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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.