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How to apply crank-nicolson method in python to a wave equation like schrodinger's

I'm trying to do a particle in a box simulation with no potential field. Took me some time to find out that simple explicit and implicit methods break unitary time evolution so I resorted to crank-nicolson, which is supposed to be unitary. But when I try it I find that it still is not so. I'm not sure what I'm missing.. The formulation I used is this:
where T is the tridiagonal Toeplitz matrix for the second derivative wrt x and
The system simplifies to
The A and B matrices are:
I just solve this linear system for using the sparse module. The math makes sense and I found the same numeric scheme in some papers so that led me to believe my code is where the problem is.
Here's my code so far:
import numpy as np
import matplotlib.pyplot as plt
from scipy.linalg import toeplitz
from scipy.sparse.linalg import spsolve
from scipy import sparse
# Spatial discretisation
N = 100
x = np.linspace(0, 1, N)
dx = x[1] - x[0]
# Time discretisation
K = 10000
t = np.linspace(0, 10, K)
dt = t[1] - t[0]
alpha = (1j * dt) / (2 * (dx ** 2))
A = sparse.csc_matrix(toeplitz([1 + 2 * alpha, -alpha, *np.zeros(N-4)]), dtype=np.cfloat) # 2 less for both boundaries
B = sparse.csc_matrix(toeplitz([1 - 2 * alpha, alpha, *np.zeros(N-4)]), dtype=np.cfloat)
# Initial and boundary conditions (localized gaussian)
psi = np.exp((1j * 50 * x) - (200 * (x - .5) ** 2))
b = B.dot(psi[1:-1])
psi[0], psi[-1] = 0, 0
for index, step in enumerate(t):
# Within the domain
psi[1:-1] = spsolve(A, b)
# Enforce boundaries
# psi[0], psi[N - 1] = 0, 0
b = B.dot(psi[1:-1])
# Square integration to show if it's unitary
print(np.trapz(np.abs(psi) ** 2, dx))

You are relying on the Toeplitz constructor to produce a symmetric matrix, so that the entries below the diagonal are the same as above the diagonal. However, the documentation for scipy.linalg.toeplitz(c, r=None) says not "transpose", but
*"If r is not given, r == conjugate(c) is assumed."
so that the resulting matrix is self-adjoint. In this case this means that the entries above the diagonal have their sign switched.
It makes no sense to first construct a dense matrix and then extract a sparse representation. Construct it as sparse tridiagonal matrix from the start, using scipy.sparse.diags
A = sparse.diags([ (N-3)*[-alpha], (N-2)*[1+2*alpha], (N-3)*[-alpha]], [-1,0,1], format="csc");
B = sparse.diags([ (N-3)*[ alpha], (N-2)*[1-2*alpha], (N-3)*[ alpha]], [-1,0,1], format="csc");

Oscillatory integral in python

I wrote the following code to plot the intensity of light exiting an optical components, which is basically a spherical Fourier integral on the incident field, so it has a Bessel function. The argument of which depends on the integrating variable (x) and the plotting variable (r).
from sympy import *
import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import quad
from scipy.special import jn
#constants
mm = 1
um = 1e-3 * mm
nm = 1e-6 * mm
wavelength = 532*nm
klaser = 2*np.pi / wavelength
waist = 3.2*mm
angle = 2 #degrees
focus = 125 * mm
ng = 1.5 # refractive index of axicon
upperintegration = 5
#integrals
def b(angle):
radians = angle* np.pi/180
return klaser * (ng-1) * np.tan(radians)
def delta(angle):
return np.pi/(b(angle)*waist)
def integrand(x, r):
return klaser/focus * waist**2 * np.exp(-x**2) * np.exp(-np.pi * 1j * x/delta(angle)) * jn(0, waist*klaser*r*x/focus) * x
def intensity1D(r):
return np.sqrt(quad(lambda x: np.real(integrand(x, r)), 0, upperintegration)[0]**2 + quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration)[0]**2)
fig = plt.figure()
ax = fig.add_subplot(111)
t = np.linspace(-3.5, 3.5, 25)
plt.plot(t, np.vectorize(intensity1D)(t))
The issue is that the plot changes drastically as I change the number of points I am using in my linspace, when I plot it.
I suspect this may be because of the oscillatory nature of the integral, so the step-size taken can dramatically change the value of the exponent and hence of the integral.
How does quad deal with this? Are there better methods to integrate numerically for this particular application?

In the call to quad, set the limit argument to a large number. This increases the maximum number subintervals that quad is allowed to use to estimate the integral. When I use
def intensity1D(r):
re = quad(lambda x: np.real(integrand(x, r)), 0, upperintegration, limit=8000)[0]
im = quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration, limit=8000)[0]
return np.sqrt(re**2 + im**2)
and compute the function with the array t defined as
t = np.linspace(1.5, 3, 1000)
I get the following plot:
(I also removed the line from sympy import *. sympy does not appear to be used in
your script.)
You should always check the error estimate that is the second return value of quad.
For example:
In [14]: r = 3.0
In [15]: val, err = quad(lambda x: np.real(integrand(x, r)), 0, upperintegration, limit=8000)
In [16]: val
Out[16]: 2.975500141416676e-11
In [17]: err
Out[17]: 1.4590630152807049e-08
As you can see, the error estimate is much larger than the approximate integral. The estimates returned by quad might be conservative, but a result with such a large error estimate should still be treated with caution. Let's take a look at the corresponding imaginary part:
In [25]: val, err = quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration, limit=8000)
In [26]: val
Out[26]: 0.0026492702707317257
In [27]: err
Out[27]: 1.4808416189183e-08
val is now orders of magnitude larger than the estimated error. So when the magnitude of the complex value is computed in intensity1D(), we end up with estimated relative error on the order of 1e-5. That may be sufficient for your calculation.
At the peak near r=2.1825, the magnitude of the error estimate is still small, and it is much smaller than the computed integral:
In [32]: r = 2.1825
In [33]: quad(lambda x: np.real(integrand(x, r)), 0, upperintegration, limit=8000)
Out[33]: (6.435730031424414, 8.801375195176556e-08)
In [34]: quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration, limit=8000)
Out[34]: (-6.583055286038913, 9.211333259956749e-08)

There are specific methods for integration of oscillatory integrands that actually increase in accuracy as the frequency increases. Filon and Levin methods are described here:
https://www.sciencedirect.com/science/article/pii/S0377042706005929
Mathematica should use one of these if you specify LevinRule as method in
NIntegrate. This is perhaps simple enough that -- if your integrand has the form, apparently common in optics calculations-- that you could even write a short program in your favorite efficient numerical programming language.
I suspect that using usual quadrature for oscillatory integrands is going to be painfully slow if you want to get accurate results.

Any suggestions for a Python Lasso solver that handles complex values?

I am looking for a Python Lasso solver that works with complex numbers to use in beamforming problems. The objective function is affine, XW - Y. I believe that there at least one such solver implemented for Matlab,
http://www.cs.ubc.ca/~schmidtm/Software/code.html
I have tried to use scikit-learn MultiTaskLasso, following a suggestion from
Is it possible to use complex numbers as target labels in scikit learn?
The matrix 21 norm in the MultiTaskLasso is the correct way to handle the L1 norm for complex numbers. However, my approach requires some gymnastics to force the solver to follow the rules of complex multiplication. Essentially, I need to minimize the L2 norm of
[Re{X}, Im{X}] * [[Re{W}, Im{W}], [-Im{W}, Re{W}]] - [Re{Y}, Im{Y}]
I attempted to enforce the relationship between the two columns of W by adding another row to the X matrix, [-Im{X}, Re{X}], and the row [-Im{Y}, Re{Y}] to Y. This ideally would equate the cost of a change a value each column of W with the corresponding value in the other column
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import MultiTaskLasso
# experiment specifications
numEl = 20
numLook = 1e2
arrayPosition = np.arange(numEl) * np.pi
thetaLook = np.r_[0 : np.pi : numLook * 1j] - (np.pi / 2)
thetaSource = -0.3758
# Make grid of look vectors
W = np.exp(1j * np.sin(thetaLook)[:,None] * arrayPosition)
data = np.exp(1j * np.sin(thetaSource) * arrayPosition).T
# Bartlet beamformer
bartlet = np.abs(np.dot(W.conj(), data))**2
B_bart = 10 * np.log10(np.abs(bartlet)); B_bart-=np.max(B_bart)
# Lasso setup
X = W.T
XSplit = np.vstack((np.hstack((X.real, X.imag)),\
np.hstack((-X.imag, X.real))))
YSplit = np.hstack((np.vstack((data.real, data.imag)),\
np.vstack((-data.imag, data.real)))).T
lasso_solver = MultiTaskLasso(alpha=0.1)
lasso = lasso_solver.fit(XSplit, YSplit).coef_
# Manipulate result back into complex values
stack1 = np.squeeze(lasso[0,:])
stack1 = np.squeeze(stack1[:numLook] + 1j * stack1[numLook:])
B_lasso = 10 * np.log10(np.abs(stack1) + np.spacing(1)); B_lasso -= np.max(B_lasso)
# stack1 ?= stack2 (Should be exact)
stack2 = np.squeeze(lasso[1,:])
stack2 = np.squeeze(-1j * stack2[:numLook] + stack2[numLook:])
np.testing.assert_almost_equal(stack1, stack2, decimal=1)
# Plot both beamformer results
_ = plt.plot(np.rad2deg(thetaLook), B_bart)
_ = plt.plot(np.rad2deg(thetaLook), B_lasso, 'r.')
_ = plt.ylim(-40,3); plt.ylabel('Beamformer Output, dB')
_ = plt.xlabel('Look Direction, deg')
While this approach seems to work for simple problems like the one above, it fails when the problems get more complicated. I define failure when the relationship between the first and second column of W no longer holds. A simple way to create small divergent behavior in the above example is to substitute a Ridge solver for the MultiTaskLasso.
Does anyone know of a Lasso solver that can solve the complex valued problem with rigorous treatment of complex numbers?

pdf estimation with scipy.stats

Say I compute the density of Beta(4,8):
from scipy.stats import beta
rv = beta(4, 8)
x = np.linspace(start=0, stop=1, num=200)
my_pdf = rv.pdf(x)
Why does the integral of the pdf not equal one?
> my_pdf.sum()
199.00000139548044

The integral over the pdf is one. You can see this by using numerical integration from scipy
>>> from scipy.integrate import quad
>>> quad(rv.pdf, 0, 1)
(0.9999999999999999, 1.1102230246251564e-14)
or by writing your own ad-hoc integration (with a trapezoidal rule in this example)
>>> x = numpy.linspace(start=0, stop=1, num=201)
>>> (0.5 * rv.pdf(x[0]) + rv.pdf(x[1:-1]).sum() + 0.5 * rv.pdf(x[-1])) / 200.0
1.0000000068732813

rv.pdf returns the value of the pdf at each value of x. It doesn't sum to one because your aren't actually computing an integral. If you want to do that, you need to divide your sum by the number of intervals, which is len(x) - 1, which is 199. That would then give you a result very close to 1.

Artefacts from Riemann sum in scipy.signal.convolve

Short summary: How do I quickly calculate the finite convolution of two arrays?
Problem description
I am trying to obtain the finite convolution of two functions f(x), g(x) defined by
To achieve this, I have taken discrete samples of the functions and turned them into arrays of length steps:
xarray = [x * i / steps for i in range(steps)]
farray = [f(x) for x in xarray]
garray = [g(x) for x in xarray]
I then tried to calculate the convolution using the scipy.signal.convolve function. This function gives the same results as the algorithm conv suggested here. However, the results differ considerably from analytical solutions. Modifying the algorithm conv to use the trapezoidal rule gives the desired results.
To illustrate this, I let
f(x) = exp(-x)
g(x) = 2 * exp(-2 * x)
the results are:
Here Riemann represents a simple Riemann sum, trapezoidal is a modified version of the Riemann algorithm to use the trapezoidal rule, scipy.signal.convolve is the scipy function and analytical is the analytical convolution.
Now let g(x) = x^2 * exp(-x) and the results become:
Here 'ratio' is the ratio of the values obtained from scipy to the analytical values. The above demonstrates that the problem cannot be solved by renormalising the integral.
The question
Is it possible to use the speed of scipy but retain the better results of a trapezoidal rule or do I have to write a C extension to achieve the desired results?
An example
Just copy and paste the code below to see the problem I am encountering. The two results can be brought to closer agreement by increasing the steps variable. I believe that the problem is due to artefacts from right hand Riemann sums because the integral is overestimated when it is increasing and approaches the analytical solution again as it is decreasing.
EDIT: I have now included the original algorithm 2 as a comparison which gives the same results as the scipy.signal.convolve function.
import numpy as np
import scipy.signal as signal
import matplotlib.pyplot as plt
import math
def convolveoriginal(x, y):
'''
The original algorithm from http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P, Q, N = len(x), len(y), len(x) + len(y) - 1
z = []
for k in range(N):
t, lower, upper = 0, max(0, k - (Q - 1)), min(P - 1, k)
for i in range(lower, upper + 1):
t = t + x[i] * y[k - i]
z.append(t)
return np.array(z) #Modified to include conversion to numpy array
def convolve(y1, y2, dx = None):
'''
Compute the finite convolution of two signals of equal length.
#param y1: First signal.
#param y2: Second signal.
#param dx: [optional] Integration step width.
#note: Based on the algorithm at http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P = len(y1) #Determine the length of the signal
z = [] #Create a list of convolution values
for k in range(P):
t = 0
lower = max(0, k - (P - 1))
upper = min(P - 1, k)
for i in range(lower, upper):
t += (y1[i] * y2[k - i] + y1[i + 1] * y2[k - (i + 1)]) / 2
z.append(t)
z = np.array(z) #Convert to a numpy array
if dx != None: #Is a step width specified?
z *= dx
return z
steps = 50 #Number of integration steps
maxtime = 5 #Maximum time
dt = float(maxtime) / steps #Obtain the width of a time step
time = [dt * i for i in range (steps)] #Create an array of times
exp1 = [math.exp(-t) for t in time] #Create an array of function values
exp2 = [2 * math.exp(-2 * t) for t in time]
#Calculate the analytical expression
analytical = [2 * math.exp(-2 * t) * (-1 + math.exp(t)) for t in time]
#Calculate the trapezoidal convolution
trapezoidal = convolve(exp1, exp2, dt)
#Calculate the scipy convolution
sci = signal.convolve(exp1, exp2, mode = 'full')
#Slice the first half to obtain the causal convolution and multiply by dt
#to account for the step width
sci = sci[0:steps] * dt
#Calculate the convolution using the original Riemann sum algorithm
riemann = convolveoriginal(exp1, exp2)
riemann = riemann[0:steps] * dt
#Plot
plt.plot(time, analytical, label = 'analytical')
plt.plot(time, trapezoidal, 'o', label = 'trapezoidal')
plt.plot(time, riemann, 'o', label = 'Riemann')
plt.plot(time, sci, '.', label = 'scipy.signal.convolve')
plt.legend()
plt.show()
Thank you for your time!

or, for those who prefer numpy to C. It will be slower than the C implementation, but it's just a few lines.
>>> t = np.linspace(0, maxtime-dt, 50)
>>> fx = np.exp(-np.array(t))
>>> gx = 2*np.exp(-2*np.array(t))
>>> analytical = 2 * np.exp(-2 * t) * (-1 + np.exp(t))
this looks like trapezoidal in this case (but I didn't check the math)
>>> s2a = signal.convolve(fx[1:], gx, 'full')*dt
>>> s2b = signal.convolve(fx, gx[1:], 'full')*dt
>>> s = (s2a+s2b)/2
>>> s[:10]
array([ 0.17235682, 0.29706872, 0.38433313, 0.44235042, 0.47770012,
0.49564748, 0.50039326, 0.49527721, 0.48294359, 0.46547582])
>>> analytical[:10]
array([ 0. , 0.17221333, 0.29682141, 0.38401317, 0.44198216,
0.47730244, 0.49523485, 0.49997668, 0.49486489, 0.48254154])
largest absolute error:
>>> np.max(np.abs(s[:len(analytical)-1] - analytical[1:]))
0.00041657780840698155
>>> np.argmax(np.abs(s[:len(analytical)-1] - analytical[1:]))
6

Short answer: Write it in C!
Long answer
Using the cookbook about numpy arrays I rewrote the trapezoidal convolution method in C. In order to use the C code one requires three files (https://gist.github.com/1626919)
The C code (performancemodule.c).
The setup file to build the code and make it callable from python (performancemodulesetup.py).
The python file that makes use of the C extension (performancetest.py)
The code should run upon downloading by doing the following
Adjust the include path in performancemodule.c.
Run the following
python performancemodulesetup.py build
python performancetest.py
You may have to copy the library file performancemodule.so or performancemodule.dll into the same directory as performancetest.py.
Results and performance
The results agree neatly with one another as shown below:
The performance of the C method is even better than scipy's convolve method. Running 10k convolutions with array length 50 requires
convolve (seconds, microseconds) 81 349969
scipy.signal.convolve (seconds, microseconds) 1 962599
convolve in C (seconds, microseconds) 0 87024
Thus, the C implementation is about 1000 times faster than the python implementation and a bit more than 20 times as fast as the scipy implementation (admittedly, the scipy implementation is more versatile).
EDIT: This does not solve the original question exactly but is sufficient for my purposes.