\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}

Find WKB solutions for $x=O(1),\ k\to +\infty$

$$y''+\left(k+\df{1}{2}-\df{1}{4}x^2\right)y=0$$

\vspace{.25in}

{\bf Answer}

$y''+\left(k+\df{1}{2}-\df{1}{4}x^2\right)y=0,\ x=O(1),\
k\to+\infty$

Try ansatz

$y \sim \exp\left\{g_0(k)\psi_0(x)+g_1(k)\psi_1(k)+\cdots\right\}$

$\{g_r(k)\}$ form asmyptotic sequence as $k \to \infty$

$\{\psi_r(x)\}=O(1)$ for $x=O(1),\ k \to +\infty$

$y' \sim
(g_0\psi_0'+g_1\psi_1'+\cdots)\exp\{g_0\psi_0+g_1\psi_1+\cdots\}$

$y'' \sim
((g_0\psi_0''+g_1\psi_1''+\cdots)+(g_0\psi_0'+g_1\psi_1'+\cdots)^2)
\times \exp\{g_0\psi_0+g_1\psi_1+\cdots\}$

Substitute into equation and simplify

$(g_0\psi_0''+g_1\psi_1''+\cdots)+(g_0\psi_0'+g_1\psi_1'+\cdots)^2+k+\df{1}{2}-\df{1}{4}x^2=0$

Now by asymptotic sequence property, we can assume that dominant
behaviour is given by

$$g_0\psi_0''+\undb{g_0^2{\psi_0'}^2+2g_0g_1\phi_0'\psi_1'}+k+\df{1}{2}-\df{1}{4}x^2=0$$

\hspace{1in} $g_0g_1=o(g_0^2)$

so

$$\undb{g_0\psi_0''+g_0^2{\psi_0'}^2}+\undb{k+\df{1}{2}-\df{1}{4}}x^2=0$$

\ assume $g_0=o(g_0^2)$\ \ \ \ $x=o(k),\ 1=o(k)$ as $k\to\infty\ \
x=O(1)$

so

$g_0^2{\psi_0'}^2=-k \Rightarrow g_0=\sqrt{k},$

${\psi_0'}^2=-1$ put $\pm \sqrt{}$ ambiguities into $\psi_0$ for
convenience

$\psi_0'=\pm i$

$\psi_\pm ix+const$ absorb the constant into the exponential
prefactor

Thus we return to the above equation:

$$\sqrt{k} \cdot 0-k+2\sqrt{k}g_1(\pm
i)\psi_1'+k+\df{1}{2}-\df{1}{4}x^2=0$$

$$\pm 2i\sqrt{kg_1}\psi_1'=-\df{1}{2}+\df{1}{4}x^2$$

To obtain balance at $O(k^0)$ we therefore need

$g_1=k^{-\frac{1}{2}} \Rightarrow \begin{array}{rcl} \pm 2i\psi_1'
& = & -\df{1}{2}+\df{1}{4}x^2\\ \psi_1' & = & \df{1}{\pm
2i}\left(-\df{1}{2}+\df{1}{4}x^2\right)\\ \psi_1 & = & \df{1}{2\pm
2i}\left(-\df{x}{2}+\df{x^3}{12}\right)\\ & = &
pm\df{i}{4}\left(x-\df{x^3}{6}\right)\end{array}$

Next order balance is given by:

\ \ =0

$(\overbrace{g_0\psi_0''}+g_1\psi_1''+\cdots)+
(2g_0g_1\psi_0'\psi_1'+2g_0g_2\psi_0'\psi_2'+\cdots)$

$+(g_0^2{\psi_0^2}'+g_1^2{\psi_1^2}'+\cdots)+k+\df{1}{2}-\df{1}{4}x^2=0$

$0=k^{-frac{1}{2}}\psi_1''+2\psi_0'\psi_1'+2k^{\frac{1}{2}}g_2\psi_0'\psi_2'+{k\psi_0^2}'
+\undb{\df{{\psi_1^2}'}{k}}+k+\df{1}{2}-\df{1}{4}x^2$

\hspace{2in} must be negligible at $O(k^{-\frac{1}{2}})$

Balance at $O(k^{-\frac{1}{2}})$ only if

$0=k^{-\frac{1}{2}}\psi_1''+2k^{\frac{1}{2}}g_2\psi_0'\psi_2'$

$\Rightarrow g_2=\df{1}{k}$ and $0=\mp\df{i}{4}x\pm2i \psi_2'$

$\Rightarrow \psi_2'=\df{x}{8} \Rightarrow
\psi_2=\df{x^2}{16}+const$ absorb the constant into the
exponential prefactor

etc...

Drawing this together we see:

$\begin{array}{rcl} Y & \sim &
A\exp\left\{+ix\sqrt{k}+\df{i}{4}\left(x-\df{x^3}{6}\right)k^{-\frac{1}{2}}
+\df{x^2}{16k}+\cdots\right\}\\ & &
+B\exp\left\{-ix\sqrt{k}-\df{i}{4}\left(x-\df{x^3}{6}\right)
k^{-\frac{1}{2}}+\df{x^2}{16k}+\cdots\right\}\end{array}$


\end{document}