<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
\parindent=0pt
\begin{document}


{\bf Question}

The Bessel functions $J_n(x)$ are the solutions of

$$x^2y''+xy'+(x^2-n^2)y=0$$

are well-behaved at the origin. Transform the equation by changing
the variables to $w=y\sqrt{x},\ t=\df{x}{\sqrt{n^2-\frac{1}{4}}}$
and hence show that the WKB solutions for large $n$ are

$$\begin{array}{l} y \sim
\df{A_\pm}{\sqrt{x}}\left(\df{x^2}{x^2-n^2}\right)^{\frac{1}{4}}\exp\left(\pm
i\left[(x^2-n^2)^\frac{1}{2}-n\arccos\left(\df{n}{x}\right)\right]\right\},\
t&gt;1\\ y \sim
\df{B_\pm}{\sqrt{x}}\left(\df{x^2}{x^2-n^2}\right)^{\frac{1}{4}}\exp\left(\pm
\left[(x^2-n^2)^\frac{1}{2}-n\rm{arccosh}\left(\df{n}{x}\right)\right]\right\},\
t&lt;1 \end{array}$$


\vspace{.25in}

{\bf Answer}

$x^2y''+xy'+(x^2-n^2)y=0$

If $t=\df{x}{\sqrt{n^2-\frac{1}{4}}},\
\pl_x=\df{1}{\sqrt{n^2-\frac{1}{4}}}\pl_t;\
\df{w(t)}{t^\frac{1}{2}(n^2-\frac{1}{4})^\frac{1}{4}}=y$

so

$\begin{array}{cl} &amp;
x^2\left(\df{1}{\sqrt{n^2-\frac{1}{4}}}\pl_t\right)
\left(\df{1}{\sqrt{n^2-\frac{1}{4}}}\right)\pl_t
\left(\df{w}{t^{\frac{1}{2}}(n^2-\frac{1}{4})^\frac{1}{4}}\right)\\
&amp;
+x\left(\df{1}{\sqrt{n^2-\frac{1}{4}}}\right)\pl_t\left(\df{w}{t^{\frac{1}{2}}
(n^2-\frac{1}{4})^\frac{1}{4}}\right)\\ &amp; +
\left[t^2\left(n^2-\df{1}{4}\right)-n^2\right]\left(\df{w}{t^{\frac{1}{2}}
(n^2-\frac{1}{4})^\frac{1}{4}}\right)=0\\ \Rightarrow &amp;
t^2\pl_t^2\left(\df{w}{t^{\frac{1}{2}}}\right)+t\pl_t\left(\df{w}{t^{\frac{1}{2}}}\right)
+\left[t^2\left(n^2-\df{1}{4}\right)-n^2\right]\left(\df{w}{t^\frac{1}{2}}\right)=0\\
\Rightarrow &amp;
t^2\pl_t\left(\df{t^{\frac{1}{2}}w'-\frac{1}{2}wt^{-\frac{1}{2}}}{t}\right)
+\left(t^{\frac{1}{2}}w'-\df{1}{2}wt^{-\frac{1}{2}}\right)\\ &amp;
+\left[t^2\left(n^2-\df{1}{4}\right)-n^2\right]\left(\df{w}{t^{\frac{1}{2}}}\right)=0\\
\Rightarrow &amp; t^2
\df{[t(\frac{1}{2}t^{-\frac{1}{2}}w'+t^{\frac{1}{2}}w''-\frac{1}{2}w't^{-\frac{1}{2}}
+\frac{1}{4}wt^{-\frac{3}{2}})-t^{\frac{1}{2}}w'+\frac{1}{2}wt^{-\frac{1}{2}}]}{t^2}\\
&amp;
+\left(t^{\frac{1}{2}}w'-\df{1}{2}wt^{-\frac{1}{2}}\right)+\left(t^2\left(n^2-\df{1}{4}\right)
-n^2\right)\left(\df{w}{t^{\frac{1}{2}}}\right)=0\\ \Rightarrow &amp;
t^{\frac{3}{2}}w''+\df{1}{4}wt^{-\frac{1}{2}}-t^{\frac{1}{2}}w'+\df{1}{2}wt^{-\frac{1}{2}}
+t^{\frac{1}{2}}w'-\df{1}{2}wt^{-\frac{1}{2}}\\ &amp; +
t^{\frac{3}{2}}\left(n^2-\df{1}{4}\right)w-n^2t^{-\frac{1}{2}}w=0\\
\Rightarrow &amp;
\un{w''+\left(n^2-\df{1}{4}\right)\left(1-\df{1}{t^2}\right)w=0}\end{array}$

$(t\ne 0)$

Phew!!

Now we have a WKB-type form of the equation. Obviously things go
wrong when $|t|=1$ (WKB-type potential $=0 \Rightarrow$ turning
point) so separate into $|t|&gt;1$ and $0 &lt; |t| &lt;1$.

$$w \sim e^{g_0^{(n)}\psi_0(t)+g_1^{(n)}\psi_1(t)+\cdots}$$

$\{g_r\}$ asymptotic sequence as $n to \infty$

$\psi_r(t)=O(1)$ for $t=O(1)$

Therefore

$(g_0\psi_0''+g_1\psi_1''+\cdots)+(g_0^2{\psi_0^2}'
+g_1^2{\psi_1^2}'+\cdots)+(2g)+\left(n^2-df{1}{4}\right)\left(1-\df{1}{t^2}\right)=0$

\un{Balance at $O(n^2)$}

$\Rightarrow g_0^2{\psi_0^2}'+n^2\left(1-\df{1}{t^2}\right) \sim
0$

$\Rightarrow g_0=n,\ {\psi_0^2}'=\df{1}{t^2}-1 \Rightarrow
\psi_0'=\pm \sqrt{\df{1-t^2}{t^2}}$ depends on whether $|t|&lt;1$ or
not.

\un{Next balance}:

$\begin{array}{cl} &amp; \pm n
\left[\left(\df{1-t^2}{t^2}\right)^{\frac{1}{2}}\right]'+n^2\left(\df{1}{t^2}-1\right)\pm
2ng_1\sqrt{\df{1-t^2}{t^2}}\psi_1'\\ &amp;
+\left(n^2-\df{1}{4}\right)\left(1-\df{1}{t^2}\right)=0\\
\Rightarrow &amp; \pm\df{1 \cdot
n}{2}\left(\df{1}{t^2}-1\right)^{-\frac{1}{2}}\left(-\df{2}{t^3}\right)\pm
2ng_1\sqrt{\df{1}{t^2}-1}\psi_1'-\df{1}{4}\left(1-\df{1}{t^2}\right)=0
\end{array}$

Must have balance at $O(n)$

$\Rightarrow
\df{n}{t^3}\left(\df{1}{t^2}-1\right)^{-\frac{1}{2}}=2ng_1\sqrt{\df{1}{t^2}-1}\psi_1'$

$\Rightarrow g_1=1(n^0)$

and $\begin{array}{rcl} \psi_1' &amp; = &amp;
\df{1}{t^3}\left(\df{1}{t^2}-1\right)^{-1}\\ &amp; = &amp;
\df{1}{t^3}\left(\df{1-t^2}{t^2}\right)^{-1}=\df{1}{2t(1-t^2)}\\
\psi_1 &amp; = &amp; \df{1}{2}\ds\int^t \df{ds}{s(1-s^2)}\\ &amp; = &amp;
\df{1}{2}\ds\int^t ds,\
\left(\df{1}{s}-\df{1}{2}\df{1}{(1+s)}+\df{1}{2}\df{1}{(1-s)}\right)\\
&amp; = &amp; \df{1}{2}\ln t-\df{1}{4}\ln(1+t)-\df{1}{4}\ln(1-t)
\end{array}$

$\Rightarrow
\psi_1=\ln\left[\left(\df{t^2}{1-t^2}\right)^{\frac{1}{4}}\right]$

Hence $w \sim A_\pm
\left(\df{t^2}{1-t^2}\right)^{\frac{1}{4}}\exp\left\{\pm in
\ds\int^t \sqrt{\df{s^2-1}{s^2}}\,ds\right\}$ $(A_\pm const)$

The $i$ has been pulled out of the $\sqrt{}$ to help what follows.
We need to evaluate the integral:

$\begin{array}{rcl}\ds\int\sqrt{\df{s^2-1}{s^2}}\,ds &amp; = &amp;
-\ds\int\df{\sin u}{\cos^2 u}\sqrt{\df{\frac{1}{\cos^2
u}-1}{\frac{1}{\cos^2 u}}}\,du\ \ s=\df{1}{\cos u}\\ &amp; = &amp;
-\ds\int \df{\sin^2 u}{\cos^2 u}\,du\\ &amp; = &amp; -\ds\int
\left(\df{1}{\cos^2 u}-1\right)\, du\\ &amp; = &amp; -\ds\int \sec^2u
\,du+u\\ &amp; = &amp; u-\tan u\\ &amp; = &amp;
\arccos\left(\df{1}{s}\right)-s\sqrt{1-\df{1}{s^2}}\\ &amp; = &amp;
\arccos\left(\df{1}{s}\right)-\sqrt{s^2-1} \end{array}$

Hence $w \sim A_\pm
\left(\df{t^2}{1-t^2}\right)^{\frac{1}{4}}\exp\left\{\pm
in\left[\arccos\left(\df{1}{t}\right)-\sqrt{t^2-1}\right]\right\}$

replace original variable $x=t\sqrt{n^2-\df{1}{4}},\
y=\df{w}{\sqrt{x}}$

$\begin{array}{rcl} y &amp; \sim &amp;
\df{A_\pm}{\sqrt{x}}\left(\df{x^2}{(n^2-\frac{1}{4})(1-\frac{x^2}{n^2-\frac{1}{4}})}\right)
^\frac{1}{4}\\ &amp; &amp; \exp\left\{\pm
in\left[\arccos\left(\df{x}{\sqrt{n^2-\frac{1}{4}}}\right)-\sqrt{\df{x^2}{n^2-\frac{1}{4}}-1}
\right]\right\}\\ &amp; \sim &amp;
\df{A_\pm}{\sqrt{x}}\left(\df{x^2}{n^2-\frac{1}{4}-x^2}\right)^{\frac{1}{4}}\\
&amp; &amp;  \exp\left\{\pm in
\left[\arccos\left(\df{x}{\sqrt{n^2-\frac{1}{4}}}\right)-\df{\sqrt{x^2-n^2+\frac{1}{4}}}
{\sqrt{n^2-\frac{1}{4}}}\right]\right\}\end{array}$

so as $n\to+\infty,\ n^2-\df{1}{4}\sim n^2$

Therefore

$y \sim
\df{A_\pm}{\sqrt{x}}\left(\df{x^2}{n^2-x^2}\right)^{\frac{1}{4}}\exp\left\{\pm
i\left[n
\arccos\left(\df{x}{n}\right)-\sqrt{x^2-n^2}\right]\right\}\ \ n
\to +\infty$

as required ($A_\pm e^{-\frac{i\pi}{4}}A_\mp$ of question)


for $|t|&lt;1$ use $w \sim B_\pm
\left(\df{t^2}{t^2-1}\right)^{\frac{1}{4}}\exp\left\{\pm
n\ds\int_1^t\sqrt{\df{1-s^2}{s^2}}\right\}\,ds$

with $s=\df{1}{\cosh u}$

$\Rightarrow y \sim
\df{B_\pm}{\sqrt{x}}\left(\df{x^2}{n^2-x^2}\right)\exp\left\{\pm
\left[\sqrt{n^2-x^2}-n \rm{arccosh}
\left(\df{n}{x}\right)\right]\right\},$

$n\to+\infty$



\end{document}
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