Колесников А.А. Синергетические методы управления сложными системами: теория системного синтеза
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˙z
1
(t)=z
2
, ˙z
2
(t)=z
3
,..., ˙z
p−1
= z
p
;˙z
p
(t)=a(x)+b(x)u.
˙z
1
(t)=z
2
, ˙z
2
(t)=z
3
,..., ˙z
p−1
= z
p
;˙z
p
(t)=a(x)+b(x)u;
˙z
p+1
(t)=ϕ
p+1
(z); ...;˙z
n
= ϕ
n
(z).
z ∈R
n
ϕ
j
(·)
z
k
k = 1,p
z
k
= γ
k
(x)=L
k
f
h = L
f
L
f
...
L
f
(h(x))
,
L
f
h = ∂h/∂xA(x) h A(x)
ϕ
i
(z) p +1 i n
ϕ
i
(z)=L
f
γ
i
Γ
−1
(z)
.
γ
k
(x) ϕ(z)
Γ(x) L
g
γ
i
(x)=0
rank Γ(x)=n,
L
g
γ
i
(x)=0.
γ
i
rank Γ(x
0
)=n
x =Γ
−1
(z) x
0
p<n Γ(x)
˙x
1
(t)=f(x
1
)+g(x
1
)u, x
1
,u∈R
u = α
1
(x
1
)
x
1
=0 V (x
1
)
∂V (x
1
)
∂x
1
f(x
1
)+g(x
1
)α
1
(x
1
)
−W (x
1
),
W (x
1
)
˙x
1
(t)=f(x
1
)+g(x
1
)x
2
,
˙x
2
(t)=u,
x
2
u = −
x
2
− α
1
(x
1
)
+
∂α
1
∂x
1
f(x
1
)+g(x
1
)x
2
−
∂V (x
1
)
∂x
1
g(x
1
)
x
1
(t) x
2
(t) x
1
→ 0 (x
2
−α
1
) → 0 t →∞
ψ
1
∆
= x
2
−α
1
(x
1
)
˙x
1
(t)=f(x
1
)+g(x
1
)
α
1
(x
1
)+ψ
1
,
˙
ψ
1
(t)=u −
∂α
1
∂x
1
f(x
1
)+g(x
1
)x
2
.
V
1
= V (x
1
)+
1
2
ψ
2
1
˙
V (t)=
∂V (x
1
)
∂x
1
[f + gα
1
]+
∂V (x
1
)
∂x
1
gψ
1
− ψ
2
1
−
∂V (x
1
)
∂x
1
gψ
1
−W (x
1
) − ψ
2
1
.
˙x
1
(t)=x
2
, ˙x
2
(t)=sinx
1
+ x
3
, ˙x
3
(t)=u.
ψ
1
= x
1
α
1
= −ψ
1
= −x
1
V
1
=1/2x
2
1
ψ
2
=
x
2
− α
1
α
2
= −ψ
2
+
∂α
1
∂x
1
x
2
−
∂V
1
∂x
1
− sin x
1
= −2x
1
− 2x
2
− sin x
1
.
V
2
= V
1
+1/2ψ
2
2
ψ
3
= x
2
−α
1
u = −ψ
3
+
∂α
2
∂x
1
x
2
+
∂α
2
∂x
2
(sin x
1
+ x
3
) −
∂V
1
∂x
1
−
∂V
1
∂ψ
2
=
= −4x
1
− (5 + cos x
1
)x
2
− 3x
3
− 3sinx
1
.
ψ
1
= x
3
− α
2
J
1
=
∞
0
m
2
1
ψ
2
1
+ c
2
1
˙
ψ
2
1
(t)
dt,
m
1
> 0 c
1
> 0
u =
∂α
2
∂x
1
x
2
−
∂α
2
∂x
2
(sin x
1
+ x
3
) −
1
T
1
α
2
,
T
1
= c
1
/m
1
ψ
1
=0
˙x
1ψ
(t)=x
2ψ
, ˙x
2ψ
(t)=sinx
1ψ
− α
2
.
ψ
2
= β
21
x
1
+ x
2
− α
2
,
β
21
> 0 α
2
α
2
=
β
21
T
2
x
1
+
β
21
+
∂α
1
∂x
1
+
1
T
2
x
2
+sinx
1
+
1
T
2
α
3
,
ψ
1
ψ
2
T
2
= c
2
/m
2
ψ
2
=0
˙x
1ψ
(t)=−β
21
x
1ψ
− u
3
,
α
3
α
2
u = −
β
21
T
1
T
2
x
1
−
β
21
T
1
+
β
21
T
2
+
1
T
1
T
2
+cosx
1
x
2
−
−
β
21
+
1
T
2
+
1
T
2
x
3
−
β
21
+
1
T
2
+
1
T
2
sin x
1
.
T
1
T
2
β
21
β
21
T
i
˙x
1
(t)=−x
1
+ x
3
1
x
2
2
,
˙x
2
(t)=x
2
(x
3
+ x
4
),
˙x
3
(t)=x
4
,
˙x
4
(t)=u.
ψ = x
3
+ x
4
+ α|x
m
2
| + βx
2
2
+ γx
4
1
,
m 1 ψ
T
˙
ψ(t)+ψ =0,
u = −x
4
−
αm|x
m
2
| +2βx
2
2
(x
3
+ x
4
) − 4γx
4
1
(x
2
1
x
2
2
− 1) −
1
T
ψ.
ψ =0
˙x
1ψ
(t)=−x
1ψ
+ x
3
1ψ
x
2
2ψ
,
˙x
2ψ
(t)=−αx
2ψ
|x
m
2
|−βx
3
2ψ
− γx
4
1ψ
x
2ψ
.
V(t)=0, 5x
2
1ψ
+0, 5x
2
2ψ
,
˙
V(t)=−x
2
1ψ
− αx
2
2ψ
|x
m
2
|−βx
4
2ψ
− (γ − 1)x
4
1ψ
x
2
2ψ
.
α>0; β>0; γ 1
˙
V(t)
x
1ψ
x
2ψ
T>0
x
1ψ
= x
2ψ
=0
x
3ψ
x
4ψ
ψ =0
x
3ψ
+ x
4ψ
= −α|x
m
2ψ
|−βx
2
2ψ
− γx
4
1ψ
,
x
1ψ
= x
2ψ
=0 x
3ψ
x
4ψ
ψ =0
˙x
3ψ
(t)=−x
3ψ
˙x
4ψ
(t)=−x
4ψ
x
3ψ
x
4ψ
m =0
x
1ψ
= x
2ψ
=0
x
3ψ
= −α
x
4ψ
=0
x
3ψ
= −α x
3
ψ
β =0
α =0
u
1
= −
1
T
x
3
−
1+
1
T
x
4
− 2βx
2
2
x
3
+ x
4
+
0, 5
T
+4γx
4
1
− 4γx
4
1
x
2
1
x
2
2
+
0, 25
T
.
u = −x
3
− 2x
4
− x
2
2
(x
3
+ x
4
+2)+4x
4
1
,
u
1
u
x
2
x
1
x
2
u
α = β = T = γ = m =1 α =10 β =0 T = m =1
γ =20 α = β = T = γ =1 m =0
˙x
1
(t)=x
2
+ x
2
3
,
˙x
2
(t)=x
3
,
˙x
3
(t)=u,
x
2
3
x
2
ψ
1
= x
3
+ γ
1
x
1
+ γ
2
x
2
+ β
1
x
3
2
.
ψ
1
T
1
˙
ψ
1
(t)+ψ
1
=0,T
1
> 0,
u
1
= −γ
1
x
2
− γ
1
x
2
3
− (γ
2
+3β
1
x
2
2
)x
3
−
1
T
1
ψ
1
.
u
1
ψ
1
=0
x
2ψ
¨x
2ψ
(t)+
γ
2
+3β
1
x
2
2ψ
+ γ
1
˙x
2ψ
(t)
˙x
2ψ
(t)+γ
1
x
2ψ
=0.
γ
1
> 0,γ
2
> 0
˙x
2
2ψ
≈ 0 x
2
2ψ
˙x
2ψ
≈ 0
γ
2
+3β
1
x
2
2
+ γ
1
x
3
0.
β
1
=0
x
30
> −
γ
2
γ
1
.
x
30
< 0
3β
1
x
2
2
u
1
τ
2
¨x
2ψ
(t)+2ξτ ˙x
2ψ
(t)+x
2ψ
=0,
ξ =
0, 5γ
2
√
γ
1
τ =
1
√
γ
1
γ
1
x
30
< 0 ξ =1
γ
1
γ
2
γ
2
=2
√
γ
1
β
1
=0 x
10
x
20
x
30
γ
2
+ γ
1
x
30
0
β =1 β =10
x
30
< 0
x
10
= −2 x
20
=2 x
30
=1 x
10
=0, 7 x
20
= −0, 8 x
30
= −2
β =1 β =10
x
3ψ
ψ
1
=0 β
1
=0
¨x
3ψ
(t)+
2γ
1
x
3ψ
+ γ
2
˙x
3ψ
(t)+γ
1
x
3ψ
=0,
x
3ψ
=0
x
3ψ0
−
γ
2
2γ
1
,
x
3
|x
3
| b =
γ
2
2γ
1
ψ
2
= x
3
+ b th(γ
1
x
1
+ γ
2
x
2
+ β
1
x
3
2
).
T
2
˙
ψ
2
(t)+ψ
2
=0
u
2
= −
b(γ
1
x
2
+ γ
1
x
2
3
+ γ
2
x
3
+3β
1
x
2
2
x
3
)
ch
2
(γ
1
x
1
+ γ
2
x
2
+ β
1
x
3
2
)
−
1
T
2
ψ
2
.
x
3
β
1
=0 x
30
< 0 β
1
x
30
< 0
β
1
=0
β
1
=1 β
1
=10
˙x
1
(t)=x
1
+ x
2
+ x
3
3
,
˙x
2
(t)=x
3
,
˙x
3
(t)=u,
x
3
2
x
1
u = k
1
x
1
+ k
2
x
2
+ k
3
x
3
+ βx
3
3
.
ψ
1
= x
3
+ γ
1
x
1
+sinγ
2
x
2
.
T
1
˙
ψ
1
(t)+ψ
1
=0,T
1
> 0,
u
1
= −γ
1
(x
1
+ x
2
+ x
3
3
) − γ
2
x
3
cos γ
2
x
2
−
1
T
1
ψ
1
.
t =(3÷ 4)T
1
x
i0
ψ
1
=0 ψ
1
=0
x
2
¨x
2ψ
(t)+
γ
2
cos γ
2
x
2ψ
− 1+γ
1
˙x
2
2ψ
(t)
˙x
2ψ
(t)+γ
1
x
2ψ
− sin γ
2
x
2ψ
=0.
γ
1
γ
2
div
˙
x
2ψ
(t)=1−γ
2
cos γ
2
x
2ψ
− 3γ
1
˙x
2
2ψ
(t).
γ
2
> 0
γ
2
< 1 γ
2
< 0
x
2ψ
˙x
2ψ
(t)
sin γ
2
x
2
∼
=
γ
2
x
2
cos γ
2
x
2
∼
=
1
ψ
1