Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 336 / 1

PutOn the specified branch, $\sqrt{z^2-1}$ is positive for $z>1$ and negative for $z<-1$. The saddle points of the phase are therefore $z=\pm a$, withandApplying the method of steepest descent at the two simple saddles givesThis formula is valid while the pole and positive saddle remain separated by much more than their $O({\lambda }^{-1/2})$ saddle width.

The pole crosses the positive saddle whenBecause the original contour passes above the pole, deformation onto the steepest-descent contour contributeswhen $z_0>a$, or $0<\theta <{\theta }_{\ast }$. It contributes no residue when $z_0<a$, or ${\theta }_{\ast }<\theta <1$. Hence, away from the transition,The residue is exponentially oscillatory and $O(1)$, whereas each ordinary saddle contribution is $O({\lambda }^{-1/2})$.

At $\theta ={\theta }_{\ast }$ the pole and saddle coalesce, so the displayed saddle formula is singular and must not be used. Passing above the coincident point gives one half of the switched residue at leading order:If $|\theta -{\theta }_{\ast }|=O({\lambda }^{-1/2})$, a uniform saddle-point approximation with a nearby pole replaces the discontinuous switch by a complementary-error-function multiplier.

As $\theta \to 1^{-}$, $s\to 0$, $a\to \infty$, and the pole lies in the no-residue regime. Each saddle coefficient is $O(s^{1/2}{\lambda }^{-1/2})$, so the leading approximation tends to

For $\theta >1$, the stationary-point equation would requireso the two saddles leave the real axis. The contour can be deformed through the appropriate complex saddle or saddles, and their contributions become exponentially small or large according to the sign of the real part of $i\Phi$. The real-axis oscillatory saddle contributions present for $0<\theta <1$ therefore disappear across the turning point $\theta =1$. The admissible deformation and the pole's residue must still respect the prescribed branch cuts and the instruction that the contour pass above $z_0$.

At $\theta =1$, the two saddles have receded to infinity and the ordinary quadratic saddle approximation is nonuniform. One should first use an asymptotic expansion of the phase and amplitude for large $z$:soThe phase varies on the scale $z=O(\lambda )$ rather than in an $O({\lambda }^{-1/2})$ neighbourhood of a finite saddle. Rescaling $z=\lambda Z$ in the positive and negative tails, retaining the corresponding large-$z$ amplitude, and matching these tail integrals to the finite part of the deformed contour produces the leading approximation. This is an endpoint at infinity problem; a uniform calculation may equivalently begin from the $\theta <1$ saddle representation and take the coalescing-at-infinity limit.