Part B · the action · branch

Integrability and chaos: determinism without predictability

Solutions exist, are unique, and depend continuously on the data — and are still useless for prediction after a few dozen orbits. Both halves are theorems, and they do not conflict.

needs 14, 02, 01~2½ h slides + 5 h problemsproblems: ./mr problems 17

Unit 01 proved that a Lipschitz force determines the future completely, and threw in continuous dependence on the initial data for free. It is tempting to read that as a promise about prediction. It is not, and this unit says exactly why: the continuous-dependence bound is $\lVert\delta(t)\rVert\le\lVert\delta(0)\rVert e^{Lt}$, and there is no $t$ in front of the $\varepsilon$–$\delta$ quantifier. Continuity in the initial data is not uniform in $t$, and that single word is the whole gap between determinism and predictability.

The unit then does the positive work. It says what the rare solvable systems have in common (Liouville–Arnold: $n$ integrals in involution, and the bounded motion lies on tori), it builds the two instruments that decide which case you are in for a system you cannot solve — the Poincaré section and the Lyapunov exponent — and it states what survives a small perturbation of an integrable system, which is the KAM theorem.

This is a branch unit: nothing later in the course needs it. It is here because a course that goes from Newton to Einstein without saying that almost nothing is solvable leaves the learner with a badly wrong picture of what mechanics is.

How to read this deck

Five sections. Section 1 defines integrability and states Liouville–Arnold. Section 2 builds the Poincaré section and proves its return map preserves area. Section 3 defines the Lyapunov exponent and measures one. Section 4 perturbs an integrable system and states KAM, Poincaré–Birkhoff and Chirikov's criterion. Section 5 spends the machinery on the asteroid belt and the solar system.

Proofs are on the slides where they are short and in these notes under Full proof where they are not. What this unit does not prove is marked cited, with the place it is proved. Terms with a dotted underline are glossary entries — click one, or press G. N opens these notes; R stacks the deck into one page.

17 · overview

Determinism was the easy part

the claim to be dismantled

"The equations determine the future, so the future is predictable." The first clause is unit 01's theorem. The second does not follow, and this unit exhibits a system where it is false — with a number attached to how false.

the two instruments

1 The Poincaré section: cut the flow with a plane and look at where it lands. An integrable system draws curves; a chaotic one fills an area. 2 The Lyapunov exponent: the asymptotic rate at which neighbouring orbits separate. Positive means chaos, and it is a number you can quote.

the running example

The Hénon–Heiles system, $H = \tfrac12(p_x^2+p_y^2) + \tfrac12(x^2+y^2) + x^2y - \tfrac13y^3$: two degrees of freedom, one known integral, and a free parameter $E$. At $E = 1/100$ its section is a family of smooth closed curves; by $E = 1/8$ half of it is a chaotic sea; at $E = 1/6$ only one pair of islands is left. We shall measure that, not admire it.

getting around

A dotted underline is a glossary term: click it, or press G. N opens the notes, where the longer proofs live. A solid underline, like the running example, shows it on hover.

Why Hénon–Heiles and not the three-body problem. The three-body problem is where the subject started and section 1 tells that story, but it has nine degrees of freedom before reduction and nothing about it can be drawn. Hénon–Heiles has two, which is the smallest number for which a bounded Hamiltonian flow can be chaotic, and two is exactly the number for which the Poincaré section is a two-dimensional picture on a page. Michel Hénon and Carl Heiles wrote it down in 1964 (Astronomical Journal 69, 73) for a real question: stars in an axially symmetric galaxy have two obvious integrals, the energy and the angular momentum about the axis, and the observed velocity distributions looked as though there had to be a third. Their cubic potential is not any particular galaxy — it is the simplest truncation with the right symmetry — and they ran it to see whether a third integral existed. The answer turned out to depend on the energy, which nobody expected.

Dimensions. Hénon–Heiles as written is dimensionless. Restoring units, the Hamiltonian is $\tfrac1{2m}(p_x^2+p_y^2) + \tfrac12 m\omega^2(x^2+y^2) + \frac{m\omega^2}{a}(x^2y - \tfrac13y^3)$; measuring lengths in $a$, times in $\omega^{-1}$ and energies in $m\omega^2a^2$ returns the form above. So the numbers in this deck read as follows: $E = 1/8$ means an energy of $m\omega^2a^2/8$, and a Lyapunov exponent of $0.044$ means $0.044\,\omega$, an inverse time. Every result in the unit will be checked against that dictionary.

The shape of the argument. Sections 1 and 2 are about what is true of a system exactly; sections 3 and 4 are about what survives when it is perturbed. The hinge between them is that the tori of section 1 are not all equally fragile: which ones survive is decided by a number-theoretic property of their frequencies, which is the strangest fact in classical mechanics and is proved, not asserted, in the sources section 4 cites.

section 1

What it means to solve a system

Integrals in involution, and the theorem that turns $n$ of them into a torus.

17 · integrability

A prize that cost more than it paid

In 1885 King Oscar II of Sweden offered a prize for the stability of the solar system. Poincaré entered with a memoir on the restricted three-body problem and won: 2500 kronor and a gold medal.

While setting the memoir for Acta Mathematica, the editor Edvard Phragmén asked about a step. Poincaré had assumed that the stable and unstable curves emerging from an unstable periodic orbit, when they met again, coincided. They do not: they cross, and having crossed once they must cross infinitely often. His conclusion of stability was wrong, and the corrected memoir — 270 pages against 158 — says the opposite.

Poincaré paid 3585 kronor to recall and destroy the printed issue. The error cost him more than the prize, and it is the origin of this subject. (cited: Barrow-Green, Poincaré and the Three Body Problem, 1997, ch. 3–5.)

what was already known

Bruns, 1887: the three-body problem has no integral algebraic in the Cartesian positions and momenta beyond the classical ten. Poincaré, 1890: the restricted problem has no integral, other than $H$, depending analytically on the mass ratio near zero. Both cited.

the lesson to carry

"No formula" is not a confession of ignorance. For most systems it is a theorem. The question then changes from what is the solution to what is true of the solutions, which is unit 02's question asked one dimension up.

The competition, precisely. Four questions were set; Poincaré answered a restricted version of the first. Weierstrass, Mittag-Leffler and Hermite judged. The prize was announced in January 1889 on Oscar's sixtieth birthday. Phragmén's query reached Poincaré in July 1889; by December Poincaré had written to Mittag-Leffler admitting the error. Volume 13 of Acta containing the flawed memoir had already been distributed to subscribers, and Mittag-Leffler spent some months quietly retrieving copies. A handful survive.

What the error became. The crossing Poincaré was forced to admit is a homoclinic point, and section 4 draws the tangle it forces. His own sentence about it is the most quoted in the subject: "one is struck by the complexity of this figure, which I shall not even attempt to draw". He is not being lazy — the figure cannot be drawn, and section 2's numerical section is the modern substitute.

Bruns and Poincaré, on what they exclude. Bruns' theorem is a statement about algebraic integrals in one particular coordinate system, and is therefore weaker than it sounds: it does not forbid an integral that is analytic but not algebraic. Poincaré's is the sharper result — it forbids analytic integrals of the perturbed system that reduce smoothly to integrals of the unperturbed one, and its proof is exactly the small-denominator computation of section 4. Neither theorem says the three-body problem is chaotic; they say the classical route to solving it is closed.

17 · integrability

Integrals in involution

conserved, and commuting

$F$ is a first integral of $H$ if $\{F,H\} = 0$, which by unit 14 is exactly $\dot F = 0$ along every orbit. Two functions are in involution if $\{F,G\} = 0$.

why involution and not merely independence

Each $F_i$ generates its own flow. $\{F_i,F_j\} = 0$ says those flows commute, so together they define an action of $\mathbb R^n$ on the level set — and it is that action, not the counting of integrals, that forces the geometry.

Independence here means: the differentials $\mathrm dF_1,\dots,\mathrm dF_n$ are linearly independent at each point of the level set, so the level set is an $n$-dimensional surface (unit 04's implicit function theorem).

Kepler, counted

Unit 14 found seven integrals for $H = p^2/2\mu - k/r$: $H$, three components of $\vec L$, three of $\vec A$ — five of them independent. But $n = 3$, so at most three can be in involution. Unit 14's brackets show $\{L_i,L_j\} = \epsilon_{ijk}L_k \ne 0$: the three components of $\vec L$ are not in involution with each other.

A commuting triple is $H$, $L_3$, $\lVert\vec L\rVert^2$. Check: $\{H,\cdot\} = 0$ as both are conserved; $\{L_3, \lVert\vec L\rVert^2\} = \sum_i 2L_i\{L_3,L_i\} = 2\sum_i L_i\,\epsilon_{3ik}L_k = 0$, since $\epsilon_{3ik}$ is antisymmetric in $i,k$ while $L_iL_k$ is symmetric.

seven is not better than three

More integrals than $n$ makes a system superintegrable: the orbits close, and lie on tori of dimension below $n$. Kepler's ellipses are one-dimensional, not three.

Full proof that the bracket of two integrals is an integral. If $\{F,H\} = \{G,H\} = 0$ then the Jacobi identity of unit 14, $\{\{F,G\},H\} + \{\{G,H\},F\} + \{\{H,F\},G\} = 0$, has its last two terms zero, so $\{\{F,G\},H\} = 0$. The conserved quantities form a Lie algebra under the bracket — that was unit 14's statement, and the reason involution is the right hypothesis is that a set of integrals in involution is an abelian subalgebra of it.

Why commuting flows matter. Write $\varphi^i_{t}$ for the flow of $F_i$. Unit 14's bracket identity gives $\frac{\mathrm d}{\mathrm dt}(F_j\circ\varphi^i_t) = \{F_j, F_i\} = 0$, so each flow preserves every $F_j$ and therefore maps a common level set to itself.

Full proof that the flows commute. Let $\{F,G\} = 0$, fix $t$, and write $\psi = \varphi^G_t$, $M = D\psi$. Unit 14 proved $M^{\!\top}JM = J$; transposing and inverting gives $MJM^{\!\top} = J$ as well. The chain rule applied to $F = (F\circ\psi^{-1})\circ\psi$ gives $\nabla F(z) = M^{\!\top}\nabla(F\circ\psi^{-1})(\psi z)$, so $M X_F(z) = MJM^{\!\top}\nabla(F\circ\psi^{-1})(\psi z) = X_{F\circ\psi^{-1}}(\psi z)$. But $F\circ\psi^{-1} = F\circ\varphi^G_{-t} = F$, because $F$ is constant along $G$'s flow. So the curve $s\mapsto\psi(\varphi^F_s z)$ has velocity $M X_F(\varphi^F_s z) = X_F(\psi\varphi^F_s z)$: it solves $\dot z = X_F$ and starts at $\psi z$. By the uniqueness of unit 01 it is $s\mapsto\varphi^F_s(\psi z)$, that is, $\varphi^G_t\circ\varphi^F_s = \varphi^F_s\circ\varphi^G_t$. $\square$

The upshot is a map $\mathbb R^n\times M_c \to M_c$, $(t_1,\dots,t_n)\cdot z = \varphi^1_{t_1}\cdots\varphi^n_{t_n}(z)$, which is the object the next theorem is really about.

The bound "at most $n$ in involution". At a point where $k$ independent integrals are in involution, their differentials span a $k$-dimensional subspace on which the symplectic form vanishes; such a subspace has dimension at most $n$ in a $2n$-dimensional symplectic vector space. The linear-algebra fact — an isotropic subspace has dimension at most half — is unit 14's $M^{\!\top}JM = J$ machinery in disguise, and is problem P3.

17 · integrability

Complete integrability, defined

completely integrable

A Hamiltonian system with $n$ degrees of freedom is completely integrable on an open set $U\subset\mathbb R^{2n}$ if there are $n$ functions $F_1 = H, F_2,\dots,F_n\in C^\infty(U)$ with $$\{F_i,F_j\} = 0 \quad\text{for all } i,j, \qquad \mathrm dF_1,\dots,\mathrm dF_n \ \text{linearly independent at every point of } U.$$

integrable

Any $n = 1$ system: $H$ alone. Central forces (unit 08): $H, L_3, \lVert\vec L\rVert^2$. $n$ uncoupled oscillators: each $E_i$. The Toda lattice. The free rigid body (unit 16).

not, provably

The restricted three-body problem (Poincaré). The double pendulum. The Sitnikov problem. Proving non-integrability is hard; the modern tool is Morales-Ramis theory, cited.

not, apparently

Hénon–Heiles. No second integral is known and none is believed to exist for general $E$, but the numerical evidence of section 2 is evidence, not proof.

the honest statement of the definition's weakness

"Integrable" is not a property of a system alone but of a system and an open set. Every Hamiltonian system is locally integrable away from equilibria — for $n = 2$ the notes prove it, for general $n$ it is cited. The content is always global: integrals defined on a large region, not on a flow box.

Why $n = 1$ is always integrable. One degree of freedom, one integral $H$, and the orbits are the level curves of $H$ in the plane — which is unit 02's phase portrait and unit 00's reduction to quadrature. Everything in this unit is therefore about $n\ge2$; two degrees of freedom is where the subject begins.

Local integrability, and why it is empty. Near a point $z_0$ where $X$ does not vanish, take a small piece $S$ of hyperplane through $z_0$ transverse to $X$, with coordinates $w\in\mathbb R^{2n-1}$, and map $(s,w)\mapsto\varphi_s(w)$. Its derivative at $(0,z_0)$ has columns $X(z_0)$ and a basis of $S$, independent by transversality, so unit 04's inverse function theorem makes it a local diffeomorphism (the flow is $C^1$ in its data — cited in unit 14). In the new coordinates the motion is $\dot s = 1$, $\dot w = 0$, so $w_1,\dots,w_{2n-1}$ are local integrals. For $n = 2$ that already gives local integrability: pick one $w_i$ whose differential is independent of $\mathrm dH$ — some must be, since the $\mathrm dw_i$ span everything orthogonal to $\mathrm ds$ — and it is automatically in involution with $H$, being an integral. For general $n$ one needs $n-1$ local integrals in involution with each other too; that they exist is the Carathéodory–Jacobi–Lie theorem, cited: Libermann and Marle, Symplectic Geometry and Analytical Mechanics, 1987. Either way this is true of every system without exception, chaotic ones included, and it shows the definition has no content until the domain is fixed. The chaos of section 2 lives in how these local boxes fail to patch together over a large region.

Morales-Ramis, in one sentence. Linearise along one particular known solution, and ask about the differential Galois group of the resulting linear equation; if that group's identity component is not abelian, the system has no meromorphic integrals. This has been used to prove Hénon–Heiles non-integrable for all but a few exceptional coefficient ratios — ours is not one of them. Cited: Morales-Ruiz, Differential Galois Theory and Non-Integrability of Hamiltonian Systems, 1999. It uses complex analysis and Galois theory well past this course.

The double pendulum. Two degrees of freedom, no second integral, and its chaos is visible on a kitchen table. It is the right thing to picture throughout this unit; Hénon–Heiles is used instead only because its section is easier to compute accurately and its energy is a single clean knob.

17 · integrability

Liouville and Arnold: the level set is a torus

Liouville–Arnold theorem

Let $F_1 = H,\dots,F_n$ be in involution with independent differentials on $M_c = \{F_1 = c_1,\dots,F_n = c_n\}$. Then:

1 $M_c$ is a smooth $n$-dimensional surface invariant under all $n$ flows. 2 If $M_c$ is compact and connected, it is diffeomorphic to the $n$-torus $\mathbb T^n = (\mathbb R/2\pi\mathbb Z)^n$ — an invariant torus. 3 There are coordinates $(\theta_1,\dots,\theta_n)$ on it in which every one of the $n$ flows is $\dot{\vec\theta} = $ constant. In particular $H$'s flow is $\theta_i(t) = \theta_i(0) + \omega_i t$. 4 In a neighbourhood there are canonical coordinates $(\vec I,\vec\theta)$, the action–angle variables, with $H = H(\vec I)$ alone and $\omega_i = \partial H/\partial I_i$.

dimensions

$I_i$ has the dimensions of $\oint p\,\mathrm dq$ — energy $\times$ time, an action — and $\theta_i$ is an angle, dimensionless. So $\omega_i = \partial H/\partial I_i$ is energy over action $=$ 1/time ✓, as a frequency must be.

the hypotheses do work

Compactness is what makes it a torus rather than a cylinder or a plane: drop it and, provided the flows are complete, $\mathbb T^k\times\mathbb R^{n-k}$ is the answer. Connectedness picks one component — a pendulum's level set at fixed energy below the separatrix has two, one per direction of swing.

proof: structure on "Why a torus, and nothing else", full proof cited — Arnold, Mathematical Methods, §49

Attribution. Liouville proved in 1855 that $n$ integrals in involution reduce the system to quadratures — an analytic statement with no geometry in it. The topological half, that the compact level set is a torus and the motion on it is linear, is Arnold's, from 1963. The modern name runs them together.

Why this is the right generalisation of unit 00. Unit 00 reduced one-dimensional motion to quadrature and found the period as an integral. Liouville–Arnold says: $n$ integrals, $n$ quadratures, $n$ periods. The actions are exactly the $n$ independent versions of unit 00's $\oint p\,\mathrm dq$, one per independent loop on the torus, and the frequencies are the derivatives of the energy with respect to them — which for $n = 1$ is the statement $T = 2\pi\,\mathrm dI/\mathrm dE$ that unit 14 proved for Halley's comet.

17 · integrability

The torus, and the straight-line flow on it

For $n = 2$: in the angles of item 3 the motion is $\theta_i(t) = \theta_i(0) + \omega_i t$. Cut the torus open along two circles and it becomes a square whose opposite edges are glued; the orbit becomes straight lines of slope $\omega_2/\omega_1$, re-entering on the opposite edge.

Left: each angle runs round one circle of the torus; cut along both and it opens into a square whose opposite edges are glued. Middle: $\omega_2:\omega_1 = 3:2$, a resonant torus — $2\omega_2 - 3\omega_1 = 0$ — on which every orbit closes after two circuits one way and three the other. Right: $\omega_2/\omega_1 = (1+\sqrt5)/2$, non-resonant, drawn for nine circuits; the lines never close, and carried on for ever they come arbitrarily close to every point.

What "the flow is linear" buys. On a torus, a straight-line flow is completely understood: if the frequency vector $\vec\omega$ is non-resonant — meaning $\vec k\cdot\vec\omega \ne 0$ for every nonzero integer vector $\vec k$ — every orbit is dense in the torus and equidistributed on it. If it is resonant, the orbit closes up or fills a subtorus. So for an integrable system the qualitative question "what does an orbit do" is answered completely by an arithmetic question about $n$ real numbers. Section 4 is about what happens to that clean dichotomy when the system is perturbed, and the answer is that the arithmetic survives, which is the last thing anyone expected.

Reading the figure. The square is the torus with coordinates $(\theta_1,\theta_2)\in[0,2\pi)^2$: the arrows say which edges are the same circle, single arrows the bottom and top, double arrows the left and right. A line leaving through the top re-enters at the bottom at the same $\theta_1$. For the $3:2$ orbit, follow the middle square from any point: after $\theta_1$ has gone round twice, $\theta_2$ has gone round exactly three times, and the line meets its start. For the golden ratio no number of circuits brings it back, because $\omega_2/\omega_1$ is irrational: returning would need $m\,\omega_2/\omega_1 = k$ for integers $m,k$. On the torus on the left, the blue circle is $\theta_2 = $ const and the teal one, partly hidden behind the tube, is $\theta_1 = $ const.

Why dense, in one line. The points where the orbit crosses the circle $\theta_1 = 0$ are $\theta_2 = \theta_2(0) + 2\pi m\,\omega_2/\omega_1 \bmod 2\pi$, $m = 1,2,\dots$: a rotation of the circle by an irrational fraction of a turn. Two of the first $N+1$ such points lie within $2\pi/N$ of each other (pigeonhole), so some power of the rotation moves every point by less than $2\pi/N$, and its iterates then visit every arc of length $2\pi/N$. Hence the crossings are dense in the circle and the orbit is dense in the torus.

17 · integrability

Why a torus, and nothing else

the geometry, in four steps

1 On $M_c$ the $n$ vector fields $X_{F_1},\dots,X_{F_n}$ are tangent, independent, and commute. So the map $$g:\mathbb R^n\times M_c\to M_c,\qquad g(\vec t, z) = \varphi^1_{t_1}\circ\cdots\circ\varphi^n_{t_n}(z)$$ is a well-defined action of the group $\mathbb R^n$ — defined for all $\vec t$ because $M_c$ is compact, so the flows last for ever, and an action because the flows commute.

2 Fix $z_0$. The orbit map $\vec t\mapsto g(\vec t,z_0)$ has independent partial derivatives, so it is a local diffeomorphism; $M_c$ connected makes it onto.

3 Its stabiliser $\Gamma = \{\vec t : g(\vec t, z_0) = z_0\}$ is a subgroup of $\mathbb R^n$, and by step 2 it is discrete. A discrete subgroup of $\mathbb R^n$ is a lattice $\mathbb Z\vec e_1\oplus\cdots\oplus\mathbb Z\vec e_k$ — this is the one substantive lemma, and it is pure group theory.

4 Hence $M_c\cong\mathbb R^n/\Gamma \cong \mathbb T^k\times\mathbb R^{n-k}$. Compact forces $k = n$: a torus.

where each hypothesis entered

involution → step 1, the action exists.
independence → step 2, no degeneracy.
connected → step 2, onto.
compact → step 1, complete flows; step 4, $k = n$.

Every hypothesis is used, and compactness twice. That is how to remember the theorem.

Full proof of step 3's lemma. Let $\Gamma\le\mathbb R^n$ be a discrete subgroup. Induct on $n$. If $\Gamma = \{0\}$ we are done. Otherwise pick $\vec e_1\in\Gamma\setminus\{0\}$ of minimal norm — a minimum exists because $\Gamma$ is discrete and closed, so $\Gamma\cap\overline{B(0,R)}$ is finite. Then $\Gamma\cap\mathbb R\vec e_1 = \mathbb Z\vec e_1$: if some $\vec t = s\vec e_1$ had $s\notin\mathbb Z$, then $\vec t - \lfloor s\rfloor\vec e_1$ is a nonzero element of $\Gamma$ of norm $\lt\lVert\vec e_1\rVert$, contradicting minimality. Now project $\Gamma$ along $\mathbb R\vec e_1$ onto its orthogonal complement $\cong\mathbb R^{n-1}$, call it $\pi$. The image is again discrete, and this is the only fiddly point: given $\vec t\in\Gamma$ with $\pi\vec t\ne0$, subtract the integer multiple of $\vec e_1$ that leaves $\vec t' = s\vec e_1 + \pi\vec t$ with $|s|\le\tfrac12$. Then $\vec t'\in\Gamma\setminus\{0\}$, so minimality gives $\lVert\vec e_1\rVert^2\le\lVert\vec t'\rVert^2 \le \tfrac14\lVert\vec e_1\rVert^2 + \lVert\pi\vec t\rVert^2$, that is, $\lVert\pi\vec t\rVert\ge\tfrac{\sqrt3}2\lVert\vec e_1\rVert$. A subgroup whose nonzero elements are bounded away from $0$ is discrete, so the induction hypothesis applies to $\pi\Gamma$. Since $\ker\pi\cap\Gamma = \mathbb Z\vec e_1$, choosing preimages of the basis of $\pi\Gamma$ and adding $\vec e_1$ gives a basis of $\Gamma$.

An example you can do by hand. The two-dimensional isotropic oscillator $H = \tfrac12(p_x^2+p_y^2) + \tfrac12(x^2+y^2)$ has $F_1 = H$ and $F_2 = \tfrac12(p_y^2 + y^2)$ in involution; the level set is $\{p_x^2+x^2 = 2(c_1-c_2)\}\times\{p_y^2+y^2 = 2c_2\}$, a product of two circles — visibly a 2-torus, with no theorem required. The actions are $I_1 = c_1 - c_2$ and $I_2 = c_2$, each a circle's enclosed area over $2\pi$, and $H = I_1 + I_2$ gives $\omega_1 = \omega_2 = 1$. Problem P2 runs the same computation for the anisotropic case.

17 · integrability

The lattice of return times

Steps 2–4 for $n = 2$. Every time vector in $\Gamma$ brings $z_0$ back to itself, so the orbit map sends the whole plane onto $M_c$ and is periodic with the periods $\vec e_1,\vec e_2$. One shaded cell, its opposite edges glued, is the torus: the glued square of the torus slide, skewed.

$\Gamma$ is the set of return times: run the flow of $F_1$ for time $t_1$ and that of $F_2$ for $t_2$, and $\vec t = (t_1,t_2)\in\Gamma$ exactly when you are back at $z_0$.

It is closed under addition and negation, and isolated (step 2), so it is a lattice $\mathbb Z\vec e_1\oplus\mathbb Z\vec e_2$; compactness rules out $k < 2$.

what is still missing

Step 4 gives the angles. Getting the actions, and the fact that $(\vec I,\vec\theta)$ is canonical, needs one more argument — that the torus is Lagrangian, so $\oint\vec p\cdot\mathrm d\vec q$ around a loop depends only on the loop's homotopy class. That is the part this unit cites.

Reading the figure. The left panel is not phase space: it is the plane of time vectors $\vec t = (t_1,t_2)$, one time for each of the two commuting flows. The map $\vec t\mapsto g(\vec t,z_0)$ wraps this plane round the level set. Two time vectors land on the same point exactly when they differ by an element of $\Gamma$, so the level set is the plane with points identified modulo $\Gamma$ — the parallelogram with opposite edges glued. The angles of the theorem are the coordinates in the basis $\vec e_1,\vec e_2$, rescaled so that each runs over $[0,2\pi)$.

What compactness rules out. If $\Gamma$ were $\mathbb Z\vec e_1$ alone ($k = 1$), the quotient would be a circle times a line — a cylinder, which is not compact. If $\Gamma$ were $\{0\}$, the orbit map would be one-to-one and $M_c$ a copy of the plane. Neither can be a compact level set, so $k = 2$.

Where the actions come from. On the torus, choose $n$ loops $\gamma_1,\dots,\gamma_n$ generating the lattice, and set $I_i = \frac1{2\pi}\oint_{\gamma_i}\vec p\cdot\mathrm d\vec q$. That this does not depend on which representative loop is chosen is the statement that the integrand's circulation vanishes on any loop that can be contracted — unit 04's criterion, applied on the torus rather than in $\mathbb R^3$, and true because the torus is a Lagrangian submanifold, on which the form $\sum\mathrm dp_i\wedge\mathrm dq_i$ restricts to zero. The vocabulary for that sentence arrives in unit 34; the fact is cited here from Arnold §49–50.

17 · integrability

Hénon and Heiles ask for a third integral

worked example 1 · the low-energy limit is integrable

Scale $H$ by $z\mapsto\epsilon z$ on all four variables and divide by $\epsilon^2$: $$\tilde H = \tfrac12(\pi_\xi^2+\pi_\eta^2) + \tfrac12(\xi^2+\eta^2) + \epsilon\left(\xi^2\eta - \tfrac13\eta^3\right),$$ with $E = \epsilon^2\tilde E$. Small energy is small perturbation — this is the whole reason the energy is the knob.

At $\epsilon = 0$, the isotropic oscillator: $F_1 = \tilde H$ and $F_2 = \tfrac12(\pi_\eta^2+\eta^2)$ are in involution, because $F_1-F_2$ involves only $(\xi,\pi_\xi)$ and $F_2$ only $(\eta,\pi_\eta)$. Each level set is a product of two circles: a 2-torus.

Frequencies. $I_1 = F_1-F_2$, $I_2 = F_2$, $\tilde H = I_1+I_2$, so $\omega_1 = \omega_2 = 1$. That is resonant — an integer combination $k_1\omega_1 + k_2\omega_2$ with $(k_1,k_2)\ne0$ vanishes, here $\omega_1-\omega_2$, so every orbit closes — and as resonant as a vector can be, which section 4 calls the worst possible starting point.

the question of 1964

Does a second integral $F_2$, independent of $H$, exist for $\epsilon\ne0$? Equivalently: do the tori survive?

what the deck will answer

Yes at small $E$, no at large $E$, partly at $E = 1/8$ — so no global $F_2$ exists, since one would force curves everywhere.

The potential that does the confining is drawn on the next slide.

The scaling argument, in full. Under $z\mapsto\epsilon z$ on all four variables, the quadratic part of $H$ scales by $\epsilon^2$ and the cubic by $\epsilon^3$. Dividing $H$ by $\epsilon^2$ leaves the quadratic part alone and multiplies the cubic by $\epsilon$. The equations of motion keep their form with no change of time: if $z = \epsilon w$ then $\dot w = \epsilon^{-1}J\nabla_zH(\epsilon w) = \epsilon^{-2}J\nabla_w\big[H(\epsilon w)\big] = J\nabla_w\tilde H(w)$, by the chain rule, since $\nabla_w[H(\epsilon w)] = \epsilon\,\nabla_zH$. The conclusion to hold onto: for Hénon–Heiles, energy is the perturbation parameter, and $\epsilon = \sqrt{E}$ up to constants. That $E = 1/8$ is already thoroughly chaotic while $E = 1/12$ is not tells you the transition is sharp in a way $\epsilon$ does not suggest.

17 · integrability · the running example

The Hénon–Heiles potential, drawn

Level curves of $V$ at the four energies of section 2: circles near the minimum (the oscillator), rounded triangles by $1/8$, and at $1/6$ a triangle through the three saddles (×). Above $1/6$ the curves open (dashed, $V = 0.21$): three ways out.

bound motion

$V$ has a minimum $0$ at the origin and three saddles at $V = 1/6$, at $(0,1)$ and $(\pm\sqrt3/2,-\tfrac12)$. Since $\tfrac12\lVert\vec p\rVert^2\ge0$, an orbit of energy $E$ stays where $V\le E$: confined for $0<E<1/6$; above, it can escape.

why a triangle

$V - \tfrac16 = -\tfrac13\,\big(y+\tfrac12\big)\big(y - 1 - \sqrt3x\big)\big(y - 1 + \sqrt3x\big)$, a product of three linear factors. Each vanishes on a line through two of the saddles, and the threefold symmetry of the cubic term is the symmetry of the triangle.

The saddles, computed. $V = \tfrac12(x^2+y^2)+x^2y-\tfrac13y^3$ gives $V_x = x(1+2y)$, $V_y = y + x^2 - y^2$. From $V_x = 0$: either $x = 0$ or $y = -\tfrac12$. With $x = 0$: $y - y^2 = 0$, so $y = 0$ (the minimum) or $y = 1$, where $V = \tfrac12 - \tfrac13 = \tfrac16$. With $y = -\tfrac12$: $-\tfrac12 + x^2 - \tfrac14 = 0$, so $x = \pm\sqrt3/2$, and $V = \tfrac12(\tfrac34+\tfrac14) - \tfrac34\cdot\tfrac12 + \tfrac13\cdot\tfrac18 = \tfrac12 - \tfrac38 + \tfrac1{24} = \tfrac{12-9+1}{24} = \tfrac16$ ✓. The three saddles sit at the corners of an equilateral triangle, which is the threefold symmetry of the cubic term.

The factorisation, checked. The last two factors multiply to $(y-1)^2 - 3x^2$. Then $(y+\tfrac12)\big(y^2 - 2y + 1 - 3x^2\big) = y^3 - \tfrac32y^2 - 3x^2y - \tfrac32x^2 + \tfrac12$, and multiplying by $-\tfrac13$ gives $-\tfrac13y^3 + \tfrac12y^2 + x^2y + \tfrac12x^2 - \tfrac16$, which is $V - \tfrac16$ ✓. So the level set $V = 1/6$ is exactly the three lines $y = -\tfrac12$ and $y = 1\pm\sqrt3x$, and the region $V < 1/6$ containing the origin is the open triangle they bound. Each line is at distance $\tfrac12$ from the origin — for $y = 1\pm\sqrt3x$ it is $1/\sqrt{1+3}$ — and the corners at distance $1$; restoring units, $a/2$ and $a$, with $a$ the length scale of the overview's notes.

Why $1/6$ is the interesting scale. Below it the level set of $V$ is a closed curve and the motion is bounded for ever; at exactly $1/6$ the three saddles join up into a triangle; above it, orbits leak out along the three channels. So the whole story of this unit has to be told in the window $0 < E < 1/6$, and the four energies of section 2 — $1/100$, $1/12$, $1/8$, $1/6$ — are $6\%$, $50\%$, $75\%$ and $100\%$ of the way up it.

section 2

Looking at a flow through a plane

A four-dimensional flow, cut twice, becomes a map of a page to itself.

17 · sections

The section and its return map

Poincaré section

Let $\dot z = X(z)$ be a flow and $\Sigma$ a hypersurface with $X\not\parallel\Sigma$ at every point of it — $\Sigma$ is transverse. For $z\in\Sigma$ let $\tau(z)$ be the first time $t>0$ with $\varphi_t(z)\in\Sigma$, if it exists. The return map is $$P:\Sigma\to\Sigma,\qquad P(z) = \varphi_{\tau(z)}(z).$$

$\tau$ and $P$ are smooth

Write $\Sigma = \{g = 0\}$. Then $\frac{\partial}{\partial t}g(\varphi_t(z)) = \nabla g\cdot X \ne 0$ by transversality, so unit 04's implicit function theorem solves $g(\varphi_t(z)) = 0$ for $t = \tau(z)$ smoothly near any $z$ with a return. $P$ is then a composition of smooth maps.

Schematic. The orbit $\varphi_t(z)$ winds round and passes through $\Sigma$ once per turn, always in the same direction. $P$ forgets the journey and keeps only where it lands: $z\mapsto P(z)\mapsto P^2(z)$. The time taken for one turn is $\tau(z)$.

Why two dimensions is the whole point. The energy surface of a two-degree-of-freedom system is three-dimensional; an orbit in it is a curve, and curves in three dimensions cannot be drawn usefully on paper. Cutting with $\Sigma$ trades the continuous time variable — the one piece of information you do not care about — for a picture. Everything qualitative survives the trade: a periodic orbit becomes a finite set of points, a torus becomes a closed curve, and a dense orbit becomes a dense set.

Reading the figure. The sheet is a piece of a hypersurface; the orbit is drawn solid in front of it and dashed behind. Each crossing is in the same direction — that is transversality plus a sign condition — so each point of $\Sigma$ has a well-defined next crossing, and the return map is a function. Had the orbit been allowed to cross both ways, each turn would give two points and $P$ would alternate between two different maps.

17 · sections · the running example

The section for Hénon–Heiles

the section used here

For Hénon–Heiles take $\Sigma_E = \{x = 0,\ \dot x>0\}$ inside $\{H = E\}$. Transversality is the condition $\dot x = p_x > 0$, imposed by hand.

Count dimensions. Phase space 4; fix $H = E$, leaving 3; impose $x = 0$, leaving 2. Coordinates $(y,p_y)$, with the fourth variable recovered as $$p_x = +\sqrt{2E - p_y^2 - y^2 + \tfrac23y^3},$$ which is why the sign condition is needed: without it each $(y,p_y)$ would name two states.

The accessible region is where that square root is real: $\tfrac12p_y^2 + \tfrac12y^2 - \tfrac13y^3 \le E$, a closed curve. Its boundary, $p_x = 0$, is the periodic orbit along the $y$-axis.

One orbit at $E = 1/8$, started on $\Sigma_E$ at $(y,p_y) = (0.10,\,0)$ and run for $60$ time units, in the $(x,y)$ plane. Of its $20$ crossings of $x = 0$, the ten moving right are its points on $\Sigma_E$, each recorded as $(y,p_y)$; all lie on one closed curve of the section.

What the section loses. Speed along the orbit, and any orbit that never crosses $\Sigma$. The $y$-axis orbit does not cross — it lies inside the closure of $\Sigma_E$, as its boundary. Whether every other orbit returns is not obvious; for this system, at these energies, experiment says yes, and the general reason is Poincaré's recurrence theorem, which follows from unit 14's Liouville theorem and is problem P4.

The transversality condition is not a technicality. Where $X$ is tangent to $\Sigma$, $\tau$ is not even continuous: an orbit grazing the surface either touches it or misses it, and an arbitrarily small change decides which. The condition $p_x>0$ keeps us away from the tangency locus $p_x = 0$, at the cost of losing exactly the boundary curve.

Reading the figure. This is configuration space, not the section: the section is what you get by keeping, at each filled dot, the height $y$ and the vertical momentum $p_y$. The dashed curve $V = 1/8$ is the wall the orbit cannot cross, since the kinetic energy would be negative beyond it. The orbit touches that wall only where its velocity momentarily vanishes. The filled dots all lie in a band of $y$, and their $(y,p_y)$ values trace one closed curve in the section — the signature of a torus, here a curve round the large island that the $E = 1/8$ panel of Four energies shows. Energy is held to $10^{-12}$ over this run.

17 · sections

The return map preserves area

area preservation

Let $H$ have two degrees of freedom and let $\Sigma_E = \{q_1 = 0,\ \dot q_1 > 0,\ H = E\}$ with coordinates $(q_2,p_2)$. Then the return map $P$ satisfies $\det DP \equiv 1$: it is an area-preserving map.

proof, in outline

Build $\Psi(E,q_2,p_2,t) = \varphi_t\big(0,\,q_2,\,p_1(E,q_2,p_2),\,p_2\big)$, a change of coordinates on phase space near $\Sigma$ (transversality makes it one).

1 $\Psi(\cdot,t+s) = \varphi_s\circ\Psi(\cdot,t)$, so $D\Psi(\cdot,t+s) = D\varphi_s\cdot D\Psi(\cdot,t)$. Unit 14's Liouville theorem gives $\det D\varphi_s = 1$, so $\det D\Psi$ does not depend on $t$.

2 At $t = 0$, expanding the determinant along the $q_1$ row leaves $\det D\Psi = \pm\,\dot q_1\cdot\dfrac{\partial p_1}{\partial E} = \pm\,\dot q_1\cdot\dfrac1{\partial H/\partial p_1} = \pm1.$

3 Repeating step 2 at the returned point gives $\det D\Psi = \pm\det DP$. Compare with step 1: $\det DP = 1$. $\square$

full proof in the notes

Full proof

Work in $\mathbb R^4$ with coordinates $(q_1,q_2,p_1,p_2)$ and suppose $\partial H/\partial p_1 \ne 0$ on the region of interest, so that $H = E$ can be solved for $p_1 = p_1(E,q_2,p_2)$ with $\partial p_1/\partial E = 1/(\partial H/\partial p_1)$ — unit 04's implicit function theorem, differentiated. Define $\sigma(E,q_2,p_2) = (0,\ q_2,\ p_1(E,q_2,p_2),\ p_2)$ and $\Psi(E,q_2,p_2,t) = \varphi_t(\sigma(E,q_2,p_2))$.

Step 1. The flow property $\varphi_{t+s} = \varphi_s\circ\varphi_t$ gives $\Psi(u,t+s) = \varphi_s(\Psi(u,t))$ for $u = (E,q_2,p_2)$. Differentiating in all four variables, $D\Psi(u,t+s) = D\varphi_s(\Psi(u,t))\cdot D\Psi(u,t)$ — the $t$-column included, since $\partial_t\Psi(u,t+s) = D\varphi_s\cdot\partial_t\Psi(u,t)$ by the chain rule. Taking determinants and using $\det D\varphi_s = 1$ (unit 14) gives $\det D\Psi(u,t+s) = \det D\Psi(u,t)$ for all $s$: the determinant is a function of $u$ alone.

Step 2. At $t = 0$ the four columns of $D\Psi$, in the row order $(q_1,q_2,p_1,p_2)$, are $\partial_E\Psi = (0,0,\partial_Ep_1,0)$, $\partial_{q_2}\Psi = (0,1,\partial_{q_2}p_1,0)$, $\partial_{p_2}\Psi = (0,0,\partial_{p_2}p_1,1)$ and $\partial_t\Psi = (\dot q_1,\dot q_2,\dot p_1,\dot p_2)$. Only the last has a nonzero first entry, so expanding along the $q_1$ row gives $\det D\Psi = \pm\dot q_1\cdot\det\!\begin{pmatrix}0&1&0\\ \partial_Ep_1&\partial_{q_2}p_1&\partial_{p_2}p_1\\ 0&0&1\end{pmatrix} = \mp\,\dot q_1\,\partial_Ep_1 = \mp1$, using $\dot q_1 = \partial H/\partial p_1$ and $\partial_Ep_1 = 1/(\partial H/\partial p_1)$.

Step 3. Let $\tau(u)$ be the return time of $\sigma(u)$ and $u' = (E, P(q_2,p_2))$ the returned point — the energy is unchanged because $H$ is conserved. For small $s$, $\Psi(u,\tau(u)+s) = \Psi(u',s)$. The map $(u,s)\mapsto(u',s)$ has Jacobian in block form: $E\mapsto E$, $(q_2,p_2)\mapsto P(q_2,p_2)$ independent of $s$, and $s\mapsto s$ up to a shift by $\tau(u)$, which contributes only off-diagonal entries. Its determinant is therefore $\det DP$. So $\det D\Psi(u,\tau(u)) = \det D\Psi(u',0)\cdot\det DP = \pm\det DP$. Step 1 says the left side is $\det D\Psi(u,0) = \pm1$ with the same sign — both are evaluated with $\dot q_1>0$ — so $\det DP = 1$. $\square$

What was used. Liouville's theorem (unit 14), the flow property (unit 02), the chain rule (unit 03) and the implicit function theorem (unit 04). Nothing else, and in particular no differential forms: the slick proof, that $P$ pulls back the symplectic form, waits for unit 34. That $\det DP = 1$ means $P$ preserves area — $\operatorname{area}P(U) = \int_U|\det DP|$ — is the change-of-variables formula for double integrals, proved in unit 05 and used here as unit 14 used it for volumes.

Why the sign condition matters here too. Step 3 compares two evaluations of the same determinant; if the section allowed both signs of $\dot q_1$, the two evaluations could differ in sign and the conclusion would weaken to $\det DP = \pm1$. Orientation is preserved precisely because the flow crosses in one direction.

17 · sections

Elliptic and hyperbolic, drawn

What $P$ does near a fixed point, to first order: $DP$ is a $2\times2$ matrix with $\det DP = 1$, so its eigenvalues satisfy $\mu_1\mu_2 = 1$. Either they are $e^{\pm i\alpha}$ and the iterates (blue) turn round ellipses, or they are real, $\mu$ and $1/\mu$, and the iterates slide along hyperbolas, out along one line and in along the other. A spiral would need $|\mu_1\mu_2| < 1$.

consequences of $\det DP = 1$

No attractors. A sink would shrink area, so nothing here ever settles down.

Recurrence. On a finite-area region, every open patch contains points that come back to it, infinitely often (problem P4).

Two kinds of fixed point. The eigenvalues of $DP$ multiply to $1$: either on the unit circle (elliptic) or real reciprocals (hyperbolic). No spirals.

Why these are the only two cases. The characteristic polynomial of a real $2\times2$ matrix with determinant $1$ is $\mu^2 - (\operatorname{tr}DP)\,\mu + 1$. If $|\operatorname{tr}DP| < 2$ the roots are complex conjugates of modulus $1$: elliptic. If $|\operatorname{tr}DP| > 2$ they are real with product $1$: hyperbolic. The boundary case $|\operatorname{tr}DP| = 2$ is parabolic — a shear — and is non-generic. A spiral sink needs complex roots of modulus less than $1$, which would make the determinant, their product, less than $1$.

What to look for. In the elliptic panel the dots step round one ellipse by a fixed angle $\alpha$ each time; the ellipse is a level curve of a quadratic form that $DP$ preserves, the linear shadow of an invariant curve round an island centre. In the hyperbolic panel the dots move away along the red line by a factor $\mu$ per step and approach along the green line by $1/\mu$: these two lines are the linear shadows of the unstable and stable sets that section 4 follows out into Poincaré's tangle. The island chains of the coming sections are built from exactly these two pieces, alternating.

Recurrence, in one line. If an open patch $U$ of finite area never returned to itself, its images $U, P(U), P^2(U),\dots$ would be disjoint — if $P^m(U)$ met $P^n(U)$ with $m<n$, then $U$ would meet $P^{n-m}(U)$ — all of the same area, inside a region of finite area. Impossible. Problem P4 makes it a proof.

17 · sections

What integrability looks like on a section

the test

Suppose a second integral $F$ exists on the region. Then every orbit stays in $\{H = E, F = f\}$, whose intersection with $\Sigma_E$ is one-dimensional: the crossings of a single orbit lie on a curve.

So: curves everywhere $\Rightarrow$ consistent with integrability; one orbit filling a patch $\Rightarrow$ no such $F$ on that patch.

the three things you see

Invariant curves — a torus cut by the section.
Island chains — $m$ closed curves visited in turn, around $m$ elliptic fixed points of $P^m$. A resonance.
A chaotic sea — one orbit's crossings spread over an area, never repeating and never settling.

the direction of the logic is one-way

A filled patch is strong evidence against a second integral, and the argument above makes that precise. Curves are not evidence for one: a chaotic layer thinner than a pixel, or an orbit that has not yet escaped, looks exactly like a curve.

This is why section 3 exists. The Lyapunov exponent is a number, not an impression, and it is what turns a picture into evidence.

a sanity check before computing

At $E\to0$ the system is the isotropic oscillator, integrable, so the section must be curves. If the computation does not show that, the integrator is wrong. It is the only case where the answer is known in advance, and it is the one to test on.

Making the test precise. "Filling a patch" needs care, because a countable set of points has zero area whatever it looks like. The honest statement is: if a smooth $F$ independent of $H$ existed on an open set $U$, then $F$ restricted to $\Sigma_E\cap U$ is a non-constant smooth function constant along orbits of $P$, so the closure of any orbit lies in one of its level curves, which has empty interior. Exhibiting an orbit whose closure contains a disc therefore rules out $F$ on $U$. Numerically one cannot exhibit a closure, only $850$ points scattered over a region — which is why the argument is completed by the Lyapunov exponent rather than by the picture.

Island chains, and where they come from. An island chain of $m$ islands is the signature of an orbit of the return map with period $m$, which means a periodic orbit of the flow that goes round $m$ times in one direction for every $q$ times in the other — a resonance $m:q$. Section 4 explains why the resonant tori are exactly the ones that break, and why what replaces them is a chain of islands.

How the pictures in this deck were made. Fixed-step fourth-order Runge–Kutta with step $0.005$, crossings located by bisection on the step to sixty iterations. The check that matters is energy drift: over the longest run here, $T = 10^6$, it never exceeds $1.2\times10^{-8}$ out of $E = 0.125$, a relative $9\times10^{-8}$, and over the $T = 20\,000$ runs of section 3 it stays below $3\times10^{-9}$ relative, so no feature of the picture larger than a pixel is an artefact of the integration. A run whose energy drifts is not evidence of anything.

17 · sections · the running example

Four energies

Horizontal $y$, vertical $p_y$. The dashed curve is the boundary $p_x = 0$; each dot is one crossing of $x = 0$ with $\dot x>0$.

Colour is the measured $\lambda_1$, the rate at which a neighbouring orbit separates (section 3 defines it). Gold: $\lambda_1<0.02$. Red: $\lambda_1>0.02$ — and in each lower panel the red points are one orbit, $850$ crossings at $E = 1/8$ and $1600$ at $E = 1/6$.

what to look at

Whether the dots of a single orbit lie on a curve or cover a patch. In the lower panels one orbit covers a patch, which no second integral permits.

This is the figure the unit exists for, and it is worth dwelling on. Each panel is the same system, differing only in the energy. Nothing has been tuned, no forcing has been added, and no noise enters: the integrator is deterministic to the last bit.

$E = 1/100$. Smooth nested curves everywhere, exactly as the integrable limit demands. Two families, around two different elliptic fixed points, separated by a curve that is the section's trace of an unstable periodic orbit.

$E = 1/12$. Still curves, but distorted, and the island structure has appeared: a resonance has opened up inside the family. A grid of $847$ starting points finds $0.1\%$ of the area chaotic — one point, and five at the lower threshold of the robustness check in the census: a thin layer, invisible at this resolution.

$E = 1/8$. The transition. A large elliptic island and a ring of satellites survive; everything else is one orbit's $850$ crossings. The measured chaotic fraction is $52\%$.

$E = 1/6$. $96\%$ chaotic. What is left is a single pair of tiny islands near $y = 0.11$ and $y = 0.55$, visited alternately — a $2$-cycle of the return map, and the last regular motion in the system.

The one thing to resist. The red points are one orbit in each panel, not a smear of many. That is what makes the picture an argument rather than a decoration: a single deterministic solution, started from one point, visits a two-dimensional set of states. No second integral can exist there, because it would have to be constant on that set.

17 · sections · example

Reading the four panels

worked example 2 · how much of the section is chaotic

Method. Lay a grid on the accessible region of $\Sigma_E$, keep the points with $\tfrac12p_y^2+\tfrac12y^2-\tfrac13y^3 < E$, start an orbit at each, measure $\lambda_1$ over $T = 2000$ — that is, $\tfrac1T\ln$ of how much a tiny separation from a neighbouring orbit has grown by time $T$, made precise in section 3 — and count the fraction above a threshold.

Threshold. A regular orbit's $\lambda_1(T)$ decays like $\ln T/T$, which at $T = 2000$ is $0.0038$; taking $0.02$ is five times that.

$E$	grid points	chaotic	$E/E_{\text{esc}}$
$1/100$	98	0%	6%
$1/12$	847	0.1%	50%
$1/8$	1315	52%	75%
$1/6$	1878	96%	100%

checks

Dimensions. A fraction of an area is dimensionless ✓; the threshold $0.02$ is in units of $\omega$, since $\lambda$ is a rate ✓.

Limit $E\to0$. Must give $0$, since the system tends to the integrable oscillator. It does ✓.

Robustness. At threshold $0.01$ the four numbers are $0\%$, $0.6\%$, $56\%$, $96\%$ — the verdict does not turn on where the line is drawn ✓.

what the table is not

A grid count cannot see islands smaller than the grid. Read it as: how much of the section a randomly thrown initial condition finds chaotic.

Why the transition looks sudden. Between $E = 1/12$ and $E = 1/8$ the chaotic fraction goes from a tenth of a per cent to half of everything, for an energy change of $50\%$. Section 4 names the mechanism: separate chaotic layers, each attached to its own resonance, grow until they touch, and at the moment they touch the regions merge and an orbit can wander through all of them. That is Chirikov's overlap picture, and it makes the transition genuinely sharp rather than merely steep.

On honest numerics. Three things were checked before the table was believed: energy drift (below $4\times10^{-10}$ relative at the end of every run), insensitivity to the integration step (doubling it to $0.01$ moves no entry by more than half a per cent), and insensitivity to the threshold (shown above). The grid has spacing $0.0254$ in both $y$ and $p_y$. A number in this course that has had none of those done to it is not a number, it is a hope.

The $E = 1/100$ row has only 98 grid points because the accessible region shrinks with $E$: its area is $2\pi E$ for small $E$, from the near-circular boundary $p_y^2+y^2\approx2E$. Integrating $2\sqrt{2E-y^2+\tfrac23y^3}$ exactly gives $0.0631$ at $E = 1/100$ against $2\pi E = 0.0628$, and $1.200$ at $E = 1/6$ against $1.047$ — the $15\%$ excess being the cubic term's distortion, visible as the triangular shape of the fourth panel's boundary. The grid counts should be in the same ratio: $1878/98 = 19.2$ against $1.200/0.0631 = 19.0$ ✓.

section 3

Measuring sensitivity

A rate, in inverse time, for how fast a mistake grows.

17 · exponents

The linearised flow

variational equation

For $\dot z = X(z)$ with flow $\varphi_t$, the derivative in the initial condition, $M(t) = D\varphi_t(z_0)$, satisfies the linear system $$\dot M(t) = DX\big(\varphi_t(z_0)\big)\,M(t),\qquad M(0) = I.$$ A single tangent vector $\delta(t) = M(t)\delta_0$ obeys the same equation.

proof

$\varphi_t$ is $C^1$ in $z_0$ (cited in unit 14: Hirsch, Smale and Devaney), so $\partial_t\partial_{z_0}\varphi_t = \partial_{z_0}\partial_t\varphi_t$ by Clairaut (unit 03). The right-hand side is $\partial_{z_0}\big[X(\varphi_t(z_0))\big] = DX(\varphi_t(z_0))\,D\varphi_t(z_0)$ by the chain rule. $\square$

this is not unit 02's linearisation

Unit 02 linearised about an equilibrium, getting $\dot\delta = A\delta$ with $A$ constant. Here the base point moves, so $DX(\varphi_t(z_0))$ is a time-dependent matrix and there is no characteristic polynomial, no eigenvalue, and no Hartman–Grobman. What replaces the eigenvalue is the Lyapunov exponent.

the matrix, for the running example

For Hénon–Heiles, $X = (p_x,\,p_y,\,-x-2xy,\,-y-x^2+y^2)$ and $$DX = \begin{pmatrix}0&0&1&0\\ 0&0&0&1\\ -1-2y&-2x&0&0\\ -2x&-1+2y&0&0\end{pmatrix}.$$ The lower-left block is $-\nabla^2V$, symmetric by Clairaut ✓. Its entries change as the orbit moves, and that is the entire mechanism of chaos in this system.

Where the eight-dimensional system comes from. To compute anything, integrate the original four equations and the four variational ones together — an eight-dimensional system whose first four components do not depend on the last four. The pictures in this deck were made that way. Integrating the variational equation separately, using a stored trajectory, is a classic way to get a wrong answer: the interpolation error in the stored orbit feeds straight into $DX$.

Why $DX$ is the whole mechanism. If $V$ were quadratic, $-\nabla^2V$ would be constant and the variational equation would be a linear system with constant coefficients, of the kind unit 02 solved in the plane — solvable exactly and bounded for a stable equilibrium. The cubic term makes $\nabla^2V$ depend on position, so the orbit visits regions where the local behaviour is a saddle and regions where it is a centre. Stretching in the saddle regions and folding back in the bounded region is, in one sentence, how a deterministic bounded flow manufactures exponential separation.

A check on the matrix. The eigenvalues of $-\nabla^2V$ at the origin are both $-1$, giving the two unit frequencies of the integrable limit. At the saddle $(0,1)$, $\nabla^2V = \begin{pmatrix}1+2y&2x\\2x&1-2y\end{pmatrix} = \operatorname{diag}(3,-1)$, one positive and one negative — a saddle of $V$ ✓, consistent with the escape channel found earlier.

17 · exponents

The Lyapunov exponent

Lyapunov exponent

For an orbit through $z_0$ and a tangent vector $\delta_0\ne0$, $$\lambda(z_0,\delta_0) \;=\; \limsup_{t\to\infty}\frac1t\,\ln\big\lVert D\varphi_t(z_0)\,\delta_0\big\rVert .$$ The largest Lyapunov exponent $\lambda_1(z_0)$ is the supremum over $\delta_0$. An orbit is called chaotic when $\lambda_1 > 0$.

independent of the norm

Two norms on $\mathbb R^{2n}$ satisfy $c\lVert\cdot\rVert_1\le\lVert\cdot\rVert_2\le C\lVert\cdot\rVert_1$, and $\tfrac1t\ln c\to0$. So $\lambda$ does not depend on which norm is used — the only thing about it that is free.

constant along an orbit

$D\varphi_{t}(\varphi_s z_0) = D\varphi_{t+s}(z_0)\,D\varphi_s(z_0)^{-1}$, and the fixed factor contributes $O(1)$ to $\ln$, hence $0$ to the limit. So $\lambda_1$ labels the orbit, not the point.

at most $2n$ values

$\{\delta_0:\lambda(z_0,\delta_0)\le c\}$ is a linear subspace, so as $c$ grows the subspaces form a flag with at most $2n$ jumps. The jump values are the spectrum.

Oseledets' multiplicative ergodic theorem — cited

For a flow preserving a finite measure, at almost every $z_0$ the $\limsup$ is a genuine limit and the flag is measurable and invariant. Oseledets 1968; a readable proof is Ruelle 1979, and Barreira and Pesin, Lyapunov Exponents and Smooth Ergodic Theory, 2002. It needs ergodic theory, which this course does not build.

Why the $\limsup$ and not the limit. Without a theorem there is no reason for $\frac1t\ln\lVert\delta(t)\rVert$ to converge: it could oscillate for ever between two values. The $\limsup$ always exists in $[-\infty,+\infty]$ and makes the definition unconditional. Oseledets says the oscillation does not happen except on a set of measure zero — with respect to an invariant measure, which for a Hamiltonian system on a compact energy surface is the Liouville measure of unit 14 restricted to it.

Full proof that the sublevel sets are subspaces. If $\lambda(\delta)\le c$ and $\lambda(\eta)\le c$ then for any $a,b$, $\lVert M(t)(a\delta+b\eta)\rVert\le|a|\lVert M\delta\rVert + |b|\lVert M\eta\rVert \le (|a|+|b|)e^{(c+\varepsilon)t}$ for $t$ large, so $\frac1t\ln\lVert M(a\delta+b\eta)\rVert\le c+\varepsilon + \frac1t\ln(|a|+|b|)$, and letting $t\to\infty$ then $\varepsilon\to0$ gives $\lambda(a\delta+b\eta)\le c$. Closure under scalars is immediate. The nesting $V_{c}\subseteq V_{c'}$ for $c\le c'$ then gives a finite flag, because a strictly increasing chain of subspaces of $\mathbb R^{2n}$ has at most $2n+1$ members.

A consequence worth stating. For a random $\delta_0$, $\lambda(z_0,\delta_0) = \lambda_1$: the vectors with a smaller exponent lie in a proper subspace, which has zero volume. This is why Benettin's algorithm, below, starts from a random vector and does not worry about the choice — and also why, if you start it exactly on the flow direction, you get $0$ and a wrong answer.

17 · exponents

What a Hamiltonian spectrum must look like

three constraints

1 $\lambda = 0$ occurs: take $\delta_0 = X(z_0)$. Then $D\varphi_t\,X(z_0) = X(\varphi_t z_0)$, which is bounded on a compact energy surface and bounded below away from equilibria, so $\tfrac1t\ln\lVert\cdot\rVert\to0$.

2 The exponents sum to zero: $\sum_i\lambda_i = \lim\tfrac1t\ln|\det D\varphi_t| = 0$ by unit 14's Liouville theorem.

3 The spectrum is symmetric: $\lambda$ an exponent $\Rightarrow$ $-\lambda$ is one, because $D\varphi_t$ is symplectic (unit 14) and a symplectic matrix's singular values come in reciprocal pairs.

two degrees of freedom, therefore

Four exponents. One is $0$ (the flow direction); one more is $0$ (the direction across the energy surface, where the separation does not grow at all). Symmetry and the zero sum then force $$\{\lambda_1,\,0,\,0,\,-\lambda_1\}.$$ There is one number to measure, and it decides everything.

Full proof of the reciprocal-pair fact. Let $M$ be symplectic, $M^{\!\top}JM = J$ (unit 14). Then $M^{-1} = J^{-1}M^{\!\top}J$, so $M^{-1}$ is similar to $M^{\!\top}$, which has the same characteristic polynomial as $M$. Hence the eigenvalue sets of $M$ and $M^{-1}$ coincide: the spectrum is closed under $\mu\mapsto1/\mu$. Now $M^{\!\top}$ is symplectic too (transpose $M^{\!\top}JM = J$ and use $J^{-1} = -J$), so $S = M^{\!\top}M$ is symplectic, symmetric and positive definite; its eigenvalues — the squared singular values of $M$ — therefore come in pairs $\mu,1/\mu$. Taking $\tfrac1{2t}\ln$ turns that into $\pm$ pairs of exponents.

Why the transverse direction gives zero. Move the initial condition to a nearby energy $E + \varepsilon$. The two orbits stay on their own surfaces for ever, and on a compact region the surfaces stay a bounded distance apart, so that component of the separation neither grows nor shrinks exponentially. The same statement for the flow direction is item 1. Together they use up two of the four exponents, which is why a two-degree-of-freedom Hamiltonian system has exactly one number in it.

17 · exponents

Stretch and contract, drawn

Schematic, in the two directions that are neither along the flow nor across the energy surface. A small disc of starting points around $z_0$ is carried to an ellipse: stretched by $e^{\lambda_1t}$ one way, squeezed by $e^{-\lambda_1t}$ the other, area unchanged.

what each case means

$\lambda_1 = 0$: separation grows at most sub-exponentially — on a torus it grows linearly, from the mismatch of frequencies.
$\lambda_1 > 0$: exponential separation in one direction, exponential contraction in another, area preserved between them. Stretch and fold, for ever.

The ellipse cannot keep growing in a bounded region: it is folded back on itself while it stretches. Stretching sets the rate; folding keeps the orbit confined.

Reading the figure. The spectrum $\{\lambda_1,0,0,-\lambda_1\}$ says what $D\varphi_t$ does to the four directions at $z_0$: along the flow, nothing exponential; across the energy surface, nothing exponential; in the remaining two-dimensional slice, one direction grows like $e^{\lambda_1t}$ and the other shrinks like $e^{-\lambda_1t}$. A disc of radius $r$ in that slice becomes, to first order, an ellipse with semi-axes $re^{\lambda_1t}$ and $re^{-\lambda_1t}$, whose area $\pi r^2$ is unchanged — the zero sum of the exponents, seen as a picture. The figure exaggerates nothing qualitatively but is not to scale: at $\lambda_1 = 0.044$ a stretch of $3.2$, as drawn, takes $26$ time units.

Linear growth on a torus, computed. On an invariant torus the motion is $\theta_i(t) = \theta_i(0) + \omega_i(\vec I)t$. Perturb the action by $\delta\vec I$: the angles drift apart by $\frac{\partial\omega_i}{\partial I_j}\delta I_j\,t$ — linear in $t$, with a coefficient that vanishes only for a degenerate Hamiltonian. So the generic behaviour on a torus is $\lVert\delta(t)\rVert\sim ct$, hence $\lambda_1(T) \approx \ln T/T\to0$. That prediction is tested on the two-orbits slide, and confirmed there to a slope of $0.997$.

17 · exponents

Computing it without overflowing

Benettin's algorithm

Choose a renormalisation interval $\Delta$ and a unit vector $\delta^{(0)}$. For $k = 1,2,\dots,N$: integrate the variational equation over $[0,\Delta]$ from $\delta^{(k-1)}$, call the result $\eta^{(k)}$, record $a_k = \lVert\eta^{(k)}\rVert$, and restart from $\delta^{(k)} = \eta^{(k)}/a_k$. Then $$\lambda_1 \;\approx\; \frac1{N\Delta}\sum_{k=1}^{N}\ln a_k .$$

why this is the same number

Linearity: $D\varphi_{N\Delta}\,\delta^{(0)} = \big(\prod_k a_k\big)\,\delta^{(N)}$, because each restart only rescales, and rescaling commutes with a linear map. Taking $\tfrac1{N\Delta}\ln$ of the norm gives the sum above plus $\tfrac1{N\Delta}\ln\lVert\delta^{(N)}\rVert = 0$. $\square$

why it is needed at all

With $\lambda_1 = 0.044$ and $T = 20\,000$, $\lVert\delta\rVert$ would reach $e^{880}$. The largest double-precision number is about $e^{709}$. Without renormalisation the computation overflows before the answer converges — and long before that, $\delta$ has aligned so exactly with the expanding direction that its short components are lost to rounding.

The trap this algorithm replaces. The naive method — integrate two nearby orbits and watch the distance — fails twice. It saturates: once the separation reaches the size of the energy surface it stops growing, and the measured rate collapses. And before saturating it is contaminated by the nonlinear terms the definition excludes. The variational equation has neither problem, because it is exactly linear and the renormalisation keeps $\delta$ at unit length for ever.

The other exponents. Benettin's algorithm generalises: carry $k$ vectors and re-orthonormalise them (Gram–Schmidt) at each step; the logarithms of the $k$ normalisation factors give $\lambda_1+\cdots+\lambda_k$. For the Hamiltonian systems of this unit the symmetry of the spectrum makes that unnecessary — one number is all there is.

17 · exponents

Stretch, record, rescale

Benettin's algorithm, schematic. Each interval $\Delta$ the unit vector is carried by the variational equation and comes out stretched (red, length $a_k$); its direction is kept and its length reset to $1$ (blue). The product $a_1a_2\cdots a_N$ is what the unrenormalised vector would have grown by, and only its logarithm is ever stored.

choosing $\Delta$

Too large and it overflows; too small and the cost is all in restarts. $\Delta\lambda_1\lesssim1$ keeps every $a_k$ modest — here $\Delta = 2$ gives $\Delta\lambda_1 = 0.09$, so $a_k\approx e^{0.09}$ on average, far inside range.

Reading the figure. The dashed curve is the base orbit, along which the variational equation is integrated; the arrows are tangent vectors at points of it, not displacements in phase space. Two things are kept from each interval: the number $\ln a_k$, and the direction of $\eta^{(k)}$. The direction matters: it carries the information about which way the flow stretches, and after a few intervals it has turned towards the most-expanding direction, which is why the sum converges to $\lambda_1$ and not to a smaller exponent.

How to know the answer has converged. Plot $\lambda_1(T)$ against $T$ and look for a plateau, not a single final value — and then distrust the plateau. For the chaotic orbit here, with step $0.005$, $\lambda_1(T)$ wanders between $0.037$ and $0.047$ for $T\in[4000,20000]$ and ends at $0.0377$. The same start with step $0.01$ — a different trajectory after a few hundred time units, since the orbit is chaotic — ends at $0.058$. Run on to $T = 10^6$, the steps $0.01$, $0.005$ and $0.0025$ give $0.0434$, $0.0455$ and $0.0442$. So the honest quotation is $\lambda_1 = 0.044\pm0.001$ at $T = 10^6$, and a budget of $20\,000$ was not enough. The fluctuation is real, not numerical: it is the orbit spending different fractions of its time stuck near the edges of islands, where the local stretching is weak.

17 · exponents · the running example

The two orbits, measured

worked example 3 · one number each

At $E = 1/8$, from the section, $\Delta = 2$, step $0.005$:

start $(y,p_y)$	$T$	$\lambda_1(T)$	verdict
$(-0.20,\,0)$	$10^6$	$0.044$	chaotic
$(0.30,\,0)$	$2\times10^4$	$3.36\times10^{-4}$	on a torus

The second number is not small, it is zero. Fit $\ln\lVert\delta(t)\rVert$ against $\ln t$ for $t>200$: slope $\mathbf{0.997}$, so the separation grows linearly and $\lambda_1(T) = (\ln T + c)/T\to0$.

Checks. Dimensions: $\lambda$ is $1/$time, i.e. $0.044\,\omega$ ✓. $1/\lambda_1 = 23$ against an orbital period near $2\pi$: $3.6$ periods per $e$-folding, the right order for a bounded orbit ✓. Both orbits have the same energy, so the difference is the orbit, not the shell ✓.

$\ln\lVert\delta(t)\rVert$ for the two orbits over $T = 20\,000$. Red: nearly a straight line, of mean slope $0.0377$ on this stretch against the long-run $0.044$. Gold: the torus, reaching $6.7$ — which its value $2.1$ at $t = 200$ plus $\ln(20000) - \ln(200) = 4.6$ of logarithmic growth accounts for entirely.

Read the figure as the argument it is. Two orbits of one system at one energy, started a quarter of the section apart. Over $T = 20\,000$ one separates by a factor $e^{754}$; the other by a factor of about $820$. The only claim being made is about the shape of the two curves: straight against flat-and-logarithmic. The values themselves are not the evidence; the straightness is.

The fit, in numbers. Both tangent vectors start as $(1,1,1,1)/2$ in the order $(x,y,p_x,p_y)$. For the torus, $\ln\lVert\delta\rVert$ at $t = 1040$ is $3.76$; at $t = 20\,000$ it is $6.71$. The difference is $2.95$. If the growth were linear in $t$, $\ln$ would grow by $\ln(20000/1040) = 2.96$. The agreement to $0.1\%$ is the content of the slope $0.997$, and it is the cleanest confirmation in this unit that a measured number is zero rather than merely small.

Why a torus gives exactly linear growth. On the torus the flow is $\dot{\vec\theta} = \vec\omega(\vec I)$. Two nearby initial conditions with slightly different actions have slightly different frequencies, so their angles drift apart at a constant rate: the separation is $\lVert\partial\vec\omega/\partial\vec I\cdot\delta\vec I\rVert\,t$ plus a bounded oscillation. Measure that with $\frac1T\ln$ and you get $(\ln T + c)/T$. The fit gives $\ln\lVert\delta\rVert = 0.997\ln t - 3.26$, so the prediction at $T = 20\,000$ is $(0.997\times9.90 - 3.26)/20\,000 = 3.31\times10^{-4}$, against the measured $3.36\times10^{-4}$ ✓.

A caution about the chaotic curve. It is not perfectly straight: there are visible plateaux where the orbit spends time in a nearly regular region before re-entering the hyperbolic one. This stickiness near the boundaries of islands is a real and much-studied phenomenon; it makes the convergence of $\lambda_1(T)$ slow, and it is why this stretch's $0.0377$ is not the number quoted: the budget that settles it is $T = 10^6$.

17 · exponents

Gronwall's revenge

what unit 01 actually proved

If $f$ is Lipschitz with constant $L$ on a region, two solutions with data $y_0$, $\tilde y_0$ satisfy $$\lVert y(t)-\tilde y(t)\rVert \;\le\; \lVert y_0-\tilde y_0\rVert\,e^{Lt}.$$ Continuous dependence follows: given $\varepsilon$ and a fixed horizon $T$, take $\delta = \varepsilon e^{-LT}$.

where the promise is void

The $\delta$ depends on $T$, and goes to zero as $T$ grows. Continuity in the data is not uniform in $t$. Nothing in unit 01 says otherwise, and no strengthening is possible: the bound is attained.

the two rates, for the running example

On the accessible region at $E = 1/8$, $\sup\lVert DX\rVert_2 = 2.35$, so Gronwall gives $e^{2.35t}$ — a guarantee, valid for every orbit, and useless: at $t = 1000$ it permits a growth factor of $e^{2350}$.

The measured rate on the chaotic orbit is $\lambda_1 = 0.044$, smaller by a factor $53$. Gronwall bounds the worst case; the Lyapunov exponent is the achieved average. Both are exponential, and that is the point.

The logical structure, stated once, because it is the checkpoint. Picard–Lindelöf: for each initial condition there is exactly one solution. Continuous dependence: for each $\varepsilon>0$ and each $T$ there is a $\delta>0$ such that data within $\delta$ stay within $\varepsilon$ up to time $T$. Positive Lyapunov exponent: $\delta$ must be taken exponentially small in $T$. The three are consistent because the second quantifies over $T$ before choosing $\delta$. A contradiction would require a single $\delta$ working for all $T$ — that is uniform continuity in the data, which unit 01 never claimed and which is false here.

The bound is attained, which is the part people skip. For $\dot y = Ly$ the Lipschitz constant is exactly $L$ and the separation is exactly $\lVert\delta_0\rVert e^{Lt}$: equality. So Gronwall cannot be improved as a general bound. What can be improved is the constant for a particular system, and the Lyapunov exponent is the best possible such constant in the asymptotic sense — problem P5 does both halves.

17 · exponents · the running example

Eight orbits per digit

An error growing at the chaotic orbit's rate $\lambda_1 = 0.044$, on a log scale: a straight line of slope $0.044/\ln10 = 0.019$ decades per time unit. Each digit of the starting data buys $\ln10/0.044 = 52$ time units. Gronwall's guaranteed rate $2.35$ (blue) uses all the digits in $19$.

the number that ends the argument

To know the state at $t = 1000$ — about $160$ orbits — to one part in $10^{2}$, the initial data must be known to $$10^{-2}e^{-\lambda_1\cdot1000} = 7.8\times10^{-22}.$$ Double precision carries $2.2\times10^{-16}$. The orbit is determined; it is not computable.

Reading the figure. Three starting errors, each a straight line until it reaches the tolerance $10^{-2}$: from $7.8\times10^{-22}$ it takes $1000$ time units, from $2.2\times10^{-16}$ — the spacing of double-precision numbers near $1$ — it takes $715$, and from $10^{-9}$, a very good physical measurement, $366$. Twelve more digits bought $634$ time units: about $52$ each, the exchange rate. The lines are $\log_{10}\lVert\delta(t)\rVert = \log_{10}\lVert\delta_0\rVert + \lambda_1t/\ln10$, which is the definition of $\lambda_1$ read as a growth law; the real separation wobbles about such a line, as the two-orbits figure shows.

The Gronwall line. It is the worst case unit 01 guarantees, $e^{2.35t}$, and it reaches the tolerance from $7.8\times10^{-22}$ at $t = \ln(1.3\times10^{19})/2.35 = 19$. Both lines are exponential; they differ in the rate by a factor $53$. The bound is true and nearly useless here, which is why the measured exponent is worth the trouble.

Why "determined but not computable" is not a cheat. The obstruction is not about floating point. Suppose the initial data were known exactly as real numbers and the integration were exact. Then the orbit is exact, and prediction is perfect. The trouble is that no physical measurement delivers a real number, and each extra digit buys only $1/\lambda_1\cdot\ln10 = 52$ more time units — about eight orbits. Eight orbits per digit is the exchange rate, and it is set by the dynamics, not by the arithmetic.

17 · exponents

What a positive exponent does not say

four things it is not

Not non-uniqueness. Unit 01's hypothesis is a Lipschitz condition on $f$; the Hénon–Heiles field is a polynomial, Lipschitz on every bounded set. Uniqueness holds, everywhere, always.

Not randomness. Nothing stochastic appears anywhere. The orbit that fills the sea is a single solution of a polynomial differential equation.

Not instability of the system. The motion stays bounded for ever, and the energy is exact. What is unstable is the labelling of orbits by initial data.

Not a property of a single point. $\lambda_1$ is an asymptotic average along an orbit; a finite-time estimate on a finite segment is a different and smaller thing.

what it does say

Information about the initial state is destroyed at a definite rate. $\lambda_1/\ln2 = 0.063$ bits per unit time is the rate at which digits of the initial condition stop mattering to the past and start mattering to the future — Pesin's identity makes that precise, and it is cited (Barreira and Pesin, 2002).

Lyapunov time

$t_\lambda = 1/\lambda_1$, the time for an error to grow by $e$. For the running example, $23$ time units. For the inner solar system, about $5$ million years — section 5.

The confusion this slide exists to prevent is the belief that chaos is a failure of determinism, or a sign that the model is incomplete, or that it somehow requires noise to get going. All three are wrong, and the first is the one worth being able to refute on demand — which is why the checkpoint asks for it with the theorem quoted precisely.

Where the bits come from. Measure the initial condition to $k$ binary digits. After time $t$ the uncertainty has grown by $e^{\lambda_1t}$, so about $\lambda_1t/\ln2$ digits have become meaningless; equivalently the orbit has revealed that many digits of information that the original measurement did not contain. Pesin's identity says that for a large class of systems the Kolmogorov–Sinai entropy — the information production rate, in the sense of Shannon — equals the sum of the positive exponents, here $\lambda_1$ alone.

Finite-time exponents are a real tool, not a poor relation. $\frac1{T}\ln\lVert\delta(T)\rVert$ for moderate $T$ varies strongly over the section, and mapping it is how the pictures in this deck were coloured. It is a different quantity from $\lambda_1$ and should be named differently; conflating them produces the common mistake of reporting a positive exponent for a regular orbit that has simply not run long enough.

section 4

Perturbing an integrable system

Which tori survive, which break, and what replaces the ones that break.

17 · perturbation

The small denominators

Take $H = H_0(\vec I) + \varepsilon H_1(\vec I,\vec\theta)$, with $\vec\omega = \partial H_0/\partial\vec I$, and remove the $\vec\theta$-dependence with unit 14's type-2 generating function $S = \vec I{\,}'\!\cdot\vec\theta + \varepsilon S_1(\vec I{\,}',\vec\theta)$, so $\vec I = \vec I{\,}' + \varepsilon\,\partial_{\vec\theta}S_1$ and

$$K = H_0(\vec I{\,}') + \varepsilon\Big[\vec\omega\cdot\partial_{\vec\theta}S_1 + H_1\Big] + O(\varepsilon^2).$$

The first-order term can be made independent of $\vec\theta$, leaving only $\langle H_1\rangle$, the average of $H_1$ over the angles (over the torus, at fixed $\vec I$), if $\vec\omega\cdot\partial_{\vec\theta}S_1 = -(H_1 - \langle H_1\rangle)$.

one harmonic, solved exactly

Take $H_1 = a\cos(\vec k\cdot\vec\theta)$, $\vec k\ne0$, and try $S_1 = b\sin(\vec k\cdot\vec\theta)$. Then $\vec\omega\cdot\partial_{\vec\theta}S_1 = b(\vec k\cdot\vec\omega)\cos(\vec k\cdot\vec\theta)$, so the equation forces $\;b = -a/(\vec k\cdot\vec\omega)$.

the obstruction, in one symbol

$\vec k\cdot\vec\omega$ is in the denominator. At a resonance, $\vec k\cdot\vec\omega = 0$ and there is no solution at all; merely near one, the "correction" exceeds what it corrects and the expansion is void.

small denominators

A general $H_1$ is a sum over all $\vec k$; for $n\ge2$ the denominators come arbitrarily close to $0$, and the series converges only if the numerators shrink faster. That is the whole difficulty of KAM.

dimensions

$a$ is an energy, $\vec k\cdot\vec\omega$ a frequency, and $S_1$ must be an action: energy over frequency ✓.

Why the infinite sum is stated rather than used. Decomposing a general periodic $H_1$ into harmonics $\sum_{\vec k}a_{\vec k}e^{\mathrm i\vec k\cdot\vec\theta}$ is a Fourier series, and the theory of those — convergence, completeness, the decay of coefficients for smooth functions — arrives later in the course. Nothing on this slide needs it: with $H_1$ a finite trigonometric sum the computation above is exact and elementary, term by term, and it already exhibits the obstruction. When the sum is infinite the same formula produces the coefficients, and the question becomes whether they are summable, which is the question KAM answers.

The averaging that motivates the whole attempt. If the transformation could be carried through, $K$ would depend on $\vec I{\,}'$ alone, so $\vec I{\,}'$ would be conserved and the system would be integrable — the perturbed motion would live on tori just like the unperturbed. Poincaré's theorem is precisely that this cannot be done globally: the formal series exists at every order but does not converge. Laplace and Lagrange had carried the first order through for the solar system and concluded stability; the small denominators are exactly what their argument swept aside.

The resonant case is not a failure of technique. When $\vec k\cdot\vec\omega = 0$, the unperturbed torus is foliated by closed orbits rather than by one dense orbit, and the perturbation's average around each closed orbit varies from orbit to orbit. There is then a real dynamical effect to capture, not a formal one to remove, and the correct treatment — secular perturbation theory near the resonance — produces a pendulum equation whose separatrix is the island chain of section 2.

17 · perturbation

Rational frequencies are the enemy, and there are few of them

Diophantine condition

$\vec\omega$ is $(\gamma,\tau)$-Diophantine if $$|\vec k\cdot\vec\omega| \;\ge\; \frac{\gamma}{\lVert\vec k\rVert^{\tau}}\qquad\text{for all }\vec k\in\mathbb Z^n\setminus\{0\}.$$ It is a quantitative statement that $\vec\omega$ is far from every resonance, with the tolerance shrinking as the resonance gets higher order.

almost every $\vec\omega$ satisfies it

For $n = 2$ write $\alpha = \omega_1/\omega_2$. For $\tau > 1$, the set of $\alpha\in[0,1]$ failing $|\alpha - p/q|\ge\gamma q^{-(\tau+1)}$ for some $p/q$ can be covered by intervals of total length at most $$\sum_{q\ge1}(q+1)\cdot\frac{2\gamma}{q^{\tau+1}} \;\le\; 4\gamma\sum_{q\ge1}q^{-\tau},$$ a convergent series. So the bad set shrinks to nothing as $\gamma\to0$. $\square$

a strange sentence, but a true one

Whether a particular orbit of a mechanical system survives a perturbation depends on the continued-fraction expansion of a ratio of its frequencies. Nothing else in classical mechanics is like this.

The covering argument, spelled out. Fix $\gamma>0$ and $\tau>1$. For each $q\ge1$ and each $p$ with $0\le p\le q$, the set of $\alpha\in[0,1]$ with $|\alpha-p/q|<\gamma q^{-(\tau+1)}$ is an interval of length at most $2\gamma q^{-(\tau+1)}$. There are $q+1$ choices of $p$, so all $\alpha$ failing the condition at denominator $q$ are covered by total length $\le 2\gamma(q+1)q^{-(\tau+1)}\le4\gamma q^{-\tau}$. Summing over $q$ and using convergence of $\sum q^{-\tau}$ for $\tau>1$ gives total length $\le4\gamma\zeta(\tau)$, which tends to $0$ with $\gamma$. Hence for every $\eta>0$ there is a $\gamma$ making the bad set coverable by length $\eta$ — the elementary substitute for "the bad set has measure zero", and all this unit needs.

The two competing intuitions, and which one is right. Rationals are dense, so every frequency is arbitrarily close to a resonance — that intuition says all tori break. Rationals are countable and can be covered by arbitrarily short intervals — that intuition says almost none break. The Diophantine condition is the referee: closeness to $p/q$ is forgiven if it is not too close relative to $q$. The second intuition wins, but only for the tori whose frequencies keep a definite distance, and the surviving set, while of nearly full measure, has empty interior. That is why the chaotic and regular regions of the section are intertwined at every scale.

17 · perturbation

The resonances, drawn on a line

The bad set of the previous slide, $\gamma = 0.08$, $\tau = 1.5$: round each $p/q$ a bar of half-width $\gamma/q^{2.5}$, one row per $q$. The bars for $q\le8$ cover $0.24$ of the line; all $q$ together, at most $4\gamma\sum_q q^{-1.5} = 0.84$. The blue line threads every gap.

the hardest number to approximate

Hurwitz: every irrational has infinitely many $p/q$ with $|\alpha - p/q| < 1/(\sqrt5q^2)$, and $\sqrt5$ cannot be improved — the obstruction being the golden ratio, whose continued fraction $[1;1,1,\dots]$ has every partial quotient as small as possible. (cited: Hardy & Wright, §11.8.)

Prediction, then: as a perturbation grows, the tori should break in order of how well their frequency ratio is approximated by rationals, and the last one to go should be the golden one. The last-torus slide shows that this is what happens.

Reading the figure. Each red bar is a set of frequency ratios too close to one particular resonance $p/q$ to pass the Diophantine test with these constants. Bars of small $q$ are wide — low-order resonances are the dangerous ones — and every further row is much thinner. The line at $\alpha = (\sqrt5-1)/2$, the reciprocal of the golden ratio, stays well clear: for each $q\le8$ its distance from the nearest $p/q$, times $q^{2.5}$, is at least $0.38$, nearly five times $\gamma$. Lower $\gamma$ and every bar shrinks in proportion, so the covered length goes to zero; that is the theorem on the previous slide, drawn.

The bound, as a number. $\sum_{q\ge1}q^{-1.5} = 2.612$, so $4\gamma\sum q^{-\tau} = 4\times0.08\times2.612 = 0.84$. It is a crude upper bound — the bars overlap and some stick out past $0$ and $1$ — and the actual covered length for $q\le8$ is $0.24$.

Continued fractions in one paragraph. $\alpha = [a_0;a_1,a_2,\dots]$ means $\alpha = a_0 + 1/(a_1 + 1/(a_2+\cdots))$; truncating gives the convergents $p_k/q_k$, the best rational approximations, with $|\alpha - p_k/q_k|\approx1/(q_kq_{k+1})$. A large partial quotient $a_{k+1}$ makes $q_{k+1}$ large and the approximation unusually good — so $\pi$, whose expansion begins $[3;7,15,1,292,\dots]$, is very well approximated by $355/113$. The golden ratio has every partial quotient equal to $1$, so it is approximated as badly as an irrational can be.

17 · perturbation

KAM, stated honestly

KAM theorem — cited

Let $H_\varepsilon = H_0(\vec I) + \varepsilon H_1(\vec I,\vec\theta)$ with $H_0,H_1$ analytic on a domain, and let $H_0$ be non-degenerate: $\det\big(\partial^2H_0/\partial I_i\partial I_j\big)\ne0$. Fix $\tau>n-1$. Then there is $\varepsilon_0>0$ such that for $|\varepsilon|<\varepsilon_0$, every torus whose frequency vector is $(\gamma,\tau)$-Diophantine with $\gamma\ge C\sqrt{|\varepsilon|}$ survives: it is deformed into a nearby invariant torus carrying the same linear flow. The surviving tori fill the phase space up to a set of measure $O(\sqrt{|\varepsilon|})$. Proof: Arnold, Mathematical Methods, appendix 8, in outline; Pöschel, A lecture on the classical KAM theorem, 2001, in full.

what "small" means

The proofs give explicit $\varepsilon_0$, and they are brutal. Hénon computed in 1966 that Arnold's version, applied to the restricted three-body problem, required a mass ratio below $10^{-333}$; Moser's, below $10^{-48}$. Jupiter's is $10^{-3}$. (cited: Dumas, The KAM Story, 2014, §6.)

what it does not give

For $n\ge3$ the surviving tori do not separate the energy surface — an $n$-torus in a $(2n-1)$-dimensional surface has codimension $n-1$. Orbits can then drift through the gaps: Arnold diffusion, cited, and still largely open.

what it does give, for $n = 2$

A $2$-torus in a $3$-dimensional energy surface does separate it. A chaotic orbit trapped between two surviving tori can never leave. Hence a genuine stability theorem — for two degrees of freedom only.

The history, briefly. Kolmogorov announced the result at the 1954 International Congress, with a sketch rather than a proof. Arnold proved the analytic case in 1963, Moser the finitely-differentiable case (originally $C^{333}$, since reduced to $C^{3+}$) in 1962. The technique is a Newton-like iteration in function space whose quadratic convergence beats the deteriorating small denominators — the same idea as Newton's method converging even when each step's linear problem is badly conditioned, provided the error is squared each time.

Why non-degeneracy is needed. $\det(\partial^2H_0/\partial I^2)\ne0$ says the map $\vec I\mapsto\vec\omega(\vec I)$ is a local diffeomorphism, so the frequencies genuinely vary from torus to torus and the Diophantine ones form a set of nearly full measure among them. For the isotropic oscillator $H_0 = I_1+I_2$ this determinant is zero: every torus has the same frequency vector $(1,1)$, which is resonant. So the standard KAM theorem does not apply to the low-energy Hénon–Heiles limit as it stands — an important honesty. Isoenergetic versions of the theorem, which require only that the frequency ratio vary along the energy surface, do apply once the cubic term shifts the frequencies apart, and those are what the picture at $E = 1/100$ illustrates.

The gap between the theorem and the numerics. Hénon's numbers are not a criticism of KAM; they are the price of a proof that must work for every system satisfying the hypotheses. The observed threshold in any particular system is enormously larger — for the standard map the proved bound and the true $K_c = 0.9716$ differ by orders of magnitude the same way. This is the normal relationship between a hard existence theorem and an experiment, and the reason section 2's pictures are not redundant.

17 · perturbation

What becomes of the resonant tori

Poincaré–Birkhoff theorem — cited

The rotation number of an invariant circle is the average fraction of a turn a point on it advances per iteration. Let $P$ be an area-preserving twist map of an annulus — rotation number varying monotonically from one edge to the other — mapping each boundary circle to itself, the two rotating in opposite directions. Then $P$ has at least two fixed points.

Applied to $P^q$: a circle of rotation number $p/q$ — every point back after $q$ steps, having gone round $p$ times — is replaced by $2mq$ points of period $q$, for some integer $m\ge1$, alternately elliptic and hyperbolic; typically $m = 1$. Poincaré 1912, Birkhoff 1913; proof: Brown and Neumann 1977.

Schematic, $q = 5$, in a frame turning with the resonant circle. Its points were all of period $5$; five survive as elliptic centres (gold), each surrounded by curves, and five as hyperbolic points (×) between them, where the chaos is born on the separatrices (red). Count the islands and you have read off the resonance.

The counting argument behind the theorem, in the spirit of unit 02. Consider the unperturbed twist map: the circle with rotation number $p/q$ consists entirely of fixed points of $P^q$. Perturb. Points just inside rotate one way under $P^q$, points just outside the other, by the twist. So on each radial line there is a point whose image lies on the same line. The curve of such points is mapped to itself and must, by area preservation, cross the original curve — and a closed curve crossing another closed curve does so an even number of times. That is the count $2mq$. This is a heuristic, not a proof: making it rigorous is a genuinely topological argument, of the same family as unit 02's index of a vector field around a closed curve, and it is the part cited.

Why alternately elliptic and hyperbolic. Between two consecutive fixed points of an area-preserving map on a curve, the map's rotation sense must reverse; the linearisation at a fixed point has determinant $1$, so its eigenvalues are either on the unit circle or real reciprocals, and the two types must alternate for the index to come out right. The elliptic ones are the island centres; the hyperbolic ones are where Poincaré's tangle lives.

17 · perturbation

Islands round islands

the hierarchy, and why it never ends

Around each new elliptic point the return map is again an area-preserving twist map, with its own resonant circles, island chains and hyperbolic points — at every scale.

Zoomed in, the big island of the $E = 1/8$ panel repeats the whole picture. So no finite computation settles whether a given orbit is regular.

the two fates, side by side

Diophantine torus: survives, slightly bent (KAM).
Resonant torus: replaced by an island chain and a thin chaotic layer (Poincaré–Birkhoff, then the tangle).
The two are interleaved at every scale.

Computed: the large island of the $E = 1/8$ section, from $54$ orbits run for $T$ up to $4000$. Near its edge, a $2/7$ resonance has become a chain of seven islands (blue) — every point of it returns after $7$ crossings, having gone twice round the centre. Enlarged, one of them is nested closed curves round its own elliptic point: the large island's structure again, one scale down.

How the figure was made. Forty orbits started on $p_y = 0$ at $y$ from $0.31$ to $0.50$, run for $T = 1500$ (gold), and fourteen started at $y$ from $0.4562$ to $0.4614$, run for $T = 4000$ (blue); fourth-order Runge–Kutta, step $0.005$, crossings of $x = 0$ located by linear interpolation on the step, energy held to $7\times10^{-11}$. The blue orbit from $y = 0.4590$ returns to within $3\times10^{-4}$ of its start after exactly $7$ crossings, having turned twice round the island centre at $y\approx0.303$: rotation number $2/7$. The two panels use different scales in $y$ and $p_y$ — the box on the left is $0.019$ wide and $0.029$ tall — so the enlarged island is stretched vertically relative to the left panel.

What to look for. On the left, the gold curves are the Liouville–Arnold tori that KAM preserved, bent but whole. The blue ring is where one resonant torus was; Poincaré–Birkhoff replaced it by seven elliptic points with their islands and seven hyperbolic points between them, which at this resolution are invisible gaps in the ring. On the right, the same theorem applies again round the new elliptic point, and with fine enough starting points one finds its own chains.

The self-similarity is not exact. The repeated structure at smaller scales is qualitatively the same but quantitatively different, and the scaling ratios near the critical golden torus are universal constants computed by renormalisation (MacKay 1983). That is a whole subject; it is named here only so that the phrase "at every scale" is not mistaken for poetry.

17 · perturbation

The tangle Poincaré found

homoclinic point

Let $z^*$ be a hyperbolic fixed point of an area-preserving map, with stable set $W^s$ (points approaching $z^*$ forwards) and unstable set $W^u$ (backwards). A point $\ne z^*$ in $W^s\cap W^u$ is homoclinic. If the crossing is transversal, it is a transversal homoclinic point.

one crossing forces infinitely many

$W^s,W^u$ are invariant, so $h\in W^s\cap W^u$ gives $P^n(h)\in W^s\cap W^u$ for every $n$ — infinitely many crossings, accumulating on $z^*$ at both ends. Area preservation forbids the enclosed lobes from shrinking, so they stretch and fold back through every other lobe. $\square$

This is the figure Poincaré declined to draw, and the correction that cost him the price of the journal issue.

Schematic. Past the first crossing $h$, the unstable set must pass through $P(h)$, $P^2(h),\dots$, closer and closer to $z^*$. Each lobe it encloses with $W^s$ is the image of the one before, so all have the same area: as they narrow, they grow taller without bound.

Reading the figure. Only half the tangle is drawn. $W^s$ is shown as the straight green line into $z^*$; the true $W^s$ oscillates about $W^u$ in the same way as it is followed backwards towards $z^*$, so the two sets cross in a mesh that thickens near $z^*$ without limit — the figure Poincaré would not draw. The crossings drawn are $P^n(h)$, approaching $z^*$ geometrically, by the factor $1/\mu$ of the incoming direction. Each lobe is mapped by $P$ onto the next, so area preservation makes their areas equal; the narrowing by $1/\mu$ is paid for by a growth in height by $\mu$. The lobes therefore stretch out across the whole region, through one another — which is how a single transversal crossing manufactures the stretching and folding that a positive exponent measures.

What is honestly not proved anywhere in this unit. That the Hénon–Heiles chaotic region at $E = 1/8$ has positive area. Numerically it plainly does; proving positive-area chaos for a specific smooth Hamiltonian system is extremely hard, and the general belief that such regions exist is not a theorem. This is the one place where the subject's standard of evidence is numerical, and saying so is part of using it honestly.

17 · perturbation

What the tangle forces, and how to detect it

Smale's horseshoe — cited

A transversal homoclinic point implies some power $P^N$ has an invariant Cantor set on which it is the shift on two symbols — hence periodic orbits of every period, a dense orbit, and $\lambda_1>0$ there. Smale 1965; Guckenheimer & Holmes §§5.1–5.3.

Melnikov function — cited

For $\dot z = X_{H_0}(z) + \varepsilon X_{H_1}(z,t)$ with an unperturbed homoclinic loop $z_0(t)$, $$M(t_0) = \int_{-\infty}^{\infty}\{H_0,H_1\}\big(z_0(t),\,t+t_0\big)\,\mathrm dt .$$ A simple zero of $M$ implies a transversal homoclinic point for small $\varepsilon$. $M$ measures, to first order, the gap between $W^s$ and $W^u$. Melnikov 1963.

Why Melnikov matters practically. It converts "is this system chaotic?" into "does this explicit integral have a simple zero?", and the integral can often be done in closed form because the unperturbed homoclinic orbit is known — for the pendulum $\ddot\theta = -\sin\theta$ it is $\theta_0(t) = 4\arctan(e^{t}) - \pi$, running from the upright position at $-\pi$ to the one at $\pi$, and the Melnikov integral can be done by residues (unit 20), producing a $\operatorname{sech}$. The result is a threshold in the perturbation parameter above which chaos is guaranteed. It is the only criterion in this unit that is both rigorous and computable by hand; it is stated rather than proved here because the proof needs the implicit function theorem in a function space and an interchange of limit and integral.

The relation to the exponent. The horseshoe gives $\lambda_1>0$ on a Cantor set, which has zero area — so it does not by itself explain a chaotic sea. What it explains is that positive exponents are present as soon as the manifolds cross at all, arbitrarily weakly. The sea appears when these thin layers, one per resonance, grow enough to merge, which is the subject of the standard map and Chirikov's criterion.

17 · perturbation

The kicked rotor and the standard map

A free rotor, $H = p^2/2$, struck once per unit time by a short torque pulse derived from the potential $K\cos\theta$. During a pulse of duration $\eta\to0$ with fixed total impulse, $\theta$ cannot change but $p$ jumps by $K\sin\theta$; between pulses $\theta$ advances by $p$. Reading $(\theta_n,p_n)$ just before the $n$-th kick:

the standard map $$\begin{aligned}p_{n+1} &= p_n + K\sin\theta_n,\\ \theta_{n+1} &= \theta_n + p_{n+1}\ (\mathrm{mod}\ 2\pi).\end{aligned}$$

it is area-preserving

$\dfrac{\partial(\theta_{n+1},p_{n+1})}{\partial(\theta_n,p_n)} = \begin{pmatrix}1+K\cos\theta_n&1\\ K\cos\theta_n&1\end{pmatrix}$, determinant $(1+K\cos\theta_n) - K\cos\theta_n = 1$ ✓ — as the previous section's theorem demands of any Hamiltonian return map.

why this map and not another

The simplest area-preserving twist map with one parameter; every two-degree-of-freedom system looks like it near a resonance; and its critical parameter is known to nine digits. It is the drosophila of the subject.

the primary resonances

$K = 0$: $p$ is conserved, and the invariant curves are the horizontal lines $p = $ const, with rotation number $p/2\pi$. Resonances sit where $p = 2\pi m$, $m\in\mathbb Z$: there, $\theta$ returns exactly and every point is fixed.

$K>0$: near $p = 0$ the map is close to the flow of the pendulum $H_{\text{eff}} = \tfrac12p^2 + K\cos\theta$, whose $\dot p = K\sin\theta$. Its separatrix leaves the hyperbolic point $(0,0)$ at energy $K$; at the island centre $\theta = \pi$, $\tfrac12p^2 - K = K$, a half-width $$\Delta p = 2\sqrt K.$$

The limit, done carefully. Replace the impulsive kick by a torque $K\sin\theta\cdot f_\eta(t) = -\partial_\theta(K\cos\theta)\,f_\eta(t)$ where $f_\eta$ is a bump of width $\eta$ and unit integral supported at the integers. Over one bump, $\Delta p = \int -\partial_\theta(K\cos\theta) f_\eta = K\sin\theta\cdot(1+O(\eta))$, since $\theta$ changes by $O(\eta p)$ during the bump; and $\Delta\theta = O(\eta)$. Letting $\eta\to0$ gives the map exactly. Nothing distributional is needed — the delta function is convenient notation for this limit, and the notation is legitimised later in the course.

Where the pendulum approximation comes from. Over one step, if $p$ and the change in $\theta$ are small, the map is the Euler step of $\dot\theta = p$, $\dot p = K\sin\theta$, which is Hamiltonian with $H_{\text{eff}} = \tfrac12p^2 + K\cos\theta$. The approximation is good when $\sqrt K\ll1$, which is exactly the regime where the widths are small — so the criterion built on it is being used at the edge of its validity, and Chirikov's factor-of-$2.5$ discrepancy should be no surprise.

The sign conventions vary. Some authors write $p_{n+1} = p_n + K\sin\theta_n$ and others $-K\sin\theta_n$, and some put the mod on $p$ as well (making the phase space a torus rather than a cylinder). The critical constant $K_c = 0.971635\dots$ is the same in all of them; the location of the elliptic fixed point is not — in this deck's convention it is $(\pi,0)$, as problem P5 finds. This deck keeps $p$ unbounded, because the question of interest — can an orbit travel from $p=0$ to $p=2\pi$? — is meaningless on the torus.

17 · perturbation

The standard map, iterated

Computed: $32$ orbits of $160$ kicks each, coloured by their exponent measured over $2000$ kicks — gold below $0.02$, red above — and one orbit of $2500$ kicks started next to the hyperbolic point $(0,0)$, red; $p$ is drawn mod $2\pi$. At $K = 0.6$ the red orbit stays in a thin layer along the separatrix of the island at $(\pi,0)$, and gold curves cross the whole square: no orbit can climb from $p\approx0$ to $p\approx2\pi$. At $K_c$ the last crossing curve is about to break. At $K = 1.4$ red covers most of the square.

What to look for. The three things of the section slides, in one map: invariant curves that run all the way across (the KAM tori of the free rotor, bent), island chains round elliptic points (the big one at $(\pi,0)$ is the primary resonance; smaller chains sit between), and chaotic layers born at the hyperbolic points. A curve that crosses the square from $\theta = 0$ to $\theta = 2\pi$ is a wall: $p$ cannot pass it, since the curve is invariant and the map is continuous and area-preserving. Global transport in $p$ is possible exactly when every such wall has gone.

Why $p$ mod $2\pi$ is allowed in the picture. The map commutes with $p\mapsto p+2\pi$: adding $2\pi$ to $p_n$ adds $2\pi$ to $p_{n+1}$ and changes $\theta_{n+1}$ by $2\pi$, which is nothing. So the phase portrait is periodic in $p$, and one period shows all of it. The question asked on the last-torus slide — can an orbit get from $p = 0$ to $p = 2\pi$? — is about the unreduced $p$, which is why the deck keeps $p$ unbounded in the dynamics and reduces it only to draw.

How it was made. Starting points $(\pi, p_0)$ for $22$ values of $p_0$ evenly across $(0,2\pi)$ and $(0,p_0)$ for $10$ more, each coloured by its exponent over $2000$ kicks, computed with the product of Jacobians as in problem P6: the regular ones measure below $0.004$ — the $\ln N/N$ decay — and the chaotic ones above $0.05$, so the threshold $0.02$ is not delicate. None of the $32$ is chaotic at $K = 0.6$, $8$ are at $K_c$ and $14$ at $K = 1.4$. The long red orbit starts at $(0.001, 0.001)$. The figure is evidence of the qualitative change only; where exactly the last curve breaks is the subject of the coming slides, and a picture of $160$ kicks cannot settle it.

17 · perturbation · example

Chirikov: when do resonances overlap?

worked example 4 · the overlap criterion, to a number

The idea. Each resonance traps orbits in a band of half-width $\Delta p$. While the bands are disjoint the motion stays local; once neighbouring bands touch, an orbit can pass from one to the next and travel across the whole cylinder.

Step 1 — the widths. $\Delta p = 2\sqrt K$, from the pendulum separatrix. Step 2 — the spacing. Primary resonances sit at $p = 2\pi m$, so $\delta p = 2\pi$.

Step 3 — the criterion. Chirikov's overlap criterion is $2\Delta p\ge\delta p$, that is $4\sqrt K\ge2\pi$, giving $K\ge K_{\text{Chirikov}} = \pi^2/4 = 2.4674$.

Step 4 — the truth. Greene's residue method gives $K_c = 0.9716354$: the criterion is high by a factor $\mathbf{2.54}$.

checks

Dimensions. $\Delta p\sim\sqrt K$ looks wrong until the kick period $T$ is restored: $\Delta p = 2\sqrt{K/T}$ against a spacing $2\pi/T$, so the criterion reads $4\sqrt{K/T}\ge2\pi/T$ and $K_c$ comes out independent of $T$ ✓.

Limit $K\to0$. Widths $\to0$, no overlap, every curve survives — the integrable rotor ✓.

Why a factor of 2.5 counts as a success. The criterion takes ten minutes, uses no computer, and applies to any system where the resonance widths can be estimated — including ones with no exact answer available. Greene's method takes a careful numerical study of periodic orbits of increasing period and applies to one map at a time. An order-of-magnitude criterion that is right within a factor of $2.5$ is the standard tool of the field, and the discrepancy is systematic and in a known direction, which makes it usable: treat $K_{\text{Chirikov}}$ as an upper bound and expect the truth at roughly $0.4$ of it.

Greene's method, in outline, since the number is quoted. For each rational $p/q$ near the golden mean, find the periodic orbit of $P$ with that rotation number and compute the residue $R = \tfrac14(2 - \operatorname{tr}DP^q)$, which is $0$ for a parabolic orbit, positive-and-small for elliptic and negative for hyperbolic. Greene's observation, later substantially proved by MacKay, is that the golden torus exists precisely when the residues of the convergent orbits tend to $0$, and is destroyed when they diverge. The threshold between the two behaviours is $K_c = 0.971635406$, obtained by bisection on $K$. Cited: Greene, J. Math. Phys. 20 (1979) 1183; MacKay, Renormalisation in Area-Preserving Maps, 1993.

17 · perturbation · example

Two resonances, about to touch

Pendulum separatrices $p = 2\pi m \pm 2\sqrt K\,\sin(\theta/2)$ for $m = 0, 1$. At $K_c$ (dashed) a gap of $2.34$ separates them, yet the real map has lost its last crossing curve. At $K = \pi^2/4$ they touch at $(\pi,\pi)$.

why it overestimates, always

It counts only the primary resonances. The higher-order ones between them, each with its own chaotic layer, help bridge the gap and pull the estimate down towards the truth. (cited: Chirikov, Phys. Rep. 52 (1979); Lichtenberg & Lieberman, §4.1b.)

Compare the $K_c$ portrait two slides back: the primary bands alone are far from touching, but the space between them is already crowded with smaller island chains and their chaotic layers.

Reading the figure. Each eye-shaped band is the region inside one pendulum separatrix, the curve $\tfrac12(p - 2\pi m)^2 + K\cos\theta = K$, i.e. $p = 2\pi m\pm2\sqrt K\sin(\theta/2)$ for $0\le\theta\le2\pi$. Its half-height at $\theta = \pi$ is $\Delta p = 2\sqrt K$, and the centres are $\delta p = 2\pi$ apart. Chirikov's criterion is the moment the two bands meet, $2\cdot2\sqrt K = 2\pi$. At the true critical value the gap between the dashed bands is $2\pi - 4\sqrt{0.9716} = 2.34$, more than a third of the spacing: the primary resonances alone are nowhere near touching.

The same criterion in the running example. The reason the Hénon–Heiles section goes from $0.1\%$ chaotic at $E = 1/12$ to $52\%$ at $E = 1/8$ is the same overlap. The island chains visible in the $E = 1/12$ panel each carry a chaotic layer too thin to see; raising the energy widens them; when they touch, the separate layers become one sea. The energy at which this happens for a given pair of resonances could in principle be estimated by Chirikov's criterion with the widths computed from secular perturbation theory — and would come out, as here, a factor of a few too high.

17 · perturbation

The last torus, numerically

the experiment

Start $400$ points on a $20\times20$ grid in the band $0<p<2\pi$ and iterate. If an invariant curve spans the cylinder, no orbit can leave the band. Record the smallest $K$ at which some orbit reaches $p>4\pi$ or $p<-2\pi$ within $N$ iterations.

$N$	threshold
$10^{5}$	$1.020$
$10^{6}$	$0.994$
$10^{7}$	$0.983$
$\infty$	$0.9716$

why it converges from above, and slowly

Above $K_c$ the broken torus does not vanish: it becomes an invariant cantorus, a Cantor set with gaps, and orbits leak through the gaps at a rate that vanishes as $K\downarrow K_c$ like $(K-K_c)^{3.01}$ (cited: MacKay, Meiss and Percival 1984). So any finite experiment sees no transport just above threshold and reports a value too high.

the check that the picture is right

$(1.020-K_c)/(0.983-K_c) = 4.26$, and $4.26^{3.01} = 78$, against the hundredfold change in $N$ the exponent predicts. The same order, within the crudeness of "first escape in $N$ steps" ✓.

The surviving curve is the one with golden rotation number, as the Diophantine slide predicted.

Why this is the right way to look for a torus. Not by staring at a plot: an orbit confined to a band looks the same whether it is confined by a curve or is simply slow. The transport test asks a yes/no question with a definite answer, and its failure mode — reporting "no transport" when the flux is merely tiny — is in a known direction and quantified by the scaling law above. Designing the experiment so that its errors all point one way is the difference between a measurement and an anecdote.

The cantorus is the object that makes the transition continuous. Below $K_c$ there is an invariant circle and the flux is exactly zero. Above, there is an invariant Cantor set with the same rotation number and the flux is positive but tiny. Nothing jumps; what changes is the topology of the invariant set. This is the picture Aubry and Mather constructed independently around 1982, and it is why the transition can be both sharp and continuous at once.

Doing this yourself. Problem P9 works with exactly this table, and the point of it is the discipline: report the iteration budget alongside the threshold, because a threshold without a budget is not a number. The table above changes by $4\%$ between $N = 10^5$ and $N = 10^7$ — a difference far larger than any plausible arithmetic error, and invisible to anyone who ran the experiment once.

section 5

Out in the solar system

Where the resonances are observable, and where the exponent has been measured.

17 · consequences · example

The Kirkwood gaps

worked example 5 · resonances you can see

The claim. An asteroid whose period is a simple rational multiple of Jupiter's sits on a resonant torus. Those break; eccentricities then grow chaotically until the body meets a planet or the Sun, so the belt should show Kirkwood gaps at the resonant semi-major axes.

Step 1. A $p\!:\!q$ mean-motion resonance means $p$ asteroid orbits to $q$ of Jupiter's: $T_a = (q/p)T_{\mathrm J}$. Step 2. Kepler's third law (unit 08), $T^2\propto a^3$, gives $a = a_{\mathrm J}(q/p)^{2/3}$. Step 3. Put in $a_{\mathrm J} = 5.2044$ AU.

$p\!:\!q$	$(q/p)^{2/3}$	$a$ / AU	observed gap
$4\!:\!1$	$0.39685$	$2.065$	$2.06$
$3\!:\!1$	$0.48075$	$2.502$	$2.50$
$5\!:\!2$	$0.54288$	$2.825$	$2.82$
$7\!:\!3$	$0.56844$	$2.958$	$2.96$
$2\!:\!1$	$0.62996$	$3.279$	$3.27$

checks

Dimensions. $(q/p)^{2/3}$ is dimensionless, $a_{\mathrm J}$ is a length ✓.

Limit. $1\!:\!1$ gives $a = a_{\mathrm J}$ — the Trojans, which are a crowd and not a gap. Resonance is not the same thing as instability ✓.

Self-consistency. $a_{\mathrm J}^{3/2} = 11.873$ yr against $T_{\mathrm J} = 11.862$ yr, $0.1\%$ ✓.

A note on which number is being predicted. The table predicts where the resonance is, and that is pure Kepler — no chaos in it. What chaos supplies is the reason a resonance should be empty rather than merely special. The agreement to three digits is therefore a check on the arithmetic, not evidence for KAM; the evidence for the mechanism is Wisdom's computation of the eccentricity's behaviour, and the fact that the gaps are at the resonances where the chaotic zone is wide and not at the ones where it is narrow.

The data. $a_{\mathrm J} = 5.2044$ AU and $T_{\mathrm J} = 11.862$ yr are Jupiter's mean elements; the observed gap positions are quoted to the precision the histogram of asteroid semi-major axes supports, which is about $0.01$ AU — the gaps have width, typically $0.02$ to $0.1$ AU, and a centre quoted to three digits is already more than the data carry.

17 · consequences · example

Where the gaps fall, and where they do not

Steps 1–2 of the worked example as a curve: $a = a_{\mathrm J}(T/T_{\mathrm J})^{2/3}$, with the commensurabilities marked. Red: the five gaps of the table, between $2.06$ and $3.28$ AU. Green: the $3\!:\!2$ and $1\!:\!1$ resonances at $3.97$ and $5.20$ AU, which hold families instead.

found 1866, explained 1983

Kirkwood found the gaps in a catalogue of fewer than a hundred asteroids. Why a resonance empties took a century: Wisdom showed the $3\!:\!1$ eccentricity jumps chaotically until the asteroid crosses Mars. (cited: Wisdom, Icarus 56 (1983).)

Kepler places every resonance; it cannot say which ones empty. That needs the dynamics near each resonance — which island the asteroids sit in.

Reading the figure. The horizontal axis is the asteroid's period as a fraction of Jupiter's, $q/p$ for a $p\!:\!q$ resonance; the curve converts it to a semi-major axis by Kepler's third law. Every rational on the axis is a resonance, as on the Diophantine slide, and only the low-order ones are strong enough to matter. The figure carries no asteroid data: it shows where the theory puts the resonances, and the colours record the observed outcome at each.

Why some resonances crowd instead of clearing. The $3\!:\!2$ resonance at $a = 5.2044\cdot(2/3)^{2/3} = 3.97$ AU holds the Hilda family, and the $1\!:\!1$ holds the Trojans. Both are protected: the resonant angle librates rather than circulating, and the libration keeps the asteroid away from Jupiter at every conjunction. The difference between a gap and a family is which island of the resonance the body sits in — elliptic centre or chaotic separatrix — which is precisely the distinction section 2 drew on the Hénon–Heiles section.

17 · consequences

Is the solar system stable?

what is actually known

1 KAM does not apply: the mass ratios are far above every proved threshold, and with $n\ge3$ surviving tori would not separate the energy surface anyway.

2 The inner solar system has a measured positive Lyapunov exponent. Laskar's integrations give a Lyapunov time near $5$ million years. (cited: Laskar, Nature 338 (1989) 237; Sussman & Wisdom 1992.)

3 Over $5$ Gyr, about $1\%$ of integrated solutions have Mercury's eccentricity grow large enough for a collision with Venus or the Sun. (cited: Laskar & Gastineau, Nature 459 (2009) 817.)

what a $5$-Myr Lyapunov time costs

Over $100$ Myr, an error is multiplied by $e^{100/5} = e^{20} = 4.85\times10^{8}$. So an initial uncertainty of one metre in a planet's position becomes $$4.85\times10^{8}\ \mathrm m = 4.85\times10^{5}\ \mathrm{km},$$ about $1.3$ times the Earth–Moon distance. Positions beyond roughly $100$ Myr are not predictable by any integration, however good.

Which claim this is. Not that the planets move away — that the phase is lost. Statistical statements survive, which is why item 3 is a frequency.

the honest summary

Marginally stable: chaotic on $10^7$ years, apparently intact on $10^9$, with a small real chance of losing a planet. Newton's laws determine all of it, and the phase of none of it can be computed.

How a Lyapunov exponent is measured for the solar system. Exactly as in section 3: integrate the variational equation alongside the equations of motion and renormalise. The difficulty is that the integration itself must be trustworthy for $10^8$ years, which is why these calculations use symplectic integrators — unit 14's remark that a good integrator for mechanics should preserve the symplectic structure, cashed in. A non-symplectic method drifts in energy and produces a spurious exponent.

What Laplace and Lagrange proved, and did not. Their secular theory shows that to first order in the masses the semi-major axes have no secular drift — the planets do not spiral in or out. That result is correct and still used. What it does not control is the eccentricities and inclinations at higher order, which is where the chaos lives, and it is the small denominators of section 4 that block the extension. The nineteenth century's confidence in stability rested on stopping the series at the order where it still looked convergent.

An honest caveat about item 3. The $1\%$ is a frequency among numerically integrated solutions with slightly different initial conditions, not a probability in any physical sense — the actual solar system has one history. It is the correct thing to compute when the Lyapunov time is a thousandth of the integration time, and the correct way to read it is: the equations plus present-day data do not exclude Mercury's loss, and they do exclude, say, the Earth's, in every solution computed.

17 · closing

The running example, closed

what was asked in 1964

Does Hénon–Heiles have a second integral? The answer this unit produced:

At $E = 1/100$ the section is nested curves; the system behaves as though it had one, and KAM explains why — most tori of the integrable limit survive a small perturbation.

At $E = 1/8$ a single orbit's crossings fill the chaotic sea, $52\%$ of the section, and its exponent is $0.044$, while an orbit a quarter of the section away has exponent zero, confirmed by a growth law with slope $0.997$.

No smooth function independent of $H$ can be constant on both. There is no second integral, and the numbers say so rather than the picture.

the three claims, side by side

claim	status
unique solutions	proved, unit 01
continuous in the data	proved, unit 01
$\lambda_1 = 0.044 > 0$	measured, this unit
not predictable past $t\sim500$	follows, arithmetically

Every row is true at once, and the only thing needed to see that is the observation that unit 01's $\delta$ depends on the horizon $T$.

what was not settled

That the chaotic region has positive area is not proved here or anywhere. Numerics show it; nobody has a proof for a specific smooth Hamiltonian system of this kind.

The $t\sim500$ figure. Taking the initial data as known to one part in $10^{9}$, as a good physical measurement might be, the uncertainty reaches order unity at $t = \ln(10^{9})/0.044 = 471$ — about $75$ orbits. Even with perfect double-precision data, uncertain at $2.2\times10^{-16}$, it is $t = 819$. Either way a few hundred time units, and no improvement in the integrator changes it: nine more digits buy $471$ more time units.

What Hénon and Heiles themselves concluded. That a third isolating integral exists for low energies and not for high ones, and that the transition is sharp. Their paper reports the fraction of the section covered by curves as a function of energy, which is the ancestor of this unit's table. They were computing on an IBM 7090 in 1964, and their pictures have the same features as the ones here.

Where this leaves the course. Part B's thesis is that a reformulation is worth exactly what it makes obvious. The Hamiltonian formulation made this unit possible: the section, its area preservation, the symmetry of the Lyapunov spectrum, action–angle variables and the whole perturbation theory are statements about the symplectic structure, and none of them is visible in Newton's equations or even in Lagrange's. That is the best argument for Part B that the course can make, and it is why this branch unit sits here rather than anywhere else.

17 · closing

What to keep

object	what it decides
$n$ integrals in involution	integrable; compact level set is a torus
action–angle variables	the flow is linear, and the frequencies are $\partial H/\partial\vec I$
Poincaré section	turns a 4-dimensional flow into an area-preserving map of a page
$\det DP = 1$	no attractors; fixed points elliptic or hyperbolic only
$\lambda_1$	$>0$ is chaos; $1/\lambda_1$ is the prediction horizon per $e$-fold
$\vec k\cdot\vec\omega$	small denominators: which tori are fragile
Diophantine condition	which survive (KAM), the golden one last
Chirikov overlap	a hand estimate of the transition, high by a factor of a few

three habits, not results

Quote the budget with the number. A Lyapunov exponent without its $T$, or a threshold without its iteration count, is not a measurement.

Check energy drift before believing a picture. Every run in this unit held its energy to a relative $10^{-7}$ or better.

Say which direction the evidence runs. A filled area refutes an integral; curves prove nothing.

the sentence to be able to say

"Unit 01 gives, for each $\varepsilon$ and each horizon $T$, a $\delta$. A positive Lyapunov exponent says $\delta$ must shrink like $e^{-\lambda T}$. Nothing in the theorem forbids that, because $\delta$ was allowed to depend on $T$."

If one line survives the year, make it the last box. It is the checkpoint, it is the reason the unit is in the course, and it is the thing that is most often got wrong in print.

What was used from earlier units. Picard–Lindelöf, Gronwall and continuous dependence from unit 01; flows, the one-parameter group property and linearisation from unit 02; the chain rule and Clairaut from unit 03; path-independence of a circulation and the inverse and implicit function theorems from unit 04; Kepler's third law from unit 08; and from unit 14 the Poisson bracket and the Jacobi identity, canonical transformations and generating functions, the symplectic condition, the variational equation it cites, Liouville's formula and Liouville's theorem. Two results are named from units this one does not need: the change-of-variables formula (unit 05), for turning $\det DP = 1$ into preserved area, and residues (unit 20), in a remark on Melnikov. Nothing else.

What this unit deliberately does not do. Ergodic theory, mixing, measure-theoretic entropy, strange attractors, dissipative chaos and the Lorenz system, bifurcation cascades and Feigenbaum's constants, renormalisation of area-preserving maps, and Arnold diffusion. Each is a unit of its own; none is needed to say honestly what chaos in a Hamiltonian system is.

17 · closing

Reading

Arnold, Mathematical Methods of Classical Mechanics, §49–50. Liouville–Arnold and action–angle variables, exactly at this level. §49 is self-contained; §50 uses differential forms from chapter 7 — read its statements and return for the proofs after unit 34.

Lichtenberg and Lieberman, Regular and Chaotic Dynamics, ch. 3–4. The working physicist's account: sections, resonances, the standard map, Chirikov's criterion with the refinements. §4.1b is where the improved overlap estimate comes from. It uses Fourier series freely, which arrive later in this course; the arguments survive reading the series as finite sums.

Ott, Chaos in Dynamical Systems, ch. 4 and 7. Lyapunov exponents and Hamiltonian chaos, with the clearest statement of what a numerical exponent does and does not establish.

Barrow-Green, Poincaré and the Three Body Problem. The prize, the error and the correction, from the archives. Chapters 3–5 need no mathematics beyond this unit.

Dumas, The KAM Story. What the theorem says, what its constants really are, and which popular statements of it are wrong. Non-technical throughout.

Deliberately not listed: texts whose first chapter is a dissipative system — logistic map, Lorenz — since the area preservation of this unit is what makes Hamiltonian chaos a different subject.

On Arnold's book. It is the best book in the subject and the least forgiving. Read §49 with a pencil and reconstruct the lattice argument yourself; it is the same argument as this unit's proof-geometry slide, at greater length. Appendix 8 on KAM is worth reading once for the shape of the proof even if the details do not land.

For the Melnikov method, Guckenheimer and Holmes, Nonlinear Oscillations, Dynamical Systems and Bifurcations of Vector Fields, §4.5–4.6, with worked examples. It uses the stable manifold theorem, which this course does not prove.

For the numerics, Hairer, Lubich and Wanner, Geometric Numerical Integration, ch. 1–2, on why a symplectic integrator behaves differently from an accurate one. That is unit 14's closing remark made precise, and it is what the solar-system calculations of section 5 rest on.

17 · checkpoint

Done when

checkpoint

You can, on paper and without the deck,

compute a largest Lyapunov exponent for a given map — write the variational equation, run Benettin's renormalisation, and quote the number with the budget it was measured over;
produce a Poincaré section that constitutes evidence for chaos rather than an impression of it: a single orbit, a stated energy drift, and an exponent to go with it; and
explain why a positive exponent does not contradict uniqueness, quoting unit 01's theorem precisely — including the place where $\delta$ is allowed to depend on the horizon.

Falsifiable: an exponent quoted without a time, a section shown without a drift figure or offered as proof of integrability, or an explanation that appeals to rounding error or to "practical" uncertainty rather than to the quantifier, each fails. Record it with ./mr done 17.

Problems

./mr problems 17 — 11 problems, about five hours, hints and solutions folded under each.

Nothing needs this unit. The spine continues at 18 the wave equation; Part B's own thread ends at 16 tensors.

Try the checkpoint cold. For the first part, take the standard map: the variational equation is the product of the Jacobians $\begin{pmatrix}1+K\cos\theta_n&1\\ K\cos\theta_n&1\end{pmatrix}$, and problem P6 walks through it. At $K = 6$ the answer is $1.14$, against the large-$K$ estimate $\ln(K/2) = 1.10$; at $K = 10$ it is $1.620$ against $1.609$. At $K = 2$ the estimate gives $0$ and the truth is about $0.44$ — a reminder that an asymptotic formula quoted outside its range is not a check on anything.

For the third part, the failure mode to watch for in your own answer is an appeal to practicality — "in practice you cannot measure the initial condition well enough". That is true and is not the point. The point is a quantifier: unit 01 promises a $\delta$ for each $T$, and never a single $\delta$ good for all $T$. If your explanation would still work with infinitely precise measurement and exact arithmetic, it is the right one.

If the second part stalls, the missing piece is usually the direction of the inference. Write the argument out: if a second integral existed on an open set, then every orbit's closure in the section would lie in a level curve, which has empty interior; an orbit whose crossings are dense in a disc therefore refutes it. Curves, by contrast, refute nothing — which is why the exponent has to be quoted alongside.

Integrability and chaos: determinism without predictability

Determinism was the easy part

What it means to solve a system

A prize that cost more than it paid

Integrals in involution

Complete integrability, defined

Liouville and Arnold: the level set is a torus

The torus, and the straight-line flow on it

Why a torus, and nothing else

The lattice of return times

Hénon and Heiles ask for a third integral

The Hénon–Heiles potential, drawn

Looking at a flow through a plane

The section and its return map

The section for Hénon–Heiles

The return map preserves area

Elliptic and hyperbolic, drawn

What integrability looks like on a section

Four energies

Reading the four panels

Measuring sensitivity

The linearised flow

The Lyapunov exponent

What a Hamiltonian spectrum must look like

Stretch and contract, drawn

Computing it without overflowing

Stretch, record, rescale

The two orbits, measured

Gronwall's revenge

Eight orbits per digit

What a positive exponent does not say

Perturbing an integrable system

The small denominators

Rational frequencies are the enemy, and there are few of them

The resonances, drawn on a line

KAM, stated honestly

What becomes of the resonant tori

Islands round islands

The tangle Poincaré found

What the tangle forces, and how to detect it

The kicked rotor and the standard map

The standard map, iterated

Chirikov: when do resonances overlap?

Two resonances, about to touch

The last torus, numerically

Out in the solar system

The Kirkwood gaps

Where the gaps fall, and where they do not

Is the solar system stable?

The running example, closed

What to keep

Reading

Done when

Problems

Next