The Logistic Growth Model
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You should be comfortable with basic calculus, specifically differentiating exponential functions and solving simple differential equations. You should also understand the concept of exponential growth and the definition of per-capita growth rate from previous units.
Why this matters
The logistic model corrects the unrealistic assumption of unlimited resources by introducing a density-dependent limit to population growth.
In nature, populations rarely grow exponentially forever. Resources such as food, space, and light are finite. As a population becomes denser, individuals compete more intensely, leading to a decline in the per-capita birth rate or an increase in the per-capita death rate. The logistic growth model captures this biological reality by positing that the growth rate slows as the population approaches a maximum sustainable size, known as the carrying capacity. This model serves as the foundational null hypothesis for understanding how populations regulate themselves in stable environments.
Intuitive Understanding of Density Dependence
Imagine a small population of bacteria introduced into a fresh petri dish. Initially, resources are abundant relative to the number of individuals. Each bacterium has ample access to nutrients, so the population grows rapidly, doubling at a constant rate. This is the exponential phase. However, as the bacterial count increases, the dish becomes crowded. Nutrients are depleted faster than they can be replenished, and waste products accumulate. The environment becomes less hospitable. In this crowded state, not every individual can reproduce successfully, and some may die due to starvation or toxicity. The key insight of the logistic model is that this reduction in growth is proportional to how close the population is to its maximum capacity. If the population is at half the capacity, the per-capita growth rate is half of what it would be at low densities. If it is at capacity, the growth rate is zero. This creates a self-regulating system where the population naturally stabilizes around a specific level determined by the environment's productivity and resource limits.
The Differential Equation Formulation
The mathematical expression of this biological intuition is the logistic differential equation. We define the per-capita growth rate, \( \frac{1}{N}\frac{dN}{dt} \), as a linear function of population size \( N \). Specifically, the per-capita rate is \( r \) when \( N \) is near zero and decreases to zero as \( N \) approaches \( K \). This relationship is written as \( \frac{1}{N}\frac{dN}{dt} = r\left(1 - \frac{N}{K}\right) \). Multiplying both sides by \( N \) yields the standard form \( \frac{dN}{dt} = rN\left(1 - \frac{N}{K}\right) \). Here, \( r \) represents the intrinsic rate of increase, which is the maximum per-capita growth rate observed at very low densities. The term \( \left(1 - \frac{N}{K}\right) \) acts as a braking mechanism. When \( N \) is small, this term is close to 1, and the equation approximates exponential growth. As \( N \) increases, the term becomes smaller, reducing the overall growth rate. When \( N = K \), the term becomes zero, and the population size remains constant. This formulation assumes that the response to density is instantaneous and that the environment is closed and well-mixed.
Analysis of the Growth Curve
The solution to the logistic equation, \( N(t) = \frac{K}{1 + \left(\frac{K-N_0}{N_0}\right)e^{-rt}} \), describes a sigmoidal or S-shaped curve. The trajectory of the population depends on the initial size \( N_0 \). If \( N_0 < K \), the population increases monotonically towards \( K \). If \( N_0 > K \), the population decreases monotonically towards \( K \). The state \( N = K \) is globally asymptotically stable, meaning that any small perturbation away from \( K \) will result in the population returning to \( K \) over time. The rate at which this convergence occurs is governed by \( r \). A larger \( r \) implies a faster approach to equilibrium. The inflection point of the curve, where the growth rate \( \frac{dN}{dt} \) is maximized, occurs at \( N = K/2 \). At this point, the population is growing at its absolute fastest rate, \( \frac{rK}{4} \). Before this point, the growth rate is accelerating; after this point, the growth rate is decelerating due to the increasing strength of density-dependent constraints.
Stability and Equilibrium States
The logistic model identifies two equilibrium states: \( N = 0 \) and \( N = K \). To determine their stability, we examine the sign of \( \frac{dN}{dt} \) near these values. Near \( N = 0 \), if the population is slightly positive, \( \frac{dN}{dt} \) is positive, causing the population to grow away from zero. Thus, \( N = 0 \) is an unstable equilibrium. Any small introduction of individuals will lead to population growth. In contrast, near \( N = K \), if the population is slightly below \( K \), \( \frac{dN}{dt} \) is positive, pushing the population up. If the population is slightly above \( K \), \( \frac{dN}{dt} \) is negative, pushing the population down. This restoring force makes \( N = K \) a stable equilibrium. The linear relaxation rate around \( K \) is \( r \), indicating that the time scale for the population to adjust to changes in \( K \) is approximately \( 1/r \). This stability analysis is crucial for predicting how populations respond to environmental changes or harvesting pressures.
The concept
Logistic population growth
Let \(N(t)\) be the size of a single closed, unstructured, well-mixed population, treated as a differentiable function of continuous time, and suppose that the per-capita growth rate declines affinely with density, \(N^{-1}\,dN/dt=r\left(1-N/K\right)\), with the intrinsic rate of increase \(r\gt 0\) and the carrying capacity \(K\gt 0\) both constant, the response to density instantaneous, and no immigration, emigration, demographic stochasticity or Allee effect. Then \(N\) obeys the logistic equation \(dN/dt=rN\left(1-N/K\right)\), whose solution through any \(N_0\gt 0\) exists, is unique, and is \(N(t)=K\left[1+\left(\left(K-N_0\right)/N_0\right)e^{-rt}\right]^{-1}\). The state \(N=K\) is globally asymptotically stable on \(\left(0,\infty\right)\) with linear relaxation rate \(r\), the state \(N=0\) is unstable with rate \(+r\); for \(0\lt N_0\lt K\) the trajectory is strictly increasing, and for \(0\lt N_0\lt K/2\) it is sigmoid, with a single inflection at \(N=K/2\), at which the absolute growth rate attains its maximum \(rK/4\) — a trajectory started at or above \(K/2\) is concave throughout and has no inflection; and the logit \(\ln\left[N/\left(K-N\right)\right]\) is exactly linear in \(t\) with slope \(r\).
A common misconception is that the logistic model implies that the per-capita growth rate is constant. In reality, the per-capita rate \( r(1 - N/K) \) changes continuously as \( N \) changes. Another subtle point is that the model assumes a linear relationship between density and per-capita growth. In many real systems, this relationship may be non-linear, or there may be a threshold (Allee effect) below which the population cannot sustain itself. The logistic model is a simplification that works best for populations with strong, immediate density dependence and no time lags in response to crowding. C
Worked example
Problem. A fish stock follows logistic growth with \(r = 0.40\ \mathrm{yr^{-1}}\) and \(K = 10\,000\) tonnes. (a) What is the largest harvest that can be taken every year indefinitely, and at what stock size? (b) After overfishing, the stock is down to 1000 t and fishing stops. How long until it recovers to the size that gives that maximum yield?
- (a) Surplus production is \(dN/dt = rN(1 - N/K)\); it is largest at \(N = K/2 = 5000\ \mathrm{t}\).
- The maximum sustainable yield is therefore \(rK/4 = 0.40 \times 10\,000/4 = 1000\ \mathrm{t\,yr^{-1}}\): harvesting exactly this at \(N = 5000\) t keeps the stock steady.
- (b) The logistic solution gives the time to grow from \(N_0\) to \(N\): \(t = \dfrac1r \ln\dfrac{N(K - N_0)}{N_0(K - N)}\).
- With \(N_0 = 1000\) and \(N = 5000\): \(t = \dfrac{1}{0.40}\ln\dfrac{5000 \times 9000}{1000 \times 5000} = 2.5\ln 9\).
- So \(t = 2.5 \times 2.197 = 5.5\) years without any fishing, which is why stocks pushed far below \(K/2\) take years to rebuild.
Units check. \(rK\) is (yr-1)(t) = t yr-1; the logarithm is dimensionless and \(1/r\) is in years.
Three quick tests
- What is the value of the per-capita growth rate when \( N = K/2 \)?
Answer
The per-capita growth rate is \( r(1 - 1/2) = r/2 \). - If \( r = 0.1 \ \mathrm{yr}^{-1} \) and \( K = 1000 \), what is the time constant for relaxation to equilibrium?
Answer
The time constant is \( 1/r = 10 \ \mathrm{yr} \). - Does the logistic model predict that a population will ever exactly reach \( K \) in finite time?
Answer
No, it approaches \( K \) asymptotically as \( t \to \infty \).
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Lecture notes checked line by line by the editor; every number in the worked example was recomputed.