7 / 15
The following function is the solution to the differential equation
˙
v
=
λv
+
+
k
S
(
t
)
k
with the initial condition
v
(
t
0
) = 0
:
v
(
t
) =
e
λt
t
Z
t
0
k
S
(
τ
)
k
e
−
λτ
dτ,
That is why, with all
t
∈
[
t
0
, t
∗
]
inequality (12) is true:
k
V
(
t
)
k
6
e
λt
Z
t
t
0
k
S
(
τ
)
k
e
−
λτ
dτ
and, consequently,
k
V
(
t
∗
)
k
6
e
λt
∗
t
∗
Z
t
0
k
S
(
t
)
k
e
−
λt
dt.
(15)
From the non-negativity of the subintegral function in the right part of
inequality (15) we have
t
∗
Z
t
0
k
S
(
t
)
k
e
−
λt
dt
6
Z
t
∗
0
k
S
(
t
)
k
e
−
λt
dt,
and with this, we get inequality (13) from (15).
Now we can prove the main result.
Theorem 3.
Let us assume the following:
1)
q
(
z
1
, . . . , z
m
, η
) =
r
P
i
=1
A
i
z
i
+
Kη
+
p
(
z
1
, . . . , z
m
)
where
A
1
, . . . , A
r
,
K
are
ρ
×
ρ
-matrixes
;
2)
matrix
M
=
A
1
+
KA
2
+
K
2
A
3
+
. . .
+
K
r
−
1
A
r
is nondegenerated
;
3)
there is such
ε >
0
, that for all
i
= 1
, r
and
(
z
1
, . . . , z
m
)
∈
R
n
−
ρ
the inequalities
k
∂p/∂z
i
k
6
ε
are fulfilled
;
4)
λ
is the largest proper number of the matrix
(
P
+
P
T
)
/
2
, where
P
=
M
−
1
KM
;
γ
=
(
k
M
−
1
k
εL
+
k
P
k
)
t
∗
,
if
λ
= 0;
(
k
M
−
1
k
εL
+
k
P
k
)
e
λt
∗
−
1
λ
,
if
λ
6
= 0
.
(16)
If
γ <
1
, then terminal problem (3), (4) for system (2) has got a solution.
J
Let us assume
c
ρ
+1
=
. . .
=
c
m
= 0
, then denote the vector with
unknown parameters by
c
= (
c
1
, . . . , c
ρ
)
T
. Then Cauchy problem (5) will
take the form of
˙
η
=
q
(
b
1
(
t
) +
c
1
d
1
(
t
)
, . . . , b
ρ
(
t
) +
c
ρ
d
ρ
(
t
)
, b
ρ
+1
(
t
)
, . . . , b
m
(
t
)
, η
);
η
(0) =
η
0
.
(17)
22
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2014. No. 5