# Bessel function approach as a trigonometric function

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

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.

• 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.