返回值odeint
是解决z(t)
你已经定义为z = [x,y,x',y']
。因此,二阶导数不是返回的解的一部分odeint
。您可以近似计算二阶导数x
and y
通过对一阶导数的返回值进行有限差分。
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,由于您已经有一个函数来计算解的二阶导数,因此您可以调用该函数:
plt.plot(time, deriv(sol.T,time)[2], label="x''")
plt.plot(time, deriv(sol.T,time)[3], label="y''")