= Inner variable
{title2=$\xi$}
An <inner variable> is a stretched coordinate such as $\xi=(x-x_0)/\delta(\epsilon)$ that remains order one inside a thin region of width $\delta(\epsilon)$. Its scale is chosen by a <dominant balance>, rather than by the size of a coefficient alone.
For the <singular perturbation> problem $\epsilon y''+y'=0$ with $y(0)=0$, $y(1)=1$, set $X=x/\epsilon$ and $Y_\epsilon(X)=y_\epsilon(\epsilon X)$. The <chain rule> gives
$$
\frac{d}{dx}=\epsilon^{-1}\frac{d}{dX},\qquad
\epsilon y_\epsilon''+y_\epsilon'=\epsilon^{-1}(Y_\epsilon''+Y_\epsilon').
$$
Thus both <derivative> terms survive in the inner <differential equation> $Y_\epsilon''+Y_\epsilon'=0$. The exact domain is $0\leq X\leq1/\epsilon$, with
$$
Y_\epsilon(X)=\frac{1-e^{-X}}{1-e^{-1/\epsilon}},\qquad
Y_0(X)=1-e^{-X}.
$$
For bounded $X$, $Y_\epsilon\to Y_0$ as $\epsilon\to0^+$: the <inner expansion> has a nonconstant limiting profile even though the <outer solution> is constant. The finite-$\epsilon$ profiles nearly collapse onto $Y_0$ after rescaling, with their remaining difference coming from the normalization at the distant right endpoint.
\Image[/singular-perturbation-inner-rescaling-balance.png]
{title=Inner rescaling reveals the common profile and derivative balance}
{description=Left: the exact normalized profiles nearly coincide as functions of the <inner variable> $X=x/\epsilon$; the $\epsilon=0.2$ curve ends at $X=5$, its actual right boundary. Right: multiplication of the original <differential equation> by $\epsilon$ gives the signed terms $\epsilon y'=Y'$ and $\epsilon^2y''=Y''$, which cancel exactly.}
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