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A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

A nullcline of a planar autonomous differential equation is a set where one component of its vector field vanishes, typically drawn as a curve. Intersections of the two component nullclines are equilibrium points. The sign of each component on either side helps determine the directions in a phase portrait.

The equilibrium points satisfy $y=0$, $x^2=\lambda$. For $\lambda <0$ there are none. For $\lambda >0$ set $a=\sqrt{\lambda }$. The Jacobian matrix isThe positive equilibrium point $(a,0)$ is always a saddle equilibrium. At $(-a,0)$ the determinant is $2a$, so it is stable for $\mu <a$ and unstable for $\mu >a$. The discriminant $(\mu -a{)}^2-8a$ distinguishes a node from a focus; its zero does not destroy hyperbolicity and is not another local bifurcation curve.

The local bifurcation curves areAt $(0,0)$ the linearization has a double zero eigenvalue and is nilpotent, giving a Bogdanov–Takens bifurcation. For $\mu \ne 0$ on $\lambda =0$, solving the local center manifold gives $\dot{x}=(\lambda -x^2)/\mu +\cdots$, displaying the saddle-node bifurcation. For $\mu <0$ the nonsaddle branch born at positive $\lambda$ is attracting; for $\mu >0$ it is repelling. On the Hopf curve the angular frequency is $\sqrt{2a}$. The permitted subcritical Hopf bifurcation creates a repelling periodic orbit on the stable-focus side $\mu <a$.

A local analysis alone does not say how far that periodic orbit extends. The global calculation below supplies the additional division of the stable-focus region. The complete nearby phase portraits and the leading global curve are shown together here. The $lambda<0$ flow has no closed periodic orbit, since a closed orbit would have index one and hence enclose an equilibrium point.

Use a real physical field, so the three positive wavevectors are accompanied by their complex-conjugate modes. Modulo lattice translations, the effective symmetry group is $G={\mathbb{T}}^2⋊C_6$. Its translation group action isA $120^{\circ }$ rotation cyclically permutes the amplitudes; a half-turn $\kappa$ acts by simultaneous complex conjugation. Together these generate the sixfold rotation group. These formulas specify the group action independently of the choice of orientation of the three wavevectors.

For a nonzero real hexagon $(h,h,h)$, its stabilizer subgroup is the rotation group $C_6$. A translation fixes it only when all three phase factors equal one. Cyclic permutation forces a fixed vector to have equal components, and the half-turn forces those components to be real. Therefore $\mathrm{Fix}(C_6)=\{(a,a,a):a\in \mathbb{R}\}$. For a roll $(r,0,0)$ with nonzero real $r$, the stabilizer subgroup consists of the translations $T_{0,\theta }$ and the half-turn, with $\kappa T_{0,\theta }\kappa =T_{0,-\theta }$. It is $S^1⋊C_2$, isomorphic to $O(2)$ as an abstract group even though spatial reflections are absent. The circle eliminates $B,C$ from its fixed-point subspace of a group action, and the half-turn makes $A$ real. ThusIf unreduced physical translations are used, the full lattice kernel is included in both stabilizer subgroups. The real representation on the six critical components is absolutely irreducible: translations distinguish the three Fourier pairs, rotations permute them, and the half-turn rules out a complex scalar in the commutant. The equivariant branching lemma consequently supplies generic steady branches of both axial types at a transverse crossing of the common linear eigenvalue. It guarantees existence, not stability or exhaustiveness of the possible branches.

Translation equivariance severely restricts the monomials in $\dot{A}$. Through cubic order they are $A$, $\overline{B}\overline{C}$, $|A{|}^{2}A$, $|B{|}^{2}A$ and $|C{|}^{2}A$: each has the same translation weight as $A$, using ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. The half-turn forces their coefficients to be real; the $120^{\circ }$ rotation gives the two cyclic counterpart equations. A spatial reflection would interchange $B,C$ and require equal cross-couplings, but it is not a symmetry here. After a real amplitude rescaling makes the generic nonzero quadratic coefficient one, the rotational hexagon amplitude equations arewith the other two equations obtained by cyclic permutation. All four coefficients are real; $\delta$ records the absence of spatial reflection. The normalization assumes the resonant quadratic coefficient is nonzero; its vanishing would be an additional degeneracy.

For the roll, $r^2=\mu /{\nu }_1$. The radial eigenvalue is $-2\mu$ and the phase of the active mode is a neutral translation. The transverse variables $(b,\overline{c})$ have matrixIts two real eigenvalues, each occurring twice in the full real linearization, are $-({\nu }_2-{\nu }_1)r^2\pm \sqrt{{\delta }^2{r}^4+r^2}$. Define $d={\nu }_2-{\nu }_1$ and $D_r=d^2-{\delta }^2$. Strict attraction modulo translation is therefore equivalent toIndeed the larger transverse eigenvalue is negative precisely when $d{r}^2>\sqrt{{\delta }^2{r}^4+r^2}$, and squaring is legitimate only after requiring $d>0$. If ${\delta }^2>d^2$, that inequality is impossible. Equality ${\delta }^2=d^2$ is also insufficient because of the additional positive term $r^2$. Any nonzero roll that exists with ${\nu }_1<0$ already has a positive radial eigenvalue; the degenerate case ${\nu }_1=0$ has no ordinary isolated nonzero roll branch.

For hexagons put $N={\nu }_1+2{\nu }_2$. A nonzero real amplitude $h$ satisfies $\mu =N{h}^2-h$, or $h=(1\pm \sqrt{1+4N\mu })/(2N)$ when $N\ne 0$. Separate real and imaginary disturbances. The real Jacobian matrix is cyclic, with diagonal $-h-2{\nu }_1{h}^2$ and off-diagonal entries $h-2({\nu }_2\pm \delta )h^2$. Its eigenvalues areThe imaginary Jacobian matrix has every entry equal to $-h$, giving eigenvalues $-3h,0,0$. The two zero modes translate the hexagonal pattern. Thus the general strict stability conditions for any nonzero root $h$ areThese conditions state orbital stability transverse to translations, not asymptotic convergence to one prescribed spatial phase. Equalities require a nonlinear bifurcation calculation.

When ${\nu }_2>{\nu }_1>0$, both $d,N$ are positive. The upper positive hexagon branch is radially stable after its saddle-node bifurcation at $\mu =-1/(4N)$, $h=1/(2N)$; it remains stable until $h=1/d$. At this second threshold the pair ${\lambda }_{\pm }$ crosses the imaginary axis with nonzero frequency if $\delta \ne 0$. The crossing is transverse, since $dh/d\mu =1/(2Nh-1)$ and the derivative of $-2h+2d{h}^2$ with respect to $h$ is two there. Hence, subject to the usual nonzero cubic coefficient, this is a Hopf bifurcation:For $\delta =0$ the same threshold has two zero real eigenvalues, so it is not a Hopf bifurcation.

A possible phase portrait when neither steady branch is stable has an attracting periodic oscillation surrounding the unstable hexagon, with the three amplitudes waxing and waning cyclically. This is realized, for example, by ${\nu }_1=1$, ${\nu }_2=2$, $\delta =2$ just above ${\mu }_H=4$. Rolls are unstable for every positive $\mu$ because ${\delta }^2=4>d^2=1$. At the hexagon Hopf point $h=1$, the real spectrum is $-9,\pm 4i\sqrt{3}$. Evaluation of the cubic Hopf normal form with a unit-norm critical eigenvector gives $\mathrm{Re}{G}_{21}=-64/27<0$, so this example has a supercritical Hopf bifurcation and an attracting small limit cycle in the real invariant subspace. The figure shows such a cycle at $\mu =4.2$ and an approaching trajectory; the positive octant is forward invariant because an amplitude's derivative at zero is the product of the other two. This is a possible portrait, rather than a claim that every parameter set with unstable rolls and hexagons has the same attractor. Stability asserted for the periodic orbit here is within the stipulated real subspace.

Use local coordinates $q=\theta -\pi /2$, $p=\varphi$ at $P_1$. The Jacobian matrix, its trace and its determinant areConsequently $P_1$ is attracting when $T<0,D>0$, repelling when $T>0,D>0$, and a saddle equilibrium when $D<0$. A transition between a node and a focus at $T^2=4D$ changes the linear geometry but is not a bifurcation of the local topological phase portrait.

The stationary bifurcation curve is $\lambda +\kappa =2$. To identify it, write $C=\cos 2\varphi$, $S=\sin 2\varphi$. At a non-equatorial equilibrium point, eliminating $S$ givesNear $(3,-1)$, if $d=2-\lambda -\kappa >0$, these yield $p=\pm \sqrt{d}/2+O(d^{3/2})$ and $q=p+o(\sqrt{d})$. The two branches are related by the local reflection $(q,p)\mapsto (-q,-p)$. Their determinants are $-4d+o(d)$, so they are saddle equilibria. This is a pitchfork bifurcation; on the portion with $T<0$ the saddle branches lie on the side where $P_1$ is stable, giving a subcritical pitchfork bifurcation.

A Hopf bifurcation requires $T=0,D>0$, henceThe oscillation frequency is $\omega =\sqrt{-2(1+\kappa )}$. At the intersection $\boxed{(\lambda ,\kappa )=(3,-1)}$, $J\ne 0$ but $J^2=0$: there is a double zero eigenvalue with one eigenvector. The map from $(\lambda ,\kappa )$ to $(T,D)$ has nonzero Jacobian determinant, so two independent parameter conditions produce this codimension-two bifurcation. Reflection removes quadratic terms; the appropriate initial comparison is with a reflection-symmetric cubic double-zero normal form, subject to the cubic degeneracy checked below.

For completeness, the printed vector field has a subcritical, rather than supercritical, Hopf bifurcation. In the mechanical coordinates $x=p$, $y=\dot{p}$, at $\lambda =3$ its cubic acceleration has $x^{2}y$ coefficient $b=6(\kappa -3)(\kappa +1)/(\kappa -1{)}^2$ and $y^3$ coefficient $d_3=-2/(\kappa -1{)}^2$. For a small harmonic oscillation, the averaged cubic change of $H=(y^2+{\omega }^2{x}^2)/2$ is ${\omega }^2{A}^4(b+3d_3{\omega }^2)/8$, whereThus a small unstable periodic orbit lies on the attracting side of the Hopf bifurcation. This calculation matters when comparing the original equations with the later assumed normal form.

For the supplied normal form, the four local topological phase portraits are as follows. When ${\mu }_2<0$, only the origin is present and it is a saddle equilibrium. When ${\mu }_2>0,{\mu }_1<0$, the origin is attracting, between the two outer saddle equilibria; inward-facing separatrices enter its basin. When ${\mu }_2>0$ and $0<{\mu }_1<h({\mu }_2)$, the origin is repelling and a stable limit cycle surrounds it, inside the separatrix boundary formed by the outer saddle equilibria. Finally, when ${\mu }_2>0,{\mu }_1>h({\mu }_2)$, the origin remains repelling but the local limit cycle has disappeared, and trajectories leave the central region through the opened separatrix channels. Whether the origin is a node or a focus within one of these regions does not change this orbit topology.

On the connection curve, the two outer saddle equilibria are joined in both directions; the period of the disappearing limit cycle diverges. This is a heteroclinic cycle, not a pair of saddle homoclinic orbits. The sketches below show the four open regions of the supplied system; they are not presented as phase portraits of the inequivalent printed spherical equations.

Here the autonomous planar system is $\dot{x}=v$, $\dot{v}=-\cos x$. Its equilibrium points areThe Jacobian matrix isFor even $n$, the eigenvalues are $1,-1$, so these points are unstable saddle equilibria. Writing $\xi =x-x_{\ast }$, their local unstable and stable directions are $v=\xi$ and $v=-\xi$ respectively. For odd $n$, the eigenvalues are $i,-i$.

To classify the latter points nonlinearly, differentiate the first integralAt $x_{\ast }=3\pi /2+2m\pi$ this energy has a strict local minimum. Its nearby level sets are closed curves, proving that the odd-$n$ points are stable center equilibria, with no attraction to the center. The phase portrait consists of closed oscillatory orbits for $-1<H<1$, heteroclinic orbits at $H=1$ joining consecutive saddle equilibria, and open rotating orbits for $H>1$. The separating curves haveAbove the horizontal axis $\dot{x}=v>0$, so flow is rightward; below it flow is leftward. The closed orbits are clockwise, and these same signs orient every separating and rotating orbit.

Introduce the velocity variable $v=\dot{x}$. The second-order differential equation becomes the autonomous planar systemA critical point of a dynamical system is a point where both components of the vector field vanish: here $v=0$ and $f(x,0)=0$. It is a stationary solution in the phase plane. The Jacobian matrix at each such point gives the local linear stability of a planar equilibrium. For the equations below, a conserved energy also determines the nonlinear phase portrait; this is important because purely imaginary eigenvalues alone do not generally prove that an equilibrium is a center equilibrium.

Write $c=(1-a-b)/a$. The admissible equilibrium points are $O=(0,0)$, $E=(1-a,0)$ when $a\le 1$, and $P=(b,c)$ when $a+b\le 1$. The Jacobian matrix isAt $O$ the eigenvalues are $(1-a)/a$ and $-1$; at $E$ they are $-(1-a)/a$ and $(1-a-b)/b$. At a positive coexistence equilibrium point,It is therefore asymptotically stable. It is a node for $b^2>4a(1-a-b)$, a focus for the reverse inequality, and has repeated real eigenvalues at equality. The node–focus change does not alter the attracting equilibrium point's topology.

There are three open physical parameter regions:

The axes are invariant. On $x=0$, $y$ decays exponentially; on $y=0$, prey obey the logistic differential equation. Positivity follows because each right-hand side has its population as a factor. Also $x$ is bounded by logistic comparison, while $W=x+by$ satisfies $\dot{W}=(1-a)x/a-x^2/a-by$, so all forward trajectories are bounded. The Dulac function $1/(xy)$ has divergence $-1/(ay)<0$ in the interior, excluding periodic orbits by the Bendixson-Dulac criterion.

In the extinction region, $\dot{x}\le -(a-1)x/a$, and then $\dot{y}/y=x/b-1$ is eventually negative. In the prey-only region, the upper bound $\mathrm{limsup}x\le 1-a<b$ implies predator decay and convergence of interior prey to $1-a$. In the coexistence region the Lyapunov functionhas $\dot{\mathcal{V}}=-(x-b{)}^2/a\le 0$. Its positive-quadrant sublevel sets are compact, and the only invariant subset with $x=b$ also has $y=c$. The LaSalle invariance principle proves convergence of every interior trajectory to $P$. These facts justify the complete qualitative phase portraits, not just their local arrows.

At $a=1$, $O$ and $E$ meet in a transcritical bifurcation; positive prey decay algebraically. At $a+b=1$ with $a<1$, $E$ and $P$ exchange stability. On its centre manifold, $x=b-ay+O(y^2)$ gives $\dot{y}=-a{y}^2/b+O(y^3)$, so the physically positive side is still attracting. There is no interior Hopf bifurcation since the coexistence trace never vanishes.

The fixed points are $(0,0)$ and $(\pm 1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, so it is a saddle with stable direction the y-axis. At either other fixed point, the characteristic polynomial is ${\lambda }^2+\lambda +2$, giving eigenvalues $(-1\pm i\sqrt{7})/2$. Thus both are stable foci. The nullclines are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.

A global argument determines the requested omega-limit sets, rather than guessing from the local phase portrait. For $x\ne 0$, use the quadratic-logarithmic Lyapunov function for a feedback systemIts sublevel sets are compact and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the omega-limit set. Every point of that set therefore has $y=1$. Invariance then requires $\dot{y}=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp ({\int }_0^t(1-y(s))ds)$ preserves its sign.

ConsequentlyOn $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The phase portrait below shows this stable axis, the nullclines and the spiral attraction in each half-plane.

The logistic equation has the constant solutions $x=0$ and $x=1$. For a nonconstant solution, separation of variables givesSolving and imposing the initial value yieldsThis expression also includes the two equilibria when $x_0=0$ or $1$. For other initial values it is taken on the maximal interval containing $t=0$ on which its denominator does not vanish.

The phase portrait follows from the sign of $x(1-x)$: arrows point left for $x<0$, right for $0<x<1$, and left for $x>1$. Thus zero is an unstable equilibrium point and one is asymptotically stable. The requested sketch is shown below.

If $0<x_0<1$, the denominator is positive for every real $t$. The solution stays between the equilibria and increases monotonically, withMore precisely $x\sim [x_0/(1-x_0)]e^t$ in the remote past and $1-x\sim [(1-x_0)/x_0]e^{-t}$ in the remote future. Outside that interval there is a finite-time pole: for $x_0<0$ it occurs in the future, and for $x_0>1$ in the past, at $t_{\ast }=\log [(x_0-1)/x_0]$. This does not affect the globally defined trajectories requested between zero and one.

Solving the two factors in each equation gives $(0,0),(3,0),(0,3),(1,1)$. The Jacobian isAt the three boundary equilibria its eigenvalues are $3,-3$. These equilibria are hyperbolic fixed points; the Hartman-Grobman theorem gives local topological conjugacy to their saddle linearizations. At $(1,1)$ the matrix is $\left(\begin{array}{cc}1& 2\\ -2& -1\end{array}\right)$, with eigenvalues $\pm i\sqrt{3}$. The linearization is a centre, but the hyperbolic theorem does not apply there.

The nonlinear classification follows from the actual HamiltonianAlong solutions $\dot{H}=H_x{H}_y-H_y{H}_x=0$. At $(1,1)$, $H=-1$ and its Hessian is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, which is positive definite. Nearby regular level curves are closed ovals around this strict minimum. Their vector field is nonzero and tangent, so each is a periodic orbit. The nonlinear equilibrium is a centre.

The separatrix level $H=0$ consists of $x=0$, $y=0$, and $x+y=3$. Within the triangle they connect the saddles as $(3,0)\to (0,0)\to (0,3)\to (3,0)$. The interior levels $-1<H<0$ are nested clockwise periodic orbits; outside the triangle the remaining level branches are unbounded. On $y=0$, flow is left for $0<x<3$ and right for $x<0$ or $x>3$; on $x=0$, it is up for $0<y<3$ and down otherwise. These directions and the invariant levels determine the full phase portrait.

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.

For $c=1$, the even equilibria are stable foci with eigenvalues $(-1\pm i\sqrt{3})/2$, and the odd equilibria are saddle equilibria with eigenvalues $r_{\pm }=(-1\pm \sqrt{5})/2$. The local saddle directions are $y=r_{\pm }[x-(2j+1)\pi ]$.

The phase portrait shows clockwise spiraling into each even equilibrium: on its right-hand horizontal axis, the flow initially points down. The stable separatrices of the saddles divide attraction basins on the unwrapped angle axis, while the unstable branches flow into neighboring sinks. Initial conditions with enough mechanical energy can cross one or more potential crests before being captured. The curves repeat under $x\mapsto x+2\pi$. They are trajectories, not conservative energy contours: energy dissipation of a damped pendulum rules out nonconstant periodic orbits.

Validity and the complete angular equation. The Lagrange planetary equations here describe a planar, circular-planet, small-orbital eccentricity, weakly perturbed exterior mean-motion resonance. Require $M_{\mathrm{pl}}/M_{\star }\ll 1$, $e\ll 1$, one isolated slowly varying resonant argument, perturbation times long compared with an orbital period, and no close planetary encounters. Other resonant harmonics and short-period terms must be negligible; coefficients may be frozen only across a narrow range of semi-major axis. Take $p>0$, $q=1$ or $2$ in a reduced integer ratio. Exactly $e=0$ makes the longitude of periapsis undefined, so the displayed angular variables must then be replaced by nonsingular eccentricity vector components.

Neglecting the perturbation to the accumulated mean longitude gives $\dot{\lambda }\simeq n(a)$; this is an accumulated-phase approximation, not differentiation of a fictitious expression $n(t)t$ while discarding $t\dot{n}$. With $n^2{a}^3=G{M}_{\star }$,Differentiating, using $\dot{n}=-3n\dot{a}/(2a)$ and the supplied orbital element equations, gives the full result within this constant-coefficient model:An equivalent formula replaces the last two terms in brackets by $q^2{C}_r{e}^{q-2}[(p+q)n-p{n}_{\mathrm{pl}}-2q{C}_s]-2q^3{C}_r^2{e}^{2q-4}\cos \varphi$. No derivatives of $C_r,C_s$ are included because they were specified as constants.

A full fixed point requires $\sin \varphi =0$ and $\dot{\varphi }=0$. Hence ${\varphi }_{\ast }=0$ or $\pi$, with its corresponding detuned semi-major axis fixed byAt either point $\dot{a}=\dot{e}=0$. The longitude of periapsis still precesses, so the fixed point refers to the reduced resonant dynamics.

Linear stability analysis and encounter geometry. Linearizing the full equation about a fixed point givesFor $q=2$, $\pi$ is a center equilibrium and zero is a saddle equilibrium. For $q=1$, $\pi$ is always a center equilibrium, while zero is a saddle equilibrium ifAt sufficiently small orbital eccentricity the full truncated equations instead admit a center at zero as well; equality is a degenerate case requiring higher-order analysis. Thus an unconditional instability claim at zero does not follow from the full equation. In the usual fixed-eccentricity weak-resonance regime the inequality holds and the stable libration center is $\pi$ for either order.

The physical explanation is resonance protection. At equal mean longitudes, the conjunction direction relative to periapsis obeys $q(\Lambda -\varpi )=\varphi$ modulo $2\pi$. For $q=1$, $\varphi =0$ places conjunction near periapsis, where the exterior particle comes closest to the planet; $\varphi =\pi$ places it near apoapsis. For $q=2$, zero includes both apsidal conjunctions, one of them near periapsis, whereas $\pi$ places the two conjunction branches near quadrature. The protected arrangements give restoring kicks in the positive-$C_r$ leading-harmonic model. This geometric explanation is conditional on its non-crossing, small-orbital eccentricity approximation; it does not override the low-$e$ term retained in the full stability calculation.

Small resonant-argument librations. Define the nominal resonant semi-major axis and the displaced center byThis follows by expanding $n=n_r[1-3(a-a_r)/(2a_r)]$ and taking $\cos (\pi +\delta )=-1+O({\delta }^2)$. It gives the requested initial semi-major axis to first order in the precession-induced detuning; $e_0=e(0)$ and $\delta =\Delta \varphi$.

For the pendulum approximation of a mean-motion resonance, also require fractional changes in $e$ small across a libration, and precession terms small compared with $n$. In particular $C_r/(n_r{e}_0^3)\ll 1$ for $q=1$ makes the extra full-equation curvature negligible; for $q=2$, small fractional eccentricity changes require $C_r/(n_r{e}_0^2)\ll 1$. Freeze $e=e_0$ in the leading restoring coefficient and setThen $u=\varphi -\pi$ obeys simple harmonic motion. Using $u(0)=\delta$, $\dot{u}(0)=0$ and integrating the remaining Lagrange planetary equations givesThe omitted apsidal modulation comes from higher perturbative orders and $u^2$. All expressions are leading resonant approximations, valid while the omitted fractional changes remain small. The small-libration phase portrait iswith period $2\pi /{\omega }_0$. The semi-major axis half-width is $2a_r{\omega }_0|\delta |/(3p{n}_{\mathrm{pl}})$ and the orbital eccentricity half-width is $q{C}_r{e}_0^{q-1}|\delta |/{\omega }_0$.

Dividing the original $\dot{e}$ and $\dot{a}$ equations gives the exact invariant of their retained terms,Thus the $e$–$a$ plot is a segment of this increasing curve traversed back and forth, not a closed ellipse. Locally its slope is $q/[2(p+q)e_0{a}_c]$.

Finite amplitude and the separatrix. Keep the same weak-resonance approximation but allow $\delta$ to be finite. The leading angular equation is $\ddot{\varphi }={\omega }_0^2\sin \varphi$, or $\ddot{u}=-{\omega }_0^2\sin u$. Multiplication by $\dot{u}$ givesThis resonant pendulum energy is constant in the pendulum approximation of a mean-motion resonance. It is not an exact integral of the earlier full equation when $e$ and its restoring coefficient vary: the $\dot{\varphi }\sin \varphi$ term alone prevents that conclusion in general.

For $|\delta |<\pi$ the phase portrait has closed libration curves around $\pi$, withwhere $K$ is the complete elliptic integral of the first kind. The largest speed is $2{\omega }_0\sin (|\delta |/2)$ and the leading semi-major axis half-width is $4a_r{\omega }_0\sin (|\delta |/2)/(3p{n}_{\mathrm{pl}})$. As $|\delta |\to \pi$, the energy approaches $2{\omega }_0^2$, the separatrix through the saddles at zero and $2\pi$; the particle spends increasingly long intervals near those saddles. With $ϵ=\pi -|\delta |$,At exactly $\delta =\pi$, $\dot{\varphi }=0$, the particle stays at the unstable fixed point in the ideal model. Nonstationary separatrix trajectories approach the saddle only in infinite time; above separatrix energy, the resonant argument circulates and resonance protection is lost. These distinctions matter when describing the limiting motion.

Use $y$ for height above the horizontal wall. In the lubrication approximation, vertical momentum is hydrostatic and the interfacial stress balance with variable surface tension givesThe horizontal equation is $\mu u_{yy}=B$, with $u(0)=0$ and $\mu u_y(h)={\gamma }_x=-A{C}_x$. ThusIntegrating across the film and adding surface diffusion to surfactant advection gives the dimensional thin-film mass flux and insoluble surfactant flux:These include Marangoni stress, hydrostatic leveling, capillarity and surface diffusion with their signs fixed by the surface traction. In particular the derivative acts on the product of surface tension and curvature, not just on curvature.

Choose $h_{\ast }=(A{C}_0/(\rho g){)}^{1/2}$ and $u_{\ast }=A{C}_0{h}_{\ast }/(\mu L)$. The dimensionless definitions areTogether with the stated concentration and horizontal scales, these giveIn steady flow $Q,J$ are constants by liquid and surfactant conservation.