(Luận án tiến sĩ) một số phương pháp hiệu chỉnh giải bài toán đặt không chỉnh luận án PTS toán học62 46 30 01

'.
-

•

4^'

BO GIAO DUG VA DAO TAO
DAI HOC QUOC GIA HA N O I
n^l/CJNG DAI HOC KHOA HOC Tl/NHIEN
&

NGUYEN VAN HUNG

M O T S O PHUONG P H A P H I E U CHJNR

GIAI BAI TO AN DAT KHONG CIliMH
Chuy^n nganh
Ma so

: Toan hoc linh (can
: 1.01.07

^\
LUAN AN PHO^ TIEN
SI KHOA HOC TOAN - LY

NGUdl HUdNG DAN KHOA HOC: TIEN SI - PIIAM KY AN13

i^u-^rz-?.>^:(ss.-:-

••:':;^v:;^.

u.y..L^/4^
fri-ij^^K

/rfa JtO

yniir^
3^

-.-4

^'%-\'"^ '^•'•.'^i•^ ' • J - v - ' ^ ' j ' •. N l l ^ . 4 •; '. j

> i^ir,

^:T:

*

Ha nOi - 1996


MUC LUC
L51 noi dau

1
CHUONG I
PHUONG PHAP COMPACT THU HEP CAI BIEN

§1. Ma dau
§2. Cac gia thi^'t cua bai toan
§3. Thuat toan compact thu hep dang Robust
§4. Tnrcmg hgp kh6ng duy nhat nghiem
§5. Tnrcmg hgp ve phai va toan tir kh6ng biet chinh xac
§6. Ap dung cho phirong trinh vi phan thuong

5
5
7
12
15
17

CHUDNG II
BAI TOAN TUYEN TINH KHONG CHINH TREN COMPACT YEU
§1 Ma dau
§2. H6i tu y^'u trong kh6ng gian Hilbert
§3. Danh gia tinh 6n dinh cua nghiem trSn compact y6u
§4. Phuang phap khai tridn ky di chat cut
§5. Phuang trinh tich phan dang tich chAp
§6. Danh gia diam Vs trong phaang phap compact thu hep cua Gaponenko

23
25
27
29
31
36

CHUONG III
M O T S O PHUONG PHAP LAP - HIEU CHINH
§1. Phirang phap Gauss - Newton hi6u chinh (RGN)
§2. Ki^m tra di^u kien B] cua Bakushinski
§3. Phuang phap hieu chinh Gasse - Newton g'an dung
§4. Phirang phap Seidel - Newton hieu chinh va bai toan phi tuyS'n cong huang
I - Phuomg phap Seidel - Newton (SN) vd phuomg phap
Seidel - Newton hieu chinh ( RSN)
II - Syc hdi tu dia phuomg cua phuomg phap RSN
III - Bdi todn Men tudn hodn cho phuomg trinh Duffing - Van derpol
nil - Bdi todn bien tudn hodn doi v&iphuomg trinh Van derpol

40
41
46
52

Phan ket luan
Phu luc
1- Bai toan - Lcfi giai
2- Thuat giai va chuong trinh

66
67
67
72

Tai lieu tham khao

gg

52
54
57
65


PHXN Ud DAU

#

Ngay nay, ciing v6i vice sur dung ph6 biC'n may tinh, loan hoc ngay cang
dirge ung dung rong rai trong cac linh virc khoa hoc va ihurc ti6n. Vice ap dung loan
hoc mot each s^u sac da thiic day manh me sir phat trien cac phirang phap tinh toan.
Trong thirc te ta thirofng gap nhiJng bai toan ma dfr ki6n ban dau chi dugc biet gan
dung, nhCmg thay ddi nho cua du* kien ban d'au c6 thd dSn dfi'n thay do! liiy y cua
nghiem, do do vi^c tun nghidm cua bai toan gap nhicu kho khan. NhCmg bai toan
khong on dinh nhu vay la m6t vi du ve bai toan dat khong chinh.
Khai niem bai toan dat khong chinh da dirge nha toan hoc Phap Hadamard J.
dira ra rfau tien cho Idrp phirang trinh vi phan [63,64]. Thoat dau nguai ta cho rang
bai toan nay kh6ng c6 y nghia toan hoc va thirc tiSn, n6n ft chu y de'n no. Nhimg den
cuo'i nhung nam 50 nguofi ta phat hien ra rang, nhicu bai toan ly thuyet va hau het
Ccic bai toan trong ki thuat va ihirc 16' deu dan de'n bai loan dat khong chinh.
Tikhonov A .N la ngirai c6 c6ng dau trong nghi^n cuti vah de nay[49,50].
6ng da d~e xuat khai niem bai toan dat khong chinh cho Icfp phirang trinh toan tu
trong kh5ng gian T6p6 va cho ra dofi mot loat cong trinh xung quanh v^n dc nay. ItCr do de'n nay, cac nha toan hoc tr6n the' gidi da tap tiling nghien cuu bai loan dat
khong chinh va da dat dirge nhung thanh tiru dang ki - Trong so do phai ke den
cac nha toan hoc Morozov

B.A , Laverientiev

M.M ,Vasin J.M, Ivanov V.K,

Bakushinski A.B, Gaponenko Yu .L, Bertero M.P, Nashed M.Z, Groetsch C M . , ...
Khai niem dat chinh theo Hadamard la:
Xet phirang trinh
Ax = y
Trong do A la toan tu dua khong gian T5p6 X vao khong gian Topo Y
1 - Vai m6i y e Ytbn tai x e X
2 - Nghiem x la duy nhat.
3 - Nghiem phu thu5c lien liic vao cac du kien cua bai loan.

(0,1)


Ne'u vi pham it nhSit m6t trong ba dieu kien tren, ihi bai toan dugc goi la bai toan dat
kh6ng chinh,
•

M6t VI du didn hinh khi A la toan tu* tuye'n Ifnh hoan toan lien tuc, con X, Y
la cac khdng gian Banach v6 ban chieu, khi do:
00

i-ImA ^ YvalmA = u {Ax:||x||< n)latap pham tru thirnha't.(dieu nay CO nghia
n=l

la bai toan (0,1) giai dugc khOng phai vdi moi y e Y).
ii - Ne'u A ' : Y-> X c6 ton tai thi cung khong lien tuc. Di^eu nay chung to nghiem
cua bai toan (0,1) khong phu thuoc lien tuc cac du" kien ban dau

Tu* dinh nghia v'e tinh chinh thay rang: Tinh chinh cua bai toan phu thu6c bo ba
{A,X,Y). Ngucri ta thucmg khac phuc tinh khong chinh bang mot trong cac phuang
phap sau day.
1 - Xet nghiem suy rong, do la phuang phap tira nghiem Ivanov va e - tua nghiem
cua Liscovets .
Tua nghiem cua (0,1) la tap M cac phan tir c6 do lech nho nha't so vai nghiem chinh
xac. Phuang phap tua nghiem la di tim cue iiin phie'm ham khong khap tren lap
chap nhan dugc nghiem [41].
2 - Thu hep mfen xac dinh cua so lieu ban d'au.
3 - Thay ddi kh6ng gian va t6p5 cua chung.
Hai phuang phap 2 va 3 ft dugc six dung vi kh5ng dap img dugc yeu cau cua thirc te.
4 - Phuang phap hieu chinh : La phuang phap thay bai toan dat kh5ng chinh bang
m6t ho bai toan dat chinh phu thu5c tham s6' ma nghiem cua bai toan dat chinh sc
d'an de'n nghiem bai toan dat kh6ng chinh khi tham so d'an tai khOng. Phuang phap
Laverenliev, phuang phap Tikhonov [54], phuang phap tua nghich dao Lattes Lions, phuang phap lap, su dung khai tridn ky di va khai tri^n ky di chat cut la
nhung phuang phap hieu chinh quen biet.
Phuang phap hieu chinh ciia Tikhonov dua tren b6 de sau:


Bo de Tikhonov A.N [54 ]
Gia su A: M -> A(M) la anh xa 1 -1, lien tuc, con M la compact khi do
• A^ : A ( M ) ^ M lien tuc.
Vao cu6'i nhung nam 50, Tikhonov[54] da de xua't khai niem phiem ham on dinh nhu sau.
i, Q : X -> R ^ a phie'm ham lien tuc va khong am.
ii, Xd e Xo = dom Q , Xo = X , (trong do Xd la nghiem cua bai loan (0,1)).
iii, Vdfi m6i r > 0, tap K(r) = (x e Xo : fi[x] < r} la tap compact tuang doi trong X
Phuang phap hieu chinh Tikhonov la lay nghiem g^an dung cua (0,1) la didm cue tieu cii
phie'm ham tran:
M "^ [x,y5 ] = p^ (Ax,y5 ) + a 0[x ] -> min
x e Xo

Vai in6t s6' gia thie't nhat dinh c6 th^ chiing minh dugc rang.
i/ 3 ! Xa= ArgminM''[ x,y5]
X e Xo
ii/ Tham s6' hieu chinh a = a (8) chon tu cac nguyen ly khong khc5rp, nguyen
ly tua tO'i iru,v.v...
iii/Xa(6)^Xd ( 6 - > 0 )

5 - Ma r6ng nglu nhien bai toan la't dinh dat kh6ng chinh.
Ban luan an nay nghien citu mot so phuang phap hieu chinh giai bai toan dat khong chin
Nhitng v^in d'e dugc quan tam trong luan an la:
1- Giai bai toan dat kh6ng chinh tren tap compact.
2- Danh gia tfnh 6n dinh ye'u cua bai loan dat khOng chinh tren compact va compact yeu
3 - Cac phuang phap lap hieu chinh giai bai toan dat kh6ng chinh.


Ban luan an gbm ba chirang, tai lieu tham khao va phan phu luc
Chuomg I: Phuomg phap compact thu hep cat Men.
Chuang nay trinh bay phuang phap compact thu hep cai bien va danh gia dugc toe do
h6i tu cua phuang phap.
Chuang II: Bdi todn tuyen tinh khong chinh tren compact yeu.
Trong bai toan (0,1) xet truong hgp X,Y la cac khOng gian Hilbert, A la toan tur tuye'n
tinh lien tuc vai khai tridn ki di da cho tru6c. Trong chuang nay chung t6i trinh bay
m6t thuat toan dang khai tri^n ki di chat cut giai phuang trinh (0,1) va danh gia tfnh on
dinh ye'u cua nghiem.
Chuang I I I : Mot so phuomg phap lap hieu chinh.
Chuang nay trinh bay phuang phap lap hieu chinh Gauss-Newton gan dung va phuang
phap Seidel - Newton hieu chinh.
Phan phu luc ; Trinh bay mot vi du so giai bdi todn Cauchy bang phuang
phdp khai trien ky di chat cut.
Noi dung chfnh cua luan an da dugc cong bo trong 6 bai bao dang tren cac tap

chf cap truong, B5, qu6'c gia, qu6'c te' va dugc bao cao tai Xeminar toan hoe tfnh toan
cua Dai Hoc Tong Hgp Ha Noi (Tien sy Pham Ky Anh chu tri). Hoi nghi khoa hoe 35
nam thanh lap khoa Toan - Co - Tin DHTHHN hoc nam 1991, Hoi nghi khoa hoc khoa
toan DHSPHN 2 1992, H5i nghi khoa hoc khoa Toan - Co - Tin hoc DHTHHN 1994,
Hoi nghi quoc te' vt bai toan ngugc 1995 ( Tai thanh pho Ho Chf Minh).


CHUONGI
PHUONG PHAP COMPACT THU HEP CAI B I C N

$1 - Ma dau:
Xet phuang trinh
Ax = y

•

(LI)

6 day A la toan tir phi tuye'n, X,, Y la cac khong gian vector t6p6. Gaponenko
Yu.L [24,25] nam 1982 da de xua't phuang phap c6 ten la "compact thu hep". Ong da
xay dung dugc cac tap Vg gbm huu han phan tir sao cho:
V xg e Vg => II X5 - Xd II < diam V5 + 5 -> 0 ( 5 -> 0)
Arsenin V.Ia , ([5], nam 1989 ), xet bai toan (1.1), trong do A : H ^ C[a,b], H la
kh6ng gian Hilbert. Thay vi biet ys e C[a,b] chi bie't m thi hien {y'sl "'1, trong do han
mOt nua y'5 thoa man dieu kien.
II / 5 - y d ||c<5

( l < j i < m ; i = l - r ,r>m/2)

(1.2)

Bang each sir dung ham Robust, Arsenin da dua ra mot phuang phap hieu chinh.
V6i ham

M" [ x,y^5 ,.•• ,y'"8 ] = ^^ (Ax) + aC)(x),

Arsenin V. la da churng minh dugc cac ke't qua sau.
l/3!Xa-ArgminM"[x,y^8

y"\]

2/ 3 a = a(5) : Xa(6) -> x* (5 -> 0 )
Cung nhu phuang phap Tikhonov AN, phuang phap Robust kh6ng cho phep danh gia
sai s6' cua nghiem g'an dung.
Y tuang cua Gaponenko va ky thuat cua Arsenin da dugc chung toi sir dung de xay
dung thuat toan giai bai toan (LI) va da danh gia dugc tO'c dO hoi tu cua nghiem g^an
dung.
52 - Cac gia thiet cua bai toan.
Xet phuang trinh (1.1), trong do A : X -> C[a,b], A la toan tir lien tuc, X la
khdng gian Banach. Goi Xi la mfen xac dinh cua phie'm ham on dinh D.[x].


-6

Xi c: X va ta c6

OO

Xi ==uK(n)

(2.1)

trong do K(n) = (v e X] ,D.{v) Chung ta da bie't m6i tap K(n) la tap compact trong X
Gia suf phuang trinh (L1) vdfi ve'phai dung yd c6 nghiem duy nhat Xd
A x d - yd

Trong thue te'ta kh6ng bie't yd(t), ma chi bie't cac ihi hien cua no y'5(t) , ...,y'"8 (t) va ,
y'6(t) e C[a,b] , (i = 1-^ m) sao cho:
II y ^ -yd II < 5 v a i j - T , r

(2.2)

c
d day m > r > m/2 1 < kj < m
Gia su vai moi vi,V2 e K(n), A la loan tir thoa man dieu kien.
(2.3)
II Avi-Av2||c
trong do r = max {n[vi ],n[v2 ]) ; 4^(l,r) > 0 la ham lien tuc, dan dieu tang theo cac
bie'n, vai 0 < t,r < +00, 4^(t,r) -> 0 khi t ^ 0
Goi S(n,h) e K(n) (Vn > 1) la (p(h) luai huu ban cua tap compact K(n), c6
nghia la:
Vxe K(n), 3 Xh e S (n,h) : ||xh - x|| < (p(h)
Trong do 0 < (p(h), la ham dan dieu tang tren ( 0,1 ]

va lim (p(h) - 0
h->0

Xet phie'm ham Ibi

b m
b
(l)6(y) = J Z P6 [ y's (t) - y(t)]dt 4 Rg (t,y(t)) dt
a

i =1

a

in

6 day

R5= E PsCy'sCO-x)

s V (2 k)

I sI < k

I s I >k
Trong d6 0 < K < C.6 , 0 < C - Const


-7-

Ham (jjg (y) c6 tinh cha't sau (xem [5])
i/ T6n tai ygCt) e C[a,b] : (t)s(y5) = Inf {(^sCy) : y e C[a,b] )
ii/ 3 V > 0

:

llyc-ysllc ^ V.5

iii/ 3 G > 0 :

I(|)8(yi)-
iiii/ 3 Q
:
aR8(t,T)

Vt < [a,b] , 5 > 0 , to
f >lkhiTo > yd(t) +Q6

5T

iiiii/

T =To I < 1 khi to < yd (t) - Q5

aR8(t,x)
<

m , V T , t e [a,b], 5 > 0

5x
S3 - Thuat toan Compact thu hep dang Robust.
Gia sir ta c6 he thiic.
b

m

(t)6(yd) = J Z P5(y'5(t)-yd(t))dt<©(8)
a

(3.1)

i^l

a day a)(5) la ham lien tuc, khong am hoi tu de'n 0 khi 5 - ^ 0 .
Nhan xet 3.1: Dfeu kien (3.1) hiin nhien duac thue hien trong tru&ng hap (1.2) thoa
man vai moi i ( i = l,m). Trong truong hap c6 cac ^§(1) khong thoa man (1.2) nhung
thoa man danh gia:
11^5-yd II

<5,

(3.2)

LPJ

Of day Pj > 1 .

,

^

Khi do di'eu kien (3.1) van dugc thue hien. Vay (3.1) dugc thue hien ngay ca trong
truong hgp c6 nhung tG[a,b] sao cho I y'5(t) -yd(i) I c6 the Ian tuy y.

De dang chiing minh dugc cac danh gia sau:
(l)5(yd)i= 1

Li


8-

||y'5-y|| < ( b - a ) | | y ^ 8 - y | | , | | y s - y | | < (b-a)^^^M|y5-y||
Li

C

Ll

Lp

d dayqj> 1 : l/qj+ 1/pj = 1, Tir he thue (2.2 ) va (3.2). suy ra he thii-c (3.1).
Gia su 5 > 0 la m6t sC c6 dinh tuy y ta chon day hn > 0 va so N = N (5) sao cho.
(p(hN-0 >5>(p(hN)
So db tinh toan theo phuang phap Compact thu hep cai bien dang Robust gbm cac
buac sau day:
Budfc 1, Chon ri = 1 va dat
V, {VE S(r,,hi) : (|)5(Av) < GT((p(h,),ri) + co(5)l
Ne'u V] = (j)

la'y

r2 = ri +1

Ne'u Vi 9^ ([)

la'y T2 = ri sau do ihirc hien cac buac tie'p theo.

Birdc n < N, Vdi r„ dugc xac dinh tu budre truac.
Vn = i ve S (rn, hn) : ^^ (Av) < GT((p (hn), rn) + CO (5)}
Ne'u Vn = (t)
Ne'u

Vn ^ ^

lay
la'y

rn+i = rn +1
Tn^-i = rn

Birdfc N, VN = V5 :=: {ve S(rN, hn) : ^^(Aw) < GT((p(hN),rN) + co(5))
Ta chutig minh su hoi tu cua phuang phap nay theo luge do sau:

Bo de 3.1 Gia sir vai v e H c6 dinh thoa man he thu"e
(j)5(Av) ^ 0 (5 -> 0 )
thi
va

I (t)8(Av) - Uyd 1 ^ 0 (5 -> 0 )
V = Xd

Chung minh day dii bd de nay xem trong [5 ].

Bo de 3,2: T6n tai s6' No = No (Xd) sao cho
V N > N o = > Q[Xd] Chvcng minh: Ta chung minh bang phan chung.
Gia sir ngugc lai: V n 3 N > n : r N < ^[Xd]


-9-

Ta dat ni = [Q[Xd] ] + 1, khi do tlm dugc Ni va mi < Ni sao cho Vmi ^ ^.
That vay, ne'u kh6ng, theo each xay dung tren ta c6 V^ = Ni > ni > n[Xd] mau thuan
vdi gia thie't.
Tuang tu, vdi ni = Ni + 1, tbn tai s6' N2 sao cho v6i mi < m2 < N2, Vm2 ^ (j) .v.v..
Tie'p tuc tie'n hanh nhu tren ta dugc day tap hgp {Vmk )"° khac r6ng. Chon day {Vn )
k= 1

ba't ky trong V m , n = l , 2 . . . taco:
n[Vn] n

Vi Tm la so nguyen nen ton tai s6' e > 0 sao cho
n[vn]Ta CO (Vn ) d D := { V e Xi : Q[v] < rd). Do D la tap hgp compact trong X, nen
tbn tai day {Vk } c (Vn} sao cho Vk —> VQ. Khong ma't tinh tong quat ta c6 the coi
n

11

Vn —> Vo ( n -^ 00 ), va do do:
Q [ V o ] < Q[Xd] - £ ^

(3.3)

Mat khac :
(l)5(AVn ) < G T ( ( p ( h ^ ) , Tn, ) + C0(5) < GT((p(hn, ),n[Xd]) + C0(5) - ^ 0 (n ^
n
n
n

O) )

Tu tinh lien tuc cua (|)s va A suy ra.
(j)5(AVo)< co(5).
Theo bd rfe (3.1) Vo = Xd e Xmau thuan vai (3.3)
Bd d'e dugc chung minh [].
Bd de 3.3: Vdfi moi sd tu nhien N > No tap VN khong rdng va chira trong tap
compact K
Chirng minh:
Tur bd d'e (3.2) suy ra vbi moi N > No, fi[Xd] < TN va do do Xd e

K{T^).

gia surxh e S(rn,hn) sao cho ||xh - Xd|| < (p(hn)
Theo tinh cha't ciia ham (t)8 va toan tii: A, ta c6 danh gia:
0 < (|)5(Axh) < I (t)5(Axh) - (1)5 (AXd) I + (|)6(Axd) < G ^ (II Xh-Xd || ,rN ) + co(5)


-10

< G ^ ( ( p ( h N ) , r N ) + 0D(5)

Suyra: Xh e

VN=V5

Bd d'e dugc chi^ng minh [].
Nhan xet 3,2: VN ^ i?, doi vdi moi N > No ta luOn c6 rw < rNo ^ No va
VN <= K (No) == K (K - Tap compact)
Bd de 3.4: Day tap compact {V5) co Ve didm Xd khi 5 -> 0
Chvcng minh: Gia sir ngugc lai di^eu do kh6ng xay ra, khi do tbn tai hai day{Vk")va
so 8 >0 dd:
||vk"-Vk-||>s>0

(3.4)

Vai Vk* e VNk c: Ko, k = 1,2 .... Do Ko la tap Compact nen khCng giam tdng quat
chung ta c6 the coi rang

Vk" -^ v"

khi k -> o)

Khi do ta cung c6 Avk" -^ Av"*". Mat khac
(|)5 (Avk") < GT((p(hNk ),rNk) + co(5).
Tir do suy ra.
0 < (1)5 (Av*) < co(5) -> 0 ( 5 -> 0)
Theo bd de (3.1) ta c6:

v* = Xd e X . Dieu nay mau thuan vox (3.4 ).

Nhu vay ta c6 diam V5 -> 0 ( 5 -> 0)
Bd de dugc chung minh [].
Ne'u ta lay x^ e V5 = VN ( N > No) tiiy y, do V5 ehira Xh:
II Xh - Xd II < (p(hN) nen la thu dugc
||x5-Xd|| < ||x6-Xh|| + ||xh-Xd|| < diam V5+ cp ( hw)
D o d o ||xs-Xd||
(3.5)

Nhu vay ta da chung minh dugc dinh ly sau:
Djnh !y 3.1: Phuang phap compact thu ht^p dang Robuts h(3i tu va ta c6 danh gia.
IIX5 - Xd 11 < diam Va + 5
Nhan xet 3.3:
Ne'u phie'm ham dn djnh Q[x] thoa man di'eu kien Q[x] > C ' | | x | Vx e X
(C-const > 0 ), thi (2.3) c6 ihi thay bang di^eu kien.


11

V Xi,X2 e X : || Ax, -Axsjl < T(||xi - X2II, r)

( 3.6)

6 day r - max (|| Xi ||,1| X2II). Khi do tat ca chung minh tren van dung, ne'u ta la'y tap
hap Vn nhu sau:
Vn = ( V e S (rn,hn) : ^^ (Av) < G ^ (cp (hn),C.r ) + co(5) )
D^ y rang phuang phap compact thu hep [24] la truong hgp rieng cua phircng
phap nay khi n = 1, Q[x ]= ||x|| 1 va ||y5 - yd|| < 5, a day ||. || 1 la chu^n nao do irong
kh6ng gian con Xo, trii mat va compact trong X
Nhan xet 3.4: Ne'u thay ddi each xay dung tap Vn mot chut thi thuat loan cua ta khong

nhat thie't phai thue hien dung N but^c nhu da trinh bay a tren.
BireJc 1: Lay r, = 1 va thanh lap tap hgp:
Vi-{v:veS(hn,ri):(|)8(Av) < G^F(5,ri ) + co(5))
Ne'u Vi ^ (|), ta la'y tuy y xs e Vi
Khi do: || X5 - Xd || < diam Vi + cp ( hN) ^ diam Vi + 8
Thuat loan dimg a day
Ne'u Vi ^ (|), ta la'y r2 = ri + 1 va thue hien buac 2 ....
Budc ( n < N) dugc thirc hien ne'u

Vi,... V„-1 = ([>

Khi do rn = n, ta thanh lap tap Vn
Vn-{v:veS(hn,rn):(l)6(Av) < GT(5,rn ) + co(5))
Ne'u Vn ^ (t>, ta la'y xs e V, tuy y, thi
II xg - Xd II < diam Vn + 5 va thuat toan dung lai.
Ngugc lai Vn = (j), dat rn = n +1 va quay lai h\x6c n cho de'n khi n < N thi dung
Nhan xet 3.5: Ne'u rn > n[Xd ], thi V„ ?^ cj) khi do thuat toan a nhan xet (3.4) dimg lai.
Dfeu nay chung to: Ne'u Qxd la mot so nguyen thi so bubc cua thuat loan la fifxd ].
Ne'u il[Xd ] khong phai la so nguyen thi so' bu6e cua thuat toan la [Q[Xd ]]+ 1. Vay so
bu(5fc ciia thuat toan khOng phu thu()c vao 5 khi 5 du be.
Nhan xet 3.6: Ne'u tien nghiem bie't rang Q[Xd] < R thi ta thanh lap tap
V5={v:veS(hN,R):(|)5(Av) < G4^(6,R)+co(5)}
va thuat toan chi c6 mot bu6e.


12

Nhdn xet 3.7: Ta c6 thd thay gia thie't (3.1) bang gia thie't i^^{y^) < (3(5), trong do
P(5) > 0 la ham lien tuc, p(8) ^ 0 khi 5 ^ 0. Khi do c6 ihi la'y tap V„ nhu sau:
•

Vn = ( V e S(rn,hn) : U^v)

< GT((p(hn),rn ) + Vs + co(5))

Cac ke't qua trinh bay b tren van con dung trong truong hgp nay

§4 - Truotig h(ifp khong duy nhat nghiem:
Trong muc nay, ta't ca cac gia thie't cua muc 52 trtr gia thie't (1.1) c6 nghiem
duy nhat dugc giu" nguyen.
Goi U = {x e Xi : Ax = Axd = yd) la tap hgp nghiem cua (1.1) Gia su trong
Utbn tai duy nhat x* sao cho
Q[x*] = minf^[x]
xe U
TrU(Je he't ta xet thuat toan cai bien sau:
Vain > l,dat:
V„ = {v:v e S(hN ,rn),: (f)8(Av) < G^(5,r„) + co(5)}
Ne'u Vn = ([), ta la'y rn +1 = rn +1 va thue hien buac tie'p theo.
Ne'u nN < N, Vn 5^ (j) va Vn = (|), v(jfi n = 1,2 .... nN - 1, thi bube tie'p theo dugc lien
hanh nhu sau:
1

LtfyTn+i = r,^

'^\~

")^^^^y<^v^gtap

Vn^ + i = { v : v eS(hN,rn^+i):(t)6(Av)Ne'u Vn^+1 = (j), thi thuat toan dimg lai va ta la'y vg e Vn tuy y. Ngoai ra

||x*-V6|| N

1
Ne'u Vn^+, ^^ftaia'y r^+2 = r„ + i
N
va tie'p tuc thue hien h\x6c n + 2
Tom lai thue hien thuat toan de'n buac nN+ m ta eo:


-13

Vi, ... , V ^ . 1 = (() , V„^ V,j^ + l , ..., V,^+ m-1

'

^^

Khi do thuat toan dimg lai va ta la'y vg e V,, + ,„-1 tuy y, ngoai ra
\^

T(5) =diamV,, . n , - i + 5
N

V<5i each xay dung thuat toan nhu tren ta c6 ke't qua sau.
Djnh ly 4.1: Phlln tir V5 dugc chon a tren c6 tinh cha't.
a) | | v s - x * | | -^ 0 khi 8 ^ 0
b) IIV5 - X* II < T(8) trong do T(8) ^ 0 khi 8 -> 0
Dd chung minh dinh ly ta can ba bd de sau:
Bd de 4.1 : 3 No = No (Xd) , VN > No => rN > 0[x*]

Chiing minh bd de nay hoan toan lap lai chung minh bd de (3.2).
Bd de 4.2 :
Vn +m-1 chiia trong tap compact Ko,
N

eon ke't luan cua bd de 4.2 suy ra tmc tiep tu each xay dung tap V„
Bd de 4.3 :
Ta CO (m > 1) lim diam V„ + „i-1 - 0
N

Chvcng minh : Gia sir ke't luan ciia bd rfe (4.3) kh5ng dung, khi do tbn tai
Vk" e Vn ^ + n^ -1 va sb e > 0

Sao cho llvk"*"-Vk'll > e > 0

vai k =: 1,2,...

Vi Vn e K (nk) - la compact, k - 1,2,... nen khong giam tdng quat ta gia sir rang:
Vk* ~ > Vo" k h i k - > 00

Tac6(i)5(Avk*) < G^((p(hnj^+ . i y i ) , r n ^ +n,^ i) + co(8)
K

K

Cho k - ^ 00, ta dugc: (j)6(AVo'^) < (x»(8)
T&bdd^e (3.1) suyra:
AVo"^ = yd = Ax*

(4.1)

-14-

Theo each xay dung thuat toan ta nhan dugc u6c lugng
m
nN •

=r
N

< Q[x*] < r
"N^"'

m
l
+ —•

= nN
"N*""'

N

N

Do nhan xet (3.5 ) thi nN = ^[x*] ne'u n[x*] la sd nguyen hoae nN = n[x*] + 1
trong tnrbng hcDfp ngugc lai.
a - Truong hcrp Cllx*] la sb nguyen thi m = 0 va
Q[x*] - [Q[x*]] = 0
b - Truong hcirp Q[x*] khong la sb nguyen thi.

m

m
< Q[x*] < [Cl[x*]] + 1 - —

N

N

[Q[x*]] + 1

I
+
N

Suy ra
m

m

1

1
+ —-

< n [ x * ] - [a[x*]] < 1 N

N

N

m

Theo each xay dimg thi 0 <

ms
m
< 1 nen tbn tai day c o n { — ) cz {
) sao cho
N
"
Ns
N

ms

ms
= a Do dinh nghia cua {

lim
ms

) ta cung eo.

nris

1

< n [ x * ] - [D.[x*]]<

1-

Ns

+
Ns

1
—
Ns

Cho s -> OO thi dugc he thiJc
1 - a < Q [ x * ] - [Q[x*]] < 1 - a

(4.2)

Theo each xay dung day ( Vk" } ta c6
mk

"N - ^ T -^f^^'' k

Suyra

Nk

mk

n
\

Nk

irik
1

Nk

mk
< n[vk* ] - [n[x*]] < 1

Nk

•+
Nk

Kh6ng mat tinh tdng quat coi k = s. Khi k -> oo ta duac

1
—
Nk


15
n[vo^ ] - [n[x']] - 1 - a
Tijr (4.2 ) suy ra

a[vo* ] = a [ X* ]

(4.3)

Til (4.1) . (4.3) va tinh duy nha't cua x* nen Vo" = x*. Difeu nay mau thuan vai gia
thie't phan chung.
Bd de dugc chung minh []
Djnh ly 4.1: Dugc suy ra true tie'p tu: bd de nay.

§5 - Truomg hgp ve phai va toan tur khong biet chinh xac
Trong muc nay chung ta gia thie't rang: Trong phuang trinh (1.1) ta khong bie't A ma
chi bie't A ^
A^ : X -> C[a,b] La toan tur lien tuc thoa man.
a - II A^v - Av|| < u(|a,Q[v]) V v e Xi, u(ja,s) la ham lien luc khOng am, kh5ng
giam theo JJ, va s, vbi mdi s cb dinh u(]a,s) —> 0 khi p ^ 0 va u(o,s) = OV s > 0
b - V v i , V 2 E X i : llA^Vi. A^V2|| < T ( | | v i - V 2 | | , r ) ;

Trong do r = max(n[vi ],Q[v2]), ham T(t,r) thoa man gia thie't trong S2
c - Ta van gia thie't (|)5( yj) = (t)5(AxT) < co(8).
Cac gia thie't khac giiJ nguyen nhu trong 52.
Chon day (hn) kh6ng tang; 0 < h„ < 1, hn —> 0 khi n —> oo, 8 va p ed dinh,
N = N(8,ii) chon tijr he thitc (p(hN. 0 > ji + 8 x p (hN ).
Budc 1 : Dat ri = 1 va xay dung tap hgp
V, = (v: V G S(h,,ri) : U^^'v) < G[T((p (hi),ri) + u(vi,ri) ]+ co(8))
Ne'u W\^^ia. la'y r2 = ri


16-

Ne'u Vi = (|) ta la'y r2 = ri + 1
Sau do thue hien budc tie'p theo
Bir6c n ( n < N )
Vn = {v: V e S(hn,rn) : UA^'v)

< G[T((p(hn),rn) + u(p,r„)] + co(8)}

Ne'u Vn ^ (|) ta la'y rn +1 = rn
Ne'u Vn = (j) ta la'y rn + i = rn + 1
BirdcN
VN - V5'' = {v:v e S(hN ,rN) : (|)6(A^v) < G[T((p(hN),rN) +u(p,rN)] + a)(S))
Ta chon

c 8 ne'u V^^ =^ ^
l^ V e Vg tuy y ne'u Vg^ ^ ^

Bo de 5.1 : T6n tai sb No = No (Xd) sao cho:
V N > No n[xd] < rN
Bd de nay dugc chung minh tuang tu bd de (3.2)
Bd de 5.2 : Vbi moi sb tu nhien N > No tap VN khong rdng va chu-a trong tap
compact Ko nao do.
Chtrng minh: Tur bd d& 5.1 suy ra v6i N > No; ^[Xd] < rN; Xd e K(rN) va do do tim
dugc Xh e S(hN,rN) sao cho
||xh

-XTII

<(p(h)< 8

Taco:
(t)5(A^Xh) < I (t)6(A^Xh) - ^5(A^Xd) I + I (t)5(A^Xd) - (t»6(AXd) I + (t)5(AXd) •
< GlJA'^Xh - A^Xdll+Gu(vi,rN)+ oj(S)
< G^(cp(hN),r) + Gu(p,rN) + 0(8)
< G T ((p(hN),rN) + Gu(p,rN) + co(8)
Trong do r = max(n[Xh],n[Xd]) < rN

Dieu nay chihig to xi, e VN hay VN ^ (j).
Bd dfe dugc chung minh.[]


17

Bd de 5.3 : Day tap compact (Vg^} co ve didm Xdkhi p, 8 -> 0
Chicng minh: Gia sir 6\tw do kh6ng xay ra, khi do tbn tai hai day{ Vk" 1 <= Vn va
sb 8 > 0 dd.
(5.1)

||vk*-Vk-||>e>0,k=l,2....

Do V c= Ko nen {Vk* ) c= Ko, kh6ng m&i tinh tdng quat chiing ta cb ihi coi rang:
Nk

Vk* ~> v"^ khi K - ^ 00

Taco:
U^x^-)

^ G[^((p( h ) , r ) + u(P, r ) ] + «(5)
k

k

k

Tur day suy ra:
lim

H-* 0

lim (|)5^(A^ Vk") = (l)6(Avo-) < co(8)
k -> 0

Theo bd d'e 3.1 ta co v = Xd
Dfeu nay mau thuSn vcfi (5.1). Nhu vay ta cb diam Vg -> 0 ( 8,|i -> 0 ).
Bd de dugc chiing minh [].
Do

Vg chiia Xh : ||xh - Xd||< q)(hN) nen ta co.
||x^5 - Xdll < 11x^^8 - Xh|| + II Xh - Xdll < diam V^g + cp(hN)

Hay

||x^g - Xdll < T(8,p) = diam V^g + 8 + u

(5.2)

Ta phat bidu ke't qua vua nhan dugc dudi dang dinh ly sau:

Dinh ly 5.1: Phuang phap compact thu hep dang robust v6i ve' phai la toan tir A bie't
g"an dung hOi tu va co danh gia (5.2) .
Nhdn xet 5.1: Ne'u chi bie't g"an dung A' va phuang trinh khong cb nghiem duy nhat,
ta cb thd ke't hgp cac phuang phap nghien cun trinh bay trong §4, §5.

§6- Ap dung cho phuang trinhvi phan thuong .
@ - Xet bai toan Cauchy dbi vai phuang trinh vi phan tuye'n tinh bac 2.
0^! HC:C QUDC GIA HA NOI
KT

T'J

^MJ^

liJiBlA^niHiiSTi?i.!r^V;


18-

Bai toan:
D[u(x)]

= f(x)

Ti u= u'(0) = (pi 0< x< 1

(6-1)

r 2 U = U(0) = (pL

Trong do D:d^^[0,l] -> C[0,1] la toan tu vi phan phi tuye'n bac 2 thoa man di^eu kien
||Dvi-Dv2||


V, -V2Lr),V vi,V2 e ^'^[0,1]

d dayr = max{|lvi|U|v2|| ) , 0 < vKt,r) la ham lien tuc, khong giam theo t e var,
0 < t,r < + 00 , va M^(0,r) = 0, Vai r > 0 .

Gia sit bai toan (6.1) eo nghiem duy nha't. Trong thue tb ta chi bie't cac thd hien cua
ve' phai y'(x)v.., y"'(x) va ditu kien bien (pi^ ,(p2^ thoa man.
a-||y^^J-f||

< 6 :V6i 1< Kj < m,

m/2
b - I (p^-(pi I < 6 : Vbi i ^ 1,2.
Ta cung gia thie't rang ( xem (3.1)) .
(t»8(Du) - <)|g (/) < co(6)
6 day co(6) -> khi 6 -> 0. Dual day ta se sur dung cac chudn || . || 2, || J ^^1. dugc xac
dinh nhu sau
2
Vbix e C'iOA]: \\ x||2 = max { || x||^|| x||^|| x||^ },aday 1| .||^aehuSn Qo.i].
v b i x e | | x | L i : | | x | L 1 =||xl|
"^2

"^2

+11x11
L2[0.1]

L2[0,l

@ - Tuong tu rai rac va cac tinh chat
Gia sir u(x) e W2^[0,l] la mCt ham cb dinh thoa man |lu||2 ^ S,

0 < p - const

Cho hai sb h, x e > 0 cb dinh, ta cb luoi:
(Xi,yk) : Xi = ih, i = 0,1,2 ..., N ; sao cho Nh=: 1
S = S (h,x,p)
yk = kT ,k = 0 , ± l , ± 2 , . . . , ± p x - '
Goi M = M (h,x,p) la tap ta't ca cac ham luoi tren S, tiJc la cac ham lien luc tren
[0,1], tuye'n tfnh tren m6i khoang (xi. 1, Xi) i = 1,2,.... , N, con tai cac didm
Xi (i = 0,1 ... ,N) chung chi nhan cac gia tri trong tap {0, ± x ,± 2x ,.... ± px''}


-19-

Goi u"h t la ham lirori trfin S thoa man dieu kiSn.
I U"h,(Xi) I

^

I u"(Xi) I

I u"i,x(Xi)-u"(Xi) I < T

i = 0, 1,2,....,N

Dinh nghia: 6.1[20] 1, Ham u"h, (x) e M(h,T,p) dirge goi la tirong tir r6i rac cua ham
u"(x)eWj[0,l]
2, Ham

xt
Uh, {x)= \ i u"aTi)dTidt + u'(0)x + u(0), 0 < x < 1
0 0
•J

dugc goi la tuang tu rbi rac cua ham u (x) e W [0,1]
Bd de 6.1 [20]: Ne'u u(x) e W ^ [0,1], eon Uhx(x) la tuang tu rai rac ctia u(x), thi tbn
tai hang sb ho = ho(u) > 0 d^ ta co danh gia.
II Uhx -u||2^ V~h~ +x

khiO
@- So do tinh toan:
Cb dinh hai day sb tuy y {hn), {Xn), hn,x„ > 0, x„ < Co h„, hn ^- 0, Xn -> 0 khi n -> co
Xet day ludl S,i = S (h,i,T,i,r,i) trong do r,i se xac dinh sau. Goi M„ = M(hn,x,i,r„) la tap ta'l
ca cac ham ludi tren S,v
Kn(r) = { a ( x ) : a ( x ) e M „ : || a || i < r }
w
2

^

Uhg vbi mbi ham a(x) e Kn (r) a C[ 0,1] ta xet ham Va (x) e €"[ 0,1]
X

V = 1(a) =

t

J J a(y) dydt + cp,^ x + (p2\ 0 < x < 1
0

0

Sb buac N = N(6) cua thuat toan dugc xac dinh tu di^eu kien
V hN-1 + TN-1 > 6 ^ V hN + TN
Budc 1: La'y ri = 1 va xay dimg tap
Vi = {ve r'(v) e Ki(ri) :^^ (Dv) < G\\f{ V"h7 + xi + 25, r, + | q),M + I q)2M ) + w (6)}
Ne'u Vi = (j), ta la'y r2 = ri + 1
Ne'u Vi 1^ (j), ta la'y r2 = ri
va thue hien budfc tie'p theo


20-

Bu6cn: ( n < N )
Vn = (v : r*(v) e Kn(rn) : ^^{Dw) < G\\f {^TK-^-ZU + 26,r„ + I (piM + ICP2H ) + co (6))
Ne'u Vn = (j), ta la'y r„ ^.i = rn + 1
Ne'u Vn ^ ^, ta la'y rn -n = rn va thue hien buac n + 1 < N.
Birdc N:

Vn = V5 = { V : r Vv) e Kn (rn + Co) ; (t)6 (Dv) < GvK (36,rN + i cpi^ I + i ipi"" I) + co (6))
La'y u^ (x) e V^ tiiy y ta cb cac ke't qua sau.
Bo de 6.1 : T6n tai sb No(Ud) sao cho:
V N > N o : ||u"|| 1 < rw
2

Chitng minh:
Gia sur ngugc lai: Vn, 3 N > n : rw < ||u" || 1
w

2

Dat m = [ II u" II 1 ] + 1 ta tim dugc Nj > ni va mi < Ni sao cho W ^ ^. That vay, ne'u

w
2

khOng, theo each xay dimg tren ta eo rNi = Ni^ ni > ||u" || 1
w

2

Dieu nay mau thuin vai gia thie't.
Tuang tu vbi n2 - Ni +1, tbn tai sb N2 sao cho vai V2 ^ N2, Vm2=?^ (j),
Tie'p tuc tie'n hanh nhu tren la dugc day tap hgp {Vmn p khac rbng. Chon day
n-l

(vnlba't ky trong V -1,2,... taco:

||v"|| , < r^n < ||u"|| iViFn,
W^

n

w

n

la sb nguyen nen tbn tai sb s > 0 sao cho.

II^'IIV -ll""llw' -^ = '•'',11=1,2,...
Vi hinh cau dong trong W2'[ 0,1] la tap hgp compact trong C[0,1] nen tir day
{v"n} = (an) CZ C[0,1], tbn tai day {ak } e {an)sao cho
n

ak -> ao khi n -> 00 trong C[ 0,1]
n

Kh6ng ma't tinh tdng quat la cb the' coi
an -> tto trong C[ 0,1] khi n ^ 00 suy ra Vn -^ Vo trong C^[0,I] khi n -^ 00,


21-

Trong db.
X t

Vn = J J an(Tl)dTldt + ( p , \ + (p2^
'

0 0

Vi Vne VtTin nen (t)6(DVn) < G\jy(V hn + Xn + 25, Vn + I (pi^ I + I (P2^ I) + co(6)
Tu* tinh lien tuc cua ^^ va D suy ra
(t)6(Dvo) < Gy (36,rN + I (piM + I ^Pi" I) + co (5)
Theo b6 de (3.1) thi Vo = u. Nhung vi
llctnll , ^rd<||u"|| ,
nen

llaoll^i = | | u " | | ^ , = II v 1 | < r.< ||u"|| ^ .
2

2

2

Bat dang thiJe cubi cung mau thuan vai gia thie't phan chung.
Bo de dugc chiing minh.[]
Bo de 6.2: Vbi sb tu nhien N > No tap VN khong rOng va chiia trong tap compact Ko
nao db.
Chieng minh: Tu bd de (6.1) suy ra:
l|u"L ^ II"11 1 ^rN va||u",,|l
< rN
Lfo.i]
"w
Lro.i]
2

2

2

Theo each xay dung, tren mOi doan [Xi,Xi +1] u"ht(x)
CO dang ajX + bi, tiif da'y suy ra
u"'hx(x) = ai
Ta cb danh gia:
||u"hx|| 1 ^ rN+ Co
^2
XN

Vi I ai I = I tga I = I u"h,(x) I =

^ Co vai x e [xi, Xi+i],
hN

Nen suy ra u"h x (x) e
Xet ham.

KN

(rw + Co), N> No

.

-


22

X t

u V = J J u"hx(r|) dridt + cpi^x + (p2^ = Uhx (x) +((pi^ -(pi )x + ((p2^ - (p2)
0 0

d day Uh t(x) la tuang tu rbi rac ciia u(x)
Ta cb danh gia:
II u \ , - U II 2 ^ II Uh X - U II 2 + I (p 1 ^ - CPI I + I Cp2^ -(P2 I
< Vhn +Xn + 5 + 6 < 35
va

(6.2)

(|)6(Du\0 < U 8 ( D u \ , ) - (|)5(Du)| + U^u) < G | | D u \ , - Du||c+ co(5)

(6.3)

d day r = max{||u\,||2,||u||2) < rN + I cpiH + I (P2H + 25

(6.4)

Tijf (6.2), (6.3) va (6.4) la cb he thu:c.
^8(Du\,) < Gvj; (35, rN + I cpi^ I + I (P2M + 25 ) + co(5)
Suy ra u\ ^ eVn , hay

VN

^ ^.

Bd de dugc chung minh.[]
Bd de 6.3 Day tap compact {V5) co ve didm u khi 5 -> 0
ChAng minh:
Gia sur ngugc lai dieu db khong xay ra. Khi do ton tai hay day {a~n 1 c: V5 va s > 0
sao cho ||a'*'n - a'n||c > £ (6.5). Vi V5 c Ko = KN (rN + Co) la tap compact, nen
khOng giam tdng quat, la cb ihi coi rang a'^n -> a*o trong C[ 0,1] khi n -> 00
Khi do:

x t
v'^n = J J a'^n(ri)dridt + (pi ^x + (p2^-> v ^ (n-^00)
0 0

trong C^[ 0,1], b day v-„ (x) G VN
Mat khac
(|)5(Dv*n) ^ Gvj/ ( VTiI7 + XN + 25, rN +" I (pi^ I + I (P2^ I) + CO (5),

nen khi n ^ 00 ta co


23

(t)6(Dv*„) < Gv|;(35,rN + I (pi^ I + ICP2'' I) + co (5)
Tu bd rfe (3.1) suy ra v*n = Vo mau thuSn vdfi (6.5). Mau ihuan nay suy ra bd de
dugc chung minh.[]
Tijr cac bd de tren suy ra dinh ly sau:
Djnh ly 6.1: Vai moi 5 > 0 tbn lai sb N(5) sao choVN = V^ chiia Uhx ,vai moi u^ e V^
ta cb danh gia:
||u^ - uh^llu"^ - u \ x II2 + II u \ , - u II2 < diam V^ + 35
Nhdn x^t 6.1: Thuat loan vijra trinh bay de dang ma rOng cho truong hgp phuang
trinh vi phan bac n.
D(u(x)) = /(x)
0< X < 1
LiU - u'(0) - cpi
Trong db D : d"^[ 0,1] -> Qo ,1] la loan tur vi phan phi tuye'n bac n.
Nhdn xet 6.2: Ta cung cb ihi xay dung thuat loan vdri s6' budfc khCng qua n
( n < N ), hoae bie't ||u" | 1 < R ta cung eo thd xay dung thuat loan mot budfc nhu
^2
nhung nhan xet a cac muc tru6c.
CHUONG II
BAI TOAN TUYfiN TINH KHONG CHINH TREN COMPACT YEU
|1 - Mo dau
Gaponenko Yu.L (1989[27]) d"e xua't phuang phap xa'p xi tuofng thich giai bai
toan (0,1) trong truong hgp X = Y = L2[ 0,1] vdfi gia thie't lien nghiem | Xd(l) I ^ R V
t e [0,1]

Pham Ky Anh [71] St xua't y tuang mdfi: Six dung co sdf Fourier thay cho

Splines cb thd nhan dugc cac ke't qua tuang tu nhu [27] cho khong gian Hilbert bat
ky, han nUa khi irb lai trubng hgp L2[0,l], la giam nhe dugc dibu kien | Xd(l) | < R.
Sur dung y tuang cua Pham Ky Anh [71], chung la xet tnrong hgp bie't khai
tri^n ky di cua loan tu tuye'n tfnh hoan toan lien luc, ta cb Ihd danh gia m6t sb ir6c
lugng dn dinh ye'u va de xua't mCt thuat loan kidu khai tridn ky di chat cut dd giai