# Numerical methods challenge: Day 23

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.

## Ritz method

Today we have the Ritz method to solve the equation:

with

The method consist in forming a functional that is *equivalent* to the
differential equation, propose an approximation as a linear combination of
a set of basis functions and find the *best* set of coefficients for that
combination. That *best* solution is found minimizing the functional.

The functional for this differential equation is

In this case, we are using the approximation

where we picked the factor \(x (1 - x)\) to enforce that the basis functions satisfy the boundary conditions. The approximated functional reads

where, in general, we will need to perform numerical integration for the second term.

Minimizing the functional

we obtain the system of equations

with

abd

We will test the implementation with the function \(f(x) = x^3\), that leads to the solution

Following are the codes.

### Python

from __future__ import division, print_function import numpy as np from scipy.integrate import quad from scipy.linalg import solve import matplotlib.pyplot as plt def ritz(N, source): stiff_mat = np.zeros((N, N)) rhs = np.zeros((N)) for row in range(N): for col in range(N): numer = (2 + 2*row + 2*col + 2*row*col) denom = (row + col + 1) * (row + col + 2) * (row + col + 3) stiff_mat[row, col] = numer/denom fun = lambda x: x**(row + 1)*(1 - x)*source(x) rhs[row], _ = quad(fun, 0, 1) return stiff_mat, rhs N = 2 source = lambda x: x**3 mat, rhs = ritz(N, source) c = solve(mat, -rhs) x = np.linspace(0, 1, 100) y = np.zeros_like(x) for cont in range(N): y += c[cont]*x**(cont + 1)*(1 - x) #%% Plotting plt.figure(figsize=(4, 3)) plt.plot(x, y) plt.plot(x, x*(x**4 - 1)/20, linestyle="dashed") plt.xlabel(r"$x$") plt.ylabel(r"$y$") plt.legend(["Ritz solution", "Exact solution"]) plt.tight_layout() plt.show()

### Julia

using PyPlot function ritz(N, source) stiff_mat = zeros(N, N) rhs = zeros(N) for row in 0:N-1 for col in 0:N-1 numer = (2 + 2*row + 2*col + 2*row*col) denom = (row + col + 1) * (row + col + 2) * (row + col + 3) stiff_mat[row + 1, col + 1] = numer/denom end fun(x) = x^(row + 1)*(1 - x)*source(x) rhs[row + 1], _ = quadgk(fun, 0, 1) end return stiff_mat, rhs end N = 2 source(x) = x^3 mat, rhs = ritz(N, source) c = -mat\rhs x = linspace(0, 1, 100) y = zeros(x) for cont in 0:N - 1 y += c[cont + 1]*x.^(cont + 1).*(1 - x) end #%% Plotting figure(figsize=(4, 3)) plot(x, y) plot(x, x.*(x.^4 - 1)/20, linestyle="dashed") xlabel(L"$x$") ylabel(L"$y$") legend(["Ritz solution", "Exact solution"]) tight_layout() show()

Both have (almost) the same result, as follows

And if we consider 3 terms in the expansion, we get

### Comparison Python/Julia

Regarding number of lines we have: 38 in Python and 38 in Julia. The comparison
in execution time is done with `%timeit` magic command in IPython and
`@benchmark` in Julia.

For Python:

%%timeit mat, rhs = ritz(5, source) c = solve(mat, -rhs)

with result

1000 loops, best of 3: 228 µs per loop

For Julia:

function bench() mat, rhs = ritz(N, source) c = -mat\rhs end @benchmark bench()

with result

BenchmarkTools.Trial: memory estimate: 6.56 KiB allocs estimate: 340 -------------- minimum time: 13.527 μs (0.00% GC) median time: 15.927 μs (0.00% GC) mean time: 17.133 μs (4.50% GC) maximum time: 2.807 ms (97.36% GC) -------------- samples: 10000 evals/sample: 1

In this case, we can say that the Python code is roughly 14 times slowe than Julia.

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