Hankel transformation

In functional analysis , a branch of mathematics , the Hankel transformation is a linear integral transformation , which is essentially based on the Bessel functions of the first kind. It is named after the mathematician Hermann Hankel . Applications include image processing for correcting imaging errors.

definition

There are different conventions for defining the Hankel transformation. Let be a complex-valued function and . Then one can do the Hankel transformation of the order of by ${\displaystyle f}$ ${\displaystyle \nu >-{\tfrac {1}{2}}}$ ${\displaystyle \operatorname {H} _{\nu }}$ ${\displaystyle \nu }$ ${\displaystyle f}$

${\displaystyle F_{\nu }(u)=\operatorname {H} _{\nu }\{f(t)\}=\int _{0}^{\infty }f(t)\cdot J_{\nu }(ut)\cdot t\,\mathrm {d} t}$

define, there are those

${\displaystyle J_{\nu }(x):=\sum _{r=0}^{\infty }{\frac {(-1)^{r}({\frac {x}{2}})^{2r+\nu }}{\Gamma (\nu +r+1)r!}}}$

Bessel functions of the first kind and is the gamma function . Insofar as the integral exists, it is called the Hankel transform of . This convention of the Hankel transform is used predominantly in this article. In individual sections, however, the variant shown below is used, which is pointed out in the corresponding sections. ${\displaystyle \Gamma (\cdot )}$ ${\displaystyle \operatorname {H} _{\nu }\{f(t)\}}$ ${\displaystyle f}$

Another possibility to define the Hankel transformation of the order of is ${\displaystyle \nu >-{\tfrac {1}{2}}}$ ${\displaystyle f}$

${\displaystyle F_{\nu }(u)=\operatorname {H} _{\nu }\{f(t)\}=\int _{0}^{\infty }f(t)\cdot {\sqrt {ut}}\cdot J_{\nu }(ut)\,\mathrm {d} t\,.}$

Here, with also the Bessel functions of the first kind referred to and also Hankel transform is, insofar as the integral exists. ${\displaystyle J_{\nu }}$ ${\displaystyle \operatorname {H} _{\nu }\{f(t)\}}$

Inverse Hankel transformation

Similar to the Fourier transformation , it is also possible with the Hankel transformation under certain circumstances to recover its output function from the Hankel transform. An important result from the theory of the Hankel transformation says that if there is a Lebesgue integrable function with limited variation , the output function from the Hankel transform with the inverse integral transformation ${\displaystyle f\in L^{1}(]0,\infty [)}$ ${\displaystyle f}$ ${\displaystyle F_{\nu }}$

${\displaystyle f(t)=\operatorname {H} _{\nu }^{-1}\{F_{\nu }(u)\}=\int _{0}^{\infty }F_{\nu }(u)\cdot J_{\nu }(ut)\cdot u\,\mathrm {d} u}$

can be recovered. So the Hankel transformation and its inverse transformation are the same. It can therefore be understood as an involutive mapping . This statement applies analogously to the alternative definition.

properties

Orthogonality

The Bessel functions form an orthogonal basis : It applies

${\displaystyle \int _{0}^{\infty }J_{\nu }(ut)\cdot J_{\nu }(u't)\cdot t\,\mathrm {d} t={\frac {\delta (u-u')}{u}}}$

for and greater than 0 and with than the delta distribution . ${\displaystyle u}$ ${\displaystyle u'}$ ${\displaystyle \delta }$

Algebraization of the Bessel differential operator

Be

${\displaystyle B_{\nu }(f):=\left(r^{2}{\frac {\mathrm {d} ^{2}}{\mathrm {d} r^{2}}}+r{\frac {\mathrm {d} }{\mathrm {d} r}}+(r^{2}-\nu ^{2})\right)(f)}$

the Bessel differential operator . The following applies to the Bessel functions . With the help of the Hankel transformation it is possible to convert this differential operator into an expression without derivatives. Precisely applies ${\displaystyle B_{\nu }(J_{\nu })=0}$

${\displaystyle \operatorname {H} _{\nu }(B_{\nu }(f))(s)=-s^{2}\operatorname {H} _{\nu }(f)(s)\,.}$

This is a central property of the Hankel transformation for solving differential equations .

Relationship to the Fourier transform

The Hankel transform has some analogies to the Fourier transform. In particular, the Hankel transform can be calculated using a two-dimensional Fourier transform . Let be a radially symmetric function. This means that the function is independent of , which is why it is only noted with the parameter in the following . The Hankel transform of this function is now described with the aid of the function and the Fourier transform. ${\displaystyle \phi \colon \mathbb {R} ^{2}\to \mathbb {C} }$ ${\displaystyle f(r,\theta ):=\phi (r\cos(\theta ),r\sin(\theta ))}$ ${\displaystyle \theta }$ ${\displaystyle r}$ ${\displaystyle f}$ ${\displaystyle \phi }$

To see this, we use the Fourier integral

${\displaystyle {\mathcal {F}}(\phi )(\xi _{1},\xi _{2})={\frac {1}{2\pi }}\int _{\mathbb {R} ^{2}}\phi (x,y)e^{-i(x\xi _{1}+y\xi _{2})}\,\mathrm {d} (x,y)}$

from transformed into polar coordinates what to ${\displaystyle \phi }$

${\displaystyle {\begin{aligned}{\mathcal {F}}(\phi )(s\cos(\sigma ),s\sin(\sigma ))=&{\frac {1}{2\pi }}\int _{r=0}^{\infty }r\int _{\theta =0}^{2\pi }\phi (r\cos(\theta ),r\sin(\theta ))e^{-isr\cos(\theta -\sigma )}\,\mathrm {d} \theta \,\mathrm {d} r\\=&{\frac {1}{2\pi }}\int _{r=0}^{\infty }rf(r)\int _{\alpha =0}^{2\pi }e^{-isr\cos(\alpha )}\,\mathrm {d} \alpha \,\mathrm {d} r\\=&\int _{r=0}^{\infty }rf(r)J_{0}(sr)\,\mathrm {d} r\end{aligned}}}$

leads. This shows that a Fourier transformation of a radially symmetric function always corresponds to the Hankel transformation of a corresponding function. In particular, it is possible to construct a corresponding radially symmetric function for a given function , with which the Hankel transform of can be calculated by Fourier transformation . ${\displaystyle f\colon {]0,\infty [}\to \mathbb {C} }$ ${\displaystyle \phi }$ ${\displaystyle f}$

Hankel transformation for distributions

As with the Fourier transformation, the Hankel transformation can be generalized to distributions in an analogous way . In contrast to the Fourier transformation, the Hankel transformation cannot be defined on the space of the tempered distributions . Therefore you define a new space and explain the Hankel transformation for distributions on your dual space . ${\displaystyle H_{\nu }(]0,\infty [)}$

Distribution room

Be , then is defined by ${\displaystyle \nu \in \mathbb {R} }$ ${\displaystyle H_{\nu }(]0,\infty [)}$

${\displaystyle H_{\nu }(]0,\infty [):=\left\{\phi \in C^{\infty }(]0,\infty [)\left|\forall k,m\in \mathbb {N} _{0}:\sup _{x\in {]0,\infty [}}\left|x^{m}\left({\tfrac {1}{x}}{\tfrac {\mathrm {d} }{\mathrm {d} x}}\right)^{k}(x^{-\nu -{\frac {1}{2}}}\phi (x))\right|<\infty \right.\right\}\,.}$

A topology in the form of a concept of convergence is also defined on this vector space . A sequence converges to zero if and only if ${\displaystyle (\phi _{j})\subset H_{\nu }(]0,\infty [)}$

${\displaystyle \lim _{j\to \infty }\sup _{x\in {]0,\infty [}}\left|x^{m}\left({\tfrac {1}{x}}{\tfrac {\mathrm {d} }{\mathrm {d} x}}\right)^{k}(x^{-\nu -{\frac {1}{2}}}\phi _{j}(x))\right|=0}$

applies to all . By forming the topological dual space , one obtains the distribution space on which the Hankel transformation can be defined. For example, all distributions with compact support in how the delta function is one in the room included. ${\displaystyle k,m\in \mathbb {N} _{0}}$ ${\displaystyle H_{\nu }'(]0,\infty [)}$ ${\displaystyle ]0,\infty [}$ ${\displaystyle H_{\nu }'(]0,\infty [)}$

Hankel transformation

For the Hankel transformation for everyone is defined by ${\displaystyle T\in H_{\nu }'(]0,\infty [)}$ ${\displaystyle \phi \in H_{\nu }(]0,\infty [)}$

${\displaystyle \operatorname {H} _{\nu }(T)(\phi ):=T(\operatorname {H} _{\nu }(\phi ))\,.}$

The expression is again a Hankel transformation of a function and is therefore defined. Due to the construction of the room , however, the convention for the transformation is used here. ${\displaystyle \operatorname {H} _{\nu }(\phi )}$ ${\displaystyle H_{\nu }(]0,\infty [)}$ ${\displaystyle \textstyle \operatorname {H} _{\nu }(\phi )=\int _{0}^{\infty }f(t){\sqrt {ut}}J_{\nu }(ut)\,\mathrm {d} t}$

As with the Fourier transform for distributions, the Hankel transform is not carried out on the distribution itself, but is calculated on the test function . ${\displaystyle \phi }$

Examples

signal ${\displaystyle f(t)\,}$	Hankel Transformed ${\displaystyle F_{0}(u):=\operatorname {H} _{0}(f)(u)\,}$
${\displaystyle 1\,}$	${\displaystyle \delta (u)/u\,}$, valid for ${\displaystyle u\neq 0}$
${\displaystyle 1/t\,}$	${\displaystyle 1/u\,}$
${\displaystyle t\,}$	${\displaystyle -1/u^{3}\,}$
${\displaystyle t^{3}\,}$	${\displaystyle 9/u^{5}\,}$
${\displaystyle t^{m}\,}$	${\displaystyle {\frac {2^{m+1}\Gamma (m/2+1)}{u^{m+2}\Gamma (-m/2)}}\,}$, valid for odd ${\displaystyle m}$
${\displaystyle {\frac {1}{\sqrt {t^{2}+z^{2}}}}\,}$	${\displaystyle {\frac {e^{-u|z|}}{u}}={\sqrt {\frac {2|z|}{\pi u}}}K_{-1/2}(u|z|)\,}$
${\displaystyle {\frac {1}{t^{2}+z^{2}}}\,}$	${\displaystyle K_{0}(uz)\,}$, ${\displaystyle z\in \mathbb {C} }$
${\displaystyle e^{\mathrm {i} at}/t\,}$	${\displaystyle \mathrm {i} /{\sqrt {a^{2}-u^{2}}}\quad (a>0,u<a)\,}$ ${\displaystyle 1/{\sqrt {u^{2}-a^{2}}}\quad (a>0,u>a)\,}$
${\displaystyle e^{-a^{2}t^{2}/2}\,}$	${\displaystyle {\frac {e^{-u^{2}/(2a^{2})}}{a^{2}}}}$
${\displaystyle -t^{2}f(t)\,}$	${\displaystyle {\frac {d^{2}F_{\nu }}{du^{2}}}+{\frac {1}{u}}{\frac {dF_{\nu }}{du}}}$

In this section, the Bessel functions of the second kind -th order, the gamma function , the imaginary unit and again the delta distribution are referred to. In the table on the right side, some pairs of Hankel transformations are also listed. ${\displaystyle K_{n}(z)}$ ${\displaystyle n}$ ${\displaystyle \Gamma }$ ${\displaystyle \mathrm {i} }$ ${\displaystyle \delta }$

The hyperbola 1 / t

For the Hankel transform of the zero-order applies ${\displaystyle {\tfrac {1}{t}}}$

${\displaystyle {\begin{aligned}\operatorname {H} _{0}\left({\frac {1}{t}}\right)(s)=&\int _{0}^{\infty }t\cdot {\frac {1}{t}}\cdot J_{0}(st)\mathrm {d} t\\=&\int _{0}^{\infty }J_{0}(st)\,\mathrm {d} t\\=&{\frac {1}{s}}\end{aligned}}}$.

The function is therefore a fixed point of the Hankel transformation. ${\displaystyle {\tfrac {1}{t}}}$

The Gaussian bell curve

This section outlines the calculation of the Hankel transformation from the Gaussian bell curve using the Fourier transformation. Since the function is analytical , it can be continued on and is even radially symmetric there. The Hankel transform can therefore be calculated using the Fourier transform via . There is a fixed point for the Fourier transformation , from which it follows that the Hankel transform of is also again . So the Gaussian bell curve is also a fixed point of the Hankel transformation. ${\displaystyle e^{-{\frac {x^{2}}{2}}}}$ ${\displaystyle \mathbb {C} \cong \mathbb {R} ^{2}}$ ${\displaystyle \mathbb {R} ^{2}}$ ${\displaystyle e^{-{\frac {x^{2}}{2}}}}$ ${\displaystyle e^{-{\frac {x^{2}}{2}}}}$ ${\displaystyle e^{-{\frac {x^{2}}{2}}}}$

The delta distribution

In this example, the zero order Hankel transformation of the delta distribution is calculated. It applies ${\displaystyle \delta }$

${\displaystyle {\begin{aligned}\operatorname {H} _{0}({\tfrac {1}{u}}\delta _{0})(\phi )=&\operatorname {H} _{0}(\delta _{0})({\tfrac {1}{u}}\phi )=\delta _{0}(\operatorname {H} _{0}({\tfrac {1}{u}}\phi ))=\operatorname {H} _{0}({\tfrac {1}{u}}\phi )(0)\\=&\int _{0}^{\infty }{\tfrac {u}{u}}J_{0}(0)\phi (u)\,\mathrm {d} u\\=&\int _{0}^{\infty }\phi (u)\,\mathrm {d} u\end{aligned}}}$.

The expression is to be understood as a distribution that is generated by the constant one function. In the field of physics, the delta distribution is often written down imprecisely as a real-valued function and not as a functional . In this case, the calculation of the Hankel transformation is shortened ${\displaystyle \textstyle \int _{0}^{\infty }\phi (u)\mathrm {d} u}$

${\displaystyle \operatorname {H} _{0}({\tfrac {1}{u}}\delta _{0})(t)=\int _{0}^{\infty }\delta (u)\cdot {\frac {1}{u}}\cdot u\cdot J_{0}(tu)\,\mathrm {d} u=J_{0}(0)=1}$.

Conversely, if one wants to calculate the Hankel transform of the constant one function, one encounters a divergent integral when inserting it into the integral representation. Due to tightness arguments , it is nevertheless possible to understand the delta distribution as a Hankel transform of the constant one function.

swell

• Larry C. Andrews, Bhimsen K. Shivamoggi: Integral Transforms for Engineers . SPIE Press, University of Central Florida, 1999, ISBN 978-0-8194-3232-2 , Chapter 7.

• Alexander D. Poularikas: The Transforms and Applications Handbook . 2nd Edition. CRC Press, 2000, ISBN 978-0-8493-8595-7 , chapter 9.

• LS Dube, JN Pandey: On the Hankel transform of distributions Tohoku Math. J. (2) Volume 27, Number 3 (1975), 337-354.

Individual evidence

1. ↑ Bernd Jähne : Digital image processing . 6th edition. Springer, ISBN 978-3-540-24999-3 , pp. 219 to 223 .
2. ↑ ^{a b} Alexander D. Poularikas: The Transforms and Applications Handbook . 2nd Edition. CRC Press, 2000, ISBN 978-0-8493-8595-7 , chapter 9.4.
3. ↑ Alexander D. Poularikas: The Transforms and Applications Handbook . 2nd Edition. CRC Press, 2000, ISBN 978-0-8493-8595-7 , chapter 9.11.

Web links

• Eric W. Weisstein : Hankel Transform . In: MathWorld (English).