
Exponential growth
\[\frac{dN}{dt} = rN\]
Logistic growth growth
\[\frac{dN}{dt} = rN\bigg(1-\frac{N}{K}\bigg)\]
Exponential growth
\[\frac{dN}{dt} = rN\]
\[\frac{1}{N}\frac{dN}{dt} = r\]
Logistic growth growth
\[\frac{dN}{dt} = rN\bigg(1-\frac{N}{K}\bigg)\]
\[\frac{1}{N}\frac{dN}{dt} = r\bigg(1-\frac{N}{K}\bigg)\]
\[\frac{dN}{dt} = rN \bigg(1-\frac{N}{K}\bigg)\]

Intuitively, we might say that the system is at equilibrium when it is at carrying capacity (\(K\))
What is special about this?
\[\frac{dN}{dt} = rN\bigg(1-\frac{K}{N}\bigg)\] When \(N = K\), \(\frac{dN}{dt} \to 0\).
This gets us back to the definition of equilibrium in ecological systems: no net change in the system (\(\frac{dN}{dt} = 0\))
\[\boxed{\frac{dN}{dt} = rN\bigg(1-\frac{K}{N}\bigg) = 0} \text{ also happens when } N = 0\]
Thus, this system has two equilibrium points:
\[N = 0 \text{ and } N = K\]
Just because a system is “at equilibrium” doesn’t mean it will never change
The dynamics depend on the stability of the equilibrium
An equilibrium is considered (locally) stable if small perturbations cause the population to return to its original state.
Alternatively, an equilibrium is unstable if small perturbations cause the system to move away from the original state
Consider a population that is at equilibrium because \(N = 0\)

A small perturbation causes \(N = 1\) (e.g. an immigration event)

This means the equilibrium \(N = 0\) is an unstable equilibrium
Consider a population that is at equilibrium because \(N = K\)

A small perturbation causes \(N\) to shift slightly lower than \(K\) (e.g. a hurricane that kills a fraction of the individuals)

Alternatively, a small perturbation causes \(N\) to shift slightly higher than \(K\) (e.g. humans release additional individuals into the system)

One of the implicit assumptions in the logistic growth model is that higher population density always results in reduced per-capita performance
We have discussed many reasons why this may not be true


\[\frac{dN}{dt} = - rN \bigg( 1-\frac{N}{T} \bigg) \bigg( 1-\frac{N}{K} \bigg)\]
\(T\) is a threshold value – when populations are below this, fitness increases with population size
\[\frac{dN}{dt} = - rN \bigg( 1-\frac{N}{T} \bigg) \bigg( 1-\frac{N}{K} \bigg)\]
This model has three equilibrium points:
\(N = 0\), \(N = T\), and \(N = K\)



\[\frac{dN}{dt} = - rN \bigg( 1-\frac{N}{T} \bigg) \bigg( 1-\frac{N}{K} \bigg)\]
This model has three equilibrium points:
\(N = 0\), \(N = T\), and \(N = K\)
What do we expect in terms of the stability for each of these equilibrium points?
\(N = 0 \to \text{stable equilibrium}\), \(N = T \to \text{unstable equilibrium}\), \(N = K \to \text{stable equilibrium}\)
\[\frac{dN}{dt} = - rN \bigg( 1-\frac{N}{T} \bigg) \bigg( 1-\frac{N}{K} \bigg)\]
This model has three equilibrium points:
\(N = 0\), \(N = T\), and \(N = K\)
What do we expect in terms of the stability for each of these equilibrium points?
\(\boxed{N = 0 \to \text{stable equilibrium}}\), \(N = T \to \text{unstable equilibrium}\), \(N = K \to \text{stable equilibrium}\)