Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 71 / 1

Write $b=g(\rho -{\rho }_0)/{\rho }_0>0$ for the reduced gravity. The Boussinesq approximation requires $|\rho -{\rho }_0|\ll {\rho }_0$: replace density by ${\rho }_0$ in inertia and retain its difference in the gravitational force. A shallow-water approximation additionally requires a small aspect ratio, negligible leading-order normal acceleration and approximately hydrostatic pressure relative to the ambient fluid:Incompressibility makes the normal velocity smaller than the downslope velocity by the aspect ratio.

Thinness alone does not force the downslope velocity to be uniform across the layer. The additional top-hat plume model assumes that turbulent mixing makes both $u(x,t)$ and $b(x,t)$ approximately uniform inside it, with a sharp transition to the ambient fluid. Then the depth-integrated volume and momentum fluxes are $hu$ and $h{u}^2$. This is the same integral closure used for a turbulent plume, applied to a shallow layer; the hydrostatic pressure term is retained because gravity has a component normal to the channel.

Assume the ambient fluid has zero downslope velocity, no bed drag, and an entrainment coefficient $\alpha >0$. The entraining shallow-water layer receives volume at rate $\alpha |u|$, but neither excess buoyancy nor downslope momentum from that ambient fluid. Consequently volume conservation, conservation of excess buoyancy, and momentum conservation giveThe pressure contribution is the integral of the hydrostatic pressure from part (a). The last source is the downslope component of the gravitational force.

With $D={\partial }_t+u{\partial }_x$, the equivalent equations areThe last term is the momentum required to accelerate newly entrained fluid. For variables $(h,u,b)$, the matrix of the spatial derivatives isThus it is a strictly hyperbolic system for $h,b>0$ and $0\le \theta <\pi /2$. At $\theta =\pi /2$ all three characteristic speeds coincide and the matrix is not diagonalizable. The exclusion of that angle presumes the physical range with $\cos \theta \ge 0$: allowing an overhanging channel with $\cos \theta <0$ would also destroy hyperbolicity.

For the steady positive-velocity branch, conservation of buoyancy flux gives $bhu=F_0$. Substituting into the other two balances givesUse a power-law ansatz $h=C{x}^p$, $u=D{x}^q$. The entrainment balance forces $p=1$ and $C(1+q)=\alpha$. The momentum powers are $x^{2q}$, $x^{-q}$ and $x^{-q}$, so the nonzero inertial branch has $q=0$. Therefore the constant-speed entraining gravity current on a slope isIts positive speed requires $\tan \theta >\alpha /2$ for $\theta <\pi /2$. In the shallow, strictly hyperbolic range this meansThe same steady integral balances have a formal vertical-plume limit at $\theta =\pi /2$, although their time-dependent system then loses strict hyperbolicity. At the lower threshold the assumed positive-flux, finite-speed solution fails.

The line source is singular: $h,uh$ and the momentum flux tend to zero, but $b$ diverges as $x\downarrow 0$. Thus this is a downstream ideal-source solution, valid only far enough away that $b/g\ll 1$ and the Boussinesq approximation and top-hat plume model hold. It does not describe the immediate source region.