编辑：
所以我发现NDSolve for ODE正在使用Runge Kutta来求解方程.
如何在我的
python代码上使用Runge Kutta方法来解决我在下面的ODE？
从我在text files with float entries的帖子中,我能够确定python和mathematica立即开始发散,公差为10到负6.
结束编辑
在过去的几个小时里,我一直试图弄清楚为什么我在Mathematica和Python中的解决方案相差5000公里.
我被认为在模拟飞行时间超过数百万秒时,一个程序具有更高的容错性.
我的问题是哪个程序更准确,如果不是python,我该如何调整精度？
使用Mathematica,距离L4不到10公里,与Python一样,我距离5947公里.
代码如下：
Python
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from numpy import linspace
from scipy.optimize import brentq
me = 5.974 * 10 ** (24) # mass of the earth
mm = 7.348 * 10 ** (22) # mass of the moon
G = 6.67259 * 10 ** (-20) # gravitational parameter
re = 6378.0 # radius of the earth in km
rm = 1737.0 # radius of the moon in km
r12 = 384400.0 # distance between the CoM of the earth and moon
d = 300 # distance the spacecraft is above the Earth
pi1 = me / (me + mm)
pi2 = mm / (me + mm)
mue = 398600.0 # gravitational parameter of earth km^3/sec^2
mum = G * mm # grav param of the moon
mu = mue + mum
omega = np.sqrt(mu / (r12 ** 3))
nu = -np.pi / 4 # true anomaly pick yourself
xl4 = r12 / 2 - 4671 # x location of L4
yl4 = np.sqrt(3) / 2 * r12 # y
print("The location of L4 is", xl4, yl4)
# Solve for Jacobi's constant
def f(C):
return (omega ** 2 * (xl4 ** 2 + yl4 ** 2) + 2 * mue / r12 + 2 * mum / r12
+ 2 * C)
c = brentq(f, -5, 0)
print("Jacobi's constant is",c)
x0 = (re + 200) * np.cos(nu) - pi2 * r12 # x location of the satellite
y0 = (re + 200) * np.sin(nu) # y location
print("The satellite's initial position is", x0, y0)
vbo = (np.sqrt(omega ** 2 * (x0 ** 2 + y0 ** 2) + 2 * mue /
np.sqrt((x0 + pi2 * r12) ** 2 + y0 ** 2) + 2 * mum /
np.sqrt((x0 - pi1 * r12) ** 2 + y0 ** 2) + 2 * -1.21))
print("Burnout velocity is", vbo)
gamma = 0.4678 * np.pi / 180 # flight path angle pick yourself
vx = vbo * (np.sin(gamma) * np.cos(nu) - np.cos(gamma) * np.sin(nu))
# velocity of the bo in the x direction
vy = vbo * (np.sin(gamma) * np.sin(nu) + np.cos(gamma) * np.cos(nu))
# velocity of the bo in the y direction
print("The satellite's initial velocity is", vx, vy)
# r0 = [x, y, 0]
# v0 = [vx, vy, 0]
u0 = [x0, y0, 0, vx, vy, 0]
def deriv(u, dt):
return [u[3], # dotu[0] = u[3]
u[4], # dotu[1] = u[4]
u[5], # dotu[2] = u[5]
(2 * omega * u[4] + omega ** 2 * u[0] - mue * (u[0] + pi2 * r12) /
np.sqrt(((u[0] + pi2 * r12) ** 2 + u[1] ** 2) ** 3) - mum *
(u[0] - pi1 * r12) /
np.sqrt(((u[0] - pi1 * r12) ** 2 + u[1] ** 2) ** 3)),
# dotu[3] = that
(-2 * omega * u[3] + omega ** 2 * u[1] - mue * u[1] /
np.sqrt(((u[0] + pi2 * r12) ** 2 + u[1] ** 2) ** 3) - mum * u[1] /
np.sqrt(((u[0] - pi1 * r12) ** 2 + u[1] ** 2) ** 3)),
# dotu[4] = that
0] # dotu[5] = 0
dt = np.linspace(0.0, 6.0 * 86400.0, 2000000.0) # secs to run the simulation
u = odeint(deriv, u0, dt)
x, y, z, x2, y2, z2 = u.T
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z, color = 'r')
# adding the Lagrange point
phi = np.linspace(0, 2 * np.pi, 100)
theta = np.linspace(0, np.pi, 100)
xm = 2000 * np.outer(np.cos(phi), np.sin(theta)) + xl4
ym = 2000 * np.outer(np.sin(phi), np.sin(theta)) + yl4
zm = 2000 * np.outer(np.ones(np.size(phi)), np.cos(theta))
ax.plot_surface(xm, ym, zm, color = '#696969', linewidth = 0)
ax.auto_scale_xyz([-8000, 385000], [-8000, 385000], [-8000, 385000])
# adding the earth
phi = np.linspace(0, 2 * np.pi, 100)
theta = np.linspace(0, np.pi, 100)
xm = 2000 * np.outer(np.cos(phi), np.sin(theta))
ym = 2000 * np.outer(np.sin(phi), np.sin(theta))
zm = 2000 * np.outer(np.ones(np.size(phi)), np.cos(theta))
ax.plot_surface(xm, ym, zm, color = '#696969', linewidth = 0)
ax.auto_scale_xyz([-8000, 385000], [-8000, 385000], [-8000, 385000])
plt.show()
# The code below finds the distance between path and l4
my_x, my_y, my_z = (xl4, yl4, 0.0)
delta_x = x - my_x
delta_y = y - my_y
delta_z = z - my_z
distance = np.array([np.sqrt(delta_x ** 2 + delta_y ** 2 + delta_z ** 2)])
minimum = np.amin(distance)
print("Closet approach to L4 is", minimum)
数学
ClearAll["Global`*"];
me = 5.974*10^(24);
mm = 7.348*10^(22);
G = 6.67259*10^(-20);
re = 6378;
rm = 1737;
r12 = 384400;
\[Pi]1 = me/(me + mm);
\[Pi]2 = mm/(me + mm);
M = me + mm;
\[Mu]1 = 398600;
\[Mu]2 = G*mm;
\[Mu] = \[Mu]1 + \[Mu]2;
\[CapitalOmega] = Sqrt[\[Mu]/r12^3];
\[Nu] = -\[Pi]/4;
xl4 = 384400/2 - 4671;
yl4 = Sqrt[3]/2*384400 // N;
Solve[\[CapitalOmega]^2*(xl4^2 + yl4^2) + 2 \[Mu]1/r12 +
2 \[Mu]2/r12 + 2*C == 0, C]
x = (re + 200)*Cos[\[Nu]] - \[Pi]2*r12 // N
y = (re + 200)*Sin[\[Nu]] // N
{{C -> -1.56824}}
-19.3098
-4651.35
vbo = Sqrt[\[CapitalOmega]^2*((x)^2 + (y)^2) +
2*\[Mu]1/Sqrt[(x + \[Pi]2*r12)^2 + (y)^2] +
2*\[Mu]2/Sqrt[(x - \[Pi]1*r12)^2 + (y)^2] + 2*(-1.21)]
10.8994
\[Gamma] = 0.4678*Pi/180;
vx = vbo*(Sin[\[Gamma]]*Cos[\[Nu]] - Cos[\[Gamma]]*Sin[\[Nu]]);
vy = vbo*(Sin[\[Gamma]]*Sin[\[Nu]] + Cos[\[Gamma]]*Cos[\[Nu]]);
r0 = {x, y, 0};
v0 = {vx, vy, 0}
{7.76974, 7.64389, 0}
s = NDSolve[{x1''[t] -
2*\[CapitalOmega]*x2'[t] - \[CapitalOmega]^2*
x1[t] == -\[Mu]1/((Sqrt[(x1[t] + \[Pi]2*r12)^2 +
x2[t]^2])^3)*(x1[t] + \[Pi]2*
r12) - \[Mu]2/((Sqrt[(x1[t] - \[Pi]1*r12)^2 +
x2[t]^2])^3)*(x1[t] - \[Pi]1*r12),
x2''[t] +
2*\[CapitalOmega]*x1'[t] - \[CapitalOmega]^2*
x2[t] == -\[Mu]1/(Sqrt[(x1[t] + \[Pi]2*r12)^2 + x2[t]^2])^3*
x2[t] - \[Mu]2/(Sqrt[(x1[t] - \[Pi]1*r12)^2 + x2[t]^2])^3*
x2[t], x3''[t] == 0, x1[0] == r0[[1]], x1'[0] == v0[[1]],
x2[0] == r0[[2]], x2'[0] == v0[[2]], x3[0] == r0[[3]],
x3'[0] == v0[[3]]}, {x1, x2, x3}, {t, 0, 1000000}];
ParametricPlot3D[
Evaluate[{x1[t], x2[t], x3[t]} /. s], {t, 0, 10*24*3600},
PlotStyle -> {Red, Thick}]
g1 = ParametricPlot3D[
Evaluate[{x1[t], x2[t], x3[t]} /. s], {t, 0, 5.75*3600*24},
PlotStyle -> {Red},
PlotRange -> {{-10000, 400000}, {-10000, 400000}}];
g2 = Graphics3D[{Blue, Opacity[0.6], Sphere[{-4671, 0, 0}, re]}];
g3 = Graphics3D[{Green, Opacity[0.6], Sphere[{379729, 0, 0}, rm]}];
g4 = Graphics3D[{Black, Sphere[{xl4, yl4, 0}, 2000]}];
Show[g2, g1, g3, g4, Boxed -> False]
(*XYdata=Flatten[Table[Evaluate[{x1[t],x2[t],x3[t]}/.s],{t,5.5*24*\
3600,5.78*24*3600,1}],1];
X1Y1data=Flatten[Table[Evaluate[{x1'[t],x2'[t],x3'[t]}/.s],{t,5.5*24*\
3600,5.78*24*3600,1}],1];
SetDirectory[NotebookDirectory[]];
Export["OrbitData.txt",XYdata,"CSV"];
Export["OrbVeloc.txt",X1Y1data,"CSV"];*)
最佳答案 如果此时您的问题已经减少到只想使用Runge-Kutta,您可以替换此行：
u = odeint(deriv, u0, dt)
用这样的东西：
#reverse the order of arguments
def deriv2(t,u):
return deriv(u,t)
# initialize a 4th order Runge-Kutta ODE solver
solver = ode(deriv2).set_integrator('dopri5')
solver.set_initial_value(u0)
u = np.empty((len(dt), 6))
u[0,:] = u0
for ii in range(1,len(dt)):
u[ii] = solver.integrate(dt[ii])
(显然用ode替换odeint导入).
请注意,对于这种类型的ODE,这个速度要慢得多.
要使用dop853,请使用solver.set_integrator(‘dop853’).
