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ME 680- Spring 2014

Geometrical Analysis of 1-D Dynamical Systems

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Logistic equation: 𝒏 = 𝒓𝒏(𝟏 − 𝒏)𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏

The length of the arrows magnitude of the velocity

(function) at that point.

Geometrical Analysis of 1-D Dynamical Systems

Equilibria or fixed points : initial conditions n* where you

start and stay without evolving for all time. They

correspond to zeros of the velocity function:

n*=0 n*=1 n

f(n) Phase diagram

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𝛼 −limit set of a point (initial condition) 𝑛0 :

It is defined as the set of limit points of the trajectory

started at 𝑛0, for t → - . Thus,

𝛼(𝑛0) = 𝑛 | lim 𝜑 (𝑛0, 𝑡) = 𝑛𝑡→−∞

𝜔 −limit set of a point 𝑛0 is the set

𝜔(𝑛0) = 𝑛 | lim 𝜑 (𝑛0, 𝑡) = 𝑛𝑡→+∞

Existence of a potential function: Consider 𝒏 = 𝒇(𝒏)

(Gradient Dynamical System)

Let there be a function V(n) such that 𝒇(𝒏) = −𝛛𝑽 𝝏 𝒏

Example: for the Logistic equation 𝒇(𝒏) = 𝒓 𝒏 (𝟏 − 𝒏),

i.e., 𝑽(𝒏) = −𝒓𝒏𝟐 𝟐 + 𝒓𝒏𝟑 𝟑

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Then, note that the equilibrium points for the system (a

Gradient Dynamical System) are at the local extrema of the

potential function. This is where the similarity with mechanical

systems with potential energy functions ends!! Considering the

Logistic equation:

𝑽(𝒏) =−𝒓𝒏𝟐

𝟐+

𝒓𝒏𝟑

𝟑

the plot of the potential function, and the equilibrium

points are as follows:

V(n)

n n*=0

n*=1

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Oscillatory behavior is not possible in 1-D autonomous Systems

Trajectories approach the equilibrium point n*=1, but

never reach it in finite time.

Invariant subspaces are regions 𝑰 ⊆ 𝕽 in phase space where

if 𝒏𝟎  ∈   𝑰, then 𝝋  𝒕, 𝒏𝟎 ∈ 𝑰 for all negative and

positive flow times (- < t < ). For the Logistic Equation,

the invariant subspaces are:

𝑰𝟏 = −∞,𝟎  , 𝑰𝟐 = 𝟎  , 𝑰𝟑 = 𝟎, 𝟏 , 𝑰𝟒 = 𝟏  , 𝑰𝟓 = {𝟏,+∞}

Thus, the state space is decomposed into:

ℝ = 𝑰𝟏 ∪  𝑰𝟐 ∪ 𝑰𝟑 ∪ 𝑰𝟒  ∪ 𝑰𝟓

𝜶 𝒂𝒏𝒅 𝝎 limit sets of any initial condition 𝑛0:

Observations:

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If 𝒏𝟎  ∈  𝑰,   𝑡ℎ𝑒   𝜶 − limit set is 0

𝝎− limit set is−∞

If 𝒏𝟎  ∈  𝑰𝟐,  𝑡ℎ𝑒𝑛  𝜶 −  𝑎𝑛𝑑 𝝎− limit sets are the same

If 𝒏𝟎  ∈  𝑰𝟑,  𝑡ℎ𝑒𝑛     𝜶 (𝒏𝟎) = {𝟎},   𝑎𝑛𝑑    𝝎(𝒏𝟎) = {𝟏} ⋯  𝑠𝑜 𝑜𝑛.

Stability of Equilibria/Fixed Points

An equilibrium point of 𝒙 = 𝒇(𝒙), say x=x*, is stable if

∀𝜺 > 𝟎 ,   ∃ 𝜹 (𝜺) > 𝟎  such that for any initial condition x0,

with | 𝒙𝟎 − 𝒙∗ | < 𝜹,   |𝝋 (𝒕, 𝒙𝟎) − 𝒙∗| < 𝜺   𝒇𝒐𝒓   𝒂𝒍𝒍  𝒕,   𝟎  < 𝒕  < +∞.

Otherwise, it is unstable.

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This definition of stability is very difficult to use directly to

deduce stability of an equilibrium point. One needs to a

priori know the solution for every given initial condition

starting inside the region of size δ. Thus, one really needs

to find other criteria that can be used to characterize stability

without solving the differential equation.

If in addition, |𝜑(𝑡, 𝑥0) − 𝑥∗| → 0 𝑎𝑠  𝑡  →  ∞,   𝑡ℎ𝑒𝑛 𝑥∗ is an

asymptotically stable equilibrium.

x*

δ

x* x0

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1. Logistic Equation

n*=0 (unstable)

n

f(n)=an2

n*=0 (stable) n

f(n) = −an3

Examples:

Observe: Any isolated stable equilibrium in 1-D

autonomous systems has to be asymptotically stable.

2. Quadratic System

n*=0

(unstable) n*=1 n

f(n)=rn(1−n)

(asymptotically stable)

3. Cubic System

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Let x* be a fixed point of 𝑥 = 𝑓(𝑥), i.e. 𝑓(𝑥∗) = 0

To linearize about x = x*, introduce a perturbation:

𝐿𝑒𝑡   𝑥 = 𝑥 − 𝑥∗ ⇒ 𝑇ℎ𝑒𝑛      𝑥 + 𝑥 ∗ = 𝑓(𝑥 + 𝑥∗)

x x0 0

dfor x x f(x ) x

dx

x 0

dfx x

dx

This is the linearized equation about x = x*

Linearization about equilibrium points

(Taylor series expansion for small 𝑋 )

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The equilibrium points are 𝑛∗ = 1   𝑎𝑛𝑑   𝑛∗ = 0.  

Let us linearize the system about 𝑛∗ = 1

Then 𝑛 = 𝑛∗ + 𝑛 = 1  + 𝑛

and 𝑛 = 𝑟(1 + 𝑛 ) (−𝑛 ) = −𝑛 𝑟 − 𝑟𝑛 2

dfr n r n

dn n 0

n 0

df0

dn

This is the linearized system near 𝒏∗ = 𝟏 . Note that 𝒏∗ = 𝟏 is linearly stable. We can make the connection between linear stability (i.e. stability of equilibrium for the linearized system) and nonlinear stability if (only if)

Example: Logistic Equation: 𝑛 = 𝑟 𝑛(1 − 𝑛)

(Hartman-Grobman theorem)

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𝑥 = 𝑓(𝑥) ;  𝑓(𝑥∗) = 0 ⇒ 𝑥∗ is an equilibrium

There are a few ways to linearize the system.

(i):

* *

*

dfx f(x ) (x x )

dx x x

* * *

*

dfx x f(x ) (x x )

dx x x

*x x x. Then

*

dfx x

dx x x

Closing Remarks on Linearization

(Taylor series expansion)

Let

linearized system around an

Equilibrium

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(ii): Let 𝑥 = 𝑥∗ + 𝑥 .      𝑇ℎ𝑒𝑛

* * ˆx x f(x x) f(x)

ˆdfˆx f(0) xdx x 0

ˆdfor x x

dx x 0

*

ˆdfx x

dx x

*

ˆdfx x(0)exp( t)

dx xGeneral solution of is

eigenvalue

if eigenvalue < 0, x=x* is asymptotically stable if eigenvalue > 0, x=x* is unstable

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If eigenvalue is ≠ 0, the equilibrium is called

“hyperbolic”. Otherwise, it is called “non-hyperbolic”.

According to the Hartman-Grobman theorem, if x* is a

hyperbolic equilibrium, stability conclusions drawn from

linearized equation (linear stability) ↔ hold also for the

nonlinear model (nonlinear stability)

if , then we have to look at higher order

terms in the Taylor series to judge stability.

,*

df0

dx x

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Interesting dynamics can occur as system (or control)

parameters vary: Equilibria can suddenly change in

number or stability type.

Ex: Consider the example of a cantilever beam with a

mass on top, with the mass being a control

parameter:

For mg < Pcr (1 equilibrium) For mg > Pcr (3 equilibriums)

mg

g g

Bifurcations of equilibria in 1-D

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A prototypical example:

• 𝒙 = 𝒓 + 𝒙𝟐 here r is some control parameter

The velocity functions for three distinct cases are as

follows:

x

x

r

r < 0

two equil

x

x

r = 0

one equil

x

x

r > 0

no equil

Saddle-node bifurcation (fold, or turning point, blue sky bifurcation)

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We can present these results in a diagram of

equilibrium solutions x* as a function of the parameter r.

This is a bifurcation diagram. (r = 0, x* = 0) is the

bifurcation point. This is called a subcritical saddle -

node bifurcation.

X*

r

stable

unstable

r =0

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Linear Stability Analysis

r > 0 : the equilibrium points are

2

Consider the system:

x r x f(x)

*x r

*

*

*

The function derivative is

x r , asympt staw be get ( le)

df

dx x x

dfFor 2 r

dx x x

X*

r

stable

unstable

r =0

Supercritical saddle node bifurcation:

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Note that both equilibria are hyperbolic

At r = 0, however, i.e., Hartman-Grobman

theorem fails!

Consider the velocity function at r = 0:

The equilibrium at x*=0

is actually unstable!

**

dfFor x r, we get 2 r

dx x(unst

xable)

df0

dx x 0

x x

df0

dx

x

x

r = 0

one equilibrium

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xConsider a system governed by : x r x e f(x)

x* **

df1 e 0 x 0

dx x 0 *r 1

r < r* r = r* r > r*

r

(r-x)

e-x

r

(r-x)

e-x

r

(r-x)

e-x 1 1 1

x x x

Another example

We can determine the equilibria and find their stability via

linearization:

What is the critical value of r? At critical value, x* and r*

must satisfy f(x*) = 0 → r* - x* - e-x* = 0 as well as

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In a sense, f(x) = r ± x2 are prototypical of all 1-D systems

undergoing a saddle-node bifurcation.

Consider the system just studied: * *xx f (x,r) with r 1, x 0.r x e

* ,x x *r r

*r r r 1 r and x x x x

2 2x 1 r x (1 x x / 2 ) r x / 2

x

f(x,r) increasing r for r=r*

Brief Introduction to Normal Forms

Near the critical point,

for small and write

higher order

terms

same form as that of super-critical saddle node bifurcation

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f(x) = a + bx2 is the “normal form” of saddle - node bifurcation,

i.e., all systems in 1-D undergoing this bifurcation must locally

possess this form.

Transcritical Bifurcation

The normal form for this bifurcation is 𝒙 = 𝒓 𝒙 − 𝒙𝟐

(similar to 𝒏 = 𝒓 𝒏 (𝟏 − 𝒏), the logistic equation).

Consider the velocity function for different parameter

values: x

x

r < 0

two equil

x

x

r = 0

one equil

x

x

r > 0

two equil

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We can display the results in the form of a bifurcation

diagram:

Example: Lasers. See notes.

x*

r

x*=r

x*=0 r=0

This is called a

transcritical bifurcation

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Active

material

pump

laser light partially reflecting

mirror

Example: Laser threshold

At low energy levels each atom oscillates acting as a little

antenna, but all atoms oscillate independently and emit

randomly phased photons. At a threshold pumping level, all

the atoms oscillate in phase producing laser! This is due to

self-organization out of cooperative interaction of atoms.

(Ref: Haken 1983, Strogatz’s book)

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Let n(t) - no. of photons

Then, 𝑛 = gain – loss (escape or leakage thru endface)

= 𝐺 𝑛 𝑁 − 𝑘 𝑛

gain coeff > 0 no. of excited atoms

Note that k > 0, a rate constant

Here 𝝉 =𝟏

𝒌  = typical life time of a photon in the laser

Note however that 𝑵(𝒕) = 𝑵𝒐 − 𝜶 𝒏

(because atoms after radiation of a photon,

are not in an excited state), i.e.,

2oo n (GN k)n Gnn Gn(N n) kn or

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The corresponding bifurcation diagram is:

n*

N0

x*=r

n*=0 N0=k/G

lamp laser

No physical meaning

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Examples:

We have already seen the example of buckling of a column

as a function of the axial load:

Another example is that of

the onset of convection in a

toroidal thermosyphan

mg g g

fluid

heating coil

Pitchfork bifurcation

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The normal form for pitchfork bifurcation is: 𝒙 = 𝒓 𝒙 − 𝒙𝟑 = 𝒇(𝒙

The behavior can be understood in terms of the velocity

functions as follows:

x

f(x)

r < 0

x

f(x)

r = 0

x

f(x)

r > 0 X*

r

stable

stable

r =0

unstable stable

Supercritical pitchfork

The bifurcation diagram is then:

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Consider 𝑥 = 𝑟 𝑥 − 𝑥3 = 𝑓(𝑥)

is stable when r < 0

is 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.

*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

Linear stability analysis

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The resulting bifurcation diagram is:

3

*

The normal form is x r x x ,

with equilibria x 0, r

X*

r

unstable

unstable

r =0

unstable

stable

Subcritical pitchfork

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Usually, the unstable behavior is stabilized by higher

order non-linear terms, e.g., 𝒙 = 𝒓 𝒙 + 𝒙𝟑 − 𝒙𝟓

The resulting bifurcation diagram can be shown to be:

X*

r

unstable

unstable

r =0

unstable stable

subcritical pitchfork, r = rP

supercritical saddle-node, r = rS

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Let 𝒇(𝒙𝟎, 𝒓𝟎) = 𝟎    𝒊. 𝒆. ,  (𝒙𝟎, 𝒓𝟎) be an equilibrium.

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

open region in the (x, r) plane containing (𝑥0, 𝑟0).

Then if in a small neighborhood of (𝑥0, 𝑟0),

we must have:

 𝑓(𝑥, 𝑟) = 0 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

Consider the system x f(x,r)

0 0

df0

(x ,r )dx

0 0

df0

(x ,r )dx

Connection between simple bifurcations and

the implicit function theorem

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The figure below illustrates the idea through two points

along a solution curve.

At (x1,r1), the derivative df/dx does not vanish, where as

at (x2,r2), the derivative df/dx vanishes.

(x1,r1)

x

r

df/dx0

df/dx=0

(x2,r2)

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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

mg g

Imperfect Bifurcations & Catastrophes

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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 equilibrium

possible for any h

3y r x x

cy h, h h (r)

ch h (r)ch h (r)

3 equilibria

in this region

ch h (r)

r > 0 , one or three

equilibria possible

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Imperfect bifurcations and catastrophes

(cont’d). Consider 𝒙 = 𝒉 + 𝒓 𝒙 − 𝒙𝟑 = 𝒇(𝒙

(h = 0 → normal form of pitchfork) 𝒚(𝒙) = 𝒓 𝒙 − 𝒙𝟑 𝒘𝒊𝒕𝒉   − 𝒉

We look for intersections of

3y r x x

y h

r 0 , only one equilibrium

possible for any h

h in

cre

as

ing

3y r x x cy h, h h (r)

ch h (r)

ch h (r)

r > 0 , one or three

equilibria possible

Cr / 3

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For critical point

Furthermore,

2 cc c c

rdy0 r 3x 0 x

dx 3

3 c cc c c c c

2r rh r x x 0 h

3 3

only one

equil soln 3 equil

solns

2 equil

solns

cusp

r

h

r=0

h=0

1

2

3

4

rC

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1.

2.

3.

r

x*

0 rC

r

x*

0

rC

r

x*

0

rC

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4.

h

x*

0 h=0

r

r=0

h x*

h=0 “catastrophic

surface”

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

follows:

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The equation of motion is:

Let 𝒎𝑳𝟐𝜽 be negligible (imagine pendulum in a vat of molasses)

2mL b mgLsin

The resulting equation is b mgLsin

or

bsin

mgL mgL

g l

θ

m

O

Re

e

1-D system on a circle:

[over-damped pendulum (acted by a constant torque )]

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(ratio of appl. torque to max.

gravitational torque)

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

Consider the system:

[0, 2 ]

dThen sin where

d

mgLLet t ;

b mgL

sin

θ

θ =0 θ =

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if 𝛾 > 1, pendulum goes around the circle albeit non-uniformly

If 𝛾 = 1,    𝜃∗ = 𝜋 2       𝑖𝑠 𝑎𝑛 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

1,

θ =0 θ =

θ =/2

*1 2* and

If 1, there are :

whi

two equilibria

opposite stability characteri

ch

s

have

tics

θ =0 θ =

θ 1* θ 2

*

Clearly, there is a saddle-node bifurcation at 𝛾 = 1.