advanced35 min studyLesson 141 in Physics

Fixed points and the geometry of motion

Locate equilibria, draw directional flow, and distinguish nodes, saddles, spirals, and centers without mistaking a sketch for a proof.

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What you’ll understand

  • Find stationary states from all components of a vector field.
  • Classify planar linear equilibria using trace, determinant, and eigenvectors.
  • Explain why local phase portraits do not determine every distant trajectory.

A pendulum can rest, circulate, or oscillate even though one differential equation describes all three behaviors. A phase portrait organizes those possibilities by plotting states rather than just position against time. Each point records enough information to predict an instantaneous direction of motion. Begin with equilibria because they anchor that geometry, then ask how nearby arrows connect rather than immediately searching for a closed formula for every trajectory.

A state is more than a position

Write a smooth autonomous system as z˙=f(z)\dot{\mathbf z}=\mathbf f(\mathbf z). Autonomous means that time does not appear explicitly; identical states have identical future directions. For a mechanical oscillator the state is position together with velocity. Two objects passing the same position with opposite velocities are different phase-space points. A fixed point requires every component of the vector field to vanish simultaneously, so a zero velocity by itself is merely a possible turning point.

Nullclines mark where one component vanishes. Their intersections are candidate fixed points, while arrows between nullclines reveal which component increases or decreases. Draw a few arrows before drawing trajectories. For a continuously differentiable field, local uniqueness prevents two distinct trajectories from crossing at an ordinary state with different directions. A portrait can therefore rule out an imagined motion even when integrating the equations is difficult; it is a constrained representation rather than freehand decoration.

x˙=y,y˙=−sin⁡x−γy;(x∗,y∗)=(nπ,0)\dot x=y,\qquad \dot y=-\sin x-\gamma y;\qquad (x_*,y_*)=(n\pi,0)
Here time and velocity have been nondimensionalized. The pendulum equilibria require both zero angular velocity and zero restoring torque. The angle can be regarded modulo 2π, or unwrapped when counting complete turns.↔ Scroll wide equations horizontally.

Classify the linear skeleton

Near a fixed point, the Jacobian matrix supplies the leading linear field. In two dimensions let τ\tau be its trace and Δ\Delta its determinant. Negative determinant gives real eigenvalues of opposite sign and therefore a saddle. Positive determinant gives eigenvalues with matching real-part signs. The discriminant separates real eigenvalues from a complex-conjugate pair: real stable directions form a node, whereas a negative discriminant with negative trace produces an inward spiral.

A saddle attracts only along its stable manifold and repels along an unstable manifold; approaching along one specially chosen line does not make it stable. Repeated real eigenvalues require inspecting eigenvectors because a defective matrix shears as well as contracts. Purely imaginary eigenvalues describe a center for the linear system, but nonlinear terms can turn that apparent center into weak attraction or repulsion. Classification is strongest for hyperbolic equilibria, whose eigenvalues all have nonzero real parts.

λ±=τ±τ2−4Δ2,τ=tr⁡J,Δ=det⁡J\lambda_{\pm}=\frac{\tau\pm\sqrt{\tau^2-4\Delta}}{2},\qquad \tau=\operatorname{tr}J,\quad \Delta=\det J
This characteristic-root formula classifies a real planar linear system. For a nonlinear system it classifies its linearization; nonhyperbolic cases need additional analysis.↔ Scroll wide equations horizontally.

Connect geometry to a physical question

A useful portrait answers an experiment: after a small displacement, does a device return to rest, ring down while returning, or depart? Node versus spiral predicts whether sufficiently small displacements oscillate locally. Stable and unstable manifolds can separate initial conditions that end in different outcomes. Those boundaries are especially valuable when a system has multiple attractors, since a stable operating point may coexist with a dangerous distant state.

Keep the scales and assumptions visible. An arrow field supplies instantaneous velocity, not equal-time spacing unless speeds are represented consistently. Local classification says nothing by itself about escape, distant cycles, or the size of an attraction basin. For a damped pendulum the mechanical energy decreases, giving a second piece of evidence beyond the Jacobian. Combining local eigenvalues with an energy argument is stronger than extrapolating a small region of a drawing to all angles.

A damped pendulum's two resting states

For x˙=y\dot x=y, y˙=−sin⁡x−y\dot y=-\sin x-y, classify the equilibria at (0,0)(0,0) and (π,0)(\pi,0).

  1. Differentiate the field: J=(01−cos⁡x−1)J=\begin{pmatrix}0&1\\-\cos x&-1\end{pmatrix}. At the downward equilibrium its trace is −1-1 and determinant is 11.

  2. At the origin the discriminant is 1−4=−31-4=-3, so λ=(−1±i3)/2\lambda=(-1\pm i\sqrt3)/2. Small perturbations spiral inward; the point is a locally asymptotically stable spiral.

  3. At the upright equilibrium the determinant becomes −1-1. Its roots are (−1±5)/2(-1\pm\sqrt5)/2, with opposite signs, so it is a saddle rather than an attractor. Exact balance is a special initial condition, not robustness.

The result

The downward rest state attracts nearby motions while ringing down. The upright state is unstable despite the existence of one attracting direction. This conclusion concerns neighborhoods of those equilibria, not every unwrapped pendulum trajectory.

Explore the model

Interactive lab

Interactive lab · oscillator

Recognize a linear ring-down

Set damping ratio ζ below one and predict a decaying oscillation before pressing Play. Compare with ζ above one at the same natural frequency ω₀. The controls solve the analytic linear oscillator from displacement one and velocity zero; they do not draw the nonlinear pendulum phase portrait or its separatrices.

Recognize a linear ring-downUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forLocal oscillatory attraction has a recognizable signature, but this linear model does not classify distant nonlinear motion.

Check your understanding

A real planar Jacobian has determinant −2. Which classification follows independently of its trace?

Try it yourself

Classify x˙=−2x\dot x=-2x, y˙=−y\dot y=-y and describe what happens to the ratio x/yx/y for an initial state with both components nonzero.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Solve each independent component and compare their decay rates.

Compare with the solution

The eigenvalues are −2-2 and −1-1, so the origin is a stable node. The solutions are x=x0e−2tx=x_0e^{-2t} and y=y0e−ty=y_0e^{-t}. Hence x/y=(x0/y0)e−tx/y=(x_0/y_0)e^{-t} tends to zero: generic trajectories align with the slower y direction near the origin. Alignment with an axis is not evidence of a spiral, since no repeated rotation occurs.

Take it with you

  • Equilibria require every state derivative to vanish.
  • Trace and determinant classify planar linear flow, with eigenvectors resolving repeated-root geometry.
  • Hyperbolic local classification is robust; attraction basins and remote trajectories need separate evidence.

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