One-dimensional shallow-water balance law with a dry-state convention

Fix gravity $g>0$ and a stationary bed $b:\mathbb{R}\to \mathbb{R}$. A shallow-water state is a pair $U=(h,m)$ with depth $h\ge 0$ and momentum $m\in \mathbb{R}$. Its admissible wet/dry state set is $K=\{(h,m):h>0,m\in \mathbb{R}\}\cup \{(0,0)\}\mathrm{.}$ For $U\in K$, define velocity by $u=m\mathrm{/}h$ when $h>0$ and $u=0$ when $h=0$. Thus a dry state has both $h=0$ and $m=0$. Define the free surface by $\eta =h+b$ and the physical flux by $F(U)=\{\begin{array}{ll}(m,m^2\mathrm{/}h+g{h}^2\mathrm{/}2),& h>0,\\ (0,0),& U=(0,0)\mathrm{.}\end{array}$ For a continuously differentiable bed, the one-dimensional shallow-water balance law is $U_t+{\mathrm{\partial }}_{x}F(U)=(0,-gh{b}_x)\mathrm{.}$ The source is evaluated as zero at a dry state.

A discrete lake-at-rest profile with level ${\eta }_0$ has $m_i=0$ and $h_i+b_i={\eta }_0$ in every wet cell; cells with $h_i=0$ are dry and satisfy $b_i\ge {\eta }_0$.

The conserved variables are depth and momentum, not depth and velocity. The convention at $h=0$ makes both $m\mathrm{/}h$ and the physical flux well-defined. No entropy inequality, convergence result, or path product for discontinuous bed steps is included in this definition.