Review / Part I — Explicit & Qualitative Methods / Ch 3

Qualitative Analysis & Long-Term Behavior

Ch 3 — Part I — Explicit & Qualitative Methods

Qualitative analysis asks what solutions look like — their long-term behavior, monotonicity, concavity — without computing closed forms.

Tools

For \(y' = f(y)\): the sign of \(f\) gives monotonicity, stationary points come from \(f(y) = 0\), and concavity from \(y'' = f(y) f'(y)\).

Long-term behavior

Between adjacent equilibria, \(f\) has constant sign, so solutions are monotone and converge to the appropriate boundary equilibrium (stable) or escape.

Concavity check

\(y'' = \tfrac{d}{dt} f(y) = f'(y) \cdot y' = f(y) f'(y).\) The sign of \(f \cdot f'\) tells you concave up (\(+\)) or down (\(-\)).

Practice Problems

Problem 1easy
For \(y' = y(1-y)\), describe the long-term behavior of \(y(t)\) for \(y(0) = 0.3\).
Hint
Equilibria at 0 and 1; \(f > 0\) on \((0,1)\).
Solution

Solution is monotone increasing on \((0,1)\); \(y(t) \to 1\) as \(t\to\infty\).

Answer: \(y(t)\to 1\).

Problem 2medium
Where is the solution of \(y' = y(1-y)\) with \(y(0) = 1/4\) concave up vs down?
Hint
\(y'' = f f' = y(1-y)(1-2y)\).
Solution

Between 0 and 1/2, \(y(1-y)(1-2y) > 0\) ⇒ concave up. Between 1/2 and 1, negative ⇒ concave down. Inflection at \(y = 1/2\).

Answer: Up on \(y\in(0,1/2)\), down on \((1/2,1)\).

Problem 3medium
Sketch the solution to \(y' = y - y^3\) starting at \(y(0) = 0.1\).
Hint
Equilibria at 0, ±1. Study signs of \(f\) on intervals.
Solution

On \((0,1)\), \(f>0\), monotone increasing toward \(y = 1\); concavity change at \(f'(y) = 1 - 3y^2 = 0\Rightarrow y = 1/\sqrt 3\).

Answer: Approaches \(y=1\), inflection at \(y = 1/\sqrt 3\).