7.6 Phase portraits, stability and linearization

An autonomous system is studied by drawing its orbits in the plane. An equilibrium is stable if nearby solutions stay near it. For a linear system the trace and the determinant of the matrix decide the picture completely. For a nonlinear system, near a hyperbolic equilibrium the picture is that of the Jacobian matrix there. The pendulum and the predator-prey equations show both the power and the limits of this idea.

A pendulum has no solution that you can write down with the functions of school: its period depends on the amplitude through an elliptic integral. Yet you know exactly what it does. It has a rest position at the bottom, which it swings around or settles into, and an upside-down balance that it leaves at once if nudged. This chapter makes that kind of knowledge precise. We draw the paths of an autonomous system in the plane, find the points where the motion stops, ask whether nearby motions stay near, and show that near such a point the nonlinear system looks like a linear one whose matrix we can read. The linear case, with its eigenvalues, is the content of 7.4; the new step is to use it as a microscope on a nonlinear equation.

Before you start

You need the solution $e^{tA}x_0$ of $x' = Ax$ and the first look at its pictures from 7.4, the Jacobian matrix of 5.3, and positive definite matrices from 3.2. Three quick checks:

Quick checkA real

2 \times 2

matrix has eigenvalues

-1 \pm 2i

. What do the solutions of

x' = Ax

do?

Quick checkWhat is the Jacobian matrix of

F(x, y) = (xy,\ x + y^2)

at the point

(1, 1)

Quick checkWhich of these symmetric matrices is positive definite?

The picture

Two pictures, one for each half of the chapter. The first is the family of phase portraits of $x' = Ax$ for a $2 \times 2$ matrix, which 7.4 introduced. Change the four entries and watch the curves reorganise around the origin. The numbers that decide the shape are the trace and the determinant, shown under the picture.

Now collect all matrices by their trace $\tau$ and determinant $\Delta$ . Each matrix is one point of the $(\tau, \Delta)$ plane, and the portrait only depends on where the point lies. Drag the dot: the left panel is the plane, with the parabola $\tau^2 = 4\Delta$ where the eigenvalues merge, and the right panel shows the orbits of one matrix with that trace and determinant.

The second picture comes later, once we have nonlinear systems to draw.

Where this came from· Poincaré, Lyapunov and the qualitative theory

Before 1880 the aim of the theory of differential equations was a formula for the solution. Poincaré changed the aim. In a long memoir on curves defined by differential equations (1881 to 1886) he studied the shapes of the solution curves in the plane, and sorted the singular points by the pictures around them, with names such as col (saddle), nœud (node) and foyer (spiral point). Lyapunov, in his doctoral thesis on the general problem of the stability of motion (1892, published by the Kharkov Mathematical Society), asked when a motion stays close to a given one, and introduced the auxiliary-function method named after him. Grobman (1959) and Hartman (1960) proved the linearization theorem that bears their names. Meanwhile biology supplied examples: Lotka’s book of 1925 and Volterra’s paper of 1926 on predator and prey, and van der Pol’s 1926 study of self-sustained oscillations in an electric circuit.

The idea, in words

Why· Do not solve the equation. Find where the arrows vanish, and look at them from close up

A system $x' = F(x)$ gives an arrow $F(x)$ at every point $x$ of the plane: the velocity of a particle that happens to be at $x$ . A solution is a path that follows the arrows. There are only two kinds of points. At most points the arrow is not zero and the particle moves on. At an equilibrium the arrow is zero and the particle stays. Since the arrows vary continuously, the equilibria organise the whole picture, in the way the points where the ground is flat organise a landscape: hilltops, valleys and passes. To learn what a landscape looks like around a flat point you ask for the second derivative, the Hessian. To learn what the arrows look like around an equilibrium, you ask for the first derivative of $F$ , the Jacobian matrix. The Jacobian is a matrix $J$ , and near the equilibrium the system behaves like the linear system $u' = Ju$ , which 7.4 solved completely. The eigenvalues of $J$ tell the story: negative real parts give a valley (everything flows in), a positive real part gives a ridge or a hill, mixed signs give a pass.

Definitions and theorems

Autonomous systems, orbits, phase portraits

Definition B

Autonomous system

A system of differential equations is autonomous if it has the form $x' = F(x)$ with $F$ defined on a region of $\mathbb R^n$ and not depending on $t$ . We assume throughout that $F$ is continuously differentiable (so it is Lipschitz on every closed ball in its region, 7.1): then every starting point gives exactly one solution. Every system can be made autonomous by adding the extra unknown $x_{n+1} = t$ with $x_{n+1}' = 1$ ; the planar case ( $n = 2$ ) is the one that can be drawn.

In one lineA system $x' = F(x)$ whose right-hand side does not depend on the time $t$ .

Why it existsThe rule of motion is the same at every moment, so a solution can be shifted in time and still be a solution, and the whole story can be told by drawing curves in space without a time axis.

Definition B

Orbit

The orbit of a solution $x(t)$ is the set of points $\{x(t)\}$ it passes through, drawn as a curve with an arrow in the direction of increasing $t$ .

In one lineThe curve in the plane traced out by one solution $x(t)$ of an autonomous system, without any record of the time.

Why it existsMany solutions (the same path, started at different times) have one orbit, and the picture of all orbits is what we study.

Definition B

Phase portrait

A phase portrait is a drawing of enough orbits, with their arrows, to show all the types of behaviour. The plane in which it is drawn is the phase plane. For the equation $y'' = G(y, y')$ the phase plane has the coordinates $(y, y')$ , as in 7.1.

In one lineA sketch of the orbits of an autonomous system in the plane, with arrows for the direction of motion and the equilibria marked.

Why it existsIt shows every possible motion at once, and the qualitative behaviour (what is attracted where) can be read from it.

Fact B MATH 376 (uniqueness, 7.1); stretch problem of 7.1

Orbits do not cross

Let $x' = F(x)$ with $F$ continuously differentiable. (a) If two solutions pass through the same point, they differ only by a shift in time. (b) If $x(t_1) = x(t_2)$ for some $t_1 \ne t_2$ , then $x$ is periodic with period $\lvert t_2 - t_1\rvert$ .

In one lineTwo orbits of an autonomous system either coincide or never meet. A solution that comes back to its starting point is periodic, and its orbit is a closed curve.

Why it is trueUniqueness of solutions for a given starting point, plus the fact that a time shift of a solution is a solution.

Proof7.1, stretch problem

The idea: Shift one solution in time to make the two meet at the same time, and apply uniqueness.

x

is a solution then so is

t \mapsto x(t + s)

, for every constant

s

Because

F

does not depend on

t

\tilde z(t) = x(t + s)

has

\tilde z'(t) = x'(t + s) = F(x(t+s)) = F(\tilde z(t))

They are solutions of the same initial value problem at time

b

Suppose

x(a) = z(b)

. Then

z(t)

and

\tilde x(t) = x(t + a - b)

are two solutions with

z(b) = \tilde x(b)

z = \tilde x

: same orbit, shifted in time.

Uniqueness (7.1, the corollary 'two solutions that meet are equal') gives

z(t) = x(t + a - b)

for all

t

. This is (a).

x

is periodic with period

\lvert t_2 - t_1\rvert

For (b) apply (a) with

z = x

a = t_2

b = t_1

x(t) = x(t + t_2 - t_1)

for all

t

\blacksquare

This is why phase portraits are pictures of curves that bend and spiral but never cross, and why any closed curve in a portrait is a periodic motion.

Equilibria and stability

Definition B

Equilibrium

A point $x^*$ is an equilibrium (also called a fixed point, a critical point or a stationary point) of $x' = F(x)$ if $F(x^*) = 0$ . For $F = -\nabla f$ the equilibria are the critical points of $f$ (5.6).

In one lineA point $x^*$ where $F(x^*) = 0$ . The constant function $x(t) = x^*$ is a solution: a particle placed there stays there.

Why it existsEquilibria are the points where the motion stops. They are the skeleton of the phase portrait.

Worked example 7.1 (logistic growth); new

The logistic equation x' = x(1 − x): find the equilibria and see which ones attract.

Solution.

Find the equilibria. $x(1 - x) = 0$ gives $x^* = 0$ and $x^* = 1$ . An equilibrium is a zero of the right-hand side.
Read the arrows on the line. For $0 < x < 1$ the right side is positive, so $x$ increases. For $x > 1$ it is negative, so $x$ decreases. For $x < 0$ it is negative, so $x$ decreases. In one dimension the phase portrait is the sign of $F$ on each interval.
Draw the conclusion. Starting near $1$ (on either side) the solution moves toward $1$ . Starting near $0$ on the right it moves away from $0$ , and on the left it moves away too. The equilibrium $1$ attracts, the equilibrium $0$ repels.
Check with the solution. The formula of 7.1 is $x(t) = \frac{1}{1 + (1/x_0 - 1)e^{-t}}$ , which tends to $1$ for every $x_0 > 0$ . It agrees with the arrows: every population that starts positive tends to the capacity $1$ .

Answer. $0$ is repelling and $1$ is attracting. The words ‘stable’ and ‘unstable’ for these will be defined next.

The three words ‘stays near’, ‘comes back’ and ‘tends to’ are different, and the definitions separate them.

Definition B

Stable

The equilibrium $x^*$ of $x' = F(x)$ is stable (in the sense of Lyapunov) if for every $\varepsilon > 0$ there is $\delta > 0$ such that

\lVert x(0) - x^*\rVert < \delta \ \Longrightarrow\ \lVert x(t) - x^*\rVert < \varepsilon \quad\text{for all } t \ge 0 .

It is unstable if it is not stable: for some $\varepsilon$ and every $\delta$ there is a solution that starts within $\delta$ and leaves the $\varepsilon$ -ball at some later time. This is the $\varepsilon$ – $\delta$ shape of continuity (1.3), with ‘for all $t \ge 0$ ’ added: the map from the starting point to the whole future path is continuous at $x^*$ .

In one lineAn equilibrium $x^*$ is stable if for every $\varepsilon > 0$ there is a $\delta > 0$ such that every solution starting within $\delta$ of $x^*$ stays within $\varepsilon$ of $x^*$ for all later times. Otherwise it is unstable.

Why it existsIt says that small causes have small effects for all time. A pencil balanced on its point is unstable; a ball in a bowl is stable.

Definition B

Asymptotically stable

The equilibrium $x^*$ is asymptotically stable if it is stable and there is $\delta_0 > 0$ such that $\lVert x(0) - x^*\rVert < \delta_0$ implies $x(t) \to x^*$ as $t \to \infty$ . (Stability is needed in the definition: there are nonlinear examples where every nearby solution eventually returns to $x^*$ but only after a large excursion.)

In one lineAn equilibrium that is stable and also attracts: there is a $\delta_0 > 0$ such that every solution starting within $\delta_0$ tends to $x^*$ as $t \to \infty$ .

Why it existsIt describes a system that forgets small disturbances. A ball in a bowl with friction ends at the bottom; without friction it keeps swinging, which is stable but not asymptotically stable.

Worked example

Which of these equilibria are stable, asymptotically stable, unstable? x' = −2x at 0; x' = x² at 0; x' = 0 at any point; x' = −x³ at 0.

Solution.

$x' = -2x$ . The solution is $x_0e^{-2t}$ , with $\lvert x(t)\rvert \le \lvert x_0\rvert$ and $x(t) \to 0$ . Take $\delta = \varepsilon$ ; the solution gets smaller and tends to $0$ . Asymptotically stable.
$x' = x^2$ . For $x_0 > 0$ the solution is $\frac{x_0}{1 - x_0t}$ , which becomes infinite at $t = 1/x_0$ . For $x_0 < 0$ it tends to $0$ . A solution starting on one side, however close, leaves every ball. Unstable (on the right it blows up, on the left it is attracted: ‘half-stable’ is still unstable).
$x' = 0$ . Every solution is constant. Starting within $\delta$ you stay within $\delta$ . Stable, not asymptotically stable: nothing attracts.
$x' = -x^3$ . The solution is $x(t) = \frac{x_0}{\sqrt{1 + 2x_0^2t}} \to 0$ , and $\lvert x(t)\rvert \le \lvert x_0\rvert$ . Check: with $u = 1 + 2x_0^2t$ we have $x = x_0u^{-1/2}$ and $x' = x_0\cdot\left(-\tfrac12\right)u^{-3/2}\cdot2x_0^2 = -x_0^3u^{-3/2} = -x^3$ . Asymptotically stable, although the linearization $x' = 0$ at this point says nothing (see below).

Answer. Asymptotically stable, unstable, stable only, asymptotically stable.

Quick checkFor the linear system

x' = Ax

with

A = \begin{bmatrix}0 & 3\\-3 & 0\end{bmatrix}

the origin is a centre: every orbit is a circle. Which statement is true?

Linear systems: the trace-determinant classification

For a linear system $x' = Ax$ the only equilibrium of interest is $x^* = 0$ (when $\det A \ne 0$ ). Everything is decided by the eigenvalues, through the solutions of 7.4. First the names of the types, in terms of eigenvalues.

Definition B

Saddle

The origin is a saddle if $A$ has real eigenvalues $\lambda_1 < 0 < \lambda_2$ . With eigenvectors $v_1, v_2$ the solutions are $c_1e^{\lambda_1t}v_1 + c_2e^{\lambda_2t}v_2$ : the line through $v_1$ is the only set of starting points that tend to the origin (the stable direction), and the line through $v_2$ is the only set of starting points that tend to it backwards in time (the unstable direction). All other orbits come near and are thrown out along $v_2$ .

In one lineThe origin of $x' = Ax$ when $A$ has two real eigenvalues of opposite signs: orbits come in along one eigendirection and leave along the other.

Why it existsExactly one direction attracts and one repels, as on a mountain pass. The system is unstable, because almost every orbit leaves.

Definition B

Node

The origin is a node if $A$ has real eigenvalues of the same sign and neither is $0$ . It is a stable node (a sink) if both are negative and an unstable node (a source) if both are positive. Two sub-cases occur when the eigenvalues are equal. A star is the case $A = \lambda I$ , where every orbit is a straight line through the origin. A degenerate node is the case of an equal eigenvalue with only one eigendirection (a Jordan block): the orbits bend around that line.

In one lineThe origin of $x' = Ax$ when $A$ has real eigenvalues of the same sign, both non-zero: every orbit runs into the origin (a stable node) or out of it (an unstable node), without turning.

Why it existsAll the motion is along or between two eigendirections, with no rotation.

Definition B

Spiral

The origin is a spiral (or focus) if $A$ has eigenvalues $a \pm ib$ with $b \ne 0$ and $a \ne 0$ . It is a stable spiral if $a < 0$ and an unstable spiral if $a > 0$ . The solutions are $e^{at}$ times a rotation, so the distance from the origin changes by the factor $e^{a\cdot\text{time}}$ while the angle turns at a rate proportional to $b$ .

In one lineThe origin of $x' = Ax$ when $A$ has non-real eigenvalues $a \pm ib$ with $a \ne 0$ : orbits wind around the origin while moving in ( $a < 0$ ) or out ( $a > 0$ ).

Why it existsThe imaginary part $b$ is the rotation speed and the real part $a$ is the rate of approach, as in $e^{at}(\cos bt, \sin bt)$ .

Definition B

Centre

The origin is a centre if the eigenvalues are $\pm ib$ with $b \ne 0$ . By 7.4 the solutions are periodic with period $2\pi/b$ , and the orbits are ellipses (circles when $A = \begin{bmatrix}0 & -b\\b & 0\end{bmatrix}$ ).

In one lineThe origin of $x' = Ax$ when $A$ has purely imaginary eigenvalues $\pm ib$ , $b \ne 0$ : every orbit is a closed curve (an ellipse) around it.

Why it existsWith no real part there is no growth or decay: the motion is a pure oscillation, like a frictionless pendulum.

Now the main theorem of the linear theory: the type and the stability are read off the two numbers $\tau = \operatorname{tr}A$ and $\Delta = \det A$ . Recall that the eigenvalues are the roots of $\lambda^2 - \tau\lambda + \Delta = 0$ , so $\lambda_1 + \lambda_2 = \tau$ , $\lambda_1\lambda_2 = \Delta$ and $\lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2}$ (2.8).

Theorem B Strogatz 5.2; the case list extends the classification theorem of 7.4

Classification of the origin by trace and determinant

For a real $2 \times 2$ matrix $A$ with trace $\tau$ and determinant $\Delta$ :

Region of the $(\tau, \Delta)$ plane	Eigenvalues	Origin of $x' = Ax$	Stability
$\Delta < 0$	real, opposite signs	saddle	unstable
$\Delta > 0$ , $\tau < 0$ , $\tau^2 > 4\Delta$	real, both negative	stable node	asymptotically stable
$\Delta > 0$ , $\tau > 0$ , $\tau^2 > 4\Delta$	real, both positive	unstable node	unstable
$\Delta > 0$ , $\tau < 0$ , $\tau^2 < 4\Delta$	$a \pm ib$ , $a < 0$	stable spiral	asymptotically stable
$\Delta > 0$ , $\tau > 0$ , $\tau^2 < 4\Delta$	$a \pm ib$ , $a > 0$	unstable spiral	unstable
$\Delta > 0$ , $\tau = 0$	$\pm ib$	centre	stable, not asymptotically
$\tau^2 = 4\Delta > 0$	equal, real, sign of $\tau$	star if $A = \frac\tau2I$ , else degenerate node	asymptotically stable if $\tau < 0$ , else unstable
$\Delta = 0$ , $\tau < 0$	$0$ and $\tau$	a line of equilibria; orbits parallel, running into it	stable, not asymptotically
$\Delta = 0$ , $\tau > 0$	$0$ and $\tau$	a line of equilibria; orbits run away from it	unstable
$\Delta = 0$ , $\tau = 0$	$0, 0$	$A = O$ (everything fixed) or a shear $x_1' = x_2$ (orbits are lines)	stable if $A = O$ , unstable if $A \ne O$

In particular every solution tends to $0$ if and only if $\tau < 0$ and $\Delta > 0$ .

In one lineReal $2 \times 2$ matrix with trace $\tau$ and determinant $\Delta$ : $\Delta < 0$ saddle (unstable); $\Delta > 0$ and $\tau^2 > 4\Delta$ node; $\Delta > 0$ and $\tau^2 < 4\Delta$ spiral if $\tau \ne 0$ , centre if $\tau = 0$ ; $\tau^2 = 4\Delta > 0$ degenerate node or star; $\Delta = 0$ a line of equilibria or worse. Asymptotically stable exactly when $\tau < 0 < \Delta$ .

Why it is trueThe eigenvalues are the roots of $\lambda^2 - \tau\lambda + \Delta$ . Their product is $\Delta$ (signs), the discriminant $\tau^2 - 4\Delta$ decides real or complex, and the sum $\tau$ (or $\tau/2$ for complex roots) decides growth or decay.

ProofStrogatz 5.2; 7.4

The idea: Compute the eigenvalues from tau and Delta, split by sign, and use the solution formulas of 7.4 in each case.

The eigenvalues are real and distinct if

\tau^2 > 4\Delta

, equal if

\tau^2 = 4\Delta

, complex conjugate if

\tau^2 < 4\Delta

. Their product is

\Delta

and their sum is

\tau

The characteristic polynomial of a

2 \times 2

matrix is

\lambda^2 - \tau\lambda + \Delta

(2.8). The quadratic formula gives

\lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2}

\Delta < 0

: a saddle, and unstable (start on the unstable direction, as close as you like).

Product

\Delta < 0

means the roots have opposite signs, so one is positive and one negative; this can only happen with real roots, since complex conjugates have product

\lvert\lambda\rvert^2 \ge 0

. By the solution

c_1e^{\lambda_1t}v_1 + c_2e^{\lambda_2t}v_2

of 7.4, the

e^{\lambda_2t}

part grows.

\Delta > 0

\tau^2 > 4\Delta

: a node, stable if

\tau < 0

and unstable if

\tau > 0

\Delta > 0

and real roots means the roots have the same sign, which is the sign of their sum

\tau

. Both exponentials

e^{\lambda_it}

then tend to

0

(

\tau < 0

) or both grow (

\tau > 0

\tau^2 < 4\Delta

: a spiral if

a = \tau/2 \ne 0

(stable if

\tau < 0

), and a centre if

\tau = 0

By the

2 \times 2

formula of 7.4,

e^{tA} = e^{at}\big(\cos bt\,I + \frac{\sin bt}{b}(A - aI)\big)

with

a = \tau/2

and

b = \sqrt{\Delta - \tau^2/4}

. The bracket is bounded and periodic in

t

, so only the factor

e^{at}

decides growth or decay. For

a = 0

the solution is periodic.

\tau^2 = 4\Delta > 0

: a star or a degenerate node, with the stability of the sign of

\tau

The repeated eigenvalue is

\lambda = \tau/2

and

A - \lambda I

is nilpotent (7.4):

e^{tA} = e^{\lambda t}(I + t(A - \lambda I))

. The polynomial factor

t

never beats the exponential if

\lambda < 0

. If

A - \lambda I = O

then

A = \lambda I

(a star), and otherwise there is a single eigendirection.

\Delta = 0

: a line (or plane) of equilibria, and the rows of the table.

\Delta = 0

0

is an eigenvalue, and every vector

v

with

Av = 0

is an equilibrium. The other eigenvalue is

\tau

. Along its eigenvector

e^{\tau t}

decays or grows. If also

\tau = 0

then

A

is nilpotent:

e^{tA} = I + tA

, which is bounded only if

A = O

So asymptotically stable if and only if

\tau < 0 < \Delta

Asymptotic stability needs all solutions to tend to

0

. From the table that happens in exactly the stable node, stable spiral and stable degenerate node cases, which together are

\tau < 0

with

\Delta > 0

(for

\Delta > 0

and

\tau < 0

the three cases cover the remaining possibilities

\tau^2 >, <, = 4\Delta

\blacksquare

In the picture above: the left panel is this table, and the parabola $\tau^2 = 4\Delta$ separates nodes (below it) from spirals (above it). A trace of $0$ is the vertical axis, a determinant of $0$ the horizontal axis. Stability needs the dot in the upper left quadrant.

Worked example MATH 376 final 5(b); worksheets 24 to 26; new

Classify the origin and decide stability for ten matrices, among them the matrix of the 376 final's problem 5(b).

Solution. Compute $\tau$ , $\Delta$ and $\tau^2 - 4\Delta$ , then look up the row of the table.

Matrix	$\tau$	$\Delta$	$\tau^2 - 4\Delta$	Eigenvalues	Origin	Stability
$\begin{bmatrix}1 & 2\\-2 & 3\end{bmatrix}$ (final 5(b))	$4$	$7$	$-12$	$2 \pm i\sqrt3$	unstable spiral	unstable
$\begin{bmatrix}-1 & -5\\1 & -3\end{bmatrix}$	$-4$	$8$	$-16$	$-2 \pm 2i$	stable spiral	asymptotically stable
$\begin{bmatrix}3 & 1\\-2 & 0\end{bmatrix}$	$3$	$2$	$1$	$2,\ 1$	unstable node	unstable
$\begin{bmatrix}2 & 1\\1 & -3\end{bmatrix}$	$-1$	$-7$	$29$	$\frac{-1 \pm \sqrt{29}}{2}$	saddle	unstable
$\begin{bmatrix}0 & 3\\-3 & 0\end{bmatrix}$	$0$	$9$	$-36$	$\pm3i$	centre	stable only
$\begin{bmatrix}-2 & 0\\0 & -2\end{bmatrix}$	$-4$	$4$	$0$	$-2, -2$	stable star	asymptotically stable
$\begin{bmatrix}-2 & 1\\0 & -2\end{bmatrix}$	$-4$	$4$	$0$	$-2, -2$	stable degenerate node	asymptotically stable
$\begin{bmatrix}-1 & 0\\0 & 0\end{bmatrix}$	$-1$	$0$	$1$	$-1,\ 0$	line of equilibria (the $x_2$ -axis)	stable only
$\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$	$0$	$0$	$0$	$0, 0$	shear	unstable
$\begin{bmatrix}-3 & 0\\0 & 1\end{bmatrix}$	$-2$	$-3$	$16$	$-3,\ 1$	saddle	unstable

Compute the trace and determinant. For the first row $\tau = 1 + 3 = 4$ and $\Delta = 1\cdot3 - 2\cdot(-2) = 7$ . The two numbers are all that the table needs.
Compute the discriminant $\tau^2 - 4\Delta$ . It separates real from complex eigenvalues. $16 - 28 = -12 < 0$ so the first matrix has complex eigenvalues, with real part $\tau/2 = 2 > 0$ .
The line of equilibria. For $\begin{bmatrix}-1 & 0\\0 & 0\end{bmatrix}$ the equation is $x_1' = -x_1$ , $x_2' = 0$ : every point $(0, x_2)$ is an equilibrium, and orbits are horizontal lines running into that axis. Stable (nearby solutions end near) but not asymptotically stable: a point near $(0, 0)$ ends at a different equilibrium, not at $(0, 0)$ .
The shear. For $\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$ , $e^{tA} = I + tA$ : a point $(x_1, x_2)$ moves to $(x_1 + tx_2, x_2)$ . Any point off the $x_1$ -axis drifts away linearly: unstable although no eigenvalue is positive. This is the ‘polynomial factor’ of the Jordan block at work.
Compare saddles with a negative trace. The fourth and last matrices have $\tau < 0$ and are still unstable. A negative trace does not give stability; $\Delta > 0$ is needed as well.

Answer. As in the last two columns. Only the second, sixth and seventh rows are asymptotically stable.

Quick checkA real

2 \times 2

matrix has trace

-3

and determinant

-2

. What is the origin of

x' = Ax

Your answerA real

2 \times 2

matrix has trace

-2

and determinant

5

. At what angular speed do its solutions turn around the origin (the imaginary part of the eigenvalues)?

Why· How to sketch each portrait in thirty seconds

Saddle: draw the two eigenlines, arrows in along the negative eigenvalue and out along the positive one, and hyperbola-like curves between them. Node: draw the two eigenlines, arrows all in (or all out); the curves hug the direction of the eigenvalue that is smaller in absolute value, since that part of the solution lasts longer, and leave tangent to the other. Spiral: a spiral turning in the direction given by the sign of the lower left entry (if $\dot x_2 > 0$ when $x_2 = 0$ , $x_1 > 0$ , it turns counter-clockwise). Centre: ellipses, in the same sense. Degenerate node: one eigenline, curves bending around it and parallel to it at the ends. Pick the starting direction by evaluating $Ax$ at one point, such as $(1, 0)$ : that gives the arrow at that point. The tangent claim for the node: for $\lambda_1 < \lambda_2 < 0$ the solution $c_1e^{\lambda_1t}v_1 + c_2e^{\lambda_2t}v_2$ is dominated by the $v_2$ term for large $t$ , so the orbits arrive tangent to $v_2$ (the slow direction).

The table is for $2 \times 2$ matrices, but the question ‘does everything tend to $0$ ?’ has an answer for any size.

Theorem B standard; follows from the Jordan form and 7.4

Stability of a linear system

(a) If every eigenvalue of $A$ has real part below $-\mu < 0$ , then there is $C$ with $\lVert e^{tA}\rVert \le Ce^{-\mu t}$ for $t \ge 0$ ; the origin is asymptotically stable. (b) If some eigenvalue has positive real part, the origin is unstable. (c) If all real parts are at most $0$ , the origin is stable if and only if every eigenvalue with real part $0$ is semisimple (its Jordan blocks are all $1 \times 1$ ); it is then not asymptotically stable.

In one lineFor $x' = Ax$ with $A$ real $n \times n$ : the origin is asymptotically stable if every eigenvalue has negative real part (and then $\lVert e^{tA}\rVert \le Ce^{-\mu t}$ ); unstable if some eigenvalue has positive real part. If all real parts are $\le 0$ it is stable exactly when every eigenvalue with real part $0$ has only $1 \times 1$ Jordan blocks.

Why it is trueThe entries of $e^{tA}$ are combinations of $t^je^{\lambda t}$ . An exponential with negative real part beats every power of $t$ , one with positive real part grows, and on the imaginary axis a power $t^j$ with $j \ge 1$ (a Jordan block of size at least $2$ ) makes the solution drift off.

Proof7.4 (Jordan blocks); standard

The idea: Write e^{tA} in Jordan form. Each entry is a sum of terms t^j e^{lambda t}. Compare the exponential with the power, in each case.

So every entry of

e^{tA}

is a combination of the functions

t^je^{\lambda t}

, with

\lambda

an eigenvalue and

j

less than the size of its block.

A = CJC^{-1}

(3.6) and

e^{tA} = Ce^{tJ}C^{-1}

(7.4). For a Jordan block of size

m

with eigenvalue

\lambda

the exponential has entries

\frac{t^j}{j!}e^{\lambda t}

for

0 \le j \le m - 1

For some

C

\lvert t^je^{\lambda t}\rvert \le Ce^{-\mu t}

for all

t \ge 0

. Hence

\lVert e^{tA}\rVert \le C'e^{-\mu t}

(a) Every eigenvalue has real part below

-\mu

, and there are finitely many, so there is

\eta > 0

with

\operatorname{Re}\lambda \le -\mu - \eta

for all of them. Then

\lvert t^je^{\lambda t}\rvert = t^je^{(\operatorname{Re}\lambda)t} \le \big(t^je^{-\eta t}\big)e^{-\mu t}

, and

t^je^{-\eta t}

is bounded on

[0,\infty)

(the exponential wins; the maximum is at

t = j/\eta

The origin is stable and every solution tends to

0

: asymptotically stable.

\lVert x(t)\rVert = \lVert e^{tA}x_0\rVert \le C'e^{-\mu t}\lVert x_0\rVert \le C'\lVert x_0\rVert

. Given

\varepsilon

, take

\delta = \varepsilon/C'

for stability; the bound also gives

x(t) \to 0

Then

\lVert x(t)\rVert \ge ce^{at}\to\infty

(b) Let

\lambda = a + ib

with

a > 0

and eigenvector

v = u + iw

(

u, w

real). If

b = 0

take

w = 0

. Then

x(t) = e^{at}(\cos bt\,u - \sin bt\,w)

is a real solution. If

b \ne 0

then

u

and

w

are linearly independent (otherwise

v

would be a complex multiple of a real vector, and

\lambda

would be real), so

\lVert\cos bt\,u - \sin bt\,w\rVert \ge c > 0

for all

t

(a continuous periodic function that is never

0

Hence the origin is unstable.

Multiplying the starting vector

x(0) = u

by a small number

\kappa

multiplies the whole solution by

\kappa

. So solutions that start as close to

0

as we like become larger than any

\varepsilon

An eigenvalue on the imaginary axis with a block of size at least

2

gives solutions growing like

t

: unstable.

\lambda = ib

has a block of size

\ge 2

there is a generalized eigenvector

z

with

(A - \lambda I)z \ne 0

and

(A - \lambda I)^2z = 0

. Then

e^{tA}z = e^{\lambda t}\big(z + t(A - \lambda I)z\big)

(nilpotent case of 7.4), whose size grows like

t

. For a real matrix take real and imaginary parts as in (b).

Otherwise

\lVert e^{tA}\rVert \le M

, so the origin is stable, and not asymptotically stable.

If all those blocks are

1 \times 1

, the entries belonging to them are

e^{ibt}

, of modulus

1

, and the others decay. So

\lVert e^{tA}\rVert

is bounded for

t \ge 0

, say by

M

, and the argument of (a) with

C' = M

gives stability. But an eigenvector for

ib

gives a solution

e^{ibt}v

that keeps its size forever, so there is no attraction.

\blacksquare

Linearization: a microscope at an equilibrium

Now a nonlinear system $x' = F(x)$ with $F$ continuously differentiable, and an equilibrium $x^*$ . Write $x = x^* + u$ , with $u$ small. Since $F(x^*) = 0$ , the definition of the total derivative (5.3) says

u' = F(x^* + u) = J\,u + g(u), \qquad J = DF(x^*), \qquad \frac{\lVert g(u)\rVert}{\lVert u\rVert} \to 0 \ \text{ as } u \to 0 . \tag{N}

Here $J$ is the Jacobian matrix of $F$ at $x^*$ (5.3) and $g$ is everything beyond the linear term. The linearization (5.3) of the system at $x^*$ is the linear system $u' = Ju$ , obtained by throwing $g$ away. Near $u = 0$ the thrown-away term is smaller than the term we keep; the question is when that survives integration over time.

Definition B

Hyperbolic equilibrium

An equilibrium $x^*$ is hyperbolic if every eigenvalue of $DF(x^*)$ has non-zero real part. In the plane that means: a saddle, a node or a spiral, but not a centre, not a line of equilibria and not a shear.

In one lineAn equilibrium $x^*$ of $x' = F(x)$ where no eigenvalue of the Jacobian matrix $DF(x^*)$ has real part $0$ .

Why it existsWith no eigenvalue on the imaginary axis, the linear part decides the local picture, and small nonlinear terms cannot change it. A zero real part is exactly where the linear terms are weak enough for the nonlinear terms to matter.

Theorem B Grobman (1959), Hartman (1960); stated without proof

Hartman–Grobman theorem

Let $x^*$ be a hyperbolic equilibrium of $x' = F(x)$ , with $F$ continuously differentiable, and let $J = DF(x^*)$ . Then there are neighbourhoods $U$ of $x^*$ and $V$ of $0$ and a homeomorphism $h : U \to V$ (continuous, with a continuous inverse, but not necessarily differentiable) that maps orbits of $x' = F(x)$ in $U$ to orbits of $u' = Ju$ in $V$ , preserving the direction of time.

In one lineNear a hyperbolic equilibrium, a nonlinear system has the same phase portrait as its linearization, up to a continuous change of coordinates that preserves the direction of time.

Why it is trueA hyperbolic linear flow is structurally stable: a small perturbation changes the orbits only a little, and a continuous distortion of the plane can undo the change.

Consequences: near a hyperbolic equilibrium, a saddle is a saddle, a sink is a sink and a source is a source, so stability is decided. Since $h$ is only continuous, the theorem does not by itself separate a node from a spiral; for stability that makes no difference. The theorem is about the neighbourhood of $x^*$ only; far away the nonlinear picture is something else (look at the pendulum: the picture near $(0,0)$ says nothing about the saddle at $(\pi, 0)$ ).

What the theorem does not cover is the non-hyperbolic case: a centre in the linearization, for example. There the thrown-away term $g$ is exactly what decides.

Worked example Strogatz 6.3; new

A linear centre that is a spiral: x' = −y + μx(x² + y²), y' = x + μy(x² + y²).

Solution.

Linearize at $(0, 0)$ . $J = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ whatever $\mu$ is, because the added terms are cubic. Eigenvalues $\pm i$ : a centre in the linearization. Cubic terms have zero derivative at the origin.
Go to polar coordinates. With $r^2 = x^2 + y^2$ : $r r' = xx' + yy' = \mu r^4$ , so $r' = \mu r^3$ . And $\theta' = \frac{xy' - yx'}{r^2} = 1$ . $xx' + yy' = -xy + \mu x^2r^2 + xy + \mu y^2r^2 = \mu r^4$ . The angular rate is $1$ whatever $\mu$ is.
Read the nonlinear picture. If $\mu < 0$ , $r' < 0$ : $r$ decreases and $\theta$ increases at a steady rate: a stable spiral, and the origin is asymptotically stable. If $\mu > 0$ , an unstable spiral. If $\mu = 0$ , a centre. The sign of the cubic term decides, and the linearization cannot see it.
Compare the rates. For $\mu < 0$ , solving $r' = \mu r^3$ gives $r(t) = \frac{r_0}{\sqrt{1 + 2\lvert\mu\rvert r_0^2t}}$ , decay like $t^{-1/2}$ , much slower than any exponential. It is the same equation as $x' = -x^3$ in the example of stability, since the linear part has no real part to produce an exponential.

Answer. The same linear centre, three different nonlinear portraits.

Quick checkFor the system

x' = -x^3

y' = -y

the Jacobian at the origin is

\operatorname{diag}(0, -1)

. What does the linearization tell you about stability of the origin?

The next theorem is the part of the story for which we can give a proof, because it uses the matrix exponential of 7.4. It gives the conclusion ‘asymptotically stable’ from the eigenvalues of $J$ alone.

Theorem B Lyapunov (indirect method); Strogatz 6.3; Hirsch–Smale–Devaney 8.2

Linearization and stability

Let $F$ be continuously differentiable with $F(x^*) = 0$ and $J = DF(x^*)$ . (a) If all eigenvalues of $J$ have negative real part, $x^*$ is asymptotically stable. (b) If some eigenvalue has positive real part, $x^*$ is unstable.

In one lineIf every eigenvalue of $J = DF(x^*)$ has negative real part, then $x^*$ is asymptotically stable for the nonlinear system. If some eigenvalue has positive real part, it is unstable.

Why it is trueBuild an energy-like function $V(u) = u^{\mathsf T}Pu$ from the linear part, with $P = \int_0^\infty e^{tJ^{\mathsf T}}e^{tJ}\,dt$ . Along the linear flow $V$ decreases like $-\lVert u\rVert^2$ , which beats the smaller nonlinear remainder $g$ .

ProofStrogatz 6.3; Hirsch–Smale–Devaney 8.2

The idea: Part (a): construct the matrix P as an integral of e^(tJ) transposed times e^(tJ). Show V = u P u is positive and its derivative along the linear flow is minus the squared length. The nonlinear remainder is smaller near zero, so V still decreases. Part (b) is quoted.

Define

P = \displaystyle\int_0^\infty e^{tJ^{\mathsf T}}e^{tJ}\,dt

. The integrand is bounded in norm by

C^2e^{-2\mu t}

, so the integral converges.

By the previous theorem, part (a):

\lVert e^{tJ}\rVert \le Ce^{-\mu t}

for

t \ge 0

with some

C, \mu > 0

P

is symmetric and positive definite (3.2).

(e^{tJ})^{\mathsf T} = e^{tJ^{\mathsf T}}

(transpose the series), so

u^{\mathsf T}Pu = \int_0^\infty\lVert e^{tJ}u\rVert^2dt

. The integrand is continuous, non-negative, and equal to

\lVert u\rVert^2 > 0

t = 0

for

u \ne 0

P

is symmetric because the integrand is.

J^{\mathsf T}P + PJ = \big[e^{tJ^{\mathsf T}}e^{tJ}\big]_0^\infty = 0 - I = -I

Differentiate

e^{tJ^{\mathsf T}}e^{tJ}

with the product rule and the derivative of the matrix exponential:

\frac{d}{dt}\big(e^{tJ^{\mathsf T}}e^{tJ}\big) = J^{\mathsf T}e^{tJ^{\mathsf T}}e^{tJ} + e^{tJ^{\mathsf T}}e^{tJ}J

. Integrate from

0

\infty

(the constant matrices

J^{\mathsf T}

and

J

come out of the integral):

J^{\mathsf T}P + PJ

equals the limit at

\infty

minus the value at

0

Let

V(u) = u^{\mathsf T}Pu

. Then

\dot V = u^{\mathsf T}(J^{\mathsf T}P + PJ)u + 2u^{\mathsf T}Pg(u) = -\lVert u\rVert^2 + 2u^{\mathsf T}Pg(u)

Product rule for

V = u^{\mathsf T}Pu

along a solution of (N):

\dot V = (u')^{\mathsf T}Pu + u^{\mathsf T}Pu'

, with

u' = Ju + g(u)

and

P

symmetric.

\dot V \le -\frac12\lVert u\rVert^2

for

\lVert u\rVert < r

Because

\lVert g(u)\rVert/\lVert u\rVert \to 0

there is

r > 0

with

\lVert g(u)\rVert \le \frac{\lVert u\rVert}{4\lVert P\rVert}

for

\lVert u\rVert < r

. Then

\lvert 2u^{\mathsf T}Pg(u)\rvert \le 2\lVert u\rVert\lVert P\rVert\lVert g\rVert \le \frac12\lVert u\rVert^2

On the set

\{V < m\}

\dot V \le -\frac{1}{2\lambda_{\max}}V

Let

\lambda_{\min}, \lambda_{\max}

be the smallest and largest eigenvalues of

P

(both positive, 3.2). Then

\lambda_{\min}\lVert u\rVert^2 \le V \le \lambda_{\max}\lVert u\rVert^2

. Put

m = \lambda_{\min}r^2

: the set

\{V < m\}

lies inside the ball

\lVert u\rVert < r

So every solution starting near

0

stays near

0

and tends to

0

x^*

is asymptotically stable. (Part (b) needs a different construction, a function built from the unstable and stable parts of

J

that grows along solutions; we quote it from Hirsch–Smale–Devaney 8.2.)

A solution starting in

\{V < m\}

cannot leave it, since

V

decreases there. So it stays in the ball, the inequality holds for all time, and

V(u(t)) \le V(u(0))e^{-ct}

with

c = \frac{1}{2\lambda_{\max}}

(compare

\frac{d}{dt}(e^{ct}V) \le 0

). Stability:

\lVert u(t)\rVert^2 \le V(u(t))/\lambda_{\min} \le \lambda_{\max}\lVert u(0)\rVert^2/\lambda_{\min}

. Attraction:

V \to 0

forces

u \to 0

\blacksquare

Worked example Strogatz 6.4 ('rabbits versus sheep', same equations)

Two species competing for food: x' = x(3 − x − 2y), y' = y(2 − x − y). Find the equilibria, classify them, and describe what happens in the long run.

Solution.

Find all equilibria. Each factor can vanish: $x = 0$ or $x + 2y = 3$ , and $y = 0$ or $x + y = 2$ . The four combinations give $(0, 0)$ ; $x = 0$ with $x + y = 2$ : $(0, 2)$ ; $y = 0$ with $x + 2y = 3$ : $(3, 0)$ ; and $x + 2y = 3$ with $x + y = 2$ : $(1, 1)$ . Missing an equilibrium is the commonest error (the 376 worksheet on critical points warned about it): list all four combinations of ‘first factor zero’ and ‘second factor zero’.
Compute the Jacobian. $DF = \begin{bmatrix}3 - 2x - 2y & -2x\\-y & 2 - x - 2y\end{bmatrix}$ . Differentiate each component with respect to $x$ and $y$ .
Evaluate at each equilibrium.

Equilibrium	$DF$	$\tau$	$\Delta$	Type	Stability
$(0, 0)$	$\begin{bmatrix}3 & 0\\0 & 2\end{bmatrix}$	$5$	$6$	unstable node	unstable (both species grow from small numbers)
$(3, 0)$	$\begin{bmatrix}-3 & -6\\0 & -1\end{bmatrix}$	$-4$	$3$	stable node	asymptotically stable ( $x$ wins)
$(0, 2)$	$\begin{bmatrix}-1 & 0\\-2 & -2\end{bmatrix}$	$-3$	$2$	stable node	asymptotically stable ( $y$ wins)
$(1, 1)$	$\begin{bmatrix}-1 & -2\\-1 & -1\end{bmatrix}$	$-2$	$-1$	saddle	unstable

Read the long-run behaviour. All four equilibria are hyperbolic, so Hartman–Grobman and the stability theorem apply. Two sinks ( $(3,0)$ and $(0,2)$ ) and a saddle between them: one species always dies out, and which one depends on the starting point. The stable direction of the saddle is the dividing curve between the two basins. The saddle’s two stable orbits form the boundary; every starting point on one side is attracted to one sink, on the other side to the other sink.

Answer. Competitive exclusion: there is no stable coexistence ( $(1, 1)$ is a saddle). Drag the dot across the dividing curve in the picture below.

Your answerAt the coexistence point

(1, 1)

x' = x(3 - x - 2y)

y' = y(2 - x - y)

, what is the determinant of the Jacobian matrix?

The pendulum, in full

Worked example Strogatz 6.7; new numbers

The pendulum θ″ = −sin θ − cθ′: all the equilibria, their Jacobians and types, the energy, and what the linearization misses.

Solution. Try the picture first: the first preset is this system, with a friction slider. Then follow the steps.

Write it as a system. $x_1 = \theta$ (angle from the hanging position) and $x_2 = \omega = \theta'$ : $x_1' = x_2$ , $x_2' = -\sin x_1 - cx_2$ , with friction $c \ge 0$ and time measured in units of $\sqrt{L/g}$ . The reduction to a first-order system from 7.1.
Find all equilibria. $x_2 = 0$ and $\sin x_1 = 0$ , so $x_1 = k\pi$ for every integer $k$ : the points $(k\pi, 0)$ . Even $k$ are the hanging position (the same physical state, since angles differ by $2\pi$ ), odd $k$ the upside-down position.
Compute the Jacobian. $DF = \begin{bmatrix}0 & 1\\-\cos x_1 & -c\end{bmatrix}$ . Differentiate $x_2$ and $-\sin x_1 - cx_2$ with respect to $x_1$ and $x_2$ .
Hanging position $(0, 0)$ . $DF = \begin{bmatrix}0 & 1\\-1 & -c\end{bmatrix}$ with $\tau = -c$ , $\Delta = 1$ and $\tau^2 - 4\Delta = c^2 - 4$ . So: for $0 < c < 2$ a stable spiral (the pendulum swings with decaying amplitude), for $c = 2$ a degenerate node, for $c > 2$ a stable node (overdamped: it creeps back without swinging). All asymptotically stable by the theorem on linearization and stability. For these values the eigenvalues have real part $-c/2 < 0$ .
The frictionless hanging position, $c = 0$ . Now $\tau = 0$ , $\Delta = 1$ : a centre in the linearization, with eigenvalues $\pm i$ . The point is not hyperbolic, so the linearization cannot tell. Real part zero: the case where the nonlinear terms decide.
Upside-down position $(\pi, 0)$ . $DF = \begin{bmatrix}0 & 1\\1 & -c\end{bmatrix}$ , $\tau = -c$ , $\Delta = -1 < 0$ : a saddle for every $c$ , so unstable. For $c = 0$ the eigenvalues are $\pm1$ , with unstable direction $(1, 1)$ . Hyperbolic, so Hartman–Grobman applies. A tiny push off the top makes the displacement grow like $e^{t}$ .
Settle the frictionless case with energy. For $c = 0$ let $E = \frac12x_2^2 - \cos x_1$ (kinetic plus potential energy per unit mass, 6.2). Then $\dot E = x_2x_2' + \sin x_1\,x_1' = x_2(-\sin x_1) + \sin x_1\,x_2 = 0$ . Energy is conserved, so each orbit lies in a level set $E =$ const.
Describe the level sets. At $(0,0)$ the function $E$ has a strict minimum $E = -1$ (its Hessian is $\operatorname{diag}(1, 1)$ , positive definite, 5.6), so for $-1 < E < 1$ the level set around it is a closed curve with no equilibrium on it: the orbit is a closed loop (the pendulum swings, periodic motion). For $E = 1$ the level set is $x_2 = \pm2\cos\frac{x_1}{2}$ through the saddles (the separatrix; the pendulum takes infinitely long to arrive at the top). For $E > 1$ , $x_2$ never vanishes and the pendulum goes round and round. $E = 1$ gives $\frac12x_2^2 = 1 + \cos x_1 = 2\cos^2\frac{x_1}2$ .
Conclude. The frictionless hanging equilibrium is a true centre of the nonlinear system and is stable (not asymptotically), by the conservation of $E$ . The linearization suggested this, and energy proves it.
The period depends on the amplitude. The small-oscillation period is $2\pi$ (units of $\sqrt{L/g}$ ) but a swing of amplitude $\theta_0$ takes $4K(\sin\frac{\theta_0}{2})$ , which is $1.0005\times$ as long at $5^\circ$ , $1.18\times$ at $90^\circ$ and $1.76\times$ at $150^\circ$ (the lab below computes these). For a pendulum of length $1$ m, $2\pi\sqrt{L/g} = 2.006$ s, and a $60^\circ$ swing takes $2.153$ s. The linearization is correct only in the limit of small amplitude.

Answer. Stable centres at $(2k\pi, 0)$ , saddles at $((2k+1)\pi, 0)$ when $c = 0$ ; with friction the centres become asymptotically stable spirals (nodes if $c \ge 2$ ) and the saddles stay.

Your answerA damped pendulum has friction

c = 0.5

(in the units above). Near the hanging position the solutions oscillate like

e^{-t/4}\cos(\omega t)

. What is

\omega

Predator and prey

Worked example Strogatz 6.4; Volterra 1926; new numbers

The Lotka–Volterra system x′ = x(3 − y), y′ = y(x − 2): equilibria, linearization, the conserved quantity, and what harvesting does.

Solution. The picture first: the second preset. The slider adds crowding among the prey.

Meaning. $x$ is the prey, $y$ the predators. Without predators the prey grow at rate $3$ per unit time. Predators die at rate $2$ when there is no prey, and gain from meeting prey. The terms $-xy$ and $+xy$ model meetings: each meeting removes prey and feeds predators.
Equilibria. $x(3 - y) = 0$ and $y(x - 2) = 0$ . If $x = 0$ the second equation gives $y = 0$ ; if $y = 3$ it gives $x = 2$ . So $(0, 0)$ and $(2, 3)$ . Each equation is a product of two factors; try the cases and discard those that contradict.
Jacobian. $DF = \begin{bmatrix}3 - y & -x\\y & x - 2\end{bmatrix}$ . Product rule on $x(3-y)$ and $y(x-2)$ .
At $(0, 0)$ . $DF = \operatorname{diag}(3, -2)$ : a saddle. Both axes are orbits: prey alone grow, predators alone die. Eigenvalues of opposite signs.
At $(2, 3)$ . $DF = \begin{bmatrix}0 & -2\\3 & 0\end{bmatrix}$ , $\tau = 0$ , $\Delta = 6$ : eigenvalues $\pm i\sqrt6$ , a centre in the linearization with period $\frac{2\pi}{\sqrt6} = 2.565$ . Not hyperbolic. Same situation as the frictionless pendulum: the linearization cannot decide.
A conserved quantity decides. Put $H(x, y) = x - 2\ln x + y - 3\ln y$ for $x, y > 0$ . Then $\dot H = (1 - \frac2x)x(3 - y) + (1 - \frac3y)y(x - 2) = (x - 2)(3 - y) + (y - 3)(x - 2) = 0$ . The proof is the next theorem. Conservation means the orbits lie on the curves $H =$ const.
Averages over a cycle. On a periodic orbit of period $T$ , $\int_0^T\frac{x'}{x}dt = \ln x(T) - \ln x(0) = 0$ . But $\frac{x'}{x} = 3 - y$ , so $\frac1T\int_0^Ty\,dt = 3$ . Likewise $\frac1T\int_0^Tx\,dt = 2$ . The time averages equal the equilibrium values $(2, 3)$ , however large the cycle.
Harvesting both species at rate $h$ per capita. The equations become $x' = x(3 - h - y)$ , $y' = y(x - 2 - h)$ , with equilibrium $(2 + h, 3 - h)$ . For $h = 0.5$ the averages change from $(2, 3)$ to $(2.5, 2.5)$ : the prey average goes UP and the predator average goes DOWN. Volterra’s principle: moderate indiscriminate harvesting favours the prey. Stopping it (as when fishing nearly ceased in the Adriatic during the First World War) lets the predators increase.
The centre is fragile. Add crowding among the prey: $x' = x(3 - y - \varepsilon x)$ . The equilibrium moves to $(2, 3 - 2\varepsilon)$ and $DF = \begin{bmatrix}-2\varepsilon & -2\\3 - 2\varepsilon & 0\end{bmatrix}$ has $\tau = -2\varepsilon < 0$ and $\Delta = 2(3 - 2\varepsilon) > 0$ : a stable spiral for small $\varepsilon$ (the discriminant $4\varepsilon^2 - 8(3 - 2\varepsilon)$ is negative while $\varepsilon < 1.16$ ). The closed orbits disappear for any $\varepsilon > 0$ . The pure Lotka–Volterra model has a special conservation law that real populations lack, so its exact cycles should not be taken literally.

Answer. A saddle at the origin and a nonlinear centre at $(2, 3)$ with period $\approx2.57$ near it; averages $(2, 3)$ ; harvesting raises the average prey, lowers the average predators; any prey crowding turns the cycles into a spiral.

Theorem B Volterra 1926; Strogatz 6.4

The Lotka–Volterra conservation law

Let $a, b, c, d > 0$ . Then $\dot H = 0$ along every solution with $x, y > 0$ ; every orbit in the open first quadrant other than the equilibrium $(c/d, a/b)$ is periodic; and over one period the averages of $x$ and $y$ are $c/d$ and $a/b$ .

In one lineFor $x' = x(a - by)$ , $y' = y(dx - c)$ with positive constants, $H = dx - c\ln x + by - a\ln y$ is constant along solutions in the open first quadrant. Every orbit there is a closed curve around the equilibrium $(c/d,\ a/b)$ , and the average of $x$ and of $y$ over a period equal its equilibrium value.

Why it is true $H$ is a sum of a function of $x$ alone and a function of $y$ alone, each with one minimum, so its level sets are loops around the minimum. The derivative along solutions is zero because the two factors in the equations are exactly the partial derivatives of $H$ , rescaled.

ProofStrogatz 6.4

The idea: Differentiate H along solutions, factor, and watch the terms cancel. Then use that H is a convex function with one minimum and compact level sets. Averages come from integrating the logarithmic derivative.

\dot H = \left(d - \frac cx\right)x(a - by) + \left(b - \frac ay\right)y(dx - c)

Chain rule:

\dot H = H_xx' + H_yy'

with

H_x = d - \frac cx

and

H_y = b - \frac ay

\dot H = (dx - c)(a - by) + (by - a)(dx - c)

\left(d - \frac cx\right)x = dx - c

and

\left(b - \frac ay\right)y = by - a

\dot H = (dx - c)\big[(a - by) + (by - a)\big] = 0

Factor

dx - c

; the brackets are negatives of each other.

H

is strictly convex, tends to

\infty

at the boundary of the quadrant, and has its minimum at the equilibrium.

H = f(x) + g(y)

with

f(x) = dx - c\ln x

and

g(y) = by - a\ln y

. Then

f'' = \frac{c}{x^2} > 0

g'' = \frac{a}{y^2} > 0

, and

f \to \infty

x \to 0^+

\infty

g \to \infty

y \to 0^+

\infty

. So

H

has a unique minimum at

(c/d, a/b)

, and every level set

H = h

is a bounded closed set inside the open quadrant.

Every orbit is a closed loop.

For

h

above the minimum, the level set is a simple closed curve around the minimum (it is the boundary of a convex region). An orbit stays on it, and there is no equilibrium on it, so the velocity is never

0

. A point moving with non-zero velocity along a closed curve goes round and returns: the orbit is periodic (by the fact that orbits do not cross).

\frac1T\int_0^Ty\,dt = \frac ab

and

\frac1T\int_0^Tx\,dt = \frac cd

\frac{x'}{x} = a - by

, so over a period

T

0 = \ln x(T) - \ln x(0) = aT - b\int_0^Ty\,dt

. Similarly

\frac{y'}{y} = dx - c

\blacksquare

Proof workshopRebuild: the Lotka–Volterra quantity is conservedVolterra 1926

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\dot H = \left(d - \frac cx\right)x' + \left(b - \frac ay\right)y'$ .
$\dot H = \left(d - \frac cx\right)x(a - by) + \left(b - \frac ay\right)y(dx - c)$ .
$\dot H = (dx - c)(a - by) + (by - a)(dx - c)$ .
$\dot H = (dx - c)\big[(a - by) + (by - a)\big]$ .
$\dot H = 0$ .

Quick checkIn the Lotka–Volterra model, hunting removes both species at the same per-capita rate

h

. How does the AVERAGE number of prey over a cycle change?

Your answerNear the equilibrium

(2, 3)

x' = x(3 - y)

y' = y(x - 2)

, small cycles have period

2\pi/\omega

with

\omega^2

equal to the determinant of the Jacobian. What is the period? (Two decimals.)

Your answerWith harvesting at per-capita rate

h = 0.5

x' = x(3 - y)

y' = y(x - 2)

(so

x' = x(3 - h - y)

y' = y(x - 2 - h)

), what is the average number of prey over a cycle?

Lyapunov functions

Conservation laws and decreasing energies are a recurring way to prove stability without solving anything. Lyapunov made it into a method.

Definition B

Lyapunov function

Let $x^*$ be an equilibrium of $x' = F(x)$ and $U$ a neighbourhood of $x^*$ . A continuously differentiable function $V : U \to \mathbb R$ is a Lyapunov function if $V(x^*) = 0$ , $V(x) > 0$ for $x \ne x^*$ , and $\dot V(x) = \nabla V(x)\cdot F(x) \le 0$ on $U$ . It is strict if $\dot V(x) < 0$ for $x \ne x^*$ . Along a solution, $\frac{d}{dt}V(x(t)) = \dot V(x(t))$ by the chain rule (5.4), so $V$ never increases.

In one lineA function $V$ with $V(x^*) = 0$ and $V > 0$ elsewhere near the equilibrium $x^*$ , whose derivative along solutions, $\dot V = \nabla V \cdot F$ , is never positive. It is called strict if $\dot V < 0$ away from $x^*$ .

Why it existsIt is a generalised energy: if it never increases, a solution that starts with little energy never gets far, and if it strictly decreases the solution runs down to the equilibrium.

Theorem B Lyapunov 1892; Strogatz 7.2; Hirsch–Smale–Devaney 9.2

Lyapunov's direct method

Let $x^*$ be an equilibrium of $x' = F(x)$ , $F$ continuously differentiable. (a) If a Lyapunov function exists near $x^*$ , then $x^*$ is stable. (b) If a strict Lyapunov function exists, then $x^*$ is asymptotically stable.

In one lineIf there is a Lyapunov function near $x^*$ , then $x^*$ is stable. If there is a strict one, $x^*$ is asymptotically stable.

Why it is trueLevel sets of $V$ are nested shells around $x^*$ . A solution cannot cross from a lower level of $V$ to a higher one, so it is trapped inside the shell it started in, and strict decrease pushes it through lower and lower shells to the centre.

ProofStrogatz 7.2; Hirsch–Smale–Devaney 9.2

The idea: Stability: on a sphere around x* the function V has a positive minimum m. Start so close that V is below m. V cannot increase, so the solution can never reach the sphere. Asymptotic stability: the limit of V along the solution must be 0, since otherwise V would decrease at a fixed speed for ever.

Let

m = \min_{S_\varepsilon}V > 0

Choose

\varepsilon > 0

small enough that the closed ball

B_\varepsilon = \{\lVert x - x^*\rVert \le \varepsilon\}

lies in

U

. The sphere

S_\varepsilon = \{\lVert x - x^*\rVert = \varepsilon\}

is compact and

V

is continuous and positive on it, so it has a positive minimum (1.3, extended to compact sets).

Choose

\delta < \varepsilon

with

V < m

on the

\delta

-ball.

V

is continuous with

V(x^*) = 0

, so

V < m

on a small ball

\lVert x - x^*\rVert < \delta

. We may take

\delta < \varepsilon

While the solution is in

B_\varepsilon

V(x(t)) < m

Start a solution in the

\delta

-ball. As long as it stays in

B_\varepsilon

\frac{d}{dt}V(x(t)) = \dot V \le 0

, so

V(x(t)) \le V(x(0)) < m

So the solution never reaches

S_\varepsilon

and stays in

B_\varepsilon

for all

t \ge 0

x^*

is stable. This proves (a).

On the sphere

S_\varepsilon

we have

V \ge m

. A solution that left

B_\varepsilon

would have to cross

S_\varepsilon

(it moves continuously), at a time when

V \ge m

, contradicting the previous step.

Take a solution starting in the

\delta

-ball, and suppose

V(x(t)) \to \ell > 0

(b) With a strict function,

V(x(t))

decreases and is bounded below by

0

, so it has a limit

\ell \ge 0

. Suppose

\ell > 0

. Then

V(x(t)) \ge \ell

for all

t

, and since

V < \ell

on some small ball

\lVert x - x^*\rVert < \rho

, the solution stays in the compact shell

K = \{\rho \le \lVert x - x^*\rVert \le \varepsilon\}

That is impossible, so

\ell = 0

On the compact set

K

the continuous function

\dot V

is negative, so

\dot V \le -\gamma

for some

\gamma > 0

. Then

V(x(t)) \le V(x(0)) - \gamma t

, which becomes negative for large

t

, contradicting

V > 0

Since

V(x(t)) \to 0

x(t) \to x^*

: asymptotically stable.

For each

\rho > 0

the minimum of

V

over the shell

\{\rho \le \lVert x - x^*\rVert \le \varepsilon\}

is positive, say

m_\rho

. Once

V(x(t)) < m_\rho

, the solution cannot be in the shell, so

\lVert x(t) - x^*\rVert < \rho

\blacksquare

Proof workshopRebuild: a Lyapunov function proves stabilityLyapunov 1892; Strogatz 7.2

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Worked example Strogatz 7.2; new

Use Lyapunov functions: the damped pendulum; x′ = −x³; and a linear system with the matrix P.

Solution.

Damped pendulum. Take $V(\theta, \omega) = \frac12\omega^2 + 1 - \cos\theta$ (energy, shifted so that $V = 0$ at rest). It is positive near $(0, 0)$ except at $(0,0)$ itself. Then $\dot V = \omega\omega' + \sin\theta\,\theta' = \omega(-\sin\theta - c\omega) + \omega\sin\theta = -c\omega^2 \le 0$ . The terms with $\sin\theta$ cancel; only the friction term remains, and it removes energy.
Conclude for the pendulum. $(0, 0)$ is stable. $\dot V = 0$ on the whole line $\omega = 0$ , so $V$ is not strict, and the theorem does not by itself give asymptotic stability. For $c > 0$ that follows from the linearization theorem (the eigenvalues have real part $-c/2$ ). Two tools: Lyapunov for stability, linearization for attraction. The extra argument that closes this gap directly is LaSalle’s invariance principle.
$x' = -x^3$ . The linearization is $x' = 0$ and says nothing. Take $V = x^2$ : $\dot V = 2x\cdot(-x^3) = -2x^4 < 0$ for $x \ne 0$ . Strict Lyapunov function, so the origin is asymptotically stable, as the explicit solution showed.
A linear system with a ready-made $V$ . $A = \begin{bmatrix}0 & 1\\-2 & -3\end{bmatrix}$ (eigenvalues $-1, -2$ , from $\lambda^2 + 3\lambda + 2$ ) and $P = \begin{bmatrix}5/4 & 1/4\\1/4 & 1/4\end{bmatrix}$ . Check: $A^{\mathsf T}P + PA = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}$ and $P$ is positive definite (its leading entries $\frac54 > 0$ , $\det P = \frac14 > 0$ ). So $V = \frac54u_1^2 + \frac12u_1u_2 + \frac14u_2^2$ has $\dot V = -u_1^2 - u_2^2$ . This is exactly the matrix $P = \int_0^\infty e^{tA^{\mathsf T}}e^{tA}dt$ of the linearization theorem, which solves the equation $A^{\mathsf T}P + PA = -I$ .

Answer. Stable (pendulum, with asymptotic stability from the linearization); asymptotically stable ( $x' = -x^3$ ); and a quadratic Lyapunov function for the linear system.

Your answerFor

A = \begin{bmatrix}0 & 1\\-2 & -3\end{bmatrix}

solve

A^{\mathsf T}P + PA = -I

for the symmetric matrix

P = \begin{bmatrix}p_{11} & p_{12}\\p_{12} & p_{22}\end{bmatrix}

. What is

p_{11}

? (A fraction.)

Quick checkA system has a function

V \ge 0

, zero only at the equilibrium, with

\dot V = -x_1^2

(no

x_2

term) near it. What does the theorem prove?

Limit cycles

A linear system has closed orbits only in the centre case, and then they come in a whole family, one through each point. A nonlinear system can have an isolated closed orbit that the neighbouring orbits wind onto.

Definition B

Limit cycle

A limit cycle is a closed orbit $\gamma$ of a planar system that is isolated: some neighbourhood of $\gamma$ contains no other closed orbit. It is stable if the orbits starting near it spiral toward it as $t \to \infty$ .

In one lineAn isolated closed orbit of a planar system: nearby orbits spiral toward it (stable) or away from it (unstable).

Why it existsIt is a self-sustained oscillation. Whatever the starting point (within its basin), the system settles into the same rhythm and amplitude, which no linear system can do.

Worked example Strogatz 7.5; van der Pol 1926

The van der Pol oscillator x″ − μ(1 − x²)x′ + x = 0: linearize at the origin, and see the limit cycle.

Solution.

System. $x' = y$ , $y' = \mu(1 - x^2)y - x$ . $y = x'$ . For $\lvert x\rvert < 1$ the damping coefficient $-\mu(1-x^2)$ is negative: the oscillator is pumped. For $\lvert x\rvert > 1$ it is damped.
The origin. $DF = \begin{bmatrix}0 & 1\\-1 & \mu\end{bmatrix}$ , $\tau = \mu$ , $\Delta = 1$ . For $0 < \mu < 2$ an unstable spiral, for $\mu \ge 2$ an unstable node. Orbits near the origin are pushed away, because there is negative damping.
Orbits far out are pulled in. With $E = \frac12(x^2 + y^2)$ , $\dot E = xy + y(\mu(1 - x^2)y - x) = \mu(1 - x^2)y^2$ . This is positive for $\lvert x\rvert < 1$ (energy is pumped in) and negative for $\lvert x\rvert > 1$ (energy is lost). So the motion neither dies out nor blows up: it is trapped in a ring between small and large circles. In the plane a bounded orbit that does not approach an equilibrium must approach a closed orbit (the Poincaré–Bendixson theorem, quoted below).
Where is the cycle for small $\mu$ ? On a circle of radius $R$ write $x = R\cos s$ , $y = -R\sin s$ . The average of $\dot E$ over the circle is $\mu R^2\left(\frac12 - \frac{R^2}{8}\right)$ , which is zero for $R = 2$ . So the cycle is close to a circle of radius $2$ with period close to $2\pi$ . $\langle\sin^2\rangle = \frac12$ and $\langle\cos^2\sin^2\rangle = \frac18$ over a full circle. A numerical solution gives amplitude $2.0001$ for $\mu = 0.1$ and $2.009$ for $\mu = 1$ .
For large $\mu$ . The motion becomes slow-fast: a long creep followed by a quick jump (a relaxation oscillation) with period about $(3 - 2\ln2)\mu = 1.61\mu$ . For $\mu = 50$ the period is $82.5$ by numerical solution, against $80.7$ from this formula.

Answer. The origin is an unstable spiral (or node), orbits from far away spiral inward, and every orbit except the equilibrium settles onto one closed curve, the limit cycle. For small $\mu$ it is nearly a circle of radius $2$ .

Beyond the course: for understanding· B: Poincaré–Bendixson, and why the plane is special

Poincaré–Bendixson theorem (stated). If a solution of a planar autonomous system stays for all $t \ge 0$ in a closed bounded region that contains no equilibrium, then it tends to a closed orbit. So in the plane, the only long-term behaviours are equilibria, closed orbits and orbits that connect them. In three dimensions orbits can wander without settling down, which is chaos (Lorenz, 1963). The proof uses the Jordan curve theorem, which is why it works in the plane only. A companion fact, Bendixson’s criterion, is problem S1 below: if $\operatorname{div}F$ has one sign throughout a simply connected region, there is no closed orbit inside it.

Van der Pol introduced the equation to describe self-sustained oscillations of a triode circuit (1926) and, with van der Mark in 1928, proposed a relaxation-oscillator model of the heartbeat.

Python labThe pendulum's period depends on its amplitude

import numpy as np

def quarter_period(theta0, h=1e-3):
    # start at rest at angle theta0; integrate theta'' = -sin(theta) until theta reaches 0
    f = lambda th, w: (w, -np.sin(th))
    th, w, t = theta0, 0.0, 0.0
    while True:
        k1 = f(th, w)
        k2 = f(th + h/2*k1[0], w + h/2*k1[1])
        k3 = f(th + h/2*k2[0], w + h/2*k2[1])
        k4 = f(th + h*k3[0], w + h*k3[1])
        th_new = th + h/6*(k1[0] + 2*k2[0] + 2*k3[0] + k4[0])
        w_new = w + h/6*(k1[1] + 2*k2[1] + 2*k3[1] + k4[1])
        if th_new <= 0:
            return t + h * th / (th - th_new)     # linear interpolation to the crossing
        th, w, t = th_new, w_new, t + h

print(" amplitude   period   period/(2 pi)   1 + theta0^2/16")
for deg in [5, 30, 60, 90, 120, 150, 170, 179]:
    th0 = np.radians(deg)
    T = 4 * quarter_period(th0)
    print(f"{deg:6d} deg  {T:8.4f}   {T / (2*np.pi):8.4f}      {1 + th0**2/16:8.4f}")

Try this. Run it. The first column shows the linearization's prediction, $2\pi = 6.2832$ : at $5^\circ$ the true period is $6.2862$ , correct to 4 digits. At $90^\circ$ it is $7.4163$ , which is 18% longer, and at $150^\circ$ it is $11.07$ , 76% longer. Near $180^\circ$ the pendulum balances at the top before falling: at $179^\circ$ the period is $24.5$ , nearly four times $2\pi$ , and it tends to infinity as the amplitude tends to $180^\circ$ (the orbit approaches the separatrix and the saddle at the top). The last column is the first-order correction $1 + \theta_0^2/16$ : good for small angles, and already far off at $90^\circ$ ( $1.154$ against $1.180$ ). Change [5, 30, ...] to [178, 179, 179.9] to see the period keep growing (it grows like $\ln\frac{1}{180^\circ - \theta_0}$ ). A linear model keeps the period fixed at $2\pi$ for all amplitudes, which is the first thing the nonlinear equation does not do.

Where it is used

Used inside maths· Inside maths: gradient systems, and why the classification of critical points of 5.6 had no spirals

Let $f$ be a $C^2$ function on the plane and consider the gradient system $x' = -\nabla f(x)$ . Its equilibria are the critical points of $f$ (5.6), and its Jacobian at one is $-H$ , minus the Hessian matrix of $f$ (5.6). The Hessian is symmetric, so its eigenvalues are real (the spectral theorem, 3.1), and so at a critical point with invertible Hessian a gradient system has only a node or a saddle: never a spiral, never a centre. A strict local minimum of $f$ (positive definite $H$ ) is a stable node, a local maximum is an unstable node, and a saddle point of $f$ is a saddle of the flow. Moreover $V(x) = f(x) - f(x^*)$ is a strict Lyapunov function at a strict minimum: $\dot V = \nabla f\cdot(-\nabla f) = -\lVert\nabla f\rVert^2 < 0$ . The flow rolls downhill and cannot circulate, and in particular a gradient system has no closed orbits.

Used in the real world· Machine learning: how large a step can gradient descent take?

Gradient descent $x_{k+1} = x_k - h\nabla f(x_k)$ is Euler’s method with step $h$ for the gradient system above. Take $f(x, y) = \frac12(x^2 + 100y^2)$ . The flow is $x' = -x$ , $y' = -100y$ : a stable node with eigenvalues $-1$ and $-100$ (the Hessian’s eigenvalues, with their condition number $100$ from 3.5). One Euler step multiplies the $x$ -coordinate by $1 - h$ and the $y$ -coordinate by $1 - 100h$ . The iteration converges only if $\lvert1 - 100h\rvert < 1$ , that is $h < \frac{2}{100} = 0.02$ , and in general $h < \frac{2}{\lambda_{\max}}$ . Starting from $(1, 1)$ and running 200 steps: with $h = 0.019$ , $y$ is $7 \times 10^{-10}$ but $x$ is still $0.022$ (the slow direction needs $\approx1/h$ steps per factor $e$ ); with $h = 0.02$ , $y$ jumps between $\pm1$ for ever; with $h = 0.021$ , $y$ reaches $1.9\times10^8$ . The slow direction wants a large step and the stiff direction forbids it, which is why ill-conditioned problems train slowly.

Used in the real world· Clocks: why a pendulum clock cares about its amplitude

The period of a pendulum of length $1$ m for small swings is $2\pi\sqrt{L/g} = 2.006$ s. For amplitude $\theta_0$ (in radians) it is longer by the factor $1 + \frac{\theta_0^2}{16} + \cdots$ , as the lab shows. A clock whose swing shrinks from $6^\circ$ to $4^\circ$ as the spring unwinds therefore changes its rate by $\frac{(0.1047)^2 - (0.0698)^2}{16} = 3.8\times10^{-4}$ , about $33$ seconds per day. Huygens, who built the first pendulum clock in the 1650s, knew this error. In his Horologium Oscillatorium (1673) he proved that a bob constrained to move along a cycloid has a period that does not depend on the amplitude at all, and that a string wrapped around cycloidal cheeks makes an ordinary pendulum follow such a curve. In practice the cheeks added errors of their own, and later clockmakers kept the swing small and steady instead.

Used in the real world· Ecology: fishing in the Adriatic

Umberto D’Ancona’s records of fish sold in the ports of the northern Adriatic showed that during the First World War, when fishing nearly stopped, the share of predatory fish (sharks, rays) rose. He asked Volterra for a mathematical explanation. Volterra’s model of 1926 (Lotka had published the same equations in a book in 1925) gives it by the harvesting computation above: fishing at per-capita rate $h$ shifts the average populations from $(c/d,\ a/b)$ to $((c+h)/d,\ (a-h)/b)$ . With the numbers $a = 3$ , $c = 2$ and $b = d = 1$ of the example, $h = 0.5$ moves the averages from $(2, 3)$ to $(2.5, 2.5)$ : less fishing means the predators are 20% more numerous and the prey 20% fewer. The model is crude, as the fragility of its centre shows, but the qualitative prediction (reversal of the effect of harvesting) is what made it famous.

Traps

Watch out· A linearization with a centre does not give a centre

For $x' = -y + \mu x(x^2 + y^2)$ , $y' = x + \mu y(x^2 + y^2)$ the Jacobian at $0$ is the same for every $\mu$ , and the nonlinear system is a centre, a stable spiral or an unstable spiral according to the sign of $\mu$ . Whenever an eigenvalue of the Jacobian has real part $0$ (a centre, a zero eigenvalue), the equilibrium is not hyperbolic and the linearization is inconclusive. Find a conserved quantity or a Lyapunov function, or compute in polar coordinates.

Watch out· 'Stable' in this chapter does not mean 'attracting'

In everyday speech a stable situation returns to rest. In the definition of Lyapunov stability, a frictionless pendulum is stable and never returns. The word for returning is ‘asymptotically stable’. A centre is stable and not asymptotically stable; a saddle is neither.

Watch out· A negative trace is not enough

Stability of a $2 \times 2$ linear system needs $\tau < 0$ AND $\Delta > 0$ . The matrix $\begin{bmatrix}-3 & 0\\0 & 1\end{bmatrix}$ has trace $-2$ and is a saddle. For larger matrices the condition is that every eigenvalue has negative real part, and the trace (the sum of the eigenvalues) can be negative while one eigenvalue is positive.

Watch out· Evaluate the Jacobian at the equilibrium, and find all the equilibria

The Jacobian is a matrix of functions. Substitute the coordinates of each equilibrium before computing trace and determinant. And solve $F = 0$ completely: for $x(3 - x - 2y) = 0$ , $y(2 - x - y) = 0$ there are four equilibria, not one. The 376 worksheet on critical points of $4xy - x^4 - y^4$ warns about exactly this for the Hessian test; the same care applies here.

Watch out· Euler's method turns a centre into an unstable spiral

For $x' = Ax$ with eigenvalues $\pm i\omega$ , one Euler step multiplies by $I + hA$ , whose eigenvalues $1 \pm ih\omega$ have modulus $\sqrt{1 + h^2\omega^2} > 1$ . For the Lotka–Volterra cycle of the example ( $\omega = \sqrt6$ ) and $h = 0.1$ this is $1.0296$ per step, and over one period (about $26$ steps) the orbit has spiralled out by the factor $2.13$ . A numerical orbit that drifts outward is an artefact; the true orbit is closed. Use a higher-order method (the labs use RK4) or one that conserves energy.

Watch out· Trajectories do not cross, and the picture has no clock

A point where two curves seem to cross in a hand-drawn portrait is an equilibrium, or an error. And an orbit does not say how fast: the same curve is traversed slowly near an equilibrium (the pendulum near the top takes forever) and quickly elsewhere. Read the arrows for direction, and compute the vector $F(x)$ at a point to read the speed.

Problems

check7.4 style

Classify the origin for $x' = Ax$ and state the stability, using the trace and determinant: (a) $A = \begin{bmatrix}-2 & -1\\2 & -2\end{bmatrix}$ ; (b) $A = \begin{bmatrix}-3 & 1\\-1 & -1\end{bmatrix}$ ; (c) $A = \begin{bmatrix}2 & 3\\1 & 0\end{bmatrix}$ ; (d) $A = \begin{bmatrix}1 & -2\\3 & -1\end{bmatrix}$ .

Solution

(a) $\tau = -4$ , $\Delta = 4 + 2 = 6$ , $\tau^2 - 4\Delta = 16 - 24 = -8 < 0$ : stable spiral (eigenvalues $-2 \pm i\sqrt2$ ), asymptotically stable. (b) $\tau = -4$ , $\Delta = 3 + 1 = 4$ , $\tau^2 = 4\Delta$ : degenerate. $A + 2I = \begin{bmatrix}-1 & 1\\-1 & 1\end{bmatrix} \ne O$ , so it is not a star: a stable degenerate node, asymptotically stable. (c) $\tau = 2$ , $\Delta = 0 - 3 = -3 < 0$ : saddle (eigenvalues $3, -1$ ), unstable. (d) $\tau = 0$ , $\Delta = -1 + 6 = 5 > 0$ : centre (eigenvalues $\pm i\sqrt5$ ), stable but not asymptotically stable.

checkStrogatz 6.5 style; new

For $x' = y$ , $y' = x - x^3$ : find all equilibria, their Jacobians and types, and decide which are genuine centres using $E = \frac12y^2 - \frac12x^2 + \frac14x^4$ .

Solution

Equilibria: $y = 0$ and $x(1 - x^2) = 0$ , so $(0, 0)$ and $(\pm1, 0)$ . Jacobian: $\begin{bmatrix}0 & 1\\1 - 3x^2 & 0\end{bmatrix}$ . At $(0,0)$ : $\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ , $\Delta = -1$ : saddle. At $(\pm1, 0)$ : $\begin{bmatrix}0 & 1\\-2 & 0\end{bmatrix}$ , $\tau = 0$ , $\Delta = 2$ : centres in the linearization, not hyperbolic. $\dot E = yy' + (-x + x^3)x' = y(x - x^3) + (-x + x^3)y = 0$ : $E$ is conserved. At $(\pm1, 0)$ , $E$ has Hessian $\operatorname{diag}(-1 + 3x^2, 1) = \operatorname{diag}(2, 1)$ , positive definite, so a strict local minimum: the level sets near it are closed curves, and $(\pm1, 0)$ are genuine nonlinear centres. The level set $E = 0$ through the saddle consists of two loops (a figure eight).

check

Decide stability of the given equilibrium: (a) $x' = \sin x$ at $0$ and at $\pi$ ; (b) $x' = x - x^3$ at $0$ , $1$ and $-1$ .

Solution

In one dimension the sign of $F'(x^*)$ decides when it is non-zero. (a) $F'(x) = \cos x$ : $F'(0) = 1 > 0$ , unstable; $F'(\pi) = -1 < 0$ , asymptotically stable. (b) $F'(x) = 1 - 3x^2$ : $F'(0) = 1$ , unstable; $F'(\pm1) = -2$ , asymptotically stable. Check with the arrows: for $0 < x < 1$ , $x - x^3 > 0$ so $x$ increases toward $1$ .

check

Which of these equilibria are hyperbolic? (a) a Jacobian with eigenvalues $-1 \pm 3i$ ; (b) eigenvalues $\pm2i$ ; (c) eigenvalues $0$ and $-5$ ; (d) eigenvalues $1$ and $-4$ .

Solution

(a) Hyperbolic (real part $-1 \ne 0$ ): a stable spiral, and the nonlinear system has the same type. (b) Not hyperbolic (real part $0$ ). (c) Not hyperbolic (eigenvalue $0$ ). (d) Hyperbolic: a saddle.

use

For $x' = y - x^2$ , $y' = x - y$ : find all equilibria, classify each with the Jacobian, and say which is asymptotically stable.

Solution

Equilibria: $y = x^2$ and $x = y$ , so $x = x^2$ , $x = 0$ or $1$ : $(0, 0)$ and $(1, 1)$ . Jacobian: $\begin{bmatrix}-2x & 1\\1 & -1\end{bmatrix}$ . At $(0, 0)$ : $\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}$ , $\tau = -1$ , $\Delta = -1$ : saddle, unstable. At $(1, 1)$ : $\begin{bmatrix}-2 & 1\\1 & -1\end{bmatrix}$ , $\tau = -3$ , $\Delta = 2 - 1 = 1 > 0$ , $\tau^2 - 4\Delta = 5 > 0$ : stable node (eigenvalues $\frac{-3 \pm\sqrt5}{2}$ ), asymptotically stable.

useStrogatz 6.4 style; new numbers

A model of rabbits $x$ (thousands) and foxes $y$ (hundreds) is $x' = 0.5x - 0.02xy$ , $y' = -0.4y + 0.01xy$ . Find the equilibrium with both species present, the period of small oscillations around it, and the average populations over a cycle.

Solution

The equations have the form $x' = x(a - by)$ , $y' = y(dx - c)$ with $a = 0.5$ , $b = 0.02$ , $c = 0.4$ , $d = 0.01$ . Equilibrium: $x^* = c/d = 40$ , $y^* = a/b = 25$ . Jacobian there: $\begin{bmatrix}a - by & -bx\\dy & dx - c\end{bmatrix} = \begin{bmatrix}0 & -0.8\\0.25 & 0\end{bmatrix}$ , so $\tau = 0$ , $\Delta = 0.2$ , eigenvalues $\pm i\sqrt{0.2}$ , $\omega = 0.447$ and the period is $\frac{2\pi}{0.447} = 14.05$ time units. By the conservation law the averages over any cycle are exactly $40$ and $25$ .

useStrogatz 7.2 style; new

Show that the origin is asymptotically stable for $x' = -x + y^2$ , $y' = -y + x^2$ , and show that every orbit starting in the open disc $x^2 + y^2 < 2$ tends to the origin.

Solution

The Jacobian at $0$ is $-I$ (the quadratic terms have zero derivative there), with eigenvalues $-1, -1$ , so the origin is asymptotically stable by the linearization theorem. For the disc let $V = x^2 + y^2 = r^2$ . Then $\dot V = 2x(-x + y^2) + 2y(-y + x^2) = -2r^2 + 2xy(x + y)$ . Since $\lvert xy\rvert \le \frac{r^2}{2}$ and $\lvert x + y\rvert \le \sqrt2\,r$ , $\dot V \le -2r^2 + \sqrt2\,r^3 = -r^2(2 - \sqrt2\,r)$ , which is negative for $0 < r < \sqrt2$ . So $V$ is a strict Lyapunov function on the disc $r < \sqrt2$ (which is the same as $x^2 + y^2 < 2$ ). Take any starting point with $r_0 < \sqrt2$ and $\rho$ with $r_0 \le \rho < \sqrt2$ : the disc $r \le \rho$ cannot be left, because $V$ decreases along solutions, and the argument of part (b) of Lyapunov’s theorem (it uses only a compact region where $\dot V < 0$ away from the origin) shows $V \to 0$ . So the whole open disc is attracted to the origin.

use

For the damped pendulum with friction $c$ , find the values of $c$ for which the hanging position is a spiral, and the time in which the amplitude of the oscillation is multiplied by $e^{-1}$ when $c = 0.5$ .

Solution

The Jacobian $\begin{bmatrix}0 & 1\\-1 & -c\end{bmatrix}$ has $\tau = -c$ , $\Delta = 1$ and discriminant $c^2 - 4$ . It is a spiral exactly when $c^2 < 4$ , that is $0 < c < 2$ (for $c = 0$ a centre). For $c = 0.5$ the real part of the eigenvalues is $-c/2 = -0.25$ , so the amplitude is multiplied by $e^{-0.25t}$ , which equals $e^{-1}$ at $t = 4$ (in units of $\sqrt{L/g}$ ; for $L = 1$ m that is $4 \times 0.319 = 1.28$ s).

proveVolterra 1926; new

Prove that for a periodic solution of $x' = x(a - by)$ , $y' = y(dx - c)$ with $x, y > 0$ and period $T$ , the time averages are $\frac1T\int_0^Tx\,dt = \frac cd$ and $\frac1T\int_0^Ty\,dt = \frac ab$ .

Solution

Plan. Divide each equation by its unknown to get a logarithmic derivative, and integrate over a period.

Since $x > 0$ , $\frac{d}{dt}\ln x = \frac{x'}{x} = a - by$ . Integrating from $0$ to $T$ : $\ln x(T) - \ln x(0) = aT - b\int_0^Ty\,dt$ . The solution is periodic, so $x(T) = x(0)$ and the left side is $0$ . Hence $\int_0^Ty\,dt = \frac{aT}{b}$ , that is $\frac1T\int y = \frac ab$ . In the same way $\frac{d}{dt}\ln y = dx - c$ gives $0 = d\int_0^Tx\,dt - cT$ , so $\frac1T\int x = \frac cd$ . $\blacksquare$

prove7.4 (the 2 by 2 theorem); new

Let $A$ be a real $2 \times 2$ matrix with $\operatorname{tr}A = 0$ and $\det A = \Delta > 0$ . Prove that $e^{tA} = \cos(\sqrt\Delta\,t)\,I + \frac{\sin(\sqrt\Delta\,t)}{\sqrt\Delta}A$ , and that every solution of $x' = Ax$ is periodic with period $\frac{2\pi}{\sqrt\Delta}$ .

Solution

Plan. Cayley–Hamilton gives $A^2 = -\Delta I$ ; then the series splits into even and odd powers, as for the rotation matrix.

The characteristic polynomial is $\lambda^2 - \tau\lambda + \Delta = \lambda^2 + \Delta$ , so by the Cayley–Hamilton theorem (3.6) $A^2 + \Delta I = O$ . Hence $A^{2k} = (-\Delta)^kI$ and $A^{2k+1} = (-\Delta)^kA$ . In $e^{tA} = \sum\frac{t^nA^n}{n!}$ , the even terms give $\sum\frac{(-\Delta)^kt^{2k}}{(2k)!}I = \cos(\sqrt\Delta\,t)I$ and the odd terms give $\sum\frac{(-\Delta)^kt^{2k+1}}{(2k+1)!}A = \frac{\sin(\sqrt\Delta\,t)}{\sqrt\Delta}A$ . Both $\cos$ and $\sin$ of $\sqrt\Delta\,t$ have period $\frac{2\pi}{\sqrt\Delta}$ , so $e^{(t + 2\pi/\sqrt\Delta)A} = e^{tA}$ , and $x(t) = e^{tA}x_0$ is periodic. $\blacksquare$

prove5.6 and 3.1

Let $x' = -\nabla f(x)$ with $f$ of class $C^2$ . Prove that $f$ strictly decreases along every non-constant solution, that there is no non-constant periodic solution, and that every equilibrium has real Jacobian eigenvalues.

Solution

Plan. Differentiate $f$ along a solution; then use symmetry of the Hessian.

Along a solution, $\frac{d}{dt}f(x(t)) = \nabla f(x)\cdot x' = -\lVert\nabla f(x)\rVert^2 \le 0$ , and it is $0$ only where $\nabla f = 0$ , an equilibrium. A non-constant solution passes through no equilibrium (uniqueness), so $f(x(t))$ is strictly decreasing. If $x$ were periodic with period $T$ , then $f(x(T)) = f(x(0))$ , contradicting strict decrease. The Jacobian of $-\nabla f$ is minus the Hessian of $f$ , which is symmetric (5.5 equality of mixed partials); a real symmetric matrix has real eigenvalues (3.1). $\blacksquare$

proveLyapunov function for a damped oscillator

For $x'' + cx' + x = 0$ with $c > 0$ , show with $V = \frac12(x^2 + (x')^2)$ that the origin is stable, and show with the table of this chapter that it is asymptotically stable.

Solution

Plan. Differentiate $V$ along solutions; then read the trace and determinant.

With $y = x'$ : $\dot V = xx' + yy' = xy + y(-x - cy) = -cy^2 \le 0$ . $V$ is positive except at the origin, so $V$ is a Lyapunov function and the origin is stable. The system matrix is $\begin{bmatrix}0 & 1\\-1 & -c\end{bmatrix}$ , with $\tau = -c < 0$ and $\Delta = 1 > 0$ , so by the table it is asymptotically stable. (Here $V$ is not strict, since $\dot V = 0$ on $y = 0$ , so the table is what proves attraction.) $\blacksquare$

stretchBendixson 1901; Strogatz 7.2

(Bendixson’s criterion.) Suppose $F = (F_1, F_2)$ is continuously differentiable on a simply connected region $D$ and $\operatorname{div}F = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y}$ is positive everywhere in $D$ , or negative everywhere. Prove that $x' = F(x)$ has no closed orbit in $D$ . Deduce that the damped pendulum with $c > 0$ has none.

Solution

Plan. On a closed orbit the field is tangent, so its flux out through the curve is zero. Green’s theorem turns the flux into the integral of the divergence over the region inside, which has one sign.

Suppose $C$ is a closed orbit in $D$ , bounding a region $R \subset D$ (simply connected). Parametrise $C$ by the solution $x(t)$ , $0 \le t \le T$ . The outward normal times arc length element is $(y'\,dt, -x'\,dt)$ for a counter-clockwise curve, so the flux of $F$ through $C$ is $\oint(F_1\,dy - F_2\,dx) = \int_0^T(F_1y' - F_2x')\,dt$ . Since $x' = F_1$ and $y' = F_2$ on the orbit, this is $\int_0^T(F_1F_2 - F_2F_1)\,dt = 0$ . By the divergence form of Green’s theorem, the flux also equals $\iint_R\operatorname{div}F\,dA$ , which is strictly positive (or negative) when $\operatorname{div}F$ has one sign. Contradiction. For the damped pendulum, $F = (x_2, -\sin x_1 - cx_2)$ has $\operatorname{div}F = 0 - c = -c < 0$ on the whole plane, so there is no closed orbit. $\blacksquare$ (Compare with $c = 0$ : $\operatorname{div}F = 0$ and there are closed orbits.)

stretchStrogatz 7.1; new

For $x' = x - y - x(x^2 + y^2)$ , $y' = x + y - y(x^2 + y^2)$ show that $r' = r - r^3$ and $\theta' = 1$ in polar coordinates, find the limit cycle, and classify the origin.

Solution

Plan. Compute $rr'$ and $r^2\theta'$ from $xx' + yy'$ and $xy' - yx'$ .

$rr' = xx' + yy' = x^2 - xy - x^2r^2 + xy + y^2 - y^2r^2 = r^2 - r^4$ , so $r' = r - r^3$ . And $r^2\theta' = xy' - yx' = x^2 + xy - xyr^2 - xy + y^2 + xyr^2 = r^2$ , so $\theta' = 1$ . The equation $r' = r(1 - r^2)$ has equilibria $r = 0$ and $r = 1$ , positive for $0 < r < 1$ and negative for $r > 1$ . Orbits therefore wind around the origin once per $2\pi$ while $r$ tends to $1$ : the circle $r = 1$ is a stable limit cycle. The Jacobian at the origin is $\begin{bmatrix}1 & -1\\1 & 1\end{bmatrix}$ , $\tau = 2$ , $\Delta = 2$ , $\tau^2 - 4\Delta = -4 < 0$ : an unstable spiral (eigenvalues $1 \pm i$ ). The explicit solution is $r(t)^2 = \frac{1}{1 + (r_0^{-2} - 1)e^{-2t}}$ . $\blacksquare$

Summary

An autonomous system $x' = F(x)$ is studied through its orbits and phase portrait. By uniqueness, orbits never cross, and a closed orbit is a periodic motion.
An equilibrium has $F(x^*) = 0$ . It is stable if nearby solutions stay near for all time, and asymptotically stable if they also tend to $x^*$ . A centre is stable but not asymptotically stable.
For $x' = Ax$ in the plane the trace $\tau$ and determinant $\Delta$ decide everything: saddle for $\Delta < 0$ , node above the parabola $\tau^2 = 4\Delta$ , spiral inside it (centre when $\tau = 0$ ), degenerate cases on the parabola and on $\Delta = 0$ . Asymptotically stable exactly when $\tau < 0 < \Delta$ ; in any dimension, when every eigenvalue has negative real part.
Near an equilibrium a nonlinear system looks like its linearization $u' = Ju$ , with $J$ the Jacobian matrix. If $J$ has no eigenvalue on the imaginary axis (hyperbolic) the phase portrait is the linear one (Hartman–Grobman), and negative real parts give asymptotic stability, a positive one instability. When an eigenvalue has real part $0$ the linearization is inconclusive.
Pendulum: hanging position a stable centre (spiral with friction), upright position a saddle, the period grows with the amplitude. Lotka–Volterra: a conserved $H$ makes the cycles around $(c/d, a/b)$ exactly closed, with averages equal to the equilibrium, and harvesting raises the prey. Both centres are non-hyperbolic and are settled by conservation laws.
A Lyapunov function $V$ (positive, $\dot V \le 0$ ) proves stability without solving anything, and a strict one proves asymptotic stability. A limit cycle (van der Pol) is an isolated closed orbit that a linear system cannot have.
In practice: find all equilibria, evaluate the Jacobian at each, read $\tau$ and $\Delta$ , and look for a conserved quantity or a Lyapunov function whenever an eigenvalue has real part $0$ . Beware Euler’s method, which breaks centres.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

7.6 Phase portraits, stability and linearization

The picture

The idea, in words

Definitions and theorems

Autonomous systems, orbits, phase portraits

Autonomous system

Orbit

Phase portrait

Orbits do not cross

Equilibria and stability

Equilibrium

Stable

Asymptotically stable

Linear systems: the trace-determinant classification

Saddle

Node

Spiral

Centre

Classification of the origin by trace and determinant

Stability of a linear system

Linearization: a microscope at an equilibrium

Hyperbolic equilibrium

Hartman–Grobman theorem

Linearization and stability

The pendulum, in full

Predator and prey

The Lotka–Volterra conservation law

Lyapunov functions

Lyapunov function

Lyapunov's direct method

Limit cycles

Limit cycle

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere