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For a force $\mathbf{F}\delta (\mathbf{x})$ acting on an unbounded incompressible fluid, the Stokeslet and its pressure areThey solve $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}\delta =0$ and $\nabla \cdot \mathbf{u}=0$. Approximate a slender body by a line distribution of these point forces on its centreline. At a point labelled by arclength $s$,To extract the large logarithm, write $s^{\prime }=s+\ell$ and approximate $\mathbf{X}(s+\ell )-\mathbf{X}(s)\simeq \ell \mathbf{t}$, $\mathbf{f}(s+\ell )\simeq \mathbf{f}(s)$, where $\mathbf{t}={\mathbf{X}}^{\prime }$ is the unit tangent vector. The near-line contribution has kernel $(I+\mathbf{tt})/(8\pi \mu |\ell |)$. Cut it off at the wire radius $R$ and a macroscopic length $L$. The two sides of the line giveThe far contribution and changes in the cutoff add terms without the leading logarithm. Since $(I+\mathbf{tt}{)}^{-1}=I-\mathbf{tt}/2$, inversion proves the leading slender-body force densityThis is force on the fluid; the fluid's force on the wire has the opposite sign. It gives twice the resistance to transverse motion as to tangential motion at this order.

The stated arclength describes the circular helix $\mathbf{X}=(a\cos \theta ,a\sin \theta ,b\theta )$. The PDF omits the second $a$ in the coordinates, so the literal printed curve and its arclength are inconsistent unless $a=1$. First use the circular interpretation consistent with the given arclength. Put $q=\sqrt{a^2+b^2}$ and $c=2\pi \mu /\ln (L/R)$. ThenRotation about $z$ has velocity $\mathbf{v}=\Omega (-a\sin \theta ,a\cos \theta ,0)$, so $\mathbf{t}\cdot \mathbf{v}=\Omega a^2/q$. The axial force density is $f_z=-c\Omega a^{2}b/q^2$. Integrating over one turn givesFor translation, $\mathbf{v}=W\hat{\mathbf{z}}$ and $\mathbf{f}=cW(2\hat{\mathbf{z}}-(b/q)\mathbf{t})$. Only the tangential term contributes to axial couple. Since $(\mathbf{X}\times \mathbf{t}{)}_z=a^2/q$, its density is $-cW{a}^{2}b/q^2$, givingThese are the off-diagonal coefficients of the axial resistance matrix of a slender helix. Their equality does not depend on this particular geometry: with symmetric local resistance $D=c(2I-\mathbf{tt})$, the rotation-induced axial force is $\int \hat{\mathbf{z}}\cdot D(\hat{\mathbf{z}}\times \mathbf{X})ds$, whereas the translation-induced axial torque is $\int (\hat{\mathbf{z}}\times \mathbf{X})\cdot D\hat{\mathbf{z}}ds$. They agree pointwise by symmetry of $D$. The full Stokes-flow statement is the Lorentz reciprocal theorem.

For completeness, retaining the literal elliptic helix instead gives $s^{\prime }=\sqrt{a^2{\sin }^2\theta +{\cos }^2\theta +b^2}$ and the axial coupling of an elliptic helixThus reciprocity holds for the literal curve as well; only the geometric coefficient and arclength change. In all these formulas the equal and opposite force or couple on the wire reverses the displayed sign.

The leading disturbance in sphere Stokes flow has size $u\sim Ua/r$. Consequently viscous diffusion has size $\nu u/r^2$, background advection has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the sphere Reynolds number $Ua/\nu$ is small, this ratio becomes order one at $r\sim \nu /U$. Thus the Stokes approximation is nonuniform at distances of order $\nu /U$. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background advection and the Oseen approximation is appropriate:Write $\mathbf{u}=\nabla \varphi +\nu \nabla \chi -\mathbf{U}\chi$. It is sufficient outside the origin to impose $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$: these give zero divergence, and substitution into momentum gives $p=-\rho \mathbf{U}\cdot \nabla \varphi$. With $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$, differentiation yieldsA decaying radial choice is $h=C{e}^{-Ur/(2\nu )}/r$. ThereforeThe exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll \nu /U$,Choosing $\varphi =-\nu C/r$ cancels the first term's gradient in $\nabla \varphi +\nu \nabla \chi$. The remaining leading velocity isMatching fixes $C=3a/2$. Thus the potential-source and wake decomposition of sphere Oseen flow isIts pressure is $p=-3\mu a(\mathbf{U}\cdot \mathbf{x})/(2r^3)$, the sign appropriate to a fixed sphere in an incident flow. The velocity decays in every direction and matches the requested leading inner disturbance; the sphere's potential-dipole term enters a higher matching order.

The scalar potential is a point source: $\nabla \varphi =(3a\nu /2)\mathbf{x}/r^3$, so integration over a centred sphere givesTo see the fluid wake, take $z$ downstream along $\mathbf{U}$ and transverse radius $R_{\perp }$. For $z\gg \nu /U$, $r-z\simeq R_{\perp }^2/(2z)$, givingThe fluid wake is concentrated within $R_{\perp }=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int \chi dA\simeq 6\pi \mu a$, balancing the outward potential-source mass flux. The corresponding momentum deficit iswhich equals the leading sphere drag. The Gaussian integral used here is ${\int }_0^{\infty }{e}^{-U{R}_{\perp }^2/(4\nu z)}2\pi R_{\perp }d{R}_{\perp }=4\pi \nu z/U$.

The PDF labels $6\pi \mu aU$ as the mass-source strength, but this has dimensions of force. The actual mass-source strength is $6\pi \mu a$; $6\pi \mu aU$ is the momentum deficit or drag. The decomposition above supplies both quantities and identifies the printed normalization error.

First solve the hinted fixed-gap problem. Let $y$ measure distance across the gap and $H(x)=h+x^2/(2a)$ be its leading local height. The lubrication approximation gives $\mu u_{yy}=p_x$, with $u=0$ at $y=0,H$, soThe constant flux per unit transverse length is $q$. Integrating its pressure gradient gives the pressure-driven flux through a parabolic lubrication gap:With $x=\sqrt{2ah}t$, the integral is $\sqrt{2a}{h}^{-5/2}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-3}dt$. The substitution $t=\tan \chi$ gives ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\chi d\chi =3\pi /8$, and therefore

Now take the sphere speed to be $U>0$ downwards and assume zero net liquid through-flow, so the displaced liquid must pass through the annular gap. Near the equator, the sphere's surface radius is $R-z^2/(2R)+\cdots$, giving $H(z)=d+z^2/(2R)$. In the sphere frame the relative volume flux is $U\pi (R+d{)}^2$; dividing by circumference $2\pi R$ gives $q\simeq UR/2$. The Couette contribution from the moving tube wall is only order $Ud$, smaller by $d/R$, so the fixed-gap pressure result givesThe pressure changes principally within axial distance $\sqrt{Rd}$ around the equator. It is nearly constant on each hemisphere, so its leading force is the pressure difference times projected area, $F_p\simeq \pi R^2\Delta p$. Direct viscous shear contributes a relative order $d/R$ and is smaller. Balance with the buoyant weight:Thus the settling speed of a nearly occluding sphere without through-flow isThe printed speed lacks the factor $1/\pi$. The parabolic-gap integral above fixes the normalization: its $3\pi /8$ cannot be dropped. With standard dynamic viscosity and Stokes drag law conventions, the requested coefficient must be corrected by dividing the printed expression by $\pi$. Also, the bypass result assumes no net through-flow; allowing a piston-like displacement of the entire fluid column would require specifying tube end conditions and a different resistance balance.

A sphere of radius $d$ in an unbounded fluid has sedimentation speed $U_d=2\Delta \rho g{d}^2/(9\mu )$. ThereforeThe large confined sphere falls much more slowly even than this much smaller unconfined sphere.

For radial offset $e<d$, let $ϵ=e/d$. To leading order the narrowest gap varies as $d(\varphi )=d(1+ϵ\cos \varphi )$. All circumferential strips share the same pressure drop and act as parallel channels. Inverting the gap-resistance formula shows that each carries flux proportional to $d(\varphi {)}^{5/2}\Delta p$. At fixed buoyant load, the leading pressure drop is fixed, so the speed ratio isThe eccentricity increases narrow-gap bypass mobility result holds because the gain through the wider side outweighs the loss through the narrower side. Strict convexity of $x^{5/2}$ and the zero mean of cosine prove the inequality. Pairing the positive- and negative-cosine contributions in the derivative proves monotonicity with offset. The broader hint's assertion for every exponent $\alpha >0$ is not correct for $0<\alpha <1$, when concavity reverses the inequality; the required exponent $5/2$ is in the valid convex range.

The Minimum-dissipation theorem for Stokes flow compares fields with fixed geometry and boundary velocities. Moving the sphere changes the geometry, while sedimentation prescribes its effective weight rather than its velocity. It therefore supplies no rule for migrating toward a position of least dissipation. Indeed, at fixed force $W$, the power is $WV$, and a lower drag resistance gives a larger settling speed and larger power, rather than a smaller power. This is a fixed-force comparison of minimum viscous dissipation issue.

The proposed migration is itself false in ideal Stokes flow. Reflect the uniform vertical tube in a horizontal plane through the sphere centre. The sphere's lateral position is unchanged, but the applied axial force reverses. Reflection leaves a transverse translational velocity unchanged. By Linearity of Stokes flow, however, reversing the force reverses that transverse velocity. It must therefore be zero. This proves no lateral migration of a single sphere in a uniform tube in Stokes flow:Rotation and a wall-dependent vertical speed are allowed. Inertial lift, particle deformation or tube-end effects belong to different problems and do not follow from the dissipation theorem.

For a fixed fluid domain, consider an incompressible Newtonian Stokes flow with prescribed velocities on its boundaries and prescribed far-field behaviour. Conservative body forces, such as gravity, may be absorbed into the pressure. Among all sufficiently regular divergence-free trial velocity fields with exactly those boundary velocities and far-field behaviour, the Stokes solution minimizes the viscous dissipationTo see this Minimum-dissipation theorem for Stokes flow, write a trial field as $\mathbf{v}=\mathbf{u}+\mathbf{w}$, with $\nabla \cdot \mathbf{w}=0$ and homogeneous velocity boundary data. The Newtonian fluid stress tensor satisfies $\nabla \cdot \mathbf{\sigma }=0$, so integration by parts givesThereforeEquality requires the difference to be a rigid motion, eliminated by the fixed wall and far-field data. The theorem varies the velocity field in one fixed geometry; it does not minimize over particle positions. If a nonconservative body force is prescribed, the appropriate variational functional also includes its work, rather than being the dissipation alone.

Let downward velocity be positive in this part, and let $W>0$ be the dense sphere's weight minus buoyancy. For the original centreline configuration, write its six-component translation-and-rotation velocity as $q$, and its diagonal positive hydrodynamic resistance matrix as $R_0$. Its single-sphere dissipation isWith only axial force and no torque, the original velocity is $V_0=W/R_{zz}$.

Now add the neutral sphere. In its actual Stokes flow, extend the velocity rigidly through the new sphere's interior. This extension is a valid incompressible trial field for the domain without that sphere, with exactly the dense sphere's actual six boundary velocities; its strain and dissipation inside the inserted ball are zero. The Minimum-dissipation theorem for Stokes flow and extra dissipation due to a rigid inclusion therefore giveThe neutral sphere has zero net external force and torque, and the dense sphere has only the axial force $W$. The dissipation equals total external power, so $D=W{V}_z$. Positivity gives $V_z>0$, and the comparison yields $V_z\le W/R_{zz}$ without assuming that either sphere translates only vertically or that either angular velocity vanishes.

For a finite nonoverlapping placement, the first inequality is strict. Equality would make the extended field the old Stokes solution and thus make that solution rigid throughout the added ball. Analyticity and connectedness then force its strain to vanish throughout the original fluid domain; the stationary tube wall and quiescent far field would force the whole motion to vanish, contradicting $V_z>0$. Hence the force-free inclusion reduces axial mobility of a centred settling sphere:The difference tends to zero as the added sphere recedes infinitely far away. The lateral and rotational contributions are nonnegative in the resistance comparison, which is precisely why they cannot invalidate the bound.

When both spheres have the same effective downward weight, the second sphere adds a force-driven disturbance instead of being merely a force-free rigid inclusion. The power now contains both $W{V}_{1z}$ and $W{V}_{2z}$, so the one-forced-sphere bound from part (iv) no longer applies.

For $a\ll R$, place the spheres in a tandem configuration with $a\ll s\ll R$. Locally the flow around the second sphere is dominated by its downward Stokeslet. Its axial incident velocity at the first sphere is downward, so the first sphere falls faster to leading order. By contrast, the zero total flux for localized tube Stokes flow requires compensating upward flow away from the downward-moving particle core in sections cutting the sphere. There are therefore positions in the return flow where the disturbance acts upwards.

One can choose the second sphere in such an upward-flow region generated by the first. The Lorentz reciprocal theorem makes the axial cross coefficients of the hydrodynamic mobility matrix equal, so the downward force on the second sphere then produces an upward contribution to the first sphere's velocity. With small well-separated spheres, this force-driven interaction dominates the higher-order self-resistance change from simply inserting the other sphere. In mobility notation, with downward positive,and the cross coefficient can have either sign in the tube. ThusThe distinction is the sign of the incident flow, not whether adding a second geometrical obstacle lowers a fixed-velocity dissipation.

Choose time $t_c$ when the passing sphere centres are at the same height, and let $S$ reflect positions in that horizontal plane. For each instantaneous configuration, Stokes flow determines particle velocities uniquely and linearly from the fixed external forces. Time reversal reverses all velocities and the dense sphere's applied force; reflecting in the horizontal plane reverses that force again. The neutral sphere remains force-free under both operations. Thus the reflected time-reversed motion obeys the same original forcing.

At $t_c$, reflection fixes both particle centres. Since the spheres have no orientational shape variables and there is no contact, uniqueness of the particle-trajectory initial-value problem makes the reflected backward motion coincide with the forward motion:This reflection symmetry of a two-sphere passing trajectory fixes the transverse coordinates and reverses the relative axial separation. Consequently, at matching large separations above and below, both spheres return to their original distances from the centreline. Their intermediate lateral excursions are undone. This uses kinematic reversibility of Stokes flow, geometric reflection symmetry and the noncontact result, not a dissipation-minimizing trajectory. It assumes the spheres pass one another; a perfectly collinear approach would not supply the side-by-side symmetry event.

One normalization of the Papkovich–Neuber representation uses a harmonic vector $\mathbf{\Phi }$ and harmonic scalar $\psi$:These formulae give $\nabla \cdot \mathbf{u}=0$ because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and give $\mu {\nabla }^2\mathbf{u}=\nabla p$. Rescaling the potentials gives the other common normalizations of the representation.

A point force requires a decaying, rotationally covariant field linear in $\mathbf{F}$, with no length scale. Thus choose the harmonic monopole $\mathbf{\Phi }=c\mathbf{F}/r$, $\psi =0$, for $r>0$. Differentiating givesThe Newtonian fluid stress tensor is ${\sigma }_{ij}=-p{\delta }_{ij}+\mu ({\partial }_i{u}_j+{\partial }_j{u}_i)$, which simplifies to ${\sigma }_{ij}=-3\mu c(\mathbf{F}\cdot \mathbf{x})x_i{x}_j/r^5$. Its traction resultant on a sphere, using an outward normal from the origin, is $-4\pi \mu c\mathbf{F}$. Momentum balance with the force $\mathbf{F}\delta (\mathbf{x})$ therefore fixes $c=1/(4\pi \mu )$. The resulting Stokeslet isAn arbitrary constant pressure may be added. The stress has purely radial traction on concentric spheres, and its integrated traction $-\mathbf{F}$ supplies the required distributional point-force normalization.

Take the upward direction as ${\mathbf{e}}_z$, with polar angle $\theta$ measured from it. The liquid is quiescent at infinity. On a translating clean inviscid bubble, the kinematic boundary condition is $(\mathbf{u}-U{\mathbf{e}}_z)\cdot \mathbf{n}=0$, and the stress-free boundary condition is zero tangential traction. The normal traction must balance uniform gas pressure and any constant-curvature surface-tension jump.

For a vertical Stokeslet of strength $F$, the radial velocity is $u_r=F\cos \theta /(4\pi \mu r)$. Its traction on a concentric sphere is purely normal, so zero tangential traction is automatic. The kinematic condition at $r=a$ givesThe singularity is only a device for generating the exterior solution; there is no liquid point force inside the actual bubble. The integrated dynamic traction on the bubble is $-F{\mathbf{e}}_z$, giving the clean-bubble Stokes drag. The upward buoyancy is $4\pi \rho g{a}^3/3$, hence force balance gives

The normal stress also shows that surface tension is not required for this particular spherical solution. Let the ambient hydrostatic pressure at the bubble centre be $p_c$, so on its surface $p_h=p_c-\rho ga\cos \theta$. Adding the Stokeslet's dynamic normal traction giveswhere the terminal buoyancy value of $F$ cancels the dipole. This is the spherical clean bubble without surface tension balance: an initially spherical bubble can remain spherical in this ideal steady Stokes solution even when surface tension is zero. With uniform surface tension $\gamma$, choose gas pressure $p_b=p_c+2\gamma /a$; with $\gamma =0$, $p_b=p_c$. This proves compatibility of the spherical shape, not stability of arbitrary deformations or validity when inertia matters.

The force scale is $F\sim \rho g{a}^3$. At the distant free surface, the Stokeslet's pressure and viscous normal-stress scales are $F/d^2$. Gravity restores a displaced surface by a pressure change $\rho gh$, soThus $h/d=O((a/d{)}^3)\ll 1$, consistent with treating the boundary as approximately flat and impermeable at leading order. This uses the stipulated gravity-dominated restoring balance.

The liquid-air interface has negligible tangential traction. The stress-free image of a normal Stokeslet places an opposite point force at the reflection point, a distance $2d$ above the bubble centre. Across the flat surface the combined normal velocity is odd and the tangential velocity is even, so the normal velocity and tangential stress vanish there. A same-radius inviscid drop of density $2\rho$ has downward excess weight $4\pi \rho g{a}^3/3$, exactly opposite the bubble's upward buoyancy. Its exterior Stokeslet therefore supplies this image, giving the asserted full-space comparison to leading order in $a/d$.

The image's incident axial velocity at the bubble isTo first reflection, the bubble translates with this locally uniform incident velocity plus its isolated rise speed $F/(4\pi \mu a)$. Hence the bubble rise near a distant stress-free free surface isOnly the first relative correction is retained; the image's spatial variation and additional reflections affect higher terms.

The customary capillary-controlled small-deformation condition is a small capillary number,equivalently a small Bond number $\mathrm{Bo}=\rho g{a}^2/\gamma$. This is a sufficient condition, not a necessary condition for the special isolated spherical solution in part (b). More precisely, the image's strain is $O(F/(\mu d^2))$, so its deformation-producing normal stress compared with $\gamma /a$ is $O(\mathrm{Bo}(a/d{)}^2)$. When exploiting the exact isolated solution, that smaller combination controls the leading boundary-induced distortion. If capillarity at the distant free surface is also retained, the gravity-dominated deflection scaling requires $\gamma /(\rho g{d}^2)\ll 1$; $a\ll \sqrt{\gamma /(\rho g)}\ll d$ allows both the usual spherical-bubble criterion and gravity-dominated surface restoration.

Let the undisturbed water surface be $z=0$, the oil's upper surface $z=\eta (x,t)$ and its lower surface $z=b(x,t)$, so $h=\eta -b$. Leading hydrostatic pressure in the nearly inviscid water is $p_w=-{\rho }_{w}gz$ relative to atmospheric pressure; in the oil it is $p_H=\rho g(\eta -z)$. Equality at the interface gives $\rho gh=-{\rho }_{w}gb$, henceThus the horizontal hydrostatic driving gradient isThis reduced gravity accounts for most of the oil thickness being below the undisturbed water level.

Because the oil is much more viscous than the water and air, tangential tractions on both faces are negligible. In a thin layer, the leading horizontal velocity is consequently independent of depth: the floating extensional viscous gravity current is a plug, rather than the parabolic profile of a no-slip gravity current. Integrating incompressibility $u_x+w_z=0$ between the moving material boundaries, and using their kinematic boundary condition, gives conservation of oil volume,To derive the factor $4$ in the extensional balance, incompressibility gives $w_z=-u_x$. The Newtonian fluid stress tensor has ${\sigma }_{zz}=-p-2\mu u_x$. Normal-stress balance relative to the hydrostatic part therefore requires $p=p_H-2\mu u_x$. The horizontal stress then satisfiesThus the longitudinal tensile force in an oil slice, relative to its hydrostatic pressure, is $4\mu h{u}_x$ per unit width. Balancing its difference across the slice with the horizontal driving force $-h{p}_{H,x}dx$ yieldsThe coefficient is the planar extensional Trouton ratio, arising from both the direct horizontal strain and the pressure correction required by the transverse normal-stress condition.

Using the pressure gradient above, integration in $x$ givesThe stated nose stress condition sets $C(t)=0$, not an arbitrary imposed inlet tension. ConsequentlyFor a material slice injected at time $t_0$, let $\tau =t-t_0$ be its age. The thickness equation along its Lagrangian trajectory is $dh/dt=-h{u}_x=-k{h}^2$, with $h=h_0$ at injection. DefineThen $h=h_0/(1+\tau /T)$. The slice injected during $d{t}_0$ has volume $h_0{u}_{0}d{t}_0$ per unit width, so at fixed observation time $t$,Since a newly injected slice has $x(t,t)=0$, integration and subsequent differentiation along a material slice give the constant-flux floating extensional gravity current:The oldest slice forms the nose. Measuring time from the start of injection, $t_0=0$ there, soIts speed ${\dot{x}}_N=u_0(1+t/T)$ is exactly the material velocity at the nose, as required.

For a fixed position $x$ behind the nose, the age needed to reach it is determined by $x/u_0=\tau +{\tau }^2/(2T)$, independently of the observation time. Hence both thickness and velocity become stationary there once the nose has passed. Eliminating $\tau$ givesTheir product is $h_0{u}_0$, so the steady profile carries precisely the inlet volume flux; integrating it to $x_N(t)$ gives the total injected volume $h_0{u}_{0}t$.

The lengthis the horizontal distance over which reduced-gravity driving causes an order-one change in inlet speed and thickness by viscous extension. The associated age is $T=8\ell /u_0$; the exact profile contains $x/(4\ell )$. It is an extensional adjustment length, not a shear-diffusion scale. The result applies while the stated thin-layer and negligible-inertia approximations remain valid.

Take positive $x$ in the snail's direction of travel, and let $y$ measure height above the rigid plane. The wave moves at laboratory velocity $U-V$. In the wave frame the plane's horizontal velocity is $V-U$ and the foot's is $V$. The foot is stationary as a shape, so its material velocity is tangent to $y=h(x)$: to leading lubrication theory order its components are $(V,V{h}_x)$. Apply the no-slip boundary condition to both surfaces.

The leading Stokes flow equations are $p_y=0$ and $\mu u_{yy}=p_x$. Their solution is the local Couette-Poiseuille flowThe normal-velocity condition on each steady surface and incompressibility make the wave-frame flux independent of $x$. Integrating the velocity givesPut $C=V-U/2$. Periodicity of pressure gives $0={\int }_0^L{p}_{x}dx=12\mu (C{I}_2-q{I}_3)$, and henceThe shear traction exerted by the fluid on the plane, positive in the $x$ direction, is

The snail is force-free horizontally: its gait consists of internal motion and it has no externally applied horizontal drive. Uniform locomotion therefore requires zero net horizontal fluid force on the foot per wavelength. The stresses on opposite periodic fluid faces cancel, so horizontal momentum balance also makes the net force on the plane zero. ThusSubstitution of $q$ proves the lubrication model of a crawling snail:The Cauchy-Schwarz inequality applied to $h^{-1/2}$ and $h^{-3/2}$ gives $0<\alpha \le 1$, with equality exactly for a uniform gap. Thus the denominator is positive, and a nonuniform waveform gives forward motion; the flat gap gives $U=0$.

For the sinusoidal shape, the given integrals yieldTherefore $\alpha =1/(1+A^2/2)$ andFor small amplitude, $U\sim 3A^{2}V$: the leading locomotion is quadratic, and there is no propulsion from a flat foot. As $A\to 1^{-}$, $U\to V$, so the wave becomes nearly stationary in the laboratory frame while the snail advances relative to it. However, the minimum gap tends to zero; finite speed in this limit does not imply finite dissipation, and $A=1$ itself is excluded by the noncontact model.

For the general waveform, use the mechanical work identity for viscous dissipation. Pressure does no boundary work because the boundary motion is tangential in the wave frame. The plane's velocity is constant, and its total force is zero, so it contributes zero net work. The upper boundary supplies, at leading lubrication order,The zero plane-force relation is $\mu U{I}_1=\frac{1}{2}\int h{p}_{x}dx$. ConsequentlyAs a direct check, integrating the squared shear rate gives $\Phi ={\int }_0^L[\mu U^2/h+h^3{p}_x^2/(12\mu )]dx=\mu I_1[U^2+12C^2(1-\alpha )]$, which reduces to the same result using the force-free speed relation. Zero horizontal force on the foot does not mean zero gait work: the tangential material velocity along its sloping surface includes vertical motion.

For prescribed speed $U>0$ and sinusoidal shape, eliminate $V=U(1+2A^2)/(3A^2)$ to obtainWriting $y=A^2$, the logarithmic derivative of the amplitude-dependent factor isThe only root in $0<y<1$ is $y=1/2$. The dissipation diverges at both endpoints, so this is the global minimum and gives the optimal sinusoidal waveform for viscous snail locomotion:The magically driven flat glider has simple Couette flow $u=Uy/h_0$ in the laboratory frame, and dissipation ${\Phi }_{\mathrm{glide}}=\mu L{U}^2/h_0$ per wavelength. HenceThe comparison is at the same $U$, mean thickness $h_0$ and wavelength $L$; the flat glider is externally propelled, whereas the snail must power its internal waveform.

Let $\chi =H(F)$ use the Heaviside step function, and ${\delta }_s=|\nabla F|\delta (F)$, so that ${\mathbf{n}}_s=\nabla F/|\nabla F|$ points from the body into the fluid. The kinematic boundary condition gives $F_t+\mathbf{u}\cdot \nabla F=0$ on the surface. In the exterior, mass conservation and the Euler equations for an inviscid fluid areProducts with $\chi$ are understood distributionally, using ${\partial }_t\chi =F_t\delta (F)$ and ${\partial }_i\chi =F_i\delta (F)$. With ${\rho }^{\prime }=\rho -{\rho }_0$, the extended continuity equation isThe corresponding extended momentum equation isbecause its advective surface term is $\rho u_i(F_t+\mathbf{u}\cdot \nabla F)\delta (F)=0$. Differentiate continuity in time and substitute the divergence of momentum. Subtracting $c_0^2{\nabla }^2({\rho }^{\prime }\chi )$ gives the Ffowcs Williams-Hawkings equationwhere the inviscid Lighthill stress tensor isIf $p$ denotes total pressure, this tensor includes the constant ambient pressure $p_0$. Its contribution $p_0{\nabla }^2\chi$ cancels the corresponding surface term exactly. For radiation integrals it is convenient to subtract $p_0$ from both terms, making $T_{ij}=\rho u_i{u}_j+(p^{\prime }-c_0^2{\rho }^{\prime }){\delta }_{ij}$ with $p^{\prime }=p-p_0$. The displayed source identity is unchanged by this consistent pressure-gauge subtraction.

Assume no incoming sound and no free initial acoustic field. Put $R_y=|\mathbf{x}-\mathbf{y}|$, ${\tau }_y=t-R_y/c_0$, $Q={\rho }_0{u}_n{\delta }_s$, and $P_i=p^{\prime }{n}_{s,i}{\delta }_s$. Convolution with the retarded acoustic Green function gives the three source integrals, for an observer outside the body:These are the surface thickness, surface loading and volume quadrupole contributions. The retarded surface Dirac delta distributions must remain inside the integrals: integrating them over a moving retarded surface produces the moving-surface retarded Jacobian, and is not generally the same as simply substituting one retarded time into an instantaneous surface integral.

For an acoustic compact-source approximation, suppose the dominant sources have size $\ell$, source frequency scale $\omega$, and $\omega \ell /c_0\ll 1$. Also require $r=|\mathbf{x}|\gg \ell$ and the acoustic far field $\omega r/c_0\gg 1$. Then $R_y\simeq r-\mathbf{n}\cdot \mathbf{y}$, $1/R_y\simeq 1/r$, and the leading source integrals use the common time $\tau =t-r/c_0$. Slow source motion relative to $c_0$ is implicit in this leading retardation approximation. DefineThus $m$ is net displaced mass per unit time, $\mathbf{f}$ is the fluid pressure force on the body, and $S_{ij}$ is the integrated excess Lighthill stress tensor. The minus sign in $\mathbf{f}$ fixes the requested loading convention; the force on the fluid is opposite. On the radiating $1/r$ terms, ${\partial }_{x_i}\simeq -(n_i/c_0){\partial }_t$, givingSpatial derivatives of $1/r$ produce near-field terms of orders $r^{-2}$ and $r^{-3}$, omitted here. Neglecting source delays is a separate approximation from discarding those observer near-field terms. If a leading integrated source vanishes by symmetry, its first nonzero delay correction can be as important as another retained multipole.