Dissipation in Runge-Kutta Methods

Copyright (C) 2020 Andreas Kloeckner

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In [1]:
import numpy as np

import matplotlib.pyplot as pt
In [2]:
def rk4_step(y, t, h, f):

    k1 = f(t, y)

    k2 = f(t+h/2, y + h/2*k1)

    k3 = f(t+h/2, y + h/2*k2)

    k4 = f(t+h, y + h*k3)

    return y + h/6*(k1 + 2*k2 + 2*k3 + k4)

Consider the second-order harmonic oscillator

$$u''=-u$$

with initial conditions $u(0) = 1$ and $u'(0)=0$.

$\cos(x)$ is a solution to this problem.

f below gives the right-hand side for this ODE converted to first-order form:

In [3]:
def f(t, y):

    u, up = y

    return np.array([up, -u])

Now, we use 4th-order Runge Kutta to integrate this system over "many" time steps:

In [4]:
times = [0]

y_values = [np.array([1,0])]



h = 0.5

t_end = 800



while times[-1] < t_end:

    y_values.append(rk4_step(y_values[-1], times[-1], h, f))

    times.append(times[-1]+h)

Lastly, plot the computed solution:

In [5]:
y_values = np.array(y_values)



pt.figure(figsize=(15,5))

pt.plot(times, y_values[:, 0])
Out[5]:
[<matplotlib.lines.Line2D at 0x7f18e0f80470>]

What do you observe?