Information Technology Reference
In-Depth Information
3 Main Results
(
tx
will be attracted by
KerW
eventually. So
we may say that
KerW
is a weak attractor for the solutions of system (1).
)
Formula (3) shows that the solutions
M
>
0
Theorem 1.
Assume that there exist matrix
and diagonal matrices
U
=
diag
(
u
,
"
,
u
)
≥
0
i
=
1
2
H
≥
H
≥
0
such that
where
i
i
1
in
1
2
⎛
−
1
T
T
⎞
MC
+
CM
−
MBM
B
M
−
2
P
MP
−
MA
−
LU
⎜
⎜
⎝
⎟
⎟
⎠
H
=
1
,
1
T
(
−
MA
−
LU
)
2
U
1
1
⎛
T
⎞
2
Q
MQ
LU
⎜
⎜
⎝
⎟
⎟
⎠
H
=
2
,
2
T
(
LU
)
M
−
2
U
2
2
n
KerW
=
{
x
∈
R
:
W
(
x
)
=
0
Then
is nonempty and formula (3) holds, where
.
T
T
T
T
W
(
x
)
=
z
(
x
)
H
z
(
x
),
z
(
x
)
=
(
x
,
f
(
x
))
H
>
H
≥
0
In particular, if
i
i
1
2
KerW
=
{
then
and the solution of system (1) will tend to zero asymptotically
with probability 1.
T
V
(
x
)
=
x
(
t
)
Mx
(
t
).
Proof.
Let
Then one obtains
LV
(
x
,
y
,
t
)
=
2
x
T
(
t
)
MF
(
x
(
t
),
y
(
t
))
+
trace
{
G
T
(
x
,
y
)
MG
(
x
,
y
)}
(4)
T
T
−
1
T
≤
2
x
(
t
)
M
[
−
Cx
(
t
)
+
Af
(
x
(
t
))]
+
x
(
t
)
MBM
B
Mx
(
t
)
+
f
T
(
y
(
t
))
Mf
(
y
(
t
))
+
2
x
T
(
t
)
P
T
MPx
(
t
)
+
y
T
(
t
)
Q
T
MQy
(
t
)].
On the other hand, one obtains
n
∑
0
≤
−
2
u
[
f
(
x
(
t
))
−
l
x
(
t
)]
f
(
x
(
t
)),
(5)
1
i
i
i
i
i
i
i
i
=
1
n
∑
0
≤
−
2
u
[
f
(
y
(
t
))
−
l
y
(
t
)]
f
(
y
(
t
))
(6)
.
2
i
i
i
i
i
i
i
i
=
1
Thus, one derives
LV
(
x
,
y
,
t
)
≤
−
W
(
x
(
t
))
+
W
(
y
(
t
)),
1
2
T
T
T
T
W
(
x
)
=
z
(
x
)
H
z
(
x
),
z
(
x
)
=
(
x
,
f
(
x
))
.
in which
From Lemma 1, one
i
i
obtains (3).
Search WWH ::
Custom Search