Comparing Population Models
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 4.4: “The Logistic Equation” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/4-4-the-logistic-equation |
| Textbook used in class | Stewart, Calculus, Section 9.4: “Models for Population Growth” (Sections on exponential and logistic) |
Quick Reference
| Feature | Exponential ($dP/dt = kP$) | Logistic ($dP/dt = kP(1-P/M)$) |
|---|---|---|
| Equilibria | $P = 0$ (unstable) | $P = 0$ (unstable), $P = M$ (stable) |
| Long-run behavior | $P \to \infty$ (if $k > 0$) | $P \to M$ |
| Growth rate max | At $t = 0$ (always decreasing) | At $P = M/2$ (inflection point) |
| Realistic? | Only for short periods | Yes, with limited resources |
Phase line: a plot of $dP/dt$ vs $P$. Where $dP/dt > 0$, arrows point right (increasing); where $dP/dt < 0$, arrows point left (decreasing).
Motivation
Comparing the two main population models side-by-side clarifies when each is appropriate. For a bacterial colony in a nutrient-rich flask, exponential growth is accurate at first. As nutrients deplete, the logistic model takes over. For wildlife management or epidemiology, the carrying capacity is often the central quantity of interest. The phase line is the tool that lets you analyze any autonomous DE $dP/dt = f(P)$ without solving it, by reading off stability from the sign of $f(P)$.
Key Concept: Phase Line Analysis
For an autonomous equation $dP/dt = f(P)$ (right side depends only on $P$, not $t$):
- Find where $f(P) = 0$: these are equilibria (constant solutions).
- For $P$ values between equilibria, determine the sign of $f(P)$.
- Draw arrows: right (increasing) where $f(P) > 0$, left (decreasing) where $f(P) < 0$.
- An equilibrium is stable if nearby solutions flow toward it; unstable if they flow away.
For $dP/dt = kP(1 - P/M)$ with $k, M > 0$:
- $f(P) = 0$ at $P = 0$ and $P = M$.
- $f(P) > 0$ for $0 < P < M$ (arrows right toward $M$).
- $f(P) < 0$ for $P > M$ (arrows right but decreasing -- actually $P$ decreasing here, so arrows left toward $M$).
Conclusion: $P = 0$ is unstable, $P = M$ is stable.
Worked Example
A fish population obeys $dP/dt = 0.2P(1 - P/500) - H$, where $H$ is a constant harvest rate (fish per year). If $H = 20$, find the equilibria and determine their stability.
Setting $dP/dt = 0$: $0.2P(1 - P/500) = 20 \implies P(1 - P/500) = 100$.
Rearrange: $P - P^2/500 = 100 \implies P^2 - 500P + 50000 = 0$.
Quadratic formula: $P = (500 \pm \sqrt{250000 - 200000})/2 = (500 \pm \sqrt{50000})/2$.
$\sqrt{50000} = 100\sqrt{5} \approx 223.6$.
$P \approx (500 + 223.6)/2 \approx 361.8$ and $P \approx (500 - 223.6)/2 \approx 138.2$.
Phase line: near $P = 361.8$, the function $f$ changes from positive to negative, so it is a stable equilibrium. Near $P = 138.2$, $f$ changes from negative to positive, so it is unstable. If the population drops below about 138 fish, it will collapse to zero.
exponential growth with $k < 0$ (decay) is the same model as logistic decay back to zero. Exponential decay $dP/dt = kP$ with $k < 0$ always drives $P \to 0$; there is only one equilibrium ($P = 0$), and every solution decays to it. Logistic growth with $P > M$ also decreases, but toward $M$, not toward zero -- the equilibrium at $M$ is stable. The two models have different long-run behaviors: one drives everything to zero, the other drives everything to $M$.
Leveled Practice
Problem 1. Draw the phase line for $dP/dt = P(3 - P)$ and classify each equilibrium.
Show answer
Equilibria: $P = 0$ and $P = 3$.
$f(P) = P(3-P)$:
- $P < 0$: $P < 0$, $3-P > 0$, so $f < 0$ (arrows left).
- $0 < P < 3$: both factors positive, so $f > 0$ (arrows right).
- $P > 3$: $P > 0$, $3 - P < 0$, so $f < 0$ (arrows left).
$P = 0$: arrows from both sides point away (unstable). $P = 3$: arrows from both sides point toward it (stable).