Lesson 2: Population Models: The Logistic Equation
Estimated time: 35-45 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Explain why the exponential model is inadequate for large populations
- State and interpret the logistic equation dP/dt = rP(1 - P/K)
- Find equilibrium solutions and determine their stability
- Solve the logistic equation using partial fractions
- Analyze logistic models with harvesting
From Exponential to Logistic
The exponential model dP/dt = rP assumes unlimited resources. To model a finite environment, we introduce a braking factor that slows growth as P approaches the environment's capacity.
Logistic Equation: dP/dt = rP(1 - P/K), where r is the intrinsic growth rate and K is the carrying capacity -- the maximum sustainable population.
When P is small relative to K, the factor (1 - P/K) ≈ 1 and growth is nearly exponential. As P approaches K, the factor approaches 0 and growth slows to a halt.
Equilibrium Analysis
Set dP/dt = 0: rP(1 - P/K) = 0 gives two equilibria:
- P = 0 (extinction) -- unstable equilibrium
- P = K (carrying capacity) -- stable equilibrium
Stability Analysis: For 0 < P < K: dP/dt > 0 (population increasing). For P > K: dP/dt < 0 (population decreasing). So P = K attracts nearby solutions from both sides -- it is asymptotically stable.
The logistic curve has a characteristic S-shape (sigmoid). The inflection point occurs at P = K/2, where growth rate dP/dt is maximized.
Solving the Logistic Equation
The logistic equation is separable. We separate and use partial fractions:
dP / [P(1 - P/K)] = r dt
Partial fractions: 1/[P(1 - P/K)] = 1/P + (1/K)/(1 - P/K)
Example 1: Solving the Logistic Equation
Solve dP/dt = 0.5P(1 - P/1000) with P(0) = 100.
Step 1: Here r = 0.5 and K = 1000. The general logistic solution is:
P(t) = K / (1 + Ae-rt) where A = (K - P0)/P0.
Step 2: A = (1000 - 100)/100 = 9.
Step 3: P(t) = 1000 / (1 + 9e-0.5t).
Check: P(0) = 1000/(1 + 9) = 100. As t → ∞, e-0.5t → 0, so P → 1000 = K.
Example 2: Finding the Inflection Point
For the solution above, when does the population grow fastest?
Solution: Maximum growth occurs at P = K/2 = 500.
500 = 1000/(1 + 9e-0.5t) ⇒ 1 + 9e-0.5t = 2 ⇒ e-0.5t = 1/9 ⇒ t = -2 ln(1/9) = 2 ln 9 ≈ 4.39 time units.
Logistic Equation with Harvesting
If a population is harvested at a constant rate h, the model becomes:
dP/dt = rP(1 - P/K) - h
Critical Harvesting Rate: The maximum sustainable harvest rate is hmax = rK/4 (the peak of the growth curve at P = K/2). If h > rK/4, the population will collapse to extinction.
Example 3: Harvesting Analysis
A fish population satisfies dP/dt = 0.4P(1 - P/5000) - h. Find the maximum sustainable harvest rate.
Solution: hmax = rK/4 = 0.4(5000)/4 = 500 fish per unit time.
At h = 500, the only equilibrium is P = K/2 = 2500 (a semi-stable equilibrium). For h > 500, no positive equilibria exist and the population goes extinct.
Comparing Exponential and Logistic
| Feature | Exponential | Logistic |
|---|---|---|
| Equation | dP/dt = rP | dP/dt = rP(1 - P/K) |
| Long-term behavior | P → ∞ | P → K |
| Growth shape | J-curve | S-curve (sigmoid) |
| Resource assumption | Unlimited | Limited (capacity K) |
| Best for | Short-term, small P | Long-term predictions |
Check Your Understanding
1. For dP/dt = 0.3P(1 - P/2000), what are the equilibria? Which is stable?
2. A population of 200 follows dP/dt = 0.5P(1 - P/1000). Write the explicit solution.
3. At what population level is the logistic growth rate maximized?
Key Takeaways
- The logistic equation dP/dt = rP(1 - P/K) models resource-limited growth with carrying capacity K.
- Two equilibria: P = 0 (unstable) and P = K (stable). Solutions form an S-shaped curve.
- The explicit solution is P(t) = K/(1 + Ae-rt) with A = (K - P0)/P0.
- Maximum growth rate occurs at P = K/2. Maximum sustainable harvest is rK/4.
- Harvesting above h = rK/4 causes population collapse.
Ready for More?
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In Lesson 3, you will learn to set up and solve mixing problems using first-order linear ODEs.
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