Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 76 / 4 / b / Solution

Retain the factor $ϵ$ in the last temporal phase, as printed in the original PDF. Let $a=U/(2\gamma )$ and put $\eta =e^{ax}b(\xi )e^{st}$ with $s=-i{\omega }_0-iϵ{\omega }_1$. The drift-removing substitution for a Dirichlet Sturm-Liouville problem cancels the first derivative:With $\xi =\sqrt{ϵ}fx$, this becomesChoose the positive scale $f=(\nu /\gamma {)}^{1/4}$. To leave an order-one spectral parameter, cancel the numerator's order-one term. ThenThe assumed decaying solutions of the Hermite differential equation require $\lambda =2j+1$, $j=0,1,\dots$. They are proportional to $e^{-{\xi }^2/2}{H}_j(\xi )$, with physicists' Hermite polynomials. Therefore the global modes of a quadratically confined Ginzburg-Landau equation haveFor every fixed positive $ϵ$, multiplying the Hermite-Gaussian by $e^{Ux/(2\gamma )}$ still gives decay at both infinities. The strongest mode is $j=0$, so the whole-line system is globally unstable precisely whenEquality is a neutral ground mode. This spectrum is exact for the stated quadratic profile, even though the notation organizes it as an expansion in $ϵ$.

The homogeneous absolute-instability threshold in part (a) is $\mu =U^2/(4\gamma )$. The global threshold approaches that value as the profile varies increasingly slowly, but at finite $ϵ$ the quadratic confinement adds the loss $ϵ\sqrt{\gamma \nu }$. A locally absolutely unstable patch at the profile maximum is necessary but must be strong and broad enough to overcome that loss. Merely having positive local temporal growth ${\mu }_0>0$, which can be only convective wave-packet instability, is insufficient for a growing decaying global normal mode. This connects the local impulse classification to the discrete global spectrum.