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\begin{document}


{\bf Question}

Find two terms in the asmyptotic solutions of the equation

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

\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}

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

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

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

Substitute:

$\Rightarrow
(\phi_0''+\phi_1''+cdots)+({\phi_0^2}'+{\phi_1^2}'+\cdots)+(2\phi_0'\phi_1'+2\phi_0'\phi_2'+\cdots)+k^2\left(\df{x^2+2}{x^2+1}\right)^2=0$

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

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

Assuming $\{\phi_r''\}$ and $\{\phi_r'\}$ are asymptotic sequences
also,

$\phi_0''+{\phi_0^2}'+k^2 \sim 0$ is the first possible balance as
$x\to+\infty\ (\phi_0'\phi_1'=o({\phi_0'}^2))$ etc...

Checking through pairwise balances, the only sensible balance is:

${\phi_0'}^2=-k^2 \Rightarrow \un{\phi_0=\pm ikx},\
(\phi_0''=0=o(k)) \surd\surd$

Next balance

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

${\phi_1'}^1=o(\phi_0'\phi_1')$ by asymptotic sequence property

Therefore $\phi_1''+2\phi_0'\phi_1'+\df{2k^2}{x^2} \sim 0$ is next
possible balance

Checking through pairwise balances, the only sensible balance is

$$\pm 2i\phi_1' k \sim \df{-2k^2}{x^2}
\begin{array}{crcl} \Rightarrow &amp; \phi_1' &amp; = &amp; \pm \df{ik}{x^2}\
\
\left(\phi_1''=O\left(\df{1}{x^3}\right)=o\left(\df{1}{x^2}\right)\right)\\
\Rightarrow &amp; \phi_1 &amp; = &amp; \mp\df{ik}{x}+const\end{array}$$

absorb the constant in the exponential prefactor

Therefore

$y \sim A_\pm e^{\pm
ikx\mp\frac{ik}{x}}=A_{+}e^{ik(x-\frac{1}{x})}+A_{-}e^{-ik(x-\frac{1}{x})}$

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

Substitute

$\lefteqn(g_0\psi_0''+g_1\psi_1''+\cdots)+(g_0^2{\psi_0'}^2
+g_1^2{\psi_1'}^2+\cdots)+(2g_0g_1\psi_0'\psi_1'+2g_0g_2\psi_0'\psi_2'
+\cdots+2g_1g_2\psi_1'\psi_2'+\cdots)+k^2\left(\df{x^2+2}{x^2+1}\right)^2=0$

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

$$g_0^2{\psi_0'}^2+k^2\left(\df{x^2+2}{x^2+1}\right)^2=0$$

$\Rightarrow g_0=k$ and
${\psi_0'}^2=-\left(\df{x^2+2}{x^2+1}\right)^2$

Therefore $\psi_0'=\pm i\left(\df{x^2+2}{x^2+1}\right),$

$\psi_0=\pm i\ds\int^x \left(\df{1}{1+x^2}+1\right)\,dx = \pm i
\left(x+\arctan x\right)+c$

absorb $c$ into the exponential prefactor

\un{Balance at $O(k)$} is next

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

For balance must have

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

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

$\begin{array}{rcl}\Rightarrow \psi_1'=\df{-\psi_0''}{2\psi_0'} &amp;
= &amp; -\df{1}{2}\df{\mp i2x}{\pm i(\frac{x^2+2}{x^2+1})(x^2+1)^2}\\
&amp; = &amp; +\df{x}{(x^2+2)(x^2+1)}\\ &amp; = &amp;
\df{x}{x^2+1}-\df{x}{x^2+2}\end{array}$

$\Rightarrow \psi_1=\df{1}{2}\ln(x^2+1)-\df{1}{2}\ln(x^2+2)+const$

absorb constant into exponential prefactor

Therefore $y\sim A_\pm\exp\left\{\pm ki(x+\arctan
x)+\ln\sqrt{\df{x^2+1}{x^2+2}}\right\}$

Therefore

$\begin{array}{rcl}y &amp; \sim &amp;
A_{+}\sqrt{\df{x^2+1}{x^2+2}}\exp\left\{+ik[x+\arctan
x]\right\}\\&amp; &amp;
A_{-}\sqrt{\df{x^2+1}{x^2+2}}\exp\left\{-ik[x+\arctan
x]\right\}\end{array}$

\un{or}

$y\sim B_+ \sqrt{\df{x^2+1}{x^2+2}}\cos[k(x+\arctan x)]$

$\ + B_- \sqrt{\df{x^2+1}{x^2+2}}\cos[k(x+\arctan x)]$
\end{description}



\end{document}
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