Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 345 / 2

The exact mixture equation of state is $\rho =(1-\varphi ){\rho }_f+\varphi {\rho }_p$. Keeping first-order terms in the small particle volume fraction and thermal expansion givesThe omitted term is ${\rho }_0\varphi \theta$. The downward solid-volume particle deposition flux is $J_v=W_s\varphi$; the corresponding particle mass flux is $J_m={\rho }_p{W}_s\varphi$. Uniform vertical mixing and constant layer depth implyFor $W_s>0$, set $r=W_s/H$. Thenand henceThe heated particle-laden layer loses its stable density stratification when this contrast vanishes. For $\beta >0$, solving for the neutral-buoyancy time givesThe layer is neutral at $T_s$ and statically unstable for $t>T_s$. The subsequent uniformly mixed lower-layer solution cannot represent the resulting overturning.

If $W_s=0$, then $\varphi ={\varphi }_0$, $\theta =\beta {\varphi }_{0}t$, and the continuous limit is $T_s=\gamma /\beta$. If $\beta =0$, the layer stays denser than its surroundings at every finite time and becomes neutral only asymptotically when $W_s>0$; thus $T_s=\infty$. These limiting cases must replace the printed formula when its denominator vanishes.

Use the shallow-water approximation, a hydrostatic vertically mixed layer, a deep motionless ambient, and no ambient fluid entrainment. Particle loss changes reduced gravity but changes total layer volume only at the discarded dilute-particle order. Likewise thermal expansion changes the equation of state while the leading Boussinesq approximation retains volume conservation. DefineDepth integration of mass conservation, horizontal momentum balance, particle transport, and the heating law givesHere the integrated excess hydrostatic pressure is $b{h}^2/2$. There is no particle source from the bed. Bed drag and the particle contribution to inertia are neglected at this order. The equivalent variable-buoyancy shallow water equations areThe term $h{b}_x/2$ is necessary when heating or particle loss makes the reduced gravity vary horizontally.

For $v=(h,u,\varphi ,\theta {)}^T$, the quasilinear system $v_t+A(v)v_x=S(v)$ hasIts characteristic polynomial factors asThus the characteristic curves have slopesOn the two repeated contact characteristic curves, the particle and heating laws are $D\varphi =-W_s\varphi /h$ and $D\theta =\beta \varphi$. Writing $c_g=\sqrt{bh}$ and $D_{\pm }={\partial }_t+(u\pm c_g){\partial }_x$, the gravity-wave compatibility relations areFor $h>0$ and $b>0$, gravity-wave right eigenvectors can be taken as $(1,\pm c_g/h,0,0{)}^T$. The contact eigenspace has dimension two: $\delta u=0$ and $b\delta h+(h/2)\delta b=0$, with independent $\delta \varphi ,\delta \theta$. Therefore $A$ has a full set of real eigenvectors.

The system is hyperbolic throughout strict static stability, but not strictly hyperbolic because the contact speed is repeated. At $b=0$ the eigenvalues coalesce and this diagonalizability is lost; when $b<0$ the gravity-wave speeds become complex, reflecting the loss of a statically stable lower layer.

Let $\mathcal{V}=H{L}_0$ be the layer volume per unit transverse width. The gravity-current box model imposes $hL=\mathcal{V}$; it approximates the front and scalar balances, rather than an exactly uniform solution of the local momentum equation. For a specified constant front Froude number $\mathrm{Fr}$, the heated particle-laden gravity current satisfiesThe initial data are $L=L_0$, $\varphi ={\varphi }_0$, and $\theta =0$. The gravity-current front condition applies only while $\gamma \varphi -\theta \ge 0$.

For heating without settling, $\varphi ={\varphi }_0$ and $\theta =\beta {\varphi }_{0}t$. Integration of the front equation givesHence the heated gravity-current runout isIt is reached at the neutral-buoyancy time $\gamma /\beta$ within this front model. A finite-momentum current could subsequently coast or lift off: the formula is the maximum predicted by the prescribed gravity-current front condition, which contains no independent front inertia.

For settling without heating, $\theta =0$. Set $K=\mathrm{Fr}\sqrt{g\gamma \mathcal{V}}$, so $\dot{L}=K\sqrt{\varphi }{L}^{-1/2}$. Eliminating time yieldsThus the runout length of a gravity current isThis length is approached as $t\to \infty$, because positive concentration cannot disappear at a finite time under the settling law. If distance means displacement from the removed barrier, the answer is $L_{\max }-L_0$ in either case.