Equilibria or fixed points : initial conditions n ... · Phase diagram The length of the arrows ......
Transcript of Equilibria or fixed points : initial conditions n ... · Phase diagram The length of the arrows ......
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Geometrical Analysis of 1-D Dynamical Systems
Logistic equation:
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:
Phase diagram
The length of the arrows magnitude of the velocity
(function) at that point.
velocity function
n rn(1 n)
n*=0 n*=1 n
f(n)
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limit set of a point (initial condition) :
It is defined as the set of limit points of the trajectory
started at , for t → - . Thus,
limit set of a point is the set
Existence of a potential function: Consider
(Gradient Dynamical System)
Let there be a function V(n)
Example: for the Logistic equation
i.e.,
0 0t
(n ) n lim (n ,t) n
n f(n)
such that f(n) V / n
0n
0n
0 0t
(n ) n lim (n ,t) n
f(n) rn(1 n) 2 3V(n) rn / 2 rn /3
0n
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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:
the plot of the potential function, and the equilibrium
points are as follows:
2 3rn rnV(n)
2 3
V(n)
nn*=0
n*=1
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Observations: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 in phase
space where if then for all negative and
positive flow times (- < t < ). For the Logistic
Equation, the invariant subspaces are:
Thus, the state space is decomposed into:
limit sets of any initial condition
0n I, 0(t,n ) I
1 2 3I { ,0} , I {0} , },I {0,1 4 5I {1} , I {1, }
1 2 3 4 5I I I I I and 0n :
I
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If limit set is 0
limit set is
If limit sets are the same
If
Stability of Equilibria/Fixed Points
An equilibrium point of say x=x*, is stable if
for any initial condition x0,
with
Otherwise, it is unstable.
0n I, the
0 2n I , then and
0 3 0n I , then (n ) {0}, and
0(n ) {1} so on.
x f(x),
0, ( ) 0 such that * *
0 0(x x ) , (t,x ) x for all t, 0 t .
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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, is an
asymptotically stable equilibrium.
* *0(t,x ) x 0 as t , then x
x*
δx*x0
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Examples:
1. Logistic Equation
2. Quadratic System
3. Cubic System
Observe: Any isolated stable equilibrium in 1-D
autonomous systems has to be asymptotically stable.
n*=0
(unstable)n*=1
(asymptotically
stable)
n
f(n)=rn(1-n)
n*=0
(unstable)
n
f(n)=an2
n*=0
(stable)
n
f(n) = -an3
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Linearization about equilibrium points
Let x* be a fixed point of , i.e.
To linearize about x = x*, introduce a perturbation:
(Taylor series expansion for small )
This is the linearized equation about x = x*
*Let x x x
x f(x)
* *Then x x f(x x )
x x0 0
dfor x x f(x ) x
dx
x
x 0
dfx x
dx
*f(x ) 0
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Example: Logistic Equation
The equilibrium points are
Let us linearize the system about
This is the linearized system near .
Note that 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)
(Hartman-Grobman theorem)
n 1 and n 0.
n rn(1 n)
dfr n r n
dn n 0
2and n r(1 n) ( n) nr rn
*Then, n n n 1 n
n 1
n 1 n 1
n 0
df0
dn
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Closing Remarks on Linearization
is an equilibrium
There are a few ways to linearize the system.
(i):
(Taylor series expansion)
Let
linearized system around an
Equilibrium
* *x f(x) ; f(x ) 0 x
* *
*
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
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(ii): Let
General solution of is
eigenvalue
if eigenvalue < 0, x=x* is asymptotically stable
if eigenvalue > 0, x=x* is unstable
* * ˆx x f(x x) f(x)
*x x x. Then
ˆdfˆx f(0) x
dx x 0
ˆdfor x x
dx x 0
*
ˆdfx x
dx x
*
ˆdfx x(0)exp( t)
dx x
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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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Bifurcations of equilibria in 1-D
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)
mgg g
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Saddle-node bifurcation
(fold, or turning point, blue sky bifurcation)
A prototypical example:
here r is some control parameter
The velocity functions for three distinct cases are as
follows:
2x r x
x
x
r
r < 0
two equil
x
x
r = 0
one equil
x
x
r > 0
no equil
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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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supercritical saddle node bifurcation:
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
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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 equil
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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
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-x1 11
x x x
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Brief Introduction to Normal Forms
In a sense, f(x) = r x2 are prototypical of all 1-D
systems undergoing a saddle-node bifurcation.
Consider the system just studied:
Near the critical point,
for small and , write
higher
order
terms
same form as that of super-critical saddle node bifurcation
* *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 rfor r=r*
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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
(similar to , the logistic equation).
Consider the velocity function for different parameter
values:
2x r x x
n rn (1 n)
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.
2
* *2
*
*
x r x x
r x x 0
x 0
x r
x*
r
x*=r
x*=0r=0
This is called a
transcritical bifurcation
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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 (Haken 1983, Strogatz’s book)
Active
material
pump
laser lightpartially reflecting
mirror
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Let n(t) - no. of photons
Then, gain - loss
(escape or leakage thru endface)
gain coeff > 0 no. of excited atoms
Note 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
oN(t) N n
1
k
2oo n (GN k)n Gnn Gn(N n) kn or
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The corresponding bifurcation diagram is:
No physical meaning
n*
N0
x*=r
n*=0N0=k/G
lamp laser
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Pitchfork bifurcation
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
mgg g
fluid
heating coil
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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 =0
unstablestable
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Linear stability analysis
Consider
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.
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
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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 =0
unstablestable
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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 =0
unstablestable
subcritical pitchfork, r = rP
supercritical saddle-node, r = rS
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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 )dx0 0(x ,r ),
0 0
df0
(x ,r )dx
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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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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
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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
(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 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 soln3 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.
Alternately, in 3-D we can visualize the solutions set as
follows:
h
x*
0h=0
r
r=0
hx*
h=0
“catastrophic
surface”
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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)
2mL b mgLsin
2mL
The resulting equation is b mgLsin
orb
sinmgL mgL
gl
θ
m
O
Re
e
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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 =0θ =
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if pendulum goes around the circle albeit
non-uniformly
If
If
Clearly, there is a saddle-node bifrucation at
1,
*1, / 2 is an equilibrium
1,
θ =0θ =
θ =/2
*1 2* and
1, there are :
whic
two equilibria
opposite stability characteris
h have
ticsθ =0θ =
θ 1*θ 2
*
1.