Linearization and Local Behavior

Linearization approximates a nonlinear system near an equilibrium by its linear part, allowing the classification of equilibria using eigenvalue analysis of the Jacobian matrix.

The Linearization

Definition6.4Jacobian linearization

For the nonlinear system $\mathbf{x}' = \mathbf{F}(\mathbf{x})$ with equilibrium $\mathbf{x}_0$ , the linearization at $\mathbf{x}_0$ is $\mathbf{u}' = A\mathbf{u}$ where $\mathbf{u} = \mathbf{x} - \mathbf{x}_0$ and $A = D\mathbf{F}(\mathbf{x}_0)$ is the Jacobian matrix:

$A = \begin{pmatrix}\partial f/\partial x & \partial f/\partial y \\ \partial g/\partial x & \partial g/\partial y\end{pmatrix}\bigg|_{(x_0, y_0)}.$

Theorem6.2Hartman-Grobman theorem

If all eigenvalues of $A = D\mathbf{F}(\mathbf{x}_0)$ have nonzero real parts (i.e., $\mathbf{x}_0$ is a hyperbolic equilibrium), then the nonlinear system is topologically conjugate to the linearization near $\mathbf{x}_0$ . The qualitative behavior (node, saddle, spiral) of the linearization correctly predicts the behavior of the nonlinear system.

RemarkNon-hyperbolic case

When eigenvalues have zero real part (centers, non-isolated equilibria), the linearization may not predict the nonlinear behavior. A linear center can become a spiral (stable or unstable) when nonlinear terms are added. Additional analysis (Lyapunov functions, center manifold theory) is needed.

Examples

ExampleLinearization of a nonlinear system

Analyze $x' = x - xy$ , $y' = -y + xy$ near $(1, 1)$ .

Jacobian: $A = \begin{pmatrix}1-y & -x\\y & -1+x\end{pmatrix}_{(1,1)} = \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$ .

Eigenvalues: $\lambda = \pm i$ (pure imaginary). This is a linear center, but since the equilibrium is non-hyperbolic, we cannot conclude the nonlinear system has a center without further analysis. (In this case, the conserved quantity $H = x + y - \ln x - \ln y$ confirms it is indeed a center.)

ExampleVan der Pol oscillator

$x'' - \mu(1-x^2)x' + x = 0$ becomes $x' = y$ , $y' = \mu(1-x^2)y - x$ .

At $(0,0)$ : $A = \begin{pmatrix}0&1\\-1&\mu\end{pmatrix}$ , eigenvalues $\lambda = \frac{\mu \pm \sqrt{\mu^2-4}}{2}$ .

For $0 < \mu < 2$ : unstable spiral (eigenvalues with positive real part). The Poincare-Bendixson theorem guarantees a limit cycle.

Stability via Trace and Determinant

RemarkTrace-determinant classification

For a $2 \times 2$ matrix $A$ with $\tau = \text{tr}(A) = \lambda_1 + \lambda_2$ and $\Delta = \det(A) = \lambda_1\lambda_2$ :

Stable node/spiral: $\tau < 0$ and $\Delta > 0$ .
Unstable node/spiral: $\tau > 0$ and $\Delta > 0$ .
Saddle: $\Delta < 0$ .
Center: $\tau = 0$ and $\Delta > 0$ .
The discriminant $\tau^2 - 4\Delta$ determines node ( $> 0$ ) vs. spiral ( $< 0$ ).

The $(\tau, \Delta)$ -plane provides a complete map of equilibrium types.