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Autonomous equations and stability

Chapter 6 - Read feedback from a phase line

An autonomous equation has no explicit time dependence:

y′=f(y). y'=f(y).

The same state always receives the same rate. This makes long-run behavior visible on a one-dimensional phase line, often before any solution formula appears.

The phase-line procedure

  1. Solve f(y)=0f(y)=0 for equilibrium states.
  2. Use those states to divide the line into intervals.
  3. Test the sign of f(y)f(y) on each interval.
  4. Draw an arrow up for f(y)>0f(y)\gt 0 and down for f(y)<0f(y)\lt 0.
  5. Classify each equilibrium by the arrows around it.

An equilibrium is:

  • asymptotically stable when nearby arrows point toward it (called stable in the phase-line examples below);
  • unstable when nearby arrows point away; and
  • semistable when one side points in and the other points out.
Phase line for y prime equals y times one minus y over four Rates are negative below zero, positive between zero and four, and negative above four. Zero is unstable and four is stable. y = 0, unstable y = 4, stable f(y) < 0 f(y) > 0 f(y) < 0 The vertical coordinate is state, not time.
A phase line compresses every solution's direction into one state axis.

Lab: logistic feedback

Logistic feedback explorerChange the environment and the initial state.

solutioncarrying capacity

Questions to test in the lab:

  • Which parameter changes the destination?
  • Which parameter changes how quickly the solution travels?
  • Does a positive initial state above KK blow up, cross KK, or approach KK?

Worked example: stability before formula

Analyze y′=y(2−y)(y−5)y'=y(2-y)(y-5)qualitative first
  1. Equilibria. The zeros are y=0,2,5y=0,2,5.
  2. Sign below 0. Test y=−1y=-1: (−)(+)(−)(-)(+)(-) is positive, so arrows point up.
  3. Sign from 0 to 2. Test y=1y=1: (+)(+)(−)(+)(+)(-) is negative, so arrows point down.
  4. Sign from 2 to 5. Test y=3y=3: (+)(−)(−)(+)(-)(-) is positive, so arrows point up.
  5. Sign above 5. All factors have signs (+)(−)(+)(+)(-)(+), so the rate is negative.
  6. Classify. Arrows point toward 0, away from 2, and toward 5. Thus 0 and 5 are stable; 2 is unstable.

Fade the phase line

For y′=(y+1)(y−2)2y'=(y+1)(y-2)^2:

  • equilibria: y=‾y=\underline{\hspace{4em}};
  • sign on (−∞,−1)( -\infty,-1): ‾\underline{\hspace{2em}};
  • sign on (−1,2)(-1,2): ‾\underline{\hspace{2em}};
  • sign on (2,∞)(2,\infty): ‾\underline{\hspace{2em}};
  • stability at −1-1: ‾\underline{\hspace{5em}}; at 2: ‾\underline{\hspace{5em}}.
Check the line

The equilibria are −1-1 and 2. Since the squared factor never changes sign, the rate is negative below −1-1 and positive on both intervals above −1-1. Thus −1-1 is unstable, while 2 is semistable: trajectories below 2 move upward toward it, while trajectories above 2 move upward away from it.

Feedback language

Near an equilibrium y∗y_*, examine whether a small displacement triggers a correcting or amplifying rate.

  • Negative feedback: a positive displacement produces a negative rate, and a negative displacement produces a positive rate. The equilibrium restores itself.
  • Positive feedback: a displacement produces a rate in the same direction. The equilibrium repels.

When ff is differentiable and f′(y∗)≠0f'(y_*)\ne0:

  • f′(y∗)<0f'(y_*)\lt 0 suggests local stability;
  • f′(y∗)>0f'(y_*)\gt 0 suggests local instability.

The sign chart remains safer when f′(y∗)=0f'(y_*)=0.

Stability from feedback. Near y=3y=3, a model is approximately y′=−0.4(y−3)y'=-0.4(y-3). What happens after a small upward displacement?

Parameters can change the portrait

Consider y′=y(r−y)y'=y(r-y). When r>0r\gt 0, y=0y=0 is unstable and y=ry=r is stable. When r<0r\lt 0, their order swaps: y=ry=r is unstable and y=0y=0 is stable. At r=0r=0, the equilibria meet.

This qualitative change as a parameter passes a threshold is a bifurcation. It matters because parameter changes can alter not only speed, but the number and stability of long-run states.

Error clinic

Error clinic: putting time on the phase line

A phase line's coordinate is the state yy. Arrows report how yy changes as time advances. A solution graph uses horizontal time and vertical state; the two pictures carry different axes.

Practice

  1. Practice - phase line

    6.1 Draw and classify the phase line for y′=y(y−1)(y−3)y'=y(y-1)(y-3).

    Detailed solution

  2. Practice - solution behavior

    6.2 For the same equation, describe the long-run behavior for y(0)=−1,1/2,2,4y(0)=-1, 1/2, 2, 4. Do not solve.

    Detailed solution

  3. Integrated - logistic interpretation

    6.3 In P′=rP(1−P/K)P'=rP(1-P/K), with r,K>0r,K\gt 0, show that the growth rate P′P' is largest at P=K/2P=K/2. Distinguish "largest growth rate" from "largest population."

    Detailed solution

  4. Challenge - threshold

    6.4 Analyze y′=y2−ay'=y^2-a for a>0a\gt 0, a=0a=0, and a<0a\lt 0. Track how the equilibria and their stability change.

    Detailed solution

  5. Practice - extension

    6.5 Analyze y′=−(y−2)3y'=-(y-2)^3. Classify the equilibrium using a sign chart, then compute f′(2)f'(2). What does this example show about concluding instability from a zero linearization derivative? Predict the limits for y(0)=1y(0)=1 and y(0)=3y(0)=3.

    Detailed solution

  6. Challenge - transfer

    6.6 A harvested population follows P′=P(1−P/8)−HP'=P(1-P/8)-H, where H≥0H\ge0 is a constant removal rate. Find and classify equilibria for 0≤H<20\le H\lt 2, H=2H=2, and H>2H\gt 2. For 0<H<20\lt H\lt 2, explain the role of the smaller equilibrium. Treat P=0P=0 as extinction: stop the physical model there instead of allowing negative population.

    Detailed solution

Retrieve and connect

Without looking, answer:

  1. Why must separation record equilibria before division?
  2. How does an integrating factor use the product rule?
  3. What do inward, outward, and one-sided arrows mean at an equilibrium?

Next: Turn second-order equations into the language of natural modes and oscillation.