Linearization and the Jacobian

The wall we just hit

By now you can read a linear planar system fluently. Given x' = A x you find the eigenvalues of A, and from their signs and whether they are real or complex you name the equilibrium at the origin a node, a saddle, a spiral, or a center — and you can sketch the whole phase portrait without solving a thing. The trace-determinant plane turned that classification into a single glance at trace(A) and det(A). That power came entirely from linearity: A x is a straight, predictable flow.

Real systems are almost never that kind. A pendulum has a sin(theta) in it; a predator-prey model has a product x*y; an oscillating circuit has a cubic term. These are nonlinear systems, written generally as x' = f(x, y), y' = g(x, y) with f and g curved functions. The honest truth that haunts this whole subject returns here in full force: such systems almost never have a closed-form solution. We cannot write down x(t). So how do we say anything at all about how they behave?

Stand at the still point and squint

The escape is the same trick single-variable calculus taught you on day one: zoom in. Near a point, a smooth curve looks like its tangent line. Near an equilibrium, a smooth vector field looks like a linear one. So we find the resting points — the equilibria where f = 0 and g = 0 at once, the places where the flow stands perfectly still — and we study the system in a tiny neighbourhood of each. The plan of linearization is to throw away everything curved and keep only the straight, leading behaviour right there.

Make this precise with a multivariable Taylor expansion. Let (x*, y*) be an equilibrium and write the small displacement as u = x - x*, v = y - y*. Expand f and g to first order around the equilibrium. The constant terms f(x*, y*) and g(x*, y*) are zero — that is exactly what 'equilibrium' means — so the leading surviving terms are the linear ones in u and v. The displacement (u, v) then obeys, to leading order, a linear system whose matrix is built entirely from the four first partial derivatives of f and g.

Meet the Jacobian

That matrix of first partials is the Jacobian matrix, the centerpiece of this guide. For a planar system it is the 2-by-2 array of how each rate responds to each variable: the top row holds the partials of f, the bottom row the partials of g. Evaluate it at the equilibrium and you get a concrete matrix J of numbers. The Jacobian is simply the multivariable derivative of the vector field — the best linear approximation to the curved flow at that point, exactly as a single derivative f'(a) is the best linear approximation to a curve at a.

          [ df/dx   df/dy ]
   J  =   [               ]      evaluated at (x*, y*)
          [ dg/dx   dg/dy ]

   near the equilibrium:   u' = J u ,   u = (x - x*, y - y*)

From the Jacobian to a verdict

Now the whole previous rung pays off at once. Once you hold J, the linearized motion u' = J u is exactly the kind of linear system you already classify in your sleep. Run the eigenvalue method on J, or simply read off trace(J) and det(J) and locate the point on the trace-determinant plane. Negative-real-part eigenvalues mean displacements shrink and the equilibrium is attracting; a positive real part means some displacement grows and it is repelling.

1. Find the equilibria: solve f(x, y) = 0 and g(x, y) = 0 simultaneously for every rest point (a nonlinear system may have several).
2. Write the Jacobian once, as the matrix of partials df/dx, df/dy, dg/dx, dg/dy.
3. Evaluate J at each equilibrium in turn to get a concrete numerical matrix there.
4. Classify that J by its eigenvalues (or by trace and determinant): node, saddle, spiral, or center.
5. Carry the local linear picture back to the nonlinear system as its behaviour near that equilibrium — with the caveat below.

A pendulum, two ways

Take the undamped nonlinear pendulum, theta'' + sin(theta) = 0, written as a planar system with x = theta and y = theta': then x' = y and y' = -sin(x). Equilibria need y = 0 and sin(x) = 0, so x = 0 (hanging straight down) and x = pi (balanced straight up). The Jacobian has top row (0, 1) and bottom row (-cos(x), 0). At the bottom rest point cos(0) = 1, giving eigenvalues +/- i — a center, the pendulum endlessly swinging. At the top, cos(pi) = -1, giving real eigenvalues +/- 1 of opposite sign — a saddle, the knife-edge instability of balancing it upright. One field, two utterly different local stories, each read straight off its J.

What linearization can and cannot see

Keep the powers and limits clearly apart. The linearization is purely local: it speaks only about a small neighbourhood of one equilibrium and says nothing about behaviour far away, where the discarded curved terms dominate. So it can tell you a rest point is a stable spiral, yet a trajectory that starts far off may never reach it — global features like which initial conditions end up where, or whether closed loops called limit cycles encircle the equilibrium, lie entirely beyond its reach.

Two more honest caveats round it out. First, linearization needs a smooth field — the partial derivatives must exist for the Jacobian to be defined at all. Second, it classifies an isolated equilibrium; if J is singular (det(J) = 0, a zero eigenvalue) the rest point is degenerate and the linear theory alone cannot resolve it. Still, within those bounds the payoff is enormous: you trade an unsolvable nonlinear system for a fistful of matrices you can classify by hand, and you get a faithful local sketch of the phase portrait without ever solving the equations. The remaining guides in this rung pick up exactly where linearization stops — Lyapunov functions for stability when J is borderline, and Poincare-Bendixson for the global loops it cannot see.