Math 334 Lecture #39
§9.5: Predator-Prey Equations
Outcome A: Apply Phase Plane Analysis to Predator-Prey Equations. Let x ≥ 0 and
y ≥ 0 denote the populations (or their densities) of the prey and the predator.
We make the following assumptions about the prey, predator, and their interaction.
1. In the absence of the predator, the prey grows exponentially, i.e., dx/dt = ax for a
constant a > 0 when y = 0.
2. In the absence of the prey, the predator dies out exponentially, i.e., dy/dt = −cy
for a constant c > 0 when x = 0.
3. The number of encounters between the predator and prey is proportional to the
product of their populations. Each encounter tends to promote the growth of
the predator, and diminish the prey. The predator grows according to γxy, for a
constant γ > 0, and the prey decreases according to −αxy, for a constant α > 0.
With this assumptions we obtain the Lotka-Volterra equations that model the predatorprey situation:
dx/dt = ax − αxy = x(a − αy),
dy/dt = −cy + γxy = y(−c + γx).
This model predicts unlimited growth of the prey in the absence of the predator.
A modification of the Lotka-Volterra equations that limits the prey to logistic growth is
dx/dt = x(a − bx − αy),
dy/dt = y(−c + γx),
for a constant b > 0, where in the absence of the predator, y = 0, gives a saturation level
of the prey as a/b.
Example. Apply phase plane analysis to the Lokta-Volterra model
dx/dt = x(1 − 0.5y),
dy/dt = y(−0.25 + 0.5x).
We first find all of the critical points in the first quadrant. These are given by
x(1 − 0.5y) = 0, y(−0.25 + 0.5x) = 0.
The are two critical points: x = 0, y = 0, and x = 1/2, y = 2.
To find the local linear systems at each of these equilibriums requires the first order
partials:
Fx = 1 − 0.5y Fy = −0.5x
Gx = 0.5y Gy = −0.25 + 0.5x.
The linearization matrix at x = 0, y = 0 is
1
0
0 −1/4
whose eigenvalues are r1 = 1, r = 2 = −1/4 with corresponding eigenvectors (1, 0)T and
(0, 1)T ; hence this equilibrium is a unstable saddle point.
The linearization matrix at x = 1/2, y = 2 is
0 −1/4
1
0
whose eigenvalues are r = ±i/2; hence this equilibrium is either center or a spiral point,
with rotation in the counterclockwise direction, but whose stability is not determined by
the linearization.
Here is a phase portrait for this Lotka-Volterra model.
It appears from this phase portrait that the equilibrium at x = 1/2, y = 2 is a stable
center. [We will learn about a technique in the next section that will determine the
stability of this equilibrium.]
Example. Apply phase plane analysis to the modified Lotka-Volterra model
dx/dt = x(1 − 0.5x − 0.5y),
dy/dt = y(−0.25 + 0.5x).
[This is the previous Example modified for logistic growth of the prey.]
The critical points are given by
x(1 − 0.5x − 0.5y) = 0, y(−0.25 + 0.5x) = 0.
The critical points in the first quadrant are x = 0, y = 0; x = 2, y = 0, and x = 1/2,
y = 3/2.
The first-order partials of F and G are
Fx = 1 − x − 0.5y, Fy = −0.5x
Gx = 0.5y, Gy = −0.25 + 0.5x.
The linearization matrix at the equilibrium x = 0, y = 0 is
1
0
0 −1/4
whose eigenvalues are r1 = 1, r2 = −1/4 with corresponding eigenvectors (1, 0)T and
(0, 1)T ; hence this equilibrium is an unstable saddle point (like before).
The linearization matrix at the equilibrium x = 2, y = 0 is
−1 −1
0 3/4
whose eigenvalues are r1 = −1, r2 = 3/4 with corresponding eigenvectors (1, 0)T and
(−4, 7)T ; hence this equilibrium is an unstable saddle point.
The linearization matrix at the equilibrium x = 1/2, y = 3/2 is
−1/4 −1/4
,
3/4
0
√
whose eigenvalues are (−1 ± i 11)/8; hence this equilibrium is an asymptotically stable
spiral point, with motion in the counterclockwise direction.
Here is a phase portrait for this modified Lotka-Volterra model.
What is the basin of attraction of the asymptotically stable spiral point x = 1/2, y = 3/2?
What are the separatrices?