Lyapunov stability — Lyapunov function, positive definiteness

3.5.31 · Physics › Guidance, Navigation & Control (GNC)

Intuition Core idea (energy bina energy ke)

Ek ball ko ek bowl mein roll karte hue imagine karo. Chahe kahaan bhi chhoddo, woh aakhir mein neeche aa jaati hai. Kyun? Kyunki uski energy hamesha neeche jaati hai jab woh move karti hai. Lyapunov ki genius yeh thi ki unhone realize kiya: system stable hai ya nahi jaanne ke liye equations of motion solve karne ki zarurat nahi — bas koi bhi "energy-jaisi" scalar dhundhni hai jo system ki trajectories ke saath hamesha decrease karti rahe.

GNC mein WHY important hai: Spacecraft attitude controller ya missile autopilot ke liye nonlinear ODEs solve karna aksar impossible hota hai. Lyapunov se hum prove kar sakte hain ki vehicle apni commanded state mein wapas aayegi — ek stability ka certificate — bina kuch bhi integrate kiye.
WHAT chahiye: ek function $V (x)$ jo ek bowl ki tarah behave kare (neeche equilibrium par) aur jiska value system evolve hone par drop kare.

1. Stage set karna: hum kya analyze kar rahe hain?

Hum ek autonomous nonlinear system study karte hain: $\dot{x} = f (x), x \in R^{n}, f (0) = 0$

Point $x = 0$ ek equilibrium hai (agar wahan zero velocity se start karo, toh wahan hi rehte ho). Hum coordinates shift karte hain taaki interest ka equilibrium origin par ho — yeh hamesha allowed hai.

WHY $ε$ – $δ$ phrasing? "Stable" ka matlab hona chahiye chhoti kick ⟹ hamesha chhota response. Nested-ball definition exactly yahi capture karti hai bina kisi solution formula ke.

2. Positive definiteness — "bowl" property

Energy function pick karne se pehle, humein jaanna hai ki koi cheez bowl-shaped kab hoti hai.

HOW to test PD for a quadratic form $V = x^{⊤} P x$ jisme $P = P^{⊤}$ : $V$ PD hai $⟺$ == $P$ ek positive-definite matrix hai== $⟺$ $P$ ki saari eigenvalues $> 0$ hain $⟺$ saare leading principal minors $> 0$ hain (Sylvester's criterion).

3. Lyapunov function aur main theorem

Crucial object hai orbital derivative — $V$ kaise change karta hai jab tum actual dynamics ke saath flow karte ho. Chain rule se: $\dot{V} (x) = \frac{d}{d t} V (x (t)) = \sum_{i} \frac{\partial V}{\partial x _{i}} \overset{x}{˙}_{i} = \nabla V (x)^{⊤} f (x) .$

WHY chain rule aur $\partial V / \partial t$ nahi? $V$ mein koi explicit time dependence nahi hai; saari time-variation $x (t)$ ke through aati hai, jiska derivative dynamics $f$ deta hai. Yahi poora trick hai: $\dot{V}$ $x (t)$ solve kiye bina compute kiya ja sakta hai.

4. Worked examples

Worked example Example 1 — Damped pendulum / spring–mass–damper

System: $\overset{q}{¨} + b \overset{q}{˙} + k q = 0$ , $b, k > 0$ . State $x = (x_{1}, x_{2}) = (q, \overset{q}{˙})$ : $\overset{x}{˙}_{1} = x_{2}, \overset{x}{˙}_{2} = - k x_{1} - b x_{2} .$ Choose mechanical energy $V = \frac{1}{2} k x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ .

Yeh step kyun? Yeh PD hai (dono coefficients $> 0$ , $V (0) = 0$ ) — ek genuine bowl — aur physical energy natural guess hai.

Orbital derivative compute karo: $\dot{V} = k x_{1} \overset{x}{˙}_{1} + x_{2} \overset{x}{˙}_{2} = k x_{1} x_{2} + x_{2} (- k x_{1} - b x_{2}) = - b x_{2}^{2} \leq 0.$

Yeh step kyun? $k x_{1} x_{2}$ terms cancel ho jaate hain — spring se fed energy wapas laut jaati hai; sirf damper $- b x_{2}^{2}$ energy remove karta hai.

$\dot{V} \leq 0$ ⟹ kam se kam stable. Yeh semi-definite hai ( $\dot{V} = 0$ jab bhi $x_{2} = 0$ ho), isliye plain Lyapunov sirf stability deta hai; ek stronger tool (LaSalle's invariance principle) isse asymptotic mein upgrade karta hai. Physics match karti hai: damping pendulum ko rest par laata hai.

5. Common mistakes (steel-manned)

Recall Feynman: 12-saal ke bacche ko explain karo

Ek marble ko ek khaane wale bowl mein socho. Chahe kahaan bhi chhoddo, woh neeche roll ho kar rok jaata hai. Ab imagine karo tumhare paas ek magic number-machine hai: tum usse marble ki taraf point karte ho aur woh tumhe marble ki "height" bowl mein batata hai. Agar woh number marble ke move karne par chhota aur chhota hota jaata hai, tum jaante ho marble neeche ki taraf ja raha hai — chahe tumne usse roll karte hua dekha bhi na ho! Lyapunov ka idea hai: kisi spacecraft ya robot ke liye aisi number-machine banao. Agar number hamesha shrink kare, toh machine safely wahan wapas ja rahi hai jahan hona chahiye. Tumhe poora safar predict nahi karna tha — bas number ko neeche jaate dekha.

Connections

LaSalle's Invariance Principle — $\dot{V} \leq 0$ ko asymptotic stability mein upgrade karta hai.
Lyapunov Equation A^TP+PA=-Q — Example 3 se linear-system machinery.
Positive Definite Matrices & Sylvester's Criterion — PD test.
Backstepping Control & Sliding Mode Control — $\dot{V} < 0$ banane ke liye by construction design kiye gaye.
Spacecraft Attitude Control (Quaternion Feedback) — Lyapunov certificates ka real GNC use.
Chetaev's Instability Theorem — sahi converse tool.
Region of Attraction Estimation — sublevel sets ${V \leq c}$ se.

Flashcards

Positive definite function

V (x)

kya hoti hai?

V (0) = 0

aur

V (x) > 0

for all

x \neq = 0

region mein.

ODE solve kiye bina trajectories ke saath

\dot{V}

kaise compute karte hain?

\dot{V} = \nabla V (x)^{⊤} f (x)

(chain rule; dynamics

\dot{x} = f

substitute karo).

Stability vs asymptotic stability ke liye Lyapunov condition?

Stability:

V

PD aur

\dot{V} \leq 0

. Asymptotic:

V

PD aur

\dot{V} < 0

(strictly,

x \neq = 0

ke liye).

V = x^{⊤} P x

(

P

symmetric) ke liye, yeh PD kab hai?

Jab

P

ki saari eigenvalues positive hon

⟺

saare leading principal minors positive hon (Sylvester).

x^{⊤} P x > 0 ⟺

saare

λ_{i} (P) > 0

kyun?

P = Q Λ Q^{⊤}

diagonalize karo;

y = Q^{⊤} x

se, form

\sum λ_{i} y_{i}^{2}

ban jaata hai, positive iff har

λ_{i} > 0

ho.

Agar valid Lyapunov function nahi milti, toh kya system unstable hai?

Nahi — theorem sufficient hai, necessary nahi.

V

na milna kuch prove nahi karta; instability prove karne ke liye Chetaev's theorem use karo.

\dot{x} = A x

V = x^{⊤} P x

ke liye

\dot{V}

derive karo.

\dot{V} = x^{⊤} (A^{⊤} P + P A) x

; set

= - x^{⊤} Q x

(Lyapunov equation).

Damped oscillator mein

\dot{V} = - b x_{2}^{2}

kyun hota hai?

Spring cross-terms

k x_{1} x_{2}

cancel ho jaate hain; sirf damper dissipate karta hai ⟹ energy loss

\propto

velocity

^{2}

Positive definite aur positive semi-definite mein kya farq hai?

PD:

V > 0

for all nonzero

x

. PSD:

V \geq 0

(nonzero points par zero ho sakta hai — "flat channels").

Ek valid Lyapunov function geometrically kya represent karta hai?

Ek "bowl" (positive walls, neeche equilibrium par) jiska height trajectories ke saath decrease kare.