Unstable Manifolds and Unpredictability in Chaotic Attractors — Epoche B2
Beyond the Pull: Unstable Manifolds and Unpredictability in Chaotic Attractors An attractor is supposed to make the long run easy. Whatever the details of the start, the system is drawn in and stays; the only thing left to know is which attractor a given start belongs to. That last clause is doing far more work than it looks. When a system has more than one attractor, its state space is carved into basins , and everything depends on how sharply the carving can be resolved. This essay is about the boundary between basins: what it is made of, why in chaotic systems it can be so convoluted that no achievable measurement decides which side of it you are on, and how to put a number on the difficulty rather than merely deplore it. The number turns out to be an exponent, and it is often close enough to zero that improving an instrument a thousandfold buys almost nothing. The set-up, and one correction to the standard example A continuous system with $n$ state variables is written $$ \frac{d\mathbf{x}}{dt} = \mathbf{f}(\mathbf{x}), \qquad \mathbf{x}\in\mathbb{R}^n, $$ where $\mathbf{f}$ is a vector field assigning a velocity to every state and the solution through $\mathbf{x}_0$ is written $\phi_t(\mathbf{x}_0)$, the flow . An attractor $A$ is a closed invariant set such that every trajectory starting in some open neighbourhood of $A$ approaches $A$ as $t\to\infty$, and such that no proper closed subset of $A$ has that property — the last condition matters, since without it the whole state space would qualify. The basin $B(A)$ is the set of all initial conditions whose trajectories approach $A$. The example always given is the damped pendulum, and it is usually given wrongly: its basin is said to include every physically possible starting state. Write the pendulum out. A bob on a light rigid rod of length $L$ under gravity $g$, with linear damping of coefficient $\gamma$, obeys $$ \ddot{\theta} + \gamma\dot{\theta} + \frac{g}{L}\sin\theta = 0 . $$ If the rod is rigid, $\theta$ is not confined to a swing: the bob can go over the top, and it will if it starts at the bottom with an angular speed exceeding a threshold that in the undamped limit is $\dot\theta_{\text{crit}} = 2\sqrt{g/L}$ — for $L=1\,$m, $6.26\ \mathrm{rad\,s^{-1}}$ — and that damping raises somewhat. Damped, the pendulum eventually stops — but at which rest position? The states $\theta = 0,\ \pm2\pi,\ \pm4\pi,\dots$ are all equilibria, all stable, and all distinct attractors. A pendulum launched at $9\ \mathrm{rad\,s^{-1}}$ goes over once; at $12$, perhaps twice; and near each threshold speed a difference of a fraction of a per cent changes the answer, because the bob arrives at the top nearly at rest and it is nearly a coin toss whether it tips forward or falls back. So a damped pendulum has infinitely many attractors and a non-trivial basin structure, and it is the standard example of the phenomenon it is usually cited to rule out. What a basin boundary is made of Boundaries are not arbitrary surfaces drawn between basins; they are made of trajectories, and they can be identified. A point $\mathbf{p}$ where $\mathbf{f}(\mathbf{p})=\mathbf{0}$ is a fixed point, and it is a saddle if the Jacobian matrix $J = \partial f_i/\partial x_j$ evaluated at $\mathbf{p}$ has eigenvalues with both positive and negative real parts — the system is pushed away along some directions and pulled in along others. The two sets of trajectories that respect those directions are its manifolds: $$ W^{s}(\mathbf{p}) = \bigl\{\mathbf{x}_0 : \lim_{t\to+\infty}\phi_t(\mathbf{x}_0)=\mathbf{p}\bigr\}, \qquad W^{u}(\mathbf{p}) = \bigl\{\mathbf{x}_0 : \lim_{t\to-\infty}\phi_t(\mathbf{x}_0)=\mathbf{p}\bigr\}. $$ The stable manifold $W^{s}$ collects the starting points that run exactly into the saddle and stay; the unstable manifold $W^{u}$ collects those that emerged from it in the infinite past. Both are invariant: a trajectory that starts on one never leaves it. Now the connection. A point on $W^{s}(\mathbf{p})$ never reaches any attractor at all — it converges on the saddle instead — so it belongs to no basin, and every neighbourhood of it contains points that miss the saddle and fall to one attractor or another. That is precisely what it means to be on a boundary. In the damped pendulum the saddles are the inverted positions $\theta = \pm\pi, \pm3\pi,\dots$, and their stable manifolds are the curves of exactly-critical launches, separating the starting states that tip over from those that fall back — the phase portrait drawn in Strogatz's Nonlinear Dynamics and Chaos [1] . In a chaotic system the same role is played by the stable manifolds of the infinitely many unstable periodic orbits embedded in the dynamics, and the boundary they collectively form need not be a smooth curve or surface. It can be a set whose box-counting dimension is not an integer: a boundary that is more than a curve without being a region, a geometry treated at length in Ott's Chaos in Dynamical Systems [2] . Why "arbitrarily close points in different basins" is not the point It is often said that on a fractal boundary, arbitrarily close initial conditions may lie in different basins. True — but that is true of any boundary, smooth ones included. Two points either side of a straight line have the same property, and no one calls a straight line unpredictable. The difference is not whether ambiguous points exist but how many of them there are. Make it precise. Fix a measurement precision $\varepsilon$. Call a state $\mathbf{x}_0$ $\varepsilon$-uncertain if some point within $\varepsilon$ of it lies in a different basin, so that knowing the state to precision $\varepsilon$ does not determine the outcome. Let $f(\varepsilon)$ be the fraction of the region that is $\varepsilon$-uncertain. For a boundary of box-counting dimension $d$ in a state space of dimension $D$, this fraction obeys a power law $$ f(\varepsilon) \propto \varepsilon^{\alpha}, \qquad \alpha = D - d, $$ where $\alpha$ is the