Ứng dụng phương pháp newton-kantorovich giải phương trình tích phân phi tuyến - Pdf 29

class="bi x0 y0 w1 h1"
class="bi x0 y1 w1 h2"
class="bi x0 y2 w1 h3"
class="bi x0 y3 w2 h4"
C
C
[a;b]
[a, b]
L
2
[a;b]
[a, b]
D
k
[a;b]
k [a, b]
l
2
x = (x
n
)


n=1
|x
n
|
2
L(X, Y ) X Y
N
N

P (P = R P = C) X → R
.
1) (∀x ∈ X) x ≥ 0 x = 0 ⇔ x = θ
2) (∀x ∈ X) (∀α ∈ P ) αx = |α|x
3) (∀x, y ∈ X) x + y ≤ x + y
x x
X 1), 2), 3)
{x
n
} X x ∈ X
lim
n→∞
x
n
− x = 0 lim
n→∞
x
n
= x x
n
→ x (n → ∞)
{x
n
}
X lim
n,m→∞
x
n
− x
m

)


n=1
|x
n
|
2
x =



n=1
|x
n
|
2
C
[a,b]
x(t) ∈ C
[a,b]
x = max
[a,b]
|x(t)| C
[a,b]
M
1
= (X, d
1
) ;

1) (∀x, y ∈ X) A (x + y) = A (x) + A (y) ;
2) (∀x ∈ X) (∀α ∈ P ) A (αx) = αA (x) .
Ax A(x) X ≡ Y A
X
ImA = {y ∈ Y |y = Ax, ∀x ∈ X} A
KerA = {x ∈ X|Ax = 0} A
A : R
n
→ R
m
A (x
1
, x
2
, , x
n
) = (y
1
, y
2
, , y
m
) y
i
=
n

j=1
a
ij

0
, a
1
, , a
k
t D
k
[a;b]
A
X ≡ Y ≡ C
[a;b]
Ax (t) =
b

a
K (t, s)x (s) ds
K (t, s) t, s a ≤ t, s ≤ b
A
X, Y
A : X → Y x
0
∈ X
∀{x
n
} ⊂ X, x
n
→ x
0
(n → ∞) Ax
n

3) A
A : X → Y A
A = sup
x≤1
Ax
A = sup
x=1
Ax
A : X → Y A
−1
α > 0
Ax ≥ α x, (∀x ∈ X) .


A
−1



1
α
X, Y f : X → Y
x
X f : X → Y x
A : X → Y :
f (x + h) − f (x) = A (h) + Φ (x, h) , (∀h ∈ X)
lim
h→0
Φ(x,h)
h

f (x + h) − f (x) = B (h) + Φ
B
(x, h)
A (h) − B (h)
h
=
Φ
B
(x, h) − Φ
A
(x, h)
h
→ θ h → 0
(∀k ∈ X) (∀ε > 0)
A(k)−B(k)
k
=
A(εk)−B(εk)
εk
ε → 0
εk → θ θ A (k) = B (k) , ∀k ∈ X
A ≡ B
X, Y, Z
g : X → Y x ∈ X f : Y → Z
y = g (x) ∈ Y φ = f ◦ g x φ

(x) =
f

(g (x)) g

, , x
n
) h = (h
1
, h
2
, , h
n
)
∈ R
n
f x df (x, h) =
n

i=1
∂f (x)
∂x
i
h
i
f x f

(x) =

∂f
∂x
1
,
∂f
∂x

n
→ R
m
x df (x, h) = (df
1
(x, h) , df
2
(x, h) , , df
m
(x, h))
f m × n
i f
i

(x) f

(x) =

∂f
i
∂x
j

f
f : X → Y
Ω ⊂ X
X Y f

: Ω → L (X, Y )
f x f

A X
Ax = f
x = λAx + f
f ∈ X λ K(K = C K =
R)
• A
• A
x

a
K[x, t, y(t)]dt = f(x)
y(x) =
x

a
K[x, t, y(t)]dt + f(x)
K[x, t, y(t)](t, s ∈ [a, b]) y(x)
[a, b]
K[x, t, y(t)]
b

a
K[x, t, y(t)]dt = f(x)
y(x) =
b

a
K[x, t, y(t)]dt + f(x)
K[x, t, y(t)](t, s ∈ [a, b]) y(x)
[a, b]

n
+ f( n ≥ 0)
x
0
x

x
n
− x

 ≤
q
n
1 − q
x
1
− x
0

x
n
− x

 ≤
q
1 − q
x
n
− x
n−1

4
y
3
(x) =
x

0

t

1
16
t
8

1
2
t
5
+ t
2

− 1

dt = −x +
1
4
x
4


K[x, t, y(t)] =
n

i=1
g
i
(x)h
i
[t, y(t)]
y(x) =
n

i=1
g
i
(x)
b

a
h
i
[t, y(t)]dt + f(x)
c
i
=
b

a
h
i

i
g
i
(t) + f(t)

dt
c
j
y(x) =
n

i=1
c
i
g
i
(x) + f(x)
c
i
y(x) =
1

−1
(x + t)y(t)dt + x
2
y(x) =
1

−1
xy(t)dt +




c
1
=
1

−1
(tc
1
+ c
2
+ t
2
)dt
c
2
=
1

−1
t(tc
1
+ c
2
+ t
2
)dt


= −
4
3
y(x) = x
2
− 2x −
4
3
class="bi x0 y82 w2 h1a"
P X
P (x) = 0
x = A (x)
A S (x
0
, r)
X
u = ϕ (u)
ϕ (u) [u
0
; u

] , (u

= u
0
+ r)
1) A (x
0
) − x
0

1
≤ u ≤ γ
2
1) ϕ (γ
1
) ≥ γ
1
{u
n
} u

2) ϕ (γ
2
) ≤ γ
2
{u
n
} u

3) γ
1
≤ ϕ (γ
1
) ≤ ϕ (γ
2
) ≤ γ
2
{u
n
} {u

u
n
≤ u

, ∀n
{u
n
} u
1
= ϕ (u
0
) = ϕ (γ
1
) ≥ γ
1
= u
0
u
n
≥ u
n−1
u
n+1
= ϕ (u
n
) ≥ ϕ (u
n−1
) = u
n
{u


]
x

x

− x
0
 ≤ u − u
0
u x

x
n
− x

 ≤ u − u
n
, n = 1, 2,
u
n
= ϕ (u
n−1
) , n = 1, 2, ; u
0
= u
0
{x
n
} ⊂ S (x

− u
k
x
n+1
− x
n
 = A (x
n
) − A (x
n−1
) =





x
n

x
n−1
A

(x) dx





x = x


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