Visual walkthrough — Linearization of nonlinear systems

We assume you know only: what a curve is, what "slope" means, and that arrows can show which way things move. Everything else we earn.

Step 1 — What a "flow" even is

WHAT. We have two rules that tell a moving dot how fast to go, depending on where it sits: $\dot{x}=f(x,y),\dot{y}=g(x,y).$

Let us decode every symbol before using it.

• $x,y$ — just the position of a dot on a flat sheet, like coordinates on graph paper.
• The little dot on top, $\dot{x}$, is shorthand for "==how fast $x$ is changing right now==" — its velocity in the $x$-direction. Read $\dot{x}$ as "x-dot".
• $f(x,y)$ — a machine: feed it a position, it hands back that $x$-velocity. $g$ does the same for $y$.

WHY these two rules. Because at every point of the sheet they hand you a little arrow $(\dot{x},\dot{y})$ — the direction and speed the dot will drift. Cover the whole sheet with these arrows and you get a vector field: a wind map. The dot just rides the wind.

PICTURE. Look at the arrows below. Each one is $(f(x,y),g(x,y))$ planted at its point. A trajectory (amber) is the path a dot traces if it always follows the local arrow.

Step 2 — The one place with no wind

WHAT. We hunt for a point where both velocities vanish: $f(x^{\ast },y^{\ast })=0\text{and}g(x^{\ast },y^{\ast })=0.$

• The star ${}^{\ast }$ marks a special, fixed location — not a variable. Think "the address of the calm spot."
• Both equations must hold at the same point. One alone is not enough.

WHY. If both velocities are zero, the arrow there has zero length — no wind. A dot placed exactly there never moves. This is an equilibrium (also called fixed or critical point). Stability is a question about behaviour near such a point, so this is where all the action lives (compare Stability of Equilibria).

PICTURE. The cyan dot below is where every surrounding arrow shrinks to nothing. Notice arrows nearby still point somewhere — the calm is exactly at the centre.

Step 3 — Rename so the calm spot sits at the origin

WHAT. Measure position relative to the equilibrium: $u=x-x^{\ast },v=y-y^{\ast }.$

• $u$ = "how far right of the calm spot am I?" $v$ = "how far above it?"
• At the equilibrium itself $u=v=0$. We slid the origin to the calm point.

WHY. Two reasons. First, a small disturbance is exactly a small $(u,v)$ — so "near equilibrium" simply means "$u,v$ tiny." Second, since $x^{\ast },y^{\ast }$ are constants, they do not change with time, so $\dot{u}=\dot{x}=f(x,y),\dot{v}=\dot{y}=g(x,y).$ The velocity of the displacement equals the velocity of the dot. Nothing lost, cleaner bookkeeping.

PICTURE. Same wind map, new grid centred on the calm point. The tiny amber displacement arrow is $(u,v)$.

Step 4 — Zoom in until the curve is a straight ramp

WHAT. Picture $f$ as a height surface: over each point $(x,y)$ the surface sits at height $f(x,y)$. Because $f(x^{\ast },y^{\ast })=0$, this surface touches zero height right above the calm spot. Now zoom in.

WHY this tool — the tangent plane. A curvy surface, magnified enough around one point, looks flat — like a ramp. That flat ramp is the best linear approximation of the surface there. Replacing surface by ramp is the entire trick of linearization: the ramp we can do algebra with; the curve we cannot. This "zoom until flat" is precisely what Taylor Series formalises.

PICTURE. Left: the true curved surface of $f$. Right: zoomed in, it is indistinguishable from a tilted flat plane through the calm point.

Step 5 — Name the two slopes (partial derivatives)

WHAT. The plane's tilt in each direction is measured by a partial derivative: $f_x=\frac{\partial f}{\partial x}{|}_{(x^{\ast },y^{\ast })},f_y=\frac{\partial f}{\partial y}{|}_{(x^{\ast },y^{\ast })}.$

• $f_x$ — "walk one step in the $x$-direction only (freeze $y$); how much does $f$ rise?" That is the steepness of the ramp along $x$.
• $f_y$ — same, but stepping in the $y$-direction with $x$ frozen.
• The bar ${|}_{(x^{\ast },y^{\ast })}$ says: measure these slopes at the calm point, so they are plain numbers, not functions.

WHY partial and not ordinary derivative. An ordinary slope $\frac{df}{dx}$ assumes one input. We have two. A partial derivative answers "steepness if I move only along one axis" — exactly the two tilts a plane needs.

PICTURE. Two cross-sections of the surface through the calm point: slicing along $x$ shows the slope $f_x$; slicing along $y$ shows $f_y$.

Step 6 — Assemble the approximation (Taylor to first order)

WHAT. The ramp's height near the calm point is base height + tilt·distance: $f(x,y)\approx \underset{=0}{\underbrace{f(x^{\ast },y^{\ast })}}+\underset{\text{tilt in }x}{\underbrace{f_x}}\underset{\text{step in }x}{\underbrace{u}}+\underset{\text{tilt in }y}{\underbrace{f_y}}\underset{\text{step in }y}{\underbrace{v}}+\underset{\text{curvature: tiny}}{\underbrace{\frac{1}{2}{f}_{xx}{u}^2+\dots }}$

Term by term:

• $f(x^{\ast },y^{\ast })=0$ — the base height vanishes because we sit on the calm point. Gone.
• $f_{x}u$ — how much height the $x$-tilt adds over a step $u$. Rise = slope × run.
• $f_{y}v$ — same for the $y$-tilt over step $v$.
• $\frac{1}{2}{f}_{xx}{u}^2+\dots$ — the curved correction.

WHY drop the curved part. If $u$ is a small fraction, say $0.01$, then $u^2=0.0001$ — a hundred times smaller. Near enough to the calm point, the squared (curvature) terms are dwarfed by the straight (linear) terms. Keeping only what dominates is the honest approximation.

PICTURE. True curve versus its straight-ramp approximation. The vertical amber gap is the discarded curvature — see how it collapses toward zero as you approach the centre.

Step 7 — Stack both ramps into one matrix, the Jacobian

WHAT. Do Step 6 for $g$ as well, and write the two straight-line rules together: $\dot{u}\approx f_{x}u+f_{y}v,\dot{v}\approx g_{x}u+g_{y}v.$ Two rows of "slope × displacement." Pack the four slopes into a grid: $\left(\begin{array}{c}\dot{u}\\ \dot{v}\end{array}\right)=\underset{J=\text{Jacobian}}{\underbrace{\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right){|}_{(x^{\ast },y^{\ast })}}}\left(\begin{array}{c}u\\ v\end{array}\right).$

Reading the matrix product row by row is literally the two equations above:

• Top row $(f_x,f_y)$ dotted with $(u,v)$ gives $\dot{u}$.

• Bottom row $(g_x,g_y)$ dotted with $(u,v)$ gives $\dot{v}$.

• $J$ — the Jacobian: the four tilts, evaluated at the calm point, so all entries are numbers.

• Because the entries are constant, this is a linear system — the kind we can crack completely with Eigenvalues and Eigenvectors.

WHY this is the finish line. The messy curved wind map, near the calm point, is now just "multiply the displacement by a fixed matrix." The whole nonlinear problem shrank to matrix multiplication.

PICTURE. The Jacobian as a machine: displacement arrow $(u,v)$ goes in, matrix $J$ acts, velocity arrow $(\dot{u},\dot{v})$ comes out — bending and stretching the little arrow.

Step 8 — Edge case: what if the linear part vanishes?

WHAT. Sometimes the two tilts conspire so the eigenvalues of $J$ are pure imaginary (a center) or zero. Then $\mathrm{Re}(\lambda )=0$: no clear growth, no clear decay.

• $\lambda$ — an eigenvalue of $J$; its real part $\mathrm{Re}(\lambda )$ says whether a mode grows ($>0$), decays ($<0$), or neither ($=0$).

WHY it's special. In Step 6 we discarded curvature because the linear terms dominated. But if the linear terms give exactly balanced circling ($\mathrm{Re}(\lambda )=0$), the ramp cannot decide the outcome — the tiny curvature we threw away is now the tie-breaker. The linearization can honestly lie (a "center" may secretly spiral).

PICTURE. Same starting arrows: the linear model draws a perfect closed loop, but the true nonlinear flow slowly spirals inward — the discarded curvature at work.

The one-picture summary

Everything above, compressed: a curved surface → tangent plane → four slopes → one matrix that turns displacements into velocities.

Active recall