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

For the broad right layer take $\zeta =(x-1)/\sqrt{\epsilon }$ and $y=Z(\zeta )$, so $\zeta \le 0$ inside the interval. Its leading equation is $Z^{\prime \prime }=\sqrt{Z}$, with $Z(0)=1$. One integration gives $(Z^{\prime }{)}^2=(4/3)(Z^{3/2}+c)$. The positive derivative branch produces the integral parametrization in the hint.

Matching to a small, nearly flat plateau requires $Z\to 0$ and $Z^{\prime }\to 0$ at the inner edge, so the leading integration constant is $c=0$. A positive order-one $c$ would leave a nonzero slope as $Z\to 0$; a negative one would prevent that limit. For $c=0$, integration with $Z(0)=1$ givesIts zero is at $\zeta =-2\sqrt{3}$, namely $x=1-2\sqrt{3\epsilon }$. Near that edge it has the fourth-power overlap obtained in solution. The neglected $\epsilon k$ reaction term becomes comparable to $\sqrt{Z}$ when $Z=O({\epsilon }^2)$, precisely where the thinner transition must restore it. The sketch in solution shows the plateau and both right-hand regions; extending the quartic profile past its inner edge would fail to match the positive plateau.