from
https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.solve_ivp.html
with
sol = scipy.integrate.solve_ivp(fun, t_span, y0, method='RK45', t_eval=None, dense_output=False, events=None, vectorized=False, **options)
t_eval is optional for store solution in specify time.
Will this override timestep select by RK45?
It does not override the timestep. One way to verify this is to use dense_output=True, which saves data at each timestep for interpolating later.
The sol attribute contains additional information about the timesteps in the ts attribute. Here, you can see that using t_eval changes the return of sol3.t but does not affect the timesteps.
import numpy as np
from scipy.integrate import solve_ivp
# To make readable
np.set_printoptions(precision=2)
method = 'RK45'
def dy(t, y):
return y
sol = solve_ivp(dy, (0, 10), [1], method=method)
print(f"No options : {sol.t}")
sol2 = solve_ivp(dy, (0, 10), [1], method=method, dense_output=True)
print(f"Dense output : {sol2.t}")
print(f" Interpolants: {sol2.sol.ts}")
t_eval = [5]
sol3 = solve_ivp(dy, (0, 10), [1], method=method, t_eval=t_eval, dense_output=True)
print(f"t_eval return : {sol3.t}")
print(f" Interpolants: {sol3.sol.ts}")
returns
No options : [ 0. 0.1 1.07 2.3 3.65 5.03 6.43 7.83 9.24 10. ]
Dense output : [ 0. 0.1 1.07 2.3 3.65 5.03 6.43 7.83 9.24 10. ]
Interpolants: [ 0. 0.1 1.07 2.3 3.65 5.03 6.43 7.83 9.24 10. ]
t_eval return : [5]
Interpolants: [ 0. 0.1 1.07 2.3 3.65 5.03 6.43 7.83 9.24 10. ]
I mildly suggest that instead of using t_eval, you should only use dense_output=True and then construct y_eval after the fact. This is a much more flexible and transparent usage.
sol = solve_ivp(dy, (0, 10), [1], method=method, dense_output=True)
y_eval = sol.sol(t_eval)