Колесников А.А. Синергетические методы управления сложными системами: механические и электромеханические системы
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˙x
1
(t)=x
2
, ˙x
2
(t)=u.
u(x
1
,x
2
)
ψ
1
= A
2
− F (x
1
) − 0, 5x
2
2
.
T
1
˙
ψ
1
(t)+x
2
2
ψ
1
=0,
u
1
= −
∂F
∂x
1
+
x
2
T
1
ψ
1
.
¨x
1
(t) −
A
2
− F (x
1
) − 0, 5˙x
2
1
˙x
1
(t)+
∂F(x
1
)
∂x
1
=0.
F (x
1
)
ψ
1
=0
F (x
1
)=0, 5ω
2
x
2
1
;
F (x
1
)=0, 5ω
2
x
2
1
+ βx
4
1
;
F (x
1
)=e
x
1
+ x
1
;
F (x
1
)=αx
3
1
− βx
2
1
F (x
1
)
¨x
1
(t) −
1
T
1
A
2
− 0, 5ω
2
x
2
1
− βx
4
1
− 0, 5˙x
2
1
˙x
1
(t)+ω
2
x
1
+4βx
3
1
=0.
ψ
1
=0
¨x
1ψ
(t)+ω
2
x
1ψ
+4βx
3
1ψ
=0,
ω β
β =0
¨x
1
(t) −
1
T
1
A
2
− 0, 5ω
2
x
2
1
− 0, 5˙x
2
1
˙x
1
(t)+ω
2
x
1
=0,
¨x
1
(t) − µ(1 − 0, 5α
1
x
2
1
− 0, 5α
2
˙x
2
1
)˙x
1
(t)+ω
2
x
1
=0,
µ
1
= µ
2
=
1
T
1
; A
2
=1; α
1
= ω
2
; α
2
=1; ω
1
= ω
2
= ω.
ω
2
1 0, 5ω
2
x
2
1
0, 5˙x
2
1
ω
2
1 0, 5˙x
2
1
A =1 ω =1 T =0, 5 x
10
=1 x
20
=0
A =1 ω =10 T =0, 5
x
1
(t) x
2
(t)
x
1
(t)
x
2
(t)
ψ
2
= A
2
+ β cos γx
1
− 0, 5x
2
2
.
ψ
2
T
2
˙
ψ
2
(t)+x
2
2
ψ
2
=0,
u
2
= −βγ sin γx
1
+
x
2
T
2
ψ
2
.
¨x
1
(t) −
1
T
2
A
2
+ β cos γx
1
− 0, 5˙x
2
1
˙x
1
(t)+βγ sin γx
1
=0.
ψ
2
=0
¨x
1
(t)+βγ sin γx
1ψ
=0.
sin γx
1
∼
=
γx
1
−
γ
3
6
x
3
1
cos γx
1
∼
=
1 − 0, 5γ
2
x
2
1
,
¨x
1
(t) −
1
T
2
A
2
+ β − 0, 5ω
2
x
2
1
− 0, 5˙x
2
1
˙x
1
(t)+ω
2
x
1
−
ω
2
γ
2
6
x
3
1
=0,
βγ
2
= ω
2
¨x
1
(t) − µ ˙x
1
cos px
1
+sinx
1
=0
(p − 1) p 1
¨x
1
(t) − µ(1 −0, 5p
2
x
2
1
)˙x
1
(t)+γx
1
−
γ
3
6
x
3
1
=0,
¨x(t) − f
µ, x, ˙x(t)
+ ω
2
x =0.
˙x
1
(t)=x
2
+ k
1
u;
˙x
2
(t)=− ω
2
x
1
+ k
2
u,
u(x
1
,x
2
)
u(x
1
,x
2
)
ψ = A
2
− 0, 5ω
2
x
2
1
− 0, 5x
2
2
.
ψ
T
˙
ψ(t)+F (x
1
,x
2
)ψ =0,
u(x
1
,x
2
)(k
1
ω
2
x
1
+ k
2
x
2
)=
F (x
1
,x
2
)
T
ψ.
F (x
1
,x
2
)=(k
1
ω
2
x
1
+ k
2
x
2
)
2
> 0
u(x
1
,x
2
)=
(k
1
ω
2
x
1
+ k
2
x
2
)
T
ψ.
˙x
1
(t)=x
2
+
k
1
T
(k
1
ω
2
x
1
+ k
2
x
2
)(A
2
− 0, 5ω
2
x
2
1
− 0, 5x
2
2
);
˙x
2
(t)=− ω
2
x
1
+
k
2
T
(k
1
ω
2
x
1
+ k
2
x
2
)(A
2
− 0, 5ω
2
x
2
1
− 0, 5x
2
2
),
ψ =0 u =0
˙x
1ψ
(t)=x
2ψ
;˙x
2ψ
(t)=−ω
2
x
1
,
T
A =1 ω =1 k
1
= k
2
=1 T =0, 5
A =1 ω =10 k
1
=10 k
2
=1 T =0, 5
x
1
(t) x
2
(t)
x
1
(t) x
2
(t)
k
1
k
2
k
1
∼ k
2
k
1
k
2
k
2
k
1
k
2
k
1
k
1
∼
=
0
¨x
1
(t) −
k
2
2
T
A
2
− 0, 5ω
2
x
2
1
− 0, 5˙x
2
1
˙x
1
(t)+ω
2
x
1
=0.
k
1
k
2
k
2
∼
=
0
¨x
2
(t) −
k
2
1
T
A
2
ω
2
− 0, 5ω
2
x
2
2
− 0, 5˙x
2
2
˙x
2
(t)+ω
2
x
2
=0.
¨x(t)+F
1
(x, z, µ
1
,...µ
k
)˙x(t)+F
2
(x, z, µ
1
,...µ
k
)=0,
˙z(t)=F
3
(x, z, µ
1
,...µ
k
),
x F
i
z F
3
x(t)
˙x(τ)=mx + y − xz − dx
3
;˙y(τ)=−x;˙z(τ )=F
3
(x, z, µ
1
,...µ
k
),
z m
F
3
F
3
= −λz + λΦ(x),
λ
Φ(x)
Φ(x)=I(x)x
2
; I(x)=
⎧
⎨
⎩
1,
x>0,
0,
x 0.
F
3
Φ(x)
y
¨x(t) − (m − z − 3dx
2
)˙x(t)+[1− λz + λΦ(x)]x =0,
˙z(t)=−λz + λΦ(x).
λ → 0
¨x(τ) − a(1 − bx
2
)˙x(τ)=0,
a = m − z
0
b =
3d
m − z
0
z
0
= z(0)
Φ(x)
z(0) = z
0
λ →∞
z =Φ(x)
¨x(τ) −
m − Φ(x) − 3dx
2
˙x(τ)+x =0.
Φ(x)=x
2
λ λ
1
λ λ
2
F
3
¨x(τ) −
m − z − 3dx
2
˙x(τ)+x + x ˙z(τ)=0,
˙z(τ)=F
3
.
ψ = z − ϕ(x).
ψ
T
˙
ψ(τ)+ψ =0,
F
3
=
∂ϕ
∂x
˙x(τ) −
1
T
ψ.
¨x(τ) −
m − z −
∂ϕ
∂x
− 3dx
2
˙x(τ)+
1+
1
T
z
x −
1
T
ϕ(x)x =0,
˙z(τ)=−
1
T
z +
1
T
ϕ(x)+
∂ϕ
∂x
˙x(τ).
(3 ÷ 4)T
x
0
˙x
0
z
0
ψ =0
¨x
ψ
(τ) −
m − ϕ(x
ψ
) − 3dx
2
ψ
− x
ψ
∂ϕ
∂x
ψ
˙x
ψ
(τ)+x
ψ
=0.
ϕ(x)
ϕ(x)=αx
2
,
¨x(τ) −
m − z − 3(α + d)x
2
˙x(τ)+
1+
1
T
z −
α
T
x
2
x =0,
˙z(τ)=−
1
T
˙z +
α
T
x
2
+2αx · ˙x(τ).
Φ(x)=x
2
2αx· ˙x(τ)
ψ =0
¨x
ψ
(τ) −
m − 3(α + d)x
2
ψ
˙x
ψ
(τ)+x
ψ
=0,
z
0
d =0
m =0, 15 T =30 α = d =1 m =1, 5 T =3
α = d =1
x(t) ˙x(t)
x(t)
˙x(t)
ϕ(x)
¨
ψ
k
(t)+α
1k
˙
ψ
k
(t)+α
2k
ψ
k
=0; k =1, 2,...,m,
m
◦