2.3 The Phase Loop

So far every picture in this chapter has plotted $x$ against $t$ . There is a second way to draw a solution that throws away the clock and gains something better. Introduce the velocity as a variable of its own, $y = x^{'}$ ; then the single second-order equation

$x^{″} + 2 a x^{'} + (a^{2} + b^{2}) x = 0$

becomes a pair of first-order equations,

$x^{'} = y, y^{'} = - (a^{2} + b^{2}) x - 2 a y,$

and a solution is now a moving point $(x (t), y (t))$ in a plane. The path that point traces is the orbit, and the plane is the phase plane. Its horizontal axis is position and its vertical axis is velocity, so a point on the positive horizontal axis is a mass displaced to the right and momentarily at rest, while a point on the positive vertical axis is a mass passing through equilibrium at speed.

For the solution $x (t) = e^{- a t} \cos b t$ with velocity $y (t) = e^{- a t} (- a \cos b t - b \sin b t)$ , the orbit is the curve in the figure below.

A red curve traced in the plane whose horizontal axis is position x and vertical axis is velocity x'. With the damping slider a at 0 the curve is a closed ellipse retraced forever, taller when the frequency slider b is larger; any positive a breaks the loop into a spiral that winds clockwise inward to the rest point at the origin. — The phase-plane orbit of the damped spring: horizontal axis is position $x$ , vertical axis is velocity $x^{'}$ . With damping $a equals 0$ the orbit closes into an ellipse; any positive $a$ turns it into a spiral winding toward the rest state at the origin.

Explore

Observe. Set $a equals 0$ , $b equals 2$ . The orbit is a closed loop, retraced forever. Where on the loop is the mass at its farthest displacement, and where is it moving fastest? Mark both on the curve.
Predict. Still with $a equals 0$ , what will raising $b$ to $4$ do to the loop — widen it, heighten it, or both? The loop is $(\cos b t, - b \sin b t)$ ; reason from that before dragging.
Verify. Drag $b$ from $1$ to $4$ . The loop's horizontal extent never changes but its vertical extent grows with $b$ . Say what that means physically about a stiffer spring released from the same displacement.
Observe. Set $b equals 2$ and nudge $a$ up to $0.15$ . The loop no longer closes: each circuit ends a little closer to the center than it began. Which direction does the point travel around the orbit — clockwise or counterclockwise — and how does the top half of the plane force the answer?
Predict, then verify. Push $a$ to $0.8$ . How many turns does the spiral complete before it is visually at the origin? Now relate that count to the $x$ -versus- $t$ picture from the first section of this chapter: what does one turn of the spiral correspond to there?

Connect

The two shapes in this figure carry the whole qualitative theory. A closed orbit means the motion is periodic — the state returns exactly to where it started, so it must repeat, forever. That happens when $a equals 0$ , which is precisely when nothing removes energy from the system: the quantity $E = \frac{1}{2} y^{2} + \frac{1}{2} b^{2} x^{2}$ is constant along the motion, and the orbit is one level set of $E$ , an ellipse. Chapter 4 makes that observation into a method.

A spiral into the origin means the motion decays to rest: energy leaks out through the damping term, the orbit crosses to lower and lower level sets of $E$ , and the origin — the equilibrium point where $x = y = 0$ , the mass sitting still at its natural length — attracts every nearby state. In the $x$ -versus- $t$ picture these two cases were "constant-amplitude wave" and "wave in a shrinking envelope," which required watching a whole time history to distinguish. In the phase plane they are "loop" and "spiral," distinguishable at a glance, and that economy is why the phase plane survives into systems too complicated to solve.

Direction of travel is fixed by the first equation, $x^{'} = y$ : above the horizontal axis the velocity is positive, so $x$ is increasing and the point moves to the right; below it, to the left. Every orbit in every phase plane of this kind therefore runs clockwise.

Example 1 From equation to orbit

Sketch the phase-plane orbit of $x^{″} + x = 0$ with $x open parenthesis 0 close parenthesis equals 2$ , $x prime open parenthesis 0 close parenthesis equals 0$ , and state its period. Then say how the sketch changes for $x^{″} + 0.2 x^{'} + x = 0$ .

With $b equals 1$ and no damping, the solution is $x equals 2 cos t$ , $y = - 2 \sin t$ , so the orbit satisfies $x^{2} + y^{2} = 4$ : a circle of radius $2$ , traversed clockwise, period $2 π$ . The mass is at maximum displacement where the orbit crosses the horizontal axis and at maximum speed where it crosses the vertical axis. Adding the damping term $0.2 x prime$ makes $a equals 0.1$ ; the circle becomes a spiral that retains only about $e^{- 0.2 π} \approx 53$ of its radius per revolution, so after three turns it is at roughly $15$ of its starting size and heading to the origin. No initial condition escapes: with damping present, the entire plane drains inward.

An original work of XYZ Homework, built around interactive XYZ 3D figures, following the standard first-course sequence (first-order equations through systems and the phase plane). Distinct from the LibreTexts-sourced Differential Equations for Engineers (Lebl) edition, which has its own attribution. License: CC-BY-NC-SA-4.0.