Solving an equation with Hankel-function

Hello,

I have an equation which contains a Hankelfunction:

exp(-i*2*pi/3) * (H1[1/3*(-2y)^(3/2)]) / (H2[1/3*(-2y)^(3/2)]) = 1/(a*sqrt(-2*y))

where H1[] = besselh(2/3,2,...) H2[] = besselh(1/3,2,...)

Is there a way to calculate y for a given a?

Greets Benny

Answer 1

[Walter Roberson]

I tried some things but could not find a symbolic method.

In Maple notion all I found was

-RootOf(4*HankelH2(1/3, (2/3)*sqrt(2)*_Z^3)*a*_Z+HankelH2(2/3, (2/3)*sqrt(2)*_Z^3)*sqrt(2)-I*HankelH2(2/3, (2/3)*sqrt(2)*_Z^3)*sqrt(2)*sqrt(3))^2

-RootOf(4*HankelH2(1/3, -(2/3)*sqrt(2)*_Z^3)*a*_Z+HankelH2(2/3, -(2/3)*sqrt(2)*_Z^3)*sqrt(2)-I*HankelH2(2/3, -(2/3)*sqrt(2)*_Z^3)*sqrt(2)*sqrt(3))^2

Here

"I" is sqrt(-1),
HankelH2(nu,z) = besselh(nu,2,z)
RootOf() indicates the value _Z for which the expression becomes zero

I tried a couple of values for "a" but got complex solutions, probably because of the "i" in your initial equation.

Answer 2

[Benjamin]

Hello Walter,

complex solutions are no problem, thats what i expected... I think i have to change from matlab to maple... just have to look if i can combine it with c#...

by the way, could you write down your maple code, please. thanks a lot!

if anybody finds out a way to calculate it in mathlab, please write it down.

greets benny

[Walter Roberson]

The Maple code was trivial,

solve(exp(-(1/3)*(I*2)*Pi)*HankelH2(1/3, (1/3)*(-2*y)^(3/2))/HankelH2(2/3, (1/3)*(-2*y)^(3/2)) = 1/(a*sqrt(-2*y)),y)

The answer was the same form even when I misspelled HankelH2, so it did not use any properties of HankelH2 in coming to the solution.

Converting the first of the pair to Bessel gives

-RootOf(4*a*_Z*BesselJ(1/3, (2/3)*sqrt(2)*_Z^3) - (4*I)*a*_Z*BesselY(1/3, (2/3)*sqrt(2)*_Z^3) + sqrt(2)*BesselJ(2/3, (2/3)*sqrt(2)*_Z^3) - I*sqrt(2)*BesselY(2/3, (2/3)*sqrt(2)*_Z^3) - I*sqrt(2)*sqrt(3)*BesselJ(2/3, (2/3)*sqrt(2)*_Z^3) - sqrt(2)*sqrt(3)*BesselY(2/3, (2/3)*sqrt(2)*_Z^3))^2

which looks to me like nothing more than a straight forward transformation based upon the definition of the HankelH2 function, with again no special properties used.

[Walter Roberson]

MATLAB equivalent with the symbolic toolbox would be

solve('exp(-(1/3)*(I*2)*Pi)*besselh(1/3, 2, (1/3)*(-2*y)^(3/2))/besselh(2/3, 2, (1/3)*(-2*y)^(3/2)) = 1/(a*sqrt(-2*y)),y)','y')

which will likely give a result that looks quite similar to what I posted earlier, but with a different syntax for RootOf().

Do you have the optimization toolbox? fsolve() should work, I think. Unfortunately, MATLAB's fzero() and fminbnd() and fminsearch() that are in basic MATLAB are not suitable as those do not handle complex solutions.