# Dissipation in Runge-Kutta Methods¶

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?