= Solution
The <square-root reaction problem with nested boundary layers> has a small positive plateau $y\sim\varepsilon^2k^2$, an $O(\varepsilon)$ left-end layer, and two distinct scales on the right. The broad right layer has width $O(\sqrt\varepsilon)$; its inner edge needs a thinner $O(\varepsilon)$ transition to the plateau. A single right-end scale cannot describe both matches.
Here is a uniformly matched leading solution. Let $L(\xi)$ solve
$$
L''=\sqrt L-k,\qquad L(0)=0,\qquad L(+\infty)=k^2,
$$
using the increasing branch below $k^2$. Define
$$
\Phi(Y)=\sqrt{2k}\log\left|\frac{\sqrt{k+2\sqrt Y}-\sqrt{3k}}{\sqrt{k+2\sqrt Y}+\sqrt{3k}}\right|+\sqrt{6(k+2\sqrt Y)}.
$$
Then $\Phi(L(\xi))=\Phi(0)-\xi$. 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 <b/i/solution> and <b/ii/solution>.
For sufficiently small $\varepsilon$, let $x_*=1-\varepsilon\Phi(\varepsilon^{-2})$. Then $R((1-x_*)/\varepsilon)=\varepsilon^{-2}$, so the right profile meets the prescribed order-one endpoint. The <composite asymptotic expansion> is
$$
\boxed{y_{\rm comp}(x)=\varepsilon^2\left[L\left(\frac x\varepsilon\right)+R\left(\frac{x-x_*}{\varepsilon}\right)-k^2\right],\qquad 0\leq x\leq1.}
$$
The 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 $x_*=1-2\sqrt{3\varepsilon}+o(\sqrt\varepsilon)$. Outside its thin transition, the broad right layer is
$$
y\sim\left[1-\frac{1-x}{2\sqrt{3\varepsilon}}\right]^4
$$
where the bracket is positive; it matches through the thin layer to $\varepsilon^2k^2$, rather than continuing as a fourth power on the wrong side. The <dominant balances> and the end-region matches are detailed below.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-79-nested-layers.png]
{title=The left layer, small plateau and two right-hand scales of the square-root reaction problem}
Back to article page
