Chapter 1

Limit Cycle and Its Perturbations

1.1 Basic notations and facts

Consider a planar system deˉned on a region G?R2 of the form

x_=f(x);(1:1:1)

where f:G ?! R2 is a Cr function,r > 1.Then for any point x02 G (1.1.1) has a

unique solution '(t;x0) satisfying '(0;x0)=x0:Let't(x0)='(t;x0):The family

of the transformations 't:G ?! R2 satisˉes the followingproperties

(i) '0=Id;

(ii) 't+s='t ± 's:

The function ' is called the °ow generated by (1.1.1) or by thevector ˉeld f.Let

I(x0) denote the maximal interval of deˉnition of '(t;x0) int.If x0 2 G is such

that '(t;x0) is constant for all t 2 I(x0),then f(x0)=0.In thiscase,x0 is called

a singular point of (1.1.1).A point that is not singular iscalled a regular point.

For any regular point x0 2 G,the solution '(t;x0) determines twoplanar curves

as follows

°+(x0)=f'(t;x0):t 2 I(x0);t > 0g;°?(x0)=f'(t;x0):t 2 I(x0);t6 0g;

which are called the positive,negative orbit of (1.1.1) throughx0 respectively.The

union °(x0)=°+(x0)[°?(x0) is called the orbit of (1.1.1) throughx0.The theorem

about the existence and uniqueness of solutions ensures thatthere is one and only

one orbit through any point in G.Thus,it is easy to prove thatany di?erent orbits

do not intersect each other.A periodic orbit of (1.1.1) is anorbit that is a closed

curve.The minimal positive number satisfying '(T;x0)=x0 is saidto be the period

of the periodic orbit °(x0).Obviously,°(x0) is a periodic orbitof period T if and

only if the corresponding representation '(t;x0) is a periodicsolution of the same

period.

Deˉnition 1.1.1 A periodic orbit of (1.1.1) is called a limitcycle if it is the

only periodic orbit in a neighborhood of it.In other words,alimit cycle is an isolated

periodic orbit in the set of all periodic orbits.

Now let us assume that (1.1.1) has a limit cycle L:x=u(t);0 6 t6 T.

Since (1.1.1) is autonomous,for any given point p 2 L we maysuppose p=u(0),

and hence,u(t)='(t;p).Further,for deˉniteness,let L be orientedclockwise.

Introduce a unit vector below

Z0 =

1

jf(p)j

(?f2(p);f1(p))T:

Then there exists a cross section l of (1.1.1) which passesthrough p and is parallel

to Z0.Clearly,a point x0 2 l near p can be written as x0=p + aZ0with a =

(x0?p)TZ0 2 R small.

Lemma 1.1.1 There exist a constant " > 0 and Cr functions Pand ?:

(?";") ?! R with P(0)=0 and?(0)=T such that

'(? (a);p + aZ0)=p + P(a)Z0 2 l;jaj < ":(1:1:2)

Proof Deˉne Q(t;a)=[f(p)]T('(t;p + aZ0)?p).We have

Q(T;0)=0;Qt(T;0)=jf(p)j2 > 0:

Note that Q is Cr for (t;a) near (T;0):The implicit functiontheorem implies that

a Cr function?(a)=T + O(a) exists satisfying

Q(? (a);a)=0 or [f(p)]T('(? (a);p + aZ0)?p)=0:

It follows that the vector '(? (a);p + aZ0)?p is parallel toZ0.Hence,it can be

rewritten as '(? (a);p + aZ0)?p=P(a)Z0,where

P(a)=ZT

0 ('(? (a);p + aZ0)?p):(1:1:3)

It is obvious that P 2 Cr for jaj small with P(0)=0.This endsthe proof.

The above proof tells us that the function?is the time of theˉrst return to l.

By Deˉnition 1.1.1,the periodic orbit L is a limit cycle if andonly if P(a) 6= a for

jaj > 0 su±ciently small.

Deˉnition 1.1.2 The function P:(?";") ! R deˉned by (1:1:2) iscalled a

Poincar?e map or return map of (1.1.1) at p 2 l.

For convenience,we sometimes use the notation P:l ?! l.

Deˉnition 1.1.3 The limit cycle L is said to be outer stable(outer unstable)

if for a > 0 su±ciently small,

a(P(a)?a) < 0(> 0):

The limit cycle L is said to be inner stable (inner unstable) ifthe inequality above

holds for ?a > 0 su±ciently small.A limit cycle is calledstable if it is both inner

and outer stable.A limit cycle is called unstable if it is notstable.

For example,if L is stable,then the orbits near it behave likethe phase portrait

as shown in Figure 1.1.1.

Let Pk(a) denote the kth iterate of a under P.It is evident thatfPk(a)g is

monotonic in k and Pk(a) > 0(< 0) for a > 0(<0).Thus,it is easy to see that L

is outer stable if and only if Pk(a) ! 0 as k ! 1 for all a >0 su±ciently small.

Similar conclusions hold for outer unstable,inner stable andinner unstable cases.

Remark 1.1.1 If the limit cycle L is oriented anti-clockwise wecan deˉne its

stability in a similar manner by using the Poincar?e map Pdeˉned by (1.1.2).For

instance,it is said to be inner stable (inner unstable) ifa(P(a)?a) < 0(> 0) for

a > 0 su±ciently small.

Deˉnition 1.1.4 The limit cycle L is said to be hyperbolic or ofmultiplicity

one if P0(0) 6= 1.It is said to have multiplicity k,2 6 k 6 r;ifP0(0)=1;P(j)(0) =

0;j=2;￠ ￠ ￠ ;k?1;P(k)(0) 6= 0:

By Deˉnition 1.1.3,one can see that L is stable (unstable) ifjP0(0)j < 1 (> 1):

Example 1.1.1 Consider a system given by

_ x1=x1?x2?x1(x2

1 + x2

2);

_ x2=x1 + x2?x2(x2

1 + x2

2):

(1:1:4)

The system has the form

r_=r(1?r2);μ_=1

in polar coordinates (r;μ) with x=(r cos μ;r sin μ).Thus,one canˉnd (1.1.4) has

a °ow of the form

'(t;x0)=r(t)(cos μ(t);sin μ(t))T;(1:1:5)

where

r(t)=r0(r2

0 + (1?r2

0)e?2t)?1

2 ;μ(t)=t + μ0;

x0=r0(cos μ0;sin μ0)T;r0 > 0;0 6 μ0 < 2:

For p=(1;0)T,we have a periodic orbit L=f(x1;x2)Tjx2

1 + x2

2=1g which has a

representation

L:x='(t;p)=(cos t;sin t)T;0 6 t 6 2;

with Z0=(?1;0)T.Then p+aZ0=(1 ?a;0)T.Hence,x0=p+aZ0 if and onlyif

r0=1?a,μ0=0.

Taking l=f(x1;0)Tjx1 > 0g.Then noting that?(0)=2,by (1.1.5)we have

'(? (a);p + aZ0) 2 l if and only if?(a)=2 for a <1.Therefore,

'(? (a);p + aZ0)=(1?a)[(1?a)2 + (2a?a2)e?4]?1

2 (1;0)T:

It follows from (1.1.3) that

P(a)=1?(1?a)[(1?a)2 + (2a?a2)e?4]?1

2

= ae?4 + O(a2)

for jaj small.By Deˉnition 1.1.3,the limit cycle L isstable.

Next,we give formulas for P0(0) and P00(0).For thepurpose,let

v(μ)=u0(μ)

ju0(μ)j

= (v1(μ);v2(μ))T;Z(μ)=(?v2(μ);v1(μ))T;

and introduce a transformation of coordinates of the form

x=u(μ) + Z(μ)b;0 6 μ 6 T;jbj < ":(1:1:6)

Lemma 1.1.2 The transformation (1:1:6) carries (1:1:1) into thesystem

dμ

dt

= 1 + g1(μ;b);

db

dt

= A(μ)b + g2(μ;b);(1:1:7)

where

A(μ)=ZT(μ)fx(u(μ))Z(μ)=trfx(u(μ)) ?

d

dμ

ln jf(u(μ))j;

g1(μ;b)=h(μ;b)[f(u(μ) + Z(μ)b)?f(u(μ))]?h(μ;b)Z0(μ)b;

g2(μ;b)=ZT(μ)[f(u(μ) + Z(μ)b)?f(u(μ))?fx(u(μ))Z(μ)b];

h(μ;b)=(jf(u(μ))j + vT(μ)Z0(μ)b)?1vT(μ);

and trfx(u(μ)) denotes the trace of the matrix fx(u(μ)),which iscalled the divergence

of the vector ˉeld f evaluated at u(μ):

Proof By (1.1.6) and (1.1.1) we have

(u0 + Z0b)

d μ

d t

+ Z

d b

d t

= f(u + Z b):(1:1:8)

In order to obtain (1.1.7) we need to solve

d μ

d t

and

d b

d t

from (1.1.8).First,multiplying

(1.1.8) by vT from the left-hand side and using

vTZ=0;vTf(u)=vTu0=ju0j=jf(u)j;

we can obtain

dμ

dt

= [jf(u)j + vTZ0b]?1vTf(u + Zb)=h(μ;b)f(u + Zb):

Note that

h(μ;b)f(u)=h(μ;b)[f(u) + Z0b]?h(μ;b)Z0b=1?h(μ;b)Z0b:

It follows that

h(μ;b)f(u + Zb)=h(μ;b)[f(u + Zb)?f(u)]?h(μ;b)Z0b + 1:

Then the ˉrst equation in (1.1.7) follows.

Now multiplying (1.1.8) by ZT from the left and using

ZTZ=1;ZTf(u)=0;ZTZ0=v1v01 + v2v02 =

1

2

(jvj2)0=0;

we obtain

db

dt

= ZT[f(u + Zb)?f(u)?fx(u)Zb] + ZTfx(u)Zb:

By writing f and Z in their components it is direct to provethat

ZTfx(u)Z=trfx(u) ?

d

dμ

ln jf(u)j:

Then the second equation of (1.1.7) follows.This ˉnishes theproof.

Set

B(μ)=[fx(u + Zb)]0bjb=0;C(μ)=vT[fx(u)Z?Z0(μ)];(1:1:9)

and

R(μ;b)=A(μ)b + g2(μ;b)

1 + g1(μ;b):

Then by Lemma 1.1.2,we can write

R(μ;b)=A(μ)b +

1

2 ?ZTBZ ?

2AC

jf(u)j?b2 + O(b3) ′ A(μ)b +

1

2A1(μ)b2 + O(b3):

(1:1:10)

For jbj small we have from (1.1.7)

db

dμ

= R(μ;b) (1:1:11)

which is a T-periodic equation.From Lemma 1.1.2 we know that thefunction R is

Cr?1 in (μ;b) and Cr in b.Let b(μ;a) denote the solution of(1.1.11) with b(0;a)=a.

Then b(T;a) deˉnes a function of a which is called a Poincar?emap of (1.1.11).For

the relationship of Poincar?e maps of (1.1.1) and (1.1.11) wehave

Lemma 1.1.3 P(a)=b(T;a):

Proof Consider the equation

dμ

dt

= 1 + g1(μ;b(μ;a)):

It has a unique solution μ=μ(t;a) satisfying μ(0;a)=0 andμ(t;0)=t.From

(1.1.10) it implies b(μ;0)=0.This yields μ(T;0)=T;

@μ

@t

(T;0)=1.Hence,by

the implicit function theorem a unique function ~? (a)=T + O(a)exists such that

μ(~? ;a)=T:

For x0=u(0) + Z(0)a,we have by (1.1.6)

'(t;x0)=u(μ(t;a)) + Z(μ(t;a))b(μ(t;a);a):

In particular,

'(~? ;x0)=u(T) + Z(T)b(T;a)=u(0) + Z(0)b(T;a)=p + Z0b(T;a) 2l:

Thus,it follows from Lemma 1.1.1 that ?=~? and P(a)=b(T;a).

The proof is completed.

For jaj small we can write

b(μ;a)=b1(μ)a + b2(μ)a2 + O(a3);

where b1(0)=1;b2(0)=0:By (1.1.10) and (1.1.11) one canobtain

b01=Ab1;b02=Ab2 +

1

2A1b2

1

which give

b1(μ)=exp Z μ

0

A(s)ds;b2(μ)=b1(μ) Z μ

0

1

2A1(s)b1(s)ds:

Then by Lemma 1.1.3 we have

P0(0)=b1(T)=exp Z T

0

A(s)ds=exp Z T

0

trfx(u(t))dt;

P00(0)=2 b2(T)=b1(T) Z T

0

A1(s)b1(s)ds:

Thus,noting (1.1.10) we obtain the following theorem.

Theorem 1.1.1 Suppose P is a Poincar?e map of (1:1:1) at p 2L.Then

(i) P0(0)=exp IL

divfdt;divf=trfx,

(ii) P00(0)=P0(0) Z T

0

eRt

0 A(s)ds ?ZT(t)B(t)Z(t) ?

2A(t)C(t)

jf(u(t))j ?dt.