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Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

The material derivative follows surfactant carried by the moving interface. Tangential surface divergence changes the local surface area and hence dilutes or concentrates insoluble material. Normal motion changes area through the curvature: an expanding sphere has positive $u_n{\nabla }_s\cdot \mathbf{n}$ and its concentration decreases. The last term is surface diffusion, which smooths concentration gradients. There is no exchange with the bulk; this is conservation of insoluble surfactant on a moving interface.

The interfacial speed is of order $a|\mathbf{F}|$. Diffusion dominates redistribution when the surface Péclet number satisfiesThen $C^{\prime }/C_0=O({\mathrm{Pe}}_s)$, and advection of $C^{\prime }$ is a second-order effect. Write $S=\mathbf{n}\cdot \mathbf{Fn}$. Since ${\mathbf{u}}_s=\mathbf{Fx}-S\mathbf{x}$, the projected divergence of the first term is $\mathrm{tr}\mathbf{F}-S=-S$, whereas the second has divergence $2S$; its gradient contracted with $\mathbf{x}$ is zero. Hence ${\nabla }_s\cdot {\mathbf{u}}_s=-3S$.

The leading steady transport equation is $D_s{\Delta }_s{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. Trace-freeness and the supplied surface Laplacian identity give ${\Delta }_{s}S=-6S/a^2$. Fixing the conserved total surfactant removes the uniform null mode and yieldsThis is quadrupolar surfactant distribution on a spherical interface; its spherical mean is zero because $⟨n_i{n}_j⟩={\delta }_{ij}/3$.

With $+$ outside, $-$ inside, and the normal directed out of the bubble, the interfacial stress balance with variable surface tension isNow ${\nabla }_{s}S=2I_s\mathbf{Fn}/a$, so $-{\nabla }_s\gamma =2A{\gamma }_1{I}_s\mathbf{Fn}/a$. Multiplying the curvature by the tension and retaining linear terms givesThe undeformed isotropic pressure jump is kept with its normal; the perturbation terms are evaluated on the reference sphere. Products of deformation and concentration perturbations are beyond the approximation.

The imposed tensor is symmetric and traceless, so $\mathbf{x}\cdot \mathbf{Ex}$ is a harmonic scalar of degree two. Its decaying partner is $\mathbf{E}:\nabla \nabla (1/r)$; the corresponding harmonic vector uses $\mathbf{E}\nabla (1/r)$. Linearity and rotational symmetry give precisely these tensorial modes, with the growing scalar chosen to reproduce the imposed strain. Thus in the unscaled convention one can takeThe coefficients multiply distinct decaying modes. No other angular harmonic is forced at this order.

No penetration in the steady state gives $P=3Q-1$. The given tangential surface velocity implies $\alpha =1+2Q$. The tangential stress balance, with $M=A{\gamma }_1/(\mu a)$, gives $1+P-8Q=M\alpha$, or $-5Q=M(1+2Q)$. ThereforeThe nonuniform normal stress is unaffected by the uniform inviscid interior pressure. Its balance isSubstitution and rearrangement giveAs $M\to \infty$, $\alpha \to 0$: the concentration-induced Marangoni stress suppresses tangential surface motion, making the exterior boundary behave like an immobile surface despite the inviscid interior. The deformation coefficient tends to $5\mu a/(2{\gamma }_0)$; the clean limit is $2\mu a/{\gamma }_0$. Both small surface Péclet number and small deformation, of order $\mu a|\mathbf{E}|/{\gamma }_0$, are required. This is Marangoni immobilization of a bubble in straining flow.

Measure pressure relative to its hydrostatic value and let $\Delta P$ be the lower-side excess pressure minus the upper-side excess pressure. The long unobstructed parts of the tube act as Poiseuille resistances in series. Their lengths add to $L$ at leading order because $l\ll L$; entrance and exterior resistances are smaller for $L\gg a$. ConsequentlyHere $Q$ is downward laboratory-frame volume flux.

Take $z$ upwards in the particle frame and let $s$ measure distance from the particle surface to the tube wall. In this frame the particle is stationary and the wall moves upwards at speed $U$. In the thin, slowly varying gap, pressure is uniform across $s$ and the leading axial Stokes equation is $\mu w_{ss}=p_z$. No slip givesThe gap's upward relative flux is constant, while conservation of displaced volume gives $2\pi aq=\pi a^{2}U-Q$. Thus $p_z=6\mu U/h^2-12\mu q/h^3$, and integration from bottom to top yieldsCurvature corrections to the gap circumference are relative order $h/a$. Rapid end regions do not contribute to the leading singular neck resistance.

For the force use a cylindrical fluid control volume containing the particle, with full-area end caps just beyond it and the particle surface as an inner boundary. The upward pressure force on the caps is $\pi a^2\Delta P$. The outer wall exerts axial traction $\mu w_s(h)$ on the fluid, whereas the particle's traction on the fluid integrates to minus its upward hydrodynamic resistance $F$. The zero-inertia momentum balance is thereforeThe velocity profile gives $w_s(h)=4U/h-6q/h^2$, provingUsing the outer-wall traction is essential; directly using the particle-side shear and a flat-end pressure force would omit the pressure on the sloping particle surface. This is control-volume drag formula for a sphere in a tube.

The four dimensionless relations areSolving them givesWithin the preceding lubrication and tube-resistance approximations, the force isSince $h/a\ll 1$, $I_3^{\ast }\gg I_2^{\ast }\gg I_1^{\ast }$. Dropping the smaller additive $-2I_2^{\ast }+4I_1^{\ast }/3$ terms in the numerator yields the requested leading formulaThe Cauchy-Schwarz inequality gives $(I_2^{\ast }{)}^2\le I_1^{\ast }{I}_3^{\ast }$, so the retained $\delta$ term is positive. The term multiplied by $\delta$ must be retained: it can compete with $I_3^{\ast }$ even though the unmultiplied lower integrals cannot.

For a sphere of radius $a(1-ϵ)$, the neck has $h=aϵ+z^2/(2a)$ at leading order. With $z=a\sqrt{2ϵ}t$, the integrals reduce to $\sqrt{2}{ϵ}^{1/2-n}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-n}dt$. For $n=1,2,3$ these last integrals are $\pi ,\pi /2,3\pi /8$, giving the full two-sided integrals. PutThe printed sphere integrals are half these values: they correspond to integrating over one side of the neck rather than the stated interval spanning both sides. The substitution above establishes the radius-deficit normalization of sphere-tube lubrication integrals. Using the two-sided values, the force becomesso the three answers areThese are the corrected sphere-tube drag regimes with a radius-deficit gap. If the supplied half-integrals are treated as stipulated data instead, the same algebra gives $3\pi {ϵ}^{-5/2}/(8\sqrt{2})$, $1/\delta$ and $\pi \sqrt{2}{ϵ}^{-1/2}/3$ in the three regimes. Thus the first and third numerical drag coefficients double when the geometrically correct integrals are used; the regime boundaries and dominant flow patterns are unchanged.

In the first regime, $Q^{\ast }\sim \delta c_3{ϵ}^{-5/2}\ll 1$, $q^{\ast }\sim 1/4$, and $\Delta P^{\ast }\sim c_3{ϵ}^{-5/2}\sim F^{\ast }$. Almost no displaced fluid escapes through the long tube; it bypasses upwards through the thin annulus. The large pressure-driven bypass resistance sets the drag, with shear smaller.

In the second regime, $Q^{\ast }\sim 1$, $\Delta P^{\ast }\sim F^{\ast }\sim 1/\delta$, andAlmost all displaced volume travels through the unobstructed tube, whose pressure resistance dominates the force. Both terms in this small gap flux are worth displaying: their crossover at $\delta \sim {ϵ}^{3/2}$ changes the balance between pressure-driven and wall-driven gap transport but does not create a fourth leading force regime.

In the third regime, $1-Q^{\ast }\sim 4ϵ/3$, $q^{\ast }\sim ϵ/3$, and $\Delta P^{\ast }\sim 1/\delta \ll F^{\ast }$. Tube throughflow still carries almost all the displaced volume, but the dominant resistance is local annular lubrication rather than the net tube pressure drop. The relative gap flux is now of order $Uh$, so wall-driven and pressure-driven components both matter locally. The formulas describe hydrodynamic resistance after subtracting hydrostatic buoyancy; terminal settling would balance it against the particle's excess weight.

The Darcy law here is $\mathbf{u}=-k\nabla p/\mu$, with hydrostatic contributions absorbed into $p$ if present. Since $k$ is uniform and the fluid is incompressible, $\nabla \cdot \mathbf{u}=0$ gives $\Delta p=0$ in the porous matrix.

Pressure is uniform across the thin crack at leading order. No-slip lubrication theory gives its flux per unit depth $q=-h^3{p}_x/(12\mu )$. Conservation in a short crack segment says $q_x=-[u_y{]}_{-}^{+}$, while Darcy flow gives $[u_y{]}_{-}^{+}=-(k/\mu )[p_y{]}_{-}^{+}$. ThereforeThe other interfacial condition is pressure continuity, $[p{]}_{-}^{+}=0$. The far pressure gradient is $-\mu U{\mathbf{e}}_x/k$; outside the slit both pressure and normal Darcy flux are continuous. The leading tapered-crack flux vanishes at its endpoints. The assumptions $h\ll a$ and $h\gg \sqrt{k}$ justify transverse pressure uniformity and neglect of porous-wall slip on scale $\sqrt{k}$. Close to the sharp tips the reduced model is an outer approximation, not a uniformly resolved tip solution. This is lubrication boundary condition for a crack in a Darcy medium.

These are planar elliptic coordinates. The curve $\xi =0$ is the slit $[-a,a]$, traced on its upper and lower sides. The curve $\xi =1$ is the ellipse with semiaxes $a\cosh 1$ and $a\sinh 1$. For fixed noncardinal $\eta$, the coordinate curve is the appropriate branch of $x^2/(a^2{\cos }^2\eta )-y^2/(a^2{\sin }^2\eta )=1$. The cardinal angles give the two exterior horizontal rays and the upper/lower vertical rays. The requested original sketch is:

On the upper face $0<\eta <\pi$, the supplied derivatives reduce to ${\partial }_x=-(a\sin \eta {)}^{-1}{\partial }_{\eta }$ and ${\partial }_y=(a\sin \eta {)}^{-1}{\partial }_{\xi }$. The same point on the lower face has angle $2\pi -\eta$, so its ${\partial }_y$ has the opposite sign. ThusInsert this and the tangential derivative into the crack condition and multiply by $12a^2\sin \eta$ to obtainPressure continuity becomes $p(0,\eta )=p(0,2\pi -\eta )$. The uniform-flow background is $p_0=-\mu Ux/k=-(\mu Ua/k)\cosh \xi \cos \eta$. Its derivatives immediately giveThe harmonic equation is $p_{\xi \xi }+p_{\eta \eta }=0$ because the metric multiplier is nonzero outside the slit.

For the specified profile, $h^3=H_0^3\sin \eta$ on the upper face. Its coefficient in the tangential operator is constant, so the imposed cosine mode separates. With $P_0=\mu Ua/k$, write $p=P_{0}F(\xi )\cos \eta$. Harmonicity gives $F^{\prime \prime }=F$, the interfacial condition gives $F^{\prime }(0)=\alpha F(0)$, and the far field fixes the growing coefficient to $-1/2$. ThereforeReflection symmetry makes the two face derivatives equal. Other homogeneous cosine modes decaying as $e^{-m\xi }$ would require $-m=\alpha m^2$ at the slit, impossible for $m>0$ and $\alpha \ge 0$. The constant pressure gauge is immaterial. This justifies the separated crack pressure in elliptic coordinates, not only a candidate satisfying selected conditions.

At the midpoint $F(0)=-1/(1+\alpha )$ and $p_x=-\mu U/[k(1+\alpha )]$. HenceFor $\alpha \ll 1$, $Q\sim H_0^{3}U/(12k)$ and the crack barely perturbs the imposed Darcy pressure gradient. For $\alpha \gg 1$, $Q\sim 2aU$ and the slit is nearly equipotential: its high axial conductance collects flow from the surrounding matrix rather than sustaining a substantial pressure gradient. The parameter compares crack conductance $H_0^3/\mu$ with matrix conductance $ka/\mu$. In fact $q(x)=Q\sqrt{1-x^2/a^2}$, so the tip flux is zero in this outer model.

Subtracting the exact uniform background gives $\delta p=P_0\alpha e^{-\xi }\cos \eta /(1+\alpha )$. Since $r\sim a{e}^{\xi }/2$ and $\eta \sim \theta$ at infinity,The printed far-field coefficient is too large by a factor of two. The factor $1/2$ is forced by the stated coordinate map. An independent check is the large-$\alpha$ solution $p=-(\mu U/k)\mathrm{Re}\sqrt{z^2-a^2}$, whose expansion has precisely $\delta p\sim \mu U{a}^2\cos \theta /(2kr)$. This is the elliptic crack pressure dipole.

The phrase “suitably chosen curve” matters for the pressure-only dissipation integral. Put $C_d=\alpha \mu U{a}^2/[2(1+\alpha )k]$ and use a rectangle with end faces $x=\pm X$ and transverse extent $|y|\le H$, taking $a\ll X\ll H$. The horizontal faces have $n_x=0$. The two end-face integrals of $\delta p{n}_x$ totalThis is a fixed-throughflow normalization: the integrated velocity perturbation on an infinite end face is zero, since it is proportional to ${\partial }_x\int \delta pdy={\partial }_x(\pi C_d\mathrm{sgn}x)=0$. Work on the transverse faces vanishes as $H/X\to \infty$. Thus the requested line functional givesA bare circular cutoff would give $-\pi U{C}_d$ instead; it does not enforce this same throughflow normalization, so its pressure-only term cannot be substituted into the fixed-throughflow dissipation comparison. Keeping the control surface explicit resolves that ambiguity. This is fixed-throughflow dissipation reduction of a conductive crack.

Let $s=\pi \varphi a^2\alpha /(1+\alpha )\ll 1$. The noninteracting average dissipation per area is $\mu U^2/k+\varphi \Delta \mathcal{D}=(\mu U^2/k)(1-s)$. Equating this with $\mu U^2/k_{xx}^{\ast }$ givesOnly the first-order density term is controlled by the noninteraction assumption; the rational form is not an all-orders result. It increases permeability along the aligned cracks and saturates as their conductance becomes large. This is dilute permeability enhancement by aligned cracks.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.

For a slowly varying mean shear flow $U(z)$, the local intrinsic frequency is $\hat{\omega }=\omega -kU(z)$. A local WKB approximation uses ${\hat{\omega }}^2=N^2{k}^2/(k^2+m(z{)}^2)$, with varying vertical wavenumber and group velocity. The appropriate transported quantity is wave action, or horizontal wave pseudomomentum $kE/\hat{\omega }$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\hat{\sigma }{|}^2$ is no longer constant.

Near a critical level of an internal gravity wave, $\hat{\omega }\to 0$, so $|m|\sim N|k|/|\hat{\omega }|$ and $|c_{gz}|\sim {\hat{\omega }}^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid WKB approximation eventually fails; viscosity and possibly internal-wave breaking regulate the small scales.