 Solution

For $Y\gg k^2$, the first integral of a square-root reaction layer reduces to $Y^{\prime }\sim (2/\sqrt{3})Y^{3/4}$. Integrating gives $Y^{1/4}\sim \xi /(2\sqrt{3})$ after a possible translation, orFor the right profile $R$ this large-positive-$\xi$ limit matches a broader scale with $y=O(1)$. Since $y={\epsilon }^{2}R$ and $\xi =(x-x_{\ast })/\epsilon$, the overlap isThus the translated $O(\epsilon )$ layer lies between the outer plateau and the $O(\sqrt{\epsilon })$ boundary layer. The large-$Y$ primitive also gives $\Phi ({\epsilon }^{-2})\sim 2\sqrt{3}/\sqrt{\epsilon }$, establishing the leading location $x_{\ast }=1-2\sqrt{3\epsilon }$.