\documentclass[a4paper,12pt]{article}
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\newcommand{\pl}{\partial}
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\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

Consider the equation

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

\begin{description}
\item[(a)]
Find the first few terms of the solutions as $x\to +\infty$ by the
method of dominant balance.

\item[(b)]
Show that one solution is exactly $x\exp(x)$, and demonstrate that
this is consistent with your answer to part (a).
\end{description}

\vspace{.25in}

{\bf Answer}

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

\begin{description}
\item[(a)]
Try ansatz $y \sim
e^{(\phi_0(x)+\phi_1(x)+\phi_2(x)+\cdots)}\{\phi_r(x)\}$

forming an asymptotic sequence as $x\to +\infty$

$y' \sim
\{\phi_0'+\phi_1'+\phi_2'+\cdots\}e^{\{\phi_0+\phi_1+\phi_2+\cdots\}}\
\ x\to +\infty$

$y'' \sim
\{(\phi_0''+\phi_1''+\phi_2''+\cdots)+(phi_0'+\phi_1'+\phi_2'+\cdots)^2\}
e^{\{\phi_0+\phi_1+\phi_2+\cdots\}}$

$x\to +\infty$

Substitute into equation and simplify

$(\phi_0''+\phi_1''+\phi_2''+\cdots)+(phi_0'+\phi_1'+\phi_2'+\cdots)^2=1+\df{2}{x}$

Assume $\{\phi_r''(x)\}$ and $\{\phi_r'(x)\}$ form an asymptotic
sequences as $x \to +\infty$ as well.

\un{First balance}

$$\phi_0''+{\phi_0'}^2=1$$

$\left\{\begin{array}{l}\phi_0\phi_r'=o({\phi_0'}^2)\\ \phi_r''=
o(\phi_0'') \end{array} \right.$ by asymptotic sequences.

The only balance which works is

$\begin{array}{rcl} {\phi_0'}^2=1 & \Rightarrow & \phi_0''=0
=o(1)\ \ x\to +\infty\\ \rm{Therefore}\ \phi_0'=\pm 1 &
\Rightarrow & \phi_0=\pm x+\undb{const} \surd\surd\end{array}$

\hspace{.5in}absorb into arbitrary prefactor of exponential.


\newpage
 \un{Second balance}

\hspace{.5in}$\phi_1'$
\begin{eqnarray*} (0+\phi_1''+\cdots)+(\overbrace{\pm
1}+\phi_1'+\cdots)^2 & = & 1+\df{2}{x}\\ \Rightarrow \phi_1''+1
\undb{\pm 2\phi_1'}+\phi_1^2+\cdots \& = &
1+\df{2}{x}\end{eqnarray*}

\hspace{.5in}$|2\phi_0 \phi_1| \gg \phi_1^2$ by asymp. sequence
assump.

$$\rm{Therefore}\ \phi_1'' \pm 2\phi_1' =\df{2}{x}$$

\newpage
The only balance which works is

$\pm 2\phi_1'=\df{2}{x} \Rightarrow \phi_1=\pm \log x$

$\left(\phi_1''=\mp \df{1}{x^2}=o\left(\df{1}{x}\right),\ x \to
+\infty\right)$

\un{Third balance}:

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

$\phi_0''$

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

$\phi_1''$\ \ \ \ ${\phi_0'}^2$\ \ $2\phi_0'\phi_1'$

By asymptotic sequence assumptions $\phi_2''=O(\phi_1'')=\mp
\df{1}{x^2}$

Thus we have

$$\mp \df{1}{x^2}+({\phi_1'}^2+\undb{2\phi_0'\phi_2'}+\cdots)=0$$

\hspace{1in}$\phi_1'\phi_2'=o(\mbox{all of this})$ so neglect it

Now we can't make much of a statement about whether ${\phi_1'}^2$
dominates $\phi_0'\phi_2'$. Why? $|\phi_0|>|\phi_1|>|\phi_2|$, but
$|phi_1^2|$ could be $>|\phi_0'\phi_2'|$ if $\phi_0',\ \phi_2'$ is
small enough, \un{or} $|{\phi_1'}^2|$ could be $<|\phi_0'\phi_2'|$
if $\phi_0',\ \phi_2'$ are \un{large} enough.

Thus we should keep \un{both}.

${\phi_1'}^2=\df{1}{x^2},\ \phi_0'=\pm 1$

Therefore $\mp \df{1}{x^2}+\df{1}{x^2} \pm 2,\ \phi_2'=0$

$\Rightarrow \phi_2'=0$ or $+\df{1}{x^2}$

$\Rightarrow \phi_2'=const$ or $-\df{1}{x}$

The constant value can be absorbed into the exp. prefactor so set
$const=0$ without loss of generality.

Therefore pulling everything together, we have

$y \sim A\exp(+x+\log x+0+o(\log x))+B\exp(-x-\log
x-\df{1}{x}+o\left(\df{1}{x}\right))\ \ x \to +\infty$

$\Rightarrow y \sim Axe^{x+o(\log
x)}+\df{B}{x}e^{-x-\df{1}{x}+o(\frac{1}{x})},\ A,\ B consts,\ x
\to +\infty$

\item[(b)]

Set $y=xe^x,\ y'=e^x(1+x),\ y''=e^x(2+x)$

Therefore $xy''=xe^x(2+x)=(x+2)y$

$\Rightarrow xy''-(x+2)y=0 \surd\surd$

So $y=xe^x$ \un{is} a solution. This is consistent with (a) as the
$\phi_2=0$ solution gives this behaviour as $x \to +\infty$. By
careful consideration of boundary data it can be established that
$\phi_r=0$ for all $r \geq 2$ is an asymptotic solution
$\Rightarrow$ an exact one as well.

\un{Moral}: asymptotic methods may give the exact result if they
truncate at some point.

($\Rightarrow$ their divergence is intimately linked to the
$\sum_{r=0}^\infty$ ($\infty$ series) representation)
\end{description}


\end{document}