Lyapunov stability — Lyapunov function, positive definiteness: Worked examples

Intuition What this page is for

You already met the idea (a bowl-shaped "energy" $V$ that only ever drops). Here we grind through every kind of case the topic can hand you: signs that work, signs that fail, degenerate bowls, boundaries where $\dot{V}$ flips, a real spacecraft, and an exam trap. By the end there is no scenario you have not seen worked start-to-finish.

Reminder of the two objects we keep using (from the parent topic):

$V (x)$ — a positive-definite scalar (written PD from here on: $V (0) = 0$ and $V (x) > 0$ for every $x \neq = 0$ ). The "bowl."
$\dot{V} (x) = \nabla V^{⊤} f (x)$ — the orbital derivative: how the bowl-height changes as the system actually moves. $\nabla V$ is the vector of partial slopes $(\frac{\partial V}{\partial x _{1}}, \dots)$ ; $f$ is the right-hand side of $\dot{x} = f (x)$ ; the $^{⊤}$ dot means "add up slope $\times$ velocity in each direction."

Throughout this page PD is shorthand for positive-definite (the two-part bowl test just above).

The scenario matrix

Every Lyapunov exercise falls into one of these cells. The examples below are labelled with the cell(s) they hit — together they cover the whole grid.

Cell	Case class	What makes it distinct	Example
A	$\dot{V} < 0$ everywhere (negative definite)	full asymptotic stability, clean	Ex 1
B	$\dot{V} \leq 0$ only (negative semi-definite)	stability only; needs LaSalle	Ex 2
C	Sign flips on a boundary	region-of-attraction estimate	Ex 3
D	Degenerate $V$ (only semi-definite)	the "bad bowl" that traps nothing	Ex 4
E	Linear / matrix case ( $V = x^{⊤} P x$ )	Sylvester + Lyapunov equation	Ex 5
F	Limiting / zero input (e.g. $b \to 0$ , no damping)	boundary between stable & unstable	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

$\overset{x}{˙}_{1} = - x_{1} + x_{2}, \overset{x}{˙}_{2} = - x_{1} - x_{2} .$ Try $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2})$ . Is the origin asymptotically stable?

Forecast: guess now — will $\dot{V}$ be $< 0$ everywhere, or only somewhere? The cross-terms look symmetric...

Check $V$ is PD (positive-definite). $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2}) > 0$ for $x \neq = 0$ , and $V (0) = 0$ . ✓
- Why this step? No orbital derivative is meaningful unless the bowl is real. Always confirm PD first.
Gradient. $\nabla V = (x_{1}, x_{2})$ .
- Why this step? $\dot{V}$ needs the slopes of $V$ in each direction.
Dot with the dynamics. $\dot{V} = x_{1} (- x_{1} + x_{2}) + x_{2} (- x_{1} - x_{2}) = - x_{1}^{2} + x_{1} x_{2} - x_{1} x_{2} - x_{2}^{2} = - x_{1}^{2} - x_{2}^{2} .$
- Why this step? $\dot{V} = \nabla V^{⊤} f$ ; the $+ x_{1} x_{2}$ and $- x_{1} x_{2}$ cancel exactly, leaving pure negatives.
Read the sign. $\dot{V} = - (x_{1}^{2} + x_{2}^{2}) < 0$ for every $x \neq = 0$ → negative definite → asymptotically stable everywhere. Cell A.

Verify: at $x = (1, 0)$ , $\dot{V} = - 1 < 0$ ; at $(1, 1)$ , $\dot{V} = - 2 < 0$ ; at $(0, 0)$ , $\dot{V} = 0$ . Height strictly drops except at the origin — the marble always rolls home.

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

Worked example Statement (damped oscillator, revisited in full)

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

Forecast: will $\dot{V}$ be strictly negative, or zero on some line?

PD check. Coefficients $2 > 0$ and $\frac{1}{2} > 0$ ; $V (0) = 0$ . ✓
- Why? Mechanical energy is the natural bowl guess for a spring–damper.
$\nabla V = (4 x_{1}, x_{2})$ .
Orbital derivative. $\dot{V} = 4 x_{1} x_{2} + x_{2} (- 4 x_{1} - 2 x_{2}) = 4 x_{1} x_{2} - 4 x_{1} x_{2} - 2 x_{2}^{2} = - 2 x_{2}^{2} .$
- Why this step? Spring terms cancel; only the damper $- b x_{2}^{2} = - 2 x_{2}^{2}$ removes energy.
Read the sign. $\dot{V} = - 2 x_{2}^{2} \leq 0$ , and $\dot{V} = 0$ on the whole line $x_{2} = 0$ , not just the origin → negative semi-definite → plain Lyapunov gives stability only. Cell B. (LaSalle's Invariance Principle upgrades it to asymptotic since the only trajectory stuck on $x_{2} = 0$ is the origin.)

Verify: at $(1, 0)$ , $\dot{V} = 0$ (marble momentarily coasts along the frictionless line); at $(1, 1)$ , $\dot{V} = - 2 < 0$ . Matches physics: damping only acts when there is velocity.

Example 3 — Cell C: sign flips on a boundary (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: everywhere stable, or only inside some strip?

PD check. Distance-squared bowl, PD. ✓
$\nabla V = (x_{1}, x_{2})$ , dot with $f$ : $\dot{V} = x_{1} (- x_{1} + x_{1} x_{2}^{2}) + x_{2} (- x_{2}) = - x_{1}^{2} + x_{1}^{2} x_{2}^{2} - x_{2}^{2} = - x_{1}^{2} (1 - x_{2}^{2}) - x_{2}^{2} .$
- Why group like this? To expose where the sign changes: the factor $(1 - x_{2}^{2})$ .
Locate the flip. $\dot{V} < 0$ requires $1 - x_{2}^{2} > 0 ⟺ ∣ x_{2} ∣ < 1$ . On the lines $x_{2} = \pm 1$ the first term vanishes; for $∣ x_{2} ∣ > 1$ it can turn positive.
- Why this step? The negative-definite region is our guaranteed basin estimate — see Region of Attraction Estimation.
Conclusion. Asymptotically stable at least for $∣ x_{2} ∣ < 1$ . Cell C.

Figure below — the $x_{1}$ – $x_{2}$ plane with $\dot{V}$ shaded: the horizontal strip $∣ x_{2} ∣ < 1$ (chalk-blue) is exactly where $\dot{V} < 0$ , our guaranteed basin; the two yellow dashed lines $x_{2} = \pm 1$ are the boundary where the first term vanishes; beyond them (pink) the guarantee is void and $\dot{V}$ can turn positive. The two marked sample points $(1, 0.5)$ and $(2, 2)$ show one case inside (negative) and one outside (positive).

Figure: sign map of $\dot{V}$ for Example 3 — negative (stable) only inside the blue strip $∣ x_{2} ∣ < 1$ , bounded by the yellow lines $x_{2} = \pm 1$ ; pink region has no guarantee.

Verify: at $(1, 0.5)$ : $\dot{V} = - 1 (1 - 0.25) - 0.25 = - 1.0 < 0$ ✓. At $(1, 2)$ : $\dot{V} = - 1 (1 - 4) - 4 = 3 - 4 = - 1$ ... still negative here, but at $(2, 2)$ : $\dot{V} = - 4 (1 - 4) - 4 = 12 - 4 = + 8 > 0$ — outside the strip the bowl can rise, confirming the boundary matters.

Example 4 — Cell D: the degenerate "bad bowl"

Worked example Statement

Same system as Ex 2, but a student picks $V = \frac{1}{2} x_{2}^{2}$ . Does it prove stability?

Forecast: it's $\geq 0$ and zero at the origin — is it a valid Lyapunov function?

PD check — and it FAILS. $V = \frac{1}{2} x_{2}^{2} = 0$ along the entire $x_{1}$ -axis ( $x_{2} = 0, x_{1}$ anything), not only at the origin. So $V$ is only positive semi-definite, not PD.
- Why this step? A Lyapunov function must be a real bowl in every direction; this one has a flat trough, so it cannot trap $x_{1}$ .
Why it's useless. Even if $\dot{V} = - 2 x_{2}^{2} \leq 0$ , the sublevel set ${V \leq c} = {∣ x_{2} ∣ \leq 2 c}$ is an infinite strip — $x_{1}$ is free to run to infinity. No trapping, no stability certificate. Cell D.

Verify: the point $(1000, 0)$ has $V = 0 = V (0)$ yet is far from the origin. A genuine PD bowl (like Ex 2's $2 x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ ) gives $V (1000, 0) = 2 \times 1 0^{6} \neq = 0$ — it does see distance in $x_{1}$ . Lesson: PD means $V > 0$ for every nonzero $x$ , all components.

Example 5 — Cell E: linear system via the Lyapunov equation

Worked example Statement

$\dot{x} = A x$ with $A = (0 - 2 1 - 3)$ . Pick $Q = I$ and solve $A^{⊤} P + P A = - Q$ for $P$ ; check $P$ is PD.

Forecast: eigenvalues of $A$ are... (roots of $s^{2} + 3 s + 2$ ). If both have negative real part, a PD $P$ must exist.

Eigenvalues of $A$ . Characteristic polynomial $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ → $s = - 1, - 2$ . Both $< 0$ → theorem promises a PD solution $P$ .
- Why this step? The Lyapunov equation has a PD solution iff all eigenvalues of $A$ have negative real part.
Set up the unknown $P$ and expand. Write $P = (p_{11} p_{12} p_{12} p_{22})$ (symmetric, so the two off-diagonal entries share one unknown $p_{12}$ ). Compute $A^{⊤} P + P A = (- 4 p_{12} p_{11} - 3 p_{12} - 2 p_{22} p_{11} - 3 p_{12} - 2 p_{22} 2 p_{12} - 6 p_{22}) = (- 1 0 0 - 1) .$
- Why this step? Two matrices are equal only if every entry matches. Matching entry-by-entry turns one matrix equation into scalar equations:
  - top-left entry $(1, 1)$ : $- 4 p_{12} = - 1$ ,
  - off-diagonal entry $(1, 2) = (2, 1)$ (equal by symmetry, and $- Q$ has $0$ there): $p_{11} - 3 p_{12} - 2 p_{22} = 0$ ,
  - bottom-right entry $(2, 2)$ : $2 p_{12} - 6 p_{22} = - 1$ .
Solve in order. From the top-left equation $p_{12} = \frac{1}{4}$ . From the bottom-right, $6 p_{22} = 2 p_{12} + 1 = \frac{1}{2} + 1 = \frac{3}{2} \Rightarrow p_{22} = \frac{1}{4}$ . From the off-diagonal, $p_{11} = 3 p_{12} + 2 p_{22} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ . $P = (\frac{5}{4} \frac{1}{4} \frac{1}{4} \frac{1}{4}) .$
- Why this step? Substitute each solved value into the next equation; the ordering (off-diagonal first, then diagonal) means no simultaneous algebra is needed.
PD check via Sylvester. Leading minors: $\frac{5}{4} > 0$ ✓ and $det P = \frac{5}{4} \cdot \frac{1}{4} - \frac{1}{4} \cdot \frac{1}{4} = \frac{5}{16} - \frac{1}{16} = \frac{4}{16} = \frac{1}{4} > 0$ ✓ → $P$ is PD. Cell E. Asymptotic stability certified.

Verify: plug back — $A^{⊤} P + P A$ equals $- I$ exactly; both leading minors ( $\frac{5}{4}$ and $det P = \frac{1}{4}$ ) are positive, so $P$ is a genuine bowl.

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

Worked example Statement

Take Ex 2 but set the damping $b = 0$ : $\overset{x}{˙}_{1} = x_{2}, \overset{x}{˙}_{2} = - 4 x_{1}$ (pure oscillator). Same $V = 2 x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ . What happens at the boundary between damped and undamped?

Forecast: with no friction, will the marble settle, or circle forever?

Orbital derivative. $\dot{V} = 4 x_{1} x_{2} + x_{2} (- 4 x_{1}) = 0$ identically.
- Why this step? No damper term survives; energy is exactly conserved.
Interpret. $\dot{V} \equiv 0 \leq 0$ → the origin is stable (marginally) — trajectories ride level sets $V = c$ (ellipses) forever — but not asymptotically stable: it never converges. Cell F: the exact boundary between the asymptotically-stable ( $b > 0$ ) and unstable regimes.
- Why this matters: it shows Lyapunov correctly distinguishes "stays near" from "goes home." A conservative system is the razor's edge.

Verify: at $(1, 0)$ : $\dot{V} = 0$ . At $(0, 2)$ : $\dot{V} = 0$ . The value $V = 2 x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ is constant along any solution — e.g. from $(1, 0)$ , energy $V = 2$ stays $2$ forever. No decay.

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

Worked example Statement

A satellite's small-angle attitude error obeys (scalar model, unit inertia) $\dot{θ} = ω, \overset{ω}{˙} = - k_{p} θ - k_{d} ω,$ where $θ$ is pointing error, $ω$ the angular rate, and $k_{p}, k_{d} > 0$ are the controller's proportional and derivative gains. Prove the controller drives the error to zero. Use $V = \frac{1}{2} k_{p} θ^{2} + \frac{1}{2} ω^{2}$ .

Forecast: which gain removes energy — $k_{p}$ or $k_{d}$ ?

PD check. $k_{p} > 0$ → $V > 0$ for $(θ, ω) \neq = 0$ , $V (0) = 0$ . ✓ ( $V$ = "elastic pointing energy" + "rotational KE".)
- Why this step? We need a valid bowl over the attitude-error state.
Orbital derivative. $\dot{V} = k_{p} θ ω + ω (- k_{p} θ - k_{d} ω) = k_{p} θ ω - k_{p} θ ω - k_{d} ω^{2} = - k_{d} ω^{2} \leq 0.$
- Why this step? The proportional gain $k_{p}$ cancels (it stores/returns energy, like a spring); only the derivative gain $k_{d}$ dissipates — it is the electronic damper.
Conclude. $\dot{V} \leq 0$ → stable; $\dot{V} = 0$ only on $ω = 0$ , and on that set the dynamics force $\overset{ω}{˙} = - k_{p} θ \neq = 0$ unless $θ = 0$ → by LaSalle the only invariant piece is the origin → asymptotically stable: the satellite settles on target. Cell G. (This is exactly quaternion PD attitude control in miniature.)

Verify: with $k_{p} = 4, k_{d} = 2$ this is literally Ex 2's algebra: $\dot{V} = - 2 ω^{2}$ . At $(θ, ω) = (0.1, 0.3)$ : $\dot{V} = - 2 (0.09) = - 0.18 < 0$ . Error energy drops — the pointing improves.

Example 8 — Cell H: the exam trap ( $\dot{V} > 0$ does NOT mean unstable)

Worked example Statement

The stable system of Ex 1 ( $\overset{x}{˙}_{1} = - x_{1} + x_{2}, \overset{x}{˙}_{2} = - x_{1} - x_{2}$ , truly asymptotically stable). A student instead tries $W = \frac{1}{2} (x_{1}^{2} - x_{2}^{2})$ and finds $\dot{W} > 0$ somewhere, then declares "unstable!" What went wrong?

Forecast: can a positive $\dot{W}$ ever prove instability by itself?

Is $W$ even a valid candidate? $W = \frac{1}{2} (x_{1}^{2} - x_{2}^{2})$ is not PD ( $W < 0$ whenever $∣ x_{2} ∣ > ∣ x_{1} ∣$ ). So it isn't a bowl at all — its $\dot{W}$ carries no stability meaning whatsoever.
- Why this step? Lyapunov's theorem only speaks about PD $V$ . A non-PD trial function is off-topic.
Even a PD $W$ with $\dot{W} > 0$ proves nothing. Lyapunov's theorem is sufficient, not necessary. Failing to make $\dot{V} \leq 0$ means your guess was bad, not that the system is unstable.
- Why this step? Ex 1 already proved the same system asymptotically stable with $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2})$ . Two different trial functions, one valid conclusion.
The right tool for instability is Chetaev's theorem, which needs a region where $V > 0$ and $\dot{V} > 0$ together touching the origin — not just one point of positive $\dot{W}$ . Cell H.

Verify: compute $\dot{W}$ for Ex 1 dynamics: $\nabla W = (x_{1}, - x_{2})$ , so $\dot{W} = x_{1} (- x_{1} + x_{2}) - x_{2} (- x_{1} - x_{2}) = - x_{1}^{2} + x_{1} x_{2} + x_{1} x_{2} + x_{2}^{2} = - x_{1}^{2} + 2 x_{1} x_{2} + x_{2}^{2}$ . At $(0, 1)$ : $\dot{W} = + 1 > 0$ — yet Ex 1's valid $V$ gave $\dot{V} = - 1 < 0$ at the same point. Same system, opposite-looking numbers ⇒ a positive $\dot{W}$ is meaningless. The system is stable, full stop.

Recall Quick self-test

Which cell needs LaSalle to reach asymptotic stability? ::: Cell B (and G) — where $\dot{V}$ is only $\leq 0$ (semi-definite). Ex 3's guaranteed basin is which region? ::: $∣ x_{2} ∣ < 1$ , where $\dot{V} = - x_{1}^{2} (1 - x_{2}^{2}) - x_{2}^{2} < 0$ . Why does $V = \frac{1}{2} x_{2}^{2}$ fail in Ex 4? ::: It is only positive semi-definite — zero along the whole $x_{1}$ -axis, so it can't trap $x_{1}$ . Does $\dot{W} > 0$ (Ex 8) prove instability? ::: No — the theorem is sufficient, not necessary; use Chetaev to prove instability. In the spacecraft (Ex 7), which gain dissipates energy? ::: The derivative gain $k_{d}$ (via the term $- k_{d} ω^{2}$ ); the proportional gain $k_{p}$ only stores and returns it, like a spring.