Keith Morse


MA 226 A1

Lab 2: The Predator/Prey System

1. BUID: U61968910 B = 1


dx/dt = x(1- (K/10)x – y)

dy/dt = y(-1+x)

Equilibrium Equations:

x(1- (K/10)x – y) = 0

y(-1+x) = 0

when y = 0, x(1-(K/10)x) = 0

x = 0, 10/K y = 0

when x = 1, (1 – K/10 – y) = 0

x = 1 , y = 1 – K/10

Equilibrium points:

(10/K ,0) (0, 0) (1, 1-K/10)


Assume K = 0

dx/dt = x(1 – y)

dy/dt = y(-1 + x)

Only two equilibrium points: (0,0), (1,1)

In the case where there are no foxes (y(0) = 0), the equation for rabbits is dx/dt = x, which solves for x (t)= Ce^t (rabbits thus increase exponentially, at a rate proportional to population). Likewise, when there are no rabbits, the fox population decreases exponentially. Otherwise, as the population of foxes (y on phase plane) increases, the population of rabbits falls to almost zero. At that point, the fox population dramatically drops off to nearly zero and remains there, as the rabbit population begins to climb again. This process repeats on without change, ad infinitum.

The Phase Plane and the graphs for x(t) and y(t), appear as follows (with a sample solution graphed):


When K becomes positive, a third equilibrium point appears, and the constant solutions change to the following:

(10/K ,0) (0, 0) (1, 1-K/10)

where K is slightly greater than 0, but less than 10 (this scenario will be covered later). This bifurcation changes the behavior of the system dramatically. Given some nonzero initial value for the two populations, the population of rabbits will begin to decline as fox population grows. When the number of rabbits approaches zero, the fox population immediately drops off, just as in the previous example, and the rabbit population begins to return.

This time, however, the rabbits don't make it back to the same population they started with before the fox population begins to return. When this happens, the fox population grows as the rabbits begin to decrease again. However, once again, they do not make it back to the previous value before the rabbit population becomes near-zero, and begin decreasing again.

This pattern results in what appears, in the phase plane, as a steady spiral down to a final non-zero equilibrium point where both populations remain constant. Looking at the graphs of x(t) and y(t), it is clear that both populations are oscillating back and forth as in the previous system, but their amplitudes are slowly decreasing until, as time approaches infinity, both graphs become straight lines, at which point the populations are no longer changing.

The point that the graphs in the phase plane tend toward is the third equilibrium point, (1, 1-K/10). Thus, as time approaches infinity, the rabbit population, x, approaches 1 and the fox population, y, approaches 1 – K/10. In the graphs shown, K = 1, so the equilibrium point lies at (1, 9/10):

For K slightly positive, the graphs appear as follows:


A third bifurcation occurs when K = 10. At this point, the system of equations simplifies to:

dx/dt = x(1- x – y)

dy/dt = y(-1+x)

With equilibrium points: (1, 0) (0, 0) and (1, 0). It is obvious that two of these points are redundant, so the only equilibrium points are (1, 0) and (0, 0).

Shortly before the bifurcation (when K is slightly less than 10), the equilibrium point found in the previous scenario comes closer and closer to the x-axis. In other words, there are always fewer and fewer of both species, but the fox population seems to be dwindling faster.

When K is greater than or equal to 10, the fox population increases a certain amount, then begins to sharply decline as before, but does not recover from its initial drop, and the system seems to behave as a damped oscillation, with the rabbit population increasing slightly after the foxes have disappeared, and then holding at a certain value. The graphs of x(t) and y(t) verify this pattern, showing the population of foxes increase slightly before they completely die out. The rabbit population drops sharply initially, nearly reaches zero, and then slowly increases to equilibrium. If we look at the equation in this situation, where foxes, y, equals 0, the equation for rabbits becomes:

dx/dt = x(1-x)

Which has equilibrium points at x = 0 and1. Thus, it is clear that the rabbit population approaches zero and then backs away from it, and begins to approach 1 as time approaches infinity.

Graph for K = 10