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

2.1.25 · D3 · Physics › Analytical Mechanics › Chaotic systems — sensitivity to initial conditions, Lyapuno

Yeh page ek drill hall hai. Parent note ne ideas build kiye; yahan hum har tarah ke cases ko cover karte hain jo Lyapunov exponent $\lambda$ de sakta hai, ek worked example per case. Shuru karne se pehle, ek reminder plain words mein:

Neeche use hone wale har symbol ke baare mein: ${\delta }_0$ = starting gap, $\delta (t)$ = time $t$ par gap, $\lambda$ = Lyapunov exponent (units: flows ke liye 1/time, maps ke liye dimensionless-per-step), $f^{\prime }(x)$ = local stretch factor (update rule ka slope), $x^{\ast }$ = ek fixed point (woh value jisko rule unchanged chhodta hai).

The scenario matrix

$\lambda$ ke baare mein har question in cells mein se kisi ek mein aata hai. Neeche ke examples sab ko fill karte hain.

Examples se pehle, teen flow cases (A, B, C) ko side by side anchor karne ke liye ek figure — gap growing, shrinking, aur coasting:

Figure s01 (words mein describe kiya gaya). Teen panels jo ek "time $t$" horizontal axis aur "gap size $|\delta (t)|$" vertical axis share karte hain. Left panel (Cell A, $\lambda >0$): ek curve jo chhotey se shuru hoti hai aur teezee se upar chadh-ti hai — gap bina bound ke badhta hai; coral mein draw kiya gaya. Middle panel (Cell B, $\lambda <0$): ek curve jo usi chhotey height se shuru hoti hai aur quickly zero ki taraf girtti hai — gap collapse ho jaata hai; mint green mein draw kiya gaya. Right panel (Cell C, $\lambda =0$): ek curve jo do fixed heights ke beech upar-neeche hoti hai, kabhi box se bahar nahi jaati — gap coast karta hai; lavender mein draw kiya gaya. Neeche har example in teen shapes mein se ek hai: grow, shrink, ya coast.

Cell A — one-dimensional blow-up

Forecast: abhi guess karo — kya ek positive growth rate ka matlab automatically chaos hai?

1. Gap equation likho. Maano $\delta =x^{\prime }-x$. $\dot{x}=kx$ ki do copies subtract karo: $\dot{\delta }=k{x}^{\prime }-kx=k\delta .$ Yeh step kyun? Chaos gap ke baare mein hai, toh hum directly $\delta$ track karte hain, $x$ nahi. Yeh growing curve hai (figure s01 ka left panel).

2. Solve karo. $\dot{\delta }=k\delta$ woh equation hai jiska ek hi solution exponential hai: $\delta (t)={\delta }_0{e}^{kt}.$ Exponential kyun? Yeh woh unique function hai jiska rate of change khud se proportional hota hai — exactly "constant relative growth" ka matlab yahi hai.

3. $\lambda$ read karo. $\delta (t)={\delta }_0{e}^{\lambda t}$ se compare karo: $\lambda =k=0.5{\text{s}}^{-1}>0.$ Yeh step kyun? $\lambda$ ko gap growth mein $t$ ke aage exponent ke roop mein define kiya jaata hai.

4. Chaos decide karo. $\lambda >0$, phir bhi trajectory $x(t)=x_0{e}^{kt}$ infinity ki taraf bhaag jaati hai — yeh unbounded hai. Chaos ke liye SDIC aur boundedness dono chahiye. Toh yeh sensitive hai lekin chaotic NAHI.

Verify karo: $t=1$ s par, gap $=0.001e^{0.5}=0.001\times 1.6487=1.6487\times 10^{-3}$. Yeh badha lekin ek gap ke roop mein raha jo do aisi numbers ke beech thi jo dono bhaag gayi — koi folding nahi, koi attractor nahi. Cell A confirmed: positive $\lambda$ zaruri hai lekin kaafi nahi.

Cell B — stable point par decay

Forecast: gap badhega ya ghategaa? Har second kitna?

1. Gap equation. $\delta =x^{\prime }-x$ ke saath, do copies subtract karke: $\dot{\delta }=-3x^{\prime }+3x=-3\delta .$ Yeh step kyun? Constant $x^{\ast }$ cancel ho jaata hai — sirf difference bachta hai, jo hum chahte hain. Yeh collapsing curve hai (figure s01 ka middle panel).

2. Solve karo. $\delta (t)={\delta }_0{e}^{-3t}$. Yeh step kyun? Same exponential logic jaise pehle, ab ek negative rate ke saath.

3. $\lambda$ read karo. $\lambda =-3{\text{s}}^{-1}<0$. Gap khatam ho jaata hai. System fixed point $x^{\ast }=2$ par converge karta hai — ek stable fixed point. Koi chaos nahi.

Verify karo: initial gap $=2.11-2.1=0.01$. $1$ s baad: $0.01e^{-3}=0.01\times 0.049787=4.98\times 10^{-4}$. ~20× ghata. Dono runs $x=2$ ki taraf march karte hain, bhool jaate hain kahan se shuru hua tha — SDIC ka bilkul ulta. Cell B confirmed.

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

Forecast: do runs identical shapes hain jo thodi shift hain. Kya unka gap explode hoga, vanish hoga, ya bounded rahega?

1. Dono runs likho. $x(t)=\cos t,x^{\prime }(t)=\cos (t+0.01).$ Yeh step kyun? Ek tiny phase difference exactly ek tiny initial-condition difference hai ek oscillator ke liye. Yeh coasting curve hai (figure s01 ka right panel).

2. Gap. $\cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}$ use karke: $\delta (t)=x^{\prime }-x=-2\sin (t+0.005)\sin (0.005).$ Yeh step kyun? Hum time ke saath gap ka actual size chahte hain, koi aisa approximation nahi jo uske behaviour ko chhupaaye.

3. Initial gap read karo. $t=0$ daalo: ${\delta }_0=\delta (0)=-2\sin (0.005)\sin (0.005)=-2{\sin }^2(0.005)\approx -5.0\times 10^{-5}.$ Yeh step kyun? Hum starting gap ko baad ki amplitude se confuse nahi kar sakte — dono alag numbers hain yahan, aur $\lambda$ unhe compare karta hai.

4. Poori curve ko bound karo. Saare baad ke $t$ ke liye, factor $\sin (t+0.005)$ size mein kabhi $1$ se zyada nahi ho sakta, isliye $|\delta (t)|=2|\sin (t+0.005)|\sin (0.005)\le 2\sin (0.005)\approx \mathrm{0.01.}$ Toh gap $5\times 10^{-5}$ se badhta hai zyada se zyada $0.01$ tak aur phir us box ke andar oscillate karta hai — kabhi bhaagta nahi. Yeh step kyun? Hum gap ko ek fixed ceiling $B=2\sin (0.005)$ ke andar pin kar rahe hain, $t$ se independent. Woh ceiling woh key ingredient hai jo next step ko chahiye.

5. Bound ko limit mein feed karo. Step 4 ki fixed ceiling $B=0.01$ lo aur use seedha Lyapunov definition mein daalo. Kyunki $|\delta (t)|\le B$ sab $t$ ke liye, log ke andar fraction cap ho jaata hai: $\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}\le \frac{1}{t}\ln \frac{B}{|{\delta }_0|}=\frac{1}{t}\underset{\text{ek fixed number}\approx 5.3}{\underbrace{\ln \frac{0.01}{5\times 10^{-5}}}}\stackrel{t\to \infty }{\to }0.$ Numerator $\ln (B/|{\delta }_0|)$ ek fixed constant hai ($t$ ke saath nahi badhta), aur kisi bhi fixed constant ko $t\to \infty$ se divide karne par woh $0$ ho jaata hai. Toh $\lambda \le 0$; aur kyunki $|\delta (t)|$ bhi zero nahi ghatta, $\lambda \ge 0$. Isliye: $\lambda ={\lim }_{t\to \infty }\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}=0.$ Yeh step kyun? Yeh woh explicit link hai jiske liye bound tha: bounded gap ⇒ constant numerator ⇒ constant / $t$ → 0. Yeh marginal case hai: periodic / quasi-periodic motion.

Verify karo: ${\delta }_0=-2{\sin }^2(0.005)=-4.9999\times 10^{-5}$; amplitude bound $2\sin (0.005)=0.0099999$. Na bina limit ke badhta hai na zero ghatta hai ⇒ $\lambda =0$. Cell C confirmed.

Cell D — exact chaotic value $\ln 2$

Forecast: compute karne se pehle value guess karo. (Hint: yeh distances double karta hai.)

Figure s02 (words mein describe kiya gaya). Do side-by-side panels. Left panel: logistic map $x_{n+1}=4x(1-x)$ ek ulta parabola (lavender mein) ke roop mein draw kiya gaya saath mein dashed diagonal line $y=x$; parabola middle mein apni peak tak uthta hai aur waapis aata hai, dikhata hai ki interval $[0,1]$ top par khud par fold hota hai — yahi folding motion ko bounded rakhti hai. Right panel: same dynamics change of variable ke baad, straight-sided doubling map $\theta \mapsto 2\theta \mathrm{mod}1$ ke roop mein draw kiya gaya (ek mint saw-tooth jisme do parallel sloping segments hain). Ek arrow dikhata hai ki do nearby $\theta$ values ke beech har chhotey gap ko exactly 2 ka factor stretch kiya jaata hai per step — yahi constant factor 2 hai jisse hum $\lambda =\ln 2$ padhte hain.

1. Maps ke liye yeh formula kyun? Har step ek chhotey gap ko local slope $|f^{\prime }(x_n)|$ se multiply karta hai. $N$ steps ke baad total stretch product ${\prod }_n|f^{\prime }(x_n)|$ hai. $\frac{1}{N}\ln$ lene se woh product ek average log-stretch mein convert ho jaata hai — $\lambda$ ki definition.

2. Ek angle variable mein change karo jo map ko seedha kare. Har point ko $x_n={\sin }^2(\pi {\theta }_n)$ ke roop mein likho jahan ${\theta }_n\in [0,1)$ — ${\theta }_n$ ko ek circle par kahan state hai aise sochon, aur $x_n$ uski height hai. Ise $x_{n+1}=4x_n(1-x_n)$ mein plug karo aur identity $4{\sin }^{2}u{\cos }^{2}u={\sin }^2(2u)$ use karo: $x_{n+1}=4{\sin }^2(\pi {\theta }_n){\cos }^2(\pi {\theta }_n)={\sin }^2(2\pi {\theta }_n)={\sin }^2(\pi \cdot 2{\theta }_n).$ $x_{n+1}={\sin }^2(\pi {\theta }_{n+1})$ se match karne par ${\theta }_{n+1}=2{\theta }_n\mathrm{mod}1$ milta hai ("mod 1" isliye kyunki ${\sin }^2$ repeat karta hai). Toh $\theta$ picture mein rule simply "angle double karo" hai. Yeh step kyun? Logistic slope $|f^{\prime }(x_n)|$ messy aur $x$-dependent hai; angle change of variable — upar callout mein stated invariance se legal — ise ek aisi rule mein convert karta hai jiska stretch ek clean constant factor 2 hai har step (mint saw-tooth, figure s02 ka right panel).

3. Average compute karo. Kyunki har step $\theta$-coordinate mein kisi bhi chhotey gap ko exactly 2 se stretch karta hai, $\ln |f^{\prime }|=\ln 2$ har step hai, toh: $\lambda =\frac{1}{N}{\sum }_{n=0}^{N-1}\ln 2=\ln 2\approx \mathrm{0.6931.}$ Yeh step kyun? $\ln 2$ ki $N$ copies ka sum, $N$ se divide karne par, sirf $\ln 2$ hai.

4. Sign ki sanity. $\lambda =\ln 2>0$ ⇒ chaos, aur map $[0,1]$ mein rehta hai toh yeh bounded hai ⇒ genuinely chaotic (Ex 1 ke unlike). The logistic map and period-doubling route to chaos dekho.

Verify karo: $\ln 2=0.693147\dots >0$. Kaafi iterates par $\frac{1}{N}\sum \ln |4(1-2x_n)|$ ka numeric average $0.693$ ke paas converge karta hai. Cell D confirmed.

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

Forecast: ek gap ka kya hota hai jo exactly zero ke slope se multiply hoti hai?

1. Fixed point find karo. $x^{\ast }=2x^{\ast }(1-x^{\ast })$ solve karo. Ya toh $x^{\ast }=0$ ya $1=2(1-x^{\ast })\Rightarrow x^{\ast }=0.5$. Yeh step kyun? Fixed point woh hai jahan map ek value ko khud pe map karta hai — woh resting state jiske stability ko hum test karte hain.

2. Slope evaluate karo. $f^{\prime }(x)=2(1-2x)$, toh $x^{\ast }=0.5$ par: $f^{\prime }(0.5)=2(1-2\cdot 0.5)=2\cdot 0=0.$ Yeh step kyun? Slope hi one-step stretch factor hai; yahi $\lambda$ ko feed karta hai.

3. $\lambda$ interpret karo. Ek step kisi bhi gap ko $0$ se multiply karta hai — yeh kuch nahi mein crush ho jaata hai. Log-average mein, $\ln |0|=-\infty$, toh is direction se $\lambda =-\infty$ milta hai: ek superstable point. Yeh step kyun? Shrink-factor-zero ka $\ln$ sabse strong possible convergence hai. Strongly stable ⇒ koi chaos nahi.

Verify karo: $2(1-2\cdot 0.5)=0$ exactly. Koi bhi nearby run ek step ki stretch mein $x=0.5$ par collapse ho jaata hai — Cell B ka extreme. Cell E confirmed.

Cell F — predictability time (word problem)

Forecast: guess karo ki 100× better sensors forecast window ko double karte hain — ya barely nudge karte hain.

1. Growth equation set karo. Error $\Delta ={\delta }_0{e}^{\lambda t}$ ki tarah badhta hai. $t$ ke liye solve karo: $t_{\text{pred}}=\frac{1}{\lambda }\ln \frac{\Delta }{{\delta }_0}.$ Yeh step kyun? Yeh exponential ko invert karta hai — ek exponent se $t$ isolate karne ke liye sirf logarithm ka algebra kaam karta hai.

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

3. Sensors 100× improve karo: ab ${\delta }_0=10^{-6}$, toh $\Delta /{\delta }_0=10^5$: $t_{\text{pred}}=4\ln (10^5)=4\times 11.513=46.05\text{days}.$ Yeh step kyun? Same formula, sirf ${\delta }_0$ badla.

4. Interpret karo. 100× better data ne humein $46.05-27.6=18.4$ extra days diye — ek $67\%$ gain, $100\times$ nahi. Yahi hai brutal logarithm: $t_{\text{pred}}\propto \ln (1/{\delta }_0)$.

Verify karo: $4\ln (10^3)=27.63$ days; $4\ln (10^5)=46.05$ days. Ratio $46.05/27.63=1.667$, yaani $100\times$ better data ke liye sirf $1.67\times$ longer. Cell F confirmed. Units: $[1/\lambda ]=$ days, $\ln (\cdot )$ dimensionless ⇒ answer days mein. ✔

Cell G — multi-dimensional spectrum

Forecast: kya ek system ek direction mein stretch kar sakta hai phir bhi volume mein simat sakta hai? Sum ka sign guess karo.

1. Chaos test. Maximal exponent ${\lambda }_1=+0.9>0$ ⇒ nearby trajectories alag hoti hain ⇒ chaotic (aur flow bounded hai, toh yeh genuine strange attractor hai). Yeh step kyun? Sirf sabse bada exponent SDIC decide karta hai; yeh ek generic gap ki growth dominate karta hai.

2. Volume test. Divergence/Liouville argument se, phase-space volume $e^{({\sum }_i{\lambda }_i)t}$ ki tarah badhta hai. Sum: ${\sum }_i{\lambda }_i=0.9+0+(-14.6)=-13.7{\text{s}}^{-1}<0.$ Yeh step kyun? Flow ki stretching ka trace (sum of ${\lambda }_i$) volume change ki exponential rate hai — Liouville's theorem (phase-space volume) dekho.

3. Paradox reconcile karo. ${\lambda }_1>0$ ⇒ ek direction mein stretch; $\sum {\lambda }_i<0$ ⇒ total volume collapse hota hai. Flow stretch karta hai, fold karta hai, aur ek zero-volume fractal set par contract karta hai. Yeh step kyun? Shrinking volume par stretch-and-fold dissipative chaos ka mechanism hai.

4. Bonus — attractor dimension (Kaplan–Yorke): $\lambda$'s ka running sum negative hone tak directions count karo. $D_{KY}=2+\frac{{\lambda }_1+{\lambda }_2}{|{\lambda }_3|}=2+\frac{0.9+0}{14.6}=\mathrm{2.062.}$ Yeh step kyun? Exponents sabse bade se add karna shuru karo: ${\lambda }_1=0.9>0$, phir ${\lambda }_1+{\lambda }_2=0.9>0$ abhi bhi positive, lekin ${\lambda }_3$ add karne par $0.9-14.6<0$ ho jaata hai. Toh attractor pehle do directions ko poori tarah fill karta hai (yahi "$2$" hai) plus teesre direction ka ek fraction $\frac{{\lambda }_1+{\lambda }_2}{|{\lambda }_3|}$ — deta hai ek fractional dimension plane ($2$) aur solid ($3$) ke beech — yeh measure karta hai strange attractor kitna "thick" hai.

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

Cell H — exam trap: SDIC ≠ chaos

Forecast: teeno "separating" ya "moving" trajectories rakhte hain — lekin sirf ek chaotic hai. Kaun?

1. System (i) $\dot{x}=x$. Gap: $\delta ={\delta }_0{e}^t$, toh $\lambda =+1>0$. Lekin $x(t)=x_0{e}^t\to \infty$: unbounded. SDIC ✔, boundedness ✘ ⇒ chaotic nahi (same lesson jaise Ex 1). Yeh step kyun? Chaos ke liye trajectory ka ek finite region mein trapped hona zaroori hai taaki woh fold back ho sake.

2. System (ii) doubling map. Har step gap double karta hai (mod 1), $\lambda =\ln 2>0$, aur yeh finite interval $[0,1)$ mein rehta hai toh yeh fold karta hai (mod). SDIC ✔, bounded ✔, mixing ✔ ⇒ chaotic. Yeh step kyun? Yeh clean archetype hai: 2 se stretch karo, wrap around karo — exactly Ex 4 ka engine.

3. System (iii) uniform rotation. Gap constant rehta hai: $\delta (t)={\delta }_0$, toh $\lambda =0$. Bounded ✔ lekin koi separation nahi ⇒ chaotic nahi (periodic, Cell C). Yeh step kyun? Koi stretching nahi matlab koi sensitivity nahi, chahe kitna bhi bounded ho.

4. Rule nikala. ==Chaos = positive $\lambda$ AND boundedness AND mixing==, jahan mixing ka matlab hai flow eventually kisi bhi chhotey starting points ke blob ko poore attractor par stir kar deta hai — toh koi bhi region kisi bhi doosre se overlap karta hai (formally: regions $U,V$ ke liye $U$ ka flow eventually $V$ ko intersect karta hai). Teen conditions mein se koi ek miss karo aur woh chaos nahi hai. Parent note ki topological mixing ki definition dekho.

Verify karo: (i) $\lambda =1>0$ lekin unbounded → nahi. (ii) $\lambda =\ln 2=0.693>0$, bounded → haan. (iii) $\lambda =0$ → nahi. Teen mein se sirf ek qualify karta hai. Cell H confirmed.

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