Predator-Prey Problem from Class
Contents
System of differential equations
Let y1 denote the number of rabbits (prey), let y2 denote the number of owls (predator). The population growth is described by
- y1' = a1*y1
- y2' = a2*y2
Here the growth rate a1 for rabbits is positive for y2=0, but decreases with increasing y2; the growth rate a2 for owls is negative for y1=0, but increases with increasing y1:
- a1 = 2 - .5*y2
- a2 = -1 + .5*y1
Hence we obtain the following system of ODEs
- y1' = (2-.5*y2)*y1
- y2' = (-1+.5*y1)*y2
Given initial conditions
The initial rabbit population is 6, the initial owl population is 2:
- y1(0) = 6
- y2(0) = 2
Definition of the function f
We have an ODE system y' = f(t,y) where y is a vector and the vector-valued function f is given by
f = @(t,y) [(2-.5*y(2))*y(1); (-1+.5*y(1))*y(2)];
Solve the initial value problem using ode45
We want to find the functions y1(t) and y2(t) for t in [0,15]:
opt = odeset('RelTol',1e-10,'AbsTol',1e-10); % use smaller tolerances [ts,ys] = ode45(f,[0,15],[6;2],opt);
Plot both y1(t) and y2(t) vs. t
It appears that the functions y1(t), y2(t) are periodic with a period of about 5.
plot(ts,ys) legend('rabbits','owls'); xlabel('time t')
Plot y2 vs. y1 (Phase plane plot)
The phase plane plot clearly shows that the solution is periodic since the trajectory is a closed curve.
The vector field given by f shows the velocity vectors with which the point (y1(t),y2(t)) moves along the trajectory. We see that the point moves along the closed curve counterclockwise as t increases.
(You first have to download the file vectfield.m)
plot(ys(:,1),ys(:,2)); hold on xlabel('rabbits'); ylabel('owls') vectfield(f,0:.8:7.5,0:.8:10.5); hold off % plot vector field
