Báo cáo nghiên cứu khoa học: "Về một điều kiện đủ cho tính ổn định mũ của một lớp phương trình vi phân ngẫu nhiên có trễ" pot

dx(t) = f(x(t), x(t − τ), t)dt + σ(t, x(t))dw(t).
dx(t) = f(x(t), x(t − τ), t)dt.
dx(t) = f(x(t), x(t−τ), t)dt+σ(t)dw(t). (1.1)
dx(t) = f(x(t), x(t−τ), t)dt+σ(t, x(t))dw(t). (1.2)
(Ω, , {
t
}
t0
, P )
{
t
}
t0
|x| x ∈ R
n
A
A = sup{|Ax| : |x| = 1} B
T
1
A = (a
ij
) Trace(A) =

a
ii
τ
C([−τ, 0]; R
d
) R
d
−

[−τ, 0] L
2

t
([−τ, 0]; R
d
) 
t
− C([−τ, 0]; R
d
)
ξ = {ξ(u) : −τ  u  0}
ξ
2
E
= sup
−τ u0
E|ξ(u)|
2
< ∞.
dx(t) = f(x(t), x(t−τ), t)dt+σ(t, x(t))dw(t); on t ≥ 0 (2.1)
x(t) = ξ(u) −τ  u  0 f : R
d
× R
d
× R
+
→ R
d
,

σ : R
d
×R
+
→ R
d×m
w ξ ∈ L
2

0
([−τ, 0]; R
d
)
x(t, ξ)
(2.1)
δ K
ξ ∈ L
2

0
([−τ, 0]; R
d
)
E|x(t, ξ)|  Kξ
2
E
e
−δt
, ∀t ≥ 0. (2.2)
δ K

u(t) v(t)
N
0
t ≥ s
u(t)  N
0
+

t
s
u(t
1
)v(t
1
)dt
1
.
t ≥ s
u(t)  N
0
exp{

t
s
v(t
1
)dt
1
}. (2.3)
c

1
− c
3
2x
T
f(x, y, t)  −c
1
|x|
2
+ c
2
|y|
2
,
T race(σ(t, x)σ
T
(t, x))  c
3
|x|
2
,
c
2
e
c
1
τ
+ c
3
< c

1
,
x, y ∈ R
d
; t  0. (2.1)
Proof. For all ξ ∈ L
2
F
0
([−τ, 0]; R
d
) Fix ξ arbitrarily and write x(t, ξ) = x(t) simple. By Ito’s
formula and assumption,
e
c
1
t
|x(t)|
2
= |x(0)|
2
+ M(t) + N(t)
for all t ≥ 0, where
M(t) = 2

t
0
e
c
1

s
x
T
(s)σ(s, x(s))dw(s).
N(t) =

t
0
e
c
1
s
(c
1
|x(s)|
2
+ 2x(s)
T
f(x(s), x(s − τ), s) + trace(σ(s, x(s))σ
T
(s, x(s)))ds.
By (i), (ii) we have
e
c
1
t
|x(t)|
2
 |x(0)|
2

+M(t)+

t
0
e
c
1
s
(c
2
|x(s−τ)|
2
+c
3
|x(s)|
2
)ds. (4.4)
But

t
0
e
c
1
s
|x(s − τ)|
2
ds 

τ

0
e
c
1
s
|x(s − τ)|
2
ds +

max {τ,t}
τ
e
c
1
s
|x(s − τ)|
2
ds


τ
0
e
c
1
s
|x(s−τ)|
2
ds+e
c

1
τ

t
0
e
c
1
s
|x(s)|
2
ds. (4.5).
From inequalities (4.5) and (4.4) follow that
e
c
1
t
|x(t)|
2
 |x(0)|
2
+ M(t) +

τ
0
e
c
1
s
c

2
|x(s − τ)|
2
ds +

t
0
(c
3
+ c
2
e
c
1
τ
)e
c
1
s
|x(s)|
2
ds
because EM(t) = 0, moreover we have

τ
0
e
c
1
s

c
2
E|x(s − τ )|
2
ds 
c
2
c
1
(e
c
1
τ
− 1)ξ
2
E
; E|x(0)|
2
 ξ
2
E
so that
e
c
1
t
E|x(t)|
2
 E|x(0)|
2

+

τ
0
e
c
1
s
c
2
E|x(s − τ )|
2
ds +

t
0
(c
3
+ c
2
e
c
1
τ
)e
c
1
s
E|x(s)|
2

ds
 (1+
c
2
c
1
(e
c
1
τ
−1))ξ
2
E
+

t
0
(c
3
+c
2
e
c
1
τ
)e
c
1
s
E|x(s)|

2
ds. (4.6).
From (4.6) and applying lemma 2.2 with
u(t) = e
c
1
t
E|x(t)|
2
; v(t) = c
3
+ c
2
e
c
1
τ
; N
0
= (1 +
c
2
c
1
(e
c
1
τ
− 1))ξ
2

E
we have
e
c
1
t
E|x(t)|
2
 (1 +
c
2
c
1
(e
c
1
τ
− 1))ξ
2
E
e
(c
3
+c
2
e
c
1
τ
)t

.
Hence we obtain
E|x(t)|
2
 (1 +
c
2
c
1
(e
c
1
τ
− 1))ξ
2
E
e
(c
3
+c
2
e
c
1
τ
−c
1
)t
By assumptions (iii) we can rewrite
E|x(t)|

2
 Kξ
2
E
e
−δt
,
where K = 1 +
c
2
c
1
(e
c
1
τ
− 1) > 0 and δ = c
1
− c
3
− c
2
e
c
1
τ
> 0.
In other words,the stochastic differential equations (2.1) is e xponential stability in mean
square. The proof is completed. 
Acknowledgement.

dx(t) = f(x(t), x(t − τ), t)dt + σ(t, x(t))dw(t).
dx(t) = f(x(t), x(t − τ), t)dt.