In this lecture, I will talk about dynamical
system analysis. We can solve equations for

particular starting point(s), but this is
often not enough to enable us to understand

the system. Therefore, we use complementary
analysis focused on finding equilibrium states

(or stationary/critical/fixed points) where
system remains unchanged over time. Analysis

determines how the system behaves over time,
in particular investigating future behaviour

of the system given any current state ie the
long-term behaviour of the system. Also we

try to classify these states/points as stable/unstable
by investigating the behaviour of the system

near them.

We learn how to find fixed points of a system
and classify them as stable/unstable for 1

(space) dimensional systems and 2 (and higher)
dimensional systems but Will NOT talk about

proofs of many of the stability theorems , many
of special cases focus will be on how to use

them to analyse your systems . These lectures
are a primer for understanding the power and

advantages of using algorithmic complexity
based tools for analysis dynamical systems.

So lets start with something you can test
right now, try to balance a coin on a table!

In how many ways you can do that, I cannot
do it in more than 2 ways but there is 3d

one, Head, tails and edge. So let assume that
you are one of the expert ones and could balance

a coin at its edge, is it stable?

… No! Small movement away (perturbation)
means that coin will end up in one of 2 different

equilibria: Heads/ Tails.

How about head and tail? Yes they are both
stable as perturbation does not result in

a change in state.

Similarly, think of a ball at rest in a dark
landscape. It’s either on top of a hill

or at the bottom of a valley. To find out
which, push it (perturb it), and see if it

comes back. (What about flat bits?)

So a set towards which a dynamical system
evolves over time called attractor . It can

be a point, a curve or more complicated structure.
A perturbation is a small change in a physical

system, most often in a physical system at
equilibrium that is disturbed from the outside.

Each attractor has a basin of attraction,
which contains all the initial conditions

that will generate trajectories joining asymptotically
this attractor.

When studying a nonlinear dynamical system,
if we are only interested in the long-time

behaviours, we will only study the attractors
of the system and determine their basin of

attractions. one of the “simplest” attractors
are the point: it is then a fixed point, i.e.

The particular points of the phase space verifying
dX/dt(x∗)= 0. The corresponding solution

of the dynamical system does not depend on
time. It is a stationary state.

So what are the fixed points of this system
dx/dt = 6x(1 – x) ?

If we solve 6x(1-x) = 0 we 
can find 2 fixed points:

Are they stable?

To test that we perturb the points and see
what happens but to perturb them we must have

a model of the system.

We can use difference equation: x(t+h) = x(t)
+h dx/dt from various different initial x’s.

If we perturb each fixed point a bit and see
if the output of system will change or not

we can see that X=0 unstable and x=1 stable.

There are a general rules regarding the stability
of one-dimensional dynamical system:

If a0 is a fixed point:

If f’(a0) > 0, a0 is unstable

if f’(a0) < 0, a0 is stable

if|f’(a0) = 0, inconclusive.

where f’(a0) means the derivative of f,
df/dx evaluated at a0. if f’(a0) = 0 need

to use higher derivatives or other methods
and can be semi-stable from above or below

or periodic …

so by now we know a fixed point is a special
point of the dynamical system which does not

change in time. It is also called an equilibrium,
steady-state, or singular point of the system.

If a system is defined by an equation dx/dt
= f(x), then the fixed point 𝑥 ̃ can

be found by examining of condition f(𝑥 ̃)=0.
We do not need to know analytic solution of

x(t).

A stable fixed point: for all starting values
x0 near the 𝑥 ̃ the system converges

to the 𝑥 ̃ as t→∞.

A marginally stable fixed point: for all starting
values x0 near 𝑥 ̃, the system stays

near 𝑥 ̃ but does not converge to 𝑥 ̃.
An unstable fixed point: for starting values

x0 very near 𝑥 ̃ the system moves far
away from 𝑥 ̃.

Now let’s finish the lecture by looking
at another biological model , bacteria Growth

Model.

We leave a nutritive solution and some bacteria
in a dish. Let b be relative rate at which

the bacteria reproduce and p be relative rate
at which they die. Then the population is

growing at the rate r = b−p. If there are
x bacteria in the dish then the rate at which

the number of bacteria is increasing is (b
− p)x, that is, dx/dt = rx. Solution of

this equation for x(0)=x0 is

However, is this mode realistic?

No, The model is not realistic, because bacteria
population goes to the infinity for r>0. But

in reality as the number of bacteria rises,
they produce more toxic products. So instead

of constant relative perish rate p, we will
assume relative perish rate dependent on their

number px. So the number of bacteria increases
by bx and their number decreases by px2 and

new differential equation will be:

How many-fixed point has this model?

To be able to find a fixed point, we have
to set the right-hand side of the differential

equation to zero. There are two possible solutions,
we have two fixed points:

Lets see what they mean and check their stability.

First fixed point means, there are no bacteria,
none can be born, none can die. However, after

small contamination (perturbation) which is
smaller than b/p, the number of bacteria will

increase by dx/dt = bx-px2>0 and will never
return to the zero state. First fixed point

is unstable.

And Second point
x= b/p, at this population level, bacteria

are being born at a rate b2/p and are dying
at the same rate, so birth and death rates

are exactly in balance.

If the number of bacteria would be slightly
increased, then dx/dt = bx-px2<0 and would

return to equilibrium. If the number of bacteria
would be slightly decreased, then dx/dt = bx-px2>0

and would return to equilibrium. Small perturbations
away from

x= b/p will self-correct back to b/p. Therefore,
second fixed point is stable. How does the

graphic solution in Mathematica look like?