6 / 15
Let us denote
V
(
t
) =
W
1
(
t
)
W
2
(
t
)
. . .
W
ρ
(
t
)
, Q
(
t
) =
P
(
t
) 0
. . .
0
0
P
(
t
)
. . .
0
...
...
. . .
...
0 0
. . . P
(
t
)
, S
(
t
) =
R
1
(
t
)
R
2
(
t
)
. . .
R
ρ
(
t
)
and put down Cauchy problem (11) in the form of
˙
V
=
Q
(
t
)
V
+
S
(
t
)
,
V
(0) = 0
.
As the Euclidean norms of the matrixes
W
(
t
)
and
R
(
t
)
coincide with the
Euclidean norms of the vectors
V
(
t
)
and
S
(
t
)
, then, to prove inequality
(12) it is sufficient to show that
k
V
(
t
∗
)
k
6
e
λt
∗
t
∗
Z
0
k
S
(
t
)
k
e
−
λt
dt.
(13)
As it follows from inequality (10), for any
t
∈
[0
, t
∗
]
and
V
=
= (
V
T
1
, . . . , V
T
ρ
)
T
∈
R
ρ
2
, where
V
j
∈
R
ρ
, the estimation
(
Q
(
t
)
V, V
) = (
P
(
t
)
V
1
, V
1
) +
. . .
+ (
P
(
t
)
V
ρ
, V
ρ
)
6
6
λ
k
V
1
k
2
+
. . .
+
λ
k
V
ρ
k
2
=
λ
k
V
k
2
.
(14)
is true.
Let us use (14) to prove inequality (13). Note that, if
V
(
t
∗
) = 0
,
then
k
V
(
t
∗
)
k
= 0
and the correctness of inequality (13) results from the
non-negativity of its right part.
If
V
(
t
∗
)
6
= 0
, then we can denote as
t
0
the exact upper boundary of
these
t
from the range
[0;
t
∗
)
, for which
V
(
t
) = 0
. Then
V
(
t
0
) = 0
, for
all
t
∈
(
t
0
;
t
∗
)
the inequality
V
(
t
)
6
= 0
is fulfilled. In the range
(
t
0
;
t
∗
)
we
can calculate and estimate
d
dt
k
V
k
, using inequality (14) and the Cauchy –
Bunyakowsky inequality:
d
dt
k
V
k
=
(
V,
˙
V
)
k
V
k
=
1
k
V
k
[(
Q
(
t
)
V, V
) + (
S
(
t
)
, V
)]
6
6
λ
k
V
k
2
k
V
k
+
S
(
t
)
,
V
k
V
k
6
λ
k
V
k
+
k
S
(
t
)
k
.
Thus, in the range
(
t
0
;
t
∗
)
the function
k
V
(
t
)
k
satisfies the differential
inequality
d
dt
k
V
k
6
λ
k
V
k
+
k
S
(
t
)
k
.
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2014. No. 5
21