Time dependent boundary conditions from an empirical function

[SpectroEric]

I want to use time-dependent boundary conditions from an empirical function (interpolate measured values). The code snippet below shows what I tried so far starting from your example for time dependent boundary conditions. Only case 1 works. What else can I try?

from pde import PDE, CartesianGrid, MemoryStorage, ScalarField, plot_kymograph
from sympy import Function, Symbol, sin
import numpy as np

def interp_nb(x_vals, x, y):
    return np.interp(x_vals, x, y)

zeiten = np.arange(0, 20, .1)
zeitwerte = np.sin(zeiten)
t = Symbol('t')
class sini(Function):
    @classmethod
    def eval(cls, t):
        cases = 1          # try the different cases
        match cases:
            case 1:
                erg = sin(t)        # ok
            case 2:
                erg = np.sin(t)     # not working
            case 3:
                erg = np.interp(t, zeiten, zeitwerte)   # not working
            case 4:
                erg = interp_nb(t, zeiten, zeitwerte)   # not working
        return erg

grid = CartesianGrid([[0, 10]], [64])  # generate grid
state = ScalarField(grid)  # generate initial condition

eq = PDE({"c": "laplace(c)"}, bc={"value_expression": sini(t)})

storage = MemoryStorage()
eq.solve(state, t_range=20, dt=1e-4, tracker=storage.tracker(0.1))

# plot the trajectory as a space-time plot
plot_kymograph(storage)

[david-zwicker]

I guess the right hand side of the expression is interpreted as a sympy expression, so it might not be surprising that only the sympy implementation of the function works. Instead of using a sympy function, you could also directly use a python function with signature func(value, dx, x, t), i.e., the following might work

def sini(value, dx, x, t):
        cases = 3          # try the different cases
        match cases:
            case 2:
                erg = np.sin(t)
            case 3:
                erg = np.interp(t, zeiten, zeitwerte)
            case 4:
                erg = interp_nb(t, zeiten, zeitwerte)
        return erg

eq = PDE({"c": "laplace(c)"}, bc={"value_expression": sini})

Note that I could not test this right now. Moreover, it might help to first get this to work without numba by using backend="numpy" in the solve call.

[SpectroEric]

I tried the three cases and they work. The perfect solution! Thank you very much for the fast reply!
From the documentation I could not figure out the syntax of func(value, dx, x, t) by myself.
backend="numpy" was not necessary.

[david-zwicker]

Great to hear that things worked out. I agree that the documentation is not very specific about this, but there is a table on all supported boundary conditions, which lists the signature in the fourth row.