Nonlinear Phase Planes and Nullclines, differential equations lesson

A prey species and a predator species share a habitat. Plenty of prey lets the predators multiply, many predators thin out the prey, and scarce prey then starves the predators. If the two population sizes are plotted against each other instead of against time, does the state circle around, settle down or drift away? This lesson reads such a model in its phase plane, sorts the directions of motion with nullclines, finds the equilibria and uses a conserved quantity to show that in the ideal model the populations repeat in closed cycles.

Key terms

Phase plane and vector field

Definition Let $t$ be the independent variable, with $x' = dx/dt$ and $y' = dy/dt$ . A planar autonomous system has the form

\begin{gathered} x' = f(x, y), \\[0.4em] y' = g(x, y), \end{gathered}

where the rates $f$ and $g$ do not depend on $t$ . Throughout this lesson they are continuously differentiable on an open set $D$ of the plane. The phase plane is the $xy$ -plane, in which each point is a state $X = (x, y)^{T}$ of the system. The vector field of the system assigns to each state the vector

F(x, y) = \begin{pmatrix} f(x, y) \\ g(x, y) \end{pmatrix}.

A solution $X(t) = (x(t), y(t))^{T}$ satisfies $X'(t) = F(X(t))$ , so the arrow $F(X)$ points in the direction in which the solution through $X$ moves, and its length is the speed. Neither axis of the phase plane is time. Because $F$ does not depend on $t$ , one picture of arrows is valid at every time.

In plain words Each point of the plane is a complete description of the system at one moment, and the arrow attached to it says where the state goes next and how fast. Time is not drawn as an axis. It is the parameter along the curves that follow the arrows.

Example In scaled, dimensionless units, let $x$ be the prey population and $y$ the predator population, with

\begin{gathered} x' = x(2 - y), \\[0.4em] y' = y(x - 1), \end{gathered}

and $x \ge 0$ , $y \ge 0$ . At the state $(3, 1)$ the arrow is $F(3, 1) = (3, 2)^{T}$ , pointing up and to the right, so prey and predators both increase there. At $(0.5, 3)$ it is $(-0.5, -1.5)^{T}$ , and both decrease.

Nullcline

Definition For the system $x' = f(x, y)$ , $y' = g(x, y)$ , the $x$ -nullcline is the set of states where $f(x, y) = 0$ , and the $y$ -nullcline is the set where $g(x, y) = 0$ . Either set may consist of several curves, and each of these curves is also called an $x$ -nullcline or a $y$ -nullcline. On the $x$ -nullcline the horizontal component of the field is zero, so the arrow there is vertical or zero. On the $y$ -nullcline the arrow is horizontal or zero. Because $f$ and $g$ are continuous, each keeps one sign on any connected region that contains no point of its nullcline. The nullclines therefore cut the plane into regions in which the arrows all point up and to the right, all up and to the left, all down and to the left, or all down and to the right.

In plain words On an $x$ -nullcline the state is not moving left or right at that instant, and on a $y$ -nullcline it is not moving up or down. Between the nullclines the signs of the two rates cannot change, so one test point in each region gives the general direction of every arrow in it.

Example For the predator-prey model, $x(2 - y) = 0$ gives the $x$ -nullclines $x = 0$ and $y = 2$ , and $y(x - 1) = 0$ gives the $y$ -nullclines $y = 0$ and $x = 1$ . In the open first quadrant the signs are as follows.

Region	Sign of $x'$	Sign of $y'$	Arrow
$x > 1$ , $0 < y < 2$	positive	positive	up and right
$x > 1$ , $y > 2$	negative	positive	up and left
$0 < x < 1$ , $y > 2$	negative	negative	down and left
$0 < x < 1$ , $0 < y < 2$	positive	negative	down and right

Two test states confirm which nullcline controls which component,

\begin{gathered} F(2, 2) = (0, 2)^{T}, \\[0.4em] F(1, 3) = (-1, 0)^{T}. \end{gathered}

The state $(2, 2)$ lies on $y = 2$ , and its arrow points straight up. The state $(1, 3)$ lies on $x = 1$ , and its arrow points straight to the left. Reading the table from top to bottom, the arrows turn counterclockwise around the point $(1, 2)$ .

Equilibrium point

Definition An equilibrium point of $x' = f(x, y)$ , $y' = g(x, y)$ is a state $X_* = (x_*, y_*)^{T}$ with $f(x_*, y_*) = 0$ and $g(x_*, y_*) = 0$ , that is, $F(X_*) = 0$ . Equivalently, it is a point where an $x$ -nullcline meets a $y$ -nullcline. The constant function $X(t) = X_*$ for all $t$ is then a solution, called an equilibrium solution, and every constant solution is of this form. A point where two $x$ -nullclines cross, or two $y$ -nullclines, is in general not an equilibrium, because there only one of the rates is known to vanish.

In plain words At an equilibrium both rates are zero at once, so a state placed there never moves. To find all equilibria, intersect each $x$ -nullcline with each $y$ -nullcline.

Example For the predator-prey model, pair the $x$ -nullclines $x = 0$ and $y = 2$ with the $y$ -nullclines $y = 0$ and $x = 1$ . The lines $x = 0$ and $y = 0$ meet at the origin, and the lines $y = 2$ and $x = 1$ meet at $(1, 2)$ . The pairs $x = 0$ , $x = 1$ and $y = 2$ , $y = 0$ are parallel and do not meet. So the model has exactly two equilibria, $(0, 0)$ , where both species are absent, and the coexistence state $(1, 2)$ , where both persist at constant levels. The point $(0, 2)$ , where the two $x$ -nullclines cross, is not an equilibrium, since $F(0, 2) = (0, -2)^{T}$ .

Trajectory

Definition Let $X(t)$ be a solution of $X' = F(X)$ on its maximal interval $I$ . Its trajectory, also called its orbit, is the curve $\{X(t) : t \in I\}$ in the phase plane, traced in the direction of increasing $t$ . At every point the trajectory is tangent to the arrow of the vector field. Since $F$ is continuously differentiable, solutions of initial-value problems are unique, so each state lies on exactly one trajectory and two different trajectories never cross or touch. An equilibrium is a trajectory consisting of a single point, a nonconstant trajectory never reaches an equilibrium at a finite time, and a trajectory that returns to a state it has already visited is a closed curve traced by a periodic solution. A nullcline is usually not a trajectory. A trajectory crosses a nullcline at every point where the arrow is nonzero and not tangent to the nullcline.

In plain words A trajectory is the path that the state actually follows, with time hidden as the parameter along it. A nullcline is a different object, a curve on which one rate happens to be zero, and trajectories usually cut straight through it.

Example In the predator-prey model the arrow $(0, 2)^{T}$ at $(2, 2)$ is vertical, while the nullcline $y = 2$ is horizontal, so the trajectory through $(2, 2)$ crosses that nullcline going up. For that instant the prey count is not changing, while the predator count is. The axes behave differently. On the line $x = 0$ the field is $(0, -y)^{T}$ , which points along the line, and the positive $y$ -axis is one trajectory, traced by $x = 0$ , $y = y_0e^{-t}$ for any $y_0 > 0$ . In the same way the positive $x$ -axis is traced by $x = x_0e^{2t}$ , $y = 0$ . Since trajectories cannot cross, a solution that starts with $x > 0$ and $y > 0$ never reaches an axis, so positive populations stay positive for as long as the solution exists.

Conserved quantity

Definition A conserved quantity, or first integral, of the system on an open set $U \subseteq D$ is a continuously differentiable function $H$ on $U$ , not constant on any open subset of $U$ , whose value does not change along any solution that stays in $U$ . By the chain rule,

\begin{aligned} &\frac{d}{dt}H(x(t), y(t)) \\[0.4em] &\quad = H_x f + H_y g, \end{aligned}

so the condition is $H_x f + H_y g = 0$ at every point of $U$ . Each trajectory in $U$ then lies in one level curve $H(x, y) = c$ , and the level curves show the shapes of the trajectories without a formula for $x(t)$ and $y(t)$ . Excluding functions that are constant on an open set rules out the useless choice of a constant $H$ .

In plain words A conserved quantity is a number computed from the state that never changes as the state moves. Each trajectory is then confined to one contour of that function, like a walker who keeps to one altitude on a hillside.

Example For the predator-prey model take, for $x > 0$ and $y > 0$ ,

H = x - \ln x + y - 2\ln y.

Here $H_x = 1 - 1/x$ and $H_y = 1 - 2/y$ , so

\begin{gathered} H_x f = (x - 1)(2 - y), \\[0.4em] H_y g = (y - 2)(x - 1), \end{gathered}

and these two products add to zero. The part $x - \ln x$ decreases on $(0, 1)$ , increases on $(1, \infty)$ and grows without bound at both ends, and $y - 2\ln y$ behaves the same way around $y = 2$ . So $H$ takes its smallest value $3 - 2\ln 2 \approx 1.614$ only at $(1, 2)$ , and every level curve $H = c$ with $c > 3 - 2\ln 2$ is a closed curve around $(1, 2)$ that stays away from both axes. The state $(2, 1)$ has $H = 3 - \ln 2 \approx 2.307$ , so the trajectory through it stays on that closed level curve forever.

Predator-prey model

Definition The Lotka-Volterra predator-prey model for a prey population $x(t) \ge 0$ and a predator population $y(t) \ge 0$ is

\begin{gathered} x' = ax - bxy, \\[0.4em] y' = -cy + dxy, \end{gathered}

with positive constants $a$ , $b$ , $c$ and $d$ . Without predators the prey grows at the per capita rate $a$ , and encounters, counted by the product $xy$ , remove prey at the rate $bxy$ . Without prey the predators die at the per capita rate $c$ , and encounters add predators at the rate $dxy$ . The origin and each positive half axis are trajectories, and the equilibria are $(0, 0)$ and $(c/d, a/b)$ . On the open first quadrant the function

\begin{aligned} H &= dx - c\ln x \\ &\quad + by - a\ln y \end{aligned}

is conserved, it has its only minimum at $(c/d, a/b)$ , and each of its level curves above the minimum value is a closed curve. As a result every solution that starts with $x > 0$ and $y > 0$ , other than the coexistence equilibrium, is periodic, and its trajectory is a closed curve around $(c/d, a/b)$ , traced counterclockwise when the prey is on the horizontal axis.

In plain words Prey alone grows, predators alone die out, and meetings between the two shift numbers from the prey to the predators. The model is an idealization. It has no limit on the food of the prey and no saturation of the appetite of the predators, and the closed cycles belong to these exact equations rather than being a claim about real populations.

Example The model of this lesson is the case $a = 2$ and $b = c = d = 1$ , so its coexistence state is $(c/d, a/b) = (1, 2)$ and its conserved quantity is $x - \ln x + y - 2\ln y$ . Follow the trajectory from $(2, 1)$ , with many prey and few predators. Both populations grow until the predators reach $y = 2$ . The prey then declines while the predators keep growing, until the prey falls to $x = 1$ . After that both decline until the predators drop below $2$ , and then the prey recovers. The trajectory is the closed level curve $H = 3 - \ln 2$ , and a numerical solution returns to $(2, 1)$ after about $4.84$ time units, after which the cycle repeats.

Check yourself

Find the nullclines and all equilibria with $x \ge 0$ and $y \ge 0$ of the system $x' = x(3 - x - y)$ , $y' = y(x - 1)$ , and give the direction of the arrow at $(2, 2)$ .

Show the answer

The rate $x(3 - x - y)$ vanishes on the $x$ -nullclines $x = 0$ and $x + y = 3$ , and the rate $y(x - 1)$ vanishes on the $y$ -nullclines $y = 0$ and $x = 1$ . Intersecting each $x$ -nullcline with each $y$ -nullcline gives the origin from $x = 0$ and $y = 0$ , the state $(3, 0)$ from $x + y = 3$ and $y = 0$ , and the state $(1, 2)$ from $x + y = 3$ and $x = 1$ . The lines $x = 0$ and $x = 1$ do not meet. So the equilibria are $(0, 0)$ , $(3, 0)$ and $(1, 2)$ . At $(2, 2)$ the field is

\begin{aligned} F(2, 2) &= \begin{pmatrix} 2(3 - 4) \\ 2(2 - 1) \end{pmatrix} \\[0.4em] &= \begin{pmatrix} -2 \\ 2 \end{pmatrix}, \end{aligned}

so the arrow points up and to the left.

For the model $x' = x(2 - y)$ , $y' = y(x - 1)$ , a classmate argues that the vertical line $x = 1$ is a trajectory, because $y' = 0$ on it. Explain what is wrong, and say which parts of the boundary of the first quadrant are trajectories.

Show the answer

On the line $x = 1$ the field is $F(1, y) = (2 - y, 0)^{T}$ . That arrow is horizontal, while the line is vertical, so for $y \ne 2$ the arrow crosses the line, to the right when $y < 2$ and to the left when $y > 2$ . For example, $F(1, 1) = (1, 0)^{T}$ . A trajectory must follow the arrows, so the only trajectory contained in the line $x = 1$ is the equilibrium $(1, 2)$ . Setting one rate to zero produces a nullcline, not a trajectory. The boundary of the quadrant is different. On $y = 0$ the field $(2x, 0)^{T}$ points along the axis, and on $x = 0$ the field $(0, -y)^{T}$ does too. So the origin, the positive $x$ -axis, traced by $x = x_0e^{2t}$ and $y = 0$ , and the positive $y$ -axis, traced by $x = 0$ and $y = y_0e^{-t}$ , are trajectories.

For the system $x' = x(3 - y)$ , $y' = y(x - 2)$ with $x > 0$ and $y > 0$ , show that $H = x - 2\ln x + y - 3\ln y$ is a conserved quantity. Can the solution that starts at $(2, 1)$ tend to the equilibrium $(2, 3)$ as $t \to \infty$ ?

Show the answer

The partial derivatives are $H_x = 1 - 2/x$ and $H_y = 1 - 3/y$ , so

\begin{gathered} H_x f = (x - 2)(3 - y), \\[0.4em] H_y g = (y - 3)(x - 2). \end{gathered}

These products add to zero, so $H$ is constant along every solution in the open first quadrant. At the start, $H(2, 1) = 3 - 2\ln 2 \approx 1.614$ , while at the equilibrium $H(2, 3) = 5 - 2\ln 2 - 3\ln 3 \approx 0.318$ . If the solution tended to $(2, 3)$ , the continuity of $H$ would force its values along the solution to tend to $0.318$ , but they stay equal to $1.614$ . So the solution cannot approach the equilibrium. Its trajectory is the closed level curve $H = 3 - 2\ln 2$ around $(2, 3)$ , and the solution is periodic.

Nonlinear Phase Planes and Nullclines

Key terms

Phase plane and vector field

Nullcline

Equilibrium point

Trajectory

Conserved quantity

Predator-prey model

Check yourself

How this lesson connects

Builds on

Used later in