Lyapunov stability — Lyapunov function, positive definiteness: Visual walkthrough

3.5.31 · D2 · Physics › Guidance, Navigation & Control (GNC) › Lyapunov stability — Lyapunov function, positive definitenes

Hum ek cheez prove karenge, dheere dheere:

Agar tumhare state space ke upar ek bowl-shaped number $V$ rakha hai, aur woh number system ke move karne par kabhi nahi chadta, toh system escape nahi kar sakta — woh stable hai.

Step 1 — "State" kya hota hai aur ek trajectory kaisi dikhti hai?

KYA. Hamara system ek aisi rule hai jo har point par kehti hai, aage kidhar jaana hai. System ki condition ko numbers ki ek list $x = (x_{1}, x_{2})$ ke roop mein likho — ek ghoomte spacecraft ke liye socho "angle error" aur "angular rate error." Rule yeh hai: $\dot{x} = f (x) .$

Term by term padho:

$x$ — floor plan par ek point (the state).
$\dot{x}$ — us point se attached ek chhota arrow jo batata hai ki aage kidhar drift karoge (the "dot" ka matlab hai time mein change ki rate).
$f$ — woh machine jo ek point ko uske arrow mein badal deti hai. Isko ek point do, ek arrow milega.

KYUN. Hum yeh reason karna chahte hain ki system kahan jaata hai bina kabhi formula se path compute kiye. Toh hum pehle arrows ko picture karte hain: woh ek field banate hain, aur ek moving point arrows ko follow karta hai jaise paani mein patta beh jaata hai.

PICTURE. Grid ke har dot par ek cyan arrow hai. Kisi bhi start se arrows follow karo aur tum ek trajectory (amber) trace karte ho jo origin ki taraf curl karti hai.

Figure — Lyapunov stability — Lyapunov function, positive definiteness

Step 2 — Bowl banao: floor ke upar ek positive-definite number

KYA. Har floor point ke upar ek height latkao: ek number $V (x)$ . Hum demand karte hain ki woh bowl ki tarah shaped ho: $V (0) = 0, V (x) > 0 for x \neq = 0 .$ Term by term:

$V (0) = 0$ — bowl ka bottom bilkul equilibrium ke upar baitha hai.
$V (x) > 0$ — baaki har jagah zameen upar hai. Koi side channels nahi, koi flat ledges nahi.

Aisa $V$ positive definite kehlata hai, jise hum aage PD abbreviate karenge.

KYUN. Hume ek aisa single number chahiye jiska chhota hona state ko ghar ke paas force kare. Bowl bilkul yahi karta hai: "low height" sirf bottom ke paas hi possible hai. Agar bowl mein ek flat trench hoti (sirf $V \geq 0$ , yaani semi-definite), toh ek point door slide kar sakta tha jabki $V$ $0$ par rehta — bekar. Isliye strict $> 0$ matter karta hai.

PICTURE. Floor ke upar paraboloid $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2})$ . Iske horizontal slices circles hain — level sets ${V = c}$ . Notice karo: ek chhoti height $c$ tumhe ek chhote circle ke andar band kar deti hai.

Recall Sirf distance kyun nahi measure karte?

Distance ::: Kar sakte ho — distance-squared ek valid PD bowl hai. Lekin system ki apni physics se matched bowl (jaise mechanical energy) zyada tar agli step ki arithmetic ko khoobsurti se cancel kar deti hai. Bowl choose karne ki freedom hi Lyapunov ki poori power hai.

Step 3 — Key question: kya moving point chadh raha hai ya doob raha hai?

KYA. Jaise state ek trajectory par slide karti hai, uski height $V$ time ke saath badal jaati hai. Hume woh rate chahiye, $\dot{V}$ . Kyunki $V$ position par depend karta hai aur position time par, chain rule use karo: $\dot{V} (x) = steepest-uphill arrow \nabla V (x) \cdot motion arrow f (x) .$ Term by term:

$\nabla V$ — gradient, ek arrow jo bowl par seedha uphill point karta hai, wahan sabse lamba jahan bowl sabse steep hai.
$f$ — Step 1 ka motion arrow, point actually kidhar jaata hai.
Beech ka dot " $\cdot$ " dot product hai: dono arrows ke matching components multiply karo aur add karo. Yeh jawab deta hai "ek arrow doosre ki kitni direction mein point karta hai?" (Tum ise $\nabla V^{⊤} f$ bhi likhte dekhoge; chhota " $⊤$ " bas hai "numbers ke column ko side pe flip karo taaki multiply-and-add kar sakein" — iska matlab wahi dot product hai.)

KYUN dot product, aur kyun yeh tool? Dot product bilkul ek hi sawaal ka jawab deta hai: arrow B, arrow A ki direction mein kitna point karta hai? Yahan yeh poochta hai "kya motion bowl par upar ja raha hai ya neeche?" Yeh precisely wahi sign hai jo hume chahiye, aur — kamaal yeh hai — yeh sirf $f$ use karta hai, woh machine jo hamare paas pehle se hai. Hum kabhi path solve nahi karte.

Agar $f$ uphill gradient ke against point kare (obtuse angle) $\Rightarrow \dot{V} < 0$ $\Rightarrow$ doob raha hai.
Agar $f$ uski taraf point kare $\Rightarrow \dot{V} > 0$ $\Rightarrow$ chadh raha hai.
Agar $f$ $\nabla V$ ke perpendicular ho $\Rightarrow \dot{V} = 0$ $\Rightarrow$ level circle ke saath glide kar raha hai.

PICTURE. Ek sample point par: amber gradient $\nabla V$ outward-uphill point karta hua, cyan motion $f$ , aur dono ke beech ka angle. Shaded projection $\dot{V}$ hai.

Step 4 — Decisive condition: $\dot{V} \leq 0$ har jagah nearby

KYA. Ek region $D$ fix karo — bas origin ke around floor plan ka ek patch (equilibrium ka ek neighbourhood, itna chhota ki hamara chosen bowl aur hamara sign check wahan valid ho). Maano ki $D$ ke andar motion arrow bowl par kabhi uphill nahi point karta: $\dot{V} (x) = \nabla V \cdot f \leq 0 for every x in D .$ Yahan $D$ exactly woh domain hai jahan humne sign check kiya hai — neeche sab kuch $D$ ke andar ke points ke baare mein claim hai.

KYUN. Yeh single inequality kehti hai ki height sirf level reh sakti hai ya gir sakti hai, kabhi nahi chadh sakti. Geometrically: kisi bhi level circle (jo $D$ ke andar fit ho) ke aas-paas har fence-post par, motion arrow inward ya tangent point karta hai — kabhi outward nahi. Uss circle ke andar shuru hone wala point kabhi bahar cross nahi kar sakta.

PICTURE. Ek level circle (dashed cyan) jiske rim par motion arrows sample kiye gaye hain. Har arrow inward/tangent jhukta hai (amber = inward component). Trajectory trap hai andar.

Step 5 — "Trapped" ka matlab stable kyun hai ( $ε$ – $δ$ picture)

KYA. Pehle, ek notation piece jo hume chahiye: $∥ x ∥$ ka matlab hai $x$ se origin tak ki straight-line distance — $x = (x_{1}, x_{2})$ ke liye yeh $x_{1}^{2} + x_{2}^{2}$ hai, centre se point tak arrow ki length. Toh " $∥ x ∥ < ε$ " bas yeh padhta hai "point $ε$ se zyada ghar ke paas hai."

Ab "stable" ka ek precise meaning hai: chhota kick in ⟹ chhota response, hamesha ke liye. Radius $ε$ ke kisi bhi target circle ke liye hume radius $δ$ ka ek start circle dhundhna hai taaki $δ$ ke andar shuru karna tumhe hamesha $ε$ ke andar rakhe.

KYUN bowl yeh deliver karta hai. Kyunki $V$ ek PD bowl hai, height $c$ choose karo us highest level-set ki jo abhi bhi target $ε$ ke andar fit ho (aur $D$ ke andar bhi). Set ${V \leq c}$ origin ke around ek closed region hai. Step 4 se, ek baar ${V \leq c}$ ke andar aane par tum wahan raho ge (height $c$ se upar nahi ja sakti). Phir $δ$ choose karo ek chhoti ball ke roop mein jo ${V \leq c}$ ke andar fit ho. Ise chain karo: $∥ x (0) ∥ < δ \Rightarrow x (0) \in {V \leq c} \Rightarrow x (t) \in {V \leq c} \forall t \Rightarrow ∥ x (t) ∥ < ε .$

PICTURE. Teen nested rings: outer target $ε$ , unke beech trapping level-set ${V \leq c}$ , inner start $δ$ . $δ$ se launch ki gayi ek amber trajectory idhar-udhar rattles karti hai lekin kabhi $ε$ ko pierce nahi karti.

Yeh stability ka ek complete proof hai — koi ODE kabhi solve nahi ki gayi.

Step 6 — Asymptotic stability tak sharpen karna ( $\dot{V} < 0$ )

KYA. Agar inequality origin se door strict hai, $\dot{V} (x) < 0 for all x \neq = 0 in D,$ toh height sirf nahi chadh paati — woh har instant strictly girti hai jab point ghar nahi hota.

KYUN. Ek quantity jo $0$ se neeche bounded hai (bowl ki heights negative nahi ho sakti) aur jo strictly decrease karti rehti hai woh kisi floor value par settle ho jaati hai — yeh monotone convergence principle hai: ek number jo sirf neeche jaata ho lekin $0$ ke through nahi gir sakta uske paas limit ki taraf jaane ke siwa koi jagah nahi. Woh limiting height $\dot{V} = 0$ ki hogi (warna abhi bhi gir rahi hoti), aur strictness kehti hai ki $\dot{V} = 0$ sirf origin par hi hota hai. Toh state sirf ghar ke paas nahi rehti — woh ghar jaati hai: $x (t) \to 0$ .

PICTURE. Height-vs-time curve $V (x (t))$ : ek monotonically falling amber line $0$ ki taraf press karti hui, saath mein trajectory origin mein spiral karti hui.

Step 7 — Degenerate case: ek flat trench ( $\dot{V} = 0$ poore set par)

KYA. Kya hoga agar $\dot{V} \leq 0$ ho lekin $\dot{V} = 0$ sirf origin se zyada jagah par ho — maano poori ek line par? Yeh exactly parent damped-pendulum case hai: $V = \frac{1}{2} k x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ deta hai $\dot{V} = - b x_{2}^{2} \leq 0,$ jo axis $x_{2} = 0$ ke saath poori tarah zero hai (koi bhi position, zero velocity).

KYUN yeh ek real gap hai. Us line par height girna band kar deti hai. Strict-decrease reasoning (Step 6) toot jaati hai — tum Lyapunov se akele convergence conclude nahi kar sakte. Lekin physically pendulum ruk jaata hai. Escape hatch yeh hai: trench $x_{2} = 0$ par dynamics abhi bhi $\overset{x}{˙}_{2} = - k x_{1} \neq = 0$ push karte hain, toh point trench par ruk nahi sakta — woh kick off ho jaata hai aur girnaa jaari rakhta hai. "Flat set par nahi reh sakta" ko formalize karna LaSalle's Invariance Principle hai.

PICTURE. Bowl mein $x_{1}$ -axis ke saath ek flat amber trench. Ek trajectory trench ko graze karti hai, lekin vertical arrows ( $\overset{x}{˙}_{2} \neq = 0$ ) use wapas dhakelte hain — woh sirf origin par rest kar sakti hai.

Step 8 — Doosri degenerate warning: ek kharab bowl kuch prove nahi karta

KYA. Maano tumne ek $V$ choose kiya aur $\dot{V} > 0$ kahi pe mila. Lalchaane wala conclusion: "unstable!" Galat.

KYUN. Lyapunov ka theorem ek one-way ticket hai: ek achha $V$ (PD bowl, non-rising) stability certify karta hai, lekin ek failed $V$ kuch certify nahi karta. Ek alag, zyada clever bowl shayad succeed kar jaaye. Sirf Chetaev's Instability Theorem hi instability prove kar sakta hai.

PICTURE. Usi stable trajectory ke upar do candidate bowls: bowl A ke arrows $\dot{V} > 0$ dikhate hain (misleading), bowl B ke $\dot{V} < 0$ dikhate hain (correct certificate). Same system, opposite readings — proof ki ek failed guess verdict nahi hai.

Ek-picture summary

Sab ek saath: arrow field (Step 1), bowl ek level ring ke saath (Step 2), gradient-vs-motion angle $\dot{V}$ ka sign deta hua (Step 3), inward arrows ring ko trap karte hue (Step 4), nested $δ$ – $ε$ rings (Step 5), aur amber trajectory bowl ke bottom ki taraf spiral karti jab uski height zero tak girti hai (Step 6).

Recall Feynman: simple shabdon mein poora walkthrough

Ek marble ko bowl mein tip karo. Mujhe iska path compute karne ki zaroorat nahi — main bas uski height dekhta hoon. Pehle main sure karta hoon ki bowl ek asli bowl hai: target spot par sabse neeche, har jagah upar (yahi positive definite, PD hai). Phir, kisi bhi jagah, main dekhta hoon ki marble ko push karna kaise compare hota hai us direction se jo use uphill le jaaye — yeh comparison dot product $\nabla V \cdot f$ hai, aur iska sign $\dot{V}$ hai. Agar push kabhi uphill nahi hai ( $\dot{V} \leq 0$ ) mere patch $D$ ke andar, toh marble ki height badh nahi sakti, toh woh ek chhoti ring ke andar fence ho jaati hai: stable. Agar push hamesha downhill hai ( $\dot{V} < 0$ ), toh height shrink karti rehti hai aur — $0$ se neeche na gir sakne ki wajah se — kisi limit pe settle ho jaani chahiye, toh marble poore ghar tak roll karti hai: asymptotically stable. Agar ek flat ledge hai jahan height momentarily girna band kar deti hai, toh main check karta hoon ki kya marble actually wahan rest kar sakti hai — zyada tar dynamics use wapas dhakelte hain (LaSalle's Invariance Principle). Aur agar mera chosen bowl ek confusing uphill reading deta hai, toh matlab sirf yeh hai ki maine ek kharab bowl choose kiya — yeh kuch prove nahi karta. Yahi Lyapunov ka direct method hai: stability ko ek girte hue number ko dekhkar judge karo, motion solve kiye bina kabhi.