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For a planar reservoir draining below a vertical lip of height one into a unit-speed stream, the potential strip $0<\mathrm{Im}w<a$ and first-quadrant unit velocity disk satisfyOn the fluid free boundary put $q=e^{-i\theta }$, $-\pi /2\le \theta <0$. Then $dz/d\theta =-2a{e}^{i\theta }\cot \theta /\pi$, so the downstream height is $1-2a/\pi$. Equating it to $a$ gives $a=\pi /(\pi +2)$.

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.

A semi-infinite horizontal plate at unit height and a floor at zero give a velocity potential strip of height $a$ and an upper unit semicircle in the complex velocity $q$. Their matching half-plane maps give $(1+e^{\pi w/a}{)}^{-1}=((q-1)/(q+1){)}^2$. Along the fluid free boundary put $q=e^{-i\theta }$, $-\pi <\theta <0$. Then $dz/d\theta =-(a/\pi )e^{i\theta }\cot (\theta /2)$ and $y=1-(a/\pi )(\theta +\pi +\sin \theta )$. Its downstream limit is $1-a$, which must equal the downstream depth $a$. Thus the contraction is exactly one half.

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.

On the fluid free boundary write $q=e^{-i\theta }$, with $-\pi <\theta <0$: the lip has $q=-1$ and the downstream limit has $q=1$. ThenSubstitution into the relation from part (ii) givesThe surface lies on the top of the velocity potential strip, so its logarithm has imaginary part $\pi$, givingFinally $dz/dw=1/q=e^{i\theta }$, and thereforeThe source's typeset $\cot \theta /2$ is interpreted as $\cot (\theta /2)$, as the derivation requires.

The lower wall has streamfunction zero. The streamline through the lip and along the upper fluid free boundary has streamfunction equal to the downstream volumetric flow rate, namely $a$ because the downstream speed is one and its depth is $a$. The complex potential of a two-dimensional potential flow therefore fillsThe chosen velocity potential origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have velocity potential tending to $-\infty$; the downstream endpoints $B,C$ have velocity potential tending to $+\infty$.

Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive real axis. Above the plate the flow goes left toward the lip, so the plate velocities run along the negative real axis from zero at $E$ to $-1$ at $D$. On the fluid free boundary Bernoulli's equation and its constant pressure give $|q|=1$. Its downward velocity has $v<0$, hence $\mathrm{Im}q>0$. Consequently the hodograph plane domain isthe upper unit semicircle. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.

Integrate real and imaginary parts from the lip value $z(-\pi )=i$:The identities $\sin s\cot (s/2)=1+\cos s$ andgive the explicit parametrizationAs $\theta \uparrow 0$, $x\to +\infty$ and $y\to 1-a$. This must be the downstream height $a$, soThe surface initially moves slightly left of the lip before turning right. Its leftmost point occurs at $\theta =-\pi /2$, with $x_{\min }=-(1-\log 2)/(2\pi )$. This is the half-contraction free streamline at a horizontal canal lip.

Write the complex potential of a two-dimensional potential flow as $w=\varphi +i\psi$, and the complex velocity as $q=w^{\prime }=u-iv$. The lower solid wall has $\psi =0$. The upper solid wall and its continuing free streamline have constant $\psi$ equal to the far-field volume flux per unit depth, namely $a$. Consequently the potential domain is the infinite stripThe points $A,E$ correspond to $\varphi \to -\infty$, $B,C$ to $\varphi \to +\infty$, and $D$ is $w=ia$ by the chosen potential normalization.

In the hodograph plane, the horizontal wall has $q$ real positive, the vertical wall has $q$ imaginary positive because the physical velocity is downward, and the fluid free boundary has $|q|=1$. The upstream reservoir velocity tends to zero, the outgoing velocity is one, and the velocity at $D$ is $i$. Thus the hodograph is the first-quadrant unit quarter disk, with $A,E$ at $q=0$, $B,C$ at $q=1$, and $D$ at $q=i$.

Scale lateral distance with $L$, transverse distance with $h$, lateral velocity with $U$, and the normal velocity with $Uh/L$. Balancing lateral pressure force with transverse viscous stress gives pressure scale $\mu UL/h^2$. The ratio of inertia to that stress is $U{h}^2/(\nu L)$, while lateral viscous derivatives are smaller by $(h/L{)}^2$. The two stated smallness conditions therefore reduce the lateral equations to $\mu u_{zz}=p_x$, $\mu v_{zz}=p_y$. The normal component of balance of linear momentum gives $p_z=0$ at leading order; continuity remains $u_x+v_y+w_z=0$. The no-slip, impermeable rigid plates require $u=v=w=0$ at both plate surfaces.

Integrating with those boundary values givesDepth-integrated continuity has no normal boundary flux, so $\nabla \cdot \bar{u}=0$ and $\boxed{{\nabla }^{2}p=0}$ away from the injection point.

Neglecting surface tension, the fluid free boundary has $p=p_{\mathrm{ext}}$, and its normal velocity is the fluid velocity: $\dot{R}={\bar{u}}_r(R)$. Any constant capillary pressure for a circular interface can instead be included in that spatially uniform boundary pressure. Conservation of injected volume gives $2\pi rh{\bar{u}}_r=Q$. HenceThe logarithmic pressure singularity represents the ideal point source, which in a physical cell is replaced by a finite inlet. The formula is an exact radial solution of the depth-averaged model for either sign of $Q$, while $R^2>0$.

Its robustness differs sharply between the signs. With viscous fluid inside a much less viscous exterior, injection displaces the less viscous phase and is stabilizing; suction draws the exterior into the viscous region and is prone to fingering. One can see this directly without relying on a slogan. Write the boundary as $R+a_n\cos n\theta$ and its regular pressure perturbation as $b_n{r}^n\cos n\theta$. Linearizing the constant-pressure condition gives $\delta p(R)=-a_n{p}_0^{\prime }(R)=a_{n}Q/(2\pi hKR)$. The perturbation radial velocity is then $-n{a}_{n}Q/(2\pi h{R}^2)$, and evaluation of the base velocity at the displaced boundary adds $-a_{n}Q/(2\pi h{R}^2)$. ThusModes decay for injection and grow for suction. Accordingly the suction circle is a mathematically possible symmetric solution but generally an unstable prediction, failing before its formal collapse in a perturbed experiment. This is Stability of an interior Hele-Shaw blob under injection and suction.

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.

For viscous fluid inside a circular fluid free boundary between parallel plates of separation $h$, the depth-averaged Hele-Shaw flow satisfies $\bar{u}=-K\nabla p$, $K=h^2/(12\mu )$. With volume injection rate $Q$, ${\bar{u}}_r=Q/(2\pi hr)$ and $R^2=R_0^2+Qt/(\pi h)$. Neglecting surface tension and exterior viscosity, pressure is constant at the interface. Perturb its radius by $a_n\cos n\theta$ and solve the interior Laplace equation with a regular pressure term proportional to $r^n\cos n\theta$. The linearized pressure boundary condition gives $\delta p(R)=a_{n}Q/(2\pi hKR)$, hence velocity perturbation $-n{a}_{n}Q/(2\pi h{R}^2)$. The displaced base velocity adds another $-a_{n}Q/(2\pi h{R}^2)$, proving the displayed growth law. Injection stabilizes the circular blob; suction destabilizes it. An exact radial solution can therefore exist while giving a fragile experimental prediction. These signs concern viscous fluid inside the blob, rather than the opposite geometry of a low-viscosity injected bubble.