Classifying 2-Cycles

You should use one of the computer algebra systems below with this module. Click on the appropriate icon for your preferred CAS and then arrange your screen so that you can easily move back-and-forth between this window and your CAS window. Click on the appropriate help button for help.

Mathematically, it is easy to classify 2-cycles because a 2-cycle of the dynamical system

p(n + 1) = f(p(n))

is just an equilibrium point of the dynamical system

q(n + 1) = f(f(q(n)).

In practice, however, the algebraic manipulations involved are fairly complicated. For this reason we use the CAS window to do the work. Be sure that you have your CAS window setup to follow the rest of this discussion. Go back-and-forth between the two windows as you continue.

We want to determine which logistic models have attracting 2-cycles.

• The first step is determining the equilibrium points and the 2-cycles for a logistic model of the form

  p(n + 1) = a [1 - b p(n)] p(n)

  Look at your CAS window to see how this is done and to see the results.

• The second step is to pick a true 2-cycle point -- that is, a point that is in a 2-cycle but is not an equilibrium point. Look at your CAS window to see how this is done.

• Next we use the CAS window to find the derivative of the function

  f(f(p))

• We are interested in the value of this derivative at the chosen 2-cycle point and, in particular, its absolute value is less than 1.
  • First we find out when this value is -1 or +1. Look at your CAS window to see how this is done. Notice the four points -- roughly

    -1.44949, -1, 3, and 3.44949

    do not involve the value of the constant b.

  • We know from our earlier work on classifying equilibrium points that the nonzero equilibrium point of our logistic model is no longer attracting when 3 < a, so we are interested in what happens in the two intervals
    • 3 < a < 3.44949 and

    • 3.44949 < a <= 4.

    We determine this by evaluating the derivative of the function f(f(p)) at our 2-cycle point using a typical value of the constant b at one point, a between 3 and 3.44949 and also at another point between a = 3.44949 and a = 4. See your CAS window.

• For 0 < a < 1 the equilibrium point zero is attracting and the population dies out.

• For 1 < a < 3 the nonzero equilibrium point is attracting and the population approaches an equilibrium where the needs of population are matched by the resources of the habitat.

• For 3 < a < 3.44949 there is an attracting 2-cycle and the population settles into a pattern going back-and-forth between years in which there is an "underpopulation" and years in which there is an "overpopulation."

Check Your Understanding

1. How would you find 3-cycles and determine if a 3-cycle was attracting?

   answer

2. How would you find 4-cycles and determine if a 4-cycle was attracting?

   answer

3. In the module on pack-hunters we looked at the model

   R(p) = 0.5 + 0.0015 p - 0.000001166667 p (p - 500)
   
   f(p) = R(p) p
   
   p(n + 1) = f(p(n))
   

   This model appeared to have an attracting 2-cycle but notice that the population will die out if the initial population is to low or too high.
   • Find the 2-cycle and verify using the methods that we have developed here that it is attracting.

   • Find an example of an initial condition that is too high and results in the population dying out.

   answer