Hankel transform

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Not to be confused with the determinant of the Hankel matrix of a sequence.

In mathematics, the Hankel transform expresses any given function f(r) as the weighted sum of an infinite number of Bessel functions of the first kind J_ν(kr). The Bessel functions in the sum are all of the same order ν, but differ in a scaling factor k along the r-axis. The necessary coefficient F_ν of each Bessel function in the sum, as a function of the scaling factor k constitutes the transformed function.

More precisely, the Hankel transform of order ν of a function f(r) is given by:

$F_{\nu }(k)=\int _{0}^{\infty }f(r)J_{\nu }(kr)\,r\operatorname {d}\!r$

where J_ν is the Bessel function of the first kind of order ν with ν ≥ −1/2. The inverse Hankel transform of F_ν(k) is defined as:

$f(r)=\int _{0}^{\infty }F_{\nu }(k)J_{\nu }(kr)k\operatorname {d}\!k$

which can be readily verified using the orthogonality relationship described below. The Hankel transform is an integral transform and was first developed by the mathematician Hermann Hankel. It is also known as the Fourier–Bessel transform. Just as the Fourier transform for an infinite interval is related to the Fourier series over a finite interval, so the Hankel transform over an infinite interval is related to the Fourier–Bessel series over a finite interval.

Domain of definition

The Hankel transform of a function f(r) is valid at every point at which f(r) is continuous provided that the function is defined in (0, ∞), is piecewise continuous and of bounded variation in every finite subinterval in (0, ∞), and the integral

$\int _{0}^{\infty }|f(r)|\,r^{{1/2}}\operatorname {d}\!r$

is finite. However, like the Fourier Transform, the domain can be extended by a density argument to include some functions whose above integral is not finite, for example $f(r)=(1+r)^{{-3/2}}$ ; this extension will not be discussed in this article.

Orthogonality

The Bessel functions form an orthogonal basis with respect to the weighting factor r:

$\int _{0}^{\infty }J_{\nu }(kr)J_{\nu }(k'r)r\operatorname {d}\!r={\frac {\delta (k-k')}{k}}$

for k and k' greater than zero.

The Plancherel theorem and Parseval's theorem

If f(r) and g(r) are such that their Hankel transforms F_ν(k) and G_ν(k) are well defined, then the Plancherel theorem states

$\int _{0}^{\infty }f(r)g(r)r\operatorname {d}\!r=\int _{0}^{\infty }F_{\nu }(k)G_{\nu }(k)k\operatorname {d}\!k.$

Parseval's theorem, which states:

$\int _{0}^{\infty }|f(r)|^{2}r\operatorname {d}\!r=\int _{0}^{\infty }|F_{\nu }(k)|^{2}k\operatorname {d}\!k.$

is a special case of the Plancherel theorem. These theorems can be proven using the orthogonality property.

Relation to other function transforms

Relation to the Fourier transform

The Hankel transform of order zero is essentially the two dimensional Fourier transform of a circularly symmetric function.

Consider a two-dimensional function f(r) of the radius vector r. Its Fourier transform is:

$F({\mathbf {k}})=\iint f({\mathbf {r}})e^{{i{\mathbf {k}}\cdot {\mathbf {r}}}}\operatorname {d}\!{\mathbf {r}}.$

With no loss of generality, we can pick a polar coordinate system (r, θ) such that the k vector lies on the θ = 0 axis. The Fourier transform is now written in these polar coordinates as:

$F({\mathbf {k}})=\int _{{r=0}}^{\infty }\int _{{\theta =0}}^{{2\pi }}f(r,\theta )e^{{ikr\cos(\theta )}}\,r\operatorname {d}\!\theta \operatorname {d}\!r$

where θ is the angle between the k and r vectors. If the function f happens to be circularly symmetric, it will have no dependence on the angular variable θ and may be written f(r). The integration over θ may be carried out, and the Fourier transform is now written:

$F({\mathbf {k}})=F(k)=2\pi \int _{0}^{\infty }f(r)J_{0}(kr)r\operatorname {d}\!r$

which is just 2π times the zero-order Hankel transform of f(r). For the reverse transform,

$f({\mathbf {r}})={\frac {1}{(2\pi )^{2}}}\iint F({\mathbf {k}})e^{{-i{\mathbf {k}}\cdot {\mathbf {r}}}}\operatorname {d}\!{\mathbf {k}}={\frac {1}{2\pi }}\int _{0}^{\infty }F(k)J_{0}(kr)k\operatorname {d}\!k$

so f(r) is 1/2π times the zero-order Hankel transform of F(k).

To generalize: If f is expanded in a multipole series,

$f(r,\theta )={\frac {1}{{\sqrt {2\pi }}}}\sum _{{m=-\infty }}^{\infty }f_{m}(r)e^{{im\theta }},$

and if $\theta _{k}$ is the angle between the direction of k and the $\theta =0$ axis,

$F({\mathbf k})=\int _{0}^{\infty }r\operatorname {d}\!r\,\int _{0}^{{2\pi }}\operatorname {d}\!\theta \,f(r,\theta )e^{{ikr\cos(\theta -\theta _{k})}}$ ,

one may substitute $\varphi =\theta -\theta _{k}$ ,

$F({\mathbf k})={\frac {1}{{\sqrt {2\pi }}}}\sum _{m}\int _{0}^{\infty }r\operatorname {d}\!r\,\int _{0}^{{2\pi }}\operatorname {d}\!\theta \,f_{m}(r)e^{{im\theta }}e^{{ikr\cos(\theta -\theta _{k})}}$

$={\frac {1}{{\sqrt {2\pi }}}}\sum _{m}e^{{im\theta _{k}}}\int _{0}^{\infty }r\operatorname {d}\!r\,f_{m}(r)\int _{0}^{{2\pi }}\operatorname {d}\!\varphi \,e^{{im\varphi }}e^{{ikr\cos \varphi }}$

$={\frac {1}{{\sqrt {2\pi }}}}\sum _{m}e^{{im\theta _{k}}}\int _{0}^{\infty }r\operatorname {d}\!r\,f_{m}(r)2\pi i^{m}J_{m}(kr)$

$={\sqrt {2\pi }}\sum _{m}i^{m}e^{{im\theta _{k}}}\int _{0}^{\infty }f_{m}(r)J_{m}(kr)r\operatorname {d}\!r.$

If $f_{m}$ is sufficiently smooth near the origin and zero outside a radius $R$ , it may be expanded into a Chebyshev series,

$f_{m}(r)=r^{m}\sum _{{t\geq 0}}f_{{mt}}[1-(r/R)^{2}]^{t},\quad 0\leq r\leq R.$

The numerically important aspect is that the expansion coefficients $f_{{mt}}$ are accessible with Discrete Fourier transform techniques. Insertion into the previous formula yields

$F(k)={\sqrt {2\pi }}\sum _{m}i^{m}e^{{im\theta _{k}}}\sum _{t}f_{{mt}}\int _{0}^{R}r^{m}[1-(r/R)^{2}]^{t}J_{m}(kr)r\operatorname {d}\!r$

$={\sqrt {2\pi }}\sum _{m}i^{m}e^{{im\theta _{k}}}R^{{m+2}}\sum _{t}f_{{mt}}\int _{0}^{1}x^{m}(1-x^{2})^{t}J_{m}(kxR)x\operatorname {d}\!x$

$={\sqrt {2\pi }}\sum _{m}i^{m}e^{{im\theta _{k}}}R^{{m+2}}\sum _{t}f_{{mt}}{\frac {t!2^{t}}{(kR)^{{1+t}}}}J_{{m+t+1}}(kR).$

This is one flavor of fast Hankel transform techniques.

Relation to the Fourier and Abel transforms

The Hankel transform is one member of the FHA cycle of integral operators. In two dimensions, if we define A as the Abel transform operator, F as the Fourier transform operator and H as the zeroth order Hankel transform operator, then the special case of the projection-slice theorem for circularly symmetric functions states that:

$FA=H.\,$

In other words, applying the Abel transform to a 1-dimensional function and then applying the Fourier transform to that result is the same as applying the Hankel transform to that function. This concept can be extended to higher dimensions.

Some Hankel transform pairs

^[1]

$f(r)\,$	$F_{0}(k)\,$
$1\,$	$\delta (k)/k\,$
$1/r\,$	$1/k\,$
$r\,$	$-1/k^{3}\,$
$r^{3}\,$	$9/k^{5}\,$
$r^{{m}}\,$	${\frac {2^{{m+1}}\Gamma (m/2+1)}{k^{{m+2}}\Gamma (-m/2)}}\,$ for -2<Re(m)<-1/2
${\frac {1}{{\sqrt {r^{2}+z^{2}}}}}\,$	${\frac {e^{{-k\|z\|}}}{k}}={\sqrt {{\frac {2\|z\|}{\pi k}}}}K_{{-1/2}}(k\|z\|)\,$
${\frac {1}{r^{2}+z^{2}}}\,$	$K_{0}(kz)\,$ , z may be any complex number
$e^{{iar}}/r\,$	$i/{\sqrt {a^{2}-k^{2}}}\quad (a>0,k<a)\,$
$\,$	$1/{\sqrt {k^{2}-a^{2}}}\quad (a>0,k>a)\,$
$e^{{-a^{2}r^{2}/2}}\,$	${\frac {e^{{-k^{2}/2a^{2}}}}{a^{2}}}$
$-r^{2}f(r)\,$	${\frac {\operatorname {d}^{2}\!F_{0}}{\operatorname {d}\!k^{2}}}+{\frac {1}{k}}{\frac {\operatorname {d}\!F_{0}}{\operatorname {d}\!k}}$
$f(r)\,$	$F_{{\nu }}(k)\,$
$r^{s}\,$	${\frac {\Gamma \left({\frac 12}(2+\nu +s)\right)}{\Gamma ({\tfrac 12}(\nu -s))}}{\frac {2^{{s+1}}}{k^{{s+2}}}}\,$
$r^{{\nu -2s}}\Gamma \left(s,r^{2}h\right)\,$	${\frac 12}\left({\frac k2}\right)^{{2s-\nu -2}}\gamma \left(1-s+\nu ,{\frac {k^{2}}{4h}}\right)\,$
$e^{{-r^{2}}}r^{\nu }U\left(a,b,r^{2}\right)\,$	${\frac {\Gamma (2+\nu -b)}{2\Gamma (2+\nu -b+a)}}\left({\frac k2}\right)^{\nu }e^{{-{\frac {k^{2}}4}}}\,_{1}F_{1}\left(a,2+a-b+\nu ,{\frac {k^{2}}4}\right)$
$-r^{2}f(r)\,$	${\frac {\operatorname {d}^{2}\!F_{\nu }}{\operatorname {d}\!k^{2}}}+{\frac {1}{k}}{\frac {\operatorname {d}\!F_{\nu }}{\operatorname {d}\!k}}-{\frac {\nu ^{2}}{k^{2}}}F_{\nu }$

$K_{n}(z)$ is a modified Bessel function of the second kind. The expression ${\frac {\operatorname {d}^{2}\!F_{0}}{\operatorname {d}\!k^{2}}}+{\frac {1}{k}}{\frac {\operatorname {d}\!F_{0}}{\operatorname {d}\!k}}$ coincides with the expression for the Laplace operator in polar coordinates $(k,\theta )$ applied to a spherically symmetric function $F_{0}(k)$ .

The Hankel transform of Zernike polynomials are essentially Bessel Functions (Noll 1976):

$R_{n}^{m}(r)=(-1)^{{(n-m)/2}}\int _{0}^{\infty }J_{{n+1}}(k)J_{m}(kr)\operatorname {d}\!k$

for even $n-m\geq 0$ .

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