使用scipy的solve_ivp解决非线性摆运动 [英] Using scipy's solve_ivp to solve non linear pendulum motion

问题描述

我仍在尝试了解resolve_ivp如何与odeint结合使用，但就在我掌握了它的本质的同时，发生了一些事情.

I am still trying to understand how solve_ivp works against odeint, but just as I was getting the hang of it something happened.

我正在尝试解决非线性摆的运动.有了odeint，一切都会像个魅力一样，在solve_ivp上发生奇怪的事情:

I am trying to solve for the motion of a non linear pendulum. With odeint, everything works like a charm, on solve_ivp hoever something weird happens:

import numpy as np
from matplotlib import pyplot as plt
from scipy.integrate import solve_ivp, odeint

g = 9.81
l = 0.1


def f(t, r):
    omega = r[0]
    theta = r[1]
    return np.array([-g / l * np.sin(theta), omega])


time = np.linspace(0, 10, 1000)
init_r = [0, np.radians(179)]

results = solve_ivp(f, (0, 10), init_r, method="RK45", t_eval=time) #??????
cenas = odeint(f, init_r, time, tfirst=True)


fig = plt.figure()
ax1 = fig.add_subplot(111)

ax1.plot(results.t, results.y[1])
ax1.plot(time, cenas[:, 1])

plt.show()

我想念什么?

推荐答案

这是一个数字问题. solve_ivp的默认相对公差和绝对公差分别为1e-3和1e-6.对于许多问题，这些值太低. odeint的默认相对公差为1.49e-8.

It is a numerical problem. The default relative and absolute tolerances of solve_ivp are 1e-3 and 1e-6, respectively. For many problems, these values are too low. The default relative tolerance for odeint is 1.49e-8.

如果将参数rtol=1e-8添加到solve_ivp调用中，则表示一致:

If you add the argument rtol=1e-8 to the solve_ivp call, the plots agree:

import numpy as np
from matplotlib import pyplot as plt
from scipy.integrate import solve_ivp, odeint

g = 9.81
l = 0.1


def f(t, r):
    omega = r[0]
    theta = r[1]
    return np.array([-g / l * np.sin(theta), omega])


time = np.linspace(0, 10, 1000)
init_r = [0, np.radians(179)]

results = solve_ivp(f, (0, 10), init_r, method='RK45', t_eval=time, rtol=1e-8)
cenas = odeint(f, init_r, time, tfirst=True)


fig = plt.figure()
ax1 = fig.add_subplot(111)

ax1.plot(results.t, results.y[1])
ax1.plot(time, cenas[:, 1])

plt.show()

情节:

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