Use $\rho=\epsilon r=\delta^2r$ in the supplied small-$\rho$ expansions. The outer approximation re-expanded in inner variables is
$$
\begin{aligned}
F\sim{}&
1-r^{-1/2}
+\delta\sqrt\pi(1-r^{-1/2})\\
&+\delta^2\left[
-\sqrt r+\log r+\log\epsilon
+\pi+\frac C4
-\left(\log\epsilon+\pi+\frac C4-1\right)r^{-1/2}
\right].
\end{aligned}
$$
This is exactly the large-$r$ re-expansion of the inner result in part 1. In particular, the identity
$$
\log\rho=\log\epsilon+\log r
$$
forces the <switchback term> $\epsilon\log\epsilon(1-r^{-1/2})$. The coefficient of the homogeneous $E_{3/2}$ contribution at outer order $\epsilon$ is fixed to $-2\sqrt\pi$, while the remaining inner homogeneous constant is fixed to $\pi+C/4-1$. Thus the two <matched asymptotic expansions> agree term by term in $1\ll r\ll\epsilon^{-1}$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 336 / 2 / 2 / 3 / Solution

A switchback term such as $\epsilon\log\epsilon$ is forced into a matched expansion when logarithms of a stretched coordinate split into parameter-dependent and local pieces in the overlap region.


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 Incoming links: Switchback term