Download Section one: Sensitive Dependence on Initial Conditions

Chapter Two
One Dimensional Chaos
In this chapter , we study many methods of describing the way in which
iterates of neighboring points separate from another : sensitive dependence on initial
conditions , Lyapunov exponent and the transitivity. These notions are fundamental
to the concept of chaos, which also will appear in the present section.
Section one: Sensitive Dependence on Initial Conditions
Before defining the sensitive dependence on initial conditions, we adopt a
notation that henceforth will facilitate our discussion. We will write 𝑓: 𝐽 → 𝐽
signifies that the domain of 𝑓 is 𝐽 and the range is contained in 𝐽.
Definition :
Let 𝐽 be an interval , and suppose that 𝑓: 𝐽 → 𝐽. Then 𝑓 has sensitive
dependence on initial conditions at x , or just sensitive dependence at 𝑥 if there is
𝜖 > 0 such that for each 𝛿 > 0, there is 𝑦 in 𝐽 and a positive integer 𝑛 such that
|𝑥 − 𝑦| < 𝛿 and |𝑓 𝑛 (𝑥) − 𝑓 𝑛 (𝑦)| > 𝜖 , that is:
∃ 𝜖 > 0 ∀ 𝛿 > 0 ∃ 𝑦 ∈ 𝐽 ∃ 𝑛 ∈ 𝑁 ∋ |𝑥 − 𝑦| < 𝛿 and |𝑓 𝑛 (𝑥) − 𝑓 𝑛 (𝑦)| > 𝜖
If 𝑓 has sensitive dependence on initial conditions at each 𝑖𝑛 𝐽 , we say that 𝑓 has
sensitive dependence on initial conditions on 𝐽 , or that f has sensitive dependence .
The “ initial conditions” in the definition refer to the given , or initial points 𝑥 and
𝑦. the definition says that f has sensitive dependence on initial conditions if
arbitrarily close to any given point 𝑥 in the domain of 𝑓 there is a point and an 𝑛 −
𝑡ℎ iterate that is farther from the 𝑛 − 𝑡ℎ iterate of 𝑥 than a distance 𝜖. This has
practical significance , because in such instance higher iterate of an approximate
value of 𝑥 may not resemble the true iterate of 𝑥.
To illustrate sensitive dependence on initial conditions, we turn to baker’s function:
Example 1 :
Consider the baker’s function B, given by:
𝐵(𝑥) = {
2𝑥
𝑓𝑜𝑟
2𝑥 − 1
0≤𝑥≤
1
𝑓𝑜𝑟
2
2
<𝑥≤1
1
Show that after 10 iterate of
1
3
and 0.333 are farther than
1
2
Solution
Notice
1
3
iterate of
1
3
2
1
2
3
3
3
3
1
1
and 𝐵2 ( ) = so that the
3
3
alternate between and .
To compare the iterate of
iterate
1
3
0.333
1
is periodic point of period 2 , that is, B( ) =
1
3
and 0.333 we make the following table:
1
2
3
4
5
6
7
8
9
10
2
3
0.666
1
3
0.332
2
3
0.664
1
3
0.328
2
3
0.656
1
3
0.312
2
3
0.624
1
3
0.248
2
3
0.496
1
3
0.992
Therefore the tenth iterate of
farther a part than a distance
1
3
1
1
and 0.333 are, respectively , and 0.992 which are
3
2
Example 2:
Show that the tent function T has sensitive dependence on initial conditions on [0,1].
Solution
Let 𝑥 be any number in [0,1]
Claim: if 𝑣 is any dyadic rational number (of the form
𝑗
2𝑚
in lowest terms) in [0,1]
and w is any irrational number in [0,1], then there is a positive integer 𝑛 such that
1
|𝑇 𝑛 (𝑣) − 𝑇 𝑛 (𝑤)| > …(1)
2
Toward that goal, if 𝑣 =
𝑗
2𝑚
then 𝑇 𝑚 (𝑣) = 1 and 𝑇 𝑚+𝑘 (𝑣) = 0 for all 𝑘 > 0
By contrast if 𝑤 is any irrational number in [0,1] then since 𝑇 doubles each number
1
in (0, ), there exists an 𝑛 > 𝑚 such that 𝑇 𝑛 (𝑤) >
2
1
2
.
1
1
2
2
Since 𝑛 > 𝑚 , it follows that 𝑇 𝑛 (𝑣) = 0 so that |𝑇 𝑛 (𝑣) − 𝑇 𝑛 (𝑤)| > |0 − | >
So the claim is valid .
Next , let 𝛿 > 0 then there exists a dyadic rational 𝑣 and an irrational number in
[0,1].
such that |𝑥 − 𝑣| < 𝛿 𝑎𝑛𝑑 |𝑥 − 𝑤| < 𝛿 therefore (1) implies that
1
< |𝑇 𝑛 (𝑣) − 𝑇 𝑛 (𝑤)| = |𝑇 𝑛 (𝑣) − 𝑇 𝑛 (𝑥) + 𝑇 𝑛 (𝑥) − 𝑇 𝑛 (𝑤)|
2
≤ |𝑇 𝑛 (𝑥) − 𝑇 𝑛 (𝑣)| + |𝑇 𝑛 (𝑥) − 𝑇 𝑛 (𝑤)|
1
1
1
4
4
4
So either |𝑇 𝑛 (𝑥) − 𝑇 𝑛 (𝑣)| > or |𝑇 𝑛 (𝑥) − 𝑇 𝑛 (𝑤)| > . thus if we let 𝜖 = then
𝑇 has sensitive dependence on initial conditions at the arbitrary number 𝑥 , and hence
on [0,1].
Basically , the reason that 𝑇 has sensitive dependence on initial conditions if 𝑥 ≠
1
1
1
then |𝑇´(𝑥)| = 2,so that distances between pairs of numbers in (0, ) or ( , 1) are
2
2
2
doubled in T