I am new here and new in programming, so excuse me if the question is not formulated clearly enough. For a uni assignment, my labpartner and I are programming a predator-prey system. In this predator-prey system, there is a certain load factor 'W0'. We want to find a load factor W0, accurate to 5 significant digits, for which applies that there will never be less than 250 predators (wnum[1] in our code). We want to find this value of W0 and we need the code to carry on further calculations with this found value of W0. Here is what we've tried so far, but python does not seem to give any response:
# Import important stuff and settings
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
print ('Results of Group 4')
def W0():
W0 = 2.0
while any(wnum[1])<250:
W0 = W0-0.0001
return W0
def W(t):
if 0 <= t < 3/12:
Wt = 0
elif 3/12 <= t <= 8/12:
Wt = W0
elif 8/12 < t < 1:
Wt = 0
else:
Wt = W(t - 1)
return Wt
# Define the right-hand-side function
def rhsf(t,y):
y1 = y[0]
y2 = y[1]
f1 = (2-2*10**-3*y2)*y1-W(t)*y1
f2 = (-3.92+7*10**-3*y1)*y2
return np.array([f1,f2])
# Define one step of the RK4 method
def RK4Step(tn,wn,Dt,f):
# tn = current time
# wn = known approximation at time tn
# Dt = the time step to use
# f = the right-hand-side function to use
# wnplus1 = the new approximation at time tn+Dt
k1 = Dt*f(tn,wn)
k2 = Dt*f(tn+0.5*Dt,wn+0.5*k1)
k3 = Dt*f(tn+0.5*Dt,wn+0.5*k2)
k4 = Dt*f(tn+Dt,wn+k3)
wnplus1 = wn + 1/6*(k1 +2*k2 +2*k3 +k4)
return wnplus1
# Define the complete RK4 method
def RK4Method(t0,tend,Dt,f,y0):
# t0 = initial time of simulation
# tend = final time of simulation
# Dt = the time step to use
# f = the right-hand-side function to use
# y0 = the initial values
# calculate the number of time steps to take
N = int(np.round((tend-t0)/Dt))
# make the list of times t which we want the solution
time = np.linspace(t0,tend,num=N+1)
# make sure Dt matches with the number of time steps
Dt = (tend-t0)/N
# Allocate memory for the approximations
# row i represents all values of variable i at all times
# column j represents all values of all variables at time t_j
w = np.zeros((y0.size,N+1))
# store the (given) initial value
w[:,0] = y0
# Perform all time steps
for n,tn in enumerate(time[:-1]):
w[:,n+1] = RK4Step(tn,w[:,n],Dt,f)
return time, w
# Set all known values and settings
t0 = 0.0
tend = 10.0
y0 = np.array([600.0,1000.0])
Dt = 0.5/(2**7)
# Execute the method
tnum, wnum = RK4Method(t0,tend,Dt,rhsf,y0)
# Make a nice table
alldata = np.concatenate(([tnum],wnum),axis=0).transpose()
table = pd.DataFrame(alldata,columns=['t','y1(t)','y2(t)'])
print('\nA nice table of the simulation:\n')
print(table)
# Make a nice picture
plt.close('all')
plt.figure()
plt.plot(tnum,wnum[0,:],label='$y_1$',marker='o',linestyle='-')
plt.plot(tnum,wnum[1,:],label='$y_2$',marker='o',linestyle='-')
plt.xlabel('$t$')
plt.ylabel('$y(t)$')
plt.title('Simulation')
plt.legend()
# Do an error computation
# Execute the method again with a doubled time step
tnum2, wnum2 = RK4Method(t0,tend,2.0*Dt,rhsf,y0)
# Calculate the global truncation errors at the last simulated time
errors = (wnum[:,-1] - wnum2[:,-1])/(2**4-1)
print('\nThe errors are ',errors[0],' for y1 and ',errors[1],' for y2 at time t=',tnum[-1])