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

The square-root reaction problem with nested boundary layers has a small positive plateau $y\sim {\epsilon }^2{k}^2$, an $O(\epsilon )$ left-end layer, and two distinct scales on the right. The broad right layer has width $O(\sqrt{\epsilon })$; its inner edge needs a thinner $O(\epsilon )$ transition to the plateau. A single right-end scale cannot describe both matches.

Here is a uniformly matched leading solution. Let $L(\xi )$ solveusing the increasing branch below $k^2$. DefineThen $\Phi (L(\xi ))=\Phi (0)-\xi$. For the right-hand branch, define $R(\xi )>k^2$ by $\Phi (R(\xi ))=\xi$; it tends to $k^2$ as $\xi \to -\infty$. These formulas follow from the first integral of a square-root reaction layer and are expanded in solution and solution.

For sufficiently small $\epsilon$, let $x_{\ast }=1-\epsilon \Phi ({\epsilon }^{-2})$. Then $R((1-x_{\ast })/\epsilon )={\epsilon }^{-2}$, so the right profile meets the prescribed order-one endpoint. The composite asymptotic expansion isThe subtraction removes the common plateau counted twice. Its boundary residuals are exponentially small: the right profile has almost reached $k^2$ at $x=0$, and the left profile has almost reached $k^2$ at $x=1$. In each active layer the opposite profile is exponentially close to its plateau, so the composite reproduces that layer's governing equation and the shared outer solution to the retained order.

The right transition location satisfies $x_{\ast }=1-2\sqrt{3\epsilon }+o(\sqrt{\epsilon })$. Outside its thin transition, the broad right layer iswhere the bracket is positive; it matches through the thin layer to ${\epsilon }^2{k}^2$, rather than continuing as a fourth power on the wrong side. The dominant balances and the end-region matches are detailed below.

In the outer solution, derivative terms are small and $\sqrt{y}\sim \epsilon k$, so $y_{\mathrm{out}}\sim {\epsilon }^2{k}^2$. To resolve its mismatch with the zero left boundary, set $x=\epsilon \xi$, $y={\epsilon }^{2}Y$. The original equation becomes exactly $Y^{\prime \prime }-\sqrt{Y}+k=0$. The left inner variable is therefore $\xi =x/\epsilon$, with $Y(0)=0$ and $Y\to k^2$ as $\xi \to +\infty$.

Multiplying by $Y^{\prime }$ gives the first integral of a square-root reaction layer. Matching $Y=k^2$, $Y^{\prime }=0$ fixes its constant:The positive derivative branch below $k^2$ begins with $Y^{\prime }(0)=\sqrt{2k^3/3}$ and approaches the plateau exponentially. Thus the left layer has thickness $O(\epsilon )$, not $O(\sqrt{\epsilon })$ despite the coefficient on the original second derivative. Its small amplitude is essential in the dominant balance.

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

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.