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
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?
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.
Fixed points of First-order autonomous
systems
dx
 x  f ( x ,  , t )
dt
nonautonomous systems
dx
 f ( x,  )
dt
autonomous systems
By a fixed point we mean that x doesn’t change as time
increases, i.e.:
dx
 f ( x,  )  0
dt
So, to find fixed points just solve above equation.
E.g.: dx/dt = Ax(1 – x) with A=6. What are the fixed
points?
Set: dx/dt = 0 ie:
so either x = 0 or x = 1
6x(1-x) = 0
Therefore 2 fixed points, and how about the stability?
Perturb the points and see what happens under the system
dynamics…
Use difference equation: x(t+h) = x(t) +h dx/dt from various
different initial values of x
Vector fields
1
0.9
(t+h, x(t)+hdx/dt)
0.8
0.7
x
0.6
0.5
(t, x(t))
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
t
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
Phase flow in 1-D phase space
dx
dt
1-D phase space
dx
 x  Ax (1  x )
dt
hdx/dt
First-order autonomous systems
dx
 f ( x,  )
dt
1. Find fixed points:
f ( xk ,  )  0
2. Identify stability
－using phase flow as above
－using the gradient at the fixed points
df ( x k ,  )
dx
 0

 0
xk is stable
xk is unstable
Exercise 1:
dx
 ax
dt
1. Fixed points: x=0
2. Stable or not?
d ( ax )
a
dx
If a<0, d(ax)/dx<0. stable
If a>0, d(ax)/dx>0. unstable
Exercise 2:
dx
 x  x
dt
1. Fixed points: x=0, +1, -1
2. Stable or not?
d ( x  x )
3
dx
 3x  1
2
3
Second-order autonomous systems
dx
Suppose we have the following system:
 x  f ( x , y )
dt
dy
 y  g ( x , y )
dt
1. Find the fixed points by setting:
f ( xk , yk )  0
g ( xk , yk )  0
2. Identify stability by Jacobian matrix (will not talk here) or
2-D phase space portrait
dx
Phase space portrait
 x  f ( x , y )
dt
dy
 y  g ( x , y )
dt
The 2-D space of possible initial conditions in which each
solution follows a trajectory given by the vector field
y
(x(t+h),y(t+h)) =(x(t)+hdx/dt, y(t)+hdy/dt)
(x(t),y(t))
x
Example 1: A damped simple pendulum
A damped simple pendulum
d 
2
dt
2
a
(a  0)
d
dt
 sin(  )  0
d 
2
dt
2
a
d
 sin(  )  0
dt
(a  0)
Second-order autonomous systems
x , y 
d
dt
First-order autonomous system in two variables
dx
 y
dt
dy
  ay  sin( x )
dt
Find the fixed points: (0, 0), (±π, 0), ……
Phase space portrait of the damped simple
pendulum
dx
 y
dt
dy
  0 . 4 y  sin( x )
dt
(Created by AI Lehnen)
Example 2: the Van der Pol oscillator
Van der Pol oscillator
2
d x
2
  ( x  1)
2
d t
dx
x0
dt
The second term is not a constant like that in the damped simple
pendulum with   a   sin(  )  0 , ( a  0 )
Second-order autonomous systems
y
dx
dt
first-order autonomous system in two variables
dx
 y
dt
dy
dt
   ( x  1) y  x
2
dx
 y
dt
dy
   ( x  1) y  x
2
dt
Find the fixed points: (0, 0). Stable or not?
Phase space portrait
with λ = 1.
The Limit Cycle
High order autonomous systems
Suppose we have the following system (imagine a neural
network …):
dx 1
dt
dx 2
dt
 x1  f 1 ( x1 , x 2 , ..., x n )
 x 2  f 2 ( x1 , x 2 , ..., x n )
......
dx n
dt
 x n  f n ( x1 , x 2 , ..., x n )
We basically need a n-dimensional phase space.
Homework
• 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