Modified midpoint method for solving the initial value problem in Python

Modified midpoint method for solving the initial value problem in Python

''' yStop = integrate (F,x,y,xStop,tol=1.0e-6)
    Modified midpoint method for solving the
    initial value problem y' = F(x,y}.
    x,y   = initial conditions
    xStop = terminal value of x
    yStop = y(xStop)
    F     = user-supplied function that returns the
            array F(x,y) = {y'[0],y'[1],...,y'[n-1]}.
'''
from numpy import zeros,float,sum
from math import sqrt
 
def integrate(F,x,y,xStop,tol):
 
    def midpoint(F,x,y,xStop,nSteps):
  # Midpoint formulas
        h = (xStop - x)/nSteps
        y0 = y
        y1 = y0 + h*F(x,y0)
        for i in range(nSteps-1):
            x = x + h
            y2 = y0 + 2.0*h*F(x,y1)
            y0 = y1
            y1 = y2
        return 0.5*(y1 + y0 + h*F(x,y2))
 
    def richardson(r,k):
  # Richardson's extrapolation      
        for j in range(k-1,0,-1):
            const = (k/(k - 1.0))**(2.0*(k-j))
            r[j] = (const*r[j+1] - r[j])/(const - 1.0)
        return
 
    kMax = 51
    n = len(y)
    r = zeros((kMax,n),dtype=float)
  # Start with two integration steps
    nSteps = 2
    r[1] = midpoint(F,x,y,xStop,nSteps)
    r_old = r[1].copy()
  # Increase the number of integration points by 2
  # and refine result by Richardson extrapolation
    for k in range(2,kMax):
        nSteps = 2*k
        r[k] = midpoint(F,x,y,xStop,nSteps)
        richardson(r,k)
      # Compute RMS change in solution
        e = sqrt(sum((r[1] - r_old)**2)/n)
      # Check for convergence
        if e < tol: return r[1]
        r_old = r[1].copy()
    print "Midpoint method did not converge"