Module IV - Population dynamics
Recurrence relations
Example 1: The number of bacteria in a colony triples every hour. If a colony begins with 10 bacteria, how many will be present after n hours?. We let denote the number of bacteria at the end of n hours. Hence , where n is a positive integer. This equation, called the recurrence relation, together with the initial condition , uniquely determines the sequence , for all nonnegative integers n. The sequence of a recurrence relation is called a solution of the recurrence relation.
Example 2: (Rabbits and the Fibonacci numbers) A pair of rabbits does not breed until they are 2 months old. After they are 2 months old, each pair produces another pair each month. Starting with a pair of rabbits, find a recurrence relation for the number of pairs of rabbits after n months, assuming that no rabbit ever dies.
Solution:
At the end of the first month, the number of pairs of rabbits is
At the end of the second month, the number of pairs of rabbits is
To find the number of pairs at the end of n months, add the number in the previous month, , and the number of newborn pairs, , since each newborn comes from a pair at least two months old. Hence,
, with and
is the desired recurrence relation.
Exercise: Assume that the population of the world in 2002 is 6.2 billion and is growing at the rate of 1.3% a year.
a) Set up a recurrence relation for the population of the world n years after 2002.
b) Find an explicit formula for the population of the world n years after 2002.
c) What will the population of the world be in 2002?
Solving recurrence relations
Definition: A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form
where are real numbers and
The recurrence relation in the definition is
· Linear since the right-hand side is a sum of multiples of the previous terms of the sequence
· Homogeneous since no terms occur that are not multiples of
· Degree k, since depends on the previous k terms.
The coefficients of the terms are constants.
Note: A sequence satisfying the recurrence relation in the definition is uniquely determined by this recurrence relation and the k initial conditions
Theorem 1: Let be real numbers. Suppose the characteristic equation
has k distinct roots . Then a sequence is a solution of the recurrence relation
if and only if
for where are constants.
Theorem 2: Let be real numbers. Suppose the characteristic equation
has t distinct roots with multiplicities respectively, so that for and . Then a sequence is a solution of the recurrence relation
if and only if
+
+
for where are constants for and
Solving a recurrence relation:
Example: Solve the recurrence relation satisfied by the Fibonacci sequence:
, with and
We look for solution of the form
Substituting into the recurrence relation and simplifying we get the characteristic equation . The characteristic roots are .
Therefore
Substituting the initial conditions we get a system of linear equation which are uniquely solvable giving
Exercise: What is the solution of the recurrence relation
, with and ?
Reference: Discrete Mathematics and its application by Rosen, 5th edition
The discrete logistic model
Consider a sequence of discrete times where for each n, and h > 0 is a fixed step size (the interval between successive breeding seasons). Let P denote the population at any time t and let The Discrete logistic model is given by the non-linear recurrence relation:
,
This equation can be rewritten as the logistic difference equation
, with , and
The substitution
simplifies the equation to
Beginning with given values of and r, the last formula generates a sequence
of values corresponding to the successive times . We may think of , the value at time , as the fraction of the maximum population that the environment can support.
Assuming
exists, we want to investigate the way in which depends on the growth parameter, r. Because , only values of r greater than 1 are pertinent to our idealized model of population growth. It turns out that the value of generally appears to stabilize on a limiting value and the number of iterations depends on the value of r and not on the initial value.
Example: with growth rate r =1.5, and , using Maple we get the following for n = 50:
> r:=1.5:
> x=array(1..50):
> x[1]:=0.5:
> for n from 2 to 50 do z:=x[n-1]:
> x[n]:=r*z*(1-z):
> od;
It can be verified that for r = 1.5, 2.0, and 2.5, 0.333333, 0.5, and 0.6 respectively.