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
- Solve for equilibrium states.
- Use those states to divide the line into intervals.
- Test the sign of on each interval.
- Draw an arrow up for and down for .
- 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.
Lab: logistic feedback
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 blow up, cross , or approach ?
Worked example: stability before formula
- Equilibria. The zeros are .
- Sign below 0. Test : is positive, so arrows point up.
- Sign from 0 to 2. Test : is negative, so arrows point down.
- Sign from 2 to 5. Test : is positive, so arrows point up.
- Sign above 5. All factors have signs , so the rate is negative.
- 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 :
- equilibria: ;
- sign on : ;
- sign on : ;
- sign on : ;
- stability at : ; at 2: .
Check the line
The equilibria are and 2. Since the squared factor never changes sign, the rate is negative below and positive on both intervals above . Thus 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 , 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 is differentiable and :
- suggests local stability;
- suggests local instability.
The sign chart remains safer when .
Stability from feedback. Near , a model is approximately . What happens after a small upward displacement?
Parameters can change the portrait
Consider . When , is unstable and is stable. When , their order swaps: is unstable and is stable. At , 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
A phase line's coordinate is the state . Arrows report how changes as time advances. A solution graph uses horizontal time and vertical state; the two pictures carry different axes.
Practice
- Practice - solution behavior
6.2 For the same equation, describe the long-run behavior for . Do not solve.
- Integrated - logistic interpretation
6.3 In , with , show that the growth rate is largest at . Distinguish "largest growth rate" from "largest population."
- Challenge - threshold
6.4 Analyze for , , and . Track how the equilibria and their stability change.
- Practice - extension
6.5 Analyze . Classify the equilibrium using a sign chart, then compute . What does this example show about concluding instability from a zero linearization derivative? Predict the limits for and .
- Challenge - transfer
6.6 A harvested population follows , where is a constant removal rate. Find and classify equilibria for , , and . For , explain the role of the smaller equilibrium. Treat as extinction: stop the physical model there instead of allowing negative population.
Retrieve and connect
Without looking, answer:
- Why must separation record equilibria before division?
- How does an integrating factor use the product rule?
- What do inward, outward, and one-sided arrows mean at an equilibrium?