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Autonomous equations and stability

Chapter 6 - Read feedback from a phase line

An autonomous equation has no explicit time dependence:

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

An equilibrium is:

  • stable when nearby arrows point toward it;
  • 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 (K) blow up, cross (K), or approach (K)?

Worked example: stability before formula

Analyze \(y'=y(2-y)(y-5)\)qualitative first
  1. Equilibria. The zeros are \(y=0,2,5\).
  2. Sign below 0. Test \(y=-1\): \((-)(+)(-)\) is positive, so arrows point up.
  3. Sign from 0 to 2. Test \(y=1\): \((+)(+)(-)\) is negative, so arrows point down.
  4. Sign from 2 to 5. Test \(y=3\): \((+)(-)(-)\) 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)^2):

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

The equilibria are \(-1\) and 2. Since the squared factor never changes sign, the rate is negative below \(-1\) and positive on both intervals above \(-1\). Thus \(-1\) 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_*), 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 (f) is differentiable and (f'(y_*)\ne0):

  • (f'(y_*)<0) suggests local stability;
  • (f'(y_*)>0) suggests local instability.

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

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

Parameters can change the portrait

Consider (y'=y(r-y)). When (r>0), (y=0) is unstable and (y=r) is stable. When (r<0), their order swaps: (y=r) is unstable and (y=0) is stable. At (r=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 \(y\). Arrows report how \(y\) 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 \(y'=y(y-1)(y-3)\).

    Detailed solution

  2. Practice - solution behavior

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

    Detailed solution

  3. Integrated - logistic interpretation

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

    Detailed solution

  4. Challenge - threshold

    6.4 Analyze \(y'=y^2-a\) for \(a>0\), \(a=0\), and \(a<0\). Track how the equilibria and their stability change.

    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.