# ## MIT License

# Permission is hereby granted, free of charge, to any person obtaining a copy
# of this software and associated documentation files (the "Software"), to deal
# in the Software without restriction, including without limitation the rights
# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
# copies of the Software, and to permit persons to whom the Software is
# furnished to do so, subject to the following conditions:
#
# The above copyright notice and this permission notice shall be included in
# all copies or substantial portions of the Software.
#
# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
# AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
# THE SOFTWARE.
#

# 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])
# What do you observe?