Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 79 / 5

Let $\mathbf{U}=(u,v)$, $\zeta =v_x-u_y$ and $D/Dt={\partial }_t+\mathbf{U}\cdot \nabla$. The inviscid shallow water equations areTaking the vertical curl of the momentum equation, with constant $f$, gives $D\zeta /Dt=-(f+\zeta )\nabla \cdot \mathbf{U}$. Combining this with the depth equation provesThus shallow-water potential vorticity is materially conserved: changes in absolute vorticity are exactly compensated by stretching or compression of the layer depth.

The relation to a Bernoulli function presupposes steady flow. In that case define $B=g(h+H)+\frac{1}{2}|\mathbf{U}{|}^2$. The vector identity $(\mathbf{U}\cdot \nabla )\mathbf{U}=\nabla (\frac{1}{2}|\mathbf{U}{|}^2)+\zeta \hat{\mathbf{z}}\times \mathbf{U}$ givesSteady volume conservation allows a transport stream function with $h\mathbf{U}=({\psi }_y,-{\psi }_x)$. Hence $\hat{\mathbf{z}}\times (h\mathbf{U})=\nabla \psi$, soIn particular, $\mathbf{U}\cdot \nabla B=0$: the Bernoulli function is constant along each streamline. Where the flow permits a single-valued dependence on the stream function, the displayed derivative describes how that constant changes between neighbouring streamlines. For an unsteady flow, shallow-water potential vorticity remains conserved but the stated steady Bernoulli function relation does not generally hold.

Take $y$ from the first wall to the second, with $f>0$. Negligible transverse velocity and cross-channel geostrophic balance give $fu=-g{\partial }_y(h+H)=-g{h}_y$, since the bed elevation $H(x)$ is independent of $y$. Therefore the volume flux isFor positive downstream volume flux, the layer is deeper at the first wall: $h_1>h_2$. This calculation does not require zero shallow-water potential vorticity.

Here $h_j$ must denote the wall depth above the local bed. If the surface heights are instead measured from a fixed vertical datum, write them as ${\eta }_j$ and use $h_j={\eta }_j-H(x)$, giving $Q=g[({\eta }_1-H{)}^2-({\eta }_2-H{)}^2]/(2f)$. The compact formula thus uses a bed-relative meaning of the wall heights.

Material conservation of shallow-water potential vorticity carries $P=0$ from the reservoir into the channel. At leading order in a long, slowly varying channel, $v$ and $v_x$ are negligible, so $\zeta \simeq -u_y$. Zero absolute vorticity therefore implies $u_y=f$. Integration from $y=0$, followed by cross-channel geostrophic balance, gives the zero-potential-vorticity rotating channel flow:In particular, $u_2=u_1+fb$ and $h_2=h_1-(fb/g)(u_1+fb/2)$. The velocity increases towards the shallower wall. Differentiating the depth-based specific energy confirmsThis cross-channel constancy is also the $P=0$ consequence of $\nabla B=-P\nabla \psi$.

For a reservoir of finite depth, exact $P=0$ requires relative vorticity $\zeta =-f$, an anticyclonic rotation cancelling the planetary vorticity. A reservoir at rest in the rotating frame has $\zeta =0$ and $P=f/h$, so it does not have exactly zero shallow-water potential vorticity. A very deep, nearly quiescent reservoir can nevertheless supply the approximation $P\simeq 0$ because $f/h$ is small. The idealization must describe vanishing absolute vorticity or this deep-reservoir limit, rather than ordinary finite-depth rest in the rotating frame.

Put $\Delta =fb$ and $v=u_1+\Delta /2$, the mean of the two wall velocities. The zero-potential-vorticity rotating channel flow has $h_2=h_1-\Delta v/g$. Substitution into the volume flux formula givesThe resulting quadratic equation has two roots, $v=g{h}_1/\Delta \pm \sqrt{g^2{h}_1^2/{\Delta }^2-2gQ/(b^{2}f)}$. The positive-depth condition $h_2>0$ selects the minus sign. ThusThe discarded sign would make $h_2<0$. On the retained branch $h_2=\sqrt{h_1^2-2fQ/g}$, which also shows the admissibility condition $h_1^2\ge 2fQ/g$.

The depth-based specific energy at the first wall isBecause ${\partial }_y(h+u^2/(2g))=0$, this same specific energy holds everywhere across the channel at fixed $x$. Its sum with the bed elevation is the Bernoulli function divided by $g$. Zero shallow-water potential vorticity makes that Bernoulli function constant throughout the connected steady channel flow, soConsequently the depth-based specific energy decreases as the bed rises and increases as it falls. It is constant along a level-bed channel, even if the width varies; that width variation instead changes the wall depths and velocities compatible with the conserved volume flux and Bernoulli function.

A hydraulic control of a zero-PV rotating channel is a critical section separating the upstream branch, where long-wave disturbances can propagate upstream, from the downstream branch, where they cannot. In the steady specific-energy description it is a stationary point of $E$ with respect to depth at fixed volume flux and local width. The control therefore fixes the discharge for a given upstream Bernoulli function and channel geometry; an arbitrary downstream condition cannot generally be imposed independently on the upstream reservoir.

Evaluate all quantities at $x=0$ and keep $Q,b$ fixed while differentiating. With $\Delta =fb$, the preceding branch givesAt the hydraulic control of a zero-PV rotating channel, this derivative vanishes. Using $h_1-h_2=\Delta v/g$ reduces it to $u_{1}v=g{h}_2$, where $v=u_1+\Delta /2$. Equivalently,Combining this with $Q=b{h}_{1}v-b\Delta v^2/(2g)$ cancels the quadratic term and leavesAt control $u_{1c}=\sqrt{g{h}_1+{\Delta }^2/16}-3\Delta /4$ and $h_{2c}=u_{1c}{v}_c/g$. Both walls must remain wet. Since $v_c>0$, the second depth is positive precisely when $u_{1c}>0$, givingEquality is the limiting case $u_{1c}=h_{2c}=0$, with the second wall just dry. For a wider channel, a dry lateral region requires a different solution and the two-wet-wall hydraulic control of a zero-PV rotating channel formula cannot be continued. The channel must also remain sufficiently long and slowly varying for the transverse geostrophic balance approximation to hold. As a check, $f\to 0$ gives $Q_c=b\sqrt{g}{h}_1^{3/2}$, the familiar rectangular-channel critical discharge.