 Solution

Use the real diffusive interpretation $\gamma >0$, $U,{\mu }_0\in \mathbb{R}$, with $\nu >0$. Without a positive real diffusion coefficient the printed ordering inequality and real oscillator coordinate would need different hypotheses. Remove the drift by $\eta =e^{Ux/(2\gamma )}v$. Direct differentiation reduces the linear complex Ginzburg-Landau equation toSet $v=b(\xi )e^{-i({\omega }_0+ϵ{\omega }_1)t}$ and $\xi =\sqrt{ϵ}fx$. The resulting equation isCanceling its order-one coefficient and normalizing the quadratic term givesThe supplied decay quantization has $\lambda =2j+1$, $j=0,1,2,\dots$. Equivalently the Hermite functions $b_j=e^{-{\xi }^2/2}{H}_j(\xi )$ solve this harmonic-oscillator problem, where $H_j$ is the physicists' Hermite polynomial. Thus ${\omega }_1=-i(2j+1)\sqrt{\nu \gamma }$ and the exact temporal growth rates areThe global modes of a quadratically confined Ginzburg-Landau equation areFor every fixed positive $ϵ$, the negative quadratic spatial exponent dominates the linear drift exponent and the polynomial at both infinities; all these normal modes are spatially bounded. The largest growth rate is $s_0$, so the least-node normal mode is temporally non-growing, and all normal modes are non-growing, whenEquality is a neutral ground-state normal mode; strict inequality gives decay. This proves the printed sufficient condition for a bounded stable global normal mode and identifies its stronger meaning: stability of the leading global normal mode. Existence of merely some stable higher-$j$ normal mode is not equivalent to the inequality, since sufficiently large $j$ gives $s_j<0$ even above the leading-mode threshold. In the quadratic-coefficient model the reduction and eigenvalues are exact, not just a first-order approximation to a more general profile.