Lyapunov stability — Lyapunov function, positive definiteness: Visual walkthrough

We will prove one thing, slowly:

If there is a bowl-shaped number $V$ sitting over your state space, and that number never climbs as the system moves, then the system cannot escape — it is stable.

Step 1 — What is "the state" and what does a trajectory look like?

WHAT. Our system is a rule that says, at every point, which way to move next. Write the system's condition as a list of numbers $x = (x_{1}, x_{2})$ — for a spinning spacecraft, think "angle error" and "angular rate error." The rule is $\dot{x} = f (x) .$

Read it term by term:

$x$ — a point on the floor plan (the state).
$\dot{x}$ — the little arrow attached to that point telling you where you drift next (the "dot" means rate of change in time).
$f$ — the machine that turns a point into its arrow. Feed it a point, out comes an arrow.

WHY. We want to reason about where the system goes without ever computing the path with a formula. So we picture the arrows first: they form a field, and a moving point just follows the arrows like a leaf in a stream.

PICTURE. Every dot on the grid has one cyan arrow. Follow the arrows from any start and you trace a trajectory (amber) that curls toward the origin.

Figure — Lyapunov stability — Lyapunov function, positive definiteness

Step 2 — Build the bowl: a positive-definite number over the floor

WHAT. Hang a height over every floor point: a number $V (x)$ . We demand it be shaped like a bowl: $V (0) = 0, V (x) > 0 for x \neq = 0 .$ Term by term:

$V (0) = 0$ — the bottom of the bowl sits exactly over the equilibrium.
$V (x) > 0$ — everywhere else the ground is higher. No side channels, no flat ledges.

Such a $V$ is called positive definite, which we'll abbreviate PD from here on.

WHY. We need a single number whose smallness forces the state to be near home. A bowl does exactly that: "low height" is only possible near the bottom. If the bowl had a flat trench (only $V \geq 0$ , i.e. semi-definite), a point could slide far away while $V$ stayed $0$ — useless. That is why the strict $> 0$ matters.

PICTURE. The paraboloid $V = \frac{1}{2} (x_{1}^{2} + x_{2}^{2})$ over the floor. Its horizontal slices are circles — the level sets ${V = c}$ . Notice: a small height $c$ pens you inside a small circle.

Recall Why not just measure distance?

Distance ::: You could — distance-squared is a valid PD bowl. But a bowl matched to the system's own physics (like mechanical energy) usually makes the next step's arithmetic cancel beautifully. The freedom to choose the bowl is Lyapunov's whole power.

Step 3 — The key question: is the moving point climbing or sinking?

WHAT. As the state slides along a trajectory, its height $V$ changes over time. We want that rate, $\dot{V}$ . Since $V$ depends on position and position depends on time, use the chain rule: $\dot{V} (x) = steepest-uphill arrow \nabla V (x) \cdot motion arrow f (x) .$ Term by term:

$\nabla V$ — the gradient, an arrow pointing straight uphill on the bowl, longest where the bowl is steepest.
$f$ — the motion arrow from Step 1, which way the point actually goes.
The centred dot " $\cdot$ " is the dot product: multiply the two arrows' matching components and add them up. It answers "how much does one arrow point along the other?" (You will also see this written $\nabla V^{⊤} f$ ; the small " $⊤$ " is just "flip the column of numbers on its side so we can multiply-and-add" — it means the same dot product.)

WHY the dot product, and why this tool? A dot product answers exactly one question: how much does arrow B point in the direction of arrow A? Here it asks "is the motion heading uphill or downhill on the bowl?" That is precisely the sign we need, and — the miracle — it uses only $f$ , the machine we already have. We never solve for the path.

If $f$ points against the uphill gradient (obtuse angle) $\Rightarrow \dot{V} < 0$ $\Rightarrow$ sinking.
If $f$ points along it $\Rightarrow \dot{V} > 0$ $\Rightarrow$ climbing.
If $f$ is perpendicular to $\nabla V$ $\Rightarrow \dot{V} = 0$ $\Rightarrow$ gliding along a level circle.

PICTURE. At one sample point: the amber gradient $\nabla V$ pointing outward-uphill, the cyan motion $f$ , and the angle between them. The shaded projection is $\dot{V}$ .

Step 4 — The decisive condition: $\dot{V} \leq 0$ everywhere nearby

WHAT. Fix a region $D$ — just a patch of the floor plan around the origin (a neighbourhood of the equilibrium, small enough that our chosen bowl and our sign check are valid there). Suppose that inside $D$ the motion arrow never points uphill on the bowl: $\dot{V} (x) = \nabla V \cdot f \leq 0 for every x in D .$ Here $D$ is exactly the domain where we have checked the sign — everything below is a claim about points inside $D$ .

WHY. This single inequality says the height can only stay level or drop, never rise. Geometrically: at every fence-post around any level circle (that fits inside $D$ ), the motion arrow points inward or tangent — never outward. A point starting inside that circle can never cross out of it.

PICTURE. A level circle (dashed cyan) with motion arrows sampled around its rim. Every arrow leans inward/tangent (amber = inward component). The trajectory is trapped inside.

Step 5 — Why "trapped" means stable (the $ε$ – $δ$ picture)

WHAT. First, one piece of notation we need: $∥ x ∥$ means the straight-line distance from $x$ to the origin — for $x = (x_{1}, x_{2})$ it is $x_{1}^{2} + x_{2}^{2}$ , the length of the arrow from the centre to the point. So " $∥ x ∥ < ε$ " just reads "the point is closer than $ε$ to home."

Now "stable" has a precise meaning: small kick in ⟹ small response, forever. For any target circle of radius $ε$ we must find a start circle of radius $δ$ so that starting inside $δ$ keeps you inside $ε$ for all time.

WHY the bowl delivers this. Because $V$ is a PD bowl, pick the height $c$ of the highest level-set that still fits inside the target $ε$ (and inside $D$ ). The set ${V \leq c}$ is a closed region around the origin. From Step 4, once inside ${V \leq c}$ you stay inside it (height can't rise past $c$ ). Then choose $δ$ as a small ball that fits inside ${V \leq c}$ . Chain it together: $∥ x (0) ∥ < δ \Rightarrow x (0) \in {V \leq c} \Rightarrow x (t) \in {V \leq c} \forall t \Rightarrow ∥ x (t) ∥ < ε .$

PICTURE. Three nested rings: outer target $ε$ , the trapping level-set ${V \leq c}$ between them, inner start $δ$ . An amber trajectory launched from $δ$ rattles around but never pierces $ε$ .

That is a complete proof of stability — no ODE was ever solved.

Step 6 — Sharpening to asymptotic stability ( $\dot{V} < 0$ )

WHAT. If the inequality is strict away from the origin, $\dot{V} (x) < 0 for all x \neq = 0 in D,$ the height doesn't just fail to rise — it strictly drops at every instant the point isn't home.

WHY. A quantity that is bounded below by $0$ (heights of a bowl can't go negative) and that keeps strictly decreasing must settle onto some floor value — this is the monotone convergence principle: a number that only ever goes down but can't fall through $0$ has nowhere to go but toward a limit. That limiting height must have $\dot{V} = 0$ (otherwise it would still be dropping), and strictness says the only place with $\dot{V} = 0$ is the origin. So the state doesn't merely stay near home — it goes home: $x (t) \to 0$ .

PICTURE. The height-vs-time curve $V (x (t))$ : a monotonically falling amber line pressing down onto $0$ , with the matching trajectory spiralling into the origin.

Step 7 — The degenerate case: a flat trench ( $\dot{V} = 0$ on a whole set)

WHAT. What if $\dot{V} \leq 0$ but $\dot{V} = 0$ on more than just the origin — say a whole line? This is exactly the parent damped-pendulum case: $V = \frac{1}{2} k x_{1}^{2} + \frac{1}{2} x_{2}^{2}$ gives $\dot{V} = - b x_{2}^{2} \leq 0,$ which is zero all along the axis $x_{2} = 0$ (any position, zero velocity).

WHY this is a real gap. On that line the height stops dropping. Strict-decrease reasoning (Step 6) breaks — you cannot conclude convergence from Lyapunov alone. But physically the pendulum does come to rest. The escape hatch: on the trench $x_{2} = 0$ the dynamics still push $\overset{x}{˙}_{2} = - k x_{1} \neq = 0$ , so the point can't linger on the trench — it gets kicked off and keeps sinking. Formalizing "can't stay on the flat set" is LaSalle's Invariance Principle.

PICTURE. The bowl with a flat amber trench along the $x_{1}$ -axis. A trajectory grazes the trench, but the vertical arrows ( $\overset{x}{˙}_{2} \neq = 0$ ) shove it back off — it can rest nowhere but the origin.

Step 8 — The other degenerate warning: a bad bowl proves nothing

WHAT. Suppose you pick a $V$ and get $\dot{V} > 0$ somewhere. Tempting conclusion: "unstable!" Wrong.

WHY. Lyapunov's theorem is a one-way ticket: a good $V$ (PD bowl, non-rising) certifies stability, but a failed $V$ certifies nothing. A different, cleverer bowl might succeed. Only Chetaev's Instability Theorem can prove instability.

PICTURE. Two candidate bowls over the same stable trajectory: bowl A's arrows show $\dot{V} > 0$ (misleading), bowl B's show $\dot{V} < 0$ (correct certificate). Same system, opposite readings — proof that a failed guess is not a verdict.

The one-picture summary

Everything at once: the arrow field (Step 1), the bowl with a level ring (Step 2), the gradient-vs-motion angle giving $\dot{V}$ 's sign (Step 3), inward arrows trapping the ring (Step 4), the nested $δ$ – $ε$ rings (Step 5), and the amber trajectory spiralling to the bottom while its height falls to zero (Step 6).

Recall Feynman: the whole walkthrough in plain words

Picture a marble tipped into a bowl. I don't need to compute its path — I just watch its height. First I make sure the bowl is a real bowl: lowest at the target spot, higher everywhere else (that's positive definite, PD). Then, at any spot, I compare the way the marble is pushed against the way that would take it uphill — that comparison is the dot product $\nabla V \cdot f$ , and its sign is $\dot{V}$ . If the push is never uphill ( $\dot{V} \leq 0$ ) inside my patch $D$ , the marble's height can't grow, so it's fenced inside a small ring: stable. If the push is always downhill ( $\dot{V} < 0$ ), the height keeps shrinking and — being unable to fall below $0$ — must settle to zero, so the marble rolls all the way home: asymptotically stable. If there's a flat ledge where height momentarily stops falling, I check whether the marble can actually rest there — usually the dynamics kick it off (LaSalle's Invariance Principle). And if my chosen bowl gives a confusing uphill reading, that just means I picked a bad bowl — it proves nothing. That is Lyapunov's direct method: stability judged by watching one falling number, never by solving the motion.