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In accordance with condition 3 of the theorem, for all
y
∈
R
ρ
the inequality
(
Py, y
)
6
λ
k
y
k
2
is fulfilled, therefore, according to Lemma 1, the solution
W
(
t
)
of the Cauchy problem
˙
W
=
PW
+
R
(
t
)
, W
(0) = 0
satisfies the inequality
k
W
(
t
∗
)
k
6
e
λt
∗
t
∗
Z
0
k
R
(
t
)
k
e
−
λt
dt.
(24)
Using the triangle inequality, condition 3 of the theorem and the inequality
0
6
D
(
t
)
6
1
, fulfilled with all
t
∈
[0
, t
∗
]
, we obtain
k
R
(
t
)
k
=
M
−
1
r
X
i
=1
∂p
∂z
i
d
(
i
−
1)
(
t
) +
PD
(
t
)
6
6
k
M
−
1
k
r
X
i
=1
∂p
∂z
i
|
d
(
i
−
1)
(
t
)
|
+
k
P
k
D
(
t
)
6
6
k
M
−
1
k
r
X
i
=1
ε
|
d
(
i
−
1)
(
t
)
|
+
k
P
k
6
k
M
−
1
k
εL
+
k
P
k
.
With regard to this estimation and notation (16), inequality (24) is taking
the form of
k
W
(
t
∗
)
k
6
e
λt
∗
t
∗
Z
0
(
k
M
−
1
k
εL
+
k
P
k
)
e
−
λt
dt
=
γ.
If
γ <
1
, then
k
v
0
(
c
)
k
=
k
W
(
t
∗
)
k
6
γ <
1
and, consequently, the mapping
v
is compressing. Thus, if the theorem conditions are satisfied, the mapping
v
is compressing and has a fixed point
c
∗
. With
c
1
=
c
1
∗
, . . . , c
ρ
=
c
ρ
∗
,
c
ρ
+1
= 0
, . . . , c
m
= 0
, the solution
η
(
t
)
of Cauchy problem (5) satisfies the
condition
η
(
t
∗
) =
η
∗
. The functions
B
1
(
t
) =
b
1
(
t
) +
c
1
∗
d
1
(
t
)
, . . . , B
ρ
(
t
) =
b
ρ
(
t
) +
c
ρ
∗
d
ρ
(
t
)
,
B
ρ
+1
=
b
ρ
+1
(
t
)
, . . . , B
m
(
t
) =
b
m
(
t
)
satisfy all the conditions of theorem 2, hence terminal problem (3), (4) for
system (2) has got a solution.
Numerical procedure.
The method of construction of terminal prob-
lem (3) solution, (4) for system (2) results from theorem 3 proving. Let us
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2014. No. 5
25