Báo cáo nghiên cứu khoa học: "Đánh giá tính ổn định nghiệm cuả bài toán Cauchy cho phương trình truyền nhiệt ngược thời gian" pot

¨o



∂u
∂t
= a(t)
∂
2
u
∂x
2
, (x, t) ∈ (−∞; +∞) × (0; 1),
u(x, 1) = ϕ(x).
1.
¨o
(1.1)



∂u
∂t
= a(t)
∂
2
u
∂x
2
, (x, t) ∈ (−∞; +∞) × (0; 1),
u(x, 1) = ϕ(x).
ϕ(x)

E
u(·, 0) :=


+∞
−∞
u
2
(x, 0)dx

1
2
 E.
1
1 u(x, t) U u(·, 0)  E, E
U
a(t)
2.
1 ( ¨o p > 1 q > 1
1
p
+
1
q
= 1
f ∈ L
p
(R), g ∈ L
q
(R) fg ∈ L

1
(R) fg
1
 f
p
g
q
.
2 ( L
1
(R) f ∈ L
1
(R)
f

f(ξ) :=
1
√
2π

+∞
−∞
e
−ix.ξ
f(x)dx (y ∈ R).
2 ( f ∈ L
1
(R) ∩L
2
(R)


f ∈ L
2
(R) f = 

f
 ·  L
2
(R)
3 ( L
2
(R)

f f ∈ L
2
(R)
{f
k
}
∞
k=1
⊂ L
1
(R) ∩ L
2
(R) f
k
→ f L
2
(R)



f
k
−

f
j
 = 

f
k
− f
j
 = f
k
− f
j
 {

f
k
}
∞
k=1
L
2
(R)

f

k
→

f L
2
(R)

f f L
2
(R)

f
{

f
k
}
∞
k=1
3 ( f, g ∈ L
2
(R)
+∞

−∞
f ¯gdx =
+∞

−∞
ˆ

f ˆgdξ

D
α
f = (iξ)
α

f α D
α
f ∈ L
2
(R)
3.
1 ( u(x, t)
(3.1)







∂u
∂t
= a(t)
∂
2
u
∂x
2

, (x, t) ∈ (−∞; +∞) × (0; 1),
u(·, 1)  ,
u(·, 0)  E, (0 <  < E),
a(t)  B > 0, t ∈ [0; 1]
u(·, t)  
µ(t)
E
1−µ(t)
, t ∈ [0; 1]
u(·, t) =


+∞
−∞
u
2
(x, t)dx

1
2
, t ∈ [0, 1]
µ(t) =
A(t)
A(1)
, ∀t ∈ [0, 1],
A(t) =

t
0
a(τ)dτ, ∀t ∈ [0, 1].

u(x, t)
∂u
∂t
= a(t)
∂
2
u
∂x
2
∂u
∂t
(ξ, t) = −ξ
2
a(t)u(ξ, t).(3.2)
u(ξ, t) = e
A(1)(1−µ(t))ξ
2
u(ξ, 1), (ξ, t) ∈ R × [0, 1].(3.3)
ˆu(·, t) ∈ L
2
(R), t ∈ [0, 1]
|u(ξ, t)|
µ(t)
= e
A(1)µ(t)(1−µ(t))ξ
2
|u(ξ, 1)|
µ(t)
, (ξ, t) ∈ R ×[0, 1].(3.4)
t = 0

u(ξ, 0) = e
A(1)ξ
2
u(ξ, 1), ξ ∈ R,(3.5)
u(ξ, 1) = e
−A(1)ξ
2
u(ξ, 0), ξ ∈ R.(3.6)
u(ξ, t) = e
−A(1)µ(t)ξ
2
u(ξ, 0), (ξ, t) ∈ R × [0, 1].(3.7)
|u(ξ, t)|
(1−µ(t))
= e
−A(1)µ(t)(1−µ(t))ξ
2
|u(ξ, 0)|
(1−µ(t))
, (ξ, t) ∈ R ×[0, 1].(3.8)
|u(ξ, t)| = | u(ξ, 1)|
µ(t)
|u(ξ, 0)|
(1−µ(t))
, ξ ∈ R.
|u(·, t)| = |u(·, 1)|
µ(t)
|u(·, 0)|
(1−µ(t))
.

t = 0 t = 1
t ∈ (0, 1)
f = |u(·, 1)|
2µ(t)
, g = |u(·, 0)|
2(1−µ(t))
, p =
1
µ(t)
, q =
1
1 − µ(t)
u(·, t)
2
= u(·, t)
2
=

+∞
−∞
|u(ξ, 1)|
2µ(t)
|u(ξ, 0)|
2(1−µ(t))
dξ
=

+∞
−∞
f(ξ)g(ξ)dξ =


+∞
−∞
|f(ξ)g(ξ)|dξ = f g
1
 f
p
g
q
=


+∞
−∞
|f(ξ)|
p
dξ

µ(t)
.


+∞
−∞
|g(ξ)|
q
dξ

(1−µ(t))
=



+∞
−∞
u
2
(ξ, 1)dξ

µ(t)
.


+∞
−∞
u
2
(ξ, 0)dξ

(1−µ(t))
= u(·, 1)
2µ(t)
.u(·, 0)
2(1−µ(t))
= u(·, 1)
2µ(t)
.u(·, 0)
2(1−µ(t))
 
2µ(t)
E

2(1−µ(t))
.
t ∈ (0, 1) 
1 a(t)
a(t)  0
a(t) = 0
a(t) ∈ L
1
(0, 1)
[1] K. A. Ames and B. Straughan, Aca-
demic Press, San Diego, 1997.
[2] Dinh Nho Hao,
Journal of Mathematical Analysis and Applications, 1996,
873-909.
¨o



∂u
∂t
= a(t)
∂
2
u
∂x
2
, (x, t) ∈ (−∞; +∞) × (0; 1),
u(x, 1) = ϕ(x).