This is an algorithm regarding the RKF method:
import numpy as np
class rkf():
def __init__(self,f, a, b, x0, atol, rtol, hmax, hmin):
self.f=f
self.a=a
self.b=b
self.x0=x0
self.atol=atol
self.rtol=rtol
self.hmax=hmax
self.hmin=hmin
def solve(self):
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 = self.a
x = np.array(self.x0)
h = self.hmax
T = np.array( [t] )
X = np.array( [x] )
while t < self.b:
if t + h > self.b:
h = self.b - t
k1 = h * self.f(t, x)
k2 = h * self.f(t + a2 * h, x + b21 * k1 )
k3 = h * self.f(t + a3 * h, x + b31 * k1 + b32 * k2)
k4 = h * self.f(t + a4 * h, x + b41 * k1 + b42 * k2 + b43 * k3)
k5 = h * self.f(t + a5 * h, x + b51 * k1 + b52 * k2 + b53 * k3 + b54 * k4)
k6 = h * self.f(t + a6 * h, x + b61 * k1 + b62 * k2 + b63 * k3 + b64 * k4 + b65 * k5)
r = abs( r1 * k1 + r3 * k3 + r4 * k4 + r5 * k5 + r6 * k6 ) / h
r = r / (self.atol+self.rtol*(abs(x)+abs(k1)))
if len( np.shape( r ) ) > 0:
r = max( r )
if r <= 1:
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( 0.94 * ( 1 / r )**0.25, 0.1 ), 4.0 )
if h > self.hmax:
h = self.hmax
elif h < self.hmin or t==t-h:
raise RuntimeError("Error: Could not converge to the required tolerance.")
break
return (T,X)
Which works just fine, but I was wondering if is it possible to make this even faster and more efficient?
class rkf():
-->class RKF:
. \$\endgroup\$