7-1

Asymptotic Techniques Introduction

Real world complex for a direct analysis.

Solution simplification.

Basis for simplification small parameter with subsequent expansions.

Often first order term of the expansion gives already good results.

For differential equations, the source of difficulties is two-fold.),( xtfx

time varying

Large dim. of x is another

Two asymptotic techniques areAveraging takes care of ; Singular perturbation takes care of

Remark:

0lim if )( oforder is

const lim if )( oforder is

00

00

o

O

7-2

Averaging

Averaging• Motivation

Ex: Consider the system

)(Q

Ts

1

wtAsin

wtAsin

sAsAxQTwds

dx

wts

wtAwtAxQT

x

sin)sin(1

sin)sin(1

7-3

Averaging (Continued)

sAsAxQds

dxTw

sin)sin(Then

11

Assume

source of difficulties

Averaging will get rid of this !

fast changessin

slowly changes

s

x

constant. is assuming averaging Take x

)0term()(

sin]sinsinsin)([

sin)sin(

342

2

0

)(H.O.T.

33!3

122!2

121

2

021

2

3

3

2

2

AAO

dssAsAsAsAxQ

dssAsAxQ

xQA

A

x

Q

x

QxQ

7-4

Averaging (Example)

Then the averaged equation is

x

QA

ds

xd

2

2

??)(&)(between iprelationsh theis But what txtx

)(xQ

x

Asymptotically stable

Ex:1),1(~ where O,0sin

sin2

yl

gyy

t

1122

21

sinsinsin xxxx

xxt

llg

ts Let

fast term

112

2

sinsin]sin[2

1

xsxx

x

llg

dsdx

dsdx

(1)

7-5

Example (Continued)

To get rid of it, we use the idea of generating equation

1sinsin

0

0Set

2

1

xsldsdx

dsdx

Solve it.

122

11

sincos)(

)(

cscsx

csx

l

Introduce the substitution.

)(sincos)()(

)()(

122

11

szsszsx

szsx

l

7-6

Example (Continued)

Then, from (1)

or

111211

12

sinsin]sinsincos[coscossinsin

]sincos[

12

1

zszzszzszs

zz

llg

ldsdz

lldsdz

ldsdz

]sincos[coscos]sinsincos[ 1211122 zszzszzsz lll

glds

dz

Thus ]}sincos[coscossinsincos{

]sincos[

121112

12

2

1

zszzszzsz

zz

lllg

ldsdz

ldsdz

Thus again, what we obtained is

10),,( xtfx

So if we will be able to analyze such systems in a simple basis, we will solve the system

7-7

Example (Continued)

The average equations are

]2sinsin[ 1412

2

2

22

1

zzz

z

llg

dszd

dszd

In the original time

1412

2

2sinsin 2

22

1

zzz

z

llg

dtzd

dtzd

or

02sinsin 2

2

4 yyyy

llg

Equation point

0

,0

2

1

z

z

7-8

Stability analysis

Stability analysis

12coscos

10

121 2

2

zzz

f

llg

(i)stable

1

10

2

2

20,0

llg

z

f

(ii)

1

10

2

2

20, llg

z

f

gll

g

l2

22

2

2

gl2 condition stability

7-9

Theory

• Theory

nn RRRfxtfx :),(

slow variable

fast variable

. allfor uniformly exists average following e that thAssume nRDx

(1) )(),(lim 01 xfdtxtfTTT

case) thecourse, of is, this),,( periodic(For xf

)0()0( and )( Consider eq. averaged

xxxfx

7-10

Theorem 1

Theorem 1 such that (i) that Assume nRD

RtDxMxtf ,,),(

RtDxxxxxtfxtf ,,,),(),(

.in uniformly exists (1) oflimit (ii) D

00 0 such that 0,,Then L

],0[,),,(),,( 0000 Lttxtxtxtx

t

)(tx

L

)(tx

].,0[, tobelongs vicinity its with together)( that provided LtDtx

7-11

Theorem 2

Theorem 2

Assume

equation. averaged theofpoint mequilibriuan is andmet

are 1 Theorem of conditions e that thAssume

sx

txtf

RtDx

x

xxtf

s

in periodic is ),( (iii)

, allfor bounded (ii)

stablelly exponentia is (i)),(

solution. periodic stableally asymptotican issolution theaddition,In

,)(

such that

),( ),(solution periodic unique a has ),(

,0 such that 0 and Then

*

*

00

txtx

ttxxtfx

s

7-12

Theorem 3

),0[ )()(

0for such that ),( ,0Then

stable.ally asymptoticglobally is system averaged

theandmet are 1 Theorem of conditions e that thAssume

00

ttxtx

D

Theorem 3

Ex: Van der Pol eq.

)oscillator nonlinear (weakly 0)1( 2 yyyy

12212

21

)1( xxxx

xx

2212

1

21

)1(

0

01

10

)()( generalIn

xxx

xx

xFxFx

2

1

2

1

01

10

equation Generating

x

x

x

x

7-13

Example (Continued)

)1cossin2sin)(coscoscos(sin

)1cossin2sin)(coscossinsin(

cossin1sincos

0

cossin

sincosThen

cossin

sincos where),( Thus

)()(

)()(

onSubstituti

cossin

sincos where)(

2122

221

22

21

2122

221

221

2

212

212

1

2

21

0

ztzttztztztzt

ztzttztztzttz

tztztztztt

tt

z

z

tt

ttezeFez

zeFzeFzAezex

tzetx

tt

ttexetx

AtAtAt

AtAtAtAt

At

AtAt

7-14

Example (Continued)

stableally asymptotic2 const

eq. unstable0 )4

11(

2Then

tan ,Let

))(1(

))(1(

cossin ,2sincossin average theTaking

2

1

2122

21

22

214

122

1

22

214

112

1

2

1

8344

812

4122

r

rrr

r

z

zzzr

zzz

zzz

z

z

ttttt

7-15

Example

.0)(solution trivial theissolution periodic the

Hence system. for thepoint mequilibriuan is 0However .0 of

odneighborho )0(in solution periodic unique a has )( Thus

stable.lly exponentia isorigin the

Hurwitz. is which11

10

yieldsion linearizat The

origin. at thepoint mequilibriuan has system average The

sin

sin

sin)cos1()sin21(2

1)(

1 where,sin)cos1()sin21(

0

12

2

12

22

012

2

122

21

tx

xx

xfx

x

f

xx

xx

xx

xdt

xtxt

xxf

xtxtx

xx

x

av

av

Ex:

7-16

Singular Perturbations

Singular Perturbations• Idea

etc. ,phenomenongain large

parameter parasitic of analysisfor model goodA

satisfied bet can' condition initial of

part a 0,on when perturbatisingular

),(

),(Then

time)slow( Introduce

satisfied) becan conditions initial all ,0(when on perturbatiRegular

)0( ),(

)0( ),(

0

0

zxZz

zxXx

t

zzzxZz

xxzxXx

7-17

Singular Perturbations (Continued)

],( )0( )()(

],[ )0( )()(

)( ),,()),(,,(

) ofsubset a is(which manifold slow

point eq. isolated stable, a assume ),(

),,(0

)( ),,(

: follows as ison simplicati of scheme The

:

10 :

)( ),,(

)( ),,(

: form in the written is system perturbed singularly a general,In

00

00

00

00

00

00

Lttttztz

Lttttxtx

xtxxtFxtzxtfx

R

xtzz

zxtg

xtxzxtfx

RRRRg

RRRRf

ztzzxtgz

xtxzxtfx

n

mmn

nmn

~

~

7-18

Example

ab

ab

ab

tta

eztz

zbza

aztzbazz

)0(

)()(

well.assolution thesolvecan wecourse, Of

0

0 ,)(

0

00Ex:

0z

ab

)(tz

thick)(3layer boundary

)O(~3

7-19

Example

it work? does when So result. wronga giveson perturbatisingular theSo

0 tr , smallly sufficientfor Otherwise, ; 0 if 0tr

011det 011tr

1111

directly.it study can weand systemlinear a is But this

1for stable )1(

0

ArArArA

rA

rxrxrxx

rxzx

zxx

xz

zxx

rz

rz

Ex:

7-20

• Theory

0

00

00

satisfy not will),( ),,(0

),,(

:equation degenerate The

10

)( : ),,,(

)( : ),,,(

ztxzzzxtg

zxtfx

ztzRRRRgzxtgz

xtxRRRRfzxtfxmmn

nmn

.0for exists )(~ so

,attraction ofdomain its tobelongs ),( and ,in uniformly

stableally asymptotic is )),,()(~,(~

oforigin The iv)(

].,0[on solution unique a has )),,(,( system The (iii)

. and offunction continuous a is ),( and ),(

solution isolatedan has 0),,( (ii)

. ,at and

in Lipschitz and in continuous are ),,( and ),,( (i)

00000

0000

z

txzztx

ttxzzxgd

zd

Tttxzxfx

txtxztxzz

zxtg

RtRRDz

xtzxtgzxtfmn

Conditions

Theory

7-21

Theorem

unstable. is systemfast the,0for

outnot work did example previous why theexplains theoremThis

],0( ),),(()()(lim

],0[ ),()(lim

Moreover ].,0[on unique is and exists

)( ),,(

)( ),,( ofsolution the

0 such that Then satisfied. are (v)-(i) that Assume

0

0

00

00

00

r

Ttttxztztz

Tttxtx

T

ztzzxtgz

xtxzxtfx

.0Re i.e., number, fixed ahan smaller t parts real have

),(),( along evaluated of seigenvalue The (v)

cz

g

txztxg

Theorem:

7-22

Example

Ex:

331

3

2

3

2

2

2

0

isequation degenerate The3

1

, 1 ; if Then,

3

11

onSubstituti Introduce

1 0)1(

: Polder Van

zzx

zx

zzxdt

dz

zdt

dx

st

yz

yys

yx

ys

yy

s

y

7-23

Linear Systems• Linear Systems

Analysis Systems Nonlinear See : Proof

Hurwitz. also is

,0t tha

such Then Hurwitz. are and Suppose : Theorem

0

Hurwitz. be should - theoremprevious hesatisfy t To

2221

1211

0

0211

22121122

0211

221211211

221211

1211

211

22

2221

22

2221

1211

AAAA

A

AAAAA

xAxAAAAxAAAxAx

zAxAx

xAAz

zAxA

A

z

x

AA

AA

z

x

7-24

NonLinear Systems• Nonlinear Systems

unstable. is (0,0) ,0 allfor such that then

ORHP, in the is or either of seigenvalue

oneleast at If stable.ally asymptotic is (1) of (0,0)point

mequilibriu the,0 allfor such that Then Hurwitz.

are and andr nonsingula is suppose and

Define

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

satisfying and abledifferentily continuous are and where

:

:

),(

),(

00

211

22121122

00

211

2212112222

0,022

0,021

0,012

0,011

AAAAA

AAAAAA

z

gA

x

gA

z

fA

x

fA

gf

gf

RRRg

RRRf

zxgz

zxfx

mmn

nmn

Theorem

Proof : See Nonlinear Systems Analysis.

7-25

Example

Ex:

.0never point whe mequilibriu stable

allyasymptotican is (0,0,0) such that 0 that conclude we

parts, real negative have & of seigenvalue theall Since

23

11

1 , 12 , 1

1 ,

35

01Then

)1()1()2(

)1()3()5(

)1(

0

0

22

211

221211

22211211

1211

2122112

22111

MA

AAAAM

AAAA

yyxxxxy

xyxxxxx

yxyxxx

7-26

Nonlinear Control

Indeed, Why do we use nonlinear control :• Modify the number and the location of the steady states.

• Ensure the desired stability properties

• Ensure the appropriate transients

• Reduce the sensitivity to plant parameters

( , ),

( , ),

n

p

x f x u x R

y h x u u R

find ( )

( )

u r x

u y

state feedbackdynamic output feedback

Remark: Consider the following problem :

stics.characteri eperformanc and stability desired

exhibits ))(,(or ))(,( system loop closed e that thso xhxfxxrxfx

7-27

Nonlinear Control Vs. Linear Control

Why not always use a linear controller ?• It just may not work.

Ex: 3x x u x R

When 0, the equilibrium point 0 is unstable.u x

Choose3 3

.

.

u kx

x x k x

We see that the system can’t be made asymptotically stable at 0.x On the other hand, a nonlinear feedback does exist :

3( )u x kx

Then(1 )x x kx k x

Asymptotically stable if 1.k

Then

7-28

Example

• Even if a linear feedback exists, nonlinear one may be better.

Ex:

y u

k

+_

v y

y u+_

vy

k

y

y ky v

1 2y k y k y v

y ky v

for 0v

y

y

1x y

2x yfor 0v

7-29

Example (Continued)

Let us use a nonlinear controller : To design it, consider again

1 2

2 1

x x

x kx

If 1k If 1k

1x

2x

1x

2x

7-30

Example (Continued)

Switch from to appropriately and obtain a variable structure system.k 1 1

1x

2x

1 2 0x x

1k

1k

1k

1k

sliding line

1

1

if 01

if 01

x sk

x s

1 2where s=x x

Created a new trajectory: the system is insensitive to disturbance in thesliding regime Variable structure control