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\begin{document}


{\bf Question}

Let

$$\begin{array}{c} y''-x^a y=0\\ y \sim
\exp\{\phi_0(x)+\phi_1(x)+\phi_2(x)+\cdots \} \end{array}$$

where $\{\phi_n(z)\}$ is an asymptotic sequence as $x\to +\infty$.
Show that the dominant terms are

$$\begin{array}{ll} {\phi_0^2}'-x^a \sim 0, & a>-2\\
{\phi_0}''+{\phi_0^2}'-x^a \sim 0, & a=-2\\ \phi_0''-x^a \sim 0, &
a<-2 \end{array}$$

Hence find the leading terms in the asymptotic expansions in each
case.

\vspace{.25in}

{\bf Answer}

$y''-x^a y=0,\ y \sim e^{\{\phi_0+\phi_1+\phi_2+\cdots\}},\
\{\phi_r\}$ asymptotic sequence as $r \to \infty$.

Substitute to get:

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

$+\left(\begin{array}{lclcl} 2\phi_0'\phi_1' & + & 2\phi_0'\phi_2'
& + & \cdots\\ & + & 2 \phi_1'\phi_2' & + & \cdots\\ & & & + &
\cdots\end{array}\right)=x^a$

($\phi_0'\phi_r'=o({\phi_0'}^2$) etc.
$\phi_1'\phi_r'=o(\phi_0'\phi_r'$) etc.)

Focus on this as dominant balance.

\begin{enumerate}
\item
$\un{\rm{If}\ \phi_0'' \sim x^a}\ (x \to +\infty) \Rightarrow
\phi_0' \sim \df{x^{a+1}}{a+1},\ \phi_0 \sim
\df{x^{a+2}}{(a+1)(a+2)}$

Consider size of ${\phi_0'}^2 = O(x^{2a+2})$

For this to be consistent we therefore need

$x^{2a+2}=o(\phi_0''\ \rm{or}\ x^a)=o(x^a)$

Therefore $2a+2<2\ \rm{as}\ x \to +\infty$

$\Rightarrow \un{a<-2}$

\item
$\un{If\ {\phi_0'}^2 \sim x^a}\ (x \to +\infty) \Rightarrow
\begin{array}{l}\phi_0' \sim \pm x^{\frac{a}{2}}\\ \phi_0 \sim
\df{\pm x^{\frac{a}{2}+1}}{(\frac{a}{2}+1)} \end{array}$

For this to be consistent we need

$$\phi-0''=o(x^a)$$

$\Rightarrow \df{\pm
(\frac{a}{2}+1)(\frac{a}{2})}{(\frac{a}{2}+1)}x^{\frac{a}{2}-1}=o(x^a)$

$\Rightarrow \df{a}{2}-1<a\ \ x \to +\infty$

$\Rightarrow \un{a>-2}$

\item
If all three are the same size,

$$\phi_0''+{\phi_0'}^2=x^a$$

Suppose $\phi_0'=\alpha x^b$,\ say,\ $\alpha,\ \beta$ consts

$$\Rightarrow \phi_0''=\alpha b x^{b-1}$$

Therefore need $\alpha b x^{b-1}+\alpha^2 x^{2b}=x^a\ \ x \to
+\infty$

or $b-1=2b=a$ balancing powers $\Rightarrow b=-1,\ a=-2$

\un{Also} need $\alpha v+\alpha^2=1 \Rightarrow
\alpha^2-\alpha-1=0$

$\alpha=\df{1\pm \sqrt{5}}{2}$

$\Rightarrow \phi_0=\left(\df{1 \pm\sqrt{5}}{2}\right)\log x$
\end{enumerate}

\un{So summary}:

$\left.\begin{array}{lcl} (i)\ {\phi_0'}^2 \sim x^a & \rm{if} &
a>-2\\ (ii)\ \phi_0''+{\phi_0'}^2 \sim x^a & \rm{if} & a=-2\\
(iii)\ \phi_0'' \sim z^a & \rm{if} & a<-2
\end{array}\right\} \Rightarrow$

$x=\left\{\begin{array}{c} \pm
\df{a^{\frac{a}{2}+1}}{(\frac{a}{2}+1)}\\ \left(\df{1\pm
\sqrt{5}}{2}\right)\log x\\ \df{x^{a+2}}{(a+1)(a+2)}
\end{array}\right\} \Rightarrow y=\left\{\begin{array}{c} e^{\pm
\df{a^{\frac{a}{2}+1}}{(\frac{a}{2}+1)}}\\ x^{\frac{1\pm
\sqrt{5}}{2}}\\ e^{\frac{x^{a+2}}{(a+1)(a+2)}}\end{array}\right\}\
\ x\to +\infty$

Note that when $a<-2$

$$y \sim e^{\frac{x^{a+2}}{(a+2)(a+1)}}\ \rm{but}\ a+2<0$$

so $y \rightarrow const$ as $x \to +\infty$

When $a=-2,\ y \sim x^{\frac{1\pm \sqrt{5}}{2}}$, algebraic growth
(modified at higher orders by an exp. which $\rightarrow const$ as
$x \to +\infty$)

\un{Question}: Where's the second solution for (iii)?

\un{Hint}: Think higher order $\cdots$



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