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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