 Solution

For $\mathbf{u}=\alpha y{\mathbf{e}}_x-\beta x{\mathbf{e}}_y$, the velocity divergence is zero and $\mathbf{u}\cdot \nabla \mathbf{u}=-\alpha \beta (x{\mathbf{e}}_x+y{\mathbf{e}}_y)$. Substituting into the steady Euler equations yieldsIntegration givesThe steady enthalpy equation also requires $\mathbf{u}\cdot \nabla Q=0$. Its $xy$ coefficient is $\alpha (3{\Omega }^2+\alpha \beta -{\beta }^2)$, so, since $\alpha >0$,For a bounded patch with positive mass density and a vacuum fluid free boundary, choose $Q_0>0$ and require both quadratic coefficients to confine the fluid. With $\Omega >0$ and $\alpha ,\beta >0$, this meansThe lower bound is imposed by $\alpha >0$; the upper bound by bounded positive $Q$. At $\beta =2\Omega$, $Q$ is constant and has no nontrivial finite vacuum boundary; $\beta \le \sqrt{3}\Omega$ violates or degenerates the stipulated elliptical flow.

For a polytropic elliptical patch in a shearing sheet, the physical fluid free boundary is $p=\rho =Q=0$. It is the ellipsewhere $a_x^2=2Q_0/[\beta (2\Omega -\beta )]$ and $a_y^2=2Q_0/[\alpha (2\Omega -\beta )]$. In the interior $\rho =[Q/((n+1)K){]}^n$. The boundary is a material streamline because $\mathbf{u}\cdot \nabla (\beta x^2+\alpha y^2)=0$. Since $\alpha <\beta$, it is elongated in the azimuthal $y$ direction.

This is a pressure-supported fluid patch in the supplied tidal model; no additional planetary self-gravity potential was assumed.