Limits to population growth

Exponential and stage-structured population growth

Exponential growth

\[\frac{dN}{dt} = rN\]

Stage-structured growth

\[N_{t+1} = \text{[transition matrix]} ~x~ \\ \text{[population vector]}\]

Exponential and stage-structured population growth

Exponential growth

\[\frac{dN}{dt} = rN\]

Stage-structured growth

\[N_{t+1} = \text{[transition matrix]} ~x~ \\ \text{[population vector]}\]

When is population at equilibrium?

(But first, what is “Equilibrium” in ecology?)

In ecological systems, “equilibrium” means no change in population size (or other relevant aspect of a system) over time

Under exponential population is at equilibrium only when \(N = 0\)

Populations with \(r = 0\) or \(\lambda = 1\) are also at equilibrium (no net births or deaths)

No immigration or emigration (Closed population)
Constant birth and death rate (\(b\) and \(d\) don’t vary with \(N\))
No variation within population (all individuals have similar \(b\) and \(d\)) (This implies that \(b\) and \(d\) don’t vary with age or stage)
Continuous population growth without time lags (e.g. no seasonality)

No immigration or emigration (Closed population)
Constant birth and death rate (\(b\) and \(d\) don’t vary with \(N\))
No variation within population (all individuals have similar \(b\) and \(d\)) (This implies that \(b\) and \(d\) don’t vary with age or stage)
We relaxed this for stage-structured growth
Continuous population growth without time lags
(e.g. no seasonality)

No immigration or emigration (Closed population)
Constant birth and death rate (\(b\) and \(d\) don’t vary with \(N\))
Today, we will relax this assumption? –> Density-dependent growth
No variation within population (all individuals have similar \(b\) and \(d\)) (This implies that \(b\) and \(d\) don’t vary with age or stage)
Continuous population growth without time lags
(e.g. no seasonality)

So far, we assumed constant birth and death rate (\(b\) and \(d\) don’t vary with \(N\))

So far, we assumed constant birth and death rate (\(b\) and \(d\) don’t vary with \(N\))

So far, we assumed constant birth and death rate (\(b\) and \(d\) don’t vary with \(N\))

We can easily imagine scenarios in which birth and death rates vary with population size.

In social species (bees, ants, wolves, some woodpeckers, etc.), birth rates (\(b\)) might increase with population size

We can easily imagine scenarios in which birth and death rates vary with population size.

Under strict resource limitation, mortality rate might increase with population size
- Could also happen without resource limitation, e.g. higher rates of disease spread

Under this assumption (\(d\) increases with \(N\)), we see that there is a point at which \(\text{growth rate} = 0\).

The population size (\(N\)) at which growth rate equals zero is called the carrying capacity (\(K\))

The population size (\(N\)) at which growth rate equals zero is called the carrying capacity (\(K\))
The population’s growth rate at \(N = 0\) is \(r\)

The per-capita population growth rate (\(\frac{1}{N}\frac{dN}{dt}\)) for a population of a given size is the Y-axis value.
- The closer a population is to \(N = 0\), the closer its actual growth rate is to \(r\)
- The closer a population is to \(N = K\), the closer its actual growth rate is to \(0\)
- Populations that are bigger than K have negative population growth (until they fall back to carrying capacity)

What is the equation of the purple line?

Recall from algebra that we can use the “point-slope” approach for the slope of a line (\(y-y_1 = m(x - x_1)\))
We know two points: \((0,r)\) and \((K,0)\)
The slope of the line turns out to be \(-\frac{r}{K}\), which we write as \(\alpha\)
And \(y\) gives the per-capita population growth rate \(\frac{1}{N}\frac{dN}{dt}\)

What is the equation of the purple line?

So, we can express the line as \[\frac{1}{N}\frac{dN}{dt} = r - \alpha N\]
Through rearranging, we get back to the equation for “realized” population growth rate (not per-capita): \[\frac{dN}{dT} = (r - \alpha N)N\]
This is equivalent to \[\frac{dN}{dT} = rN(1-\alpha N)\]
On your own: Convince yourself through algebra that this is equal to the canonical form of the equation: \[\frac{dN}{dt} = rN \bigg(1-\frac{N}{K}\bigg)\]

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

When population size is very small (\(N \to 0\)), \(\frac{N}{K} \to 0\), and the population grows ‘exponentially’
When population size is close to carrying capacity (\(N \to K\)), \(\frac{N}{K} \to 1\), and the population remains constant
When population size exceeds carrying capacity (\(N > K\)), \(\frac{N}{K} > 1\), and the population growth rate is negative (shrinks)
“Logistic growth” model

The population will converge to the carrying capacity no matter the initial size