Linear Stability Analysis

We can spot a fixed point of \dot{x} = f(x) and, in one dimension, eyeball its stability from the flow arrows. But "eyeball the arrows" won't survive contact with higher dimensions, and it gives no number — no rate. Linear stability analysis replaces the eyeball with a derivative. The idea is the oldest trick in calculus: zoom in. Near a fixed point the messy nonlinear system looks, to first order, exactly like a linear system — and a linear system is completely decoded by its eigenvalues.

One dimension: the slope is the rate

Let x^\* be a fixed point, so f(x^\*) = 0. Watch a small disturbance: write x(t) = x^\* + \eta(t) with \eta tiny. Since x^\* is constant, \dot{\eta} = \dot{x} = f(x^\* + \eta). Taylor-expand about x^\*:

\dot{\eta} = f(x^\*) + f'(x^\*)\,\eta + \tfrac12 f''(x^\*)\,\eta^2 + \cdots = f'(x^\*)\,\eta + O(\eta^2).

The constant term is zero (that's what "fixed" means), and near the fixed point \eta^2 is negligible beside \eta. So the disturbance obeys the linear equation \dot{\eta} = f'(x^\*)\,\eta, whose solution is the pure exponential

\eta(t) = \eta_0\, e^{f'(x^\*)\,t}.

Everything hinges on the sign of the single number f'(x^\*). If it is negative the disturbance decays and the point is stable; if positive it grows and the point is unstable. The magnitude |f'(x^\*)| even gives the rate: the characteristic time to relax is 1/|f'(x^\*)|.

For \dot{x} = f(x) with a fixed point f(x^\*)=0:

The linearization is a tangent line

Geometrically, f'(x^\*) is just the slope of f where it crosses the axis. Below is f(x) = x - x^3 again, now with its tangent lines drawn at the three fixed points. At x^\* = 0 the tangent has slope f'(0) = 1 > 0 (an upward crossing — unstable); at x^\* = \pm1 the tangent has slope f'(\pm 1) = -2 < 0 (downward crossings — stable). The tangent line is the linearized flow.

The steeper the crossing, the faster the flow reacts nearby. At \pm 1 the slope is -2, so disturbances halve roughly twice as fast as they would with slope -1. Stability is a sign; the slope's size is the speed.

Many dimensions: the Jacobian

For a system \dot{\mathbf{x}} = \mathbf{f}(\mathbf{x}) in n variables, the single slope becomes a matrix of partial derivatives — the Jacobian — evaluated at the fixed point:

J = D\mathbf{f}(\mathbf{x}^\*) = \begin{pmatrix} \partial f_1/\partial x_1 & \cdots & \partial f_1/\partial x_n \\ \vdots & & \vdots \\ \partial f_n/\partial x_1 & \cdots & \partial f_n/\partial x_n \end{pmatrix}_{\mathbf{x}=\mathbf{x}^\*}.

The same Taylor argument gives, for the disturbance \boldsymbol{\eta} = \mathbf{x} - \mathbf{x}^\*, the linear system \dot{\boldsymbol{\eta}} = J\,\boldsymbol{\eta}. And you already know how such a system behaves: decompose \boldsymbol{\eta} along the eigenvectors of J, and along eigenvector \mathbf{v}_k the disturbance evolves as e^{\lambda_k t}. Stability is therefore governed by the real parts of the eigenvalues.

Let \lambda_1,\dots,\lambda_n be the eigenvalues of the Jacobian J = D\mathbf{f}(\mathbf{x}^\*).

The middle case has a name worth knowing: when no eigenvalue sits on the imaginary axis the fixed point is called hyperbolic, and the Hartman–Grobman theorem promises that near a hyperbolic fixed point the true nonlinear flow is a smoothly deformed copy of its linearization — the linear picture is not just a guess, it is topologically exact. All the danger lives in the non-hyperbolic case, where an eigenvalue grazes the imaginary axis; that borderline is precisely where bifurcations are born.

Worked example: a 2-D nonlinear system

Classify the origin for

\dot{x} = -x + 2y + x^2, \qquad \dot{y} = -3y.

Step 1 — confirm the fixed point. At (0,0) both right sides vanish, so the origin is a fixed point.

Step 2 — build the Jacobian. With f_1 = -x + 2y + x^2 and f_2 = -3y,

J = \begin{pmatrix} -1 + 2x & 2 \\ 0 & -3 \end{pmatrix}, \qquad J(0,0) = \begin{pmatrix} -1 & 2 \\ 0 & -3 \end{pmatrix}.

Step 3 — eigenvalues. The matrix is upper-triangular, so the eigenvalues are the diagonal entries: \lambda_1 = -1, \lambda_2 = -3.

Step 4 — classify. Both eigenvalues are real and negative, so the origin is a stable node. The quadratic term x^2 was irrelevant to the verdict — near the fixed point it is swamped by the linear terms, exactly as Hartman–Grobman guarantees. Two partial derivatives and a glance at a triangular matrix settled the fate of a nonlinear system.

For a 2\times2 Jacobian you rarely need the eigenvalues themselves. They satisfy \lambda^2 - (\operatorname{tr}J)\lambda + \det J = 0, so the product of the eigenvalues is \det J and their sum is \operatorname{tr}J. Both eigenvalues have negative real part — i.e. the fixed point is stable — iff

\operatorname{tr}J < 0 \quad\text{and}\quad \det J > 0.

A negative determinant forces one positive and one negative eigenvalue — a saddle, always unstable — with no need to solve anything. This "trace–determinant plane" is the map we colour in fully in the next lesson.

Two traps snap shut here. First, the linearization is only local. It tells you what happens to small disturbances near the fixed point and nothing about far-away starting points, which may head somewhere else entirely. A "stable" fixed point can have a tiny basin of attraction.

Second, a zero real part is a genuine cliff-edge, not a rounding issue. If the leading eigenvalue is -0.0001 the point is (barely) stable; if it is exactly 0 the linear test says nothing and the neglected quadratic or cubic terms take over. Never round a real part to zero and declare victory — a purely imaginary pair (a "centre" in the linear system) can become a slow inward or outward spiral once nonlinear terms are restored.