Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 84 / 5

Let $P$ denote total gravitational potential energy, $P_b$ the background potential energy of the adiabatically sorted density field, and $A=P-P_b$ the available potential energy. Define $B$ positive for reversible transfer from kinetic energy to $P$, $ϵ$ as viscous kinetic-energy loss, $F$ as imposed mechanical work, $W$ as molecular transfer from internal energy to potential energy, and $M$ as the irreversible conversion of $A$ to $P_b$. With boundary contributions removed, the stratified mixing energy budget isFor an unforced flow set $F=0$. In an alternative notation, if ${\Phi }_d={\dot{P}}_b$ is the full diffusive increase of background energy and ${\Phi }_i=W$, then $M={\Phi }_d-{\Phi }_i$ and $\dot{A}=B+{\Phi }_i-{\Phi }_d$. This distinguishes irreversible density mixing from reversible buoyancy exchange. The internal reservoir obeys $\dot{I}=ϵ-W$, so summing the kinetic, potential and internal budgets gives $d(K+P+I)/dt=F$ as required by conservation of energy.

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.

Use the model's normalization in which $K$ has velocity-squared units: it is kinetic energy per fixed reference mass, or a correspondingly normalized total energy. LetFor $K>0$, setting $s=\sqrt{K}$ reduces the equation to $2\dot{s}=a-s^2/L_v$. Suppose first that $L_f<L_{\rho }$, so $a>0$. Put $s_{\infty }=\sqrt{a{L}_v}$, $y=s/s_{\infty }$, $y_0=\sqrt{K_0/(a{L}_v)}$ and $\gamma =s_{\infty }/(2L_v)$. Separating variables gives $\dot{y}=\gamma (1-y^2)$ and therefore the forced stratified mixing with square-root power input solutionThis expression satisfies $K(0)=K_0$ and the differential equation directly.

For $0<K_0<K_{\infty }$, one may equivalently write $y=\tanh [\gamma t+\mathrm{artanh}{y}_0]$; the energy increases monotonically to $K_{\infty }$. For $K_0>K_{\infty }$, $y=\coth [\gamma t+\mathrm{arccoth}{y}_0]$, and the energy decreases monotonically to the same value. $K_0=K_{\infty }$ is stationary. For a very large initial energy, the initial evolution is approximately $s=s_0/[1+s_{0}t/(2L_v)]$, until forcing becomes comparable with dissipation. For a very small positive seed, $s$ initially grows approximately linearly, before leveling off. Thus both paragraphs labelled (c) in the PDF are answered by the solution and its two branches.

The positive-seed condition is material. If $K_0=0$, the original equation also admits $K\equiv 0$, and its square-root right-hand side is not Lipschitz at zero. It admits delayed-start solutions $K=0$ for $t\le t_d$ followed by $K=K_{\infty }{\tanh }^2[\gamma (t-t_d)]$. Therefore convergence to the positive steady state is not guaranteed from exactly zero turbulence without a startup prescription. For completeness, if $a=0$ a positive seed decays as $K=K_0/[1+\sqrt{K_0}t/(2L_v){]}^2$. If $a<0$, it reaches zero in finite time, $t_{\ast }=2\sqrt{L_v/|a|}\arctan \sqrt{K_0/(|a|L_v)}$, and the nonnegative continuation is at rest.

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.