9 / 15
we can write the inequality
k
v
0
(
c
)
k
6
γ <
1
. Therefore, we can prove that
the mapping
v
is compressing. We calculate
˙
W
with the help of (18):
˙
W
=
d
(
t
)
E
+
r
X
i
=2
d
(
i
−
1)
(
t
)
N
i
−
1
−
M
−
1
˙
ν
=
=
d
(
t
)
E
+
r
X
i
=2
d
(
i
−
1)
(
t
)
N
i
−
1
−
M
−
1
"
Kν
+
r
X
i
=1
A
i
+
∂p
∂z
i
d
(
i
−
1)
(
t
)
#
.
(20)
Taking (19) into consideration, we express
ν
(
t
)
in terms of
W
(
t
)
:
ν
(
t
) =
M
D
(
t
)
E
+
r
−
1
X
i
=1
d
(
i
−
1)
(
t
)
N
i
−
W
(
t
)
!
,
substitute the received relation in (20). As a result, we have the equality
˙
W
=
PW
+ [
E
−
M
−
1
A
1
−
PN
1
]
d
(
t
)+
+
r
−
1
X
i
=2
[
N
i
−
1
−
M
−
1
A
i
−
PN
i
]
d
(
i
−
1)
(
t
)+
+[
N
r
−
1
−
M
−
1
A
r
]
d
(
r
−
1)
(
t
)
−
M
−
1
r
X
i
=1
∂p
∂z
i
d
(
i
−
1)
(
t
)
−
PD
(
t
)
.
(21)
Now we choose matrices
N
1
, . . . , N
r
−
1
from the condition
E
−
M
−
1
A
1
−
PN
1
= 0;
N
i
−
1
−
M
−
1
A
i
−
PN
i
= 0
, i
= 2
, r
−
1;
N
r
−
1
−
M
−
1
A
r
= 0
.
(22)
With a direct substitution, we can show that the solution to system (22) of
the matrix equations is the matrices
N
1
=
M
−
1
(
A
2
+
KA
3
+
. . .
+
K
r
−
2
A
r
);
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
N
r
−
2
=
M
−
1
(
A
r
−
1
+
KA
r
);
N
r
−
1
=
M
−
1
A
r
.
(23)
Let us assume that the matrices
N
i
are specified by formulae (23). Then
equality (21) will take the form of
˙
W
=
PW
+
R
(
t
)
,
where
R
(
t
) =
−
M
−
1
r
X
i
=1
∂p
∂z
i
d
(
i
−
1)
(
t
)
−
PD
(
t
)
.
24
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2014. No. 5