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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.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.

For the requested sedimentation drift of a weighted two-rod body sketch, a full parametric trajectory follows by also integrating $U_v/\dot{\theta }$. Put $\psi =\theta -\pi /4$ and set the initial height to zero:It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda <1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda >1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda )mg/(3CL)$.

For the motionless uniform layer, the heat equation reduces to $T_{zz}=0$. The lower temperature and upper heat-transfer condition determine the quasistatic temperature of a conductively cooled film:When $\alpha h/\kappa \ll 1$, $T_s=T_0-\Delta T\alpha h/\kappa +O(\Delta T(\alpha h/\kappa {)}^2)$. For a varying film let its vertical velocity scale as $Uh/L$, as required by incompressible flow. Relative to vertical thermal diffusion, horizontal thermal diffusion is smaller by $(h/L{)}^2$, and either horizontal or vertical thermal advection is smaller by $U{h}^2/(\kappa L)$. Time variation on the film's transport timescale $L/U$ has the same small factor. Thus the leading heat equation is again $T_{zz}=0$ at each $x,t$ when both specified dimensionless parameters are small. This argument retains the small $\alpha h/\kappa$ assumption. If thickness were externally varied on a faster timescale $t_h$, one would additionally need $h^2/(\kappa t_h)\ll 1$; the spatial conditions alone do not control arbitrary rapid time dependence.

Define $b={\gamma }^{\prime }\Delta T\alpha /\kappa >0$. The surface tension at the surface is ${\gamma }_s={\gamma }_0+bh$, so the Marangoni stress is ${\gamma }_{s,x}=b{h}_x$. With capillary pressure $p=-{\gamma }_0{h}_{xx}$ to leading order, horizontal Stokes flow in the film solves $\mu u_{zz}=p_x$, $u(0)=0$, $\mu u_z(h)={\gamma }_{s,x}$. HenceThe continuity equation gives the thermocapillary thin-film equationTo justify using constant surface tension in the capillary pressure, compare the two volume fluxes on horizontal scale $L$: their magnitudes are ${\gamma }_0{h}^4/(3\mu L^3)$ and $b{h}^3/(2\mu L)$. When both mechanisms are retained at leading order, their balance gives $bh/{\gamma }_0=O((h/L{)}^2)\ll 1$. Thus replacing ${\gamma }_s$ by ${\gamma }_0$ in the curvature term incurs only a relative small correction. Equivalently, small $\alpha h/\kappa$ ensures this if ${\gamma }^{\prime }\Delta T/{\gamma }_0$ is bounded; if it were parametrically large, the extra small surface-tension-variation condition would have to be imposed separately.

The capillary pressure is higher beneath a crest than beneath a trough, so pressure-driven volume flux drains the crest and smooths thickness. The Marangoni effect acts oppositely: a thicker region has a colder surface and therefore larger surface tension. Surface velocity is pulled towards that region, supplying more liquid and amplifying the thickness variation. The arrows below represent these two mechanisms separately.

Put $a_c={\gamma }_0/(3\mu )$ and $a_m=b/(2\mu )$. Substitution of $h=\hat{h}H$, $x=\hat{h}X/ϵ$, $t=\hat{t}\tau$ gives coefficients $a_m{ϵ}^2\hat{t}$ and $a_c{ϵ}^4\hat{t}/\hat{h}$. Making both equal to one yieldsIn these scales the fractional surface tension change is $b\hat{h}/{\gamma }_0=2{ϵ}^2/3$, making the previous approximation explicit.

The linear stability analysis about $H=1$ gives ${\eta }_{\tau }+{\eta }_{XX}+{\eta }_{XXXX}=0$. For a normal mode $e^{s\tau +ikX}$,Long waves with $0<|k|<1$ grow, $|k|>1$ decay, and $k=0,\pm 1$ are neutral. The neutral $k=0$ mode changes the mean film thickness. The dispersion relation is even in $k$, with its two maxima at $\pm 1/\sqrt{2}$.

For a steady positive film with zero volume flux, divide $H^2{H}_X+H^3{H}_{XXX}=0$ by $H^3$ and integrate to obtain $H_{XX}+\ln H=C$. Choosing $\hat{h}$ as specified fixes $C=0$. Multiplication by $H_X$ and another integration then give the zero-flux thermocapillary film profiles first integralFor $H>0$, $V^{\prime }=\ln H$ and $V^{\prime \prime }=1/H>0$: the unique minimum is $V(1)=-1$, while $V(0^{+})=0$ and $V(e)=0$. This potential energy interpretation classifies the profiles without assuming a sinusoidal shape at finite amplitude.

For $E=-1$, the only profile is the uniform film $H=1$. For $-1<E<0$, there are two positive turning heights $H_{-}<1<H_{+}<e$ and a smooth periodic film oscillates between them. Its period isFor $E=0$, the maximum is $e$ and the lower endpoint is zero. A drop reaches zero thickness in finite distance, with zero limiting slope but unbounded curvature since $H_{XX}=-\ln H$. Its half-width isThus the formal limiting profile consists of drops of peak height $e$ and footprint width $2\sqrt{\pi e}$, with zero limiting contact angle. Identical drops can touch, or a zero-thickness region can separate them in the degenerate zero-flux model. These are limiting wet-region solutions, not everywhere positive twice differentiable films; the singular contact region is not resolved by lubrication theory. A useful parametric drop profile isFinally let $E=-1+\delta E$ with small positive $\delta E$. Writing $H=1+\eta$ gives $V=-1+{\eta }^2/2+O({\eta }^3)$ and $H_{XX}=-\eta +O({\eta }^2)$. The small-amplitude steady profile is $H=1+\sqrt{2\delta E}\cos (X-X_c)+O(\delta E)$, with period tending to $2\pi$. Its wavenumber tends to the neutral boundary $|k|=1$ of the unstable band, not to the fastest-growing wavenumber.

Let $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ be the rate-of-strain tensor of an incompressible flow, and let $s(t)$ be the supremum over the conductor of its largest eigenvalue. State Backus' necessary condition for dynamo action with its magnetic boundary conditions: an isolated bounded conductor of uniform positive magnetic diffusivity $\eta$, surrounded by an electrical insulator with a decaying potential exterior field, and no imposed magnetic field or boundary energy input. For definiteness take a no-slip boundary condition on the fluid, which eliminates the stretching surface term. If the conductor lies within a sphere of radius $R$ and $S={\sup }_{t}s(t)$, a necessary condition for a nondecaying dynamo isThe constant is the free-decay spectral bound for an insulating exterior, not a universal constant for every magnetic boundary condition. The condition is necessary, not sufficient, and involves maximum stretching rather than an rms velocity.

To see both the condition and the growth-rate bound, include exterior magnetic energy:This follows from the resistive induction equation and integration by parts, with the stated boundary assumptions. The magnetic free-decay spectral bound is ${\int }_D|\nabla \times \mathbf{B}{|}^{2}dV\ge ({\pi }^2/R^2){\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^{2}dV$. For a sphere its lowest mode is the dipolar poloidal free-decay mode; enclosing a smaller conductor gives the same valid lower bound. Since $B_i{e}_{ij}{B}_j\le s(t)|\mathbf{B}{|}^2$, we obtainIntegrating this differential inequality gives decay whenever $S<\eta {\pi }^2/R^2$. More generally the exponential rate of the field norm, rather than of its squared energy, satisfiesSimply discarding the nonnegative resistive dissipation already proves the requested maximum-strain bound. The energy exponent is twice the field-amplitude exponent.

For the alpha-Omega dynamo model, write $a=|{\alpha }_0|$, $w=|\Omega |$, $k\ge 0$ and $d=\eta k^2$. Direct differentiation givesFor $a,kw>0$, use the weighted energy estimate for two coupled modes and form the positive weighted norm $N=kw|A{|}^2+a|B{|}^2$. The inequality $N\ge 2\sqrt{akw}|A||B|$ givesFor each fixed $k$ and model parameters, this norm is equivalent to the amplitude norm; its square-root exponential rate is therefore bounded by $h(k)=\sqrt{awk}-\eta k^2$, uniformly over all admissible $f(t)$. Maximizing over $k$ givesThe exponent is also achievable in order of magnitude. Choose the admissible constant $f=1$. The growing eigenvalue of the two-component system has real part $-\eta k^2+\sqrt{awk/2}$. Its maximum occurs at $k=(aw/(32{\eta }^2){)}^{1/3}$ and equals $32^{-10/3}(a^2{w}^2/\eta {)}^{1/3}$. Thus the bounded-modulation alpha-Omega growth estimate has the scalingThis means the maximum over allowed modulations and wavenumbers, not that every bounded modulation grows; $f=0$ supplies no regenerating alpha coupling.

The Omega effect rapidly makes toroidal field from poloidal field, but exponential dynamo action also requires the slower alpha effect to regenerate the poloidal component. The coupled amplification rate is of order $\sqrt{a\Omega k}$ rather than $\Omega$; shortening the wavelength to increase it also increases magnetic diffusion as $\eta k^2$. Their optimal balance gives $k\sim {\Omega }^{1/3}$ and growth $\sim {\Omega }^{2/3}$. The Backus' necessary condition for dynamo action estimate controls stretching alone and does not incorporate this regeneration bottleneck. A shear without regeneration can give transient amplification but not this sustained exponential feedback.

Use the inviscid, nonrotating Boussinesq approximation, with a stable background mass density $\bar{\rho }(z)$ and reference density ${\rho }_{\ast }$. Write $B=-g{\rho }^{\prime }/{\rho }_{\ast }$ for the perturbation buoyancy and $\pi =p^{\prime }/{\rho }_{\ast }$ for the kinematic pressure. The buoyancy frequency is $N_0^2=-(g/{\rho }_{\ast }){\bar{\rho }}_z>0$. Dropping products of perturbations in the Boussinesq equations gives the Linearized Boussinesq equationsThe last equation expresses incompressible flow; the second follows by advecting the background mass density gradient. Neglecting rotation and viscosity is part of this internal gravity wave model.

For a plane internal gravity wave proportional to $\exp [i({\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t)]$, put $\kappa =|{\mathbf{k}}_h|>0$ and $K^2={\kappa }^2+m^2$. Eliminating the horizontal velocity, pressure and buoyancy from the linear equations givesFor example, the horizontal momentum and continuity equations give $\pi =-\omega mw/{\kappa }^2$; substituting $B=-i{N}_0^{2}w/\omega$ into vertical momentum yields the displayed dispersion relation. Thus an IGW has $0<|\omega |\le N_0$ in this model.

On the positive-frequency branch, the phase velocity normal to a constant-phase plane and the group velocity areConsequently $({\mathbf{k}}_h,m)\cdot {\mathbf{c}}_g=0$, and phase and group velocity are perpendicular. Equivalently, the dispersion relation is homogeneous of degree zero in the wave vector, so differentiating with respect to its scale proves the same orthogonality. Energy travels with the group velocity, along the phase planes. At the degenerate limit $m=0$, $\omega =N_0$ and the group velocity vanishes; the orthogonality statement then has this limiting interpretation.

Write the conservative body force per unit mass as $-\nabla \Phi$. For a fluid of constant density $\rho$, the Euler equations for an inviscid fluid and the incompressible flow condition areUsing part (a), these becomeFor irrotational flow, locally choose a velocity potential with $\mathbf{u}=\nabla \varphi$. Then $\nabla ({\partial }_t\varphi +H)=0$, so the unsteady Bernoulli integral isThe constant is spatially uniform on a connected region where the velocity potential is defined; its time dependence can be absorbed into the additive time-dependent gauge of $\varphi$. A global single-valued velocity potential requires an appropriate topological condition, such as a simply connected domain; without it, the integral is locally valid. In a connected steady irrotational region, $\nabla H=0$ directly gives one spatial Bernoulli constant even without choosing a global potential.

Use the standard ideal-fluid assumptions of incompressible flow, inviscid flow, gravity $-g{\mathbf{e}}_y$, and no surface tension. Irrotational flow gives ${\mathbf{u}}_j=\nabla {\phi }_j$ in each layer, and incompressibility gives Laplace equationThe rigid-wall boundary conditions are ${\phi }_{1y}=0$ at $y=h_1$ and ${\phi }_{2y}=0$ at $y=-h_2$. At the moving interface, each fluid has the same normal velocity as the interface:The dynamic condition is continuity of pressure, $p_1=p_2$ at $y=\eta$. Bernoulli equation in each fluid reads ${\phi }_{jt}+\frac{1}{2}|\nabla {\phi }_j{|}^2+p_j/{\rho }_j+gy=C_j(t)$. Choosing the potential gauges to subtract a common reference pressure, the interface condition becomesTangential velocities need not agree across an inviscid interface.

By definition the vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of the curl identity yieldsFor incompressible flow, $\nabla \cdot \mathbf{u}=0$, so

Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of $z$ alone. Incompressible flow and zero normal velocity at the walls give $u_z^{\prime }=0$ and hence $u_z=0$. The zero divergence of the magnetic field makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/({\mu }_0\rho )$, integrating by parts and adding givesThus $u_y=b_y=0$, establishing the asserted forms on this fully developed flow branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.

The current density and Lorentz force density areThe advective acceleration vanishes because $\mathbf{u}\cdot \nabla =u(z){\partial }_x$ acts on fields independent of $x$. The $x$-component of the magnetohydrodynamic momentum equation therefore givesThe $z$-component must also balance: it requires ${\partial }_z(p+b^2/(2{\mu }_0))=0$. Hence a compatible pressure is $p(x,z)=p_{\ast }-\rho Gx-b(z{)}^2/(2{\mu }_0)$; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in $z$. This accounts for the magnetic pressure of the induced field.

The resistive induction equation with constant magnetic diffusivity gives ${\partial }_t\mathbf{B}=\nabla \times (\mathbf{u}\times \mathbf{B})+\eta {\nabla }^2\mathbf{B}$. Since $\mathbf{u}\times \mathbf{B}=(0,-B_{0}u,0)$, its curl has $x$-component $B_0{u}^{\prime }$. Steadiness consequently givesFinally the no-slip boundary condition gives $u(\pm L)=0$, and the normal magnetic field boundary condition gives $b(\pm L)=0$. The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.

Let $e=(\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)/2$ be the rate-of-strain tensor and $\mathbf{\sigma }=-pI+2\mu e$ the Newtonian fluid stress tensor. The viscous dissipation is $D[\mathbf{u}]=2\mu {\int }_{V}e:edV$. The Minimum-dissipation theorem for Stokes flow compares the Stokes flow with every sufficiently regular incompressible flow in the same domain having the same prescribed boundary velocity, with no body force or with a conservative body force absorbed into the fluid pressure. A trial field need not satisfy the Stokes flow equations. Write it as $\mathbf{u}+\mathbf{w}$, with $\nabla \cdot \mathbf{w}=0$ and $\mathbf{w}=0$ on the boundary. Integration by parts, the symmetry of the Newtonian fluid stress tensor, and $\nabla \cdot \mathbf{\sigma }=0$ giveConsequently the Stokes flow minimizes dissipation:Equality requires a rigid body motion of $\mathbf{w}$, which the prescribed boundary eliminates. Extending the flow inside the inserted particle as its rigid body velocity produces an admissible comparison field with zero internal rate-of-strain tensor; hence the extra dissipation due to a rigid inclusion is nonnegative.

Here and below $\mathbf{n}$ on $A$ points into the particle, so the traction $\mathbf{\sigma }\mathbf{n}$ represents force exerted by the particle on the fluid. Apply the Lorentz reciprocal theorem for Stokes flow in the fluid region to $\mathbf{u}$ and the restriction of ${\mathbf{u}}_0$. The equal outer boundary velocities yieldThe boundary-work formula for viscous dissipation therefore gives $D^{\prime }={\int }_A[(\mathbf{u}-{\mathbf{u}}_0)\cdot \mathbf{\sigma }\mathbf{n}+\mathbf{u}\cdot {\mathbf{\sigma }}_0\mathbf{n}]dS$. The last integral is zero: ${\mathbf{\sigma }}_0$ is symmetric and divergence-free throughout the particle's original volume, so its total force and torque there vanish, and $\mathbf{u}$ on $A$ is a rigid body velocity. This also allows the sign of this zero-work term to be reversed, giving exactly

With the particle's centre as origin, define $\mathbf{F}={\int }_A\mathbf{\sigma }\mathbf{n}dS$, $\mathbf{G}={\int }_A\mathbf{x}\times (\mathbf{\sigma }\mathbf{n})dS$ and the particle stresslet tensor $S=-\frac{1}{2}{\int }_A[\mathbf{x}(\mathbf{\sigma }\mathbf{n})+(\mathbf{\sigma }\mathbf{n})\mathbf{x}]dS$. Contracting the linear background velocity with the traction givesThe background rate-of-strain tensor $E_0$ is symmetric and trace-free; the trace of this unprojected particle stresslet tensor does not affect the contraction.

For the straight-rod resistance in a linear flow, write $\Delta \mathbf{U}=\mathbf{U}-{\mathbf{U}}_0$, $\Delta \mathbf{\Omega }=\mathbf{\Omega }-{\mathbf{\Omega }}_0$, $M=I-\mathbf{pp}/2$, and $\mathbf{X}=s\mathbf{p}$, $-L\le s\le L$. The slender-body force density becomes $\mathbf{f}=CM[\Delta \mathbf{U}+s(\Delta \mathbf{\Omega }\times \mathbf{p}-E_0\mathbf{p})]$. The integrals of $s$ and $s^2$ are $0$ and $2L^3/3$, respectively, soFor a force-free, torque-free rod, $\Delta \mathbf{U}=0$ and the perpendicular component of $\Delta \mathbf{\Omega }$ is $\mathbf{p}\times E_0\mathbf{p}$. Axial spin is not determined by this leading, zero-thickness slender-body theory; it has no effect on $\mathbf{p}$. Put $a_p=\mathbf{p}\cdot E_0\mathbf{p}$. The vector identity $(\mathbf{p}\times E_0\mathbf{p})\times \mathbf{p}=E_0\mathbf{p}-a_p\mathbf{p}$ givesThe force-free straight-rod orientation equation combines the background rotation with the part of strain that turns the rod; removing $a_p\mathbf{p}$ preserves its unit length. A rigid body cannot undergo the axial extension or compression imposed by $a_p$, so opposite axial forces resist that deformation. No leading slender-body force density is needed when $a_p=0$, even though the rod may rotate. Its particle stresslet tensor and viscous dissipation are

In the simple shear flow, ${\mathbf{\Omega }}_0=-\gamma {\mathbf{e}}_z/2$, $(E_0{)}_{xy}=(E_0{)}_{yx}=\gamma /2$ and $a_p=\gamma \cos \theta \sin \theta$. Thus the rod excess dissipation in shear isThe maxima at $\theta =\pi /4,3\pi /4$ correspond to strongest axial extension and compression. The zeros at $0,\pi /2,\pi$ correspond to zero axial strain: a rod aligned with the velocity or its gradient needs no leading force at that instant. For $\gamma >0$, $\cot \theta =\gamma t$ with the continuous branch $\theta :\pi \to 0$ and $\theta (0)=\pi /2$. Reversing the shear flow reverses the traversal. Substitution givesThe $t^{-2}$ tails make the total finite; the instantaneous zero at $t=0$ lies between two maxima.

For a dilute rod suspension, the added bulk stress is the number density times the orientation average of the particle stresslet tensor. In this ideal infinitely slender, non-interacting model, rods approach alignment with the shear flow and their excess viscous dissipation decays. A real finite-aspect-ratio rod continues to tumble; rotational diffusion and interactions also maintain a spread of orientations. One therefore expects increased effective viscosity, with alignment offering a mechanism for reduced excess viscosity at strong shear flow. This is a qualitative expectation, not a claim that a finite rod suspension has zero steady excess stress.

Take $z$ positive downwards. The leading axial velocity of a slender viscous thread is independent of radius. Incompressible flow then gives $u_r=-r{w}^{\prime }/2$, while steady mass conservation gives $(a^{2}w{)}^{\prime }=0$. The Newtonian fluid stress tensor has ${\sigma }_{rr}=-p-\mu w^{\prime }$ and ${\sigma }_{zz}=-p+2\mu w^{\prime }$. The Young–Laplace equation, neglecting axial curvature at slender order, gives ${\sigma }_{rr}=-p_e-\gamma /a$ and hence $p=p_e+\gamma /a-\mu w^{\prime }$.

The axial cut transmits both the integrated Newtonian fluid stress tensor and the circumferential surface tension. After subtracting the ambient pressure, its tensile force isThe coefficient three is the Trouton ratio for axial extension. The last term is essential: the capillary tensile force of a slender thread is $\pi \gamma a$, not $2\pi \gamma a$, because the capillary pressure subtracts half the interfacial line contribution. Axial force balance is $T^{\prime }+\pi \rho g{a}^2=0$. Together the governing equations areDefine the dimensionless variablesThen $a=a_0{W}^{-1/2}$ and the axial equation becomesFor the increasing power-law branch, all three terms have the same dependence when $\alpha =2$. Substitution leaves $1-6k^2-\Gamma k=0$. ThusThis remains valid at $\Gamma =0$, when $k=1/\sqrt{6}$. The other root is negative and describes a decreasing local branch up to its singular endpoint, not a thread accelerating downwards indefinitely. Positive surface tension decreases $k$, so it slows the fall at a given height on this branch. Thinning reduces the capillary tensile force $\pi \gamma a$. Its negative axial derivative supplies an upward capillary force on a thread segment, balancing part of the weight. The viscous tensile force therefore supplies a smaller remaining balance; on the quadratic family this reduces the stretching rate.

The condition $W(0)=1$ alone does not select a unique solution of this second-order ordinary differential equation. The boxed power law is the requested verified branch, with tensile force tending to zero at large height; an additional nozzle tension or downstream condition is required to select it physically. The slender, negligible-inertia approximation must remain valid over the height range considered.

Write $D=R_e-R_i$, $S=R_e+R_i$, $A=R_e^2-R_i^2=DS$, and $\Delta p=p_i-p_e$. Incompressible flow with axial strain rate $E$ givesIts radial Stokes equation gives $p_r=0$, because $u_{r,rr}+u_{r,r}/r-u_r/r^2=0$. The radial Newtonian fluid stress tensor component is ${\sigma }_{rr}=-p-\mu E-2\mu C/r^2$. The outer and inner Young–Laplace equation conditions have opposite curvature signs:Subtracting and eliminating $C$ givesIn particular, the boxed pressure is equivalent to the printed identity for $(-p-\mu E)A$. For equal gas pressures, $C=-\gamma R_i{R}_e/(2\mu D)$, so surface tension draws both interfaces inward in addition to the imposed extension.

The kinematic boundary condition is ${\dot{R}}_j=-E{R}_j/2+C/R_j$ at either interface. Taking the difference, and then the difference of squared radii, givesThe cancellation of $C$ in the area equation is mass conservation: axial stretching reduces the liquid area while radial redistribution changes the hole size. For $E=0$, equal pressures, and $R_i(0)=b$, $R_e(0)=2b$, one has $A=3b^2$ and $D=b+\gamma t/(2\mu )$. Since $S=A/D$,Closure occurs at $D=\sqrt{A}=\sqrt{3}b$, soThis capillary collapse of an annular viscous cylinder has finite closure time for $\gamma >0$ and $b>0$. For $\gamma =0$ it does not close in this case. The annular formula is used only up to closure; the singular cylindrical inner curvature is not a boundary condition on the subsequent solid cylinder.