Calculate the equilibrium/stationary state, to two decimal places, of the difference equation
Answers
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
=