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?