Community Ecology

Nature of species interactions

Sp 1 effect
on Sp 2		 Sp 2 effect
on Sp 1		 Shorthand
Benefit (+) Benefit (+) Mutualism
Harm (-) Harm (-) Competition
Benefit (+) Harm (-) Predation
Herbivory
Parasitism
Neutral (0) Benefit (+) Commensalism
Neutral (0) Harm (-) Ammensalism

Population dynamics of two competing species

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

\[\frac{dN_2}{dT} = r_2N_2(1-\alpha_{21}N_1 - \alpha_{22}N_2)\]

\(\alpha_{11}\) and \(\alpha_{22}\): competitive effect of each species on itself (e.g. how strongly do ospreys compete with one another, and how strongly do eagles compete with one another?)

\(\alpha_{12}\) and \(\alpha_{21}\): competitive effect of each species on the other (e.g. how strongly do ospreys compete with eagles, and how strongly do eagles compete with ospreys?)

Population dynamics of two competing species

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

What happens if species 1 is growing alone?

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1)\]

Growth to Species 1’s carrying capacity (\(\frac{1}{\alpha_{11}} = K\))

Similarly, if Species 2 is growing alone, it will grow to its carrying capacity \(\frac{1}{\alpha_{22}} = K_2\)

But what happens if both species are present in the system?

Possible outcomes of two species competing:

  • Both species can have stable coexistence
  • Species 1 can win, and exclude Species 2
  • Species 2 can win, and exclude Species 1
  • “It depends” – whichever species comes first, wins in competition.

How to predict the outcome for any given pair of species?

  • Or, given that species coexistence is something we are trying to explain – what enables coexistence?

Population dynamics of two competing species

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

\[\frac{dN_2}{dT} = r_2N_2(1-\alpha_{21}N_1 - \alpha_{22}N_2)\]

Under what conditions do both species co-exist?

Signature of stable coexistence:

  • System is at equilibrium
  • Both species are present (abundance > 0)

Under what conditions do both species co-exist?

Approach: Graphical analysis of the competition model

  • New type of visualization: phase space (AKA “state space”)
  • New type of analysis: null-cline analysis
    (AKA “zero net-growth isocline”)

Community ecology
Coexistence of competing species

Population dynamics of two competing species

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

\[\frac{dN_2}{dT} = r_2N_2(1-\alpha_{21}N_1 - \alpha_{22}N_2)\]

Possible outcomes of two species competing:

  • Both species can have stable coexistence
  • Species 1 can win, and exclude Species 2
  • Species 2 can win, and exclude Species 1
  • “It depends” – whichever species comes first, wins in competition.

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

\[\frac{dN_2}{dT} = r_2N_2(1-\alpha_{21}N_1 - \alpha_{22}N_2)\]

Given that we frequently see similar species coexisting, but are now in an era of biodiversity declines, an important question is: under what conditions do both species co-exist?

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

\[\frac{dN_2}{dT} = r_2N_2(1-\alpha_{21}N_1 - \alpha_{22}N_2)\]

Signature of stable coexistence:

  • System is at equilibrium (\(\frac{dN_1}{dt} = 0\) and \(\frac{dN_2}{dt} = 0\))

  • Both species are present (abundance > 0)

When are the conditions above satisfied?

Evaluating coexistence through phase space analysis

  • Graph with each axis representing a state variable

  • Any point on the graph represents a possible state of the system

  • Lines on the graph show how a system changes through time (trajectory)

Null-cline analyses (AKA zero net growth isocline analysis)

  • At what points in the state space does the system not change? (i.e. is at equilibrium)

  • Analyzed one axis at at time

  • Key questions:
    At what points does the abundance of species 1 (\(N_1\)) not change?
    At what points does the abundance of species 2 (\(N_2\)) not change?

At what points does the abundance of species 1 (N1) not change?

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

Two “extreme” cases…

  • When species 1 is at its carrying capacity, and species 2 is not around
    • \(N_1 = \frac{1}{\alpha_{11}}, N_2 = 0\)
  • When there are so many individuals of Species 2 that Species 1 cannot begin to grow
    • When does that happen?
    • \(N_1\) is low (0), and \(N_2\) is… some high number

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

Solve for \(\frac{dN_1}{dT} = 0, N_1 = 0, N_2 > 0\)

At what points does the abundance of species 1 (N1) not change?

  • When species 2 is absent, and species 1 is at its carrying capacity
    • \(N_2 = 0, N_1 = \frac{1}{\alpha_{11}}\)
  • When there are so many individuals of Species 2 that Species 1 cannot begin to grow
    • When does that happen?
  • \(N_1 = 0, N_2 = \frac{1}{\alpha_{12}}\)

Summary of the null-cline analysis so far:

  • We set out to find cases where \(N_1\) doesn’t change, i.e. \(dN_1/dt = 0\)
  • We identified two extreme scenarios:
    • \(N_1 = 1/\alpha_{11}, N_2 = 0\)
    • \(N_1 = 0, N_2 = 1/\alpha_{12}\)
  • We can put these two extreme points on the state space graph.

Draw state space with Species 1 equilibrium points

Growth of species 1 is also zero for intermediate combinations between these extremes

  • These intermediate combinations are defined by the line connecting the two extremes.

  • We can solve for the equation of this line.

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

\[N_2^* = \frac{1-\alpha_{11}N1}{\alpha_{12}} \]

\[N_2^* = \frac{1-\alpha_{11}N1}{\alpha_{12}} = \frac{1}{\alpha_{12}} + \frac{\alpha_{11}}{\alpha_{12}}N_1\]

Growth of species 1 is also zero for intermediate combinations between these extremes

\[\frac{dN_1}{dT} = r_1N_1(1-\alpha_{11}N_1 - \alpha_{12}N_2)\]

\[N_2^* = \overbrace{\frac{1}{\alpha_{12}}}^{\text{y-intercept}} - \overbrace{\frac{\alpha_{11}}{\alpha_{12}}}^{\text{slope}}N_1\]

This is the equation of the null-cline for species 1 (AKA zero net-growth isocline, or ZNGI)

Draw state space with Species 1 equilibrium points, plus intermediate line

  • What happens on either side of the null-cline?

Recall the key questions of null-cline analysis:

  • Key questions:
    At what points does the abundance of species 1 (\(N_1\)) not change?
    • \(N_1 = 1/\alpha_{11}, N_2 = 0\);
      \(N_1 = 0, N_2 = 1/\alpha_{12}\);
      \(N_2^ = \frac{1}{\alpha_{12}} - \frac{\alpha_{11}}{\alpha_{12}}N_1\)
  • At what points does the abundance of species 2 (\(N_2\)) not change?

At what points does the abundance of species 2 (N2) not change?

  • When species 1 is not around, and species 2 is at its carrying capacity
    • \(N_1 = 0, N_2 = 1/\alpha_{22}\)
  • When there are so many individuals of Species 1 that Species 2 cannot begin to grow
    • \(N_1\) is some high number… and \(N_2\) is low (0)
    • Following algebra, \(N_1 = 1/\alpha_{21}, N_2 = 0\)
  • We can add these two extremes to the state space plot

Growth of species 2 is also zero for intermediate combinations between these extremes

\[\frac{dN_2}{dT} = r_2N_2(1-\alpha_{21}N_1 - \alpha_{22}N_2)\]

\[N_2^* = \frac{1}{\alpha_{22}} - \frac{\alpha_{21}}{\alpha_{22}}N_1\]

This is the equation of the null-cline for species 2 (AKA zero net-growth isocline, or ZNGI)

Add null-cline to state space

What happens on either side of the nullcline?

Graphical analysis of the Lotka-Volterra competition model

Review:

  • Our goal is to identify conditions under which both species can coexist at equilibrium
  • Being ‘at equilibrium’ means \(dN_1/dt = dN_2/dt = 0\)
  • Null-cline analysis lets us find the conditions at which \(dN_1/dt = 0\), and the conditions at which \(dN_2/dt\)
  • There are a couple of ‘extreme’ cases (e.g. one species at carrying capacity, and the other absent), and a whole bunch of in-between cases that result in \(dN_1/dt = 0\) or \(dN_2/dt = 0\)

Test your recollection

  • Consider a pair of species that interact with the following strength:
    • \(\alpha_{11} = 0.01\), \(\alpha_{12} = 0.005\), \(\alpha_{22} = 0.02\), \(\alpha_{21} = 0.001\)
    • (\(\frac{1}{0.01} = 100,\ \frac{1}{0.01} = 200,\ \frac{1}{0.02} = 50,\ \frac{1}{0.001} = 1000\))
  • On separate graphs, draw the isoclines for species 1 and 2.