= Solution
Here the drift vanishes at the left endpoint. The outer equation is $xy_0'+y_0=0$, giving $y_0=1/x$ after applying the right boundary value. The ordinary $x=O(\epsilon)$ layer does not balance diffusion and drift: their balance selects $x=O(\sqrt\epsilon)$ instead.
For the <square-root boundary layer at a vanishing drift>, put $X=x/\sqrt\epsilon$. The inner equation is
$$
Y''+XY'+Y=0,\qquad (Y'+XY)'=0.
$$
Matching $1/x$ requires a leading inner <amplitude> of order $\epsilon^{-1/2}$, so an inner expansion with an everywhere order-one leading term would be incorrect. This can be seen directly by integrating the original equation once:
$$
\epsilon y'+xy=C,\qquad
y(x)=e^{-x^2/(2\epsilon)}\left[1+\frac C\epsilon\int_0^x e^{s^2/(2\epsilon)}\,ds\right].
$$
The right boundary value determines
$$
C=\frac{\epsilon(e^{1/(2\epsilon)}-1)}{\int_0^1e^{s^2/(2\epsilon)}ds}=1+O(\epsilon).
$$
In the inner variable the exact representation becomes
$$
Y(X)=e^{-X^2/2}+\frac C{\sqrt\epsilon}e^{-X^2/2}\int_0^X e^{s^2/2}\,ds.
$$
The first term enforces $Y(0)=1$, while the second is zero there and matches $C/(\sqrt\epsilon X)=C/x$ for large $X$. One would expand this integral in the overlap region, determine successive outer terms from the right boundary, and subtract their common expansion. \b[The endpoint layer has width $\sqrt\epsilon$ and an $O(\epsilon^{-1/2})$ interior peak], while its value at the endpoint remains one. This distinguishes it from the ordinary exponential layers in the two preceding cases.
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