= Solution
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 $\epsilon=K^{3/2}/L_v$, so for $K(t)>0$ the <instantaneous mixing efficiency> is
$$
\boxed{\eta(t)=\frac1{1+K(t)L_\rho/(L_vU_0^2)}
=\frac1{1+(L_\rho/L_f-1)y(t)^2}.}
$$
For a positive initial seed and $L_f<L_\rho$, both loss rates approach positive constants. Their time integrals grow linearly at those rates, implying
$$
\boxed{\lim_{t\to\infty}\eta_c(t)=\frac{B_\infty}{B_\infty+\epsilon_\infty}
=\frac{L_f}{L_\rho}.}
$$
If mixing efficiency instead denotes the flux coefficient, then $\Gamma(t)=L_vU_0^2/[L_\rho K(t)]$ and $\Gamma_{c,\infty}=L_f/(L_\rho-L_f)$. During the transient, $\int(B+\epsilon)\,dt=\int F\,dt+K_0-K(t)$; 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.
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