Limit Cycle• Isolated closed trajectories

Analysis is, in general, quite difficult. For nonlinear systems, two simple results exist.

Limit Cycle

2 - 1

Bendixson’s Theorem

• Bendixson’s Theorem

B

A

Connected(but not simple set)

(angular region)

BA, :set connected a is

set ithin theentirely wlying curve aby connected becan , BA

AB

:set connectedsimply a is D

set connected a is

and connected is

D

D

#

)( of ies trajectorclosed no has Then . ofsign change

not does and ofregion - subany over zeroy identicallnot is

)(

assume weand abledifferentily continuous is

, ,)(Consider set. connectedsimply a be Let

2

2

1

1

22

xfxDD

D

x

f

x

fxf

f

RxxfRD

: Theorem

2 - 2

.in sign its changes )( or

,0)(

either if indeed, This,

0)()()(

theoremdivergence By the

0)()( 0))()((

,0)()(

Then .at tonormal outward thebe )(Let .for

tol tangentiais )(Then ).( ofy trajectorclosed a be Let

Sxf

Sxxf

dsxfdlxxf

dlxxfdlxxf

xxxf

xxx

xfxfx

S

T

T

)(

0)(

22112

22121

xxxx

xxxx

Ex:

2

2122

21

2

2

1

1

in ies trajectorclosed No

)0,0(),(0)()(

R

xxxxx

f

x

fxf

).( of ies trajectorclosedany contain not does Hence

not true. is then this, ofsubset a is If

xfxD

DS

: Proof

2 - 3

Poincare-Bendixson Theorem

• Poincare-Bendixson Theorem

. as and )(y with analogousl defined

isset limit Negative ).( of , set,limit (positive) thecalled is )( of

pointslimit all ofset The . as )( and as such that

in )( sequence a if )( ofpoint limit (positive) a is point A

. ),( of y)(trajectorsolution a be )(Let

1

12

2

ntt

txLtx

nztxnt

RttxRz

Rxxfxtx

nnn

nn

nn

:Definition

1x

2x

))0(,( xtx ))0(,( xtx

positive limit set

1x

2xpositive limit set

1x

2x

Negative limit set exists.Positive limit set does not.

2 - 4

Theorem

set.limit negativefor the validis same The orbit. closed a is or orbit closed a is )(either Then ).( of equilibria no contains and region bounded

closed ain contained is that Assume ).( ofset limit positive thebe Let 2

LtxxfxMRM

LtxL

:Theorem

curve. closed a of seperation plane offact on the Based: Proof

1x

2x

Systems.Nonlinear in A.10Appendix see proof, detailed For the

.region thefind tohaving ofnecessity theis theorem thisofpoint weak The M

2 - 5

Example for Poincare-Bendixson Theorem

)1(

)1(22

21212

22

21121

xxxxx

xxxxx

Ex :

22

21 xxV

))(1)((2)1)((2

)1(22)1(22

22

22

21

22

21

22

21

2221

22

21

2121

221122

11

xVxVxxxx

xxxxxxxxxx

xxxxfx

Vf

x

VV

1)(for01)(for0

xVVxVV

sin cos

Indeed

2

1

rxrx

1),1( 2 rrr

2 - 6

Index Theorem

Index Theorem2

2 2:

D R

f R R D

f J

D

Suppose is an open, simply connected subset of and

is a vector field. Suppose contains only

isolated equilibria of . Let be a simple, closed, positively

oriented Jordan curve in that

( ) ( )ff x x

x J

f

doesn't pass through any

equilibria of and let denote the direction of vector

field at . Then the index of the curve with respect to the

vector field is

1( ) ( )

2f fJI J d x

:Definition

2 - 7

• Facts

Isolated Equilibrium

. ,

( ), ( )s s

f s f

s

x f x

I x I J J

x J J

Let be an isolated equilibrium point of Then, the index of

is equal to when is any Jordan curve such that

is enclosed by and doesn't enclose any other equilibria.

( ) 0 .fI J J f if doesn't enclose any equilibria of

1

1.The indices of a center, focus, and node are . The index

of a saddle point is

:J fSuppose encloses a finite number of equilibria of

1( ) ( )

ki

f f si

I J I x

Then

+1 -1

.,,1 kss xx

:Definition

2 - 8

Facts (Continued.)

( ).J x f xLet be a simple closed, positively oriented trajectory of

( ) 1.fI J Then

( )x f xSuppose has only isolated equilibria. Then every closed

trajectory (if any) encloses at least one equilibrium point. Moreover,

the sum of indices of all equillibria enclosed by the closed traj 1.ectory is

Either (1) center, focus, node or (2) 2 node + 1 saddle

Ex : 1 2 1

2 1

cos

sin

x x x

x x

( ,0) 0,1,2,n n eq. point for

no periodic solution !

2 1 1

1 ( ,0)

sin cos 0 11 (

cos 0 1 0n

x x xA

x

linearization : saddle point)

2 - 9

Volterra predator – prey model

Ex: : modelprey -predator Volterra

2122

2111

xxxxxxxx

1

1

0

00)1(0)1(

22

21

12

11

12

21

s

s

s

s

x

x

x

xxxxxeq)

11

00 21

ss xx

saddle center

1)( 1 sf xI 1)( 2 sf xI

. enclose

not must and enclosemust y trajectorclosedevery Thus1

2

s

s

x

x

2 - 10

Introduction and motivation

3. Mathematical Foundations existence, uniqueness, continuously dependent on initial conditions

],0[,)0(:,),(

0 TtxxRRfRxxfx nnn

or

],0[,)0(:,),,(

0 TtxxRRRfRxtxfx nnn

Obviously, we can write (for simplicity, we assume a time invariant case)

t Ttdxfxtx 00 ],0[,))(()(

To analyze the fundamental properties, use the simple idea of successive approximations : Choose the initial approximation

],0[,const)( 00 Ttxtx

not initial condition but the initial approximation2 - 11

Introduction and motivationThen, we can write

dxfxtx t 0 001 ))(()(

first approximation

In general,

],0[,))(()( 001 Ttdxfxtx tii

So, the procedure generates the sequence

),(,),(),( 10 txtxtx i

Questions:

?))(()(or )(

satisfies that )( toconverge procedure theDoes (i)

00 dxfxtxxfx

txt

? uniquesolution theIs (ii)

? place takeseconvergenc thefor which is What (iii) T

2 - 12

Introduction (Continued)

The idea to provide the answers is quite simple. We write again,

.))](())(([)()( 01 dxfxftxtx tii

i.e., function, Lipschitz globally)or (locally a is )( Assume xf

such that 0K

xxKxfxf ii )()(

Lipschitzglocally ,

Lipschitzlocally ,

ni

i

Rxx

Dxx

Then we can write

dxxKdxfxftxtx it

ii 001 )()())(())(()()(

Processing this and analogous inequalities, one can arrive at the conclusions ofexistence, uniqueness, etc.

2 - 13

Introduction (Continued)

The modern approach to the problem is based on a more general principle,useful for many other applications – the contraction mapping theorem(functional analysis technique).

Introduce an operator)())(( 1 txtTx ii

T)(txi )())(( 1 txtTx ii

Define T as follows :

dxfxtTx tii 00 ))(())((

Then, the recursion has the form

,1,0),)(()(1 itTxtx ii

2 - 14

Introduction and motivationIt is possible to show that under the Lipschitz condition, T has a fixed point

))(()( tTxtx

which says that x is a solution. It is also possible to find conditions under whichthis fixed point is unique. To develop, more or less carefully, the machinery forsuch a proof we need several metrics :

Linear Space NormLinear

NormedSpace

+ =

Set of elementsclosed w.r.t.addition and multiplicationby a scalar

Measure of distance betweentwo elements(after generatedby the projection -inner product of elements - )

InnerProduct

InnerProductSpace

=

2 - 15

Introduction (Continued)

space.Banach a is ),(Then . ofelement

an toconverges ),( spacelinear normed ain sequenceCauchy every that Assume

XX

X

Complete linearnormed space

Complete linearnormed space

OperatorT

everyconvergent sequence hasthe limit in this set )(txi ))(( tTxi

unique is andexists ofsolution the

space, normed complete on thencontractio a is If

1 nn Txx

T

2 - 16

Preliminaries•

Preliminaries

nR spaceEuclidean

vectordim- all ofset nT

nxxx ][ 1 (real number) n

n

Rx

Ryx

spaceLinear

Norm real valued function of x such that

RRxxx

Ryxyxyx

xxRxx

n

n

n

,,

,,

0 ifonly and if 0,,0

P-norm

ii

ppn

p

p

xx

pxxx

max

1)(1

1

2 - 17

Preliminaries(continued)

Most used norms

xxx and,,21

Euclidean norm

xcxxc 21

(Example) xnxx

1

(proof)

xnxxx

xxxx

n

ii

i

n

ii

n

iii

111

11

max

max

., norm,other any for holds normin provedproperty spaces, ldimensiona finitein So,

212

1

pppp

All finite dimensional norms are equivalent.

2 - 18

Preliminaries(continued)

mnnm RRARA : wherematrix a of norm Induced

matrix norm

pxp

p

xp

Axx

AxA

p 10supsup

For instance,

m

iij

ja

11

maxA

m

jij

ia

1maxA

(max column sum) (max row sum)

21

max2)]([ AAA T

AAT of eigenvalue maximum

2 - 19

Preliminaries(continued)

pxp

p

xp

Axx

AxA

p 10supsup

pxαxu whereconsider this,show To

Then

p

p

p

p

p

p

p

p

u

Au

u

Au

x

Au

x

Ax

.0 all of instead considered be toneed torslength vecunity only Thus x

2 - 20

• Linear infinite dimensional (function) spaces

Preliminaries(continued)

: norm inducedmatrix between bridges are There

AAA12

Proof

1111

2

2

2

21

1

)(

1)(

AAvAAA

AvAA

vvvAA

T

T

T

constant aby tion multiplica andaddition : functions tworespect to with closed properties specific a with functions all ofset

2 - 21

Preliminaries(Examples)

properties three thesatisfyingfunction enonnegativ - Norm

Xyxyxyx

Xxxx

Xxx

, ,

,

0,

pLdxtxbac ppb

a

p

p1),())(()(],[

11

)(max)(],[

txtxbat

first],[ bacn

nitxpi ,,1)(

pn

i

p

pi txp1

1))(( vectors thisof norm- Take

:Ex to],[ fromfunction continuous all],[ Rbabac

:Ex

:Ex

2 - 22

Preliminaries(Examples)

S :Ex

,2,1 max

1 )(1

1

ixx

pxx

ii

p

i

pip

.equivalentnot are spaces ldimensiona infinitein Normsnorm.or then vectnorm,component :y analogousl defined is on Norm nS

),( pair a as denoted becan spacelinear Normed

X

: Definition

)(,

such that )( ,0 if toconvergent is

sequence This .in elements of sequence a be }{Let ).,(Consider

0

0

0

Nixx

NXx

XxX

i

ii

2 - 23

Cauchy Sequence

)(,,

such that )(

0 if sequenceCauchy a is },{in element of }{ sequenceA 0

Njixx

N

Xxx

ji

ii

: sequenceCauchy a is sequence convergentEvery

jixxxxxxxxxx jijiji , as 00000

.convergent is sequenceCauchy every spaces dimesional finiteIn

. of outside bemight , element, limiting thefact that the

todue happens This : y truenecessarilnot is thisspaces ldimemsiona infiniteIn 0 Xx

: Definition

2 - 24

Cauchy Sequence(Examples)

Rx

l

ii

il

ii

where define

and for which },,{ scalars of sequences all ofset thebe Let

2/1

1

2

1

2

212 Ex :

norm. defined wella is that Showl

x

Solution :

.for },,{ },,,{ Define. tobelong },,{ and },,{Let

space.tor linear vec is that showFirst

212211

22121

2

Rxyxlyx

l

2

1

22

1

2

1

2

2

Then,

lyxi

ii

iii

iii

2

1

22

1

2

and,

lxi

ii

i

2 - 25

Cauchy Sequence(Examples)

ll

l

l

l

xx

xx

x

lx

l

and

0 iff 0

,0

clear that isIt

.on norm a is that show sLet'

space.tor linear vec a is Therefore,

2

2

1

2

1

2

1

,inequality Schwartz-Cauchy from know we,inequalityr triangula theshow To

ii

ii

iii

llllll

iiiii

iiil

yxyxyx

yx

2

)2( Now

22

1

22

1

2

2 - 26

Cauchy Sequence(Examples)

space.Banach a is norm with the that Show 2 lxl

(sol.)

mn

mnxx

xlx

mi

ni

i

mi

nilmn

nnnnn

, as 0

, as 0Then

},,{:in sequencecahchy a be }{let First

1

2

2120

n

RRi

ini

ni

as

complete is Since.in sequencecauchy a is }{ sequence the,each for Hence,

21

2

1

2

1

Hence

.in uniformly , such that

constant aexist therecauchy, is }{ Since}.,,{Let

lxM

nM

Mxx

ii

i

ni

n

i

2 - 27

Cauchy Sequence(Examples)

.in as }{ that showswhich

.' where ,' Thus

,

letting

1 , ,

have we, Letting

1 ,, ,

such that integer positive a ,0Given

.in toconverges }{ that show Now

2

1

2

1

2

2

1

2

2

lnxx

Nnxx

Nn

k

kNn

m

kNmnxx

N

lxx

n

ln

ii

ni

k

ii

ni

lmn

k

i

mi

ni

n

2 - 28

Cauchy Sequence(Examples)

)|||| ],1,0[( is have what weSo

|||||||| ]1,0[

1

1

c

cXEx :

jitxtx

i

t

t

t

ittx

jiji

i

ii

i

, as 0)()( Here

,3,2

0

for

for

for

1

1

0

)(

sequence heConsider t

1121

1

21

211

21

121

2

x

t6

14

12

1

1

2i

3i4i

2 - 29

Banach Space

space.Banach aor complete a called is

convergent is sequenceCauchy each where||)||,( spacelinear Normed : Definition X

spaces.-

integrable absolutely are which functions measurable all precisely, moreor

function ousdiscontinu addingby space completecan weexample, above In the

pL

step.unit a isfunction

limiting thesince convergentnot but Cauchy is seqeunce theThus

points). ofnumber countablealmost an at except (i.e., zero measure ofset aon except

everywhere continuous isit if measurable be tosaid is :)(function A RRf

2 - 30

Banach Space(continued)

TOperator

)||||,( XX )||||,( YY

T

YXT :

),( when )()(

such that ),( , if at continuous is

000

00

xxxxTxT

xXxT

XY

. oft independen is and continuous isit if continousuniformly is

allfor continuous isit if continuous is

0xT

XxT

manner standard in the introduced is of norm induced The T

X

Y

xXxXY

i tx

tTxT

)(

))((Sup

0by

2 - 31

Banach Space(continued)

spacelinear : Recollect To X

SzxXzxN

X

SxXS

}:{ ),(

, of odneighborho-an such that

0 , ifopen is

open. is in complement its if closed is XXS

Sxrx

rXS

,

such that 0 if bounded is

bounded. and closed isit ifcompact is XS

2 - 32

Contraction Mapping

Contraction Mapping•

XxxxxTxTx

XXT

X

212121 , ,

such that 1 Suppose .:operator and

||)||,( spaceBanach aConsider : theoremmappingn contractio Global

**

*

such that unique a exists Then there

xTx

Xx

00*

*

1

0

1

is econvergenc of speed eFinally th . toconverge

Moreover

xTxxx

x

Txx

Xx

k

k

kk

not necessarily linear

2 - 33

0001

212

21

111

0

cauchy. is }{ that showFirst

xTxxx

xxTxTx

xxTxTxxx

x

kk

kkkk

kkkkkk

kk

Proof :

kxTx

xTxxTx

xTx

xxxxxx

xxxxxxxx

k

i

ikr

i

ik

krkrk

kkrkrkrkrk

krkrkrkrkrkkrk

as 01

][

Now

00

000

1

000

0021

1211

2211

Cauchy

Banach. is ||)||,( assumptionby since,

*say to,Converge

X

x

space.Banach a isit that

fact the toDue

2 - 34

**

*

satisfies that thisshow tohave weNow

xTx

x

*1

* limlimlim Indeed xxTxxTTx kk

kk

kk

continuousuniformly therefore,

and, Lipschitz is fact that the todue T

**

*

say point, fixedanother be Let there :ion contradictBy

: unique ispoint that thisshow weNow . ofpoint fixed theis ,So

x

Tx

*********Then xxTxTxxx

*** ifonly holds this,1 since xx

00

*

1 lim

lim

: econvergenc of speed theshow weFinally,

xTxxx

xxxx

k

krkr

krkr

k

2 – 35

Banach Space(Examples)

Ex :

Rxxf

x

xfx

xfTx

RX

kk

,1)(

;

)(

)(

)||||,(||)||,(

0

1

p

.)(

such that unique a is there,by theorem Thus

1 ,)(' )()( So

),)((')()(

theoremmean value by the Indeed,

n.contractio a is )( ,1)( Since

**

*

xxf

x

yxyxzfyfxf

yzxyxzfyfxf

xfxf

0x1x

)( 0xf

)( 1xf

)(xf

x

yxy

sequence econvergenc

2 – 36

Local Contraction mapping theorem

• Local Contraction mapping theorem

XSX ||)||,(

10 ,,

:

212121

SxxxxTxTx

SST

00*

1*

****

1

and wherelim

unique is ,such that

xTxxx

Txxxx

xTxxSx

k

k

kkkk

Proof : same as before.

2 - 37