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

Take the spatial Fourier transform of the causal Green function. The transformed equation is ${\hat{G}}_t=(\mu -iUk-\gamma k^2)\hat{G}+\delta (t)$, with zero response for $t<0$. Integration across the impulse sets $\hat{G}(k,0^{+})=1$, soInverting the Gaussian Fourier transform givesThis solves the homogeneous equation for $t>0$ and tends to the Dirac delta distribution as $t\downarrow 0$. Its spatial integral is $e^{\mu t}$, as required by the zero-wavenumber growth law.

Along a ray $x=vt$, the exponential growth rate isThe impulse grows in the moving interval of rays $U-2\sqrt{\gamma \mu }<v<U+2\sqrt{\gamma \mu }$, while its peak at $x=Ut$ grows as $t^{-1/2}{e}^{\mu t}$. At a fixed position,Thus, under the printed assumption $\mu >0$, $0<\mu <U^2/(4\gamma )$ gives convective wave-packet instability, because the growing disturbance is swept away from each fixed observation point. $\mu >U^2/(4\gamma )$ gives absolute wave-packet instability, with exponential growth at fixed position. At equality the fixed-position exponential rate is zero and the impulse decays algebraically as $t^{-1/2}$, even though its moving peak grows. The two bounding rays have the same algebraic prefactor.

These statements classify the Green function and localized initial disturbances propagated by its convolution. They do not assert that every possible initial condition decays at fixed position in the convective case: spatially uniform initial data instead produce $e^{\mu t}$. Persistent external forcing can likewise change the observed late-time response. The distinction between convective wave-packet instability and absolute wave-packet instability concerns localized disturbances in a specified frame.

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.