= Solution
Let $\boldsymbol\zeta=\nabla\times\boldsymbol u$ and $\boldsymbol\omega=\boldsymbol\zeta+2\Omega\boldsymbol e_z$ be the <relative vorticity> and <absolute vorticity>. The nonlinear acceleration identity rewrites the momentum equation as
$$
\partial_t\boldsymbol u-\boldsymbol u\times\boldsymbol\omega
=-\nabla\left(\frac P{\rho_0}+\frac12|\boldsymbol u|^2-\frac32\Omega^2x^2\right)-N^2\theta\boldsymbol e_x.
$$
Taking the <curl> removes the <pressure>, <kinetic energy> gradient and <shearing-sheet tidal potential>. Thus
$$
\partial_t\boldsymbol\omega-\nabla\times(\boldsymbol u\times\boldsymbol\omega)
=-N^2\nabla\theta\times\boldsymbol e_x.
$$
Both $\nabla\cdot\boldsymbol u$ and $\nabla\cdot\boldsymbol\omega$ vanish. The <vorticity equation> is consequently
$$
\partial_t\omega_j+u_i\partial_i\omega_j-\omega_i\partial_i u_j
=-N^2(\nabla\theta\times\boldsymbol e_x)_j.
$$
For the <absolute-vorticity flux tensor> $T_{ij}=u_i\omega_j-\omega_i u_j$, use the flux convention $(\nabla\cdot\boldsymbol T)_j=\partial_iT_{ij}$, with the first index denoting transport direction. This gives
$$
\boxed{\partial_t\boldsymbol\omega+\nabla\cdot\boldsymbol T=-N^2\nabla\theta\times\boldsymbol e_x.}
$$
The scalar $\theta$ is a <buoyancy displacement variable>, so dimensional consistency gives it units of length; the printed potential-temperature terminology presupposes this normalization. Its <curl> source directly forces only the $y,z$ components of <absolute vorticity>. Hence <radial vorticity conservation in a shearing sheet> takes the local flux form
$$
\boxed{\partial_t\omega_x+\partial_i(u_i\omega_x-\omega_i u_x)=0.}
$$
Its volume integral is constant when the net boundary flux vanishes. However $D_t\omega_x=\boldsymbol\omega\cdot\nabla u_x$ still permits <vortex stretching>; the radial component is not generally a materially conserved scalar in three dimensions. Using divergence on the second index of the stated tensor would reverse the transport sign, so the tensor convention matters.
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