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Solving an ODE with a forcing termยค

This example demonstrates how to incorporate an external forcing term into the solve. This is really simple: just evaluate it as part of the vector field like anything else.

This example is available as a Jupyter notebook here.

import jax
import jax.numpy as jnp
import matplotlib.pyplot as plt
from diffrax import diffeqsolve, ODETerm, SaveAt, Tsit5

def force(t, args):
    m, c = args
    return m * t + c

def vector_field(t, y, args):
    return -y + force(t, args)

def solve(y0, args):
    term = ODETerm(vector_field)
    solver = Tsit5()
    t0 = 0
    t1 = 10
    dt0 = 0.1
    saveat = SaveAt(ts=jnp.linspace(t0, t1, 1000))
    sol = diffeqsolve(term, solver, t0, t1, dt0, y0, args=args, saveat=saveat)
    return sol

y0 = 1.0
args = (0.1, 0.02)
sol = solve(y0, args)
plt.plot(sol.ts, sol.ys)

Now let's consider a more complicated example: the forcing term is an interpolation, and what's more we would like to differentiate with respect to the values we are interpolating.

from diffrax import backward_hermite_coefficients, CubicInterpolation

def vector_field2(t, y, interp):
    return -y + interp.evaluate(t)

def solve(points):
    t0 = 0
    t1 = 10
    ts = jnp.linspace(t0, t1, len(points))
    coeffs = backward_hermite_coefficients(ts, points)
    interp = CubicInterpolation(ts, coeffs)
    term = ODETerm(vector_field2)
    solver = Tsit5()
    dt0 = 0.1
    y0 = 1.0
    sol = diffeqsolve(term, solver, t0, t1, dt0, y0, args=interp)
    (y1,) = sol.ys
    return y1

points = jnp.array([3.0, 0.5, -0.8, 1.8])
grads = solve(points)

In this example, we computed the interpolation in advance (not repeatedly on each step!), and then just evaluated it inside the vector field.