Bounds of Hankel/Bessel function

Question

Is it possible to prove that when $p > z$, the Bessel functions satisfy $$ \vert H_p(z) \vert \leq 2|Y_p(z)|. $$ or even stronger $$ \vert J_p(z) \vert \leq |Y_p(z)|. $$

I saw some author used this in paper without any references or proof.

Answer 1

Assume $0<z<p$ and consider $$ z \mapsto \frac{{J_p (z)}}{{Y_p (z)}}. $$ Note that the first zeros of $J_p(z)$ and $Y_p(z)$ are larger than $p$ (cf. $(10.21.3)$), so this mapping is well defined and it takes negative values (because $J_p(z)>0$ and $Y_p(z)<0$ near $z=0$ when $p>0$). By the Wronskian for the Bessel functions (see $(10.5.2)$), $$ \frac{d}{{dz}}\frac{{J_p (z)}}{{Y_p (z)}} = - \frac{2}{{\pi zY_p^2 (z)}} < 0. $$ Thus, our function is strictly decreasing on $0<z<p$. Therefore, it remains to prove that $J_p(p)/Y_p(p)>-1$ for all $p>0$, since this will imply $$ -1<\frac{{J_p (z)}}{{Y_p (z)}} = - \frac{{\left| {J_p (z)} \right|}}{{\left| {Y_p (z)} \right|}} \Longleftrightarrow \left| {J_p (z)} \right| < \left| {Y_p (z)} \right| $$ for $0<z<p$. By the results of the this paper, $\mathrm{i}H_{\mathrm{i}t}^{(1)} (\mathrm{e}^{\pi \mathrm{i}/2} t) > 0$ for $t>0$ and $$ J_p (p) + \frac{1}{{\sqrt 3 }}Y_p (p) = - \frac{1}{{\sqrt 3 \pi p^{5/3} }}\int_0^{ + \infty } {\frac{{t^{2/3} \mathrm{e}^{ - 2\pi t} }}{{1 + (t/p)^2 }}\mathrm{i}H_{\mathrm{i}t}^{(1)} (\mathrm{e}^{\pi \mathrm{i}/2} t)\mathrm{d}t} < 0. $$ Consequently, $$ \frac{{J_p (p)}}{{Y_p (p)}} > - \frac{1}{{\sqrt 3 }} > - 1 $$ which completes the proof.

Note that in fact we obtained the stronger inequality $$ \left| {J_p (z)} \right| < \frac{1}{{\sqrt 3 }}\left| {Y_p (z)} \right| $$ for $0<z<p$. The constant $\frac{1}{{\sqrt 3 }}$ is the best possible, since $\left| {J_p (p)} \right| \sim \frac{1}{{\sqrt 3 }}\left| {Y_p (p)} \right|$ as $p\to +\infty$.

From the above analysis it can be inferred that the phase function $\theta_p(z)$ satisfies $$ - \frac{\pi }{2} < \theta _p (z) < - \frac{\pi }{3}, $$ when $0<z<p$. Accordingly $$ \left| {H_p^{(1,2)} (z)} \right| = \left| {Y_p (z)} \right|\left| {\csc \theta _p (z)} \right| \le \frac{2}{{\sqrt 3 }}\left| {Y_p (z)} \right| $$ provided $0<z<p$. The constant $\frac{2}{{\sqrt 3 }}$ is again sharp.