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For a fixed geometry, the steady Stokes equation maps $(\mathbf{u},p^{\prime },\mathbf{f})$ to $(-\mathbf{u},-p^{\prime },-\mathbf{f})$ when all imposed boundary velocities and disturbance forces are reversed; $p^{\prime }$ is pressure relative to any unchanged hydrostatic pressure. Linearity of Stokes flow and Uniqueness of Stokes flow therefore reverse the entire velocity field. If the driving schedule and evolving geometry are retraced, fluid elements retrace their paths too. Reversing just one drive while leaving another applied force unchanged does not meet these conditions.

This reversal concerns the paths, while viscous dissipation remains positive: for a sphere, the required power $6\pi \mu a{U}^2$ is unchanged by $U\mapsto -U$. A reversed experiment must supply fresh energy. Inertia, Brownian motion and irreversible contact can prevent perfect retracing. Edward M. Purcell illustrated the mechanism with a viscous fluid flow demonstration in Life at low Reynolds number, American Journal of Physics 45, 3–11 (1977).

The scalar linear partial differential equation $u_t+u_x+u_{xxx}=0$ is an Airy equation with a first-order transport term. It is a linearization of the Korteweg-De Vries equation with an advective term retained. It is distinct from the viscous incompressible-flow Stokes equation. On the whole line its spatial Fourier transform evolves by $e^{i(k^3-k)t}$.

The Stokes equation is linear in velocity, pressure, body force, and boundary data. Solutions may therefore be superposed, and rigid-body velocities depend linearly on applied forces and torques.

Subtract the ambient velocity and stress, writing ${\mathbf{u}}^{\prime }=\mathbf{u}-{\mathbf{u}}_{\infty }$. This perturbation decays at infinity and on the sphere has velocity $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}-{\mathbf{u}}_{\infty }$. The ambient flow is regular through the ball and has no body force. Its net torque on the imaginary spherical surface is zero: the divergence theorem converts the torque to the volume integral of the stress divergence plus its antisymmetric part, both zero. Thus the perturbation has zero torque too.

Reciprocity with the auxiliary rotlet givesUsing ${\mathbf{\sigma }}_G{\mathbf{n}}_D=3(\mathbf{G}\times \mathbf{x})/(8\pi a^4)$ yieldsOn this sphere $\mathbf{x}=a\mathbf{n}$, so the vector divergence theorem gives $\int \mathbf{x}\times {\mathbf{u}}_{\infty }dS=a{\int }_{r<a}\nabla \times {\mathbf{u}}_{\infty }dV$. Taking the curl of the homogeneous Stokes equation makes ${\mathbf{\omega }}_{\infty }$ harmonic. Its ball average is its central value: the derivative of each spherical mean is proportional to the volume integral of its Laplacian, hence zero, and its small-radius limit is its value at the center. ThereforeThis proves Faxén rotation law directly from the reciprocal theorem, including why no finite-radius Laplacian correction is needed for a body-force-free ambient flow. Regularity of the ambient field throughout the sphere's prospective interior is part of this argument.

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.

For the given harmonic potentials, ${\nabla }^2(\mathbf{\Phi }\cdot \mathbf{x})=2\nabla \cdot \mathbf{\Phi }$. Taking the divergence of the velocity representation therefore gives $\nabla \cdot \mathbf{u}=0$. Taking its Laplacian gives ${\nabla }^2\mathbf{u}=-2\nabla (\nabla \cdot \mathbf{\Phi })$, so the Stokes equations hold withThis is the Papkovich–Neuber representation in the sign and normalization of this problem. The potentials used in other conventions can differ by an overall factor and sign; the velocity and pressure must be scaled together.

(a) Point force. Put the singularity at the origin, $r=|\mathbf{x}|$, and choose $\mathbf{\Phi }=\mathbf{F}/(8\pi \mu r)$, $\chi =0$. The resulting Stokeslet isThe harmonic calculation applies away from the force. Distributionally this is the Green function for $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}{\delta }^{(3)}(\mathbf{x})=0$, with zero divergence.

(b) Point source. Set $\mathbf{\Phi }=0$, $\chi =-Q/(4\pi r)$. The three-dimensional point source field isIts outward volume flux across every enclosing sphere is $Q$. The divergence is zero away from the source and equals $Q{\delta }^{(3)}(\mathbf{x})$ distributionally, so the singular point represents volume injection rather than an incompressible material point.

(c) Stresslet. Let $G_{ij}(\mathbf{x})=({\delta }_{ij}/r+x_i{x}_j/r^3)/(8\pi \mu )$ and use a symmetric trace-free stresslet tensor $S$ with the convention $u_i=-S_{jk}{\partial }_k{G}_{ij}$. Differentiating the Stokeslet and using symmetry and zero trace givesFor an axial force-dipole flow with axis $\mathbf{n}$, write $S=s(\mathbf{nn}-I/3)$. Then the radial factor is proportional to $s[3(\mathbf{n}\cdot \hat{\mathbf{x}}{)}^2-1]$. Specifying the derivative convention is essential for interpreting the signs of image strengths.

(d) Source dipole. A positive source and negative source separated by a vector $\mathbf{\ell }$ have moment $\mathbf{M}=Q\mathbf{\ell }$. Their dipole limit is minus the directional derivative of the unit source field. Hence the source dipole hasIt carries no net volume flux, and unlike a stresslet it is an irrotational potential flow.

Now set $\mathbf{r}=\mathbf{x}-d\mathbf{n}$ and $\mathbf{R}=\mathbf{x}+d\mathbf{n}$, with $d>0$. Define the unscaled kernels $K(\mathbf{R})=\mathbf{n}/R+(\mathbf{n}\cdot \mathbf{R})\mathbf{R}/R^3$, $H(\mathbf{R})=\mathbf{R}/R^3$, and let ${\partial }_n=\mathbf{n}\cdot {\nabla }_{\mathbf{R}}$. The no-slip image system of a normal Stokeslet isThe reflected Stokeslet has force $-F\mathbf{n}$. With the conventions above, the remaining image strengths areThe isotropic part of $S$ produces zero velocity, because ${\partial }_j{G}_{ij}=0$. Its inclusion simply puts the tensor in standard trace-free form. The pressure follows by superposition:

To verify the no-slip boundary condition explicitly, write $\mathbf{x}=\mathbf{\rho }$ on the plane, with $\mathbf{\rho }\cdot \mathbf{n}=0$, and let $b^2={\rho }^2+d^2$. ThenThe $b^{-5}$ terms cancel, while the remaining image contribution is $2d(\mathbf{R}-d\mathbf{n})/b^3=2d\mathbf{\rho }/b^3$. Thus both normal and tangential velocity vanish. All image singularities lie outside the fluid, so the physical half-space contains only the prescribed force.

For the far field, expand about the wall origin. The pair of opposite forces starts with $-2d{\partial }_{n}K$; the image stresslet cancels this entire $r^{-2}$ term. The first surviving field isWriting $z=\mathbf{x}\cdot \mathbf{n}$ and now $r=|\mathbf{x}|$, this isTherefore the generic fixed-angle far-field velocity decays as $r^{-3}$, rather than the $r^{-1}$ free-space Stokeslet or an uncancelled $r^{-2}$ stresslet. Along the wall it is exactly zero; at fixed height and large horizontal distance the leading tangential field instead scales as ${\rho }^{-4}$, consistent with the angular zeros of the displayed far field.

Take $x$ downslope and $z$ normal to the plane. To leading order in the lubrication approximation, Newtonian gravity gives the hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. The downslope Stokes equation is therefore $\mu u_{zz}=\rho g(h_x-\theta )$. Imposing the no-slip boundary condition at $z=0$ and the stress-free boundary condition at $z=h$ givesHere $\nu =\mu /\rho$ is the kinematic viscosity; $u$ is the downslope velocity. Applying mass conservation to this lubrication gravity-current flux,For typical thickness $H=\hat{h}$ and length $L=\hat{x}$, the transverse velocity is $O(UH/L)$ by incompressibility. Both convective terms are then $O(U^2/L)$, while transverse viscous acceleration is $O(\nu U/H^2)$. Thus the inertia criterion for an inclined viscous film is $U{H}^2/(\nu L)\ll 1$, in addition to $H/L\ll 1$. With evolution time $T\sim L/U$ the same condition controls unsteady inertia. If a faster time is independently imposed, also require $H^2/(\nu T)\ll 1$. Estimating $U$ from the velocity profile gives the two requested limits:

At long times, $h\sim A/x_N$ and the ratio of hydrostatic-gradient forcing to downslope Newtonian gravity in the bulk is $h_x/\theta =O(A/(\theta x_N^2))\ll 1$. Put $K=g\theta /(3\nu )$; the bulk conservation law is $h_t+(K{h}^3{)}_x=0$. The finite-volume gravity-driven thin-film current scales as $x\sim t^{1/3}$, $h\sim t^{-1/3}$. More explicitly, writing $h=t^{-1/3}f(\xi )$, $\xi =x/t^{1/3}$, givesThe constant of integration is zero for the zero-flux dry tail. ConsequentlyThe reduced bulk solution is zero outside this interval. Its thin nose region is resolved below. Integrating to impose the conserved area,The bulk front height is $h_N=3A/(2x_N)$, with $h_N^2=x_N/(3Kt)$. Differentiation of the front law givesEquivalently, this is the Rankine-Hugoniot condition for the bulk shock wave, whose speed is flux divided by height.

In moving coordinate $y=x-x_N(t)$, the full thin-film equation becomes $h_t{|}_y-{\dot{x}}_N{h}_y+q_y=0$. A nose thickness $h_N$ and slope $\theta$ give axial scale ${\ell }_N=h_N/\theta$. Since ${\ell }_N/x_N=O(A/(\theta x_N^2))\ll 1$, the explicit time derivative is smaller than translation by that ratio. The travelling front of a gravity-driven thin film is therefore quasisteady. Integrate once with dry-tip flux zero to obtain $q={\dot{x}}_{N}h$, orWith $f=h/h_N$, separation of variables and the convention $y=0$ at $f=0$ giveThe nose approaches $h_N$ exponentially far behind its tip, then falls to zero with a vertical tangent in this small-slope model. The requested sketch shows this monotone profile:

Expanding the inverse hyperbolic tangent, $f-\mathrm{atanh}f=-f^3/3+O(f^5)$, so $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. Its slope has magnitude $\theta /f^2$ near the tip. The small-slope breakdown at an inclined-current nose occurs at $f=O(\sqrt{\theta })$, which givesFinally, surface tension produces a pressure-gradient scale $\gamma h_N/{\ell }_N^3$, whereas downslope Newtonian gravity gives $\rho g\theta$. Their ratio is $\gamma {\theta }^2/(\rho g{h}_N^2)$. Thus surface tension changes the principal nose length whenThis is a bulk-nose crossover; capillarity and contact-line physics can already affect a smaller region very close to the dry tip even when the bulk nose is Newtonian gravity dominated.

Let $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ be two Stokes flows in the same fluid domain, with equal dynamic viscosity and zero body force. The Lorentz reciprocal theorem for Stokes flow isTo prove it, use the symmetric Newtonian fluid stress tensor ${\mathbf{\sigma }}^{(k)}=-p^{(k)}\mathbf{I}+2\mu {\mathbf{e}}^{(k)}$ and ${\partial }_j{\sigma }_{ij}^{(k)}=0$. Incompressibility removes the pressure contraction, while symmetry removes the antisymmetric velocity gradient. ThereforeThe divergence theorem proves the boundary identity. It applies to an exterior domain by a limit at infinity; the boundary integral there vanishes for decaying rigid-body Stokes flows.

Use a normal directed outward from the fluid, into the rigid body. Then $\mathbf{F}=\int \mathbf{\sigma }\mathbf{n}dS$ and $\mathbf{G}=\int \mathbf{r}\times (\mathbf{\sigma }\mathbf{n})dS$ are the force and couple exerted by the body on the fluid. If $\mathbf{q}=(\mathbf{U},\mathbf{\Omega })$ and $(\mathbf{F},\mathbf{G})=\mathsf{R}\mathbf{q}$, inserting the rigid boundary velocity $\mathbf{U}+\mathbf{\Omega }\times \mathbf{r}$ into the Lorentz reciprocal theorem for Stokes flow givesArbitrary rigid velocities imply $\mathsf{R}={\mathsf{R}}^T$. A separate integration by parts, using the Stokes equation, givesEquality requires zero rate-of-strain tensor everywhere. Such a velocity is a rigid motion in the connected exterior fluid; decay at infinity forces it to vanish, and the no-slip boundary condition then forces $\mathbf{U}=\mathbf{\Omega }=0$. Hence the resistance matrix is symmetric positive definite. Defining force as fluid-on-body would reverse its sign, so the positive convention matters.

For the rods, choose body axes along $OA$, $OB$, and their normal. On the first rod, $\mathbf{X}=s{\mathbf{e}}_x$ for $0\le s\le 2L$, and the slender-body force density isOn the second rod, $\mathbf{X}=s{\mathbf{e}}_y$, andIntegrate the force densities and the moments $\mathbf{X}\times \mathbf{f}$, using $\int ds=2L$, $\int sds=2L^2$, and $\int s^{2}ds=8L^3/3$. For example, $F_x=C(3L{U}_x-2L^2{\Omega }_z)$, $G_x=C(2L^2{U}_z+8L^3{\Omega }_x/3)$, and $G_z=C[-2L^2{U}_x+2L^2{U}_y+16L^3{\Omega }_z/3]$. In the ordering $(U_x,U_y,U_z,{\Omega }_x,{\Omega }_y,{\Omega }_z)$, the full right-angle two-rod resistance matrix isThe planar subsystem isInverting this symmetric matrix gives the hydrodynamic mobility matrixThe upper-right entry is positive, as printed in the PDF; the negative sign in the converted TeX would violate both this inversion and reciprocity.

Let laboratory $X$ point along the initial $OA$ and laboratory $Y$ point upward. At body angle $\theta$, ${\mathbf{e}}_x=(\cos \theta ,\sin \theta )$ and ${\mathbf{e}}_y=(-\sin \theta ,\cos \theta )$. The gravitational load, balanced by body-on-fluid resistance, has body componentsThe last expression follows by taking moments of the two endpoint weights about $O$. Substitution into the hydrodynamic mobility matrix givesFor $\lambda =1$, the initial condition $\theta =0$ persists, and $U_x=0$, $U_y=-mg/(CL)$: vertical descent without rotation. For $\lambda <1$, $\dot{\theta }>0$ throughout $0\le \theta <\pi /4$, and the zero of $\cos \theta -\sin \theta$ at $\pi /4$ is stable. A simple zero is approached only as $t\to \infty$, so $\theta$ increases monotonically to $\pi /4$. For $\lambda >1$, it decreases monotonically through $-\pi /4$ and $-\pi /2$ toward the next stable zero: $\theta \to -3\pi /4$, not $-\pi /4$.

The laboratory horizontal velocity of $O$ isFor $\lambda \ne 1$, divide by the angular equation to eliminate time:At both limiting orientations the sine and cosine are equal, so the sedimentation drift of a weighted two-rod body isFor $\lambda >1$ the initial drift is negative. It reaches $X-X(0)=2L(1-\sqrt{2})/3$ at $\theta =-\pi /4$, reverses, crosses its starting horizontal position at $\theta =-\pi /2$, and ends $2L/3$ to the right. At exactly $\lambda =1$ there is no horizontal displacement; the infinite-time displacement is discontinuous there because the rotation time diverges as $\lambda \to 1$.

For a Newtonian fluid, write $\mathbf{\sigma }=-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation without body force gives ${\partial }_j{\sigma }_{ij}=0$. Since the stress tensor is symmetric, its contraction with the antisymmetric part of the velocity gradient is zero. ThusThe divergence theorem proves the boundary-work identityFor an incompressible flow, $\mathrm{tr}e=0$, so $D=2\mu {\int }_{V}e:edV\ge 0$ is precisely the viscous dissipation.

Apply this identity to the particle-free volume and then to the fluid outside the added particle. On the rigid particle, the velocity has the form $V+\Omega \times x$. Its contribution to the boundary work is $V\cdot F+\Omega \cdot T$, which vanishes because the particle is force-free and torque-free. The outer velocity is unchanged. Subtracting the two boundary-work identities therefore givesTo move this expression to the particle surface, apply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the punctured fluid domain. For completeness, its local proof isOn the outer boundary $u^{\prime }=0$. Let $N=-n$ be the particle-outward normal, opposite to the fluid-outward normal there. Integrating the preceding divergence identity givesThis expression includes the removal of the original fluid inside the particle; that removal must not be counted separately again.

For a sphere in a uniform straining Stokes flow, $E$ is symmetric and trace free, and $u_0=Ex$. Symmetry gives zero rigid translation and rotation. Put $H=x\cdot Ex$ and $r=|x|$. Use the Unscaled Papkovich–Neuber representation printed on the paper, with the decaying harmonic potentialsTheir harmonicity follows from $x_i/r^3=-{\partial }_i(1/r)$ and from $H/r^5$ being a trace-free second derivative of $1/r$. Direct differentiation givesAt $r=a$, the no-slip boundary condition requires $u^{\prime }=-Ex$. Matching the independent $Ex$ and $Hx$ terms fixes $B=-1/2$ and $C=5/6$. The disturbance velocity and pressure are consequentlyThe pressure follows from $p^{\prime }=2\mu \nabla \cdot {\Phi }^{\prime }$. The disturbance decays, the total velocity vanishes at the sphere, and the harmonic representation verifies the Stokes equation. The boundary traction has zero total force by odd symmetry, and zero total torque because $E$ is symmetric, confirming the assumed force-free and torque-free motion.

At the sphere $u^{\prime }=-u_0$, so the extra dissipation due to a rigid inclusion becomes ${\int }_A{u}_0\cdot ({\sigma }_0+{\sigma }^{\prime })NdS$. Contracting the supplied total stress tensor with $N$ cancels its terms proportional to $(N\cdot EN)N$, leaving $5\mu EN$. ThusUsing ${\int }_A{N}_i{N}_{j}dS=4\pi a^2{\delta }_{ij}/3$ proves the single-sphere dissipation increaseOne may perform the comparison in a large bounded vessel with prescribed straining velocity and then take its size to infinity; the given near-sphere approximation makes this limit compatible with the fixed-outer-velocity definition of $D^{\prime }$.

The sphere number density is $n_s=\varphi /(4\pi a^3/3)=3\varphi /(4\pi a^3)$. Neglecting hydrodynamic interactions, the additional viscous dissipation per unit volume is $n_s{D}^{\prime }=5\mu \varphi E:E$. Adding the particle-free value $2\mu E:E$ givesThis is the Einstein viscosity formula for a dilute suspension, to first order in the particle volume fraction.

Use the Unscaled Papkovich–Neuber representation with $e_k=e^{ikx-ky}$ and harmonic potentialsTo see how these coefficients arise, start with $\Phi =(0,A{e}_k)$ and $\chi =B{e}_k$. The representation givesAt $y=0$, the prescribed tangential velocity requires $ikB=U$, while zero normal velocity requires $A=-kB=iU$. Hence the decaying half-space flow isThe pressure follows from $2\mu {\partial }_y{\Phi }_y$. These harmonic potentials verify the Stokes equation and incompressibility; the polynomial factors do not prevent decay as $y\to \infty$.

For the Newtonian fluid stress tensor, ${\sigma }_{xy}=\mu (u_y+v_x)$ and ${\sigma }_{yy}=-p+2\mu v_y$. On the boundary $u_y=-2kU{e}^{ikx}$, $v_x=0$ and $v_y=-ikU{e}^{ikx}$. ThereforeTake real parts for a physical sinusoidal flow. Together with the normal-velocity result supplied in the question, these formulas give the Fourier traction map for a viscous half-space; for signed wave numbers its resistance coefficient is $2\mu |k|$.

Applying the stated long-wave expression formally in the intended high-viscosity regime, the numerator is dominated by $2\lambda {\kappa }^2/3$ and the displayed denominator by $\lambda$. It givesThe mechanism is inner transverse shear: the outer fluids almost immobilize the interfaces tangentially, so the middle layer has a pressure-driven parabolic velocity instead of the plug-like extensional velocity of part (b). Indeed, with nearly stationary faces at $z=\pm h$, its integrated flux is $-2h^3{p}_x/(3\mu )$. With $p_x=\Delta \rho g{h}_x$, conservation of mass gives precisely the displayed decay rate. The leading viscous dissipation is then in the middle layer, even though its neighbours have much greater dynamic viscosity.

There is a nonuniformity in taking this conclusion literally for every $\lambda \gg {\kappa }^{-1}$. A long-wave remainder written for fixed $\lambda$ cannot be assumed uniform as $\lambda$ diverges. Here this matters physically because sufficiently viscous outer fluids also resist normal interface motion. The distinction can be resolved directly, without an unspecified remainder, from the Stokes equation. For a symmetric thickness mode, the middle vertical velocity has the formIt is odd in $z$, and the middle horizontal velocity is even. At $z=h_0$, the exact outer traction map is ${\sigma }_{xz}^{+}=-2\lambda \mu ku$ and ${\sigma }_{zz}^{+}=-2\lambda \mu kw$. Tangential continuity and the normal hydrostatic jump therefore giveThe latter uses $p=\mu (w_{zzz}-k^2{w}_z)/k^2$, obtained from horizontal force balance. The kinematic condition is $w(h_0)=-s\eta$. Writing $S=\sinh \kappa$ and $C=\cosh \kappa$, tangential continuity fixesEliminating $A,B$ in the remaining two equations gives the exact symmetric-layer rateFor $\kappa \ll 1$ and $\lambda \kappa \gg 1$, this proves the uniform high-viscosity limit of viscous-layer relaxationThus the intended inner-shear answer is valid in the distinguished range ${\kappa }^{-1}\ll \lambda \ll {\kappa }^{-3}$. If $\lambda {\kappa }^3\gg 1$, the main resistance is instead outer normal motion andBoth ranges satisfy the printed $\lambda \gg {\kappa }^{-1}\gg 1$. For example $\lambda ={\kappa }^{-4}$ lies in the latter range and has $s\sim [\Delta \rho g{h}_0/(2\mu )]{\kappa }^3$, rather than a rate proportional to ${\kappa }^2$. The exact denominator contains a ${\lambda }^2{\kappa }^3$ contribution hidden by a fixed-parameter remainder. The inner-shear limit is the intended result, but an additional upper restriction on $\lambda$ is needed to make it a uniform conclusion.