Mathematics.

systems and stability

Equilibrium and Stability

Differential Equations30 minDifficulty7 out of 10

You should know: systems of differential equations

Overview

An equilibrium (or fixed point) of a system x′=f(x) is a state x* where f(x*)=0, so a trajectory starting exactly there stays there forever. Stability asks what happens to trajectories that start near, but not exactly at, x*: they may return (stable), stay nearby without necessarily converging (neutrally stable), or move away (unstable). For a system linearized near an equilibrium, x′=Ax with A the Jacobian at x*, the eigenvalues of A determine stability — negative real parts mean decay toward the equilibrium, positive real parts mean growth away from it. This eigenvalue criterion (the linearization or Hartman–Grobman theorem, informally) is the standard first tool for classifying equilibria of nonlinear systems.

Intuition

Picture a marble on a landscape: a bowl bottom is a stable equilibrium (nudge the marble and it rolls back), a hilltop is unstable (nudge it and it rolls away), and a perfectly flat plateau is neutrally stable (it stays wherever you put it). The eigenvalues of the linearized system play the role of the landscape's curvature at that point — negative real parts mean the 'bowl' curves upward in every direction (restoring), while a positive real part means the landscape falls away in at least one direction (a way to escape), just like a saddle point.

Formal Definition

Definition

Near an equilibrium x*, the system is approximated by its Jacobian, and stability is read off the eigenvalues of that Jacobian:

x′=Ax,A=Df(x∗) (the Jacobian at the equilibrium)\mathbf{x}' = A\mathbf{x}, \qquad A = Df(\mathbf{x}^*)\ \text{(the Jacobian at the equilibrium)}
Linearization
Re(λi)<0 ∀i  ⟹  x∗ asymptotically stable\text{Re}(\lambda_i) < 0 \ \forall i \implies \mathbf{x}^* \text{ asymptotically stable}
Stability criterion
Re(λi)>0 for some i  ⟹  x∗ unstable\text{Re}(\lambda_i) > 0 \ \text{for some } i \implies \mathbf{x}^* \text{ unstable}
Instability criterion

Worked Examples

  1. This system is already diagonal, so the Jacobian is constant and its eigenvalues are read directly from the diagonal.

    A=(−300−2),λ1=−3, λ2=−2A = \begin{pmatrix}-3&0\\0&-2\end{pmatrix}, \quad \lambda_1=-3,\ \lambda_2=-2

  2. Both eigenvalues are negative, so both decoupled solutions decay exponentially.

    x(t)=x0e−3t→0,y(t)=y0e−2t→0x(t)=x_0e^{-3t}\to 0, \qquad y(t)=y_0e^{-2t}\to 0

Answer: The origin is asymptotically stable (a stable node) since both eigenvalues are negative (verify: x′=-3x₀e^{-3t}=-3x ✓, y′=-2y₀e^{-2t}=-2y ✓, and both terms → 0 as t→∞).

Practice Problems

Difficulty 7/10

Classify the equilibrium (0,0) for x′=5x, y′=-y.

Difficulty 6/10

For a Jacobian with trace -4 and determinant 4 at an equilibrium, are the eigenvalues real or complex, and is the equilibrium stable?

Difficulty 7/10

A chemical reactor's steady state has Jacobian eigenvalues -1±3i. Is the steady state stable, and what qualitative behavior do nearby trajectories show?

Quiz

An equilibrium x* of x′=f(x) satisfies:
If every eigenvalue of the Jacobian at an equilibrium has negative real part, the equilibrium is:
Complex eigenvalues α ± iβ with α > 0 at an equilibrium indicate:

Summary

  • An equilibrium x* satisfies f(x*)=0; stability describes whether nearby trajectories converge to, stay near, or diverge from x*.
  • Linearizing at x* gives a Jacobian A whose eigenvalues classify stability: all negative real parts means asymptotically stable, any positive real part means unstable.
  • Complex eigenvalue pairs α±iβ give spiraling behavior, decaying if α<0 and growing if α>0.

References

  1. WebsiteWikipedia — Stability theory