Lyapunov stability — Lyapunov function, positive definiteness: Question bank

3.5.31 · D5 · Physics › Guidance, Navigation & Control (GNC) › Lyapunov stability — Lyapunov function, positive definitenes

Definition Is poore bank mein use ki gayi notation (zero se bani)

$x \in R^{n}$ state hai — un numbers ki list jo system ko abhi describe karti hai (jaise ek spacecraft ka angle aur angular rate).
System ek rule $\dot{x} = f (x)$ ke zariye evolve karta hai. Yahan $\dot{x}$ state ki velocity hai (har number kitni tezi se change hota hai), aur $f$ dynamics map hai: ek machine jo current state $x$ dekar state ke move karne ki direction aur speed return karta hai. $f$ ko plane ke har point par laga ek arrow samjho — state in arrows ke saath slide karti hai (neeche vector field dekho).
$f (0) = 0$ ka matlab hai origin ek equilibrium hai: wahan koi arrow nahi, isliye agar tum origin par rest mein start karo toh wahi rehte ho.
$V (x)$ ek scalar "height" hai (bowl), $\nabla V$ iska gradient (steepest-uphill vector), aur $\dot{V} = \nabla V^{⊤} f$ iska rate of change hai jab state $f$ ke arrows ke saath flow karti hai.
PD = positive definite: $V (0) = 0$ aur $V (x) > 0$ har $x \neq = 0$ ke liye.

Neeche di gayi picture in teeno objects ko anchor karti hai — bowl $V$ , arrows $f$ , aur woh invariant set jahan $\dot{V} = 0$ hai — traps try karne se pehle.

Figure — Lyapunov stability — Lyapunov function, positive definiteness

True or false — justify

Agar

\dot{V} \leq 0

har jagah ho, toh origin asymptotically stable hai.

False.

\dot{V} \leq 0

(negative semi-definite) sirf itna guarantee karta hai ki height kabhi badhegi nahi, yaani plain stability/boundedness; convergence ke liye tumhe

\dot{V} < 0

chahiye

x \neq = 0

ke liye, ya LaSalle's Invariance Principle, origin par convergence ke liye.

Ek positive-definite

V

V (0) = 0

satisfy karna chahiye.

True. Definition ke anusaar bowl ka bottom equilibrium par baithta hai; agar

V (0) > 0

ho toh woh PD nahi hai (halankि kabhi-kabhi constant se shift karke theek kiya ja sakta hai).

V = \frac{1}{2} x_{2}^{2}

ek 2D system ke liye valid Lyapunov candidate hai.

False. Yeh sirf positive semi-definite hai:

V = 0

poori

x_{1}

-axis ke saath hota hai, sirf origin par nahi, isliye yeh

x_{1}

component ko trap nahi kar sakta aur PD requirement fail karta hai.

Agar koi Lyapunov function nahi milta, toh system unstable hai.

False. Lyapunov ka theorem sufficient hai, necessary nahi; working

V

na milna sirf itna dikhata hai ki tum kaafi clever nahi the. Instability actually prove karne ke liye Chetaev's Instability Theorem use karo.

V = x^{⊤} P x

ke liye symmetric

P

ke saath, "

P

ke saare eigenvalues

> 0

" aur "

P

ke saare leading principal minors

> 0

" equivalent tests hain.

True. Dono positive-definite matrix ke equivalent characterizations hain; minor test Sylvester's criterion hai aur usually haath se check karna aasaan hota hai.

Lyapunov equation

A^{⊤} P + P A = - Q

kisi bhi PD

Q

ke liye hamesha ek PD solution

P

rakhta hai.

False. Ek unique PD

P

tab aur sirf tab exist karta hai jab

A

ke har eigenvalue ka strictly negative real part ho; agar

A

unstable hai, toh koi PD

P

ise solve nahi karta. Dekho Lyapunov Equation A^TP+PA=-Q.

Non-symmetric

P

wale quadratic form

x^{⊤} P x

ki koi well-defined definiteness nahi hoti.

False. Sirf symmetric part

\frac{1}{2} (P + P^{⊤})

x^{⊤} P x

ki value ko affect karta hai, isliye definiteness usi symmetric part se decide hoti hai; hum conventionally

P

ko symmetric lete hain.

Agar ek chosen

V

ke liye

\dot{V} > 0

ho, toh system unstable hona chahiye.

False. Ek bura candidate kuch prove nahi karta — woh sirf galat bowl ho sakta hai. Sirf woh

V

jo Chetaev ki hypotheses meet kare (suitable cone mein positive

V

aur positive

\dot{V}

) instability certify karta hai.

Positive semi-definite

V

ab bhi LaSalle ke saath milkar asymptotic stability de sakta hai.

True. Agar

\dot{V} \leq 0

aur

{\dot{V} = 0}

mein rehne wali ek matra trajectory origin ho, toh LaSalle's Invariance Principle stability ko asymptotic stability mein upgrade kar deta hai — exactly isi tarah damped pendulum handle hota hai.

Spot the error

\dot{V} = \nabla V

, isliye main sirf

V

ko uske arguments se differentiate karta hoon."

Error.

\nabla V

gradient hai (partials ka vector); orbital derivative hai

\dot{V} = \nabla V^{⊤} f (x)

— tumhe gradient ko dynamics map

f

ke saath dot karna hoga.

\cdot f

step skip karne se ek meaningless quantity milti hai.

"Main

\dot{V} = \partial V / \partial t

compute karunga kyunki

V

time mein change hota hai."

Error. Autonomous

V (x)

ke liye koi explicit

t

-dependence nahi hai; saara time variation

x (t)

ke through aata hai, isliye chain rule deta hai

\dot{V} = \nabla V^{⊤} \dot{x} = \nabla V^{⊤} f

, na ki

\partial V / \partial t = 0

P

ka positive determinant hai, isliye

P

positive definite hai."

Error. Positive determinant sirf itna dikhata hai ki negative eigenvalues ki even count hai; jaise

diag (- 1, - 1)

ka determinant

+ 1

hai lekin woh negative definite hai. Tumhe saare leading principal minors positive chahiye (Sylvester).

\dot{V} = - x_{1}^{2} (1 - x_{2}^{2}) - x_{2}^{2} < 0

har jagah hai, isliye system globally asymptotically stable hai."

Error. Pehla term sirf tab negative hai jab

∣ x_{2} ∣ < 1

; us strip ke bahar

\dot{V}

positive ho sakta hai, isliye tum sirf

∣ x_{2} ∣ < 1

par local asymptotic stability claim kar sakte ho. Dekho Region of Attraction Estimation.

"Damped pendulum ki energy,

V = \frac{1}{2} k x_{1}^{2} + \frac{1}{2} x_{2}^{2}

, deti hai

\dot{V} < 0

, isliye directly asymptotic stability."

Error.

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

sirf

\leq 0

hai (zero jab bhi

x_{2} = 0

ho), yaani semi-definite. Plain Lyapunov sirf stability deta hai; asymptotic stability ke liye LaSalle chahiye.

V

PD hai aur

\dot{V} \leq 0

hai, isliye trajectories kisi nonzero point par converge karti hain."

Error.

\dot{V} \leq 0

sirf ek sublevel set ke andar boundedness guarantee karta hai, kisi specific point par convergence nahi; trajectory

{\dot{V} = 0}

level set par oscillate ya drift kar sakti hai bina settle kiye.

Why questions

V

continuously differentiable kyun hona chahiye?

Kyunki poora method chain-rule identity

\dot{V} = \nabla V^{⊤} f

par tika hai; bina well-defined gradient ke orbital derivative — woh object jo theorem test karta hai — exist hi nahi karta.

Damped pendulum ke liye

\dot{V}

mein spring term cancel kyun ho jaata hai?

Spring conservative hai: jo energy store karti hai woh return karti hai, isliye uska contribution

k x_{1} x_{2}

coupling term se exactly annihilate ho jaata hai, sirf dissipative damper

- b x_{2}^{2}

bachta hai, jahan energy actually leave karti hai.

Hum interest ke equilibrium ko hamesha origin par kyun rakh sakte hain?

Stability ek fixed point ke baare mein local property hai; coordinate shift

z = x - x_{e q}

kisi bhi equilibrium ko

0

par le aata hai bina dynamics ke qualitative behaviour ko change kiye, har "bowl at the bottom" statement ko simplify karta hai.

Hum

\dot{V}

saari directions mein ki bajaye trajectories ke saath kyun test karte hain?

Stability isse concern karti hai ki system actually flow kaisa karta hai, arbitrary displacements se nahi; sirf direction

f (x)

jo dynamics map deta hai woh realize hoti hai, isliye

\dot{V} = \nabla V^{⊤} f

woh actual rate of change of height measure karta hai jo system experience karta hai.

Yeh prove karne ke liye diagonalization argument kyun kaafi hai ki

x^{⊤} P x > 0 ⟺

eigenvalues

> 0

Orthogonal diagonalization

P = Q Λ Q^{⊤}

form ko

\sum λ_{i} y_{i}^{2}

mein convert kar deta hai jahan

y = Q^{⊤} x

ek pure rotation hai, isliye pure sum ki positivity sab

y \neq = 0

ke liye exactly har

λ_{i}

ki positivity hai.

Lyapunov ka method ODE solve karne se kyun bachata hai?

Kyunki

\dot{V} = \nabla V^{⊤} f

dynamics map

f

ko directly use karta hai — hum

x (t)

find karne ke liye kabhi integrate nahi karte; hum sirf ek scalar expression ka sign check karte hain, yahi cheez is method ko unsolvable nonlinear GNC problems par usable banati hai.

Physical energy ko

V

ke liye pehla guess kyun banana achha hai?

Real systems energy dissipate karte hain, isliye mechanical energy automatically rest par centered ek PD bowl hai aur damping ke saath decrease hoti rehti hai — yeh almost bina kisi engineering ke dono conditions satisfy karta hai, jaise Sliding Mode Control aur Backstepping Control designs mein hota hai.

Edge cases

Kya hoga agar

\dot{V} = 0

sirf origin par nahi, poori surface par ho?

Plain Lyapunov sirf stability deta hai; tumhe check karna hoga ki koi trajectory us surface par reh sakti hai ya nahi. Agar ek matra invariant subset origin hai, toh LaSalle's Invariance Principle asymptotic stability recover kar leta hai.

Kya hoga agar

f (0) \neq = 0

Tab origin ek equilibrium nahi hai aur theorem wahan apply nahi hota; tumhe pehle ek actual equilibrium dhundna hoga aur coordinates shift karne honge taaki

f (0) = 0

ho jaye.

Kya hoga agar

V

locally PD ho lekin

V \to

finite as

∥ x ∥ \to \infty

Local asymptotic stability hold kar sakti hai, lekin tum global stability conclude nahi kar sakte: global results ke liye radial unboundedness bhi chahiye (

V \to \infty

∥ x ∥ \to \infty

) taaki sublevel sets bounded rahein.

Zero matrix

P = 0

ki definiteness kya hai

V = x^{⊤} P x

mein?

V \equiv 0

, jo sirf positive semi-definite hai (practically identically zero) — yeh PD nahi hai, isliye yeh ek useless Lyapunov candidate hai jo kuch bhi trap nahi karta.

Nonlinear system ke liye

\dot{V} = - x_{1}^{2} (1 - x_{2}^{2}) - x_{2}^{2}

mein boundary

∣ x_{2} ∣ = 1

par kya hota hai?

Wahan

\dot{V} = - x_{1}^{2} (1 - 1) - x_{2}^{2} = - x_{2}^{2} \leq 0

sirf semi-definite ho jaata hai (poori line

x_{2} = 0

par zero); woh strict negativity jo asymptotic stability de rahi thi exactly edge par khatam ho jaati hai, guaranteed basin ki frontier mark karte hue.

Kya hoga agar

A

ka eigenvalue zero real part wala ho (marginal case)?

Lyapunov equation ka koi PD solution nahi hoga aur linearization inconclusive hai; linear method stability certify nahi kar sakta, aur tumhe nonlinear terms directly analyze karne honge (centre-manifold ya ek custom

V

Kya system stable ho sakta hai lekin asymptotically stable nahi?

Haan — ek undamped oscillator (

b = 0

pendulum mein) ka

\dot{V} = 0

hota hai: trajectories constant-energy orbits par hamesha circle karti rehti hain, bounded rehti hain (stable) lekin origin par kabhi converge nahi karti.

Kya PD

V

ke saath

\dot{V} < 0

guarantee karta hai ki state apne starting sublevel set ke andar rahe?

Haan — kyunki height strictly girti hai, trajectory kabhi kisi higher level set mein re-enter nahi kar sakti, isliye woh

{V \leq V (x (0))}

ke andar rehti hai; yahi trapping property Region of Attraction Estimation ke neeche kaam karti hai.

Recall Har trap ka ek-line summary

Ek positive definite (PD) $V$ ka matlab hai == $V > 0$ har nonzero $x$ ke liye== $V (0) = 0$ ke saath; stability ke liye == $\dot{V} \leq 0$ chahiye, asymptotic stability ke liye $\dot{V} < 0$ ya LaSalle== chahiye; $V$ ki failed search ya ek candidate jiska == $\dot{V} > 0$ ho kuch prove nahi karta== (instability prove karne ke liye Chetaev use karo); hamesha gradient ko dynamics map $f$ ke saath dot karo; aur har "global" claim ke peeche ek radial-unboundedness assumption chupi hoti hai.