7-1 asymptotic techniques introduction real world complex for a direct analysis. solution...
TRANSCRIPT
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