Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 76 / 5

Assume $\gamma >0$, the positive diffusivity requirement for the forward heat kernel. A normal mode $e^{ikx-i\omega t}$ of the linear complex Ginzburg-Landau equation givesFor real $k$, the maximum temporal growth rate is $\mu$. The Briggs-Bers criterion starts the inverse time Laplace transform above the temporal growth spectrum and follows the two spatial roots as its frequency contour is lowered. A contributing spatial pinch point occurs when roots continued from opposite sides of the spatial contour coalesce and prevent its deformation. An arbitrary algebraic saddle is not sufficient without this branch selection.

Here $D=D_k=0$ givesTo check the pinch, start at $\omega =i\Omega$ with $\Omega$ large. The roots are $k=-iU/(2\gamma )\pm i\sqrt{U^2+4\gamma (\Omega -\mu )}/(2\gamma )$ and initially lie in opposite half-planes. They meet at $\Omega =\mu -U^2/(4\gamma )$, giving the causal saddle. Positive imaginary absolute frequency means fixed-position exponential growth. ThereforeFor $\mu \le 0$ there is no positive temporal growth. At $\mu =U^2/(4\gamma )$ the fixed-position exponential rate is zero; its algebraic prefactor, derived below, decays. When $U=0$ the convective interval is empty.

The source specifies only real coefficients. For $\gamma <0$, temporal growth is unbounded at high real wavenumber, so the usual causal forward-diffusion construction is ill posed and the above criterion does not apply. For $\gamma =0$ the equation reduces to transport with amplification; its point impulse remains a moving delta, rather than a spreading heat kernel. These cases cannot be included by division by $\gamma$.

Select the causal Green function, zero for $t<0$; otherwise the impulse equation permits arbitrary homogeneous additions. Use the spatial Fourier transform convention $e^{-ikx}$ and the temporal Laplace transform boundary convention $e^{i\omega t}$. The transformed impulse equation isIf $\mu >0$, a literal real-frequency Fourier transform of the growing response need not exist. Its causal inverse instead starts on $\mathrm{Im}\omega =c>\max (\mu ,0)$ and is continued from there. For $t>0$ close the frequency contour below; its clockwise orientation and the pole residue $1/(-i)=i$ give $e^{-i\omega (k)t}$. For $t<0$ the upper closure contains no pole. The remaining inverse spatial transform is a Gaussian integral:The unit-mass Gaussian heat kernel tends to $\delta (x)$ at $t\downarrow 0$, giving the required impulse normalization.

At the advected center $x=Ut$, its amplitude is $e^{\mu t}/\sqrt{4\pi \gamma t}$. At a fixed spatial point,Thus $\mu >0$ permits growth of the moving wave packet, but it is convective hydrodynamic instability if the fixed-position rate is negative. A positive fixed-position rate is absolute hydrodynamic instability. At its equality threshold the $t^{-1/2}$ factor decays, explaining the marginal exponential classification. This directly verifies the physically selected Briggs-Bers criterion saddle rather than assuming its accessibility.

Write $\mathbf{c}=(U,V{)}^T$ and $\Gamma =\left(\begin{array}{cc}{\gamma }_{11}& {\gamma }_{12}\\ {\gamma }_{12}& {\gamma }_{22}\end{array}\right)$. A well-posed anisotropic diffusion problem requires $\Gamma$ to be symmetric positive definite: ${\gamma }_{11}>0$ and $\Delta ={\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}^2>0$. For the normal mode in the source the dispersion relation isThe group velocity vanishes when $\mathbf{c}-2i\Gamma \mathbf{k}=0$. ConsequentlyThe requested anisotropic absolute-instability threshold for positive diffusion isFor positive definite diffusion this is also sufficient: the anisotropic Gaussian heat kernel has exponential factor $\exp [\mu t-(\mathbf{x}-\mathbf{c}t{)}^T{\Gamma }^{-1}(\mathbf{x}-\mathbf{c}t)/(4t)]$ and a $t^{-1}$ prefactor, so its fixed-position growth rate is precisely $\mu -f$.

Let the prescribed speed be $q$ and let $d_{\min }\le d_{\max }$ be the positive eigenvalues of $\Gamma$. The Rayleigh quotient for ${\Gamma }^{-1}$ givesAbsence of absolute hydrodynamic instability for every direction therefore requiresThis is an upper bound on $\mu$, not a minimum required value. The source's last sentence reverses this monotonicity: reducing the growth parameter always helps, so there is no finite lower threshold guaranteeing absence. If it had asked for instability in every direction, the corresponding condition would be $\mu >q^2/(4d_{\min })$. Without the positive-definite assumption, negative diffusion directions make the forward problem ill posed; a singular diffusion tensor requires a separate transport analysis and does not support the displayed inverse-matrix formula.

Assume again positive diffusion $\gamma >0$. Remove the drift with $A(x,t)=e^{Ux/(2\gamma )-i\omega t}f(x)$. Differentiating explicitly cancels the first derivative and givesBalance the second derivative with the linear confinement by taking $x=O({ϵ}^{-1/3})$, so the requested exponent is $\sigma =1/3$. More precisely, putThe eigenvalue problem reduces to $f_{\xi \xi }=(\xi -c)f$. Its two solutions are Airy functions. The growing $\mathrm{Bi}(\xi -c)$ is excluded by the condition at infinity: its growth proportional to $e^{2{\xi }^{3/2}/3}$ dominates any linear-in-$x$ drift factor. The decaying $\mathrm{Ai}(\xi -c)$ remains admissible even after multiplication by $e^{Ux/(2\gamma )}$. The wall imposes $\mathrm{Ai}(-c)=0$, so $-c=z_n$, giving the Airy global modes of a linearly confined Ginzburg-Landau equation:The least negative zero, $z_1\simeq -2.338107$, gives the largest temporal growth rate. HenceAt equality the leading global eigenmode is neutral, and below it all these eigenmodes decay. In contrast, freezing the coefficients locally gives absolute hydrodynamic instability wherever ${\mu }_0-ϵ\lambda x>U^2/(4\gamma )$. Such a pocket exists iff ${\mu }_0>U^2/(4\gamma )$ and extends to $x_a=[{\mu }_0-U^2/(4\gamma )]/(ϵ\lambda )$. Thus global instability implies a locally absolutely unstable pocket, while the converse fails: a pocket shorter than $|z_1|(\gamma /(ϵ\lambda ){)}^{1/3}$ does not overcome the wall and confinement losses. The additional threshold is order ${ϵ}^{2/3}$, precisely the loss produced by the Airy function spatial scale.