= Solution
Within the <reduced equations for an anisotropic rotating fluid>, a flow that remains purely horizontal has $u_\parallel=0$ and $\partial_tu_\parallel=0$. The second reduced equation therefore requires
$$
0=-2\Omega\partial_z\Phi,\qquad
\boxed{\partial_z\Phi=0,\quad\partial_z\mathbf u_\perp^{(0)}=0.}
$$
Thus the retained horizontal motion is two-dimensional. This is a <Taylor–Proudman theorem> consequence in the slow balanced ordering; setting $u_\parallel=0$ at only one instant would not suffice, because vertical motion could subsequently develop.
The word “exactly” applies to the reduced model. It is not a theorem for every finite-<Rossby number> solution satisfying only $\Omega\gg kU$. For example, the full rotating <Euler equation> admits
$$
\mathbf u=A\cos(kz)(\cos2\Omega t,-\sin2\Omega t,0),\qquad p=\text{constant}.
$$
It is an <incompressible flow>, has zero nonlinear advection, and obeys $\partial_t\mathbf u=2\mathbf u\times\boldsymbol\Omega$. It is purely horizontal but depends on $z$, and $\Omega\gg kA$ can hold. This fast inertial oscillation lies outside the slow anisotropic <critical balance> used in part <1/c>; it shows why that ordering cannot be dropped from the conclusion.
Back to article page
