ordinary di erential equations: a constructive approach

Ordinary Differential Equations: A Constructive Approach

J.B. van den Berg, M. Gameiro, J.-P. Lessard, J.D. Mireles James, K. Mischaikow

March 3, 2017

Contents

1 Motivation 7

1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Fundamental Results: Existence and Uniqueness 13

2.1 Contraction Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Existence and Uniqueness of Solutions to ODEs . . . . . . . . . . . . . . . . 15

2.2.1 Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Equilibria and Radii Polynomial Approach in Finite Dimension 35

3.1 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2 Radii Polynomial Approach in Finite Dimension . . . . . . . . . . . . . . . 37

3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 Linear Theory and Stability of Equilibria 57

4.1 Homogeneous Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.2 Constant Coefficient Linear Systems . . . . . . . . . . . . . . . . . . . . . . 60

4.3 Hyperbolic Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

4.4 Linear Approximations of Nonlinear Systems . . . . . . . . . . . . . . . . . 69

4.5 Rigorous Computation of Eigenvalues and Eigenvectors . . . . . . . . . . . 71

4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

5 Continuation of Equilibria 77

5.1 Parameterized Families of Equilibria . . . . . . . . . . . . . . . . . . . . . . 77

5.2 Computing Branches of Equilibria . . . . . . . . . . . . . . . . . . . . . . . 79

5.3 Saddle-Node Bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6 Banach Spaces and Radii Polynomials in Infinite Dimension 95

6.1 Banach Spaces, Duals and Operator Norms . . . . . . . . . . . . . . . . . . 95

6.2 The Main Examples: Banach Spaces of Infinite Sequences . . . . . . . . . . 105

6.3 Calculus on Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

3

4 CONTENTS

6.4 Banach Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246.5 The Radii Polynomial Approach on Banach spaces: global version . . . . . 1286.6 The Radii Polynomial Approach on Banach spaces: local version . . . . . . 1306.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

6.7.1 Validated numerics for reciprocals and square roots in `1ω,N . . . . . 132

6.7.2 Validated numerics for reciprocals and square roots in `1ω,Z . . . . . 1326.7.3 A radii polynomial proof of the implicit function theorem . . . . . . 1326.7.4 A radii-polynomial proof of the Hartman-Grobman theorem . . . . . 140

6.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

7 Appendix 1497.1 Topological and Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 1497.2 Normed Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1517.3 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1547.4 Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

Chapter 1

Motivation

As indicated earlier in this introduction this book develops the theory of differential equa-tions in a manner that facilitates computer assisted proofs. In particular, there are manyoccasions in which the most practical means of obtaining necessary explicit inequalities isby using the computer. While conceptually this may appear to be a simple task, the realityis that these are done using floating point arithmetic which is only capable of representinga finite set of rational numbers. Therefore, it is a nontrivial task to ensure that the boundsreported by the computation are, in fact, valid. A common approach to dealing with thisissue is to make use of interval arithmetic [?] where real numbers are represented by inter-vals bounded by rational numbers that have floating point representation. Furthermore,there is software that performs interval arithmetic. For further information on the softwarereader is referred to [?]. With this in mind, in this text we will often use phrases of theform ”we can compute the following bounds . . . .” As a reader this should be interpretedas shorthand for the statement ”using interval arithmetic we can compute the followingbounds . . . .”

1.1 Exercises

Exercise 1.1.1. (Frictionless spring) Consider a time varying population of organismsP (t) living in a closed environment (possibly a petri dish). The initial population at timet = 0 is P (0) = p0. The population in the dish at future times is governed by the logisticmodel

d

dtP (t) = kP (t)− k

KP (t)2,

where k > 0 is the reproductive rate of the species and K > 0 is the carrying capacity ofthe environment.

• Solve the equation analytically.

5

6 CHAPTER 1. MOTIVATION

• What is the limiting value of P (t)?

• If the initial population in the dish is p0 = 10, the reproductive rate is k = 2 andthe carrying capacity is K = 1000, at what time t > 0 does the population double?Triple? Ten times its initial value?

• Suppose that the initial population is very small compared to the carrying capacity.Derive a simplified model of the population growth. Explain what happens to thepopulation size as time goes on (i.e. why was the Reverend Dr. Malthus so worriedafter he forgot to include the P 2 term in his analysis?)

Exercise 1.1.2. (Frictionless spring) Recall Hook’s law, which describes the force exertedon a mass attached to the end of a stretched spring

F (x) = −Kx,

where K is the “stiffness” constant of the spring, and x is the signed displacement fromthe equilibrium position of the spring.

• Let the object have mass m > 0. Use Newton’s laws to derive the equation ofmotion for the system consisting of a mass connected to a spring, moving freely ona frictionless table top (perhaps a surface of ice or perhaps the mass is attached towell oiled wheels/bearings).

• Suppose that the mass is held motionless at an initial position x(0) = a, and thatat time t = 0 the mass is released. Describe the motion of the system for all futuretime. What is the frequency of the systems oscillation?

• Suppose that the mass is initially at rest in the equilibrium position x = 0, andthat at time t = 0 the system is “kicked”, imbuing the mass with an initial velocityof x′(0) = b. Describe the motion of the system for all future time. What is thefrequency of the systems oscillation?

Exercise 1.1.3. (Spring with friction) Consider the same setup as in the previous problem,but now suppose that we account for friction between the table and the mass (or friction inthe wheels). A typical model of friction is to assume that the force of friction is proportionalto the magnitude of the velocity, acting against the direction of motion. (So the faster theobject moves the more friction “tries to slow it down”).

• Let the object have mass m > 0. Use Newton’s laws to derive the equation of motionfor the system consisting of a mass connected to a spring, moving on a table top,subject to the force of friction described above. Suppose that the proportionalitycoefficient of the friction is C > 0.

1.1. EXERCISES 7

• Suppose that the mass is held motionless at an initial position x(0) = a, and thatat time t = 0 the mass is released. Describe the motion of the system for all futuretime.

• Suppose that the mass is initially at rest in the equilibrium position x = 0, and thatat time t = 0 the system is “kicked”, imbuing the mass with an initial velocity ofx′(0) = b. Describe the motion of the system for all future time.

Exercise 1.1.4. Derive Kepler’s laws from Newton’s laws and the law of universal gravi-tation. Consider only two bodies. (Try this problems for a few minutes/a few hours beforeyou look up the solution online).

Exercise 1.1.5. (The pendulum)

• Use Newton’s laws to derive the equations of motion for a frictionless pendulum.Suppose that the pendulum moves close to the surface of the Earth, so that theacceleration of the gravitational field can be assumed constant.

• Rewrite the equation as a system of first order differential equation, i.e. as a vectorfield. Sketch the vector field, first by hand, and then using a computer program.What kinds of behaviour do you expect from the pendulum? Compare this to whatyou see in the vector field drawings.

• Repeat the problem assuming that the pendulum is subject to the force of friction.

• Suppose that the initial displacement of the pendulum is “small” compared to thelength of the pendulum. Simplify the equations of motion accounting this the “small-ness” assumption. Now justify Galileo’s observation that the period of a pendulumdepends only on the length of the pendulum, and not on the initial displacement ofthe pendulum (again for small displacements).

Exercise 1.1.6. (Meet the van der Pol equation) The Van der Pol equation is given bythe second-order differential equation

d2

dt2x(t)− µ(1− x(t)2)

d

dtx(t) + x(t) = 0,

and describes the voltage versus time behaviour in a certain electrical circuit consisting ofresistors, inductors, capacitors, a vacuum tube (triode), and a constant external voltagesource.

• Re-write the system as a vector field.

• Write a computer program which can numerically solve the equation (you can useor modify one of mine). Take µ = 0.1. Pick 25 initial conditions in the plane andintegrate them “for a long time”. Describe what happens.


• Same problem as above but with µ = 0.25.

• Now repeat with µ = 0.5, . . . , µ = 10, taking at least 5 different values of µ. Describewhat happens.

Exercise 1.1.7. (Meet the Lorenz equations) Consider the system of three nonlinear cou-pled first order ordinary differential equations

d

dtx = σ(y − x)

d

dty = x(ρ− z)− y

d

dtz = xy − βz

These are called the Lorenz equations, and they were derived by the Meteorologist EdwardLorenz in the early 1960’s as a simple model of weather (more precisely as a model of a apair of coupled convection cells in the atmosphere). In the model σ, β, ρ > 0 are positivereal parameters.

• Write a program (or use one of mine) which can simulate solutions for the Lorenzequations. Take σ = 10, β = 8/3 and ρ = 0.75. Choose 25 different initial conditionsand integrate them for a “long time”. Describe what happens.

• Take σ = 10, β = 8/3 and ρ = 2. Choose 25 different initial conditions and integratethem for a long time. Describe what happens. Repeat this experiment with ρ =5, 8, 10, 12.

• Repeat with ρ = 14, 18, 20, 22, 24.

• Repeat with ρ = 28. This is the so called “classic parameter” for the Lorenz system.

• Explore what happens for ten different values of ρ between ρ = 145.96 and ρ = 166.07.

Exercise 1.1.8. (Strange attractor scavenger hunt) Write computer programs (by modi-fying the Lorenz code I gave you) which reproduce all 7 of the strange attractors found onthe posters in the second floor hall way, outside the main math office in the FAU ScienceBuilding.

1.1. EXERCISES 9

Notation

Given x, y ∈ Rn,

x y if and only if xk < yk for all k = 1, . . . , n.

The default norm on Rn is the 2-norm:

‖x‖ :=

√√√√ n∑k=1

|xk|2.

Let ‖ · ‖X denote a norm on a space X. Then

Bε(x0) := x ∈ X | ‖x− x0‖X ≤ ε

Let J be an interval and α : J → X where X is a normed space with norm ‖ · ‖X . TheC0 norm of α is given by

‖α‖C0(J) := sup ‖α(t)‖X | t ∈ J .

Let Mn(R) the set a n × n matrices with real coefficients and Mn(C) the set a n × nmatrices with complex coefficients.

Given a matrix A = ai,ji,j and a vector v = vii (real or complex valued), then theelementwise absolute value is represented by |A| = |ai,j |i,j and |v| = |vi|i, where | · |denotes the absolute value.

The sup norm on Rn is defined as follows. Given x = (x1, . . . , xn) ∈ Rn

‖x‖∞ := maxk=1,...,n

|xk| .

Given a matrix A, the matrix norm induced by ‖ · ‖∞ is defined to be

‖A‖∞ = max‖x‖∞≤1

‖Ax‖∞.

If T : X → X, then its n-th iterate Tn is recursively defined by Tn(x) = T (Tn−1(x)).

Chapter 2

Fundamental Results: Existenceand Uniqueness

In this chapter we provide fundamental results concerning the existence, uniqueness, andcontinuity of solutions to ODEs. The results of this section are classical and can be foundin any graduate level text on ordinary differential equations. We have chosen not to presentthe most general results, but rather the minimal results necessary for this book. This isdone because we can directly use the contraction mapping theorem which is suggestive ofmany of the proofs in this class. See [4, Section 1.12] or [8, Chapter 1] for alternativeand/or more general proofs of existence and uniqueness. Our presentation follows that of[13, Section V.5.3].

2.1 Contraction Mapping Theorem

Consider a function T : X → X where X is a topological space. An element x ∈ X is afixed point of T , if T (x) = x. A fixed point is globally attracting if limn→∞ T

n(x) = x forall x ∈ X.

Definition 2.1.1. Let (X,d) denote a metric space. A function T : X → X is a contractionif there is a number κ ∈ [0, 1), called a contraction constant, such that

d(T (x), T (y)) ≤ κd(x, y)

for all x, y ∈ X.

Theorem 2.1.2 (Contraction Mapping Theorem). Let (X,d) be a complete metricspace. Assume that T : X → X is a contraction with contraction constant κ. Then thereexists a unique, globally attracting fixed point x ∈ X. Furthermore, for any x ∈ X,

d(Tn(x), x) ≤ κn

1− κd(T (x), x). (2.1)

11

12 CHAPTER 2. FUNDAMENTAL RESULTS: EXISTENCE AND UNIQUENESS

Proof. Choose x0 ∈ X and recursively define

xn+1 := T (xn).

By the assumption that T is a contraction we have that

d(xn+1, xn) = d(T (xn), T (xn−1)) ≤ κd(xn, xn−1).

Thus, by inductiond(xn+1, xn) ≤ κnd(x1, x0).

Applying the triangle inequality, we have for n < m

d(xn, xm) ≤m−1∑j=n

d(xj+1, xj)

≤m−1∑j=n

κjd(x1, x0)

≤ κn( ∞∑k=0

κk

)d(x1, x0)

≤ κn 1

1− κd(x1, x0). (2.2)

This implies that xn is a Cauchy sequence. Since X is complete there exists x ∈ Xsuch that

limn→∞

xn = x.

Hence, by continuity of T ,

x = limn→∞

xn = limn→∞

T (xn−1) = T(

limn→∞

xn−1

)= T (x).

This establishes the existence of a fixed point.We prove by contradiction that the fixed point is unique. Assume x is as above and y

is another fixed point of T , that is T (y) = y. Then

d(y, x) = d(T (y), T (x))

≤ κd(y, x).

Assume that y 6= x, so that d(y, x) > 0. Dividing by d(y, x) gives

κ ≥ 1,

which is the desired contradiction.

2.2. EXISTENCE AND UNIQUENESS OF SOLUTIONS TO ODES 13

Turning to (2.1), note that since x0 ∈ X was arbitrary, the estimates of (2.2) areindependent of the base point x0. Applying the triangle inequality we have

d(Tn(x), x) ≤ d(Tn(x), Tm(x)) + d(Tm(x), x)

≤ κn

1− κd(T (x), x) + d(Tm(x), x).

for any x ∈ X, and all m > n. Taking the limit as m → ∞ yields the bound claimed in(2.1). Finally, (2.1) and the fact that κ ∈ [0, 1) proves that x is a globally attracting fixedpoint.

2.2 Existence and Uniqueness of Solutions to ODEs

Definition 2.2.1. Let f : U → Rn be a continuous function defined on an open set U ⊂ Rn.A solution to the differential equation

x :=dx

dt= f(x) (2.3)

on an interval J ⊂ R is a differentiable function ϕ : J → U such that

dϕ

dt(t) = f(ϕ(t))

for all t ∈ J .

In this section we focus on solutions to the initial value problem (IVP)

x = f(x), x(t0) = x0,

that is, the existence of a solution x : J → U such that t0 ∈ J and x(t0) = x0.

Definition 2.2.2. Consider metric spaces (X,dX) and (Y,dY ). A function f : X → Y isLipschitz continuous if there exists a real constant K ≥ 0 such that, for all x1, x2 ∈ X,

dY (f(x1), f(x2)) ≤ KdX(x1, x2).

The smallest K satisfying this inequality is denoted by Lip(f) := K and is called theLipschitz constant of f .

More generally, f is locally Lipschitz continuous if every point in X has a neighborhoodsuch that f restricted to that neighborhood is Lipschitz continuous.

Note that Lipschitz continuity implies continuity, and moreover that a function which isLipschitz continuous on a bounded set is itself bounded. The following proposition, whoseproof is left to the reader, indicates that smooth functions are Lipschitz continuous.


Proposition 2.2.3. Let U ⊂ Rn be an open set and f : U → Rn. If f ∈ C1(U), then f islocally Lipschitz continuous.

The first goal of this section is the proof of the following theorem which guaranteeslocal existence and uniqueness of solutions.

Theorem 2.2.4. Assume f : U → Rn is a locally Lipschitz continuous function defined onan open set U ⊂ Rn. If x0 ∈ U , then there exists an open interval J ⊂ R, containing t0,over which a solution to the initial value problem

x = f(x), x(t0) = x0 (2.4)

is defined. Furthermore, any two solutions to the initial value problem agree on the inter-section of their domains of definition.

The proof of Theorem 2.2.4 is obtained via a series of propositions and lemmas. Thefirst step in the proof is the observation that by the fundamental theorem of calculus asolution to an ODE can be recast as a solution to an integral equation.

Lemma 2.2.5. Assume f : U → Rn is a continuous function defined on an open setU ⊂ Rn, and let x0 ∈ U . A continuous function ϕ : [a, b] → U solves the the initial valueproblem x = f(x), x(t0) = x0 on (a, b) with t0 ∈ (a, b) if and only if

ϕ(t) = x0 +

∫ t

t0

f(ϕ(s)) ds,

for all t ∈ [a, b].

Observe that since f is independent of t, there is no loss of generality in assuming thatt0 = 0, which for the sake of simplicity of expression will be done henceforth. The followingproposition provides for existence of solutions.

Proposition 2.2.6. If f : U → Rn is a locally Lipschitz continuous function defined on anopen set U ⊂ Rn and x0 ∈ U , then there exists a solution ϕ : (−a, a)→ U for some a > 0to the initial value problem

x = f(x), x(0) = x0. (2.5)

Moreover the solution ϕ(t) is Lipschitz continuous on (−a, a).

Proof. Since f is locally Lipschitz continuous, there exists an open neighborhood V0 ⊂ Uof x0 so that f is Lipschitz continuous on V0. Choose ε > 0 so that Bε(x0) ⊂ V0. SinceBε(x0) is compact, it follows that f is bounded on Bε(x0). Therefore, there are constantsK and M with

‖f(x)− f(y)‖ ≤ K‖x− y‖ and ‖f(x)‖ ≤Mfor all x, y ∈ Bε(x0).


By Lemma 2.2.5 it is sufficient to prove the existence of ϕ satisfying

ϕ(t) = x0 +

∫ t

0f(ϕ(s)) ds, (2.6)

for all t in some closed interval [−a, a] ⊂ R. The strategy is to define a function space Xwithin which we expect to find a solution, and define a contraction T : X → X such thatthe fixed point of T is a solution to (2.6).

Since solutions are functions of time the domain of elements of the function space shouldbe an interval J ⊂ R. Choose a > 0 with

a < min

ε

M,

1

K

, (2.7)

and set J := [−a, a]. Define the set X ⊂ C0(J) by

X := BaM (x0) =

α : J → Rn

∣∣∣∣ α is continuous and supt∈J‖α(t)− x0‖ ≤ aM

,

i.e. X is the closed ball in C0(J) of radius aM about the constant function x0. Note thatX is a complete metric space, as X is a closed subset of the complete metric space C0(J).We define the operator T : X → C0(J) by

T (α)(t) := x0 +

∫ t

0f(α(s)) ds,

with α ∈ X. Observe that a function satisfies (2.6) if and only if it is a fixed point of T .Application of the contraction mapping theorem (Theorem 2.1.2) requires two things: (i)T : X → X and (ii) T is a contraction.

We first prove (i). Note that for any α ∈ X and t ∈ J we have that

‖α(t)− x0‖ ≤ aM.

Then, since aM < ε, we have that

α(t) ∈ Bε(x0) ⊂ V ⊂ U ⊂ Rn,

for all t ∈ J . So, for any α ∈ X one has

supt∈J‖f(α(t))‖ ≤ sup

x∈Bε(x0)

‖f(x)‖ ≤M,


from which follows

‖T (α)(t)− x0‖C0(J) = supt∈J

∥∥∥∥(x0 +

∫ t

0f(α(s)) ds

)− x0

∥∥∥∥≤ sup

t∈J

∥∥∥∥∫ t

0f(α(s)) ds

∥∥∥∥≤ sup

t∈J

∣∣∣∣∫ t

0‖f(α(s))‖ ds

∣∣∣∣≤ sup

t∈J

∣∣∣∣∫ t

0M ds

∣∣∣∣≤ sup

t∈J|t|M

≤ aM,

and we see that T (α) ∈ X. But α ∈ X was arbitrary, so that T : X → X as desired.

We now prove (ii). Observe that given α, β ∈ X

‖T (α)− T (β)‖C0(J) = supt∈J‖T (α)(t)− T (β)(t)‖

= supt∈J

∥∥∥∥∫ t

0f(α(s)) ds−

∫ t

0f(β(s)) ds

∥∥∥∥≤ sup

t∈J

∣∣∣∣∫ t

0‖f(α(s))− f(β(s))‖ ds

∣∣∣∣≤ sup

t∈J

∣∣∣∣∫ t

0K‖α(s)− β(s)‖ ds

∣∣∣∣≤ sup

t∈J

∣∣∣∣∫ t

0K sup

τ∈J‖α(τ)− β(τ)‖ ds

∣∣∣∣≤ sup

t∈J

∣∣∣∣∫ t

0K‖α− β‖C0(J) ds

∣∣∣∣≤ sup

t∈J|t|K‖α− β‖C0(J)

≤ Ka‖α− β‖C0(J).

By (2.7), Ka < 1, and therefore T : X → X is a contraction with contraction constantκ = Ka < 1. Denote by ϕ : J → U the unique fixed point of T within X and observe thatthis implies, by Lemma 2.2.5, that ϕ is a solution of the initial value problem (2.5) on J .

Finally, suppose that ϕ : J → U is the fixed point of T obtained above. Again, we have


that f(ϕ(t)) ∈ Bε(x0) for all t ∈ J . Then for any t1, t2 ∈ J , one has that

‖ϕ(t1)− ϕ(t2)‖ = ‖T (ϕ)(t1)− T (ϕ(t2))‖

=

∥∥∥∥∫ t2

t1

f(ϕ(s)) ds

∥∥∥∥≤M |t2 − t1|.

Thus, ϕ(t) is Lipschitz continuous.

Returning to the details of the proof of Theorem 2.2.6, observe that if δ = ε/2, thenfor any x ∈ Bδ(x0),

Bδ(x) ⊂ U, ‖f(z)− f(y)‖ ≤ K‖z − y‖, and ‖f(z)‖ ≤M

for all z, y ∈ Bδ(x). This leads to the following corollary.

Corollary 2.2.7. Let f : U → Rn be a locally Lipschitz continuous function defined on anopen set U ⊂ Rn. For every x0 ∈ U there exists a neighborhood V of x0 and a constanta = a(V ), such that for every y ∈ V there exists a solution ϕ(·, y) : (−a, a) → U to theinitial value problem

x = f(x), x(0) = y.

Remark 2.2.8. It is worth noting that while we have proven the existence of a solution, it islimited to a time interval of length 2a where a < min

εM ,

1K

, which could be very small.

Theorem 2.2.20 provides information concerning existence over longer intervals of time.

Observe that we have only proven uniqueness of solutions over the family of continuousfunctions on [−a, a] with an a priori bound aM , as opposed to all differentiable functions.Thus the proof of Theorem 2.2.4 remains to be completed. For the moment we turn ourattention to a different question and demonstrate that solutions to an IVP are Lipschitzcontinuous as a function of the initial value. The following inequality will be used to provethe previous statement and is fundamental to the study of differential equations.

Theorem 2.2.9 (Gronwall’s Inequality). Let α, β : (a, b)→ [0,∞) be continuous non-negative functions. Assume

α(t) ≤ C +

∣∣∣∣∫ t

t0

α(s)β(s) ds

∣∣∣∣ , t0, t ∈ (a, b)

for some constant C ≥ 0. Then,

α(t) ≤ C exp

(∣∣∣∣∫ t

t0

β(s) ds

∣∣∣∣) .


Proof. Assume first that a < t0 ≤ t < b. Define

G(t) := C +

∫ t

t0

α(s)β(s) ds.

Then

G′(t) =dG

dt(t) = α(t)β(t) ≤ G(t)β(t).

Hence

exp

(−∫ t

t0

β(s)ds

)(G′(t)− β(t)G(t)

)≤ 0

d

dt

(G(t) exp

(−∫ t

t0

β(s)ds

))≤ 0

G(t) exp

(−∫ t

t0

β(s)ds

)≤ G(t0)

G(t) ≤ C exp

(∫ t

t0

β(s)ds

)α(t) ≤ C exp

(∫ t

t0

β(s)ds

).

Now assume a < t ≤ t0 < b. Define

G(t) := C +

∫ t0

tα(s)β(s)ds

and repeat the argument.

We now use Gronwall’s Inequality to show that solutions to an IVP with differentinitial data diverge at most at an exponential rate.

Proposition 2.2.10. Let U ⊂ Rn be an open set and assume f : U → Rn is a Lipschitzcontinuous function with Lip(f) = K. If ϕ(·, x0) : Jx0 → U and ψ(·, y0) : Jy0 → U aresolutions to the initial value problem x = f(x) with x(0) = x0 and x(0) = y0, respectively,then

‖ϕ(t, x0)− ψ(t, y0)‖ ≤ ‖x0 − y0‖eK|t| (2.8)

for all t ∈ Jx0 ∩ Jy0.

Proof. Let ϕ(·, x0) : Jx0 → U and ψ(·, y0) : Jy0 → U be solutions to the initial value prob-lems x(0) = x0 and x(0) = y0, respectively. By Proposition 2.2.6 the interior of theinterval Jx0 ∩ Jy0 contains 0. Let J be any open interval such that 0 ∈ J ⊂ Jx0 ∩ Jy0 . ByLemma 2.2.5, for all t ∈ J ,

ϕ(t, x0)− ψ(t, y0) = x0 − y0 +

∫ t

0f(ϕ(s, x0))− f(ψ(s, y0)) ds.


Let α(t) := ‖ϕ(t, x0)− ψ(t, y0)‖. Then

α(t) ≤ ‖x0 − y0‖+

∥∥∥∥∫ t

0f(ϕ(s, x0))− f(ψ(s, y0)) ds

∥∥∥∥≤ ‖x0 − y0‖+

∣∣∣∣∫ t

0‖f(ϕ(s, x0))− f(ψ(s, y0))‖ ds

∣∣∣∣≤ ‖x0 − y0‖+

∣∣∣∣∫ t

0K‖ϕ(s, x0)− ψ(s, y0)‖ ds

∣∣∣∣≤ ‖x0 − y0‖+

∣∣∣∣∫ t

0Kα(s) ds

∣∣∣∣since Lip(f) = K. Applying Gronwall’s Inequality with C = ‖x0 − y0‖ and β(t) = K, weobtain

α(t) ≤ ‖x0 − y0‖ exp

(∣∣∣∣∫ t

0K ds

∣∣∣∣)≤ ‖x0 − y0‖eK|t|

‖ϕ(t, x0)− ψ(t, y0)‖ ≤ ‖x0 − y0‖eK|t|,for all t ∈ J , where J was an arbitrary open subinterval of Jx0 ∩ Jy0 . A limit argumentthen implies that the final inequality holds for all t ∈ Jx0 ∩ Jy0 .

To see that solutions are locally Lipschitz continuous with respect to initial conditions,observe that given initial values x0 and y0 and the interval J = (−a, a) as defined above

‖ϕ(·, x0)− ψ(·, y0)‖C0(J) = supt∈J‖ϕ(t, x0)− ψ(t, y0)‖

≤ ‖x0 − y0‖ supt∈J

eK|t|

= ‖x0 − y0‖eKa.Remark 2.2.11. Inequality (2.8) provides Lipschitz continuity of the dependence on theinitial data. Suppose that f : U → Rn is a locally Lipschitz continuous function, andlet x0 ∈ U . Recall that by Corollary 2.2.7 there is an a > 0, an open neighborhoodx0 ∈ V0 ⊂ U , and a function ϕ : (−a, a)×V0 → U so that for all y ∈ V0 the function ϕ(·, y)solves that initial value problem x = f(x) and x(0) = y on (−a, a). Choosing V0 and asufficiently small and arguing as in Corollary 2.2.7, Proposition 2.2.10 implies that ϕ(t, y)is Lipschitz continuous in y for t ∈ (−a, a).

Proof of Theorem 2.2.4. As is remarked above, Proposition 2.2.6 guarantees the existenceof solutions. To show that two solutions to the same IVP agree on the intersection of theirdomains of definition let ϕ : J0 → Rn and ψ : J1 → Rn denote two solutions to the initialvalue problem x = f(x), x(0) = x0. By (2.8) for all t ∈ J0 ∩ J1

‖ϕ(t)− ψ(t)‖ = 0.


The following Corollary provides a summary of the results obtained so far.

Corollary 2.2.12. Assume f : U → Rn is a locally Lipschitz continuous function definedon an open set U ⊂ Rn. For every x0 ∈ U and t0 ∈ R there exists a neighborhood V0 ⊂ U ofx0, an interval J0 ⊂ R containing t0, and a Lipschitz continuous function ϕ : J0×V0 → Rnso that for each x ∈ V0 the function ϕ(·, x) is the solution to the IVP x = f(x), x(t0) = xon J0.

Observe that Proposition 2.2.3 guarantees existence and uniqueness of solutions forevery C1 vector field.

Note that since ϕ(·, x0) is a solution to the IVP it is differentiable in t. One canalso prove differentiability with respect to the initial conditions [4, Theorem 1.261]. Thisquestion is considered in Exercise 2.3.16 of this Chapter.

Models that arise in applications typically depend on set of parameters Λ and often aretime dependent. Thus, we are interested in solutions to differential equations that appearto take a more general form. Let J ⊂ R, U ⊂ Rn and Λ ⊂ Rm be open sets and letf : J ×U ×Λ→ Rn be a continuous function. For fixed λ ∈ Λ a solution to the differentialequation

x = f(t, x, λ) (2.9)

is a differentiable function ϕ : J0 → U defined on an open interval J0 ⊂ J such that

dϕ

dt(t) = f(t, ϕ(t), λ)

for all t ∈ J0. For t0 ∈ J , x0 ∈ U and λ0 ∈ Λ, the initial value problem associated with (2.9)requires finding a solution ϕ(t) = ϕ(t; t0, x0, λ0) to x = f(t, x, λ0) satisfying ϕ(t0) = x0.

The corresponding existence and uniqueness theorem is as follows.

Theorem 2.2.13. Let J ⊂ R, U ⊂ Rn and Λ ⊂ Rm be open sets, and assume f : J ×U ×Λ → Rn is a locally Lipschitz continuous function. If (t0, x0, λ0) ∈ J × U × Λ, then thereexists an open neighborhood of the form J0×U0×Λ0 of (t0, x0, λ0) and a unique Lipschitzcontinuous function ϕ : J0×J0×U0×Λ0 → Rn such that for every (t1, x1, λ1) ∈ J0×U0×Λ0

ϕ(·, t1, x1, λ1) : J0 → U

is the solution to the initial value problem

x = f(t, x, λ1), x(t1) = x1. (2.10)

Proof. The proof follows from the realization that result can be viewed as a special case ofTheorem 2.2.4. Define F : U × J × Λ→ Rn+1+m by

F (x, s, λ) = (f(s, x, λ), 1, 0).


By Theorem 2.2.4, there exists a function ϕ : J0 → Rn+1+m which satisfies the initial valueproblem

x = f(s, x, λ)

s = 1

λ = 0

(x(t0), s(t0), λ(t0)) = (x0, s0λ0) ∈ U × J × Λ.

It is left to the reader to check that the first n components of ϕ solve (2.10).

The most significant restriction on the assumptions of Theorem 2.2.4 is that f is Lip-schitz. As the following example indicates, existence is possible in more general settings,however uniqueness of the solution can no longer be assumed.

Example 2.2.14. Consider the initial value problem

x = 3x23 , x(0) = 0. (2.11)

Observe that

ϕ(t) ≡ 0 and ψ(t) :=

t3 if t ≥ 0

0 if t ≤ 0

are two (of infinitely many) distinct solutions to (2.11) on R.

Having established existence and uniqueness we turn to the question of the maximaltime interval on which a solution is defined.

Example 2.2.15. Consider the initial value problem

x = x2, x(0) = x0, (2.12)

where we assume that x0 > 0. By Theorem 2.2.4 we know that this IVP has a locallyunique solution. It is straightforward to check that

ϕ(t, x0) :=x0

1− tx0

is a solution and that the maximal interval in time over which ϕ is defined is −∞ < t <1/x0. As motivation for the next theorem it is also worth observing that

limt→1/x0

ϕ(t, x0) =∞.

Definition 2.2.16. Let ϕ : I → U and ψ : J → U be solutions to x = f(x), wheref : U → Rn is a continuous function defined on an open set U ⊂ Rn. We say that ψ is anextension of ϕ if I ⊂ J and ϕ(t) = ψ(t) for all t ∈ I. If we also have that I is a propersubset of J , then we say that ψ is a proper extension of ϕ.


Definition 2.2.17. A solution ϕ : J → U to x = f(x) is called a maximal solution if it hasno proper extensions. In this case the interval J is called the maximal interval of existenceof ϕ.

Theorem 2.2.18. If f : U → Rn is a locally Lipschitz continuous function defined on anopen set U ⊂ Rn and x0 ∈ U , then the initial value problem

x = f(x), x(0) = x0

has a unique maximal solution ϕ : J → U , and the maximal interval of existence J is open.

Proof. Let J be the union of all intervals I for which there is a solution ψ : I → U for theinitial value problem x = f(x), x(0) = x0. Since all the intervals I contain t0 = 0 we havethat J is an interval. Define ϕ : J → U by ϕ(t) := ψ(t) if t ∈ I and ψ : I → U is a solution.Since any two solutions agree on the intersection of their domains of definition, we havethat ϕ is well defined. The function ϕ : J → Rn just defined is clearly the unique maximalsolution to x = f(x), x(0) = x0. It remains to show that the interval J is open. Let t−and t+ be the left and right end points of J , respectively. Assume that J is closed at t+.Then, since U is open, we can by Theorem 2.2.4 extend the solution to an interval aboutt+, which contradicts the fact that J is the maximal interval. Likewise if we assume thatJ is closed at t−. Therefore J is the open interval (t−, t+).

The proof of the following Lemma is left as an exercise.

Lemma 2.2.19. Let f : U → Rn be a locally Lipschitz continuous function defined on anopen set U ⊂ Rn. Let Q be some compact subset of U , and let ϕ : (a, b)→ Q be a solutionto the initial value problem x = f(x), x(t0) = x0. Define the one-sided limits

ϕ(a) = x0 + limt→a

∫ t

t0

f(ϕ(s)) ds and ϕ(b) = x0 + limt→b

∫ t

t0

f(ϕ(s)) ds.

Then ϕ : [a, b]→ Q is differentiable, and

dϕ

dt(a) = f(ϕ(a)) and

dϕ

dt(b) = f(ϕ(b)).

Theorem 2.2.20. Let f : U → Rn be a locally Lipschitz continuous function defined onan open set U ⊂ Rn. Consider the initial value problem

x = f(x), x(0) = x0

with maximal solution ϕ(t, x0). Let J = (t−, t+) be the maximal interval of existence of ϕ.

(i) If t+ <∞, then given any compact set C ⊂ U there is a time t+C ∈ (0, t+) such thatϕ(t+C , x0) 6∈ C. Similarly, if t− > −∞, then given any compact set C ⊂ U there is atime t−C ∈ (t−, 0) such that ϕ(t−C , x0) 6∈ C.


(ii) If U = Rn and ‖f(x)‖ is bounded, then (t−, t+) = R.

Proof. (i) Assume, by contradiction, that C ⊂ U is compact and ϕ([0, t+), x0) ⊂ C. SinceC is compact and f is continuous, there are positive constants K1 and K2 such that

‖f(x)− f(y)‖ ≤ K1‖x− y‖ and ‖f(x)‖ ≤ K2

for all x, y ∈ C. Indeed, a locally Lipschitz continuous function is Lipschitz continuouson any compact subset. Thus the proof of Theorem 2.2.4 implies that the solution ϕsatisfies Lip(ϕ) ≤ K2 for all t ∈ [0, t+).

By Lemma 2.2.19 we can extend our solution to ϕ : [0, t+] → U such that dϕdt (t+) =

f(ϕ(t+)).

Theorem 2.2.4 guarantees the existence of δ > 0 and a solution ψ : (t+ − δ, t+ + δ) tothe initial value problem x = f(x), x(t+) = x+. Furthermore, because dϕ

dt (t+) = f(ϕ(t+))we can construct a solution to the original initial value problem by

φ(t) :=

ϕ(t) if t ≤ t+ψ(t) if t ≥ t+.

This contradicts the assumption that J = (t−, t+) is the maximal interval of existence ofϕ.

The argument for t− is similar.

(ii) Letting K := supx∈Rn ‖f(x)‖ <∞, note that

‖ϕ(t, x0)− x0‖ = ‖∫ t

0f(ϕ(s, x0))ds‖ ≤

∣∣∣∣∫ t

0‖f(ϕ(s, x0))‖ds

∣∣∣∣ ≤ K|t|. (2.13)

This implies that ϕ(t, x0) ∈ BR(x0) for all |t| < R/K. Suppose that t+ < ∞. Considerthe compact set C := BR0(x0) with R0 := Kt+. From part (1), there is a time t+C ∈ (0, t+)such that ϕ(t+C , x0) 6∈ C. Now, t+C < t+ = R0/K. By (2.13), we get that ϕ(t+C , x0) ∈BR0(x0) ⊂ C. This is a contradiction, implying that t+ =∞. Similarly, one can show thatt− = −∞. We conclude that J = R.

Based on Theorem 2.2.20, from now on we will interpret ϕ as being naturally definedon a set of the form (t, x0) | x0 ∈ U, t ∈ Jx0, where Jx0 is the maximal interval ofexistence.

We now prove that ϕ is locally Lipschitz continuous with respect to the initial data.

Theorem 2.2.21. Suppose that f : U → Rn is a locally Lipschitz continuous function, andlet x0 ∈ U . Let J be a closed subinterval of the maximal interval of existence Jx0. Thenthere is an open neighborhood V0 ⊂ U of x0 such that J ⊂ Jx for all x ∈ V0. The mapΨ : V0 → C0(J) given by Ψ(x)(t) = ϕ(t, x) is Lipschitz continuous.


Proof. For ε sufficiently small the set Q =⋃t∈J Bε(ϕ(t, x0)) is contained in U . Since Q is

compact, f is Lipschitz continuous on Q with Lipschitz constant, say, K. We can applyProposition 2.2.10 on Q. Let t0 = maxt∈J |t| and let ε0 = εe−Kt0 . Then inequality (2.8)implies that for x ∈ Bε0(x0) the solution ϕ(t, x) cannot leave the compact set Q for t ∈ J ,hence by Theorem 2.2.20(i) the interval of existence Jx contains J . Set V0 = Bε0(x0). Themap Ψ : V0 → C0(J) given by Ψ(x)(t) = ϕ(t, x) is Lipschitz continuous, since we inferfrom (2.8) that

‖ϕ(·, x1)− ϕ(·, x2)‖C0(J) ≤ eKt0‖x1 − x2‖, for all x1, x2 ∈ V0.

2.2.1 Flows

For the most part the focus of the previous sections is on the local existence of solutionsto differential equations. In this section we take a more global perspective.

Definition 2.2.22. Let f : U → Rn be a Lipschitz continuous function defined on an openset U ⊂ Rn. Let ϕ(·, x0) : J → Rn be the maximal solution to the IVP

x = f(x), x(0) = x0. (2.14)

The orbit through x0 is the set γ(x0) := ϕ(J, x0) ⊂ Rn.

The dynamical systems perspective on ODEs is to focus on the geometry of orbitsand/or sets of orbits. A fundamental concept that extends beyond the setting of ODEs isthe following.

Definition 2.2.23. Let X be a topological space. A continuous map ϕ : R×X → X is aflow if

(i) ϕ(0, x) = x

(ii) ϕ(t, ϕ(s, x)) = ϕ(t+ s, x)

for all x ∈ X and t, s ∈ R. The space X is called the phase space for the flow.

To put this into the context of differential equations consider the following Corollarythat arises as a special case of Theorem 2.2.4 and an application of Theorem 2.2.20(ii).

Corollary 2.2.24. Assume f : Rn → Rn is a bounded locally Lipschitz continuous function.Define ϕ : R× Rn → Rn by ϕ(·, x0) : R→ Rn to be the unique solution to the initial valueproblem x = f(x), x(0) = x0. Then, ϕ is a locally Lipschitz flow.


Using [4, Theorem 1.261] and extensions thereof one can prove that if f ∈ Cr, then theflow ϕ : R× Rn → Rn is also Cr.

The assumption that the vector field f is bounded is used to guarantee that the domainof the maximal solution is all of R for every initial condition. Boundedness is an extremelystrong constraint. However, as mentioned above in dynamical systems the focus is on thegeometry of orbits as opposed to their parameterization, which allows us to circumventthis constraint. To be more specific, in Definition 2.2.22 it is possible that the maximalsolution is ϕ : J → Rn where J 6= R. However, if we consider the following modified IVP,

x =f(x)

1 + ‖f(x)‖2 , x(0) = x0, (2.15)

then by Theorem 2.2.20(ii) the maximal solution takes the form ψ(·, x0) : R→ Rn. Observethat if ϕ(t0, x0) = x1 and ψ(t1, x0) = x1, then

dϕ(t, x0)

dt(t0) =

1

1 + ‖f(x1)‖2dϕ(t, x0)

dt(t1),

which means that the tangent vectors to the two solutions are the scalar multiples of eachother at every point that they have in common. As is shown in the following theorem thisimplies that both differential equations produce the same orbits.

Theorem 2.2.25 (Reparameterization of time). Let U ⊂ Rn be an open set. Assumef : U → Rn is Ck (with k = ∞ allowed). Given x0 ∈ U , let ϕ(·, x0) : Jx0 → U be themaximal solution for the IVP (2.14). Denote by γ(x0) the orbit through x0. Then thereexists a Ck function g : U → (0, 1] such that R is the maximal domain of existence for anysolution to

x = g(x)f(x).

Moreover, for each x0 ∈ U , denote by ψ(·, x0) : R → U the solution for the initial valueproblem x = f(x)g(x), x(0) = x0. Then,

ψ(t, x0) = ϕ(τ(t, x0), x0),

where τ(·, x0) : R→ Jx0 satisfies τ(0, x0) = 0, and solves the differential equation

τ(t, x0) = g(ϕ(τ(t, x0), x0)

)> 0.

Hence, ψ(·, x0) is a reparameterization of ϕ(·, x0) with the same oriented solution curves,yielding the same orbit γ(x0) = ϕ(Jx0 , x0) = ψ(R, x0).

Proof. If U = Rn, let g(x) := 11+‖f(x)‖2 , with ‖ · ‖ the Euclidean norm (hence ‖ · ‖2 is a C∞

function from Rn to [0,∞)). Then g : U → (0, 1] is a Ck function, and ‖g(x)f(x)‖ ≤ 1 forall x ∈ U . From Theorem 2.2.20(ii), any solution to x = g(x)f(x) is defined on R.


In case U 6= Rn, then consider a function G : U → (0, 1] to be a C∞ function such thatG(x) approaches 0 as x goes to the boundary of U or as ‖x‖ goes to infinity (the functionG should be thought of as the square distance to the boundary of U). Moreover, wescale G so that it satisfies the bound supx∈U ‖DG(x)‖ ≤ 1, where the row vector DG(x)is interpreted as a linear functional on Rn, hence the bound on the norm implies that

|DG(x)y| for any y ∈ Rn and any x ∈ U . In this case, let g(x) := G(x)2

1+‖f(x)‖2 , which

is a Ck function, and let F (x) := g(x)f(x). Then, ‖F (x)‖ ≤ |G(x)|2 ≤ 1 for all x ∈ U .Now, consider x0 ∈ U and denote by ψ(t, x0) the unique solution of the IVP x = F (x),x(0) = x0. From Theorem 2.2.20(i), to show that ψ(t, x0) is defined on R, it is enough toshow that G(ψ(t, x0)) does not go to 0 in finite time, or equivalently that 1

G(ψ(t,x0)) doesnot go to infinity in finite time. Now,

d

dt

1

G(ψ(t, x0))= − 1

G(ψ(t, x0))2DG(ψ(t, x0))F (ψ(t, x0))

and therefore,

1

G(ψ(t, x0))− 1

G(ψ(0, x0))= −

∫ t

0

1

G(ψ(s, x0))2DG(ψ(s, x0))F (ψ(s, x0)) ds

= −∫ t

0DG(ψ(s, x0))

f(ψ(s, x0))

1 + ‖f(ψ(s, x0))‖2 ds.

This implies that ∣∣∣∣ 1

G(ψ(t, x0))

∣∣∣∣ ≤ ∣∣∣∣ 1

G(x0)

∣∣∣∣+

∫ |t|0

ds =

∣∣∣∣ 1

G(x0)

∣∣∣∣+ |t|,

and therefore 1G(ψ(t,x0)) does not go to infinity in finite time. Hence, ψ(t, x0) is defined for

all t ∈ R.

Now, let τ(t, x0) ∈ R be the unique solution of the IVP

τ(t, x0) = g(ϕ(τ(t, x0), x0)

), τ(0, x0) = 0.

Then, ϕ(τ(t, x0), x0) satisfies ϕ(τ(0, x0), x0) = ϕ(0, x0) = x0 and

d

dtϕ(τ(t, x0), x0) = f

(ϕ(τ(t, x0), x0)

)τ(t, x0)

= f(ϕ(τ(t, x0), x0)

)g(ϕ(τ(t, x0), x0)

)= F

(ϕ(τ(t, x0), x0)

).

By unicity of the solutions of the IVP x = F (x), x(0) = x0, we obtain that ψ(t, x0) =ϕ(τ(t, x0), x0).

2.3. EXERCISES 27

The following definition formalizes the comparison between vector fields performed inthe previous theorem.

Definition 2.2.26. Two flows ϕ : R × X → X and ψ : R × Y → Y are topologicallyequivalent if there exists a homeomorphism h : X → Y such that orbits of ϕ are mappedonto orbits of ψ preserving the direction of time, that is, there exists a continuous andstrictly increasing time-rescaling map τ : R×X → R such that

h(ϕ(t, x)) = ψ(τ(t, x), h(x)),

for all (t, x) ∈ R×X.

2.3 Exercises

Exercise 2.3.1. Let (X, d) be a metric space. Prove that the metric is a continuousfunction (in the metric space topology).

Exercise 2.3.2. Use the contraction mapping theorem to prove that the function f(x) =cos(x) has a positive real fixed point.

Exercise 2.3.3. Suppose that f : R → R is a continuously differentiable function (i.e.that f ′ is continuous on R). Suppose that there is an x0 ∈ R so that f(x0) = x0, and that|f ′(x0)| < 1. Prove that f is a contraction mapping in some neighborhood of x0.

Exercise 2.3.4. Suppose that f : R → R is a continuously differentiable mapping (i.e.that the function f ′ is continuous on R). Assume that there is an x0 ∈ R and positiveconstants Y0 and Z0 so that

|f(x0)− x0| ≤ Y0,

and

|f ′(x0)| ≤ Z0 < 1.

• Prove that there exists an ε > 0 so that

|f ′(x)| ≤ 1 + Z0

2,

for all x ∈ [x0 − ε, x0 + ε].

• Prove that if2Y0

1− Z0< ε,

then there exists a unique fixed point for f in [x0 − ε, x0 + ε].


Exercise 2.3.5. Use the contraction mapping theorem to prove that the function

f(x, y) =

(x2 + y

2 + xy25 + x3

5032 −

y2 −

xy30 + y4

100

)

has a fixed point near x = 1, y = 1. Write a MatLab program which computes the fixedpoint to 8 decimal places.

Exercise 2.3.6. In Chapter 1 you worked with the following differential equations:

•d

dtx(t) = kx(t)− k

Kx(t)2, (Logistic Equation)

• (x′

y′

)=

(y

−cy − ω sin(x)

)(The pendulum)

• (x′

y′

)=

(µ(x− 1

3x3 − y

)1µx

)(van der Pol oscillator)

• x′

y′

z′

=

σ(y − x)x(ρ− z)− yxy − βz

(Lorenz Equation)

• w′

x′

y′

z′

=

x−w

(w2+y2)3/2

z−y

(w2+y2)3/2

(The “one body problem”)

In each case, determine the largest open set U ⊂ Rn on which the vector field is locallyLipschitz. Repeat this problem for some of the vector fields from the scavenger hunt.

Exercise 2.3.7. Let A be an n× n matrix and define the linear map T : Rn → Rn by

T (x) = Ax,

for all x ∈ Rn. Find conditions on A which guarantee that T is a contraction.

Exercise 2.3.8. Give an example of a metric space X and a contraction mapping T : X →X so that T does not have any fixed points. (Of course X cannot be complete).

2.3. EXERCISES 29

Exercise 2.3.9. Consider the metric space

X = x ∈ R : 1 ≤ x <∞ ,

with the metric inherited from R, and let T : X → X be the map

T (x) = x+1

x.

Prove that

• |T (x)− T (y)| < |x− y| for all x, y ∈ X with x 6= y,

• and the T has no fixed points.

Exercise 2.3.10 (Existence via Picard’s method). An alternative proof for the exis-tence of solutions is given by Picard’s method. In this exercise we use Picard’s method toprove the existence of a solution to the non-autonomous IVP

x = f(x, t), x(t0) = x0.

We begin with some hypotheses. Consider an open set U ⊂ Rn and an open intervalI ⊂ R which contains t0. Denote D := U × I. Assume that f : D → Rn is continuous andthat for all t ∈ I, f(·, t) : U → Rn is a locally Lipschitz continuous function.

The goal is to show that there exists a solution ϕ : J = (t0 − a, t0 + a)→ Rn for somea > 0 to the integral equation

ϕ(t) = x0 +

∫ t

t0

f(ϕ(s), s) ds, for all t ∈ J. (2.16)

Let ε, δ > 0 small enough so that Dε,δ := (x, t) | ‖x− x0‖ ≤ ε, |t− t0| ≤ δ ⊂ D andso that

‖f(x, t)− f(y, t)‖ ≤ K‖x− y‖, for all (x, t), (y, t) ∈ Dε,δ. (2.17)

LetMε,δ := max

(x,t)∈Dε,δ‖f(x, t)‖, (2.18)

and let a > 0 such that

a ≤ min

δ,

ε

Mε,δ

. (2.19)

This existence time a is independent of the Lipschitz constant, cf. (2.7). Define a timeinterval by

J := (t0 − a, t0 + a). (2.20)

For any t ∈ J , define the Picard operator by

T (x)(t) = x0 +

∫ t

t0

f(x(s), s) ds. (2.21)


(i) Show that (T (x)(t), t) ∈ Dε,a, for every (x, t) ∈ Dε,a.

This allows us to define a sequence of functions xnn≥0 with xn : J → Rn by

x0(t) ≡ x0, xn+1(t) = x0 +

∫ t

t0

f(xn(s), s) ds, n ≥ 0. (2.22)

The iterations (2.22) define what is known as the Picard iterative process.

(ii) Show that for every t ∈ J = (t0 − a, t0 + a),

‖xn(t)− xn−1(t)‖ ≤Mε,δKn−1 |t− t0|n

n!.

(iii) Show that xnn≥0 is a Cauchy sequence in the space C0(J) endowed with the supre-mum norm.

(iv) Show that there exists a continuous function ϕ : J → Rn that is solution of theintegral equation (2.16).

Exercise 2.3.11. Consider the differential equation

dx

dt= x,

with the initial condition x(0) = 1. Compute the Picard iteration for this differentialequation starting with the initial function

x0(t) = 1.

Exercise 2.3.12. Consider the differential equation(x′(t)y′(t)

)=

(0 −11 0

)(x(t)y(t)

)with the initial condition x(0) = 1, y(0) = 1. Compute the Picard iteration for thisdifferential equation starting with the initial function(

x0(t)y0(t)

)=

(11

)Exercise 2.3.13. Consider the differential equation

dx

dt= 2x+ x2

with the initial condition x(0) = 1/2. Compute the Picard iteration for this differentialequation starting with the initial function

x0(t) = 1/2.

2.3. EXERCISES 31

Exercise 2.3.14. Prove Proposition 2.2.3.

Exercise 2.3.15 (Existence and uniqueness for linear differential equations).Consider J ⊂ R an open interval. Suppose that g : J → Rn and ai,j : J → R, 1 ≤ i, j ≤ n,are continuous functions. Set

A(t) =

a1,1(t) . . . a1,n(t)...

. . ....

an,1(t) . . . an,n(t)

.

Prove that for any t0 ∈ J the initial value problem

x(t) = A(t)x+ g(t), x(t0) = x0, (2.23)

has a unique solution x : J → Rn.

Hint: Define an integrating factor C(t) by writing down a solution of the equation

C ′(t) = −C(t)A(t),

that is write down an explicit formula for such a C(t). Show directly that

• C(t) is well defined, continuous, and differentiable on J .

• C(t0) = Id.

• C(t) is invertible for every t ∈ J .

Now multiply both sides of Equation (2.23) by C(t), apply the product rule on the left,integrate both sides from t0 to t with t ∈ J , and solve for x(t). This provides an explicitformula for the solution x(t). Show that the formula gives a well defined, continuous,differentiable function for each t ∈ J .

Exercise 2.3.16 (Differentiability of the flow with respect to initial conditions).Let ϕ : R × Rn → Rn be the flow generated by x = f(x), where f : Rn → Rn is C1 andbounded on Rn. Proposition 2.2.10 guarantees that ϕ(t, x) is continuous with respect toinitial conditions. The following approach shows that ϕ is actually differentiable withrespect to x.

Observe that differentiability of ϕ with respect to x is equivalent to the statement thatfor any t ∈ R, x0 ∈ Rn, there exists a unique n× n matrix Dϕ(t, x0) satisfying

lim‖h‖→0

‖ϕ(t, x0 + h)− ϕ(t, x0)−Dϕ(t, x0)h‖‖h‖ = 0. (2.24)


Let x0 ∈ Rn and γ(t) ≡ ϕ(t, x0) denote the orbit segment through x0. Define thematrix Dϕ(t, x0) to be the solution of the first variation equation,

d

dtDϕ(t, x0) = Df(γ(t))Dϕ(t, x0), Dϕ(0, x0) = Id.

Use the results of problem 2.3.15 to establish that the matrix solving this equation existsassuming only that φ(t, x0) ≡ γ(t) exists and is continuous. Now show that Dϕ(t, x0)satisfies (2.24).

Exercise 2.3.17 (Flow Box Theorem). Consider the differential equation x = f(x)where f : Rn → Rn is a Cr vector field for r ≥ 1. A point x ∈ Rn is called a ordinarypoint if f(x) 6= 0. Prove that there exists a Cr change of coordinates x = g(y) defined ona neighborhood of x such that

y1 = 1

y2 = 0...

yn = 0

Hint: Without loss of generality assume that x = 0 and f(0) = (1, 0, . . . , 0). Let U be asufficiently small neighborhood of the origin. Consider the map h : U → Rn defined by

h(y1, . . . , yn) := ϕ(y1, (0, y2, . . . , yn))

where ϕ : R × Rn → Rn is the flow associated with x = f(x). Make use of the inversefunction theorem Theorem 7.3.1.

Exercise 2.3.18. Let J = [a, b] ⊂ R. Show that X = x : J → Rn | x is continuous withthe C0(J) norm

‖x− y‖C0(J) := supt∈J‖x(t)− y(t)‖

is a complete metric space

Exercise 2.3.19. Let f : Rn → Rn be a globally Lipschitz continuous function. Considerthe initial value problem

x = f(x), x(0) = x0

with maximal solution ϕ(t, x0). Let J = (t−, t+) be the maximal interval of existence of ϕ.Show that J = R. Hint: use Gronwall’s inequality (Theorem 2.2.9).

Exercise 2.3.20. Prove Lemma 2.2.19.

Chapter 3

Equilibria and Radii PolynomialApproach in Finite Dimension

Consider an open set U ⊂ Rn and a locally Lipschitz continuous function f : U → Rn.Theorem 2.2.13 guarantees the existence of a unique maximal solution to the initial valueproblem problem

x = f(x), x(0) = x0

for any x0 ∈ U . Therefore, we can now turn our attention to studying solutions withspecific properties. The simplest, and hence the starting point for our investigations, isthat of an equilibrium point, also called a fixed point, steady state, or critical point, i.e., asolution that is constant in time. One of our goals is to develop a constructive method toprove the existence of equilibria. Observe that x0 ∈ Rn is an equilibrium point if and onlyif f(x0) = 0. Thus, it is sufficient for us to provide a constructive approach for provingthe existence of zeros of a function defined on a finite dimensional space X (in our case Rnor Cn). More precisely, the goal of this section is twofold: prove the existence of a pointx ∈ X such that f(x) = 0 and provide bounds on the location of x. This is done using theradii polynomial approach which is a variant on Newton’s method. With this in mind werecall Newton’s method in Section 3.1 and then introduce the radii polynomial approachin finite dimension in Section 3.2.

3.1 Newton’s Method

We begin with a trivial proposition that sets the stage for our strategy for finding zeros ofa function.

Proposition 3.1.1. Let X be be a topological vector space and let U, V ⊂ X be open sets.Consider f : U → V . Assume that A : X → X is an injective linear map. Let T : U → X

33

34CHAPTER 3. EQUILIBRIA AND RADII POLYNOMIAL APPROACH IN FINITE DIMENSION

be defined byT (x) := x−Af(x). (3.1)

If T (x) = x, then f(x) = 0.

If X is a finite dimensional vector space, then the assumption that A is injective isequivalent to A being invertible. Thus, for the remainder of this chapter we will, withoutloss of generality, assume that A is invertible.

Proposition 3.1.1 allows us to replace the problem of directly finding a zero of f withthat of proving the existence of a fixed point of T . As is indicated at the beginningof Chapter 2 the contraction mapping theorem (Theorem 2.1.2) provides existence anduniqueness if T is a contraction. Furthermore, it gives bounds on the location of x as afunction of an initial guess (recall (2.1)). Thus the problems we are trying to address arereduced to finding an injective linear map A that makes T a contraction. This leads us toNewton’s method. We begin with a simple example.

Example 3.1.2. Consider f ∈ C1(R,R). Recall that in Newton’s method the fixed pointmap T is given by

T (x) := x− f(x)

f ′(x)

and that T is used iteratively to find an approximate value of a root of f . More explicitly,given an initial guess x0 ∈ R, inductively define xk+1 := T (xk). Assume

limk→∞

xk = x.

Observe that if T is continuous at x (a sufficient condition for this is that f ′(x) 6= 0), thenT (x) = x and hence f(x) = 0. Thus the problem of proving the existence of a zero of f isessentially reduced to finding and/or identifying whether an initial guess x0 ∈ R leads toconvergence of Newton’s method.

Anticipating that fast convergence of the Newton iterations is related to the contractionrate of T , let us assume for simplicity that f ∈ C2 and, more importantly, that the zerox of f is non-degenerate: f ′(x) 6= 0. In that case we can compute the derivative

T ′(x) = 1− (f ′(x))2 − f(x)f ′′(x)

(f ′(x))2= 0.

This implies that T ′ is small near the fixed point x of T , hence T has a small contractionconstant in a neighborhood of x.

The intended take away message from Example 3.1.2 is that in a sufficiently smallneighborhood of a nondegenerate zero of f the associated Newton operator is an extremelystrong contraction. In particular, returning to the question concerning the choice of aninvertible linear map A, a naive interpretation of this example suggests setting A = A(x) =(f ′(x))−1.

3.2. RADII POLYNOMIAL APPROACH IN FINITE DIMENSION 35

Next, consider f : U → Rn a C1 function defined on an open set U ⊂ Rn. In this casethe Newton operator is given by

T (x) := x− (Df(x))−1f(x). (3.2)

An argument similar to that presented in Example 3.1.2 demonstrates that if f(x) = 0and Df(x) is invertible, then in a small neighborhood of x, T is a contraction mappingwith small contraction constant. Again, this suggests the choice of A(x) = (Df(x))−1.However, the cost of computing the inverse of a n × n matrix is of order n3 and thus forhigh dimensional problems repeatedly computing the inverse is prohibitively expensive.

The second, and perhaps more substantial reason, is that for most of this book, i.e.,Chapters 6 onward, we are interested in applying these techniques to maps that are definedon infinite dimensional Banach spaces. In this setting, finding an explicit representationof (Df(x))−1 is typically not possible. Thus the approach we adopt is as follows. Weassume that we are given an initial guess x for a zero of f . In practice x is obtained usinga standard numerical method. We then use some, typically problem dependent, form ofapproximation of (Df(x))−1 to choose A.

3.2 Radii Polynomial Approach in Finite Dimension

The discussion of the previous section indicates that we intend to use the contractionmapping theorem applied to functions of the form T (x) = x− Af(x) to identify zeroes off . Two key issues remain to be resolved: how does one choose the domain of T and howdoes one prove that T is a contraction?

Consider f(x) = x2 − 1. The associated Newton operator T : R \ 0 → R is given byT (x) = x− (2x)−1(x2 − 1). Since T has two distinct fixed points, T (±1) = ±1, T cannotbe a contraction mapping over its entire domain. However, the analysis of Example 3.1.2suggests that there exists a neighborhood U+ of 1 such that T : U+ → R defined byT (x) := x− 1

2(x2− 1) is a contraction mapping. The theorem below provides a mechanismfor rigorously identifying a domain on which T is a contraction mapping.

The theorem is formulated for a general norm on Rn (or Cn). To obtain explicitestimates, in the examples we choose to work with the ∞-norm.

Theorem 3.2.1. [Radii polynomial approach in finite dimension] Consider a C1

map f : Rn → Rn. Let x ∈ Rn and A ∈ M(Rn,Rn). Let Y0 and Z0 be non-negativeconstants and Z2 : (0,∞)→ [0,∞) a non-negative function satisfying

‖Af(x)‖ ≤ Y0 (3.3)

‖Id−ADf(x)‖ ≤ Z0 (3.4)

‖A[Df(c)−Df(x)]‖ ≤ Z2(r)r, for all c ∈ Br(x) and all r > 0. (3.5)


Definep(r) := Z2(r)r2 − (1− Z0)r + Y0. (3.6)

If there exists r0 > 0 such that p(r0) < 0, then the matrix A is invertible and there existsa unique x ∈ Br0(x) satisfying f(x) = 0.

The same result holds when Rn is replaced by Cn.

We remark that no constant Z1 appears in the theorem. It will become clear why it is“missing” here once we discuss infinite dimensional generalizations in Section 6.6.

Since in practice the bound Z2(r) is usually chosen to be a polynomial with nonnegativecoefficient, the function p(r) is called the radii polynomial.

Proof. We write Z2 = Z2(r0). Note that the assumption that p(r0) < 0 is equivalent to

κ := Z0 + Z2r0 < 1− Y0

r0≤ 1. (3.7)

Define the mapping T byT (x) = x−Af(x).

Now, given c ∈ Br0(x) and applying the bounds (3.3), (3.4), (3.5), and (3.7), we obtain

‖DT (c)‖ = ‖Id−ADf(c)‖≤ ‖Id−ADf(x)‖+ ‖A[Df(c)−Df(x)]‖≤ Z0 + Z2r0

= κ < 1. (3.8)

In particular, ‖Id − ADf(x)‖ < 1 and hence ADf(x) is invertible (see Exercise 3.3.11).This implies that A is invertible.

We now show that T maps Br0(x) into itself. Let x ∈ Br0(x), apply the Mean ValueInequality (Corollary 3.3.8) and (3.8) to obtain

‖T (x)− x‖ ≤ ‖T (x)− T (x)‖+ ‖T (x)− x‖≤ sup

c∈Br0 (x)

‖DT (c)‖‖x− x‖+ ‖Af(x)‖

≤ (Z0 + Z2r0)r0 + Y0

< r0,

where the last inequality follows from the assumption that p(r0) < 0. Hence, T (Br0(x)) ⊂Br0(x).

To see that T is a contraction on Br0(x), let a, b ∈ Br0(x) and apply (3.8) to obtain

‖T (a)− T (b)‖ ≤ supc∈Br0 (x)

‖DT (c)‖‖a− b‖

= κ‖a− b‖.


Hence T : Br0(x) → Br0(x) is a contraction with contraction constant κ < 1. Thecontraction mapping theorem yields the existence of a unique x ∈ Br0(x) such thatT (x) = x−Af(x) = x. The invertibility of A yields the existence of a solution x ∈ Br0(x)of f(x) = 0.

Remark 3.2.2. Assume that Z2(r) is a polynomial with non-negative coefficients, or moregenerally a non-decreasing function of r. Observe that if there exists r > 0 such that p(r) <0, then there exists a range of values I = (r−, r+) ⊂ [0,∞) over which the inequalities aresatisfied. Since x is the unique zero of f in Br(x) for all r ∈ I, r− provides tight boundson the location of x, while r+ provides information about the domain of isolation of x.The maximal such interval is called the existence interval for the radii polynomials. If theexistence interval for the radii polynomials is nonempty, then one can present an explicitdomain in which there exists a unique zero of f .

Remark 3.2.3. We claim that, while choices need to be made to apply Theorem 3.2.1, thereis considerable lattitude in how these choices can be made to achieve successful applicationsfor concrete problems. For the sake of simplicity, let us assume that Z2 > 0 is independentof r. A necessary and sufficient condition for the existence interval of p (defined as in (3.6))to be non empty is

Z0 + 2√Y0Z2 < 1. (3.9)

Indeed, (3.9) amounts to saying that Z0 < 1 and the discriminant of p is positive, in whichcase p(r) must be negative for some positive r, since at its vertex 1−Z0

2Z2> 0 the radii

polynomial p is negative.Observe that if x = x, then f(x) = 0 and one can take Y0 = 0, hence the lower bound

of the existence interval is 0. If A := (Df(x))−1, then Z0 can be taken equal to 0. Alsonotice that, once some (possibly coarse) Lipschitz bound β on the derivative Df has beenfound, i.e.,

‖Df(b)−Df(c)‖ ≤ β‖b− c‖,it is, at least theoretically, always possible to use the radii polynomial approach to provethe existence of and provide bounds on the location of the zero x of f , provided it isnon-degenerate (i.e. Df(x) is invertible). Indeed, in that case Df(x) is invertible for x ina neighborhood of x. By choosing x in this neighborhood and choosing an approximateinverse A of DF (x), the condition (3.9) becomes

‖Id−ADf(x)‖+ 2‖A‖√β‖f(x)‖ < 1. (3.10)

Sufficient conditions for this inequality to be satisfied are having a small residue ‖f(x)‖and a good approximation A ≈ (Df(x))−1.

However, (3.10) also indicates that these are not necessary conditions. The residuef(x) may differ significantly from 0 if either ‖A‖ or β is sufficiently small. Similarly,A ≈ (Df(x))−1 is not a necessary condition, since Z0 = ‖Id− ADf(x)‖ can be close to 1if f(x) is sufficiently small.


To demonstrate how the radii polynomials are used in practice we consider severalsimple examples.

Example 3.2.4 (The radii polynomial approach for a one-dimensional example).Consider the simplest nonlinear function f(x) = x2−2. The first step is to choose an initialguess x for the zero of f . In a typical application x is taken as the output from a standardnumerical procedure for finding zeros of a function. The next step is to fix a value forA ∈ R. Example 3.1.2 suggests that A be chosen based on Df(x)−1. We leave it generalfor now. To prove that

T (x) := x−A(x2 − 2)

is a contraction mapping via Theorem 3.2.1 we need to determine bounds Y0, Z0 and Z2

satisfying (3.3), (3.4) and (3.5), respectively.Consider any Y0 and Z0 such that

|A(x2 − 2)| := Y0,

|1−ADf(x)| = |1−A(2x)| := Z0.

If c ∈ Br(x) = [x− r, x+ r], then

|A[Df(c)−Df(x)]| = |A(2c− 2x)| = 2|A||c− x| ≤ 2|A|r = Z2r,

that is Z2 := 2|A|. Hence the radii polynomial (3.6) is given by

p(r) = Z2r2 − (1− Z0)r + Y0

= 2Ar2 − (1− |1− 2xA|) r + |A(x2 − 2)|.

Let us now make some explicit choices.

(a) The best possible approximations

Let x =√

2 (this is an exact solution) and A = Df(x)−1 = 12x (the exact inverse).

In this case, the radii polynomial is p(r) = 2Ar2 − (1− |1− 2xA|) r + |A(x2 − 2)| =√2

2 r2−r. Then, the existence interval for the radii polynomial is given by I =

(0,√

2).

(b) Not the best approximations

For the purpose of applications it is not practical to assume that x is the exactsolution, nor is it reasonable to assume that A = Df(x)−1. However, the proposedapproach works well even with coarse approximations.

Choose x = 1.3. Since Df(x)−1 = (2x)−1 ≈ 0.384615, we set the approximate inverseto be A = 0.38. To prove that

T (x) := x− 0.38(x2 − 2)


is a contraction mapping we need to determine bounds Y0, Z0 and Z2. To obtain Y0

observe that| −Af(x)| = | − 0.38(1.32 − 2)| = 0.1178.

Since we only need a bound we choose Y0 := 0.12. As above, we have Z0 = |1−2xA| =0.012 and Z2 = 0.76r.

For the choices of bounds, the associated radii polynomial is given by

p(r) = Z2r2 − (1− Z0)r + Y0 = 0.76r2 − 0.988r + 0.12.

See Figure 3.1 for a geometrical interpretation of the radii polynomial p(r). Usingthe quadratic formula we see that p(r) < 0 for all r ∈ I = [0.136, 1.164]. By The-

r

p(r)

I

0 0.2 0.4 0.6 0.8 1 1.2

−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

Figure 3.1: The radii polynomial p(r) = 0.76r2 − 0.988r + 0.12 of Example 3.2.4 andI = [0.136, 1.164], where the radii polynomial is strictly negative.

orem 3.2.1 and Remark 3.2.2 we can conclude that x ∈ B0.136(1.3) = [1.164, 1.436].Given that the actual root contained in this interval is

√2 ≈ 1.414 we see that the

relative error for this bound is

1.414− 1.164

1.414≈ 0.1768.

We can also conclude that there is a unique root in the interval x ∈ B1.164(1.3) =[0.136, 2.464]. Given that −

√2 ≈ −1.414 is also a root, this is a reasonably accurate

statement.

Example 3.2.4 demonstrates how the radii polynomial can be employed. It also makesclear that the choice of A plays a central role in the values of Y0, Z0 and Z2, which in turndetermines whether the radii polynomial inequality can be satisfied. With this in mind wereturn to this example to gain some insight into how broad a range of A’s is possible.


•Br(x)

•xx

• •xx

Br(x)

x1

• •xx

Br(x)•

Figure 3.2: For different values of r > 0, we can have different scenarios. On the left,the radius r is too small and p(r) ≥ 0. Hence the method based on the radii polynomialscannot conclude about the existence of a unique solution x∗ ∈ Br(x). In the context ofExample 3.2.4, that would correspond for instance to r = 0.1 /∈ I = [0.136, 1.164]. In themiddle, the radius r ∈ I is chosen so that p(r) < 0, which implies that the ball Br(x)contains a unique solution. In the context of Example 3.2.4, that would correspond forinstance to r = 0.15 ∈ I. On the right, the radius r is too large and p(r) ≥ 0. Hence,it could happen that there exist more than one solution in the ball Br(x), that is thereexist x∗, x∗1 ∈ Br(x) both solutions of f = 0. In the context of Example 3.2.4, that wouldcorrespond for instance to r = 3, where x∗ =

√2 and x∗1 = −

√2.

Example 3.2.5. As a slight generalization of Example 3.2.4, let f(x) = x2 − λ. Recallthat the philosophy of the approach we are taking is that we begin with a guess x > 0of the location of a zero of f and from that attempt to prove that a zero exists. UsingNewton’s method as a guide we set A = (f ′(x))−1 = (2x)−1 and hence

T (x) = x− x2 − λ2x

.

Applying the same analysis as in Example 3.2.4, set Z0 = 0

Y0 =|x2 − λ|

2x= |Af(x)|

Z2 = 2A = 21

2x=

1

x.

Thus, the radii polynomial inequality takes the form

p(r) =1

xr2 − r +

|x2 − λ|2x

< 0⇐⇒ r2 − xr +|x2 − λ|

2< 0.

This in turn implies that the interval of convergence I = (r−, r+) for λ ≥ 0 is given by

r± =x±

√x2 − 2 |x2 − λ|

2.

This in turn tells us that if

x ∈(√

2λ

3,√

2λ

),

then the radii polynomials will provide a positive answer to the question of existence of azero of f .


Example 3.2.6 (The radii polynomial approach for a two-dimensional example).Consider the problem of looking for equilibria of

x1 = x2 + 4x21 − λ

x2 = x1 + x22 − 1,

(3.11)

where λ ∈ R is a parameter. An equilibrium solution x = (x1, x2) is a solution of f(x) = 0where f : R2 → R2 is given by the rand-hand side of (3.11). At some given parametervalues λ ∈ R, there are up to four real solutions.

Given an initial guess x

Df(x) =

(8x1 11 2x2

),

and the exact formula for the inverse is

Df(x)−1 =1

16x1x2 − 1

(2x2 −1−1 8x1

).

Set A := Df(x)−1, and let T (x) := x−Af(x). To apply the radii polynomial approach wecompute the bounds Y0, Z0 ad Z2 satisfying (3.3), (3.4) and (3.5), respectively. To obtainY0, observe that

Af(x) = Df(x)−1f(x) =1

16x1x2 − 1

(2x2 −1−1 8x1

)(x2 + 4x2

1 − λx1 + x2

2 − 1

).

Using this expression, we can choose Y0 := ‖Df(x)−1f(x)‖∞. Since A = Df(x)−1, then‖1−ADf(x)‖∞ = 0 so that we set Z0 = 0.

Given c ∈ Br(x), denote b = c− x ∈ Br(0). Consider h ∈ B1(0). Then

(Df(c)−Df(x))h = (Df(x+ b)−Df(x))h

=

[(8x1 + 8b1 1

1 2x2 + 2b2

)−(

8x1 11 2x2

)](h1

h2

)=

(8b1h1

2b2h2

)and hence

A (Df(c)−Df(x))h = − 1

16x1x2 − 1

(2x2 −1−1 8x1

)(8b1h1

2b2h2

).

By assumption b ∈ Br(0) and h ∈ B1(0), hence |b1|, |b2| ≤ r, |h1|, |h2| ≤ 1. Therefore,

|(A (Df(c)−Df(x))h)1| ≤(

16|x2|+ 2

|16x1x2 − 1|

)r


and

|(A (Df(c)−Df(x))h)2| ≤(

16|x1|+ 8

|16x1x2 − 1|

)r.

Setting

Z2 := max

(16|x2|+ 2

|16x1x2 − 1| ,16|x1|+ 8

|16x1x2 − 1|

),

we get that

‖A[Df(c)−Df(x)]‖∞ ≤ Z2r, for all c ∈ Br(x).

Finally, the radii polynomial is defined by

p(r) := max

(16|x2|+ 2

|16x1x2 − 1| ,16|x1|+ 8

|16x1x2 − 1|

)r2 − r + ‖Df(x)−1f(x)‖∞. (3.12)

Observe that the definition of the radii polynomial (3.12) is the same for any approx-imate solution x ∈ R2. This means that we have derived an explicit formula of the radiipolynomial that can be applied to any initial approximation x. If the initial approxima-tion is reasonable, then we expect to be able to prove the existence of a true solutionx. As an example, set λ = 3, and using some numerical scheme, e.g., Newton’s method,find x = (x1, x2) ∈ R2 such that ‖f(x)‖ < tol for some fixed tolerance tol. We chosetol = 10−15, and computed four approximate solutions with Newton’s method obtaining

x(1) =

(−0.6545436118927946

1.286290640521338

), x(2) =

(0.79863337536104250.4487389270377125

),

x(3) =

(0.9086121587039679−0.3023042197787387

), x(4) =

(−1.052701922172216−1.432725347780312

).

(3.13)

For each i = 1, 2, 3, 4, we find that the following intervals I(i) are contained in theexistence intervals for the associated radii polynomial:

I(1) = [1.609× 10−16, 0.640]

I(2) = [6.469× 10−16, 0.227]

I(3) = [1.219× 10−15, 0.239]

I(4) = [4.856× 10−16, 0.927].

(3.14)

Figure 3.3 shows the largest enclosures for each equilibrium of (3.11) for λ = 3. For eachi = 1, 2, 3, 4, the radius around x(i) is the largest value of I(i).

In Example 3.2.6, we defined A as the exact inverse of Df(x). However, as the followingexample demonstrates, having the exact inverse is not necessary.


x1

-3 -2 -1 0 1 2

x2

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Figure 3.3: Largest existence and uniqueness enclosures for each equilibrium of (3.11) forλ = 3. For each i = 1, 2, 3, 4, the radius around x(i) is the largest value of I(i). Thequantities x(i) and I(i) are found in (3.13) and (3.14) respectively. The smallest enclosureis too small to (re)present graphically, which implies that the dots representing x(i) alsorepresent the true equilibria x(i).

Example 3.2.7. The Lorenz system is given byx1 = σ(x2 − x1)

x2 = ρx1 − x2 − x1x3

x3 = −βx3 + x1x2

(3.15)

For any β > 0 and ρ > 1, the set of equilibria of (3.15) is given by(0, 0, 0),

(±√β(ρ− 1),±

√β(ρ− 1), ρ− 1

)which is obtained by solving f(x) = 0, where f : R3 → R3 is given by the right-hand sideof (3.15). At the classical parameter values, σ = 10, ρ = 28 and β = 8/3,

(√72,√

72, 27)

is an equilibrium solution, which is approximated by

x :=

8.48538.4853

27

. (3.16)


Using that

Df(x) =

−σ σ 0ρ− x3 −1 −x1

x2 x1 −β

we compute

Df(x)−1 ≈ A :=

−0.051851843722001 −0.018518437220007 0.0589254357535970.048148156277999 −0.018518437220007 0.058925435753597−0.011785087150719 −0.117850871507195 0

.

We can compute a bound Y0 such that

‖Af(x)‖∞ ≤ Y0. (3.17)

Denote B := I −ADf(x), and compute Z0 such that

‖I −ADf(x)‖∞ = ‖B‖∞ = maxi=1,2,3

3∑j=1

|Bi,j | ≤ Z0. (3.18)

Given c ∈ Br(x) denote b := c− x ∈ Br(0). Given h ∈ B1(0), note that

(Df(c)−Df(x))h = (Df(b+ x)−Df(x))h

=

−σ σ 0ρ− (x3 + b3) −1 −(x1 + b1)x2 + b2 x1 + b1 −β

− −σ σ 0ρ− x3 −1 −x1

x2 x1 −β

h1

h2

h3

=

0 0 0−b3 0 −b1b2 b1 0

h1

h2

h3

=

0−b3h1 − b1h3

b2h1 + b1h2

and using that |bi| ≤ r and |hi| ≤ 1 for i = 1, 2, 3, |A (Df(c)−Df(x))h| |A|

(022

)r.

Setting

Z2 := maxk=1,2,3

(|A|(

022

))k, (3.19)

we obtain that ‖A[Df(c) − Df(x)]‖∞ ≤ Z2r for all c ∈ Br(x). Combining (3.17), (3.18)and (3.19), we define the radii polynomial as

p(r) = Z2r2 − (1− Z0) r + Y0.

Applying Theorem 3.2.1 we conclude that there is a unique equilibrium to (3.15) in theball Br(x) for any

r ∈ I =[1.8626× 10−5, 4.2426

]⊂ r > 0 | p(r) < 0 .


We emphasize that careful choices of approximate inverses and approximate solutionsare necessary, but they do not always need to be very accurate. Keeping x as in (3.16)consider the crude approximation of Df(x)−1 given by

A =

−0.05 −0.01 0.050.04 −0.01 0.05−0.01 −0.11 0

.

The radii polynomial approach (as stated in Theorem 3.2.1) succeeded with this choice ofA. As a consequence, the matrix A is invertible. Using the same analysis as above we getthat the interval

I =[2.0486× 10−5, 3.5068

]is a subset of the existence interval. In fact, with this latest choice of A, if instead of xgiven in (3.16), we let

x :=

8.485281374238578.48528137423857

27

,

then

I =[9.2097× 10−16, 3.5069

].

The functions f : Rn → Rn considered in all examples so far are polynomials. We nowconsider a non polynomial problem.

Example 3.2.8 (The radii polynomial approach for a non polynomial problem).Consider the following model

x1 = 3x1(1− x1)− x1x2 − λ(1− e−5x1)

x2 = −x2 + 3x1x2.(3.20)

At some given parameter values λ ∈ R, there are up to four real equilibrium solutions.

Let f : R2 → R2 be the right hand-side of (3.20). Hence,

Df(x) =

(3− 6x1 − x2 − 5λe−5x1 −x1

3x2 −1 + 3x1

).

Since we are working with a two dimensional example, it is easy to obtain an explicitexpression for Df(x)−1 and thus we can set A := Df(x)−1, so that Z0 satisfying (3.4) canbe taken to be 0. To apply the radii polynomial approach we compute the bounds Y0 andZ2. The bound Y0 ≥ ‖Af(x)‖∞ can be easily be computed. To compute the bound Z2,


consider c ∈ Br(x), set b := c− x ∈ Br(0) and consider h ∈ B1(0)

(Df(c)−Df(x))h = (Df(x+ b)−Df(x))h

=

[(3− 6(x1 + b1)− (x2 + b2)− 5λe−5(x1+b1) −(x1 + b1)

3(x2 + b2) −1 + 3(x1 + b1)

)

−(

3− 6x1 − x2 − 5λe−5x1 −x1

3x2 −1 + 3x1

)]h

=

(−6b1 − b2 − 5λe−5x1(e−5b1 − 1) −b1

3b2 3b1

)h.

Applying the mean value theorem to the function t 7→ e−5b1t over the interval t ∈ [0, 1] weconclude that there exists ξ ∈ [0, 1] such that

e−5b1 − 1 = −5b1e−5b1ξ.

Hence since |b1| ≤ r, |e−5b1 − 1| ≤ 5e5rr. Using that |b1|, |b2| ≤ r and |h1|, |h2| ≤ 1, weget

|(Df(c)−Df(x))h| =∣∣∣∣(−6b1 − b2 − 5λe−5x1(e−5b1 − 1) −b1

3b2 3b1

)h

∣∣∣∣(

8 + 25|λ|e−5x1e5r

6

)r.

Setting

Z2(r) := maxk=1,2

(|A|(

8 + 25|λ|e−5x1+5r

6

))k

,

we get that ‖A (Df(c)−Df(x))h‖∞ ≤ Z2(r), for all c ∈ Br(x). From this, we define

p(r) := Z2(r)r2 − r + Y0.

Although clearly Z2(r) and p(r) are not polynomial, Theorem 3.2.1 applies.To simplify the analysis somewhat, let us observe that Z2(r) is nondecreasing in r.

Hence, if we choose an a priori bound r∗ such that we restrict attention to

r ≤ r∗, (3.21)

then we may replace the function Z2(r) by the constant Z2(r∗), leading to the quadraticradii polynomial

p(r) := Z2(r∗)r2 − r + Y0, (3.22)

which can only be interpreted for r ≤ r∗. In particular, Theorem 3.2.1 can be appliedto (3.22) for r0 ≤ r∗ only.


To demonstrate the use of this quadratic radii polynomial (3.22) we fix λ = 1/2 andconsider x = (x1, x2) ∈ R2 such that ‖f(x)‖ < tol for some fixed tolerance tol. In thisexample, we choose tol = 10−15, and compute four approximate solutions with Newton’smethod to obtain

x(1) =

(0.7940710596186642

0

), x(2) =

(0.33333333333333340.7833134042563428

),

x(3) =

(00

), x(4) =

(−0.1095121750607109

0

).

(3.23)

For each i = 1, 2, 3, 4, we compute an interval I(i) on which each radii polynomial isnegative:

I(1) = [1.534× 10−16, 0.15]

I(2) = [7.393× 10−16, 0.029]

I(3) = (0, 0.022]

I(4) = [5.845× 10−16, 0.0197].

(3.24)

To obtain I(1), we set r∗ = 0.15, to obtain I(2), we set r∗ = 0.03, to obtain I(3), we setr∗ = 0.03 and to obtain I(4), we set r∗ = 0.03.

x1

-0.2 0 0.2 0.4 0.6 0.8 1

x2

-0.2

0

0.2

0.4

0.6

0.8

Figure 3.4: Largest existence and uniqueness enclosures for each equilibria of (3.20) forλ = 1/2. For each i = 1, 2, 3, 4, the radius around x(i) is the largest value of I(i). Thequantities x(i) and I(i) are found in (3.23) and (3.24) respectively.


Let us make some observation about the examples discussed above. In the case of aquadratic nonlinearity (Examples 3.2.4, 3.2.5, 3.2.6 and 4.5.2) the bound Z2 reduced toa constant (independent of r) automatically. For the nonpolynomial nonlinearity inExample 3.2.8 we first obtained a transcendental function p(r), which we then “reduced”to a quadratic polynomial by putting an a priori restriction on r. This strategy will beformalized in the Theorem below.

In all the examples discussed so far the nonlinearities f were defined on the whole ofRn. In general we want to require that f is defined on some open set U ⊂ Rn. This leadsto the following reformulation of Theorem 3.2.1. As a by-product we choose the bound Z2

to be a constant rather than a function of r. This is at the cost of restricting the domainof r.

Theorem 3.2.9. [Radii polynomial approach in finite dimension] Consider a C1

map f : U → Rn where U ⊂ Rn is open. Let x ∈ U and A ∈ M(Rn,Rn). Let r∗ > 0 besuch that Br∗(x) ⊂ U . Let Y0, Z0, and Z2 be non-negative constants satisfying

‖Af(x)‖ ≤ Y0 (3.25)

‖Id−ADf(x)‖ ≤ Z0 (3.26)

‖A[Df(c)−Df(x)]‖ ≤ Z2r, for all c ∈ Br(x) and all r ∈ (0, r∗]. (3.27)

Define the radii polynomial

p(r) := Z2r2 − (1− Z0)r + Y0. (3.28)

If there exists r0 ∈ (0, r∗] such that p(r0) < 0, then the matrix A is invertible and thereexists a unique x ∈ Br0(x) satisfying f(x) = 0.

Proof. The proof is identical to the one of Theorem 3.2.1.

The choice of r∗ has a certain level of arbitrariness. In particular, it is not necessary tochoose r∗ close to the distance of x to the boundary of U . In practice we usually choosea rather loose a priori bound r∗ on r. After checking that p(r0) < 0 and r0 ≤ r∗ we canthen invoke Theorem 3.2.9 to conclude that a (unique) zero x of f lies within distance r0

of x. One may subsequently vary r∗ to optimise the existence interval, which in this caseis given by

r | p(r) < 0 ∩ [0, r∗]. (3.29)

A smaller r∗ will allow a sharper bound Z2, but such a more restrictive choice for r∗ maynevertheless shrink the existence interval (3.29). When r∗ is chosen very small, it canhappen that the existence interval is empty even though there is an r0 > 0 such thatp(r0) < 0 (but r0 > r∗). This indicates that the choice for r∗ needs to be increased.

3.3. EXERCISES 49

Remark 3.2.10. In the ∞-norm it is straightforward to evaluate (with interval arithmetic)the left hand sides of (3.25) and (3.26) to obtain the bounds Y0 and Z0. This holds for anarbitrary function f . Furthermore, the following second derivative estimate provides (viathe mean value theorem) a general bound Z2 satisfying (3.5) in the case of the ∞-norm:

Z2 = supb∈Br∗ (x)

(max

1≤i≤n

∑1≤k,m≤n

∣∣∣ ∑1≤j≤n

AijD2kmfj

(b)∣∣∣). (3.30)

For any fixed r∗ this bound can be computed directly using interval arithmetic, whenexpressions for the second derivatives of f are available.

3.3 Exercises

Exercise 3.3.1. Recall the definition of the derivative:

Definition 3.3.2. Suppose that f : U → Rm with U ⊂ Rn open. We say that f isdifferentiable at x0 ∈ U if there exists an m× n matrix A having that

limh→0

‖f(x0 + h)− f(x0)−Ah‖Rm‖h‖Rn

= 0.

The matrix A is called the derivative of f at x0 and we write

Df(x0) = A.

Prove the following basic facts about the derivative.

• If f is differentiable at x0 then the derivative is unique.

• If f is constant then the derivative is zero (the zero matrix).

• If L : Rn → Rm is a linear map then L is differentiable at each x0 ∈ Rn and

DL(x0) = L.

• (Linearity) Suppose that α, β ∈ R and f, g : U → Rm with U ⊂ Rn open. Assumethat f, g are differentiable at x0 ∈ U . Then αf + βg is differentiable at x0 and

D(αf + βg)(x0) = αDf(x0) + βDg(x0).

• Suppose that f is differentiable at x0 ∈ U . Then f is continuous at x0.

• (Challenge) State and prove “the chain rule”.


Exercise 3.3.3. Let a, b, x0 ∈ R with a < x0 < b. Prove that a function f : (a, b) → R isdifferentiable at x0 in the sense of Definition 3.3.2 if and only if f is differentiable at x0 inthe sense of elementary calculus, so that

A = limh→0

f(x0 + h)− f(x0)

h.

Exercise 3.3.4. Let U ⊂ Rn be open and f : U → Rm. Define the Jacobian matrix of fat x0 ∈ U , which we denote by J(f, x0), to be the matrix with entries

J(f, x0)ij =∂

∂xjfi(x0).

i.e. J(f, x0) is the m×n matrix whose ij-entry is the partial derivative of the i-th compo-nent of f with respect to the variable xj with 1 ≤ i ≤ m and 1 ≤ j ≤ n.

Theorem 3.3.5. If f is differentiable at x0 ∈ U then Df(x0) = J(f, x0). Conversely,if each partial derivative of f exists and is continuous in a neighborhood of x0, then f isdifferentiable at x0.

The following examples show that this statement cannot be weakened.

• Let f : R→ R be defined by

f(x) =

x2 sin

(1x

)if x 6= 0,

0 if x = 0.

Prove that f is differentiable everywhere (including at x = 0) but that f ′ is notcontinuous at x = 0.

• Define f : R2 → R by

f(x, y) =

(x2 + y2) sin

(1√x2+y2

)if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0).

Prove that f is differentiable at (x, y) = (0, 0), and that the first partial derivativesof f with respect to both x and y exist, but that the partial derivatives are notcontinuous. Compare the derivative at (0, 0) to the matrix of first partial derivatives(0, 0).

• (Challenge) Prove Theorem 3.3.5.

Exercise 3.3.6 (Mean values). Prove the following version of the mean value theorem.Recall that f : U → Rm is continuously differentiable on U if for all x0 ∈ U the each firstpartial derivative of f exists and and is continuous at x0.

3.3. EXERCISES 51

Theorem 3.3.7 (The mean value theorem). Suppose that U ⊂ Rn is open and thatf : U → Rm is continuously differentiable. Assume that x, y ∈ U are such that the linesegment from x to y is contained in U , i.e. that

(1− t)x+ ty ∈ U,

for all t ∈ [0, 1]. Then

f(y)− f(x) =

(∫ 1

0Df((1− t)x+ ty) dt

)(y − x).

(Hint: recall that if the first partial derivatives of f exist and are continuous at x0,then f is differentiable and Df(x0) is the matrix of partial derivatives). Now prove theimportant corollaries.

Corollary 3.3.8 (The mean value inequality). Suppose that f : V → Rm is continu-ously differentiable with V ⊂ Rn open and convex. Assume further that that there existsan M > 0 so that

supy∈V‖Df(y)‖B(Rn,Rm) ≤M.

Then

‖f(x)− f(y)‖Rm ≤M‖x− y‖Rn .

Corollary 3.3.9. Suppose U ⊂ Rn is open and that f : U → Rm is continuously differen-tiable. Then f is locally Lipschitz.

Exercise 3.3.10. Prove Remark 3.2.10.

Exercise 3.3.11. Let A : Rn → Rn be a linear map. If ‖Id−A‖∞ < 1, then A is invertible.

Exercise 3.3.12. Use Newton’s method to compute√

7. How many correct digits do youthink you have? Why do you think this?

Exercise 3.3.13. Use Newton’s method to compute the zero of xex−1 = 0. Use the radiipolynomial approach to prove your result.

Exercise 3.3.14. Use Newton’s method in the complex variable setting to compute thethree zeros of z3 + z + 1 = 0. Use the radii polynomial approach to prove your result.

Exercise 3.3.15. Consider the function f : R2 → R2 given by

f(x, y) =

(y + 4x2 − 3x+ y2 − 1

)• Explain why there are at most four real solution of this equation.


• Write a MatLab program which solves the equation f(x, y) = (0, 0) using Newton’Method. Compute four distinct solutions.

Exercise 3.3.16. Consider the function f : R2 → R2 given by

f(x, y) =

(3x(1− x)− xy − 1

2(1− e−5x)−y + 3xy

).

Write a MatLab program which solves the equation f(x, y) = (0, 0) via Newton’s Method.Use you program to compute four distinct solutions.

Exercise 3.3.17. Choose several of your favorite example vector fields from the scavengerhunt and use Newton’s method to find as many equilibrium solutions as you can. Explainwhy you think you have found all the equilibrium solutions or why you think there may bemore.

Exercise 3.3.18. Suppose that f : Rn → Rn is twice continuously differentiable, and thatfor some x∗ ∈ Rn we have that

f(x∗) = 0,

and that Df(x∗) is invertible. Prove that there exists a neighborhood Br(x∗) of x∗ so that,for all x ∈ Br(x∗) the Newton method with x0 = x converges. Moreover, prove that theconvergence is “quadratically fast”, i.e. that if

εn = ‖f(xn)‖Rn ,

then

εn+1 ≤Mε2n,

for some M > 0.

Exercise 3.3.19. Consider the equation z5 − 1 = 0 with z ∈ C. The equation has 5 rootsin C, namely the 5 complex roots of unity. Use Newton’s method to compute all 5.

Now study the “domain of attraction for each root”. In other words, choose severalthousand points in the complex plane and start Newton from each. Keep track of whichpoints converge to which of the 5 roots. (Maybe color points red, blue, green, orange, oryellow depending on which root they converge to under Newton). Examine the boundarybetween the regions and see if you notice anything interesting.

Exercise 3.3.20. Show that the bound Z2 as defined in (3.30) satisties (3.5).

Exercise 3.3.21. Let f : R→ R be differentiable with at least two real roots. Prove thatthere exist points x0 ∈ R such that Newton’s method applied to x0 does not converge.

3.3. EXERCISES 53

Exercise 3.3.22. Consider the FitzHugh-Nagumo equation

f(x) =

(x1(x1 − a)(1− x1)− x2

ε(x1 − γx2)

)(3.31)

at the parameter values (a, ε, γ) = (5, 1, 2). Consider the initial point x0 = (1, 0)T ∈ R2 andapply Newton’s method to converge to a point x ∈ R2. What is the point x you obtain?Decrease the parameter γ by .05 and apply Newton’s method with initial point x0 = x,where x is the value obtained at the previous step. This initial guess is called a 0th orderpredictor. Applying Newton’s method, you should converge to a new point that we denoteagain by x. Repeat this procedure 40 times. What do you observe? This procedure is thesimplest form of a predictor-corrector method.

Exercise 3.3.23. Consider the problem f(x) = 0, where f is the right hand side of(3.31). Given an approximate solution x such that f(x) ≈ 0, consider an approximateinverse A that is A ≈ Df(x)−1. Define T (x) = x − Af(x) and compute the boundsY0, Z0 and Z2 satisfying (3.3), (3.4) and (3.5), respectively. Define the radii polynomialsp(r) = Z2 − (1 − Z0)r + Y0. Use Theorem 3.2.1 to find r > 0 such that Br(x) containsa unique solution of f(x) = 0. Write a MATLAB program that verifies the hypotheses ofTheorem 3.2.1. Prove the existence of the fixed points close to the 40 numerical valuesobtain in Exercise 3.3.22.

Chapter 4

Linear Theory and Stability ofEquilibria

We begin this section by introducing the notion of stability for equilibrium solutions.

Definition 4.0.24. An equilibrium x of x = f(x) is stable if for any ε > 0, there existsδ > 0 such that if ‖x0 − x‖ < δ, then ‖ϕ(t, x0)− x‖ < ε for all t > 0. An equilibrium thatis not stable is called unstable.

A stronger notion of stability is the following.

Definition 4.0.25. An equilibrium x of x = f(x) is asymptotically stable if it is stableand there exists ρ > 0 such that if ‖x0 − x‖ < ρ, then limt→∞ ϕ(t, x0) = x.

4.1 Homogeneous Linear Systems

In this section we study the homogeneous linear system

x = A(t)x (4.1)

where A : R → Rn×n is a continuous matrix valued function. More general and morein-depth results can be found in [4, 8].

Proposition 4.1.1. Given the homogeneous linear system (4.1) the solution to the initialvalue problem x(t0) = x0 is defined for all t ∈ R.

Proof. The proof is by contradiction. The existence of a solution ϕ : J → Rn follows fromTheorem 2.2.13. Assume that J = (α, β), is the maximal interval of existence and β <∞.By Lemma 2.2.5

ϕ(t) = ϕ(t0) +

∫ t

t0

A(s)ϕ(s) ds. (4.2)

55

56 CHAPTER 4. LINEAR THEORY AND STABILITY OF EQUILIBRIA

By assumption A is continuous over R, hence there exists M such that

supt∈[t0,β]

‖A(t)‖ ≤M.

Applying this bound to (4.2) we obtain

|ϕ(t)| ≤ |x0|+∫ t

t0

M‖ϕ(s)‖ ds,

which by Gronwall’s inequality implies

|ϕ(t)| ≤ |x0|eM(t−t0).

In particular, ϕ(t) is bounded on [t0, β]. This contradicts Theorem 2.2.20. Thereforeβ =∞.

A similar argument shows that α = −∞.

The principle of superposition, which follows directly from the linearity, is fundamentalto our ability to characterize all solutions to (4.1).

Proposition 4.1.2 (Principle of superposition). If ϕ and ψ are solutions to (4.1),then αϕ+ βψ is also a solution for all scalars α and β.

The principle of superposition allows us to think of linear combinations of solutions,which in turn can be codified as follows. A matrix valued function X : R → Rn×n is amatrix solution to (4.1) if each column is a solution to (4.1). Observe that if X is a matrixsolution, then

X = A(t)X.

Definition 4.1.3. A fundamental matrix solution to (4.1) is a matrix solution X(t) of(4.1) such that detX(t) 6= 0. The principal fundamental matrix solution X(t) at initialtime t0 is a fundamental matrix solution satisfying X(t0) = I, the identity matrix. Thisprincipal matrix solution is denoted by X(t, t0).

Proposition 4.1.4. Let X(t) be a matrix solution to (4.1). Then, either detX(t) ≡ 0 ordetX(t) 6= 0 for all t ∈ R.

Proof. If there exists τ such that detX(τ) = 0, then X(τ) is not invertible and hencethere exists a nonzero vector v such that X(τ)v = 0. Since X(t) is a matrix solution to(4.1), X(t)v is a solution to (4.1). In particular, it is a solution with initial value x(τ) = 0.However, 0 is an equilibrium for (4.1) and hence X(t)v = 0, for all t ∈ R. Therefore,detX(t) = 0 for all t ∈ R.

Corollary 4.1.5. If X(t) is a matrix solution of (4.1) satisfying X(t0) = X0 where X0 isa non-singular matrix, then X(t) is a fundamental matrix solution.

4.1. HOMOGENEOUS LINEAR SYSTEMS 57

Theorem 4.1.6 (Liouville’s formula). If X(t) is a matrix solution to (4.1) satisfyingthe initial condition X(t0) = X0, then

d

dt(detX) = (trA(t)) detX

and hencedetX(t) = detX0 e

∫ tt0

trA(s) ds. (4.3)

The last equation (4.3) is typically called Liouville’s formula.

Proof. Let

X(t) =

R1(t)R2(t)

...Rn(t)

where Rj(t) denotes the j-th row. Since det is a multilinear function of its rows,

d

dt(detX) =

n∑i=1

det

R1(t)

...ddtRi

...Rn(t)

(4.4)

Writing out (4.1) in matrix form gives the expression

d

dtxi,j =

∑k

ai,kxk,j

from which we can deduce that

dRidt−∑k 6=i

ai,kRk = ai,iRi.

Thus (4.4) becomes

d

dt(detX) =

n∑i=1

det

R1(t)

...ai,iRi

...Rn(t)

= (trA) detX

where the last equality follows from the multilinearity of det.


4.2 Constant Coefficient Linear Systems

In this section we study properties of the linear system

x = Ax (4.5)

where A ∈Mn(R).

We know that if n = 1, then A ∈ R and the solution to (4.5) takes the form eAt.Observe that if x0 = 1, then eAt is the principal fundamental matrix solution. In analogyto this we define

eAt := X(t)

where X(t) = X(t, 0) is the principal fundamental matrix solution to (4.5).

The following proposition provides essential properties of the function eAt and indicatesthat it is equivalent to the standard definition of the exponential map.

Proposition 4.2.1. Let A,B ∈Mn(R). The function eAt satisfies the following properties:

(i) eA(t+s) = eAteAs;

(ii)(eAt)−1

= e−At;

(iii) ddte

At = AeAt = eAtA;

(iv) eAt =∑∞

k=01k!A

ktk;

(v) If AB = BA, then e(A+B)t = eAteBt;

(vi) If B is an invertible matrix, then

eB−1ABt = B−1eAtB.

Proof. (i) Let s ∈ R. Observe that eA(t+s) and eAteAs are fundamental matrix solutionsto (4.5) that agree at t = 0. Therefore, the desired equality follows from uniqueness ofsolutions.

(ii) In (i) let s = −t.(iii) The first equality follows from the definition of eAt. The second equality follows

from the observation that both sides are matrix solutions to (4.5) that agree at t = 0.

(iv) Let ‖A‖ = M and let E(K)(t) :=∑K

k=01k!A

ktk. Then

‖E(K)(t)x‖ ≤K∑k=0

1

k!‖A‖ktk‖x‖ ≤ eMt‖x‖.

4.2. CONSTANT COEFFICIENT LINEAR SYSTEMS 59

Furthermore,

‖E(K+1)(t)x− E(K)(t)x‖ ≤ 1

(K + 1)!MK+1tK+1‖x‖.

Therefore, for all x ∈ Rn and t ∈ R, the sequence E(K)(t)x converges uniformly asK →∞. Define

ϕ(t, x) := limK→∞

E(K)(t)x.

It remains to be shown that ϕ(t, x) = eAtx.Observe that

E(K+1)(t)x = E(0)(t)x+

∫ t

0AE(K)(s)x ds.

Therefore,

limK→∞

E(K+1)(t)x = x+ limK→∞

∫ t

0AE(K)(s)x ds,

hence

ϕ(t, x) = x+

∫ t

0limK→∞

AE(K)(s)x ds = x+

∫ t

0Aϕ(s, x) ds,

which implies that ϕ(t, x) is a solution to (4.5) with initial condition x. By uniqueness ofsolutions this implies the desired equality eAtx = ϕ(t, x).

(v) Using (iv)

BeAt = B

∞∑k=0

1

k!Aktk =

∞∑k=0

1

k!AktkB = eAtB.

By definition e(A+B)t is the principal fundamental matrix solution to x = (A+B)x. Observethat

d

dt(eAteBt) = AeAteBt + eAtBeBt = AeAteBt +BeAteBt = (A+B)(eAteBt).

Thus the desired equality follows from uniqueness of solutions.(vi) Since B is invertible it acts as a linear change of coordinates. In the coordinate

system y = B−1x, (4.5) becomesy = B−1ABy

with principal fundamental matrix solution eB−1ABt. Let y0 = B−1x0 be an initial condi-

tion, thenx(t) = By(t) = BeB

−1ABty0 = BeB−1ABtB−1x0.

By uniqueness of solutionseAt = BeB

−1ABtB−1

or equivalentlyB−1eAtB = eB

−1ABt.


The following result, which is a direct consequence of Proposition 4.2.1, provides a flowfor constant coefficient linear systems.

Corollary 4.2.2. Given A ∈Mn(R), the function

ϕ(t, x) := etAx

is a solution of x = Ax and is a flow (as defined in Definition 2.2.23).

Recall that the real Jordan normal form is a real valued block diagonal matrix wherethe diagonal blocks take one of the following four forms

[λ],

[α −ββ α

],

λ 1

. . .

. . . 1λ

,D I

. . .

. . . ID

where I is the 2× 2 identity matrix and

D =

[α −ββ α

].

Furthermore, up to ordering of the diagonal blocks any real valued matrix A has a uniquereal Jordan normal form J and there exists an invertible matrix P such that

J = PAP−1.

Definition 4.2.3. Two flows ϕ : R×X → X and ψ : R×Y → Y are topologically conjugateif there exists a homeomorphism h : X → Y such that

h ϕ(t, x) = ψ(t, h(x))

for all t ∈ R and x ∈ X.

Observe that if two flows are topologically conjugate, then they are topologically equiv-alent, but the converse need not be true.

Example 4.2.4. Let ϕ(t, x) = e−tx and ψ(t, x) = e−2ty the two flows associated to theproblems x = −x and y = −2y, respectively. Define h : R→ R the homeomorphism by

h(x) =

x2, x ≥ 0

−x2, x < 0

Then, for all x ∈ R and t ∈ R, h(ϕ(t, x)) = h(e−tx) = e−2th(x) = ψ(t, h(x)).

4.2. CONSTANT COEFFICIENT LINEAR SYSTEMS 61

Corollary 4.2.5. Consider two n-dimensional constant coefficient linear systems

x = Ax and x = Bx.

If A and B have the same real Jordan normal form then the corresponding flows ϕA(t, x) :=eAtx and ϕB(t, x) := eBtx are topologically equivalent.

Proof. Since A and B have the same real Jordan normal form, there exist P1 and P2

invertible matrices such that

P1AP−11 = J = P2BP

−12 .

This implies that

P1eAtP−1

1 = eP1AP−11 t = eP2BP

−12 t = P2e

BtP−12 ,

and letting S := P−12 P1 which is invertible, we get

eBt = SeAtS−1.

Recall from Definition 2.2.26 that two flows ϕ : R×Rn → Rn and ψ : R×Rn → Rn are topo-logically equivalent if there exists a homeomorphism h : Rn → Rn and a reparameterizationfunction τ : R× Rn → R such that

ψ(t, h(x)) = h(ϕ(τ(t, x), x)),

for all x ∈ Rn and t ∈ R. In this case, letting ϕ(t, x) = ϕA(t, x) = eAtx, ψ(t, x) =ϕB(t, x) = eBtx, h = S and τ(t, x) = t for all x ∈ Rn, we get the result. In fact, since thereparameterization function is τ(t, x) = t for all x ∈ Rn, we have even more: the flows ϕAand ϕB are topologically conjugate, as defined in Definition 4.2.3.

As is made clear in Section 4.3 the converse of Corollary 4.2.5 is not true.The following two propositions allow us to give an explicit formula for solutions to a

homogeneous linear differential equation.

Proposition 4.2.6. Let A be a matrix with real entries.

(i) λ ∈ R is an eigenvalue of A with associated eigenvector v if and only if t 7→ eλtv isa solution to (4.5).

(ii) If λ = α+ iβ is an eigenvalue of A with associated eigenvector v = u+ iw, then

t 7→ eαt(cos(βt)u− sin(βt)w) and t 7→ eαt(sin(βt)u+ cos(βt)w)

are solutions to (4.5) and if β 6= 0, then they are linearly independent.


The proof of Proposition 4.2.6 is left as an exercise.

Proposition 4.2.7. Let A = λI+N where I denotes the identity matrix and N is nilpotent,i.e. Nk = 0, for some k ≥ 1. Then,

eAt = eλt

k−1∑j=0

tj

j!N j

.

Proof. Since IN = NI, by Proposition 4.2.1(v)

eAt = e(λI+N)t = eλIteNt = eλt

k−1∑j=0

tj

j!N j

.

Given the homogeneous linear differential equation (4.5), let J = PAP−1 denote aJordan normal form for A. Then, Propositions 4.2.6 and 4.2.7 can be used to express eJt

and by Proposition 4.2.1(vi)

eAt = P−1eJtP. (4.6)

4.3 Hyperbolic Linear Systems

Definition 4.3.1. A matrix is hyperbolic if all of its eigenvalues have nonzero real parts.

The focus of this section is on characterizing the dynamics of constant coefficient linearsystems under the assumption that the matrix is hyperbolic. We begin by studying thecase where all eigenvalues have negative real part.

Theorem 4.3.2. Consider the differential equation x = Ax under the assumption thatA ∈Mn(R). The following statements are equivalent.

(i) There is a norm ‖ · ‖a on Rn and a real number λ > 0 such that for all x ∈ Rn andall t ≥ 0, ∥∥eAtx∥∥

a≤ e−λt‖x‖a.

(ii) If ‖ · ‖ is an arbitrary norm on Rn, then there is a constant C > 0 and a real numberλ > 0 such that for all x ∈ Rn and all t ≥ 0,∥∥eAtx∥∥ ≤ Ce−λt‖x‖.

(iii) The real part of every eigenvalue of A is negative.

4.3. HYPERBOLIC LINEAR SYSTEMS 63

Proof. (i) ⇒ (ii) By Proposition 7.2.4 all norms on Rn are equivalent, thus there existpositive constants K0 and K1 such that

K0‖x‖ ≤ ‖x‖a ≤ K1‖x‖.

Therefore,

‖eAtx‖ ≤ 1

K0‖eAtx‖a ≤

e−λt

K0‖x‖a ≤

K1

K0e−λt‖x‖.

(ii) ⇒ (iii) The proof is by contradiction. Observe that by (ii) limt→∞ ϕ(t, x) =limt→∞ e

Atx = 0, for every x ∈ Rn. Let µ = α + iβ be an eigenvalue of A and assumeα ≥ 0. Let v = u+ iw 6= 0 be a corresponding eigenvector. By Proposition 4.2.6(ii),

t 7→ eαt(cos(βt)u− sin(βt)w)

is a solution.But ϕ(t, u) does not tend to 0 as t goes to infinity, a contradiction.(iii) ⇒ (i) Let

vj | j = 1, . . . , n

be a set of linearly independent generalized eigenvec-

tors for A. Let νj : R→ Rn denote the solution to x = Ax satisfying νj(0) = vj . Given aninitial condition x =

∑nj=1 cjv

j ,

eAtx =n∑j=1

cjνj(t).

Explicitly writing out (4.6) (that is eAt = P−1eJtP ) allows us to conclude that thecomponents of eAt can be written as a sum of terms of the form p(t)eαt sin(βt) andp(t)eαt cos(βt), where α, β ∈ R are the real and imaginary parts of an eigenvalue of Aand the polynomial p(t) has degree bounded by the geometric dimension of the corre-sponding eigenspace. Choose a > 0 such that α < −a for all eigenvalues α+ iβ of A. Thenfor each x there exists τ(x) (continuous in x) such that∥∥eAtx∥∥ ≤ e−at‖x‖for all t ≥ τ(x). By compactness τ := sup‖x‖=1 τ(x) < ∞. Thus, for ‖x‖ = 1, if t > τ ,then ∥∥eAtx∥∥ ≤ e−at.Finally, for arbitrary x ∥∥eAtx∥∥ = ‖x‖

∥∥∥∥eAt x‖x‖∥∥∥∥ ≤ ‖x‖e−at (4.7)

for all t ≥ τ .At this point we have proved that given the norm ‖ · ‖ the flow eventually produces a

contraction. We claim that

‖x‖a :=

∫ τ

0eas∥∥eAsx∥∥ ds


defines a norm that contracts for all positive time. It is left to the reader to check that‖ · ‖a is a norm.

Given t ≥ 0 choose a positive integer n and 0 ≤ T < τ such that t = nτ + T . Observethat ∥∥eAtx∥∥

a=

∫ τ

0eas∥∥eAseAtx∥∥ ds

=

∫ τ

0eas∥∥∥eA(nτ+T+s)x

∥∥∥ ds=

∫ τ−T

0eas∥∥∥eAnτeA(T+s)x

∥∥∥ ds+

∫ τ

τ−Teas∥∥∥eA(n+1)τeA(T−τ+s)x

∥∥∥ ds≤∫ τ−T

0ea(s−nτ)

∥∥∥eA(T+s)x∥∥∥ ds+

∫ τ

τ−Tea(s−(n+1)τ)

∥∥∥eA(T−τ+s)x∥∥∥ ds

=

∫ τ

Tea(u−T−nτ)

∥∥eAux∥∥ du+

∫ T

0ea(u+τ−T−(n+1)τ)

∥∥eAux∥∥ du= e−at

∫ τ

0eau∥∥eAux∥∥ du

= e−at ‖x‖a ,where the inequality is obtained using (4.7) and we use the substitutions u = T + s in thefirst integral and u = T −τ+s in the second integral. Thus (i) holds for the norm ‖·‖a.

Corollary 4.3.3. If every eigenvalue of A has negative real part, then 0 is an asymptoticallystable equilibrium for x = Ax.

The fundamental conclusion of Theorem 4.3.2 is that if every eigenvalue of A hasnegative real part then every solution to x = Ax converges to the origin exponentiallyfast and furthermore there is a uniform lower bound on the exponential rate. To extendthis theorem and its conclusions to hyperbolic linear equations it is useful to introduce thefollowing subspaces.

Definition 4.3.4. Let A ∈ Mn(R). The stable, unstable, and center eigenspaces of A aredefined by

EsA := span v | v a generalized eigenvector associated with eigenvalue α+ iβ, α < 0EuA := span v | v a generalized eigenvector associated with eigenvalue α+ iβ, α > 0EcA := span v | v a generalized eigenvector associated with eigenvalue α+ iβ, α = 0

respectively.

By definition if A is a hyperbolic matrix, then EcA = ∅. This implies that

Rn = EuA ⊕ EsA,

which is called a hyperbolic splitting.

4.3. HYPERBOLIC LINEAR SYSTEMS 65

Theorem 4.3.5. Consider the differential equation x = Ax where A ∈Mn(R).

(i) The subspace EuA, EsA and EcA are invariant sets under the flow eAt.

(ii) x ∈ EsA if and only if there exist a > 0 and C ≥ 1 such that

‖eAtx‖ ≤ Ce−a|t| ∀ t ≥ 0.

(iii) x ∈ EuA if and only if there exist a > 0 and C ≥ 1 such that

‖eAtx‖ ≤ Ce−a|t| ∀ t ≤ 0.

(iv) x ∈ EcA if and only if for all a > 0

limt→±∞

e−a|t|‖eAtx‖ = 0.

Proof. (i) The proof is left as an exercise.(ii) By (i) EsA is invariant. Restricting the differential equation to EsA implies that

the conditions of Theorem 4.3.2(iii) are satisfied and hence the result follows from Theo-rem 4.3.2(ii).

(iii) This is essentially the same as (ii) after applying the transformation t 7→ −t.(iv) If x ∈ EcA, then by (4.6) the growth rate in forward or backward time is subexpo-

nential.So assume x 6∈ EcA. Since Rn = EsA⊕EuA⊕EcA, we can write x = xs + xu + xc where

x∗ ∈ E∗A, ∗ ∈ s, u, c and xs 6= 0 or xu 6= 0. By (i), (ii) and (iii), ‖xs(t)‖ and/or ‖xu(t)‖grows at an exponential rate for negative or positive time, respectively. This contradictslimt→±∞ e

−a|t|‖eAtx‖ = 0.

Theorem 4.3.6. Let A,B ∈Mn(R) be hyperbolic matrices.

(i) If all eigenvalues of A and B have negative real parts, then the flows eAt and eBt aretopologically conjugate.

(ii) If the number of eigenvalues with negative real parts are the same for A and B, thenthe flows eAt and eBt are topologically conjugate.

(iii) If A ∈Mn(R) is hyperbolic, then there exists ε = ε(A) such that if B ∈ Bε(A) = M ∈Mn(R) : ‖A−M‖ < ε, then the flows eAt and eBt are topologically conjugate.

Theorem 4.3.7. Let A,B ∈Mn(R).

(i) If A and B are hyperbolic and the number of eigenvalues with negative real parts arethe same for A and B, then the flows eAt and eBt are topologically conjugate.


(ii) If A is hyperbolic, then there exists ε = ε(A) such that if B ∈ M ∈Mn(R) : ‖A−M‖ < ε, then B is hyperbolic and the flows eAt and eBt are topologically conjugate.

Proof. (i) Let ‖ · ‖a and ‖ · ‖b be the norms for x = Ax and y = By guaranteed byTheorem 4.3.2(i), respectively, that is

‖eAtx‖a ≤ e−λat‖x‖a, ‖eBty‖b ≤ e−λbt‖y‖b,

for some λa, λb > 0 and for all x, y ∈ Rn and t ≥ 0.Let SA := x ∈ Rn | ‖x‖a = 1 and SB := y ∈ Rn | ‖y‖b = 1. Define h0 : SA → SB

by

h0(x) =x

‖x‖b.

Because ‖ · ‖a is strictly decreasing along the orbits eAtx we can define τ : Rn \ 0 → R tobe the solution to

eAτ(x)x ∈ SA.Note that

τ(x) > 0, if ‖x‖a > 1τ(x) = 0, if ‖x‖a = 1τ(x) < 0, if ‖x‖a < 1.

Define h : Rn → Rn by

h(x) :=

e−Bτ(x)h0

(eAτ(x)x

)if x 6= 0

0 if x = 0.

By construction h is continuous except perhaps at 0. To show continuity at this point,consider a sequence xk → 0 and observe that this implies that τ(xk) → −∞. Since bydefinition ‖h0(eAτ(xk)xk)‖b = 1, limk→∞ h(xk) = 0 = h(0). Thus h is continuous.

That h defines a conjugation follows from the following calculations. If x = 0, then

h(eAt0) = 0 = eBt0 = eBth(0).

If x 6= 0, then

h(eAtx) = e−Bτ(eAtx)h0

(eAτ(eAtx)eAtx

)= e−B(τ(x)−t)h0

(eA(τ(x)−t)eAtx

)= eBte−Bτ(x)h0

(eAτ(x)e−AteAtx

)= eBte−Bτ(x)h0

(eAτ(x)x

)= eBth(x).

4.4. LINEAR APPROXIMATIONS OF NONLINEAR SYSTEMS 67

It is left to the reader to prove that h is a homeomorphism (see Exercise 4.6.4).

(ii) Since A and B are hyperbolic we write them as A = As ⊕Au and B = Bs ⊕Bu byrestricting them to their stable and unstable eigenspaces Es and Eu, respectively. AssumedimEs = k. Let hs : Rk → Rk and hu : Rn−k → Rn−k be defined as in (i) using As and Bs,and Au and Bu, respectively. The reader can check that h := hs ⊕ hu is a conjugacy.

(iii) This follows from Exercise 4.6.5 where it is shown that hyperbolic matrices aregeneric.

4.4 Linear Approximations of Nonlinear Systems

The goal of this section is to prove that if x is an equilibrium to x = f(x), and the realparts of all the eigenvalues of Df(x) are negative, then x is asymptotically stable. We alsodiscuss the case where Df(x) is a hyperbolic matrix.

Proposition 4.4.1 (Variation of Constants Formula). Let x(t) be the unique solutionof

x = Ax+ g(t), x(t0) = x0. (4.8)

Then

x(t) = eAt(e−At0x0 +

∫ t

t0

e−Asg(s) ds

). (4.9)

Equation (4.9) is called the variation of constants formula.

Proof. Set

y(t) := e−Atx(t)

and observe that

y = −Ae−Atx+ e−Atx = −Ae−Atx+ e−At(Ax+ g(t)) = e−Atg(t). (4.10)

An initial condition x(t0) = x0 to (4.8) is equivalent to an initial condition y(t0) = e−At0x0

to (4.10). We can solve (4.10) by direct integration, i.e.

y(t) = e−At0x0 +

∫ t

t0

e−Asg(s) ds.

Rewriting this in the original coordinates results in equation (4.9).

Theorem 4.4.2. Let f : Rn → Rn be a C1 function and let x be an equilibrium solutionto

x = f(x), x ∈ Rn. (4.11)


If all the eigenvalues of Df(x) have negative real part strictly less than −a, for some a > 0,then for any norm ‖ · ‖ on Rn there exists a neighborhood U of x and a constant C ≥ 1such that for any initial condition x0 ∈ U the solution ϕ(·, x0) : [0,∞)→ Rn satisfies

‖ϕ(t, x0)− x‖ ≤ Ce−at‖x0 − x‖, ∀t ≥ 0.

Therefore x is asymptotically stable.

Proof. Let y(t) = x(t) − x. We are interested in the dynamics in a neighborhood of x.Observe that using a Taylor approximation (4.11) can be written as

y = f(x+ y) = f(x) +Df(x)y + g(y) = Df(x)y + g(y) (4.12)

where g(0) = 0 and Dg(0) = 0.Let y(t) be a solution to (4.12) with initial condition y(0) = y0 and view g(y(t)) as a

function of time. Using the variation of constants formula (4.9)

y(t) = eAt(y0 +

∫ t

0e−Asg(y(s)) ds

)= eAty0 +

∫ t

0eA(t−s)g(y(s)) ds.

Since the negative real parts of the eigenvalues of A are strictly less than −a, there existsλ = a+ δ for some δ > 0 and a constant C ≥ 1 which satisfy Theorem 4.3.2(ii), that is∥∥eAtx∥∥ ≤ Ce−λt‖x‖, (4.13)

for all x ∈ Rn and all t ≥ 0.Choose m > 0 such that mC < δ. Because g(0) = 0 ∈ Rn and Dg(0) = 0 ∈ Mn(R)

there exists ε > 0 such that

‖g(y)‖ ≤ m‖y‖ for all ‖y‖ < ε.

By (4.13),

‖y(t)‖ ≤ Ce−λt‖y0‖+

∫ t

0Ce−λ(t−s)m‖y(s)‖ ds

as long as ‖y(s)‖ < ε for s ∈ [0, t]. Thus,

eλt‖y(t)‖ ≤ C‖y0‖+

∫ t

0Ceλsm‖y(s)‖ ds.

By Gronwall’s inequality (with α(t) = eλt‖y(t)‖ and β(t) = Cm),

eλt‖y(t)‖ ≤ C‖y0‖eCmt,or equivalently

‖y(t)‖ ≤ C‖y0‖e−(λ−Cm)t < C‖y0‖e−at. (4.14)

Observe that if ‖y0‖ ≤ ε/C, then (4.14) guarantees that ‖y(t)‖ < ε for all t > 0.

4.5. RIGOROUS COMPUTATION OF EIGENVALUES AND EIGENVECTORS 69

The following theorem is fundamental. The proof can be found in [4, 13]. Consider

x = Ax+ g(x) (4.15)

where A ∈Mn(R) and g ∈ Cr(Rn,Rn) satisfies g(0) = 0.

Theorem 4.4.3. [Hartman-Grobman Theorem] Consider (4.15) and assume A ishyperbolic. Let ϕ be the flow generated by (4.15) and let ψ be the flow generated by x = Ax.Then, there exists µ > 0 such that for any g ∈ B1

µ(Rn,Rn) there exists a homeomorphism

h = h(g) = I + ρ(g)

where ρ : B1µ → C0

b is continuous, unique, and ρ(0) = 0, and for all x ∈ Rn, t ∈ R

h(ϕ(t, x)) = ψ(t, h(x)).

4.5 Rigorous Computation of Eigenvalues and Eigenvectors

As is made clear in the previous section, determining the linear stability of an equilibriumsolution requires knowledge of the eigenvalues and determining a linear approximation ofthe stable and unstable manifolds requires knowledge of the eigenvectors of a matrix. Inpractice, for all but the simplest systems, computation of eigenvalues and eigenvectors mustbe done numerically and there are a variety of techniques for doing so [?, Chapter 9]. Withthis in mind we present Theorem 4.5.1 which provides for the existences of solutions (λ, v)of the equation

Mv = λv, (4.16)

for a given matrix M ∈Mn(C) under the assumption that an approximate eigenpair (λ, v)has been determined numerically. For sake of simplicity we do not present how to verifygeneralized eigenvectors.

A remark on notation: given a vector x = (x1, . . . , xn), set

xk := (x1, . . . , xk−1, xk+1, . . . , xn).

Theorem 4.5.1. Let v ∈ Cn, λ ∈ C, and M ∈ Mn(C). Choose k such that |vk| =max |vi| | i = 1, . . . , n. Define f : Cn → Cn by

f(z) = (M − z1I)

z2...zkvkzk+1

...zn

. (4.17)


Choose A ∈Mn(C) and positive constants Y0, Z0 and Z2 such that

‖A(M − λI)v‖ ≤ Y0 (4.18)

‖I −ADf(λ, vk)‖ ≤ Z0 (4.19)

2‖A‖ ≤ Z2. (4.20)

Set p(r) = Z2r2 − (1− Z0)r + Y0.

If there exists r0 > 0 such that p(r0) < 0, then A is invertible and there exists uniquev and λ such that ‖v − v‖ < r0, ‖λ− λ‖ < r0, and

Mv = λv.

Perhaps a few words of motivation are in order before turning to the details of theproof. Theorem 4.5.1 is a special case of Theorem 3.2.1. In particular, we can rewrite(4.16) as

g(λ, v) = (M − λI)v = 0.

This turns the problem of finding an eigenpair into the problem of finding a zero of g whichsuggests that we consider the fixed point problem

T (z) = z −ADg(z).

However, we cannot hope to use a radii polynomial to find a fixed point of T . To under-stand why, observe that given a solution (λ, v) of (4.16) and any θ ∈ C \ 0, (λ, θv) is alsoa solution. This implies that the solution (λ, v) is not isolated, i.e. that there is no neigh-borhood of (λ, v) in Cn+1 on which (λ, v) is the unique solution to g(λ, v) = 0. To addressthis issue, we introduce the notion of a phase condition which will ensure that solutionsare isolated. Notice that uniqueness fails along a two dimensional parameter space C\0.Thus, one expects that if we can find an appropriated reduction by two of the dimension ofthe space on which g acts, then the reduced function will have a unique fixed point. Withthis in mind we choose a phase condition that involves fixing one of the components of vto be a given constant.

To be more precise, suppose that an approximate eigenpair (λ, v) of M has been com-puted, i.e., Mv ≈ λv (note that this is the first line of the hypothesis of Theorem 4.5.1).Choose k such that

|vk| = max |vj | | j = 1, . . . , n .


Since v is a numerical eigenvector we can assume that vk 6= 0. Define f : Cn → Cn by

f(z) := (M − z1I)

z2...zkvkzk+1

...zn

. (4.21)

By definition, a solution z of f(z) = 0 corresponds to an eigenpair (λ, v) of M withthe eigenvalue λ = z1 and the eigenvector v = (z2, . . . , zk, vk, zk+1, . . . , zn). Since the k-thcoordinate of v is fixed and there are finitely many eigenvectors up to scalar multiplication,if (λ, v) is an eigenpair, then it is isolated as a point in Cn+1.

Proof of Theorem 4.5.1. To proof follows as an application of Theorem 3.2.1 where f isgiven by labeleq:feigs. Observe that f(λ, vk) = (M − λI)v, hence satisfying (4.18) impliesthat (3.3) is satisfied. Similarly, (4.19) is a restatement of (3.4). Thus we turn our attentionto showing that (4.20) verifies (3.5).

Observe that

Df(z) =

−z2...−zk−vk−zk+1

...−zn

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣(M − z1I)

k

where (M − z1I)

kdenotes the n× (n− 1) matrix obtained by deleting the k-th column of

M − z1I. Therefore,

Df(c)−Df(λ, vk) =

−c2 + v2...

−ck + vk−1

0−ck+1 + vk+1

...−cn + vn

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣((−c1 + λ)I)

k


We are interested in c ∈ Br((λ, vk)) or equivalently c = (λ+ z1, vk + z1) where ‖z‖∞ ≤ r.Thus, by (7.1)

‖Df(c)−Df(λ, vk)‖ =

∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥

−z2...−zk

0−zk+1

...−zn

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣(−z1I)

k

∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥≤ 2r.

Therefore,‖A[Df(c)−Df(λ, vk)‖ ≤ 2‖A‖r,

and hence (4.20) verifies (3.5). The conclusion now follows from Theorem 3.2.1.

Example 4.5.2. Recall that the Lorenz system is given byx1 = σ(x2 − x1)x2 = ρx1 − x2 − x1x3

x3 = −βx3 + x1x2.

For any β > 0 and ρ > 1, the set of equilibria of (3.15) is given by(0, 0, 0),

(±√β(ρ− 1),±

√β(ρ− 1), ρ− 1

).

The stability of each hyperbolic equilibrium solution x = (x1, x2, x3) is determined by the

eigenvalues of Df(x). For instance, at the positive eye x =(√

β(ρ− 1),√β(ρ− 1), ρ− 1

),

the Jacobian matrix is given by

Df(x) =

−σ σ 0

1 −1 −√β(ρ− 1)√

β(ρ− 1)√β(ρ− 1) −β

.

For the classical parameter values σ = 10, β = 83 and ρ = 28, numerical calculations lead

to the following candidates for eigenvalues:

λ1 = −13.854577914596042, λ2,3 = 0.093955623964686± 10.194505220927850i

and their associated eigenvectors

v1 =

0.855665024602210−0.329822750612395−0.398816146677856

, v2,3 =

−0.266119316830765± 0.295010166256352i0.032128610535763± 0.569077429163104i

−0.719213558699417

.


Consider the approximation (λ1, v1). The component of v1 with largest magnitude is v1

1.Defining the radii polynomials as

pk(r) = (Z1)kr2 + ((Z0)k − 1)r + Yk, k = 1, 2, 3

where Y is defined by (??), and Z0 and Z1 are given by (??), we obtained an interval ofexistence I =

[9.3438× 10−15 , 0.64912

].

For λ2,3 we obtain the interval of existence[5.6641× 10−15 , 0.45322

]. This altogether

yields a proof that the true eigenvalues λ1, λ2, λ3 of Df(x) satisfy

λ1 < 0 < Re(λ2) = Re(λ3)

and therefore that the positive eye x =(√

β(ρ− 1),√β(ρ− 1), ρ− 1

)is an unstable

equilibrium solution.

As discussed above, to analyze the stability of an equilibrium solution x its is necessaryto compute one or all the eigenvalues of Df(x). In Example 4.5.2 this was done using theexact formula for the equilibrium to compute Df(x). However in general we will not knowthe equilibria exactly, and hence we need to compute the eigenvalues without knowingDf(x) exactly. One way to do this is to compute the equilibrium and the eigenpairs at thesame time as discussed next.

Let f : Rn → Rn, and assume we want to compute an equilibrium solution x and keigenpairs (λi, vi), i = 1, . . . , k, of Df(x). For each eigenpair (λi, vi) consider the mapgi : Rn × Cn → Cn, corresponding to (4.21), given by

gi(x, z) := (Df(x)− λiI)

z2...zk

(vi)kzk+1

...zn

,

where z = (λi, z2, . . . , zn), and consider the map F : Rn×Cn×· · ·×Cn → Rn×Cn×· · ·×Cngiven by

xz1

...zk

7→ F

xz1

...zk

=

f(x)

g1(x, z1)...

gk(x, zk)

.Applying the radii polynomials to compute a zero of F will provide the equilibrium

solution x of x = f(x) and the k eigenpairs (λi, vi), i = 1, . . . , k, of Df(x) as desired.


4.6 Exercises

Exercise 4.6.1.

Prove that if M(v1(t), v2(t), . . . , vn(t) is a multilinear function, then

dM

dt(t) =

n∑k=1

M

(v1(t), . . . , vk−1(t),

dvkdt

(t), vk+1, . . . , vn(t)

)Exercise 4.6.2. Prove Proposition 4.2.6.

Exercise 4.6.3. Consider x = Ax and assume the real parts of all eigenvalues of A arenegative.

(i) Prove that x | ‖x‖ ≤ 1 is an attracting neighborhood for 0.

(ii) Prove that there exists a trapping region containing 0.

Exercise 4.6.4. Complete the proof of Theorem 4.3.7(i) and show that the conjugacyh : Rn → Rn is a homeomorphism. Hint: you can write down the inverse explicitly.

Exercise 4.6.5. Let X be a topological space. A set G ⊂ X is generic if G is open anddense in X. Prove that hyperbolic matrices are open and dense in the space of linear mapson Rn.

Hint: Given A ∈Mn(R), show that ‖A‖ := max1≤i,j≤n |ai,j | defines a metric onMn(R).

Exercise 4.6.6. Prove part (i) of Theorem 4.3.5. (Hint: use (4.6).)

Chapter 5

Continuation of Equilibria

5.1 Parameterized Families of Equilibria

Radii polynomials are introduce in Chapter 3 as a means of rigorously verifying the ex-istence of an equilibrium to an ODE of the form x = f(x). However, in applicationsparameters play a fundamental role and thus it is important to be able to find equilibriafor

x = f(x, λ),

where f : Rn × Rm → Rn takes the form (x, λ) 7→ f(x, λ). Assume that f(x0, λ0) = 0. Anobvious question is under what conditions does there exists a neighborhood V ⊂ Rm ofλ0 and a continuous function α : V → Rn such that α(λ0) = x0 and f(α(λ), λ) = 0. Theimplicit function theorem provides an answer.

Theorem 5.1.1. [Implicit Function Theorem] Let U ⊂ Rn and V ⊂ Rm be opensets. Assume that f : U × V → Rn is C1 and that f(x0, y0) = 0 for (x0, y0) ∈ U × V . IfDxf(x0, y0) is invertible, then there exists neighborhoods U0 ⊂ U and V0 ⊂ V of x0 andλ0, respectively, and a C1 function α : V0 → U0 such that f(α(y), y) = 0. Furthermore, if(x, y) ∈ U0 × V0 and f(x, y) = 0, then x = α(y).

We take a slightly different, more restrictive approach in this Chapter and assume thatλ ∈ R. Consider the following initial value problem

dα

dλ= F (α, λ) := − [Dxf(α, λ)]−1 ∂f

∂λ(α, λ), α(λ0) = x0, (5.1)

and assume that f(x0, λ0) = 0. By Theorem 2.2.13, if Dxf(α, λ) is invertible and if F issmooth, then there exists a smooth solution α : J → Rn where J ⊂ R is a neighborhood of

75

76 CHAPTER 5. CONTINUATION OF EQUILIBRIA

λ0. To understand the significance of this observe that (5.1) can be rewritten as

Dxf(α, λ)dα

dλ= −∂f

∂λ(α, λ)

Dxf(α, λ)dα

dλ+∂f

∂λ(α, λ) = 0

d

dλf(α, λ) = 0

f(α, λ) = C

for some constant C. However, since we are assuming that f(α(λ0), λ0) = 0, C = 0 andhence

f(α(λ), λ) = 0. (5.2)

Notice that we have essentially recovered a simplified form of the inverse function theorem.While we cannot expect to obtain an exact closed form solution to (5.2) we can use the

fact that α is the solution to a differential equation to compute an approximate solution.The most straightforward approach is to make use of the Euler approximation and set

x1 := x0 + hF (x0, λ0).

For sufficiently small step size |h| > 0 it is reasonable to assume that f(x1, λ1) ≈ 0 whereλ1 = λ0 + h. In fact, it is also reasonable to assume that the Newton operator

T (x) := x−Dxf(x1, λ1)−1f(x, λ1)

is a contraction mapping and thus that 0 ≤ ‖f(T (x1), λ1)‖∞ < ‖f(x1, λ1)‖∞, i.e. T (x1)is a better approximation of a root of f at λ1 than x1. This technique of approximatingequilibria is a standard numerical approach and is often referred to as a predictor-correctormethod. The Euler step is used to predict a new root and Newton’s method is used toimprove upon the prediction.

Example 5.1.2. To make the above discussion more concrete consider

x = f(x, λ) = x2 − λ.

Example 3.2.4 allows us to conclude that f(x(2), 2) = 0 for a unique value of x(2) ∈B1.3(0.136). Let us choose (x0, λ0) = (x(λ0), λ0) = (1.3, 2) as an initial condition for

dx

dλ= F (x, λ) = − [Dxf(x, λ)]−1 ∂f

∂λ(x, λ) =

1

2x.

Applying the Euler approximation with h = 1 we obtain

x1 = x0 + hF (x0, λ0) = x0 +1

2x0= 1.3 +

1

2(1.3)≈ 1.68.

5.2. COMPUTING BRANCHES OF EQUILIBRIA 77

For the associated Newton operator at λ1 = λ0 + h = 2 + 1 = 3, we choose

T (x) = x− 0.3f(x, 3)

since Dxf(x1, λ1)−1 = (2 · 1.68)−1 ≈ 0.3. Set

x(3) = 1.73 ≈ T (x1).

We would like to know if we have found a reasonable representative of an equilibrium atλ = 3. To this we apply the method of radii polynomials. We use the mapping

T (x) := x−Af(x, 3)

where A = 0.3 since Dxf(x(3), 3)−1 = (2 · 1.73)−1 ≈ 0.3. The same analysis as in Exam-ple 3.2.4 leads to the radii polynomial

p(r) =∣∣0.3(1.732 − 3)

∣∣+ (|1− 2(0.3)(1.73)| − 1)r + 2(0.3)r2

= 0.6r2 − 0.962r + 0.00213

which is negative over the interval I = [0.003, 1.6] and hence there exists x(3) ∈ B1.73(0.003)such that f(x(3), 3) = 0.

Given that√

3 ≈ 1.73205080756888 the computation in Example 5.1.2 is a fairly goodresult. The problem is that we have only established the existence and bounds on the rootsx(2) and x(3), whereas we would like to have existence and bounds for x(λ) for λ ∈ [2, 3].

5.2 Computing Branches of Equilibria

Consider the differential equationx = f(x, λ),

where f : Rn × Λ → Rn and Λ ⊂ R. Assume that f ∈ C1(Rn × Λ,Rn). Assume we haveapproximate solutions to f(x, λ) = 0 given by (x0, λ0) and (x1, λ1). Let

λs := (1− s)λ0 + sλ1 and xs := (1− s)x0 + sx1. (5.3)

We want to use radii polynomials to conclude about the existence of a smooth curve ofequilibria xs lying in a region centered about the segment xs : s ∈ [0, 1].Theorem 5.2.1. Consider the differential equation

x = f(x, λ),

where f ∈ Ck(Rn × Λ,Rn), k ≥ 2, and Λ ⊂ R. Let xs and λs as in (5.3). Set

Br :=⋃

s∈[0,1]

Br(xs)× λs .


Choose A ∈Mn(R) and positive constants Y0, Z0, and Z2 such that

‖Af(x, λs)‖ ≤ Y0 (5.4)

‖I −ADf(x, λs)‖ ≤ Z0 (5.5)

‖A [Df(c, λs)−Df(x, λs)] ‖ ≤ Z2(r)r, for all c ∈ Br(x) and all r > 0. (5.6)

Set p(r) = Z2r2 − (1− Z0)r + Y0.

If there exists r0 > 0, such that p(r0) < 0 and Dxf(x, λ) is invertible for all (x, λ) ∈ Br0 ,then there exists a Ck−1 function α : [λ0, λ1]→ Br0 such that

f(α(λ), λ) = 0

and if (x, λ) ∈ Br0 and f(x, λ) = 0, then x = α(λ).

Proof. Because pk(r0) < 0, for all k = 1, . . . , n, by continuity, there exists 0 < r− < r0 < r+

such that if r ∈ (r−, r+), then pk(r) < 0, for all k = 1, . . . , n. Furthermore, by Corollary ??for each s ∈ [0, 1], there exists a unique xs ∈ Br−(xs) such that f(xs, λs) = 0 and for allx ∈ Br+(xs)\Br−(xs), f(x, λs) 6= 0. It remains to be shown that xs lies on a smooth curve.

Let α : (λ−, λ+)→ Rn be the maximal solution to the initial value problem

dα

dλ= − [Dxf(α, λ)]−1 ∂f

∂λ(α, λ), α(λ1/2) = x1/2.

Observe that by (5.2), f(α(λ), λ) = 0. Our goal is to show that λ− < λ0 < λ1 < λ+ andthat α([λ0, λ1]) ⊂ Br.

Assume λ+ < λ1. By assumption Dxf(x, λ) is invertible on Br0 , and hence, by conti-nuity, Dxf(x, λ) is invertible on a neighborhood of Br0 . By Theorem 2.2.20, this impliesthat there exists λs ∈ (λ1/2, λ+) such that α(λs) ∈ Br+(xs) \ Br−(xs). However, by (5.2)f(α(λs), λs) = 0, a contradiction. Thus, λ1 < λ+. Observe, that the same argument alsoimplies that α([λ1/2, λ1]) ⊂ Br0 . Yet another application of this argument allows us toconclude that α([λ0, λ1]) ⊂ Br0 .

Example 5.2.2. Consider the Lorenz system, fix the values of σ and β, and leave ρ as aparameter. Denote the right-hand side of the system by

f(x, ρ) :=

σ(x2 − x1)ρx1 − x2 − x1x3

−βx3 + x1x2

.

Assume that at ρ = ρi, we have an approximate solution xi for i = 0, 1. As in (5.3), defineρs = (1− s)ρ0 + sρ1, and xs = (1− s)x0 + sx1.

5.2. COMPUTING BRANCHES OF EQUILIBRIA 79

Following (??), we first find a bound for |T (xs, s) − xs| = |Af(xs, ρs)|. To simplifynotation define g : [0, 1]→ R3 by g(s) = f(xs, ρs). Since g is quadratic in s,

g(s) = g(0) + g′(0)s+1

2g′′(0)s2

= f(x0, ρ0) + [Dxf(x0, ρ0)(x1 − x0) +Dρf(x0, ρ0)(ρ1 − ρ0)] s

+1

2

[D2xf(x0, ρ0)(x1 − x0, x1 − x0)

+Dρ,xf(x0, ρ0)(x1 − x0)(ρ1 − ρ0) +D2ρf(x0, ρ0)(ρ1 − ρ0)2

]s2.

Since f is linear in ρ, D2ρf(x0, ρ0)(ρ1 − ρ0)2 = 0. To further simplify the notation set

y(0) := f(x0, ρ0)

y(1) := Dxf(x0, ρ0)(x1 − x0) +Dρf(x0, ρ0)(ρ1 − ρ0)

y(2) :=1

2

[D2xf(x0, ρ0)(x1 − x0, x1 − x0) +Dρ,xf(x0, ρ0)(x1 − x0)(ρ1 − ρ0)

].

Then,

|T (xs, s)− xs| = |Af(xs, ρs)| ∣∣∣Ay(0)

∣∣∣+∣∣∣Ay(1)

∣∣∣+∣∣∣Ay(2)

∣∣∣ .Choose Y = (Y1, Y2, Y3) such that

∣∣∣Ay(0)∣∣∣+∣∣∣Ay(1)

∣∣∣+∣∣∣Ay(2)

∣∣∣ Y. (5.7)

In practice this is usually done using an interval arithmetic computation. This providesthe first bound of (??).

Turning to the second bound of (??), observe that

DxT (xs + b, s)c = [I −ADxf(xs + b, ρs)]c

= [I −ADxf(x0 + sv + b, ρ0 + s(ρ1 − ρ0))]c

where v := x1 − x0. Making use of the fact that f is quadratic in x and linear in ρ, theTaylor series expansion of Dxf(x, ρ) at (x0, ρ0) takes the form

Dxf(x0+sv+b, ρ0+s(ρ1−ρ0)) = Dxf(x0, ρ0)+D2xxf(x0, ρ0)(sv+b)+D2

xρf(x0, ρ0)(s(ρ1−ρ0)).


Note that

Dxf(x0, ρ0) =

−σ σ 0ρ0 − x0,3 −1 −x0,1

x0,2 x0,1 −β

Dxxf(x0, ρ0)(sv + b) =

0 0 0−sv3 − b3 0 −sv1 − b1sv2 + b2 sv1 + b1 0

D2xρf(x0, ρ0)(s(ρ1 − ρ0)) =

0 0 0s(ρ1 − ρ0) 0 0

0 0 0

To explicitly express the powers of r, let b = b/r and c = b/r. Then using the Taylor seriesexpansion we obtain

DxT (xs + b, s)c = [I −ADxf(xs + br, ρs)]cr

= [I −ADxf(x0, ρ0)]cr −A

0

(−sv3 − b3r)c1 + (−sv1 − b1r)c3

(sv2 + b2r)c1 + (sv1 + b1r)c2

r−A

0s(ρ1 − ρ0)c1

0

r= [I −ADxf(x0, ρ0)]cr −A

0(ρ1 − ρ0)c1 − v3c1 − v1c3

v2c1 + v1c2

sr−A

0

−b3c1 − b1c3

b2c1 + b1c2

r2.

Thus we define

Z(1) := |I −ADxf(x0, ρ0)|13 + |A|

0|ρ1 − ρ0|+ |v3|+ |v1|

|v2|+ |v1|

(5.8)

and

Z(2) := 2|A|

011

, (5.9)

and leave it to the reader to check that for all s ∈ [0, 1],

supb,c∈Br(0)

|DxT (xs + b, s)c| Z(r) := Z(1)r + Z(2)r2.

5.3. SADDLE-NODE BIFURCATION 81

To do an explicit computation fix σ = 10 and β = 8/3 and at ρ0 = 28 and 29 considerthe numerical approximations

x0 :=

8.4852813742385708.485281374238570

27

and x1 :=

8.6409875978771468.640987597877146

28

.

Using the numerically computed inverse

Dxf(x0, ρ0)−1 ≈ A :=

−0.051851851851852 −0.018518518518519 0.0589255650988790.048148148148148 −0.018518518518519 0.058925565098879−0.011785113019776 −0.117851130197758 0

.

and the bounds (5.7), (5.8) and (5.9), we obtain the radii polynomials pk(r) = Z(2)k r2 +

(Z(1)k − 1)r + Yk, for k = 1, 2, 3.The existence interval contains I = [0.0886, 3.0762]. Choosing r0 = 0.0886, we get that

for each s ∈ [0, 1], there is a true equilibrium solution x(s) ∈ Br(xs), that is f(x(s), 28+s) =0, for all s ∈ [0, 1].

For the Lorenz equations we have a closed form expression for the this branch of equi-libria

x(s) =(√

β(ρs − 1),√β(ρs − 1), ρs − 1

), s ∈ [0, 1]

and thus we can check that

sups∈[0,1]

‖x(s)− xs‖∞ ≈ 3.539× 10−4.

This is 2 orders of magnitude smaller than r0 the optimal radius obtained from the existenceinterval for the radii polynomials.

5.3 Saddle-Node Bifurcation

Consider a differential equation f : Rn × Rm → Rn which induces a continuous family offlows

ϕ : R× Rn × Rm → Rn × Rm

(t, x, λ) 7→ (ϕ(t, x, λ), λ) = (ϕλ(t, x), λ).

A parameter value λ ∈ Rm is a bifurcation point if given any ε > 0 there exists λ ∈ Rmsuch that ‖λ− λ‖ < ε and ϕλ is not topologically equivalent to ϕλ.

There is no hope of classifying all bifurcation points. However there are some bifurca-tions that can be detected by changes in the eigenvalue structure of the linearized equationat equilibria. The equilibria for the differential equation x = f(x, λ), f : Rn×Rm → Rn are


solutions to f(x, λ) = 0. By the implicit function theorem, if f(x, λ) = 0 and Dxf(x, λ) isinvertible then there exists a smooth family of equilibria x(λ) defined in a neighborhoodof λ. This raises the question of what happens when Dxf(x, λ) is not invertible. The lackof invertibility is equivalent to the existence of a zero eigenvalue. The simplest possiblesetting is a unique zero eigenvalue and a one dimensional parameter space, m = 1. Thisleads to the following definitions.

Definition 5.3.1. Consider f : Rn×R→ Rn. A saddle-node for f is a point (x, λ) ∈ Rn×Rsuch that

(i) f(x, λ) = 0, and

(ii) 0 is an eigenvalue of Dxf(x, λ) with algebraic multiplicity one and all other eigenval-ues have non-zero real parts.

Definition 5.3.2. Given f : Rn ×R→ Rn a saddle-node bifurcation occurs at the saddle-node (x, λ) ∈ Rn × R if the following conditions are met.

SNB1 There exists a smooth curve g : (−δ, δ) → Rn × R denoted by s 7→ (g1(s), g2(s))such that g(0) = (x, λ) and f(g1(s), g2(s)) = 0.

SNB2 The curve defined by g has a quadratic tangency with Rn × λ at (x, λ); that is,

g2(0) = λ, g′2(0) = 0, and g′′2(0) 6= 0.

SNB3 If s 6= 0 then Dxf(g1(s), g2(s)) is hyperbolic and if σ(s) is the eigenvalue ofDxf(g1(s), g2(s)) that satisfies σ(0) = 0, then σ′(0) 6= 0.

Before stating a general theorem we consider the simplest setting where the phase spaceis one dimensional.

Proposition 5.3.3. Consider a C2 function f : R× R→ R satisfying

f(0, 0) = 0, Dxf(0, 0) = 0, D2xf(0, 0) 6= 0, and Dλf(0, 0) 6= 0.

Then, there is a saddle-node bifurcation at (0, 0).

Proof. Since Dλf(0, 0) 6= 0 the implicit function theorem implies the existence of a curveg2 : (−δ, δ) → R such that f(s, g2(s)) = 0. Observe that by defining g1(s) = s, SNB1 issatisfied.

Differentiating f(s, g2(s)) = 0 with respect to s gives

Dxf(s, g2(s)) +Dλf(s, g2(s))g′2(s) = 0 (5.10)

and in particularDxf(0, 0) +Dλf(0, 0)g′2(0) = 0.


By assumption Dxf(0, 0) = 0 and Dλf(0, 0) 6= 0, hence g′2(0) = 0. Differentiating (5.10)gives

D2xf(s, g2(s)) + 2DλDxf(s, g2(s))g′2(s) +D2

λf(s, g2(s))(g′2(s))2 +Dλf(s, g2(s))g′′2(s) = 0.

At the saddle-node we have

D2xf(0, 0) +Dλf(0, 0)g′′2(0) = 0

and hence

g′′2(0) = −D2xf(0, 0)

Dλf(0, 0)6= 0.

This implies SNB2.

To check that SNB3 is satisfied observe that

d

dsDxf(s, g2(s)) = D2

xf(s, g2(s)) +DλDxf(s, g2(s))g′2(s)

and hence at s = 0,d

dsDxf(0, 0) = D2

xf(0, 0) 6= 0

which implies that for |δ| sufficiently small Dxf(s, g2(s)) 6= 0.

The following result, [4, Theorem 8.12], provides sufficient conditions in the setting ofan n-dimensional phase space for the occurrence of a saddle-node bifurcation.

Theorem 5.3.4. Assume f : Rn × R → Rn is C1, (x, λ) ∈ Rn × R is a saddle-node, andthe kernel of Dxf(x, λ) is spanned by the non-zero vector v ∈ Rn. If

Dλf(x, λ) 6= 0 and D2xf(x, λ)(v, v) 6= 0

and both are not in the range of Dxf(x, λ), then there is a saddle-node bifurcation at(x, λ). Moreover, among all C∞ one parameter families that have a saddle-node, thosethat undergo a saddle-node bifurcation form an open and dense subset.

The following lemma provides an elegant means of checking that if u1, u2 6= 0, thenthey are not in the range of Dxf(x, λ).

Lemma 5.3.5. Let D : Rn → Rn a linear transformation with dim(ker(D)) = 1. ThenkerDT = 〈w〉, for some w ∈ Rn \ 0. Also, u is in the range of D if and only if u ·w = 0.

Proof. By the rank-nullity theorem, rank(D) := dim(image(D)) = n − dim(ker(D)) =n−1. Since rank(D) = rank(DT ) the rank-nullity theorem implies that dim(ker(DT )) = 1.Hence, there is a non-zero vector w ∈ Rn such that kerDT = 〈w〉.


Now, if u ∈ Rn is in the range of D, there exists y ∈ Rn such that u = Dy, and thenu · w = wTu = wT (Dy) = (wTD)y = (DTw)T y = 0. This implies that

image(D) ⊂ (ker(DT ))⊥ = (〈w〉)⊥ := u ∈ Rn | u · w = 0. (5.11)

Conversely, assume u · w = 0, that is u ∈ (〈w〉)⊥ = (ker(DT ))⊥. Since (ker(DT ))⊥ is an(n−1)-dimensional subspace and dim(image(D)) = n−1, we use (5.11) to get image(D) =(ker(DT ))⊥. Hence, u ∈ (ker(DT ))⊥ = image(D), that is u is in the range of D.

Example 5.3.6. Consider the FitzHugh-Nagumo equation

x = f(x, γ) =

(x1(x1 − a)(1− x1)− x2

ε(x1 − γx2)

), (5.12)

with parameter values (a, ε) = (5, 1) and where γ is left as a free parameter. We useTheorem 5.3.4 to show the existence of a saddle-node bifurcation.

The first step is to identify a saddle-node, which we do numerically by using a predictor-corrector approach to compute two branches of hyperbolic equilibria, as portrayed in Fig-ure 5.1. The computations suggest this occurs at (x, γ), with x = (3, 12)T and γ = 1

4 .Direct computation gives

f(x, γ) =

(3(3− 5)(1− 3)− 12

1(3− 1

4 · 12) )

=

(00

), and D := Dxf(x, γ) =

(4 −11 −1

4

),

and the eigenvalues of D are given by 0 and 154 . Therefore, (x, γ) is a saddle-node.

v = (1, 4)T is an eigenvector associated with eigenvalue 0 and hence kerD = 〈v〉.Since dim kerDT = dim kerD = 1, we can solve for the eigenpair (0, w) for DT to

obtain w = (1,−4)T 6= 0. In particular, kerDT = 〈w〉.Again, direct computation shows that

u1 := Dγf(x, γ) =

(0−εx2

)=

(0−12

)u2 := D2

xf(x, γ)(v, v) =

(2v2

1(a− 3x1 + 1)0

)=

(2(5− 3 · 3 + 1)

0

)=

(−60

)and thus,

u1 · w = (0,−12)(1,−4)T = 48 6= 0 and u2 · w = (−6, 0)(1,−4)T = −6 6= 0.

Lemma 5.3.5 implies that both u1, u2 are not in the range of Dxf(x, λ).

Therefore, by Theorem 5.3.4 that there is a saddle-node bifurcation at (x, γ).


γ

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x1

1

1.5

2

2.5

3

3.5

4

4.5

5

Figure 5.1: The below (red) branch was obtained by applying Newton to theinitial point (1, 0)T at γ = 2. After Newton converged to the point x =(1.129171306613029 , 0.564585653306515)T , we decreased the parameter γ up to 1

4 , wherethe branch seemed to disappear. The top (blue) branch was obtained by applyingNewton to the initial point (4, 0)T at γ = 2. After Newton converged to the pointx = (4.870828693386970 , 2.435414346693485)T , we decrease the parameter γ up to 1

4 ,where the branch seems to disappear.

Our success in directly applying Theorem 5.3.4 to demonstrate the existence of a saddle-node bifurcation for the FitzHugh-Nagumo equation made use of the fact that we hadexplicit values for the saddle-node (x, γ) and that the system is two dimensional and sowe could perform the linear algebra computations by hand. Keeping with the spirit of thisbook we are interested in recasting Theorem 5.3.4 in such a way that we can rigorouslyverify the existence of a point at which a saddle-node bifurcation occurs for more generalproblems.

Algorithm 5.3.7. Let f : Rn × R → Rn be smooth. Assume there exists (x, λ) ∈ Rn × Rand µ ∈ σ(Dxf(x, λ)) such that f(x, λ) ≈ 0 and µ ≈ 0. Perform the following steps.

(i) Using the techniques of Section 4.5 compute eigenpairs

(µi, vi) | i = 1, . . . , N

such

that µi ∈ Bri(µi) where

0 ∈ Br1(µ1) and Bri(µi) ∩ iR = ∅, i = 2, . . . , n.

(ii) Verify that µ1 = 0.


(iii) Define u1 := Dλf(x, λ) and u2 := D2xf(x, λ)(v1, v1) and verify that

wTui 6= 0, i = 1, 2, (5.13)

where w ∈ Rn \ 0 such that wTDxf(x, λ) = 0.

If steps (i)-(iii) are carried out successfully, then a saddle-node bifurcation occurs at the

saddle-node (x, λ) ∈ Br(x, λ) where r is obtain explicitly in the verification of step (ii).

There are two parts to the justification of this algorithm. The first part is to showthat if steps (i)-(iii) are carried out, then the hypotheses of Theorem 5.3.4 are satisfied andhence a saddle-node bifurcation occurs. We leave this to the reader.

The second part is to demonstrate that there are rigorous numerical procedures capableof carrying out steps (i)-(iii). As indicated in the statement of Algorithm 5.3.7 step (i) isto be carried out using the techniques of Section 4.5. We assume that this computation issuccessful.

Step (ii) requires us to rigorously verify that µ1 = 0, which in turn requires us tomodify the problem being address in Section 4.5. By Step (i) there is at most one zeroeigenvalue, thus it is sufficient to prove the existence of an eigenvector v that lies in thekernel of Dxf(x, λ). Observe that if it exists, then v ∈ Rn, and thus to obtain isolation it issufficient to require that ‖v‖ = 1. This leads to the following problem: prove the existenceof X = (x, λ, v) ∈ R2n+1 satisfying

F (X) :=

f(x, λ)‖v‖2 − 1Dxf(x, λ)v

= 0. (5.14)

This problem can be addressed using the radii polynomial approach described in Sec-tion 3.2. This solution to (5.14) identifies the existence of (x, λ) ∈ Br(x, λ) for some r > 0.We now assume that step (ii) has been successfully completed.

Let D := Dxf(x, λ). Observe that step (iii) requires the existence of a vector w ∈Rn\0 such that wTD = 0. Let us now introduce how to use the radii polynomial approachto compute rigorously a non-zero vector w such that kerDT = 〈w〉. At this point, it isknown rigorously that dim ker(D) = 1, which implies that rank(D) = rank(DT ) = n − 1.Therefore, when looking for a non-zero w satisfying DTw = 0, we can get rid of one rowof DT without changing the solution space. Now the question is which equation to get ridof? As usual, we use a numerical approximation to answer that question. Assume thatv 6= 0 satisfies Dv ≈ 0. Let k the component of v with the largest magnitude, that is

|vk| = maxi=1,...,n

|vi| 6= 0.

Denote by C1, . . . , Cn the columns of D and R1, . . . , Rn the corresponding rows of DT that


is Ri = CTi for i = 1, . . . , n. Then since Dv ≈ 0,

Ck ≈1

vk

n∑i=1i6=k

viCi =⇒ Rk = CTk ≈1

vk

n∑i=1i 6=k

viCTi =

1

vk

n∑i=1i 6=k

viRi.

Since the k-th row Rk of DT is a linear combination of the other rows, we get rid of it,

or equivalently we get rid of the k-th column Ck of D. Denote M :=(Dk

)T, with Dk the

n × (n − 1) matrix defined by D without its k-th column Ck. A non-zero unit vector wsuch that kerDT = 〈w〉 is an isolated solution of

g(w) :=

(‖w‖2 − 1Mw

)= 0, (5.15)

which we solve using the radii polynomial approach as introduced in Section 3.2.The final step is to verify (5.13) which, in practice, is done using interval arithmetic.

Example 5.3.8. The Hodgkin-Huxley model for the action potential of a space-clampedsquid axon is defined by the four dimensional vector field

x1

x2

x3

x4

=

−20− 120x32x3

(x1 − 25.1 ln(

23

1350(550− λ))

)− 36x4

4

(x1 − 25.1 ln(

λ

400)

)− 0.3(x1 + 24.3)

9

25(1− x2)

x1 −∆(λ) + 35

1− exp(−x1−∆(λ)+35

10

) − 72

5x2 exp

(−x1 −∆(λ) + 60

18

)63

250(1− x3) exp

(−x1 −∆(λ) + 60

20

)− 18

5x3

1

exp(−x1−∆(λ)+30

10

)+ 1

9

250(1− x4)

x1 −∆(λ) + 50

1− exp(−x1−∆(λ)+50

10

) − 9

20x4 exp

(−x1 −∆(λ) + 60

80

)

(5.16)

where

∆(λ) := 9.32 ln

(11

10− λ

500

).

The variable x1 is the membrane potential, x2 is the activation of a sodium current, x3 isthe activation of a potassium current, x4 is the inactivation of the sodium current and theparameter λ is the external potassium concentration.

Using a standard pseudo-arclength continuation technique (see Exercise ??), we obtainthe bifurcation diagram of Figure 5.2. From this numerical simulation, we can conjecturethe existence of two saddle-node bifurcations at λ ≈ 426.42 and at λ ≈ 53.61.

Denote by f(x, λ) the right-hand side of (5.16). Let X = (x, λ, v) ∈ R9, and defineF : R9 → R9 as in (5.14). Applying Newton’s method to problem (5.14), we compute two


λ: external potassium concentration (mM)0 50 100 150 200 250 300 350 400 450

x1: m

em

bra

ne p

ote

ntial (m

V)

-100

-80

-60

-40

-20

0

20

Figure 5.2: A branch of equilibria which undergoes two saddle-node bifurcations.

approximate solutions

X1 :=

−87.25156054399080.00890348887480.92318576319590.1363687823857

426.41597255550500.99989820430070.0011653907919−0.01055188172910.0095331365494

and X2 :=

−49.2705162821502260.1708949161547820.2419466852503400.48785290593769553.6071084136836970.9994270220352320.016569867126847−0.0251523842303170.015441006985167

(5.17)

corresponding to the two possible saddle-node points.

Theorem 5.3.9. There is a saddle-node bifurcation at a point

(x1, λ1) ∈ Br1((x1, λ1)), r1 := 1.715× 10−12,

where (x1, λ1) ∈ R5 is given by the first five components of X1 in (5.17).

Theorem 5.3.10. There is a saddle-node bifurcation at a point

(x2, λ2) ∈ Br2((x2, λ2)), r2 := 2.034× 10−12,

where (x2, λ2) ∈ R5 is given by the first five components of X2 in (5.17).


The proofs of Theorem 5.3.9 and Theorem 5.3.10 are both obtained by applying Algo-rithm 5.3.7.

Proof. (a) Let X one of the two points X1 or X2 given in (5.17). We already have thatF (X) ≈ 0, with F given by (5.14). Note that

DXF (X) =

Dxf(x, λ) Dλf(x, λ) 00 0 2vT

Dx

(Dxf(x, λ)v

)DxDλf(x, λ)v Dxf(x, λ)

. (5.18)

Using INTLAB in MATLAB, compute the exact inverse A = DXF (X)−1. Note that inpractice, the so obtained A will have interval entries. Define T : R9 → R9 by

T (X) = X −AF (X).

Using interval arithmetic, compute the upper bound Y such that∣∣AF (X)

∣∣ Y .Now, thanks to the (perfect) choice of A

DT (X + b)c = −A(DF (X + b)−DF (X)

)c.

Defining h : [0, 1]→ R9 by h(s) = DXF (X+sb)c, h(1)−h(0) =(DXF (X + b)−DXF (X)

)c.

For each k ∈ 1, . . . , 9, there exists sk ∈ [0, 1] such that(DXFk(X + b)−DXFk(X)

)c = hk(1)− hk(0) = h′k(sk) = D2

XFk(X + skb)(b, c).

Now, let b, c ∈ B1(0) such that b, c ∈ Br(0) are given by b = br and c = cr. In this case,(DXFk(X + b)−DXFk(X)

)c = D2

XFk(X + skb)(b, c)r2.

Set r∗ = 10−4 an a-priori upper bound for the left point of the existence interval of the radiipolynomials. We will have to show a-posteriori that r ≤ r∗. Denote by b∗ = [−r∗, r∗]9a vector in R9 whose entries are given by the interval [−r∗, r∗]. Denote by X∗ = X + b∗

a vector in R9 with its k-th entry given by the interval [Xk − r∗, Xk + r∗]. Denote byδ = [−1, 1]9 a vector in R9 whose entries are given by the interval [−1, 1]. Then, for eachb, c ∈ Br(0), it is left to the reader to verify that∣∣A (DF (X + b)−DF (X)

)c∣∣ ∈ ∣∣AD2

XF (X∗)(δ, δ)∣∣ .

Using interval arithmetic, compute Z1 ∈ R9 such that∣∣AD2XF (X∗)(δ, δ)

∣∣ Z(2).

Using the previous bounds, define the radii polynomials pk(r) = Z(2)k r2 − r+ Yk. For each

of the point X1 and X2 given in (5.17), we computed the radii polynomials and obtainedthe existence intervals

I1 =[1.715× 10−12 , 8.052× 10−6

]and I2 =

[2.034× 10−12 , 1.974× 10−5

],


respectively. Since 8.052× 10−6, 1.974× 10−5 < r∗ = 10−4, then the existence intervals arevalid. Let r1 := 1.715×10−12 and r2 := 2.034×10−12. Recall (5.17), then by Corollary ??,

there exists a unique X1 = (x1, λ1, v1) ∈ Br1(X1) such that F (X1) = 0 and there exists a

unique X2 = (x2, λ2, v2) ∈ Br2(X2) such that F (X2) = 0. Hence, for j = 1, 2, f(xj , λj) = 0and the kernel of Dxf(xj , λj) must be one dimensional, as otherwise we would not havethat Xj isolated solution in R9.

(b) Choose j ∈ 1, 2 and let I := Ij the existence interval associated to X := Xj . Let

r the smallest radius of the existence interval I. Define B = Br((x, λ)) ⊂ R5, that is

B =4∏

k=1

[xk − r, xk + r]×[λ− r, λ+ r

].

Let D := Dxf(x, λ) and D := Dxf(B) a 4× 4 interval matrix computed with intervalarithmetic. Note that D ⊂ D. Using the radii polynomial approach as introduced inSection 4.5, we now show that σ(D) ⊂ ⋃n

j=1Bj , for some small balls Bj ∈ C. The onlymodification from the theory of Section 4.5 is that we now have a matrix whose entriesare intervals. Hence, the bounds Y in (??) and the bonds Z0, Z1 in (??) have to boundall possible error coming from D. Interval arithmetic can be used to do this. Using theabove procedure and INTLAB, we proved that the eigenvalues of Dxf(x1, λ1) are enclosedin ∪4

j=1Bj , where

B1 =z ∈ C : |z + 32.02633660454969| ≤ 3.394274197807681× 10−11

B2 =

z ∈ C : |z − 9.305978062941147× 10−16| ≤ 1.069542699059249× 10−11

B3 =

z ∈ C : |z + 0.9317141708275124| ≤ 9.975910428070430× 10−12

B4 =

z ∈ C : |z + 0.5558951614569074| ≤ 1.774408327300926× 10−12

.

Therefore, we obtain that 0 ∈ Bk for the unique k = 2 ∈ 1, 2, 3, 4, and that Bj ∩ iR = ∅,for all j ∈ 1, 3, 4. This shows that (x1, λ1) is a saddle-node. We repeated the sameprocedure to show that (x2, λ2) is also a saddle-node.

(c) Let

w1 :=

0.0191791615230120.0016716237612230.0003095602023970.999814617621566

and w2 :=

−0.009242057198078−0.253654297666144−0.9378905136254270.236506799280007

, (5.19)

numerical approximations satisfying Dxf(x1, λ1)T w1 ≈ 0 and Dxf(x2, λ2)T w2 ≈ 0.

Choose j ∈ 1, 2, let D = Dxf(xj , λj) and w = wj . Based on D, construct the3 × 4 matrix M as above in order to define the problem g(w) = 0 as in(5.15). Using theradii polynomial approach of Section 3.2 applied on (5.15), we showed the existence of (i)


w1 ∈ B7.246×10−13(w1) such that Dxf(x1, λ1)w1 = 0 and (ii) w2 ∈ B1.595×10−11(w2) suchthat Dxf(x2, λ2)w2 = 0.

(d) For j = 1, 2, define

u(j)1 = Dγf(xj , λj) and u

(j)2 = D2

xf(xj , λj)(v,j vj).

With interval arithmetic, we showed that for both j = 1, 2, u(j)1 · wj 6= 0 and u

(j)2 · wj 6= 0.

This concludes the proofs of Theorem 5.3.9 and Theorem 5.3.10.

Chapter 6

Banach Spaces and RadiiPolynomials in Infinite Dimension

In previous chapters we studied the theory and practice of solving f(x) = 0 problemsformulated on the finite dimensional vector spaces Rn and Cn. In the remainder of thebook we will direct our attention to f(x) = 0 problems where x can be an unknown function(or collection of functions). Then we will need to consider a more general collection of vectorspaces in an infinite dimensional setting. The spaces we need are referred to as Banachspaces.

This chapter presents some general background material on Banach spaces. We brieflystudy linear operators between Banach spaces and extends some of the basic notions ofdifferential calculus to this new setting. We consider f(x) = 0 problems for differentiablemaps between Banach spaces and extend the earlier notions of Newton-liker operators andradii-polynomials. This material is the basis of the computer assisted proofs on functionspaces developed in the remainder of the book.

6.1 Banach Spaces, Duals and Operator Norms

Definition 6.1.1. A Banach space is a normed vector space which is complete in themetric defined by the norm.

Example 6.1.2. The vector spaces Rn and Cn as well as the collection of all m×n matricesof real (or complex) numbers provide familiar examples of Banach spaces.

Example 6.1.3. Let [a, b] ⊂ R and f : [a, b] → R be a continuous function. Recall thatthe usual supremum norm of f is

‖f‖C0([a,b]) := supt∈[a,b]

|f(t)|,

93

94CHAPTER 6. BANACH SPACES AND RADII POLYNOMIALS IN INFINITE DIMENSION

and that this norm endows the space

C0([a, b]) =f : [a, b]→ R | f is continuous, and ‖f‖C([a,b]) <∞

,

with a Banach space structure.

Example 6.1.4. Let [a, b] ⊂ R and f : R be a continuously differentiable function, i.e.suppose that f ′ exists and is continuous on [a, b]. Define the norm

‖f‖C1([a,b]) = ‖f‖C0([a,b]) + ‖f ′‖C0([a,b]).

Then

C1([a, b]) :=f : [a, b]→ R | f, f ′are continuous, and ‖f‖C1([a,b]) <∞

.

is a Banach space.

Example 6.1.5. It is an exercise to show that C0([a, b]) is not a Banach space whenendowed with the norm

‖f‖1 =

∫ b

a|f(t)| dt,

(to see this, find a sequence of continuous functions which are Cauchy in this norm butwhich do not converge pointwise to a continuous function). The set

L1([a, b]) = f : [a, b]→ R | f is a measurable function, and ‖f‖1 <∞ ,

is the completion of C0([a, b]) under the ‖·‖1 norm. Under this norm, L1([a, b]) is a Banachspace.

Example 6.1.6. More generally, let Ω ⊂ Rn be a measurable set and define

‖f‖1,Ω :=

∫Ω|f(x)| dx,

where the integral is taken in the sense of Lebesgue. Then

L1(Ω) = f : Ω→ Rn|f is measurable, and ‖f‖1,Ω <∞ ,

is a Banach space.

The collection of all linear maps from a normed vector space V to a normed vectorspace W is itself a vector space using the operations defined by

(αA)v := αAv

(A+B)v := Av +Bv

for any scalar α, vector v ∈ V , and linear maps A,B : V →W .

6.1. BANACH SPACES, DUALS AND OPERATOR NORMS 95

Definition 6.1.7. Let V and W be normed vector spaces and A : V →W be a linear map.We say that A is a bounded linear map if

sup‖x‖V =1

‖Ax‖W <∞.

Building on the definition above, we define for a linear map A : V → W the operatornorm of A by

‖A‖ := sup‖x‖V =1

‖Ax‖W ,

(see also Definition 7.2.2). Denote by

L(V,W ) := A : V →W |A is a linear map, and ‖A‖ <∞ ,

the normed linear space of all bounded linear maps from V to W . The following boundfrequently is useful.

Proposition 6.1.8. Suppose that A ∈ L(V,W ). Then

‖Ax‖W ≤ ‖A‖‖x‖V ,

for all x ∈ V .

Proof. Let x ∈ V . If x = 0 then Ax = 0 and we have the equality

‖Ax‖W = ‖0‖W = 0 = ‖A‖‖x‖V .

On the other hand suppose that x 6= 0. Then 0 < ‖x‖V and we have

‖Ax‖W =

∥∥∥∥Ax‖x‖V‖x‖V

∥∥∥∥W

= ‖x‖V∥∥∥∥A x

‖x‖V

∥∥∥∥W

≤ ‖x‖V sup‖y‖V =1

‖Ay‖W

= ‖A‖‖x‖V .

Example 6.1.9. Define the linear operator A : C0([a, b])→ C0([a, b]) by

(Af)(t) =

∫ t

af(x) dx,


for t ∈ [a, b] and f ∈ C0([a, b]). Note that A is well defined by the fundamental theorem ofCalculus (i.e. Af is continuous). Then

‖A‖ = sup‖f‖=1

‖Af‖C0([a,b])

= sup‖f‖=1

supt∈[a,b]

|(Af)(t)|

= sup‖f‖=1

supt∈[a,b]

∣∣∣∣∫ t

af(x) dx

∣∣∣∣≤ sup‖f‖=1

supt∈[a,b]

∫ t

a|f(x)|dx

≤ sup‖f‖=1

supt∈[a,b]

∫ t

asupy∈[a,b]

|f(y)|dx

≤ sup‖f‖=1

∫ b

a‖f‖C0([a,b])dx

≤ sup‖f‖=1

|b− a|‖f‖C0([a,b])

≤ |b− a| <∞,

so that A is a bounded linear operator.

Example 6.1.10. Fix g ∈ L1(R), and recall that for any other f ∈ L1(R) the convolutionof f and g is defined by

(f ∗ g)(t) :=

∫Rf(x)g(t− x) dx.

Define the linear operator A : L1(R)→ L1(R) by

(Af)(t) = (f ∗ g)(t).

It is an exercise to prove that A is a bounded linear operator.

Since any two normed linear spaces V,W are a metric spaces with the metric inducedby the norm, it is reasonable to ask “what are the continuous linear maps from V to W?”The next theorem provides the answer.

Theorem 6.1.11. Let (V, ‖ · ‖V ) and (W, ‖ · ‖W ) be normed linear spaces and A : V →Wbe a linear map. Then A is a continuous map from V to W if and only if A ∈ L(V,W ).

Proof. Suppose that A ∈ L(V,W ) and x ∈ V . Since V and W are metric spaces we havethat continuity is equivalent to sequential continuity. So, choose xn∞n=0 ⊂ V having


xn → x as n→∞. Using the bound of Proposition 6.1.8 we have that

‖Ax−Axn‖W = ‖A(x− xn)‖W≤ ‖A‖‖x− xn‖V

so that ‖Ax−Axn‖W → 0 as n→∞, i.e.

limn→∞

Axn = Ax,

i.e. A is sequentially continuous, and hence continuous at x.Suppose on the other hand that A : V → W is continuous. Then A is continuous at

0 ∈ V , and for ε = 1 we can find a δ > 0 so that ‖x − 0‖V = ‖x‖V ≤ δ implies that‖Ax‖W = ‖Ax−A0‖W ≤ 1. Now let y ∈ V have that ‖y‖V = 1. Then

‖Ay‖W =1

δ‖A(δy)‖W

≤ 1

δ,

where we used that ‖δy‖ = δ implies ‖A(δy)‖ ≤ 1. Since this bound is uniform in y, takingthe supremum over all y with ‖y‖V = 1 gives that

‖A‖ ≤ 1

δ<∞,

so that A ∈ L(V,W ).

Example 6.1.12. If V and W are finite dimensional and A : V → W is a linear map,then A ∈ L(V,W ). This can be proven using the fact that in finite dimensions ever linearoperator has a matrix representation.

It is important to understand when L(V,W ) is itself a Banach space.

Theorem 6.1.13. Let V be a normed vector space. If W is a Banach space, then L(V,W )is a Banach space.

Proof. Let Ann∈N ⊂ L(V,W ) be a Cauchy sequence of bounded linear operators.We begin by proving that there exists a function A : V →W defined by

Av := limn→∞

Anv, v ∈ V.

Since An is Cauchy, given ε > 0, there exists N > 0 such that ‖An − Am‖ < ε, for anyn,m > N . By Definition 7.2.2,

‖Anv −Amv‖W < ε‖v‖V ,


for all n,m > N . Thus, for each v ∈ V , Anv is Cauchy. Since W is a Banach space,limn→∞Anv converges.

We now prove that A ∈ L(V,W ). Given v ∈ V , a scalar α and ε > 0, choose nsufficiently large such that

‖Av −Anv‖ < ε and ‖Aαv −Anαv‖ < ε.

Then

‖Aαv − αAv‖ ≤ ‖Aαv −Anαv‖+ ‖Anαv − αAv‖≤ ε+ ‖αAnv − αAv‖≤ ε(1 + |α|).

Since this is true for any ε, Aαv = αAv. A similar argument shows that A(v + v′) =Av +Av′.

Finally, we need to show that A is bounded. Using the fact that An is Cauchy thereexists n > 0 such that if m ≥ n, then

‖An −Am‖ < 1.

Choose v ∈ V such that ‖v‖V = 1 and observe that

‖Amv‖W ≤ ‖Amv −Anv‖W + ‖Anv‖W ≤ 1 + ‖An‖.

Thus,‖A‖ ≤ lim sup

m‖Am‖ ≤ 1 + ‖An‖

and hence A is bounded.

Recall that an operator A : V →W is invertible if there exists and operator B : W → Vso that

AB = IdW , and BA = IdV ,

and that we write A−1 := B.

Definition 6.1.14. We say that A : V →W is boundedly invertible if A−1 exists and

‖A−1‖ <∞.

Note that in the definition above it is not necessary that A itself is bounded. Indeed,there are important applications where an unbounded linear operator has a bounded inverse(think of differentiation versus integration). That being said, if A is bounded we have thefollowing theorem, which is a deep result from functional analysis whose proof we omit (seefor example REFERENCE KREYSZIG).


Theorem 6.1.15 (Bounded Inverse Theorem). Suppose that A : V →W is a bijectivelinear map with V and W Banach spaces. If A ∈ L(V,W ) then A−1 ∈ L(V,W ), i.e. if Ais bounded and A−1 exists, then the inverse is bounded as well.

If L : V → V then we say that L is a linear operator on V . We write L(V ) to denotethe set of all bounded linear operators on V . It is an exercise to show that if A,B ∈ L(X)then AB ∈ L(X) and

‖AB‖L(X) ≤ ‖A‖L(X)‖B‖L(X).

Moreover, if A and B are invertible, then AB is invertible and

(AB)−1 = B−1A−1.

(The proof is again left as an exercise). More generally, in infinite dimensions assumingthat AB is invertible does not imply that A and B are invertible seperately. What one hasis the following.

Proposition 6.1.16. Let A ∈ L(Y,X) and B ∈ L(X,Y ) with X and Y Banach spaces.Suppose that AB is invertible. Then B is one-to-on and A is onto.

Proof. Suppose that u ∈ X has that Bu = 0. Then

(AB)u = A(Bu) = A0 = 0,

i.e. u is in the kernel of AB. But AB is one-to-one, and hence ker(AB) = 0. Sou ∈ ker(B) implies that u = 0. Then the kernel of B is trivial and B is one-to-one.

Suppose now that b ∈ X. Since, AB is invertible there is a unique x ∈ X so that

ABx = b.

Let Bx = y ∈ Y . ThenAy = A(Bx) = b.

Since b was arbitrary A is onto.

Now we are especially interested in the invertible elements of L(V ), and the followingtheorem is a fundamental “perturbation” result for linear operators.

Theorem 6.1.17 (The Neumann Series). Let V be a Banach space and A : V → V bea bounded linear operator. If ‖A‖ < 1 then

• I −A is invertible,

•(I −A)−1 =

∞∑n=0

An,


• and

‖(I −A)−1‖ ≤ 1

1− ‖A‖ .

Proof. First note that, since we assume that V is a Banach space, the space L(V ) is alsoa Banach space by Theorem 6.1.13. Define the sequence of partial sums Bn∞n=0 ⊂ L(V )by Bn =

∑nk=0A

k. We will show that the sequence Bn∞n=0 is Cauchy. To see this letε > 0 be given. Choose N ∈ N so that

‖A‖N+1

1− ‖A‖ < ε.

Note that this N exist by the hypothesis that ‖A‖ < 1. Now for m,n ≥ N we assumewithout loss of generality that n > m and have that

Bn −Bm =

n∑k=0

Ak −m∑k=0

Ak =

n∑k=m+1

Ak.

Then

‖Bn −Bm‖ =

∥∥∥∥∥n∑

k=m+1

Ak

∥∥∥∥∥≤

n∑k=m+1

‖Ak‖

≤n∑

k=m+1

‖A‖k

≤∞∑

k=m+1

‖A‖k

≤∞∑k=N

‖A‖k

≤ ‖A‖N+1

1− ‖A‖< ε,

and indeed the sequence is Cauchy. Since L(V ) is complete, there exists a B ∈ L(V ) towhich the partial sums converge, i.e.

B = limN→∞

BN = limN→∞

N∑k=0

Ak =∞∑k=0

Ak,


i.e. the infinite series converges to a bounded linear operator on V .We now claim that B inverts I − A. To see this note first that ‖An‖ ≤ ‖A‖n → 0 as

n→∞, so that An → 0 as n→∞. Now observe that

(I −A)Bn = Bn −ABn =

n∑k=0

Ak −n+1∑k=1

Ak = I −An+1,

and similarly that

Bn(I −A) = Bn −BnA =

n∑k=0

Ak −n+1∑k=1

Ak = I −An+1.

Then taking limits yields that(I −A)B = I.

and thatB(I −A) = I.

So (I −A) is invertible with inverse B =∑∞

n=0An.

Finally we note that

‖(I −A)−1‖ =

∥∥∥∥∥∞∑n=0

An

∥∥∥∥∥ ≤∞∑n=0

‖An‖ ≤∞∑n=0

‖A‖n =1

1− ‖A‖ ,

by applying the geometric series for the number ‖A‖ < 1, yielding the desired estimate.

As in finite dimensions, the Neumann series is useful when we want to show that asmall perturbation of a boundedly invertible linear operator is boundedly invertible. Thefollowing gives a general result which follows from the Neumann series.

Corollary 6.1.18. Let X be a Banach space and L(X) denote the set of all bounded linearoperators on X (i.e. linear maps from X into itself). Define the set

G(X) :=A ∈ L(X) |A is invertible and A−1 ∈ L(X)

.

Then G(X) ⊂ L(X) is open.

Proof. Let A : X → X be a bounded linear operator. Suppose in addition that A isboundedly invertible, i.e. that A−1 exists and is a bounded linear operator. Let

R =1

2‖A‖−1<∞,

and define the open ball

BR(A) =B ∈ L(X) | ‖A−B‖L(X) < R

,


and note that any B ∈ BR(A) can be written as B = A + H, where ‖H‖L(X) < R. ThenA−1H and (−1)A−1H are bounded linear operators, and we have

‖(−1)A−1H‖L(X) ≤ ‖A−1‖L(X)‖H‖L(X) ≤ ‖A−1‖L(X)R ≤1

2< 1.

The Neumann series shows that the operator I + A−1H = I − ((−1)A−1H) is boundedlyinvertible as

‖I +A−1H‖L(X) = ‖I − ((−1)A−1H)‖L(X)

≤ 1

1− ‖(−1)A−1H‖L(X)

≤ 1

1− ‖A−1‖L(X)‖H‖L(X)

≤ 1

1− 1/2

= 2 <∞.

So

B = A+H

= A+ IH

= A+(AA−1

)H

= A(I +A−1H).

Since A and I +A−1H are both invertible we have that[A(I +A−1H)

]−1= (I +A−1H)−1A−1 = B−1,

and moreover that

‖B−1‖L(X) ≤‖A−1‖L(X)

1− ‖A−1‖L(X)‖H‖L(X)< 2‖A−1‖L(X) <∞.

Then B ∈ G(X). But B ∈ BR(A) was arbitrary, so

BR(A) ⊂ G(X).

As A was arbitrary we have that G(X) is open.

An important special case of linear maps from V to W arises when W = R (or W = Cif V is a vector space over C). In this case the linear map is called a linear functional on V .The set of all bounded linear functionals on V is called the dual space of V and is denotedby V ∗. By Theorem 6.1.13 V ∗ is a Banach space.

6.2. THE MAIN EXAMPLES: BANACH SPACES OF INFINITE SEQUENCES 103

Example 6.1.19. Consider the linear functional ` : C0([a, b])→ R defined by

`(f) =

∫ b

af(x) dx.

It is an exercise to show that ` ∈ C0([a, b])∗, i.e. that ` is bounded.

We are often interested in building new Banach spaces out of known ones. Especiallyimportant is the direct product of Banach spaces. If V1, . . . , Vn are Banach spaces we willbe interested in the Cartesian product of these spaces, which it turns out is again a Banachspace. More precisely let

V n := V1 ⊕ . . .⊕ Vn = (v1, . . . , vn)|vj ∈ Vjfor1 ≤ j ≤ n ,

For v = (v1, . . . , vn) ∈ V n define

‖v‖ = max1≤j≤n

‖vj‖Vj .

We leave it as an exercise to verify that V n is a Banach space when endowed with thisnorm.

Suppose that Aij : Vj → Vi are bounded linear operators and consider the linear oper-ator A : V n → V n defined by

(Av)i =n∑j=1

Aijvj ,

for 1 ≤ i ≤ n. We can think of this as a matrix of operators

A =

A11 . . . A1n...

. . ....

An1 . . . Ann

,

and the reader can verify that

‖A‖ ≤ max1≤i≤n

n∑j=1

‖Aij‖B(Vj ,Vi).

6.2 The Main Examples: Banach Spaces of Infinite Sequences

While there are a wide variety of Banach spaces, those that arise from infinite sequencesare of particular interest to us. We write a = ann∈N and a = ann∈Z to to denoterespectively one and two sided infinite sequences of complex numbers. We are interestedin sequences with various growth or decay properties. More precisely we will measure


growth or decay by asking wether or not the sequence is absolutely summable, perhapsafter assigning appropriate “growth rates” to the terms of the sequence.

Let SN(C) denote the set of all one sided infinite sequences of complex numbers. Byone sided, we mean that the sequence is indexed by the natural numbers N. A one sidedsequence ω = ωn∞n=0 ⊂ R is said to be a sequence of admissible weights if

ωn > 0,

for all n ∈ N. With a ∈ SN(C) let

‖a‖1,ω,N :=

∞∑n=0

|an|ωn,

denote the weighted sum and

‖a‖∞,ω,N := supn≥0|an|ωn,

denote the weighted supremum of the sequence a. Similarly, if SZ(C) denotes the set ofall two sided infinite sequences of complex numbers, we say that a two sided sequenceω = ωnn∈Z ⊂ R is again an admissible sequence of weights if ωn > 0, for all n ∈ Z. Witha ∈ SZ(C) let

‖a‖1,ω,Z :=∑n∈Z|an|ωn,

denote the weighted sum, and

‖a‖∞,ω,Z := supn∈Z|an|ωn,

the weighted supremum. Note that SN(C) and SZ(C) are both vector spaces. In fact wehave the following.

Lemma 6.2.1. Suppose that ω is an admissible weight sequence. The weighted sum ‖·‖1,ω,Ndefines a norm on SN(C).

Proof. We must verify the three properties of a norm. Namely we must show that

•‖a‖1,ω,N ≥ 0,

and that ‖a‖ = 0 if and only if a = 0.

• That‖αa‖1,ω,N = |α|‖a‖1,ω,N,

for all a ∈ SN(C) and all α ∈ C,


• and that

‖a+ b‖1,ω,N ≤ ‖a‖1,ω,N + ‖b‖1,ω,N,

for all a, b ∈ SN(C).

For the first property let a ∈ SN(C). Then

‖a‖1,ω,N =∞∑n=0

|an|ωn ≥ 0,

simply because this is a sum of positive terms. Moreover, if a = 0, i.e. if an = 0 for alln ∈ N, then

‖a‖1,ω,N =∞∑n=0

|0|ωn = 0.

On the other hand, if a 6= 0 then there is a j ∈ N so that aj 6= 0, in which case |aj |ωj > 0.Then

‖a‖1,ω,N =

∞∑n=0

|an|ωn > |aj |ωj > 0,

which establishes the first property of the norm. (Note that this is precisely the momentwhere admissibility of the sequence ω is used).

For the second property, let a ∈ SN(C) and α ∈ C. Then

‖αa‖1,ω,N =∞∑n=0

|αan|ωn

=

∞∑n=0

|α||an|ωn

= |α|∞∑n=0

|an|ωn

= |α|‖a‖1,ω,N.

Note that these equalities holds even if ‖a‖1,ω,N =∞.

For the final property, let a, b ∈ SN(C). Suppose that one of the sequences has divergentweighted sum, so without loss of generality assume that ‖a‖1,ω,N =∞. Then

‖a+ b‖1,ω,N =∞∑n=0

|an + bn|ωn ≤ ∞ = ‖a‖1,ω,N ≤ ‖a‖1,ω,N + ‖b‖1,ω,N.


If instead both of the sums converge then we have

‖a+ b‖1,ω,N =∞∑n=0

|an + bn|ωn

≤∞∑n=0

|an|ωn + |bn|ωn

=∞∑n=0

|an|ωn +∞∑n=0

|bn|ωn

= ‖a‖1,ω,N + ‖b‖1,ω,N,

by the elementary properties of convergent sums.

Remark 6.2.2. The weighted supremum ‖ · ‖∞,ω,N defines a norm on SN(C), as do theweighted sums and supremums ‖ · ‖1,ω,Z, and ‖ · ‖∞,ω,Z on SZ(C). We leave the proofs asexercises.

We are especially interested in linear subspaces of SN(C) and SZ(C) comprised of se-quences which have finite norm. Then for any admissible sequence ω we write

`1ω,N := a ∈ SN(C) | ‖a‖1,ω,N <∞ ,

and`∞ω,N := a ∈ SN(C) | ‖a‖∞,ω,N <∞ ,

for the normed linear spaces of sequences with finite weighted sums and supremums, andwrite

`1ω,Z := a ∈ SZ(C) | ‖a‖1,ω,Z <∞ ,and

`∞ω,Z := a ∈ SZ(C) | ‖a‖∞,ω,Z <∞ ,for two sided sequences.

Example 6.2.3. In Chapter (REFERENCE CH 7) we are especially interested in thefollowing examples. Choose any ν > 0 and, for one sided sequences, consider the weights

ωn = νn,

and

ω−1n =

1

νn.

Note that both sequences are admissible. In a slight abuse of notation we write ω to denotethe first sequence and ω−1 to denote the second. We write

‖a‖1,ν,N := ‖a‖1,ω,N =

∞∑n=0

|an|νn,


to denote the weighted sum, and similarly

‖a‖∞,ν,N := ‖a‖∞,ω−1,N = supn≥0

|an|νn

,

to denote the weighted supremum. We write

`1ν,N := `1ω,N,

and

`∞ν,N := `∞ω−1,N,

to denote the spaces associated with these norms. Note that the left hand side of theprevious line is a definition, and we choose not to write ν−1 in this definition.

Example 6.2.4. The following examples play a key role in Chapter (REFERENCE CH8). For ν ≥ 1 consider the weights

ωn = ν|n|,

and

ω−1n =

1

ν|n|.

These are admissible weights. As above write ω to denote the first sequence and ω−1 todenote the second. Let

‖a‖1,ν,Z := ‖a‖1,ω,Z =∑n∈N|an|ν|n|,

denote the weighted sum, and similarly

‖a‖∞,ν,Z := ‖a‖∞,ω−1,N = supn∈Z

|an|ν|n|

,

to denote the weighted supremum.

We write

`1ν,Z := a ∈ SZ(C) | ‖a‖1,ν,Z <∞ ,and

`∞ν,Z := a ∈ SZ(C) | ‖a‖∞,ν,Z <∞ ,to denote the spaces associated with these norms.

Example 6.2.5. The following examples are considered in Chapter (REFERENCE BVP).For K ∈ N consider the admissible weights

ωn = (nK + 1),


and

ω−1n =

1

nK + 1.

As above write ω to denote the first sequence and ω−1 to denote the second. Let

‖a‖1,K,Z := ‖a‖1,ω,Z =∑n∈N

(|n|K + 1)|an|,

denote the weighted sum, and similarly

‖a‖∞,K,Z := ‖a‖∞,ω−1,N = supn∈Z

|an||n|K + 1

,

to denote the weighted supremum.

We write

`1K,Z := a ∈ SZ(C) | ‖a‖1,K,Z <∞ ,

and

`∞K,Z := a ∈ SZ(C) | ‖a‖∞,K,Z <∞ ,

to denote the spaces associated with these norms.

The weighted sequences spaces defined so far are the main examples of Banach spacesused throughout the remainder of the book. We include the proof of the following theoremfor the sake of completeness. Other cases are left as exercises.

Theorem 6.2.6. Suppose that ω is an admissible one sided sequence of weights. Then`1ω,N is a Banach space.

Proof. Note that `1ω,N ⊂ SN(C) is closed under addition and scalar multiplication. Then

`1ω,N is a vector subspace of SN(C), hence inherits the normed linear structure. Whatremains is to show completeness under ‖ · ‖1,ω,N norm.

To this end, choose any Cauchy sequence an∞n=0 ⊂ `1ω,N. We want to prove thatthis sequence has a limit in our space. The proof has three steps. First, we constructa candidate sequence i.e. a guess as to the value of the limit. Next, we prove that ourcandidate is indeed in `1ω,N. Finally, we show that the Cauchy sequence an∞n=0 converges

to our guess in the `1ω,N norm.

To begin we remark that, since an∞n=0 is an infinite sequence consisting of points


which are themselves infinite sequences, a little notation is in order. We write

a0 = a(0)k ∞k=0 = a(0)

0 , a(0)1 , a

(0)2 , . . .

a1 = a(1)k ∞k=0 = a(1)

0 , a(1)1 , a

(1)2 , . . .

a2 = a(2)k ∞k=0 = a(2)

0 , a(2)1 , a

(2)2 , . . .

...

an = a(n)k ∞k=0 = a(n)

0 , a(n)1 , a

(n)2 , . . .

...

to denote the individual terms of the sequence. Note that for any k ∈ N we have that

|a(m)k ωk − a(n)

k ωk| = |a(m)k − a(n)

k |ωk ≤∞∑j=0

|a(m)j − a(n)

j |ωj = ‖am − an‖1,ω,N,

with the later Cauchy. Then for each fixed k ∈ N the sequence a(n)k ωk∞n=0 is a Cauchy

sequence of numbers. Since ωk is constant, it follows that the sequence a(n)k ∞n=0 is itself

Cauchy in C. Then a(n)k ∞n=0 has a limit in C, which we denote by

ak := limn→∞

a(n)k .

Collecting these results, one for each k ∈ N, defines a new infinite sequence a = ak∞k=0 ∈SN(C). This is our candidate.

To see that the norm of a is bounded let ε = 1. Since an∞n=0 ⊂ `1ω,N is Cauchy, thereis a K ∈ N so that for all n,m ≥ K we have

‖an − am‖1,ω,N < 1.

Choose n ≥ K. Then for any fixed N ∈ N we have that

N∑k=0

|a(n)k |ωk ≤

∞∑k=0

|a(n)k − a

(K)k |ωk +

∞∑k=0

|a(K)k |ωk

= ‖an − aK‖1,ω,N + ‖aK‖1,ω,N< 1 + ‖aK‖1,ω,N<∞,


as aK ∈ `1ω,N, by hypothesis. The estimate is uniform in n, and since each of the sequences

a(n)k ∞n=0 converge, we have also that

limn→∞

N∑k=0

|a(n)k |ωk =

N∑k=0

limn→∞

|a(n)k |ωk

=N∑k=0

|ak|ωk

< 1 + ‖aK‖1,ω,N.

Here there is no issue commuting the limit with the sum, as the sum is finite. Noting thatthe terms in the sum are positive, we have that the partial sums satisfy

sN :=N∑k=0

|ak|ωk < 1 + ‖aK‖1,ω,N,

for eacd N ∈ N; hence sN∞N=0 is a bounded and increasing sequence of real numbers, sothat (by the least upper bound axiom of R) sN∞N=0 converges to its least upper bound.Moreover, since the the limit/least upper bound is less than or equal to any upper bound,and we retain the estimate

‖a‖1,ω,N =∞∑k=0

|ak|ωk < 1 + ‖aK‖1,ω,N <∞,

i.e. a ∈ `1ω,N as desired.

The last step in the proof is to show that the Cauchy sequence actually converges to a.So, given ε > 0 choose K ∈ N so that ‖an − am‖1,ω,N < ε for all m,n ≥ K. Then, for eachfixed N ≥ 0 we have that

N∑k=0

|a(m)k − a(n)

k |ωk ≤∞∑k=0

|a(m)k − a(n)

k |ωk = ‖an − am‖1,ω,N < ε,

a bound uniform in N , as well as in m,n ≥ K. Since the sequencesa

(m)k

∞m=0

converge


for every k ∈ N we take the limit and have that

limm→∞

N∑k=0

∣∣∣a(m)k − a(n)

k

∣∣∣ωk =N∑k=0

limm→∞

∣∣∣a(m)k − a(n)

k

∣∣∣ωk=

N∑k=0

∣∣∣ limm→∞

a(m)k − a(n)

k

∣∣∣ωk=

N∑k=0

∣∣∣ak − a(n)k

∣∣∣ωk≤ ε.

Again, these limit operations are justified as the sum is finite and the absolute value is acontinuous function. Once more, the partial sums

sN :=N∑k=0

∣∣∣ak − a(n)k

∣∣∣ωk < ε,

form an increasing and bounded sequence of positive real numbers, so that the limit asN →∞ exists and satisfies the same bound. In other words we have that

‖a− an‖1,ω,N < ε,

so that

limn→∞

an = a,

as desired. Since the Cauchy sequence was arbitrary, it follows that `1ω,N is complete.

Recall that a basis for a vector space V is a collection of vectors vααI (here I is someunspecified index set) with the property that for any vector v ∈ V there are constantsc1, . . . , cN and indices α1, . . . , αN ∈ I (where N may depend on v) so that

v = c1vα1 + . . .+ cNvαN ,

i.e. any v ∈ V can be expressed as a finite linear combination of elements of vααI . Thecollection vααI is also referred to as a Hamel basis for V . While it can be shown (usingthe Axiom of choice) that every vector space has a Hamel basis, in specific applications itis often difficult if not impossible to describe a Hamel basis explicitly.

Instead one typically weakens the notion of a basis, and asks only that vectors can beexpressed as countable sum of basis vectors. The following definition makes this notionprecise for Banach spaces.


Definition 6.2.7. Let V be a complex Banach space. A countable collection vn∞n=0 ⊂ Vis called a Schauder basis for V if for each v ∈ V there is a unique collection of complexnumbers an∞n=0 so that

limn→∞

∥∥∥∥∥v −n∑k=0

anvn

∥∥∥∥∥V

= 0.

In this case we write

v =∞∑n=0

anvn.

The next theorem constructs an explicit Schauder basis for `1ω,N.

Theorem 6.2.8. Define the standard basis vectors en ∈ `1ω,N by

(en)k =

1 if k = n

0 otherwise.

Then the collection en∞n=0 is a Schauder basis for `1ω,N.

Proof. Note that

e0 = 1, 0, 0, . . .e1 = 0, 1, 0, . . .e2 = 0, 0, 1, . . .,

and so on. Then for any a = an∞n=0 ∈ `1ν,N we can write

N∑k=0

akek = a0 1, 0, 0, . . . , + a1 0, 1, 0, . . .+ . . .+ aN 0, 0, 0, . . . , 0, 1, 0

= a0, . . . , aN , 0, 0, 0, . . . ,

so that

a−N∑k=0

anen = 0, . . . , 0, aN+1, aN+2, . . . . (6.1)

Now,∞∑n=0

|an|ωn <∞,

as a ∈ `1ω,N, which implies that

limN→∞

∞∑n=N

|an|ωn = 0.


So, let ε > 0 be given and choose N ∈ N so that n ≥ N implies

∞∑m=n

|am|ωm < ε, (6.2)

Now, taking n ≥ N and combining Equation (6.1) with Equation (6.2), we have that∥∥∥∥∥a−n−1∑k=0

akek

∥∥∥∥∥1,ω,N

= ‖0, . . . , 0, an, an+1, an+2, . . .‖1,ω,N

=∞∑k=n

|ak|ωk ≤ ε,

which is to say that

limn→∞

∥∥∥∥∥a−n∑k=0

akek

∥∥∥∥∥1,ω,N

= 0.

But (by definition 6.2.7) this is exactly what it means to say that

∞∑n=0

anen = a.

Remark 6.2.9. Consider the two sided sequences en ∈ `1ω,Z defined by

(en)k =

1 if k = n

0 otherwise.

Then the countable collection enn∈Z is a Schauder basis for `1ω,Z. We leave the proof asan exercises.

Remark 6.2.10. Consider the set E ⊂ `1ω,N defined by

E :=

a ∈ `1ω,N | a =

∞∑n=0

qnen, with qn∞n=0 ⊂ Q

.

It is an exercise to show that E is countable and dense in `1ω,N. Then `1ω,N is a separable

metric space for any choice of admissible weights ωn∞n=0. A similar result holds for `1ω,Z,thanks to Remark 6.2.9.


Remark 6.2.11. It is worth remarking that the spaces `∞ω,N and `∞ω,Z are not always separable.Choose for example the admissible sequence of weights ω = ωn∞n=0 with constant termsgiven by ωn = 1, for n ∈ N. The resulting weighted sequence space is usually denoted by`∞ω,N =: `∞, and we write ‖ · ‖∞ to denote the norm.

To see that `∞ is not separable, consider the subset V ⊂ `∞ given by

V = a ∈ `∞ | an = 0, 1 ,

i.e. let V be the collection of all infinite binary sequences. Now, define the map R : V → Rby

R(a) =

∞∑n=0

an2n.

Note that the value R(a) ∈ [0, 1] and has binary expansion

R(a) = (0.a0a1a2 . . .),

i.e. R associates with any a ∈ V the real number x ∈ [0, 1] whose binary representation isgiven by the sequence an∞n=0. Note that every x ∈ [0, 1] has a binary representation, andthat any x, y ∈ [0, 1] with x 6= y have different binary representations. Then since [0, 1] isuncountable the set R(V ) is uncountable. Moreover, if a, b ∈ V and a 6= b then

‖a− b‖∞ = 1.

Now, for each a ∈ V we construct a small ball, say for example

B1/10(a) = b ∈ `∞ | ‖b− a‖∞ < 1/10 .

Note that none of the balls intersect, and that there are uncountable many of them.Finally, take M any dense subset of `∞. Then each B1/10(a), being open, contains a

point in M . Since there are uncountable many balls, none of them intersect, and eachcontains a point in M we have that

card(M) ≥ card (V ) = card(R(V )) ≥ card(R) > card(N),

i.e. M cannot be countable. Hence `∞ has no countable dense subset, i.e. it is not aseparable Banach space.

More generally, suppose that ωn∞n=0 is a sequence of admissible weights with theproperty that supn∈N ωn < ∞. Then, by imitating the construction above, one can showthat `∞ω,N is not separable.

If on the other hand ω is an admissible sequence of weights with ωn → ∞ as n → ∞,then

‖a‖∞,ω,N <∞,implies that an → 0 as n → ∞, and one can now show that `∞ω,N is separable. Similarremarks hold for two sided sequences in `∞ω,Z. We leave the details for the exercises.


Definition 6.2.12. Two Banach spaces (X, ‖ · ‖X) and (X, ‖ · ‖X) are isomorphic if thereexists a linear bijection J : X → Y such that J and its inverse J−1 are continuous. Wesay that X and Y are isometrically isomorphic if in addition, J is an isometry, that is‖J(x)‖Y = ‖x‖X for every x ∈ X.

The following Theorem provides a useful characterization of the Banach space dual ofa space of summable sequences. We state the theorem for one sided sequences and leavethe generalization to two sided sequences as an exercise.

Theorem 6.2.13. Let ω = ωn∞n=0 be an admissible sequence of weights and write ω−1 :=ω−1

n ∞n=0 (which is itself an admissible sequence of weights). Then `∞ω−1,N and (`1ω,N)∗ areisometrically isomorphic.

In less precise language, we say that `∞ω−1,N is the dual of `1ω,N. The following estimateis used in the proof of Theorem 6.2.13.

Lemma 6.2.14. If a ∈ `1ω,N and c ∈ `∞ω−1,N, then∣∣∣∣∣∞∑n=0

cnan

∣∣∣∣∣ ≤ ‖c‖∞,ω−1,N‖a‖1,ω,N.

Proof. Given c ∈ `∞ω−1,N and k ≥ 0 we have, by definition of the weighted supremum norm,that

|ck|ωk≤ ‖c‖∞,ω−1,N.

Therefore, ∣∣∣∣∣∞∑k=0

ckak

∣∣∣∣∣ ≤∞∑k=0

|ck||ak|

≤∞∑k=0

(‖c‖∞,ω−1,N ωk

)|ak|

= ‖c‖∞,ω−1,N

∞∑k=0

|ak|ωk

= ‖c‖∞,ω−1,N‖a‖1,ω,N.

Proof of Theorem 6.2.13. We begin by constructing an invertible linear map J : `∞ω−1,N →(`1ω,N)∗. Given an element c ∈ `∞ω−1,N, define a linear functional on `1ω,N by

Qc(a) :=∞∑k=0

ckak.


Then letJc := Qc.

We leave it as an exercise to check that J is an linear transformation.To see that Qc is bounded (and hence that J is well defined) note that

‖Qc‖ := sup‖a‖1,ω,N=1

|Qc(a)|

= sup‖a‖1,ω,N=1

∣∣∣∣∣∞∑k=0

ckak

∣∣∣∣∣≤ sup‖a‖1,ω,N=1

‖c‖∞,ω−1,N‖a‖1,ω,N

≤ ‖c‖∞,ω−1,N, (6.3)

where we have used Lemma 6.2.14. Then Qc ∈ (`1ω,N)∗ as desired.The argument that J is continuous is similar. We check that J is a bounded linear

operator, and find that

‖J‖L(`∞ω−1,N

,(`1ω,N)∗) = sup

‖c‖=1‖Jc‖(`1ω,N)∗

= sup‖c‖=1

‖Qc‖(`1ω,N)∗

= sup‖c‖=1

sup‖a‖=1

|Qc(a)|

≤ sup‖c‖=1

sup‖a‖=1

‖c‖∞,ω−1,N‖a‖1,ω,N

≤ 1, (6.4)

where we used the bound of Equation (6.3) to obtain the second to last inequality.To see that J is injective, we show that the kernel of J is trivial. So, suppose that

c ∈ `∞,ω−1,N has that

Jc = 0 ∈ (`1ω,N)∗.

This means that Jc : `1ω,N → C is the zero map, or to be more precise

(Jc)(a) =

∞∑k=0

akck = 0,

for all a ∈ `1ω,N. By choosing an = en the standard basis vectors for `1ω,N we obtain theinfinitely many equations

(Jc)(en) =∞∑k=0

(en)kck = encn = cn = 0.


Then c ∈ ker(J) implies that cn = 0 for all n ∈ N, i.e. c = 0 ∈ `∞ω−1,N.

Next, we need to show that J is surjective, i.e. given l ∈ (`1ω,N)∗ we need to producec ∈ `∞ω−1,N such that Jc = l. Here we adapt the useful idea from linear algebra that a linear

operator is determined by its action on a basis. Since ek∞k=0 is a Schauder basis for `1ω,N(by Theorem 6.2.8) we can, given given a ∈ `1ω,N, write

l(a) = l

( ∞∑k=0

akek

)

= l

(limN→∞

N∑k=0

akek

)

= limN→∞

l

(N∑k=0

akek

)

= limN→∞

N∑k=0

akl(ek)

=

∞∑k=0

akl(ek).

Here the Schauder basis lets us write a as the limit of a sequence (of partial sums) in`1ω,N, and the fact that l is bounded, hence continuous, hence sequentially continuous, letsus move the limit outside the argument of l. Linearity of l then lets us distribute l overthe finite sum. Note also that the convergence of the infinite sum in the last line of thecalculation is part of the conclusion, i.e. since the second to last line is equal to l(a) in thefirst line, the limit necessarily exists. We define c = cn := l (en)n∈N, and hope to showthat Jc = l.

First we need to check that c ∈ `∞ω−1,N. The assumption that l ∈ (`1ω,N)∗, combined withthe bound of Proposition 6.1.8, implies that

|l(a)| ≤ ‖l‖‖a‖1,ω,N <∞,

for all a ∈ `1ω,N, with ‖l‖ the operator norm. Note also that for any n ∈ N, we have

‖en‖1,ω,N =

∞∑k=0

|enk |ωk =

∞∑k=0

|δnk |ωk = ωn <∞.

Then|cn|ωn

=|l(en)|ωn

≤ ‖l‖‖en‖1,ω,Nωn

=‖l‖ωnωn

= ‖l‖ <∞, (6.5)


so that

‖c‖∞,ω−1,N = supn∈N

|cn|ωn≤ ‖l‖ <∞, (6.6)

i.e. c ∈ `∞ω−1,N. Then Jc = l. But l ∈ (`1ω,N)∗ arbitrary, and it follows that J is surjective.This proves that J is a continuous linear isomorphism. But then, by the bounded inversetheorem 6.1.15, J−1 is bounded and hence continuous as well.

The final step is to show that J is an isometry. First, note that

‖Jc‖ ≤ ‖J‖‖c‖∞,ω−1,N ≤ ‖c‖∞,ω−1,N, (6.7)

as, recalling the bound in Equation (6.4), ‖J‖ ≤ 1. Next, let c ∈ `∞ω−1,N and Qc = Jc ∈(`1ω,N)∗. Then, since for each k ∈ N we have that

Qc(ek) = ck,

repeating the estimates of Equations (6.5) and (6.6) with l(a) = Qc(a) leads to

‖c‖∞,ω−1,N ≤ ‖Qc‖ = ‖Jc‖. (6.8)

Combining the estimates of Equation (6.7) and Equation (6.8) we have

‖c‖∞,ω−1,N ≤ ‖Qc‖ = ‖Jc‖ ≤ ‖c‖∞,ω−1,N,

and hence‖Jc‖ = ‖c‖∞,ω−1,N.

When working with Banach spaces of infinite sequences, the dual space provides aconvenient means for studying more general linear operators. Let πn : `1ω,N → C denote the

canonical projection maps; that is, given b = bn∞n=0 ∈ `1ω,N, πnb = bn. Then a linear map

A : `1ω,N → `1ω,N can be expressed as

(πn A)bn∈N = [Ab]nn∈N .If we now assume that A is bounded, then each individual linear functional πn A isbounded, and can be expressed uniquely as

[Ab]n = an · b =∞∑k=0

an,kbk, (6.9)

where an := an,kk≥0 ∈ `∞ω−1,N.

Proposition 6.2.15. Let A : `1ω,N → `1ω,N a bounded linear operator. For any n ≥ 0,consider an = an,kk≥0 ∈ `∞ω,−1N satisfying (6.9). Then

‖A‖ ≤∞∑n=0

(‖an‖∞,ω−1,N

)ωn. (6.10)


Proof. For b = bnn∈N ∈ `1ω,N, observe that

‖Ab‖1,ω,N =∞∑n=0

|[Ab]n|ωn =∞∑n=0

∣∣∣∣∣∞∑k=0

an,kbk

∣∣∣∣∣ωn≤∞∑n=0

‖an‖∞∞,ω−1,N‖b‖1,ω,Nωn = ‖b‖1,ω,N∞∑n=0

‖an‖∞,ω−1,Nωn,

where the inequality follows from Lemma 6.2.14. Thus

‖A‖ = sup‖b‖1,ω,N=1

‖b‖1,ω,N∞∑n=0

‖an‖∞,ω−1,Nωn =∞∑n=0

‖an‖∞,ω−1,Nωn.

Proposition 6.2.16. Consider a linear operator A : `1ω,N → `1ω,N of the form

A =

AN 0

cN+1

cN+2

0. . .

where AN =[am,n

]m,n∈0,1,...,N ∈MN+1(C) and cn ⊂ C.

If there exists a constant C such that |cn| < C, for all n ≥ N + 1, then

‖A‖ ≤ max(K,C)

where

K := max0≤n≤N

1

ωn

N∑m=0

|am,n|ωm. (6.11)


Proof. Given b = bnn∈N ∈ `1ω,N, let bF = (b0, . . . , bN )T . By definition

‖A‖ = sup‖b‖1,ω,N=1

‖Ab‖1,ω,N

= sup‖b‖1,ω,N=1

∥∥∥∥∥∥∥∥∥

AN

cN+1

cN+2

. . .

bFbN+1

bN+2...

∥∥∥∥∥∥∥∥∥

1,ω,N


(N∑m=0

∣∣∣∣∣N∑n=0

am,nbn

∣∣∣∣∣ωm +∞∑

n=N+1

|cnbn|ωn)


(N∑n=0

∣∣∣∣∣N∑m=0

am,nωmbn

∣∣∣∣∣+∞∑

n=N+1

|cnbn|ωn)

≤ sup‖b‖1,ω,N=1

N∑n=0

(N∑m=0

|am,n|ωm)|bn|+

∑n≥N+1

|cn|ωn|bn|

. (6.12)

For n ≥ 0, define

cn =

∑Nm=0 |am,n|ωm if n ≤ N

|cn|ωn if n ≥ N + 1.

Observe that c := cnn∈N ∈ `∞ω−1,N since

‖c‖∞,ω−1,N := supn∈N

|cn|ωn

= max(K,C).

By (6.12)

‖A‖ ≤ sup‖b‖1,ω,N=1

∑n∈N|cn||bn| ≤ sup

‖b‖1,ω,N=1‖c‖∞,ω−1,N‖b‖1,ω,N = ‖c‖∞,ω−1,N

where the latter inequality follows from Lemma 6.2.14.

6.3 Calculus on Banach Spaces

Definition 6.3.1. Let V,W be Banach spaces. A function F : V → W is Frechet differ-entiable at x0 ∈ V if there exists a bounded linear operator A : V →W satisfying

limh→0

‖F (x0 + h)− F (x0)−Ah‖W‖h‖V

= 0.

The linear operator A is called the derivative of F at x0 and denoted by A = DF (x0).

6.3. CALCULUS ON BANACH SPACES 121

Example 6.3.2. Consider the Banach space V = C([a, b],R) and define the mappingF : V → V by

F [u](t) =

∫ t

au2(x) dx.

Note that for any u ∈ V the function F [u] is also continuous. Then F maps the set ofcontinuous functions back into itself. The important point is that we are considering thevariable in the function F to be u. But u is itself a continuous function, and the result ofcomputing F [u] is a new continuous function.

We claim that for any u0 ∈ V the derivative of F at u0 is given by

DF (u0)h = 2

∫ t

au0(x)h(x) dx.

To see this consider

F [u0 + h]− F [u0] =

∫ t

a[u0(x) + h(x)]2 dx−

∫ t

au2

0(x) dx

=

∫ t

a

(u2

0(x) + 2u0(x)h(x) + h2(x))dx−

∫ t

au2

0(x) dx

= 2

∫ t

au0(x)h(x) dx+

∫ t

ah2(x) dx.

This computation suggests the correct guess. Now we check that

limh→0

‖F (u0 + h)− F (u0)−DF (u0)h‖‖h‖ = lim

h→0

1

‖h‖

∥∥∥∥∫ t

ah2(x) dx

∥∥∥∥≤ lim

h→0

1

‖h‖(b− a)‖h‖2

≤ 0.

The proof of the following result is left as an Exercise.

Proposition 6.3.3. Let X be a Banach space. If A : X → X is a bounded linear operatorand F : X → X is Frechet differentiable at x0 ∈ X, then the mapping G : X → X definedby G(x) = AF (x) is Frechet differentiable at x0 and DG(x0) = ADF (x0).

An important theorem is the following Mean Value Theorem, which is used in order toobtain Lipschitz bounds for smooth functions.

Theorem 6.3.4. [Mean Value Theorem] Let V,W be Banach spaces. Let u0 ∈ Vand suppose that F : Br(u0) ⊂ V → W is a C1 mapping, i.e. we assume that for everyv ∈ Br(u0) the Frechet derivative DF (v) exists and is a bounded linear operator. Let

K := supv∈Br(u0)

‖DF (v)‖.


Then for any u, v ∈ Br(u0) we have that

‖F (u)− F (v)‖W ≤ K‖u− v‖V .

Proof. Mimic the proof of Theorem 1.225 and Theorem 1.226 on pages 113 and 114 ofChicone.

6.4 Banach Algebra

Definition 6.4.1. A Banach algebra is a Banach space X with a multiplication operation∗ : X ×X → X that satisfies

x ∗ (y ∗ z) = (x ∗ y) ∗ z (6.13)

(x+ y) ∗ z = x ∗ z + y ∗ z, x ∗ (y + z) = z ∗ y + x ∗ z, (6.14)

α(x ∗ y) = (αx) ∗ y = x ∗ (αy) (6.15)

and

‖x ∗ y‖ ≤ ‖x‖‖y‖, (6.16)

for all x, y, z ∈ X and all scalars α. The Banach algebra is commutative if x ∗ y = y ∗ x,for all x, y ∈ X.

A unit element e of a Banach algebra X satisfies

x ∗ e = e ∗ x = x

for all x ∈ X. Observe that if a unit element exists, then it is unique.

The importance of (6.16) is that it implies that ∗ is a continuous operation.

Proposition 6.4.2. Let X be a Banach algebra. If xn → x and yn → y, then xn∗yn → x∗y.

Proof. Observe that

limn→∞

‖xn ∗ yn − x ∗ y‖ = limn→∞

‖(xn − x) ∗ yn + x ∗ (yn − y)‖≤ lim

n→∞‖(xn − x) ∗ yn‖+ lim

n→∞‖x ∗ (yn − y)‖

≤ limn→∞

‖xn − x‖‖yn‖+ limn→∞

‖x‖‖yn − y‖= 0.

In addition to continuity, we have that certain operations in a Banach algebra arealways Frechet differentiable.

6.4. BANACH ALGEBRA 123

Theorem 6.4.3. Let X be a Banach algebra and fix c ∈ X. Define the maps F,G : X → Xby

F (x) = c ∗ x,and

G(x) = x ∗ x.Then F and G are Frechet differentiable on X. Moreover

DF (x)h = c ∗ h, (6.17)

and

DG(x)h = x ∗ h+ h ∗ x, (6.18)

for all h ∈ X.

Proof. The formula for the derivative of F given in Equation (6.17) follows as soon as weobserve that, for fixed c ∈ X, F is a linear map on X.

For the derivative of G start by letting x, h ∈ X, and let A : X → X be the map definedby

Ah = x ∗ h− h ∗ x.Note that A is linear in h and that A ∈ L(X) (by the Banach algebra bound). Moreover,we have that

G(x+ h)−G(x)−Ah = G(x+ h)−G(x)− (x ∗ h+ h ∗ x)

= (x+ h) ∗ (x+ h)− (x ∗ x)− x ∗ h− h ∗ x= h ∗ h.

Then for any h 6= 0 we have that

‖G(x+ h)−G(x)−Ah‖‖h‖ =

‖h ∗ h‖‖h‖

≤ ‖h‖‖h‖‖h‖= ‖h‖.

Noting that

limh→0‖h‖ = 0,

(by the continuity of the norm), and that for h 6= 0

0 ≤ ‖G(x+ h)−G(x)−Ah‖‖h‖ ≤ ‖h‖,


we have that

limh→0

‖G(x+ h)−G(x)−Ah‖‖h‖ = 0,

by the squeeze theorem. ThenDG(x) = A,

as desired.

Remark 6.4.4. Note that if X is a commutative Banach algebra then Theorem 6.4.3 givesthat

D(x ∗ x)h = 2(x ∗ h),

generalizing the formulad

dxx2 = 2x,

from Calculus of a single real variable.

Example 6.4.5. Let X be a Banach space and L(X) be the set of all bounded linearoperators on X. Note that if A,B ∈ L(X) then the composition AB ∈ L(X) and

‖AB‖L(X) ≤ ‖A‖L(X)‖B‖L(X).

Define the product ∗ : L(X)× L(X)→ L(X) by

(A ∗B)x = ABx,

for all x ∈ X, i.e. A ∗ B is operagtor composition. Then the pair (L(X), ∗) is a Banachalgebra.

Example 6.4.6. Define a product ∗ : L1(R) × L1(R) → L1(R) by convolution, i.e. forf, g ∈ L1(R) let

(f ∗ g)(t) =

∫Rf(x)g(t− x) dx,

be their product. Then (L1(R)), ∗) is a commutative Banach algebra.

Theorem 6.4.7. Given a, b ∈ `1ω,N, define the Cauchy product a ∗ b = (a ∗ b)nn∈N by

(a ∗ b)n =n∑k=0

an−kbk. (6.19)

Then for all a, b ∈ `1ω,N we have that a ∗ b = b ∗ a ∈ `1ω,N and the bound

‖a ∗ b‖1,ω,N ≤ ‖a‖1,ω,N‖b‖1,ω,N,

i.e. the pair (`1ω,N, ∗) is a commutative Banach algebra.

6.4. BANACH ALGEBRA 125

Proof. We begin by showing that a ∗ b ∈ `1ω,N, i.e. that

‖a ∗ b‖1,ω,N =∞∑n=0

|(a ∗ b)n|ωn <∞.

Consider the partial sums

cN :=N∑n=0

|(a ∗ b)n|ωn =N∑n=0

∣∣∣∣∣n∑k=0

bn−kak

∣∣∣∣∣ωn= |a0b0|ω0 + |a1b0 + a0b1|ω1 + . . .+ |aNb0 + . . .+ a0bN |ωN≤ (|a0||b0|ω0 + |a0||b1|ω1 + |a0||b2|ω2 + . . .+ |a0||bN |ωN )+

(|a1||b0|ω1 + |a1||b1|ω2 + |a1||b2|ω3 + . . .+ |a1||bN |ωN+1) + . . .

(|aN ||b0|ωN + |aN ||b1|ωN+1 + |aN ||bn|ωN+2 + . . .+ |aN ||bN |ω2N )

= (|a0|ω0 + |a1|ω1 + . . .+ |aN |ωN ) (|b0|ω0 + |b1|ω1 + . . .+ |bN |ωN )

=

(N∑k=0

|an|ωn)(

N∑k=0

|bn|ωn)≤( ∞∑k=0

|an|ωn)( ∞∑

k=0

|bn|ωn)

= ‖a‖1,ω,N‖b‖1,ω,N.

Since the partial sums cN are positive, monotonically increasing, and bounded from above,it follows that the series converges absolutely and therefore,

‖a ∗ b‖1,ω,N ≤ ‖a‖1,ω,N‖b‖1,ω,N.

To prove commutativity we simply observe that

n∑k=0

an−kbk = anb0 + an−1b1 + . . .+ a1bn−1 + a0bn

= a0bn + a1bn−1 + . . .+ an−1b1 + anb0

= bna0 + bn−1a1 + . . .+ b1an−1 + b0an =

n∑k=0

bn−kak.

We have the following two sided analogy of Theorem 6.4.7, whose proof we leave as anexercise.

Theorem 6.4.8. Given a, b ∈ `1ω,Z, define the discrete convolution product by a ∗ b =(a ∗ b)nn∈Z by

(a ∗ b)n =∑k∈Z

an−kbk. (6.20)

The pair (`1ω,Z, ∗) is a commutative Banach algebra.


6.5 The Radii Polynomial Approach on Banach spaces: globalversion

Consider the problem of finding a fixed point of a nonlinear map T : X → X with X aBanach space, or of finding a zero of a function F : X → Y with X and Y Banach spaces.In the first half of the book, when considering finite dimensional problems, we found thatthe contraction mapping theorem and Newton’s method were the appropriate tools. Weeven developed a-posteriori analysis which facilitated computer assisted proofs based onthese ideas. In this Section we aim to extend these tools to the setting of Banach spaces.

The following theorem facilitates a-posteriori analysis of fixed point problems.

Theorem 6.5.1. Radii-polynomial approach for the contraction mapping the-orem. Suppose that X is a Banach space, that T : X → X is a Frechet differentiablemapping, and that x ∈ X. Let Y0 ≥ 0 and Z : (0,∞) → [0,∞) a non-negative functionsatisfying that

‖T (x)− x‖X ≤ Y0, (6.21)

and

supx∈Br(x)

‖DT (x)‖L(X,X) ≤ Z(r). (6.22)


p(r) := Z(r)r − r + Y0.

If there exists r0 > 0 such that p(r0) < 0, then there exists a unique x ∈ Br0(x) so thatT (x) = x.

Proof. Suppose that r0 > 0 and that p(r0) < 0. Then (Z(r0)−1)r0 ≤ (Z(r0)−1)r0 +Y0 =p(r0) < 0, and so Z(r0) < 1. We now show that T : Br0(x) → Br0(x) is a contractionmapping with contraction constant Z(r0) < 1.

Let x1, x2 ∈ Br0(x). By the Mean Value Theorem (Theorem 6.3.4),

‖T (x1)− T (x2)‖X ≤ supx∈Br0 (x)

‖DT (x)‖L(X,X)‖x1 − x2‖X

≤ Z(r0)‖x1 − x2‖X . (6.23)

Now for any y ∈ Br0(x) we use (6.23) and that p(r0) = Z(r0)r0 − r0 + Y0 < 0 to get

‖T (y)− x‖X ≤ ‖T (y)− T (x)‖X + ‖T (x)− x‖X≤ Z(r0)‖y − x‖X + ‖T (x)− x‖X≤ Z(r0)r0 + Y0

< r0,

6.5. THE RADII POLYNOMIAL APPROACHON BANACH SPACES: GLOBAL VERSION127

which shows that T maps the complete metric space Br0(x) into Br0(x) ⊂ Br0(x). By thecontraction mapping theorem (Theorem 2.1.2), there exists a unique x ∈ Br0(x) so thatT (x) = x.

Consider a Frechet differentiable map F : X → X. In line with the philosophy ofChapter 3 assuming that a ∈ X is an approximate zero of F , i.e. ‖F (a)‖X ≈ 0, andexploiting the fact that F is Frechet differentiable construct the Newton-like map

T (a) = a−DF (a)−1F (a).

If, as is explicitly assumed, DF (a) is invertible, then fixed points of T are in one-to-onecorrespondence with zeros of F . In this case we can attempt to show that T is a contractionmapping in a neighborhood of a.

As is observed repeatedly in the examples of Sections 3.2 and 4.5 and Chapter 5 in-verting DF (a) is in some sense “too hard.” Thus we make use of an approximate inverse,typically obtained by computing a numerical inverse. However, we are now working withinfinite dimensional operators, thus finding an exact expression for DF (a)−1 is even morechallenging and making use of a numerical inverse is no longer an option. With thisin mind we adopt a slightly different approach and approximate DF (a) by an operatorA† ∈ L(X,X) specifically chosen to be easier to work with. We then choose A ∈ L(X,X)to be an approximate inverse of A† and define the Newton-like operator

T (a) = a−AF (a).

Observe that if A is injective, then fixed points of T correspond to zeros of F . The followingtheorem provides conditions under which we can guarantee the existence of a fixed pointof T and hence a zero of F . The assumptions are more general in the sense that we do notrequire that F : X → X but instead that AF : X → X.

Theorem 6.5.2. Radii polynomial approach for a-posteriori analysis in infinitedimensions Let X and Y be Banach space and F : X → Y be a Frechet differentiablemapping. Suppose that x ∈ X, A† ∈ L(X,Y ) and A ∈ L(Y,X). Moreover assume that Ais a one-to-one operator. Let Y0, Z0, and Z1 be positive constants and Z2 : (0,∞)→ [0,∞)a non-negative function satisfying

‖AF (x)‖Y ≤ Y0, (6.24)

‖IdX −AA†‖L(X,X) ≤ Z0, (6.25)

‖A[DF (x)−A†]‖L(X,X) ≤ Z1, (6.26)

and

‖A[DF (c)−DF (x)]‖L(X,X) ≤ Z2(r), for all c ∈ Br(x) for all r > 0. (6.27)


Definep(r) := Z2(r)r2 − (1− Z0 − Z1)r + Y0.

If there exists r0 > 0 such that p(r0) < 0, then the linear operator A is one to one, andthere exists a unique x ∈ Br0(x) satisfying F (x) = 0.

Proof. To begin we define the “Newton-like operator” T : X → X by

T (x) = x−AF (x).

Since A is one-to-one we have that x ∈ X has F (x) = 0 if and only if T (x) = x. Now wewill use the contraction mapping theorem to show that T has a fixed point.

First note that for any x ∈ X, the operator T is Frechet differentiable with

DT (x) = IdX −ADF (x).

(...UNDER CONSTRUCTION...)

6.6 The Radii Polynomial Approach on Banach spaces: localversion

Theorem 6.6.1. [Radii Polynomial Approach in Banach Spaces] Let X and Y beBanach spaces. Denote the norm on X by ‖ · ‖X . Consider the bounded linear operatorsA† ∈ L(X,Y ) and A ∈ L(Y,X). Assume F : X → Y is C1, that A is injective and that

AF : X → X. (6.28)

Let Y0, Z0, Z1, and Z2 be positive constants satisfying

‖AF (x)‖X ≤ Y0 (6.29)

‖I −AA†‖L(X,X) ≤ Z0 (6.30)

‖A[DF (x)−A†]‖L(X,X) ≤ Z1, (6.31)

‖A[DF (b)−DF (x)]‖L(X,X) ≤ Z2r, for all b ∈ Br(x). (6.32)


p(r) := Z2r2 − (1− Z1 − Z0)r + Y0. (6.33)

If there exists r0 > 0 such thatp(r0) ≤ 0,

then there exists a unique x ∈ Br0(x) := x ∈ X | ‖x− x‖X ≤ r0 satisfying

F (x) = 0.

6.6. THE RADII POLYNOMIAL APPROACHON BANACH SPACES: LOCAL VERSION129

Proof. Define the operator T : X → X by

T (x) = x−AF (x).

The goal is to show that T is a uniform contraction mapping Br0(x) into itself, in whichcase the result follows from the contraction mapping theorem and the assumption that Ais injective.

By Proposition ??DT (x) = I −ADF (x),

for all x ∈ X. Now, given x ∈ Br0(x) and applying the bounds (6.29), (6.30), (6.31), and(6.32), we obtain

‖DT (x)‖L(X,X) = ‖I −ADF (x)‖L(X,X)

≤ ‖I −AA†‖L(X,X) + ‖A[A† −DF (x)]‖L(X,X)

+ ‖A[DF (x)−DF (x)]‖L(X,X)

≤ Z0 + Z1 + Z2r0 (6.34)

We now show that T maps Br0(x) into itself. Let x ∈ Br0(x) and apply the Mean ValueTheorem (Theorem 6.3.4) to obtain

‖T (x)− x‖X ≤ ‖T (x)− T (x)‖X + ‖T (x)− x‖X≤ sup

b∈Br0 (x)

‖DT (b)‖L(X,X)‖x− x‖X + ‖AF (x)‖X

≤ (Z0 + Z1 + Z2r0)r0 + Y0

where the last inequality follows from (6.34). Applying (6.33) and the assumption thatp(r0) ≤ 0 implies that ‖T (x)− x‖X ≤ r0, the desired result.

To see that T is a contraction on Br0(x), let a, b ∈ Br0(x) and apply (6.34) to obtain

‖T (a)− T (b)‖X ≤ supb∈Br0 (x)

‖DT (b)‖L(X,X)‖a− b‖X

≤ (Z0 + Z1 + Z2r0)‖a− b‖X .Again, from the assumption that p(r0) ≤ 0, it follows that

Z0 + Z1 + Z2r0 ≤ 1− Y0

r0.

Since Y0/r0 > 0 we conclude that κ := Z0 +Z1 +Z2r0 < 1 and hence T : Br0(x)→ Br0(x)is a contraction with contraction constant κ < 1.

Remark 6.6.2. The preceding theorem also goes through if we replace the condition givenby (6.32) with the condition

‖A[DF (b)−DF (c)]‖ ≤ Z2‖b− c‖X ,for all b, c ∈ Br(a) with the bound Z2 being a polynomial function of r. (See the exercises).


6.7 Applications

6.7.1 Validated numerics for reciprocals and square roots in `1ω,N

(UNDER CONSTRUCTION...)

6.7.2 Validated numerics for reciprocals and square roots in `1ω,Z

Consider b ∈ `1ω,Z and define 1 ∈ `1ω,Z by

1n =

0 if n < 0

1 if n = 0

0 if n > 0

We are interested in computing the multiplicative inverse of b (if it exists), i.e. we hope tofind an a ∈ `1ω,Z so that

a ∗ b = 1.

To this end define the map F : `1ω,Z → `1ω,Z by

F (a) = a ∗ b− 1.

Under construction...

...

For b ∈ `1ω,Z we are interested in computing a “square root” of b. More precisely, we

seek an a ∈ `1ω,Z so that a ∗ a = b. As in Chapter REFERENCE CH 3, we define the map

F : `1ω,Z → `1ω,Z byF (a) = a ∗ a− b,

and look for a zero of F (think of b as fixed). Since `1ω,Z is a Banach algebra we have that

DF (a)h = 2a ∗ h.

UNDER CONSTRUCTION...

6.7.3 A radii polynomial proof of the implicit function theorem

We would like to illustrate the use of the machinery developed in the previous section ina familiar context. To this end we return again to the implicit function theorem. To keepthings simple we consider the two dimensional cases, but extensions are in the exercises.

So, let f : R2 → R be a smooth function and suppose that (λ0, x0) ∈ R2 is a zero, i.e.

f(λ0, x0) = 0.

6.7. APPLICATIONS 131

We will give a new proof of the existence of a continuous branch of zeros, i.e. a continuousfunction x(λ) so that

f(λ, x(λ)) = 0. (6.35)

A few observations will help us set up the proof. Suppose that x(λ) existed and wassmooth. Then we could differentiate Equation (6.35) with respect to λ to get

∇f(λ, x(λ))

[1

x′(λ)

]=

∂

∂λf(λ, x(λ)) +

∂

∂xf(λ, x(λ))x′(λ) = 0.

Evaluating at λ = λ0 and solving for x′(λ0) gives

x′(λ0) = − ∂

∂λf(λ0, x(λ0))

(∂

∂xf(λ0, x(λ0))

)−1

,

assuming that∂

∂xf(λ0, x(λ0)) 6= 0.

Letting

b = − ∂

∂λf(λ0, x(λ0))

(∂

∂xf(λ0, x(λ0))

)−1

,

we see that a first order approximation of x(λ) is

x(λ) = x0 + b(λ− λ0).

Taking this as out “initial guess” for the desired branch of zeros x(λ), we try to improvethe guess via a Newton Method.

Then define the nonlinear operator

F (x)(λ) = f(λ, x(λ)),

which maps C1 functions to C1 functions. In order to solve F (x) = 0 with a Newtonmethod we first need to find the Frechet derivative. Consider

F (x+ h)(λ)− F (x)(λ) = f(λ, x(λ) + h(λ))− f(λ, x(λ))

= f(λ, x(λ)) +∂

∂xf(λ, x(λ))h(λ) +R(λ, h(λ))− f(λ, x(λ))

=∂

∂xf(λ, x(λ))h(λ) +R(λ, h(λ)),

from which we can show that

DF [x(λ)]h(λ) =∂

∂xf(λ, x(λ))h(λ).


(See Exercise 6.8.17). It follows that

DF [x(λ)]h(λ) =∂

∂xf(λ, x0 + b(λ− λ0))h(λ).

Then, if λ− λ0 is small (and f is C2) a good “approximate derivative” is

A† :=∂

∂xf(λ0, x0),

which is on the one hand a bounded linear operator from C1 to C1, but on the other handis just a number. Moreover, this operator is “easy to invert”, i.e. we define

A :=1

∂∂xf(λ0, x0)

.

Then a Newton-like method for improving our approximate branch is given by

xn+1(λ) = xn(λ)−AF (xn(λ)),

withx0(λ) = x(λ) = x0 + b(λ− λ0).

This method can be iterated to obtain better and better approximations. Moreover, wecan actually use Theorem 6.6.1 to prove the implicit function theorem. First we recall thefollowing version of Taylor’s theorem, as well as a theorem about derivatives of uniformlimits of differentiable functions.

Theorem 6.7.1 (A linear approximation theorem). Let f : R2 → R be differentiableat (λ0, x0) ∈ R. Then there exists a function R(λ0,x0) : R2 → R so the

f(λ, x) = f(λ0, x0) +∇f(λ0, x0)

[λ− λ0

x− x0

]+R(λ0,x0)(λ− λ0, x− x0),

andlim

(λ,x)→(λ0,x0)R(λ0,x0)(λ− λ0, x− x0).

Moreover, if f is twice continuously differentiable on D := [λ0−r∗, λ0+r∗]×[x0−r∗, x0+r∗]then

sup(λ,x)∈D

∣∣R(λ0,x0)(λ− λ0, x− x0)∣∣ ≤ 2Mr2

∗,

where M > 0 is a bound of the form

max

(sup

(λ,x)∈D

∣∣∣∣ ∂2

∂λ2f(λ, x)

∣∣∣∣ , sup(λ,x)∈D

∣∣∣∣ ∂2

∂λ∂xf(λ, x)

∣∣∣∣ , sup(λ,x)∈D

∣∣∣∣ ∂2

∂x2f(λ, x)

∣∣∣∣)≤M.


Theorem 6.7.2 (Two dimensional implicit function theorem). Suppose that f : R2 →R is a continuous function, and that

• (λ0, x0) ∈ R2 is a zero of f , i.e.

f(λ0, x0) = 0.

• That f is differentiable at (λ0, x0) ∈ R2 and that

∂

∂xf(λ0, x0) 6= 0.

• That there is an r∗ > 0 so that f is twice continuously differentiable on

D = [λ0 − r∗, λ0 + r∗]× [x0 − r∗, x0 + r∗].

Let M > 0 have

max

(sup

(λ,x)∈D

∣∣∣∣ ∂2

∂λ2f(λ, x)

∣∣∣∣ , sup(λ,x)∈D

∣∣∣∣ ∂2

∂λ∂xf(λ, x)

∣∣∣∣ , sup(λ,x)∈D

∣∣∣∣ ∂2

∂x2f(λ, x)

∣∣∣∣)≤M.

Then there exists an ε > 0, and a C1 function x : (λ0 − ε, λ0 + ε)→ R so that

f(λ, x(λ)) = 0

for all λ ∈ (λ0 − ε, λ0 + ε).

Proof. Let

K =1∣∣ ∂

∂xf(λ0, x0)∣∣ ,

b = −∂∂λf(λ0, x0)∂∂xf(λ0, x0)

and choose

ε <r∗

2|b| .

Let I = (λ0 − ε, λ0 + ε) and let

X = Y = C0(I,R).

Note that X is a Banach space. The function x(λ) defined by

x(λ) = x0 + b(λ− λ0),


has x ∈ X. Moreover, the correspondence

F (x)(λ) := f(λ, x(λ)),

gives a well defined mapping F : X → Y . Let x ∈ C0(I,R) and note that the correspon-dences

A†x(λ) :=∂

∂xf(λ0, x0)x(λ),

and

Ax(λ) :=1

∂∂xf(λ0, x0)

x(λ),

define bounded linear operators on X, i.e. A,A† ∈ B(C0(I,R)).Step 1: First we will use Theorem 6.6.1 to prove the existence of a continuous branch

of zeros. This this end we define the Y0, Z0, Z1, and Z2 bounds needed for the Theorem.

Y0 bound: The Y0 bound measures how well x approximates a zero of F . Then note thatfor any λ ∈ I we have

F (x(λ)) = f(λ, x0 + b(λ− λ0))

= f(λ0 + (λ− λ0), x0 + b(λ− λ0))

= f(λ0, x0) +∂

∂λf(λ0, x0)(λ− λ0) +

∂

∂xf(x0, λ0)b(λ− λ0) +R(λ− λ0, b(λ− λ0))

= 0 +∂

∂λf(λ0, x0)(λ− λ0) +

∂

∂xf(x0, λ0)

[−

∂∂λf(λ0, x0)∂∂xf(x0, λ0)

](λ− λ0)

+R(λ− λ0, b(λ− λ0))

= R(λ− λ0, b(λ− λ0)).

It follows that

AF (x(λ)) =1

∂∂xf(x0, λ0)

R(λ− λ0, b(λ− λ0)),

and

‖AF (x)‖C0 ≤ K supλ∈I|R(λ− λ0, b(λ− λ0))|

≤ 2KM(1 + |b|)2|λ− λ0|2≤ 2KM(1 + |b|)2ε2.

Z0 bound: The Z0 bound measures how close A is to being an inverse of A†. But since Ais the exact inverse of A†, we have that

‖Id−AA†‖B(C0(I,R)) = 0,


i.e. we take Z0 = 0.

Z1 bound: The Z1 bound measures how well A† approximates the Frechet derivative of Fat the approximate solution x. So, note that for any λ ∈ I we have

∂

∂xf(λ, x0 + b(λ− λ0))− ∂

∂xf(λ0, x0) = ∇ ∂

∂xf(λ∗, x∗)

[λ− λ0

b(λ− λ0)

]=

∂2

∂λ∂xf(λ∗, x∗)(λ− λ0) + b

∂2

∂x2f(λ∗, x∗)(λ− λ0),

for some (λ∗, x∗) ∈ [λ0 − r, λ0 + r]× [x0 − r, x0 + r] by the mean value theorem. Then∥∥∥DF (x(λ))−A†∥∥∥B(C0(I,R))

= sup‖h‖C0=1

∥∥∥[DF (x(λ))−A†]h∥∥∥C0

= sup‖h‖C0=1

supλ∈I

∣∣∣∣ ∂2

∂λ∂xf(λ∗, x∗)(λ− λ0)h(λ) + b

∂2

∂x2f(λ∗, x∗)(λ− λ0)h(λ)

∣∣∣∣≤ (1 + |b|)Mε.

Then we take

Z1 = (1 + |b|)Mε.

Z2 bound: The Z2 bound is a local Lipschitz bound for the Frechet derivative of F nearx. Here we choose any 0 < r < r∗/2 and consider the ball

Br(x) =x ∈ C0(I,R)|‖xx‖C0 ≤ r

.

First note that for any x ∈ Br(x) and λ ∈ I we have that

|x(λ)− x0| ≤ |x(λ)− x0|+ |(x(λ)− x(λ))|≤ |x0 + bε− x0|+ r

≤ |b| r∗|b|2 +r∗2

≤ r∗,

as ε < r∗/2|b|. In other words, (λ, x(λ)) ∈ D whenever λ ∈ I. So, again taking x ∈ Br(x)and λ ∈ I we have

∂

∂xf(λ, x(λ))− ∂

∂xf(λ, x(λ)) =

∂2

∂x2f(λ, x∗)(x(λ)− x(λ)),


for some x∗ ∈ [x0 − r, x0 + r], again by the Mean Value Theorem. It follows that

‖DF (x)−DF (x)‖B(C0(I,R)) = sup‖h‖C0=1

‖(DF (x)−DF (x))h‖C0

= sup‖h‖C0=1

supλ∈I|[DF (x(λ))−DF (x(λ))]h(λ)|

= sup‖h‖C0=1

supλ∈I

∣∣∣∣ ∂∂xf(λ, x(λ))h(λ)− ∂

∂xf(λ, x(λ))h(λ)

∣∣∣∣= sup

‖h‖C0=1supλ∈I

sup(λ∗,x∗)∈D

∣∣∣∣ ∂2

∂x2f(λ∗, x∗)

∣∣∣∣ |h(λ)||x(λ)− x(λ)|

≤ M‖x− x‖C0

≤ Mr.

Then, taking

Z2 = KM,

we have that

‖A[DF (x)−DF (x)]‖B(C0(I,R)) ≤ Z2r,

whenever x ∈ Br(x).

Construction of the radii-polynomial: Consider the polynomial

p(r) := Z2r2 − (1− Z1 − Z0)r + Y0.

with

Y0 := 2KM(1 + |b|)2ε2, Z0 := 0,

Z1 := (1 + |b|)Mε, and Z2 := KM.

The roots of p are

r± =1− Z1 ±

√(1− Z1)2 − 4Z2Y0

2Z2,

i.e.

r− =1− (1 + |b|)Mε−

√(1− (1 + |b|)Mε)2 − 8K2M2(1 + |b|)2ε2

2KM

Multiplying and dividing by the conjugate gives

r− =1

2KM

(1− (1 + |b|)Mε)2 − ((1− (1 + |b|)Mε)2 − 8K2M2(1 + |b|)2ε2)

1− (1 + |b|)Mε+√

(1− (1 + |b|)Mε)2 − 8K2M2(1 + |b|)2ε2

=4KM(1 + |b|)2

1− (1 + |b|)Mε+√

(1− (1 + |b|)Mε)2 − 8K2M2(1 + |b|)2ε2ε2


To get a real positive result we need

(1 + |b|)Mε < 1,

and

2√

2KM(1 + |b|)ε < 1− (1 + |b|)Mε,

which reduce to the single condition

ε <1

M(1 + |b|)1

1 + 2√

2K. (6.36)

Since the quantity on the right is strictly positive we can always choose an ε > 0 satisfyingthe condition of Equation (6.36). Then for any 0 < r− < r < r+ we have that p(r) < 0,i.e. we have satisfied the hypothesis of Theorem 6.6.1. We conclude that there exists ax ∈ Br(x) so that F (x) = 0, i.e. a continuous function x : I → R with

f(λ, x(λ) = 0

for all λ ∈ I = (λ0 − ε, λ0 + ε), and having

supλ∈I|x(λ)− x(λ)| ≤ r,

for all λ ∈ I.

Step 2: Now we will show that x : I → R is actually C1. This part of the proof isstandard but we include it for the sake of completeness. First, choose λ, h ∈ R having thatλ, λ+ h ∈ interior(I). Note that

0 = f(λ+ h, x(λ+ h))− f(λ, x(λ))

= ∇f(λ∗, x∗)

[h

x(λ+ h)− x(λ)

]=

∂

∂λf(λ∗, x∗)h+

∂

∂xf(λ∗, x∗)(x(λ+ h)− x(λ))

with λ∗ between λ and λ+ h, and x∗ between x(λ+ h) and x(λ). Then

x(λ+ h)− x(λ) = −∂∂λf(λ∗, x∗)h∂∂xf(λ∗, x∗).

Now

limh→0

x∗ = x(λ), and limh→0

λ∗ = λ,


as x is continuous and λ∗, x∗ squeezed between λ, λ + h and x(λ), x(λ + h) respectively.Then by the continuity of the partial derivatives we have that

limh→0

∂

∂λf(λ∗, x∗) =

∂

∂λf(λ, x(λ)),

and

limh→0

∂

∂xf(λ∗, x∗) =

∂

∂xf(λ, x(λ)).

Putting it all together gives

limh→0

x(λ+ h)− x(λ)

h= − lim

h→0

∂∂λf(λ∗, x∗)∂∂xf(λ∗, x∗)

= −∂∂λf(λ, x(λ))∂∂xf(λ, x(λ))

. (6.37)

This shows that x is differentiable on the interior of I and gives the formula for the deriva-tive.

Remark 6.7.3. Consider the Newton sequence

x0(λ) := x(λ) = x0 + b(λ− λ0),

xn+1(λ) = xn(λ)−Af(λ, xn(λ)).

The sequence converges uniformly to our C1 branch x(λ). Observe that

x0(λ0) = x0.

Now assume that xn(λ0) = x0 for some n ≥ 0. Then

xn+1(λ0) = xn(λ0)−Af(λ0, xn(λ0))

= x0 −Af(λ0, x0)

= x0.

Now since xn converges (uniformly) to x we see that

x(λ0) = limn→∞

xn(λ0) = x0.

By Equation (6.37) it now follows that

x(λ0) = −∂∂λf(λ0, x0)∂∂xf(λ0, x0)

,

i.e. we recover that x(λ) equals x(λ) to first order without ever imposing this constraint.

6.7.4 A radii-polynomial proof of the Hartman-Grobman theorem

6.8. EXERCISES 139

6.8 Exercises

Exercise 6.8.1. Show by example that C0([0, 1]) is not a Banach space when endowedwith the norm

‖f‖1 =

∫ 1

0|f(x)| dx.

Exercise 6.8.2. (This exercise involves the Lebesgue integral) Choose a fixed g ∈ L1(R)and define the map A : L1(R)→ L1(R) by

(Af)(t) = (f ∗ g)(t),

where f ∈ L1(R).

• Prove that A is a well defined linear operator.

• Prove that A is a bounded linear operator and provide a bound on the norm.

Exercise 6.8.3. Let ω be a one or two sided sequence of admissible weights (dependingon the context).

• Prove that the one sided weighted supremum

‖a‖∞,ω,N := supn≥0|an|ωn,

defines a norm on SN(C).

• Prove that the two sided weighted sum

‖a‖1,ω,N :=∑n∈Z|an|ωn,

defines a norm on SZ(C).

• Prove that the two sided weighted supremum

‖a‖∞,ω,N := supn∈Z|an|ωn,

defines a norm on SZ(C).

Exercise 6.8.4. Prove that `∞,ω,N, `1,ω,Z, and `∞,ω,Z are Banach spaces.

Exercise 6.8.5. • Recall the definition of the standard two sided basis vectors fromRemark 6.2.12. Prove that these vectors provide a Schauder basis for `1,ω,Z.


• Let ω = ωn∞n=0 be an admissible sequence of weights and suppose that

limn→∞

ωn =∞.

Prove that the standard one sided basis vectors of Theorem 6.2.11 form a Schauderbasis for `∞,ω,N.

• Let ω = ωn∞n∈Z be an two sided admissible sequence of weights and suppose that

limn→±∞

ωn =∞.

Prove that the standard one sided basis vectors of Remark 6.2.12 form a Schauderbasis for `∞,ω,Z.

Exercise 6.8.6. Let X,Y be Banach spaces and suppose that φn∞n=0 ⊂ X be a Schauderbasis for X. Let A : X → Y be a bounded linear operator and x ∈ X. Show that if

x =∞∑n=0

anφn,

is the expansion of x in the Schauder basis then

Ax =∞∑n=0

anA(φn).

Explain how this is like “writing A as an infinite matrix”.

Exercise 6.8.7. Recall that a metric space is separable if it has a countable dense subset.Prove the following general theorem about Banach spaces.

Theorem 6.8.8. Let X be a Banach space. If X has a Schauder basis then X is separable.

Use the theorem to justify the following claims.

• `1,ω,N is a separable Banach space.

• `1,ω,Z is a separable Banach space.

• Suppose that ω = ωn∞n=0 is an admissible sequence of weights with

limn→∞

ωn =∞.

Then `∞,ω,N is a separable Banach space.

6.8. EXERCISES 141

• Suppose that ω = ωnn∈Z is an admissible sequence of weights with

limn→±∞

ωn =∞.

Then `∞,ω,Z is a separable Banach space.

• Suppose that ω = ωnn∈Z is an admissible sequence of weights and that

limn→±∞

ωn = 0.

Prove that `∞,ω,Z is not separable.

Exercise 6.8.9. Let ω = ωnn∈Z be a two sided admissible sequence of weights and defineω−1 := ω−1

n n∈Z . Prove that `∞,ω−1,Z is isometrically isomprphic to (`1,ω,Z)∗.

Exercise 6.8.10. Calculus on Banach spaces: the basics. Let X,Y be Banach spaces,x0 ∈ X, and suppose that F,G : X → Y are Frechet differentiable at x0.

• Prove that the Frechet derivative of F at x0, if it exists, is unique.

• Suppose that A : X → Y is a bounded linear map. Prove the A is Frechet differen-tiable for all x0 ∈ X and that

DA(x0) = A,

i.e. for all h ∈ X the differential has the action

DA(x0)h = Ah.

• Suppose that A : Y → Y is a bounded linear operator. Prove that AF is Frechetdifferentiable at x0 and that

D(AF )(x0) = ADF (x0).

• Let α, β ∈ C. Prove that αF + βG is Frechet differentiable at x0 and that

D(αF + βG)(x0) = αDF (x0) + βDG(x0).

• (A chain rule) Let X,Y , F : X → Y and x0 be as above. Suppose that Z is a thirdBanach space and that H : Y → Z is Frechet differentiable at y0 = F (x0). Provethat H F : X → Z is Frechet differentiable at x0 and that

D(H F )(x0) = DH(y0)DF (x0).


Exercise 6.8.11. (This problem uses the Lebesgue integral) Consider the functionN : L1(R)→L1(R) defined by

N [f ](t) := (f ∗ f)(t) =

∫Rf(s)f(t− s) ds,

for f ∈ L1(R).

• Prove that N is Frechet differentiable for each f ∈ L1(R) and compute the derivative(don’t use any theorems about Banach algebras. Instead, work directly from thedefinition).

• Consider the “triple” convolution

T [f ](t) = (f ∗ (f ∗ f))(t).

Explain why T is well defined. Moreover, show that

((f ∗ f) ∗ f)(t) = (f ∗ (f ∗ f))(t).

Because of this associativity it makes sense to drop the parenthesis and write T (f)(t) =(f ∗ f ∗ f)(t).

• Prove that N [f ](t) is Frechet differentiable for all f ∈ L1(R), and compute thederivative.

• Extend the previous result by induction to any number of “monomial” convolutionsof f with itself, i.e. define

(fn)(t) = (f ∗ . . . ∗ f)(t), (convolution n times).

Prove that the repeated convolution makes sense (associavitity), that the result isFrechet differentiable, and that compute the derivative.

• Suppose that g ∈ L1(R) is fixed. Give a short explanation (one or two sentences) forwhy the map

(Af)(t) = (g ∗ f)(t),

is Frechet differentiable. What is the derivative?

• Let g0, g1, . . . , gn ∈ L1(R) be a collection of Lebesgue integrable functions. Definethe “polynomial” operator P : L1(R)→ L1(R) by

P [f ](t) = (gn ∗ fn)(t) + . . .+ (g2 ∗ f2)(t) + (g1 ∗ f)(t) + g0(t).

where f ∈ L1(R). Prove that P is Frechet differentiable and compute the derivative.

6.8. EXERCISES 143

• How much of the preceding work generalizes if you work instead with a commutativeBanach algebra (X, ∗)?

Exercise 6.8.12. Given a, b ∈ `1ω,Z define the discrete convolution product of a and b tobe the new sequence a ∗ b = (a ∗ b)nn∈Z given by

(a ∗ b)n =∑k∈Z

an−kbk.

• Prove that ∗ : `1ω,Z × `1ω,Z → `1ω,Z is a well defined binary product (the challenge here

is to show that the result a ∗ b is actually an element of `1ω,Z). Show that the productcommutes.

• Prove that‖a ∗ b‖1,ω,Z ≤ ‖a‖1,ω,Z‖b‖1,ω,Z,

so that (`1ω,Z, ∗) is a commutative Banach algebra.

• Think of all the calculus you now get for free!

Exercise 6.8.13. Let X be a normed linear space and Y be a Banach space. For T : X →Y recall that the operator define norm is defined by

‖T‖L(X,Y ) := sup‖x‖X=1

‖T (x)‖Y ,

and that the set

L(X,Y ) :=T : X → Y |T is linear and ‖T‖L(X,Y ) <∞

,

is a Banach space. Let T, S ∈ L(X,Y ) and define the product T ∗ S by composition, i.e.

(T ∗ S)(x) = T (S(x)),

for any x ∈ X. Prove that this product endows L(X,Y ) with a Banach algebra structure.What can you say about calculus in this Banach algebra?

Exercise 6.8.14. Let X and Y be Banach spaces and T : X → Y be a linear operator.Prove that the sets

image(T ) = y ∈ Y : y = T (x) for some x ∈ X ,

andker(T ) = x ∈ X : T (x) = 0 ,

are Banach (sub)spaces. Prove that T is one-to-one if and only if ker(T ) = 0.


Exercise 6.8.15. Let X,Y be Banach spaces and T : X → Y be a one-to-one and ontolinear map. Prove that the inverse of T is also linear.

Exercise 6.8.16. LetX be a Banach space andA,B : X → X be bounded linear operators.Suppose that A,B are invertible. Prove that AB is invertible and

(AB)−1 = B−1A−1.

Exercise 6.8.17. Suppose that f : R2 → R is a smooth function. Consider the operatorF : C0([a, b],R)→ C0([a, b],R) defined by

F (x)(λ) := f(λ, x(λ)),

for λ ∈ [a, b] and x ∈ C0([a, b],R). Prove that the Frechet derivative of F is given by

DF (x(λ))h(λ) =∂

∂xf(λ, x(λ)).

Exercise 6.8.18. Imitate the proof of the implicit function theorem of this chapter andprove Theorem 5.1.1 from Chapter 5.

Exercise 6.8.19. Use the implicit function theorem to prove that the eigenvalues of amatrix depend smoothly on the coefficients of the matrix.

Exercise 6.8.20. Imitate the proof in the text and prove the following infinite dimensionalversion of the implicit function theorem.

Theorem 6.8.21 (The implicit function theorem). Let X,Y, Z be Banach spaces andsuppose that F : Z×X → Y is a C1 mapping. Let y0 = f(z0, x0). Suppose that the boundedlinear operator DxF (z0, x0) : X → Z is boundedly invertible, i.e. that DxF (z0, x0)−1 : Z →X exists and is a bounded linear operator.

Then there are open sets W ⊂ Z, U ⊂ X and V ⊂ Y with z0 ∈W , x0 ∈ U and y0 ∈ Vand a unique mapping

g : W × V → U,

so thatF (z, g(z, y) = y,

for all (z, y) ∈W × V . Moreover g is C1 and, letting x = g(z, y) we have that

Dg(z, y)(η, ζ) = DxF (z, x)−1(ζ −DzF (z, x)η),

for all (z, y) ∈W × V and all (η, ζ) ∈ Z × Y .

Exercise 6.8.22. The following is a typical abstract perturbation theory result. You canprove it with the implicit function theorem. Try also to prove it using Theorem 6.6.1.

6.8. EXERCISES 145

Theorem 6.8.23. Let X and Y be Banach spaces, and suppose that F,G : X → Y areC1. Define

H(x, ε) = F (x) + εG(x),

with ε ∈ R. If H(x0, 0) = 0 and DF (x0) is boundedly invertible, show that there exists anx ∈ X so that H(x, ε) = 0 for ε sufficiently small.

Consider the operator ...

Chapter 7

Appendix

In this chapter we present some definitions, results and notations that are used throughoutthe book. The results from this chapter can be found in the following references [?, ?, ?,?, ?, ?]

7.1 Topological and Metric Spaces

Throughout this chapter F will denote either the real field R or the complex field C.

Definition 7.1.1. A topology on a set X is a collection T of subsets of X with the followingproperties:

(1) ∅ and X are in T .

(2) The union of the elements of any subcollection of T is in T .

(3) The intersection of the elements of any finite subcollection of T is in T .

A topological space is a pair (X, T ) where X is a set and T is a topology on X. We oftenrefer to X as a topological space when there is no need to explicitly indicate the topologyT .

Definition 7.1.2. Given a topological space X with topology T we say that a subset Uof X is open if U belongs to T . A subset F of X is closed if its complement F c := X \ Fis open. Hence a topological space is a set X together with a collection T of open subsetsof X. Given a point x ∈ X, if x ∈ U and U is open, we say that U is a neighborhood of x.

Proposition 7.1.3. Let X be a topological space. Then we have the following:

(1) ∅ and X are closed.

(2) Arbitrary intersections of closed sets are closed.

147

148 CHAPTER 7. APPENDIX

(3) Finite unions of closed sets are closed.

Definition 7.1.4. Let A be a subset of a topological space X. We define the interior of A,denoted int (A), as the union of all open sets contained in A, and the closure of A, denotedA, as the intersection of all closed sets containing A. In other words, int (A) is the largestopen set contained in A and A is the smallest closed set containing A.

Definition 7.1.5. Let A be a subset of a topological space X. A collection C of subsets ofX is said to cover A, or to be a covering of A, it the union of the elements of C containsA. It is called an open covering of A if its elements are open subsets of X.

Definition 7.1.6. A subset K of a topological space X is said to be compact if every opencovering of K contains a finite subcollection that also covers K.

Definition 7.1.7. A topological space X is said to be locally compact if for every x ∈ Xthere is a compact set K that contains a neighborhood of x.

Definition 7.1.8. A metric space is a pair (X,d) where X is a set and d is a metricon X, i.e., a function d : X ×X → [0,∞) such that for any x, y, z ∈ X, the following holds

(1) d(x, y) = 0 if and only if x = y.

(2) d(x, y) = d(y, x).

(3) d(x, z) ≤ d(x, y) + d(y, z) (triangle inequality).

The value d(x, y) is called the distance from x to y.

If (X,d) is a metric space, given a point x0 ∈ X and an r > 0, the open ball of radiusr centered at x0 is given by

Br(x0) := x ∈ X | d(x0, x) < r ,

and the closed ball of radius r centered at x0 is given by

Br(x0) := x ∈ X | d(x0, x) ≤ r .

Definition 7.1.9. Let (X,d) be a metric space and A ⊂ X. We say that A is open in Xif, for every x ∈ A there is an ε > 0 such that Bε(x) ⊂ A.

It is easy to see that the collection of all open sets as above define a topology on X.Hence a metric space is a topological space. This topology is called the metric topology orthe topology induce by the metric.

Definition 7.1.10. A metric space X is said to be complete if every Cauchy sequenceconverges in X, i.e., if d(xn, xm) → 0 as n,m → ∞, then there is some y ∈ X withd(xn, y)→ 0, as n→∞.

7.2. NORMED VECTOR SPACES 149

Definition 7.1.11. Let (X,d) denote a metric space. A function T : X → X is a contrac-tion if there is a number λ ∈ [0, 1), called a contraction constant, such that

d(T (x), T (y)) ≤ λd(x, y)

for all x, y ∈ X.

7.2 Normed Vector Spaces

Definition 7.2.1. Let X be a vector space over F. A function ‖ · ‖ : X → [0,∞) is a normon X if it satisfies the following conditions:

(1) ‖x‖ = 0 if and only if x = 0.

(2) ‖αx‖ = |α|‖x‖ for all vectors x ∈ X and scalars α ∈ F.

(3) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all vectors x, y ∈ X.

A normed vector space is a vector space X equipped with a norm ‖ · ‖.

Definition 7.2.2. Let V and W be normed vector spaces with norms denoted by ‖ · ‖Vand ‖ · ‖W , respectively. Let A : V →W be a linear map. The norm of A is defined by

‖A‖ := supx 6=0

‖Ax‖W‖x‖V

= sup‖x‖V =1

‖Ax‖W .

Observe that

‖Ax‖W ≤ ‖A‖‖x‖V .Using different norms on vector spaces results in different norms on operators. It is left

to the reader the check that given a matrix A = aiji,j

Vector Norm Matrix Norm

‖x‖1 :=n∑k=1

|xk| ‖A‖1 = max1≤j≤n

n∑i=1

|ai,j |

‖x‖2 :=

√√√√ n∑k=1

x2k ‖A‖2 =

√rσ(A∗A)

‖x‖∞ := max1≤k≤n

|xk| ‖A‖∞ = max1≤i≤n

n∑j=1

|ai,j |

(7.1)

Here rσ(M) denotes the maximum of the magnitudes of the eigenvalues of the matrix M .


Definition 7.2.3. Let X be a linear space. Two norms ‖ · ‖a and ‖ · ‖b are equivalent ifthere exists constants K0,K1 > 0 such that

K0‖x‖a ≤ ‖x‖b ≤ K1‖x‖a.

It is left to the reader to check that this defines an equivalence relation and furthermorethat an alternative statement of the definition is that

K0 ≤ ‖x‖b ≤ K1.

for all x ∈ X such that ‖x‖a = 1.

Proposition 7.2.4. If X is a finite dimensional linear space, then all norms are equivalent.

Proof. Assume the dimension of X is N . Let en | n = 1, . . . , N be a basis for X. Then,for every x ∈ X, there are a unique set of coefficients xn | n = 1, . . . , N such that

x =N∑n=1

xnen.

Observe that it is sufficient to prove all norms are equivalent to a single norm. It is leftto the reader to check that

‖x‖1 :=

N∑n=1

|xn|

is a norm. Let ‖·‖ be an arbitrary norm on X we need to show that there are fixed positiveconstants K0 and K1 such that

K0 ≤ ‖x‖ ≤ K1 (7.2)

for all x ∈ X such that ‖x‖1 = 1. Observe that demonstrating (7.2) is equivalent toshowing that ‖ · ‖ is a bounded function on the unit sphere Sn−1 (under the ‖ · ‖1 norm).

Since Sn−1 is compact it is sufficient to show that ‖ · ‖ : V → R is a continuous functionunder the topology on V induced by ‖ · ‖1. Fix ε > 0 and define

δ =ε

maxn ‖en‖.

Choose x, y ∈ X such that ‖x− y‖1 < δ. Then

|‖x‖ − ‖y‖| ≤ ‖x− y‖ ≤N∑n=1

|xn − yn|‖en‖ ≤ ‖x− y‖1 maxn‖en‖ ≤ ε.

Hence ‖ · ‖ : V → R is a continuous function.

7.2. NORMED VECTOR SPACES 151

If X is a normed vector space, it is easy to check that d(x, y) = ‖x−y‖ defines a metricon X.

Definition 7.2.5. A Banach space is a normed vector space which is complete with respectto the metric defined by the norm.

Definition 7.2.6. Let X be a vector space over F. A map 〈·, ·〉 : X ×X → F is an innerproduct (or scalar product) on X if for all x, y, z ∈ X and all α ∈ F, the following holds:

(1) 〈x+ y, z〉 = 〈x, z〉+ 〈y, z〉.

(2) 〈αx, y〉 = α 〈x, y〉.

(3) 〈x, y〉 = 〈y, x〉.

(4) 〈x, x〉 > 0 if x 6= 0.

Here, for α ∈ F, we use the convention that α = α if F = R and α is the complexconjugate of α if F = C. Also, if α ∈ C, the statement α > 0 means that α ∈ R and α ispositive.

A vector space X equipped with an inner product is called and inner product space. Itis easy to see that an inner product defines a norm on X given by

‖x‖ =√〈x, x〉.

It is left to the reader to check the Schwarz inequality

| 〈x, y〉 | ≤ ‖x‖‖y‖

which can be used to prove the triangle inequality for the norm defined above.

Definition 7.2.7. An inner product space H is called a Hilbert space if H is a Banachspace with respect to the norm defined by its inner product.

Definition 7.2.8. Let J ⊂ R be an interval. A weight is a non-negative integrable functionw defined on J (it may assume the value of infinity) such that if g ∈ C0(J,R) and∫

Jw(t)g(t) dt = 0

then g ≡ 0.

We are primarily interested in the weights

w(t) ≡ 1 0 ≤ t ≤ pw(t) =

1√1− t2

− 1 ≤ t ≤ 1.


Example 7.2.9. Given 0 < p <∞ and f : X → C a Lebesgue measurable function, define

‖f‖p :=

(∫X|f |p dx

)1/p

and let Lp(X) be defined to be the set of all measurable functions f : X → C for which‖f‖p < ∞. If we identify functions which differ on a set of measure 0 then Lp(X) is aBanach space.

As the following example indicates there are different inner products that can be im-posed on L2([a, b],R).

Example 7.2.10. The Banach space L2([a, b],R) with inner product

(f, g) :=

∫ b

a

f(t)g(t)√1− t2

dt

is a Hilbert space

7.3 Calculus

The Inverse Function theorem.

Theorem 7.3.1. Let U ⊂ Rn be an open set. Let f : U → Rn be a Ck function with k ≥ 1.If Df(x) is invertible at a point x ∈ U , then there exits an open neighborhood V ⊂ U of xsuch that f : V → f(V ) is invertible with Ck inverse.

7.4 Analytic Functions

Here we will define analytic functions and introduce the multi-index notation. (Still workingon this)

The proof of this theorem is analogous to that of the one-dimensional ODE, howeverthe higher dimensionality leads to the need for more sophisticated notation. An n-tuple ofnonnegative integers α = (α1, . . . , αn) is called a multi-index. We define

|α| :=n∑k=1

αk and α! := α1!α2! · · ·αn!.

We extend this notation to vector valued functions. Given x ∈ Rn

xα := xα11 xα2

2 · · ·xαnn .

7.4. ANALYTIC FUNCTIONS 153

As an exercise in use of multi-index notation we leave it to the reader to check that

(x1 + . . .+ xn)k =∑|α|=k

|α|!α!

xα. (7.3)

We use multi-indices to express higher order derivatives, in particular

∂α :=

(∂

∂x1

)α1(

∂

∂x2

)α2

· · ·(

∂

∂xn

)αn.

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ordinary di erential equations: a constructive approach

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