Foundations — Linearization of nonlinear systems

This page assumes nothing. If the parent note Linearization used a symbol, we build it here from the ground up, in the order that lets each idea lean on the one before it.

1. The dot, the plane, and a position

Everything starts with a single point moving on a flat sheet of paper.

The whole map — every possible $(x,y)$ — is called the phase plane (built properly in Phase Plane Analysis). One dot = one complete snapshot of the system.

2. The dot in time and the dot-notation $\dot{x}$

Now let the dot move. Its address changes as time $t$ ticks forward.

Why an overdot instead of $dx/dt$? Pure laziness that pays off: our equations will be littered with rates, and $\dot{x}$ keeps them readable. The dot always means "rate of change with respect to time", nothing else.

3. Functions of position: $f(x,y)$ and $g(x,y)$

Where does the velocity arrow come from? A rule.

So at every point of the map, $f$ and $g$ jointly paint an arrow. The map covered in arrows is a vector field — the rulebook of the system.

Nonlinear simply means these machines contain products or curves — terms like $x^2$, $xy$, or $\sin x$ — so the arrows bend. If $f,g$ were just $ax+by$ (straight sums), the system would be linear and already solvable.

4. The resting spot: equilibrium $(x^{\ast },y^{\ast })$

Some special addresses have a zero arrow.

5. Small displacements $u$ and $v$

We never look far from a resting spot; we look just beside it.

We will keep $u,v$ tiny. That tininess is the engine of the whole method (next section).

6. Slopes: partial derivatives $f_x,f_y$

To zoom in on the curved arrow-field, we need to measure how steeply each push changes as we nudge $x$ or $y$.

Why partial (not ordinary) derivatives? Because $f$ depends on two inputs. We must isolate the effect of one nudge at a time — freeze the other, ask "which way does the ground tilt?", then repeat. The full tilt is captured by the two answers together.

7. Taylor's approximation: straightening a curve

Here is the single move that powers linearization: replace a curve by its tangent line near a point.

Why is this allowed, and why stop here? Because near the equilibrium the steps $u,v$ are tiny. The dropped terms carry $u^2,uv,v^2$ — a tiny number squared is vastly smaller still ($0.01^2=0.0001$). So the straight-line part dominates; the curvature is negligible. This is the built-from-scratch version of "a smooth curve looks straight if you zoom in enough" — the full machinery lives in Taylor Series.

At an equilibrium the first term is zero (that's the definition of ${}^{\ast }$!), so the approximation collapses to a pure straight-line rule in $u,v$: $\dot{u}\approx f_{x}u+f_{y}v.$

8. The Jacobian matrix $J$

Do section 7 for both $f$ and $g$, and stack the four slopes into a grid.

A matrix is just a machine that turns one arrow into another. $J$ is that machine for our zoomed-in world.

9. Eigen-things: $\lambda$ and $\mathbf{v}$

The linear machine $J$ has favourite directions in which it acts by pure stretching.

Full construction lives in Eigenvalues and Eigenvectors. Why do we care here? Because along an eigen-direction the future is dead simple: the displacement follows $e^{\lambda t}$.

The symbols $\tau =f_x+g_y$ (trace, the sum of the diagonal) and $\Delta =f_x{g}_y-f_y{g}_x$ (determinant) are just quick handles for finding $\lambda$ via $\lambda =\frac{\tau \pm \sqrt{{\tau }^2-4\Delta }}{2}$.

How the foundations feed the topic

Read it top to bottom: each box is a symbol this page defined, and each arrow is "you needed the box above to understand the box below." Miss one and the topic stalls.

Equipment checklist

Test yourself — cover the right side and answer aloud.

Once every answer comes easily, you're ready for the full Linearization derivation.