Results
Steady States
There are three families of steady state solutuions of this model.
 x=0, y=0, corresponding to extinction of both populations.
 x=0, y = 1/c2, corresponding to extinction of the predators, and
the equilibrium population of the prey in the absence of
predators.
 x= 1 + a1 c2, y= a1, corresponding to coexistence of the prey
and the predator populations. This steady state makes physical
sense only when x and y are both positive. Since a1 is negative,
y is always positive. For the value of x to be positive, we
require a1 > 1/c2.
Steady State Stability
The Jacobian matrix for the model has the following form.
Jacobian matrix entries
a1+y  x

y  1x2 c2y

When the three different families of steady state solutions are
substituted into the Jacobian, the following results are obtained.
 For family 1, x=0, y=0, the Jacobian matrix becomes as follows.
Jacobian matrix entries
a1  0

0  1

The eigenvalues are a1, which is negative for this model, and 1 so the
origin is a saddle.
 For family 2, x=0, y=1/c2, the Jacobian becomes
Jacobian matrix entries
a1+1/c2  0

1/c2  1

The eigenvalues are 1, and a1+1/c2, so this steady state is stable if
a1 < 1/c2 and is a saddle if a1 > 1/c2.
 For family 3, x=1+c1a2, y=a1, the Jacobian matrix is
Jacobian matrix entries
0  1+c1a2

a1  c2a1

The eigenvalues can be calculated, but it is easier to determine
stability from the characteristic polynomial, which has the form
r^{2}  a1c2 r a1(1+c2a1) = 0
Since a1 is negative, and c2 is positive, the trace is always
positive. The determinant is positive if a1 > 1/c2 and is negative
if a1 < 1/c2. This means that this family is stable if a1 >
1/c2 and is a saddle if a1 < 1/c2.
Phase Portraits
Given the stability information from the previous section, It isn't
too hard to predict the phase portraits. Before we look at some of
them, however, it will help to look at the dynamics restricted to the
coordinate axes.
If y=0, then the dynamics on the x axis is governed by
x' = a1x
Since a1 is negative, any solution starting on the positive x axis
will approach the origin asymptotically. This corresponds to the
predators becoming extinct in the absence of prey.
If x=0, then the dynamics on the y axis is governed by
y' = y(1c2y)
which is just the logistic equation. That is, solutions starting on the
y axis will approach the solution y=1/c2 asymptotically.
The phase portraits can be divided up into 2 classes, depending on the
values of a1 and c2. Phase portraits
within a class are qualitatively similar.
Phase portraits for a1 > 1/c2
For this class, solution 1 is a saddle and solution 3 is stable. All
trajectories are attracted to solution 3. Examples can be seen by
following the links below. The value of c2=1 is used throughout.
 a1=1/4 Stable spiral at (3/4,1/4), saddle
at (0,1).
 a1=1/2 Stable spiral at (1/2,1/2), saddle
at (0,1).
 a1=3/4 Stable node at (1/4,3/4), saddle
at (0,1).
Note that as a1 is decreased, the stable fixed point moves toward the
saddle at (0,1). Note also that the stable fixed point changes from a
stable spiral to a stable node as a1 is decreased.
Phase portraits for a1 < 1/c2
At a1=1/c2, the fixed point from family 3 collides with the saddle at
(0,1/c2). For a1 < 1/c2, there is only the stable fixed point from
family 2.
Bifurcation diagram
The results of the stability analysis can be summarized in the form
of a bifurcation diagram. For 0 > a1 > 1/c2, solution 3 is
stable and solution 2 is unstable. For a1 < 1/c2, solution 2 is
stable and solution 3 does not exist in the physical regime. A sample
bifurcation diagram can be found here.Stable
solutuions are shown in green, and unstable solutions in red.
Relation to the LotkaVolterra model
If the parameter c2 is set equal to zero, the LotkaVolterra model
results. In this case, solution 2 goes off to infinity and solution 3
occurs at x=1, y= a1. The characteristic equation is
r^{2} a1 = 0
So that the equilibrium solution is a center of the linearized system
for every value of a1 < 0.
Bill Farr < bfarr@wpi.edu>
Last modified: Tue Oct 8 12:31:27 EDT 1996