I was trying to find the eigenvalues, which is the positive roots of the equation bellow: $$J_{1}(a\lambda)Y_{1}(c\lambda) -J_{1}(c\lambda)Y_{1}(a\lambda) =0$$ I was presented with this trigonometric approach that worked fine. However, I need to know where that approach came from. Could anyone tell me how I get to this approach? $$J_{1}(a\lambda)Y_{1}(c\lambda) -J_{1}(c\lambda)Y_{1}(a\lambda) \approx -\frac{2\sin[(a-c)\lambda]}{\pi \lambda\sqrt{ac}}$$ and the zeros are $$\lambda_n \approx n\pi/ \vert a-c\vert$$ $$ a = 0.02$$ $$c=0.05$$
2026-04-03 04:36:25.1775190985
Bessel function approach as a trigonometric function
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It comes from the asymptotic form of the Bessel functions for $x\gg|\alpha^2-\frac{1}{4}|$:
$J_{\alpha}(x) = \sqrt{\frac{2}{\pi x}}\left[\cos\left(x-\frac{\alpha\pi}{2}-\frac{\pi}{4}\right)+O\left(\frac{1}{x}\right)\right]$,
$Y_{\alpha}(x) = \sqrt{\frac{2}{\pi x}}\left[\sin\left(x-\frac{\alpha\pi}{2}-\frac{\pi}{4}\right)+O\left(\frac{1}{x}\right)\right]$.
It follows that
$J_{\alpha}(x)Y_{\alpha}(y)-J_{\alpha}(y)Y_{\alpha}(x)= -\frac{2}{\pi\sqrt{xy}} \left[\sin(x-y)+O\left(\frac{1}{x},\frac{1}{y}\right)\right]$
if both $x$ and $y$ are sufficiently large.
Source: https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms
(Edited to incorporate suggestion made by Gary in the comments.)