WEBVTT
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So, today we will study stability of bioreactors
which stressed up on it a little bit in the
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last class, but we will start of fresh today.
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So, if you go to this screen so, just hope
such process is called stability analysis
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of bioreactors by which we test the stability
of the bioreactors. Now, the method that we
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used is called the Liapunov method of stability
analysis of chemical reactors. It was started
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in 1955 by Bilous and Amundson, and the way
to do it as I said, is to start to be able
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to write the unsteady state equation given
over here del C del t equals f unsteady state
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equation of the system.
Now, then first thing we do is actually find
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the steady state of the that, we will see
here we find the steady state f C s s , P
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equals 0. So, that is the steady state so,
why do we find the steady state because what
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we want to find is that, we want to give some
small perturbations into the system and small
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perturbations are always available in the
system because in a real system, you all know
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that we discussed and we want to see the how
is the system behaves with respect to this
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small perturbation.
It has the do the perturbation increase with
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time do the decrease with time or die wave
if they decrease with time and die die wave
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then this system is stable, if they stay with
time and increase over time then the system
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is unstable. Now, what we have to do is we
have to write everything in a vectorial form
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over here as the we have written. So, that
to retain the generality of the process or
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the analysis and we write it as d d t of C
vector equals f vector which is a function
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of C vector mouse.
You can see C vector and P vector. So, C vector
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for the chemostat as we have discussed in
the previous classes consist of two variables.
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One is the substrate S and the other one is
the cell X and the P vector is a vector of
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parameter which consist of the following elements
D, the delusion rate Y the yield ratio of
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X over S, k s the Michaelis kinetic constant
mu max is a maximum growth rate and S 0, is
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a initial concentration of the substrate.So,
these are the variable that are concerned
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that P consist of now, x is define as a deviation
from this steady state. Why?
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Do we define deviation from the steady state
is because we want to study how this deviation
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would change? As you keep you know as you
change the steady state, as you go away from
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the steady state. If this deviation is going
to increase then it then with time then, if
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the system is unstable if this deviation x
of t is going to decrease with time then the
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system is stable. So, that is the basic definition
that is what we are aiming towards that all
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points of time. So, x of t why is that x of
t because x is the deviation of the function
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of time and you want to study the dynamics
of it as a function of time. So, that is written
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as C vector as a function of time minus this
steady state.
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So, what we did we realize in the last class
is that we have to do a Taylor series expansion.
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So, C t is written as C s s plus x. So, C
s s we have to do a Taylor series expansion
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of C t about the steady state C s s and so,
what we do is we take the function f which
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is the function of the steady state of C t.
Now, C t is written as the steady state plus
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the deviation and a function of the parameters.
Now, if we are to do a scalar, Taylor series
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expansion for a certain scalar g this is the
way we write it g C s about the steady state
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let us say, g C s C s s plus x is g C s that
is zeroth order term, then you have the first
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order term g del C evaluated at C s s. I reminded
you the other day that this is not g del g
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del C s s, but this is del g del C evaluated
at C s s and then del two g del C square evaluated
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as C s s time the x square over to vectorial
plus the higher order terms, which be del
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three g del C cubed x cubed over three vectorial
and so on.
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Now, for a linear stability analysis first
question is that why do we do a linear stability
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analysis? We do the linear stability analysis
because we at this point are not interested
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to find how the system behaves, far away from
the equilibrium linear stability analysis
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is valid. How valid for systems close to the
equilibrium and we are trying to find out,
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how the system is going to behave close to
the equilibrium on the application of small
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perturbation or noise. The reason we are not
interested in how the system, behaves far
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away from the equilibrium using a Taylor series
expansion is because far away from the equilibrium
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this Taylor series expansion is not valid.
Why is that?
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Hm
What ?
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Is that is a question is Taylor series expansion
valid for all values effects no.
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What you think Liza is it valid for all values
of x, what kind of values of x is it valid
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for what do you think is it valid for all
values of x ?
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Yes, it is Taylor series expansion is valid
for all values of x then there is no constraint,
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on x for Taylor series expansion to evaluate,
but if you want to truncate it. Then and then
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only it valid for small values of x, do you
see why because if x is large for example,
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then as you go to x square, x cubed, x to
the power 4 these terms are dominate. So,
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leaving that those terms out would be very
stupid thing to do.
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So, x you know if you see look at that series
over here, if x is much smaller than one then,
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what happens this term for example, x square
over vectorial two is very small the next
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term x cubed over vectorial three is even
small, x to the x to the power four over vectorial
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four is even smaller. So, you can leave all
those terms out.
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So, that is a point so, if you want Taylor
series if you want to truncate, the Taylor
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series at the first order and the second order
then x has to be small, but otherwise this
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is like a mantra you need to remember because
most of the students do not get it, they all
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say that Taylor series is valid for small
values of x, which is absolutely wrong Taylor
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series is valid for all values of x provided,
you take into account up to infinite term.
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A very large number of terms say for example,
if you have large value of x you know reasonable
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large value of say almost close to one and
0.9999 and you took into account 200 terms.
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Then Taylor series is fine it is a pretty
good approximation fine. So, here we are concerned
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so. What is our concerned? Our concerned is
small values of x, why is that because it
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is a linear stability analysis, we are doing
a, b. We are trying to figure out that how
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the system behaves in the vicinity of the
steady state, we are trying to figure out
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how the system behaves in the vicinity of
the steady state.
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If you are trying to figure out how the system
behaves in the vicinity of the steady state.
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What is the deviation from the steady state
x therefore, x has to be much smaller than,
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what x has to be much smaller than what, just
by saying that its small does not make any
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sense whenever, you say something is small
you have to compared it with something else.
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So, x has to be much smaller than.
(C s s)
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C s s so, that is my criteria if I want to
truncate it over the first after the first,
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order term that is right here. So, g g C s
plus del g del C c a valid at s s times x
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then, x has to be much smaller than one much
smaller than C s s. So, which is criteria
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that will take and we want to do a linear
stability analysis and I am trying to impress
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upon you the fact why we are doing a linear
stability analysis is because we want to qualitatively
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understand which direction, the system is
going to go see a linear stability analysis
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will only give you a good qualitative understanding
near the steady state.
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So, we want to say we want to figure out that
this is a steady state, I have a steady state
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over here I perturb the system slightly. So,
how is the system going to behave when I perturb
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it, just after I give the perturbation not
too far in time. I mean in the vicinity of
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the steady state means, not too far away in
time is small time from the time you gave
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the perturbation. How the system is going
to behave because morning shows a day right,
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you know if you give for a certain perturbation,
you see the system behaves in a certain way
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you kind of sort of figure out from that,
how the system is going to behave on the long
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term and we will look at that you know in
great detail.
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So, this is the scalar now what would be the
vector form of it. So, we want to for the
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system for the reactor are the chemostat,
you saw that it is a vectorial system right
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need to consist of two many parameters, parameters
and two variable the substrate and the and
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the excuse me and the substrate and the cells.
Now, what is the problem in this system ? Our
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major problem is that you cannot, it is a
coupled system over you cannot decouple, the
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system remember if you could decouple then
you can treat them as two scalar equation,
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which you cannot right because look at the
substrate equation go back and look at the
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substrate equation I will show you, if you
need to it is a here.
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This equation for example, it the both the
substrate equation, and the equation for the
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cells; so look at the equation for the cell
it has X over here and substrate concern through
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mu right similarly, the substrate equation
S over here and the cell comes in through
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X. So, both these equations are coupled and
there is no way to decouple then, if you decouple
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these equations then you can solve individually
for these equations using scalar expansion,
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but there is no way to decouple them. So,
what we need to do is kind of divide this
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strategy of how to do the Taylor series expansion
for a vectorial system. I am not sure if you
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have done that, but even if you haven’t
we will go through it and if there are questions
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you can stop me, if you do not understand
anything.
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So, this is a Taylor series expansion what
you see on the screen is a Taylor series expansion
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for a vectorial system; so f here being vector
f C s s plus x and A being matrix and x again
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being a vector; so plus h o t higher order
term which we are ignoring fine. Now, let
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us first look at the look at the orders of
these. So, f is a, n cross one vector both
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f a is a n cross n and x is a n cross one
vector of variables.
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So, as a result this is of course, n cross
one so, everything is matched out both in
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the left and right hand side. The n cross
one thing is kind of matched out right dimensionally
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consistent fine. Now, what does A what is
A composed out for to intuitive I will show
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you in a minute, but I want you to into it.
(Jacobian)
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Jacobian another function why is that because
if you go back here, look at the stuff. So,
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this is for a scalar its del g del C.
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So, you can you tell me what the see if f
consist of let us, let me write it here if
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f consist of f equals f 1, f 2 do you do you
know how to write it this way. So, if you
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are writing a vector and you do not want to
waste your space in the vertical direction,
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this is how you write f 1, f 2 transpose so
and x is x 1, x 2 transpose and my A is going
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to be what this is very simple case.
Del f 1 del X 1
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Del f 1 del X 2
Del f 2 del X 1
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Del f 2 del X 2
So, this my A matrix and similarly, for n
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cross one n cross n matrix you can write it.
So, A is n cross n or in other words f is
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say f 1 to f n transpose and x is x 1 to x
n transpose. So, this would be del f 1 del
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x 1 up to del f n del f 1 del f x del x n
and this direction, it will be del f n del
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x one and this direction it will be del f
n del x n right, fine with everybody. So,
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this is my Jacobian this is how we define
the Jacobian.
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Now, I can approximate as you can see in the
screen, I can approximate this as this. Now,
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you have to tell me what are the, what are
the assumptions involved in that approximation.
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Let us put lets go back to the screen here,
what is the what is the approximation involved
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in that, from here to here there is straight
forward I mean, it is not a rocket science
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A the higher order terms are ignored.
And then f C s s P is a steady states so that
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0. So, we just put that over here so, f C
s s P is 0 and this is ignored. So, this is
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approximately equal to approximation sign
is not for f C s s P, f C s s P is equals
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0, but for ignoring the because we ignored
the higher order term that is why we get.
13:55.290 --> 14:02.290
So, f C s s plus x , P equals A x fine and
what was if you, if you let me remind you
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what was f C s s plus x , P, that was d X
d t. so, essentially what we have is d X d
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t equals x is it right.
Now, this is what I just said, so the magnitude
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of the higher order term. If this is x is
very small then the magnitude of this is going
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to be small, you know magnitude of the higher
order terms is going to be smaller than the
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first, first order right this is what I explained
in the scalar case. So, because higher order
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X square X cube is obviously going to be larger.
So, n over here the order of the matrices
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is going to be same as the number of species
that is there in the system fine, if you have
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two species the just that you have in the
chemostat.
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So, most of chemostat equation will be a two
gross two matrix and which also equals the
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number of equation, in the system it has to
equal the number of equation because for a
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system which is less for a system, which is
has to be complete the numbers equation has
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to equal for well defined system, the number
of equations has to equal the number of species
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right is that clear up to this point.
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So, next we go to the next step which is that
if x is small then x, but this is also, something
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I discussed x bar its much, much less than
the steady state the deviation is much, much
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less than the steady state. Now, somebody
has to tell me that I discussed this just
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now, with respect to the scalar concentration.
Now, how does it fit into the vector form?
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So, let us look at the screen so, if x bar
just now I discussed that x for deviation
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for this whole thing to be valid the deviation
x has to be much smaller than, the steady
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state concentration, but that was for a scalar
for a single species. Now, what if I have
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a vector what, how do I define this it could
be that, you know concentration of one is
15:53.800 --> 16:00.800
smaller the other one is not smaller, but
how do I define this you know how do I satisfy
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this condition.
16:03.620 --> 16:10.620
Each of these are smaller no that is not the
way we do it in mathematically, that is not
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the way we do it what you are saying is X
1 is smaller much much smaller than C s s
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1, X 2 and no is its very hard to you know
it will be a lot of comparisons, involve then
16:23.910 --> 16:30.910
how. So, what is the way we do it mathematically
you know you should some of you might have
16:31.110 --> 16:38.110
taken that advanced math course. So, you should
know if you had when you have a vector and
16:40.160 --> 16:45.660
you want to compare the magnitude the vectors.
So, how do you do?
16:45.660 --> 16:52.660
What determinant vector does not have determinant.
What is determinant of vector, how do you
16:57.000 --> 17:04.000
compare magnitude of vectors you want to go
back to class twelve or what just before the.
17:06.699 --> 17:13.699
How do you differ, how do you compare magnitudes
of vectors 2, vectors I give you three you
17:16.260 --> 17:23.260
know each of them say consist of 3 variables.
How do you compare the magnitudes the 2 variables,
17:24.120 --> 17:31.120
3 variables, 5 variables whatever anything
larger than one variable, how do you compare
17:33.440 --> 17:34.899
the magnitude?
17:34.899 --> 17:41.899
Norms, yes by comparing the norms of the vectors
so that, that is the way to compare and so
17:43.590 --> 17:48.490
there are different kinds of norms, there
are there is a two norms there is infinity
17:48.490 --> 17:49.120
norms and so on.
17:49.120 --> 17:56.120
So, we have the two norms for example, here
it will be it will be x 1 square plus x 2
17:57.080 --> 18:03.740
square plus you know x 3 square. Let us say
if you have 3 square root of that then you
18:03.740 --> 18:08.269
can the infinity norm you can the absolute
norm. What is the absolute norm? May be absolute
18:08.269 --> 18:15.269
is be the absolute values of this x 1 plus.
So any of these norms the most popular is
18:17.490 --> 18:22.629
this one norm and this is, say I believe you
know how to calculate the length of a vector.
18:22.629 --> 18:26.820
So, this is the this is defined as a length
of the vector that norm and that is the most
18:26.820 --> 18:31.269
usually, norm that you use to compare to vectors
you can use other forms of norms also like
18:31.269 --> 18:36.539
the absolute norm for example, the infinite
norm and so on, but typically what you use
18:36.539 --> 18:43.539
is this is this norm, where you know just
compare the lengths of those lengths of those
18:43.850 --> 18:50.320
vectors take the square of them and sum, sum
them up and take the square root. So, if I
18:50.320 --> 18:51.879
am to going back to the screen.
18:51.879 --> 18:58.320
So, if I am to decide whether as my x is small,
I just compared the length of x as compared
18:58.320 --> 19:05.320
to the length of the other vector. So, when
I have done that and when that turns out to
19:05.610 --> 19:12.610
be small then, I can make this approximation
that d x d t equals A x clear. So, where a
19:13.320 --> 19:20.320
i j that is the elements of the matrix A are
given by just as, I showed you know f del
19:20.600 --> 19:25.029
f. So, this is a Jacobian and given by this.
So, del a i j could be written in the compact
19:25.029 --> 19:31.720
form as del f i del C j is that clear to everybody
any question, you ask me please. So, this
19:31.720 --> 19:38.720
is this is the compact form of writing a i
j. So, a i j is the elements of the matrix
19:39.470 --> 19:46.399
A. Now, what I want you to sort of you know
tell me if not completely, but at least to
19:46.399 --> 19:53.399
some extent what would be the solution of
this if I am allowed to go back to this.
19:53.940 --> 19:59.230
No, no. What is the solution of this equation?
Let us not go that part d x d t equals A x
19:59.230 --> 19:59.639
what would be the solutions?
19:59.639 --> 19:59.889
Yeah,
19:59.700 --> 20:06.700
How do you know that. So, this is my equation
del x A what is the solution ?
20:12.889 --> 20:17.389
Eigenvector into.
20:17.389 --> 20:22.309
Some constant some constant alpha I times.
20:22.309 --> 20:25.409
Exponentially mean.
20:25.409 --> 20:28.509
Hm
Minus.
20:28.509 --> 20:30.059
Minus.
20:30.059 --> 20:31.610
Yeah
20:31.610 --> 20:32.259
Into into
20:32.259 --> 20:37.519
Values into t and what sigma.
20:37.519 --> 20:44.519
Sigma of that so how do you, how did you get
this?
20:53.340 --> 20:53.799
How do you know this?
20:53.799 --> 21:00.799
See there is a easy way of doing it, which
is very straightforward to do it. Now, this
21:05.950 --> 21:12.950
is the equation and the solution of this is
if X 0, X equals X 0 at t equals 0. What is
21:14.710 --> 21:21.710
the easy way? The easy way is X equals X 0
exponential A t. Now, you have been taught
21:24.429 --> 21:31.429
how to break up A t, if A is know A is given
then any f of A it could be written as a function
21:33.049 --> 21:40.049
of the Eigenvectors, Eigenvectors and the
Eigen values. You aware of that I think its
21:40.210 --> 21:47.210
you know f of lambda f of lambda times the
Eigenvalue, Eigenvectors also coming there.
21:47.639 --> 21:52.749
And if you use that you can go back to that
other option is let us look at the screen
21:52.749 --> 21:58.190
other option is this is the solution. Other
option is to actually breakdown A x, there
21:58.190 --> 22:03.289
when you have the matrix in terms of it Eigen
values and Eigenvectors and then you can decouple
22:03.289 --> 22:08.029
the system and solve it, solve it in this
way that is like, but just coupling the system
22:08.029 --> 22:13.240
that is one way other way is to straight away
go and solve this in the matricial form and
22:13.240 --> 22:18.700
then convert the matricial form into this
form, but whatever it is this is the final
22:18.700 --> 22:22.389
solution that you have to remember this solution
and now, this is a well known solution.
22:22.389 --> 22:29.389
So, d a d d t of x equals excuse me a matrix
of times x the solution of that would be x
22:30.869 --> 22:37.869
is given as sigma times some constant alpha
i times beta i, beta i being the Eigenfunction
22:38.889 --> 22:43.759
times e to the lambda i t you said minus lambda
i t, but that is fine, but that depends only
22:43.759 --> 22:48.889
how you define you Eigenvalue. If you define
your Eigenvalue as you know lambda t and its
22:48.889 --> 22:53.659
one and if you define it as minus lambda t
that is another, but that is all right. So,
22:53.659 --> 22:59.259
if you define A x equals lambda x then it
comes out to be lambda i t fine, that is that
22:59.259 --> 23:03.409
is how it is and if you define A x equals
minus lambda even is minus lambda i t, but
23:03.409 --> 23:10.009
typically you define A x lambda, lambda x.
So, let us not confuse it so given the definition
23:10.009 --> 23:14.350
that A x equals lambda x then, this is your
solution there are two ways of reaching the
23:14.350 --> 23:18.309
solution. One is as I said to use the Eigen
values and the Eigenvectors right, away and
23:18.309 --> 23:23.080
decuple the system and solve it there the
other one is to not decouple, it use a matricial
23:23.080 --> 23:28.179
solution form first get the matricial solution
and then use the Cayley-Hamilton and other
23:28.179 --> 23:33.679
theorem to break up that matrix say.
So, exponential this is something might want
23:33.679 --> 23:38.100
to do in Matlab or mathematic has a good thing.
So, if you give your exponential if you give
23:38.100 --> 23:42.389
a matrix in three by three matrix and you
give exponential of that matrix then mathematic
23:42.389 --> 23:46.649
will immediately spit out what. So, you put
exponential so you put a three by three matrix
23:46.649 --> 23:52.179
A input it then you say exponential of a and
then mathematic will immediately, spit out
23:52.179 --> 23:59.179
what that exponential of A is in terms of
just another matrix you see what I am saying.
23:59.600 --> 24:01.570
So, basically it is like this.
24:01.570 --> 24:08.570
So, A let us say is two by two; so a 1 1,
a 1 2, a 2 1, a 2 2. Now, exponential of a
24:12.369 --> 24:19.369
would be some other matrix b 1 1, b 1 2, b
2 1, b 2 2, but these are related these, these
24:23.259 --> 24:28.679
are related through the Eigen values and the
Eigenfunction, Eigen functions and Eigen values
24:28.679 --> 24:35.679
of the previous matrix.
Now, give you quick problem say A equals simple
24:36.749 --> 24:43.749
A equals a 1, 0, 0, a 2 what would be my exponential
of a how do you make exponential of a does
24:55.350 --> 25:02.350
not say or whatever three by three whatever
any idea no it would be e to by a 1, 0, 0,
25:04.909 --> 25:11.909
e to by a 2 and this is valid for any value.
So, if you have diagonal matrix then any exponential
25:14.559 --> 25:21.559
or anything that you do on that is same as
doing on its individual elements is it clear
25:21.679 --> 25:26.009
Liza. If you have a diagonal if you have diagonal
matrix anything any operation that you do
25:26.009 --> 25:31.659
on the matrix is same as doing on the individual
elements of the matrix fine. So, I hope you
25:31.659 --> 25:38.659
agree with me on this and I hope you remember
this also those of you do not that, the vector
25:39.989 --> 25:44.399
something that we will need all the time that
if d x d t equals x then x is given by this
25:44.399 --> 25:50.320
form alpha i beta i into the lambda i t where
lambda i is the Eigen values of the system,
25:50.320 --> 25:56.570
given as you know x equals lambda x and beta
i are the Eigen functions. Now, how what alpha
25:56.570 --> 25:59.110
i is a certain constant how am I going to
define?
25:59.110 --> 26:05.100
How I going to obtain those constants is through
defining the initial values. So, if my initial
26:05.100 --> 26:07.159
value is given at x t equals 0.
26:07.159 --> 26:14.159
Let us say so my solution over here is x t
equals sigma alpha i beta i e to the power
26:16.409 --> 26:23.409
lambda i t, i going from say 1 to n then my
initial value is given x 0 equals x 0 fine.
26:26.669 --> 26:33.669
So, I put my x 0, t equals sigma i equals
i going from 1 to n alpha i beta i right.
26:36.320 --> 26:43.320
So, this is how I define my now my Eigen functions
are known. So, there has to be a unique there
26:44.690 --> 26:50.509
has to be there have to be unique values of
alpha i with such that this will work out
26:50.509 --> 26:54.649
right.
So, if your beta i has you know three elements
26:54.649 --> 26:58.559
and with each of these three deductions. So,
let us say if you have two by two matrix and
26:58.559 --> 27:05.559
beta i has three, three elements each of these.
So, let us say alpha 1 times beta 1 1 plus
27:08.549 --> 27:15.549
alpha 2 times beta 2 1 should equal x 0 1
first element similarly, you will have alpha
27:17.929 --> 27:24.619
1 times beta 1 2 plus alpha 2 times beta 2
2 should equal the second element of this
27:24.619 --> 27:29.200
and you have. Now, two unknowns two equations
alpha one and alpha 2 2 unknowns equations
27:29.200 --> 27:36.200
you can solve for them and get the values
fine. So, you just need to invert it.
27:36.659 --> 27:43.029
using the Eigenvector.
(That is transpose of the matrix
27:43.029 --> 27:50.029
You can solve what alpha i.
(Alpha a
27:51.249 --> 27:57.950
Eigenvectors of the transpose let us call
it.
27:57.950 --> 28:01.109
Y i.
It’s called Y i.
28:01.109 --> 28:04.169
beta i.
That is correct. So, you can do it that way
28:04.169 --> 28:09.119
also, but any how this is a system of equations
with alpha i is as unknown and these two as
28:09.119 --> 28:12.059
known. So, you can solve them any which way.
28:12.059 --> 28:18.859
So, the Eigenvectors are solved and you know.
So, the Eigen vectors and Eigen values you
28:18.859 --> 28:21.869
know how to solve them and I do not need to
repeat them, but still I will just quickly
28:21.869 --> 28:26.019
go through. So, this is the Eigen values are
solved using determinant of a minus lambda
28:26.019 --> 28:30.499
i equals 0 and then Eigenvectors that obtained
by solving a minus lambda i times beta equals
28:30.499 --> 28:33.499
0 fine.
So, these are some basic things you learned
28:33.499 --> 28:37.590
in matrix algebra, but these are things you
should be able to do because when a problem
28:37.590 --> 28:44.590
is given to you. You have to be able to actually
find the lambda i then the beta i fine. So,
28:44.739 --> 28:48.249
and just as I said the initial condition just
as I showed is shown by that. So, alpha i
28:48.249 --> 28:52.559
depends on the initial conditions so, alpha
i times beta i equals X 0.
28:52.559 --> 28:58.970
So, there are n for n by n nonsingular matrix,
there is n Eigen values and n Eigen vector
28:58.970 --> 29:05.970
n corresponding Eigen vectors. So, which means
that the system is uniquely defined if there
29:06.580 --> 29:10.779
are n Eigen values and n Eigenvectors in the
system is uniquely defined if there is a,
29:10.779 --> 29:17.779
if there is a less than n Eigen value n minus
1 Eigen values then what is what does it imply,
29:20.559 --> 29:27.559
if there is.
Unique solution is not possible solution is
29:28.570 --> 29:32.299
possible, but unique solution is not possible
and in other words the system is ill-defined
29:32.299 --> 29:37.619
is it is an ill-defined problem. So, essentially
what is happening is that you have n variables,
29:37.619 --> 29:42.409
but n minus one equations like we said if
the system has to be completely defined then
29:42.409 --> 29:47.809
you have n invariables and n equation.
Now, if you have if the matrix is singular
29:47.809 --> 29:51.850
then you know it what it means is that, if
the for example, is less than the order then
29:51.850 --> 29:56.499
what it means is that the system is ill-defined
or in other words the number of equations
29:56.499 --> 30:00.769
that, you have from the system is less than
the number of variables. When that happens
30:00.769 --> 30:04.049
what are you suppose, to do you are actually
not suppose to settle around with the algebra,
30:04.049 --> 30:07.139
you are suppose to go back to the model and
check if there is something wrong with the
30:07.139 --> 30:10.989
model. You are missing out an equation with
the model it is there is no point fiddling
30:10.989 --> 30:17.210
around with the algebra or the mathematic
when there is a when the system is nonsingular.
30:17.210 --> 30:22.679
Now, these Eigen values now we come towards
the stability to you. Now, we are kind of
30:22.679 --> 30:28.919
walking towards the stability equation these
Eigen values that you get are complex numbers
30:28.919 --> 30:33.369
in general, they are complex numbers then
they will have a imaginary part and a real
30:33.369 --> 30:39.350
part for example, here any lambda i am sorry
this should be lambda i excuse me is written
30:39.350 --> 30:46.350
as a i plus minus b i, b i times i. So, i
b being the imaginary part and a i being the
30:47.269 --> 30:50.840
real part and now, we will look at what the
effects of these imaginary and real parts
30:50.840 --> 30:52.440
are on the system.
30:52.440 --> 30:59.440
So, if my Eigen values let us see so basically
what is happening is that. I need to have
30:59.720 --> 31:06.149
the solution out here. So, the what is happening
is this is my solution. So, let us write lambda
31:06.149 --> 31:13.149
i as a i plus b i times t. Now, what happens
is let me write this over here and then let
31:16.429 --> 31:17.669
us see.
31:17.669 --> 31:24.669
So, X equals sigma alpha i beta i e to the
power just write it like this a i plus b i
31:28.950 --> 31:35.950
times i times t i from 1 to infinity. So,
if this is our basic thing and now all stability
31:36.109 --> 31:43.109
results will depend on this. Now, you tell
me that if my a i is what, what are the effects
31:44.009 --> 31:49.889
that you think of a i and b i. So, look at
this what we are interested in is how it varies
31:49.889 --> 31:54.869
with time right how it varies with time. So,
what are the two factors that are going to
31:54.869 --> 31:59.109
decide how it varies with time obviously a
i and b i fine.
31:59.109 --> 32:03.999
Now, let us try and you know try to discuss
what would be the effect of different values
32:03.999 --> 32:10.999
of a i and b i. So, a i is the real part b
i is the imaginary part. So, a i is a real
32:11.929 --> 32:18.929
and b i is the imaginary part. So, you have
to tell me that what could you think would
32:19.149 --> 32:25.289
be the effects of these.
(Real part should be negative)
32:25.289 --> 32:26.859
What?
Real part should be negative.
32:26.859 --> 32:32.109
Real part should be negative for all kinds
of stability. Now, the imaginary part may
32:32.109 --> 32:37.580
or may not be negative. So, what do you think
if it is negative fine if it is not negative,
32:37.580 --> 32:42.690
if the imaginary part is not negative then
what will happen?
32:42.690 --> 32:48.789
System will be stable, but something else
will happen.
32:48.789 --> 32:52.320
(Oscillations)
Oscillations see what does this imaginary
32:52.320 --> 32:55.960
part, you know e to the power i theta i to
the power i what does this imaginary part
32:55.960 --> 33:00.659
is composed of when cos and sin. When you
write e to the power i x, what is it composed
33:00.659 --> 33:06.669
of cos and sin. So, if the real part say a
is the real part and b is a b is a imaginary
33:06.669 --> 33:11.619
part. So, here over here if a is the real
part is negative then definitely means, that
33:11.619 --> 33:16.639
the system is stable. If the real part is
positive then the system is always unstable.
33:16.639 --> 33:23.639
If the real part is 0, what is it called you
know what is it called, if the real part is
33:26.649 --> 33:27.690
0.
33:27.690 --> 33:34.049
No, the system is called neutrally stable
that is just stable slight perturbation from
33:34.049 --> 33:40.720
this side, slight increase in a will lead
to growth or growth of noise slight decrease
33:40.720 --> 33:44.989
will lead to d k. So, it is called neutrally
stable just stable so that, that are those
33:44.989 --> 33:49.759
are my explanations on the real part. Now,
let us look at the imaginary part b i. So,
33:49.759 --> 33:51.799
what do you think are the effects of imaginary
part?
33:51.799 --> 33:56.440
If the imaginary part is 0 it means, that
there are no oscillations in the system, if
33:56.440 --> 34:01.559
the imaginary part is nonzero that is positive
or negative then there could be oscillations
34:01.559 --> 34:05.239
in the system. So, what you have to do when
you looking at the total system you have to
34:05.239 --> 34:07.929
couple the effects of the real part and the
imaginary part.
34:07.929 --> 34:12.700
So, what we did over here is in the let us
go to the screen we divided into 5 different
34:12.700 --> 34:19.530
parts. So, first case is where the real part
is negative and the imaginary part is 0, this
34:19.530 --> 34:25.290
system is completely stable fine. Next, is
the real part is positive and the imaginary
34:25.290 --> 34:31.220
part is 0, the system is completely unstable
third is real part is 0, but the imaginary
34:31.220 --> 34:35.950
part is nonzero positive or negative, what
it means that there will be oscillation, there
34:35.950 --> 34:41.109
will be cos and sin part as a result of which
there will be oscillations and sustained oscillations.
34:41.109 --> 34:46.559
sustained oscillations means that, these oscillations
are not growing with time or decaying with
34:46.559 --> 34:50.520
time they remain what they are, but there
are sustained oscillations I will show you
34:50.520 --> 34:57.520
pictures in a minute and you will appreciate
this better next, is that with a negative,
34:58.400 --> 35:03.510
negative real part you have the imaginary
part nonzero which means that, that would
35:03.510 --> 35:05.700
going to be oscillation, but these damped
oscillation.
35:05.700 --> 35:11.010
So, oscillations are large at the beginning
and then the slowly decay and the last one
35:11.010 --> 35:16.230
is the real part is positive and the imaginary
part is nonzero which mean that the oscillations
35:16.230 --> 35:23.230
start and then they grow with time. So ,out
of this how many stable a is stable first
35:24.190 --> 35:31.190
case is stable a number 1, number 2 unstable
number three neutrally stable number 3 is
35:31.250 --> 35:38.250
neutrally stable just stable number 4, stable
number 5 is unstable. So, two of them stable
35:38.940 --> 35:41.130
two of them unstable one of them neutrally
stable.
35:41.130 --> 35:46.970
Let’s look at the pictures. Now, so this
is the case first case where it the real part
35:46.970 --> 35:53.970
is negative and the imaginary part is 0, second
case real part is positive imaginary part
35:55.319 --> 36:00.569
is 0. So, first case as you see x decreases
x is a deviation right. So, you started you
36:00.569 --> 36:04.900
gave a little perturbation started with a
little deviation at t equals 0 and the perturbation
36:04.900 --> 36:09.000
or the deviation decreases with time and goes
back to the steady state.
36:09.000 --> 36:13.380
So, this is this is your steady state value
at 0 is at the bottom of the x your steady
36:13.380 --> 36:18.059
state value, what do you see and we increased
it to this extend and it goes back to the
36:18.059 --> 36:23.549
steady state value fine. So, that is why this
is stable node and then the next one is a
36:23.549 --> 36:30.549
is larger than 0 and b is a 0. So, this you
start with a small deviation let us say whatever,
36:30.740 --> 36:35.380
the deviation value is and it grows over time.
So, the system even if it is a very small
36:35.380 --> 36:40.470
you give a large so, that is a difference
for a stable system you gave even whenever,
36:40.470 --> 36:47.470
I mean sort of significant perturbation and
still it goes back to 0, for unstable system
36:49.299 --> 36:54.160
you gave a small perturbation and that blows
up and these are the most dangerous ones as
36:54.160 --> 36:58.119
far as biochemical reactors or any kind of
reactors are concerned and I think I discussed
36:58.119 --> 37:02.299
this the concept of hotspots here.
In this class at one point of time and this
37:02.299 --> 37:08.190
is what it is related to systems, where you
have hotspots you give them a small perturbation
37:08.190 --> 37:13.490
and they growth and let us look at other things
then I will come back to the hotspot question.
37:13.490 --> 37:19.049
This is the case where the vortex point it
is called neutrally stable which, where the
37:19.049 --> 37:25.329
real part is 0 and the unreal part that the
imaginary part is nonzero and you have sustained
37:25.329 --> 37:27.680
perturbations.
So, whatever perturbation you gave remains,
37:27.680 --> 37:33.520
it does not go away and does not increase
either then, this is a stable one it is called
37:33.520 --> 37:39.829
the stable focus. Stable node with a imaginary
part or with a perturbation or with a oscillation.
37:39.829 --> 37:45.410
So, you start with an osculation and you give
a perturbation it starts oscillating and then
37:45.410 --> 37:50.440
it decrease all of this in the control theory,
I believe this starts decreasing and, but
37:50.440 --> 37:54.869
it keep keeps oscillating went through the
oscillation it goes to 0 it decays, but decays
37:54.869 --> 37:56.240
in an oscillating way.
37:56.240 --> 38:03.240
And the last one is an unstable focus which
is that you start with a little give a little
38:03.470 --> 38:08.510
perturbation it starts to oscillate and the
oscillation start to increase. Now, this is
38:08.510 --> 38:13.240
one I am coming back to the hotspot question,
this is the one that actually is most seen
38:13.240 --> 38:17.920
in hotpots. So, there are two possibilities
of instability one is the this one this is
38:17.920 --> 38:22.720
the example the second one here, where it
grows uniformly and this one where it grows
38:22.720 --> 38:25.369
the oscillation.
Now, what is seeing in most reactor than what
38:25.369 --> 38:29.940
we are studying biochemical reactors that
a chemical and biochemical reactors. What
38:29.940 --> 38:35.559
happens when the system goes unstable, it
goes unstable through this through this unstable
38:35.559 --> 38:41.240
focus the reason being that it typically that
is an oscillatory part and if it is a especially
38:41.240 --> 38:45.859
this happens, when there is a what kind of
coordinates do you think this will happen,
38:45.859 --> 38:52.859
what kind of systems will happen in circular
or spherical systems in cylindrical or spherical
38:52.950 --> 38:55.780
systems, why because you have the radial direction
over here.
38:55.780 --> 39:01.980
Let us see in this in this cylindrical system
and these oscillations happens in theta. So,
39:01.980 --> 39:06.680
the hotpots starts to rotate so, these oscillations
will actually lead to rotation of the hotspots.
39:06.680 --> 39:10.770
So, you have a hotspot over here and these
are in practical system this is what happen.
39:10.770 --> 39:14.280
When you study the systems theoretically,
we kind of assume there is a hotspot over
39:14.280 --> 39:18.569
here and there is another hotspot emerging
and another hotspot emerging and so on.
39:18.569 --> 39:22.930
But in a real system that does not happen,
what happens is that these hotspots start
39:22.930 --> 39:27.980
to travel you know because there is a imaginary
wave, there is a cos and sin part of it and
39:27.980 --> 39:34.039
these are oscillation, these hotspots start
to travel and as they travel they gain in
39:34.039 --> 39:38.520
size and then and then it will lead to a huge
hotspots and then, whole explosion the reactor
39:38.520 --> 39:42.829
blows up and that is how reactor you know
in industry for example, all these accidence
39:42.829 --> 39:45.750
in the plants that you hear of actually happen
this way.
39:45.750 --> 39:50.640
So, hotspot means a temperature a small region
in the in the reactor, where the temperature
39:50.640 --> 39:56.430
grows suddenly and then it starts to gather
you know gather in space like a for example,
39:56.430 --> 39:59.990
if you start started over here and then it
will goes over here it becomes like this it
39:59.990 --> 40:04.460
goes over here it becomes like this. So, this
is our hotspot is generated small increase
40:04.460 --> 40:09.339
in temperature is a major problem lesser concentration
temperature is a major problem.
40:09.339 --> 40:14.329
You can have concentrations spots also you
know, but concentration hotspots, but temperature
40:14.329 --> 40:18.240
hotspots is a major problem because it lead
to the reactor blow up. So, you have you start
40:18.240 --> 40:23.319
with a small hotspot over here and because
there is a if there is no see, if there is
40:23.319 --> 40:28.960
no unstable focus if it is just simply unstable
or in other words, the imaginary part is 0
40:28.960 --> 40:33.640
then what happens is that hotspot grows in
one space. So, it just stays there and keeps
40:33.640 --> 40:40.640
growing, keeps growing which is bad also,
but this is lot worse. Why it lot worse? Because
40:41.240 --> 40:46.869
it is it propagates you know it helps in propagation.
So, it goes to another regional space and
40:46.869 --> 40:51.549
spreads it over there with that is why it
is if you, if it just increases in one space
40:51.549 --> 40:55.770
you have an idea and you can cool and through
cooling and stuffs like that, but then this
40:55.770 --> 41:00.859
propagates and as it propagates at it continues
to infect other regions to hotspot, where
41:00.859 --> 41:07.329
regions wherever there are no hotspot it continues
to infect therefore, the unstable focus so
41:07.329 --> 41:11.910
unstable focus means, why if there is a focus
of hotspot. So, this is my focus of hotspot
41:11.910 --> 41:15.039
and that itself is unstable.
So, there is a focus and the focus itself
41:15.039 --> 41:20.260
is moving whereas, this is a stable focus
you know the system is unstable overall, but
41:20.260 --> 41:24.240
the focus that is there is not unstable that
is a its unstable in one place is that clear
41:24.240 --> 41:28.069
this mathematics and how it translates to
the physics of the problem and these are very
41:28.069 --> 41:31.640
real problem, which is not it is just not
something that is of mathematical interest.
41:31.640 --> 41:35.150
You know this is these are things that happen
in the reactor and you will read in a papers
41:35.150 --> 41:38.829
here and there every now and then, that reactors
blow up and this is the reason why reactors
41:38.829 --> 41:43.559
blow up because you have certain noises or
perturbations in the system that lead to because
41:43.559 --> 41:47.210
why is this, what is the root cause of all
of this is because the nonlinearity of the
41:47.210 --> 41:50.770
reaction and as we do a problem, you know
just in the next few minutes you will see
41:50.770 --> 41:55.059
that how the nonlinearity try start to infect
the system.
41:55.059 --> 41:58.789
So, because there are nonlinearity in the
system these hotspots are generated to start
41:58.789 --> 42:03.180
with it its a completely linear system nothing
like that would ever happen, you have hotspot
42:03.180 --> 42:06.380
it will go away you know you increase the
introduce the perturbation, you can try this
42:06.380 --> 42:11.779
with a linear system introduce a small perturbation
to the system, it would be decayed. You know
42:11.779 --> 42:15.010
it will damp out and go away, but with a non-linear
system you introduce the perturbation. The
42:15.010 --> 42:20.890
system it starts to grow there are possibility
that it grows and what we here or today trying
42:20.890 --> 42:24.960
to understand is under, what conditions that
will happen? Why are you trying to do that?
42:24.960 --> 42:29.200
Because as a engineers we want to predict
a priory whether there is a possibility of
42:29.200 --> 42:33.730
hotspot or not, you do not want to have the
hotspot blow up into an accident and then
42:33.730 --> 42:37.819
you know. So, you will get fire if that happens
so, you want to take care of your job and
42:37.819 --> 42:41.720
you want to figure out how the, whether the
hotspot is going to happen or not and that
42:41.720 --> 42:45.260
is why? We are doing all these analysis I
am think some of you might have done similar
42:45.260 --> 42:46.210
things.
42:46.210 --> 42:53.210
So, as we said that a i plus b i t is the
whole thing, you know b sorry a i times plus
42:54.119 --> 42:57.609
b i times, I is a whole thing and what is
a whole thing the Eigenvalue of the system
42:57.609 --> 43:02.220
right. So, what we need to find out is what
is the, what is the nature of that Eigenvalue
43:02.220 --> 43:07.319
is it positive is the real part positive or
negative and is a imaginary part positive
43:07.319 --> 43:11.400
or negative, but what is a prime importance
for us more than imaginary part, the imaginary
43:11.400 --> 43:14.680
part only leads to oscillations.
So, what is a prime importance to us is a
43:14.680 --> 43:18.589
real part whether, the real part is positive
negative or 0. If it’s positive its unstable,
43:18.589 --> 43:23.369
if its negative its stable and if it is 0
its neutrally stable. So, we want to do that
43:23.369 --> 43:27.089
now, but we want to little clever in what
we want to do is we do not want to compute
43:27.089 --> 43:30.770
all the Eigen values. Why do we not want to
compute the Eigen values is because if you
43:30.770 --> 43:35.299
have a system like a two by two system for
example, or three by three system is still
43:35.299 --> 43:38.930
possible to go and compute the Eigen values,
but in industries you will network of reaction.
43:38.930 --> 43:42.869
You do not have reactions where you just with
three species you know and I showed you the
43:42.869 --> 43:47.539
examples of this multiple reactions. What
happens is when you start a reactor with two
43:47.539 --> 43:52.529
or three species a whole network of reactions
emerge because you know, you cannot control
43:52.529 --> 43:56.099
once you put the thing in you do not have
control over the entire system. So, you want
43:56.099 --> 44:01.160
a going to b going to c or a going a reacting
with b producing c, but then d e and e and
44:01.160 --> 44:04.140
f and all will come up and it will lead to
a network of reactions.
44:04.140 --> 44:07.940
Now, when is there is a network of reaction,
this matrix a that we are talking of will
44:07.940 --> 44:13.339
be a large matrix it will not be a 2 by 2
matrix n is going to be large number maybe
44:13.339 --> 44:17.770
10 or 20 or 50 there are you know most of
the industries many of the industries in petroleum
44:17.770 --> 44:23.859
industry for example, the number of n value
of n is very, very large it could be 100,
44:23.859 --> 44:28.020
200, 500.
So, how I going to 5, 500 Eigen values each
44:28.020 --> 44:31.640
time it is very hard. So, you do not want
to find all the Eigen values and that is a
44:31.640 --> 44:36.140
lot of work. So, what you want is a shortcut
to figuring out so your aim is not to find
44:36.140 --> 44:40.430
the Eigen values you know it is we are not
doing a course applied mathematics over here,
44:40.430 --> 44:44.069
that we our job is to find the Eigen values.
Your aim is to find out whether the reactor
44:44.069 --> 44:47.960
is going to remain stable or blow up and for
that you do not need the Eigen values, but
44:47.960 --> 44:54.140
what you need is the science of the real part
of the Eigen values and that is what you concentrated
44:54.140 --> 44:57.710
on.
So, what all we need to do is we need to determine
44:57.710 --> 45:03.440
if the real part of the Eigenvalue is positive
or negative. Now, so this is your equation
45:03.440 --> 45:10.440
for obtaining the Eigenvalue a minus lambda
equals 0. Now, this equation you for finding
45:10.500 --> 45:17.000
the Eigen values could be written as this
right. So, you can just open this up and write,
45:17.000 --> 45:23.390
write this in this form fine is that clear
if it say n by n matrix, then your Eigenvalue
45:23.390 --> 45:26.589
in this characteristic equation is called
the characteristic equation for Eigenvalue.
45:26.589 --> 45:32.089
So, characteristic equation for the Eigenvalue
will have n cross. You know it is it could
45:32.089 --> 45:38.420
e to be off of order n right is that clear
why because is that clear Liza why or do you
45:38.420 --> 45:39.529
want me to go through that.
45:39.529 --> 45:46.529
So, if you have a i 1 1 minus lambda a 2 2
minus, minus lambda and so on. Say a n n minus
45:48.940 --> 45:52.549
lambda then, what is the order of this. So,
lambda, lambda, lambda, lambda to the n and
45:52.549 --> 45:57.089
then the other terms this is the highest order
and then the all the lower order terms will
45:57.089 --> 46:02.799
be there is that clear completely. So, when
you arrange these terms you collect terms
46:02.799 --> 46:06.299
of lambda to the n and so on.
Then divide the numerator and everything and
46:06.299 --> 46:10.549
get it in the cleanest form possible, this
is the characteristic equation you get where
46:10.549 --> 46:16.430
lambda n is given by lambda n plus b 1 lambda
n minus 1 plus b 2 to the lambda n minus 2
46:16.430 --> 46:22.520
and so on, b n minus 1 to the lambda to times
lambda plus b n this is how it is given. Now,
46:22.520 --> 46:26.609
all you need to do is find and you know what
is a criteria called do you remember?
46:26.609 --> 46:32.549
Routh root criteria and this is the Hurwitz
criteria, if you remember so of finding the
46:32.549 --> 46:38.069
two different ways of finding what the signs
of this a would be.
46:38.069 --> 46:44.589
So, the Hurwitz states that all what this
is what we do and you can do the routh criteria
46:44.589 --> 46:49.319
also it is either it is mine they states that
all roots of an equation will have negative
46:49.319 --> 46:56.319
real roots. If one is here B 1 the first coefficient
here is greater than 0 and what is the other
46:59.210 --> 47:06.210
one and if you remember the determinant of
this B 1, B 3, 1, B 2 that is you go this
47:07.809 --> 47:14.809
way 1, B 1, B 2, B 3 like this is greater
than 0. Now, if this is I n by n matrix if
47:16.230 --> 47:17.400
it is not a 2 by 2 matrix.
47:17.400 --> 47:24.400
Then, this is how it is 1, B 1, B 2, B 3,
B 4, B 5 this to see how it is going, its
47:26.940 --> 47:33.940
going zigzag like this; so, one this reduction,
then this reduction and so on. Fine it is
47:40.670 --> 47:46.480
going like this so, if I look here let us
go back to the screen; so 1, B 1, then B 2
47:46.480 --> 47:53.480
then B 3 then B 4 then and so on. And then
you start at B 3 again here and then go on
47:53.579 --> 47:58.390
in the same direction, you just move and then
go on in the same direction and then up to
47:58.390 --> 48:05.390
B n. So, luckily for our case you know for
the chemostat case, we have how many components
48:07.960 --> 48:09.910
what will be the order of this matrix.
Two
48:09.910 --> 48:15.480
Two yes typically two by two, but it could
be three by three if you know multiple substrate
48:15.480 --> 48:20.019
or if it is a inhibitor that is involved then
it could be three by three, but typically
48:20.019 --> 48:24.410
it is two by two. Why? Because even if there
is multiple substrate its different even,
48:24.410 --> 48:27.450
if there is a inhibitor. What we do? We get
rid of the inhibitor out of equate problem.
48:27.450 --> 48:31.720
So, we come up with two by two system.
If there are multiple substrate then yes you
48:31.720 --> 48:35.990
know if s 1 and s 2 are two substrate then,
how many what would be the order of the equation
48:35.990 --> 48:38.970
three. So, it would be three by three. So,
either it would be two by two or it would
48:38.970 --> 48:42.910
be a three by three matrix for us. So, this
matrix is this either if it is a 2 by 2 then
48:42.910 --> 48:47.109
this is going to be the form, if it is a three
by three then this is going to be the form
48:47.109 --> 48:50.369
of it.
So, let us remember this and you can also
48:50.369 --> 48:54.740
prepare for the three by three and I do not
think you need the generalize form, but I
48:54.740 --> 48:59.319
still gave you the generalize form. So, these
two criteria have to be satisfied and they
48:59.319 --> 49:05.569
are they are joined by the logical which means,
that both of these criteria have to be satisfied.
49:05.569 --> 49:10.250
So, let us now go and look at the stability
of this chemostat fine.
49:10.250 --> 49:14.460
So, this is the stability analysis that we
did, and this is generalized this is for valid
49:14.460 --> 49:19.450
for everything. Now, what we are going to
do is we are going to try and apply it to
49:19.450 --> 49:22.839
the chemostat. So, the first equation for
the chemostat is what we have done remember
49:22.839 --> 49:29.839
dx dt equals dilution rate time x 0 minus
x plus mu times x this is mu the whole thing,
49:31.160 --> 49:36.380
the initial calculation that we will do we
will assume it to be mu. Why would we assume
49:36.380 --> 49:41.670
it to be mu? We own put the whole thing mu
max S over K s plus x we just assume, it to
49:41.670 --> 49:45.630
be mu, why would you do that would assume
it to be mu as a function of as, but why would
49:45.630 --> 49:52.630
you do that, what is the growth rate of this
system?
50:00.849 --> 50:07.849
What is it, what is the ideal typical growth
rate of the system? It is mu times x, what
50:11.559 --> 50:18.559
is the growth of the system its mu times.
So, I think there is a basic problem here
50:20.349 --> 50:26.170
so, let me go to it. So what is a difference
between mu times x, and just you know the
50:26.170 --> 50:33.170
concept of mu; what is the fundamental difference,
growth is mu times x so, is the growth rate
50:38.539 --> 50:44.220
is what is seeing seen on the screen over
here mu, mu max S x over K s plus K plus S
50:44.220 --> 50:51.220
is that the growth rate at all time. Yes or
no, I do not understand you not if.
50:52.119 --> 50:59.119
So, I do not know why you could not you know
we are not so, what is the growth rate? The
51:01.420 --> 51:06.680
growth rate is mu times x at all times, but
the specific growth rate is not the Monod
51:06.680 --> 51:11.109
growth kinetics at all time see, there is
fundamental difference growth rate is mu times
51:11.109 --> 51:16.230
x is all time there is no confusion about
that then. Next, question comes what model
51:16.230 --> 51:20.240
do I use for the specific growth rate and
I can use different kinds of model for specific
51:20.240 --> 51:23.529
growth rate.
So, as a result when I am going doing this
51:23.529 --> 51:28.640
analysis I want to do it for mu. So, that
I can plug in any kind of growth rates at
51:28.640 --> 51:34.250
a later time and the model analysis would
be still valid is that clear because I think
51:34.250 --> 51:38.930
that is the major thing that, you are not
differentiating between the growth rate as
51:38.930 --> 51:43.460
a whole and the fact that you are using just
a Monod growth kinetics. You can use modified
51:43.460 --> 51:47.970
Monod growth kinetics, you can use you know
the Malthusian kinetics or different types
51:47.970 --> 51:51.230
of kinetics.
So anyhow, so these are the coupled equations
51:51.230 --> 51:56.660
that you have; and so my f 1; so I need to
define my two functions in order to be able
51:56.660 --> 52:01.700
to do all these calculations. So, what I need
to do first I need to define my two functions
52:01.700 --> 52:06.279
and then I need to obtain my Jacobian obtain
my A using these two functions and then I
52:06.279 --> 52:12.279
need to look at the Eigen values of a not
specifically, but we need to look at it through
52:12.279 --> 52:18.480
the characteristic equation fine. So, first
step in this is therefore, looking finding
52:18.480 --> 52:23.329
the functions so, this is my first function
and this is my second function. Next thing,
52:23.329 --> 52:27.630
I need to do is find the Jacobian. So, why
would you work on your copies to find the
52:27.630 --> 52:34.630
Jacobian so, these are my f 1 and f 2. So,
I want you to find the A matrix.
54:35.029 --> 54:40.309
This is minus mu over Y, you got plus minus
or plus.
54:40.309 --> 54:47.309
Minus and this is minus D can we see other
fine. So, these are my calculations we will
55:07.000 --> 55:11.569
stop here and continue with the rest in the
next class.
55:11.569 --> 55:17.980
So, I just quickly capture back say f 1 and
f 2 are my functions, these are my Jacobian
55:17.980 --> 55:24.980
and I take the derivates in each step and
get this.