Chapter 27 · Regimes, bistability, and tipping points

Part VI — Predictability of the second kind Knob: tilt, noise amplitude live notebook

Overview

Chapter 25 offered a reassurance: when the trajectory is unpredictable the statistics need not be, so a projection can be confident about a distribution it cannot resolve in detail. That reassurance assumes the attractor deforms smoothly as the forcing changes. If a slowly changing parameter destroys the state the system currently occupies, the distribution moves abruptly, and it does so exactly where a smooth-response argument is least willing to look.

So: can you see a tipping point coming? The literature says yes in principle — a state about to be destroyed relaxes more slowly, and slow relaxation shows up in a record as rising variance and rising autocorrelation. This chapter verifies that theory against exact identities, and then does what the theory does not: runs the indicator as a detection problem with a calibrated false-alarm rate.

The model

The normal form for two competing states, tilted by a parameter and kicked by noise:

\dot x = x - x^3 + \mu + \sigma\,\xi(t), \qquad V(x) = -\tfrac12 x^2 + \tfrac14 x^4 - \mu x ,

so $\dot x = -V'(x)$ — a ball in a double well. Being a gradient system buys two exact results that everything else is checked against. The stationary density is exactly Boltzmann, $p \propto e^{-2V/\sigma^2}$ ; and the fold sits at $\mu_c = 2/(3\sqrt3)$ , $x_c = -1/\sqrt3$ , where the right-hand side and its derivative vanish identically rather than to tolerance.

Measured as the ratio of the two lobes’ populations, the Boltzmann form holds to 4.6 % across a ratio spanning a factor of fourteen. That single comparison tests the drift and the integrator’s $\sigma\sqrt{\Delta t}$ noise convention together, which is worth having, since a factor of two in the diffusion coefficient is the standard error and it is invisible in a trajectory plot.

Prefer to run it yourself? Download as a Jupyter notebook — it installs its own dependencies in the first cell. The sliders are static in Jupyter; edit value= and re-run to change a parameter.

Tipping without a tipping point

A system can leave its state with the parameter held perfectly still: an unlucky run of kicks carries the ball over the barrier. Kramers’ waiting time is exponential in $1/\sigma^2$ with slope exactly $2\Delta V$ , and the measurement is unambiguous — fitted slope 0.2333 against an exact 0.2350, an error of 0.8 %.

The prefactor is another matter. It sits at 0.61 of the formula, so Kramers' expression overestimates the waiting time by about 60 % at these barrier heights. It is asymptotic in $2\Delta V/\sigma^2$ and a tipping problem lives at moderate values of that ratio. Use it for scaling, not for a date.

One noise level is deliberately left in the figure as an open square: only 84 % of its members escaped within the run, so its mean is censored, and censoring biases a mean escape time low because the slow escapes are the ones missing. Including that single point drags the fitted slope to 0.2065 — an error of 12.1 % rather than 0.8 %. It is the deepest barrier in the sweep, which is to say the most interesting one, and it is also what a finite computing budget or a finite observational record will always hand you. chaoslib.earlywarning.escape_times returns nan for a member that never escaped rather than quietly omitting it.

Critical slowing down is exactly true, and stops being useful before the fold

Near a stable state the dynamics linearise to an Ornstein–Uhlenbeck process, for which $\operatorname{var} x = \sigma^2/2|\lambda|$ and the lag-1 autocorrelation is $e^{\lambda\Delta t}$ , both exact. And $\lambda \to 0$ at the fold at a known rate: expanding about $(x_c,\mu_c)$ , where $f_{xx} = 2\sqrt3$ ,

\lambda \simeq -2 \cdot 3^{1/4} \sqrt{\mu_c - \mu}, \qquad \Delta V \simeq \tfrac43 3^{-1/4} (\mu_c - \mu)^{3/2}.

Up to $\mu = 0.25$ the measured standard deviation matches its exact value to 3.5 % and the autocorrelation to 1.6 %, both rising as advertised.

Then the indicator stops describing its own system, well short of the fold. At $\mu = 0.32$ — against a fold at $0.385$ — 46 % of members have already left the well and the measured spread is 3.6 times the theory. Closer in it is worse than useless: by $\mu = 0.382$ every member has escaped and the measured autocorrelation has fallen to 0.47 while the theory says 0.96. An analyst watching that number would conclude the system was becoming more stable at the moment it finished tipping.

The reason is not numerical. The predicted fluctuation grows as $(\mu_c-\mu)^{-1/2}$ while the distance from the well to the saddle shrinks as $(\mu_c-\mu)^{1/2}$ . They cross, and once the predicted fluctuation is the size of the basin, the fluctuation is the escape.

Two timing laws, with opposite signs

A noiseless system leaves late. In the fold normal form $\dot u = \sqrt3 u^2 + \gamma\tau$ the substitution $u = -w'/w$ gives Airy’s equation, so departure is the first zero of $w$ :

\mu_{\text{tip}} - \mu_c \simeq |a_1|\,3^{-1/6}\,\gamma^{2/3} = 1.9469\,\gamma^{2/3},

with $|a_1| = 2.3381\ldots$ . Measured over a factor of sixteen in sweep rate, the ratio to that law climbs monotonically to 0.988 as the sweep slows — approaching from below, as an asymptotic law should.

A noisy system leaves early, and by more than the naive estimate. The barrier vanishes as $(\mu_c-\mu)^{3/2}$ , faster than linearly, so escape becomes certain strictly before the fold. Integrating Kramers’ rate along the sweep is elementary under $w = d^{3/2}$ :

d^* = \left[\frac{\sigma^2}{\tfrac83 3^{-1/4}} \ln \frac{2\cdot3^{1/4}\,\sigma^2} {3\pi\,\tfrac83 3^{-1/4}\,\gamma\,\ln 2}\right]^{2/3}.

With the sweep slow enough that the deterministic delay cannot contaminate the answer, this predicts the measured median tipping tilt to 3.3 % across a factor of five in $\sigma$ .

The logarithm is the whole content. Drop it and you get the $\sigma^{4/3}$ scaling usually quoted, which is wrong here by a factor rising from 1.81 to 3.18 across that range — not an offset, a trend, which is why fitting a power law to the measurements returns $\sigma^{1.69}$ rather than $\sigma^{1.333}$ . There is no pure power law: the logarithm carries the sweep rate, and a slower sweep gives noise more time to find the barrier.

Which says something about real systems. The distance-to-fold at which a system commits is set by its noise and its rate of change, not by the fold alone. Two systems with identical bifurcation structure tip at different forcings if their internal variability differs. A tipping threshold quoted as a property of the system — a temperature, a freshwater flux — is incomplete without the variability and the rate of approach attached.

Early warning: good recall, no specificity

The recipe: slide a window along the record, compute variance and lag-1 autocorrelation in each, and test for an upward trend with Kendall’s $\tau$ . How large a $\tau$ ? That is the only part that determines whether the method has skill, and it is where most presentations stop. $\tau$ computed on overlapping windows of an autocorrelated record has a null distribution far wider than the independent-sample one, so a threshold from a table is meaningless. Here it is calibrated against a null run — the same system with the parameter held fixed — so the false-alarm rate is 5 % by construction.

Four scenarios, 200 realisations each:

scenario	tipped	alarm (variance)	alarm (autocorrelation)
null, $\mu$ fixed at 0	0 %	5 %	5 %
ramp through the fold	100 %	100 %	89 %
ramp stopping at $\mu = 0.30$	0 %	100 %	89 %
noise-induced, $\mu$ fixed at 0.25	80 %	13 %	18 %

What works. Against a genuine sweep the indicator fires on every realisation, and not at the last moment: the alarm still fires on more than 90 % of cases when the decision must be taken 1200 time units early — a third of the record discarded, the call made at a tilt of 0.28 when tipping comes at 0.36. Lead time is not the problem.

What fails. The scenario that ramps to $\mu = 0.30$ and stops never tips, and the alarm fires on 100 % of those realisations — indistinguishable from the runs that do. The indicator is not broken; it is answering a different question. Rising variance is evidence that the system is approaching a bifurcation, true in both cases. It carries no information about whether the approach will continue, because that is a fact about the forcing and not about the system’s dynamics. No statistic computed from a record of $x$ can supply it.

And the transitions with no precursor at all. The noise-induced scenario tips 80 % of the time with the parameter perfectly still, and the alarm fires on 13 % against a 5 % baseline — essentially no skill. Four realisations in five tip with no warning whatever. That is not an estimator failing: there is genuinely nothing to detect, because the potential never changed.

The honest summary is that early-warning indicators have good lead time and good recall, and no specificity. In a real record you do not know which scenario you are in — that is the entire problem — and the indicator does not tell you. Which is worth being precise about rather than cynical about: a rising-variance alarm is real information, saying the system is closer to a fold than it was. What it is not is a forecast.

Exercises

The noise slider changes $\sigma$ without changing the potential. At what noise does “which state the system is in” stop being a useful description, and what does that correspond to in the detection problem?
The Kramers prefactor is 60 % optimistic here. Estimate the barrier height at which it would be good to 10 %, and say whether a system with that barrier is one anybody would worry about tipping.
The two failure conditions above — the fluctuation reaching the basin width, and members escaping — coincide. Show that they must, using the near-fold scalings.
The “stops short” scenario defeats the indicator. Design a measurement that would distinguish it from a genuine sweep, and say what extra information your measurement requires.