# Results

There are three families of steady state solutuions of this model.
1. x=0, y=0, corresponding to extinction of both populations.
2. x=0, y = 1/c2, corresponding to extinction of the predators, and the equilibrium population of the prey in the absence of predators.
3. 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.

The Jacobian matrix for the model has the following form.
 a1+y x -y 1-x-2 c2y

When the three different families of steady state solutions are substituted into the Jacobian, the following results are obtained.

1. For family 1, x=0, y=0, the Jacobian matrix becomes as follows.
 a1 0 0 1
The eigenvalues are a1, which is negative for this model, and 1 so the origin is a saddle.
2. For family 2, x=0, y=1/c2, the Jacobian becomes
 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.
3. For family 3, x=1+c1a2, y=-a1, the Jacobian matrix is
 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

r2 - 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(1-c2y)

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 Lotka-Volterra model

If the parameter c2 is set equal to zero, the Lotka-Volterra model results. In this case, solution 2 goes off to infinity and solution 3 occurs at x=1, y= -a1. The characteristic equation is

r2 -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>