Numerical ODE solving in Python

Question

How do I numerically solve an ODE in Python?

Consider

\\ddot{u}(         


        

Answer 1

The code from your other question is really close to what you want. Two changes are needed:

• You were solving a different ODE (because you changed two signs inside function deriv)
• The y component of your desired plot comes from the solution values, not from the values of the first derivative of the solution, so you need to replace u[:,0] (function values) for u[:, 1] (derivatives).

This is the end result:

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

def deriv(u, t):
    return np.array([u[1], -u[0] + np.sqrt(u[0])])

time = np.arange(0.01, 7 * np.pi, 0.0001)
uinit = np.array([1.49907, 0])
u = odeint(deriv, uinit, time)

x = 1 / u[:, 0] * np.cos(time)
y = 1 / u[:, 0] * np.sin(time)

plt.plot(x, y)
plt.show()

However, I suggest that you use the code from unutbu's answer because it's self documenting (u, udot = z) and uses np.linspace instead of np.arange. Then, run this to get your desired figure:

x = 1 / u * np.cos(phi)
y = 1 / u * np.sin(phi)
plt.plot(x, y)
plt.show()

Answer 2

scipy.integrate() does ODE integration. Is that what you are looking for?

Answer 3

You can use scipy.integrate.ode. To solve dy/dt = f(t,y), with initial condition y(t0)=y0, at time=t1 with 4th order Runge-Kutta you could do something like this:

from scipy.integrate import ode
solver = ode(f).set_integrator('dopri5')
solver.set_initial_value(y0, t0)
dt = 0.1
while t < t1:
    y = solver.integrate(t+dt)
    t += dt

Edit: You have to get your derivative to first order to use numerical integration. This you can achieve by setting e.g. z1=u and z2=du/dt, after which you have dz1/dt = z2 and dz2/dt = d^2u/dt^2. Substitute these into your original equation, and simply iterate over the vector dZ/dt, which is first order.

Edit 2: Here's an example code for the whole thing:

import numpy as np
import matplotlib.pyplot as plt

from numpy import sqrt, pi, sin, cos
from scipy.integrate import ode

# use z = [z1, z2] = [u, u']
# and then f = z' = [u', u''] = [z2, -z1+sqrt(z1)]
def f(phi, z):
    return [z[1], -z[0]+sqrt(z[0])]


# initialize the 4th order Runge-Kutta solver
solver = ode(f).set_integrator('dopri5')

# initial value
z0 = [1.49907, 0.]
solver.set_initial_value(z0)

values = 1000
phi = np.linspace(0.0001, 7.*pi, values)
u = np.zeros(values)

for ii in range(values):
    u[ii] = solver.integrate(phi[ii])[0] #z[0]=u

x = 1. / u * cos(phi)
y = 1. / u * sin(phi)

plt.figure()
plt.plot(x,y)
plt.grid()
plt.show()

Answer 4

import scipy.integrate as integrate
import matplotlib.pyplot as plt
import numpy as np

pi = np.pi
sqrt = np.sqrt
cos = np.cos
sin = np.sin

def deriv_z(z, phi):
    u, udot = z
    return [udot, -u + sqrt(u)]

phi = np.linspace(0, 7.0*pi, 2000)
zinit = [1.49907, 0]
z = integrate.odeint(deriv_z, zinit, phi)
u, udot = z.T
# plt.plot(phi, u)
fig, ax = plt.subplots()
ax.plot(1/u*cos(phi), 1/u*sin(phi))
ax.set_aspect('equal')
plt.grid(True)
plt.show()