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Lesson 94 — Population models: Malthus and Verhulst

Exponential growth (Malthus) and logistic growth (Verhulst). Equilibria, stability, inflection at K/2.

Used in: Spécialité Maths (France, Terminale) · AP Calculus BC (EUA) · Leistungskurs alemão

\dot P = rP\!\left(1 - \frac{P}{K}\right) \;\Longrightarrow\; P(t) = \frac{K}{1 + \left(\frac{K - P_0}{P_0}\right)e^{-rt}}

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Rigorous notation, full derivation, hypotheses

Malthus, Verhulst and equilibrium analysis

Malthus Model (1798)

"If the rate of change of the population is proportional to the population itself, we get the Malthusian model." — Lebl, Notes on Diffy Qs, §1.8

Logistic Model (Verhulst, 1838)

"The logistic equation is another separable equation... The assumption is that the rate of growth of the population is proportional to the current population, but decreases as the population approaches the carrying capacity." — OpenStax Calculus Volume 2, §4.4

Closed-form solution

Via partial fractions:

$P(t) = \frac{K}{1 + \left(\dfrac{K - P_0}{P_0}\right)e^{-rt}}$

Equilibrium analysis

Phase diagram

1D phase diagram: arrows indicate the direction of variation of $P$ . $P = 0$ repels; $P = K$ attracts.

Solved examples

Example— 1· Malthusian growth with doubling

Problem. A bacterial colony doubles every 20 minutes. How many bacteria are there after 2 hours if $P_0 = 1000$ ?

Strategy. Malthus model: $P(t) = P_0 e^{rt}$ . Calculate $r$ using the doubling period $T_2 = 20$ min.

Solution.

$P(20) = 2P_0 \Rightarrow e^{20r} = 2 \Rightarrow r = \ln 2 / 20 \approx 0.0347\,\text{min}^{-1}$ .

At $t = 120$ min: $P(120) = 1000 \cdot e^{120 \cdot \ln 2/20} = 1000 \cdot 2^6 = 64,000$ .

Verification. In 2 hours = 6 doubling periods: $1000 \times 2^6 = 64,000$ . Correct.

Source. Lebl, Notes on Diffy Qs, §1.3, Example 1.3.1 — CC-BY-SA.

Example— 2· Logistic model with carrying capacity

Problem. A deer population in a reserve has $K = 1200$ , $r = 0.4$ /year, $P(0) = 200$ . Find $P(t)$ and calculate when $P = 900$ .

Strategy. Closed-form logistic formula. For $P = 900$ : solve for $t$ .

Solution.

$A = (K - P_0)/P_0 = (1200 - 200)/200 = 5$ .

$P(t) = \frac{1200}{1 + 5e^{-0.4t}}$

For $P = 900$ : $900(1 + 5e^{-0.4t}) = 1200 \Rightarrow 1 + 5e^{-0.4t} = 4/3 \Rightarrow e^{-0.4t} = 1/15$ .

$t = \ln(15)/0.4 \approx 2.71/0.4 \approx 6.8$ years.

Verification. $P(6.8) = 1200/(1 + 5e^{-2.71}) \approx 1200/(1 + 5 \times 0.0667) \approx 1200/1.333 \approx 900$ . Correct.

Source. OpenStax Calculus Volume 2, §4.4, Example 4.14 — CC-BY-NC-SA.

Example— 3· Derivation of the logistic solution by partial fractions

Problem. Derive the closed-form logistic formula from $\dot P = rP(1 - P/K)$ , $P(0) = P_0$ .

Strategy. Separate variables and decompose into partial fractions.

Solution.

$\frac{dP}{P(1-P/K)} = r\,dt \;\Rightarrow\; \frac{K\,dP}{P(K-P)} = r\,dt$

Partial fractions: $\dfrac{K}{P(K-P)} = \dfrac{1}{P} + \dfrac{1}{K-P}$ .

Integrating: $\ln P - \ln(K-P) = rt + C_0 \Rightarrow \ln\!\left(\dfrac{P}{K-P}\right) = rt + C_0$ .

Let $A = e^{C_0}$ : $\dfrac{P}{K-P} = A e^{rt}$ .

Solving: $P = Ae^{rt}(K-P) \Rightarrow P(1 + Ae^{rt}) = KAe^{rt} \Rightarrow P = \dfrac{KAe^{rt}}{1+Ae^{rt}} = \dfrac{K}{1 + A^{-1}e^{-rt}}$ .

For $t = 0$ : $P_0 = K/(1+1/A) \Rightarrow A = (K-P_0)/P_0$ .

$P(t) = \frac{K}{1 + \frac{K-P_0}{P_0}\,e^{-rt}}$

Verification. $P(0) = K/(1+(K-P_0)/P_0) = KP_0/K = P_0$ . $\lim_{t\to\infty} P(t) = K$ . Correct.

Source. Lebl, Notes on Diffy Qs, §1.7, Example 1.7.1 — CC-BY-SA.

Example— 4· Inflection point and maximum sustainable yield

Problem. For the logistic model with $r = 0.3$ /year, $K = 500$ , $P_0 = 50$ : (a) find the inflection time $t^*$ ; (b) calculate the maximum sustainable yield (MSY).

Strategy. Inflection when $P = K/2 = 250$ . MSY = maximum value of $\dot P$ .

Solution.

(a) With $A = (500-50)/50 = 9$ : $P(t^*) = 250 \Rightarrow 500/(1+9e^{-0.3t^*}) = 250 \Rightarrow 9e^{-0.3t^*} = 1 \Rightarrow t^* = \ln(9)/0.3 \approx 7.3$ years.

(b) $\dot P_{\max} = rK/4 = 0.3 \times 500/4 = 37.5$ individuals/year (occurs at $P = 250$ ).

Verification. $\dot P(250) = 0.3 \times 250 \times (1-250/500) = 75 \times 0.5 = 37.5$ . Correct.

Source. OpenStax Calculus Volume 2, §4.4, Exercise 186 — CC-BY-NC-SA.

Example— 5· Comparison Malthus vs logistic on IBGE data

Problem. Brazil: $P_0 = 95$ M (1970), $r = 2.5\%$ /year. Using $K = 250$ M, compare the Malthusian prediction with the logistic one for 2024 ( $t = 54$ years). Actual data: 214 M.

Strategy. Calculate both formulas for $t = 54$ .

Solution.

Malthus: $P(54) = 95 \cdot e^{0.025 \times 54} = 95 \cdot e^{1.35} \approx 95 \times 3.857 \approx 366$ M.

Verhulst: $A = (250-95)/95 \approx 1.632$ . $P(54) = 250/(1 + 1.632\,e^{-1.35}) = 250/(1 + 1.632 \times 0.259) \approx 250/1.423 \approx 176$ M.

Relative error: Malthus +71%; Verhulst +18%.

Observation. Neither model captures migration, urbanization, or the decline in fertility rates. The logistic is better, but $r$ changed over time.

Source. Lebl, Notes on Diffy Qs, §1.7, Exercise 1.7.4 — CC-BY-SA.

Exercise list

23 exercises · 5 with worked solution (25%)

Application 10Understanding 3Modeling 5Challenge 3Proof 2

Ex. 94.1ApplicationAnswer key
Solve $\dot P = 0.03P$ , $P(0) = 500$ .
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Ex. 94.2Application
Bacterial colony starts with 500, doubles every 30 min. How many bacteria after 3 hours? Find $r$ .
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Ex. 94.3Application
Write the logistic solution for $r = 0.2$ , $K = 5000$ , $P_0 = 200$ .
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Ex. 94.4Application
For the logistic model from the previous exercise ( $K = 5000$ , $r = 0.2$ , $P_0 = 200$ ): when does the inflection occur?
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Ex. 94.5Application
For the logistic model with $r = 0.2$ , $K = 5000$ : identify the equilibria and calculate the maximum sustainable yield (MSY).
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Ex. 94.6Application
Endangered species: $\dot P = -0.015 P$ . Calculate the half-life of the population.
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Ex. 94.7Application
Logistic: $K = 8000$ , $r = 0.3$ , $P(0) = 1000$ . Calculate $P(5)$ .
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Ex. 94.8Application
Logistic: $K = 1000$ , $r = 0.5$ , $P(0) = 100$ . Calculate $P(8)$ .
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Ex. 94.9Application
Determine $r$ knowing $P(0) = 100$ , $P(5) = 300$ , $K = 1000$ .
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Ex. 94.10Application
Carbon-14 has a half-life of 5730 years. A sample retains 70% of the original carbon. How old is it?
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Ex. 94.11Understanding
What is the maximum growth rate $\dot P_{\max}$ of the logistic equation $\dot P = rP(1-P/K)$ ?
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Ex. 94.12Understanding
For the logistic model with $r, K > 0$ : which values of $P_0$ lead to $P(t) \to K$ ?
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Ex. 94.13Modeling
Deer reserve: $K = 1200$ , $r = 0.4$ /year. What is the maximum sustainable annual harvest? At what population level should the herd be maintained?
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Ex. 94.14Modeling
World population: $P_0 = 6$ billion (year 2000), $r = 1.2\%$ /year, $K = 10$ billion. Predict the population for 2050 using the logistic model.
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Ex. 94.15ModelingAnswer key
Logistic with constant harvest: $\dot P = 0.3P(1-P/1500) - 50$ . Find the equilibria and their stability.
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Ex. 94.16ModelingAnswer key
Product diffusion: market of 50,000 customers, 500 in the first month, $r = 0.6$ /month. When have 90% of the market adopted?
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Ex. 94.17ModelingAnswer key
At the start of an epidemic ( $I$ small, $S \approx N$ ), show that $\dot I \approx (\beta N - \gamma)I$ . For $\beta = 0.3$ , $\gamma = 0.1$ , $N = 1000$ : is there an epidemic?
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Ex. 94.18Understanding
Gompertz model: $\dot P = rP\ln(K/P)$ . Compare the position of the inflection point with the logistic model.
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Ex. 94.19ChallengeAnswer key
Logistic with harvest: $\dot P = 0.4P(1-P/1200) - H$ . For what value of $H$ does no positive equilibrium exist? What happens to the population in this case?
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Ex. 94.20Challenge
Allee effect: $\dot P = rP(P/A - 1)(1-P/K)$ with $0 < A < K$ . Find the equilibria and classify them. What happens if $P_0 < A$ ?
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Ex. 94.21Challenge
Lotka-Volterra: $\dot x = 2x - xy$ , $\dot y = -y + xy$ . Find the equilibria and show that the trajectories satisfy $y - \ln y + x - 2\ln x = C$ .
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Ex. 94.22Proof
Demonstrate that the logistic solution $P(t)$ has an inflection point exactly at $P = K/2$ .
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Ex. 94.23Proof
Demonstrate via linearization that $P^* = K$ is a stable equilibrium and $P^* = 0$ is unstable for the logistic equation with $r, K > 0$ .
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Sources

Notes on Diffy Qs — Jiří Lebl · v6.6 · §1.3, §1.7–1.8 · EN · CC-BY-SA. Primary source.
Calculus Volume 2 — OpenStax · §4.4 · EN · CC-BY-NC-SA.
Elementary Differential Equations — William F. Trench · §1.3 · EN · open.