# Numerical methods challenge: Summary

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

## Summary

This post is a summary of that challenge. For the source code you can check the repository.

### The verdict

Since the challenge is with myself, and the main purpose was to learn some Julia the verdict is: success. Nevertheless, I failed during day 26 with the Boundary Element Method.

### The list of methods

Day

Numerical method

01

Bisection

02

Regula falsi

03

Newton

04

Newton multivariate

05

Broyden

06

07

08

Newton for optimization

09

Lagrange interpolation

10

Lagrange interpolation with Lobatto sampling

11

Lagrange interpolation using Vandermonde matrix

12

Hermite interpolation

13

Spline interpolation

14

15

16

17

Euler integration

18

Runge-Kutta integration

19

Verlet integration

20

Shooting method

21

Finite differences with Jacobi method

22

Finite differences for eigenvalues

23

Ritz method

24

Finite element method in 1D

25

Finite element method in 2D

26

Boundary element method

27

Monte-Carlo integration

28

LU factorization

29

Cholesky factorization

30

31

Finite element method with solver

### Conclusions

• This was an exercise of code-kata to learn some of the details of Julia for scientific computing. As such, it was really useful for me to get my hands dirty with Julia.

• Implementing the Boundary Element Method in one day seems to be rough. I knew this beforehand, but I gave it a try anyways … without succcess.

• Julia syntax is somewhat in a sweetspot between Python and MATLAB. This makes it really easy to use, although the documentation of some packages is at an arcane stage right now.

• I won't take a challenge like this in a while. It takes a lot of atttention to get it done.

# Numerical methods challenge: Day 31

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

## Putting some things together

Today, I am putting some things together, namely, I am going to solve the system of equations that results in the finite element using the conjugate gradient.

Following are the codes

### Python

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

def FEM1D(coords, source):
N = len(coords)
stiff_loc = np.array([[2.0, -2.0], [-2.0, 2.0]])
eles = [np.array([cont, cont + 1]) for cont in range(0, N - 1)]
stiff = np.zeros((N, N))
rhs = np.zeros(N)
for ele in eles:
jaco = coords[ele] - coords[ele]
rhs[ele] = rhs[ele] + jaco*source(coords[ele])
for cont1, row in enumerate(ele):
for cont2, col in enumerate(ele):
stiff[row, col] = stiff[row, col] +  stiff_loc[cont1, cont2]/jaco
return stiff, rhs

r = b - A.dot(x)
p = r
rsq_old = r.dot(r)
for cont in range(len(b)):
Ap = A.dot(p)
alpha = rsq_old / p.dot(Ap)
x = x + alpha*p
r = r - alpha*Ap
rsq_new = r.dot(r)
if np.sqrt(rsq_new) < tol:
break
p = r + (rsq_new / rsq_old) * p
rsq_old = rsq_new
return x, cont, np.sqrt(rsq_new)

fun = lambda x: x**3
N_vec = np.logspace(0.5, 3, 50, dtype=int)
err_vec = np.zeros_like(N_vec, dtype=float)
niter_vec = np.zeros_like(N_vec)
plt.figure(figsize=(8, 3))
for cont, N in enumerate(N_vec):
x = np.linspace(0, 1, N)
stiff, rhs = FEM1D(x, fun)
sol = np.zeros(N)
sol[1:-1], niter, _ = conj_grad(stiff[1:-1, 1:-1], -rhs[1:-1], rhs[1:-1])
err = np.linalg.norm(sol - x*(x**4 - 1)/20)/np.linalg.norm(x*(x**4 - 1)/20)
err_vec[cont] = err
niter_vec[cont] = niter

plt.subplot(121)
plt.loglog(N_vec, err_vec)
plt.xlabel("Number of nodes")
plt.ylabel("Relative error")
plt.subplot(122)
plt.loglog(N_vec, niter_vec)
plt.xlabel("Number of nodes")
plt.ylabel("Number of iterations")
plt.tight_layout()
plt.show()


### Julia

using PyPlot

function FEM1D(coords, source)
N = length(coords)
stiff_loc = [2.0 -2.0; -2.0 2.0]
eles = [[cont, cont + 1] for cont in 1:N-1]
stiff = zeros(N, N)
rhs = zeros(N)
for ele in eles
jaco = coords[ele] - coords[ele]
rhs[ele] = rhs[ele] + jaco*source(coords[ele])
stiff[ele, ele] = stiff[ele, ele] +  stiff_loc/jaco
end
return stiff, rhs
end

r = b - A * x
p = r
rsq_old = dot(r, r)
niter = 1
for cont = 1:length(b)
Ap = A * p
alpha = rsq_old / dot(p, Ap)
x = x + alpha*p
r = r - alpha*Ap
rsq_new = dot(r, r)
if sqrt(rsq_new) < tol
break
end
p = r + (rsq_new / rsq_old) * p
rsq_old = rsq_new
niter += 1
end
return x, niter, norm(r)
end

fun(x) = x.^3
N_vec = round.(logspace(0.5, 3, 50))
err_vec = zeros(N_vec)
niter_vec = zeros(N_vec)
figure(figsize=(8, 3))
for (cont, N) in enumerate(N_vec)
x = linspace(0.0, 1.0,N)
stiff, rhs = FEM1D(x, fun)
sol = zeros(N)
sol[2:end-1], niter, _ = conj_grad(stiff[2:end-1, 2:end-1],
-rhs[2:end-1], rhs[2:end-1])
err = norm(sol - x.*(x.^4 - 1)/20)/norm(x.*(x.^4 - 1)/20)
err_vec[cont] = err
niter_vec[cont] = niter
end
subplot(121)
loglog(N_vec, err_vec)
xlabel("Number of nodes")
ylabel("Relative error")
subplot(122)
loglog(N_vec, niter_vec)
xlabel("Number of nodes")
ylabel("Number of iterations")
tight_layout()
show()


In this case, we are analyzing the error of the solution as a function of the number of nodes. This, and the number of iterations required in the conjugate gradient are shown in the following image

# Numerical methods challenge: Day 30

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.

Today we have the conjugate gradient method. This method is commonly used to solve positive-definite linear systems of equations. Compared with gradient descent, we choose as descent direction a direction that is conjugated with the residual, that is, it is orthogonal with the matrix as weighting.

Following are the codes

### Python

from __future__ import division, print_function
import numpy as np

r = b - A.dot(x)
p = r
rsq_old = r.dot(r)
for cont in range(len(b)):
Ap = A.dot(p)
alpha = rsq_old / p.dot(Ap)
x = x + alpha*p
r = r - alpha*Ap
rsq_new = r.dot(r)
if np.sqrt(rsq_new) < tol:
break
p = r + (rsq_new / rsq_old) * p
rsq_old = rsq_new
return x, cont, np.sqrt(rsq_new)

N = 1000
A = -np.diag(2*np.ones(N)) + np.diag(np.ones(N-1), -1) +\
np.diag(np.ones(N-1), 1)
b = np.ones(N)
x0 = np.ones(N)
x, niter, accu = conj_grad(A, b, x0)


### Julia

function conj_grad(A, b, x; tol=1e-8)
r = b - A * x
p = r
rsq_old = dot(r, r)
niter = 1
for cont = 1:length(b)
Ap = A * p
alpha = rsq_old / dot(p, Ap)
x = x + alpha*p
r = r - alpha*Ap
rsq_new = dot(r, r)
if sqrt(rsq_new) < tol
break
end
p = r + (rsq_new / rsq_old) * p
rsq_old = rsq_new
niter += 1
end
return x, niter, norm(r)
end

N = 1000
A = -diagm(2*ones(N)) + diagm(ones(N-1), -1) + diagm(ones(N-1), 1)
b = ones(N)
x0 = ones(N)
x, niter, accu = conj_grad(A, b, x0)


In this case, we are testing the method with a matrix that comes from the discretization of the second order derivative using finite differences.

### Comparison Python/Julia

Regarding number of lines we have: 27 in Python and 27 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia.

For Python:

%timeit conj_grad(A, b, x0)


with result

10 loops, best of 3: 107 ms per loop


For Julia:

@benchmark conj_grad(A, b, x0)


with result

BenchmarkTools.Trial:
memory estimate:  27.13 MiB
allocs estimate:  3501
--------------
minimum time:     128.477 ms (0.54% GC)
median time:      294.407 ms (0.24% GC)
mean time:        298.208 ms (0.30% GC)
maximum time:     464.223 ms (0.30% GC)
--------------
samples:          17
evals/sample:     1


In this case, we can say that the Python code is roughly 2 times faster than Julia code.

# Numerical methods challenge: Day 29

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.

## Cholesky decomposition

Today we have Cholesky decomposition. It is a factorization of a Hermitian, positive-definite matrix into a lower and upper matrix, the main difference with LU decomposition is that it the lower matrix is the Hermitian transpose of the upper one.

Following are the codes

### Python

from __future__ import division, print_function
import numpy as np

def cholesky(mat):
m, _ = mat.shape
mat = mat.copy()
for col in range(m):
print(mat[col, col] -
mat[col, 0:col].dot(mat[col, 0:col]))
mat[col, col] = np.sqrt(mat[col, col] -
mat[col, 0:col].dot(mat[col, 0:col]))
for row in range(col + 1, m):
mat[row, col] = (mat[row, col] -
mat[row, 0:col].dot(mat[col, 0:col]))/mat[col, col]
for row in range(1, m):
mat[0:row, row] = 0
return mat

A = np.array([
[4, -1, 1],
[-1, 4.25, 2.75],
[1, 2.75, 3.5]], dtype=float)
B = cholesky(A)


### Julia

function cholesky(mat)
m, _ = size(mat)
mat = copy(mat)
for col = 1:m
mat[col, col] = sqrt(mat[col, col] -
dot(mat[col, 1:col-1], mat[col, 1:col-1]))
for row = col + 1:m
mat[row, col] = (mat[row, col] -
dot(mat[row, 1:col-1], mat[col, 1:col-1]))/mat[col, col]
end
end
for row = 2:m
mat[1:row-1, row] = 0
end
return mat
end

A = [4 -1 1;
-1 4.25 2.75;
1 2.75 3.5]
B = cholesky(A)


As an example we have the matrix

\begin{equation*} A = \begin{bmatrix} 4 &-1 &1\\ -1 &4.25 &2.75\\ 1 &2.75 &3.5 \end{bmatrix} \end{equation*}

And, the answer of both codes is

2.0  0.0  0.0
-0.5  2.0  0.0
0.5  1.5  1.0


### Comparison Python/Julia

Regarding number of lines we have: 23 in Python and 22 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia.

For Python:

%timeit cholesky(np.eye(100))


with result

100 loops, best of 3: 13 ms per loop


For Julia:

@benchmark cholesky(eye(100))


with result

BenchmarkTools.Trial:
memory estimate:  4.01 MiB
allocs estimate:  20303
--------------
minimum time:     1.010 ms (0.00% GC)
median time:      1.136 ms (0.00% GC)
mean time:        1.370 ms (17.85% GC)
maximum time:     4.652 ms (40.37% GC)
--------------
samples:          3637
evals/sample:     1


In this case, we can say that the Python code is roughly 10 times slower than Julia code.

# Numerical methods challenge: Day 28

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.

## LU factorization

Today we have LU decomposition. That is a factorization of a matrix into a lower (L) and upper (U) matrix. Basically it stores de steps of a Gauss elimination in matrices.

Following are the codes

### Python

from __future__ import division, print_function
import numpy as np

def LU(mat):
m, _ = mat.shape
mat = mat.copy()
for col in range(0, m - 1):
for row in range(col + 1, m):
if mat[row, col] != 0.0:
lam = mat[row, col]/mat[col, col]
mat[row, col + 1:m] = mat[row, col + 1:m] -\
lam * mat[col, col + 1:m]
mat[row, col] = lam
return mat

A = np.array([
[1, 1, 0, 3],
[2, 1, -1, 1],
[3, -1, -1, 2],
[-1, 2, 3, -1]], dtype=float)
B = LU(A)


### Julia

function LU(mat)
m, _ = size(mat)
mat = copy(mat)
for col = 1:m - 2
for row = col + 1:m
if mat[row, col] != 0.0
lam = mat[row, col]/mat[col, col]
mat[row, col + 1:m] = mat[row, col + 1:m] -
lam * mat[col, col + 1:m]
mat[row, col] = lam
end
end
end
return mat
end

A = [1.0 1.0 0.0 3.0;
2.0 1.0 -1.0 1.0;
3.0 -1.0 -1.0 2.0;
-1.0 2.0 3.0 -1.0]
B = LU(A)


As an example we have the matrix

\begin{equation*} A = \begin{bmatrix} 1 &1 &0 &3\\ 2 &1 &-1 &1\\ 3 &-1 &-1 &2\\ -1 &2 &3 &-1 \end{bmatrix} = \begin{bmatrix} 1 &1 &0 &0\\ 2 &1 &0 &0\\ 3 &4 &1 &2\\ -1 &-3 &0 &1 \end{bmatrix} \begin{bmatrix} 1 &1 &0 &3\\ 0 &-1 &-1 &-5\\ 0 &0 &3 &13\\ 0 &0 &0 &-13 \end{bmatrix} \end{equation*}

And, the answer of both codes is

[[  1.,   1.,   0.,   3.],
[  2.,  -1.,  -1.,  -5.],
[  3.,   4.,   3.,  13.],
[ -1.,  -3.,   0., -13.]]


### Comparison Python/Julia

Regarding number of lines we have: 23 in Python and 22 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia.

For Python:

%timeit LU(np.random.rand(10, 10))


with result

1000 loops, best of 3: 303 µs per loop


For Julia:

@benchmark LU(rand(10, 10))


with result

BenchmarkTools.Trial:
memory estimate:  29.25 KiB
allocs estimate:  310
--------------
minimum time:     9.993 μs (0.00% GC)
median time:      11.725 μs (0.00% GC)
mean time:        14.943 μs (15.90% GC)
maximum time:     3.285 ms (95.64% GC)
--------------
samples:          10000
evals/sample:     1


In this case, we can say that the Python code is roughly 30 times slower than Julia code.

# Numerical methods challenge: Day 27

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.

## Monte Carlo integration

Today we have Monte Carlo integration. Where we use random sampling to numerically compute an integral. This method is important when we want to evaluate higher-dimensional integrals since common quadrature techniques imply an exponential growing in the number of sampling points.

The method computes an integral

\begin{equation*} I = \int_\Omega f(x) dx \end{equation*}

where $\Omega$ has volume $V$.

The integral is approximated as

\begin{equation*} I \approx \frac{V}{N} \sum_{i=1}^{N} f(x_i) \end{equation*}

where the $x_i$ distribute uniform over $\Omega$.

Following are the codes

### Python

from __future__ import division, print_function
import numpy as np

def monte_carlo_int(fun, N, low, high, args=()):
ndims = len(low)
pts = np.random.uniform(low=low, high=high, size=(N, ndims))
V = np.prod(np.asarray(high) - np.asarray(low))
return V*np.sum(fun(pts, *args))/N

return 0.5*(1 - np.sign(x[:, 0]**2 + x[:, 1]**2 - rad**2))

N = 1000000
low = [-1, -1]
high = [1, 1]
inte = monte_carlo_int(circ, N, low, high, args=(rad,))


### Julia

using Distributions

function monte_carlo_int(fun, N, low, high; args=())
ndims = length(low)
pts = rand(Uniform(0, 1), N, ndims)
for cont = 1:ndims
pts[:, cont] = pts[:, cont]*(high[cont] - low[cont]) + low[cont]
end
V = prod(high - low)
return V*sum(fun(pts, args...))/N
end

return 0.5*(1 - sign.(x[:, 1].^2 + x[:, 2].^2 - rad^2))
end

N = 1000000
low = [-1, -1]
high = [1, 1]
inte = monte_carlo_int(circ, N, low, high, args=(rad,))


One of the most common examples is the computation of $\pi$, this is illustrated in the following animation. ### Comparison Python/Julia

Regarding number of lines we have: 20 in Python and 24 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia.

For Python:

%timeit monte_carlo_int(circ, N, low, high, args=(rad,))


with result

10 loops, best of 3: 53.2 ms per loop


For Julia:

@benchmark monte_carlo_int(circ, N, low, high, args=(rad,))


with result

BenchmarkTools.Trial:
memory estimate:  129.70 MiB
allocs estimate:  46
--------------
minimum time:     60.131 ms (5.15% GC)
median time:      164.018 ms (55.64% GC)
mean time:        204.366 ms (49.50% GC)
maximum time:     357.749 ms (64.04% GC)
--------------
samples:          25
evals/sample:     1


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

# Numerical methods challenge: Day 26

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.

## Boundary element method

Today we have the Boundary element method , or, an attempt. We want so solve the equation

\begin{equation*} \nabla^2 u = -f(x) \end{equation*}

with

\begin{equation*} u(x) = g(x),\quad \forall (x, y)\in \partial \Omega \end{equation*}

For this method, we need to use an integral representation of the equation, that in this case is

\begin{equation*} u(\xi) = \int_{S} [u(x) F(x, \xi) - q(x)G(x, \xi)]dS(x) + \int_{V} f(x) G(x, \xi) dV(x) \end{equation*}

with

\begin{equation*} G(x,\xi)= -\frac{1}{2\pi}\ln|x- \xi| \end{equation*}

and

\begin{equation*} F(x,\xi) = -\frac{1}{2\pi |x- \xi|^2}(x - \xi)\cdot\hat{n} \end{equation*}

Then, we can form a system of equations

\begin{equation*} [G]\{q\} = [F]\{u\}\, , \end{equation*}

that we obtain by discretization of the boundary. If we take constant variables over the discretization, the integral can be computed analytically by

\begin{equation*} G_{nm} = -\frac{1}{2\pi}\left[r \sin\theta\left(\ln|r| - 1\right) + \theta r\cos\theta\right]^{\theta_B, r_B}_{\theta_A, r_A} \end{equation*}

and

\begin{equation*} F_{nm} = \left[\frac{\theta}{2\pi}\right]^{\theta_B}_{\theta_A} \end{equation*}

for points $n$ and $m$ in different elements, and the subindices $A,B$ refer to the endpoints of the evaluation element. For diagonal terms the integrals evaluate to

\begin{equation*} G_{nn} = -\frac{L}{2\pi}\left(\ln\left\vert\frac{L}{2}\right\vert - 1\right) \end{equation*}

and

\begin{equation*} F_{nn} = - \frac{1}{2\pi} \end{equation*}

with $L$ the size of the element.

Below is the Python code. In this case, I did not succeed and the code is not working right now.

from __future__ import division, print_function
import numpy as np
from numpy import log, sin, cos, arctan2, pi, mean, arctan
from numpy.linalg import norm, solve
import matplotlib.pyplot as plt

def assem(coords, elems):
nelems = elems.shape
Gmat = np.zeros((nelems, nelems))
Fmat = np.zeros((nelems, nelems))
for ev_cont, elem1 in enumerate(elems):
print(coords[elem1], coords[elem1])
for col_cont, elem2 in enumerate(elems):
pt_col = mean(coords[elem2], axis=0)
if ev_cont == col_cont:
L = norm(coords[elem1] - coords[elem1])
Gmat[ev_cont, ev_cont] = - L/(2*pi)*(log(L/2) - 1)
Fmat[ev_cont, ev_cont] = - 0.5
else:
Gij, Fij = influence_coeff(elem1, coords, pt_col)
Gmat[ev_cont, col_cont] = Gij
Fmat[ev_cont, col_cont] = Fij
return Gmat, Fmat

def influence_coeff(elem, coords, pt_col):
r_A = coords[elem] - pt_col
r_B = coords[elem] - pt_col
theta_A = arctan2(r_A, r_A)
theta_B = arctan2(r_B, r_B)
G_coeff = r_B*(log(norm(r_B)) - 1) + theta_B*r_B -\
(r_A*(log(norm(r_A)) - 1) + theta_A*r_A)
F_coeff = theta_B - theta_A
return -G_coeff/(2*pi), F_coeff/(2*pi)

def eval_sol(ev_coords, coords):
nelems = elems.shape
Gmat = np.zeros((nelems, nelems))
Fmat = np.zeros((nelems, nelems))
for ev_cont, elem1 in enumerate(elems):
L = norm(coords[elem1] - coords[elem1])
for col_cont, elem2 in enumerate(elems):
pt_col = mean(coords[elem2], axis=0)
if ev_cont == col_cont:
Gmat[ev_cont, ev_cont] = - L/(2*pi)*(log(L/2) - 1)
Fmat[ev_cont, ev_cont] = - 0.5
else:
Gmat[ev_cont, col_cont], Fmat[ev_cont, col_cont] = \
influence_coeff(elem1, coords, pt_col)

nelems = 3
theta =  np.linspace(0, 2*pi, nelems, endpoint=False)
coords = rad * np.vstack((cos(theta), sin(theta))).T
elems = np.array([[cont, (cont + 1)%nelems] for cont in range(nelems)])
Gmat, Fmat = assem(coords, elems)
u_boundary = np.ones_like(theta)
q_boundary = solve(Gmat, Fmat.dot(u_boundary))


# Numerical methods challenge: Day 25

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.

## Finite element method

Today we have the Finite element method to solve the equation:

\begin{equation*} \nabla^2 u = -f(x) \end{equation*}

with

\begin{equation*} u(\sqrt{3}, 1) = 0 \end{equation*}

As in the Ritz method we form 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

\begin{equation*} \Pi[u] = -\int_\Omega \left(\nabla u\right)^2 d\Omega -\int_\Omega u f(x) d\Omega \end{equation*}

Using linear triangles, the local stiffness matrices read

\begin{equation*} K_\text{local} = \frac{1}{2|J|} \begin{bmatrix} 2 & -1 &1\\ -1 & 1 &0\\ -1 & 0 &1\\ \end{bmatrix} \end{equation*}

and

\begin{equation*} b_\text{local} = -|J| f(x_m) \mathbf{1}_3\, , \end{equation*}

where $|J|$ is the Jacobian determinant of the transformation and $x_m$ is the centroid of the triangle. I am skipping a great deal about assembling, but it would be just too extensive to describe the complete process.

We will solve the problem in a mesh with 6 triangles forming a regular hexagon

Following are the codes.

### Python

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

def assem(coords, elems, source):
ncoords = coords.shape
stiff = np.zeros((ncoords, ncoords))
rhs = np.zeros((ncoords))
for el_cont, elem in enumerate(elems):
stiff_loc, jaco = local_stiff(coords[elem])
rhs[elem] += jaco*np.mean(source[elem])
for row in range(3):
for col in range(3):
row_glob, col_glob = elem[row], elem[col]
stiff[row_glob, col_glob] += stiff_loc[row, col]
return stiff, rhs

def local_stiff(coords):
dHdr = np.array([[-1, 1, 0], [-1, 0, 1]])
jaco = dHdr.dot(coords)
stiff = 0.5*np.array([[2, -1, -1], [-1, 1, 0], [-1, 0, 1]])
return stiff/np.linalg.det(jaco), np.linalg.det(jaco)

sq3 = np.sqrt(3)
coords = np.array([
[sq3, -1],
[0, 0],
[2*sq3, 0],
[0, 2],
[2*sq3, 2],
[sq3, 3],
[sq3, 1]])
elems = np.array([
[1, 0, 6],
[0, 2, 6],
[2, 4, 6],
[4, 5, 6],
[5, 3, 6],
[3, 1, 6]])
source = -np.ones(7)
stiff, rhs = assem(coords, elems, source)
free = range(6)
sol = np.linalg.solve(stiff[np.ix_(free, free)], rhs[free])
sol_c = np.zeros(coords.shape)
sol_c[free] = sol
plt.tricontourf(coords[:, 0], coords[:, 1], sol_c, cmap="hot")
plt.colorbar()
plt.axis("image")
plt.show()


### Julia

using PyPlot

function assem(coords, elems, source)
ncoords = size(coords)
nelems = size(elems)
stiff = zeros(ncoords, ncoords)
rhs = zeros(ncoords)
for el_cont = 1:nelems
elem = elems[el_cont, :]
stiff_loc, jaco = local_stiff(coords[elem, :])
rhs[elem] += jaco*mean(source[elem])
for row = 1:3
for col = 1:3
row_glob = elem[row]
col_glob = elem[col]
stiff[row_glob, col_glob] += stiff_loc[row, col]
end
end
end
return stiff, rhs
end

function local_stiff(coords)
dHdr = [-1 1 0; -1 0 1]
jaco = dHdr * coords
stiff = 0.5*[2 -1 -1; -1 1 0; -1 0 1]
return stiff/det(jaco), det(jaco)
end

sq3 = sqrt(3)
coords =[sq3 -1;
0 0;
2*sq3 0;
0 2;
2*sq3 2;
sq3 3;
sq3 1]
elems =[2 1 7;
1 3 7;
3 5 7;
5 6 7;
6 4 7;
4 2 7]
source = -ones(7)
stiff, rhs = assem(coords, elems, source)
free = 1:6
sol = stiff[free, free] \ rhs[free]
sol_c = zeros(size(coords))
sol_c[free] = sol
tricontourf(coords[:, 1], coords[:, 2], sol_c, cmap="hot")
colorbar()
axis("image")
show()


Both have the same result, as follows ### Comparison Python/Julia

Regarding number of lines we have: 51 in Python and 56 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia. For the test we are just comparing the time it takes to generate the matrices.

For Python:

%timeit assem(coords, elems, source)


with result

1000 loops, best of 3: 671 µs per loop


For Julia:

@benchmark assem(coords, elems, source)


with result

BenchmarkTools.Trial:
memory estimate:  13.30 KiB
allocs estimate:  179
--------------
minimum time:     7.777 μs (0.00% GC)
median time:      8.934 μs (0.00% GC)
mean time:        10.810 μs (14.54% GC)
maximum time:     797.432 μs (95.85% GC)
--------------
samples:          10000
evals/sample:     4


In this case, we can say that the Python code is about 70 times slower than Julia code.

# Numerical methods challenge: Day 24

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.

## Finite element method

Today we have the Finite element method to solve the equation:

\begin{equation*} \frac{d^2 u}{dx^2} = f(x) \end{equation*}

with

\begin{equation*} u(0) = u(1) = 0 \end{equation*}

As in the Ritz method we form 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

\begin{equation*} \Pi[u] = -\int_{0}^{1} \left(\frac{d u}{d x}\right)^2 dx -\int_{0}^{1} u f(x) dx \end{equation*}

The main difference is that we use a piecewise interpolation for the basis functions,

\begin{equation*} \hat{u}(x) = \sum_{n=0}^{N} u_n N_n(x)\, , \end{equation*}

this leads to the system of equations

\begin{equation*} [K]\{\mathbf{c}\} = \{\mathbf{b}\} \end{equation*}

where the local stiffness matrices read

\begin{equation*} K_\text{local} = \frac{1}{|J|}\begin{bmatrix} 2 & -2\\ -2 &2\end{bmatrix} \end{equation*}

and

\begin{equation*} b_\text{local} = -|J|\begin{bmatrix} f(x_m)\\ f(x_{n})\end{bmatrix}\, , \end{equation*}

where $|J|$ is the Jacobian determinant of the transformation. I am skipping a great deal about assembling, but it would be just too extensive to describe the complete process.

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

\begin{equation*} u(x) = \frac{x (x^4 - 1)}{20} \end{equation*}

Following are the codes.

### Python

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

def FEM1D(coords, source):
N = len(coords)
stiff_loc = np.array([[2.0, -2.0], [-2.0, 2.0]])
eles = [np.array([cont, cont + 1]) for cont in range(0, N - 1)]
stiff = np.zeros((N, N))
rhs = np.zeros(N)
for ele in eles:
jaco = coords[ele] - coords[ele]
rhs[ele] = rhs[ele] + jaco*source(coords[ele])
for cont1, row in enumerate(ele):
for cont2, col in enumerate(ele):
stiff[row, col] = stiff[row, col] +  stiff_loc[cont1, cont2]/jaco
return stiff, rhs

N = 100
fun = lambda x: x**3
x = np.linspace(0, 1, N)
stiff, rhs = FEM1D(x, fun)
sol = np.zeros(N)
sol[1:-1] = np.linalg.solve(stiff[1:-1, 1:-1], -rhs[1:-1])

#%% Plotting
plt.figure(figsize=(4, 3))
plt.plot(x, sol)
plt.plot(x, x*(x**4 - 1)/20, linestyle="dashed")
plt.xlabel(r"$x$")
plt.ylabel(r"$y$")
plt.legend(["FEM solution", "Exact solution"])
plt.tight_layout()
plt.show()


### Julia

using PyPlot

function FEM1D(coords, source)
N = length(coords)
stiff_loc = [2.0 -2.0; -2.0 2.0]
eles = [[cont, cont + 1] for cont in 1:N-1]
stiff = zeros(N, N)
rhs = zeros(N)
for ele in eles
jaco = coords[ele] - coords[ele]
rhs[ele] = rhs[ele] + jaco*source(coords[ele])
stiff[ele, ele] = stiff[ele, ele] +  stiff_loc/jaco
end
return stiff, rhs
end

N = 100
fun(x) = x.^3
x = linspace(0, 1, N)
stiff, rhs = FEM1D(x, fun)
sol = zeros(N)
sol[2:end-1] = -stiff[2:end-1, 2:end-1]\rhs[2:end-1]

#%% Plotting
figure(figsize=(4, 3))
plot(x, sol)
plot(x, x.*(x.^4 - 1)/20, linestyle="dashed")
xlabel(L"$x$")
ylabel(L"$y$")
legend(["FEM solution", "Exact solution"])
tight_layout()
show()


Both have the same result, as follows ### Comparison Python/Julia

Regarding number of lines we have: 37 in Python and 35 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia. For the test we are just comparing the time it takes to generate the matrices.

For Python:

%timeit FEM1D(x, fun)


with result

100 loops, best of 3: 2.15 ms per loop


For Julia:

@benchmark FEM1D(x, fun)


with result

BenchmarkTools.Trial:
memory estimate:  183.73 KiB
allocs estimate:  1392
--------------
minimum time:     60.045 μs (0.00% GC)
median time:      70.445 μs (0.00% GC)
mean time:        98.276 μs (25.64% GC)
maximum time:     4.269 ms (96.70% GC)
--------------
samples:          10000
evals/sample:     1


In this case, we can say that the Python code is roughly 30 times slower than Julia code.

# 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:

\begin{equation*} \frac{d^2 u}{dx^2} = f(x) \end{equation*}

with

\begin{equation*} u(0) = u(1) = 0 \end{equation*}

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

\begin{equation*} \Pi[u] = -\int_{0}^{1} \left(\frac{d u}{d x}\right)^2 dx -\int_{0}^{1} u f(x) dx \end{equation*}

In this case, we are using the approximation

\begin{equation*} \hat{u}(x) = x (1 - x)\sum_{n=0}^{N} c_n x^n\, , \end{equation*}

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

\begin{align*} \Pi[\hat{u}] = -\sum_{n=0}^{N} \sum_{m=0}^{N} c_n c_m \left[\frac{2 + 2m + 2n + 2mn}{(n + m + 1)(n + m + 2)(n + m +3)}\right] -\\ \sum_{n=0}^{N} c_n\int_{0}^{1} x^{n + 1}(1 - x) f(x) dx \end{align*}

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

Minimizing the functional

\begin{equation*} \frac{\partial \Pi[\hat{u}]}{\partial c_m} = 0\, , \end{equation*}

we obtain the system of equations

\begin{equation*} [K]\{\mathbf{c}\} = \{\mathbf{b}\} \end{equation*}

with

\begin{equation*} K_{mn} = \frac{2 + 2m + 2n + 2mn}{(n + m + 1)(n + m + 2)(n + m +3)} \end{equation*}

and

\begin{equation*} b_m = -\int_{0}^{1} x^{m + 1}(1 - x) f(x) dx\, . \end{equation*}

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

\begin{equation*} u(x) = \frac{x (x^4 - 1)}{20} \end{equation*}

Following are the codes.

### Python

from __future__ import division, print_function
import numpy as np
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 slower than Julia.