Predator-Prey Dynamics

Computer Methods in Chemical Engineering


Problem Statement: The following set of ordinary differential equations (ODEs) describe the dynamics of a predator-prey system.

   dr/dt = a*r - b*r*f  r(0)=3   ...Rabbits (Prey)
   df/dt =-c*f + d*r*f  f(0)=2   ...Foxes (Predator)

   where
     a = 2
     b = 2
     c = 1
     d = 1
Find r(0.1) and f(0.1) by manually advancing these ODEs from the initial position (at t=0) to the next step (at t=0.1) with the Euler's method and the Runge-Kutta's 4th-order method. Show your work.

Solution:

Euler's Method for Numerical Integration of ODEs.

With a=2, b=2, c=1, and d=1, the dynamic equations are: dr/dt = 2*r - 2*r*f = 2*r*(1-f) df/dt = -f + r*f = f*(r-1) Start with r=3 and f=2, and the slopes at this starting point are: dr/dt = 2*3*(1-2) = -6 df/dt = 2*(3-1) = 4 With h=0.1 r (at t=0.1) = r(0) + dr/dt*h = 3 + -6*0.1 = 2.4 f (at t=0.1) = f(0) + df/dt*h = 2 + 4*0.1 = 2.4

Runge-Kutta's Method for Numerical Integration of ODEs.

In the following, we repeat with Runge-Kutta's 4th-Order Method.
  1st evaluation is the same as Euler's Method.
    r1 = 3
    f1 = 2
  1st slope for r and f (from Problem 1) ...
    kr1= fr(r1,f1) = -6
    kf1= ff(r1,f1) =  4
  2nd evaluation is at:
    r2 = 3 + kr1*h/2 = 3 + -6*0.05 = 2.7
    f2 = 2 + kf1*h/2 = 2 +  4*0.05 = 2.2
  2nd slope for r and f ...
    kr2= fr(r2,f2) = 2*2.7*(1-2.2) = -6.48
    kf2= ff(r2,f2) = 2.2*(2.7-1)   =  3.74
  3rd evaluation is at:
    r3 = 3 + kr2*h/2 = 3 + -6.48*0.05 = 2.676
    f3 = 2 + kf2*h/2 = 2 +  3.74*0.05 = 2.187
  3rd slope for r and f ...
    kr3= fr(r3,f3) = 2*2.676*(1-2.187) = -6.35
    kf3= ff(r3,f3) = 2.187*(2.676-1)   =  3.67
  4th evaluation is at:
    r4 = 3 + kr3*h = 3 + -6.35*0.1 = 2.365
    f4 = 2 + kf3*h = 2 +  3.67*0.1 = 2.367
    u  = 9.8116/(1+9.8116) = 0.9075
  4th slope for r and f ...
    kr4= fr(r4,f4) = 2*2.365*(1-2.367) = -6.47
    kf4= ff(r4,f4) = 2.367*(2.365-1)   =  3.23
  Weighted average slope for r and f ...
    krave = (kr1+2*kr2+2*kr3+kr4)/6 = (-6-2*6.48-2*6.35-6.47)/6 = -6.36
    kfave = (kf1+2*kf2+2*kf3+kf4)/6 = ( 4+2*3.74+2*3.67+3.23)/6 =  3.67
  Update the values of r and f at t=h=0.1
    r = r + krave*h = 3 + -6.36*0.1 = 2.364
    f = f + kfave*h = 2 +  3.67*0.1 = 2.367


Problem Statement: Write a program to integrate the coupled set of ODEs from t=0 to t=10.

Solution:


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Computer Methods in Chemical Engineering -- Predator-Prey Dynamics
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