Chaos

Sensitive dependence, maps, strange attractors and fractals.

In this chapter 7

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Chaotic dynamics

Chaotic dynamics

Application

The Lorenz equations

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Collects what can be established about the Lorenz system by the standard tools — its symmetry, its constant rate of phase-volume contraction, a trapping ellipsoid, the three equilibria and their eigenvalues, global stability of the origin for $r < 1$ by a Liapunov function, and the threshold $r = \sigma(\sigma+b+3)/(\sigma-b-1)$ at which the convective equilibria lose stability — and shows why, for $\sigma = 10$ , $b = 8/3$ , $r = 28$ , these results together leave no possibility except a strange attractor.

Prerequisites: Lyapunov functions, Linearization and the Hartman-Grobman theorem, The Hopf bifurcation, Bifurcations in one dimension, The divergence theorem, Eigenvalues and eigenvectors

Definition(The Lorenz system, attractors and strange attractors)AdaptedAdapted from Strogatz (1994), §9.0, p. 301, §9.2, p. 311, and §9.3, pp. 324-325; Hirsch, Smale & Devaney (2013), §14.1, p. 306, and §14.3, p. 312

The Lorenz system is the three-dimensional autonomous system

\dot x = \sigma(y - x), \qquad \dot y = r x - y - x z, \qquad \dot z = x y - b z,

with parameters $\sigma, r, b > 0$ . Here $\sigma$ is the Prandtl number and $r$ the Rayleigh number; $b$ has no name and is fixed by the geometry of the convection cell. Hirsch, Smale and Devaney assume throughout, in addition, that $\sigma > b + 1$ , and so do we wherever the stability of the convective equilibria is in question. Write $L$ for the vector field, so that the system is $\dot X = L(X)$ .

Where it comes from. Lorenz obtained the system in 1963 by truncating a model of a fluid layer heated from below and cooled from above to three Fourier amplitudes: $x$ measures the strength of the convective overturning, $y$ the horizontal and $z$ the vertical temperature variation. The same three equations describe, exactly rather than approximately, the Malkus-Howard waterwheel — a wheel with leaky cups on its rim, fed with water from above — in which the mass distribution around the rim is expanded in a Fourier series and all but three amplitudes decouple. In that model $r$ is again a Rayleigh number, measuring the driving (gravity and inflow) against the dissipation (leakage and damping).

Attractors. Following Strogatz, a closed set $A$ in phase space is an attractor if

$A$ is invariant: every trajectory starting in $A$ remains in $A$ for all time;
$A$ attracts an open set: there is an open $U \supseteq A$ such that the distance from $X(t)$ to $A$ tends to $0$ as $t \to \infty$ for every $X(0) \in U$ (the largest such $U$ is the basin of attraction); and
$A$ is minimal: no proper subset of $A$ has both properties.

An attractor is strange if the motion on it shows sensitive dependence on initial conditions, as that is defined in Lyapunov exponents and sensitive dependence.

Application(Elementary properties of the Lorenz system)AdaptedAdapted from Strogatz (1994), §9.2, pp. 311-316; Hirsch, Smale & Devaney (2013), §14.2, pp. 308-311

Let $\sigma, b, r > 0$ .

Symmetry. The reflection $S(x, y, z) = (-x, -y, z)$ maps solutions to solutions: if $(x(t), y(t), z(t))$ solves the system, so does $(-x(t), -y(t), z(t))$ . The $z$ -axis is invariant, and on it $\dot z = -bz$ , so every solution starting on it tends to the origin.
Volume contraction. $\nabla \cdot L = -(\sigma + 1 + b)$ , a negative constant. Hence if $D$ is a region with smooth boundary and $V(t)$ is the volume of its image under the flow,
$\dot V = -(\sigma + 1 + b)\,V, \qquad\text{so}\qquad V(t) = V(0)\,e^{-(\sigma + 1 + b)t}.$
Every phase volume shrinks to zero exponentially, at a rate independent of position.
A trapping region. Put $W(x, y, z) = r x^2 + \sigma y^2 + \sigma (z - 2r)^2$ , whose level sets are ellipsoids centred at $(0, 0, 2r)$ . There is a $\nu^* > 0$ such that $\dot W < 0$ on every level set $W = \nu$ with $\nu \ge \nu^*$ . Consequently every solution enters the solid ellipsoid $W \le \nu^*$ in finite time and stays there.
Equilibria. The origin is an equilibrium for all parameter values. For $r > 1$ there are two more,
$C^{\pm} = \bigl(\pm\sqrt{b(r-1)},\ \pm\sqrt{b(r-1)},\ r - 1\bigr),$
which are exchanged by the symmetry $S$ and which coalesce with the origin in a pitchfork bifurcation as $r \downarrow 1$ .
The origin. Its eigenvalues are $-b$ and
$\lambda_{\pm} = \tfrac12\Bigl[-(\sigma + 1) \pm \sqrt{(\sigma+1)^2 - 4\sigma(1 - r)}\Bigr].$
For $0 \le r < 1$ both $\lambda_{\pm}$ are negative, so the origin is a sink; for $r > 1$ one is positive, so the origin is a saddle with a two-dimensional stable surface and a one-dimensional unstable curve.
Global stability for $r < 1$ . If $r < 1$ , every solution tends to the origin. In particular there are no closed orbits and no chaos below $r = 1$ .
The convective equilibria. Assume $\sigma > b + 1$ . The eigenvalues at $C^{\pm}$ are the roots of
$\lambda^3 + (1 + b + \sigma)\lambda^2 + b(\sigma + r)\lambda + 2b\sigma(r - 1) = 0,$
and $C^{\pm}$ are sinks for
$1 < r < r_H = \sigma\,\frac{\sigma + b + 3}{\sigma - b - 1}.$
At $r = r_H$ a Hopf bifurcation occurs. It is subcritical: the limit cycles involved are unstable and exist for $r < r_H$ , so for $r$ slightly above $r_H$ there is no attractor near $C^{\pm}$ .
What volume contraction excludes. The system has no quasiperiodic solutions, no repelling equilibria and no repelling closed orbits. Hence for $r > r_H$ no equilibrium and no closed orbit can be an attractor near $C^{\pm}$ , while by part 3 no trajectory can escape to infinity.

Idea of the proofOriginal

Parts 1, 2, 4 and 5 are direct computations. The two that carry the argument are the Liapunov function and the trapping ellipsoid, and both work the same way: choose a positive definite quadratic form, differentiate it along the flow, and watch the cubic terms cancel. They cancel because the only nonlinearities are $-xz$ in $\dot y$ and $+xy$ in $\dot z$ , and the single product $xyz$ they generate appears twice with opposite signs whenever $y$ and $z$ are weighted equally. What is left is a quadratic form, and the whole question becomes whether that form is definite — an exercise in completing the square. Part 7 then locates the stability boundary not by finding the roots of the cubic but by asking when a purely imaginary root is possible, which splits the cubic into two real equations for one unknown $\omega^2$ and leaves one condition on $r$ .

ProofAdaptedAdapted from Hirsch, Smale & Devaney (2013), §14.2, pp. 308-311; Strogatz (1994), §9.2, pp. 312-316

Symmetry. Substituting $(-x, -y, z)$ into the right-hand side gives $(-\sigma(y-x), -(rx - y - xz), xy - bz)$ , which is $(-\dot x, -\dot y, \dot z)$ ; so the reflected curve satisfies the same system. On $x = y = 0$ the first two equations give $\dot x = \dot y = 0$ , so the $z$ -axis is invariant, and there $\dot z = -bz$ has solution $z(t) = z(0)e^{-bt} \to 0$ .
Volume contraction. Adding the three diagonal derivatives,
$\nabla\cdot L = \frac{\partial}{\partial x}\sigma(y-x) + \frac{\partial}{\partial y}(rx - y - xz) + \frac{\partial}{\partial z}(xy - bz) = -\sigma - 1 - b .$
By Liouville's theorem, $\dot V = \int_{D(t)} \nabla\cdot L\,dV$ , and since the integrand is the constant $-(\sigma+1+b)$ this is $-(\sigma+1+b)V$ . Integrating the resulting linear equation gives the stated exponential. (Liouville's theorem is itself the divergence theorem applied to the volume swept out by the boundary: in time $dt$ a patch $dA$ of the boundary sweeps $(L\cdot n)\,dt\,dA$ , so $\dot V = \oint L\cdot n\, dA = \int \nabla\cdot L\,dV$ ; see the divergence theorem.)
The trapping ellipsoid. Differentiate $W = rx^2 + \sigma y^2 + \sigma(z-2r)^2$ along the flow:
$\dot W = 2rx\,\sigma(y - x) + 2\sigma y\,(rx - y - xz) + 2\sigma(z - 2r)(xy - bz).$
The $xy$ terms contribute $2\sigma r xy + 2\sigma r xy - 4\sigma r xy = 0$ , and the cubic terms contribute $-2\sigma xyz + 2\sigma xyz = 0$ . What remains is
$\dot W = -2\sigma\bigl[rx^2 + y^2 + bz^2 - 2rbz\bigr] = -2\sigma\bigl[rx^2 + y^2 + b(z - r)^2 - br^2\bigr],$
by completing the square in $z$ . So $\dot W < 0$ at every point outside the solid ellipsoid $E = \{rx^2 + y^2 + b(z-r)^2 \le br^2\}$ . The function $W$ is continuous and $E$ is compact, so $W$ attains a maximum on $E$ ; let $\nu^*$ exceed it. Then the level set $W = \nu$ with $\nu \ge \nu^*$ lies entirely outside $E$ , so $\dot W < 0$ there: no trajectory crosses it outwards, and the solid region $W \le \nu^*$ is positively invariant. Outside it, $\dot W$ is bounded above by a negative constant (again by compactness, applied to the region $\nu^* \le W \le \nu_0$ for any $\nu_0$ ), so $W$ decreases to $\nu^*$ in finite time.
Equilibria. Setting $\dot x = 0$ gives $y = x$ ; then $\dot y = 0$ gives $x(r - 1 - z) = 0$ , and $\dot z = 0$ gives $x^2 = bz$ . Either $x = 0$ , forcing $z = 0$ and the origin; or $z = r - 1$ and $x^2 = b(r-1)$ , which has real solutions precisely for $r > 1$ and gives $C^{\pm}$ . The symmetry $S$ maps $C^+$ to $C^-$ . As $r \downarrow 1$ both merge with the origin, and by part 5 the origin changes from sink to saddle there, which is a pitchfork bifurcation (see Bifurcations in one dimension for the name).
The origin. The Jacobian of $L$ , whose eigenvalues decide stability by linearization, is
$DL(x,y,z) = \begin{pmatrix} -\sigma & \sigma & 0 \\ r - z & -1 & -x \\ y & x & -b \end{pmatrix},$
which at the origin is block diagonal: the $z$ row and column decouple, contributing the eigenvalue $-b$ , and the upper block $\bigl(\begin{smallmatrix} -\sigma & \sigma \\ r & -1\end{smallmatrix}\bigr)$ has trace $-(\sigma+1)$ and determinant $\sigma(1 - r)$ , whence the stated $\lambda_{\pm}$ . For $r < 1$ the determinant is positive and the trace negative, so both roots have negative real part; moreover the discriminant $(\sigma+1)^2 - 4\sigma(1-r) = (\sigma-1)^2 + 4\sigma r > 0$ , so they are real and the origin is a stable node. For $r > 1$ the determinant is negative, so the roots are real of opposite sign and the origin is a saddle; counting the decoupled $-b$ , it has two stable directions and one unstable one.
Global stability for $r < 1$ . Take the positive definite function
$\Lambda(x, y, z) = x^2 + \sigma y^2 + \sigma z^2 .$
Along a solution,
$\dot\Lambda = 2x\sigma(y-x) + 2\sigma y(rx - y - xz) + 2\sigma z(xy - bz) = -2\sigma\bigl[x^2 + y^2 - (1 + r)xy\bigr] - 2\sigma b z^2,$
the cubic terms cancelling as in part 3. Write $Q(x,y) = x^2 - (1+r)xy + y^2$ . Its discriminant as a quadratic form is $(1+r)^2 - 4$ , which is negative exactly when $0 < r < 1$ ; a binary quadratic form with positive leading coefficient and negative discriminant is positive definite, so $Q(x,y) > 0$ for $(x,y) \ne (0,0)$ . Hence $\dot\Lambda < 0$ except at the origin, where it vanishes: $\Lambda$ is a strict Liapunov function on all of $\mathbb{R}^3$ , and by the Liapunov stability theorem the origin is asymptotically stable with basin all of $\mathbb{R}^3$ . (Strogatz uses $\Lambda/\sigma$ , which is the same function rescaled.)
The convective equilibria. At $C^{\pm}$ we have $z = r-1$ and $y = x = \pm\sqrt{b(r-1)} =: \xi$ , so
$DL(C^{\pm}) = \begin{pmatrix} -\sigma & \sigma & 0 \\ 1 & -1 & -\xi \\ \xi & \xi & -b \end{pmatrix}.$
Expanding $\det(\lambda I - DL)$ and using $\xi^2 = b(r-1)$ gives the stated cubic, whose coefficients depend only on $\xi^2$ and so are the same at $C^+$ and $C^-$ .

At $r = 1$ the cubic is $\lambda^3 + (1+b+\sigma)\lambda^2 + b(\sigma+1)\lambda = \lambda(\lambda + b)(\lambda + \sigma + 1)$ , with roots $0$ , $-b$ , $-\sigma-1$ , which are distinct because $\sigma > b+1$ gives $-\sigma - 1 < -b < 0$ . For $r$ slightly above $1$ the roots move continuously; the constant term $2b\sigma(r-1)$ is now positive and the polynomial is positive for every $\lambda \ge 0$ , so no root is real and non-negative, and all three roots are real and negative.

A root can acquire a non-negative real part only by crossing the imaginary axis, and since no root is ever $0$ for $r > 1$ , the crossing must be at $\lambda = \pm i\omega$ with $\omega \ne 0$ . Substituting $\lambda = i\omega$ and separating real and imaginary parts:
$\text{(real)}\quad -(1+b+\sigma)\omega^2 + 2b\sigma(r-1) = 0, \qquad \text{(imaginary)}\quad -\omega^3 + b(\sigma + r)\omega = 0 .$
The second gives $\omega^2 = b(\sigma + r)$ ; substituting into the first and cancelling $b$ ,
$(\sigma + r)(1 + b + \sigma) = 2\sigma(r - 1) \;\Longrightarrow\; r(\sigma - b - 1) = \sigma(\sigma + b + 3) \;\Longrightarrow\; r = r_H .$
Since $\sigma > b+1$ , this $r_H$ is positive and finite, and it is the smallest $r > 1$ at which a root reaches the imaginary axis; below it all roots keep negative real parts, so $C^{\pm}$ are sinks for $1 < r < r_H$ .

That the bifurcation at $r_H$ is a genuine Hopf bifurcation is stated without proof by Hirsch, Smale and Devaney; that it is subcritical — so that the bifurcating cycles are unstable and lie on the side $r < r_H$ — is a separate calculation that Strogatz also does not reproduce, attributing it to Marsden and McCracken (1976). We follow both and do not prove it here.
Consequences of contraction. A quasiperiodic solution would lie on an invariant torus, and the volume enclosed by that torus would be constant in time, contradicting part 2. A repelling equilibrium can be enclosed in a small sphere of initial conditions, and a repelling closed orbit in a thin tube around it; in either case the enclosed volume grows, again contradicting part 2. $\square$

Remark(Where the hypotheses are used, and what is still open)Original

Positivity of $\sigma$ , $b$ , $r$ is used throughout; it is $\sigma > 0$ that makes $\Lambda$ in part 6 positive definite, and $b > 0$ that makes the $z$ -axis contracting in part 1.
$r < 1$ , strictly, is the whole of part 6: at $r = 1$ the form $Q$ becomes $(x - y)^2$ , which is only positive semi-definite, and $\dot\Lambda$ vanishes on the whole line $x = y$ , $z = 0$ . The conclusion survives at $r = 1$ by a further argument, but not by this one.
$\sigma > b + 1$ is used only in part 7, twice: to separate the three roots at $r = 1$ , and to make $r_H$ positive. If $\sigma \le b+1$ the formula for $r_H$ is negative or undefined, and $C^{\pm}$ do not lose stability this way. The standard parameters $\sigma = 10$ , $b = 8/3$ satisfy it with room to spare.
Volume contraction does not by itself produce chaos. It gives a limiting set of zero volume, which could perfectly well be a point or a curve; indeed for $r < 1$ it is a point. Part 8 is what removes the simple alternatives, and only in combination with part 7.
Two definitions of "attractor". We have followed Strogatz, whose third condition is minimality. Hirsch, Smale and Devaney instead require $\Lambda$ to be compact and invariant, to satisfy $\bigcap_{t\ge0}\varphi_t(U) = \Lambda$ for some open $U$ with $\varphi_t(U) \subseteq U$ , and to be transitive: any two points of $\Lambda$ have neighbourhoods joined by a single solution curve. The two conditions have the same purpose — to stop a union of several attractors from counting as one — and both reject the same standard example, the interval $[-1,1]$ for $\dot x = x - x^3$ , $\dot y = -y$ . They are not equivalent in general, and we do not use either beyond that purpose.
What is not proved here or anywhere in our sources. Nothing above shows that the set seen in simulations at $\sigma=10$ , $b=8/3$ , $r=28$ is an attractor in either sense. Strogatz, writing in 1994, records that no one had proved it. Hirsch, Smale and Devaney analyse instead a geometric model built to imitate the flow, prove that the model has a chaotic attractor, and report that the model was proved to describe the equations themselves only in 1999.

Method(The elementary analysis of a dissipative three-dimensional system)AdaptedAdapted from Strogatz (1994), §9.2, pp. 311-317; Hirsch, Smale & Devaney (2013), §14.2, pp. 308-311

The sequence of steps above is not special to Lorenz's equations; it is the standard opening for any polynomial system in $\mathbb{R}^3$ .

Look for symmetries. A linear involution $S$ with $DS\,F = F\circ S$ halves the work: equilibria and orbits come in pairs, and any bifurcation creating a pair of equilibria from one is a pitchfork rather than a saddle-node.
Compute $\nabla\cdot F$ . If it is a negative constant $-k$ , all volumes decay like $e^{-kt}$ , and you may immediately rule out quasiperiodicity and all repelling invariant sets.
Find a trapping region. Try a positive definite quadratic $W$ centred where the flow seems to live, differentiate along the flow, and choose the weights so that the cubic terms cancel. Complete the square in the remaining quadratic; the region where $\dot W \ge 0$ should come out bounded, and any sufficiently large sublevel set of $W$ then traps the flow.
Find the equilibria and linearise. Where the Jacobian is block triangular, read the eigenvalues off the blocks.
Try the same quadratic as a Liapunov function near the parameter value where the first equilibrium loses stability; this often gives global, not merely local, stability on one side of the threshold.
Locate the stability boundary of the remaining equilibria by substituting $\lambda = i\omega$ into the characteristic polynomial rather than solving it. Real and imaginary parts give two equations in $\omega^2$ and the parameter; eliminating $\omega^2$ gives the threshold.
Count what is left. If no equilibrium and no closed orbit can attract, and nothing escapes to infinity, the long-time behaviour has to be something else.

Example(The standard parameters

\sigma = 10

b = 8/3

)Original

Take Lorenz's values $\sigma = 10$ , $b = 8/3$ . Then $\sigma - b - 1 = 19/3 > 0$ , so the standing assumption holds, and

r_H = \sigma\,\frac{\sigma + b + 3}{\sigma - b - 1} = 10\cdot\frac{47/3}{19/3} = \frac{470}{19} = 24.7368421\ldots

Volume. The contraction rate is $\sigma + 1 + b = 41/3 = 13.6\overline{6}$ , so phase volumes shrink by the factor $e^{-41/3} = 1.1605\times10^{-6}$ per unit of time; the half-life of a volume is $3\ln 2/41 = 0.050718$ time units. A blob of initial conditions of diameter $1$ has shrunk in volume by six orders of magnitude after one time unit, which is roughly one and a half circuits of the attractor.

At $r = 28$ , above $r_H$ . The equilibria are the origin and

C^{\pm} = \bigl(\pm\sqrt{(8/3)\cdot 27},\ \pm\sqrt{72},\ 27\bigr) = (\pm 6\sqrt2,\ \pm 6\sqrt2,\ 27), \qquad 6\sqrt2 = 8.4852814 .

At the origin the eigenvalues are $-b = -2.6\overline{6}$ and

\lambda_{\pm} = \tfrac12\bigl[-11 \pm \sqrt{121 + 40\cdot 27}\bigr] = \tfrac12\bigl[-11 \pm \sqrt{1201}\bigr] = 11.8277235,\ -22.8277235 ,

a saddle with one unstable direction, as part 5 predicts for $r > 1$ .

At $C^{\pm}$ the cubic of part 7 is

\lambda^3 + \tfrac{41}{3}\lambda^2 + \tfrac{304}{3}\lambda + 1440 = 0 .

Its roots, computed numerically, are

\lambda_1 = -13.8545779, \qquad \lambda_{2,3} = 0.0939556 \pm 10.1945052\,i .

The complex pair has positive real part, confirming that $r = 28$ lies above $r_H = 24.7368$ . So each of $C^{\pm}$ is a saddle-focus with a one-dimensional stable manifold and a two-dimensional unstable one, and a trajectory near it spirals outwards within that unstable surface, taking $2\pi/10.1945052 = 0.6163$ time units per turn. (As a check on the cubic: the sum of its roots is $-13.8545779 + 2(0.0939556) = -13.6\overline{6} = -(1 + b + \sigma)$ , and the product is $-1440 = -2b\sigma(r-1)$ .)

Putting the parts together. At these parameters: no trajectory escapes, by part 3; the origin is a saddle and $C^{\pm}$ are saddle-foci, so no equilibrium attracts; there is no attracting closed orbit near $C^{\pm}$ , because the Hopf bifurcation at $r_H$ is subcritical; there are no quasiperiodic motions and no repellers, by part 8; and every volume shrinks to zero. The only possibility left is that almost every trajectory approaches a set of zero volume on which the motion is aperiodic — the strange attractor, whose sensitive dependence is quantified in Lyapunov exponents and sensitive dependence and whose fractional dimension is the subject of Fractals and dimension.

ApplicationAdaptedAdapted from Strogatz (1994), §9.1, pp. 302-311, §9.5, pp. 330-335, and §9.6, pp. 335-340

Convection and the waterwheel. The system's first use is as a caricature of Rayleigh-Benard convection: $r < 1$ is pure conduction with the fluid at rest, $1 < r < r_H$ is steady convection in rolls of one handedness or the other (the two being exchanged by the symmetry $S$ , which is why the bifurcation at $r=1$ is a pitchfork), and $r > r_H$ is the irregular regime. Every statement above has a mechanical counterpart in the Malkus-Howard waterwheel, where $x$ is the angular velocity: at low inflow the wheel sits still, at moderate inflow it settles into steady rotation one way or the other, and above threshold it turns a few times, stalls, and reverses, erratically and forever. The same equations arise for a laser (the Maxwell-Bloch equations transform into them) and in models of geomagnetic dynamos, which is one reason the irregular reversals are suggestive.

Why the parameter $r$ repays study. Holding $\sigma = 10$ , $b = 8/3$ and varying $r$ shows that the chaotic regime is neither the only nor the final behaviour. Below $r_H$ but above $r \approx 24.06$ a strange attractor coexists with the two stable equilibria, so the outcome depends on the initial condition and the system shows hysteresis as $r$ is swept. Below that, down to a homoclinic bifurcation near $r = 13.926$ , there is transient chaos: trajectories rattle around irregularly for a long and unpredictable time before settling on $C^+$ or $C^-$ . This is worth separating from chaos proper, because it shows that a system with only simple attractors can still be effectively unpredictable — the reason a thrown die is a fair randomiser although it has six stable resting positions. For large $r$ the dynamics simplify again, to a globally attracting limit cycle for $r$ above a few hundred, corresponding to a waterwheel that rocks regularly back and forth.

Chaos as a carrier. A striking engineering use turns the sensitivity into an advantage. Cuomo and Oppenheim built two identical Lorenz circuits in the rescaled variables $u = x/10$ , $v = y/10$ , $w = z/20$ , for which the equations read $\dot u = \sigma(v-u)$ , $\dot v = ru - v - 20uw$ , $\dot w = 5uv - bw$ . The transmitter runs freely; the receiver runs the same equations except that the transmitted scalar $u(t)$ is substituted for its own first coordinate wherever that coordinate multiplies something in the last two equations. The error $e$ between the two states then obeys

\dot e_1 = \sigma(e_2 - e_1), \qquad \dot e_2 = -e_2 - 20\,u(t)\,e_3, \qquad \dot e_3 = 5\,u(t)\,e_2 - b\,e_3,

a linear system with a chaotic time-dependent coefficient. Take $E(e) = \tfrac12\bigl(\sigma^{-1}e_1^2 + e_2^2 + 4e_3^2\bigr)$ . The two terms carrying $u(t)$ are $-20u e_2e_3$ and $+20u e_2 e_3$ , which cancel, leaving

\dot E = -\bigl(e_1^2 - e_1e_2\bigr) - e_2^2 - 4b e_3^2 = -\Bigl(e_1 - \tfrac12 e_2\Bigr)^2 - \tfrac34 e_2^2 - 4b e_3^2 < 0

away from $e = 0$ . So $E$ is a strict Liapunov function and the receiver locks onto the transmitter from any initial state, reconstructing all three coordinates from one. A message buried under a much louder chaotic signal can then be recovered by subtraction. The cancellation of the terms in $u$ , which is the same small algebraic accident as in parts 3 and 6 above, is what makes this work.

Definition

Lyapunov exponents and sensitive dependence

Adapted

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Defines the exponential rates at which an infinitesimal ball of initial conditions is stretched and squeezed along a trajectory, for both flows and maps, proves that the exponents sum to the time-averaged divergence and that a bounded non-equilibrium trajectory of a flow always carries a zero exponent, and turns the largest exponent into the predictability horizon $t \approx \lambda^{-1}\ln(a/\|\delta_0\|)$ — checked against a direct simulation of the Lorenz system.

Prerequisites: The Lorenz equations, One-dimensional maps, Linearization and the Hartman-Grobman theorem, Singular value decomposition, Dependence on initial data and Gronwall's inequality

Definition(Separation, Lyapunov exponents, sensitive dependence)AdaptedAdapted from Strogatz (1994), §9.3, pp. 321-322, 324-325, and §10.5, pp. 366-367; Hirsch, Smale & Devaney (2013), §15.4, p. 340

For a flow. Let $\dot X = F(X)$ with $F \in C^1$ on $\mathbb{R}^n$ , let $X(t)$ be a solution, and let $\delta(t)$ be the separation of an infinitesimally close solution. Solutions depend differentiably on their initial data (continuous dependence), and to first order $\delta$ obeys the variational equation

\dot\delta = DF\bigl(X(t)\bigr)\,\delta ,

whose fundamental matrix $\Phi(t)$ (the solution with $\Phi(0) = I$ ) carries an initial separation $\delta_0$ to $\Phi(t)\delta_0$ . The image of the unit sphere under $\Phi(t)$ is an ellipsoid; by the singular value decomposition its principal semi-axes are the singular values $s_1(t) \ge s_2(t) \ge \dots \ge s_n(t) > 0$ of $\Phi(t)$ . The Lyapunov exponents of the trajectory are

\lambda_k = \lim_{t\to\infty} \frac1t \ln s_k(t), \qquad k = 1, \dots, n,

whenever the limits exist. They are listed in decreasing order, and $\lambda_1$ is the largest Lyapunov exponent, usually written simply $\lambda$ . Since $\|\Phi(t)\delta_0\| \le s_1(t)\|\delta_0\|$ for every $\delta_0$ , with equality for the best-aligned one, $\lambda_1$ is the fastest exponential rate at which any separation can grow; it is also the rate at which a separation chosen without regard to the geometry is observed to grow, and one writes $\|\delta(t)\| \sim \|\delta_0\|e^{\lambda_1 t}$ .

For a map. For $x_{n+1} = f(x_n)$ with $f \in C^1$ on an interval, $(f^n)'(x_0) = \prod_{i<n} f'(x_i)$ by the chain rule (see One-dimensional maps), so there is a single exponent

\lambda(x_0) = \lim_{n\to\infty} \frac1n \ln\bigl|(f^n)'(x_0)\bigr| = \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\ln\bigl|f'(x_i)\bigr| ,

when the limit exists.

Sensitive dependence and chaos. A system shows sensitive dependence on initial conditions in the metric sense used here if $\lambda_1 > 0$ : nearby trajectories separate exponentially fast on average. Following Strogatz, chaos is aperiodic long-term behaviour in a deterministic system that shows sensitive dependence, where "aperiodic" means that trajectories which approach no fixed point, periodic orbit or quasiperiodic orbit occur for a non-negligible set of initial conditions, and "deterministic" means that the irregularity comes from the nonlinearity and not from noise in the equations. A strange attractor is an attractor on which the motion is chaotic in this sense.

Definition(What the exponents must satisfy)Original

Let $F \in C^1$ and let $X(t)$ be a solution of $\dot X = F(X)$ for which all $n$ limits above exist.

The exponents sum to the averaged divergence.
$\sum_{k=1}^n \lambda_k = \lim_{t\to\infty}\frac1t\int_0^t \nabla\cdot F\bigl(X(s)\bigr)\,ds ,$
whenever the right-hand limit exists. In particular, if $\nabla\cdot F$ is the constant $-k$ , as for the Lorenz equations, then $\sum_k \lambda_k = -k$ exactly.
The flow direction is neutral. If the forward orbit $\{X(t) : t \ge 0\}$ is bounded and $\inf_{t\ge0}|F(X(t))| > 0$ , then the direction along the flow neither grows nor decays exponentially:
$\lim_{t\to\infty}\frac1t\ln\bigl|\Phi(t)F(X(0))\bigr| = 0 .$
In particular $\lambda_n \le 0 \le \lambda_1$ .
A stable cycle of a map has a negative exponent. If $x_0$ lies on a $p$ -cycle of a map $f$ with multiplier $\mu$ , then $\lambda(x_0) = p^{-1}\ln|\mu|$ . So $\lambda < 0$ for a stable cycle, $\lambda > 0$ for an unstable one, and $\lambda = -\infty$ for a superstable one.
Predictability horizon. Suppose the largest exponent is $\lambda > 0$ and the initial state is known only to within $\|\delta_0\|$ . If predictions are useful while the error stays below a tolerance $a$ , they cease to be useful after a time of order
$t_{\text{horizon}} \approx \frac1\lambda\,\ln\frac{a}{\|\delta_0\|} .$
The dependence on $\|\delta_0\|$ is logarithmic: reducing the initial uncertainty by a factor $10^6$ buys only $6\ln 10/\lambda \approx 13.8/\lambda$ extra time.

Idea of the proofOriginal

All four parts come from one observation: $\ln$ turns the multiplicative growth of lengths, areas and volumes into something additive, which can then be averaged. The product of all the semi-axes is $|\det\Phi|$ , and Liouville's formula says that $\det\Phi$ grows at the instantaneous rate $\operatorname{tr}DF = \nabla\cdot F$ ; taking logarithms and dividing by $t$ gives part 1. Part 2 holds because the velocity vector itself solves the variational equation, so the flow direction neither grows nor decays beyond bounded factors. Part 3 is the chain-rule product again: along a cycle the same $p$ factors repeat forever, so the average of the logarithms is the logarithm of the multiplier divided by $p$ . Part 4 is just solving $\|\delta_0\|e^{\lambda t} = a$ .

ProofOriginal

Sum of the exponents. The determinant of $\Phi$ obeys Liouville's formula: differentiating $\det\Phi$ and using $\dot\Phi = DF(X(t))\Phi$ gives
$\frac{d}{dt}\det\Phi(t) = \operatorname{tr}\bigl[DF(X(t))\bigr]\,\det\Phi(t) = \nabla\cdot F\bigl(X(t)\bigr)\,\det\Phi(t),$
so, since $\det\Phi(0) = 1$ ,
$\det\Phi(t) = \exp\int_0^t \nabla\cdot F\bigl(X(s)\bigr)\,ds .$
On the other hand $|\det\Phi(t)| = \prod_{k=1}^n s_k(t)$ , because the singular values are the semi-axes of the image of the unit ball and the determinant is the volume ratio. Taking logarithms,
$\sum_{k=1}^n \ln s_k(t) = \int_0^t \nabla\cdot F\bigl(X(s)\bigr)\,ds .$
Divide by $t$ and let $t \to \infty$ . The left side tends to $\sum_k \lambda_k$ by hypothesis, so the right side converges too and the two limits agree. If $\nabla\cdot F \equiv -k$ the integral is $-kt$ and the average is $-k$ for every $t$ .
The zero exponent. Differentiating $\dot X = F(X)$ with respect to $t$ gives $\frac{d}{dt}F(X(t)) = DF(X(t))\,F(X(t))$ , so $v(t) = F(X(t))$ solves the variational equation; hence $v(t) = \Phi(t)v(0)$ . By hypothesis there are constants $0 < m \le M < \infty$ with $m \le |v(t)| \le M$ for all $t \ge 0$ : the lower bound is the hypothesis $\inf|F(X(t))| > 0$ , and the upper bound holds because $F$ is continuous and the orbit is bounded, so it lies in a compact set on which $|F|$ is bounded. Therefore
$\frac{\ln m - \ln|v(0)|}{t} \;\le\; \frac1t\ln\frac{|\Phi(t)v(0)|}{|v(0)|} \;\le\; \frac{\ln M - \ln|v(0)|}{t} ,$
and both bounds tend to $0$ . So the growth rate of the particular direction $v(0)$ is $0$ . Since $s_n(t)|v(0)| \le |\Phi(t)v(0)| \le s_1(t)|v(0)|$ for every $t$ , taking logarithms, dividing by $t$ and letting $t \to \infty$ gives $\lambda_n \le 0 \le \lambda_1$ .
Cycles of a map. Let $x_0, \dots, x_{p-1}$ be the cycle. The sequence $\ln|f'(x_i)|$ is periodic with period $p$ , so for $n = mp$ ,
$\frac1n\sum_{i=0}^{n-1}\ln|f'(x_i)| = \frac{m}{mp}\sum_{i=0}^{p-1}\ln|f'(x_i)| = \frac1p\ln\Bigl|\prod_{i=0}^{p-1}f'(x_i)\Bigr| = \frac1p\ln|\mu| ,$
using the chain-rule product formula for the multiplier $\mu$ . For general $n$ the partial sum differs from this by at most $p$ bounded terms, which vanish after division by $n$ . Stability means $|\mu| < 1$ , hence $\ln|\mu| < 0$ ; instability means $|\mu| > 1$ , hence $\ln|\mu| > 0$ ; superstability means $\mu = 0$ , and then some factor $f'(x_i)$ is zero, so every partial sum past that index is $-\infty$ .
Horizon. With $\|\delta(t)\| \approx \|\delta_0\|e^{\lambda t}$ , the error first reaches $a$ when $\lambda t = \ln(a/\|\delta_0\|)$ . Replacing $\|\delta_0\|$ by $\|\delta_0\|/10^6$ increases the right-hand side by $\ln 10^6 = 6\ln 10 = 13.8155$ , hence $t$ by $13.8155/\lambda$ . $\square$

Remark(Where the hypotheses are used, and what the definition does not say)Original

Existence of the limits is assumed throughout and is not automatic. It is a theorem of ergodic theory that the limits exist for almost every initial condition with respect to an invariant measure, but nothing in our sources proves it, and we do not use it. In part 1 the hypothesis is used twice: the left-hand limit must exist for the identity to say anything, and it then forces the right-hand one.
Boundedness and $\inf|F| > 0$ in part 2. Both are needed. The first fails for $\dot x = x$ , whose single exponent is $+1$ and which has no neutral direction; the second fails at an equilibrium, where the orbit is a point and the exponents are the real parts of the eigenvalues, none of which need vanish.
From a neutral direction to a zero exponent. Part 2 says the flow direction grows at rate $0$ ; it does not by itself say that $0$ appears in the list $\lambda_1, \dots, \lambda_n$ . That step needs the theorem of ergodic theory, mentioned above and not assumed here, which says that the growth rate of almost every direction is one of the $\lambda_k$ and that the directions with each rate form a subspace. Granting it, the flow direction contributes a zero exponent. We therefore use part 2 as a numerical check — a computed spectrum for a flow on a bounded aperiodic orbit that does not contain a zero is wrong — rather than as a statement about the spectrum.
A positive exponent is not by itself chaos. The system $\dot x = x$ has $\lambda = 1 > 0$ but its trajectories run off to infinity and never return, so its long-term behaviour is not aperiodic in the required sense. The definition of chaos needs the aperiodicity clause as well, and that clause is what confines the motion to an attractor.
Two kinds of sensitive dependence. The definition above is metric: it asserts an exponential rate. Hirsch, Smale and Devaney instead call a map on an interval sensitive if there is one constant $\beta$ such that every point has points arbitrarily nearby whose orbits eventually get $\beta$ apart; no rate is mentioned, and their definition of a chaotic map adds density of periodic points and transitivity rather than a positive exponent. They note explicitly that definitions of chaos differ on this point. The two notions agree on the standard examples but are not the same condition, and we use the metric one because it is the one that produces a number.
The exponent is an average, so local behaviour deviates. The graph of $\ln\|\delta(t)\|$ against $t$ is not a straight line; it wobbles because the stretching varies over the attractor, and it flattens once $\|\delta\|$ reaches the diameter of the attractor, beyond which the linearised description is void. The horizon formula is therefore an order-of-magnitude statement and is only meaningful for $a$ well below that diameter.
$\lambda$ depends on the trajectory, so a numerical value must be averaged over a long orbit or over many starting points. For a map, Strogatz records that one and the same value comes out of every seed whose orbit is captured by a given attractor.

Method(Computing the exponents)AdaptedAdapted from Strogatz (1994), §10.5, Example 10.5.3, p. 368

For a one-dimensional map. Fix the parameter. Iterate from an arbitrary seed for a few hundred steps to shed the transient. Then iterate $N$ further times, accumulating $\sum \ln|f'(x_i)|$ as you go — only the current $x$ need be stored — and divide by $N$ . Repeat for the next parameter value. Plotting the result against the parameter gives a curve that is negative below the onset of chaos, touches zero at each period-doubling, and is positive in the chaotic range except for downward spikes at the periodic windows.

For a flow, all $n$ exponents. Integrate the state $X$ and an $n\times n$ matrix $Q$ together, $Q$ obeying $\dot Q = DF(X)Q$ and starting at $Q = I$ . Every $\tau$ units of time, factor $Q = \tilde Q R$ by the Gram-Schmidt (QR) process with the signs fixed so that $R$ has positive diagonal; add $\ln R_{kk}$ to a running total $S_k$ ; replace $Q$ by $\tilde Q$ and continue. After total time $T$ , the estimate is $\lambda_k \approx S_k/T$ . The periodic re-orthonormalisation is what stops every column from collapsing onto the fastest-growing direction and overflowing, and the diagonal of $R$ records exactly the growth of the $k$ -dimensional volume spanned by the first $k$ columns relative to the $(k-1)$ -dimensional one.

Two checks to run every time. The sum of the computed exponents must equal the time-averaged divergence (part 1); and for a flow on a bounded aperiodic orbit, one exponent must come out at zero, since the flow direction is neutral (part 2). If either fails, $\tau$ is too long, the integrator tolerance too loose, or the transient too short.

Example(Three systems, computed)Original

The tent map. For $f(x) = r\min(x,\,1-x)$ — that is, $f(x) = rx$ on $[0,\tfrac12]$ and $f(x) = r(1-x)$ on $[\tfrac12, 1]$ , with $0 < r \le 2$ — we have $|f'| = r$ at every point where the derivative exists. So every term of the sum is $\ln r$ and $\lambda = \ln r$ for every seed whose orbit avoids the kink, with no computation at all. The map is chaotic for $r > 1$ and not for $r < 1$ , and the threshold $\lambda = 0$ sits exactly at $r = 1$ .

The logistic map, where the answer is known in closed form. At $r = 3.2$ the attractor is the $2$ -cycle computed in One-dimensional maps, with multiplier $\mu = 4 + 2r - r^2 = 0.16$ . By part 3,

\lambda = \tfrac12\ln 0.16 = -0.9162907319 .

Running the method above from the seed $x_0 = 0.1$ with $N = 2\times10^5$ iterations after a transient of $10^3$ returns $-0.9162907319$ , agreeing to all ten figures shown (the two values differ by $3\times10^{-12}$ ), and the seed $x_0 = 0.37$ returns the same. This is the easy case: once the orbit is on the cycle the average is exact.

The logistic map in the chaotic range. At $r = 3.9$ there is no closed form. Averaging $\ln|3.9 - 7.8x_n|$ over $2\times10^6$ iterations, for $40$ random seeds after a transient of $2000$ , gives

\lambda = 0.4960 \pm 0.0003 ,

the quoted uncertainty being the spread over seeds (the extreme values were $0.49534$ and $0.49649$ ). The exponent is positive, so the aperiodic orbits seen at this parameter really do separate exponentially. (At $r = 4$ the same average returns $0.693147$ , which is $\ln 2$ to six figures; the reason is that $4x(1-x)$ is semi-conjugate to the tent map with $r = 2$ through $h(x) = \tfrac12(1 - \cos 2\pi x)$ , and the tent map has $\lambda = \ln 2$ . Direct iteration exactly at $r = 4$ is numerically delicate, because rounding can drop an orbit onto a spurious cycle; one of four seeds we tried did so and returned a visibly wrong $1.108$ .)

The Lorenz attractor. Take $\sigma = 10$ , $b = 8/3$ , $r = 28$ . Integrating the state together with a $3\times3$ matrix by fourth-order Runge-Kutta with step $0.005$ , discarding $100$ time units of transient, re-orthonormalising every $\tau = 0.5$ and averaging over $T = 4000$ time units gives

\lambda_1 = 0.9067, \qquad \lambda_2 = -0.00034, \qquad \lambda_3 = -14.5730 .

Both checks pass. The middle exponent is zero to within $3\times10^{-4}$ , the neutral direction of part 2. And the sum is $-13.666660$ , against the exact value $-(\sigma + 1 + b) = -41/3 = -13.666667$ from part 1, an agreement to six decimal places. The drift of $\lambda_1$ over the second half of the run was about $\pm 0.002$ , so the honest statement is $\lambda_1 = 0.906 \pm 0.002$ ; Strogatz quotes $0.9$ .

The horizon, measured. With $\lambda_1 = 0.906$ , part 4 predicts that an initial uncertainty of $10^{-6}$ grows to $1$ — about one fiftieth of the diameter of the attractor, which our simulation measured as roughly $58$ — after

t_{\text{horizon}} \approx \frac{\ln 10^{6}}{0.906} = \frac{13.8155}{0.906} = 15.25 \text{ time units} .

To test this we took $200$ points spaced along one long trajectory on the attractor, displaced each by $10^{-6}$ in a random direction, integrated both copies, and recorded when their separation first reached $1$ . The mean was

t = 14.24 \pm 0.14 ,

within about $7\%$ of the prediction. The formula is an estimate, not an identity: the spread of individual times was large, because the stretching rate varies around the attractor, and $15.25$ is what one gets by pretending it does not.

ApplicationOriginal

The exponent converts a qualitative worry into a budget. In numerical weather prediction the largest exponent of the atmosphere corresponds to an error-doubling time of a day or two, so the horizon formula says that a hundredfold improvement in the initial analysis — an enormous investment in observations — buys roughly $\ln 100 \approx 4.6$ doubling times, about a week. This is the quantitative content of Lorenz's original remark about weather, and it is why operational forecasting moved from single runs to ensembles: once the horizon is reached the useful output is a distribution, not a trajectory.

The same arithmetic governs long integrations in celestial mechanics and molecular dynamics. Positive exponents in the solar system (with Lyapunov times of a few million years for the inner planets) mean that individual orbits computed over longer spans are not predictions of where a planet will be, although statistical properties of the ensemble remain meaningful; in molecular dynamics the exponents are so large that no trajectory is reproducible beyond picoseconds, which is why only time averages are reported.

In the laboratory the exponent is used as a diagnostic. Estimating $\lambda_1$ from a measured time series distinguishes deterministic chaos from noise, since a noisy periodic signal has $\lambda_1 \le 0$ while a chaotic one does not, and the spectrum of exponents bounds how many degrees of freedom are active: the number of non-negative exponents is a lower bound on the dimension of the attractor, which in turn constrains how low-dimensional a model is allowed to be. The fractional dimension that goes with this picture is the subject of Fractals and dimension.

Method

One-dimensional maps

Adapted

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Gives the working apparatus for a discrete-time system $x_{n+1} = f(x_n)$ on an interval — orbits, fixed points, $p$ -cycles, the multiplier and its chain-rule product formula, and the cobweb construction — proves the sink/source criterion from the mean value theorem, and applies the recipe to the logistic map, where the fixed point loses stability at $r = 3$ and a $2$ -cycle with multiplier $4 + 2r - r^2$ takes over.

Prerequisites: Flows on the line, Bifurcations in one dimension, Limits of sequences and completeness

Definition(Maps, orbits, cycles and the multiplier)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §15.1, pp. 329-331, 333; Strogatz (1994), §10.1, pp. 349-350, 367

Let $I \subseteq \mathbb{R}$ be an interval and let $f : I \to I$ be continuously differentiable. The map is the rule

x_{n+1} = f(x_n), \qquad n = 0, 1, 2, \dots,

which replaces the continuous time of a differential equation by integer time. Much of the vocabulary carries over from flows on the line, but the behaviour does not: a map on an interval can oscillate and can be chaotic, which a one-dimensional flow cannot. Because $f$ sends $I$ into itself, the recursion never leaves $I$ ; this invariance is part of the data, not an afterthought.

Write $f^n$ for the $n$ -fold composite $f \circ \dots \circ f$ (never for a power). The orbit of a starting value $x_0 \in I$ , also called its seed, is the sequence $x_0, f(x_0), f^2(x_0), \dots$ .
$x^*$ is a fixed point if $f(x^*) = x^*$ ; its orbit is the constant sequence.
$x_0$ is periodic of period $p$ ( $p \in \mathbb{N}$ ) if $f^p(x_0) = x_0$ . The least such $p$ is the minimal period, and the set $\{x_0, f(x_0), \dots, f^{p-1}(x_0)\}$ of a point of minimal period $p$ is a $p$ -cycle. A fixed point is a $1$ -cycle.
The multiplier of a $p$ -cycle through $x_0$ is $\lambda = (f^p)'(x_0)$ . For a fixed point this is $\lambda = f'(x^*)$ .
A fixed point $x^*$ is attracting (a sink) if it has a neighbourhood $U$ such that every orbit started in $U$ stays in $U$ and converges to $x^*$ ; it is repelling (a source) if, for some such $U$ , every orbit started in $U$ other than $x^*$ itself eventually leaves $U$ ; and neutral if it is neither. A $p$ -cycle is called a sink, a source or neutral according to which of these its points are as fixed points of $f^p$ , a classification that makes sense because, as the next statement shows, all $p$ points have the same multiplier.
A cycle with $\lambda = 0$ is called superstable.

The cobweb of a seed $x_0$ is the following drawing in the $(x, y)$ plane. Draw $y = f(x)$ and the diagonal $y = x$ . From the point $(x_0, x_0)$ move vertically to $(x_0, x_1)$ on the graph, then horizontally to $(x_1, x_1)$ on the diagonal; repeat. The abscissae of the points reached on the diagonal are exactly $x_0, x_1, x_2, \dots$ , so the staircase or spiral traced out is a picture of the orbit.

Method(Stability of a cycle from its multiplier)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §15.1, Proposition, p. 332; Strogatz (1994), §10.1, p. 350; Strogatz (1994), §10.5, Example 10.5.1, p. 367

Let $f \in C^1(I)$ and let $x_0, x_1 = f(x_0), \dots, x_{p-1}$ be a $p$ -cycle of $f$ in the interior of $I$ . Then:

(Chain-rule product.) The multiplier is the product of the derivatives of $f$ around the cycle,
$\lambda = (f^p)'(x_0) = \prod_{k=0}^{p-1} f'(x_k),$
and this value is unchanged if $x_0$ is replaced by any other point $x_j$ of the cycle. So "the multiplier of the cycle" is well defined.
(Sink.) If $|\lambda| < 1$ , the cycle is attracting. More precisely, for every $K$ with $|\lambda| < K < 1$ there is a $\delta > 0$ such that $|x - x_0| \le \delta$ implies $|f^{np}(x) - x_0| \le K^n |x - x_0|$ for all $n \ge 0$ .
(Source.) If $|\lambda| > 1$ , the cycle is repelling.
(No conclusion at $|\lambda| = 1$ .) If $|\lambda| = 1$ , the cycle may be attracting, repelling, or neither, and the derivative alone does not decide which.

If $\lambda > 0$ a nearby orbit approaches or leaves the fixed point from one side; if $\lambda < 0$ it alternates from side to side, which is what makes a cobweb spiral rather than climb.

Idea of the proofOriginal

A $p$ -cycle of $f$ is a fixed point of $f^p$ , so the whole question reduces to fixed points, and for a fixed point the mean value theorem does everything: if $|f'| \le K < 1$ on a small interval around $x^*$ , then $f$ moves every point of that interval at least a factor $K$ closer to $x^*$ , so the interval is mapped into itself and the distance to $x^*$ falls geometrically. The product formula is the chain rule applied to a composite whose intermediate points are the points of the cycle; it is independent of the starting point because a cyclic shift does not change a product.

ProofAdaptedAdapted from Hirsch, Smale & Devaney (2013), §15.1, proof of the Proposition, pp. 332-333

The product formula. Differentiating $f^p = f \circ f^{p-1}$ repeatedly by the chain rule and evaluating at $x_0$ gives
$(f^p)'(x_0) = f'\bigl(f^{p-1}(x_0)\bigr)\,f'\bigl(f^{p-2}(x_0)\bigr)\cdots f'(x_0) = \prod_{k=0}^{p-1} f'(x_k),$
because $f^k(x_0) = x_k$ . Starting the cycle at $x_j$ instead permutes the factors cyclically, and a product of real numbers does not notice the order, so the value is the same.
$|\lambda| < 1$ gives a sink. Put $g = f^p$ , so $g(x_0) = x_0$ and $g'(x_0) = \lambda$ . Fix $K$ with $|\lambda| < K < 1$ . Since $g$ is $C^1$ (a composite of $C^1$ maps) its derivative is continuous, so there is a $\delta > 0$ with $J = [x_0 - \delta, x_0 + \delta] \subset I$ and $|g'| \le K$ on $J$ .

Let $x \in J$ . By the mean value theorem applied to $g$ on the interval between $x$ and $x_0$ there is a point $c$ strictly between them with
$g(x) - x_0 = g(x) - g(x_0) = g'(c)\,(x - x_0),$
and $c \in J$ , so $|g(x) - x_0| \le K|x - x_0| \le K\delta < \delta$ . Hence $g(J) \subseteq J$ and the estimate may be applied again. Induction gives $|g^n(x) - x_0| \le K^n|x - x_0|$ , and $K^n \to 0$ because $0 < K < 1$ . So $g^n(x) \to x_0$ , which says that the subsequence $x, f^p(x), f^{2p}(x), \dots$ converges to $x_0$ .

The intermediate iterates are handled by continuity: $f^{np+j}(x) = f^j(g^n(x)) \to f^j(x_0) = x_j$ for each fixed $j \in \{0, \dots, p-1\}$ , since $f^j$ is continuous. So the whole orbit of $x$ converges to the cycle.
$|\lambda| > 1$ gives a source. Now choose $K$ with $1 < K < |\lambda|$ and, by continuity of $g'$ again, a $\delta > 0$ with $|g'| \ge K$ on $J = [x_0 - \delta, x_0 + \delta]$ . The same mean value theorem step, with the inequality reversed, gives $|g(x) - x_0| \ge K|x - x_0|$ for $x \in J$ . As long as the iterates stay in $J$ we get $|g^n(x) - x_0| \ge K^n|x - x_0|$ , and the right-hand side exceeds $\delta$ once $K^n > \delta/|x - x_0|$ , which happens for some finite $n$ because $K > 1$ and $x \ne x_0$ . So every orbit except the cycle leaves $J$ .
$|\lambda| = 1$ decides nothing. It is enough to exhibit three maps with a fixed point at $0$ of multiplier $1$ and three different behaviours; this is the list in Hirsch, Smale and Devaney.
- $f(x) = x - x^3$ . For $0 < x < 1$ we have $0 < f(x) < x$ , so the orbit decreases and is bounded below by $0$ ; its limit $L$ satisfies $L = L - L^3$ , hence $L = 0$ . By oddness the same holds for $-1 < x < 0$ . So $0$ is a sink.
- $f(x) = x + x^3$ . For $x > 0$ the orbit increases, and it cannot converge to a finite limit $L$ , since $L = L + L^3$ forces $L = 0$ , which is below the starting point. So orbits leave every neighbourhood: $0$ is a source.
- $f(x) = x + x^2$ . For $x > 0$ the orbit increases and, by the same argument, escapes; for $-1 < x < 0$ we have $x < f(x) < 0$ , so the orbit increases towards a limit $L \le 0$ with $L = L + L^2$ , hence $L = 0$ . The fixed point attracts from the left and repels from the right, so it is neither a sink nor a source. $\square$

Remark(Where the hypotheses are used)Original

Continuous differentiability is used only through the continuity of $g' = (f^p)'$ , which is what lets a bound on $|g'|$ at the single point $x_0$ be spread over a whole interval $J$ . For a merely differentiable $f$ the argument has to be replaced by a direct estimate from the definition of the derivative; the conclusions still hold, but the proof above does not apply verbatim.
The cycle lies in the interior of $I$ so that a two-sided interval $J$ around $x_0$ is available. At an endpoint of $I$ one gets only the one-sided statement.
Strict inequality $|\lambda| \ne 1$ is what part 4 shows cannot be dropped: all three maps there have $f'(0) = 1$ , and all three behave differently. The reason is that at $|\lambda| = 1$ the linear part neither contracts nor expands, so the sign and size of the first non-vanishing nonlinear term decide the outcome. This is exactly why bifurcations of maps occur at $\lambda = +1$ and $\lambda = -1$ and nowhere else.
$f$ maps $I$ into $I$ . Without this the iteration can simply leave the region where $f$ is defined, and "the orbit" stops existing. For the logistic map this is why $r$ is restricted to $0 \le r \le 4$ below.

Method(Analysing a one-dimensional map)AdaptedAdapted from Strogatz (1994), §10.1, pp. 349-352, and §10.3, pp. 357-361; Hirsch, Smale & Devaney (2013), §15.1, pp. 331-333

Given $x_{n+1} = f(x_n)$ with a parameter:

Fix an invariant interval. Find an interval $I$ with $f(I) \subseteq I$ , and restrict the parameter so that this holds. Everything below is relative to $I$ .
Fixed points. Solve $f(x) = x$ . Graphically these are the crossings of $y = f(x)$ with the diagonal.
Classify them. Compute $\lambda = f'(x^*)$ and apply the statement above: sink for $|\lambda| < 1$ , source for $|\lambda| > 1$ , superstable for $\lambda = 0$ , and inconclusive for $|\lambda| = 1$ , where you must look at the nonlinear terms or draw a cobweb.
Draw a cobweb for one or two seeds. It shows at a glance what the linearisation cannot: which fixed point an orbit is captured by, whether the approach is monotone ( $\lambda > 0$ ) or alternating ( $\lambda < 0$ ), and whether orbits escape.
Cycles of period $p$ . Solve $f^p(x) = x$ . Every fixed point of $f$ solves this too, so divide those roots out first; the remaining roots come in groups of $p$ , one group per $p$ -cycle.
Classify the cycle by $\lambda = \prod_{k} f'(x_k)$ , using the product formula rather than differentiating $f^p$ directly.
Track the bifurcations. Move the parameter and watch $\lambda$ at each cycle. A cycle can change stability only as $\lambda$ passes through $+1$ or $-1$ . Passage through $+1$ is where cycles are created or destroyed in pairs or exchanged (saddle-node, transcritical, pitchfork, as for flows on the line); passage through $-1$ is a flip bifurcation, at which the cycle keeps existing but its period doubles.

Example(Fixed points and the

2

-cycle of the logistic map)AdaptedAdapted from Strogatz (1994), §10.2, pp. 353-355; Strogatz (1994), §10.3, Examples 10.3.1-10.3.3, pp. 357-361

Take the logistic map

f_r(x) = r\,x\,(1 - x),

the standard test case for everything above, and run the recipe on it with the parameter kept general.

Step 1: an invariant interval. The graph is a downward parabola vanishing at $x = 0$ and $x = 1$ , with apex at $x = \tfrac12$ of height $r/4$ . Hence $f_r([0,1]) = [0, r/4]$ , which is contained in $[0,1]$ precisely when $0 \le r \le 4$ . Take $I = [0,1]$ and $r$ in that range.

Step 2: fixed points. $rx(1-x) = x$ holds at $x = 0$ and at $x^* = 1 - 1/r$ ; the second belongs to $I$ only once $r \ge 1$ .

Step 3: their multipliers. Since $f_r'(x) = r(1 - 2x)$ ,

f_r'(0) = r, \qquad f_r'(x^*) = r\Bigl(1 - 2 + \tfrac2r\Bigr) = 2 - r .

So the origin is a sink for $r < 1$ and a source for $r > 1$ , while $x^*$ is a sink exactly when $|2 - r| < 1$ , that is on $1 < r < 3$ . The two fixed points swap stability where they collide at $r = 1$ (a transcritical bifurcation), and at $r = 3$ the multiplier of $x^*$ arrives at $-1$ , which by step 7 is a flip.

Step 5: the $2$ -cycle, without solving a quartic. The equation $f_r^2(x) = x$ is a quartic whose roots include the two fixed points, and one route is to divide those out. It is quicker to work from the defining relations. Let $p \ne q$ be the two points of a $2$ -cycle, so $q = f_r(p)$ and $p = f_r(q)$ . Subtracting the relations,

p - q = f_r(q) - f_r(p) = r(q - p) - r(q^2 - p^2) = r(q-p)\bigl[1 - (p+q)\bigr],

and dividing by $q - p \ne 0$ leaves $-1 = r[1 - (p+q)]$ , so the sum is pinned down:

s := p + q = \frac{r+1}{r} .

Adding the relations instead, and using $p^2 + q^2 = s^2 - 2pq$ ,

s = r s - r\bigl(s^2 - 2pq\bigr), \qquad\text{so}\qquad 2r\,pq = s\,(1 - r + rs) = 2s ,

because $rs = r+1$ makes the bracket equal to $2$ . Hence $pq = s/r = (r+1)/r^2$ . Knowing the sum and the product, $p$ and $q$ are the roots of $t^2 - st + pq = 0$ , that is of

r^2 t^2 - r(r+1)\,t + (r+1) = 0, \qquad p, q = \frac{r + 1 \pm \sqrt{(r-3)(r+1)}}{2r} ,

whose discriminant is $s^2 - 4pq = (r+1)(r-3)/r^2$ . For $r > 0$ this is positive exactly when $r > 3$ : the $2$ -cycle exists only beyond $r = 3$ , born at the flip bifurcation, at the very parameter where the fixed point lost its stability.

Step 6: the multiplier of the cycle. The product formula needs only the two symmetric functions, so $p$ and $q$ never have to be evaluated:

\lambda = f_r'(p)\,f_r'(q) = r^2 (1 - 2p)(1 - 2q) = r^2\bigl[1 - 2s + 4pq\bigr] = r^2 - 2r(r+1) + 4(r+1) = 4 + 2r - r^2 .

At $r = 3$ this gives $\lambda = 1$ , as it must, since the cycle is just emerging from the fixed point there; $\lambda$ then decreases, and reaches $-1$ when $r^2 - 2r - 5 = 0$ , that is at

r = 1 + \sqrt6 = 3.449489\ldots

So the $2$ -cycle is a sink exactly for $3 < r < 1 + \sqrt6$ , and at the right-hand end it flips in turn and a $4$ -cycle appears. That is the first step of the cascade treated in Period doubling and universality.

Example(The logistic map at

r = 3.2

, in numbers)Original

Fix $r = 3.2$ in the family above and put every step to the test.

Both fixed points repel. $x^* = 1 - 1/3.2 = 0.6875$ with multiplier $\lambda = 2 - 3.2 = -1.2$ , and the origin has multiplier $3.2$ ; both exceed $1$ in absolute value, so no orbit other than the two constant ones can settle at a fixed point.

The $2$ -cycle. The discriminant is $(r-3)(r+1) = (0.2)(4.2) = 0.84$ , so

p = \frac{4.2 + \sqrt{0.84}}{6.4} = 0.799455490\ldots, \qquad q = \frac{4.2 - \sqrt{0.84}}{6.4} = 0.513044510\ldots,

and its multiplier is $\lambda = 4 + 6.4 - 10.24 = 0.16$ , comfortably inside $(-1,1)$ . The cycle is therefore a sink, and because $\lambda > 0$ the iterates of $f^2$ approach each cycle point from one side rather than alternating about it.

Check by iterating. Take the seed $x_0 = 0.2$ and iterate $x_{n+1} = 3.2\,x_n(1 - x_n)$ :

\begin{aligned} x_1 &= 0.512000000, & x_2 &= 0.799539200, & x_3 &= 0.512884057,\\ x_4 &= 0.799468803, & x_6 &= 0.799457619, & x_8 &= 0.799455831,\\ x_{12} &= 0.799455499, & x_{16} &= 0.799455491, & x_{20} &= 0.799455490 . \end{aligned}

The even iterates run to $p = 0.799455490467$ and the odd ones to $q = 0.513044509533$ , and the error should shrink by the factor $\lambda = 0.16$ every two steps. It does: $x_4 - p = 1.3313 \times 10^{-5}$ and $x_6 - p = 2.1280 \times 10^{-6}$ , a ratio of $0.15985$ against the predicted $0.16$ . (All the numbers in this block were computed at $30$ -digit precision and then rounded.)

A superstable cycle nearby. The $2$ -cycle is superstable where its multiplier vanishes, $4 + 2r - r^2 = 0$ , that is at

r = 1 + \sqrt5 = 3.236067977\ldots

By the product formula $\lambda = r^2(1-2p)(1-2q)$ can vanish only if one factor does, so one point of the cycle must be $x = \tfrac12$ , the apex of the parabola. And indeed at this $r$ ,

f\bigl(\tfrac12\bigr) = \frac{r}{4} = \frac{1 + \sqrt5}{4} = 0.809016994\ldots, \qquad f^2\bigl(\tfrac12\bigr) = r \cdot \frac r4 \cdot \Bigl(1 - \frac r4\Bigr) = \frac{r^2(4-r)}{16} = \frac12 ,

exactly, because $r^2 = 2r + 4$ turns $r^2(4-r)$ into $(2r+4)(4-r) = 16 + 4r - 2r^2 = 16 + 4r - 2(2r+4) = 8$ . This is what makes superstable parameters easy to locate numerically: a superstable cycle of a map with a single critical point always contains that critical point, so one solves $f_r^p(c) = c$ for $r$ with $c$ the critical point, instead of hunting for the cycle first.

ApplicationOriginal

Maps of an interval reach physics from two directions.

As reductions of flows. A continuous-time system is turned into a map by recording a trajectory only at selected moments: the crossings of a surface in phase space (Poincare maps), or the successive maxima of one coordinate, as Lorenz did for his equations. A periodic orbit of the flow becomes a fixed point or a cycle of the map, and its multiplier is exactly what decides the orbit's stability, so the apparatus above is the standard tool for the stability of periodic solutions. Because the map inherits the dimension of the surface minus nothing, the reduction is only one-dimensional when the attractor of the flow is nearly two-dimensional, as it is for strongly dissipative systems.

As models in their own right. Discrete time is the natural setting wherever the state is updated once per cycle rather than continuously: non-overlapping generations in population biology, where $f_r(x) = rx(1-x)$ was popularised as the simplest model with a carrying capacity; sampled-data and switching circuits; impact and kick maps in mechanics, where a particle is struck once per revolution; and the period-to-period return of a driven oscillator. The same map also appears in numerical analysis: Newton's iteration for $g(x) = 0$ is the map $f(x) = x - g(x)/g'(x)$ , and a simple root is a superstable fixed point of it, $f'(x) = g(x)g''(x)/g'(x)^2$ vanishing where $g$ does, which is the reason for the method's rapid convergence.

Theorem

Period doubling and universality

Adapted

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States Feigenbaum's universality — that every smooth one-parameter family of unimodal maps with a quadratic maximum reaches chaos through an infinite cascade of period doublings whose parameter spacings shrink by the same factor $\delta = 4.6692$ and whose cycles shrink by the same factor $\alpha = -2.5029$ — explains it by renormalisation, carries out an approximate renormalisation in closed form to get $\delta \approx 1 + \sqrt{17}$ , and treats intermittency, the other way a periodic window ends.

Prerequisites: One-dimensional maps, Bifurcations in one dimension

Definition(Cascades, the Feigenbaum constants, intermittency)AdaptedAdapted from Strogatz (1994), §10.2, pp. 353-355, §10.4, pp. 363-364, §10.6, pp. 370, 373, and §10.7, p. 380; Hirsch, Smale & Devaney (2013), §15.2, pp. 336-337

Let $f(x, r)$ be a smooth one-parameter family of maps of an interval into itself.

A map is unimodal on $[0,1]$ if it is smooth, vanishes at both endpoints, and has a single interior maximum. The maximum is quadratic if $f''$ is non-zero there; write $x_m$ for its location. The logistic map $rx(1-x)$ and the sine map $r\sin\pi x$ are the standard examples.
A cycle undergoes a flip (or period-doubling) bifurcation at $r = r_0$ if its multiplier passes through $-1$ there. By the multiplier criterion, this and passage through $+1$ are the only ways a cycle can change stability, so these are the only parameter values at which a bifurcation of cycles can occur.
A period-doubling cascade is an infinite sequence of flip bifurcations in which a $2^{n}$ -cycle loses stability and a $2^{n+1}$ -cycle takes over. Write $r_n$ for the parameter at which the $2^n$ -cycle is born, and $r_\infty = \lim_n r_n$ for the accumulation point of the cascade.
A cycle is superstable if its multiplier is $0$ . Because a superstable cycle of a unimodal map must contain the critical point $x_m$ , the superstable parameter values are easy to find numerically; write $R_n$ for the value at which the $2^n$ -cycle is superstable. Then $r_n < R_n < r_{n+1}$ , so the $R_n$ and the $r_n$ accumulate at the same point.
The Feigenbaum constants are
$\delta = \lim_{n\to\infty}\frac{r_n - r_{n-1}}{r_{n+1} - r_n}, \qquad \alpha = \lim_{n\to\infty}\frac{d_n}{d_{n+1}},$
where $d_n$ is the signed distance from $x_m$ to the nearest other point of the $2^n$ -cycle, measured at $r = R_n$ . The same limits are obtained with $R_n$ in place of $r_n$ in the first.
Intermittency is long stretches of nearly periodic motion interrupted at unpredictable times by short chaotic bursts. The kind considered here, type I, occurs just outside a periodic window, where a stable and an unstable cycle have just annihilated in a tangent (saddle-node) bifurcation.

Theorem(Feigenbaum universality)AdaptedAdapted from Strogatz (1994), §10.6, pp. 370, 373-374, and §10.7, pp. 383-384

Consider families of unimodal maps undergoing a period-doubling route to chaos.

Qualitative universality (the U-sequence). Metropolis, Stein and Stein proved that for all families $x_{n+1} = r f(x_n)$ with $f$ unimodal and $f(0) = f(1) = 0$ — the precise conditions are in their paper — the order in which stable cycles appear as $r$ increases does not depend on $f$ . Up to period $6$ that order is
$1,\ 2,\ 2^2,\ 6,\ 5,\ 3,\ 2\cdot3,\ 5,\ 6,\ 4,\ 6,\ 5,\ 6 .$
Only the shape of $f$ matters, not its algebraic form; the parameter values at which the cycles appear are not universal.
Quantitative universality. For every such family whose maximum is quadratic, the limits $\delta$ and $\alpha$ exist and take the same values,
$\delta = 4.669\ldots, \qquad \alpha = -2.5029\ldots,$
independent of the family. The negative sign of $\alpha$ records that the nearest cycle point lies alternately above and below $x_m$ . A maximum of a different order — quartic, say — gives a different pair of constants.
The functional equation. The mechanism is renormalisation. With the origin of $x$ moved to $x_m$ , the rescaled iterates converge,
$g_i(x) = \lim_{n\to\infty}\alpha^n f^{(2^n)}\!\Bigl(\frac{x}{\alpha^n},\ R_{n+i}\Bigr),$
to universal functions $g_i$ independent of $f$ . At the accumulation point of the cascade no shift of parameter is needed, and the limit $g = g_\infty$ satisfies
$g(x) = \alpha\, g\bigl(g(x/\alpha)\bigr), \qquad g'(0) = 0, \quad g(0) = 1 ,$
from which $\alpha = 1/g(1)$ . No closed form for $g$ is known; a power series $g(x) = 1 + c_1x^2 + c_2x^4 + \cdots$ substituted into the functional equation and truncated at seven terms gave Feigenbaum $c_1 = -1.5276$ , $c_2 = 0.1048$ and $\alpha = -2.5029$ . The corresponding account of $\delta$ requires the spectrum of the linearisation of the renormalisation operator on a space of functions, which is beyond what we develop here; neither Strogatz nor Hirsch, Smale and Devaney proves it, and neither do we.

Lemma(An approximate renormalisation that can be done in closed form)AdaptedAdapted from Strogatz (1994), §10.7, pp. 384-387

Near a flip bifurcation every smooth family looks, after translating $x$ and $\mu$ and rescaling $x$ , like

f_\mu(x) = -(1+\mu)x + x^2 ,

whose fixed point at $x = 0$ has multiplier $-(1+\mu)$ and so flips at $\mu = 0$ . For this family:

For $\mu > 0$ there is a $2$ -cycle $\{p, q\}$ with $p + q = \mu$ and $pq = -\mu$ , namely $p, q = \tfrac12\bigl(\mu \pm \sqrt{\mu^2+4\mu}\bigr)$ .
Writing $x = p + \eta$ and discarding terms of order $\eta^3$ , the second iterate obeys
$\eta_{n+1} = \bigl(1 - 4\mu - \mu^2\bigr)\eta_n + C(\mu)\,\eta_n^2, \qquad C(\mu) = \mu^2 + 4\mu - 3\sqrt{\mu^2+4\mu} .$
Rescaling by $\tilde x = C(\mu)\eta$ puts this back into the original form with the renormalised parameter
$\bar\mu = T(\mu) = \mu^2 + 4\mu - 2 .$
Consequently the parameter $\mu_k$ at which a $2^k$ -cycle is born satisfies $T(\mu_{k+1}) = \mu_k$ with $\mu_1 = 0$ ; the sequence increases to the fixed point
$\mu^* = \tfrac12\bigl(-3 + \sqrt{17}\bigr) = 0.5615528 ,$
which is the accumulation point of the cascade in this approximation, and the convergence is geometric with ratio
$T'(\mu^*) = 2\mu^* + 4 = 1 + \sqrt{17} = 5.1231 ,$
the approximation's value of $\delta$ . The rescaling factor at the accumulation point,
$C(\mu^*) = \frac{1+\sqrt{17}}{2} - 3\sqrt{\frac{1+\sqrt{17}}{2}} = -2.2399 ,$
is the approximation's value of $\alpha$ . Both are within about $10\%$ of the true constants.

Idea of the proofOriginal

The orbit diagram of a unimodal map repeats itself: each branch of the figtree, magnified in the $x$ direction and stretched in the $r$ direction, looks like the whole tree. The reason is that the second iterate of the map, restricted to a small interval around the critical point and with the parameter shifted to the next superstable value, is again a unimodal map with a quadratic maximum — an inverted and shrunken copy of the original. So one step of the cascade for $f$ is one step of the cascade for $f^2$ , which is the same problem over again. The whole of the cascade therefore comes from iterating a single operation on maps: take the second iterate, rescale $x$ by a factor $\alpha$ , and shift the parameter. The constants are the properties of that operation at its fixed point — $\alpha$ is the rescaling the fixed point demands, and $\delta$ is the rate at which the parameter shift converges, so they cannot depend on the family one started with. The Lemma carries out exactly this programme with the smallest possible model of a map, a quadratic kept only to second order, where every step can be done by hand.

ProofAdaptedAdapted from Strogatz (1994), §10.7, pp. 384-387

We prove the Lemma. Throughout, $S = \sqrt{\mu^2 + 4\mu}$ and $a = -(1+\mu)$ , so that $f_\mu(x) = ax + x^2$ . Assume $\mu > 0$ , so $S$ is real and positive.

The $2$ -cycle. A $2$ -cycle $\{p,q\}$ satisfies $f(p) = q$ and $f(q) = p$ . Subtracting the two equations,
$p - q = a(q - p) + q^2 - p^2 = (q-p)\bigl[a + (p+q)\bigr],$
and since $p \ne q$ we may divide by $p - q$ to get $-1 = a + (p+q)$ , that is $p + q = -1 - a = \mu$ . Adding the two equations instead,
$p + q = a(p+q) + p^2 + q^2 = a\mu + \mu^2 - 2pq ,$
so $\mu = -(1+\mu)\mu + \mu^2 - 2pq = -\mu - 2pq$ , giving $pq = -\mu$ . Hence $p$ and $q$ are the roots of $t^2 - \mu t - \mu = 0$ , which are real and distinct for $\mu > 0$ . Take $p$ to be the root with the $+$ sign.
The second iterate near $p$ . Put $x = p + \eta$ . Then
$f(p + \eta) = a(p+\eta) + (p+\eta)^2 = \underbrace{ap + p^2}_{=\,q} + (a + 2p)\eta + \eta^2 ,$
so $f(p+\eta) = q + u$ with $u = (a+2p)\eta + \eta^2$ , and applying $f$ again,
$f^2(p+\eta) = aq + q^2 + (a+2q)u + u^2 = p + (a+2q)(a+2p)\,\eta + \bigl[(a+2q) + (a+2p)^2\bigr]\eta^2 + O(\eta^3),$
where the $\eta^2$ terms come from $(a+2q)\eta^2$ and from the leading part of $u^2$ . Now $2p = \mu + S$ and $2q = \mu - S$ , and $a = -1-\mu$ , so
$a + 2p = -1 + S, \qquad a + 2q = -1 - S .$
Hence the coefficient of $\eta$ is $(-1-S)(-1+S) = 1 - S^2 = 1 - 4\mu - \mu^2$ , and the coefficient of $\eta^2$ is
$C = (-1 - S) + (-1 + S)^2 = -1 - S + 1 - 2S + S^2 = S^2 - 3S = \mu^2 + 4\mu - 3\sqrt{\mu^2+4\mu},$
which is part 2. (That the linear coefficient is the multiplier of the $2$ -cycle is the chain-rule product formula of One-dimensional maps; here it has been obtained directly.)
Rescaling. Multiply the recursion of part 2 by $C$ and set $\tilde\eta = C\eta$ :
$\tilde\eta_{n+1} = \bigl(1 - 4\mu - \mu^2\bigr)\tilde\eta_n + \tilde\eta_n^2 .$
This has exactly the algebraic form $-(1+\bar\mu)\tilde\eta + \tilde\eta^2$ of the original map, provided we define $\bar\mu$ by $-(1+\bar\mu) = 1 - 4\mu - \mu^2$ , that is $\bar\mu = \mu^2 + 4\mu - 2 = T(\mu)$ .
Iterating the renormalisation. The renormalised map describes the dynamics of $f^2$ near $p$ , so the renormalised family flips — at $\bar\mu = 0$ — precisely when the $2$ -cycle of the original family doubles into a $4$ -cycle. More generally, the renormalised family gives birth to a $2^k$ -cycle exactly when the original gives birth to a $2^{k+1}$ -cycle. Therefore
$T(\mu_{k+1}) = \mu_k , \qquad \mu_1 = 0 .$
Solving for the forward iteration, $T(\mu) = \mu_{k}$ means $(\mu+2)^2 = 6 + \mu_k$ , and the relevant root is
$\mu_{k+1} = -2 + \sqrt{6 + \mu_k} .$
The map $\mu \mapsto -2 + \sqrt{6+\mu}$ is increasing, is a contraction on $[0, 1]$ (its derivative there is $\tfrac12(6+\mu)^{-1/2} \le \tfrac12\cdot6^{-1/2} < 1$ ), and maps $[0,1]$ into itself since it sends $0$ to $0.4495$ and $1$ to $0.6458$ . So the sequence converges monotonically to the unique fixed point in $[0,1]$ , which satisfies $\mu = \mu^2 + 4\mu - 2$ , that is $\mu^2 + 3\mu - 2 = 0$ , with positive root $\mu^* = \tfrac12(-3 + \sqrt{17}) = 0.5615528$ .

For the rate: $\mu_k - \mu^* = T(\mu_{k+1}) - T(\mu^*)$ , so by the mean value theorem $\mu_k - \mu^* = T'(c_k)(\mu_{k+1} - \mu^*)$ with $c_k$ between $\mu_{k+1}$ and $\mu^*$ , hence $c_k \to \mu^*$ and
$\frac{\mu_k - \mu^*}{\mu_{k+1} - \mu^*} = T'(c_k) \longrightarrow T'(\mu^*) = 2\mu^* + 4 = 1 + \sqrt{17} = 5.1231 .$
The constant $\delta$ was defined by the spacings rather than by the distances to $\mu^*$ , but the two give the same number. Write $e_k = \mu_k - \mu^*$ and $\rho_k = e_{k-1}/e_k$ , so $\rho_k \to T'(\mu^*)$ by the display above. Then
$\frac{\mu_k - \mu_{k-1}}{\mu_{k+1} - \mu_k} = \frac{e_k - e_{k-1}}{e_{k+1} - e_k} = \frac{\rho_{k+1} - \rho_k\rho_{k+1}}{1 - \rho_{k+1}} \longrightarrow \frac{T'(\mu^*) - T'(\mu^*)^2}{1 - T'(\mu^*)} = T'(\mu^*),$
on dividing numerator and denominator by $e_{k+1}$ . So the spacings shrink by the same factor $1+\sqrt{17}$ , which is the approximation's $\delta$ . Finally, $C$ was the factor by which $\eta$ had to be rescaled at each renormalisation, which is the role played by $\alpha$ ; evaluating at the fixed point, with $\mu^{*2} + 4\mu^* = \mu^* + 2 = \tfrac12(1+\sqrt{17})$ ,
$C(\mu^*) = \frac{1+\sqrt{17}}{2} - 3\sqrt{\frac{1+\sqrt{17}}{2}} = 2.5615528 - 4.8014557 = -2.2399027 . \qquad\square$

Remark(Where the hypotheses are used)Original

A quadratic maximum is what fixes the values of the constants. In the Lemma this enters at the very first line: the model map was truncated after $x^2$ , and it is the $x^2$ term that supplied $C$ and hence $\alpha$ . A map whose maximum is quartic renormalises to a different fixed function and has its own pair of constants. This is also the sense in which $g$ is universal: the argument $x/\alpha^n$ shrinks to zero, so only the behaviour of $f$ in a vanishing neighbourhood of the maximum survives, and all that survives of that is the order of the maximum.
Unimodality is what makes the renormalised map a map of the same kind, with one critical point in the interval being magnified. A map with two humps can renormalise to something else, and the U-sequence does not apply to it.
Where the Lemma is only an approximation. Two things were discarded: the $O(\eta^3)$ terms at each renormalisation, and the dependence of the family on $\mu$ beyond linear order. Both are genuinely present, and the price is the $10\%$ error in $\delta$ and $\alpha$ . What survives exactly is the structure: that there is a renormalisation operation at all, that it has a fixed point, and that the constants are a derivative and a scale factor at that fixed point. The exact theory replaces the one-parameter map $T$ by an operator on a space of functions; its fixed point is $g$ , and $\delta$ is the one eigenvalue of modulus greater than $1$ of its linearisation.
What the cascade does not require. Nothing above needs the map to be the logistic map, or even to have a formula. It needs a family of maps of an interval with one quadratic maximum. That is why the same constants appear in flows: if a flow is dissipative enough that its return map is nearly one-dimensional and unimodal — as it is for the Rossler system, and apparently for convection in mercury — the theory applies to the flow.
The constants are limits. The ratios approach $\delta$ and $\alpha$ only in the limit, and the early ratios are noticeably off, as the computation below shows. An experiment that resolves four doublings is measuring something still several percent from $\delta$ , which is one reason measured values cluster near $4.4$ rather than $4.669$ .

Remark(Intermittency: the other way a window ends)AdaptedAdapted from Strogatz (1994), §10.4, pp. 361-365

A periodic window in the orbit diagram begins with a tangent bifurcation and ends with a cascade. For the logistic map's period- $3$ window, which begins at $r = 1 + \sqrt8 = 3.8284271\ldots$ — a value due to Myrberg and, as Strogatz notes, surprisingly hard to derive — the graph of $f^3$ becomes tangent to the diagonal at three points simultaneously, creating a stable and an unstable $3$ -cycle there. Raising $r$ further, the stable $3$ -cycle period-doubles through its own cascade, with the same $\delta$ , and a miniature copy of the whole orbit diagram appears inside the window.

Lowering $r$ below $1+\sqrt8$ instead produces intermittency. The three tangency points become three narrow channels between the graph of $f^3$ and the diagonal. An orbit that enters a channel takes many iterations to squeeze through it, and while it does, $f^3(x_n) \approx x_n$ , so the motion looks like the $3$ -cycle that no longer exists. On leaving the channel the orbit wanders chaotically until it falls into a channel again, at an unpredictable time.

How long the laminar phase lasts. Near the tangency the third iterate is, to leading order, $f^3(x) \approx x + a(x - x_c)^2 + c\,\epsilon$ with $a, c > 0$ and $\epsilon = r_c - r$ . Treating $n$ as continuous, $dx/dn \approx a(x-x_c)^2 + c\epsilon$ , so the number of iterations to cross the channel is

n \approx \int_{-\infty}^{\infty}\frac{ds}{as^2 + c\epsilon} = \frac{\pi}{\sqrt{ac\,\epsilon}} \propto \epsilon^{-1/2} .

So the laminar stretches lengthen without bound as the window is approached, and the bursts become rarer; moving away from the window they become more frequent until the motion is fully chaotic. This progression is the intermittency route to chaos, and it has been observed in lasers, in fluids and in chemical reactions. The $\epsilon^{-1/2}$ law is checked numerically below.

Method(Measuring

\delta

)AdaptedAdapted from Strogatz (1994), §10.7, pp. 380-381, and §10.6, pp. 374-376

In a simulation, do not chase the bifurcation values $r_n$ : they are the parameters at which a multiplier passes through $-1$ , which is hard to localise because nothing visible happens at the cycle itself. Use the superstable values $R_n$ instead, and exploit the fact that a superstable cycle contains the critical point $x_m$ .

Locate $x_m$ , the critical point of the family.
For each $n$ , solve $f^{(2^n)}(x_m, r) = x_m$ for $r$ . This is one scalar equation in one unknown and is solved by bisection or by a secant method.
Bracket the root: the next $R_n$ lies close to $R_{n-1} + (R_{n-1} - R_{n-2})/\delta$ , so use that as the prediction and search a small interval around it, taking the root nearest the prediction — the equation has many other roots, belonging to cycles of other periods.
Work in extended precision. Each level needs about $\log_{10}\delta \approx 0.67$ more decimal digits in $r$ than the last, and $2^n$ iterations of the map amplify rounding error, so double precision runs out near $n = 9$ or $10$ .
Form the ratios $(R_n - R_{n-1})/(R_{n+1} - R_n)$ and extrapolate; the accumulation point follows from $R_\infty \approx R_n + (R_n - R_{n-1})/(\delta - 1)$ .

In an experiment, step 1 is unavailable, so one measures the $r_n$ directly from time series and accepts a few percent of error. Strogatz tabulates the results: $4.3 \pm 0.8$ from convection in water (Giglio and others, 1981), $4.4 \pm 0.1$ from convection in mercury (Libchaber and others, 1982, over four doublings), $4.5 \pm 0.6$ and $4.3 \pm 0.1$ from diode circuits, $4.7 \pm 0.3$ from a transistor circuit, and $4.5 \pm 0.3$ from a simulated Josephson junction. Since each further doubling needs about a fivefold improvement in the resolution of the control parameter, and noise blurs high-period orbits, about five doublings is the practical limit.

Example(The logistic cascade, computed)Original

We carried out the method above for $f(x,r) = rx(1-x)$ , with $x_m = \tfrac12$ , in $60$ -digit arithmetic. Solving $f^{(2^n)}(\tfrac12, r) = \tfrac12$ gives the superstable parameters

\begin{array}{llll} R_0 = 2 & R_1 = 3.236067977499790 & R_2 = 3.498561699327702 & R_3 = 3.554640862768825 \\ R_4 = 3.566667379856269 & R_5 = 3.569243531637110 & R_6 = 3.569795293749945 & R_7 = 3.569913465422349 \\ R_8 = 3.569938774233305 & R_9 = 3.569944194608065 & & \end{array}

The first two are exact: $R_0 = 2$ because $f(\tfrac12) = r/4 = \tfrac12$ , and $R_1 = 1 + \sqrt5 = 3.2360679774997897$ , as computed in One-dimensional maps.

The constant $\delta$ . The successive ratios $(R_n - R_{n-1})/(R_{n+1} - R_n)$ are

4.70894,\quad 4.68077,\quad 4.66296,\quad 4.66840,\quad 4.66895,\quad 4.66916,\quad 4.66919,\quad 4.669199 ,

for $n = 1, \dots, 8$ . They settle on $4.66920$ , against the accepted $\delta = 4.6692016$ . Note that the first ratio is off by nearly $1\%$ and the third actually overshoots: the convergence is not monotone, which is worth knowing before reading much into an experiment that resolves four doublings.

The accumulation point. Extrapolating, $R_\infty \approx R_9 + (R_9 - R_8)/(\delta - 1) = 3.5699456719$ , in agreement with the value $3.569946$ that Strogatz obtains from the $r_n$ .

The constant $\alpha$ . At each $R_n$ we took $d_n$ to be the displacement from $x_m = \tfrac12$ to the point of the $2^n$ -cycle half a period away, having checked for every $n \le 8$ that this is indeed the nearest cycle point to $x_m$ . The values alternate in sign:

d_1 = 0.3090170,\quad d_2 = -0.1164018,\quad d_3 = 0.0459752,\quad d_4 = -0.0183262,\quad d_5 = 0.0073184,

and so on down to $d_8 = -4.666897\times10^{-4}$ . The ratios $d_n/d_{n+1}$ are

-2.65474,\quad -2.53184,\quad -2.50872,\quad -2.50411,\quad -2.50316,\quad -2.50296,\quad -2.50292 ,

converging to $-2.50292$ against the accepted $\alpha = -2.5029079$ . The sign alternation predicted in the statement is visible in the $d_n$ themselves.

Comparison with the Lemma. The approximate renormalisation gave $\delta \approx 1 + \sqrt{17} = 5.1231$ (high by $9.7\%$ ), $\alpha \approx -2.2399$ (low in magnitude by $10.5\%$ ), and an accumulation point at $\mu^* = 0.5616$ , which corresponds to $r = 3 + \mu^* = 3.5616$ for the logistic map since $\mu = 0$ was placed at the first flip, $r = 3$ . The true value is $3.5699$ : an error of $0.2\%$ , far better than the error in $\delta$ , which is what one expects from a calculation that approximates the renormalisation map $T$ well near its fixed point but its derivative there less well.

Example(Intermittency near the period-3 window, measured)Original

Set $r = r_c - \epsilon$ with $r_c = 1 + \sqrt8 = 3.8284271$ , iterate $f^3$ for $4\times10^6$ steps after discarding a transient of $10^4$ , and call a step laminar when $|f^3(x) - x| < 10^{-2}$ . The mean length $L$ of a run of consecutive laminar steps, in iterations of $f^3$ , came out as

\begin{array}{lllll} \epsilon = 10^{-3}: & L = 4.70 & \quad & L\sqrt\epsilon = 0.1486 \\ \epsilon = 3\times10^{-4}: & L = 8.87 & & L\sqrt\epsilon = 0.1537 \\ \epsilon = 10^{-4}: & L = 15.43 & & L\sqrt\epsilon = 0.1543 \\ \epsilon = 3\times10^{-5}: & L = 28.25 & & L\sqrt\epsilon = 0.1547 \\ \epsilon = 10^{-5}: & L = 48.57 & & L\sqrt\epsilon = 0.1536 \end{array}

Across the whole table a hundredfold decrease in $\epsilon$ multiplies $L$ by $48.57/4.70 = 10.33$ , against the predicted $10$ ; over the last four rows, where $\epsilon$ falls by a factor $30$ , it multiplies $L$ by $48.57/8.87 = 5.476$ , against the predicted $\sqrt{30} = 5.477$ . The product $L\sqrt\epsilon$ is constant to within $1\%$ over the last four rows, which is the $\epsilon^{-1/2}$ law. (The value of the constant $0.154$ is not universal: it depends on the threshold $10^{-2}$ chosen to call a step laminar, and on $a$ and $c$ , which are properties of this particular map at this particular tangency. The exponent is what the argument predicts.)

ApplicationOriginal

Feigenbaum's constants are the reason a one-dimensional toy map is taken seriously as a model of a laboratory system. A convecting layer of mercury has an enormous number of degrees of freedom and nothing in its equations resembles $rx(1-x)$ ; yet if the dissipation is strong enough that the attractor is barely more than two-dimensional, the return map of the flow is nearly one-dimensional and unimodal, and the ratio of successive bifurcation intervals must then be $4.669$ . That is a parameter-free prediction about a real fluid, and it is why Libchaber's group went to the trouble of stabilising the roll structure with a magnetic field: increasing the damping is exactly what pushes a system into the regime where the theory applies.

The practical uses follow from that. Observing a few period doublings with roughly the right ratio in a measured signal — from a laser, a chemical reactor, a cardiac preparation, a nonlinear circuit — is evidence that the irregularity to come is low-dimensional chaos rather than noise or high-dimensional turbulence, and therefore that a model with two or three active variables may be worth building. Conversely, when the ratios come out wrong, the one-dimensional picture is the wrong picture. Where the behaviour is unwanted, the cascade also says where to intervene: since the route is a sequence of flips, keeping a multiplier away from $-1$ — by adding damping, or by small feedback corrections applied once per cycle — suppresses the whole cascade at its first step.

Intermittency plays the same diagnostic role for the other common route. Nearly periodic behaviour interrupted by irregular bursts, with the burst rate growing as a control parameter is moved, points to a saddle-node bifurcation of cycles rather than to a cascade; and the $\epsilon^{-1/2}$ law for the laminar lengths gives a quantitative test and, if it holds, a way of locating the critical parameter from data taken away from it.

Method

Poincare maps

Adapted

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Replaces a closed orbit of a flow by a fixed point of the first-return map on a transverse surface, shows that this map is well defined and continuously differentiable near the orbit, identifies the eigenvalues of its linearisation with the nontrivial Floquet multipliers of the orbit — so that they decide its stability and their product is $\exp\int_0^T \nabla\cdot f$ — and works the machinery through two systems whose return maps can be written down.

Prerequisites: Limit cycles and the Poincare-Bendixson theorem, One-dimensional maps, Floquet theory, Dependence on initial data and Gronwall's inequality, Implicit and inverse function theorems, Jordan normal form

Definition(Surface of section, return map, multipliers)AdaptedAdapted from Strogatz (1994), §8.7, pp. 278-281; Hirsch, Smale & Devaney (2013), §10.2, p. 217, and §10.3, p. 219

Let $f$ be continuously differentiable on an open set $U \subseteq \mathbb{R}^n$ and let $\phi_t$ be the flow of $\dot x = f(x)$ (Existence and uniqueness of solutions).

A surface of section is a set $S \subseteq U$ of dimension $n-1$ that the flow crosses cleanly. Concretely we take

S \;=\; \{\, x \in W : g(x) = 0 \,\}, \qquad W \subseteq U \text{ open},

where $g : W \to \mathbb{R}$ is continuously differentiable and

\nabla g(x) \cdot f(x) \;\neq\; 0 \qquad\text{for every } x \in S .

This last condition is transversality: the velocity never lies in the tangent plane $\ker \nabla g(x)$ of $S$ , so no trajectory grazes $S$ or runs along it, and all trajectories pierce $S$ in the same direction, the one in which $g$ increases. For $n = 2$ this is the local section already used in the Poincare-Bendixson theorem, where $g(x) = (x - x_0)\cdot f(x_0)$ and $S$ is a short straight segment.

Fix $x \in S$ . If there is a smallest $t > 0$ with $\phi_t(x) \in S$ , write $\tau(x)$ for it — the return time — and set

P(x) \;=\; \phi_{\tau(x)}(x) .

The map $P$ , defined on the set of points of $S$ that do return, is the Poincare map, or first-return map, of $S$ . It turns the continuous time of the flow into the integer time of a map of the $(n-1)$ -dimensional set $S$ into itself, so the whole apparatus of one-dimensional maps and its higher-dimensional analogue applies to it.

The point of the construction is the dictionary it sets up. A closed orbit $\gamma$ of least period $T$ that meets $S$ at $x^*$ and nowhere else in one circuit gives $P(x^*) = x^*$ with $\tau(x^*) = T$ ; conversely a fixed point of $P$ is a point whose trajectory returns to itself, hence a closed orbit. A $p$ -cycle of $P$ is a closed orbit that threads $S$ $p$ times before closing up.

Let $\gamma$ be such a closed orbit and $x^*$ its intersection with $S$ . The derivative $DP(x^*)$ , restricted to the tangent plane $T_{x^*}S = \ker\nabla g(x^*)$ , is the linearised Poincare map: an $(n-1)\times(n-1)$ matrix once a basis of that plane is fixed. Its eigenvalues $\lambda_1, \dots, \lambda_{n-1}$ are the characteristic multipliers, or Floquet multipliers, of $\gamma$ . The name is the one used for the eigenvalues of the monodromy matrix of the $T$ -periodic variational equation $\dot v = Df(\gamma(t))v$ (Floquet theory); the statement below shows that the two lists agree once the monodromy matrix's unavoidable eigenvalue $1$ — the trivial multiplier, belonging to displacements along the orbit, which are time shifts and not real perturbations — is deleted. Strogatz records the same convention.

A time-periodic system $\dot x = F(x,t)$ with $F(x, t+T) = F(x,t)$ is covered by this: adjoin $\dot\theta = 1$ on $\mathbb{R}^n \times \mathbb{R}/T\mathbb{Z}$ and take $S = \{\theta = 0\}$ , which is transverse because $\nabla\theta \cdot (F,1) = 1$ . Every point returns, with $\tau \equiv T$ , and $P$ is the stroboscopic map: photograph the state once per drive period.

Method(The return map exists, is smooth, and decides stability)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §10.2, Proposition, p. 217, and §10.3, p. 219; Strogatz (1994), §8.7, pp. 281-282

Let $f \in C^1(U)$ , let $\gamma$ be a closed orbit of least period $T > 0$ , let $x^* \in \gamma$ , and let $S$ be a surface of section containing $x^*$ , as above.

(Well defined and $C^1$ .) There is a relatively open neighbourhood $V$ of $x^*$ in $S$ and a continuously differentiable $\tau : V \to \mathbb{R}$ with $\tau(x^*) = T$ such that $\phi_{\tau(x)}(x) \in S$ for every $x \in V$ , and $\tau(x)$ is the first positive return time for $x$ close enough to $x^*$ . Hence $P(x) = \phi_{\tau(x)}(x)$ is continuously differentiable on a neighbourhood of $x^*$ in $S$ , with $P(x^*) = x^*$ .
(Attracting case.) If every eigenvalue of $DP(x^*)$ satisfies $|\lambda_j| < 1$ , then $\gamma$ is asymptotically orbitally stable: there is a neighbourhood $N$ of $\gamma$ in $U$ such that every trajectory starting in $N$ stays in $N$ and its distance to $\gamma$ tends to $0$ .
(Repelling case in the plane.) For $n = 2$ , $DP(x^*)$ is the single number $P'(x^*)$ , and the criterion is exactly the sink/source criterion for a fixed point of a map on an interval (One-dimensional maps): $|P'(x^*)| < 1$ gives an attracting closed orbit, $|P'(x^*)| > 1$ a repelling one, and $|P'(x^*)| = 1$ settles nothing.
(Borderline.) If $\max_j|\lambda_j| = 1$ the linearisation does not decide the stability of $\gamma$ ; this is where closed orbits bifurcate.

Theorem(The multipliers, and what their product must be)Original

In the setting above, let $\Phi(t)$ be the matrix solution of the variational equation along $\gamma$ ,

\dot\Phi = Df\bigl(\gamma(t)\bigr)\,\Phi, \qquad \Phi(0) = I ,

with $\gamma(0) = x^*$ , so that $M = \Phi(T)$ is the monodromy matrix of $\gamma$ (Floquet theory). Then:

(The two lists of multipliers agree.) The characteristic polynomial of $M$ factors as
$\det(\mu I - M) \;=\; (\mu - 1)\,\det\bigl(\mu I - DP(x^*)\bigr) .$
So the eigenvalues of $M$ are $1$ together with the eigenvalues $\lambda_1, \dots, \lambda_{n-1}$ of the linearised Poincare map, counted with multiplicity.
(No choice matters.) Consequently $\lambda_1, \dots, \lambda_{n-1}$ depend only on the orbit $\gamma$ : not on the surface of section, not on the point of $\gamma$ at which it is placed, and not on the coordinates used on $S$ .
(Their product is fixed by the divergence.)
$\prod_{j=1}^{n-1}\lambda_j \;=\; \det DP(x^*) \;=\; \exp\int_0^T \nabla\cdot f\bigl(\gamma(t)\bigr)\,dt .$
In particular, in the plane there is only one multiplier and it is given by this formula, with no linearised map to compute; and in a system of constant divergence $-k$ the product of the multipliers is $e^{-kT}$ , so at least one multiplier has modulus below $1$ whenever $k > 0$ .

Idea of the proofOriginal

Everything rests on one observation: hitting $S$ is a scalar equation, $g(\phi_t(x)) = 0$ , and transversality says that its $t$ -derivative does not vanish at the solution we already know. The implicit function theorem therefore solves it for $t$ as a smooth function of the starting point, and that function is the return time.

Once $P$ is differentiable, stability is the fixed-point question for a map, and a map whose derivative has spectral radius below $1$ contracts in a suitably weighted norm — which is what the Jordan form provides.

For the multipliers, differentiate the definition $P(x) = \phi_{\tau(x)}(x)$ by the product rule. The first term is $\Phi(T)$ acting on the perturbation; the second is a multiple of the velocity $f(x^*)$ , because wobbling the return time only slides the point along its own trajectory. So in the splitting $\mathbb{R}^n = \mathbb{R}f(x^*)\oplus T_{x^*}S$ the monodromy matrix is block triangular with the blocks $1$ and $DP(x^*)$ , and both the factorisation of the characteristic polynomial and the determinant formula fall out, the latter combined with Liouville's formula for $\det\Phi$ .

Proof(Part 1: existence and smoothness of the return map)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §10.2, proof of the Proposition, p. 218

Define $G(x,t) = g(\phi_t(x))$ on a neighbourhood of $(x^*, T)$ in $U\times\mathbb{R}$ . The flow is continuously differentiable in both arguments, by smooth dependence on initial data, and so is $g$ ; hence $G$ is continuously differentiable.

The equation holds at the known point. $G(x^*,T) = g(\phi_T(x^*)) = g(x^*) = 0$ , because $x^*$ has period $T$ and lies on $S$ .
The $t$ -derivative is nonzero there. By the chain rule and $\frac{d}{dt}\phi_t(x) = f(\phi_t(x))$ ,
$\frac{\partial G}{\partial t}(x^*,T) \;=\; \nabla g\bigl(\phi_T(x^*)\bigr)\cdot f\bigl(\phi_T(x^*)\bigr) \;=\; \nabla g(x^*)\cdot f(x^*) \;\neq\; 0 ,$
which is exactly the transversality hypothesis at $x^*$ . This is the only place it is used, and without it the argument stops here.

By the implicit function theorem there are a neighbourhood $V_0$ of $x^*$ in $U$ and a continuously differentiable $\tau : V_0 \to \mathbb{R}$ with $\tau(x^*) = T$ and $G(x,\tau(x)) \equiv 0$ , that is, $\phi_{\tau(x)}(x) \in S$ . Put $V = V_0 \cap S$ .

It remains to check that $\tau(x)$ is the first return, not a later one. Two facts combine. First, trajectories leave $S$ immediately: since $\frac{d}{dt}g(\phi_t(x)) = \nabla g\cdot f \neq 0$ on $S$ and this quantity is continuous, there is $\sigma > 0$ and a neighbourhood of $x^*$ on which $t \mapsto g(\phi_t(x))$ is strictly monotone for $|t| \le \sigma$ ; so no point near $x^*$ meets $S$ at a time in $(0,\sigma]$ . Second, on the compact time interval $[\sigma, T-\sigma]$ the orbit of $x^*$ stays at a positive distance $d$ from $S$ , because it does not meet $S$ there and both sets are closed in the relevant compacta; by continuity of $(x,t)\mapsto\phi_t(x)$ on the compact set $\{x^*\}\times[\sigma,T-\sigma]$ , every $x$ close enough to $x^*$ keeps its orbit within $d/2$ of that of $x^*$ on $[\sigma,T-\sigma]$ and so also misses $S$ there. Shrinking $V$ to the intersection of these neighbourhoods, $\tau(x)$ is the first positive return time, and $P = \phi_{\tau(\cdot)}(\cdot)$ is continuously differentiable on $V$ as a composite of continuously differentiable maps. $\square$

Proof(Parts 2-4: stability)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §10.3, p. 219; Strogatz (1994), §8.7, pp. 281-282

An adapted norm. Write $A = DP(x^*)$ , acting on the $(n-1)$ -dimensional space $E = T_{x^*}S$ , and let $\rho(A) = \max_j|\lambda_j|$ be its spectral radius. We use: for every $\rho > \rho(A)$ there is a norm $\|\cdot\|_*$ on $E$ with $\|Av\|_* \le \rho\|v\|_*$ for all $v$ . To see this, put $A$ in Jordan form, $A = Q J Q^{-1}$ , and rescale the basis vectors of each Jordan block by successive powers of $\delta = \rho - \rho(A) > 0$ ; in the rescaled basis the superdiagonal entries become $\delta$ instead of $1$ , and the resulting matrix has every row sum of moduli at most $\rho(A)+\delta = \rho$ , so the induced maximum norm of $A$ is at most $\rho$ . (For complex eigenvalues work over $\mathbb{C}$ and take the real part of the resulting norm; the bound is unchanged.)

Part 2. Suppose $\rho(A) < 1$ and fix $\rho$ with $\rho(A) < \rho < 1$ ; take $\|\cdot\|_*$ as above and set $\rho' = (1+\rho)/2 < 1$ . Since $P$ is continuously differentiable at $x^*$ ,

P(x) - x^* \;=\; A\,(x - x^*) + o\bigl(\|x - x^*\|_*\bigr) ,

so there is $\delta > 0$ such that $\|P(x) - x^*\|_* \le \rho' \|x - x^*\|_*$ whenever $\|x-x^*\|_* \le \delta$ and $x \in S$ . The ball $B = \{x \in S: \|x - x^*\|_* \le \delta\}$ is therefore mapped into itself, and by induction

\|P^k(x) - x^*\|_* \;\le\; (\rho')^k\,\|x - x^*\|_* \longrightarrow 0 .

So $x^*$ attracts every point of $B$ under $P$ , geometrically.

This is a statement about the discrete times at which a trajectory crosses $S$ ; the continuous-time statement follows from it. The return times $\tau$ are continuous and $\tau(x^*) = T$ , so after shrinking $\delta$ we may assume $T/2 \le \tau \le 2T$ on $B$ . Every trajectory starting at a point $x \in B$ therefore consists of the arcs $\{\phi_t(P^k(x)) : 0 \le t \le \tau(P^k(x))\}$ , $k = 0,1,2,\dots$ , laid end to end. Each arc starts within $(\rho')^k\delta$ of $x^*$ and lasts at most $2T$ ; by continuous dependence on initial data over the bounded time interval $[0,2T]$ , for every $\varepsilon > 0$ there is $\eta > 0$ such that starting within $\eta$ of $x^*$ keeps the whole arc within $\varepsilon$ of the corresponding arc of $\gamma$ , hence within $\varepsilon$ of $\gamma$ . Since $(\rho')^k\delta \to 0$ , all but finitely many arcs are within $\varepsilon$ of $\gamma$ , and the finitely many early ones are within $\varepsilon$ too once $\delta$ is chosen below the $\eta$ belonging to $\varepsilon$ . Thus the trajectory stays in the $\varepsilon$ -neighbourhood $N$ of $\gamma$ and its distance to $\gamma$ tends to $0$ .

Finally, a point of $N$ not on $S$ flows to $S$ in time at most $2T$ provided $N$ is thin enough, by the flow-box argument of part 1 applied around each point of $\gamma$ and compactness of $\gamma$ ; so the conclusion covers a full neighbourhood of $\gamma$ and not merely of $x^*$ .

Part 3. For $n = 2$ the section is a segment, $E$ is one-dimensional and $A$ is the number $P'(x^*)$ . Part 2 gives the attracting case. For $|P'(x^*)| > 1$ , the source half of the multiplier criterion of One-dimensional maps says that every orbit of $P$ other than $x^*$ itself leaves a fixed neighbourhood of $x^*$ in $S$ ; since points of $S$ near $x^*$ are points near $\gamma$ , $\gamma$ is not stable.

Part 4. The logistic examples recorded in One-dimensional maps already show, for $n = 2$ , that a multiplier of modulus $1$ is compatible with an attracting, a repelling and a one-sided closed orbit, so no criterion phrased in $DP(x^*)$ alone can decide. $\square$

Proof(The multiplier theorem)Original

Write $v = f(x^*)$ , which is nonzero because $\gamma$ is a nonconstant periodic orbit, and $E = T_{x^*}S = \ker\nabla g(x^*)$ . Transversality says $\nabla g(x^*)\cdot v \neq 0$ , that is, $v \notin E$ ; since $\dim E = n-1$ , this gives the direct sum

\mathbb{R}^n \;=\; \mathbb{R}v \,\oplus\, E .

The monodromy matrix fixes the flow direction. Differentiating $\dot\gamma = f(\gamma)$ in $t$ shows that $t\mapsto f(\gamma(t))$ solves the variational equation; it equals $v$ at $t=0$ , so $\Phi(t)v = f(\gamma(t))$ and in particular

Mv \;=\; \Phi(T)v \;=\; f(\gamma(T)) \;=\; f(x^*) \;=\; v .

The monodromy matrix on $E$ . Let $w \in E$ , and differentiate $P(x) = \phi_{\tau(x)}(x)$ at $x^*$ in the direction $w$ by the chain rule, using $\tau(x^*) = T$ and $\partial_t\phi_t(x^*)|_{t=T} = f(x^*) = v$ :

DP(x^*)\,w \;=\; \underbrace{D_x\phi_T(x^*)}_{=\;\Phi(T)\;=\;M}\,w \;+\; v\,\bigl(D\tau(x^*)w\bigr) .

(The identity $D_x\phi_T(x^*) = \Phi(T)$ is the definition of the variational equation: the derivative of the flow with respect to its initial point solves it with the identity as initial value.) Writing $c(w) = D\tau(x^*)w$ , a linear functional on $E$ , this rearranges to

M w \;=\; -c(w)\,v \;+\; DP(x^*)\,w , \qquad w \in E .

Reading off the matrix. In a basis consisting of $v$ followed by a basis of $E$ , the two displays say that $M$ has the block form

M \;=\; \begin{pmatrix} 1 & -c \\ 0 & DP(x^*)\end{pmatrix} ,

upper block triangular, because $Mv = v$ has no $E$ -component and $Mw$ has $E$ -component $DP(x^*)w$ . The characteristic polynomial of a block triangular matrix is the product of those of its diagonal blocks, which is part 1; taking determinants gives $\det M = \det DP(x^*)$ .

Part 2. By Floquet theory (Floquet theory) the eigenvalues of the monodromy matrix of a periodic linear system are unchanged if the fundamental matrix is normalised at a different time: replacing the base point $\gamma(0)$ by $\gamma(s)$ replaces $M$ by a matrix similar to it. The eigenvalues of $M$ are therefore an attribute of $\gamma$ alone. Part 1 identifies the eigenvalues of $DP(x^*)$ as those of $M$ with one copy of $1$ removed, and nothing in that list refers to $S$ or to the coordinates on it. Hence the $\lambda_j$ are attributes of $\gamma$ alone. (That one copy of $1$ is removed and not more is the content of the factorisation: if $1$ happens to be an eigenvalue of $DP(x^*)$ as well, then $M$ has $1$ as an eigenvalue of algebraic multiplicity at least two, and the trivial multiplier is still a single factor $\mu - 1$ .)

Part 3. Liouville's formula for the variational equation — derived in Lyapunov exponents — gives

\det\Phi(t) \;=\; \exp\int_0^t \operatorname{tr}Df\bigl(\gamma(s)\bigr)\,ds \;=\; \exp\int_0^t \nabla\cdot f\bigl(\gamma(s)\bigr)\,ds ,

since the trace of the Jacobian is the divergence. At $t = T$ the left side is $\det M = \det DP(x^*) = \prod_j \lambda_j$ . For $n = 2$ the product is a single multiplier. If $\nabla\cdot f \equiv -k$ then the product is $e^{-kT}$ , which for $k>0$ is less than $1$ in modulus, so not all $|\lambda_j|$ can be $\ge 1$ . $\square$

Remark(Where the hypotheses are used)Original

Transversality is used exactly twice, and both uses are essential. It makes $\partial_t G \neq 0$ , which is what the implicit function theorem needs to produce the return time; and it makes $f(x^*) \notin T_{x^*}S$ , which is what splits $\mathbb{R}^n$ into the flow direction and the section and so block-triangularises $M$ . A section tangent to the flow at some point gives no return map near that point: trajectories can touch it and turn back, so "the next crossing" is not even locally well defined.
$f(x^*) \neq 0$ , that is, the orbit is a genuine closed orbit and not an equilibrium. Without it $v = 0$ , the splitting collapses, and there is no trivial multiplier to remove.
Least period $T$ . If $T$ is taken as a multiple $kT_0$ of the least period, the orbit crosses $S$ $k$ times in a circuit, $x^*$ is a $k$ -cycle of $P$ rather than a fixed point, and the multipliers come out as $k$ -th powers. The statements above are for a section crossed once per circuit.
$C^1$ and no more. Only continuous differentiability of $f$ and $g$ is used; $P$ is then $C^1$ , which is all the stability argument needs. If $f$ is $C^r$ so is $P$ , by the same implicit function theorem.
What part 2 does not say. It gives attraction when all $|\lambda_j| < 1$ . The converse at the nonlinear level — that some $|\lambda_j| > 1$ forces instability — is true but needs an invariant-manifold argument that we do not give; what is immediate is the linear statement, that the recursion $v_{k+1} = DP(x^*)v_k$ then has geometrically growing solutions. In the plane there is no gap, because part 3 reduces the question to a map of an interval, where both directions are proved.
Orbital, not pointwise. Even for a strongly attracting limit cycle, a nearby trajectory does not converge to $\gamma(t)$ for the same $t$ ; it converges to the set $\gamma$ , generally with a persistent phase lag. That is why the trivial multiplier is $1$ and not something smaller: a phase shift is neither damped nor amplified.

Method(Using a section)AdaptedAdapted from Strogatz (1994), §8.7, pp. 278-283; Hirsch, Smale & Devaney (2013), §10.3, pp. 219-220

To study a suspected closed orbit of $\dot x = f(x)$ :

Choose the section. Pick a hypersurface $S = \{g = 0\}$ that the orbit must cross and check $\nabla g\cdot f \neq 0$ on the part of $S$ you will use. Natural choices: a half-plane $\{\theta = \theta_0\}$ for a system that circulates; the set where one coordinate is at a maximum, $\{\dot z = 0,\ \ddot z<0\}$ ; and, for a $T$ -periodic forcing, the stroboscopic section $\{t \equiv 0 \bmod T\}$ , which is transverse automatically.
Integrate to the next crossing. From $x \in S$ , integrate $\dot x = f(x)$ until $g$ changes sign, then locate the crossing time accurately by solving $g(\phi_t(x)) = 0$ — for instance by Newton's method on $t$ , whose derivative $\nabla g\cdot f$ is the transversality quantity and so is nonzero. This evaluates $P(x)$ .
Find the fixed point. Solve $P(x) = x$ on $S$ , numerically if need be; $n-1$ equations in $n-1$ unknowns. Each solution is a closed orbit.
Get $DP$ . Either integrate the variational equation $\dot\Phi = Df(\phi_t(x))\Phi$ alongside the trajectory and project, or difference $P$ : form $[P(x^*+h e_i) - P(x^*-h e_i)]/2h$ for a basis $e_i$ of $T_{x^*}S$ .
Read off the stability from the eigenvalues of $DP(x^*)$ : attracting if all lie strictly inside the unit circle, bifurcating when one reaches it. A multiplier leaving through $+1$ signals a saddle-node or pitchfork of cycles, through $-1$ a period doubling of the cycle, and a complex pair crossing together a torus (Neimark-Sacker) bifurcation.
Check the determinant. Compare $\prod_j\lambda_j$ with $\exp\int_0^T\nabla\cdot f\,dt$ computed along the orbit. For a system of constant divergence this is a one-line check that catches most integration errors; in the plane it replaces step 4 entirely.

Example(A return map in closed form, and its multiplier two ways)Original

On the punctured plane $U = \mathbb{R}^2\setminus\{0\}$ , in polar coordinates, take

\dot r \;=\; -\,r\ln r, \qquad \dot\theta \;=\; 1 .

The field is continuously differentiable on $U$ (it is not defined at the origin, which is why the origin is excluded). Take $S$ to be the positive $x$ -axis, $g(x,y) = y$ restricted to $x>0$ ; then $\nabla g\cdot f = \dot y = \dot r\sin\theta + r\dot\theta\cos\theta = r > 0$ on $S$ , so $S$ is transverse and every trajectory returns to it after exactly $\tau \equiv 2\pi$ , since $\dot\theta = 1$ .

The return map. Substituting $u = \ln r$ turns the radial equation into $\dot u = -u$ , so $u(t) = u_0e^{-t}$ and $r(t) = r_0^{\,e^{-t}}$ . Therefore, with $\kappa = e^{-2\pi}$ ,

P(r) \;=\; r^{\,\kappa}, \qquad \kappa \;=\; e^{-2\pi} \;=\; 0.00186744273\ldots

The closed orbit. $P(r) = r$ forces $r^{\kappa-1} = 1$ , hence $r = 1$ : the unit circle is the only closed orbit, and it has period $T = 2\pi$ .

The multiplier, from the map. $P'(r) = \kappa r^{\kappa-1}$ , so

\lambda \;=\; P'(1) \;=\; \kappa \;=\; e^{-2\pi} \;=\; 0.00186744273 .

Since $0 < \lambda < 1$ , the circle is an attracting limit cycle, approached monotonically from each side.

The multiplier, from the divergence. The divergence in polar coordinates is $\nabla\cdot f = r^{-1}\partial_r(rF_r) + r^{-1}\partial_\theta F_\theta$ with $F_r = -r\ln r$ and $F_\theta = r\dot\theta = r$ , so

\nabla\cdot f \;=\; \frac1r\,\frac{d}{dr}\bigl(-r^2\ln r\bigr) \;=\; -2\ln r - 1 ,

which equals $-1$ on the orbit $r=1$ . Part 3 of the multiplier theorem then predicts

\lambda \;=\; \exp\int_0^{2\pi}(-1)\,dt \;=\; e^{-2\pi} ,

the same number, obtained without knowing $P$ . This is the check of step 6, passed exactly.

Iterating the map. The cobweb recipe of One-dimensional maps applies to $P$ verbatim. Because $\ln P(r) = \kappa\ln r$ , the logarithms of the successive crossings are exactly geometric, $\ln r_k = \kappa^k\ln r_0$ . From $r_0 = 3$ :

r_1 = 1.0020537015,\quad r_2 = 1.0000038312,\quad r_3 = 1.0000000072,\quad r_4 = 1.0000000000134 ,

each excess over $1$ being about $\kappa = 1.87\times10^{-3}$ times the previous one, as the multiplier says. Integrating the Cartesian form of the system numerically from $(3,0)$ over $[0,2\pi]$ returns $(1.00205370150, -7\times10^{-13})$ , agreeing with $r_1$ to eleven digits and confirming the return time.

Example(A forced oscillator: the multiplier without the orbit)Original

Now a case where the closed orbit cannot be written down but the multiplier can. Consider the $2\pi$ -periodically forced equation

\dot x \;=\; (1 + \cos t)\,x \;-\; x^3 , \qquad x > 0 ,

a logistic-type growth whose growth rate is modulated once per period. Use the stroboscopic section $S = \{t \equiv 0 \bmod 2\pi\}$ , on which the return time is $\tau \equiv T = 2\pi$ .

Finding the fixed point. The equation is a Bernoulli equation: $u = x^{-2}$ obeys the linear equation

\dot u \;=\; -2(1+\cos t)\,u + 2 ,

whose integrating factor is $e^{\mu(t)}$ with $\mu(t) = 2t + 2\sin t$ . Since $\mu(2\pi) = 4\pi$ and $\mu(0)=0$ ,

u(2\pi) \;=\; e^{-4\pi}u(0) + K, \qquad K = e^{-4\pi}\!\!\int_0^{2\pi}\!\! 2\,e^{2s+2\sin s}\,ds \;=\; 0.52781657316 .

So in the variable $u$ the stroboscopic map is the affine map $u \mapsto e^{-4\pi}u + K$ , with the unique fixed point

u^* \;=\; \frac{K}{1 - e^{-4\pi}} \;=\; 0.52781841384 , \qquad x^* \;=\; (u^*)^{-1/2} \;=\; 1.37644141493 .

There is therefore exactly one $2\pi$ -periodic solution with $x>0$ , and it passes through $x = 1.37644141493$ at $t = 0$ .

The multiplier. The map $x\mapsto u = x^{-2}$ is a diffeomorphism of $(0,\infty)$ , so it conjugates the stroboscopic map in $x$ to the affine map in $u$ , and a multiplier is unchanged by a smooth change of variable. Hence

\lambda \;=\; e^{-4\pi} \;=\; 3.4873423562\times10^{-6} ,

far inside the unit circle: the periodic solution is violently attracting, which is why a numerical integration from almost any positive initial value looks periodic after one cycle.

Three checks. (i) Integrating the original equation from $x(0) = 1.37644141493$ over one period returns $1.3764414149302$ , reproducing the starting value to thirteen digits. (ii) Integrating the variational equation $\dot v = \bigl[(1+\cos t) - 3x(t)^2\bigr]v$ along that solution with $v(0)=1$ gives $v(2\pi) = 3.48734235622\times10^{-6}$ , matching $e^{-4\pi}$ to eleven digits — this is the divergence formula of part 3, since for the one-dimensional nonautonomous equation the bracket is the divergence. (iii) The periodic solution oscillates between $0.51956$ and $1.39773$ over the cycle, so the forcing is far from a small perturbation, yet the multiplier is exactly $e^{-4\pi}$ : the two $e^{-2\pi}$ -sized factors come from the mean growth rate $1$ and the mean of $-3x^2$ along the orbit, and the modulation $\cos t$ averages away.

ApplicationOriginal

A surface of section is the standard way to see a flow of three or more dimensions, because it trades one dimension of picture for one of integer time.

In mechanics. For a Hamiltonian system of two degrees of freedom the energy surface is three-dimensional, and a section of it is a two-dimensional picture in which a periodic orbit is a point, a quasiperiodic orbit an invariant curve, and a chaotic orbit a scatter of dots. This is the construction behind every Poincare-section plot of the Henon-Heiles system or of the restricted three-body problem; the multipliers of a periodic orbit come in reciprocal pairs there, because a Hamiltonian flow has zero divergence and part 3 forces the product to be $1$ .

In forced oscillators and circuits. The stroboscopic map turns a driven system into a map, and the bifurcations of the driven response are bifurcations of its fixed points: a multiplier leaving through $-1$ is the period doubling that halves the observed frequency, and a cascade of these is the route to chaos described in Period doubling and universality. The same reduction applied to a strongly dissipative flow, where all but one multiplier are tiny, is what makes a one-dimensional map a faithful model of a three-dimensional flow — the justification for studying the logistic map at all, and the step Lorenz took when he plotted successive maxima of $z$ for his equations.

In experiment and in numerics. Shaft-angle triggers in rotating machinery, the once-per-orbit crossings recorded in accelerator physics, and the fixed-step sampling of an experimental time series are all Poincare sections taken by the apparatus rather than by the analyst; the measured return map is then fitted directly. On the numerical side, steps 2-5 of the method are exactly how a periodic-orbit solver works: Newton's method on $P(x)-x$ , with the Jacobian supplied by the variational equation, converges quadratically to closed orbits that a plain time integration would take thousands of periods to reveal — or would never reveal at all, if the orbit is unstable.

Definition

Fractals and dimension

Adapted

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Defines the similarity, box and correlation dimensions, proves that a set built by keeping $m$ of the $r^D$ subcubes of a cube at every stage has box dimension $\ln m/\ln r$ — so the middle-thirds Cantor set, which is also shown to be compact, totally disconnected, uncountable and of measure zero, has dimension $0.6309$ — and tests the theorem against a box count of a planar dust of dimension $\ln 6/\ln 4 = 1.2925$ .

Prerequisites: One-dimensional maps, Sets, functions and countability, Compactness, Lebesgue measure

Definition(Self-similarity, and three dimensions)AdaptedAdapted from Strogatz (1994), §11.3, pp. 406-408, and §11.4, pp. 409-410, and §11.5, pp. 411-412; Folland (1999), §11.3, pp. 355-356, 359

A similitude of ratio $s \in (0,1)$ is a map $S : \mathbb{R}^D \to \mathbb{R}^D$ of the form $S(x) = s\,O x + c$ with $O$ orthogonal. A nonempty compact set $F$ is self-similar with $m$ pieces of ratio $s$ if there are similitudes $S_1,\dots,S_m$ of ratio $s$ with

F \;=\; S_1(F)\cup\dots\cup S_m(F) ,

the pieces overlapping only negligibly. Iterating, $F$ is a union of $m^k$ copies of itself at ratio $s^k$ , so it has detail at every scale: magnifying a small piece reproduces the whole. Writing $s = 1/r$ , the similarity dimension of $F$ is the exponent $d$ for which $m$ copies at ratio $1/r$ make up one copy at ratio $1$ , that is $m = r^d$ :

d_{\text{sim}} \;=\; \frac{\ln m}{\ln r} .

For a segment ( $m=r$ ), a square ( $m=r^2$ ) and a cube ( $m=r^3$ ) this returns $1$ , $2$ and $3$ , which is the reason for the definition; for the sets below it returns a number that is not an integer.

Self-similarity is special, so one wants a definition that applies to any bounded set. Let $F \subseteq \mathbb{R}^D$ be bounded and nonempty and, for $\varepsilon > 0$ , let

N(\varepsilon) \;=\; \text{the least number of cubes of side } \varepsilon \text{ whose union contains } F .

A smooth curve of length $L$ needs $N \approx L/\varepsilon$ cubes and a smooth surface of area $A$ needs $N \approx A/\varepsilon^2$ : the exponent in $N(\varepsilon) \propto \varepsilon^{-d}$ is the dimension. Taking that as the definition, the box dimension (also called the capacity) is

d_{\text{box}} \;=\; \lim_{\varepsilon\to0^+} \frac{\ln N(\varepsilon)}{\ln(1/\varepsilon)} ,

when the limit exists. Counting instead the cubes of a fixed mesh of side $\varepsilon$ that contain a point of $F$ changes $N(\varepsilon)$ by at most the factor $2^D$ — each mesh cube that meets $F$ is covered, and each covering cube meets at most $2^D$ mesh cubes — and a bounded factor does not survive division by $\ln(1/\varepsilon)$ . So the mesh count, which is what a computer does, gives the same dimension.

Finally, a definition for a set that carries a dynamics. Let $A$ be an attractor and $x_0, x_1, \dots, x_{n-1}$ a long trajectory on it. Fix a point $x$ of $A$ and let $N_x(\varepsilon)$ be the number of trajectory points within distance $\varepsilon$ of $x$ ; the pointwise dimension at $x$ is the exponent in $N_x(\varepsilon)\propto\varepsilon^{d}$ . Averaging over $x$ gives the correlation sum

C(\varepsilon) \;=\; \frac{\#\{(i,j) : i\neq j,\ |x_i - x_j| < \varepsilon\}}{n(n-1)} ,

and the correlation dimension $d_{\text{corr}}$ is the exponent in $C(\varepsilon)\propto\varepsilon^{\,d_{\text{corr}}}$ , read off as the slope of $\ln C$ against $\ln\varepsilon$ . Unlike the box dimension it weights a region by how often the trajectory visits it, so it describes the invariant measure on $A$ and not only the set; Strogatz records that in general $d_{\text{corr}} \le d_{\text{box}}$ , usually with little between them.

A closed set $F$ is a topological Cantor set if it is totally disconnected — its only connected subsets are single points, which for a subset of $\mathbb{R}$ means it contains no interval — and has no isolated point: every point of $F$ has other points of $F$ arbitrarily close. The two conditions pull in opposite directions, and that tension is what a Cantor set is. Neither mentions dimension or self-similarity, so the notion survives the continuous deformations under which those are destroyed.

Definition(Grid constructions)Original

All the explicit sets below are built the same way. Fix an integer $r \ge 2$ , a dimension $D \ge 1$ , and divide the unit cube $Q_0 = [0,1]^D$ into $r^D$ closed subcubes of side $1/r$ in the obvious grid. Choose $m$ of them, say the cubes with lower corners $c_1/r, \dots, c_m/r$ where each $c_i \in \{0,1,\dots,r-1\}^D$ , and let

S_i(x) \;=\; \frac{x + c_i}{r}, \qquad i = 1,\dots,m ,

so that $S_i(Q_0)$ is the $i$ -th chosen cube. Put $F_0 = Q_0$ and $F_{k+1} = S_1(F_k)\cup\dots\cup S_m(F_k)$ , and define the grid set

F \;=\; \bigcap_{k\ge0} F_k .

Thus $F_k$ is a union of $m^k$ closed cubes of side $r^{-k}$ — call them the stage- $k$ cubes — each stage- $k$ cube contains exactly $m$ stage- $(k+1)$ cubes, and distinct stage- $k$ cubes have disjoint interiors. The middle-thirds Cantor set $C$ is the case $D=1$ , $r=3$ , $m=2$ , $c_1 = 0$ , $c_2 = 2$ : remove the open middle third of $[0,1]$ , then of each remaining interval, and so on.

Definition(Grid sets and their dimension)Original

Let $F$ be a grid set as above, with $1 \le m \le r^D$ and the $c_i$ distinct.

( $F$ is a self-similar compact set.) $F$ is nonempty and compact, $F = S_1(F)\cup\dots\cup S_m(F)$ , and the $S_i$ are similitudes of ratio $1/r$ whose images meet only in sets of Lebesgue measure zero. Every stage- $k$ cube contains at least one point of $F$ .
(Box dimension.) The limit defining $d_{\text{box}}$ exists and
$d_{\text{box}}(F) \;=\; \frac{\ln m}{\ln r} \;=\; d_{\text{sim}}(F) .$
(Measure.) The Lebesgue measure of $F$ is $\lim_k m^k r^{-kD}$ , which is $0$ if $m < r^D$ and $1$ if $m = r^D$ (in which case $F = Q_0$ ).
(Cardinality and topology, for $m \ge 2$ .) $F$ is uncountable. If moreover no two chosen cubes touch — as happens for the middle-thirds Cantor set, where the two chosen intervals $[0,\frac13]$ and $[\frac23,1]$ are disjoint — then $F$ is a topological Cantor set.

Specialising to $D=1$ , $r=3$ , $m=2$ : the Cantor set $C$ is a compact topological Cantor set of Lebesgue measure zero, is uncountable, and has

d_{\text{box}}(C) \;=\; \frac{\ln 2}{\ln 3} \;=\; 0.6309297536 .

Idea of the proofOriginal

The covering number is handed to us at the special scales $\varepsilon = r^{-k}$ : the $m^k$ stage- $k$ cubes cover $F$ , so $N(r^{-k}) \le m^k$ , and since each of them contains a point of $F$ and a cube of side $r^{-k}$ can straddle at most $2^D$ cells of the stage- $k$ grid, $N(r^{-k}) \ge m^k/2^D$ . Both bounds give $\ln N/\ln(1/\varepsilon) = \ln m/\ln r$ up to a term of order $1/k$ . Intermediate scales are squeezed between two consecutive special ones, and the squeeze costs only a factor $m$ and a shift of one in $k$ , neither of which survives the limit.

The measure and the cardinality need no more than the construction: the areas multiply by $m/r^D$ at each stage, and the stage- $k$ cube containing a point of $F$ records a choice of one of $m$ branches, so $F$ is coded by infinite words in $m$ letters.

ProofOriginal

Part 1. Each $F_k$ is a finite union of closed cubes, hence compact, and $F_{k+1}\subseteq F_k$ : indeed $F_1 \subseteq F_0$ because the chosen subcubes lie in $Q_0$ , and applying the (inclusion-preserving) operation $E \mapsto \bigcup_i S_i(E)$ repeatedly keeps the inclusion. So $F$ is an intersection of a nested sequence of nonempty compact sets, hence compact and nonempty (Compactness). The same argument inside a fixed stage- $k$ cube $Q$ shows that $Q\cap F \neq \emptyset$ : the sets $Q \cap F_{k+j}$ , $j \ge 0$ , are nonempty — each stage- $(k+j)$ cube inside $Q$ contains $m\ge1$ stage- $(k+j+1)$ cubes — compact and nested.

For self-similarity, $S_i(F) = S_i\bigl(\bigcap_k F_k\bigr) = \bigcap_k S_i(F_k)$ , because $S_i$ is injective; hence

\bigcup_{i=1}^m S_i(F) \;=\; \bigcup_i \bigcap_k S_i(F_k) \;=\; \bigcap_k \bigcup_i S_i(F_k) \;=\; \bigcap_k F_{k+1} \;=\; F ,

where the middle exchange of union and intersection is legitimate because the sequences $S_i(F_k)$ are nested in $k$ and the union is finite. Each $S_i$ is a similitude of ratio $1/r$ with $O = I$ . Two distinct images $S_i(F)$ , $S_j(F)$ lie in distinct stage- $1$ cubes, which meet at most in a common face, a set of Lebesgue measure zero.

Part 2. Fix $k$ and put $\varepsilon = r^{-k}$ .

Upper bound. The $m^k$ stage- $k$ cubes have side $\varepsilon$ and cover $F$ , so $N(\varepsilon) \le m^k$ .
Lower bound. Let cubes $B_1,\dots,B_N$ of side $\varepsilon$ cover $F$ . Consider the grid of all $r^{kD}$ cells of side $\varepsilon$ in $Q_0$ . In each coordinate, an interval of length $\varepsilon$ meets at most two of the grid intervals of length $\varepsilon$ , so each $B_l$ meets at most $2^D$ grid cells. The $m^k$ stage- $k$ cubes are grid cells and each contains a point of $F$ by part 1, so each must be met by at least one $B_l$ . Counting the pairs (stage- $k$ cube, covering cube that meets it) in two ways gives $m^k \le 2^D N$ , that is $N(\varepsilon) \ge 2^{-D}m^k$ .

Therefore

\frac{k\ln m - D\ln 2}{k\ln r} \;\le\; \frac{\ln N(r^{-k})}{\ln(r^{k})} \;\le\; \frac{k \ln m}{k\ln r} ,

and both ends tend to $\ln m/\ln r$ as $k\to\infty$ .

Now let $\varepsilon$ be arbitrary and let $k$ be the integer with $r^{-(k+1)} < \varepsilon \le r^{-k}$ . Covering numbers are non-increasing in the size of the covering cube, so $N(r^{-k}) \le N(\varepsilon) \le N(r^{-(k+1)})$ , while $k\ln r \le \ln(1/\varepsilon) < (k+1)\ln r$ . Combining with the two displayed bounds,

\frac{k\ln m - D\ln 2}{(k+1)\ln r} \;\le\; \frac{\ln N(\varepsilon)}{\ln(1/\varepsilon)} \;\le\; \frac{(k+1)\ln m}{k\ln r} ,

and as $\varepsilon \to 0^+$ we have $k \to \infty$ and both bounds tend to $\ln m/\ln r$ . So the limit exists and has that value. It coincides with $\ln m/\ln r = d_{\text{sim}}$ by part 1.

Part 3. $F_k$ is a union of $m^k$ cubes of side $r^{-k}$ with disjoint interiors, so its Lebesgue measure is exactly $m^kr^{-kD} = (m/r^D)^k$ (Lebesgue measure). The $F_k$ decrease to $F$ and have finite measure, so the measure of $F$ is $\lim_k (m/r^D)^k$ , which is $0$ when $m < r^D$ and $1$ when $m = r^D$ .

Part 4. Assume $m \ge 2$ . Each point of $F$ lies in exactly one nested sequence of cubes — stage $1$ , stage $2$ , and so on — unless it lies on a shared face, in which case it lies in more than one; conversely every infinite word $i_1i_2i_3\cdots$ in the alphabet $\{1,\dots,m\}$ determines a nested sequence of stage- $k$ cubes whose intersection is a single point of $F$ (single, because the diameters $\sqrt{D}\,r^{-k}$ tend to $0$ ), and distinct words give distinct points whenever the corresponding cubes are eventually disjoint. To avoid the face-sharing nuisance, note simply that the map

\text{word } (i_1,i_2,\dots) \;\longmapsto\; \text{the point } \bigcap_k S_{i_1}\cdots S_{i_k}(Q_0)

is onto $F$ and is at most $2^D$ -to-one (a point lies in at most $2^D$ cubes of a grid). The set of words is uncountable, being $\{1,\dots,m\}^{\mathbb{N}}$ with $m\ge2$ (Sets, functions and countability); a set that is the image of an uncountable set under an at-most- $2^D$ -to-one map is uncountable, since a countable image would force the domain to be a countable union of finite sets. Hence $F$ is uncountable.

Suppose now that no two chosen cubes touch, and let $\delta>0$ be the least distance between two of them. Then two distinct stage- $k$ cubes inside a common stage- $(k-1)$ cube are at distance at least $\delta r^{-(k-1)} > 0$ .

No isolated points. Let $x\in F$ and $\eta>0$ . Choose $k$ with $\sqrt D\,r^{-k} < \eta$ and let $Q$ be a stage- $k$ cube containing $x$ . Its $m \ge 2$ stage- $(k+1)$ children each contain a point of $F$ by part 1, and at least one of those children does not contain $x$ ; the point of $F$ in it is distinct from $x$ and within $\operatorname{diam}Q < \eta$ . So $x$ is not isolated.
Totally disconnected. Let $x \ne y$ both lie in $F$ and choose $k$ with $\sqrt D\,r^{-k} < |x-y|$ , so $x$ and $y$ lie in different stage- $k$ cubes $Q_x \neq Q_y$ . Walk back to the first stage $j \le k$ at which their cubes differ; those two stage- $j$ cubes sit inside one stage- $(j-1)$ cube and are therefore at distance at least $\delta r^{-(j-1)} > 0$ . The open slab strictly between them (take any hyperplane separating the two cubes) misses $F_j \supseteq F$ entirely, so it disconnects $F$ into two relatively open pieces, one containing $x$ and the other $y$ . Hence no connected subset of $F$ contains two points. $\square$

Remark(Where the hypotheses are used, and what the box dimension is not)Original

Separation of the chosen cubes is used only in part 4, for the topological conclusions. Part 2 needs nothing of the kind: the chosen cubes may be adjacent, in which case $F$ is connected and still has dimension $\ln m/\ln r$ . What saves the dimension computation is that distinct cubes of a grid have disjoint interiors, which is enough for the two-way count.
The grid form is used in both bounds of part 2: that the stage- $k$ cubes are cells of one mesh of side $r^{-k}$ is what makes "at most $2^D$ cells per covering cube" available. For general similitudes with a common ratio — rotations allowed, as in the von Koch curve — the same conclusion holds under a separation condition, but the counting needs the volume argument of Folland's Lemma 11.32 rather than a grid.
Equal ratios. Nothing above covers pieces of different ratios $s_1,\dots,s_m$ . There the dimension is the root $p$ of $\sum_i s_i^{\,p} = 1$ , which reduces to $\ln m/\ln r$ when all $s_i = 1/r$ ; Folland states this and attributes it to Hutchinson, and we do not prove it.
The box dimension can be wrong. It is not countably additive and does not vanish on all countable sets: the rationals in $[0,1]$ have box dimension $1$ , as Strogatz notes, because any $\varepsilon$ -cover of a dense subset of $[0,1]$ is an $\varepsilon$ -cover of $[0,1]$ . Hausdorff's definition repairs this by allowing covers with sets of different small sizes: $H_p(A) = \lim_{\delta\to0}\inf\{\sum_j(\operatorname{diam}B_j)^p : A\subseteq\bigcup B_j,\ \operatorname{diam}B_j<\delta\}$ , and the Hausdorff dimension is the value of $p$ at which $H_p(A)$ drops from $\infty$ to $0$ . For the sets here the two agree: Folland's Theorem 11.33 gives Hausdorff dimension $\log_{1/s}m$ for the compact invariant set of $m$ similitudes of ratio $s$ admitting a separating set — an open bounded $U$ with $\bigcup_iS_i(U)\subseteq U$ and the $S_i(U)$ disjoint — and for a grid set the open unit cube is such a $U$ whenever the chosen cubes are distinct. In general $d_{\text{Hausdorff}} \le d_{\text{box}}$ and the inequality can be strict, the rationals again being the example.
The limit may not exist for a general bounded set, in which case one keeps the upper and lower limits separately. Part 2 proves existence for grid sets.
$d_{\text{corr}}$ is not a property of the set. It depends on the invariant measure, so two systems with the same attractor but different statistics can return different correlation dimensions. It is nevertheless the one that is measurable in practice, because the correlation sum needs only a list of points.

Method(Measuring a dimension)AdaptedAdapted from Strogatz (1994), §11.4, pp. 409-411, and §11.5, pp. 411-413

For a self-similar set given by a rule. Identify $m$ , the number of copies, and $r$ , the factor by which each is shrunk; then $d = \ln m/\ln r$ , with no computation. Check the rule really does reproduce the whole set, not merely a stage of the construction.

By box counting, for a set given as a list of points or by a construction.

Choose a sequence of scales $\varepsilon$ spaced evenly in $\ln\varepsilon$ , spanning the range between the resolution of your data and the diameter of the set. If the set is built on a grid of ratio $r$ , the scales $r^{-k}$ give the cleanest numbers, but any sequence converges to the same limit.
At each $\varepsilon$ , lay a mesh of side $\varepsilon$ and count the occupied cells $N(\varepsilon)$ — in practice by binning the coordinates, dividing by $\varepsilon$ , rounding down, and counting distinct index tuples.
Plot $\ln N$ against $\ln(1/\varepsilon)$ and fit a straight line over the scaling region: scales small enough that the mesh resolves the structure, but large enough that each occupied cell still holds many points. The slope is $d_{\text{box}}$ .
Expect the local slope to oscillate rather than settle, if the set has a preferred ratio $r$ : the count is exactly on the line only at $\varepsilon = r^{-k}$ , and in between it overshoots. Average over at least one full period in $\ln\varepsilon$ , or use only the scales $r^{-k}$ .

By the correlation sum, for a trajectory on an attractor. Discard the transient, store $n$ points, and compute $C(\varepsilon)$ as the fraction of pairs closer than $\varepsilon$ for a range of $\varepsilon$ ; the slope of $\ln C$ against $\ln\varepsilon$ over the scaling region is $d_{\text{corr}}$ . Two cautions: exclude pairs whose indices differ by less than a few correlation times, or consecutive points on one trajectory will inflate $C$ at small $\varepsilon$ and bias the slope upward; and expect saturation at $\varepsilon$ near the diameter of the attractor, where $C \to 1$ , and noise at $\varepsilon$ below the typical nearest-neighbour distance.

Example(A planar dust, by rule and by count)Original

Divide the unit square into a $4\times4$ grid of $16$ squares of side $\tfrac14$ , index them by $(i,j)$ with $i$ the column and $j$ the row, $i,j\in\{0,1,2,3\}$ , and keep the six squares

(0,0),\quad (0,3),\quad (1,1),\quad (2,2),\quad (3,0),\quad (3,3) .

Repeat inside each kept square for ever. Call the limit set $F$ : a dust of points scattered along a bent diagonal with two outlying clumps at each end.

By the rule. This is a grid set with $D = 2$ , $r = 4$ , $m = 6$ , and the six chosen squares are distinct, so part 2 of the theorem gives

d_{\text{box}}(F) \;=\; \frac{\ln 6}{\ln 4} \;=\; \frac{1.791759469}{1.386294361} \;=\; 1.2924812504 .

The open unit square is a separating set for the six maps $S_i(x) = (x+c_i)/4$ — their images are disjoint open squares inside it — so by Folland's theorem the Hausdorff dimension is the same number. The area is $(6/16)^k = (3/8)^k \to 0$ , so $F$ is a planar set of zero area and dimension strictly between $1$ and $2$ . Two of the chosen squares touch at a corner, $(0,0)$ and $(1,1)$ for instance, so part 4 does not apply and we do not claim that $F$ is totally disconnected.

By count. To test the theorem rather than trust it, we generated the $6^9 = 10\,077\,696$ stage- $9$ squares, of side $4^{-9} = 3.81\times10^{-6}$ , took their lower corners as a point set, and counted occupied cells of a mesh of side $\varepsilon$ for $\varepsilon = 10^{-j/2}$ :

\begin{array}{llll} \varepsilon = 10^{-1.5}: & N = 280 \qquad & \varepsilon = 10^{-3}: & N = 25\,140\\ \varepsilon = 10^{-2}: & N = 1\,230 & \varepsilon = 10^{-3.5}: & N = 112\,403\\ \varepsilon = 10^{-2.5}: & N = 5\,775 & \varepsilon = 10^{-4}: & N = 484\,884 \end{array}

A least-squares fit of $\ln N$ against $\ln(1/\varepsilon)$ over these six scales gives a slope of

1.2979 ,

and the chord from the first point to the last gives $1.2954$ , both within $0.4\%$ of the exact $1.2924812504$ . The residual wobble is the effect predicted in step 4 of the method: the local slope between consecutive scales ranges from $1.27$ to $1.34$ and returns to the exact value only at the scales $4^{-k}$ , where the count is exactly $6^k$ — we checked that it is, for $k = 1,\dots,6$ : $6, 36, 216, 1296, 7776, 46656$ . Below $\varepsilon \approx 10^{-5}$ the count saturates at $10\,077\,696$ , because the mesh has become finer than the stage- $9$ approximation; this is the lower end of the scaling region, and including those scales drags the fitted slope down to $1.05$ , which is the single most common way of getting a dimension wrong.

The one-dimensional case for comparison. The same theorem with $D = 1$ , $r = 3$ , $m = 2$ gives $d_{\text{box}}(C) = \ln2/\ln3 = 0.6309297536$ for the middle-thirds Cantor set, and with $D=1$ , $r=5$ , $m=3$ — keep the first, third and fifth fifths of every interval — gives $\ln3/\ln5 = 0.6826061944$ . The von Koch curve, four copies of itself at ratio $\tfrac13$ , has similarity dimension $\ln4/\ln3 = 1.2618595071$ ; it is not a grid set, since its copies are rotated, so our theorem does not cover it, and we quote it from Strogatz, with Folland's Corollary 11.34 giving the same value for its Hausdorff dimension.

ApplicationOriginal

Fractal dimension entered dynamics because the attractors of dissipative chaotic systems are not surfaces.

Counting the active degrees of freedom. A dissipative flow contracts phase-space volume, so its attractor has measure zero; but an attractor on which the motion is chaotic cannot be a point, a curve or a torus either. What is left is a set of non-integer dimension, and that number is the most compact summary of how complicated the long-term motion is. For an experimental time series, the correlation dimension computed from a reconstructed trajectory is the standard estimate: a low value, stable as the reconstruction dimension is increased, is evidence that a few variables suffice, while a value that keeps rising with the reconstruction dimension indicates noise and not low-dimensional chaos. Strogatz reports the value $2.05\pm0.01$ obtained by Grassberger and Procaccia for the Lorenz attractor at $\sigma=10$ , $b=8/3$ , $r=28$ , and $0.500\pm0.005$ for the attractor of the logistic map at the accumulation point of the period-doubling cascade, against a box dimension of about $0.538$ for that set — an attractor that is a Cantor-like set but, because the gaps in the orbit diagram scale differently in different places, is not strictly self-similar.

Why the cross section is a Cantor set. A strange attractor is made by stretching a sheet of phase space, folding it, and laying it back down. After $k$ rounds there are $2^k$ layers inside the original thickness, so a transverse cut through the attractor meets a set of points separated by gaps of every size — exactly the construction of this unit, run in the transverse direction. The attractor is then locally the product of a smooth sheet with a Cantor set, and its dimension exceeds the dimension of the sheet by the dimension of that set. This is the mechanism made explicit in The Henon map and strange attractors.

Outside dynamics. The same arithmetic is used for the measured roughness of surfaces, for the dimension of fracture and porous networks, and for percolation clusters, in each case to answer the question "how does the amount of the object within a box scale with the size of the box?" — the only question the box dimension answers, but one that controls transport, scattering and adsorption on such objects.

Example · advanced

The Henon map and strange attractors

Adapted

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Shows how repeated stretching, folding and re-injection builds a strange attractor, writes Henon's map as the composition of a fold, a contraction and a flip, proves it is invertible with constant area ratio $-b$ , locates its two saddles and an exact trapping quadrilateral at $a = 1.4$ , $b = 0.3$ , and measures the attractor there — Lyapunov exponents $0.4191$ and $-1.6231$ , correlation dimension $1.21$ , box dimension $1.25$ , and a transverse cut resolving into sheets in groups of one, two and three.

Prerequisites: Fractals and dimension, One-dimensional maps, Lyapunov exponents and sensitive dependence, Eigenvalues and eigenvectors, Compactness

Definition(Stretching, folding, re-injection, and Henon's map)AdaptedAdapted from Strogatz (1994), §12.0-12.1, pp. 423-425, and §12.2, pp. 429-430

A strange attractor — an attractor carrying chaotic motion, in the sense of Lyapunov exponents and sensitive dependence — has to do two incompatible-looking things at once. Nearby states must separate exponentially, yet the whole motion must stay inside a bounded set. The way out is stretching and folding. Take a small blob of initial conditions. Dissipation squeezes it flat in some directions while the instability draws it out in others; the stretching cannot continue indefinitely inside a bounded region, so the elongated blob is bent back over itself and laid down beside where it started. That last step is the re-injection. Rolling and folding dough is the standard picture: each pass doubles the number of layers and halves their thickness, so after many passes a cut across the slab meets a set of layers separated by gaps of every size, and a speck of colouring placed in the dough is drawn out and distributed through it. In the limit the layers accumulate on a set that is, transverse to the sheets, a Cantor set, and along the sheets a smooth curve or surface. That product structure — smooth one way, Cantor the other — is what "strange" means geometrically, and the exponential separation is the stretching seen in the tangent space.

Henon built the smallest concrete model of this mechanism. He wanted the stretch-and-fold action of the Lorenz system with the amount of dissipation under the author's control: Lorenz's flow squashes a phase volume by a factor of roughly $14{,}000$ in one circuit of the attractor, which packs the sheets far too tightly to be resolved numerically. A map, besides, iterates faster and more accurately than a differential equation is integrated. The Henon map is

T(x, y) \;=\; \bigl(\,y + 1 - a x^2,\ b x\,\bigr), \qquad a > 0,\ \ 0 < |b| < 1 ,

written as a difference scheme $x_{n+1} = y_n + 1 - a x_n^2$ , $y_{n+1} = b x_n$ . It is assembled from three elementary moves, each of which is one word of the mechanism above. Begin with a region elongated along the $x$ -axis.

Fold. $T'(x, y) = (x,\ 1 + y - a x^2)$ bends the region over a parabola; horizontal lines become downward parabolas, and $a$ sets how sharp the bend is.
Contract. $T''(x, y) = (b x,\ y)$ squeezes it horizontally by the factor $|b|$ . This is the dissipation, and because the region is now bent it also tightens the fold.
Flip. $T'''(x, y) = (y,\ x)$ reflects in the diagonal, returning the long axis to the horizontal so that the next pass can repeat the first two moves.

Composing, $T = T''' \circ T'' \circ T'$ , which is the formula above. The Henon attractor is the attractor obtained at the standard parameters $a = 1.4$ , $b = 0.3$ , which lie well inside the chaotic range: as $a$ is increased at fixed $b = 0.3$ the map passes from a stable fixed point through a stable $2$ -cycle and the rest of a period-doubling cascade (Period doubling and universality) into chaos interleaved with periodic windows.

Example(Four properties the map was designed to have)AdaptedAdapted from Strogatz (1994), §12.2, pp. 430-432, Examples 12.2.1-12.2.2

Let $a > 0$ and $0 < |b| < 1$ .

$T$ is invertible, so every state has exactly one past as well as one future. This is the feature that distinguishes $T$ from the logistic map, which also stretches and folds but is two-to-one, and it is what makes $T$ an honest analogue of a flow.
$T$ contracts areas, at the same rate at every point: the Jacobian determinant is the constant $-b$ . This is the counterpart of the constant negative divergence of the Lorenz field.
For suitable parameters there is a bounded trapping region $Q$ , a set with $T(Q) \subseteq Q$ ; the attractor lies inside it.
Unlike the Lorenz flow, whose orbits are all bounded, $T$ has orbits that run to infinity: far out, the term $-a x^2$ dominates and throws the orbit away.

Proposition(The two saddles, the escape radius and a trapping quadrilateral)Original

Let $a > 0$ and $0 < |b| < 1$ , and write $T(x,y) = (y + 1 - ax^2,\ bx)$ .

Inverse. $T$ is a bijection of $\mathbb{R}^2$ , with
$T^{-1}(X, Y) \;=\; \Bigl(\frac{Y}{b},\ X - 1 + \frac{a}{b^2}Y^2\Bigr) .$
Area. $DT(x,y) = \begin{pmatrix} -2ax & 1 \\ b & 0\end{pmatrix}$ , so $\det DT \equiv -b$ and every measurable set $E$ satisfies $\operatorname{area}(T(E)) = |b|\operatorname{area}(E)$ . Consequently any compact set $A$ with $T(A) = A$ has area zero.
Fixed points. $T$ has a fixed point $(x, bx)$ exactly when $a x^2 + (1-b)x - 1 = 0$ . Since $a > 0$ the discriminant $(1-b)^2 + 4a$ is positive, so there are always two of them,
$x_\pm \;=\; \frac{-(1-b) \pm \sqrt{(1-b)^2 + 4a}}{2a} ,$
with $x_- < 0 < x_+$ . The multipliers at $(x, bx)$ are the roots of $\mu^2 + 2ax\,\mu - b = 0$ , namely
$\mu_{\pm}(x) \;=\; -ax \pm \sqrt{a^2x^2 + b} \qquad (b > 0),$
and they satisfy $\mu_+\mu_- = -b$ , so $|\mu_+\mu_-| < 1$ : a fixed point is never a repeller.
Escape. Put
$R \;=\; \frac{(1 + |b|) + \sqrt{(1+|b|)^2 + 4a}}{2a} ,$
the positive root of $a R^2 - (1+|b|)R - 1 = 0$ . If an orbit satisfies $|x_{n}| > R$ and $|x_{n-1}| \le |x_{n}|$ for some $n$ , then $|x_m|$ increases strictly for $m \ge n$ and $|x_m| \to \infty$ . Consequently every compact set $A$ with $T(A) = A$ lies in the strip $|x| \le R$ .
A trapping region at the standard parameters. Let $a = 1.4$ , $b = 0.3$ and let $Q$ be the closed convex quadrilateral with vertices
$(-1.33,\ 0.41), \quad (1.33,\ 0.15), \quad (1.25,\ -0.13), \quad (-1.09,\ -0.46)$
taken in this (clockwise) order. Then $T(Q) \subseteq Q$ . Hence $A = \bigcap_{n \ge 0} T^n(Q)$ is a nonempty compact set with $T(A) = A$ , of area zero, and every orbit starting in $Q$ has $\operatorname{dist}(T^n(x,y), A) \to 0$ .

Idea of the proofOriginal

Everything here is elementary once one notices that the $y$ -equation is trivial: $y_{n+1} = bx_n$ merely stores the previous $x$ . So the map is really the scalar second-order recurrence $x_{n+1} = 1 - ax_n^2 + bx_{n-1}$ , and inverting it, finding its fixed points and bounding its orbits are all one-variable problems. Invertibility is reading that recurrence backwards; the fixed-point equation and the characteristic polynomial are quadratics; and escape follows because the quadratic term beats the linear ones once $|x|$ passes the root $R$ . The area statement is the chain rule plus the change-of-variables formula, and it gives the attractor zero area at once, since an invariant set would have to keep an area that is multiplied by $|b| < 1$ at every step. Only part 5 needs more than algebra, and even there the work collapses: because $Q$ is convex and $T$ moves points along horizontal lines in the image (the image height $bx$ depends only on $x$ ), it is enough to check that the upper and lower boundary curves of $Q$ land in $Q$ , and each such check is one quadratic inequality on an interval.

ProofOriginal

1. Inverse. Given $(X, Y)$ , the equations $X = y + 1 - ax^2$ and $Y = bx$ determine $(x,y)$ uniquely: the second gives $x = Y/b$ , which is legitimate because $b \neq 0$ , and substituting into the first gives $y = X - 1 + a(Y/b)^2$ . Both operations are defined for every $(X,Y)$ , so $T$ is a bijection with the stated inverse, and both $T$ and $T^{-1}$ are polynomial, hence smooth.

2. Area. Differentiating the two components of $T$ gives the matrix displayed, whose determinant is $(-2ax)\cdot 0 - 1\cdot b = -b$ for every $(x,y)$ . The change-of-variables formula for the diffeomorphism $T$ therefore gives $\operatorname{area}(T(E)) = \int_E |\det DT| = |b|\operatorname{area}(E)$ . If now $T(A) = A$ with $A$ compact, then $\operatorname{area}(A) = |b|\operatorname{area}(A)$ with $\operatorname{area}(A) < \infty$ by compactness, and since $|b| < 1$ this forces $\operatorname{area}(A) = 0$ .

3. Fixed points and multipliers. A fixed point must have $y = bx$ , and then $x = bx + 1 - ax^2$ , that is $ax^2 + (1-b)x - 1 = 0$ . The discriminant is $(1-b)^2 + 4a > 0$ because $a > 0$ , so there are two real roots; their product is $-1/a < 0$ , so one is negative and one positive. The Jacobian at $(x,bx)$ has trace $-2ax$ and determinant $-b$ , so its characteristic polynomial is $\mu^2 + 2ax\mu - b$ ; the quoted roots are the quadratic formula, real because $a^2x^2 + b > 0$ when $b > 0$ . Their product is the determinant $-b$ , whose modulus is $|b| < 1$ , so the two multipliers cannot both have modulus $\ge 1$ .

4. Escape. Write $t_m = |x_m|$ . The recurrence $x_{m+1} = 1 - ax_m^2 + bx_{m-1}$ gives the lower bound

t_{m+1} \;\ge\; a t_m^2 - 1 - |b|\,t_{m-1} .

Let $p(t) = at^2 - (1+|b|)t - 1$ ; this is the quadratic whose positive root is $R$ , and since $a > 0$ it is strictly increasing on $[R,\infty)$ with $p(t) > 0$ there. Suppose $t_m > R$ and $t_{m-1} \le t_m$ . Substituting $t_{m-1} \le t_m$ into the bound above,

t_{m+1} \;\ge\; a t_m^2 - |b| t_m - 1 , \qquad\text{so}\qquad t_{m+1} - t_m \;\ge\; a t_m^2 - (1+|b|) t_m - 1 \;=\; p(t_m) \;>\; 0 ,

the last inequality because $t_m > R$ . So $t_{m+1} > t_m > R$ , and the pair $(t_m, t_{m+1})$ again satisfies the hypothesis. By induction $t_m$ is strictly increasing for $m \ge n$ and stays above $R$ . It cannot converge to a finite limit $L$ : letting $m \to \infty$ in $t_{m+1} - t_m \ge p(t_m)$ would give $0 \ge p(L) > 0$ . An increasing sequence that does not converge tends to $+\infty$ . Now let $A$ be compact with $T(A) = A$ , and let $M = \max\{|x| : (x,y) \in A\}$ , which exists because $A$ is compact and the coordinate function is continuous. Choose $(x^*, y^*) \in A$ attaining it. Because $T(A) = A$ , every point of $A$ has a preimage in $A$ , so there is a bi-infinite orbit inside $A$ passing through $(x^*, y^*)$ ; along it $t_m \le M$ for every $m$ , and $t_n = M$ at the index of $(x^*,y^*)$ . In particular $t_{n-1} \le t_n$ . If $M > R$ the previous paragraph would make $t_m \to \infty$ , contradicting $t_m \le M$ . Hence $M \le R$ .

5. The trapping quadrilateral. Write $Q$ as the intersection of four closed half-planes $\{\,n_i \cdot p \le c_i\,\}$ , $i = 1, \dots, 4$ , one per edge, with the normals $n_i$ pointing out of $Q$ ; this is possible because the vertices listed are in convex position and in clockwise order, as the four cross products of consecutive edge vectors, all negative, confirm.

Reduction to the boundary. Let $x_\ell = -1.33$ and $x_r = 1.33$ be the smallest and largest abscissas of $Q$ . For $x \in [x_\ell, x_r]$ the vertical slice $Q \cap (\{x\}\times\mathbb{R})$ is the segment $\{x\}\times[\varphi_-(x), \varphi_+(x)]$ , where $\varphi_-$ and $\varphi_+$ are the piecewise-affine lower and upper boundary curves of $Q$ . The image of that slice is

T\bigl(\{x\}\times[\varphi_-(x), \varphi_+(x)]\bigr) \;=\; \bigl[\,1 - ax^2 + \varphi_-(x),\ 1 - ax^2 + \varphi_+(x)\,\bigr] \times \{bx\} ,

a horizontal segment, because the second coordinate of $T$ does not involve $y$ and the first is affine in $y$ with slope $1$ . A convex set contains a segment as soon as it contains both endpoints. So $T(Q) \subseteq Q$ holds if and only if $T$ maps the graphs of $\varphi_-$ and $\varphi_+$ into $Q$ .

The boundary check. Each of $\varphi_\pm$ is affine on each of the (four in total) edges of $Q$ : say $y = \alpha x + \beta$ for $x$ in an interval $[u, v]$ . Along such an edge,

T(x, \alpha x + \beta) \;=\; \bigl(\,1 + \beta + \alpha x - a x^2,\ b x\,\bigr) ,

so the requirement $n_i \cdot T(x, \alpha x + \beta) \le c_i$ reads $g(x) \le 0$ with

g(x) \;=\; -a\,n_{i,1}\,x^2 \;+\; \bigl(\alpha\,n_{i,1} + b\,n_{i,2}\bigr)x \;+\; n_{i,1}(1+\beta) - c_i ,

a quadratic in $x$ . Its maximum over $[u,v]$ is attained at $u$ , at $v$ , or (only if the leading coefficient is negative) at the vertex $-\,(\alpha n_{i,1} + b n_{i,2})/(-2an_{i,1})$ , so each condition is decided by evaluating $g$ at no more than three points. With $a = 7/5$ , $b = 3/10$ and the given vertices, all four edges and all four half-planes have rational data, and the $4 \times 4 = 16$ inequalities were evaluated in exact rational arithmetic. All sixteen hold; the largest value of $g/\|n_i\|$ over all of them is $-4.873 \times 10^{-4}$ , so $T(Q)$ lies inside $Q$ with a clearance of about $4.9 \times 10^{-4}$ at the tightest point, on the image of the top edge against the left edge.

The attractor. $Q$ is compact and $T$ continuous, so each $T^n(Q)$ is compact, and $T(Q) \subseteq Q$ gives $T^{n+1}(Q) \subseteq T^n(Q)$ : the sets are nested. A nested sequence of nonempty compact sets has nonempty compact intersection $A$ (compactness), and $T(A) = \bigcap_{n\ge1}T^n(Q) = A$ because the sequence is nested and $T$ is a homeomorphism onto its image. By part 2, $\operatorname{area}(A) = 0$ . For attraction, note that $T^n(Q)$ has area $|b|^n\operatorname{area}(Q) \to 0$ ; more directly, $\operatorname{dist}(p, A) \le \sup\{\operatorname{dist}(q, A) : q \in T^n(Q)\} =: d_n$ for $p \in T^n(Q)$ , and $d_n \to 0$ because the $T^n(Q)$ are nested compact sets with intersection $A$ : if $d_n \ge \varepsilon$ for all $n$ , pick $q_n \in T^n(Q)$ with $\operatorname{dist}(q_n, A) \ge \varepsilon$ ; a subsequence converges to some $q \in \bigcap_n T^n(Q) = A$ with $\operatorname{dist}(q, A) \ge \varepsilon$ , a contradiction. $\square$

Remark(Where the hypotheses are used)Original

$b \neq 0$ is what makes the map invertible. At $b = 0$ the second coordinate collapses to $0$ and $T$ degenerates to the one-dimensional quadratic map $x \mapsto 1 - ax^2$ , which is two-to-one. Part 1 divides by $b$ at exactly the step where this would fail. So the Henon map is a genuinely two-dimensional object only for $b \neq 0$ , and the quadratic family is its singular limit.
$|b| < 1$ is used only for dissipation. Parts 1, 3 and 4 hold for any $b \neq 0$ . Part 2's conclusion that an invariant compact set has zero area needs $|b| < 1$ ; at $|b| = 1$ the map preserves area, there can be no attractor at all, and the phase portrait is of the conservative kind discussed in Conservative and reversible systems.
$b > 0$ in the multiplier formula. For $b < 0$ the discriminant $a^2x^2 + b$ can be negative and the multipliers are then a complex-conjugate pair of modulus $\sqrt{|b|} < 1$ , a stable spiral. The displayed real formula is stated for $b > 0$ for that reason; the relation $\mu_+\mu_- = -b$ holds in all cases.
$a > 0$ in part 3. It is what makes the discriminant $(1-b)^2 + 4a$ positive, so that the two fixed points exist for every $a > 0$ . For $a < 0$ they can disappear: the fixed points are created in a saddle-node bifurcation at $a = -(1-b)^2/4$ .
The hypothesis $|x_{n-1}| \le |x_n|$ in part 4 cannot be dropped as stated. A single large $|x_n|$ does not by itself force escape, because the term $bx_{n-1}$ could in principle be large and negative and cancel the growth; the proof uses the inequality $|b||x_{n-1}| \le |b||x_n|$ at the first step. The last paragraph of the proof shows that the hypothesis costs nothing where it is needed: on a compact invariant set the orbit through a point of largest $|x|$ supplies it automatically.
Part 5 is for the stated parameters only. Nothing in the construction of $Q$ survives a change of $a$ or $b$ ; the clearance is only about $5\times10^{-4}$ , so even $a = 1.41$ would need a different quadrilateral. This is not fussiness but a feature of the map: the exact check above already fails for $Q$ at $a = 1.41$ , and at $b = 0.3$ there can be no trapping region at all once $a$ passes about $1.427$ , where in twenty runs from random starting points near the origin eighteen escaped within $2\times10^4$ iterations while none escaped at $a = 1.426$ .
What the Proposition does not prove. It gives a compact invariant attracting set $A$ of zero area. It does not prove that $A$ is a strange attractor: that the motion on it is chaotic, that $A$ is not simply a stable cycle, or that $A$ has the fractal structure seen in the example below. Those are genuinely hard; the chaotic behaviour at $a = 1.4$ , $b = 0.3$ is established here only numerically.

Remark(What the attractor looks like, and why)AdaptedAdapted from Strogatz (1994), §12.1, p. 425, and §12.2, pp. 433-434

Plotting ten thousand iterates from almost any starting point inside $Q$ draws a thin boomerang-shaped arc. Magnifying a small square of it does not smooth it out: the arc resolves into several nearly parallel curves, and magnifying one of those shows the same pattern again, at every scale that numerics can reach. Strogatz describes Henon's own sequence of magnifications as showing six curves in groups of one, two and three, with each curve of a group resolving into the same arrangement at the next level. This is the dough picture made visible: each application of $T$ folds the strip once more, so after $n$ steps a transverse cut meets a rapidly growing number of intervals — $2^n$ of them for the idealised dough — whose widths shrink geometrically, which is exactly the construction of a Cantor set. Locally the attractor is a smooth arc times a Cantor set, and its dimension is therefore between $1$ and $2$ .

There is a reason the longitudinal direction is smooth. The saddle fixed point lies on the attractor, and its unstable manifold is a smooth curve that the dynamics stretches along the attractor and folds back; Strogatz reports the theorem of Benedicks and Carleson that the Henon attractor is the closure of one branch of that unstable manifold. So the "curves" seen under magnification are successive passes of a single immersed curve, and the Cantor structure is the pattern in which those passes accumulate.

Method(Measuring a two-dimensional attractor)Original

Given a planar map $T$ depending on parameters, this is the sequence of computations that turns a picture into numbers. Each step has a check attached; skipping the checks is how wrong dimensions get published.

Fix the invariant structure first. Solve $T(p) = p$ and diagonalise $DT(p)$ at each fixed point. Classify: the saddles are the organising centres, and their unstable manifolds will trace the attractor.
Find a trapping region. Guess a convex polygon enclosing the plotted orbit, then verify $T(Q) \subseteq Q$ by the boundary reduction used above — for a polynomial map the verification is a finite set of polynomial inequalities and can be done exactly. If no polygon works, enlarge the parameter search: orbits may be escaping.
Discard the transient. Iterate a few thousand steps before recording anything, and confirm the recorded orbit stays in $Q$ .
Lyapunov exponents. Carry a tangent vector $u_{n+1} = DT(p_n)u_n$ alongside the orbit, renormalising $u$ to unit length each step and accumulating the logarithms of the norms; the average is $\lambda_1$ . Check: when $\det DT$ is a constant $c$ , the exponents must satisfy $\lambda_1 + \lambda_2 = \ln|c|$ exactly, so $\lambda_2$ comes free and any discrepancy is an error. Repeat from several seeds; the spread is the honest error bar.
Correlation dimension. Collect $N$ points, keeping only every $k$ th iterate so that successive stored points are dynamically decorrelated, and compute the correlation sum $C(\varepsilon)$ of Fractals and dimension with a $k$ -d tree. Plot $\ln C$ against $\ln\varepsilon$ and fit the slope over the middle of the range — above the scale at which only a handful of pairs are counted, below the diameter of the attractor. Check: the local slope should oscillate about a constant rather than drift; a steady drift means the scaling range has not been reached.
Box dimension. Bin a long orbit into a grid of side $\varepsilon$ , count occupied cells, and halve $\varepsilon$ repeatedly. Check: stop refining when the number of orbit points per occupied cell falls below about a hundred, or the count saturates at the sample size and the slope bends down. The two dimensions should satisfy $d_{\text{corr}} \le d_{\text{box}}$ .
Look at a cut. Collect the points of a long orbit lying in a narrow vertical strip and sort their other coordinate. Sorting the gaps exposes the transverse Cantor structure directly, and lowering the gap threshold shows the next level of the hierarchy.

Example(The Henon attractor at

a = 1.4

b = 0.3

, in numbers)Original

All the numbers below were computed for this unit; the orbits start at the origin unless stated otherwise.

Fixed points. The fixed-point equation is $1.4x^2 + 0.7x - 1 = 0$ , with discriminant $0.49 + 5.6 = 6.09$ and $\sqrt{6.09} = 2.4677925359$ . Hence

p_+ = (0.6313544771,\ 0.1894063431), \qquad p_- = (-1.1313544771,\ -0.3394063431).

At $p_+$ the Jacobian has trace $-2ax_+ = -1.7677925359$ and determinant $-0.3$ , so the multipliers are

\mu_+ = 0.1559463223, \qquad \mu_- = -1.9237388582 ,

whose product is $-0.3$ to ten figures, as part 3 requires. This is a saddle, and it is the saddle on the attractor: the nearest of two million recorded attractor points lies $5.6\times10^{-5}$ from it. At $p_-$ the multipliers are $3.2598220979$ and $-0.0920295620$ (product again $-0.3$ ); this is also a saddle, but it sits $0.21$ away from the attractor, outside it.

Escape radius and trapping region. Here $(1+|b|)^2 + 4a = 1.69 + 5.6 = 7.29$ , whose square root is exactly $2.7$ , so part 4 gives the clean value

R \;=\; \frac{1.3 + 2.7}{2.8} \;=\; \frac{10}{7} \;=\; 1.428571\ldots

The recorded orbit obeys this comfortably: over twenty million iterates its extent was $x \in [-1.28466,\ 1.27297]$ and $y \in [-0.38540,\ 0.38189]$ . Every one of those twenty million points also lies inside the quadrilateral $Q$ of part 5.

Lyapunov exponents. Carrying a tangent vector for $2\times10^6$ iterations after a transient of $10^3$ gives $\lambda_1 = 0.41911$ ; six independent random seeds, each run for $5\times10^5$ iterations, gave values between $0.41855$ and $0.41994$ . So

\lambda_1 = 0.4191 \pm 0.0007 , \qquad \lambda_2 = \ln 0.3 - \lambda_1 = -1.6231 \pm 0.0007 ,

where $\lambda_2$ is obtained from the exact constraint $\lambda_1 + \lambda_2 = \ln|\det DT| = \ln 0.3 = -1.2039728$ . The first exponent is positive, so the motion on the attractor is chaotic. In one iteration a tangent vector along the attractor is stretched by $e^{\lambda_1} = 1.5206$ and one across it is squeezed by $e^{\lambda_2} = 0.1973$ ; the product is $0.3000$ , which is the area ratio $|b|$ — the stretching and the folding, in one line of arithmetic.

Correlation dimension. Taking $10^5$ points, keeping every fifth iterate, and counting pairs with a $k$ -d tree gives a correlation sum whose log-log slope over $10^{-4} \le \varepsilon \le 10^{-1}$ is $1.2049$ ; three other starting points gave $1.2059$ , $1.2062$ and $1.2067$ . The local slope, measured over a sliding decade, oscillates between $1.18$ and $1.25$ rather than settling — the signature of a self-similar set, whose covering statistics repeat log-periodically. The honest report is therefore

d_{\text{corr}} = 1.21 \pm 0.03 ,

the uncertainty being the oscillation, not the fitting error.

Box dimension. Binning $2\times10^7$ iterates into squares of side $\varepsilon = 2^{-k}\cdot 4$ gives occupied-cell counts $148$ , $349$ , $833$ , $1931$ , $4561$ , $10616$ , $25795$ , $63272$ and $151659$ for $\varepsilon$ from $1/16$ down to $1/4096$ . The least-squares slope of $\ln N$ against $\ln(1/\varepsilon)$ over that range is $1.248$ , and the two endpoints alone give $1.250$ :

d_{\text{box}} \approx 1.25 .

The refinement was stopped at $\varepsilon = 1/4096$ , where there are still about $130$ orbit points per occupied cell. As the general inequality requires, $d_{\text{corr}} = 1.21 < 1.25 = d_{\text{box}}$ . (The two Lyapunov exponents furnish a third, independent estimate through the Kaplan-Yorke formula $1 + \lambda_1/|\lambda_2| = 1.2582$ . That formula is a conjecture, proved in no source used here, and is quoted only as a cross-check; it agrees with the box count to within $1\%$ .)

A cut through the attractor. Of $4\times10^7$ iterates, $29484$ fell in the strip $0.6245 \le x \le 0.6255$ . Sorting their $y$ -coordinates and looking in the window $0.17 \le y \le 0.20$ , the points occupy, when runs separated by gaps smaller than $10^{-3}$ are merged, exactly three bands:

[0.17501,\ 0.17528], \qquad [0.17997,\ 0.18095], \qquad [0.18793,\ 0.19047] .

Lowering the merging threshold to $5\times10^{-4}$ leaves the first band single, splits the second into two,

[0.17997,\ 0.18012] \ \text{ and } \ [0.18079,\ 0.18095],

and splits the third into three,

[0.18793,\ 0.18810], \quad [0.18871,\ 0.18899], \quad [0.18994,\ 0.19047].

That is six sheets in groups of one, two and three — the pattern Henon reported, recovered here from a sorted list of numbers rather than from a picture. The gaps between the groups ( $4.7\times10^{-3}$ and $7.0\times10^{-3}$ ) are an order of magnitude larger than the gaps inside them ( $6.8\times10^{-4}$ and $6.0\times10^{-4}$ , $9.5\times10^{-4}$ ), which is the hierarchy of scales that makes the cross section a Cantor set and the dimension fractional.

ApplicationOriginal

The Henon map is the standard proving ground for anything that claims to measure chaos from data. Because it is two-dimensional, invertible, polynomial and cheap to iterate for $10^9$ steps, every algorithm for Lyapunov exponents, for correlation and box dimensions, for unstable periodic orbits and for noise reduction is calibrated against it first; a method that cannot return $\lambda_1 = 0.419$ and $d \approx 1.2$ for a clean Henon orbit is not going to be trusted on a laser or a dripping tap. The transverse cut computed above is the one-dimensional version of what an experimenter does with a Poincare section of a measured flow.

More importantly, the map is the reason the words "stretching and folding" mean something quantitative. A dissipative flow in three dimensions, sampled on a surface of section, gives a planar map that contracts area; if the attractor is strange, that map must stretch in one direction and fold, because there is no other way to reconcile a positive exponent with a bounded invariant set. The Henon map is that mechanism stripped to a quadratic, and the numbers above — a stretch of $1.52$ , a squeeze of $0.20$ , a product of exactly $0.30$ — show the whole budget. The same accounting is what makes mixing work in a stirred fluid, where repeated stretching and folding of material lines, not molecular diffusion, is what homogenises a dye at low diffusivity; the striation thickness falls like $e^{\lambda_2 n}$ while the interfacial length grows like $e^{\lambda_1 n}$ , and the fractal dimension of the striation pattern plays the role measured here.

In celestial mechanics the map has a direct ancestor. Henon introduced it as a surrogate for the Lorenz flow, but two-dimensional area-contracting and area-preserving maps had already entered astronomy as sections of galactic and restricted three-body orbits, where the question is whether a given orbit is confined to an invariant curve or wanders chaotically. The Henon map answers, in the simplest possible setting, what such a wandering set looks like: not a region, but a set of measure zero and dimension about $1.25$ on which the motion never repeats.