dynamical systems analysis
DESCRIPTION
LECTURE 5. Dynamical Systems Analysis. Why?. A dynamical system (e.g. a neuron or a neural system) is usually described by a set of nonlinear differential equations What is ‘analysis’ here? To determine how the system behaves over time given any current - PowerPoint PPT PresentationTRANSCRIPT
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Dynamical Systems Analysis
LECTURE 5
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Why?
A dynamical system (e.g. a neuron or a neural system) is usually described by a set of nonlinear differential equations
What is ‘analysis’ here?
To determine how the system behaves over time given any current state (the future or long-term behavior ) before solving it numerically
For instance, are there equilibrium states (physics) or fixed
points (mathematics)? Are these states stable or unstable?
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Equilibria of a dynamical system
Coin balanced on a table - How many equilibria? (Face Up, Face Down, Edge) - Is it stable?
A ball in the track:
It’s either on top of a hill or at the bottom of the track. To find out which, push it (perturb it), and see if it comes back.
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Fixed points of First-order autonomous systems
0) ,( xfdt
dx
) , ,( txfxdt
dx
By a fixed point we mean that x doesn’t change as time increases, i.e.:
So, to find fixed points just solve above equation.
) ,( xfdt
dx
autonomous systemsnonautonomous systems
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E.g.: dx/dt = Ax(1 – x) with A=6. What are the fixed points?
Set: dx/dt = 0 ie: 6x(1-x) = 0so either x = 0 or x = 1
Use difference equation: x(t+h) = x(t) +h dx/dt from various different initial values of x
Therefore 2 fixed points, and how about the stability?
Perturb the points and see what happens under the system
dynamics…
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Vector fields
0 0.5 1 1.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
t
x
(t, x(t))
(t+h, x(t)+hdx/dt)
So x=0 is unstable and x=1 seems to be stable. Change the value of parameter A to see the influence of parameters on systems
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Phase flow in 1-D phase space
)1( xAxxdt
dx
hdx/dt
1-D phase space
dt
dx
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First-order autonomous systems
0
0) ,(
dx
xdf k
) ,( xfdt
dx
1. Find fixed points:
2. Identify stability - using phase flow as above - using the gradient at the fixed points
xk is stable
xk is unstable
0),( kxf
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Exercise 1:
adx
axd
)(
1. Fixed points: x=0
2. Stable or not?
If a<0, d(ax)/dx<0. stable
axdt
dx
If a>0, d(ax)/dx>0. unstable
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Exercise 2: 3xxdt
dx
13)( 2
3
xdx
xxd
1. Fixed points: x=0, +1, -1 2. Stable or not?
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Second-order autonomous systems
),(
),(
yxgydt
dy
yxfxdt
dx
Suppose we have the following system:
1. Find the fixed points by setting:
0),(
0),(
kk
kk
yxg
yxf
2. Identify stability by Jacobian matrix (will not talk here) or 2-D phase space portrait
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Phase space portrait
The 2-D space of possible initial conditions in which each solution follows a trajectory given by the vector field
(x(t+h),y(t+h)) =(x(t)+hdx/dt, y(t)+hdy/dt)
(x(t),y(t))
x
y
),(
),(
yxgydt
dy
yxfxdt
dx
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Example 1: A damped simple pendulum
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A damped simple pendulum
)0(
0)sin(2
2
adt
da
dt
d
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Second-order autonomous systems
First-order autonomous system in two variables
)sin(xaydt
dy
ydt
dx
dt
dyx
,
Find the fixed points: (0, 0), (±π, 0), ……
)0(
0)sin(2
2
adt
da
dt
d
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Phase space portrait of the damped simple pendulum
)sin(4.0 xydt
dy
ydt
dx
(Created by AI Lehnen)
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Example 2: the Van der Pol oscillator
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Van der Pol oscillator
0)1( 22
2
xdt
dxx
td
xd
The second term is not a constant like that in the damped simple pendulum with )0( ,0)sin( aa
Second-order autonomous systems
first-order autonomous system in two variables
xyxdt
dy
ydt
dx
)1( 2
dt
dxy
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Find the fixed points: (0, 0). Stable or not?
xyxdt
dy
ydt
dx
)1( 2
Phase space portrait with λ = 1.
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High order autonomous systems
) ..., ,,(
......
) ..., ,,(
) ..., ,,(
21
21222
21111
nnnn
n
n
xxxfxdt
dx
xxxfxdt
dx
xxxfxdt
dx
Suppose we have the following system (imagine a neural network …):
We basically need a n-dimensional phase space.
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• Find fixed points of a simple system
• Classify fixed points as stable/unstable. Especially, use graphical methods (vector field plots in phase space) to analyze and elucidate the behaviour of simple systems
Homework