]

( baa dong cua mi€n

B]

= B; ) ta co th€ ve -them hai du'ong trOll phg:

Du'ong troll thu nha't la C (W)'1)) gioi h<;ln mOt hlnh troll dong chua thanh ph~n


41

bien (}j; Duong troll thu hai la C (w2'r2) chua trong Bl va baa bQc C (WI'1j). GQi

B2 la mi€n nhi lien gioi h~n bdi C(w],1j) va C(W2'r2).
z=g(w)

~
B

c:::::>
Q
/

.

~

.. ..

.;..
..f

A

.

~

/.

I::i'l

...\

\

w

///

AI

~~
/

.'

/>3
fI\;\ B, )) B,
0 ., \jJ)
~~

.........

--./

A,

.

..~

z

Hinh 3.3: PBHKABG z = g( w)bie'n mi€n A leD mi€n A voi p = 4
Thea h~ qua 2.3, ta co:
mod(B2) ~ mod(B]).
D<)t

R.

= min{lwllw

E C(w2,r2)}'

~ = max{lwllwE C(w2,r2)}'
RI=lw21-r2

tuc

(O<)~

(3.10)

=lw21+r2

Sau do ta tie'p t\lCve hai duong troll C(0, RI) va C(0, ~), tuc C(w2'r2)n~m trong

phftngiaa cua

BJ

va hlnh vanh khan B3={wiR, < Iwl< ~} .

Ta tinh tie'n va quay mi€n

B2

r6i ap d\lng b6 d€ 2.11, mi€n B2 co th€ bie'n baa

giac don di~p len hlnh vanh khan B4 = {slr < Isl< I} saG rho C (W2'r2) tuong ung
voi Isl= 1,


42

2

,

va r = r(lj, r2,h) =

h 2 +lj

r2 -

2

-\I

I 2 h2 2 2 4h 2 2
(r2 -lj ) lj

I

2ljr2

f)~t Al =g(B1XcA),

A2 =g(B2XcA]),

~.

, VOl h = W2 -

W]

I

.

(3.11)

gEG.

M~t khac t6n t(,li PBHBG don di<%p; =; (z) mi~n

A2

ten hinh vanh khan

Bs ={;lr'Vi phep bi€n hinh hQp ;ogos-]mi~n A2 ten Bs la mQt PBHKABG nen ta co:
r'<_r 7c.

Thea tinh don di<%u
(1.17) cua ham phv T(p,r,s), ta co:
T(2,r',0)~T(2,r7c,0) vdi r xacdinhnhu'(3.11).
GQi D la du'ong kinh cua mQt nhat cat cling trOll Li' D' la du'ong kinh cua anh

du'ong troll C (WI'r]) bdi z = g (w), g E G tilc du'ong kinh bien trong cua A2. R6
rang ta co D ~ D' .
f)~t m=m(Rpg)

, M=M(R2,g),

gEG.

Thea (3.4), ta co:
m 2::4 -:

R] K

=m

, M ~ 4 ~M* (0, g)

R; = M .

Thea b6 d~ 2.6, ta co
- N€u p = 1 , d€ co quail h<%D = 2Rsin ~ c~n thi€t cha vi<%c
tim c~n tren
cua P(g) ta c~n them gia thi€t 2p ~ n vi n€u 2p > nthi D = 2R.
Gia thi€t nay du'Qcthai n€u

(D ~ D)

= 2T( 2,rt,0)M

(R2, g) < 2.41 r7c4~ M* (O,g )R; < 2.4-~ dK « 2R),

tilc
1>:1-1.

4P+'rKM

*

-1>:

(O,g)R;

<4

pdK.

Luc nay ta mdi co th€ ap dvng du'Qcquail h<%D = 2R sin ~ .

(3.12)


43

- Ne'u p ~ 2 thi du'dng nhien 213s 2n S 1[do do ta luau co D = 2R sin p .
p

/

?

,

Ap dlJng ba de 2.6, ta du'Qc:

D's 2MF( 2,r-K,0).
Mi,Hkhac, ta co:
DsD'

va

M=M(Rz,g)sM.

Suy fa
DSD's2T(2,r-K,0)M.
Tildo

Ds2T(2,r-K,0)4~ M*(O,g)R~
s2.41r-K4~M*(O,g)R~ =4~+lr-KM*(0,g)R~.

M~t khac ta co D = 2R sin p
Suy fa
sin 13 = D < 4~+lr1-M* (O,g)RK
2R 2R
2

0

-K

Vi R chu'a bie't, ta thay R b~ng c~n du'oi, nghla la R ~ E = 4p dK .

Suy fa
0

4~+1r1-M*

sIn 13 <
-

(0' g ) RK2<

4~+1r1-M*
-

2R

(0 g ) RK2

K '

2A-1i dK

V~y
2.42: r1-M* (O,g)RK
f3(g)sarcsin

(

dK

2

J

=f31(g),

.


44

2K

ydi di€uki~n

2.4pr1-M*(O,g)R;
dK

::;1

0

M~t khac, ap dl:mgb6 d€ 2.8, ta co:

(

IS In 1-

D'::;

(2

-7r

)

(

ydi (= T 1,r t ,0) ,

7r

[

S::;SI = P M

2

(

2

)

R2,g -m

(

7r

Rl'g

K

*

K

)] ::;P [ 4p M (O,g ) R2 -4

_K

K

PRJ

V~y
D ::;D '::; ,I SIn (1- (2)
-7r

<=>

.

2Rsinp ~ ~SIn~~t2)

Sur ra
SIn (1-

(2

)

-7r

13 (g) ::;arcsin

2R

K

Thay R = R = 4-P dK ta duQc

fJ(g)::; arcsin

'0

O;:;

YOI d leu

'
lln~~t')
ki <:fn
K I

4-P+2dK

::;1.

~s In (

J-t'

I

_K~7r

4

p+2

dK

)

1= 132 (g),

-

]=S.

(3.13)


45

Nhu' vi;tyta da tlm du'Qcci;tntren cua fJ(g) du'oi d~ng:
Dinh ly 3.5:
Du'oi cac ki hi~u va gia thie't trong m\,lc1.2, va nhu'moi lieU (j tren, Vg
Ne'u p

= 1 va

E

G ta co

thoa (3.12) hoi;lc p'2 2 thl fJ(g):::;mill {fJI(g), fJ2(g)},

trong d6

~ (g) = arcsin

[

2.41f r+M* (0, g) R:
dK

S In (1fJ2(g) = arcsin

(2

J

(3.14)

'

)

-7(

(3.15)
4-K+1

dK

voi r,~ va S xac dinh nhu' d (3.10), (3.11) va (3.13).
Vi d1} 3.2:

~

gEG

A

(

~

~
Y4

w

z

Hinh 3.4

46

Gia sa mi€n B c6 p

= 4 thanh

ph~n bien O"j' j = 1"",4 la cac du'ong troll

c (aJ'&) voi aj = euta; & du'dng, du be, Trang d6 thanh ph~n bien 0"1la du'ong
troll C(ap&)voi

al =a,&=fj

> 0 (ffinh 3.4),

Ta ve du'ong cong Jordan rl la du'ong phan giac cua g6c ph~n tu' thil nha't.
Sau d6 dung phep quay ta xac dinh du'Qcr2' r3 va r4 la 3 du'ong phan giac cua 3
g6c ph~n tu'con I~i, Cac du'ong phan giac nay chia B thanh 4 ph~n b~ng nhau,
Ta ve them du'ong troll C (a, r2)saD cho ban kinh r2 Ia khoang cach tu a d€n
du'ong phan giac cua g6c ph~n tu' thil nha't. Khi d6 mi€n B2 chinh la mi€n nhi
lien gioi h~n bdi hai du'ong troll C (a, &) va C (a, r2) ,
Mi€n

c6 th€ bi€n baa giac ddn di~p leu hinh vanh khan r < Isl< 1, Theo h~

B2

,;:,
1 r2
qua. 2 2 , tIllh b at blen cua mo dun mIen n h ~ Ilen, ta co - = -,

?

/

"'"

' "'"

?

'

~

' ~

/

r

&

M~t khac , ta c6:
ff

a

r2= a sin 4" = 12'
r = &12
a

V~y
Ap dvng (3,14), ta c6:

2.4':'2' EKM' (O,gJ( a+ ;JK

PI(g) = arcsin
I

Cho a c6 dinh kha lOn, &~ 0 , M* (0, g)

aK

t

(a + &

s const

thi PI (g ) ~ 0,

/

tilc

&

r = -,

r2

47

Tu'ongtv , ap dvng (3,15)
Sln(1-t2)

)

/32

voi t = +(
va

RJ

t

I

(g = arcsm

4K+!

<:)',0). s = : [M2 (R"g)-rn'
a

-Jr

V

,

I'

(a + &

(R"g)] ~: [4' M' (O,g)R;' -4-' R,'

r

a

=a- .[2,R2 =a+ .[2 .

Theo (1.24), ta co:
I

I

&.[2 K

&.[2 K

t T [ {--;;~

V~ykhicho

a c6dinhkhalOn,

J ' 0) ~ {--;;&~O,

J

.
->

0

kh1&

M*(O,g)~const

->

0.

thlln(1-t2)~O,tuc

ta

cling co /32(g) ~ o.
Nhu' v~y trong tru'ong h<;1pnay cae cong thuc (3.14), (3.15) Ia khong hi€n nhien

va ti~m c~n dung.