Math 427

Introduction to Dynamical Systems

Winter 2012

Lecture Notes

by

Benjamin C. Wallace

Instructor

Prof. Dan Offin

Queen’s University

Dept. of Mathematics and Statistics

1 Introduction

The goal of this course is to understand the long term behaviour of systems that function accord-

ing to deterministic laws. For example, given a family of functions fµ : Rn → Rn parametrized

by µ ∈ Rl and given an initial condition x0 = (x01, . . . , x

0n) ∈ Rn, we may study the forward orbit

of x0 under fµ, that is, the sequence (xm)∞m=0, where xm+1 = fµ(xm) = fmµ (x0). Similarly, we

may wish to study the backward orbit (xm)−∞m=0. Note that when fµ is not invertible, the inverse

image x−m is actually the set f−mµ (x0) = {y ∈ Rn : fmµ (y) = x0}. In this case, there are many

possible backwards orbits we could study. We call xm the state of the system at time m. Since

m runs over a discrete set, this is a discrete-time system.

In the continuous case, rather than defining a system via iteration of some function, we use a

differential equation. We will be dealing with first-order ordinary differential equations (ODEs).

For example, for x, x0, and fµ as before, we may study the system given bydx

dt= fµ(x(t)). If

we choose an initial condition x0, then we get an initial value problem or IVP. Such a problem is

known to always have a unique solution t 7→ Φµ(t, x0). In continuous time, we define the forward

orbit of x0 as the set {Φµ(t, x0) : t ≥ 0} and the backwards orbit as the set {Φµ(t, x0) : t < 0}.We will also discuss equilibrium points of such systems, i.e. points x∗ at which f(x∗) = 0.

Example 1. Let fµ(x) = 1 + µx for µ 6= 0 and consider the simple discrete-time dynamical

system given by the recurrence relation

xn+1 = fµ(xn).

It is easy to see that xn = fnµ (x) = 1 + µ+ . . .+ µn−1 + µnx. First we notice that when µ 6= 1,

we have fµ

(1

1− µ

)=

1

1− µ. Let us look at some other possibilities.

1. For |µ| < 1, we see that fnµ (x)→ 1

1− µ.

2. For µ = 1, fnµ (x)→∞.

3. For µ = −1,

fnµ (x) =

{x− 1 n even

−x n odd.

So if −x = x− 1 = −1

2, the sequence (fnµ (x))∞n=1 converges to −1

2. Otherwise, it diverges

but contains two convergent subsequences, one converging to x− 1, the other −x.

4. For µ > 1, the limit depends on the value of x.

(a) When x >1

1− µ, fnµ (x)→∞.

1

(b) When x <1

1− µ, fnµ (x)→ −∞.

5. For µ < −1, the limit again diverges.

The following definitions apply to a discrete-time dynamical system xn+1 = f(xn), where

f : R→ R.

Definition. A point x0 is periodic (with period n) if fn(x0) = x0 and f j(x0) 6= x0 for j =

1, . . . , n− 1. If n = 1, x is called a fixed point.

So when µ 6= 1 in the previous example,1

1− µis a fixed point of fµ.

Definition. We say that a point x is forward asymptotic to a fixed point p when |fn(x)−p| → 0.

The set of points forward asymptotic to p is called the stable set of p and is denoted W s(p).

In the previous example, for |µ| < 1, every point is forward asymptotic to the fixed point1

1− µ, so W s

(1

1− µ

)= R.

Definition. A point x is backwards asymptotic for a fixed point p when there is a sequence

(xn)0n=−∞, with each xn ∈ f−n(x), such that |x−n − p| → 0. The set of backwards asymptotic

points to p is called the unstable set of p and is denoted W u(p).

2 Linear Systems

In this section, we consider systems of linear differential equations of the form x(t) = A(t)x(t),

where x(t) = (x1(t), . . . , xn(t))T ∈ Rn and A(t) is an n × n matrix, for each t. We shall begin

by focusing on the case of autonomous systems, i.e. ones in which A(t) = A. In particular, we

shall study the system {x(t) = Ax(t)

x(0) = x0

. (1)

Recall that such systems always have a unique solution.

First, suppose A has n distinct real eigenvalues λ1, . . . , λn with associated linearly indepen-

dent eigenvectors v1, . . . , vn. Recall that in this situation, these eigenvectors form a basis of Rn

and can be used to diagonalize A via the change of basis

B−1AB =

λ1

. . .

λn

,where B =

[v1 . . . vn

]is an invertible matrix. Thus, letting y = B−1x, so that x = By, we

get

y(t) = B−1x(t) = B−1Ax(t) = B−1ABy(t).

2

That is, if y = (y1, . . . , yn), we have yi(t) = λiyi(t) for i = 1, . . . , n. Thus, for each i, we can

solve for yi(t) = eλity(0) or

y(t) =

eλ1t

. . .

eλnt

B−1x(0).

But then it follows that

x(t) = B

eλ1t

. . .

eλnt

B−1x(0).

Alternately setting x(0) = Be1, . . . , Ben, where e1, . . . , en is the standard basis of Rn, we get n

linearly independent solutions x1, . . . , xn to the system given by

xi(t) = B

eλ1t

. . .

eλnt

ei = eλitvi.

We can form the non-singular matrix of these solutions

X(t) =[v1 . . . vn

]eλ1t

. . .

eλnt

.Definition. A fundamental matrix solution to the system x(t) = Ax(t) ∈ Rn is a mapping

t 7→ Φ(t) ∈ L(Rn,Rn) such that Φ = AΦ.

So X, as defined above, is a fundamental matrix solution to (1).

The fundamental matrix solution X of a linear system x = Ax has the following properties,

which are easily seen from the definition.

1.d

dtX(t) = AX(t).

2. The solution to the IVP x = Ax with initial condition x(0) = x0 is given by x(t) =

X(t)X−1(0)x0.

3. If Y(t) = X(t)X−1(0), then Y(t) is a fundamental matrix solution to x = Ax and Y(t+s) =

Y(t)Y(s).

4. With Y(t) as above, (Y(t))−1 = Y(−t).

3

2.1 Complex Eigenvalues

Next, suppose A has a complex eigenvalue λj = aj + ibj with eigenvector vj = uj + iwj . As

before, we can diagonalize A and obtain the solution

xj(t) = eλjtvj

= e(aj+ibj)t(uj + iwj)

= eajt((cos(bjt)uj − sin(bjt)wj) + i(cos(bjt)wj + sin(bjt)uj)).

But by linearity of our system,

x1j (t) = eajt(cos(bjt)uj − sin(bjt)wj)

and

x2j (t) = eajt(sin(bjt)uj + cos(bjt)wj)

are also solutions, and moreover, they are real and independent. We can write these two solutions

together as [x1j x2

j

]= eajt

[uj wj

] [ cos(bjt) sin(bjt)

− sin(bjt) cos(bjt)

].

More generally, if A has both real and complex eigenvalues (but nevertheless has n linearly

independent eigenvectors), we get the fundamental matrix solution

X(t) =[v1 . . . vn

]A1(t)

. . .

An(t)

,where Ai(t) = eλit when λi ∈ R and

Ai(t) = eait

[cos(bit) sin(bit)

− sin(bit) cos(bit)

]

and vi =[ui wi

]when λi = ai + ibi ∈ C \ R and vi = ui + iwi.

2.2 The Matrix Exponential

Here we shall give an alternative formulation of the fundamental matrix solution. First, let

us review some basic notions from analysis while establishing notation. We will denote the

Euclidean norm by | · |.

Definition. We define the operator norm

‖ · ‖ : L(Rn,Rn)→ R

by

‖T‖ = max|x|≤1

|Tx|.

4

One can check that ‖ · ‖ indeed satisfies the properties of a norm.

Lemma 2.1. For any X ∈ Rn,

|Tx| ≤ ‖T‖|x|.

Proof. The case x = 0 is trivial so we assume x 6= 0. Letting y =1

|x|x, we see that since |y| = 1,

‖T‖ ≥ |Ty| = 1

|x||Tx|,

from which the result follows.

We can now define the matrix exponential.

Definition. We define the matrix exponential by

eT =∞∑n=0

Tn

n!.

Notice that

e‖T‖ −N∑n=0

‖T‖n

n!= ‖

∞∑n=N+1

Tn

n!

≤∞∑

n=N+1

‖T‖n

n!

→ 0,

so eT is absolutely and uniformly convergent for any T ∈ L(Rn,Rn).

Proposition 2.2 (Properties of the matrix exponential).

1. e0 = I.

2. If AB = BA, then eA+B = eAeB.

3. eA(t+s) = eAteAs.

4. det(eAt) 6= 0.

5. If B is invertible, then eBAB−1

= BeAB−1.

6.d

dteAt = AeAt.

Proof. Part (2) follows from the fact that the binomial theorem applies to commuting matrices.

Part (6) follows from (2) and the uniform convergence of the matrix exponential. The other

parts follows quite directly from the definition of the matrix exponential.

Corollary 2.3. The matrix exponential eAt is the fundamental matrix solution to x = Ax.

5

2.3 Generalized Eigenvectors

Recall the following definitions.

Definition. Let A be a matrix. The algebraic multiplicity of an eigenvalue λ of A is its mul-

tiplicity as a root of the characteristic polynomial p(x) = det(A − xI) of A. The geometric

multiplicity of λ is the dimension its associated eigenspace, i.e. dim(Ker(A− λI)).

In this section we consider the case where A does not have n linearly independent eigenvec-

tors. For instance, let λ1, . . . , λn be the eigenvalues of A, but suppose one of these, call it λ, has

geometric multiplicity less than its algebraic multiplicity. In this case, we must “enlarge” the

eigenspace of λ.

Definition. If m is the algebraic multiplicity of an eigenvalue λ of a matrix A, then we call a

vector v ∈ Ker((A − λI)m) a generalized eigenvector. The set of all generalized eigenvectors of

λ is called the generalized eigenspace of λ and is a vector space.

The following lemma is quite easy to see.

Lemma 2.4. For any n× n matrix S, we have

Ker(S) ⊂ Ker(S2) ⊂ . . . ⊂ Ker(Sk) = Ker(Sk+1) = . . .

for some k.

So for some k, we have

Ker((A− λI)k−1) ⊂ Ker((A− λI)k) = Ker((A− λI)k+1).

For such k, we can find k linearly independent non-zero vectors ξ1, . . . , ξk satisfying

ξi ∈ Ker(A− λI)i, i = 1, . . . , k.

We call such ξ1, . . . , ξk a Jordan chain. Let’s assume for now that λ ∈ R so that the ξi ∈ Rn.

Notice that

Aξ1 = λξ1

Aξ2 = λξ2 + ξ1

...

Aξk = λξk + ξk−1

or in matrix form,

A[ξ1 . . . ξk

]=[ξ1 . . . ξk

]λ 1

λ. . .

. . . 1

λ

.

6

Now suppose λ = a+ ib ∈ C so that ξi = ui + iwi for i = 1, . . . , k. Proceeding as before, we

get

Au1 + iAw1 = (a+ ib)(u1 + iw1)

= (au1 − bw1) + i(bu1 + aw1)

Auj + iAwj = (a+ ib)(uj + iwj) + (uj−1 + iwj−1)

= (auj − bwj + uj−1) + i(buj + awj + wj−1),

so that

Au1 = au1 − bw1

Aw1 = bu1 + aw1

Auj = auj − bwj + uj−1

Awj = buj + awj + wj−1

for j = 2, . . . , k. In matrix form, this is

A[u1 w1 . . . uk wk

]=[u1 w1 . . . uk wk

]D I2

D. . .

. . . I2

D

,

where

D =

[a b

−b a

].

Matrices of the form ofD I2

D. . .

. . . I2

D

or λI +N =

λ 1

. . .. . .

λ 1

λ

are known as elementary Jordan blocks.

Theorem 2.5 (The Jordan normal form). For any matrix A, there is a basis v1, . . . , vn of Rn

made up of real generalized eigenvectors of A and real and imaginary parts of complex generalized

eigenvectors of A such that letting

B =[v1 . . . vn

]

7

we get

B−1AB =

J1

. . .

Jk

,a block diagonal matrix with each Ji an elementary Jordan block.

Corollary 2.6. Let A be a matrix and let v1, . . . , vn be a matrix of Rn such that B−1AB is in

the Jordan normal form, where B =[v1 . . . vn

].

eAt = B

eJ1t

. . .

eJlt

B−1.

This corollary is very useful to us because each eJi can in fact be computed very easily. For

instance, Jordan blocks of real eigenvalues are of the form

Ji = λiI +Ni,

where Ni is a nilpotent matrix. That is, Nmi = 0 for some m. Thus, eJi can be computed by a

finite sum.

2.4 Invariant Subspaces

Definition. Let T : Rn → Rn be a linear map. A subspace U ⊆ Rn is said to be invariant

under T if TU ⊆ U .

For instance, any eigenspace or generalized eigenspace of an eigenvalue of a matrix A is

clearly invariant under A.

Definition. The flow of a linear system x = Ax is the linear mapping given by the fundamental

matrix solution eAt.

We will be especially interested in subspaces invariant under the flow of a linear system.

Theorem 2.7. A subspace U ⊆ Rn is invariant under eAt if and only if it is invariant under

A.

Example 2. Let

A =

−2 −1 0

1 −2 0

0 0 3

.

8

This matrix has eigenvalues λ1 = −2 + i and λ2 = 3. Corresponding to λ1 are is the complex

eigenvector

v = v1 + iv2 =

0

1

0

+ i

1

0

0

and corresponding to λ2 is the real eigenvector

v3 =

0

0

1

.We thus get invariant subspaces

U1 = span(v1, v2)

and

U2 = span(v3).

Also,

eAt =

0 1 0

1 0 0

0 0 1

[eJ1t 0

0 e3t

]0 1 0

1 0 0

0 0 1

,where

eJ1t = e−2t

[cos t sin t

− sin t cos t

].

Now we can see that eAt acts on the invariant subspace U1 by performing a counterclockwise

rotation combined with an exponential contraction. Hence, the orbit of a point in U1 under

eAt forms a spiral asymptotically approaching the origin. Meanwhile, its action on U2 is an

exponential dilation. That is, points on U2 get “stretched” away from the origin. The action on

any other point in R3 is then a linear combination of these two actions; an exponential stretch

upwards or downwards combined with an exponential contraction and rotation inwards, towards

U2.

Definition. We call a function I : Rn → R a conserved quantity or an integral of motion for the

system x = Ax (where x ∈ Rn) if I is a constant function of time, i.e. the composed function

I ◦ x : R→ R is constant.

Definition. A set S is called an invariant set for the system x = Ax if it is invariant under the

action of eAt.

Notice that if I is a conserved quantity, then its level sets are invariant sets.

9

Definition. Let A be a matrix with eigenvalues λj and corresponding eigenvectors vj (when

real) or vj + iuj (when complex). We define the stable subspace

Es = span(vj , uj : Re(λj) < 0),

the unstable subspace

Eu = span(vj , uj : Re(λj) > 0),

and the center subspace

Ec = span(vj , uj : Re(λj) = 0)

of A.

Theorem 2.8. We have the subspace decomposition

Rn = Es ⊕ Eu ⊕ Ec.

Proof. This follows directly from the definitions of Es, Eu, and Ec and the fact that the vj and

uj form a basis of Rn.

Theorem 2.9. The following are equivalent:

1. for all x0 ∈ Rn, limt→∞

eAtx0 = 0 (respectively, limt→−∞

eAtx0 = 0);

2. all eigenvalues of A have negative (respectively, positive) real part; and

3. there are positive constants a,m,M , and k such that

m|tk|e−at|x0| ≤ |eAtx0| ≤Me−at|x0|

for all t ≥ 0 (respectively, change e−at to eat).

3 Nonlinear Systems

We turn to autonomous nonlinear differential equations. That is, we consider equations of the

form x = f(x), where f : E → Rn is continuously differentiable on the open subset E ⊆ Rn. We

begin by studying the continuity of solutions to an initial value problem in the initial value. So

consider the initial value problem {x = f(x)

x(0) = y. (2)

Theorem 3.1 (Existence/uniqueness). Let E ⊆ Rn be open and let f : E → Rn be C1 on E.

Given y ∈ E, there exists a > 0 such that the IVP (2) has a unique solution x : [−a, a]→ Rn.

We do not prove this theorem here but note that the proof relies on Banach’s fixed-point

theorem, which we recall below.

10

Theorem 3.2 (Banach’s fixed-point theorem). Let (X, d) be a complete metric space and sup-

pose T : X → X is a contraction mapping, i.e. there exists a constant 0 ≤ k < 1 such that

d(x, y) ≤ kd(x, y). Then T has a unique fixed point, i.e. there exists x∗ ∈ X such that T (x) = x

if and only if x = x∗. Moreover, for any x0 ∈ X,

x∗ = limn→∞

xn,

where xn = T (xn−1).

Let u(t, y) = y +

∫ t

0f(u(s, y)) ds be the solution to the IVP (2). We know that u(t, y)

is continuous in t for each initial condition y. However, we wish to show next that it is also

continuous in y. That is, a given point on the curve traced by the solution to the IVP should

change very little as a result of a small change in the initial condition. We first prove the

following lemma.

Lemma 3.3 (Gronwall’s inequality). Assume that g : R→ R is continuous on an interval [0, a]

and suppose that for some C,K > 0,

g(t) ≤ C +

∫ t

0Kg(s) ds

for all t ∈ [0, a]. Then g(t) ≤ CeKt for t ∈ [0, a].

Proof. Let G(t) = C +

∫ t

0Kg(s) ds. So g(t) ≤ G(t) for t ∈ [0, a] and G′(t) = Kg(t). We thus

get, for all t ∈ [0, a],

G′(t)

G(t)=Kg(t)

G(t)≤ K ⇔ d

dtln(G(t)) ≤ K

⇒ ln(G(t)) ≤ ln(G(0)) +Kt

⇒ g(t) ≤ G(t) ≤ CeKt.

Recall that a function f : E → Rn, where E ⊆ Rm is said to be differentiable at a point

x0 ∈ E if there is a linear map Df(x0) : Rm → Rn such that

limh→0

f(x0 + h)− f(x0)−Df(x0)h

|h|= 0.

In this case, we call the matrix representative of Df(x0), which consists of partial derivatives of

f at x0, the Jacobian matrix of f at x0. We are now ready to state the conditions under which

the solutions of a nonlinear IVP vary smoothly in the initial conditions.

11

Theorem 3.4 (Smooth dependence on initial data). Let E ⊆ Rn be open and let f : E → Rn

be C1 on E. Given x0 ∈ E, there exist a, δ > 0 such that for all y ∈ Nδ(x0) and all t ∈ [−a, a],

the solution u(t, y) to the IVP (2) is C1 on some interval G = [−a, a]×Nδ(x0) ⊆ Rn+1.

Proof. We know that u(t, y) = y +

∫ t

0f(u(s, y)) ds is C2 in t, since f is C1, and that u(t, y) =

f(u(t, y)), so u(t, y) = Df(u(t, y))f(u(t, y)). Let y ∈ Nδ/2(x0) and h ∈ Rn with |h| < δ/2 so

that y + h ∈ Nδ(x0). We have

u(t, y + h)− u(t, y) = h+

∫ t

0(f(u(s, y + h))− f(u(s, y))) ds,

and so letting g(t) = |u(t, y + h)− u(t, y)|, we get

g(t) ≤ |h|+∫ t

0K|u(s, y + h)− u(s, y)| ds,

wher K = maxx∈Nδ(x0)

‖Df(x)‖. Thus, by Gronwall’s inequality,

g(t) ≤ |h|eKt ≤ |h|eKa, t ∈ [−a, a]. (3)

Hence, we see that y 7→ u(t, y) is continuous for y ∈ Nδ/2(x0).

Next, we need to show that

limh→0

|u(t, y0 + h)− u(t, y0)− lh||h|

= 0

for some linear transformation l : Rn → Rn. Let Φ(t, y0) denote the unique fundamental matrix

solution of the linear system

Φ(t, y) = Df(u(t, y0))Φ(t, y)

Φ(0, y0) = I.

We claim that Φ is the required linear map l. Letting u1(s) = u(s, y0 + h) and u0(s) = u(s, y0),

12

we see that

|u1(t)− u0(t)− Φ(t, y0)h| =∣∣∣∣h− Φ(t, y0)h+

∫ t

0(f(u1(s))− f(u0(s))) ds

∣∣∣∣=

∣∣∣∣∫ t

0(f(u1(s))− f(u0(s))−Df(u0(s))Φ(s, y0)h) ds

∣∣∣∣≤∫ t

0|(f(u0(s))− f(u0(s))−Df(u0(s))Φ(s, y0)h)| ds

=

∫ t

0|R(s)(u1(s)− u0(s)) +Df(u0(s))(u1(s)− u0(s)− Φ(s, y0)h)| ds

≤∫ t

0(ε0|h|eKa + |Df(u0(s))(u1(s)− u0(s)− Φ(s, y0)h)| ds, (by (3))

≤ ε0|h|eKat+

∫ t

0‖Df(u0(s))‖|u1(s)− u0(s)− Φ(s, y0)h| ds

≤ ε0|h|eKaa+

∫ t

0‖Df(u0(s))‖|u1(s)− u0(s)− Φ(s, y0)h| ds,

where R(s) is a remainder term in the Taylor series expansion of f about u0(s) evaluated at

u1(s). But by continuity, we have that ‖Df(u0(s))‖ = ‖Df(u(s, y0))‖ ≤M for some M and for

all (s, y0) ∈ [−a, a]×Nδ(x0). Thus,

|u1(t)− u0(t)− Φ(t, y0)h| ≤ ε0|h|eKaa+

∫ t

0M |u1(s)− u0(s)− Φ(s, y0)h| ds

≤ ε0|h|eKaaeMt

≤ ε0|h|e(K+M)aa.

This shows that y 7→ u(t, y) is differentiable and has derivative

∂

∂y

∣∣∣∣y=y0

u(t, y) = Φ(t, y0).

Moreover, since Φ(t, y0) is continuous in y0, we have that u(t, y) is C1 over [−a, a]×Nδ(x0).

Corollary 3.5 (Corollary (to the proof)). Let u(t, y) be the solution to the IVP{x(t) = f(x(t))

x(0) = y.

Then Φ(t, y) =∂u

∂y(t, y) if and only if Φ(·, y) is the solution of the IVP

d

dtΦ(t, y) = Df(u(t, y))Φ(t, y)

Φ(0, y) = In×n

.

13

Definition. Let u(t, y) be as above. Then the LTV system

ξ(t) = Df(u(t, y))ξ(t),

where ξ(t) ∈ Rn, is called the variational equation for u(t, y).

To finish off, let us adapt some definitions from the introduction to the context of continuous-

time nonlinear systems.

Definition. Let u(t, y) denote the solution to the systemx = f(x)

x(0) = y

evaluated at time t. If for some x0 ∈ Rn there exists T > 0 such that u(T, x0) = x0, then x0 is

called a periodic point of x, the corresponding trajectory u(t, x0) is said to be periodic, and the

minimal such T is referred to as the period of x0 or of u(t, x0). If u(t, x0) = x0 for all t, then x0

is called a fixed point of x.

Definition. If f(x0) = 0, then x0 is called an equilibrium point of the vector field f .

Notice that fixed-points and equilibrium points are the same; however, it is common to see a

fixed-point in terms of the solution curves to a given differential equation while when we talk of

equilibrium points, we are thinking in terms of vector fields used to define differential equations.

Definition. If A and B are two subsets of Rn, define their distance d(A,B) = infa∈A,b∈B

|a − b|.

A periodic orbit u(t, x0) is said to be stable if for each ε > 0, there is a neighbourhood U of the

trajectory Γ = {u(t, x0) ∈ Rn : t ∈ R} of u(t, x0) (i.e. an open set U containing Γ) such that

for all x ∈ U and all t ≥ 0, d(u(t, x),Γ) < ε. If, moreover, there is a neighbourhood V of Γ

such that for all x ∈ V , limt→∞

d(u(t, x),Γ) = 0, then u(t, x0) is said to be asymptotically stable. If

u(t, x0) is not stable, then we say it is unstable.

Definition. Let U ⊆ Rn and consider the trajectory u(t, x). We define the stable and unstable

sets of P respectively by

W s(U) = {x ∈ Rn : limt→∞

d(Φt(x), U) = 0}

W u(U) = {x ∈ Rn : limt→−∞

d(Φt(x), U) = 0}.

3.1 Example of a Non-Autonomous System

Consider the systemdy

dt= (a cos t+ b)y − y3, (4)

14

where a, b > 0. This nonlinear system is non-autonomous, so the results we have so far cannot

be directly applied to it. However, we can transform the above system into

dy

dt= (a cos τ + b)y − y3 (5)

dτ

dt= 1 (6)

or

x(t) = f(τ, y) = (1, (a cos τ + b)y − y3),

where x = (τ, y) and the planar vector field f ∈ C1(R2) is periodic in τ with period 2π. We

thus identify τ with τ + 2kπ for all k ∈ Z. This is akin to folding a vertical rectangular strip of

the plane R2 of width 2π into a cylinder and considerin the vector field f over this cylinder, i.e.

the set T× R.

Now notice that the curve y = 0 (a circular cross-section of our cylinder) is invariant invariant

under the flow defined by the system, i.e. if we set the initial condition y(0) = 0, then y(t) = 0

for all t. Now consider the curve y = a+ b+ 1 (another circular cross-section). Along this curve,

y(t) = (a cos t+ b)(a+ b+ 1)− (a+ b+ 1)3 < 0

as can be verified. Thus, the vector field points downward along this curve, i.e. this curve

repels the flow downward. But the lower circle y = 0 is fixed, hence by continuity, the compact

cylindrical region

C = {(τ, y) ∈ T× R : 0 ≤ y ≤ a+ b+ 1}

is positively invariant under the flow, i.e. it is invariant for t ≥ 0.

Next, we will show that there is a unique attracting periodic cycle within C. To do this, we

study iterations of the map

P : [0, a+ b+ 1]→ [0, a+ b+ 1]

y0 7→ y(2π, 0, y0),

where we denote by y(t, τ0, y0) the solution to the system (5) with initial condition (τ(0), y(0)) =

(τ0, y0). Clearly, P (0) = 0, so 0 is a fixed point of P . We claim the following:

1. The fixed point 0 is repelling.

2. There exists a unique attracting fixed point ξ ∈ [0, a+ b+ 1] for P .

3. ξ = W s(0, a+ b+ 1].

To prove these claims, let Φ(t, y0) =∂y

∂y0(t, 0, y0). Then Φ is the solution to the IVP

Φ(t, y0) = ((a cos t+ b)− (y(t, y0))2)Φ(t, y0)

Φ(0) = 1.

15

Then since y(t, 0, 0) = 0 and

P ′(0) =∂y

∂y0(2π, 0, 0) = Φ(2π, 0),

we get

Φ(t, 0)

Φ(t, 0)= a cos t+ b⇒ ln Φ(t, 0)− ln Φ(0, 0) =

∫ t

0(a cos t+ b) dt

⇒ ln Φ(t, 0) =

∫ t

0(a cos t+ b) dt

⇒ P ′(0) = exp

(∫ 2π

0(a cos t+ b) dt

)= e2πb > 1.

So 0 is a repelling fixed point. It follows directly from the intermediate value theorem that

[0, a + b + 1] contains some other fixed point of P . By similar reasoning, if we can show that

every fixed point in the interior of [0, a+ b+ 1] must be stable, then it will follow that the stable

fixed point we have found is unique. Otherwise, in a manner similar to the way in which the

unstable fixed point 0 forced the existence of a stable fixed point, another stable fixed point

would lead to another unstable fixed point.

To prove this claim, begin by noticing that P (ξ) = y(2π, 0, ξ) is C1 so that P ′(ξ) = yξ(2π, 0, ξ)

is continuous. Moreover, P ′(ξ) is the unique solution to the variational equationd

dtyξ(t, 0, ξ) = [(a cos t+ b)− 3y2(t, 0, ξ)]yξ(t, 0, ξ)

yξ(0, 0, ξ) = 1.

Dividing both sides of the above differential equation by yξ(t, 0, ξ) and evaluating the result at

a point ξ0 gives usd

dtln yξ(t, 0, ξ0) = (a cos t+ b)− 3y2(t, 0, ξ0).

Integrating over [0, 2π], we get

ln yξ(2π, 0, ξ0) =

∫ 2π

0((a cos t+ b)− 3y2(t, 0, ξ0)) dt

= 2πb−∫ 2π

03y2(t, 0, ξ0) dt,

and thus

P ′(ξ0) = yξ(2π, 0, ξ0) = e2πbe−∫ 2π0 3y2(t,0,ξ0) dt.

To simplify this further, integrate

d

dtln y(t, 0, ξ0) = (a cos t+ b)− y2(t, 0, ξ0),

16

to get

ln y(2π, 0, ξ0)− ln y(0, 0, ξ0) = 2πb−∫ 2π

0y2(t, 0, ξ0) dt.

So when ξ0 is a fixed point, the left hand side of this expression is 0 by periodicity, so∫ 2π

0y2(t, 0, ξ0) dt = 2πb

and so

P ′(ξ0) = e2πbe−2πb = e−4πb < 1,

since b > 0.

3.2 Extensibility of Solutions

In this section, we wish to identify the maximal interval on which an initial value problem has

a well-defined solution.

Lemma 3.6. Suppose x1, defined on the open interval I1 and x2, defined on the open interval

I2, are solution to the initial value problemx = f(x),

x(0) = y, f : R→ E ⊆ Rn.

Then if there is a point t0 ∈ I1∩I2 such that x1(t0) = x2(t0), then x1(t) = x2(t) for all t ∈ I1∩I2.

Proof. By existence and uniqueness, if x1(t0) = x2(t0), then there exists a closed interval [−a+

t0, t0 + a] ⊆ I1 ∩ I2 on which x1(t) = x2(t). In particular, x1 and x2 agree on the open interval

(−a+ t0, t0 + a) obtained by removing the endpoints t0 ± a. Let I∗ be the maximal such open

interval and suppose, by way of contradiction, that I∗ 6= I1 ∩ I2. Writing I∗ = (α, β), we

must have by continuity that x1 and x2 agree at the endpoints α and β of this interval. But

then taking α and β as initial conditions to the initial value problem with differential equation

x = f(x), it follows from existence and uniqueness of solutions that x1 and x2 agree on an

open interval about α and β. But this contradicts the maximality of I∗. Hence, we must have

I∗ = I1 ∩ I2.

3.3 The Flow Operator

Definition. Let f ∈ C1(E), where E ⊆ Rn is open. The flow of the vector field f is the unique

solution Φ(t, y) to the initial value problemx = f(x)

x(0) = y ∈ E

defined on the maximal interval of existence J(y) to this problem. We often write Φt : E → E,

where Φt(y) = Φ(t, y). This mapping is sometimes called the flow operator.

17

Note that the flow operator is in general nonlinear.

Proposition 3.7. The flow operator satisfies the group composition property

Φ(t,Φ(s, y)) = Φ(t+ s, y).

Proof. This follows directly from existence and uniqueness and from the previous lemma.

A familiar example is found in the case of linear systemsx = Ax

x(0) = y,

where the flow operator is simply the matrix exponential Φ(t, y) = eAty.

Definition. Suppose U, V ⊆ Rn are open and F : U → V is continuous on U . Then F is called

a homeomorphism if there exists a continuous function G : V → U such that F ◦G = idV and

G ◦ F = idU . If both F and G are C1 on their respective domains, then F (and by symmetry,

G) is called a diffeomorphism.

For instance, x1/3 is a homeomorphism on R but not a diffeomorphism because its derivative

is undefined at 0. Of greater significance is the fact that for f ∈ C1(E), the flow operator

Φt : E → E of x = f(x) is a diffeomorphism. We have already shown that Φ is continuously

differentiable. Since the inverse of Φt is simply the flow operator Φ−t, it follows that it too is

C1.

The following familiar theorem serves as an excellent tool for proving that a given mapping

is a diffeomorphism.

Theorem 3.8 (Inverse function theorem). Suppose that F : Rn → Rn is C1 and let p ∈ Rn be

such that F (p) = q ∈ Rn and such that DF (p) is non-singular. Then there exist neighborhoods

U and V of p and q, respectively, and a C1 function G : V → U such that G ◦ F = idV and

F ◦G = idU .

We do not prove the inverse function theorem. A proof can be obtained via Banach’s fixed

point theorem.

Corollary 3.9. If F (p) = q and DF (p) is non-singular, then with G as in the above, DG(q) =

[DF (p)]−1.

Proof. Since G ◦ F (x1, . . . , xn) = (x1, . . . , xn),

I = D(G ◦ F )(p)

= DG(F (p)) ·DF (p)

= DG(q) ·DF (p),

from which our claim follows.

18

Let f ∈ C1(E) and let u : W ′ →W be a diffeomorphism for some open sets W ′ and W . We

can “pull back” the differential equation x = f(x) to the domain W ′ via the function u by a

simple change of coordinates. Supposing x = u(y), we get

f(x) =dx

dt=∂u

∂y(y)

dy

dt.

The resulting expression for y is given in terms of a specially named vector field.

Definition. Let f and u be as above. We define the pullback of f to be the vector field defined

by

f∗(y) =

(∂u

∂y(y)

)−1

f(u(y)).

Let us recall another important theorem.

Theorem 3.10 (Implicit function theorem). Let f : Rn×Rm → Rm be C1 on an open set U× V

and suppose f(x0, y0) = 0 for some x0 ∈ Rn and y0 ∈ Rm. Moreover, assume that∂f

∂y(x0, y0)

(i.e the matrix of partial derivatives of f with respect to the y variables only) is non-singular.

Then there exist neighbourhoods U of x0 and V of y0 and a C1 function g : U → V such that

f(x, y) = 0 if and only if y = g(x) for all (x, y) ∈ U × V .

Proof. Let us define F : Rn × Rm → Rn × Rm by F (x, y) = (x, f(x, y)) = (x, y′), where

y′ = f(x, y). Then F (x0, y0) = (x0, f(x0, y0)) = (x0, 0). Now notice that since

DF =

In×m 0∂f

∂x

∂f

∂y

,then by the assumption that

∂f

∂yis non-singular,

det(DF ) = det(In×m) det

(∂f

∂y

)= det

(∂f

∂y

)6= 0.

So by the inverse function theorem, there are open sets U × V about (x0, y0) and U × V ′ about

(x0, 0) and a smooth function G : U×V ′ → U×V such that F ◦G = idU×V ′ and G◦F = idU×V .

Moreover, writing G(x, y′) = (G1(x, y′), G2(x, y′)), where G1(x, y′) = x and G2(x, y′) = y, we

see that y′ = f(x, y) = 0 if and only if y = G2(x, 0) = g(x), where g : U → V is smooth and

g(x0) = y0.

3.4 The Poincare Map

Let us consider a vector field f : E → Rn which is C1 on the open subset E ⊆ Rn. Also, let

S = {x ∈ Rn : l(x) = h} be a hyperplane in Rn, where h ∈ R and l : Rn → R is a linear map.

19

We will call an open1 subset Σ ⊆ S of such a hyperplane a section of Rn and say that f is

transverse to Σ if f(x) is never tangent to Σ for x ∈ Σ. Let us rephrase this a little bit.

We denote a vector at a point x ∈ Rn by a pair (x, v) ∈ Rn × Rn and define the tangent

plane of Σ at x to be the set

TxΣ = {(x, v) : l(v) = 0}

of vectors at x that are tangent to Σ. Thus, for f to be transverse to Σ means that (x, f(x)) /∈TxΣ for any x ∈ Σ. Equivalently, for any x ∈ Σ, we must have l(f(x)) 6= 0. Since f is C1, we

can assume without loss of generality that l(f(x)) > 0 for x ∈ Σ. Clearly, f(x) cannot change

sign on Σ otherwise it would be 0 somewhere on Σ.

Now suppose that for some x ∈ Σ and T > 0 we have ΦT (x) ∈ Σ, where Φt : E → E is

the flow of the vector field f . Then we can find some open neighbourhood N ⊆ Σ of x (in the

relative topology of Σ) such that for all x′ ∈ N we have |ΦT (x)− ΦT (x′)| < ε and l(f(x′)) > 0.

Shrinking Σ down to the size of this neighbourhood, we can see that there is a return mapping

T : Σ→ R such that ΦT (x′)(x′) ∈ Σ for all x′ ∈ Σ.

Definition. In the above scenario, we call the mapping P : Rn → Rn defined by P (x) = ΦT (x)(x)

the Poincare mapping on the section Σ.

Theorem 3.11. The return map T : Rn → R is continuously differentiable.

Proof. Let F : R×Rn be the C1 function defined by F (t, ξ) = l(Φt(ξ))−h, so that F (T, x0) = 0.

Moreover,∂F

∂t(T, x0) = l

(d

dtΦt(ξ)

)∣∣∣∣ξ=x0t=T

= l(f(ΦT (x0))) 6= 0,

by transversality. Thus, we can apply the inverse function theorem to get an open neighbourhood

Σ′ ⊆ Σ of x0 and an open neighbourhood V ⊆ R of T and a C1 function T : Σ′ → V such that

for all (t, ξ) ∈ V × Σ′ we have F (t, ξ) = 0 if and only if t = T (ξ). In addition, T (x0) = T . In

other words, the first return time map T is C1.

We thus have the follow important property of the Poincare map.

Corollary 3.12. The Poincare map P as above is continuously differentiable.

Proof. With F (t, ξ) = l(Φt(ξ)) − h as above, we have that P (x) = F (T (x), x) is a composition

of C1 functions and so is itself C1.

Theorem 3.13. If v ∈ Tx0Σ, then the directional derivative of the Poincare map at x0 in the

direction v is given by

DP (x0)v = f(x0)〈∇T (x0), v〉+ uξ(T, x0)v,

1If S is a subspace of a topological space (X, τ), the relative topology on S is the topology {S ∩ U : U ∈ τ}consisting of intersections of S with the open sets of X.

20

where u(t, ξ) is the solution to the IVPx(t) = f(x(t))

x(0) = ξ.

Proof. This follows by a straightforward application of the chain rule:

DP (x0)v =d

dε

∣∣∣∣ε=0

P (x0 + εv)

= ut(T (x0), x0)〈∇T (x0), v〉+ uξ(T (x0), x0)v

= f(x0)〈∇T (x0), v〉+ uξ(T, x0)v.

3.5 Stability of Periodic Solutions

Theorem 3.14. Let P be the Poincare map corresponding to a periodic trajectory u(t, x0) in

the plane R2. Then u(t, x0) is asymptotically stable if and only if P ′(x) < 1 for some x along

this trajectory.

Recall that if u(t, ξ) is the solution to the IVPx(t) = f(x(t))

x(0) = ξ,

then uξ(t, x0) is the solution to the variational equationd

dtuξ(t, x0) = Df(u(t, x0))uξ(t, x0)

uξ(0, x0) = In×n

.

Let us denote Φ(t, x0) = uξ(t, x0).

Lemma 3.15. f(u(t, x0)) = Φ(t, x0)f(x0).

Proof. By evaluating the derivative of f with respect to t along the solution u(t, x0), we get

d

dtf(u(t, x0)) = Df(u(t, x0))

d

dtu(t, x0)

= Df(u(t, x0))f(u(t, x0)).

It is also clear that f(u(0, x0)) = f(x0). Thus, since Φ(t, x0) is the fundamental matrix solution

to the equationd

dtf(u(t, x0)) = Df(u(t, x0))f(u(t, x0)),

it follows that f(u(t, x0)) = Φ(t, x0)f(x0).

21

Corollary 3.16. f(x0) = Φ(T, x0)f(x0).

Theorem 3.17. If Γ is the periodic orbit of x0 and Σ is a section such that x0 = Γ ∩ Σ, then

Φ(T, x0) has eigenvalues 1, λ1, . . . , λn−1, where λ1, . . . , λn−1 are the eigenvalues of DP (x0).

Proof. That 1 is an eigenvalue is evidenced directly by the above corollary. To show that the

remaining n− 1 eigenvalues of Φ(T, x0) are identical to those of DP (x0), we choose

α = {f(x0), s1, . . . , sn−1}

as a basis of Rn, where s1, . . . , sn−1 is a basis of the tangent space Tx0Σ of Σ at x0. The vectors

in α are indeed linearly independent by transversality of f(x0) to the tangent space. Under this

basis, the matrix representative of Φ(T, x0) is given in block form by

[Φ(T, x0)]α =

1 · · · a · · ·...

. . .

0 b...

. . .

,

where a = (a1, . . . , an−1) and b = (bij)n−1i,j=1. Also in the basis α, a tangent vector v ∈ Tx0Σ has

the form

[v]α =

0

v1

...

vn−1

=

0...

vΣ

...

,

where vΣ =

v1

...

vn−1

. Thus,

[Φ(T, x0)]α[v]α =

avΣ

...

bvΣ

...

.Also, since DP (x0)v ∈ Tx0Σ, it follows from Theorem 3.13 that

[DP (x0)v]α =

0...

bvΣ

...

.So the eigenvalues of b and of DP (x0) are the same. But these are precisely the remaining

eigenvalues of Φ(T, x0).

22

Definition. The matrix Φ(T, x0) as above is called the monodromy matrix of the periodic

solution u(t, x0).

Next, we present a useful formula for computing the derivative of the Poincare map at its

corresponding periodic point. We will use Liouville’s formula for linear homogeneous systems

of the form dΦ

dt= A(t)Φ(t)

Φ(0) = In×n

.

Liouville’s formula tells us that

det Φ(t) = exp

∫ T

0tr(A(s)) ds.

Theorem 3.18. Suppose f is a planar vector field with periodic point x0 of period T . Let Σ be

a local section such that x0 = Γ ∩ Σ, where Γ is the periodic orbit of x0. For a corresponding

Poincare map P : Σ→ Σ, we have

P ′(x0) = exp

∫ T

0div f(u(t, x0)) dt.

Proof. Let Φ(T, x0) = ux0(t, x0), where u(t, x0) is the solution ofx(t) = f(x(t))

x(0) = x0

.

Since P is simply a function from R to itself, P ′(x0) is its own (and only) eigenvalue. Thus, the

monodromy matrix M = Φ(T, x0) has eigenvalues 1 and P ′(x0). Therefore,

P ′(x0) = detM = exp

∫ T

0tr(Df(u(t, x0))) dt.

But writing f = (f1, f2) so that Df =

∂f1

∂x1

∂f1

∂x2∂f2

∂x1

∂f2

∂x2

, we see that

tr(Df(u(t, x0))) =∂f1

∂x1(u(t, x0)) +

∂f2

∂x2(u(t, x0)) = div(f(u(t, x0))).

Example 3. Consider the systemx = f1(x, y) = x− y − x(x2 + y2)

y = f2(x, y) = x+ y − y(x2 + y2).

23

We can pull the vector field f = (f1, f2) back to a domain that uses polar coordinates defined

by (x, y) = u(r, θ) = (r cos θ, r sin θ), obtaining the vector field g(r, θ) via the formula

g(r, θ) = [Du(r, θ)]−1f(u(r, θ))

=

[cos θ −r sin θ

sin θ r cos θ

]−1 [r cos θ − r sin θ − r3 cos θ

r cos θ + r sin θ − r3 sin θ

]

=1

r

[r cos θ r sin θ

− sin θ cos θ

][r cos θ − r sin θ − r3 cos θ

r cos θ + r sin θ − r3 sin θ

]

=

[r − r3

1

].

Hence, our system takes on the much simpler formr = r − r3

θ = 1.

It is no longer even coupled. Moreover, it is clear that setting with the initial condition r(0) = 1,

we get the solution curve u(t, 1, θ(0)) = (1, θ(0) + t), which is clearly periodic, with period 2π

and orbit Γ = S1. We can let θ(0) = 0 for convenience. Notice then that u(t, 1, θ) = (1, t), so

the angle θ will always be equal to the time t at any point on this particular trajectory. Hence,

we can safely ignore θ.

To study the periodic solution u(t, 1) = u(t, 1, 0), we let Σ = {(r, 2π) : 0 < r < ∞} and let

τ(r) be the return time function to Σ. Thus, τ(1) = 2π. The corresponding Poincare map is

then P (r) = u1(τ(r), 1) and has derivative (evaluated at r = 1)

P ′(1) =∂P

∂r

∣∣∣∣r=1

u1(2π, r, 0).

In order to compute this, we shall study the variational equationd

dtuξ(t, r) = Dg(u(t, 1))uξ(t, 1)

uξ(0, 1) = I,

where ξ = (r0, θ0). Writing

uξ =

∂r∂r0

∂r

∂θ0∂θ

∂r0

∂θ

∂θ0

,where u = (r, θ), and since

Dg(u(t, r0)) =

[1− 3r2(t, r0) 0

0 0

],

24

the variational equations reduce to the initial value problemd

dt

∂r

∂r0(t, 1) = −2

∂r

∂r0(t, 1)

∂r

∂r0(0, 1) = 1

,

which has solution∂r

∂r0(t, 1) = e−2t. Thus, P ′(1) =

∂r

∂r0(2π, 1) = e−4π < 1, so the periodic

solution u(t, 1) is asymptotically stable.

3.6 Lotka-Volterra System

The Lotka-Volterra system is described by the differential equationsx = f1(x, y) = x(A−By)

y = f2(x, y) = y(Cx−D),

where A,B,C,D > 0. This system is important in modeling two populations, one of prey (x)

and one of predator (y). It is easily seen that (x∗, y∗) = (D/C,A/B) is an equilibrium point

for this system. Let us consider the transversal section Σ = {(x, y∗) : x∗ < x <∞} with return

map τ : Σ → R. We shall denote a generic point on Σ by q = (x, y∗). The Poincare map on Σ

is defined by

(P1(q), P2(q)) = P (q) = Φτ(q)(q).

Note that P really only depends on x, so we will abuse notation by writing (P1(x), P2(x)) =

P (x) = P (q).

We let Φ(t, x, y) = DΦt(x, y) be the derivative with respect to the initial condition (x, y) of

the solution curve Φt(x, y). Then we know that Φ(t, x, y) satisfies the variational equationd

dtΦ(t, x, y) = Df(Φt(x, y))Φ(t, x, y)

Φ(0, x, y) = I2×2

,

where f = (f1, f2).

Proposition 3.19. P ′1(x) =f2(q)

f2(P (q))exp

∫ τ(q)

0div f(Φt(q)) dt.

Proof. Begin by writing

P ′1(x)f2(P (q)) = f2(Φτ(q)(q)) = det

[P ′1(x) f1(P (q))

0 f2(P (q))

].

Recall also that

DP (q) · v = f(Φτ(q)(q))〈∇τ(q), v〉+DΦτ(q)(q) · v,

25

for v ∈ R2. Letting v = (1, 0), we get[P ′1(x)

0

]= f(P (q))

∂τ

∂x(q) + Φ(τ(q), q)

[1

0

].

Thus,

P ′1(x)f2(P (q)) = det

[f(P (q))

∂τ

∂x(q) f(P (q))

]+ det

[Φ(τ(q), q)

[1

0

]f(P (q))

]

= det

[Φ(τ(q), q)

[1

0

]f(P (q))

]

= det

[Φ(τ(q), q)

[1

0

]Φ(τ(q), q)f(q)

]

= det Φ(τ(q), q) det

[1 f1(q)

0 f2(q)

]

= f2(q) exp

∫ τ(q)

0div f(Φt(q)) dt.

where the equalities follow respectively from the fact that the determinant of a matrix is mul-

tilinear in its columns, linear dependence of the columns of

[f(P (q))

∂τ

∂x(q) f(P (q))

], Lemma

(3.15), commutativity of the determinant and matrix multiplication, and Liouville’s formula.

Since for the Lotka-Volterra system, div f =x

x+y

y, it follows from this proposition that

f2(q)

x= P ′1(x)

f2(P1(x))

P1(x).

Integrating both sides of this equation from x∗ to x gives us

F (x)− F (x∗) =

∫ x

x∗

f2(q)

xdx

=

∫ x

x∗

f2(P (x))

P (x)P ′(x) dx

=

∫ P (x)

P (x∗)

f2(u, y∗)

udu

= F (P (x))− F (P (x∗)),

where F (x) is the antiderivative off2(x, y∗)

x. This implies that F (x) = F (P (x)). But since

f2(x, y∗) is easily seen to be monotone on Σ, it must be the case that x = P (x) for x > x∗, i.e.

the points x > x∗ are all periodic.

26

3.7 Continuation of Periodic Solutions

Consider a system x = f(x, ε), where f : E × U → Rn is C1 and E ⊆ Rn and U ⊆ Rm are

open. We wish to study the qualitative effects of changing the parameter ε ∈ U . In this section,

in particular, we will study the conditions under which periodic solutions for some parameter

ε = ε0 can be continued as ε changes. More precisely, suppose Φt(x, ε) is the flow of this system

and suppose that for some x0 ∈ E and T > 0 we have ΦT (x0, ε0) = x0. Then we wish to ask

whether we can find functions x0 : U → Rn and T : U → R such that ΦT (ε)(x0(ε), ε) = x0(ε).

Another way of phrasing this is by asking what happens to the periodic solution with initial

point x0 as ε is perturbed.

Before we answer this question, we should take care of the more basic concern involving

the smooth (i.e. C1) dependence of solutions to this system in the parameter ε. However,

we can easily reduce the question of smooth dependence in the parameter to our previous

understanding of smooth dependence in the initial condition. Simply let u =

(x

ε

)∈ Rn+m so

that u =

(f(x, ε)

0

)= g(u). Then we know that the flow Ψt corresponding to the vector field g

is smooth in all variables. But Ψt(u) = Ψt(x, ε) =

(Φt(x, ε)

ε

), so Φt(x, ε) is smooth in (x, ε).

Theorem 3.20. If f ∈ C1(E × U) for E ⊆ Rn and U ⊆ Rm open, then the flow Φt(x, ε)

associated to f is smooth in (x, ε).

In what follows, let us take ε0 = 0 for simplicity. If x0 is periodic for this parameter, we can

construct a transversal section Σ to the periodic orbit of x0. It would seem then, by continuity,

that small perturbations in ε should not affect the fact that orbits originating in Σ should return

to Σ. Of course, their respective return times and return locations may change. However, we

are mainly interested in the qualitative dynamics, so in this case the existence of periodic orbits.

We make this discussion more precise in the theorem that follows.

Definition. If (x0, ε0) is an equilibrium point of f , then it is called elementary if Df(x0, ε0) has

no zero eigenvalues. If (x0, ε0) is a periodic point of f with period T , then it is called elementary

if the monodromy matrix evaluated at (x0, ε0), i.e. DxΦT (x0, ε0) has 1 (which we know to be an

eigenvalue) as a simple eigenvalue (i.e. 1 has multiplicity one in the characteristic polynomial of

the monodromy matrix).

Theorem 3.21. An elementary equilibrium (respectively, periodic) point (x0, ε0) can always be

continued smoothly, i.e. there exist neighbourhoods V and W of x0 and ε0, respectively, and a

smooth function g : W → V such that g(ε0) = x0 and (g(ε), ε) is an equilibrium (respectively,

periodic) point for all ε ∈W .

Proof. This follows from the implicit function theorem and Theorem 3.17.

27

3.8 Van der Pol Oscillator

The van der Pol oscillator, an important example of a dynamical system, is defined by the

second-order differential equation

x+ ε(1− x2)x+ ω2x = 0.

We can convert this into the systemx = −y

y = −ε(1− x2)y + x.

More generally, we can consider any system given by

u = Au+ εg(u),

where u = (x, y)T , A =

[0 −1

1 0

], and g = (g1, g2)T . This last form makes clear the fact that

when ε = 0, the system is linear. Thus, if Φt(u, ε) denotes the flow of the system with parameter

ε evaluated at u, then

Φt(u, 0) = eAtu =

[cos t − sin t

sin t cos t

][x0

y0

].

So for ε = 0, all solutions to the system are periodic with period 2π. We wish to extend this

result to the case where ε 6= 0. However, for ε = 0, the monodromy matrix is simply the identity,

which does not have 1 as a simple eigenvalue.

To get around this problem, consider the section Σ = {(x, 0) : x > 0} and define the

parametrized Poincare map P (ξ, ε) = ΦT (ξ,ε)(ξ, ε) for ξ = (x, 0) ∈ Σ. So we know that P (ξ, 0) =

ξ. Also, define the displacement function δ(ξ, ε) = x(T (ξ, ε), ξ, ε) − ξ. Since δ has no simple

zeros, let’s take a look at the Taylor expansion of δ about ε = 0, which is given by

δ(ξ, ε) = εδε(ξ, 0) +O(ε2) = ε∆(ξ, ε),

where ∆(ξ, ε) = δε(ξ, 0)+O(ε) (∆ is known as a reduced displacement function). If ∆ has simple

zeros, then they can be continued by the implicit function theorem. So we want to find ξ0 such

that δε(ξ0, 0) = 0 and δξε(ξ0, 0) 6= 0. To begin, we take the derivative

δε(ξ, 0) =d

dt

∣∣∣∣t=T (ξ,0)

x(t, ξ, 0)Tε(ξ, 0) + xε(T (ξ, 0), ξ, 0).

Since ξ ∈ Σ, we getd

dt

∣∣∣∣t=T (ξ,0)

x(t, ξ, 0) = −y(t, ξ, 0) = 0 since ξ ∈ Σ and so

δε(ξ, 0) = xε(T (ξ, 0), ξ, 0) = π1

(∂

∂εΦt(ξ, 0, 0)

).

28

To evaluate this, note thatd

dtΦt(ξ, 0, 0) = f(Φt(ξ, 0, 0)), so

d

dt

∂

∂εΦt(ξ, 0, 0) = Df(Φt(ξ, 0, 0))

∂

∂εΦt(ξ, 0, 0).

But we have the initial condition Φt(ξ, 0, 0) = ξ + fε(Φt(ξ, 0)), and so∂

∂εΦt(ξ, 0, 0) = 0.

[[Incomplete]]

3.9 Hopf Bifurcation

Consider the system {x = µx− y − (x2 + y2)x

y = x+ µy − (x2 + y2)y

It can be seen that as µ pass through 0 (from the left), a periodic orbit is born at r =√µ. This

kind of phenomenon can be found in a more general setting under some assumptions.

1. Assume vector field F (x, µ) = A(µ)x+ f(x, µ) has an equilibrium point at x = 0 for all µ,

i.e. f(0, µ) = 0.

2. Suppose the linear map A(µ) has eigenvalues λ1(µ), λ1(µ), λ3, . . . , λn. We assume that

α(0) = 0, α′(0) 6= 0, and β(0) 6= 0.

3. Finally, we assume the non-resonance conditionsλkλ1

/∈ Z for k = 3, . . . , n.

Theorem 3.22 (Hopf-Andronov). Under the above assumptions, there exists a smooth family

of periodic solutions x(t) = u(t, x0(ε), ε), µ = µ(ε), and T (ε) on 0 ≤ ε ≤ ε0 such that

u(t, x0(ε), ε)→ 0

µ(ε)→ 0

T (ε)→ 2π

β(0).

[[Proof]]

29