Unit 6, lesson 51

Nonlinear Systems and Phase-Plane Models

How to find the equilibria of a nonlinear system, classify them with the Jacobian when they are hyperbolic, and use conserved quantities or radial equations when the linear picture says nothing.

  • Theory part 101
  • Practice part 102
  • Systems of Equations

Two species compete for the same food, or foxes and rabbits rise and fall in turn. The equations for such models are nonlinear and almost never have formulas for their solutions. Still, you can draw the velocity arrows in the plane, find where they vanish, and zoom in on those points, where the system looks almost linear. That picture answers most questions about long-term behavior without solving anything.

Key terms

Autonomous systems

Definition An autonomous system in the plane is

x′=f(x,y),y′=g(x,y),x' = f(x, y), \qquad y' = g(x, y),

where ff and gg do not depend on tt and have continuous first partial derivatives on an open set UU. In vector form x′=F(x)\mathbf{x}' = \mathbf{F}(\mathbf{x}) with F=⟨f,g⟩\mathbf{F} = \langle f, g \rangle. Under these hypotheses, through each point of UU and each starting time there passes exactly one solution, defined on some open time interval. If x(t)\mathbf{x}(t) is a solution, so is x(t−c)\mathbf{x}(t - c) for every constant cc.

In plain words The vector field F\mathbf{F} assigns a velocity to every point, and the rule never changes with time. Starting the same state later just shifts the whole motion in time.

Example The competition model

x′=x(3−x−2y),y′=y(2−x−y)\begin{aligned} x' &= x(3 - x - 2y), \\ y' &= y(2 - x - y) \end{aligned}

is autonomous, with polynomial right sides, so the hypotheses hold on all of R2\mathbb{R}^2. At the point (2,2)(2, 2) the velocity is ⟨2(−3),2(−2)⟩=⟨−6,−4⟩\langle 2(-3), 2(-2) \rangle = \langle -6, -4 \rangle.

Equilibria

Definition A point (x∗,y∗)(x^*, y^*) in UU is an equilibrium if f(x∗,y∗)=0f(x^*, y^*) = 0 and g(x∗,y∗)=0g(x^*, y^*) = 0. Then the constant function x(t)=(x∗,y∗)\mathbf{x}(t) = (x^*, y^*) is a solution. The curves f=0f = 0 and g=0g = 0 are the nullclines, and the equilibria are exactly the points where an xx nullcline meets a yy nullcline.

In plain words An equilibrium is a point where both velocity components vanish at once. A point where only one of them is zero is just a place where the path is momentarily vertical or horizontal.

Example For the competition model, x′=0x' = 0 when x=0x = 0 or x+2y=3x + 2y = 3, and y′=0y' = 0 when y=0y = 0 or x+y=2x + y = 2. Pairing them gives four equilibria, (0,0)(0, 0), (0,2)(0, 2), (3,0)(3, 0), and the solution of x+2y=3x + 2y = 3 and x+y=2x + y = 2, which is (1,1)(1, 1).

Jacobian linearization

Definition Let F\mathbf{F} have continuous first partial derivatives near an equilibrium x∗\mathbf{x}^*. Its Jacobian matrix there is

J=[fxfygxgy](x∗,y∗).J = \begin{bmatrix} f_x & f_y \\ g_x & g_y \end{bmatrix}_{(x^*, y^*)}.

Writing u=x−x∗\mathbf{u} = \mathbf{x} - \mathbf{x}^*, differentiability of F\mathbf{F} gives u′=Ju+r(u)\mathbf{u}' = J\mathbf{u} + \mathbf{r}(\mathbf{u}) with ∥r(u)∥/∥u∥→0\lVert \mathbf{r}(\mathbf{u}) \rVert / \lVert \mathbf{u} \rVert \to 0 as u→0\mathbf{u} \to \mathbf{0}. The linear system u′=Ju\mathbf{u}' = J\mathbf{u} is the linearization at x∗\mathbf{x}^*.

In plain words Near an equilibrium the vector field is almost its best linear approximation. The linear system keeps the leading behavior and drops terms that are tiny compared with the distance to the equilibrium.

Example For the competition model, at a point (x,y)(x, y)

J=[3−2x−2y−2x−y2−x−2y].J = \begin{bmatrix} 3 - 2x - 2y & -2x \\ -y & 2 - x - 2y \end{bmatrix}.

At (1,1)(1, 1) this is [−1−2−1−1]\begin{bmatrix} -1 & -2 \\ -1 & -1 \end{bmatrix}, with trace −2-2 and determinant −1-1.

Hyperbolic equilibrium classification

Definition An equilibrium x∗\mathbf{x}^* is hyperbolic if no eigenvalue of JJ has zero real part. Linearization theorem (Hartman-Grobman). If F\mathbf{F} has continuous first partial derivatives on an open set containing a hyperbolic equilibrium x∗\mathbf{x}^*, then there is a homeomorphism, a continuous map with continuous inverse, from a neighborhood of x∗\mathbf{x}^* onto a neighborhood of 0\mathbf{0} that carries trajectories of the nonlinear system onto trajectories of the linearization and preserves the direction of time. In particular

  • if every eigenvalue has negative real part, x∗\mathbf{x}^* is asymptotically stable (a sink),
  • if every eigenvalue has positive real part, x∗\mathbf{x}^* is a source, and solutions starting near it move away,
  • if the eigenvalues are real with opposite signs, x∗\mathbf{x}^* is a saddle, which is unstable.

A homeomorphism can bend straight approaches into spirals, so this theorem alone does not separate a node from a spiral. A finer result in the plane does. Suppose F\mathbf{F} has continuous second partial derivatives near x∗\mathbf{x}^*. If the eigenvalues of JJ are real, distinct and of the same sign, then x∗\mathbf{x}^* is a node, meaning every trajectory that tends to it, as t→∞t \to \infty for a sink or as t→−∞t \to -\infty for a source, does so tangent to a line through it. If they are α±iβ\alpha \pm i\beta with α≠0\alpha \ne 0 and β≠0\beta \ne 0, then x∗\mathbf{x}^* is a spiral, meaning nearby trajectories wind around it infinitely often. A repeated real eigenvalue still gives a sink or a source, but if F\mathbf{F} is only C1C^1 the trajectories near it may spiral, so the linearization does not settle the finer type in that case.

In plain words If the linearization has no neutral directions, the small nonlinear terms cannot change the picture close to the equilibrium. Sinks stay sinks, sources stay sources and saddles stay saddles.

Example The four equilibria of the competition model classify as follows.

PointEigenvalues of JJType
(0,0)(0, 0)3,23, 2unstable node
(3,0)(3, 0)−3,−1-3, -1stable node
(0,2)(0, 2)−1,−2-1, -2stable node
(1,1)(1, 1)−1±2-1 \pm \sqrt{2}saddle

At (1,1)(1, 1) the determinant is −1<0-1 < 0, so the eigenvalues are real with opposite signs, approximately 0.4140.414 and −2.414-2.414.

Inconclusive nonhyperbolic cases

Definition If some eigenvalue of JJ has zero real part, the equilibrium is nonhyperbolic, and the linearization alone does not determine stability or type. In particular a linear center may be a nonlinear center, a stable spiral or an unstable spiral. Other tools are then needed, such as a conserved quantity, a polar or radial equation, or a Lyapunov function.

In plain words A center is balanced on a knife edge. Tiny nonlinear terms that the Jacobian ignores can tip it either way.

Example For each constant aa, the system

x′=−y+ax(x2+y2),y′=x+ay(x2+y2)\begin{aligned} x' &= -y + a x(x^2 + y^2), \\ y' &= x + a y(x^2 + y^2) \end{aligned}

has Jacobian [0−110]\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} at the origin, a center with eigenvalues ±i\pm i. With r2=x2+y2r^2 = x^2 + y^2, the identity rr′=xx′+yy′=ar4r r' = x x' + y y' = a r^4 gives the radial equation r′=ar3r' = a r^3. For a=−1a = -1 the solution is

r(t)=r01+2r02t→0,r(t) = \frac{r_0}{\sqrt{1 + 2 r_0^2 t}} \to 0,

so the origin is asymptotically stable, and since the angle obeys θ′=1\theta' = 1 it is a stable spiral. For a=1a = 1 the radius grows and the origin is unstable, and for a=0a = 0 it is a true center.

Trajectories

Definition A trajectory, or orbit, is the set of points {x(t)}\{\mathbf{x}(t)\} traced by one solution over its interval of existence, drawn with an arrow for increasing tt. Under the hypotheses above, two distinct trajectories never meet, and each trajectory is an equilibrium point, a closed curve (a periodic solution), or a curve that never crosses itself. Where f≠0f \ne 0, trajectories satisfy

dydx=g(x,y)f(x,y).\frac{dy}{dx} = \frac{g(x, y)}{f(x, y)}.

In plain words The solution is a moving point, and the trajectory is the path it leaves behind. Paths cannot cross, because at a crossing point the velocity would have to point two ways at once.

Example For the pendulum x′=yx' = y and y′=−sin⁡xy' = -\sin x, the slope equation dydx=−sin⁡xy\frac{dy}{dx} = -\frac{\sin x}{y} separates to y dy=−sin⁡x dxy\, dy = -\sin x\, dx. Integrating gives the energy

E(x,y)=y22−cos⁡x,E(x, y) = \frac{y^2}{2} - \cos x,

which is constant on every trajectory. The trajectory through (0,1)(0, 1) lies on the curve E=−12E = -\frac{1}{2}.

Predator-prey models

Definition The Lotka-Volterra model for prey x≥0x \ge 0 and predators y≥0y \ge 0 is

x′=ax−bxy,y′=−cy+dxy,\begin{aligned} x' &= ax - bxy, \\ y' &= -cy + dxy, \end{aligned}

with constants a,b,c,d>0a, b, c, d > 0. Its equilibria are (0,0)(0, 0), a saddle, and (cd,ab)\bigl(\frac{c}{d}, \frac{a}{b}\bigr), where the linearization has eigenvalues ±iac\pm i\sqrt{ac}, a nonhyperbolic center. In the open quadrant x>0x > 0, y>0y > 0 the function

H(x,y)=dx−cln⁡x+by−aln⁡yH(x, y) = dx - c\ln x + by - a\ln y

is constant along every trajectory and has a strict minimum at the interior equilibrium. Hence every trajectory in the open quadrant other than that equilibrium is a closed curve, and the populations are periodic.

In plain words Prey grow until predators catch up, predators then crash for lack of food, and the cycle repeats. The conserved quantity is what proves the cycles close up exactly, since the linearization cannot.

Example Take x′=x(2−y)x' = x(2 - y) and y′=y(x−3)y' = y(x - 3), so a=2a = 2, b=1b = 1, c=3c = 3, d=1d = 1, and the interior equilibrium is (3,2)(3, 2). With H=x−3ln⁡x+y−2ln⁡yH = x - 3\ln x + y - 2\ln y,

dHdt=(1−3x)x(2−y)+(1−2y)y(x−3)=(x−3)(2−y)+(y−2)(x−3)=0.\begin{aligned} \frac{dH}{dt} &= \Bigl(1 - \frac{3}{x}\Bigr)x(2 - y) \\ &\quad + \Bigl(1 - \frac{2}{y}\Bigr)y(x - 3) \\[0.4em] &= (x - 3)(2 - y) \\ &\quad + (y - 2)(x - 3) = 0. \end{aligned}

Small cycles around (3,2)(3, 2) have period close to 2π6≈2.57\frac{2\pi}{\sqrt{6}} \approx 2.57.

Local versus global behavior

Definition Linearization at an equilibrium describes trajectories only in some neighborhood of it, whose size the theorem does not give. Global questions, such as which equilibrium a given starting point approaches, whether solutions exist for all t≥0t \ge 0, or whether closed orbits exist far from equilibria, need the whole phase portrait. The basin of attraction of an asymptotically stable equilibrium is the set of initial points whose solutions tend to it.

In plain words A stable equilibrium attracts the points near it, not every point. Far away, other equilibria, cycles or blow-up can take over.

Example The system x′=−x+x2x' = -x + x^2, y′=−2yy' = -2y has Jacobian [−100−2]\begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} at the origin, a stable node. But from (2,0)(2, 0) the solution is

x(t)=11−12et,y(t)=0,x(t) = \frac{1}{1 - \frac{1}{2}e^{t}}, \qquad y(t) = 0,

which blows up as t→ln⁡2t \to \ln 2. In the competition model both (3,0)(3, 0) and (0,2)(0, 2) are sinks, and the two trajectories that approach the saddle (1,1)(1, 1) separate their basins. Which species survives depends on where the populations start.

Check yourself

  1. Find and classify the equilibria of x′=yx' = y, y′=x−x3y' = x - x^3. Where linearization is inconclusive, use the conserved quantity H=y22−x22+x44H = \frac{y^2}{2} - \frac{x^2}{2} + \frac{x^4}{4}.
Show the answer

The equilibria are (0,0)(0, 0), (1,0)(1, 0) and (−1,0)(-1, 0). The Jacobian is [011−3x20]\begin{bmatrix} 0 & 1 \\ 1 - 3x^2 & 0 \end{bmatrix}. At the origin the eigenvalues are ±1\pm 1, a saddle. At (±1,0)(\pm 1, 0) they are ±i2\pm i\sqrt{2}, which is nonhyperbolic. Along solutions dHdt=(x3−x)y+y(x−x3)=0\frac{dH}{dt} = (x^3 - x)y + y(x - x^3) = 0, and HH has a strict local minimum at (±1,0)(\pm 1, 0), so nearby trajectories are closed curves and both points are centers.

  1. For x′=y+x(1−x2−y2)x' = y + x(1 - x^2 - y^2) and y′=−x+y(1−x2−y2)y' = -x + y(1 - x^2 - y^2), find the radial equation and describe the behavior.
Show the answer

From rr′=xx′+yy′=r2(1−r2)r r' = x x' + y y' = r^2(1 - r^2) the radial equation is r′=r(1−r2)r' = r(1 - r^2). The radius grows when 0<r<10 < r < 1 and shrinks when r>1r > 1, so every nonzero solution approaches the circle r=1r = 1, a periodic orbit. The angle obeys θ′=−1\theta' = -1, so motion is clockwise. At the origin the Jacobian has eigenvalues 1±i1 \pm i, an unstable spiral, which matches.

  1. Show that H=x−ln⁡x+2y−4ln⁡yH = x - \ln x + 2y - 4\ln y is conserved for x′=x(4−2y)x' = x(4 - 2y), y′=y(x−1)y' = y(x - 1) in the open first quadrant, and find the eigenvalues of the linearization at the interior equilibrium.
Show the answer

Along solutions

dHdt=(x−1)(4−2y)+(2y−4)(x−1)=0.\begin{aligned} \frac{dH}{dt} &= (x - 1)(4 - 2y) \\ &\quad + (2y - 4)(x - 1) = 0. \end{aligned}

The interior equilibrium is (1,2)(1, 2), where the Jacobian is [0−220]\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} with eigenvalues ±2i\pm 2i. The linearization alone is inconclusive, but the conserved quantity has a strict minimum at (1,2)(1, 2), since each of x−ln⁡xx - \ln x and 2y−4ln⁡y2y - 4\ln y has a strict minimum there, so the orbits near it are closed and the equilibrium is a center.

How this lesson connects

It also uses Vector fields and trajectories, Jacobian matrices and Linear-system eigenvalues and stability.