Дыхта В.А. Динамические системы в экономике. Введение в анализ одномерных моделей
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ρ
C
t
= wL
t
w > 0
L
t
= βX
t
X C L
t ∈
+
x y
x
t+1
= f(x
t
). (2.39)
f(x)
f : →
x
x
∗
∈
f
x
t
≡ x
∗
f(x
∗
) = x
∗
f x
∗
x
∗
x
t
≡ ¯x x
t
≡ ¯x x
∗
x
t+1
= ax
t
+ b
|a| < 1 y = ax + b
|a|
¯
¯
¯
∆y
∆x
¯
¯
¯
=
|a(x
2
− x
1
)|
|x
2
− x
1
|
< 1 ∀x
1
, x
2
, x
1
6= x
2
,
|y
0
| = |a| < 1
f(x)
|f(x
2
) − f(x
1
)|
|x
2
− x
1
|
< 1 ∀x
1
, x
2
∈ , x
1
6= x
2
, (2.40)
|f
0
(x)| < 1 ∀x ∈ , (2.41)
f
f : → R
|f(x
2
) − f(x
1
)| < |x
2
− x
1
| ∀x
1
, x
2
∈ (2.42)
x
1
x
2
→ x
1
f(x
2
) → f(x
1
) f
f : → R
f
x
∗
, x
∗∗
∈
x
∗
= f(x
∗
), x
∗∗
= f(x
∗∗
).
|x
∗∗
− x
∗
| = |f(x
∗∗
) − f(x
∗
)| < |x
∗∗
− x
∗
|.
x
∗
= x
∗∗
f : →
x
∗
x
∗
x
t+1
= f(x
t
)
x
0
∈ x
∗
t → ∞
y
t
= |x
t
−x
∗
| x
0
= x
∗
x
t
≡ x
∗
y
t
≡ 0 x
0
6= x
∗
y
t
> 0 ∀t
f
y
t+1
= |x
t+1
− x
∗
| = |f(x
t
) − f(x
∗
)| < |x
t
− x
∗
| = y
t
.
{y
t
}
a ≥ 0
a = 0 x
t
→ x
∗
{x
t
}
|x
t
| = |x
t
− x
∗
+ x
∗
| ≤ |x
t
− x
∗
| + |x
∗
| =
= y
t
+ |x
∗
| < y
0
+ |x
∗
|
{y
t
}
{x
t
}
b
{x
t
} b
a = 0 |x
t
− x
∗
| → 0
x
∗
{x
t
}
b 6= x
∗
a 6= 0
a = lim
t→∞
y
t
= lim
t→∞
|x
t
− x
∗
| = |b − x
∗
|,
|x − x
∗
|
a = lim
t→∞
y
t+1
= lim
t→∞
|x
t+1
− x
∗
| =
= lim
t→∞
|f(x
t
) − f(x
∗
)| = |f(b) − f(x
∗
)| < |b − x
∗
| = a,
|f(x)−f(x
∗
)| f
a = 0 x
t
→ x
∗
x
t+1
= e
−x
t
f(x) =
e
−x
≥ 0 ∀x f
R
+
f x
∗
> 0
x
f(x)
0 x
∗
x
0
y = x
y = e
−x
f R
+
0 ≤ x
1
≤ x
2
f
0
(x) = −e
−x
e
−x
2
− e
−x
1
= −e
−c
(x
2
− x
1
), c ∈ (x
1
, x
2
).
|e
−c
| < 1
|e
−x
2
− e
−x
1
| = |e
−c
| · |x
2
− x
1
| < |x
2
− x
1
|.
x
∗
2
f : → R
k ∈ (0, 1)
|f(x
2
) − f(x
1
)| ≤ k|x
2
− x
1
| ∀x
1
, x
2
∈ . (2.43)
k f
x
1
x
2
k x
1
x
2
k
f(x)
|f
0
(x)| ≤ k < 1 ∀x ∈ R. (2.44)
f k R
f : →
R
x
1
, x
2
∈ R x
1
< x
2
[x
1
, x
2
]
f(x
2
) − f(x
1
) = f
0
(c)(x
2
− x
1
), c ∈ (x
1
, x
2
).
|f(x
2
) − f(x
1
)| = |f
0
(c)||x
2
− x
1
| ≤ k|x
2
− x
1
|, k ∈ (0, 1).
x
1
, x
2
∈ R f
R R 2
f(x) =
1
2
cos x R
|f
0
(x)| =
1
2
|sin x| ≤
1
2
f(x) = cos x [0, b]
b < π/2
|f
0
(x)| = |sin x| = sin x ≤ sin b < 1,
f : [0, b] → R k = sin b
b = π/2 f : [0, π/2] → R
|f
0
(x)| = sin x ≤ 1 [0, π/2]
[0, π/2]
2
f
x
∗
∈
f f(x
∗
) = x
∗
x
0
∈ x
t+1
= f(x
t
)
x
∗
t → ∞ x
∗
f
x
0
∈ S
R R
+
[a, b]
x
0
∈ S
|x
2
− x
1
| = |f(x
1
) − f(x
0
)| ≤ k|x
1
− x
0
|,
|x
3
− x
2
| = |f(x
2
) − f(x
1
)| ≤ k|x
2
− x
1
| ≤ k
2
|x
1
− x
0
|,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
|x
t+1
− x
t
| = |f(x
t
) − f(x
t−1
)| ≤ k|x
t
− x
t−1
| ≤ k
t
|x
1
− x
0
|, (2.45)
|x
n+p
− x
n
| =
= |(x
n+p
− x
n+p−1
) + (x
n+p−1
− x
n+p−2
+ ··· +
+ (x
n+1
− x
n
)| ≤
≤ |x
n+p
− x
n+p−1
| + |x
n+p−1
− x
n+p−2
| + ··· +
+ |x
n+1
− x
n
| ≤
≤ k
n+p−1
|x
1
− x
0
| + k
n+p−2
|x
1
− x
0
| + ··· +
+ k
n
|x
1
− x
0
| =
=
k
n
− k
n+p
1 − k
|x
1
− x
0
| ≤
k
n
1 − k
|x
1
− x
0
|.
p n → ∞
{x
t
}
x
∗
x
t
∈ ∀t x
∗
∈
x
∗
2
f(x)
[a, b]
|f
0
(x)| < 1 ∀x ∈ [a, b].
[a, b]
x
t+1
= f(x
t
)
{x
t
}
p |x
n+p
− x
n
| → 0 n → ∞
{x
t
}
f(x) = cos x [0, 1]
x
t+1
=
cos x
t
[0, 1]
f(x)
x
∗
∈
|f
0
(x
∗
)| < 1 x
∗
|f
0
(x
∗
)| > 1 x
∗
|f
0
(x
∗
)| = 1 x
∗
f(x)
f(x) = f(x
∗
) + (x − x
∗
)
£
f
0
(x
∗
) + α(x)
¤
,
lim
x→x
∗
,
x∈S
α(x) = 0 α(x
∗
) = 0. (2.45)
|f
0
(x
∗
)| < 1 k
|f
0
(x
∗
)| < k < 1 g(x) =
|f
0
(x
∗
)+α(x)| x
∗
g(x
∗
) = |f
0
(x
∗
)| <
k
x
∗
δ x
∗
g(x) = |f
0
(x
∗
) + α(x)| < k ∀x ∈ ∩ (x
∗
− δ, x
∗
+ δ).
² > 0 δ
∗
= min{², δ} I = ∩
(x
∗
− δ, x
∗
+ δ) I g(x) < k
|f(x) − x
∗
| = |f(x) − f(x
∗
)| = |x − x
∗
|g(x) ≤ k|x − x
∗
| < δ
∗
. (2.46)
f I
x
0
∈ I
{x
t
} x
t+1
= f(x
t
) x
∗
t → ∞
|x
1
− x
∗
| = |f(x
0
) − f(x
∗
)| ≤ k|x
0
− x
∗
|,
|x
2
− x
∗
| = |f(x
1
) − f(x
∗
)| ≤ k|x
0
− x
∗
| ≤ k
2
|x
0
− x
∗
|,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
|x
n+1
− x
∗
| = |f(x
n
) − f(x
∗
)| ≤ k|x
n
− x
∗
| ≤ k
n+1
|x
0
− x
∗
|,
n →
∞ k ∈ (0, 1) |x
n+1
−x
∗
| → 0
{x
t
} x
∗
t → ∞
|f
0
(x
∗
)| > 1 q
1 < q < |f
0
(x
∗
)|
δ > 0
g(x) = |f
0
(x
∗
) + α(x)| > q
x ∩ (x
∗
− δ, x + δ)
ε > 0 δ
∗
= min{ε, δ} I = ∩(x
∗
−δ
∗
, x
∗
+δ
∗
)
x
0
∈
I x
0
6= x
∗
|x
1
− x
∗
| = |f(x
0
) − f(x
∗
)| = |x
0
− x
∗
0
|g(x) ≥ q|x
0
− x
∗
|,
|x
2
− x
∗
| = |f(x
1
) − f(x
∗
)| = |x
1
− x
∗
|g(x) ≥ q
2
|x
0
− x
∗
|,
|x
n+1
− x
∗
| = |f(x
n
) − f(x
∗
)| = |x
n
− x
∗
|g(x) ≥ q
n+1
|x
0
− x
∗
|.
{q
n
} → +∞ q > 1 |x
n+1
− x
∗
| → +∞
x
0
∈ I x
0
6= x
∗
x
∗
|f
0
(x
∗
)| = 1
x
t+1
= x
t
+ ax
3
t
x
∗
= 0 a 6= 0
f
0
(x
∗
) = 1 a
• a > 0 t = 0, 1, . . .
|x
t+1
− x
∗
| = |x
t+1
| = |x
t
|(1 + ax
2
t
) > |x
t
|, (2.47)
|x
t
| ≥ |x
0
| ∀t
|x
1
| ≥ (1 + ax
2
0
)|x
0
|,
|x
2
| ≥ (1 + ax
2
0
)|x
1
| ≥ (1 + ax
2
0
)
2
|x
0
|,
|x
t+1
| ≥ (1 + ax
2
0
)|x
t
| ≥ (1 + ax
2
0
)
t+1
|x
0
|.
|x
t+1
− x
∗
| = |x
t+1
| → +∞ x
0
= 0 x
∗
= 0
• a = 0 x
∗
= 0
• a < 0
|x
t+1
| = |x
t
||1 + ax
2
t
| < |x
t
|,
x
0
6= 0 < 1
t = 0
−1 < 1 + ax
2
0
< 1
x
0
|x
0
| < b b =
p
2/(−a)
|x
1
| < |x
0
| < b
|x
t
| < b |1 + ax
2
t
| < 1
t y
t
= |x
t
|
c ∈ [0, b)
y
t+1
= y
t
|1 + ay
2
t
|
c
c = c|1+ac
2
|
c = 0 y
t
= |x
t
| → 0 = x
∗
x
∗
x
0
∈ (−b, b)
x
t+1
= 2x
3
t
/(1 + x
2
t
).
f(x) = 2x
3
/(1 + x
2
)
R
f
0
(x) = 2x
2
(x
2
+ 3)(1 + x
2
)
−2
.
f(x) = x x
∗
= 0
¯x = 1 x
e
= −1 f
0
(0) = 0 < 1 ⇒ x
∗
= 0