Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 76 (Asymptotic Methods in Fluid Dynamics)

For $0<\lambda <1$ fixed, the denominator stays away from zero, so dominated convergence givesThis approximation remains appropriate on the left of the transition when $1-\lambda \gg ϵ$. To resolve the Lorentzian peak approaching an integration endpoint, let $\lambda -1=ϵ\tau$ with bounded $\tau$, and put $x=ϵX$. The cosine is $1+O({ϵ}^2{X}^2)$ in the contributing region. ThereforeThus the distinguished transition has width $|\lambda -1|=O(ϵ)$. For large negative $\tau$ this expression matches $1/(1-\lambda )$ to leading order; for large positive $\tau$ it gives the full interior-peak contribution.

For $\lambda >1$ away from the transition, the denominator has a narrow minimum at $x_{\ast }=1-1/\lambda$. Set $c_{\ast }=\cos (\pi x_{\ast }/2)=\sin (\pi /(2\lambda ))$ and $x-x_{\ast }=(ϵc_{\ast }/\lambda )s$. The slowly varying cosine may be frozen over the peak and the integration limits extended, givingThe small parameter controlling the left-end truncation near $\lambda =1$ is $ϵ/(\lambda -1)$, so this formula requires $\lambda -1\gg ϵ$ there.

For $\lambda \gg 1$, the peak approaches $x=1$, but its width is $ϵ\pi /(2{\lambda }^2)$ whereas its distance from that endpoint is $1/\lambda$. Their ratio is $O(ϵ/\lambda )$, and the relative variation of the cosine across the peak is also $O(ϵ/\lambda )$. Hence proximity to the endpoint does not destroy the approximation: the peak becomes narrower still. Equivalently, $s=\lambda (1-x)$ gives $J={\lambda }^{-1}{\int }_0^{\lambda }[(1-s{)}^2+{ϵ}^2{\sin }^2(\pi s/(2\lambda )){]}^{-1}ds$, with its peak at $s=1$. The large-$\lambda$ limit of the leading result is $J\sim 2/ϵ$.

Take the usual Heaviside step function, $H(x)=1$ for $x>0$ and $0$ for $x<0$. Its Fourier transform must be interpreted as a tempered distribution. For $a>0$, exponential damping gives an ordinary integral:Against any Schwartz function $\phi$, the first term is an approximate identity of total mass $\pi$: changing variables $k=au$ gives the limit $\pi \phi (0)$. For the second term, pair positive and negative $k$ near zero. The difference $\phi (k)-\phi (-k)=O(k)$ removes the singularity and gives the Cauchy principal value of $1/k$. The tails converge by dominated convergence. Since the damped functions also approach $H$ as tempered distributions, continuity of the Fourier transform yields the Fourier transform of the Heaviside step function:The printed plus sign is inconsistent with the printed $e^{-ikx}$ convention for this $H$. In fact $H^{\prime }=\delta$ requires $ik\hat{H}=1$; the printed expression would give $-1$. Its plus sign would instead be correct for $H(-x)$.

To find the Fourier transform of a principal-value reciprocal, exponential damping of its odd part givesThe integral vanishes at $k=0$ and its $k$-derivative is $a/(a^2+k^2)$, so it equals $\arctan (k/a)$. Taking the distributional convergence limit givesAn ordinary improper integral of $1/x$ across zero is not a substitute for the specified Cauchy principal value.

The locally integrable function $\log |x|$ defines a tempered distribution. Its distributional derivative is $\mathrm{PV}(1/x)$: perform integration by parts outside $(-a,a)$, and observe that the boundary contribution $\log a[\phi (a)-\phi (-a)]$ tends to zero. By the Fourier transform of a derivative and the preceding result,Division by $|k|$ needs a finite-part inverse-absolute-value distribution, since $1/|k|$ is not locally integrable at zero. One possible normalization isIt satisfies $k\mathrm{Pf}(1/|k|)=\mathrm{sgn}k$. The general solution of $kT=0$ is a multiple of the Dirac delta distribution: a test function vanishing at zero can be written $k\psi (k)$, so $T$ depends only on its value at zero. Consequently the Fourier transform of the absolute logarithm isChanging the finite-part normalization changes $C$. The printed reciprocal expression is meaningful with this interpretation, not as an ordinary function at $k=0$.

The barred integral in the PDF denotes a Cauchy principal value. With $p(x)=\mathrm{PV}(1/x)$ its left side is the convolution $p\ast f$, whose Fourier transform is $-i\pi \mathrm{sgn}(k)\hat{f}(k)$. The right side is $-1$ on $(-1,1)$ and zero outside; either direct integration or the Fourier transform shift rule givesThus, away from the irrelevant single point $k=0$ in the $L^2$ formulation,By the Fourier transform of the absolute logarithm, the difference $\log |x-1|-\log |x+1|$ has transform $2i\pi \sin k/|k|$; its two delta terms cancel. ThereforeIts endpoint logarithms are locally square integrable, and $f(x)=2/({\pi }^{2}x)+O(x^{-3})$ at infinity, so it belongs to L2 space. The Hilbert transform multiplier is invertible almost everywhere on this space, proving uniqueness there. Without a decay or function-space condition, one may also add a constant under the symmetric principal-value convention. If the reversed step function were used to retain the printed sign in part (i), this answer would acquire the opposite sign.

Write $K_0(z)=-F(z)\log z+G(z)$ locally, with analytic $F,G$ and $F(0)=1$. The coefficient of $\log z$ in the Modified Bessel differential equation requires $z^2{F}^{\prime \prime }+z{F}^{\prime }-z^{2}F=0$. If $F=\sum a_m{z}^m$, its recurrence is $m^2{a}_m=a_{m-2}$, with $a_1=0$. HenceThe analytic part is even: an odd-power term in $G$ would produce an uncancelled odd-power term under the same differential operator, and the recurrence removes all such terms. Insert a smooth even cutoff equal to one near zero. The smooth part and the exponentially decaying tails give no algebraic large-wavenumber terms after repeated integration by parts. Therefore the logarithmic singularities determine the requested terms. For $k\ne 0$,Thus the Fourier transform of the modified Bessel function K0 has expansionAn independent check follows from the integral representation of the modified Bessel function of the second kind. Integrating $e^{-|x|\cosh t}$ first and then setting $u=\sinh t$ gives $2{\int }_0^{\infty }du/(u^2+1+k^2)=\pi /\sqrt{1+k^2}$, with exactly these first two terms.

Use the Gaussian pole transition for a large gamma parameter. With $t=1+s/\sqrt{n}$ and $\lambda =n+i\mu \sqrt{n}+\nu +o(1)$, Taylor expansion givesThe upper semicircle is traversed from left to right, so $\int ds/s=-i\pi$ on the indentation. The minus sign in $dt/(1-t)$ therefore makes its contribution $+i\pi$. The remaining leading integral isAt $\mu =0$ its Cauchy principal value vanishes by oddness. Differentiating with respect to $\mu$ removes the pole, leaving the Fourier transform of a Gaussian:Integrating from zero givesThis is uniform for bounded real $\mu ,\nu$ (and extends analytically over bounded complex values where the local contour is continued consistently). Contributions away from the saddle are exponentially small after localization. The bounded shift $\nu$ enters at the next order, not the leading error function. For example, retaining the displayed cubic correction gives $\sqrt{2\pi /n}{e}^{-{\mu }^2/2}[\nu +(2+{\mu }^2)/3]$ as the next term. This also checks the constant and contour sign.

For positive integer $n$ and any positive integer $N$, the finite geometric series identitygives an exact decomposition along the indented contour. The polynomial terms have no pole and may be returned to the positive real axis; their Gamma function integrals areEquivalently, expansion about the endpoint $t=0$ in the Laplace transform integral gives the same coefficients. The resulting formal asymptotic expansion isThe ratio of consecutive terms is $r/\lambda$, so the infinite series is factorially divergent. It should be truncated, not treated as a convergent sum; the pole prescription controls the contribution beyond its algebraic orders. Although the source initially calls $n$ an integer, the ordinary lower endpoint integral requires $n>0$, as is already implicit in the large-positive-$n$ limit of part (i).

By Stirling's approximation, for fixed $a$,Apply this with $a=1/6,5/6$ and use $\Gamma (r+1)=r\Gamma (r)$. Their exponents add to one, and henceMore explicitly, the ratio is $1-5/(36r)+O(r^{-2})$. The exact ratio of neighboring terms isThus the Airy function expansion is factorially divergent, and optimal truncation retains about $|\sigma |$ terms.

The printed summation starts at $r=1$, but its leading $r=0$ term is missing. The Gamma reflection formula gives $\Gamma (1/6)\Gamma (5/6)=2\pi$, so $Y_0=1$. The correctly normalized dominant Airy function expansion starts at zero. This correction is essential for the exponentially small remainder in the next part; if the printed series is used literally, its remainder also contains the entire missing leading term.

Let $P(z)=1/(2\sqrt{\pi }{z}^{1/4})$, choose a consistent branch, and truncate the corrected series after $r=n-1$, with $n=|\sigma |+O(1)$. The leading factorial tail in part (iii) is represented by the same indented integral as in part (ii):Here the divergent sum means its lateral Borel summation, not ordinary summation. To see why the pole approximation controls this remainder, write $a_r={\sigma }^r{Y}_r$ and insert $\Gamma (r)={\int }_0^{\infty }{e}^{-v}{v}^{r-1}dv$ in the leading late terms. Summing the geometric tail creates $1/(1-v/\sigma )$; $v=\sigma t$ produces $I(\sigma ,n)$. The relative $O(r^{-1})$ corrections to the late coefficients give lower-order contributions on its saddle scale. More concretely, the Borel transform of the coefficient series isThe simple pole supplies the leading Stokes phenomenon; its weaker logarithmic singularity supplies the smaller corrections. This is the integral justification for retaining the factorial tail.

For $\arg \sigma =\varphi /\sqrt{|\sigma |}$, expand $\sigma =n+i\varphi \sqrt{n}+O(1)$. Part (i), with $\mu =\varphi$, now givesThis formula uses the upper-pole prescription and the upper Airy function switching ray: near $\arg z=2\pi /3$, take $\arg \sigma =3\arg z/2-\pi$ near zero. The coefficient of the subdominant exponential changes smoothly from zero to $i$ across this ray and has value $i/2$ on it. The angular transition width is $O(|\sigma {|}^{-1/2})=O(|z{|}^{-3/4})$. The conjugate lower ray has the conjugate contour sign, rather than the same $+i$ on both rays. This is error-function smoothing of a Stokes multiplier, not a discontinuity of the analytic Airy function. With the literal printed $r=1$ series, the missing $P(z)e^{\sigma /2}$ would invalidate this remainder formula.

The outer solution satisfies $y(x)[x+y(x)]=0$, so on a connected interior interval the nonoscillatory branches are $y=0$ and $y=-x$. Both happen to solve the differential equation exactly away from any endpoint mismatch. For a fixed positive endpoint coordinate $q$ ($q=a$ or $q=1$), introduce the inward distance $\xi =(x-a)/\sqrt{ϵ}$ on the left or $\xi =(1-x)/\sqrt{ϵ}$ on the right. The leading nonlinear endpoint layer with a quadratic reaction obeysMatching to zero fixes $E=0$, so $(Y^{\prime }{)}^2=-q{Y}^2-2Y^3/3$. It is negative for every sufficiently small nonzero $Y$. Thus no real nonconstant layer can approach zero: the zero outer branch requires $\alpha =\beta =0$, and then $y=0$ is an exact solution.

Matching to $-q$ fixes $E=q^3/6$. The first integral factors asConsequently an endpoint value $\eta$ can match this branch only if $\eta \le q/2$. For $-q<\eta <q/2$, there are two choices: a direct approach to $-q$, and an initial excursion to $q/2$ followed by a return to $-q$. Both are translations ofThe negative translation gives the direct layer; the positive translation gives the excursion. At $\eta =q/2$ the two coincide, with the maximum at the wall. For $\eta <-q$, the unique admissible direction is a monotone approach from below:where ${\xi }_0$ is fixed by $Y(0)=\eta$. At $\eta =-q$ only the constant leading layer matches at a finite inward coordinate; a pulse translated infinitely far from the wall is outside this endpoint-layer classification.

Thus the leading admissibility condition for the negative outer branch is $\alpha \le a/2$ and $\beta \le 1/2$. Under strict inequalities, each endpoint has two choices when its datum lies between $-q$ and $q/2$, and one choice below $-q$. Their independent combinations give one, two or four formal matched profiles. In particular, for $\alpha =\beta =0$ there are four negative-bulk profiles in addition to the exact zero solution, so uniqueness certainly cannot be inferred from the outer equation. The endpoint equalities and the constant-layer cases are degenerate; higher-order endpoint matching is needed to decide exact finite-$ϵ$ counts at these cutoffs. These conditions and counts concern leading families under the stated exclusion of interior layers and rapid oscillations.

A matched asymptotic expansion for the negative branch isThe following sketch shows the four choices for zero endpoint data, together with the exact zero solution. It plots leading composite profiles, not numerical claims of exact finite-$ϵ$ multiplicity.

Additional interior layers cannot be excluded merely by counting the outer roots. Freezing $x=q$ gives a homoclinic orbit to $-q$ with the same hyperbolic-secant pulse. Its location would require a further matching or solvability condition; it is not an arbitrary parameter in the original variable-coefficient problem. Indeed, put $z=y+x$. The exact equation and an energy identity areAn isolated order-one pulse with exponentially small tails on both sides would decrease this energy by an algebraic amount, whereas its two tail energies would be exponentially small. This obstructs a freely standing isolated pulse well away from the walls. It does not by itself classify oscillatory solutions or pulse interactions with endpoint layers, so a complete global uniqueness claim would exceed this leading construction.

When $a=\alpha =0$, the two outer roots meet at the left endpoint. Let $x={ϵ}^{p}X$, $y={ϵ}^{q}Y$. Balancing $xy$ with $y^2$ gives $p=q$, and balancing $ϵy^{\prime \prime }$ with them gives $1-q=2q$. The one-third-power scaling at a nonlinear turning endpoint is thereforeIts left condition is $Y(0)=0$. Matching to the negative branch requires $Y(X)+X\to 0$ and $Y^{\prime }(X)+1\to 0$ as $X\to \infty$; the right endpoint is then handled by the earlier $q=1$ layer. Write $Y=-X+u$. In the intermediate matching region a small correction satisfies $u^{\prime \prime }-Xu=0$, so $u=C\mathrm{Ai}(X)+D\mathrm{Bi}(X)$. The growing Airy function excludes $D$; the decaying Airy function supplies a permissible stable matching direction. In fact $Y=-X$ is an exact inner solution satisfying $Y(0)=0$, establishing existence of an inner match without solving a nonlinear shooting problem. Since $\mathrm{Ai}(0)\ne 0$, the wall condition excludes an infinitesimal nonzero decaying correction about this solution; it does not prove absence of other finite-amplitude inner solutions. The right wall still requires $\beta \le 1/2$ at leading order.

Matching instead to zero requires $Y\to 0$. Its linearized equation is $Y^{\prime \prime }+XY=0$, with solutions $\mathrm{Ai}(-X)$ and $\mathrm{Bi}(-X)$. Their large-$X$ forms have envelope $X^{-1/4}$ and phase $2X^{3/2}/3$; although their values decay, they generate rapid spatial oscillations in the overlap region. The imposed nonoscillatory assumption removes these nonzero tails. The exact $Y=0$ remains, and a globally zero outer branch again requires $\beta =0$. The PDF's integration hint uses an inverse hyperbolic tangent; the converted TeX's inverse tangent would give a different and incorrect primitive.

Put $T=ϵt$ and use the method of multiple scales with $x=x_0(t,T)+ϵx_1+\cdots$. At leading order, $x_{0,tt}+x_0=k$, so write $x_0=k-r(T)\cos (t+\varphi (T))$ with $r(0)=k$, $\varphi (0)=0$. The order-$ϵ$ equation isUsing ${\cos }^{2}t\sin t=(\sin t+\sin 3t)/4$, its fundamental sine coefficient is $-2r_T-cr(1-k^2-r^2/4)$, and its fundamental cosine coefficient forces ${\varphi }_T=0$. Removing these resonant forcings is the solvability condition in the method of multiple scales. Thus the shifted Van der Pol amplitude evolution isWith $a=k^2-1$ and $R=r^2$, solve $R_T=caR+c{R}^2/4$ by putting $W=1/R$. This givesThe uniformly valid leading approximation is $x(t)=k-r(ϵt)\cos t+O(ϵ)$ on bounded slow-time intervals for which $r$ remains bounded. The exact initial derivative is recovered by the first-order correction; the displayed leading expression already satisfies the leading initial data.

There is an important limitation to the source's unqualified time-range wording. Starting from $r(0)=k$, the amplitude decays for $k^2<4/5$, is constant at leading order for $k^2=4/5$, and grows for $k^2>4/5$. In the latter case the denominator vanishes atA bounded-amplitude multiple-scale expansion is uniform for $0\le T\le C<T_b$, not through its predicted large-amplitude breakdown. For the decaying case every fixed bounded slow-time interval is available; the leading unstable equilibrium at $k^2=4/5$ also has bounded amplitude over such intervals. The envelope singularity is a breakdown warning, not a proof that the exact original solution has precisely this blow-up time. The coefficient in the PDF is $ϵc$, correcting the converted TeX's duplicated $ϵ$.

Again put $T=ϵt$. To leading order the oscillator has form $x_0=k-r(T)\cos (t+\varphi (T))$, while the second variable has a slow part $Y(T)$. Averaging its equation over the fast period givesThe oscillatory forcing of the second variable produces only an order-$ϵ$ correction: its leading fast derivative is $-r(T)\cos t$, whose bounded primitive is $-r(T)\sin t$. Thus it creates no secular term.

The oscillator solvability condition is the same as in part (i), with $c$ replaced by $Y(T)$:Introduce accumulated slow timeThe shifted Van der Pol amplitude evolution therefore gives, for $a=k^2-1$,ConsequentlyAs before, uniformity for $t=O({ϵ}^{-1})$ means a fixed bounded slow-time interval on which the envelope stays bounded. For $k^2>4/5$ it fails when $S(T)$ reaches $\log (5-4/k^2)/(k^2-1)$, with limiting value $4$ at $k=1$. There is no bounded leading approximation covering all such slow times for arbitrary positive $k$.

Use the Burgers reduction of weakly nonlinear rightgoing acoustics, with $\theta =t-x$, $X=Mx$, ${\partial }_t={\partial }_{\theta }$ and ${\partial }_x=-{\partial }_{\theta }+M{\partial }_X$. The equation of state expands asAt order one, mass conservation and momentum conservation give $({\rho }_0-u_0{)}_{\theta }=0$ and $(u_0-p_0{)}_{\theta }=0$. For the pure rightgoing acoustic wave with no independent leading mean density or velocity shift, ${\rho }_0=u_0=p_0$. This is the background choice implicit in the requested traveling-wave expansion; additional mean components would modify its phase convention. The source's duplicated phrase about expanding $p$ should refer to expanding $\rho$ as well as $u$.

At order $M$, the mass conservation equation becomeswhereas the momentum conservation equation becomesIn the latter equation, the two nonlinear products ${\rho }_0{u}_{0,\theta }$ and $-u_0{u}_{0,\theta }$ cancel. Substitute the equation of state relation into the former equation and add the two equations to eliminate the first corrections. The result isThe same calculation gives $u_1-p_1=-(\gamma +1)p_0^2/4+C(X)$, so the first corrections may be chosen without an explicit secular factor of $\theta$. This is why the retarded-time description can remain uniform for $|\theta |=O(M^{-1})$, provided the leading profile and its relevant derivatives remain bounded over that range and the slow propagation interval.

This Inviscid Burgers equation can develop a shock, so the source's uniformity assertion needs a pre-shock qualification. If $p_0(\theta ,0)=F(\theta )$ and $a=(\gamma +1)/2$, its method of characteristics givesThe smooth construction ends when this denominator vanishes; if $\max F^{\prime }>0$, the first such distance is $X_s=[a\max F^{\prime }{]}^{-1}$.

For viscosity write $\beta =M\nu$ with bounded positive $\nu$. Since $u_{0,xx}=p_{0,\theta \theta }+O(M)$, the order-$M$ momentum conservation equation instead reads $(u_1-p_1{)}_{\theta }+p_{0,X}=\nu p_{0,\theta \theta }$. Addition now yields the viscous Burgers equationIts positive diffusion resolves the inviscid steepening rather than supporting an unqualified smooth inviscid expansion through shock formation.

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.