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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.

Along the downward gradient flow, the gradient-flow dissipation identity, the chain rule and the defining property of the Riemannian gradient giveA smooth vector field on a compact manifold is complete, so the trajectory exists for every $t\ge 0$. In particular, its function value never increases.

If the trajectory is still in $C_{[a,b]}$ at time $t$, its monotone function value has stayed in $[a,b]$ throughout $[0,t]$. ConsequentlyThus a uniform exit bound isThe trajectory is outside the closed band for every $t>T$. Once it leaves below level $a$, monotonicity prevents its return. The argument also covers the zero-width band $a=b$.

For the total occupation-time bound for a gradient flow, let $m={\min }_{M}f$ and $F={\max }_{M}f$, which exist by compactness. For every $R>0$, the same gradient flow identity givesTaking $R\to \infty$, by monotone convergence theorem, gives the total occupation-time boundThis allows arbitrarily many visits to $C_{\epsilon }$; it bounds their total duration. Neither of these two estimates needs the Morse function hypothesis.

For exponential convergence of a Morse gradient flow, first observe that a limiting point $z=x_{\infty }$ is a critical point. Indeed, writing ${\varphi }_s$ for the flow and using continuous dependence on initial data,Differentiation at $s=0$ gives $\nabla f(z)=0$.

We use the Morse lemma: near a nondegenerate critical point of Morse index $\lambda$, there are coordinates $(u,v)\in {\mathbb{R}}^{\lambda }\times {\mathbb{R}}^{n-\lambda }$ in whichChoose a Riemannian metric equal to the Euclidean metric on a smaller ball in these coordinates. Such a global Riemannian metric exists by a partition of unity: patch this metric to any background metric outside a slightly larger ball. One can do this simultaneously at every critical point, since a Morse function on a compact manifold has only finitely many critical points. The metric is chosen before forming its gradient flow; changing the metric can change trajectories.

For this metric, any trajectory converging to $z$ eventually remains in the smaller coordinate ball. From some time $t_0$ onward its equations are exactlyConvergence forces $u(t_0)=0$, and thenThe straight radial segment stays in the coordinate ball, so its length bounds the Riemannian distance from above:Increase $C$ beyond both $e^{2t_0}|v(t_0)|$ and ${\max }_{0\le t\le t_0}{e}^{2t}d({\gamma }_x(t),z)$. Then the requested strict estimate holds for all $t\ge 0$:This proof uses the Morse lemma, existence and uniqueness for smooth ordinary differential equations, continuous dependence on initial data, and the existence of a partition of unity. The expanding coordinates explain why convergence to a saddle occurs only along its stable manifold.

For a counterexample, use an arc coordinate $x$ around a point of the circle, its flat local metric, and a globally smooth function that equals $x^4$ on this arc. Such a function can be made by a smooth cutoff, agreeing with a positive constant outside a larger arc. For $x_0>0$ sufficiently small the downward gradient flow remains in the arc and isIt converges to the degenerate critical point $0$, and its Riemannian distance from $0$ is $x(t)$ for sufficiently large $t$. For every $k>0$, $e^{kt}x(t)\to \infty$, so no exponential bound is possible. This counterexample is on a compact manifold as well.

The equilibrium equations are $x(-1+x^2+\beta y^2)=y(-1+\beta x^2+y^2)=0$. For $\beta \ne 1$ the fixed points are the origin, four axis points $(\pm 1,0),(0,\pm 1)$, and four diagonal points $(\pm s,\pm s)$ with independent signs and $s=(1+\beta {)}^{-1/2}$. The hypothesis $\beta >-1$ ensures that $s$ is real. The Jacobian matrix has diagonal entries $-1+3x^2+\beta y^2$, $-1+\beta x^2+3y^2$ and off-diagonal entries $2\beta xy$.

At the origin both eigenvalues are $-1$, so it is a stable node. At each axis point the eigenvalues are $2,\beta -1$. At each diagonal point they are $2,2(1-\beta )/(1+\beta )$, along the radial and transverse diagonal directions respectively. ThusThe axes and diagonals are invariant lines. For $\beta =1/2$ the axis saddle points have stable directions transverse to their axes and unstable directions along them; the diagonal points are unstable nodes. For $\beta =2$ these roles switch: the diagonal saddle points have stable transverse directions and unstable radial directions, and the axis points are unstable nodes. Their stable manifolds separate trajectories attracted to the origin from outward-escaping trajectories. Symmetry gives the other quadrants.

For $\beta =1$, the entire unit circle consists of equilibria. In polar coordinates $\dot{r}=r(r^2-1)$ and $\dot{\theta }=0$: every ray inside the circle moves toward the origin, and every ray outside moves outward. The circle has neutral tangential and unstable radial directions; its points are not isolated saddle points or unstable nodes.

All three phase portraits are shown below. The potential $\Phi =-(x^2+y^2)/2+(x^4+y^4)/4+\beta x^2{y}^2/2$ satisfies $(\dot{x},\dot{y})=\nabla \Phi$ and $\dot{\Phi }=|\nabla \Phi {|}^2$, ruling out nonconstant periodic orbits. Blue curves show the saddle stable separatrices, integrated backwards from their local stable directions. The flow arrows and invariant lines, together with the equilibrium classification, specify the topology of the sketches.

At an equilibrium, the second equation gives $x=y=:s$. The first equation then givesHence the fixed-point branches in the $(a,x)$ plane areandThe three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt{a}$, so the second bifurcation value is

The Jacobian matrix at a general point isOn the branch $P_0=(a^2,a^2)$ it is triangular, with eigenvaluesThus $P_0$ is unstable for $a<0$, asymptotically stable for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm \sqrt{a}$. Thenwhose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt{a}$ is therefore a saddle equilibrium for every $a>0$. The positive branch is a saddle for $0<a<1$ and asymptotically stable for $a>1$; close to $a=1$ it is a stable node.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitutionand append $\dot{a}=0$. The equations becomeThe centre subspace is $Y=X$. Seek the extended centre manifold for a parameter asThe centre-manifold invariance equation isTo second order, $\dot{X}=-aX+O(3)$, so comparison of coefficients givesand hence $A=C=0$, $B=1$. ThereforeSubstitution into the $X$ equation gives the reduced flowIts central branch $X=0$ is unstable for $a<0$ and stable for $a>0$, while the two nonzero branches $X\sim \pm \sqrt{a}$ for $a>0$ are unstable. Thus the bifurcation at zero is a subcritical pitchfork bifurcation with reversed normal-form parameter $\mu =-a$, exactly as recorded by the extended centre manifold of the 2023 Cambridge quadratic-product system.

The complete bifurcation diagram is therefore as follows. For $a<0$, only $P_0$ exists and is unstable. For $0<a<1$, $P_0$ is stable while both $P_{+}$ and $P_{-}$ are saddles. For $a>1$, $P_{+}$ is stable while $P_0$ and $P_{-}$ are saddles. At $a=1$, $P_0$ and $P_{+}$ cross and exchange stability, so the bifurcation is transcritical. This gives the bifurcation diagram of the 2023 Cambridge quadratic-product system.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll 1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt{a},\sqrt{a})$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt{a},\sqrt{a})$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional stable and unstable manifolds; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.

The fixed points are $(0,0)$ and $(\pm |\mu |,{\mu }^2)$. At the origin the eigenvalues are ${\mu }^2$ and $-1$, so it is a saddle. At either nonzero point the characteristic polynomial isThey are stable nodes for $0<|\mu |<1/(2\sqrt{2})$, degenerate nodes at equality, and stable spirals above it.

The stable manifold of the origin is exactly $x=0$. Write the stable manifold in the unstable direction as $y=a{x}^2+b{x}^4+O(x^6)$. Graph invariance givesThus $W^u$ has

At a hyperbolic equilibrium, the unstable manifold is formed by nearby trajectories approaching that equilibrium as time tends to minus infinity. It is tangent to the span of eigenvectors whose eigenvalues have positive real part. Its dimension counts departing amplitudes; translation along an autonomous trajectory can remove one parameter when comparing orbit shapes. The stable manifold instead approaches the equilibrium forward in time.