 Solution

Set $x=\delta X$ and write $y(x)=Y(X)$. The equation becomesA dominant balance must compare all three coefficients, rather than just the two derivative terms.

For $-1<\alpha <1$, balancing diffusion and drift gives $\delta ={ϵ}^{1/(\alpha +1)}$. Their common coefficient tends to infinity, so the reaction term is smaller and the leading inner equation isThis derivative is integrable at infinity, allowing an ordinary localized endpoint adjustment. Thus the thickness is ${ϵ}^{1/(\alpha +1)}$ in this range.

For $\alpha =1$, $\delta =\sqrt{ϵ}$ makes all three terms comparable. Keep all of them:One decaying solution is $e^{-X^2/2}$, and reduction of order gives the other. This is a square-root-thickness layer with all terms retained.

For $\alpha >1$, the first variation scale is $\delta =\sqrt{ϵ}$. Keep diffusion and reaction, and drop drift at this scale:This leading motion is oscillatory, not a monotone decaying boundary-layer profile. To describe localization one must also resolve the slower damping envelope. Removing the first derivative in the original equation bygivesNear the endpoint the wavelength is $O(\sqrt{ϵ})$, while the envelope changes on $x=O({ϵ}^{1/(\alpha +1)})$. Hence an oscillatory endpoint region is possible, but it has distinct wavelength and envelope scales; calling the diffusion-drift balance alone its leading equation would incorrectly discard the reaction term. This generalizes the oscillatory endpoint with a quadratically vanishing drift.

For $\alpha \le -1$, the diffusion-drift balance would require ${\delta }^{\alpha +1}=ϵ$, which gives no shrinking positive power scale. At every shrinking scale the singular drift dominates diffusion, so no regular finite-amplitude diffusion boundary layer of the preceding kind occurs. The coefficient itself is singular at zero; singular endpoint solutions are a different issue from a bounded boundary-layer correction.