BỘ GIÁO DỤC VÀ ĐÀO TẠO
TRƯỜNG ĐẠI HỌC VINH NGUYỄN VĂN TÁM
VỀ MỘT PHƯƠNG PHÁP CHỈNH HÓA
CHO PHƯƠNG TRÌNH PARABOLIC TUYẾN TÍNH
NGƯỢC THỜI GIAN VỚI HỆ SỐ PHỤ THUỘC
THỜI GIAN TRONG KHÔNG GIAN BANACH
LUẬN VĂN THẠC SĨ TOÁN HỌC
NGHỆ AN - 2014
NGHỆ AN - 2014
···
du
dt
= a(t)Au(t) 0 s t < T
u(s) = χ
X −A
θ ∈
0,
π
2
X a ∈ C([0, T ] : R
+
)
C
0
θ
class="bi x3 y46 w5 h1a"
X
X X T (t), 0 t < ∞
X
T (0) = I, ( )
T (t + s) = T (t)T (s) t, s 0
T (t)
=
∞
n=0
(tA)
n
n!
t 0
T (t) T (0) = I
T (t + s) = T (t)T (s)
T (t) − I tAe
tA
T (t) − I
t
− A
A max
0st
T (s) − I.
T (t)
X A
T (t)
X
ρ
0
T (s)ds
= h
−1
ρ+h
ρ
T (s)ds −
h
0
T (s)ds
.
h
−1
(T (h) − I) =
h
−1
ρ+h
0
T (s)ds − h
−1
h
−1
T (t)
T (t) C
0
A
x ∈ X lim
h→0
1
h
t+h
t
T (s)xds = T(t)x.
x ∈ X
t
0
T (s)xds ∈ D(A)
A
t
0
T (s)xds
= T (t)x − x.
x ∈ D(A) T (t)x ∈ D(A)
d
dt
T (t)x = AT(t)x = T (t)Ax.
n
→ x Ax
n
→ y n → ∞
T (t)x
n
− x
n
=
t
0
T (s)Ax
n
ds.
T (s)y
n → ∞
T (t)x − x =
t
0
T (s)yds.
t > 0 t ↓ 0
x ∈ D(A) Ax = y
T (t) C
0
ω 0 M 1
T (t) Me
ωt
0 t < ∞.
n
M
n+1
MM
t/n
= Me
ωt
.
A C
0
T (t), t 0
A D(A) = X
ρ(A) A R
+
λ > 0
R(λ : A)
1
λ
,
R(λ : A) = (λI −A)
−1
I X
θ ∈ (0;
π
2
]
T (t) t > 0 X
θ
T (t)
T (z)
π
/
2
−
θ
1
ω ∈ ρ(A)
(ω − A)
−1
≤
M
1
dist(ω, S
π/
2
−θ
1
)
.
−A
θ 0 ∈ ρ(A) σ > 0
A
A
−σ
=
1
2πi
Γ
ω
) σ > σ
> 0
(A
σ
) X σ ≥ 0
A
σ
1
+σ
2
x = A
σ
1
A
σ
2
x σ
1
, σ
2
∈ R x ∈ (A
σ
)
σ = max{σ
1
, σ
2
, σ
A
u
t
= Au(t) + h(t), u(0) = x, (0 t < T ).
u
t
+ Au(t) = 0, (0 t < T), u(T ) = x, A
−A
X x ∈ X
du
dt
= A(t, D)u(t), 0 s t < T,
u(s) = χ
X −D
θ ∈ (
π
4
,
π
2
] χ ∈ X
u(t) u(t)
L
p
(R), 1 p < ∞
X
du
dt
β > 0 f
β
(t; A), 0 ≤ t ≤ T a(t)A
θ ∈ (0,
π
2
]
f
β
(t, A) =
a(t)A − βA
σ
θ ∈ (0,
π
4
]
a(t)A(I + βA)
−1
θ ∈ (
π
4
,
π
2
]
σ > 1 θ ∈ (0;
π
4
]
2
− θ) <
π
2
A(I + βA)
−1
θ ∈ (
π
4
;
π
2
]
−A
θ ∈ (0; π/
4
] 0 ∈ ρ(A) 0 < β < 1 σ > 1
σ(π/
2
− θ) < π/
2
f
β
(t; A); 0 ≤ t ≤ T
f
β
(t, A) = a(t)A −βA
σ
.
v
−θ
φ
σ(π/
2
− θ) < π/
2
.
V
β
(t; s) 0 ≤ s ≤ t ≤ T
0 ≤ s < t ≤ T 0 ∈ ρ(A)
d ∈ (0; 1)
ρ(A)
(t − s)
−1/
σ
≤ d
Γ
φ
ρ(A)
Γ
1
= {re
iφ
: r ≥ (t −s)
−1/
σ
},
Γ
2
(ω, S
π/
2
−θ
1
) = |ω|sin(φ −(π/
2
−θ
1
))
(ω − A)
−1
≤
M
1
|ω|sin(φ − (π/
2
− θ
1
))
.
M
1
= M
1
|e
t
s
(a(τ)ω−βω
σ
)dτ
||ω|
−1
dω
= 2M
1
∞
(t−s)
−1/
σ
e
t
s
(a(τ)τ cos φ−βτ
σ
cos σφ)dτ
τ
−1
dτ
≤ 2M
1
∞
1
e
BT
1−1/σ
x cos φ−βx
σ
cos σφ
dx ≤ K,
K t s σ > 1 π/
2
σ >
φ > π/
2
− θ 0 < φ < σφ < π/
2
cos φ > 0
cos(σφ) > 0 ω ∈ Γ
2
Γ
2
−1
cos σθ
)dτ
(t − s)
−1/
σ
dθ
≤ dM
d
φ
−φ
e
B(t−s)
1−1/σ
cos θ
−β cos σθ
dθ
≤ dM
d
φ
−φ
e
BT
iφ
: r ≥ (t −s)
−1/σ
},
Γ
2
= {(t −s)
−1/σ
e
iθ
: −φ ≤ θ
≤ π},
Γ
3
= {re
iπ
: d ≤ r ≤ (t −s)
−1/σ
},
Γ
4
= {de
−iθ
: −π ≤ θ
≤ π},
Γ
3
Γ
1
∪Γ
7
=
Γ
1
∪Γ
3
≤ K
ω ∈ Γ
2
ω ∈ Γ
2
(ω, S
π/2−θ
1
) ≥ ((t −s)
−1/σ
e
iφ
, S
π/2−θ
1
).
π
φ
e
t
s
(a(τ)(t−s)
−1/σ
cos θ
−β(t−s)
−1
cos σθ
)dτ
dθ
≤ M
1
π
φ
e
BT
1−1/σ
cos φ−βcosσθ
dθ
Γ
5
=
(t−s)
−1/σ
d
(e
t
s
(−a(τ)τ−βr
σ
e
−iπσ
)dτ
− e
t
s
(−a(τ)τ−βτ
σ
e
iπσ
)dτ
)(−τ − A)
−1
dτ
(t−s)
−1/σ
d
e
−(
t
s
(a(τ)dτ)τ
|(e
−β(t−s)τ
σ
cos σπ
2i sin(β(t −s)τ
σ
sin σπ)|τ
−1
dτ
≤ M
1
(t−s)
−1/σ
d
e
−β(t−s)τ
σ
cos σπ
2|sin(β(t −s)τ
−1/2
e
−βx
σ
cos σπ
{4x
−1
sin
2
(βx
σ
sin σπ)}
1/2
dx
= M
1
1
(t−s)
−1/σ
d
x
−1/2
e
−βx
σ
cos σπ
{2x
−1
e
−βx
σ
cosσπ
dx ≤ M
1
e
1
0
x
−1/2
dx = M
1
2e
0 < β < 1
Γ
4
≤ M
≤ dM
d
π
−π
e
BTd
e
−β(t−s)d
σ
cos σθ
dθ
≤ dM
d
e
BTd
(1 + e
T d
σ
)2π,
M
d
= max
|ω|≤d
(ω − A)
−1
V (t; s) 0 ≤ s ≤ t ≤ T
β
(t; s)A
−σ
− V
β
(t
0
, s
0
)A
−σ
→ 0 (t, s) → (t
0
, s
0
).
x ∈ (A
σ
)
V
β
(t; s)x − V
β
(t
0
, s
0
)x
≤ V
β
t → V
β
(t; s)χ [s; T ]
∂
∂t
V
β
(t, s)χ = f
β
(t, A)V
β
(t, s)χ t ∈ (s, T )
∂
∂t
V
β
(t, s)χ =
1
2πi
Γ
φ
(
∂
∂t
e
t
s
f
t
s
f
β
(τ,ω)dτ
)a(t)ω(ω − A)
−1
χdω
+
1
2πi
Γ
φ
e
t
s
f
β
(τ,ω)dτ
)(−βω
σ
)(ω − A)
−1
χdω.
1
2πi
2πi
Γ
φ
e
t
s
f
β
(τ,ω)dτ
dω)χ
+ a(t)
1
2πi
Γ
φ
e
t
s
f
β
(τ,ω)dτ
A(ω − A)
−1
χdω
= a(t)AV
β
−σ
G = V
β
(t; s) A
σ
= (A
σ
)
−1
(V
β
(t; s)) ⊆ (A
−σ
) = (A
σ
) G = A
σ
V
β
(t; s)
1
2πi
Γ
φ
e
t
s
f
β
(t; s)χ
t → f
β
(t, A)V
β
(t, s)χ
(s; T ) t → e
t
s
f
β
(τ,ω)dτ
f
β
(t, ω)
t → V
β
(t; s)χ (s; T ) t → V
β
(t; s)χ
t → V
β
(t; s)χ
0 < β < 1 f
β
(t, A) 0 ≤ t ≤
T V
β
V
β
(t, s) ≤ K
1
+
M
1
π
∞
1
e
BT
1−1/σ
x cos φ−βx
σ
cos σφ
dx,
K
1
β
β q(x) = 2BT
1−1/σ
x cos φ − βx
σ
cos(σφ)
[1; ∞) x
0
= (
.
∞
1
e
BT
1−1/σ
x cos φ−βx
σ
cos σφ
dx ≤ e
K
2
β
1/(σ−1)
∞
1
e
BT
1−1/σ
x cos φ
dx
=
K
2
β
1/(σ−1)
BT
1−1/σ
K
2
β
−1/(σ−1)
0 ≤ s ≤ t ≤ T
K
3
K
3
−A
θ ∈ (π/4; π/2] 0 ∈ ρ(A) β ∈ (0, 1)
f
β
(t; A) 0 ≤ t ≤ T
f
β
(t, A) = a(t)A(I + βA)
−1
.
v
β
(t) =
V
β
(t; s)χ, χ ∈ X V
β
(t; s) 0 ≤ s ≤ t ≤ T
V
β
−1
)
≤
B(1 + C)
β
,
B = max
t∈[0;T ]
|a(t)| t → f
β
(t; A)
A(I + βA) a(t)
v
β
(t) χ ∈ X v
β
(t)
v
β
(t) = V
β
(t; s)χ V
β
(t; s)
V
β
(t, s) ≤ e
t
s