The Lotka–Volterra system lets prey grow, predators die, and every meeting move one to the other. It has an equilibrium with both populations present, where the Jacobian has trace zero: a centre of the linearisation, which settles nothing. A quantity that stays constant along solutions does, and the drawn trajectory is a closed loop that matches it.
Core ideas
Three ideas. The model states its assumptions, the Jacobian gives a borderline answer, and a conserved quantity finishes the argument.
Let x be the prey and y the predators. Prey would grow at a rate proportional to x alone, predators would die at a rate proportional to y alone, and each meeting, which happens at a rate proportional to x·y, takes prey away and adds predators. With the numbers of this lesson, x′ = 2x − x·y and y′ = −y + x·y.
Setting x·(2 − y) = 0 and y·(x − 1) = 0 gives (0, 0) and (1, 2). At (1, 2) the Jacobian has rows 0 −1 and 2 0, so τ = 0 and Δ = 2: roots ±i·√2. That is the borderline case from the last lesson, and the type tile says the linearisation is inconclusive.
The function V = x − ln x + y − 2·ln y has the same value at every point of a solution. So a solution stays on one level curve of V, and near (1, 2) those level curves are closed loops. The populations then return to where they started, which neither settling nor spiralling away would do.
The material
The equations, both equilibria, the quantity the true solutions keep, and what the lab draws and what it cannot.
Prey x and predators y obey x′ = α·x − β·x·y and y′ = −γ·y + δ·x·y, where α, β, γ and δ are positive constants.
The assumptions are named, not defended: prey have unlimited food and grow without bound alone, predators starve alone, and a meeting is as likely as the product of the two populations.
Take α = 2 and β = γ = δ = 1. The x-rate factors as x·(2 − y) and the y-rate as y·(x − 1). The prey grow while there are fewer than 2 predators, and the predators grow while there are more than 1 prey. Each nullcline is a pair of lines, x = 0 or y = 2 for the first and y = 0 or x = 1 for the second, and the crossings that matter are (0, 0) and (1, 2).
The Jacobian has rows 2 − y, −x and y, x − 1. At (0, 0) the rows are 2 0 and 0 −1: determinant −2, a saddle. Prey grow along the x-axis and predators die along the y-axis.
At (1, 2) the rows are 0 −1 and 2 0: trace 0, determinant 2, roots ±i·√2. The lab prints a centre of the linearisation, and that is all it can claim.
Along every solution with x > 0 and y > 0, V = x − ln x + y − 2·ln y is constant.
Differentiate along a solution. The rate of ln t is 1/t: that is the claim “The Integral of 1/t” made in defining ln x as the integral of 1/t from 1 to x, and the one Separable Equations, Growth and Decay used to integrate 1/y. With the chain rule, applied here as a claim beyond the polynomials it was verified on, ln x changes at the rate x′/x and ln y at the rate y′/y.
V′ = (1 − 1/x)·x′ + (1 − 2/y)·y′, and substituting x′ = x·(2 − y) and y′ = y·(x − 1) gives V′ = (x − 1)·(2 − y) + (y − 2)·(x − 1).
The two terms are negatives of each other, so V′ = 0. The value of V never changes, and a solution is confined to a level curve.
□
The proof is algebra the reader has; what it does not show is that the level curves are closed. That is a fact about the shape of V, which has its smallest value at (1, 2) and rises on all sides, and the lab does not prove it. It draws it. The trajectory view steps the system in floating point from the start and the legend says so, and the curve closes on the picture. A drawing is evidence. The reason is the conserved quantity.
Start at (1, 2) itself and the arrow is zero, so the point stays for ever, and Euler's exact steps stay at (1, 2) too. Start anywhere else in the first quadrant and the solution runs round the point. Because y′ > 0 only when x > 1, the predator peak comes after the prey peak, and the loop is traversed anticlockwise.
Interactive
Checked exactly, linearised, and classified
Method
Find the points, linearise, and when the linearisation is silent go to the conserved quantity.
x′ = x·(2 − y) and y′ = y·(x − 1). Each factor is a family of points where that rate is zero.
x = 0 with y = 0, and y = 2 with x = 1. The other pairings give the same two points. Check each in both equations.
The rows 2 − y, −x and y, x − 1 become 2 0 and 0 −1 at the origin, a saddle, and 0 −1 and 2 0 at (1, 2).
τ = 0 and Δ = 2 > 0 at (1, 2): a centre of the linearisation, which the Jacobian cannot decide.
Show V′ = 0 by substitution. Then read the drawn loop as agreeing with the level curves of V, and say that the drawing illustrates the argument rather than proving it.
The first preset, linearised at the interior equilibrium.
The linearisation at (1, 2) is a centre, and the lab says so. The reason a closed loop is still the right answer is that V′ = 0, not that the Jacobian predicted it.
The value of V at the start (1, 1) is 2, while at the equilibrium it is 3 − 2·ln 2 ≈ 1.61371, rounded. The start is on a bigger loop around the point, and the two never meet: the populations do not tend to (1, 2).
You should be able to write the Lotka–Volterra equations, find both equilibria exactly, show the interior one has trace zero, verify by substitution that the stated quantity is conserved, and say why the drawn trajectory illustrates the result without proving it.
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