Arshak Grigoryan
Project for Math.
Modeling
Predator-Prey model (By Odell)
Lets consider one of the variations of the Odell`s model
where x, y are population of the prey and predator
respectively and a and b are positive control
parameters.
x'  x(bx  x  y )
2
y '  y ( x  a)
Bifurcation
We are already familiar with the bifurcation in
one dimensional systems. In this example we
consider two dimensional system with two
parameters a and b. And on this model I
would like to present Hopf bifurcation.
Bifurcation
Definition. If the phase portrait changes its
topological structure as parameters are
varied, we say that a bifurcation has
occurred. Examples include changes in the
number or stability of fixed points, closed
orbits or saddle connections as a parameter
is varied.
Closed Orbit and Limit Cycle
Closed Orbit: If a phase point starting
anywhere else would circulate around the
origin and eventually return to its starting
point.
Limit cycle: A limit cycle is an isolated closed
trajectory. Isolated means that neighboring
trajectories are not closed; they spiral either
toward or away from the limit cycle.
Hopf Bifurcation
Suppose a two-dimensional system has a stable fixed
point. How possibly it could lose stability as
parameter vary? The eigenvalues of the Jacobean
are the key.
If the fixed point is stable then eignevalues are
negative.( or their real parts are negative in complex
case). To destabilize the fixed point we need one or
both of the eigenvalues to change their sign(s).
Predator-Prey
Now let's investigate our system.
x'  x(bx  x 2  y )
y '  y ( x  a)
Lets look at the equilibrium solutions of the system.
To find them we need solutions of the system
x(bx  x 2  y )  0
y ( x  a)  0
Predator-Prey
After simple calculations we find out that fixed
points are the following.
1. (0,0)
2. (b,0)
3. (a, ab-a^2)
Predator-Prey
Now we will need Jacobean of the system in order to
analyze stability of equilibrium solutions.
 2bx  3x  y
J  
y

2
x 


x  a
Predator-Prey
i)
Solution at the origin.
J 0, 0
0 0 

 
0  a
analyzing the Jacobean we can see that this is
an unstable solution and that can be seen on
the graph.
Equilibrium at (b,0)
ii) Thus eigenvalues are  b
and b  a
.
We can see that first eigenvalue always negative and
when b<a second eigenvalue is negative also and for that
reason solution will be stable.
2
J b,0
  b2
 
 0
b 

b  a 
Last equilibrium soultion
And the Jacobean of the last equilibrium solution is
J a ,aba 2
 ab  2a 2
 
2
 ab  a
 a

0 
And conditions for stability are
2
Trace= ab  2a
Det=
And
a 2 (b  a)
a(b  2a)  a 2 (b  2a) 2  4a 2 (b  a)

2
Last equilibrium solution
Now we clearly can see that change of the sign
of eigenvalues occur at b=2a and that is the
the point where Hopf bifurcation happens. So
the stability of the solution we need a<b<2a.
We can see all this on the graphs.