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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 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.