11 / 15
take a random number
c
(0)
∈
R
ρ
and build a sequence of approximations
{
c
(
j
)
}
according to the rule
c
(
j
+1)
=
c
(
j
)
−
M
−
1
(Ψ(
c
(
j
)
)
−
η
∗
)
, j
= 0
,
1
, . . .
(25)
In order to determine the value
Ψ(
c
(
j
)
)
, it is necessary to find the solution
η
(
t, c
(
j
)
)
to the Cauchy problem
˙
η
=
q
(
b
1
(
t
) +
c
(
j
)
1
d
1
(
t
)
, . . . , b
ρ
(
t
) +
c
(
j
)
ρ
d
ρ
(
t
)
, b
ρ
+1
(
t
)
, . . . , b
m
(
t
)
, η
);
η
(0) =
η
0
.
Then
Ψ(
c
(
j
)
) =
η
(
t
∗
, c
(
j
)
)
.
As the mapping
v
is compressing, the sequence
{
c
(
j
)
}
converges to the
fixed point
c
∗
of the mapping
v
. Thereby, the estimation is true
k
c
(
j
)
−
c
∗
k
6
γ
j
1
−
γ
k
c
(1)
−
c
(0)
k
.
(26)
It follows from (25) that
Ψ(
c
(
j
)
)
−
η
∗
=
M
(
c
(
j
+1)
−
c
(
j
)
)
,
hence, using the triangle inequality and estimation (26), we obtain
k
Ψ(
c
(
j
)
)
−
η
∗
k
6
k
M
kk
c
(
j
+1)
−
c
(
j
)
k
=
k
M
kk
c
(
j
+1)
−
c
∗
+
c
∗
−
c
(
j
)
k
6
6
k
M
kk
c
(
j
+1)
−
c
∗
k
+
k
M
kk
c
∗
−
c
(
j
)
k
6
k
M
k
1
−
γ
(
γ
j
+1
+
γ
j
)
k
c
(1)
−
c
(0)
k
=
=
(1 +
γ
)
γ
j
1
−
γ
k
M
kk
c
(1)
−
c
(0)
k
.
Having chosen the number
J
from the condition
(1 +
γ
)
γ
J
1
−
γ
k
M
kk
c
(1)
−
c
(0)
k
6
σ,
where
σ >
0
is a given accuracy, we try to obtain the inequality fulfilment
k
Ψ(
c
(
J
)
)
−
η
∗
k
6
σ.
(27)
The vector-functions
z
1
=
b
1
(
t
) +
c
(
J
)
1
d
1
(
t
)
, . . . , z
ρ
=
b
ρ
(
t
) +
c
(
J
)
ρ
d
ρ
(
t
)
,
z
ρ
+1
=
b
ρ
+1
(
t
)
, . . . , z
m
=
b
m
(
t
);
η
=
η
(
t, c
(
J
)
)
, t
∈
[0
, t
∗
]
,
26
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2014. No. 5