Exercises — Chaotic systems — sensitivity to initial conditions, Lyapunov exponents

• (a) $\lambda >0\Rightarrow$ chaotic. Nearby trajectories separate exponentially — the gap grows like $e^{0.7t}$.
• (b) $\lambda =0\Rightarrow$ marginal (regular/periodic/quasi-periodic). The gap stays roughly constant — neither grows nor shrinks exponentially.
• (c) $\lambda <0\Rightarrow$ convergent. Nearby trajectories collapse together onto a stable fixed point or limit cycle; the gap dies like $e^{-1.2t}$. Mnemonic: Positive Pulls aparT → chaoS.

Growth factor $=\frac{10^{-2}}{10^{-4}}=100$. The gap grew by 100× in finite time, so the local separation rate is positive $\Rightarrow$ this stretch is chaotic (exponentially separating). (We'll compute the actual $\lambda$ in L2.1.)

Invert $|\delta (t)|=|{\delta }_0|e^{\lambda t}$: take $\ln$ of both sides, $\lambda =\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}=\frac{1}{5}\ln \frac{10^{-2}}{10^{-4}}=\frac{1}{5}\ln (100).$ Since $\ln 100=2\ln 10\approx 4.605$, $\lambda \approx \frac{4.605}{5}\approx 0.921{\text{s}}^{-1}.$ WHAT we did: solved the growth law for $\lambda$. WHY: $\ln$ undoes the exponential, and dividing by $t$ turns "total growth" into "growth per second."

$t_{\text{pred}}=\frac{1}{\lambda }\ln \frac{\Delta }{{\delta }_0}=\frac{1}{0.5}\ln \frac{1}{10^{-3}}=2\ln (10^3)=2\cdot 3\ln 10\approx 2\times \mathrm{6.908.}$ $t_{\text{pred}}\approx 13.8\text{days}.$

$t_{\text{pred}}=\frac{1}{0.5}\ln \frac{1}{10^{-6}}=2\ln (10^6)=2\cdot 6\ln 10\approx 2\times 13.816\approx 27.6\text{days}.$ Improvement: $27.6-13.8=13.8$ days. A thousand-fold better data only doubled the horizon — the brutal logarithm at work.

$f^{\prime }(0.5)=2(1-2\cdot 0.5)=2(1-1)=0.$ The stretch factor per step is $0$, so a small gap is crushed to zero each iteration — super-strong contraction. Then $\ln |f^{\prime }|=\ln 0=-\infty$, giving $\lambda =-\infty$ at that point: strongly convergent, not chaotic. (Perturbations don't just shrink — they vanish faster than any exponential near this "super-stable" point.)

$f^{\prime }(0.1)=4(1-2\cdot 0.1)=4(0.8)=3.2,|f^{\prime }(0.1)|=\mathrm{3.2.}$ Local stretch $=3.2>2=e^{\lambda }$. So at this point the gap is stretched more than average. Analysis: $\lambda$ is the average of $\ln |f^{\prime }(x_n)|$ over the whole orbit. Some points stretch harder ($x$ near $0$ or $1$, factor up to $4$), some stretch less (near $x=0.5$, factor near $0$). The average log-stretch lands at exactly $\ln 2$. No single point equals the average — that's the whole point of averaging along the orbit.

• (a) All ${\lambda }_i<0$: every error dies, trajectory settles to the rest point (a stable fixed point / attractor). Not chaotic.
• (b) $\lambda =0$: motion is periodic; a small error stays bounded and neither grows nor shrinks exponentially. Marginal.
• (c) ${\lambda }_1>0$ (stretching in one direction ⇒ SDIC ⇒ chaos) but ${\sum }_i{\lambda }_i<0$ (total phase-space volume contracts due to friction). How both at once? Dissipation shrinks the volume of a blob of initial conditions, but the flow simultaneously stretches the blob in one direction and folds it back to stay bounded. Stretch (positive ${\lambda }_1$) makes it chaotic; net contraction (negative sum) squeezes it onto a strange attractor. See Strange attractors and fractal dimension and Driven damped pendulum.

From $t_{\text{pred}}=\frac{1}{\lambda }\ln (\Delta /{\delta }_0)$ with fixed $\Delta$, the difference in horizons is $t_2-t_1=\frac{1}{\lambda }\left(\ln \frac{\Delta }{{\delta }_0^{\prime }}-\ln \frac{\Delta }{{\delta }_0}\right)=\frac{1}{\lambda }\ln \frac{{\delta }_0}{{\delta }_0^{\prime }}.$ Solve for the shrink factor ${\delta }_0/{\delta }_0^{\prime }$: $\ln \frac{{\delta }_0}{{\delta }_0^{\prime }}=\lambda (t_2-t_1)=0.4\times 10=4\Rightarrow \frac{{\delta }_0}{{\delta }_0^{\prime }}=e^4\approx \mathrm{54.6.}$ You must make the initial error about 54.6× smaller. To add 10 more seconds, error must shrink by $e^{\lambda \cdot 10}$ — exponential cost for linear gain in horizon.

Continuous estimate: $\Delta ={\delta }_0{e}^{\lambda N}\Rightarrow N=\frac{1}{\lambda }\ln \frac{\Delta }{{\delta }_0}=\frac{1}{\ln 3}\ln (10^6).$ $N=\frac{6\ln 10}{\ln 3}=\frac{6\times 2.302585}{1.098612}\approx \frac{13.8155}{1.0986}\approx \mathrm{12.58.}$ Since $N$ must be an integer and we need the gap to exceed 1, round up: $N=13$ steps. Check: after 12 steps gap $\approx 10^{-6}\cdot 3^{12}=10^{-6}\cdot 531441\approx 0.531<1$; after 13 steps $\approx 10^{-6}\cdot 3^{13}\approx 1.594>1$. ✓

• Inner limit ${\delta }_0\to 0$ first keeps the linearisation $\dot{\delta }\approx f^{\prime }(x)\delta$ honest. That equation is only valid while $\delta$ is infinitesimal; a finite $\delta$ would feel the nonlinear folding of the attractor.
• Then $t\to \infty$ averages the local stretch rate over the whole attractor, making $\lambda$ a property of the system, not of one lucky starting point. If you swap the order (fix a finite ${\delta }_0$, then $t\to \infty$): the gap $|\delta (t)|$ cannot exceed the attractor's diameter — it saturates at some bounded value. Then $\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}\to \frac{1}{t}\ln (\text{const})\to 0$. You'd wrongly measure $\lambda =0$ for every bounded system, chaotic or not. Taking ${\delta }_0\to 0$ first is what lets the exponential separation run "forever" before saturation ruins the measurement.

(a) ${\lambda }_1=+0.9>0\Rightarrow$ chaotic (SDIC in one direction). (b) Volume grows as $e^{({\sum }_i{\lambda }_i)t}$. Here $\sum {\lambda }_i=0.9+0-1.4=-0.5{\text{s}}^{-1}<0$, so volume shrinks at rate $0.5{\text{s}}^{-1}$ (dissipative, consistent with Liouville's theorem (phase-space volume) being violated by friction — Liouville holds only for conservative Hamiltonian flow). (c) Find $k$: running sums are ${\lambda }_1=0.9\ge 0$ ✓; ${\lambda }_1+{\lambda }_2=0.9\ge 0$ ✓; ${\lambda }_1+{\lambda }_2+{\lambda }_3=-0.5<0$ ✗. So the largest $k$ with non-negative sum is $k=2$. Then $D_{KY}=2+\frac{{\lambda }_1+{\lambda }_2}{|{\lambda }_3|}=2+\frac{0.9+0}{1.4}=2+0.642857\dots \approx 2.643.$ A fractal dimension between 2 and 3 — the signature of a strange attractor (Strange attractors and fractal dimension).

Solve: $\dot{x}=x\Rightarrow x(t)=x_0{e}^t$, and the gap $\delta (t)=(x_0^{\prime }-x_0)e^t={\delta }_0{e}^t$, so indeed $\lambda =1>0$: it is sensitive (nearby starts separate exponentially). But chaos needs three things, not one: SDIC + boundedness + topological mixing. Here the trajectory $x_0{e}^t\to \infty$ is unbounded — nothing folds it back. Positive $\lambda$ implies chaos only for a bounded system on an attractor. So "positive $\lambda \Rightarrow$ chaos" is shorthand that silently assumes boundedness. Stretching without folding is mere explosion; chaos is stretch-and-fold in a confined region.

Cover the answers. Growth law solved for $\lambda$ ::: $\lambda =\frac{1}{t}\ln (|\delta (t)|/|{\delta }_0|)$ Predictability time formula ::: $t_{\text{pred}}=\frac{1}{\lambda }\ln (\Delta /{\delta }_0)$ To double the horizon, shrink ${\delta }_0$ by ::: a factor $e^{\lambda \Delta t}$ (exponential cost) Three ingredients of chaos ::: SDIC + boundedness + topological mixing Kaplan–Yorke idea ::: fractional dimension from where cumulative $\lambda$ crosses zero