<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\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}

Solve the equation

$$y''+(1-\varepsilon x)y=0,\ y(0)=1,\ y'(0)=0,\ \varepsilon \to
0^+$$

by regular perturbation theory up to order $\varepsilon$. By
consideration of the relative size of the terms in the expansion,
show that the regular perturbation ceases to be an accurate
approximation for large value of $x$ such that
$x=O(\varepsilon^{-\frac{1}{2}})$. Show by substitution that when
$x=O(\varepsilon^{-\frac{1}{2}})$, the solution then behaves like
$y(x)=\cos\left(x-\df{\varepsilon
x^2}{4}\right)+O(\varepsilon^{-\frac{1}{2}})$. Deduce that the
solution cannot be regular over the full range of $0&lt;x&lt;+\infty$ as
$\varepsilon\to 0^+$.



\vspace{.25in}

{\bf Answer}

$y''+(1-\varepsilon x)y=0;\ y(0)=1,\ y'(0)=0 \ \ \varepsilon \to
0^+$

Try $y(x;\varepsilon)=y_0(x)+\varepsilon y_1(x)+O(\varepsilon^2)$

Therefore ${y_0}''+\varepsilon{y_1}''+(1-\varepsilon
x)(y_0+\varepsilon y_1)+O(\varepsilon^2)=0$

$\un{O(\varepsilon^0)}: y_0''+y_0=0 \Rightarrow y_0=A\sin x+B \cos
x$

Boundary conditions $\Rightarrow \begin{array}{rclcl}1 &amp; = &amp; A
\cdot 0+B &amp; \Rightarrow &amp; B=1\\ 0 &amp; = &amp; A \cdot 1+B\cdot 0 &amp;
\Rightarrow &amp; A=0\end{array}$

Therefore $y_0=\cos x$

$\un{O(\varepsilon^1)}: y_1''+y_1-xy_0=0 \Rightarrow
y_1''+y_1=x\cos x$

$y=CF+PI$

$y_{CF}=C\cos x+D\sin x$

\un{TRY} $y_{PI}=(\alpha^2+\beta x)\cos x+(\phi x^2+\gamma x)\sin
x$

Substitution of $y_{PI}$ in equation gives

$$2\alpha \cos x-2(2\alpha x+\beta)\sin x+2\phi \sin x+2(2\phi
x+\gamma )\cos x=x\cos x$$

Comparison of like terms gives

$$\alpha=0,\ \beta=\df{1}{4},\ \phi=\df{1}{4},\ \gamma=0$$

so $y_{PI}=\df{1}{4}x\cos x+\df{1}{4}x^2 \sin x$

Therefore $y=C\cos x+\df{1}{4}x \cos x+\df{1}{4}x^2\sin x$.

Use boundary conditions to find $C$ and $D$:

$\left\{\begin{array}{ll} y_1(0)=0: &amp; 0=C\\ \undb{y_1'(0)=0}: &amp;
0=\df{1}{4}+D\end{array}\right\} \Rightarrow y_1=\df{1}{4}x^2\sin
x+\df{1}{4}x \cos x-\df{1}{4}\sin x$

from perturbation of boundary conditions $y'(0)=0$.

Thus $y=\cos x+\varepsilon\left[\df{1}{4}x^2 \sin x+\df{1}{4}x
\cos x-\df{1}{4}\sin x\right]+O(\varepsilon^2)$

Clearly when $\varepsilon x^2=O(1)$ the $\df{\varepsilon x^2}{4}
\sin x$ term amplitude has grown to be the same size as the $\cos
x$, initial term. Thus the oscillations from the \lq\lq small"
$\varepsilon$ correction are of the same size as the leading order
behaviour. Hence the perturbation series breaks down here, i.e.,
where $x=O(\varepsilon^{-\frac{1}{2}})$. Therefore it can't be
regular over $0&lt;x&lt;+\infty$.

Given $y=\cos\left(x-\df{\varepsilon
x^2}{4}\right)+O(\varepsilon^{\frac{1}{2}}),$

$y' \sim -\left(1-\df{\varepsilon x}{2}\right)\sin
\left(x-\df{\varepsilon x^2}{4}\right)$

$y'' \sim \df{\varepsilon}{2}\sin \left(x-\df{\varepsilon
x^2}{2}\right)-\left(1-\df{\varepsilon x}{2}\right)^2\cos x$ and
$y(0)=1,\ y'(0)=0 \Rightarrow y''+y(1-\varepsilon
y)=O(\varepsilon)$ as $\varepsilon \to 0$

so leading order \un{uniform} expansion.



\end{document}
</pre></body></html>