Equilibrium Points

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.

We begin this section by looking at an example of an exponential model with immigration.

p(n + 1) = 0.80 p(n) + 100

This particular model might describe a population living in a very inhospitable habitat. Left alone the population would fall by 20% each generation. There is, however, immigration into this habitat from a nearby habitat at the rate of 100 individuals per generation. We can picture this as shown below.

Each generation 100 new individuals enter our inhospitable habitat and the population already there declines by 20%.

• If the current population is less than 500 then the 20% decline will the less than the immigration of 100 and the net result will be rising population.

• If the current population is greater than 500 then the 20% decline will be greater than the immigration of 100 and the net result will be declining population.

• If the current population is exactly 500 then the 20% decline will be exactly 100 and will be exactly offset by the immigration of 100. The net result will be no change in the population.

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

then an equilibrium point is a point p such that

p = f(p)

In our example above we had

p(n + 1) = 0.80 p(n) + 100

That is,

f(p) = 0.80 p + 100

so we can find any equilibrium points by solving the equation

 
     p = 0.80 p + 100
0.20 p = 100
     p = 100/0.20
       = 500

As another example consider the logistic model

p(n + 1) = 2.8[1 - .001 p(n)] p(n)

We find the equilibrium points of this model by solving the equation

 
p = 2.8 (1 - .001 p) p
p = 2.8 p - .0028 p^2
0 = 1.8 p - .0028 p^2
0 = p(1.8 - .0028 p)

p = 0  or 1.8 - .0028 p = 0
                .0028 p = 1.8
                      p = 1.8/.0028
                        = 642.86
 

We can use algebraic, numeric, and graphic tools to find equilibrium points by solving the equation

p = f(p)

We have seen two examples using algebraic tools above. We can look for solutions graphically by drawing the two curves

y = f(p) and

y = p

on the same set of axes as shown below using the example

f(p) = 2.8 (1 - .001 p) p

from above.

An equilibrium point is a point where the two curves

y = p and

y = f(p)

intersect.

We often use a combination of graphic techniques and numeric techniques. Graphic techniques give us rough estimates for any equilibrium points and then numeric techniques like the bisection method or Newton's method can be used to find more precise estimates. This one-two approach is particularly effective with a computer algebra system.

Check Your Understanding

Find all the equilibrium points for each of the following models.

1. p(n + 1) = 1.05 p(n)

   answer

2. p(n + 1) = p(n) + 250

   answer

3. p(n+1) = 3.2 (1 - .001 p(n)) p(n)

   answer

4. p(n + 1) = 0.80 p(n) + 3000

   answer

5. p(n + 1) = 3.4 (1 - .001 p(n)) p(n)

   answer

6. p(n + 1) = R p(n) + M

   answer

Stretch Your Understanding