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Cumulative mixing efficiency is the ratio of total irreversible density-mixing conversion to total irreversible mechanical-energy loss over an interval. It is a loss-rate-weighted average of instantaneous mixing efficiency, not an unweighted time average. Stored kinetic energy, changes in available potential energy and boundary work must be accounted for before replacing its denominator by integrated forcing.

For $A=U_0^2(1/L_f-1/L_{\rho })>0$, define $K_{\infty }=A{L}_v$ and $y=\sqrt{K/K_{\infty }}$. The positive-energy solution is $y(t)=[y_0+\tanh (\sqrt{K_{\infty }}t/(2L_v))]/[1+y_0\tanh (\sqrt{K_{\infty }}t/(2L_v))]$. Energies above and below $K_{\infty }$ approach it from their respective sides. The zero-energy equation is not Lipschitz and admits both permanent rest and delayed startup, so a unique positive steady limit needs a positive seed or an additional startup prescription. With negligible available and internal energy exchange, the long-time instantaneous mixing efficiency fraction is $L_f/L_{\rho }$, independent of $L_v$.

An instantaneous mixing efficiency defined as a fraction of irreversible mechanical-energy loss isThe cumulative mixing efficiency over $[0,t]$ isIt is weighted by the loss rate, rather than being the arithmetic time average of $\eta$. The flux coefficient $\Gamma =M/ϵ$ is also sometimes called mixing efficiency; it obeys $\eta =\Gamma /(1+\Gamma )$, and the corresponding cumulative coefficient is ${\Gamma }_c=\int M/\int ϵ$. These conventions should be stated rather than interchanged. Only when $\dot{A}$ is negligible does $M\simeq B$, permitting the frequently used buoyancy-flux form $B/(B+ϵ)$. Input-power efficiency $M/F$ is another quantity and generally differs during changes in stored energy.

With negligible internal-to-potential exchange and negligible change in available potential energy, the budgets give $M=B$. The model supplies $B=\sqrt{K}{U}_0^2/L_{\rho }$ and $ϵ=K^{3/2}/L_v$, so for $K(t)>0$ the instantaneous mixing efficiency isFor a positive initial seed and $L_f<L_{\rho }$, both loss rates approach positive constants. Their time integrals grow linearly at those rates, implyingIf mixing efficiency instead denotes the flux coefficient, then $\Gamma (t)=L_v{U}_0^2/[L_{\rho }K(t)]$ and ${\Gamma }_{c,\infty }=L_f/(L_{\rho }-L_f)$. During the transient, $\int (B+ϵ)dt=\int Fdt+K_0-K(t)$; equating irreversible loss with the instantaneous forcing would omit changing kinetic-energy storage. A permanently resting zero-seed branch has zero numerator and denominator, so it has no defined loss-based efficiency.

Keep the forcing ratio $r=L_f/L_{\rho }$ fixed in $(0,1)$. ThenFor $L_v\ll L_{\rho }$, dissipation is strong at a given $K$ and $K_{\infty }\ll U_0^2$ when $r$ is not extreme. If the initial energy is of order $U_0^2$, it decays toward a small steady value and starts with a small instantaneous mixing efficiency. This is a regime of short velocity-fluctuation scales and efficient viscous loss relative to the modeled buoyancy conversion at that initial energy.

For $L_v\gg L_{\rho }$, dissipation is weak at that same energy and $K_{\infty }\gg U_0^2$. A moderate initial energy grows and initially has a large instantaneous mixing efficiency; increasing stored energy eventually makes dissipation comparable with the remaining forcing. The energy-relaxation time is of order $L_v/\sqrt{K_{\infty }}$, which grows as $\sqrt{L_v}$ at fixed forcing and density lengths.

Within this closure, both limits have the same long-time loss fraction ${\eta }_{c,\infty }=L_f/L_{\rho }$. The steady energy changes with $L_v$ so as to compensate for its explicit effect on dissipation. Different initial conditions and finite observation windows give different transient efficiencies; one cannot infer a universal long-time efficiency from $L_v/L_{\rho }$ alone. In a real experiment the length scales need not remain independent of the changing flow, so this conclusion belongs to the stated model.