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Section 7.6 Logistic Models

Subsection The Logistic Model

One of the most famous autonomous differential equations is the logistic model, which describes population growth under limited resources. The equation is:
\begin{equation*} \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right), \end{equation*}
where \(P\) is the population at time \(t\text{,}\) \(r\) is the growth rate, and \(K\) is the carrying capacity, the maximum population that the environment can sustain.
This equation has two equilibrium points:
  • \(P = 0\) (extinction),
  • \(P = K\) (population stabilizes at carrying capacity).
The phase line for this model shows:
  • At \(P = 0\text{,}\) the equilibrium is unstable; any small population increase leads to growth.
  • At \(P = K\text{,}\) the equilibrium is stable; populations near this value settle toward it.
  • When \(0 < P < K\text{,}\) populations increase.
  • When \(P > K\text{,}\) populations decrease.
This model captures how populations grow rapidly when small, slow down as they approach the carrying capacity, and eventually level off.

🌌 Example 107. Logistic Model Analysis.

Consider the logistic equation
\begin{equation*} \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)\text{.} \end{equation*}
  • Identify the equilibrium points.
  • Sketch the phase line and classify the stability of each equilibrium.
  • Explain how this model describes long-term population behavior.
Solution.
The equilibrium points are \(P = 0\) and \(P = K\text{.}\)
On the phase line:
  • \(P = 0\) is a source (unstable).
  • \(P = K\) is a sink (stable).
Populations grow when between \(0\) and \(K\text{,}\) and shrink when above \(K\text{.}\) The model predicts eventual stabilization at the carrying capacity.
The logistic model is sometimes written in a slightly different form:
\begin{equation*} \frac{dP}{dt} = kP\left(1 - \frac{P}{N}\right), \end{equation*}
where \(N\) is the carrying capacity and \(k\) is the growth-rate constant. This form emphasizes that growth is proportional to both the current population and the remaining resource capacity \((1 - P/N)\text{.}\)
This is still a first-order autonomous differential equation, but unlike the exponential model, it is nonlinear since the right-hand side is not a linear function of \(P\text{.}\)
The analysis is the same: the equilibria are \(P = 0\) and \(P = N\text{.}\) The phase line indicates their stability, as before.

Subsection A Modified Logistic Model

Let us now consider a variation of the logistic model that incorporates an important ecological detail: what happens to very small populations. The standard logistic model assumes that populations always grow when small, but in reality, sparse populations can have difficulty finding mates, making growth negative at low levels.
As an example, consider pine squirrels in the Rocky Mountains. These animals are territorial, and both overcrowding and underpopulation can limit their growth:
  • If the population is too large, territorial conflict slows growth or even causes decline.
  • If the population is too small, individuals may not find mates, also resulting in decline.
  • If the population is exactly zero, it stays at zero forever.
Let’s model this situation. We define:
  • \(t\) = time,
  • \(S(t)\) = squirrel population at time \(t\text{,}\)
  • \(k\) = growth-rate coefficient,
  • \(N\) = carrying capacity (too large),
  • \(M\) = sparsity threshold (too small).
We want the growth rate \(g(S)\) to satisfy the following:
  • \(g(S) < 0\) if \(S < M\) or \(S > N\text{,}\)
  • \(g(S) > 0\) if \(M < S < N\text{,}\)
  • \(g(0) = 0\text{.}\)
A function with this behavior might look like:
\begin{equation*} \frac{dS}{dt} = kS\left(1 - \frac{S}{N}\right)\left(\frac{S}{M} - 1\right). \end{equation*}
This is the modified logistic model. It adjusts the standard model to account for low-population risks.
Let’s analyze it. The equilibrium points occur where the right-hand side equals zero:
  • \(S = 0\) (extinction),
  • \(S = M\) (sparsity threshold),
  • \(S = N\) (carrying capacity).
The phase line tells us:
  • \(S = 0\) is a stable equilibrium (a sink).
  • \(S = M\) is an unstable equilibrium (a source).
  • \(S = N\) is a stable equilibrium (a sink).
This model better captures real-world behavior by showing that small populations might go extinct if they fall below the sparsity threshold.

Checkpoint 108. Modified Logistic Model Analysis.

Consider the differential equation:
\begin{equation*} \frac{dS}{dt} = kS\left(1 - \frac{S}{N}\right)\left(\frac{S}{M} - 1\right). \end{equation*}
  • Find all equilibrium solutions.
  • Sketch the phase line and label the stability of each equilibrium.
  • Explain how this model improves on the standard logistic model.
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