1

not, it would have to converge to a ¬xed point of L2 di¬erent from 0 and 1 ’ µ .

µ

We shall show that there are no such points. Indeed, a ¬xed point of L2 is a

µ

zero of

L2 (x) ’ x = µLµ (x)(1 ’ Lµ (x)) = µ[µx(1 ’ x)][1 ’ µx(1 ’ x)] ’ x.

µ

We know in advance two roots of this quartic polynomial, namely the ¬xed

1

points of Lµ ,which are 0 and 1 ’ µ . So we know that the quartic polynomial

1

factors into a quadratic polynomial times µx(x ’ 1 + µ ). A direct check shows

that this quadratic polynomial is

’µ2 x2 + (µ2 + µ)x ’ µ ’ 1. (2.2)

The b2 ’ 4ac for this quadratic function is

µ2 (µ2 ’ 2µ ’ 3) = µ2 (µ + 1)(µ ’ 3)

which is negative for ’1 < µ < 3 and so (2.2) has no real roots.

We thus conclude that the iterates of any point in ( µ , µ ] oscillate about the

1

4

1

¬xed point, 1 ’ µ and converge in towards it, eventually with the geometric rate

of convergence a bit less than µ ’ 2. The graph of Lµ is strictly above the line

1

y = x on the interval (0, µ ] and hence the iterates of Lµ are strictly increasing

so long as they remain in this interval. Furthermore they can™t stay there, for

this would imply the existence of a ¬xed point in the interval and we know that

1 1

there is none. Thus they eventually get mapped into the interval [ µ , 1 ’ µ ]

and the oscillatory convergence takes over. Finally, since Lµ is decreasing on

32 CHAPTER 2. BIFURCATIONS

1 1 1

[1 ’ µ , 1], any point in [1 ’ µ , 1) is mapped into (0, 1 ’ µ ] and so converges to

the non-zero ¬xed point.

In short, every point in (0, 1) is in the basin of attraction of the non-zero

1

¬xed point and (except for the points µ and the ¬xed point itself) eventually

converge toward it in a “spiral” fashion.

µ = 3.

2.1.4

Much of the analysis of the preceding case applies here. The di¬erences are: the

2 1

quadratic equation (2.2) now has a (double) root. But this root is 3 = 1 ’ µ .

So there is still no point of period two other than the ¬xed points. The iterates

2

continue to spiral in, but now ever so slowly since Lµ ( 3 ) = ’1.

For µ > 3 we have

1

Lµ (1 ’ ) = 2 ’ µ < ’1

µ

1

so both ¬xed points, 0 and 1 ’ µ are repelling. But now (2.2) has two real roots

which are

1 1 1

± (µ + 1)(µ ’ 3).

p2± = +

2 2µ 2µ

Both roots lie in (0, 1) and give a period two cycle for Lµ . The derivative of

2

Lµ at these periodic points is given by

(L2 ) (p2± ) = Lµ (p2+ )Lµ (p2’ )

µ

= (µ ’ 2µp2+ )(µ ’ 2µp2’ )

= µ2 ’ 2µ2 (p2+ + p2’ ) + 4µ2 p2+ p2’

1 1

= µ2 ’ 2µ2 (1 + ) + 4µ2 — 2 (µ + 1)

µ µ

2

= ’µ + 2µ + 4.

This last expression equals 1 when µ = 3 as we√already know. It decreases as µ

increases reaching the value ’1 when µ = 1 + 6.

√

3<µ<1+ 6.

2.1.5

In this range the ¬xed points are repelling and both period two points are

attracting. There will be points whose images end up, after a ¬nite number of

iterations, on the non-zero ¬xed point. All other points in (0, 1) are attracted

to the period two cycle. We omit the proof.

Notice also that there is a unique value of µ in this range where

1

p2+ (µ) = .

2

Indeed, looking at the formula for p2+ we see that this amounts to the condition

that (µ + 1)(µ ’ 3) = 1 or

µ2 ’ 2µ ’ 4 = 0.

33

2.1. THE LOGISTIC FAMILY.

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 2.4: µ = 3.3, graphs of y = x, y = Lµ (x), y = L2 (x).

µ

34 CHAPTER 2. BIFURCATIONS

The positive solution to this equation is given by µ = s2 where

√

s2 = 1 + 5.

At s2 , the period two points are superattracting, since one of them coincides

with 1 which is the maximum of Ls2 .

2

3.449499... < µ < 3.569946....

2.1.6

√