Asymptotic behavior of a Bessel function on a sequence on zeros with a shifted parameter of type

Question

Let $J_\nu$ be a Bessel function of the first kind and let $\{\lambda_{n, \nu}\}_{n\ge 1}$ be a sequence of its zeroes. I claim that $$ \inf_{n\ge 1}\bigg|\sqrt{\lambda_{n,\nu}} J_{\nu+1}(\lambda_{n,\nu})\bigg|>0. $$ The reason I believe this is true is that:

(1) we have $|J_{\nu+1}(x)|\lesssim x^{-1/2},\qquad x\rightarrow \infty$,

(2) there's an asymptotic formula $$ \lambda_{n,\nu}=\pi n+\frac{\pi(2\nu-1)}{4}+O(n^{-1}), $$ which shows in particular that zeroes of $J_{\nu+1}$ are separated from zeroes of $J_\nu$, thus the only decay $J_{\nu+1}(\lambda_{n,\nu})$ is only due to (1) (i.e. oscillations should not be relevant).

I don't know how to make this argument rigorous. Any hints are appreciated!

Answer 1

You have, using the asymptotics for $\lambda_{n,\nu}$, \begin{align*} & \cos\left(\lambda_{n,\nu} - \frac{2\nu+3}{4}\pi \right) \\ &= \cos\left(\lambda_{n,\nu} - \frac{2\nu-1}{4}\pi+\pi \right) \\ &= \cos\left(\lambda_{n,\nu} - \frac{2\nu-1}{4}\pi-n\pi \right) (-1)^{n-1}\\ &= \cos\left(O(n^{-1}) \right) (-1)^{n-1} \\ &= (-1)^{n-1} +O(n^{-1}). \end{align*} Inserting this in the asymptotic formula for $J_{\nu+1}(x)$, we obtain $$ \sqrt{\lambda_{\nu,n}}J_{\nu+1}(\lambda_{\nu,n})= (-1)^{n-1}\sqrt{\frac2\pi} +O(n^{-1})+O(\lambda_{\nu,n}^{-1})=(-1)^{n-1}\sqrt{\frac2\pi} +O(n^{-1}). $$ Which shows that $$ \lim_{n\to\infty} \lambda_{\nu,n}J_{\nu+1}^2(\lambda_{\nu,n})=\frac2\pi. $$ As $J_{\nu}$ and $J_{\nu+1}$ have distinct zeros, you have a proof.

Answer 2

To see that the given infimum is positive one needs to combine the asymptotics $$ J_{\nu}(x) = \sqrt{\frac{2}{\pi x}} \bigg( \cos\Big( x - \frac{2\nu+1}4 \pi \Big) + \mathcal{O}\big( x^{-1} \big) \bigg), \qquad x \to \infty. $$ with the asymptotic formula for zeros, which I stated in the question, and with the fact that zeros of Bessel functions with shifted parameters interlace, i.e. one has $$ 0 < \lambda_{1,\nu} < \lambda_{1,\nu+1} < \lambda_{2,\nu} < \lambda_{2,\nu+1} < \lambda_{3,\nu} < \ldots $$