PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
Twice iterated tent map : find corresponding map M 2 (x)
1)
Note there are points where the map M ( x ) goes to zero or 1.
M (x)
also,
2)
M
( )=1
M (0) = 0
M
( ) = 12
M
1
2
1
4
M (1 ) = 0
( ) = 12 .
3
4
There are points where M 2 ( x ) goes to zero or 1.
Clearly,
M2
( ) = M ( M ( )) = M ( ) = 1
( ) = M ( M ( )) = M ( ) = 1
M2
( ) = M ( M ( ) ) = M ( 1) = 0
M2
and
1
4
3
4
1
2
1
4
3
4
1
2
1
2
1
2
M2 (0) = M ( M (0)) = M (0) = 0
Folding points – another method.
1
0
1
1
2
1
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
M2( x) = 1
1
2
M (x) =
1
4
1
iterates as M
4
1
iterates as M
8
1
iterates as M
16
( ) = 12
1
4
M2
1
2
( )=1
1
4
( ) = 14
and M
( ) = 14
so similarly
1
8
1
16
M5
Now consider all the segments between, ie
1 1 1 1 3 3
0! ,
! ,
! ,
! 1.
4 4 2 2 4 4
2
( )=0
1
4
M2
1
4
M3(x) = 0
3
4
M3
( ) = 12
1
2
( )=1
1
4
( )=0
1
16
so M 4
( )=0
1
8
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
Notation
Write M ( x ) as
M < ( x ) = 2x "left half"
M> ( x ) = 2(1 ! x )
1
2
1
4
1
2
1%
(
4 '&
1%
"
$0 ! 2 '
#
&
0
"
$0 !
#
"
$0 !
#
Clearly,
and
only need
0!x!
1
4
1%
( [ 0 ! 1]
2 '&
3
4
M< ( x )
M< ( x )
M < ( x ) here
M 2 ( x ) = M <2 ( x )
Now
"1 1%
"1
%
$ 4 ! 2 ' ( $ 2 ! 1'
#
&
#
&
"1
%
$ 2 ! 1' ( [1 ! 0 ]
#
&
M< ( x )
M> ( x )
so,
1
1
!x!
4
2
M 2 ( x ) = M> ( M< ( x ) )
similarly
3
1
"right half"
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
1
3
!x!
2
4
3
! x !1
4
M 2 ( x ) = M >2 ( x )
M 2 ( x ) = M< ( M> ( x ) )
Then plug in M < , M > to get the result.
1
M 2 ( x ) = 2 [ 2x ] = 4x
4
1
1
!x!
M 2 ( x ) = 2 ( 1 " [ 2x ] ) = 2 " 4x
4
2
1
3
!x!
M 2 ( x ) = 2(1 " [ 2(1 " x ) ] )
2
4
= 2 ( 1 " 2 + 2x )
0!x!
= 4x " 2
3
! x ! 1 M 2 ( x ) = 2[ 2 ( 1 " x ) ]
4
= 4 " 4x
Sketch
M
1/4
1/4
1/2
1/2
x
3/4
3/4
x
Note – neighbouring points on x move apart as length of M ( x ) line increases as we iterate –
stretch/fold.
4
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
1 triangle
2 fixed points
•
1/2
M2
2 triangles
4 fixed points
•
•
•
•
•
1/2
1/4
M3
•
•
•
•
•
•
•
•
5
1/8
1/4
4 triangles
8 fixed points
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
p th iterate of map
2 p !1 triangles
2 p fixed points
2! p
2! p +1
- base length
Therefore, any points in x separated by more than 2! p can be anywhere in [ 0, 1 ] after p iterates.
-
chaos
follows from inevitability of map.
Note, gradient is always 2 p > 1
therefore, always globally unstable.
6
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
local gradient and stability
Note that this shows if a fixed point is unstable.
eg: tent map
xn +1 = 2 xn , xn +1 = 2 (1 ! xn )
xn +1 = xn
M
•
•
x1
unstable
stability just depends on gradient
locally
2
x1 x0
x
here gradient >
1
M
1
2
here gradient <
stable
1x
7
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
Mathematically, for some map M ( x ) consider small displacement ! xn from fixed point x and
Taylor expand xn+1 = M ( xn ) at xn = x + ! xn .
xn+1 = M ( x + ! xn ) ! M ( x ) + ! xn
but
xn+1 = x + ! xn+1
so
x + ! xn+1 = x + ! xn
(
)
dM
dM
x ) + 0 ! x 2 ! x + ! xn
(
(x)
dx
dx
dM
(x)
dx
ie:
! xn+1 dM
=
(x ).
! xn
dx
dM
( x ) < 1.
dx
Must hold for all iterates.. Lyapunov exponents
So, for stability
Measure of divergence of trajectories.
Lyapunov exponents (or Lyapunov number)
Consider a general map
xn+1 = f ( xn )
how fast are two initially neighbouring points converging/diverging?
Have initial condition x0 .
This then has iterates x1 , x2 ...xn
x1 = f ( x0 ) ,
x2 = f ( x1 ) , etc.
Consider an initially neighbouring point x!0 = x0 + ! 0 separated from x0 by ! 0 ! 1 .
After we iterate x!1 = f ( x!0 ) = f ( x0 + ! 0 ) = f ( x0 ) + ! 0
But, after one iterate x!1 = x1 + !1
then
df
( x ) + ... , by Taylor expansion.
dx 0
Now two points are separated by !1
df
( x ) + ...
dx 0
to first order in ! 0 .
x!1 = x1 + !1 = f ( x0 + ! 0 ) = f ( x0 ) + ! 0
ie:
!1 = ! 0 f " ( x0 )
So, generally for j th iterate have
x! j = x j + ! j
8
(
! j = ! j"1 f # x j"1
)
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
provided ! j ! 1 0 < j < n .
x!n = xn + ! n = xn + ! n"1 f # ( xn"1 )
= xn + ! n"2 f # ( xn"2 ) f # ( xn"1 )
Then
= xn + ! 0 f # ( x0 ) f # ( x1 ) ... f # ( xn"1 )
x!n+1 = xn+1 + ! 0 f " ( x0 ) ... f " ( xn )
n
( )
x!n+1 = xn+1 + ! 0 " f # x j
j=0
Now use a trick (handy to turn a product into a sum)
( )
f ! xj = e
( ).
ln f ! n j
( ) (and we don't need to know – just interested in
In addition, we do not know the signs of all f ! x j
whether the product is getting bigger or not in magnitude).
So, write
% n#1
(
x!n = xn + ! 0 exp ' $ ln f " x j * ,
& j=0
)
and hence, define Lyapunov number (exponent)
( )
n%1
!=
NB: after written with
n!1! n
n"#
( )
1
& ln f $ x j
Lim
n j=0
n"#
which is a measure of exponential divergence of trajectories
x!n ! xn = " 0 en# .
What about approximation ! j ! 1 ?
Can always take ! 0 (and thus ! j ) infinitesimally small – still same dynamics.
So
!<0 !>0 -
x is at an attractor – trajectories converge.
x is at a repellor → chaos
trajectories diverge.
Note, there is a ! for each degree of freedom (coordinate) here one (x).
( )
Note that generally if all f ! x j
9
>1
all iterates
PX391 Nonlinearity, Chaos, Complexity SUMMARY Lecture Notes 10- S C Chapman
( )
ln f ! x j
then all
" ln f ! ( x j )
then
>0
>0
and ! > 0
chaotic
( )
Similarly, if all f ! x j
<1
then all
ln f ! x j
<0
!<0
and
*
( )
-
stable attractor
-
if our graphical result that for each iterate if gradient > 1 unstable
-
now generalised this for all iterates of the map – characterised the dynamics.
*
Tent map after p iterates:
gradient = ±2 p
gradient >
2
! p+1
* Therefore, tent map globally chaotic *
Next, consider polynomial maps. More complicated f ! ( x )
10
all iterates