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# Numerical methods challenge: Day 19

During October (2017) I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.

## Verlet integration

Today we have the Verlet integration technique. This method is widely used to integrate equations of motion of several systems, such as orbital mechanics or molecular dynamics. One of the main reasons for choosing this method is that it is a symplectic integrator.

The stepping is done with the formula

\begin{equation*} \mathbf{x}_{n+1}=2 \mathbf{x}_n- \mathbf{x}_{n-1}+ A(\mathbf{x}_n)\,\Delta t^2. \end{equation*}

where $A(\mathbf{x}_n)$ is the aceleration for that time-step.

Following are the codes.

### Python

from __future__ import division, print_function
import numpy as np
import matplotlib.pyplot as plt

def verlet(force, x0, v0, m, t, args=()):
ndof = len(x0)
ntimes = len(t)
x = np.zeros((ndof, ntimes))
dt = t - t
x[:, 0] = x0
F = force(x0, v0, m, t, *args)
x[::2, 1] = x0[::2] + v0[::2]*dt + 0.5*F[::2]*dt**2/m
x[1::2, 1] = x0[1::2] + v0[1::2]*dt + 0.5*F[1::2]*dt**2/m
for cont in range(2, ntimes):
dt = t[cont] - t[cont - 1]
v = (x[:, cont - 1] - x[:, cont - 2])/dt
acel = force(x[:, cont - 1], v, m, t, *args)
acel[::2] = acel[::2]/m
acel[1::2] = acel[1::2]/m
x[:, cont] = 2*x[:, cont - 1] - x[:, cont - 2] + acel*dt**2
return x

spring_force = lambda x, v, m, t, k: -k*x
x0 = np.array([1.0, 0.0])
v0 = np.array([0.0, 1.0])
m = np.array([1.0])
t = np.linspace(0, 10.0, 1000)
k = 1.0
x = verlet(spring_force, x0, v0, m, t, args=(k,))

#%% Plot
plt.figure(figsize=(6, 3))
plt.subplot(121)
plt.plot(t, x[0, :])
plt.plot(t, x[1, :])
plt.subplot(122)
plt.plot(x[0, :], x[1, :])
plt.show()


### Julia

using PyPlot

function verlet(force, x0, v0, m, t; args=())
ndof = length(x0)
ntimes = length(t)
x = zeros(ndof, ntimes)
dt = t - t
x[:, 1] = x0
F = force(x0, v0, m, t, args...)
x[1:2:end, 2] = x0[1:2:end] + v0[1:2:end]*dt + 0.5*F[1:2:end]*dt^2./m
x[2:2:end, 2] = x0[2:2:end] + v0[2:2:end]*dt + 0.5*F[2:2:end]*dt^2./m
for cont = 3:ntimes
dt = t[cont] - t[cont - 1]
v = (x[:, cont - 1] - x[:, cont - 2])/dt
acel = force(x[:, cont - 1], v, m, t, args...)
acel[1:2:end] = acel[1:2:end]./m
acel[2:2:end] = acel[2:2:end]./m
x[:, cont] = 2*x[:, cont - 1] - x[:, cont - 2] + acel*dt^2
end
return x
end

spring_force(x, v, m, t, k) = -k*x
x0 = [1.0, 0.0]
v0 = [0.0, 1.0]
m = [1.0]
t = linspace(0, 10.0, 1000)
k = 1.0
x = verlet(spring_force, x0, v0, m, t, args=(k,))

#%% Plot
figure(figsize=(6, 3))
subplot(121)
plot(t, x[1, :])
plot(t, x[2, :])
subplot(122)
plot(x[1, :], x[2, :])
show()


In both cases the result is the following plot ### Comparison Python/Julia

Regarding number of lines we have: 40 in Python and 40 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia.

For Python:

%timeit -n 100 verlet(spring_force, x0, v0, m, t, args=(k,))


with result

100 loops, best of 3: 26.5 ms per loop


For Julia:

@benchmark verlet(spring_force, x0, v0, m, t, args=(k,))


with result

BenchmarkTools.Trial:
memory estimate:  4.36 MiB
allocs estimate:  101839
--------------
minimum time:     73.159 ms (0.00% GC)
median time:      74.883 ms (0.00% GC)
mean time:        75.464 ms (1.02% GC)
maximum time:     80.017 ms (4.87% GC)
--------------
samples:          67
evals/sample:     1


In this case, we can say that the Python code is roughly 3 times faster than Julia.