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For thickness $h(r,t)$, radial velocity $u(r,t)$ and external pressure $p_{\mathrm{ext}}$, the leading Newtonian fluid stress tensor and incompressibility giveRadial force balance on an annular sector includes the inward projection of hoop traction and external pressure on the sloping broad faces. Together with conservation of mass it yields

For cylindrical rotation independent of height, write $W_{r\varphi }=\int T_{r\varphi }dz$. Define the effective kinematic viscosity by $W_{r\varphi }=\bar{\nu }\Sigma r{\Omega }^{\prime }$, where $\Sigma$ is the surface density of a disk. For Keplerian rotation, $\bar{\nu }=-2W_{r\varphi }/(3\Sigma \Omega )$, and the outward viscous torque in an accretion disk is $-2\pi r^2{W}_{r\varphi }=3\pi \bar{\nu }\Sigma \sqrt{GMr}$. A Newtonian fluid stress tensor gives the density-weighted viscosity of a disk; the stress definition can also parameterize an effective turbulent or magnetic transport stress.

// Target: astrophysics.bigb

A slender axisymmetric column of a Newtonian fluid, with radius $a(z,t)$, axial velocity $w$, external pressure $p_e$ and axial body-force density $f$, obeys $(a^2{)}_t+(a^{2}w{)}_z=0$ and $3\mu (a^2{w}_z{)}_z/a^2=p_{e,z}-f$ when inertia, surface tension and leading external shear are negligible. The factor three is the Trouton ratio. Radial mass conservation gives $u_r=-r{w}_z/2$; the Newtonian fluid stress tensor gives $p=p_e-\mu w_z$ and axial excess stress $3\mu w_z$. This reduces a three-dimensional Stokes flow to one-dimensional evolving geometry.

The Navier slip boundary condition allows a fluid to slide tangentially past a solid wall while resisting that motion through wall friction in the Navier slip boundary condition. For parallel flow past a flat, impermeable wall translating uniformly with tangential velocity $U_w$, write the fluid's tangential velocity as $u$ and its relative tangential velocity as $u_{\mathrm{rel}}=u-U_w$. Thenwhere $b$ is the slip length, also written ${\ell }_s$. The unit normal $\mathbf{n}$ points from the wall into the fluid, and ${\partial }_{n}u=\mathbf{n}\cdot \nabla u$ is the normal derivative. The separate no-penetration boundary condition requires $(\mathbf{u}-{\mathbf{U}}_w)\cdot \mathbf{n}=0$: tangential sliding does not allow flow through the wall.

The sign convention matters. In a channel $0<z<H$, the lower wall has inward unit normal $+{\mathbf{e}}_z$ and the upper wall has inward unit normal $-{\mathbf{e}}_z$. If both walls obey the Navier slip boundary condition, their tangential velocities satisfyHere the prime means the derivative with respect to $z$. Using $+b_H{u}^{\prime }(H)$ at the upper wall would reverse the friction law. The examples Couette flow with a slipping lower wall and planar Poiseuille flow with equal wall slip use different upper-wall boundary conditions.

This flat-wall formula is a local linear model. On a curved wall the Newtonian fluid stress tensor must supply the tangential traction; the normal derivative alone generally misses curvature terms. The author-written review by Lauga, Brenner and Stone states the full strain-based condition with the unit normal directed into the liquid. ^[ref]

The same review credits Navier's 1823 treatise with the linear slip law. Its history is a reminder that the no-slip boundary condition is an experimentally successful model, whose validity depends on the interface and scale. The history here follows that review rather than a new reading of Navier's original memoir. ^[ref]

A Newtonian fluid has viscous stress proportional to the local rate-of-strain tensor, with coefficients independent of that rate. For an isotropic fluid these coefficients are viscosities. Its Newtonian fluid stress tensor specifies the corresponding constitutive relation; the material and its stress tensor are distinct concepts.

Use the same constant dynamic viscosity $\mu$ for both incompressible Stokes flows, and let ${\mathbf{f}}^{(a)}$ be body force per unit volume. Their equations and Newtonian fluid stress tensor arewhere $e_{ij}^{(a)}=({\partial }_i{u}_j^{(a)}+{\partial }_j{u}_i^{(a)})/2$ is the strain-rate tensor. Consider the cross-work fluxBy the product rule,The pressure terms vanish by incompressibility. Symmetry of the stress tensor removes antisymmetric velocity gradient contributions, leaving $2\mu e^{(1)}:e^{(2)}$ in both cross terms; they cancel. Apply the divergence theorem, with $\mathbf{n}$ outward from $V$, and rearrange:This is the Lorentz reciprocal theorem with body forces. It depends on reciprocal linear viscous response in the same fluid; arbitrary unrelated constitutive laws would not supply the cross-term cancellation.

Write the supplied velocity in the radial formAt $r=a$, $f=1$ and $g=0$, giving the no-slip boundary condition $\mathbf{u}=\mathbf{U}$. At infinity $\mathbf{u}\to 0$ and the supplied gauge-fixed pressure tends to zero. These are the laboratory-frame conditions for a sphere translating through otherwise quiescent fluid.

The exterior Stokes equations can also be checked directly:andTo compute surface traction in translating-sphere Stokes flow, let $\mathbf{n}=\mathbf{x}/a$ point from the sphere into the fluid. At the surface, $f^{\prime }(a)=-3/(2a)$, $g^{\prime }(a)=3/(2a^3)$ and $g(a)=0$. HenceSubstitute into the Newtonian fluid stress tensor. Its viscous contribution to the surface traction is $3\mu [-\mathbf{U}+(\mathbf{U}\cdot \mathbf{n})\mathbf{n}]/(2a)$, so the normal part cancels the pressure contribution:The vector is constant over the entire sphere in the stated pressure gauge. Integrating it gives the fluid's drag force on the sphere,This is Stokes's law. On an exterior fluid domain the outward domain normal at its inner boundary is $-\mathbf{n}$; that distinction will matter in the reciprocal calculation.

Work in the laboratory frame, with the fluid at rest at infinity. SetAt $r=a$, $f=1$ and $g=0$, so the no-slip boundary condition is $\mathbf{u}=\mathbf{U}$ on the translating sphere. As $r\to \infty$, $\mathbf{u}=O(r^{-1})\to 0$ and the gauge-fixed pressure tends to zero.

It is also possible to check the exterior Stokes equations directly. Differentiating the radial functions givesandThus the Newtonian fluid stress tensor $\sigma =-pI+\mu (\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)$ satisfies $\nabla \cdot \sigma =0$ throughout $r>a$.

Let $\mathbf{n}$ point away from the common centre on each concentric sphere. Applying the divergence theorem to the intervening fluid shell, whose inner normal is $-\mathbf{n}$, yieldsThis is the force flux at infinity in Stokes flow. It is legitimate even though the stress tensor tends to zero pointwise, because the area of the large sphere grows as $R^2$.

For an explicit evaluation, put $A=3a/4$ and $\mathbf{n}=\mathbf{x}/R$. The leading terms are $\mathbf{u}=A[\mathbf{U}+(\mathbf{U}\cdot \mathbf{n})\mathbf{n}]/R$ and $p=2\mu A(\mathbf{U}\cdot \mathbf{n})/R^2$. Their velocity gradient isAfter symmetrizing and subtracting the pressure contribution,By rotational symmetry, $\int n_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$. HenceThis is Stokes's law, with the force on the sphere opposite its velocity. For a component calculation with $\mathbf{U}=U{\mathbf{e}}_z$, the same integral is $F_z=-(9\mu aU/2)\int {\cos }^2\theta d\Omega$.

Use the force flux at infinity in Stokes flow. With an outward normal from a large sphere, the force on the body is $\mathbf{F}={\int }_{S_R}\sigma \mathbf{n}dS$. Let $C=3a/4$. The leading far-field velocity and pressure areTheir rate-of-strain tensor isIn the Newtonian fluid stress tensor ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}$, the isotropic pieces cancel. Consequently,The second spherical moment integral $\int n_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$ follows from rotational symmetry and ${\sum }_i{n}_i^2=1$. Since $dS=R^{2}d\Omega$, taking $R\to \infty$ gives the Stokes drag law:The omitted dipole contributes a surface integral of order $R^{-2}$ and vanishes in this limit. The sign refers to the fluid's drag on the moving sphere.

For two Stokes flows in the same fluid domain with the same dynamic viscosity, symmetry of the Newtonian fluid stress tensor and incompressibility give the local reciprocal identityIn index notation, its remaining cross-gradient terms are both $2\mu e:e^{\prime }$. On the remote sphere the integrated reciprocal flux tends to zero: the velocity is $O(R^{-1})$, the stress tensor is $O(R^{-2})$ and the area is $O(R^2)$. On the body, the no-slip boundary condition makes the two velocities the constants $\mathbf{U},{\mathbf{U}}^{\prime }$. The Lorentz reciprocal theorem therefore givesBy the Linearity of Stokes flow, the sum of the two velocity-pressure fields satisfies the boundary conditions for body velocity $\mathbf{U}+{\mathbf{U}}^{\prime }$. Uniqueness of the exterior Stokes flow identifies it with the third solution. Since the stress tensor and its surface integral are linear too, $\mathbf{F}(\mathbf{U}+{\mathbf{U}}^{\prime })=\mathbf{F}(\mathbf{U})+\mathbf{F}({\mathbf{U}}^{\prime })$; scalar multiplication follows in the same way. Thus $\mathbf{F}=-A\mathbf{U}$ for a fixed-orientation hydrodynamic resistance matrix $A$. Substituting arbitrary $\mathbf{U},{\mathbf{U}}^{\prime }$ in the reciprocal identity yieldsThe minus sign is natural because viscous dissipation gives $-\mathbf{F}\cdot \mathbf{U}=2\mu \int e:edV>0$ for nonzero rigid translation in an otherwise stationary fluid; $A$ is a positive-definite matrix.

Differentiating the axisymmetric stream function givesThe normal kinematic boundary condition fixes $C=Ua/2$, soBoth velocity components vanish at infinity, and the axisymmetric stream function is regular at the polar axes. The tangential component of the strain-rate tensor vanishes everywhere:Thus the stress-free boundary condition is satisfied. The vorticity is ${\omega }_{\varphi }=2C\sin \theta /r^2=Ua\sin \theta /r^2$, and the supplied dipolar pressure is consistent with the Stokes equation.

Use the Newtonian fluid stress tensor $\mathbf{\sigma }=-p\mathbf{I}+2\eta \mathbf{e}$. At the bubble surface,The vertical force exerted by the liquid is the integral of the vertical traction:Since the integral is $2/3$, the drag is downward and has magnitudeThe constant pressure integrates to zero force. The pressure perturbation contributes $4\pi \eta aU/3$ and the normal viscous stress contributes $8\pi \eta aU/3$; there is no tangential traction contribution.

For the free-interface qualification in part (b), the calculated normal traction at $\theta =0$ and $\theta =\pi$ differs by $6\eta U/a$. It cannot equal a single uniform gas pressure if every restoring effect is discarded. Restoring the external-minus-internal variation of hydrostatic pressure supplies a term $({\rho }_w-{\rho }_a)ga\cos \theta$. It cancels the dynamic normal traction at the terminal velocity $U=({\rho }_w-{\rho }_a)g{a}^2/(3\eta )$, consistent with a spherical clean bubble without surface tension. This distinguishes the drag calculation from the extra conditions needed for a freely rising bubble.

Assume axisymmetry in a thin disk orbiting a fixed central mass $M$. To leading order in thickness, the Keplerian rotation is independent of height, $\Omega =(GM/r^3{)}^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial pressure support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface torques. The Cauchy stress tensor can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the conservation of angular momentum equation.

Define the surface density and its mass-weighted radial velocity byand set $W_{r\varphi }=\int T_{r\varphi }dz$. Integrating mass conservation vertically givesThe azimuthal equation in cylindrical coordinates, multiplied by $r$ and combined with mass conservation, is the conservative equation for specific angular momentum $l=r^2\Omega$. The cylindrical stress tensor divergence contributes $r^{-1}{\partial }_r(r^2{T}_{r\varphi })+{\partial }_z(r{T}_{\varphi z})$. With vanishing surface torque, vertical integration therefore givesBecause $l(r)$ is fixed in time, subtracting $l$ times the mass equation yieldsDefine the effective viscosity defined by disk stress byThis sign convention treats $T_{r\varphi }$ as the shear stress appearing with $+\nabla \cdot \mathbf{T}$. For the Newtonian fluid stress tensor it reduces to the density-weighted viscosity of a disk, $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$, rather than an unweighted height average.

Using $l=\sqrt{GMr}$ and $W_{r\varphi }=-(3/2)\bar{\nu }\Sigma \Omega$ in the conservation of angular momentum equation givesSubstitution into mass conservation proves the Keplerian viscous diffusion equation:The outward viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^2{W}_{r\varphi }=3\pi \sqrt{GM}\sqrt{r}\bar{\nu }\Sigma$.

For $\bar{\nu }=Ar$ with $A>0$, make the heat-equation transform for linear-radius disk viscosityWriting $y=\sqrt{r}$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The torque is $3\pi \sqrt{GM}g$, so the zero-torque inner boundary condition at the origin becomes $g(0,t)=0$ on the half-line $x>0$.

A positive similarity solution obeying this boundary condition isFor example, substitution of $g=t^{-1}F(x/\sqrt{t})$ reduces the heat equation to $F^{\prime \prime }+(\eta /2)F^{\prime }+F=0$, and $F=C\eta e^{-{\eta }^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.

To determine the global integrals, transform the mass and angular momentum:The Gaussian integrals giveThus the accreting similarity disk with viscosity proportional to radius loses mass as $t^{-1/2}$ while conserving angular momentum; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-{\dot{M}}_d=2\pi C\sqrt{3/A}{t}^{-3/2}$. Its radial velocity ${\bar{u}}_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner density singularity is part of the idealized extension to $r=0$.

For the other profile, direct differentiation givesIt also satisfies the heat equation, but has $g(0,t)=t^{-1/2}\ne 0$ and $g_x(0,t)=0$. It therefore has a nonzero inner torque and zero mass flow at the origin, not the previous zero-torque inner boundary condition. HereAn arbitrary positive amplitude multiplies these quantities. This constant-mass disk with viscosity proportional to radius spreads outward under a central torque that supplies angular momentum, with no ongoing central mass supply. It has the reflecting Neumann boundary condition for the heat equation and can be regarded as the spreading of an initially central mass distribution driven by an applied torque. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.

Put $r=|\mathbf{x}|$. In the Unscaled Papkovich–Neuber representation, choose the componentwise harmonic vector fieldBoth $\mathbf{\Omega }\times \mathbf{x}$ and $(\mathbf{\Omega }\times \mathbf{x})/r^3$ are harmonic away from the origin: the latter is a linear combination of derivatives of the harmonic function $1/r$. Also $\mathbf{x}\cdot \mathbf{\Phi }=0$, so the representation gives $\mathbf{u}=f(r)\mathbf{\Omega }\times \mathbf{x}$. The no-slip boundary condition requires $f(a)=1$ and $f(b)=0$, givingThus the Rotational Stokes flow between concentric spheres isThe divergence of $\mathbf{\Phi }$ is zero: the cross-product field itself has zero divergence, and $\nabla f$ is radial and perpendicular to $\mathbf{\Omega }\times \mathbf{x}$. Hence $p=2\mu \nabla \cdot \mathbf{\Phi }=0$ in this pressure gauge.

There is also a symmetry argument. Take the rotation axis as the polar axis. Linearity of Stokes flow makes velocity and pressure perturbations change sign when the rotation reverses. Reflection in any plane containing that axis reverses the axial rotation vector. Combined with axial symmetry, this excludes meridional velocity components and forces the pressure perturbation to be both unchanged and sign-reversed by that reflection. Equivalently a purely azimuthal axisymmetric Stokes flow has no meridional component of its vector Laplacian, so its meridional pressure gradient vanishes. Thus the pressure is constant; setting that arbitrary constant to zero gives the same result without solving the radial coefficients.

For the Newtonian fluid stress tensor, let $\mathbf{q}=\mathbf{\Omega }\times \mathbf{x}$. The symmetric part of the velocity gradient isbecause the gradient of rigid rotation is antisymmetric. Consequently the full stress tensor, with $p=0$, isFor $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, its only nonzero spherical-coordinate components are ${\sigma }_{r\varphi }={\sigma }_{\varphi r}=-3\mu B\Omega \sin \vartheta /r^3$. On the inner sphere the traction exerted by the fluid, using the sphere's outward normal $\mathbf{n}$, isIts torque isusing $\int n_i{n}_{j}d\omega =4\pi {\delta }_{ij}/3$. Therefore the applied Torque in rotational Stokes flow between concentric spheres isFor $b\gg a$, the relevant speed and length are $\Omega a$ and $a$, so negligible inertia requires the rotational Reynolds numberFor a narrow gap $d=b-a\ll a$, the viscous gradients are controlled by $d$, not $a$. A centrifugal acceleration of scale ${\Omega }^{2}a$ is then compared with viscous scale $\mu \Omega a/d^2$, rather than $\mu \Omega /a$. Their ratio is $\rho |\Omega |d^2/\mu$. This explains why the large-gap condition is not a necessary narrow-gap criterion; a thin gap greatly strengthens viscous resistance. This base-flow scaling is distinct from a criterion for stability to arbitrary short-scale disturbances.

The Minimum-dissipation theorem for Stokes flow compares a Stokes solution with all sufficiently regular incompressible trial velocities in the same fluid domain that have the same prescribed boundary velocities. With no body-force work, the dissipation functional isIf $\mathbf{v}=\mathbf{u}+\mathbf{w}$ and $\mathbf{w}=0$ on the prescribed boundary, integration by parts with $\nabla \cdot \mathbf{\sigma }(\mathbf{u})=0$ and $\nabla \cdot \mathbf{w}=0$ makes the cross term vanish. ThusThe comparison fixes velocities, rather than applied torques or forces.

Now let the particles take their actual rigid translations and rotations in the particle-containing flow. Extend the fluid velocity through each particle by its rigid velocity. No-slip boundary conditions make this extension continuous; rigid motion has zero rate-of-strain tensor and zero divergence. The extension is therefore an admissible trial field in the original particle-free annulus, and its dissipation equals that of the fluid outside the particles. Hence $D\ge D_0$.

The Stokes boundary-work identity gives $D=\mathbf{G}\cdot \mathbf{\Omega }$. The outer sphere is stationary, the maintained inner sphere has zero translational velocity, and every added particle has zero total force and torque, so none of those other boundaries contributes power. Likewise $D_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor nonzero rotation and a particle of positive volume the inequality is strict: equality in the dissipation identity would force the extension to be the unique original flow, but that original flow has nonzero strain on every open particle region, whereas a rigid interior has zero strain. This proves that force-free inclusions increase rotational resistance:No dilute-suspension approximation or assumption about the particle shapes was used.

In a slender symmetric viscous sheet, the two broad faces have zero tangential traction. A cross-thickness variation of $w$ on the thickness scale would produce leading shear $\mu w_x$, whereas $\mu u_z$ is smaller by the square of the aspect ratio. The leading transverse momentum balance and the shear-free conditions therefore give $w_x=0$: write $w=w(z,t)$.

Incompressibility gives $u_x=-w_z$. Symmetry about $x=0$ sets the integration constant to zero, soThe normal traction on either nearly vertical face matches the ambient hydrostatic pressure, giving ${\sigma }_{xx}=-p_a$ to leading order. The Newtonian fluid stress tensor then yieldsThe factor four is the planar extensional resistance, as in the planar viscous-sheet stretching equations, rather than the factor three for uniaxial extension with two contracting transverse directions.

The following slice diagram shows all axial-force contributions per unit $y$-width; the end arrows indicate positive tensile stress and reverse if that signed stress is negative.

The two end tractions have net $z$-component ${\partial }_z(h{\sigma }_{zz})\delta z$. The normal pressure on the inclined side faces contributes $p_a{h}_z\delta z$, and gravitational acceleration gives the weight $\rho gh\delta z$. Hence force balance isSubstitute the stress expression and $p_a^{\prime }={\rho }_{a}g$:soKeeping the side-pressure component is essential: dropping it would retain an incorrect dependence on the arbitrary ambient pressure level. The remaining equation from conservation of mass isIt follows either by integrating incompressibility across the moving faces or directly from the face kinematic boundary condition.

When effective gravitational acceleration vanishes, $4\mu h{w}_z$ is independent of $z$. It equals the applied excess tensile force $F(t)$, with ambient pressure removed from the end loading. Along a material element labeled by its initial position $z_0$,Integration gives uniform material thinning under planar-sheet tension:A material interval conserves its area: $h(z,t)dz=h_0(z_0)d{z}_0$. ThusWith the specified fixed material origin, integrate the latter relation to obtainThese formulas hold while the material thickness remains positive.

For the quadratic initial profile, put $q=H-\Delta >0$. If $k\ne 0$, evaluating the integral at $z_0=L_0$ givesFor $k=0$ it instead givesFor a constant positive pulling force, $\Delta =Ft/(4\mu )$ first reaches the minimum initial thickness $H$ atBoth expressions for $L$ diverge there. Write $q=F(t^{\ast }-t)/(4\mu )$. If $k\ne 0$, the inverse tangent tends to $\pi /2$ and $\Delta \to H$, givingFor $k=0$,This is finite-time extension of a quadratically thickened sheet. With nonzero $k$, the dominant extension comes from the small material region $|z_0|=O(\sqrt{q/H}/|k|)$ near the thickness minimum, where the integrand is proportional to $1/(q+H{k}^2{z}_0^2)$. Its width shrinks like $\sqrt{q}$, giving an integrated divergence $q^{-1/2}$. The fixed ends are asymptotically far away in those local coordinates, so $A$ does not depend on $L_0$. For a uniform sheet the entire material interval has denominator $q$, giving the stronger $q^{-1}$ divergence and retaining its length in $A$. The singularity is a prediction of the ideal slender-sheet model as its minimum thickness tends to zero.

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.