Question

我正在尝试使用scipy.ode和zvode积分器解决python中耦合复杂ODE的系统。但是，一旦我运行代码，就会显示此错误消息。

ZVODE--  At T(=R1) and step size H(=R2), the error test failed repeatedly or with abs(H) = HMIN. In above,  R1 =  0.1018805870139D-15   R2 =  0.2392554739952D-22

我确实看过FORTRAN源代码，但无法弄清楚它的含义。

对此有任何帮助。

编辑：已包含代码。我还尝试打印出一些值，并且还为使用简单的Euler方法的集成编写了单独的代码。从这些我感觉到错误可能是由于值超出范围，即大于10 ^ 308。（可能是由于某些参数错误）。有人可以确认吗？

import numpy as np
import matplotlib.pyplot as plt 
from scipy.integrate import ode

# Constants
h_cross = 6.5821e-16 
charge = 1 
a = 5.65e-10
gamma = 1/50e-15 
E_g = 1.43  
temp = 300
k_b = 8.6173e-5  



k = np.linspace(-1, 1, 32) * np.pi / a 
k = k[1:]
grad_k = k[1] - k[0]  

delta_c_1 = 6.9
delta_v = 1

e_c_1 = delta_c_1/2.0 * (1 - np.cos(np.abs(k * a))) + E_g/2.0
e_v = -delta_v/2.0 * (1 - np.cos(np.abs(k * a))) - E_g/2.0
t_0 = 1e-14 # Initial time
dt = 5e-15 # Time interval 
t_f = t_0 + 50e-14 # Final time
t_mid = (t_0 + t_f) / 2.0
steps = int((t_f - t_0) / dt) 
t = np.linspace(t_0,t_f,steps)

d = np.ones(k.size) * 3.336e-30 


w_0 = 0.1 / h_cross
f_0 = w_0 / (2*np.pi)

y_0 = np.zeros([k.size,3],dtype = complex)
y_input = y_0.flatten()

solution = y_input # Inserting initial condition as the first entry in the solution

def E(times):
    pulse = np.cos(w_0 * times ) 
    fwhm = 10e-14
    sigma = fwhm / 2.35 
    envelope = ( 1 / (2 * np.pi *sigma**2)**0.5 ) * np.exp( -((times-t_mid)/sigma)**2 / 2.0)
    waveform = pulse * envelope 
    return waveform

# NORMALIZE VALUE OF E(t) USING VALUE AT PEAK VALUE OF E(t)
E_peak_req = 1e8
E_peak = E(t).max()
normalisation = np.abs(E_peak_req / E_peak) * (1/1.6e-19)


def dynamics(t,y):

    dydt = np.zeros([k.size,3],dtype = complex)


    if(solution.size == (k.size*3)):
        prev_y = solution
        prev_y = np.reshape(prev_y,(k.size,3))
        prev_prev_y = prev_y
        prev_prev_y = np.reshape(prev_prev_y,(k.size,3))
    else:
        last_step = solution.shape[0] - 1
        prev_y = solution[last_step,:]
        prev_y = np.reshape(prev_y,(k.size,3)) # Extracting the latest values of the density matrix elements obtained in the last time step 
        prev_prev_y = solution[last_step - 1, :] 
        prev_prev_y = np.reshape(prev_prev_y,(k.size,3))

    for index in range(k.size):

        grad_p = prev_y[index][0] - prev_prev_y[index][0]
        grad_f_c = prev_y[index][1] - prev_prev_y[index][1] 
        grad_f_v = prev_y[index][2] - prev_prev_y[index][2]

        dipole_contr =  d[index] * (E(t) * normalisation)
        grad_contr_1 = 1j * charge * (E(t) * normalisation) * grad_p / grad_k 
        grad_contr_2 = charge * (E(t) * normalisation) * grad_f_c / grad_k 
        grad_contr_3 = charge * (E(t) * normalisation) * grad_f_v / grad_k 

        dpdt = (-1j  / h_cross) * ( (e_c_1[index] + e_v[index] - 1j*h_cross*gamma) * prev_y[index][0] - (1 - prev_y[index][1] - prev_y[index][2]) * dipole_contr + grad_contr_1 )
        dfcdt = (1 / h_cross) * ( -2 * np.imag( dipole_contr * np.conjugate(prev_y[index][0]) ) + grad_contr_2 )
        dfvdt = (1 / h_cross) * ( -2 * np.imag( dipole_contr * np.conjugate(prev_y[index][0]) ) + grad_contr_3 )

        dydt[index] = np.array([dpdt, dfcdt, dfvdt])
    dydt = dydt.flatten()
    return dydt



solver = ode(dynamics, jac = None).set_integrator('zvode', method ='bdf')
solver.set_initial_value(y_input, t_0) #.set_f_params()
while (solver.successful() and solver.t + dt <= t_f):
            solver.integrate(solver.t + dt)
            solution = np.vstack((solution,solver.y))

sol = np.reshape(solution,(solution.shape[0],k.size,3))

Answer 1

这意味着您的系统很僵硬，步长控制器的启发式计算得出它需要非常小的步长才能保证所需的误差范围，但是步长变得很小，因此所需的步数也很大浮点噪声的积累变得更加占主导地位，这意味着控制器将失去对误差积累的控制。似乎要避免这种情况，控制器将步长限制为2e-7的一部分，约为sqrt(mu)，比t的值要小一些。

Scipy.ode“错误测试反复失败”

1 个答案: