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For liquid between spatially uniform parallel plates with gap $h(t)$ and circular material rim $R(t)$, lubrication theory gives ${\overline{u}}_r=-h^2{p}_r/(12\mu )$ and $\dot{h}+r^{-1}(rh{\overline{u}}_r{)}_r=0$. Bounded pressure at zero excludes the logarithmic solution, and $p(R)=0$ gives the displayed pressure. Then ${\overline{u}}_r=-r\dot{h}/(2h)$ and rim advection implies $(R^{2}h{)}^{\prime }=0$, expressing volume conservation. Closing plates have positive gauge pressure; opening requires suction, whose validity can be limited by cavitation or air invasion. Negative gauge pressure is not in itself a contradiction.

A thin liquid lining inside a rigid cylindrical tube has pressure perturbation $p_1=-\sigma (a^{-2}{h}_1+h_{1,xx})$ at leading order in $h_0/a$. Lubrication theory gives flux $q=-h_0^3{p}_x/(3\mu )$, so mass conservation gives the displayed growth rate. Long waves with $0<k^2{a}^2<1$ grow, while short waves are stabilized by curvature. The maximum occurs at $k^2{a}^2=1/2$.

A thin axisymmetric coating of physical thickness $d=ϵah$ on a rigid cylinder of radius $a$ has leading curvature $a^{-1}-d/a^2-d_{\bar{z}\bar{z}}$. With constant surface tension $\gamma$, negligible Newtonian gravity, no-slip boundary condition at the wall and negligible exterior stress, lubrication theory gives $q=-d^3{p}_{\bar{z}}/(3\mu )$. Set $z=\bar{z}/a$ and $t=\gamma {ϵ}^3\bar{t}/(3\mu a)$ to obtain the displayed thin-film equation. A perturbation of a uniform film $h_0$ proportional to $e^{st+ikz}$ hasThus the fastest Rayleigh–Plateau instability has $|k|=1/\sqrt{2}$, or physical wavelength $2\pi \sqrt{2}a$.

For a sphere of radius $R$ moving through a membrane tube with a uniform narrow gap $\delta \ll R$, lubrication theory givesIt exceeds the unbounded-fluid Stokes drag law coefficient by $2R^2/{\delta }^2$.

With $K=\rho g/(3\mu )$ and formal edge conditions $h(\pm L,t)=0$, the lubrication theory equation is $h_t=K(h^3{h}_x{)}_x$. Its positive separated long-time profile hasWriting $C=B(4/5,1/2)$ in terms of the beta function, the centre height is $h_0^3=10\mu L^2/(\rho g{C}^{2}t)$. The edge region of a draining viscous film determines how the singular outer profile matches the flow over the physical edge.

The outer similarity solution approaches a plate edge as $h/h_0\sim [(4C/5)(1-x/L){]}^{1/4}$, where $C=B(4/5,1/2)$. Its slope becomes order one atThis region lies outside lubrication theory because its vertical and horizontal scales are comparable. The outer zero-thickness boundary condition is consistent at leading order since the matching height is $O(\ell )\ll h_0$, while the actual edge flow requires an inner solution. A finite outward flux survives even though the outer thickness vanishes, because the slope diverges.

In a dimensionless thin bearing with lower-wall speed one, stationary upper wall and gap $h(x)$, lubrication theory gives $u=1-Y/h+(p^{\prime }/2)Y(Y-h)$ and constant volume flux $q=h/2-h^3{p}^{\prime }/12$. Equal endpoint pressures, chosen as zero at $x=0,1$, imply $(h^3{p}^{\prime }-6h{)}^{\prime }=0$.

For a converging exponential gap $h=e^{-\alpha x}$, $\alpha >0$, integration gives $p^{\prime }=6e^{3\alpha x}(e^{-\alpha x}-M)$ with $M=3(e^{2\alpha }-1)/(2(e^{3\alpha }-1))$. This constant is the positive weighted average $\int h{h}^{-3}dx/\int h^{-3}dx$, so $e^{-\alpha }<M<1$. It follows that the pressure first rises, reaches a unique maximum at the displayed $x_m$, and falls back to zero. The interior pressure is positive and supports a load. Expansion gives $x_m=1/2+5\alpha /24+O({\alpha }^3)$ as $\alpha \to 0$.

For inner radius $a$, mean gap $\Delta \ll a$, fixed eccentricity parameter $|\alpha |<1$, gap $h=\Delta (1+\alpha \sin \theta )$, and inner angular velocity $\Omega$, the constant circumferential flux isIt follows by integrating the lubrication theory relation $p_s=6\mu \Omega a/h^2-12\mu q/h^3$ over a full period. The leading holding force is horizontal:Integration by parts computes this force without a closed-form pressure. The fluid couples about the separate cylinder axes obey $C_i+C_o=\alpha \Delta F_x$; moments about a common axis balance exactly at this order.

An icicle ripple is a corrugation of the ice-water interface produced while a thin flowing water film freezes. Coupling lubrication theory, surface tension, thermal conduction, and the Stefan condition determines its growth and migration.

For a film of thickness $H$, axial length $L\gg H$, and velocity $U$, the ratio of convective inertia to transverse viscous stress in lubrication theory is $U{H}^2/(\nu L)$. On a small incline of slope $\theta$, lubrication gravity-current flux gives $U\sim g\theta H^2/\nu$ when $H\ll \theta L$, and $U\sim g{H}^3/(\nu L)$ when $H\gg \theta L$. The respective criteria are $g\theta H^4/({\nu }^{2}L)\ll 1$ and $g{H}^5/({\nu }^2{L}^2)\ll 1$. Unsteady inertia also requires $H^2/(\nu T)\ll 1$ if the evolution timescale $T$ is independent of $L/U$.

A long-wave approximation assumes that a layer thickness $H$ is much smaller than its streamwise length $L$. The small aspect ratio $H/L$ makes transverse gradients dominate longitudinal gradients and is the geometric basis of lubrication theory.

In a thin fluid-filled crack, lubrication theory gives flux per unit depth $q=-h^3{p}_x/(12\mu )$. The surrounding Darcy flow has normal velocity $-k{p}_y/\mu$. Conservation in the crack gives $q_x=-[u_y{]}_{-}^{+}$, yielding the displayed condition. Pressure is continuous across the crack. This reduced model neglects transverse pressure variation and assumes wall slip lengths are small compared with $h$.

For smooth no-slip spheres of radius scale $a$ with minimum gap $\delta \ll a$, the lateral gap scale is $(a\delta {)}^{1/2}$. Squeeze-film lubrication theory gives pressure $p\sim \mu a(-\dot{\delta })/{\delta }^2$ and the displayed normal resistance. Bounded forcing therefore gives $-\dot{\delta }=O(\delta )$, so the gap cannot vanish in finite time. The conclusion uses smooth surfaces, no slip and continuum creeping flow; it is not a contact model for rough particles.

Lubrication theory predicts large viscous resistance when fluid must be squeezed through a narrow gap. For normal approach of smooth curved surfaces with curvature scale $a$ and minimum separation $h\ll a$, the resisting force scales as $\mu a^{2}U/h$. Under a bounded normal force this gives $|\dot{h}|=O(h)$, so ideal smooth surfaces do not make contact in finite time. Tangential and rolling motions have different resistance laws; the normal squeeze law should not be applied to them.

For an incompressible Newtonian fluid at low Reynolds number, the quasi-steady Stokes equations areThe first equation expresses local conservation of volume for constant density; the second balances pressure, viscous stresses and body force with negligible inertia. Convective inertia is negligible when $\mathrm{Re}=U\ell /\nu \ll 1$ at the relevant flow scale. Quasi-steady flow also requires the viscous diffusion time ${\ell }^2/\nu$ to be short compared with the evolution time. Low convective Reynolds number alone does not justify neglecting a rapidly oscillating unsteady acceleration.

For a thin layer, slenderness further changes the comparison of longitudinal and transverse derivatives; these scalings, as well as low inertia, are needed for lubrication theory.

Let $H$ be the typical thickness, ${\ell }_r$ the radial variation length, $U$ the horizontal speed, and $W$ the vertical speed. For a thin smooth profile, $\epsilon =H/{\ell }_r\ll 1$. The axisymmetric continuity equation isAway from singular boundary regions, balancing its terms gives $U/{\ell }_r\sim W/H$, hence $\boxed{W/U\sim \epsilon \ll 1}$. Near $r=0$, regularity gives $v_r=O(r)$ and the same conclusion follows without a divergent $v_r/r$ term.

Vertical viscous derivatives dominate horizontal ones by ${\epsilon }^{-2}$. The ratio of convective acceleration to vertical viscous acceleration isFor thickness-evolution time $T$, the local inertial ratio is $H^2/(\nu T)$. When both ratios are small, lubrication theory reduces the Navier-Stokes equation toThinness must therefore accompany a slow enough motion; it is not a justification for dropping arbitrarily rapid imposed acceleration. The small vertical velocity also makes the leading vertical balance hydrostatic balance, $p_z=-\rho g$. The neglected normal viscous stress at the interface is smaller than the characteristic lubrication pressure by order ${\epsilon }^2$.

Use the source's curvature sign: $p_{\mathrm{air}}-p_{\mathrm{liquid}}=-\gamma \kappa$. The Young–Laplace equation and atmospheric pressure $p_0$ thus give $p(r,h,t)=p_0+\gamma \kappa$. Integrating the hydrostatic balance yieldsConsequently $p_r=\gamma {\kappa }_r+\rho g{h}_r$, independent of $z$ at this order. The radial Stokes equation and boundary conditions areThe first is the no-slip boundary condition; the second is the leading stress-free boundary condition, since the atmosphere has negligible dynamic viscosity and the surface tension is constant. Integrating twice givesThis is the horizontal velocity only. A smaller vertical velocity is generally required by the continuity equation.

The gravity-capillary drop with a meridional curvature approximation uses the stipulated $\kappa =-h^{\prime \prime }$ for this calculation. At equilibrium the thin-film equation givesRegularity at $r=0$ forces $C=0$. In the wetted interior $h>0$, the equilibrium pressure gradient therefore vanishes, givingThis is the capillary length. Integrating the constant-coefficient ordinary differential equation givesThe central boundary condition $h^{\prime }(0)=0$ sets $C_2=0$, and contact at $r=R$ gives $C_0=-C_1\cosh (R/l)$. Write $C_1=-\alpha l$. The prescribed contact angle fixesHere $\alpha$ is the dimensionless coefficient of this drop solution, unrelated to the hinge angle in question 1.

The axisymmetric volume integral is $V=2\pi {\int }_0^{R}rh(r)dr$. Substituting $\xi =r/l$ givesFor numerical evaluation, with $s=R/l$, the integral isTogether with $\alpha =\tan {\theta }_c/\sinh s$, this determines $R$ for fixed $V$ and ${\theta }_c$.

The Bond number $R^2/l^2=\rho g{R}^2/\gamma$ distinguishes the dominant balances. For $R\ll l$, surface tension dominates gravity and the surface is nearly a small-slope constant-curvature cap:For $R\gg l$, gravity flattens the interior, while surface tension bends the edge over the capillary length. The stipulated model givesup to relative corrections of order $l/R$. The contact angle fixes the edge slope and therefore the height scale; conservation of mass then fixes the lateral extent. Consistency of lubrication theory requires small slopes, in particular $\tan {\theta }_c\ll 1$.

The stated approximation differs from axisymmetric free-surface curvature. For an axisymmetric graph with the same sign convention, the sum of principal curvatures isThe omitted $h^{\prime }/r$ term is generally of the same order as $h^{\prime \prime }$. For example, $h=H(1-r^2/R^2)$ gives stipulated $\kappa =2H/R^2$ but axisymmetric small-slope $\kappa =4H/R^2$. Thus the required hyperbolic-cosine profile solves the explicitly prescribed reduced model, not the full axisymmetric capillary equilibrium. Keeping both small-slope terms instead giveswhere $I_0,I_1$ are modified Bessel functions. The qualitative capillary length and Bond number balances remain valid.

For a stationary film of uniform thickness, thermal conduction gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:Consequently the surface temperature isThis is the quasistatic temperature of a conductively cooled film, with a weak-cooling expansion when $\alpha h/\kappa \ll 1$.

Let the thickness and longitudinal scales be $h_{\ast },L$, with $ϵ=h_{\ast }/L\ll 1$. Incompressibility gives transverse speed of order $ϵU$, and film evolution has the kinematic time scale $L/U$. For the temperature variation scale $\Theta$, longitudinal diffusion divided by transverse diffusion is $h_{\ast }^2/L^2={ϵ}^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse diffusion isThus ${ϵ}^2\ll 1$ and $Pe\ll 1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_{\ast }^2/(\kappa t_{\mathrm{evol}})$ to be small.

Define the positive surface-tension gradient coefficientThe weak-cooling approximation gives ${\gamma }_s\simeq {\gamma }_0+Bh$, so the Marangoni stress is ${\gamma }_{s,x}=B{h}_x$. The Young–Laplace equation gives the leading liquid pressure $p=p_a-{\gamma }_0{h}_{xx}$, with constant ambient pressure. In lubrication theory, $p_z=0$ and $\mu u_{zz}=p_x$. The no-slip boundary condition at the wall and the tangential surface condition $\mu u_z(h)={\gamma }_{s,x}$ giveHence the volume flux isConservation of mass gives the thermocapillary thin-film equationThe replacement of surface tension by ${\gamma }_0$ in the capillary term needs $B{h}_{\ast }/{\gamma }_0\ll 1$. For comparable capillary and thermocapillary fluxes,so the fractional tension change is only $O({ϵ}^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution ${\gamma }_x{h}_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|{\gamma }^{\prime }|\Delta T/{\gamma }_0$ were allowed to grow without bound.

The capillary term smooths the film: a crest has negative curvature and higher liquid pressure, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher surface tension; Marangoni stress pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.

Use $H=h/\hat{h}$, $X=ϵx/\hat{h}$, and $\tau =t/\hat{t}$. The two dimensionless flux-divergence coefficients are $B{ϵ}^2\hat{t}/(2\mu )$ and ${\gamma }_0{ϵ}^4\hat{t}/(3\mu \hat{h})$. Setting both equal to one gives the balanced scales for a thermocapillary film:These choices yield $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The predicted $ϵ$ must be small for the assumed long-wave approximation.

Linearize about $H=1$ using a normal mode of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the dispersion relation isModes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbersThe physical fastest wavenumber is $ϵ/(\sqrt{2}\hat{h})$.

For a positive steady film with zero volume flux, $H^2{H}_X+H^3{H}_{XXX}=0$. Divide by $H^3$ and integrate once:Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the zero-flux thermocapillary film profiles relationThus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.

The Darcy law here is $\mathbf{u}=-k\nabla p/\mu$, with hydrostatic contributions absorbed into $p$ if present. Since $k$ is uniform and the fluid is incompressible, $\nabla \cdot \mathbf{u}=0$ gives $\Delta p=0$ in the porous matrix.

Pressure is uniform across the thin crack at leading order. No-slip lubrication theory gives its flux per unit depth $q=-h^3{p}_x/(12\mu )$. Conservation in a short crack segment says $q_x=-[u_y{]}_{-}^{+}$, while Darcy flow gives $[u_y{]}_{-}^{+}=-(k/\mu )[p_y{]}_{-}^{+}$. ThereforeThe other interfacial condition is pressure continuity, $[p{]}_{-}^{+}=0$. The far pressure gradient is $-\mu U{\mathbf{e}}_x/k$; outside the slit both pressure and normal Darcy flux are continuous. The leading tapered-crack flux vanishes at its endpoints. The assumptions $h\ll a$ and $h\gg \sqrt{k}$ justify transverse pressure uniformity and neglect of porous-wall slip on scale $\sqrt{k}$. Close to the sharp tips the reduced model is an outer approximation, not a uniformly resolved tip solution. This is lubrication boundary condition for a crack in a Darcy medium.

These are planar elliptic coordinates. The curve $\xi =0$ is the slit $[-a,a]$, traced on its upper and lower sides. The curve $\xi =1$ is the ellipse with semiaxes $a\cosh 1$ and $a\sinh 1$. For fixed noncardinal $\eta$, the coordinate curve is the appropriate branch of $x^2/(a^2{\cos }^2\eta )-y^2/(a^2{\sin }^2\eta )=1$. The cardinal angles give the two exterior horizontal rays and the upper/lower vertical rays. The requested original sketch is:

On the upper face $0<\eta <\pi$, the supplied derivatives reduce to ${\partial }_x=-(a\sin \eta {)}^{-1}{\partial }_{\eta }$ and ${\partial }_y=(a\sin \eta {)}^{-1}{\partial }_{\xi }$. The same point on the lower face has angle $2\pi -\eta$, so its ${\partial }_y$ has the opposite sign. ThusInsert this and the tangential derivative into the crack condition and multiply by $12a^2\sin \eta$ to obtainPressure continuity becomes $p(0,\eta )=p(0,2\pi -\eta )$. The uniform-flow background is $p_0=-\mu Ux/k=-(\mu Ua/k)\cosh \xi \cos \eta$. Its derivatives immediately giveThe harmonic equation is $p_{\xi \xi }+p_{\eta \eta }=0$ because the metric multiplier is nonzero outside the slit.

For the specified profile, $h^3=H_0^3\sin \eta$ on the upper face. Its coefficient in the tangential operator is constant, so the imposed cosine mode separates. With $P_0=\mu Ua/k$, write $p=P_{0}F(\xi )\cos \eta$. Harmonicity gives $F^{\prime \prime }=F$, the interfacial condition gives $F^{\prime }(0)=\alpha F(0)$, and the far field fixes the growing coefficient to $-1/2$. ThereforeReflection symmetry makes the two face derivatives equal. Other homogeneous cosine modes decaying as $e^{-m\xi }$ would require $-m=\alpha m^2$ at the slit, impossible for $m>0$ and $\alpha \ge 0$. The constant pressure gauge is immaterial. This justifies the separated crack pressure in elliptic coordinates, not only a candidate satisfying selected conditions.

At the midpoint $F(0)=-1/(1+\alpha )$ and $p_x=-\mu U/[k(1+\alpha )]$. HenceFor $\alpha \ll 1$, $Q\sim H_0^{3}U/(12k)$ and the crack barely perturbs the imposed Darcy pressure gradient. For $\alpha \gg 1$, $Q\sim 2aU$ and the slit is nearly equipotential: its high axial conductance collects flow from the surrounding matrix rather than sustaining a substantial pressure gradient. The parameter compares crack conductance $H_0^3/\mu$ with matrix conductance $ka/\mu$. In fact $q(x)=Q\sqrt{1-x^2/a^2}$, so the tip flux is zero in this outer model.

Subtracting the exact uniform background gives $\delta p=P_0\alpha e^{-\xi }\cos \eta /(1+\alpha )$. Since $r\sim a{e}^{\xi }/2$ and $\eta \sim \theta$ at infinity,The printed far-field coefficient is too large by a factor of two. The factor $1/2$ is forced by the stated coordinate map. An independent check is the large-$\alpha$ solution $p=-(\mu U/k)\mathrm{Re}\sqrt{z^2-a^2}$, whose expansion has precisely $\delta p\sim \mu U{a}^2\cos \theta /(2kr)$. This is the elliptic crack pressure dipole.

The phrase “suitably chosen curve” matters for the pressure-only dissipation integral. Put $C_d=\alpha \mu U{a}^2/[2(1+\alpha )k]$ and use a rectangle with end faces $x=\pm X$ and transverse extent $|y|\le H$, taking $a\ll X\ll H$. The horizontal faces have $n_x=0$. The two end-face integrals of $\delta p{n}_x$ totalThis is a fixed-throughflow normalization: the integrated velocity perturbation on an infinite end face is zero, since it is proportional to ${\partial }_x\int \delta pdy={\partial }_x(\pi C_d\mathrm{sgn}x)=0$. Work on the transverse faces vanishes as $H/X\to \infty$. Thus the requested line functional givesA bare circular cutoff would give $-\pi U{C}_d$ instead; it does not enforce this same throughflow normalization, so its pressure-only term cannot be substituted into the fixed-throughflow dissipation comparison. Keeping the control surface explicit resolves that ambiguity. This is fixed-throughflow dissipation reduction of a conductive crack.

Let $s=\pi \varphi a^2\alpha /(1+\alpha )\ll 1$. The noninteracting average dissipation per area is $\mu U^2/k+\varphi \Delta \mathcal{D}=(\mu U^2/k)(1-s)$. Equating this with $\mu U^2/k_{xx}^{\ast }$ givesOnly the first-order density term is controlled by the noninteraction assumption; the rational form is not an all-orders result. It increases permeability along the aligned cracks and saturates as their conductance becomes large. This is dilute permeability enhancement by aligned cracks.

Let $\delta =H/L\ll 1$. Incompressibility makes the transverse velocity $O(\delta U)$. In the axial Navier-Stokes equation, transverse viscous stress scales as $\mu U/H^2$, longitudinal viscous stress as $\mu U/L^2$, and convective inertia as $\rho U^2/L$. Thus lubrication theory requiresUnsteady inertia is also negligible if $\rho H^2/(\mu T)\ll 1$; this coincides with the preceding condition when $T=L/U$. The transverse pressure variation is smaller than this pressure scale by $O({\delta }^2)$, and viscous normal stress at the free surface is similarly smaller. Consequently pressure is uniform across the film to leading order, while axial viscous shear balances its axial gradient. These estimates justify the lubrication approximation with small surface slopes and a slowly varying rigid wall.

Write the physical cylindrical-film thickness as $d=ϵ\bar{h}$, so that $\bar{h}=O(a)$, and use $\bar{z},\bar{t}$ for dimensional coordinates. For a surface of radius $a+d$, its outward mean curvature, with the convention giving cylindrical curvature $1/a$, isThe Young–Laplace equation gives $p=p_{\mathrm{ext}}+\gamma \kappa$. With local normal coordinate $y$ measured from the rigid cylinder, the axial velocity solves $\mu u_{yy}=p_{\bar{z}}$, with no-slip boundary condition at $y=0$ and stress-free boundary condition at $y=d$. ThereforeThe factors of circumference in axisymmetric mass conservation cancel at leading thin-film order, giving $d_{\bar{t}}+q_{\bar{z}}=0$. IntroduceThis produces the capillary instability of a thin film on a rigid cylinder governed byThe curvature term $h_z$ destabilizes long waves; the axial-curvature term $h_{zzz}$ smooths short waves.

Linearize about $h_0$ using $h=h_0+\eta e^{st+ikz}$ with small $\eta$. The linear equation is $s+h_0^3(-k^2+k^4)=0$, soThe growth rate is positive for $0<|k|<1$, zero at $k=0,\pm 1$, and negative for $|k|>1$. Its derivative is $2h_0^{3}k(1-2k^2)$, givingThe requested sketch is even in $k$:

For the Rayleigh–Plateau instability of an entire liquid cylinder in the purely viscous limit with negligible exterior viscosity, the fastest mode tends to infinite wavelength. The rigid-wall coating instead selects a finite wavelength. The difference comes from the flow resistance: a coating must shear across its thickness, so its long-wave growth vanishes as $k^2$; a free liquid cylinder can undergo nearly plug-like axial extension. For example, the long-wave free-cylinder balances ${\eta }_t+(a/2)u_z=0$ and $3\mu u_{zz}=-\gamma ({\eta }_z/a^2+{\eta }_{zzz})$ give $s\sim \gamma (1-a^2{k}^2)/(6\mu a)$, which does not vanish as $k\to 0$. This comparison is for the creeping-flow, negligible-exterior-viscosity limit; finite inertia or exterior viscosity changes the wavelength selection. The full free-cylinder dispersion relation is discussed in Plateau-Rayleigh Instability of a Cylinder of Viscous Liquid.

The collar is a quasisteady sliding collar on a coated cylinder. It must slowly grow as well as translate: an exact travelling wave would obey $q-ch=\text{constant}$, whereas its stationary uniform far films have $q=0$ and would require $-c{h}_{+}=-c{h}_{-}$. Thus the stipulated unequal films require the stated slow evolution of the collar amplitude. In its $h=O(1)$ core, $c\ll 1$ and the volume flux is small, so the leading pressure gradient vanishes: $h^{\prime }+h^{\prime \prime \prime }=0$. Integration gives a constant plus sine and cosine terms. Choose the collar centre at its maximum to remove the sine term, and parameterize its edge height as $B$. ThenThe core is a local capillary equilibrium. In the moving frame, swept-up volume minus deposited volume gives the growth rate $c(h_{+}-h_{-})$. Its leading excess volume is $2\pi A$, and for constant $B$,Thus the printed rate is the rate of collar volume, or an order-of-magnitude amplitude rate; interpreted as an exact equation for $\dot{A}$, it lacks $1/(2\pi )$. This slow change is smaller than the translational core terms by the small far-film thickness and does not alter the leading local matching below.

Near either edge $x_f=\pm \pi$, let $h_f$ denote the adjacent film thickness, $h_{-}$ at the rear and $h_{+}$ at the front. Across the short transition region the collar's slow growth is higher order, so local mass conservation integrates to $q=c(h-h_f)$. The axial third derivative dominates the first derivative in the capillary flux, givingFor collar-to-film capillary matching, chooseThe two sides become $c{h}_f{H}^3{H}^{\prime \prime \prime }$ and $c{h}_f(H-1)$ respectively, and henceThe resulting transition length is small compared with the core length: the matching relations below imply $h_f=O(c^{2/3}/A)$, so $h_f/c^{1/3}=O(c^{1/3}/A)$ for an order-one core amplitude.

To count admissible branches, put $H=1+\zeta$ and linearize. The characteristic equation is $r^3=1$, with rootsAt the rear, decay as $X\to -\infty$ permits only $\zeta \sim D{e}^X$. For the nontrivial branch growing into a positive collar, $D>0$; translation rescales $D$, leaving one orbit. This branch has positive $H^{\prime },H^{\prime \prime },H^{\prime \prime \prime }$. It exists for all forward $X$, since $0<H^{\prime \prime \prime }=(H-1)/H^3\le 4/27$ for $H\ge 1$, and once $H^{\prime \prime }>0$, $H$ grows at least quadratically. Then $H^{\prime \prime \prime }=O(X^{-4})$, so successive integrations give a quadratic asymptote with a finite constant term; translation removes the linear term. The constant solution is not a rear-to-collar match. On the downward branch, $H<1$ gives $H^{\prime \prime \prime }<0$ while $H^{\prime }$ and $H^{\prime \prime }$ are initially negative; the slope remains negative and $H$ reaches zero, so it cannot match a positive collar. Uniqueness here refers to the positive, nontrivial rear matching branch, modulo translations.

At the front, decay as $X\to +\infty$ permits the two real amplitudes inTranslation removes one degree of freedom, leaving one parameter $\lambda$. These oscillatory branches account for the small capillary ripples ahead of the collar. This local stable manifold count gives the family of front branches; admissible global branches must also remain positive and reach a parabolic core asymptote. It does not, by itself, prove the global uniqueness of the matching parameter supplied in the question.

Write the rear and front core asymptotes, after translation removes their linear terms, as $a_{-}{X}^2+b_{-}$ and $a_{+}(\lambda )X^2+b_{+}(\lambda )$. Expanding the outer collar near either edge giveswhereas the inner expansion givesMatching the constant and quadratic coefficients therefore yieldsEliminate $h_{-}/h_{+}$ to obtain $a_{+}(\lambda )b_{+}(\lambda )=a_{-}{b}_{-}$. Using the specified selected root ${\lambda }_0$ and writing $\alpha =a_{+}({\lambda }_0)$, $\beta =b_{+}({\lambda }_0)$,The physical branch has $\alpha >a_{-}$, as required by $h_{-}<h_{+}$. No numerical values of these universal matching coefficients are needed for the requested relations.

Finally, the energy source is decreasing capillary surface energy, not Newtonian gravity. This can be checked through the volume-constrained capillary energy of a cylindrical coating. The exact liquid volume per axial length is $\pi [(a+d{)}^2-a^2]$, and the free-surface area is $2\pi \int (a+d)\sqrt{1+d_{\bar{z}}^2}d\bar{z}$. Expanding the latter and eliminating $\int d$ with fixed exact volume gives the nonconstant quadratic partApart from its positive dimensional factor, this is the surface energy $\mathcal{E}=\frac{1}{2}\int (h_z^2-h^2)dz$. Its variational derivative is $-h-h_{zz}$, and the thin-film equation is a conserved gradient flow. With periodic or zero-energy-flux boundaries, integration by parts givesFor an isolated collar between unequal infinite films, use the corresponding local energy balance, including far-film fluxes and subtracting background energies; the unsubtracted total energy is infinite. Collecting film into the growing collar lowers surface energy, which supplies the viscous dissipation sustaining its slow motion.