Chaotic systems — sensitivity to initial conditions, Lyapunov exponents: Visual walkthrough

We only assume you can read a graph and that you have met the idea of a trajectory (a path a system traces over time) from Phase space and flows.

Step 1 — Two dots that start almost together

WHAT. Picture a rule that tells a dot where to go next. Drop two dots, almost on top of each other. Call the first dot's position at time $t$ the value $x (t)$ , and the second dot's position $x^{'} (t)$ . The little arrow-gap between them is what we watch.

WHY. Chaos is never about one trajectory — a single path can't be "sensitive" to anything. Sensitivity is a statement about two neighbours. So the very first object we build is the gap, not the path.

PICTURE. Look at Step Figure 1. The teal dot is $x (t)$ , the plum dot is $x^{'} (t)$ . At $t = 0$ they sit a hair apart; the burnt-orange segment between them is the gap.

Figure — Chaotic systems — sensitivity to initial conditions, Lyapunov exponents

The symbol $δ$ (Greek "delta") here means "a small difference." $∣ \cdot ∣$ means "length of" — always a positive number, because a distance can't be negative.

Step 2 — Ask how the gap changes, not where it is

WHAT. We don't care about the dots' absolute positions. We care whether the gap $∣ δ ∣$ is growing or shrinking as time ticks. So we ask about the rate of change of $δ$ .

WHY. "Rate of change" is exactly what a derivative answers — and it is the right tool because our whole question is "how fast?", not "how much?". Writing $\dot{δ}$ (a dot over a symbol means "its rate of change per unit time") turns the vague phrase "the gap grows" into something we can compute.

PICTURE. Step Figure 2 shows the same two dots one instant later. Both moved along their own arrows (their velocities), and the gap arrow got longer. The change in the gap is the difference of the two velocities.

Each dot obeys the same rule $\overset{x}{˙} = f (x)$ — read this as "the velocity of a dot is a fixed function $f$ of where it currently is." So:

Step 3 — Zoom in until the curve becomes a straight line

WHAT. The two dots are extremely close, so $δ$ is tiny. Over a tiny gap, the rule $f$ can't do anything fancy — it looks like a straight ramp. We replace the curve of $f$ near $x$ by its tangent line.

WHY. This is the linear approximation (the first term of a Taylor expansion). We use it because a tiny separation only ever feels the local slope of $f$ — not its bends far away. The slope of $f$ at $x$ is written $f^{'} (x)$ ; the prime $^{'}$ means "derivative of $f$ ", i.e. how steeply $f$ rises as $x$ moves right.

PICTURE. Step Figure 3 zooms into the graph of $f$ . Over the tiny window $[x, x + δ]$ the curve (plum) and its tangent (burnt orange) are indistinguishable. The height difference across the window is slope × width.

Step 4 — A constant stretch rate forces an exponential

WHAT. Pretend for a moment the slope is a fixed number, $f^{'} (x) = λ$ . Then the gap obeys $\dot{δ} = λ δ$ : "the gap's growth rate is proportional to the gap." The only function whose rate of change is a constant multiple of itself is the exponential $e^{λ t}$ .

WHY. We reach for $e^{λ t}$ because it is the unique answer to "what grows in proportion to its own size?" — not because it looks fancy. Doubling money at fixed interest, bacteria dividing, and separating trajectories are all the same equation.

PICTURE. Step Figure 4 plots the gap length versus time for three slopes: $λ > 0$ (plum, explodes), $λ = 0$ (teal, flat), $λ < 0$ (burnt orange, decays).

Step 5 — Extract the rate: take a log, divide by time

WHAT. We have $∣ δ (t) ∣ = ∣ δ_{0} ∣ e^{λ t}$ but we want the number $λ$ hiding inside it. Divide by $∣ δ_{0} ∣$ , take the natural log to peel off the exponential, then divide by $t$ .

WHY. The logarithm $ln$ is the exact tool that undoes $e$ : $ln (e^{λ t}) = λ t$ . That's the whole reason it appears — it is the question "what exponent produced this?". Dividing by $t$ then converts the total stretch into a per-unit-time rate.

PICTURE. Step Figure 5 shows the same three curves on a log vertical axis. Now they are straight lines, and $λ$ is literally the slope of each line — the picture of "average stretch per unit time."

$\frac{∣ δ ( t ) ∣}{∣ δ _{0} ∣} = e^{λ t} \Rightarrow ln \frac{∣ δ ( t ) ∣}{∣ δ _{0} ∣} = λ t \Rightarrow λ = \frac{1}{t} ln \frac{∣ δ ( t ) ∣}{∣ δ _{0} ∣}$

$\frac{∣ δ ( t ) ∣}{∣ δ _{0} ∣}$ — how many times bigger the gap grew (a pure ratio, no units).
$ln (\dots)$ — turns that "×-factor" into a "+-amount" (the exponent that made it).
$\frac{1}{t}$ — spreads that exponent evenly across the elapsed time → a rate.

Step 6 — Why two limits, and in this exact order

WHAT. Real slopes are not constant along a wiggly orbit, and $δ$ won't stay infinitesimal forever. So the final definition wraps the boxed result in two limits: $λ = lim_{t \to \infty} lim_{δ_{0} \to 0} \frac{1}{t} ln \frac{∣ δ ( t ) ∣}{∣ δ _{0} ∣} .$

WHY.

$δ_{0} \to 0 lim$ first — keeps the Step-3 straight-line picture exact. If the gap ever grew big, our tangent-line approximation would break and the number would be junk.
$t \to \infty lim$ second — averages the ever-changing local slope over the whole orbit, so $λ$ describes the system, not one lucky starting point.

PICTURE. Step Figure 6: the local slope $f^{'} (x)$ varies wildly as the dot rides the orbit (jagged burnt-orange line); the long-time average of all that jitter is the single flat teal level — that level is $λ$ .

Step 7 — The edge cases the picture must also cover

WHAT & PICTURE. Step Figure 7 collects the degenerate behaviours as four mini-panels, each a gap-vs-time sketch.

Case	What the gap does	Sign of $λ$	Example
Stable fixed point	dot & neighbour both fall to the same rest point	$λ < 0$	damped pendulum at rest
Periodic / quasi-periodic	gap oscillates, never grows	$λ = 0$	frictionless pendulum
Chaotic but bounded	gap grows then folds back	$λ_{1} > 0$ , $\sum λ_{i} < 0$	Driven damped pendulum
Unbounded blow-up	gap grows forever, no folding	$λ > 0$ but not chaos	$\overset{x}{˙} = x$

The one-picture summary

Step Figure 8 compresses all seven steps into one frame: two dots split apart (Step 1–2), the local slope stretches the gap (Step 3), the exponential grows it (Step 4), the log-slope reads off $λ$ (Step 5), the orbit-averaging flattens the jitter (Step 6), and the fold keeps it bounded (Step 7).

Recall Feynman retelling — the whole walkthrough in plain words

Put two toy cars almost on the same spot of a bumpy track. First, forget the cars — watch the gap between them (Step 1). Ask only one thing: is the gap growing? That's a "how fast" question, so we use a rate — a derivative (Step 2). Because the cars are so close, the track under the gap looks like a straight ramp, and the ramp's steepness is all that matters (Step 3). "Grows in proportion to its own size" is precisely the exponential rule, so the gap swells like $e^{λ t}$ (Step 4). To read the growth rate $λ$ , undo the exponential with a log and divide by time — on log-paper $λ$ is just the slope of a straight line (Step 5). But the ramp's steepness keeps changing as the cars ride the track, and the straight-line trick only works while the gap is tiny — so we take the gap tiny first, then average over a long, long ride (Step 6). Finally: a positive $λ$ means the gap keeps doubling, but if the track loops back on itself the cars stay on the board even while their gap explodes — stretch locally, fold globally. That, and only that, is chaos (Step 7).

Recall Quick self-checks

Why is $δ$ (not $x$ ) the star of the show? ::: Chaos is a statement about neighbours separating; a single trajectory can't be "sensitive," only the gap between two can. Why does the exponential appear? ::: $\dot{δ} = λ δ$ says the gap grows in proportion to itself, and $e^{λ t}$ is the unique function with that property. Why take $ln$ then divide by $t$ ? ::: $ln$ undoes the exponential to expose the exponent $λ t$ ; dividing by $t$ turns it into a per-unit-time rate. Why $δ_{0} \to 0$ before $t \to \infty$ ? ::: Tiny-gap-first keeps the straight-line (linear) approximation exact; long-time-second averages $λ$ over the whole attractor. Does $λ > 0$ alone prove chaos? ::: No — $\overset{x}{˙} = x$ has $λ > 0$ but is unbounded. Chaos also needs boundedness and folding.

Related builds: Fixed points and linear stability analysis (the $λ < 0$ case), The logistic map and period-doubling route to chaos (discrete version of this derivation), Hamiltonian chaos and the KAM theorem (the $λ = 0$ regular islands).