IAPT Workshop
2nd August ISP, CUSAT
V P N Nampoori
nampoori@gmail.com
Simple pendulum
Oscillate
Rotate
Simple pendulum
Oscillate
Rotate
Double pendulum
Taking the amplitude
small
d 2
2


 0
2
dt
   0 Sin t
1
T 
2
θ
L
g
  2

g
 

2
L
2nd order linear DE
solutions 1 and  2
  a1  b 2
 sup erposition principle
The hallmark of linear equations.
We can predict state of the simple pendulum at any future time
Linearlty helps us to predict future.
Alternate representation of damped motion of simple pendulum
Phase space plot
x
1 3
sin     
3
d 2
1 3
2
  (   )  0
2
dt
3
Second order nonlinear differential equation
Superposition principle is not valid
Prediction is not possible
Linear dynamics – prediction possible
Traffic jam –Nonlinear dynamics - prediction is impossible
One-dimensional maps
One-dimensional maps, definition:
- a set V (e.g. real numbers between 0 and 1)
- a map of the kind f:VV
Linear maps:
- a and b are constants
- linear maps are invertible with no ambiguity
Non-linear maps: The logistic map
One-dimensional maps
Non-linear maps: The logistic map
with
Motivation: Discretization of the logistic equation for the dynamics
of a biological population x
b: birth rate (assumed constant)
cx: death rate depends on population (competition for food, …)
How do we explore the logistic map?
Geometric representation
1
Evolution of a map:
1) Choose initial conditions
2) Proceed vertically until you hit f(x)
3) Proceed horizontally until you hit y=x
4) Repeat 2)
5) Repeat 3)
.
:
x
f(x)
0
0.5
1
Evolution of the logistic map
fixed point ?
Phenomenology of the logistic map
a)
b)
1
1
y=x
y=x
fixed point
c)
f(x)
f(x)
0
0.5
fixed point
0
1
1
d)
0.5
1
1
chaos?
2-cycle?
0
0.5
1
What’s going on?
0
0.5
1
Analyze first a)  b)
b)  c) , …
Geometrical representation
1
1
x
x
f(x)
0
0.5
f(x)
1
Evolution of the logistic map
0
0.5
1
fixed point
How do we analyze the existence/stability of a fixed point?
Fixed points
- Condition for existence:
- Logistic map:
- Notice: since
the second fixed point exists only for
Stability
- Define the distance of
from the fixed point
Taylor expansion
- Consider a neighborhood of
- The requirement
implies
Logistic map?
Stability and the Logistic Map
- Stability condition:
- First fixed point: stable (attractor) for
- Second fixed point: stable (attractor) for
- No coexistence of 2 stable fixed points for these parameters
(transcritical biforcation)
1
1
x
x
f(x)
0
0.5
What about
f(x)
0
1
?
0.5
1
Period doubling
1
Observations:
x
1) The map oscillates
between two values of x
f(x)
0
0.5
1
0
0.5
1
Evolution of the logistic map
2) Period doubling:
What is it happening?
Period doubling
- At
since
the fixed point
becomes unstable,
>
-Observation: an attracting 2-cycle starts 
(flip)-bifurcation
The points are found solving the equations
0
0.5
1
and thus:
These points form a 2-cycle for
However, the relation
suggests
they are fixed points for the iterated map
Stability analysis for
and thus:
For
Why do these
points appear?
:
, loss of stability and bifurcation to a 4-cycle
Now, graphically..
Bifurcation diagram Plot of fixed points vs
International Relations and Logistic Map
Let A and B are two neighbouring countries
Both countries look each other with enmity.
Country A has x1 fraction of the budget for the Defence for year 1
Country B has same fraction in its budget as soon as A’s budget session
Is over
Next year A has increased budget allocation x2
Budget allocation goes on increasing.
If complete budget is for Defence , it is not possible since no funds for
Other areas
Fund allocation for subsequent years for the country A
xn1  f ( xn )   xn (1  xn )
1
x
f(x)
0
0.5
1
As time progresses, budget allocation for defense decreases.
Peace time . A and B are friends.
xn1  f ( xn )   xn (1  xn )
Parameter μ is called enmity parameter.
Now let a third country C intervenes
In the region to modulate the enmity parameter and
μ = μ(t)
Phenomenology of the logistic map
a)
b)
1
1
y=x
y=x
fixed point
c)
f(x)
f(x)
0
0.5
fixed point
0
1
1
d)
0.5
1
1
chaos?
2-cycle?
0
0.5
1
What’s going on?
0
0.5
1
Analyze first a)  b)
b)  c) , …
1
1
x
x
f(x)
0
0.5
f(x)
1
Budget allocation decreases
and goes to zero
Full peace time Green
0
0.5
1
Budget allocation stabilises
To a fixed value.
Caution time. Yellow
Period doubling
1
Observations:
x
1) The map oscillates
between two values of x
f(x)
0
0.5
1
0
0.5
1
Evolution of the logistic map
2) Period doubling:
What is it happening?
Bifurcation diagram Plot of fixed points vs
Tension
builds up
Peace time
WAR!!!
Evolution of International Relationships between three countries
Two countries are at enmity and the third is the controlling country
From Peace time to War time
Interaction leads to modification of dynamics.
A, B and C are three components of a system
with two states YES (1) or NO ( 0)
Case 1 A, B and C are non interacting
Following are the 8 equal probable states of the system
1=( 000), 2=( 001), 3=(010), 4=( 100),5=(110),6=(101),7=(011), 8=(111)
Probability of occurrence is 1/8 for all the states.
States evolve randomly.
Case II Let A obeys AND logic gate while B and C obey OR logic gate
T1
(000)
(010)
T2
(000)
(001)
(001)
(010)
(100)
(011)
(110)
(011)
T3
(000)
(010)
(001)
(111)
(111)
(101)
(011)
(111)
(011)
(111)
(111)
(111)
(111)
(111)
(000)
1/8
I
(001)
(010)
II
bistable
2/8
(111)
(111)
(111)
(111)
(111)
III
5/8
Evolution to
Fixed state
Blissful state!!!
Bistable picture
Bistable picture
Rabbit and duck – bistable state
Photograph of melting ice landscape –
Face of Jesus Christ - evolution leading to fixed state
Evolution – Esher’s painting
Nonlinearity in Optics
Linear Optics
Maxwell’s Equations : Light -- Matter Interaction
Maxwell’s equations for charge free,
nonmagnetic medium
.D = 0 .B = 0 XE = -B/t
XH = D/t
D = 0E + P and B = 0H
In vacuum, P = 0 and on combining above eqns
2E - 00 2 E/t2 = 0 or 2E - 1/c2 2 E/t2 = 0
In a medium, 2E - 00 2 E/t2 - 02P/t2 = 0
writing P = 0 E,
2E - 1/v2 2 E/t2 = 0 where, v = (0 )-1/2
Defining c/v = n, 2E - n2/c2 2 E/t2 = 0
we get a second order linear diffl eqn describing
what is called, Linear Optics
P=(1) E + (2) E2 + (3)E3 +…..
(n)/ (n+1) << 1
For isotropic medium, (n) will be scalar.
(n) represents nth order non linear optical coefficient
Polarisation of a medium P = PL +PNL
where,PL = (1) E and PNL = (2) E2 + (3)E3 +…..
On substituting P in the Maxwell’s eqns, we get
2E - n2/c2 2 E/t2 = 02PNL/t2
This is nonlinear differential equation and describes
various types of Nonlinear optical phenomena.
Type of NL effects exhibited by the medium depend the
order of nonlinear optical coefficient.
E
Polarisation
Maxwell’s eqns
feedback

2
Optical second harmonic generation
1
3
2
Sum (difference) frequency generation
OPC by DFWM
Consequence of 3rd order optical nonlinearity
intensity dependent complex refractive index
P   0 1 E   0  3 E 3
 ( 0  1   0  3 E 2
)E
 E
   0 1   0  3 E 2
refract ive index  ( 0 1   0  3 E 2 )1 / 2
which gives

n  0

 0


1 1/ 2

(1  1 / 2

 1 / 2(
1 1/ 2
  0 1
3 1/ 2
(1  1 I)

1/ 2
n ( I)  n 1  n 2 I
3
I)
1

 0 1/ 2 3
)  I
1

One of the consequences of 3rd order NL
PNL = (1)E + (3)E3 =( (1) + (3)E2)E = (n1 + n2 I)E
We have
n(I) = (n1 + n2 I)
n2< 0
n
n1
n2> 0
I
Intensity dependent ref.index has applications in
self induced transparency, self focussing, optical limiting
and in optical computing.
Saturable absorbers– Materials which become
transparent above threshold intense light pulses
Materials become transparent at high intensity
α
T
Absorption decreases with intensity
Is
I
Optical Limiters : Materials which are opaque above a threshold laser
intensity
It
Materials become opaque at higher intensity
Is
I
Materials become opaque at higher pump intensity – optical limiter
Materials become transparent at higher pump intensity- saturable absorber
Optical limiters are used in……
Protection of eyes and sensitive devices from intense
light pulses
Laser mode locking
Optical pulse shaping
Optical signal processing and computing
.
Intensity dependent refractive index
Laser beam :Gaussian beam
I(r )
I(r ) = I0exp(-2r2/w2)
Beam cross section
r
I( r) = I0exp(-2r2/w2)
T ( r) =T0exp(-2r2/w2)
n(r )=n0exp(-2r2/w2)
n = n(I)
n= n1-n2I
Fabry - Perot etalon as NLO device
n=n1+n2I
L
Condition for transmission peak
L = m / 2n = m  / 2(n1+n2I)
m = 2nL/m is the resonant
condition
It

It

i/p#2
Ii=I1+I2
It o/p
Saturable abspn
i/p#1
It
Ii
Materials become transparent at higher pump intensity- saturable absorber
i/p#2
Ii=I1+I2
It o/p
Optical limiting
( extra abspn at high intensity)
i/p#1
It
Materials become opaque at higher pump
intensity – optical limiter
Ii
Threshold logic
I/p 1
+
0 or 1
0
I/p 2
AND
NOT/NOR
1
Binary o/p
nonlinearity
0 or 1
O/p
1
0,1 or 2
2
I/p
OR
1
2
XOR
1
1
2
1
2
Z-scan Experiment
Aperture
Detector
B
Beam splitter
Sample
z
A method to vary the intensity of pump beam incident on a sample:
The sample is moved along a focused gaussian beam. Laser Intensity can be
varied continuously with maximum at the focal point.
Open aperture z-scan : Sensitive to nonlinear absorption of sample
Closed aperture z-scan : Sensitive to both nonlinear absorption & refraction
M S Bahae et.al; “High-sensitivity, single-beam n2 measurements”, Opt Lett, 14, 955 (1989)
(a) Open aperture z scan
The propagation through the sample is given by
dI
  I I
dz
1. Reverse Saturable Absorption
RSA- transmittance valley
Application: Optical limiting
2. Saturable Absorption
SA- transmittance peak
Application: Optical pulse compression
self assembled films of ZnO
Self assembled nano
films :
Drop casting & solvent evaporation
 Reaction solution dropped onto a
substrate @ 120oC
 Solvent evaporation helps particles
to spontaneously assemble into
periodic structures
Temperature and solvent – significantly
influence the process of self-assembly
Open aperture
RSA in Colloids
The observed nonlinear absorption is
attributed to TPA followed by FCA
SA in Self assembled films
self assembled films of ZnO
Size dependence
Intensity-220 MW/cm2 and Irradiation wavelength-532 nm
Colloids
Self assembled thin films
Normalized
Transmittance
Normalized
Transmittance
0
z
Negative nonlinearity : peak-valley
Closed Aperture Z scan
0
z
Positive nonlinearity: valley-peak
For a positive (self-focusing) nonlinearity,
 The positive lensing in the sample placed before the focus moves the focal
position closer to the sample resulting in a greater far field divergence and a
reduced aperture transmittance corresponds to valley
 When the sample is after focus, the same positive lensing reduces the far field
divergence allowing for a larger aperture transmittance corresponds to peak
The opposite occurs for a self-defocusing nonlinearity
To Conclude
Nature is Nonlinear
Linearity is an approximation for our convenience