Hankel function, mathematical definition

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Kevin ROUARD
Kevin ROUARD el 11 de Dic. de 2021
Comentada: Rik el 20 de Dic. de 2021
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 eiZ, 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.
  3 comentarios
Stephen23
Stephen23 el 20 de Dic. de 2021
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 eiZ, 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
Rik el 20 de Dic. de 2021
Regarding your flag, why should this question be deleted?

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David Goodmanson
David Goodmanson el 12 de Dic. de 2021
Editada: David Goodmanson el 12 de Dic. de 2021
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.
  1 comentario
Kevin ROUARD
Kevin ROUARD el 12 de Dic. de 2021
Hi David,
thank you for the answer. Effectively, I speak about spherical Hankel functions. Thanks for the scaling explanation.

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