CODEWITHSUNDEEP
scipyodeint
Esash
Esash

Reputation: 137

python scipy odeint return values

I have been playing around with odeint in scipy and I could not understand what the function returns as return values. For example,

# -*- coding: utf-8 -*-
"""
Created on Sat Feb 04 20:01:16 2017

@author: Esash
"""

from scipy.integrate import odeint
import matplotlib.pyplot as plt
import numpy as np

def MassSpring(state,t):
  # unpack the state vector
  x = state[0]
  xd = state[1]

  # these are our constants
  k = -5.5 # Newtons per metre
  m = 1.5 # Kilograms
  g = 9.8 # metres per second

  # compute acceleration xdd
  xdd = ((k*x)/m) + g

  # return the two state derivatives
  return [xd, xdd]

state0 = [0.0, 0.0]
t = np.arange(0.0, 10.0, 0.1)

state = odeint(MassSpring, state0, t)

plt.plot(t, state)
plt.xlabel('TIME (sec)')
plt.ylabel('STATES')
plt.title('Mass-Spring System')
plt.legend(('$x$ (m)', '$\dot{x}$ (m/sec)'))

In the above code, I have set the two parameters as 0.0 and 0.0 and the xd in the function is just 0.0 which I return as well. But the return value is not just 0.0, it varies. 

In [14]: state
Out[14]: 
array([[ 0.        ,  0.        ],
       [ 0.04885046,  0.97402207],
       [ 0.19361613,  1.91243899],
       ..., 
       [ 0.10076832, -1.39206172],
       [ 0.00941998, -0.42931942],
       [ 0.01542821,  0.54911655]])

Also, if I have one differential equation for which I need to send many parameters, then I cannot send M parameters in the odeint call as a list or tuple and return only the solution of the ODE as a single array. It expects that the number of parameters sent should be equal to the number of parameters returned form the function. Why is this?

I am not able to understand how this function works. Can someone please explain this to me? My apologies if I sound too confusing.

Thanks a lot.

Upvotes: 1

Views: 2988

Answers (1)

Warren Weckesser
Warren Weckesser

Reputation: 114921

I could not understand what the function returns as return values.

The return value of odeint is the computed solution at the requested time values. That is, after this call

state = odeint(MassSpring, state0, t)

state[0] is [x(t[0]), x'(t[0])], state[1] is [x(t[1]), x'(t[1])], etc. If you wanted to plot just the x coordinate, you could call plt.plot(t, state[:, 0]) to plot the first column of state.

I have set the two parameters as 0.0 and 0.0 [...]

What you are calling the "parameters" are usually called the initial conditions. They are the values of x(t) and x'(t) at t=0.

But the return value is not just 0.0, it varies.

That is because (0, 0) is not an equilibrium of the system. Look at the equation

xdd = ((k*x)/m) + g

When x is 0, you get xdd = g, so xdd is initially positive. That is, there is a nonzero force (gravity) acting on the mass, so it accelerates.

The equilibrium state is [-g*m/k, 0].

Also, if I have one differential equation for which I need to send many parameters, then I cannot send M parameters in the odeint call as a list or tuple and return only the solution of the ODE as a single array. It expects that the number of parameters sent should be equal to the number of parameters returned form the function. Why is this?

odeint only solves the system for one set of initial conditions at a time. If you want to generate several solutions (corresponding to different initial conditions), you'll have to call odeint multiple times.

Upvotes: 1

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