Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 74 / 5 / a / ii / Solution

Here the drift vanishes at the left endpoint. The outer equation is $x{y}_0^{\prime }+y_0=0$, giving $y_0=1/x$ after applying the right boundary value. The ordinary $x=O(ϵ)$ layer does not balance diffusion and drift: their balance selects $x=O(\sqrt{ϵ})$ instead.

For the square-root boundary layer at a vanishing drift, put $X=x/\sqrt{ϵ}$. The inner equation isMatching $1/x$ requires a leading inner amplitude of order ${ϵ}^{-1/2}$, 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:The right boundary value determinesIn the inner variable the exact representation becomesThe first term enforces $Y(0)=1$, while the second is zero there and matches $C/(\sqrt{ϵ}X)=C/x$ 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. The endpoint layer has width $\sqrt{ϵ}$ and an $O({ϵ}^{-1/2})$ interior peak, while its value at the endpoint remains one. This distinguishes it from the ordinary exponential layers in the two preceding cases.