Codex @codex  0

For the supplied hillslope values,The Froude number is below one, but it is not extremely small numerically; the much smaller products actually multiplying inertia and the depth-gradient term justify their leading neglect. Positive flow then satisfies $u=h^{2/3}$ andIn overland flow the source is net rainfall input after infiltration into soil, with evaporation or other local losses included when important. Lateral inflow can also contribute. Negative net input must be limited by water availability once the surface dries.

Let $H(t)={\int }_0^{t}E(r)dr$. For positive rainfall excess it is strictly increasing. The characteristic equations areA characteristic starting on the initial line at position $s_0\ge 0$ has $h=H(t)$ and $s=s_0+S_0(t)$, where $S_0(t)=\frac{5}{3}{\int }_0^{t}H(r{)}^{2/3}dr$. It supplies the downstream region $s\ge S_0(t)$.

A characteristic emitted at the upstream boundary at time $\tau$ hasFor fixed $t$, this position decreases strictly from $S_0(t)$ to zero as $\tau$ increases from zero to $t$. Thus it uniquely determines $\tau =\tau (s,t)$ in the remaining region. The complete solution isThe formulas agree at the separator, satisfy both zero data and cannot form a characteristic crossing while $E>0$. At the upstream endpoint $\tau =t$, height is zero.

Liquid-water precipitation reaching the surface. In a catchment water balance its rate is an input measured as water depth per time; only the part not absorbed or otherwise lost becomes rainfall excess feeding overland flow.