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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 tension-dominated cylindrical lipid vesicle of fixed length $L$ and volume, a real axisymmetric sine amplitude $u_q$ has stiffness $K_q=\pi \sigma L[(q{R}_0{)}^2-1]/R_0$. Fixed volume requires a mean-radius shift $-u_q^2/(4R_0)$. Stable modes have $⟨u_q^2⟩=k_{B}T/K_q$ by the equipartition theorem. Modes with $q{R}_0<1$ have the Rayleigh–Plateau instability, rather than a negative equilibrium variance. Bending modifies very short wavelengths and near-marginal modes.

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.

Take a long periodic section of length $L$, or neglect end effects and choose an integer number of wavelengths. The amplitude $u_q$ here is a single real sine amplitude, fixing the normalization of the thermal membrane undulation spectrum. For a lipid vesicle with an axisymmetric membrane deformation, conservation of the enclosed volume givesThe mean radius therefore decreases at second order. This adjustment is essential: fixing the mean radius instead of the volume would miss the unstable term in the fixed-volume capillary spectrum of a cylindrical membrane.

The surface area of an axisymmetric graph is $\mathcal{S}=2\pi \int R\sqrt{1+R_z^2}dz$. Expanding for $|u_q|\ll R_0$ and $q|u_q|\ll 1$ givesMultiplication by the membrane tension yields a quadratic potential energy $\Delta E=K_q{u}_q^2/2$, whereFor $q{R}_0>1$, the equipartition theorem gives the stable-mode fixed-volume capillary spectrum of a cylindrical membraneHere $k_B$ is the Boltzmann constant and $T$ is the temperature. The cosine amplitude has the same variance and is an independent real coordinate at quadratic order. If instead $R-{\rho }_0={\sum }_{q\ne 0}{a}_q{e}^{iqz}$ with $a_{-q}=a_q^{\ast }$, the corresponding positive-$q$ complex coefficient has $⟨|a_q{|}^2⟩=k_{B}T{R}_0/[2\pi \sigma L((q{R}_0{)}^2-1)]$. This is the same thermal membrane undulation spectrum in a different Fourier mode normalization.

For $q{R}_0<1$, the cylinder is unstable rather than having a negative fluctuation variance. The quadratic surface area change is negative: a sufficiently long-wavelength modulation reduces area while retaining volume. This is the Rayleigh–Plateau instability, expressed here as pearling of a tension-dominated lipid vesicle. A cylinder supporting such modes has no unconstrained harmonic thermal equilibrium about the uniform state. The $q=0$ uniform radius change is forbidden by fixed volume, and $q{R}_0=1$ is marginal in the tension-only approximation. The finite length and endpoint constraints determine which nonzero Fourier modes are allowed.

The large membrane tension assumption controls modes with $q{R}_0$ of order one. For large $q$ the omitted membrane bending modulus contribution grows approximately as $\kappa q^4$ and overtakes the tension term when $\kappa q^2\gtrsim \sigma$. Thus the displayed tension-only spectrum requires $q\ll \sqrt{\sigma /\kappa }$ as well as small amplitudes and a positive stiffness. Close enough to the marginal wavelength, bending corrections must also be retained; their relative scale at $q{R}_0$ of order one is $\kappa /(\sigma R_0^2)$.

For the circular phase boundary in a lipid bilayer, the energy is its perimeter times the line tension $\gamma$. Let $\phi$ be the polar angle and write $r(\phi )={\rho }_0+u_m\sin (m\phi )$, where the periodicity requires an integer $m=q{R}_0\ge 1$. Fixed enclosed area impliesThe perimeter, including the mean-radius change, isThus the fixed-area capillary spectrum of a circular boundary, again for a single real sine or cosine amplitude, isThe complex Fourier mode coefficient convention again divides this result by two. The mode $m=0$ is excluded by the fixed-area constraint, so the negative-stiffness interval $0<m<1$ is not an allowed mode of a closed circular boundary.

At $q{R}_0=1$, the zero stiffness is the translation mode of a circular boundary, not a shape instability. Displacing the centre by a small distance $h$ changes the radius to first order by $h\cos \phi$ or $h\sin \phi$ while leaving area and perimeter unchanged. For example, an exactly translated circle hasThe higher harmonics complete a true translation mode of a circular boundary. The $m=1$ denominator therefore represents free centre motion: an unconfined domain's centre has no restoring force or finite equilibrium variance in an infinite membrane. Radius fluctuations about a recentered domain omit this mode; external confinement would give it a separate restoring stiffness.

Let the actual dimensional thickness be $d=ϵah$, and let $\bar{z}$ increase downward. The leading curvature of the outer free surface isThe Young–Laplace equation gives $p_{\bar{z}}=-\gamma (d_{\bar{z}}/a^2+d_{\bar{z}\bar{z}\bar{z}})$. In the lubrication approximation, the downward velocity at distance $y$ from the wall is $(\rho g-p_{\bar{z}})(dy-y^2/2)/\mu$, from no slip and zero free-surface shear. Integrating this profile givesCircumferential corrections are higher order in $d/a$, so volume conservation is $d_{\bar{t}}+q_{\bar{z}}=0$ at leading order. DefineThen the thin liquid film on a vertical cylinder satisfiesThe circumferential-curvature term $h_z$ is retained along with axial curvature.

The perturbation calculation givesLong waves with $0<|k|<1$ grow, while $|k|>1$ decay; the fastest growth is $1/12$ at $|k|=1/\sqrt{2}$. Disturbances drift downward at speed $G$. The dimensional unstable wavelengths exceed $2\pi a$, with most unstable wavelength $2\pi \sqrt{2}a$. This is the thin-film version of the Rayleigh–Plateau instability driven by circumferential curvature and opposed by short-wave axial curvature.

For a travelling wave, one integration fixed by $H\to 1$ givesor $H^{\prime \prime \prime }+H^{\prime }=3c(H-1)/H^3-G(1-H^{-3})$. In the large positive-speed core, substitute the expansion using the supplied assumptions. The left side at order $c^{2/3}$ is $H_2^{\prime \prime \prime }+H_2^{\prime }$, while the gravity term is order one and $3c/H^2$ is only order $c^{-1/3}$. Thus $H_2^{\prime \prime \prime }+H_2^{\prime }=0$. Its general solution is a constant plus a sine and cosine. The vanishing leading height and slope at both endpoints selectThe core length is $2\pi$ in the axial coordinate; its amplitude is fixed by transition matching.

At either edge set $X=(3c{)}^{1/3}(x-x_e)$, with $x_e=0$ at the trailing edge and $x_e=2\pi$ at the leading edge. Keeping the height $H$ of order one, the exact wave equation becomesThus the transition equation is $H^3{H}_{XXX}=H-1$. Its linearization about $H=1$ has exponents $1$ and $-1/2\pm i\sqrt{3}/2$. Behind the pulse, approaching the uniform film as $X\to -\infty$ permits only the positive real exponent. The positive-growing, nontrivial branch has one amplitude that is removed by translation, giving the relevant unique $H_{-}$. In front, decay as $X\to +\infty$ permits the two real oscillatory amplitudes; translation removes one parameter and leaves a one-parameter family $H_{+}$. These are the unstable manifold and stable manifold dimensions for the transition system. The literal boundary condition also admits the constant solution and an opposite departing branch, so uniqueness here is specifically for the branch matching a raised positive pulse.

Match the supplied trailing quadratic asymptotic to the core near zero. It gives $K_2=\kappa 3^{2/3}$, because $\frac{1}{2}{c}^{2/3}{K}_2{x}^2=\frac{1}{2}\kappa (3c{)}^{2/3}{x}^2$. The leading maximum occurs at $x=\pi$, soThe front tail has the form $H-1\sim C{e}^{-X/2}\cos (\sqrt{3}X/2+\theta )$. Its capillary waves occur ahead of the descending pulse, not behind it; the trailing tail is monotone to leading order. Converting its frequency to physical distance gives

This large solitary pulse on a cylindrical film sketch shows the leading core and the front-tail eigenmodes; the edge transitions are schematic and the ringing is enlarged for visibility. The thin-film asymptotics require both $c\gg 1$ and $ϵc^{2/3}\ll 1$, rather than taking speed to infinity at fixed film slenderness.

At the next core order, $H_0^{\prime \prime \prime }+H_0^{\prime }=-G$. Integrating and imposing zero $H_0^{\prime }$ at both endpoints givesThe constants in the two quadratic transition asymptotics match the core endpoint heights: $A_0+B_0={\beta }_{-}$ and $A_0+B_0-2\pi G={\beta }_{+}$. Hence $2\pi G={\beta }_{-}-{\beta }_{+}$, and the dimensional uniform thickness selected by matching isWith the supplied constants this fixes $G=3.75/(2\pi )$; $B_0$ is not separately fixed by these height conditions. The thickness selection and capillary wavelength follow from matched asymptotic expansion, not from the linear instability calculation.

The expression in braces in the first equation is the net axial force. The term $-\pi a^2\gamma /a$ is the compressive force from the capillary pressure $\gamma /a$, $3\pi \mu a^2{w}_z$ is the Newtonian extensional tension with Trouton ratio three, and $2\pi a\gamma$ is the axial pull of surface tension around the circumference. Its $z$-derivative vanishes because axial force is conserved.

The final equation is conservation of an insoluble surfactant. The surface divergence $w_z+u/a$ is the sum of axial and circumferential extension rates. Positive surface divergence increases interfacial area and dilutes $C$; negative divergence concentrates it. The absence of a diffusion term expresses the assumption of negligible surface diffusion.

Linearizing the area and surfactant equations givesThusand hencewhere $\psi$ is fixed by the initial data.

Linearizing the displayed net axial force and using ${\gamma }^{\prime }=-A{C}^{\prime }$ givesEliminating $w_z$ and $C^{\prime }$ therefore gives directlyThus the three displayed evolution equations implyThe target formula printed later in the paper contains an additional factor of $\pi$ in both denominators. That factor does not follow from the displayed equations because every term in the axial-force balance contains the same factor $\pi$. If the target formula is adopted as the intended normalization, its corresponding value is $s=({\gamma }_0-A{C}_0)/(6\pi \mu a_0)$.

Initially $\eta =0$ and $C^{\prime }>0$, so $\psi >0$ and ${\eta }_t<0$: a surfactant-rich, low-tension region begins to neck as neighboring higher tension pulls fluid away. The accompanying axial extension dilutes the surfactant.

If $A<{\gamma }_0/C_0$, then $s>0$. The Rayleigh–Plateau instability overwhelms the weak surface-elastic response: the necking perturbation grows in the linear model while the original concentration excess is diluted and eventually changes sign.

If $A>{\gamma }_0/C_0$, then $s<0$. Strong surface elasticity arrests the disturbance atThe concentration perturbation becomes negative, raising the local surface tension until its axial force balances that of the wider regions. This is stabilization by a surfactant-induced Marangoni stress.

For an exterior thin film, circumferential and axial curvature together give $h_t+(h^3[h_{zzz}+h_z+G]{)}_z/3=0$ in appropriate dimensionless variables. The $h_z$ term destabilizes long waves through the Rayleigh–Plateau instability, whereas $h_{zzz}$ damps short waves. The linear growth rate is $(k^2-k^4)/3-iGk$, with fastest growth at $|k|=1/\sqrt{2}$. The gravity parameter is $G=\rho g{a}^2/(\gamma ϵ)$ when film thickness is scaled as $ϵah$ and axial distance by cylinder radius $a$.