= Solution
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> give
$$
\boxed{\begin{aligned}
h_t+(hu)_x&=\alpha|u|,\\
(bh)_t+(bhu)_x&=0,\\
(hu)_t+\left(hu^2+\tfrac12bh^2\cos\theta\right)_x&=bh\sin\theta.
\end{aligned}}
$$
The 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 are
$$
Dh+hu_x=\alpha|u|,\qquad
Db=-\frac{\alpha|u|}{h}b,\qquad
Du+b\cos\theta\,h_x+\frac h2\cos\theta\,b_x
=b\sin\theta-\frac{\alpha|u|u}{h}.
$$
The last term is the momentum required to accelerate newly entrained fluid. For variables $(h,u,b)$, the matrix of the spatial derivatives is
$$
A=\begin{pmatrix}
u&h&0\\
b\cos\theta&u&h\cos\theta/2\\
0&0&u
\end{pmatrix},
\qquad
\boxed{\lambda=u,\quad u\pm\sqrt{bh\cos\theta}.}
$$
Thus it is a strictly <hyperbolic system> for $h,b>0$ and $0\leq\theta<\pi/2$. 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 $\cos\theta\geq0$: allowing an overhanging channel with $\cos\theta<0$ would also destroy <hyperbolicity>.
Back to article page
