Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2009 / A / 2

The cross-sectional area $A(s,t)$ multiplied by the mean downstream speed $u(s,t)$ is the volume flux $Q=Au$. The term $A_t$ is local storage change per unit river length, $(Au{)}_s$ is net downstream export per unit length, and $M>0$ is lateral volume input per unit length per unit time, for example distributed tributary or runoff supply.

For a fixed reach $[s_1,s_2]$, incompressibility and constant density giveApply the fundamental theorem of calculus and localize the resulting integral identity to obtain $A_t+Q_s=M$. The density cancels because every term transports the same water; no sediment source is included in this balance.

To obtain a power-law river kinematic wave, let $P_b$ be the wetted perimeter and $I$ the small bed slope. A uniform-flow force balance per unit river length is $\rho gAI={\tau }_b{P}_b$. With a Chézy friction law ${\tau }_b=\rho f{u}^2$, it gives $u=(gI/f{)}^{1/2}(A/P_b{)}^{1/2}$. Thus the hydraulic radius determines the power of area.

For a channel family with $P_b=P_0{A}^q$, where $q<1$, we have $u=C{A}^{(1-q)/2}$, with $C=(gI/(f{P}_0){)}^{1/2}$. Set $m=(1-q)/2>0$ and $c=(m+1)C$. ThenFor example a broad fixed-width rectangular channel has approximately constant wetted perimeter, giving $m=1/2$; a geometrically similar channel family has $P_b\propto A^{1/2}$, giving $m=1/4$. The exponent is therefore a geometry-dependent consequence of the friction balance, not a universal arbitrary value fixed by Chezy's law alone.

At equilibrium $Q_s=M$, so $Q=Q_{\mathrm{in}}+Ms$ for an upstream boundary flux $Q_{\mathrm{in}}$. HenceWithout a specified upstream condition this integration constant must be retained. On a region of positive area, the characteristic speed is $Q^{\prime }(A)=c{A}^m>0$, increasing downstream along the equilibrium. Small disturbances move downstream, spread in space as this speed increases, and decrease in area amplitude in the manner derived next. They are transported away from a fixed location; the model has no diffusion giving instantaneous smoothing. Finite-amplitude compressive disturbances can steepen into shocks because the flux is convex, $Q^{\prime \prime }(A)=cm{A}^{m-1}>0$, so the small-disturbance picture is not a global no-shock assertion.

Taylor expansion gives $Q(A_{\mathrm{eq}}+a)=Q(A_{\mathrm{eq}})+c{A}_{\mathrm{eq}}^{m}a+O(a^2)$. Subtract the equilibrium equation and write $V(s)=c{A}_{\mathrm{eq}}^m(s)$ to obtain $a_t+(Va{)}_s=0$. Set $b=Va$. Since $V$ is time independent, $b_t+V{b}_s=0$. The travel coordinate for a conservative transport perturbation $\xi (s)={\int }_0^{s}V(r{)}^{-1}dr$ transforms this into $b_t+b_{\xi }=0$, and thereforeThe conserved quantity along a characteristic is the flux perturbation $Va$, explaining the diminishing area amplitude as $V$ grows.

Differentiate the equilibrium relation to get $A_{\mathrm{eq}}^{\prime }=M/(c{A}_{\mathrm{eq}}^m)=M/V$. Thus the arrival time is explicitlyFor zero upstream flux the integral remains finite, giving the limiting formula with its second term zero. However $V(0)=0$ then makes the displayed amplitude formula singular at the head; a uniformly small arbitrary area disturbance imposed exactly there is outside this linearization. A positive upstream flux avoids that degeneracy.