i have this code, that should calculate the speed and distance with given acceleration. The boundrary condition says that the speed in the beginning or and in the end has to be zero. but i cant get it to work code: from scipy.integrate import solve_bvp, cumulative_trapezoid from scipy.interpolate import interp1d,UnivariateSpline import numpy as np import pandas as pd data = pd.read_csv("Linear Accelerometer.csv", header=None, skiprows=1) t_raw = data.iloc[:, 0].astype(str).str.replace(',', '.').astype(float).values a_raw = data.iloc[:, 2].astype(str).str.replace(',', '.').astype(float).values n_points = 1000 t_coarse = np.linspace(t_raw[0], t_raw[-1], n_points) a_coarse = np.interp(t_coarse, t_raw, a_raw) a_spline = UnivariateSpline(t_coarse, a_coarse, s=0.1) # s > 0 glättet a_func = a_spline v_guess = cumulative_trapezoid(a_func(t_coarse), t_coarse, initial=0) s_guess = cumulative_trapezoid(v_guess, t_coarse, initial=0)""" y_initode = np.zeros((2, t_coarse.size)) DGL-System def fun(t, y): return np.array([y[1], a_func(t)]) Randbedingungen v(0)=0, v(end)=0 def bc(ya, yb): return np.array([ya[1] , yb[1] ]) Solver aufrufen sol = solve_bvp(fun, bc, t_coarse, y_initode, max_nodes=100000, verbose=2) Prüfen print(sol.message) Ergebnis extrahieren s = sol.sol(t_coarse)[0] v = sol.sol(t_coarse)[1] import matplotlib.pyplot as plt plt.figure(figsize=(12, 6)) plt.subplot(2, 1, 1) plt.plot(t_coarse, s, label="s(t) (Strecke)") plt.ylabel("Strecke [m]") plt.legend() plt.grid() plt.subplot(2, 1, 2) plt.plot(t_coarse, v, label="v(t) (Geschwindigkeit)", color="orange") plt.xlabel("Zeit ") plt.ylabel("Geschwindigkeit [m/s]") plt.legend() plt.grid() plt.tight_layout() plt.show() like this it gives me the error: A singular Jacobian encountered when solving the collocation system. I expected the speed to be 0 in the beginning and in the end but it doesnt work as explained. But when i say the distance has to be zero at beginning and the end it works and looks like this:Plot for d(0) = d(T) = 0 Continue reading...
