tóm tắt luận án một số phương pháp hiệu chỉnh giải hệ phương trình toán tử đặt không chỉnh

BỘ GIÁO DỤC VÀ ĐÀO TẠO
VIỆN HÀN LÂM KHOA HỌC
VÀ CÔNG NGHỆ VIỆT NAM
VIỆN CÔNG NGHỆ THÔNG TIN



NGUYỄN ĐÌNH DŨNG


MỘT SỐ PHƯƠNG PHÁP HIỆU
CHỈNH GIẢI HỆ PHƯƠNG TRÌNH
TOÁN TỬ ĐẶT KHÔNG CHỈNH

Chuyên ngành: Toán học tính toán
Mã số: 62.46.30.01

TÓM TẮT LUẬN ÁN TIẾN SĨ TOÁN HỌC



Hà Nội – 2014
Công trình được hoàn thành tại Viện Công nghệ Thông tin – Viện Hàn lâm
Khoa học và Công nghệ Việt Nam.


Tập thể hướng dẫn khoa học:
1. GS.TS. Nguyễn Bường
2. TS. Nguyễn Công Điều

Phản biện 1:…………………………………………………………………….

Phản biện 2:…………………………………………………………………….

Phản biện 3:…………………………………………………………………….





Luận án sẽ được bảo vệ trước Hội đồng chấm luận án cấp Viện họp
tại:………………………………………………………………………………
………………………………………………………………………………….
………………………………………………………………………………….
vào hồi …… giờ… ngày……tháng……năm……
N
C
j
, j = 1, 2, , N H X
C
j
N
N = 1
N > 1
X X
∗
X
H
N = 1

A
j
: H → H
N > 1 U−
H
U−
A
j
(x) = f
j
, j = 1, 2, , N.
f
j
f
δ
j
A
j
: X → Y
j
, j =
1, 2, , N,
N

j=1
A
j
(x) −f
δ
j


2
+ αx −x
∗

2
→ min
X
,
A
1
A
j
: X → X U−
A
1
(x) + α
˜µ
N

j=2
(A
j
(x) −f
δ
j
) + α(x −x
∗
) = f
δ

1
α = α(δ) ˜µ ∈ (0, 1)
A
1
U−
U−
U−
A
j
(x) = f
j
, ∀j = 1, 2, , N
x
0
∈ D (1.11) A
j
, j = 1, 2, , N
D X
Y
j
f
j
∈ Y
j
S
j
= {¯x ∈ D : A
j
(¯x) = f
j

}, j = 1, 2, , N, S = ∩
N
j=1
S
j
S = ∅ A
j
S
j
S X
f
j
f
δ
j
j
f
j
− f
δ
j
j

Y
j
≤ δ
j
.
A
j

, j = 1, 2, , N
(1.11) f
δ
j
j
N

j=1
A
j
(x) −f
δ
j
j

2
Y
j
+ αx −x
∗

2
X
→ min
D
,
α x
∗
∈ X\S
j

A
j
(2.2)
(2.2)
(i)
(ii) (2.2) (1.11) α, δ
j
→ 0
α > 0, f
δ
jk
j
→ f
δ
j
j
δ
j
≥ 0 x
k
(2.2) f
δ
j
j
f
δ
jk
j
{x
k

} ˜x
(2.2)
(2.2) (1.11)
δ
j
= δ
α(δ) α(δ) → 0,
δ
2
α(δ)
→ 0 δ → 0 {x
δ
k
α
k
}
(2.2) δ
k
→ 0, α
k
= α(δ
k
)
x
∗
x
∗
x
0
l

im
δ→0
x
δ
α(δ)
= x
0
x
δ
α(δ)
→ x
0
δ → 0
{x
δ
α(δ)
}
A
j
, j = 1, 2, , N
x
0
A
1
L > 0 A

1
(x
0
) −A


1
(z)
Y
1
≤ Lx
0
− z
X
z
x
0
x
0
z U
ω ∈ Y
1
x
0
− x
∗
= A

1
(x
0
)
∗
ω
Lω

Y
1
< 1
α ∼ δ
p
, 0 < p < 2 x
δ
α(δ)
− x
0

X
= O(δ
1−
p
2
)
(1.11) f
j
A
j
f
δ
j
j
A
h
j
f
δ

j
j
(2.1) A
h
j
A
h
j
(x) −A
j
(x)
Y
j
≤ hg (x
X
) ,
g(t), t ≥ 0 A
h
j
A
j
(1.11)
N

j=1
A
h
j
x −f
δ

j
j

2
Y
j
+ αx −x
∗

2
X
→ min
D
.
(2.15) x
h,δ
j
α(h,δ
j
)
(2.15) x
0
(1.11) α, h, δ
j
→ 0
α > 0 h
k
→ h > 0 A
h
k

j
A
h
j
(2.14) f
δ
jk
j
→ f
δ
j
j
δ
j
≥ 0 x
k
(2.15) f
δ
j
j
A
h
j
f
δ
jk
j
A
h
k

j
{x
k
} ˜x
(2.15)
(2.15) (1.11)
δ
j
= δ
α(h, δ) α(h, δ) → 0,
δ
2
α(h,δ)
→ 0,
h
2
α(h,δ)
→ 0 h → 0, δ →
0

x
h
k
δ
k
α
k

(2.15) h
k

→ 0, δ
k
→ 0, α
k
=
α(h
k
, δ
k
) x
0
x
∗
− x
0
lim
h,δ→0
x
hδ
α(h,δ)
= x
0
x
hδ
α(h,δ)
→ x
0
h, δ → 0
A
1

L > 0 A

1
(x
0
) −A

1
(z)
Y
1
≤ Lx
0
− z
X
z
x
0
x
0
z U
ω ∈ Y
1
x
0
− x
∗
= A

1

(x
0
)
∗
ω
Lω
Y
1
< 1
α ∼ (h + δ)
p
, 0 < p < 2
x
hδ
α(h,δ)
− x
0

X
= O((h + δ)
1−
p
2
).
x
0
∈ X (1.11) A
j
j = 1, , N,
X Y

j
f
j
∈ Y
j
S
j
= {¯x ∈ X : A
j
¯x = f
j
}, j = 1, , N, S = ∩
N
j=1
S
j
, S = ∅.
A
j
S
j
X
S f
δ
j
j
δ
j
(2.1)
j

A
j
x −f
δ
j
j

2
Y
j
+ αx −x
∗

2
X
.
x
δ
j
α
x
δ
j
α
→ x
0
δ
j
, α → 0
α = α (δ

j
) x
δ
j
α(δ
j
)
→ x
0
δ
j
→ 0
x
δ
j
α(δ
j
)
− x
0

X
x
0
x
∗
(1.11)
min
x∈X
N


j=1
A
j
x −f
δ
j
j

2
Y
j
+ αx −x
∗

2
X
,
A
j
j = 1, 2, , N (2.28)
(2.28)
α > 0, δ
j
≥ 0, f
δ
jk
j
→ f
δ

j
j
k → ∞ x
k
(2.28)
f
δ
j
j
f
δ
jk
j
{x
k
} (2.28)
α(δ) α(δ) → 0,
δ
α(δ)
→ 0 δ → 0

x
δ
α

x
0
x
∗
(1.11) x

δ
α
(2.28)
ω ∈ Y
1
x
0
− x
∗
= A
∗
1
ω
α ∼ δ
p
, 0 < p < 1
x
δ
α(δ)
− x
0

X
= O(δ
p/2
).
A
j
A
h

j
A
j
x −A
h
j
x
Y
j
≤ hCx
X
,
C A
h
j
A
j
(1.11)
N

j=1
A
h
j
x −f
δ
j
j

2

Y
j
+ αx −x
∗

2
X
→ min
D
.
A
j
, j = 1, , N (2.36)
(2.36)
α > 0 h
k
→ h > 0 A
h
k
j
A
h
j
(2.35) f
δ
j,k
j
→ f
δ
j

j
A
h
k
j
→ A
h
j
k → ∞ δ
j
≥ 0 x
k
(2.36) f
δ
j
j
A
h
j
f
δ
j,k
j
A
h
k
j
{x
k
} (2.36)

δ
j
= δ α(h, δ)
α(h, δ) → 0,
δ
α(h, δ)
→ 0,
h
α(h, δ)
→ 0
h, δ → 0 {x
h,δ
α(h,δ)
} (2.36) x
0
x
∗
−
(1.11)
ω ∈ Y
1
x
0
−x
∗
= A
∗
1
ω
α ∼ (h + δ)

p
x
h,δ
α(h,δ)
− x
0

X
=

O

(h + δ)
1−p/2

1 ≤ p < 2
O

(h + δ)
p/2

0 < p ≤ 1.
Bx = f,
B : H → H
B
Bx −By, x −y ≥ 0, ∀x, y ∈ H;
B
∃L > 0 : Bx − By ≤ Lx −y, ∀x, y ∈ H.
(2.41)
x

(k+1)
= x
(k)
− β
k
(Bx
(k)
+ α
k
(x
(k)
− x
(0)
) −f), x
(0)
= x
∗
,
f
˜
f
δ
f −
˜
f
δ
 ≤ δ,
x
(k+1)
= x

(k)
− β
k
(Bx
(k)
+ α
k
(x
(k)
− x
(0)
) −
˜
f
δ
) , x
(0)
= x
∗
.
Bx
(K)
−
˜
f
δ

2
≤ τδ < Bx
(k)

−
˜
f
δ

2
, 0 ≤ k < K, τ > 1.
x
(K(δ))
→ x
0
δ → 0 x
0
x
∗
−
(2.41)
B {α
k
} {β
k
}
α
k
> 0, α
k
 0,
α
k
− α

k+1
α
3
k
α
k+1
≤ 1, β
k
:= cα
k
, c > 0
(1 −cλ)cα
2
0
≤ 1, λ :=
α
2
0
+ L
2
2
.
x
0
τ > 1
ω := x
0
− x
(0)
 ≤

[1 −cλ −(2α
0
/c)] (
√
τ −1)
2
α
0
(1 + α
3
0
) (1 + α
2
0
) + 2α
0
,
k = 0, 1, , K(δ)
l :=

c(1 −cλ −2ω/(
√
τ −1)
2
) −2α
0

(
√
τ −1)

2
(1 + α
3
0
) (1 + α
2
0
) c
,
z
(k)
− x
(k)

α
k
≤ l,
z
(k)
Bz − f + α
k
(z − x
(0)
) = 0, α
k
> 0,
K = K(δ)
B(x
(K)
) −

˜
f
δ

2
≤ τδ < B(x
(k)
) −
˜
f
δ

2
, 0 ≤ k ≤ K, τ > 1,
lim
δ→0
x
(K(δ))
− x
0
 = 0 lim
δ→0
K(δ) = +∞
A
j
x = f
j
, j = 1, 2,
A
1

=



1 2 −1
2 0 1
3 2 0



, f
1
=



2
3
5



,
A
2
=



1 −2 −1

−2 1 0
−1 −1 −1



, f
2
=



−2
−1
−3



.
(2.54) x
0
= (1; 1; 1) det(A
j
) = 0 j = 1, 2
(2.54)
(2.54)
A
1
x −f
1


2
R
3
+ A
2
x −f
2

2
R
3
+ αx −x
∗

2
R
3
→ min
R
3
,
x
R
3
=

x
2
1
+ x

2
2
+ x
2
3
x = (x
1
; x
2
; x
3
) ∈ R
3
. (2.55)
Bx + α(x −x
∗
) =
˜
f,
B = A
∗
1
A
1
+ A
∗
2
A
2
,

˜
f = A
∗
1
f
1
+ A
∗
2
f
2
, x
∗
∈ R
3
,
x
∗
= (0; 0; 0) A
∗
1
=



1 2 3
2 0 2
−1 1 0




, A
∗
2
=



1 −2 −1
−2 1 −1
−1 0 −1



.
A
2
ω = (−1; −1; 0)
x
0
− x
∗
= A
∗
2
ω det(B) = 996 (2.56)
x
(0)
= (2; 2; 2)
α x

α
1
x
α
2
x
α
3
x
α
− x
0

R
3
A
h
j
= A
j
+ H, f
δ
j
= f
j
+ ∆
H =




0 h/3 h/2
h/2 h h/3
h h/4 h/4



, ∆ =



δ/
√
3
δ/
√
3
δ/
√
3



.
x
h,δ
α(h,δ)
= (x
α,h,δ
1
; x

α,h,δ
2
; x
α,h,δ
3
) (2.54)
A
h
1
x −f
δ
1

2
R
3
+ A
h
2
x −f
δ
2

2
R
3
+ α(h, δ)x −x
∗

2

R
3
→ min
R
3
,
B
h
x + α(h, δ)(x −x
∗
) =
˜
f
δ
,
x
∗
∈ R
3
, B
h
= (A
h
1
)
∗
A
h
1
+ (A

h
2
)
∗
A
h
2
,
˜
f
δ
= (A
h
1
)
∗
f
δ
1
+ (A
h
2
)
∗
f
δ
2
.
x
(0)

= (0; 0; 0)
h δ α x
α,h,δ
1
x
α,h,δ
2
x
α,h,δ
3
x
h,δ
α(h,δ)
− x
0

R
3
10
−1
10
−1
0.2000 0.985296 0.977224 0.961036 0.047467
10
−2
10
−2
0.0200 0.998553 0.997783 0.996333 0.004523
10
−3

10
−3
0.0020 0.999856 0.999779 0.999636 0.000450
10
−4
10
−4
0.0002 0.999986 0.999978 0.999964 0.000045
α = (h + δ), x
∗
= (0; 0; 0)
det(B) = 0 Rank(B) = 3
A
j
f
j
j = 1, 2
A
1
=






0.1 −0.2 0.1 −0.1
0.2 −0.1 0 0.2
0.3 −0.3 0.1 0.1
0.1 0.1 −0.1 0.3







, f
1
=






−0.1
0.3
0.2
0.4






A
2
=







0.1 0.2 −0.1 0.1
0.2 −0.1 0 0.2
0 0.3 −0.2 0.4
−0.1 0.5 −0.3 0.5






, f
2
=






0.3
0.3
0.5
0.6







(2.54) x = (1; 1; 1; 1) x

=
(1; 3; 6; 2) (2.54) x
0
= x

+ t (x −x

) x
0
x
∗
t
t
x
0
− x
∗

R
4
→ min,
x
∗
= (x
∗

1
; x
∗
2
; x
∗
3
; x
∗
4
) t = (8 − 2x
∗
2
− 5x
∗
3
− x
∗
4
)/30.
f
j
f
δ
j
= f
j
+ ∆, j = 1, 2, ∆ = (δ/2; δ/2; δ/2; δ/2) ,
˜
f

δ
= A
∗
1
f
δ
1
+ A
∗
2
f
δ
2
.
(2.46)
x
(k+1)
= x
(k)
− β
k
(Bx
(k)
+ α
k
(x
(k)
− x
(0)
) −

˜
f
δ
) , x
(0)
= x
∗
∈ R
4
,
β
k
= cα
k
, α
k+1
=
α
k
1 + α
3
k
, k = 0, 1, 2, ,
(2.47)
Bx
(K)
−
˜
f
δ


2
R
4
≤ τδ < Bx
(k)
−
˜
f
δ

2
R
4
, τ > 1, k = 0, 1, , K − 1.
(2.48)
(2.49) (2.50)
L = B
R
4
= 1.1847, α
0
= 0.1, λ =
α
2
0
+ L
2
2
= 0.7068, c =

1
2λ
= 0.7075.
τ (2.50)
τ =


x −x
(0)

R
4
[(1 + α
3
0
) (1 + α
2
0
) + 2α
0
]

1 −cλ −2
α
0
c

α
0
+ 1


2
= 133.57463.
x
(0)
= x
∗
= (0; 0; 0; 0) t =
4
15
x
0
=

1;
7
15
; −
1
3
;
11
15

x
∗
−
n K x
(K)
1

x
(K)
2
x
(K)
3
x
(K)
4
1 0 0.000000 0.000000 0.000000 0.000000
2 0 0.000000 0.000000 0.000000 0.000000
3 16 0.177032 0.245830 −0.207462 0.545650
4 77 0.435608 0.225849 −0.259445 0.845527
5 10821 0.678054 0.280364 −0.286240 0.870472
6 439824 0.842983 0.367841 −0.310054 0.814587
7 15936490 0.940141 0.427954 −0.324409 0.766137
x
0
δ = 10
−n
, n = 1, 2,
n K Bx
(K)
−
˜
f
δ
 τδ x
(K)
− x

0

1 0 1.089943 13.357463 1.366260
2 0 1.004651 1.335746 1.366260
3 16 0.361763 0.133575 0.881541
4 77 0.114956 0.013357 0.628154
5 10821 0.036548 0.001336 0.399228
6 439824 0.011567 0.000134 0.203875
7 15936490 0.003655 0.000013 0.078987
x
0
δ = 10
−n
, n = 1, 2,
A
j
f
j
A
h
j
f
δ
j
A
h
j
= A
j
+ H, f

δ
j
= f
j
+ ∆, j = 1, 2,
H =






h/2 h/3 h/4 h/5
h/3 h/4 h/5 h/6
h/4 h/5 h/6 h/7
h/5 h/6 h/7 h/8






, ∆ =






δ/2

δ/2
δ/2
δ/2






B
h
= (A
h
1
)
∗
A
h
1
+ (A
h
2
)
∗
A
h
2
,
˜
f

δ
= (A
h
1
)
∗
f
δ
1
+ (A
h
2
)
∗
f
δ
2
.
x
(k+1)
= x
(k)
− β
k
(B
h
x
(k)
+ α
k

(x
(k)
− x
(0)
) −
˜
f
δ
)
L = B
h

R
4
α
0
= 0.05 λ =
α
2
0
+L
2
2
c =
1
2λ
τ (2.50)
τ =



x−x
(0)

R
4
[(1+α
3
0
)(1+α
2
0
)+2α
0
]
(
1−cλ−2
α
0
c
)
α
0
+ 1

2
= 145.8741.
n K x
(K)
1
x

(K)
2
x
(K)
3
x
(K)
4
1 0 0.000000 0.000000 0.000000 0.000000
2 0 0.000000 0.000000 0.000000 0.000000
3 29 0.167796 0.240586 −0.200835 0.525637
4 96 0.393307 0.251553 −0.267696 0.835876
5 7009 0.670939 0.277273 −0.285218 0.871628
6 380309 0.837765 0.364742 −0.309287 0.816961
7 13912613 0.937705 0.426402 −0.324047 0.767432
x
0
h = δ = 10
−n
, n = 1, 2,
n K B
h
x
(K)
−
˜
f
δ
 τδ x
(K)

− x
0

1 0 1.155791 14.587413 1.366260
2 0 1.009850 1.458741 1.366260
3 29 0.381565 0.145874 0.896866
4 96 0.120050 0.014587 0.655113
5 7009 0.038193 0.001459 0.406930
6 380309 0.012078 0.000146 0.210430
7 13912613 0.003819 0.000015 0.082164
x
0
h = δ = 10
−n
, n = 1, 2,
x
∗
A
j
(x) = f
j
, j = 1, 2,
A
1
(x) = f
1
⇔






x
4
1
+ x
4
2
= 2
x
2
1
− x
2
2
= 0
x
1
x
3
= 0
,
A
2
(x) = f
2
⇔






x
1
− x
2
+ x
3
= 0
2x
1
− 2x
2
+ x
3
= 0
x
1
− x
2
− x
3
= 0
A
1
(x) = f
1
s
1
= (−1; −1; 0), s

2
= (−1; 1; 0), s
3
= (1; −1; 0), s
4
= (1; 1; 0),
A
2
rank(A
2
) = 2 (2.57)
s
1
= (−1; −1; 0), s
4
= (1; 1; 0).
f
j
f
δ
j
= (f
δ
j,1
; f
δ
j,2
; f
δ
j,3

) f
δ
j
− f
j

R
3
≤ δ
f
δ
j
= f
j
+ ∆, ∆ = (δ/
√
3; δ/
√
3; δ/
√
3), j = 1, 2. (2.57)
F (x) = A
1
x −f
δ
1

2
R
3

+ A
2
x −f
δ
2

2
R
3
+ αx −x
∗

2
R
3
→ min
x
∗
= (x
∗
1
; x
∗
2
; x
∗
3
) ∈ R
3
α

(2.58)
∂F
∂x
i
= 0, i = 1, 2, 3.
(2.60)
x
(k+1)
= x
(k)
− J(x
(k)
)
−1
G(x
(k)
), x
(0)
= x
∗
= (x
∗
1
; x
∗
2
; x
∗
3
),

J =



∂g
1
∂x
1
∂g
1
∂x
2
∂g
1
∂x
3
∂g
2
∂x
1
∂g
2
∂x
2
∂g
2
∂x
3
∂g
3

∂x
1
∂g
3
∂x
2
∂g
3
∂x
3



, G =



g
1
g
2
g
3



, g
i
=
∂F

∂x
i
, i = 1, 2, 3,
α = δ
α = δ x
α,δ
1
x
α,δ
2
x
α,δ
3
x
δ
α(δ)
− s
1

R
3
x
δ
α(δ)
− s
4

R
3
x

(0)
= x
∗
= (5; 5; 5)
α = δ x
α,δ
1
x
α,δ
2
x
α,δ
3
x
δ
α(δ)
− s
1

R
3
x
δ
α(δ)
− s
4

R
3
x

(0)
= x
∗
= (−5; −5; −5)
(2.57)
A
h
1
(x) =





(1 + h)x
4
1
+ x
4
2
x
2
1
− (1 − h)x
2
2
(1 + h)x
1
x
3

A
h
2
(x) =





(1 + h)x
1
− x
2
+ x
3
2x
1
− (2 + h)x
2
+ (1 + h)x
3
x
1
− x
2
− (1 + h)x
3
A
h
j

(x) = f
δ
j
, j = 1, 2,
F
h
(x) = A
h
1
x −f
δ
1

2
R
3
+ A
h
2
x −f
δ
2

2
R
3
+ αx −x
∗

2

R
3
→ min .
h = δ x
α
1
x
α
2
x
α
3
x
hδ
α(h,δ)
− s
1

R
3
x
hδ
α(h,δ)
− s
4

R
3
x
(0)

= x
∗
= (5; 5; 5) α = h + δ
h = δ x
α
1
x
α
2
x
α
3
x
hδ
α(h,δ)
− s
1

R
3
x
hδ
α(h,δ)
− s
4

R
3
x
(0)

= x
∗
= (−5; −5; −5) α = h + δ
x
∗
x
(0)
= x
∗
= (5; 5; 5) x
(0)
− s
1

R
3
=
9.8489, x
(0)
− s
4

R
3
= 7.5498, s
4
x
∗
x
(0)

= x
∗
= (−5; −5; −5)
x
(0)
− s
1

R
3
= 7.5498, x
(0)
− s
4

R
3
= 9.8489,
s
1
x
∗
U−
U−
(1.11)
A
j
U− U−
A
j

f
j
A
h
j
, f
δ
j
(2.1) (2.14)
(1.11)
A
h
1
(x) + α
˜µ
N

j=2
(A
h
j
(x) −f
δ
j
) + α(x −x
∗
) = f
δ
1
,

˜µ ∈ (0, 1) α
X
A
1
U− A
j
U−
γ
j
X j = 2, , N
α > 0 f
δ
j
∈ X
A
1
(x) + α
˜µ
N

j=2
(A
j
(x) −f
δ
j
) + α(x −x
∗
) = f
δ

1
x
δ
α
S = θ f
δ
j
(2.1) j = 1, , N α
α, δ/α → 0 x
δ
α
x
0
∈ S
x
0
− x
∗
, U(x
0
− z) ≤ 0, ∀z ∈ S.
X
A
h
j
U− X (2.14)
g(t) h > 0
α > 0 f
δ
j

∈ X (3.1) x
hδ
α
S = θ f
δ
j
(2.1) j = 1, , N α
α, (δ + h)/α → 0 x
hδ
α
x
0
∈ S (3.4)
(1.11) N = 1
A(x
δ
α
) −f
δ
 = Kδ
p
, K > 1, 0 < p ≤ 1.
(3.1)
ρ(α) = K(h + δ)
p
, K > 2, 0 < p ≤ 1,
ρ(α) ≡ αx
hδ
α
− x

∗
.
ρ(α) ≡ αx
hδ
α
− x
∗

ρ(α) (α
0
, +∞) α
0
> 0

N

j=2
(A
h
j
(x
∗
) −f
δ
j
) > 0, h, δ > 0,
lim
α→+∞
ρ(α) = +∞ A
0

j
≡ A
j
, f
0
j
≡ f
j
x
∗
∈ X (3.10)
¯α = α(h, δ)
¯α ≥
(K −2)(δ + h)
p
2x
∗
− z
, z ∈ S
ρ(¯α) = [K + 2g(x
hδ
¯α
)](δ + h)
p
, K > 2, p ∈ (0, 1],
x
hδ
¯α
(3.1) α = ¯α
h, δ → 0 ¯α → 0 p ∈ (0, 1) (δ +h)/α → 0 x

hδ
α
→ x
0
∈ S
p = 1 U S = {x
0
} x
hδ
α
 x
0
(δ + h)/α ≤
C, C > 0
A
1
(y) −A
1
(x
0
) −QA

1
(x
0
)
∗
U(y − x
0
) ≤ γA

1
(y) −A
1
(x
0
),
y S γ > 0 Q
X
∗
A
1
(3.21)
ω ∈ X x
∗
− x
0
= A

1
(x
0
)ω
α
0 < p < 1
x
hδ
α
− x
0
 = O((h + δ)

ν
), ν = min{1 −p; ˜µp/2}.
U−
F (
1
2
B
j
x, x)B
j
(x) = f
j
, j = 1, 2, 3,
F : R → R
F (t) =





0, t ≤ a
0
t−a
0
ε
, a
0
< t ≤ a
0
+ ε

1, t > a
0
+ ε
a
0
ε B
j
: L
2
[0, 1] → L
2
[0, 1]
B
j
x(t) =
1

0
k
j
(t, s)x(s)ds, j = 1, 2, 3,
k
j
(t, s), j = 1, 2, 3
k
1
(t, s) =

t(1 −s), t ≤ s
s(1 −t), s < t

k
2
(t, s) =

(1−s)
2
st
2
2
−
(1−s)
2
t
3
(1+2s)
6
+
(t−s)
3
6
, t ≤ s
s
2
(1−s)(1−t)
2
2
+
s
2
(1−t

3
)(2s−3)
6
+
(s−t)
3
6
, s ≤ t
k
3
(t, s) = ts.
f
j
f
δ
j
= f
j
+ δ, δ > 0
(3.24)
F (
1
2
B
j
x, x)B
j
(x) = f
δ
j

, j = 1, 2, 3.
(3.25)
A
1
(x) + α
˜µ
(A
2
(x) −f
δ
2
+ A
3
(x) −f
δ
3
) + α(x −x
∗
) = f
δ
1
,
A
j
(x) = F (
1
2
B
j
x, x)B

j
(x), j = 1, 2, 3,
B(x) + α(x −x
∗
) =
˜
f
δ
,
B(x) = A
1
(x) + α
˜µ
(A
2
(x) + A
3
(x)),
˜
f
δ
= f
δ
1
+ α
˜µ
(f
δ
2
+ f

δ
3
).
(3.26) B
j
x(t)
B
j
x(t
i
) ≈
˜
B
j
x
i
= h

k
j
(t
i
, t
0
)x
0
+ k
j
(t
i

, t
M
)x
M
2
+
M−1

q=1
k
j
(t
i
, t
q
)x
q

, i = 0, 1, , M,
x(t) ≈ ˜x = (x
0
; x
1
; ; x
M
), x
i
≈ x(t
i
), i = 0, 1, , M,

t
0
= 0, t
M
= 1, t
i
= i/M, t
q
= q/M, h = 1/M,
˜
B
j
˜
B
j
= (b
i,q
)
M
i,q=0
; b
i,q
= hk
j
(t
i
, t
q
), i, q = 1, , M −1;
b

i,0
= hk
j
(t
i
, t
0
)/2; b
i,M
= hk
j
(t
i
, t
M
)/2.
(3.26)
˜
B˜x + α(˜x − ˜x
∗
) =
˜
f
δ
,
˜
B =
˜
A
1

+ α
˜µ
(
˜
A
2
+
˜
A
3
),
˜
A
j
˜x = F (
1
2

˜
B
j
˜x, ˜x)
˜
B
j
˜x, j = 1, 2, 3.
(3.24)
(3.24) x(t) = 1
(3.27)
x

(k+1)
= x
(k)
− β
k
(
˜
Bx
(k)
+ α
k
(x
(k)
− x
(0)
) −
˜
f
δ
) , x
(0)
= ˜x
∗
∈ R
M+1
,
β
k
= cα
k

, α
k+1
=
α
k
1 + α
3
k
, k = 0, 1, 2, ,

˜
Bx
(K)
−
˜
f
δ

2
R
M+1
≤ τδ < 
˜
Bx
(k)
−
˜
f
δ


2
R
M+1
, τ > 1, k = 0, 1, , K − 1.
(2.48) (2.49)
(2.50)
L = 
˜
B
R
M+1
, α
0
= 0.1, λ =
α
2
0
+ L
2
2
, c =
1
2λ
,
τ =


x −x
(0)


R
M+1
[(1 + α
3
0
) (1 + α
2
0
) + 2α
0
]

1 −cλ −2
α
0
c

α
0
+ 1

2
.
x
(0)
= (0.9; 0.9; ; 0.9)
∈ R
M+1
a
0

=
10
−3
3
ε = 10
−2
M = 50 ˜µ =
1
2
n K 
˜
Bx
(K)
−
˜
f
δ
 τδ x
(K)
− x
0

x
0
= (x
0
0
; x
0
1

; ; x
0
M
) = (1; 1; ; 1) δ = 10
−n
B
h
j
x(t) =
1

0
k
h
j
(t, s)x(s)ds, j = 1, 2, 3,
k
h
j
(t, s) = k
j
(t, s) + h(t, s), j = 1, 2, 3, 0 < h(t, s) ≤ h, ∀t, s h → +0
h(t, s) = h
n K 
˜
B
h
x
(K)
−

˜
f
δ
 τδ x
(K)
− x
0

x
0
= (x
0
0
; x
0
1
; ; x
0
M
) = (1; 1; ; 1)
h = δ = 10
−n
x(t) = 1
(3.27)
˜
B
δ
x
(0)