Колесников А.А. Синергетические методы управления сложными системами: теория системного синтеза
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X Y
˙
X(t)=
a
0
sX
k
s
+ s
− βX −γXY,
˙
Y (t)=
a
0
sY
k
s
+ s
− βY −γXY,
˙s(t)=−
αa
0
s
k
s
+ s
(X + Y )+U.
s
U
τ =βt; x =
γX
β
; y =
γY
β
; z =
γs
β
;
u =
γU
β
2
; f(z)=
a
0
z
k
z
+ z
; k
z
=
γk
s
β
.
˙x(τ)=f(z)x − x − xy,
˙y(τ)=f(z)y − y − xy,
˙z(τ )=− αf(z)(x + y)+u.
z
˙z(τ )=0
f(z)=
u
α(x + y)
=
u
0
x + y
.
˙x(τ)=x
u
0
x + y
− (1 + y)
,
˙y(τ)=y
u
0
x + y
− (1 + x)
.
• x =0 y =0
• x =0 y = u
0
• x = u
0
y =0
•
x = y =
√
1+2u
0
− 1
2
.
x y
α
ψ
1
= f (z) −
u
0
x + y
.
T
1
˙
ψ
1
(t)+ψ
1
=0
u
1
= αf(z)(x + y) −
(k
z
+ z)
2
k
z
a
0
(x + y)
2
u
0
(x + y)f(z) −2u
0
xy − u
0
(x + y)+
1
T
1
ψ
1
(x + y)
2
,
3T
1
ψ
1
=0
ψ
1
=0
T
1
a
0
=2 α =10 k
z
=1 u
0
=1 T
1
=1 u
1
(x, y, z)
ψ(t)
ψ
2
= f(z)+λx + λy −2,
λ>0
T
2
˙
ψ
2
(t)+ψ
2
=0
u
2
= αf(z)(x + y) −
(k
z
+ z)
2
k
z
a
0
λ(x + y)f(z) −2λxy − λ(x + y)+
1
T
2
ψ
2
,
3T
2
ψ
2
=0
˙x
ψ
(t)=x
ψ
− (λ +1)x
ψ
y
ψ
− λx
2
ψ
,
˙y
ψ
(t)=y
ψ
− (λ +1)x
ψ
y
ψ
− λy
2
ψ
.
• x
ψ
= y
ψ
=0
• x
ψ
= y
ψ
=
1
2λ +1
• x
ψ
=
1
λ
y
ψ
=0
• y
ψ
=
1
λ
x
ψ
=0
x =
1
λ
y =
1
λ
ψ
2
=0
λ =1
T
2
=1
S
u
1
(x, y, z)
u
2
(x, y, z)
|u(t)| u
max
|x
i
| x
i max
|u(t)| u
max
u
i
u
u = arctg by u = A th by.
˙x
j
(t)=f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
,j=1, 2,...,n− 1;
˙x
n
(t)=f
n
(x
1
,...,x
n
)+u;
˙y(t)= ˙x
n+1
(t)=u
1
,
x
1
,...,x
n
u y = x
n+1
u
1
˙x
j
(t)=f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
,j=1, 2,...,n− 1;
˙x
n
(t)=f
n
(x
1
,...,x
n
)+A arctg bx
n+1
;
˙x
n+1
(t)=u
1
,
ψ
1
= A arctg x
n+1
+ u
2
(x
1
,...,x
n
)
u
1
(x
1
,...,x
n
)
T
1
˙
ψ
1
(t)+ψ
1
=0.
ψ
1
Ab ˙x
n+1
(t)
1+b
2
x
2
n+1
+
n
k=1
∂u
2
(x
1
,...,x
n
)
∂x
k
˙x
k
(t)+
A
T
1
arctg bx
n+1
+
1
T
1
u
2
(x
1
,...,x
n
)=0.
u
1
= −
1+b
2
x
2
n+1
Ab
n−1
j=1
∂u
2
∂x
j
f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
+
∂u
2
∂x
n
f
n
(x
1
,...,x
n
)+
+ A arctg bx
n+1
+
A
T
1
arctg bx
n+1
+
1
T
1
u
2
(x
1
,...,x
n
)
.
u
1
ψ
1
=
0
˙x
j
(t)=f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
,j=1, 2,...,n− 1;
˙x
n
(t)=f
n
(x
1
,...,x
n
) − u
2
;
n
m = n −1
ψ
2
= x
n
+ u
3
(x
1
,...,x
n−1
);
ψ
3
= x
n−1
+ u
4
(x
1
,...,x
n−r
),...,;
ψ
m
= x
r
+ u
m+1
(x
1
).
u
2
,u
3
,...,u
n+1
u
1
u = arctg bx
n+1
|u(t)| <A
˙x
1
(t)=x
2
;˙x
2
(t)=u = A arctg bx
3
;˙x
3
(t)=u
1
.
ψ
1
= A arctg bx
3
+ u
2
(x
1
,x
2
).
u
1
= −
1+b
2
x
2
3
Ab
∂u
2
∂x
1
x
2
+ A
∂u
2
∂x
2
+
1
T
1
arctg bx
3
+
1
T
1
u
2
(x
1
,x
2
)
.
ψ
1
=0
˙x
1ψ
1
(t)=x
2ψ
1
, ˙x
2ψ
1
(t)=−u
2
(x
1ψ
1
,x
2ψ
1
).
u
2
T
2
˙
ψ
2
(t)+ψ
2
=0
ψ
2
= α
1
x
1
+ x
2
.
ψ
2
u
2
=
α
1
T
2
x
1
+
α
1
+
1
T
2
x
2
,
ψ
2
=0 ψ
2
=0
˙x
1ψ
2
(t)=−α
1
x
1ψ
2
.
α
1
> 0 u
2
u
1
= −
1+b
2
x
2
3
Ab
α
1
T
1
T
2
x
1
+
α
1
T
1
+
α
1
T
2
+
1
T
1
T
2
x
2
+ A
α
1
+
1
T
1
+
1
T
2
arctg bx
3
.
T
1
= T
2
=
1
α
1
= T
u
1
= −
1+b
2
x
2
3
Ab
x
1
T
3
+
3x
2
T
2
+
3A
T
arctg bx
3
.
T
u
1
t
p
(3 ...4)
T
1
+ T
2
+
1
α
1
≈ 10T.
A =0, 6366
b =20 T =0, 5 x
10
=1 x
20
= x
30
=0 u = A arctg bx
3
|u(t)| A
π
2
=1
x
1
(t) x
2
(t)
u(t)
max
= −0, 86 |u| 1 t
p
≈ 4 < 5
T =1 x
1
(t) x
2
(t)
t
p
< 10
u u(t)
max
= −0, 23
T =0, 5 T =1
T =1, 2 T =1, 5
T =2
x
10
=5 x
20
= x
30
=0 T u<1
T =1, 2
x
1
(t) x
2
(t) t
p
≈ 9
u(t)
max
= −0, 8 |u| < 1
x
10
=5 x
20
= x
30
=0 T =1, 5
t
p
≈ 11
u(t)
max
= −0, 51 u(t)
max
= −0, 28
T =2
T
|u| u
max
y(t) u(t)
˙u(t)=k
1
y − k
1
ϕ(u).
ϕ(u)=A
1
tg b
1
u
u =
1
b
1
arctg
1
A
1
y −
1
k
1
A
1
˙u(t)
.
u
u
max
=
π
2b
1
b
1
=
π
2
|u| < 1
A
1
tg b
1u
u
˙x
j
(t)=f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
,j=1, 2,...,n− 1;
˙x
n
(t)=f
n
(x
1
,...,x
n
)+u;
˙u(t)=k
1
y − k
1
A
1
tg b
1
u;
˙y(t)=k
2
u
1
,
x
1
,...,x
n
u y
u
1
u = x
n+1
y = x
n+2
˙x
j
(t)=f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
,j=1, 2,...,n− 1;
˙x
n
(t)=f
n
(x
1
,...,x
n
)+x
n+1
;
˙x
n+1
= k
1
x
n+2
− k
1
A
1
tg b
1
x
n+1
;
˙x
n+2
= k
2
u
1
.
ψ
1
= x
n+2
+ u
2
(x
1
,...,x
n−1
)
ψ
2
= x
n+1
+ u
3
(x
1
,...,x
n
) ψ
3
= x
n
+ u
4
(x
1
,...,x
n−1
)