I have written very simple Python code to solve the simple harmonic oscillator using Euler method, but I am not sure if the program is correct or not. I would be very grateful if anyone can look at my code and suggest further improvements since I am very new to this programming thing.

import matplotlib.pyplot as plt
v=0.0   #initial velocity
h=0.01  #time step
x=5.0   #initial position


while t<1.0:

    v=v-(10.0/1.0)*x*h    #k=10.0, m=1.0
  • \$\begingroup\$ In what sense is this the "Euler" method? It just looks like a discrete-timestep simulation to me. \$\endgroup\$ – 200_success May 29 '17 at 2:09
  • \$\begingroup\$ For dy/dx=f(x,y), we can find the value of y at x1=x0+h as y(x1)=y(x0)+hf(x0,y0)$. Isn't this what the Euler's method precisely about ? \$\endgroup\$ – Roshan Shrestha May 29 '17 at 2:22
  • \$\begingroup\$ I don't see what role does g play. Besides, h is really dt. That said, it is far from harmonic oscillator. \$\endgroup\$ – vnp May 29 '17 at 4:37
  • \$\begingroup\$ Yes, it seems I did a mistake by calling variable g which is not needed for this one since I was doing free fall motion earlier. So, haven't I used the Euler method correctly in here. \$\endgroup\$ – Roshan Shrestha May 29 '17 at 5:50

The code is quite monolithic: calculation and presentation are separate concerns, and while I appreciate the context given by showing the presentation, it would be better to make clear that they're separate concerns by separating them with whitespace and commenting briefly.

Also on structure: separate constants (h) from variables (v, x, t).

    v=v-(10.0/1.0)*x*h    #k=10.0, m=1.0

Two things: firstly, what are k and m? It seems that they should be constants with comments explaining their physical significance.

Secondly, this isn't Euler's method. Euler's method is \$\vec{x}_{n+1} = \vec{x}_n + hf(t_n, \vec{x}_n)\$ where \$f = \frac{d\vec{x}}{dt}\$. Here \$\vec{x} = (v, x)\$ and \$f(t, v, x) = (\frac{dv}{dt}, \frac{dx}{dt}) = (-\frac{k}{m}x, v)\$. In other words, to be Euler's method you should update x according to the previous value of v, not the updated value of v. Since you're using Python, you can take advantage of simultaneous assignment:


(As it happens your buggy implementation works better than Euler's method, but if it was intended to implement Euler's method then it's still technically buggy).

| improve this answer | |
  • \$\begingroup\$ Thanks @Taylor. I would definitely take that into consideration, and I should have updated x first. \$\endgroup\$ – Roshan Shrestha May 29 '17 at 10:34
  • \$\begingroup\$ @RoshanShrestha, no, that would be equally wrong. You should update them simultaneously, which depending on the language may mean using temporary variables. \$\endgroup\$ – Peter Taylor May 29 '17 at 10:52
  • \$\begingroup\$ But, I have used temporary variables for x,v and t outside the while loop. So, does it mean that if I order my code for Euler method x=x+vh v=v-(10.0/1.0)*xh will still be wrong ? \$\endgroup\$ – Roshan Shrestha May 29 '17 at 10:56
  • \$\begingroup\$ @RoshanShrestha, I wouldn't call those temporary variables. Also, I've realised that the code is Python (I had initially misread and thought it was Matlab, which I don't know), so that means it supports simultaneous assignment. I've edited sample code into the answer. \$\endgroup\$ – Peter Taylor May 29 '17 at 11:22

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