 Solution

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.