Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2008 / A / 2

Here $s$ is downstream distance, $A$ the wetted cross-sectional area, $u$ the section-averaged downstream velocity, and $M$ the lateral water input per channel length and time. Thus $A_t$ is local water storage and $(Au{)}_s$ the divergence of discharge. Their balance is mass conservation.

The momentum terms $u_t$ and $u{u}_s$ are local and advective acceleration. The bed slope $S$ gives the downstream gravitational acceleration $gS$. Bed shear stress $\tau$ acting over wetted perimeter $l$ produces resisting force per unit mass $\tau l/(\rho A)$; the hydraulic radius is $R=A/l$. Finally $-g{\overline{h}}_s$ is the hydrostatic free-surface pressure-gradient term in the chosen bed-following description. The Saint-Venant approximation depth-averages the flow, assumes a long shallow geometry and hydrostatic pressure, and closes the boundary shear through a roughness law.

Manning's uniform-flow law implies the friction slope $S_f=n^2{u}^2/R^{4/3}$ for positive downstream flow. In a wide rectangular section $A\simeq wh$ and $R\simeq h$, so $\tau l/(\rho A)=g{n}^2{u}^2/h^{4/3}$.

Scale distance with $L$, depth with $h_0$, velocity with $u_0=h_0^{2/3}{S}^{1/2}/n$, and time with $L/u_0$. Divide momentum by $gS$ to obtainThe factor $ϵ$ compares depth change over $L$ with the mean bed drop $SL$, rather than with $L$ alone. Its product with $F^2$ measures inertial acceleration relative to bed-slope forcing.

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.

The physically meaningful extension requires $h\ge 0$: negative net input removes water where wet and cannot keep removing nonexistent surface water where dry. Keeping the algebraic source $E(t)<0$ unchanged in a dry zone would instead produce negative depths, so the drying-front description implicitly includes this availability condition.

Use the same $H(t)$, increasing until $t^{\ast }$ and decreasing afterwards. While $H(t)>0$, there is one earlier wetting time ${\tau }_d(t)\in [0,t^{\ast }]$ such that $H({\tau }_d(t))=H(t)$. The boundary-emitted characteristic with this label has just exhausted its water. Its position is the drying front:The integrand is nonnegative because $H(r)\ge H(t)$ between these two equal-height times. At $t=t^{\ast }$ the interval collapses and $s_d=0$; as net input becomes negative the front moves downstream. Differentiating gives $s_d^{\prime }=-(10/9)E(t){\int }_{{\tau }_d(t)}^t[H(r)-H(t){]}^{-1/3}dr>0$, where endpoint singularities are integrable away from the turning-time limit.

The wet solution ahead of the front still follows unexhausted characteristics. With $S_0(t)=\frac{5}{3}{\int }_0^{t}H(r{)}^{2/3}dr$,where $0\le \tau <{\tau }_d(t)$ is determined by the same characteristic-position integral as in (iv). If cumulative loss reaches $H(t)=0$, all remaining initial-line water has also vanished and the surface is dry thereafter while $E<0$. If the total later loss is smaller, the formula applies for all later times with $H(t)>0$. This is the drying front in a rainfall-driven Manning kinematic wave.