Lec 1: Introductionese617/lec01.pdf · 1 Introduction to nonlinear dynamical systems 1.1 Linear...

CONTENTS 1 INTRODUCTION TO NONLINEAR DYNAMICAL SYSTEMS

Lecture 1: Introduction

Cameron Nowzari

August 25, 2015

Abstract

This lecture states the goals of the course and provides an introduction to basic concepts. We discussgeneral dynamical properties of nonlinear systems and explore its differences with linear systems. Wealso introduce some interesting nonlinear examples. The treatment corresponds to Chapters 1-2 in [2].

Contents

1 Introduction to nonlinear dynamical systems 1

1.1 Linear versus nonlinear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Goals of this course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Basic concepts 3

3 Examples of nonlinear systems 4

3.1 Bowing of a violin string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3.2 The buckling beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

4 Planar dynamical systems 6

4.1 Linear systems on the plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4.2 Linearization of nonlinear systems around an equilibrium point . . . . . . . . . . . . . . . . . 7

5 Closed orbits and limit cycles 9

5.1 Bendixson’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5.2 Poincare-Bendixson theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1 Introduction to nonlinear dynamical systems

1.1 Linear versus nonlinear

There is a huge body of work in the analysis and control of linear systems, which are of the form

x(t) = A(t)x(t) +B(t)u(t) , x(0) = x0

1

ESE 617 – Nonlinear Systems and ControlThese notes have been adapted from Professor Jorge Cortes’ notes for MAE 281A at the University of California, San Diego.

1.1 Linear versus nonlinear 1 INTRODUCTION TO NONLINEAR DYNAMICAL SYSTEMS

where x ∈ Rn (the state), u ∈ Rm (the control inputs), A(t) ∈ Rn×n (the dynamics) and B(t) ∈ Rn×m (thecontrol action).

Most models currently available are linear - especially in industry. However, most real systems are nonlinear...

Why nonlinear theory? Because the dynamics of linear systems are not rich enough to describe manycommonly observed phenomena like

• Finite escape time. The state of an unstable linear system goes to infinity as time approaches infinity;however, a nonlinear system’s state can go to infinity in finite time.

• Multiple (finite) equilibria. Consider the linear differential equation

x(t) = Ax(t) (1)

For the initial condition x0 ∈ Rn, the solution of the system in x(t) = exp(At)x0.

The point x = 0 is an equilibrium. If A is non-singular, then x = 0 is the only equilibrium of thesystem. If A is singular, then kerA is the set of equilibria. However, an infinitesimal perturbation“almost surely” makes any singular matrix non-singular.

Therefore, problems in nature and engineering which have a finite number of equilibria cannot bedescribed by means of a linear system model.

Example: a pendulum

• (Finite number of) limit cycles. A limit cycle is a nontrivial periodic solution. Consider thelinear system (1). If A has eigenvalues in the imaginary axis, then the system admits a continuum ofperiodic solutions. However, infinitesimal perturbations will cause the eigenvalues to be displaced ofthe imaginary axis, and hence destroy the existence of periodic solutions.

Therefore, problems in nature and engineering which have a finite number of limit cycles cannot bedescribed by means of a linear system model.

Example: the pumping heart

• Bifurcations. Many real systems exhibit qualitative changes as the parameters of the model are varied.These include changes in the number of equilibrium points, limit cycles and their stability properties.

However, linear systems are very simple in this regard. As the parameters of the model change, thesystem can only change from stable to unstable (or vice versa). The number of equilibria goes to infinityif one of the eigenvalues of the matrix A goes through the origin. Limit cycles only exist (and in aninfinite number) when the matrix A has eigenvalues in the imaginary axis.

Example: buckling of a beam

• Chaos. The dynamical behavior of linear systems is quite simple. The responses are sums of exponen-tials, with exponents given by the eigenvalues of the matrix A. They decay to zero or blow up as t goesto infinity when there are no imaginary eigenvalues - independently of the initial condition! When Ahas eigenvalues in the imaginary axis, the solutions neither decay nor blow up for any initial condition.

On the contrary, many physical systems exhibit very complex dynamical behavior. Small changes inthe initial condition can make the trajectories vastly differ over time.

Example: population models, climatic models, turbulent fluid flow models

• Many more reasons we will see along the course...

Linear techniques are good (and, in some cases, useful for the analysis of nonlinear systems!), but often timesnot sufficient.

2


1.2 Goals of this course 2 BASIC CONCEPTS

1.2 Goals of this course

(i) To learn and use various notions and tools for the analysis and control of nonlinear systems.

(ii) To get a feeling and gain insight into the complexity of nonlinear systems.

(iii) To introduce a wide variety of interesting, inherently nonlinear examples.

2 Basic concepts

Basic notions from differentiable topology on Rn that by now you should be familiar with are the following:

• B(x, δ) denotes a ball centered at x of radius δ, i.e., B(x, δ) = {y ∈ Rn | ‖y − x‖2 < δ}

• a neighborhood N of a point x ∈ Rn is a set such that there exists δ > 0 with B(x, δ) ⊂ N

• a limit point x ∈ Rn of a set S ⊂ Rn is a point such that every neighborhood of x contains a point ofS \ {x}

• a closed set S is a set that contains all its limit points. A set is closed if and only if it contains itsboundary

• an open set S is a set such that for any x ∈ S, there exists a neighborhood of x contained in S

• a bounded set S is such that there exists K > 0 such that ‖x‖ ≤ K for all x ∈ S

• a compact set S is a bounded and closed set.

Consider the dynamical system

x = f(x, t),

where f : Rn × R→ Rn. Let us define the following notions:

(i) the system is called autonomous if f does not explicitly depend on time, i.e., f(x, t) = f(x);

(ii) a point x0 ∈ Rn is called an equilibrium point if f(x0, t) = 0 for all t;

(iii) an equilibrium x0 is called isolated if there exists δ > 0 such that there is no other equilibrium point inthe ball B(x0, δ)

Let x0 be an equilibrium of the autonomous system x = f(x). Assume f is C1. Writing down the Taylorseries about x0, we get

f(x) = f(x0) +∂f

∂x(x0)(x− x0) + r(x) =

∂f

∂x(x0)(x− x0) + r(x),

where r(x) is the tail of the Taylor series corresponding to the higher-order terms, i.e.,

lim‖x−x0‖2→0

‖r(x)‖2‖x− x0‖2

= 0

Denoting y = x− x0, near x0, we can approximate the nonlinear system by its linearization

y =∂f

∂x(x0)y.

The linearization of a nonlinear system tells us how the system behaves around x0 in some cases.

3


3 EXAMPLES OF NONLINEAR SYSTEMS

Theorem 2.1 If the Jacobian matrix ∂f∂x (x0) is non-singular, then x0 is an isolated equilibrium of the non-

linear system.

Another situation when the linearization might be useful is when deciding the stability properties of anequilibrium point. We will discuss this at depth in the chapter on Lyapunov stability – it is intrinsicallyrelated to the following (famous) result.

Theorem 2.2 (Hartman-Grobman) If the Jacobian matrix ∂f∂x (x0) has no eigenvalues in the imaginary

axis, then there exists δ > 0 and a continuous map h : B(x0, δ) → Rn, with a continuous inverse (i.e., ahomeomorphism) mapping the trajectories of the nonlinear system onto the trajectories of the linearization.

3 Examples of nonlinear systems

3.1 Bowing of a violin string

The following example is the mechanical analog of the bowing of a violin string (see Figure 1(a)). Bowing ismodeled by the friction between the body and the conveyor belt. The conveyor belt is moving at a constantvelocity b. Consider the second-order system

(a) Mechanical analog

-6 -4 -2 2 4 6

-15

-10

-5

5

10

(b)

0 1 2 3 4

-2

-1

0

1

2

(c) Phase portrait

Figure 1: Bowing of a violin string.

Mx+ kx+ Fb(x) = 0,

where the function Fb is of the form plotted in Figure 1(b). Defining x1 = x and x2 = x, we can rewrite itas a system of first-order differential equations

x1 = x2

x2 =1

M

(− kx1 − Fb(x2)

)The equilibrium points of the system are determined by

x2 = 0, −kx1 − Fb(x2) = 0

Therefore, there is only one equilibrium point, x0 = (−Fb(0)k , 0), which is obviously isolated. To assess its

stability properties, we linearize the system around it. The Jacobian at x0 is

∂f

∂x(x0) =

[0 1

− kM −F

′b(0)M

]4


3.2 The buckling beam 3 EXAMPLES OF NONLINEAR SYSTEMS

For instance, for k = 3, M = 3 and Fb defined by

Fb(z) =

{−((z − b) + c)2 − d, x < b

((z − b)− c)2 − d, x > b

with b = 1, c = 2 and d = 3, we get x0 = ( 43 , 0) and

∂f

∂x(x0) =

[0 1−1 2

3

]which has eigenvalues 1

3 ±2√2

3 i.

3.2 The buckling beam

Consider the following model

mx+ dx− µx+ λx+ x3 = 0

Here x represents the one-dimensional deflection of the beam normal to the axial direction, µx stands for theapplied axial force, λx + x3 models the restoring spring force in the beam, and dx is the damping. We canrewrite the system as

x1 = x2

x2 = − 1

m

(dx2 − µx1 + λx1 + x31

)The equilibrium points are determined by

x2 = 0, dx2 − µx1 + λx1 + x31 = 0

Therefore, for λ ≤ µ, the system has three equilibrium points (0, 0), (±√µ− λ, 0). For λ > µ, the only

equilibrium point is (0, 0). Figure 2 shows a phase portrait for the undamped case. The linearization of thesystem is

∂f

∂x(x) =

[0 1

1m

(µ− λ− 3x21

)− dm

]For the values d = 1, m = 1, µ = 2 and λ = 1, we get the equilibrium points (0, 0), (−1, 0), and (1, 0). Thecorresponding eigenvalues of the linearization at (0, 0) are

−1±√

5

2

and at (±1, 0),

−1

2±√

7

2i

Therefore, (0, 0) is unstable and the other two equilibria are stable.

5


4 PLANAR DYNAMICAL SYSTEMS

x ’ = y y ’ = − (d y − mu x + lambda x + x3)/m

m = 1lambda = 1

d = 1mu = 2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

y

Figure 2: Phase portrait of the buckling beam for d = 1, m = 1, µ = 2 and λ = 1.

4 Planar dynamical systems

A planar dynamical system is represented by two scalar differential equations

x1 = f1(x1, x2)

x2 = f2(x1, x2) (2)

The (x1, x2)-plane is usually called the phase plane. The ‘bowing of a violin string’ and the ‘buckling beam’above are both planar examples.

If t 7→ x(t) = (x1(t), x2(t)) ∈ R2 denotes the solution of the system that starts at x0 ∈ R2, the velocity vectorof this curve in the (x1, x2)-plane at time t is precisely the vector

(x1(t), x2(t)) = f(x1(t), x2(t)) = (f1(x1(t), x2(t)), f2(x1(t), x2(t))).

The vector field x ∈ R2 → f(x) ∈ R2 can be plotted with Mathematica/Matlab/Maple to give us an idea ofthe behavior of the solutions of the system. In Mathematica, use the command PlotVectorField. In Matlab,use the routine pplane provided by Prof. Polking at Rice University at http://math.rice.edu/~dfield

Example 4.1 (Tunnel-diode circuit) Consider the following equations describing the behavior of a tunnel-diode circuit

x1 = .5(−h(x1) + x2)

x2 = .2(−x1 − 1.5x2 + 1.2)

with h(x) = 17.76x− 103.79x2 + 229.62x3 − 226.31x4 + 83.72x5. Solving for the equilibrium points, we get

Q1 = (.063, .758), Q2 = (.285, .61), Q3 = (.884, .21).

Figure 3 shows the phase portrait of this system. •

6


http://math.rice.edu/~dfield

4.1 Linear systems on the plane 4 PLANAR DYNAMICAL SYSTEMS

x ’ = .5 ( − 17.76 x + 103.79 x2 − 229.62 x3 + 226.31 x4 − 83.72 x5 + y)y ’ = .2 ( − x − 1.5 y + 1.2)

−0.4 −0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

x

y

Figure 3: Phase portrait of the tunnel-diode circuit example.

4.1 Linear systems on the plane

Consider the linear time-invariant system

x = Ax (3)

where A is a 2 × 2 matrix. From linear algebra, we know that there exits an invertible matrix Q such thatQ−1AQ = J , where J may take one of the following forms[

λ1 00 λ2

],

[λ 00 λ

],

[λ 10 λ

],

[α −ββ α

]. (4)

The solution of (3) starting from x0 is given by x(t) = Q exp(Jt)Q−1x0. The first case (4) corresponds tothe case when the eigenvalues λ1 and λ2 are real and distinct. The second and third case corresponds to thecase when the eigenvalues are real and equal (second case, two linearly independent eigenvectors, third caseonly one). Finally, the fourth case corresponds to the case of complex eigenvalues α± iβ.

(Un)stable (improper) node, saddle point, (un)stable focus, center: follow discussion in [2, Chapter 2.1]

4.2 Linearization of nonlinear systems around an equilibrium point

Consider the linearization of the nonlinear system (2) around an equilibrium point x0 ∈ R2,

y = Ay

where y ∈ R2 and

A =

[a11 a12a21 a22

]=∂f

∂x(x0).

Recall the Hartman-Grobman theorem, cf. Theorem 2.2. From our previous discussion on linear systems, wededuce

7


4.2 Linearization of nonlinear systems around an equilibrium point4 PLANAR DYNAMICAL SYSTEMS

(i) if the origin of the linearization is a stable (respectively unstable) node with distinct eigenvalues, thenthe equilibrium of the nonlinear system is a stable (respectively unstable) node;

(ii) if the origin of the linearization is a stable (respectively unstable) focus, then the equilibrium of thenonlinear system is a stable (respectively unstable) focus;

(iii) if the origin of the linearization is a saddle point, then the equilibrium of the nonlinear system is asaddle point.

If the origin of the linearization is a center point, then we cannot make any claims about the nonlinear system.

Example 4.2 (Tunnel-diode circuit revisited) Consider again the tunnel-diode circuit example. TheJacobian matrix of the system is

∂f

∂x=

[−.5h′(x1) .5−.2 −.3

]Evaluating this at each of the equilibrium points, we get

∂f

∂x(Q1) =

[−3.598 .5−.2 −.3

], eigenvalues −3.57,−.33⇒ stable node

∂f

∂x(Q2) =

[1.82 .5−.2 −.3

], eigenvalues 1.77,−.25⇒ saddle point

∂f

∂x(Q3) =

[−1.427 .5−.2 −.3

], eigenvalues −1.33,−.4⇒ stable node

Figure 4 shows various solutions of this system. •

x ’ = .5 ( − 17.76 x + 103.79 x2 − 229.62 x3 + 226.31 x4 − 83.72 x5 + y)y ’ = .2 ( − x − 1.5 y + 1.2)

−0.4 −0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

x

y

Figure 4: Phase portrait of the tunnel-diode circuit example.

Example 4.3 Consider the dynamical system

x1 = x2

x2 = −x1 − εx21x2

8


5 CLOSED ORBITS AND LIMIT CYCLES

The linearization around the equilibrium point (0, 0) is

y1 = y2

y2 = −y1

with eigenvalues ±i. Therefore, it is inconclusive about the stability properties of the equilibrium point ofthe nonlinear system (we cannot apply the Hartman-Grobman theorem). In fact, checking out the evolutionof the function V (x1, x2) = 1

2 (x21 + x22) along the solutions of the system, we get

d

dt(V (x1(t), x2(t))) = −εx21x22.

Therefore, for ε > 0, the equilibrium is stable, and for ε < 0 it is unstable (see Figure 5). Note how thelinearization around the equilibrium is completely blind to this fact, i.e., no matter the value of ε, it alwaysgives the same answer (“the equilibrium of the linear system is a center”). We will learn how to invoke thiskind of argument systematically when we discuss Lyapunov stability analysis. •

x ’ = y y ’ = − x − eps x2 y

eps = 1

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

y

x ’ = y y ’ = − x − eps x2 y

eps = − 1

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

y

Figure 5: Phase portraits for ε = 1 and ε = −1.

Example 4.4 The dynamical systems

x = x3, x = −x3

have the same linearization around 0, simply y = 0. However, their behaviors are very different. •

5 Closed orbits and limit cycles

A non-trivial periodic solution of (2) satisfies x(t+T ) = x(t) for all t ≥ 0 and some period T > 0. Non-trivialexcludes the possibility of constant solutions corresponding to equilibrium points. A closed orbit γ is thetrace of the trajectory of a nontrivial periodic solution. An isolated closed orbit is called a limit cycle. Alllimit cycles are closed orbits, but not all closed orbits are limit cycles.

Example 5.1 (Duffing’s equation) Consider the following dynamical system

x+ δx− x+ x3 = 0

9


5.1 Bendixson’s theorem 5 CLOSED ORBITS AND LIMIT CYCLES

Writing this as a system of first-order differential equations, we get

x1 = x2

x2 = x1 − x31 − δx2

Figure 6(a) shows the phase portrait for δ = 0. Note the infinitely many number of closed orbits. •

-2 -1 0 1 2

-4

-2

0

2

4

(a) δ = 0

-2 -1 0 1 2

-4

-2

0

2

4

(b) δ = 1

Figure 6: Phase portraits of the Duffing example.

5.1 Bendixson’s theorem

Theorem 5.2 (Bendixson’s theorem) Let D ⊂ R2 be a simply connected region. Consider the vectorfield f : R2 → R2 and assume that its divergence

div(f) =∂f1∂x1

+∂f2∂x2

is not identically zero and does not change sign on D. Then the dynamical system x = f(x) does not possessany closed orbits.

Proof: Let us reason by contradiction. Assume there exists a closed orbit γ. Along any orbit of f , wehave dx2

dx1= f2/f1. Therefore, ∫

γ

f2(x1, x2)dx1 − f1(x1, x2)dx2 = 0

By Green’s theorem, this implies that ∫ ∫S

( ∂f1∂x1

+∂f2∂x2

)dx1dx2 = 0

where S is the interior of γ. This contradicts the hypothesis that div(f) is not identically zero and does notchange sign on D, and this concludes the proof.

Example 5.3 (Duffing’s equation revisited) Writing Duffing’s equation (cf. Example 5.1) as a systemof first-order differential equations, we get

x1 = x2,

x2 = x1 − x31 − δx2,

10


5.2 Poincare-Bendixson theorem 5 CLOSED ORBITS AND LIMIT CYCLES

and recognize f(x1, x2) = (x2, x1 − x31 − δx2). We compute div(f) = −δ. So for δ 6= 0, there are no closedorbits on R2. Figure 6(b) shows the phase portrait for δ = 1. Note that the closed orbits of Figure 6(a) haveall disappeared. •

5.2 Poincare-Bendixson theorem

Definition 5.4 (Positively invariant sets) A set M is called positively invariant for the dynamical sys-tem (2) if every trajectory starting in M stays in M for all future time.

Theorem 5.5 (Poincare-Bendixson theorem) Let M ⊂ R2 be a non-empty, compact, positively invari-ant set. Then M contains an equilibrium point or a closed orbit.

The intuition behind this result is the trajectories in the set M will have to approach closed orbits orequilibrium points as time tends to infinity.

Remark 5.6 (How to find invariant sets) Consider a simple closed curve defined by the equation V (x) =c, where V : Rn → R is continuously differentiable. The derivative of V along the flow field f (Lie derivative)is then given by

∇V · f =∂V

∂x1f1(x) +

∂V

∂x2f2(x)

If this is less than or equal to zero at each point in the curve defined by V (x) = c, that means that thevector field f is pointing inward or is tangent to the curve, and therefore, the trajectories of the dynamicalsystem are trapped inside the closed region defined by the curve. The set M =

{x ∈ R2 | V (x) ≤ c

}is then

positively invariant with respect to f . •

Example 5.7 (Harmonic oscillator) Consider the dynamical system

x1 = x2

x2 = −x1

and the annular region M ={x ∈ R2 | c1 ≤ x21 + x22 ≤ c2

}, with c2 > c1 > 0. The set M is closed, bounded,

and does not contain any equilibrium point. Moreover, since f(x) · ∇V (x) = 0 everywhere, trajectories aretrapped inside M , i.e., the set is positively invariant. Hence, by Poincare-Bendixson theorem, we know thatthere is at least one periodic orbit in M . •

Example 5.8 (Lotka-Volterra predator-prey model with limited growth) Motivated by the studyof the cyclic variation of the populations of small fish in the Adriatic, Count Vito Volterra proposed thefollowing model for the population evolution of two species: prey x and predators y,

x = (a− by − λx)x,

y = (cx− d− µy)y,

where a, b, c, d, λ, µ > 0. The equilibria of the system are

e1 = (0, 0), e2 = (a

λ, 0), e3 = (

bd+ aµ

bc+ λµ,ac− dλbc+ λµ

), e4 = (0,− dµ

)

We neglect the last one, since our attention is restricted to positive values x, y ≥ 0 for the populations. Forthe same reason, the third one is not considered if ac − dλ < 0. The Jacobian of the vector field takes theform

∂f

∂x=

[a− by − 2λx −bx

cy cx− d− 2µy

]11


5.2 Poincare-Bendixson theorem 5 CLOSED ORBITS AND LIMIT CYCLES

From here, we get

∂f

∂x(e1) =

[a 00 −d

], eigenvalues are a,−d⇒ saddle point

∂f

∂x(e2) =

[a −abλ0 ca

λ − d

], eigenvalues are −a, ac

λ− d⇒ saddle point or stable node

∂f

∂x(e3) = . . .

Consider the following two cases

Case ac < dλ: In this case, we only have the equilibria e1 and e2. The first one is a saddle point and thesecond one is a stable node. A sample phase portrait is plotted in Figure 7. Note that the region A∪B

c x - d - μ y = 0

a - by - λx = 0

A

B

C

x ’ = (a − b y − lambda x) xy ’ = (c x − d − mu y) y

b = .5mu = 2

c = 1lambda = 2

a = 2d = 2

0 1 2 3 4 5 6

0

1

2

3

4

5

6

x

y

Figure 7: Phase portrait of Lotka-Volterra model for ac < dλ

is non-empty, compact, and positively invariant. By Poincare-Bendixson, this region must contain anequilibrium point or a closed orbit. Note that y ≤ 0 in this region, therefore there are no closed orbits.Also, the trajectories starting in C eventually reach B (why?). Therefore, all trajectories eventuallyconverge to either e1 or e2.

Case ac > dλ: In this case, we have the equilibria e1, e2 and e3. The first two ones are saddle points and thethird one is a stable focus/node (check!). A sample phase portrait is plotted in Figure 8. The regionplotted in the figure is non-empty, closed, bounded and positively invariant. By Poincare-Bendixson,it contains an equilibrium point or a closed orbit. By index theory, if such an orbit exists, it mustenclose the equilibrium e3. All trajectories starting outside this region eventually enter it. Therefore,all trajectories will eventually converge to the equilibria e1, e2, e3, or the limit cycle if it exists.

This example shows how, without actually computing the explicit solutions of a dynamical system, we candeduce a lot of things about its behavior with the qualitative tools that we have introduced. •

12


REFERENCES REFERENCES

c x - d - μ y = 0

a - by - λx = 0

x ’ = (a − b y − lambda x) xy ’ = (c x − d − mu y) y

b = .5mu = .5

c = 1lambda = .5

a = 2d = 2

0 1 2 3 4 5 6

0

1

2

3

4

5

6

x

y

Figure 8: Phase portrait of Lotka-Volterra model for ac > dλ

References

[1] S. S. Sastry, Nonlinear Systems: Analysis, Stability and Control, ser. Interdisciplinary Applied Mathematics.Springer, 1999, no. 10.

[2] H. K. Khalil, Nonlinear Systems, 3rd ed. Prentice Hall, 2002.

[3] D. V. Zenkov, A. M. Bloch, and J. E. Marsden, “The energy-momentum method for the stability of nonholonomicsystems,” Dynamics and Stability of Systems, vol. 13, pp. 123–166, 1998.

[4] S. H. Strogatz, Nonlinear Dynamics and Chaos. Cambridge: Westview Press, 1994.

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