12 / 15
specify the
t
-parameter curve in the range of system (2) conditions,
connecting states (3) and (4). The control implementing this trajectory as
system (2) trajectory, can be found using formula (7), if we assume that
B
1
(
t
) =
b
1
(
t
) +
c
(
J
)
1
d
1
(
t
)
, . . . , B
ρ
(
t
) =
b
ρ
(
t
) +
c
(
J
)
ρ
d
ρ
(
t
)
,
B
ρ
+1
=
b
ρ
+1
(
t
)
, . . . , B
m
(
t
) =
b
m
(
t
);
η
(
t
) =
η
(
t, c
(
J
)
)
.
Example
. Let us consider the system
˙
z
i
1
=
z
i
2
;
˙
z
i
2
=
u
i
,
i
= 1
,
2;
˙
η
1
=
−
0
.
1
η
2
+
z
1
1
+
z
2
2
+ 0
.
08 cos
z
1
2
;
˙
η
2
= 0
.
1
η
1
+
z
2
1
+
z
1
2
−
0
.
08 sin
z
2
2
(28)
with the following boundary conditions:
z
1
1
(0) = 0
, z
1
2
(0) = 0
, z
2
1
(0) = 0
, z
2
2
(0) = 0
, η
1
(0) = 0
, η
2
(0) = 0
,
z
1
1
(2) =
−
4
, z
1
2
(2) =
−
8
, z
2
1
(2) = 0
, z
2
2
(2) = 4
, η
1
(2) =
−
5
, η
2
(2) = 4
.
For this task
t
∗
= 2
,
m
= 2
,
ρ
= 2
,
r
1
=
r
2
= 2
,
z
1
= (
z
1
1
, z
2
1
)
T
,
z
2
= (
z
1
2
, z
2
2
)
T
,
A
1
=
1 0
0 1
, A
2
=
0 1
1 0
, K
=
0
−
0
,
1
0
,
1 0
,
p
(
z
1
, z
2
) =
0
.
08 cos
z
1
2
−
0
.
08 sin
z
2
2
, M
=
A
1
+
KA
2
=
0
.
9 0
0 1
.
1
,
M
−
1
=
10
/
9 0
0 10
/
11
, P
=
M
−
1
KM
=
0
−
11
/
90
9
/
110 0
,
∂p
∂z
1
=
0 0
0 0
,
∂p
∂z
2
=
−
0
.
08 sin
z
1
2
0
0
−
0
.
08 cos
z
2
2
.
Since
k
∂p/∂z
1
k
= 0
, а
k
∂p/∂z
2
k
6
0
.
08
√
2
, we can assume
ε
= 0
.
08
√
2
as the number
ε
from condition 3 of theorem 3. Matrix
(
P
+
P
T
)
/
2
has
the form of
1
2
(
P
+
P
T
) =
0
−
2
/
99
−
2
/
99 0
,
Its largest proper value
λ
= 2
/
99
.
Let us check the fulfilment of theorem 3 conditions. The function
d
(
t
)
,
built with formula (9), has the form of
d
(
t
) =
15
16
t
2
(2
−
t
)
2
, therefore,
d
0
(
t
) =
15
4
t
(
t
−
1)(
t
−
2);
L
= max
[0
,
2]
{
d
(
t
) +
|
d
0
(
t
)
|}
6
1
.
95
.
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2014. No. 5
27