Lesson · Systems and the Phase Plane · Nonlinear systems

Competing Species

In a competition model each species would grow logistically alone, and the other subtracts from its growth. Four equilibria matter: both extinct, either species alone, and a mixed state. The Jacobian at each sorts them, and the mixed state is a saddle when competition is strong, so the winner depends on the start, and a stable node when it is weak enough, so they coexist.

Open the interactive lesson Practice

x′ = x·(3 − x − 2y) y′ = y·(2 − x − y) equilibria: origin, two axis points, (1, 1) classify each by its own Jacobian mixed point a saddle: no coexistence mixed point a node and stable: coexist

Four equilibria, four Jacobians

Core ideas

What this lesson establishes

Three ideas. The equations come from a product of two factors, the factors give the four points, and the determinant at the mixed point decides the story.

Each rate is a factor times a factor

In x′ = x·(3 − x − 2y) the species x grows logistically towards 3 when y = 0, and each unit of y costs it twice as much as a unit of itself. The rate is zero when x = 0 or when the bracket is zero, and the same is true of y.

Pairing the factors gives four equilibria

Either factor of the first rate with either factor of the second: (0, 0), then x = 0 with 2 − y = 0 for (0, 2), then y = 0 with 3 − x = 0 for (3, 0), then both brackets zero: x + 2y = 3 and x + y = 2 give (1, 1). Only points with both coordinates not negative are populations.

The determinant at the mixed point decides coexistence

The mixed point is the only equilibrium where both species are present. If its Jacobian is a stable node or spiral, nearby starts end with both species. If the determinant is negative it is a saddle, almost every start leaves it, and one species ends alone.

The material

Who survives when they share

The model, the four points and four Jacobians worked out, and the version in which competition is weak enough for both to live.

Definition · A two-species competition model

Competing species x and y obey x′ = x·(r − α·x − β·y) and y′ = y·(s − γ·x − δ·y), with positive constants.

The assumptions are named: each species grows logistically alone, each crowds its own kind and the other kind, and nothing else enters. This lesson uses the integer values below.

x′ = x·(3 − x − 2y) y′ = y·(2 − x − y)

The two brackets are lines, x + 2y = 3 and x + y = 2, and each rate is zero on its own line or where its own species is zero. The crossings are the four equilibria, and one of them, (1, 1), is where the lines cross. At each the Jacobian is the same formula, with different numbers.

J rows: 3 − 2x − 2y, −2x and −y, 2 − x − 2y (0, 0): rows 3 0 and 0 2 unstable node (3, 0): rows −3 −6 and 0 −1 stable node (0, 2): rows −1 0 and −2 −2 stable node (1, 1): rows −1 −2 and −1 −1 saddle

Example · The mixed point is a saddle

At (1, 1) the rows are −1 −2 and −1 −1. The trace is −2 and the determinant is 1 − 2 = −1 < 0, so it is a saddle.

A saddle has one direction that comes in and the rest go out. A start exactly on the incoming curve settles at (1, 1). Every other start near it leaves, and ends at (3, 0) or (0, 2), both stable nodes. Which one depends on which side of the incoming curve the start is on.

So in this model coexistence is an equilibrium that exists and does not last, and the lab's tile for it reads saddle. The claim about where nearby starts end is from the saddle and the two nodes, and the trajectory view, drawn in floating point, shows one example from the start the preset gives.

Weaker competition changes the answer

Now halve the cost of the other species: x′ = x·(2 − x − y/2) and y′ = y·(2 − y − x/2). The mixed point is (4/3, 4/3), since 2 − 4/3 − 2/3 = 0. The Jacobian rows there are −4/3 −2/3 and −2/3 −4/3: trace −8/3, determinant 16/9 − 4/9 = 4/3, positive, and the roots are −2/3 and −2. A stable node: both species survive from every nearby start. Competition is the same in kind in both models. The strength of the cross terms decided whether the mixed point is a saddle or a sink.

Interactive

Equilibria and the Jacobian

Checked exactly, linearised, and classified

Classify all four equilibria

Each preset lists four points and selects one. Use the Linearise at control to go through all four, and record the determinant and type of each. The tile for the selected point is exact; the portrait behind it is the linear system, drawn by stepping in floating point. Switch the view to a trajectory to see where one start ends.

System

x′ = f(x, y)

y′ = g(x, y)

Points to check (x y; x y)

Linearise at

Start for the trajectory (may be empty)

Search a grid of fractions

View

At the selected point—

Jacobian J—

Trace of J—

Determinant of J—

Eigenvalues of J—

Linearised type—

Grid search—

Epidemic threshold—

Polynomials of degree 3 at most: y - x^2, x(3 - x - 2y).

Method

Classifying a competition model

Factor, pair, linearise four times, and then answer the question about the species.

Factor both rates

Write each as a species times a bracket: x·(3 − x − 2y) and y·(2 − x − y). Do not multiply out.

Pair the factors

Four pairings give four points: the origin, two axis points and the crossing of the brackets. Check each in both equations.

Write the Jacobian once

Differentiate the products: f_x = 3 − 2x − 2y, f_y = −2x, g_x = −y, g_y = 2 − x − 2y.

Evaluate and classify at each point

Substitute, then read Δ, the discriminant and τ in that order. The result at each point is a node or a saddle.

Answer the question about the species

Coexistence needs the mixed point to be stable: Δ > 0 and τ < 0. A saddle there means the outcome depends on the start.

Four points and a verdict

Score 0 / 0

Choose the answer you can justify.

x′ = x(3 − x − 2y), y′ = y(2 − x − y): the mixed point

The first preset: find the mixed equilibrium from the two brackets and classify it.

x + 2y = 3 and x + y = 2: y = 1, x = 1 f_x = 3 − 2x − 2y f_y = −2x g_x = −y g_y = 2 − x − 2y at (1, 1): rows −1 −2 and −1 −1 τ = −2, Δ = 1 − 2 = −1 Δ < 0: a saddle (3, 0) and (0, 2) are stable nodes

The lab's tiles for the first preset confirm the point, read the determinant −1 and name the saddle. Selecting (3, 0) and then (0, 2) in the control reads two stable nodes, and the grid search on this system reports found 4: (0, 0), (0, 2), (1, 1), (3, 0).

The model predicts that, from a start near (1, 1), one species ends alone, and which one is the saddle's decision, not the model's luck.

Common mistakes

Thinking two competing species always drive one to extinctionThe strong-competition model does: its mixed point has Δ = −1, a saddle. A reader generalises. With the cross terms halved the mixed point (4/3, 4/3) has τ = −8/3 and Δ = 4/3, a stable node, and both survive. Whether competition excludes depends on the numbers, and the determinant at the mixed point is where the lab shows it.

Equating an equilibrium with both species present with coexistenceThe point (1, 1) is an equilibrium of the strong-competition model, and both populations are positive there. But it is a saddle, so a population that reaches it is knocked off by any disturbance. Coexistence in the long run needs the equilibrium to be stable.

Dividing by x or y and losing the axis equilibriaDividing x′ = 0 by x gives 3 − x − 2y = 0 and no more: the points (0, 0) and (0, 2) are lost, and with them both outcomes in which the first species is extinct. Both factors of each rate are zero in turn.

Check your understanding

✓

Finish when you can find and classify the equilibria of a competition model and say whether the species coexist.

You should be able to factor the rates, list the four equilibria, compute the Jacobian at each, read the type from trace and determinant, and state whether the mixed point is stable and what that means for the species.

Note: One more model, and the one where the equilibria are not isolated: “An Epidemic Model”.