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

• (a) $\lambda >0\Rightarrow$ chaotic. Nearby trajectories exponentially separate hoti hain — gap $e^{0.7t}$ ki tarah badhta hai.
• (b) $\lambda =0\Rightarrow$ marginal (regular/periodic/quasi-periodic). Gap roughly constant rehta hai — na exponentially badhta hai na ghatta hai.
• (c) $\lambda <0\Rightarrow$ convergent. Nearby trajectories ek stable fixed point ya limit cycle par collapse hoti hain; gap $e^{-1.2t}$ ki tarah khatam hota hai. Mnemonic: Positive Pulls aparT → chaoS.

Growth factor $=\frac{10^{-2}}{10^{-4}}=100$. Gap finite time mein 100× bada ho gaya, toh local separation rate positive hai $\Rightarrow$ yeh stretch chaotic hai (exponentially separating). (Actual $\lambda$ hum L2.1 mein compute karenge.)

$|\delta (t)|=|{\delta }_0|e^{\lambda t}$ ko invert karo: dono sides ka $\ln$ lo, $\lambda =\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}=\frac{1}{5}\ln \frac{10^{-2}}{10^{-4}}=\frac{1}{5}\ln (100).$ Kyunki $\ln 100=2\ln 10\approx 4.605$, $\lambda \approx \frac{4.605}{5}\approx 0.921{\text{s}}^{-1}.$ KYA kiya humne: growth law ko $\lambda$ ke liye solve kiya. KYUN: $\ln$ exponential ko undo karta hai, aur $t$ se divide karna "total growth" ko "growth per second" mein convert karta hai.

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

$t_{\text{pred}}=\frac{1}{0.5}\ln \frac{1}{10^{-6}}=2\ln (10^6)=2\cdot 6\ln 10\approx 2\times 13.816\approx 27.6\text{days}.$ Improvement: $27.6-13.8=13.8$ days. Hazaar-guna better data ne horizon sirf double kiya — yahi brutal logarithm ka kaam hai.

$f^{\prime }(0.5)=2(1-2\cdot 0.5)=2(1-1)=0.$ Stretch factor per step $0$ hai, toh ek chhoti gap har iteration mein crush ho ke zero ho jaati hai — super-strong contraction. Tab $\ln |f^{\prime }|=\ln 0=-\infty$, giving $\lambda =-\infty$ us point par: strongly convergent, chaotic nahi. (Perturbations sirf shrink nahi hote — woh is "super-stable" point ke paas kisi bhi exponential se tezi se gayab ho jaate hain.)

$f^{\prime }(0.1)=4(1-2\cdot 0.1)=4(0.8)=3.2,|f^{\prime }(0.1)|=\mathrm{3.2.}$ Local stretch $=3.2>2=e^{\lambda }$. Toh is point par gap average se zyada stretch hota hai. Analysis: $\lambda$ poore orbit mein $\ln |f^{\prime }(x_n)|$ ka average hai. Kuch points zyada stretch karte hain ($x$ near $0$ ya $1$, factor up to $4$), kuch kam stretch karte hain ($x=0.5$ ke paas, factor near $0$). Average log-stretch exactly $\ln 2$ par aata hai. Koi bhi single point average ke barabar nahi hota — orbit ke saath average karne ka yahi poora point hai.

• (a) Sab ${\lambda }_i<0$: har error khatam ho jaata hai, trajectory rest point par settle ho jaati hai (ek stable fixed point / attractor). Chaotic nahi.
• (b) $\lambda =0$: motion periodic hai; ek chhoti error bounded rehti hai aur na exponentially badhti hai na ghatti hai. Marginal.
• (c) ${\lambda }_1>0$ (ek direction mein stretching ⇒ SDIC ⇒ chaos) lekin ${\sum }_i{\lambda }_i<0$ (friction ki wajah se total phase-space volume contract hota hai). Dono ek saath kaise? Dissipation initial conditions ke ek blob ka volume shrink karta hai, lekin flow simultaneously blob ko ek direction mein stretch karta hai aur use bounded rehne ke liye fold wapas karta hai. Stretch (positive ${\lambda }_1$) use chaotic banata hai; net contraction (negative sum) use ek strange attractor par squeeze karta hai. Dekho Strange attractors and fractal dimension aur Driven damped pendulum.

$t_{\text{pred}}=\frac{1}{\lambda }\ln (\Delta /{\delta }_0)$ se, fixed $\Delta$ ke saath, horizons mein difference hai $t_2-t_1=\frac{1}{\lambda }\left(\ln \frac{\Delta }{{\delta }_0^{\prime }}-\ln \frac{\Delta }{{\delta }_0}\right)=\frac{1}{\lambda }\ln \frac{{\delta }_0}{{\delta }_0^{\prime }}.$ Shrink factor ${\delta }_0/{\delta }_0^{\prime }$ ke liye solve karo: $\ln \frac{{\delta }_0}{{\delta }_0^{\prime }}=\lambda (t_2-t_1)=0.4\times 10=4\Rightarrow \frac{{\delta }_0}{{\delta }_0^{\prime }}=e^4\approx \mathrm{54.6.}$ Initial error ko 54.6× chhota banana padega. 10 aur seconds add karne ke liye, error ko $e^{\lambda \cdot 10}$ se shrink karna hoga — horizon mein linear gain ke liye exponential cost.

Continuous estimate: $\Delta ={\delta }_0{e}^{\lambda N}\Rightarrow N=\frac{1}{\lambda }\ln \frac{\Delta }{{\delta }_0}=\frac{1}{\ln 3}\ln (10^6).$ $N=\frac{6\ln 10}{\ln 3}=\frac{6\times 2.302585}{1.098612}\approx \frac{13.8155}{1.0986}\approx \mathrm{12.58.}$ Kyunki $N$ integer hona chahiye aur hume gap ka 1 exceed karna hai, upar round karo: $N=13$ steps. Check: 12 steps ke baad gap $\approx 10^{-6}\cdot 3^{12}=10^{-6}\cdot 531441\approx 0.531<1$; 13 steps ke baad $\approx 10^{-6}\cdot 3^{13}\approx 1.594>1$. ✓

• Inner limit ${\delta }_0\to 0$ pehle linearisation $\dot{\delta }\approx f^{\prime }(x)\delta$ ko honest rakhta hai. Yeh equation tabhi valid hai jab $\delta$ infinitesimal ho; ek finite $\delta$ attractor ki nonlinear folding feel karega.
• Phir $t\to \infty$ local stretch rate ko poore attractor par average karta hai, jisse $\lambda$ system ki property ban jaata hai, na kisi ek lucky starting point ki. Agar order swap karo (finite ${\delta }_0$ fix karo, phir $t\to \infty$): gap $|\delta (t)|$ attractor ki diameter se zyada nahi ho sakta — woh kisi bounded value par saturate ho jaata hai. Tab $\frac{1}{t}\ln \frac{|\delta (t)|}{|{\delta }_0|}\to \frac{1}{t}\ln (\text{const})\to 0$. Tum galti se har bounded system ke liye $\lambda =0$ measure kar lete, chahe chaotic ho ya nahi. ${\delta }_0\to 0$ pehle lena hi woh hai jo exponential separation ko saturation barbad karne se pehle "hamesha ke liye" run karne deta hai.

(a) ${\lambda }_1=+0.9>0\Rightarrow$ chaotic (ek direction mein SDIC). (b) Volume $e^{({\sum }_i{\lambda }_i)t}$ ki tarah badhta hai. Yahaan $\sum {\lambda }_i=0.9+0-1.4=-0.5{\text{s}}^{-1}<0$, toh volume rate $0.5{\text{s}}^{-1}$ se shrink karta hai (dissipative, consistent hai Liouville's theorem (phase-space volume) ke violate hone se friction ki wajah se — Liouville sirf conservative Hamiltonian flow ke liye hold karta hai). (c) $k$ nikalo: running sums hain ${\lambda }_1=0.9\ge 0$ ✓; ${\lambda }_1+{\lambda }_2=0.9\ge 0$ ✓; ${\lambda }_1+{\lambda }_2+{\lambda }_3=-0.5<0$ ✗. Toh non-negative sum ke saath largest $k$ hai $k=2$. Phir $D_{KY}=2+\frac{{\lambda }_1+{\lambda }_2}{|{\lambda }_3|}=2+\frac{0.9+0}{1.4}=2+0.642857\dots \approx 2.643.$ 2 aur 3 ke beech fractal dimension — strange attractor ki pehchaan (Strange attractors and fractal dimension).

Solve karo: $\dot{x}=x\Rightarrow x(t)=x_0{e}^t$, aur gap $\delta (t)=(x_0^{\prime }-x_0)e^t={\delta }_0{e}^t$, toh indeed $\lambda =1>0$: yeh hai sensitive (nearby starts exponentially separate hote hain). Lekin chaos ko teen cheezein chahiye, na ek: SDIC + boundedness + topological mixing. Yahaan trajectory $x_0{e}^t\to \infty$ unbounded hai — kuch bhi use fold wapas nahi karta. Positive $\lambda$ chaos imply karta hai sirf ek bounded system par attractor ke saath. Toh "positive $\lambda \Rightarrow$ chaos" shorthand hai jo silently boundedness assume karta hai. Folding ke bina stretching sirf explosion hai; chaos confined region mein stretch-and-fold hai.

Answers cover karo. Growth law solved for $\lambda$ ::: $\lambda =\frac{1}{t}\ln (|\delta (t)|/|{\delta }_0|)$ Predictability time formula ::: $t_{\text{pred}}=\frac{1}{\lambda }\ln (\Delta /{\delta }_0)$ Horizon double karne ke liye, ${\delta }_0$ ko shrink karo ::: factor $e^{\lambda \Delta t}$ se (exponential cost) Chaos ke teen ingredients ::: SDIC + boundedness + topological mixing Kaplan–Yorke idea ::: fractional dimension wahan se jahan cumulative $\lambda$ zero cross kare