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

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.