This python code can solve one non- coupled differential equation:
import numpy as np
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
@numba.jit()
# A sample differential equation "dy / dx = (x - y**2)/2"
def dydx(x, y):
return ((x - y**2)/2)
# Finds value of y for a given x using step size h
# and initial value y0 at x0.
def rungeKutta(x0, y0, x, h):
# Count number of iterations using step size or
# step height h
n = (int)((x - x0)/h)
# Iterate for number of iterations
y = y0
for i in range(1, n + 1):
"Apply Runge Kutta Formulas to find next value of y"
k1 = h * dydx(x0, y)
k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1)
k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * k2)
k4 = h * dydx(x0 + h, y + k3)
# Update next value of y
y = y + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4)
# Update next value of x
x0 = x0 + h
return y
def dplot(start,end,steps):
Y=list()
for x in np.linspace(start,end,steps):
Y.append(rungeKutta(x0, y, x , h))
plt.plot(np.linspace(start,end,steps),Y)
print("Execution time:",time.clock() - start_time, "seconds")
plt.show()
start,end = 0, 10
steps = end* 100
x0 = 0
y = 1
h = 0.002
dplot(start,end,steps)
This code can solve this differential equation:
dydx= (x - y**2)/2
Now I have a system of coupled differential equations:
dydt= (x - y**2)/2
dxdt= x*3 + 3y
How can I implement these two as a system of coupled differential equations in the above code? Is there any more generalized way for system of n-number of coupled differential equations?
With the help of others, I got to this:
import numpy as np
from math import sqrt
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
a=1
b=1
c=1
d=1
# Equations:
@numba.jit()
#du/dt=V(u,t)
def V(u,t):
x, y, vx, vy = u
return np.array([vy,vx,a*x+b*y,c*x+d*y])
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([1., 0., 1. , 0.]) , 0. , 10. , 100000)
x,y, vx,vy = u.T
# plt.plot(t, x, t,y)
plt.semilogy(t, x, t,y)
plt.grid('on')
print("Execution time:",time.clock() - start_time, "seconds")
plt.show()