6. The logistic map, period-doubling to chaos
One of the simplest, yet most intriguing maps is the logistic map. The wealth of information gathered on this map in the last twenty years revealed its very complicated dynamics and universality. We discuss it in this section.

The logistic map looks very innocent

ƒ(x) = rx(1 - x),     x is in [0,1],    r > 0

Question: for given x0is in[0,1], what happens to ƒn(x) as n → ∞?

1. 0 < r < 1:

In this case, ƒ(x) = rx(1 - x) = x gives us a single fixed point x = 0.

Since ƒ(0) = r implies x = 0 is attracting.

Actually, we can show that it is also globally attracting, i.e.

ƒ(x0) → 0 as n → ∞ for any x0is in[0,1].

Proof: |ƒ(x) < r|x| implies |ƒ2(x)| < r|ƒ(x)| < r2|x|, ...

implies |ƒn(x)| < rn |x| implies |fn(x)| → 0 as n → ∞ .
2. 1 < r < 3:

In this case, solving ƒ(x) = x gives us two fixed points x = 0 and x = 1 - 1/r.

ƒ′(0)= r > 1 implies x = 0 is repelling

|ƒ′(1 - 1/r)| = |2-r| < 1 implies x = 1 - 1/r is attracting.

Are there periodic points in this case? No.
In fact, we can show that,

ƒ(x0) → 1 - 1/r as n → ∞ for any x0 isin (0,1).

Sketch of the proof:

|ƒ′(x)| = |r - 2r| < 1 implies |x - 1/2| < 1/2r implies x is in (1/2 - 1/2r, 1/2 + 1/2r),

Note that 1 - 1/r is in (1/2 - 1/2r, 1/2 + 1/2r) and |ƒ(x)| < 1 in this interval

implies for any x0 is in (1/2 - 1/2r, 1/2 + 1/2r),    ƒn(x0) → 1 - 1/r as n → ∞,

Then we can show that all the preimages of this interval are (0,1).

figure 84
3. r > 3: In this case, both of the two fixed points are repelling. But the periodic points appear.

(a) 2-cycle
Solving ƒ2(x) = x gives us four roots   x = 0, 1-1/r, figure 85

The first two are clearly the repelling fixed points. The second two are two-periodic points.

Stability: (ƒ2)′(x0) == ƒ′(x1) ƒ′(x0)     where x1 = ƒ(x0).

When x0 = figure 85,   (f2)′(x0) = 4 + 2r - r2.

Therefore the 2-cycle is attracting for |4 + 2r - r2 < 1.

i.e.     3 < r < 1 + √6 ≈ 3.449...

When y > 1 + √6, it is repelling.

figure 86
When 3 - r - 1 + √6, are there other periodic orbits besides the 2-cycle? No, actually it can be shown that, except for a countable number of points in [0,1] (which are 0, 1 and the preimages of the repelling fixed point 1 - 1/r), every point is eventually attracted to this 2-cycle. Numerically, one usually finds that every point in [0,1] is attracted to this 2-cycle because both 0 and 1 - 1/r are repelling.

What happens when r > 1 + √6?

In this case, the 2-cycle is repelling, but a 4-cycle appears. Clearly the analysis gets more and more awkward. Hence we switch to numberical exploration. But keep in mind that numerics can find attractors but miss the unstable structures.
The following is what happens when 4 > r > 1 + √6 ≈ 3.449.

3.449... < r < 3.54409...:  there is an attracting 4-cycle. Numerically every pointis eventually attracted to it.
3.54409... < r < 3.5644...:  there is an attracting 8-cycle which numberically attracts every point in (0,1).
3.5644... < r < 3.568759...:  there is an attracting 16-cycle.

This is the so-called period-doubling bifurcation. Note that the successive bifurcations come faster and faster.
Ultimately the intervals of 2n-cycles converge to a point r∞ ≈ 3.569946... as n → ∞

What happens if r > r∞? Here, the orbit {ƒn(x0)} is chaotic for most r values. But periodic windows mysteriously appear here and there. The largest window is near (3.8284, 3.8415) which is 3-periodic. The complet orbit diagram, which is the plot of the map's attractor as a function of r, is given below. This amazing diagram is as beautiful as it is mysterious. If you look at it more closely, you will find that lying just above the periodic windows in the chaotic region are small copies of the whole orbit diagram. Thus this picture has fine structures at arbitrarily small scales.

figure 87

figure 88


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