Lyapunov stability — Lyapunov function, positive definiteness: Worked examples

3.5.31 · D3 · Physics › Guidance, Navigation & Control (GNC) › Lyapunov stability — Lyapunov function, positive definitenes

Intuition Yeh page kis liye hai

Aap idea se already mil chuke ho (ek bowl-shaped "energy" $V$ jo sirf girta hi rehta hai). Yahan hum har tarah ka case grind karte hain jo yeh topic de sakta hai: signs jo kaam karte hain, signs jo fail karte hain, degenerate bowls, boundaries jahan $\dot{V}$ flip karta hai, ek real spacecraft, aur ek exam trap. Iske baad koi bhi scenario aisa nahi bachega jo tumne start-to-finish worked na dekha ho.

Un do objects ki reminder jo hum baar baar use karte hain (from the parent topic):

$V (x)$ — ek positive-definite scalar (yahan se PD likhenge: $V (0) = 0$ aur $V (x) > 0$ har $x \neq = 0$ ke liye). Yeh "bowl" hai.
$\dot{V} (x) = \nabla V^{⊤} f (x)$ — orbital derivative: bowl-height kitna change hota hai jab system actually move karta hai. $\nabla V$ partial slopes ka vector hai $(\frac{\partial V}{\partial x _{1}}, \dots)$ ; $f$ hai $\dot{x} = f (x)$ ka right-hand side; $^{⊤}$ dot ka matlab hai "har direction mein slope $\times$ velocity add karo."

Is poore page mein PD shorthand hai positive-definite ke liye (upar wala two-part bowl test).

Scenario matrix

Har Lyapunov exercise inhi cells mein se ek mein aata hai. Neeche ke examples cell(s) ke saath labeled hain — milake poora grid cover hota hai.

Cell	Case class	Kya isko distinct banata hai	Example
A	$\dot{V} < 0$ har jagah (negative definite)	full asymptotic stability, clean	Ex 1
B	Sirf $\dot{V} \leq 0$ (negative semi-definite)	sirf stability; LaSalle chahiye	Ex 2
C	Sign ek boundary par flip karta hai	region-of-attraction estimate	Ex 3
D	Degenerate $V$ (sirf semi-definite)	"bad bowl" jo kuch trap nahi karta	Ex 4
E	Linear / matrix case ( $V = x^{⊤} P x$ )	Sylvester + Lyapunov equation	Ex 5
F	Limiting / zero input (jaise $b \to 0$ , no damping)	stable aur unstable ke beech ki boundary	Ex 6
G	Real-world word problem	spacecraft attitude	Ex 7
H	Exam twist ( $\dot{V} > 0$ ≠ instability)	sufficient-not-necessary trap	Ex 8

Example 1 — Cell A: clean asymptotic stability

Worked example Statement

$\overset{x}{˙}_{1} = - x_{1} + x_{2}, \overset{x}{˙}_{2} = - x_{1} - x_{2} .$ $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2})$ try karo. Kya origin asymptotically stable hai?

Forecast: abhi guess karo — kya $\dot{V}$ har jagah $< 0$ hoga, ya sirf kahin kahin? Cross-terms symmetric lagte hain...

Check karo ki $V$ PD (positive-definite) hai. $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2}) > 0$ for $x \neq = 0$ , aur $V (0) = 0$ . ✓
- Yeh step kyun? Koi bhi orbital derivative meaningful nahi hai jab tak bowl real na ho. Hamesha pehle PD confirm karo.
Gradient. $\nabla V = (x_{1}, x_{2})$ .
- Yeh step kyun? $\dot{V}$ ke liye $V$ ke har direction mein slopes chahiye.
Dynamics ke saath dot karo. $\dot{V} = x_{1} (- x_{1} + x_{2}) + x_{2} (- x_{1} - x_{2}) = - x_{1}^{2} + x_{1} x_{2} - x_{1} x_{2} - x_{2}^{2} = - x_{1}^{2} - x_{2}^{2} .$
- Yeh step kyun? $\dot{V} = \nabla V^{⊤} f$ ; $+ x_{1} x_{2}$ aur $- x_{1} x_{2}$ exactly cancel ho jaate hain, sirf pure negatives bachte hain.
Sign padho. $\dot{V} = - (x_{1}^{2} + x_{2}^{2}) < 0$ har $x \neq = 0$ ke liye → negative definite → har jagah asymptotically stable. Cell A.

Verify: $x = (1, 0)$ par, $\dot{V} = - 1 < 0$ ; $(1, 1)$ par, $\dot{V} = - 2 < 0$ ; $(0, 0)$ par, $\dot{V} = 0$ . Height strictly drop karta hai origin ko chhod kar — marble hamesha ghar roll karta hai.

Example 2 — Cell B: sirf stability (semi-definite $\dot{V}$ )

Worked example Statement (damped oscillator, poori tarah revisited)

$\overset{x}{˙}_{1} = x_{2}, \overset{x}{˙}_{2} = - k x_{1} - b x_{2}$ with $k = 4, b = 2$ . Energy $V = \frac{1}{2} (4) x_{1}^{2} + \frac{1}{2} x_{2}^{2} = 2 x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ use karo.

Forecast: kya $\dot{V}$ strictly negative hoga, ya kisi line par zero?

PD check. Coefficients $2 > 0$ aur $\frac{1}{2} > 0$ ; $V (0) = 0$ . ✓
- Kyun? Mechanical energy spring–damper ke liye natural bowl guess hai.
$\nabla V = (4 x_{1}, x_{2})$ .
Orbital derivative. $\dot{V} = 4 x_{1} x_{2} + x_{2} (- 4 x_{1} - 2 x_{2}) = 4 x_{1} x_{2} - 4 x_{1} x_{2} - 2 x_{2}^{2} = - 2 x_{2}^{2} .$
- Yeh step kyun? Spring terms cancel ho jaate hain; sirf damper $- b x_{2}^{2} = - 2 x_{2}^{2}$ energy remove karta hai.
Sign padho. $\dot{V} = - 2 x_{2}^{2} \leq 0$ , aur $\dot{V} = 0$ poori line $x_{2} = 0$ par, sirf origin par nahi → negative semi-definite → plain Lyapunov sirf stability deta hai. Cell B. (LaSalle's Invariance Principle ise asymptotic tak upgrade karta hai kyunki $x_{2} = 0$ par stuck rehne wali akeli trajectory origin hai.)

Verify: $(1, 0)$ par, $\dot{V} = 0$ (marble momentarily frictionless line par coast karta hai); $(1, 1)$ par, $\dot{V} = - 2 < 0$ . Physics se match karta hai: damping tabhi kaam karta hai jab velocity ho.

Example 3 — Cell C: boundary par sign flip (basin estimate)

Worked example Statement

$\overset{x}{˙}_{1} = - x_{1} + x_{1} x_{2}^{2}, \overset{x}{˙}_{2} = - x_{2}$ , with $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2})$ .

Forecast: har jagah stable, ya sirf kisi strip ke andar?

PD check. Distance-squared bowl, PD. ✓
$\nabla V = (x_{1}, x_{2})$ , $f$ ke saath dot karo: $\dot{V} = x_{1} (- x_{1} + x_{1} x_{2}^{2}) + x_{2} (- x_{2}) = - x_{1}^{2} + x_{1}^{2} x_{2}^{2} - x_{2}^{2} = - x_{1}^{2} (1 - x_{2}^{2}) - x_{2}^{2} .$
- Aise group kyun kiya? Yeh expose karne ke liye ki sign kahan change hota hai: factor $(1 - x_{2}^{2})$ .
Flip locate karo. $\dot{V} < 0$ ke liye chahiye $1 - x_{2}^{2} > 0 ⟺ ∣ x_{2} ∣ < 1$ . Lines $x_{2} = \pm 1$ par pehla term vanish ho jaata hai; $∣ x_{2} ∣ > 1$ ke liye woh positive turn ho sakta hai.
- Yeh step kyun? Negative-definite region hi hamaara guaranteed basin estimate hai — dekho Region of Attraction Estimation.
Conclusion. Asymptotically stable kam se kam $∣ x_{2} ∣ < 1$ ke liye. Cell C.

Figure neeche — $x_{1}$ – $x_{2}$ plane jisme $\dot{V}$ shaded hai: horizontal strip $∣ x_{2} ∣ < 1$ (chalk-blue) exactly woh jagah hai jahan $\dot{V} < 0$ , hamaara guaranteed basin; do yellow dashed lines $x_{2} = \pm 1$ woh boundary hain jahan pehla term vanish hota hai; unke baad (pink) guarantee void ho jaati hai aur $\dot{V}$ positive turn ho sakta hai. Do marked sample points $(1, 0.5)$ aur $(2, 2)$ dikhate hain ek case inside (negative) aur ek outside (positive).

Figure: Example 3 ke liye $\dot{V}$ ka sign map — negative (stable) sirf blue strip $∣ x_{2} ∣ < 1$ ke andar, yellow lines $x_{2} = \pm 1$ se bounded; pink region mein koi guarantee nahi.

Verify: $(1, 0.5)$ par: $\dot{V} = - 1 (1 - 0.25) - 0.25 = - 1.0 < 0$ ✓. $(1, 2)$ par: $\dot{V} = - 1 (1 - 4) - 4 = 3 - 4 = - 1$ ... yahan bhi negative, lekin $(2, 2)$ par: $\dot{V} = - 4 (1 - 4) - 4 = 12 - 4 = + 8 > 0$ — strip ke bahar bowl rise kar sakta hai, confirm karta hai ki boundary matter karti hai.

Example 4 — Cell D: degenerate "bad bowl"

Worked example Statement

Same system as Ex 2, lekin ek student $V = \frac{1}{2} x_{2}^{2}$ pick karta hai. Kya yeh stability prove karta hai?

Forecast: yeh $\geq 0$ hai aur origin par zero — kya yeh valid Lyapunov function hai?

PD check — aur yeh FAIL karta hai. $V = \frac{1}{2} x_{2}^{2} = 0$ poore $x_{1}$ -axis par ( $x_{2} = 0, x_{1}$ kuch bhi), sirf origin par nahi. Toh $V$ sirf positive semi-definite hai, PD nahi.
- Yeh step kyun? Ek Lyapunov function ko har direction mein real bowl hona chahiye; isme ek flat trough hai, isliye yeh $x_{1}$ ko trap nahi kar sakta.
Yeh kyun useless hai. Chahe $\dot{V} = - 2 x_{2}^{2} \leq 0$ ho, sublevel set ${V \leq c} = {∣ x_{2} ∣ \leq 2 c}$ ek infinite strip hai — $x_{1}$ infinity tak jaane ke liye free hai. Koi trapping nahi, koi stability certificate nahi. Cell D.

Verify: point $(1000, 0)$ par $V = 0 = V (0)$ hai lekin origin se bahut door hai. Ek genuine PD bowl (jaise Ex 2 ka $2 x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ ) deta hai $V (1000, 0) = 2 \times 1 0^{6} \neq = 0$ — woh $x_{1}$ mein distance dekhta hai. Lesson: PD ka matlab hai $V > 0$ har nonzero $x$ ke liye, sabhi components mein.

Example 5 — Cell E: Lyapunov equation se linear system

Worked example Statement

$\dot{x} = A x$ with $A = (0 - 2 1 - 3)$ . $Q = I$ pick karo aur $A^{⊤} P + P A = - Q$ ko $P$ ke liye solve karo; check karo ki $P$ PD hai.

Forecast: $A$ ke eigenvalues hain... ( $s^{2} + 3 s + 2$ ke roots). Agar dono ke negative real part hain, toh ek PD $P$ exist karna chahiye.

$A$ ke eigenvalues. Characteristic polynomial $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ → $s = - 1, - 2$ . Dono $< 0$ → theorem ek PD solution $P$ ka promise karta hai.
- Yeh step kyun? The Lyapunov equation ka PD solution tab aur sirf tab hota hai jab $A$ ke sabhi eigenvalues ka negative real part ho.
Unknown $P$ set up karo aur expand karo. Likho $P = (p_{11} p_{12} p_{12} p_{22})$ (symmetric, isliye dono off-diagonal entries ek unknown $p_{12}$ share karte hain). Compute karo $A^{⊤} P + P A = (- 4 p_{12} p_{11} - 3 p_{12} - 2 p_{22} p_{11} - 3 p_{12} - 2 p_{22} 2 p_{12} - 6 p_{22}) = (- 1 0 0 - 1) .$
- Yeh step kyun? Do matrices equal tabhi hote hain jab har entry match kare. Entry-by-entry match karna ek matrix equation ko scalar equations mein badal deta hai:
  - top-left entry $(1, 1)$ : $- 4 p_{12} = - 1$ ,
  - off-diagonal entry $(1, 2) = (2, 1)$ (symmetry se equal, aur $- Q$ mein wahan $0$ hai): $p_{11} - 3 p_{12} - 2 p_{22} = 0$ ,
  - bottom-right entry $(2, 2)$ : $2 p_{12} - 6 p_{22} = - 1$ .
Order mein solve karo. Top-left equation se $p_{12} = \frac{1}{4}$ . Bottom-right se, $6 p_{22} = 2 p_{12} + 1 = \frac{1}{2} + 1 = \frac{3}{2} \Rightarrow p_{22} = \frac{1}{4}$ . Off-diagonal se, $p_{11} = 3 p_{12} + 2 p_{22} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ . $P = (\frac{5}{4} \frac{1}{4} \frac{1}{4} \frac{1}{4}) .$
- Yeh step kyun? Har solved value ko agle equation mein substitute karo; ordering (pehle off-diagonal, phir diagonal) ka matlab hai ki koi simultaneous algebra nahi chahiye.
PD check via Sylvester. Leading minors: $\frac{5}{4} > 0$ ✓ aur $det P = \frac{5}{4} \cdot \frac{1}{4} - \frac{1}{4} \cdot \frac{1}{4} = \frac{5}{16} - \frac{1}{16} = \frac{4}{16} = \frac{1}{4} > 0$ ✓ → $P$ PD hai. Cell E. Asymptotic stability certified.

Verify: plug back karo — $A^{⊤} P + P A$ exactly $- I$ barabar hai; dono leading minors ( $\frac{5}{4}$ aur $det P = \frac{1}{4}$ ) positive hain, toh $P$ ek genuine bowl hai.

Example 6 — Cell F: limiting case (undamped, $b \to 0$ )

Worked example Statement

Ex 2 lo lekin damping $b = 0$ set karo: $\overset{x}{˙}_{1} = x_{2}, \overset{x}{˙}_{2} = - 4 x_{1}$ (pure oscillator). Same $V = 2 x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ . Damped aur undamped ke beech ki boundary par kya hota hai?

Forecast: bina friction ke, marble settle karega, ya hamesha circle karta rahega?

Orbital derivative. $\dot{V} = 4 x_{1} x_{2} + x_{2} (- 4 x_{1}) = 0$ identically.
- Yeh step kyun? Koi damper term survive nahi karta; energy exactly conserved hai.
Interpret karo. $\dot{V} \equiv 0 \leq 0$ → origin stable (marginally) hai — trajectories level sets $V = c$ (ellipses) par hamesha ke liye ride karti hain — lekin asymptotically stable nahi: yeh kabhi converge nahi karta. Cell F: asymptotically-stable ( $b > 0$ ) aur unstable regimes ke beech ki exact boundary.
- Yeh kyun matter karta hai: yeh dikhata hai ki Lyapunov sahi tarah distinguish karta hai "paas rehta hai" aur "ghar jaata hai." Ek conservative system razor's edge hai.

Verify: $(1, 0)$ par: $\dot{V} = 0$ . $(0, 2)$ par: $\dot{V} = 0$ . Value $V = 2 x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ kisi bhi solution ke along constant rahta hai — jaise $(1, 0)$ se, energy $V = 2$ hamesha $2$ rehti hai. Koi decay nahi.

Example 7 — Cell G: real-world word problem (spacecraft attitude)

Worked example Statement

Ek satellite ka small-angle attitude error yeh obey karta hai (scalar model, unit inertia) $\dot{θ} = ω, \overset{ω}{˙} = - k_{p} θ - k_{d} ω,$ jahan $θ$ pointing error hai, $ω$ angular rate hai, aur $k_{p}, k_{d} > 0$ controller ke proportional aur derivative gains hain. Prove karo ki controller error ko zero par drive karta hai. $V = \frac{1}{2} k_{p} θ^{2} + \frac{1}{2} ω^{2}$ use karo.

Forecast: kaun sa gain energy remove karta hai — $k_{p}$ ya $k_{d}$ ?

PD check. $k_{p} > 0$ → $V > 0$ for $(θ, ω) \neq = 0$ , $V (0) = 0$ . ✓ ( $V$ = "elastic pointing energy" + "rotational KE".)
- Yeh step kyun? Hume attitude-error state par ek valid bowl chahiye.
Orbital derivative. $\dot{V} = k_{p} θ ω + ω (- k_{p} θ - k_{d} ω) = k_{p} θ ω - k_{p} θ ω - k_{d} ω^{2} = - k_{d} ω^{2} \leq 0.$
- Yeh step kyun? Proportional gain $k_{p}$ cancel ho jaata hai (woh energy store/return karta hai, spring ki tarah); sirf derivative gain $k_{d}$ dissipate karta hai — yeh electronic damper hai.
Conclude karo. $\dot{V} \leq 0$ → stable; $\dot{V} = 0$ sirf $ω = 0$ par, aur us set par dynamics force karti hain $\overset{ω}{˙} = - k_{p} θ \neq = 0$ jab tak $θ = 0$ na ho → LaSalle se akela invariant piece origin hai → asymptotically stable: satellite target par settle karta hai. Cell G. (Yeh exactly quaternion PD attitude control hai miniature mein.)

Verify: $k_{p} = 4, k_{d} = 2$ ke saath yeh literally Ex 2 ki algebra hai: $\dot{V} = - 2 ω^{2}$ . $(θ, ω) = (0.1, 0.3)$ par: $\dot{V} = - 2 (0.09) = - 0.18 < 0$ . Error energy drop karti hai — pointing improve hoti hai.

Example 8 — Cell H: exam trap ( $\dot{V} > 0$ ka matlab unstable NAHI hai)

Worked example Statement

Ex 1 ka stable system ( $\overset{x}{˙}_{1} = - x_{1} + x_{2}, \overset{x}{˙}_{2} = - x_{1} - x_{2}$ , truly asymptotically stable). Ek student instead $W = \frac{1}{2} (x_{1}^{2} - x_{2}^{2})$ try karta hai aur paata hai ki kahin $\dot{W} > 0$ hai, phir declare karta hai "unstable!" Kya galat hua?

Forecast: kya ek positive $\dot{W}$ kabhi akele instability prove kar sakta hai?

Kya $W$ ek valid candidate bhi hai? $W = \frac{1}{2} (x_{1}^{2} - x_{2}^{2})$ PD nahi hai ( $W < 0$ jab bhi $∣ x_{2} ∣ > ∣ x_{1} ∣$ ). Toh yeh bowl hi nahi hai — iska $\dot{W}$ koi bhi stability meaning carry nahi karta.
- Yeh step kyun? Lyapunov ka theorem sirf PD $V$ ke baare mein baat karta hai. Ek non-PD trial function off-topic hai.
Ek PD $W$ bhi $\dot{W} > 0$ ke saath kuch prove nahi karta. Lyapunov ka theorem sufficient hai, necessary nahi. $\dot{V} \leq 0$ banana fail karna matlab hai tumhara guess bura tha, system unstable nahi hai.
- Yeh step kyun? Ex 1 ne already same system ko $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2})$ se asymptotically stable prove kar diya. Do alag trial functions, ek valid conclusion.
Instability ke liye sahi tool Chetaev's theorem hai, jisme ek region chahiye jahan $V > 0$ aur $\dot{V} > 0$ saath mein origin ko touch karein — sirf $\dot{W}$ ka ek positive point nahi. Cell H.

Verify: Ex 1 dynamics ke liye $\dot{W}$ compute karo: $\nabla W = (x_{1}, - x_{2})$ , toh $\dot{W} = x_{1} (- x_{1} + x_{2}) - x_{2} (- x_{1} - x_{2}) = - x_{1}^{2} + x_{1} x_{2} + x_{1} x_{2} + x_{2}^{2} = - x_{1}^{2} + 2 x_{1} x_{2} + x_{2}^{2}$ . $(0, 1)$ par: $\dot{W} = + 1 > 0$ — lekin Ex 1 ke valid $V$ ne same point par $\dot{V} = - 1 < 0$ diya tha. Same system, opposite-looking numbers ⇒ ek positive $\dot{W}$ meaningless hai. System stable hai, full stop.

Recall Quick self-test

Asymptotic stability tak pohonchne ke liye kaun si cell ko LaSalle chahiye? ::: Cell B (aur G) — jahan $\dot{V}$ sirf $\leq 0$ ho (semi-definite). Ex 3 ka guaranteed basin kaun sa region hai? ::: $∣ x_{2} ∣ < 1$ , jahan $\dot{V} = - x_{1}^{2} (1 - x_{2}^{2}) - x_{2}^{2} < 0$ . Ex 4 mein $V = \frac{1}{2} x_{2}^{2}$ fail kyun karta hai? ::: Yeh sirf positive semi-definite hai — poore $x_{1}$ -axis par zero, isliye $x_{1}$ ko trap nahi kar sakta. Kya $\dot{W} > 0$ (Ex 8) instability prove karta hai? ::: Nahi — theorem sufficient hai, necessary nahi; instability prove karne ke liye Chetaev use karo. Spacecraft (Ex 7) mein kaun sa gain energy dissipate karta hai? ::: Derivative gain $k_{d}$ ( $- k_{d} ω^{2}$ term ke through); proportional gain $k_{p}$ sirf ise store aur return karta hai, spring ki tarah.