Solving a system of odes (with changing constant!) using scipy.integrate.odeint?

Question

I currently have a system of odes with a time-dependent constant. E.g.

def fun(u, t, a, b, c):
    x = u[0]
    y = u[1]
    z = u[2]
    dx_dt = a * x + y *         


        

Answer 1

Yes, this is possible. In the case where a is constant, I guess you called scipy.integrate.odeint(fun, u0, t, args) where fun is defined as in your question, u0 = [x0, y0, z0] is the initial condition, t is a sequence of time points for which to solve for the ODE and args = (a, b, c) are the extra arguments to pass to fun.

In the case where a depends on time, you simply have to reconsider a as a function, for example (given a constant a0):

def a(t):
    return a0 * t

Then you will have to modify fun which computes the derivative at each time step to take the previous change into account:

def fun(u, t, a, b, c):
    x = u[0]
    y = u[1]
    z = u[2]
    dx_dt = a(t) * x + y * z # A change on this line: a -> a(t)
    dy_dt = b * (y - z)
    dz_dt = - x * y + c * y - z
    return [dx_dt, dy_dt, dz_dt]

Eventually, note that u0, t and args remain unchanged and you can again call scipy.integrate.odeint(fun, u0, t, args).

A word about the correctness of this approach. The performance of the approximation of the numerical integration is affected, I don't know precisely how (no theoretical guarantees) but here is a simple example which works:

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

tmax = 10.0

def a(t):
    if t < tmax / 2.0:
        return ((tmax / 2.0) - t) / (tmax / 2.0)
    else:
        return 1.0

def func(x, t, a):
    return - (x - a(t))

x0 = 0.8
t = np.linspace(0.0, tmax, 1000)
args = (a,)
y = sp.integrate.odeint(func, x0, t, args)

fig = plt.figure()
ax = fig.add_subplot(111)
h1, = ax.plot(t, y)
h2, = ax.plot(t, [a(s) for s in t])
ax.legend([h1, h2], ["y", "a"])
ax.set_xlabel("t")
ax.grid()
plt.show()

I Hope this will help you.