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 $$\lambda_n \approx n\pi/ \vert a-c\vert$$ $$ a = 0.02$$ $$c=0.05$$


1 Answer 1


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.)

  • 3
    $\begingroup$ It is better to replace $\sim$ with $=$ and put an extra $+\mathcal{O}\!\left( {\frac{1}{{x^{3/2} }}} \right)$ on the RHSs. This is because at zeros of the trigonometric functions, these would tell you that the Bessel functions are asymptotic to $0$ which makes no sense. See dlmf.nist.gov/10.17.i for the complete asymptotic expansions. $\endgroup$
    – Gary
    Apr 9, 2021 at 7:18

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