Báo cáo nghiên cứu khoa học: "Đánh giá tính ổn định và chỉnh hóa nghiệm của phương trình truyền nhiệt ngược thời gian" pdf

¨o



∂u
∂t
= a
∂
2
u
∂x
2
(x, t) ∈ (−∞; +∞) × (0; 1),
u(·, 1) − ϕ(·)  ε,
u(·, 0)
H
s
(R)
 E, ϕ(·) ∈ L
2
(R) E > ε > 0
s  0 a > 0
1.
¨o



∂u
∂t
= a
∂

2
u
∂x
2
(x, t) ∈ (−∞; +∞) × (0; 1),
u(·, 1) − ϕ(·)  ε,
(1.1)
u(·, 0)
H
s
(R)
 E, (1.2)
 · ,  · 
H
s
(R)
L
2
(R)
H
s
(R)(s  0)
L
2
(R)
x u(x, t), (x, t) ∈ (−∞; +∞) × [0; 1]
u(x, t) ∈ L
2
(R) t ∈ [0, 1]
−

−
s = 0 s = 0 H
s
(R) = H
0
(R) =
L
2
(R)
s s  0
1
v(x, t) ∈ L
2
(R) t ∈ [0; 1+β]



∂v
∂t
= a
∂
2
v
∂x
2
(x, t) ∈ (−∞; +∞) × (0; 1 + β),
αv(x, 0) + v (x, 1 + β) = ϕ(x), x ∈ (−∞; +∞), α > 0, β > 0
(1.3)
−
¨o

t ∈ (0, 1]
t = 0 s > 0
2.
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
.
2.1 ( L
1
(R)
f ∈ L
1
(R)

f(ξ) :=

1
√
2π

+∞
−∞
e
−ix.ξ
f(x)dx (y ∈ R)
f
f
∨
(ξ) :=
1
√
2π

+∞
−∞
e
ix.ξ
f(x)dx (y ∈ R).
2.2 ( f ∈ L
1
(R) ∩ L
2
(R)

f, f
∨

∈ L
2
(R)
f = 

f = f
∨

2.2 ( 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
∨
2.3 ( f, g f ∗g
(f ∗g)(x) =

+∞
−∞
f(y)g(x − y)dy.
2.3 ( f, g ∈ L
2
(R)
+∞

−∞
f ¯gdx =
+∞

−∞
ˆ
f ˆgdξ


D
α
f = (iξ)
α

f α D
α
f ∈ L
2
(R)

f ∗g =
√
2π

fg
f = (

f)
∨
.
2.4 f ∈ H
s
(R) f
f
H
s
(R)
:=



+∞
−∞
|

f(ξ)|
2
(1 + ξ
2
)
s
dξ

1/2
< +∞. (2.1)
H(η)
H(η) =

η
η
(1 − η)
1−η
, η ∈ (0, 1),
1, η = 0 1.
(2.2)
H(η) ≤ 1
C(x, y) 1 > x  0, y > 0
C(x, y) =

y

1 − x

y
e
1−x−y
. (2.3)
2.4 x, y z
x + zy  (z + 1)x
1/(z+1)
y
z/(z+1)
.
2.5 0  p  q < ∞, q = 0 α > 0
αe
−p
α + e
−q
 H

p
q

α
p
q
.
p = 0 p = q
0 < p < q < ∞
x = α, z =
p

q −p
, y =
q −p
p
e
−q
,
α + e
−q
= x + zy  (z + 1)x
1/(z+1)
y
z/(z+1)
,
=
1

p
q

p
q

1 −
p
q


1−
p

q

α
1−
p
q
e
−p
.

2.6 u(x, t) ∈ L
2
(R) t ∈ [0; 1]
∂u
∂t
= a
∂
2
u
∂x
2
, (x, t) ∈ (−∞; +∞) × (0; 1). (2.4)
u(·, t)  u(·, 1)
t
u(·, 0)
1−t
, t ∈ [0; 1]
u(x, t)
∂u
∂t

= a
∂
2
u
∂x
2
x
∂u
∂t
(ξ, t) = −ξ
2
au(ξ, t). (2.5)
u(ξ, t) = e
a(1−t)ξ
2
u(ξ, 1), (ξ, t) ∈ R × [0, 1]. (2.6)
ˆu(·, t) ∈ L
2
(R), t ∈ [0, 1]
|u(ξ, t)|
t
= e
at(1−t)ξ
2
|u(ξ, 1)|
t
, (ξ, t) ∈ R × [0, 1]. (2.7)
t = 0
u(ξ, 0) = e
aξ

2
u(ξ, 1), ξ ∈ R, (2.8)
u(ξ, 1) = e
−aξ
2
u(ξ, 0), ξ ∈ R. (2.9)
u(ξ, t) = e
−atξ
2
u(ξ, 0), (ξ, t) ∈ R ×[0, 1]. (2.10)
|u(ξ, t)|
(1−t)
= e
−at(1−t)ξ
2
|u(ξ, 0)|
(1−t)
, (ξ, t) ∈ R × [0, 1]. (2.11)
|u(ξ, t)| = |u(ξ, 1)|
t
|u(ξ, 0)|
(1−t)
, ξ ∈ R.
t = 0 t = 1
t ∈ (0, 1)
p =
1
t
, q =
1

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

−∞
|f(ξ)g(ξ)|dξ = f g
1
 f
p
g
q
=


+∞
−∞
|f(ξ)|
p
dξ

t
.


+∞
−∞
|g(ξ)|
q
dξ

(1−t)
=



+∞
−∞
u
2
(ξ, 1)dξ

t
.


+∞
−∞
u
2
(ξ, 0)dξ

(1−t)
= u(·, 1)
2t
.u(·, 0)
2(1−t)
= u(·, 1)
2t
.u(·, 0)
2(1−t)

3.
3.1 u
1
(x, t) u

2
(x, t) −
t ∈ [0, 1]
u
1
(·, t) − u
2
(·, t)  2C
1
(t, a, s, β)ε
t
E
1−t

ln
E
ε

−
s
2
(1−t)
(1 + o(1)), ε → 0
+
,
C
1
(t, a, s, β) =

1 + C(a, β, s)a

s
2

C(a, β, s) =



e
aβ
s = 0,
max

1, e
aβ

s
2aeβ

s/2

s > 0.
3.2
3.3 u(x, t) − v
α
(x, t)
u(·, t) − v
α
(·, t + β)  C(a, β, s)α
t
1+β


a(1 + β)
ln
1
α

s/2
E + α
t+β
1+β
−1
ε, ∀t ∈ [0, 1], ∀α ∈ (0, 1).
α =

ε
E

ln
E
ε

s
2

1+β
,
t ∈ [0; 1]
u(·, t) − v
α
(·, t + β)  C

1
(t, a, s, β)ε
t
E
1−t

ln
E
ε

−
s
2
(1−t)
(1 + o(1)), ε → 0
+
.
3.1 t = 0
u(·, 0) − v
α
(·, β)  C
1
(0, a, s, β)E

ln
E
ε

−
s

2
(1 + o(1)), ε → 0
+
.
s > 0
3.4 ε < ϕ(·) τ > 1 τ ε < ϕ(·)
α
ε
> 0
v
α
ε
(·, 1 + β) − ϕ(·) = τε. (3.1)
u(x, t) v
α
(x, t)
α = α
ε
. t ∈ [0, 1]
u(·, t) − v
α
ε
(·, t + β) 

C(τ, a, β, s)ε
t
E
1−t

ln

E
ε

−s(1−t)/2
(1 + o(1)) khi ε → 0
+

C(τ, a, β, s) τ, a, β, s
3.2 s
E t = 0
u(·, 0) − v
α
ε
(·, β) 

C(τ, a, β, s)E

ln
E
ε

−s/2
(1 + o(1)) khi ε → 0
+
.
s > 0
4.
v
x ∈ R




dv
dt
= −aξ
2
v, 0 < t < 1 + β,
αv(ξ, 0) + v(ξ, 1 + β) = ϕ(ξ), ξ ∈ (−∞; +∞), α > 0, β > 0.
(4.1)
v(ξ, t) = e
−atξ
2
v(ξ, 0), ∀t ∈ [0, 1 + β]. (4.2)
t = 1 + β v(ξ, 1 + β) = e
−a(1+β)ξ
2
v(ξ, 0), ϕ(ξ) = αv(ξ, 0) +
v(ξ, 1 + β) = (α + e
−a(1+β)ξ
2
)v(ξ, 0)
v(ξ, 0) =
ϕ(ξ)
α + e
−a(1+β)ξ
2
. (4.3)
v(ξ, t) =
e
−atξ

2
α + e
−a(1+β)ξ
2
ϕ(ξ) t ∈ [0, 1 + β]
v(x, t) =

e
−atξ
2
α + e
−a(1+β)ξ
2
ϕ(ξ)

∨
, ∀t ∈ [0, 1 + β]. (4.4)
v =
F ∗ ϕ
√
2π
(4.5)
F

F =
e
−atξ
2
α + e
−a(1+β)ξ

2
. (4.6)
|

F | =

F =
e
−atξ
2
α + e
−a(1+β)ξ
2
 H

t
1 + β

α
t
1+β
−1
, ∀t ∈ [0, 1 + β]. (4.7)
v(·, t) =





F ϕ


∨



=




F ϕ



 H

t
1 + β

α
t
1+β
−1
ϕ = H

t
1 + β

α
t

1+β
−1
ϕ
v(·, t)  H

t
1 + β

α
t
1+β
−1
ϕ, ∀t ∈ [0, 1 + β]. (4.8)

u(·, t) − v
α
(·, t + β) = u(·, t) − v
α
(·, t + β)
=





e
a(1−t)ξ
2
u(ξ, 1) −
e

−a(t+β)ξ
2
α + e
−a(1+β)ξ
2
ϕ(ξ)





=





e
a(1−t)ξ
2
u(ξ, 1) −
e
−a(t+β)ξ
2
α + e
−a(1+β)ξ
2
u(ξ, 1) +
e
−a(t+β)ξ

2
α + e
−a(1+β)ξ
2
(u(ξ, 1) − ϕ(ξ))












e
a(1−t)ξ
2
−
e
−a(t+β)ξ
2
α + e
−a(1+β)ξ
2

u(ξ, 1)






+





e
−a(t+β)ξ
2
α + e
−a(1+β)ξ
2
(u(ξ, 1) − ϕ(ξ))











αe
a(1−t)ξ

2
α + e
−a(1+β)ξ
2
u(ξ, 1)





+ sup
ξ∈R
e
−a(t+β)ξ
2
α + e
−a(1+β)ξ
2
u(ξ, 1) − ϕ(ξ)
=





αe
−atξ
2
α + e
−a(1+β)ξ

2
u(ξ, 0)





+ sup
ξ∈R
e
−a(t+β)ξ
2
α + e
−a(1+β)ξ
2
u(ξ, 1) − ϕ(ξ)






αe
−atξ
2
(1 + ξ
2
)
−s/2
α + e

−a(1+β)ξ
2
(1 + ξ
2
)
s/2
u(ξ, 0)





+ H

t + β
1 + β

α
t+β
1+β
−1
ε
 sup
ξ∈R
αe
−atξ
2
(1 + ξ
2
)

−s/2
α + e
−a(1+β)ξ
2
u(ξ, 0)
H
s
+ H

t + β
1 + β

α
t+β
1+β
−1
ε
 sup
ξ∈R
αe
−atξ
2
(1 + ξ
2
)
−s/2
α + e
−a(1+β)ξ
2
E + α

t+β
1+β
−1
ε. (4.9)
A = sup
ξ∈R
αe
−atξ
2
(1 + ξ
2
)
−s/2
α + e
−a(1+β)ξ
2
. e
−a(1+β)ξ
2
= αz
0 < z 
1
α
αe
−atξ
2
(1 + ξ
2
)
−s/2

α + e
−a(1+β)ξ
2
=
α(αz)
t
1+β

1 −
ln(αz)
a(1 + β)

−s/2
α(1 + z)
= α
t
1+β
z
t
1+β
(1 + z)

1 −
ln(αz)
a(1 + β)

−s/2
= α
t
1+β


a(1 + β)
ln
1
α

s/2
z
t
1+β
(1 + z)

−ln α
−ln(αz) + a(1 + β)

s/2
= α
t
1+β

a(1 + β)
ln
1
α

s/2
z
t
1+β
(1 + z)


−ln α
−ln α − ln z + a(1 + β)

s/2
B =
z
t
1+β
(1 + z)

−ln α
−ln α − ln z + a(1 + β)

s/2
C(a, β, s)
0 < z  1 0 < α < 1 0  −ln α < −ln α − ln z + a(1 + β)
B =
z
t
1+β
(1 + z)

−ln α
−ln α − ln z + a(1 + β)

s/2
<
z
t

1+β
(1 + z)
< 1.
z > 1
0 <
−ln α
−ln α − ln z + a(1 + β)
< 1 +
ln z
a(1 + β)
. (4.10)
0 < a(1 + β) −
ln z
a(1 + β)
ln(αz) = a(1 + β) + ξ
2
ln z. (4.11)
z > 1
z > 1
B =
z
t
1+β
(1 + z)

−ln α
−ln α − ln z + a(1 + β)

s/2
<

z
t
1+β
(1 + z)

1 +
ln z
a(1 + β)

s/2

z
1
1+β
(1 + z)

1 +
ln z
a(1 + β)

s/2
<
z
1
1+β
z

ln(e
a(1+β)
z)

a(1 + β)

s/2
= z
−β
1+β

ln(e
a(1+β)
z)
a(1 + β)

s/2
= e
aβ

1
a(1 + β)

s/2
(e
a(1+β)
z)
−β
1+β

ln(e
a(1+β)
z)


s/2
= e
aβ

1
a(1 + β)

s/2
y
−β
1+β
(ln y)
s/2
= e
aβ

1
a(1 + β)

s/2
g(y),
g(y) = y
−β
1+β
(ln
y
)
s/2
, y = e
a(1+β)

z > e
a(1+β)
> 1
g(y) y > 1
g

(y) =
−β
1 + β
y
−β
1+β
−1
(ln y)
s/2
+ y
−β
1+β
s
2
(ln y)
s
2
−1
1
y
= y
−β
1+β
−1

(ln y)
s
2
−1

s
2
−
β
1 + β
ln y

,
g

(y) = 0 ⇔ y = e
s(1+β)
2β
.
sup
y>1
g(y) = 1 s = 0 sup
y>1
g(y) = g

e
s(1+β)
2β

=


s(1+β)
2eβ

s/2
s > 0
B  max

1, e
aβ

= e
aβ
s = 0,
B  max

1, e
aβ

s
2aeβ

s/2

s > 0.
A  C(a, β, s)α
t
1+β

a(1 + β)

ln
1
α

s/2
E.
u(·, t) − v
α
(·, t + β)  C(a, β, s)α
t
1+β

a(1 + β)
ln
1
α

s/2
E + α
t+β
1+β
−1
ε, ∀t ∈ [0, 1], ∀α ∈ (0, 1).
α =

ε
E

ln
E

ε

s
2

1+β
u(·, t) − v
α
(·, t + β)  ε
t
E
1−t

ln
E
ε

−
s
2
(1−t)


1 + C(a, β, s)a
s
2

ln
E
ε

ln
E
ε
−
s
2
ln ln
E
ε

s
2


 ε
t
E
1−t

ln
E
ε

−
s
2
(1−t)

1 + C(a, β, s)a
s

2

(1 + o(1)) ε → 0
+
lim
→0
+

ln
E
ε
ln
E
ε
−
s
2
ln ln
E
ε

s
2
= 1.


ρ(α) = v
α
(·, 1 + β) − ϕ 0 < ε < ϕ
ρ

lim
α→0
+
ρ(α) = 0,
lim
α→+∞
ρ(α) = ϕ,
ρ
ρ(α) = v
α
(·, 1 + β) − ϕ = v
α
(ξ, 1 + β) − ϕ(ξ)
=





e
−a(1+β)ξ
2
α + e
−a(1+β)ξ
2
ϕ(ξ) − ϕ(ξ)






=




−
α
α + e
−a(1+β)ξ
2
ϕ(ξ)




=




α
α + e
−a(1+β)ξ
2
ϕ(ξ)





=


+∞
−∞

α
α + e
−a(1+β)ξ
2

2
|ϕ(ξ)|
2
dξ

1
2
. (4.12)
δ ϕ
2
=
ϕ
2
=

+∞
−∞
|ϕ(ξ)|
2

dξ n
δ

|ξ|>n
δ
|ϕ(ξ)|
2
dξ <
δ
2
2
α
0 < α <
δ
√
2ϕ
e
−a(1+β)n
2
δ
ρ
2
(α) =

+∞
−∞

α
α + e
−a(1+β)ξ

2

2
|ϕ(ξ)|
2
dξ
=

|ξ|n
δ

α
α + e
−a(1+β)ξ
2

2
|ϕ(ξ)|
2
dξ +

|ξ|>n
δ

α
α + e
−a(1+β)ξ
2

2

|ϕ(ξ)|
2
dξ


|ξ|n
δ

α
α + e
−a(1+β)n
2
δ

2
|ϕ(ξ)|
2
dξ +

|ξ|>n
δ
|ϕ(ξ)|
2
dξ
 α
2
e
2a(1+β)n
2
δ


|ξ|n
δ
|ϕ(ξ)|
2
dξ +
δ
2
2
 α
2
e
2a(1+β)n
2
δ

∞
−∞
|ϕ(ξ)|
2
dξ +
δ
2
2
= α
2
e
2a(1+β)n
2
δ

ϕ
2
+
δ
2
2
< δ
2
.
lim
α→0
+
ρ(α) = 0
ρ
2
(α) =

+∞
−∞

α
α + e
−a(1+β)ξ
2

2
|ϕ(ξ)|
2
dξ 


+∞
−∞
|ϕ(ξ)|
2
dξ = ϕ
2
(4.13)
ρ
2
(α) =

+∞
−∞

α
α + e
−a(1+β)ξ
2

2
|ϕ(ξ)|
2
dξ 

+∞
−∞

α
α + 1


2
|ϕ(ξ)|
2
dξ
=

α
α + 1

2
ϕ
2
. (4.14)
ρ(α)
α
α + 1
ϕ  ρ(α)  ϕ. lim
α→+∞
α
α + 1
ϕ =
ϕ lim
α→+∞
ρ(α) = ϕ.
0 < α
1
< α
2
ρ(α
1

) < ρ(α
2
)
0 <
α
1
α
1
+ e
−a(1+β)ξ
2
<
α
2
α
2
+ e
−a(1+β)ξ
2
, ∀ξ ∈ R. (4.15)
ϕ = ϕ > 0 ϕ
2
=

+∞
−∞
|ϕ(ξ)|
2
dξ n
0


+n
0
−n
0
|ϕ(ξ)|
2
dξ > 0. (4.16)
ρ(α
1
) < ρ(α
2
)
α
0
ρ α
0
α
0
> 0 ϕ > 0 ρ
2
(α
0
) > 0
ρ(α
0
) > 0 ρ(α
0
) α
ρ(α

0
)|ρ(α) − ρ(α
0
)|  (ρ(α) + ρ(α
0
))|ρ(α) − ρ(α
0
)| = |ρ
2
(α) − ρ
2
(α
0
)|


+∞
−∞






α
α + e
−a(1+β)ξ
2

2

−

α
0
α
0
+ e
−a(1+β)ξ
2

2





|ϕ(ξ)|
2
dξ. (4.17)






α
α + e
−a(1+β)ξ
2


2
−

α
0
α
0
+ e
−a(1+β)ξ
2

2





=




α
α + e
−a(1+β)ξ
2
−
α
0
α

0
+ e
−a(1+β)ξ
2








α
α + e
−a(1+β)ξ
2
+
α
0
α
0
+ e
−a(1+β)ξ
2




 2





α
α + e
−a(1+β)ξ
2
−
α
0
α
0
+ e
−a(1+β)ξ
2




= 2





(α − α
0
)e
−a(1+β)ξ
2

(α + e
−a(1+β)ξ
2
)(α
0
+ e
−a(1+β)ξ
2
)





 2
|α − α
0
|
αα
0
, ∀n ∈ N
∗
. (4.18)
ρ(α
0
)|ρ(α) − ρ(α
0
)|  2
|α − α
0

|
αα
0

+∞
−∞
|ϕ(ξ)|
2
dξ = 2
|α − α
0
|
αα
0
ϕ
2
.
0  |ρ(α) − ρ(α
0
)|  2
|α − α
0
|
αα
0
ρ(α
0
)
ϕ
2

.
lim
α→α
0
2
|α − α
0
|
αα
0
ρ(α
0
)
ϕ
2
= 0 lim
α→α
0
|ρ(α) −
ρ(α
0
)| = 0 lim
α→α
0
ρ(α) = ρ(α
0
) ρ α
0
ρ α
ε

z(·, t) = u(·, t) −v
α
ε
(·, t + β) t ∈ [0, 1]
z
t
= az
xx
, (x, t) ∈ (−∞, +∞) × (0; 1) (4.19)
z(·, 1) = u(·, t) −v
α
ε
(·, t + β)
 u(·, t) − ϕ + v
α
ε
(·, t + β) − ϕ
 ε + τ ε = (1 + τ )ε. (4.20)
z(·, 0)
z(·, 0) = u(·, 0) −v
α
ε
(·, β) = u(·, 0) − v
α
ε
(·, β)
=






e
aξ
2
u(ξ, 1) −
e
−aβξ
2
α
ε
+ e
−a(1+β)ξ
2
ϕ(ξ)





=





e
aξ
2
u(ξ, 1) −

e
−aβξ
2
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1) +
e
−aβξ
2
α
ε
+ e
−a(1+β)ξ
2
(u(ξ, 1) − ϕ(ξ))













e
aξ
2
−
e
−aβξ
2
α
ε
+ e
−a(1+β)ξ
2

u(ξ, 1)





+





e
−aβξ
2
α
ε

+ e
−a(1+β)ξ
2
(u(ξ, 1) − ϕ(ξ))











α
ε
e
aξ
2
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1)






+ sup
ξ∈R
e
−aβξ
2
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1) − ϕ(ξ)
=




α
ε
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 0)




+ sup

ξ∈R
e
−aβξ
2
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1) − ϕ(ξ)






α
ε
(1 + ξ
2
)
−s/2
α
ε
+ e
−a(1+β)ξ
2
(1 + ξ
2
)

s/2
u(ξ, 0)





+ H

β
1 + β

α
−1
1+β
ε
ε
 sup
ξ∈R
α
ε
(1 + ξ
2
)
−s/2
α
ε
+ e
−a(1+β)ξ
2

u(ξ, 0)
H
s
+ H

β
1 + β

α
−1
1+β
ε
ε
 sup
ξ∈R
α
ε
(1 + ξ
2
)
−s/2
α
ε
+ e
−a(1+β)ξ
2
E + α
−1
1+β
ε

ε. (4.21)
A t = 0
sup
ξ∈R
α
ε
(1 + ξ
2
)
−s/2
α
ε
+ e
−a(1+β)ξ
2
 C
2
(a, β, s)

a(1 + β)
ln
1
α
ε

s/2
(4.22)
C
2
(a, β, s) =


e
a(1+β)
s = 0,
max

1, e
a(1+β)

s
2

s/2

s > 0.
(4.23)
z(·, 0)  C
2
(a, β, s)

a(1 + β)
ln
1
α
ε

s/2
E + α
−1
1+β

ε
ε (4.24)





α

α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1)

− ( ϕ − v
α
ε
(ξ, 1 + β))




=






α
ε
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1)

− α
ε
v
α
ε
(ξ, 0)




= α
ε





1
α
ε
+ e

−a(1+β)ξ
2
u(ξ, 1)

−
1
α
ε
+ e
−a(1+β)ξ
2
ϕ




=




α
ε
α
ε
+ e
−a(1+β)ξ
2
(u(ξ, 1) − ϕ)





 sup
ξ∈R
α
ε
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1) − ϕ  u(ξ, 1) − ϕ = u(ξ, 1) − ϕ  ε.
ϕ − v
α
ε
(ξ, 1 + β) 




α
ε
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1)





+





α
ε
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1)

− ( ϕ − v
α
ε
(ξ, 1 + β))









α
ε
α
ε
+ e
−a(1+β)ξ
2
u(·, 1)




 τε − ε = (τ −1)ε
ε 
1
τ −1




α
ε
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 1)





. (4.25)
z(·, 0)  C
2
(a, β, s)

a(1 + β)
ln
1
α
ε

s/2
E + α
−1
1+β
ε
1
τ −1




α
ε
α
ε
+ e
−a(1+β)ξ

2
u(ξ, 1)




= C
2
(a, β, s)

a(1 + β)
ln
1
α
ε

s/2
E + α
−1
1+β
ε
1
τ −1





α
ε

e
−aξ
2
α
ε
+ e
−a(1+β)ξ
2
u(ξ, 0)





 C
2
(a, β, s)

a(1 + β)
ln
1
α
ε

s/2
E + α
−1
1+β
ε
1

τ −1
C(a, β, s)α
1
1+β
ε

a(1 + β)
ln
1
α
ε

s/2
E
=

C
2
(a, β, s) +
1
τ −1
C(a, β, s)


a(1 + β)
ln
1
α
ε


s/2
E
(τε)
2
= v
α
ε
(·, 1 + β) − ϕ
2
= v
α
ε
(·, 1 + β) − ϕ
2
=




α
ε
α
ε
+ e
−a(1+β)ξ
2
ϕ





2


α
ε
α
ε
+ 1

2
ϕ
2
=

α
ε
α
ε
+ 1

2
ϕ
2
.
τε 
α
ε
α
ε

+ 1
ϕ
ε
α
ε

ϕ − τε
τ
(4.26)
a(1 + β)
ln
1
α
ε
=
a(1 + β)
ln

E
ε
ε
α
ε
1
E


a(1 + β)
ln


E
ε
ϕ − τε
τ
1
E

=
a(1 + β)
ln
E
ε
ln
E
ε
ln

E
ε
ϕ − τε
τ
1
E

. (4.27)
lim
ε→0
+
ln
E

ε
ln

E
ε
ϕ − τε
τ
1
E

= 1. (4.28)
ε → 0
+
z(·, 0) 

C
2
(a, β, s) +
1
τ −1
C(a, β, s)


a(1 + β)
ln
E
ε

s/2
E(1 + o(1)). (4.29)

z(·, t)  z(·, 1)
t
z(·, 0)
1−t
 C
3
(τ, t, a, β, s)ε
t
E
1−t

ln
E
ε

−s(1−t)/2
(1 + o(1)) khi ε → 0
+
C
3
(τ, t, a, β, s) = (τ + 1)
t
(a(1 + β))
s(1−t)/2

C
2
(a, β, s) +
1
τ −1

C(a, β, s)

1−t
.

C(τ, a, β, s) = sup
t∈[0,1]
C
3
(τ, t, a, β, s) < +∞

[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.
[3] Dinh Nho H`ao, Nguyen Van Duc and D. Lesnic,
, Inverse problems, 25 (2009),055002,
27pp.
[4] , ,
¨o



∂u
∂t
= a
∂
2
u

∂x
2
(x, t) ∈ (−∞; +∞) × (0; 1),
u(·, 1) − ϕ(·)
L
2
(R)
 ε,
u(·, 0)
H
s
(R)
 E, E > ε > 0 s  0 a > 0 ϕ(·) ∈ L
2
(R)