Shallow Water PV

The rotating shallow water equations are

\begin{align*} u_t + \vec{u} \cdot\nabla u -fv &= -gh_x \\
v_t + \vec{u} \cdot\nabla v +fu &= -gh_y \\
h_t + \nabla \cdot (\vec{u}h) &= 0
\end{align*}

where \(\vec{u}=(u,v)\) is the horizontal velocity, \(h=h(x,y,t)\) is the depth of the fluid, \(g\) is the gravitational acceleration, and \(f\) is the Coriolis parameter.

  1. Define the potential vorticity for the rotating shallow water equations (SWPV).
  2. Show that the SWPV is conserved.
  3. Show that for a parcel initially at rest in the Northern hemisphere, there is an absolute lower bound on its relative vorticity if it remains in the same hemisphere, but no upper bound.
Answer


  1. Define the potential vorticity for the rotating shallow water equations (SWPV).

We can derive the potential vorticity by first taking the curl of the momentum equations. First, let's re-write the non-linear term as:

$$\vec{u} \cdot \nabla \vec{u} = \vec{\omega} \times \vec{u} - \frac{1}{2}\nabla |\vec{u}|^2 $$

So when we take the curl of this, the second term will disappear. The vertical component of the curl of the first term can be written as:\(\require{cancel}\)
\begin{align*}
&\nabla \times \left(\vec{\omega} \times \vec{u}\right)\cdot\hat{k} = \\
& (\nabla \cdot \vec{u})\vec{\omega}\cdot\hat{k} + \vec{u} \cdot \nabla \vec{\omega}\cdot\hat{k} \\
&- \cancel{(\nabla \cdot \vec{\omega})\vec{u}\cdot\hat{k}} - \cancel{\vec{\omega} \cdot \nabla \vec{u}\cdot\hat{k}}
\end{align*}

The first cancelled term disappears because we are taking the divergence of a curl (the vorticity), while the second cancelled term disappears because the vertical component of the horizontal velocity vector is zero. So we arrive at the following equation:
\begin{align*} \frac{\partial \zeta}{\partial t} + \vec{u} \cdot \nabla \zeta + f\nabla \cdot \vec{u} &= -(\nabla \cdot \vec{u})\zeta \\
\frac{D \zeta}{D t} &= -(\nabla \cdot \vec{u})(\zeta+f)
\end{align*}

where \(\zeta\) is the vertical component of vorticity. We can now use the mass conservation equation to get an alternative expression for the horizontal velocity divergence:

\begin{align*}\frac{Dh}{Dt} + h\nabla \cdot \vec{u} &= 0 \\
\implies \nabla \cdot \vec{u} &= - \frac{1}{h}\frac{Dh}{Dt}
\end{align*}

Substituting this into the vorticity evolution equation,
\begin{align*}
\frac{D \zeta}{D t} &=\frac{1}{h}\frac{Dh}{Dt}(\zeta+f)\\
h\frac{D \zeta}{D t} -(\zeta+f)\frac{Dh}{Dt} &= h^2\frac{D}{Dt}\left( \frac{\zeta+f}{h}\right) = 0
\end{align*}

The conserved quantity \(q = \frac{\zeta+f}{h}\) is the shallow water PV.

  1. Show that the SWPV is conserved.

We just showed this in part (1). Alternatively, if you remembered the SWPV, you could work backwards: take the advective derivative and show that it is equal to zero using the shallow water equations.

  1. Show that for a parcel initially at rest in the Northern hemisphere, there is an absolute lower bound on its relative vorticity if it remains in the same hemisphere, but no upper bound.

If we are at rest in the Northern hemisphere, then \(\zeta = 0\) and \(f>0\), so

\begin{align*}\frac{f_0}{h_0}  & =\frac{\zeta_1 + f_1}{h_1}\\
\zeta_1 &= f_0\frac{h_1}{h_0} - f_1 \geq - f_1
\end{align*}

As \(h_1 \to 0\), \(\zeta_1 \to -f_1\), so \(-f_1\) is a lower bound for relative vorticity. But as \(h_0 \to 0\), \(\zeta_1 \to \infty\), so there is no upper bound.