Đánh giá các phép biến hình á bảo giác những miền nội tiếp trong hình vành khăn 4

ChU'O11g. DINH NGIDA PREP BIEN HINH K-A BAo
2

GIAC vA MOT s6 CONG CD
.
.

.

2.1 Mo dun cua mi~n nhi lien
B6 d~ 2.1.(T. Carleman [4, tr.212])
,
Gia sir w=fl(z) la mot PBHBG dan diep hinh vanh khan ~ = K ~

.

.

b'

b

,

len mot mien
.

nhi lien G khong chua di~m 0Ci sao cho bien trong c cua G tuang irng v6i duemg
trim Izl= r. GQi s la di~n tich (ngoai) cua t~p dong do c baa bQc, S la di~n dch
(trong) cua t~p ma do C baa bQc. Khi do ta co quail h~ sau:

s~(:

(2.1)
)'s,

trong do ding thuc xay ra khi va chi khi f(z)= az + b v6i a, b la cac hfuIgs6 va
a:;tO.

Tir b& dS ta suy ra cac h~ qua sau:

H~ qua 2.1. (Binb ngbia mo dun mi~n nbi lien)
N@umiSn nhi lien G qua cac PBHBG dan di~pf va fl IAnluQ'tthanh hai hinh
VaMkhan: H: r thi:

R=RI
f

(2.2)

fl

Dlililuang R duQ'cdinh nghla la ma dun cua miSn nhi lien noi tren kYhi~u la
r
m(G)= R.
r

7

Ron mra nSu Gila anh cua G b6i mQtPBHBG don di~p thi m (G1)= m (G), do
la tinh b~t biSn cua ma dun mien nhi lien.

H~ qua 2.2(tinb don di~u ctia mo dun mi~n nbi lien)
NSu cac mien nhi lien G va G' v6i cac ma dun lftn luot R va R' co tinh ch~t
.
. r
r'
G c G' va G ngfm cach hai thanh phftn bien cua G'thi R::;;R' , D~ng thuc xay
r r'
ra khi va chi khi: G == G' .

2.2 Binb ngbia pbep bi~n binb K-a baGgiac
Binb ngbia 2.1
Phep biSn hinh mQt mQtw = f (z) tir mien A trong m~t ph~g z len mien B trong
m~t ph~ng w'lien t\lC hai chieu va baa toan chieu duong tren bien duQ'c g9i la

PBHKABG nSu m9i ill giac cong Q c A co madun la m(Q) thi ill giac cong

Q'=f(Q) comadunla m(Q') se th6a b~t d~ngthuc:
~m(Q)
K

::;;

m(Q')

::;;

Km(Q) (K ~ 1)

(2.3)

Sail day la mQt s6 tinh ch~t cua phep biSn hinh K-a baa giac:
1. K= 1 thi phep biSn hinh tr6 thanh baa giac
2. RQ'Pcua hai phep biSnhinh K}-a baa giac va phep biSnhinh K2-a baa
giac la mQt phep biSn hinh K}.K2-a baa giac.
3. Phep biSn hinh nguQ'c cua phep biSn hinh K-a baa giac la phep biSn hinh
K-a baa giac.

8


4. NSu Q la hinh chfrnh~tABCDco cac c~

la a, b, Q' la hinh chfrnh~t

A'B'C'D' co cac c~nh a', b'song song v6i cac tf\lCto~ dQ. Gia sir [ la
PBHKABG mi~n Q leu Q' saDcho b6n dinh tuang Ungv6i Mall. Khi do:
,

:' = K: khi va chikhi [co d~g [(x + iy)= a(Kx + iy)+ p,
a'
b'

-

1
=-- a

,

'

.

'

,

1

'

~

'

kh1va ch 1 kh1 [ co dan g f( x + lY) =a (- x + lY + 1-',
) A
Kb
.
K

v6i cac h~ng s6

a > 0, f3 phuc.

Sau day la mQt s6 b~t d~ng thuc ma rQng cho PBHKABG.

B8 d~2.2
Gia sir G la mi~n nhi lien v6i modun la m(G) duQ'cbiSn K-a baa giac leu mi~n
G' v6i modun m(G'). Khi do:
1

K

(2.4)

[m(G)JK ~m(G')~[ m(G)J .

Trang truang hQ'P G = {z,q < Izi< I} va G' = {w,q' < Iwl< I} thi (2.4) tra thanh:
1
qK

~ q' ~ qK .

(2.4a)
1

Ran nfra q' = q K

<=>

[( z) = a IzIK-l , la! = 1, va q = q K
z

1
<=>

f ( z) = a IzIK-l,lal = 1.

B8 d~ 2.3 (Soy rQng bAtding thue Carleman eho PBHKABG)
Gia sir [la PBHKABG hinh VaM khan (0 <)r khong chua diSm

00

v6i bien trong c, bien ngoai C saD cho z 1= R tuang Ung
1

9


v6i bien ngoai C. GQi S la di~n tich (trong) cua t~p ma do bien ngoai C bao
bQc,s la di~n tich (ngoai) cua t~p dong do c bao bQc.Khi do:
(2Ab)

s~(~)% s,
trong do dfuIg thuc xiiy ra khi va chi khi:

.,

~~

w=f(z)=alzIK

z+b v6i cachang so a:;t:O vab.

ChUng minh:(Xem Thao [14, tr.55-56] ho~c Luong [11, tr23]).

B&d~ 2.4 (Bit ding thu-c di~n tich cho mi~n da lien)

Giii sir A la hinh vanh khan (0 <) r
Lj(j = O,I-,...,p n~m tren dUOngtrOlldang tam z 1= Rl (r < Rl < R). GQiflit
-1)
I

PBHKABG miSn A len miSn B cua m~t phfuIg w sao cho duOng trOll z 1= R
I

tuong Ung v6i bien ngoai C cua B, cac nhat c~t Lj biSn thanh cac bien
o)j = O,I,...,p-1). Khi do ta co:

S(R, f) > S(r, f)

~

( r J

+

Yrc

i> ( ~
R
.
J=1

J

'

~
1

J

'

(2.5)

trong do S(R,f) 1ftdi~n tich (trong) cua t~p ma do bien ngoai bao bQc, S(r,f) la
di~n tich (ngofti) cua t~p dong do bien trong bao bQc, Sj la di~n tich (ngoai) cua
t~p dong do cac IT bao bQc.
j

.

Bang thuc xiiy ra khi va chi khi fez)

1

--1.

,

= a z IK +b vai a, b la cac hang so,
I

a:;t:O.

10


ChUng minh: Xem Thao[14, tr.56] ho~c Thao[13, tr.522 ].

2.3 Cae ham sa pht} T(p,r,s) va R(p,t,s)
Gia su hiOO vaOO khan r s Pj={wl s~lwl~t,argw=j21t},(j=O,1,...,p-l)(xem
p

hiOO 2.1), do tiOO don

di~u cua ma dun miSn OOilien (h~ qua 2.2) ta co 0 ~ s ~ r < t < 1. Han nua t la
d~i luang xac diOOduy oofit theo r va s, r la m<)td~i lugng xac diOOduy oofit

theo t va s.
BG

Or

1

1

z

HiOO2.1 (v6i P =2)

Tir do ta diOOnghia cae ham s6:

t=T(p,r,s)

v6ip= 1,2,...; O~s
r=R(p,t,s)

v6ip= 1,2,...; O~s
Cling do tinh don di~u cua madun miSn nhi lien ta co cae tiOOchfit cua cae ham
T(p,r,s) va R(p, t,s) OOusau:
1. r
(2.6)


2. T(p,r,sl) > T(p,r,sz) (0 ~ Sl < Sz< r < 1)

(2.7)

3. T(p,rps)
(2.8)

4. s
(2.9)

(O~s

5. R(p, tpS) < R(p, tz,s) (0:5;t1 < tz < 1)

(2.10)

6. R(P,t,SI) < R(p,t,sz) (O~SI
(2.11)

7. R(p, t,s) > R(1,t,s) (0 ~ s < t < 1,p ~ 2)

(2.12)

Nha cae cong thuc trong Nehari [10, tr.280-295], Thao[12, tr.100-107 ] va
Luang [11,tr.15-18] dii chi ra cong thuc cua ham R va T OOusau:
R(P,t,s)=exp
,

-JrKf(U»

(

)

2pK(u)

h = (l-k)(l-ak)

, voi u=1+h-.Jh(2+h),

1+s4pj

fI[ 1+
00

, k = 4sp

k( 1+ a)

j=l

i2pb

trongd6:

"

S4PJ-Zp

]

t

a = sn ( b + ;-In;,k

) , b = K(k),

6 day sn(z,k) chi sin eliptic v6i tham sa k.
.!.

T (p,r,O ) =4Pr

00

1+ r4pj

fI [ 1 + r PJ P]
j=l
4 '-2

.

T(p,r,s) = sexp (

a

~

(O
dx

2;:;k)! x')( 1- k'x') J ,
~(1-

0 < s < r < 1,pEN, voi K(k) 0011'
tren

12

,


2

a= I-m m= k(l-h ) h=4rP
k +m '
2h (1- k)'

00

l+r4pj

U[ 1+ R 4pj-2p .
]

Tir d6 sur ra cae t£OO
eh~t khae eua ham R(p,t,s), T(p,r,s):
-1

8. R(p,t,s)~4Ptkhit~0,

(2.13)
1{2

9. l-R(p,t,s)~

2pln

8

khi t~l

(2.14)

p(1- t)

-1
10.4Pt
O:5;s
\:fp,

(2.15)

1

11. T(p,r,0)~4Pr,khi

12. 1- T(p,r,s)

~

(2.16)

r~O

8
-exp

P

1{2

( 2p(l- r) J

khi r ~ 1,

(2.17)

1

(2.18)

13. r
2.4 Cac b6 d~khac:
B6 d~ 2.5 (BAtding thtfc Grotzschl)
Gia sir trong m~t ph~ng z eho truae mien Eo eho truae, giai h~ b6i Izl= 1,
Izl=q,O
va p(l:5;p
q < Izi< 1. Gia sir ham w

=f(z)

trong

bi~n baa giac dan di~p mien Eo len mien Bo

ehua trong mQt hiOOvaOOkhan q' < Iwl< 1 sao eho Izi= 1 ehuy~n thanh Iwl 1,
=
Izi= q ehuy~n thanh Iwl= q', crj thanh crj'(j = 1,2,...,n). N~u t~t ea crj' la OOung

13


nhat c~t theo ban kinh thi ta viSt fo,qo thay vi f, q', trai l~i nSu t~t ca crj' la
nhUng nhat c~t theo cac cung iron d6ng tam t~i 0 ta viSt fl,Qo thay vi f, q' .
GrOtzsch dff chi ra quan h~ sail:
qo < q '< Q 0,
trong do q' = qo <=> f

= fo va

q'=Qo

"

<=>

f

= fl

(2.19a)

.

ChUng minh: xem [6, tr.372].

=
Bay gia nSu miSn Eo noi tren duQ'c ham w

baa giac len miSn

BI chua trong q" < Iwl< 1 sao cho Izl= 1 chuySn thanh Iwl= I, Izi= q chuySn
thanh Iwl= q:'. Khi do
I

(2.19b)

q~ ~q"
trong

do

q" = q~

<=>


=(

I
q"

= Q K <=>

v6i

(

=alulK-Iu, lal= I,

va

I

= (

ChUng minh:Xem Thao[14,tr.58] ho~c Luang[ll,

lal = 1.

tr.29].

BB d~ 2.6 (Mo- rQng bAt ding thuc Grotzsch2)
Gia sir D la hinh vanh khan R
nhat c~t n~m

tren cac duang d6ng tam 0 sao cho D trimg v6i chinh no b6i phep quay
.21t
1-,

'"

Z =e P z, f la phep bien hinh K-a baa giac mien D len mien El nam trong
0 1sao cho duang iron

1

z 1= R tuang lIng v6i bien trong CI, sao cho t~p

dong gi6i h~ b6i CIchua g6c tQa dQ, duang iron

14

I

z 1= 1 tuang lIng v6i bien


.27t
1-

ngoai C2. Han 111la sir El trimg v6'i chinh no qua phep quay W = e
giil

PW

. Khi

do ta co danh gia dung
(2.20)

M} ::;T(p,RX ,m}),
v6'i M} =max{lwl,WEC}},

m} =min{lwl,wEC}}(~O),va

T(p,r,s) 1ahamph\l

du<;ycdinh nghTa trong 2.3.

.

~

Dang thuc xily ra a (2.20)

<=>

fez)

= fo(z) = ah(t),

1

a 1= 1, t = biz IK z,1b 1=1, h 1a

1

phep bi@nhinh bilo giac hinh vanh khan RK }

Itl= 1 tuang' Ung v6'i Iwl= 1 va t 1= R K tuang Ung v6'i c trong do c du<;yc
dinh
1

nghTanhu sau:
w M
c = {w ,1w 1=m}} u {w,m} ::;1 I::; pargw = j 21t},j = O, ,p -1.

P

ChUng minh: Xem Thao[14,tr.63], ho~c LuO'ng[19,tr.33].

DBd~ 2.7 (M6' rQng bit ding thuc Grotzsch2)
Giil sir Dl 1a hinh vanh khan Q< Izi n&mtren cac duang trOll d6ng tam 0 sao cho Dl trUng v6'i chinh no qua phep
.2J1:

quay Z = e 1--; , f 1aPBHKABG miSn D} 1en miSn Ez n&mtrong 0 < Iwl< 00 sao
z

cho duang iron Izi=Q thanh bien trong C}bao gBc tQa dQ, duang trOll Izi= R
thanh bien ngoai C2, Han nua E2 trimg v6'i chinh no qua phep quay
.2J1:

1-

W

=e

P

w . Khi do ta co danh gia dUng

15


rnt

rn>
2

-

(2.21)

'

t

Q K rnt

T[ P.(R) 'M, )
rnj

=min{lwllw E Cj},j

= 1,2 va M2 = rnax{lwllw E Cz}' v6i T(p,r,s) la ham ph\!

duQ'cdjnh nghi'a trong 2.3.
t

D~ng thuc Kay ra

~f(z)=fo(z)=aH(t),lal=l,t=blzIK-tz,lbl=l,
t

H la

t

1

PBHKABG hinh VaM khan QK < It 1 < RK ten mien nhj lien P sao cho

ItI =RK

tuang li'ng v6i

c = {wllwl = M,} u {ill, ,;;1wI:,; argw = j 2; }, j = O,...,p-1.
M"

It I

= Qk

wang

t'mg v6i c = { w w 1= rnt} .
II

ChUng minh: B6 de 2.7 sur ill b6 de 2.6 nha cac phep biSn d6i Z = Q

z va

w = rnt , xern Thao [14, tr.64].
w

2.5 Ly thuy~t dQ diti c1}'c
trj

B8d~2.8
Trong rn~t phkg

z cho hinh chft nh~t D

= {z = x + iy

I

0 < x < a,O < y < b} .

Gia sir ham s6 w = f1 (z) th\lc hi~n PBHKABG hinh chft nh~t D ten rnQtill giac
cong H cua rn~t phkg w sao cho cac dinh 0, a, a+ib va ib cua D lfu1lugt tuang

16


tmg v6i cac dinh WI' W2' W3' va W4 cua H. GQi r la hQcac cling y trong H n6i
c~nh WIW2 v6i c~nh W3W4 cua H.

Gia

sir co

ham

dQ do

p = p(w) ~ 0 lien

t\1c trong

H

sao

cho

Oy

0 < Sp(H)

= ffp2dudv
H

< 00,W = u + iv t6n t~i theo nghia Lebesgue

D~t Ip = inf Ip(y) Khi do ta co Sp(H) ~
YEr

~~

Kb

(2.22)

l~

Chu y : DAngthuc co thS xay ra.
ChUng minh: Trong truemg hqp K = 1

D~tcrx =Dn{zliRz=x}vayx

=f(crx),O
Theo gia thiSt ta co
a

Sp

(H) = II p2dudv = II Dp2If;(z)12 dxdy = I dx Iax p2 If;(z)12Idyl.
H

0

Theo bfit dAngthuc Schwarzl ta nh~n duqc Vx E (O,a)

Lx P

2 If'(z)12

Idy

I Lx Idy I ~

(Lx If'(z)lldy If

i

va do axIdyl = b > 0 nen co
2
Lx p 2 If' (z) 12Id YI ~ ~ (L xcr If' (z) lid YI)
dS Y Y E r ta co
x

17


1

a

2

(

)

1

a

2

(

)

Sp(H)~-f 0 icrx plf'(Z)lldyl dX=-f 0 IpldWI dx
b
b
Yx
~ ~12
b

f
a

P 0

dX

= ~12.
b

(2.22a)

P

Truemg hQ'P K> 1.

Xet hI (w) 1aPBHBG H 1en hinh chfr nh~t D' , v6i D' dugc dinh nghia nhu sau:
D' = {11 = s + itlO < s < a', 0 < t < b'}, hiSn nhien h1il 1a PBHKABG ill D 1en D'
dodo
a'
1
->-- a

(2.22b)

b' - K b .

Ap d\mg2.22a cho phep biSn hinh h~1ta co:
a '
S p (H

)~

(2.22c)

b'1~

Thay kSt qua tfong (2.22b) vao (2.22c) ta dugc Sp(H) ~ ~

:

1~.

,
Dfiu bimg co thS Kay fa tfong truemg hQ'P H trimg vdi D', pew) = 1,va

khi do

:

= K :'

.

~~ 12 =~~ (b,)2 =a'b' = S (H).
KbP
Kb
P

BB d~ 2.9
Trang m~t ph~ng z cho tu giac cong E

= {zIfI
18

z

1< f2'

( -7t <)<1>1 arg z < <1>2 7t) }
<
(<


Gia sir ham s6 W= fez) thgc hi~n PBHKABG miSn E Ien mQtill ghic cong H cua

mat Phtm w saDcho cac dinh z I
g
.

= r1'21'32'42Z = 1", z
e ieicp\

= r eicp\z = r eiCP2 E
cua

I~n luqt tuang (rug v6i cac dinh WI' w 2' W3' W4cua H.

V6i cac ky hi~u p, Ip(Y), YE r,Sp(H)nhu

Sp(H)2

trong b6 dS 2.8 ta co:

1 </>2-</>112

K
r
p'
In ---L
rl

(2.23)

Chu y: f)~ng thuc co thS xay fa.
ChWzg minh:

Dung phep biSn hinh t = Inz biSn baa giac miSn E Ien hinh chu nh~t D v6i cac
dinh tuang (rugsau do ap d\lng b6 dS 2.8 ta dugc kSt qua (2.23).

B8 d~ 2.10
Gia sir trong m~t ph~ng z cho tru6c miSn A2 tuy Y saD cho v6i mQi r ma
(0 <) < r1< r < r2« 00) co

0 < OCr) = II dq>1<~(~

2n), trong do
= argz

va Yr = A2 (\ {z:1 z 1=r}.

Yr

Ham s6 w=f(z) thgc hi~n mQt PBHBG don di~p miSn A2 Ien miSn B2 cua m~t
ph~ng w. Ta d~t:

A21

B 21

= A2

n {z: rl
= f (A

21) va y

= f (y),

rl

< r < r2 .

Han lllia gia sir p = p(w ) ~ 0 dugc xac dinh trong B21saD cho

19


fP

2

dud

v

«

00 ),

r1

y

(0 <) S p (B 21) =

ff P dud v «
2

< r < r2
00 ),

w

=u

+ iv, t6n

t~i

theo

nghia

BZ1

Lebegue va v6i mQi r, r1 < r < r2 c 6

1 p dw
I

I

~ 1p

Yr

.

khi do

ta

co: (1 p )

TZ

2

J ra

TI

d
~ r)

<

ChUng minh: xem [12, tr 124-125].

20

s

p

(B

2 1

)

(2.24)