 Solution

Use the horizontal streamfunction convention $(u,v)=(-{\psi }_y,{\psi }_x)$, and set $a=\pi /L$, $k_d^2=f_0^2/(g{H}_0)$. Geostrophic balance gives $f_0\bar{u}=-g{\bar{\eta }}_y$, henceThe constant is fixed by the total volume or the reference surface height. To obtain shallow-water quasi-geostrophic potential vorticity, require a small Rossby number, nearly geostrophic horizontal motion and $|\eta |/H_0\ll 1$. Expanding the exact shallow-water potential vorticity givesAt leading nontrivial order $\zeta ={\nabla }_h^2\psi$, so after removing the constant background and multiplying by $H_0$,The mean potential vorticity and its gradient areWith $\psi =\bar{\psi }+{\psi }^{\prime }$ and $q=\bar{q}+q^{\prime }$, expand the Jacobian. Since both mean fields depend only on $y$, their Jacobian vanishes. The two mixed terms are $\bar{u}{q}_x^{\prime }$ and $v^{\prime }{\bar{q}}_y$, givingThe PDF has this correct mean-advection term; the TeX aid incorrectly inserts a division by $\partial x$.

Drop the quadratic perturbation Jacobian. Freeze $\bar{u}$ and ${\bar{q}}_y$ locally for a short-wave plane wave, with $K^2=k^2+l^2$ and $D=K^2+k_d^2$. Then $q^{\prime }=-D{\psi }^{\prime }$, $v^{\prime }=ik{\psi }^{\prime }$, and the local Rossby-wave dispersion relation in a zonal jet isThe intrinsic phase moves against an eastward mean flow because $\omega -k\bar{u}=-k{\bar{q}}_y/D$.

The printed wavelength bound is not a consequence of these equations. The local short-wave assumption is $K/a\gg 1$, equivalently wavelength $2\pi /K\ll 2L$; a particular cutoff $L/2$ is an optional scale-separation criterion, not a universal derived threshold. If one requires the ground-frame phase to follow the mean current, the dispersion relation only gives $K>a$, or wavelength less than $2L$. For example $K=2a,l=0$ has wavelength $L$ and a phase moving with the mean. More decisively, the stationary condition requested next fixes wavelength $2L$, contradicting a bound $L/2$ if both are imposed on the same waves.

For nonzero zonal $k$ and mean velocity, a ground-stationary wave satisfiesIt has the background length scale, so its existence should not be justified solely by the short-wave approximation. It is nevertheless an exact stationary linear solution of the Rossby-wave equation for a sheared zonal current: writing ${\psi }^{\prime }=F(y)e^{ikx}$ gives $\bar{u}[F^{\prime \prime }-(k^2+k_d^2)F]+{\bar{q}}_{y}F=0$, which reduces to $F^{\prime \prime }+(a^2-k^2)F=0$. Thus $F=e^{ily}$ with $k^2+l^2=a^2$ works globally. Indeed the total field then has $q=-(a^2+k_d^2)\psi +$ a constant, so its full Jacobian vanishes as well.

Differentiating the local dispersion relation givesOn the stationary circle this becomesFor a purely zonal wavevector, $l=0$, this is the printed expression ${\mathbf{c}}_g=2\bar{u}{K}^2\hat{\mathbf{x}}/(K^2+k_d^2)$. With $l\ne 0$ the printed expression needs a directional qualification; the general vector is the one above. Here stationarity means $\omega =0$ in the ground frame, not vanishing intrinsic frequency in the moving fluid.

A stationary obstacle can therefore anchor the phase pattern while wave energy travels away from it. The zonal group component has the mean-flow sign, so an eastward jet carries the stationary wave response downstream; tilted wavevectors also carry energy meridionally. This is the usual stationary Rossby wave wake distinction between fixed phase crests and propagating wave energy.