Một số phương pháp hiệu chỉnh giải bài toán đặt không chỉnh

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GIAO
DUG
VA
DAO
TAO
DAI HOC QUOC
GIA
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n^l/CJNG
DAI
HOC KHOA HOC
Tl/NHIEN
&
NGUYEN
VAN
HUNG
MOT
SO
PHUONG
PHAP
HIEU
CHJNR
GIAI BAI TO
AN
DAT
KHONG CIliMH

Chuy^n nganh
:
Toan
hoc
linh
(can
Ma
so :
1.01.07
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LUAN AN PHO TIEN
SI
KHOA HOC TOAN
- LY
NGUdl HUdNG
DAN KHOA HOC: TIEN SI -
PIIAM KY AN13
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-
1996
MUC
LUC
L51
noi
dau 1
CHUONG
I
PHUONG PHAP COMPACT
THU
HEP
CAI BIEN
§1.
Ma dau 5
§2.
Cac gia
thi^'t
cua bai toan
5

§3.
Thuat
toan compact thu hep dang Robust 7
§4.
Tnrcmg
hgp kh6ng duy nhat nghiem 12
§5.
Tnrcmg hgp ve phai va toan
tir
kh6ng biet
chinh
xac 15
§6.
Ap dung cho
phirong trinh
vi
phan thuong
17
CHUDNG II
BAI TOAN TUYEN
TINH
KHONG CHINH TREN COMPACT YEU
§1
Ma
dau 23
§2.
H6i
tu
y^'u
trong kh6ng gian Hilbert 25

§3.
Danh gia
tinh
6n dinh cua
nghiem trSn
compact
y6u
27
§4.
Phuang
phap khai
tridn
ky di chat cut 29
§5.
Phuang trinh tich phan
dang
tich chAp
31
§6.
Danh gia diam
Vs
trong
phaang
phap compact thu hep cua Gaponenko 36
CHUONG III
MOT
SO
PHUONG PHAP LAP - HIEU CHINH
§1.
Phirang

phap Gauss - Newton
hi6u chinh
(RGN) 40
§2.
Ki^m
tra
di^u
kien
B]
cua Bakushinski 41
§3.
Phuang
phap hieu chinh Gasse - Newton
g'an
dung 46
§4.
Phirang phap Seidel - Newton hieu chinh va bai toan phi
tuyS'n
cong
huang
52
I -
Phuomg
phap Seidel - Newton (SN)
vd phuomg
phap
Seidel - Newton hieu chinh ( RSN) 52
II -
Syc hdi
tu

dia
phuomg cua phuomg phap RSN 54
III -
Bdi
todn Men tudn hodn cho phuomg trinh Duffing - Van
derpol
57
nil
- Bdi todn bien tudn hodn doi
v&iphuomg
trinh Van
derpol
65
Phan ket
luan
66
Phu
luc
67
1-
Bai toan -
Lcfi
giai 67
2-
Thuat
giai va
chuong
trinh 72
Tai lieu
tham

khao gg
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.
I<c
tCr
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 CM.,
Khai niem dat chinh theo Hadamard la:
Xet phirang trinh
Ax
=
y (0,1)
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.
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
III:
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
BICN
$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-yd
||c<5
(l<ji<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].
OO
-6
Xi
c:
X va ta c6
Xi ==uK(n)
(2.1)
trong do K(n)
=
(v e
X] ,D.{v)
<n )
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
Axd-
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
vaij-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.
II
Avi-Av2||<T(||v,
-V2||,r)
(2.3)

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
s
I
< 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/
3V>0
:
llyc-ysllc ^
V.5
iii/
3G>0

:
I(|)8(yi)-<l)5(y2)l
<G
||y,-y2||,
Vyi,
y2
,
Vg
iiii/
3
Q : Vt < [a,b] , 5 > 0 ,
to
aR8(t,T) f >lkhiTo
>
yd(t)
+Q6
iiiii/
5T
aR8(t,x)
5x
T =To
I <
1
khi
to
<
yd
(t) - Q5
< m , V
T

, t
e
[a,b],
5>0
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
i^l
(3.1)
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||y'8-yd||
i=
1
Li
8-
||y'5-y||
<(b-a)||y^8-y||,||ys-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)

lay
rn+i = rn
+1
Ne'u
Vn
^ ^
la'y
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
I (t)8(Av)
-
Uyd
1^0
(5
->
0 )
va
V
=
Xd
Chung
minh day
dii
bd de nay xem trong [5 ].
Bo de
3,2:
T6n tai s6'
No = No (Xd)
sao cho
VN>No=> Q[Xd] <rN
Chvcng

minh: Ta chung minh bang phan chung.
Gia
sir
ngugc
lai:
Vn3N>n:rN< ^[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] <r,T,< n[Xd]
,
n

Vi
Tm
la so nguyen nen ton tai s6'
e
> 0 sao cho
n[vn]<Q[Xd]-s =
rd,n-
1,2
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[Vo]
<
Q[Xd]
-
£
^
(3.3)
Mat khac :
(l)5(AVn
) <
GT
((p(h^),
Tn,
)
+
C0(5) <
GT((p(hn,
),n[Xd])
+

C0(5)
-^
0 (n
^
O)
)
n
n n
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(hN),rN) + 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)
Dodo
||xs-Xd|| <diam
V5 + 5
(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)<GT(5,rn^+ i + co(5)
}
Ne'u
Vn^+1 =
(j),
thi
thuat toan
dimg
lai va ta la'y
vg
e
Vn
tuy y. Ngoai ra
||x*-V6||
<dimVn
+ 8
-T(8)
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)
||vs-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"
khi
k
->
00
Tac6(i)5(Avk*)
<
G^((p(hnj^+
.iyi),rn^
+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
ml
nN
-
=r
<
Q[x*]
< r

= nN +
—•
•
N
"N^"' "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 I
[Q[x*]]
+
1
<
Q[x*]
<
[Cl[x*]]
+ 1
-
— +
N N N
Suy ra
m m 1
1 <n[x*]
-
[a[x*]]
< 1
- + —-
N N N
m
ms
m
Theo
each xay dimg
thi 0 <
<
1 nen tbn tai day
con{—
)

cz {
) sao cho
N "
Ns
N
ms ms
lim =
a Do dinh nghia cua
{
) ta cung eo.
ms nris
1
1
<n[x*]-
[D.[x*]]<
1-
+
—
Ns Ns Ns
(4.2)
Cho s
-> OO
thi dugc he
thiJc
Theo each
1 -a<
Q[x*]-
[Q[x*]]
xay dung day (
Vk"

}
ta c6
mk
"N
- ^T
-^f^^''
-
k
Nk
<
n
1 -
\
a
mk
Nk
Nk
Suyra irik mk
1
1
<
n[vk*
] -
[n[x*]]
<
1 •
+
—
Nk Nk Nk
Kh6ng mat

tinh
tdng quat coi k
=
s. Khi k
->
oo ta duac
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-Vvi,V2EXi
:
llA^Vi.
A^V2||
<
T(||vi
-V2||,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 xp
(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)
<
GT ((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.
||vk*-Vk-||>e>0,k=l,2
(5.1)
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:
Tur
day suy ra:
U^x^-)
^ G[^((p( h
)
, r)
+
u(P,
r
) ] +
«(5)
k k k
lim lim
(|)5^(A^
Vk") = (l)6(Avo-)
<
co(8)
H-*
0

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)

r2
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||
<M;(||
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<j<m
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].
vbixe||x|L
i:||x|L
1
=||xl|
+11x11
"^2 "^2
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, ,±px-'
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 x t
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<h<ho,
0<x< +
co
@- 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)eM„:
|| 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 t
V
=
1(a)
= 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:
VN>No:
||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
v1|
<
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
KN
(rw +
Co),
N>
No . -

Xet ham.
22
X t
uV =
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 (6.2)
va
(|)6(Du\0
<
U8(Du\,)-
(|)5(Du)|
+
U^u)
<
G||Du\,
- Du||c+ co(5)
<Gv|/(||u\,-u||2,r)
+
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