Linearization and Eigenvalue Stability

1Introduction

This lecture explains how to extract the first-order approximation of a nonlinear system near an equilibrium and use the eigenvalues of the resulting linear system to determine local stability.

2Standard Form

Consider

An equilibrium $$x_{*}$$\boldsymbol{x}_* satisfies $$\mathit{F}\left(x_{*}\right)=0$$\boldsymbol{F}(\boldsymbol{x}_*)=\boldsymbol{0}. Assume that $$\mathit{F}$$\boldsymbol{F} is $$C^1$$C^1 near the equilibrium. Local existence and uniqueness then hold, so the solution determined by an initial value is well defined.

3Definitions of Stability

Fix an initial time $$t_0$$t_0 and write the solution as $$\mathit{x}(t;t_0,x_0)$$\boldsymbol{x}(t;t_0,\boldsymbol{x}_0). For every initial value in the neighborhoods below, forward existence for all $$t\text{[PARSE ERROR: Undefined("Command("ge")")]}{t}_0$$t\ge t_0 is included in the definition.

The equilibrium $$x_{*}$$\boldsymbol{x}_* is Lyapunov stable if, for every $$\epsilon >0$$\varepsilon>0, there exists $$\delta >0$$\delta>0 such that

Thus, solutions starting sufficiently near the equilibrium remain near it; convergence is not required. The equilibrium is locally attractive if some $$r>0$$r>0 satisfies

An equilibrium that is both Lyapunov stable and locally attractive is locally asymptotically stable. It is unstable if it is not Lyapunov stable. It is globally asymptotically stable if it is Lyapunov stable, every solution exists forward for all initial values, and every solution converges to it.

For a linear system, Jordan blocks associated with eigenvalues on the imaginary axis determine whether solutions remain bounded. We now transfer the part of that criterion away from the spectral boundary to nonlinear equilibria.

4The Linear Approximation

Set $$\mathit{u}=\mathit{x}-x_{*}$$\boldsymbol{u}=\boldsymbol{x}-\boldsymbol{x}_*. Taylor expansion gives

Here $$D\mathit{F}\left(x_{*}\right)$$D\boldsymbol{F}(\boldsymbol{x}_*) is the Jacobian matrix. The $$C^1$$C^1 assumption gives the displayed little-o remainder; an $$O(\parallel \mathit{u}{\parallel }^2)$$O(\|\boldsymbol{u}\|^2) estimate requires stronger smoothness. The Jacobian does not make the entire nonlinear map linear. It extracts the linear map

acting on increments at the fixed equilibrium.

5Eigenvalue Criteria

5.1Complete Criterion for the Linear System

For the exact linear system $$u^{\prime }=A\mathit{u}$$\boldsymbol{u}'=A\boldsymbol{u}:

• The origin is globally asymptotically stable if and only if every eigenvalue satisfies $$\mathrm{Re}\lambda <0$$\operatorname{Re}\lambda<0.
• The origin is Lyapunov stable if and only if every eigenvalue satisfies $$\mathrm{Re}\lambda \text{[PARSE ERROR: Undefined("Command("le")")]}0$$\operatorname{Re}\lambda\le0 and every Jordan block belonging to an eigenvalue on the imaginary axis has size one.
• The origin is unstable if an eigenvalue has positive real part or if an eigenvalue on the imaginary axis has a Jordan block of size at least two.

This is a complete statement about the linear system itself.

5.2Linearization Theorem for a Nonlinear System

Suppose that $$\mathit{F}$$\boldsymbol{F} is $$C^1$$C^1 near $$x_{*}$$\boldsymbol{x}_* and that

has no eigenvalue on the imaginary axis. Such an equilibrium is hyperbolichyperbolic. If all eigenvalues of $$A$$A have negative real part, then $$x_{*}$$\boldsymbol{x}_* is locally asymptotically stable. If $$A$$A has an eigenvalue with positive real part, then $$x_{*}$$\boldsymbol{x}_* is unstable.

The instability conclusion from a positive real part remains valid even when other eigenvalues lie on the imaginary axis. If no eigenvalue has positive real part but at least one has zero real part, linearization alone is generally inconclusive. Nonlinear remainder terms then become essential. This theorem concerns local behavior only and does not establish global stability.

6Example 1: Computing a Jacobian

For

the equilibria are $$(0,0),(1,0),(0,2)$$(0,0),(1,0),(0,2), and

At $$(1,0)$$(1,0),

whose eigenvalues are $$-1$$-1 and $$1$$1. The positive eigenvalue proves that $$(1,0)$$(1,0) is unstable. Locally, one direction approaches the equilibrium and another departs from it.

7Example 2: An Inconclusive Linearization

For $$x^{\prime }=x^2$$x'=x^2, the equilibrium $$x=0$$x=0 has linearization $$u^{\prime }=0$$u'=0. Nevertheless,

so every $$x_0>0$$x_0>0 produces finite-time blow-up and the origin is Lyapunov unstable. By contrast, $$x^{\prime }=-x^3$$x'=-x^3 has the same linearization, but $$\left|x\right(t\left)\right|$$|x(t)| decreases to zero, so the origin is locally asymptotically stable. Equal boundary linearizations can therefore lead to opposite conclusions.

8Summary Table for Nonlinear Equilibria

Node, saddle, focus, and center are two-dimensional geometric classifications treated in the next lecture.

9Scope

Linearization is local. Far-field behavior, periodic orbits, and global stability require additional analysis. At a nonhyperbolic equilibrium, the original equations, monotone quantities, or conserved quantities must be examined directly.

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