Дыхта В.А. Динамические системы в экономике. Введение в анализ одномерных моделей
Подождите немного. Документ загружается.
x
t
= x
0
a
t
= (y
0
− y
∗
)a
t
,
y
t
= x
t
+ y
∗
= (y
0
− y
∗
)a
t
+ y
∗
(a 6= 1).
a = 1 y
t+1
=
y
t
+ b f = y + b
b
y
t
= y
0
+ tb (a = 1) .
t → ∞
a 6= 1
• |a| < 1 y
t
→ y
∗
∀y
0
y
∗
• |a| > 1 |a
t
| → ∞
y
∗
• a = −1
a > 1
a < 0
f = ay + b
ay |a| b
< 1
2
a 6= 1 y
∗
y
0
−y
∗
a ≥ 1
t
y
0
y
0
y = y
∗
0 < a < 1
t
y
0
y
0
y = y
∗
a = 0
t
y
0
y
0
= b y
t
≡ y
∗
= b
a
a = 1
y(t, C) = Ca
t
+ y
∗
,
t ∈
+
C
a 6= 1
y
0
−1 < a < 0
t
y
0
y
0
y = y
∗
a = −1
t
y
0
y
0
y = y
∗
a < −1
t
y
0
y
0
y = y
∗
a < 0
y
t+1
= 0, 5y
t
+ 10 a = 0, 5
b = 10 y
∗
= 20
x
t+1
= 0, 5x
t
x
t
= C(0, 5)
t
y
t
= y(t, C) = C(0, 5)
t
+ 20.
y|
t=0
= y
0
C = y
0
− y
∗
= y
0
− 20
y
t
= (y
0
− 20)(o, 5)
t
+ 20.
y
∗
= 20
y
t+1
= ay
t
+ b(t)
a
{¯y
t
}
y
0
y
t
= (y
0
− ¯y
0
)a
t
+ ¯y
t
, t ∈
+
y
0
{y
t
} y
0
x
t
= y
t
− ¯y
t
x
0
= y
0
− ¯y
0
x
t+1
= y
t+1
− ¯y
t+1
(2.16)
= ay
t
+ b(t) − ¯y
t+1
=
= ax
t
+ a¯y
t
+ b(t) − ¯y
t+1
= ax
t
,
{¯y
t
} x
t
x
t
= (y
0
− ¯y
0
)a
t
2
b(t) = P (t) t
¯y
t
= Q(t) Q(t)
• P a 6= 1
• P a = 1
y
t+1
=
ay
t
+ t b(t) = t
¯y
t
= Q(t) = At + B A B
¯y
t
B + A(t + 1) = a(At + B) + t.
t
A B
aA + 1 = A, A + B = aB.
a 6= 1
A =
1
1 − a
, B = −
1
(1 − a)
2
.
¯y
t
=
t
1 − a
−
1
(1 − a)
2
, ¯y
0
= −
1
(1 − a)
2
a = 1 ¯y
t
=
At
2
+ Bt + C
2A + B = B + 1, A + B = 0, C = 0
A = 1/2 B = −1/2
¯y
t
=
t
2
2
−
t
2
, ¯y
0
= 0.
b(t) = λ
t
P (t) λ 6= 0
t
¯y
t
= λ
t
Q(t) Q(t)
• P λ 6= a
• P λ = a
¯y
t
y
t+1
= ay
t
+ λ
t
P (t)
λ
t
λQ(t + 1) = aQ(t) + P (t).
Q(x)
x ∈
λQ(x + 1) − aQ(x) = P (x).
n P
λ 6= a n = 0 P
P (x) ≡ c Q(x) = c
0
=
c
0
= c/λ −a n = 0
P (x)
≤ n − 1 P (x) n
P (x) = Ax
n
+ R(x) (A, R ),
R(x) ≤ n −1
Q(x)
Q(x) = Bx
n
+ S(x) (B, S ),
S(x) ≤ n−1
λS(x + 1) − aS(x) = R(x) + Ax
n
−
−B[λ(x + 1)
n
− ax
n
].
B x
n
A − B(λ − a)
B = A/λ − a
S(x)
≤ n −1
λ 6= a
λ = a n = 0 P ≡ c Q(x) = cx/λ
Q
n P ≤ n −1
n P
Q
Q(x) = Cx
n+1
+ S(x),
S ≤ n
aS(x + 1) − aS(x) = R(x) + Ax
n
−
−aC[(x + 1)
n+1
− x
n+1
].
n + 1
x
n
A −aC(n + 1) A a
C = A/a(n + 1)
≤ n − 1
2
y
t+1
= ay
t
+tλ
t
λ 6= a P (t) = t
¯y
t
= λ
t
(At + B)
λ
t+1
[A(t + 1) + B] = aλ
t
(At + B) + tλ
t
,
λ
t
λAt + λA + λB = (aA + 1)t + aB.
t
A B
λA = aA + 1, λA + λB = aB.
A = 1/λ − a, B = −λ/(λ − a)
2
,
λ = a ¯y
t
= λ
t
(Ct
2
+Dt)
y
t
= (y
0
− Q(0))a
t
+ λ
t
Q(t),
Q(t)
t → ∞
• |a| < 1, |λ| < 1 y
t
→ 0 ∀y
0
• a = 1, |λ| < 1 y
t
→ y
0
− Q(0) ∀y
0
• a = −1, |λ| < 1
y = 0
•
y
t+1
= −3y
t
+ 4, y
0
= 4
y
t+1
− 0, 2y
t
= 6, y
0
= 4
2y
t+1
− y
t
= 6, y
0
= 7
y
n+1
= 2y
n
− 10, n = 0, 1, . . . , y|
n=0
= y
0
y
n+1
= y
n
, n = 0, 1, . . .
y
t
= 0, 5y
t−1
+ 1, t = 1, 2, . . . , y|
t=1
= y
0
y
t+1
− 0, 1y
t
= 9
y
t
− y
t−1
= −1
y
t+1
− 5y
t
= −2
y
t+1
=
1
3
y
t
+ 6, y
0
= 1
y
t+1
= −
1
4
y
t
+ 5, y
0
= 2
y
t+1
= −2y
t
+ 9, y
0
= 4
y
t+1
− y
t
= 3, y
0
= 5
y
t+1
= b(t) b(t)
+
y
t+1
= a(t)y
t
y
t
= a(0)a(1) . . . a(t − 1)y
0
=
t−1
Y
i=0
a(i)y
0
,
Q
y
t+1
= a(t)y
t
+ b(t)
y
t
=
t−1
Y
i=0
a(i)y
0
+
t−1
X
j=0
b(j)
t−1
Y
i=j+1
a(i).
{y
t
}
t → ∞
y
t+1
=
α
t+1
y
t
; y
t+1
= ay
t
+ t;
y
t+1
= (t + 1)y
t
+ b; y
t+1
= ay
t
+ β
t
;
y
t+1
= α
t
y
t
+ b.
Q
D
= f(P ) Q
S
= g(P )
Q
D
= Q
S
, f(P )
|{z}
= g(P )
|{z}
P
∗
E (Q
∗
, P
∗
) Q
∗
=
f(P
∗
) = g(P
∗
)
P
∗
E P
∗
P
∗
t = 0 P
0
P
0
6= P
∗
P
t
→ P
∗
t → ∞
P
t
P
t−1
t − 1
t − 1 t
Q
Dt
Q
St
t
P
t
Q
Dt
= f(P
t
),
P
t−1
Q
St
= g(P
t−1
).
Q
Dt
= Q
St
f(p
t
) = g(P
t−1
).
P
0