Step-by-step explanation:

What is the solution of the logistic equation when r PASSES through the value 3? We know from

the previous section that when 3 < r ≤ 4 the fixed points x

∗ = 0 and x

∗ = (r − 1) /r are both

unstable. So the solution isn’t a period-1 solution. In exploring the behaviour of the logistic

map over the range 3 < r ≤ 4 it is useful to INTRODUCE the following notation for the iterative

procedure:

X1 = f (x0) (47)

x2 = f (x1) = f (f (x1)) = f

2

(x0) (48)

.

.

.

un =f (xn−1) =f (f (xn−2)) = f

n

(x0) (49)

Figure 7 shows an example of iterations of the logistic map, starting from the initial value x0 =

0.7, with the parameter choice r = 3.4. The solution is converging to a sequence p, q, p, q, p, . . .,

where

✞

✝

☎

✆

f(p) = and

✞

✝

☎

✆

f(q) = . This is a

✞

✝

☎

✆solution. Note th

64

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.2 0.4 0.6 0.8 1

Population at generation (t+1)

Population at generation (t)

y=x

y=f(x)

y=f(f(x))

O

O

O

A

B

C

Figure 8: Second iteration xn+2 = f

2

(xn) as a function of xn for the logistic map when r = 3.18.

The dotted line reproduces the first iteration curve of xn+1 as a function of xn; it passes through

the origin and B, the unstable period-1 steady states.

✞

✝

☎

✆

f

2

(p) = and

✞

✝

☎

✆

f

2

(q) = . Another way to say this is that p and q are

fixed points of the map

✞

✝

☎

✆.

Definition 2 (Period-2 solution) A period-2 solution is a pair x

∗

0

, x∗

1 with f (x

∗

0

) = x

∗

1 and

f (x

∗

1

) = x

∗

2

but x

∗

0 6= x

∗

1

.

For the logistic model (16) we have

xn+1 = rxn (1 − xn) (50)

xn+2 = rxn+1 (1 − xn+1) (51)

= (52)

We now look at the second iteration (52) and ask if it has any fixed points, i.e. are there any

values x

∗

2

for which xn+2 = xn = x

∗

2

? The function xn+2 = f (xn) is shown in figure 8.

Question 18 How many fixed points does the map xn+2 = f

2

(xn) have in figure 8?

Fixed points of the map xn+2 = f

2

(xn) satisfy the equation

x

∗

2 = f

2

(x

∗

2

), (53)

x

∗

2 = , (54)

0 = x

∗

2 {r [r (1 − x

∗

2

)] [1 − rx∗

2

(1 − x

∗

2

)] − 1} . (55)

Fixed points of the map xn+2 = f

2

(xt) OBVIOUSLY include fixed points of the map

✞

✝

☎

✆.

Thus two factors of equation (55) are

✞

✝

☎

✆and

✞

✝

☎

✆. Using this knowledge

equation (55) can be factored

x

∗

2

[rx∗

2 − (r − 1)] h

r

2x

∗

2

2 − r (r + 1) x

∗

2 + (r + 1)i

=