Lyapunov stability — Lyapunov function, positive definiteness

Intuition The core idea (energy without energy)

Imagine a ball rolling in a bowl. No matter where you drop it, it eventually settles at the bottom. Why? Because its energy only ever goes down as it moves. Lyapunov's genius was to realize: ==you don't need to solve the equations of motion to know a system is stable — you just need to find any "energy-like" scalar that always decreases along the system's trajectories.==

WHY it matters (GNC): For a spacecraft attitude controller or a missile autopilot, solving the nonlinear ODEs is often impossible. Lyapunov lets us prove the vehicle returns to its commanded state — a certificate of stability — without integrating anything.
WHAT we need: a function $V(\mathbf{x})$ that behaves like a bowl (bottom at the equilibrium) and whose value drops as the system evolves.

1. Setting the stage: what are we analyzing?

We study an autonomous nonlinear system: $\dot{\mathbf{x}} = \mathbf{f}(\mathbf{x}), \qquad \mathbf{x}\in\mathbb{R}^n, \quad \mathbf{f}(\mathbf{0}) = \mathbf{0}$

The point $\mathbf{x}=\mathbf{0}$ is an equilibrium (if you start there with zero velocity, you stay). We shift coordinates so the equilibrium of interest sits at the origin — always allowed.

WHY the $\varepsilon$ – $\delta$ phrasing? "Stable" must mean small kick ⟹ small response, forever. The nested-ball definition captures exactly that without needing a solution formula.

2. Positive definiteness — the "bowl" property

Before we can pick an energy function, we need to know what makes something bowl-shaped.

HOW to test PD for a quadratic form $V = \mathbf{x}^\top P \mathbf{x}$ with $P=P^\top$ : $V$ is PD $\iff$ == $P$ is a positive-definite matrix== $\iff$ all eigenvalues of $P$ are $>0$ $\iff$ all leading principal minors are $>0$ (Sylvester's criterion).

Formula Why

$\mathbf{x}^\top P\mathbf{x}>0 \iff$ eigenvalues $>0$ (derivation) Since $P$ is symmetric, it is orthogonally diagonalizable: $P = Q\Lambda Q^\top$ with $Q^\top Q = I$ and $\Lambda=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ . Let $\mathbf{y} = Q^\top\mathbf{x}$ (a rotation, so $\mathbf{y}=\mathbf{0}\iff\mathbf{x}=\mathbf{0}$ ). Then $\mathbf{x}^\top P\mathbf{x} = \mathbf{x}^\top Q\Lambda Q^\top\mathbf{x} = \mathbf{y}^\top\Lambda\mathbf{y} = \sum_{i=1}^n \lambda_i y_i^2.$ This sum is $>0$ for every $\mathbf{y}\neq\mathbf{0}$ iff every $\lambda_i>0$ . Done — no hand-waving. ∎

3. The Lyapunov function and the main theorem

The crucial object is the orbital derivative — how $V$ changes as you flow along the actual dynamics. By the chain rule: $\dot V(\mathbf{x}) = \frac{d}{dt}V(\mathbf{x}(t)) = \sum_{i}\frac{\partial V}{\partial x_i}\dot x_i = \nabla V(\mathbf{x})^\top \mathbf{f}(\mathbf{x}).$

WHY the chain rule and not $\partial V/\partial t$ ? $V$ has no explicit time dependence; all time-variation comes through $\mathbf{x}(t)$ , whose derivative the dynamics $\mathbf f$ supply. This is the whole trick: $\dot V$ is computable without solving for $\mathbf x(t)$ .

4. Worked examples

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

System: $\ddot q + b\dot q + kq = 0$ , $b,k>0$ . State $\mathbf x=(x_1,x_2)=(q,\dot q)$ : $\dot x_1 = x_2,\qquad \dot x_2 = -kx_1 - b x_2.$ Choose the mechanical energy $V = \tfrac12 k x_1^2 + \tfrac12 x_2^2$ .

Why this step? It is PD (both coefficients $>0$ , $V(\mathbf 0)=0$ ) — a genuine bowl — and physical energy is the natural guess.

Compute the orbital derivative: $\dot V = kx_1\dot x_1 + x_2\dot x_2 = kx_1x_2 + x_2(-kx_1 - bx_2) = -b\,x_2^2 \le 0.$

Why this step? The $kx_1x_2$ terms cancel — energy fed by the spring exactly returns; only the damper $-bx_2^2$ removes energy.

$\dot V\le0$ ⟹ at least stable. It's semi-definite ( $\dot V=0$ whenever $x_2=0$ ), so plain Lyapunov gives only stability; a stronger tool (LaSalle's invariance principle) upgrades it to asymptotic. Physics matches: damping brings the pendulum to rest.

Worked example Example 2 — A purely nonlinear system

$\dot x_1 = -x_1 + x_1 x_2^2,\qquad \dot x_2 = -x_2.$ Choose $V = \tfrac12(x_1^2 + x_2^2)$ (PD).

Why this step? No physical energy available, so we try the simplest "distance-squared" bowl. $\dot V = x_1(-x_1+x_1x_2^2) + x_2(-x_2) = -x_1^2 + x_1^2x_2^2 - x_2^2 = -x_1^2(1-x_2^2) - x_2^2.$
Why this step? We grouped to expose the sign. For $|x_2|<1$ , the term $-x_1^2(1-x_2^2)<0$ , so $\dot V<0$ : locally asymptotically stable in the region $|x_2|<1$ .

Lesson: the region of validity comes straight from where $\dot V$ stays negative — Lyapunov gives us a guaranteed basin estimate.

Worked example Example 3 — Linear system, matrix Lyapunov equation

For $\dot{\mathbf x}=A\mathbf x$ , try $V=\mathbf x^\top P\mathbf x$ with $P$ PD. $\dot V = \dot{\mathbf x}^\top P\mathbf x + \mathbf x^\top P\dot{\mathbf x} = \mathbf x^\top(A^\top P + PA)\mathbf x.$

Why this step? Product rule; substitute $\dot{\mathbf x}=A\mathbf x$ . Set $A^\top P + PA = -Q$ . If we can solve this Lyapunov equation for a PD $P$ given a chosen PD $Q$ , then $\dot V=-\mathbf x^\top Q\mathbf x<0$ ⟹ asymptotic stability. Theorem: such a $P$ exists $\iff$ all eigenvalues of $A$ have negative real part. This is how GNC software auto-certifies linear autopilots.

5. Common mistakes (steel-manned)

Recall Feynman: explain to a 12-year-old

Think of a marble in a mixing bowl. Wherever you drop it, it rolls to the bottom and stops. Now imagine you had a magic number-machine: you point it at the marble and it tells you the marble's "height" in the bowl. If that number keeps getting smaller and smaller as the marble moves, you know the marble is heading for the bottom — even if you never watched it roll! Lyapunov's idea is: build such a number-machine for a spacecraft or robot. If the number always shrinks, the machine is safely returning to where it should be. You never had to predict the whole journey — you just watched the number go down.

Connections

LaSalle's Invariance Principle — upgrades $\dot V\le0$ to asymptotic stability.
Lyapunov Equation A^TP+PA=-Q — the linear-system machinery from Example 3.
Positive Definite Matrices & Sylvester's Criterion — the PD test.
Backstepping Control & Sliding Mode Control — designed by construction to make $\dot V<0$ .
Spacecraft Attitude Control (Quaternion Feedback) — real GNC use of Lyapunov certificates.
Chetaev's Instability Theorem — the correct converse tool.
Region of Attraction Estimation — from sublevel sets $\{V\le c\}$ .

Flashcards

What is a positive definite function

V(\mathbf x)

V(\mathbf 0)=0

and

V(\mathbf x)>0

for all

\mathbf x\neq\mathbf 0

in the region.

How do you compute

\dot V

along trajectories without solving the ODE?

\dot V = \nabla V(\mathbf x)^\top \mathbf f(\mathbf x)

(chain rule; substitute the dynamics

\dot{\mathbf x}=\mathbf f

Lyapunov condition for stability vs asymptotic stability?

Stability:

V

PD and

\dot V\le0

. Asymptotic:

V

PD and

\dot V<0

(strictly, for

\mathbf x\neq\mathbf 0

For

V=\mathbf x^\top P\mathbf x

(

P

symmetric), when is it PD?

When all eigenvalues of

P

are positive

\iff

all leading principal minors positive (Sylvester).

Why is

\mathbf x^\top P\mathbf x>0 \iff

all

\lambda_i(P)>0

Diagonalize

P=Q\Lambda Q^\top

; with

\mathbf y=Q^\top\mathbf x

, the form becomes

\sum\lambda_i y_i^2

, positive iff each

\lambda_i>0

If you cannot find a valid Lyapunov function, is the system unstable?

No — the theorem is sufficient, not necessary. Failing to find

V

proves nothing; use Chetaev's theorem to prove instability.

Derive

\dot V

for

\dot{\mathbf x}=A\mathbf x

V=\mathbf x^\top P\mathbf x

\dot V=\mathbf x^\top(A^\top P+PA)\mathbf x

; set

=-\mathbf x^\top Q\mathbf x

(Lyapunov equation).

For the damped oscillator, why does

\dot V=-bx_2^2

Spring cross-terms

kx_1x_2

cancel; only damper dissipates ⟹ energy loss

\propto

velocity

^2

Difference between positive definite and positive semi-definite?

PD:

V>0

for all nonzero

\mathbf x

. PSD:

V\ge0

(can be zero at nonzero points — "flat channels").

What geometric object does a valid Lyapunov function represent?

A "bowl" (positive walls, bottom at equilibrium) whose height decreases along trajectories.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek ball ko bowl (katori) mein daala — chahe kahin se bhi chhodo, wo ghoomte ghoomte bottom pe aakar ruk jaata hai. Kyun? Kyunki uski energy har time kam hoti jaati hai. Lyapunov ka genius idea yahi hai: system ki equations solve kiye bina, agar hum koi bhi "energy jaisa" scalar function $V(\mathbf x)$ dhoondh lein jo trajectory ke saath ghatta jaata hai, to hum prove kar sakte hain ki system stable hai. GNC mein — jaise spacecraft attitude control ya missile autopilot — nonlinear equations solve karna almost impossible hota hai, isliye Lyapunov method ek "stability certificate" de deta hai bina kuch integrate kiye.

Do cheezein zaroori hain. Pehli: $V$ positive definite ho — matlab origin (equilibrium) pe zero, aur baaki har jagah strictly positive. Yeh hi "bowl" shape deta hai. Test simple hai: agar $V=\mathbf x^\top P\mathbf x$ hai to $P$ ke saare eigenvalues positive hone chahiye. Doosri cheez: $V$ ka time derivative trajectory ke saath — jise hum orbital derivative kehte hain — nikaalo chain rule se: $\dot V=\nabla V^\top \mathbf f$ . Yahan magic hai — $\mathbf f$ dynamics hai, to $\dot V$ bina solve kiye mil jaata hai.

Rule yaad rakho: agar $V$ PD hai aur $\dot V\le 0$ , to system stable hai (bounded rahega). Agar $\dot V<0$ strictly hai, to asymptotically stable — ghar (origin) wapas pahunch jaayega. Damped spring-mass example mein $\dot V=-bx_2^2$ nikalta hai — spring ke cross terms cancel ho jaate hain, sirf damper energy nikaalta hai, bilkul physics ke hisaab se.

Ek badi galti se bacho: agar tumhe koi valid $V$ nahi milta, iska matlab system unstable hai — galat! Theorem sirf sufficient hai, necessary nahi. Shaayad tumne galat $V$ choose kiya. Aur "PD bowl, downhill roll" mantra yaad rakhna — bowl positive definite, aur ball hamesha neeche jaaye to stable. Bas itna samajh lo to Lyapunov ka poora concept clear hai.

Lyapunov stability — Lyapunov function, positive definiteness

1. Setting the stage: what are we analyzing?

2. Positive definiteness — the "bowl" property

3. The Lyapunov function and the main theorem

4. Worked examples

5. Common mistakes (steel-manned)

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Guidance, Navigation & Control (GNC)

Connections