 Solution

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.