Review / Part I — Explicit & Qualitative Methods / Ch 4

Phase Line & Stability

Ch 4 — Part I — Explicit & Qualitative Methods

For a scalar autonomous ODE \(y' = f(y)\), the phase line is a 1-D picture of \(y\)-values with arrows showing sign of \(f\). Equilibria are marked points; arrows show direction of motion between them.

Classifying equilibria

An equilibrium \(y = c\) with \(f(c)=0\) is

Concavity change test

On the phase line, solution curves in the \((t,y)\) plane change concavity where \(f'(y) = 0\) between equilibria (and \(f(y) \ne 0\)).

y=0 semi-stable y=1 unstable y=2 stable
Phase line for \(y' = -2y(y-1)(y-2)\).

Practice Problems

Problem 1medium
Draw the phase line and classify equilibria for \(y' = -2y(y-1)(y-2)\).
Hint
Zeros of \(f\) at 0, 1, 2. Pick test values in each interval and record the sign.
Solution

Signs: \(f(-1) = -12 < 0\), \(f(0.5) = -0.75 < 0\), \(f(1.5) = 0.75 > 0\), \(f(3) = -12 < 0\). So the sign pattern across 0, 1, 2 is \(-,\,-,\,+,\,-\). At \(y=0\) the sign doesn't flip (−→−) so it's semi-stable. At \(y=1\) it flips − to + (arrows pointing away) so unstable. At \(y=2\) it flips + to − (arrows pointing in) so stable.

Answer: \(y=0\) semi-stable, \(y=1\) unstable, \(y=2\) stable.

Problem 2easy
If \(f\) passes through zero with \(f'(c) < 0\), classify \(y = c\).
Hint
\(f\) is crossing from + to −.
Solution

Arrows converge toward \(c\); stable.

Answer: Stable.

Problem 3medium
If \(f(c) = 0\) and \(f'(c) = 0\) but \(f''(c) \ne 0\), classify \(y = c\).
Hint
Graph of \(f\) touches axis without crossing.
Solution

\(f\) doesn't change sign, so \(y=c\) is semi-stable.

Answer: Semi-stable.

Problem 4medium
For \(y' = y(y - 3)^2\), give the phase-line classification.
Hint
Zeros at 0 (simple) and 3 (double).
Solution

\(f(-1) < 0\), \(f(1) > 0\), \(f(4) > 0\). So \(y=0\) is unstable (\(-\to+\)); \(y=3\) has same sign on both sides: semi-stable.

Answer: 0 unstable, 3 semi-stable.