Simulation of Predator-Prey Ecological Model
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The predator prey system below has a term for the carrying capacity of the prey species. Initially the prey population N(0) = 330 and the predator population P(0) = 270.
dN/dt = 0.07N(700-N) - 0.05NP
dP/dt = 0.04PN - 4P
(a) In the absence of predators, what is the carrying capacity of the prey's environment?
(b) Use a step size of delta(t) = 0.05 and Euler's method, to model the population numbers over the next 5 years.
(c) i) Plot prey and predator population versus time on one graph.
ii) Plot prey population size versus predator population size.
iii) Do the results suggest a stable situation develops in time?
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Solution Summary
The solution numerically and graphically solves a problem pertaining to a predator-prey ecological model governed by a system of two coupled differential equations.
Solution Preview
See the attached file(s).
(a) The carrying capacity of the prey's environment is 700, as can be seen by the term 700-N in the first equation. As N approaches 700, dN/dt approaches zero.
(b) See the first attachment.
(c), i. - ii. See the second and third attachments.
iii. As you can see, the predator and prey populations show oscillatory behavior with a period of approximately 0.55 years.
// Predator-prey numerical simulation.
import java.awt.Color;
import java.awt.Graphics;
import javax.swing.JPanel;
public class PredatorPrey extends JPanel
{
public void paintComponent( Graphics g )
{
// call paintComponent to ensure the panel displays correctly
super.paintComponent( g );
int width = getWidth(); // total width
int height = getHeight(); // total height
int k;
double t, dt = 0.05, tmax = 5.0, n = 330.0, p = 270.0, dndt, dpdt, n0, p0;
Color c1 = new Color( 255, 0, 0 );
Color c2 = new Color( 0, 0, 255 );
// Output predator and prey population vs. time.
for ( k = 0; k < 100; k ++ )
{
t = 0.05 * ( double ) k;
System.out.printf( "t = %5.2f; N = ...
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