point is negative. On this range of µ, the point 0 is an attracting ¬xed point

since 0 < Lµ (0) < 1. Under iteration, all points of [0, 1] tend to 0 under the

iteration. The population “dies out”.

For µ = 1 we have

L1 (x) = x(1 ’ x) < x, ∀x > 0.

Each successive application of L1 to an x ∈ (0, 1] decreases its value. The limit

of the successive iterates can not be positive since 0 is the only ¬xed point. So

all points in (0, 1] tend to 0 under iteration, but ever so slowly, since L1 (0) = 1.

25

26 CHAPTER 2. BIFURCATIONS

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 2.1: µ = .5

27

2.1. THE LOGISTIC FAMILY.

In fact, for x < 0, the iterates drift o¬ to more negative values and then tend

to ’∞.

For all µ > 1, the ¬xed point, 0, is repelling, and the unique other ¬xed

1

point, 1 ’ µ , lies in [0, 1]. For 1 < µ < 3 we have

1

|Lµ (1 ’ )| = |2 ’ µ| < 1,

µ

so the non-zero ¬xed point is attractive.

1

We will see that the basin of attraction of 1 ’ µ is the entire open interval

(0, 1), but the behavior is slightly di¬erent for the two domains, 1 < µ ¤ 2 and

2 < µ < 3:

In the ¬rst of these ranges there is a steady approach toward the ¬xed point

from one side or the other; in the second, the iterates bounce back and forth

from one side to the other as they converge in towards the ¬xed point. The

graphical iteration spirals in. Here are the details:

1 < µ ¤ 2.

2.1.2

1

For 1 < µ < 2 the non-zero ¬xed point lies between 0 and 2 and the derivative

at this ¬xed point is 2 ’ µ and so lies between 1 and 0.

1

Suppose that x lies between 0 and 1 ’ µ . For this range of x we have

1

<1’x

µ

so, multiplying by µx we get

x < µx(1 ’ x) = Lµ (x).

1

Thus the iterates steadily increase toward 1 ’ µ , eventually converging geomet-

rically with a rate close to 2 ’ µ. If

1

1’ <x

µ

then

Lµ (x) < x.

If, in addition,

1

x¤

µ

then

1

Lµ (x) ≥ 1 ’ .

µ

To see this observe that the function Lµ has only one critical point, and that is

1 1 1

a maximum. Since Lµ (1 ’ µ ) = Lµ ( µ ) = 1 ’ µ , we conclude that the minimum

11

value is achieved at the end points of the interval [1 ’ µ , µ ].

28 CHAPTER 2. BIFURCATIONS

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 2.2: µ = 1.5

29