Modeling PredatorPrey Relations:
Abstract
Our goal is to model predatorprey relations, assuming that
both the predator and the prey
populations satisfy logistic
growth models. Here, we assume
that the per-capita growth rate is
linear. If this work were new, I
might say something about my
results here, briefly.
A Population Model
 Assume only two populations: a predator and a
prey
 Assume logistic growth of both predator and prey
dx
 x ( k  ax  by )
dt
dy
 y ( m  cx  dy )
dt
where a, b, c, d , k , m are positive constants
Stationary Points
Used to analyze the behavior of the
population models
bm  dk
x
ad  bc
am  ck
y
ad  bc
Other Stationary Points
 For completeness
 Somewhat degenerate in nature
( x , y )  (0,0),
( x , y )  (  k / a ,0),
( x , y )  (0, m / c )
Observations
• The x-coordinate of the first
stationary point is positive only
if bm > dk
• The point (0,0) is only
semistable
• There is always one stationary
point calculated with an xcoordinate less than zero
(which means it is physically
impossible).
Type of Stationary
Points
  ax
L  
  cy
bx 

- dy 
• We need to find the eigenvalues of
the above matrix
• Positive (negative) eigenvalues
imply unstable (stable) stationary
point
The Eigenvalues
• We determine the eigenvalues of
the matrix L to be:
 (ax  dy )  (ax  dy )2  4 x y (ad  bc)
1,2 
2
Implications of
Eigenvalues
When  ( ax  dy )  0, the stationary points must
be stable (this happens for x  0).
If (a x  dy) 2  4 x y ( ad  bc )  0, then the stationary
points are a stable node; otherwise, they are a stable
focus.
Phase Plot
Stationary Points
( x , y )  (0,0),
( x , y )  (0,2),
( x , y )  ( 1 / 3,0),
( x , y )  (3 / 7,8 / 7)
Conclusions
• For both plots, (0,2) is a stable
node
• In the second plot, we also
have a stable focus, (3/7,8/7)