如何在Python中绘制耦合的二阶非线性常微分方程的二阶导数？

Question

我是Python的新手，我写了这段代码来模拟弹簧摆的运动：

import numpy as np
from scipy.integrate import odeint
from numpy import sin, cos, pi, array
import matplotlib.pyplot as plt

init = array([0,pi/18,0,0]) 

def deriv(z, t):
    x, y, dxdt, dydt = z
    dx2dt2=(4+x)*(dydt)**2-5*x+9.81*cos(y)
    dy2dt2=(-9.81*sin(y)-2*(dxdt)*(dydt))/(0.4+x)

    return np.array([dxdt, dydt, dx2dt2, dy2dt2])

time = np.linspace(0.0,10.0,1000)
sol = odeint(deriv,init,time)

plt.xlabel("time")
plt.ylabel("y")
plt.plot(time, sol)
plt.show()

但是它给了我x，dxdt，y和dydt的图形，而不是dx2dt2和dy2dt2 (分别是x和y的二阶导数)。我如何改变我的代码来绘制二阶导数呢？

Answer 1

odeint的返回值是您定义为z = [x,y,x',y']的z(t)的解决方案。因此，二阶导数不是odeint返回的解的一部分。您可以通过对一阶导数的返回值进行有限差来近似x和y的二阶导数。

例如：

import numpy as np
from scipy.integrate import odeint
from numpy import sin, cos, pi, array
import matplotlib.pyplot as plt

init = array([0,pi/18,0,0]) 

def deriv(z, t):
    x, y, dxdt, dydt = z
    dx2dt2=(4+x)*(dydt)**2-5*x+9.81*cos(y)
    dy2dt2=(-9.81*sin(y)-2*(dxdt)*(dydt))/(0.4+x)

    return np.array([dxdt, dydt, dx2dt2, dy2dt2])

time = np.linspace(0.0,10.0,1000)
sol = odeint(deriv,init,time)

x, y, xp, yp = sol.T

# compute the approximate second order derivative by computing the finite
# difference between values of the first derivatives
xpp = np.diff(xp)/np.diff(time)
ypp = np.diff(yp)/np.diff(time)

# the second order derivatives are now calculated at the midpoints of the
# initial time array, so we need to compute the midpoints to plot it
xpp_time = (time[1:] + time[:-1])/2

plt.xlabel("time")
plt.ylabel("y")
plt.plot(time, x, label='x')
plt.plot(time, y, label='y')
plt.plot(time, xp, label="x'")
plt.plot(time, yp, label="y'")
plt.plot(xpp_time, xpp, label="x''")
plt.plot(xpp_time, ypp, label="y''")
plt.legend()
plt.show()

或者，由于您已经有了一个函数来计算解的二阶导数，所以您可以直接调用该函数：

plt.plot(time, deriv(sol.T,time)[2], label="x''")
plt.plot(time, deriv(sol.T,time)[3], label="y''")