这里 scipy.integrate.odeint 被调用了六个不同的标准 ode 问题, rtol = atol1E-061E-13 。我查看了所有较大公差的结果之间的最大差异减去最小公差的结果之间的最大差异,以获得某种“错误”的表示。我很好奇为什么对于给定的容差,一个问题 (D5) 的错误率比另一个问题 (C1) 严重一百万倍,即使步数的范围相当窄(在 10 倍以内)。

脚本中给出了 ode 问题的引用。所有的问题都很好地标准化了,所以我对 rtolatol 的处理方式类似。

重申 - 我的问题是为什么错误在不同问题之间的差异几乎是 1E+06 ,尽管错误会随着容差而变化。当然,C1 是“最软的”,而 D5 在“近日点”处具有戏剧性的峰值,但我认为例程会在内部调整步长,以便误差相似。

编辑: 我添加了“错误”的时间演变,这可能会有所启发。

python - 需要更好地了解 rtol、atol 在 scipy.integrate.odeint 中的工作方式-LMLPHP

python - 需要更好地了解 rtol、atol 在 scipy.integrate.odeint 中的工作方式-LMLPHP

python - 需要更好地了解 rtol、atol 在 scipy.integrate.odeint 中的工作方式-LMLPHP

# FROM: "Comparing Numerical Methods for Ordinary Differential Equations"
# T.E. Hull, W.H. Enright, B.M. Fellen and A.E. Sedgwidh
# SIAM J. Numer. Anal. vol 9, no 4, December 1972, pp: 603-637

def deriv_B1(y, x):
    return [2.*(y[0]-y[0]*y[1]), -(y[1]-y[0]*y[1])] # "growth of two conflicting populations"

def deriv_B4(y, x):
    A = 1./np.sqrt(y[0]**2 + y[1]**2)
    return [-y[1] - A*y[0]*y[2],  y[0] - A*y[1]*y[2],  A*y[0]]  # "integral surface of a torus"

def deriv_C1(y, x):
    return [-y[0]] + [y[i]-y[i+1] for i in range(8)] + [y[8]] # a radioactive decay chain

def deriv_D1toD5(y, x):
    A = -(y[0]**2 + y[1]**2)**-1.5
    return [y[2],  y[3],  A*y[0],  A*y[1]] # dimensionless orbit equation

deriv_D1, deriv_D5 = deriv_D1toD5, deriv_D1toD5

def deriv_E1(y, x):
    return [y[1], -(y[1]/(x+1.0) + (1.0 - 0.25/(x+1.0)**2)*y[0])] # derived from Bessel's equation of order 1/2

def deriv_E3(y, x):
    return [y[1], y[0]**3/6.0 - y[0] + 2.0*np.sin(2.78535*x)] # derived from Duffing's equation

import numpy as np
from scipy.integrate import odeint as ODEint
import matplotlib.pyplot as plt
import timeit

y0_B1 = [1.0, 3.0]
y0_B4 = [3.0, 0.0, 0.0]
y0_C1 = [1.0] + [0.0 for i in range(9)]
ep1, ep5 = 0.1, 0.9
y0_D1 = [1.0-ep1, 0.0, 0.0, np.sqrt((1.0+ep1)/(1.0-ep1))]
y0_D5 = [1.0-ep5, 0.0, 0.0, np.sqrt((1.0+ep5)/(1.0-ep5))]
y0_E1 = [0.6713968071418030, 0.09540051444747446] # J(1/2, 1), Jprime(1/2, 1)
y0_E3 = [0.0, 0.0]

x  = np.linspace(0, 20, 51)
xa = np.linspace(0, 20, 2001)

derivs = [deriv_B1, deriv_B4, deriv_C1, deriv_D1, deriv_D5, deriv_E3]
names  = ["deriv_B1", "deriv_B4", "deriv_C1", "deriv_D1", "deriv_D5", "deriv_E3"]
y0s    = [y0_B1, y0_B4, y0_C1, y0_D1, y0_D5, y0_E3]

timeit_dict, answer_dict, info_dict = dict(), dict(), dict()

ntimes = 10
tols   = [10.**-i for i in range(6, 14)]

def F():           # low density of time points, no output for speed test
    ODEint(deriv, y0, x, rtol=tol, atol=tol)
def Fa():           # hight density of time points, full output for plotting
    return ODEint(deriv, y0, xa, rtol=tol, atol=tol, full_output=True)

for deriv, y0, name in zip(derivs, y0s, names):
    timez = [timeit.timeit(F, number=ntimes)/float(ntimes) for tol in tols]
    timeit_dict[name] = timez
    alist, dlist = zip(*[Fa() for tol in tols])
    answer_dict[name] = np.array([a.T for a in alist])
    info_dict[name] = dlist

plt.figure(figsize=[10,6])

for i, name in enumerate(names):
    plt.subplot(2, 3, i+1)
    for thing in answer_dict[name][-1]:
        plt.plot(xa, thing)
    plt.title(name[-2:], fontsize=16)
plt.show()

plt.figure(figsize=[10, 8])
for i, name in enumerate(names):
    plt.subplot(2,3,i+1)
    a = answer_dict[name]
    a13, a10, a8 = a[-1], a[-4], a[-6]
    d10 = np.abs(a10-a13).max(axis=0)
    d8  = np.abs(a8 -a13).max(axis=0)
    plt.plot(xa, d10, label="tol(1E-10)-tol(1E-13)")
    plt.plot(xa, d8,  label="tol(1E-08)-tol(1E-13)")
    plt.yscale('log')
    plt.ylim(1E-11, 1E-03)
    plt.title(name[-2:], fontsize=16)
    if i==3:
        plt.text(3, 1E-10, "1E-10 - 1E-13", fontsize=14)
        plt.text(2, 2E-05, "1E-08 - 1E-13", fontsize=14)
plt.show()

fs = 16
plt.figure(figsize=[12,6])

plt.subplot(1,3,1)
for name in names:
    plt.plot(tols, timeit_dict[name])
plt.title("timing results", fontsize=16)
plt.xscale('log')
plt.yscale('log')
plt.text(1E-09, 5E-02, "D5", fontsize=fs)
plt.text(1E-09, 4.5E-03, "C1", fontsize=fs)

plt.subplot(1,3,2)
for name in names:
    a = answer_dict[name]
    e = a[:-1] - a[-1]
    em = [np.abs(thing).max() for thing in e]
    plt.plot(tols[:-1], em)
plt.title("max difference from smallest tol", fontsize=16)
plt.xscale('log')
plt.yscale('log')
plt.xlim(min(tols), max(tols))
plt.text(1E-09, 3E-03, "D5", fontsize=fs)
plt.text(1E-09, 8E-11, "C1", fontsize=fs)

plt.subplot(1,3,3)
for name in names:
    nsteps = [d['nst'][-1] for d in info_dict[name]]
    plt.plot(tols, nsteps, label=name[-2:])
plt.title("number of steps", fontsize=16)
plt.xscale('log')
plt.yscale('log')
plt.ylim(3E+01, 3E+03)
plt.legend(loc="upper right", shadow=False, fontsize="large")
plt.text(2E-12, 2.3E+03, "D5", fontsize=fs)
plt.text(2E-12, 1.5E+02, "C1", fontsize=fs)

plt.show()

最佳答案

自从我发布了这个问题,我学到了更多。不能只是将每一步的数值精度乘以步数,而希望得到整体精度。

如果解决方案出现分歧(附近的起点导致路径随着时间的推移变得更远),那么数值误差可能会被放大。每个问题都会有所不同 - 一切都应该如此。

赫尔等人。是学习 ODE 求解器的好起点。 (问题中显示的问题的来源)

关于python - 需要更好地了解 rtol、atol 在 scipy.integrate.odeint 中的工作方式,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33748601/

10-12 21:32