# Numerical methods challenge: Day 20

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.

## Shooting method

Today we have the shooting method. This is a method for solving boundary value problems by turning them into a sequence of initial value problems.

Loosely speaking, for a second order equation we have

and we solve the sequence of problems

until the function \(F(a) = y(x_1; a) - y_1\) has a root.

We will test the method with the boundary value problem

Following are the codes.

### Python

from __future__ import division, print_function import numpy as np import matplotlib.pyplot as plt from scipy.optimize import newton from scipy.integrate import odeint def shooting(dydx, x, x0, xf, shoot=None): if shoot is None: shoot = np.random.uniform(-20, 20) F = lambda s, x0, xf, x: odeint(dydx, [x0, s], x)[-1, 0] - xf shoot = newton(F, shoot, args=(x0, xf, x)) y = odeint(dydx, [x0, shoot], x) return y[:, 0], shoot func = lambda y, t: [y[1], 1.5*y[0]**2] x = np.linspace(0, 1, 1000) y, shoot = shooting(func, x, 4, 1, shoot=-5) plt.plot(x, y) plt.xlabel(r"$x$") plt.ylabel(r"$y$") plt.show()

### Julia

using PyPlot, DifferentialEquations, Roots function shooting(dydx, x, y0, yf; shoot=nothing) if shoot == nothing shoot = rand(-20:0.1:20) end function F(s) prob = ODEProblem(dydx, [y0, s], (x[1], x[end])) sol = solve(prob) return sol(x[end])[1] - yf end shoot = fzero(F, shoot) prob = ODEProblem(dydx, [y0, shoot], (x[1], x[end])) sol = solve(prob, solveat=x) return sol(x)[1, :], shoot end func(x, y) = [y[2], 1.5*y[1]^2] x = linspace(0, 1, 1000) y, shoot = shooting(func, x , 4.0, 1.0, shoot=-5.0) plot(x, y)

In both cases the result is the following plot, and the slope is -7.9999999657800833.

We should mention that the convergence of the method relies on the selection of initial guesses. For example, if we choose as initial parameter -50 in the previous problem, we obtain a completely differente solution.

And the obtained slope is -35.858547970130971.

### Comparison Python/Julia

Regarding number of lines we have: 20 in Python and 23 in Julia. The comparison
in execution time is done with `%timeit`

magic command in IPython and
`@benchmark`

in Julia.

For Python:

with result

For Julia:

with result

BenchmarkTools.Trial: memory estimate: 4.18 MiB allocs estimate: 78552 -------------- minimum time: 10.065 ms (0.00% GC) median time: 10.593 ms (0.00% GC) mean time: 11.769 ms (9.28% GC) maximum time: 22.252 ms (48.58% GC) -------------- samples: 425 evals/sample: 1

In this case, we can say that the Python code is roughly 5 times faster than
Julia. Although, the codes are more different than in the other challenges.
For example, I am not using `newton`

in Julia.

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