tống quan các vấn đề : lý thuyết mô hình toán tử

CHUaNG
1

,

-

1
-

,

TONG QUAN vB CAC VAN DE BAT RA
TRONG LUAN AN

Giai tich ham noi chung, va d~c bi~t 1a 1y thuy~t mo hiOOtoan tlf co r~t
OOi~uling d\illg trong cac 1Inh VL;fckhac OOau cua toan hQC va v~t 1y. Ly
thuy~t cac toan tlf k~t h<;5pvdi mo hiOO cac h~ d9ng llJC tuy~n tiOO da dfu1 d~n
cac k~t qua thll vi trong vi~c nghien coo cac tiOO ch~t cac h~ tuy~n tiOOvo h?ll

chi~u. N 9i dung cua 1u~ an 1a nghien coo cac tiOOch~t cua cac h~ d9ng llJC
tuy~n tioo vo h?ll chi~u bfulg cong Clfmo hiOOtoan tlf va mo hiOOh~ tuy~n
tiOO.

1. Ly tbuy~t fiG binb tmin tn.
Nho vao OOilng k~t qua n6i ti~ng cua Hilbert v~ ph6 toan tlf, ta da co
du<;5C
mo hiOOcua cac toan tlf tlJ lien h<;5pva toan tlf ddn nguyen, chUng du<;5c
bi€u di~n d d?llg tich phan
T

=

fAdP).
a(T)

d m9t hudng khac, ta da bi~t cac k~t qua cua Shill v~ dua m9t ma tr~
tlJ lien h<;5pv~ d?llg duong cheo 000 phep bi~n d6i ddn nguyen.
Phat tri€n hai hudng tren, vao dftu OOilngnam 50, cac OOatoan hQc Xo
Vi~t b~t dftu xay d1..fng1y thuy~t mo hiOO cho cac toan tlf k.hong tlJ lien h<;5p
, ho~c khong ddn nguyen va nguoi di tien phong trong 1Inh VL;fcnay 1a
M.S.Livsis. Nam 1946, Livsis da cong b6 cong trioo qUail trQng [38], trong do


l§.nd~u tien ham toan tit d~c tnfng cua toan tit dutoan tit d~c tnfng saDnay trd thanh mQtcong clf quan trQng trong nghien cUu
cua r~t nhi~u nha toan hQC.Qua trinh ti~n boa cua ly thuy~t ham toan tit d~c
trlfng di~n ra khcong thuc ham toan tit d~c tnfng cua toan tit co d?Ilg kh
eA(Z)= -A sign(I-AA *)+z I-AA * 11/2(I-zA *f1 II-A *A 1112
I

(1.1)

D?Ilg ham d~c tnfng nhu tren kha ti~n dlfng cho toan tit "g§.n"don nguyen.
D6i vdi toan tit "g§.n" tV lien h
saD
W(z)=I+2i (SignAl)

1

AI

I

(A *-zIfl

112

I

AI 1112

*

vdi

A = A-A
I

2i

Dinh nghla tren da dud?Ilg saD
W(z)=I-2iK*(A-zI)

trong do

-lKJ

J=J*, J2=I, KJK*=Ar vdi J, K la cac toan tit bi ch~ lien h~,vdi

toan tit A bdi cac cong thuc tren.
Sl} cai ti~n ham d~c tnfng d?Ilg (1.1) cho toan tit "g§.n" don nguyen mai
d~n nam 1972 mdi due A(z)=D+zC(I-zAflB
trong do B, C, D la cac toan tit b6 tr
I-A *A=C*C , I-AA *=BB*, I-D*D=B*B , I-DD*=CC* , -A *B=C*D .

4


Cilia khoa d€ su d\lIlg cac ham toan tV d~c tning d tren duejc th€ hi~n qua
dinh ly co ban sau day.
Dinh ly Livsis [37]. N~u hai toan tV bi ch~, don gian co clIng ham toan tV
d~c tning thi Wong duong d6n nguyen.
Ti~p sau, nho nhi1ng nghien cdu sau s~c ve ham toan tV cua Potapov

[32], Ginzburg [18]; Brodski va Livsis [37] da khai tri€n ham toan tV d~c
tning vao d~g
"J
1

1

e A(Z)=7 eel-aCt)

de:( t)

0

trong

00

IT(I k=l

"J
1
q~qk)
Ak - Z

do cac toan tVaCt),set), ~ thoa man mQts6 h~ thuc.
Nho vao bi€u di~n nay, Livsis da xay dvng mo hinh cua toan tV "gfu1"tl!

lien hejp, day chinh la budc d~u tien qUailtrQngtrong ly thuy~t mo hinh toan
tV khong don nguyen ho~c khong tl! li~n hejp. Mo hinh cu~ Livsis da ducac d6ng nghi~p va cac hQc tro cua ong cM.ti~n , md rQng V3.0nhUng nam 60

va 70 [41]. Mo hinh nay trong tniong h(Af)k=AJk + i I, fjqjJq~;
j=k+l

trong tniong h1
(Af)(x)=a(x)f(x)+i f f(t)q(t)Jq *(x)dt.
x

D~u nhvng nam 60, d mQt hudng khac, cac nha toan hQc Dong Au, Nagy
va Foias da ti~n hanh nhUng nghien cilil r~t sau s~c ve cac toan tV co trong

khong gian Hilbert ma mQtv~ de trQng tam la phep gian (dilation) toan tV.

5


Trong qua trinh xtly dlJng phep giffil cac toan tv, cac nha toan hQcnay da thi~t
l~p m9t d<:lih1Qngd~c tnfng cua toan tv, va th~t thu vi la d<:lilv(jng d~c tnfng

nay l<:litrimg vdi khai ni~m v~ ham d~c tnfng cua Livsis. DlJa vao khai ni~m
ham toan

cl d~c tnfng do, Nagy va Foias da xtly dlJng m9t ma hinh rftt ti~n

d1Jngma san nay lien h;1cxuftt hi~n tren cac cang trinh cua cac nha toan hQc
tren th~ gidi.
Ly thuy~t ma hinh toan cl cua Nagy va Foias co th~ tom t~t nhv san.

Cho A la toan cl co tren khang gian Hilbert, ham toan

cl d~c tnfng cua A

dvQc dinh nghla b6i

eA(z) = -A + z(I - AA *)1/2(I - zA *)-1 (I - A * A)1/2
Ngv(jc l<:li,cho trlfdc ham e(Z)E $ (U,V), Nagy-Foias xtly dlJng ma hinh toan

clco nhv san.

x = [L~(V) Ef)&2(U) ]e{(eO) Ef)ilO))/ 0) E L~(U)},
A(

\jf)

= e -it

«p( eit) -
trong do il( eit) = (I - e( eit) * e( eit) )1/2.
Toan

cl A nay ddn gian va co ham toan cl d~c tnfng trimg vdi e(z).

San do,

vao nam 1972 Brodski da xtly dlJngth~m cac toan cl
Bu

= e -it

(e( eit) - e(o))u Ef)e -it il( eit)u ,

C(
Du = e(O)u ,
d~ dVa v~ ma hinh cua h~. H~ tuy~n tinh dv(jc xtly dlJng nhv tren la ddn gian,
ddn nguyen va co ham truy~n la e(z). Ta co k~t qua (diM ly Livsis- Brodski)

6

L1hai h~ don gian, don nguyen co cUng ham truy~n thi Wong duong don
nguyen. Nhu v~y mo hinh cua Nagy-Foias da co m9t vai tro qUailtr<;mgd~
nghien cUu cac h~ don nguyen, mo hinh nay co th~ cho ta nhi~u thu~ 1Qivi
cac toan tUd~u duQcxay dlJIlgtlfdng minh.
Hudng thll ba trong 1ythuy~t mo hinh toan tU duQcphat tri~n b6i cac
nha toan hQc My, De Branges, Rovnyak [13]. SaD day 1a mo hinh cua De
Branges va Rovnyak cho toan tU co duQcxay dlJIlgtheo ham 8(Z)E $ (U,V)
cho trudc. GQi Be 1a khong gian g6m cac phfu1 tU (f(z),g(z)) vdi f(z)EH2(V),
g(z) EH2(U) sao cho
«f(z),g(z)),
vdi

K ew,x,/z) 1aham

K ew,X,/Z»Be

= <few), X>y+ <g(w), y>u

cac toan tU duQc dinh nghia b6i

K ew,x,y(z) =
I - 8(z)8(w) *
8(z) - 8(w)
x+
y,
z-w
( 1-zw

8(z) - 8(w)
1- 8(z)8(w) *
.
x+
y
z-w
1-zw
J

trong do 8(z) = 8("2)*, WEq}),XEV, YEU.
Toan tU mo hinh tren Be duQc diM nghia b6i

A : (f(z),g(z)) H (zf(z)-8(z)g(O), g(z) - g(O))
z

co ham toan tU d~c trlfng 1a8(z). Vi cac toan tU A trong cac mo hiOOcua
Nagy-Foias va cua De Branges- Rovnyak d~u don gian lien chUng tlfong
duc5ngdon nguyen. Trong mo rJlli~cua De Branges- Rovnyak, tuy khong gian
Be khong co bi~u di~n tlfdng minh nhung couu di~m 1acac phfu1tU fez), g(z)
d~u 1acac ham gild tich.
7


Ngoai cac Gong triOOchu y~u tren, ly thuy~t mo hiOOtoan tV con duQc
md r<)ngcho cac lop toan tV khac, kS ca toan tV khong bi ch~ [6],[8].
Trong lu?n an nay, chUng Wi S11dlJIlg chu y~u mo hiOOcua Nagy-Foias.
LUll

Y

la cac mo hiOOtren la mo hiOOham. Ngoai ra con co huang xfty

dlJIlg mo hiOOd~g tich phan t:.heocac khong gian con bftt bi~n duQc xfty d\fng
bdi Brodski [14], Gohberg, Krein [19].
2. Ly thuy~t h~ dQng hfe tuy~n tinh.
BM d~u tU nam 1960, sail cac Gong trioo cua Kalman [23], m<)ts6 huang
nghien coo cac tiOOchftt diOOtiOOcua h~ d<)ngh.Jctuy~n tiOOphat triSn m~.
Kalman da dUa fa cac khai ni~m rftt qUail tr<;mg: tiOOdieD khiSn duQc, qUail
sat duQc, xfty dlJng mo hiOOcac h~ (ly thuy~t thS hi~n), sq d6ng d~g cua cac
h~ tuy~n tiOO[23], ...
Xet h~ d<)ng lqc tuy~n tiOOa=(X,U,V,A,B,C,D)

duQc mo hiOOhoa bdi

h~ phuong trioo sail
dx = Ax(t) + Bu(t) ,
dt
vet) = Cx(t) + Du(t) ;

x(t), u(t), vet), la cac ham vecto voi gia tri la cac vectd l~ luQtthu<)ccac
khong gian Hilbert kha tach X, U, V. Ham x(t) duQc gQi la ham tr~g thai,
u(t) duQc gQi Ia ham dieD khiSn va vet) duQc gQi Ia ham qUail sat.
H~ a duQc gQi Ia dieD khiSn duQc tU tr~g thai Xod~n tr~g thai Xl trong
khoang thdi gian [to,tl] n~u t6n t(;lim<)tham dieD khiSn u(t) xac diOOtren [to,t1]
sao cho n~u h~ b~t d~u tU tr~g thai Xo(tUc la x(to)=Xo)thi t(;lithdi diSm tl no

8


co tr?llg thai Xl' tUGla x(tl)=Xl' Di~u do d6i vdi h~ tuy~n t1nh xet d tren co

nghia la
X(tl) = e A(tl-tO)Xo+ ftt~eA(tl-S)Bu(s)ds.

H~ a duQc gqi la di~u khi€n duQc hoan toan n~u a di~u khi€n duQc tU tr?llg
thai bM 10' Xov~ tr?llg thai b~t 10' Xl trong khO
Trong di~u ki~n X,U,V la cac khong gian hitu h?ll chi~u thi h~ di~u
khi€n duQckhi va cm khi
rang (B, AB, ..., An-lB)= n = dimX.
D6i vdi h~ vo h?ll chi~u, khai ni~m di~u khi€n duQc thuang duQc hi€u d
d~lg di~u ~~i€n duQc x~p xi, nghia 1.1vdi ill<)tIan c~ cho trVdc cua Xb luon
t6n t?i m9t ham di~u khi€n u(t) di~u khi€n quy d?o cua h~ tU tr?llg thai Xod~n
Ian c~ cua tr?llg thai Xl trong m9t thai gian hitu h?ll, Khi ~y di~u ki~n cfu1va
du d€ h~ di~u khi€n duQc la

:AkBU=X
0

D6i ng~u vdi khai ni~m di~u khi€n duQc, Kalman dua ra khai ni~m qUail

sat duQc.V~ d~ d~t ra la khi bi~t ham qUailsat vet) (t 2 to) thi tr?llg thai ban
d~u Xo= x(to) co duQcxac dinh duy nh~t kh6ng ? N~u h~ a co tr?llg thai x(to)
= Xo"*0, ham di~u khi€n u(t) = 0 (t 2 to) l'iti co ham qUail sat vet) = 0 (t 2 to) thi
tr?llg thai Xo gqi la kt~ong qUail sat duQC t'iti thai di€m to. H~ duQC gqi la qUail

sat duQc hoan toan n~u t'itimqi thai di€m , khong co vecto nao khong qUail sat
duQc. Khi do ta co k~t qua d6i ng~u cho tinh qUail sat duQc. H~ hitu h?ll chi~u
,

,,',

l'

qUail sat duQc hoan toan neu va chi neu

9


rang (C*, A *C*,.. .,A *n-1C*)=n=dimX ;

\trong tru6ng h<;>p
vo h~ chi~u thi di~u ki~n c§n va du d~ h~ qUailsat duQc
hoantoanla
00

vA *kc*v=x.
0

MQt khai ni~m qUailtr(;mgtrong h~ tuyen tioo dUngla khai ni~m ham
truy~n, ham nay du<;>c
xac dinh b6i cong thuc
eaCz)=D+zC(I-zAr1B:U

~ V.

Ly thuyet h~ dQng l\fC tuyen tinh d\fa tren ham truy~n va ly thuyet mo hiOO
toan tU trong giai tich phat tri~n dQc l~p song song OOungco OOi~udi~m Wong
d6ng th1.ivi. D6i vdi mQt s6 ldp cac h~ thi ham truy~n trUng vdi ham d~c trung
?

,

?

cua toan ill A .
Ham truy~n mang

y nghia

OOusail :gia sit h~ a co vecto tr~g thai xCi) =

XoeZ\vecto vao u(t) = uoezt,vecto ra vet) = voezt,thi vet) = e(l(z)u(t). Nhu v~y

hai h~ co cling ham truy~n co th~ coi la Wong duong vi tr~g thai ben trong
cua hai h~ co th~ khac OOaunhung khi cho cling tin hi~u vao u(t), ta d~<;>c
cling tin hi~u ra vet). TiOO qUail trQng cua ham truy~n con duQc th~ hi~n

6

dinh ly Kalman [23] : neu "hai h~ hitu h~ chi~u ai, ~ di~u khi~n duQc, qUail

sat duQCco cling ham truy~n thi chUngd6ng d~g, nghia la khi do t6n t?i mQt
toan tUkha nghich lien 1:\1c
W: xi~ X2sao cho
A2

= WA1W-1

,

B2 = WE1 ,
C2 =C1W-1 ,

10


D2 =Dl

;

va ra rang hai h~ d6ng d~g till chUng co clIng mQt s6 cac tinh ch~t qUail tn;mg
nhl1 tinh di~u khi€n dl1W la don nguyen till ng116ita noi hai h~ la Wong dl1ong don nguyen.
Tren co s6 dinh ly d6ng d~g, Kalman da xay dljIlg cac mo hinh cua h~
tuy~n tinh IDeo mQt ham truy~n 8(z) cho trUoc ma ong ta gQi la cac th€ hi~n
(realization) cua ham 8(z) [23]. Ly thuy~t th€ hi~n da pilat tri€n kha m~,
khong nhltng cho h~ tuy~n tinh dung, h~ khong dUng ma Celh~ phi tuy~n.
Di Sail hon nlta d6i voi h~ tuy~n tinh, cac nha lOan hQc My (Brockett,
Barass... [10], Israel (Gohberg [11]), ... da nghien CUllSlJlien k~t cae h~. Cac
h~ tuy~n tinh khi lien k~t n6~ ti~p nhau IDeo nghia : Cho hai h~ a k =
(Xk,Uk,VbAbBbCbDk),

k = 1,2, sao cho U2 = Vl . H~ a = (X,U,V,A,B,C,D )

dl1
la a = a2al n~u :
U = Vl ; V = V2 ; X = Xl EBX2 ,
A = A1P1 + A2P2 + B2ClP1 ,

B = Bl + B2Dl ,
C = C2P2+ D2C1P1,
D = D2D 1 ,

trong do Pk la phep chi~u vuong goc tU khong gian X len khong gian Xk ,
k=12.
, ,


thi cac tiOOdi~u khi€n dUdC,qUail sat dUdC, t6i thi€u, ddn g ian, t6i ULl... co th€

.

~.

khong du\Jc baa toan. Cac taG gia tren da co mQt s6 k@t qua v~ di~u ki~n d€
baa toan cac tiOO ch~t do khi lien k@tcac h~. Cac k@t qua nay du\Jc phat bi€u

tren ligon ngv b~c Mac Milan cila ham ma tr~. Trong t~t ca cac th€ hi~n cila
ham 8(z), th€ hi~n co s6 chi~u cila khong gian tr~g thai la 006 OO~tdu\Jc gqi
la th€ hi~n t6i thi€u. S6 chi~u cila khong gian tr~g thai trong th€ hi~n t6i
thi€u cila 8(z) du\Jc gqi la b~c Mac Milan cila 8(z) va du\JCkYhi~u la deg8(z).
MQt k@tqua v~ di~u ki~n d€ baa toan tiOOch~t t6i thi€u du\Jc phat bi€u trong
diOOly Gohberg : Lien k@tn6i ti@pa cila hai h~ t6i thi€u aj va a2 la t6i thi€u
.

n@uva chi n@udeg8a(z) = degeaj (z) + degea2 (z).

Cong Clfcila hudng nghien coo nay la GongClfd?i s6 ma tr~, r~t kho

phat tri€n cho tr1.fdngh\Jpvo h~ chi~u.

3. Ly thuy~t h~ dQng I1fctuy~n tlnh tren khong gian Hilbert.
Livsis la ngudi d~u tien nghien coo ly thuy@th~ tuy@ntiOOtrong khong
gian vo h?il chi~u [40]. Ong da khao sat cac h~ dQng h.;tctuy@ntiOOdissipative

d~g
x = Ax+Bu ,
v = Cx+Du ;
trong do C=B* , D=I ,

A- A*

~=

,

,,'
,. h '
BB *
, VOl am truyen

co

d

~g

S(z)=I+2iB*(zI-AflB; va nghien coo nhi~u ling d\lilg cila chUngtrong v~t ly.

12


~,
,cac hQctra cua ang cling d3:tien hanh cac nghien cUuve cac h~ ngau nhien
?

(Iancevich [30]), v~ th€ hi~n cua cac ham phan hinh (meromorphicD.C.Khanh

[42]). D~c bi~t Arov d3: nghien cUu sail v~ cac h~ bi dQng

(passive). Do la h~ rai r~c d~g
~+l =~+B~,

vn =C~+D~;

vcii (~

~): XEBU~XEj)V 1aloan ttJ co va mUlltruy~n 8(z)=D+zC(I-zArIB.

Ong d3:xay dl,fngcac h~ ma hinh cua lop cac h~ bi dQng t6i liu, xay dl,fng cac
th€ hi~n bi dQng khac nhau cua cac lop ham toan hf trong khang gian Hilbert

voi nhifng y nghla v~t ly hfong ling, d6ng thai lien h~ voi phep gian cac h~.
Arov d3:xay dlJIlgphep gian cua ham toan hf co gi::iitich S(z), hIc la tim ma
tr~ kh6i
~

S(z)

=

Sll (z)

S(z)

"

( S21(z) S22(z))

don nguyen tren yang trOll don vi fj}Jj va thoa di~u ki~n t6i thi€u
KerSll(z)={O} h~u kh~p tren fj}Jj.sv dl)11gcac k~t qua cua Arov, D.C.Khanh
d3:khao sat cac bai toan v~ lien k~t cac h~, slJ baa toan cac tinh chM diOOtiOO
cua h~ trong qua triOOlien k~t [24],[44],[45],[46]. Phuong phap nghien CUlld
day la OOanhf hoa cac ham toan hf co va ly thuy~t ma hiOOtoan hf. Dua vao
cac khai ni~m moi (:t) nhan hf hoa chiOOquy cua ham toan hf, D.C .Khanh d3:
thi~t l~p cac di~u ki~n c~ va du d€ baa toan tioo di~u khi€n du<;jc,qUail sat
du<;jc,t6i thi€u khi lien k~t cac h~ don nguyen ho~c cac h~ bi dQng.

13


Cho 8(z) E ,%\U,V-j, 8k(z) E $(UbVk), k=l, 2, U)= U, V)= U2, V?= V.
Nhan 111hoa ham toan 1118(z) = 8iz)8)(z) dllQCgQila (+) chiOOquy n~u toan
111
Z+ : Llli ~ .1.28)hEB.1.)h, Vh E H2(U)
sail iliac tri€n tuy~n t£OOlien h;1c1a toan 111ddn nguyen tU khong gian .1.H2(U)
1en khong gian .1.2H2(U 2) EB .1.1H2(U1);

trong trlidng hQp toan 111

*
Z-: .1.*h~.1.2*h EB.1.1*82h, hE 1"2(V)
sail iliac tri€n tuy~n tiOO lien h;1c 1a toan 111ddn nguyen tU khong gian
.1.*1"2(V) 1en .1.2*1"2(V 2) EBL11*L2 (VI) ; vdi .1.*(eit) = (I - 8(eit)8 (eit)*)112,
.1.

k*

*
( eit ) = ( 1-8 k (eit ) 8 k ( eit ) ) 112k=12'
"
,

thi OOan 111hoa ham toan 111d tren dllQCgQi la (-) chiOOquy.

.
Sail day la mQt vai k~t qua dllQCdUng trong 1u~ an.

DiM 1y 1. Cho h~ a 1a lien k~t n6i ti~p cua hai h~ ddn gian, ddn nguyen, di~u
khi€n dllQCa) va 0.2' Khi do h~ a la di~u khi€n dllQCn~u va chi n~u OOan111
hoa ham truy~n 8a(z) = 8aj (Z)8a2 (z) 1a(-) chiOOquy.

DiM 1y 2. Cho a) va a2 1a cac h~ bi dQng t6i thi€u. N~u OOan111hoa ham
truy~n 8a(z) = 8aj (Z)8a2(z) 1a(:1:)chiOOquy thi h~ a=~aj1a h~ t6i thi€u.
4. Cae viin d~ nghien etfu trong lu~n an.
Ti~p h;1chlldng nghien coo tren, d6i vdi cac h~ dQng h;fc tuy~n tioo rdi
f?C bi dong vdi ham truy~n 1aham cac toan 111co giai tich tren dra trim ddn vi,

14

m9t 10<;1t
cac bai loan mai duqc d~t ra ho~c dj thi~n cac k~t qua cua cac lac
gia neu tren. Nguqc l<;1i
vai vfu1d~ lien k~t cac h~, chUngtoi xet bai loan tach
m9t h~ thanh n6i cua hai h~ ddn gian d~ nghien coo tUngh~ rieng. Trong bai
loan nay, chung toi da tim duqc d?llg Wongminh cua cac h~ thanh phftn va da
tach h~ theo hai hl1dng : hudng thll OO~tla tach h~ theo rinh chiOOquy cua
ham truy~n, huang thll hai la tach h~ theo khong gian con b~t bi~n cua loan tV
chiOOA. ChUng toi cling da tim duqc m6i lien h~ gi11ahai cach khai tri~n noi
tren. Cac k~t qua nay duqc triOObay trong chudng 2 cua lu~ an va da duqc
cong b6 trong [25]. K~ d~n, chUng Wi cling xet d~n cac riOOch~t dinh riOO
cua cac h~ vo h<;1nchi~u va vfu1d~ lien k~t cac h~ Dhung phudng phap nghien
coo d day chu y~u la dUng khai ni~m ham non t6t OO~tcua ham truy~n. Cho

8(z) : U ~ V la ham cac loan tV co giai rich tren dla trim ddn vi qj).NagyFoias da chUng miOOduqc t6n t<;1im9t ham ngoai cac loan tV co tU khong gian U vao khong gian F sao cho
,

,

va
n~u ~(z) la ham giai rich cac loan tV co sao cho

~(eit)*~ (eit) < 1- 8(eit)*8 (eit) a.e. thi ~(eit)*~ (eit) < Ham v~ bell trai va duqc gQi la ham non t6t OO~tcua ham I - 8(z)*8(z).

15

Theo dinh nghla, ham cac toan tU co giai rich tren dla trim ddn vi
cp(z):U~F dl1<;5c
gqi la ham ngoai n~u ~H2(F), (cpu)(z)= cp(z)u(z),uEH2(u); va cp(z)dl1<;5c
gQi la ham trong n~u
cpla mQt d&ng clJ.
Ham non t6t nhftt cp(z) nay da co nhi~u vai tra trong vi~c khao sat cac h~ ddn

nguyen : khao sat cac khong gian con b:lt bi~n, thanh phfu1 khong qUail sat
dl1<;5c,
khong di~u khiSn dl1<;5c
cua h~ ...[9J,[dinh ly 3.4,chl1dng 3J; xfly dt;fng
mo hinh cac h~ bi dQng t6i 00, t6i thiSu [35J... Do k~t qua: n~u h~ a la rich

n6i ti~p cua hai h~ exIva a2 thi ham truy~n cua h~ exla eiz) = eal (Z)8a2(z).
Nen mQt bai toan dl1<;5c
d~t ra cho chUng Wi la xfly dt;fngham non t6t nhftt cua
ham 8(z)=8iz)8I(z)

tU cac ham non t6t nhftt cua 8I(z) va 8iz). Cac k~t qua

thu dl1<;5c
dl1<;5c
trinh bay trong mQt phfu1 nQi dung cua chl1dng 3 va da dl1<;5c
cong b6 trong [26J. K~t qua nay da dl1<;5c
sit dlJIlg dS chUng minh dinh ly baa

toan tinh t6i 00 cua h~ n6i. DlJa tren khai ni~m ham non t6t nh:lt, chUng Wi

cling tim dl1<;5c
di~u ki~n dS mQth~ bi dQngla t6i 00, d6i t6i 00; di~u ki~n dS
mQt h~ bi dQng la ddn nguyen... Cac k~t qua nay dl1<;5c
trinh bay trong chl1dng
4 va da dl1<;5C
cong b6 trong [28J. Cling tren cd sd ham non t6t nh:lt, chUng toi

da thi~t l~p cac di~u ki~n cfu1va du dS baa toan tinh t6i 00, tinh t6i thiSu, tinh
hoan toan khong qUail sat dl1<;5c.
.. trong qua trinh lien k~t cac h~ bi dQng. Cac
k~t qua thu dl1<;5C
dl1<;5ctrinh bay trong chl1dng 5 cua lu~

an va da dl1<;5ccong

b6 mQt phfu1 trong [27J va mQt phfu1 se dl1<;5C
cong b6 trong [20 J.

16


Gia sV CP1(z), CP2(z) va cp(z) Ifm lu,

P2(Z)8I (Z)

,

81(z), 8iz) va 8(z), 8(z)= 82(z)81(z).Ta luon co
,

,.

Vng VOl8(z.)

,
. ,
:.
~
v an
de dX<:;t
t ra 1a kh1 nao

P2(Z)8I (Z»


«

)


«

)

-

,

'

,

,

la ham non
:.

:.

se 1a ham non tot nh at,

nghla la khi nao trong bftt d~g thUG
'"

P2(Z)8I (Z)
«


P2(Z)8I (Z»

J

«

::;;cp(z)*cp(z)

)

co dftu bfu1gxay fa. Bfu1gcac phuong phap khac Mati, chUng toi thu dus6 k~t qua v~ vftn d~ nay xet tren cac h~ mo hinh khac Mati. ChUng du
bay trong chuang 3 va 5 cila lu~ an va da duNgoai fa, chUng Wi cling tim dutruy~n; k~t qua co du
trong [20].

17