Created
February 27, 2020 16:11
-
-
Save CryogenicPlanet/04f4f580581488516662795373e5d859 to your computer and use it in GitHub Desktop.
Single Time Step for Runge Kutta 4th Order
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
#Runge-Kutta Method Single Time Step | |
def rk4(func,t,a,b,c,dt): | |
""" | |
Peforms a single time step of the Runge-Kutta 4th Order Method. | |
The below function finds the ki value for [dx,dy,dz] and return the value to move Yn+1 | |
func is an input of functions, for the Lorenz system this is [dx,dy,dz] | |
Recall Rk4 Equations : | |
k1 = h*f(xn,yn) | |
k2 = h*f(xn+h/2,yn+k1/2) | |
k3 = h*f(xn+h/2,yn+k2/2) | |
k4 = h*f(xn,yn+k3) | |
Where f is a function [dx,dy,dz] | |
Yn+1 = Yn + 1/6*(k1+k2+k3+k4) | |
""" | |
k1,k2,k3,k4 = [],[],[],[] | |
for f in func: | |
k1.append(dt*f(t,a,b,c)) | |
for f in func: | |
k2.append(dt*f(t+dt/2,a+k1[0]/2,b+k1[1]/2,c+k1[2]/2)) | |
for f in func: | |
k3.append( dt*f(t+dt/2,a+k2[0]/2,b+k2[1]/2,c+k2[2]/2)) | |
for f in func: | |
k4.append( dt*f(t+dt/2,a+k3[0],b+k3[1],c+k3[1])) | |
k1,k2,k3,k4 = np.array(k1),np.array(k2),np.array(k3),np.array(k4) | |
return (1/6)*(k1+k2+k3+k4) |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment