131 lines
3.1 KiB
Python
131 lines
3.1 KiB
Python
import numpy as np
|
|
from scipy.optimize import least_squares
|
|
|
|
# Эталонные значения параметров
|
|
K1_REFERENCE = 0.577
|
|
K2_REFERENCE = 0.422
|
|
|
|
K1_BOUNDS = (1e-5, 1e2)
|
|
K2_BOUNDS = (1e-2, 1.0)
|
|
|
|
T_BEGIN = 0.0
|
|
T_END = 10.0
|
|
H = 0.1
|
|
|
|
|
|
def analytical_solution(t, k1, k2):
|
|
"""
|
|
Аналитическое решение задачи Коши.
|
|
"""
|
|
t = np.asarray(t, dtype=float)
|
|
|
|
x1 = np.exp(-k1 * t)
|
|
|
|
if np.isclose(k1, k2):
|
|
x2 = k1 * t * np.exp(-k1 * t)
|
|
else:
|
|
x2 = (
|
|
k1 / (k2 - k1)
|
|
* (np.exp(-k1 * t) - np.exp(-k2 * t))
|
|
)
|
|
|
|
x3 = 1.0 - x1 - x2
|
|
|
|
return np.column_stack((x1, x2, x3))
|
|
|
|
|
|
def derivative(state, k1, k2):
|
|
"""
|
|
Правая часть системы дифференциальных уравнений.
|
|
"""
|
|
x1, x2, x3 = state
|
|
|
|
return np.array([
|
|
-k1 * x1,
|
|
k1 * x1 - k2 * x2,
|
|
k2 * x2
|
|
])
|
|
|
|
|
|
def rk4_step(state, h, k1, k2):
|
|
"""
|
|
Один шаг метода Рунге--Кутты четвёртого порядка.
|
|
"""
|
|
q1 = derivative(state, k1, k2)
|
|
q2 = derivative(state + h * q1 / 2.0, k1, k2)
|
|
q3 = derivative(state + h * q2 / 2.0, k1, k2)
|
|
q4 = derivative(state + h * q3, k1, k2)
|
|
|
|
return state + h / 6.0 * (q1 + 2.0 * q2 + 2.0 * q3 + q4)
|
|
|
|
|
|
def rk4_solution(t, k1, k2):
|
|
"""
|
|
Решение задачи Коши методом RK4.
|
|
"""
|
|
solution = np.zeros((len(t), 3), dtype=float)
|
|
solution[0] = [1.0, 0.0, 0.0]
|
|
|
|
for index in range(1, len(t)):
|
|
h = t[index] - t[index - 1]
|
|
|
|
solution[index] = rk4_step(
|
|
solution[index - 1],
|
|
h,
|
|
k1,
|
|
k2
|
|
)
|
|
|
|
return solution
|
|
|
|
time = np.linspace(T_BEGIN, T_END, 101)
|
|
|
|
# Эталонные данные
|
|
reference_solution = analytical_solution(time, K1_REFERENCE, K2_REFERENCE)
|
|
|
|
def residuals(parameters):
|
|
"""
|
|
Вектор невязок между моделью и эталонными данными.
|
|
"""
|
|
k1, k2 = parameters
|
|
|
|
numerical_solution = rk4_solution(time, k1, k2)
|
|
|
|
# Получается вектор из 101 * 3 = 303 невязок
|
|
return (numerical_solution - reference_solution).ravel()
|
|
|
|
|
|
# Начальное приближение
|
|
initial_parameters = np.array([0.3, 0.7])
|
|
|
|
result = least_squares(
|
|
residuals,
|
|
initial_parameters,
|
|
bounds=(
|
|
[K1_BOUNDS[0], K2_BOUNDS[0]],
|
|
[K1_BOUNDS[1], K2_BOUNDS[1]]
|
|
)
|
|
)
|
|
|
|
# result.x - solution
|
|
found_k1, found_k2 = result.x
|
|
|
|
absolute_error_k1 = abs(found_k1 - K1_REFERENCE)
|
|
absolute_error_k2 = abs(found_k2 - K2_REFERENCE)
|
|
|
|
relative_error_k1 = absolute_error_k1 / K1_REFERENCE * 100.0
|
|
relative_error_k2 = absolute_error_k2 / K2_REFERENCE * 100.0
|
|
|
|
print("Results")
|
|
print(f"Found k1:{found_k1:.4f}")
|
|
print(f"Theoretical k1:{K1_REFERENCE:.4f}")
|
|
print(f"Absolute error of k1:{absolute_error_k1:.5e}")
|
|
print(f"Relative error of k1:{relative_error_k1:.5e} %")
|
|
print()
|
|
|
|
print(f"Found k2:{found_k2:.5f}")
|
|
print(f"Theoretical k2:{K2_REFERENCE:.5f}")
|
|
print(f"Absolute error of k2:{absolute_error_k2:.5e}")
|
|
print(f"Relative error of k2:{relative_error_k2:.5e} %")
|
|
print()
|