# Python - Using A Kronecker Delta With ODEINT

## 10 October 2019 - 1 answer

I'm trying to plot the output from an ODE using a Kronecker delta function which should only become 'active' at a specific time = t1. This should give a sawtooth like response where the initial value decays down exponentially until t=t1 where it rises again instantly before decaying down once again. However, when I plot this it looks like the solver is seeing the Kronecker delta function as zero for all time t. Is there anyway to do this in Python?

from scipy import KroneckerDelta
import scipy.integrate as sp
import matplotlib.pyplot as plt
import numpy as np

def dy_dt(y,t):

dy_dt = 500*KroneckerDelta(t,t1) - 2y

return dy_dt

t1 = 4
y0 = 500
t = np.arrange(0,10,0.1)

y = sp.odeint(dy_dt,y0,t)

plt.plot(t,y)

I think the problem could be internal rounding errors, because 0.1 cannot be represented exactly as a python float. I would try

import math

def dy_dt(y,t):
if math.isclose(t, t1):
return 500 - 2*y
else:
return -2y

Also the documentation of odeint suggests using the args parameter instead of global variables to give your derivative function access to additional arguments and replacing np.arange by np.linspace:

import scipy.integrate as sp
import matplotlib.pyplot as plt
import numpy as np
import math

def dy_dt(y, t, t1):
if math.isclose(t, t1):
return 500 - 2*y
else:
return -2*y

t1 = 4
y0 = 500
t = np.linspace(0, 10, num=101)
y = sp.odeint(dy_dt, y0, t, args=(t1,))
plt.plot(t, y)

I did not test the code so tell me if there is anything wrong with it.

EDIT:

When testing my code I took a look at the t values for which dy_dt is evaluated. I noticed that odeint does not only use the t values that where specified, but alters them slightly:

...
3.6636447422787928
3.743098503914526
3.822552265550259
3.902006027185992
3.991829287543431
4.08165254790087
4.171475808258308
...

Now using my method, we get

math.isclose(3.991829287543431, 4) # False

because the default tolerance is set to a relative error of at most 10^(-9), so the odeint function "misses" the bump of the derivative at 4. Luckily, we can fix that by specifying a higher error threshold:

def dy_dt(y, t, t1):
if math.isclose(t, t1, abs_tol=0.01):
return 500 - 2*y
else:
return -2*y

Now dy_dt is very high for all values between 3.99 and 4.01. It is possible to make this range smaller if the num argument of linspace is increased.

TL;DR

Your problem is not a problem of python but a problem of numerically solving an differential equation: You need to alter your derivative for an interval of sufficient length, otherwise the solver will likely miss the interesting spot. A kronecker delta does not work with numeric approaches to solving ODEs.