A Family of Predator–Prey Equations
Differential Equations (MATH 3310) Project
This project (found on page 496 of the Blanchard–Devaney–Hall textbook) concerns a
study of the family of differential equations
dx  x9  x  3xy
dt
dy
 2y  xy.
dt
This is a predator–prey model with predator population y and prey population x. The
parameter  is assumed to be non–negative. The goal of the project is to study how
the value of  determines the behavior of solutions.
To help us get started on the project, let us study the model in the case   1. In this
case, our model is
dx  x9  x  3xy
dt
dy
 2y  xy.
dt
The first thing to do is to determine the equilibrium points of the model. We do this by
solving the system of equations
x9  x  3xy  0
 2y  xy  0
which can be written as
x9  x  3y  0
y2  x  0.
The only solutions of this system of equations are 0, 0, 9, 0, and 2, 7/3, so we have
three equilibrium points, all of which are biologically meaningful.
The equilibrium point 0, 0 corresponds to the constant solution
xt  0
yt  0
and the meaning of this equilibrium point is that if we start with no prey and no
predators, there will never be any prey or predators (no spontaneous generation)
since our model does not take immigration into account.
The equilibrium point 9, 0 corresponds to the constant solution
xt  9
yt  0
and the meaning of this equilibrium point is that if we start with 9 prey and no
predators, there will always be 9 prey and no predators. This makes sense because 9
is the carrying capacity of the prey population in the logistic differential equation
dx  x9  x  9x 1  x
9
dt
and we don’t have spontaneous generation of predators.
The equilibrium point 2, 7/3 corresponds to the constant solution
1
xt  2
yt  7
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and the meaning of this equilibrium point is that if we start with 2 prey and 7/3
predators, there will always be 2 prey and 7/3 predators. These population levels are
“just right” for the predator and prey to survive together at constant population levels.
The next thing we would like to do is to determine the behavior of solutions of our
model with initial conditions near the equilibrium points. To do this, we must consider
the linearized models at each of the equilibrium points. Since
fx, y  9x  x 2  3xy
and
gx, y  2y  xy,
the Jacobian matrix for our model is
Jx, y 
9  2x  3y
3x
y
2  x
.
At the equilibrium point 0, 0, we have
J0, 0 
9
0
.
0 2
This matrix has eigenvalues 9 and 2, so 0, 0 is a local saddle for our model.
At the equilibrium point 9, 0, we have
9 27
J9, 0 
0
7
.
This matrix has eigenvalues 9 and 7, so 9, 0 is a local saddle for our model.
At the equilibrium point 2, 7/3, we have
J 2, 7
3

2 6
7
3
0
.
This matrix has eigenvalue 1  13 i, so 2, 7/3 is a local spiral sink for our model.
Based on our linearization analysis, we suspect that if we take any initial populations
x 0 , y 0  with x 0  0 and y 0  0, the long term limiting behavior of the populations will be
lim xt  2
t
lim yt  7
t
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and the approach to equilibrium (for both predator and prey) will be oscillatory. To
verify this conjecture, we compute a few numerical solutions. The picture below shows
a numerical solution (in the phase plane) for the initial condition x0  2, y0  1/2.
2
4
3
y2
1
0
1
2
3x
4
5
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This picture can be generated using X(plore) with the sequence of commands
Load (F8) solve2d.xpl
f(t,x,y)x*(9-x)-3*x*y
g(t,x,y)y*(-2x)
window(0,6,0,4)
axis(1,1,1,1)
solve2d(f,g,0,2,1/2,0.01,10,’RK’)
To see the prey and predator graphs separately (as functions of time), we use
Load (F8) xivst.xpl
window(0,10,0,6)
axis(1,1,1,1)
xivst(f,g,0,2,1/2,0.01,10,’RK’,1) (to obtain the prey graph)
erase
window(0,10,0,4)
axis(1,1,1,1)
xivst(f,g,0,2,1/2,0.01,10,’RK’,2) (to obtain the predator graph).
These graphs are shown below.
6
5
4
x3
2
1
0
2
4
t
6
8
10
3
4
3
y2
1
0
2
4
t
6
8
10
As we suspected, we see that
lim xt  2
t
lim yt  7  2 1
t
3
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and these limiting values are approached in an oscillatory fashion.
Finally, let us examine the nullclines of our system. (This is probably not really
necessary in this case, since we already know how solutions behave, but it might be
useful when other values of  are considered, so we will do it for practice.)
The x nullcline consists of all points satisfying the equation
x9  x  3y  0.
Thus, the x nullcline consists of the line
x0
and the line
9  x  3y  0.
Solution curves cross this nullcline with vertical tangents.
The y nullcline consists of all points satisfying the equation
y2  x  0.
Thus, the y nullcline consists of the line
y0
and the line
x  2.
Solution curves cross this nullcline with horizontal tangents.
The picture below shows the x nullcline in red and the y nullcline in green. The
numerical solution of our model with initial condition x0  2, y0  1/2 is also shown.
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The above plot can be generated in X(plore) (assuming that the subroutine
solve2d.xpl has already been loaded and the functions f and g have already been
defined) by using the sequence of commands
window(0,6,0,4)
axis(1,1,1,1)
color(red)
line(0,0,0,4)
graph(3-(1/3)*x,x)
color(green)
graph(0,x)
line(2,0,2,4)
solve2d(f,g,0,2,1/2,0.01,10,’RK’).
On with the Project!
Repeat the above analysis for the model
dx  x9  x  3xy
dt
dy
 2y  xy
dt
for the parameter values   0, 2, 3, 4, and 5. For each value of , your analysis
should include
 a discussion of the equilibrium points of the model (including a discussion of
which equilibrium points are biologically meaningful and why)
 a local analysis (i.e., linearization analysis) of each of the equilibrium points
 a few sample numerical solutions of the model (including phase plane, prey
versus time, and predator versus time graphs).
 an analysis of the nullclines of the model including pictures of the nullclines
(preferably in different colors) and showing a sample solution curve or two.
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Needless to say, all figures should be carefully labeled and properly referenced in your
writing to make things as easy as possible for the reader.
Finally, discuss (including mathematical details) how the behavior of solutions of the
family of predator–prey models changes qualitatively as we vary the value of  from
  0 to   5. Values of  at which the qualitative nature of solutions changes
abruptly are called bifurcation values for the family of predator–prey models. You can
summarize this information by making a table like the following:
 range
ab
Equilibrium Points Classification of Equilibrium Points
0, 0
saddle (or whatever)
eq. pt. 2
type
eq. pt 3
type
bc
0, 0
etc.
etc.
.
etc.
Your report should be written in a narrative “reader–friendly” style (at least as
reader–friendly as mathematics can be) and should conclude with a discussion of your
overall findings explained (as best possible) in such a way that somebody not familiar
with differential equations (such as perhaps an ecologist) could understand the results.
For example, say things such as “If the parameter  is in the range a    b, then, no
matter what the initial prey and predator populations are, the predator will eventually
become extinct and the prey will settle down at the equilibrium population level of 2.”
Good luck and have fun!
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