example: laser threshold · example: laser threshold at low energy levels each atom oscillates...

1

�

Example: Laser threshold

At low energy levels each atom oscillates acting asa little antenna, but all atoms oscillateindependently and emit randomly phased photons. At a threshold pumping level, all the atoms oscillatein phase producing laser! This isdue to self-organization out of cooperativeinteraction of atoms (Haken 1983, Strogatz’s book)

Activematerial

pump

���� laser lightpartially reflectingmirror

�

Let n(t) - no. of photonsThen, gain - loss

(escape or leakage thru endface)

� gain coeff > 0 no. of excited atomsNote that k > 0, a rate constant

Here typical life time of a photon in the laser

Note however that (because atoms after radiation of a photon,are not in an excited state), i.e.,

=�n= −GnN kn

= − αoN(t) N n

τ = =1k

= − − α= − α − �� 2oo n (GN k)n Gnn Gn(N n) kn or

2

�

The corresponding bifurcation diagram is:

No physical meaning

n*

N0

x*=r

n*=0N0=k/G

lamp laser

�

Pitchfork bifurcationExamples:We have already seen the example of buckling of a columnas a function of the axial load:

Another example is that ofthe onset of convection in a toroidal thermosyphan

mgg g

fluid

heating coil

3

�

The normal form for pitchfork bifurcation is:

The behavior can be understood in terms of the velocity functions as follows:

The bifurcation diagram is then:

Supercritical pitchfork

= − =� 3x r x x f(x)

x

f(x)

r < 0

x

f(x)

r = 0

x

f(x)

r > 0

X*

r

stable

stable

r =0unstablestable

�

Linear stability analysis

Consider

is stable when r < 0is unstable r > 0

� what about when r = 0? The linear analysis fails!!

For the non-zero equilibria:

���� eigenvalue is negative if r > 0

i.e., these bifurcating equilbiria are asymp. stable.

= − =� 3x r x x f(x)

= ±*The equilibria are at x 0, r

� − ==

2*

dfr 3(0) r

dx x 0

� =*x 0

� − ± = −= ±

2*

dfr 3( r ) 2r

dx x r

4

�

Subcritical pitchfork

The resulting bifurcation diagram is:

= +

= ± −

� 3

*

The normal form is x r x x ,

with equilibria x 0, r

X*

r

unstable

unstable

r =0unstable

stable

�

Usually, the unstable behavior is stabilized by higher order non-linear terms, e.g.,The resulting bifurcation diagram can be shown to be:

= + −� 3 5x r x x x

X*

r

unstable

unstable

r =0unstable

stable

subcritical pitchfork, r = rP

supercritical saddle-node, r = rS

5

Connection between simple bifurcations and the implicit function theorem

Let be an equilibrium.

Let f be continuously differentiable w.r.t. x and r in some

open region in the (x, r) plane containing

Then if in a small neighborhood of

we must have:� has a unique solution x=x(r) such that

f(x(r),r)=0� furthermore, x(r) is also continuously differentiable.���� No bifurcations arise so long as

=0 0 0 0f(x ,r ) 0 i.e., (x ,r )

=�Consider the system x f(x,r)

=f(x,r) 0

0 0(x ,r ).

≠0 0

df0

(x ,r )dx 0 0(x ,r ),

≠0 0

df0

(x ,r )dx

�

The figure below illustrates the idea through two points along a solution curve. At (x1,r1), the derivative df/dx does not vanish, where asat (x2,r2), the derivative df/dx vanishes.

(x1,r1)

x

r

df/dx≠0

df/dx=0

(x2,r2)

6

��

Imperfect Bifurcations & Catastrophes Consider the buckling example. If the load does not coincide with the axis of the column, what happens?

� Real physical systems have imperfections and mathematical imposition of reflection symmetry is an idealization.

� Do the bifurcation diagrams change significantly if imperfections or “perturbations” are added to the model (velocity function)? This is related to the concept of “structural stability” or robustness of models.

g

symmetric loading

mg

asymmetric loading

mgg

��

Consider the two-parameter normal form:

� we now have the two parameters, h and r. Note that it is a perturbation of the normal form for pitchfork bifurcation (h=0)

3x h r x x= + −�

3y r x x= −

y h= −

r ≤≤≤≤ 0 , only one equilibriumpossible for any h

3y r x x= −cy h, h h (r )= − >

ch h (r )>ch h (r )

3 equilibriain t his region

<

ch h (r)=

r > 0 , one or threeequilibria possible

7

��

Imperfect bifurcations and catastrophes (cont’d). Consider(h = 0 � normal form of pitchfork)We look for intersections of

== + −� 3 f(x)x h r x x

= − −3y(x) r x x with h

3y r x x= −

y h= −

r ≤≤≤≤ 0 , only one equilibriumpossible for any h

h in

crea

sing

3y r x x= − cy h, h h (r )= − >

ch h (r)>

ch h (r)=

r > 0 , one or threeequilibria possible

Cr / 3

��

For critical point

Furthermore,

= � − = � = ±2 cc c c

rdy 0 r 3x 0 xdx 3

+ − = � = ±3 c cc c c c c

2r rh r x x 0 h3 3

only oneequil soln

3 equilsolns

2 equilsolns

cusp

r

h

r=0

h=0

1

2

3

4

rC

8

��

1.

2.

3.

r

x*

0 rC

r

x*

0

rC

r

x*

0rC

��

4.

Alternately, in 3-D we can visualize the solutions set as follows:

h

x*

0h=0

r

r=0

hx*

h=0

“catastrophicsurface”

9

��

�

1-D system on a circle: over-damped pendulum(acted by a constant torque )The equation of motion is:

�

Let be negligible (imagine pendulum in a vat of molasses)

�

�

�

θ + θ + θ = Γ�� �2mL b mgLsin

�2mL

� θ + θ = Γ�The resulting equation is b mgLsinΓθ = − θ�or

b sinmgL mgL

gl�

m

O

Re�

eθ�

ΓΓΓΓ

ΓΓΓΓ

��

(ratio of appl. torque to max.gravitational torque)

We say that i.e., the phase space is a circle

� Consider the system:

θ∈ π[0, 2 ]

θ′ ′� θ = γ − θ θ =τ

dThen sin whered

Γτ = γ =mgLLet t ;b mgL

′θ = γ − θsin

�

� =0� =ππππ

10

�

if pendulum goes around the circle albeit non-uniformly

If

If

Clearly, there is a saddle-node bifrucation at

γ > 1,

γ = θ = π*1, / 2 is an equilibrium

1,γ <

� =0� =ππππ

� =ππππ/2

γ <

θ θ *1 2* and

1, there are :

whic

two equilibria

opposite stability characterish have

tics� =0� =ππππ

� 1*� 2

*

γ = 1.

�

1-D Non-autonomous Systems:Consider the system with explicit

dependence on timeGeneral propertiesMonotonicity w.r.t. initial data

If

If for all times ‘t’, then the velocity function is T-periodic ���� T-periodic systems

=�x f (x,t)

= +f(x,t) f(x,t T)

< ϕ < ϕ0 0 0 0 0 0x y then (t,t ,x ) (t,t ,y )(follows from uniqueness)

x

tt0

x0

y0ϕϕϕϕ(t,t0,x0)

ϕϕϕϕ(t,t0,y0)

11

��

Consider the T-periodic system

Now, consider the following:

� obeys the same equation as x(t + T), only they have different initial conditions. Thus,

� or

+ = + +�x(t T) f(x(t T),t T)

=�x(t) f(x,t)

� x(t)

= +f(x(t T),t)

x

tt0

x0

ϕϕϕϕ(t,t0,x0)

t0+T t0+2T

x

tt0

x0

ϕϕϕϕ(t,t0,x0)

t0+T

��

Steady state solutions of the T-periodic system:A period-T solution of the system with T-periodic velocity function is one where � Pictorially

Stability of a steady state solution: Intuitively

ϕ + =0 0 0 0(t T,t ,x ) x

x

tt0

x0

ϕϕϕϕ*(t,t0,x0)

t0+T

x0

x

tt0

x0

t0+Tϕϕϕϕ*(t,t0,x0) is asymptotically stable

x

tt0

x0

t0+Tϕϕϕϕ*(t,t0,x0) is unstable

12

��

An example with multiple periodic solutions

> with(DEtools):dfieldplot(diff(x(t),t)=x(t)*x(t)-0.9+0.3*(sin(5*Pi*t)),x(t),t=0...10,x=-3..3,\title=`One-D Periodic NonAutonomous System`,color=x(t)*x(t)-0.9+0.3*(sin(5*Pi*t)));

= − + π� 2x x 0.9 0.3sin(5 t)

��

> with(DEtools):phaseportrait(diff(x(t),t)=x(t)*x(t)-0.9+0.3*(sin(5*Pi*t)), x(t),t=0...5,[[x(0)=0],[x(0)=0.8],[x(0)=0.928],[x(0)=-2],[x(0)=0.9298]], stepsize=0.01,\title=`One-D Periodic Non Autonomous System`,colour=magenta,linecolor=[gold,yellow,black,blue,green]);

13

��

Consider the periodic system:We can think of the stability of a periodic solution in a different way.

We define ‘Poincare map’

It is a mapping that relates a point at time t0 to the

solution at the end of the period, time t0+T. Then,

This is an example of a Poincare map:

= = +�x f(x,t) f(x,t T)

∏ → ∏ ∏ −20 0: x ( (x )) iterate of the mapping

∏ → ϕ +0 0 0 0: x (t T,t ,x )

x

tt0

x0

t0+T

x2x3

x1

periodic soln

��

We begine the study with linear non-autonomous systems (non-homogenous)The solution can be written as:

homogenous soln

The particular solution can be constructed using variation of constants approach:

= +�x a(t)x b(t)

= +p hx(t) x (t) x (t)

� �� �= τ τ� �� �� �0

t

h 0t

x (t) x(t )exp a( )d

�= τ τ0

t

pto be determined t

Let x (t) A(t) exp a( )d . Substituting

�� � τ τ + τ τ� � �

�

0 0

t t

t t

Aexp a( )d A(t) a(t)exp a( )d

14

��

Simplifying, we get

or

= τ τ +0

t

t

a(t)A(t)exp a( )d b(t)

� �� �

� = − τ τ� �� �� ��

0

t

t

A b(t)exp a( )d

τ� �� �= τ − τ� �� �� �

0 0

t

t t

A(t) b( )exp a(s)ds d

0 0 0

t t

h p 0t t t

x(t) x x exp a( )d x(t ) b( )exp a(s)ds dτ� � �

� �� = + = τ τ + τ − τ� �� � � �� �� �

��

Now consider the special case of a homogenous system:Then,

One can write this as

It depends on initial state transition functiontime & current time or propagator

b(t) 0 x a(t)x= � =�

0

t

0t

x(t) x(t )exp a( )d� �� �

� = τ τ� �� �� �

0 0x(t) k(t, t ) x(t )� =���

15

�

The properties of the propagator include:

and

These can be pictorially depicted as:

0 0 2 0 2 1 1 0k(t ,t ) 1, k(t ,t ) k(t , t ) k(t ,t )= =

=1 22 1

1k(t ,t )

k(t ,t )

xx2x1x0

1 0k(t ,t ) 2 1k(t ,t )

2 0k(t ,t )

0 2k(t ,t )

�

Linear Periodic SystemsConsider the system

Define a period propagator:

Now consider

T

0

k(0, T ) exp a( )d= τ τ

= + =�x a(t)x a(t T) a(t) for all time

0 0 0

0

0 0 0

0

0

0

t t T t TT

0 0 t T

t t T t T

0 t T

t T

0 0t

exp a( )d exp a( )d a( )d a( )d

exp a( )d exp a( )d exp a( )d

exp a( )d k(t T, t )

+ +

+ +

+

� �� �� � � �τ τ = τ τ + τ τ − τ τ� � � �� � � �� � � �

� �� � � �� � � � � �= τ τ τ τ − τ τ� � � � � �� � � � � �� � � �� �

� �� �= τ τ = +� �� �� �

16

��

Having defined the period propagator, now consider solutions at the end of every period:

gives iterates of the mapping

�+ =

��������0 0xkx 01

x(t T) k(T,0)x(t ) or �=1 0periodpropagator

x k x

= + = + = =2 0 0 1 0x x(t 2T) k x(t T) k x kk x(t )

= + = + =3 0 0 0x x(t 3T) k x(t 2T) kkk x(t )

= + = + − = nn 0 0 0

.

.

x x(t nT) k x(t (n 1)T) k x(t )

xx2x1x0

k

x3

k k

��

Consider stability of a fixed point.� x = 0 is a fixed point of the t-periodic systemThen, for some initial condition x0

Following the iterates, we have

If k > 1, x = 0 is unstable (xn � ∞∞∞∞ as n � ∞∞∞∞).

�=1 0periodpropagator

x k x=�x a(t)x

= =

=

= = τ τ

2 1 03

3 0T

0

x k x kk x

x k x ........

Note that k k(T,0) exp a( )d

xx2x1x0

k

x3

k kx=0x2 x1

x0

kk

17

��

x = 0 is asymptotically (strongly) stable

If -1 < k < 0, (not physically possible for 1-D systems)

x = 0 is asymptotically stable (in an oscillatory manner)

What do you think is the time response of the system?

< <If 0 k 1

xx1x2x2

k

x0

k kx=0x0 x1

x3

kk

k

x3

k x=0x2x1 x0

kx

��

If - ∞∞∞∞ <k< – 1, x=0 is unstable (in an oscillatory manner)

2nd Order or 2-Dimensional Systems (autonomous)Every system is of the form

Ex: The damped pendulum+ + =�� �x f(x,x ) g(x ) 0

k

x0

k x=0x1x2 x3

kx

gl�

m

O

Re�

eθ�

RKinematics : r e= �� �

Rr e eθ= = θ�� �� �� � �

2Rr e e e eθ θ θ= θ + θ = − θ + θ� �� � ���� �� � � �

� � � � �

18

��

Applying Newton’s second law to the particle

Let us define two nondimensional variables:

= ���� �

mr F

e : m mgsin cθ θ = − θ − θ�� �� �

− + θ = − θ�� 2re : T mgcos m

2n n2 sin 0� θ + γω θ + ω θ =�� �ω =

�

2n

g γ =

�

cg2m

( mgsin c ) eθ− θ − θ��

RF (m g cos T ) e= θ −�

g�

m

Re�

eθ�

mg

T

c l θ�

��

Write the equations in first-order form:

���� (1)

We first study the Undamped problem in detail.

1. Equilibria of the system (i.e., when

2. Now, linearize (1) about an equilibrium:

θ = θ =�1 2x x=�1 2x x= − ζω − ω� 2

2 n 2 n 1x 2 x sin(x )

( 0)ζ == =� �1 2x x 0)

2 1x 0 and x 0, n , n 1, 2, 3� = = ± π =

19

��

Let are small perturbations

Substituting into equation (1) ����

and

1 1x n y= ± π += +2 2x 0 y

= n 0,1, 2, 3

1 2n y 0 y± π + = +��

(with 0)ζ =

22 n 10 y sin( n y )+ = −ω ± π +� �

1 2Here y and y

� =�1 2y y2

2 n 1 12 nn 1

y (sin( n )cos y cos( n )siny )

( 1) siny

= −ω ± π + ± π

= −ω −

�

��

Assume small variations from equilibrium point: (Taylor Series expansion for small y1 and y2, and retain only linear terms) ����

Case 1: n = even

+� �� � � �

= � �� � � �ω −� �� � � �� �

�

�1 1

2 n 12 2n

0 1y yy y( 1) 0

= = −ω� � 21 2 2 n 1y y y y

1 22

2 n 1

We can int egrate equations in phase planedy y

first write asdy y

� = −ω

2n 1 1 2 2y dy y dy 0� ω + =

2 2 2n 1 2Integrating, this gives ( y y ) / 2 C (cons

defines an ellipse around the equilib u

t)

( )ri m

ω + =

20

�

Phase portrait around a center> with(DEtools):phaseportrait([D(x1)(t)=x2(t),D(x2)(t)=-x1(t)],\[x1(t),x2(t)],t=-7...7,[[x1(0)=0,x2(0)=1],[x1(0)=0,x2(0)=0.5],[x1(0)=0,x2(0)=0.2]],stepsize=0.02,title=`Phase portrait around a center`,colour=magenta,linecolor=[gold,blue,red]);

�

Case 2: n = odd ����

Note that there are two special curves:These curves (straight lines for C=0) pass through the equilibrium with specified slope

=�1 2y y = ω� 22 n 1y y

1 22

2 n 1

Pr oceeding along thesame lines, we can int egratedy y

as follows : first writedy y

=ω

� ω =2n 1 1 2 2y dy y dy

ω − =2 2 2n 1 2

defines a hyperbola around the equilibrium,C de

In

pe

tegrating ( y y ) /

nds on initial co

2 C (c

ndit

o

i

nst)

(ons)

= ±ω2 n 1y y

n±ω

21

��

Phase portrait around a Saddle> with(DEtools):phaseportrait([D(x1)(t)=x2(t),D(x2)(t)=x1(t)],\[x1(t),x2(t)],t=-1.5...1.5,[[x1(0)=0,x2(0)=0.5],[x1(0)=0.5,x2(0)=0],[x1(0)=0,x2(0)=-0.5],[x1(0)=-0.5,x2(0)=0]],stepsize=0.02,title=`Phase portrait around a saddle`,colour=magenta,linecolor=[green,gold,blue,red]);

��

From last time: The equations for pendulum motion are

(undamped pendulum)The equilibria are at

Linearized EOM about each equilibrium:n = even

n = odd

�����

= = θ= θ= − ζω − ω

�

��

1 2 12

2n n2 2 1

x x x

xx 2 x sinx

ζ = 0

=�� = π = ± ±�

2

1

x 0x n ,n 0, 1, 2

= = −ω� � 21 2 2 n 1y y y y

= = ω� � 21 2 2 n 1y y y y

22

��

Phase portrait for the undamped pendulum> with(DEtools):phaseportrait([D(x1)(t)=x2(t),D(x2)(t)=-sin(x1(t))],\[x1(t),x2(t)],t=-7...7,[[x1(0)=0,x2(0)=1],[x1(0)=0,x2(0)=2],[x1(0)=0,x2(0)=2.1],[x1(0)=0,x2(0)=-2],[x1(0)=0,x2(0)=-2.1],[x1(0)=6.28,x2(0)=1],[x1(0)=-6.28,x2(0)=1],[x1(0)=9.424777962,x2(0)=0],[x1(0)=-2*Pi,x2(0)=2],[x1(0)=-2*Pi,x2(0)=-2],[x1(0)=2*Pi,x2(0)=2],[x1(0)=2*Pi,x2(0)=-2]],stepsize=0.02,title=`Phase portrait for the undamped Pendulum`,colour=magenta,linecolor=[blue,blue,blue,blue,blue,blue,blue,blue,blue,blue,blue,blue]);

ππππ-ππππ 0-2ππππ 2ππππ

separatrix

saddle

center or libration

rotation

��

The equation for nonlinear, undamped pendulum is:

As seen on the last page, the phase portrait is symmetric about x1 axis. It is governed by the system:

(1)This equation governs the locus of points along a trajectory. Note it is invariant under the transformation , and invariant under the transformation

(Normal form all conservative (Particular to the pendulum

problem)

=�1 2x x = −ω� 2n2 1x sinx

= −ω

1 222 n 1

dx xdx sinx

→ −�����2 2x x

→ −�����1 1x x

oscillators)

23

��

We can integrate (1) along a solution trajectory:

For different constants, we get different trajectories in

the phase plane.

� Consider a trajectory passing

“through” - a saddle point.From (2):

� ω = − 2n 1 1 2 2sinx dx x dx

= π1 2,x )(x ( ,0)

� − ω =2 2n2 1/ 2The result is x cosx constant (2)

� − ω = = − ω π = ω2 2 2 2 2n n n2 1/ 2 / 2x cosx C (0) cos( )

2 n2 2

n2 1 1/ 2 ) or 2 )

equation for trajectory through the saddThus, the

x (1 cosx x

le is

(1 cosx� = ω + = ±ω +

��

This trajectories passing through saddles look like:

x1

x2

(�,0)(-�,0)

(0,2)

24

��

� Now consider a trajectory passing throughFrom equation (2), we get

ωn(0,4 )

� − ω = = ω − ω = ω2 2 2 2 22 n 1 n n nx / 2 cosx C (4 ) / 2 cos(0) 7

2 22 n 1 2 n 1

:

x /

The equa

2 (7

tion for traject

cosx ) or x

ory is

2(7 c

the

x

n

os )= ω + = ±ω +

x1

x2

(�,0)(-�,0)

(0,4)

(0,-4)

these representrotations of the pendulum

��

As before, the composite phase portrait then is:

Note that separatrices can also be generated by starting with initial conditions on the trajectories through the saddles; e.g., starting with ( ±±±± εωεωεωεωn,εεεε)

ππππ-ππππ 0-2ππππ 2ππππ

separatrix

saddle

center or libration

rotation

(0,-2)

(0,2)

25

�

Remarks: The phase space for the pendulum can be viewed as a cylinder. Then, the phase portrait is on the surface of a cylinder.

-� �

fold θ�

θθθθ

�

Time period of oscillations: when initial conditions lead to oscillatory motions, the period depends on amplitude

To determine period of motion, consider the initial condition (ααααs,0), where ααααS is the amplitude of max excurtion. Then one can integrate the equation along a closed curve as shown next.

T ≈≈≈≈ 2ππππT ≈≈≈≈ ∞∞∞∞

26

��

Details are provided in the notes; however, in principle:The energy equation is

212 n 1

dxor x 2( cos x C)

dt= = ± ω +

2 22 n 1

1x cos x C

2= ω +

π

�

=± ω +

= φ − φ = αω

1 1

0 0

x(t ) t1

2x(t ) tn 1

/ 22 2

n 0

we int egrate over quarter of the per

Integrating along a trajec tory

dxdt

2( cos x C)

If , we get

4T d / (1 k sin ) where k si

od

n(

i

/ 2)

��

The resulting curve for period as a function of amplitude ααααS is:

period vs modulus

0

5

10

15

20

25

0 0.2 0.4 0.6 0.8 1 1.2

T

modulus ‘k’

27

��

Conservative Systems & First Integrals of MotionConsider the system in first-order form:

A scalar quantity H(x,y) such that along any solution, is called an integral of motion. As an example, consider mechanical oscillators where

-

=�x f(x,y) =�y g(x,y)=�H 0

−∂= =∂

��V(x)

x Fx

∂� + =

∂��

Vx 0x

−= �

� = +�−∂= �∂ �

�

�

2

Let us write the system in first order form :

x y1

. (3) The

int egral of motion

n, H(x,y) y V(x) is anV 2yx

o conserved quanr a tity.integral of motion (conserved quantity)

��

Differentiating H(x,y) gives:

However, are solutions of (3) �

Note that is the total energy of the system.

Example (pendulum): The equation of motion is:

� �x and y

∂= +∂

� � �1 V

H 2y y x2 x

−∂ ∂ �= + =� ∂ ∂� �� 1 V VH 2y (y) 0

2 x x

� �= +2

potential energyKin. energy

H y / 2 V(x)

θ + ω θ =−∂ ∂θ

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2n sin 0

V /

θ

= θ − ω θ =

=

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�

2 2nV( )

1Then, H cos total energy

2

and verify that H 0