Дыхта В.А. Динамические системы в экономике. Введение в анализ одномерных моделей
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S(t + 1) =
h
1 + r
³
1 −
S(t)
S
max
´i
S(t), (1.28)
S
0
= r
³
1 −
S
S
max
´
S, (1.29)
S
max
r, S
max
r
t → ∞
2
y
1 − y
y
0
= ay(1 − y)
y(t + 1) = y(t) + ay(t)(1 − y(t))
a > 0
(−cy) c > 0
D = D(I)
E(D , I) = I
D
0
(I)
D(I)
: (1.30)
E(D , I) > 0
E
0
D
(D, I) < 0
E
0
I
(D, I) > 0
E(D , I) = aI(M − D), (1.31)
a, M > 0 0 < D < M
E > 0
dD
dI
= aD(M − D), (1.32)
D(I) I
v
y
0
= ay − by
2
− v. (1.33)
y
v
W
y
0
= ay
y
a a − by + cW/y
y
0
= ay − by
2
+ cW.
y v
p(t) t D(p), S(p)
z(p) = D(p) −S(p)
p
0
= αF
¡
z(p)
¢
(1.34)
p(t + 1) = p(t) + αF
³
z
¡
p(t)
¢
´
.
α > 0
F (z)
F (0) = 0, F (z) > 0 z > 0 F (z) <
0 z < 0 F = z
D, S
p(t + 1) = p(t)e
αz(p(t))
, α > 0, (1.35)
D, S
y
t
y(t)
Y
t
= C
t
+ I
t
, Y
t
= F (K
t
, L
t
)
K
t+1
= K
t
+ I
t
− bK
t
=
= I
t
+ (1 − b)K
t
, (1.36)
b
S
t
= I
t
= sY
t
L
t
= (1 + n)
t
L
0
n
Y C I
K L S s
F (λK, λL) = λF (K, L) ∀λ > 0
y =
Y
L
, c =
C
L
, k =
K
L
, i =
I
L
.
f(k) = F (
K
L
, 1),
y
t
= c
t
+ i
t
, y
t
= f(k
t
),
k
t+1
= v
£
i
t
+ (1 − b)k
t
¤
,
v = 1/1 + n y
t
, i
t
k
p
t+1
= v
£
f(k
t
) + (1 − b)k
t
− c
t
¤
, (1.37)
c
t
˙
k = f(k) − (b + n)k − c. (1.38)
Y
t
L
t
= f(k
t
) = Bk
β
t
(M − k
t
)
γ
, k
t
≤ M,
β, γ, M > 0
β = γ = 1
Y = AL
α
, α ∈ (0, 1), A > 0
L
0
L
= a − b
L
Y
L/Y Y
L
0
= aL −
b
A
L
2−α
. (1.39)
c = c(t)
c S
c(S) ≥ 0 S ≥ 0;
c(S)
C > 0
C
∀t
Z
t+θ
t
c
¡
S(τ)
¢
dτ ≤ C.
c(S) =
A(S + a
1
)
S + a
1
+ a
2
, (1.40)
A, a
1
, a
2
A = lim
S→∞
c(S) = sup
S≥0
c(S)).
A cos(at + ϕ),
ϕ
y
00
+ a
2
y = 0
t
C
t
= cY
t−1
+ c
0
, (1.41)
Y c c
0
I
ind
t
= v(Y
t−1
− Y
t−2
); (1.42)
v > 0
Y
t
= C
t
+ I
t
,
I
t
= I
ind
t
+ I
a
t
,
(1.43)
I
a
t
t
Y
Y
t
= (c + v)Y
t−1
− vY
t−2
+ A
t
, (1.44)
A
t
Y
Y
0
Y
1
c v
c = 0, 5 v = 1, 5 A
0
= 100 A
t
= 110 t ≥ 1
Y
0
= 200 Y
1
= 210
c = 0, 55 v = 0, 8
c = 0, 5 v = 1
t
Y
Y
0
0
t
Y
Y
0
0
t
Y
Y
0
0
t
Y
Y
0
0
C(t) = cY (t) + c
0
,
dK(t)
dt
= α[vY (t) − K(t)],
dY (t)
dt
= β[ C(t) +
dK(t)
dt
− Y (t)], (1.45)
α, β, v > 0
K
I K
0
= I
K
0
= dK/dt
K/Y = v
I = 0 K
0
= 0 K/Y > v
I < 0 K
0
< 0 K/Y < v
I > 0 K
0
> 0
K Y
Y
0
K
0
(t) = βv
Z
∞
0
e
−βτ
Y
0
(t − τ)dτ. (1.46)
v > 0 β > 0
β
Z
∞
0
e
−βτ
dτ = 1.
Y
0
(t −τ)
τ
t
s =
t − τ
K
0
(t) = βv
Z
t
−∞
e
−β(t−s)
Y
0
(s)ds =
= βve
−βt
Z
t
−∞
e
βs
Y
0
(s)ds
e
βt
K
0
(t) = βv
Z
t
−∞
e
βs
Y
0
(s)ds.
t
βe
βt
K
0
(t) + e
βt
K
00
(t) = βve
βt
Y
0
(t)
exp(βt)
βK
0
(t) + K
00
(t) = βvY
0
(t).
βK(t) + K
0
(t) = βvY (t) + D,
D D = 0
K
0
= I
Y + R = C + K + R
K = vY