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Adiabatic elimination replaces a rapidly relaxing variable by its instantaneous stable quasistatic value. If $\dot{R}=-\Gamma (t)R+P(t)$ with $\Gamma >0$, write $R=R_0+\delta R$, $R_0=P/\Gamma$. Then $\dot{\delta R}+\Gamma \delta R=-{\dot{R}}_0$. After the transient, $\delta R\simeq -{\dot{R}}_0/\Gamma$ when coefficients vary slowly compared with ${\Gamma }^{-1}$. This gives both the leading reduction and its time-scale error; rapid relaxation alone does not justify an additional Taylor expansion in another parameter.

Let $n=|\Psi {|}^2$. Two independent approximations are required for the reservoir reduction of a polariton condensate. First the exciton reservoir must follow the changing condensate number density rapidly: after its initial transient, its relaxation time $[{\gamma }_R+R_{R}n{]}^{-1}$ must be small compared with the local density-evolution time. A sufficient modewise condition is that the relevant number density frequencies and growth rates are much smaller than ${\gamma }_R+R_{R}n$; pump variations must be comparably slow. There is no reservoir diffusion in the given model. Adiabatic elimination then givesSecond, to obtain the particular cubic equation by expansion about the empty condensate, require $\epsilon =R_{R}n/{\gamma }_R\ll 1$. ThenKeeping the full denominator instead gives a saturable-gain equation, not exactly the requested cubic complex Ginzburg–Landau equation. A near-threshold condensate is one natural regime in which the second condition holds. Rapid reservoir relaxation alone does not justify this number density expansion.

Multiplying the condensate equation by $-i$ and inserting the expanded reservoir separates real gain from the conservative frequency shift:These are local functions if the pump is spatially varying. Their signs are important: the reservoir blueshift gives negative $s$ in the printed convention $+is\Psi$, while reservoir depletion subtracts from the effective interaction $g$. Positive rates and pump give $\beta >0$; the effective $g$ need not be positive merely because the original interactions are repulsive.

For reservoir kinetics ${\mathcal{R}}_t=-({\gamma }_R+R_{R}n)\mathcal{R}+P$, adiabatic elimination gives $\mathcal{R}=P/({\gamma }_R+R_{R}n)$. Expanding for $R_{R}n/{\gamma }_R\ll 1$ and inserting into the condensate equation gives a complex Ginzburg–Landau equation withFast reservoir response and weak depletion are separate assumptions. Without the number density expansion the gain is saturable, not cubic. The coefficients depend on position when the pump does.