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\begin{document}


{\bf Question}

Find two terms in the asymptotic solutions of the equation

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

\begin{description}
\item[(a)]
for $x\to +\infty,\ k=O(1)$

\item[(b)]
for $k\to +\infty,\ x=O(1)$
\end{description}

\vspace{.25in}

{\bf Answer}

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

\begin{description}
\item[(a)]
\un{$x\to +\infty,\ k=O(1)$}:

Try $y \sim e^{\psi_0(x)+\psi_1(x)}$

$\phi_r\}=$ asymptotic sequence as $x\to +\infty$

Substitute:

$\Rightarrow
(\phi_0''+\phi_1''+\cdots)+({\phi_0'}^2+{\phi_1'}^2+\cdots)+(2\phi_0'\phi_1'
+2\phi_0'\phi_2'+\cdots)-\df{k^4(x^2+1)}{k^2+x^2} \sim 0,\ \ x \to
+\infty$

Now as $x \to +\infty,\ k=O(1)$

$\begin{array}{rcl} \df{k^4(x^2+1)}{k^2+x^2} & \sim &
k^4\left(1+\df{1}{x^2}\right)\left(1+\df{k^2}{x^2}\right)^{-1}\\ &
\sim &
k^4\left(1+\df{1}{x^2}\right)\left(1-\df{k^2}{x^2}+O\left(\df{1}{x^4}\right)\right)\\
& \sim & k^4
\left(1+\df{(1-k^2)}{x^2}+O\left(\df{1}{x^4}\right)\right)
\end{array}$

Assuming that $\{\phi_r''\}$ and $\{\phi_r'\}$ are asymptotic
sequences also, the first possible balance is

$$\phi_0''+{\phi_0'}^2-k^4 \sim 0$$

Checking through pairwise balances, the only sensible balance is

${\phi_0'}^2=k^4 \Rightarrow \phi_0\pm k^2,\ \phi_0=\pm
k^2x+const$

Absorb the constant into the exponential prefactor

($\phi_0''=o(k^4)\surd\surd$)

Next balance

$(0+\phi_1''+\cdots)+)k^4+{\phi_1'}^2+\cdots)+(2\phi_0\phi_1'+\cdots)-k^4-\df{k^4(1-k^2)}{x^2}
\sim 0$

${\phi_1'}^2=o(\phi_0'\phi_1')$ by a symmetric sequence property

Therefore $\phi_1''+2\phi_0'\phi_1'-\df{k^4(1-k^2)}{x^2} \sim 0$
is the next possible balance.

Checking through pairwise, the only sensible balance is

$\begin{array}{rcl} \pm 2 k^2\phi_1' & = & \df{k^4(1-k^2)}{x^2}\\
\Rightarrow \phi_1' & = & \pm \df{k^2(1-k^2)}{2x^2} \Rightarrow
\phi_1''=O\left(\df{1}{x^3}\right)=o\left(\df{1}{x^2}\right)
\surd\surd\\ \phi_1 & = & \mp \df{k^2(1-k^2)}{2x}\end{array}$

Therefore $y \sim A_{+} e^{+k^2 x-\frac{k^2(1-k^2)}{2x}}+A_{-}
e^{-k^2 x+\frac{k^2(1-k^2)}{2x}}$

\item[(b)]
\un{$k \to +\infty,\ x=O(1)$}:

Try $y \sim e^{g_0(k)\psi_0(x)+g_1(k)\psi_1(x)+\cdots}$

$\{g_r\}$ asymptotic sequence as $k\to \infty,\ x=O(1),\
\{\psi+r\}=O(1)$

Substitute

$(g_0\psi_0''+g_1\psi_1''+\cdots)+(g_0^2{\psi_0'}^2+g_1^2{\psi_1'}^2+\cdots)+(2g_0g_1\psi_0'\psi_1'+2g_0g_2\psi_0'\psi_2'
+\cdots)-\df{k^4(x^2+1)}{k^2+x^2}=0$

As $k \to +\infty,\ x=O(1)$

$\begin{array}{rcl} \df{k^4(x^2+1)}{k^2+x^2} & \sim &
\df{k^4(x^2+1)}{k^2}\left(1+\df{x^2}{k^2}\right)^{-1}\\ & \sim &
k^2(x^2+1)\left(1-\df{x^2}{k^2}+O\left(\df{1}{k^4}\right)\right)\\
& \sim & k^2(x^2+1)-x^2(x^2+1)+O\left(\df{1}{k^2}\right)
\end{array}$

\un{Balance at $O(k^2)$} (is most denominators as $k \to +\infty$)

$\ g_0^2{\psi_0'}^2-k^2(x^2+1) \sim 0$

$ \Rightarrow g_0=k$ and ${\psi_0'}^2=x^2+1$

$\psi_0'=\pm \sqrt{x^2+1} \Rightarrow \pi_0=\pm
\df{1}{2}[\rm{arcsinh} x+x\sqrt{1+x^2}]$


\newpage
\un{Next balance at $O(k)$}:

$(k\psi_0''+\cdots)+(k^2{\psi_0'}^2
+g_1^2{\psi_1'}^2+\cdots)+(2kg_1\psi_0'\psi_1'+\cdots)-k^2(x^2+1)+x^2\undb{(x^2+1)}
\sim 0$

\hspace{.4in} $=O(k^0)$ Therefore neglect

For balance must have:

$\ k\psi_0''+2kg_1\psi_0'\psi_1' \sim 0$

$\Rightarrow g_1=1(=k^0)$ and $\psi_0''+2\psi_0'\psi_1' \sim 0$

$\Rightarrow \psi_1'=\df{-\psi_0''}{2\psi_0'}=\mp
\df{1}{2}\df{\frac{1}{2}(x^2+1)^{-\frac{1}{2}}}{\sqrt{x^2+1}}
\cdot 2x=\mp \df{x}{2(x^2+1)}$

$\ \psi_1 =\mp\df{1}{4}\ln(x^2+1)$

Therefore

$\begin{array} {rcl} y & \sim & \df{A_+}{(x^2+1)^{\frac{1}{4}}}
\exp \left\{\df{1}{2} \rm{arcsinh}\
x+\df{x}{2}\sqrt{1+x^2}\right\}\\ & &
+A_{-}(x^2+1)^{\frac{1}{4}}\exp\left\{-\df{1}{2} \rm{arcsinh}\
x-\df{x}{2}\sqrt{1+x^2}\right\} \end{array}$
\end{description}



\end{document}
