Worked examples — Chaotic systems — sensitivity to initial conditions, Lyapunov exponents

This page is a drill hall. The parent note built the ideas; here we hit every kind of case the Lyapunov exponent $\lambda$ can throw at you, one worked example per case. Before we start, one reminder in plain words:

Every symbol used below: ${\delta }_0$ = starting gap, $\delta (t)$ = gap at time $t$, $\lambda$ = Lyapunov exponent (units: 1/time for flows, dimensionless-per-step for maps), $f^{\prime }(x)$ = the local stretch factor (slope of the update rule), $x^{\ast }$ = a fixed point (a value the rule leaves unchanged).

The scenario matrix

Every question about $\lambda$ falls into one of these cells. The examples below fill all of them.

Before the examples, one figure to anchor the three flow cases (A, B, C) side by side — the gap growing, shrinking, and coasting:

Figure s01 (described in words). Three panels sharing a "time $t$" horizontal axis and a "gap size $|\delta (t)|$" vertical axis. Left panel (Cell A, $\lambda >0$): a curve that starts small and climbs steeply upward — the gap widens without bound; drawn in coral. Middle panel (Cell B, $\lambda <0$): a curve that starts at the same small height and falls quickly toward zero — the gap collapses; drawn in mint green. Right panel (Cell C, $\lambda =0$): a curve that bobs up and down between two fixed heights, never escaping the box — the gap coasts; drawn in lavender. Every example below is one of these three shapes: grow, shrink, or coast.

Cell A — one-dimensional blow-up

Forecast: guess now — does a positive growth rate automatically mean chaos?

1. Write the gap equation. Let $\delta =x^{\prime }-x$. Subtract the two copies of $\dot{x}=kx$: $\dot{\delta }=k{x}^{\prime }-kx=k\delta .$ Why this step? Chaos is about the gap, so we track $\delta$ directly, not $x$. This is the growing curve (left panel of figure s01).

2. Solve it. $\dot{\delta }=k\delta$ is the equation whose only solution is an exponential: $\delta (t)={\delta }_0{e}^{kt}.$ Why the exponential? It is the unique function whose rate of change is proportional to itself — exactly what "constant relative growth" means.

3. Read off $\lambda$. Compare with $\delta (t)={\delta }_0{e}^{\lambda t}$: $\lambda =k=0.5{\text{s}}^{-1}>0.$ Why this step? $\lambda$ is defined as the exponent in front of $t$ in the gap growth.

4. Decide chaos. $\lambda >0$, yet the trajectory $x(t)=x_0{e}^{kt}$ runs off to infinity — it is unbounded. Chaos requires SDIC and boundedness. So this is sensitive but NOT chaotic.

Verify: at $t=1$ s, gap $=0.001e^{0.5}=0.001\times 1.6487=1.6487\times 10^{-3}$. It grew but stayed a gap between two numbers that both fly off — no folding, no attractor. Cell A confirmed: positive $\lambda$ is necessary but not sufficient.

Cell B — decay to a stable point

Forecast: will the gap grow or shrink? By how much per second?

1. Gap equation. With $\delta =x^{\prime }-x$, subtracting the two copies: $\dot{\delta }=-3x^{\prime }+3x=-3\delta .$ Why this step? The constant $x^{\ast }$ cancels — only the difference survives, which is what we want. This is the collapsing curve (middle panel of figure s01).

2. Solve. $\delta (t)={\delta }_0{e}^{-3t}$. Why this step? Same exponential logic as before, now with a negative rate.

3. Read $\lambda$. $\lambda =-3{\text{s}}^{-1}<0$. Gap dies. The system converges to the fixed point $x^{\ast }=2$ — a stable fixed point. No chaos.

Verify: initial gap $=2.11-2.1=0.01$. After $1$ s: $0.01e^{-3}=0.01\times 0.049787=4.98\times 10^{-4}$. Shrank ~20×. Both runs march toward $x=2$, forgetting where they started — the opposite of SDIC. Cell B confirmed.

Cell C — marginal / periodic, $\lambda =0$

Forecast: the two runs are identical shapes shifted slightly. Will their gap explode, vanish, or stay bounded?

1. Write both runs. $x(t)=\cos t,x^{\prime }(t)=\cos (t+0.01).$ Why this step? A tiny phase difference is exactly a tiny initial-condition difference for an oscillator. This is the coasting curve (right panel of figure s01).

2. Gap. Using $\cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}$: $\delta (t)=x^{\prime }-x=-2\sin (t+0.005)\sin (0.005).$ Why this step? We want the actual size of the gap over time, not an approximation that hides its behaviour.

3. Read off the initial gap. Put $t=0$: ${\delta }_0=\delta (0)=-2\sin (0.005)\sin (0.005)=-2{\sin }^2(0.005)\approx -5.0\times 10^{-5}.$ Why this step? We must not confuse the starting gap with the later amplitude — the two are different numbers here, and $\lambda$ compares them.

4. Bound the whole curve. For all later $t$, the factor $\sin (t+0.005)$ can never exceed $1$ in size, so $|\delta (t)|=2|\sin (t+0.005)|\sin (0.005)\le 2\sin (0.005)\approx \mathrm{0.01.}$ So the gap grows from $5\times 10^{-5}$ up to at most $0.01$ and then oscillates inside that box — it never runs away. Why this step? We are pinning the gap inside a fixed ceiling $B=2\sin (0.005)$, independent of $t$. That ceiling is the key ingredient the next step needs.

5. Feed the bound into the limit. Take the fixed ceiling $B=0.01$ from step 4 and drop it straight into the Lyapunov definition. Because $|\delta (t)|\le B$ for all $t$, the fraction inside the log is capped: $\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}\le \frac{1}{t}\ln \frac{B}{|{\delta }_0|}=\frac{1}{t}\underset{\text{a fixed number}\approx 5.3}{\underbrace{\ln \frac{0.01}{5\times 10^{-5}}}}\stackrel{t\to \infty }{\to }0.$ The numerator $\ln (B/|{\delta }_0|)$ is one fixed constant (it does not grow with $t$), and dividing any fixed constant by $t\to \infty$ sends it to $0$. So $\lambda \le 0$; and since $|\delta (t)|$ does not shrink to zero either, $\lambda \ge 0$. Hence: $\lambda ={\lim }_{t\to \infty }\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}=0.$ Why this step? This is the explicit link the bound was for: bounded gap ⇒ constant numerator ⇒ constant / $t$ → 0. This is the marginal case: periodic / quasi-periodic motion.

Verify: ${\delta }_0=-2{\sin }^2(0.005)=-4.9999\times 10^{-5}$; amplitude bound $2\sin (0.005)=0.0099999$. Neither grows without limit nor decays to zero ⇒ $\lambda =0$. Cell C confirmed.

Cell D — the exact chaotic value $\ln 2$

Forecast: guess the value before computing. (Hint: it doubles distances.)

Figure s02 (described in words). Two side-by-side panels. Left panel: the logistic map $x_{n+1}=4x(1-x)$ drawn as an upside-down parabola (in lavender) together with the dashed diagonal line $y=x$; the parabola rises to its peak at the middle and comes back down, showing how the interval $[0,1]$ is folded back on itself at the top — this folding is what keeps the motion bounded. Right panel: the same dynamics after the change of variable, drawn as the straight-sided doubling map $\theta \mapsto 2\theta \mathrm{mod}1$ (a mint saw-tooth with two parallel sloping segments). An arrow shows that each small gap between two nearby $\theta$ values is stretched by exactly a factor of 2 per step — this constant factor 2 is what we read $\lambda =\ln 2$ from.

1. Why this formula for maps? Each step multiplies a small gap by the local slope $|f^{\prime }(x_n)|$. After $N$ steps the total stretch is the product ${\prod }_n|f^{\prime }(x_n)|$. Taking $\frac{1}{N}\ln$ turns that product into an average log-stretch — the definition of $\lambda$.

2. Change to an angle variable that straightens the map. Write each point as $x_n={\sin }^2(\pi {\theta }_n)$ with ${\theta }_n\in [0,1)$ — think of ${\theta }_n$ as where on a circle the state sits, and $x_n$ as its height. Plug this into $x_{n+1}=4x_n(1-x_n)$ and use the identity $4{\sin }^{2}u{\cos }^{2}u={\sin }^2(2u)$: $x_{n+1}=4{\sin }^2(\pi {\theta }_n){\cos }^2(\pi {\theta }_n)={\sin }^2(2\pi {\theta }_n)={\sin }^2(\pi \cdot 2{\theta }_n).$ Matching $x_{n+1}={\sin }^2(\pi {\theta }_{n+1})$ gives ${\theta }_{n+1}=2{\theta }_n\mathrm{mod}1$ (the "mod 1" is because ${\sin }^2$ repeats). So in the $\theta$ picture the rule is simply "double the angle." Why this step? The logistic slope $|f^{\prime }(x_n)|$ is messy and $x$-dependent; the angle change of variable — legal by the invariance stated in the callout above — converts it into a rule whose stretch is a clean constant factor 2 every step (the mint saw-tooth, right panel of figure s02).

3. Compute the average. Since each step stretches any small gap by exactly a factor of 2 in the $\theta$-coordinate, $\ln |f^{\prime }|=\ln 2$ every step, so: $\lambda =\frac{1}{N}{\sum }_{n=0}^{N-1}\ln 2=\ln 2\approx \mathrm{0.6931.}$ Why this step? The sum of $N$ copies of $\ln 2$, divided by $N$, is just $\ln 2$.

4. Sanity of sign. $\lambda =\ln 2>0$ ⇒ chaos, and the map lives in $[0,1]$ so it is bounded ⇒ genuinely chaotic (unlike Ex 1). See The logistic map and period-doubling route to chaos.

Verify: $\ln 2=0.693147\dots >0$. A numeric average of $\frac{1}{N}\sum \ln |4(1-2x_n)|$ over many iterates converges near $0.693$. Cell D confirmed.

Cell E — degenerate slope, $\lambda \to -\infty$

Forecast: what happens to a gap that gets multiplied by a slope of exactly zero?

1. Find the fixed point. Solve $x^{\ast }=2x^{\ast }(1-x^{\ast })$. Either $x^{\ast }=0$ or $1=2(1-x^{\ast })\Rightarrow x^{\ast }=0.5$. Why this step? A fixed point is where the map maps a value to itself — the resting state whose stability we test.

2. Evaluate the slope. $f^{\prime }(x)=2(1-2x)$, so at $x^{\ast }=0.5$: $f^{\prime }(0.5)=2(1-2\cdot 0.5)=2\cdot 0=0.$ Why this step? The slope is the one-step stretch factor; that's what feeds $\lambda$.

3. Interpret $\lambda$. A single step multiplies any gap by $0$ — it is crushed to nothing. In the log-average, $\ln |0|=-\infty$, so this direction gives $\lambda =-\infty$: a superstable point. Why this step? $\ln$ of a shrink-factor-zero is the strongest possible convergence. Strongly stable ⇒ no chaos.

Verify: $2(1-2\cdot 0.5)=0$ exactly. Any nearby run collapses onto $x=0.5$ in one step's worth of stretch — the extreme of Cell B. Cell E confirmed.

Cell F — predictability time (word problem)

Forecast: guess whether 100× better sensors doubles your forecast window — or barely nudges it.

1. Set the growth equation. Error grows as $\Delta ={\delta }_0{e}^{\lambda t}$. Solve for $t$: $t_{\text{pred}}=\frac{1}{\lambda }\ln \frac{\Delta }{{\delta }_0}.$ Why this step? This inverts the exponential — the only algebra that isolates $t$ from an exponent is the logarithm.

2. Plug in. $t_{\text{pred}}=\frac{1}{0.25}\ln \frac{0.1}{10^{-4}}=4\ln (10^3)=4\times 6.9078=27.6\text{days}.$ Why this step? $\Delta /{\delta }_0=10^{-1}/10^{-4}=10^3$; $\ln 10^3=3\ln 10=6.9078$.

3. Improve sensors 100×: now ${\delta }_0=10^{-6}$, so $\Delta /{\delta }_0=10^5$: $t_{\text{pred}}=4\ln (10^5)=4\times 11.513=46.05\text{days}.$ Why this step? Same formula, only ${\delta }_0$ changed.

4. Interpret. 100× better data bought us $46.05-27.6=18.4$ extra days — a $67\%$ gain, not $100\times$. That is the brutal logarithm: $t_{\text{pred}}\propto \ln (1/{\delta }_0)$.

Verify: $4\ln (10^3)=27.63$ days; $4\ln (10^5)=46.05$ days. Ratio $46.05/27.63=1.667$, i.e. only $1.67\times$ longer for $100\times$ better data. Cell F confirmed. Units: $[1/\lambda ]=$ days, $\ln (\cdot )$ dimensionless ⇒ answer in days. ✔

Cell G — multi-dimensional spectrum

Forecast: can a system stretch in one direction yet shrink in volume? Guess the sign of the sum.

1. Chaos test. The maximal exponent ${\lambda }_1=+0.9>0$ ⇒ nearby trajectories separate ⇒ chaotic (and the flow is bounded, so it's a genuine strange attractor). Why this step? Only the largest exponent decides SDIC; it dominates the growth of a generic gap.

2. Volume test. By the divergence/Liouville argument, phase-space volume grows as $e^{({\sum }_i{\lambda }_i)t}$. Sum: ${\sum }_i{\lambda }_i=0.9+0+(-14.6)=-13.7{\text{s}}^{-1}<0.$ Why this step? The trace of the flow's stretching (sum of ${\lambda }_i$) is the exponential rate of volume change — see Liouville's theorem (phase-space volume).

3. Reconcile the paradox. ${\lambda }_1>0$ ⇒ stretch along one direction; $\sum {\lambda }_i<0$ ⇒ total volume collapses. The flow stretches, folds, and contracts onto a zero-volume fractal set. Why this step? Stretch-and-fold on a shrinking volume is the mechanism of dissipative chaos.

4. Bonus — attractor dimension (Kaplan–Yorke): count directions until the running sum of $\lambda$'s turns negative. $D_{KY}=2+\frac{{\lambda }_1+{\lambda }_2}{|{\lambda }_3|}=2+\frac{0.9+0}{14.6}=\mathrm{2.062.}$ Why this step? Start adding exponents largest-first: ${\lambda }_1=0.9>0$, then ${\lambda }_1+{\lambda }_2=0.9>0$ still positive, but adding ${\lambda }_3$ would make $0.9-14.6<0$. So the attractor fills the first two directions completely (that's the "$2$") plus a fraction $\frac{{\lambda }_1+{\lambda }_2}{|{\lambda }_3|}$ of the third — giving a fractional dimension between the plane ($2$) and a solid ($3$). This measures how "thick" the strange attractor is.

Verify: $\sum {\lambda }_i=-13.7<0$ (contracts) while ${\lambda }_1=0.9>0$ (chaos). $D_{KY}=2+0.9/14.6=2.0616$. Cell G confirmed.

Cell H — the exam trap: SDIC ≠ chaos

Forecast: all three have "separating" or "moving" trajectories — but only one is chaotic. Which?

1. System (i) $\dot{x}=x$. Gap: $\delta ={\delta }_0{e}^t$, so $\lambda =+1>0$. But $x(t)=x_0{e}^t\to \infty$: unbounded. SDIC ✔, boundedness ✘ ⇒ not chaotic (same lesson as Ex 1). Why this step? Chaos needs the trajectory trapped in a finite region so it must fold back.

2. System (ii) doubling map. Each step doubles the gap (mod 1), $\lambda =\ln 2>0$, and it lives in the finite interval $[0,1)$ so it folds (the mod). SDIC ✔, bounded ✔, mixing ✔ ⇒ chaotic. Why this step? This is the clean archetype: stretch by 2, wrap around — exactly Ex 4's engine.

3. System (iii) uniform rotation. Gap stays constant: $\delta (t)={\delta }_0$, so $\lambda =0$. Bounded ✔ but no separation ⇒ not chaotic (periodic, Cell C). Why this step? No stretching means no sensitivity, no matter how bounded.

4. Rule extracted. ==Chaos = positive $\lambda$ AND boundedness AND mixing==, where mixing means the flow eventually stirs any little blob of starting points across the whole attractor — so any region overlaps any other (formally: for regions $U,V$ the flow of $U$ eventually intersects $V$). Miss any one of the three conditions and it's not chaos. See the parent note's definition of topological mixing.

Verify: (i) $\lambda =1>0$ but unbounded → no. (ii) $\lambda =\ln 2=0.693>0$, bounded → yes. (iii) $\lambda =0$ → no. Only one of three qualifies. Cell H confirmed.

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