Phase Plane Analysis — Mathematics Knowledge System

Overview

Phase plane analysis studies a two-variable system x′=f(x,y), y′=g(x,y) by plotting trajectories in the (x,y) plane rather than as functions of time. Each point in the plane has an associated velocity vector (f,g), so the phase plane is filled with a vector field whose flow lines are the system's solution curves. Qualitative features — spirals, closed loops, straight-line convergence — reveal the long-term dynamics without needing an explicit formula for x(t) and y(t). This graphical/qualitative approach is indispensable for nonlinear systems, which rarely have closed-form solutions but whose phase portraits can still be classified near equilibrium points.

Intuition

Imagine leaves floating on a pond with unseen currents pushing them around — at every point of the pond there's a definite current direction, and a leaf dropped anywhere just follows the local flow. The phase plane is exactly this: at each (x,y), the vector (f,g) tells you which way the system state is currently moving, and the trajectory through any starting point is the path a 'leaf' dropped there would trace. Equilibrium points are the still spots where the current vanishes, and whether nearby leaves drift toward, away from, or circle around such a spot classifies the system's qualitative behavior.

Formal Definition

Definition

A planar autonomous system and its associated direction field, together with the trajectory's tangent-slope relation used to sketch it:

x' = f(x,y), \qquad y' = g(x,y)

Planar autonomous system

\frac{dy}{dx} = \frac{g(x,y)}{f(x,y)}

Slope of trajectories in the phase plane

f(x^*,y^*) = 0,\ g(x^*,y^*) = 0

Equilibrium (fixed) points

Worked Examples

This system decouples completely: solve each equation independently.
$x(t) = x_0e^{-t}, \qquad y(t) = y_0e^{-4t}$
Both components decay to 0, but y decays 4× faster, so trajectories flatten onto the x-axis as t grows.
$\frac{y}{y_0} = \left(\frac{x}{x_0}\right)^4 \quad (\text{eliminating } t)$
The origin is an equilibrium where both derivatives vanish, and every trajectory approaches it — a stable node.
$x'(0,0) = 0,\ y'(0,0) = 0$

Answer: The origin is a stable node; all trajectories flow into it, approaching along the slower-decaying x-direction (verify: x=x₀e^{-t} solves x′=-x since d/dt[x₀e^{-t}]=-x₀e^{-t}=-x; similarly y′=-4y checks for y=y₀e^{-4t}).

Practice Problems

Difficulty 7/10

Find the equilibrium point(s) of x′ = y - x, y′ = x + y - 2.

Difficulty 6/10

For the decoupled system x′=2x, y′=-3y, describe the long-term behavior of trajectories starting away from the origin.

Difficulty 7/10

The linearized pendulum system is θ′ = ω, ω′ = -θ (small angle, unit constants). Show trajectories in the phase plane are circles.

Quiz

Summary

The phase plane plots trajectories of a planar system (x′,y′)=(f,g) directly in the (x,y) plane, exposing qualitative behavior without an explicit time solution.
Equilibrium points are where f=g=0; the local flow pattern around them (node, saddle, spiral, center) classifies the system's dynamics.
This qualitative approach is essential for nonlinear systems that rarely admit closed-form solutions.

References

WebsiteWikipedia — Phase plane