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Work in the body frame and approximate the narrow annulus locally by a planar gap. Let $y=0$ be the cylindrical wall, moving at $-U$ in this frame, and $y=h(x)$ the stationary cell surface. The red blood cell is represented by the given compliant body; the surrounding fluid is modeled as incompressible and Newtonian. Assume $h\ll a$, small local gap slope, and negligible inertia. The lubrication equations are $p_y=0$ and $\mu w_{yy}=p_x$, with no-slip conditions $w(0)=-U$, $w(h)=0$. Integration givesMass conservation makes the body-frame flux constant. With the sign convention of the problem,so the lubrication model of a compliant translating plug givesThe circumferential multiplier is $2\pi a$ to leading order, giving the relative volume flux $-2\pi aQ$.

Let $p_d$ be the pressure in front of the cell. Match the film to the reservoirs by $p(L)=p_d$ and $p(-L)=p_d+\Delta p$. Use $h=a-r$ and symmetric end matching for the parabolic-gap reduction. The elastic law givesFor $r_0(-L)=r_0(L)$ this gives $h(-L)-h(L)=\Delta p/\alpha$. If the outer unstressed shape is not symmetric at the matching ends, the general relation instead includes $h(-L)-h(L)=\Delta p/\alpha -[r_0(-L)-r_0(L)]$. The proposed reduced end equation assumes the symmetric matching used here. The gap must remain positive. To determine the unknown flux and translating speed one also needs zero net axial force on the freely moving body, and a specified absolute downstream pressure relative to $p_0$, not merely a pressure difference. If an external force pulls the body, it must replace the force-free condition. The stated elasticity law fixes $p_0$; treating it as an unknown internal pressure would require a further cell-volume constraint.

The force-free condition is most easily obtained from a control volume containing both body and surrounding film. The pressure force is $\pi a^2\Delta p$, and the resisting wall traction is $2\pi a\mu {\int }_{-L}^L{w}_y(0)dx$. The local velocity solution gives $w_y(0)=4U/h-6Q/h^2$, soThis is the force-free pressure drop of an annular lubrication plug. It includes the film pressure and shear contributions without separately approximating the deformed body's surface normal.

Put $s=2Q/U>0$, $\ell =\sqrt{s/\kappa }$, $h=sH$ and $x=\ell X$. The local parabolic elastic law is $p=p_0+\alpha [r_{00}-a-\kappa x^2/2+h]$, whence $p_x=\alpha \sqrt{\kappa s}(H_X-X)$. Substitution into the local lubrication equation givesThe derivative is with respect to the scaled coordinate $X$; a derivative with respect to dimensional $x$ would be inconsistent with this nondimensional equation.

Let $C=s/a$ and $\tilde{L}=L/\ell$. The elastic end condition gives $\Delta p=\alpha s[H(-\tilde{L})-H(\tilde{L})]$. Scaling the force integral and eliminating $\Delta p$ then givesThis completes the requested formulation and both dimensionless equations. The local parabolic approximation is used in the region contributing most strongly to the lubrication integrals; matching to the outer shape and reservoirs supplies the stated end conditions.

Take an attractive nonretarded molecular potential $U(R)=-\beta /R^6$, with molecular number densities $n_1,n_2$. Its force magnitude is $6\beta /R^7$. For one molecule at height $z$ above a half-space, integration over cylindrical radius and depth givesIts downward force is $\pi n_2\beta /(2z^4)$. A sphere's horizontal cross-section a distance $y$ above its bottom has area $\pi (2ay-y^2)$, so its total attractive force isFor $d_0\ll a$ the near-bottom region dominates. Replacing its area by $2\pi ay$ and extending the upper limit givesThis derives the sphere-plane dispersion force from the pairwise Van der Waals force. Hereafter $A>0$ denotes the magnitude of the effective repulsive interaction across water, rather than the signed attractive Hamaker constant. Its planar pressure magnitude is $\Pi (d)=A/(6\pi d^3)$.

For a nearly planar freezing front, the spherical gap is $d(r)=d_0+r^2/(2a)$. In the particle frame the incoming normal motion has speed $V$. Axisymmetric mass conservation gives the outward gap flux per circumference $q_r=Vr/2$. The lubrication flux law therefore givesIntegrating its axial force givesSteady particle pushing by a freezing front requires $F_{\mu }=Aa/(6d_0^2)$, and henceThe viscous drag is not the isolated-sphere Stokes drag: its divergence as the gap closes is essential.

For the deformed interface let $c=\cos {\theta }_0$ and measure $\theta$ from the sphere's downward pole, as in the original geometry. The height below the remote melting isotherm is $a(\cos \theta -c)$. Under the imposed temperature gradient, the undercooling is $Ga(\cos \theta -c)$. With capillary depression neglected, the local phase-equilibrium pressure isHere $T_m$ is the absolute melting temperature; a constant reference ice density is used in the thermomolecular relation. ThusThe numerical value of $K$ will cancel from the speed relation.

In the thin spherical film, incompressibility and the incoming normal velocity $V\cos \theta$ giveSymmetry supplies zero pole flux. The tangential Couette contribution is smaller by $d/a$ and is negligible at this order. Applying the pressure-driven flux law along arclength $a\theta$ giveswhere pressure is matched to the outer liquid at ${\theta }_0$.

A cap area element is $2\pi a^2\sin \theta d\theta$ and its vertical normal component is $\cos \theta$. Put $y=\cos \theta$. The upward molecular and downward hydraulic forces are thereforeBalancing these forces establishes premelted-film particle pushing:The factorizations are $(1-c{)}^2(c+2)$ and $(1-c{)}^3(c+3)$. They confirm positivity on the physical cap range and provide a numerically better-conditioned shallow-cap expression. Since $h=a(1-c)$, the shallow-cap limit within $d\ll h$ is $V\sim A/(12\pi \mu ah)$; it is not a limit in which the film can be identified with the flat-front gap $d_0$.

The local film formula diverges right at the rim ${\theta }_0$. Accordingly the thin-film calculation is an outer approximation away from a small matching fringe, not a claim of a uniformly thin film at its contact with bulk water. Both integrated stress densities vanish at that rim; in the thin-film limit the fringe contributes only a higher-order correction. Surface-tension curvature, altered thermal gradients or appreciable density-change fluxes would require a different balance.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.

For the basic axisymmetric flow, ${\mathbf{u}}_0=\Omega R{\mathbf{e}}_{\varphi }$. The steady radial Euler equation for fluid motion gives $-{\Omega }^{2}R=-p_{0,R}/\rho$; the other components give $p_{0,z}=p_{0,\varphi }=0$. Thus, in the absence of an additional body force,The additive constant is undetermined without an imposed reference pressure; the rigid boundary does not prescribe a uniform boundary pressure. Its shape is fixed by the container rather than by a free-surface pressure condition.

Let $v_R,v_{\varphi },v_z$ be the small velocity amplitudes and $W=p_1/\rho$ the pressure amplitude. Linearizing the cylindrical Euler equations for an inviscid fluid about uniform rotation, the material derivative is ${\partial }_t+\Omega {\partial }_{\varphi }$. The intrinsic frequency for the stated phase convention is thereforeThe radial and azimuthal curvature terms supply the Coriolis acceleration terms, givingFor $s\ne 0,\pm 2\Omega$, solving these equations givesThe incompressible flow constraint isOn substitution, the terms proportional to $m\Omega W_R$ cancel. The remaining equation iswhich provesThis is the pressure equation for inertial modes in a rigid rotating sphere. The exceptional frequencies used in the divisions must instead be treated in the original momentum system; they are not needed for the mode below.

A fixed rigid sphere is impermeable, so its normal velocity vanishes:Using the solved amplitudes and multiplying by $s(s^2-4{\Omega }^2)/i$ givesEquivalently divide by ${\bar{\sigma }}^2$ for $\bar{\sigma }\ne 0$. Smoothness means tangential slip is allowed: there is no no-slip condition on these inviscid normal modes. Nor is there a condition $W=0$, since the wall supplies the required varying normal pressure.

For an axisymmetric oscillation, $m=0$ and $s=\sigma$. Write $\beta =1-4{\Omega }^2/{\sigma }^2$. For $W=z(A{R}^2+B{z}^2+C)$ the radial and vertical operators areThe interior pressure equation therefore requiresThe boundary equation becomesSubstituting $R^2=r_0^2-z^2$, this polynomial must vanish for every boundary height. Its constant and $z^2$ coefficients consequently giveTogether with the interior relation the latter condition is $(4+\beta )A=0$. A nontrivial nonsingular oscillatory solution needs $A\ne 0$: if $A=0$ and $\beta \ne 0$, the other equations force $B=C=0$. Hence $\beta =-4$, givingThe cubic axisymmetric inertial mode of a rigid sphere is explicitlyIts cubic pressure is regular at the axis and centre, and its reconstructed velocity is divergence free and tangent to the sphere by the preceding identities. The frequency lies within the inertial wave band $|\sigma |<2|\Omega |$: Coriolis acceleration, rather than compression or a moving surface, supplies the restoring dynamics. The amplitude $A$ is arbitrary in this linear problem.