Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 90 / 2

The shallow-water approximation assumes a horizontal length scale $\ell$ much greater than the current depth, $h/\ell \ll 1$. Vertical acceleration is then small compared with gravity and pressure is given by the hydrostatic approximation. A depth-averaged model also assumes that the horizontal velocity and excess mass density can be represented by nearly uniform cross-sectional values. Here the ambient fluid is deep or sufficiently undisturbed that its motion and interface-pressure perturbations can be neglected.

The Boussinesq approximation assumes $|\rho -{\rho }_0|/{\rho }_0\ll 1$: mass density differences are retained in gravitational forces but a common reference mass density is used for inertia. A dense bottom current has the positive reduced gravityThus the pressure excess over the ambient hydrostatic value at height $z$ within the current is ${\rho }_0{g}^{\prime }(h-z)$. We neglect bed drag, molecular diffusion and horizontal turbulent stress in the leading balances, but retain the prescribed turbulent entrainment across the interface. Negligible bed drag does not imply that entrainment has no effect on the current's velocity or mass density.

The occupied cross-sectional area and interface width areThe incoming volume flux per unit channel length is $h{w}_e$. Volume conservation therefore gives $A_t+(Au{)}_x=h{w}_e$. Ambient fluid carries no excess mass density, so integrating the excess-mass density balance gives $(A{g}^{\prime }{)}_t+(Au{g}^{\prime }{)}_x=0$. These two equations imply dilution of the reduced gravity, even though total integrated buoyancy is conserved.

The integrated horizontal hydrostatic pressure force, divided by the reference mass density, isThe prismatic walls contribute no axial component of pressure force. Since ambient fluid enters with zero horizontal velocity, it supplies no horizontal momentum flux. Thus the entraining shallow-water current in a triangular channel obeys the conservative equationsTo put them in primitive form, let $D={\partial }_t+u{\partial }_x$ be the material derivative. The volume equation gives $Dh+(h/2)u_x=w_e$. Subtracting $u$ times the volume equation from the momentum equation, and $g^{\prime }$ times it from the buoyancy equation, yieldsThe momentum source is the acceleration of newly entrained ambient fluid, not bed drag. The source in the reduced gravity equation describes mixing with ambient fluid of zero excess mass density. In particular, the rectangular-channel coefficients cannot be used: the triangular area produces both $h/2$ in the depth equation and $h/3$ in the buoyancy-gradient force.

For the state vector $q=(h,u,g^{\prime }{)}^T$, the primitive equations are $q_t+\mathsf{A}{q}_x=S$, withThe characteristic polynomial isConsequently the three characteristic speeds areFor $h>0$ and $g^{\prime }>0$, the hyperbolic system has three distinct real eigenvalues. The long-wave speed relative to the current is $c$, while $u$ transports a buoyancy/contact disturbance.

A left eigenvector for the material characteristic is $(0,0,1)$, givingFor $u\pm c$, left eigenvectors can be chosen asMultiplying $q_t+\mathsf{A}{q}_x=S$ by these rows gives the characteristic compatibility for an entraining triangular-channel current. With all differentials evaluated along the indicated characteristic,The source follows directly from ${\ell }_{\pm }S=-2u{w}_e/h\pm 2c{w}_e/h\mp 2g^{\prime }{w}_e/(3c)$ and $g^{\prime }/c=2c/h$.

Since $4dc=(2c/h)dh+(h/c)d{g}^{\prime }$, an equivalent form isCombining it with the material buoyancy equation also givesOnly when entrainment vanishes and $g^{\prime }$ is spatially constant do $u\pm 4c$ become constant Riemann invariants on these characteristic curves. Dropping the buoyancy differential in the general problem would lose one of the required transport couplings. The characteristic formulas with $1/c$ apply to the nondegenerate region $g^{\prime }h>0$; dry or neutrally buoyant limits require their conservative form instead.