Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 79 / 3 / b

Put $w=\sqrt{Y}$. The separated first integral of a square-root reaction layer gives a primitive $\Phi (Y)$ as defined in solution. One can verify it directly: above the plateau,while below it the derivative has the opposite sign. Accordingly the increasing left profile uses $\Phi (L)=\Phi (0)-\xi$, and the increasing right profile uses $\Phi (R)=\xi$ after choosing its translation constant.

As $w\to k$, the logarithm dominates:Hence the left profile approaches from below with $|\sqrt{L}-k|\propto e^{-\xi /\sqrt{2k}}$, while the right profile approaches from above with $|\sqrt{R}-k|\propto e^{\xi /\sqrt{2k}}$ as $\xi \to -\infty$. These are precisely the required two plateau matching conditions. The absolute value in the logarithm is necessary to use the same real primitive on both sides.

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 $x_{\ast }=1-2\sqrt{3\epsilon }$.