Numerical integration: Why does my orbit simulation yield the wrong result?

问题

I read Feynman's Lecture on Physics Chapter 9 and tried to my own simulation. I used Riemann integrals to calculate velocity and position. Although all start-entry is same, my orbit look's like a hyperbola. Here is lecture note: https://www.feynmanlectures.caltech.edu/I_09.html (Table 9.2)

import time
import matplotlib.pyplot as plt
x=list()
y=list()
x_in=0.5
y_in=0.0
x.append(x_in)
y.append(y_in)

class Planet:
    def __init__(self,m,rx,ry,vx,vy,G=1):
        self.m=m
        self.rx=rx
        self.ry=ry
        self.a=0
        self.ax=0
        self.ay=0
        self.vx=vx
        self.vy=vy
        self.r=(rx**2+ry**2)**(1/2)
        self.f=0
        self.G=1
        print(self.r)
    def calculateOrbit(self,dt,planet):
        self.rx=self.rx+self.vx*dt
        self.vx=self.vx+self.ax*dt
        self.ax=0
        for i in planet:
            r=((((self.rx-i.rx)**2)+((self.ry-i.ry)**2))**(1/2))
            self.ax+=-self.G*i.m*self.rx/(r**3)

        self.ry=self.ry+self.vy*dt
        self.vy=self.vy+self.ay*dt
        self.ay=0
        for i in planet:
                self.ay+=-self.G*i.m*self.ry/(r**3)
        global x,y
        x.append(self.rx)
        y.append(self.ry)

        #self.showOrbit()
    def showOrbit(self):
        print(""" X: {} Y: {} Ax: {} Ay: {}, Vx: {}, Vy: {}""".format(self.rx,self.ry,self.ax,self.ay,self.vx,self.vy))

planets=list();
earth = Planet(1,x_in,y_in,0,1.630)
sun =   Planet(1,0,0,0,0)
planets.append(sun)
for i in range(0,1000):
    earth.calculateOrbit(0.1,planets)

plt.plot(x,y)
plt.grid()
plt.xlim(-20.0,20.0)
plt.ylim(-20.0,20.0)
plt.show()

回答1:

dt is supposed to be infinitly small for the integration to work.

The bigger dt the bigger the "local linearization error" (there is probably a nicer mathematical term for that...).

0.1 for dt may look small enough, but for your result to converge correctly (towards reality) you need to check smaller time steps, too. If smaller time steps converge towards the same solution your equation is linar enough to use a bigger time step (and save comptation time.)

Try your code with

for i in range(0, 10000):
    earth.calculateOrbit(0.01, planets)

and

for i in range(0, 100000):
    earth.calculateOrbit(0.001, planets)

In both calculations the overall time that has passed since the beginning is the same as with your original values. But the result is quite different. So you might have to use an even smaller dt.

More info:

https://en.wikipedia.org/wiki/Trapezoidal_rule

And this page states what you are doing:

A 'brute force' kind of numerical integration can be done, if the integrand is reasonably well-behaved (i.e. piecewise continuous and of bounded variation), by evaluating the integrand with very small increments.

And what I tried to explain above:

An important part of the analysis of any numerical integration method is to study the behavior of the approximation error as a function of the number of integrand evaluations.

There are many smart approaches to make better guesses and use larger time steps. You might have heared of the Runge–Kutta method to solve differential equations. It seems to become Simpson's rule mentioned in the link above for non-differential equations, see here.

回答2:

The problem is the method of numerical integration or numerical solution for differential equations. The method you're using(Euler's Method for numerical solutions to differential equations), although it gives very close to the actual value but still gives a very small error. When this slightly errored value is used over multiple iterations(like you have done 1000 steps), this error gets larger and larger at every step which gave you the wrong result.

There can be two solution two this problem:

1. Decrease the time interval to an even smaller value so that even after amplification of errors throughout the process it doesn't get largely deviated from the actual solution. Now the thing to note is that if you decrease the time interval(dt) and not increase the number of steps then you can see the evolution of the system for a shorter period of time. Therefore you'll have to increase the number of steps too along with a decrease in time interval(dt). I checked your code and it seems if you put dt = 0.0001 and put number of steps as 100000 instead of just 1000, you'll get your beautiful elliptical orbit you're looking for. Also, delete or comment out plt.xlim and plt.ylim lines to see your plot clearly.

2. Implement Runge-Kutta method for the numerical solution of differential equations. This method has better convergence per iteration to the actual solution. Since, it will take much more time and changes to your code, that's why I'm suggesting it as second option otherwise this method is superior and more general than Euler's method.

Note: Even without any changes, solution it is not behaving like a hyperbola. For the initial values that you've provided for the system, solution is giving a bounded curve but just because of the error amplification it is spiraling into a point. You'll notice this spiraling in if you just increase the steps to 10000 and put dt = 0.01.

来源：https://stackoverflow.com/questions/62367842/numerical-integration-why-does-my-orbit-simulation-yield-the-wrong-result