Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 336 (Perturbation Methods)

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.

Away from $x=1/2$, the outer solution is obtained by setting $\epsilon =0$:It already satisfies both endpoint conditions, but its derivative jumps at $x=1/2$. Introduce the inner variableand write $y=\epsilon Y(\xi )$. The inner equation isMatching to the outer cusp requires $Y\sim |\xi |$ as $|\xi |\to \infty$. The even solution isSubtracting the common part $\epsilon |\xi |$ gives the composite asymptotic expansionIts endpoint errors are exponentially small.

At $x=1/2$, the first and second derivatives from the two sides agree, but the third derivatives have opposite signs. The composite expansion is therefore $C^2$ but not $C^3$.

For $\sigma =-1$, the reduced first-order outer equation isThe condition at $x=0$ selectsIndeed $z_0\to 1$ as $x\to 0^{+}$, while $z_0\to 0$ as $x\to 1^{-}$. The boundary condition at $x=1$ must therefore be supplied by a boundary layer.

SetSince $x{\log }^{2}x=O({\epsilon }^2{X}^2)$ in this layer, the convection term is lower order. The leading inner equation and matching conditions areThusThe leading uniformly valid expression isNear $x=0$ diffusion regularizes the divergent derivative of the outer approximation on the thinner scale $x|\log x|=O(\epsilon )$, but the leading value there remains the already matched constant $1$.

For $\sigma =1$, the formal outer family isEvery nonzero member diverges as $x\to 1^{-}$, so bounded matching selects the outer solution $z_0=0$. Boundary layers are now required at both endpoints.

The $x=1$ layer still has width $O(\epsilon )$ and leading profileNear $x=0$, diffusion and the logarithmically vanishing convection balance on the thinner scaleAt leading order the reaction term is smaller there. Using the supplied first integral with $\mu ={\epsilon }^2$, defineThe left layer that equals $1$ at the endpoint and matches zero isConsequently a leading composite description isIts sketch has value $1$ at each endpoint, drops sharply to an almost-zero outer plateau just to the right of $x=0$, and rises through an $O(\epsilon )$ layer just before $x=1$.

The linear $\cos x$ mode has frequency squared $k^2-1$, so $k_c(0)=1$. To resolve the small frequency near threshold, writeThe leading equationand the initial data give

At $O({\epsilon }^2)$, the equation for ${\theta }_2$ has a forcing whose $\cos x$ component isThe Fredholm solvability condition removes this secular term and yields the slow amplitude equationIt has the conserved energyThe potential has maxima atA periodic orbit launched from $A(0)=1$, $A^{\prime }(0)=0$ exists only when that turning point lies inside the two maxima, namelyAt equality the orbit is the separatrix; below it, the assumed real periodic oscillation is lost. Therefore