Mathematics.

qualitative theory

Phase Portraits and Stability

Differential Equations50 minDifficulty7 out of 10

Overview

A phase portrait is a geometric picture of the trajectories of an autonomous ODE system dx/dt = F(x) drawn in the phase plane (or phase space). Instead of tracking solutions as functions of time, we visualize the collection of all possible trajectories simultaneously. Stability theory classifies equilibrium points (where F(x*)=0) by their behavior: stable (nearby solutions stay close), asymptotically stable (nearby solutions converge), or unstable. For linear systems, eigenvalues of the Jacobian determine the type: nodes, spirals, saddles, and centers.

Intuition

Think of the phase portrait as a map of 'currents' in a river: each point tells you the velocity of a particle placed there. Trajectories are the paths particles follow. An equilibrium is a point of still water. A stable equilibrium (like the bottom of a bowl) attracts nearby particles; an unstable one (like the top of a hill) repels them. The eigenvalues of the Jacobian tell you whether solutions spiral in/out, slide in/out along axes (node), or split apart (saddle).

Formal Definition

Definition

For the planar autonomous system, equilibria and their stability are determined by the linearization:

x˙=F(x),x∈R2\dot{\mathbf{x}} = \mathbf{F}(\mathbf{x}),\quad \mathbf{x} \in \mathbb{R}^2
Autonomous system
F(x∗)=0 (equilibrium point)\mathbf{F}(\mathbf{x}^*) = \mathbf{0} \text{ (equilibrium point)}
Equilibrium condition
J=DF(x∗)=(∂F1/∂x1∂F1/∂x2∂F2/∂x1∂F2/∂x2)J = DF(\mathbf{x}^*) = \begin{pmatrix} \partial F_1/\partial x_1 & \partial F_1/\partial x_2 \\ \partial F_2/\partial x_1 & \partial F_2/\partial x_2 \end{pmatrix}
Jacobian at equilibrium
λ1,2=tr(J)±tr(J)2−4det⁡(J)2\lambda_{1,2} = \frac{\text{tr}(J) \pm \sqrt{\text{tr}(J)^2 - 4\det(J)}}{2}
Eigenvalues determining stability type

Worked Examples

  1. The system is already linear: J = [[1,−1],[1,1]].

    J=(1−111)J = \begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}

  2. tr(J) = 2, det(J) = 1+1 = 2. Eigenvalues: λ = (2 ± √(4−8))/2 = 1 ± i.

    λ=1±i\lambda = 1 \pm i

  3. Complex eigenvalues with positive real part Re(λ)=1>0: unstable spiral (trajectories spiral outward).

    Re(λ)=1>0⇒unstable spiral\text{Re}(\lambda) = 1 > 0 \Rightarrow \text{unstable spiral}

Answer: The origin is an unstable spiral: solutions rotate outward in the phase plane.

Practice Problems

Difficulty 6/10

Classify the equilibrium of ẋ = y, ẏ = −x (simple harmonic oscillator).

Difficulty 7/10

Find and classify all equilibria of ẋ = x(1−x−y), ẏ = y(2−x−3y).

Difficulty 7/10

State the Hartman–Grobman theorem and explain when linearization accurately predicts stability.

Common Mistakes

Common Mistake

A center in the linearization means the nonlinear system has a center.

Centers are non-generic: higher-order nonlinear terms can turn a linearized center into a stable or unstable spiral. Only for Hamiltonian systems can one often confirm a true center.

Common Mistake

Phase portraits only exist for 2D systems.

Phase portraits generalize to any dimension, though visualization requires projecting onto 2D subspaces for higher-dimensional systems.

Quiz

If the Jacobian at an equilibrium has eigenvalues λ = 2 ± 3i, the equilibrium is:
A saddle point occurs when:

Summary

  • Phase portraits show all trajectories of an autonomous system simultaneously in phase space.
  • Equilibria x* satisfy F(x*)=0; their stability is determined by eigenvalues of J=DF(x*).
  • Negative real parts → stable (spiral or node); positive real parts → unstable; opposite signs → saddle.
  • Purely imaginary eigenvalues give a center for linear systems (may differ for nonlinear).
  • The Hartman–Grobman theorem justifies linearization at hyperbolic (Re≠0) equilibria.

References

  1. WebsiteWikipedia — Phase portrait