hankel 0.3.0

Setup

This implementation is set up to allow efficient calculation of multiple functions f(x). To do this, the format is class-based, with the main object taking as arguments the order of the Bessel function, and the number and size of the integration steps (see Limitations for discussion about how to choose these key parameters).

For any general integration or transform of a function, we perform the following setup:

from hankel import HankelTransform     # Import the basic class

ht = HankelTransform(nu= 0,            # The order of the bessel function
                     N = 120,          # Number of steps in the integration
                     h = 0.03)         # Proxy for "size" of steps in integration

Alternatively, each of the parameters has defaults, so you needn’t pass any. The order of the bessel function will be defined by the problem at hand, while the other arguments typically require some exploration to set them optimally.

Integration

A Hankel-type integral is the integral

Having set up our transform with nu = 0, we may wish to perform this integral for f(x) = 1. To do this, we do the following:

f = lambda x : 1   # Create a function which identically 1.
ht.integrate(f)    # Should give (1.0000000000003544, -9.8381428368537518e-15)

The correct answer is 1, so we have done quite well. The second element of the returned result is an estimate of the error (it is the last term in the summation). The error estimate can be omitted using the argument ret_err=False.

We may now wish to integrate a different function, say x/(x^2 + 1). We can do this directly with the same object, without re-instantiating (avoiding unnecessary recalculation):

f = lambda x : x/(x**2 + 1)
ht.integrate(f)               # Should give (0.42098875721567186, -2.6150757700135774e-17)

The analytic answer here is K_0(1) = 0.4210. The accuracy could be increased by creating ht with a higher number of steps N, and lower stepsize h (see Limitations).

Transforms

The Hankel transform is defined as

We see that the Hankel-type integral is the Hankel transform of f(r)/r with k=1. To perform this more general transform, we must supply the k values. Again, let’s use our previous function, x/(x^2 + 1):

import numpy as np              # Import numpy
k = np.logspace(-1,1,50)        # Create a log-spaced array of k from 0.1 to 10.
ht.transform(f,k,ret_err=False) # Return the transform of f at k.

Fourier Transforms

One of the most common applications of the Hankel transform is to solve the radially symmetric n-dimensional Fourier transform:

We provide a specific class to do this transform, which takes into account the various normalisations and substitutions required, and also provides the inverse transform. The procedure is similar to the basic HankelTransform, but we provide the number of dimensions, rather than the Bessel order directly. Say we wish to find the Fourier transform of f(r) = 1/r in 3 dimensions:

from hankel import SymmetricFourierTransform
ft = SymmetricFourierTransform(ndim=3, N = 200, h = 0.03)

f = lambda r : 1./r
ft.transform(f,k, ret_err=False)

To do the inverse transformation (which is different by a normalisation constant), merely supply inverse=True to the .transform() method.

Efficiency

An implementation-specific limitation is that the method is not perfectly efficient in all cases. Care has been taken to make it efficient in the general sense. However, for specific orders and functions, simplifications may be made which reduce the number of trigonometric functions evaluated. For instance, for a zeroth-order spherical transform, the weights are analytically always identically 1.

Lower-Bound Convergence

In terms of limitations of the method, they are very dependent on the form of the function chosen. Notably, functions which tend to infinity at x=0 will be poorly approximated in this method, and will be highly dependent on the step-size parameter, as the information at low-x will be lost between 0 and the first step. As an example consider the simple function f(x) = 1/sqrt(x) with a 1/2 order bessel function. The total integrand tends to 1 at x=0, rather than 0:

f = lambda x: 1/np.sqrt(x)
h = HankelTransform(0.5,120,0.03)
h.integrate(f)  #(1.2336282286725169, 9.1467916948046785e-17)

The true answer is sqrt(pi/2), which is a difference of about 1.6%. Modifying the step size and number of steps to gain accuracy we find:

h = HankelTransform(0.5,700,0.001)
h.integrate(f)   #(1.2523045156429067, -0.0012281146007910256)

This has much better than percent accuracy, but uses 5 times the amount of steps. The key here is the reduction of h to “get inside” the low-x information. This limitation is amplified for cases where the function really does tend to infinity at x=0, rather than a finite positive number, such as f(x) = 1/x. Clearly the integral becomes non-convergent for some f(x), in which case the numerical approximation can never be correct.

Upper-Bound Convergence

If the function f(x) is monotonically increasing, or at least very slowly decreasing, then higher and higher zeros of the Bessel function will be required to capture the convergence. Often, it will be the case that if this is so, the amplitude of the function is low at low x, so that the step-size h can be increased to facilitate this. Otherwise, the number of steps N can be increased.

For example, the 1/2-order integral supports functions that are increasing up to f(x) = x^0.5 and no more (otherwise they diverge). Let’s use f(x) = x^0.4 as an example of a slowly converging function, and use our “hi-res” setup from the previous section:

h = HankelTransform(0.5,700,0.001)
f = lambda x : x**0.4
h.integrate(f)   # (0.53678277933471386, -1.0590954621246349)

The analytic result is 0.8421449 – very far from our result. Note that in this case, the error estimate itself is a good indication that we haven’t reached convergence. We could try increasing N:

h = HankelTransform(0.5,10000,0.001)
h.integrate(f,ret_err=False)/0.8421449 -1     ## 7.128e-07

This is very accurate, but quite slow. Alternatively, we could try increasing h:

h = HankelTransform(0.5,700,0.03)
h.integrate(f,ret_err=False)/0.8421449 -1     ## 0.00045616

Not quite as accurate, but still far better than a percent for a hundredth of the cost!

There are some notebooks in the devel/ directory which toy with some known integrals, and show how accurate different choices of N and h are. They are interesting to view to see some of the patterns.