Competing species · Differential Equations

Intuition

Two species eating the same food hold each other back as well as themselves. Each alone would settle at its carrying capacity, as in the logistic model; together, the outcome depends on how strongly each crowds the other compared with how strongly it crowds itself. Weak competition lets them coexist at a stable equilibrium; strong competition makes coexistence a saddle, and one species drives the other out — which one depends on where they start.

Two shops on one street selling the same goods. If each has its own loyal customers, both survive; if every customer buys purely on price, the first to gain an edge takes the whole street.

Strong competition, $x'=x(1-x-2y)$ , $y'=y(1-y-2x)$ . The coexistence point $\bigl(\tfrac13,\tfrac13\bigr)$ is a saddle; starts below the diagonal end with $x$ alone at $(1,0)$ and starts above it with $y$ alone at $(0,1)$ .

Crowding each other

$x'=x(1-x-\alpha y)$ and $y'=y(1-y-\beta x)$ with : each species alone grows logistically to , and and measure how much each crowds the other. The equilibria are , , and, when it lies in the quadrant, the coexistence point below. If and it is a stable node and the species coexist; if and it is a saddle, and one species wins, decided by which side of the separatrix the start lies.

\bigl(x^{*},y^{*}\bigr)=\Bigl(\frac{1-\alpha}{1-\alpha\beta},\ \frac{1-\beta}{1-\alpha\beta}\Bigr)

Weak and strong competition

$\alpha=\beta=\tfrac12$ : coexistence at $\bigl(\tfrac23,\tfrac23\bigr)$ , where the Jacobian has eigenvalues and : a stable node.

Where the coexistence point lies

\begin{aligned}&1-x-\alpha y=0=1-y-\beta x\\&\Longrightarrow\ \bigl(x^{*},y^{*}\bigr)=\Bigl(\frac{1-\alpha}{1-\alpha\beta},\ \frac{1-\beta}{1-\alpha\beta}\Bigr)\end{aligned}

Away from the axes, each species is at rest on a straight nullcline. Substituting the first line into the second gives a linear equation for y, solved when αβ is not 1; the first line then gives x. Both coordinates are positive exactly when the two numerators have the sign of the denominator, which happens when α and β are both below one or both above.

Proof steps

1-x-\alpha y=0,\qquad 1-y-\beta x=0

Away from the axes, both species are at rest on their nullcline lines.

x=1-\alpha y\ \Longrightarrow\ y=1-\beta+\alpha\beta y

Substitute the first into the second.

y^{*}=\frac{1-\beta}{1-\alpha\beta},\qquad x^{*}=\frac{1-\alpha}{1-\alpha\beta}

Solve for y, then use the first line for x.

x^{*},y^{*}>0\ \Longleftrightarrow\ \alpha,\beta<1\ \text{or}\ \alpha,\beta>1

Both are positive exactly when the two crowdings lie on the same side of one.

Applications

Practice

Crowding Coefficients

α measures how much species y holds back species x, compared with how much x holds itself back.

x'=x(1-x-\alpha y),\qquad y'=y(1-y-\beta x)

Try it

In $x'=x(1-x-\alpha y)$ , $y'=y(1-y-\beta x)$ , what does $\alpha$ measure?