Amirhossein Rezaei
Amirhossein Rezaei

Reputation: 364

Issue on Runge Kutta Fehlberg algorithm

I have wrote a code for Runge-Kutta 4th order, which works perfectly fine for a system of differential equations:

import numpy as np
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
 

@numba.jit()
def V(u,t):
    x1,dx1, x2, dx2=u
    ddx1=-w**2 * x1 -b * dx1
    ddx2=-(w+0.5)**2 * x2 -(b+0.1) * dx2
    return np.array([dx1,ddx1,dx2,ddx2])


@numba.jit()
def rk4(f, u0, t0, tf , n):
    t = np.linspace(t0, tf, n+1)
    u = np.array((n+1)*[u0])
    h = t[1]-t[0]
    for i in range(n):
        k1 = h * f(u[i], t[i])    
        k2 = h * f(u[i] + 0.5 * k1, t[i] + 0.5*h)
        k3 = h * f(u[i] + 0.5 * k2, t[i] + 0.5*h)
        k4 = h * f(u[i] + k3, t[i] + h)
        u[i+1] = u[i] + (k1 + 2*(k2 + k3) + k4) / 6
    return u, t

u, t  = rk4(V,np.array([0,0.2,0,0.3]) ,0,100, 20000)

print("Execution time:",time.clock() - start_time, "seconds")
x1,dx1,x2,dx2 = u.T
plt.plot(x1,x2)
plt.xlabel('X1')
plt.ylabel('X2')
plt.show()

The above code, returns the desired result: enter image description here

And thanks to Numba JIT, this code works really fast. However, this method doesn't use adaptive step size and hence, it is not very suitable for a system of stiff differential equations. Runge Kutta Fehlberg method, solves this problem by using a straight forward algorithm. Based on the algorithm (https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta%E2%80%93Fehlberg_method) I wrote this code which works for only one differential equation :

import numpy as np

def rkf( f, a, b, x0, tol, hmax, hmin ):

    a2  =   2.500000000000000e-01  #  1/4
    a3  =   3.750000000000000e-01  #  3/8
    a4  =   9.230769230769231e-01  #  12/13
    a5  =   1.000000000000000e+00  #  1
    a6  =   5.000000000000000e-01  #  1/2

    b21 =   2.500000000000000e-01  #  1/4
    b31 =   9.375000000000000e-02  #  3/32
    b32 =   2.812500000000000e-01  #  9/32
    b41 =   8.793809740555303e-01  #  1932/2197
    b42 =  -3.277196176604461e+00  # -7200/2197
    b43 =   3.320892125625853e+00  #  7296/2197
    b51 =   2.032407407407407e+00  #  439/216
    b52 =  -8.000000000000000e+00  # -8
    b53 =   7.173489278752436e+00  #  3680/513
    b54 =  -2.058966861598441e-01  # -845/4104
    b61 =  -2.962962962962963e-01  # -8/27
    b62 =   2.000000000000000e+00  #  2
    b63 =  -1.381676413255361e+00  # -3544/2565
    b64 =   4.529727095516569e-01  #  1859/4104
    b65 =  -2.750000000000000e-01  # -11/40

    r1  =   2.777777777777778e-03  #  1/360
    r3  =  -2.994152046783626e-02  # -128/4275
    r4  =  -2.919989367357789e-02  # -2197/75240
    r5  =   2.000000000000000e-02  #  1/50
    r6  =   3.636363636363636e-02  #  2/55

    c1  =   1.157407407407407e-01  #  25/216
    c3  =   5.489278752436647e-01  #  1408/2565
    c4  =   5.353313840155945e-01  #  2197/4104
    c5  =  -2.000000000000000e-01  # -1/5

    t = a
    x = np.array(x0)
    h = hmax

    T = np.array( [t] )
    X = np.array( [x] )
    
    while t < b:

        if t + h > b:
            h = b - t

        k1 = h * f( x, t )
        k2 = h * f( x + b21 * k1, t + a2 * h )
        k3 = h * f( x + b31 * k1 + b32 * k2, t + a3 * h )
        k4 = h * f( x + b41 * k1 + b42 * k2 + b43 * k3, t + a4 * h )
        k5 = h * f( x + b51 * k1 + b52 * k2 + b53 * k3 + b54 * k4, t + a5 * h )
        k6 = h * f( x + b61 * k1 + b62 * k2 + b63 * k3 + b64 * k4 + b65 * k5, \
                    t + a6 * h )

        r = abs( r1 * k1 + r3 * k3 + r4 * k4 + r5 * k5 + r6 * k6 ) / h
        if len( np.shape( r ) ) > 0:
            r = max( r )
        if r <= tol:
            t = t + h
            x = x + c1 * k1 + c3 * k3 + c4 * k4 + c5 * k5
            T = np.append( T, t )
            X = np.append( X, [x], 0 )

        h = h * min( max( 0.84 * ( tol / r )**0.25, 0.1 ), 4.0 )

        if h > hmax:
            h = hmax
        elif h < hmin:
            raise RuntimeError("Error: Could not converge to the required tolerance %e with minimum stepsize  %e." % (tol,hmin))
            break

    return ( T, X )

but I'm struggling to convert it to a function like the first code, where I can input a system of differential equations. The most confusing part for me, is how can I vectorize everything in the second code without messing things up. In other words, I cannot reproduce the first result using the RKF algorithm. Can anyone point me in the right direction?

Upvotes: 1

Views: 2004

Answers (1)

Lutz Lehmann
Lutz Lehmann

Reputation: 26040

I'm not really sure where your problem lies. Setting the not given parameters to w=1; b=0.1 and calling, without changing anything

T, X = rkf( f=V, a=0, b=100, x0=[0,0.2,0,0.3], tol=1e-6, hmax=1e1, hmin=1e-16 )

gives the phase plot

enter image description here

The step sizes grow as the system slows down as

enter image description here

which is the expected behavior for an unfiltered step size controller.

Upvotes: 1

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