Pass args for solve_ivp (new SciPy ODE API)

Question

For solving simple ODEs using SciPy, I used to use the odeint function, with form:

scipy.integrate.odeint(func, y0, t, args=(), Dfun=None, col_deriv=0, full_         


        

Answer 1

Relatively recently there appeared a similar question on scipy's github. Their solution is to use lambda:

solve_ivp(fun=lambda t, y: fun(t, y, *args), ...)

And they argue that there is already enough overhead for this not to matter.

Answer 2

For completeness, I think you can also do this but I'm not sure why you would bother since the other two options posted here are perfectly fine.

from functools import partial
fun = partial(dy_dt, arg1=arg1, arg2=arg2)
scipy.integrate.solve_ivp(fun, t_span, y0, method='RK45', t_eval=None, dense_output=False, events=None, vectorized=False, **options)

Answer 3

It doesn't seem like the new function has an args parameter. As a workaround you can create a wrapper like

def wrapper(t, y):
    orig_func(t,y,hardcoded_args)

and pass that in.

Answer 4

Adding to Cleb's answer, here's an example for using the lambda t,y: fun(t,y,args) method. We set up the function handle that returns the rhs of a second order homogeneous ODE with two parameters. Then we feed it to our solver, along with a couple options.

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


def rhs_2nd_order_ode(t, y, a, b):
    """
    2nd order ODE function handle for use with scipy.integrate.solve_ivp
    Solves u'' + au'+ bu = 0 after reducing order with y[0]=u and y[1]=u'.

    :param t: dependent variable
    :param y: independent variables
    :param a: a
    :param b: b
    :return: Returns the rhs of y[0]' = y[1] and y[1]' = -a*y[1] - b*y[0]
    """
    return [y[1], -a*y[1] - b*y[0]]


if __name__ == "__main__":
    t_span = (0, 10)
    t_eval = np.linspace(t_span[0], t_span[1], 100)
    y0 = [0, 1]
    a = 1
    b = 2
    sol = integrate.solve_ivp(lambda t,y: rhs_2nd_order_ode(t,y,a,b), t_span, y0, 
                              method='RK45', t_eval=t_eval)

    fig, ax = plt.subplots(1, 1)
    ax.plot(sol.t, sol.y[0])
    ax.set(xlabel='t',ylabel='y')

Answer 5

According to Javier-Acuna's ultra-brief, ultra-useful answer, the feature that you (as well as I) desire has recently been added. This was announced on Github by none other than the great Warren Weckesser (See his Github, StackOverflow) himself. Anyway, jokes aside the docstring of solve_ivp has an example using it in for the `Lotka-Volterra equations

solve_ivp( fun, t_span, y0, method='RK45', t_eval=None, dense_output=False, events=None, vectorized=False, args=None, **options, )

So, just include args as a tuple. In your case

args = (arg1, arg2)

Please don't use my answer unless your scipy version >= 1.4 . There is no args parameter in solve_ivp for versions below it. I have personally experienced my answer failing for version 1.2.1.

The implementation by zahabaz would probably still work fine in case your scipy version < 1.4

Answer 6

Recently the 'args' option was added to solve_ivp, see here: https://github.com/scipy/scipy/issues/8352#issuecomment-535689344