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

Linear Stability Test

Ch 4 — Part I — Explicit & Qualitative Methods

The linear stability test uses the sign of \(f'(c)\) at an equilibrium \(c\) to classify it — faster than plotting and often sufficient.

The test

For \(y' = f(y)\) with \(f(c) = 0\):

Why it works (proof sketch)

Near \(c\), \(f(y) \approx f'(c) (y - c)\). Let \(u = y - c\); the linearized ODE is \(u' = f'(c) u\), exponential with rate \(f'(c)\). Decay if \(f'(c) < 0\), growth if \(f'(c) > 0\).

Practice Problems

Problem 1easy
For \(y' = y - y^3\), classify all equilibria using the linear test.
Hint
\(f'(y) = 1 - 3y^2\).
Solution

Equilibria: 0, ±1. \(f'(0) = 1 > 0\) (unstable). \(f'(\pm 1) = -2 < 0\) (stable).

Answer: 0 unstable, ±1 stable.

Problem 2easy
If \(f(y) = \sin y\), classify \(y = 0\) and \(y = \pi\).
Hint
\(f'(y) = \cos y\).
Solution

\(f'(0) = 1 > 0\) (unstable). \(f'(\pi) = -1 < 0\) (stable).

Answer: 0 unstable, π stable.

Problem 3hard
For \(y' = y^3 - y^5\), classify \(y = 0\).
Hint
Linear test is inconclusive since \(f'(0) = 0\). Inspect higher behavior.
Solution

Near 0, \(f(y) \approx y^3\). Sign of \(f\) flips across 0 from negative (for \(y<0\)) to positive (for \(y>0\)): unstable.

Answer: Unstable.

Problem 4easy
A population model \(P' = rP(1 - P/K)\) with \(r,K>0\). Use the linear test on the two equilibria.
Hint
\(f(P) = rP - rP^2/K\), \(f'(P) = r - 2rP/K\).
Solution

\(P=0\): \(f'(0) = r > 0\) (unstable). \(P=K\): \(f'(K) = -r < 0\) (stable).

Answer: Extinction unstable, carrying capacity stable.