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Work in the frame in which the distant fluid is stationary, and take the pressure at infinity as zero. In the Unscaled Papkovich–Neuber representation, choose the harmonic potentialsBoth are harmonic away from the sphere's centre: $1/r$ is harmonic there and $\mathbf{V}\cdot \mathbf{x}/r^3=-\mathbf{V}\cdot \nabla (1/r)$. Substitution into $\mathbf{u}=\nabla (\mathbf{x}\cdot \mathbf{\Phi }+\chi )-2\mathbf{\Phi }$, $p=2\mu \nabla \cdot \mathbf{\Phi }$, gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. At $r=a$ the coefficients of $\mathbf{I}$ sum to one and those of $\mathbf{nn}$ sum to zero, so the no-slip boundary condition is satisfied. The field decays at infinity. The Papkovich–Neuber representation enforces the Stokes equation and incompressibility, and Uniqueness of Stokes flow identifies the physical solution.

Differentiate radially at fixed direction $\mathbf{n}$:On the surface, multiplying the specified Newtonian fluid stress tensor by the outward normal from the solid givesThus the surface traction in translating-sphere Stokes flow is uniform and the fluid's force on the sphere isThis is the Stokes drag law; an external force of the opposite sign maintains the motion.

Let $\mathbf{F}$ denote the applied force, and take the normal ${\mathbf{n}}_S$ outward from the solid. At leading order the Stokes drag law and the torque on a rotating sphere in Stokes flow giveThe exact no-slip boundary condition and the exact resultant-load conditions areThe stress condition specifies the total force and torque, not a pointwise prescribed traction.

For the boundary perturbation of a nearly spherical particle, evaluate the no-slip condition at $\mathbf{x}=(a+\epsilon f)\mathbf{n}$. Taylor expansion givesThere is no $f{\mathbf{\Omega }}_0\times \mathbf{n}$ term because ${\mathbf{\Omega }}_0=0$. Therefore

To expand the integral conditions correctly, use surface independence of Stokes force and torque integrals. The Stokes equation implies $\nabla \cdot \mathbf{\sigma }=0$, and symmetry of the stress tensor implies that the angular-momentum flux is divergence-free as well. Evaluate the exact force and torque on a fixed sphere of radius $R$ enclosing the perturbed particle. There are then no geometric terms: the $O(\epsilon )$ integrals of ${\mathbf{\sigma }}_1$ on that fixed sphere vanish because the external force and torque have no $O(\epsilon )$ corrections. Each perturbation field solves the homogeneous exterior Stokes equations, so the divergence theorem transfers those integrals back to $r=a$. HenceIf the integrals are expanded directly on the displaced surface, terms involving the displaced evaluation of ${\mathbf{\sigma }}_0$, the changed normal and the changed area do appear individually. Their sum vanishes by the same surface-independence identity. Thus their absence in the final equations is a cancellation, not permission to omit geometric terms independently.

First estimate transverse relaxation without assuming a thin cross-section. With both depth and width of order $h_0$, a transverse surface-height variation of order $h_0$ gives a pressure gradient of order $\rho g$. The Stokes equation then gives transverse horizontal and vertical velocities of order $g{h}_0^2/\nu$. Moving the surface through distance $h_0$ takesFor longitudinal variations over $L\gg h_0$, the hydrostatic pressure gradient is of order $\rho g{h}_0/L$, so the longitudinal velocity is $u\sim g{h}_0^3/(\nu L)$. Incompressibility gives a vertical velocity $w\sim h_{0}u/L$, henceThus cross-channel variations relax rapidly compared with the longitudinal evolution. These are viscous estimates, requiring inertial relaxation to be negligible; the slow-flow regime in particular excludes a large transverse inertial response. After that initial relaxation the free surface is nearly horizontal across the channel, so its height is $h_0(x,t)$ and its local vertical depth is $h_0-|y|\cot \alpha$.

Retain the full cross-sectional geometry for a fixed $0<\alpha <\pi /2$; no small-angle approximation is used. With $|h_{0x}|\ll 1$, the leading hydrostatic pressure is $p=p_a+\rho g(h_0-z)$. Axial viscous derivatives are small relative to cross-sectional ones. The axial Stokes equation is thereforeIntroduce $Y=y/h_0$, $Z=z/h_0$, and the triangular domainLet the dimensionless cross-sectional Poisson equation beThe two lower sloping sides impose the no-slip boundary condition; the upper horizontal side imposes the stress-free boundary condition. Symmetry would additionally give ${\varphi }_Y=0$ at $Y=0$ if only half the domain were used. The full domain needs no extra midline boundary condition. Define $Q(\alpha )={\int }_{{\mathcal{D}}_{\alpha }}\varphi dYdZ$. ThenThe cross-sectional area is $A=h_0^2\tan \alpha$. Conservation of mass, $A_t+q_x=0$, gives the V-shaped channel gravity current equationIn particular, neither replacing this triangular Poisson problem by a vertically parabolic film nor taking an extreme angle is necessary.

Put $C=gQ(\alpha )/(\nu \tan \alpha )$. For a symmetric fixed-volume similarity solution, write $h_0=t^{-b}f(x{t}^{-d})$. Volume conservation gives $d=2b$, and balancing $(h_0^2{)}_t=C(h_0^4{h}_{0x}{)}_x$ gives $2b+1=5b+2d$. Hence $b=1/7$, $d=2/7$. Integrating the similarity equation once, with zero flux at the centre, yieldsIntegrating again and imposing $h_0=0$ at $x=\pm X(t)$ givesThe constant-volume similarity in a V-shaped channel has a compact support. If $H(t)=h_0(0,t)$, its volume isEliminating $H$ therefore givesThis describes the large-time spreading from a localized release; the point-release profile is singular at $t=0$, and the idealized front has a narrow region where its divergent slope violates the long-wave approximation. Neither feature changes the leading bulk similarity law.

At a fixed nonzero $x^{\ast }$, the depth is initially zero until $X(t)=|x^{\ast }|$. After arrival it rises and then falls. Because $X^2\propto t^{4/7}$, differentiation of $h_0(x^{\ast },t{)}^3=3(X^2-x^{\ast 2})/(7Ct)$ givesThus its unique maximum occurs at $X^2=7x^{\ast 2}/3$; thereafter the depth tends to zero as $t^{-1/7}$. If $t_a$ is the arrival time, $t_{\max }/t_a=(7/3{)}^{7/4}$. At the singular release point $x^{\ast }=0$, the point-source similarity profile instead decreases from the start; a finite initial release regularizes its early behaviour.

The right-hand sketch shows the triangular bulk at half its previous maximum depth, with thin draining films on the previously wetted portions of both walls. These residual wall films in a receding gravity current are not captured by a perfectly dry-wall triangular section. They drain rapidly compared with bulk longitudinal spreading. For an order-one angle, a wall film of thickness $b$ and wetted length $\ell$ has speed $O(g{b}^2/\nu )$ by falling film flow; its drainage time gives $b\sim (\nu \ell /(gt){)}^{1/2}$. At a fixed nonzero station the previous maximum wetted height is finite, so film area decays as $t^{-1/2}$, faster than the bulk area $h_0^2\sim t^{-2/7}$. In stations whose arrival and peak times are of order the observation time, the typical residual film fraction is $O(\sqrt{{\tau }_y/t})\to 0$. Therefore the films modify the detailed receding cross-section but do not hold a leading-order fraction of the volume or invalidate the large-time nose law. Microscopic contact-line physics is beyond this ideal viscous model.

Let the longitudinal variation scale be $\ell \gg h$. The stress-free boundary condition gives $\mu (u_z+w_x)=0$ at each face. Incompressibility implies $w=O(hu/\ell )$, so $w_x$ is smaller than a putative $u_z=O(u/h)$ by $O(h^2/{\ell }^2)$. The leading Stokes equation and the two shear-free conditions therefore make $u$ independent of $z$; its thickness-dependent correction is smaller-order. Symmetry gives $w=0$ at $z=0$, and incompressibility integrates toVertical momentum makes the leading normal stress independent of $z$. With the normal directed from liquid into gas, the upper-face curvature is $-h_{xx}/2$ to leading order. The Young–Laplace equation givesThe Newtonian fluid stress tensor also gives ${\sigma }_{zz}=-p+2\mu w_z=-p-2\mu u_x$. Hence $p=p_a-\gamma h_{xx}/2-2\mu u_x$ andThe factor four is the planar extension factor: both the longitudinal viscous stress and the pressure adjustment from transverse contraction contribute.

For force balance on the illustrated slice, the vertical cuts give $(h{\sigma }_{xx}{)}_x\delta x$. Pressure on the sloping broad faces gives $p_a{h}_x\delta x$ horizontally. Each of the two interfaces exerts a tangential surface tension pull at each cut, with horizontal component $\gamma \cos \theta$, where $\tan \theta =h_x/2$. Thus the horizontal force balance is the derivative of the effective tensionto the retained small-slope order. The ambient-pressure contributions cancel in this excess tension. Differentiating cancels the two $h_x{h}_{xx}$ terms, leaving the planar viscous-sheet stretching equationsThe upper-face kinematic condition is $w(x,h/2,t)=(h_t+u{h}_x)/2$. Combining it with $w=-z{u}_x$ gives the second equation,which is conservation of mass per unit transverse width.

For the two bubbles, let $p_g$ be the common gas pressure and $p_e$ the surrounding liquid pressure away from the sheet. A cylindrical bubble has one nonzero principal curvature, so the Young–Laplace equation gives $p_g-p_e\simeq \gamma /a$. In the nearly flat sheet the normal stress is $-p_g$, giving liquid pressure $p_{\mathrm{film}}=p_g-2\mu u_x$. The extensional correction is small compared with $\gamma /a$ because the transition scale will satisfy $\sqrt{a{h}_0}/L\ll 1$. Thus $p_{\mathrm{film}}\simeq p_g>p_e$, and this capillary pressure difference drives liquid out through both ends.

For a spatially uniform flat sheet, $h_x=h_{xxx}=0$, so the longitudinal equation gives $u_{xx}=0$. Reflection symmetry at $x=0$ selectsHere $U$ is the outward speed at the positive end. The mass equation then gives $h_t=-Uh/L$, independent of $x$, so initially uniform thickness remains uniform. Writing it as $h_0(t)$,The end transitions determine $U$, as derived in the following parts.

Take $p$ as excess pressure relative to the fluid far from the gap, and let ${\dot{h}}_0<0$ for approach. The lower surface of the sphere is $h_0+a-\sqrt{a^2-r^2}$ above the plane. In the dominant region $r=O(\sqrt{a{h}_0})\ll a$, this is the parabolic lubrication gapIts slope is $O(\sqrt{h_0/a})\ll 1$, justifying lubrication theory. Both rigid surfaces have zero radial velocity, so the no-slip boundary condition and radial Stokes equation give the radial volume flux per unit circumferenceAxisymmetric conservation of mass gives $h_t+r^{-1}(rq{)}_r=0$. Symmetry at $r=0$ therefore yields $q=-r{\dot{h}}_0/2$, hence $p_r=6\mu {\dot{h}}_{0}r/h^3$. Matching to $p\to 0$ outside the pressure-bearing region givesExtending the parabolic profile to infinity in this integral is a leading-order matching device; the main pressure and force come from $r=O(\sqrt{a{h}_0})$.

The radial scale $\sqrt{a\hat{h}}$ balances the two contributions to the gap. The time scale $\hat{h}/\hat{v}$ is the time for a typical gap change, and the lubrication pressure scale $\mu a\hat{v}/{\hat{h}}^2$ follows either from the formula above or from $p\sim \mu \hat{v}{\ell }^2/{\hat{h}}^3$ with ${\ell }^2=a\hat{h}$. The elastic surface's upward displacement is $-\eta p$, so the actual gap is increased by $+\eta p$. Thus the dimensionless gap in weakly compliant sphere-plane squeeze flow isA dot now means differentiation with respect to $T$. The dimensionless Reynolds lubrication equation isEven without fluid inertia, the elastic surface has a pressure-dependent normal velocity, so ${\mathcal{H}}_T$ includes $\epsilon P_T$; omitting it would wrongly discard the acceleration term below.

Write $d=H_0+R^2/2$ and abbreviate $H=H_0$, $v={\dot{H}}_0$, $b={\ddot{H}}_0$ during the calculation. The rigid term isAt first order the Reynolds equation givesBecause $RdR=dd$, one integration givesSet $\xi =R^2/(2H)$, so $d=H(1+\xi )$. Dividing and changing variables yieldsThe first three terms are derivatives of $I_3,I_4,I_5$, respectively. Integrating from infinity to enforce $P_1\to 0$ givesThe rational expansion of $I_n$ follows by differentiating $\log [\xi /(1+\xi )]+{\sum }_{j=1}^{n-1}[j(1+\xi {)}^j{]}^{-1}$, obtaining $1/[\xi (1+\xi {)}^n]$, and fixing its value to zero at infinity.

Near $\xi =0$ every $I_n$ has the same singular term $\log \xi$. Its coefficient must vanish to keep the pressure regular, soEquivalently, the integrated radial flux above must vanish at $R=0$; otherwise $P_{1R}$ would have a $1/R$ singularity. Using $I_4=I_3+[3(1+\xi {)}^3{]}^{-1}$ and $I_5=I_4+[4(1+\xi {)}^4{]}^{-1}$ cancels all logarithms and gives the simplified regular expression

The upward hydrodynamic force has scale $\mu a^2\hat{v}/\hat{h}$. Its leading contribution is pressure integrated over horizontal projected area; vertical shear forces are smaller in the lubrication limit. Thus $F=2\pi {\int }_0^{\infty }(P_0+\epsilon P_1+\cdots )RdR$. Use $RdR=H_{0}d\xi$ and ${\int }_0^{\infty }(1+\xi {)}^{-n}d\xi =1/(n-1)$. The two contributions areTherefore

For weight-controlled sedimentation, let the fixed effective load be $F=6\pi A$, $A>0$, and neglect particle inertia as well as fluid inertia. At leading order ${\dot{H}}_0=-A{H}_0$ and ${\ddot{H}}_0=A^2{H}_0$. The first-order bracket evaluated on this trajectory is $-(7/5)A^2/H_0^2$. Solving the force balance perturbatively therefore givesHence weak compliance initially increases the sedimentation speed in the regime where the perturbation is valid. The squeeze pressure indents the coating away from the sphere, widens the flow passage and lowers resistance. This is a load-controlled conclusion on the slow sedimenting branch, not a claim about an arbitrary externally imposed acceleration at startup.

The load-controlled breakdown of weak squeeze-flow compliance occurs when the relative correction $O(\epsilon A/H_0^2)$ becomes order one:The central rigid pressure has size $P_0(0)\sim A/H_0$, so the relative elastic gap change is also $\epsilon P_0(0)/H_0\sim \epsilon A/H_0^2$. Thus breakdown means that elastic deflection is comparable with the nominal undeformed gap. The regular expansion about a rigid plane then ceases to be uniform and the coupled elastic-lubrication problem must be retained. It does not by itself imply contact, fluid inertia or failure of the thin-gap geometry.

The lubrication approximation applies in a gap thin compared with its horizontal scale, with small slopes, no slip at the walls, negligible inertia, and approximately constant pressure across the gap. Continuity makes the horizontal velocity much larger than the normal velocity. At $t=0$, write $h(x)=h_0+\delta x/L$, where $\delta =\Delta h$ and $|\delta |<h_0$ keeps the gap open.

The horizontal Stokes equation becomes $\mu u_{yy}=p_x$. Both plates have zero horizontal velocity, soIntegrated incompressibility with the moving upper wall gives $h_t+Q_x=0$. Since $h_t=-V$, $Q_x=V$, and $Q(x)=Vx+C$, giving $Q(L)-Q(-L)=2VL$.

Equal end pressures impose ${\int }_{-L}^L(Vx+C)/h^{3}dx=0$. The two elementary integrals areHence $C=VL\delta /h_0$. In endpoint form this pressure condition is $(h_0-\delta )Q(L)+(h_0+\delta )Q(-L)=0$. Combining it with the flux difference givesThe limits $\delta =0$ recover equal outward fluxes.

The horizontal velocity scale is $U\sim VL/h_0$. The ratio of convective inertia $\rho U^2/L$ to viscous force $\mu U/h_0^2$ is $\rho V{h}_0/\mu$; the unsteady term on time scale $h_0/V$ has the same ratio. Therefore, alongside $h_0/L\ll 1$, the required condition is $\boxed{V\ll \mu /(\rho h_0)}$.

For axisymmetric incompressible flow without swirl, the Stokes streamfunction satisfiesSubstitution into the divergence in spherical coordinates makes incompressibility an identity. Its only vorticity component isTaking the curl of the Stokes equation eliminates pressure and gives ${\nabla }^2\mathbf{\omega }=0$. For an axisymmetric azimuthal vector, its scalar component obeys $(\Delta -1/(R^2{\sin }^2\theta )){\omega }_{\varphi }=0$. Direct differentiation verifiesApply this to $h={\mathcal{D}}^2\Psi$ to obtain

Write $D={\partial }_n{A}_n$ and use repeated-index summation. Because every $A_i$ is a harmonic function,andThus $\mu {\nabla }^2{u}_i-{\partial }_{i}p=0$, the incompressible Stokes equation. MoreoverThe antisymmetric first two terms cancel in the sum with ${\partial }_i{u}_j$, giving the Newtonian fluid Cauchy stress tensorFor $A_i=V_i(1-a/(2r))$, harmonic outside the origin,At $r=a$ this gives $\mathbf{u}=\frac{1}{2}[\mathbf{V}-(\mathbf{V}\cdot \mathbf{n})\mathbf{n}]$, where $\mathbf{n}=\mathbf{x}/a$, so the normal velocity is zero. AlsoHence on the surface,a purely normal traction, appropriate to a shear-free sphere rather than a no-slip solid sphere. Finally rotational symmetry gives ${\int }_{S^2}{n}_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$, so the fluid's force on the sphere isThis is the clean-bubble Stokes drag in the far-field velocity $\mathbf{V}$.

For an incompressible flow, write the Newtonian fluid stress tensor as $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation gives ${\partial }_j{\sigma }_{ij}=0$. Symmetry of the Cauchy stress tensor therefore givesIntegrating and applying the divergence theorem expresses the viscous dissipation as boundary power:At a moving rigid body, the surface velocity is $U+\Omega \times (x-x_c)$. Its boundary-power contribution is $U\cdot F+\Omega \cdot G$, with the force and torque evaluated using the fluid's outward normal vector. Both resultants vanish for a force-free, torque-free inclusion. Thus the new viscous dissipation comes entirely from the unchanged outer boundary velocity. Subtracting the particle-free boundary power givesApply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the fluid outside the inclusion. On the outer boundary $u^{\prime }=0$, soWriting $n_p=-n$ for the particle's outward normal vector gives the required extra dissipation due to a rigid inclusion:This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local viscous dissipation.

For the sphere, the ambient rate-of-strain tensor $E$ is symmetric and trace free. The sphere in a uniform straining Stokes flow has zero translational and rotational velocity: inversion symmetry eliminates its force, and symmetry of $E$ eliminates its torque. To derive the disturbance, put $r=|x|$, $Q=x\cdot Ex$ and seekThe incompressible flow condition and Stokes equation reduce toA decaying family satisfying these equations is $f=A{r}^{-5}$, $g=B{r}^{-5}-5A{r}^{-7}/2$, $j=2B{r}^{-5}$. The no-slip boundary condition requires $f(a)=-1$ and $g(a)=0$, so $A=-a^5$ and $B=-5a^3/2$. ConsequentlyThe total velocity is zero at $r=a$, and the disturbance decays as $r^{-2}$. The pressure constant has been set to zero. These boundary and far-field conditions, together with Uniqueness of Stokes flow, establish the solution. There is no rotational background in a pure strain flow.

On the sphere $u^{\prime }=-u_0=-aE{n}_p$. Contracting the supplied Newtonian fluid stress tensor with $n_p$ makes its two terms proportional to $(n_p\cdot E{n}_p)n_p$ cancel, leaving $\sigma n_p=5\mu E{n}_p$. Hence the extra dissipation due to a rigid inclusion isHere ${\int }_{S^2}{n}_j{n}_{k}d\Omega =4\pi {\delta }_{jk}/3$. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.

If $N_v$ denotes the number of spheres per unit volume, $N_v(4\pi a^3/3)=\varphi$. ThusAdd this to the particle-free viscous dissipation density $2\mu E:E$. Comparing with $2{\mu }_{\mathrm{eff}}E:E$ gives the Einstein viscosity formula for a dilute suspension:to first order in $\varphi$. The hydrodynamic interactions neglected here contribute beyond that dilute order.

For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.