Báo cáo nghiên cứu khoa học: "Đánh giá tính ổn định cho ph­ương trình dạng Burgers ng­ược thời gian" pps






u
t
= (a(x, t)u
x
)
x
− uu
x
, (x, t) ∈ (0, 1) × (0, 1),
u(0, t) = u(1, t) = 0, 0  t  1,
u(x, 1) = ϕ(x), 0  x  1.
1.
(1.1)





u
t
= (a(x, t)u
x
)
x
− uu
x

, (x, t) ∈ (0, 1) × (0, 1),
u(0, t) = u(1, t) = 0, 0  t  1,
u(x, 1) = ϕ(x), 0  x  1.
ϕ(x)
C
2,1
1
D = {(x, t) : 0 < x < 1, 0 < t < 1}. a(x, t) C
2,1
(D)
C
2,1
(D) x
t D D β > 0 a(x, t) 
β > 0, (x, t) ∈ D
1 u(x, t) V
u(x, t)
C
2,1
(D)
 M,
M
V
u
xx
a(t)
u
xx
2.
u

1
(x, t) u
2
(x, t)
ϕ
1
(x) ϕ
2
(x) z(x, t) = u
1
(x, t) − u
2
(x, t) f(t) =

1
0
z
2
(x, t)dx.
1 u
i
(x, t) ∈ V, i = 1, 2, c
1
c
2
u
i
(x, t)
f(t)  e
(c

2
/c
1
)[t−α]
[f(0)]
(1−α)
[f(1)]
α
, ∀t ∈ [0, 1],(2.1)
α =
exp{c
1
t} − 1
exp{c
1
} − 1
.(2.2)
u
i
(x, t), i = 1, 2,
z
t
= (az
x
)
x
− u
2
z
x

− u
1x
z.(2.3)
f(t) =

1
0
z
2
dx
f

(t) = 2

1
0
zz
t
dx
= 2

1
0
z{(az
x
)
x
− u
2
z

x
− u
1x
z}dx
= 2

1
0
z(az
x
)
x
dx − 2

1
0
u
2
zz
x
dx − 2

1
0
u
1x
z
2
dx
= 2az

x
z



1
0
− 2

1
0
a(z
x
)
2
dx − u
2
z
2



1
0
+

1
0
{u
2x

− 2u
1x
}z
2
dx
= −2

1
0
a(z
x
)
2
dx −

1
0
u
1x
z
2
dx −

1
0
z
2
z
x
dx

= −2

1
0
a(z
x
)
2
dx −

1
0
u
1x
z
2
dx
f

(t) = −2

1
0
a
t
(z
x
)
2
dx − 4


1
0
az
x
z
xt
dx
−

1
0
u
1xt
z
2
dx − 2

1
0
u
1x
zz
t
dx
= −2

1
0
a

t
(z
x
)
2
dx − 4az
x
z
t



1
0
+ 4

1
0
z
t
(az
x
)
x
dx
−

1
0
u

1xt
z
2
dx − 2

1
0
u
1x
zz
t
dx
= −2

1
0
a
t
(z
x
)
2
dx + 4

1
0
z
t
{z
t

+ u
2
z
x
+ u
1x
z}dx −

1
0
u
1xt
z
2
dx − 2

1
0
u
1x
zz
t
dx
= 4

1
0
z
2
t

dx − 2

1
0
a
t
(z
x
)
2
dx + 4

1
0
u
2
z
x
z
t
dx + 2

1
0
u
1x
zz
t
dx −


1
0
u
1xt
z
2
dx
= 4

1
0
z
2
t
dx − 2

1
0
a
t
(z
x
)
2
dx + 2

1
0
(2u
2

z
x
+ u
1x
z)z
t
dx +

1
0
2u
1t
zz
x
dx
= 4

1
0
z
2
t
dx − 2

1
0
a
t
(z
x

)
2
dx + 2

1
0
(2u
2
z
x
+ u
1x
z){(az
x
)
x
− u
2
z
x
− u
1x
z}dx
+

1
0
2u
1t
zz

x
dx
= 4

1
0
z
2
t
dx +

1
0
{4u
2
(a
x
− u
2
) − 2(au
1x
) −
1
2
(au
2
)
x
− 2a
t

}(z
x
)
2
dx
+2

1
0
(u
1x
a
x
− (au
1x
)
x
− 3u
1x
u
2
+ 2u
1t
)(zz
x
)dx − 2

1
0
(u

1x
)
2
z
2
dx.
u ∈ V a ∈ C
2,1
(
¯
D) c
3
u
1
, u
2
|u
1x
a
x
− (au
1x
)
x
− 3u
1x
u
2
+ 2u
1t

|  c
3
, ∀(x, t) ∈ D.
|bc| 
εb
2
2
+
c
2
2ε
ε > 0 b, c
2

1
0
(u
1x
a
x
− (au
1x
)
x
− 3u
1x
u
2
)(zz
x

)dx  −c
3

ε

1
0
(z
x
)
2
dx +
1
ε

1
0
z
2
dx

.
f

(t)  4

1
0
z
2

t
dx +

1
0
{4u
2
(a
x
− u
2
) − 2(au
1x
) −
1
2
(au
2
)
x
− 2a
t
− c
3
ε}(z
x
)
2
dx
−


1
0
(
c
3

+ 2u
2
1x
)z
2
dx =
= 4

1
0
z
2
t
dx − 2

1
0
{−2u
2
(a
x
− u
2

) + (au
1x
) +
1
4
(au
2
)
x
+ a
t
+
1
2
c
3
ε}(z
x
)
2
dx
−

1
0
(
c
3

+ 2u

2
1x
)z
2
dx =
= 4

1
0
z
2
t
dx − 2

1
0
{−2u
2
(a
x
− u
2
) + (au
1x
) +
1
4
(au
2
)

x
+ a
t
+
1
2
c
3
ε}
a
.a(z
x
)
2
dx
−

1
0
(
c
3

+ 2u
2
1x
)z
2
dx.
c

1
u
1
, u
2
{−2u
2
(a
x
− u
2
) + (au
1x
) +
1
4
(au
2
)
x
+ a
t
+
1
2
c
3
ε}
a
 c

1
, ∀(x, t) ∈ D.
f

(t)  4

1
0
z
2
t
dx − 2c
1

1
0
a(z
x
)
2
dx −

1
0
(
c
3

+ 2u
2

1x
)z
2
dx
= 4

1
0
z
2
t
dx + c
1
{f

(t) +

1
0
u
1x
z
2
dx} −

1
0
(
c
3


+ 2u
2
1x
)z
2
dx
= 4

1
0
z
2
t
dx + c
1
f

(t) +

1
0
{c
1
u
1x
− (
c
3


+ 2u
2
1x
)}z
2
dx.
c
2
u
1
, u
2
|c
1
u
1x
− (
c
3

+ 2u
2
1x
)|  c
2
, ∀(x, t) ∈ D.
f

(t)  4


1
0
z
2
t
dx + c
1
f

(t) − c
2

1
0
z
2
dx
= 4

1
0
z
2
t
dx + c
1
f

(t) − c
2

f(t).(2.4)
f(t)  0
f

(t).f(t)  4

1
0
z
2
t
dx.

1
0
z
2
dx + c
1
f

(t)f(t) − c
2
f
2
(t)


2


1
0
zz
t
dx

2
+ c
1
f

(t)f(t) − c
2
f
2
(t)
 (f

(t))
2
+ c
1
f

(t)f(t) − c
2
f
2
(t), ∀t ∈ (0, 1).(2.5)
f


(t).f(t) − (f

(t))
2
 c
1
f

(t)f(t) − c
2
f
2
(t), ∀t ∈ (0, 1).(2.6)
t
0
∈ (0, 1) f(t
0
) = 0
f(t) 0
f(t) >
0, t ∈ (0, 1)
f
2
(t)
f

(t).f(t) − (f

(t))

2
f
2
(t)
 c
1
f

(t)
f(t)
− c
2
, ∀t ∈ (0, 1).(2.7)

f

(t)
f(t)


 c
1
f

(t)
f(t)
− c
2
, ∀t ∈ (0, 1)
⇔ (

f

(t)
f(t)
)

− c
1
f

(t)
f(t)
 −c
2
, ∀t ∈ (0, 1)
⇔ e
−c
1
t
{(
f

(t)
f(t)
)

− c
1
f


(t)
f(t)
}  −c
2
e
−c
1
t
, ∀t ∈ (0, 1)
⇔ {e
−c
1
t
f

(t)
f(t)
}

 −c
2
e
−c
1
t
, ∀t ∈ (0, 1).(2.8)
s = e
c
1
t

d
ds

1
f
df
ds

 −
c
2
c
2
1
1
s
2
.
ψ
1
(s) = ln(f s
−
c
2
c
2
1
)
(ψ


1
(s)  0)
ln(exp{−
c
2
c
1
t}f(t)) 
exp{c
1
} − exp{c
1
t}
exp{c
1
} − 1
ln(exp{−
c
2
c
1
0}f(0))
+
exp{c
1
t} − 1
exp{c
1
} − 1
ln(exp{−

c
2
c
1
1}f(1)).(2.9)

1 u
i
(x, t) ∈ V, i = 1, 2,





u
t
= u
xx
− uu
x
, (x, t) ∈ (0, 1) × (0, 1),
u(0, t) = u(1, t) = 0, 0  t  1,
u(x, 1) = ϕ
i
(x), 0  x  1, (i = 1, 2)
ϕ
1
− ϕ
2
 = {


1
0
(ϕ
1
(x) − ϕ
2
(x))
2
dx}
1
2
 ε
u
1
(·, t) − u
2
(·, t)  e
(c
2
/c
1
)[t−α]
[2M]
(1−α)
ε
α
, ∀t ∈ [0, 1],(2.10)
 ·  L
2

(0, 1)
α =
exp{c
1
t} − 1
exp{c
1
} − 1
;
c
1
= 3M
2
+
9
4
M
c
2
= 3M
3
+
35
4
M
2
+
9
2
M.

2 u
i
(x, t) ∈ V, i = 1, 2,





u
t
= ((x + t + 1)u
x
)
x
− uu
x
, (x, t) ∈ (0, 1) × (0, 1),
u(0, t) = u(1, t) = 0, 0  t  1,
u(x, 1) = ϕ
i
(x), 0  x  1, (i = 1, 2)
ϕ
1
− ϕ
2
 = {

1
0
(ϕ

1
(x) − ϕ
2
(x))
2
dx}
1
2
 ε
u
1
(·, t) − u
2
(·, t)  e
(c
2
/c
1
)[t−α]
[2M]
(1−α)
ε
α
, ∀t ∈ [0, 1],(2.11)
 ·  L
2
(0, 1)
α =
exp{c
1

t} − 1
exp{c
1
} − 1
;
c
1
= 3M
2
+
21
2
M + 1
c
2
= 3M
3
+ 14M
2
+
7
2
M.
a(x, t) = 1, (x, t) ∈
D
c
3
= 3M
2
+ 3M

 =
2
3
c
1
= 3M
2
+
9
4
M
c
2
= 3M
3
+
35
4
M
2
+
9
2
M
a(x, t) = x + t + 1, (x, t) ∈ D
c
3
= 3M
2
+ 5M

 = 2
c
1
= 3M
2
+
21
2
M + 1
c
2
= 3M
3
+ 14M
2
+
7
2
M.
ε L
2
(0, 1)
L
2
(0, 1)
t ∈ (0, 1] V
a(x, t)
c
1
c

2
M
L
2
[1] K. A. Ames and B. Straughan, Aca-
demic Press, San Diego, 1997.
[2] A. Carasso,

Journal of Mathematical Analysis and Applications, 59 (1977), 169-209.
[3] L. E. Payne and B. Straughan,
, Int. J. Nonlinear Mech., 24 (1989), 209–214.
[4] S. M. Ponomarev, Soviet Math.
Dokl., 33 (1986), 621-624.
[5] A. Friedman, Prentice-Hall, Engle-
wood Cliffs, N. J., 1964.





u
t
= (a(x, t)u
x
)
x
− uu
x
, (x, t) ∈ (0, 1) × (0, 1),
u(0, t) = u(1, t) = 0, 0  t  1,

u(x, 1) = ϕ(x), 0  x  1.