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A fluid free boundary is an initially unknown material interface separating a fluid from its exterior. For a polytropic elliptical patch in a shearing sheet bounded by vacuum, it is the level set $p=\rho =Q=0$, and the velocity is tangent to it in a steady 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.

The steady planar velocity $\mathbf{u}=\alpha y{\mathbf{e}}_x-\beta x{\mathbf{e}}_y$ supports a bounded polytropic patch when $\alpha =({\beta }^2-3{\Omega }^2)/\beta$ and $\sqrt{3}\Omega <\beta <2\Omega$. Its polytropic enthalpy is $Q=Q_0-(2\Omega -\beta )(\beta x^2+\alpha y^2)/2$ with $Q_0>0$. Its fluid free boundary is the material ellipse $\beta x^2+\alpha y^2=2Q_0/(2\Omega -\beta )$. No additional self-gravity potential is part of this model.