Hankel function, mathematical definition

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Kevin ROUARD 2021-12-11

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评论： Rik 2021-12-20

采纳的回答： David Goodmanson

Hello everyone,

I'm wonder about the besselh(.) function.

The definition given is,

H = besselh(nu,K,Z,scale) specifies whether to scale the Hankel function to avoid overflow or loss of accuracy. If scale is 1, then Hankel functions of the first kind H(1)ν(z) are scaled by e−iZ, and Hankel functions of the second kind H(2)ν(z) are scaled by e+iZ.

But I found that (in eq. 12.140-2, Weber & Arfken, 2003)

Hankel first kind:

Hankel second kind:

That mean H(1)ν(z) correspond to

and H(2)ν(z) correspond to

? and why that is inverted so ?

Thank you.

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Stephen23 2021-12-20

https://web.archive.org/web/20211220041046/https://www.mathworks.com/matlabcentral/answers/1609065-hankel-function-mathematical-definition

Hankel function, mathematical definition

Hello everyone,

I'm wonder about the besselh(.) function.

The definition given is,

H = besselh(nu,K,Z,scale) specifies whether to scale the Hankel function to avoid overflow or loss of accuracy. If scale is 1, then Hankel functions of the first kind H(1)ν(z) are scaled by e−iZ, and Hankel functions of the second kind H(2)ν(z) are scaled by e+iZ.

But I found that (in eq. 12.140-2, Weber & Arfken, 2003)

Hankel first kind:

Hankel second kind:

That mean H(1)ν(z) correspond to and H(2)ν(z) correspond to ? and why that is inverted so ?

Thank you.

Rik 2021-12-20

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Answer 1

David Goodmanson 2021-12-12

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https://ww2.mathworks.cn/matlabcentral/answers/1609065-hankel-function-mathematical-definition#answer_853290

编辑：David Goodmanson 2021-12-12

在 MATLAB Online 中打开

Hi Kevin,

The hankel functions h that you cited are spherical hankel functions, which have half-integer order and are related to the regular hankel function H by

h(n,1,z) = const/sqrt(z)*H(n+1/2,1,z)       % first kind
h(n,2,z) = const/sqrt(z)*H(n+1/2,2,z)       % second kind
where
H(m,1,z) = besselh(m,1,z) 
H(m,2,z) = besselh(m,2,z)

To the best of my knowledge (I have 2019b), spherical bessel functions still are not a part of core Matlab.

Those details do not change the basic question about normalization. For large z,

besselh(m,1,z) --> const/sqrt(z)*exp(i*z)    as |z| --> inf
besselh(m,2,z) --> const/sqrt(z)*exp(-i*z)   as |z| --> inf

so the first kind goes like exp(i*z) and the second kind goes like exp(-i*z) as you said.

For larger but not overly large z, the factor in front is a slowly varying function that goes over to const/sqrt(z) in the limit.

Including scaling just means that the bessel function of the first kind is multiplied by exp(-i*z) to make the known exponential factor go away, leaving the slowly varying function. Similarly for the second kind.