¨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
∂

−
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

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
∨

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

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)
:=


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

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

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

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

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


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

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



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
. (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

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

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)


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

−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
)

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


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)

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
<

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)


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


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)

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

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

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

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

|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

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 


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


+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

−

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


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 + 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

.
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

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


+ 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

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

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) =

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

−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)


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



. (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

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

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




+ 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


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)

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