Lyapunov stability — Lyapunov function, positive definiteness: Question bank

Definition Notation used throughout this bank (built from zero)

$x \in R^{n}$ is the state — the list of numbers describing the system right now (e.g. angle and angular rate of a spacecraft).
The system evolves by a rule $\dot{x} = f (x)$ . Here $\dot{x}$ is the velocity of the state (how fast each number changes), and $f$ is the dynamics map: a machine that, given the current state $x$ , returns the direction and speed the state moves. Picture $f$ as an arrow attached to every point of the plane — the state slides along these arrows (see the vector field below).
$f (0) = 0$ means the origin is an equilibrium: no arrow there, so if you start at rest at the origin you stay.
$V (x)$ is a scalar "height" (the bowl), $\nabla V$ its gradient (steepest-uphill vector), and $\dot{V} = \nabla V^{⊤} f$ its rate of change as the state flows along the arrows of $f$ .
PD = positive definite: $V (0) = 0$ and $V (x) > 0$ for every $x \neq = 0$ .

The picture below anchors these three objects — the bowl $V$ , the arrows $f$ , and the invariant set where $\dot{V} = 0$ — before you attempt the traps.

Figure — Lyapunov stability — Lyapunov function, positive definiteness

True or false — justify

\dot{V} \leq 0

everywhere, the origin is asymptotically stable.

False.

\dot{V} \leq 0

(negative semi-definite) only guarantees the height never rises, i.e. plain stability/boundedness; you need

\dot{V} < 0

for

x \neq = 0

, or LaSalle's Invariance Principle, to get convergence to the origin.

A positive-definite

V

must satisfy

V (0) = 0

True. By definition the bottom of the bowl sits at the equilibrium; if

V (0) > 0

it is not PD (though shifting by a constant can sometimes repair it).

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

is a valid Lyapunov candidate for a 2D system.

False. It is only positive semi-definite:

V = 0

along the entire

x_{1}

-axis, not just the origin, so it cannot trap the

x_{1}

component and fails the PD requirement.

If no Lyapunov function can be found, the system is unstable.

False. Lyapunov's theorem is sufficient, not necessary; failing to find a working

V

means only that you haven't been clever enough. Use Chetaev's Instability Theorem to actually prove instability.

For

V = x^{⊤} P x

with symmetric

P

, "all eigenvalues of

P > 0

" and "all leading principal minors of

P > 0

" are equivalent tests.

True. Both are equivalent characterizations of a positive-definite matrix; the minor test is Sylvester's criterion and is usually easier to check by hand.

The Lyapunov equation

A^{⊤} P + P A = - Q

always has a PD solution

P

for any PD

Q

False. A unique PD

P

exists iff every eigenvalue of

A

has strictly negative real part; if

A

is unstable, no PD

P

solves it. See Lyapunov Equation A^TP+PA=-Q.

A quadratic form

x^{⊤} P x

with a non-symmetric

P

has no well-defined definiteness.

False. Only the symmetric part

\frac{1}{2} (P + P^{⊤})

affects the value of

x^{⊤} P x

, so definiteness is decided by that symmetric part; we conventionally take

P

symmetric.

\dot{V} > 0

for one chosen

V

, the system must be unstable.

False. A bad candidate proves nothing — it may just be the wrong bowl. Only a

V

meeting Chetaev's hypotheses (positive

V

and positive

\dot{V}

in a suitable cone) certifies instability.

Positive semi-definite

V

can still yield asymptotic stability when combined with LaSalle.

True. If

\dot{V} \leq 0

and the only trajectory staying in

{\dot{V} = 0}

is the origin, LaSalle's Invariance Principle upgrades stability to asymptotic stability — exactly how the damped pendulum is handled.

Why questions

Why must

V

be continuously differentiable?

Because the whole method rests on the chain-rule identity

\dot{V} = \nabla V^{⊤} f

; without a well-defined gradient the orbital derivative — the object the theorem tests — does not exist.

Why does the spring term cancel in

\dot{V}

for the damped pendulum?

The spring is conservative: energy it stores it returns, so its contribution

k x_{1} x_{2}

is exactly annihilated by the coupling term, leaving only the dissipative damper

- b x_{2}^{2}

, which is where energy truly leaves.

Why can we always place the equilibrium of interest at the origin?

Stability is a local property about a fixed point; a coordinate shift

z = x - x_{e q}

moves any equilibrium to

0

without changing the dynamics' qualitative behaviour, simplifying every "bowl at the bottom" statement.

Why do we test

\dot{V}

along trajectories rather than in all directions?

Stability concerns how the system actually flows, not arbitrary displacements; only the direction

f (x)

given by the dynamics map is realized, so

\dot{V} = \nabla V^{⊤} f

measures the true rate of change of height that the system experiences.

Why is a diagonalization argument enough to prove

x^{⊤} P x > 0 ⟺

eigenvalues

> 0

Orthogonal diagonalization

P = Q Λ Q^{⊤}

turns the form into

\sum λ_{i} y_{i}^{2}

with

y = Q^{⊤} x

a pure rotation, so positivity of the whole sum for all

y \neq = 0

is exactly positivity of each

λ_{i}

Why does Lyapunov's method avoid solving the ODE?

Because

\dot{V} = \nabla V^{⊤} f

uses the dynamics map

f

directly — we never integrate to find

x (t)

; we only check the sign of a scalar expression, which is what makes the method usable on unsolvable nonlinear GNC problems.

Why is picking physical energy a good first guess for

V

Real systems dissipate energy, so mechanical energy is automatically a PD bowl centred at rest and tends to decrease under damping — it frequently satisfies both conditions with almost no engineering, as in Sliding Mode Control and Backstepping Control designs.

Edge cases

What if

\dot{V} = 0

on an entire surface, not just the origin?

Plain Lyapunov gives only stability; you must check whether any trajectory can stay on that surface. If the only invariant subset is the origin, LaSalle's Invariance Principle recovers asymptotic stability.

What if

f (0) \neq = 0

Then the origin is not an equilibrium and the theorem does not apply there; you must first find an actual equilibrium and shift coordinates so that

f (0) = 0

What if

V

is PD locally but

V \to

finite as

∥ x ∥ \to \infty

Local asymptotic stability may hold, but you cannot conclude global stability: global results also require radial unboundedness (

V \to \infty

∥ x ∥ \to \infty

) so that sublevel sets stay bounded.

What is the definiteness of the zero matrix

P = 0

V = x^{⊤} P x

V \equiv 0

, which is only positive semi-definite (indeed identically zero) — it is not PD, so it is a useless Lyapunov candidate that traps nothing.

What happens at the boundary

∣ x_{2} ∣ = 1

for the nonlinear system with

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

There

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

becomes only semi-definite (zero on the whole line

x_{2} = 0

); the strict negativity that gave asymptotic stability is lost exactly at the edge, marking the frontier of the guaranteed basin.

What if

A

has an eigenvalue with zero real part (marginal case)?

The Lyapunov equation has no PD solution and linearization is inconclusive; the linear method cannot certify stability, and you must analyze the nonlinear terms directly (centre-manifold or a custom

V

Can a system be stable but not asymptotically stable?

Yes — an undamped oscillator (

b = 0

in the pendulum) has

\dot{V} = 0

: trajectories circle forever on constant-energy orbits, staying bounded (stable) but never converging to the origin.

Does a PD

V

with

\dot{V} < 0

guarantee the state stays inside its starting sublevel set?

Yes — since height strictly drops, the trajectory can never re-enter a higher level set, so it remains inside

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

; this trapping property underlies Region of Attraction Estimation.

Recall One-line summary of every trap

A positive definite (PD) $V$ means == $V > 0$ for every nonzero $x$ == with $V (0) = 0$ ; stability needs == $\dot{V} \leq 0$ , asymptotic stability needs $\dot{V} < 0$ or LaSalle==; a failed search for $V$ or a candidate whose == $\dot{V} > 0$ proves nothing== (use Chetaev to prove instability); always dot the gradient with the dynamics map $f$ ; and every "global" claim hides a radial-unboundedness assumption.

Question bank — Lyapunov stability — Lyapunov function, positive definiteness

True or false — justify

Spot the error

Why questions

Edge cases