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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.

For $\omega \ne 0$, put $\mu =\hat{\mu }{\omega }^2$, $C=\omega z$ and $\tau =\omega T$, where $z=\hat{C}$. Then $C_T={\omega }^2{z}_{\tau }$ and division by ${\omega }^2$ givesWe may take $\omega >0$ without loss of the dynamics by conjugating the original equation if necessary; otherwise $\tau$ reverses the time orientation. The scaling is singular at $\omega =0$ and is not a transformation for that exactly zero-frequency case.

The Hamiltonian limit of three-to-one forcing drops the terms proportional to $\omega$. For $z=x+iy$ the remaining real system isThe proposed first integral isIndeed $x_{\tau }=H_y/2$ and $y_{\tau }=-H_x/2$, hence $H_{\tau }=0$. The unperturbed equation is a planar Hamiltonian system, with a center equilibrium at the origin and three saddle equilibria atAll saddle equilibria have $H=1/3$. The factorizationshows that their central separatrix consists of the three sides of an equilateral triangle. Each level $0<H<1/3$ inside this triangle is a closed periodic orbit. Indeed, inside the triangle $0<r<1$ and ${\partial }_{r}H=2r(1+r\sin 3\theta )>0$. Each ray from the origin therefore meets each such level once, giving a compact simple closed contour with no equilibrium point on it. The nonzero vector field traverses this contour periodically; the period grows without bound as the separatrix is approached. This supplies an infinite family, not a claim that every level outside the central region is closed.

Restore the small radial perturbation. Its exact effect on the first integral isConsequently the continuum of Hamiltonian system orbits generally does not persist. The origin becomes a weak attracting focus for $\hat{\mu }<0$ or a repelling focus for $\hat{\mu }>0$, and the three hyperbolic saddle equilibria persist with perturbed stable manifold and unstable manifold. For a small positive $\hat{\mu }$, outward drift on very small orbits balances cubic damping on somewhat larger ones, selecting a stable limit cycle rather than an arbitrary energy level. Near the center equilibrium $H\simeq r^2$, so its leading radius is $r\simeq \sqrt{\hat{\mu }}$ when $\hat{\mu }$ is also small.

For a more general closed unperturbed orbit ${\Gamma }_h$, the averaged area criterion for perturbed Hamiltonian cycles says that persistence requires the averaged energy drift ${\oint }_{{\Gamma }_h}{H}_{\tau }d\tau$ to vanish. Since the unperturbed speed is $|\nabla H|/2$, the planar divergence theorem converts this leading drift towhere ${\mathcal{D}}_h$ is the enclosed region. Isolated zeros select candidate periodic orbits; a drift changing from positive inside to negative outside gives an attracting limit cycle. The separatrix triangle has mean $r^2=1/4$, so its leading flux changes sign at $\hat{\mu }=1/2$. This marks the leading possible heteroclinic transition, with higher-order corrections needed to locate it precisely.

As a cycle approaches the saddle equilibria, long residence times and splitting of the heteroclinic cycle become important. Orbits can instead drift inward to the equilibrium point at the origin or leave the periodic island and approach one of the stable states with phase locking of the full canonical equation. Those upper-branch threefold phase-locked equilibria have $C=O(1)$, so they lie outside the local $C=O(\omega )$ scaling. Thus the small perturbation gives energy selection, attracting or repelling oscillations, and possible switching/locking transitions; it does not preserve a conserved $H$ or an infinite family of neutral periodic solutions. This qualitative picture does not assume all global parameter values have the same attractor.